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Riccati Differential Equation Help Show that $y(x)=\frac{1}{x}$ is a solution of the Ricatti differential equation:
$$y'+\frac{1-3xy}{x^2}+3y^2=0$$
Hence using a suitable transformation of the dependant variable, solve the differential equation when $y(1)=3$.
Any help appreciated.
| Multiply by $x^2$ and substitute $y=Q\frac{w'}{w}$
$$x^2Q'\frac{w'}{w}+x^2Q\frac{w''}{w}-x^2Q\frac{(w')^2}{w^2}+1-3xQ\frac{w'}{w}+3x^2Q^2\frac{(w')^2}{w^2}=0$$
We choose what $Q$ is so we choose it conveniently:
$$-x^2Q\frac{(w')^2}{w^2}+3x^2Q^2\frac{(w')^2}{w^2}=0$$
We get $Q=\frac 13$. Now the equation reads:
$$\frac13x^2\frac{w''}{w}+1-x\frac{w'}{w}=0$$
Multiply by $w$
$$x^2 w''(x)-3 x w'(x)+3 w(x)=0$$
This is a Cauchy Euler equation. Try the solution $x^r$ (I am assuming you can do this)
The solutions are $w=c_2 x^3+c_1 x$
Bring it all together:
$$y=\frac 13\frac{3c_2x^2+c_1}{c_2 x^3+c_1 x}$$
Divide the num and denum by $c_2$ and call the new constant $c$
$$y=\frac{3x^2+c}{ 3x^3+3c x}$$
And the final solution:
$$\frac{4 x^2-1}{x \left(4 x^2-3\right)}$$
Your answer is wrong. It is a solution to the equation, but it does not satisfy the initial condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How do I prove this claim? Claim :Let $p$ be a prime and $m \geq 2$ be an integer. Prove that the equation $ \frac{ x^p + y^p } 2 = \left( \frac{ x+y } 2 \right)^m $ has a positive integer solution $(x, y) \neq (1, 1)$ if and only if $m = p$.
Thank you for your help .
| $ p > 2$ $ \implies$ $ (x + y)|x^{p} + y^{p}$
$ q > 2$ ,$ q$ is prime,$ q^{a}\parallel{}x,q^{b}\parallel{}y$ ,$ a\leqq$
$ \implies$ $ ap = am$ $ \implies$ $ p = m$
$ \implies$ $ (x,y) = 1$ or $ (x,y) = 2^{\alpha}$
$ (x,y) = 2^{\alpha}$,$ x = 2^{\alpha}.x_{1},y = 2^{\beta}.y_{1}$ ,$ {\beta}\geq {\alpha}$
$ x + y = 2^{\alpha}[x_{1} + y_{1}.2^{{\beta} - {\alpha}}]$
$ r$ is prime, $ r|[x_{1} + y_{1}.2^{{\beta} - {\alpha}}]$ $ \implies$ $ (r,x) = (r,y) = 1$
because if $ (r,y) > 1$ $ \implies$ $ (r,x) > 1$
and $ r|[x_{1} + y_{1}.2^{{\beta} - {\alpha}}]$ if $ {\beta} > {\alpha}$ $ \implies$ $ r > 2$
and $ r|x_{1},r|y_{1}$ Contradiction!
$ \implies$ $ {\alpha} = {\beta}$
$ \implies$ $ x = 2^{\alpha}.x_{1},y = 2^{\alpha}.y_{1}$
$ r|x_{1} + y_{1}$, if $ r > 2$ $ \implies$ $ r\nmid x,r\nmid y$
$ (r,x) = (r,y) = 1$ $ \implies$ $ v_{r}(x^{p} + y^{p}) = v_{r}(x + y) + v_{r}(p) = m.v_{r}(x + y)$
if $ p\not = r$ $ \implies$ $ m = 1$ Contradiction
if $ p = r$ $ \implies$ $ m = 2$
$ \implies$ $ x^{p} + y^{p} = \frac {(x + y)^{2}}{2}$ $ \implies$ $ p\leq 2$ $ \implies$ $ p = 2$ for $ x = y$
$ \implies$ $ p = m = 2$
if $ x^{p} + y^{p} = 2^{c}$,and $ v_{2}(x) = v_{2}(y) = {\alpha}$
$ \implies$ $ x_{1}^{p} + y_{1}^{p} = 2^{s}$
$ x_{1} = 2^{k} - y_{1}$
$ (2^{k} - y_{1})^{p} + y_{1}^{p} = 2^{s}$
$ 2^{kp} - 2^{k(p - 1)}.y_{1} + ... + 2^{k}y_{1}^{p - 1} = 2^{s}$
$ \implies$ $ 2^{k}\parallel{}2^{s}$ $ \implies$ $ k = s$
$ \implies$ $ (2^{s} - y_{1})^{p} + y_{1}^{p} = 2^{s}$
$ 2^{sp} - 2^{s(p - 1)}.y_{1} + .. + 2^{s}y_{1}^{p - 1} = 2^{s}$
$ \implies$ $ 2^{sp}\geq2^{s(p - 1)}.y_{1}$
$ \implies$ $ 2^{s}\geq2^{s}.y_{1}^{p - 1}$
$ \implies$ $ y_{1} = 1$
if $ (x,y) = 1$ $ \implies$,$ r$ is prime $ r|x + y$ and $ (r,x) = (r,y) = 1$
the rest is directly tricky lemma .
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof of $\sum_{x = 1}^\infty \frac{1}{x}$'s divergence by absurdity? (From this site.)
The following argument purports to show that the series $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \dots = 0$. It begins with the harmonic series.
$$
\begin{aligned}
\sum \frac{1}{x} &= \sum \left( \frac{1}{2x - 1} + \frac{1}{2x} \right)\\
&= \sum \left( \frac{2(2x - 1)}{2x(2x - 1)} + \frac{1}{2x(2x - 1)} \right)\\
&= \sum \left(\frac{1}{x} + \frac{1}{2x(2x - 1)}\right)\\
&= \sum \left(\frac{1}{x}\right) + \sum\left(\frac{1}{2x(2x - 1)}\right)\\
\end{aligned}
$$
So $\sum \frac{1}{x}$ equals itself plus something, which implies that the second series is $0$. Of course, this argument is false because the harmonic series diverges.
My question: suppose we didn't know the harmonic series diverged, but we did know that $\sum\left(\frac{1}{2x(2x - 1)}\right)$ converged to a nonzero value -- it is in fact $\ln{(2)}$.
Would we then be able to conclude, from the above argument, that the harmonic series diverges? That is, if we arrive at $A = A + c$, where $c$ is constant, can we conclude that $A$ is infinite?
| $$A = A + c$$ so subtract A from both sides. $$c = 0$$ but $c$ is not $0$, it's $\ln 2$. Contradiction. $\therefore$ the harmonic series diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ To find the above minimal polynomial, let
$$x=\sqrt{2}+\sqrt{3}+\sqrt{5}$$
$$x^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}$$
Subtracting 10 and squaring gives
$$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$
$$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$
$$x^4-20x^2-24=40\sqrt{6}+24\sqrt{10}+16\sqrt{15}$$
$$x^4-20x^2-24=8(2\sqrt{6}+2\sqrt{10}+2\sqrt{15})+24\sqrt{6}+8\sqrt{10}$$
$$x^4-20x^2-24=8(x^2-10)+24\sqrt{6}+8\sqrt{10}$$
$$x^4-28x^2-104=24\sqrt{6}+8\sqrt{10}$$
Again, squaring both sides
$$x^8-56x^6+576x^4+5428x^2+10816=4096+765\sqrt{6}$$
But if I square again, I will get a degree 16 polynomial. Mathematica says the minimal polynomial is degree 8, which would make sense since elements of $\mathbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5}]$ look like
$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5}+e\sqrt{6}+f\sqrt{10}+g\sqrt{15}+h\sqrt{30}$$
Where am I making mistakes?
| In general, if $p_1,\dotsc,p_n$ is any list of integers, then the polynomial
$$f=\prod_{e_1,\dotsc,e_n \in \{\pm 1\}} (x+e_1 \sqrt{p_1}+\dotsc+e_n \sqrt{p_n})$$
has coefficients in $\mathbb{Z}$, which follows by induction from the observation
$p \in \mathbb{Z},\,g \in \mathbb{Z}[x] \Rightarrow g(\sqrt{p}) \cdot g(-\sqrt{p}) \in \mathbb{Z}[x]$.
You can prove this using the automorphism group of $\mathbb{Z}[\sqrt{p}]$. But you can also just verify it via some direct calculation, which has the advantage that you really see why the $\sqrt{p}$-terms vanish and that you may compute $f$ faster.
Clearly, $f$ is monic, has degree $2^n$, and has $\sqrt{p_1}+\dotsc+\sqrt{p_n}$ as a root. So this is quite elementary and for any fixed $n$, you can also compute $f$. What is not so easy to prove is that if $p_1,\dotsc,p_n$ are square-free and pairwise coprime integers, then $f$ is irreducible. Equivalently, the degree of $\sqrt{p_1}+\dotsc+\sqrt{p_n}$ equals $2^n$. You can find the proof here ("square-root-extension.pdf").
For the numbers $p_1,p_2,p_3 = 2,3,5$, we get $f = x^8 - 40 x^6 + 352 x^4 - 960 x^2 + 576$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Let $\mathbf{r}=(x,y,z)$,$r=||\mathbf{r}||$. Show the following equation on $B\cdot \nabla (A\cdot \nabla (\frac{1}{r}))$ Let $\mathbf{r}=(x,y,z)$ and let $r=||\mathbf{r}||$. If $A$ and $B$ are constant vectors show that:
$$B\cdot \left(\nabla \left (A\cdot \nabla \left(\frac{1}{r}\right)\right)\right)=\frac{3A\cdot \mathbf{r}B\cdot \mathbf{r}}{r^5}-\frac{A\cdot B}{r^3}$$
I've found so far that $A\cdot \nabla\left(\frac{1}{r}\right)=-\frac{A\cdot \mathbf{r}}{r^3}$.
However, I have not been able to show the equation above.
I would greatly appreciate any solutions, suggestions, or hints.
| Let's start from what you have found so far, $A\cdot \nabla(\frac{1}{r})=-\frac{A\cdot \mathbf{r}}{r^3}$, and denote vector constant vector $A$ as $(a_x,a_y,a_z)$. The expression can be re-written as
$$A\cdot \nabla\left(\frac{1}{r}\right)=-\frac{A\cdot \mathbf{r}}{r^3}=-\frac{a_xx+a_yy+a_zz}{(x^2+y^2+z^2)^{3/2}}$$
Evaluating the partial derivative with respect to $x$, $y$ and $z$ (using the product rule of differentiation for each partial derivative) results in
$$\begin{align}\frac{\partial}{\partial x}A\cdot \nabla\left(\frac{1}{r}\right)&=\frac{3x(a_xx+a_yy+a_zz)}{(x^2+y^2+z^2)^{5/2}}-\frac{a_x}{(x^2+y^2+z^2)^{3/2}}\\
\frac{\partial}{\partial y}A\cdot \nabla\left(\frac{1}{r}\right)&=\frac{3y(a_xx+a_yy+a_zz)}{(x^2+y^2+z^2)^{5/2}}-\frac{a_y}{(x^2+y^2+z^2)^{3/2}}\\
\frac{\partial}{\partial z}A\cdot \nabla\left(\frac{1}{r}\right)&=\frac{3z(a_xx+a_yy+a_zz)}{(x^2+y^2+z^2)^{5/2}}-\frac{a_z}{(x^2+y^2+z^2)^{3/2}}\end{align}$$
Combining all three partial derivatives leads to
$$\nabla (A\cdot \nabla\left(\frac{1}{r}\right))=\frac{3(A\cdot\mathbf{r})\mathbf{r}}{r^5}-\frac{A}{r^3}$$
The final stage is more straightforward, leading to the final result:-
$$B\cdot \nabla (A\cdot \nabla (\frac{1}{r}))=B\cdot \left(\frac{3(A\cdot\mathbf{r})\mathbf{r}}{r^5}-\frac{A}{r^3}\right)=\frac{3(A\cdot\mathbf{r})(B\cdot\mathbf{r})}{r^5}-\frac{A\cdot B}{r^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1314288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplifying the product $\prod\limits_{k=2}^n \left(1-\frac1{k^2}\right)$ Can we simplify the given product to a general law?
$$\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{n^2}\right)$$
| We can write, from the difference of squares,
\begin{gather*}
\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dots\left(1-\frac{1}{n^2}\right)\\
=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\dots\left(1-\frac{1}{n}\right)\cdot \left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\dots\left(1+\frac{1}{n}\right)\\
=\frac{1}{2}\frac{2}{3}\dots\frac{n-1}{n}\cdot\frac{3}{2}\frac{4}{3}\dots\frac{n+1}{n}=\frac{1}{n}\frac{n+1}{2}=\frac{n+1}{2n}.
\end{gather*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1314371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sqrt{2} + \sqrt[3]{2}$ is irrational I solved the problem by way of contradiction. Suppose $x = \sqrt{2} + \sqrt[3]{2}$ is rational. Then we have
$$2 = (x - \sqrt{2})^3 = x^3 - 3\sqrt{2}x^2 + 6x - 2\sqrt{2} = (x^3 + 6x) - \sqrt{2}(3x^2 - 2)$$
I've already shown that $\sqrt{2}$ is irrational, and it's easy to show a rational number plus an irrational number is irrational and that the product of an irrational and a rational is irrational. Given those facts, we have $(x^3 + 6x) \in \mathbb{Q}$ and $\sqrt{2}(3x^2 - 2) \not\in \mathbb{Q}$, so
$$2 = a\in\mathbb{Q} + b\not\in\mathbb{Q} \implies 2 \not\in\mathbb{Q}$$ a contradiction.
However, the book does it differently: it says
Show that $x$ satisfies an equation of the type $$x^6 + a_1x^5 +
\ldots + a_6 = 0$$ where $a_1,\ldots,a_6$ are integers; prove that $x$
is then either irrational or an integer.
*
*Is my way correct as well?
*I don't understand how the book does it. How does one obtain that equation, and once obtained, how does it prove $x$ is irrational?
| You've done half the job for their way: from
$$2 = (x - \sqrt{2})^3 = x^3 - 3\sqrt{2}x^2 + 6x - 2\sqrt{2} = (x^3 + 6x) - \sqrt{2}(3x^2 - 2)$$
you deduce:$$(x^3 + 6x-2)^2=2(3x^2+2)^2,$$
whence $$x^6-6x^4-4x^3+12x^2-24x-4=0.$$
By the Rational roots theorem, we know that a rational root has to be an integer, and a divisor of $4$, i. e. it can be only $\pm 1, \pm 2, \pm 4$. Just test them to check none is a root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Evaluate $\int\frac{d\theta}{1+x\sin^2(\theta)}$ We have to evaluate $$\int\frac{d\theta}{1+x\sin^2(\theta)}$$
Here is all my steps:
$$\tan\frac{\theta}{2}=\omega\Rightarrow \sin(\theta)=\frac{2\omega}{1+\omega^2}\Rightarrow 1+x\sin^2{(\theta)}=1+\frac{x 4\omega^2}{(\omega^2+1)^2}$$
Therefore: $$\int\frac{d\theta}{1+x\sin^2(\theta)}=2\int\frac{\omega^2+1\:\ d \omega}{(\omega^2+1)^2+x(2\omega)^2}=2\left(\int\frac{\omega^2\:\ d \omega}{(\omega^2+1)^2+x(2\omega^2)}+\int\frac{d\omega}{(\omega^2+1)^2+x(2\omega)^2}\right)$$
I don't know if I'm wrong somewhere but I don't have ideea how can I continue...
| $$
\tan\frac{\theta}{2}=\omega,\qquad \sin(\theta)=\frac{2\omega}{1+\omega^2},\qquad d\theta=\frac{2\,d\omega}{1+\omega^2}
$$
\begin{align}
& \int\frac{d\theta}{1+x\sin^2(\theta)} = \int \frac{\dfrac{2\,d\omega}{1+\omega^2}}{1+x\left(\dfrac{2\omega}{1+\omega^2}\right)^2} = \int \frac{2(1+\omega^2)\,d\omega}{(1+\omega^2)^2 + 4\omega^2 x} \\[10pt]
= {} & \int \frac{2(1+\omega^2)\,d\omega}{\omega^4 + (2+4x)\omega^2 + 1}
= \int \frac{2(1+\omega^2)\,d\omega}{\Big(\omega^4 + (2+4x)\omega^2 + (1+2x)^2 \Big) + 1 - (1+2x)^2} \\[10pt]
= {} & \int \frac{2(1+\omega^2)\,d\omega}{\Big(\omega^2 + 1+2x \Big)^2 + 1 - (1+2x)^2} = \int \frac{2(1+\omega^2)\,d\omega}{\Big(\omega^2 + 1+2x \Big)^2 - (4x+4x^2)} \\[10pt]
= {} & \int \frac{2(1+\omega^2)\,d\omega}{\Big(\omega^2 + 1+2x - 2\sqrt{x+x^2}\Big)\Big( \omega^2 + 1+2x + 2\sqrt{x+x^2} \Big)} \\[10pt]
= {} & \int \left( \frac A {\omega^2 + 1+2x - 2\sqrt{x+x^2}} + \frac B{\omega^2 + 1+2x + 2\sqrt{x+x^2}} \right) \,d\omega.
\end{align}
We do not need $A\omega+B$ and $C\omega+D$ in the numerators because in every place where $\omega$ appears it's $\omega^2$.
If $x\ge 0$ then $1+2x\pm\sqrt{x+x^2}$ is positive regardless of whether it's plus or minus. Therefore we should get some arctangents.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can this recurrence relation be solved with generating functions? I have this recurrence relation,
$$a_{n+1}=\frac{n+2}{n}a_n$$
with $a_1=1$.
I've already solved this using a substitution approach by letting $a_n=\dfrac{(n+1)!}{(n-1)!}b_n$. This means $a_{n+1}=\dfrac{(n+2)!}{n!}b_{n+1}$, and so
$$\begin{align*}
\frac{(n+2)!}{n!}b_{n+1}&=\frac{n+2}{n}\frac{(n+1)!}{(n-1)!}b_n\\
b_{n+1}&=b_n\\
\sum_{n=1}^{k-1}(b_{n+1}-b_n)&=0\\
b_k&=b_1\\
b_k&=\frac{(1-1)!}{(1+1)!}a_1\\
b_k&=\frac{1}{2}
\end{align*}$$
Finally, $a_n=\dfrac{n(n+1)}{2}$.
I'm wondering if there's a way to use generating functions to solve this relation? I've been having trouble working with the $\dfrac{2}{n}a_n$ term. Are there only certain cases of relations for which generating functions are useful?
| Induction is easier, but to use generating functions let's assume that
$$
f(x)=\sum_{n=1}^\infty a_nx^n
$$
Then
$$
\begin{align}
\sum_{n=1}^\infty na_{n+1}x^n
&=x\left(\frac{f(x)}x\right)'\\
&=f'(x)-\frac{f(x)}x
\end{align}
$$
and
$$
\begin{align}
\sum_{n=1}^\infty(n+2)a_nx^n
&=\frac1x\left(f(x)x^2\right)'\\
&=2f(x)+xf'(x)
\end{align}
$$
Equating these functions gives
$$
(1-x)f'(x)=\left(2+\frac1x\right)f(x)
$$
Therefore, dividing both sides by $(1-x)f(x)$ and integrating, we get
$$
\begin{align}
\log(f(x))
&=\int\frac{2x+1}{x(1-x)}\,\mathrm{d}x\\
&=\int\left(\frac1x+\frac3{1-x}\right)\,\mathrm{d}x\\
&=\log\left(\frac x{(1-x)^3}\right)+C
\end{align}
$$
So that if $a_1=1$,
$$
\begin{align}
f(x)
&=\frac{x}{(1-x)^3}\\
&=\sum_{n=1}^\infty(-1)^{n-1}\binom{-3}{n-1}x^n\\
&=\sum_{n=1}^\infty\binom{n+1}{n-1}x^n\\
&=\sum_{n=1}^\infty\binom{n+1}{2}x^n\\
\end{align}
$$
Therefore,
$$
a_n=\binom{n+1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Closed form of power series First day on the site and this is an amazing place!
this is my question:
knowing that: $$1+\frac{x^2}{2}+\frac{x^4}{4}+\frac{x^6}{6}+\frac{x^8}{8}+\cdots $$
is a power series, how can I obtain its closed form?
Thank you.
| David's answer is certainly the way to go. I will just expand a little about the result (for $|x|<1$)
$$
S =x + x^3 + x^5 + \cdots = \frac{x}{1-x^2}
$$
Notice that in such case both series:
$$
S_1 =1 + x + x^2 +\cdots = \frac{1}{1-x}
$$
$$
S_2 = 1 + x^2 + (x^2)^2 + (x^2)^3 + \cdots = \frac{1}{1-x^2}
$$
converge absolutely, so we may write $S = S_1-S_2$ and substract termwise, one gets:
$$
S = \frac{1}{1-x} -\frac{1}{1-x^2} =\frac{x}{1-x^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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A unusual inequality about function $\ln$ These day,I met a unusual inequality when I solve a difficult problem, and proving the inequality means I have done the work!
Could you show me how to prove it or deny it? By the way, I believe that it's true!
Prove that, for all $t > 0$,
\begin{align*}
&4\ln t\ln (t + 2) - \ln t\ln (t + 1) - 3\ln t\ln (t + 3)\\
+ &4\ln (t + 1)\ln (t + 3) - 3\ln (t + 1)\ln (t + 2) - \ln (t + 2)\ln \left( {t + 3} \right)>0.
\end{align*}
Let
$$f\left( t \right) = 4\ln t\ln \left( {t + 2} \right) - \ln t\ln \left( {t + 1} \right) - 3\ln t\ln \left( {t + 3} \right) + 4\ln \left( {t + 1} \right)\ln \left( {t + 3} \right) - 3\ln \left( {t + 1} \right)\ln \left( {t + 2} \right) - \ln \left( {t + 2} \right)\ln \left( {t + 3} \right),$$
We have
$$f'\left( t \right) = \frac{{2\left[ {{t^2}\ln t - 3{{\left( {t + 1} \right)}^2}\ln \left( {t + 1} \right) + 3{{\left( {t + 2} \right)}^2}\ln \left( {t + 2} \right) - {{\left( {t + 3} \right)}^2}\ln \left( {t + 3} \right)} \right]}}{{t\left( {t + 1} \right)\left( {t + 2} \right)\left( {t + 3} \right)}}.$$
Let
$$g\left( t \right) = {t^2}\ln t - 3{\left( {t + 1} \right)^2}\ln \left( {t + 1} \right) + 3{\left( {t + 2} \right)^2}\ln \left( {t + 2} \right) - {\left( {t + 3} \right)^2}\ln \left( {t + 3} \right),$$
we got
$$g'\left( t \right) = 2\left[ {t\ln t - 3\left( {t + 1} \right)\ln \left( {t + 1} \right) + 3\left( {t + 2} \right)\ln \left( {t + 2} \right) - \left( {t + 3} \right)\ln \left( {t + 3} \right)} \right].$$
And let
$$h\left( x \right) = t\ln t - 3\left( {t + 1} \right)\ln \left( {t + 1} \right) + 3\left( {t + 2} \right)\ln \left( {t + 2} \right) - \left( {t + 3} \right)\ln \left( {t + 3} \right),$$
we have
\begin{align*}
h'\left( x \right) &= \ln t - 3\ln \left( {t + 1} \right) + 3\ln \left( {t + 2} \right) - \ln \left( {t + 3} \right)\\
&= \ln \frac{{t{{\left( {t + 2} \right)}^3}}}{{{{\left( {t + 1} \right)}^3}\left( {t + 3} \right)}} = \ln \left[ {1 - \frac{{2t + 3}}{{{{\left( {t + 1} \right)}^3}\left( {t + 3} \right)}}} \right] < 0.
\end{align*}
However, it seems that there are no use!
| The desired inequality is written as
$$3ac - ab - b^2 - bc \ge 0$$
where
$a = \ln (t + 1) - \ln t, ~
b = \ln(t + 2) - \ln(t + 1)$ and
$c = \ln(t + 3) - \ln(t + 2)$.
Clearly, $a > b > c > 0$.
Fact 1: It holds that, for all $u \ge 0$,
$$\frac{2u}{2 + u} \le \ln(1 + u) \le \frac{u}{2}\cdot \frac{2 + u}{1 + u}.$$
(Note: See bounds for logarithm. It is also easily proved by taking derivative.)
Using Fact 1, we have
$$c \ge \frac{2}{2t + 5}, \quad a \ge \frac{2}{2t + 1}, \quad b \le \frac{2t + 3}{2(t + 1)(t + 2)}. \tag{1}$$
(Note: For example, we have $a = \ln(1 + \frac{1}{t})$. Let $u = \frac{1}{t}$.)
Using (1) and $a > b > 0$, we have
\begin{align*}
& 3ac - ab - b^2 - bc\\
=\,& (3a - b) c - ab - b^2\\
\ge\,& (3a - b)\cdot \frac{2}{2t + 5} - ab - b^2 \\
=\,& \left(\frac{6}{2t + 5} - b\right)a - \frac{2b}{2t+5} - b^2 \\
\ge\,& \left(\frac{6}{2t + 5} - \frac{2t + 3}{2(t + 1)(t + 2)}\right)a - \frac{2b}{2t+5} - b^2\\
=\,& \frac{8t^2 + 20t + 9}{2(2t+5)(t+1)(t+2)} a - \frac{2b}{2t+5} - b^2\\
\ge\,& \frac{8t^2 + 20t + 9}{2(2t+5)(t+1)(t+2)} \cdot \frac{2}{2t+1} - \frac{2}{2t+5}\cdot \frac{2t+3}{2(t+1)(t+2)} - \left(\frac{2t+3}{2(t+1)(t+2)}\right)^2\\
=\,& \frac{3}{4(2t+5)(t+1)^2(t+2)^2(2t+1)}\\
>\,& 0.
\end{align*}
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
How to integrate$ I=\int\ln\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy $ I am stuck with the integration
$$
I=\int\ln\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy
$$
I got this from the question from the book
"Field and wave electromagnetics, Cheng, 2nd, Problem 3-18.
I tried to solve this equation using method of integration by parts, but my equation got worse.
I know the answer by Wolfram Alpha, but I can't get how.
| Considering $$I=\int\log\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy$$ what I would first do is to get rid of the logarithm by a first integration by parts $$u=\log\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)$$ $$du=\frac{y\, dy}{\sqrt{\frac{L^2}{4}+y^2+z^2}
\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)}$$ $v'=dy$, $v=y$ which makes $$I=y \log\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)- J$$ where $$J=\int\frac{y^2\, dy}{\sqrt{\frac{L^2}{4}+y^2+z^2}
\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)}$$ Now, an apparent change of variable could be $$\frac{L^2}{4}+y^2+z^2=t^2$$ $$y=\frac{1}{2} \sqrt{4 t^2-(L^2+4 z^2)}$$ $$dy=\frac{2 t}{\sqrt{4 t^2-(L^2+4 z^2)}}$$ which makes $$J=\int \frac{\sqrt{4 t^2-(L^2+4 z^2)}}{2t+L}\,dt$$ Again, $2t+L=w$, $t=\frac{w-L}{2}$, $dt=\frac{dw}{2}$ make $$J=\frac 12 \int\frac{\sqrt{w^2-2 L w-4 z^2}}{ w} dw$$ At this point, we find integrals which are given in the Table of Integrals, Series, and Products by I.S. Gradshteyn and I.M. Ryzhik (in the $7^{th}$ edition, look at section $2.267$).
It is effectively quite tedious !
| {
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"source": "stackexchange",
"question_score": "6",
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How can I apply Newton's sums to solve this problem? Given $x_1,x_2,x_3,x_4$ real numbers such that $x_1+x_2+x_3+x_4 = 0$ and $x_1^7+x_2^7+x_3^7+x_4^7 = 0,$ how can I use symmetric functions and Newton's sums to prove that $x_1(x_1+x_2)(x_1+x_3)(x_1+x_4)=0$?
This is what I have so far:
Note that $x_1 = -(x_2+x_3+x_4)$ and the relation between Newton's sums, elementary symmetric polynomials, and power sums gives us that $P_7-S_1 P_6+S_2 P_5-S_3 P_4+S_4P_3=0,$ where $P_k$ denotes the $kth$ power sum and $S_k$ denotes the $kth$ elementary symmetric polynomial of $x_1,x_2,x_3,x_4.$
We want to show that $(x_2+x_3+x_4)(x_2+x_3)(x_2+x_4)(x_3+x_4) = 0.$
But how?
| Here is another useful application of the AM-GM inequality(didn't think of it, did you)
Tricky factorization:
$$0=-(a^7+b^7+c^7+d^7)$$$$=7(b+c)(c+d)(d+a)E$$ where $E=(b^2+c^2+d^2+bc+cd+da)^2+bcd(b+c+d)$. However, $$4E=((b+c)^2+(c+d)^2+(b+d)^2)^2)-4abcd$$$$=((b+a)^2+(c+a)^2+(a+d)^2)^2)-4abcd$$$$=(3a^2+2a(b+c+d)+b^2+c^2+d^2)^2-4abcd$$ The last expression reduces to $$(a^2+b^2+c^2+d^2)^2-4abcd$$ However we see that $$(a^2+b^2+c^2+d^2)^2-4abcd \geq 12|abcd| \geq 0$$ from the AM-GM inequality. We conclude that $4E \geq 0$ with equality if and only if $|a|=|b|=|c|=|d|=0$. Otherwise one of $b+c,b+d,c+d$ must be $0$. Hence, the result follows.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Differential equation $ y' = \frac {1 +py}{1 + p^2} $ I have a problem with the equation which is a part of the solution:
$$ y' = \frac {1 +py}{1 + p^2} $$
I tried to use $y=uv, y'=u'v + uv'$ substitutions (Bernouilli's method):
$$ u'v + uv' - \frac {p}{1 + p^2} uv = \frac {1} {1+p^2}$$
By solving $v' - \frac {p}{1 +p^2}v = 0 $ I have found $v = \sqrt {1 + p^2}$. Than I have:
$$u'\sqrt{1 +p^2} = \frac {1}{1+p^2} \implies u = \int \frac{dp}{\sqrt{(1+p^2)^3}} $$
Here I'm stuck but I know the solution should be: $$y = \sqrt{1+p^2}\left(\frac {p}{\sqrt{1+p^2}} + C \right) $$
| This is a linear differential equation, so you can use integrating factors (should be essentially equivalent to what you've done): you have
$$ y' - \frac{p}{1+p^2}y = \frac{1}{1+p^2}. $$
The integrating factor is
$$ \exp{\left( \int \frac{-p}{1+p^2} \, dp \right)} = \exp{\left( -\frac{1}{2}\log{(1+p^2)} \right)} = \frac{1}{\sqrt{1+p^2}}. $$
Therefore
$$ \left( \frac{y}{\sqrt{1+p^2}} \right)' = \frac{1}{(1+p^2)^{3/2}}. $$
We need to integrate the right-hand side: we use the substitution $p=\tan{x}$, so $dp = \sec^2{x} \, dx $
$$ \int \frac{dp}{(1+p^2)^{3/2}} = \int \frac{\sec^2{x} dx}{\sec^3{x}} = \int \cos{x} \, dx = \sin{x}+C = \frac{p}{\sqrt{1+p^2}}+C, $$
where in the last bit I have used
$$ \sin{x} = \frac{\tan{x}}{\sec{x}} = \frac{\tan{x}}{\sqrt{1+\tan^2{x}}} $$
Hence it comes out as
$$ \frac{y}{\sqrt{1+p^2}} = \frac{p}{\sqrt{1+p^2}}+C, $$
which is the form you want.
So really the answer was that you had to push on and try to do the integral, even if it didn't seem doable to begin with.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a rule for $\sqrt{a+b}$? You learn in algebra that
$$\sqrt{ab}=\sqrt{a} \sqrt{b}$$
and that
$$\sqrt{\frac ab}=\frac {\sqrt a}{\sqrt b}$$
You also learn to never make the fatal mistake of thinking
$$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$$
However, I am wondering if there is a rule for $\sqrt{a+b}$. I would think it would be pretty complex, if it exists at all.
$$\sqrt{a+b}=\text{?}$$
| There is no finite rule but there is the binomial series:
$$
(1 + x)^\alpha = \sum_{k=0}^{\infty} \; {\alpha \choose k} \; x^k
$$
where
$$
{\alpha \choose k} = \frac{\alpha (\alpha-1) (\alpha-2) \cdots (\alpha-k+1)}{k!}
$$
This series converges for $|x|<1$.
Taking $\alpha=1/2$, we get
$$
(1+x)^{1/2} =
1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5 x^4}{128}+\frac{7 x^5}{256}+\cdots
$$
We can apply this to $\sqrt{a+b}$ as follows. Assume wlog that $a>b$. Then
$$
\sqrt{a+b}=(a+b)^{1/2} = \sqrt{a}(1+x)^{1/2}
$$
with $x=b/a<1$.
We then have
$$
\sqrt{a+b} = \sqrt{a}\left(1+\frac{b}{2a}-\frac{b^2}{8a^2}+\frac{b^3}{16a^3}-\frac{5 b^4}{128a^4}+\frac{7 b^5}{256a^5}+\cdots\right)
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How would you prove this converges? $\sum_{1}^{\infty} \frac{(-1)^n}{[n-(-1)^n]^{\frac{2}{3}}}$ How would you check if this converges or not?
$$\sum_{1}^{\infty} \frac{(-1)^n}{[n-(-1)^n]^{\frac{2}{3}}}$$
It looks like a telescopic sequence so I thought I'd first write the beginning values:
$$S_n=\frac{-1}{2^{\frac{2}{3}}} + \frac{1}{1^{\frac{2}{3}}} + \frac{-1}{4^{\frac{2}{3}}} ... $$
But I got no conclusions from it...
What am I missing here?
Thanks.
| First, note that your series is not absolutely convergent: placing a modulus inside the sum you get $\sum \frac 1 {\big( n - (-1)^n \big)^{\frac 2 3}}$ which, keeping only the dominant term and applying the limit comparison test, behaves like $\sum \frac 1 {n^{\frac 2 3}}$ which is clearly divergent.
Now, is your series convergent? One cannot apply Leibniz's test because the sequence $\frac 1 {\big( n - (-1)^n \big)^{\frac 2 3}}$ does not decrease, even though it tends to $0$.
Let us examine once more the definition of the concept of convergent series: $x_1 + x_2 + \cdots + x_n + \cdots$ converges if and only if the sequences $(x_1 + x_2) + (x_3 + x_4) + \cdots + (x_{2N-1} + x_{2N})$ and $(x_1 + x_2) + (x_3 + x_4) + \cdots + (x_{2N-1} + x_{2N}) + x_{2N+1}$ converge, both, to the same limit, when $N \to \infty$.
In our case, $x_{2n+1} \to 0$, so we just have to show that $(x_1 + x_2) + (x_3 + x_4) + \cdots + (x_{2N-1} + x_{2N}) = \sum \limits _{n=1} ^N (x_{2n-1} + x_{2n})$ converges. In order to study this convergence we shall transform the summand as follows:
$$x_{2n-1} + x_{2n} = \\ -\frac 1 { {\sqrt[3] {2n-1 +1}} ^2} + \frac 1 { {\sqrt[3] {2n-1}} ^2} = \\ \Big( \frac 1 {\sqrt[3] {2n-1}} + \frac 1 {\sqrt[3] {2n}} \Big) \Big( \frac 1 {\sqrt[3] {2n-1}} - \frac 1 {\sqrt[3] {2n}} \Big) \sim \\ \frac 1 {\sqrt[3] n} \frac {\sqrt[3] {2n} - \sqrt[3] {2n-1}} {\sqrt[3] {2n} \sqrt[3] {2n-1}} \sim \\ \frac 1 n \frac {\sqrt[3] {2n} - \sqrt[3] {2n-1}} 1 = \\ \frac 1 n \frac {2n - (2n-1)} { {\sqrt[3] {2n}}^2 + {\sqrt[3] {2n}} {\sqrt[3] {2n-1}} + {\sqrt[3] {2n-1}}^2 } \sim \\ \frac 1 {n^{\frac 5 3}} ,$$
where the notation $a_n \sim b_n$ means $a_n, b_n >0$ and $\lim \limits _{n \to \infty} \frac {a_n} {b_n} \in (0, \infty)$.
Since $\sum \frac 1 {n^{\frac 5 3}}$ converges, the convergence of your series follows now easily.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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The number of ways to write $10$ as the sum of five natural numbers not equal to $3$ How many answers are there for the equation
$$x_1+x_2+x_3+x_4+x_5=10$$
given that $x_1,x_2\dots x_5\in\Bbb{Z^{0+}}\setminus\{3\}$.
| The generating function for the number of ways to sum to $k$ with these numbers is
$$
\begin{align}
&\left(\frac1{1-x}-x^3\right)^5\\
&=\frac1{(1-x)^5}-\frac{5x^3}{(1-x)^4}+\frac{10x^6}{(1-x)^3}-\frac{10x^9}{(1-x)^2}+\frac{5x^{12}}{1-x}-x^{15}\\
&=\sum_{k=0}^\infty\left[\binom{k+4}{k}x^k-5\binom{k+3}{k}x^{k+3}+10\binom{k+2}{k}x^{k+6}\right]\\
&+\sum_{k=0}^\infty\left[-10\binom{k+1}{k}x^{k+9}+5\binom{k}{k}x^{k+12}-\binom{k-1}{k}x^{k+15}\right]\\[3pt]
&\small=\sum_{k=0}^\infty\left[\binom{k+4}{k}-5\binom{k}{k-3}+10\binom{k-4}{k-6}-10\binom{k-8}{k-9}+5\binom{k-12}{k-12}-\binom{k-16}{k-15}\right]x^k
\end{align}
$$
Thus, the number of ways to solve $x_1+x_2+x_3+x_4+x_5=k$ where each $x_j$ is a non-negative integer not equal to $3$ is
$$
\small\binom{k+4}{k}-5\binom{k}{k-3}+10\binom{k-4}{k-6}-10\binom{k-8}{k-9}+5\binom{k-12}{k-12}-\binom{k-16}{k-15}
$$
Setting $k=10$ gives
$$
\binom{14}{10}-5\binom{10}{7}+10\binom{6}{4}-10\binom{2}{1}=531
$$
Note that the second to last term is $0$ for $k\lt12$ and the last term is $0$ for $k\ne15$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is this Function differentiable and continuous at x=0? Is $f(x)$ continuous and differentiable at $x = 0$ ?
$$f(x) = x(\sqrt{x} - \sqrt{x+1})$$
| First $f$ is defined only for $x \ge 0$.
Then, you have $$f(x) = x(\sqrt{x} - \sqrt{x+1})=x(\sqrt{x} - \sqrt{x+1})\frac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} + \sqrt{x+1}} = \frac{-x}{\sqrt{x} + \sqrt{x+1}}$$
which is continuous at $x=0$ with $f(0)=\lim\limits_{x \to 0^+} f(x) = 0$. Moreover $$\frac{f(x)}{x}=\frac{-1}{\sqrt{x} + \sqrt{x+1}}$$ has a right hand side limit as $x$ goes to $0$ equal to $-1$. So $f^\prime(0)=-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Greatest integer function in infinite sum of geometric series I am not sure how to approach this infinite sum. I can see that it's a geometric series and that has a straightforward solution, but I am not sure how to address the alternating signs with the greatest integer function. Any ideas?
$\sum_{n=0} ^\infty (-1)^{\lfloor n/m \rfloor}r^n$, such that $0<r<1$ and $m \in N$.
We have $0<r<1$, so a general geometric series $\sum_{n=0} ^\infty ar^n$ will converge to $\frac{a}{1-r}$, but our $a$ here is $(-1)^{\lfloor n/m \rfloor}$.
| Let us do a concrete calculation, say with $m=4$. Our sum is equal to
$$1+r+r^2+r^3-r^4-r^5-r^6-r^7+r^8+r^9+r^{10}+r^{11}-r^{12}-r^{13}-r^{14}-r^{15}+\cdots,$$
which can be rearranged as
$$(1+r+r^2+r^3)-r^4(1+r+r^2+r^3)+r^8(1+r+r^2+r^3)-r^{12}(1+r+r^2+r^3)+\cdots.$$
This is an infinite geometric series, first term "$a$" equal to $1+r+r^2+r^3$ and common ratio $-r^4$. By the standard formula, the sum is equal to
$$\frac{1+r+r^2+r^3}{1+r^4}.$$
Exactly the same reasoning shows that in the general case, the sum is equal to
$$\frac{1+r+r^2+\cdots+r^{m-1}}{1+r^m}.\tag{1}$$
We can simplify a bit to get a closed form. Note that
$$1+r+r^2+\cdots+r^{m-1}=\frac{1-r^m}{1-r}=\frac{2-(1+r^m)}{1-r}.$$
Divide by $1+r^m$. We find that (1) can be rewritten as
$$\frac{2}{(1-r)(1+r^m)}-\frac{1}{1-r}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all $x$ such that $2^x,2^{x^2}$ and $2^{x^3}$ form $3$ terms of an A.P. I know that if $a,b,c$ are in Arithmetic Progression, then $2b=a+c$, but in this case, I am unable to solve for $x$. Hints are appreciated. Thanks.
| For $x>0\land x\ne1$,
$$0<x(x-1)^2\implies x^2<\frac{x+x^3}2.$$
Then by monotonicity and convexity of the exponential
$$2^{x^2}<2^{\frac{x+x^3}2}<\frac12\left(2^x+2^{x^3}\right).$$
For $x<0$,
$$\dfrac12\left(2^x+2^{x^3}\right)<1<2^{x^2}.$$
Two cases remain, $x=0$ and $x=1$, that happen to be solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$ Let $\varphi=\frac{1+\sqrt5}2$ (the golden ratio).
How can I simplify the following expression?
$$7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$$
| We will use a well-known$^{[1]}$ formula for sum of arctangents:
$$\arctan u + \arctan v = \arctan \left( \frac{u+v}{1-uv} \right) \pmod \pi\tag1$$
The exact equality holds for $uv<1$, for other values there is additional term — an integer multiple of $\pi$ (it is easy to determine in each case).
Using this formula, we can establish the following identities:
$$\begin{align}
\arctan\varphi\phantom{^1}&=\frac\pi2-\frac12\,\arctan2,\tag2\\
\arctan\varphi^3&=\frac\pi4+\frac12\,\arctan2,\tag3\\
\arctan\varphi^5&=\pi-\frac32\,\arctan2.\tag4
\end{align}$$
Plugging these into the original expression in the question and expanding parentheses, we can see that all $\arctan2$ terms cancel, and we get the result
$$7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5=\frac{7\pi^2}8.\tag5$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$r! \equiv (−1)^k \pmod p$ Suppose that p ≡ 3 (mod 4)
and $r = \frac {p-1}2$
Show that $r! \equiv (−1)^k \pmod p$
where k is the number of non-quadratic residues modulo p which are smaller than $\frac p2$
I know from Wilson's Theorem for a prime p:
$(p−1)! \equiv −1 \pmod p$
I don't know where to start from ..
I managed to show that
$r! \equiv \pm 1 \pmod p$
any help or hint will be appreciated
| First, since $p \equiv 3$ (mod $4$), $-1$ is not a QR. $-1 \equiv g^{\frac{p-1}{2}}$ for any primitive root $g$ and $\frac{p-1}{2}$ is odd. The QRs have even exponent.
So now take $1 \leq a \leq \frac{p-1}{2}$. Then exactly one of $a$ and $-a$ is a QR. Suppose both were QRs. Then $a \equiv x^2$ (mod $p$) and $-a \equiv y^2$ (mod $p$). What can we conclude about $-1$? Then, remember that there are $\frac{p-1}{2}$ non-zero QRs. So that ensures one of $a$ and $-a$ is a QR.
Now $r! \equiv 1 \cdot 2 \cdot 3 \cdot \ldots \cdot \frac{p-3}{2} \cdot \frac{p-1}{2}$ (mod $p$). For every non-quadratic residue $a$ in the product, replace $a$ with $-(-a)$ so that $-a$ is a QR. This gives:
$r! \equiv (-1)^k \cdot 1^2 \cdot 2^2 \cdot 3^2 \cdot \ldots \cdot (\frac{p-3}{2})^2 \cdot (\frac{p-1}{2})^2$ (mod $p$) because we have $\frac{p-1}{2}$ distinct quadratic residues in our product, i.e. all the quadratic residues. So $r! \equiv (-1)^k (r!)^2$ (mod $p$) and $r!$ is invertible.
| {
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"timestamp": "2023-03-29T00:00:00",
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$ \cos ^2\left(x\right)+\cos ^2\left(2x\right)+\cos ^2\left(3x\right)=\frac{3}{2} $ $$ \cos ^2\left(x\right)+\cos ^2\left(2x\right)+\cos ^2\left(3x\right)=\frac{3}{2} $$
How can I solve this one, I mean I get something like this:
$-3+\left(-1+2\cos ^2\left(x\right)\right)^22+2\left(-3\cos \left(x\right)+4\cos ^3\left(x\right)\right)^2+2\cos ^2\left(x\right)=0$
This equation seems rather hard to solve from here, any tips or other ways to come to an solution?
| Linearise the left-hand side: we start from formula:
$$\cos^2u=\frac12(1+\cos 2u)$$
from which we obtain the equation:
$$\frac12(3+\cos 2x+\cos 4x+\cos 6x)=\frac32\iff\cos 2x+\cos 4x+\cos 6x=0.$$
Now factorise the left-hand side:
\begin{align*}
\cos 2x+\cos 4x+\cos 6x&=\cos(4x- 2x)+\cos 4x+\cos(4x+2x)\\
&=2\cos4x\cos2x+\cos 4x=\cos 4x(2\cos2x+1)
\end{align*}
Thus the equation is equivalent to the system of standard equations:
$$\begin{cases}
\cos 4x=0\\\cos2x=-\dfrac12\end{cases}$$
Solutions:
$$\begin{cases}
4x\equiv \dfrac\pi2&\bmod\pi\\
2x\equiv \pm\dfrac{2\pi}3&\bmod2\pi
\end{cases}\iff \begin{cases}
x\equiv \dfrac\pi8&\bmod\dfrac\pi4\\
x\equiv \pm\dfrac\pi3&\bmod\pi
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341096",
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"source": "stackexchange",
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I need help with a Finite Series Problem:
Find the sum to $n$ terms of
\begin{eqnarray*}
\frac{1}{1\cdot 2\cdot 3} + \frac{3}{2\cdot 3\cdot 4} + \frac{5}{3\cdot 4\cdot 5} +
\frac{7}{4\cdot 5\cdot 6}+\cdots \\
\end{eqnarray*}
Answer:
The way I see it, the problem is asking me to find this series:
\begin{eqnarray*}
S_n &=& \sum_{i=1}^{n} {a_i} \\
\text{with } a_i &=& \frac{2i-1}{i(i+1)(i+2)} \\
\end{eqnarray*}
We have:
\begin{eqnarray*}
S_n &=& S_{n-1} + a_n \\
S_n &=& S_{n-1} + \frac{2n-1}{n(n+1)(n+2)} \\
\end{eqnarray*}
I am tempted to apply the technique of partial fractions
but I believe there is no closed formula for a series of the of the form:
\begin{eqnarray*}
\sum_{i=1}^{n} \frac{1}{i+k} \\
\end{eqnarray*}
where $k$ is a fixed constant. Therefore I am stuck. I am hoping that somebody
can help me.
Thanks Bob
| Setting
$$\frac{2n-1}{n(n+1)(n+2)}=\frac{An+B}{n(n+1)}-\frac{A(n+1)+B}{(n+1)(n+2)}$$
gives us $A=2,B=-\frac 12$, i.e.
$$\frac{2n-1}{n(n+1)(n+2)}=\frac{2n-\frac 12}{n(n+1)}-\frac{2(n+1)-\frac 12}{(n+1)(n+2)}.$$
Hence, we have
$$\begin{align}\sum_{i=1}^{n}\frac{2i-1}{i(i+1)(i+2)}&=\sum_{i=1}^{n}\left(\frac{2i-\frac 12}{i(i+1)}-\frac{2(i+1)-\frac 12}{(i+1)(i+2)}\right)\\&=\frac{2\cdot 1-\frac 12}{1\cdot (1+1)}-\frac{2(n+1)-\frac 12}{(n+1)(n+2)}\\&=\color{red}{\frac 34-\frac{4n+3}{2(n+1)(n+2)}}\end{align}$$
| {
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What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$?
What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$ ?
$7 \equiv 3 \pmod 4$
$7^2 \equiv 9 \pmod 4\equiv 1 \pmod 4$
$(7^2)^{16} \equiv 1^{16} \pmod 4$
i.e $7^{32} \equiv 1 \pmod 4$
Similarly $9 \equiv 1 \pmod 4$ implies $9^{45} \equiv 1 \pmod 4$.
But the problem arise with the coefficients and addition sign.
what to do?
| Reduction modulo an integer is a homomorphism of rings ie $$a+b \pmod n=a \pmod n+b\pmod n$$ and $$a\times b\pmod n=a\pmod n\times b \pmod n$$
These facts are easy to check directly e.g. $(a+pn)(b+qn)=ab+(aq+bp+pqn)n$ and $(a+pn)+(b+qn)=a+b+(p+q)n$.
This means that it doesn't matter whether you do your reductions first and the arithmetic after, or do the arithmetic first and then the reduction, or a combination of the two.
Note also that the reduction $7\equiv -1$ is allowed. So I would be doing this as $$6\times 7^{32}+7\times 9^{45}$$ and do all the reductions you can to get something like $$2 \times (-1)^{32}+(-1)\times 1^{45}=2-1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof by induction for golden ratio and Fibonacci sequence I have to prove the following equation by induction for $$x = \phi$$ I am stuck and I don't know how to proceed.
This is the equation
$$
\phi ^n = f_n\phi + f_{n-1}
$$
where $f_n$ is the nth term of the Fibonacci sequence and $f_{n-1}$ is the (n-1)st term.
| More insight:
One way to consider the basic $x^2 - x - 1 = 0$ starting point in the above answer is to consider the initial golden ratio itself, i.e., $a + b$ is to $a$ as $a$ is to $b$,
or
\begin{align*}
\frac{a + b}{a} = \frac{a}{b} = \varphi.
\end{align*}
Now, if $b$ is of length $1$ and $a$ is $x$, we have $a + b = 1 + x$. Then we have
\begin{align*}
\frac{x + 1}{x} = \frac{x}{1} = \varphi
\end{align*}
so that
\begin{align}
x^2 - x - 1 = 0.
\end{align}
We then can plug this into the quadratic equation
\begin{align*}
\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\end{align*}
which gives
\begin{align*}
\varphi = \frac{1 + \sqrt{5}}{2} = 1.6180339887498948482\dots
\end{align*}
but also
\begin{align*}
\varphi = \frac{1 - \sqrt{5}}{2} = -0.6180339887498948482\dots
\end{align*}
but since the golden ratio is the ratio of positives, we discard the second solution$-$initially, at least. See this for a use of the conjugate.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the general values of $x$ satisfying the trigonometric equation
Find the general values of $x$ satisfying
$$
\frac{\tan^2 x \sin^2 x}{1-\sin^2 x \cos2x}+\frac{\cot^2 x \cos^2 x}{1-\cos^2 x \cos2x}+\frac{2\sin^2 x}{\tan^2 x+\cot^2 x}=\frac{3}{2}
$$
It seems to me just some equality case of an inequality. But I am unable to find the inequality. Thanks.
| since
$$\dfrac{\tan^2{x}\sin^2{x}}{1-\sin^2{x}\cos{2x}}=\dfrac{\tan^2{x}}{\csc^2{x}-\cos{2x}}=\dfrac{\tan^2{x}}{\cot^2{x}+2\sin^2{x}}$$
and simaler other
$$\dfrac{\cot^2{x}\cos^2{x}}{1-\cos^2{x}\cos{2x}}=\dfrac{\cot^2{x}}{\sec^2{x}-\cos{2x}}=\dfrac{\cot^2{x}}{\tan^2{x}+2\sin^2{x}}$$
so your equation is
$$\dfrac{\tan^2{x}}{\cot^2{x}+2\sin^2{x}}+\dfrac{\cot^2{x}}{\tan^2{x}+2\sin^2{x}}+\dfrac{2\sin^2{x}}{\tan^2{x}+\cot^2{x}}=\dfrac{3}{2}$$
let
$$a=\tan^2{x},b=\cot^2{x},c=2\sin^2{x}$$
then
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{3}{2}$$
But we known well inequality
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\tag{1}$$
$=$ iff$a=b=c$
so $\tan^2{x}=\cot^2{x}=2\sin^2{x}$
ADD $(1)$ proof
By Cauchy-Schwarz inequality we have
$$\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\left(a(b+c)+b(c+a)+c(a+b)\right)\ge(a+b+c)^2$$
it is enought to prove
$$(a+b+c)^2\ge 3(ab+bc+ac)$$ it is clear
$"="$ iff $a=b=c$
| {
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What is wrong in my $f'(x)$? We have $f:\mathbb{R}\rightarrow\mathbb{R}, f(x)=\frac{x^2-x+1}{x^2+x+1}$ and we need to find $f'(x)$.
Here is all my steps:
$$\begin{align}f'(x)&=\frac{(2x-1)(x^2+x+1)-(x^2-x+1)(2x+1)}{(x^2+x+1)^2}\\&=\frac{(2x-1)(x^2+1)-(2x+1)(x^2+1)}{(\cdots)^2}\\&=\frac{(x^2+1)(2x-1-2x-1)}{(\cdots)^2}\\&=\frac{2(1-x)(1+x)}{(\cdots)^2},\forall x\in\mathbb{R}\end{align}$$
But in my book they say that $f'(x)=\frac{2(x-1)(x+1)}{(x^2+x+1)^2}$. What is wrong in my method ?
| You are missing the red parts in your "method".
$$(2x-1)(x^2+x+1)-(x^2-x+1)(2x+1)$$
$$=(2x-1)(x^2+1)+(2x-1)x-(x^2+1)(2x+1)+x(2x+1)$$
$$=(x^2+1)(2x-1-2x-1)\color{red}{+2x^2-x+2x^2+x}$$
$$=(x^2+1)(-2)\color{red}{+4x^2}$$
| {
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Prove that $\frac{ 5^{125}-1}{ 5^{25}-1}$ is a composite number Prove that $\dfrac {\left( 5^{125}-1\right)}{\left( 5^{25}-1\right)}$ is composite number using number theory. Do not use calculator or Wolfram alpha or anything like that.
| Hint $\ $ The factorization arises by applying a variation of cyclotomic factorization known as Aurifeuillian factorization (aka Aurifeuillean). For the OP we can employ the following one
$$\begin{align}\frac{(5x^2)^5-1}{5x^2-1} =\, (25x^4\!+15x^2+1)^2 - (5x(5x^2\!+1))^2\\[4pt]
\overset{\large x\, =\, 5^{\large 12}}\Longrightarrow\ \ \frac{5^{125}-1}{5^{25}-1}\, =\, (5^{50}+3\cdot 5^{25}+1)^2 - (5^{13}(5^{25}+1))^2\end{align}$$
Remark $\ $ There are many known Aurifeuillian factorizations (e.g. see below for a few more). For more see this answer and see Aurifeuillian Factorization by A. Granville and P. Pleasants.
$$\begin{align}
\frac{(3x^2)^3+1}{3x^2+1} &=\, (3x^2\!+1)^2-(3x)^2\\[4pt]
\frac{(5x^2)^5-1}{5x^2-1} &=\, (25x^4\!+15x^2+1)^2 - (5x)^2(5x^2\!+1)^2\\[4pt]
\frac{(7x^2)^7+1}{7x^2+1} &=\, (7x^2\!+1)^6-(7x)^2(49 x^4\!+7x^2+1)^2
\end{align}$$
| {
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Find the smallest possible value for: $a+b$ If $a,b$ are positive integers with $a, b > 1$, and
$$\sqrt{a\sqrt{a\sqrt{a}}}=b,$$
find the smallest possible value for $a+b$.
| We have: $a^7 = b^8 \to a = b\cdot b^{\frac{1}{7}}$. Take $c = b^{\frac{1}{7}} \to a = c^7\cdot c = c^8, b = c^7$. Thus: $a+b = c^7+c^8$. Clearly the minimum occurs if $c = 2$, for $a = 2^8 = 256, b = 2^7 = 128$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all numbers that have 30 factors and have 30 as one of their factors. Find all numbers that have 30 factors and have 30 as one of their factors.
Thank you.
Note: please show way if possible.
| Let $n \in \mathbf N$. Writing the primefactor decomposition as
$$ n = \prod_p p^{\alpha_p(n)}, $$
we know that $n$ has
$$ \tau(n) = \prod_p \bigl(\alpha_p(n)+ 1\bigr) $$
divisors. (For any $p$ in a divisor we can choose $p$ appearing form $0$ to $\alpha_p(n)$ times), we have as $30 \mid n$, that $\alpha_p(n) \ge 1$, for $p \in \{2,3,5\}$. On the other hand, we must have that $2$, $3$ and $5$ appear in some $\alpha_p(n)+1$, as three of them are greater or equal to two, these must equal $2$, $3$ and $5$. So we are left with six possibilities
$$ 2^1 3^2 5^4, 2^1 3^4 5^2, 2^2 3^1 5^4, 2^2 3^4 5^1, 2^4 3^1 5^2, 2^4 3^2 5^1 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Three variable, second degree diophantine equation I am trying to solve this diophantine equation:
$x^2 + yx + y^2 = z^2$
In other words, I am trying to find integers $x$ and $y$ such that $x^2 + yx + y^2$ is a perfect square.
So far, the only methods to solve quadratic diophantine equations I am familiar with are Pythagorean triples and Pell equations. The $yx$ term has a coefficient of one, so I can't complete the square to reduce it to a Pell equation somehow, and I am wondering if there are other methods to solve this kind of equation.
Some insight would be highly appreciated.
| Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical.
Write the formula can someone come in handy. the equation:
$Y^2+aXY+X^2=Z^2$
Has a solution:
$X=as^2-2ps$
$Y=p^2-s^2$
$Z=p^2-aps+s^2$
more:
$X=(4a+3a^2)s^2-2(2+a)ps-p^2$
$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$
$Z=(2a^3+a^2-8a-8)s^2+2(a^2-2)ps-p^2$
more:
$X=(a+4)p^2-2ps$
$Y=3p^2-4ps+s^2$
$Z=(2a+5)p^2-(a+4)ps+s^2$
more:
$X=8s^2-4ps$
$Y=p^2-(4-2a)ps+a(a-4)s^2$
$Z=-p^2+4ps+(a^2-8)s^2$
For the particular case: $Y^2+XY+X^2=Z^2$ You can draw more formulas.
$X=3s^2+2ps$
$Y=p^2+2ps$
$Z=p^2+3ps+3s^2$
more:
$X=3s^2+2ps-p^2$
$Y=p^2+2ps-3s^2$
$Z=p^2+3s^2$
In the equation: $X^2+aXY+bY^2=Z^2$ there is always a solution and one of them is quite simple.
$X=s^2-bp^2$
$Y=ap^2+2ps$
$Z=bp^2+aps+s^2$
$p,s$ - integers asked us.
| {
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Can the roots of $f(x)=x^4-x^3+2x^2-x-1$ be found algebraically? Can the roots of $f(x)=x^4-x^3+2x^2-x-1$ be found algebraically? Are there multiple methods for doing so?
| If you read French, you can have a description of Ferrari's method, and links to a bunch of methods for algebraic equations (some exist in other languages).
In the present case, however, as $1$ is a root,we have a factorisation by $x-1$:
$$ x^4-x^3+2x^2-x-1=(x-1)(x^3+2x+1). $$
So we have a cubic equation in standard form. Its discriminant is $\Delta=-4\cdot 2^3-27\cdot 1^2=-59<0$, hence it has only $1$ real root (we could as well check this point with the derivative), and we can use Cardano's method:
Set $x=u+v$. The equation rewrites as:
$$u^3+v^3+(3uv+2)(u+v)=1=0.$$
As we've replaced $1$ unknown with $2$, we can impose a condition: $3uv+2=0$. The equation is now equivalent to the system:
$$\begin{cases}uv=-\frac23\\u^3+v^3=-1\end{cases}\iff\begin{cases}u^3v^3=-\frac8{27}\\u^3+v^3=-1\end{cases} $$
Thus the problem boils down to the standard problem on quadratic equations: find two numbers, given their sum and their product. They are the roots of the equation:
$$t^2+t-\frac8{27}=0.$$
The discriminant of this equation is $\;1+\dfrac{32}{27}=\dfrac{59}{27}\;$ so
$$u^3,v^3=\frac{-1\pm\sqrt{\frac{59}{27}}}2$$
and finally:
$$ x=u+v= \sqrt[3]{\frac{-1+\sqrt{\frac{59}{27}}}2} + \sqrt[3]{\frac{-1-\sqrt{\frac{59}{27}}}2} $$
| {
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Find coefficients of polynomials $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d$ $(a,b,c,d \in \mathbb{R})$ Roots of polynomial $f(x)=x^2+ax+b$ are cubes of the roots of polynomial $g(x)=x^2+cx+d$. Sum and product of roots of polynomial $g(x)$ are equal. Find coefficients $a,b,c,d$ so that polynomial $f(x)$ has double root, and $g(x)$ doesn't have double root.
From the condition that $f(x)$ has double root, discriminant of quadratic equation $x^2+ax+b=0$ must be zero, $$D=a^2-4b=0,b=\frac{a^2}{4}$$
Sum and product of roots of polynomial $g(x)$ are equal, $$x_1^3+x_2^3=x_1^3x_2^3$$
Vieta's formulas for two polynomials,
$$x_1+x_2=-a$$
$$x_1x_2=b$$
$$x_1^3+x_2^3=-c$$
$$x_1^3x_2^3=d$$
If we cube first equation, $$(x_1^3+x_2^3)+3x_1x_2(x_1+x_2)=-a^3$$
$$-c-3ba=-a^3$$
From $b=\frac{a^2}{4}$ $$a^3-4c=0,c=\frac{a^3}{4}$$
From $x_1^3+x_2^3=x_1^3x_2^3$ implies $d=-c$
If we cube second equation, $$x_1^3x_2^3=b^3=\frac{a^6}{64}=d=-c$$
I don't know how to find these coefficients.
Thanks for replies.
| Hint: Let $\alpha, \beta$ be the roots of $g(x)=0$ then the roots of $f(x)$ are $\alpha^3=\beta^3$ which means that $\alpha^3-\beta^3=0=(\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2)$ and since $\alpha\neq \beta$ you have $\alpha^2+\alpha\beta+\beta^2=0$
Then plug in the condition $\alpha \beta=\alpha+\beta=r$ so that $\alpha^2+\alpha\beta+\beta^2=r^2-r=0$
Since others have completed the solution, here's my take on a simpler answer using the hint.
You can't have $r=0$, because this would give a double root of $0$, so it must be that $r=1$. Then $g(x)=x^2-x+1$. It is clear that $b=\alpha^3\beta^3=r^3=1$, and then $-a=\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)=r(r^2-3r)=-2$ so that $a=2$ so that $f(x)=x^2+2x+1$
| {
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Mathematical Induction getting the right side So I 've been doing Mathematical Inductions but I seem to have a issue in simplify and getting the right side.
So I have this on the L.H.S
$$\frac{k(k + 1)(2k +1)}{6} + (k + 1)^2 $$
And I'm trying to make it equal to
$$\frac{(k+1)(k+2)(2k+3)}{6}$$
I've check the book for help but for this it seems to be skipping some steps.
And I might as well post this as well which I had the same issue as above.
$$\frac{3(5^{k+1} −1)}{4}+ 3·5^{k+1}$$
to
$$\frac{3(5^{k+2} −1)}{4}.$$
| (1)$\frac{k(k+1)(2k+1)}{6} + (k+1)^2 =\frac{k(k+1)(2k+1) + 6(k+1)^2}{6} =
(k+1)\frac{k(2k+1)+6(k+1)}{6} = (k+1)\frac{k(2k+1)+6k+6}{6} = (k+1)\frac{2k^2 + 7k + 6}{6} = \frac{(k+1)(k+2)(2k+3)}{6} $
(2) Is your second statement even true? for $k=1$ I get: $34 = 93$...
Let $5^k = x$
$\frac{3(5x-1)}{4} + 3x + 1 = \frac{15x - 3 + 12x + 4}{4} = \frac{27x+1}{4}$
which doesn't seem to be equal to your given RHS
| {
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Proving $n^3 + 3n^2 +2n$ is divisible by $6$ The full question is: Factorise $n^3 + 3n^2 + 2n$. Hence prove that when $n$ is a positive integer, $n^3 + 3n^2 + 2n$ is always divisible by $6$.
So i factorised and got $n(n+1)(n+2)$ which i think is right? I'm not sure how to actually prove this is divisible by $6$ though. Thanks for help and i apologise if someone has already asked this, i couldn't find it. Also i've not been told whether i have to do induction or not.
| Keep going with your idea. $n(n+1)$ is the product of two consecutive integer so one of them is even. If $n$ is even we are done and if not $n+1$ is and the product is therefore divisible by $2$.
Similarly $n(n+1)(n+2)$ is the product of three consecutive integers so one of them is divisible by $3$. Let's test the residues modulo $3$
$$\begin{array}{c | c c c}
n & n+1 & n+2 & n(n+1)(n+2)\\
\hline
0 & 1 & 2 & 0\\
1 & 2 & 0 & 0\\
2 & 0 & 1 & 0
\end{array}$$
So the product is also divisible by $3$ and therefore the product $n(n+1)(n+2)\equiv 0\pmod 6$
One could also test directly the congruences modulo $6$
$$\begin{array}{c|c c c c}
n & n^3 & 3n^2 & 2n & n^3+3n^2+2n\\
\hline
0 & 0 & 0 & 0 & 0\\
1 & 1 & 3 & 2 & 0\\
2 & 2 & 0 & 4 & 0\\
3 & 3 & 3 & 0 & 0\\
4 & 4 & 0 & 2 & 0\\
5 & 5 & 3 & 4 & 0
\end{array}$$
And we have proven that $n^3+3n^2+2n\equiv 0\pmod 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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} |
When are we permitted to multiply or divide both sides of an equation by a variable? As it is said in the mathematics books (at least the one I have), we are not permitted to divide or multiply both sides of an equation by a variable, because it is possible to lose some answers. For example, in the following equation
$$x^2=x$$
if we divide both sides by $x$, we would have $x = 1$, but the original equation has two answers, $0$ and $1$, and we've lost $x = 0$ by dividing it by a variable.
But in the same book, in order to solve a rational equation, the author multiplies both sides of the equation by $x(x-2)$, solving the equation:
$$\begin{align*}
\frac {x+2}{x-2} - \frac1x &= \frac{2}{x(x-2)}\\\\
x(x-2)\frac {x+2}{x-2} - x(x-2)\frac 1x &= x(x-2)\frac{2}{x(x-2)}\\\\
x(x+2)-(x-2)&=2\\\\
x^2+x&=0\\\\
x(x+1)&=0\\\\
x&=0\\
x&=-1
\end{align*}$$
Can someone please explain when we are allowed to do this and when we are not? I got a bit confused!
| When $c\ne0$, $a=b$ and $ac=bc$ are equivalent, that's about all you need to know.
Applications:
*
*$x^2=x$ is equivalent to $x=1$ when $x\ne0$. Obviously, $0^2=0$ so that $x=0$ is also a solution (and there are no others).
*$\dfrac {x+2}{x-2} - \dfrac1x = \dfrac{2}{x(x-2)}$ and $x(x+2) - (x-2)= 2$ are equivalent when $x(x-2)\ne0$. Then $x=0$ and $x=2$ are known not to be solutions, as the expressions are undefined for them. So if they appear later in the computation as possible roots, these must be discarded.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Help with this limit? I am trying to focus on the limits of functions with similar series expansions and I stumbled on this.
$$\lim_{x\to\infty}\left({\left(\frac{x^2+5}{x+5}\right)}^{1/2}\sin{\left({\left(\frac{x^3+5}{x+5}\right)}^{1/2}\right)}-(x-5)^{1/2}\sin{\left({\left(x^2-5x+25\right)}^{1/2}\right)}\right)=0$$
I heard the mean value is possible but the entire function is not bounded. I can take the taylor series at infinity however the terms would be undefined. I could use substitution with the taylor series but it would become a complicated mess.
$$\lim_{x\to\infty}\left(\left(\frac{x^2+5}{x+5}\right)^{1/2}{\left(\frac{x^3+5}{x+5}\right)}^{1/2}-\frac{\left(\frac{x^2+5}{x+5}\right)^{1/2}\left({\left(\frac{x^3+5}{x+5}\right)}^{1/2}\right)^3}{3!}+\frac{\left(\frac{x^2+5}{x+5}\right)^{1/2}\left({\left(\frac{x^3+5}{x+5}\right)}^{1/2}\right)^5}{5!}...-\left(\left({x-5}\right)^{1/2}{\left(x^2-5x+25\right)}^{1/2}-\frac{\left(x-5\right)^{1/2}\left({\left(x^2-5x+25\right)}^{1/2}\right)^3}{3!}+\frac{\left(x-5\right)^{1/2}\left({\left(x^2-5x+25\right)}^{1/2}\right)^5}{5!}....\right)\right)$$
I only have limited knowledge of series expansion so I am not so sure how to approach this. Is their an easier way?
| Since $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$, we have $|\sqrt{a}-\sqrt{b}|\le\frac{|a-b|}{2\sqrt{\min(a,b)}}$. Therefore
$$
\begin{align}
\left|\left(\frac{x^3+5}{x+5}\right)^{1/2}-\left(\frac{x^3+125}{x+5}\right)^{1/2}\right|
&\le\overbrace{\frac12\left(\frac{x^3+5}{x+5}\right)^{-1/2}}^{\large\frac1{2\sqrt{\min(a,b)}}}\overbrace{\frac{120}{x+5}\vphantom{\left(\frac{x^3}{x}\right)^{1/2}}}^{|a-b|}\\
&\le\frac{60}{x^2}\tag{1}
\end{align}
$$
Since $|\sin(x)-\sin(y)|\le|x-y|$, we get that
$$
\left|\sin\left(\left(\frac{x^3+5}{x+5}\right)^{1/2}\right)-\sin\left(\left(\frac{x^3+125}{x+5}\right)^{1/2}\right)\right|\le\frac{60}{x^2}\tag{2}
$$
Similarly,
$$
\begin{align}
\left|\left(\frac{x^2+5}{x+5}\right)^{1/2}-\left(\frac{x^2-25}{x+5}\right)^{1/2}\right|
&\le\overbrace{\frac12(x-5)^{-1/2}}^{\large\frac1{2\sqrt{\min(a,b)}}}\overbrace{\frac{30}{x+5}}^{|a-b|}\\
&\le\frac{15}{(x-5)^{3/2}}\tag{3}
\end{align}
$$
Using $(2)$ and $(3)$ and $ab-cd=(a-c)b+(b-d)c$, we get
$$
\begin{align}
&\left|\left(\frac{x^2+5}{x+5}\right)^{1/2}\sin\left(\left(\frac{x^3+5}{x+5}\right)^{1/2}\right)-\left(\frac{x^2-25}{x+5}\right)^{1/2}\sin\left(\left(\frac{x^3+125}{x+5}\right)^{1/2}\right)\right|\\
&\le\left(\frac{x^2+5}{x+5}\right)^{1/2}\frac{60}{x^2}+1\cdot\frac{15}{(x-5)^{3/2}}\\[12pt]
&\to0\tag{4}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Expected number of coin flips for heads The expected number of coin flips to get one heads is $2$. What is wrong with this argument?
There is a $1/2$ chance of getting $H$, $1/4$ chance of getting $TH$,
$1/8$ chance of getting $TTH$, etc. so the expected value of flips is
$$\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + ...\approx \ ?$$
Edit: incorrect value
| Here is an alternate method for computing the sum (not better, just different).
To compute the sum, note that $$\frac{1}{1-x} = 1 + x + x^2 + x^3+...$$
Differentiate to get $$\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3+...$$
Whence $$\frac{x}{(1-x)^2} = x + 2x^2 + 3x^3 + 4x^4+...$$
In particular, taking x = $\frac{1}{2}$ we have
$$ 2 = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + ...$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do you find the value of $f(x)$ for this trig function satisfying all values of $x$? If $ f(x) = 3[\sin^4(\frac{3\pi}{2} - x) + \sin^4(3\pi+x)] -2[\sin^6(\frac{\pi}{2} + x) + \sin^6(5\pi-x)] $ then, for all permissible values of $x$, $f(x)$ is:-
Here's how I attempted it-
$ f(x) = 3[\sin^4(\frac{3\pi}{2} - x) + \sin^4(3\pi+x)] -2[\sin^6(\frac{\pi}{2} + x) + \sin^6(5\pi-x)] $
$ f(x) = 3[\cos^4x + \sin^4x] -2[\cos^6x + \sin^6x] $
$ f(x) = 3\cos^4x-2\cos^6x+ 3\sin^4x - 2\sin^6x $
$ f(x) = \cos^4x(3-2\cos^2x)+ \sin^4x(3 - 2\sin^2x) $
$ f(x) = \cos^4x(1+2\sin^2x)+ \sin^4x(1+2\cos^2x) $
$ f(x) = \cos^4x+2\sin^2x\cos^4x+ \sin^4x+2\cos^2x\sin^4x $
$ f(x) = (\sin^2x)^2+ (\cos^2x)^2 +2\sin^2x\cos^2x(\cos^2x+\sin^2x) $
$ f(x) = (\sin^2x+\cos^2x)^2$
I don't know what to do next. Thanks in advance to anyone who's willing to help. The final answer should come up to '$-1$'.(According to the textbook).
| Since $$\sin \left(\frac{3\pi}{2} - x\right) = \cos x \implies \sin^4 \left(\frac{3\pi}{2} - x\right) = \cos^4 x$$ and $$\sin^4 (3\pi + x) = \sin^4 x$$
We have $f(x) = f(x) = 3(\sin ^4 x + \cos ^4 x)-2[\sin^6(\frac{\pi}{2} + x) + \sin^6(5\pi-x)]$. Let's work on simplifying the second part of the function now. We have $$\sin \left(\frac{\pi}{2} + x\right) = \cos x \implies \sin^6 \left(\frac{\pi}{2} + x\right) = \cos^6 x$$ and $$\sin^6 (5\pi-x) = \sin^6 x$$ so that your function becomes $$f(x) = 3(\sin^4 x + \cos^4 x) - 2(\sin^6 x + \cos^6 x)$$
But we know that $$\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x$$ and $$\sin^4 x + \cos^4 x = 1 - \frac{1}{2}\sin^2 2x = 1 - 2\sin^2 x \cos^2 x$$ Putting this all together yields $$f(x) = 3 - 6 \sin^2 x \cos^2 x - 2 + 6\cos^2 x \sin^2 x = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit of definite integral sequence This is taken from an admission exam textbook at our local construction university :
$$ \lim_{n\to\infty} n\int_{1}^{2} \frac{1}{x^2(1+x^n)}dx = ?$$
I tried finding tight bounds in order to use the sandwich theorem, but that got me nowhere. I also tried using Lebesgue's Dominated Convergence Theorem, but I ended up with nothing once more.
Do you have any ideas?
| Using
\begin{align}
\frac{1}{1+x^{n}} = \frac{1}{x^{n}} \, \frac{1}{1 + \frac{1}{x^{n}}} = \frac{1}{x^{n}} \, \sum_{k=0}^{\infty} \frac{(-1)^{k}}{x^{kn}}
\end{align}
then
\begin{align}
I_{n} &= \int_{1}^{2} \frac{dx}{x^{2} \, (1+x^{n})} \\
&= \int_{1}^{2} \frac{1}{x^{n+2}} \, \left( 1 - \frac{1}{x^{2n+2}} + \cdots \right) \, dx \\
&= - \frac{1}{n+1} \left[ \frac{1}{x^{n+1}} \right]_{1}^{2} + \frac{1}{2n+1} \, \left[ \frac{1}{x^{2n+1}} \right]_{1}^{2} + \cdots \\
&= \frac{1}{n+1} \, \left(1 - \frac{1}{2^{n+1}} \right) + \frac{1}{2n+1} \, \left( \frac{1}{2^{2n+1}} - 1 \right) + \cdots \\
&= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{nk+n+1} \, \left( 1 - \frac{1}{2^{nk+n+1}} \right)
\end{align}
Now,
\begin{align}
\lim_{n\to\infty} \left\{ n \, I_{n} \right\} = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k+1} = \ln 2.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Math contest question - prove unsolvability of equations. Prove that the system of equations has no real solutions:
$$\begin{cases} y=\sqrt{x+\sqrt{1-x}} \\ x=\sqrt{y-\sqrt{1+y}}\end{cases}. $$
This is a former problem from a national math contest which I´ve solved already. However, since my solution was very similar to the solution manual, I´m interested in other solving approaches.
My idea, in short, was as follows:
We first notice that from equation (1), we get the condition $y < \sqrt{2}$.
From equation (2) we have that $y \geq \sqrt{1+y}$, but for $y<1$ this can't be true since then $\sqrt{1+y}>1$. So $y \geq 1$ and thus $$0 \leq y-\sqrt{1+y}\leq y-\sqrt{2},$$
which is a contradiction to our first condition $y < \sqrt{2}$.
| Suppose that there exists a set of real solution.
First of all, $x$ has to satisfy the followings :
$$1-x\ge 0\ \ \ \text{and}\ \ \ x+\sqrt{1-x}\ge 0\ \ \ \text{and}\ \ \ x\ge 0.$$
This is equivalent to $0\le x\le 1$.
Similarly, $y$ has to satisfy the followings :
$$1+y\ge 0\ \ \ \text{and}\ \ \ y-\sqrt{1+y}\ge 0\ \ \ \text{and}\ \ \ y\ge 0.$$
This is equivalent to $y\ge\frac{1+\sqrt 5}{2}$.
Now let $f(x)=\sqrt{x+\sqrt{1-x}}\ \ \ (0\le x\le 1)$. Since we have
$$f'(x)=\frac{2\sqrt{1-x}-1}{4\sqrt{1-x}\cdot\sqrt{x+\sqrt{1-x}}},$$
we have $f'(x)=0\iff x=\frac 34$, and so $1=f(0)=f(1)\le f(x)\le f(3/4)=\frac{\sqrt 5}{2}$.
However, this contradicts $y\ge\frac{1+\sqrt 5}{2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Arc length of $f(x)=x^2-\ln x$ over [1,e] This is how I learned to solve for arc length:
$$ \frac d{dx} f(x)=2x - \frac 1x$$
$$ \left(\frac d{dx}f(x)\right)^2 = \left(2x - \frac 1x\right)^2 = 4x^2-4+\frac1{x^2} $$
$$ 1+\left(\frac d{dx}f(x)\right)^2 = 4x^2-3+\frac1{x^2} $$
$$ \sqrt{1+\left(\frac d{dx}f(x)\right)^2} = \sqrt{4x^2-3+\frac1{x^2}}$$
So to solve for arc length, I have:
$$\int_1^e \sqrt{4x^2-3+\frac1{x^2}} ~dx $$
I already have the answer (from Wolfram alpha) but I cannot figure out how to solve this integral. It seems complicated since Wolfram cannot generate the steps for this integral.
Any suggestion is highly appreciated.
| After substituting $t=(8x^2-3)/\sqrt{7}$, we get
$$\int\sqrt{4x^2-3+\frac{1}{x^2}}dx=\frac{7}{8}\int\frac{\sqrt{t^2+1}}{\sqrt{7}t+3}dt.$$
Making another substitution, $t=\sinh s$, we obtain
$$\frac{7}{8}\int\frac{\sqrt{t^2+1}}{\sqrt{7}t+3}dt=
\frac{\sqrt{7}}{8}\int\frac{\sinh^2 s+1}{\sinh s+\displaystyle\frac{3}{\sqrt{7}}}ds=\frac{\sqrt{7}}{8}\int\sinh s~ds-\frac{3}{8}\int ds+\frac{2}{\sqrt{7}}\int\frac{1}{\sinh s+\displaystyle\frac{3}{\sqrt{7}}}ds.$$
The first two terms are trivially integrable; as to the third term, I cheated and checked my Gradshteyn ($\int (a+b\sinh x)^{-1}dx=2(a^2+b^2)^{-1/2}{\rm atanh}\big([a\tanh(x/2)-b][a^2+b^2]^{-1/2}\big)$). After simplification, we get
$$\int\sqrt{4x^2-3+\frac{1}{x^2}}dx=\frac{1}{2}\sqrt{4x^4-3x^2+1}-\frac{3}{8}{\rm asinh}~\frac{8x^2-3}{\sqrt{7}}+{\rm atanh}\left(\frac{3-\sqrt{7}}{4}-\frac{3\sqrt{7}}{16x^2-6+2\sqrt{7}+8\sqrt{4x^4-3x^2+1}}\right).$$
Up to an integration constant, this agrees with the result given by Maple.
| {
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The number $2^{2^n} + 2^{2^{n - 1}} + 1$ can be expressed as the product of at least $n$ prime factors
Prove that the number $2^{2^n} + 2^{2^{n - 1}} + 1$ can be expressed as the product of at least $n$ prime factors, not necessarily distinct.
Doing what the hint has suggested, I have done the same where $n>1$, and took $\bmod 7$. I then found that $2^{2^n} + 2^{2^{n - 1}} + 1=7k$.
However, I got stuck when trying to simplify the case of $n=3$, since $$2^8+2^4+1=273=3\cdot 7\cdot 13$$
I do not understand where the 13 is from, since I thought that it would be something like $$2^{2^n} + 2^{2^{n - 1}} + 1 \equiv 0 \pmod {2^n-1}$$
| We start with $x^4+x^2+1=\left(x^2+1\right)^2-x^2=\left(x^2-x+1\right)\left(x^2+x+1\right)$. Thus, we can proceed by induction (on $n$) that $2^{2^n}+2^{2^{n-1}}+1$ has at least $n$ prime factors. For $n=1$, there is nothing to prove. Suppose the statement is true for $n=k$ for some $k\in\mathbb{N}$. Then, for $n=k+1$,
$$2^{2^{k+1}}+2^{2^k}+1=\left(2^{2^k}-2^{2^{k-1}}+1\right)\left(2^{2^k}+2^{2^{k-1}}+1\right)\,.$$
By the induction hypothesis, $2^{2^k}+2^{2^{k-1}}+1$ has $k$ prime factors. Furthermore, $2^{2^k}-2^{2^{k-1}}+1>1$ has at least one prime factor. Hence, $2^{2^{k+1}}+2^{2^k}+1$ has at least $k+1$ prime factors. We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Iterated Integral and Sign Change in Answer Given the iterated integral $\int_0^1\int_x^{2-x}(x^2-y) \, dy \, dx$, the value for the type I integral is,
\begin{align*}
& \int_0^1\int_x^{2-x}(x^2-y)\,dy\,dx \\
= {} & \int_0^1 x^2y\Big|_x^{2-x} \, dx - \int_0^1 \left.\frac{y^2}{2}\right|_x^{2-x} \, dx \\
= {} & \int_0^1 x^2(2-2x) \, dx - \int_0^1(2-2x) \, dx \\
= {} & \int_0^1 2x^2 \, dx - \int_0^1 2x^3 \, dx - \int_0^1 2 \, dx + \int_0^1 2x \, dx \\
= {} & 2 \left.\frac{1}{3}x^3\right|_0^1 - \left.2 \frac{x^4}{4}\right|_0^1 - 2x\Big|_0^1 + 2\frac{1}{2} x^2\Big|_0^1 \\
= {} & \frac{2}{3} - \frac{1}{2} -2 + 1 \\
= {} & \frac{2}{3} - \frac{3}{2} = -\frac{5}{6}
\end{align*}
We then calculate the type II integral, thus,
$$
\int_0^1\int_y^{2-y}(x^2-y) \, dx \, dy = \frac{5}{6}
$$
The two integrals differ by the presence of a negative sign. Does this mean that these integrals are not iterated?
EDIT: It would seem that the issue has to deal with the order of the integration limits. Why are the limits reversed when you take the iterated integral with respect to $dx \, dy$?
Thank you for your time.
| The integral for $dxdy$ has to be split. You are integrating a different region in your actual set up. You should be integrating
$$\int_0^1\int_0^y(x^2-y) \, dx\,dy+\int_1^2\int_0^{2-y}(x^2-y)\,dx\,dy$$
instead.
The area you were integrating for $dx \, dy$ was the triangle below $y=x$ and $y=2-x$ and above $y=0$. The one you were integrating for $dy \, dx$ was to the left of $y=x$ and $y=2-x$ and to the right of $x=0$. (the first is the lower triangle, and the second is the upper triangle)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Median of triangle I know that a median of a triangle is a line joining one
of the vertices to the mid-point of the opposite side.
For example, in a triangle $OAB$, $O$ is the origin, $A$ is the point $(0,6)$ and $B$ is the point $(6,0)$.
How can I show that the point $(2,2)$ lies on all three medians?
| $\color{red}{\text{Method 1}}$: All three medians of a triangle pass through the centroid of the triangle hence the centroid of $\triangle OAB$ having vertices $O(0, 0)$, $A(0, 6)$ & $B(6, 0)$ is given as $$\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\equiv\left(\frac{0+0+6}{3}, \frac{0+6+0}{3}\right)\equiv(2, 2)$$
Thus the given point $(2, 2)$ (coincident with the centroid) lies on all three medians of $\triangle OAB$.
$\color{red}{\text{Method 2}}$: Calculating the mid-points of all the sides of $\triangle OAB$ & finding the equations of all three medians of $\triangle OAB$ as follows
$\color{blue}{\text{Median 1}}$: drawn from the vertex $A(0, 6)$ to the mid-point of side $OB$ i.e. $\left(\frac{0+6}{2}, \frac{0+0}{2}\right) $ or $(3, 0) $ given as $$y-6=\frac{0-6}{3-0}(x-0)$$ $$\implies \color{blue}{2x+y-6=0}$$
Substituting the coordinates of the given point $(2, 2)$ in the above equation of the median, as follows $$2(2)+2-6=0$$ $$\implies 0=0$$ Hence, point $(2, 2)$ lies on the median through the vertex $A$. Similarly,
$\color{blue}{\text{Median 2}}$: drawn from the vertex $B(6, 0)$ to the mid-point of side $OA$ i.e. $\left(\frac{0+0}{2}, \frac{0+6}{2}\right) $ or $(0, 3) $ given as $$y-0=\frac{3-0}{0-6}(x-6)$$ $$\implies \color{blue}{x+2y-6=0}$$
Substituting the coordinates of the given point $(2, 2)$ in the above equation of the meridian, as follows $$2+2(2)-6=0$$ $$\implies 0=0$$ Hence, point $(2, 2)$ lies on the median through the vertex $B$.
$\color{blue}{\text{Median 3}}$: drawn from the vertex $O(0, 0)$ to the mid-point of side $AB$ i.e. $\left(\frac{0+6}{2}, \frac{6+0}{2}\right) $ or $(3, 3) $ given as $$y-0=\frac{3-0}{3-0}(x-0)$$ $$\implies \color{blue}{y=x}$$
Substituting the coordinates of the given point $(2, 2)$ in the above equation of the median, as follows $$2=2$$ Hence, point $(2, 2)$ lies on the median through the vertex $O$.
Thus the given point $(2, 2)$ lies on all three medians of $\triangle OAB$
| {
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} |
Unfairish Probability
Charles has two six-sided dice. One of the dice is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$ Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers. Find $p + q$.
Case 1: Fair die chosen:
$$P = \binom{3}{3} \frac{1}{6^3} \cdot \frac{5^0}{6^0} = \frac{1}{216}$$
Case 2: Unfair die chosen:
$$P_2 = \binom{3}{3} \frac{2^3}{3^3} = \frac{8}{27}$$
$$P + P_2 = \sum P = \frac{1728 + 27}{5832} = \frac{1755}{5832}$$
But that is incorrect, why?
| The answer you wrote down doesn't capture the information gained from getting two sixes. This problem requires conditional probability. Let die $A$ be the event that the fair die was selected and $B$ be the event that the unfair die was selected. Then we have
\begin{align} \mathbb{P}(\text{another } 6 \text{ is rolled }| \text{ two } 6\text{'s were rolled}) &= \mathbb{P}(\text{another } 6 \text{ is rolled }|A)\cdot \mathbb{P}(A| \text{ two } 6\text{'s were rolled}) \\ &+ \mathbb{P}(\text{another } 6 \text{ is rolled }|B)\cdot \mathbb{P}(B| \text{ two } 6\text{'s were rolled}) \\
&= \frac{1}{6}\cdot \mathbb{P}(A| \text{ two } 6\text{'s were rolled}) \\ &+ \frac{2}{3}\cdot \mathbb{P}(B| \text{ two } 6\text{'s were rolled}) .\end{align}
We now use Bayes' theorem to calculate $ \mathbb{P}(A| \text{ two } 6\text{'s were rolled})$ and the similar probability for $B$. We have \begin{align} \mathbb{P}(A| \text{ two } 6\text{'s were rolled})&= \frac{\mathbb{P}(A)\mathbb{P}(\text{ two } 6\text{'s were rolled } | A)}{\mathbb{P}(A)\mathbb{P}(\text{ two } 6\text{'s were rolled } | A) + \mathbb{P}(B)\mathbb{P}(\text{ two } 6\text{'s were rolled } | B)} \\
&= \frac{\frac{1}{36}\cdot\frac{1}{2}}{\frac{1}{36}\cdot\frac{1}{2} + \frac{4}{9}\cdot\frac{1}{36}} \\
&= \frac{1}{17}.\end{align}
Thus, we also have $ \mathbb{P}(B| \text{ two } 6\text{'s were rolled}) =\frac{16}{17}.$ Putting it all together gives $$ \mathbb{P}(\text{another } 6 \text{ is rolled }| \text{ two } 6\text{'s were rolled}) = \frac{1}{6}\cdot \frac{1}{17} + \frac{2}{3}\cdot\frac{16}{17} = \frac{65}{102}.$$
Thus, your answer is $65 + 102 = 167$.
| {
"language": "en",
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} |
Proof by induction that $3^{2n} + 7$ is divisible by $4$ Demonstrate by induction: $3^{2n} + 7 = 4k$ is true, for any $n\in \mathbb N$. I need to demonstrate this using the induction principle.
So far I have:
$n = 1$
$$3^{2\cdot 1} + 7 = 4\cdot k $$
$$9 + 7 = 4k$$
$$16 = 4k$$
$$k = 4$$
So it checks for $n=1$.
$n = h$
$$3^{2\cdot h} + 7 = 4\cdot k$$
$n = h +1$
$$3^{2\cdot (h + 1)} + 7 = 4\cdot k'$$
(I use $k'$ to note that it's not the same $k$ as in $n = h$)
And I don't know how to continue.
| Just to be contrary,
here is a non-induction proof.
$\begin{array}\\
3^{2n}+7
&=3^{2n}-1+8\\
&= (3^n+1)(3^n-1)+8
\end{array}
$
Since $3^n$ is odd
(for $n \ge 1$),
$3^n-1$ and
$3^n+1$
are consecutive even numbers,
so one is exactly divisible by $2$
and the other is divisible by
(at least) $4$.
Therefore,
their product is divisible
by at least 8,
so
$3^{2n}+7$
is divisible by 8.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Induction Proof using factorials Recall that for $n \in N$, $n! = 1 \cdot 2 \cdots n$.
Prove the following for each $n \in N$:
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$$
I understand how to do the proof, but in the inductive step I am facing some difficulty proving the left-hand side is equivalent to the right-hand side.To be direct I am facing some difficulty with the algebra required to make LHS = RHS.
Here is what I have done so far:
1) Base Case
$n = 1$
LHS:
$1/2$
and RHS is $1/2$ $\checkmark$
2) Inductive Step
For $k \geq 1$, Assume $n = k$
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{k}{(k+1)!} = 1 - \frac{1}{(k+1)!}$$
$$n = k + 1$$
$$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{n}{(n+1)!} + \frac{k+1}{(k+2)!} = 1 - \frac{1}{(n+1)!}$$
$$\implies 1-\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} = 1 - \frac{1}{(k+2)!}$$
Here is where i do not know how to make the LHS = RHS.
| For $n=1,2,3$ we have:
\begin{eqnarray}
\frac{1}{2!}&=&\frac12=1-\frac12=1-\frac{1}{2!}\\
\frac{1}{2!}+\frac{2}{3!}&=&1-\frac{1}{2!}+\frac13=1-\frac16=1-\frac{1}{3!}\\
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}&=&1-\frac16+\frac{1}{8}=1-\frac{1}{24}=1-\frac{1}{4!}.
\end{eqnarray}
If we assume that up to $n=3$, we have
$$
\sum_{k=1}^n\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}.
$$
Then
\begin{eqnarray}
\sum_{k=1}^{n+1}\frac{k}{(k+1)!}&=&\sum_{k=1}^n\frac{k}{(k+1)!}+\frac{n+1}{(n+2)!}=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}\\
&=&1-\frac{n+2-(n+1)}{(n+2)!}=1-\frac{1}{(n+2)!}.
\end{eqnarray}
| {
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} |
If $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ then.. If $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ then
(A) $a=c$
(B) either $a=c$ or $a+b+c+d=0$
(C) $a+b+c+d=0$
(D) $a=c$ and $b=d$
I solved $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ and got $a(a+b+d)=c(c+b+d)$ and so I thought that (A) is the correct option. But the correct answer is (B).
I'm how $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ if $a+b+c+d=0$. Please help.
| Hint: we get by cross multiplication
$$a^2+ab+ad+bd=bc+bd+c^2+dc$$from here we get
$$ab-bc+ad-cd+a^2-c^2=0$$
or
$$(a-c)(a+b+c+d)=0$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Wanted : for more formulas to find the area of a triangle? I know some formulas to find a triangle's area, like the ones below.
*
*Is there any reference containing most triangle area formulas?
*If you know more, please add them as an answer
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=\frac{1}{2}bc\sin(A)\\s=2R^2\sin A \sin B \sin C$$
Another symmetrical form is given by :$$(4s)^2=\begin{bmatrix}
a^2 & b^2 & c^2
\end{bmatrix}\begin{bmatrix}
-1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -1
\end{bmatrix} \begin{bmatrix}
a^2\\
b^2\\
c^2
\end{bmatrix}$$
Expressing the side lengths $a$, $b$ & $c$ in terms of the radii $a'$, $b'$ & $c'$ of the mutually tangent circles centered on the triangle's vertices (which define the Soddy circles)
$$a=b'+c'\\b=a'+c'\\c=a'+b'$$gives the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$
If the triangle is embedded in three dimensional space with the coordinates of the vertices given by $(x_i,y_i,z_i)$ then $$s=\frac{1}{2}\sqrt{\begin{vmatrix}
y_1 &z_1 &1 \\
y_2&z_2 &1 \\
y_3 &z_3 &1
\end{vmatrix}^2+\begin{vmatrix}
z_1 &x_1 &1 \\
z_2&x_2 &1 \\
z_3 &x_3 &1
\end{vmatrix}^2+\begin{vmatrix}
x_1 &y_1 &1 \\
x_2&y_2 &1 \\
x_3 &y_3 &1
\end{vmatrix}^2}$$
When we have 2-d coordinate $$ s=\frac{1}{2}\begin{vmatrix}
x_a &y_a &1 \\
x_b &y_b &1 \\
x_c &y_c & 1
\end{vmatrix}$$
In the above figure, let the circumcircle passing through a triangle's vertices have radius $R$, and denote the central angles from the first point to the second $q$, and to the third point by $p$ then the area of the triangle is given by:
$$ s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$
| Consider $\triangle ABC$ and inscribed circle with radius $r$.
Three smaller similar triangles
$\triangle AcBcC,\ \triangle AbBCb,\ \triangle ABaCa$
are cut by
tangent lines to the incircle parallel to
the corresponding sides of $\triangle ABC$.
Given the radii $r_a,\ r_b,\ r_c$ of the incircles of
$\triangle ABaCa,\ \triangle AbBCb,\ \triangle AcBcC$,
respectively,
the area of $\triangle ABC$ is found as
\begin{align}
\bbox[5px,border:2px solid #C0A000]{
S=
\sqrt{\frac{(r_a+r_b+r_c)^7}{r_a\,r_b\,r_c}}
}
\tag{1}
.
\end{align}
Also, inradius $r$, semiperimeter $\rho$, and circumradius $R$
of $\triangle ABC$ can be found as
\begin{align}
r&=r_a+r_b+r_c
\tag{2}
,\\
\rho&=
\sqrt{\frac{(r_a+r_b+r_c)^5}{r_a\,r_b\,r_c}}
\tag{3}
,\\
R&=
\tfrac14\,\frac{r(r-r_a)(r-r_b)(r-r_c)}{r_a r_b r_c}
\\
&=
\tfrac14\,\frac{(r_a+r_b+r_c)(r_a+r_b)(r_b+r_c)(r_c+r_a)}{r_a r_b r_c}
\tag{4}
.
\end{align}
And the side lengths of $\triangle ABC$ can be found explicitly as
\begin{align}
a&=r\,(r_b+r_c)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
,\\
b&=r\,(r_c+r_a)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
,\\
c&=r\,(r_a+r_b)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
.
\end{align}
Related answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 10
} |
The entrance law of a Brownian motion with absorbing boundary In the article "Construction of Diffusion processes with Wentzell's Boundary conditions by means of poisson point processes of Browninan excursions" one reads:
I tried to compute it for $n=1$
Then I guess we have:
\begin{align}
K(t,x) &= \frac{\sqrt{\pi t^3}}{\sqrt{2}} x \exp \bigg(-\frac{x^2}{2t}\bigg) \\
p^0_K(t,x,y) &= \frac{1}{\sqrt{2\pi t}} \Bigg(\exp \bigg(-\frac{x^2-y^2}{2t}\bigg)-\exp \bigg(-\frac{x^2+y^2}{2t}\bigg) \Bigg)
\end{align}
The question is how do we prove that
$$\int_0^\infty K(t,x) p^0_K(s,x,y)\, dx = K(t+s, y)$$
| I doubt that the claimed identity holds true for $n=1$:
By the very definition of $K$ and $p_K^0$, we have
$$\begin{align*} \int_0^{\infty} K(t,x) p_K^0(s,x,y) \, dx = I_1+I_2 \end{align*}$$
where
$$\begin{align*} I_1&:= \frac{1}{2} \sqrt{\frac{t^3}{s}} \exp \left( \frac{y^2}{2s} \right) \int_{0}^{\infty} x \exp \left(- \frac{x^2}{2} \left[ \frac{1}{t}+ \frac{1}{s} \right] \right) \, dx \\ &= - \frac{1}{2} \sqrt{\frac{t^3}{s}} \left[ \frac{1}{t}+\frac{1}{s} \right]^{-1} \exp \left( \frac{y^2}{2s} \right) \int_0^{\infty} \frac{d}{dx} \exp \left(- \frac{x^2}{2} \left[ \frac{1}{t}+ \frac{1}{s} \right] \right) \, dx \\ &= \sqrt{t^5 s} \frac{1}{t+s} \exp \left( \frac{y^2}{2t} \right) \end{align*}$$
and
$$\begin{align*}I_2 &:= - \frac{1}{2} \sqrt{\frac{t^3}{s}} \exp \left( - \frac{y^2}{2t} \right) \int_0^{\infty} x \exp \left(- \frac{x^2}{2} \left[ \frac{1}{t}+ \frac{1}{s} \right] \right) \, dx \\ &= \dots = -\sqrt{t^5 s} \frac{1}{t+s} \exp \left(- \frac{y^2}{2t} \right) \end{align*}$$
Hence,
$$ \int_0^{\infty} K(t,x) p_K^0(s,x,y) \, dx = 2\sqrt{t^5 s} \frac{1}{t+s} \sinh \left( \frac{y^2}{2t} \right) \neq K(t+s,y).$$
If I'm not mistaken, then the density of (one-dimensional) absorbed Brownian motion equals
$$\frac{1}{\sqrt{2\pi t}} \left( \exp \left( - \frac{(x-y)^2}{2t} \right) - \exp \left( - \frac{(x+y)^2}{2t} \right) \right),$$
which does not agree with the given formula for $n=1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
| $${1\over(1-x)^2}={1\over 1-x}\cdot{1\over 1-x}=\sum_{j\geq0} x^j\cdot\sum_{k\geq0}x^k
=\sum_{r\geq0} x^r\left(\sum_{j+k=r}1\right)=\sum_{r\geq0}(r+1)x^r\ .$$
| {
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"source": "stackexchange",
"question_score": "27",
"answer_count": 15,
"answer_id": 5
} |
Solve $\begin{cases} x + y + z = 2 \\ 2xy - z^2 = 4 \\ \end{cases} $ for x, y, z. It came to my mind to rewrite the expression above as
$$\begin{cases}
x + y = 2 - z \\
2xy = (2 - z)^2 + 4z \\
\end{cases}
$$
and see if there any restrictions on the values of the variables occur.
What I can see is that $(2 - z)^2 + 4z \gt 0$ for all $z$ (since $z^2 > -4$) but when I express $y$ in terms of $z$ and $x$
$$\begin{cases}
y = 2 - z - x\\
x(x - (2 - z)) = (2 - z)^2 + 4z \\
\end{cases}
$$
I see that $x(x - (2 - z)) \gt 0$ not for all $z$. This is all I got for now.
The answer is $(2, 2, -2)$.
| You have:
$$
x + y= 2-z\\
xy = 2 + \frac{z^2}{2}.
$$
From Vieta's formulas, $x$ and $y$ are roots of
$$
t^2 - (2-z)t + 2 + \frac{z^2}{2} = 0;
$$
hence
$$
t = \frac{2 - z \pm \sqrt{(2-z)^2 - 8 - 2z^2}}{2} = \frac{2 - z \pm \sqrt{-(z+2)^2}}{2}.
$$
So, we have $z=-2$, and $t=2$; therefore $t=x=y=2$ and $z=-2$.
Greate thanks to @Shailesh for notes.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Evaluate the integral $\int \frac{x}{a+bx^3}\ dx$ How do I solve integral at this form
$\displaystyle\int \frac{x}{a+bx^3}\ dx$ ?
| partial fraction : in general form ,write it like below
and find A,B,C
$$\frac{x}{a+bx^3}= \\ \frac{x}{(\sqrt[3]{a}+\sqrt[3]{b}x)((\sqrt[3]{a})^2+(\sqrt[3]{b}x)^2-(\sqrt[3]{a}\sqrt[3]{b}x))}=\\ \space \\ \frac{A}{(\sqrt[3]{a}+\sqrt[3]{b}x)}+\frac{Bx+C}{((\sqrt[3]{a})^2+(\sqrt[3]{b}x)^2-(\sqrt[3]{a}\sqrt[3]{b}x))} $$ then you will have logarithm part + log or arctan part (depends on a,b)
| {
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Suppose that $A$ be a square, invertible matrix such that $A^{4} = A$. Find all real eigenvalues of $A$. Suppose that $A$ be a square, invertible matrix such that $A^4=A$. Find all real eigenvalues of $A$.
| Since
$A^4 = A, \tag{1}$
we have
$A(A^3 - I) = A^4 - A = 0, \tag{2}$
and since $A$ is invertible we may write
$A^3 - I = I(A^3 - I) = (A^{-1}A)(A^3 - I)$
$= A^{-1}(A(A^3 - I)) = A^{-1}(A^4 - A) = 0; \tag{3}$
now if $\lambda$ is an eigenvalue of $A$ there is a vector $v \ne 0$ such that
$Av = \lambda v, \tag{4}$
whence
$A^2v = A(Av) = A(\lambda v) = \lambda(Av)$
$= \lambda(\lambda v) = \lambda^2 v; \tag{5}$
likewise
$A^3 v = A(A^2 v) = A(\lambda^2 v)$
$= \lambda^2(Av) = \lambda^2 (\lambda v) =\lambda^3 v;.\tag{6}$
combining (6) with (3) we find
$(\lambda^3 - 1)v = \lambda^3 v - v = A^3 v - Iv$
$= (A^3 - I)v = 0, \tag{7}$
and since $v \ne 0$ we conclude that
$\lambda^3 - 1 = 0 \tag{8}$
for any eigenvalue $\lambda$ of $A$; the eigenvalues must then lie among the roots of the polynomial $\lambda^3 - 1$, and since
$(\lambda - 1)(\lambda^2 + \lambda + 1) = \lambda^3 - 1 = 0, \tag{9}$
we see that either $\lambda = 1$ or
$\lambda^2 + \lambda + 1 = 0; \tag{10}$
by the quadratic formula, the roots of (10) are
$\lambda = \dfrac{1}{2}(-1 \pm i\sqrt{3}); \tag{11}$
we thus conclude that the only possible real eigenvalue of $A$ is $1$.
The question remains, is $1$ necessarily an eigenvalue of $A$? The answer is "no"; to see this, observe that
$(A - I)(A^2 + A + I) = A^3 - I = 0; \tag{12}$
if now there exists a vector $y$ such that
$z = (A^2 + A + I)y \ne 0, \tag{13}$
then
$(A - I)z = (A - I)(A^2 + A + I)y$
$= (A^3 - I)y = 0, \tag{14}$
that is,
$Az = z; \tag{15}$
$1$ is in fact an eigenvalue in this case; the other option is
$(A^2 + A + I)y = 0 \tag{16}$
for all $y$, or
$A^2 + A + I = 0; \tag{17}$
certainly such matrices exist; an example is
$A = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix}. \tag{18}$
The final conclusion is thus
$1$ is an eigenvalue of $A$ if and only if $A^2 + A + I \ne 0$, and no other real eigenvalues are possible.
| {
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Proving that $1\cdot3+3\cdot5+5\cdot7+\cdots+(2n-1)(2n+1)={n(4n^2+6n-1) \over 3}$ by induction for $n\geq 1$
Prove using mathematical induction that
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2n-1)(2n+1)= {n(4n^2+6n-1) \over 3}.$$
Step 1: If we assume that the equation is true for a natural number, $n=k$, then we get
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2k-1)(2k+1)= {k(4k^2+6k-1) \over 3}$$
Step 2: When a statement is true for a natural number $n = k$,
then it will also be true for its successor, $n=k+1$. Hence, we have to prove that it is also true for $n=k+1$.
$$1\cdot3+3\cdot5+5\cdot7+\cdots+(2k-1)(2k+1)+(2k+1-1)(2k+1+1) = {k(4k+1^2+6k+1-1) \over 3}$$
I replace the LHS by step 1.
$${k(4k^2+6k-1) \over 3} + 2(k+1-1)(2k+1+1)={k(4k+1^2+6k+1-1) \over 3}$$
Now I need to make LHS equal to RHS.
| For what it's worth, I think Mario's answer is definitely the most elegant, but I know you are trying to prove the relation by induction. Thus, I will outline a proof by induction.
Proof. For $n\geq 1$, let $S(n)$ denote the statement
$$
S(n) : \sum_{i=1}^n(2i-1)(2i+1)=\frac{n(4n^2+6n-1)}{3}.
$$
Base step ($n=1$): $S(1)$ says that $(2-1)(2+1)=3=\frac{4+6-1}{3}$, and this is true.
Induction step ($S(k)\to S(k+1)$): Fix some $k\geq 1$ and assume that
$$
S(k) : \sum_{i=1}^k(2i-1)(2i+1)=\frac{k(4k^2+6k-1)}{3}
$$
holds. To be shown is that
$$
S(k+1) : \sum_{i=1}^{k+1}(2i-1)(2i+1)=\frac{(k+1)[4(k+1)^2+6(k+1)-1]}{3}
$$
follows.
Note: Later in the proof, it may help to first observe the following:
$$
(k+1)[4(k+1)^2+6(k+1)-1]=(k+1)(4k^2+14k+9)=4k^3+18k^2+23k+9\tag{$\dagger$}
$$
Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\sum_{i=1}^{k+1}(2i-1)(2i+1)&= \sum_{i=1}^k(2i-1)(2i+1)+\bigl(2(k+1)-1\bigr)\bigl(2(k+1)+1\bigr)\\[1em]
&= \frac{4(k^2+6k-1)}{3}+(2k+1)(2k+3)\tag{by $S(k)$}\\[1em]
&= \frac{k(4k^2+6k-1)+(12k^2+24k+9)}{3}\\[1em]
&= \frac{4k^3+18k^2+23k+9}{3}\tag{simplify}\\[1em]
&= \frac{(k+1)[4(k+1)^2+6(k+1)-1]}{3},\tag{by $(\dagger)$}
\end{align}
one arrives at the right-hand side of $S(k+1)$, completing the inductive proof.
By mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Cauchy's Integral Formula, Evaluate Evaluate,
$$
\oint_{C}\frac{\cos2z}{z^{2}(z^{2}-z+1)}dz
$$
where $C$ is the circle of radius $2$ centred at the origin. Answer should be in the format $J=A\cos(2z+)+B\cos(2z-)+C$.
Really would appreciate if someone could show me how to do this. I was able to find the singularities but I have never come across when all the singularities are in the unit circle.
The singularities I got were $z=0$, $z=(1\pm i\sqrt3)/2$.
Any help is welcome, Thanks!
| If you want to apply the Cauchy integral formula directly, you'll need to use partial fraction decomposition:
$$
\frac{1}{z^2(z^2-z+1)}=\frac{1}{z^2}\left(z+1-\frac{z^3}{z^2-z+1}\right)
=\frac{1}{z}+\frac{1}{z^2}-\frac{z}{z^2-z+1}
$$
Let $a=e^{\frac{\pi i}{3}}=\frac{1+i\sqrt{3}}{2}$. Then
\begin{align}
\frac{z}{z^2-z+1}&=\frac{1}{3}\left(\frac{3z-(\overline{a}-a)-(a-\overline{a})}{z^2-z+1}\right)\\&=
\frac{1}{3}\left(\frac{2z+2z\Re a-(\overline{a}+\overline{a}^2)-(a+a^2)}{z^2-z+1}\right)\\&=
\frac{1}{3}\left(\frac{2z+z(a+\overline{a})-\overline{a}(1+\overline{a})-a(1+a)}{(z-a)(z-\overline{a})}\right)\\&=
\frac{1}{3}\left(\frac{z(1+a)+z(1+\overline{a})-\overline{a}(1+\overline{a})-a(1+a)}{(z-a)(z-\overline{a})}\right)\\&=
\frac{1}{3}\left(\frac{1+\overline{a}}{z-a}+\frac{1+a}{z-\overline{a}}\right)
\end{align}
Thus the contour integral we are interested in is
$$
\oint_C\cos 2z\left(\frac{1}{z}+\frac{1}{z^2}-\frac{1+\overline{a}}{3(z-a)}-\frac{1+a}{3(z-\overline{a})}\right)dz=I_1+I_2+I_3+I_4
$$
Let $f(z)=\cos 2z$ and $g(z)=\sin 2z$, so that $f(z)=\frac{1}{2}g'(z)$. Note that each of $0,a,\overline{a}$ is in the interior of $C$, which justifies our use of the Cauchy integral formula.
The first integral is
$$
I_1=\oint_C\frac{\cos 2z}{z}dz=\oint_C\frac{f(z)}{z-0}dz=2\pi i f(0)=2\pi i
$$
The second is
$$
I_2=\oint_C\frac{\cos 2z}{z^2}dz=\frac{1}{2}\oint_C\frac{g'(z)}{(z-0)^2}dz=\frac{1}{2}\frac{2\pi i}{1!}g(0)=0
$$
The third is
$$
I_3=-\frac{1+\overline{a}}{3}\oint_C\frac{f(z)}{z-a}dz=-\frac{1+\overline{a}}{3}\cdot 2\pi if(a)
$$
The fourth is
$$
I_4=-\frac{1+a}{3}\oint_C\frac{f(z)}{z-\overline{a}}dz=-\frac{1+a}{3}\cdot 2\pi if(\overline{a})
$$
Finally the total contour integral is
$$
I_1+I_2+I_3+I_4=2\pi i\left(1-\frac{1+\overline{a}}{3}\cos(2a)-\frac{1+a}{3}\cos(2\overline{a})\right)
$$
which you can simplify somewhat if you like.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify Square Root Expression $\sqrt{125} - \sqrt{5}$ $\sqrt{125}-\sqrt5$ simplify it.
I thought it would be $\sqrt {5\cdot5\cdot5}-\sqrt 5$ which would be the square root of 25 which is 5 but it is not.
Can you show how to simplify this?
| \begin{align}
\sqrt{125}-\sqrt{5}&=\sqrt{5^3}-\sqrt{5}\\
&=\sqrt{5^2\cdot 5}-\sqrt{5}\\
&=\sqrt{5^2}\sqrt{5}-\sqrt{5}\\
&=5\sqrt{5}-\sqrt{5}\\
&=4\sqrt{5}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
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} |
Calculus Optimization problem
Help with the problem??
The cost of building the pipeline is $\$3$ million per mile in the water, and $\$4$ million per mile on land. Hence, the cost of the pipeline depends on the location of point $P$. where it meets shore. What would be the most economical route of the pipeline?
| Using the figure in the post, let us define the coordinates of the points : $A (0,2)$ , $P (0,x)$, $B (d,-1)$.
Now, for computing $d$, the square of the distance between points $A$ and $B$ is given by $$D_{AB}^2=(x_A-x_B)^2+(y_A-y_B)^2=(0-d)^2+(2+1)^2=d^2+9=5^2=25$$ which makes $d=4$. Similarly $$D_{AP}^2=(x_A-x_P)^2+(y_A-y_P)^2=(0-x)^2+(2-0)^2=x^2+4$$ $$D_{BP}^2=(x_B-x_P)^2+(y_B-y_P)^2=(4-x)^2+(-1-0)^2=1+(4-x)^2$$ So, the cost function is $$C=3 \sqrt{x^2+4}+4\sqrt{1+(4-x)^2}$$ and this is what we need to minimize.
The derivative of $C$ with respect to $x$ is given by $$C'=\frac{3 x}{\sqrt{x^2+4}}-\frac{4 (4-x)}{\sqrt{1+(4-x)^2}}$$ Uisng $C'=0$, we can then rewrite $$\frac{\sqrt{1+(4-x)^2}}{\sqrt{x^2+4}}=\frac{4 (4-x)}{3x}$$ Squaring, reducing to same denominator and simplifying leads to $$7 x^4-56 x^3+167 x^2-512 x+1024=0$$ which is not pleasant since it is a quartic. Plotting it between $x=0$ and $x=4$ which are our bounds shows that the root is close to $x=3$. So, we can start Newton method using $x_0=3$ and the iterates will be given by $$x_{n+1}=\frac {21 x_n^4-112 x_n^3+167 x_n^2-1024}{28 x_n^3-168 x_n^2+334 x_n-512}$$ that is to say $3.17293$, $3.17846$ , $3.17847$ which is the solution for six significant figures.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find x in $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
Which one satisfies the equation $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
(A)$27$ (B)$32$ (C)$45$ (D)$52$ (E)$63$
let $a = 6+\sqrt x , b=6-\sqrt x$
cube each side
\begin{align}
(\sqrt[3]a + \sqrt[3]b)^3 &= (\sqrt[3]3)^3 \\
(\sqrt[3]{a^2} + 2\sqrt[3]{ab} + \sqrt[3]{b^2})(\sqrt[3]a + \sqrt[3]b) &= 3 \\
\sqrt[3]{a^3} + \sqrt[3]{3a^2b} + \sqrt[3]{3ab^2} + \sqrt[3]{b^3} &= 3 \\
a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3
\end{align}
There's still had cube root, how do I remove it?
| Let $ a = \sqrt[3]{6+\sqrt x} $ and $ b = \sqrt[3]{6-\sqrt x}$, then we have:
$$ \left\{\begin{matrix}
a^3 + b^3 = 12 \\
a + b = \sqrt[3]{3}
\end{matrix}\right. $$
Therefore,
$$ ab = \sqrt[3]{-9}$$
or,
$$ \sqrt[3]{36 - x} = \sqrt[3]{-9} $$
then,
$$ x = 45$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Find the maximum value of the fraction
Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b}<\frac{3}{2}$. The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
Trial and error makes the job very easy, but it isn't rigorous.
I used factoring:
$$= \frac{(ab + 1)(a^2b^2 - ab + 1)}{(a+b)(a^2 - ab + b^2)} < \frac{3}{2} \cdot \bigg( \frac{a^2b^2 -ab + 1}{a^2 -ab + b^2} \bigg)$$
But that doesnt get you anywhere either.
Hints only please!
| Without loss of generality, we assume that $a\leq b$. Since $a=2\left(\frac{ab}{2b}\right)\leq2\left(\frac{ab}{a+b}\right)< 2\left(\frac{ab+1}{a+b}\right)<3$, we have $a=1$ or $a=2$. If $a=1$, then $b$ can be any natural number and $\frac{a^3b^3+1}{a^3+b^3}=1$. If $a=2$, then $\frac{2b+1}{b+2}=\frac{ab+1}{a+b}<\frac{3}{2}$ gives $4b+2<3b+6$, or $b<4$. The rest is only checking with $(a,b)\in\big\{(2,2),(2,3)\big\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\iint dy\,dx;\frac{\pi}{4}\leq\theta \leq\frac{3\pi}{4};0\leq r\leq2$ I need to evaluate
$\displaystyle\iint \color{red}{dydx}\;\;\;,\frac{\pi}{4}\leq\theta \leq\frac{3\pi}{4}\;\;\;\;,0\leq r\leq2$
$\color{blue}{\text{without using polar coordinates}}$.
My attempt:
$$\int_{x=-2}^{x=0}\int_{y=-x}^{\sqrt{4-x^2}}dydx+\int_{x=0}^{x=2}\int_{y=x}^{y=\sqrt{4-x^2}}dydx$$
Is the correct?
| At $\theta=\pi/4$ and $y<\sqrt{2}$, $x=y$. Similarly, at $\theta=3\pi/4$ and $y<\sqrt{2}$, $x=-y$.
At $y\ge\sqrt{2}$, $-\sqrt{4-y^2}\le x\le\sqrt{4-y^2}$.
So you can write your integral as
$$\int_0^{\sqrt{2}}\int_{-y}^ydxdy + \int_{\sqrt{2}}^2\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}dxdy=\int_0^{\sqrt{2}}2ydy+\int_{\sqrt{2}}^22\sqrt{4-y^2}dy=2+\pi-2=\pi.$$
EDIT (in response to comment): If you wish to evaluate the integral with respect to $y$ first... well, $x$ goes from $-\sqrt{2}$ to $\sqrt{2}$. For any given $x$, the lower bound on $y$ is $x$; the upper bound, $\sqrt{4-x^2}$. So then, your integral reads
$$\int_{-\sqrt{2}}^{\sqrt{2}}\int_x^{\sqrt{4-x^2}}dydx=\int_{-\sqrt{2}}^{\sqrt{2}}\sqrt{4-x^2}-x~dx=2+\pi-2=\pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value of trigonometric function The minimum value of the expression $\left|\sin x+\cos x+\tan x+\cot x+\sec x+\mathrm{cosec} x\right|$ can be expressed as $(\sqrt a-\sqrt b)$ where a and b are natural number then find the value of $(a-b)?$
My attempt:I applied AM-GM inequality (but AM GM inequality can be applied when all quantities are positive and all trigonometric ratios are not positive except when in first quadrant.)
$$\frac{\sin x+\cos x+\tan x+\cot x+\sec x+\mathrm{cosec} x}{6}\geq\sqrt[6]{\sin x.\cos x.\tan x.\cot x.\sec x.\mathrm{cosec} x}$$
$\sin x+\cos x+\tan x+\cot x+\sec x+\mathrm{cosec} x\geq6 $
but this is the minimum value of $\sin x+\cos x+\tan x+\cot x+\sec x+\mathrm{cosec}x$ ,not minimum value of $\left|\sin x+\cos x+\tan x+\cot x+\sec x+\mathrm{cosec} x\right|$ and it not expressible as $\sqrt a -\sqrt b$
Someone help me find $a$ and $b$.Thanks in advance..
| Let $y=x+\frac{\pi}{4}$. Then:
$$\begin{eqnarray*} f(x)&=&\sin x+\cos x+\tan x+\cot x+\sec x+\text{cosec } x\\&=&\sqrt{2}\,\sin\left(x+\frac{\pi}{4}\right)+\frac{2+2\sqrt{2}\,\sin\left(x+\frac{\pi}{4}\right)}{\sin(2x)}\\&=&\sqrt{2}\,\sin(y)-\frac{2+2\sqrt{2}\sin(y)}{\cos(2y)}\end{eqnarray*} $$
hence by setting $u=\sqrt{2}\sin y$ we just need to find the stationary points of:
$$ g(u) = u-\frac{2+2u}{1-u^2} = u-\frac{2}{1-u}$$
over $[-\sqrt{2},\sqrt{2}]$. $g'(u)$ vanishes at $u=1-\sqrt{2}$, and since:
$$ g(1-\sqrt{2})=1-2\sqrt{2} $$
it follows that:
$$ \min_{x\in\mathbb{R}}\left|f(x)\right| = 2\sqrt{2}-1 = \sqrt{8}-\sqrt{1} $$
hence the answer is $\color{red}{7}$. I bet it is a problem from Brilliant, am I right?
| {
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"timestamp": "2023-03-29T00:00:00",
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polynomial factorization when exponent is not given How can I factorise this equation, given i already know some of its factors which are: $(a-b)(b-c)(c-a).$
Equation is :
$$a^nb^{n-1} + a^{n-1}c^n - a^nc^{n-1} - a^{n-1}b^n - b^{n-1}c^n + b^nc^{n-1}$$
| Denote the above polynomial by $f(a,b,c)$. Then $f(a,b,c)=0$ for $a=b$ or $a=c$ or $b=c$. Hence we can write
$$
f(a,b,c)=g(a,b,c)(a-b)(a-c)(b-c).
$$
For $n=2$ we have $g(a,b,c)=1$. For $n\ge 3$ the homogeneous polynomial $g$ is given by
$$
g(a,b,c)=a^{n-2}b^{n-2}+a^{n-2}b^{n-3}c+a^{n-2}b^{n-4}c^2+\ldots +b^{n-2}c^{n-2},
$$
where all monomials $a^ib^jc^k$ appear with $i+j+k=2n-4$. For example, for $n=3$ we have $g(a,b,c)=ab+ac+bc$, and for $n=4$ we have $$g(a,b,c)=a^2b^2+a^2bc+a^2c^2+ab^2c+abc^2+b^2c^2.$$
However, in general the polynomial $g(a,b,c)$ has no further nontrivial factors.
| {
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"timestamp": "2023-03-29T00:00:00",
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how to solve $\displaystyle \frac{a^3}{a^2+2b+c}+\frac{b^3}{b^2+2c+a}+\frac{c^3}{c^2+2a+b}\geq\frac{3}{4}$
$a,b,c>0,a+b+c=3,$
prove that:
$$\frac{a^3}{a^2+2b+c}+\frac{b^3}{b^2+2c+a}+\frac{c^3}{c^2+2a+b}\geq\frac{3}{4}$$
| Starting from the Cauchy-Schwarz inequality:
$$(x_1^2+y_1^2+z_1^2)( x_2^2+y_2^2+z_2^2)\geqslant (x_1x_2+y_1y_2+z_1z_2)^2;x_i,y_i\in R$$
Setting
$$x_1^2=a^2+2b+c, y_1^2=b^2+2c+a, z_1^2=c^2+2a+b $$
$$x_2^2=\frac{a^3}{a^2+2b+c}, y_2^2=\frac{b^3}{b^2+2c+a}, z_2^2=\frac{c^3}{c^2+2a+b} $$
Which gives
$$(a^2+b^2+c^2+3(a+b+c))( \frac{a^3}{a^2+2b+c}+\frac{b^3}{b^2+2c+a}+\frac{c^3}{c^2+2a+b}))\geqslant (a^{\frac{3}{2}}+ b^{\frac{3}{2}}+ c^{\frac{3}{2}})^2$$
Therefore, if the following inequality holds, the original problem is solved.
$$\frac{(a^{\frac{3}{2}}+ b^{\frac{3}{2}}+ c^{\frac{3}{2}})^2}{ a^2+b^2+c^2+3(a+b+c)}\geqslant\frac{3}{4} ;\quad s.t\quad a+b+c=3$$
Since $a,b,c>0$ we can use the following change of variables;
$$u=a^\frac{1}{2}, v=b^\frac{1}{2}, w=c^\frac{1}{2}$$
$$\Rightarrow\frac{(u^3+v^3+w^3)^2}{u^4+v^4+w^4+3(u^2+v^2+w^2)}\geqslant\frac{3}{4}$$
$$\quad s.t\quad u^2+v^2+w^2=3$$
In order to prove the above inequality, it suffices to show that the optimum value of the following optimization problem is zero;
$$minf(u,v,w)=4(u^3+v^3+w^3)^2-3(u^4+v^4+w^4+3(u^2+v^2+w^2))$$
$$\quad s.t\quad u,v,w>0; u^2+v^2+w^2=3$$
Using the Lagrange multiplier method, the solution is
$$u=v=w=1,f_{min}=0$$
Therefore
$$f(u,v,w)\geqslant0\Rightarrow\frac{a^3}{a^2+2b+c}+\frac{b^3}{b^2+2c+a}+\frac{c^3}{c^2+2a+b}\geqslant\frac{(a^{\frac{3}{2}}+ b^{\frac{3}{2}}+ c^{\frac{3}{2}})^2}{ a^2+b^2+c^2+3(a+b+c)}\geqslant\frac{3}{4}$$
Q.E.D.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $ f: R \to [\frac{1}{2} , 1]$ and $f(x+2) = \frac{1}{2} +\sqrt{f(x) -f(x)^2}$ Problem :
Let $ f: R \to [\frac{1}{2} , 1]$ and $f(x+2) = \frac{1}{2} +\sqrt{f(x) -f(x)^2}$
Then which of the following is always true
$(a) f(2) = f(7)$
$(b) f(4) = f(10) $
$(c) f(2) =f(4) $
Please suggest how to proceed in this problem , will be of great help please guide thanks
| We first note that the function that takes us from $f(x)$ to $f(x+2)$, namely
$$g(x)=\frac 12+\sqrt{x-x^2}$$
is the upper-right quadrant of an circle, and if its domain is $\left[\frac 12,1\right]$ then so is its range. We can see that $g(x)$ is a decreasing function, a bijection from $\left[\frac 12,1\right]$ to itself. (These statements are easily confirmed if you know conic sections, or are also confirmed by graphing.)
The important point is that if $f(x)$ is defined, so is $f(x+2)$. That is also the only restriction on $f$, so we can choose any values at all for a domain $[a,a+2)$ and then the values of $f$ are then defined in a sort-of-periodic way for all real numbers.
Choice (a) cannot always be true, because we can define $f(2)$ and $f(3)$ any way we want, then $f(3)$ alone determines $f(3+2)$ and $f(3+2+2)$. So there is no relationship between $f(2)$ and $f(7)$.
We can check (c) with the formula:
$$f(4)=f(2+2)=\frac 12+\sqrt{f(2)-f(2)^2}$$
If we let $f(2)=1$ then $f(4)=\frac 12+\sqrt{1-1^2}=\frac 12$, so (c) is certainly not always true.
Now we'll set $f(4)=\frac 12$, then we get
$$f(6)=f(4+2)=\frac 12+\sqrt{\frac 12-\left(\frac 12\right)^2}=1$$
$$f(8)=f(6+2)=\frac 12+\sqrt{1-1^2}=\frac 12$$
$$f(10)=f(8+2)=\frac 12+\sqrt{\frac 12-\left(\frac 12\right)^2}=1$$
So here $f(4)\ne f(10)$, and (b) is not always true.
Interestingly, if we do the calculations we will see that $g(g(x))=x$. We can also see that from the graph of $g(x)$, which is symmetric with the line $y=x$ and is therefore its own inverse. So we can say that
$$f(x+4)=f(x+2+2)=g(f(x+2))=g(g(f(x)))=f(x)$$
So we do get that $f(2)=f(6)=f(10)$ and $f(4)=f(8)$. But your question does not ask about those.
| {
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Expressing the integral in terms of the original variable In evaluating the integral:
$$ \int{dx\over(a^2-x^2)^{3/2}} $$ or $$ \int{dx\over(a^2-x^2)^{1/2}\ (a^2-x^2)}$$
Let $ x=a\sin\theta $ and $ dx=a\cos\theta\ d\theta $. Then
$$ \int{{a\cos\theta\ d\theta}\over{a\cos\theta\ (a^2-a^2\sin^2\theta)}} = {1\over a^2}\int {{d\theta}\over{1-\sin^2\theta}} = {1\over a^2}\int {{d\theta}\over{\cos^2\theta}} = {1\over a^2}\int \sec^2\theta\ d\theta $$
$$ ={{\tan \theta} \over {a^2}}$$
My question has to do with the method of expressing the transformed variable $\theta$ in terms of the original variable $x$. Taking the initial substitution $x=a\sin\theta$, I am able to derive the following:
$$ \theta=\arcsin{x \over a}$$
And substitute into the answer in the following manner:
$$ \tan(\arcsin {x\over a})\over a^2$$
However, the book expresses the integral as such:
$$ {1 \over a^2}{{x} \over {\sqrt {a^2-x^2}}} + C$$
Although the term $1 \over a^2$ remains the same, how do I express the trigonometric functions in this particular form?
| Ultimately you want to re-express your answer $\frac{\tan\theta}{a^2}$ in terms of $\sin\theta$ and then convert back using the substitution you already did. Observe that $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and that $\cos\theta = \sqrt{1 - \sin^2\theta}$. Then
$$
\tan\theta \;\; =\;\; \frac{\sin\theta}{\cos\theta} \;\; =\;\; \frac{\sin\theta}{\sqrt{1 - \sin^2\theta}} \;\; =\;\; \frac{1}{a} \frac{x}{\sqrt{1 - \frac{x^2}{a^2}}} \;\; =\;\; \frac{x}{\sqrt{a^2 - x^2}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Expressing polynomial as linear combinaion I found these questions in Adams Introduction to Groebner bases
Let $f=x^6-1$ and $g=x^4+2x^3+2x^2-2x-3$. Let $I=\langle f,g\rangle$. Calculate the polynomial that generates $I$ alone. After a couple of polynomial divisions I came to the answer $h=x^2-1$.
Is it correct?
Furthermore, is $x^5+x^3+x^2-7\in I$? I found that it is not, is that result correct?
Furthermore show that $p:=x^4+2x^2-3\in I$. Again reduction by $h$ shows that it is. But then it goes like: write $p$ as a linear combination of $f$ and $g$. Is this done by the "usual backward Euclidean algorithm"? I mean expressing the remainders and "going backwards"? Is there another method?
Thanks and sorry for the lack of fancy characters, I am in Indonesia and have a hard time with the keyboard.
| According to WolframAlpha the gcd between $f$ and $g$ is indeed $x^2 - 1 = (x+1)(x-1)$. The easiest way to see this by hand is to use the Euclidean algorithm. Namely, by polynomial long division we have
$$
\begin{gather}
x^6-1 = (x^2-2 x+2) (\color{green}{x^4+2x^3+2x^2-2x-3}) + \color{red}{2 x^3-5 x^2-2 x+5} \\
\color{green}{x^4+2x^3+2x^2-2x-3} = \left(\frac{x}{2}+\frac{9}{4}\right) (\color{red}{2 x^3-5 x^2-2 x+5}) + \frac{57}{4} x^2 - \frac{57}{4}
\end{gather}
$$
Furthermore, substituting one equation into the other and multiplying by $\frac{4}{57}$ we get
$$
\begin{align}
x^2-1 &= \frac{1}{57} (2 x^3+5 x^2-14 x+22) (x^4+2x^3+2x^2-2x-3) \\
&-\,\frac{1}{57} (2 x+9) (x^6-1) \label{eq:bezout} \tag{1}
\end{align}
$$
which will be useful later.
You can easily see that $x^5 + x^3 + x^2 - 7 \notin I$ because its degree isn't even, but every polynomial in $I = \langle x^2-1 \rangle$ has even degree.
Now, note that $x^4+2x^2−3$ is a quadratic polynomial in $x^2$ and by the usual "sum and product" rule you can see that it factors as $(x^2+3)(x^2-1)$, hence $x^4+2x^2−3 \in I$. Finally, multiplying \eqref{eq:bezout} by $x^2+3$ it immediately follows that
$$
\begin{align}
x^4+2x^2−3 &= \frac{1}{57} (2 x^5+5 x^4-8 x^3+37 x^2-42 x+66) (x^4+2x^3+2x^2-2x-3) \\
&-\,\frac{1}{57} (2 x^3+9 x^2+6 x+27) (x^6-1)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Harmonic progression based question
If $a,b,c,d$ are distinct positive numbers in harmonic progression, then
(A) $a+b>c+d$
(B) $a+c>b+d$
(C) $a+d>b+c$
(D) none of these
I tried $\frac{1}{a},\frac{1}{b},\frac{1}{c},\frac{1}{d}$ are in AP so either $\frac{1}{a}>\frac{1}{b}>\frac{1}{c}>\frac{1}{d}$ or $\frac{1}{a}<\frac{1}{b}<\frac{1}{c}<\frac{1}{d}$.That means either $a>b>c>d$ or $a<b<c<d$.But i could not judge correct answer based on this.Is my approach not correct?What is correct answer?Can someone help me in solving this question?
| $$a=\dfrac1{x-3y},b=\dfrac1{x-y},c=\dfrac1{x+y},d=\dfrac1{x+3y}$$
$\implies a+d=\dfrac{2x}{x^2-9y^2}$ and $b+c=\dfrac{2x}{x^2-y^2}$
$\implies a+d>b+c$ as $9y^2>y^2$
| {
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"url": "https://math.stackexchange.com/questions/1384013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a point using complex geometry
In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$. Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$. Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$, where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$.
I have seen solutions as follows:
Consider: $A = 1 + 0i$ and $B = 2 + i2\sqrt{3}$
Obviously inside a triangle each angle is $60$ degrees.
Really we need to consider $B$ since that is the angle of the ray anyway.
So: $\theta = \arctan(2/2 = 1) = \frac{\pi}{4}$.
The third point must then be $\theta_2 = \frac{\pi}{4} + \frac{\pi}{3} = \frac{7\pi}{12}$
But this cant be since it isnt in the first quadrant.
$\frac{\pi}{4} - \frac{\pi}{3} = -\frac{\pi}{12}$.
but that isnt the first quadrant either.
| The point $P(X,Y)$ such that $\triangle{ABX}$ is an equilateral triangle satisfies
$$X+Yi-(1+0i)=(2+2\sqrt 3i-(1+0i))(\cos(\pm 60^\circ)+i\sin(\pm 60^\circ))$$
$$\Rightarrow (X,Y)=(-3/2,3\sqrt 3/2),(9/2,\sqrt 3/2).$$
So, $C(9/2,\sqrt 3/2)$ because $C$ lies in the first quadrant. Hence, the center of $\triangle{ABC}$ is
$$\left(\frac{1+2+(9/2)}{3},\frac{0+2\sqrt 3+(\sqrt 3/2)}{3}\right)=\left(\frac{5}{2},\frac{5}{6}\sqrt 3\right)$$
Thus,
$$\frac{5}{2}\cdot\frac{5}{6}\sqrt 3=\frac{25}{12}\sqrt 3\Rightarrow p+q+r=25+3+12=40.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How can I find an ODE equation from $dy/dx$ What is the ODE satisfied by $y=y(x)$
given that $$\frac{dy}{dx} = \frac{-x-2y}{y-2x}$$
I understand that I need to get it in some form of $\int \cdots \;dy = \int \cdots \; dx$, but am not sure how to go about it.
| \begin{align}
\frac{dy}{dx} = \frac{ -x-2y }{ y-2x } \\
\frac{ dy }{ dx } =\frac{ -1-2\frac{ y }{ x } }{ \frac{ y }{ x } -2 } \\ \frac{ y }{ x } = t \to \quad y=xt \quad \Rightarrow \quad dy=t+x \frac{ dt }{ dx } \\
t+x\frac{ dt }{ dx } =\frac{ -1-2t }{ t-2 } \\
x\frac{ dt }{ dx } =\frac{ -1-2t }{ t-2 } -t=\frac{ -1-{ t }^{ 2 } }{ t-2 } \\
\int { \frac{ 2-t }{ t^{ 2 }+1 } dt } =\int { \frac{ dx }{ x } } \\
\int { \left( \frac{ 2 }{ { t }^{ 2 }+1 } -\frac{ t }{ { t }^{ 2 }+1 } \right) dt= } \int { \frac { dx }{ x } } \\ 2\int { \frac { 1 }{ { t }^{ 2 }+1 } dt-\frac { 1 }{ 2 } \int { \frac { d\left( { t }^{ 2 }+1 \right) }{ { t }^{ 2 }+1 } } =\int { \frac { dx }{ x } } } \\ 2\arctan { \left( t \right) -\frac { 1 }{ 2 } } \ln { \left( 1+{ t }^{ 2 } \right) =\ln { \left| x \right| +C } } \\ \arctan { \left( t \right) =\frac { 1 }{ 2 } \ln { \left| x\sqrt { 1+{ t }^{ 2 } } \right| +C } } \\ \arctan { \left( \frac { y }{ x } \right) =\frac { 1 }{ 2 } \ln { \left| x\sqrt { 1+{ \left( \frac { y }{ x } \right) }^{ 2 } } \right| +C } }
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A puzzling step in a solution for "Find $\sin(x)$ and $\cos(x)$, if $a\sin(x)+b\cos(x)=c$" A textbook problem:
Find $\sin(x)$ and $\cos(x)$, if $a\sin(x)+b\cos(x)=c$
The solution from the textbook:
Let's divide each term of this equation by $\sqrt{a^2+b^2}$:
$$\frac{a}{\sqrt{a^2+b^2}}\sin(x)+\frac{b}{\sqrt{a^2+b^2}}\cos(x)=\frac{c}{\sqrt{a^2+b^2}}$$
Since the sum of squares of $\frac{a}{\sqrt{a^2+b^2}}$ and $\frac{b}{\sqrt{a^2+b^2}}$ equals 1, then there will always exist an angle, let's call it $\phi$, for which
$$\sin(\phi)=\frac{a}{\sqrt{a^2+b^2}};$$
$$\cos(\phi)=\frac{b}{\sqrt{a^2+b^2}}$$
Using this, let's transform our equation to
$$\sin(\phi)sin(x)+cos(\phi)cos(x)=\frac{c}{\sqrt{a^2+b^2}}$$
This will bring us to
$$\cos(x-\phi)=\frac{c}{\sqrt{a^2+b^2}}$$
This I understand. But the next step is:
It is evident from this that
$$\sin(x-\phi)=\pm\frac{\sqrt{a^2+b^2-c^2}}{\sqrt{a^2+b^2}}$$
Could you give me a hint regarding this last transformation?
| Let's say you posted like this:
~~~~~
for which
$$\cos(\phi)=\frac{a}{\sqrt{a^2+b^2}};$$
$$\sin(\phi)=\frac{b}{\sqrt{a^2+b^2}}$$
Using this, let's transform our equation to
$$\cos(\phi)\sin(x)+\sin(\phi)\cos(x)=\frac{c}{\sqrt{a^2+b^2}}$$
This will bring us to
$$ \sin(x+\phi)=\frac{c}{\sqrt{a^2+b^2}}$$
This I understand. But the next step is:
It is evident from this that
$$\cos(x+\phi)=\pm\frac{\sqrt{a^2+b^2-c^2}}{\sqrt{a^2+b^2}}$$
Could you give me a hint regarding this last transformation?
~~~~
You can see not immediately catching the $\sin, \cos $ Pythagorean relationship.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Finding fomulas for hyperbolic functions I'm trying to find formulas for hyperbolic functions, starting with this image
Knowing that the area between the origin, vertex and a point on hyperbola (enclosed by x-axis and hyperbola itself) is equal to $\frac{\alpha}{2}$, I wrote
$$
x^2 - y^2 = 1
$$
$$
x = \sqrt{1 + y^2}
$$
... and started integrating this expression for $x$.
$$
\begin{align}
\frac{\alpha}{2}
& = \int^{y}_0{\sqrt{1 + y^2}\,\textrm{d}y}
\\ & = \int^{\arctan y}_0{\sqrt{1 + \tan^2\theta}\sec^2\theta\,\textrm{d}\theta} && \text{let }y = \tan\theta \text{ and }\textrm{d}y = \sec^2\theta\,\textrm{d}\theta
\\ & = \int^{\arctan y}_0{\sec\theta\sec^2\theta\,\textrm{d}\theta}
\end{align}
$$
Here I used integration by parts to find $\int{sec^3\theta\,\textrm{d}\theta}$
$$
\begin{align}
\int{\sec^3\theta\,\textrm{d}\theta}
& = \sec\theta\tan\theta - \int{\sec\theta\tan^2\theta \,\textrm{d}\theta}
\\ & = \sec\theta\tan\theta - \int{\sec\theta(\sec^2\theta - 1) \,\textrm{d}\theta}
\\ & = \sec\theta\tan\theta - \int{\sec^3\theta \,\textrm{d}\theta} + \int{\sec\theta \,\textrm{d}\theta}
\\ & = \frac{1}{2}\bigg(\sec\theta\tan\theta + \int{\sec\theta \,\textrm{d}\theta}\bigg)
\\ & = \frac{1}{2}\bigg(\sec\theta\tan\theta + \ln{\bigg|\sec\theta + \tan\theta \,\bigg|}\bigg)
\end{align}
$$
... and by substituting $\tan\theta = y$, $\sec\theta = \sqrt{1 + y^2}$ back I got
$$
\alpha = y\sqrt{1 + y^2} + \ln{\bigg|\sqrt{1 + y^2} + y \,\bigg|}
$$
Which doesn't make much sense, because this is not what an inverse hyperbolic sine looks like. Where have I made mistake? What should I do different?
Thanks.
| The integral you have calculated $\int_0^y\sqrt{1+y^2}dy$ does not represent the red area alone, but includes the area of the right angled triangle immediately above it. Therefore you need to subtract $\frac 12 xy$ and then you will get the answer you expected
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $(1 + \frac{1}{n})^n < n$ for natural numbers with $n \geq 3$. Prove with induction on $n$ that \begin{align*} \Bigl(1+ \frac{1}{n}\Bigr)^n < n \end{align*} for natural numbers $n \geq 3$.
Attempt at proof: Basic step. This can be verified easily.
Induction step. Suppose the assertion holds for $n >3$, then we now prove it for $n+1$. We want to prove that \begin{align*} \big( 1+ \frac{1}{n+1})^{n+1} < n+1. \end{align*} So we have \begin{align*} \big( 1+ \frac{1}{n+1}\big)^{n+1} &= \big( 1+ \frac{1}{n+1} \big)^n \cdot \big( 1 + \frac{1}{n+1} \big) \\ & < \big (1 + \frac{1}{n} \big)^n \cdot \big( 1 + \frac{1}{n+1} \big) \\ & = n \cdot \big( 1 + \frac{1}{n+1} \big) \qquad \text{(Induction hypothesis)} \\ &= n \cdot \big( \frac{n+1+1}{n+1} \big) \\ &= \frac{n^2 + 2n}{n+1} \\ &= \frac{(n+1)^2 -1}{(n+1)}
\end{align*}
And now I'm stuck. I don't know how to get $n+1$ on the RHS. Please help!
| First: When using the induction hypothesis it should read as $$\left(1+\frac{1}{n}\right)^n\cdot\left(1+\frac{1}{n+1}\right)\color{red}<\color{black}n\cdot\left(1+\frac{1}{n+1}\right)$$
Then: $$(n+1)^2-1<(n+1)^2,$$ so we get...can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Vectors as linear combinations I want to express v=$(\sqrt2,1/2)$ as linear combination of p=(1,2) and q=(2,1).
My answer:
$(\sqrt2,1/2)$=c(1,2)+d(2,1). So solve for c and d:
$\begin{pmatrix} \sqrt2 \\ 1/2 \end {pmatrix}=\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} c \\ d\end{pmatrix}$
$\begin{pmatrix} c \\ d \end {pmatrix}=\begin{pmatrix} -\sqrt2/3 +1/3 \\ 2\sqrt2/3+1/6 \end{pmatrix}$
Is this right? It doesn't feel right for some reason.
| We have
$$
\binom{c}{d}=\begin{pmatrix} 1 & 2 \\ 2 & 1\end{pmatrix}^{-1}\binom{\sqrt{2}}{1/2}=\frac{1}{3}\begin{pmatrix} -1 & 2 \\ 2 & -1\end{pmatrix}\binom{\sqrt{2}}{1/2}=\frac{1}{3}\binom{1-\sqrt{2}}{(1-4\sqrt{2})/2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiate $y=a(x+b)^2 -8$, then find $a$ and $b$ if the parabola: a) passes through the origin with the gradient $16$
b) has tangent $y=2x$ at the point $P =(4,8)$
for a) $y'= 2a(x+b)$ then $16=2a(0 +b)$
I'm not sure what to do after this or if it's even right?
| Notice, we have $$y=a(x+b)^2-8$$ $$\frac{dy}{dx}=2a(x+b)$$
a) Since, the parabola: $y=a(x+b)^2-8$ passes through the origin hence, substituting the coordinates of $(0, 0)$ in the equation, we get
$$0=a(0+b)^2-8\iff ab^2=8 $$ $$ ab=\frac{8}{b}\tag 1$$
Now, gradient at the origin $(0, 0)$ is $16$, hence we have
$$ \frac{dy}{dx}=2a(0+b)=2ab$$$$\iff 2ab=16\iff ab=8\tag 2$$ Now, equating (1) & (2), we get $$\frac{8}{b}=8\iff b=1$$ $$\implies a=\frac{8}{a}=\frac{8}{1}=8$$ Hence, we have
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{a=8,\ b=1}}$$
b) Since, the parabola: $y=a(x+b)^2-8$ passes through the point $(4, 8)$ hence, substituting the coordinates of $(4, 8)$ in the equation of parabola, we get
$$8=a(4+b)^2-8\iff a(4+b)^2=16 \tag 3$$
Now, gradient at the origin $(4, 8)$ will be equal to the slope of line: $y=2x$
$$\implies \frac{dy}{dx}=2a(4+b)=8a+2ab$$$$\iff 8a+2ab=2$$$$\iff a(4+b)=1$$ $$\iff a=\frac{1}{4+b}\tag 4$$
From (3) & (4), we get $$\frac{1}{4+b}(4+b)^2=16 $$$$\iff 4+b=16\iff b=16-4=12$$
Now, $$a=\frac{1}{4+b}=\frac{1}{4+12}=\frac{1}{16}$$
Hence, $$\bbox[5px, border:2px solid #C0A000]{\color{red}{a=\frac{1}{16},\ b=12}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Question about congruence modulo n Is there any sort of algorithm to calculate the remainder when $10^{515}$ is divided by $7$?
Could the same algorithm be applied to finding the remainder of $8^{391}$ divided by $5$?
Is it even necessary to use congruence here?
| By Fermat's Little Theorem you know that
$$10^6\equiv 1 \mod 7$$
because $7$ is not a divisor of $10$. Now, since
$$10^{515}=10^{6\cdot 85+5}=(10^6)^{85}\cdot 10^5,$$
you get
$$
10^{515}\equiv
(10^6)^{85}\cdot 10^5\equiv
1^{85}\cdot 10^5\equiv
10^5\pmod 7.
$$
Finally, dividing $10^5=100000$ by $7$ the quotient is $14285$ and the remainder is $5$, so $5$ is the solution.
Same thing for $8^{391}\mod 5$: this time $8^4\equiv 1\mod 5$ because $5$ is not a divisor of $8$, then $8^{391}=8^{4\cdot 97+3}$ and so on.
| {
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"timestamp": "2023-03-29T00:00:00",
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Different approaches for evaluating a series I was playing around with the Fourier series of $\cos x , \; x \in (0, \frac{\pi}{2})$ and a series came up. First of all the fourier series of $\cos x$ at the given interval is:
$$\cos x =\frac{8}{\pi} \sum_{n=1}^{\infty} \frac{n \sin 2nx}{4n^2-1}$$
Calculations are pretty straightforward. By applying Parseval we get that:
$$\sum_{n=1}^{\infty} \frac{n^2}{(4n^2-1)^2}= \frac{\pi^2}{64}$$
What other approaches can we use in order to evaluate this series? Perhaps partial decomposition? One notes that:
$$\frac{n^2}{\left ( 4n^2-1 \right )^2}= \frac{1}{16}\left [ \frac{1}{\left ( 2n+1 \right )^2}- \frac{1}{(2n-1)^2} \right ]+ \frac{1}{16}\left [ \frac{1}{2n-1} - \frac{1}{2n+1} \right ]$$
Hence:
$$\sum_{n=1}^{\infty}\frac{n^2}{\left ( 4n^2-1 \right )^2}= \frac{1}{16}\sum_{n=1}^{\infty}\left [ \frac{1}{\left ( 2n+1 \right )^2}- \frac{1}{(2n-1)^2} \right ]+ \frac{1}{16}\sum_{n=1}^{\infty}\left [ \frac{1}{2n-1} - \frac{1}{2n+1} \right ]$$
However, we all know where the first two sums evaluate to. They are quite famous:
$\displaystyle {\color{gray} \bullet} \;\;\; \sum_{n=1}^{\infty}\frac{1}{(2n+1)^2}= \frac{\pi^2}{8}- 1$
$\displaystyle {\color{gray} \bullet} \;\;\; \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}= \frac{\pi^2}{8}$
Hence the first bracket is evaluated to $-\frac{1}{16}$. The second bracket, is easily noted that it can be expressed in terms of digamma. However, I am having a difficult time converting to the digamma.
Any help on that point would be nice. Also I am interest in other approaches, such as contour integration.
| There's a slight mistake in your decomposition; it should be:
$$\frac{n^2}{\left ( 4n^2-1 \right )^2}= \frac{1}{16}\left [ \frac{1}{\left ( 2n+1 \right )^2} \color{red}{+} \frac{1}{(2n-1)^2} \right ]+ \frac{1}{16}\left [ \frac{1}{2n-1} - \frac{1}{2n+1} \right ]$$
Note that:
$$\sum_{k=1}^n \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right) = -\sum_{k=1}^n (a_{k} - a_{k-1}) = -(a_n - a_0) = a_0 - a_n = 1 - \frac{1}{2n+1} \to 1$$
Where $a_k = \frac{1}{2k+1}, k=0,1,2,...$
| {
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Can someone show me HOW to do this, I don't just want the answer $4n$ to the power of $3$ over $2 = 8$ to the power of negative $1$ over $3$
Written Differently for Clarity:
$$(4n)^\frac{3}{2} = (8)^{-\frac{1}{3}}$$
EDIT
Actually, the problem should be solving $4n^{\frac{3}{2}} = 8^{-\frac{1}{3}}$. Another user edited this question for clarity, but they edited it incorrectly to add parentheses around the right hand side, as can be seen above.
| So the real problem we are trying to complete is to solve $4n^{\frac{3}{2}} = 8^{-\frac{1}{3}}$.
The way to do this is to first divide both sides by $4$ and get:
$n^{\frac{3}{2}} = \dfrac{8^{-\frac{1}{3}}}{4}$
Now, since $8 = 2^{3}$ and $4 = 2^{2}$, we can rewrite this as:
$n^{\frac{3}{2}} = \dfrac{(2^{3})^{-\frac{1}{3}}}{2^{2}}$
and in the numerator, since we have something with a power raised to another power, we can multiply the two powers to get:
$n^{\frac{3}{2}} = \dfrac{2^{3(-\frac{1}{3})}}{2^{2}}$
which gives
$n^{\frac{3}{2}} = \dfrac{2^{-1}}{2^{2}}$
Now, raising something to a negative exponent means you move it through the fraction either up or down depending on where it started. So since $2^{-1}$ is in the numerator, we move it down to the denominator and remove the negative in the exponent, to get:
$n^{\frac{3}{2}} = \dfrac{1}{2^{1}2^{2}}$
Since we are multiplying two things raised to exponents with the same base in the denominator (the base is $2$), we can just add the exponents to get:
$n^{\frac{3}{2}} = \dfrac{1}{2^{3}}$
Finally, we can raise both sides to the $\frac{2}{3}$ power to get:
$(n^{\frac{3}{2}})^{\frac{2}{3}} = \left (\dfrac{1}{2^{3}} \right )^{\frac{2}{3}}$.
When you raise something with an exponent to another exponent, you can just multiply the exponents together to get:
$n^{\frac{3}{2} \cdot \frac{2}{3}} = \left (\dfrac{1}{2^{3}} \right )^{\frac{2}{3}}$
or
$n = \left (\dfrac{1}{2^{3}} \right )^{\frac{2}{3}}$
or
$n = \dfrac{1^{\frac{2}{3}}}{(2^{3})^{\frac{2}{3}}}$
Now, $1$ raised to any power is just $1$, so the numerator is $1$. With the denominator, we again multiply exponents to get:
$n = \dfrac{1}{2^{3 \cdot {\frac{2}{3}}}}$
or
$n = \dfrac{1}{2^{2}}$
or
$n = \dfrac{1}{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of Interesting Quadruples
Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$, and a+d>b+c. How many interesting ordered quadruples are there?
This is a bit of trouble here actually, I am to use $a + d \gt b + c$ as a constraint.
Without any restrictions (no $a + d \gt b + c$) there are: $\binom{10}{4} = 210$ possible values for $a, b, c, d$.
We could have three cases: $a + d \gt b + c$ or $a + d < b + c$ or $a + d = b + c$.
We need to take out $a + d = b + c$ cases first: It is possible that:
$a + b = \{1 + \sum_{k=4}^{10}k, 2 + \sum_{k=5}^{10}, 3 + \sum_{k=6}^{10}, 4 + \sum_{k=7}^{10}k, ..., 7 + 10 \}$
Total (incl. overcounting): $7 + 6 + 5 + 4 + 3 + 2 + 1 = 28$ possible.
But I think I have messed the whole problem.
Hints Please!
| Let $t_1=a,\; t_2=b-a,\; t_3=c-b,\; t_4=d-c$;
so $t_1+t_2+t_3+t_4=d$ and $a+d>b+c\implies
t_1+t_2+t_3+t_4\le10$ and $t_4>t_2$.
Now let $t_5=t_4-t_2$ and $t_6=10-(t_1+t_2+t_3+t_4)$ to get
$\hspace{.3 in}2t_2+t_1+t_3+t_5+t_6=10$ with $t_i\ge1$ for $i<6$ and $t_6\ge0$.
Letting $y_i=t_i-1$ for $i<6$ and $y_6=t_6$ gives
$\hspace{.3 in}2y_2+y_1+y_3+y_5+y_6=5$ with $y_i\ge0$ for each $i$, so taking $y_2=0, y_2=1, \text{ and }y_2=2$
gives a total of $\displaystyle\binom{8}{3}+\binom{6}{3}+\binom{4}{3}=80$ solutions.
| {
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Prove that if $|x-2|<0.001$, then $|\frac{1}{x}-\frac{1}{2}|<3\times 10^{-3}$ I still have difficulties with absolute value, and even if I manage to solve questions and problems, I do that awkwardly.
So, please show me if this is the way to answer this question.
Thank you in advance.
$|x-2|<0.001 \iff -0.001<x-2$ or $x-2<0.001$
$$\begin{align}\\-0.001<x-2 & \iff 2-0.001<x\\ &\iff1.999<x\\ &\iff\frac{1}{x}<\frac{1}{1.999}\\ &\iff\frac{1}{x}-\frac{1}{2}<\frac{0.001}{3.998}\\ &\iff \frac{1}{x}-\frac{1}{2}<3\times 10^{-3}\end{align}$$
$$\begin{align}\\ x-2<0.001 &\iff x<2.001\\ &\iff\frac{1}{x}>\frac{1}{2.001}\\ &\iff\frac{1}{x}-\frac{1}{2}>\frac{3.002}{2}\\ &\iff\frac{1}{x}-\frac{1}{2}>-3\times10^{-3}\end{align}$$
We have: $-3\times 10^{-3}<\frac{1}{x}-\frac{1}{2}<3\times10^{-3}$
Then, $|\frac{1}{x}-\frac{1}{2}|<3\times10^{-3}$
| Your work is correct. Also, you can simply write:
We have by the triangle inequality $$2-|x|\le |2-x|<0.001\implies 3.998<|2x|$$
so
$$\left|\frac1x-\frac12\right|=\frac{|2-x|}{|2x|}<\frac{0.001}{3.998}<3\cdot 10^{-3}$$
| {
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Prove that $\frac{1}{1999} < \prod_{i=1}^{999}{\frac{2i−1}{2i}} < \frac{1}{44}$
Prove that $$\dfrac{1}{1999} < \prod_{i=1}^{999}{\dfrac{2i−1}{2i}} < \dfrac{1}{44}$$
from the 1997 Canada National Olympiad.
I have been able to prove the left half of the inequality using induction. Need help with the second part.
Prove $\prod_{i=1}^{n}{(1-\frac{1}{2i})} \ge \frac{1}{2n}$
Define $$p(n)=\prod_{i=1}^{n}{\left(1-\frac{1}{2i}\right)} \tag{1}$$
We wish to show $$p(n)\ge\frac{1}{2n},\quad\forall{n}\in\mathbb{Z^+} \tag{2}$$
For the base case, $n=1$, we have $p(1)=1-\frac{1}{2}=\frac{1}{2}\ge\frac{1}{2}$ which holds.
Assume the induction hypothesis (2) for $n$. Then for $n+1$, we can write:
$$\begin{align}
p(n+1) = \prod_{i=1}^{n+1}{\left(1-\frac{1}{2i}\right)} &= \left(1-\frac{1}{2(n+1)}\right) \cdot \prod_{i=1}^{n}{\left(1-\frac{1}{2i}\right)} \\
&\ge \left(1-\frac{1}{2(n+1)}\right) \cdot \frac{1}{2n} &(\text{by IH}) \\
&\ge \left(\frac{2n+1}{2(n+1)}\right) \cdot \frac{1}{2n} \\
&\ge \left(\frac{1}{2(n+1)}\right) \cdot \frac{2n+1}{2n} \\
&> \left(\frac{1}{2(n+1)}\right) &(\text{for }n\ge1)
\end{align}$$
This proves the induction step, establishing (2) for all positive $n$. Hence, we conclude
$$\boxed{\prod_{i=1}^{999}{\dfrac{2i−1}{2i}} = p(999) \ge \frac{1}{2\times999} = \frac{1}{1998} > \frac{1}{1999}}$$
Prove $\prod_{i=1}^{999}{\left(1-\frac{1}{2i}\right)} < \frac{1}{44}$
For this part, I have applied AM-GM to get an upper bound with a harmonic sum.
$$\left(\prod_{i=1}^{999}{\left(1-\frac{1}{2i}\right)}\right)^{1/999} \le \frac{1}{999} \cdot \sum_{i=1}^{999}{\left(1-\frac{1}{2i}\right)} \tag{3}$$
I can't see a way of evaluating the summation on the RHS.
| $${ P }_{ n }^{ 2 }\left( 999 \right) =\frac { { 1 }^{ 2 }\cdot { 3 }^{ 2 }...\cdot { \left( 2\cdot 999-1 \right) }^{ 2 } }{ { 2 }^{ 2 }\cdot { 4 }^{ 2 }...\cdot { \left( 2\cdot 999 \right) }^{ 2 } } =\frac { 1\cdot 3 }{ { 2 }^{ 2 } } \cdot \frac { 3\cdot 5 }{ { 4 }^{ 2 } } \cdot ...\cdot \frac { \left( 2\cdot 999-1 \right) \left( 2\cdot 999+1 \right) }{ { \left( 2\cdot 999 \right) }^{ 2 } } \cdot \frac { 1 }{ 2\cdot 999+1 } <\frac { 1 }{ 2\cdot 999+1 } =\frac { 1 }{ 1999 } <\frac { 1 }{ 1936 } =\frac { 1 }{ { 44 }^{ 2 } } $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Using right-hand Riemann sum to evaluate the limit of $ \frac{n}{n^2+1}+ \cdots+\frac{n}{n^2+n^2}$ I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:
\begin{align*}
\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)&=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{n}{n^2+k^2}\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{n^2}{n^2+k^2})\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{1}{1+(k/n)^2})\\
&=\lim_{n \to \infty} \sum_{k=1}^{n}f(\frac{k}{n})(\frac{k-(k-1)}{n})\\
&=\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4}
\end{align*}
where $f(x)=\frac{1}{1+x^2}$. Is this correct, are there any steps where I am not clear?
| Yes, what you've written is correct.
I think your write-up would be cleaner if you explicitly mentioned the antiderivative
$$ \int \frac{1}{1 + x^2} dx = \arctan x,$$
which makes it completely clear where $\frac{\pi}{4}$ comes from at the end.
| {
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Another integral $\int \frac{3 x^2+2 x+1}{ \left(x^3+x^2+x+2\right) \sqrt{1+\sqrt{x^3+x^2+x+2}}} \, dx$ Here is an indefinite integral that is similar to an integral I wanna propose for a contest. Apart from
using CAS, do you see any very easy way of calculating it?
$$\int \frac{1+2x +3 x^2}{\left(2+x+x^2+x^3\right) \sqrt{1+\sqrt{2+x+x^2+x^3}}} \, dx$$
EDIT: It's a part from the generalization
$$\int \frac{1+2x +3 x^2+\cdots n x^{n-1}}{\left(2+x+x^2+\cdots+ x^n\right) \sqrt{1\pm\sqrt{2+x+x^2+\cdots +x^n}}} \, dx$$
Supplementary question: How would you calculate the following integral using the generalization above? Would you prefer another way?
$$\int_0^{1/2} \frac{1}{\left(x^2-3 x+2\right)\sqrt{\sqrt{\frac{x-2}{x-1}}+1} } \, dx$$
As a note, the generalization like the one you see above and slightly modified versions can be wisely used for calculating very hard integrals.
| Let $\sinh^4(t)=x^3+x^2+x+2$. Then $\cosh(t)=\sqrt{1+\sqrt{x^3+x^2+x+2}}$ and
$$
\begin{align}
&\int\frac{3x^2+2x+1}{\left(x^3+x^2+x+2\right)\sqrt{1+\sqrt{x^3+x^2+x+2}}} \,\mathrm{d}x\\
&=\int\frac{\mathrm{d}\sinh^4(t)}{\sinh^4(t)\cosh(t)}\\
&=4\int\frac{\mathrm{d}t}{\sinh(t)}\\
&=4\int\frac{\mathrm{d}\cosh(t)}{\cosh^2(t)-1}\\
&=2\int\left(\frac1{\cosh(t)-1}-\frac1{\cosh(t)+1}\right)\mathrm{d}\cosh(t)\\[2pt]
&=2\log\left(\frac{\cosh(t)-1}{\cosh(t)+1}\right)+C\\[6pt]
&=-4\operatorname{arccoth}(\cosh(t))+C\\[6pt]
&=-4\operatorname{arccoth}\left(\sqrt{1+\sqrt{x^3+x^2+x+2}}\right)+C
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Inequality in proof of SLLN This comes from theorem 5.1.2 of KL Chung's A Course in Probability Theory.
Suppose ${X_n}$ are uncorrelated and their second moments have a common bound.
Then For each $n \ge 1 $, $D_n:= \max_{n^2 \le k < (n+1)^2 } |S_k - S_{n^2}| $, we have
$$E[|D_n|^2] \le 2n E\left[\left|S_{(n+1)^2}- S_{n^2}\right|^2\right] .$$
This is the only part of the proof I couldn't figure out. I'm sure its some silly trick that I somehow couldn't see.
| Since
\begin{align}
& D_n^2 = \max_{n^2 \leq k < (n + 1)^2 } \left|S_k - S_{n^2}\right|^2 \\
\leq & \sum_{k = n^2}^{(n + 1)^2 - 1} \left|S_k - S_{n^2}\right|^2 \\
= & \sum_{k = n^2 + 1}^{(n + 1)^2 - 1} \left|S_k - S_{n^2}\right|^2
\end{align}
It follows that
\begin{align}
& E\left(D_n^2\right) \leq \sum_{k = n^2 + 1}^{(n + 1)^2 - 1}E\left(\left|S_k - S_{n^2}\right|^2\right) \\
\leq & [(n + 1)^2 - 1 - (n^2 + 1) + 1] E\left(\left|S_{(n + 1)^2} - S_{n^2}\right|^2\right) \\
= & 2nE\left(\left|S_{(n + 1)^2} - S_{n^2}\right|^2\right)
\end{align}
where we used for every $k \in \{n^2 + 1, \ldots, (n + 1)^2 - 1\}$, \begin{align}
& E\left(\left|S_k - S_{n^2}\right|^2\right) = E\left[\left(X_{n^2 + 1} + \cdots + X_k\right)^2\right] = \sum_{i = n^2 + 1}^k E\left(X_i^2\right) \\
\leq & \sum_{i = n^2 + 1}^{(n + 1)^2} E\left(X_i^2\right) = E\left(\left|S_{(n + 1)^2} - S_{n^2}\right|^2\right)\end{align}
in which we used uncorrelated condition.
| {
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"timestamp": "2023-03-29T00:00:00",
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an interesting q-series and a certain continued-fraction My aim is to find a rigorous proof of the following conjectured identity.Given
$$1+q+q^2-q^4-q^5+q^7+q^8-q^{10}-q^{11}+\ddots=\cfrac{1}{1-q+\cfrac{(q^3)}{1-q^3+\cfrac{q^2(1+q)(1+q^3)}{1-q^5+\cfrac{q^3(1+q^2)(1+q^4)}{1-q^7+\cfrac{q^4(1+q^3)(1+q^5)}{1-q^9+\ddots}}}}}$$
The exponents on the left-hand side $${1,2,4,5,7,8,\ddots}$$ are positive integers not divisible by $3$.
How do we show that, the identity is true?
Independent verification and attempts at proving/disproving the result are very welcome.
| Regarding R. Israel's remark, three of your continued fractions, while not exactly the same, are variants of a common form discussed here for $|q|<1$,
$$\frac{1}{1-q} =\cfrac{1}{1+q-\cfrac{\color{brown}{2q(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}\tag0$$
First, the one for the Fibonacci numbers in this post,
$$\chi(q)=\cfrac{1}{1+q-\cfrac{\color{brown}{(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}\tag1$$
Second, the one for the Narayana numbers in this post,
$$\psi\Big(q\Big)=\cfrac{1}{1+q-\cfrac{\color{brown}{q^2}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}\tag2$$
Third, the one for this post with $q \to -q$,
$$M(q) = \cfrac{1}{1-q+\cfrac{\color{brown}{q^3}}{1-q^3+\cfrac{q^2(1+q)(1+q^3)}{1-q^5+\cfrac{q^3(1+q^2)(1+q^4)}{1-q^7+\cfrac{q^4(1+q^3)(1+q^5)}{1-q^9+\ddots}}}}}\tag3$$
Arranged in this way, one can see their common form. Notice that the brown part is the only "level" that changes, as well as letting $q\to-q$. (Based on this observation, I experimented with using $\color{brown}{-1+q^2, -q+q^2, q+q^2, q^4}$ instead of the ones above, and they also yield generating functions of known sequences in the OEIS.)
P.S. Transforming $(0)$ to $(3)$, I get,
$$\begin{aligned}M(q)
&=\frac{1+q^2}{1-q+q^2}\\
&= 1+q+q^2-q^4-q^5+q^7+q^8-q^{10}-q^{11}+\dots
\end{aligned}$$
and the sequence is A163806 (the Gf is corrected by B. Berselli). It states that,
$$a(3n) = 0$$
thus your observation that the exponents with non-zero coefficients are the integers not divisible by 3 is correct. (Note that this formula is only valid for $|q|<1$.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $ \int_0^\infty \frac{x^5}{e^x+1} \, dx $ This improper integral has stumped me and not many integration problems give me problems. However, this one made me think over my limit. Finally, as I could not even get started on this problem, I thought to post it here, and have some talented mathematicians look at it. The problem is to integrate:
$$ \int_0^\infty \frac{x^5}{e^x+1} \, dx $$
| $$I=\int _{ 0 }^{ \infty }{ \cfrac { { x }^{ 5 } }{ { e }^{ x }+1 } dx= } \int _{ 0 }^{ \infty }{ \cfrac { { e }^{ -x } }{ { 1+e }^{ -x } } { x }^{ 5 }dx } =\int _{ 0 }^{ \infty }{ \left( \sum _{ r=1 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx } } { x }^{ 5 } \right) dx } \\ =\sum _{ r=1 }^{ \infty }{ \left\{ \int _{ 0 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx }{ x }^{ 5 }dx } \right\} }$$
Let $rx=t$:
$$\int _{ 0 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx }{ x }^{ 5 }dx } ={ \left( -1 \right) }^{ r-1 }\int _{ 0 }^{ \infty }{ { e }^{ -t }{ \left( \cfrac { t }{ r } \right) }^{ 6-1 }\cfrac { dt }{ r } } =\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } \Gamma (6)=120\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } \\ \\ \therefore I=\sum _{ r=1 }^{ \infty }{ 120\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } } =120\eta \left( 6 \right) =120\left\{ \left( 1-{ 2 }^{ 1-6 } \right) \right\} \zeta \left( 6 \right) \\ \\ =120\times \cfrac { 31 }{ 32 } \zeta \left( 6 \right) =\cfrac { 465 }{ 4 } \zeta \left( 6 \right)$$
$$\zeta \left( 6 \right) =\zeta \left( 2\times 3 \right) =\cfrac { { \left( -1 \right) }^{ 3+1 }{ B }_{ 6 }{ \left( 2\pi \right) }^{ 6 } }{ 2\left( 6 \right) ! } =\cfrac { { \pi }^{ 6 } }{ 945 }$$
Note that $B_6$ refers to the $6$th Bernoulli Number $\left(B_6=\frac{1}{42}\right)$:
$$\therefore I=\cfrac { 465 }{ 4 } \times \cfrac { { \pi }^{ 6 } }{ 945 } =\cfrac { { 31\pi }^{ 6 } }{ 252 }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the longest side of the triangle. The sides $a,b,c$ of a $\triangle ABC$ are in $GP$ whose common ratio is $\frac{2}{3}$ and the circumradius of the triangle is $6\sqrt{\frac{7}{209}}$.Find the longest side of the triangle.
I used law of sines $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$
I took $a=a,b=\frac{2}{3}a,c=\left(\frac{2}{3}\right)^2a$
which gives $$\frac{\sin A}{\sin B}=\frac{\sin B}{\sin C}$$
I am stuck now,how to find the longest side $a$.
| Let $a,b,c$ be the side lengths. Then, $b,c$ can be written as $$b=\frac 23a,\quad c=\left(\frac 23\right)^2a$$
where $a$ is the longest side. So, we have, by the law of cosines,
$$a^2=\left(\frac 23a\right)^2+\left(\left(\frac 23\right)^2a\right)^2-2\cdot\frac 23a\cdot\left(\frac 23\right)^2a\cos A$$
$$\Rightarrow \cos A=-\frac{29}{48}.$$
Hence,
$$a=2R\sin A=2R\sqrt{1-\cos^2A}=2\cdot 6\sqrt{\frac{7}{209}}\sqrt{1-\left(-\frac{29}{48}\right)^2}=\frac 74.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integer solutions to $x^2=2y^4+1$. Find all integer solutions to $x^2=2y^4+1$.
What I tried
The only solutions I got are $(\pm 1 ,0)$, I rewrote the question as : is $a_{n}$ a perfect square for $n>0$ were
$$a_0=0,\quad a_1=2, \quad a_{n+2}=6a_{n+1}-a_n.$$
I tried taking $\pmod{4}$ and $\pmod{12}$ but that lead me nowhere.
| There are no solutions other than those trivial ones. First we note that $x$ must be odd. Let $x = 2k+1$. This gives $2k^2+2k = y^4$. Thus $y = 2j$ and we have $k(k+1) = 8j^4$. Note that $\gcd(k,k+1) = 1$ so we have two cases.
Case 1: $k+1 = 8a^4, k = b^4$.
This case is impossible since this implies $b^4 \equiv -1 \pmod 4$ but $-1$ is a not a quadratic residue $\pmod 4$.
Case 2: $k+1 = b^4, k = 8a^4$.
This case is equivalent to solving the diophantine equation $8a^4+1 = b^4$. An elementary solution is given here. In the referenced paper, it is prove that the only non trivial solution to this equation is $(a,b) = (1,3)$ which gives $k = 8, k+1 = 9$ but $9 = b^4$ is impossible.
| {
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Why does the diophantine equation $x^2+x+1=7^y$ have no integer solutions? This following Problem is from Pell equation chapters exercise
Let $y>3$ positive integer numbers, show that following diophantine equation
$$x^2+x+1=7^y\tag{1}$$ has no integer solutions.
I tried write the equation
$$(2x+1)^2+3=4\cdot 7^y$$
if $y=2k$ then we have
$$(2\cdot 7^k+2x+1)(2\cdot 7^k-2x-1)=3$$
this case has no integer.
But $y$ is odd number, How to prove equation (1) has no integer solutions for $x,y (y>3)$? Any help would be appreciated.
| We work in the Eisenstein integers. The equation then becomes$$(x + 1 + \omega)(x - \omega) = 6^y.$$Suppose $d\, |\, (x + 1 + \omega)$ and $\text{ }d\,|\,(x - \omega)$.Then we have $d\,|\,(1 + 2\omega)$, so $N(d)\,|\,3$. However, we also have $N(d)\,|\,7^{2y}$, so $N(d) = 1$. On the other hand, $7 = (-2 - 3\omega)(1 + 3\omega)$ and $-2 - 3\omega$, $1 + 3\omega$ are relatively prime. Thus, it follows that $x + 1 + \omega$ and $x - \omega$ are $e(-2-3\omega)^y$ and $f(1 + 3\omega)^y$ in some order, where $e$ and $f$ are units $\pm1$, $\pm\omega$, $\pm1\pm\omega$ with $ef = 1$. We just need to show that if$$a + b\omega = f(1 + 3\omega)^y,$$then $|b|$ can never be $1$, which would solve the problem. Indeed, let$$A = 1 + 3^3\binom{y}{3} + \dots,\text{ }B = 3\binom{y}{1} + \dots,\text{ }C = 3^2\binom{y}{2} + \dots.$$We have$$a + b\omega = f(A - C + \omega(B - C)),$$so we want to show none of $|A - B|$, $|B- C|$, $|C - A|$ are one. However, this is a relatively easy task for $y > 3$, so we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 4
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Determine a positive integer $n\leq5$,such that $\int_{0}^{1}e^x(x-1)^ndx=16-6e$ Determine a positive integer $n\leq5$,such that $\int_{0}^{1}e^x(x-1)^ndx=16-6e$.
I tried to solve it.But since $n$ is given to be $\leq$ 5,my calculations went lengthy.
Applying integration by parts repeatedly,we get
\begin{align}
\int e^x(x-1)^n \, dx &= \left[(x-1)^ne^x-n(x-1)^{n-1}e^x+n(n-1)(x-1)^{n-2}e^x \right. \\
& \hspace{5mm} \left.-n(n-1)(n-2)(x-1)^{n-3}e^x+n(n-1)(n-2)(n-3)(x-1)^{n-4}e^x \right. \\
& \hspace{5mm} \left.-n(n-1)(n-2)(n-3)(n-4)e^x\right]
\end{align}
\begin{align}
\int_{0}^{1}e^x(x-1)^n \, dx &= -n(n-1)(n-2)(n-3)(n-4)e-(-1)^n+n(-1)^{n-1}-n(n-1)(-1)^{n-2} \\
& \hspace{5mm} +n(n-1)(n-2)(-1)^{n-3}-n(n-1)(n-2)(n-3)(-1)^{n-4} \\
& \hspace{5mm} +n(n-1)(n-2)(n-3)(n-4) \\
&=16-6e
\end{align}
Now solving this is very difficult,is there another simple and elegant method to find $n=3.$
| HINT:
Let $x-1=y$
$$\int_0^1e^x(x-1)^n\ dx=\int_{-1}^0e^{y+1}y^n\ dy=e\int_{-1}^0e^yy^n\ dy$$
Now integrating by parts, $$I_n=\int_{-1}^0e^yy^n\ dy=[y^n\int e^y\ dy]_{-1}^0-\int_{-1}^0\left(\dfrac{d(y^n)}{dy}\int e^y\ dy\right)dy$$
$\implies I_n=-(-1)^n\dfrac1e-nI_{n-1}$
Now $I_0=\int_{-1}^0e^y\ dy=1-\dfrac1e$
$\implies I_1=\dfrac1e-I_0$ and so on
| {
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"url": "https://math.stackexchange.com/questions/1405064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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How to find range of a logarithmic function? How do I find the range of these logarithmic functions?
\begin{align}
& \ln(3x^2 -4x +5), \\
& \log_3(5+4x-x^2).
\end{align}
how should I approach questions like this ?
What I did: I found out the roots of quadratic: i.e. $(5-x)(1+x) >0$
so $(x-5)(1+x) <0$ so it lies between $-1$ and $5$, and then I took log on both sides of the inequality.
| You can only take a logarithm of a number greater than zero.
So you need $3x^2-4x+5 > 0$ in the first case. Completing the square give you $\left(x-\frac{2}{3}\right)^2+\frac{11}9$. We see that the quadratic is always greater than $\frac{11}{9}$ and goes to infinity. Therefore the range is $[\ln\left(\frac{11}{9}\right), \rightarrow \rangle$
For the second one, you want $-x^2+4x+5 > 0$. We first solve $-x^2+4x+5 =0$. This gives $x=-1$ or $x=5$ as you found. Because the coefficient for $x^2$ is negative, this means that the quadratic is positive when $-1<x<5$. The maximum is attained at $x=-\frac{b}{2a}=2$, whith a value of $9$.
So we can make the argument of the log very close to zero but never greater than 9. As $\log_3(9)=2$, the range is $\langle \leftarrow , 2]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Maximum of $\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}....\cos \alpha_{n}.$
Maximum value of $\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdot \cos \alpha_{4}....\cos \alpha_{n}.$ If it is given that
$\cot \alpha_{1}\cdot \cot \alpha_{2}\cdot \cot \alpha_{3}.......\cot \alpha_{n} = 1$ and $\displaystyle 0 \leq \alpha_{1},\alpha_{2},\alpha_{3},.......,\alpha_{n}\leq \frac{\pi}{2}.$
$\bf{My\; Try::}$ Using $\bf{A.M\geq G.M}$
$$\displaystyle \frac{\sin^2 \alpha_{1}+\sin^2 \alpha_{2}+\sin^2 \alpha_{3}+........+\sin^2 \alpha_{n}}{n}\geq \sqrt[n]{\sin^2 \alpha_{1}\cdot \sin^2 \alpha_{2}...\sin^2 \alpha_{n}}$$
So we get
$$\displaystyle \sin^2 \alpha_{1}+\sin^2 \alpha_{2}+\sin^2 \alpha_{3}+........+\sin^2 \alpha_{n}\geq n\cdot \sqrt[n]{\sin^2 \alpha_{1}\cdot \sin^2 \alpha_{2}...\sin^2 \alpha_{n}}.......(1)$$
Similarly
$$\displaystyle \frac{\cos^2 \alpha_{1}+\cos^2 \alpha_{2}+\cos^2 \alpha_{3}+........+\cos^2 \alpha_{n}}{n}\geq \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}$$
So we get
$$\displaystyle \cos^2 \alpha_{1}+\cos^2 \alpha_{2}+\cos^2 \alpha_{3}+........+\cos^2 \alpha_{n}\geq n\cdot \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}.......(2)$$
Now Adding $(1)$ and $(2)\;,$ We get $$n\geq 2n\cdot \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}$$
Because above it is given that $$\sin \alpha_{1}\cdot \sin\alpha_{2}\cdot\sin \alpha_{3}.........\sin\alpha_{n}=\cos\alpha_{1}
\cdot\cos\alpha_{2}\cdot \sin\alpha_{3}......\cos\alpha_{n}$$
So we get $$\displaystyle \cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdot \cos \alpha_{4}....\cos \alpha_{n}\leq \frac{1}{2^{\frac{n}{2}}}$$
My question is that can we solve it using any other short method? If yes then please explain it here.
Thanks.
| $\cot\alpha_1.\cot\alpha_2......\cot\alpha_n=1$
$\cos\alpha_1.\cos\alpha_2......\cos\alpha_n=\sin\alpha_1.\sin\alpha_2......\sin\alpha_n$....................(1)
Let $y=\cos\alpha_1.\cos\alpha_2......\cos\alpha_n$(to be maximized)
$\Rightarrow y^2=\cos^2\alpha_1.\cos^2\alpha_2......\cos^2\alpha_n$
$y^2=(\cos\alpha_1.\cos\alpha_2......\cos\alpha_n)(\sin\alpha_1.\sin\alpha_2......\sin\alpha_n)$
$y^2=\frac{1}{2^n}[\sin(2\alpha_1).\sin(2\alpha_2)......\sin(2\alpha_n)]$
As $0\leq\alpha_1,\alpha_2,.....,\alpha_n\leq\frac{\pi}{2}$
$0\leq2\alpha_1,2\alpha_2,.....,2\alpha_n\leq\pi$
$\Rightarrow 0\leq \sin(2\alpha_1),\sin(2\alpha_2),......,\sin(2\alpha_n)\leq 1$
$y^2\leq\frac{1}{2^n}\times 1\Rightarrow y\leq \frac{1}{2^{n/2}}$
Therefore maximum value of $y$ is $\frac{1}{2^{n/2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Integers which are the sum of non-zero squares Lagrange's four-square theorem states that every natural number can be written as the sum of four squares, allowing for zeros in the sum (e.g. $6=2^2+1^2+1^2+0^2$). Is there a similar result in which zeros are not allowed in the sum? For example, does there exist $n\in\mathbb{N}$ such that every natural number greater than $n$ can be written as the sum of five non-zero squares, or six non-zero squares, for example?
| This question is answered in "Introduction to Number
Theory" by Niven, Zuckerman& Montmogery (pp.318-319 of
the fifth edition). I summarize their proof below.
Every integer $\geq 34$ is a sum of five positive squares
(while $33$ is not). The number five is optimal, because
the only representation of $2^{2r+1}$ as a sum of four
squares is $0^2+0^2+(2^r)^2+(2^r)^2$ (easy exercice by induction
on $r$).
One can check by hand by noting all the numbers between $34$ and $169$
are sums of five positive squares. Now, let $n\geq 169$ and let us
show that $n$ is a sum of five positive squares.
We know that $n-169$ is a sum of four not necessarily positive
squares, $n-169=x_1^2+x_2^2+x_3^2+x_4^2$ and we can assume
$x_1 \leq x_2 \leq x_3 \leq x_4$.
If $x_1>0$, writing $n=13^2+x_1^2+x_2^2+x_3^2+x_4^2$ we are done. So assume
$x_1=0$.
If $x_2>0$, writing $n=5^2+12^2+x_2^2+x_3^2+x_4^2$ we are done. So assume
$x_2=0$.
If $x_3>0$, writing $n=3^2+4^2+12^2+x_3^2+x_4^2$ we are done. So assume
$x_3=0$.
If $x_4>0$, writing $n=2^2+4^2+7^2+12^2+x_4^2$ we are done. So assume
$x_4=0$.
So now all the $x_i$ are zero, and $n=169=5^2+6^2+6^2+6^2+6^2$. This concludes
the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Proving $\frac{n^n}{3^n} < n!$ for $n\ge6$ by induction How would I prove this using mathematical induction:
$\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$.
Here is what I have tried:
$\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$
Base case: $\dfrac{6^6}{3^6} < 6!$. Yes, this is true.
Assume for $k$ it's true so $\dfrac{k^k}{3^k} < k!$.
Now prove for $n=k+1$,
$\dfrac{(k+1)^{k+1}}{3^{k+1}} < (k+1)!$.
By assumption, we have that
$\dfrac{k^k}{3^k} < k!$.
Multiplying by $k+1$, we get
$$
\frac{k^k(k+1)}{3^k} < (k+1) k!
$$
and then
$$
\frac{k^k(k+1)}{3^k} < (k+1) !.
$$
I tried lots of things after this, but I am unable to figure out the proof.
| We will use induction to show that $\;\;\color{red}{3^n n!>n^n}$ for all $\color{blue}{n\ge1}$:
1) This is true for $n=1$, since $3>1$.
2) Let $n\in\mathbb{N}$ with $3^n n!>n^n$. Then $3^{n+1}(n+1)!=3(n+1)\big(3^n n!\big)>3(n+1)\big(n^n\big)$,
$\;\;\;$and $3(n+1)(n^n)>(n+1)^{n+1}\iff3n^n>(n+1)^{n}\iff3>\big(1+\frac{1}{n}\big)^n,\;$ and
$\hspace{.25 in}(1+\frac{1}{n})^n=\sum_{k=0}^{n}\binom{n}{k}(\frac{1}{n})^k=\sum_{k=0}^{n}\frac{n(n-1)(n-2)\cdots(n-(k-1))}{k!n^k}=\sum_{k=0}^{n}\frac{(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k-1}{n})}{k!}$
$\hspace{.87 in}\le\sum_{k=0}^{n}\frac{1}{k!}=2+\sum_{k=2}^n\frac{1}{k!}\le2+\sum_{k=2}^n\frac{1}{k(k-1)}=2+\sum_{k=2}^n(\frac{1}{k-1}-\frac{1}{k})=3-\frac{1}{n}<3$
$\;\;\;$Therefore $3^{n+1}(n+1)!>(n+1)^{n+1}$.
Thus $\color{red}{3^n n!>n^n}$ for all $\color{blue}{n\ge1}$ by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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trying to grasp disphenoid tetrahedral honeycomb, what are the dihedral angles? What are the dihedral angles in a disphenoid with four identical triangles, each having one edge of length $2$ and two edges of length $\sqrt{3}$? Tried to look it up, but couldn't find it...
| Say the four vertices of the shape are $ABCD$, with $\overline{AB} = \overline{CD} = 2$, and the rest of the segments having length $\sqrt{3}$. Let $O$ denote the midpoint of $AB$, then the triangle $AOC$ is a right triangle with hypotenuse $\sqrt{3}$ and one of the legs $1$. We get that $\overline{OC} = \sqrt{2}$. Similarly, $\overline{OD} = \sqrt{2}$. So the triangle $COD$ is an isosceles triangle with two sides of length $\sqrt{2}$ and the third side of length $2$, which implies that it is a right triangle(!).
So one of the dihedral angles is $90^{\circ}$. Now we need to compute the angle between two faces meeting on a side of length $\sqrt{3}$- for example, between $ABC$ and $DBC$. One way to do this is to assign coordinates to the vertices. From the previous analysis, a particularly nice choice is $A = (-1,0,0)$, $B = (1,0,0)$, $C = (0,\sqrt{2},0)$, $D = (0,0,\sqrt{2})$.
The plane containing $ABC$ is given by the equation $z = 0$, and the plane containing $DBC$ is given by $x\sqrt{2} + y + z = \sqrt{2}$. Then the angle between the planes is equal to the angle between the normal vectors $\mathbf{n}_1 = (0,0,1)$ and $\mathbf{n}_2 = (\sqrt{2},1,1)$:
$$\cos(\theta) = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{1}{1 \cdot 2} = \frac{1}{2}$$
It follows that the other dihedral angle is $60^{\circ}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Integrate $\int_{0}^{2}{\int_{y^2}^{4}{y\cos(x^2)dxdy}}$ I'm asked to evaluate $\int_{0}^{2}{\int_{y^2}^{4}{y\cos (x^2)dxdy}}$. Letting $f(x)=\cos (x^2)$ We have have that $\int_{y^2}^{4}{\cos (x^2)dx} = F(4)-F(y^2)$ by the FTC. This gives us
\begin{align*}
\int_{0}^{2}{\int_{y^2}^{4}{y\cos(x^2)dxdy}} &= \int_{0}^{2}{(yF(4)-yF(y^2))dy}\\
&=2F(4)-\int_{0}^{2}yF(y^2)dy\\
&=2F(4)-(\frac{y^2}{2}F(4) \big|_0^2 - \int_{0}^{2}{\frac{y^2}{2} \cdot 2y \cdot f(y^2)dy})\\
&= \int_{0}^{2}{y^3\cos(y^4)dy} = \frac{\sin(16)}{4}
\end{align*}
Is this a valid way to evaluate this integral? The solution given to me changed the order of integration instead and did it a bit differently. Thanks.
| hint: Switch the order is better: $I = \int_{0}^ 4\int_{0}^{\sqrt{x}} y\cos(x^2)dydx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the condition on $a$ and $b$ so that the two tangents drawn to the parabola $y^2=4ax$ from a point are normals to the parabola $x^2=4by$ Find the condition on $a$ and $b$ so that the two tangents drawn to the parabola $y^2=4ax$ from a point are normals to the parabola $x^2=4by.$
The required condition is $a^2>8b^2$.I dont know how to prove it.I tried.
Let $(h,k)$ be the point from where tangents are drawn to the parabola $y^2=4ax$.Let $m_1,m_2$ be the slopes of the tangents.Equations of tangents are $y=m_1x+\frac{a}{m_1}$ and $y=m_2x+\frac{a}{m_2}$ and equations of normal to the parabola $x^2=4by$ is $y=m_1x-2bm_1-bm^3_1$ and $y=m_2x-2bm_2-bm^3_2$.
As $y=m_1x+\frac{a}{m_1}$ and $y=m_1x-2bm_1-bm^3_1$ are the same lines.Therefore,$\frac{a}{m_1}=-2bm_1-bm^3_1$
$bm^4+2bm^2+a=0$
And I am stuck.I cannot get the desired condition.Please help me.
|
Equations of tangents are $y=m_1x+\frac{a}{m_1}$ and $y=m_2x+\frac{a}{m_2}$
I think this is correct.
and equations of normal to the parabola $x^2=4by$ is $y=m_1x-2bm_1-bm^3_1$ and $y=m_2x-2bm_2-bm^3_2$.
I don't think this is correct.
Let $(s,t)$ be a point on $x^2=4by$, i.e. $s^2=4bt$. Since $2x=4by'\Rightarrow y'=\frac{x}{2b}$, the equation of normal to the parabola at $(s,t)$ will be $y-t=-\frac{2b}{s}(x-s)\iff y=-\frac{2b}{s}x+2b+t$. Let $k=-\frac{2b}{s}$. Then, since $s=-\frac{2b}{k}$, the equation of the normal can be written as
$$y=kx+2b+\frac{s^2}{4b}=kx+2b+\frac{1}{4b}\left(-\frac{2b}{k}\right)^2=kx+2b+\frac{b}{k^2}.$$
So, in our case, we have
$$y=m_1x+2b+\frac{b}{m_1^2}\qquad\text{and}\qquad y=m_2x+2b+\frac{b}{m_2^2}.$$
Thus, since $y=m_1x+\frac{a}{m_1}$ and $y=m_1x+2b+\frac{b}{m_1^2}$ are the same lines,
$$\frac{a}{m_1}=2b+\frac{b}{m_1^2}\iff 2bm_1^2-am_1+b=0.$$
Now, the discriminant is positive, so $(-a)^2-4\cdot 2b\cdot b\gt 0$, i.e. $a^2\gt 8b^2$.
As harry pointed out in the comments, one has to consider the case where one of the tangents is the $y$-axis.
Two tangents drawn to the parabola $y^2=4ax$ from $(0,k)$ where $k\not=0$ are $x=0$ and $y=\frac akx+k$. It follows from $u^2=4bv,\frac ak=-\frac{2a}{u}$ and $k=2b+v$ that $bk^2-a^2k+2a^2b=0$. So, one has to have $a^2\geqslant 8b^2$. If $a^2=8b^2$, then two tangents drawn to the parabola $y^2=4ax$ from a point $(0,4b)$ are $x=0$ and $ax-4by+16b^2=0$ (at $(2a,8b)$) which are normals to the parabola $x^2=4by$ (at $(-a,2b)$).
In conclusion, the answer is $a^2\geqslant 8b^2$.
| {
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"url": "https://math.stackexchange.com/questions/1414507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve $\sum_{k=1}^n\frac{k}{n^2+k}$? Can someone show me what is wrong with the expression I got for evaluating $\sum_{k=1}^n\frac{k}{n^2+k}$?
Steps:
$\sum_{k=1}^n\frac{k}{n^2+k} = \frac{\sum_{k=1}^nk}{\sum_{k=1}^{n}n^2+k} = \frac{\frac{1}{2}n(n+1)}{\sum_{k=1}^{n}n^2+\sum_{k=1}^{n}k} = \frac{\frac{1}{2}n(n+1)}{n^3+\frac{1}{2}n(n+1)} = \frac{n(n+1)}{2n^3+n(n+1)} = \frac{n+1}{2n^2+n+1}$
However, when I plug in something like $\sum_{k=1}^{10}\frac{k}{10^2+k}$ into this online summation calculator, the answer I get is $\sim0.514$, which is not the same answer I get with $\frac{10+1}{2(10)^2+10+1} = \frac{11}{211} \approx 0.0521$
| Your first step is wrong:
$$\sum_{k=1}^n \frac{k}{n^2+k} \neq \frac{\sum_{k=1}^n k}{\sum_{k=1}^n n^2+k}$$
It's like saying $\frac ab + \frac cd=\frac{a+c}{b+d}$ which is false...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1415120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.