Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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I made an inequality to solve myself, but someone pointed out my solution is wrong. I made the following inequality for myself to solve, but my friend found out an mistake:
If $a,b,c\in\mathbb{R^{+}}$ and $abc=1$, prove that $$\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}\ge\frac32$$
I tried to substitute $a=\frac xy, b=... | By C-S $$\sum_{cyc}\frac{a}{b+1}=\sum_{cyc}\frac{a^2}{ab+a}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+a)}.$$
Thus, it's enough to prove that
$$2(a+b+c)^2\geq3\sum_{cyc}(ab+a)$$ or
$$\sum_{cyc}(2a^2+ab)\geq3(a+b+c).$$
Now, by AM-GM $$\sum_{cyc}ab\geq3\sqrt[3]{a^2b^2c^2}=3$$ and by C-S again
$$a^2+b^2+c^2=\frac{1}{3}(1+1+... | {
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"url": "https://math.stackexchange.com/questions/3383742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<2$ Prove that $$\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<2 \tag{1} $$ $\forall$ $n \gt 1$
I tried using Induction:
For th... | By the generalized binomial formula, for $x\ne0$,
$$(1+x)^{1/n}+(1-x)^{1/n}
\\=2+\frac2n\left(\frac1n-1\right)\frac{x^2}2+\frac2n\left(\frac1n-1\right)\left(\frac1n-2\right)\left(\frac1n-3\right)\frac{x^4}{4!}+\cdots\\<2.$$
That's all folks.
Alternatively, the first derivative of $(1+x)^{1/n}+(1-x)^{1/n}$ is odd and m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3383865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Determine $x+y$ where $x$ and $y$ are real numbers such that $(2x+1)^2+y^2+(y-2x)^2=\frac{1}{3}$ Determine $x+y$ where $x$ and $y$ are real numbers such that $(2x+1)^2+y^2+(y-2x)^2=\frac{1}{3}$
I used the quadratic equation to get $$x=\frac{y-1\pm\sqrt{-2y-3y^2-\frac{5}{3}}}{4}$$
But I don’t see how that helps, hints a... | Multiply out and collect terms $$8x^2+4x-4xy+2y^2+\frac{2}{3}=0.$$
Multiply by $6$ and complete the square for $x$ terms $$3(4x+1-y)^2+9y^2+6y+1=0$$
$$3(4x+1-y)^2+(3y+1)^2=0$$
Therefore $x=y=-\frac{1}{3}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In how many ways can 10 blankets be given to 3 beggars such that each recieves at least one blanket? The question was to find the number of ways in which 10 identical blankets can be given to 3 beggars such that each receives at least 1 blanket. So I thought about trying the multinomial theorem...this is the first time... | While generating functions are all well and good to understand, it is unnecessary here and a more direct approach is usually preferred.
The technique of stars and bars leads to a direct solution of $\binom{10-1}{3-1}=\binom{9}{2}=\binom{9}{7}$ outcomes where each person receives at least one item.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $x^2 + y^2 = z^2$, then $ xyz \equiv 0\pmod{60} $ (Pythagorean triples)
Prove that, if $x$, $y$ and $z$ are integers, and if $x^2 + y^2 = z^2$, then $ xyz \equiv 0\pmod{60} $.
Attempt:
$ xyz \equiv 0\pmod{60} \iff 60\mid xyz$. Now notice that $60 = 3 \cdot 4 \cdot 5 $ so the statement $3 \cdot 4 \cdot 5|xyz $ is e... | The easiest way to do this is to note:
If $m \equiv -1,0,1 \pmod 3$ then $m^2 \equiv 0,1\pmod 3$. If none of them $x, y$ or $z$ are divisible by $3$ then $x^2,y^2,z^2 \equiv 1 \pmod 3$ and $1+1\equiv 1\pmod 3$ is a contradiction. So at least one of $x,y,z$ is divisible by $3$.
Similarly if $m\equiv -2,-1,0,1,2 \pmod ... | {
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"timestamp": "2023-03-29T00:00:00",
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Why doesn't this converge to $\pi$? $\lim_{n\to\infty}\frac12\sqrt{2-2\cos( \frac{2 \pi}{n} )} \times n$ Suppose a regular polygon with $n$ side has radius (line from center to point that connect sides *I don't know how to call it) of length $r$.
From cosine law, the side would has length of $$\sqrt{2r^2-2r^2\cos( \fr... | The limit set up is correct, indeed by standard limit we have that
$$\frac{ \sqrt{ 2-2cos( \frac{2 \pi}{n} )}n }{2}=\sqrt 2 \pi \sqrt{ \frac{1-cos( \frac{2 \pi}{n} )}{\left( \frac {2\pi} n\right)^2} }\to \pi$$
therefore the problem could be a numerical issue with desmos.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Solve $0.9^n+0.8^n \leq 0.1$ For students who studied the logarithmic function, it is easy to solve the equation
$$0.9^n \leq 0.1$$ in $\mathbb{N}$, which has as solutions $n\geq \frac{\ln0.1}{\ln 0.9} \approx 21.85 $. That is all natural numbers starting from
$22$.
Now how can we solve the following equation
$$0.9^n... | Ask a computer
Asking a numerical solver to find $n$ such that $0.8^n + 0.9^n = 0.1$, yields $n = 22.5020{\dots}$. Checking that the function decreases as $n$ increases, the solution set is $n \geq 23$.
OP has stated they do not want to use this method. But at least we know what answer we should be getting.
Basic Ans... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Show that $(1+x)^{(1+x)}>e^x$ How can I prove that $(1+x)^{(1+x)}>e^x$ for all $x>0$?
The problem arose as I tried to prove the well-known & intuitive econometric principle that the more often you compound your interest, the more interest you ultimately get (in maths, that $\frac{d}{dn}((1+\frac{1}{n})^n)>0$ for $n>0$)... | $$
(1+x)^{1+x} = 1 + x + x^2 + \frac{x^3}{2} + \frac{x^4}{3} + \cdots
$$
is clearly greater than
$$
e^{x} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{24} + \cdots
$$
for $x > 0.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
} |
Find all the values of the parameter 'a' for which the given inequality is satisfied for all real values of x. Find all the values of the parameter 'a' for which the inequality is satisfied for all real values of x.
$$a\cdot 9^x+4\cdot \left(a-1\right)\cdot 3^x+\left(a-1\right)>0$$
My attempt is as follows:-
$$a\cdot \... | I think I got the point where I did mistake, thanks to @copper.hat
It is important to note that when substituting $3^x$ with $t$, it means t will also be greater than $0$.
So we have to find the condition where at least one root is positive, so there can be multiple cases to it.
Case 1: When both the roots are positive... | {
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"timestamp": "2023-03-29T00:00:00",
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Roots over $\mathbb{C}$ equation $x^{4} - 4x^{3} + 2x^{2} + 4x + 4=0 $. I need roots over $\mathbb{C}$ equation $$x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = 0$$
From Fundamental theorem of algebra we have statement that the equation have 4 roots over complex.
But I prepare special reduction:
$$ \color{red}{ x^{4} - 4x^{3} +... | Answer to the revised question
You have correctly obtained $x=1\pm \sqrt{2\pm i\sqrt3}$.
By the standard method find the complex square roots. This gives one form of the full answer as $$x=1\pm \sqrt[4]7(\frac{2+\sqrt7\pm i\sqrt3}{\sqrt{14+4\sqrt7}}).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Help with $\lim\limits_{x \to 0} \frac{x^2 \sin(2x)}{\log (1+(\sin3x)^3)}$ I'm preparing for my first exam in university (just recently enrolled in computer science) and I'm having difficulties working out this limit. I either currently lack the proper reasoning process to get it done or they haven't yet explained us a... | Hint:
By L'Hôpital ("$\frac{0}{0}$"),
\begin{align}
\lim_{x \to 0} \frac{x^2 \sin(2x)}{\log(1 + \sin^3(3x))}
& = \lim_{x \to 0} \frac{2(x^2 \cos(2x)+ x \sin(2x))(\sin^3(3x) + 1)}{9 \cos(3x) \sin^2(3x)} \\
& = \frac{2}{9} \left( \lim_{x \to 0} \left(\frac{x}{\sin(3x)}\right)^2 + \lim_{x \to 0} \left(\frac{x \sin(2x)}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Range of $k$ in $x^2+y^2-6x-10y+k=0$
The circle $x^2+y^2-6x-10y+k=0$ does not touch or intersect the x-axis and the point $(1,4)$ lies inside the circle, then find the range of $k$
$$
C(3,5)\;\&\;r=\sqrt{34-k}
$$
Attempt 1
$$
d=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}=|3|>r=\sqrt{34-k}\implies k>25
$$
Attempt 2
$$
y=0\implie... | The distance from center $C(3,5)$ to the $x-$ axis is $d=5$ .
It must be greater than the radius.
thus
$$5 >\sqrt{34-k}$$
the distance from the center to the point $(1,4)$ is
$D=\sqrt{(3-1)^2+(5-4)^2}=\sqrt{5}$
It must be smaller than the radius. thus
$$\sqrt{5}<\sqrt{34-k}$$
So
$$5<34-k<25$$
or
$$9<k<29.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396663",
"timestamp": "2023-03-29T00:00:00",
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Write the expression in terms of $\sin$ only $\sin(4x)-\cos(4x)$ I am currently taking a Precalc II (Trig) course in college. There is a question in the book that I can't figure out how to complete it. The question follows:
Write the expression in terms of sine only:
$\sin(4x)-\cos(4x)$
So far I have $A\sin(x)+B\cos(x)... | Use the two double angle identities that you know:
\begin{align}
\sin2\theta&=2\sin\theta\cos\theta\\
\cos2\theta&=1-2\sin^2\theta
\end{align}
You will also need to know
$$\cos x=\sqrt{1-\sin^2x}$$
As such, we get
\begin{align}
\sin4x-\cos4x&=\sin2(2x)-\cos2(2x)\\
&=2\sin2x\cos2x-(1-2\sin^22x)\\
&=2(2\sin x\cos x)(1-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3399498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factor $x^{35}+x^{19}+x^{17}-x^2+1$ I tried to factor $x^{35}+x^{19}+x^{17}-x^2+1$ and I can see that $\omega$ and $\omega^2$ are two conjugate roots of $x^{35}+x^{19}+x^{17}-x^2+1$. So I divide it by $x^2+x+1$ and the factorization comes to the following
$$(x^2+x+1)(x^{33}-x^{32}+x^{30}-x^{29}+x^{27}-x^{26}+x^{24}-x^{... | I think if we simply add and then subtract $x^{18}$ and x this factorization becomes easier.
$x^{35} + x^{19} + x^{17} - x^{2} +1 = x^{35}- x^{18}+ x^{17}+ x^{19}- x^{2}+x+ x^{18}- x+ 1$ = $x^{17}(x^{18}-x+1)+ x(x^{18}-x +1)+(x^{18}- x +1)$= $(x^{18}-x+1)(x^{17}+x+1)$.
Now it is easy to see that omega is a root of $x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Dealing with degrees in decomposition into partial fractions I had to decompose $ \frac{2x^2}{x^4-1} $ into partial fractions in order to determine its antiderivative.
So, I said:
$$
\frac{2x^2}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}
$$
However, in the answer key, they said:
$$
\frac{2x^2}{x^4-1} =... | In your very first example, notice that the function
$$f(x)=\frac{2x^2}{x^4-1}$$
is an even function (a function that satisfies $f(-x)=f(x)$ for all $x$). So, the partial fraction descomposition
$$g(x)=\frac{a}{x-1}+\frac{b}{x+1}+\frac{cx+d}{x^2+1}$$
also satisfies the same. Thus
$$\begin{align}
\frac{a}{x-1}+\frac{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400562",
"timestamp": "2023-03-29T00:00:00",
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If $3\sec^4\theta+8=10\sec^2\theta$, find the values of $\tan\theta$ If $3\sec^4\theta+8=10\sec^2\theta$, find the values of $\tan\theta$.
$$3\sec^4\theta-10\sec^2\theta+8=0$$
$$3\sec^4\theta-6\sec^2\theta-4\sec^2\theta+8=0$$
$$(3\sec^2\theta-4)(\sec^2\theta-2)=0$$
$$\sec^2\theta=2 \text { or } \sec^2\theta=\frac{4}{3}... | Your method is correct. Whatever source told you only the positive roots are right either made a mistake or imposed (or meant to impose) a range on $\theta$ that ensures $\tan\theta>0$, such as $\theta\in\left(0,\,\frac{\pi}{2}\right)$.
| {
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Calculate the smallest possible value to measure the ∠BAC angle Let $ABC$ be a right triangle acutangle and let $\overline{ AD}$, with D in $\overline{BC}$, be a height relative to point $A$. Let $Γ_1$
and $Γ_2$ as circumferences circumscribed to triangles $ABD$ and $ACD$, respectively. The circumference
$Γ_1$ crosses ... |
In this optimized drawing we have:
BA=BC=CY
AX=AY
$\angle BPA=90^o$
$\angle BQY=90^o$
BC=a, AC=b, AB=c
In triangle ACX, AC is diameter of circle so $\angle CXA=90^o$ and we can write:
$2AX^2=AC^2=b^2$
$AX \times YX=AX \times 2 AX=2AX^2$
⇒ $AX \times YX= b^2$
$QY \times QC = AX \times YX$
$QC=b Cos(\angle QCA)=b Sin(\a... | {
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"url": "https://math.stackexchange.com/questions/3402011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find real parameter $a$, such that the solution of the linear system lies in the second quadrant
For which real parameter $a$ lies the solution of the system of equations
$$\begin{aligned} \frac{x}{a+1} + \frac{y}{a-1} &= \frac{1}{a-1}\\ \frac{x}{a+1} - \frac{y}{a-1} &= \frac{1}{a+1} \end{aligned}$$
in the second quad... | Subtracting the two equations you get
$$\frac{2y}{a-1} = \frac{2}{a^2-1} \Leftrightarrow y = \frac{1}{a+1}, \qquad \text{whenever $a\neq -1$}.$$
Therefore, setting $y$ in the second equation,
$$\frac{x}{a+1} = \frac{1}{a+1}+\frac{y}{a-1} = \frac{1}{a+1} + \frac{1}{a^2-1}=\frac{a}{a^2-1} \Leftrightarrow x = \frac{a}{a-1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find x from the equation $\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$ For m are real number,
Find x from the equation
$$\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$$
I tried to multiply $\sqrt{x + \sqrt{x}}$ to the both sides and I get
$$x + \sqrt{x} -... | After dividing by $\sqrt{x}$ you may use the substitution $\boxed{y = \sqrt{x}}$. Then, after a bit rearranging you get
$$y+1 - m = \sqrt{y^2-1}$$
Squaring removes the $y^2$ and you can solve directly for $y$:
$$y = \frac{2-2m+m^2}{2(m-1)} \Rightarrow \boxed{x = \frac{(2-2m+m^2)^2}{4(m-1)^2}}$$
Note, that there are sol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Solving indefinite integral $\int \frac{1+\sqrt{1+x^2}}{x^2+\sqrt{1+x^2}}dx$ How can we solve this integration?
$$\int \frac{1+\sqrt{1+x^2}}{x^2+\sqrt{1+x^2}}dx$$
I tried to make the following substitution
$$1+x^2=w^2$$
but this substitution complicated the integral.
| Substitute $x=\sinh u$,
$$I=\int \frac{1+\sqrt{1+x^2}}{x^2+\sqrt{1+x^2}}dx=
\int \frac{\cosh u+\cosh^2 u}{\cosh^2 u+\cosh u - 1}du = u+I_1$$
where,
$$I_1=\int \frac{du}{\cosh^2 u+\cosh u-1}$$
Next, use the substitution $\cosh u = \frac{1+t^2}{1-t^2}$, along with $d u = \frac{2dt}{1-t^2}$,
$$I_1=2\int \frac{t^2-1}{t^4-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\sum_{j=1}^{n-1} M_j(M_{j+1} - M_j) = \frac{1}{2}\left( \sum_{j=1}^n X_j \right)^2 - \frac{n}{2}$ The following question is taken from Steven Shreve Stochastic Calculus for Finance Volume 1, question $2.5$
Toss a coin repeatedly. Assume the probability of head on each toss is $\frac{1}{2},$ as is the probab... | Note that since $X_j^2 = 1$ and $M_1^2 = X_1^2 = 1$,
$$I_n = \sum_{j=1}^{n-1} \frac{1}{2}(M_j+M_{j+1})(M_{j+1} - M_j) - \sum_{j=1}^{n-1} \frac{1}{2}X_{j+1}(M_{j+1} - M_j) \\= \frac{1}{2}\sum_{j=1}^{n-1}(M_{j+1}^2- M_j^2)- \frac{1}{2}\sum_{j=1}^{n-1}X_{j+1}^2 = \frac{1}{2} M_n^2 - \frac{1}{2} M_1^2 -\frac{n-1}{2} \\ ... | {
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"url": "https://math.stackexchange.com/questions/3406742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Demonstrate that there are no perfect squares ending with $8$ A number n will always end in some digit of the set {$0,1,2,3,4,5,6,7,8,9$}. The last digit of $n^2$ is the last digit of its last squared digit. Like this:
$$\ldots 0^2 = \ldots 0$$
$$\ldots 1^2 = \ldots 1$$
$$\ldots 2^2 = \ldots 4$$
$$\ldots 3^2 = \ldots 9... | It's good but you can tweak it. Notice the palindromic symmetry: 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0. If $n \equiv 0 \pmod{10}$, then $n^2 \equiv 0 \pmod{10}$; if $n \equiv \pm 1 \pmod{10}$, then $n^2 \equiv 1 \pmod{10}$; if $n \equiv \pm 2 \pmod{10}$, then $n^2 \equiv 4 \pmod{10}$; etc. By proving $n^2 \equiv 8 \pmod{10}$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Definite integration evaluation of $\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}~dx$. $$\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}~dx$$
how to proceed please help
The answer given is $\dfrac{\pi}{4a^3b}$.
| Let $I(a,b)=\int_0^{\pi/2}\frac{dx}{a^2\sin^2x+b^2\cos^2x}=\frac{\pi}{2ab}$(try proving this yourself)
partially differentiating wrt $a$ and applying Leibnitz rule,
$\int_0^{\pi/2}\frac{-2a \sin^2x dx}{(a^2\sin^2x+b^2\cos^2x)^2}= \frac{- \pi}{2ba^2}$.
So $\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A conjecture about the number of divisors of a natural number Conjecture:
$\tau(n)\mid\tau(n^2)\iff$ $n$ is a perfect square and $\sqrt n=p^2s$,
where $p$ is prime and $s$ is a non prime squarefree number such that $\gcd(p,s)=1$.
$\tau(n)$ is the number of factors of $n$.
See this question on MSE:
Number theory ... | If $n=p^4s^2$ with $p$ prime, $s$ is a product of $m\ge1$ distinct primes $q_1,\ldots,q_m$ and $p\nmid s$, then
$$\tau(n) =\tau(p^4)\tau(q_1^2)\cdots \tau(q_m^2)=5\cdot 3\cdots 3=5\cdot 3^m$$
and
$$\tau(n^2) =\tau(p^8)\tau(q_1^4)\cdots \tau(q_m^4)=9\cdot 5\cdots 5=3^2\cdot 5^m.$$
It follows that for such $n$, we have $... | {
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If equation rep $x^2+2y^2-5z^2+2kyz+2zx+4xy=0$ represent pair of plane. then $k$ is The values of $k$ for which the equation
$x^2+2y^2-5z^2+2kyz+2zx+4xy=0$
represents a pair of plane passing Through origin,is
what i try
$x^2+2y^2-5z^2+2kyz+2zx+4xy=(ax+by+cz)(px+qy+rz)$
and camparing coefficients
but it is very tediou... | ADDED: conclusion without the matrices:
When $k^2 - 4 k - 8 = 0,$ we get factoring over the reals because
$$ \color{magenta}{ (x+2y+z)^2 - \frac{1}{2} \left( 2y + (2-k) z \right)^2 } $$
is your quadratic form, and then we can factor as
$$ T^2 - \frac{1}{2} U^2 = \left(T + \frac{U}{ \sqrt 2}\right) \left(T - \frac{U}{... | {
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Pointwise convergence of a series of functions using the ratio test. I am asked to calculate the pointwise convergence of the series of functions
$$\sum_{n\geq 0} a_n=\sum_{n\geq 0}\frac{(2n)^{\frac{n+1}{2}}}{\sqrt{n!}}x^ne^{-nx^2}$$
Since it is a series of positive terms, I apply the ratio test and after simplifying ... | By ratio test we obtain
$$\left|\frac{[2(n+1)]^{\frac{n+2}{2}}}{\sqrt{(n+1)!}}\frac{\sqrt{n!}}{[2n]^{\frac{n+1}{2}}}\frac{x^{n+1}e^{-(n+1)x^2}}{x^ne^{-nx^2}}\right|=\sqrt 2|x|e^{-x^2}\left(1+\frac{1}{n}\right)^\frac{n+1}2\to \sqrt {2e}|x|e^{-x^2}\le 1$$
for $\sqrt {2e}|x|e^{-x^2}=1 \implies |x|e^{-x^2}=\frac1{\sqrt{2e}... | {
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Prove that $(11 \cdot 31 \cdot 61) | (20^{15} - 1)$ Prove that
$$ \left( 11 \cdot 31 \cdot 61 \right) | \left( 20^{15} - 1 \right) $$
Attempt:
I have to prove that $20^{15}-1$ is a factor of $11$, $31$, and $61$. First, I will prove
$$ 20^{15} \equiv 1 \bmod11 $$
Notice that
$$ 20^{10} \equiv 1 \bmod 11$$
$$ 20^{5} ... | $20\equiv3^2\pmod{11}$ therefore $20^{15} \equiv 3^{30}\equiv 1 \bmod11$.
$20\equiv12^2\pmod{31}$ therefore $20^{15} \equiv 12^{30}\equiv 1 \bmod31$.
$20\equiv3^4\pmod{61}$ therefore $20^{15} \equiv 3^{60}\equiv 1 \bmod61$.
| {
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The number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$ I'm trying to find the number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$.
Let $2^x=a$
$$|a-1|=a^2-2a-5$$
Then there are two cases
$$a-1=a^2-2a-5$$
And $$a-1=-a^2+2a+5$$
Solving both equations
$$a=1,-4,-2,3$$
Now -4 and -2 can be neglected so th... | Your way is right indeed for $2^x \ge1$
$$5+|2^x-1|=2^x(2^x-2)\implies 5+2^x-1=2^{2x}-2\cdot 2^{x} \iff2^{2x}-3 \cdot 2^x-4=0$$
for $0<2^x <1$
$$5+|2^x-1|=2^x(2^x-2)\implies 5-2^x+1=2^{2x}-2\cdot 2^{x} \iff2^{2x}- 2^x-6=0$$
then let $2^x=t>0$ and solve keeping only the solutions which agree with the assumptions.
Note t... | {
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General formula for $e^x+\cos(x)$, $e^x+\sin(x)$, $e^x-\sin(x)$, $e^x-\sin(x)$ I have been able to derive the formal series for these four functions:
$e^x+\sin(x) = 1+2x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{2x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{2x^9}{9!}+...$
$e^x+\cos(x) = 2+\dfrac{x^3}{3!}+\dfrac{2x^4}{4... | Take the series for $e^x+\sin x$ for example. This is obtained by taking the series for $e^x$, doubling the terms $\dfrac{x^n}{n!}$ for $n=4k+1$ and deleting the terms $\dfrac{x^n}{n!}$ for $n=4k+3$. So you could write it as
$$\sum_{k=0}^{\infty} \left( \dfrac{x^{4k}}{(4k)!} + \dfrac{2x^{4k+1}}{(4k+1)!} + \dfrac{x^{4k+... | {
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Inequality $\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\leq \frac{\sqrt{3}}{k+1}$ Hi it's related to this If $a+b+c = 3abc$ and $\frac17 \leq k \leq 7$ prove $ \frac1{ka+b}+\frac1{kb+c}+\frac1{kc+a} \leq \frac3{k+1} $
I propose this :
Let $a,b,c>0$ and $a+b+c=abc$ and $a\geq b \geq c $ then we have :
$$\frac{1}{ka... | The Buffalo Way works.
After homogenization, it suffices to prove that, for $a\ge b\ge c > 0$ and $8 \le k\le \alpha$,
$$\frac{3}{(k+1)^2}\ge \frac{abc}{a+b+c}\left(\frac{1}{ka+b} + \frac{1}{kb+c} + \frac{1}{kc+a}\right)^2.$$
WLOG, assume that $c = 1$. Let $a = 1 + s + t, b = 1 + s$ for $s, t\ge 0$. It suffices to prov... | {
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Series expansion for $\arctan(1-x)$ Series expansion for $\arctan(1-x)$
I try to expand this function into its Taylor series by means of differentiating it and then integrating it terms by terms but I fail to obtain the correct result.
The derivative of $\arctan(1-x)=-\dfrac{1}{x^2-2x+2}$. By using long division, I can... | For $n\in\mathbb{N}$, the derivative of $\arctan z$ is
\begin{equation*}
(\arctan z)^{(n)}
=\frac{(n-1)!}{(2z)^{n-1}}\sum_{k=0}^{n-1}(-1)^k\binom{k}{n-k-1}\frac{(2z)^{2k}}{(1+z^2)^{k+1}}.
\end{equation*}
The function $\frac{\arctan z}{z}$ has Taylor's series expansion
\begin{equation}\label{arctan-pi-4-ser-eq}
\frac{\a... | {
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prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.
I have tried induction as follows.
Step 1:
Try n = 0, we get: $5 - 3 - 2 = 0$, which is divisible by 30.
Try n = 1, we get: $5^{3} - 3^{3} - 2^... | Much the simplest method is to use arithmetic modulo 2,3 and 5.
$$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 1^{2n+1} - 1^{2n+1} - 0^{2n+1}\equiv 0 \pmod 2$$
$$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 2^{2n+1} - 0^{2n+1} - 2^{2n+1}\equiv 0 \pmod 3$$
$$5^{2n+1} - 3^{2n+1} - 2^{2n+1}\equiv 0^{2n+1} - (-2)^{2n+1} - 2^{2n+1}\equiv 0... | {
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Problematic inequality & hint I would like to ask for hint for proving following inequality:
$$x^3(1+x)+y^3(1+y)+z^3(1+z)\geq \frac{3}{4}(1+x)(1+y)(1+z)$$
for all $x>0$, $y>0$, $z>0$ such that $xyz=1$.
Generally, I tried to find some elementary solution, but even with some calculus I didn't solve it.
Edit. My attempt:
... | We need to prove that:
$$\sum_{cyc}(4x^4+4x^3)\geq\sum_{cyc}(2+3x+3xy),$$ which is true by Muirhead.
Indeed, we need to prove that:
$$\sum_{cyc}(3x^4+3x^3+x^4+x^3)\geq\sum_{cyc}(3xy+3x+1+1).$$
Now, $$\sum_{cyc}x^4\geq\sum_{cyc}xy$$ it's
$$\sum_{cyc}x^4\geq\sum_{cyc}xy\sqrt[3]{x^2y^2z^2}$$ or
$$\sum_{cyc}x^4\geq\sum_{cy... | {
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Find the limit $ \cos x / (1 - \sin x)^{2/3}$ My problem:
Find the following limit
$$\lim\limits_{x \to \pi/2} \frac{\cos x}{\left(1 - \sin x\right )^{2/3}}$$
I tried $(1 - \sin x)^2 = 1 - 2\sin x + \sin^{2}x$ but can go as far as that.
| We have that by
*
*$\cos x=\sin\left(\frac \pi 2 -x\right)$
*$\sin x=\cos\left(\frac \pi 2 -x\right)$
and $\frac \pi 2 -x =y \to 0$
$$\frac{\cos x}{\sqrt[3]{(1-\sin x)^2}} =\frac{\sin y}{\sqrt[3]{(1-\cos y)^2}}=\frac{\sin y}{y}\frac{\sqrt[3]{y^4}}{\sqrt[3]{(1-\cos y)^2}}\frac1{\sqrt[3]y} \to \pm \infty $$
indeed
... | {
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$x^4 + 1020x^3 - 4x^2 + 2039x+1$ is divisible by 1019. Give all integers x in the range [1,2018] such that $x^4 + 1020x^3 - 4x^2 + 2039x+1$
is divisible by 1019.
Applying mod 1019 to the coefficients, we have
$x^4 + x^3 - 4x^2 + x + 1$ which is equal to $(x-1)^2(x^2+3x+1)$
I tried making 1 of the factors equal to a mul... | $(x-1)^2(x^2+3x+1)\equiv0\bmod1019\implies$
$ x-1\equiv0\bmod1019$ or $x^2+3x+1\equiv0\bmod1019$.
The former implies $x\equiv1\bmod 1019$; solutions in $[1,2018]$ are $1$ and $1020$.
The latter implies $x^2+3x+\dfrac94\equiv\dfrac94-1\bmod1019$, so $\left(x+\dfrac32\right)^2\equiv9\times255-1\equiv256\bmod1019$,
so $... | {
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A refinement of a famous inequality on the forum . It's related to a big problem Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$ .I have this (For one time I take the time to check it )
Let $a,b,c>0$ such that $a+b+c=1$ then we have :
$$\sum_{cyc}\frac{a^4}{8a^3+5b^3}> \sum_{... | The inequality
$$
f(a,b)=\frac{a^4}{8a^3+5b^3}>\frac{\tan(a^4)}{8\tan(a^3)+5\tan(b^3)}
$$
reduces to
$$
5 a^4 \tan \left(b^3\right)-\left(8 a^3+5 b^3\right) \tan \left(a^4\right)+8 a^4 \tan \left(a^3\right) \ge 0
$$
in $$T=\{(a,b):a>0,b>0,a+b \le 1\}.$$
One way to do it is to cut $T$ into pieces and approximate $tan(x... | {
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Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$. Suppose that $a,b$ are non-negative numbers such that $a+b=2$. I want to prove $$\sqrt{2 a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4.$$
My attempt: I tried proving $$2a^2+b+1\geq 4$$ but this seems wrong (try $a=0,b=1$).
How can I prove the above result?
| Suppose without loss of generality that $a\geqslant b$ so that $0\leqslant b \leqslant 1$ and write $a = 2- b$.
Our expression becomes
$$\sqrt{2(2-b)^2 + b + 1} + \sqrt{2b^2 - b + 3}$$
and this can be minimized for $b\in[0,1]$ with usual calculus techniques.
| {
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Prove that $x^2+px+p^2$ is a factor $(x+p)^n-x^n-p^n$, if $n$ be odd and not divisible by $3$. Question:
Prove that $x^2+px+p^2$ is a factor of $(x+p)^n-x^n-p^n$, if $n$ is odd and is not divisible by $3$.
My approach:
$$(x+p)^n-x^n-p^n=\sum_{r=0}^n\limits {n\choose r} x^{n-r}p^r-x^n-p^n$$ What can I do after that?
| First of all substitute $x$ with $py$
Observe $y^3-1=(y-1)(?)=0$
So, $y=w,w$ is a complex cube root of unity
Now $(x+p)^n-x^n-p^n=p^n((1+y)^n-1-y^n)$
Now $f(n)=(1+w)^n-1-w^n=(-w^2)^n-1-w^n$
which is $=-1-w^n-w^{2n}$ if $n$ is odd
If $3\mid n,f(n)=-1-1^n-1^{2n}=?$
Else
$w^n\ne1,f(n)=-\dfrac{1-(w^3)^n}{1-w^n}=0$
Check fo... | {
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Solve system of equations $ y + 3x = 4x^3$ and $x + 3y = 4y^3$ Problem: In the set of real numbers find all $x,y$ that satisfy
$ y + 3x = 4x^3,x + 3y = 4y^3$.
What I don't understand, is how do I know that a condition satisfies both equations(I will show later exactly).
The solution goes like this: By addition and subs... | Here is a natural way to solve the system. Continue with,
$$x + y = (x + y)(x^2 − xy + y^2),\>\>\>\>\>
x − y = 2(x − y)(x^2 + xy + y^2)$$
or,
$$(x + y)(x^2 − xy + y^2-1)=(x − y)(x^2 + xy + y^2-\frac12)=0$$
So, there are four cases to be considered,
Case 1: Substitute $x-y=0$ into one of the original equations, say, $y... | {
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Proving $\sum_{cyc} \sqrt{a^2+ab+b^2}\geq\sqrt 3$ when $a+b+c=3$ Good evening everyone, I want to prove the following:
Let $a,b,c>0$ be real numbers such that $a+b+c=3$. Then $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geq 3\sqrt 3.$$
My attempt: I try noting that $$\sum_{cyc} \sqrt{a^2+ab+b^2}=\sum_{cyc}... | Lemma For all $x$ we have $$\sqrt{x^2+x+1}\geq {\sqrt{3}\over 2}(x+1)$$
Proof After squaring and clearing the denominator we get $$4x^2+4x+4\geq 3(x^2+2x+1)$$
which is the same as $$x^2-2x+1\geq 0$$
Using lemma we get $$\sqrt{a^2+ab+b^2}= b\sqrt{\Big({a\over b}\Big)^2+{a\over b}+1}\geq b\cdot {\sqrt{3}\over 2}({a\ov... | {
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"url": "https://math.stackexchange.com/questions/3450882",
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How can I evaluate $\lim_{x\to \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$ using L'Hospital's rule? How can I find the following limit using the L'Hospital's rule?
$$\lim_{x\rightarrow \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$$
I have ... | By Taylor expansion, we see that for $x>1$,
\begin{align*}
\dfrac{1}{x}-\dfrac{1}{2}\dfrac{1}{x^{2}}+\dfrac{1}{3}\dfrac{1}{x^{3}}-\dfrac{1}{4}\dfrac{1}{x^{4}}\leq\log\left(1+\dfrac{1}{x}\right)\leq \dfrac{1}{x}-\dfrac{1}{2}\dfrac{1}{x^{2}}+\dfrac{1}{3}\dfrac{1}{x^{3}},
\end{align*}
so
\begin{align*}
x^{2}-\dfrac{1}{2}x... | {
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Proving iff statement about polynomials I have the following question:
Let $\mathbb F$ be a field and $ x^3 +ax+b, x^2+cx-1 $ $\in \mathbb F$[x].
Prove that $(x^2+cx-1) | (x^3+ax+b) $if and only if $c=b$ and $ a= -1-b^2$.
Attempt:
If $(x^2+cx-1) | (x^3+ax+b) $ then there exists $(x+d)$ such that $(x^2+cx-1)(x+d) = (x... | Just equate the coefficients and eliminate $d$. We have $c+d=0, cd-1=a$ and $-d=b$. So $d=-c$. Plug this into the other equations to get $a=cd-1=-1-c^{2}$ and $b=-d=c$. Hence $b=c$ and $a =-1-b^{2}$. Now check that the converse is also true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the period of the $f(x)=\sin x +\sin3x$?
What is the period of the $f(x)=\sin x +\sin3x$?
$f(x)=\sin x+\sin 3x=2\frac{3x+x}{2}\cos\frac{x-3x}{2}=2\sin2x\cos x=4\sin x\cos^2x\\f(x+T)=4\sin(x+T)\cos^2(x+T)=4\sin x\cos^2 x$
how can I deduct this I have no idea
| The period of $~\sin x~$ is $~2\pi~$ and that of $~\sin 3x~$ is $~\frac{2\pi}{3}~$, because $~\sin\left(3\frac{2\pi}{3}\right)=\sin 2\pi~$.
Now given that $~f(x)=\sin 3x+\sin x~$
Let $~a~$be a period $~f(x)~$, then by the definition $~f(x+a)=f(x)~$.
Here $~f(x+a)=\sin(3x+3a)+\sin(x+a)~$
From $~\sin(3x+3a)~$, we have $... | {
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"timestamp": "2023-03-29T00:00:00",
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What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5? What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5?
My work:
1^5 ends with 1.
2^5 ends with 2.
3^5 ends with 3.
And so on.
Do I simply add the ending digits to get my answer?
| Using Faulhaber's formula:
$$S_{99}=\frac{4a^3-a^2}{4}=a^2\cdot \frac{4a-1}{4}=\frac{9900^2}{4}\cdot \frac{2\cdot 9900-1}{3}=33\cdot 25\cdot 9900\cdot 19799,$$
where:
$$a=\frac{n(n+1)}{2}.$$
Hence, the answer is $00$.
| {
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"answer_id": 3
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$\frac{d}{dx} \frac{1}{x+\frac{1}{x+\frac{1}{ \ddots}}}$, the derivative of an infinite continued fraction. I computed the first few derivatives in the sequence $\frac{1}{x+\frac{1}{x}}, \frac{1}{x+\frac{1}{x+\frac{1}{x}}} \dots$ and eventually lost feeling in my fingers and also realized I was seeing nothing meaningfu... | Let $y:=\frac{1}{x+\frac{1}{x+\cdots}}=\frac{1}{x+y}$ so$$y^2+xy-1=0\implies y=\frac{-x\pm\sqrt{x^2+4}}{2}.$$If $x>0$, $y>0$ so$$y=\frac{-x+\sqrt{x^2+4}}{2}\implies\frac{dy}{dx}=\frac{-1+\frac{x}{\sqrt{x^2+4}}}{2}.$$Your $f(x,\,t)$ are odd for each integer $t\ge0$, so if we are to extend this to $x\le0$ we need$$y=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
calculate:$\int \frac{x+1}{-x(x+1)+\sqrt{x^2+1}}dx$ calculate:$$\int \frac{x+1}{-x(x+1)+\sqrt{x^2+1}}dx$$
I tried:
$$\int \frac{x+1-\frac{\sqrt{x^2+1}}{x}+\frac{\sqrt{x^2+1}}{x}}{-x(x+1)+\sqrt{x^2+1}}dx=\int\frac{-dx}{x}+\int \frac{\frac{\sqrt{x^2+1}}{x}}{-x(x+1)+\sqrt{x^2+1}}dx$$
I can not continue from here
| Let $x = \tan u $
$$\int \frac{(\tan u +1) \sec^2 u}{\sec u - \tan u ( \tan u +1 )} \; du $$ multiply with $\cos ^2 u $
$$\int \frac{(\tan u +1 )}{\cos u - \sin u ( \sin u + \cos u ) } $$
$$\int \frac{\sin u + \cos u }{\cos ^2 u - \sin u \cos u(\sin u + \cos u) } $$
$$\int \frac{2 \sqrt{2} \sin (0.25 \pi + u ) }{ 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $f$ is continuous in $(0,0)$ and check if $f$ differentiable in $(0,0)$ and if $f$ is uniformly continuous on $\mathbb R^2$
Let $f: \mathbb R^2 \to \mathbb R$ such that:
$$f(x,y)= \begin{cases} \frac{(x+y)^3}{\sqrt{1+x^2+y^2}-1}, (x,y)\neq (0,0) \\ 0, (x,y)=(0,0) \end{cases}$$
a) Prove that $f$ is conti... | Hints:
a) No, $(x+y)^3\le (x^2+y^2)^3$ is false. Take $x=1/2,y=0$ to see this. But this is true:
$$|f(x,y)| \le \frac{(|x|+|y|)^3}{\sqrt {1+x^2+y^2}-1}=\frac{(|x|+|y|)^3}{x^2+y^2}(\sqrt {1+x^2+y^2}+1).$$
b) If $f$ is differentiable at $(0,0),$ then
$$f(x,y) = \nabla f(0,0)\cdot (x,y) + o\left(\sqrt {x^2+y^2}\right).$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the equation of the line $ r $ Find the equation of the line $ r $ that goes through $ (1, -2,3) $, concurrent with the line
$$\begin{cases}
x=2+3t \\
y=1+2t \\
z= -1 \\
\end{cases}$$
and has orthogonal director vector $ (1, -3,1) $
Solution: line r is contained in the plane $(x-1)-3(y+2)+(z-3)=0$
$x-3y+z=10$
Next... | A straight $r$ going through point $(1,-2,3)$ has this parametric form:
$$x = 1 + a·u$$
$$y = -2 + b·u$$
$$z= 3 + c·u$$
If it has intersection with the other given line, then
$$ x = 1 + au = 2+ 3t$$
$$ y= -2 + bu = 1 +2t$$
$$ z= 3 + cu = -1$$
If it has a orthogonal director vector (1,−3,1) then the dot product with $(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Is there a way to guss the sign of 1 in the expressions (2^8-1)/17=15 or (3^8+1)/17 =386 I found some expressions like above that are divisible by 17 as bellow:
(4^8-1)/17 ,(5^8+1)/17 ,(6^8+1)/17 ,(7^8+1)/17 ,(8^8-1)/17 ,(9^8-1)/17,(10^8+1)/17
then I sum and get (1+2^8+3^8+...+10^8) that are divisible by 17
My questio... | We know that $a^{16}=1\mod17$ (for $17\nmid a$) by Fermat's Little Theorem. So $a^8=\pm1\bmod 17$.
It is not hard to check that $2^8=1,3^8=-1\bmod17$. Hence $4^8=8^8=1,6^8=(2^8)(3^8)=-1\bmod17$ and $9^8=(3^8)^2=1\bmod17$. Similarly $12^8=(2^8)^23^8=-1\bmod17$, so $5^8=(17-5)^8=-1\bmod17$ and $10^8=(2^8)(5^8)=-1\bmod17$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$(a+bi)$ of $\frac{3+4i}{5+6i}$ and $i^{2006}$ and $\frac{1}{\frac{1}{1+i}-1}$ How can one get the Cartesian coordinate form $(a+bi)$ of the following complex numbers?
$$\frac{3+4i}{5+6i}$$
$$i^{2006}$$
$$\frac{1}{\frac{1}{1+i}-1}$$
Regarding $\frac{3+4i}{5+6i}$ I tried expanding it with $\frac{5+6i}{5+6i}$ and got $\f... | $$ \frac{z_1}{z_2}=\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}} $$
$$\frac{z_1}{z_2}=(\frac{r_1}{r_2})e^{i(\theta_1-\theta_2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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${2n \choose 1} + {2n \choose 3} + ... {2n \choose 2n-1}$ $(1) \ \ \ \ {2n \choose 1} + {2n \choose 3} + ... {2n \choose 2n-1} = ?$
I know that $\sum_{k=0}^n {n \choose k} = 2^n$ and it is pretty easy to obtain, as it's an amout of all possible sets we can get from ${1,2,...,n}$. There is also a formula for $k = k + at... | We have the following binomial expansion:
$$(1+1)^{2n}=\binom{2n}{0}+\binom{2n}{1}+\dots+\binom{2n}{2n}$$
and
$$(1-1)^{2n}=\binom{2n}{0}-\binom{2n}{1}\pm\dots+\binom{2n}{2n}.$$
Summing both gives
$$2^{2n}=2\binom{2n}{0}+2\binom{2n}{2}+\dots+2\binom{2n}{2n},$$
So dividing throughout by $2$ we get
$$2^{2n-1}=\binom{2n}{0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find area bounded by $y=\frac 3x, y=\frac 5x, y=3x, y=6x$ Let S be the area of the region bounded by the curves
$$y=\frac 3x,\>\>\> y=\frac 5x,\>\>\> y=3x,\>\>\> y=6x$$
Need to find $S$.
The coordinates of the vertices of the resulting figure were found. The problem with the transition from a double integral to a repe... | 1) Subtract the two integrals of $5/x$ and $3/x$ in the range of $a$ which is the solution of $\frac{3}{x}=6x$ (the intersection of two lines) to $b$ which is the solution of $\frac{5}{x}=3x$.
2) One would obtain $area=[ln(2x)]_{a}^{b}$.
3)Now, to subtract the area of the two hyperbolic triangles, simply obtain $c$, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding sum of series $\sum^{\infty}_{k=0}\frac{(k+1)(k+2)}{2^k}$
Finding sum of series $\displaystyle \sum^{\infty}_{k=0}\frac{(k+1)(k+2)}{2^k}$
what i try
Let $\displaystyle x=\frac{1}{2},$ Then series sum is $\displaystyle \sum^{\infty}_{k=0}(k+1)(k+2)x^k$
$\displaystyle \sum^{\infty}_{k=0}x^k=\frac{1}{1-x}\Righta... | Another way:
Let $f(m)=\dfrac{a+bm+cm^2}{2^m}$
and $\dfrac{(k+1)(k+2)}{2^k}=f(k)-f(k+1)$
so that $$\sum_{k=0}^n(f(k)-f(k+1))=f(0)-f(n+1)$$
$$\implies2(k+1)(k+2)=2(a+bk+ck^2)-[a+b(k+1)+c(k+1)^2]$$
$$\iff4+6k+2k^2=k^2(2c-c)+k(2b-b-2c)+2a-a-b-c$$
Compare the constants and the coefficients of $k,k^2$ to find $a,b,c$
Now w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
particular solution of $(D^2+4)y=4x^2\cos 2x$
Find a particular solution to the equation $$(D^2+4)y=4x^2\cos 2x$$
\begin{align}
y_p&=\frac{1}{D^2+4}(4x^2\cos 2x)\\
&=\frac{1}{D^2+4}\left[2x^2(e^{2ix}+e^{-2ix})\right]\\
&=2\left[e^{2ix}\frac{1}{(D+2i)^2+4}x^2+e^{-2ix}\frac{1}{(D-2i)^2+4}x^2\right]\\
&=2\left[e^{2ix}\f... | $$E=\frac 1 {D+4i}=\frac 1 {4i}\frac 1 {(1+D/4i)}$$
$$E=\frac 1 {4i}({1+D/4i})^{-1}$$
Now you can use this ( after a certain number the derivative is zero for $x^2$:
$$\frac 1 {1-x}=\sum_{n=0}^{\infty} x^n$$
So we have that:
$$E=\frac 1 {4i}(1-\frac D {4i}-\frac {D^2}{16}+........)$$
Apply the operator on $x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Solve $x^3+3y-xy^\prime=0$
Solve $x^3+3y-xy^\prime=0$
Divided by $x$ to get $x^2+3\frac{y}{x}-y^\prime=0$
Let $v=\frac{y}{x}$ then $y^\prime=v^\prime x+v$
then $x^2+3v=v^\prime x+v$
then $v^\prime=x+2\frac{v}{x}$
Used integrating factor $e^{\int -2/x dx}=e^{-x^2}$
then $e^{-x^2}v=\int e^{-x^2}\cdot x dx$
But when I s... |
Divided by $x$ to get $x^2+3\frac{y}{x}-y^\prime=0$
From this step, multiply by negative one to form
$$y'-\frac{3}{x}y=x^2$$
therefore the integrating factor is
$$\mu(x)=\text{exp}\left(-{\int {\frac{3}{x}}dx}\right)=x^{-3}$$
hence the equation can be written as
$$\frac{d}{dx}\left(\frac{y}{x^3}\right)=\frac{1}{x}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the recurrence relation solution. $a_n$ = $a_{n-1} + 3n - 5$ $a_n$ = $a_{n-1} + 3n - 5$ $,$ $a_0 = 7$
So far what I got is:
$a_0 = 7$
$a_1 = 7+3(1)-5$
$a_2 = 7+3(1)-5 + 3(2)-5$
$a_3 = 7+3(1)-5 + 3(2)-5 + 3(3)-5$
$a_4 = 7+3(1)-5 + 3(2)-5 + 3(3)-5 + 3(4)-5$
This is where I'm stuck I'm having trouble finding a patter... | By telescoping sum, $a_n-7=a_n-a_0=\sum\limits_{k=1}^n(a_k-a_{k-1})=\sum\limits_{k=1}^n(3k-5)=3\sum\limits_{k=1}^n k-\sum\limits_{k=1}^n 5=3\frac{n(n+1)}2-5n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Integrating $\frac1{a+b\cos(x)}$ using $e^{ix}$ In calculus class our teacher demonstrated us the evaluation of the definite integral $\int_0^\pi \dfrac{1}{a+b\cos(x)}dx=\dfrac{\pi}{\sqrt{a^2-b^2}}$, for $a\ne 0, b \ne 0, |\dfrac ba \lt 1|$, but there's a part I could not grasp.
Our teacher started out by setting up an... | A simpler method:
Let us recall the fact that when $A, B, C\in{\mathbb R}$ and $\Delta=B^2-4 A C < 0$ then
\begin{equation}
\int \frac{d t}{A t^2 + B t + C} = \frac{2}{\sqrt{-\Delta}} \arctan\left(\frac{2 A t + B}{\sqrt{- \Delta}}\right)
\end{equation}
The substitution $t = \tan(x/2)$ in the original integral gives
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Count a triple integral on a sphere $\int\int\int_{x^2+y^2+z^2 \le 1} |x+y|e^{2xy-z^2}dxdydz$
I have tried to change coordinates to spherical, but it didn't help. Can you give me a hint, please?
| Use the substitution
$$\begin{cases} u = \frac{x+y}{\sqrt{2}} \\ v = \frac{x-y}{\sqrt{2}} \\ w = z\\ \end{cases} \implies u^2-v^2 = 2xy$$
(chosen so that the Jacobian is $1$) to get the following integral
$$\sqrt{2}\iiint_{u^2+v^2+w^2\leq 1} |u|e^{u^2-v^2-w^2}\:du \:dv \:dw = \sqrt{2} \iint_{v^2+w^2\leq 1} e^{1-2(v^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3470745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Without calculating the square roots, determine which of the numbers:$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$ is greater.
Without calculating the square roots, determine which of the numbers:
$$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$$
is greater.
My work (I was wondering if there are other way... | Another way of doing it. Using the result that,
$$\sqrt{1+\frac{n}{m}}-\sqrt{1-\frac{n}{m}} ≥ \frac{n}{m}$$
$$\sqrt{1+\frac{3}{32}}-\sqrt{1-\frac{3}{32}} >\frac{3}{32}$$
$$\sqrt{1+\frac{3}{32}} > \frac{3}{32}+\sqrt{1-\frac{3}{32}}$$
$$8\sqrt{1+\frac{3}{32}} > \frac{3}{4}+8\sqrt{1-\frac{3}{32}}$$
$$\sqrt{70} > \frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Find the minimal polynomial for $\cos(\frac{2\pi}{5})$ and $\sin(\frac{2\pi}{5})$ Let $\omega$ be the primitive 5th root of $1$, then $\cos(\frac{2\pi}{5}) = \frac{w+w^{-1}}{2}$ and $\sin(\frac{2\pi}{5}) = \frac{w-w^{-1}}{2i}$. How to find the minimal polynomial of $\frac{w+w^{-1}}{2}$ then? (without using the Chebysh... | $$
\alpha = \frac{\omega+\omega^4}{2}\\
\alpha^2 = \frac{\omega^2 + \omega^3 + 2}{4}\\
2 \alpha + 4 \alpha^2 = \omega+\omega^4 + \omega^2 + \omega^3 + 2\\
4 \alpha^2 + 2 \alpha -1 = 0\\
$$
Do $\alpha^n$ enough times that you see all the powers of $\omega^k$. Then you can see how to combine them that gives you the minim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$ The question states:
Show that: $$2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$$
This is what I have done
$2(\sin y + 1)(\cos y + 1) = 2(\sin y + \cos y + 1)^2$
L. H. S. = R. H. S.
From L. H. S.
$2(\sin y +1)(\cos y + 1) = 2(\sin y... | If you have difficulty finding clever groupings, you could brute-force the problem by expanding everything and seeing what happens:
Writing $c := \cos y$ and $s := \sin y$ to save typing, we have
$$\require{cancel}\begin{align}
2(s+1)(c+1) &= (s+c+1)^2 \\
2(s c + s + c + 1) &= s^2 + c^2 + 1 + 2 s c + 2 s + 2 c \\
\canc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Harmonic series bounded above by $\sqrt{n}$ Question:
(a) Prove that $$\sqrt{n+1}-\sqrt{n} > \frac{1}{2\sqrt{n+1}}$$
(b) Prove by induction, for $n \geqslant 7$, that $$1 + \frac{1}{2} + \frac13 + \cdots + \frac1n<\sqrt{n}$$
| Answer:
(a) Rationalising the numerator, we have:
\begin{align}
\sqrt{n+1}-\sqrt{n} &= \frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}} \\
&=\frac{1}{\sqrt{n+1}+\sqrt{n}} \\
&> \frac{1}{2\sqrt{n+1}} \qquad (\text{since $\sqrt{n}<\sqrt{n+1}$})
\end{align}
(b) The base case is satisfied, since $H_7 = 2.592\ldots < 2.645\ldots = \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $m$ and $n$ are integers, show that $\left|\sqrt{3}-\frac{m}{n}\right| \ge \frac{1}{5n^{2}}$ If $m$ and $n$ are integers, show that $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr| \ge \dfrac{1}{5n^{2}}$.
Since $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr|$ is equivalent to $\biggl|\dfrac{ \sqrt{3}n-m}{n}\biggr|$
So I performed the fol... | Using a technique similar to Liouville's theorem, $\sqrt{3}$ is a root of $P_2(x)=x^2-3$. Then, for any $\frac{m}{n}$ we have an $\varepsilon$ in between $\sqrt{3}$ and $\frac{m}{n}$ such that (this is MVT)
$$\left|P_2\left(\frac{m}{n}\right)\right|=
\left|P_2(\sqrt{3})-P_2\left(\frac{m}{n}\right)\right|=
|P_2'(\vareps... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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If $\cos(A+B+C)=\cos A\cos B\cos C\neq 0$, then evaluate $\left|\frac{8\sin(A+B)\sin(B+C)\sin(C+A)}{\sin 2A\sin 2B\sin 2C}\right|$
If
$$\cos(A+B+C)=\cos A\cos B\cos C, \quad\text{with}\;A,B,C\neq \frac{k\pi}{2}$$
then
$$\left|\frac{8\sin(A+B)\sin(B+C)\sin(C+A)}{\sin 2A\sin 2B\sin 2C}\right|$$ is what integer?... | Using $\sin(2x)=2\sin(x)\cos(x)$ along with $\sin(A+B)=\sin(\pi-C)=\sin(C)$ and $\cos(A)\cos(B)\cos(C)=\cos(A+B+C)=\cos(\pi)=-1$ should give the value as $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that $f'\left(\frac13\right)$ does not exist? Consider the function $f(x)=\begin{cases}
x^2\left|\cos\dfrac{\pi}{2x}\right|, &x\ne 0\\
0, & x=0
\end{cases}$
$$f(x)=\begin{cases}
x^2\cos\dfrac{\pi}{2x}, &\dfrac{\pi}{2x}\in\left[2n\pi-\dfrac{\pi}{2},2n\pi+\dfrac{\pi}{2}\right]\\
-x^2\cos\dfrac{\... | For a derivative to exist, the right hand derivative must be equal to the left hand derivative.
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{f(x)-f(x-h)}{h}$$
Also the limits for defining functions will be $[\frac{1}{4n+1},\frac{1}{4n-1}] \text{ and }(\frac{1}{4n+3},\frac{1}{4n+1})$
For $x=1/3$, the le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does $49\cdot\left(\frac{1}{7}\right)^{x}-\left(\frac{1}{7}\right)^x$ become $\left(\frac{1}{7}\right)^{x}\left(49-1\right)$? Rewrite $\left(\frac{1}{7}\right)^{x-2}-\left(\frac{1}{7}\right)^x$ as $A\cdot\left(\frac{1}{7}\right)^x$ is a question I got.
Now the answer to A is 48, but I don't really get how they got ... | Note that, for all $a$ and $b$, due to the distributive property, you have
$$(a-1)b = ab - b \tag{1}\label{eq1A}$$
Thus, with $a = 49$ and $b = \left(\frac{1}{7}\right)^x$, you get what you're asking about, i.e.,
$$49\left(\frac{1}{7}\right)^x - \left(\frac{1}{7}\right)^x = (49 - 1)\left(\frac{1}{7}\right)^x \tag{2}\la... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Find the smallest positive number k For any positive number $n$, let $a_n = \sqrt{2+\sqrt{2+{...+\sqrt{2+\sqrt 2}}}}$ ($2$ appear $n$) and let $k$ is positive number such that $\displaystyle\frac{1}{k}\leq\frac{3-a_{n+1}}{7-a_n}$ for any positive number $n$, then find the smallest positive number $k$.
I have $a_1=\sqrt... | You have given a lower & upper bound for $a_n$. With this, you've determined a possible value for $k$, but you haven't shown it's necessarily the smallest such value of $k$.
Instead, note that $a_{n+1} = \sqrt{2 + a_n}$. Thus, you have
$$\begin{equation}\begin{aligned}
\frac{3-a_{n+1}}{7-a_n} & = \frac{3-\sqrt{2 + a_n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the value of $a+b+c+d$ with following conditions?
$$
\begin{split}
a^2 &+b &+c &+d&=10\\
a &+b^2 &+c &+d&=12\\
a &+b &+c^2 &+d&=16\\
a &+b &+c &+d^2&=22\\
a &+b &+c &+d&=?
\end{split}
$$
By adding these equations I get that $3(a+b+c+d)=60-(a^2+b^2+c^2+d^2)$.
So can I get some hint that... | Subtract the first equation from the second to get $b^2 + a - a^2 - b = 2$, which becomes $2 = (b - a)(b + a - 1)$.
Assuming that $a,b,c,d$ are all positive integers, then $b - a = 1$ and $b + a - 1 = 2$ OR $b - a = 2$ and $b + a - 1 = 1$. The restriction that all four numbers are positive integers gives $a = 1$ and $b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\int\frac{x^2-1}{x^4+x^2+1}dx$
Find
$$
\int\frac{x^2-1}{x^4+x^2+1}dx
$$
$x^4+x^2+1>0$, so not possible to factor, so I guess there is no direct way or through partial fraction decomposition.
$$
\int\frac{x^2-1}{x^4+x^2+1}dx=\int\frac{(x^2-1)^2}{(x^2)^3-1}dx\\
$$
I have no clue of how to solve it, is there an... | At the bottom, write $ x^2 + 1/x^2 = (x + 1/x)^2 -2$. Then assume $ x+ 1/x = t$. Then you're done.
Let me right the detailed solution. Divide the numerator and denominator of the integrand by $x^2.$ Then we have $$ \frac{1-1/x^2}{(x+1/x)^2 - 1}$$. Then do the substitution business. Then just integrate the following ni... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How can I calculate the limit $\lim\limits_{n \to \infty} \frac1n\ln \left( \frac{2x^n}{x^n+1} \right)$. I have the following limit to find:
$$\lim\limits_{n \to \infty} \dfrac{1}{n} \ln \bigg ( \dfrac{2x^n}{x^n+1} \bigg)$$
Where $n \in \mathbb{N}^*$ and $x \in (0, \infty)$.
I almost got it. For $x > 1$, I observed th... | The term reduces to
\begin{align*}
\dfrac{1}{n}\log\left(2-\dfrac{2}{x^{n}+1}\right)&=\dfrac{1}{n}\log 2+\dfrac{1}{n}\log\left(1-\dfrac{1}{x^{n}+1}\right).
\end{align*}
We do the L'Hopital to the second term, it becomes
\begin{align*}
&\lim_{n\rightarrow\infty}\dfrac{\dfrac{1}{1-\dfrac{1}{x^{n}+1}}\dfrac{1}{(x^{n}+1)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Evaluating $\int \sqrt{\frac{5-x}{x-2}}\,dx$ with two different methods and getting two different results I tried Evaluating $\int \sqrt{\dfrac{5-x}{x-2}}dx$ using two different methods and got two different results.
Getting two different answers when tried using two different methods:-
M-$1$:
$$\int \dfrac{5-x}{\sqrt{... | Both your answers are correct. Your two functions differ by $3\pi/4$, so they have the same derivative.
Let $b=\sqrt{\frac{x-2}3}$. For your integral to make sense you need $0\leq b\leq1$ (this comes from $2\leq x\leq5$). Let $a=\arcsin b$. Note $0\leq\arcsin b\leq \tfrac\pi2$, and so the sine is injective when applie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Finding the volume of the tetrahedron with vertices $(0,0,0)$, $(2,0,0)$, $(0,2,0)$, $(0,0,2)$. I get $8$; answer is $4/3$. The following problem is from the 7th edition of the book "Calculus and Analytic Geometry Part II". It can be found in section 13.7. It is
problem number 5.
Find the volume of the tetrahedron who... | your tetrahedron is also a pyramid. with the volume of $\frac{1}{3}\cdot \frac{2\cdot 2}{2}\cdot 2=\frac{4}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
On sums such as $\sum_{k=0}^\infty \binom{2k}{k}\frac{1}{8^k}=\sqrt{2}$ This identity is a special case of a more general formula found here, and known at least since 1972 (published in Abramowitz and Stegun 1972, p. 555.) Many remarkable series involving the inverse of binomial coefficients are known and listed in the... | The Lagrange Inversion Formula provides an appropriate method to derive
\begin{align*}
\sum_{k=0}^\infty \binom{3k}{k}\frac{1}{8^k}=\frac{4\sqrt{10}}{5}\cos\Big(\frac{1}{3}\arcsin \frac{3\sqrt{6}}{8}\Big)\tag{1}
\end{align*}.
Let a formal power series $w=w(t)$ be implicitely defined by a relation $w=t\Phi(w)$, where $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Inequality $\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2} \geqslant \frac{x+y+z}{2}$ Help to prove this Inequality:
If x,y,z are postive real numbers then:
$\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}+\dfrac{z^3}{z^2+x^2} \geqslant \dfrac{x+y+z}{2}$
I tied to use analytic method with convex function but no... | We apply AM-GM Inequality with $x^2 + y^2 \geq 2xy$.
\begin{align*}
&\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}+\dfrac{z^3}{z^2+x^2} \\
&= \dfrac{x^3 + xy^2}{x^2+y^2}+\dfrac{y^3 + yz^2}{y^2+z^2}+\dfrac{z^3 + zx^2}{z^2+x^2} - \left(\dfrac{xy^2}{x^2+y^2}+\dfrac{yz^2}{y^2+z^2}+\dfrac{zx^2}{z^2+x^2}\right) \\
&= x + y + z -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Formula for probability of sums of rolling 2 six-sided dice I'm learning about expected probability and have got stuck.
In summing the outcomes of rolling 2 six-sided dice, how is the probability for each event given by $p_i = \frac{1}{36} (6 - abs(7 - i))p$ please?
Obviously I can check by counting instances of each s... | The best way is indeed to look at the table of the outcomes. Each outcome has a probability of $\frac1{36}$.
$$\begin{array}{|c|c|c|c|c|c|}\hline \text{die 1 / die 2 } & 1 &2 &3 &4 &5 &6 \\ \hline\hline \hline 1 & 2&3 &4 &5 &6 &7 \\ \hline 2 & 3 &4 &5 &6 &7&8 \\ \hline 3 &4 &5 &6 &7&8&9 \\ \hline 4 &5 &6 &7&8&9&10 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that..........? If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that:
$$(\frac{1}{\beta^3}+\frac{1}{\gamma^3}-\frac{1}{\alpha^3})(\frac{1}{\gamma^3}+\frac{1}{\alpha^3}-\frac{1}{\beta^3})(\frac{1}{\alpha^3}+\frac{1}{\beta^3}-\frac{1}{\gamm... | Notice that if we multiplly the polynomial with $x-1$ we get $$x^3(x-1)+(x-1)(x^3+x^2+x+1)=0$$
so $$2x^4-x^3-1=0\implies \boxed{{1\over x^3}= 2x-1}$$
So your expression is $$E=(2\beta + 2\gamma -2\alpha -1)(2\beta -2\gamma +2\alpha -1)(-2\beta + 2\gamma +2\alpha -1)$$
Now notice that $$\beta + \gamma +\alpha =-{1\over ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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How do you calculate the following limit of a standard Brownian motion $lim_{n\rightarrow \infty} E\left[\sum_{k=1}^{n}\left(B(\frac{kt}{n})-B(\frac{(k-1)t}{n}) \right)^2\right]$, where $\{B(t); t\geq 0\}$ is a standard Brownian motion.
Here is my attempt to solve it. I'd appreciate any feedback on it:
Recall that a s... | By linearity of expectation and expanding the quadratic term, we have
\begin{align}
\mathbb E\left[ \sum_{k=1}^n \left(B\left(\frac{kt}n\right) - B\left(\frac{(k-1)t}n\right)\right)^2\right] &= \sum_{k=1}^n \mathbb E\left[B\left(\frac{kt}n\right) - B\left(\frac{(k-1)t}n\right) \right]^2\\
&= \sum_{k=1}^n\mathbb E\left[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3493596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the correlation coefficient between $2$ variables Value $A$ is the product of $2$ integers $X$ and $Y$. Each of the last two integers takes values from the set {$1,2,5$} with probabilities: $0.2$, $0.5$, $0.3$. Determine the correlation coefficient between $X$ and $A$.
Any ideas on how to solve this? I would ap... | We have
\begin{align}
\mathbb P(X = 1) &= \frac15\\
\mathbb P(X = 2) &= \frac12\\
\mathbb P(X = 5) &= \frac3{10}\\
\mathbb P(A = 1) &= \frac1{25}\\
\mathbb P(A = 2) &= \frac15\\
\mathbb P(A = 4) &= \frac14\\
\mathbb P(A = 5) &= \frac3{25}\\
\mathbb P(A = 10) &= \frac3{10}\\
\mathbb P(A = 25) &= \frac9{100},
\end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Why must absolute value be used in this equation? I have a simple equation:
$(x-2)^2 < 3$
My first solution was:
$x-2 < \sqrt{3}$
But this gives me only:
$x < \sqrt{3} + 2$
Which is only one solution, so it's not enough. I've figured it out that I must use:
$|x-2| < \sqrt{3}$
Then, solutions are:
$-\sqrt{3} + 2 < x < \... | If you are not comfortable using $|x-2|$ then you can divide your derivation into two cases as follows:
Case 1: $x-2\ge 0$ :
$(x-2)^2 < 3 \\ \Rightarrow 0 \le x-2 < \sqrt 3 \\ \Rightarrow 2 \le x < 2+\sqrt{3}$
Case 2: $x-2< 0$ :
$(x-2)^2 < 3 \\ \Rightarrow -\sqrt 3 < x-2 < 0\\ \Rightarrow 2-\sqrt 3 < x < 2$
Taking the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How does $\tan 70^\circ - \sec 10^\circ$ have the exact value of $\sqrt{3}$? Recently, while trying to solve a problem posed by the Youtube video here, I found the following relation:
$$\tan 70^\circ - \sec 10^\circ = \sqrt{3}$$
This relation is exact, and that can be proved by combining the purely geometric arguments ... | Great answer by Michael! Since his answer was a bit terse and I had to spend some time figuring it out, here is an expansion. Since $\tan 60^\circ = \sqrt{3}$, and because $\sec 10^\circ = \frac {1} {\cos 10^\circ}$ by definition, we can say that we need to prove
$$
\frac {1} {\cos 10^\circ} = \tan 70^\circ - \tan 60^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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If both roots of the equation $ax^2-2bx+5=0$ are $\alpha$ and roots of the equation $x^2-2bx-10=0$ are $\alpha$ and $\beta$.
find $\alpha^2+\beta^2$
Both equations have a common root
$$(-10a-5)^2=(-2ab+2b)(20b+10b)$$
$$25+100a^2+100a=60b^2(1-a)$$
Also since the first equation has equal roots
$$4b^2-20a=
0$$
$$b^2=... | From the first equation, $\alpha^2=\frac{5}{a},2\alpha=\frac{2b}{a}$ and we obtain $\alpha =\frac{5}{b}$.
From the second equation
$\alpha + \beta=2b, \alpha\beta=-10$. Then $\beta=-2b$ and $\frac{5}{b}=4b$ i.e.
$b^2=\frac{5}{4}.$
Then $\alpha^2+\beta^2=(\alpha+\beta)^2 -2\alpha\beta=4b^2+20=25.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim\limits_{n \to \infty} \int\limits_0^n \frac1{1 + n^2 \cos^2 x} dx$. I have to find the limit:
$$\lim\limits_{n \to \infty} \displaystyle\int_0^n \dfrac{1}{1 + n^2 \cos^2 x} dx$$
How should I approach this?
I kept looking for some appropriate bounds (for the Squeeze Theorem) that I could use to determine the ... | Let
$$ S_n=\int_0^\pi\frac{1}{1+n^2\cos^2x}\,dx $$
Then
$$\lim_{n\to\infty}S_n=0 $$
Assume that
$$ L=\lim_{n\to \infty} \int_0^n \dfrac{1}{1 + n^2 \cos^2 x} dx $$
Then
$$ \left\lfloor\frac{n}{\pi} \right\rfloor S_n\le L\le\left(\left\lfloor\frac{n}{\pi} \right\rfloor+1\right)S_n $$
So
$$ L=\lim_{n\to\infty}\left\lflo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Linear transformation with respect to basis problem Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ such that $T(\begin{bmatrix} 3 \\ 1 \end{bmatrix}) = \begin{bmatrix} 1 \\2 \end{bmatrix}$ and $T(\begin{bmatrix} -1 \\ 0 \end{bmatrix}) = \begin{bmatrix} 1 \\1 \end{bmatrix}$. Find the matrix $A$ representing $T$.
I understand t... | $\begin {bmatrix} 1\\0 \end{bmatrix} = - \begin {bmatrix} -1\\0 \end{bmatrix}$
$T(\begin {bmatrix} 1\\0 \end{bmatrix} ) = -T(\begin {bmatrix} -1\\0 \end{bmatrix}) = -\begin {bmatrix} 1\\1 \end{bmatrix} = \begin {bmatrix} -1\\-1 \end{bmatrix}$
$\begin {bmatrix} 0\\1 \end{bmatrix} = \begin {bmatrix} 3\\1 \end{bmatrix} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Given positives $a, b, c$, prove that $\frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$.
Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$
Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2... | Let $a+b+c=p$, then consider a function $f(a)=\dfrac{a}{(p-a)^2}$, then $f''(a)=\dfrac{2(a+p)}{(p-a)^4} >0$. So from the Jensen's in equality it follows that
$$\frac{f(a)+f(b)+f(c)}{3} \ge f\left[\frac{a+b+c}{3}\right]$$
So we get $$\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+c)^2} \ge 3*\frac{p/3}{(p-p/3)^2}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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An old and interesting problem in combinatorics from Russia Mathematics Olympiad Can the numbers from $1$ to $81$ be written on a $9 \times 9$ board, so that the sum of the numbers in each $3\times 3$ square is the same?
I believe I have not made much progress and am missing the key insight here. Any advice?
| I believe if you take the following matrix:
$$\left[\begin{array}{ccccccccc}0&3&6&0&3&6&0&3&6\\1&4&7&1&4&7&1&4&7\\2&5&8&2&5&8&2&5&8\\3&6&0&3&6&0&3&6&0\\4&7&1&4&7&1&4&7&1\\5&8&2&5&8&2&5&8&2\\6&0&3&6&0&3&6&0&3\\7&1&4&7&1&4&7&1&4\\8&2&5&8&2&5&8&2&5\end{array}\right]$$
it already satisfies the conditions, and so does the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 0
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Show: For every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.
Show that for every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.
What I did:
Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix... | Pick some $X$ that maps some vector $v$ to another (independent) vector $w$ and that $w$ maps to $-\alpha v$.
*
*What does that imply for $X^2$?
*How many different $X$ can you obtain this way? (Say, let $v=e_1$, then how many choices for $w$ are possible?)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove the positive sequence $a_{n+1} = \sqrt{1+\frac{a^2_n}{4}} $ is strictly increasing for $0 \leq a_0<\frac{2}{\sqrt{3}}$ My attempt:
$a_{n+1} - a_n= \sqrt{1+\frac{a^2_n}{4}} -a_n > \frac{a_n}{2}-a_n = -\frac{1}{2}a_n$, which doesn't tell me any thing. How do I prove that this sequence is strickly increasing? Than... | $a_n^2=1+\frac{a_{n-1}^2}{4}=\frac{1}{4}\left(4+a_{n-1}^2\right)=\frac{1}{4^2}\left(4^2+1\cdot4+a_{n-2}^2\right)=\frac{1}{4^3}\left(4^3+4^2+4+a_{n-3}^2\right)$
Hence, a recursive formula can be written as follows: $$ a_{n+1}^2=\frac{1}{4^n}\left(4^n+4^{n-1}+...4+a_1\right)=\frac{4}{3}\cdot\frac{4^n-1}{4^n}+\frac{a_1^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a^3 + b^3 + c^3 = 3abc$ , can this be true only when $a+b+c = 0$ or $a=b=c$, or can it be true in any other case? Since
$$
a^3 + b^3 + c ^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc ),
$$ from this it is clear that if $(a+b+c) = 0$ , then $a^3 + b^3 + c ^3 - 3abc = 0$ and
$a^3 + b^3 + c ^3 = 3abc$ . Also if $a=... | Without loss of generality, suppose that $a\ge b$ and $a\ge c$. Then $$a^2+b^2+c^2-ab-bc-ca=(a-b)(a-c)+(b-c)^2\ge 0.$$
Equality can only occur if $a=b=c$ and so there are no other solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Can any polynomials in the rational field be decomposed like this I' ve learned that the following examples can be used to decompose a
factor in this way:
x^5 - 5 x + 12 = (x - a) (a^4 - (5 a^3)/8 + (7 a^2)/8 + 1/8 (5 a^2 - 12 a) +
1/8 (5 a^3 - 12 a^2) + ((5 a^4)/16 - a^3/8 + (7 a^2)/16 -
3/16 (5 a^2 - 12 a... | Every (monic) polynomial $f(x)\in \Bbb{Q}[x]$ factorizes into $$f(x) = \prod_{j=1}^{\deg(f)} (x-\beta_j)=\prod_{j=1}^{\deg(f)} (x-g_j(\alpha)), \qquad g_j\in \Bbb{Q}[x]$$
$\beta_j$ are the roots, they are algebraic numbers, they generate $f$'s splitting field $$K=\Bbb{Q}(\beta_1,\ldots,\beta_{\deg(f)})$$
The primitive ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Why does this would eventually simplify into the original circle equation? I was trying to solve this problem:
The point $A$ has coordinates $(5, 16)$ and the point $B$ has coordinates $(-4,4)$. The variable $P$ has coordinate $(x,y)$ and moves on a path such that $ AP = 2BP$. Show that the Cartesian equation of the pa... | I'll suppose you use the usual euclidian distance between two points. Then the distance between $A$ and $P$ is
$$d(A, P) =\sqrt{(x-5)^2+(y-16)^2}$$
Same goes for the distance between $B$ and $P$.
$$d(B, P)=\sqrt{(x-(-4))^2+(y-4)^2}$$
Since $d(A, P)=D(B, P)$, we have
$$\sqrt{(x-5)^2+(y-16)^2} =2\times\sqrt{(x+4)^2+(y-4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3522446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$I_n=\int_0^1{\frac{x^n}{x^n+1}}$. Prove $I_{n+1} \le I_n$ for any $n \in \mathbb N$ $$I_n=\int_0^1{\frac{x^n}{x^n+1}}$$
Prove $\lim_{n\to\infty}{I_n} = 0$
Here is what I tried.
First, I rewrite $I_n$.
$$I_n=\int_0^1{1-\frac{1}{x^n+1}}=1 - \int_0^1{\frac{1}{x^n+1}}$$
Now the limit becomes:
$$L=1-\lim_{n\to\infty}\int_... | $I_n
=\int_0^1{\frac{x^n}{x^n+1}}dx
=\int_0^1{\frac{x^n+1-1}{x^n+1}}dx
=1-\int_0^1{\frac{1}{x^n+1}}dx
=1-J_n$.
Need to show that
$J_{n+1} \ge J_n
$.
$\begin{array}\\
J_{n+1} - J_n
&=\int_0^1{\frac{1}{x^{n+1}+1}}dx-\int_0^1{\frac{1}{x^n+1}}dx\\
&=\int_0^1(\frac{1}{x^{n+1}+1}-\frac{1}{x^n+1})dx\\
&=\int_0^1\frac{x^n+1-(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Special Eigenvalues of a Matrix in Strang p.368 This question arises from Strang's Linear Algebra p.368.
It concerns the matrix
$$A = \left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 \\
1 & 1 & 1 & 1 \\
\end{array}
\right)$$.
Some straightforward computation shows that
$$AA^T = \left(
\... | If we're looking for the fact that each eigenvalue can be expressed as $2 - 2 \cos \theta$ for some $\theta$, which is to say that the eigenvalues fall on the interval $[0,4]$, then there is indeed a straightforward way to confirm that this is the case.
It suffices to note that the eigenvalues are real because $(AA^T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\frac{ 1- \sin\frac{5\pi}{18}}{\sqrt{3} \sin \frac{5\pi}{18}}= \tan\frac{\pi}{18} $ It's easy to verify on WolframAlpha, but I have difficulty deriving it.
It's easy to see
$$
\tan\left(\frac{\pi}{18}\right)=\frac{\sqrt{3}-\tan(5/18 \pi)}{1+\sqrt{3}\tan(5/18 \pi)}
$$
| Evaluate,
$$\frac1{\sin\frac{5\pi}{18}}-\sqrt3 \tan\frac{\pi}{18}
=\frac{\cos\frac{\pi}{18}-\sqrt3 \sin\frac{5\pi}{18}\sin\frac{\pi}{18}}{\sin\frac{5\pi}{18}\cos\frac{\pi}{18}}\tag 1$$
Examine the numerator,
$$\cos\frac{\pi}{18}-\frac{\sqrt3}2\cdot2 \sin\frac{5\pi}{18}\sin\frac{\pi}{18}$$
$$= \cos\frac{\pi}{18}-\cos\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Factor pairs problem Is it possible to find a number $N$ such that it has two factor pairs {a,b}, {c,d} such that (a+1)(b+1)=(c+1)(d+1)
My intuition tells me this is impossible in the case where a,b,c,d are positive integers. However I’m curious if this is still true for negative $N$ too. For negative $N$, assume $a$ a... | Let $a, b, c, d, N \in \mathbb{Z}\setminus\{0\}$ such that $N = ab = cd$. Suppose that $(a+1)(b+1) = (c+1)(d+1)$.
Expanding $(a+1)(b+1) = (c+1)(d+1)$ we get that $a+b = c + d$.
Now substituting $b = \frac{N}{a}$ and $d = \frac{N}{c}$ we get $a+\frac{N}{a} = c + \frac{N}{c}$.
Rearranging we have $(a-c)(ac-N)=0$ and usin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sin x + \arcsin x > 2x$ using Maclaurin series My teacher asked us to solve this problem using the Maclaurin series, but I could not figure out how to approach..
Prove that the inequality sin x + arcsin x > 2x holds for all values of x such
that 0 < x ≤ 1.
I know that the Maclaurin series of
sin(x) = x - $\fra... | Hint: A nicer representation of the arcsin series is
$$\arcsin(x) = \sum_{n = 0}^\infty \frac{1}{4^n} {2n \choose n} \cdot \frac{x^{2n + 1}}{2n + 1}$$
Since both $\sin(x)$ and $\arcsin(x)$ are both odd, we can look at the coefficients on all the odd-degree terms. Can you prove that for all $n \geq 1$,
$$\frac{1}{4^n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Finding power series representation of the fucntion $f(x) = \frac{1+x^2}{1-x^2}$. So I want to find the power series representation of the function $f(x) = \frac{1+x^2}{1-x^2}. $ So how do I solve this?
I can start by reducing the function to the known form of geometric series but I have a summation at the numerator. H... | I would go straight using with $X=x^2$
$${1\over 1-X}=\sum_{n=0}^\infty X^n$$
This leads to
$${1+x^2\over 1-x^2}=\left(1+x^2\right)\sum_{n=0}^\infty x^{2n}$$
Now rearranging the sum
$${1+x^2\over 1-x^2}=1+2\sum_{n=1}^\infty x^{2n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Bernoulli First Order ODE I want to know if my answer is equivalent to the one in the back of the book. if so what was the algebra? if not then what happened?
$$x^2y'+ 2xy = 5y^3$$
$$y' = -\frac{2y}{x} + \frac{5y^3}{x^2}$$
$n = 3$
$v = y^{-2}$
$-\frac{1}{2}v'=y^{-3}$
$$\frac{-1}{2}v'-\frac{2}{x}v = \frac{5}{x^2}$$
$$v'... | You switched one sign too many in
$$
-\frac12v'+\frac2xv=\frac5{x^2}
$$
Then
$$
\left(\frac{v}{x^4}\right)'=\frac{v'}{x^4}-\frac{4v}{x^5}=-\frac{10}{x^6}
\implies
\frac{v}{x^4}=\frac2{x^5}+C
$$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find $a$ such that the minimum and maximum distance from a point to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ Context: I have to solve the following problem: find $a\in \mathbb{R},\, a>0\,\, /$ the minimum and maximum distance from $(4,2)$ to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ respectively.... | The point $(4,2)$ lies on the line $y=\frac{1}{2}x$ which passes through the center of the circle.
Now the minimum and maximum is happened in the two intersection points of the line and the circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Integral with binomial to a power $\int\frac{1}{(x^4+1)^2}dx$ I have to solve the following integral:
$$\int\frac{1}{(x^4+1)^2}dx$$
I tried expanding it and then by partial fractions but I ended with a ton of terms and messed up. I also tried getting the roots of the binomial for the partial fractions but I got complex... | We first decrease the power 2 using integration by parts.
$$\begin{aligned} \int \frac{d x}{\left(1+x^{4}\right)^{2}} &=-\frac{1}{4}\int \frac{1}{x^{3}} d\left(\frac{1}{1+x^{4}}\right) \\ &=-\frac{1}{4}\left[\frac{1}{x^{3}\left(1+x^{4}\right)}+3 \int \frac{d x}{x^{4}\left(1+x^{4}\right)}\right] \\ &=-\frac{1}{4}\left[\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Prove that if $ (x+y)$ is even, then $(x−y)$ is even, for integers. Did I prove correctly? is this a direct proof? For $x+y$ to be even, either $x$ and $y$ are both even, or $x$ and $y$ are both odd.
If $x$ and $y$ are both even we obtain: $x=2k$ and $y=2j$.
substituting into $x-y$ we get $2k-2j$.
$2(k-j)$ is even, so ... | Note that if the difference between $p^2$ and $q^2$ is a multiple of $4,$ then $p$ and $q$ must be of equal parity, for otherwise, we have wlog that $(2m+1)^2-(2n)^2=(2m+1-2n)(2m+1+2n)=(2M+1)(2N+1),$ which is never even.
Now since $(x+y)^2-(x-y)^2=4xy,$ it follows that $x+y$ and $x-y$ are of equal parity, and in partic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
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.