Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the quadratic equation with real coefficients and solutions $x_1,x_2$ if you know that $\Delta =b^2-4ac=-36$ and $x_1+x_2=6m$ What I've done so far is:
We know that $\Delta \lt 0$, which means that $x_1$ and $x_2$ are two conjugate complex numbers with the form: $x_1=r+n\cdot i \space\text{ and }\space x_2=r-n\cdo... | Alternative approach:
Without loss of generality, the equation is
$x^2 + Bx + C = 0.$
Since $x_1 + x_2 = 6m$,
and since you must have that
$(x - x_1) \times (x - x_2) = x^2 + Bx + C$,
you must have that
$B = -6m.$
Here, there is some ambiguity involved. Taking the constraint of (in effect)
$B^2 - 4C = -36$
at ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4318399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Rolling a die - Conditional Probability A die is thrown repeatedly.
Let $X$ ~ First 5 is thrown and $Y$ ~ First 6 is thrown
Calculate $\mathbb{E}(X|Y=3)$
You may use the identity: $\sum_{n=k}^\infty nz^{n-k} = \frac{1}{(1-z)^2}+\frac{k-1}{1-z}$
I know from the definition of expectation, we have:
$\mathbb{E}(X|Y=3) = (1... | While it is not necessary, if the question expects you to use the given identity then here is how you would go about it -
$ \small \displaystyle E(X|Y = 3) = 1 \cdot \frac{1}{5} + 2 \cdot \frac{4 \cdot 1}{5^2} + 4 \cdot \frac{4 \cdot 4}{5^2} \cdot \frac{1}{6} + 5 \cdot \frac{4 \cdot 4}{5^2} \cdot \frac{5 \cdot 1}{6^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4320165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Express an integer $m$ as sum of 1, 0 and -1 for fixed number of summands $j$ Let $j \in \mathbb{N}_0$. I'm looking for the number of all possible combinations, such that for a given $|m| \leq j, m \in \mathbb{Z}$
$m = \sum \limits_{k=1}^{j} a_k\quad$ where $a_k \in {-1, 0, 1}$
Examples:
$j=3;m=3 \Rightarrow a \in A=\l... | I'll use $n$ instead of $j$. We can assume $m$ is nonnegative, since replacing $m$ with $-m$ gives the same answer (by negating all the solutions).
We can count the solutions in groups according to how many $1$s there are. In order for the sum $m$ to be possible, we need at least $m$ $1s$ and at most $\left\lfloor (m+n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find inverse of element in a binary field The question states:
Let us consider the field $GF(2^4)$ with multiplication modulo $x^4+ x^3+1$
Find all y such that $1010(y + 0011) = 1111$, in other words find y that satisfies $(x^3+x)(y +x+1) = x^3+x^2+x+1$
I tried to find the inverse for element $1010$ to multiply both si... | Here are some methods to compute this using a computer. The software is sagemath.
First, if the program supports defining finite fields with a given polynomial, you can just use that:
K.<a> = GF(2^4, modulus=x^4 + x^3 + 1)
(a^3 + a)^-1
=> a^3 + a + 1
You can also use the extended GCD which gives polynomials $u, v$ s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the sum of geometric progression Evaluate the sum:
$$\sum_{x=0}^\infty x(x-1) {2+x \choose x}(0.008)(0.8)^x $$
I was able to make this into:
$$0.004\sum_{x=0}^\infty x(x-1) (x+1)(x+2)(0.8)^x $$
Let $x=n-2$ then $n=x+2$:
$$0.004\sum_{n=2}^\infty n(n-1)(n-2)(n-3)(0.8)^{n-2} $$
$$0.004\sum_{n=4}^\infty (n)_3~(0.8)... | There are a couple ways to go about this. An important identity is that $\frac{1}{(1-x)^k}=\sum_{n=0}^\infty \binom{n+k-1}{k-1}x^n$ for $|x|<1$. This can be proven with just binomial theorem and the negative binomial coefficients. It can also be proven by noting that $\frac{1}{(1-x)^k}=(1+x+x^2+\ldots)^k$. The coeffici... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4329796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve $\sqrt{x-5}-\sqrt{9-x}\gt1,x\in\mathbb Z$
Solve $\sqrt{x-5}-\sqrt{9-x}\gt1,x\in\mathbb Z$
The statement tells us that $x\in[5,9]$. Also,
$$\sqrt{x-5}\gt1+\sqrt{9-x}$$
Since both sides are positive, we can square
$$x-5>1+9-x+2\sqrt{9-x}\\2x-15\gt2\sqrt{9-x}$$
$\implies 2x-15\gt0\implies x\gt7.5$
Since $x\in\math... | Hint:
Let $\sqrt{x-5}=a\ge0$ and $\sqrt{9-x}=b\ge0$
$$\implies a^2+b^2=4$$
and $a-b>1\iff a> b+1$
$$4=a^2+b^2>(b+1)^2+b^2\iff 2b^2+2b-3<0$$
Now for $(x-a)(x-b)<0, a<b;$ $$a<x<b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4331630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given two distinct intersecting circles, length of that chord of larger circle which is bisected by the smaller circle is equal to? Two circles whose centres lie on the x axis, whose radii are $\sqrt2cm$ and $1cm$ and whose centres are 2 cm apart intersect at a point A.The chord AC of the larger circle cuts the smalle... | Drop a perp from $C_1$ to $AB$ and extend. Then $D$ is the midpoint of $AB$. Drop a perp from $C_2$ to $C_1D$ extend.
If $AC = 4a$, $C_2E = BD = a$ and $C_1E = C_1D + C_2B$
$C_1D = \sqrt{1-a^2}, C_2B = \sqrt{2 - 4a^2}$
Using Pythagoras in $\triangle C_1EC_2$,
$ (\sqrt{1-a^2} + \sqrt{2 - 4a^2})^2 + a^2 = 4$
$ 3 - 4a^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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relation between roots and coefficient in a cubic polynomial If $\alpha,\beta,\gamma$ are roots of the cubic equation
$$2x^{3}+3x^2-x-1=0$$ then I want to find the equation whose roots are $\frac{\alpha}{\beta+\gamma}, \frac{\beta}{\gamma+\alpha}, \frac{\gamma}{\alpha+\beta}$.
I have $\alpha+\beta+\gamma$= - $\frac{3}{... | If
$$
2x^3+3x^2-x-1=0
$$
has roots $\alpha,\beta,\gamma$, then substituting $x\mapsto\frac1x$ (and multiplying by $-x^3$ to clear denominators)
$$
\begin{align}
&-x^3\left(\frac2{x^3}+\frac3{x^2}-\frac1x-1\right)\\
&=x^3+x^2-3x-2=0
\end{align}
$$
has roots $\frac1\alpha,\frac1\beta,\frac1\gamma$. Then substituting $x\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve the ODE $y'=1-\frac{y}{x}$ I substituted $u=\frac{y}{x}$ then tried to solve the ODE $$\frac{u'}{2u-1}= -\frac{1}{x}$$ and I came this far $$\frac{1}{2}\ln |{2u-1}|=- \ln |{x}| + c_1$$ but then in the solution there was the step $$c_1=\ln c_2 \in \mathbb{R}, c_2 > 0$$ to get $$\ln |2u-1|=\ln{(\frac{c_2}{x})^2}$$ ... | $y' = 1 - \dfrac{y}{x}; \tag 1$
$y' + \dfrac{y}{x} = 1; \tag 2$
$xy' + y = x; \tag 3$
$(xy)' = y + xy'; \tag 4$
$(xy)' = x; \tag 5$
$xy = \dfrac{x^2}{2} + C, \; C \; \text{the arbitrary constant}; \tag 6$
$y = \dfrac{x}{2} + \dfrac{C}{x}; \tag 7$
We Check:
$y' = \dfrac{1}{2} - \dfrac{C}{x^2}; \tag 8$
$\dfrac{y}{x} = \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $(2x^2-3x+1)(2x^2+5x+1)=9x^2$ Solve the equation $$(2x^2-3x+1)(2x^2+5x+1)=9x^2$$
The given equation is equivalent to $$4x^4+4x^3-11x^2+2x+1=9x^2\\4x^4+4x^3-20x^2+2x+1=0$$ which, unfortunately, has no rational roots. What else can I try?
| Hint:
As $x\ne0$ we can divide both sides by $x^2$ to find
$$(a-3)(a+5)=9\iff a^2+2a-24=0$$ where $2x+\dfrac1x=a$
Can you take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation $x^4-8x^3+23x^2-30x+15=0$ Solve the equation $$x^4-8x^3+23x^2-30x+15=0$$
As $x=0$ is obviously not a solution, we can consider $x\ne0$, so I have tried to divide both sides by $x^2$ to get $$x^2-8x+23-\dfrac{30}{x}+\dfrac{15}{x^2}=0$$ Clearly this does not help. I have also tried to find the rational... | Since you tried the rational root theorem and didn't find any rational roots, then next thing to try is factoring as a product of two quadratics. Set your polynomial equal to
$$(x^2+ax+b)(x^2+cx+d)$$ $$=x^4 + (a+c)x^3+(d+ac+b)x^2+(ad+bc)x+bd$$
and equate coefficients. Since $bd=15$ you can try $b=3$, $d=5$ and see wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4343240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Closed-form for $\sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right)$ I'm looking for a closed-form for the following sum:
$$\sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right)$$ for $|m|>1.$
In a previous question of mine, the following similar sum was determined:
$$\sum_{n=1}^{\inft... | EDIT: There was a question asked a long time ago about the related infinite series $$\sum_{n=1}^{\infty} \left( n \operatorname{arccot}(n)-1 \right). $$
The approach used in the accepted answer is similar to the approach I used for this question.
For $|x| >1 $, the inverse hyperbolic cotangent function has the Laurent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4343353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How many ways to deal with the integral $\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}$? I tackle the integral by rationalization on the integrand first.
$$
\frac{1}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{2 x}
$$
Then splitting into two simpler integrals yields $$
\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\frac{1}{... | The answer is no.
There are only two different representations of the result at all.
$ 1/2 (2 \sqrt{1-x}+2 \sqrt{x+1}+ln(\sqrt{1-x}-1)-ln(\sqrt{1-x}]+1)+ln(\sqrt{x+1}-1)-ln(\sqrt{x+1}+1))+constant$
or alternatively:
$ \sqrt{1-x}+\sqrt{x+1}+1/2 ln(((\sqrt{1-x}-1) (\sqrt{x+1}-1))/((\sqrt{1-x}+1) (\sqrt{x+1}+1)))+constant... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove $\sum_{cyc}{\sqrt{\frac{x+1}{x^2+16x+1}}}\geqslant 1$ and $ \sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\leqslant 1$ for $x,y,z>0,xyz=1$ Source: https://artofproblemsolving.com/community/c6t243f6h2745656_inequalities_with_3
Let $x,y,z>0,xyz=1$ then $$
\sum_{cyc}{\sqrt{\frac{x+1}{x^2+16x+1}}}\geqslant 1 \ \ \ \ (1)
... | Hint :
For $x\leq 0$ we have :
$$\frac{d}{dx}\left(\frac{d}{dx}f\left(e^{x}\right)\right)<0$$
Where :
$$f(x)=\sqrt{\frac{x+1}{4x^{2}+4+10x}}$$
Next we show that for $x>0$ and $n= 3$ a natural number :
$$f\left(x\right)+\left(n-1\right)f\left(\frac{1}{x^{\frac{1}{n-1}}}\right)-nf\left(1\right)\leq 0$$
In the case $n=3$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4346877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Let $X$ be a random variable with Cauchy distribution, compute the density function of $Y=\frac{1}{1+X^2}$
Let $X$ be a random variable with Cauchy distribution, if $$Y=\frac{1}{1+X^2}$$ then compute the density function of $Y$
I have tried to use some random variable transformation theorem using the functions $\frac... | The problem can also be solved via the change of the variable in PDF.
$$X:\,f(x)=\frac{1}{\pi}\frac{1}{1+x^2}; \,\,Y=\frac{1}{1+X^2}; \,\,y=f(x);\,\,g(y)-?$$
$$1+x^2=\frac{1}{f(x)}\,\Rightarrow x=\sqrt{\frac{1-f(x)}{f(x)}}$$
$$ dx=-\frac{1}{2}\bigg(\frac{1}{\sqrt{1-f(x)}\sqrt f(x)}+\frac{\sqrt{1-f(x)}}{(\sqrt f(x))^3}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $1If $a$, $b$, $c$ are three positive real numbers such that:
$a+b>c$
$b+c>a$
$c+a>b$
$a+b+c=2$
$a$, $b$ and $c$ might be or might not be equal in value
Show that:
$$1<ab+bc+ca-abc<\frac{28}{27}$$
From the first half of the question, I realised that $a$, $b$ and $c$ are the sides of a triangle whose perimeter is ... | I'll write up my comment as an answer. Consider the polynomial
$$
\begin{split}
f(x)&=(x-a)(x-b)(x-c)\\
&=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc\\
&=x^3-2x^2+(ab+bc+ca)x-abc.
\end{split}
$$
Then $f(1)=1-2+ab+bc+ca-abc=ab+bc+ca-abc-1$, so we need to show that
$0<f(1)\le 1/27$.
Note that $1=(a+b+c)/2=p$, the semiperimeter of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is there a pattern to the roots of continued-fraction equations with depth $n$? I've found what looks like the start of a pattern, but I don't have the tools or background to know if one really exists here. Does anyone know how to analyze this mathematical behavior and continue the pattern?
Consider the continued fract... | Let $\,f_n = \frac{p_n}{q_n}\,$ then for $\,n \ge 1\,$:
$$
\frac{p_{n+1}}{q_{n+1}}= f_{n+1} = \frac{1}{x+f_n}=\frac{1}{x+\frac{p_n}{q_n}}=\frac{q_n}{xq_n+p_n} \;\;\implies\;\;\\ \begin{cases}p_{n+1}=q_n \\ q_{n+1}=xq_n+p_n=xq_n+q_{n-1}\end{cases}
$$
It follows that $\,p_{n+1}\,$ satisfies $\,p_{n+1}=xp_n+p_{n-1}\,$ wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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how this simplification was done in the integral? I'm stuck at simplification
Here is the Integral ,
$$I=\int\frac{x^2+x+1}{\sqrt{x^2+2x+3}} \ dx$$
To solve this ,
First let's Substitute, we got
$$x^2+2x+3=t^2$$
$$\implies {x}=\sqrt{(t^2-2)}-1$$
$$\implies dx=\frac{t}{\sqrt{t^2-2}}dt$$
Putting this back into the Integr... | Substitution is not the way to go here,
Write the integral as
$$=\int\sqrt{x^2+2x+3}dx-\int\frac{x+1}{\sqrt{x^2+2x+3}}dx-\int\frac{1}{\sqrt{x^2+2x+3}}dx$$
We have three types of integrals. The middle is a straightforward substitution.
The first can be done by parts,
$$\int\sqrt{x^2+2x+3}dx=x\sqrt{x^2+2x+3}-\int\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4355890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Transformation Matrix with respect to a basis and the General Linear Group I need some help going over this problem because I'm not entirely sure if my solution is sound.
Let $$\phi:\mathbb{R^2} \rightarrow \mathbb{R^2},\begin{pmatrix} x\\y
\end{pmatrix} \mapsto \begin{pmatrix} \frac{3}{2}x- \frac{1}{2}y \\ -\frac{1}{... | c) If $T\in Z$ and $\lambda\in\Bbb R\setminus\{0\}$, then\begin{align}(\lambda T)\begin{bmatrix}\frac32&-\frac12\\-\frac12&\frac32\end{bmatrix}(\lambda T)^{-1}&=\lambda T\begin{bmatrix}\frac32&-\frac12\\-\frac12&\frac32\end{bmatrix}\lambda^{-1}T^{-1}\\&=\lambda\lambda^{-1}T\begin{bmatrix}\frac32&-\frac12\\-\frac12&\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove $\left(\sqrt{x}+\sqrt{y}\right)^2\ge 2\sqrt{2\left(x+y\right)\sqrt{xy}}$, with $x,\:y\in \mathbb{R}$ and $x\ne 0$ My thinking:
Starting with the basic fact $\left(\sqrt{x}-\sqrt{y}\right)^4\ge 0$ which is trivial.
$\left(\sqrt{x}-\sqrt{y}\right)^4\ge 0\:$
$\rightarrow \:x^2-4\sqrt{yx}\sqrt{x}+6xy-4\sqrt{xy}\sqrt{... | Your proof works, but there is a much more direct proof.
First, just write $a=\sqrt x,b=\sqrt y$ and rewrite it as $$(a+b)^2\geq 2\sqrt{(a^2+b^2)(2ab)}$$
Letting $u=a^2+b^2, v=2ab,$ use AM/GM:
$$\frac{u+v}{2}\geq\sqrt{uv}$$
That first step is important from an aesthetic point of view, as much as anything - all those s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Determine the greatest of the numbers $\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$ Determine the greatest of the numbers $$\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$$ The least common multiple of $2,3,4,5$ and $6$ is $LCM(2,3,4,5,6)=60$, so $$\sqrt2=\sqrt[60]{2^{30}}\\\sqrt[3]3=\sqrt[60]{3^{20}}\\\sqrt[4]4=\sq... | The following fills-in the remaining step in OP's approach.
Now how do we compare $2^{30},3^{20},4^{15},5^{12}$ and $6^{10}$?
*
*$3^{20} = 9^{10} \gt 8^{10}=2^{30}\,$ which excludes $\,\sqrt{2} = \sqrt[4]{4}\,$ as possible maximums;
*$3^{20} = 9^{10} \gt 6^{10}$ which excludes $\,\sqrt[6]{6}\,$ as a possible maxi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Using De Moivre's theorem to solve $(z−3+2i)^4 = z^4$ What are all the solutions to: $(z−3+2i)^4 = z^4$?
I know I have to use De Moivre's theorem which states:
$$(\cos\theta + i\sin\theta)^n=\cos\theta n + i\sin\theta n$$
| There's not realy a need to use De Moivre's Theorem to solve this.
Let a=3-2i:
\begin{align}
(z-a)^4=z^4 &\Leftrightarrow (z-a)^4 - z^4= 0 \\
& \Leftrightarrow \big((z-a)^2-z^2 \big)\big( (z-a)^2 + z^2\big) =0 \\
& \Leftrightarrow z^2 -2az +a^2 - z^2=0 \quad \vee \quad z^2-2az + a^2 +z^2=0\\
& \Leftrightarrow z=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Problem of finding values of $ a $ for which two matrices are similar Problem: Let $A=\left(\begin{array}{ccc}1 & 2 & 3 \\ 1 & 2 & 7-a^{2} \\ 2 & 2+a & 6\end{array}\right) $ , $ B=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 9\end{array}\right)$ where $ a \in \mathbb{R} $. Find all the values of $ a $ for ... | You have $\det(A)=a^3-2a^2-4a+8$. On the other hand, if $a=\pm2$, then the first two line of $A$ will be equal, and therefore $\det(A)=0$. And if you divide $\det(A)$ by $(a+2)(a-2)$, then you will get $a-2$, and therefore $\det(A)=0\iff a=\pm2$. So, since $\det(B)=0$, we only have to consider the cases $a=2$ and $a=-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove that $\frac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\frac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=2\sqrt[4]{ax}$ Prove that $$\dfrac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\dfrac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=2\sqrt[4]{ax}, a>0, x>0,x\ne a.$$ My try $$\dfrac{a-x}{\sqrt{a}-\sqrt{x}... | $$
\begin{aligned}
\frac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\frac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right) &= \frac{\left(\sqrt{a}\right)^2-\left(\sqrt{x}\right)^2}{\sqrt{a}-\sqrt{x}}-\frac{\left(\sqrt[4]{a}\right)^4+\sqrt[4]{a}\left(\sqrt[4]{x}\right)^3}{\left(\sqrt[4]{a}\right)^2+\sqrt[4]{a}\sqrt[4]{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Indefinite integral with geometric series I was trying to calculate the following integral
$$\int \frac{\text{d}x}{(1 + x^n)^n}$$
for $n > 0$.
I tried this road:
$$\int \frac{\text{d}x}{\left(x^n\left(1 + \frac{1}{x^n}\right)\right)^n} = \int \frac{\text{d}x}{x^{n^2}} \sum_{k = 0}^{+\infty} (-1)^k \left(\frac{1}{x^n}\r... | There are mistakes in the proposed derivation, since the geometric series should have been taken with an exponent $n$.
To obtain the quoted result, one can use the generalized binomial expansion
\begin{equation}
\frac{1}{(1-z)^{s}}=\sum _{k=0}^{\infty }\binom{s+k-1}{k}z^{k}
\end{equation}
to express
\begin{align}
f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving the system $\tan x + \tan y = 1$ and $\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$ How can I solve this system of trigonometric equations:
$$\tan x + \tan y = 1$$
$$\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$$
I tried to write tangent as $\sin/\cos$ and then multiply the first equation with the second one but it is ... | $$\cos x \sin y=\frac{\sqrt2}{2}$$
$$\cos x \sin y=\frac{1}{\sqrt2}$$
$\frac{1}{\sqrt2}$ can be written as $1*\frac{1}{\sqrt2}$.
This gives rise to two cases-->
*
*$\cos x=1$ and $\sin y=\frac{1}{\sqrt2}$. So, $x=0$ and $y=\frac{\pi}{4}$.
*$\cos x=\frac{1}{\sqrt2}$ and $\sin y=1$. So, $x=\frac{\pi}{4}$ and $y=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Finding bounds for a function I would like to show that
$$\frac{1}{\pi^2}<\int_{\pi/2}^\pi\frac{\sin x}{x^3}<\frac{3}{2\pi^2}.$$
We know that $\frac{\sin x}{x^3}\leq\frac{1}{x^3}$ and integrating gives $\int_{\pi/2}^\pi\sin x/x^3\leq3/(2\pi^2)$. I don't know how to make $\leq$ into $<$. On the other hand, $\frac{1}{\pi... | Just for the fun.
Without any special function, we can have a more than decent approximation of the integral using, as an approximation,
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed, more than $\large 1,400$ years ago, by Mahabhaskariya of Bhaskara I, a seventh-century ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4368515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Trying to understand the sigma expansion in an answer Iteration 2: Reassign cluster labels according to the smallest squared distances:
$\begin{array}{ccccc}\text { Data point } X_{i} & \left\|X_{i}-(0,1)\right\|^{2} & \left\|X_{i}-(4,4)\right\|^{2} & \left\|X_{i}-(6,1)\right\|^{2} & \text { Cluster Label } \\ (6,-1) &... | You have three cluster centroids $\left\{\mu_{l_{1}}, \mu_{l_{2}}, \mu_{l_{3}}\right\}=\{(0,1),(4,4),(6,1)\}$. Now you (re-)assign 6 points to one of the 3 clusters with the regard of the centroid. You choose the smallest (squared) distance between the point and the three centroids and assign the corresponding point t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Plane passing through a given point so that the tetrahedron is of the minimum possible positive volume
Find the plane passing through the point $(1,1,2)$ s. t. the volume of the tetrahedron enclosed by the plane and the coordinate axes is of the minimum possible positive volume.
$$A(x-1)+B(y-1)+C(z-2)=0\\Ax+By+Cz=A+B... | Let the normal to the plane be $N = (1, A, B) $, then the equation of the plane is
$N \cdot (r - (1, 1, 2) ) = 0 $
which simplifies to
$ (x - 1) + A (y - 1) + B (z - 2) = 0 $
Setting $y=z=0$, gives us
$ a = 1 + A + 2 B $
Setting $x = z = 0 $ gives us
$ b = 1 + \dfrac{1}{A} ( 1 + 2 B ) $
Setting $ x = y = 0$ gives us
$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In a triangle ABC, if certain areas are equal then P is its centroid Let $P$ be a point in the interior of $\triangle ABC$. Extend $AP$, $BP$, and $CP$ to meet $BC$, $AC$, and $AB$ at
$D$, $E$, and $F$, respectively. If $\triangle APF$, $\triangle BPD$, and $\triangle CPE$, have equal areas, prove that $P$ is the centr... |
Let us give name $a$ for the area of triangles with common area and $x,y,z$ for the other areas. We have (using the following result: the areas of triangles sharing a same altitude with collinear bases are in the ratio of the lengths of these bases):
$$\begin{cases}\dfrac{EA}{EC}&=&\dfrac{x}{a}&=&\dfrac{a+z}{a+y}\\ \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to compute $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n^2}}$ Trying to prove $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n^2}} = 2e^{-1/2}$
I tried using ratio-root criteria with $a_n= \frac{1}{n^n} \l... | We have
\begin{align*}
& \frac{1}{n}\left( {\prod\limits_{k = 1}^n {(n^2 + k^2 )^k } } \right)^{1/n^2 } = \frac{1}{n}\exp \left( {\frac{1}{{n^2 }}\sum\limits_{k = 1}^n {k\log (n^2 + k^2 )} } \right) \\ & = \frac{1}{n}\exp \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}\log \left( {1 + \left( {\frac{k}{n}} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4376764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving that $\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}$ when $n$ is very large This is an example from Mathematical Methods in the Physical Sciences, 3e, by Mary L. Boas.
My question is,
\begin{equation}
\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}
\end{equation}
can also be written as,
\begin... | It's simpler than that:
$$\frac{1}{n^2}-\frac{1}{(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}=\frac{2n+1}{n^2(n+1)^2}=\frac{2+1/n}{n(n+1)^2}\approx \frac{2}{n^3}$$
as for large $n$ one has $1/n\approx 0$ and $n+1\approx n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve tangent of $y=\left(\log_{a}{x}\right)^2$ and $y=-ax+2$ How can I solve the tangent point and $a$ when $f(x)=\left(\log_{a}{x}\right)^2$ is tangent to $g(x)=-ax+2$?
Although this can be solved by substituting $a=e^2$ and $x=e^{-2}$, then
$f\left(e^{-2}\right)=g\left(e^{-2}\right)$ and $f^\prime\left(e^{-2}\right)... | for real $a,x$
$$\frac{d}{dx}\left(\log_a(x)^2\right)=\frac{d}{dx}\left(-ax+2\right)$$
$$\frac{2\ln(x)}{\ln(a)^2x}=-a$$
$a\to e^t$:
$$t^2e^t=-\frac{2\ln(x)}{x}$$
$$\sqrt{t^2e^t}=\sqrt{-2\frac{\ln(x)}{x}}$$
$$te^{\frac{1}{2}t}=\pm\sqrt{-2\frac{\ln(x)}{x}}$$
$$\frac{1}{2}te^{\frac{1}{2}t}=\pm\frac{1}{2}\sqrt{-2\frac{\ln(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
From homogeneous to non-homogeneous linear recurrence relation I'm trying to do the following exercise:
Find a non-homogeneous recurrence relation for the sequence whose general term is
$$a_n = \frac{1}{2}3^n - \frac{2}{5} 7^n$$
From this expression we can obtain the roots of the characteristic polynomial $P(x)$, which... | Write the relations for two consecutive terms:
$$
\begin{align}
\begin{cases}
a_n &=\, \dfrac{1}{2}\,3^n - \dfrac{2}{5}\, 7^n
\\ a_{n+1} &=\, \dfrac{1}{2}\,3^{n+1} - \dfrac{2}{5}\, 7^{n+1} \,=\, \dfrac{3}{2}\,3^n-\dfrac{14}{5}\,7^n
\end{cases}
\end{align}
$$
Eliminate (for example) $\,7^n\,$ between the two:
$$
\requir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x$. Latest Edit
By the contributions of the writers, we finally get the closed form for the integral as:
$$\boxed{\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x =-\frac{\pi(2 n-3) ! !}{2^{n}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}}$$
I first evaluat... | Inspired by metamorphy, I do want to try his wonderful trick
$$
I_{n}=\frac{1}{4} F^{\prime} _n\left(\frac{1}{2}\right) \text { where } F_{n}(\alpha)=2 \int_{0}^{\infty} \frac{x^{2 \alpha-1}}{\left(x^{2}+1\right)^{2}} d x
$$
In my post, I had found $$ \displaystyle \int_{0}^{\infty} \frac{x^{r} d x}{\left(x^{m}+a\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 2
} |
$3$-digit numbers are formed by arranging three of the digits $1$, $2$, $3$, $4$ and $5$
Three-digit numbers are formed by arranging three of the digits $1$, $2$, $3$, $4$ and $5$. No digit may be used more than once. Find the number of three-digit numbers if there exists at least $1$ prime number in the $3$ digits.
... | Three of the five numbers in the set $S = \{1, 2, 3, 4, 5\}$ are prime numbers. Therefore, any three-digit number formed using three distinct digits of the set will contain at least one prime number. Thus, the number of three-digit numbers which contain at least one prime number in the three digits is $$5 \cdot 4 \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$f(f(x - y)) = f(x) f(y) - f(x) + f(y) - xy$ Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that
$$f(f(x - y)) = f(x) f(y) - f(x) + f(y) - xy$$for all $x,$ $y.$ Find the sum of all possible values of $f(1).$
I tried plugging in combinations of $0$'s and $1$'s, but it didn't really get me anywhere. I would appr... | Hint: try plugging $x=y$ in the formula.
Full answer: This yields $f(f(0)) = f(x)^2-x^2$. Let $c=f(f(0))$. Then $$f(x)^2 = x^2+c.$$ Therefore
$$c^2=f(f(0))^2 = f(0)^2+c = 0^2+2c=2c.$$
Thus $c$ is either $0$ or $2$.
If $c=0$, then $f(1)^2=1$ and therefore $f(1)$ is either $1$ or $-1$. As the function $f(x)=-x$ is a solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How is this differential equation solved? I have a differential equation as follows:
$$ y \cdot y'' +(y')^2 +1=0.$$
I'm interested in how to solve it. So far I have found a few solutions like $y=\sqrt{r^2-(x-k)^2}$ and $y=x+k.$ [In these, $r$ and $k$ are real constants.]
It came up while looking at a property of surfac... | You can observe that
$(y^2)’’=(2yy’)’=2(yy’’+(y’)^2)$
Thus you get
$(y^2)’’=-2 $
$(y^2)’-(y^2)’(0)=-2x$
$y^2=y^2(0)+ (y^2)’(0)x-x^2$
Set $a= y^2(0)$, $b= (y^2)’(0)$, then
$y^2=a+bx-x^2$ so that all the possible solutions are of kind
$y=+/-\sqrt{a+bx-x^2}$ for every $a,b\neq 0$
The function it makes sense only for $ \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4390550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Joint PDF of $U=XY$ and $V=\frac{X}{Y}$ Given that $X$ and $Y$ are jointly distributed with pdf:
$$f(x, y)=\left\{\begin{array}{cc}
\frac{1}{x^{2} y^{2}}, & x \geqslant 1, y \geqslant 1 \\
0 & \text { else }
\end{array}\right\}$$
Find the joint pdf of $U=XY, V=\frac{X}{Y}$
My try:
First i found the Jacobian of the tran... | $x \geq 1$ and $y \geq 1$ become $\sqrt {uv} \geq 1$ and $\sqrt {\frac u v} \geq 1$. So the limits are $\frac 1 u \leq v \leq u$ and $u \geq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Using diagonalisation of a symmetric on quadratic form Let $f (x, y) = 7x_2 + 4y_2 + 4xy$. Use the diagonalisation of a symmetric matrix to write this quadratic form in the form $λ_1u_2^2 +λ_2v_2^2$, with u and v linear combinations of x and y.
Here's what I have tried:
step 1. get the symmetric matrix
$$f (x, y) = 7x_... | You start with the quadratic form
$$
7x^2 + 4y^2 + 4xy=
\left(\begin{matrix}x&y\end{matrix}\right)A
\left(\begin{matrix}x\\y\\\end{matrix}\right)=
\left(\begin{matrix}x&y\end{matrix}\right)
\left(\begin{matrix}7&2\\2&4\\\end{matrix}\right)
\left(\begin{matrix}x\\y\\\end{matrix}\right)
$$
Write the eigenvalues decompos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4397322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Closed form for $T_n$ if $T_{n+3}=2 T_{n+2}+2 T_{n+1}-T_{n}$ Consider the recurrence relation $$T_{n+3}=2 T_{n+2}+2 T_{n+1}-T_{n}$$ with first three terms as:
$T_{1}=20, T_{2}=12, T_{3}=70$
Find the closed form expression for $T_n$.
My try:
Since it is a constant coefficient difference equation, the auxiliary equation ... |
$$T_{n}=A(-1)^{n}+B\left(\frac{3+\sqrt{5}}{2}\right)^{n}+C\left(\frac{3-\sqrt{5}}{2}\right)^{n}$$
Where the constants $A,B,C$ to be determined by values of $T_1,T_2,T_3$ but its becoming too tedious to solve. Any alternate approach?
The approach is correct, and can be made less tedious by noting that:
*
*It is easi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum of $a^4+2b^4+3c^4$ Suppose that $a, b,$ and $c$
are positive real numbers satisfying
$a+b+c=3$. Find the minimum of $$a^4+2b^4+3c^4$$
We know $f(x)=x^4$ is convex on the positive reals so by Jensen's Inequality, we have $$a^4+b^4+c^4\ge \frac{1}{27}$$
Hence $$a^4+2b^4+3c^4\ge \frac{1}{27}+b^4+2c^4$$
W... | Using Holder's inequality:
$$\left(a^4+2b^4+3c^4\right)\cdot\left(1+\frac1{\sqrt[3]2}+\frac1{\sqrt[3]3}\right)^3 \geqslant (a+b+c)^4=3^4$$
Equality is possible when $a^3:b^3:c^3 =1:\frac12:\frac13$, so this gives the minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4402274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
inverse trigonometric differentiation - 2 approaches, 2 different answers I have the following question:
$\arctan x^2+\arctan y^2=\frac{\pi}{4}$
Find $\frac{dy}{dx}$ in terms of $x$ and $y$.
I am aware we can differentiate implicitly straight away to get $\frac{2x}{1+x^4}+\frac{2y}{1+y^4}\cdot\frac{dy}{dx}=0$
then $\fr... | From $x^2 + y^2 = 1 - x^2y^2$, we get:
$$x^2 = \frac{1-y^2}{1+y^2} $$
Then,
$$\frac{1+y^2}{1+x^2} = \frac{1+y^2}{1+\frac{1-y^2}{1+y^2}} = \frac12(1+y^2)^2$$
And,
$$\frac{1+y^4}{1+x^4} = \frac{1+y^4}{1+ \left( \frac{1-y^2}{1+y^2} \right)^2} = \frac12 (1+y^2)^2$$
The two expressions are equal and there's nothing wrong wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4403733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove : $\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$ Let $a,b,c>0$ satisfy $abc=1$, prove that:
$$\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$$
My attempt:
Let $a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac... | Use this inequality:
If a, b and c are all positive, show that:
$$\frac a{b+c}+\frac b{a+c}+\frac c{b+a}\geq\frac 32$$
Solution:
Since a, b and c are positive we have a+b, b+c, and c+a are also positive. Let b+c<c+a<a+b, so that $\frac 1{b+c}\geq \frac 1{c+a} \geq \frac 1{a+b}$
Using:
$(a_1+a_2+a_3+...+a_n)(b_1+b_2+b_3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is it possible to prove algebraically that these formulas for calculating a rotated point produce the same result? The following shows 2 formulas to calculate a point $(x',y')$, which is $(x,y)$ rotated $\theta$ degrees.
$$\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\e... | Following about the study of the geometric transformations of the plane it's known that the counterclockwise rotations of center $O$ are described by the linear equations:
$$
\begin{cases}
x' = x\,\cos\theta - y\,\sin\theta \\
y' = x\,\sin\theta + y\,\cos\theta \\
\end{cases}\,.
$$
At this point we can decline two case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Pythagorean triples and primes.
Determine whether there are any right-angled triangles with integer lengths such that the
lengths of both of the sides adjacent to the right angle are primes.
This was the question I was asked by my teacher. My first answer is no, which I still believe is correct, but then she asked m... | If $a=2$ and $b\ge 2$ is prime, we need an integer $c$ with $c^2=b^2+2^2$. As $(b+1)^2=b^2+2b+1>b^2+4$, we’d need $b^2<c^2<(b+1)^2$, which is impossible.
Hence we must have odd sides $a=2r+1$ and $b=2s+1$, so
$$\tag1c^2=(2r+1)^2+(2s+1)^2
=4(r^2+r+s^2+s)+2,$$
a number that is even, but not a multiple of $4$. If $c$ is o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Fair price to play a dice game We play a game with a pot and a single die. The pot starts off empty.
*
*if the die roll is 1, 2 or 3, I put 1 pound in the pot, and the die is thrown again
*if its 4 or 5, the game finishes, and you win whatever is in the pot
*if its 6, you leave with nothing
What is the fair price... | This question is equivalent to asking the expected value of the price in the pot when the game ends.
There’s a $\frac{1}{2}\cdot \frac{1}{3}$ chance the total is 1 (1, 2, or 3 and then 4 or 5), $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{3}$ chance the total is 2 (1, 2, or 3 and then 1, 2, or 3 and then 4 or 5), and so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum{\frac{1}{(k+1)^\alpha}\frac{1}{(n+1-k)^\beta}}\le K\frac{1}{(n+1)^\alpha}$
I want to prove that
$$\sum_{k=0}^{n}{\frac{1}{(k+1)^\alpha}\frac{1}{(n+1-k)^\beta}}\le K\frac{1}{(n+1)^\alpha}$$
for all $n\ge 0$, where $1<\alpha\le\beta$ and $K$ a constant.
Everything I tried to do, I always arrived at the... | Complement to Diger's proof:
According to Diger's idea, we have
\begin{align*}
&\sum_{k=0}^{n}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\beta}}\\
\le\,& \sum_{k=0}^{n}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\alpha}}\\
\le\,& \sum_{0\le k\le n/2}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Finding the slope of line intersecting the parabola A line $y=mx+c$ intersects the parabola $y=x^2$ at points $A$ and $B$. The line $AB$ intersects the $y$-axis at point $P$. If $AP−BP=1$, then find $m^2$. where $m > 0$.
so far I know $x^2−mx−c=0,$ and $P=(0,c)$.
$x = \frac{m \pm \sqrt{m^2 + 4c}}{2}$
$A_x = \frac{m + \... | We may also work from the disposition of the points of interest toward obtaining the slope of the line. We then need to treat this question as two separate cases.
The "easier" of the two to set up geometrically occurs for $ \ c \ < \ 0 \ \ , \ $ in which $ \ P \ , \ $ the $ \ y-$intercept of the line $ \ y \ = \ mx + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find equation of the plane that this $x=\frac{1+t}{1-t}, y=\frac{1}{1-t^2}, z=\frac{1}{1+t}$ curve lies
Prove that all points of the given curve lie in one plane, and find
the equation of that plane:
$$x=\frac{1+t}{1-t}, y=\frac{1}{1-t^2}, z=\frac{1}{1+t}.$$
If the given curve lies in one plane, then
$$a\left(\frac{... | Observe that (for $t\neq\pm1$)
$$\begin{align}
x&=\frac{2}{1-t}-1\iff\frac{1/2}{1-t}=\frac{x+1}{4},\\
y&=\frac{1/2}{1-t}-\frac{1/2}{1+t},\\
z&=\frac{1}{1+t}\iff\frac{1/2}{1+t}=\frac{z}{2}.
\end{align}$$
From here
$$y=\frac14(x+1)-\frac12z\iff x-4y+2z=-1.$$
Alternatively, the curves equation is
$$\begin{pmatrix}
-1\\ 0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Quadrilateral ABCD satisfies $\overline{2AB}=\overline{AC}$, $\overline{BC}=\overline{\sqrt{3}}$, $\overline{BD}=\overline{DC}$ and $Quadrilateral ABCD is inscribed in a circle satisfies $\overline{2AB}=\overline{AC}$, $\overline{BC}=\overline{\sqrt{3}}$, $\overline{BD}=\overline{DC}$ and $<BAC=60$
I've never been good... | Let $x = \overline{AB}$ , then $\overline{AC} = 2 x$, and we are given that $\overline{BC} = \sqrt{3} $ and that $\angle BAC = 60^\circ$
Applying the law of cosines to $\triangle ABC$, we get
$ (\sqrt{3})^2 = x^2 + (2 x)^2 - 2 x (2 x) (\frac{1}{2}) $
Hence,
$ 3 = x^2 + 4 x^2 - 2 x^2 = 3 x^2 $
From which, $x = 1$. Ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Rewriting $\cos^4 x \sin^2 x $ with exponent no higher than $1$ I'm having some trouble finishing this one off.
Rewrite with exponent no higher than $1$:
$$\cos^4 x \sin^2 x$$
The answer is:
$$\frac{2 + \cos(2x) - 2\cos(4x) - \cos(6x)}{32}$$
So I started like this:
$$\cos^4 x \sin^2 x = \frac{1+\cos(2x)}{2}\frac{1+... | Since $\cos(3\theta) = 4\cos^{3}(\theta) - 3\cos(\theta)$, you can also proceed backwards:
\begin{align*}
\frac{2 + \cos(2x) - 2\cos(4x) - \cos(6x)}{32} & = \frac{2 + \cos(2x) - 2(2\cos^{2}(2x) - 1) - (4\cos^{3}(2x) - 3\cos(2x))}{32}\\\\
& = \frac{-4\cos^{3}(2x) - 4\cos^{2}(2x) + 4\cos(2x) + 4}{32}\\\\
& = \frac{-\cos^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to? If $S_n=\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to?
So, the most obvious course of action in my mind is to find a closed form for the partial summations, but alas, this task eludes me. I started doing this by hand.... | This is a simple telescoping series trick
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{1}{k}-\frac{1}{k+1} \\ $$
By choosing values of k for $k = 1, \cdots, n$ then we have
$$\begin{align}k&=1 \implies 1-\frac{1}{2} \\ k&=2 \implies \frac{1}{2} - \frac{1}{3} \\\vdots \\ k&= n-1 \implies \frac{1}{n-1}- \frac{1}{n} \\ k&=n \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to prove by induction $\frac 12 + \frac 14 + ... + \frac {1}{2^n} = 1- \frac {1}{2^n} $ I'm trying to solve this problem about mathematical induction but every time I try to solve it I end up with an incorrect answer.
The problem is as follows:
Prove by induction : $$\frac 12 + \frac 14 + ... + \frac {1}{2^n} = 1- ... | Be careful with the minus sign.
$$-\frac{1}{2^k} + \frac{1}{2^{k+1}} = \frac{-2^{k+1} + 2^k}{2^k 2^{k+1}}
= \frac{-2 + 1}{2^{k+1}} = -\frac{1}{2^{k+1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}} \ge \frac{{55}}{{12\left( {a + b + c} \right)}}$
Let $a,b,c>0,\left( {a + b + c} \right)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) =10.$ Prove that
$$\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}} \ge \frac{{55}}{{12\left( {... | It suffices to prove that
$$\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} \ge \frac{19}{12}.$$
Since the condition and the inequality are both homogeneous and symmetric, assume that $c = \max(a, b, c) = 1$.
Let $p = a + b, q = ab$. We have $0 < p \le 2$.
It suffices to prove that
$$\frac{p^2 - 2q + p}{1 + p + q}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4422755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find: $\lim\limits _{n \rightarrow \infty} n^{3} \int\limits_{-\pi}^{\pi} \frac{\cos (2 n x)}{1+x^{8}} d x$ Problem:
What is the following limit?
$\lim _{n \rightarrow \infty} n^{3} \int_{-\pi}^{\pi} \frac{\cos (2 n x)}{1+x^{8}} d x$
Attempt:
Denote $ f(x) = \frac{1}{1+x^8} $ for $ x \in [ -\pi , \pi] $ ( Note t... | Too long for a comment
Using the recommendation of @Gribouillis, integrating by part several times, using the fact that the integrand is even and denoting $\,\Big[\frac{1}{1+x^8}\Big]^{(k)}=\frac{d^k}{dx^k}\frac{1}{1+x^8}$
$$I(n)=2\int_0^\pi \frac{\cos (2 n x)}{1+x^{8}} d x=\frac{1}{n}\frac{\sin 2nx}{1+x^8}\bigg|_{x=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
If $(1+x+x^2)^n=a_0+a_1x+a_2x^2+\dots+a_{2n}x^{2n}$ then prove that $a_0^2-a_1^2+a_2^2-a_3^2+\dots+(-1)^{n-1}a_{n-1}^2=\frac12a_n(1-(-1)^na_n)$
If $(1+x+x^2)^n=a_0+a_1x+a_2x^2+\dots+a_{2n}x^{2n}$ then prove that $a_0^2-a_1^2+a_2^2-a_3^2+\dots+(-1)^{n-1}a_{n-1}^2=\frac12a_n(1-(-1)^na_n)$
My Attempt:
Replacing $x$ by $... | Your approach is absolutely the way to go. Just a tiny step is missing.
So, you let $f(x)=(1+x+x^2)^n=\sum_{k=0}^{2n}a_kx^k$ and compute the coefficient of $x^{2n}$ in $$f(x)f(-x)=f(x^2)$$ in two ways: obviously $[x^{2n}]f(x^2)=a_n$; on the other hand, $$[x^{2n}]f(x)f(-x)=\sum_{k=0}^{2n}(-1)^k a_k a_{2n-k}=\sum_{k=0}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Straight line $y = x + 2$ cuts circle at $x^{2} + y^{2} = 20$ at $A$ and $B$ determine the values of $A$ and $B$ Im currently studying circles so now that I have two equations I'm confused. I'm asked to determine the values of $A$ and $B$ given straight line $y = x+2$ that cuts circle at $x^{2} + y^{2} = 20$.
| If $(x,y)$ is in the intersection of the line $y=x+2$ and the circle $x^2+y^2=20$, then\begin{align*}x^2+(x+2)^2 & =20 \\
x^2+x^2+4x+4 & =20 \\
2x^2+4x-16 & =0 \\
x^2+2x-8 & =0 \\
(x-2)(x+4) & =0
\end{align*}hence $x=2$ or $x=-4$. This gives the points $(2,4)$ and $(-4,-2)$, because $y=x+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4433181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating $\iint_D\frac{\sin y}{y}dxdy$, where $D$ is the region enclosed by $y=2$, $y=1$, $y=x$, $y=2x$ I have been given this exercise: Calculate the double integral:
$$\iint_D\frac{\sin(y)}{y}dxdy$$
Where $D$ is the area enclosed by the lines: $y=2$, $y=1$, $y=x$, $2y=x$ (not $y = 2x$).
Visualising $D$ is easy. Y... | Note that the integrand does not depend on $x$:
$$
\int\int_D \frac{\sin y}{y} dx dy
= \int_1^2 \left( \int_{y/2}^y dx\right) \frac{\sin y}{y} dy
= \int_1^2 \frac{y}{2}\frac{\sin y}{y} dy
$$
$$
= \int_1^2 \frac{\sin y}{2} dy = \frac{\cos 1 - \cos 2}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Compute the following integral: $\lim_{n\rightarrow\infty}\int_0^{\infty}(1+\frac{x}{n})^{-n}\sin(\frac{x}{n})dx$ Question: Compute: $\lim_{n\rightarrow\infty}\int_0^{\infty}(1+\frac{x}{n})^{-n}\sin(\frac{x}{n})dx$. This is from Folland's Real Analysis book.
If we can find an integrable majorant, then the integral will... | Let
$$I_1(n) := \int_0^n (1 + x/n)^{-n} \sin \frac{x}{n} \mathrm{d} x$$
and
$$I_2(n) := \int_n^\infty (1 + x/n)^{-n} \sin \frac{x}{n} \mathrm{d} x.$$
Clearly, $I_1(n) \ge 0$.
Using $\sin u \le u$ for all $u\ge 0$, we have
$$I_1(n) \le \int_0^n (1 + x/n)^{-n} \frac{x}{n} \mathrm{d} x = \frac{n(1 - n 2^{1 - n})}{(n - 1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4438721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Evaluate $\int_0^1 \left(\{ax\}-\frac{1}{2}\right)\left(\{bx\}-\frac{1}{2}\right) dx$ for $a,b\in\mathbb{Z}^+$ Evaluate $$\int_0^1 \left(\{ax\}-\frac{1}{2}\right)\left(\{bx\}-\frac{1}{2}\right) dx$$
where $a$ and $b$ are positive integers and $\{x\}:=x-\lfloor x\rfloor$
WLOG, we may assume that $a\le b$. Via the substi... | This is a brute force approach:
$$\int_0^1 \left(\{ax\}-\frac{1}{2}\right)\left(\{bx\}-\frac{1}{2}\right) dx$$
Break the interval up into $ab$ parts to avoid the switch:
$$\sum_{k=1}^{ab}\int_{(k-1)/(ab)}^{k/(ab)} \left(\{ax\}-\frac{1}{2}\right)\left(\{bx\}-\frac{1}{2}\right) dx$$
Now $\{ax\}$ is just a straight line s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Let $z_1$ and $z_2$ be the roots of $az^2+bz+c=0;\,a,b,c\in \mathbb{C}, a\ne 0$ and $w_1,w_2$ be roots of $(a+\bar{c})z^2+(b+\bar{b})z+(\bar{a}+c)=0$ Let $z_1$ and $z_2$ be the roots of $az^2+bz+c=0;\,\,a,b,c\in \mathbb{C}, a\ne 0$ and $w_1,w_2$ be roots of $(a+\overline{c})z^2+(b+\overline{b})z+(\overline{a}+c)=0$. If... | Let $a+\overline{c} = t$ and $b+\overline{b} = 2s$ with $s$ real.
Claim: $t\neq 0\text{ and }s^2-|t|^2 <0$
proof:
(i) $t=0\implies a=-\bar{c}\implies |z_1z_2|=1$ contradiction!!
(ii) let $\frac{\overline{a}}{a}=\alpha.$
\begin{align*}
4(s^2-|t|^2)&= |a(z_1+z_2)+\overline{a(z_1+z_2)}|^2-4|a+\overline{az_1z_2}|^2\\
&=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4445610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find all solutions to the equation:$3^{\sin x} \cos x-3^{\cos x} \sin x=0$ Find all solutions to the equation:
$3^{\sin x} \cos x-3^{\cos x} \sin x=0$
My attempt:
$3^{\sin x} \cos x-3^{\cos x} \sin x=0\implies 3^{\sin x} \cos x=3^{\cos x} \sin x\implies3^{\sin x-\cos x}=\tan x.$
I have found that $x=45^{\circ}$ or $225... | Assume $x \in [0,2\pi]$. Rewrite it : $3^{\cos x}\sin x = 3^{\sin x}\cos x$. If $\sin x > 0 \implies \cos x > 0$ and the same for the case negative sign. Thus you might consider $x \in (0,\pi/2)\cup(\pi, 3\pi/2)$. For the first case, rewrite further: $\dfrac{3^{\sin x}}{\sin x}= \dfrac{3^{\cos x}}{\cos x}$. Consider $f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4446279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
My Attempt:$$y^2=\frac{x^2+2x-3}{x^2+4x+4}=z\\\implies x^2(z-1)+x(4z-2)+4z+3=0$$
Discriminant greater than equal to zero, so, $$(4z-2)^2-4(z-1)(4z+3)\ge0\\\implies z\le\frac43\\\implies -\frac2{\sqrt3}\le y\le\frac2{\... | First notice that
$${\rm Dom}(f)=\left]-\infty,-3\right]\cup \left[1,+\infty\right[$$
Now, setting
$$y=f(x)\implies y=\frac{\sqrt{(x-1)(x+3)}}{x+2}\implies y^{2}=\frac{(x-1)(x+3)}{(x+2)^{2}}$$
Then,
$$y^{2}(x+2)^{2}=(x-1)(x+3)$$
Expanding and collecting all in terms of $x$, we have
$$(1-y^{2})x^{2}+(2-4y^{2})x+(-3-4y^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
$\sum_{k=0}^\infty\frac{1}{k+1}\binom{3k+1}{k}\left(\frac{1}{2}\right)^{3k+2}$ converges to $\frac{3-\sqrt{5}}{2}$? I stumble upon the expression
$$
\sum_{k=0}^\infty
\frac{1}{k+1}
\binom{3k+1}{k}
\left( \frac{1}{2} \right)^{3k+2}
$$
and it seems to converge to
$$
\frac{3-\sqrt{5}}{2}
$$
Do they equate ? How to prove ... | Let
$$S(z)=\sum_{k=0}^\infty {{3k+1}\choose{k}}z^{k}(1-z)^{2k+1}.$$
One has
\begin{align}S(z)&=\sum_{k=0}^\infty {{3k+1}\choose{k}}z^{k}\sum_{l=0}^{2k+1}(-1)^l{{2k+1}\choose{l}}z^l \\
&=\sum_{n=0}^\infty\sum_{k+l=n}(-1)^l{{3k+1}\choose{k}}{{2k+1}\choose{l}}z^n\\
&=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^{n-k}{{3k+1}\choose{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Asymptotic for $\sum_{k=1}^n k^n$ Consider the OEIS sequence A031971, which is defined as:
$$a_n=\sum\limits_{k=1}^n k^n\quad\color{gray}{(1,\,5,\,36,\,354,\,4425,\,67171,\,1200304,\,.\!.\!.\!)}\tag{1}$$ I'm interested in the asymptotic behavior of $a_n$ for $n\to\infty$.
Empirically, it appears that
$$a_n\stackrel{\co... | Let us write
$$
\sum\limits_{k = 1}^n {k^n } = n^n \sum\limits_{k = 0}^{n - 1} {\left( {1 - \frac{k}{n}} \right)^n } .
$$
Now by power series expansions
\begin{align*}
\left( {1 - \frac{k}{n}} \right)^n & = \exp \left( {n\log \left( {1 - \frac{k}{n}} \right)} \right) = \exp \left( { - n\sum\limits_{j = 1}^\infty {\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Solutions of PDE I am looking for solutions $g \colon \mathbb R^2 \times \mathbb R^2 \to \mathbb C$ of the PDE
$$\partial_{x_1} g(x,y) + \partial_{y_1} g(x,y) = a(y_1+x_1+i(x_2-y_2))g(x,y), \\ \partial_{x_2} g(x,y) + \partial_{y_2} g(x,y) = a(y_2+x_2+i(y_1-x_1))g(x,y),$$
where $a>0$ and $(x,y) = (x_1,x_2,y_1,y_2) \in \... | As the system of PDEs strongly decouples, with one equation only dependent on the $(x_{1}, y_{1})$ coordinates while the other is only dependent on the $(x_{2}, y_{2})$ coordinates, it makes sense to solve the problems separately.
For the first PDE, the method of characteristics implies (labelling $g \to g_{1}$)
$$\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $f(x)=\sqrt{1+\sqrt{x+\sqrt{x^2 +\sqrt{x^3 +\cdots}}}},$ then find the value of $f(4).$ If $f(x)=\sqrt{1+\sqrt{x+\sqrt{x^2 +\sqrt{x^3 +\cdots}}}},$ then find the value of $f(4).$
My attempt:
$f(x) = \sqrt{1+{\sqrt{x}} f(x)} \implies f(4) =\sqrt{1+2f(4)} \implies
f(4)(f(4)-2)=1.$
I don't know how to proceed from her... | For $y>0$,
\begin{eqnarray}
1+y & = & \sqrt{y^2+(2y+1)}\\
& = & \sqrt{y^2+\sqrt{(2y+1)^2}}\\
& = & \sqrt{y^2+\sqrt{4y^2+(4y+1)}}\\
& = & \sqrt{y^2+\sqrt{4y^2+\sqrt{(4y+1)^2}}}\\
& = & \sqrt{y^2+\sqrt{4y^2+\sqrt{16y^2+(8y+1)}}}\\
& \vdots & \\
& = & \sqrt{y^2+\sqrt{4y^2+\sqrt{4^2y^2+\sqrt{4^3y^2+\sqrt{4^4y^2+\cdots}}}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality involving sums with binomial coefficient I am trying to show upper- and lower-bounds on
$$\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i)$$
(where $n\geq 1$) in order to show that it basically grows as $\Theta(n)$.
The upper-bound is easy to get since $\min(i, n-i)\leq i$ for $i\in\{0, \dots n\}$ so that
$... | We start with
$$
\sum_{k=0}^{\left\lfloor\frac{n}2\right\rfloor}\binom{n}{k}
=\left\{\begin{array}{}
2^{n-1}&\text{if $n$ is odd}\\
2^{n-1}+\frac12\left(\raise{2pt}{n}\atop\frac{n}2\right)&\text{if $n$ is even}
\end{array}\right.\tag1
$$
Substitute $n\mapsto n-1$:
$$
\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate $\int_{0}^{\infty} \frac{1}{x^{n}+1} d x?$ I first investigate the integral
$$\int_{0}^{\infty} \frac{1}{x^{6}+1} d x$$
using contour integration along the semicircle $\gamma=\gamma_{1} \cup \gamma_{2},$
$\textrm{ where }$ $$ \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} ... | Let $f(z) = \frac{1}{z^n + 1}$.
The trick is to pick a different contour. We choose the ray parametrised by $z(t) = e^{2 \pi i /n} t$ for $t \in [0, \infty)$.
We then have, after taking a bit of care with limits, $\int\limits_0^\infty f(t) dt - e^{2 \pi i / n} \int\limits_0^\infty f(t) dt = 2 \pi i Res(f(z), z = e^{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$\frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7}+\cdots+\frac{500^2}{999\cdot1001} = ?$ I found this problem in a high school text book.
Let $ \displaystyle s = \frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7}+\cdots+\frac{500^2}{999\cdot1001}$. Find $s$.
How I tried:
Observe that $T_n... | I don't think your $T_n$ is correct. Here's how I broke it down:
$$\frac{1^2}{1.3} + \frac{2^2}{3.5} + \frac{3^2}{5.7}+ ... = \frac{1^2}{1 + 0.1 \cdot 3} + \frac{2^2}{3 +0.1 \cdot 5} + \frac{3^2}{5+ 0.1 \cdot 7} + ...
\\ = \sum_{n=1}^N \frac{n^2}{(2n-1)+0.1(2n+1)}
\\ = \sum_{n=1}^N \frac{10n^2}{10(2n-1)+(2n+1)}
\\ = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why is the triangle inequality equivalent to $a^4+b^4+c^4\leq 2(a^2b^2+b^2c^2+c^2a^2)$? Consider the existential problem of a triangle with side lengths $a,b,c\geq0$. Such a triangle exists if and only if the three triangle inequalities
$$a+b\geq c,\quad b+c\geq a\quad\text{and}\quad c+a\geq b\tag{0}$$
are all satisfie... | For non-negative $a,b,c$:
$$a+b\geq c; a+c\geq b;b+c\geq a \Leftrightarrow |b-a|\leq c \leq b+a\Leftrightarrow\\
(b-a)^2\leq c^2 \leq (b+a)^2 \Leftrightarrow (c^2-(b+a)^2)(c^2-(b-a)^2)\leq 0 \Leftrightarrow\\
c^4-((b+a)^2+(b-a)^2)c^2+(b^2-a^2)^2\leq 0\Leftrightarrow
c^4+b^4+a^4\leq 2(a^2b^2+a^2c^2+b^2c^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove or disprove $\sum\limits_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18$ for $\sum\limits_{i=1}^n x_i = \frac12$($x_i\ge 0, \forall i$)
Problem 1: Let $x_i \ge 0, \, i=1, 2, \cdots, n$ with $\sum_{i=1}^n x_i = \frac12$. Prove or disprove that
$$\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac1... | Partial proof with hint :
Let us consider the inequality for $x,y>0$ and $x,y\leq 0.5$ :
$$l\left(x,y\right)=g\left(2xy+\frac{4}{3}xy\left(x-y\right)^{2}\right)-\frac{x^2y^2}{1-x^2-y^2}\ge 0$$
Where $g\left(x\right)=\frac{\frac{x^{2}}{4}}{1-x}$
As $g$ is convex on $[0,0.5]$ we use Jensen-Mercer inequality
We need to s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
What is $\sum_{k = 1}^n (k \log k)\binom{n}{k}$? If the exact answer is difficult to find, what is the tightest asymptotic upper bound? While trying to solve the complexity of my program I came across the the following summation:
$$\sum_{k = 1}^n (k \log k)\binom{n}{k}$$
Could you please provide a solution to this sum.... | a) rewriting the sum
First of all we have better to rewrite the sum as
$$
\eqalign{
& S(n) = \sum\limits_{k = 1}^n {k\ln k\left( \matrix{
n \cr k \cr} \right)} = \sum\limits_{k = 1}^n {\ln \left( {k^k } \right)\left( \matrix{
n \cr k \cr} \right)} = \sum\limits_{k = 0}^n {\ln \left( {k^k } \right)\left( \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Proving contour integral equal to zero Let $G$ be the path traversed once as shown:
Show that $\displaystyle{\int_{G}{\dfrac{1}{v^4-1} \text{d}v} = 0}$.
By partial fraction decomposition,
$\dfrac{1}{v^4 -1} = \dfrac{1}{4} \left( \dfrac{1}{v-1} - \dfrac{1}{v+1} + \dfrac{i}{v-i} - \dfrac{i}{v+i} \right)$
The singular p... | You could do it easier by using the special symmetry your function and your path have: They both behave nicely under rotations of $e^{i\frac{\pi}{2}}$. For the path it is evident that $C_k = e^{k\cdot i\frac{\pi}{2}}C_1$, and for the function it is the same:
\begin{align}f(e^{i\frac{\pi}{2}}z) &= \frac{1}{(e^{i\frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4465771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Distribution of random variable $Y = (X + 1)^2$. Could you please help to solve the following problem:
The random variable $X$ has a uniform distribution on the segment $[0,2]$. Find the density of the distribution of a random variable $Y = (X + 1)^2$.
My attempt (but it is obviously wrong):
$$ f(x) = \begin{cases}
(... | $$ P (Y \leq u) = P((X+1)^2 \leq u) = P(X \leq \sqrt{u}-1)$$
Thus, using $F$ for the cummulative distribution function we have
$$F_Y (u) = F_x (\sqrt{u}-1)$$
The density $f$ is a derivative of $F$:
$$f_Y(u) = \frac{d}{du} F_X(\sqrt{u}-1) = F_X^\prime (\sqrt{u}-1) \frac{1}{2\sqrt{u}}$$
$$ = \frac{f_x\left(\sqrt{u}-1 \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$? I first wrote $y$ in terms of $x$. in $5x + 3y = 100$, I subtracted $5x$ from both sides to get $3y = 100 - 5x$. Therefore, $y = \frac{100 - 5x}{3}$. I substituted this into $xy$ to get $x(\frac{100 - 5x}{3}$) – You... | A not very strict approach using a rather standard geometric reuslt is as follows.
$5x+3y=100$ depicts a line on the plane, specifically the $y=-\frac{5}{3}x-\frac{100}{3},$ which intersects the $y-$axis on $A=(0,33,333)$ and the $x-$axis on $B=(20,0)$.
Thus the right triangle $\triangle AOB$ is formed, where $O=(0,0)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Product of $n$ terms of sequence where the $n^{th}$ term is of the form $(x^{a^n}+1)$ While practicing from a book I found a product in the form $$(x^{a^1}+1)\cdot(x^{a^2}+1)\cdot(x^{a^3}+1)\cdot(x^{a^4}+1)$$ and was immediately curious if I could a formula to solve the product for $n$ terms, that is, a single formula ... | Using $[n]=\{1,2,\ldots,n\}$ we can write the product as
\begin{align*}
\prod_{j=1}^n\left(x^{a^j}+1\right)=\sum_{S\subseteq [n]}x^{\sum_{j\in S}a^j}
\end{align*}
where $S$ runs over all subsets of $[n]$. A special case is $S=\emptyset$, which gives the empty sum. The empty sum is zero per definition, resulting in $x^0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to show that $\forall n \geq 1$, $\frac{n^{2n}}{n!^2} \geq (\frac{n+1}{n})^{n^2-n}$? I tried using the fact that $(n+1)/n \leq 2$ and thus $\frac{n^{2n}}{n!^2} \geq 2^{n^2-n}$ but this does not seem to be true.
| We can proceed by induction on $n$. The case $n=1$ is easy to check. Suppose that the inequality holds for $1,2,\ldots,n$. Then
\begin{align*}
\frac{{(n + 1)^{2n + 2} }}{{(n + 1)!^2 }} = \frac{{(n + 1)^{2n} }}{{n!^2 }} &= \frac{{n^{2n} }}{{n!^2 }}\left( {\frac{{n + 1}}{n}} \right)^{2n} \ge \left( {\frac{{n + 1}}{n}} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve $2^x \cdot 3^y = 1 + 5^z$ in positive integers
Solve $2^x \cdot 3^y = 1 + 5^z$ in positive integers.
I think a useful idea for tackling this question is considering congruences. If we consider the equation modulo 4, we get that the RHS is congruent to 2, so $x=1.$ If $y > 1,$ the LHS is congruent to 0 modulo 9.... | If $z=6k+3,$ then $5^z+1$ is divisible by $7,$ so it cannot yield a solution.
That finishes your solution, since it means that $x=y=1$ are the only solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the smallest $n\in \mathbb Z^+$ that makes $\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$ integer. Find the smallest $n\in \mathbb Z^+$ that makes $\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$
Clearly if $n=0$ then we will have $20$ but I couldn't decide that how can I find the other integers. Any hint?
If I say that $x=\sqrt{... | Alternative approach:
You want
$$M^2 = \left(\sqrt{100 + \sqrt{n}} + \sqrt{100 - \sqrt{n}}\right)^2 = 200 + 2\sqrt{10000 - n}.$$
To minimize $n$, you need $M^2$ as large as possible,
and still less than $200 + 2\sqrt{10000} = 400$.
The largest such square is
$19^2 = 361 \implies 2\sqrt{10000 - n} = 161 \implies $
$\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$. $$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$$
Edit : $D> 0$.
My work:
Let $x = D\tan \theta$
$$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}dx$$
$$=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2... | Here is another method to solve. Let
\begin{eqnarray}
I&=&\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx=\frac{2}{D^3}\int_{0}^\infty \frac{1}{\bigg[\big(\frac{x}{D}\big)^2 + 1)\bigg]^2}d(\frac xD)\\
&\stackrel{x/D\to x}{=}&\frac{2}{D^3}\int_{0}^\infty \frac{1}{(x^2 + 1)^2}dx\tag1\\
&\stackrel{1/x\to x}{=}&\frac{2}{D^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4477389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Where is my mistake in the computation of $\int \frac{1}{\sin x} \text{d}x$ Here is my attempt:
$$\quad \ \sin x = \frac{2 \tan(x/2)}{1 + \tan^2 (x/2)} \\
\implies \int \frac{\text{d}x}{\sin x}=\int \frac{1+\tan^2(x/2)}{2 \tan(x/2)} \ \text{d}x$$
I do the following variable change: $u = \tan (x/2)$
$$\quad \ \frac{\tex... | The derivative is precisely $$\frac{d}{dx}\bigg(\ln\big(\tan(\frac{x}{2})\big)\bigg)=\frac{1}{2}\cdot \frac{\sec^{2}(\frac{x}{2})}{\tan(\frac{x}{2})}=\frac{1}{2\cdot\sin(\frac{x}{2})\cos(\frac{x}{2})}=\frac{1}{\sin(x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4477985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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About a Conjecture: $-\left(\frac{x^{n}+1}{x^{n-1}+1}\right)^{n}-\left(\frac{x+1}{2}\right)^{n}+\left(x^{\frac{x}{x+1}}+\sqrt{x}-1\right)^{n}+1\leq 0$ Hi it's a conjecture wich refine for $0< x\leq 1$ the inequality Refinement of a famous inequality :
Problem/Conjecture
Let $0<x\leq 1$ then for $n\geq 3$ a natural numb... | Not an answer just a useful remark :
We can split the problem because it seems we have :
Let $\exists x\in(0,1]$ and $\exists n\geq 3$ a natural number such that
$$\frac{\left(1+\frac{1}{2^{n}}+\frac{1}{n^{2}}-nx\right)\left(x^{n}+1\right)^{n}}{\left(x^{\left(n-1\right)}+1\right)^{n}}+nx-\frac{1}{n^{2}}-\frac{1}{2^{n}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Why can't we solve $\lim\limits_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}}$ through dividing by xy in both numerator and denominator? When solving this limit, we use methods like $\epsilon$ - $\delta$ definition and polar coordinate convertion. And I wonder if we can solve it by dividing by $xy$ in both numerator ... | You can actually divide by $x$ and $y$, with a grain of salt:$\def\sgn{\operatorname{sgn}}$
$$\begin{align}
f(x,y) = \frac{xy}{\sqrt{x^2 + y^2}}
&= \frac{xy}{|x||y|\sqrt{\frac1{x^2} + \frac1{y^2}}} \\
&= \frac{\sgn(x)\sgn(y)}{\sqrt{\frac{1}{y^2} + \frac{1}{x^2}}} \tag 1
\end{align}$$
which is due to $\sqrt {x^2}=|x|$.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Convergence of $\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$ How can I prove the following sequence converges?
$$\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$$
I tried everything.
Could not find any candidates for comparison test, and failed to find an upper bound.
Any hints will be app... | $$\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^2}\left(\sum_{i=1}^n \frac{1}{\sqrt{i}}\right) &= \sum_{i=1}^{\infty} \frac{1}{\sqrt{i}}\left(\sum_{n=i}^{\infty} \frac{1}{n^2}\right) \\
&< \frac{\pi^2}{6} + \sum_{i=2}^{\infty}\frac{1}{\sqrt{i}}\left(\sum_{n=i}^{\infty} \frac{1}{n(n-1)}\right) \\
&=\frac{\pi^2}{6} + \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Proving $(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$ with various solutions.
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
Solutions in the answers.
$\ \\ \ \\ \ \\ \ \\$
Edit) Since this question is closed, I'll add more contexts for this question.
This identity is called "Brahmagupta-Fibonacci identity", which the comment... | Simpler Solution.
\begin{align} & (ac+bd)^2+(ad-bc)^2 \\ = \; & (ac)^2+2abcd+(bd)^2+(ad)^2-2abcd+(bc)^2 \\ = \; & a^2c^2+b^2d^2+a^2d^2+b^2c^2 \\ = \; & (a^2+b^2)(c^2+d^2) \end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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Integrate $\int\frac{3x}{x^5+x^4+1}dx$ I have an integral which I solved. But, I am not sure whether my answer is right or not. The integral is $$\int\frac{3x}{x^5+x^4+1}dx$$ My answer $$3\left(-\dfrac{\displaystyle\sum_{\left\{Z:\>Z^3-Z+1=0\right\}}\frac{\left(2Z^2-3Z-1\right)\ln\left(\left|x-Z\right|\right)}{3Z^2-1}}... | $x^3-x+1= (x-a)(x^2+ax-a^{-1}) $ has a single real root $a=-1.3247$, or
$$a =-\sqrt[3]{\frac{1}2+\frac16\sqrt{\frac{23}3}}-\sqrt[3]{\frac{1}2-\frac16\sqrt{\frac{23}3}}
$$
Then
\begin{align}
&\int\frac{3x}{x^5+x^4+1}dx\\
=& \ \frac17\int
\frac{3(2x-1)}{x^2+x+1}-\frac{2(3x^2-1)}{x^3-x+1} + \frac{9x+1}{(x-a)(x^2+ax-a^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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What is wrong with my proof that $\int 2x dx= 2x^2$ by writing $2x=\underbrace{2+2+\cdots+2}_{x\;\text{times}}$? I know $\int 2x \,dx = x^2 + C$ (by the power rule) but why does the following proof not give the same answer?
\begin{align*}
\int 2x \,dx &= \int \underbrace{(2 + 2 + 2 + \dots + 2)}_{x \text{ times}} \, dx... | An integral might run from $x=0$ to $x=X$.
In $\int2xdx$, the function is increasing as you go along. So the first $2$ is there all the way as x goes from 1 to X, but the second $2$ doesn't start until x is 2, and the final $2$ doesn't appear until x has already reached X.
In $\int2dx+\int2dx+...$, all the $2$s are th... | {
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"url": "https://math.stackexchange.com/questions/4482632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Limit of sequence given by $x_{n} = \frac{x_{n-2}+x_{n-3}}{2}$? Let $x_1,x_2,x_3\in\mathbb{R}$ be three distinct real numbers. I am interested in the convergence of the sequence
$$
x_{n} = \frac{x_{n-2}+x_{n-3}}{2},\quad n = 4,5,\ldots
$$
ie,
$$
x_{4} = \frac{x_{2}+x_{1}}{2},\quad x_{5} = \frac{x_{3}+x_{2}}{2},\quad x_... | The characteristic equation of the linear recurrence is
$$\lambda^3 - \frac{1}{2}\lambda-\frac{1}{2} = 0$$ with roots $\lambda_1=1$, $\lambda_{2,3} = \frac{1}{\sqrt{2}}\cdot(-\frac{1}{\sqrt{2}} \pm \frac{i}{\sqrt{2}})$. There exist unique $a$, $b$, $c$ such that for all $n \ge 0$ we have
$$x_n = a \cdot 1^n + b \cdot ... | {
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"source": "stackexchange",
"question_score": "6",
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Proving $\int_0^\infty \frac{x^k}{(x^2 + b^2)^l}\,dx=\frac1{2b^{2l-k-1}}\Gamma(\frac{k+1}2)\Gamma(\frac{2l-k-1}{2})/\Gamma(l)$ . My quantum mechanics textbook states the following integral without proof:
$$
\int_0^\infty \frac{x^k}{(x^2 + b^2)^l} \mathrm dx = \frac1{2b^{2l-k-1}}\frac{\Gamma(\frac{k+1}2)\Gamma(\frac{2l-... | Letting $x=b \tan \theta, \quad$ then $d x=b \sec ^{2} \theta d \theta$ and
$$
I=\int_{0}^{\frac{\pi}{2}} \frac{b^{k} \tan ^{k} \theta}{b^{2 l} \sec ^{2 l} \theta} \cdot b \sec ^{2} \theta d \theta =\frac{1}{b^{2 l-k-1}} \int_{0}^{\frac{\pi}{2}} \sin ^{k} \theta \cos ^{2 l-k-2} \theta d \theta
$$
Using $$
B(x, y)=2 \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find the sum of constants given the following systems of equations? Given that $q,r,s$, and $t$ are different constant values in the following systems of equations containing $a,b,c$, and $d$. Find the sum $q+r+s+t$.
$\frac{1}{qa+1} + \frac{1}{qb+1} + \frac{1}{qc+1}+ \frac{1}{qd+1} = 1$
$\frac{1}{ra+1} + \frac{1... | Hint: $\,$ the equations are of the form:
$$
\frac{1}{x+a} + \frac{1}{x+b} + \frac{1}{x+c}+ \frac{1}{x+d} = \frac{1}{x} \quad\quad\text{for}\;\; x = \frac{1}{q}, \frac{1}{r}, \frac{1}{s}, \frac{1}{t} \tag{1}
$$
Let $P(x) = (x+a)(x+b)(x+c)(x+d)$ then $(1)$ can be written as:
$$
P(x) - x P'(x) = 0 \quad\quad\text{for}\;\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Which interval is correct here$?$ The equation $$2\textrm{sin}^2\theta x^2-3\textrm{sin}\theta x+1=0$$ where $\theta \in \left(\frac{\pi}{4},\frac{\pi}{2}\right)$ has one root lying in the interval
$(0,1)$
$(1,2)$
$(2,3)$
$(-1,0)$
I know that if $f(a)$ and $f(b)$ are of opposite signs then at least $1$ or in ge... | $\begin{align}
2\sin^2 \theta x^2 − 3 \sin \theta x +1 & = 0 \\
x & = \frac{3 \sin \theta \pm \sqrt{9 \sin^2 \theta - 8 \sin^2 \theta}}{4 \sin^2 \theta} \\
x & = \frac{3 \pm 1}{4 \sin \theta} \\
x & = \frac{1}{\sin\theta} \; \; \text{or} \; \; x = \frac{1}{2\sin\theta}
\end{align}$
$$ \therefore \, \theta = (2n+1)\pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4495411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Computing the value of $(x+y)^4$ if $x^4+y^4=5$ and $x^2+xy+y^2=10$
Let $x^4+y^4=5$ and $x^2+xy+y^2=10.$ Find $(x+y)^4.$
First, I tried expanding $(x+y)^4$ using the binomial theorem to get $5+4x^3y+6x^2y^2+4xy^3,$ so simplifying I got $5+4xy(x^2+y^2)+6(xy)^2.$ Then I rearranged the given equation to get $x^2+y^2=10-... | An alternative approach is to rearrange the second equality and square:
$$
x^2 + y^2 = 10 - xy \implies \underbrace{x^4 + y^4}_{=\, 5} + 2 x^2y^2 = 100 - 20 xy +x^2 y^2 \implies x^2y^2 + 20 xy = 95
$$
The last quadratic could be solved for $\,xy\,$, then $\,x,y\,$ determined and $\,(x+y)^4\,$ calculated by brute force.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Computing the integral of trig function under square root How can we solve this integral
$$\int \sqrt{\csc^2x -2}
\mathrm{d}x$$
My idea was substituting $\csc^2x=2\csc^2\theta$. Then the integral became $$\sqrt{2}\int \frac{\csc^2\theta-1}{\sqrt{2\csc^2\theta-1}} \mathrm{d}\theta$$ after a few simplifications. I don'... | Letting $u=\sqrt{\csc ^{2} x-2}$ transforms the integral into
$$\begin{aligned}
I&=-\frac{1}{2} \underbrace{\int \frac{u^{2} d u}{\left(u^{2}+2\right) \sqrt{u^{2}+1}}}_{J}\end{aligned}
$$
For the integral $J$, letting $t=\frac{u}{\sqrt{u^{2}+1}} $, we have $$ u^{2}=\frac{t^{2}}{1-t^{2}} \Rightarrow u d u=\frac{t d t}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Solve $\sin{(70^{\circ}-x)}=\frac{1}2 + \frac{1}{4\cos{(20^{\circ})}}$ $$\text{Solve} \ \ \sin{(70^{\circ}-x)}=\frac{1}2 + \frac{1}{4\cos{(20^{\circ})}}$$
I came across this equation while solving a problem. The computer provides $x=20^{\circ}$, and I tried to draw a shape with that in mind:
Then I got the equation $\... | $$\sin{(70^{\circ}-x)}=1/2 + \frac{1}{4\cos{(20^{\circ})}}$$
Start from RHS:
$$\begin{align}
\frac{1}{2} + \frac{1}{4\cos{(20^{\circ})}}&=\frac{2\cos(20^{\circ})+1}{4\cos(20^{\circ})}=\frac{2\sin(20^{\circ})\cos(20^{\circ})+\sin(20^{\circ})}{4\sin(20^{\circ})\cos(20^{\circ})}\\
\\
&=\frac{\sin(40^{\circ})+\sin(20^{\cir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is... | $\begin{align}p(x)&=x^{20}+x^{19}+\dots+x^{10}\\&=x^{17}(x^3+x)+x^{16}(x^3+x)+x^{13}(x^3+x)+x^{12}(x^3+x)+x^9(x^3+x)+\color{red}{x^{11}}\end{align}$
Again $x^{11}=x^8(x^3+x) -x^6(x^3+x) +x^4(x^3+x) -x^2(x^3+x) +(x^3+x) -x$
Hence remainder is $=-x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 3
} |
The least possible number of tokens on a $n×n$ board Tokens are placed on the squares of a
$2021
×
2021$
board in such a way that each square contains at most one token. The token set of a square of the board is the collection of all tokens which are in the same row or column as this square. (A token belongs to the tok... | On an $N\times N$ board, we can do this with $\frac32N$ tokens if $N$ is even: fill the main diagonal with tokens, and a parallel sub-diagonal of length $\frac12N$. And if $N$ is odd, we only need $\frac12(3N-1)$ tokens, because the parallel sub-diagonal can be of length $\frac12(N-1)$. For example ($N=10$ and $N=11$):... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$
Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$ .
Here is a solution by someone:
\begin{align*} f(x)&=(1+\sqrt{x})^{2n+2}=\sum_{k=0}^{2n+2}\binom{2n+2}{k}x^{\frac{k}{2}}\\ &=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\sum_{j=0}^{\infty}\binom{\frac{k}{2}}{j}(x-1)^j\\ &=\sum_{j=0}^... | This is just a supplement to the nice answer of @Cathedral. Here we close a gap and show
\begin{align*}
\color{blue}{\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{1}{2r+1}=\frac{(2r)!!}{(2r+1)!!}}\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{r=0}^n}&\color{blue}{(-1)^r\binom{n}{r}\frac{1}{2r+1}}\\
&=\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Proving $\sum_{j=1}^{n+1} {2n+2 \choose 2j-1} {j-1/2 \choose n}=2(n+1)^2$ I have given a partial answer to a very interesting recent post Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$, where it remains to prove that
$$\sum_{j=1}^{n+1} {2n+2 \choose 2j-1} {j-1/2 \choose n}=2(n+1)^2.$$
Wolfarm Mathematica supports this... | We seek to prove that
$$\sum_{j=1}^{n+1} {2n+2\choose 2j-1} {j-1/2\choose n}
= 2(n+1)^2.$$
The LHS is
$$\sum_{j=0}^{n} {2n+2\choose 2j+1} {j+1/2\choose n}
= \sum_{j=0}^{2n+2} {2n+2\choose j} {j/2\choose n}
\frac{1-(-1)^j}{2}.$$
The first piece $A$ is
$$\frac{1}{2}
\sum_{j=0}^{2n+2} {2n+2\choose j} {j/2\choose n}
= \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.