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Cross product and inverse of a matrix I would like to show that $\left(\begin{array}{ccc}
1 & s & s^2 \\
1 & t & t^2 \\
1 & u & u^2
\end{array}\right)$ has an inverse provided $s$, $t$ and $u$ are distinct.
I have tried to prove $A\cdot B\times C \neq 0$ without success.
I computed $A\cdot B\times C = tu^2-ut^2+st^2-su^2+s^2u-s^2t$. What to do next ?
| $\det\begin{pmatrix} 1 & s & s^2 \\ 1 & t & t^2 \\ 1 & u & u^2\end{pmatrix}$
$=\det\begin{pmatrix} 1 & s & s^2 \\ 1-1 & t-s & t^2-s^2 \\ 1-1 & u-s & u^2-s^2\end{pmatrix}$ (applying $R_2'=R_2-R_1$ and $R_3'=R_3-R_1$)
$=\det\begin{pmatrix} 1 & s & s^2 \\ 0 & (t-s) & (t+s)(t-s) \\ 0 & (u-s) & (u+s)(u-s)\end{pmatrix}$
$=(t-s)(u-s)\det\begin{pmatrix} 1 & s & s^2 \\ 0 & 1 & (t+s) \\ 0 & 1 & (u+s)\end{pmatrix}$
$=(t-s)(u-s)(u+s-t-s)=(t-s)(u-s)(u-t)\ne 0$ if $t,s,u$ are distinct.
Now ,we can use this to find the inverse.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An inequality for all natural numbers Prove, using the principle of induction, that for all $n \in \mathbb{N}$, we have have the following inequality:
$$1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n} \leq 2\sqrt n$$
| Try this:
The inequality hold for $k=1$, assume it holds for $k=n$. Now look at $k=n+1$:
$$
S_{n+1}=S_{n} + \frac{1}{\sqrt{n+1}} < 2\sqrt{n} + \frac{1}{\sqrt{n+1}}
$$
The last step is the assumption. So LHS is $2\sqrt{n} + \frac{1}{\sqrt{n+1}}$ and RHS is $2 \sqrt{n+1}$ (induction). Clearly
$$
2 \sqrt{n+1} - 2 \sqrt{n}=2(\sqrt{n+1} - \sqrt{n})=\frac{2}{\sqrt{n+1}+\sqrt{n}}
>\frac{1}{\sqrt{n+1}}
$$
Therefore,
$$
S_{n+1}=S_{n} + \frac{1}{\sqrt{n+1}} < 2\sqrt{n} + \frac{1}{\sqrt{n+1}}<2 \sqrt{n+1}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the sequence is convergent How can we show that the sequence $$a_n=\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3-n^2}$$ is convergent?
| HINT
The sequence converges.
Use the identity $$a- b = \dfrac{a^3 - b^3}{a^2 + b^2 + ab}$$
Move your cursor over the gray area below for the answer.
\begin{align}\sqrt[3]{n^3 + n^2} - \sqrt[3]{n^3 - n^2} & = \dfrac{(n^3 + n^2)- (n^3-n^2)}{(n^3+n^2)^{2/3} + (n^3-n^2)^{2/3} + (n^6-n^4)^{1/3}}\\ & = \dfrac{2n^2}{n^2 \left(\left(1+1/n \right)^{2/3} + \left(1-1/n \right)^{2/3} + (1-1/n^2)^{1/3}\right)}\\& = \dfrac2{\left(1+1/n \right)^{2/3} + \left(1-1/n \right)^{2/3} + (1-1/n^2)^{1/3}}\end{align}
| {
"language": "en",
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"source": "stackexchange",
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Find the volume of the solid under the plane $x + 2y - z = 0$ and above the region bounded by $y=x$ and $y = x^4$. Find the volume of the solid under the plane $x + 2y - z = 0$ and above the region bounded by $y = x$ and $y = x^4$.
$$
\int_0^1\int_{x^4}^x{x+2ydydx}\\
\int_0^1{x^2-x^8dx}\\
\frac{1}{3}-\frac{1}{9} = \frac{2}{9}
$$
Did I make a misstep? The answer book says I am incorrect.
| $\int_0^1\int_{x^4}^x{(x+2y)\mathrm{d}y\mathrm{d}x}$ You have set correctly the integral but you did not integrate correctly.
Be careful in $\int_{x^4}^x{(x+2y)}\mathrm{d}y$ you integrate in respect of $y$ so you should find
$$\int_{x^4}^x{(x+2y)}\mathrm{d}y=\left | xy+y^2 \right |_{x^4}^{x}=x^2+x^2-(x^5+x^{8}) $$
And finally you get $\int_0^1 2x^2-x^5-x^{8}\mathrm{d}y=\frac{2}{3}-\frac{1}{6}-\frac{1}{9}=\frac{7}{18}$
You could verify your results here
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Understanding what this probability represents Say I want the probability that a five card poker hand contains exactly two kings. This would be
$$\frac{{4\choose 2}{48 \choose 3}}{52\choose 5}$$
Now if I drop the $48 \choose 3$, which represents the 3 non king cards, what can
the probability $\frac{4\choose 2}{52\choose 5}$ be taken to represent?
Is it the number of hands containing at least 2 kings?
| In this context the fraction $$\frac{\binom42}{\binom{52}5}$$ has no very natural interpretation. The original fraction $$\frac{\binom42\binom{48}3}{\binom{52}5}$$ is another story: the numerator is the number of $5$-card hands containing exactly two kings, and the denominator is the number of $5$-card hands, so the fraction is the probability of being dealt a hand containing exactly two kings.
Just as there are $\binom42\binom{48}3$ hands with exactly two kings, there are $\binom43\binom{48}2$ hands with exactly three kings and $\binom44\binom{48}1$ hands with exactly four kings. Thus, the number of hands with at least two kings is $$\binom42\binom{48}3+\binom43\binom{48}2+\binom44\binom{48}1\;,$$ and the probability of being dealt such a hand is $$\frac{\binom42\binom{48}3+\binom43\binom{48}2+\binom44\binom{48}1}{\binom{52}5}\;.\tag{1}$$
Note: The number of hands with no kings is $\binom40\binom{48}5$, and the number with exactly one king is $\binom41\binom{48}4$, so the number with at most one king is $$\binom40\binom{48}5+\binom41\binom{48}4\;.$$ Thus, the number of hands with at least two kings is $$\binom{52}5-\left(\binom40\binom{48}5+\binom41\binom{48}4\right)\;,$$ the total number of possible hands minus the number having fewer than two kings. Thus, we could also have computed the probability in $(1)$ as
$$\frac{\binom{52}5-\left(\binom40\binom{48}5+\binom41\binom{48}4\right)}{\binom{52}5}=1-\frac{\binom40\binom{48}5+\binom41\binom{48}4}{\binom{52}5}\;.$$ This has a perfectly good intuitive significance: it’s $1$ minus the probability of getting a hand with fewer than two kings.
| {
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"url": "https://math.stackexchange.com/questions/222928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determining a limit I'm having troubles showing that
$$
\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = -1.
$$
In particular, why is the following derivation wrong?
$$
\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = \lim_{x\rightarrow -\infty} \frac{x\sqrt{1+2/x}}{x} = \lim_{x\rightarrow -\infty} \sqrt{1+2/x} = \sqrt{1+\lim_{x\rightarrow -\infty}(2/x)} = 1.$$
| Your mistake is here
$$
\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = \lim_{x\rightarrow -\infty} \frac{x\sqrt{1+2/x}}{x} = \ldots = 1.
$$
The correct is
$$
\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2x}}{x} = \lim_{x\rightarrow -\infty} \frac{-x\sqrt{1+2/x}}{x} = \ldots = -1
$$
since $x<0$ ($\sqrt{x^2}=|x|$, for example $\sqrt{(-1254)^2}=1254$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/225317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Using induction to prove $3$ divides $\left \lfloor\left(\frac {7+\sqrt {37}}{2}\right)^n \right\rfloor$ How can I use induction to prove that $$\left \lfloor\left(\cfrac {7+\sqrt {37}}{2}\right)^n \right\rfloor$$ is divisible by $3$ for every natural number $n$?
| Let $a=\dfrac{7+\sqrt{37}}{2}$, and $b=\dfrac{7-\sqrt{37}}{2}$. We first show that $a^n+b^n$ is always an integer.
We have
$$a^{n+1}+b^{n+1}=(a+b)(a^n+b^n)-ab(a^{n-1}+b^{n-1})=7(a^n+b^n)-3(a^{n-1}+b^{n-1}).\tag{$1$}$$
It is easy to check that $a^0+b^0$ and $a^1+b^1$ are integers. The others are dealt with by induction using Equation $(1)$.
Note that $b\lt 1/2$, and $b$ is positive. Since $a^n+b^n$ is an integer, it follows that $\lfloor a^n\rfloor=a^n+b^n-1$.
So we want to prove that $a^n+b^n\equiv 1\pmod{3}$. This is true for $n=0$ and $n=1$. Now from Equation $(1)$, on the assumption that $a^n+b^n \equiv 1\pmod{3}$, we obtain that $a^{n+1}+b^{n+1}\equiv 1\pmod{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/226177",
"timestamp": "2023-03-29T00:00:00",
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How to integrate $\frac{x^2}{1-x^2}$ I want to integrate
\[\frac{x^2}{1-x^2},\]
what I have try is trigonometric substitution and partition function and integration by part
but still cannot solve it
Thx for your reading!
| $$\frac{x^2}{1-x^2} = -\frac{x^2}{x^2 - 1} = -\frac{x^2 - 1 + 1}{x^2 - 1} = -1 - \frac{1}{x^2 - 1}$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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$5^{2^{a-3}} \equiv 1+2^{a-1}\pmod{2^a}$ Here's a tricky question I have in a homework problem:
Via induction, show
$5^{2^{a-3}} \equiv 1+2^{a-1}\pmod{2^a}$
and then deduce that $5\pmod{2^a}$ has order $2^{a-2}$ for $a \ge 3$
Help please I'm so lost with this question!!!
| Let $a^{2^b}=1+c2^{b+2}$ where $a=2d+1$ is odd
So,$a^{2^{b+1}}=(a^{2^b})^2=(1+c2^{b+2})^2=1+c2^{b+3}+2^{2(b+2)}c^2\equiv1\pmod {2^{b+3}}$ if $2b+4\ge b+3$ or if $b\ge 0$
Now, for $b=1,a^{2^1}=a^2=(2d+1)^2=1+8\frac{d(d+1)}2\equiv 1\pmod{2^{1+2}}$--> this is base case for the induction.
So, using induction we can prove the proposition for any odd number with the power of $2>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/229890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving a reduction formula for the antiderivative of $\cos^n(x)$ I want to show that for all $n\ge 2$, it holds that
$$
\int \cos^n x\ dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x\ dx.
$$
I'm not even getting the result for the induction base $(n=2)$:
Using integration by parts, I only get
$$
\int \cos^2 x\ dx = \cos x \sin x + \int \sin^2 x\ dx.
$$
I'm suspecting that I need to use some trigonometric identity here.
| You only need that $$\sin^2 x+\cos^2 x=1$$
Let $$\varphi(n)=\int \cos^n x dx$$
Integrate by parts
$$\begin{cases}\cos^{n-1} x =u\\ \cos x dx =dv\end{cases}$$
Then
$$\begin{align}\varphi(n)&=\int \cos^n x dx\\ &=uv-\int v du\\&=\cos^{n-1}x\sin x+\int (n-1)\cos^{n-2}x \sin ^2x d x\end{align}$$
But $\sin^2 x=1-\cos^2 x $
$$\begin{align}&=\cos^{n-1}x\sin x+\int (n-1)\cos^{n-2}x \sin ^2x d x \\&=\cos^{n-1}x\sin x+\int (n-1)\cos^{n-2}x \left(1-\cos^2 x\right) d x \\ &=\cos^{n-1}x\sin x+\int (n-1)\cos^{n-2}x dx-\int (n-1)\cos^{n-2}x \cos^2 x d x \\&=\cos^{n-1}x\sin xdx+(n-1)\int \cos^{n-2}x dx-(n-1)\int \cos^{n }x dx \\&=\cos^{n-1}x\sin xdx+(n-1)\int \cos^{n-2}xdx -(n-1)\varphi(n)\end{align}$$
Now solve for $\varphi(n)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ Prove that
$$
\frac{1}{\sqrt{1-4t}} \left(\frac{1-\sqrt{1-4t}}{2t}\right)^k = \sum\limits_{n=0}^{\infty}\binom{2n+k}{n}t^n,
\quad
\forall k\in\mathbb{N}.
$$
I tried already by induction over $k$ but i have problems showing the statement holds for $k=0$ or $k=1$.
| We proceed by induction on $k$.
For $k = 0$, we have $\displaystyle \sum\limits_{m=0}^{\infty} \binom{2m}{m}t^m = \frac{1}{\sqrt{1-4t}}$
and for $k = 1$, we have
\begin{align*} \sum\limits_{m=0}^{\infty} \binom{2m+1}{m+1}t^m &= \sum\limits_{m=0}^{\infty} \left(2 - \frac{1}{m+1}\right)\binom{2m}{m}t^m \\&= \frac{2}{\sqrt{1-4t}} - C(t)\\&= \frac{2}{\sqrt{1-4t}} -\frac{2}{1+\sqrt{1-4t}} \\&= \frac{2}{\sqrt{1-4t}(1+\sqrt{1-4t})}\end{align*}
Where, $\displaystyle C(t) = \sum\limits_{m=0}^{\infty} \frac{1}{m+1}\binom{2m}{m}z^m = \frac{2}{1+\sqrt{1-4t}}$ is the generating function of Catalan Numbers.
Assuming the result holds for $k$ we prove for $k+1$:
\begin{align*}\sum\limits_{m=0}^{\infty} \binom{2m+k+1}{m+k+1}t^m &= \sum\limits_{m=0}^{\infty} \left(2 - \frac{k+1}{m+k+1}\right)\binom{2m+k}{m+k}t^m\\
&= \frac{2^{k+1}(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}} - \frac{k+1}{t^{k+1}}\int_0^t \sum\limits_{m=0}^{\infty} \binom{2m+k}{m+k}t^{m+k}\,\mathrm{d}t\\
&= \frac{2^{k+1}(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}} - \frac{2^k(k+1)}{t^{k+1}}\int_0^t \frac{t^k(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}}\,\mathrm{d}t\\
&= \frac{2^{k+1}(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}} - \frac{k+1}{2^kt^{k+1}}\int_0^t \frac{(1-\sqrt{1-4t})^{k}}{\sqrt{1-4t}}\,\mathrm{d}t\\
&= \frac{2^{k+1}(1+\sqrt{1-4t})^{-p}}{\sqrt{1-4t}} - \frac{k+1}{2^{k+1}t^{k+1}} \frac{(1-\sqrt{1-4t})^{k+1}}{k+1}\\
&= \frac{2^{k+1}(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}} - 2^{k+1}(1+\sqrt{1-4t})^{-(k+1)}\\
&= \frac{2^{k+1}(1+\sqrt{1-4t})^{-(k+1)}}{\sqrt{1-4t}}\\
\end{align*}
| {
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Solve the recurrence relation:$ T(n) = \sqrt{n} T \left(\sqrt n \right) + n$ $$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$
Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.
The answer is $Θ (n \log \log n)$.
Can anyone arrive at the solution.
| Use substitution method:
$$\Large\begin{align*} \text{T}(n) &= \sqrt{n}\ \text{T}(\sqrt{n})+n\\ &= n^{\frac{1}{2}}\ \text{T}\left(n^{\frac{1}{2}} \right )+n\\ &= n^{\frac{1}{2}}\left( n^{\frac{1}{2^2}}\ \text{T}\left(n^{\frac{1}{2^2}} \right )+n^{\frac{1}{2}} \right )+n\\ &= n^{\frac{1}{2}+\frac{1}{2^2}}\ \text{T}\left(n^{\frac{1}{2^2}}\right ) +n^{\frac{1}{2}+\frac{1}{2}}+n\\ &= n^{\frac{1}{2}+\frac{1}{2^2}}\ \text{T}\left(n^{\frac{1}{2^2}}\right ) +2n\\ &= n^{\frac{1}{2}+\frac{1}{2^2}}\left(n^{\frac{1}{2^3}}\ \text{T}\left(n^{\frac{1}{2^3}}\right ) +n^{\frac{1}{2^2}} \right )+2n\\ &= n^{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}}\ \text{T}\left(n^{\frac{1}{2^3}}\right ) +n^{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^2}} +2n\\ &= n^{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}}\ \text{T}\left(n^{\frac{1}{2^3}}\right ) +3n\\ \vdots \\ &= n^{\sum_{i=1}^{k}\frac{1}{2^i}}\ \text{T}\left(n^{\frac{1}{2^k}}\right ) +kn\\ \end{align*}$$
assuming $\text{T}(2) = 2$, which is the least value of n that could be. So, $$\begin{align*} n^{\frac{1}{2^k}} &= 2\\ \frac{1}{2^k}\log_2(n) &= \log_2(2) \\ \log_2(n) &= {2^k} \\ \log_2\log_2(n) &= k\log_2(2) \\ \log_2\log_2(n) &= k \end{align*}$$
therefore, the recurrence relation will look like:
$$\large \begin{align*} \text{T}(n)&=n^{\sum_{i=1}^{k}\frac{1}{2^i}}\ \text{T}\left(n^{\frac{1}{2^k}}\right ) +kn\\ &=n^{\sum_{i=1}^{\log_2\log_2(n)}\frac{1}{2^i}}\ \text{T}\left(n^{\frac{1}{2^{\log_2\log_2(n)}}}\right ) +n \log_2\log_2(n)\\ \end{align*}$$
where, $$\sum_{i=1}^{\log_2\log_2(n)}\frac{1}{2^i} = 1-\frac{1}{\log_2(n)} = \text{fraction always, as }n\geq 2$$
so, $$\large \text{T}(n) = \mathcal{O}\left( n \log_2 \log_2 n \right)$$
| {
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"url": "https://math.stackexchange.com/questions/239402",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "13",
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Find the sum of the first $n$ terms of $\sum^n_{k=1}k^3$ The question:
Find the sum of the first $n$ terms of $$\sum^n_{k=1}k^3$$
[Hint: consider $(k+1)^4-k^4$]
[Answer: $\frac{1}{4}n^2(n+1)^2$]
My solution:
$$\begin{align}
\sum^n_{k=1}k^3&=1^3+2^3+3^3+4^3+\cdots+(n-1)^3+n^3\\
&=\frac{n}{2}[\text{first term} + \text{last term}]\\
&=\frac{n(1^3+n^3)}{2}
\end{align}$$
What am I doing wrong?
| Let $f(n)=\Big(\frac{n(n+1)}2\Big)^2$ , then$\space$$f(n-1)=\Big(\frac{n(n-1)}2\Big)^2$ ; now
we know that
$(n+1)^2 - (n-1)^2=4n$ ,
this implies
$\space$ $n^2\Big((n+1)^2 - (n-1)^2\Big)=4n^3=$$\big(n(n+1)\big)^2 - \big(n(n-1)\big)^2$
$\implies$ $\frac{\big(n(n+1)\big)^2}4 - \frac{\big(n(n-1)\big)^2}4=$ $\Big(\frac{n(n+1)}2\Big)^2-\Big(\frac{n(n-1)}2\Big)^2=n^3$$=f(n)-f(n-1)$
$\implies$ $\sum_{n=1}^m \Big(f(n)-f(n-1)\Big) =$$\sum_{n=1}^mn^3$$=f(m)-f(0)=f(m)=\Big(\frac{m(m+1)}2\Big)^2$ , where we
have used $f(0)=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the recurrence $T(n) = 2T(n-1) + n$ Solve the recurrence $T(n) = 2T(n-1) + n$
where $T(1) = 1$ and $n\ge 2$.
The final answer is $2^{n+1}-n-2$
Can anyone arrive at the solution?
| This recurrence $$T(n) = 2T(n-1) + n$$ is difficult because it contains $n$. Let $D(n) = T(n) - T(n-1)$ and compute $D(n+1) = 2D(n) + 1$ this recurrence is not so difficult. Of course $D(1) = 4 - 1 = 2 + 1$.
The sequence $D(n)$ goes: $2 + 1$, $2^2 + 2 + 1$, $2^3 + 2^2 + 2 + 1$. $D(n) = 2^{n+1}-1$.
Now $$T(n) = \sum_{i=1}^n D(i) = 2 \sum_{i=1}^n 2^{i} - \sum_{i=1}^n 1 = 2^{n+1}-2-n.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of an improper integral with trig function $$\int_0^{ + \infty } {\left( {\frac{x}{{1 + {x^6}{{\sin^2 x}}}}} \right)dx}$$
I'd like your help with see why does??
I tried to compare it to other functions and to change the variables, but it didn't work for me.
Thanks a lot!
| We can divide up the ray $[0,\infty)$ into subintervals $[(n-\frac{1}{2})\pi,(n+\frac{1}{2})\pi]$, $n=1,2,\ldots$ (there a little piece $[0,\pi/2]$ left over, which we may ignore). On each subinterval, we compute
$$
\begin{eqnarray*}
\int_{(n-\frac{1}{2})\pi}^{(n+\frac{1}{2})\pi}\frac{x}{1+x^6\sin^2 x}\,dx&\leq& (n+\frac{1}{2})\pi\int_{(n-\frac{1}{2})\pi}^{(n+\frac{1}{2})\pi}\frac{1}{1+x^6\sin^2 x}\,dx\\
&\leq&(n+\frac{1}{2})\pi\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{1}{1+(n-\frac{1}{2})^6\pi^2\sin^2 u}\,du\\
&\leq&(n+\frac{1}{2})\pi\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{1}{1+4(n-\frac{1}{2})^6u^2}\,du\\
&\leq&(n+\frac{1}{2})\pi\int_{-\infty}^{\infty}\frac{1}{1+4(n-\frac{1}{2})^6u^2}\,du\\
&=&\frac{\pi^2(n+\frac{1}{2})}{2(n-\frac{1}{2})^3}
\end{eqnarray*}
$$
These integrals are decaying like $\frac{1}{n^2}$, so the sum over all $n$ converges.
| {
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Example 2, Chpt 4 Advanced Mathematics (I)
$$\int \frac{x+2}{2x^3+3x^2+3x+1}\, \mathrm{d}x$$
I can get it down to this:
$$\int \frac{2}{2x+1} - \frac{x}{x^2+x+1}\, \mathrm{d}x $$
I can solve the first part but I don't exactly follow the method in the book.
$$ = \ln \vert 2x+1 \vert - \frac{1}{2}\int \frac{\left(2x+1\right) -1}{x^2+x+1}\, \mathrm{d}x $$
$$= \ln \vert 2x+1 \vert - \frac{1}{2} \int \frac{\mathrm{d}\left(x^2+x+1\right)}{x^2+x+1} + \frac{1}{2}\int \dfrac{\mathrm{d}x}{\left(x+\dfrac{1}{2}\right)^2 + \frac{3}{4}} $$
For the 2nd part:
I tried $ u = x^2+x+1 $ and $\mathrm{d}u = 2x+1\, \mathrm{d}x$ that leaves me with $\frac{\mathrm{d}u - 1}{2} = x\, \mathrm{d}x$ which seems wrong.
because $x^2+x+1$ doesn't factor, I don't see how partial fractions again will help.
$x = Ax+B$ isn't helpful.
| Let
$$x+\frac12=\frac{\sqrt3}2\tan t,dx=\frac{\sqrt3}2\sec^2tdt$$
$$\frac12\int\frac{dx}{\left(x+\frac12\right)^2+\frac34}=\frac{\sqrt3}4\int\frac{\sec^2tdt}{\frac34\tan^2t+\frac34}=$$
$$\frac{\sqrt3}4\int\frac{\sec^2tdt}{\frac34\sec^2t}=\frac{\sqrt3}3\int dt=\frac{\sqrt3}3t+C$$
Now just solve for $t$ in terms of $x$.
| {
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Showing $\sqrt{2}\sqrt{3} $ is greater or less than $ \sqrt{2} + \sqrt{3} $ algebraically How can we establish algebraically if $\sqrt{2}\sqrt{3}$ is greater than or less than $\sqrt{2} + \sqrt{3}$?
I know I can plug the values into any calculator and compare the digits, but that is not very satisfying. I've tried to solve $$\sqrt{2}+\sqrt{3}+x=\sqrt{2}\sqrt{3} $$ to see if $x$ is positive or negative. But I'm just getting sums of square roots whose positive or negative values are not obvious.
Can it be done without the decimal expansion?
| $$\sqrt{3}\sqrt{2}-\sqrt{3}-\sqrt{2}+1=(\sqrt{3}-1)(\sqrt{2}-1) < 1$$
The last inequality follows from the fact that $(\sqrt{3}-1)$ and $(\sqrt{2}-1)$ are in $(0,1)$.
Second solution
By AM-GM you have
$$2\sqrt[4]{6} \leq \sqrt{2}+\sqrt{3}$$
Combine this with $\sqrt[4]{6} < 2$, which is easy to prove, and you are done.
And a non-algebraic one, which is an overkill :)
Let $\theta$ be the angle so that $\cos(\theta)=-\frac{1}{2\sqrt{6}}$. Plot a point $A$ draw two rays with an angle of $\theta$ between them, and pick points $B$ respectively $C$ on these ray so that $AB=\sqrt{2}$ and $AC=\sqrt{3}$. By the cosine law
$$BC^2=2+3+2\sqrt{2}\sqrt{3}\frac{1}{2\sqrt{6}}=6$$
Thus the triangle $ABC$ has the edges of length $\sqrt{2}, \sqrt{3}$ and $\sqrt{6}$, and your inequality is exactly the triangle inequality.
| {
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How to solve $ \sum_{i=1}^{n-1} i^2 \equiv \;? \pmod n$ I solved expressions below . it is ok?
My attempt:
$$1)\;\;\;\;\; ?\cong\pmod n\sum_{i=1}^{n-1} i^2 \stackrel{?}= {(n-1)^3\over3}+{(n-1)^2\over2}+{(n-1)\over6}={n^3\over3}-{n^2\over6}+{n\over6}=n\left({n^2\over3}-{n\over2}+{1\over6}\right)
$$
(continue?)how to solve for all n?
$$2)\;\;\;\;\;?\cong\pmod n \sum_{i=1}^{n-1} i^3 = \left( {(n-1)n\over2} \right)^2$$
(continue?)how to solve for all n?
Thank you.
| You know that
First, since $n^2 =0 \pmod n$ and $n^3 =0 \pmod n$, you can calculate the sum up to $n$.
(i)
$$\sum_{i=1}^{n-1} i^2 =\sum_{i=1}^{n} i^2 =\frac{n(n+1)(2n+1)}{6} \pmod n$$
Thus, you have to decide which of the terms is divisible by 2,3. CRT tells you this is a problem modulo 6.
Now a case by case analysis solves it:
Case 1: $n=6k$. Then
$$ \frac{n(n+1)(2n+1)}{6} \equiv k (n+1)(2n+1) \equiv k \equiv \frac{n}{6} \pmod n$$
Case 2: $n=6k+1$. Then $n+1=2(3k+1)$ and $n+2=3(2k+1)$
$$ \frac{n(n+1)(2n+1)}{6} \equiv n(3k+1)(2k+1) \equiv 0 \pmod n$$
the other cases are similar.
(ii) This is easier
$$\sum_{i=1}^{n-1} i^3 =\sum_{i=1}^{n} i^3 =\frac{n^2(n+1)^2}{4} \pmod n$$
We only have two cases now, odd or even.
Case 1: $n=2k$. Then
$$ \frac{n^2(n+1)^2}{4} \equiv k^2 (n+1)^2 \equiv k^2 \equiv \frac{n^2}{4} \pmod n$$
Case 2: $n=2k+1$. Then $n+1=2(k+1)$
$$ \frac{n^2(n+1)^2}{2} \equiv n^2(k+1)^2 \equiv 0 \pmod n$$
| {
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Generating Pythagorean triples for $a^2+b^2=5c^2$? Just trying to figure out a way to generate triples for $a^2+b^2=5c^2$. The wiki article shows how it is done for $a^2+b^2=c^2$ but I am not sure how to extrapolate.
| This is one of those CW answers. Country and Western.
I did Gerry's recipe and I quite like how it works. Educational, you might say.
I took the slope $t = \frac{q}{r}$ and starting rational point $(2,1).$ The other point works out to be
$$ x = \frac{2 t^2 - 2 t - 2}{t^2 + 1}, \; \; \; y = \frac{- t^2 - 4 t + 1}{t^2 + 1}, $$ so multiply everything by $r^2$ to arrive at
$$ x = \frac{2 q^2 - 2qr - 2r^2}{q^2 + r^2}, \; \; \; y = \frac{- q^2 - 4 qr + r^2}{q^2 + r^2}. $$
So far $x^2 + y^2 = 5.$ Multiply through by $q^2 + r^2$ to get
$$ a = 2 q^2 - 2 q r - 2 r^2 $$
$$ b = -q^2 - 4 q r + r^2 $$
$$ c = q^2 + r^2 $$
$$ a^2 + b^2 = 5 q^4 + 10 r^2 q^2 + 5 r^4 $$
and
$$ c^2 = q^4 + 2 r^2 q^2 + r^4 $$
and
$$ a^2 + b^2 = 5 c^2 $$
| {
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How to simplify polynomials I can't figure out how to simplify this polynominal $$5x^2+3x^4-7x^3+5x+8+2x^2-4x+9-6x^2+7x$$
I tried combining like terms
$$5x^2+3x^4-7x^3+5x+8+2x^2-4x+9-6x^2+7x$$
$$(5x^2+5x)+3x^4-(7x^3+7x)+2x^2-4x-6x^2+(8+9)$$
$$5x^3+3x^4-7x^4+2x^2-4x-6x^2+17$$
It says the answer is $$3x^4-7x^3+x^2+8x+17$$ but how did they get it?
| Group the terms with the same power of $x$ together.
$5x^2+3x^4−7x^3+5x+8+2x^2−4x+9−6x^2+7x$
$=3x^4−7x^3+5x^2+2x^2−6x^2+5x−4x+7x+8+9$
$=3x^4−7x^3+x^2+8x+17$
| {
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Advanced integration, how to integrate 1/polynomial ? Thanks I have been trying to integrate a function with a polynomial as the denominator.
i.e, how would I go about integrating
$$\frac{1}{ax^2+bx+c}.$$
Any help at all with this would be much appreciated, thanks a lot :)
ps The polynomial has NO real roots
$${}{}$$
| The details depend on $a$, $b$, and $c$.
Assume $a\ne 0$. If there are two distinct real roots, use partial fractions.
If there are two identical real roots, we are basically integrating $\dfrac{1}{u^2}$.
If there are no real roots, complete the square. With the right substitution, you basically end up integrating $\dfrac{1}{1+u^2}$, and get an $\arctan$.
For polynomials of higher degree, factor as a product of linear terms and/or quadratics with no real roots. (In principle this can be done. In practice, it may be very unpleasant).
Then using partial fractions and substitutions you end up with integrals of $\dfrac{1}{u^{n}}$, and/or $\dfrac{u}{(1+u^2)^{n}}$ and/or $\dfrac{1}{(1+u^2)^{n}}$. All of these are doable.
Added: It turns out the OP was interested in the irreducible case. Will write a bit on that, because I want to advocate a procedure slightly different from the standard one. Assume that $a$ is positive.
Rewrite $\dfrac{1}{ax^2+bx+c}$ as
$\dfrac{4a}{4a^2x^2+4abx+4ac}$, and then, completing the square, as $\dfrac{4a}{(2ax+b)^2+4ac-b^2}$.
Note that $4ac-b^2$ is positive. Call it $k^2$, with $k$ positive.
Make the change of variable $2ax+b=ku$. Substitute. There is some cancellation, and we end up integrating $\dfrac{2}{k}\dfrac{1}{1+u^2}.$
I would suggest going through this procedure in any individual case. As an example, with $\dfrac{1}{x^2+x+1}$ we write $\dfrac{4}{4x^2+4x+4}$, then $\dfrac{4}{(2x+1)^2+3}$, make the change of variable $2x+1=\sqrt{3} u$.
Another addition: The OP has expressed a wish to see the particular problem $\int\frac{dx}{x^2+10x+61}$. The numbers here are particularly simple, designed for the "standard" style, so we will do it that way. Also, we will use more steps than necessary.
First we complete the square. We get $x^2+10x+61=(x+5)^2-25+61=(x+5)+36$.
Now let $u=x+5$. Then $du=dx$ and
$$\int\frac{dx}{x^2+10x+61}\int \frac{dx}{(x+5)^2+36}=\int\frac{du}{u^2+36}.$$
Now maybe think: it would be nice if we had $u^2=36w^2$, because the $36$ could then "come out." So let $u=6w$. Then $du=6\,dw$, and we get
$$\int\frac{du}{u^2+36}=\int \frac{6\,dw}{36w^2+36}=\int\frac{1}{6}\frac{dw}{w^2+1}=\frac{1}{6}\arctan(w) +C.$$
Finally, we undo our substitution. We have $w=\frac{u}{6}=\frac{x+5}{6}$.
| {
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If $x$,$y$ are odd integers $a$,$b$ and $c$ such that $a^4-b^4=c^4$, argue that... If $x$,$y$ are odd integers $a$,$b$ and $c$ such that $a^4-b^4=c^4$, argue that $\gcd(a,b,c)=1$ implies that $\gcd(a,b)=1$
What I know:
A Pythagorean Triple is a triple of positive integers $a$,$b$,$c$ such that $a^2+b^2=c^2$
A Primitive Pythagorean Triple is a Pythagorean triple $a$,$b$,$c$ with the constraint that $\gcd(a,b)=1$, which implies $\gcd(a,c)=1$ and $\gcd(b,c)=1$.
I'm not sure how to use this, since we are dealing with $a^4-b^4=c^4$.
| Hint: Suppose to the contrary that $a$ and $b$ are not relatively prime. Let $p$ a common prime divisor of $a$ and $b$. Show that $p$ must divide $c$.
Remark: We do not need anything about Pythagorean triples to solve this problem. And we do not need to use the fact that $a$ and $b$ are odd.
But if we do want to borrow facts about Pythagorean triples, we can note that $a^4 -b^4=c^4$ if and only if $(b^2)^2 +(c^2)^2=(a^2)^2$, so $(b^2, c^2, a^2)$ is a Pythagorean triple. Then knowledge about triples becomes useful in the further analysis of $a^4-b^4=c^4$.
| {
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Inequality for Gamma functions Let $k, n ,m \in N$ and such that $0\leq k \leq n \leq m$. When the following ineuality is true?
$$
\frac{2^{m-k}\Gamma(n+1)\Gamma\left(\left[\frac{m+1-k}{2}\right]\right)\Gamma(m+1-n)}{\Gamma(m+1)\Gamma\left(\left[\frac{m+2-k}{2}\right]\right)}\leq \sqrt{\pi}
$$
Thank you for your help.
| I'm assuming that "$[\cdots]$" is the floor function, sometimes written "$\lfloor \cdots \rfloor$".
Since $m$ and $n$ are integers we have
$$\Gamma(n+1) = n! \qquad \Gamma(m+1) = m! \qquad \Gamma(m+1-n) = (m-n)!$$
This allows us to rewrite the left-hand side of the inequality as
$$
2^{m-k} \frac{\Gamma\left(\left[\frac{m+1-k}{2}\right]\right)}{\Gamma\left(\left[\frac{m+2-k}{2}\right]\right)} \cdot \frac{n!(m-n)!}{m!} = 2^{m-k} \frac{\Gamma\left(\left[\frac{m+1-k}{2}\right]\right)}{\Gamma\left(\left[\frac{m+2-k}{2}\right]\right)} \cdot \frac{1}{\binom{m}{n}}.
$$
Now if $m-k$ is even then $m+2-k$ is even, which implies that $(m+2-k)/2$ is an integer and
$$
\left[\frac{m+2-k}{2}\right] = \frac{m+2-k}{2} = \frac{m-k}{2} + 1.
$$
This tells us that
$$
\Gamma\left(\left[\frac{m+2-k}{2}\right]\right) = \Gamma\left(\frac{m-k}{2} + 1\right) = \frac{m-k}{2} \,\Gamma\left(\frac{m-k}{2}\right).
$$
Then since
$$
\frac{m+1-k}{2} = \frac{m+2-k}{2} - \frac{1}{2},
$$
we have
$$
\left[\frac{m+1-k}{2}\right] = \frac{m+2-k}{2} - 1 = \frac{m-k}{2},
$$
which gives us
$$
\Gamma\left(\left[\frac{m+1-k}{2}\right]\right) = \Gamma\left(\frac{m-k}{2}\right).
$$
Some cancellation happens in the inequality and we are left with
$$
\frac{2^{m+1-k}}{(m-k)\binom{m}{n}} \leq \sqrt{\pi}.
$$
After multiplying by the denominator and taking logarithms this is equivalent to
$$
k \geq m+1 - \log_2\left[\sqrt{\pi}(m-k)\binom{m}{n}\right] \qquad \text{if } m-k \text{ is even.}
$$
If $m-k$ is odd then similar considerations reveal that
$$
\Gamma\left(\left[\frac{m+1-k}{2}\right]\right) = \Gamma\left(\left[\frac{m+2-k}{2}\right]\right),
$$
which leaves the inequality in the form
$$
\frac{2^{m-k}}{\binom{m}{n}} \leq \sqrt{\pi}.
$$
By again multiplying by the denominator and taking logarithms we see that this is equivalent to
$$
k \geq m - \log_2\left[\sqrt{\pi}\binom{m}{n}\right] \qquad \text{if } m-k \text{ is odd.}
$$
| {
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greatest common divisor is 7 and the least common multiple is 16940 How many such number-pairs are there for which the greatest common divisor is 7 and the least common multiple is 16940?
| Theorem. The number of unordered pairs $(x,y)$ such that $\gcd(x,y) = G$ and $\operatorname{lcm}(x,y) = L$ is $2^{n-1}$ where $n$ is the number of distinct prime factors of $\frac LG$.
Proof
If $\gcd(x,y) = G$, and $\operatorname{lcm}(x,y) = L$, then
\begin{array}{ll}
1. & u = \dfrac xG \; \text{and} \; v = \dfrac yG \; \text{are integers} \\
2. & \gcd(u,v) = 1 \\
3. & uv = \dfrac LG
\end{array}
It should be pretty easy to see that first two properties are true. The third property is a consequence of the fact that
$xy = \gcd(x,y) \operatorname{lcm}(x,y) = GL$:
\begin{align}
L
&= \dfrac{xy}{G} \\
&= \dfrac{(Gu)(Gv)}{G} \\
&= Guv\\
\end{align}
It follows that the number of unordered pairs $(x,y)$ such that $\gcd(x,y) = G$ and $\operatorname{lcm}(x,y) = L$ is equal to the number of unordered pairs (u,v) such that $\gcd(u,v) = 1$ and $uv = \dfrac LG$.
Suppose that $\displaystyle \frac LG = \prod_{i=1}^n p_i^{a_i}$ where the $p_i$ are distinct prime numbers. For $i=1..n$, let $P_i = p_i^{a_i}$, so that $\displaystyle \frac LG = \prod_{i=1}^n P_i$. We see that finding a pair of relatively prime $u$ and $v$ such that $\dfrac LG = uv$ corresponds to expressing the set $\{ P_1, P_2, \dots, P_n\}$ as the union of a pair of disjoint subsets (where the empty set must correspond to the number 1). There are $2^{n-1}$ such subsets.
EXAMPLE
For this problem, $G = 7$ and $L = 16940\;$,
$\dfrac LG = 2420 = 2^2 \times 5 \times 11^2 = 4 \times 5 \times 121$.
Note that $2420 = 2^2 \times 5 \times 11^2 = 4 \times 5 \times 121$. So there are $2^{3-1} = 4$ unordered pairs.
In order to enumerate those pairs, we seek all $u, v$ such that $gcd(u,v) = 1$ and $uv = 2420$. We find
\begin{array}{c|cc|cc}
\text{Partition} & u & v & x & y\\
\hline
\{\},\{4,5,121\} & 1 & 2420 & 7 & 14640\\
\{4\}, \{5,121\} & 4 & 605 & 28 & 4235\\
\{5\}, \{4,121\} & 5 & 484 & 35 & 3388\\
\{121\}, \{4,5\} & 121 & 20 & 847 & 140\\
\hline
\end{array}
| {
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Two series questions I'm trying to correct an old quiz. I want to see if I have sufficiently corrected the second problem. The first I am still a bit unsure what to do.
(1) Show that the following series is divergent if $\alpha \in \mathbb{R}$ such that $|\alpha|<1$.
$$\sum_{k=1}^{\infty} \frac{1}{1+\alpha^k}$$
The above I wasn't sure what to do so I left it blank.
(2) Use the root test to decide whether or not the following series converges:
$$\frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+...$$
First note: $a_n =\begin{cases} \frac{1}{2^n} \text{ if n is odd}\\ \frac{1}{3^n} \text{ if n is even } \end{cases}$.
So $\liminf a_n = \sqrt[2n]{\frac{1}{3^n}}=\frac{1}{3}$ and $\limsup a_n = \sqrt[2n-1]{\frac{1}{2^n}}=1$ So the sequence diverges. I just want to make sure I got the values correct on this one.
| Note that if $\alpha \in (-1,1)$, then $1 + \alpha^k \leq 2$ so that
$$
\frac{1}{1 + \alpha^k} \geq \frac{1}{2}.
$$
For the second one, since $\frac{1}{3} \leq \frac{1}{2}$, then:
$$
\frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^3} + \dots \leq \sum_{k=1}^{\infty} \frac{1}{2^k} = 1
$$
If you want to use the root test: Note that
\begin{align}
\limsup_{n \to \infty} \sqrt[n]{a_n} = \limsup_{n \to \infty} \sqrt[n]{\frac{1}{2^n}} = \frac{1}{2} < 1
\end{align}
so the series converges absolutely.
| {
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The particular solution of the recurrence relation I cannot find out why the particular solution of $a_n=2a_{n-1} +3n$ is $a_{n}=-3n-6$
here is the how I solve the relation
$a_n-2a_{n-1}=3n$
as $\beta (n)= 3n$
using direct guessing
$a_n=B_1 n+ B_2$
$B_1 n+ B_2 - 2 (B_1 n+ B_2) = 3n$
So $B_1 = -3$, $B_2 = 0$
the particular solution is $a_n = -3 n$
and the homo. solution is $a_n = A_1 (-2)^n$
Why it is wrong??
| Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence without subtractions in indices:
$$
a_{n + 1} = 2 a_n + 3 n + 3
$$
Multiply by $z^n$, sum over $n \ge 0$ and recognize the resulting sums, particularly:
\begin{align}
\sum_{n \ge 0} z^n
&= \frac{1}{1 - z} \\
\sum_{n \ge 0} n z^n
&= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \\
&= \frac{z}{(1 - z)^2}
\end{align}
to get:
$$
\frac{A(z) - a_0}{z} = 2 A(z) + \frac{3}{(1 - z)^2} + \frac{3}{1 - z}
$$
As partial fractions this gives:
$$
A(z)
= \frac{a_0 + 6}{1 - 2 z}
- \frac{3}{1 - z}
- \frac{3}{(1 - z)^2}
$$
Using the generalized binomial theorem, in particular:
$$
\binom{-m}{n} = \binom{m + n - 1}{m - 1} (-1)^n
$$
you read off the coefficients:
\begin{align}
a_n &= (a_0 + 6) 2^n - 3 - 3 \binom{n + 2 - 1}{2 - 1} \\
&= (a_0 + 6) 2^n - 3 - 3 (n + 1) \\
&= (a_0 + 6) 2^n - 3 n - 6
\end{align}
| {
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} |
$\sqrt{(a+b-c)(b+c-a)(c+a-b)} \le \frac{3\sqrt{3}abc}{(a+b+c)\sqrt{a+b+c}}$ Suppose $a, b, c$ are the lengths of three triangular edges. Prove that:
$$\sqrt{(a+b-c)(b+c-a)(c+a-b)} \le \frac{3\sqrt{3}abc}{(a+b+c)\sqrt{a+b+c}}$$
| Here is another way to proof $ sinA+sinB+sinC\leq \dfrac{3\sqrt{3}}{2}$ in case some one don't know the Jensen inequality.
in general ,let $A \geq B \geq C$,then $C< \dfrac{\pi}{2}$ and $cos\dfrac{C}{2}>0$
$ sinA+sinB+sinC=2sin\dfrac{A+B}{2}cos\dfrac{A-B}{2}+2sin\dfrac{C}{2}cos\dfrac{C}{2}=2cos\dfrac{C}{2}(cos\dfrac{A-B}{2}+sin\dfrac{C}{2}) \leq 2cos\dfrac{C}{2}(1+sin\dfrac{C}{2})$ ,
since $cos\dfrac{A-B}{2} \leq 1$,when $A=B$, it equals 1.
from $x_1*x_2 \leq (\dfrac{x_1+x_2}{2})^2$, we can get $x_1x_2x_3x_4 \leq (\dfrac{x_1+x_2+x_3+x_4}{4})^4$
$ 2cos\dfrac{C}{2}(1+sin\dfrac{C}{2})$$=2\sqrt{1-sin^2\dfrac{C}{2}}(1+sin\dfrac{C}{2})$$=2\sqrt{(1-sin^2\dfrac{C}{2})(1+sin\dfrac{C}{2})^2}=2\sqrt{(1-sin\dfrac{C}{2})(1+sin\dfrac{C}{2})(1+sin\dfrac{C}{2})(1+sin\dfrac{C}{2})}$$=2\sqrt{\dfrac{1}{3}(3-3sin\dfrac{C}{2})(1+sin\dfrac{C}{2})(1+sin\dfrac{C}{2})(1+sin\dfrac{C}{2})}$ $ \leq 2\sqrt{\dfrac{1}{3}(\dfrac{3-3sin\dfrac{C}{2}+1+sin\dfrac{C}{2}+1+sin\dfrac{C}{2}+1+sin\dfrac{C}{2}}{4})^4}$=$2\sqrt{\dfrac{1}{3}(\dfrac{6}{4})^4}$=$2\sqrt{\dfrac{1}{3}(\dfrac{3}{2})^4}=$$\dfrac{3\sqrt{3}}{2} $
when and only when $3-3sin\dfrac{C}{2}=1+sin\dfrac{C}{2}$,ie $sin\dfrac{C}{2}=\dfrac{1}{2}$,you get the equals. put $A=B$ together, we get when and only when $A=B=C$, the LHS=$\dfrac{3\sqrt{3}}{2} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/262619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Legendre Polynomials Triple Product I have to solve the following integral:
\begin{align}
\int_{-1}^{1} \left(x^2 -1\right)^3 P_k(x)\,P_l(x)\, P_m(x) \;dx
\end{align}
where $P_{k,l,m}$ are Legendre Polynomials
The triple product
\begin{align}
\int_{-1}^{1} P_k(x)\,P_l(x)\, P_m(x) \;dx = 2 \begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix}^2
\end{align}
using the special case of $3j$ symbol form
\begin{align}
\begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix}
&= (-1)^s \sqrt{(2s-2k)! (2s-2l)! (2s-2m)! \over (2s+1)!}
{s! \over (s-k)! (s-l)! (s-m)!}
\\ & \mbox{for $2s=k+l+m$ even} \\[3pt]
\begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix}
&= 0
\quad\mbox{for $2s=k+l+m$ odd} \\
\end{align}
I'm sure you should be able to solve this by doing integration by parts but can't seem to get it to work. Any tips?
So using the answer below I think you get the following for step 1 of 3
\begin{multline}
\int_{-1}^{1}(x^2-1)^3 P_k P_l P_m
= \overbrace{(x^2-1)^3\frac{(P_{k+1} - P_{k-1})}{2k+1} P_l P_m \Big]_{-1}^1}^\text{ = 0}\\
-\int_{-1}^{1} \frac{(P_{k+1} - P_{k-1})}{2k+1}(x^2-1)^2\Big(
6xP_l P_m \\
+ (1+l) P_m(P_{l+1} - P_{l-1}) +
(1+m) P_l(P_{m+1} - P_{m-1}) \Big) \; dx\\
\end{multline}
Not sure if the formula for integration works as I think the $6xP_lP_m$ term might cause problems?
| I think the best way to approach this is as follows,
note that
\begin{align}
(x^2 -1 ) = \frac{P_2 - 2}{3}
\end{align}
You can then use the following definition
\begin{align}
P_kP_l = \sum_{m=|k-l|}^{k+l}
\begin{pmatrix}
k & l & m \\
0 & 0 & 0
\end{pmatrix}^2
(2m+1)P_m
\end{align}
This allows the integral to be written as follows
\begin{align}
\int_{-1}^{1} (x^2-1)^3P_iP_jP_k \; dx &= \int_{-1}^{1} \frac{1}{9}\left(P_2^3 + . . .-8 \right) P_i P_j P_k \; dx
\end{align}
The most difficult term to deal with is the $ P_2^3 P_i P_j P_k$
\begin{align}
P_2^3 P_i P_j P_k &= \sum_{m=0}^{4}
\begin{pmatrix}
2 & 2 & m \\
0 & 0 & 0
\end{pmatrix}^2 (2m+1)P_m P_2 P_i P_j P_k \\
&= \sum_{m=0}^{4}
\begin{pmatrix}
2 & 2 & m \\
0 & 0 & 0
\end{pmatrix}^2 (2m+1)
\sum_{n=|m-2|}^{m+2}
\begin{pmatrix}
2 & m & n \\
0 & 0 & 0
\end{pmatrix}^2 (2n+1)P_n P_i P_j P_k \\
&=
\sum_{m=0}^{4}
\begin{pmatrix}
2 & 2 & m \\
0 & 0 & 0
\end{pmatrix}^2 (2m+1)
\sum_{n=|m-2|}^{m+2}
\begin{pmatrix}
2 & m & n \\
0 & 0 & 0
\end{pmatrix}^2 (2n+1)
\sum_{l=|n-i|}^{n+i}
\begin{pmatrix}
n & i & l \\
0 & 0 & 0
\end{pmatrix}^2 (2l+1)
P_l P_j P_k
\end{align}
Which can then make use of the usual triple integral.
All other terms can be solved for in a similar manner.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/262973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Prove the following inequality: for
$a,b,c>0$
$$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$$
What I tried is using substitution:
$p=a+b+c$
$q=ab+bc+ca$
$r=abc$
But I cannot reduce $a^2(b+c)(c+a)+b^2(a+b)(c+a)+c(a+b)(b+c) $ interms of $p,q,r$
| The Cauchy-Schwarz inequality implies that
$$\left(\frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+a}\right)\left((a+b)+(b+c)+(c+a)\right)\geq(a+b+c)^2,$$
i.e.
$$\left(\frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+a}\right)\times 2\geq a+b+c,$$
which gives the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/264931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Find the number of pairs $(m, n)$ of positive integers such that $\frac{ m}{n+1} < \sqrt{2} < \frac{m+1}{n}$ Find the number of pairs $(m, n)$ of positive integers such that $\frac{ m}{n+1} < \sqrt{2} < \frac{m+1}{n}$ Constraint: $m$ and $n$ are both less than or equal to 1000
I toiled over this problem for a bit. I tested the first few positive integers of m and found the corresponding values of $n$, but no real pattern seemed to emerge. Clearly $m \ge n$ . I believe that for each value of $m$, there were either two or one value(s) for $n$ in no clear order, which is the part that is messing me up. This is meant to be done by hand so I have not used any software on it.
| You can rewrite this as:
$$\frac{m}{\sqrt{2}}-1<n<\frac{m+1}{\sqrt{2}}$$
Since $\sqrt{2}$ is irrational, we know that $\frac{m}{\sqrt{2}}-1$ is not an integer, so for an integer $n>\frac{m}{\sqrt{2}}-1$ iff $n\geq \left\lceil \frac{m}{\sqrt{2}}-1\right\rceil =\left\lfloor \frac{m}{\sqrt{2}}\right\rfloor$.
Similarly, $n<\frac{m+1}{\sqrt{2}}$ iff $n\leq \left\lfloor \frac{m+1}{\sqrt{2}}\right\rfloor$. So altogether, we are seeking $m,n$ such that $$\left\lfloor\frac{m}{\sqrt 2}\right\rfloor\leq n\leq \left\lfloor\frac{m+1}{\sqrt{2}}\right\rfloor$$
For a particular $m$, then, the number of possible $n$ is: $\left\lfloor\frac{m+1}{\sqrt{2}}\right\rfloor-\left\lfloor\frac{m}{\sqrt 2}\right\rfloor +1$.
Summing over all $m$, the result is $$-1+\sum_{m=1}^{1000}\left(\left\lfloor\frac{m+1}{\sqrt{2}}\right\rfloor-\left\lfloor\frac{m}{\sqrt 2}\right\rfloor +1\right)$$
(The $-1$ is because we don't want to count $n=0$ when $m=1$.)
But this is just $999$ plus a telescoping sum, and we see that the result is:
$$999+\left\lfloor\frac{1001}{\sqrt{2}}\right\rfloor$$
Actually, even more specifically, it is:
$$1000-\lfloor\sqrt{2}\rfloor + \left\lfloor\frac{1000+1}{\sqrt{2}}\right\rfloor$$
This will work for any irrational number $\alpha>1$ and any upper bound, $M>\alpha$, yielding a total: $$M-\lfloor\alpha\rfloor + \left\lfloor\frac{M+1}{\alpha}\right\rfloor$$
which counts the pairs $(m,n)$ with $1\leq m,n\leq M$ and
$$\frac{m}{n+1}<\alpha<\frac{m+1}n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/265214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
} |
A non-linear maximisation We know that $x+y=3$ where x and y are positive real numbers. How can one find the maximum value of $x^2y$? Is it $4,3\sqrt{2}, 9/4$ or $2$?
| $x^2y=x^2(3-x)=3x^2-x^3$
So, $\frac{x^2y}{dx}=3(2x)-3(x^2)=3x(2-x)$
For the extreme values of $x^2y,\frac{x^2y}{dx}=0\implies x=0$ or $x=2$
Now, $\frac{d^2(x^2y)}{dx^2}=6-3(2x)=6(1-x)$
At $x=0,\frac{d^2(x^2y)}{dx^2}=6>0$ so $x^2y$ will have minimum value at $x=0$
At $x=2,\frac{d^2(x^2y)}{dx^2}=-6<0\implies x=2$ will make $x^2y$ the maximum
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/267055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
how many number like $119$ How many 3-digits number has this property like $119$:
$119$ divided by $2$ the remainder is $1$
119 divided by $3$ the remainderis $2 $
$119$ divided by $4$ the remainder is $3$
$119$ divided by $5$ the remainder is $4$
$119$ divided by $6$ the remainder is $5$
| Observe that $119$ leaves remainder $n-1$ when divided by $n$ where $2\le n\le 6$
So, $119+1$ is divisible by $n$ for $2\le n\le 6$
So, we numbers of the form $\operatorname{lcm}(2,3,4,5,6)m-1=60m-1$ where $m$ is any positive integer.
As the required number is of three digits $100\le 60m-1\le 999\implies 2\le m\le 16$
Hence there are $16-2+1=15$ such numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/268619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Equation I simply don't understand I am working on some math problems from math books, and I have seen this in one math problem.
$x_1^2+x_2^2= (x_1+x_2)^2 - 2x_1x_2$
I simply don't understand this. Can you please explain this to me.
The whole math problem is:
You have this equation
$x^2 - 3x + m^2 -1 = 0$
For what parameter m is:
$x_1^2 + x_2^2 > 3$
| I think what's going on is that you want to find all values of the constant $m$ so that, if $x_1$ and $x_2$ are the two (not necessarily distinct) solutions to the equation $x^2 - 3x + m^2 -1 = 0,$ then $x_{1}^{2} + x_{2}^{2} > 3.$
Recall Vieta's formulas for the sum and product of the solutions of the quadratic equation $ax^2 + bx + c = 0.$ They say that the sum of the solutions equals $-\frac{b}{a}$ and the product of the solutions equals $\frac{c}{a}.$
Thus, in the case of your quadratic equation, we have $x_{1} + x_{2} = 3$ and $x_{1}x_{2}=m^2 - 1.$ Since $x_{1}^{2} + x_{2}^{2} = \left(x_1 + x_2\right)^2 - 2x_{1}x_{2},$ it follows that $x_{1}^{2} + x_{2}^{2} = 3^2 - 2(m^2 - 1) = 11 - 2m^2.$ Therefore, $x_{1}^{2} + x_{2}^{2} > 3$ becomes $11 - 2m^2 > 3,$ or $m^2 < 4,$ or $-2 < m < 2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/269970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
$ \displaystyle\lim_{n\to\infty}\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$ $$ \ X_n=\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$$ Find $\displaystyle\lim_{n\to\infty} X_n$ using the squeeze theorem
I tried this approach:
$$
\frac{1}{\sqrt{n^3+1}}\le\frac{1}{\sqrt{n^3+1}}<\frac{n}{\sqrt{n^3+1}}
$$
$$
\frac{1}{\sqrt{n^3+1}}<\frac{2}{\sqrt{n^3+2}}<\frac{n}{\sqrt{n^3+1}}$$
$$\vdots$$
$$\frac{1}{\sqrt{n^3+1}}<\frac{n}{\sqrt{n^3+n}}<\frac{n}{\sqrt{n^3+1}}$$
Adding this inequalities:
$$\frac{n}{\sqrt{n^3+1}}\leq X_n<\frac{n^2}{\sqrt{n^3+1}}$$
And this doesn't help me much. How should i proced?
| From the inequality $$n^3 \leqslant n^3+k \leqslant n^3+n \leqslant 2n^3, \;\; (1 \leqslant k \leqslant n)$$
we have
$$
\dfrac{1}{\sqrt{2n^3}} \leqslant \dfrac{1}{\sqrt{n^3+k}} \leqslant \dfrac{1}{\sqrt{n^3}},$$
therefore
$$\dfrac{n(n+1)}{2\sqrt{2n^3}} = \dfrac{1}{\sqrt{2n^3}} \sum\limits_{k=1}^{n}{k} \leqslant \sum\limits_{k=1}^{n} \dfrac{k}{\sqrt{n^3+k}} \leqslant \dfrac{1}{\sqrt{n^3}}\sum\limits_{k=1}^{n}{k}= \dfrac{n(n+1)}{2\sqrt{n^3}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/270216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Calculate value of expression $(\sin^6 x+\cos^6 x)/(\sin^4 x+\cos^4 x)$
Calculate the value of expresion:
$$
E(x)=\frac{\sin^6 x+\cos^6 x}{\sin^4 x+\cos^4 x}
$$
for $\tan(x) = 2$.
Here is the solution but I don't know why $\sin^6 x + \cos^6 x = ( \cos^6 x(\tan^6 x + 1) )$, see this:
Can you explain to me why they solved it like that?
| $$E(x)=\frac{\sin^6x+\cos^6x}{\sin^4x+\cos^4x}=\frac{\sin^6x+\cos^6x\over\cos^6x}{\sin^4x+\cos^4x\over\cos^6x}=\frac{\tan^6x+1}{\frac{1}{\cos^2x}(\tan^4x+1)}=$$
$$=\cos^2x\cdot\frac{2^6+1}{2^4+1}=\frac{1}{1+\tan^2x}\frac{65}{17}=\frac{1}{2^2+1}\frac{65}{17}=\frac{1}{5}\cdot\frac{5\cdot13}{17}=\frac{13}{17}$$
$$\sin^6x+\cos^6x=\cos^6x\left(\frac{\sin^6x}{\cos^6x}+1\right)=\cos^6x(\tan^6x+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/271458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that $x = 2$ is the unique solution to $3^x + 4^x = 5^x$ where $x \in \mathbb{R}$ Yesterday, my uncle asked me this question:
Prove that $x = 2$ is the unique solution to $3^x + 4^x = 5^x$ where $x \in \mathbb{R}$.
How can we do this? Note that this is not a diophantine equation since $x \in \mathbb{R}$ if you are thinking about Fermat's Last Theorem.
| For all $x_j>x_i$ and $0<a<1$, $a^{x_i}>a^{x_j}$ .
Hence
\begin{align}
\left(\frac{3}{5}\right)^{x} + \left(\frac{4}{5}\right)^{x} - 1 < \left(\frac{3}{5}\right)^{2} + \left(\frac{4}{5}\right)^{2} - 1 = 0
\end{align}
for all $x>2$. Hence, there is no solution for $x>2$.
Similarly
\begin{align}
\left(\frac{3}{5}\right)^{x} + \left(\frac{4}{5}\right)^{x} - 1 > \left(\frac{3}{5}\right)^{2} + \left(\frac{4}{5}\right)^{2} - 1 =0
\end{align}
for all $x<2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/272114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Check convergence $\sum\limits_{n=1}^\infty\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$ Please help me to check convergence of $$\sum_{n=1}^{\infty}\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$$
| Update. Compare
$$\sum_{n=1}^{\infty}\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)
$$
with the serie
$$
\sum_{n=1}^{\infty} \frac{1}{n^a}
$$
for a convenient $a\geq 1$.
Note that if $a_n=\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$, $b_n=\frac{1}{n^a}$ and
$$
0<\lim_{n\to\infty} \frac{a_n}{b_n}<\infty.
$$
then $\sum_{n=1}^{\infty} a_n<\infty$ if, only if, $\sum_{n=1}^{\infty} b_n<\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If x,y,z are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what If $x,y, z$ are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what?
$108$ , $216$ , $405$ , $1048$
| $$(x^2+8y^2+27z^2)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 \ge (1+2+3)^3 = 216$$
by Holder's inequality. It's clear that the equality can hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
How to show that $\frac{x^2}{x-1}$ simplifies to $x + \frac{1}{x-1} +1$ How does $\frac{x^2}{(x-1)}$ simplify to $x + \frac{1}{x-1} +1$?
The second expression would be much easier to work with, but I cant figure out how to get there.
Thanks
| $$\frac{x^2}{x-1}=\frac{x^2-x+x-1+1}{x-1} = \frac{x(x-1) + (x-1)+1}{x-1}=\frac{x(x-1)}{x-1}+\frac{x-1}{x-1}+\frac{1}{x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/278481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
prove the divergence of cauchy product of convergent series $a_{n}:=b_{n}:=\dfrac{(-1)^n}{\sqrt{n+1}}$ i am given these series which converge. $a_{n}:=b_{n}:=\dfrac{(-1)^n}{\sqrt{n+1}}$ i solved this with quotient test and came to $-1$, which is obviously wrong. because it must be $0<\theta<1$ so that the series converges. my steps:
$\dfrac{(-1)^{n+1}}{\sqrt{n+2}}\cdot \dfrac{\sqrt{n+1}}{(-1)^{n}} = - \dfrac{\sqrt{n+1}}{\sqrt{n+2}} = - \dfrac{\sqrt{n+1}}{\sqrt{n+2}}\cdot \dfrac{\sqrt{n+2}}{\sqrt{n+2}} = - \dfrac{(n+1)\cdot (n+2)}{(n+2)\cdot (n+2)} = - \dfrac{n^2+3n+2}{n^2+4n+4} = -1 $
did i do something wrong somewhere?
and i tried to know whether the cauchy produkt diverges as task says:
$\sum_{k=0}^{n}\dfrac{(-1)^{n-k}}{\sqrt{n-k+1}}\cdot \dfrac{(-1)^{k}}{\sqrt{k+1}} = \dfrac{(-1)^n}{nk+n-k^2+1} = ..help.. = diverging $
i am stuck here how to show that the produkt diverges, thanks for any help!
| Here is another solution: Observe that
\begin{align*}
\sum_{k=0}^{n}\frac{(-1)^{k}}{\sqrt{k+1}} \frac{(-1)^{n-k}}{\sqrt{n-k+1}}
&= (-1)^{n} \sum_{k=0}^{n}\frac{1}{\sqrt{(k+1)(n-k+1)}} \\
&= (-1)^{n} \sum_{k=0}^{n}\frac{1}{n} \frac{1}{\sqrt{\left(\frac{k+1}{n}\right)\left(1-\frac{k-1}{n}\right)}}.
\end{align*}
But since
$$ \sum_{k=0}^{n}\frac{1}{n} \frac{1}{\sqrt{\left(\frac{k+1}{n}\right)\left(1-\frac{k-1}{n}\right)}} \xrightarrow[]{n\to\infty} \int_{0}^{1} \frac{dx}{\sqrt{x(1-x)}} = \pi,$$
the general term of the Cauchy product diverges. Therefore the series itself diverges. (Here, the specific value of the integral is irrelevant. It suffices to observe that the integral is positive.)
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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Solving lyapunov equation, Matlab has different solution, why? I need to solve the lyapunov equation i.e. $A^TP + PA = -Q$. With $A = \begin{bmatrix} -2 & 1 \\ -1 & 0 \end{bmatrix}$ and $Q = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$.
Hence...
$$
\begin{bmatrix}
-2 & -1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
P_{11} & P_{12} \\
P_{12} & P_{22}
\end{bmatrix}
+
\begin{bmatrix}
P_{11} & P_{12} \\
P_{12} & P_{22}
\end{bmatrix}
\begin{bmatrix}
-2 & 1 \\
-1 & 0
\end{bmatrix} +
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} = 0
$$
$$
\begin{bmatrix}
-4P_{11} - 2P_{12} + 1 & P_{11} -2P_{12} - P_{22} \\
P_{11} - 2P_{12} - P_{22} & 2P_{12} + 1
\end{bmatrix} = 0
$$
$$
\begin{bmatrix}
-4 & -2 & 0 \\
1 & -2 & -1 \\
1 & -2 & -1 \\
0 & 2 & 0
\end{bmatrix}
\begin{bmatrix}
P_{11} \\
P_{12} \\
P_{22}
\end{bmatrix} =
\begin{bmatrix}
-1 \\
0 \\
0 \\
-1
\end{bmatrix}
\Rightarrow
\begin{bmatrix}
-4 & -2 & 0 \\
1 & -2 & -1 \\
0 & 2 & 0
\end{bmatrix}
\begin{bmatrix}
P_{11} \\
P_{12} \\
P_{22}
\end{bmatrix} =
\begin{bmatrix}
-1 \\
0 \\
-1
\end{bmatrix}$$
And such we get that $P = \begin{bmatrix} 1/2 & -1/2 \\ -1/2 & 3/2\end{bmatrix}$.
This is the same solution as given by my professor. I wanted to check however if I can also find the solution using Matlab. I entered the following:
A = [-2 1; -1 0];
Q = [1 0; 0 1];
P = lyap(A,Q)
This however tells me that $P = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 3/2\end{bmatrix}$.
What is going on here? Is Matlab correct or wrong? Or is my solution wrong? Or are we both correct?
| The matlab definition must be different from yours. With Matlab's $P$, one has $AP+PA^T=-Q$. Doing things your way with matlab's $P$ gives
$$
A^TP+PA=
\begin{bmatrix}
-3 & -2 \\
-2 & 1
\end{bmatrix}
$$
Maybe look in matlab's documentation to see what equation their lyap(A,Q) is solving; I'd guess it had the transpose switched.
EDIT: It seems Mathematica9 has the other $AP+PA^T=-Q$ definition, or close to it.
Their command LyapunovSolve[a,c] gives solution to $ax+xa^T=c$, which has the transposed matrix mentioned last, opposite to your version. Reference page:
http://reference.wolfram.com/mathematica/ref/LyapunovSolve.html
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $(a+b+c)^3 = a^3 + b^3 + c^3+ (a+b+c)(ab+ac+bc)$ As stated in the title, I'm supposed to show that $(a+b+c)^3 = a^3 + b^3 + c^3 + (a+b+c)(ab+ac+bc)$.
My reasoning:
$$(a + b + c)^3 = [(a + b) + c]^3 = (a + b)^3 + 3(a + b)^2c + 3(a + b)c^2 + c^3$$
$$(a + b + c)^3 = (a^3 + 3a^2b + 3ab^2 + b^3) + 3(a^2 + 2ab + b^2)c + 3(a + b)c^2+ c^3$$
$$(a + b + c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc$$
$$(a + b + c)^3 = (a^3 + b^3 + c^3) + (3a^2b + 3a^2c + 3abc) + (3ab^2 + 3b^2c + 3abc) + (3ac^2 + 3bc^2 + 3abc) - 3abc$$
$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3a(ab + ac + bc) + 3b(ab + bc + ac) + 3c(ac + bc + ab) - 3abc$$
$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3(a + b + c)(ab + ac + bc) - 3abc$$
$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3[(a + b + c)(ab + ac + bc) - abc]$$
It doesn't look like I made careless mistakes, so I'm wondering if the statement asked is correct at all.
| $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)$ with $3(a+b)(b+c)(c+a) = 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc$
I guess the right factorization
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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} |
Factorization problem Find $m + n$ if $m$ and $n$ are natural numbers such that: $$\frac {m+n} {m^2+mn+n^2} = \frac {4} {49}\;.$$
My reasoning:
Say: $$m+n = 4k$$ $$m^2+mn+n^2 = 49k$$
It follows:$$(m+n)^2 = (4k)^2 = 16k^2 \Rightarrow m^2+mn+n^2 + mn = 16k^2 \Rightarrow mn = 16k^2 - 49k$$
Since: $$mn\gt0 \Rightarrow 16k^2 - 49k\gt0 \Rightarrow k\gt3$$
Then no more progress.
| $4m^2+4mn+4n^2=49m+49n$
$4m^2+4mn+n^2+3n^2-49m-49n=0$
$(2m+n)^2+3n^2-(49/2)(2m+n)-(49/2)n=0$
$16(2m+n)^2-392(2m+n)+48n^2-392n=0$
$(4(2m+n)-49)^2+48n^2-392n=2401$
$3(4(2m+n)-49)^2+144n^2-1176n=7203$
$3(4(2m+n)-49)^2+(12n-49)^2=9604$
$3(8m+4n-49)^2+(12n-49)^2=9604$.
$3x^2+y^2=98^2$
Now the last equation has only finitely many integer solutions, with a finite procedure for finding them, then for each solution you can check whether $8m+4n-49=x,12n-49=y$ has integer solutions, and that should do it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/290166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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no. of subsets which have at least $r+1$ element is A Set has $2r+1$ elements. Then the no. of subsets which have at least $r+1$ element is
My Try::
selecting $r+1$ element from $2r+1$ which is $\displaystyle = \binom{2r+1}{r+1}$
selecting $r+2$ element from $2r+1$ which is $\displaystyle = \binom{2r+1}{r+2}$
..........................................................
..........................................................
selecting $2r+1$ element from $2r+1$ which is $\displaystyle = \binom{2r+1}{2r+1}$
So Total $ = \binom{2r+1}{r+1}+\binom{2r+1}{r+2}+..........+\binom{2r+1}{2r+1}$
and answer given is $ = 2^{2r}$
How can i Calculate it. Thanks
| Substitute $x=1$ into $(1+x)^{2r+1}=\sum_{k=0}^{2r+1}\binom{2r+1}{k}x^k$ to get
$2^{2r+1}$
$=\sum_{k=0}^{2r+1}\binom{2r+1}{k}$
$=\sum_{k=0}^{r}\binom{2r+1}{k}+\sum_{k=r+1}^{2r+1}\binom{2r+1}{k}$
$=\sum_{k=0}^{r}\binom{2r+1}{2r+1-k}+\sum_{k=r+1}^{2r+1}\binom{2r+1}{k}$
$=2\sum_{k=r+1}^{2r+1}\binom{2r+1}{k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/291096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof of inequality involving surds Determine whether $ \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}$ is greater or less than $\frac{3}{10}$
So what I did was:
Add the fraction with surds to get $\frac{\sqrt{6}-\sqrt{2}}{\sqrt{12}}$
Then squaring both sides results in $\frac{8-4\sqrt{3}}{12} = \frac{2-\sqrt{3}}{3}$
Therefore we have $\frac{2-\sqrt{3}}{3} ? \frac{9}{100}$ (using question mark in place of < or > therefore $200-100\sqrt{3} = 27$, but $1\times1 = 1$ and $2\times2 = 4$ thefore $1<\sqrt{3}<2$ and since $100-100\times1 = 0 <27$, $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}} < \frac{3}{10}$
I checked the answer and they had a method but same result, however would this suffice?
| Assume the LHS is $<$ the RHS. Square the LHS
$$\frac{1}{2} + \frac{1}{6} - \frac{2}{\sqrt{12}} = \frac{2}{3} - \frac{1}{\sqrt{3}} < \frac{9}{100}$$
Subtract $2/3$ from the LHS and the RHS and we are assuming that
$$ \frac{1}{\sqrt{3}} > \frac{2}{3} - \frac{9}{100} = \frac{173}{300}$$
Square both sides again:
$$\frac{1}{3} > \frac{29929}{90000}$$
which is true, and therefore the assumption is true. Therefore
$$\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}} < \frac{3}{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/293485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limits calculus very short question? Can you help me to solve this limit? $\frac{\cos x}{(1-\sin x)^{2/3}}$... as $x \rightarrow \pi/2$, how can I transform this?
| As $\cos x=\cos^2\frac x2-\sin^2\frac x2$ and $1-\sin x=(\cos\frac x2-\sin \frac x2)^2$
$$\lim_{x\to\frac\pi2}\frac{\cos x}{(1-\sin x)^\frac23}$$
$$=\lim_{x\to\frac\pi2}\frac{(\cos\frac x2-\sin\frac x2)(\cos\frac x2+\sin\frac x2)}{(\cos\frac x2-\sin \frac x2)^\frac43}$$
$$=\lim_{x\to\frac\pi2}\frac{(\cos\frac x2+\sin\frac x2)}{(\cos\frac x2-\sin \frac x2)^\frac13} \text{ which is of the form } \frac{\sqrt2}0$$
as $x\to \frac\pi2, \frac x2\to \frac\pi4\implies \tan \frac x2\to1 \implies \tan \frac x2\ne1\implies \cos \frac x2\ne \sin\frac x2$
Alternatively,
putting $t=\tan\frac x2$ so that $x\to\frac\pi2,t\to1$ and $$\cos x=\frac{1-\tan^2\frac x2}{1+\tan^2\frac x2}=\frac{1-t^2}{1+t^2}\text {and } \sin x=\frac{2\tan\frac x2}{1+\tan^2\frac x2}=\frac{2t}{1+t^2},1-\sin x=\frac{(1-t)^2}{1+t^2}$$
So, $$\lim_{x\to\frac\pi2}\frac{\cos x}{(1-\sin x)^\frac23}$$
$$=\lim_{t\to1}\frac{(1-t^2)}{(1+t^2)}\cdot \frac{(1+t^2)^\frac23}{(1-t)^\frac43}$$
$$=\lim_{t\to1}\frac{(1+t)}{(1+t^2)^\frac13(1-t)^\frac13} \text{ which is of the form } \frac10$$
as $t\ne1$ as $t\to1$
| {
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"timestamp": "2023-03-29T00:00:00",
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when $f(x)^n$ is a degree of $n$ why is useful to think $\sqrt{f(x)^n}$ as $n/2$? I have come across this question when doing problems from "Schaum's 3000 Solved Calculus Problems". I was trying to solve $$\lim_{x\rightarrow+\infty}\frac{4x-1}{\sqrt{x^2+2}}$$ and I couldn't so I looked the solution and solution said
Can someone please explain to me why this is and exactly how it works? Also, the next question is as such $$\lim_{x\rightarrow-\infty}\frac{4x-1}{\sqrt{x^2+2}}$$ and there the author has suggested that $x= -\sqrt{x^2}$. Why is that?
Thanks
EDIT
Can someone use the above technique and solve it, to show it works? Because I understand the exponent rules, I am aware of that but what I don't understand is why you want to do that?
Here is the solution that book shows:
| As you may know $(\alpha^n)^m = \alpha^{n.m}$, since the action of raising something to the power of $n$ can be thought of as an inverse action of taking the $n$-th root.
So, we define $\sqrt[m]{a^n} = a^{\frac{n}{m}}$ (this is because, division is the inverse operator of multiplication), where $\frac{m}{n}$ is in lowest terms. And in fact, this definition is valid.
Examples
Some other examples:
*
*$\sqrt{2} = 2^{\frac{1}{2}}$
*$\sqrt[5]{a^3} = a^{\frac{3}{5}}$
*$\sqrt[4]{a^2} = a^{\frac{2}{4}}$, this is incorrect, as $\frac{2}{4}$ can be further reduced.
*$\sqrt[2]{a^2} = a^{\frac{2}{2}} = a$, this is wrong too, the same reason as above. It should read $\sqrt[2]{a^2} = |a|$ instead.
Applying it here, we have: $\sqrt{a^n} = a^{\frac{n}{2}}$, which means, if we take the square root of some $n$ degree term, we'll have a term of degree $\frac{n}{2}$.
Secondly, $sqrt{x^2}$ is always non-negative. Like this:
*
*$\sqrt{2^2} = \sqrt{4} = 2$
*$\sqrt{(-2)^2} = \sqrt{4} = 2$
*$\sqrt{3^2} = \sqrt{9} = 3$
*$\sqrt{(-3)^2} = \sqrt{9} = 3$
so, if x is nonnegative, we'll have $x = \sqrt{x^2}$, but when x is negative, then we'll have to put a minus sign in front of $\sqrt{x^2}$ to make it negative too, so $x = -\sqrt{x^2}$.
As x tends to $+\infty$, x will be positive, and vice versa, when x tends to $\infty$, it'll be negative.
Remember that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ as long as $a; b \ge 0$.
Example
Evaluate $\lim\limits_{x \rightarrow -\infty} \frac{-5x}{\sqrt{x^2+x}}$.
$\begin{align}\lim\limits_{x \rightarrow -\infty} \dfrac{-5x}{\sqrt{x^2+x}} &= \lim\limits_{x \rightarrow -\infty} \dfrac{-5x}{\sqrt{x^2 \left(1+\dfrac{1}{x} \right)}}\\
&= \lim\limits_{x \rightarrow -\infty} \dfrac{-5x}{\sqrt{x^2} \sqrt{\left(1+\dfrac{1}{x} \right)}}\\
&= \lim\limits_{x \rightarrow -\infty} \dfrac{-5x}{-x \sqrt{\left(1+\dfrac{1}{x} \right)}}\\
&= \lim\limits_{x \rightarrow -\infty} \dfrac{5}{\sqrt{\left(1+\dfrac{1}{x} \right)}} \quad \mbox{cancel } -x\\
&= \dfrac{5}{\sqrt{\left(1+0 \right)}}\\
&= 5 \end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factoring $3x^2 - 10x + 5$ How can $3x^2 - 10x + 5$ be factored? FOIL seemingly doesn't work (15 has no factors that sum to -10).
| $3x^2 - 10x + 5$
Testing the polynomial using the discriminant (the $\;\sqrt{b^2 - 4ac}\;$ portion of the quadratic formula,),
with $$b = -10, \,a=3,\,c = 5$$ gives you $$\sqrt{100 - 4\cdot 3 \cdot 5} = \sqrt{100 - 60} = \sqrt{40} = \sqrt{4\cdot 10} = \sqrt{4}\cdot \sqrt{10} = 2\sqrt{10},$$ which tells you that there aren't any rational, let alone integer, roots.
So we need to find the roots using the entire quadratic formula: For $ax^2 + bx + c$, we get two roots: $$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a},\;\; x_2 =\frac{-b - \sqrt{b^2 - 4ac}}{2a}$$ and we already know that $\sqrt{b^2 - 4ac} = 2\sqrt{10}$ for our polynomial:
$$x=\dfrac {10 \pm 2\sqrt{10}}{6} =\dfrac 16 (2\cdot 5 \pm 2\sqrt{10})= 2\cdot\frac 16(5 \pm \sqrt{10}) = \frac13 (5 \pm \sqrt{10})$$
In general, $ax^2+bx+c=a(x-x_1)(x-x_2)$:
So, to express your polynomial in factored form you need $$3x^2 - 10x + 5\quad = \quad 3\left(x-\frac 13(5 + \sqrt{10})\right)\left(x-\frac 13(5 - \sqrt{10})\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Quadratic Diophantine Equations I note that the Diophantine equation, $x^2 + y^2 = z^2$, with $x, y, z \in \mathbb{N}$, has infinitely many solutions. Indeed, $(x, y, z) = (3,4,5)$ provides a solution, and for any $k \in \mathbb{N}$ : $(kx, ky, kz ) = (3k, 4k, 5k)$ provides a solution.
However, assuming $x, y, z \in \mathbb{N}$ with $x, y > 1$, is the same true for the Diophantine equations,
$x^2 +y^2 = z^2 + 1$,
$x^2 + y^2 = z^2 + 2$,
$x^2 + y^2 = z^2 + 3$
and more generally, for $x^2 + y^2 = z^2 + n$, for any $n \in \mathbb{N}$?
In particular, are there infinitely-many triples $(x, y, z) \in \mathbb{N}^3$ for which $x^2 + y^2 = z^2 + n$ is true for infinitely-many values of $n \in \mathbb{N}$?
| $x^2+y^2 = z^2+n$ is equivalent to $x^2-n = (z-y)(z+y)$.
Any composite odd number can be written as $(z-y)(z+y)$ for some integers $z$ and $y$, so it is enough to show that $x^2-n$ contains infinitely many composite odd numbers.
If $n$ is odd, then you can simply pick $x = 2kn$ for any $k$. Then $x^2-n$ is a multiple of $n$ and odd, which gives you two integers $y$ and $z$ satisfying the equation. You end up with the family of solutions $(2kn, 2k^2n-(n+1)/2, 2k^2n-(n-1)/2)$
If $n$ is even, then you can simply pick $x=2k(n-1)+1$ for any $k$. Then $x^2-n \equiv 1^2-1 = 0 \pmod {n-1}$, and it is odd so again this gives you two integers $y$ and $z$ satisfying the equation. You end up with the family of solutions $(2k(n-1)+1,2k^2(n-1)+2k-n/2,2k^2(n-1)+2k-1+n/2)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the ratio How to find the ratio between the area of the big regular pentagon $ABCDE$ and the small regular pentagon $PQRST$
| By angle-chasing one can show that $EP=AP=AQ$.
Without loss of generality we can assume that $PQ=1$. Let $x=EA$. Then $EP=AP=AQ=x-1$.
Note that triangles $EQA$ and $APQ$ are similar. It follows that $\dfrac{EA}{AQ}=\dfrac{AQ}{PQ}$. Thus
$$\frac{x}{x-1}=\frac{x-1}{1}.$$
This simplifies to $x^2-3x+1=0$, which has roots $\frac{3\pm\sqrt{5}}{2}$. One of these is less than $1$. So $x=\dfrac{3+\sqrt{5}}{2}$.
Now the ratio of the area of the big pentagon to the little one is $x^2$ to $1$.
Compute. The number $x^2$ simplifies to $\dfrac{7+3\sqrt{5}}{2}$.
Remark: Note that $\frac{3+\sqrt{5}}{2}$ is the square of the famous "Golden Ratio" $\frac{1+\sqrt{5}}{2}$. The Golden Ratio is often denoted by $\varphi$. So the ratio of the areas is $\varphi^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/298143",
"timestamp": "2023-03-29T00:00:00",
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How can we show the convergence of $x_n = \sin(2\pi (n^3-n^2+1)^{\frac{1}{3}})$? Show that the sequence $(x_n)_{n\geq 1}$ defined by $$x_n = \sin(2\pi (n^3-n^2+1)^{\frac{1}{3}})$$ converges and compute its limit.
| We have
$$
2\pi\sqrt[3]{n^3-n^2+1}=2\pi n\sqrt[3]{1-1/n+1/n^3}.
$$
Now use $(1+u)^\alpha=1+\alpha u+O(u^2)$ as $u$ approaches $0$, to get:
$$
2\pi n\sqrt[3]{1-1/n+1/n^3}=2\pi n\left(1-\frac{1}{3n}+O\left(\frac{1}{n^2}\right)\right)=2\pi n-\frac{2\pi}{3}+O\left(\frac{1}{n}\right)
$$
Then
$$
x_n=\sin\left(2\pi n-\frac{2\pi}{3}+O\left(\frac{1}{n}\right) \right)=\sin\left(-\frac{2\pi}{3}+O\left(\frac{1}{n}\right) \right)
$$
so the limit is
$$
\sin(-2\pi/3)=-\frac{\sqrt{3}}{2}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of definite Integral Evaluate
$$
\int_{\ln(0.5)}^{\ln(2)}\left(
\frac{\displaystyle\sin x
\frac{\sqrt{\sin^2(\cos x)+\pi
e^{(x^4)}}}{1+(xe^{\cos x}\sin x)^2}+
2\sin(x^2+2)\arctan\left(\frac{x^3}{3}\right)
}
{\displaystyle 1+e^{-\frac{x^2}{2}}+x^7 \sin(-\pi
x)+\frac{12}{11}|x|^{2\pi+1}} \,d x\right)
$$
This is my solution, it's correct?
First we observe that
$\ln (0.5)=\ln \frac{1}{2}=-\ln 2$
therefore, the integral is, said $ f (x) $ the integrand,
$\int_{-\ln 2}^{\ln 2} f(x)\,\,dx$
that is, an integral over an interval symmetrical about the origin; without taking roads for the search of all the primitives, and by exploiting the symmetry of the interval, we check if the function is odd, in that case one can immediately conclude that the value of 'integral is $ 0 $, then we have: therefore, the integral is, that $ f (x) $ the integrand,
\begin{align}
f(-x)&= \frac{\displaystyle\sin (-x)\frac{\sqrt{\sin^2(\cos (-x))+\pi e^{((-x)^4)}}}{1+((-x)e^{\cos (-x)}\sin (-x))^2}+2\sin((-x)^2+2)\arctan\left(\frac{(-x)^3}{3}\right)}{\displaystyle 1+e^{-\frac{(-x)^2}{2}}+(-x)^7 \sin(-\pi (-x))+\frac{12}{11}|(-x)|^{2\pi+1}}\\
&= \frac{\displaystyle-\sin x\frac{\sqrt{\sin^2(\cos x)+\pi e^{(x^4)}}}{1+( x e^{\cos x}\sin x )^2}-2\sin(x^2+2)\arctan\left(\frac{ x ^3}{3}\right)}{\displaystyle 1+e^{-\frac{x^2}{2}}+x^7 \sin(-\pi x )+\frac{12}{11}|x|^{2\pi+1}}\\
&= -\frac{\displaystyle \sin x\frac{\sqrt{\sin^2(\cos x)+\pi e^{(x^4)}}}{1+( x e^{\cos x}\sin x )^2}+2\sin(x^2+2)\arctan\left(\frac{ x ^3}{3}\right)}{\displaystyle 1+e^{-\frac{x^2}{2}}+x^7 \sin(-\pi x )+\frac{12}{11}|x|^{2\pi+1}}\\
&=-f(x)
\end{align}
the function is therefore odd and therefore the integral is equal to $ 0 $
| Yes everything is fine. ${}{}{}{}{}{}{}{}{}{}{}$
| {
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Finding the smallest positive integer $N$ such that there are $25$ integers $x$ with $2 \leq \frac{N}{x} \leq 5$
Find the smallest positive integer $N$ such that there are exactly $25$ integers $x$ satisfying $2 \leq \frac{N}{x} \leq 5$.
| If $x > 0$, $2 \le N/x \le 5$ iff $\frac{N}{5} \le x \le \frac{N}{2}$. Thus you need
$\lfloor N/2 \rfloor - \lceil N/5 \rceil = 24$. If $N = 10 q + r$, $r \in [0,1,\ldots,9]$,
$$ \lfloor N/2 \rfloor - \lceil N/5 \rceil = 3 q + \lfloor r/2 \rfloor - \lceil r/5 \rceil$$
For $r = 0,1,\ldots,9$, $ \lfloor r/2 \rfloor - \lceil r/5 \rceil = 0, -1, 0, 0, 1, 1, 1, 1, 2, 2$. We need a value divisible by $3$ since $24$ and $3q$ are divisible by $3$, so $r = 0, 2$ or $3$; $q = 24/3 = 8$, and thus $N$ is either $80$, $82$ or $83$.
| {
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Evaluate $\sum\limits_{k=1}^{\infty} \frac{k^2-1}{k^4+k^2+1}$ How to find $$\sum_{k=1}^{\infty} \frac{k^2-1}{k^4+k^2+1}$$
I try something like this:
$$\begin{align*}\sum_{k=1}^{\infty} \frac{k^2-1}{k^4+k^2+1}=\sum_{k=1}^{\infty}\frac{k^2}{k^4+k^2+1}-\sum_{k=1}^{\infty}\frac{1}{k^4+k^2+1}.\end{align*}$$
Using fact that $$\sum_{k = 1}^{n}{\frac{1}{k^4+k^2+1}}=\frac{1}{2}\cdot\frac{n+1}{n^2+n+1}+\frac{1}{2}\cdot\sum_{k = 1}^{n-1}{\frac{1}{k^2+k+1}}$$ we find that $$\begin{align*}\sum_{k=1}^{\infty}\frac{1}{k^4+k^2+1}
&=\frac{1}{2}\cdot\sum_{k=1}^{\infty}{\frac{1}{k^2+k+1}}\\
&=\frac{1}{6}\left(\sqrt{3}\pi \tanh{\left(\frac{\sqrt{3}\pi}{2}\right)}-1\right).\end{align*}$$
But I don't know how to find $\displaystyle\sum_{k=1}^{\infty}\frac{k^2}{k^4+k^2+1}.$
If someone want to know how to evaluate $\displaystyle\sum_{k=0}^{\infty}\frac{1}{k^2+k+1}$:
First, $$\displaystyle\sum_{k=0}^{\infty} \frac{1}{k^2+k+1}=\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}.$$ Now, using "well-know" formula $$\displaystyle\cos(\phi)=\prod_{k=0}^{\infty}{\left( 1-\frac{4\phi^2}{(2k+1)^2\pi^2}\right)}$$ we find that $$\displaystyle\log (\cos(\phi))=\sum_{k=0}^{\infty}{\log\left( 1-\frac{4\phi^2}{(2k+1)^2\pi^2}\right)}$$ and then we attack with $\dfrac{d}{d\phi}$ and find $$\displaystyle\tan(\phi)=\sum_{k=0}^{\infty}{\frac{8\phi}{(2k+1)^2\pi^2-4\phi^2}}.$$ Let $\phi=\pi\alpha\cdot i$, then we get $$\displaystyle\tan(\pi\alpha\cdot i)=i\cdot\tanh(\pi\alpha)=i\cdot\sum_{k=0}^{\infty}{\frac{8\pi\alpha}{(2k+1)^2\pi^2+4\pi^2\alpha^2}}=\frac{2\alpha i}{\pi}\cdot\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\alpha^2}}.$$ So, we find that $$\displaystyle\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\alpha^2}}=\frac{\pi}{2\alpha}\cdot\tanh(\pi\alpha).$$ Let $ \alpha=\dfrac{\sqrt{3}}{2}.$ We get $$\displaystyle\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}=\frac{\sqrt{3}\pi}{3}\cdot\tanh\left(\frac{\sqrt{3}\pi}{2}\right)$$ or $$\displaystyle\sum_{k=0}^{\infty} \frac{1}{k^2+k+1}=\frac{\sqrt{3}\pi}{3}\cdot\tanh\left(\frac{\sqrt{3}\pi}{2}\right).$$
| Using Partial Fraction Decomposition $$\frac{k^2-1}{k^4+k^2+1}=\frac{Ak+B}{k^2-k+1}+\frac{Ck+D}{k^2+k+1}$$
So, $k^2-1=k^3(A+C)+k^2(A+B-C+D)+k(A+B+C-D)+B+D$
Comparing the coefficients of different powers of $k$ in the above identity,
$A+C=0$
From $A+B+C-D=0,B+D=0$ and $B+D=-1\implies B=-D=-\frac12$
From $A+B-C+D=1\implies A-C=2$ and $A+C=0\implies A=-C=1$
$$\implies\frac{k^2-1}{k^4+k^2+1}=\frac{k-\frac12}{k^2-k+1}-\frac{k+\frac12}{k^2+k+1}$$
$$\frac{2(k^2-1)}{k^4+k^2+1}=\frac{2k-1}{k^2-k+1}-\frac{2k+1}{k^2+k+1}=T(k)\text{ say}$$
$$\implies T(n)=\frac{2n-1}{n^2-n+1}-\frac{2n+1}{n^2+n+1}$$
If we set $U(m)=\dfrac{2m-1}{m^2-m+1},U(m+1)=\dfrac{2(m+1)-1}{(m+1)^2-(m+1)+1}=\dfrac{2m+1}{m^2+m+1}$
$$\implies T(n)=U(n)-U(n+1)$$
Clearly, the first part of any term except the first term is cancelled by the last part of the previous term.
$$\implies2\sum_{k=1}^{\infty} \frac{k^2-1}{k^4+k^2+1}=U(1)=\cdots$$
| {
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A improper integral with Glaisher-Kinkelin constant Show that :
$$\int_0^\infty \frac{\text{e}^{-x}}{x^2} \left( \frac{1}{1-\text{e}^{-x}} - \frac{1}{x} - \frac{1}{2} \right)^2 \, \text{d}x = \frac{7}{36}-\ln A+\frac{\zeta \left( 3 \right)}{2\pi ^2}$$
Where $\displaystyle A$ is Glaisher-Kinkelin constant
I see Chris's question is a bit related with this Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$
| Define $$f(s) = \int_0^{\infty} x^{s - 1} \frac{e^{-x}}{x^2} \left(\frac1{1 - e^{-x}} - \frac1x - \frac12\right)^2 dx.$$ This defines an analytic function on the domain $\operatorname{Re}(s) > 0$ and the problem is to evaluate $f(1)$.
We have $$f(s) = \int_0^{\infty} x^{s - 3} e^{-x} \left(\frac1{(1 - e^{-x})^2} + \frac1{x^2} + \frac14 - \frac2{x(1 - e^{-x})} - \frac1{1 - e^{-x}} + \frac1x\right) dx$$
$$\quad\ = \int_0^{\infty} \left(\frac{x^{s - 3} e^{x}}{(e^x - 1)^2} + x^{s - 5} e^{-x} + \frac{x^{s - 3} e^{-x}}4 - \frac{2x^{s - 4}}{e^x - 1} - \frac{x^{s - 3}}{e^x - 1} + x^{s - 4} e^{-x}\right) dx.$$
For $\operatorname{Re}(s) > 4$, integrating by parts on the first term gives $$f(s) = \int_0^{\infty} \left(\frac{(s - 3)x^{s - 4}}{e^x - 1} + x^{s - 5} e^{-x} + \frac{x^{s - 3} e^{-x}}4 - \frac{2x^{s - 4}}{e^x - 1} - \frac{x^{s - 3}}{e^x - 1} + x^{s - 4} e^{-x}\right) dx.$$
$$= \int_0^{\infty} \left(\frac{(s - 5)x^{s - 4}}{e^x - 1} + x^{s - 5} e^{-x} + \frac{x^{s - 3} e^{-x}}4 - \frac{x^{s - 3}}{e^x - 1} + x^{s - 4} e^{-x}\right) dx.$$
Again assuming $\operatorname{Re}(s) > 4$, this gives us $$f(s) = (s - 5)\Gamma(s - 3)\zeta(s - 3) + \Gamma(s - 4) + \frac14 \Gamma(s - 2) - \Gamma(s - 2)\zeta(s - 2) + \Gamma(s - 3),$$ but, by analytic continuation, the equation must be valid (where the right side is defined) for $\operatorname{Re}(s) > 0$ and the apparent singularities of the right side at $s = 1$, $2$, $3$, and $4$ must be removable.
We may write $$f(s) = \frac{(s - 4)(s - 5)\zeta(s - 3) + 1 + \frac14 (s - 4)(s - 3) - (s - 4)(s - 3)\zeta(s - 2) + s - 4}{(s - 4)(s - 3)(s - 2)(s - 1)}\Gamma(s)$$ and so $$f(1) = \lim_{s \to 1} \left(\frac{(s - 5)\zeta(s - 3)}{(s - 3)(s - 2)(s - 1)} - \frac{\zeta(s - 2) + \frac1{12}}{(s - 2)(s - 1)} + \frac{\frac13 (s - 4)(s - 3) + s - 3}{(s - 4)(s - 3)(s - 2)(s - 1)}\right)$$
$$= \lim_{s \to 1} \left(\frac{(s - 5)\zeta(s - 3)}{(s - 3)(s - 2)(s - 1)} - \frac{\zeta(s - 2) + \frac1{12}}{(s - 2)(s - 1)} + \frac1{3(s - 4)(s - 2)}\right)$$
$$= -2\zeta'(-2) + \zeta'(-1) + \frac19$$
$$= \frac{\zeta(3)}{2\pi^2} + \frac1{12} - \ln A + \frac19$$
$$= \frac7{36} - \ln A + \frac{\zeta(3)}{2\pi^2}.$$
| {
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Exercise of complex variable, polynomials. Calculate the number of zeros in the right half-plane of the following polynomial:
$$z^4+2z^3-2z+10$$
Please, it's the last exercise that I have to do. Help TT. PD: I don't know how do it.
| Consider the polynomial
$$
p(z) = z^4 + 2z^3 - 2z + \epsilon.
$$
When $\epsilon = 0$ we can solve for the zeros explicitly using the cubic formula. They are $z=0$ and
$$
z \in \left\{\begin{array}{c}
-\frac{2}{3}+\frac{1}{3} \sqrt[3]{19-3 \sqrt{33}}+\frac{1}{3} \sqrt[3]{19+3 \sqrt{33}}, \\
-\frac{2}{3}-\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{19-3 \sqrt{33}}-\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{19+3 \sqrt{33}}, \\
-\frac{2}{3}-\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{19-3 \sqrt{33}}-\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{19+3 \sqrt{33}}
\end{array}\right\}.
$$
Now,
$$
\sqrt[3]{19-3 \sqrt{33}} > \sqrt[3]{19-3 \sqrt{36}} = 1,
$$
so that the first zero in this set satisfies
$$
\begin{align*}
& -\frac{2}{3}+\frac{1}{3} \sqrt[3]{19-3 \sqrt{33}}+\frac{1}{3} \sqrt[3]{19+3 \sqrt{33}} \\
&\qquad > -\frac{2}{3}+\frac{1}{3} \sqrt[3]{19-3 \sqrt{33}}+\frac{1}{3} \sqrt[3]{19-3 \sqrt{33}} \\
&\qquad = -\frac{2}{3}+\frac{2}{3} \sqrt[3]{19-3 \sqrt{33}} \\
&\qquad > 0.
\end{align*}
$$
The real part of the next two zeros is clearly negative:
$$
-\frac{2}{3}-\frac{1}{6} \sqrt[3]{19-3 \sqrt{33}}-\frac{1}{6} \sqrt[3]{19+3 \sqrt{33}} < 0.
$$
So when $\epsilon = 0$ we have one zero at $z=0$, one zero with $\Re(z) > 0$, and two zeros with $\Re(z) < 0$.
Next,
$$
p(iy) = y^4 + \epsilon - i2(y+y^3),
$$
so that $p(z)$ has no zeros on the imaginary axis when $\epsilon > 0$.
We now consider the zero of $p(z)$ which is located at $z=0$ when $\epsilon = 0$ as an analytic function $z_0 = z_0(\epsilon)$ with $z_0(0) = 0$. Expand $z_0$ as a Taylor series
$$
\begin{align*}
z_0(\epsilon) &= z_0(0) + z_0'(0)\epsilon + O(\epsilon^2) \\
&= z_0'(0)\epsilon + O(\epsilon^2)
\end{align*}
$$
(valid for small $\epsilon$) and substitute this into the equation $p(z_0) = 0$ to get
$$
[1-2z_0'(0)]\epsilon = O(\epsilon^2).
$$
Divide both sides by $\epsilon$ and let $\epsilon \to 0$ to find that
$$
1-2z_0'(0) = 0
$$
or $z_0'(0) = 1/2$. Thus for small $\epsilon > 0$ there are two zeros satisfying $\Re(z) > 0$. Since $p(z)$ has no zeros with $\Re(z) = 0$ when $\epsilon > 0$ and the zeros of polynomials are continuous functions of the coefficients we conclude that this is true for all $\epsilon > 0$.
In particular, when $\epsilon = 10$ there are exactly two zeros with $\Re(z) > 0$ and two zeros with $\Re(z) < 0$.
| {
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Solve $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$ for $x$ Is there any smart way to solve the equation: $$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$
Use Maple I can find $x \in \{1;ab+bc+ca\}$
| Multiplying the original equation out is
$-a^3b^2x + a^3c^2x + a^2b^3x - a^2c^3x - b^3c^2x + b^2c^3x + a^3b^2 - a^3c^2 - a^2b^3 +a^2bx^2 + a^2c^3 - a^2cx^2 - ab^2x^2 + ac^2x^2 + b^3c^2 - b^2c^3 + b^2cx^2 - bc^2x^2 a^2bx + a^2cx + ab^2x - ac^2x - b^2cx + bc^2x=0$
Taking all the coefficients $a$, $b$ and $c$ in the quadratic $ax^2+bx+c=0$ grouping them and then factoring leads to the following quadratic $$(b-c)(a-c)(a-b)x^2-(b-c)(a-c)(a-b) (ab+ac+bc+1)x+(b-c)(a-c)(a-b)(ab+ac+bc)=0$$ Dividing both sides of the equation by $$(b-c)(a-c)(a-b)$$ gives $$x^2-(ab+ac+bc+1)x+(ab+ac+bc)=0$$ Using the quadratic formula
$$x=\frac{(ab+ac+bc+1)\pm{\sqrt{(ab + ac + bc + 1)^2-4(ab + ac + bc)}}}{2}$$ or $$x=\frac{(ab+ac+bc+1)\pm\sqrt{(ab + ac + bc-1)^2}}{2}$$ So the roots are $$x=ab+ac+bc$$ and $$x=1$$
| {
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Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet
First, I showed that $(m^2 - n^2, 2mn, m^2 + n^2)$ is in fact a Pythagorean triplet.
$$\begin{align*} (m^2 - n^2)^2 + (2mn)^2 &= (m^2 + n^2)^2 \\
&= m^4 -2m^2n^2 + n^4 + 4m^2n^2 \\
&= m^4 + 2m^2n^2 + n^4 \\
&= 1\end{align*}$$ which shows that it respect $a^2+b^2 = c^2$
let p be a prime number, $ p|(m^2 + n^2) \text { and } p|(m^2 - n^2) $
if $gcd(m^2 + n^2, (m^2 - n^2)) = 1$
$p | (m^2 + n^2) , \text { so, } p |m^2 \text { and } p |n^2$
that means $ (m^2 + n^2) \text { and } (m^2 - n^2) $ are prime together
I'm kind of lost when I begin to show the gcd = 1... I think I know what to do, just not sure how to do it correctly.
Thanks
| The question is wrong.
It is not always primitive.
As when we put m=3, n=1
The triplet comes to be 8,6,10, which is not a primitive Pythagorean triplet.
| {
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Permutation question. $qpq^{-1}$, $q,p,r,s \in S_{8}$. Let $p,r,s,q \in S_{8}$ be the permutation given by the following products of cycles:
$$p=(1,4,3,8,2)(1,2)(1,5)$$
$$q=(1,2,3)(4,5,6,8)$$
$$r=(1,2,3,8,7,4,3)(5,6)$$
$$s=(1,3,4)(2,3,5,7)(1,8,4,6)$$
Compute $qpq^{-1}$ and $r^{-2}sr^{2}.$
thanks for your help.
I want to write the following permutations like :
$p=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\4&1&8&3&?&?&?&2\end{pmatrix}$
$q=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\2&3&1&5&6&8&7&4\end{pmatrix}$
$r=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 2&3&8&3&6&5&4&7\end{pmatrix}$
$s=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 3&?&4&1&?&?&?&?\end{pmatrix}$
can you help me plese to fill the $?$ mark. Is there another method to compute $qpq^{-1}$?
thanks:)
| There is a quick way to compute both of your permutations, and it is by using this fact:
Suppose $\tau, \sigma \in S_n$ and in cycle notation, $\sigma = (a_1, a_2, \cdots a_p) (b_1, b_2, \cdots, b_q) \cdots (z_1, z_2, \cdots, z_l)$ say. Then $$ \tau \sigma \tau^{-1} = (\tau(a_1), \tau(a_2), \cdots \tau(a_p)) (\tau(b_1), \tau(b_2), \cdots, \tau(b_q)) \cdots (\tau(z_1), \tau(z_2), \cdots, \tau(z_l))$$
That is, to conjugate by $\tau$, replace the elements in the cycles of $\sigma$ by their images under $\tau.$ Exercise: Prove this.
| {
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How to rewrite $\sin^4 \theta$ in terms of $\cos \theta, \cos 2\theta,\cos3\theta,\cos4\theta$? I need help with writing $\sin^4 \theta$ in terms of $\cos \theta, \cos 2\theta,\cos3\theta, \cos4\theta$.
My attempts so far has been unsuccessful and I constantly get developments that are way to cumbersome and not elegant at all. What is the best way to approach this problem?
I know that the answer should be:
$\sin^4 \theta =\frac{3}{8}-\frac{1}{2}\cos2\theta+\frac{1}{8}\cos4\theta$
Please explain how to do this.
Thank you!
| $$\begin{align}\sin^4 \theta &= (\sin^2\theta)^2\\ &= \left(\frac12-\frac12\cos(2\theta)\right)^2\\ &= \frac14 \left(1 - \cos(2\theta)\right)^2\\ &= \frac14\left(1 - 2 \cos(2\theta) + \cos^2(2 \theta)\right)\\ &= \frac14\left(1 - 2 \cos(2 \theta) + \frac12(\cos (4\theta) + 1)\right)\\ &= \frac14\left(\frac32 - 2\cos(2\theta) + \frac12\cos(4 \theta)\right)\\ &= \frac38 - \frac12\cos(2\theta) + \frac18\cos(4\theta).\end{align}$$
| {
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If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then find the largest possible value of $|a-b|$ I came across the following problem that says:
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then the largest possible value of $|a-b|$ is which of the following?
$(1)\sqrt 7$, $(2)\sqrt{7/2}$ , $(3)\sqrt {35}$ $(4)7$
My Attempt: We notice, $(a-b)^2=7-(4ab+b^2)$ and hence $|a-b|=\sqrt {7-(4ab+b^2)}$ and so $|a-b|$ will be maximum whenever $(4ab+b^2)$ will be minimum. But now I am not sure how to progress further hereon.Can someone point me in the right direction? Thanks in advance for your time.
| You can solve the given equation as a quadratic to get $a=-b \pm \sqrt {b^2-2b^2+7}=-b \pm \sqrt {7-b^2}$. To minimize $4ab+b^2$ you want $a$ as negative as possible, so take the minus sign. Now you have $|a-b|=\sqrt{7-4b(-b-\sqrt{7-b^2})-b^2}=\sqrt{7+3b^2+4b\sqrt{7-b^2}}$. Now you can take the derivative, set to zero, etc.
Added: Wolfram Alpha shows the maximum at $b=2\sqrt{\frac 75}, a=-3\sqrt {\frac 75}$ with $|a-b|=\sqrt {35}$
| {
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Probability of choosing two equal bits from three random bits Given three random bits, pick two (without replacement). What is the probability that the two you pick are equal?
I would like to know if the following analysis is correct and/or if there is a better way to think about it.
$$\Pr[\text{choose two equal bits}] = \Pr[\text{2nd bit} = 0 \mid \text{1st bit} = 0] + \Pr[\text{2nd bit} = 1 \mid \text{1st bit} = 1]$$
Given three random bits, once you remove the first bit the other two bits can be: 00, 01, 11, each of which occurring with probability $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$. Thus,
$$\Pr[\text{2nd bit} = 0] = 1\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{4} + 0\cdot\frac{1}{4} = \frac{3}{8}$$
And $\Pr[\text{2nd bit} = 1] = \Pr[\text{2nd bit} = 0]$ by the same analysis.
Therefore,
$$\Pr[\text{2nd bit}=0 \mid \text{1st bit} = 0] = \frac{\Pr[\text{1st and 2nd bits are 0}]}{\Pr[\text{1st bit}=0]} = \frac{1/2\cdot3/8}{1/2} = \frac{3}{8}$$
and by the same analysis, $\Pr[\text{2nd bit} = 1 \mid \text{1st bit} = 1] = \frac{3}{8}$.
Thus, $$\Pr[\text{choose two equal bits}] = 2\cdot\frac{3}{8} = \frac{3}{4}$$
| I hope I'm not repeating anyone else's answer here.
Since there are 3 bits, there are $2^3$ possible combinations. We can break them down into 2 categories:
Category 1: all bits are the same, i.e. $\{000,111 \}$.
Category 2: two bits are 0, one is 1.
Category 3: two bits are 1, one is 0.
Clearly the last category is the same as 2.
Hence your law of total probability should be something like
$$
P(\text{sample two same bits})=P(\text{sample 2 same bits}|C1)P(C1) +2P(\text{sample 2 same bits}|C_2)P(C_2)\\
=1 \cdot \frac{2}{2^3} + 2 \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot \binom{3}{1}\cdot \frac{1}{2^3}=\frac{1}{2}
$$
Expanation: $\binom{3}{1} \cdot \frac{1}{2^3}$ is the probability of the outcome 'two bits are the same 1, one is different' (since you have 3 slots). $\frac{2}{3} \cdot \frac{1}{2}$ is the probability to sample two of those equal bits without replacement. 2 is of course due to 2 options (0 or 1).
| {
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} |
Prove that $ \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$.
Given $a,b,c>0$, prove that $\displaystyle \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$.
I expanded the LHS, and realized I have to prove $\displaystyle\frac a b +\frac a c +\frac b c +\frac b a +\frac c a +\frac c b \geq \frac{2(a+b+c)}{\sqrt[3]{abc}}$, but I don't know how. Please help. Thank you.
| The same inequality mentioned by @Sanchez used three times takes us to:
$$\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+3\geq \frac{3(a+b+c)}{\sqrt[3]{abc}}$$
On the other hand, AM-GM gives us:
$$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$
so that:
$$-3 \geq -\frac{a+b+c}{\sqrt[3]{abc}}$$
And adding the first and third inequalities gives the result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Need help integrating $\frac{(t-1)^2-2t(t-1)}{t^2+(a(t-1)^2)^2}$ I need to integrate $$a\int_1^2 \frac{(t-1)^2-2t(t-1)}{t^2+(a(t-1)^2)^2} dt$$ $$=a\int_1^2 \frac{-t^2+1}{t^2+(a(t-1)^2)^2} dt $$
$$=-a\int_1^2 \frac{t^2-1}{t^2+(a(t-1)^2)^2} dt $$ with the hint that two trigonomic substitutions would be necessary and to consider using arctan. I have tried to tackle this a few different ways and get stuck every time. Can anyone help me start off?
| Looks like we could make use of this
$$ a\int \frac{-t^2+1}{t^2+(a(t-1)^2)^2} \; dt =
-a \int \frac{1 - \frac{1}{t^2}}{1 + \left( a \left( \sqrt{t } - \frac{1}{\sqrt t } \right )^2 \right )^2 }dt =
-a \int \frac{1 - \frac{1}{t^2}}{1 + \left( a \left( t + \frac{1}{ t } - 2 \right ) \right )^2 }dt $$
Substitute $\displaystyle a\left( t + \frac{1}{ t } - 2 \right ) = u$, you get $\displaystyle - \int \frac{ 1}{1 + (u)^2} du $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/315964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding :$x_1^8+x_2^8+\cdots+x_8^8$ If $x_1,x_2,\ldots,x_8$ are roots for the equation :
$$x^8-13x^2+7x-6=0$$ then how to find
$$x_1^8+x_2^8+\cdots+x_8^8$$
| $x^8-13x^2+7x-6=\prod_{k=1}^8(x-x_k)$, in which the coefficient of $x^7$ is $-\sum_{k=1}^8x_k$, so $\sum_{k=1}^8x_k=0$. The coefficient of $x^6$ is $\sum_{1\le i<k\le 8}x_ix_k$, so
$$0=\left(\sum_{k=1}^8x_k\right)^2=\sum_{k=1}^8x_k^2+2\sum_{1\le i<k\le 8}x_ix_k=\sum_{k=1}^8x_k^2+0\;,$$
and $\sum_{k=1}^8x_k^2=0$. Thus,
$$\begin{align*}
\sum_{k=1}^8x_k^8&=\sum_{k=1}^8\left(13x_k^2-7x_k+6\right)\\
&=13\sum_{k=1}^8x_k^2-7\sum_{k=1}^8x_k+48\\
&=48\;.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
Conditions that $\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$ is rational Motivation
I am working on one of the questions from Hardy's Course of Pure Mathematics and was wondering if I could get some assistance on where to go next in my proof. I have attempted rearranging the expression in numerous ways from the step I am at, but seem to get no-where.
Question
If $a^2-b>0$, then the necessary and sufficient conditions that $\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$ is rational are $a^2-b$ and $\dfrac{1}{2} (a+ \sqrt{a^2-b})$ be squares of rational numbers.
Attempt
Suppose that $\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$ is rational. Then it can be written as the ratio of two integers, p and q, that have no common factor. Write this as:
$\dfrac{p}{q}=\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$
Then by squaring both sides we have:
$\dfrac{p^2}{q^2} = (\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}})^2=2a+ 2\sqrt{a^2-b}$
-Note sure where to go from here.
| The assertion that "If $a^2−b>0$, then the necessary and sufficient conditions that $\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}$ is rational are $a^2−b$ and $\frac{1}{2}(a+\sqrt{a^2-b})$ be squares of rational numbers." is false. Take $a=2-\sqrt[4]{2}, b=4-4\sqrt[4]{2}, a^2-b=\sqrt{2}, \sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}=2$.
I believe the assertion should read: "If $a^2−b>0$, then the necessary and sufficient condition that $\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}$ is rational is that $\frac{1}{2}(a+\sqrt{a^2-b})$ is a square of a rational number."
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How does $\frac{1}{2} \sqrt{4 + 4e^4} = \sqrt{1 + e^4}$ My understanding would lead me to believe that:
$$\frac{1}{2} \sqrt{4 + 4e^4} = \frac{1}{2}(2 + 2e^4) = 1 + e^4$$
But it actually equals: $\sqrt{1 + e^4}$
Can you explain why?
| You have
$$
\frac{1}{2}\sqrt{4 + 4e^4} = \frac{1}{2}\sqrt{4(1+e^4)} = \frac{1}{2}\sqrt{4}\sqrt{1 + e^4} = \frac{1}{2}2\sqrt{1+e^4} = \sqrt{1+e^4}
$$
It looks like you where thinking that
$$
\sqrt{a + b} = \sqrt{a} + \sqrt{b}.
$$
But that is not true (try to check this with $a=b=2$). And even if you did that it looks like you forgot that $\sqrt{4e^4}= 2 e^2$ (as pointed out by TMM in the comment below.)
| {
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Property of $10\times 10 $ matrix Let $A$ be a $10 \times 10$ matrix such that each entry is either $1$ or $-1$. Is it true that $\det(A)$ is divisible by $2^9$?
| Alternatively, we can choose $A$ as below:
$$A = {\left[ {\begin{array}{*{20}{c}}
1&1& \ldots &1 \\
{ - 1}&1& \ldots &1 \\
{ - 1}&{ - 1}& \ldots &{ - 1} \\
\vdots &{}&{}&{} \\
{ - 1}&{ - 1}& \ldots &{ - 1}
\end{array}} \right]_{10 \times 10}}$$
If we add fist row to another rows, we get:
$$A = {\left[ {\begin{array}{*{20}{c}}
1&1& \ldots & \ldots &1 \\
0&2& \ldots & \ldots &2 \\
0&0& \ldots & \ldots &2 \\
\vdots &{}&{}&{}&{} \\
0&0& \ldots &0&2
\end{array}} \right]_{10 \times 10}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\det (A) = {2^9}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove $a\sqrt[3]{a+b}+b\sqrt[3]{b+c}+c\sqrt[3]{c+a} \ge 3 \sqrt[3]2$ Prove $a\sqrt[3]{a+b}+b\sqrt[3]{b+c}+c\sqrt[3]{c+a} \ge 3 \sqrt[3]2$ with $a + b+c=3 \land a,b,c\in \mathbb{R^+}$
I tried power mean inequalities but I still can't prove it.
| Hint, you can try to use calculus. Let $$f(a, b, c, \lambda )=a\sqrt[3]{a+b}+b\sqrt[3]{b+c}+c\sqrt[3]{c+a}-\lambda (a+b+c-3)=0.$$
Then, the extrema occurs at $\nabla f=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to calculate the asymptotic expansion of $\sum \sqrt{k}$? Denote $u_n:=\sum_{k=1}^n \sqrt{k}$. We can easily see that
$$ k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + O(k^{-1/2}),$$
hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + O(n^{1/2})$, because $\sum_1^n O(k^{-1/2}) =O(n^{1/2})$.
With some more calculations, we get
$$ k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + \frac{1}{2} (k^{1/2}-(k-1)^{-3/2}) + O(k^{-1/2}),$$
hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + \frac{1}{2} n^{1/2} + C + O(n^{1/2})$ for some constant $C$, because $\sum_n^\infty O(k^{-3/2}) = O(n^{-1/2})$.
Now let's go further. I have made the following calculation
$$k^{1/2} = \frac{3}{2} \Delta_{3/2}(k) + \frac{1}{2} \Delta_{1/2}(k) + \frac{1}{24} \Delta_{-1/2}(k) + O(k^{-5/2}),$$
where $\Delta_\alpha(k) = k^\alpha-(k-1)^{\alpha}$. Hence :
$$\sum_{k=1}^n \sqrt{k} = \frac{2}{3} n^{3/2} + \frac{1}{2} n^{1/2} + C + \frac{1}{24} n^{-1/2} + O(n^{-3/2}).$$
And one can continue ad vitam aeternam, but the only term I don't know how to compute is the constant term.
How do we find $C$ ?
| As shown in this answer, we can use the Euler-Maclaurin Sum Formula to get
$$
\sum_{k=1}^n\sqrt{k}=\frac23n^{3/2}+\frac12n^{1/2}+\zeta\left(-\frac12\right)+\frac1{24}n^{-1/2}-\frac1{1920}n^{-5/2}+\frac1{9216}n^{-9/2}+O\left(n^{-13/2}\right)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\frac{1}{x-1}+ \frac{2}{x-2}+ \frac{3}{x-3}+\cdots+\frac{10}{x-10}\geq\frac{1}{2} $ I would appreciate if somebody could help me with the following problem:
Q: find $x$
$$\frac{1}{x-1}+ \frac{2}{x-2}+ \frac{3}{x-3}+\cdots+\frac{10}{x-10}\geq\frac{1}{2} $$
| Let $x$ be any number such that $10 \lt x \lt 21$. Then for each $k\in [1,10]$ we have $21k \gt x \gt k$, hence $20k \gt x-k \gt 0$, so $\frac{k}{x-k} \gt \frac{1}{20}$. Summing from $k=1$ to $10$, we obtain the desired inequality.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to create a generating function / closed form from this recurrence? Let $f_n$ = $f_{n-1} + n + 6$ where $f_0 = 0$.
I know $f_n = \frac{n^2+13n}{2}$ but I want to pretend I don't know this. How do I correctly turn this into a generating function / derive the closed form?
| $$\begin{array}{rcl}
G(x) &=& \sum_{n=1}^\infty f_n x^n \\
&=& \sum_{n=1}^\infty (f_{n-1} + n + 6) x^n \\
&=& \sum_{n=1}^\infty f_{n-1} x^n + \sum_{n=1}^\infty n x^ n + \sum_{n=1}^\infty 6 x^n \\
&=& x G(x) + x \frac{d}{dx}\left(\frac{x}{1-x}\right) + 6 \frac{x}{1-x}
\end{array}$$
So $$G(x) = \frac{6x^2 - 7x}{x^3 - 3x^2 + 3x - 1} = 7x + 15x^2 + 24x^3 + 34x^4 + 45x^5 + \cdots$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Equilateral triangle geometric problem I have an Equilateral triangle with unknown side $a$. The next thing I do is to make a random point inside the triangle $P$. The distance $|AP|=3$ cm, $|BP|=4$ cm, $|CP|=5$ cm.
It is the red triangle in the picture. The exercise is to calculate the area of the Equilateral triangle (without using law of cosine and law of sine, just with simple elementary argumentation).
The first I did was to reflect point $A$ along the opposite side $a$, therefore I get $D$. Afterwards I constructed another Equilateral triangle $\triangle PP_1C$.
Now it is possible to say something about the angles, namely that $\angle ABD=120^{\circ}$, $\angle PBP_1=90^{\circ} \implies \angle APB=150^{\circ}$ and $\alpha+\beta=90^{\circ}$
Now I have no more ideas. Could you help me finishing the proof to get $a$ and therefore the area of the $\triangle ABC$. If you have some alternative ideas to get the area without reflecting the point $A$ it would be interesting.
| You can solve this without any trig if you consider the properties of a equilateral triangle, and the fact that you've created six right triangles in which you know the length of the hypotenuse and the relationship:
$a+b = c+d = e+f$
$a^2 + g^2 = |AP|^2$
$b^2 + g^2 = |BP|^2$
$c^2 + h^2 = |BP|^2$
$d^2 + h^2 = |CP|^2$
$e^2 + i^2 = |CP|^2$
$f^2 + i^2 = |AP|^2$
Then:
$a+b = c+d = e+f$
$a^2 + g^2 = 9$
$b^2 + g^2 = 16$
$c^2 + h^2 = 16$
$d^2 + h^2 = 25$
$e^2 + i^2 = 25$
$f^2 + i^2 = 9$
Then:
$a+b = c+d = e+f = s$
$b^2 - a^2 = 7$
$d^2 - c^2 = 9$
$e^2 - f^2 = 16$
Then:
$b^2 - (s-b)^2 - 7 = s^2 + 2sb - 7 = 0$
$d^2 - (s-d)^2 - 9 = s^2 + 2sd - 9 = 0$
$e^2 - (s-e)^2 - 16 = s^2 + 2se - 16 = 0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to simplify a square root How can the following:
$$
\sqrt{27-10\sqrt{2}}
$$
Be simplified to:
$$
5 - \sqrt{2}
$$
Thanks
| If you're faced with a question that says "Prove that $\sqrt{27-10\sqrt{2}}$ $=5 - \sqrt{2}$", then it's just a matter of squaring $5 - \sqrt{2}$ and seeing that you get $27-10\sqrt{2}$. But suppose the question your faced with is to find a square root of $27-10\sqrt{2}$ of the form $a+b\sqrt{2}$, where $a$ and $b$ are rational. Then you have
$$
27-10\sqrt{2}=\left(a+b\sqrt{2}\right)^2 = a^2 + 2ab\sqrt{2} + 2b^2
$$
so
\begin{align}
27 & = a^2+2b^2 \\[8pt]
-10 & = 2ab
\end{align}
From the second equation we get $a=-5/b$, then the first equation becomes
$$
27 = \frac{25}{b^2} + 2b^2
$$
or
$$
2(b^2)^2 -27b^2 + 25 = 0.
$$
A solution is $b^2=1$, and you can go on from there to find $b$ and then $a$.
(And remember that the number will have two square roots.)
Later note: In order for all this to work, we have to rely on the fact that $\sqrt{2}$ is irrational. That enables us to conclude that the rational parts are equal and the irrational parts are equal, so we have two equations.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$ Find the value of
$$I=\displaystyle\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$$
We have the information that
$$J=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)\ \mathrm dx=\dfrac{\pi^2}{8}\ln^2(2)-\dfrac{\pi^4}{192}$$
| I'm still struggling with this integral, but I guess the following result may have a chance to be helpful:
\begin{align*}
\int_{0}^{\frac{\pi}{2}} x^2 \log^2 \cos x \, dx
&= \frac{11 \pi^5}{1440} + \frac{\pi^3}{24} \log^2 2 + \frac{\pi}{2}\zeta(3) \log 2 \tag{1} \\
&\approx 4.2671523609840988652 \cdots.
\end{align*}
To prove this, let us consider the following identity
$$ \int_{0}^{\frac{\pi}{2}} \cos^{z}x \cos wx \, dx = \frac{\pi}{2^{z+1}} \binom{z}{\frac{z+w}{2}}.$$
You can find the proof of this identity at here. Thus it follows that
$$ \int_{0}^{\frac{\pi}{2}} x^2 \log^2 \cos x \, dx = - \left. \frac{\partial^4}{\partial z^2 \partial w^2} \frac{\pi}{2^{z+1}} \binom{z}{\frac{z+w}{2}} \right|_{(z, w) = (0, 0)}. $$
Performing a bunch of calculations, we obtain $(1)$. Similar idea shows that
$$ \int_{0}^{\frac{\pi}{2}} \log^2 \cos x \, dx = \left. \frac{\partial^2}{\partial z^2} \frac{\pi}{2^{z+1}} \binom{z}{\frac{z+w}{2}} \right|_{(z, w) = (0, 0)} = \frac{\pi^3}{24} + \frac{\pi}{2}\log 2. \tag{2} $$
Indeed, starting from the identity
$$ \log^2 \left( \frac{\sin 2x}{2} \right) = \log^2 \cos x + \log^2 \sin x + 2\log \cos x \log \sin x, $$
I obtained
\begin{align*}I
&= -\frac{7}{8}\int_{0}^{\frac{\pi}{2}} x^2 \log^2 \cos x \, dx
+ \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} x \log^2 \cos x \, dx
- \frac{3\pi^2}{32} \int_{0}^{\frac{\pi}{2}} \log^2 \cos x \, dx \\
&\quad -\frac{\log 2}{8}\int_{0}^{\pi} x^2 \log \sin x \, dx
+ \frac{\pi^3}{48} \log^2 2 \\
&\approx 0.077821979372293864338\cdots.
\end{align*}
From the identity
$$ \log \sin x = -\log 2 - \sum_{n=1}^{\infty} \frac{\cos 2nx}{n}, $$
we obtain
$$\int_{0}^{\pi} x^2 \log \sin x \, dx = -\frac{\pi}{2} \zeta (3) - \frac{\pi^3}{3} \log 2. \tag{3}$$
Putting $(1)$, $(2)$ and $(3)$ together, I was able to derive
\begin{align*}I
&= -\frac{61 \pi^5}{5760} - \frac{3\pi}{8} \zeta (3) \log 2 -\frac{\pi^3}{48} \log^2 2
+ \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} x \log^2 \cos x \, dx.
\end{align*}
I'm not sure if this formula will be helpful, since the last remaining integral seems to defy my techniques.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "60",
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How do I transform the left side into the right side of this equation? How does one transform the left side into the right side?
$$
(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2
$$
| $(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2$
If $(a^2+b^2)\ne 0$ and $(c^2+d^2)\ne 0$ (If either of them is $0$ then the statement is vacuously true).
Let $\sin \alpha =\displaystyle \frac{a}{\sqrt{a^2+b^2}}\Rightarrow \cos \alpha\displaystyle \frac{b}{\sqrt{a^2+b^2}}$ and $\sin \beta =\displaystyle \frac{c}{\sqrt{c^2+d^2}}\Rightarrow \cos \beta\displaystyle \frac{d}{\sqrt{c^2+d^2}}$
So we have ,
$\displaystyle \frac{(ac-bd)^2 + (ad+bc)^2}{(a^2+b^2)(c^2+d^2)}=(-\cos (\alpha+\beta))^2+(\sin(\alpha+\beta))^2 =1$
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Prove inequality: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ Prove: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ with $a, b, c \ge0$
I can do this by: $EAT^2$ (expand all of the thing)
*
*$(x+y+z)^4={x}^{4}+{y}^{4}+{z}^{4}+4\,{x}^{3}y+4\,{x}^{3}z+6\,{x}^{2}{y}^{2}+6\,{
x}^{2}{z}^{2}+4\,x{y}^{3}+4\,x{z}^{3}+4\,{y}^{3}z+6\,{y}^{2}{z}^{2}+4
\,y{z}^{3}+12\,x{y}^{2}z+12\,xy{z}^{2}+12\,{x}^{2}yz$
*$(x+y-z)^4={x}^{4}+{y}^{4}+{z}^{4}+4\,{x}^{3}y-4\,{x}^{3}z+6\,{x}^{2}{y}^{2}+6\,{
x}^{2}{z}^{2}+4\,x{y}^{3}-4\,x{z}^{3}-4\,{y}^{3}z+6\,{y}^{2}{z}^{2}-4
\,y{z}^{3}-12\,x{y}^{2}z+12\,xy{z}^{2}-12\,{x}^{2}yz$
...
$$28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4\\
\iff a^4 + b^4 + c^4 \ge a^2b^2+c^2a^2+b^2c^2 \text{(clearly hold by AM-GM)}$$
but any other ways that smarter ?
| This inequality is true for all reals $a$, $b$ and $c$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Hence, our inequality is a linear inequality of $w^3$, which says that it's enough to prove our inequality for an extremal value of $w^3$, which happens for equality case of two variables.
Since our inequality is homogeneous and fourth degree, we can assume $b=c=1$,
which gives $(a-1)^2(a+1)^2\geq0$.
Done!
| {
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"answer_id": 1
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Evaluating $\int \frac{1}{{x^4+1}} dx$ I am trying to evaluate the integral
$$\int \frac{1}{1+x^4} \mathrm dx.$$
The integrand $\frac{1}{1+x^4}$ is a rational function (quotient of two polynomials), so I could solve the integral if I can find the partial fraction of $\frac{1}{1+x^4}$. But I failed to factorize $1+x^4$.
Any other methods are also wellcome.
| Hint:
$$x^4+1=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1) \tag{1}$$
You can integrate using partial fraction decomposition. Since
$$x^4+1=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1),$$
then
$$\frac{1}{x^4+1}=\frac{Ax+B}{x^2-\sqrt{2}x+1}+\frac{Cx+D}{x^2+\sqrt{2}x+1}=\frac{(Ax+B)(x^2+\sqrt{2}x+1)+(Cx+D)(x^2-\sqrt{2}x+1)}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)} = \\
= \frac{x^3(A+C)+x^2(A\sqrt{2}+B+D-C\sqrt{2})+x(B\sqrt{2}-D\sqrt{2})+B+D}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}$$
$$\begin{cases}
A+C=0;\\
B+D+\sqrt{2}(A-C)=0; \\
B-D=0; \\
B+D=1.
\end{cases}$$
$$
A=-C=-\frac{1}{2\sqrt{2}}; \\
B=D=\frac{1}{2}$$
$$
\frac{1}{x^4+1}=\frac{1}{2\sqrt{2}}\left(\dfrac{-x+\sqrt{2}}{x^2-\sqrt{2}x+1}+ \dfrac{x+\sqrt{2}}{x^2+\sqrt{2}x+1} \right).
$$
Added
Decomposition (1) can be done using one of the following ways:
*
*Completion to the full square
$$x^4+1=x^4 +2x^2+1 -2x^2=(x^2+1)^2-\left(\sqrt{2}x\right)^2 = \\ =\big(x^2-\sqrt{2}x +1 \big)\big( x^2+\sqrt{2}x +1 \big).$$
*Let $\omega_i, \ i\in\{1,\, 2,\,3,\,4\} $ are roots of the equation $x^4+1=0$ over $\mathbb{C}:$ $\omega_1=\frac{\sqrt{2}}{2}(1-i), \ \omega_2=\frac{\sqrt{2}}{2}(1+i) \ \omega_3=\frac{\sqrt{2}}{2}(-1+i), \ \omega_4=\frac{\sqrt{2}}{2}(-1-i). $ Then use the decomposition into prime factors and multiply pairwise complex conjugate:
$x^4+1= \left( x-\frac{\sqrt{2}}{2}(1-i) \right)\left( x-\frac{\sqrt{2}}{2}(1+i) \right)\left( x-\frac{\sqrt{2}}{2}(-1+i) \right)\left( x-\frac{\sqrt{2}}{2}(-1-i) \right) = \\ =\big(x^2-\sqrt{2}x +1 \big)\big( x^2+\sqrt{2}x +1 \big).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find $\int_0^\infty \frac{\ln ^2z} {1+z^2}{d}z$ How to find the value of the integral $$\int_{0}^{\infty} \frac{\ln^2z}{1+z^2}{d}z$$ without using contour integration - using usual special functions, e.g. zeta/gamma/beta/etc.
Thank you.
| With $ \int_0^\infty \frac{2y\ln y }{(y^2+z^2)(y^2+1)}{dy}= \frac{\ln^2 z}{z^2-1}$
\begin{align}
&\int_0^\infty\frac{\ln^2 z}{1+z^2}dz\\
=&\ \int_0^\infty \frac1{1+z^2} \left(\int_0^\infty\frac{2(z^2-1)y\ln y}{(y^2+z^2)(y^2+1)}\ dy \right) dz\\
=& \int_0^\infty\int_0^\infty
\frac{2y\ln y}{(y^2+z^2)(y^2-1)}-\frac{4y\ln y}{(1+z^2)(y^4-1)}\ dz \ dy\\
=& \ \pi\int_0^\infty \frac{\ln y}{y^2-1} {dy}- 2\pi\int_0^\infty \frac{y\ln y}{y^4-1}\overset{y^2\to y} {dy}\\
=&\ \frac\pi2 \int_0^\infty \frac{\ln y}{y^2-1}\ dy
= \frac\pi4 \int_0^\infty \int_0^1 \frac{x}{1+(y^2-1)x^2}dx\>dy\\
= &\ \frac{\pi^2}4\int_0^1 \frac1{\sqrt{1-x^2}}dx=\frac{\pi^3}{8}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Variation of Pythagorean triplets: $x^2+y^2 = z^3$ I need to prove that the equation $x^2 + y^2 = z^3$ has infinitely many solutions for positive $x, y$ and $z$.
I got to as far as $4^3 = 8^2$ but that seems to be of no help.
Can some one help me with it?
| Set $z=a^2+b^2=(a+bi)(a-bi)$, $(i=\sqrt{-1})$
$x^2+y^2=(x+yi)(x-yi)=z^3=(a+bi)^3(a-bi)^3,$
$(1)\quad x+yi=(a+bi)^3=a^3+3a^2bi-3ab^2-b^3i,$
$x=a^3-3ab^2,y=3a^2b-b^3,z=a^2+b^2.$
$(2)\quad x+yi=(a+bi)^2(a-bi)=(a^2+b^2)(a+bi),$
$x=(a^2+b^2)a,y=(a^2+b^2)b,z=a^2+b^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 8,
"answer_id": 6
} |
Help me to prove this integration Where the method used should be using complex analysis.
$$\int_{c}\frac{d\theta}{(p+\cos\theta)^2}=\frac{2\pi p}{(p^2-1)\sqrt{p^2-1}};c:\left|z\right|=1$$
thanks in advance
| On $|z| = 1$, we have $z = e^{i\theta}$, so $\frac{dz}{d\theta} = i e^{i\theta}$, and hence $d\theta = \frac{dz}{iz}$.
Also on $|z| = 1$, we have $\cos \theta = \frac12\left(z+\frac{1}{z}\right)$.
Putting this together, you get
$$\frac{1}{i} \int_{|z| = 1} \frac{1}{z\left(p+\frac12\left(z+\frac{1}{z}\right)\right)^2}\ dz.$$
Expand the denominator,
$$\begin{align*}
\frac{1}{i} \int_{|z| = 1} \frac{1}{z\left(p+\frac12\left(z+\frac{1}{z}\right)\right)^2}\ dz &= \frac{1}{i} \int_{|z| = 1} \frac{1}{z\left(p^2 + p(z+1/z) + \frac14(z+1/z)^2\right)} \\
&= \frac{1}{i} \int_{|z| = 1} \frac{1}{\left(p^2z + pz^2+p + \frac{z}{4}(z^2+2+1/z^2)
\right)} \\
&= \cdots
\end{align*} $$
Then, apply the residue theorem.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Prove $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$.
Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$
Here's my idea:
$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$
And I'm stuck here.
I need to prove that:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ or
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)$, because $a+b+c = 3$
In the first case using Cauchy-Schwarz Inequality I prove that:
$(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$
$3(a^2 + b^2 + c^2) \ge (a+b+c)^2$
Now I need to prove that:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)$
$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2$
I need I don't know how to continue.
In the second case I tried proving:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ and
$a^2 + b^2 + c^2 \ge a+b+c$
Using Cauchy-Schwarz Inequality I proved:
$(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$
$(a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2$
$a^2 + b^2 + c^2 \ge a+b+c$
But I can't find a way to prove that $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$
So please help me with this problem.
P.S
My initial idea, which is proving:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$
maybe isn't the right way to prove this inequality.
| Hint:
What lower bound does AM-GM give you when you consider $a^2 + \sqrt{a} + \sqrt{a}$?
Your hope that $\sum \sqrt{a} \ge \sum a = 3$ is false, by using Cauchy Schwarz: $9 = 3(\sum a) \ge (\sum \sqrt{a})^2$. In fact, when $a+b+c = 3$, we have
$$\sum a^2 \ge \sum a = 3 \ge \sum \sqrt{a}$$
all by Cauchy-Schwarz, so your hope to split the inequality up is thwarted. This also signals us that we should try to "mix" $a^2$ and $\sqrt{a}$ together in some way, hence the hint.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 3
} |
Calculate:$y'$ for $y = x^{x^{x^{x^{x^{.^{.^{.^{\infty}}}}}}}}$ and $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....\infty}}}}$ (1) If $y = x^{x^{x^{x^{x^{.^{.^{.^{\infty}}}}}}}}$
(2) If $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....\infty}}}}$
then find $y'$ in both cases
(3)If $ y= \sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+....\infty}}}}\ $ then find$\ (2y-1)\frac{dy}{dx}$
| *
*As $y=x^y = e^{y \log{x}}$, take derivatives of both sides:
$$y' = \left( \frac{y}{x} + y' \log{x} \right ) x^y$$
Solve for $y'$.
*
*As $y^2 = x+y$, then a similar implicit differentiation as above yields
$$2 y \,y' = 1 + y'$$
Solve for $y'$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
How can find this sequence $ a_{n+1}=a_{n}+na_{n-1},$ let $a_{n+1}=a_{n}+na_{n-1},a_{1}=1，a_{2}=2$.
find the $a_{n}=?$
my ideas: $\dfrac{a_{n+1}}{(n+1)!}=\dfrac{1}{n+1}\dfrac{a_{n}}{n!}+\dfrac{1}{n+1}\dfrac{a_{n-1}}{(n-1)!},$
and let $b_{n}=\dfrac{a_{n}}{n!}$,then we have
$(n+1)b_{n+1}=b_{n}+b_{n-1},b_{1}=1,b_{2}=1$,but I failure,
so let $f(x)=\displaystyle\sum_{n=0}^{\infty}b_{n}x^n$,and assume that $b_{0}=0$
$f'(x)=\displaystyle\sum_{n=0}^{\infty}nb_{n}x^{n-1}=1+\displaystyle\sum_{n=1}^{\infty}(n+1)b_{n+1}x^{n-1}=1+\displaystyle\sum_{n=1}^{\infty}(b_{n}+b_{n-1})x^{n-1}=1+\dfrac{f(x)}{x}+f(x)$
then $f'(x)-(1+1/x)f(x)=1,f'(0)=1$
so
$f(x)=xe^x(\displaystyle\int \dfrac{e^{-x}}{x}dx+c)$
so $c=0$
and $f(x)=\displaystyle\sum_{n=0}^{\infty}\dfrac{x^{n+1}}{n!}\displaystyle\sum_{n=0}^{\infty}(-1)^n\dfrac{x^n}{(n+1)!}$
and my problem：How can prove that:
if:$f(x)=\displaystyle\sum_{n=0}^{\infty}b_{n}x^n=\displaystyle\sum_{n=0}^{\infty}\dfrac{x^{n+1}}{n!}\displaystyle\sum_{n=0}^{\infty}(-1)^n\dfrac{x^n}{(n+1)!}$
then we have
$b_{n}=\displaystyle\sum_{k=0}^{[n/2]}\dfrac{1}{（n-2k)!2^kk!}?$
| Let $f(x)$ be the exponential generating function of $(a_n)$:
$$ f(x) = \sum_{n=1}^{\infty} \frac{a_n}{n!} x^n = \sum_{n=1}^{\infty} b_n x^n, $$
where $(b_n)$ is as in OP's notation. Then
\begin{align*}
f'(x)
&= 1 + 2x + \sum_{n=1}^{\infty} (n+2) b_{n+2} x^{n+1} \\
&= 1 + 2x + \sum_{n=1}^{\infty} (b_{n+1} + b_n) x^{n+1} \\
&= 1 + 2x + (f(x) - x) + x f(x).
\end{align*}
Thus we have
$$ f'(x) - (x+1) f(x) = 1 + x. $$
Multiplying the integration factor $e^{-(x+1)^2 / 2}$, we have
$$ \left\{ e^{-(x+1)^2 / 2} f(x) \right\}' = (1 + x) e^{-(x+1)^2 / 2}. $$
Integrating, we have
$$ e^{-(x+1)^2 / 2} f(x) = - e^{-(x+1)^2 / 2} + c, $$
or equivalently
$$ f(x) = c e^{(x+1)^2 / 2} - 1. $$
By noting that $f(0) = 0$, we must have $c = e^{-1/2}$ and therefore
\begin{align*}
f(x)
= e^{\frac{x^2}{2}+x} - 1
&= \sum_{n=1}^{\infty} \frac{1}{n!} \left( \frac{x^2 + 2x}{2} \right)^n \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} \binom{n}{k} 2^{k-n} x^{2n-k} \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \sum_{l=n}^{2n} \binom{n}{2n-l} 2^{n-l} x^{l} \\
&= \sum_{l=1}^{\infty} \left( \sum_{\frac{l}{2}\leq n \leq l} \frac{1}{n!} \binom{n}{2n-l} 2^{n-l} \right) x^{l}
\end{align*}
Therefore we have
$$ a_n = n! b_n = \sum_{\frac{n}{2}\leq k \leq n} \frac{n!}{(2k-n)!(n-k)!2^{n-k}}. $$
Finally, substituting $k \mapsto n-k$, we have
$$ a_n = \sum_{0\leq k \leq \frac{n}{2}} \frac{n!}{(n-2k)!k!2^k} = \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{n!}{(n-2k)!k!2^k}. $$
The followings are computer calculations:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
} |
Easier way of calculating the determinant for this matrix I have to calculate the determinant of this matrix:
$$
\begin{pmatrix}
a&b&c&d\\b&c&d&a\\c&d&a&b\\d&a&b&c
\end{pmatrix}
$$
Is there an easier way of calculating this rather than the long regular way?
| A pedestrian's solution (experimentX's suggestion below the question).
Add the first three columns to the fourth:
\begin{align*}
\begin{vmatrix}
a&b&c&d\\b&c&d&a\\c&d&a&b\\d&a&b&c
\end{vmatrix} &
=(a+b+c+d)\begin{vmatrix}
a&b&c&1\\b&c&d&1\\c&d&a&1\\d&a&b&1
\end{vmatrix}
\end{align*}
Subtract the second row from the first row, the third row from the second row and the fourth row from the third row; develop after the fourth column:
\begin{align*}
&=(a+b+c+d)
\begin{vmatrix}
a-b&b-c&c-d \\
b-c&c-d&d-a \\
c-d&d-a&a-b
\end{vmatrix} \\
\end{align*}
Add the first column to the third column:
\begin{align*}
&=(a+b+c+d)(a-b+c-d)
\begin{vmatrix}
a-b&b-c&1 \\
b-c&c-d&-1 \\
c-d&d-a&1
\end{vmatrix}
\end{align*}
Add the second row to the first row and the third row to the second row:
\begin{align*}
&=(a+b+c+d)(a-b+c-d)
\begin{vmatrix}
a-c&b-d&0 \\
b-d&c-a&0 \\
c-d&d-a&1
\end{vmatrix} \\
&=-(a+b+c+d)(a-b+c-d)[(a-c)^2+(b-d)^2].
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
} |
What are some useful tricks/shortcuts for verifying trigonometric identities? What "tricks" are there that could help verify trigonometric identities?
For example one is:
$$a\cos\theta+b\sin\theta = \sqrt{a^2+b^2}\,\cos(\theta-\phi)$$
| Note that $\cos(x-y)=\cos x\cos y+\sin x\sin y$. This is obtained from the more familiar formula for $\cos(x+y)$ by replacing $y$ by $-y$.
Note also that
$$a\cos\theta+b\sin\theta=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos\theta+ \frac{b}{\sqrt{a^2+b^2}}\sin\theta \right).$$
So if $\phi$ is the angle whose cosine is $\frac{a}{\sqrt{a^2+b^2}}$ and whose sine is $\frac{b}{\sqrt{a^2+b^2}}$, then in the formula above we can replace $\frac{a}{\sqrt{a^2+b^2}}$ by $\cos\phi$, and $\frac{b}{\sqrt{a^2+b^2}}$ with $\sin\phi$, and obtain
$$a\cos\theta+b\sin\theta=\sqrt{a^2+b^2} \cos(\theta-\phi).$$
Remark: As to tricks and shortcuts, mostly it is a question of experience and practice. Already, I am sure, you recognize certain patterns and know how to exploit them. After a while, you will have used most of the common devices a dozen times, and then things get easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Calculate:$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$ How to calculate following with out using L'Hospital rule
$$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$$
| \begin{eqnarray*}
\lim_{x\rightarrow\left(-1\right)^{+}}\frac{\sqrt{\pi}-\sqrt{\arccos x}}{\sqrt{x+1}} & = & \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\pi-\arccos x}{2\sqrt{\pi}\sqrt{x+1}}\\
& = & \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\sin\left(\pi-\arccos x\right)}{2\sqrt{\pi}\sqrt{x+1}}\\
& = & \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\sin\arccos x}{2\sqrt{\pi}\sqrt{x+1}}\\
& = & \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\sqrt{1-x^{2}}}{2\sqrt{\pi}\sqrt{x+1}}\\
& = & \frac{1}{\sqrt{2\pi}}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Surface integral over ellipsoid I've problem with this surface integral:
$$
\iint\limits_S {\sqrt{ \left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)}}{dS}
$$, where
$$
S = \{(x,y,z)\in\mathbb{R}^3: \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}= 1\}
$$
| Note that the outward unit normal ${\bf n}$ to the ellipsoid is given by ${\displaystyle {({x \over a^2}, {y \over b^2}, {z \over c^2}) \over \sqrt{ \left(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\right)}}}$.
So if ${\bf F} = ({x \over a^2}, {y \over b^2}, {z \over c^2})$, your integral is $\int_S {\bf F} \cdot {\bf n}\,dS$. By the divergence theorem, this is equal to $\int_E div\,\mathbf{F}$, where
$E$ is the ellipsoid's interior. But $ div \,\mathbf{F}$ is the constant ${1 \over a^2} + {1 \over b^2} + {1 \over c^2}$ and the ellipsoid has volume ${4\pi \over 3}abc$, so the integral will evaluate to
$${4\pi \over 3}abc \times \bigg({1 \over a^2} + {1 \over b^2} + {1 \over c^2}\bigg)$$
$$= {4\pi \over 3} \bigg({bc \over a} + {ac \over b} + {ab \over c}\bigg)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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If $2x = y^{\frac{1}{m}} + y^{\frac{-1}{m}}(x≥1) $ then prove that $(x^2-1)y^{"}+xy^{'} = m^{2}y$ How do I prove following?
If $2x = y^{\frac{1}{m}} + y^{\frac{-1}{m}},(x≥1)$, then prove that $(x^2-1)y^{"}+xy^{'} = m^{2}y$
| $2x = y^{\frac{1}{m}} + y^{\frac{-1}{m}}(x≥1) $ then prove that $(x^2-1)y^{"}+xy^{'} = m^{2}y.........(1)$
Differntiating w.r.t x
$2 = \frac{1}{m}y^{\frac{1}{m}-1}.y^{'} -\frac{-1}{m}y^{\frac{-1}{m}-1}.y^{'}$
$2 = \frac{1}{m}\left(\frac{y^\frac{1}{m}}{y}-\frac{y^{\frac{-1}{m}}}{y}\right)y^{'}$
$2my = \left(y^\frac{1}{m}-y^\frac{-1}{m}\right)y^{'}$
$4m^2y^2 = \left(y^\frac{1}{m}-y^\frac{-1}{m}\right)^2(y^{'})^2$
$4m^2y^2 = (y^{'})^2\left((y^\frac{1}{m}+y^\frac{-1}{m})^2-4\right)$
$4m^2y^2 = (y^{'})^2\left(4x^2-4\right)$......(By -(1))
$m^2y^2 = (y^{'})^2\left(x^2-1\right)$
Differentiating w.r.t x
$m^2(2yy^{'}) = (y^{'})^2(2x) + \left(x^2-1\right)(2y^{'}y^{''})$
canceling $2y^{'}$ on both side
$(x^2-1)y^{"}+xy^{'} = m^{2}y$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$x$ is rational, $\frac{x}{2}$ is rational, and $3x-1$ is rational are equivalent How do we prove that the three statements below about the real number $x$ are equivalent?
(i) $\displaystyle x$ is rational
(ii) $\displaystyle \frac{x}{2}$ is rational
(iii) $\displaystyle 3x-1$ is rational
| If $x$ is rational, then $x=\frac{p}{q}$ for some integer $p$ and some nonzero integer $q$. Then $\frac{x}{2}=\frac{p}{2q}$ with $p$ an integer and $2q$ a nonzero integer.
If $\frac{x}{2}$ is rational, then $\frac{x}{2}=\frac{m}{n}$ for some integer $m$ and some nonzero integer $n$. Then $3x-1=\frac{6m-n}{n}$ with $6m-n$ an integer and $n$ a nonzero integer.
If $3x-1$ is rational, then $3x-1=\frac{a}{b}$ for some integer $a$ and some nonzero integer $b$. Then $x=\frac{a+b}{3b}$ with $a+b$ an integer and $3b$ a nonzero integer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factor $(a^2+2a)^2-2(a^2+2a)-3$ completely I have this question that asks to factor this expression completely:
$$(a^2+2a)^2-2(a^2+2a)-3$$
My working out:
$$a^4+4a^3+4a^2-2a^2-4a-3$$
$$=a^4+4a^3+2a^2-4a-3$$
$$=a^2(a^2+4a-2)-4a-3$$
I am stuck here. I don't how to proceed correctly.
| Often, a problem is handed to us in a slightly convenient form. Here, we may note the quadratic form:
$$(a^2+2a)^2−2(a^2+2a)−3$$
We can guess it will factor into four factors, so let's find the four roots, via setting the equation to zero and solving.
$$(a^2+2a)^2−2(a^2+2a)−3=0$$
Lets complete the square: $[(a^2+2a)-1]^2=4$, then $a^2+2a-1=\pm2$. We solve the two equations, $a^2+2a-3=0$ and $a^2+2a+1=0$. This will give real roots, so we can completely factor the above polynomial. I'll let you finish.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\displaystyle\lim_{x\rightarrow 0}\frac{5^x-4^x}{x}=\log_e\left({\frac{5}{4}}\right)$
*
*Show that $\displaystyle\lim_{x\rightarrow 0}\frac{5^x-4^x}{x}=\log_e\left({\frac{5}{4}}\right)$
*If $0<\theta < \frac{\pi}{2} $ and $\sin 2\theta=\cos 3\theta~~$ then find the value of $\sin\theta$
| $$\sin2\theta=\cos3\theta$$
$$\implies 2\sin\theta\cos\theta=4\cos^3\theta-3\cos\theta--->(1)$$
As $0<\theta<\frac\pi2,\cos\theta\ne0\implies 2\sin\theta=4\cos^2\theta-3$
$=4(1-\sin^2\theta)-3=1-4\sin^2\theta$
$$\text{So, }4\sin^2\theta+2\sin\theta-1=0--->(2)$$
$$\implies \sin\theta=\frac{-2\pm\sqrt{4-4(4)(-1)}}{2\cdot4}=\frac{-1\pm\sqrt5}4$$
As $0<\theta<\frac\pi2,\sin\theta>0 \implies \sin\theta=\frac{-1+\sqrt5}4 $
Determination of $\theta$:
$$\cos3\theta=\sin2\theta=\cos\left(\frac\pi2-2\theta\right)$$
$$\implies 3\theta=2n\pi\pm\left(\frac\pi2-2\theta\right)\text{ where } n \text{ is any integer}$$
Taking the '+' sign, $\theta=\frac{(4n+1)\pi}{10}$
$2\pi$ will divide $\frac{(4n_1+1)\pi}{10}-\frac{(4n_2+1)\pi}{10}=\frac{2(n_1-n_2)\pi}5$ if $5\mid (n_1-n_2)$
So, $5$ the in-congruent values of $n$ will give us $5$ the in-congruent values of $\theta$
Those values of $\theta$ are $\frac\pi{10},\frac{5\pi}{10}=\frac\pi2,\frac{9\pi}{10}=\pi-\frac{\pi}{10},\frac{13\pi}{10}=\pi+\frac{3\pi}{10},\frac{17\pi}{10}=2\pi-\frac{3\pi}{10}$
Taking the '-' sign, $\theta=2n\pi-\frac\pi2$ which clearly gives us exactly one in-congruent value of $\theta$, namely $-\frac\pi2$
Clearly, for $\cos\theta=0$ in $(1), \theta=\pm\frac\pi2$
Now, $\sin\frac{9\pi}{10}=\sin\left(\pi-\frac{\pi}{10}\right)=\sin\frac{\pi}{10}$
$\sin\frac{13\pi}{10}=\sin\left(\pi+\frac{3\pi}{10}\right)=-\sin\frac{3\pi}{10}$
and
$\sin\frac{17\pi}{10}=\sin\left(2\pi-\frac{3\pi}{10}\right)=-\sin\frac{3\pi}{10}$
So, $\sin\frac{\pi}{10},-\sin\frac{3\pi}{10}$ are the roots of $(2)$
As $0<\frac{\pi}{10}<\frac\pi2, \sin\frac{\pi}{10}>0\implies \sin\frac{\pi}{10}=\frac{-1+\sqrt5}4$
As $0<\frac{3\pi}{10}<\frac\pi2, -\sin\frac{3\pi}{10}<0\implies -\sin\frac{3\pi}{10}=-\frac{1+\sqrt5}4$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Separating $\frac{1}{1-x^2}$ into multiple terms I'm working through an example that contains the following steps:
$$\int\frac{1}{1-x^2}dx$$
$$=\frac{1}{2}\int\frac{1}{1+x} - \frac{1}{1-x}dx$$
$$\ldots$$
$$=\frac{1}{2}\ln{\frac{1+x}{1-x}}$$
I don't understand why the separation works. If I attempt to re-combine the terms, I get this:
$$\frac{1}{1+x} \frac{1}{1-x}$$
$$=\frac{1-x}{1-x}\frac{1}{1+x} - \frac{1+x}{1+x}\frac{1}{1-x}$$
$$=\frac{1-x - (1+x)}{1-x^2}$$
$$=\frac{-2x}{1-x^2} \ne \frac{2}{1-x^2}$$
Or just try an example, and plug in $x = 2$:
$$2\frac{1}{1-2^2} = \frac{-2}{3}$$
$$\frac{1}{1+2} -\frac{1}{1-2} = \frac{1}{3} + 1 = \frac{4}{3} \ne \frac{-2}{3}$$
Why can $\frac{1}{1-x^2}$ be split up in this integral, when the new terms do not equal the old term?
| you wrote wrong fraction :
$$\dfrac{1}{1-x^2}=\dfrac{1}{2}(\dfrac{1}{1-x}-\dfrac{1}{1+x})$$
instead of :$$\dfrac{1}{1-x^2}=\dfrac{1}{2}(\dfrac{1}{1+x}-\dfrac{1}{1-x})$$
Now check it by plug in $x=2$
LHS=$\dfrac {1}{-3}\,$ ; RHS=$\dfrac{1}{2}(\dfrac{1}{3}-{1})\implies \dfrac {1}{-3}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Question about limit at infinity I have questions which I need to solve:
1) $\lim\limits_{n\to\infty}\dfrac{\sqrt{n^2+1}}{\sqrt n}=\infty$
2) $\lim\limits_{n\to\infty}(\sin(n)-n)=-\infty$
Using this definition:
$X_n\rightarrow\infty \iff$ for all $\alpha>0$, these exists an $N$ in $\Bbb N$ such that if $n>N$, then $X_n > \alpha$
$X_n\rightarrow-\infty \iff$ for all $\beta<0$, these exists an $N$ in $\Bbb N$ such that if $n>N$, then $X_n < \beta$
Please solve it by using these definitions. What are the values for $N$?
Thank you.
| In (1),
$$
X_n = \frac{\sqrt{n^2 + 1}}{\sqrt n} = \sqrt{n + \frac 1n} > \sqrt n
$$
for $n > 0$. Also, $X_{n+1} > X_n$, as can be seen from
$$
\begin{align}
\frac{X_{n+1}}{X_n} &= \sqrt{\frac{n+1+\frac{1}{n+1}}{n+\frac 1n}} \\
&= \sqrt{\left(\frac{(n+1)^2 + 1}{n^2 + 1}\right)\frac{n}{n + 1}} \\
&= \sqrt{\frac{n^3 + 2n^2 + 2n}{n^3 + n^2 + n + 1}} \\
&= \sqrt{1 + \frac{n^2 + n - 1}{n^3 + n^2 + n + 1}} \\
&= \sqrt{1 + \frac{(n + \frac{1 + \sqrt{5}}{2})(n + \frac{1 - \sqrt{5}}{2})}{n^3 + n^2 + n + 1}} \\ &> 1
\end{align}
$$
for $n > \frac{\sqrt 5 - 1}2$ (which obviously holds for $n \ge 1$).
Therefore, for any given $\alpha > 0$, the choice $N = \lceil \alpha \rceil^2$ makes $X_n > X_N > \sqrt N = \lceil\alpha\rceil \ge \alpha$ for all $n > N$.
For (2),
$$
X_n = -n + \sin n \le -n + 1
$$
for all $n > 0$. For any given $\beta < 0$, pick $N = -\lfloor \beta \rfloor + 1$. Then, for any $n > N$, we have
$$
X_n \le -n + 1 < -N + 1 = \lfloor \beta \rfloor \le \beta.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to compute the area of the shadow?
If we can not use the integral, then how to compute the area of the shadow?
It seems easy, but actually not?
Thanks!
| Assume a square of side length $2$ for simplicity.
Denote the various sub-areas as $A, B, \ldots, H$ from top to bottom, left to right. So, the shaded area is $C$, the small disk is $C+D=\pi$, the big quarter disk is $\pi = A+E+G+H$.
Note that $A+B=F+H=D=G=1-\frac\pi 4$.
We locate the right vertex of $A$, measured from the center by solving $x^2+y^2=1$ and $(1+y)^2+(1+x)^2=4$, which implies
$$x+y=\frac{(1+y^2)+(1+x)^2-(x^2+y^2)-2}2=\frac12$$
and then $x^2+(\frac12-x)^2=1$, hence $x=\frac{1\pm\sqrt 7}4$ and by sign consideration,
$$ x=\frac{1-\sqrt 7}4, \quad y=\frac{1+\sqrt 7}4.$$
Now $B$ can be computet as the difference of a rectangle $1\times (1-y) =\frac{3-\sqrt 7}4$ and thow halves of circular segments.
But for these segments (using e.g. the formula $ r^2\arccos(1-\frac hr)-(r-h)\sqrt{(2r-h)h}$ for the area of a segment), we need to compute some angles!
I doubt that anything better than the $\arccos$ expression can be found
for these angles (i.e. they are not a rational multiple of $\pi$).
Having thus obtained an expression
$$B=a+b\sqrt 7+c\arccos\frac{1+\sqrt 7}4+d\arccos\frac{5+\sqrt 7}8$$
with $a,b,c,d$ rational, we finally find
$$C = A+G+H = 3G-2B = 3-\frac34\pi -2B. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve recursive equation $ f_n = \frac{2n-1}{n}f_{n-1}-\frac{n-1}{n}f_{n-2} + 1$ Solve recursive equation:
$$ f_n = \frac{2n-1}{n}f_{n-1}-\frac{n-1}{n}f_{n-2} + 1$$
$f_0 = 0, f_1 = 1$
What I have done so far:
$$ f_n = \frac{2n-1}{n}f_{n-1}-\frac{n-1}{n}f_{n-2} + 1- [n=0]$$
I multiplied it by $n$ and I have obtained:
$$ nf_n = (2n-1)f_{n-1}-(n-1)f_{n-2} + n- n[n=0]$$
$$ \sum nf_n x^n = \sum(2n-1)f_{n-1}x^n-\sum (n-1)f_{n-2}x^n + \sum n x^n $$
$$ \sum nf_n x^n = \sum(2n-1)f_{n-1}x^n-\sum (n-1)f_{n-2}x^n + \frac{1}{(1-z)^2} - \frac{1}{1-z} $$
But I do not know what to do with parts with $n$. I suppose that there can be useful derivation or integration, but I am not sure. Any HINTS?
| Did you try difference equations? The original expression can be rewritten as (denote $\Delta f_k=f_{k}-f_{k-1}$)
$$
\Delta f_{k}=\frac{k-1}{k}\Delta f_{k-1}+1 =\frac{k-1}{k} \cdot \frac{k-2}{k-1} \Delta f_{k-2} + \frac{k-1}{k}+1= \ldots =\frac{1}{k} \Delta f_1 + \frac{k-1}{k} \\
+ \frac{k-2}{k-1}+\ldots + 1 = \frac{1}{k} \Delta f_1 + k - H_k
$$
By summing both sides over $k$ and using boundary values you should get something like (I haven't checked the algebra!)
$$
f_n= H_n + \frac{n(n+1)}{2}-(n+1)(H_n-1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Integral solutions of hyperboloid $x^2+y^2-z^2=1$ Are there integral solutions to the equation $x^2+y^2-z^2=1$?
| Of course for the equation $X^2+Y^2=Z^2+t$
There is a particular solution:
$X=1\pm{b}$
$Y=\frac{(b^2-t\pm{2b})}{2}$
$Z=\frac{(b^2+2-t\pm{2b})}{2}$
But interessuet is another solution:
$X^2+Y^2=Z^2+1$
If you use the solution of Pell's equation: $p^2-2s^2=\pm1$
Making formula has the form:
$X=2s(p+s)L+p^2+2ps+2s^2=aL+c$
$Y=(p^2+2ps)L+p^2+2ps+2s^2=bL+c$
$Z=(p^2+2ps+2s^2)L+p^2+4ps+2s^2=cL+q$
number $L$ and any given us.
The most interesting thing here is that the numbers $a,b,c$ it Pythagorean triple. $a^2+b^2=c^2$
This formula is remarkable in that it allows using the equation $p^2-2s^2=\pm{k}$
Allows you to find Pythagorean triples with a given difference.
$a=2s(p+s)$
$b=p(p+2s)$
$c=p^2+2ps+2s^2$
$b-a=\pm{k}$
Pretty is not expected relationship between the solutions of Pell's equation and Pythagorean triples.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots+ \frac{1}{\sqrt{24} + \sqrt{25}}$ Simplify$$\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \frac{1}{\sqrt{24} + \sqrt{25}}.$$
I know you can solve this using generating functions but I'm not totally sure.
| Here, we simply use André Nicolas' hint to observe the "collapsing house of cards":
We first represent the $k$th term of the sum, to simplify matters.
$$\sum_{k = 1}^{24} \color{blue}{\bf \frac{1}{\sqrt k + \sqrt{k+1}}}\quad = \quad\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \frac{1}{\sqrt{24} + \sqrt{25}}$$
$$ {\bf NOTE:}\quad \frac{1}{\sqrt k + \sqrt{k+1}}\cdot \frac{\sqrt k - \sqrt{k+1}}{\sqrt k - \sqrt{k+1}} = \frac{\sqrt k - \sqrt{k+1}}{k - (k + 1)} = \color{blue}{\bf \sqrt {k+1} - \sqrt{k}}$$
$$ $$
$$
\begin{align}\sum_{k = 1}^{24} \frac{1}{\sqrt k + \sqrt{k+1}} \quad
& = \quad\sum_{k=1}^{24} \sqrt {k+1} - \sqrt{k} \\ \\
& = (\sqrt 2 - 1) + (\sqrt 3 - \sqrt 2) + (\sqrt 4 - \sqrt 3) \cdots + (\sqrt{25} -\sqrt {24}) \\ \\
& = \sqrt{25} - 1= 4 \\
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find minimum in a constrained two-variable inequation I would appreciate if somebody could help me with the following problem:
Q: find minimum
$$9a^2+9b^2+c^2$$
where $a^2+b^2\leq 9, c=\sqrt{9-a^2}\sqrt{9-b^2}-2ab$
| $ 9a^2+9b^2+c^2=9a^2+9b^2+(9-a^2)(9-b^2)-4ab\sqrt{9-a^2}\sqrt{9-b^2}+4a^2b^2$
=$81+5a^2 b^2-4ab \sqrt{9^2-9(a^2+b^2)+a^2b^2} $
$\ge 81+5a^2 b^2-4|ab|\sqrt{9^2-9(a^2+b^2)+a^2 b^2} $ $ (-ab \ge -|ab|)$
$\ge 81+5a^2 b^2-4|ab|\sqrt{9^2-9(2|ab|)+a^2 b^2}$ $(a^2+b^2 \ge 2|ab|)$
=$81+5a^2 b^2-4|ab|(9-|ab|)$=$9a^2b^2-36|ab|+81$
$=9(|ab|-2)^2+45 \geq 45$ , $ (9(|ab|-2)^2 \ge 0)$
first "=", $ab \ge 0$, 2nd "=", $|a|=|b|$, last "=", $|ab|=2$, so we got the min is 45 when $a=b=\pm \sqrt{2}$
with same method, we can get max also.
Edit: I add max in same way:
$81+5a^2 b^2-4ab \sqrt{9^2-9(a^2+b^2)+a^2b^2} $
$\leq 81+5a^2 b^2+4|ab| \sqrt{9^2-9(a^2+b^2)+a^2b^2}$ $ (-ab \leq |ab|)$
$ \leq 81+5(\dfrac{a^2+ b^2}{2})^2+4*\dfrac{a^2+b^2}{2} \sqrt{9^2-9(a^2+b^2)+( \dfrac{a^2+b^2}{2} )^2} $ $ ( |ab| \leq \dfrac{a^2+b^2}{2}, a^2 b^2 \leq (\dfrac{a^2+b^2}{2})^2)$
$=81+5x^2+4x(9-x)=81+x^2+36x $ ......... $ here: x=\dfrac{a^2+b^2}{2} \leq \dfrac{9}{2} $
$\leq 81+(\dfrac{9}{2})^2+36*\dfrac{9}{2}=\dfrac{567}{4}$
when $a=-b=\pm \dfrac{3\sqrt{2}}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Identity for $e$ in terms of the Fibonacci sequence. The following identity appears in Martin Gardner's paper, "Dr. Matrix on Little Known Fibonacci Curiosities:
$$e = \frac{1 + 1 + \frac{2}{2!} + \frac{3}{3!} + \frac{5}{4!} + \frac{8}{5!} + \frac{13}{6!} + \frac{21}{7!} + \frac{34}{8!} + \frac{55}{9!} + \cdots}{1 + 0 + \frac{1}{2!} + \frac{1}{3!} + \frac{2}{4!} + \frac{3}{5!} + \frac{5}{6!} + \frac{8}{7!} + \frac{13}{8!} + \frac{21}{9!} + \cdots} $$
How can we prove this?
| The correct form of this expression seems to be:
$$e=\frac{1+1+\frac{2}{2!}+\frac{3}{3!}+\frac{5}{4!}+\frac{8}{5!}+\frac{13}{6!}+\ldots}{1-0+\frac{1}{2!}-\frac{1}{3!}+\frac{2}{4!}-\frac{3}{5!}+\frac{5}{6!}-\ldots} = \frac{\sum\limits_{k=0}^\infty \frac{F_{k+1}}{k!}}{\sum\limits_{k=0}^\infty (-1)^k\frac{F_{k-1}}{k!}}$$
There are two useful observations that can be used to prove this statement:
*
*$\sum\limits_{k=0}^\infty \frac{x^k}{k!}=e^x$ (the definition of $e^x$)
*$F_n=\frac{\varphi^n-\psi^n}{{\varphi-\psi}}$, where $\varphi=\frac{1+\sqrt{5}}{2}$ and $\psi=\frac{1-\sqrt{5}}{2}$ (the Euler-Binet formula)
Now we are all set for the proof!
$$\begin{eqnarray}
\sum\limits_{k=0}^\infty \frac{F_{k+1}}{k!} & = & \frac{1}{\varphi-\psi}\sum\limits_{k=0}^\infty \frac{\varphi^{k+1}-\psi^{k+1}}{k!} \\
& = & \frac{1}{\varphi-\psi}\left(\varphi\sum\limits_{k=0}^\infty \frac{\varphi^k}{k!}-\psi\sum\limits_{k=0}^\infty \frac{\psi^k}{k!}\right) \\
& = & \frac{1}{\varphi-\psi}\left(\varphi e^\varphi - \psi e^\psi\right) \\
\end{eqnarray}$$
$$\begin{eqnarray}
\sum\limits_{k=0}^\infty (-1)^k\frac{F_{k-1}}{k!} & = & \frac{1}{\varphi-\psi}\sum\limits_{k=0}^\infty \frac{(-1)^k\varphi^{k-1}-(-1)^k\psi^{k-1}}{k!} \\
& = & \frac{1}{\varphi-\psi}\left(
\frac{1}{\varphi}\sum\limits_{k=0}^\infty \frac{(-\varphi)^k}{k!} -
\frac{1}{\psi}\sum\limits_{k=0}^\infty \frac{(-\psi)^k}{k!}\right) \\
& = & \frac{1}{\varphi-\psi}\left(\frac{1}{\varphi} e^{-\varphi} - \frac{1}{\psi} e^{-\psi}\right) \\
\end{eqnarray}$$
Dividing these two yields: $$\frac{\sum\limits_{k=0}^\infty \frac{F_{k+1}}{k!}}{\sum\limits_{k=0}^\infty (-1)^k\frac{F_{k-1}}{k!}} = \frac{\varphi e^\varphi - \psi e^\psi}{ \frac{1}{\varphi} e^{-\varphi} - \frac{1}{\psi} e^{-\psi} }$$
Since $\varphi + \psi = 1$ and $\varphi\psi = -1$, we this can be simplified as
$$\frac{\varphi e^\varphi - \psi e^\psi}{ \frac{1}{\varphi} e^{-\varphi} - \frac{1}{\psi} e^{-\psi} } =
\frac{\varphi e^\varphi - \psi e^\psi}{ e^{-1}\left(\varphi e^{\varphi} - \psi e^{\psi}\right)} = e$$
Q.E.D.
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits