Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the generating function or closed form for the recurrence relation $a_n = a_{n-1} + 4a_{n-2} + 2a_{n-3}$ I was trying to solve this recurrence relation using generating function
$a_n = a_{n-1} + 4a_{n-2} + 2a_{n-3} \qquad : \quad a_0 =1,a_1 =1,a_2 =5, $
I did in the following way
$
\begin{align*}
&G(x) = \sum_{n... | Hint: factorise the denominator and thereby rewrite the fraction as a linear combination of factors of the form $\frac{1}{1-\lambda x}$. And to get you started, one root by inspection is $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $x^2 + y^2 + z^2 + 2xyz = 1$ then $x+ y+ z \leq \frac{3}{2}?$ True or false?
If $x \geq 0, y \geq 0, z \geq 0 $ and $x^2 + y^2 + z^2 + 2xyz = 1$ then $x+ y+ z \leq \frac{3}{2}.$
I want to know if there is a way to demonstrate this conditional inequality.
I know I can make a connection with two properties known in a ... | If $z=0$ so $x+y+z=x+y\leq\sqrt{2(x^2+y^2)}=\sqrt2<\frac{3}{2}$.
Thus, it remains to prove our inequality for $xyz\neq0$.
Let $x=\frac{a}{\sqrt{(a+b)(a+c)}}$ and $y=\frac{b}{\sqrt{(a+b)(b+c)}}$,where $a$, $b$ and $c$ are positives.
Hence, $z=\frac{c}{\sqrt{(a+c)(b+c)}}$ and we need to prove that $\sum\limits_{cyc}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the closed form for a recurrence relation I'm having trouble finding a closed form for a geometric recurrence relation where the term being recursively multiplied is of the form (x+a) instead of just (x).
Here's the recursive sequence:
$a_{n} = 4a_{n-1} + 5$ for $n \geq 1$ with the initial condition $a_{0} = 2... | You have
\begin{aligned}
a_n-4a_{n-1}&=5\\\\
4a_{n-1}-4^2a_{n-2}&=5\cdot4\\\\
4^2a_{n-2}-4^3a_{n-3}&=5\cdot4^2\\\vdots\\
4^{n-1}a_1-4^na_0&=5\cdot4^{n-1}
\end{aligned}
so
$a_n-4^na_0=5\sum\limits_{k=0}^{n-1}4^k=5\cdot\dfrac{4^n-1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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If $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$, $\lim_{n\to\infty}\frac{a_{2n}}{a_n}=\frac{1}{2}$ then $\lim_{n\to\infty}\frac{a_{3n}}{a_n}=\frac{1}{3}$ Let $\{a_n\}$ be a decreasing sequence and $a_n>0$ for all $n$.
If $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ and $\displaystyle\lim_{n\to\infty}\frac{a_{2n}}... | The limit of $\frac{a_{3n}}{a_n}$ need not be $\frac13$, because it need not exist.
Each $n>0$ can be written uniquely as $2^k + r$, where $2^k \le n < 2^{k+1}$. Define $a_n$ (in terms of these $k$ and $r$) piecewise as follows: $$a_n = \begin{cases} \frac{2^{1 - r/2^{k-1}}}{2^k} & \mbox{if }r < 2^{k-1} \\
\frac{1}{2^k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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Solve $\int \frac{1}{\sin ^{\frac{1}{2}}x \cos^{\frac{7}{2}}x}dx$ So it's given this indefinite integral $$\int \frac{1}{\sin ^{\frac{1}{2}}x \cos^{\frac{7}{2}}x}dx$$
Is there anyone could solve this integral? Thanks in advance.
| With $u=\tan(x)$, you get $du=\frac{dx}{\cos^2(x)}$ : $$\int \frac{1}{\sin ^{\frac{1}{2}}x \cos^{\frac{7}{2}}x}dx=\int \frac{u^2+1}{\sqrt u}du=\int u^{\frac{3}{2}}du+\int \frac{1}{\sqrt u}du\\=\frac{2u^{\frac{5}{2}}}{5}+2\sqrt u + C=\frac{2\tan^{\frac{5}{2}}(x)}{5}+2\sqrt{\tan(x)} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Prove that the gcd is 1. I need to show that for $x$ odd the $\gcd\bigg(x,\dfrac{x^2-1}{2},\dfrac{x^2+1}{2}\bigg)=1$. Im trying it doing it pairwise. I already show that $\gcd\bigg(\dfrac{x^2-1}{2},\dfrac{x^2+1}{2}\bigg)=1$, but I dont have an idea how to show that $\gcd\bigg(x,\dfrac{x^2-1}{2}\bigg)=1$ and $\gcd\bigg(... | Hint: Since you already proved that $\gcd\left(\frac{x^2-1}{2},\frac{x^2+1}{2}\right)=1$, assume that $\gcd\left(x,\frac{x^2-1}{2},\frac{x^2+1}{2}\right)=d>1$ which means $d \mid \frac{x^2+1}{2}$ and $d \mid \frac{x^2-1}{2}$ or $\gcd\left(\frac{x^2-1}{2},\frac{x^2+1}{2}\right)\geq d>1$ a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Number of ways to get a sum 11 using numbers 1,3 and 5? How can we find the number of ways to get sum $11$ using numbers $\{1,3,5\}$? Repetition of numbers is allowed. For example, one way is to take number $1$ and sum it $11$ times. Another is $5+5+1$. Ordering of numbers does not matter.
| One method is to use generating functions:
\begin{align}
& (1+x+x^2+x^3+\cdots)(1+x^3+x^6+x^9+\cdots)(1+x^5+x^{10}+x^{15}+\cdots) \\
& = \frac{1}{1-x} \cdot \frac{1}{1-x^3} \cdot \frac{1}{1-x^5} \\
& = 1 + x + x^2 + 2x^3 + 2x^4 + 3x^5 + 4x^6 + 4x^7 + 5x^8 + 6x^9 + 7x^{10} \\
& \quad + 8x^{11} + 9x^{12} + 10x^{13} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Series Convergence / Divergence: $\sum \frac{3^{1/\sqrt{n}}-1}{n}$ Does the following series converge or diverge:
$$\sum_{n=1}^\infty \frac{3^{1/\sqrt{n}}-1}{n}$$
Many thanks!
| For any $a > 1$ and $b > 0$,
consider
$\sum_{n=1}^\infty \frac{a^{1/n^b}-1}{n}
$.
I will show that this sum converges.
OP's problem is
$a=3$ and $b = \frac12$.
$a^{1/n^b}
=e^{\ln a/n^b}
$.
If $0 < x < \frac12$,
$\begin{array}\\
e^x
&=\sum_{k=0}^{\infty} \frac{x^k}{k!}\\
&<\sum_{k=0}^{\infty} x^k\\
&=\frac1{1-x}\\
&< 1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to find constant term in two quadratic equations Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - x + p=0$ and let $\gamma$ and $\delta$ be the roots of the equation $x^2 -4x+q=0$. If $\alpha , \beta , \gamma , \delta$ are in Geometric progression then what is the value of $p$ and $q$?
My approach:
From... | $\alpha + \beta = 1$
Also,
$\alpha + \beta = \frac{a}{r^3} + \frac{a}{r^1}$
So we have,
$\frac{a}{r^3} + \frac{a}{r^1} = 1$
Put value of $r$,
$\frac{a}{2\sqrt2} + \frac{a}{\sqrt 2}= 1$
$a = \frac{2\sqrt2}{3}$
Now you can find roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why is $5^{n+1}+2\cdot 3^n+1$ a multiple of $8$ for every natural number $n$? I have to show by induction that this function is a multiple of 8. I have tried everything but I can only show that is multiple of 4, some hints? The function is
$$5^{n+1}+2\cdot 3^n+1 \hspace{1cm}\forall n\ge 0$$, because it is a multiple of... | By induction:
It is true for case $n=1$, Let it be true for $n=k$ then
$5^{k+2} +2.3^{k+1} +1 = 5.5^{k+1}+3.2.3^k+1 = 2(5^{k+1} -1) + 3(5^{k+1}+2.3^k+1)$
The second part is a multiple of $8$ and one can easily show that $(5^{k+1} -1)$ is a multiple of $4$.
Hint for showing $(5^{k+1} -1)$ is a multiple of $4$:
$5^{k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Use series to approximate the definite integral I. (Round your answer to three decimal places.) **Use series to approximate the definite integral $I$. (Round your answer to three decimal places.)
$$I=\int _0^1x \cos\left(x^4\right) \, dx$$
I understand:
$$\int _0^1x \cos(x^2) \, dx=\frac{\sin(1)} 2 \quad \left(\text{De... | It said "Use series".
You have
$$
\cos x = 1 - \frac{x^2} 2 + \frac{x^4}{24} - \frac{x^6}{720} + \cdots.
$$
So
\begin{align}
\cos(x^4) & = 1 - \frac{(x^4)^2} 2 + \frac{(x^4)^4}{24} - \frac{(x^4)^6}{720} + \cdots \\[10pt]
& = 1 - \frac{x^8} 2 + \frac{x^{16}}{24} - \frac{x^{24}}{720} + \cdots
\end{align}
and then
$$
x\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that $\sum_{n=1}^\infty \frac{n^2}{(n+1)!}=e-1$ Show that:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=e-1$$
First I will re-define the sum:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!} = \sum^\infty_{n=1} \frac{n^2-1+1}{(n+1)!} - \sum^\infty_{n=1}\frac{n-1}{n!} + \sum^\infty_{n=1} \frac{1}{(nm)!}$$
Bow I will define e:
$$e^... | $$\frac{n^2}{(n+1)!}=\frac{(n+1)^2-2(n+1)+1}{(n+1)!}=\frac{n+1}{n!}-2\frac{1}{n!}+\frac{1}{(n+1)!}=\frac{1}{(n-1)!}-\frac{1}{n!}+\frac{1}{(n+1)!}$$
so
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=\sum^\infty_{n=1}\frac{1}{(n-1)!}-\sum^\infty_{n=1}\frac{1}{n!}+\sum^\infty_{n=1}\frac{1}{(n+1)!}=(e)-(e-1)+(e-2)=e-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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How prove this inequality $\sum\limits_{cyc}\sqrt{\frac{yz}{x^2+2016}}\le\frac{3}{2}$ Given $x,y,z$ are positive real number satisfy $xy+yz+xz=2016$. Prove that $$\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{z^2+2016}}+\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{3}{2}$$
I tried
$\sqrt{\frac{yz}{x^2+2016}}=\sqrt{\frac{yz}{x^... | By AM-GM
$$\sum_{cyc}\sqrt{\frac{xy}{z^2+2016}}=\sum_{cyc}\sqrt{\frac{xy}{z^2+xy+xz+yz}}=\sum_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\leq$$
$$\leq\frac{1}{2}\sum_{cyc}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{x}{x+z}+\frac{z}{z+x}\right)=\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Calculate $\int^1_{-1}\sqrt{(1-x^2)^3} dx$ using Integration by parts or substitution technique How can I calculate $\int^1_{-1}\sqrt{(1-x^2)^3} dx$?.
I'm trying to solve it by using substitutions or integrate by parts but none of those leads me to the solution.
My attempt (with the help of a friend):
first since $(1-... | "$1-x^2$" should immediately make you think about the substitution $x=\sin u$, as $1-x^2 = 1-\sin^2 u = \cos^2 u$. So let $x = \sin u$, then $dx = \cos u \,du$ and note that $\sin (-\tfrac{\pi}{2}) = -1$ and $\sin (\tfrac{\pi}{2}) = 1$, so
\begin{align}
\int_{-1}^{1} (1-x^2)^{3 / 2}\,dx
&= \int_{-\pi/2}^{\pi/2} (1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to prove an upper bound on polynomial $x + \sqrt{2} x^2 + \sqrt{3} x^3 + \ldots + \sqrt n x^n$ If $P(x) = x + \sqrt 2 x^2 + \sqrt{3}x^3+\ldots + \sqrt n x^n$ how can I prove that if $x > 1$ then $$ P(x) \leq \sqrt {\frac{n(n+1)}{2} \frac{x^{2n+1}-1}{x^2-1}}$$
given that $ 1 + 2 + \ldots + n = n(n+1)/2$ and some geo... | As didgogns pointed out, this is an application of the Cauchy-Schwarz inequality. The idea is as follows.
$$(\sum_{k=1}^n u_k v_k)^2\leq(\sum_{k=1}^n u_k^2)(\sum_{k=1}^n v_k^2)$$
Substituting $u_k=\sqrt{k}$ and $v_k = x^k$ gives you
$$P(x) \leq \sqrt {(\sum_{k=0}^n k)*(\sum_{k=0}^n x^{2n})}$$
And simplifying gives the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2217615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Polynomial and Function I have been give a question if
$ f(x) = x^3-1 $
then show
$$ \frac {f(b) - f(a)} {b-a} =b^2+ab+a^2 $$
how to show that the above fraction is equal to that polynomial if $ f(X) = x^3-1 $ ?
| $$ \frac {f(b) - f(a)} {b-a}$$
$$ = \frac {b^3 - 1 - a^3 + 1} {b-a}$$
$$ = \frac {b^3 - a^3} {b-a}$$
$$ = \frac {(b-a)(b^2+ab+a^2)} {b-a} $$
$$ = b^2 + ab + a^2 $$
Formula used.
$$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to solve $uu_{x_1}+u_{x_2}=1$ with characteristic method? $$uu_{x_1}+u_{x_2}=1\text{ with initial condition }u(x_1,x_1)=\frac{1}{2}x_1$$
I have problem to use following characteristic method to solve it.
Let $$x(s) = (x_1(s), x_2(s))$$$$z(s) = u(x(s))$$$$p(s) = (u_{x_1}(x(s)), u_{x_2}(x(s))$$$$F(x,z,p) = zp_1 + p_2... | $uu_{x_{1}} + u_{x_{2}} = 1, u(x_{1}, x_{1}) = \frac{1}{2} x_{1}$
Let
\begin{cases}
x(s) = (x_1(s), x_2(s))\\
z(s) = u(x(s))\\
p(s) = (u_{x_1}(x(s)), u_{x_2}(s))
\end{cases}
Need to solve the equation:
\begin{cases}
\frac{dx_1}{ds} = z(s) \\
\frac{dx_2}{ds} = 1 \\
\end{cases}
Observe that,
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Checking whether solution of differential curve inersects another curve.
The differential equation is $(x^2+xy+y^2+4x+2y+4)\dfrac {\mathrm dy}{\mathrm dx}=y^2$ passing through $(1,3) $. I have already solved it. The solution of differential curve is:
$$\ln \left|\frac {y}{3e}\right|=-\frac {y}{x+2},x>0$$
I have to c... | Notice $\ln\left|\frac{y}{3e}\right| = \ln|y|- \ln(3e)$
Let $$f(x,y) = \ln|y| + \frac{y}{x+2} -\ln(3e)$$ $$x>0$$
First lets start with simply $y=x+2$. Substitute this:
$$f(x) = \ln|x+2| + \frac{x+2}{x+2} -\ln(3e)$$
If the above expression equals 0 at some point for a positive $x$ then a solution exists.
Notice the modu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Tangent Points for Common Tangent to Two Ellipses This is somewhat similar to my other question here.
Consider the two ellipses given by the equations
\begin{equation}
\frac{x^2}{2^2} + \frac{(y-1)^2}{1^2} = 1
\end{equation}
and
\begin{equation}
\frac{x^2}{1^2} + \frac{(y-4)^2}{(1/2)^2} = 1.
\end{equation}
How do I... | Let such a tangent line be $y=mx+c$
Solving simultaneously with the ellipses gives two quadratics in $x$ which are $$(4m^2+1)x^2+8mx(c-1)+4c^2-8c=0$$ and $$(4m^2+1)x^2+8mx(c-4)+4c^2-32c+63=0$$
Each of these must have double roots at the point of tangency and therefore the discriminant is zero, so this leads to two equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
recurrence relation concrete way to solve it I have the following recurrence relation:
$$a_n = 2a_{n-1} + 2^n; a_0 = 0$$ I used the characteristic equation method and some method I found online by calculating the $n+1$ th term and subtracting accordingly the equation with $a_{n+1}$ minus the equation with $a_{n}$:
$$a_... | Given: $a_n = 2a_{n-1} + 2^n; a_0 = 0$ and you can find $a_n = 2^n A_1 + n \cdot 2^n A_2$. Now we have to find $A_1$ and $A_2$.
We put $n=1$ in $a_n = 2a_{n-1} + 2^n$. So $a_1 = 2a_0 + 2^1=2$.
Now, let's put $n=0$ and $n=1$ in $a_n = 2^n A_1 + n\cdot 2^n A_2$. Hence
$2^0 A_1 + 0\cdot 2^0 A_2 = 0 \implies A_1 =0$
$ 2^1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove $\prod_1^\infty (1+p_n)$ converges Let $p_{2n-1} = \frac{-1}{\sqrt{n}}$, and $p_{2n} = \frac{1}{n}+\frac{1}{\sqrt{n}}$.
Prove $\prod_1^\infty (1+p_n)$ converges.
By numerical simulations, it appears to converge (to something around $0.759$). However, I'm not sure how to prove this. I know we can skip the first t... | I'm not 100% sure if this can be made rigorous, but maybe you can take it as a starting point.
\begin{align*}\log \Pi(1+p_n)= \sum \log((1+\frac{1}{2n}+\frac{1}{\sqrt{2n}})(1-\frac{1}{\sqrt{2n+1}}))\\\approx\sum \log((1+\frac{1}{\sqrt{2n}})(1-\frac{1}{\sqrt{2n}}))\approx\sum \log(1-\frac{1}{2n})\approx \sum -\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the solution of the $x^2+ 2x +3=0$ mod 198 Find the solution of the $x^2+2x+3 \equiv0\mod{198}$
i have no idea for this problem i have small hint to we going consider $x^2+2x+3 \equiv0\mod{12}$
| We can rewrite the equation as \begin{align}x^2+2x+3&\equiv0\mod198\\
(x+1)^2-1+3&\equiv0\mod 198\\
(x+1)^2&\equiv-2\mod198\\
(x+1)^2&\equiv196\mod198\end{align}
We let $y=x+1$ and we now need to solve \begin{align}y^2&\equiv-2\mod 198\\
&\equiv196\mod198\end{align}
We can see that $196=14^2$ and so we can say that $y\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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If condition on $a$ is given, then find the domain of the function If $\log _{\frac{1}{3}} (|a|+1)>-1$, then find the domain of
$$f(x)=\sqrt{2x^4+ax^3-6x^2-4ax-8}$$
Using $\log _{\frac{1}{3}} (|a|+1)>-1$, I got $-2<a<2$ but to use it in $\sqrt{2x^4+ax^3-6x^2-4ax-8} \geq0$
Could someone help me with this.
| HINT:
$$2x^4+ax^3-6x^2-4ax-8=2x^4-6x^2-8+ax(x^2-4)=\\=2(x^4-3x^2-4)+ax(x^2-4)=2(x^2-4)(x^2+1)+ax(x^2-4)=(x^2-4)(2x^2+ax+2)\\$$
NOTE:For factoring $x^4-3x^2-4$ you can use the change of variables $t=x^2$ and use a quadratic.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometric Inequality I have a problem about this trigonometric inequality, which I cannot completely solve. In particular, I cannot get the whole solution the book provides and what a bad luck: I don't have the book with me, because this problem arose from one of my student's problem during a private lesson.
$$\si... | $$\sin\left(\frac{x}{2}\right)\left(2\cos(x) - \sqrt{3}\right) >0 \iff\sin\left(\frac{x}{2}\right)\left(\cos(x) - \frac{\sqrt{3}}{2}\right) >0\\\iff\left(\sin\left(\frac{x}{2}\right)>0\text{ and }\cos(x) - \frac{\sqrt{3}}{2}>0\right)\text{ or } \left(\sin\left(\frac{x}{2}\right)<0\text{ and }\cos(x) - \frac{\sqrt{3}}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving three lines are concurrent
In the triangle $ABC$ the segment $AD$ is altitude. $M$ is the midpoint of $BC$ and $N$ is the reflection of $M$ in $D$. The circumcircle of $ANM$ meets $AB$ at $P$ and $AC$ at $Q$. Show that $AN, BQ, CP$ are concurrent.
All I have tried to do is angle chasing and hopefully get the ... | By Ceva's theorem $AN,BQ,CP$ are concurrent iff
$$ BN\cdot CQ\cdot AP = CN\cdot AQ\cdot BP\tag{1}$$
and the involved lengths are not difficult to compute. From $BD=c\cos B=\frac{a^2+c^2-b^2}{2a}$ and $BM=\frac{a}{2}$ we get $DM=\frac{b^2-c^2}{2a}$ and $NM=\frac{b^2-c^2}{a}$, from which $CN=\frac{a^2+2b^2-2c^2}{2a}$. Si... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Algebra rules when finding inverse modulo The idea is to find an inverse modulo for two numbers, $660$ and $43$ in this case. I find that the GCD is easy to calculate, but the step after that, calculating the inverse modulo when calculating back trough the GCD calculation.
The thing I do not get, is that 'by algebra' t... | I was looking for one of my other answers on the extended Euclidean algorithm to link you to, but since I couldn't find one, here's a table for your question:
$$\begin{array}{|c|c|} \hline
n & s & t & q & \text{equation; }n=\\ \hline
660 & 1 & 0 & & 1\cdot 660 + 0\cdot 43\\ \hline
43 & 0 & 1 & 15 & 0\cdot 660 + 1\cdo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$ We have to find the integration of
$$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$$
In this I tried to do substitution of $x=e^t$
After that got stuck .
| Let $$I = \int^{\infty}_{0}\frac{\ln x}{x^2+2x+4}dx = \int^{\infty}_{0}\frac{\ln x}{(x+1)^2+(\sqrt{3})^2}dx$$
put $\displaystyle x=\frac{1^2+(\sqrt{3})^2}{y} = \frac{4}{y}$ and $\displaystyle dx = -\frac{4}{y^2}dy$
$$I = \int^{0}_{\infty}\frac{\ln (4)-\ln (y)}{16+8y+4y^2}\cdot -\frac{4}{y^2}\cdot y^2 dy = \int^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Find the probability that $m^2-n^2$ is divisible by $4$ Two integers are $m$ , $n$ chosen at random with replacement from the set of integers $1,2,3..9$ . Find the probability that $m^2-n^2$ is divisible by $4$ .
my solution : Two integers can be chosen (with replacement ) in $9\times 9=81$ ways .
$m^2-n^2 $ is divisib... | Everything is fine, the only thing wrong is :
When order is considered, the total ways will be $9 \times 8$ rather than $2 \times 9 \times 8$.
You have counted repeatedly, consider selecting $m$ and $n$ from set {$1,2$}.
Total ordered pairs are : (1,2), (2,1).
Id est $2 \times 1$, not $2 \times 2 \times 1$.
Although ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Solve$\int\frac{x^4}{1-x^4}dx$ Question: Solve $\int\frac{x^4}{1-x^4}dx.$
My attempt:
$$\int\frac{x^4}{1-x^4}dx = \int\frac{-(1-x^4)+1}{1-x^4}dx = \int 1 + \frac{1}{1-x^4}dx$$
To integrate $\int\frac{1}{1-x^4}dx,$ I apply substitution $x^2=\sin\theta.$ Then we have $2x \frac{dx}{d\theta} = \cos \theta.$ which implies... | Hint. Alternatively, one may use a partial fraction decomposition to get
$$
\frac{1}{x^4-1}=\frac{1}{4(x-1)}-\frac{1}{4(x+1)}-\frac{1}{2 \left(x^2+1\right)}
$$ then one may integrate each term easily.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the constant term in the expansion of $(x^2+1)(x+\frac{1}{x})^{10}$ I can't solve this problem. How to solve it?
The Problem is
"Find the constant term in the expansion of $
\left({{x}^{2}\mathrm{{+}}{1}}\right){\left({{x}\mathrm{{+}}\frac{1}{x}}\right)}^{\mathrm{10}}
$"
| We want to find the coefficients of the binomial expansion of $(x+\frac{1}{x})^{10}$ of the constant term and of the term with $\frac{1}{x^2}$.
$$\left(x+\frac{1}{x}\right)^{10}=\sum_{k=0}^{10}\binom{10}{k}x^{10-k}\frac{1}{x^k}$$
To find the constant term, want $10-k=k$, ie: $k=5$. This term is $\binom{10}{5}x^5\cdot x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Triangle - length of the sides - proof a, b and c are the lengths of the sides of a triangle. Prove that
$$a^2+b^2 \ge \frac{1}{2}c^2$$
Let $\gamma$ be the angle between sides a and b. then:
$$a^2 + b^2 - 2ab\cos(\gamma) = c^2$$
Hence we need to prove that
$$a^2+b^2 \ge \frac{1}{2}c^2$$
$$2a^2+2b^2 \ge a^2+b^2 -... | How about this:
Triangle inequality theorem: a + b > c .
The sum of the lengths of two sides of a triangle is greater than the length of the third side.
1) $(a + b)^2 > c^2$ ; or
2)$ a^2 + 2ab + b^2 > c^2$.
Now:
3)$ (a -b)^2 \ge 0$; hence : $a^2 + b^2 \ge 2ab$.
Putting together:
$ 2a^2 + 2b^2 \ge a^2 + 2ab + b^2 \gt c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help calculating a math limit Can anyone help with this limit?
\begin{equation*}
\lim_{x \rightarrow 4}
\frac{16\sqrt{x-\sqrt{x}}-3\sqrt{2}x-4\sqrt{2}}{16(x-4)^2}
\end{equation*}
I've tried a variable change of \begin{equation*} y=\sqrt{x} \end{equation*} but this didn't help.
| HINT: we have $$\frac{16\sqrt{x-\sqrt{x}}-\sqrt{2}(3x+4)}{16(x-4)^2}$$
multiply numerator and denominator by
$$16\sqrt{x-\sqrt{x}}+\sqrt{2}(3x+4)$$
the numerator is given after this multiplication as
$$-2 \left(\sqrt{x}-2\right)^2 \left(9 x+36 \sqrt{x}+4\right)$$
and write $$x-4$$ as $$(\sqrt{x}-2)(\sqrt{x}+2)$$
after ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\int x \sqrt{1- x^2 \over 1 + x^2}dx$.
$$J =\int x \sqrt{1- x^2 \over 1 + x^2}dx$$
Substituting $u = \sqrt{1 + x^2}$
$$J = \int \sqrt{2 - u^2} du = \sqrt{2}\int \sqrt{1 - \left({u \over \sqrt{2}}\right)^2} du$$
Now substituting $\sin z = u/\sqrt{2}$
$$J = 2\int \cos^2 z dz = z + (\sin 2z)/2 + C = \arcsin\l... | $u = \sqrt {1+x^2}$
$u^2 = 1 + x^2$
$2u du = 2x dx$
$u du = x dx$
Here you are wrong.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is $x$ in the formula? What is $x$ (assume it's integer) in $512p+ 1 = x^3$, where $p$ is a prime number.
Attempt:
$$a^3 - b^3 = (a-b) (a^2 + ab+b^2) \Longrightarrow 512p = x^3 -1 = (x-1)(x^2+x+1).$$
Here I got stuck. Am I suppose to plug in $p$ and try it one by one?
What about $16p + 1 = x^3$??
| $512p=2^9p=x^3-1=(x-1)(x^2+x+1)$. Note that $x^3$ is odd (and hence $x$ is odd) since it's an even number plus $1$. When $x$ is odd, then $x^2+x+1$ is also odd and $x-1$ is the only even factor which implies that $p=x^2+x+1$ and $2^9=x-1$ (otherwise $p$ would not be prime or odd). Hence, the only possible solution coul... | {
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"source": "stackexchange",
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Proving $\lim_{(x,y)\to (0,0)} (x^4+y^4)/(x^2+y^2)=0$ by definition I need to prove by definition (and nothing else) that
$$\lim_{(x,y) \to (0,0)}\frac{x^4+y^4}{x^2+y^2} = 0.$$
I've been stuck on this for almost an hour with no luck, and ran out of ideas.
Can anyone help or give a hint?
| Take any $\epsilon > 0$. Suppose that $\|(x,y) - (0,0)\|<\delta(\epsilon) = \sqrt{\epsilon/2}$, which is to say that $x^2 + y^2 <\epsilon/2$. We have
$$
\left|\frac{x^4 + y^4}{x^2 + y^2} - 0 \right| =
\left|\frac{x^4}{x^2 + y^2} + \frac{y^4}{x^2 + y^2}\right| \leq
\left|\frac{x^4}{x^2 + y^2}\right| + \left|\frac{y^4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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For the function $f(x)= \begin{cases}1, & \text{for } -2For the function $f(x)= \begin{cases}1, & \text{for } -2<x<0 \\ x , & \text{for } 0<x < 2, \end {cases}$ find the Fourier series of $f(x)$.
Here, $f(-x)=-f(x)$. So $f(x)$ is Odd function, hence $a_n=0$ and the Fourier series looks like $ f(x) \sim \sum b_n \sin \f... | Define problem
Piecewise function: Resolve $f(x)$ into a left and right piece
$$
\begin{align}
%
l(x) &= 1, \quad -2 \le x < 0 \\
%
r(x) &= x, \quad \ \ \ 0 \le x \le 2
%
\end{align}
$$
Find the Fourier expansion
$$
f(x) = \frac{1}{2}a_{0} +
\sum_{k=1}^{\infty} \left(
a_{k} \cos \left( \frac{k \pi x}{2} ... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction that $\forall n\geq 1,\ 7\mid 3^{2n+1} + 2^{n-1}$ Prove by induction that $$7 \mid 3^{2n+1} + 2^{n-1},\ \forall n\geq 1$$
Base case $n=1$:
$$3^{2 × 1+1} + 2^{1-1} = 28.$$
Induction:
$$P(k): 3^{2k+1} + 2^{k-1},\ P(k+1): 3^{2(k+1)+1} + 2^{(k+1)-1}.$$
$$3^{2k+3} + 2^k = 9 \times 3^{2k+1} + 2^{k-1} \tim... | Note that from your assumption you have:
$3^{2k+1} +2^{k-1} = 7p,\quad p\in\mathbb{Z}$
Then you can change the following line you have:
$9 \cdot3^{2k+1} + 2^{k-1} \cdot 2$
into:
$2(3^{2k+1}+2^{k-1})+7\cdot3^{2k+1} = 2\cdot 7p+7\cdot3^{2k+1}=7(2p+3^{2k+1})$
which is divisible by $7$ as required.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some ... | The desired inequality follows by noting that from the convexity of $f(t) = t^{2},$ we have $ f\left(\dfrac{x+y+z}{3}\right) \leq \dfrac{f(x)+f(y)+f(z)}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 11
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Tangents concurent at (1,1/2)
I got $y'=\frac{1-x^2}{(1+x^2)^2}$..
after tht I got stuck
| If $(s,t)=\left(s,\dfrac{s}{1+s^2}\right)$ is a point of tangency of the graph then the slope there is $m=\dfrac{1-s^2}{(1+s^2)^2}$ so the equation of the line of tangency is
$$ y-\dfrac{s}{1+s^2}= \dfrac{1-s^2}{(1+s^2)^2}(x-s)$$
In order for the line of tangency to contain the point $\left(1,\dfrac{1}{2}\right)$ the f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of product of mixed root indexes I am unable to recover the limit of this function.
$$
\lim_{x \to +\infty} \left (\frac{1}{\sqrt{x}} \times \sqrt[3]{x+1} \right )
$$
I have tried many ways of solving it. These are the two simplest:
*
*Variable substitution
Using $y = \sqrt[3]{x+1} $, with which I got a comm... | Your calculations are right!
$$
\begin{aligned}
\lim _{x\to \infty }\left(\frac{\sqrt[3]{1+x}}{\sqrt{x}}\right)
& = \lim _{t\to 0\:}\left(\frac{\sqrt[3]{1+\frac{1}{t}}}{\sqrt{\frac{1}{t}}}\right)
\\& = \lim _{t\to 0\:}\left(\sqrt[6]{t}\sqrt[3]{t+1}\right)
\\& =0^{\frac{1}{6}}\sqrt[3]{0+1} = \color{red}{0}
\end{aligned}... | {
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"url": "https://math.stackexchange.com/questions/2248708",
"timestamp": "2023-03-29T00:00:00",
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why$2 (-1 + 2^{1 + n})$ is the answer to the recurrence relation $a_{n}=2a_{n-1}+2$? $a_{0}=2$
$a_{1}=2(2)+2$
$a_{2}=2(2(2)+2)+2$
$a_{3}=2(2(2(2)+2)+2)+2$
$a_{4}=2(2(2(2(2)+2)+2)+2)+2$
$a_{5}=2(2(2(2(2(2)+2)+2)+2)+2)+2$
To simplifiy
$a_{6}=2^{6}+2^{5}...2^{1}$
so my answer is
$a_{n}=2^{n+1}+2^{n}+...2^{1}$
The correct... |
It's equivalent to
$a_{n} + 2 = 2\left(a_{n - 1} + 2\right) = 2^{2}\left(a_{n - 2} + 2\right) = \cdots =
2^{n}\left(a_{0} + 2\right) = 4 \times 2^{n}\implies
\bbox[10px,border:1px dotted navy]{\displaystyle a_{n} = 2\left(\,2^{n + 1} - 1\right)}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}$ without L'Hospital's rule How to show that without using L'Hospital's rule
\begin{align}
\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}
\end{align}
I was able to show the upper bound by using th... | Note that below we have used well-known limits $$\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ \frac { 1 }{ x } } } =e\\ \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } } =1$$
$$\lim _{ x\to 0 } \frac { \log \left( \cosh \left( x^{ 2 }-xc \right) \right) }{ x^{ 2 } } =\lim _{ x\to 0 } \frac { \log ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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The line $y=mx+c$ is a tangent to $x^2+y^2=a^2$ if: The line $y=mx+c$ is a tangent to $x^2+y^2=a^2$ if:
$1$. $c=a\sqrt {1+m^2}$
$2$. $c=\pm a\sqrt {1+m^2}$
$3$. $c^2=\pm a\sqrt {1+m^2}$
$4$. $\textrm {None}$
My Attempt:
The tangent to circle $x^2+y^2=a^2$ at point $(x_1,y_1)$ is given by :
$$xx_1+yy_1=a^2$$
Now, what s... | You have $x^2 + (mx+c)^2 = a^2 \iff x^2 + m^2x^2 + 2mxc + c^2 = a^2$ which simplifies to $$(1+m^2)x^2 + 2mcx + (c^2-a^2) = 0$$
But as $y$ must be tangent, there must be a single solution to the above quadratic. Which means its discriminant must be $0$, i.e: $$4m^2c^2-4(1+m^2)(c^2-a^2) = 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\lim_{n \rightarrow \infty} (a_{n+1}-\frac{a_n}{2})=0$ then show $a_n$ converges to $0$. I have been stuck on this question for a while now. I have tried many attempts. Here are two that I thought looked promising but lead to a dead end:
Attempt 1:
Write out the terms of $b_n$:
$$b_1=a_{2}-\frac{a_{1}}{2}$$
$$b_2=a... | Let $b_n=a_{n+1}-(a_{n}/2)$ so that $b_{n} \to 0$ as $n\to\infty$. Now we have $a_{2}=b_{1}+(a_{1}/2),a_{3}=b_{2}+b_{1}/2+a_{1}/4$ and continuing in this manner we get $$a_{n+1}=b_{n}+b_{n-1}/2+\cdots+b_{1}/2^{n-1}+a_{1}/2^{n}$$ Let $\epsilon>0$ be given then we can see that there is a positive integer $m$ such that $|... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many ways to create and select 2 lines from a set of 5 points We have 5 points: A, B, C, D, E
Any two unique points define a line.
How many ways can we select two unique lines?
example) AB-AC form a set of 2 lines. AB-AD, AB-AE, AB-BC, AB-BD, etc.
I tried brute force counting, and I got 40.
However, I was expecting... | Yes, your calculation is correct: There are
$${5\choose 2} = 10$$ lines, and so there are
$${10\choose 2} = 45$$ pairs of lines.
So you must have missed some when doing this by brute force.
And yes, this clearly generalizes for $n$ points:
$${n\choose 2} = \frac{n(n-1)}{2}$$ lines and thus
$${{\frac{n(n-1)}{2}}\choo... | {
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"timestamp": "2023-03-29T00:00:00",
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Find a non-zero integer matrix $X$ such that $XA=0$ where $X,A,0$ are all $4 \times 4$ Let $A$ be the following $4 \times 4$ matrix.
\begin{bmatrix}1&2&1&3\\1&3&2&4\\2&5&3&7\\1&4&1&5\end{bmatrix}
How can we find a non-zero integer matrix $C$ such that $CA = 0_{4 \times 4}$
Note that $0$ is a $4 \times 4$ matrix.
| Solve the equation $A^Tx = 0$. If $A$ is not invertible then it must have a nonzero solution $\vec x_0$.
Then set $B = \begin{pmatrix} \vec x_0 \\ \vec x_0 \\ \dots \\ \vec x_0\end{pmatrix}$ -- the matrix having $\vec x_0$ as rows. Since $\vec x_0^T A = 0$, it means that $BA = 0$ and $B \neq 0$.
Since $\vec x_0$ is a v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the maximum and minimum value of $P =x+y+z+xy+yz+zx$ Let $x^2+y^2+z^2\leq27$ and $P = x+y+z+xy+yz+zx$. Find the value of $x, y, z$ such that $P$ is the maximum value and minimum value.
My attempt :
$$(x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0$$
$$27 \geq x^2+y^2+z^2 \geq xy+yz+zx\tag{1}$$
$$(x+y+z)^2 \leq 3(x^2+y^2+z^2) \... | A minimum is $-14$.
Prove that $$(x+y+z)\sqrt{\frac{x^2+y^2+z^2}{27}}+xy+xz+yz+\frac{14}{27}(x^2+y^2+z^2)\geq0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Average amount of appearances of given set inside another set. So, I was studying a little bit, and I realised that the questions in my book(applied mechanics) which had an answer at the end of the book were of the form $[1,5,9,13,17,...]$, or $[2,6,10,...]$ or $[3,7,11,...]$ which is pretty obviously an arithmetical p... | If all you know is $n$ (the number of problems in the chapter), and you are flipping to the back of the book, you can expect there to be
$$A = \frac{1}{4} \sum_{j=1}^{4} \left(\left\lfloor\frac{n-j}{4}\right\rfloor+ 1\right) = \frac{1}{4}\left(\left\lfloor \frac{n-1}{4}\right\rfloor + \left\lfloor \frac{n-2}{4}\right\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the exact value of $A(\beta)=8\pi-16\sin(2\beta)$ with $\tan(\beta)= \frac{1}{2}$
The picture below represents a semi-circumference of diameter [AB] and
center C. Point D belongs to the semi-circumference and it's one of
the vertices of the triangle $ABC$. Consider that BÂD = $\beta (\beta
\in ]0,\frac{\pi}{... | Note that $\sin 2\beta = 2\sin \beta \cos \beta$ and $1 + \tan^2 \beta = \frac{1}{\cos^2 \beta}$ so $\cos^2 \beta = \frac{4}{5}$. Also $\sin^2 \beta + \cos^2 \beta = 1$ so $\sin^2 \beta = \frac{1}{5}$ (alternatively you know $\sin \beta = \frac{1}{2}\cos \beta$ from what you've shown).
Hence $\sin 2\beta = 2 \sqrt{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find naturals numbers $p$ and $q$ such that $\frac{\sqrt{p^2+7}}{\sqrt{q^2-3}}$ is rational Find naturals numbers $p$ and $q$ such that $\frac{\sqrt{p^2+7}}{\sqrt{q^2-3}}$ is rational
i have taken $$\frac{\sqrt{p^2+7}}{\sqrt{q^2-3}}=\frac{a}{b}$$ where $a$ and $b$ are positive integers which are co prime then after si... | If $\dfrac{p^2+7}{q^2-3} = \dfrac{a^2}{b^2}$, where $\gcd(a,b) = 1$, then there is a positive integer $r$ such that
$$\begin{cases}p^2 - ra^2 = -7& \\ q^2 - rb^2 = 3. & \end{cases}$$
For each $r$ we have two independent generalized Pell's Equations. There are several algorithms to find the solutions to this equations.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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eliminate A,B and C How to eliminate $A$, $B$, $C$ from the given three equations:
$$X \cos A+ Y \cos B + Z \cos C=0$$
$$X \sin A+ Y \sin B + Z \sin C=0$$
$$X \sec A+ Y \sec B + Z \sec C=0$$
Answer is
$$(X^2+ Y^2 - Z^2 )^2 = 4 X^2 Y^2$$
I was squaring and adding 1st and 2nd eqn .But could not use the 3rd one.Please an... | Since $X\cos{A}=-Y\cos{B}-Z\cos{C}$ and $X\sin{A}=-Y\sin{B}-Z\sin{C}$, we obtain
$$X^2(\sin^2A+\cos^2A)=(Y\cos{B}+Z\cos{C})^2+(Y\sin{B}+Z\sin{C})^2$$ or
$$X^2=Y^2+Z^2+2YZ\cos(B-C).$$
In another hand, $X\cos{A}=-Y\cos{B}-Z\cos{C}$ and $\frac{X}{\cos{A}}=-\frac{Y}{\cos{B}}-\frac{Z}{\cos{C}}$, which gives
$$X^2=\left(Y\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $x^p+y^p \equiv (x+y)^p \pmod{x^2+xy+y^2}$
Let $p > 3$ be a prime, and let $x,y$ be integers such that $\gcd(x,y) = 1$. Prove that $$x^p+y^p \equiv (x+y)^p \pmod{x^2+xy+y^2}.$$
I thought about expanding $(x+y)^p$ using the binomial theorem, but I didn't see how that would help because the modulus is an alg... | Expanding on previous comment: $\,(x+1)^p-x^p-1\,$ is divisible by $\,x^2+x+1\,$ as polynomials.
Let $\,\omega_1,\omega_2\,$ be the roots of $x^2+x+1\,$, and note that they are the two complex cube roots of unity since $x^3-1=(x-1)(x^2+x+1)$. By Vieta's relations $1+\omega_1+\omega_2=0\,$. It also follows that $1+\omeg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of infinite series: $\frac{1}{3^2+1} + \frac{1}{4^2+2} + \frac{1}{5^2+3} + \cdots \text{ad inf.}$ How to find the sum of the following infinite series:
$$ S = \frac{1}{3^2+1} + \frac{1}{4^2+2} + \frac{1}{5^2+3} + \cdots \text{ad inf.} $$
I only need to prove that $\left[\frac{1}{S}\right] = 2$.
| Hint: Use $\dfrac{1}{n^2+n-2} = \dfrac{1}{(n-1)(n+2)} = \dfrac{1}{3}\Big(\dfrac{1}{n-1}-\dfrac{1}{n+2} \Big).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$ If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$
My Attempt:
$$\sin x + \sin^2 x=1$$
$$\sin x = 1-\sin^2 x$$
$$\sin x = \cos^2 x$$
Now,
$$\cos^8 x + 2\cos^6 x + \cos^4 x$$
$$=\sin^4 x + 2\sin^3 x +\sin^... | \begin{align}
& \cos^8 x + 2\cos^6 x + \cos^4 x \\[8pt]
= {} & \sin^4 x + 2\sin^3 x + \sin^2 x \\[8pt]
= {} & (\sin^2 x)^2 + 2(\sin^2 x)(\sin x) + \sin^2 x \\[8pt]
= {} & (1-\cos^2 x)^2 + 2(1-\cos^2 x)(\sin x) + (1-\cos^2 x) \\[8pt]
= {} & (1-\sin x)^2 + 2(1 - \sin x)(\sin x) + (1-\sin x) \\[8pt]
= {} & (1-2\sin x + \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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$A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ $A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ and $\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$. Then $A+2B$ is equal to:
$1$. $\dfrac {\pi}{4}$
$2$. $\dfrac {\pi}{3}$
$3$. $... | From the first equation
$$\cos 2B = 2\cos ^2B - 1=\left(4-3\cos ^2A\right)-1=3\sin ^2A$$
From the second equation
$$\sin 2B = \frac{3}{2}\sin 2A$$
Replace these values in
$$\cos (A+2B)=\cos A \cos 2B - \sin A \sin 2B=$$
$$=\cos A \left(3\sin ^2A\right)-\sin A \left(\frac{3}{2}\sin 2A\right)=0$$
$$A+2B=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Calculate $\sqrt{2i}$ I did:
$\sqrt{2i} = x+yi \Leftrightarrow i = \frac{(x+yi)^2}{2} \Leftrightarrow i = \frac{x^2+2xyi+(yi)^2}{2} \Leftrightarrow i = \frac{x^2-y^2+2xyi}{2} \Leftrightarrow \frac{x^2-y^2}{2} = 0 \land \frac{2xy}{2} = 1$
$$\begin{cases}
\frac{x^2-y^2}{2} = 0 \\
xy = 1\\ \end{cases} \\
=\... | I'll stick with your solving algorithm. You obtained (apart from a mistake):
$$\frac{1}{y^2}-y^2 = 0, \implies \frac{1}{y^2} = y^2$$
You can check that both $1$ and $-1$ satisfy this equation! So your chain of systems splits in two: knowing $x = \frac{1}{y}$,
$$y=1 \implies x = 1 \qquad y=-1 \implies x= -1 $$
So you ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Domain of $f(x)=\sqrt{\lfloor x\rfloor-1+x^2}$ I drew the number line and tested with different values, getting the correct domain $(-\infty,-\sqrt3)\cup[1,\infty)$. However, how do I solve this faster by manipulating the function?
| $f(x)$ is not real-valued if the expression $g(x)$ inside the square root is negative; i.e., $$g(x) = \lfloor x \rfloor - 1 + x^2 < 0.$$ Since $$x-1 < \lfloor x \rfloor \le x,$$ we find $$(x+2)(x-1) < g(x) \le x^2+x-1.$$ Hence we are assured to be outside the domain if $x^2 + x - 1 < 0$, and assured to be inside the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2274330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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What is the largest $a$ for which all the solutions to the equation $3x^2+ax-(a^2-1)=0$ are positive and real? Problem: What is the largest $a$ for which all the solutions to the equation $3x^2+ax-(a^2-1)=0$ are positive and real?
Attempt: Solving the equation for $x$ I get $$x_{1,2}=-\frac{a}{6}\pm\sqrt{\frac{13a^2-12... | First of all, the inequality you end up with is wrong: you need either
$$a \leq -\frac{2\sqrt{39}}{13} \;\;\;\text{ or }\;\;\; a \geq \frac{2\sqrt{39}}{13}$$
Since the root
$$x_2 = -\frac{a}{6} - \sqrt{\frac{13a^2 - 12}{36}} = \frac{1}{6}\left(-a - \sqrt{13a^2 - 12}\right)$$
is always smaller than $x_1$ (when ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to find period of a real function $f$ given the functional equation $\sqrt{3}f(x) = f(x-1) + f (x+1) $? If a periodic function satisfies the equation $\sqrt{3}f(x) = f(x-1) + f (x+1) $ for all real $x$ then prove that fundamental period of the function is $12$.
Here fundamental period means the smallest positive r... |
I tried replacing $x$ by $x \pm 1$ then try to find $f(x)$ in terms of other but always end up with it in terms of sum of other two arguments in the function eg $f(x-2)$ + $f (x+2)$ etc.
Start with:
$f(x+1) = \sqrt{3}\,f(x) - f(x-1) \tag{1}$
Using $(1)\,$:
$$
f(x+3) = \sqrt{3}\,\big(\sqrt{3}\,f(x+1) - f(x)\big) - f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Area of triangle and determinant The area of a $\vartriangle ABC$ with given vertices $(a,a^2),(b,b^2),(c,c^2)$ is $\frac{1}{4}$ $sq. units$ and area of another $\vartriangle PQR$ with given vertices $(p,p^2),(q,q^2),(r,r^2)$ is $3$ $sq. units$.
Then what is the value of
$$
\begin{vmatrix}
(1+ap)^2 & (1... | The oriented area of a triangle with vertices
$$A(x_A,y_A), B(x_B,y_B), C(x_C,y_C)$$
is known to be given by the following formula:
$$\tag{1}Area(ABC)=\tfrac12 \begin{vmatrix}x_A & x_B & x_C\\y_A & y_B & y_C\\1 & 1 & 1\end{vmatrix}$$
(this area is positive if the orientation of ABC is direct, negative otherwise).
See ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ x^4+x^3+x^2+x+1=0$ then what is the value of $x^5$ If $$x^4+x^3+x^2+x+1=0$$ then what's the value of $x^5$ ??
I thought it would be $-1$ but it does not satisfy the equation
| The left side of the equation is $P= (x^4 +x^2) + (x^3 + x) +1 = 2$
In the equation $x^2(x^2+1) +1 > 0$, there is no such real number for $x$. Logically, a false statement implies anything, so $x^5$ = anything you wish .
M Andrews' correct argument above can be modified to see that $x^5-1 =(x-1)(x^4+x^3+x^2+x+1)$, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How to find the maximum of the value $\frac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$ Find the maximun of the value
$$f(x)=\dfrac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$$
I use wolframpha this found this maximum is $\dfrac{4\sqrt{2}}{5}$,But How to prove and how to find this value?(without derivative)
idea 1
let $\tan{\df... | Hint:
Assume that $f(x)\leq 4\sqrt2/5$
Now cross multiply, do squares on both sides and ultimately if it's the maximum, you will get an inequality where the equality condition can hold, then write the whole sum backwards!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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Calculate $\frac{i^2-i^{31}+(4+3i)(5-2i)}{\frac{1}{2}i}$ without a calculator I tried:
$$\frac{i^2-i^{31}+(4+3i)(5-2i)}{\frac{1}{2}i} = \\
\frac{-1-(-1)+(4+3i)(5-2i)}{\frac{1}{2}i}= \\
\frac{26+7i}{\frac{1}{2}i} = \\
\frac{52+14i}{i} = \\
\frac{(52+14i)\cdot-i}{i\cdot -i} = \\ \\
14-52i$$
But my book states the solutio... | Numerator: $-1+i+20-8i+15i+6$, which simplifies to $25+8i$. Can you divide this by $\frac{i}{2}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equation of circumcircle From a point $(a,b)$ two tangents $\overset{\leftrightarrow}{PQ}$ and $\overset{\leftrightarrow}{PR}$ are drawn to a circle
$x^2 + y^2 - a^2=0$
Find the equation of the circumcircle of $\triangle PQR$.
My attempt:
The circumcircle and the given circle have a common chord $\overline{QR}$.
Apart... | $$x^2+y^2-a^2=0\ \ \ \ (0)$$
Using The equation of a pair of tangents to a circle from a point.
tangents-to-a-circle-from-a-point,
the equation of the pair of tangents, $$(a^2+b^2-a^2)(x^2+y^2-a^2)=(ax+by-a^2)^2\ \ \ \ (1)$$
Now the equation of any conic passing through the intersection of $(0),(1)$
will be $$(a^2+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Having trouble using my usual method of partial fraction decomposition for $\frac{9 + 3s}{s^3 + 2s^2 - s - 2}$. I'm having trouble using my usual method of partial fraction decomposition for $\dfrac{9 + 3s}{s^3 + 2s^2 - s - 2}$.
We can factor such that $$\dfrac{9 + 3s}{s^3 + 2s^2 - s - 2} = \dfrac{A}{s - 1} + \dfrac{B}... | I would prefer a simpler method for this problem, known as Heaviside cover-up method:
$$\dfrac{9+3s}{s^3+2s^2-s-2}=\dfrac{9+3s}{s^2(s+2)-(s+2)}
=\dfrac{9+3s}{(s-1)(s+1)(s+2)}$$
Now,$$\dfrac{9+3s}{(s-1)(s+1)(s+2)}\\=\left.\dfrac{9+3s}{(s+1)(s+2)}\right|_{s=1}\cdot\dfrac{1}{s-1}+\left.\dfrac{9+3s}{(s-1)(s+2)}\right|_{s=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Is $f(x)=\left(\frac{x^2+1}{x^2}\right)^{x^2-x}$ an increasing function for $x \ge1$? The answer seems to me to be yes. Here's my reasoning:
$\frac{x^2+1}{x^2}$ is a decreasing function since for $a > 0, x \ge 1$, $\frac{(x+a)^2+1}{(x+a)^2} < \frac{x^2+1}{x^2}$ since:
$$x^2[(x+a)^2+1]=x^2(x+a)^2+ x^2 < (x^2+1)(x+a)^2... | I think it's an increasing function.
Let $g(x)=(x^2-x)\ln\left(1+\frac{1}{x^2}\right)$.
Thus, $$g'(x)=(2x-1)\ln\left(1+\frac{1}{x^2}\right)-\frac{(x^2-x)\cdot\frac{2}{x^3}}{1+\frac{1}{x^2}}=(2x-1)\ln\left(1+\frac{1}{x^2}\right)-\frac{2(x-1)}{x^2+1}.$$
We'll prove that $g'(x)>0$ for all $x\geq1$.
Indeed, let $h(x)=\ln\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2}$ fine $x,y$ :
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2} \ \ \ : x ,y \in \mathbb{Z}$$
My Try :
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2} \\ x^2-x+1+y^2-y+1+2\sqrt{(x^2-x+1)(y^2-y+1)}=x^2+xy+y^2 \\ 2\sqrt{(x^2-x+1)(y^2-y+1)}=xy +x+y-2$$
Now ?
| Applying Minkowski's inequality, we have that $$ \sqrt{x^{2}-x+1}+\sqrt{y^{2}-y+1}=\big((x-\frac{1}{2})^{2}+\big(\frac{\sqrt{3}}{2}\big)\big)^{\frac{1}{2}}+\big((y-\frac{1}{2})^{2}+\big(\frac{\sqrt{3}}{2}\big)\big)^{\frac{1}{2}} \geq \big(\big(x-\frac{1}{2}+y-\frac{1}{2}\big)^{2}+(\sqrt{3})^{2}\big)^{\frac{1}{2}}=\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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$\lim_{n\to \infty} \frac{3}{n}(1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+...+\sqrt{\frac{n}{4n-3}})$ fine the limits :
$$\lim_{n\to \infty} \frac{3}{n}(1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+...+\sqrt{\frac{n}{4n-3}})$$
My Try :
$$\lim_{n\to \infty} \frac{3}{n}(1+\frac{1}{\sqrt{1+\frac{3}{n}}}+\frac{1}{\sqrt{1+... | Note that $$\lim _{ n\to \infty } \frac { 3 }{ n } \left( 1+\frac { 1 }{ \sqrt { 1+\frac { 3 }{ n } } } +\frac { 1 }{ \sqrt { 1+\frac { 6 }{ n } } } +...+\frac { 1 }{ \sqrt { 1+\frac { 3(n-1) }{ n } } } \right) =\\ =3\lim _{ n\to \infty } \sum _{ k=1 }^{ n }{ \frac { 1 }{ n } \left( \frac { 1 }{ \sqrt { 1+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a point on an ellipsoid Find a point on the ellipsoid $x^2+4y^2+z^2=9$ where the tangent plane is perpendicular to the line with parametric equations \begin{align}x&=2+2t\\y&=1+2t\\z&=3-t\end{align}
The answer to this question is: $$\left(\frac{6\sqrt{13}}{13}, \frac{6\sqrt{13}}{13}, -\frac{3\sqrt{13}}{13}\righ... | Projective geometry to the rescue.
The points on the line (in homogeneous coordinates) are
$${\bf q} =\left[ \matrix{ {\bf r} + {\bf e}\, t \\ 1}\right] =\left[ \matrix{ \pmatrix{2\\1\\3}+\pmatrix{2\\2\\-1} t \\ 1}\right] $$
The planes normal to the line (in homogeneous coordinates) are
$${\bf w} =\left[ \matrix{ {\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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let $x ,y ,z \in \mathbb{N} ,x \leq y\leq z$ and : $\frac{100}{336}=\frac{1}{x}+\frac{1}{xy}+\frac{1}{xyz}$then : $x+y+z =?$ let $x ,y ,z \in \mathbb{N} ,x \leq y\leq z$
and :
$$\frac{100}{336}=\frac{1}{x}+\frac{1}{xy}+\frac{1}{xyz}$$
then :
$$x+y+z =?$$
my try :
$$\frac{100}{336}=\frac{1}{x}+\frac{1}{xy}+\frac{1}{xyz}... | Let $x,y,z\in\mathbb{N}$ with $x<y<z$. First, you can rewrite the equation as
$$ \dfrac{100}{336}=\dfrac{25\lambda}{84\lambda}=\dfrac{yz+z+1}{xyz}. $$
for a positive integer $\lambda$. Since we know that $xyz$ and that $yz+z+1$ are integers, we can hope to find a solution for the system of equations:
$$\begin{cases}84... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve Quadratic Congruence Equation How to solve $3x^2 - 5x + 5 \equiv 0 \pmod 7$? In general, how to approach this kind of problem? Any help is appreciated.
| If $a,b,c,x,n\in\mathbb Z$, $n\ge 3$, $\gcd(2a,n)=1$, then
$$ax^2+bx+c\equiv 0\pmod{n}$$
$$\stackrel{\cdot 4a}\iff (2ax+b)^2\equiv b^2-4ac\pmod{n}$$
You're left with finding the modular square roots of $b^2-4ac$ mod $n$.
First, you may want to find a Legendre symbol, use Quadratic reciprocity.
Here you're solving $3x^2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $n^5+n^4+1$ is composite for $n>1.$
Prove that $f(n)=n^5+n^4+1$ is composite for $n>1, n\in\mathbb{N}$.
This problem appeared on a local mathematics competition, however it looks like there is no simple method to solve it. I tried multiplying it by $n+1$ or $n-1$ and then tried factorizing it, but it was t... | $$n^5+n^4+1 = n^5 - n^2 + n^4 + n^2 + 1 = n^2(n^3-1) + (n^2+1)^2 - n^2 = n^2(n-1)(n^2+n+1) + (n^2-n+1)(n^2+n+1) = (n^3-n+1)(n^2+n+1)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\frac{d^{100}}{dx^{100}}\left(\frac{p(x)}{x^3-x}\right)$ I am given that $\dfrac{d^{100}}{dx^{100}}\left(\dfrac{p(x)}{x^3-x}\right) = \dfrac{f(x)}{g(x)}$ for some polynomials $f(x)$ and $g(x).$ $p(x)$ doesn't have the factor $x^3-x$ and I need to find the least possible degree of $f(x)$.
My Attempt: I am de... | Using the chain rule we write:
\begin{eqnarray}
&&\frac{d^n}{d x^n} \left(\frac{p(x)}{x^3-x}\right) =
\sum\limits_{p=0}^n \binom{n}{p} d^{n-p} p(x) \cdot d_x^p
\underbrace{\left[
\frac{1}{2} \frac{1}{x-1} + \frac{1}{2} \frac{1}{x+1} - \frac{1}{x}
\right]}_{(-1)^p\left(\frac{1}{2} \frac{1}{(x-1)^{p+1}} + \frac{1}{2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the general solution of recurrence relation of order four Find the general solution of the recurrence relation.
$a_n = 4a_{n - 1} - 5a_{n - 2} + 4a_{n - 3} - 4a_{n - 4}$
I've found the roots, but I don't really understand how does general solution look like.
$$r^n = 4r^{n - 1} - 5r^{n - 2} + 4r^{n - 3} - 4r^{n - 4... | Generating functions make it transparent. Define $A(z) = \sum_{n \ge 0} a_n z^n$, shift your recurrence by 4, sum over $n \ge 0$, recognize resulting sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 4} z^n
&= 4 \sum_{n \ge 0} a_{n + 3} z^n
- 5 \sum_{n \ge 0} a_{n + 2} z^n
+ 4 \sum_{n \ge 0} a_{n + 1}... | {
"language": "en",
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PDF of $\frac{X}{1+X^2}$ in terms of the PDF of $X$ Given a r.v. $X$ with a know probability density $g$, I would like to find the probability density of $\frac{X}{1+X^2}$. Below I write my calculations:
denote
$f(X) := \frac{X}{1+X^2}$
Now $f$ is not monotone so we can't apply the known theorem, we proceed by finding... | You might be interested to know that, if $X$ has PDF $f_X$ and if $Y=h(X)$ for some function $h$ regular enough then, the PDF $f_Y$ of $Y$ is given by
$$f_Y(y)=\sum_{x:h(x)=y}\frac1{|h'(x)|}f_X(x)$$
In your case, $$h(x)=\frac{x}{1+x^2}$$ hence $f_Y(y)=0$ for $|y|\geqslant\frac12$. For every $0<|y|<\frac12$, $h(x)=y$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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There exists a $2 \times 2$ matrix $R$ such that $r = R v$ for all 2-dimensional vectors $v$. Find $R$. For a vector $v$, let $r$ be the reflection of $v$ over the line
$$x = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}.$$
There exists a $2 \times 2$ matrix $R$ such that
$$r = R v$$
for all 2-dimensional vectors $v$. Find $... | Finding the matrix that performs a reflection over a line through the origin.
Consider: a) A vector $\vec{v}$ ,
b) a straight line passing through the origin with a normalized direction vector $\vec{w}$ = $(1/√5)\begin{pmatrix} 2 \\ -1\end{pmatrix}$, and
c) the reflected vector $\vec{v'}$.
We have:
1) $\vec{v} + \ve... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate arctan from $0$ to $3\pi$ How do I integrate
$$\int_0^{3\pi} \frac 1 {\sin^4x + \cos^4 x} \,dx$$
I tried with Weierstrass and obtained:$$
\int\frac 1 {u^2+2}\, du
$$ I think it's correct but how do I integrate this given that I cant integrate arctan for $3\pi$ and $0$
| $\sin^4(x)+\cos^4(x)=(\sin^2(x)+\cos^2(x))^2-2\sin^2(x)\cos^2(x) = 1-\frac{1}{2}\sin^2(2x) $ leads to:
$$ I = \frac{1}{2}\int_{0}^{6\pi}\frac{dz}{1-\frac{1}{2}\sin^2(z)}=6\int_{0}^{\pi/2}\frac{dz}{1-\frac{1}{2}\sin^2(z)}=6\int_{0}^{\pi/2}\frac{dz}{1-\frac{1}{2}\cos^2(z)} $$
by periodicity and symmetry. By setting $z=\a... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $T(n) = T(n/3) + T2n/3) + 5n$ is $O(n log n)$ I'm doing some research about time complexity of algorithms and stumbled upon the following problem that I'm not able to solve:
Let $T(n) = T(n/3) + T(2n/3) + 5n$. prove that $T(n) = O(n log n)$
First, I made a recursion tree, which is the same as the one in the ... | One approach that can tackle recurrences like this is the Akra-Bazzi method:
$T(n) = T(n/3) + T(2n/3) + 5n$ means we must solve for $\left(\frac{1}{3}\right)^p + \left(\frac{2}{3}\right)^p = 1$, which is true for $p=1$. Then we have:
$$T(n) \in \Theta\left(n^p\left(1 + \int_{1}^{n} \frac{5u}{u^{p+1}} \, du\right)\right... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\sin 9^{\circ}$ or $\tan 8^{\circ} $ which one is bigger? $\sin 9^{\circ}$ or $\tan 8^{\circ} $ which one is bigger ?
someone ask me that , and said without using calculator !!
now my question is ,how to find which is bigger ?
Is there a logical way to find ?
I s there a mathematical method to show which is greater... | In terms of radian, $9^\circ = \frac{\pi}{20} \approx 0.157$ is reasonable small. We can use Taylor series expansion to estimate the value of $\sin$ and $\tan$. For small $\theta$, we have
$$\sin\theta \approx \theta - \frac{\theta^3}{6}
\quad\text{ and }\quad
\tan\theta = \frac{\sin\theta}{\cos\theta}
\approx \frac{\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find $\sin(A)$ and $\cos(A)$ given $\cos^4(A) - \sin^4(A) = \frac{1}{2}$ and $A$ is located in the second quadrant.
Question: Find $\sin(A)$ and $\cos(A)$, given $\cos^4(A)-\sin^4(A)=\frac{1}{2}$ and $A$ is located in the second quadrant.
Using the fundamental trigonometric identity, I was able to find that:
• $\cos^... | We can use the double angle identities for cosine to solve the problem. They are
\begin{align*}
\cos(2x) & = \cos^2x - \sin^2x\\
& = 2\cos^2x - 1\\
& = 1 - 2\sin^2x
\end{align*}
You correctly found that
$$\cos^2A - \sin^2A = \frac{1}{2}$$
where $A$ is a second-quadrant angle. Substituting $2\cos^2A... | {
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"timestamp": "2023-03-29T00:00:00",
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From $a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$ to $a_n$, Case 2
Find and prove by induction an explicit formula for $a_n$ if $a_1=1$ and, for $n \geq 1$,
$$P_n: a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$$
Checking the pattern:
$$a_1=1 $$
$$a_2= \frac{3}{4 \cdot 5}$$
$$a_3= \frac{3^2} { 4 \cdot 5 \cdot 6 \cdot 7}$$
$$a_4= \frac{... | $$a_{n+1} = \frac{n!.3^n}{(2n+3)!} = \frac{n!.3.3^{n-1}}{(2n+3)(2n+2)(2n+1)!}=\frac{3!3^{n-1}}{(2n+1)!} \frac{3}{(2n+2)(2n+3)} = a_n \frac{3}{(2n+2)(2n+3)} = \frac{3a_n}{(2n+2)(2n+3)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}} $
Find the real roots of the equation $$x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}} $$ $$(0<a<\frac{1}{4})$$
My attempt,
$(x^2+2ax+\frac{1}{16}+a)^2=a^2+x-\frac{1}{16}$
I did an expansion which becomes $x^4+4ax^3(4a^2+2a+\frac{1}{8})x^2+(4a^2+\frac{1}{4}... | Note that the right-hand side is a solution to the equation
$$x=y^{2}+2 a y+\displaystyle\frac{1}{16}$$
Furthermore, the solution to this equation gives
$$y=x^{2}+2 a x+\displaystyle\frac{1}{16}$$
That is, we have an equation for points where a function equal to its inverse function. The points in such a case lie along... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Limit of function calculations I must solve limit of next function:
$$\lim_{x\to \infty}\frac{2x^3+x-2}{3x^3-x^2-x+1}$$
Does my calculations are proper? If not where is my mistake?
$$=\lim_{x\to \infty}\frac{x^3\left(2+\frac{1}{x^2}-\frac{2}{x^3}\right)}{x^3\left(3-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}\right)} \\
\ =... | Exactly correct! In the future you can also check your work on Wolfram Alpha.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving Trigonometric Equality I have this trigonometric equality to prove:
$$\frac{\cos x}{1-\tan x}-\frac{\sin x}{1+\tan x}=\frac{\cos x}{2\cos^2x-1}$$
I started with the left hand side, reducing the fractions to common denominator and got this:
$$\frac{\cos x+\cos x\tan x-\sin x+\sin x\tan x}{1-\tan^2x}\\=\frac{\cos... | Multiplying numerator/denominator by $\cos x$,
$$\frac{c}{1-t}-\frac{s}{1+t}=c\left(\frac c{c-s}-\frac s{c+s}\right)=c\frac{c^2+cs-cs+s^2}{c^2-s^2}=\frac c{2c^2-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2299015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Find limit $a_{n + 1} = \int_{0}^{a_n}(1 + \frac{1}{4} \cos^{2n + 1} t)dt,$ Find the limit of the sequence:
$$a_{n + 1} = \int_{0}^{a_n}(1 + \frac{1}{4} \cos^{2n + 1} t)dt,$$
such that $a_0 \in (0, 2 \pi)$
That was one of the tasks in the Olympiad.
Here is my approach.
First, I wanted to simplify the integral:
$\int... | The limit is $\pi$.
First note that
$$a_{n+1} = a_n + \frac{1}{4}\int_0^{a_n} \cos^{2n+1} t\,dt.$$
$\cos^{2n+1} t$ is positive on $(0,\pi/2)$, negative on $(\pi/2,3\pi/2)$, and again positive on $(3\pi/2,2\pi)$. By the symmetries of the cosine, we have
$$\int_0^{\pi} \cos^{2n+1} t\,dt = 0 = \int_0^{2\pi} \cos^{2n+1} t\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating $\sum_{k=1}^\infty \frac{k^2}{2^k}=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots+\frac{k^2}{2^k}+\cdots$ I want to know the value of $$\sum_{k=1}^\infty \frac{k^2}{2^k}=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots+\frac{k^2}{2^k}+\cdots$$
I added up to $k=5... | Start with the geometric series $\frac{1}{1-x} = \sum_i x^i$. Differentiate it once to get $\frac{d}{dx} \left[ \frac{1}{1-x} \right] = \frac{d}{dx} \sum_i x^i = \sum_i i x^{i-1}$. Differentiate again to get $\frac{d^2}{dx^2} \left[ \frac{1}{1-x} \right] = \frac{d^2}{dx^2} \sum_i x^i = \sum_i i (i-1) x^{i-2}$.
Now, pl... | {
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Solve the Inequality $x+\frac{x}{\sqrt{x^2-1}} \gt \frac{35}{12}$ Solve the Inequality $$x+\frac{x}{\sqrt{x^2-1}} \gt \frac{35}{12}$$
First of all the Domain of LHS is $(-\infty \:\: -1) \cup (1 \:\: \infty)$
So i assumed $x=\sec y$ since Range of $\sec y$ is $(-\infty \:\: -1) \cup (1 \:\: \infty)$
So
$$\sec y+ |\cs... | The inequality is:$$\frac{x}{\sqrt{x^2-1}} > \frac{35}{12} - x$$
For $x > 1$, LHS is positive and so is RHS unless $x \geqslant \frac{35}{12}$. But if $x \geqslant \frac{35}{12}$ the inequality is trivially true. So we need to consider $x \in (1, \frac{35}{12})$, where we may square and simplify$^\dagger$ to get $(1, ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is $a^4+b^4$ factorable in two different ways, and why are the solutions not the same? I am working on factoring $a^4+b^4$ and I have found two different solutions to this.
First, I have factored it to $$a^4+b^4=(a+b)[a^3-a^2b+ab^2+b^3]-2ab^3$$
But then I also found that the equation is factorable to $$a^4+b^4=(a^2... | A factorization is always a pure product, e.g.,
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$
or
$$a^4-b^4=(a-b)(a+b)(a^2+b^2)$$
So the problem is that $(a+b)(a^3-a^2b+ab^2+b^3)-2ab^3$, is not a pure product. Instead, it's the sum of the product $(a+b)(a^3-a^2b+ab^2+b^3)$ and $-2ab^3$.
I.e., what tilper just said!
| {
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"source": "stackexchange",
"question_score": "2",
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Ratios of $HM:GM=12:13$
If the ratio of $HM$ between two positive numbers $a$ and $b$ ($a>b$) is to their $GM$ as $12$ to $13$, find $a:b$
My attempts:
$\dfrac{HM}{GM}=\dfrac{12}{13}\implies \dfrac{\dfrac{2ab}{a+b}}{\sqrt{ab}}=\dfrac{12}{13}\implies 13\sqrt{ab}=6(a+b)$
$\dfrac{13}{6}=\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{... | HINT:
Let $\sqrt{\dfrac ab}=p\implies p+\dfrac1p=\dfrac{13}6\iff0=6p^2-13p+6=(3p-2)(2p-3)$
If $p=\dfrac23,\dfrac ab=\dfrac49,\dfrac a4=\dfrac b9=k$(say)
$a=4k,b=?$
| {
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Find the least integer $n$ such that the fractions are irreducible
Find the least integer $n$ with the property that the fractions $$\dfrac{7}{n+9},\dfrac{8}{n+10},\dfrac{9}{n+11},\ldots,\dfrac{31}{n+33}$$ are irreducible.
In order for all the fractions to be irreducible, we must have $\gcd(n+k,k-2) = 1$ for $k = 9,1... | You just need to have $n\not \equiv -2 \bmod p$ for all prime $p$ not exceeding $31$.
In other words $n+2$ must not be divisible by any prime not exceeding $31$.
In other words we can take $n+2=37$ and so $n=35$ is the smallest options (because $377=35+2$ is the prime after $31$ (I am assuming you wanted $n$ to be posi... | {
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How to find the number of solution pairs for this equation? Question: Consider the equation $(1+a+b)^2=3(1+a^2+b^2)$ where $a$ and $b$ are real numbers. How many solution pair(s), ($a$,$b$), are possible for this equation?
My attempt:
$(1+a+b)^2=3(1+a^2+b^2)$
=>$1+a^2+b^2+2(a+ab+b)=3+3a^2+3b^2$
=>$1+a^2+b^2+2a+2ab+2b=... | Although, dxiv showed you a standard solution, but you have a good result there. Let $x=1-a$ and $y=b-1$ then $xy=(x+y)^2$ or $-xy=x^2+y^2$. Notice that $$2|x||y| \le x^2+y^2=-xy \le |x||y|,$$which gives you $|x||y| \le 0$. This is possible only when $|x||y|=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ellipse in a Rectangle What is the equation for an ellipse (or rather, family of ellipses) which has as its tangents the lines forming the rectangle $$x=\pm a, y=\pm b\;\; (a,b>0)$$?
This question is a modification/extension of this other question here posted recently.
| This solution is inspired by the Lissajous diagram.
Consider two signal inputs $x, y$ which are out of phase, with magnitudes $a,b$ respectively, fed into an oscilloscope.
The resulting Lissajous diagram will be a tilted ellipse with tangents $x=\pm a, y=\pm b$ and parametric equations
$$\begin{cases}
x=a\sin(t+u)\\
y... | {
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"url": "https://math.stackexchange.com/questions/2308198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A difficult problem on factorization algebra Factorize $(a+b+c)^7 - a^7-b^7-c^7$.
| When $a=-b$,
$$(a+b+c)^7-a^7-b^7-c^7=c^7-a^7-(-a)^7-c^7=0$$
By Factor Theorem, $a+b$ is a factor of $(a+b+c)^7-a^7-b^7-c^7$.
Similarly, $b+c$ and $c+a$ are also factors of $(a+b+c)^7-a^7-b^7-c^7$.
As $(a+b+c)^7-a^7-b^7-c^7$ is cyclic and symmetric,
$$(a+b+c)^7-a^7-b^7-c^7=(a+b)(b+c)(c+a)[p(a^4+b^4+c^4)+q(a^2b^2+b^2c^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What does the group of the geometric images of z, such that $z+\overline{z}+|z^2|=0$ define?
What does the group of the geometric images of z, such that
$z+\overline{z}+|z^2|=0$ define?
A) A circumference of center (0,-1) and radius 1.
B) Two lines, of equations $y=x+2$ and $y=x$.
C)A circumference of center (1,0) a... | Hint: using $|z^2|=|z|^2= z \bar z\,$:
$$
0 = z+\bar{z}+|z^2| = z + \bar z + z \bar z \color{red}{+1-1}=(z+1)(\bar z + 1) -1 = |z+1|^2 - 1 \iff |z+1|^2 = 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
solving ${x + y + z = 4, x^2 + y^2 + z^2 = 4, x^3 + y^3 + z^3 = 4}$ Knowing that :
$$x + y + z = 4$$
$$x^2 + x^2 + z^2 = 4$$
$$x^3 + y^3 + z^3 = 4 $$
Solving the equation with complex numbers :
I know that:
$$4 = x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy + xz + yz)$$
So:
$$ (xy + xz + yz) = 6$$
$$4 = x^3 + y^3 + z^3 = (x+... | From just the first two equations and the QM-AM inequality:
$$\sqrt{\frac43}= \sqrt{\frac{x^2+y^2+z^2}3} \geqslant \frac{x+y+z}3 = \frac43$$
we have the fact that there are no real solutions...
If looking for complex solutions, note
we have $2(xy + yz + zx) = (x+y+z)^2-(x^2+y^2+z^2) \implies xy+yz+zx = 6 $
and $3xyz =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate the value of the series $\,\sum_{n=1}^\infty\frac{1}{2n(2n+1)(2n+2)}$ Calculate the infinite sum
$$\dfrac{1}{2\cdot 3\cdot 4}+ \dfrac{1}{4\cdot 5\cdot 6}+\dfrac{1}{6\cdot 7\cdot 8}+\cdots$$
I know this series is convergent by Comparison Test, but I can't understand how can I get the value of the sum.
Is th... | Let us use partial fraction decomposition and the identity
$$\frac{1}{n+1}=\int_0^1 x^n dx$$
to evaluate
$$\begin{align}S&=\frac{1}{2·3·4}+\frac{1}{4·5·6}+\frac{1}{6·7·8}+...\\
\\
&=\sum_{k=0}^\infty \frac{1}{(2k+2)(2k+3)(2k+4)}=\sum_{k=0}^\infty \frac{1}{2} \left(\frac{1}{2k+2}-\frac{2}{2k+3}+\frac{1}{2k+4}\right)\\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Show that if $\frac1a+\frac1b+\frac1c = a+b+c$, then $\frac1{3+a}+\frac1{3+c}+\frac1{3+c} \leq\frac34$
Show that if $a,b,c$ are positive reals, and
$\frac1a+\frac1b+\frac1c = a+b+c$, then
$$\frac1{3+a}+\frac1{3+b}+\frac1{3+c} \leq\frac34$$
The corresponding problem replacing the $3$s with $2$ is shown here:
How... | By C-S
$$\sum_{cyc}\frac{1}{3+a}\leq\frac{1}{(3+1)^2}\sum_{cyc}\left(\frac{3^2}{2+a}+\frac{1^2}{1}\right)=$$
$$=\frac{9}{16}\sum_{cyc}\frac{1}{2+a}+\frac{3}{16}\leq\frac{9}{16}+\frac{3}{16}=\frac{3}{4}$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculate: $\sin9°$ I found this question in the olympiad book. But I could not find the solution.
The question is to calculate the following real number: $$\sin{9°}$$
| Let $x = 18$ then $5 x = 90$ so $2x = 90 - 3x$.
Now
\begin{align}
\sin(2x) &= \sin(90 - 3x)\\
2 \sin x \cos x &= \cos 3x\\
2\sin x \cos x &= 4\cos^3x - 3\cos x\\
2\sin x &= 4\cos^2x - 3\\
2\sin x&= 4 - 4\sin^2x-3\\
4\sin^2x + 2\sin x - 1 &=0
\end{align}
Solving this quadratic equation, we get
$$\sin x = \frac{-1+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2313450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Integration of $1/(1+a \csc^2(x))$
Integration of $$\int_{0}^{\frac{(M-1)\pi}{M} }\frac{1}{1+\alpha \csc^2(x)} dx,$$ where $ \alpha $ is a constant.
I tried taking $\cot(x) = t$, then differentiating it w.r.t $dx$ we get, $-\csc^2(x)dx = dt$. And as we know that, $\csc^2(x)= \cot^2(x) +1$, so tried substituting the... | Hint
Considering the problem of the antiderivative, using $$x=\tan ^{-1}\left(u\,\sqrt{\frac{a}{a+1}} \right)\implies dx=\frac{ \sqrt{a(a+1)}}{a u^2+a+1}\,du$$ This makes
$$\int\frac{dx}{1+a \csc^2(x)}=\sqrt{\frac{a}{a+1}}\int\frac{ u^2}{\left(u^2+1\right) \left(a u^2+a+1\right)}\,du$$ Now, using partial fraction deco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve for $x$ in $\cos(2 \sin ^{-1}(- x)) = 0$
Solve $$\cos(2 \sin ^{-1}(- x)) = 0$$
I get the answer $\frac{-1}{ \sqrt2}$
By solving like this
\begin{align}2\sin ^{-1} (-x )&= \cos ^{-1} 0\\
2\sin^{-1}(- x) &= \frac\pi2\\
\sin^{-1}(- x) &= \frac\pi4\\
-x &= \sin\left(\frac\pi4\right)\end{align}
Thus $x =\frac{-1}{\... | Trig functions are not one-to-one. For example if $\sin x = \frac 12$ then $x$ might be equal to $\frac {\pi}6$, but $x$ could also be $\frac {5\pi}6$ or $\frac {13\pi}6$ or it could be any $2k\pi + \frac {\pi}6$ or $(2k+1)\pi - \frac {\pi} 6$.
So when we say $\sin^{-1} x = \theta$ we are not just saying that $\theta$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"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.