Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
why does the quotient become messy when we switch the terms of the divisor when using long division? If we were to use long divison to divide $x^2 - y^2$ by $x + y$, the long division easily gives $x-y$. But if I were to use long division to divide $x^2 - y^2$ by $y + x$ instead, the quotient becomes lengthy and fruitl... | Dividing $x^2-y^2$ by $y+x$ gives:$$\frac{x^2}{y}-y-\frac{x^3}{y^2}+x+\frac{x^4}{y^3}-\frac{x^2}{y}-\frac{x^5}{y^4}+\frac{x^3}{y^2}+\frac{x^6}{y^5}-\frac{x^4}{y^3}-\frac{x^7}{y^6}+\frac{x^5}{y^4}+\frac{x^8}{y^7}-\frac{x^6}{y^5}-\cdots$$ The pattern of signs is + - - +, and every fractional term cancels with the fifth ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Closed form for $\prod_{k=1}^{n-1} \sin(\frac{\pi k}{n})^{n-k}$ Question: I am hoping to analytically continue the function:
$F(n) = \prod_{k=1}^{n-1} \sin(\frac{\pi k}{n})^{n-k}$
to $n=1/2$. My understanding is that means I will need a closed form for this product. What is a closed form of the above product? (Or are t... | We break up the product in a different way (analogous to changing the order of products):
$$\begin{align}
F(n) &= \sin{\left(\frac{\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \sin{\left(\frac{2\pi}{n}\right)} \\
&\quad \times \sin{\left(\frac{\pi}{n}\right)} \sin{\left(\frac{2\pi}{n}\right)} \sin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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exponential equation: $6^x+8^x+15^x=9^x+10^x+12^x$ What are the solutions of this equation? Or at least in which interval are they? $$6^x+8^x+15^x=9^x+10^x+12^x$$ I tried to find an increasing function, or use some inequalities but I got nothing out of it...
| Let $a=3^x$ and $b=2^x$ and $c=5^x$.
Then we have that
$$ab+b^3+ac=a^2+bc+ab^2$$
$$ab+b^3+ac-a^2-bc-ab^2=0$$
$$a(b-a)+b^2(b-a)-c(b-a)=0$$
$$(b-a)(a+b^2-c)=0$$
Now $a=b \implies x=0$ and $a+b^2-c=0 \implies 3^x+4^x=5^x$ which gives solution for $x=2$ only and not for any higher integer $x$ by Fermat's Last Theorem.
So ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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write the given matrix as a product of elementary matrices Please explain this in detail because i simply cannot understand previous explanations I have read for this type of problem.
By the way this is from elementary linear algebra 10th edition section 1.5 exercise #29. There is a copy online if you want to check the... | It took me a good 20 minutes to type this, so I'm gonna be pissed af if you don't read it.
Take the matrix $\begin{pmatrix}-3 & 1 \\ 2 & 2 \end{pmatrix}$ and add $2/3$ times the first row to the second. You get $\begin{pmatrix}-3 & 1 \\ 0 & \frac{8}{3} \end{pmatrix}$. Let $E_1$ be the elementary row matrix correspond... | {
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"url": "https://math.stackexchange.com/questions/2157090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Geometric series and sequences $t_n$ is the $n^{th}$ term of an infinite sequence and $s_n$ is the partial sum of the first $n$ terms. Given that $t_1 = \frac{2}{4}$, $s_2 = \frac{4}{7}$, $t_3 = \frac{1}{35}$, $s_4 = \frac{8}{13}$. Determine a formula for $s_n$ and calculate its limit.
| $s_2 = t_1 + t_2, s_4 = s_2 + t_3 + t_4$ this is enough information to find $t_2, t_4$
$t_1,t_2,t_3, t_4 =$$ \frac 12, \frac 1{14}, \frac 1{35}, \frac 1{65}\\
\frac {1}{2\cdot 1}, \frac {1}{2\cdot 7}, \frac {1}{5\cdot 7},\frac {1}{5\cdot 13}$
Just a hunch, but it appears that the first factor in the denominator increas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove an identity involving factorials inside summations Prove that:
\begin{align*}\sum _{k=0}^N\left(k\cdot \frac{N!}{k!\left(N-k\right)!}\cdot \frac{\left(X+k-1\right)!}{\left(X-1\right)!}\cdot \frac{\left(L-X+N-k-1\right)!}{\left(L-X-1\right)!}\frac{\left(L-1\right)!}{\left(L+N-1\right)!}\right)&= \frac{XN}{L}
\end... | Here is an approach without using induction. We start with the left-hand side, do some simplifications to finally obtain the right-hand side.
We obtain
\begin{align*}
\sum _{k=0}^N& k\cdot \frac{N!}{k!\left(N-k\right)!}\cdot \frac{\left(X+k-1\right)!}{\left(X-1\right)!}\cdot \frac{\left(L-X+N-k-1\right)!}{\left(L-X-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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For $a, b, c$ is the length of three sides of a triangle. Prove that $\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{8}$ For $a, b, c$ is the length of three sides of a triangle. Prove that $$\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{8}$$
| HINT: use that we get for the sum $$\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=\frac{(a-b) (a-c) (b-c)}{(a+b) (a+c) (b+c)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show $|-\ln(\cos (x))-\frac{x^2}2| \le \frac23|x^3| \quad\forall x \in[-\frac{\pi}{4},\frac{\pi}{4}] $
Let
$$f:(-\frac{\pi}{2},\frac{\pi}{2}) \to \Bbb R$$
$$f(x)=-\ln(\cos (x))$$
Show that
$$\left| f(x)-\frac{x^2}2 \right| \le \frac23\left| x^3 \right| \qquad x
\in[-\frac{\pi}{4},\frac{\pi}{4}]$$
My attempt:
We... | Applying Taylor's theorem with remainder to
$$ \begin{aligned}
f(x) &= -\ln(\cos(x)) & f(0) &= 0 \\
f'(x) &= \tan(x) & f'(0) &= 0\\
f''(x) &= 1 + \tan^2(x) & f''(0) &= 1 \\
f'''(x) &= 2 \tan(x) (1+\tan^2(x))
\end{aligned}$$
gives
$$
f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f'''(\xi)}{3!} x^3
= \frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For $x,y,z>0$. Minimize $P=\frac{1}{4x^{2}-yz+2}+\frac{1}{4y^{2}-zx+2}+\frac{1}{4z^{2}-xy+2}$ For $x,y,z>0$ and $xy+yz+xz=1$, minimize $$P=\frac{1}{4x^{2}-yz+2}+\frac{1}{4y^{2}-zx+2}+\frac{1}{4z^{2}-xy+2}$$
My try: Let $xy=a; yz=b;zx=c \Rightarrow a+b+c=1$
$\Rightarrow x^2=\frac{ac}{b};y^2=\frac{ab}{c};z^2=\frac{bc}{... | If $x=y=z=\frac{1}{\sqrt3}$ we get $P=1$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{1}{4x^2-yz+2}\geq1$$ or
$$\sum_{cyc}\left(\frac{1}{4x^2-yz+2}-\frac{1}{3}\right)\geq0$$ or
$$\sum_{cyc}\frac{1-4x^2+yz}{4x^2-yz+2}\geq0$$ or
$$\sum_{cyc}\frac{xy+xz+2yz-4x^2}{4x^2-yz+2}\geq0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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If $\frac{1}{n}A^n = \frac{1}{n+1}A^{n+1} = \frac{1}{n+2}A^{n+2}$ then any eigenvalue of $A$ is $0$, please check my proof I'd like to show that for a square matrix A with the property that there exists a positive integer n such that $\dfrac{1}{n}A^n = \dfrac{1}{n+1}A^{n+1} = \dfrac{1}{n+2}A^{n+2}$, any eigenvalue of A... | Essentially correct. You should stick to mathematical english or to $\Rightarrow$ language, though and not mix them up as you do in the first block of the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Polynomial system If there are 3 numbers $x,y,z$ satisfying
$f=x+y+z=3$ , $g=x^2+y^2+z^2=5$ , $h=x^3+y^3+z^3=7$ then prove that they also satisfy
$x^4+y^4+z^4=9$ but not $x^5+y^5+z^5=11$
I dont know how to tackle this to be honest, i have started trying to write $x^4+y^4+z^4-9$ as $r_1\times f + r_2\times g+ r_3\time... | $$2(xy+yz+zx) = (x+y+z)^2 - (x^2+y^2+z^2) = 4$$
$$-3xyz = (x+y+z)^3 - (x^3+y^3+z^3) - 3(xy+yz+zx)(x+y+z) = 27 - 25 = 2$$
Thus $x,y,z$ are the roots of
$$t^3 -3t^2 + 2t + \frac{2}{3} = 0$$
Now
\begin{align*}
t^4 = 3t^3 -2t^2 -\frac{2}{3}t
\end{align*}
Putting $t=x,y,z$ in the above and adding, we get
$$x^4 + y^4 + z^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Find the value of a given integral sequence: $I_n=\int_{0}^{1}(1-x^2)^ndx$ Let $(I_n)_{n \geq 1}$ be a sequence such that:
$$I_n = \int_0^1 (1-x^2)^n dx$$
Find the value of $I_n$ (The solution is $\frac{2}{3} \cdot \frac{4}{5} \cdot ... \cdot \frac{2n}{2n+1}$).
I've tried using the binomial expansion of $(1-x^2)^n$ but... | Here is an approach that relies on the Beta function $\operatorname{B}(m,n)$, namely
$$\operatorname{B} (m,n) = \int_0^1 x^{m - 1} (1 - x)^{m - 1} \, dx.$$
In the integral for $I_n$ we begin by enforcing a substitution of $x \mapsto \sqrt{x}$. Thus
\begin{align}
I_n &= \frac{1}{2} \int_0^1 x^{-1/2} (1 - x)^n \, dx\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Quadratic equation find all the real values of $x$ Find all real values of $x$ such that
$\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$
I tried sq both sides by taking 1 in RHS but it didn't worked out well...
| Squaring both sides:
$x^{2} - x +\frac{2}{x} - 1 = 2\sqrt{(x-\frac{1}{x})(1-\frac{1}{x})}$
Squaring again we get:
$x^{4} - 2x^{3} - x^{2} + \frac{4}{x^{2}} + 6x -\frac{4}{x} - 3 = 4(\frac{1}{x^{2}} + x - \frac{1}{x} - 1)$
We can reduce this to:
$x^{4} -2x^{3} -x^{2} +2x + 1 = 0$
Notice that this factorises into $(x^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Regarding complex roots of a polynomial I am having difficulty in finding the roots of the following polynomial when it is given that all the roots are complex:-
$$f(x)= x^4+4x^3+8x^2+8x+4$$
How can I factorize the polynomial to get its roots?
| Hint:
$$
\begin{align}
f(x)= x^4+4x^3+8x^2+8x+4 & = x^2\left(x^2+\frac{4}{x^2} + 4\left(x + \frac{2}{x}\right) +8\right) \\
& = x^2\left(\left(x+\frac{2}{x}\right)^2 + 4\left(x + \frac{2}{x}\right) +4\right) \\
& = x^2\left(x+\frac{2}{x} + 2\right)^2 = \;\cdots
\end{align}
$$
[ EDIT ] The above is based on the obs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve an integral $\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$ Solve an integral $$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$$
Using the third substitution of Euler,
$$\sqrt{x^2-1}=(x-1)t\Rightarrow x=\frac{1+t^2}{1-t^2},dx=\frac{4t}{(1-t^2)^2}$$
we get an integral $$\int\frac{1-t^2}{1+t^4}dt=\frac{1}{2}\int\frac{1-\sqrt 2 t}{t^2-\... | Setting $$x=\sinh(t)$$ then we have $$dx=\cosh(t)dt$$ and our integral will be $$\int \frac{dt}{1+\sinh^2(t)}$$
then use $$\sinh(t)=-2\,{\frac {\tanh \left( t/2 \right) }{ \left( \tanh \left( t/2
\right) \right) ^{2}-1}}
$$
note that $$x^2-1=\sinh^2(t)-1=\cosh^2(t)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum of series $\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$ Find the radius of convergence and the sum of power series
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$$
Radius of convergence is $R=1$, and the interval of convergence is $-1<x<1$.
I am having trouble in finding the sum.
Here is what I have tried.
$$\sum... | Hint:
$$f(x)=\sum_{n=0}^\infty\frac{n+1}{2n+1}x^{2n+1}$$
$$\implies f'(x)=\sum_{n=0}^\infty(n+1)x^{2n}$$
And now use the geometric series
$$\frac1{1-r}=\sum_{n=0}^\infty r^n$$
$$\frac d{dr}\frac1{1-r}=\sum_{n=0}^\infty(n+1)r^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to find limit of $\lim_{x\to 1} \frac{f(x) - f(1)}{x-1}$ if $f(x)=-\sqrt{25-x^2}$ I have a question , if then find $$\lim_{x\to 1} \frac{f(x) - f(1)}{x-1}$$
I got $f(1)=-\sqrt {24}$. Should I get limit = $\frac{\sqrt{24} - \sqrt{25-x^2}}{x-1}$ but answer is $\frac{1}{\sqrt{24}}$ . Should I use $f'(x)$? If I use ... | Method $1$. One may recall that, for any differentiable function $f$ near $a$, one has
$$
\lim_{x \to a}\frac{f(x)-f(a)}{x-a}=f'(a),
$$
Here we have
$$
f(x)=-\sqrt{25-x^2},\qquad f'(x)=\frac{x}{\sqrt{25-x^2}},\qquad a=1.
$$
Can you finish it?
Method $2$. (without derivatives) One has, a $x \to 1$,
$$
\begin{align}
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the sum of series $\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$ Evaluate the sum of series
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$$
I have tried two methods:
1) using power series
2) using partial sums
but I can't find the sum.
1) Using power series:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{... | An alternative approach: since $\frac{1}{k(4k^2-1)} = \frac{1}{2k-1}-\frac{2}{2k}+\frac{1}{2k+1}$ is the integral over $(0,1)$ of $x^{2k-2}-2x^{2k-1}+x^{2k} = x^{2k-2}(1-x)^2$, we have:
$$ \sum_{k\geq 1}\frac{(-1)^{k+1}}{k(4k^2-1)}=\int_{0}^{1}(1-x)^2\sum_{k\geq 1}(-1)^{k+1} x^{2k-2}\,dx = \int_{0}^{1}\frac{(1-x)^2}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Ackermann function $A(x,y)$: Prove that $A(4,y) = \underbrace{2^{2^{.^{\cdot^2}}}}_\textrm{y+3 times} -3$. Prove that $A(4,y) = \underbrace{2^{2^{.^{\cdot^2}}}}_\textrm{y+3 times} -3$.
I'm having trouble with proving that there are indeed $y+3$ exponentiations:
\begin{align*}
A(4,y) &= A(3, A(4,y-1))\\
&= 2^{A(4,y-1)+3... | Prove by induction that
$$A(1,n)=n+2$$
Use this to prove by induction that
$$A(2,n)=2(n+3)-3$$
Use this to prove by induction that
$$A(3,n)=2^{n+3}-3$$
And use this to prove by induction that
$$A(4,n)=^{n+3}2-3$$
where we used tetration notation.
$$\begin{align}A(4,n+1)&=A(3,A(4,n))\\&=2^{(^{n+3}2-3)+3}-3\\&=2^{(^{n+3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Count number of exact matching sequences Consider all pairs of binary strings $P$ and $T$. Let the length of $P$ be $n$ and the length of $T$ be $2n-1$. For each such pair, we can check if $P$ is exactly equal to each of the $n$ substrings of $T$ in order from left to right and output a sequence representing these res... | Lemma. Let $R = R(P, T)$ be an output sequence for given strings $P$ and $T$ of length $n$ and $2n - 1$ correspondingly. Then
$$R = \underbrace{00\ldots\ldots\ldots\ldots\ldots\ldots0}_{\text{non-negative number of zeros}}\underbrace{1\underbrace{0\ldots0}_{\text{$k - 1$ zeros}}1\underbrace{0\ldots0}_{\text{$k - 1$ zer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve a system of two congruences with modules not pairwise coprime I have the following system of congruences:
$\left \{ \begin{array}{rcl} x & \equiv & 225 & \mbox{ (mod 250) } \\ x & \equiv & 150 & \mbox{ (mod 1225) } \end{array} \right .$
Since $\gcd(250, 1225) \ne 1$ the Chinese Remainder Theorem is not applicable... | $x \equiv 225 \pmod {250}$ so $x=25(9+10j)$ for some $j \in \mathbb{Z} $.
$x \equiv 150 \pmod {1225}$ so $x=25(6+49k)$ for some $k \in \mathbb{Z} $. So we need to find $y$ satisfying
\begin{eqnarray*}
y \equiv 9 \pmod {10} \\
y \equiv 6 \pmod {49}
\end{eqnarray*}
$10$ and $49$ are coprime so we can use the Chinese rema... | {
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"url": "https://math.stackexchange.com/questions/2187100",
"timestamp": "2023-03-29T00:00:00",
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If $a+b+c=x+y+z=1$ then $\frac{a}{y+z}+\frac{b}{x+z}+\frac{c}{x+y}\geq1+\frac{27}{2}abc$
Let $a$, $b$, $c$, $x$, $y$ and $z$ be positives such that $a+b+c=x+y+z=1$. Prove that:
$$\frac{a}{y+z}+\frac{b}{x+z}+\frac{c}{x+y}\geq1+\frac{27}{2}abc$$
I tried to make the homogenization:
We need to prove that
$$(x+y+z)\left... | I used Lagrange Multipliers here. I'd love to see a way that doesn't rely on them, because I feel there should be a much snappier solution. I'm a bit rusty with contest math, but I think I've checked my own work well enough. That's no excuse for holes, but I offer it as a caveat. It's likely there can be improvements i... | {
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"url": "https://math.stackexchange.com/questions/2190546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve the integral $\int\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}\,dx$ .
$$\large \int \frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}~dx$$
In this I tried to substitute $x^2 +1$ by $t$.
after that got stuck.
| HINT:
$$x^2-1=x^2\left(1-\dfrac1{x^2}\right)$$
As $\displaystyle\int\left(1-\dfrac1{x^2}\right)dx=x+\dfrac1x$
write
$$x^2+1=x\left(x+\dfrac1x\right)$$
$$x^4+1=x^2\left(x^2+\dfrac1{x^2}\right)\implies\sqrt{x^4+1}=|x|\sqrt{x^2+\dfrac1{x^2}}$$
Finally $ x^2+\dfrac1{x^2}=\left(x+\dfrac1x\right)^2-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Characterize Numbers that can be written as the Sum Of Three Consecutive Positive Integers I am looking to characterize the numbers that can be characterized as the sum of 3 consecutive integers, but that can only be written in one way
so for example $12 = 3 + 4 + 5$
but not $15 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5.$
So in... | -----------This answer is incomplete now due to the recent clarification in the edit specifying looking only for those numbers which can be written as the sum of 3 consecutive integers but not by any amount larger -----------
Original answer for when interpretation of question was just about whether or not it could be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $ \sqrt{5} $ is irrational geometrically Hardy and Wright I am reading Hardy and Wright's Intro to Number Theory chapter about irrational number and there is a section proving that $ \sqrt{5} $ is irrational geometrically, but the proof is confusing me for some parts. The proof, extracted verbatim from the b... |
$ \textbf{at each stage the ratios of the number divided,} $
$ \textbf{the divisor, and the remainder are the same.} $ I am confused about this bold sentence.
They mean that
$$\color{red}{\text{red}}:\color{blue}{\text{blue}}:\color{green}{\text{green}}=1:x:x^2$$
always holds in
$$\color{red}{1}=1\times \color{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\sin (\theta+\alpha)=a$ and $\sin(\theta+\beta)=b$, prove that, $\cos [2(\alpha-\beta)]-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$ If $\sin (\theta+\alpha)=a$ and $\sin(\theta+\beta)=b$, prove that, $\cos [2(\alpha-\beta)]-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$.
My Attempt:
.$$\sin (\theta+\alpha)=a$$
$$\sin \theta. \cos \alph... | Set $\alpha-\beta=\gamma$, $\varphi=\theta+\alpha$, $\psi=\theta+\beta$; then $\gamma=(\theta+\alpha)-(\theta+\beta)=\varphi-\psi$, so
$$
\cos\gamma=
\cos\varphi\cos\psi+
\sin\varphi\sin\psi=
ab+\cos\varphi\cos\psi
$$
Also
$$
\cos2\gamma=2\cos^2\gamma-1=
2(a^2b^2+2ab\cos\varphi\cos\psi+\cos^2\varphi\cos^2\psi)-1
$$
so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Determine all pairs $(x,y)$ of integers
Determine all pairs $(x,y)$ of integers such that
$$1+2^{x}+2^{2x+1}=y^2.$$
I tried:
$$1+2^x+2^{2x+1}=y^2$$
$$2^x+2^{2x+1}=y^2-1$$
$$2^x+2^x2^x2=(y+1)(y-1)$$
$$2^x(1+ 2^{x+1})=(y+1)(y-1).$$
| Completing the square gives
\begin{equation}
\left(2^{x+2}+1\right)^2=8y^2-7
\end{equation}
so $8y^2-7$ must be a perfect square of a number which is one more than four times a power of two. So two solutions are $(0,2)$ and $(0,-2)$.
Two more solutions are $(4,\pm23)$.
The steps to completing the square are as follows:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find condition for $ a $ so that $ x^3 - 3ax + 1$ is separable Let $ K $ be an algebraically closed field of characteristic $ 0 $. What is a polynomial condition for $ a \in K $ such that $ f(x) = x^3 - 3ax + 1$ has distinct roots?
Here is my attempt: note that $ Df(x) = 3x^2 - 3a $. If $ a = 0 $, then $ (x^3 + 1, 3x^... | Your computation of the greatest common divisor is wrong.
The division of $x^3-3ax+1$ by $x^2-a$ has remainder $-2ax+1$.
Now we can divide $4a(x^2-a)$ by $2ax-1$, which has remainder $1-4a^3$.
(The case $a=0$ is trivial, so multiplying by a constant doesn't change the greatest common divisor, which is determined up to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show: $\cos \left( \frac{ 3\pi }{ 8 } \right) = \frac{1}{\sqrt{ 4 + 2 \sqrt{2} }}$ I'm having trouble showing that:
$$\cos\left(\frac{3\pi}{8}\right)=\frac{1}{\sqrt{4+2\sqrt2}}$$
The previous parts of the question required me to find the modulus and argument of $z+i$ where $z=\operatorname{cis{\theta}}$. Hence, I found... | Problem statement
$$
\cos \left( \frac{3\pi}{8} \right) = \cos \left( \frac{\pi}{4} - \frac{\pi}{8} \right)
$$
Basic formulas
Use the $\color{blue}{angle \ addition}$ formula
$$
\cos \left( \alpha + \beta \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta,
$$
and the $\color{blue}{half\ angle}$ formula
$$
\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
$a_{n+1}=\frac{a_n}{\sqrt{a_n^2+1}}$. Find $a_n$. Given $\{a_n\}$: $a_{n+1}=\frac{a_n}{\sqrt{a_n^2+1}}$, $a_1=1$.
Find $a_n$ like a function of $n$.
My trying.
Let $a_n=\tan\alpha_n$, where $\alpha_n\in\left(0,\frac{\pi}{2}\right)$.
Hence, $a_{n+1}=\sin\alpha_n$ and what is the rest?
Thank you!
| Note that $${ a }_{ 2 }=\frac { 1 }{ \sqrt { 2 } } \\ { a }_{ 3 }=\frac { \frac { 1 }{ \sqrt { 2 } } }{ \sqrt { \frac { 3 }{ 2 } } } =\frac { 1 }{ \sqrt { 3 } } \\ { a }_{ 4 }=\frac { \frac { 1 }{ \sqrt { 3 } } }{ \sqrt { \frac { 4 }{ 3 } } } =\frac { 1 }{ \sqrt { 4 } }
$$
so $$\\ { a }_{ n }=\frac { 1 }{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to prove this condition if three vectors are colinear? So I was given this problem:
Let $\vec{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\vec{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\vec{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$.
Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_... | Hint: Collinear vectors have the same direction vector.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Find answer of a system of equations Let $a,b,c\in \mathbb{R}$ that $a^{2}+b^{2}+c^{2}=1$. We want to find $x,y,z,w$ in the following equations:
$$\begin{cases}
\begin{align}
x^{2}+y^{2}&=\frac{1}{2}(1+a) \tag{1}\\
w^{2}+z^{2}&= \frac{1}{2}(1-a) \tag{2}\\
xw+yz&= \frac{1}{2} b \tag{3}\\
yw-xz&= \frac{1}{2} c.\tag{4}
\e... | Hint (assuming $x,y,z,w$ are reals): let $u=y+ix$ and $v=w+iz$ then the system becomes:
$$
\begin{cases}
\begin{align}
|u|^2 &= \frac{1}{2}(1+a) \\
|v|^2 &= \frac{1}{2}(1-a) \\
uv & = \frac{1}{2}(c+ib)
\end{align}
\end{cases}
$$
From the first two equations $|u|^2|v|^2= \frac{1}{4}(1-a^2)=\frac{1}{4}(b^2+c^2)\,$. From ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $|w_{n+1}-w_n| \leq (2/3)^2 |w_n - w_{n-1}|$? If a sequence is given by $S_{n+1}=S_n + S_{n-1}$ where $S_1=1$ and $S_2=2$ and we let $a_n=S_n/S_{n-1}$ for all $n \geq 2$. For each $n \geq 2$, prove that $|a_{n+1}-a_n| \leq (2/3)^2 |a_n - a_{n-1}|$?
This is what I'm thinking...
Let n=3. Then $|a_{n+1}... | Like user dezdichado already mentioned, this is the fibonacci sequence, but we can approach this independently without necessarily using known facts about the sequence.
You also have a typo on your final inequality which it should instead read as:
$$\vert \frac{S_{k+1}S_{k-1}-S_k^2}{S_kS_{k-1}} \vert \leq (2/3)^2\ve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solving equations with floor function and square roots? I was solving a problem to discover n and after I transformed the problem it gave me this equation:
\begin{equation*}
\left\lfloor{\frac{2}{3}\sqrt{10^{2n}-1}}\right\rfloor = \frac{2}{3}(10^{n}-1)
\end{equation*}
So I tried to simplify it by defining:
\begin... | Hint: $10^n-1$ is a multiple of $9$, so $\frac{2}{3}k$ is an integer, then:
$$
\begin{align}
\left\lfloor{\frac{2}{3}\sqrt{k(k+2)}}\right\rfloor = \frac{2}{3}k \;\;&\iff\;\; \frac{2}{3}k \le \frac{2}{3}\sqrt{k(k+2)} \lt \frac{2}{3}k \,+\, 1 \;\; \\
&\iff\;\; \frac{4}{9}k^2 \le \frac{4}{9}(k^2+2k) \lt \frac{4}{9}k^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integration of rational functions How would you integrate a rational function like this:
$$\frac{2x+3}{x^2+2x+10}$$
| Notice that:
$$\frac{2x + 3}{x^2 + 2x + 10} = \frac{2x + 2}{x^2 + 2x + 10} + \frac{ 1}{x^2 + 2x + 10} = \frac{2x + 2}{x^2 + 2x + 10} + \frac{ 1}{(x+1)^2 + 3^2}$$
Therefore,
$$\int \frac{2x + 3}{x^2 + 2x + 10} dx = \int\frac{2x + 2}{x^2 + 2x + 10}dx+ \int\frac{ 1}{(x+1)^2 + 3^2}dx$$
$$ = \ln|x^2 + 2x + 10| + \int\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluating the indefinite integral $\int \frac{x^3}{\sqrt{4x^2 -1}}~dx$.
Find $$\int {x^3\over \sqrt{4x^2 -1}}\,dx.$$
Let
$2x = \sec u$, $2 =\sec (u) \tan(u) u^{'}(x).$ Then
$$
\begin{align*}
\int {x^3\over \sqrt{4x^2 -1}}\,dx &= \frac1{16}\int {\sec^3 u\over \tan u}\tan u \sec u \, du\\
&= \frac1{16}\int {\sec^4... | $$\tan\left(\sec^{-1}(2x)\right)=\sqrt{4x^2-1}$$
Make this substitution into your answer and you will get the result you expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Continuity of some function related to the Hopf fibration of $S^3$ By the identification $\mathbb R^4$ with the field of quaternions $\mathbb H$ by $(x_0,x_1,x_2,x_3) \sim x_0+x_1 i+x_2 j+x_3k$ and $S^3$ with the set of unit quaternions and $S^2$ with the set of purely imaginary unit quaternions, let's consider the Ho... | We can get the explicite formula on $f_1^{-1}$. From the system of equations $(a,b,c,d)=\sqrt{\frac{1+v_1}{2}}(a_1, a_2, \frac{v_2 a_2-v_3 a_1}{1+v_1},\frac{v_2 a_1+v_3 a_2}{1+v_1})$,
we get
$v_1=2(a^2+b^2)-1$,
$v_2=2\sqrt{a^2+b^2}(c a_2+da_1)$,
$v_3=2\sqrt{a^2+b^2}(da_2-ca_1)$,
$a_1=\frac{a}{\sqrt{a^2+b^2}}$,
$a_2=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Mathematical Olympiad Treasures Problem 1.11 Let $n$ be a positive integer, prove that:
$3^{3^{n}}(3^{3^{n}}+1) + 3^{3^{n} + 1} - 1$
is not prime
The solution states:
Observe:
$3^{3^{n}}(3^{3^{n}}+1) + 3^{3^{n} + 1} - 1 = a^{3} + b^{3} + c^{3} - 3abc$
for $a = 3^{3^{n-1}}$, $b = 9^{3^{n-1}}$, $c = -1$
After that main ... | $$3^{3^{n}}(3^{3^{n}}+1) + 3^{3^{n} + 1} - 1\equiv 1\cdot 0 +1-1\equiv 0\bmod 2$$ hence it's not a prime
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show that $\lim_{n \to \infty} \sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \ln(2)$ Show that
$$\lim_{n \to \infty} \sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \ln(2)$$
How many ways are there to prove it ?
Is there a standard way ?
I was thinking about making it a Riemann sum.
Or telescoping.
What is the easiest way ?
What is the sho... | Note that
$$\sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \sum_{k=1}^n \frac{2k}{k^2+n^2+1} - \frac{2}{2+n^2} - \frac{4}{5+n^2}.$$
We can ignore the last two terms since they converge to $0$.
Consider
$$\sum_{k=1}^n \frac{2k}{k^2+n^2+1} = \frac{1}{n}\sum_{k=1}^n \frac{2(k/n)}{1+(k/n)^2 + (1/n^2)}. $$
This is almost a Riemann sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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How to find all possible solutions to $a^2 + b^2 = 2^k$? I'm working on the following problem, this is for an introductory discrete mathematics class.
Find all possible solutions to the equation $a^2 + b^2 = 2^k, k\geq1$ and $a$ and $b$ positive integers.
I've observed that the following are answers:
$2^1 = 1^2 + 1^2$
... | Your statement that $2^{k+1} = 2^k + 2^k$ is proven correct by the observation on the right-hand side that for any integer $c$, $c + c = 2c$. My first thought is that $a,b$ must have the same parity for this to ever work.
If both are even, let $a = 2A$ and $b = 2B$, so that $a^2 + b^2 = 4A^2 + 4B^2 = 4(A^2 + B^2)$. Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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If $a_n+b_n\sqrt{3}=(2+\sqrt{3})^n$, then what's $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}$? Let $a_n$ and $b_n$ be integers defined in the following way: $$a_n+b_n\sqrt{3}=(2+\sqrt{3})^n.$$ Compute $$\lim_{n\rightarrow\infty}\frac{a_n}{b_n}.$$
I tried expanding using binomial theorem:
$$ (2+\sqrt{3})^n=\binom{n}{0}2+\... | Note that
$$a_n - b_n\sqrt{3} = (2-\sqrt{3})^n$$
Hence,
$$2a_n = (a_n - b_n\sqrt{3}) + (a_n + b_n\sqrt{3}) \Rightarrow a_n = \dfrac{(2+\sqrt{3})^n + (2-\sqrt{3})^n}{2}$$
Expressing $b_n$ we obtain:
$$b_n = \dfrac{(2+\sqrt{3})^n - a_n}{\sqrt{3}} = \dfrac{(2+\sqrt{3})^n - (2-\sqrt{3})^n}{2\sqrt{3}}$$
Now it should be eas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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Show that the sequence {$a_n$}, such that $a_1 =4$ and $a_{n+1}=3-{{2}\over\ a_n}$ is convergent to 2. Show that the sequence {$a_n$}, such that
$a_1 =4$ and $a_{n+1}=3-{{2}\over\ a_n}$ is convergent to 2.
I show that the sequence is bounded but I cannot show that is monotone.
| Let's solve the difference equation.
Find the first few terms:
$a_2=3-\frac{2}{4}=\frac{5}{2}=\left(2+\frac{1}{2}\right)$
$a_3=3-\frac{4}{5}=\frac{11}{5}=\left(2+\frac{1}{5}\right)$
$a_4=3-\frac{10}{11}=\frac{23}{11}=\left(2+\frac{1}{11}\right)$
$a_5=3-\frac{22}{23}=\frac{47}{23}=\left(2+\frac{1}{23}\right)$
...
$a_{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I solve $5^{2x} + 4(5^x) - 5 = 0$? This is a math problem I'm currently working on.
$$5^{2x} + 4(5^x) - 5 = 0$$
I've used logarithm to try solve the problem. Here's what I've done so far:
\begin{align}5^{2x} + 4(5^x) - 5 &= 0\\
5^{2x} + 5^x &= \frac{5}{4}\\
\log_5{2x} + \log_5{x} &= \log_5\left(\frac{5}{4}\rig... | $\require{enclose}5^{2x}+4(5)^{x}-5=0 \enclose{updiagonalstrike}{\implies} 5^{2x}+5^{x}=\frac{5}{4}$
Instead:
Let $y=5^x\quad$ noting that $y\geq 0$, then we have:
$y^2+4y-5=0\implies (y+5)(y-1)=0$
$\implies y=-5\quad\text{reject since}\quad y\geq 0\quad \text{or}\quad y=1\implies 5^x=1\implies x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Maximization of $f(\theta)=\frac{1}{\sqrt{2\pi c\theta}}e^{-\frac{1}{2c\theta}(x-\theta)^2}$ with inequality constraints I want to maximize the following function:
$$f(\theta)=\frac{1}{\sqrt{2\pi c\theta}}e^{-\frac{1}{2c\theta}(x-\theta)^2}$$
with respect to $\theta\in[0,1]$. Assume the rest of the variables known and ... | Hint. What you have done is right. Then solving the quadratic equation gives two potential solutions
$$
\theta_0=-\frac12 \left(\sqrt{c^2+4 x^2}+c\right), \qquad \theta_1=\frac12 \left(\sqrt{c^2+4 x^2}-c\right).
$$ Which $\theta_i$ satisfies $0\le\theta_i \le1$ ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int \frac{\cos\pi z}{z^2-1}\, dz$ inside rectangle with vertices $2+i,2-i,-2+i,-2-i$ My attempt: The poles are $z=1,-1$, both lie inside the rectangle.
Residue at the poles are $-\frac12$ each, since residue at $$f(a)=\left[\frac{\cos \pi z}{\frac{d}{dz}(z^2+1)}\right]_{z=a}=\frac{\cos\pi a}{2a}.$$
So, by re... | The residue of $\frac{\cos(\pi z)}{z^2-1} =\frac{h(z)}{z-1}, \ h(z) = \frac{\cos(\pi z)}{z+1}$ at $z = 1$ is $h(1)=\frac{\cos(\pi)}{2} = -1/2$.
The residue of $\frac{\cos(\pi z)}{z^2-1} =\frac{H(z)}{z+1}, H(z) = \frac{\cos(\pi z)}{z-1}$ at $z = -1$ is $H(-1) = 1/2$.
The proof of the residue theorem in the case of fini... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Roots of Quadratic lies in $(0, 1)$ Given that the Quadratic equation $f(x)=ax^2-bx+c=0$ has two roots in $(0 \: 1)$
if $a,b,c \in \mathbb{N}$ Find Minimum values of $a$ and $b$
Since $a$ is Natural number graph of parabola will be open upwards.
Now $f(0) \gt 0$ and $f(1) \gt 0$ so we get
$c \gt 0$ and $a+c \gt b$ and ... | For $f(x) = ax^2 - bx + c$ , we get $f^{'}(x) = 2ax - b$.
Thus the minima is at $\displaystyle f^{'}(x) = 0 \iff 2ax-b = 0 \iff x = {b\over 2a}$.
Therefore a root must be in $[0, b/2a]$ and one in $[b/2a, 1]$.
And since $f(0) > 0$ therefore $f(b/2a) < 0$
therefore $f(b/2a) < f(0) \implies c < b^2/4a$
Similarly, $\dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given functions $f,g$ find the direct form for $f(x)=g(x)+f(x-1)$ Let $g(x)=1+x+\frac{x(x+1)}{2}$ and let $f(x)=g(x)+f(x-1)$. Find the closed form for $f(x)$ and $x\geq0$ given that $f(0)=0$.
So what we are asked are is a direct form for $$\sum_{n=1}^{k}g(n)$$
I want to show that $$\sum_{n=0}^{k}g(n)=\frac{1}{2}(\sum_{... | Use generating functions.
Define $f(-n) = 0$ and say that for all $n \geq 1$ we have $$f(n) = 1+n+\frac{1}{2} n(n+1) + f(n-1)$$
Then $$F(y) := \sum_{n=1}^{\infty} f(n) y^n = \sum_{n=1}^{\infty} [1+n+\frac{1}{2}n(n+1) + f(n-1)] y^n \\
= \sum_{n=1}^{\infty} y^n + \sum_{n=1}^{\infty} n y^n + \sum_{n=1}^{\infty} \frac{n(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2234008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Taylor series expansion of $\frac{(1- \cos x)}{x^2} $ According to my calculations, the Taylor series expansion for $1- \cos x$ around $x=\pi$ is
$$2 - \frac{(x-\pi)^2}{2} + \frac{(x-\pi)^4}{24} - \frac{(x-\pi)^6}{720} + \frac{(x-\pi)^8}{40320} -\dotsb
$$
In solving the Taylor series expansion of $\frac{(1- \cos x)}{x... | As $x \to \pi$, you may express $\dfrac1{x^2}$ as
$$
\begin{align}
\dfrac1{x^2}&=\dfrac1{(\pi+\color{red}{x-\pi})^2}
\\\\&=\dfrac1{\pi^2} \cdot \dfrac1{\left(1+\frac{\color{red}{x-\pi}}\pi\right)^2}
\\\\&=\dfrac1{\pi^2} \cdot \sum_{n=0}^\infty\frac{(-1)^n(n+1)}{\pi^n}\left({\color{red}{x-\pi}}\right)^n
\end{align}
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show that there is a unique polynomial $\int_{x}^{x+1} b_{n}{(t)} dt = x^n$ For each n = 1, 2, 3, . . . show that there is a unique polynomial $ b_{n}{(x)} $ that
satisfies the equation
$\int_{x}^{x+1} b_{n}{(t)} dt = x^n$
n = 1, I find $ b_{1}{(t)} = t + x_{0} $ and find $x_{0} = -\frac{1}{2}$
n = 2, $ b_{2}{(t)} = t^... | First observe that
$$
\int_x^{x + 1} t^n \,dt = \frac{(x + 1)^{n + 1} - x^n}{n+1} = \frac{1}{n + 1} \sum_{k = 0}^n {n + 1 \choose k} x^k.
$$
Therefore
$$\int_x^{x+1} t^n \,dt = x^n + \text{lower order terms}.$$
Therefore the matrix which writes $\{\int_{x}^{x+1} t^n \,dt : n = 0, 1, 2, 3\dots \}$ in terms of $\{1,x,x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to turn $12x^2-8x+1$ into $(2x-1)(6x-1)$ without quadratic equation? After almost seven years I recently started again learning math and have few holes in my algebra knowledge, so I apologize for the beginner question.
My question is:
Is there any simple trick to turn $12x^2-8x+1$ into $(2x-1)(6x-1)$ without using ... | Not sure what you mean by "without quadratic equation" but I'll assume it means without using the quadratic formula. So we can factorise it by completing the square:
Let $f(x) = 12x^2-8x+1$, then:
$f(x) = 12(x^2-\frac{8}{12}x+\frac{1}{12})=12(x^2-\frac{2}{3}x+\frac{1}{12})$
$f(x) = 12[(x-\frac{1}{2}\cdot\frac{2}{3})^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 2
} |
Mapping of a horizontal line $y = c$ by $w = \frac{1}{z}$ onto which region? I was trying to solve this problem ,
so $y = c$. Suppose for ease let us take $c > 0$.
so $w = \frac{1}{z} = \frac{1}{x + iy} = \frac{x - iy}{x^2 + y^2} = \frac{x - ic}{x^2 + c^2} = \frac{x}{x^2 + c^2} - i( \frac{c}{x^2 + c^2})$.
Now as $w = ... | Hint: Consider the absolute value of $w+\frac i{2c}$.
Also look at Inversive Geometry.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If matrix $A$ in $\mathbb{R}^3 $ such that, $A^3 = I$, $\det A = 1$. Is there a such matrix which is not orthogonal, rotation and identity? I tried to use Cayley–Hamilton theorem to learn something about the matrix.
Using the theorem we have:
$p(A)=0 = - A^3 + \text{tr} A\cdot A^2 -\left(\begin{vmatrix}a_{11} && a_{12}... | Here's a quick example:
$$
A = \pmatrix{\cos(2\pi/3) & -10\sin(2 \pi /3) & 0\\
\frac 1{10}\sin(2 \pi /3) & \cos (2 \pi /3) & 0\\0 & 0 & 1}
$$
Is like the usual rotation, but with the off-diagonal entries slightly altered. Show that $A^3 = I$, but $A$ is not orthogonal (or a rotation) since $A^TA \neq I$.
Another eas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15. Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15.
My instinct was to find the primitive root and then use a theorem to directly show the number of incongruent solutions, which follows from knowing the primitive root. B... | $x^2\equiv 1 \pmod 3$ and $x^2\equiv 1 \pmod 5$. Now, $x\equiv 1 \text{ or } 2 \pmod 3$. There are $2$ solutions. Also $x\equiv 1 \text{ or } 4 \pmod 5$. There are $2$ solutions. By Chinese remainder theorem: $x^2\equiv 1 \pmod {15}$ has $2\cdot 2=4$ incongruent solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$ Here is my question: Prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$.
They gave me a hint that I should consider the triple $3x+2z+1, 3x+2z+2, 4x+3z+2$, but I honestly do not know h... | Start with the hint . . .
Let $a,b,c$ be given by
\begin{align*}
a &= 3x+2z+1\\[4pt]
b &= 3x+2z+2\\[4pt]
c &= 4x+3z+2
\end{align*}
where $x,z$ are unknown positive integers.
If we were lucky, the equation $a^2 + b^2 = c^2$ would hold identically, for all $x,z$.
Let's try . . .
\begin{align*}
a^2 + b^2 - c^2 &= (3x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Let $( \sqrt{2} + 1)^{1000} = a + b \sqrt{2}$, where a and b are integers. Find the greatest common factor of b and 81. Let $( \sqrt{2} + 1)^{1000} = a + b \sqrt{2}$, where $a$ and $b$ are integers. Find the greatest common factor of $b$ and $81$.
How would one solve this question? I tried to use the binomial theorem b... | It is a bit annoying, but certainly feasible, to compute the answer by repeated squaring in $(\mathbb{Z}/81\mathbb{Z})[\sqrt{2}]$.
$$(1+\sqrt{2})^2 \equiv 3+2\sqrt{2}$$
$$(1+\sqrt{2})^4 \equiv 17+12\sqrt{2}$$
$$(1+\sqrt{2})^8 \equiv 10+3\sqrt{2}$$
$$(1+\sqrt{2})^{16} \equiv 37+60\sqrt{2}$$
$$(1+\sqrt{2})^{32} \equiv 64... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Least possible polynomial deegree of complex roots What is the least possible deegree of polynomial with real coefficients having roots $2\omega , 2+3\omega , 2+3\omega ^2 , 2-\omega -\omega ^2$
As there are four roots so the deegree should be four but the answer is given as five . how ?
| The least possible degree of polynomial with real coefficients having these roots is the product of $(x-\alpha)(x-\bar\alpha)$, where $\alpha$ are the given roots, unless $\alpha$ is real, in which case the second factor is omitted.
Since $\bar\omega=\omega^2$ and $1+\omega+\omega^2=0$, the polynomial is
$$
(x-2\omega)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding $\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}$ Finding $$\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}$$
Attempt: $$\lim_{n\rightarrow \infty}\bigg[\frac{1}{n^n}+\frac{2^2}{n^n}+\frac{3^3}{n^n}+\cdots \cdots +\frac{n^n}{n^n}\bigg] = 1$$
because all terms are approaching to z... | Bounding by a geometric series,
$$
\begin{align}
&\frac{n^n}{n^n}+\frac{(n-1)^{n-1}}{n^n}+\frac{(n-2)^{n-2}}{n^n}+\cdots+\frac{1^1}{n^n}\\
&\le1+\frac1n+\frac1{n^2}+\frac1{n^3}+\cdots\\
&=\frac{n}{n-1}
\end{align}
$$
Since the sum is obviously always $\ge1$, and $\le\frac{n}{n-1}$, the Squeeze Theorem says that the lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Number of integer partitions $p(n,m)$ In "Integer partitions" by Andrews and Eriksson, the authors provide formulas to compute $p(n,m)$, i.e., the number of partitions of $n$ into parts less than or equal to $m$, for $m=1,2,3,4,5$. As discussed in this question, it seems that no formula for $m>5$ is known.
Howevere, th... | Via Flajolet-Sedgewick: $p_m(z)=\frac{1}{\left(1-z\right) \left(1-z^2\right)\text{...} \left(1-z^m\right)}$ is an ordinary generating function. For general k, the coefficients are "polynom + some small (O(1)) periodic function". Using Mathematica,
$$ p(n,5) = \frac{30 n \left(15 \left(3 (-1)^n+85\right)+n (n (n+30)+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2244719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve $2^{x+5} - 2^{x+2} = 7$ I have a logarithms related problem at hand which I need to know its related formulas.
$$2^{x+5} - 2^{x+2} = 7$$
I already know the answer which is: $-2$.
So far, I have reached this form:
$$x = \frac{\ln(7 + 2^x+2)}{\ln (2)} - 5.$$
Any tips or hints will be appreciated. Thanks in advance... | Note that $2^{x+5} = 2^x\cdot 2^5 = 32\cdot 2^x$ and $2^{x+2} = 2^x\cdot 2^2 = 4\cdot 2^x$ so your expression becomes
$$ 32\cdot 2^x - 4\cdot 2^x = 7.$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve $2^{33} \mod 4725$? Basically what I should do is:
$$2^{33} \equiv x \pmod{4725}.$$
I need to find $x$ which will give same result as $2^{33}$ when modulo $4725$.
I should find prime factors of $4725$ which are $4725 = 3^3 × 5^2 × 7^1$, and use them to calculate
\begin{align*}
2^{33} &\equiv x \pmod{3^3}\\... | I will compute $2^{33}\pmod {27}$ as the OP is ok with the rest of the calculation.
In what follows $\equiv $ denotes congruence $\pmod {27}$.
Method I (iterated squaring): This works unusually well in this case because $33=32+1$ is so near a power of $2$. We write:
$$2^2=4\implies 2^4=16\implies 2^8=16^2\equiv 13 \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Equation with integrals and absolute complex log function problem (physics related) I was working on a problem that I described in a previous question. The problem is the following: I need to solve this for $\theta$:
$$\frac{\text{n}}{\text{A}}\int_0^{2\pi}\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\tex... | Using Wolfram :
$$\int_0^\theta\frac{1}{\left(1+\epsilon\cos\left(x\right)\right)^2}\space\text{d}x=\frac{\epsilon}{\epsilon^2-1}\cdot\frac{\sin\left(\theta\right)}{1+\epsilon\cos\left(\theta\right)}-\frac{2}{\sqrt{\left(\epsilon^2-1\right)^3}}\cdot\tanh^{-1}\left({\frac{\epsilon-1}{\sqrt{\epsilon^2-1}}\cdot\tan\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Limit of trigonometric function $\lim_{x\to\pi/3} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})}$ I want to compute this limit:
$\displaystyle \lim_{x\to\pi/3} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})}$
Using L'Hopital, is easy to get the result, which is $\sqrt{3}$
I tried using linear approximation (making $u = x - ... | Note that $\cos (\pi/3) = 1/2$ and hence we can write $$\lim_{x \to \pi/3}\frac{1 - 2\cos x}{\sin (x - \pi/3)} = -2\lim_{x \to \pi/3}\frac{\cos x - \cos (\pi/3)}{x - \pi/3}\cdot\frac{x - \pi/3}{\sin (x - \pi/3)} = -2(-\sin \pi/3)\cdot 1 = \sqrt{3}$$ here we have used the fact that $(\cos x)' = -\sin x$ so that $$\lim_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
If $\frac{a}{b} = \frac{c}{d}$ why does $\frac{a+c}{b + d} = \frac{a}{b} = \frac{c}{d}$? Can anyone prove why adding the numerator and denominator of the same ratios result in the same ratio? For example, since $\dfrac{1}{2}=\dfrac{2}{4}$ then $\dfrac{1+2}{2+4}=0.5$.
| An alternative solution, not to disparage the other answers.
$$\begin{aligned}
\frac{a}{b}=\frac{c}{d}\quad&\Rightarrow\quad\frac{ad}{b}=c&\text{solve for $c$}\\
\frac{a+c}{b+d}&=\frac{a+\left(\frac{ad}{b}\right)}{b+d}&\text{substitute $c$}\\
&=a\cdot\frac{1+\left(\frac{d}{b}\right)}{b+d}&\text{factor $a$ from numerato... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 2
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Expressing partial fractions in ascending powers of $x$ Let $f(x) = \frac {5x^2 + x + 6}{(3-2x)(x^2 + 4)}$
$i) $ Express $f(x)$ in partial fractions.
$ii)$ Hence obtain the expansion of $f(x)$ in ascending powers of $x$, up to and including the term in $x^2$.
I got part $i)$ but how do I do part $ii)$, I know how to th... | Note that we can expand $\frac{3}{3-2x}$ as a geometric series given by
$$\begin{align}
\frac{3}{3-2x}&=\frac{1}{1-(2/3)x}\\\\
&=\sum_{n=0}^\infty \left(\frac23\right)^n\,x^n\\\\
&=1+\frac23x+\frac49x^2+O(x^3)
\end{align}$$
for $|x|<3/2$.
Analogously, we can expand $\frac{-x-2}{x^2+4}$ as a geometric series given by
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Polynomials $P(x)$ such that $P(x-1)$ $=$ $P(-x)$ Are there an infinite number of polynomials $P(x)$ of even degree such that
$P(x-1) = P(-x)$
$P(x) = x^2+x+1$ is a good example because $P(x-1) = (x-1)^2+(x-1)+1 = x^2-x+1 = P(-x)$.
$P(x) = x^4+2x^3+4x^2+3x+1$ is another example because $P(x-1) = (x-1)^4+2(x-1)^3+4(x-1... | One easy way to generate infinitely many such polynomials is to start with your example $P(x)=x^2+x+1$ and consider $P(x)^2,P(x)^3,P(x)^4,\dots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$\cos^2(\frac{\pi}{101})+\cos^2(\frac{2\pi}{101})+\cos^2(\frac{3\pi}{101})+...+\cos^2(\frac{100\pi}{101})=?$
Find the value: $$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cd... | $$\cos\left(\frac{k\pi}{101}\right)= \frac{1}{2} \left(e^{i\frac{k\pi}{101}}+e^{-i\frac{k\pi}{101}} \right) \\
\cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \left(e^{2i\frac{k\pi}{101}}+e^{-2i\frac{k\pi}{101}} +2\right) \\
\sum_{k=1}^{100}\cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \sum_{k=1}^{100}\left(e^{2i\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Prove $f(x) =\ \frac{x^5}{5!} +\frac{x^4}{4!} +\frac{x^3}{3!}+\frac{x^2}{2!} +x+1$ has only one root. We have to prove that the equation $\displaystyle \frac{x^5}{5!} +\frac{x^4}{4!} +\frac{x^3}{3!}+\frac{x^2}{2!} +x+1=0$ have exactly one real root .
My sir told me it is just an application of derivative .
But I could... | With a little inspection you can note that you have a truncated exponential function, which is monotonically increasing, and as pointed out can only have one real root. However, letting $f(x) = \frac{x^5}{120} + \frac{x^4}{24} + \frac{x^3}{6} + \frac{x^2}{2} + x + 1$, and noting that
$$f'(x) = \frac{x^4}{24} + \frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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show that $\sum_{i = 1}^{2k}\frac{ (-1)^{i+1}}{i} = \sum_{i = k+1}^{2k} \frac{1}{i}$ I have a proof but it does not seem elegant. Is there a more elegant solution? Thanks.
Consider $X = \sum_{i = 1}^{2k}\frac{(-1)^{i+1}}{i} = X_1 + X_2$ where
$X_1 = 1 - \frac{1}{2} + ... + \frac{1}{k-1} - \frac{1}{k}$
$X_2 = \frac{1}{k... | Let $S_k=\sum_{k=1}^{2k}(-1)^{i+1}/i$ and $T_k=\sum_{i+1}^{2i}1/i$. Then
$$T_k-T_{k-1}=\frac1{2k-1}+\frac1{2k}-\frac1k=\frac1{2k-1}-\frac1{2k}=
S_k-S_{k-1}.$$
Obviously, $S_0=T_0$ so by induction....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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How to find two vectors orthogonal to the gradient space of the feasible set? After finding the solution to the minimization of $f(x,y,z)=\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2+\left(z-\frac{1}{2}\right)^2$ constrained by $h(x,y,z)=x^2+y^2+z^2-1=0$ by using the Lagrange multiplier method:
$$\nabla f(... | The optical inspection method works. The target vector is of the form
$$
u =
\left[ \begin{array}{c}
1 \\ 1 \\ 1
\end{array} \right]
$$
Two orthogonal vectors are
$$
v_{1} =
\left[ \begin{array}{r}
1 \\ 0 \\ -1
\end{array} \right],
\quad
v_{2} =
\left[ \begin{array}{r}
1 \\ -1 \\ 0
\end{array} \right].
$$
The pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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X and Y intercept basic issue Finding the Intercepts of the Graph of an Equation
Given an equation involving x and y, we find the intercepts of the graph as follows
*
*x-intercepts have the form (x,0); set y = 0 in the equation and solve for x.
*y-intercepts have the form (0,y); set x = 0 in the equation and solve f... | For the $x$-intercept: You need to isolate the square root before you square both sides. So when you have
$$ 0 = 2\sqrt{x+4}-2,$$
you don't want to square both sides here because you will have to FOIL out the right side like
$$0^{2} = (2\sqrt{x+4}-2)^{2}$$
and you will end up with a mess.
If you get the square root by... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int \frac {1}{(x+2)(x+3)} \textrm {dx}$ Integrate $\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$
My Attempt:
$$\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$$
$$\int \dfrac {1}{x+2} \textrm {dx} . \int \dfrac {1}{x+3} \textrm {dx}$$
$$\dfrac {\textrm {log (x+2)}}{1} . \dfrac {\textrm {log (x+3)}}{1} + C$$
$$\textrm ... | Nope! That's not how integration works.
You want to split the denominator using partial fractions.
$\displaystyle\int \frac{1}{(x+3)(x+2)}\,dx =\int \left(\frac{1}{x+2}-\frac{1}{x+3}\right)\,dx=\ln|x+2|-\ln|x+3|+C=\boxed{\ln \left|\frac{x+2}{x+3}\right|+C}$
| {
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"source": "stackexchange",
"question_score": "3",
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Finding all real values of $a$ for which $\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$ Find all real values of $a$ for which $\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$ for all $x>0$
Attempt: Let $$\sqrt{9-(x-a)^2}> \sqrt{16-x^2}$$
So $$x^2-(x-a)^2>7$$
So $$a(2x-a)>7\Rightarrow a^2-2ax+7<0$$
could some help me how to find range of $a,... | For the square roots to be defined, we need $|x|\leq 4$ and $|x-a|\leq 3$.
To solve a quadratic inequality, it's useful to find its roots. We have
$$a^2-2ax+7=0\iff a=\frac{2x\pm\sqrt{4x^2-28}}{2}=x\pm\sqrt{x^2-7}$$
In particular, if $|x|\leq\sqrt{7}$ the polynomial is always nonnegative in $a$.
If $\sqrt{7}<|x|\leq 4$... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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geometric series calculating the sum for first 45 terms I am able to calculate up to the step in blue, but I cant understand how do I simply the terms from -3^45 to 10^20 ?
Can anyone explain?
| $$S_{45}=-\frac{1}{8}(1-(-3)^{45})=-\frac{1}{8}+\frac{1}{8}(-3)^{45}$$
$$=-\frac{1}{8}-\frac{1}{8}(3)^{45}=-\frac{1}{8}-\frac{1}{8}(3^5)^9$$
Since $ $ $3^5=3\times3\times3\times3\times3=243$, we have
$$-\frac{1}{8}-\frac{1}{8}(243)^9=-\frac{1}{8}-\frac{1}{8}(2.43\times10^2)^9=-\frac{1}{8}-\frac{1}{8}(2.43^9\times10^{18... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Induction proof of divisibility I am trying to solve a proof by induction, which says
For all $n \in \mathbb{N}_0$,
$$ ((2n+1)^2 - 2^{n+1}(2n+1) + 2^{2n+1}) \mid ((2n+1)^4 + 4^{2n+1}). $$
Every try ended in a dead end. Does anybody have a hint?
| Notice that we can expand the right hand side as
\begin{equation}
\begin{split}
(2n+1)^4+4^{2n+1} &= \left((2n+1)^2+2^{2n+1}\right)^2-2^{2n+2}\cdot(2n+1)^2 \\&= \left((2n+1)^2-2^{n+1}\cdot(2n+1)+2^{2n+1}\right)\left((2n+1)^2+2^{n+1}\cdot(2n+1)+2^{2n+1}\right)
\end{split}
\end{equation}
Hence $$\left((2n+1)^2-2^{n+1}\cd... | {
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"source": "stackexchange",
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How do you prove that the sum of 1 to k divided by k is a multiple of 1/2? Expressed mathematically: $\frac{1}{k}\sum_{i=1}^k n = \frac{x}{2}$ where x is an integer. I tested this series mathematically and I found this trend, but I don't know how to actually prove it. Thanks in advance.
| Notice that $\frac 1k \sum\limits_{n=1}^k n$ is, by definition the average of $1,2,3,4......, .... k$. Convince yourself that the average of $1....n$ (equally distributed) is $\frac {k+1}2$.
.... or ...
Let $N = \sum\limits_{n=1}^k n = \sum\limits_{n=k;-1}^1 n = \sum\limits_{n=1}^k ((k+1) -n)$
So $2N = (\sum\limits_... | {
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"source": "stackexchange",
"question_score": "2",
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$p$,$q$ and $r$ are roots of $x^3+2x^2+3x+3=0$ $p$,$q$ and $r$ are roots of $x^3+2x^2+3x+3=0$. Find value of
$$\sum_{cyc}\left(\frac{p}{p+1}\right)^3$$
I have modified the equations as $$(x+1)^3=x^2-2$$ so
$$ \frac{x^3}{(x+1)^3}=\frac{x^3}{x^2-2}=x+\frac{1}{x-\sqrt{2}}+\frac{1}{x+\sqrt{2}}$$ So
$$\sum_{cyc}\left(\fr... | There is an easier way if $y=\frac{1}{x}$ then $y$ is a root of
$$3y^3+3y^2+2y+1=0$$ if $z=y+1$ then $z$ is a root of
$$3z^3-6z^2-6z-2=0$$ and if $w=\frac{1}{z}$ then $w$ is a root of
$$2w^3+6w^2+6w-3=0$$ and in the highly unlikely event I have not made a mistake, you can use Newton formula to find the sum of the cu... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding minimum of trigonometric function Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Then find difference between maximum and minimum of $v^2$.
I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?
I tried guessing, and got max... | We can restrict to $x\in[0,\pi/2]$, because of symmetries.
The maximum and minimum of $v$ are attained at the same values as the maximum and minimum of $v^2$:
$$
v^2=a^2+b^2+
2\sqrt{(a^2\cos^2(x)+b^2\sin^2(x))(b^2\cos^2(x)+a^2\sin^2(x))}
$$
We can also remove $a^2+b^2$, then the factor $2$ and square again, so we reduc... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Find all functions $f$ such that $f(x)+f(\frac{1}{1-x})=x$ I would like to find all functions $f:\mathbb{R}\backslash\{0,1\}\rightarrow\mathbb{R}$ such that
$$f(x)+f\left( \frac{1}{1-x}\right)=x.$$
I do not know how to solve the problem. Can someone explain how to solve it?
In one of my attempts I did the following, ... | make $x:= \frac{1}{1-x}$ then
$$f\left( \frac{1}{1-x}\right)+f\left( \frac{1}{1-\frac{1}{1-x}}\right)=\frac{1}{1-x}\to
f\left( \frac{1}{1-x}\right)+f\left(1- \frac{1}{x}\right)=\frac{1}{1-x}\quad (1)$$
do it again in the last equation:
$$f\left( \frac{1}{1-\frac{1}{1-x}}\right)+f(x)=\frac{1}{1-\frac{1}{1-x}}\to f\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Is there a possible explanation in plain English how to use the Chinese Reminder Theorem? For example, if it is the problem of
Find the smallest integer that leave a remainder of 3 when divided by
5, a remainder of 5 when divided by 7, and a remainder of 7 when
divided by 11
In some explanation such as in this ar... | It's definitely worthwhile learning a little bit about modular arithmetic for questions such as this.
But if you want to know in very concrete terms how to solve questions such as this, here's an attempt.
Suppose you want a number $n$ (not necessarily the least, we'll handle that later) with remainder $3$ when divided ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Problem in Rationalization The Question is :
If $1 < a <2$ then
$\sqrt{a - 2 \sqrt{a - 1}} - \sqrt{a + 2 \sqrt{a - 1}}$ can be:
A) $2$
B)$-2 \sqrt{a-1}$
C) $0$
D) $\sqrt{a-1}$
I have tried solving it a number of times but failed every time.
I took $\sqrt{a-1}$ as b then tried to rationalise by multiplying with unity... | $$k=\sqrt{a-2\sqrt{a-1}}-\sqrt{a+2\sqrt{a-1}}$$
square both sides and get:
$$k^2=2a-2\sqrt{a^2-4(a-1)}=2a-2|a-2|$$
once $1<a<2$ then $|a-2|=2-a$, so
$$k^2=2a-2(2-a)=4a-4\to k=\pm2\sqrt{a-1}$$
but $k<0$ because $\sqrt{a-2\sqrt{a-1}}<\sqrt{a+2\sqrt{a-1}}$, so
$$k=-2\sqrt{a-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find these singular matrices? Let $X$ and $Y$ be two matrices different from $I$ such that $XY=YX$ and $X^n-Y^n$ is invertible for some natural number $n$ . If $X^n-Y^n=X^{n+1}-Y^{n+1}=X^{n+2}-Y^{n+2}$, then:
A) $I-X$ is singular
B) $I-X$ is singular
C) $X+Y=XY+I$
D) $(I-X)(I-Y)$ is non-singular
Like everyone, I... | Our assumptions are as follows:
$1.) \: X,Y \neq I;$
$2.) \:X \: \text{and} \:Y \:\text{commute;}$
$3.) \:\text{there is some }n \text{ such that }X^n - Y^n \:\text{is invertible;}$
$4.) \: \text{for that same }n \text{, we have that } X^n - Y^n = X^{n+1} - Y^{n+1} = X^{n+2}-Y^{n+2}.$
$\text{--} $
By 2.), it is easily ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrals with Taylor expansion? Can I use the Taylor series to find this integrals?
$$ \lim_{n\to \infty } \int _{0}^n \frac{\arctan x}{x^2 + x + 1}dx $$
$$\lim _{n\to \infty }n \int_{-1}^0(x + e^x)^{n}dx = \: ?$$
| Here is a relatively simple standard approach to evaluating the first of your integrals.
Let
$$I = \int_0^\infty \frac{\arctan x}{x^2 + x + 1} \, dx.$$
Dividing up the interval of integration as follows
$$I = \int_0^1 \frac{\arctan x}{x^2 + x + 1} \, dx + \int_1^\infty \frac{\arctan x}{x^2 + x + 1} \, dx.$$
In the se... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What is the result of the following binomial sum? When computing the expectation of $S_n^{-1}I_{[S_n > 0]}$ with $S_n \sim \text{binomial}(n, p)$, I need to evaluate the sum:
\begin{align*}
\sum_{k = 1}^n \frac{1}{k}\binom{n}{k}p^{k}(1 - p)^{n - k}
\end{align*}
for $p \in (0, 1)$.
Is there any well-known binomial ... | Here is a transformation which simplifies the sum somewhat by extracting Harmonic numbers $H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$.
We obtain for $n\geq 1$
\begin{align*}
\color{blue}{f_n}&\color{blue}{=\sum_{k = 1}^n \frac{1}{k}\binom{n}{k}p^{k}(1 - p)^{n - k}}\\
&=\sum_{k = 1}^n \frac{1}{k}\left(\binom{n-1}{k}+\bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2271526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Factorial expression of $\frac{n!}{(n-r)!}$ If $n \gt r,$
Then how is the following valid?
$$n \cdot (n-1) \cdot (n-2) \cdots (n-(r-1))=\frac{n!}{(n-r)!}?$$
I thought it the list of multiplications would be equal to $\frac{n!}{(n-(r-1))!}?$
| $\require{cancel}$Maybe a concrete example will make things clearer:
\begin{align}
& \frac{9!}{(9-4)!} = \frac{9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{5\cdot4\cdot3\cdot2\cdot1} = \frac{9\cdot8\cdot7\cdot6\cdot\cancel{5\cdot4\cdot3\cdot2\cdot1}} {\cancel{5\cdot4\cdot3\cdot2\cdot1}} = 9\cdot8\cdot7\cdot6 \\[1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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$y''+\frac{y'}{x^2}-\frac{a^2}{x^2}y = 0$ by Frobenius, finding $s$ in $\sum a_n x^{n+s}$ I need to solve:
$$y''+\frac{y'}{x^2}-\frac{a^2}{x^2}y = 0, \ \ \ a\ge 0$$
by Frobenius method. So I did:
$$y = \sum_{n=0}^{\infty}a_nx^{n+s}$$
$$y' = \sum_{n=0}^{\infty}a_n(n+s)x^{n+s-1}$$
$$y'' = \sum_{n=0}^{\infty}a_n(n+s)(n+s-... | The coefficient of $y'$ is $1/x^2$, so $x=0$ is not a regular singular point. The Frobenius method won't work here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the sum of the power series $\sum\limits_{n=1}^\infty \frac{(n+2)!}{(2!)(n!)}x^n$
Find the sum of the power series $\sum\limits_{n=1}^\infty \frac{(n+2)!}{(2!)(n!)}x^n$
frist I use $$\sum_{n=1}^\infty x^n = \frac{x}{1-x}$$
and multiple two side by $x^2$
can get $$\sum_{n=1}^\infty x^{n+2} = \frac{x^3}{1-x}$$
the... | Starting from zero,
$\begin{array}\\
\sum\limits_{n=0}^\infty \frac{(n+2)!}{(2!)(n!)}x^n
&=\sum\limits_{n=0}^\infty \binom{n+2}{2}x^n\\
&=\frac1{x^2}(1-x)^{-3}\\
\end{array}
$
by the generalized binomial theorem
(since
$\binom{-n}{k}
=(-1)^n \binom{n+k-1}{k}
$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Error in Polynomial Factoring
From Ramanujan's Notebooks IV:
Let $\alpha,\beta$ and $\gamma$ be the roots of$$x^3-ax^2+bx-1=0\tag1$$Now, choose cube roots such that $(\alpha\beta\gamma)^{1/3}=1$ and then let$$z^3-\theta z^2+\varphi z-1=0\tag2$$Denote the cubic polynomial with roots $\alpha^{1/3},\beta^{1/3},\gamma^{1/... | The equation (4) is correct. After both sides of equation (3) are cubed and the terms on the right get moved to the left there is only zero on the right. Now the cube of the right side of (4) is expanded by binomial theorem to four terms. Two of them are combined and factored with one factor being the right side of (3)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Maximum value for $M^4$
$$M=|\sqrt{x^2+4x+5}-\sqrt{x^2+2x+4}|$$
Find the maximum value of $M^4$.
I think it could have a geometric solution. Because it looks like the difference between two points formula. Please help.
| Assuming $x \in \mathbb{R}$,
$$M(x) = \left \lvert \sqrt{x^2 + 4 x + 5} - \sqrt{x^2 + 2 x + 4} \right \rvert$$
and
$$M(x)^4 = (M(x)^2)^2 = g(x)^2$$
where
$$g(x) = \left ( \sqrt{x^2 + 4 x + 5} - \sqrt{x^2 + 2 x + 4} \right )^2$$
As Jaideep Khare and Abishanka Saha already succintly answered, we can write
$$M(x) = \left... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all primes such that $a^2+b^2=9ab-13$.
Let $a,b$ be primes. Find all primes such that $a^2+b^2=9ab-13$.
Whatever I've done is from parity checking. But I can't proceed with the case when both $a,b$ are odd primes.
I tried with some modulo-chasing but couldn't complete.
| Not a complete answer, just a few restrictions...
Proposition 1. For any prime $p>3 \Rightarrow 3 \mid p^2-1$ (from LFT) and $8 \mid p^2-1$ (from $(2k+1)^2 \equiv 1 \pmod{8}$). As a result $24 \mid p^2-1$.
For $a>3,b>3 \Rightarrow 24 \mid a^2-1$, $24 \mid b^2-1$
and $$24 \mid a^2 + b^2 -2=9ab-15$$
or
$$8 \mid 3ab-5 \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Find integer part of the maximum value of $x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$ for $x+y=3$
If $M$ is the maximum value of $x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$. Subject to $x+y=3$. Find the value of $\lfloor M \rfloor$.
I think the maximum of $M$ occurred when $x=y=3/2$. And I just put the value of that in M and found 34.... | Extended hint: let $s=x+y, \,p=xy\,$, then $\,x^2+y^2=s^2-2p, \;x^3+y^3=s^3-3sp\,$ so:
$$
\begin{align}
x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4 &= xy(x^3+x^2+x+1+y+y^2+y^3) \\
&= xy(1+x+y+x^2+y^2+x^3+y^3) \\
&= p(1+s+s^2-2p+s^3-3sp) \\
&= p(40-11p)
\end{align}
$$
The maximum value is attained for $p=\cfrac{20}{11}\,$ and ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to find the range of unknowns from a logarithmic equation? If the equation $\log(ax)\cdot \log(bx) + 1 = 0$ with $a>0, b>0$ constants has a solution $x>0$, it follows that $$b/a \geq (?)$$ or $$(?)\geq b/a >(?)$$
Hints maybe (for both :p)?
EDIT:
Answer $$b/a \ge 100$$ $$1/100 \ge b/a \ > 0$$
| $\begin{array}\\
-1
&=\log(ax)\cdot \log(bx) \\
&=(\log(a)+\log(x))(\log(b)+\log(x))\\
&=(A+X)(B+X)
\qquad\text{with } A=\log a, B=\log b, X = \log x\\
&=AB+X(A+B)+X^2\\
\end{array}
$
so
$X^2+X(A+B)+AB+1 = 0$.
Solving,
$\begin{array}\\
X
&=\dfrac{-A-B\pm\sqrt{A^2+2AB+B^2-4AB-4}}{2}\\
&=\dfrac{-A-B\pm\sqrt{A^2-2AB+B^2-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{(10^4+324)(22^4+324)\cdots(58^4+324)}{(4^4+324)(16^4+324) \cdots (52^4+324)}$
From AIME 1987, compute $$\frac{(10^4+324)(22^4+324)\cdots (58^4+324)}{(4^4+324)(16^4+324) \cdots (52^4+324)}$$
So basically the way used to solve this is by Sophie Germain's Identity which is $a^4+4b^4=(a^2+2b^2-2ab)(a^2+2b^2+2ab)$
... | Noticing that $x^4+324=0$ implies $x=-3\pm3i$ and $x=3\pm 3i$ each root corresponds to one factor. Now lets try to generalize our problem with placing $x$ instead of $7$.
$$\frac{((x+3)^4+324)((x+15)^4+324)((x+27)^4+324)((x+39)^4+324)((x+51)^4+324)}{((x-3)^4+324)((x+9)^4+324)((x+21)^4+324)((x+33)^4+324)((x+45)^4+324)}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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What are the integration limits for this volume? I had a test today and want to know whether I set the limits right for this volume computation. The solid is the one defined by $$z= 2-2x^2-y^2, x + y + z = 3, 2x^2 + y^2 = 2$$, and the limits I set are (in adapted cylindrical coordinates in which
$$x=r\cos t , y = \sq... | We can visualize the three surfaces as follows:
Looking at the image, we want the volume that is above the upside-down paraboloid, inside the elliptic cylinder, and below the plane.
What you are doing is not really converting to cylindrical coordinates, because the coordinate transformation is not:
$$ x = r \cos \thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cubic equation with unknown coefficients given roots I was given this equation. $x^3 + 3px^2 + qx + r=0$. The roots are $1, -1$, and $3$.
Ive tried dividing the equation by $(x-1)$ to get a quadratic to make it easier for me. But that ended up really badly.
I also inputted the different roots into the equation to get ... | Subbing in these roots to the equation gives us three equations.
$1 + 3p + q + r = 0$
$-1 + 3p -q + r = 0$
$27 + 27p + 3q + r = 0$
This can become a matrix equation
$\begin{bmatrix}1 \\ -1 \\ 27 \end{bmatrix} + \begin{bmatrix}3 & 1 & 1\\3 & -1 & 1 \\27 & 3 & 1\end{bmatrix} \begin{bmatrix}p \\ q \\ r \end{bmatrix} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the minimum value of $xy+yz+xz$ Question:
Find the minimum value of $xy+yz+xz$, given that $x,y,z$ are real and $x^2+y^2+z^2=1$
My attempt,
Since $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
$=1+2(xy+xz+yz)$
$(x+y+z)^2\geq0$
So that $1+2(xy+xz+yz)\geq0$
$(xy+xz+yz)\geq -\frac{1}{2}$
So the minimum value is $-\frac{1}{2}... | You are almost done. You are right that, since $(x+y+z)^2 \geq 0$, we have $xy+xz+yz \geq -\frac{1}{2}$. Now, to find the actual minimum, turn the inequality sign into an equality sign:
$$x+y+z = 0 \Rightarrow xy+xz+yz=-\frac{1}{2}$$
Now, just find values of $x, y, z$ such that
$$\begin{cases} x^2+y^2+z^2=1 \\ x+y+z=0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the smallest cosntant $k>0$ such that $\frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b} \leq k(a+b+c)$ for every $a,b,c>0$. In the book 'Putnam and beyond', page $173$, it has the following problem:
Find the smallest cosntant $k>0$ such that
$$\frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b}... | The inequality is equivalent to $f(a,b,c) \leq k$ where
$$
f(a,b,c)=\frac{\frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b}}{a+b+c}
$$
Now, if we simplify $f(ta,tb,tc)$ we find that it is equal to $f(a,b,c)$ : it "remains
unchanged". For any $p_0=(a_0,b_0,c_0)$ with $a_0,b_0,c_0>0$, if we put $p_1=(a_1,b_1,c_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2288816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Infinite sum of rationals is irrational
a) Let $x$ be a number between $0$ and $1$. Let $a_1$ be the smallest positive integer such that $x_1=x-a_1^{-1}\geq 0$, let $a_2$ be the smallest positive integer such that $x_2=x_1-a_2^{-1}\geq 0$, etc. Show that this leads to a finite expansion $$x=\frac{1}{a_1}+\frac{1}{a_... | To solve (b), we can use (a). Let $x=\sum b_k^{-1}$, and for $n\ge 1$, let $x_n=\sum_{k>n} b_k^{-1}$.
We have to show $b_{n + 1}$ is the smallest integer s.t. $x_n - b_{n+1}^{-1}\ge 0$ for all $n$.
Let $n\in\mathbb{N}$. On the one hand, $x_n - b_{n+1}^{-1}=x_{n+1}$ is indeed positive. On the other hand
\begin{eqnarray... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Can anyone help to proof convergence and find the sum of such series? May be correct my mistakes. $$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$
I personally have such an idea: try to make geometric series like this
$$1-\frac{1}{\sqrt{10}}(1-\frac... | My idea: suppose that $$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$ is a convergent serie. Let $S$ the sum of the serie $$S= 1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}\left(1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\cdot\cdot\cdot\right)=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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$F(\alpha, \beta):F$ is a simple extension Let $K:F$ be a extension, $F$ characteristic other than $2$, $\alpha, \beta \in K-F$ with $\alpha \neq \beta$ and $\alpha \neq -\beta$ such that $\alpha^2, \beta^2 \in F$. I have to prove extension $F(\alpha, \beta):F$ is simple.
So as $F$ characteristic other than $n$, then $... | As with the hint of @Chickenmancer, consider $F(a+b)$. Then clearly $F(a+b) \subset F(a, b)$. We'll show we have the reverse inclusion and hence equality.
Since $(a+b) \in F(a+b)$ we must have that $(a+b)^2 = a^2 + 2ab + b^2 \in F(a+b)$.
Now, $a^2, b^2 \in F \subset F(a+b)$ gives us that $2ab \in F(a+b)$.
Now the ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Calculate $\lim_{x\to a}(2-\frac{x}{a} )^{\tan( \frac{\pi x}{2a})}$
I would like to calculate
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)},\quad a \in\mathbb{R}^* \,\,\text{fixed} $$
we've
$$\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}=e^{\tan\left... | Hint:
$$\left(2-\frac xa\right)^{\tan(\pi x/2a)}=\left[\left(1+\left(1-\frac xa\right)\right)^{1/(1-x/a)}\right]^{(1-x/a)\tan(\pi x/2a)}\\\stackrel{x\to a}\longrightarrow e^{2/\pi}$$
Since,
$$e=\lim_{u\to0}(1+u)^{1/u}$$
$$\frac2\pi=\lim_{u\to1}(1-u)\tan(\pi u/2)$$
for those who want to make this rigorous, use squeeze t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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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.