Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$? I'm trying to find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$ in $\mathbb{Z}[\sqrt[3]{2}]$. I know it is a unit, so there is an inverse, but I feel like I may be doing too much work in the wrong direction. Here's what I have so far:
Let $\alpha = 5+4\sqrt[3]{2}+3\sqr... | $$
\left(
\begin{array}{ccc|c}
5 & 6 & 8 & 1 \\
4 & 5 & 6 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
1 & 1 & 2 & 1 \\
4 & 5 & 6 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
1 & 1 & 2 & 1 \\
1 & 1 & 1 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Diophantine Equation on squares and cubes Find all integer solutions to: $(a^2+b)(b^2+a)=(a-b)^3$
I've found some of the trivial cases, just finding difficulty proving the existence (or not existence) of others. Perhaps taking $mod5$ or something?
| (EDITED)
$$(a^2+b)(b^2+a)-(a-b)^3 = b(a^2 b+3 a^2-3 a b+2 b^2+a)$$
The factor $b$ means $a=arbitrary, b=0$ are solutions.
The curve $a^2 b+3 a^2-3 a b+2 b^2+a=0$ has genus $0$ and rational parametrization
$$ a = {\frac {-2{s}^{2}}{ \left( 2\,s+1 \right) \left( s+1 \right) }}, b = -{
\frac { \left( 4\,s+1 \right) s}{(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $a + b$ is prime and $a < -1$, show $a^2 + ab + b < 2$ I want to show $a^2 + ab + b < 2$ given $a, b \in \mathbb{Z}$, $a + b$ is prime and finally $a < -1$.
I've reduced the bounds a little. As $a + b$ is prime, $a + b \ge 2$. Since $a \in \mathbb{Z}$ the first integer less than $-1$ is $-2$, and thus $a \le -2$.... | Let $a+b = p$. $p$ prime so $p\geq 2$. Then $a^2+ab+b = (p-1)a+p$. What can you conclude now?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluation of a certain polynomial on a matrix. Given $A=\left[\begin{matrix}\dfrac{-1+i\sqrt{3}}{2i}&\dfrac{-1-i\sqrt{3}}{2i}\\\dfrac{1+i\sqrt{3}}{2i}&\dfrac{1-i\sqrt{3}}{2i}\end{matrix}\right]$ where $i=\sqrt{-1}$.
Also Given$f(x)=x^2+2.$
Find $f(A)$
My approach:
A can be written as
$A=i\left[\begin{matrix}w&w^2\\w^... | Your expression for $A$ is not correct. You have
$$
A=-i\begin{bmatrix} w&w^2\\ -w^2&-w\end{bmatrix}=-iw\begin{bmatrix} 1&w\\-w&-1\end{bmatrix}
$$
Then
$$
A^2+2I=-w^2\begin{bmatrix} 1-w^2&0\\0&1-w^2\end{bmatrix}+2I=\begin{bmatrix} w-w^2+2&0\\0&w-w^2+2\end{bmatrix}=(2+i\sqrt 3)I,
$$
since
$$
w-w^2+2=-\frac12+\frac i2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if the matrix $\begin{bmatrix}a & b\\b & c\end{bmatrix}$ is nonnegative definite, then it has a factorization $LL^{T}$ Prove that if the matrix $\begin{bmatrix}a & b\\b & c\end{bmatrix}$ is nonnegative definite, then it has a factorization $LL^{T}$ in which $L$ is lower triangular.
I am using the following ... | The (semi)definedness of the matrix tells you that $a\ge0$, $c\ge0$ and $ac-b^2\ge0$.
In particular, if $a=0$, then also $b=0$, so the matrix is
$$
\begin{bmatrix} 0 & 0 \\ 0 & c\end{bmatrix}=
\begin{bmatrix} 0 & 0 \\ 0 & \sqrt{c}\end{bmatrix}
\begin{bmatrix} 0 & 0 \\ 0 & \sqrt{c}\end{bmatrix}^T
$$
Suppose $a>0$. We ca... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Alternate solutions to similar integrals? So, when I started this problem I couldn't see any reasonable trig substitutions that would make it simpler, and I don't know of a better method, so I attempted to solve it using repeated use of integration by parts. Is there a much easier, better, alternate solution to integra... | $\sin^2 x = \frac 12 (1 - \cos 2x)\\
\sin^4 x = \frac 14 (1 - \cos 2x)^2 = \frac 14 (1 - 2\cos 2x + \cos^2 2x)\\
\cos^2 2x = \frac 12 (1+\cos 4x)\\
\sin^4 x = \frac 14 (1 - \cos 2x)^2 = \frac 14 (1 - 2\cos 2x + \frac 12(1+\cos 4x))$
and simplify.
This is a nifty trick you can use if you know Euler's identity.
$e^{ix} =... | {
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Flux of a Sphere without using Divergence Theorem I have solved the problem below. my answer is $2\pi a^3$ but in the text book the answer is $4\pi a^3$. Please I would appreciate if someone can tell me where i made a mistake.
PROBLEM
Find the Flux of F= $xi+yj+zk$ outward across the sphere $x^2+y^2+z^2=a^2$
SOLUTION
... | You are considering the flux over an hemisphere, indeed we have
$$2\int_0^{2\pi} \int_0^{a} \frac{a^2}{\sqrt{a^2-r^2}}r\mathrm dr\mathrm d\theta = 4\pi\left[-a^2(a^2-r^2)^\frac12\right]_0^a =4\pi a^3$$
Note that $4\pi a^3$ is the correct result since the $|F|=a$ and $F$ is parallel to $n$ thus the flux is $4\pi a^2\cd... | {
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"url": "https://math.stackexchange.com/questions/2743221",
"timestamp": "2023-03-29T00:00:00",
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How to factor $a^{3} + b^{3} + c^{3} - 3abc$ into a product of polynomials The question is in the title.
This question is from "Algebra" by Gelfand.
My initial thought is that if $a$, $b$ and $c$ are $1$ or $-1$, then the polynomial evaluates to $0.$ So, maybe two of the factors will be $(a + b + c - 3)$ and $(a + b + ... | I suggests that you use $(a+b)^3=a^3+b^3+3ab(a+b)\Rightarrow a^3+b^3=(a+b)^3-3ab(a+b)$ instead, you will need to use it twice like this:
$a^3+b^3+c^3-3abc$
$=(a+b)^3+c^3-3ab(a+b)-3abc$
$=(a+b+c)^3-(3c(a+b)^2+3(a+b)c^2)-3ab(a+b+c)$
$=(a+b+c)^3-3c(a+b)(a+b+c)-3ab(a+b+c)$
$=(a+b+c)^3-(a+b+c)(3ab+3bc+3ca)$
$=(a+b+c)(a^2+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Lines tangent to graph of equation So I have the following problem and can't seem to arrive at the official answer:
Consider the curve whose equation is $$(2-x)y^2 = x^3$$
Obtain the equations of the lines tangent to the graph of the curve at the points where $x=3/2$.
My attempt
My first thought is that you can iso... | Note that fo $x\ne 2$
$$(2-x)y^2 = x^3\implies y^2=\frac{x^3}{2-x}\implies 2ydy=-\frac{2(x-3)x^2}{(2-x)^2}dx \implies\frac{dy}{dx}=-\frac{2(x-3)x^2}{2y(2-x)^2}$$
and for $x=\frac32$
$$y^2=\frac{x^3}{2-x}=\frac{\frac{27}{8}}{\frac{1}{2}}=\frac{27}{4}\implies y={\pm} \frac{3\sqrt{3}}2 $$
and we obtain
*
*$(x,y)=\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Computing determinant of a specific matrix: This question might seem very easy to some. But I am having a very tough time solving it.
$$
A = \begin{pmatrix}
1 + x^2 - y^2 - z^2& 2(xy + z) & 2(zx-y) \\
2(xy - z) & 1 + y^2 - z^2 - x^2 & 2(yz + x) \\
2(zx + y) & 2(yz - x) & 1 + z^2 - x^2 - y^2
\end{pmatrix}
$$
Then $ ... | Let $B = \begin{bmatrix}0 & z & -y\\ -z & 0 & x \\ y & -x & 0\end{bmatrix}$ be half of the anti-symmetric part of $A$. Since its square
$$B^2 = \begin{bmatrix}-y^2-z^2 & xy & xz\\ xy & -x^2-z^2 & yz\\xz & yz & - x^2-y^2\end{bmatrix}$$
has same off-diagonal elements as half of the symmetry part of $A$, we obtain followi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Generalizing $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}$ If $a+b+c=0$ as discussed in this, this, and this post, then,
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}\tag1 $$
$$ \frac{a^3+b^3+c^3}{3} \times \frac{a^4+b^4+c^4}{2} = \frac{a^7+b^7+c^7}{7}\tag... | When denote
$$
s_k = \dfrac{a^k+b^k+c^k}{k},
$$
then some expressions for $s_{17}$ and $s_{19}$:
$$
s_5 \left(6s_{12}-7s_5s_7-\frac{1}{2}{s_3^4}\right) = s_{17},
$$
$$
s_7 \left(6s_{12}-5s_5s_7+\frac{3}{2}{s_3^4}\right) = s_{19}.
$$
Or
$$
s_5 \left(s_{12}+3s_6^2+4s_4s_8\right) = s_{17},
$$
$$
s_7 \left(3s_{12}+9s_6^2-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find $U+V$ given $U\alpha^{-5} + V\beta^{-5} = 22$
Find $U+V$ given $U\alpha^{-5} + V\beta^{-5} = 22$, when $\alpha, \beta$ are roots
of $x^2-x-1$ and $U,V$ are integers.
For simplicity I only show the U term:
$$U+V = U\alpha^{-5}\alpha^{5}+ ... \\ = U \alpha^{-5}(\alpha^4 \alpha)\\
= U \alpha^{-5}((\alpha+1)^2 \... | Start with $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$ since these satisfy $x^2 = x+1$. Clearly, both $\alpha$ and $\beta$ are non-zero.
Now, $\alpha^{-1} = \alpha - 1$. Square both sides : $\alpha^{-2} = \alpha^2 - 2\alpha + 1 = 2 - \alpha$. Squaring again, $\alpha^{-4} = (2-\alpha)^2 =$ $ 4 - 4 \alpha + \alph... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Figure out all positive integers n with consecutive + integers a,b,c. When $2018^n$ = $a^4$ + $b^4$ + $({b^2+c^2})^2$,
then what is the possible positive integers n be?
| Since we have consecutive integers $a,b,c$, take $a=b-1$ and $c=b+1$. Then, we have $2^n 547^n=(b-1)^4+b^4+(b^2+(b+1)^2)^2=(b-1)^4+2b^4+2b^2(b+1)^2+(b+1)^4=(b-1)^4+(b+1)^4+4b^4+4b^3+2b^2$. If $b=2k$, then expanding the braces, we get the RHS is divisible by $2$, not $4$. This means $n$ is at most 1.
If $b=2k+1$, then ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I derive the formula for the reciprocal of a hypotenuse? Given $a^2 + b^2 = c^2$,
Why is it that the equation below yields the reciprocal of the hypotenuse c, ($\frac{1}c$)?
$\sqrt{(\frac{a}{a^2+b^2})^2 + (\frac{b}{a^2+b^2})^2}$
Worked example:
$3^2 + 4^2 = c^2$
$c = 5$
$\sqrt{(\frac{3}{3^2+4^2})^2 + (\fra... | Observe that, for $ab\ne0$,
$$
\left(\frac{a}{a^2+b^2}\right)^2 + \left(\frac{b}{a^2+b^2}\right)^2=\frac{a^2}{(a^2+b^2)^2}+\frac{b^2}{(a^2+b^2)^2}=\frac{a^2+b^2}{(a^2+b^2)^2}=\frac{1}{a^2+b^2}.
$$ Hope you can take it from here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limits of a product $N$ is a positive integer.
I want to calculate this limit, but i couldn't get anywhere when i tried.
$$P[n]=\left(1+\frac {1}{n^2}\right)\left(1+\frac {2}{n^2}\right)\cdots\left(1+\frac {n-1}{n^2}\right)$$ as $n\to \infty$
I tried to apply $\ln()$ at both sides to transform into a sum etc..
tried to... | First realize that, $$\prod_{i=1}^{n} 1+ \frac i{n^2} = \prod_{i=1}^n \frac {n^2+i}{n^2} = \prod_{i=n^2+1}^{n^2+n} \frac {i}{n^2} $$
Then,
$$\prod_{i=n^2+1}^{n^2+n} \frac {i}{n^2} = \exp \left ( \sum_{i=n^2+1}^{n^2+n} \log(i) - 2n\log(n) \right )$$
Then we use a Riemann inequality since $\log$ is an increasing fun... | {
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How prove $\cos(\frac{2\pi}{17}) + \cos(\frac{18\pi}{17})+\cos(\frac{26\pi}{17})+\cos(\frac{30\pi}{17}) = \frac{\sqrt{17}-1}{4}$
Prove that $\cos(\frac{2\pi}{17}) + \cos(\frac{18\pi}{17})+\cos(\frac{26\pi}{17})+\cos(\frac{30\pi}{17}) = \frac{\sqrt{17}-1}{4}$
Regards that value of $\cos(2\pi/17)$, I can't find the eas... | Let $p$ be an odd prime number. Then
$$g_p=\sum_{k=0}^{p-1}\exp(2\pi i k^2/p)$$
is a quadratic Gauss sum. Gauss proved that $g_p=\sqrt p$ or
$i\sqrt p$ according to whether $p\equiv1$ or $p\equiv3\pmod 4$.
It is quite easy to prove this up to sign, but hard to prove the sign.
So $g_{17}=\sqrt{17}$. Therefore
\begin{ali... | {
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Simplification of $\frac{\sqrt{n}}{\sqrt{4n+\sqrt{n}} + 2\sqrt{n}}$ Can somebody explain how pass to this : $\sqrt{n} / \left(\sqrt{4n+\sqrt{n}} + 2*\sqrt{n}\right)$
to this :
$1 \big/ \left(2 + 2\sqrt{1+ \frac{\sqrt{n}}{4n}}\right)$
Thanks in advance
| \begin{align}
\frac{\sqrt{n} }{ \sqrt{4n+\sqrt{n}} + 2\sqrt{n}}
=&\;\frac{\frac{\sqrt{n}}{\sqrt{n}} }{ \frac{\sqrt{4n+\sqrt{n}}}{\sqrt{n}} + 2\frac{\sqrt{n}}{\sqrt{n}}}\\[2ex]
=&\;\frac{1 }{ \sqrt{\frac{4n+\sqrt{n}}{n}} + 2}\\[2ex]
=&\;\frac{1 }{ \sqrt{4+\frac{\sqrt{n}}{n}} + 2}\\[2ex]
=&\;\frac{1 }{ \sqrt{4\left(1+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757818",
"timestamp": "2023-03-29T00:00:00",
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Removing a discontinuity? How would you "remove the discontinuity" of $f$ ? In other words, how would you define $f(4)$ in order to make $f$ continuous at $x=4$?
$$f(x) = \dfrac{x^2-x-12}{x-4}$$
| if $ x \neq 4 $
$f(x)= \dfrac{x^2-x-12}{x-4}= \dfrac{(x-4)(x+3)}{x-4}=\dfrac{1(x+3)}{1}= x+ 3$
so we have
f(x)=x+3
(for $ x \neq 4$)
$\lim_{x\to4}f(x) = \lim_{x\to4}(x+3) $= 7
we want to make f continuous for this reason
the ammount of function at 4 should be equall to $$\lim_{x\to4}f(x)=7$$
so we define a functio... | {
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"url": "https://math.stackexchange.com/questions/2763300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof check modulo arithmetic what is wrong with this proof,
so I am supposed to show $7x^2-15y^2=1$ has no integer solutions...
so since $7x^2=1+3(5y^2)$ so $7x^2\equiv 1 \pmod{3}$
hence in mod $3$, $x\equiv 0, 1$ or $2$ so $x^2\equiv 0, 1$ and $7x^2\equiv0$ or $7$
and since $7$ is congruent to $1 \bmod3$ I end up wi... | There are also no integer solutions to $$ 13 x^2 - 17 y^2 = 1. $$
We know that there are rational solutions, therefore $p$-adic, as we can solve
$$ 13 x^2 - 17 y^2 = 25 $$
in integers.
$$ 13 \cdot 7^2 - 17 \cdot 6^2 = 637 - 612 = 25 $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Number of terms in a Polynomial Expansion For a binomial $(a + b)^n$, the number of terms is n + 1.
For a trinomial $(a + b + c)^n$, the number of terms is $\frac{(n+1)(n+2)}{(2)}$.
For a multinomial $(a + b + c +d)^n$, the number of terms is $\frac{(n+1)(n+2)(n+3)}{(6)}$.
I'm guessing that for $(a + b + c + d + e)^n$... | for (a+b+c+d+e)^n it would be [(n+4)(n+3)(n+2)(n+1)]/24
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Whenever $a+b+c=1$, $\frac{bc+a+1}{a^2+1} + \frac{ac+b+1}{b^2+1} + \frac{ab+c+1}{c^2+1} \le \frac{39}{10}$ Prove that for a + b + c =1 and a,b,c are positive real numbers,
then
$$\frac{bc+a+1}{a^2+1} + \frac{ac+b+1}{b^2+1} + \frac{ab+c+1}{c^2+1} \le \frac{39}{10}$$
My try:
if one term is proven to be $\le \frac{13}{1... | This is wrong. If $a=1$ and $b=c=0$ then the left-hand side is $3$, which is less than $\frac{39}{10}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Maximum power of $2$ which divides $3^{1024}-1$ What is the maximum power of $2$ which completely divides $3^{1024}-1$?
I proceeded thus:
$\phi(2^n)=2^{n-1}$ for all $n\ge1$
$$3^{1024}=3^{2^{10}}\equiv1\pmod {2^{11}}$$
$$3^{1024}-1\equiv0\pmod {2^{11}}$$
Since $\phi(2^{11})=2^{10}$. So, maximum power of $2$ must be $1... | As requested in the comments:
We start by factoring the polynomial $x^{1024}-1$. To do that, we note that $1$ is a root, as is every $2^k-$root of $1$ for $k=0,\cdots 10$. For $k>0$ such a root of unity is also a $2^{k-1}-$st root of $-1$ so our polynomial is divisible by $$(x-1)\times \prod_{k=0}^9(x^{2^k}+1)$$
Comp... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability of exactly three of a kind in a roll of 5 dice The answer according to edx course HarvardX: FC1x Fat Chance: Probability from the Ground Up
is $$\frac{6*5*5* \left(^5_2\right)}{6^5} =\frac{1500}{6^5}$$
The remaining two dice can be same say this is valid favourable outcome 4,4,4,5,5.
But according to my r... | Assume the dice are distinguishable. Then there are $6^5$ possible outcomes since there are six possible outcomes for each of the five dice.
Three of a kind: There are $\binom{5}{3}$ ways for three of the five dice to display the same outcome and six possible outcomes those three dice could display. There are $\bino... | {
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"url": "https://math.stackexchange.com/questions/2772282",
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"answer_id": 0
} |
Sum of $4$ dice rolls greater than the product Let us roll a fair die $4$ independent times, and denote the outcomes as $X_1, X_2, X_3$ and $X_4$.
What is the probability of $X_1+X_2+ X_3+X_4 > X_1X_2 X_3X_4$?
My try:
I could get the answer for $2$ rolling case by enumerating possibilities, but couldn't get this larger... | A solution to the equation has at least one $1$.
When you roll all 2s, the equation is false.
$2 + 2 + 2 + 2 > 2 \times 2 \times 2 \times 2$
$8 > 16 \implies false$
Adding one to any of the numbers always maintains the falsehood of this equation, or the truth of the reverse. Assuming the opposite equation:
$x_1 + x_2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 2
} |
If I draw 4 cards from a deck of 52 cards, what is the probability that the cards are 2 of one suit and 2 of another? 4 cards are drawn from a stack of 52, what is the probability that 2 cards are of one suit and the other 2 cards are of a different suit?
I.e. 2 Diamonds and 2 Clubs or 2 Spades and 2 Diamonds etc..
I ... | In my opinion "2 cards are of one suit and the other 2 cards are of a different suit" is a bit ambiguous. Let us consider the following two possible interpretations.
2 different suits: two cards of one suit and two of another suit
The total number of hands of $4$ cards is $\binom{52}{4}.$
We have $\binom{4}{2}$ ways ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Logarithmic expression how to simplify $$ \log_{3}24 - 3\log_{3}5\times \log_{5}2$$
What I can get is:
$$ \log_3{24} - \log_3{5^3} \times \log_{5}2$$
Change of base rule to get it all in base 3:
$$ \log_5{2} = \frac{\log_3{2}}{\log_3{5}} $$
Now I have:
$$\log_3{24} - \frac{\log_3{5^3}\times \log_3{2}}{\log_3{5}}$$
How ... | Continuing from what you have,
\begin{align}
\log_3{24} - \frac{\log_3{5^3}\times \log_3{2}}{\log_3{5}} &= \log_3{24} - \frac{3\cdot\log_3{5}\cdot \log_3{2}}{\log_3{5}} \\
&=(\log_3{3} + \log_3{8}) - 3\cdot \log_3{2} \\
&= 1 + 3 \log_3 2 - 3 \log_3 2 \\
&= 1.
\end{align}
In questions like these, it's often a good idea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Expected number of die rolls - conditional probability The aim here is to roll a fair die until a 3 is rolled twice in a row, where the number of rolls that this takes (including the final two where threes are consecutive) is given by the random variable X. We are also told of the random variable Y, and this is the num... | In fact, you may compute the entire distribution of $X$. Let's denote $a_n = \mathbb{P}(X = n)$, where $n=1,2,...$ . We have $a_1 = 0$ and $a_2 = 1/6^2$.
Let also $X_1$ and $X_2$ be the first and the second rolls of the die.
Then, using the law of total probability, for $n>2$ we have
$$
a_n = \frac 56 \mathbb{P}(X = n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2774670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Binomial Inequality in which Binomial coefficient is in square root.
$$1\cdot \sqrt{C_{1}}+2\cdot \sqrt{C_{2}}+\cdots \cdots +100\cdot \sqrt{C_{100}}\leq \frac{1}{2}\cdot (2^{100}-1)+\frac{20301}{12}$$
where $\displaystyle C_{r}=\binom{n}{r}$
Try: Using Cauchy Schwarz Inequaity
$$\bigg(1^2+2^2+\cdots \cdots +100^2\bi... | By C-S,
as you have done,
$\begin{array}\\
\left(\sum_{k=1}^n k\sqrt{\binom{n}{k}}\right)^2
&\le \sum_{k=1}^n k^2\sum_{k=1}^n\binom{n}{k}\\
&=\dfrac{n(n+1)(2n+1)}{6}(2^n-1)\\
&\lt\dfrac{2(n+1)^3}{6}2^n\\
&=\dfrac{(n+1)^3}{3}2^n\\
\text{so}\\
\sum_{k=1}^n k\sqrt{\binom{n}{k}}
&\lt\dfrac{(n+1)^{3/2}}{\sqrt{3}}2^{n/2}\\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$2^{x}+ 5^{y}= 7^{z}$ Help me solve this integers exponents equation:
$$2^{x}+ 5^{y}= 7^{z}$$
We have: $z= \frac{i(2\,\pi\,n-i\,\log(2^{x}+ 5^{y}))}{\log(7)}$
I ask because I read that a slightly complication of the equation occuring in Fermat's last theorem can lead to an undecidable case and I wonder whether the give... | Only $x=y=z=1$, corresponding to $2+5=7$, can work. Most of this proof rehashes the comments, the real purpose here is to consolidate all the steps into one place.
1) $x$ is odd, or else $2^x \equiv 7^z \equiv 1 \bmod 3$ and $5^y$ can't be divisible by $3$.
2) $y$ is odd, or else $5^y \equiv 7^z \equiv 1 \bmod 3$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Complex Number Series If
$$
z + 1/z = -1
$$
where $z$ is a complex number then value of the sum from r = 1 to r = 99 of $$
( z^r + 1/z^r)^2
$$
is equal to
A) 198
B) 3
C) 99
D) 0
I tried by putting z = x + iy but that gives, |z| = 1
| HINT: observe that
$$
\left(z^r+\frac1{z^r}\right)^2=\left(z^{2}\right)^r+\left(\frac1{z^2}\right)^r+2
$$
then split the sum and using the geometric series
$$
\sum_{n=1}^Nq^n=\frac{q-q^{N+1}}{1-q}\;,\;\;\;\;\;q\in\Bbb C
$$
we can evaluate the first two sums:
\begin{align*}
\sum_{r=1}^{99}\left(z^r+\frac1{z^r}\right)^2
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
My attempt:
We have: $f'(x)=-\frac{(a-1)(x^2-1)}{(x^2+x+1)^2}\implies x = 1$ and $x = -1$ points of extrema.
then ... | The derivative of $f(x)$ is equal to $\dfrac{(a-1)(x^2-1)}{(x^2+x+1)^2}$ so the extremes are independent of a and taken at $x=\pm 1$. This extremes are $\dfrac{2\pm a}{3}$ It follows $0\le a\le 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
$\sum_{m=0}^{\infty}x^m\sum_{k=0}^{\infty}W_{m,k}f_k=\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\sum_{k=0}^{\infty}\left(\frac{n+x}{n(n+1)}\right)^k f_k$? How do we solve
$\sum_{m = 0}^{\infty} x^m \sum_{k = 0}^{\infty} W_{m, k} f_k = \sum_{n =
1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{n + x}{n (n
+ 1)... | $\sum_{m = 0}^{\infty} x^m \sum_{k = 0}^{\infty} W_{m, k} f_k = \sum_{n =
1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{n + x}{n (n
+ 1)} \right)^k f_k
$
I'll naively
expand the right side,
not worrying about convergence.
$\begin{array}\\
\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2784351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find $C\in\mathbb R$ such that $(2,3,5)$ be in $\text{Im}(F)$ Let $$\color{red}{\underbrace{F:\mathbb R^3\to\mathbb R^3\;|}_{\text{Added by me}}}\; F(x,y,z)=(x+Cz,\;x+y+2z,\;x+Cy)$$ a linear transformation. Find $C\!\!\!\!\!\color{red}{\underbrace{\in\mathbb R}_{\text{Added by me}}}\!\!\!\!\!\!$ such that $(2,3,5)$ be ... | $$(x+Cz, x+y+2z, x+Cy)=(2,3,5)$$
$$\implies x+Cz=2;~~~ x+y+2z=3;~~~~ x+Cy=5$$
$$\begin{bmatrix}1&0&C\\1&1&2\\1&C&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\3\\5\end{bmatrix}$$
applying row transformation, $R_2\leftarrow R_2-R_1$, $R_3\leftarrow R_3-R_1$
$$\begin{bmatrix}1&0&C\\0&1&2-C\\0&C&-C\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Suppose $a,b,c\geq 0$ and $ab+bc+ca=1$, prove that $a\sqrt{a^2+1}+b\sqrt{b^2+1}+c\sqrt{c^2+1}\geq 2$. Suppose $a,b,c\geq 0$ and $ab+bc+ca=1$, prove that
$$a\sqrt{a^2+1}+b\sqrt{b^2+1}+c\sqrt{c^2+1}\geq 2.$$
In my opinion, I do this problem by th following:
$$\sum_{cyc}a\sqrt{a^2+1}=\sum_{cyc}a\sqrt{(a+b)(c+a)},$$
but I... | By C-S, Schur and AM-GM we obtain:
$$\sum_{cyc}a\sqrt{a^2+1}=\sum_{cyc}a\sqrt{(a+b)(a+c)}\geq\sum_{cyc}a(a+\sqrt{bc})=$$
$$=\sum_{cyc}(a^2-\sqrt{a^3b}-\sqrt{a^3c}+a\sqrt{bc})+\sum_{cyc}(\sqrt{a^3b}+\sqrt{ab^3})\geq2\sum_{cyc}ab=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$. Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$.
First, observe that $\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!}=\frac{(2n)!}{(n!)^2}=\frac{1\cdot 2\cdot 3\cdot...\cdot 2n-1\cdot 2n}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 4\... | Another approach: the expression equals
$$\frac{2n(2n-1)\cdots (2n-(n-1))}{n!}=\frac{2n}{n}\frac{2n-1}{n-1}\frac{2n-2}{n-2}\cdots \frac{2n-(n-1)}{n-(n-1)}.$$
Since $2n-k \ge 2(n-k),$ $ k=0,\dots ,n-1,$ each fraction on the right is at least $2.$ Since there are $n$ fractions, we have the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
This function has two different integrals? $f(x)=∫\frac{1}{x^2}dx$
Integrating by u-substitution:
$u=x^2$
$du=2dx$
$\frac{1}{2}du = dx$
$∫\frac{1}{x^2}dx=$ $∫\frac{1}{u}\times\frac{1}{2}du$
$\frac{1}{2}$∫ $\frac{1}{u}du$
$=\frac{1}{2}ln u+c$
$=\frac{1}{2}ln x^2+c$
$=lnx+c$
Another way:
$∫\frac{1}{x^2}dx=∫x^{-2}dx $
... | It’s not $du=2dx$, it’s $du=2xdx$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Indefinite integral of $\sqrt{x^2-x}$ i was trying to compute the indefinite integral:
$$
\int\sqrt{x^2-x}dx
$$
but i got stuck: after a few (unsuccessful) attempts for some $u$-substitution, i tried integration by parts:
$$
\int\sqrt{x^2-x} \ dx=\int(x)'\sqrt{x^2-x} \ dx= \\ x\sqrt{x^2-x}-\int x(\sqrt{x^2-x})'dx= \\
=... | hint
$$\sqrt {x^2-x}=\frac {1}{2}\sqrt {(2x-1)^2-1} $$
then put $$2x-1=\cosh (t) $$
and use
$$\cosh^2(t)-1=\sinh^2(t)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Evaluating definite integral $\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$ Question: $$\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$$ is equals to
(a) $-\frac\pi6$
(b) $-\frac{\pi}{12}$
(c) $\frac\pi{12}$
(d) $\frac\pi6$
My attempt: Denoting given integral by $I$ and letting $z=e^{iθ}$ then giv... | Use periodicity to rewrite your integral
$$I=\int_{-\pi}^{\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$$
With the change of variable $t=\tan\frac\theta2$,
$$I=\int_{-\infty}^{+\infty}\frac{2\mathrm dt}{(1+t^2)\left(13-5\dfrac{2t}{1+t^2}\right)}=\int_{-\infty}^{+\infty}\frac{2\mathrm dt}{13t^2-10t+13}$$
Then
$$I=\dfr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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How to solve the following trigonometrical equation? I have the following equations
\begin{align*}
R_1\cos(\omega T_1-\phi_{1})& =Q_1-R_2\cos(\omega T_2-\phi_{2})\\
R_1\sin(\omega T_1-\phi_{1})& =-Q_2-R_2\sin(\omega T_2-\phi_{2}).
\end{align*}
From these equations how can I obtain the following solution
$$\omega T_2=\p... | From $$a\sin\theta+b\cos\theta+c=0,$$
$$a^2\sin^2\theta=a^2(1-\cos^2\theta)=(b\cos\theta+c)^2$$
and
$$(a^2+b^2)\cos^2\theta+2bc\cos\theta+c^2-a^2=0.$$
You solve the quadratic equation for $\cos\theta$, and $\sin\theta$ follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Show that the annulus $1 \leq x^2 + y^2 \leq 3$ is positively invariant
Given the planar dynamical system
$$\dot x = x - y - x^3, \\ \dot y = x + y - y^3$$
show this is positively invariant in the annulus $1 \leq x^2 + y^2 \leq 3$. Hints:
$$x^4 + y^4 = (x^2 + y^2)^2 − 2x^2y^2$$
$$x^4 + y^4 = \frac 12 (x^2 + y^2)^2 + \... | You found that
$$
2V(1-V)\le \dot V\le V(2-V)
$$
Treating these differential inequalities in the Bernoulli fashion now divide by $V^2$ to get for $U=1/V$
$$
2U-2\le -\dot U\le 2U-1\\
1\le\dot U+2U\le 2\\
U_0e^{-2t}+\frac12(1-e^{-2t})\le U(t)\le U_0e^{-2t}+(1-e^{-2t})\\
\frac{1}{1+(1/V_0-1)e^{-2t}}\le V(t)\le \frac2{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How many solutions are there to the sum $x+y+z+w = 12$ when $x \geq 4$ and $y \leq 3$? I am currently trying to work out how many integer solutions there are to the sum
$$x+y+z+w = 12$$
when $x \geq 4$ and $y \leq 3$ (and $z,w \geq 0$).
I have worked out that
*
*With $x,y,z,w \geq 0$ the number of solutions is $45... | We wish to solve the equation
$$x + y + z + w = 12 \tag{1}$$
in the nonnegative integers subject to the restrictions $x \geq 4$ and $y \leq 3$.
To handle the restriction $x \geq 4$, let $x' = x - 4$. Then $x'$ is a nonnegative integer. Substituting $x' + 4$ for $x$ in equation 1 yields
\begin{align*}
x' + 4 + y + ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Prove that $\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}\geq \frac{x}{1+x^2},~x\geq 0.$
Prove that for all $x\geq 0$:
$$\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}\geq \frac{x}{1+x^2}.$$
Attempt. I have tried algebra, by using $f(x)=\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}- \frac{x}{1+x^2}$, where $$f'(x)=-\frac{1}{1+... | You can express the right hand side as a definite integral:
$$\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}\geq \frac{x}{1+x^2} \iff \int\limits_{x}^{+\infty}\frac{1}{1+s^2}ds\geq \int\limits_{x}^{+\infty}\frac{s^2-1}{(1+s^2)^2}ds \iff \\
\int\limits_{x}^{+\infty}\frac{1}{1+s^2}-\frac{s^2-1}{(1+s^2)^2}ds\geq 0 \iff \int\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Can we say that $\frac{a+b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $ If $a,b,c,d$ are positive real numbers, can we say that $\frac{a+b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $ is always true? If no, can you please give insignts on under which conditions this might be true.
Any references to a similar type of inequalities ... | Solution to new version:
Since $$\frac{a}{c+d} \leq \frac{a}{c}$$
and
$$\frac{b}{c+d} \leq \frac{b}{d}$$
we have
$$\frac{a+b}{c+d} = \frac{a}{c+d} + \frac{b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find primes $p$, for which the $x^2 \equiv 7 \pmod{p}$ has a solution
I want to determine for which primes $p$ the following congruence has a solution.
$$x^2 \equiv 7 \pmod{p}.$$
I have thought the following.
We are looking for the primes $p$ for which the Legendre symbol $\left( \frac{7}{p}\right)=1$. Then $(7,p)=1$... | It's a little longer. From law of quadratic reciprocity, we have
$$\biggl(\frac7p\biggr)=\biggl(\frac p7\biggr)(-1)^{\tfrac{p-1}2},$$
\begin{align}&\text{On the other hand, }&&\begin{cases}(-1)^{\tfrac{p-1}2}=1&\text{if }\;p\equiv 1\mod 4,\\(-1)^{\tfrac{p-1}2}=-1&\text{if }\;p\equiv 3\;(\text{or }-1)\mod 4.
\end{cases... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Showing $f(x,y):=(y^2-x^2)(y^2-2x^2)$ has no extrema how can I show that $f(x,y):=(y^2-x^2)(y^2-2x^2)$ has no extrema?
I already computed the first derivatives for $x$ and $y$ but that doesn't really help me either
$f_{1_x}=8x^3-6xy^2$
$f_{2_x}=24x^2-6y^2$
$f_{1_y}=-6x^2y+4y^3$
$f_{2_y}=-6x^2+12y^2$
| We need to find the critical points by
*
*$f_x=-2x(y^2-2x^2)-4x(y^2-x^2)=0$
*$f_y=2y(y^2-2x^2)+2y(y^2-x^2)=0$
which has solution for $x=y=0$ and for $x,y\neq 0$ we have
*
*$f_x=-2x(y^2-2x^2)-4x(y^2-x^2)=0\\\implies -2(y^2-2x^2)-4(y^2-x^2)=-6y^2+8x^2=0\implies3y^2=4x^2$
*$f_y=2y(y^2-2x^2)+2y(y^2-x^2)=0\implies (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find $x^5+x^{-5}$ given the value of $x^2+x^{-2}$.
Find $x^5+\dfrac1{x^5}$ in its simplest form given that $x^2+\dfrac1{x^2}=a$ for $a,x>0$.
Attempt:
We write $$x^2+\frac1{x^2}=a\implies x^4-ax^2+1=0\implies x^5=ax^3-x$$ and $$\frac1{x^2}=a-x^2\implies \frac1{x^4}=a^2-2ax^2+x^4\implies \frac1{x^5}=\frac{a^2}x-2ax+x^3... | With $S_n:=x^n+x^{-n}$, you can establish the recurrence
$$S_{n+1}=S_nS_1-S_{n-1}.$$
From this, $$S_2=S_1^2-2,
\\S_3=S_1(S_2-1),
\\S_4=S_2^2-2,
\\S_5=S_1(S_2^2-S_2-1).
$$
You draw $S_1=\sqrt{S_2+2}$ from the first identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 7
} |
Prove or disprove that, for any $n \in \mathbb{N_+}$, there exist $a,b \in \mathbb{N_+} $ such that $\frac{a^2+b}{a+b^2}=n.$ Problem
Prove or disprove that, for any $n \in \mathbb{N_+}$,
there exist $a,b \in \mathbb{N_+} $ such that $$\frac{a^2+b}{a+b^2}=n.$$
My Thought
Assume that the statement is ture. Then, the equa... | Solution for non-square $n$ is provided in @Oldboy's answer and in linked questions. This answer handles the case for square $n$.
Case 1: $n=k^2,k \equiv 0 \pmod {2}$
Choose
\begin{align}
a=\frac{k^2(k^3+2)}{4},
b=\frac{k^4}{4}.
\end{align}
Conditions imply that $k^2 \equiv 0 \pmod {4}$ and so both $a$ and $b$ are int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 1,
"answer_id": 0
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Solve the initial value problem $ (x+1)^2 dx+(2xy+x^2-1)dy =0 , \ y(0)=1 \ $ Solve the initial value problem $$ (x+1)^2 dx+(2xy+x^2-1)dy =0 \ , \ \ y(0)=1 $$
Answer:
Consider the above equation with $ \ M(x,y)dx+N(x,y)dy=0 \ $ , then we get
$ M(x,y)=(x+1)^2, \ N(x,y)=2xy+x^2-1 \ $
Since $ \ \frac{\partial M}{\partial ... | $$ (x+1)^2 dx+(2xy+x^2-1)dy =0 \ , \ \ y(0)=1$$
Yes, the integrating factor is $\mu=x^ay^b$. There's a formula for this case
$$\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=a\frac{N(x,y)}{x}-b\frac{M(x,y)}{y}\\ -2x-2y=a\left(\frac{x^2+2xy-1}{x}\right)-b\left(\frac{x^2+2x+1}{y}\right)\\ -2x-2y=ax+2ay-ax^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding $f(\sqrt 3 )$
Let $$x^2 + \left( f(x)-2 \right) x + 2\sqrt 3 - 3 - \sqrt 3 f(x) = 0$$ for all $x \in \mathbb R$ and let $f(x)$ be continuous for all $x \in \mathbb R$. Find $f(\sqrt 3)$.
When I substituted $x= \sqrt 3$, I got $0=0$. How else do I solve this problem? I don't think there are many more obvious c... | If $x\ne \sqrt{3}$ then you can write $f$ explicitly:
$$f(x)= {-x^2+2x+3-2\sqrt{3}\over x - \sqrt 3 }$$
So you can only calculate $$\lim_{x\to \sqrt{3}}f(x) =\lim_{x\to \sqrt{3}} {-x^2+2x+3-\color{red}{2}\sqrt{3}\over x - \sqrt 3 }$$ $$= \lim_{x\to \sqrt{3}} {(\sqrt{3}-x)(\sqrt{3}+x)+2(x-\sqrt{3})\over x - \sqrt 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric equation: $2\arcsin \left(\frac{2x}{1+x^2}\right)- \pi x^3 = 0$
The number of solutions of the equation $$2\arcsin \left(\dfrac{2x}{1+x^2}\right)- \pi x^3 = 0$$ is?
Let $x= \tan \theta$
$\implies \sin 2\theta = \sin(\dfrac \pi 2 \tan^3\theta)$
I had to delete the rest of my attempt because it was total... | Consider
$$
f(x)=\arcsin\frac{2x}{1+x^2}
$$
Then
$$
f'(x)=\frac{1}{\sqrt{1-\dfrac{4x^2}{(1+x^2)^2}}}\frac{2(1+x^2)-4x^2}{(1+x^2)^2}
=\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}
$$
Therefore
$$
f'(x)=\begin{cases}
\dfrac{2}{1+x^2} & |x|<1 \\[6px]
-\dfrac{2}{1+x^2} & |x|>1
\end{cases}
$$
which implies
$$
f(x)=\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\sum_{n=1}^{\infty}\frac{4n^2+26n+34}{n^4+10n^3+35n^2+50n+24}$ convergent? serie is convergent?
$$\sum_{n=1}^{\infty}\frac{4n^2+26n+34}{n^4+10n^3+35n^2+50n+24}$$
Factor: $$\frac{4n^2+26n+34}{(n+1)(n+2)(n+3)(n+4)}$$
What test can i Use?
| $$
\frac{4 n^2+26 n+34}{(n+1) (n+2) (n+3) (n+4)} = \frac{1}{n+2}-\frac{4}{n+3}+\frac{1}{n+4}+\frac{2}{n+1}
$$
Then
$$
\sum_{n=1}^{\infty}\frac{4 n^2+26 n+34}{(n+1) (n+2) (n+3) (n+4)} = \lim_{n\to\infty}\left(2\sum_{k=2}^{n}\frac{1}{k}+\sum_{k=3}^{n}\frac{1}{k}-4\sum_{k=4}^{n}\frac{1}{k}+\sum_{k=5}^{n}\frac{1}{k}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate the product of two or more polynomials raised to some powers, not necessarily integral This question is inspired by my own answer to a question which I tried to answer and got stuck at one point.
The question was:
HI DARLING.
USE MY ATM CARD, TAKE ANY AMOUNT OUT, GO SHOPPING AND TAKE YOUR FRIENDS FO... | $\require{begingroup}\begingroup$This should help to get closer to the final result (if you want to calculate this manually):
$$\newcommand{\dd}{\; \mathrm{d}} I=\int_0^1 \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \dd x =
\int_0^1 \frac{3x^3 - x^2 + 2x - 4}{\sqrt{(x-\frac32)^2 -\frac14}} \dd x$$
It will take some... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Prove $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$ when $a=b+c$ I want to prove this identity:
$$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) \qquad\text{when}\;a=b+c$$
Can somebody give a hint in the easiest way possible? I am debugging this for hours and can't get the left side to be t... | If $a = b+ c$ then
$\cos a = \cos b\cos c - \sin b\sin c$
$\sin a = \sin b\cos c + \cos b\sin c$
So solving $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$
involves solving
$\sin^2 b + \sin^2 c + \sin^2 b\cos^2 c + 2\cos b\sin b\cos c\sin c + \cos^2 b \sin^2 c = 2(1 - \cos b^2 \cos^2 c + \cos b\cos c\sin b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Convergence of $\sum_{n=1}^{\infty} 3^n \sin(\frac{1}{4^nx})$ I wish to prove the convergence of:
$$\sum_{n=1}^\infty 3^n \sin\left(\frac 1 {4^nx}\right)$$
for $1\le x \lt \infty$, using Cauchy's criterion.
Here is what I tried:
\begin{align}
|S_{n+p}-S_n| & = \left| 3^{n+1} \sin\left(\frac 1 {4^{n+1}x}\right)
+ \cd... | Note that
$$3^n \sin\left(\frac{1}{4^nx}\right)\sim \frac{3^n}{4^nx}$$
and by ratio test $\sum \frac{3^n}{4^nx}$ converges $\forall x\neq 0$, therefore the given series converges by limit comparison test with $\sum \frac{3^n}{4^nx}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluate $ \lim_{x \to -0.5^{-}} \left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor$ Evaluate $$L=\lim_{x \to -0.5^{-}} \left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor $$
My try:
Let $t=\frac{1}{x}$ Now when $ t \to -0.5^{-}$ we have $t \to -2^{+}$ we get
$$L=\lim_... | Assume $x=-0.5-\epsilon$ with $\epsilon >0$, then we have that
$$\frac1x=\frac1{-0.5-\epsilon}=-\frac{2}{1+2\epsilon}=-2(1-2\epsilon)=-2+4\epsilon$$
therefore for $\epsilon$ sufficiently small we have
$$\left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor=\left\lfloor (-2+4\epsilon) \left\lfloor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Linear equation help $$\frac{1-x}{4} + \frac{5x+1}{2} = 3 - \frac{2(x+1)}{8}$$
I got x=1 but the book says x=4/5 and I don't understand how to get to that, I tried working backwards too but I just can't figure it out. Any help would be appreciated, thanks.
Here's my work
using 8 as the lcm
$$
\begin{align}
\frac 81(\f... | You have a sign mistake here
$$2(1-x) + 4(5x+1) = 24 - 2x \color{red}{+ 2}$$
It should be
$$2(1-x) + 4(5x+1) = 24 - 2x \color{blue}{- 2}$$
Because
$$-\frac {2(x+1)}8 =-\frac x4-\frac 14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$ where $a,b$ are finite natural numbers
Evaluate $$\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$$ where $a,b$ are finite natural numbers
I have spent about a day thinking over this problem. I tried int... | Let $I(a,b)=\int_0^{\pi/2} \log(a^2\cos^2(\theta)+b^2\sin^2(\theta))\,d\theta$. Differentiating under the integral with respect to $a^2$ reveals
$$\begin{align}
\frac{\partial I(a,b)}{\partial (a^2)}&=\int_0^{\pi/2}\frac{1}{a^2+b^2\tan^2(\theta)}\,d\theta\\\\
&=\frac{\pi/2}{a(a+b)}\tag1
\end{align}$$
Integrating $(1)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2817172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplify the polynomial $xy(x+y)+(x+y)+(x+y)^2=13xy$ to the form $y^2=4x^3-g_{2}x-g_{3}$
I try to simplify a polynomial to the form: $y^2=4x^3-g_{2}x-g_{3}$, which is the elliptic curves. And the polynomial is $xy(x+y)+(x+y)+(x+y)^2=13xy$.
I try to let the $u=x+y$ and $v=x-y$, then I get the $u^3-uv^2+4u-9u^2+13v^2=... | You made a good start but there was a complication you did not anticipate.
The homogeneous version of your equation is:
$\, 0 = -W X Y Z + (X+Y)(X+Z)(Y+Z), \,$ where $\, W=13 \,$ is a constant. Now substitute $\, X = 1 + c_3 x + \sqrt{c_1} y, $
$\, Y = 1 + c_3 x - \sqrt{c_1} y, $ $\, Z = 2 x + c_2, \,$
where $\,c_1,c_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Determining rational functions that simplify to a simple expression
Question: Is there a way to find, if possible, another way to write $(\sqrt[3]2-1)^2$ and $(\sqrt[3]4-1)^2$, in the form of $\frac {a+b}{c+d}$?
What I meant was that, let's take the second expression $(\sqrt[3]4-1)^2$ as an example. The expression is... | As for $(\sqrt[3]{2}-1)^2$:
$$
\dfrac{4-3\sqrt[3]{2}}{2+\sqrt[3]{2}} = (\sqrt[3]{2}-1)^2. \tag{1}
$$
We can easily check it: denote $x = \sqrt[3]{2}$, then $$(x-1)^2(2+x)=(x^2-2x+1)(2+x)= \\ 2x^2-4x+2+x^3-2x^2+x=\\ x^3-3x+2 = \\ 4-3x.$$
Let's try to find $(x-1)^2$ in the form
$$
\dfrac{a+bx}{c+dx}, \qquad a,b,c,d\in\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $\frac{4x}{y+1} + \frac{16y}{z+1} + \frac{64z}{x+1}$
Let $x,y,z$ be positive real numbers such that $x+y+z = 1$ and $xy+yz+zx = \frac{1}{3}$. Find the vlaue of $$\frac{4x}{y+1} + \frac{16y}{z+1} + \frac{64z}{x+1}.$$
So far using the fact that $(x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx)$, I was able to ge... | The point is that $x=y=z=1/3$.
Too see this consider
$$0=(x+y+z)^2-3(xy+xz+yz)=\frac{(x-y)^2+(x-z)^2+(y-z)^2}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Functional Equation satisfying f(2x)=f(x) Determine all contimouse functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f\left(2x\right)=f\left(x\right)$
Attempt
Let $a$ and $b$ be two distinct points on $\mathbb{R}$.
Then consider the sequences of real numbers
$$\{x_n\}=\frac{a}{2},\frac{a}{2^2},\frac{a}{2^3},\frac{a}... | Let $x\ne 0$. for $n\in\mathbb N,$
$$f (x)=f (\frac {x}{2^n} )$$
$f $ is continuous at $0$ and $$\lim_{n\to+\infty}\frac {x}{2^n}=0$$
$$\implies \lim_{n\to+\infty}f (\frac {x}{2^n})=f (0) $$
thus
$$(\forall x\in\mathbb R) \;\;f (x)=f (0) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find $y$ in $-\frac12+\frac y2=1$ without combining $-\frac12+\frac y2$? I am trying to find the value of a variable. The problem given was$$
\frac{1}{2}(-1+y) = 1.
$$
I understand that $\dfrac{1}{2}(-1+y) = 1$ is the same as $-\dfrac{1}{2} + \dfrac{y}{2} =1$. How do I get $y$ by itself starting from $-\dfrac{1... | As Ethan Bolker pointed out in the comments, without having to divide each summand by $2$ in the expression $$\frac{y-1}{2}=1,$$
one can multiply both sides by two to get $$\require{cancel}\frac{y-1}{\cancel{2}}\cdot\cancel{2}=1\cdot 2\iff y-1=2\iff y=3.$$
Note: I use the symbol $\iff$ to mean “if and only if,” or “th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find the finite limit of this function?
Let $f(x) = \dfrac{1-\cos \{x\}}{(x^4 + ax^3 +bx^2 +cx)^2}$. If $l= \lim_{x\to 1^+}f(x), m = \lim_{x\to 2^+}f(x) $ and $n= \lim_{x\to 3^+}f(x),$ where $l,m$ and $n$ non-zero finite then:
$a+b+c=? $
$l+m+n=?$
$\lim_{x\to 0^+}f(x)=? $
where {} denotes the fractional part... | Try breaking the fraction part function into different intervals, like $(1,2)$ etc. For $x\in (1,2) $ we have $\{x\} = x-1$ and so
$$\begin{align}
l=\lim_{x\to1^+} \frac{1-\cos\{x\}}{(x^4+ax^3+bx^2+cx)^2} &= \lim_{x\to1^+}\frac{2\sin^{2}(\frac{x-1}{2})}{(\frac{x-1}{2})^2}\cdot \frac{(\frac{x-1}{2})^2}{(x^4+ax^3+bx^2+cx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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The ring $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$ is a UFD. Contradiction? Considering the ring $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$, I know that this ring is an euclidian domain and therefore a unique factorization domain.
Now, $23=(3+4\sqrt{2})(3-4\sqrt{2})=(11+7\sqrt{2})(11-7\sqrt{2... | There is no contradiction because $3+4\sqrt{2}$ and $11+7\sqrt{2}$ are associates:
$$
\frac{11+7\sqrt{2}}{3+4\sqrt{2}} = 1+\sqrt{2}
$$
and $1+\sqrt{2}$ is a unit, since it has norm $-1$. Or note that
$$
\frac{3+4\sqrt{2}}{11+7\sqrt{2}} = -1+\sqrt{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Want to Confirm Answer for the Sum of Fractions with Triangular Number Sequence as the Denominators I am trying to compute the sum of these fractions:
$$\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3} + \dots + \frac{3}{1+2+3+\dots+100}.$$
I believe the denominators are a triangular number sequence, therefore the expression... | Call the $n$th Triangular number $T_n$. Then, noting $T_{100}=5050$, the reciprocal sum of the first $100$ triangular numbers is:
\begin{align*}
\frac{1}{1}&+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\dotsb+\frac{1}{5050}\\
&=2\left[\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Minimizing $c(x-2)(y-2)(z-2)$ subject to $xyz =V$ Given constants $c > 0$ and $V > 0$, how can I solve the following optimization problem?
$$\begin{array}{ll}
\text{minimize} & c \, (x-2) (y-2) (z-2)\\
\text{subject to} & xyz = V\\
& x,y,z > 2
\end{array}$$
My attempt
Since $V=xyz$, we get $z = \frac{V}{xy}$
Let $S ... | Let$f(x,y,z):=c (x - 2) (y - 2) (z - 2)$, and $g(x,y,z)=x y z - V$. Then solve for $x,y,z,\lambda$ in:
$$\frac{\partial f(x,y,z)}{\partial x}=\lambda \frac{\partial g(x,y,z)}{\partial x}\\
\frac{\partial f(x,y,z)}{\partial y}=\lambda \frac{\partial g(x,y,z)}{\partial y}\\\frac{\partial f(x,y,z)}{\partial z}=\lambda ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Rationalize this I'm having trouble rationalizing the denominator of this fraction. Would you kindly explain this to a fellow self-learning math student?
$$\frac{10}{\sqrt[4]{3}-1}$$ knowing that $a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}\cdot b+a^{n-3}\cdot b^2+...+a\cdot b^{n-2}+b^{n-1}\right)$
$a,b\in \mathbb{R... | $$\dfrac{10}{\sqrt[4]{3}-1} = \dfrac{5\cdot (\sqrt[4]{3^4}-1^4)}{\sqrt[4]{3}-1} = 5\left(\sqrt[4]{3^3}\cdot 1^0+\sqrt[4]{3^2}\cdot 1^1+\sqrt[4]{3^1}\cdot 1^2+\sqrt[4]{3^0}\cdot 1^3\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Cube root of numbers such as $2+11i$ How can I find powers and roots of complex numbers with ugly argument such as cube root of 2+11i? I saw Find the solutions to $z^3 = 2 + 11i$., but the answer includes some guessing and I would like to have some algorithm for such a tasks, which I can program. I would also like to s... | Note that $\tan 3\alpha=\dfrac {t(3-t^2)}{1-3t^2}$, where $t=\tan\alpha$.
Putting $\theta=3\alpha$ where $\tan\theta=\tan3\alpha=\dfrac {11}2$, we have
$$\begin{align}
\frac {11}2&=\frac {t(3-t^2)}{1-3t^2}\\
11(1-3t^2)&=2t(3-t^2)\\
2t^3-33t^2-6t+11&=0\end{align}$$
By inspection, putting $t=\frac 12$ gives $\text{LHS}=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Eliminating $\theta$ from $a\cos(\theta-\alpha)=x$ and $b\cos(\theta-\beta)=y$ Eliminate $\theta$ in following equations
$$\begin{align}
a \cos(\theta-\alpha) &= x \\
b \cos(\theta- \beta) &=y
\end{align}$$
I am trying to solve this problem but still I am unable to get the perfect answer
I added both the equations bu... | For simplicity, define $u:=x/a$ and $v:=y/b$, so that we have
$$\begin{align}
u &= \cos(\theta-\alpha) = \cos\theta \cos\alpha + \sin\theta\sin\alpha \\
v &= \cos(\theta-\beta) = \cos\theta\cos\beta + \sin\theta\sin\beta
\end{align}$$
This is a linear system in $\cos\theta$ and $\sin\theta$. Solving, we obtain
$$\cos\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding the determinant of a $2\times 2$ matrix Let $R$ be a commutative ring with 1. Show with the Leibniz signature formula that $\det\biggl(\begin{bmatrix}
\lambda & A_{12}\\
0 & A_{22}\\
\end{bmatrix}\biggl) = \lambda \cdot \det(A_{22})$ applies whereby $\lambda \in R \land A_{22} \in R^{n,n}$.
My idea:
$\sigma_{1}... | While you are right to be concerned that it's a little more tricky if $A_{22}$ is a larger matrix, since $A$ is only a $2\times 2$ matrix, $A_{22}$ is just a number and your work is correct!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Remainder on division with $22$ What is the remainder obtained when $14^{16}$ is divided with $22$?
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $22$, excep... | Using the Euclidean Algorithm
Note that
$$
\begin{align}
14^{16}&\equiv0&\pmod2\tag1
\end{align}
$$
Reducing mod $11$ and using Fermat's Little Theorem, we get
$$
\begin{align}
14^{16}
&\equiv3^6&\pmod{11}\\
&\equiv3&\pmod{11}\tag2
\end{align}
$$
Solving $11a+2b=1$ using the Euclidean Algorithm as implemented in this a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Minimum $e$ where $a,b,c,d,e$ are reals such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ I have a question about this 1978 USAMO problem:
Given that $a,b,c,d,e$ are real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the maximum value that $e$ can attain.
I had the following solution:
Let $a+... | Given $a+b+c+d+e=8$ and $a+b+c+d=8-e$
$$(8-e)^2=(a+b+c+d)^2$$
Now expand $(a+b+c+d)^2$
$$=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)$$
$$\le4(a^2+b^2+c^2+d^2)$$
And Now, $a^2+b^2+c^2+d^2+e^2=16$
$a^2+b^2+c^2+d^2=16-e^2$ $$(16-e^2)=a^2+b^2+c^2+d^2$$$$4(a^2+b^2+c^2+d^2)\ge(a+b+c+d)^2$$$$4(16-e^2)\ge(8-e)^2$$$$64-4e^2\ge64-16e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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prove that the polynomial $x^8 -x^5 +x^2 -x +1$ is positive for all real values of x is there any factorization possible of the above expression or can it be shown that is a a sum of two or three squares?
i tried various factorizations but none of them were conclusive.
| $$x^8-x^5+x^2-x+1 = \frac{1}{2}\bigg[2x^8-2x^5+2x^2-2x+2\bigg]$$
So $$ = \frac{1}{2}\bigg[x^8+(x^8-2x^5+x^2)+(x^2-2x+1)+1\bigg]$$
So $$ = \frac{1}{2}\bigg[x^8+(x^4-x)^2+(x-1)^2+1\bigg]>0\forall x \in \mathbb{R}.$$
Added:: For $x=0,x^8-x^5+x^2-x+1>0$
Using Arithmetic Geometric Inequality $(x\neq 0)$
$$\frac{x^8}{2}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding the inverse of $A$
Find the inverse of $$A =\left[\begin{matrix}0 & 1 & 0 & 0\\ 0& 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ a & b & c & d\end{matrix}\right]$$
My attempts:
$$A^{-1} = \frac {\operatorname{adj}A}{\det A}$$
As I can find the $\det A$ that is $\det A = -b$. Here, how can I find the inverse? Is there... | You're wrong: expanding by the first column, you obtain at once $\;\det A=-a$. We'll suppose $a\ne0$.
To find the inverse you can perform column reduction:
\begin{align}
\begin{bmatrix}
\begin{array}{cccc|cccc}
0&1&0&0&1&0&0&0 \\
0&0&1&0 &0&1&0&0\\
0&0&0&1&0&0&1&0\\
a&b&c&d&0&0&0&1
\end{array}
\end{bmatrix}&\xrightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Series Convergence of Harmonic Means
Let $\{x_n\}$ be a sequence of real numbers such that $0< x_1 <x_2$. If $$x_n= \frac{2}{\frac{1}{x_{n-1}}+\frac{1}{x_{n-2}}}, $$then show that
$$\lim_{n\to\infty}x_n=\frac{3x_1x_2}{2x_1+x_2}.$$
| Note that
\begin{align}
& x_n = \frac{2}{\frac{1}{x_{n-1}} + \frac{1}{x_{n-2}}} \\
\iff & \frac{1}{x_{n-1}} + \frac{1}{x_{n-2}} = \frac{2}{x_{n}} \\
\iff & a_n = \frac{1}{2}a_{n-1} + \frac{1}{2}a_{n-2},
\end{align}
where $a_i = \frac{1}{x_i}$.
It's easy to verify that $$a_n = \alpha + \beta\left(-\frac{1}{2}\right)^n.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Asymptotes of an implicit curve Per the method described at How to find asymptotes of implicit function? , I proceeded to find the asymptotes of
$$
x^3 + 3x^2y - 4y^3 - x + y + 3 = 0
$$
Whilst, it correctly generates the asymptote $$ y=x $$ , the remaining asymptote(s):-
$$ 2y + x = ±1 $$ can't be deduced.
Instead ano... | Giving
$$
f(x,y) = x^3 + 3 x^2 y - 4 y^3 - x + y - 3 = 0
$$
the asymptotic directions can be explored by substituting $ y = a x + b\;\; $ into $f(x,y)$ giving
$$
f(x,ax+b) = 3 x^2 (a x+b)-4 (a x+b)^3+a x+b+x^3-x-3 = (1+3a-4a^3)x^3+3b(1-4a^2)x^2 +(a(1-12b^2)-1)x+b(1-4b^2)-3
$$
Now the conditions for $f(x,ax+b)$ to hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Reference request: Transformations under which the discriminant is invariant I had an occasion to think about this quadratic equation:
$$
ax^2 +bx(1-x) + c(1-x)^2 = 0.
$$
Its solution is
$$
x = \frac{2c-b\pm\sqrt{b^2-4ac\,}}{2(a-b+c)}.
$$
The thing under the radical is the same thing we were all taught in eighth or nin... | Let $P(x)=ax^2+bx(1-x)+c(1-x)^2$. Then
\begin{align}
P(x)&=ax^2+bx(1-x)+c(1-x)^2\\
&=x^2\big [a+b(\frac{1}{x}-1)+c(\frac{1}{x}-1)^2\big ]
\end{align}
Hence, $P(x)$ is obtained from $a+bx+cx^2$ by a projective transformation. Also,
\begin{align}
a+bx+cx^2 &= x^2\big [a\frac{1}{x^2}+b\frac{1}{x}+c\big ]\\
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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What is the probability that both tubes are good?
This a conditional probability where we calculate:
Probability that both are good knowing that one of them is good= prob that both are good/ probability that one of them is good
Which gives me: (1/3) / (6/10)=5/9
But the correct answer is: 5/13
How should I proceed?
| You presumably calculated the probability both are good as $$\frac{6}{10}\times\frac{5}{9}=\frac{1}{3}$$
Similarly the probability both are bad is $$\frac{4}{10}\times\frac{3}{9}=\frac{2}{15}$$
And so the probability at least one is good $$1-\frac{2}{15} = \frac{13}{15}$$
which you could alternatively calculate as $\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the value of this product: $\prod_{n=1}^\infty \;\frac{3}{1+2 \cos(\frac{\pi}{3^n})} \;=\;? $ emphasized text$$\left[\, \prod_{n=1}^\infty \;\frac{3}{1+2 \cos\left(\frac{\pi}{3^n}\right)}\, \right]\, =\;? $$
Where $\;[\, .]\;$ denotes the integral part function.
$\mathbf {My Attempt}$
I tried to confine the n... | $$1+2\cos2x=1+2(1-2\sin^2x)=\dfrac{\sin3x}{\sin x}=\dfrac{f(n+1)}{f(n)}$$
where $f(m)=\sin(3^mx)$
Here $3^mx=?$
Related:$\cos x(2\cos2x-1)=\cos3x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
An inequality with substitution If $a,b,c$ positive real numbers, then I have to prove $ \frac {1}{18} \sum\limits_{cycl}^{} \frac{a^2}{b^2} + \sum\limits_{cycl}^{} \frac {a}{2a+b+c} \ge \frac {11}{12}$
We have that $\frac {1}{18} \sum\limits_{cycl}^{} \frac{a^2}{b^2} \ge \frac {3}{18} = \frac {1}{6}$
If we assume that... | Let $c=\max\{a,b,c\}$.
Since by C-S $$\sum_{cyc}\frac{a}{2a+b+c}=\sum_{cyc}\frac{a^2}{2a^2+ab+ac}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2a^2+ab+ac)}=\frac{(a+b+c)^2}{2\sum\limits_{cyc}(a^2+ab)},$$
it's enough to prove that
$$\frac{1}{18}\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)+\frac{(a+b+c)^2}{2\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$ Solve:
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$
This is taken from one of the TAU entry tests (I have one in 2 weeks :) )
So, I don't really recognize anything speical here ... | Hint: your equation is equivalent to $$\frac{4x(6+x)}{5(1+x)(5+x)}=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How do I prove the following inequality? How do I proceed to solve the inequality $$\frac{(a^2+b^2)}{(a+b)} + \frac {(b^2+c^2)}{(b+c)} + \frac{(a^2+c^2)}{(a+c)} \geq (a+b+c)$$ where $a , b , c > 0$
I have thought of taking the terms on the $LHS$ and converting them to $AM$ and then use the $AM-GM$ theorem , but I can... | Using Cauchy Schwarz in Engel form, we have
$$\!\!\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{a+c}+
\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{a+c}\geq \frac{(a+b+c)^2}{2(a+b+c)}+\frac{(b+c+a)^2}{2(a+b+c)}= a+b+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Conic Section Equation from Michael Spivak's Book So i've been reading Michael Spivak's Calculus lately and now i feel im stuck in his conic section equation, page 81.
What i dont understand is, how can the first equation becomes the second? After squaring it, i got different coefficient especially in the x² part.
| There is indeed a typo. From:
$$M (\alpha x + \beta) + B = \pm C \sqrt{(\alpha x + \beta)^2 + y^2}$$
square both sides:
$$\left [M (\alpha x + \beta) + B \right ]^2 = \left [\pm C \sqrt{(\alpha x + \beta)^2 + y^2} \right]^2$$
Now on the left use $(a + b)^2 = a^2 + 2 a b + b^2$:
$$M^2(\alpha x + \beta)^2 + 2 B M (\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the remainder when $x^{10}+1$ is divided by $(x^2+1)(x^2+x+1)$ Find the remainder when $x^{10}+1$ is divided by $(x^2+1)(x^2+x+1)$
I have done it until the the divisor is of second degree.
But here the degree of the remainder is $4$
This means the remainder will be of the form. : $ax^3+bx^2+cx+d$ which makes ... | Note that:
$$\frac{x^{10}+1}{(x^2+1)(x^2+x+1)}=\frac{(x^2+1)(x^8-x^6+x^4-x^2+1)}{(\color{green}{x^2+1})(x^2+x+1)}$$
Using long division:
$$
\require{enclose}
\begin{array}{r}
\color{blue}{x^6-x^5-x^4+2x^3-2x+1} \\[-3pt]
x^2+x+1 \enclose{longdiv}{x^8-x^6+x^4-x^2+1} \\[-3pt]
\underline{x^8+x^7+x^6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $B$ if $B=A-{{1}\over{2}} A^2+{{1}\over{3}} A^3 -{{1}\over{4}} A^4+...$
Let $$
\ A=\begin{bmatrix} 0 & a & a^2 & a^3 \\ 0 & 0 & a & a^2 \\ 0 & 0 & 0 & a \\ 0 & 0 & 0 & 0 \end{bmatrix}
$$ and
$B=A-{{1}\over{2}} A^2+{{1}\over{3}} A^3 -{{1}\over{4}} A^4+...$
$i)$ Find the matrix $B$
$ii)$ Prove that $A=B+ {{1}\ove... | You are on the right track. For (ii), just compute $B^2$, $B^3$, $B^4$, and observe that $B^4 = O$. Afterwards, verify that
$$A = B + \frac{1}{2!}B^2 + \frac{1}{3!} B^3.$$
For your information,
$$B^2 = \begin{bmatrix}0 & 0 & a^2 & a^3 \\ 0 & 0 & 0 & a^2 \\0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix},$$ and
$$B^3 = \be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sequences and Sums There is a list of numbers $a_{1} , a_{2} , …, a_{2010}$ . For $1 \leq n \leq 2010$, where $n$ is positive
integer, let $a_1+a_2+ \ldots +a_n = S_n$ . If $a_1 = 2010$ and $S_n = a_nn^2$ for all n, what is
the value of $a_{2010}$ ?
I've been trying to manipulate the formula but I cant seem to find a g... | We have $a_n\cdot n^2=a_1+a_2+\ldots+a_n=S_n$. Then
$$
\begin{array}{cc}
\begin{array}{rlrl}
\frac{\quad}{\quad}a_1\cdot 1^2=&2010\\
\frac{\quad}{\quad}a_2\cdot 2^2=&a_1+a_2\\
\frac{\quad}{\quad}a_3\cdot 3^2=&a_1+a_2+a_3\\
\frac{\quad}{\quad}a_4\cdot 4^2=&a_1+a_2+a_3+a_4\\
\end{array}
&
\begin{array}{rl}
\implies a_1=&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$
T... | $$\frac{a\sqrt{b}+c}{d}~=~2\sqrt{2}~~~~\text{with}~~~~a+b+c+d~=~10$$
The numerator of the $abcd$-fraction contains one square root plus a number. Hence the sum of the four variable equals an integer the easiest way to get rid of $c$ is by setting it $0$. From thereone we not got $a+b+d=10$. Now $b$ has to be $2$ so tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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The sides of a triangle are in Arithmetic progression If the sides of a triangle are in Arithmetic progression and the greatest and smallest angles are $X$ and $Y$, then show that
$$4(1- \cos X)(1-\cos Y) = \cos X + \cos Y$$
I tried using sine rule but can't solve it.
| The law of sines helps!
From the given we obtain
$$\sin X+\sin Y=2\sin(X+Y)$$ or
$$2\sin\frac{X+Y}{2}\cos\frac{X-Y}{2}=4\sin\frac{X+Y}{2}\cos\frac{X+Y}{2}$$ or
$$\cos\frac{X-Y}{2}=2\cos\frac{X+Y}{2}$$ or
$$\cos\frac{X}{2}\cos\frac{Y}{2}=3\sin\frac{X}{2}\sin\frac{Y}{2}$$ or
$$\cos\frac{X-Y}{2}=4\sin\frac{X}{2}\sin\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Help summing the telescoping series $\sum_{n=2}^{\infty}\frac{1}{n^3-n}$. I know a priori that the series $$\sum_{n=2}^{\infty}\frac{1}{n^3-n}$$ converges. However, I am tasked with summing the series by treating it as a telescoping series.
By partial fraction decomposition, the series can be written as: $$\sum_{n=2}^... | Your partial fraction decomposition is fine, but you need not to break up into separate positive and negative sums. Instead, you need to cancel terms within the summation.
$$\sum_{n=2}^{\infty}\bigg(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\bigg)=\\\frac 12\sum_{n=2}^{\infty}\bigg(\frac{1}{(n+1)}+\frac{1}{(n-1)}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
For a given $D\in \mathbb{N}$ are there infinite solutions to $D=b^2-4ac$ with $a,b,c \in \mathbb{N}$ I am wondering if for given $D\in \mathbb{N}$ we can find an infinite amount of solutions to $D=b^2-4ac$. Obvious $D$ is the discriminant of a binary quadratic form or a polynomial of degree $2$. Maybe the answer can b... |
I am wondering if for given $D\in \mathbb{N}$ we can find an infinite amount of solutions to $D=b^2-4ac$
No, not in general.
Consider $D=1$. Then $b^2-1=4ac$. Hence $(b+1)(b-1)=4ac$.
This obviously has four immediate solutions. $b=\pm 1$ and $a=0\vee c=0$.
Since $4|(b+1)(b-1)$, $b$ has to be odd. Which gives $b=2k+1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to solve $\sqrt{49-x^2}-\sqrt{25-x^2}=3$? I recognize the two difference of squares: $49-x^2$ and $25-x^2$.
I squared the equation to get:
${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$
However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.
| First express one root: $$\sqrt{49-x^2}=3+\sqrt{25-x^2}$$
Now square:
$$ 49-x^2=9+6\sqrt{25-x^2}+ 25-x^2\implies \boxed{5=2\sqrt{25-x^2}}$$
and square again...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $... | I would use Taylor polynomials at order $2$.
$$\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)}=\frac{1-\frac{x^2}{2}-\left(1-\frac{(3x)^2}{2}\right)+o(x^2)}{3x^2-x^2+o(x^2)}=\frac{4x^2+o(x^2)}{2x^2+o(x^2)}=2+o(1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 2
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Maximum minus minimum of $c$ where $a+b+c=2$ and $a^2+b^2+c^2=12$ Let $a,b,$ and $c$ be real numbers such that
$a+b+c=2 \text{ and } a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{... | Hint:
$$(2-c)^2 = (a+b)^2 = (a \cdot 1 + b \cdot 1)^2 \le (a^2 + b^2) (1^2 + 1^2) = 2(12-c^2)$$
Finding $c$ satisfying this inequality amounts to solving a quadratic.
$$c^2 - 4c + 4 \le 24 - 2 c^2$$ $$3 c^2 - 4 c - 20 \le 0$$ So $c$ is between the two roots $-2$ and $10/3$. Note that you should check that for each of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
$\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$? On a problem book solution I was faced with the following step:
$$\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$$
I identified $\frac{1}{n+1}+\frac{1}{(n+1)^2}...$ as a geometric series so the sum would be $\frac{1}{1-r}$ so that $\frac{1}{1-\frac{1}{n+1}}=1+\frac{1}... | If you know that $a+ar+ar^2+ar^3+\cdots = \dfrac{a}{1-r}$ for $|r|\lt 1$
then here you have $a= \dfrac1{n+1}$ and $r= \dfrac1{n+1}$
so the sum is $\dfrac{\frac1{n+1}}{{1-\frac1{n+1}}}=\dfrac{1}{n+1-1}=\dfrac{1}{n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
last 2 digits of a sequence $x+\frac{1}{x} = 3$, what are the last 2 digits of $x^{2^{2013}}+\frac{1}{x^{2^{2013}}}$?
Getting the next value, we have to square then subtract by 2, I am clueless in getting to the next step
| Definitely you see the pattern : $x^{2m} + \frac 1{x^{2m}} = \left(x^m + \frac 1{x^m}\right)^2 - 2$ for all $m \geq 1$.
So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.
Since we are only looking at the last two digits of the number, it is also enough to trac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to find the limit of the sequence $a_n=\frac{n^n}{3^n\cdot n!}$ as $n$ tends to infinity.
Let $a_n=\dfrac{n^n}{3^n\cdot n!}.$ Show that $a_n\to0$ as $n\to\infty$.
I know that I can use the ratio test for sequences, and $n^n$ increases faster than $3^n \cdot n!$ so it will tend towards infinity so I invert the seq... | A Third Proof
Since $$\frac{a_{n}}{a_{n-1}}=\frac{1}{3}\left(1+\frac{1}{n-1}\right)^{n-1}$$ for $n=2,3\cdots$
then $$a_{n}=a_1\cdot\prod_{k=2}^{k=n}\frac{a_{k}}{a_{k-1}}=a_1\cdot \prod_{k=2}^{k=n}\frac{1}{3}\left(1+\frac{1}{k-1}\right)^{k-1}.$$
Notice that $$\left(1+\frac{1}{k-1}\right)^{k-1}<e$$ for $k=2,3,\cdots$
Thu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\int_{0}^{\infty}\frac{dx}{(x+\sqrt{1+x^2})^{2n}}\cdot \frac{1}{1+x^2}$ $$F(n)=\large \int_{0}^{\infty}\frac{\mathrm dx}{(x+\sqrt{1+x^2})^{2n}}\cdot \frac{1}{1+x^2}$$
$\large x=\tan u$
$\large \mathrm dx=\sec^2 u\mathrm du$
$$F(n)=\large \int_{0}^{\pi/2}\frac{\mathrm du}{(\tan u+\sec u)^{2n}}$$
$\large \tan u... | Alternatively:
$$F(n)=\int_{0}^{\infty}\frac{dx}{(x+\sqrt{1+x^2})^{2n}}\cdot \frac{1}{1+x^2}=
\int_{0}^{\infty}\frac{(x-\sqrt{1+x^2})^{2n}}{1+x^2}dx$$
Change: $x-\sqrt{1+x^2}=t \Rightarrow x=\frac{1-t^2}{2t}, dx=-\frac{1+t^2}{2t^2}dt$, then:
$$F(n)= \int_{-1}^{0}\frac{t^{2n}}{\frac{(1+t^2)^2}{4t^2}}\cdot \left(-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
2 ways to find a Laurent series? A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 8.17,19,36 --> These exercises involve possibility of computing multiple Laurent series.
-
(Q1) For Exer 8.17, do we obtain multiple Laurent series depending on how we rewrit... | Note that isolated singularities of a rational function specify different regions of convergence when expanding the function around a point in a Laurent series. So, for each of these different regions there is a specific representation of $f$ as Laurent series.
Ad Q.1: Yes, we obtain two different Laurent series expan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
symmetric polynomial recursion to solve the system, $x^5+y^5=33$, $x+y=3$ I was just reading on symmetric polynomials and was given the system of equations$$x^5+y^5=33 \text{ , } x+y=3$$
In the text they said to denote $\sigma_1=x+y$ and $\sigma_2=xy$, and to use recursion. I understand how the formula below works
$$S... | Let $x+y=3=a$, $xy=b$, $x^5+y^5=33=c$.
\begin{align}
x^5+y^5&=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
\\
c&=
a(x^4+y^4-x^2xy+(xy)^2-xyy^2)
\\
c&=
x^4+y^4-b(x^2+y^2)+b^2
,\\
x^4+y^4&=(x^2+y^2)^2-2x^2y^2=(x^2+y^2)^2-2b^2
,\\
c&=
a((x^2+y^2)^2-2b^2-b(x^2+y^2)+b^2)
=
a(x^2+y^2)(x^2+y^2-b)-b^2)
,\\
x^2+y^2&=(x+y)^2-2xy=a^2-2b
,\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$ I am trying to calculate $$I=\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$$ So far I have done the following. First I tried to let $\sin x= t^2$ then:
$$I=2\int_0^1 \frac{x\arcsin x}{\sqrt{1-x^4}}dx =\int_0^1 (\arcsin^2 x)'\frac{x}{\sqrt{1+x^2}}dx $$
$$=\frac... | Mathematica gives:
$$\frac{1}{24} \left(-6 \text{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)+6
\text{Li}_2\left(3-2 \sqrt{2}\right)+4 \pi ^2-3 \log ^2(2)+3 \log
^2\left(\sqrt{2}-1\right)+3 \log ^2\left(3+2 \sqrt{2}\right)+ \log (64) \log
\left(\sqrt{2}-1\right)+6 \sinh ^{-1}(1)^2-12 \log \left(2
\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
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Fractions in Questions and Answers
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