Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Product of repeated cosec.
$$P = \prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$$
I realize that there must be some sort of trick in this.
$$P = \csc^2(1)\csc^2(3).....\csc^2(89) = \frac{1}{\sin^2(1)\sin^2(3)....\sin^2(89)}$$
I noticed that: $\sin(90 + x) = \cos(x)$ hence,
$$\sin(89) = \cos(-1) = \cos(359)$$
$$\sin(1) = \co... | $$\sin(2n+1)x=(2n+1)\sin x+\cdots+(-1)^n2^{2n}\sin^{2n+1}x$$
If we set $2n+1=45,2^{44}\sin^{45}x-\cdots+45\sin x-\sin45x=0$
If we set $\sin45x=\sin45^\circ,45x=360^\circ m+45^\circ$ where $m$ is any integer
$\implies x=8^\circ m+1^\circ$ where $0\le m\le44$
$\implies Q=\prod_{m=0}^{44}\sin(8^\circ m+1^\circ)=\dfrac1{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Parabolic sine approximation Problem Find a parabola ($f(x)=ax^2+bx+c$) that approximate the function sine the best on interval [0,$\pi$].
The distance between two solutions is calculated this way (in relation to scalar product): $\langle u,v \rangle=\int_0^\pi fg$.
My (wrong) solution I thought that I would get the s... | Is not the direct modern equivalent of Bhaskara's parabola approximation for Sine curve good enough as a start?
$$ y(x) = \dfrac{x(\pi-x)}{2},$$
$$ y(\pi/2)= \pi^2/8;y^{'}(\pi/2) =0; y^{''}(\pi/2) =-1. $$
And cannot the amplitude alone be improved/adjusted from here further to minimize error by least squares?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Finf $f(x)$ which is a second degree polynomial, such that $f(1)=0$ and $f(x) = f(x-1)$ I must find a function $f(x) = ax^2+bx+c$ such that:
$$f(1) = a+b+c=0\\f(x)=f(x-1)\implies ax^2+bx+c = a(x-1)^2+b(x-1)+c\implies\\ax^2+bx+c = ax^2+(-2a+b)x+a-b+c\implies\\a = a, b = -2a+b, c = a-b+c$$
but this results fo $a=b=c=0$. ... | If $f(x) = f(x-c)$
for all $x$,
where $c$ is a constant,
and $f$ is a polynomial,
then
$f$ is constant.
Proof:
Let
$g(x) = f(x)-f(0)$.
Then $g(0) = 0$.
Also,
$g(-nc) = g(0) = 0$
for all positive integers $n$.
Since a polynomial of
degree $d$ has at most $d$
distinct zeros,
and $g$ has an infinite
number of zeros,
$g$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational
How can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?
I know that $\sqrt{3}, \sqrt{5}$ and $\sqrt{7}$ are all irrational and that $\sqrt{3}+\sqrt{5}$, $\sqrt{3}+\sqrt{7}$, $\sqrt{5}+\sqrt{7}$ are all irrational, too. But how can I prove that $\sqrt{... | Let $\alpha=\sqrt{3}+ \sqrt{5}+ \sqrt{7}$.
Then $\alpha$ is a root of $x^8-60 x^6+782 x^4-3180 x^2+3481=0$.
Since this is a monic polynomial with integer coefficients, the rational root theorem tells you that $\alpha$ is either irrational or an integer.
Now
$\quad 1.7 < \sqrt 3 < 1.8 $
$\quad 2.2 < \sqrt 5 < 2.3 $
$\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to show that $zw=1\implies w = z^{-1}, z = w^{-1}$? I need to show that
$$zw=1\implies w = z^{-1}, z = w^{-1}$$
But I don't know how to start. I've tried to consider $z=a+bi, w = c+di$ then:
$$zw = ac+adi+bci-bd\implies zw = ac-bd + (ad+bc)i = 1\implies \\ac-bd = 1\\ad+bc = 0$$
$$ac = 1+bd\\ad = -bc$$
$$c = \frac{1... | Hint Multiply both $zw=1$ by $z^{-1}$. Do the same for $w$.
Also, how do you define $z^{-1}$?
Since complex multiplication is commutative, the definition of $z^{-1}=w$ is $zw=1$. So there is absolutely nothing to prove....
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $x^n + x < (x^n)x$ using induction. I need to prove $x^n$ + x < $x^n\cdot$ x, n $\in$ N, x $\in$ R>2 using induction. I started by
$x^n$ + x + (x^(n+1)+x) < ($x^n\cdot$ x) + (x^(n+1)+x)
I simplified to this:
< 2x^(n+1) + x
I am stuck here. All help is appreciated.
| Let $P(n): x^n+x< x^{n+1}, n \in \mathbb{N}, x > 2$.
Check $P(1)$ is true. $x^1+x = 2x < x^2 \iff x(x-2) > 0$ is true since $x > 2$.
Assume $P(n)$ is true, i.e. $x^n + x < x^{n+1}$, prove $P(n+1)$ is true.
$x^{n+1} + x = x\cdot x^n + x < x(x^{n+1}-x) + x = x^{n+2} - x^2 + x < x^{n+2} - x^2 + x^2 = x^{n+2}$ since $x^2 >... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration of $\int\log(\sqrt{1-x}+\sqrt{1+x}) \, dx$ Integration of
$$\int\log\left(\sqrt{1-x}+\sqrt{1+x}\right) \, dx$$
Please help to go through this problem as i have started with putting $x$= $cos2y$.
| It is shorter to integrate by parts, setting $ u = \ln\bigl(\sqrt{1-x} + \sqrt{1 + x}\bigr) $, hence:
\begin{align*}
\mathrm d\mkern1.5mu u & = \frac{\frac12\Bigl(\dfrac{-1}{\sqrt{1-x}} + \dfrac{1}{\sqrt{1 + x}}\Bigr)}{\sqrt{1-x} + \sqrt{1 + x}}\,\mathrm d\mkern1.5mu x= \frac12\frac{\sqrt{1-x}-\sqrt{1 + x}}{\sqrt{(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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When does the curve $y = {x^p}\cos \frac{\pi }{x}(0 < x \leqslant 1)$ have finite length? When does the curve $y = {x^p}\cos \frac{\pi }{x}(0 < x \leqslant 1)$ have finite length?
I cannot figure out how to solve this problem. Thanks for your help.
| Thanks for everyone! With your hints I have figured out a solution.
We need to check if the integral
$\begin{gathered}
\int_0^1 {\sqrt {1 + y'{{(x)}^2}} dx} = \int_1^{ + \infty } {\frac{1}{{{t^2}}}\sqrt {1 + y'{{(\frac{1}{t})}^2}} dt} \hfill \\
= \int_1^{ + \infty } {\frac{1}{{{t^2}}}\sqrt {1 + \frac{1}{{{t^{2(p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
permutation & combinations How many odd three digit numbers are there when tens digit is greater than units digit and hundreds digit is greater than tens digit?
*
*$225$
*$ 45$
*$ 50$
*$230$
My attempt:
The units digit can be $1$ or $3$ or $5$ so $3$ ways ($9$ cannot be taken)
units digit when = 1 _ _ _1... | For a number to be odd, it must end with $1$ or $3$ or $5$ or $7$
Ending with 1: Let us fix $1$ at one's place now we have to fill ten's and hundred's place such that tens digit is greater than units digit and hundreds digit is greater than tens digit.
Placing $2$ at ten's place leaves us with $[3,4,5,6,7,8,9]$ i.e, $7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Number of solution of Frobenius equation Oke I am trying to find all presentable of a number $n$ as sum $ax+(a+1)y$ where $a=0,1,\ldots$ and $x,y\geq0$ are integers.
I find that
$5=1+1+1+1+1=1+1+1+2=1+2+2=2+3=5$ so we have $5$ ways.
$7=1+1+1+1+1+1+1=1+1+1+1+1+2=1+1+1+2+2=1+2+2+2=2+2+3=3+4$ so we have $7$ ways.
I suppo... | This should hold for any $n$, not just $n$ prime. For the induction step from $n-1$ to $n$, note that a generic sum is
$$n-1 = ax + (a+1)y,$$
or
$$n = ax + (a+1)y +1 = a(x-1) + (a+1)(y+1),$$
assuming $x$ is not zero. So in this case, a sum for $n-1$ corresponds to one for $n$. If x is zero, then use
$$n-1= ay,$$
or
$$... | {
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"timestamp": "2023-03-29T00:00:00",
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Modular Arithmetic with large exponents! Decide whether each of the following is true or false without using a calculator:
The problem is:
$$11^{99}\equiv 1\pmod{5}$$
Now I know I can break the $11$ into $(10+1)^{99}$ and maybe rewrite it as $(10+1)^{98}\times (10+1)$
and then realize that $10+1\equiv 1\pmod{5}$ and th... | There are many ways to approach this.
One way is to note that for all $n$: if $\gcd(a,n) = 1$:
$a^{\phi(n)} = 1$ (mod $n$). This is known as Euler's Theorem.
Here, we have $a = 11, n = 5$, and $\phi(5) = 4$, so we have:
$11^{99} = (11^{96})(11^3) = (11^{4})^{24}(11^3) = (1^{24})(11^3) = 11^3$ (mod $5$).
Then it is easy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Riemann integral problem 4 How can I evaluate the following limit with Riemann integral not with telescoping series:
$$ u_n=\frac{1}{n}\sum_{k=1}^n n\cdot \arctan\left(\frac{1}{k^2+k+1}\right),$$
$$\lim _{n\to \infty }u_n=\lim _{n\to \infty }\frac{1}{n}\sum_{k=1}^n n\cdot \arctan\left(\frac{1}{k^2+k+1}\right)=\lim_{n\t... | Integration by parts gives:
$$\int_{0}^{1}\arctan\left(\frac{1}{1+x+x^2}\right)\,dx = \arctan\frac{1}{3}+\int_{0}^{1}\frac{x(1+2x)}{(1+x^2)(1+(x+1)^2)}\,dx$$
and the last integral equals:
$$ \int_{0}^{1}\left(\frac{x}{1+x^2}-\frac{x+1}{1+(x+1)^2}+\frac{1}{1+(x+1)^2}\right)\,dx =\frac{2\log 2-\log 5}{2}+\arctan 2-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Simultaneusly solving $2x \equiv 11 \pmod{15}$ and $3x \equiv 6 \pmod 8$ Find the smallest positive integer $x$ that solves the following simultaneously.
Note: I haven't been taught the Chinese Remainder Theorem, and have had trouble trying to apply it.
$$
\begin{cases}
2x \equiv 11 \pmod{15}\\
3x \equiv 6 \pmod{8}
\en... | Let's proceed completely naively and see where it takes us.
The first congruence is equivalent to $x \equiv 13 \pmod{15}$. This is the same as $x = 13 + 15n$ for any integer $n$. Let's use this in the second congruence.
$$\begin{align}
3x &\equiv 6 \pmod 8 \\
3(13 + 15n) &\equiv 6 \pmod 8 \\
39 + 45n &\equiv 6 \pmod 8 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $\left(\cos{\frac{5\pi}{9}}\right)^{11}+\left(\cos{\frac{7\pi}{9}}\right)^{11}+\left(\cos{\frac{11\pi}{9}}\right)^{11}$ How to evaluate $$\left(\cos{\frac{5\pi}{9}}\right)^{11}+\left(\cos{\frac{7\pi}{9}}\right)^{11}+\left(\cos{\frac{11\pi}{9}}\right)^{11}?$$
I found the problem on this page.
| Starting from Frieder's final expression, there is a terrible closed form since $$\cos \left(\frac{\pi }{9}\right)=\frac{\sqrt[3]{1-i \sqrt{3}}+\sqrt[3]{1+i \sqrt{3}}}{2 \sqrt[3]{2}}$$ $$\cos \left(\frac{2\pi }{9}\right)=2\cos^2 \left(\frac{\pi }{9}\right)-1$$ $$\sin \left(\frac{\pi }{18}\right)=\sqrt{\frac{1-\cos \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Residue Integration I am attempting to calculate the integral of $\frac{(1+sin(\theta))}{(3+cos(\theta))}$ from $0$ to $2\pi$. I have already changed $sin$ and $cos$ into $\frac{1}{2i(z-z^{-1})}$ and $\frac{1}{2(z+z^{-1})}$. I am really stuck now. Can anyone please guide me?
| If direct way can be considered, just as Dr.MW already answered, tangent half-angle substitution $t=\tan \frac \theta 2$ makes the problem simple since $$I=\int\frac{(1+sin(\theta))}{(3+cos(\theta))}d\theta=\int\frac{(t+1)^2}{t^4+3 t^2+2}dt=\int \frac{2 t}{t^2+1} dt+\int\frac{1-2 t}{t^2+2}dt$$ $$I=\log \left(1+t^2\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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how to solve $2(2y-1)^{\frac{1}{3}}=y^3+1$ how can I solve $2(2y-1)^{\frac{1}{3}}=y^3+1$
I got by $x=\frac{y^3+1}{2}$ that $y=\frac{x^3+1}{2}$ but I was told earlier that can't say that $x=y$.
So I can I solve this equation?
Thanks.
| You correctly set $x^3=2y-1$, so the equation becomes
$$
\begin{cases}
2y=x^3+1\\
2x=y^3+1
\end{cases}
$$
Subtract them:
$$
2(y-x)=x^3-y^3
$$
or
$$
(x-y)(x^2+xy+y^2)+2(x-y)=0
$$
This factors as
$$
(x-y)(x^2+xy+y^2+2)=0
$$
but the second factor is positive for all $x$ and $y$, because $x^2+xy+y^2\ge0$.
Therefore you con... | {
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Prove that $1-\frac{1}{1+\frac{\alpha}{nm}}\leq \sqrt{1-\frac{1}{1+\frac{\alpha}{n}}}\sqrt{1-\frac{1}{1+\frac{\alpha}{m}}}$. For what values of the real parameter $\alpha$ the following inequality is true?
$$1-\frac{1}{1+\frac{\alpha}{nm}}\leq \sqrt{1-\frac{1}{1+\frac{\alpha}{n}}}\sqrt{1-\frac{1}{1+\frac{\alpha}{m}}}$$... |
$$1-\frac{1}{1+\frac{\alpha}{nm}}\leq \sqrt{1-\frac{1}{1+\frac{\alpha}{n}}}\sqrt{1-\frac{1}{1+\frac{\alpha}{m}}}$$
Coverts to:
$$\frac{\alpha}{\alpha+mn}\le\sqrt{\frac{\alpha}{\alpha +m}}\sqrt{\frac \alpha{\alpha+n}}$$
Or if $\alpha\ne0$ we can cancel it and get:
$$(\alpha+m)(\alpha+n)\le(\alpha+mn)^2\\
\alpha^2+(m+n... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving $ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ for all $n\geq 2$ by induction Question:
Let $P(n)$ be the statement that $1+\dfrac{1}{4}+\dfrac{1}{9}+\cdots +\dfrac{1}{n^2} <2- \dfrac{1}{n}$. Prove by mathematical induction. Use $P(2)$ for base case.
Attempt at solution:
So I plugged in $... | For $n\geq 2$, let $S(n)$ denote the statement
$$
S(n) : 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}.
$$
Base step ($n=2$): $S(2)$ says that $1+\frac{1}{4}=\frac{5}{4}\leq\frac{3}{2}= 2-\frac{1}{2}$, and this is true.
Inductive step: Fix some $k\geq 2$ and suppose that $S(k)$ is true. It remains to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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diophantine equation $x^3+x^2-16=2^y$ Solve in integers: $x^3+x^2-16=2^y$.
my attempt:
of course $y\ge 0$, then $2^y\ge 1$, so $x\ge 1$.
for $y=0,1,2,3$ there is no good $x$.
so $y\ge 4$ and we have equation $x^2(x+1)=16(2^z+1)$, where $z=y-4\ge 0$.
what now?
| Not my answer. Copied.
$x^2(x+1) = 2^y + 16$
Since LHS is an integer, then we must have $y \ge 0$.
Since RHS is a positive integer, then we must have $x \ge 1$.
$x^2(x+1)$ is strictly increasing for $x \ge 0$.
$2^y + 16$ is strictly increasing for $y \ge 0$.
$$\begin{array}{n|c|c|}
\hline
n & n^2(n+1) & 2^n + 16 \\ \h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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How to calculate sum of combinations with different n and k Input: $[X,Y]$ and $L$
Output : no of increasing sequence of length L and all elements should be $X\le i \le Y$ e.g: for $[6,7]$ and $2$ sequences are $6,66,67,7,77.$
For the above question my answer is the following sequence
$a+b+c+d+..$ where
$a=N$
$b=\sum_... | Notice that $${n \choose m}={n-1 \choose m-1}+{n-1 \choose m}$$
Hence
\begin{align*}
\sum_{k=1}^{D} {n+k-1 \choose k} &= \sum_{k=1}^{D} {n+k-1 \choose n-1} \\
&= {n \choose n} + \sum_{k=1}^{D} {n+k-1 \choose n-1} - {n \choose n} \\
&= {n \choose n} + {n+1-1 \choose n-1}+ \cdots +{n+D-1 \choose n-1} - {n \choose n} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find $n$ such that $(m-1)(m+3)(m-4)(m-8)+n$ is a perfect square for all $m$ Find $n$ such that $(m-1)(m+3)(m-4)(m-8)+n$ is a perfect square for all $m$
I am thinking of starting like this
$(m-1)(m+3)(m-4)(m-8)+n = k^2 \implies (m-1)(m+3)(m-4)(m-8) = k^2-n$
Honestly somewhat scared of expanding the products on left han... | The trick is to do $(m-1)(m+3)(m-4)(m-8)=\underbrace{(m-1)(m-4)}\underbrace{(m+3)(m-8)}$ since $-1-4=-5=3-8$ giving same coefficient of m:
$$(m-1)(m+3)(m-4)(m-8)+n=(m^2-5m+4)(m^2-5m-24)+n\stackrel{y=m^2-5m+4}=y(y-28)+n=y^2-28y+n=(y-14)^2+n-14^2\implies n=14^2=196$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to deduce that $1\cdot 1 + 2\cdot 1 + 2\cdot 2 + 3\cdot 1+3\cdot 2+3\cdot 3 +...+(n\cdot n) = n(n+1)(n+2)(3n+1)/24$ I know how to reason $$1\cdot2 + 2\cdot3 + 3\cdot4 + n(n-1) = \frac{1}{3}n(n-1)(n+1)$$
However, I'm stuck on proving $$1\cdot1 + (2\cdot1 + 2\cdot2) + (3\cdot1+3\cdot2+3\cdot3) + \cdots +(n\cdot 1+...... | A similar but slightly different variation of Jack's proof above.
$$\begin{align}
&1\cdot (1)\\
+&2\cdot(1+2)\\
+&3\cdot(1+2+3)\\
+&4\cdot(1+2+3+4)\\
+&\vdots\qquad\vdots \; \quad\ddots\quad \ddots\\
+&n\cdot(1+2+3+\cdots+n)\\
&=\sum_{i=1}^ni\sum_{j=1}^i j=\sum_{i=1}^ni\sum_{j=1}^i {j\choose 1}\\
&=\sum_{i=1}^ni{i+1\ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Find the last three digits of $17^{256}$ Find the last three digits of
$ 17 ^{256} $
We have to check mod $1000$
I tried to check some patterns but in vain.!
| As $1000=8\times 125$ comptute first $17^{256}$ modulo $8$ and modulo $256$, then use the Chinese Remainder Theorem to recover $17^{256}\mod 1000$.
Modulo $8$: $\enspace 17\equiv 1\mod 8$, hence $17^{256}\equiv 1 \mod 8$.
Modulo $125$:
By Euler's theorem, $n^{\varphi(125)}\equiv 1\mod125$ for all $n$. As $\varphi(125)=... | {
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How can i calculate Total no. of digit in $2^{100}\cdot 5^{75}$
How can i calculate Total no. of digit in $2^{100}\cdot 5^{75}$
$\bf{My\; Try::}$ I have used $$\log_{10}(2) = 0.3010$$.
Now Total no. of digit in $$x^y = \lfloor \log_{10}x^y\rfloor +1$$
Now $$\log_{10}(2^{100}\cdot 5^{75}) = 100\cdot \log_{10}(2)+75\l... | I will note that your answer went from $100 \log_{10}(2) + 75 - 75\log_{10}(20)$ to $100 \log_{10}(2) + 75$. The general form of argument is correct, but taking this mistake into account the answer is $\lfloor 25\log_{10}(2) + 75\rfloor+1 = \lfloor 75 + 7.525...\rfloor+1 = 83$.
Doing this without logarithms is fairly ... | {
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"source": "stackexchange",
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Proving $\lim_{x\to c}x^3=c^3$ for any $c\in\mathbb R$ using $\epsilon$-$\delta$ definition
$\lim_{x\to 3}x^3=c^3$ for any $c\in\mathbb R$
Let $\epsilon>0$. Then
$$|x^3-c^3|=|x-c||x^2+xc+c^2|.$$
Let $\delta=\min\{1,\epsilon/x^2+xc+c^2\}$.
Then if $0<|x-c|<\delta$ and therefore since $|x-c|<\epsilon/(x^2+xc+c^2)$,
$$|... | Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-c|<\delta$, then $|x^3-c^3|<\epsilon$.
Now,
$
|x^3-c^3| = |x-c||x^2+cx+c^2|
$
If $|x-c|<1$, then
$||x|-|c|| \leq |x-c|<1$
i.e. $||x|-|c||<1 \implies -1<|x|-|c|<1 \implies |x|<|c|+1$
so that
$$
|x^2+cx+c^2|\leq |x|^2+|c||x|+|c|^2 < (|c|+1)^2+|c|(|c|+1)+|c|^2=3|c|... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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evaluate $\int \frac{\tan x}{x^2+1}\:dx$ $$\int \frac{\tan x}{x^2+1}\, \mathrm dx$$
I used By-parts method setting $u=\tan x$ and $\, \mathrm dv=\frac{1}{x^2+1}\, \mathrm dx$, but then I got an integral that's more complicated
I also thought of trigonometric substitution, setting $x=\tan\theta$, but how am going to ... | The Laurent series of tan(x) is
$$\sum_{n=1,3,5..}^{\infty }\frac{8x}{(n\pi )^2-4x^2}$$
so
$$\frac{\tan(x)}{1+x^2}=\sum_{n=1,3,5..}^{\infty }\frac{8x}{\left [(n\pi )^2-4x^2 \right ](1+x^2)}$$
use the partial fraction to get
$$\sum_{n=1,3,5,..}^{\infty }\frac{8x}{((n\pi)^2+4 )(1+x^2)}+\frac{8}{((n\pi)^2+4 )(n\pi -2x)}-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Maximum value of $ x^2 + y^2 $ given $4 x^4 + 9 y^4 = 64$ It is given that $4 x^4 + 9 y^4 = 64$.
Then what will be the maximum value of $x^2 + y^2$?
I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .
| If $64= (3y^2)^2 + 4x^4$, then $3y^2 =\sqrt{ 64 - 4x^4}$. Plugging that to $x^2+y^2$ we get:
$$x^2+\frac{\sqrt{64 - 4x^4}}{3}.$$
Now you have to maximalize a function of one variable. Remember that if $64 = 4x^4+9y^4$, then (setting $y = 0$) we obtain $16 \ge x^4$, therefore $|x| \le 2$.
| {
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"source": "stackexchange",
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Prove ${4n \choose 2n} = {\frac{1\cdot3\cdot5\cdots(4n-1)}{(1\cdot3\cdot5\cdots(2n-1))^{2}}}{2n \choose n}$ Prove that prove $\dbinom{4n}{2n} = \dfrac{1\cdot3\cdot5\cdots(4n-1)}{(1\cdot3\cdot5\cdots(2n-1))^2} \dbinom{2n}{n}$ using mathematical induction. I have looked all over the internet, been able to prove a simila... | Hints:
*
*$\dfrac{1\times 3\times 5\times \cdots \times (4n-1)}{(1\times 3\times 5\cdots \times (2n-1))^{2}} $ $= \dfrac{1\times 2\times 3\times 4\times 5\times \cdots \times (4n-1)\times 4n}{(1\times 2\times 3\times 4\times 5\times \cdots \times (2n-1)\times 2n)^{2}} \times \dfrac{(2\times 4\times \cdots \times 2n)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the function that equals to $1-x^3+x^6-x^9+ \cdots$
Find the function that equals to $1-x^3+x^6-x^9+ \cdots$ for all $|x| < 1$
I know that $\frac{1}{1+x} = 1-x+x^2-x^3+...$ But I couldn't find the pattern here
| Substitute $y=x^{3}$
$$S = 1-y+y^{2}-y^{3}+\cdots$$
$$S = \frac{1}{1+y} = \frac{1}{1+x^{3}}$$
| {
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How do I solve these questions using Diophantine equations? I have been told that it is easier to solve the below 2 questions using Diophantine equations instead of simply trial and error.
1) Find the smallest positive integer which, when divided by 6, gives a remainder of 1 and when divided by 11, gives a rema... | One should maybe say that using this approach replaces the guesswork by some modeling and applying a solution algorithm, so it is more systematic.
Problem 1):
From the equations for the divisions
$$
u = 6x+1 =11y+6
$$
we get a Diophantine linear equation
$$
6x-11y=5
$$
which is the name for a linear equation with inte... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\lim\limits_{n \to \infty} \frac{n^2+1}{5n^2+n+1}=\frac{1}{5}$ directly from the definition of limit.
Prove $\lim\limits_{n \to \infty} \frac{n^2+1}{5n^2+n+1}=\frac{1}{5}$ directly
from the definition of limit.
So far ive done this:
Proof:
It must be shown that for any $\epsilon>0$, there exists a positive i... | $$|\frac{4-n}{25n^2+5n+5}|< ϵ\\\frac{|4-n|}{|25n^2+5n+5|}< ϵ\\\frac{|n-4|}{|25n^2+5n+5|}< ϵ\\$$ now $n∈N ,n \rightarrow \infty $so $n-4>0 ,|n-4|=n-4 $ also $25n^2+5n+5>0 \rightarrow |25n^2+5n+5|=25n^2+5n+5$ $$\frac{n-4}{25n^2+5n+5} <\frac{n-4}{25n^2+5n}<\frac{n}{25n^2+5n}=\frac{1}{5n+1} < ϵ\\5n+1 > \frac{1}{ϵ}\\5n>\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$ Could anybody help me by checking this solution and maybe giving me a cleaner one.
Prove by mathematical induction:
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.
So after I check special cases for $n=2,3$, I have to prove... | Your reasoning is a little overbearing on the sums, so here's a simplification although it looks like it's basically the same thing. Define $F(n)=\sum_{k=n}^{n^2}\frac{1}{k}$. Then
$$F(n+1)=F(n)-\frac{1}{n}+\frac{1}{n^2+1}+\cdots +\frac{1}{n^2+2n+1}.$$
By the induction hypothesis, $F(n)>1$. So it suffices to prove:
$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inverse function. A function $h$ is defined by $h:x\rightarrow 2-\frac{a}{x}$, where $x\neq 0$ and $a$ is a constant. Given $\frac{1}{2}h^2(2)+h^{-1}(-1)=-1$, find the possible values of $a$.
Can someone give me some hints? Thanks
| Thank you for this problem. I was not familiar with the notation $h^2(x)=h(h(x))$, but this is clearly what must be meant in this problem since interpreting $h^2(x)=(h(x))^2$, yields no real solutions.
If $h(x)=2-\frac{a}{x}$, then we can solve for $h^{-1}(x)$ in the following manner.
$x=2-\frac{a}{h^{-1}(x)} \Right... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove equation $(ad-bc)(a-c)^2 = (b-d)^3$, if polynomials has common root
$$\begin{split}
W(x) &= x^3 + ax + b \wedge a,b \in \mathbb{R}
&\wedge \mathbb{D}_W &= \mathbb{R}\\
G(x) &= x^3 + cx + d \wedge c,d \in \mathbb{R}
&\wedge \mathbb{D}_G &= \mathbb{R}
\end{split}
$$ Prove:
$$ \left(\exists p \in \mathbb... | Let $y$ be the common root. We then have
$$y^3 + ay+b = 0 \text{ and }y^3+cy+d = 0$$
This means we have
$$ay+b = cy+d \implies y = \frac{d-b}{a-c}$$
Hence,
$$\left(\frac{d-b}{a-c}\right)^3 + a \left(\frac{d-b}{a-c}\right) + b = 0$$
This gives us
$$(d-b)^3 + a(d-b)(a-c)^2 + b(a-c)^3 = 0 \implies (b-d)^3 = (a-c)^2(ad-ab+... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$
Calculate $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$. accurate upto two decimal places or in surds .
$\begin{align}\sin 69^{\circ}&=\sin (60+9)^{\circ}\\~\\
&=\sin (60^{\circ})\cos (9^{\circ})+\cos (60^{\circ})\sin (9^{\circ})\\~\\
&=\dfr... | You may exploit:
$$ \sin(60^\circ)=\frac{1}{2}\sqrt{3},\quad \sin(18^\circ)=\frac{1}{4}\left(\sqrt{5}-1\right),\quad \tan(22^\circ 30')=\sqrt{2}-1$$
then use some form of interpolation. For instance, in a neighbourhood of $x=\frac{\pi}{3}$:
$$ \sin(x)=\frac{1}{2}\sqrt{3}+\frac{1}{2}\left(x-\frac{\pi}{3}\right)-\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I find the sum of the series $\sum_{n=0}^\infty {(-1)^n \over 4^n}$ or show that it diverges using the geometric series test? First, I reindexed it:
$$\sum_{n=0}^\infty {(-1)^n \over 4^n} = \sum_{n=1}^\infty {(-1)^{n-1} \over 4^{n-1}} = \sum_{n=1}^\infty {\left(-1 \over 4\right)}^{n-1} $$
So now I'm pretty sure... | If we break the summations down we get:
$$S = \sum_{i=0}^{n}\frac{1}{4^{2(n-1)}} - \sum_{i=0}^{n}\frac{1}{4^{2n-1}}$$
That is the sum of the reciprocals of 4 to the power of the even positive numbers, then you subtract the reciprocals of 4 to the power of the odd positive numbers. $2(n-1)$ is the $n_{th}$ even number,... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Volume of a hole through cylinder (from the side) I need to calculate the volume of a circular hole in a cylinder and I've come across a problem. The problem is finding the "cap-volume", which is needed to complete the volume of the hole. I created a quick example of the problem.
Cylinder which will be drilled
Hole.... | Let $R$ and $r$ be the radius of the cylinder and the hole. We will assume $k = \frac{r}{R} < 1$.
We will further assume the hole is drilled toward the center and perpendicular to the axis of the cylinder. Under these assumptions, the volume of the cap is given by
$$\begin{align}
\verb/Vol/_{cap}
&= 4
\int_0^r \int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1246464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate volume of oblique frustum of right circular cone with elliptical section?
Let there be an oblique frustum, with an elliptical section, of a right circular cone with apex point O & cone angle $2\alpha=60^{o}$. It is obtained by cutting the cone by a plane at a normal distance $OM=h=20 cm$ & making an an... | We will not only find the volume but prove that the cross-section is indeed an ellipse.
We look for the intersection (see figure) of the cone:
$$
x^2+y^2=z^2\,\tan^2\alpha\tag1
$$
with the plane:
$$
z=\frac d{\sin\theta}+\frac x{\tan\theta}\tag2
$$
Substituting $(2)$ into $(1)$ results in:
$$\begin{align}
&x^2+y^2=\le... | {
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"timestamp": "2023-03-29T00:00:00",
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Integration: $\int\frac{1}{(x^2+x+1)^{1/2}} dx$ Find the value of $$\int\frac{1}{(x^2+x+1)^{1/2}} dx$$
Anyone can provide hint on how to integrate this, and how you know what method to use? (I mean, is there any general guideline to follow for solving?)
Thank you!
| If you make the Euler substitution $t=x+\sqrt{x^2+x+1}$, then $\displaystyle x=\frac{t^2-1}{2t+1}$ and $\displaystyle dx=\frac{2(t^2+t+1)}{(2t+1)^2}dt$;
so $\displaystyle\int\frac{1}{\sqrt{x^2+x+1}}dx=\int\frac{1}{t-\frac{t^2-1}{2t+1}}\cdot\frac{2(t^2+t+1)}{(2t+1)^2}dt=\int\frac{2}{2t+1}dt=\ln\lvert2t+1\lvert+C$
$\disp... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve for $x:1 + \tan^2(x) = 8\sin^2(x)$ I have a tricky problem , I tried several methods and I can't seem to get a definite answer.
$1 + \tan^2(x) = 8\sin^2(x), x \in [\frac{\pi}{6} , \frac{\pi}{2}]$
I got to $8\cos^4(x)-8\cos^2(x)+1=0$ and found that $\cos^2(x) = \frac{1}{4}[2-\sqrt{2}]$ but that is not too useful... | If $\sin^2y=\sin^2A\iff\cos2y=1-2\sin^2y=\cdots=\cos2A$
$\iff2y=2m\pi\pm2A\iff y=m\pi\pm A$ where $m$ is any integer
We have $\sin^2(2x)=\dfrac12=\left(\sin\dfrac\pi4\right)^2$
$\implies2x=r\pi\pm\dfrac\pi4=\dfrac\pi4\left(4r\pm1\right)$ where $r$ is any integer
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Write a function as $\sum _{n=0} ^{\infty} a_n x^n$ We have $f(x) = (x+ x^2 + x^3 + x^4 + x^5 + x^6)^4$.
Now I want to write this as $\sum _{n=0} ^{\infty} a_n x^n$.
What I got:
$f(x) = (x+ x^2 + x^3 + x^4 + x^5 + x^6)^4 = x^4 (1+ x + x^2 + x^3 + x^4 + x^5)^4$
We know: $\frac{1}{1-x} = \sum _{n=0} ^{\infty} x^n$.
Note ... | (can't leave a comment)
You'll want the multinomial series formula for this purpose:
$$\left(\sum_{j=1}^k a_j\right)^n=\sum_{\stackrel{n_j \geq 0}{\sum_{j=1}^k n_j=n}}\frac{n!}{\prod_{j=1}^k n_j!}\prod_{j=1}^k a_j^{n_j}$$
where the condition on the sum on the right means to sum all over the integer partitions of the po... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$a^2 = 2b^3 = 3c^5$ Find the smallest value of $abc$. We have following equation:
$a^2 = 2b^3 = 3c^5$
Where $a, b, c$ are natural numbers.
Find the smallest possible value of product $abc$.
| Introducing any prime factors apart from $2$ and $3$ makes things worse. Therefore we may assume $a=2^\alpha3^\beta$, which leads to
$$b^3=2^{2\alpha-1}3^{2\beta},\qquad c^5=2^{2\alpha}3^{2\beta-1}\ .$$
therefore we have to find the smallest $\alpha\geq0$, $\beta\geq0$ such that
$$2\alpha-1=0\quad(3),\qquad 2\alpha=0\q... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of times $2^k$ appears in factorial
For what $n$ does: $2^n | 19!18!...1!$?
I checked how many times $2^1$ appears:
It appears in, $2!, 3!, 4!... 19!$ meaning, $2^{18}$
I checked how many times $2^2 = 4$ appears:
It appears in, $4!, 5!, 6!, ..., 19!$ meaning, $4^{16} = 2^{32}$
I checked how many times $2^3 = ... | A simple trick to compute $k$ such that $2^k|n!$ is to compute $\sum_{i=1}^\infty \left\lfloor\frac{n}{2^i}\right\rfloor$, this is because $n$ has $[n/2]$ numbers divided by $2$, if we pick out these numbers and find out that there're $[n/4]$ numbers divided by $4$.. If we continue this procedure, we see that $$k=1\cdo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Am I misinterpreting this matrix determinant property? I was reading matrix determinant properties from wikipedia.
The property reads
$\det(cA) = c^n \det(A)$ for $n \times n$ matrix.
However I am not able to realize it. What I find is $\det(cA) = c\det(A)$
For example, multiplying matrix by 2 and then taking the dete... | $$2 \begin{bmatrix} 4 & 5 & 6 \\ 6 & 5 & 4 \\ 4 & 6 & 5 \end{bmatrix} =
\begin{bmatrix} 8 & 10 & 12 \\ 12 & 10 & 8 \\ 8 & 12 & 10 \end{bmatrix}.$$
You only multiplied the first row by 2.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Factorize Trigonometric Equation: $ 3\sin(x)^2 - 2\sin(x)\cos(x) - \cos(x)^2 = 0 $ I have a problem with the following trigonometric equation:
$$ 3\sin(x)^2 - 2\sin(x)\cos(x) - \cos(x)^2 = 0 $$
It's from the book Engineering Mathematics 7th edition by Stroud.
The book is giving the answer, but I can't seem to be able t... | Let $u = \sin x$; let $v = \cos x$. Then the equation
$$3\sin^2x - 2\sin x\cos x - \cos^2x = 0$$
becomes
$$3u^2 - 2uv - v^2 = 0$$
To split the linear term, we must find two numbers with product $3 \cdot -1 = -3$ and sum $-2$. They are $-3$ and $1$. Hence,
\begin{align*}
3u^2 - 2uv - v^2 & = 0\\
3u^2 - 3uv + uv - v^2... | {
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Sum limit. Please tell if correct. I just solved this limit, and the result provided by the book is different.
$$ \lim_{x\to 1} \frac{x+x^2+x^3 + ... + x^n - n}{x-1} $$
I turned this into:
$$ \lim_{x\to 1} \frac{x-1}{x-1} + \frac{x^2-1}{x-1} + \frac{x^3-1}{x-1} + ... + \frac{x^n-1}{x-1} $$
$$ \lim_{x\to 1} 1 + x + (x ... | Your method of solving is correct and nice, but there are some mistakes. First
$$\frac{x^k-1}{x-1}=x^{k-1}+x^{k-2}+\ldots+x+1$$
and thus
$$\frac{x+x^2+\ldots+x^n-n}{x-1}=\sum_{k=1}^n\frac{x^k-1}{x-1}=$$
$$=\sum_{k=1}^n\left(1+x+\ldots+x^{k-1}\right)=1+2+\ldots+n=\frac{n(n+1)}2$$
One can also do the following, though pe... | {
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How can a binomial coefficient can be approximated by using Stirling's formula? I've met some difficulties with such question:
How can we approximate a binomial coefficient by using a Stirling's factorial approximation.
I've evaluate a little bit and got this
How can I transform the right side of this equation for gett... | Thank you all for your answers, but I mentioned that I have to use slightly different factorial approximation:
$$
n!=\sqrt{2\pi n}\cdot(\frac{n}{e})^n\cdot\left(1+\frac{1}{12n}+\frac{1}{288n^2}+O(\frac{1}{n^3})\right)
$$
So, by using this approximation it's necessary to obtain approximation for the binomial coefficient... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Show that if $p$ is an odd prime, then the congruence $x^2\equiv1\pmod{p^{\alpha}}$ has only two solutions, $x\equiv1,x\equiv-1\pmod{p^{\alpha}}$.
Show that if $p$ is an odd prime, then the congruence $x^2 \equiv 1 \pmod{p^{\alpha}}$ has only two solutions, which are $x \equiv 1, x \equiv -1 \pmod{p^{\alpha}}$.
Clear... | IMHO there is something missing. Your last implication says, in effect, that if $p\mid x-1$ then $p^\alpha\mid x-1$, which is certainly not true.
The point you need to make (earlier in the proof) is this: since $x+1$ and $x-1$ differ by $2$, they cannot both be multiples of $p$ (because $p$ is greater than $2$). Ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1258616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible
Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible.
I am trying to prove it with Eisenstein's criterion but without success: for p=2, it divides -4 and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the cons... | You've seen that $f(x)$ has no roots, so you want to exclude factorizations of the form
$$f(x) = (x^2 + ax + b)(x^2 + cx + d)$$
Since $f(x) = f(-x)$, the above implies
$$f(x) = (x^2 - ax + b)(x^2 - cx + d)$$
Here $a,b,c$, and $d$ are integers by Gauss's Lemma.
So a given root $r$ of $x^2 - ax + b$ is a root of $x^2 + a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 9,
"answer_id": 0
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Is Vieta the only way out? Let $a,b,c$ are the three roots of the equation $x^3-x-1=0$. Then find the equation whose roots are $\frac{1+a}{1-a}$,$\frac{1+b}{1-b}$,$\frac{1+c}{1-c}$.
The only solution I could think of is by using Vieta's formula repeatedly which is no doubt a very messy solution. Is there any easier and... | Transform the equation. Since all the roots are symmetric, say $$y=\frac {1+x}{1-x}\implies x(y+1)=y-1\implies x=\frac {y-1}{y+1}$$ Substitute this expression in place of $x$ in the original equation and simplify. $$\require{cancel}\begin{align}f(y)&=\biggl(\frac {y-1}{y+1}\biggr)^3-\frac {y-1}{y+1}-1=0\\&\implies(y-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
} |
Computing the complex integral? I am dealing with the following:
$$\int_{0}^{\infty}\frac{x\sin(x)}{(x^2+a^2)(x^2+b^2)}dx$$
Furthermore, I know $a,b>0$ and I know $a\neq b$. I believe this is using Jordan's Lemma? I see that the singularities in the upper half of the plane are $ai$ and $bi$ where $R>a$. I'm stuck writi... | $x\sin(x)$ is an even function, so your integral is equal to
$$
\int_0^\infty\frac{x\sin(x)}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x
=\frac12\int_{-\infty}^\infty\frac{x\sin(x)}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x
\tag{1}
$$
We can break up $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$ and compute
$$
\begin{align}
\frac12\int_{-\infty}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1263745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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What is the connection between the discriminant of a quadratic and the distance formula? The $x$-coordinate of the center of a parabola $ax^2 + bx + c$ is $$-\frac{b}{2a}$$
If we look at the quadratic formula
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
we can see that it specifies two points at a certain offset from the cen... | Suppose that $a\ne 0$ and $b^2-4ac\geq 0$. Let $f(x)=ax^2+bx+c$ and
$$
x_0= -\frac{b}{2a},\; x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}, \;x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}.
$$
It follows that
$$
y_0=f(x_0)=\frac{-b^2+4ac}{4a},\;y_1=f(x_1)=0, \; y_2=f(x_2)=0.
$$
The distances from vertex to roots:
$$
d_1=\sqrt{(x_1-x_0)^2+(y_1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 8,
"answer_id": 0
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How to evaluate the $\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5}$, using power series? How to evaluate the $\displaystyle\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5}$, using power series?
It made sense to first try and build the numerator using power series that are commonly used... | if you add up the expansions you have given us, you get 1/2 +0. You take a common factor of $x^5$ from the numerator and divide through you get 1/2+ x(...) --> 1/2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Computing $\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2 + x + 2} \, dx$ I wish to compute
$$\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2 + x + 2} \, dx$$
The singularities are $\pm \frac{\sqrt{7}}{2}i - \frac{1}{2}$
I then make a half-circle contour $C = \{Re^{it}, t \in [0,\pi], R \in [0,\infty) \}$
$$ \int_{-\infty}^{\inf... | We have $x^2+x+2 = (x+a)(x+b)$, where $a+b=1$ and $ab=2$. Hence, we have
$$\dfrac{\cos(x)}{x^2+x+2} = \dfrac{\cos(x)}{(x+a)(x+b)} = \dfrac1{b-a}\left(\dfrac{\cos(x)}{x+a} - \dfrac{\cos(x)}{x+b}\right)$$
Our integral is of the form
$$I = \dfrac1{b-a}\int_{-\infty}^{\infty}\left(\dfrac{\cos(x)}{x+a} - \dfrac{\cos(x)}{x+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Solve the following recurrence relation: $S(1) = 2$; $S(n) = 2S(n-1)+n2^n, n \ge 2$ Solve the following recurrence relation:
$$\begin{align}
S(1) &= 2 \\
S(n) &= 2S(n-1) + n 2^n, n \ge 2
\end{align}$$
I tried expanding the relation, but could not figure out what the closed relation is:
+-------+----------------------... | We can write $S(n)=2+2(2^2)+3(2^3)+\cdots+n(2^n)$. However, we can simplify the problem slightly by generalizing: replace $2$ with $x$ to get
$$f_n(x)=x+2x^2+3x^3+\cdots+ nx^n = x(1+2x+3x^2+\cdots+nx^{n-1})=x \frac{d}{dx}(1+x+\cdots+x^n)$$
Now, using the formula for a geometric sum, $1+x+\cdots+x^n=\frac{1-x^{n+1}}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Convert Riemann sum to definite integral $ \lim_{n\to\infty}\sum_{i=1}^{n} \frac{21 \cdot \frac{3 i}{n} + 18}{n} $ The limit
$\quad\quad \displaystyle \lim_{n\to\infty}\sum_{i=1}^{n} \frac{21 \cdot \frac{3 i}{n} + 18}{n} $
is the limit of a Riemann sum for a certain definite integral
$\quad\quad \displaystyle \int_a^... | You were so close with $\frac {3}{n} (7 \cdot \frac {3i}{n} + 6)$. You can simply factor the $7$ outside the sum as follows:
Given the Riemann sum definition of the definite integral,
$$\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(a + i \, \mathrm{d}x) \, \mathrm{d}x = \int_a^b f(x)\, \mathrm{d}x$$
we manipulate to fin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Integral-derivative issue The derivative of $sin^2(x)$ is $2sin(x)cos(x)$. You can also write it as $sin(2x)$.
If we integrate $\sin(2x)$ we get $-0.5\cos(2x)$ and according to calculator does not equal $\sin^2(x)$. Help?
| Both functions ($\sin^2$ and $-\frac 12 \cos(2\cdot)$) differ by a constant, as any two function $\mathbf R \to \mathbf R$ having the same derivative do. Note that
\begin{align*}
\cos(2x) &= \cos^2 x - \sin^2 x\\
&= 1 - \sin^2 x - \sin^2 x\\
&= 1 - 2\sin^2 x\\
\iff \sin^2 x &= \frac 12 - \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluate $ \lim_{x\to 0} \frac{\tan(4x)}{\sin(7x)}$ Evaluate $$ \lim_{x\to 0} \ \frac{\tan(4x)}{\sin(7x)}$$
I am stuck after I convert tan(4x) into sin (4x) / cosine(4x)
| Recall that $\lim_{x\to0} \sin(x)/x = 1$ and $\lim_{x\to 0} \tan(x)/x = 1$.
Then $$\lim_{x\to 0} \frac{\tan(4x)}{\sin(7x)} = \lim_{x\to 0} \frac{7}{7} \cdot \frac{4x}{4x} \cdot \frac{\tan(4x)}{\sin(7x)} = \lim_{x\to 0} \frac{4}{7} \frac{\tan(4x)}{4x} \cdot \frac{7x}{\sin(7x)} = \frac47 \cdot 1 \cdot 1 = \frac47.$$
Onc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Maclaurin series for a function Provided I have the function
\begin{equation*}
f(x)=(1+x)^{1/x},
\end{equation*}
and I want to calculate a 3rd order Maclaurin series, how can that be done without taking direct derivatives (as this seems hard..). I know that
\begin{equation*}
(1+x)^{1/x}=e^{ln(1+x)/x},
\end{equation*... | $$\frac{\ln(1+x)}x=1-\frac x2+\frac{x^2}3-\frac{x^3}4+o(x^3)$$
Substitute $-\frac x2+\frac{x^2}3-\frac{x^3}4$ to $u$ in the development of $\mathrm e^u$, first computing the succesive powers of $u$:
\begin{align*}
u^2&=\frac{x^2}4-\frac{x^3}3+o(x^3),\\
u^3&-\frac{x^3}8+o(x^3),
\end{align*}
so that $$(1+x)^{\tfrac 1x}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Let $a,b$ be relative integers such that $2a+3b$ is divisible by $11$. Prove that $a^2-5b^2$ is also divisible by $11$. The divisibility for $11$ of $a^2 - 5b^2$ can be easily verified; in fact: $$a \equiv \frac {-3}{2}b \pmod {11}$$ therefore $$\frac {9}{4}\cdot b^2 - 5b^2 = 11(-\frac{b^2}{4}) \equiv 0 \pmod {11}.$$
T... | My first thought was to think about a difference of squares: $(2a+3b)(2a-3b) = 4a^2 - 9b^2$. So if $2a+3b$ is divisible by 11, then $4a^2-9b^2$ as well. And therefore $4a^2-9b^2+11b^2 = 4a^2+2b^2$ is divisible by 11.
That's not what you asked about. But is there some number $k$ such that if $k(4a^2+2b^2)$ is divisib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve 3 variable in 2 equation? This paper is abstracted from 2007 British Mathematics Olympiad Round 1 Question 2.
I am currently practicing grade 8 (Singapore Secondary 2) for the upcoming Singapore Mathematics Olympiad(SMO). Even before solving the question,is it even possible to solve 3 variable of 2 equatio... | $$x+y-12=z \Rightarrow x^2+y^2-(x+y-12)^2=12$$
This leads to
$$xy-12x-12y+78=0$$
or
$$(x-12)(y-12)=66$$
Now check all the possible ways of writing $66$ as a product of 2 integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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How many $n-$digit number that contain only digits $ 1,2,3,4,5,6$ How many $n- $digit numbers can be formed from the digits $1,2,3,4,5$ and 6, which contains the numbers $1$ and $2$ as neighbours.
Let $p_n$ be the number of n-digit numbers which consist only of the digits 1,2,3,4,5,6, which contains the numbers $1$ a... | Start by making $n$-symbol words from the five symbols $[12],3,4,5,6$; there are $\binom{n+4}{4}$ ways to do this (use "stars and bars".) For each such word, you have two possible orderings of the $1$ and $2$. So
$$p_n=2\binom{n+4}{4}.$$
And $p_{n+1}-p_n=2\binom{n+5}{4}-2\binom{n+4}{4}=2\binom{n+4}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Trouble constructing $\mathbb Z_3[x]/(x^2+1)$ If I have $\mathbb Z_3[x]/(x^2+2x+2)$, I can construct a field by letting $x^2=x+1$. The reps are:
$0$
$1$
$x$
$x^2=x+1$
$x^3=x^2+x=x+1+x=2x+1$
$x^4=2x^2+x=2x+2+x=2$
$x^5=2x$
$x^6=2x^2=2x+2$
$x^7=2x^2+2x=2x+2+2x=x+2$
However, when I try this for $\mathbb Z_3[x]/(x^2+1)$, l... | If I understand you correctly, the issue is that there's no guarantee that $x$ will be a multiplicative generator for the group of units for the resulting field. You might to choose another element. One choice might be $x+2$, which squares to $x^2+4x+4 \equiv x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Extracting real and imaginary numbers from a complex number How can I get the real number and the imaginary number from: $$\frac{3+i}{5-12i}$$
| $$\frac{3+i}{5-12i}=$$
$$\left|\frac{3+i}{5-12i}\right|e^{\arg\left(\frac{3+i}{5-12i}\right)i}=$$
$$\frac{\left|3+i\right|}{\left|5-12i\right|}e^{\arg\left(\frac{3}{169}+\frac{41}{169}i\right)i}=$$
$$\frac{\sqrt{3^2+1^2}}{\sqrt{5^2+12^2}}e^{\tan^{-1}\left(\frac{41}{3}\right)i}=$$
$$\frac{\sqrt{10}}{\sqrt{169}}e^{\tan^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$ Question:
Show that:
$$\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$$
then go on to prove the general case that:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k=1}^n \binom{n}{k}x^{k-1}$$
Attempted solution:
It might be doable to first prove... | Try induction in $n$, and then use $\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For which values does the Matrix system have a unique solution, infinitely many solutions and no solution? Given the system:
$$\begin{align}
& x+3y-3z=4 \\
& y+2z=a \\
& 2x+5y+(a^2-9)z=9
\end{align}$$
For which values of a (if any) does the system have a unique solution, infinitely many solutions, and no solution?... | Note that your system is equivalent to the matrix equation
$$
\begin{bmatrix}
1 & 3 & -3\\ 0 & 1 & 2 \\ 2 & 5 & a^2-9
\end{bmatrix}
\begin{bmatrix}
x\\ y\\ z
\end{bmatrix}
=
\begin{bmatrix}
4\\ a\\9
\end{bmatrix}
$$
Since
$$
\det\begin{bmatrix}
1 & 3 & -3\\ 0 & 1 & 2 \\ 2 & 5 & a^2-9
\end{bmatrix}=a^2-1
$$
this system ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Inequality for sides and height of right angle triangle
Someone recently posed the question to me for the above, is c+h or a+b greater, without originally the x and y lengths. I used this method: (mainly pythagorus)
$a^2+b^2=c^2=(x+y)^2=x^2+y^2+2xy$
$a^2=x^2+h^2$ and $b^2=y^2+h^2$
therefore $x^2+h^2+y^2+h^2=x^2+y^2+2x... | Perhaps easier, in the right triangle, once you note $ab=ch$, the smaller sum is when the terms are closer to each other, and clearly $c>a,b>h$, so $c+h>a+b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Compute power of a matrix $A$ as $n\rightarrow \infty$ We are given $A^p=A ...A$(p times)
And we are given matrix A:
$A=\begin{vmatrix}0.6&-0.4&0\\-0.4&0.6&0\\0&0&0.5\end{vmatrix}$
I need to compute $A^p$ as p approach Infinity.
By some calculations results that the values of matrix A get always close to $0$ as n grows... | Everything is in the passage you quoted.
THe matrix $A$ has three distinct real eigenvalues, hence it is diagonalisable:
$A=PDP^{-1}$, where $D$ is a digonal matrix with eigenvalues of $A$ and $P$ is a matrix whose columns are eigenvectors of $A$.
In your case it easy to find these eigenvectors: $(1,-1,0)^T$ correspon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integral giving wrong values Given following function:
$x^3 + 1$
(source: gyazo.com)
Find the area that is selected in red lines.
So I solved the root $\sqrt[3]{-1} = -1$ so $x = -1$
So now I have to create the 3 integrals:
$S_1 = \int_{-2}^{-1}\begin{pmatrix}0 - (x^3 + 1)\end{pmatrix}dx = \begin{bmatrix} \dfrac{-x^4... | Hint: If you have $\begin{bmatrix} -\frac{x^4}{4}-x \end{bmatrix}_a^b$, then you have to calculate $-\frac{b^4}{4}-b-\left(-\frac{a^4}{4}-a \right)=-\frac{b^4}{4}-b+\frac{a^4}{4}+a$
You have to substract the whole term, if you insert the lower bound. And both signs change, if you remove the brackets.
Additionally you h... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding Limit Using Taylor Polynomial
Find the limit $$\displaystyle{\lim_{x\to4}}\frac{\left(1-\cos \left( x-4 \right)\right)^4\,\ln \left( x-3 \right)}{\left(e^{\left(x-4\right)^2}-1\right)^2\,\sin ^5\left(\pi \,x\right)}$$
Using Taylor Polynomial and Peano reminder
So first thing would be to move $\lim_{x\to 4}$... | Using Taylor for this simple limit is not a good idea. However if one does wish to use the Taylor series it can be done by a little simplification at first (similar to what I present below) and then using first term (i.e. upto $x^{1}$) of the Taylor series for $\sin x, \log(1 + x), e^{x}$ and for $\cos x$ you need to g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Triangular Array's Recursive Formula Breakdown I have the following polynomials:
$$1$$
$$z-1$$
$$z^2-2z+3$$
$$z^3-3z^2+9z-15$$
$$z^4-4z^3+18z^2-60z+93$$
$$z^5-5z^4+30z^3-150z^2+465z-725$$
$$...$$
They are generated both recursively and explicitly. The recursive formula is
$$p_n(z)=z^n-\sum_{k=0}^{n-1}\binom{n}{k}\fra... | $$p_n(z)=z^n-\sum_{k=0}^{n-1}\binom{n}{k}\frac{1+(-1)^{n-k}+2(n-k)(-1)^{n-k-1}}{2}p_k(z)$$
$T(n, k) = [z^{n-k}]p_n(z)$ and you're particularly interested in $T(n, n) = [z^0]p_n(z)$
$$T(n, n) = [n = 0] - \sum_{k=0}^{n-1}\binom{n}{k}\frac{1+(-1)^{n-k}+2(n-k)(-1)^{n-k-1}}{2}T(k, k)$$
*
*$T(0, 0) = 1$ is a special case.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Perpendicular vectors in 3d Suppose a vector $v$ in $\mathbb{R}^3 $
How can I find two arbitrary unit vectors $u$ and $u^*$, that are perpendicular to each other and $v$ ?
There are infinitely many solutions, but I cannot hand pick them. I need some function $Q(v)$ = $(u, u^*)$ that deterministically finds a solutio... | $$u^\star = a \times u$$
Pick an arbitrary vector $a$ which is not parallel to $u$ and do a cross product. The result is perpendicular to both vectors. You can use a fixed vector such as $a=\hat{x}$, $a=\hat{y}$ or $a=\hat{z}$ by selecting the least parallel (lowest $a\cdot u$ value).
Alternatively pick any point is sp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the length of chord $BC$.
On a semicircle with diameter $AD$. Chord $BC$ is parallel to the diameter.Further each of the chords $AB$ and $CD$ has length of $2$ cm while $AD$ has the length $8$ cm.Find the length of $BC$.
$a.)7.5\quad cm\\
\color{green}{b.)7\quad cm}\\
c.)7.75\quad cm\\
d.)\text{cannot be determi... | Drop altitudes from $B$ and $C$ to $AD$, and call them $X, Y$ respectively.
Then, by symmetry, $AX = DY = 4 - \frac{BC}{2}$.
Furthermore $OX = \frac{BC}{2}$.
Also, $OB = 4$ since $AD = 8$.
Now use pythagorean theorem twice:
$AX^2 + BX^2 = AB^2$
$OX^2 + BX^2 = OB^2$
$OX^2 - AX^2 = OB^2 - AB^2$
$\frac{1}{4}BC^2 - (4 - \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Proving $\sqrt{2}(a+b+c) \geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}$ I've been going through some of my notes when I found the following inequality for $a,b,c>0$ and $abc=1$:
$$
\begin{equation*}
\sqrt{2}(a+b+c) \geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}
\end{equation*}
$$
This was what I attempted, but that ... | Hint
$$\sqrt{2}x-\sqrt{x^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{x},x>0$$
it is easy to prove by derivative.
so
$$\sqrt{2}a-\sqrt{a^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{a}\tag{1}$$
$$\sqrt{2}b-\sqrt{b^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{b}\tag{2}$$
$$\sqrt{2}c-\sqrt{c^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{c}\tag{3}$$
$(1)+(2)+(3)$
$$\sqrt{2}(a+b+c)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
A set of numbers Problem: Let $E(x)$ be the number defined by the following expression
\begin{equation*}
E(x)=\sqrt[3]\frac{x^3-3x+(x^2-1)\sqrt{x^2-4}}{2}+\sqrt[3]\frac{x^3-3x-(x^2-1)\sqrt{x^2-4}}{2}
\end{equation*}
where $x$ is a real number and
$\sqrt[3]{Z}$ denotes the real cubic root of the real number $Z$.
Det... | Let: $t=E(n)$ for some $n\gt 1$. So: $$t=\sqrt[3]{\frac {n^3-3n+(n^2-1)\sqrt{n^2-4}}{2}}+\sqrt[3]{\frac {n^3-3n-(n^2-1)\sqrt{n^2-4}}{2}}$$ From here: $$t^3=\frac {n^3-3n+(n^2-1)\sqrt{n^2-4}}{2}+\frac {n^3-3n-(n^2-1)\sqrt{n^2-4}}{2}+3t\sqrt[3]{\frac {(n^3-3n+(n^2-1)\sqrt{n^2-4})(n^3-3n-(n^2-1)\sqrt{n^2-4})}{4}}$$ But: $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What mistake have I made when trying to evaluate the limit $\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b}$? Suppose $a$ and $b$ are positive constants.
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = ?$$
What I did first:
I rearranged $\sqrt{n+a} \sqrt{n+b} = n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b... | Another approach:
$$\sqrt{a+n}\sqrt{b+n}=\sqrt{ab+(a+b)n+n^2}$$ and
$$\begin{align}n-\sqrt{n+a}\sqrt{n+b}&=\frac{n^2-(n+a)(n+b)}{n+\sqrt{n^2+(a+b)n+ab}}\\
&=\frac{-(a+b)n-ab}{n+\sqrt{n^2+(a+b)n+ab}}\\
&=\frac{-(a+b)-\frac{ab}{n}}{1 + \sqrt{1+\frac{a+b}{n} + \frac{ab}{n^2}}}
\end{align}$$
The numerator converges to $-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 1
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What can the various ways of integrating $\int \frac { x ^2 }{ (x \sin (x) +\cos(x))^2} \, dx$ $$
\int \frac {x^2}{(x\sin(x) + \cos(x))^2} \, \mathrm{d}x
$$
Well I found a method for solving this sum in a book saying that : We can multiply and divide the expression by $x\cos(x)$ and then apply integration by parts.
... | $\bf{My\; Solution::}$ Let $$\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)^2}dx$$
We can write
$$\displaystyle (x\sin x+\cos x) = \sqrt{1+x^2}\left\{\frac{x}{\sqrt{1+x^2}}\cdot \sin x+\frac{1}{\sqrt{1+x^2}}\cdot \cos x\right\} = \sqrt{1+x^2}\cdot \cos \left(x-\phi \right)$$
So Here $$\displaystyle \cos x\ \phi =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determining the limit of this series $$ \sum_{n=0}^\infty \frac{(-2)^n + 2^{3n}}{3^n4^n} = $$
$$ \sum_{n=0}^\infty \frac{(-1)^n2^n}{3^n4^n} + \sum_{n=0}^\infty \left(\frac{2}{3}\right)^n = $$
$$ \sum_{n=0}^\infty (-1)^n\frac{1}{6^n} + \sum_{n=0}^\infty \left(\frac{2}{3}\right)^n = $$
The second sum is a geometric seri... | Both are geometric series:
$$
\sum_{n=0}^\infty \frac{(-2)^n}{3^n \cdot 4^n} = \sum_{n=0}^\infty \left(\frac{-2}{3\cdot 4}\right)^n = \sum_{n=0}^\infty \left(\frac{-1}{6}\right)^n
$$
and
$$
\sum_{n=0}^\infty \frac{2^{3n}}{3^n\cdot 4^n} =
\sum_{n=0}^\infty \frac{8^n}{3^n\cdot 4^n} =
\sum_{n=0}^\infty \left(\frac{8}{3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determinant of matrices without expanding Show that $$\begin{array}{|ccc|}
-2a & a + b & c + a \\
a + b & -2b & b + c \\
c + a & c + b & -2c
\end{array} = 4(a+b)(b+c)(c+a)\text{.}$$
I added the all rows but couldn't get it.
| We can deduce from the structure of the matrix that (a) the determinant will be a symmetric polynomial in $a,b,c$ with every monomial with non-zero coefficient having degree $3$, and (b) that the coefficient of $a^3$, $b^3$, and $c^3$ will be $0$. So
$$
\det= k\,(a^2 b+ a^2 c+ b^2 a+ b^2 c+ c^2 a+ c^2 b)+m\,abc
$$
fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Simple trigonometry equation The previous class we were doing trigonometry exercises. Before the class finished, our teacher wrote exercises on the table. I am stuck with the following one:
$$
\cos(2x) + 1 + 3\sin x = 0
$$
I have come up with this:
$$
1= \sin^2 x + \cos^2 x
$$
$$
\cos(2x) = \cos^2 x - \sin^2 x
$$
Whe... | Although the procedure is almost same as given but I will explain it to the answer $$\cos(2x)+1+3\sin x=0$$ $$2-2\sin^2 x+3\sin x=0$$
$$2\sin^2 x-3\sin x-2=0$$ $$\implies (\sin x-2)(2\sin x+1)=0$$ $$ \implies \sin x-2=0 \implies\sin x=2\quad \text{but}\quad -1\leq\sin x\leq 1 $$ Hence we have $$ 2\sin x+1=0 \implies\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How can I show this inequality: $-2 \le \cos \theta (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$
Show that $$-2 \le \cos \theta ~ (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$$ for all value of $\theta$.
Trial: I know that $0\le \sin^2 \theta \le1 $. So, I have $\sqrt3 \le \sqrt{\sin ^2 \theta +3} \le 2 $. After tha... | Use this well known inequality
$$-\dfrac{a^2+b^2}{2}\le ab\le\dfrac{a^2+b^2}{2},a,b\in R$$
so
$$-\dfrac{\cos^2{\theta}+4\sin^2{\theta}}{2}\le\cos{\theta}\cdot 2\sin{\theta}\le\dfrac{\cos^2{\theta}+4\sin^2{\theta}}{2}\tag{1}$$
$$-\dfrac{4\cos^2{\theta}+\sin^2{\theta}+3}{2}\le2\cos{\theta}\cdot\sqrt{\sin^2{\theta}+3}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Sum of binomial coefficients from $\binom{2m+1}{0}$ to $ \binom{2m+1}{m} $ Prove that: $\displaystyle 4^m = \binom{2m+1}{0}+\binom{2m+1}{1}+\binom{2m+1}{2}+\ldots + \binom{2m+1}{m} $
From the Binomial Theorem:
$\displaystyle (a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$
If $\displaystyle a = b = 2$, then we have
$\d... | Your sum is the first half of the binomial expansion of $(1+1)^{2m+1}$ -- and since $\binom{2m+1}n = \binom{2m+1}{2m+1-n}$, the sum is equal to the other half, so the sum is
$$\frac{(1+1)^{2m+1}}2 = \frac{2^{2m+1}}2 = 2^{2m} = 4^m $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to prove that this matrix is positive definite? Let $\mathbf{A}=\begin{pmatrix}a^2+b^2 & b^2 & b^2 & ... & b^2 \\ b^2 & a^2+b^2 & b^2 & ... & b^2\\ \vdots & b^2 & \ddots & & b^2 \\ b^2 & \dots & & & a^2+b^2 \end{pmatrix}$, where $a,b\ne 0$. How can I be sure that this matrix is positive definite?
Any help would ... | Since $\mathbf{A}=a^2 \mathbf{I} + b^2 \mathbf{J}$ where $\mathbf{I}$ is an identity matrix and $\mathbf{J}$ has all ones, $$z^T \mathbf{A} z = z^T(a^2 \mathbf{I} + b^2 \mathbf{J})z = a^2 z^T \mathbf{I} z + b^2 z^T \mathbf{J} z.$$ If $z = (z_1, \ldots, z_n) \neq (0, \ldots, 0)$ then $$z^T I z = z_1^2 + \ldots z_n^2 > 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Roots of the equation $x^2 + px + q = 0$ If $\tan A$ and $\tan B$ are the roots of the equation $x^2 + px + q = 0 $ , show that $$\sin^2(A+B) + p \sin(A+B)\cos(A+B) + q \cos^2(A+B) = q$$
I tried using the result that $\tan A + \tan B = -p $ and $(\tan A)(\tan B) = q$ and tried substituting in the original equation but... | As $\tan(A+B)=\cdots=\dfrac p{q-1}$
Method $\#1:$
$\implies(1-q)\sin(A+B)+p\cos(A+B)=0$
Multiplying both sides by $\sin(A+B),$
$(1-q)\sin^2(A+B)+p\cos(A+B)\sin(A+B)=0$
$\iff\sin^2(A+B)+p\cos(A+B)\sin(A+B)+q(\cos^2(A+B)-1)=0$
$\iff\sin^2(A+B)+p\cos(A+B)\sin(A+B)+q\cos^2(A+B)=q$
Method $\#2:$
Dividing the numerator & the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What is the principal cubic root of $-8$? According to my book it should be a real number, and according to WolframAlpha it should be $1+1.73i$
What is the correct answer?
| Consider $x^3 = - 8 $. It "seems" we can just say $x = -2$ because $(- 2)^3 = -8$ but $- 2$ is not the official Primary Solution!
Using De Moivre's theorem again:
$$x^3 = - 8$$
$$( r \operatorname{cis}(\theta) )^3 = -8$$
$$r^3 \operatorname{cis}(3\theta) = 8\operatorname{cis}( 180^\circ + 360^\circ n)$$
$$r^3 = 8 \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
integral $\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt$ I want to compute this integral
$$\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt$$
where $0<b \leq a$.
I have this results
$$I_1=-\frac{ab}{2\pi}\int_0^\pi \frac{\cos(2t)}{a^2sin^2(t)+b^2cos^2(t)}dt=\frac{a}{a+b}-\frac{1}{2}$$
But I don't know ... | One trick to facilitate analysis is to write
$$\sin^2x=\frac{1-\cos 2x}{2}$$
and
$$\cos^2x=\frac{1+\cos 2x}{2}$$
Thus,
$$a^2\sin^2(t)+b^2\cos^2(t)=\frac{b^2+a^2}{2}+\frac{b^2-a^2}{2}\cos 2t$$
For the integral of interest, we can write
$$\begin{align}
I_1&=\int_0^{\pi} \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt\\\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Showing $ \int_0^{2 \pi } \frac{dt}{a^2 \cos^2 t + b^2 \sin^2 t} = \frac{2 \pi}{ab}$ The question:
Let $\gamma$ be a contour such that $0 \in I(\gamma),$ where $I$ is the interior of the contour. Show that
$$\int_\gamma z^n \, \text{d}z = \begin{cases} 2\pi i & \text{if } n = -1 \\ 0 & \text{otherwise} \end{cases}$$
... | Let us consider the ellipsis in the complex plane $z=a\cos t+ib\sin t$, $t \in [0,2\pi]$, from which $dz=(-a\sin t+ib\cos t)dt$, then
\begin{align}
&2\pi i=\int_\gamma z^{-1}dz=\int_0^{2\pi}\frac{-a\sin t+ib\cos t}{a\cos t+ib\sin t}dt=\\
&\qquad=\int_0^{2\pi}\frac{(b^2-a^2)\sin t \cos t+iab}{a^2\cos^2 t+b^2\sin^2 t}dt=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer... | Suppose $\sqrt2$ is arational, then there exist $p,q$ two natural numbers such that $$\color{Red}{p^2=q^2+q^2.}$$ Then by the parametric solution of Pythagoras Equation there exist two natural numbers $a,b$ such that $a\gt b\ge 1$ and $p=a^2+b^2,$ $$\color{Green}{q=a^2-b^2=2ab.}$$
Now, if $r=a+b$ and $s=a,$ then $$\col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 6
} |
Prove that $f=(x+i)^{10}+(x-i)^{10}$ have all real roots We have $f=(x+i)^{10}+(x-i)^{10}$ and we need to prove that $f$ have all the roots in $\mathbb{R}$.
Here is all my steps:
*
*Suppose that $z\in\mathbb{R}$ is a root of $f\Rightarrow (z+i)^{10}+(z-i)^{10}=0$
Therefore: $f(z)=\sum_{k=0}^{10}\left[\left(\dbino... | $x=a+bi$
$$(x+i)^{10}+(x-i)^{10}=0 \Rightarrow (x+i)^{10}=-(x-i)^{10} \Rightarrow (\frac{x+i}{x-i})^{10}=-1 \\ \Rightarrow \left (\frac{a+(b+1)i}{a+(b-1)i}\right )^{10}=i^{10} \Rightarrow \frac{a+(b+1)i}{a+(b-1)i}=i \\ \Rightarrow \frac{(a+(b+1)i)(a-(b-1)i)}{(a+(b-1)i)(a-(b-1)i)}=i \Rightarrow \frac{a^2-a(b-1)i+a(b+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
integer solution of $(x-y)(x+y)xy=z^2$ By wolfram alpha, integer solution of $(x-y)(x+y)xy=z^2$ is $x=y=z=0$.
How to show that there are not another solutions with $z \neq 0$.
Thanks.
| If $(x,y)=k$ then $k^4(x_1-y_1)(x_1+y_1)x_1y_1= z^2$ which implies an equation (we use the same notation) $(x-y)(x+y)xy=z^2$ with $(x,y)=1$; it follows that $(x+y,x-y)=1$ ($x+y=mX$ and $x-y=mY$ give clearly a contradiction). Hence $x-y, x+y, x, y$ are four coprime integers and each of them must be a square, $x, y$ be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Tangent numbers are divisible by $2^{n}$ Let us consider a $$\tan(z) = \sum_{n=1}^{\infty}{T_{2n-1} \cdot \frac{z^{2n-1}}{(2n-1)!}}$$.
So, it can be shown that $$T_{2n+1}=\frac{(-1)^{n} 4^{n+1}(4^{n+1}-1) B_{2n+2}}{2n+2} $$ where $B_{2n+2}$ is the $2n+2$ th Bernoulli number.
How to prove that $T_{2n+1}$ is divisible by... | Take a look at the first several derivatives of $\tan(z)$:
$$
\begin{align}
\tan(z)&=\tan(z)\\
\sec^2(z)&=\tan(z)^2+1\\
2\sec^2(z)\tan(z)&=2\tan^3(z)+2\tan(z)\\
6\tan^2(z)\sec^2(z)+2\sec^2(z)&=6\tan^4(z)+8\tan^2(z)+2\\
24\tan^3(z)\sec^2(z)+16\tan(z)\sec^2(z)&=24\tan^5(z)+40\tan^3(z)+16\tan(z)\\
\end{align}
$$
This sug... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Finding streamlines from complex potential. I'm currently studying a Fluid Dynamics module and mock exam question has me completely stumped, I have been given the complex potential and shown it to be in the form given, however when trying to separate the complex potential into velocity potential and stream function I c... | You'll have to accept or justify the definition that
$$(1) \quad \ln(a+b \cdot i)={{\ln(a^2+b^2)} \over 2}+\left({{\pi \cdot sign(b)} \over 2}-\tan^{-1}{ a \over b} \right) \cdot i$$
$$W=C \cdot \ln(z^2+a^2)$$
Where $C={m \over {2 \cdot \pi}}$
The stream function is given by the imaginary part of the complex potential.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent
First i subbed numbers in
$$\lim_{n \to \infty} \frac{(-1)^n}{1+\sqrt{n}} = \frac{-1}{1+\sqrt{1}} + \frac{1}{1+\sqrt{2}} - \frac{-1}{1+\... | There are so many mistakes (and so severe) in your "solution" that it is a superhuman task to discuss them. Your series is convergent, Leibniz's test solves the problem in a second. The series is not absolutely convergent by an application of the limit comparison test (compare your series with $\sum \frac 1 {\sqrt n}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x^2-y^2=196$, can we find the value of $x^2+y^2$? $x$ and $y$ are positive integers.
If $x^2-y^2=196$, can we know what the value of $x^2+y^2$ is?
Can anyone explain this to me? Thanks in advance.
| It is possible to find out $x^2+y^2$, as $196$, which will be $(x-y)(x+y)$, can be factorised as five possible integer products, $14\times 14$, $7\times28$, $4\times 49$, $2\times 98$, $1\times 196$.
Of these only one product leads to both positive and integer solutions for $x$ and $y$. Can you work out which of the p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this inequality, with the hypothesis more complicated than the conclusion? Given $x,y,z \in \mathbb{R}$ and $x,y,z>2,$ I want to show that if,
$$\frac{1}{x^2-4}+\frac{1}{y^2-4}+\frac{1}{z^2-4} = \frac{1}{7}$$
then,
$$\frac{1}{x+2} + \frac{1}{y+2} + \frac{1}{z+2} \leq \frac{3}{7}.$$
I follow the solution h... | Here is an adaptation of hyperbolictangent's answer in your given link.
Let $S:= \frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}$. Note that
\begin{align*}
3-4S &= \frac{x^2-4}{x^2-4} + \frac{y^2-4}{y^2-4} + \frac{z^2-4}{z^2-4} - 4 \left(\frac{x-2}{x^2-4}+\frac{y-2}{y^2-4}+\frac{z-2}{z^2-4}\right)\\
&= \frac{(x-2)^2}{x^2-4}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Why is $\sqrt{-i} \neq i\sqrt{i}$? I wanted to figure out the square root of $-i$.
Since $\sqrt{-x} = i\sqrt{x}$,
$\sqrt{-i}$ should equal $i\sqrt {i}$, however, WolframAlpha said it was false.
However, if I do say that $\sqrt{-i} = i\sqrt{i}$, I can replace $\sqrt{i}$ with $\dfrac {1+i} {\sqrt{2}}$, leaving me w... | You are running up against a complication that arises with roots of complex numbers: they are not single-valued, but multi-valued. If you express the question "what is $ \ \sqrt{-i} \ $ ?" as "what are the roots of the equation $ \ x^2 \ - \ (-i) \ = \ 0 \ ? $ " , there are two solutions. DeMoivre's Theorem tells us t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Please help to find term's coefficient in the following example I trying find the number of all solutions in the following:
$ x_1 + x_2 + x_3 + x_4 + x_5 = 24 $
where:
2 of variables are natural odd numbers
3 of variables are natural even numbers
none of variables are equal to $0$ or $1$
(all the variables are $>= 2$)... | Let the numbers be $2a + 1, 2b + 1, 2c, 2d, 2e$, where $a, b, c, d, e$ are natural. Note that this makes the odd numbers greater than or equal to $3$, and the even numbers greater than or equal to $2$.
Then you have $2a + 1 + 2b + 1 + 2c + 2d + 2e = 24 \rightarrow a + b + c + d + e = 11$. You can solve this with a stan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is I got this task two days ago, quite difficult for me, since I have not done applications of Vieta's formulas and Bezout's Theorem for a while. If can someone solve this and add exactly how I am supposed to use these two theorem's on this task, I would ... | There is some quadratic $x^2 + cx + d$ such that:
$$(x^2 + 4x + 6)(x^2 + cx + d) = x^4 + ax^2 + b$$
Multiply it out:
$$x^4 + (c + 4)x^3 + (d + 4c + 6)x^2 + (6c + 4d)x + 6d = x^4 + ax^2 + b$$
Equate coefficients:
$$\begin{cases}
c + 4 = 0 \\
d + 4c + 6 = a \\
4d + 6c = 0 \\
6d = b
\end{cases}$$
Use the first equation to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Alternative way to count the number of solutions to the equation $x^2 + y^2 = -1$ over $\Bbb Z /p$
$x^2 + y^2 = -1$ is a weird equation because it has no solutions over $\Bbb R$. I want to count the number of solutions it has over $\Bbb Z / p$ where $p$ is prime.
If $p = 2$ then it has $p$ solutions. This is to do wi... | Given a finite field $F$, let $C = F[i]/(i^2+1)$. Define the mapping $N:C \to F$ by $N(x+iy) = x^2 + y^2$ and note that it's a multiplicative homomorphism from $C^*$ to $F^*$.
Now let $X = \{x \in C \mid N(x) = 1\}$ and $Y = \{x \in C \mid N(x) = -1\}$. The aim is to determine $|Y|$.
Note that there exists an $e \in Y$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Michael Spivak's Calculus - Chapter 1, Problem 19 Problem 19. The fact that ${a^2}\ge{0}$ for all the numbers a, elementary as it may seem, is nevertheless the fundamental idea upon which most important inequalities are ultimately based. The great-grandaddy of all inequalities is the Schwarz inequality:
${x_{1}y_{1}+x_... | We have
\begin{align}2x_iy_i &\le \frac{x_i^2 \sqrt{y_1^2 + y_2^2}}{\sqrt{x_1^2 + x_2^2}} + \frac{y_i^2 \sqrt{x_1^2 + x_2^2}}{\sqrt{y_1^2 + y_2^2}}\\
&\le \frac{x_i^2 (y_1^2 + y_2^2) + y_i^2 (x_1^2 + x_2^2)}{\sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}}\\
&\le \frac{x_i^2 (y_1^2 + y_2^2) + y_i^2 (x_1^2 + x_2^2)}{\sqrt{x_1^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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