Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find all irreducible characters of a matrix group on finite field $\mathbb F_5$ Find all irreducible characters of matrix group $G =\left\{ \left( \begin{array}{cc}
a & b \\0 & a^{-1}\end{array} \right)|\,\,\, a,b \in\mathbb F_5, a\not=0 \right\}$.
The former question is to find irreducible characters of subgroup of $G... | From Nicky's answer it is fairly straight forward to obtain the full table. Assuming the reader has calculated the conjugacy classes. We know that $C_4$ is a quotient so we can lift the nontrivial characters from it. We should determine which classes are mapped where. Since we know that the copy of $C_{10}$ in the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Solutions of $2^a+5^b=c^2$ I tried to solve this question using this way:
$a$ must be even, because if it's odd the equation have no solution. Let $a=2n$, so
\begin{align*}
2^{2n}+5^b &= c^2 \\
(2^n)^2+5^b &= c^2 \\
5^b &= c^2-(2^n)^2 \\
5^b &= (c-2^n)(c+2^n)
\end{align*}
Only one of $c-2^n$ and $c+2^n$ can be divided ... | Continue from your work, we solve $$1+2^{n+1} = 5^b$$
This special case of Catalan equation is easy: $2$ is a primitive root of $25$, hence a primitive root for any $5^b$. Therefore
$$2^{n+1} \equiv -1 \pmod{5^b}$$
implies $n+1$ is divisible by $\varphi(5^b)/2 = 2(5^{b-1})$, so $n+1 \geq 2(5^{b-1})$.
Hence $$5^b \geq 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Function Optimization with Non-linear Constraint, Lagrange Multipliers Fails I am trying to maximize the function $A(x,y)=\frac{1}{2}(x(12-x)+y(13-y))$ subject to the constraint $x^2+(12-x)^2-y^2-(13-y)^2=0$.
My attempt:
$\begin{align*} \nabla A=\frac{1}{2}\langle 12-2x,\,13-2y\rangle &= \lambda\langle4x-24,\, -4y+26\r... | First, let's re-explain why the first two equations are contradictory. The first equation gives us:
$$\lambda=\frac{-x+6}{4x-24}=-\frac 1 4$$
While the second equation gives us:
$$\lambda=\frac{-y+\frac{13} 2}{-4y+26}=\frac 1 4$$
However, what's important to realize here is that these two equations only work when $4x-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$I_{m,n}=\int\frac{x^m}{(ax^2+bx+c)^n}dx$ Reduction Formula I'm having trouble proving the following reduction formula:
If
$$I_{m,n}=\int\frac{x^m}{(ax^2+bx+c)^n}dx$$
then
$$\int\frac{x^m}{(ax^2+bx+c)^n}dx=-\frac{x^{m-1}}{a(2n-m-1)(ax^2+bx+c)^{n-1}}-\frac{b(n-m)}{a(2n-m-1)}I_{m-1,n}+\frac{c(m-1)}{a(2n-m-1)}I_{m-2,n}$$
... | Let $f(x)=\frac{x^{m-1}}{(ax^2+bx+c)^{n-1}}$ and differentiate
$$f’(x)
=\frac{a(m-2n-1)x^m+b(m-n)x^{m-1}+c(m-1)x^{m-2}}{(ax^2+bx+c)^{n}}
$$
Integrate both sides
$$f(x) = a(m-2n-1)I_{m,n} +b(m-n) I_{m-1,n} +c(m-1) I_{m-2,n}
$$
and then rearrange to obtain the reduction formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Algebraic Closed Form for $\sum_{n=1}^{k}\left( n- 3 \lfloor \frac{n-1}{3} \rfloor\right)$ Let's look at the following sequence:
$a_n=\left\{1,2,3,1,2,3,1,2,3,1,2,3,...\right\}$
I'm trying to calculate:
$$\sum_{n=1}^{k} a_n$$
Attempts:
I have a Closed Form for this sequence.
$$a_n=n- 3 \bigg\lfloor \frac{n-1}{3} ... | If you consider$$b_k=\sum_{n=1}^{k}\left( n- 3 \bigg\lfloor \frac{n-1}{3} \bigg\rfloor\right)$$ you could see, after some simple manipulationd that it corresponds to the recurrence relation
$$b_k=b_{k-1}+b_{k-3}-b_{k-4}\qquad \text{with}\qquad b_1=1, b_2=3, b_3=6, b_4=7$$ Using the conventional method you should end ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
show this inequality with $\sum_{i=1}^{n}a_{i}=n$
Let $n\ge 3$ be postive number,$a_{i}>0,i=1,2,\cdots,n$,and $\displaystyle\sum_{i=1}^{n}a_{i}=n$,show that
$$a^3_{1}a_{2}+a^3_{2}a_{3}+\cdots+a^3_{n}a_{1}+n\ge 2(a_{1}a_{2}\cdots a_{n-1}+a_{2}a_{3}\cdots a_{n}+a_{n}a_{1}\cdots a_{n-2})$$
it seem can use indution to... | A dull proof for $n<5$ by BW(Buffalo Way).
For $n=4$, it is enough to prove$$64(a^3b+b^3c+c^3d+d^3a)+(a+b+c+d)^2 \ge 32(a+b+c+d)(abc+abd+acd+bcd)$$and WLOG we can assume $a$ is smallest of the four numbers.
That is, $b=a+u$, $c=a+v$ and $d=a+w$ for some positive numbers $u,v,w$. Expanding gives
$$64a^2(3 u^2 - u v - 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Rolling $4$ dice and multiplying the results. What is the probability that the product is divisible by $5$ or has $5$ as the least significant digit? Four fair dice are rolled and the four numbers shown are multiplied together. What is the probability that this product
(a) Is divisible by $5$?
(b) Has last digit $5$?
M... | Your answer for part $(a)$ is correct. For part $(b)$, observe that a number $N$ has last digit $5$ iff it's divisible by $5$ and odd. This means that if we denote our rolls by $a,b,c,d$ then $N=abcd$ where at least one of $a,b,c,d$ is a $5$ and none of $a,b,c,d$ is even. This means we have to avoid $2,4,6$ on all roll... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Solving for $x$ in $y=N\cdot\left(\frac{10}{x}\right)^{-2.6}$ I want to confirm my solution of $x$ from
$$y=N\cdot\left(\frac{10}{x}\right)^{-2.6}$$
My answer is:
$$x=\frac{N^{2.6}}{10\cdot y^{2.6}}$$
Is this right? How would you solve?
| Let $$y = N \cdot (\frac{10}{x})^{-2.6}$$
Then,
$$y \cdot (\frac{10}{x})^{2.6} = N$$
$$ y \cdot 10 ^{2.6} = N \cdot x^{2.6}$$
$$ (y \cdot 10^{2.6})^{1/2.6} = (N \cdot x^{2.6})^{1/2.6}$$
$$ y^{1/2.6} \cdot 10 = N^{1/2.6} \cdot x$$
$$ x = y^{1/2.6} \cdot 10 \cdot N^{-1/2.6} = (\frac{y}{N})^{1/2.6} \cdot 10 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Suppose $a, b\in G$ such that $\lvert a\rvert$ is odd and $aba^{-1}=b^{-1}$. Show that $b^2=e$. I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.30.
Suppose $a$ and $b$ belong to a group, $a$ has odd order, and $aba^{-1}=b^{-1}$. Show that $b^2=e$.
My Attempt:
Let $\lvert a\rvert=2n+1$, $n\... | We have
$$\begin{align}
aba^{-1}=b^{-1}&\implies ab=b^{-1}a=x\\
&\implies (ab) \cdot (b^{-1}a)=x^2 \\
&\implies a^2=x^2.
\end{align}$$
Now:
$$\begin{align} x^{2m+1}&= (x^2)^m \cdot x\\ &= (a^2)^m \cdot (ab)\\ &= a^{2m+1}\cdot b\\ &=b.\end{align}$$
and
$$\begin{align} x^{2m+1}=b &\implies (x^{2m+1})^2=b^2 \\ &\implies (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
How to apply CRT to a congruence system with moduli not coprime? $x=1 \pmod 8$
$x=5 \pmod{12}$
8 and 12 are not coprime, I could break it to:
$x=1 \pmod 2$
$x=1 \pmod 4$
and
$x=5 \pmod 3$
$x=5 \pmod 4$
But what are the next steps to solve it? By the way, $x$ should be $17$ not sure how to get that number ...
Thanks in... | No point in figuring it out to any lesser power of $2$ than $2^3$.
Leave it at $x\equiv 1 \pmod 8$ but $x \equiv 5 \pmod {12}$ can be reduced to $x\equiv 5 \equiv 2 \pmod 3$.
So CRT says there is a unique solution $\pmod {28}$; $x \equiv 17 \pmod{24}$.
$x = A \pmod {n_1}$ $(A = 1; n_1 = 8)$
$x = B \pmod {n_2}$ $(B = 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3075823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Proof for area of an equilateral triangle with respect to one side? I'm trying to find out the area of an equilateral triangle with respect to one side. Anything wrong with my proof?
An equilateral triangle with sides of length $a$ can be divided in half along the base to create two triangles with right angles. The sid... | Your proof is fine, but in the spirit of visual geometry, let me present a cute alternative: (You might want to break out paper/pencil to follow along with the diagram, but maybe not since the figure is so simple.)
Draw a regular hexagon made from 6 equilateral triangles, then draw the equilateral triangle formed by ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integration by substitution to take out square root
Find $$\int (x+1)\sqrt{x^2+1}\,dx .$$
In order to not bother with the square root I thought of doing this:
$let$ $ x^2+1=(x+t)^2$ $\Rightarrow$ $x=\frac{1-t^2}{2t}$ $\Rightarrow x-t=\frac{1+t^2}{2t}$ $(1)$
$\Rightarrow$ $t=\sqrt{x^2+1}-x$ $(2)$
And
$dx=-\frac{t^2+1}... | Before trying the substitution, it is worth to simplify the integrand, by observing that
$$(x^2+1)'=2x$$ so that
$$\int(x+1)\sqrt{x^2+1}\,dx=\frac13(x^2+1)^{3/2}+\int\sqrt{x^2+1}\,dx.$$
For the last integral, substitution with $x=\sinh t$ will allow to get rid of the square root. By parts intégration also works:
$$I:=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Intuitively, why should I expect a circle in the complex plane from the equation $\left|\frac{z-1}{z+1}\right| = c$? I know how to prove that : ($c \in [0,1[$)
$$C = \{z \in \mathbb{C}: \left|\frac{z-1}{z+1}\right| = c \}$$
is circle in the complex plane. To do so we can for example write $z = x+iy$ and use the brute f... | Let's explore.
$$\bbox{
\left\lvert \frac{z - 1}{z + 1} \right\rvert = c
} \quad \iff \quad \bbox{
\frac{\lvert z - 1 \rvert}{\lvert z + 1 \rvert} = c
} \quad \iff \quad \bbox{
\left\lvert z - 1 \right\rvert = c \left\lvert z + 1 \right\rvert
} \tag{1}\label{NA1}$$
This only makes sense if $0 \lt c \in \mathbb{R}$.
Sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How to evaluate: $\lim\limits_{n\to\infty} \frac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$ How to evaluate: $$\lim_{n\to\infty} \dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$$
when
$i)$ $p\in\mathbb R,p\neq0$
$ii)\space p=0$
So for $i)$ I tried using Stolz–Cesàro theorem and Binomial theorem and If I didn't mess up I got $1$. But... | We can apply the Stolz-Cesàro theorem for positive real values $p, p\ne 1$
and consider other values of $p$ separately.
We consider for $p\in\mathbb{R}$:
\begin{align*}
\lim_{n\to \infty}\frac{1^{p-1}+2^{p-1}+\cdots+n^{p-1}}{n^p}\tag{1}
\end{align*}
Case $p>1, 0<p<1$:
If $p>1$ resp. $0<p<1$ the sequence $(n^p)_{n\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Finding the closed form of a generating function Given that $k$ is a positive integer and $f(x)$ is the generating function of the sequence $(b_0,b_1,b_2,...)$ where $b_n = {n \choose k}\;\, \forall \;n$, show that: $$f(x)=\frac{x^k}{(1-x)^{k+1}}$$
I tried writing a few terms of $f(x)$:$$f(x)=x^k+{k+1 \choose k}x^{k+1}... | HINT:
Inductive hypothesis
\begin{eqnarray*}
f_k(x)= \sum_{n=k}^{\infty} \binom{n}{k} x^n = \frac{x^k}{(1-x)^{k+1}}
\end{eqnarray*}
and use
\begin{eqnarray*}
\binom{n}{k} + \binom{n}{k+1}= \binom{n+1}{k+1}.
\end{eqnarray*}
More detail available on request.
Further Hint :
\begin{eqnarray*}
f_{k+1}(x) &=& \sum_{n=k+1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Idea for $\lim\limits_{x \to \frac{\pi}{2}} \left( \tan \left( \tfrac{\pi}{4} \sin x\right)\right)^{1/ ( \tan (\pi \sin x))}$ $$\lim_{x \to \frac{\pi}{2}} \tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right).$$
Need help for this. I tried using trigonometric identities but nothing seems to fi... | Alternatively:
$$\lim_{x \to \frac{\pi}{2}} \tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right)=\\
\lim_{x \to \frac{\pi}{2}} \left[1+\tan \left(\frac{\pi}{4}\sin x\right)-1\right]^\left({\dfrac 1{\tan(\pi \sin x)}}\right)=\\
\exp \left[\lim_{x \to \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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How to prove binomial coefficient $ {2^n \choose k} $ is even number? Prove:
${2^n \choose k}$ (for integers $k$ & $n$ : $0<k<2^n$) is even number.
I have tried induction but was unable to get any useful results.
| The following solution (copypasted from my old coursework) is purely
elementary number theory:
We set $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $.
Lemma 1. Let $n$ be an integer. Let $m$ be a positive integer. Then,
$\dbinom{n}{m}=\dfrac{n}{m}\cdot\dbinom{n-1}{m-1}$.
Proof of Lemma 1. We have
\begin{align*}
\dbino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
What is the dimension of $V$? Let $$A=\begin{pmatrix} 1 & 1& 1\\2 &2&3\\x&y&z\end{pmatrix}$$
and let $V=\{(x,y,z)\in \mathbb{R}^3: \det(A)=0\}$ Then the dimension of $V$ equals,
Efforts:
Applying row transformation $R_2\to R_2-2R_1$, we get $$A=\begin{pmatrix} 1 & 1& 1\\0 &0&1\\x&y&z\end{pmatrix}$$
Now expanding along ... | You should be more precise with the language: “informally speaking” is not enough.
You can go on with the row reduction:
$$
\begin{pmatrix} 1 & 1& 1\\2 &2&3\\x&y&z\end{pmatrix}
\xrightarrow{\begin{aligned}R_2&\gets R_2-2R_1 \\ R_3&\gets R_3-xR_1\end{aligned}}
\begin{pmatrix} 1 & 1& 1\\0 &0&1\\0&y-x&z-x\end{pmatrix}
\xr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove that $\sum_{k=0}^n 2^k \binom{n}{k} \binom{n-k}{\lfloor (n-k)/2 \rfloor}=\binom{2n+1}{n}$ Where the thing that looks like a floor function is the floor function. This is an interesting result which I hoped to prove by induction, but ran into trouble applying the inductive hypothesis. The base case is trivial. Her... | Seeking to verify
$$\sum_{k=0}^n 2^k {n\choose k}
{n-k\choose \lfloor (n-k)/2 \rfloor} = {2n+1\choose n}$$
we observe that
$${n\choose k}
{n-k\choose \lfloor (n-k)/2 \rfloor}
= \frac{n!}{k! \times \lfloor (n-k)/2 \rfloor!
\times (n-k-\lfloor (n-k)/2 \rfloor)!}
\\ = {n\choose \lfloor (n-k)/2 \rfloor}
{n-\lfloor (n-k)/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3089150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find solutions of the given equation in the form of power series Find solutions of the given equation in the form of power series
$$y'' +2xy' = 0$$
My approach:
$$2xy' = 2a_1x + 2\times2a_2x^2 + 2\times3a_3x^3 + ... = \sum 2ia_{i}x^{i}$$
$$y'' = 2a_2 + 3\times2a_3x + 4\times3a_4x^2 + ... = \sum (i+1)(i+2)a_{i+2}x^{i}$$... | You already did most of the job.
We continue
\begin{align*}
\color{blue}{a_{2n+1}}&= -\frac{2(2n-1)}{(2n+1)2n}a_{2n-1}\\
&=+\frac{2(2n-1)2(2n-3)}{(2n+1)2n(2n-1)(2n-2)}a_{2n-3}\\
&=\cdots\\
&=(-1)^n\frac{2^n(2n-1)!!}{(2n+1)!}a_1\tag{1}\\
&=(-1)^n\frac{2^n}{(2n+1)(2n)!!}a_1\tag{2}\\
&\,\,\color{blue}{=\frac{(-1)^n}{(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3089497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}, x \in \mathbb{R}$ if $\sqrt{49-x^2}-\sqrt{25-x^2}=3$ Suppose that real number $x$ satisfies $$\sqrt{49-x^2}-\sqrt{25-x^2}=3$$What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$?
This is what I did:
I try to multiply by the conjugate. Its value I believe is technically ... | Another way:
$$1) \ \sqrt{49-x^2}=\sqrt{25-x^2}+3 \Rightarrow \\
49-x^2=25-x^2+9+6\sqrt{25-x^2} \Rightarrow \\
2.5=\sqrt{25-x^2};\\
2) \ \sqrt{25-x^2}=\sqrt{49-x^2}-3 \Rightarrow \\
25-x^2=49-x^2+9-6\sqrt{49-x^2} \Rightarrow \\
\sqrt{49-x^2}=5.5.$$
Hence:
$$\sqrt{49-x^2}+\sqrt{25-x^2}=5.5+2.5=8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding an $\epsilon-\delta$ proof for a multivariable limit. Suppose $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is defined as $$(x,y) \longmapsto
\left\{
\begin{array}{cl}
\dfrac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2} & \mbox {if } (x,y) \neq (0,0) \\
\\
0 & \mbox {if } (x,y) = (0,0)
\end{array}
\right.
$$
I ha... | The essential point is that $r:=\sqrt{x^2+y^2}$ governs the distance from some point $(x,y)$ to $(0,0)$. In particular $|x|\leq \sqrt{x^2+y^2}=r$, and similarly $|y|\leq r$. It follows that
$$\left|{4x^2y^3+x^4y-y^5\over(x^2+y^2)^2}\right|\leq(4+1+1){r^5\over r^4}=6r\ .\tag{1}$$
Given an $\epsilon>0$ choose $\delta:={... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Reciprocal of $x_{n-1}^2(1-\frac{x_{n-1}^2}{3}+o(x_{n-1}^2))$ where $x_n\searrow0$
Suppose that $\{x_n\}$ is a decreasing sequence having limit $0$. How to find the reciprocal of $x_{n-1}^2(1-\frac{x_{n-1}^2}{3}+o(x_{n-1}^2))$ as $n\to\infty$.
In my textbook it says that
\begin{equation}\frac{1}{x_{n-1}^2(1-\frac{x_... | This comes out from the geometric series ($x\rightarrow 0$)
$$
\frac{1}{1-x}=1+x+O(x^2).
$$
In your case, you have to evaluate
$$
\frac{1}{x_n^2\left(1-\frac{x^2_n}{3}+O(x^4_n)\right)}
$$
Using the geometric series you will get
$$
\frac{1}{x_n^2}\left(1+\frac{x_n^2}{3}+O(x_n^4)\right)=\frac{1}{x_n^2}+\frac{1}{3}+O(x_n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{x^2}{x-1} \,dx$ Evaluate $\int \frac{x^2}{x-1} \,dx$
(A) $2x^2+x+\ln|x-1|+C$
(B) $\frac{x^2}{2}+x+\ln|x+1|+C$
(C) $\frac{x^2}{2}+x+\ln|x-1|+C$
(D) $x^2+x+\ln|x-1|+C$
My attempt :
Let $u=x-1$, so : $du=dx$ and $u+1$
$\int \frac{(u+1)^2}{u}\,du\\
=\int u +2+\frac{1}{u}\,du\\
=\frac{u^2}{2}+2u+\ln|u|+... | Your answer is right except for the constant I changed.
$$\frac{x^2-3}{2}+x+\ln|x-1|+C_1 = \frac{x^2}{2}+x+\ln|x-1|+C_1-\frac{3}{2} = \frac{x^2}{2}+x+\ln|x-1|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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$\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ if $a+b+c=ab+bc+ca$? For $a,b,c>0 $ and $a+b+c=ab+bc+ca$ . Prove or disprove that : $\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$
I checked in very many cases. Example :$c=1,
a=2,b=\frac{1}{2}...$ then it’s true, but cannot prove that
My attempts:
I conside... | What are the allowed values of $a,b,c$? If they are integers greater than zero then how about this:
Firstly $a = b = c = 1$ meets your constraint, and with these values $3\sqrt{24 +25} = 21$
By my assumption that they are positive integers any valid $a,b$ would satisfy $\sqrt{24ab + 25} \ge \sqrt{24 + 25} = 7$
And sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Differentiation uner the integral sign - help me find my mistake This is my integral:
$$I(a)=\int_0^\infty\frac {\ln(a^2+x^2)}{(b^2+x^2)}dx.$$
Taking the first derivative with respect to a:
$$I'(a)=\int_0^\infty \frac {2adx} {(a^2+x^2)(b^2+x^2)}.$$
This is how I did the partial fraction decomposition:
$\frac {2a} {(a^2... | $$I(a)=\int_0^\infty\frac{\ln(a^2+x^2)}{(b^2+x^2)}dx$$
$$I'(a)=\int_0^\infty\frac{2a}{(a^2+x^2)(b^2+x^2)}dx$$
now we want to look at simplifying this expression:
$$\frac{2a}{(a^2+x^2)(b^2+x^2)}$$
and we know (due to both factors being quadratic) that it will be of the form:
$$\frac{2a}{(a^2+x^2)(b^2+x^2)}=\frac{Ax+B}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3101376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Use the definition of limit to show that $\lim_{x \to -1} \frac{x+5}{2x+3}=4$
Use the definition of limit to show that
$\lim_{x \to -1} \frac{x+5}{2x+3}=4$
We have $|\frac{x+ 5}{2x+3} -4|= |\frac{x+5-8x-12}{2x+3}|=|\frac{-7x-7}{2x+3}|=\frac{7}{|2x+3|}|x+1|$
To get a bound on the coefficient of $|x+1|$, we restrict $x... | The neighborhood you have chosen is too large, it contains a root of the denominator. Let us try with the bounds
$$-1\pm\frac13$$ (any deviation smaller than $\dfrac12$ can do).
The function is monotonous in that range and the extreme values of $\left|\dfrac7{2x+3}\right|$ are
$$21, \frac{21}{5}$$ so you can write
$$\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
remainder in double sum having binomial coefficients find the remainder when
$\displaystyle \sum^{2014}_{r=0}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{2019}{r-k}$ is divided by $64$
what i try
$$\sum^{2014}_{r=0}\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{2019}{r-k}$$
$$\sum^{2014}_{r=0}(-1)^k\sum^{r}_{k=0}(k+1)(k+2)\binom{2019}{r... | Let us use the followings :
$$\sum^{r}_{k=0}(-1)^k(k+1)(k+2)\binom{n}{r-k}=\frac{-2(r-n)(r-n+1)(r-n+2)}{n(n-1)(n-2)}\binom{n}{r}\tag1$$
$$\sum_{r=0}^{n}r^3\binom nr=2^{n-3}n^2(n+3)\tag2$$
$$\sum_{r=0}^{n}r^2\binom nr=2^{n-2}n(n+1)\tag3$$
$$\sum_{r=0}^{n}r\binom nr=2^{n-1}n\tag4$$
$$\sum_{r=0}^{n}\binom nr=(1+1)^n=2^n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove by induction: $C(2n, 2) = 2C(n, 2) + n^2$
Show that if $n$ is a positive integer, then $C(2n, 2) = 2C(n, 2) + n^2$. Here, $C(a, b)$ means the binomial coefficient $\dbinom{a}{b}$.
Prove this by induction.
Here is my calculation:
$n$ cannot be $1$ because $n$ should be equal or larger than $2$ for $C(n, 2)$.
If ... | We assume for our inductive hypothesis that $\binom{2n}{2}=2\binom{n}{2}+n^2$ is true for all positive integers $n$ which are less than or equal to some $k$. We then try to prove that this will further imply it is true for $k+1$ as well.
$$\begin{array}{rl}\binom{2(k+1)}{2}&=\binom{2k+2}{2}=\binom{2k+1}{1}+\binom{2k+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Is it true that $\int_0^\infty \frac{f(x)}{x} \sin \bigl(\frac{\pi x}{a}\bigr) \,\mathrm{d}x = \frac{\pi}{2} \int_0^{a/2} f(x) \,\mathrm{d}x$? I dont recall where, but I found this interesting identity a few years ago. It was shown in one of Victor Moll's papers about elliptic integrals.
Corollary 3.1. Let $f$ be an e... | Corollary 3.1
An even function with period $a$ means $f(a-x)=f(x)$.
$$
\begin{align}
\int_0^\infty\frac{f(x)}x\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x
&=\sum_{k=0}^\infty\int_0^af(x)\left(\frac1{x+2ka}-\frac1{x+(2k+1)a}\right)\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag1\\
&=\sum_{k=0}^\infty\int_0^af(x)\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
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Better proof for $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ Prove this $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$
My attempt
Proof - by using [axiomdistributive] and [axiommulcommutative]:
$$\begin{split}
&(x+y)(x^2 - xy + y^2)\\
&= (x+y)x^2 - (x+y)xy + (x+y)y^2\\
&= (x^3+x^2y) - (x^2y+xy^2) + (xy^2+y^3)\\
&= x^3 + x... | Just let $k=-y$ to obtain $$x^3-y^3=(x-y)(x^2+xy+y^2)\iff x^3+k^3=(x+k)(x^2-xk+k^2)$$
Note that the same works whenever $n$ is odd. Given $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dots + xy^{n-2}+y^{n-1})$$
Let $k=-y$ and ...
See how it works? Try to explain why it doesn't work whenever $n$ is even!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
a,b,c are three real numbers where $a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}$. Now $abc$ = ? Here will the answer be a number? I want to know whether it is possible to get a real number (not an algebraic expression) as the product of $a$, $b$ and $c$.
I tried for a long time and this is what I got.
$$3abc =... | Hint
\begin{align}
a-b&= \frac1c-\frac 1b \\ &=\frac{b-c}{bc}\\ &=\frac{c-a}{abc^2} \\& =\frac{a-b}{a^2b^2c^2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate : $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
Evaluate: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
My solution:
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}} & = \lim_{x\to -\infty} \frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\\
& = \fra... | Don't forget that $\sqrt{x^2}=|x|$ and when $x<0$ ($x$ is going to negative infinity), $|x|$ is equivalent to $-x$:
$$
\frac{\sqrt{x^2}\sqrt{4-\frac{2}{x}}}{\sqrt[3]{x^3}\sqrt[3]{1+\frac{1}{x^3}}}=
\frac{|x|\sqrt{4-\frac{2}{x}}}{x\sqrt[3]{1+\frac{1}{x^3}}}=
\frac{-x\sqrt{4-\frac{2}{x}}}{x\sqrt[3]{1+\frac{1}{x^3}}}=\\
-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Evaluating $\int_{} \frac{xe^{2x}}{(1+2x)^2}dx$ via integration by parts $\int_{} \frac{xe^{2x}}{(1+2x)^2}dx$
I am having trouble picking the correct $u/dv$ before integrating by parts. I felt like L.I.A.T.E. did not really help me here...
This is what I tried, but it ended up with integration spiraling into an endless... | Hint:
Let $u=xe^{2x}$, $\text dv=\dfrac{\text dx}{(1+2x)^2}$.
Result:
$$\dfrac{\mathrm{e}^{2x}}{8x+4} + C$$
EDIT: More steps.
\begin{eqnarray*}
\int \frac{xe^{2x}}{(1+2x)^2} \ \text dx &=& \left|\begin{array}{2}
u=xe^{2x} & \text dv=\dfrac{\text d x}{(1+2x)^2} \\
\text du = (1+2x)e^{2x}\ \text dx & v=-\dfrac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solving $(x^2+4x+3)^x+(2x+4)^x=(x^2+4x+5)^x$ with $x\in(-1,\infty)$ I've been struggling for a few hours on the below pre-calculus olympiad equation to which I still don't have an answer:
$$(x^2+4x+3)^x+(2x+4)^x=(x^2+4x+5)^x$$
where $x \in (-1,\infty)$.
Now, according to WolframAlpha, this has a unique solution, $x... | First, the trivial solution is $x=2$.
Then, consider the fact $$(x^2+4x+3)^2+(2x+4)^2=(x^2+4x+5)^2 \tag{1}$$
And the fact $\dfrac{x^2+4x+3}{x^2+4x+5}$ and $\dfrac{2x+4}{x^2+4x+5}$ are both in $(0,1)$ if $x \in (-1,\infty)$. (I believe you can figure this out yourself.)
Consider that $a^x$ is strictly decreasing if $a\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Integral problem. Unsure of the approach. I have this integral:
$$\int_0^1 \frac{1 + 12t}{1 + 3t}dt$$
I can split this up into:
$$\int_0^1 \frac{1}{1+3t} dt + \int_0^1\frac{12t}{1+3t}dt$$
The left side:
$u = 1+3t$ and $du = 3dt$ and $\frac{du}{3} = dt$
so $$\frac{1}{3} \int_0^1 \frac{1}{u} du = \frac{1}{3} \ln |u| + C... | My first attempt is to split the integral in another way:
$$\int_0^1{\frac{1+12t}{1+3t}}dt = \int_0^1{\frac{1+3t+9t}{1+3t}} dt = \int_0^1 1 \, dt + \int_0^1 \frac{9t}{1+3t} dt$$
Now observe that
$$\int_0^1 \frac{9t}{1+3t} dt = \int_0^1{\frac{3t}{1+3t}} \cdot 3dt = \int_0^1{\frac{3t}{1+3t}}d(3t) = \int_0^3 \frac{u}{1+u}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
The intersection of a sphere and the xy-plane is the circle $(x-1)^2+(y-2)^2=5$ and point $(1,2,1)$ is on the sphere The intersection of a sphere and the $xy$-plane is the circle $$(x-1)^2+(y-2)^2=5$$ and point $(1,2,1)$ is on the sphere. Find the center and radius of the sphere.
When the sphere and $xy$-plane intersec... | One more :
Equation of the sphere:
$(x-a)^2+(y-b)^2+(z-c)^2=R^2.$
Cut with plane $z=0:$
$(x-a)^2+(y-b)^2 +c^2=R^2.$
1) Compare with the given circle:
$a=1,b=2, R^2-c^2= (√5)^2$
2) $(1,2,1)$ lies on the sphere:
$0+0+(1-c)^2=R^2.$
Hence: $1-2c+c^2=R^2;$
$1-2c= R^2-c^2=(√5)^2.$
$c= (1-5)/2=-2$, and $R^2 =9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
prove that the triangle is isosceles
In a $\triangle ABC$, If
$\begin{vmatrix}
1 & \;\;1\;\;&\;\; 1\;\;\\\\
\displaystyle \cot \frac{A}{2} & \displaystyle \cot \frac{B}{2} & \displaystyle \cot \frac{C}{2} \\\\
\displaystyle \tan\frac{B}{2}+\tan \frac{C}{2} &\;\;\displaystyle \tan \frac{C}{2}+\tan\frac{A}{2} & \... | I have got this here for your determinant: $$\tan \left(\frac{A}{2}\right) \cot \left(\frac{B}{2}\right)-\cot \left(\frac{A}{2}\right)
\tan \left(\frac{B}{2}\right)-\tan \left(\frac{A}{2}\right) \cot
\left(\frac{C}{2}\right)+\cot \left(\frac{A}{2}\right) \tan \left(\frac{C}{2}\right)+\tan
\left(\frac{B}{2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove that $\ x_{n}=\int_{n}^{2n} \frac{x+a}{x^{3}+2a}dx$ is decreasing Let $\ a>0$ and the sequence $(x_{n})_{n>=0}$ defined by $\ x_{n}=\int_{n}^{2n} \frac{x+a}{x^{3}+2a}dx$. Prove the sequence is monotonically decreasing and $0<x_n<\frac{4+3a}{8}$, for any $n>0$. I've made little progress towards proving that $x_{n}... | Let $x_n=\int_{n}^{2n}\frac{x+a}{x^3+2a}\,dx$. Then, the first difference $x_{n+1}-x_n$ is given by
$$\begin{align}
x_{n+1}-x_n&=\int_{n+1}^{2n+2}\frac{x+a}{x^3+2a}\,dx-\int_{n}^{2n}\frac{x+a}{x^3+2a}\,dx\\\\
&=\int_{2n}^{2n+2}\frac{x+a}{x^3+2a}\,dx-\int_n^{n+1}\frac{x+a}{x^3+2a}\,dx\\\\
&=\int_n^{n+1}\left(\frac{4x+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3127499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Coefficients of $1,x,x^2$ in $((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
Finding coefficients of $x^0,x^1,x^2$ in
$((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
where there are $k$ parenthesis in the left side
Try:
Let $P(x)=((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
Let we assume
$P(x)=a_{k}+b_{k}x+c_{k}x^2+d_{k}x^3+\cdots\c... | Lets consider the cases $k=1,2,3,4$:$$P_1(x)=(x-2)^2=x^2-4x+4,\\ P_2(x)=(P_1(x)-2)^2=P_1(x)^2-2P_1(x)+4=x^4-8x^3+20x^2-16x+4,\\ P_3(x)=(P_2(x)-2)^2=x^8+\dots+336x^2-64x+4,\\ P_4(x)=x^{16}+\dots+5440x^2-256x+4.$$
From this it is reasonable to conjecture that the coefficient of $x$ in $P_k(x)$ is $4^k$. Moreover, a quick... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Prove that the product of the $n$ roots of $e^{-i\cdot \frac{\pi }{2}}$ is given by $e^{i\cdot \frac{\left(2n-3\right)\pi }{2}}$ Let $w=e^{-i\cdot \frac{\pi }{2}}$.
Here's how to tried to solve this:
$\sqrt[n]{w}=e^{i\cdot \left(-\frac{\pi }{2n}+\frac{2\pi }{n}k\right)},\:k\in \left(0;\:1;\:2;\:...;\:n-1\right)$
Fact... | Your approach seems correct, except for a small mistake in the "product". I have solved it as follows.
The $n$ roots of $e^{- \iota \frac{\pi}{2}}$ are given by $e^{\iota \left( - \frac{\pi}{2n} + \frac{2 \pi k}{n} \right)}$ for $k = 0, 1, 2, \cdots, n - 1$
\begin{align*}
\prod\limits_{k = 0}^{n - 1} e^{\iota \left( - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Condition when angle between two lines is ${\pi\over 3}$ The whole question is-
Show that if the angle between the lines whose direction cosines are
given $l+m+n=0$ and $fmn+gnl+hlm=0$ is ${\pi\over 3}$ then ${1\over
f}+{1\over g}+{1\over h}=0$.
I'm trying to solve the problem in the following manner-
From the fir... | In general if a, b, c and d, e, f are components of 2 vectors then the direction cosines are for the first vector :
$\frac{a}{\sqrt{a^2 + b^2 + c^2}}$ and $\frac{b}{\sqrt{a^2 + b^2 + c^2}}$ and $\frac{c}{\sqrt{a^2 + b^2 + c^2}}$ similarly for the second vector.
Then the angle $\theta$ between these vectors is
$$cos\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3129344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Finding lengths when circles and squares tangents.
Should one approach by coordinates or by euclidean geometry?
By pure geometry, I am not able to solve.
|
As we see in the above diagram, $ABCD$ is a square and two semi circles with its center $E$ and $F$ respectively. Let place the point $Q$ in such that $\triangle QEF$ is a right angled triangle and draw two altitude lines $MH$ and $GL$ from two vertices $H$ and $G$ respectively.
Again, denote the side of the sqaure = ... | {
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"url": "https://math.stackexchange.com/questions/3129843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}$ = $21\sqrt{6}$ but I get $207\sqrt{6}$ I'm asked to simplify $6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}$
The provided solution is $21\sqrt{6}$ but I arrive at a different amount.
Here is my working, trying to understand where I went wrong:
First expression:
$6\sqrt{24}$ = $6\sqrt{4}$ *... | Notice while simplifying the radicals you are multiplying 6,7 twice...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving a system of equations with trig How do you solve a system of the following form:
$$a = 2\sin(x) - \sin(y) + \sin(x+y)\tag1$$
$$b = 2\sin(y) - \sin(x) + \sin(x+y)\tag2$$
where $a,b$ are constants, and $x,y$ the variables I'd like to solve for.
Subtracting $(1)-(2)$ gives an expression for $\sin(y)$. However, re... | You are on the right track. Subtracting $(2)$ from $(1)$ gives $\sin x=c+\sin y$ where $c=\frac{a-b}3$ and putting this into $(1)$ gives $$a=2c+\sin y+\sin x\cos y+\sin y\cos x$$ so $$\cos x=-1-\cos y+\frac{b+c-c\cos y}{\sin y}$$ Thus $$\small \sin^2x+\cos^2x\\=\\\small c^2+2c\sin y+\sin^2y+1+\cos^2y+\left(\frac{b+c-c\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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What is the maximum value of $n$ if $4^n$ divides $1000!$ without a remainder?
If $1000!$ is divided by $4^n$ with a remainder 0, what is the highest
possible value of $n$?
I placed 2, 3, 4, etc value in $n$ but didn't found any possible $4^n$. Moreover I have seen that only $4^1$ can divide 1000! without remainder... | For a solution by hand: $1000!$ has
$500$ multiples of $2^{1}$
$250$ multiples of $2^{2}$
$125$ multiples of $2^{3}$
$62$ multiples of $2^{4}$
$31$ multiples of $2^{5}$
$15$ multiples of $2^{6}$
$7$ multiples of $2^{7}$
$3$ multiples of $2^{8}$
$1$ multiples of $2^{9}$
So $2^{1+3+7+15+...+500}=2^{994}=4^{497}|1000!$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
... | The probability generating function of a Binomiall random variable $X\sim \text{Bin}(n, 1/2)$ with probability of success $1/2$ is given by
$$
g_{X}(t)=Et^X=\sum_{k=0}^nP(X=k)t^k=\sum_{k=0}^n\binom{n}{k}\frac{t^k}{2^n}=\frac{1}{2^n}(1+t)^n
$$
In particular the probability that $X$ is even is given by
$$
\sum_{0\leq k\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
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"answer_id": 1
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Finding the sum of first $N$ squares with generating functions I'm trying to prove the well-known result
$$\sum_1^Nn^2=\frac{N(N+1)(2N+1)}6$$
using generating functions. The obvious choice is to set $f(x)=\sum_{n=1}^N n^2x^n$, so that we have $$f(x)=(xD)^2\sum_1^N 1x^n=(xD)^2\left(\frac{x^{N+1}-1}{x-1}\right).$$
It see... | \begin{align}
f(x)
&=(xD)^2\left(\frac{x^{N+1}-1}{x-1}\right)\\
&=(xD)(xD)\left(\frac{x^{N+1}-1}{x-1}\right)\\
&=x\left(xD\left(\frac{x^{N+1}-1}{x-1}\right)+D^2\left(\frac{x^{N+1}-1}{x-1}\right)\right)\\
&=x^2D\left(\frac{x^{N+1}-1}{x-1}\right)+xD^2\left(\frac{x^{N+1}-1}{x-1}\right)
\end{align}
If $t=x-1$:
\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the sum $S_n = \sum\limits_{k=1}^{\infty}\left\lfloor \frac{n}{2^k} + \frac{1}{2}\right\rfloor $ I am doing tasks from Concrete Mathematics by Knuth, Graham, Patashnik for trainning, but there are a lot of really tricky sums like that:
Calculate sum $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \f... | Since $S_1=1$ try to prove that $S_n=n$ by induction. Note that if $n=2m$ is even
$$\begin{align*}
S_n=\sum_{k=1}^{\infty}\left\lfloor\frac{n}{2^k}+\frac12\right\rfloor
&=\sum_{k=1}^{\infty}\left\lfloor\frac{2m}{2^k}+\frac12\right\rfloor
=\sum_{k=1}^{\infty}\left\lfloor\frac{m}{2^{k-1}}+\frac12\right\rfloor\\
&=\left\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
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Finding the general formula of $a_{n+1}=\frac{a_n^2+4}{a_{n-1}}$ with $a_1=1$ and $a_2=5$
Find the general formula of $a_{n+1}=\dfrac{a_n^2+4}{a_{n-1}}$ with $a_1=1$, $a_2=5$.
I have tried to write the recursion as a product, make summations, tried to look at patterns but its value grows very fast: $1,5,29,169,985,57... | If we start from $a_{n+1}^2+(a_n^2+4) = a_n^2+(a_{n+1}^2+4)$ we will get:
$$a_{n+1}^2+(a_n^2+4) = a_n^2+(a_{n+1}^2+4)\\ \Leftrightarrow\\ a_{n+1}^2+a_{n-1}a_{n+1} = a_n^2+a_{n}a_{n+2}\\ \Leftrightarrow\\ \frac{a_{n-1}+a_{n+1}}{a_n} = \frac{a_{n}+a_{n+2}}{a_{n+1}}$$
Therefore we have $\frac{a_{n-1}+a_{n+1}}{a_n} = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the least possible value of perimeter of $\triangle ABC$ with given ranges of angle
In $\triangle ABC$,$\angle A >2\angle B$ and $\angle C > 90^\circ$. If the length of all side of triangle $\triangle ABC$ are positive integers, then what is the least possible value of perimeter of $\triangle ABC$?
However, I... | In the standard notation we have:
$$a+b>c,$$
$$c^2>a^2+b^2$$ and
$$\frac{\sin\beta}{b}=\frac{\sin\alpha}{a}>\frac{\sin2\beta}{a}=\frac{2\sin\beta\cos\beta}{a},$$ which gives
$$\frac{a}{b}>\frac{a^2+c^2-b^2}{ac}$$ or
$$a^2(c-b)>b(c^2-b^2)$$ or
$$a^2>bc+b^2.$$
If $b=1$ we obtain $$c<a+1$$ or
$$c-a<1,$$ which is impossib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I find ordered pair, given slope of the tangent line? The function is $f(x) = x^3 + 9x^2 + 36x + 10$ and the slope given is $9$.
I found the derivative and set it equal to $9$, but I ended up with $x = (-9,-33)$ and the answer is $(-3,-44)$.
I've asked two Math majors and neither knows how to find it.
Where did ... | You got $3x^2+18x=-27.$
This is equivalent to $3x^2+18x+27=0$ or $x^2+6x+9=0$ or $(x+3)^2=0$.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In how many ways can $4$ boys and $6$ girls be put into $3$ groups of $3$ people so that there is a boy in each group? In how many ways can $4$ boys and $6$ girls be put into $3$ groups of $3$ people so that there is a boy in each group?
I tried by inclusion exclusion principle where: $$ A_{i} = \left\{\text{there is a... | I would condition on whether the one left out is a boy or a girl. If the one left out is a boy, there is one boy in each group. You can choose the first group in $4{6 \choose 2}$ ways, the second in $3{4 \choose 2}$ and the third in $2{2 \choose 2}$ ways, giving $$4{6\choose 2}3{4\choose 2}2{2 \choose 2}=2160$$
If a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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A logarithmic integral, generalization of a result of Shalev As many of you are already aware, I and Marco Cantarini are currently working on the applications of fractional operators to hypergeometric series, extending the class of $\phantom{}_{p+1} F_p$s whose closed form is provided by FL-expansions (like the ones ap... | I'll use the ingenious method of user @FDP from
the IntegralsAndSeries forum.
The idea is to make the following chain of substitutions:
$$ x = \frac{ 1 + \sqrt{2} \cos y }{ 1 - \sqrt{2} \cos y} $$
$$ z = \frac{y}{2} - \frac{\pi}{8} $$
$$ t = \tan z $$
Together with the following observations:
$$ \frac{ 1 + \sqrt{2} \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Is $n^2+3n+6$ divisible by 25, where $n$ is a integer? If we want $n^2+3n+6$ to be divisible by $25$, it firstly has to be divisible by $5$.
So, let's take a look at couple of cases:
$n=5k: 25k^2+15k+6$. The remainder is $6$, so it's not divisible by $25$.
$n=5k+1: 25k^2+25k+10$. The remainder is $10$, so it is not div... | If $25$ divides $n^2+3n+6,$ the later will be divisible by $5$
$n^2+3n+6\equiv n^2-2n+1\pmod5$
So, we need $n\equiv1\pmod5\implies n=1+5m$
$(5m+1)^2+3(5m+1)+6=25m^2+25m+10\not\equiv0\pmod{25}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove closed form for $\sum_{n\in\Bbb N}\frac1{5n(5n-1)}$ While looking for solutions to a difficult geometric problem, I encountered this sum:
$$
\sum_{n\in\Bbb N}\frac1{5n(5n-1)} = \frac1{4\cdot 5} + \frac1{9\cdot 10} + \frac1{14\cdot 15} + \ldots
$$
A bit of numerical exploring has convinced me that the answer is
$$... | Start with $$f(x) = \sum_{n=2}^\infty \frac{x^n}{n(n-1)} = x + (1-x) \log (1-x)$$ If $w$ is a primitive $5$-th root of unity then $$\frac{1}{5}\sum_{k=0}^4 w^{nk} = 1$$ if $5 | n$ and $0$ otherwise. Thus
$$\frac{1}{5}\sum_{k=0}^4 f(w^k) =\sum_{n> 2,\, 5 | n} \frac{1}{n(n-1)} = \sum_{n=1}^\infty \frac{1}{5n(5n-1)}$$
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
If rank of a given matrix of order $3 \times 4$ is $2$ then the value of $b$ is Q) Suppose the rank of the matrix
$\begin{pmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{pmatrix}$
is $2$ for some real numbers $a$ and $b$. Then $b$ equals
$(A)$ $1\;\;\;$ $(B)$ $3\;\;\;$ $(C)$ $1/2\;\;\;$ $(D)$ $1/3\;\;\;$
My Approach... | After getting$$\begin{bmatrix}1&1&2&1\\0&0&-1&1\\0&b-a&b-2a&1-2a\end{bmatrix},$$you can't just say (as you did) that the all the elements of the last row must be zero simultaneously. At this point, you should consider the cases $b=a$ and $b\neq a$. In the first case, you now have the matrix$$\begin{bmatrix}1&1&2&1\\0&0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3144398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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What is the value of $\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1}$? $$\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = ?$$
I have done these steps to find the answer:
*
*$x^2-1=(x+1)(x-1)$
*$\sqrt{4x-4}=2\sqrt{x-1}$
*$\displaystyle\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}... | The expression is not of the form $0/0$ notice carefully the denominator goes to $0$ where the numerator is finite.
$$\lim_{x\to 1^+}\dfrac{\sqrt{x^2-1}+x+1}{\sqrt{4x-4}+x^2-1}\to \infty$$
This graph confirms the result: https://www.desmos.com/calculator/ejj9xiyjcf
After the OP's edit: $$\dfrac{\sqrt{x^2-1}+x-1}{2\sqr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find values of $a$ such that $x^2+ax+a^2+6a \lt 0$ $\forall$ $x \in (1,2)$ Find values of $a$ such that $x^2+ax+a^2+6a \lt 0$ $\forall$ $x \in (1,2)$
My try: Since $y=x^2+ax+a^2+6a$ is an open upward Parabola, the roots $\alpha,\beta$ should be distinct and satisfy $1 \lt \alpha \lt 2$ and $1 \lt \beta \lt 2$ implies b... | Hint: if you reverse the role of $a$ and $x$ you get $$a^2+a(x+6)+x^2<0$$
Solution of coresponding equation (on $a$) is $$a_{1,2} = {-x-6\pm \sqrt{D}\over 2}$$
where $D= -3(x-6)(x+2)>0$ for $x\in(1,2)$. So $$a_1<a<a_2$$ and now you have to find the maximum and minimum of expression(s):
$${-x-6\pm \sqrt{D}\over 2}$$ whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$\sum_{k=0}^\infty\frac{ (-1)^{\left\lfloor\frac kn\right\rfloor}}{k+1}$ While testing a program I came across an interesting equality:
$$
\sum_{k=0}^\infty\frac{ (-1)^{\left\lfloor\frac kn\right\rfloor}}{k+1}
=\frac1n\left[\log2+\frac\pi2\sum_{k=1}^{n-1}\tan\frac{k\pi}{2n}\right],
$$
which generalises the well-known i... | For $n \in \mathbb{N}$ write $k = m n + l - 1$ with $m \in \mathbb{N}_0$ and $l \in \{1,\dots,n\}$. Then $\left\lfloor \frac{k}{n} \right\rfloor = m$, so
\begin{align}
S_n &\equiv \sum \limits_{k=0}^\infty \frac{(-1)^{\left\lfloor \frac{k}{n} \right\rfloor}}{k+1} \stackrel{1.}{=} \sum \limits_{l=1}^n \sum \limits_{m=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3153111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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Prove that $\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$ Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
$$\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$$
I write
$$a^2+b^2+c^2+d^2=16-2\left(ab+cd+\left(a+b\right)\left(c+d\right)\right)$$. Then the inequality is equival... | Another way.
$$\frac{1}{ab}+\frac{1}{cd}-\frac{a^2+b^2+c^2+d^2}{2}=$$
$$=\left(\frac{1}{\sqrt{ab}}-\sqrt{ab}\right)^2+\left(\frac{1}{\sqrt{cd}}-\sqrt{cd}\right)^2+4-\left(ab+cd+\frac{a^2+b^2+c^2+d^2}{2}\right)=$$
$$=\left(\frac{1}{\sqrt{ab}}-\sqrt{ab}\right)^2+\left(\frac{1}{\sqrt{cd}}-\sqrt{cd}\right)^2+\frac{(a+b+c+d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$ Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$
I know that final answer is 377, but how?
Edit:
Drawing f... | This is $f(6)$ where $f(n)$ for integers $n\ge0$ is the number of lattice
points in the polyhedron $nP$. This is the $n$-fold dilate of $P$, the polyhedron with vertices $(\pm1,0,0)$, $(0,\pm1,0)$ and $(0,0,\pm1)$. By Ehrhart's theorem, $f$ is a degree $3$ polynomial in $n$. Moreover, its leading coefficient is the vol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Sum of all real numbers $x$ such that $(\text{A quadratic})^\text{Another quadratic}=1$.
What is the sum of all real numbers $x$ such that
$(x^2-5x+5)^{(x^2-7x+12)}=1$?
So I know that $x^0=1$ and $1^x=1$. So, I can solve for them and find $x$, and add them up.
Solving $x^2-7x+12=0$ for $x^0=1$ gives $x=3, 4$.
Solv... | Hint
You have missed $$(-1)^{\text{even number}}=1$$
Now if $x^2-5x+5=-1,x=?$
Which values of $x$ make the exponent even?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $\frac{a-b}{\sqrt{1+a^2}\sqrt{1+b^2}} < \arctan{a}- \arctan{b}$ when $0$$\frac{a-b}{\sqrt{1+a^2}\cdot\sqrt{1+b^2}} < \arctan{a}- \arctan{b}$$ when $0<b<a$
This might relate to the mean value theorem, but I just can't prove it.
This question was put on hold as off-topic, I couldn't understand the reas... | You can write right side as
$$ \arctan a - \arctan b = \int_b^a \frac{1}{1+x^2}dx $$
and left side as
$$ \frac{a-b}{\sqrt{1+a^2}\sqrt{1+b^2}} = \int_b^a \frac{\partial}{\partial x}\left(\frac{x-b}{\sqrt{1+x^2}\sqrt{1+b^2}}\right) dx = \int_b^a \frac{xb+1}{(1+x^2)^\frac32\sqrt{1+b^2}} dx$$
Therefore the inequality you w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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A box has $4$ red and $20$ white balls. A person takes $10$ balls. What is the probability that all or none of the red balls were taken? A box has $24$ balls, $4$ red and $20$ white. One person takes $10$ balls and the second the remaining $14$. What is the probability that one of the two people picked up the $4$ red o... | Because $\tfrac{\binom 44\binom{20}6\binom{14}{14}}{\binom{24}{10}\binom{14}{14}}$ is the probability that the first person select 4 red and 6 white and the second person receives the remaining 14 white, while $\tfrac{\binom{20}{10}\binom{4}{4}\binom{10}{10}}{\binom{24}{10}\binom{14}{14}}$ is the probability that the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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series sum $\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots $ Finding value of $\displaystyle \int^{\pi}_{0}\ln(1+k\cos x)dx$ for $0<k<1$
what I try
Let $\displaystyle I =\int^{\pi}_{0}\ln(1+k\cos x)dx$
put $\displaystyle x\rightarrow ... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Evaluate $\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$ By accident, I find this summation when I pursue the particular value of $-\operatorname{Li_2}(\tfrac1{2})$, which equals to integral $\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x}$.
Notice this observation
$$\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x... | Here is an approach that avoids knowing the value of $\operatorname{Li}_2 (\frac{1}{2})$.
Let
$$S = \sum_{n = 1}^\infty \frac{1}{n} \left (H_{2n} - H_n - \ln 2 \right ).$$
Observing that
$$\int_0^1 \frac{x^{2n}}{1 + x} \, dx = H_n - H_{2n} + \ln 2,$$
your sum can be rewritten as
\begin{align}
S &= -\int_0^1 \frac{1}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Number of solutions to the equation $x_1 + x_2 + \ldots + x_n = k$ when $0 \leq x_i \leq m$ and $m + 1 \leq k \leq 2m + 1$ The problem:
Find the number of solutions to the equation: $x_1+x_2+...+x_n=k$ when $0\leq x_i\leq m$ and $m+1\leq k\leq 2m+1$.
The answer in the book is ${n+k-1\choose k}-{n\choose 1}{n+k-(m+1)-1\... | A solution of the equation in the nonnegative integers
$$x_1 + x_2 + \ldots + x_n = k \tag{1}$$
corresponds to the placement of $n - 1$ addition signs in a row of $k$ ones. For instance, if $n = 4$ and $k = 10$,
$$1 1 + + 1 1 1 1 1 + 1 1 1$$
corresponds to the solution $x_1 = 2$, $x_2 = 0$, $x_3 = 5$, $x_4 = 3$. The... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the average value of $f$ on the given interval $f(x) = (x-3)^2$, $[2,5]$ After finding the average value of $f$, I need to find $c$ such that $f_\text{avg}= f(c)$
$$f(x) = (x-3)^2 \textrm{, }[2,5]$$
$$\begin{align}
f_\text{avg} &= \frac{1}{5-2} \int_{2}^{5} (x-3)^2dx \\
u = x-3, du = dx & ~~~~~ \frac{1}{3} \int_{2... | Just the comments have pointed out, you miscalculated the last line. It should be
$$\frac{1}{3}\cdot\Big(\frac{8}{3}+\frac{1}{3}\Big)=\frac{1}{3}\cdot\frac{9}{3}=1\neq3$$
To solve the second part, it should be an easy substitution
$$(x-3)^2=1 \iff x-3=\pm1 \iff x=4,2$$
| {
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show that if $\frac{2(m^2+n^2+n)+1}{m(2n+1)}=k$, k is the multiple of 3 $$\frac{2(m^2+n^2+n)+1}{(2n+1)m}=k$$
where k, m, and n are all integer.
show that $3|k$
| Write it as
$$
2(m^2+n^2+n)+1=k(m(2n+1))
$$
Compute both sides mod $3$, that is, for $m,n=0,1,2$.
It turns out that there are two cases:
*
*The LHS is not zero and the RHS is zero. This cannot happen.
*The LHS is zero and the RHS is not zero. This implies that $k \equiv 0 \bmod 3$.
$$
\begin{array}{cccc}
m & n & 2(... | {
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Understanding a proof that, if $xy$ divides $x^2+y^2+1$ for positive integers $x$ and $y$, then $x^2+y^2+1=3xy$ This is a Worked Example from Brilliant.org's entry on Vieta Root Jumping.
Let $x$ and $y$ be positive integers such that $xy$ divides $x^2+y^2+1$. Prove that $$x^2+y^2+1=3xy$$
The solution proposes that $x... | The proof you've given is incomplete.
You should start assuming $k \neq 3$, so you get $x \neq y$ (otherwise $k=3$ which is absurd).
Then, we have
$x \geq y+1$,
$x^2 \geq (y+1)^2 > y^2+1$,
$x>\frac{y^2+1}{x}=t_2$.
So $(t_2, y)$ is another solution to $\frac{x^2+y^2+1}{xy}=k$ with $k \neq 3$. But $t_2+y<x+y$, which cont... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\cos^{2}\frac{\pi}{2n} \cos^{2}\frac{3\pi}{2n}\cdots \cos^{2}\frac{\left(n-2\right)\pi}{2n}=n2^{1-n}$ using complex numbers Using $$z^{2n}+1 = \prod_{k=1}^{n}\left(z^2-2z\cos{\left(\frac{\left(2k-1\right)\pi}{2n}\right)}+1\right)$$
and $$z^{2n}+1 = \left(1+z^2\right)\left(1-z^2+z^4-\cdots+z^{2n-2}\right)$$
pro... | There is no problem; you can view the identity as an equality of polynomials in $\Bbb{C}[z]$. This is an integral domain, meaning that you can cancel non-zero factors; the factor $1+z^2$ is a non-zero polynomial, so you can cancel it.
This gives you a simpler equality of polynomials. You can then evaluate these two po... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Degree of splitting field of $X^4+2X^2+2$ over $\mathbf{Q}$
Find the degree of splitting field of $f=X^4+2X^2+2$ over $\mathbf{Q}$.
By Eisenstein, $f$ is irreducible. By setting $Y=X^2$, we can solve for the roots: $Y=-1\pm i \iff X=\sqrt[4]{2}e^{a\pi i/8}$, $a\in\{3,5,11,13\}$. Clearly $f$ splits in $\mathbf{Q}(\sqr... | Let $\Omega$ be the OP's splitting field.
We have a chain of 3 field extensions,
$$\tag 1 \mathbb Q \subset \mathbb Q(\sqrt 2) \subset Q(\sqrt 2)\,(\sqrt{1 + \sqrt 2}) \subset Q(\sqrt 2)\,(\sqrt{1 + \sqrt 2})\,(i) = \mathbb F$$
It is easy to explain why the first and last extensions have degree two. For the $2^\text{n... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Exercise XIX number 15 - Calculus Made Easy $$
\text{Use substitution}\quad\frac{1}{x}=\frac{b}{a}cosh\;u\quad\text{to show that}\quad\\
\int\;\frac{dx}{x\sqrt{a^2-b^2x^2}}=\frac{1}{a}\ln\frac{a-\sqrt{a^2-b^2x^2}}{x}\;+\;C.\\
\text{My way:}\\
\text{Let}\quad\frac{1}{x}=\frac{b}{a}cosh\;u\quad\\
x=\frac{a}{b}\frac{1}{co... | You only have a minor error: $\cosh^-(\frac{a}{bx})= \ln\left(\frac{a}{bx} + \sqrt{\frac{a^2}{b^2x^2} - 1}\right)$, the minus-sign is not correct. This gives you
$$\int\;\frac{dx}{x\sqrt{a^2-b^2x^2}}\quad = \quad\frac{1}{a}\ln\frac{bx}{a+\sqrt{a^2-b^2x^2}}+ C \\ = \quad \frac{1}{a}\ln\frac{b^2x}{a+\sqrt{a^2-b^2x^2}} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$ then $y=2^k$ and $x=1$ Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. Is it necessarily the case that $x=1$ and $y=2^k$?
Equivalently (I prove equivalence at the end): Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two posit... | I've finally solved it!
My solution to this question proves a generalized version.
To deduce this special case, set $z=2^k$, note that $z$ is divisible by $4$, and use Theorem 9 to conclude that $y=z$ and $yz-1=y^2-x^2$.
Then $y=2^k$ and $x=1$ as desired.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "10",
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Is The Series $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$ Divergent This is my solution for $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$. First I let the sequence in the series be labeled $A$, then I constructed a new sequence ($B$) that would have the similar behavior as $A$. This gives be the following:
$$A=\frac{... | So, not quite. Suppose you are considering two series $\sum a_n$ and $\sum b_n$ consisting of non-negative terms. If you know that $\sum a_n$ is divergent and it happens that $ a_n \leq b_n$ for every $n$ then you may conclude that $\sum a_n \leq \sum b_n$. However, since you know the first series is divergent and con... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\tan 9\theta = 3/4$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
If $\tan9\theta=\dfrac{3}{4}$, where $0<\theta<\dfrac{\pi}{18}$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
My approach:-
$$\begin{align*} \tan9\theta &=\frac{3}{4} \\[6pt] \implies \theta & = \frac{37^{\circ}}{3} \end{align... | Set $3\theta=t$
$\tan3t=\dfrac34\implies\dfrac{\sin3t}3=\dfrac{\cos3t}4=\dfrac1k$
$\implies(k\sin3t)^2+(k\cos3t)^2=3^2+4^2\implies k=5$ as $0<t<\dfrac\pi6$
$$F(t)=\dfrac3{\sin t}-\dfrac4{\cos t}=\dfrac{k\sin3t}{\sin t}-\dfrac{k\cos3t}{\cos t}$$
Method$\#1:$
As $\sin t\cos t\ne0,$ using $\sin3t,\cos3t$ formula,
$$F(t)=k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$ Let $x,y,u,v \in \mathbb{R}.$ Prove that
$$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$$
Proof 1:
$$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$$
Square both side, we have
$$... |
$$
\overline{AB} = \sqrt{(x+u)^{2} + (x+u)(y+v) + (y+v)^{2}} \\
\overline{BC} = \sqrt{x^{2} + xy + y^{2}} \\
\overline{CA} = \sqrt{u^{2} + uv + v^{2}}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$
Prove $$ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$$
So I started by combining the two fractions, which gave me:
$$ \frac{\sin\theta(1+\cos\theta) - \sin\theta(1-\cos\theta)}{(1-\cos\theta)... | It's $$\frac{\sin2\theta}{\sin^2\theta}=\frac{2\sin\theta\cos\theta}{\sin^2\theta}=\frac{2\cos\theta}{\sin\theta}=2\cot\theta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Value of $\lim\limits_{n\rightarrow \infty}(a_{1}+a_{2}+\cdots +a_{n})$
If $\displaystyle a_{n}=\bigg(\frac{n!}{1\cdot 3 \cdot 5 \cdot 7\cdot...\cdot (2n+1)}\bigg)^2.$
Then $\displaystyle \lim_{n\rightarrow \infty}\bigg(a_{1}+a_{2}+...+a_{n}\bigg)$ is
Options:
$(a)$ Does not exists
$(b)$ Greater than $\displaystyle ... | Why, why bounty when @gt6989b gave you an awesome hint :) ?
$$a_n=\left(\left(\frac{1}{3}\right)\cdot\left(\frac{2}{5}\right)\cdot...\cdot\left(\frac{n}{2n+1}\right)\right)^2 <
\frac{1}{2^{2n}}=\frac{1}{4^n}$$
and (calculating the first 3 terms and using infinite geometric progression)
$$\sum\limits_{n=1}a_n =
\frac... | {
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"url": "https://math.stackexchange.com/questions/3191145",
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find : $\int_0^{\pi/4}x\ln(\sin x)\mathrm dx$ I'm try to find this integral
$$\int_0^{\pi/4}x\ln(\sin x)\mathrm dx$$
My try use :
$\ln(\sin x)=-\ln2-\sum\limits_{n=1}^{\infty}\frac{\cos (2nx)}{n}$
But I don't know how to complete summation ...
I will happy if someone help me
Thanks!
| Your approach works perfectly well:
We can use the Fourier series and integrate by parts to obtain
$$ I \equiv \int \limits_0^{\pi/4} x [- \ln(\sin(x))] \, \mathrm{d} x = \frac{\pi^2}{32} \ln(2) + \frac{1}{4} \sum \limits_{n=1}^\infty \frac{1}{n^2} \left[\frac{\pi}{2} \sin\left(\frac{\pi}{2} n \right) - \frac{1}{n} \le... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solving Pell's Equation for $x^2 -7y^2 = 1$ for the first three integral solutions. Like the title states my goal is the find the first three integral solutions of the Diophantine equation. I know $x^2 -7y^2 = 1$ is a Pell's equation where $d = 7$. I found the minimal solution to be $(x,y) = (8,3)$ through brute force.... | Remember that $x^2-7y^2=1$, so you can write $x^6-21x^4y^2=-14x^4y^2+7x^4$ and keep going
$$x^6 -21x^4y^2+147x^2y^4-343y^6 = 1\\
-14x^4y^2+7x^4+147x^2y^4-343y^6=1\\
7x^4+49x^2y^4-14x^2y^2-343y^6=1\\
7x^4-14x^2y^2+49y^4=1$$
Now you can plug in your fourth degree solution and be done. You can also use the Brahmgupta-Fer... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\lim_{n\to \infty} \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$ $$\lim_{n\to \infty} \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$$
$$\lim_{n\to \infty} \sum_{r=0}^{n-1} \left[\frac{1}{\sqrt{n^... | Hint
$$\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+n}}+...+\frac{1}{\sqrt{n^2+n(n-1)}}=\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\sqrt{1+\frac{k}{n}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $3x^2 - 4x -2 = 0$ by completing the square I can't understand the solution from the textbook (Stroud & Booth's "Engineering Mathematics" on a problem that involves solving a quadratic equation by completing the square.
The equation is this:
$$
\begin{align}
3x^2 - 4x -2 = 0 \\
3x^2 - 4x = 2
\end{align}
$$
Now,... | Try going backward; expand the square $(x-\tfrac{2}{3})^2$ to find that
$$\left(x-\frac{2}{3}\right)^2=\left(x-\frac{2}{3}\right)\left(x-\frac{2}{3}\right)=x^2-\frac43x+\frac49.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
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Showing $\lvert\cos z\rvert^2=a\cos2x+b\cosh2y$, where $z=x+iy$, for numbers $a$ and $b$ to be determined
Show that, for $z=x+iy$,
$$\lvert\cos z\rvert^2=a \cos2x+b\cosh2y$$
where $a$ and $b$ are numbers to be determined.
My incomplete solution:
$$\begin{align}
\lvert z\rvert &= \sqrt{z^*z} \\
\lvert\cos(z)\rvert... | \begin{align}\lvert\cos z\rvert^2&=\cos(z).\overline{\cos(z)}\\&=\cos(z).\cos\left(\overline z\right)\\&=\cos(x+yi)\cos(x-yi)\\&=\bigl(\cos(x)\cosh(y)-\sin(x)\sinh(y)i\bigr)\bigl(\cos(x)\cosh(y)+\sin(x)\sinh(y)i\bigr)\\&=\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)\\&=\cos^2(x)\cosh^2(y)+\bigl(1-\cos^2(x)\bigr)\bigl(\cosh^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a,b \in \Bbb [0,2]$ and $a+b=2$, maximize $ab(a^2+b^2)$ without differentiation. If $a,b \in \Bbb [0,2]$ and $a+b=2$, maximize $ab(a^2+b^2)$ without differentiation.
My try
$a+b=2 \iff a=2-b \implies \max \{ab(a^2+b^2)\} \iff \max \{[(2-b)b][(2-b)+b^2)\}$
$\iff \max \{-2 b^4 + 8 b^3 - 12 b^2 + 8 b\}$
At this point... | You can use the fact that\begin{align}ab(a^2+b^2)&=ab\bigl((a+b)^2-2ab\bigr)\\&=ab(4-2ab)\\&=2\bigl(2ab-(ab)^2\bigr)\\&=2\left(1-\left(ab-1\right)^2\right)\\&=2-2(ab-1)^2.\end{align}So, the maximum is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A modular equation of 23rd degree of Dedekind’s $\eta$ function. Regarding the Post Additional values of Dedekind's $\eta$ function in radical form
I wrote the equation that has as root the value $\frac{\eta(23i)}{\eta(i)}$ that is missing.
Can someone help me solve (in radical form) the following equation, whose sol... | After substitution $x^4\to x,$ we get a 12-degree equation that factors into cubics over the rationals extended with $\sqrt{23}$ and $\sqrt2\cdot\sqrt[4]{23\,}$. Solving the cubic equation, we get
$$\small\begin{align}
\!\!\frac{\eta(23i)}{\eta(i)}&=\\
&\!\!\!\!\!\!\sqrt[4]{\frac{2\cdot 2^{2/3} \sqrt[3]{\alpha }+\sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove Smaller Distance from Hyperbola to Asymptote There is a canonical hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 $$ and the asymptote $$ y= \pm \frac{b}{a}x $$
Let us say that the value of hyperbola at $x$ is given as
$$y=\frac{b}{a}\sqrt{x^2 - a^2}$$ and the value of asymptote at the same $x$ is given as
$$ y = ... | OK, by using comment from Blue, I will try to answer the question myself. Since the difference is $$\frac{b}{a}(x-\sqrt{x^2 - a^2}) $$ then the derivative will be $$\frac{b}{a}(1-\frac{x}{\sqrt{x^2-a^2}}) $$
Since I am only interested to the value of $x > a$ where $x$ and $a$ is always positive then $$\frac{x}{\sqrt{x... | {
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"url": "https://math.stackexchange.com/questions/3206331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove with using mathematical induction $\prod_{i=0}^n 1+q^{2^i} = \frac{1-q^{2^{n+1}}}{1-q}$?
Prove the identity
\begin{align}
\prod_{i=0}^n \left(1+q^{2^i}\right) = \frac{1-q^{2^{n+1}}}{1-q}
\end{align}
for each nonnegative integer $n$.
To begin with, I cannot verify the equality itself.
$\prod_{i=0}^n 1+q^... | For the base case $n=1$, prove that $$(1+q)(1+q^2)=\frac{1-q^4}{1-q}\iff \underbrace{(1-q)(1+q)}_{=(1-q^2)}(1+q^2)=1-q^4$$
Which is true since $(a-b)(a+b)=a^2-b^2$.
For the inductive step $\color{blue}{\text{assume that the equation holds for some $n$}}$. Therefore
$$\begin{align*}(1+q)(1+q^2)(1+q^4)...(1+q^{2^{n+1}})&... | {
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"url": "https://math.stackexchange.com/questions/3207238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ Show that $\alpha = \sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ by presenting a polynomial $p$ from $\mathbb{Q}[X]$ with $p(\alpha)$.
It seems like $$\mathbb{Q}(\sqrt[3]{3} + \sqrt[3]{9}) = \{a + b\sqrt[3]{3} + c\sqrt[3]{9},~a,b,c\in\mathbb{Q} \}... | Let's have $q=\sqrt[3]{3}\quad$ and $\quad\alpha=q+q^2$.
$$q+q^2+q^3=q(1+q+q^2)\iff3+\alpha=q(1+\alpha)$$
Now since $q^3=3$ then $\alpha$ is solution of $$(3+\alpha)^3=3(1+\alpha)^3\iff 12+9\alpha-\alpha^3=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Integration by substitution using $u^2=3-x$ The question says to integrate $\frac x{\sqrt{3-x}}$ using the substitution $u^2=3-x$.
What I did was first rearrange $u^2=3-x$ into $ 3 - u^2 =x$ and then subbed it into the integral .
I then rearranged $u^2=3-x$ to $u=\sqrt{3-x} $ and then also subbed that in.
Next I found... | Using $u^2 = 3-x$, we have the following
$$\int \dfrac{x}{\sqrt{3-x}} dx = \int \dfrac{3-u^2}{\sqrt{u^2}} dx = \int \dfrac{3-u^2}{u} dx$$
Now, like you did, solve the transformation for $u$ such that $u=(3-x)^\frac{1}{2}$. Then, $du = -\dfrac{1}{2}(3-x)^{-\frac{1}{2}} dx = -\dfrac{1}{2}(u^2)^{-\frac{1}{2}}dx = -\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find $\sum_{k=1}^{2n} k {2n\choose k}^2 (-1)^{k+1}$.
I need to find $$S = {2n\choose 1}^2 -2 {2n\choose 2}^2 + ... - 2n{2n\choose 2n}^2= \sum_{k=1}^{2n} k {2n\choose k}^2 (-1)^{k+1}$$, given $$\sum_{k=1}^{2n} k {2n\choose k} x^{k-1} = 2n(1+ x)^{2n -1}$$
Using $$-(1-x)^{2n} + 1 = \sum_{k=1}^{2n} {2n \choose k} (-1)^{k... | $[x^n]:f(x)$ denote the coefficient of $x^n$ in the function $f(x)$
\begin{eqnarray*}
\binom{2n}{k} =\binom{2n}{2n-k}=[x^{2n-k}]: (1+x)^{2n}=[x^{2n}]:x^k(1+x)^{2n}.
\end{eqnarray*}
So your sum can be written as
\begin{eqnarray*}
\sum_{k=1}^{2n} k \binom{2n}{k}^2 (-1)^{k+1} &=& [x^{2n}]: (1+x)^{2n} \sum_{k=1}^{2n} k \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
The exponential generating function for the central binomial coefficients I am having difficulty proving the following result
\begin{align}
\sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^{n}}{n!} = e^{2x} \ I_{0} (2x) \text{,}
\end{align}
where
\begin{align}
I_{0}(y) = \sum_{n=0}^{\infty} \frac{\Big( \frac{y}{2} \Big)^{2n}... | Extracting the coefficient we seek to show
$${2n\choose n} \frac{1}{n!}
= [z^n] \sum_{k\ge 0} \frac{2^k}{k!} z^k
\sum_{k\ge 0} \frac{1}{k! \times k!} z^{2k}.$$
This is
$${2n\choose n} \frac{1}{n!}
= \sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{1}{q! \times q!} \frac{2^{n-2q}}{(n-2q)!}.$$
or
$${2n\choose n}
= \sum_{q=0}^{\lfl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How can I compute this integral in closed form : $\int_0^{\frac{π}{4}}\ln^2(\tan x)dx$ How can I compute this integral in closed form :
$$\displaystyle\int_{0}^{\displaystyle \tfrac{π}{4}}\ln^{2}\left(\tan x\right)dx$$
How can use Fourier series here ?
$$-2\displaystyle \sum_{n=0}^{\infty}\frac{\cos((4n+2)x)}{2n+1}$$
$... | An elementary solution. Utilize
$$ J=\int_0^\infty \frac{(x^2-1)\ln y }{(y+x^2)(y+1)}
\overset{y\to \frac{x^2}y}{dy}=\int_0^\infty \frac{(x^2-1)\ln (x^2) }{(y+x^2)(y+1)}dy- J= {2\ln^2 x}$$
and $x=\tan t $
\begin{align}
&\int_0^{\pi/4}\ln^2(\tan t)\ dt\\
=& \int_0^1\frac{\ln^2x}{1+x^2}\overset{x\to 1/x}{dx}
=\int_1^\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Is my solution correct? (trigonometry) $$2\cos x\left(\cos x-\sqrt{8}\tan x\right)<5$$
After expanding I get $2\sin^2 x+4\sqrt{2}\sin x-5>0$. I then factorized $(2\sin x+3\sqrt{2})(2\sin x+\sqrt{2})>0$. The first one is always positive so $\sin x>\sin(-\pi/4)$. How to continue?
| $\sin x>-\sqrt{2}/2$ and you have $\sin(-\pi/4)=\sin(5\pi/4)=-\sqrt{2}/2$. The big arc between $-\pi/4$ and $5\pi/4$ is the solution :
$$\left[\frac{-\pi}{4}, \frac{5\pi}{4}\right]~~~\text{(mod }2\pi\text{)}$$
Another way to denote this solution is
$$\bigcup\limits_{n\in\mathbb{Z}}\left[\frac{-\pi}{4}+2n\pi, \frac{5\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $a-b$ from $(\cos(a)+\cos(b))^{2}+(\sin(a)+\sin(b))^{2}=2\cos^{2}(\frac{a-b}{2})$ $a,b\in \mathbb{R}$ and these numbers satisfy this egality:
$(\cos(a)+\cos(b))^{2}+(\sin(a)+\sin(b))^{2}=2\cos^{2}(\frac{a-b}{2})$
I need to find $a-b$ (The right answer is $a-b=2k\pi+\pi|k\in \mathbb{Z}$)
My try: I know that the exp... | Clayton has the answer for what is wrong with your answer.
An alternative answer is to remember $2\cos^2{x}-1=\cos 2x$ so:
$$2\cos^2\left(\frac{a-b}{2}\right)=\cos(a-b)+1$$
And you have the left side is, after expanding and canceling $\cos^2 a+\sin^2 a = 1=\cos^2b +\sin^2 b$
$$2+2(\cos a \cos b + \sin a \sin b)=2+2\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Series that evaluates to different values I am trying to find a sequence $x_{pq}$ such that $\sum_{p=1}^\infty\sum_{q=1}^\infty x_{pq}$ and $\sum_{q=1}^\infty\sum_{p=1}^\infty x_{pq}$ both exist and evaluate to different values. I am thinking of this as a matrix, but I can't seem to see how adding all the elements of i... | The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence.
Consider, on the other hand, $a_{jk} = 1/(k^2-j^2)$ if $j\neq k$ and $a_{jk} = 0$ if $j=k$.
Here we have,
$$\frac{\pi^2}{8}=\tag{*}\sum_{j=1}^\infty \sum_{k=1\\j\neq k}^\infty \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\lim\limits_{n\rightarrow \infty}\left( \frac{2^{-n^2}}{\sum_{k=n+1}^{\infty}2^{-k^2}}\right)$
Find the value of $$\lim_{n\rightarrow \infty}\left( \frac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty}2^{-k^2}}\right)$$
My answer:
Take $r_{n+1} = \sum\limits_{k=n+1}^{\infty}2^{-k^2}$.
If $\lim\limits_{n\r... | For $k \geq n+1$ it is easy to see that $n^{2}-k^{2} \leq -k$. Hence $\sum\limits_{k=n+1}^{\infty} 2^{n^{2}-k^{2}} \leq \sum\limits_{k=n+1}^{\infty} 2^{-k} \to 0$. Hence the required limit is $\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve $99x^2 \equiv 1 \mod 125$ Solve $$x^{98} \equiv 99 \mod 125$$
Is there any easy way to solve equations like that? My observation is that from Euler's theorem we know that
$$ x^{100} \equiv 1 \mod 125 $$
so
$$x^{98} \equiv 99 \mod 125 \\
x^{100} \equiv 99x^2 \mod 125 \\
99x^2 \equiv 1 \mod 125$$
but what is gener... | Start mod $5$, and then lift...
$$99 x^2 \equiv 4 x^2 \equiv (2x)^2 \equiv 1 \mod 5$$
so $2 x \equiv \pm 1 \mod 5$, i.e. $x \equiv 2$ or $3 \mod 5$.
If $x \equiv 2 \mod 5$, $x \equiv 2 + 5 y \mod 25$, and then
$$ 99 x^2 - 1 \equiv 5 y + 20 \equiv 0 \mod 25$$
$$ y + 4 \equiv 0 \mod 5$$
$$ y \equiv 1 \mod 5$$
So now $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$ Calculate $$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$$
Here is my attempt:
$$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}= \left(\frac{4\infty^2+5\infty-6}{4\infty^2+3\infty-10}\right)^{3-4\i... | Hint:
$$\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)=\frac{(n+2)(4n-3)}{(n+2)(4n-5)}=\frac{4n-3}{4n-5}=\frac{1}{(1-\frac{2}{4n-3})}=(1-\frac{2}{4n-3})^{-1}$$
so
$$\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}=(1-\frac{2}{4n-3})^{4n-3}$$
then use
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3222029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.