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Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$ Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.
We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\sqrt{\dfrac{\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha}{\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha}}=\sqrt{\dfrac{(\sin\alpha+\cos\alpha)^2}{(\sin\alpha-\cos\alpha)^2}}=\sqrt{\left(\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right)^2}.$
Using the fact that $\sqrt{a^2}=|a|$ the given expression is equal to $\left|\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right|.$ I think that in the inverval $\left(45^\circ;90^\circ\right) \sin\alpha>\cos\alpha$ but how can I prove that? What to do next?
| Apparently your second question hasn't been answered, about what to do next. Here's what to do:
$$\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\equiv\frac{\frac{1}{\cos \alpha}}{\frac{1}{\cos \alpha}}\times\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\equiv\dots$$
I hope that's helpful.
If you need any more help please don't hesitate to ask.
| {
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"url": "https://math.stackexchange.com/questions/3972195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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difference of recursive equations Lets have two recursive equations:
\begin{align}
f(0) &= 2 \\
f(n+1) &= 3 \cdot f(n) + 8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot g(n) + 12
\end{align}
We want a explicit equation for f(x) - g (x).
I firstly tried to do in manually for first $n$ numbers
\begin{array}{|c|c|c|c|}
\hline
n & f(n) & g(n) & f(n) - g(n) \\ \hline
0 & 2 & -2 & 4 \\ \hline
1 & 6 & 6 & 0 \\ \hline
2 & 26 & 30 & -4 \\ \hline
3 & 94 & 102 & -8 \\ \hline
4 & 306 & 318 & -12 \\ \hline
\end{array}
We can deduce that $f(n) - g(n) = 4 - 4n$
But now we have to prove it.
Lets extend recursive equation $f(n)$:
$$f(n) = 3^n \cdot f(0) + 8 \cdot (3^{n+1} \cdot 0 + \dots + 3^{0}(n-1))$$
for $g$ we get $$g(n) = 3^n \cdot g(0) + 12 \cdot (3^{n-1} + \dots + 3^0)$$
We can simply check this by induction but I will skip it, so the question won't be so long.
Now lets put it together:
$$
f(n) - g(n) = 2 \cdot 3^n + 8 * 3^{n-1} \cdot 0 + ... + 3^0 \cdot (n-1) + 2 \cdot 3^n - 4 \cdot 3 \cdot ( 3^{n-1}+ ... + 3^0)= \\
= 4 \cdot 3^n + 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^n + 3^{n-1} + ... + 3^1) = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4 \cdot 3^0 = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4
$$
As we can see, we already got the $4$, so to get $-4n + 4$, the rest of the equation must equal $-4n$. But this is where I don't know how to continue.
How to prove that:
$$8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = -4n$$
All I could do is this:
\begin{align}
&8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = \\
&= 4 \cdot (\frac{0}{2}3^{n-1} + \dots + \frac{1}{2} \cdot (n-1) - 4 * (\frac{2}{2}3^{n-1} + \dots + \frac{2}{2}3^0) = \\
&= 4 \cdot (3^{n-1} \cdot (\frac{0- 2}{2}) + \dots + 3^0 \cdot \frac{(n-1)-1}{2}) = \\
&= 4 \cdot (-\frac{2}{2}3^{n-1} + \dots + \frac{n-2}{2})
\end{align}
And I made sum function out of it:
$\sum^{n-1}_{i=0}{\frac{i - 2}{2}\cdot 3^{n-1-i}}$
What to do next? Did I go the wrong direction anywhere?
Thank you for your fast responses.
| This is an elegant problem. You can use induction on the difference and skip the brute force computation.
Proof by induction:
Let S(n) be the claim that $f(n)-g(n)=4-4n$.
Base case: trivial.
Inductive case: Suppose S(n-1) is true. That is, suppose $f(n-1)-g(n-1)=4-4(n-1)$. Then $f(n)-g(n)=[3f(n-1)+8(n-1)]-[3g(n-1)+12]=3[f(n-1)-g(n-1)]+8n-20.
$
By induction, that equals:$3[4-4(n-1)]+8n-20=12-12(n-1)+8n-20=12-12n+12+8n-20=24-4n-20=4-4n$
and you're done.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Change of variable with g(u,v) = (u-v+1,u+v) I want compute the integral of f(x,y) = 1 over $R = \{ (x,y) : 0\leq x \leq 1 \land 0\leq y \leq x^2 \} $ with the transformation $g(u,v) = (u-v+1,u+v)$ and check that is the same that compute without change of variable. i.e. $ \int_{0}^{1} \int_{0}^{x^2} dy dx = \frac{1}{3} $
I have problems for determinate the new region, I know that if I get the inverse function of $g(u,v)$, then I can map the vertices as follow
$$ g^{-1}(x,y) = \left( \frac{x}{2} + \frac{y}{2} - \frac{1}{2} , \frac{-x}{2} + \frac{y}{2} + \frac{1}{2} \right) $$
$$ g^{-1}(0,0) = \left(\frac{-1}{2}, \frac{1}{2} \right), \ g^{-1}(1,1) = \left(\frac{1}{2}, \frac{1}{2} \right), \ g^{-1}(1,0) = (0,0) $$
But, how I determine the ranges of $u$ and $v$? for the integral $\int\int J_g(u,v) du dv$
| $0 \leq x \leq 1, 0 \leq y \leq x^2$
Transformation $x = u - v + 1, y = u + v$
$J = \begin{vmatrix} 1 & -1 \\ 1 & 1\end {vmatrix} = \, 2$
For the upper bound of $v$,
$y = x^2 \implies u + v = (u-v+1)^2$ (which is equation of another parabola)
$v^2 - (2u+3)v + (u^2+u+1) = 0$ which is a quadratic in $v$.
$(v - \frac{1}{2}\big(2u+3)\big)^2 = \big(\frac{2u+3}{2}\big)^2 - (u^2+u+1)$
$v = \frac{1}{2} (2u+3 \, \pm \sqrt{8u+5})$. As we have to find area below the parabola, we take the lower part of the curve.
So your integral becomes,
$\displaystyle \int_{-1/2}^0 \int_{-u}^{\frac{1}{2} (2u+3 - \sqrt{8u+5})} 2 \,dv \, du + \int_0^{1/2} \int_u^{\frac{1}{2} (2u+3 - \sqrt{8u+5})} 2 \,dv \, du = \frac{1}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $\iint_D (x^4-y^4) \,dx\,dy$ How should I proceed? the question is :
$$\iint_D (x^4-y^4) \,dx\,dy$$
$$D= \left\{(x,y):1<x^2-y^2<4, \quad \sqrt{17}<x^2+y^2<5,\quad x<0,\ \ y>0\right\}$$
I've tried to solve it with change of variable:
$$u=x^2-y^2, \quad v=x^2+y^2$$ $$|J|=\frac{1}{8xy}, \quad \frac{1}{8}\iint uv\frac{1}{xy} \,du\,dv$$
How should I proceed?
Any suggestion would be great, thanks
| Oddly enough polar coordinates will do the trick and does it well as an alternative. The region is one integral if done angular first (which is easy to see because the region is between two circular arcs)
$$1<x^2-y^2<4 \implies \pi-\frac{1}{2}\cos^{-1}\left(\frac{1}{r^2}\right)<\theta<\pi-\frac{1}{2}\cos^{-1}\left(\frac{4}{r^2}\right)$$
in the second quadrant. Which makes the integral
$$\int_{\sqrt[4]{17}}^{\sqrt{5}}\int_{\pi-\frac{1}{2}\cos^{-1}\left(\frac{1}{r^2}\right)}^{\pi-\frac{1}{2}\cos^{-1}\left(\frac{4}{r^2}\right)}r^5\cos2\theta\:d\theta dr = \int_{\sqrt[4]{17}}^{\sqrt{5}} \frac{r^3}{2}\left(\sqrt{r^2-1}-\sqrt{r^2-4}\right)dr$$
$$ = \int_{\sqrt{17}}^{5} \frac{t}{4} \left(\sqrt{t-1}-\sqrt{t-4}\right)dt$$
Integration by parts takes care of the rest.
$$ = \boxed{\frac{113}{15}-\frac{\sqrt{17}}{6}\left[(\sqrt{17}-1)^{\frac{3}{2}}-(\sqrt{17}-4)^{\frac{3}{2}}\right]+\frac{1}{15}\left[(\sqrt{17}-1)^{\frac{5}{2}}-(\sqrt{17}-4)^{\frac{5}{2}}\right]}$$
As a side note, this answer would have come out a lot more cleanly had the lower bound been $4$ instead of $\sqrt{17}$ giving
$$\frac{113}{15}-\frac{7\sqrt{3}}{5}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to minimise the area of the folded part of a piece of paper when it touches the other side I am studying maths purely out of interest and have come across this question in my text book:
A rectangular piece of paper ABCD is folded about the line joining points P on AB and Q on AD so that the new position of A is on CD. If AB = a and AD = b, where $a \ge\frac{2b}{\sqrt3}$, show that the least possible area of the triangle APQ is obtained when the angle AQP is equal to $\frac{\pi}{3}$. What is the significance of the condition $a \ge\frac{2b}{\sqrt3}$?
I realise I need to get an equation for the area of the fold, which I can then differentiate. Even then, I am not sure how to relate this to angle AQP.I have looked at solutions on the internet but they tend to look at the length of the crease, not the area of the fold.
I have taken A in my diagram to represent its position before the fold. I could be mistaken.
This is how I visualise it:
I have said:
$x^2 = m^2 + (b - x)^2 = m^2 + b^2 - 2bx +x^2 \implies m^2 = 2b(x - \frac{b}{2})$
$L^2 = (L - m)^2 + b^2 \implies L = \frac{m^2 + b^2}{2m}$
$y^2 = x^2 + L^2 = x^2 + \frac{(m^2 + b^2)^2}{4m^2} = x^2 + \frac{(2bx - b^2 + b^2)^2}{4m^2} = x^2 + \frac{4b^2x^2}{8b(x - \frac{b}{2})} = \frac{bx^2}{2x - b}$
After this I am not sure how to proceed.
| If $\angle AQP = \theta$, $\angle APQ = 90^0 - \theta$.
Draw a perp from $A_1$ to line segment $AP$ and say it is point $A_2$ on line segement $AP$.
Then $A_1A_2 = b$ and $\angle A_2PA_1 = 2 \ \angle APQ = 180^0 - 2\theta \ $.
So $AP = L = \frac{A_1 A_2}{\sin(180^0 - 2 \theta)} = \frac{b}{\sin (180^0 - 2 \theta)} = \frac{b}{\sin 2 \theta} \ $ (in $\triangle A_1PA_2$)
$AQ = x = L \cot \theta = \frac{b}{2 \sin^2 \theta}$
Area of $\triangle APQ = \frac{1}{2} \ L \ x = \frac{b^2}{8 \sin^3 \theta \cos \theta}$
So to minimize area of $\triangle APQ$, we need to maximize $f(\theta) = \sin ^3 \theta \ \cos \theta$.
$f'(\theta) = 3 \sin^2 \theta \cos^2 \theta - \sin^4 \theta = 0$
$\implies \sin \theta = 0, \tan \theta = \pm \sqrt3$. This leads to $\theta = \frac{\pi}{3}$.
Which also tells that $L = \frac{b}{\sin 2 \theta} = \frac{2b}{\sqrt3}$
Also note that for us to be able to fold such that we obtain minimum area,
$a \geq L = \frac{2b}{\sqrt3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $e^{2z} - 2e^z + 2= 0$ So I've started by looking at
\begin{align} e^{z} = x:\\
x^2 - 2x + 2 = 0
\end{align}
Whose solutions should be: \begin{align}\ 1+i, 1-i \end{align}
Then I did the following:
\begin{align}
e^z=e^{x+iy}=e^x(\cos y+i\sin y)&=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{1}{2}+\frac{i \cdot 1}{2}\\
e^x= {\sqrt{2}} \quad \text{and}\quad y&=\frac{\pi}{4}+2\pi k\quad k \in \Bbb{Z}\\
x&= \sqrt{2} \quad \text{and} \quad y=\frac{\pi}{4}\\
\text{Also,} \quad e^x(\cos y+\sin y)&=\cos(\frac{7\pi}{4})+i\sin\frac{7\pi}{4})=\frac{1}{2}-\frac{i \cdot 1}{2}\\
e^x= \sqrt{2} \quad \text{and}\quad y&=\frac{7\pi}{4}+2\pi k\quad k \in \Bbb{Z}\\
x&= \sqrt{2} \quad \text{and} \quad y=\frac{7\pi}{4}\\
\end{align}
Are the solutions the following? \begin{align} \sqrt{2} + \frac{\pi}{4} \quad \text{and} \quad \sqrt{2} + \frac{7\pi}{4} \end{align}
Forgive the shoddy formatting as this is my first ever post here. Thanks.
| Yes, the solutions of $x^2-2x+2=0$ are $1+i$ and $1-i$. So, the solutions of $e^{2z}-2e^z+2=0$ are all those numbers $z$ such that $e^z=1+i$ or that $e^z=1-i$. But$$1+i=\sqrt2\left(\frac1{\sqrt2}+\frac i{\sqrt2}\right)=\sqrt2e^{\pi i/4}. $$Therefore$$e^z=1+i\iff z=\log\sqrt2+\frac{\pi i}4+2\pi in,$$for some $n\in\Bbb Z$. The case of the equality $e^z=1-i$ is similar.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $p+\frac{2p-1}{ \left(\frac{1-p}{p}\right)^n -1} \leq \frac{1}{2} - \frac{1}{2n}$, if $0
Prove that $p+\frac{2p-1}{ \left(\frac{1-p}{p}\right)^n -1} \leq \frac{1}{2} - \frac{1}{2n}$, if $0<p<1, n\in \mathbb N$
This is from a recently closed question.
Notice that the fraction on the LHS is well defined. The denominator can be written as
$$\frac{1}{p^n} \left( (1-p)^n - p^n \right)=\frac{1-2p}{p^n} g(p)$$
where $g(p)$ is a polynomial of $p$ taking positive values. When $p=\frac 12$, the LHS becomes
$p-\frac{1}{2^n g(\frac 12)}$.
My effort was shown here with variable substitution and taking derivatives. I'd like to know if there's a simpler, elementary proof without calculus. Thanks.
| No calculus is required.
Using your substitution in your linked effort:
Let $x=\frac{1-p}{p}$ then $p=\frac{1}{1+x}, x>0$. The inequality becomes
$$\frac{1}{1+x}+\frac{1}{1+x}\cdot \frac{1-x}{x^n-1} = \frac{x-x^n}{(1+x)(1-x^n)} \le \frac1 2-\frac{1}{2n}$$
$$\iff g(x) = \frac{x-x^n}{(1+x)(1-x^n)} \le \frac 12 - \frac{1}{2n}\tag 1$$
From here, notice that
$$ \begin{array}{ll} g(x) & = &\frac{ x}{1+x} \times \frac{ 1 - x^{n-1 }} { 1 - x^n} \\
&=& \frac{x}{1+x} \times \frac{ 1 + x + \ldots + x^{n-2} } {1 + x + \ldots + x^{n-1} } \\
& =& \frac{ x + \ldots + x^{n-1}} { 1 + 2x + 2x^2 + \ldots + 2x^{n-2} + 2x^{n-1} + x^{n} } \\
& =& \frac{1}{2} - \frac{1}{2} \times \frac{ 1+x^n} {1 + 2x + 2x^2 + \ldots + 2x^{n-2} + 2x^{n-1} + x^{n}} \end{array}$$
So, it remains to show that
$$ 1 + 2x + 2x^2 + \ldots + 2x^{n-2} + 2x^{n-1} + x^{n} \leq n ( 1 + x^n). $$
This is obvious, since $ x^k + x^{n-k} \leq 1 + x^n$ for $ 0 \leq k \leq n$.
(EG by factoring, if you want to avoid convexity-calculus arguments ).
As you mentioned, equality holds iff $ x =1$ or $ n = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the product of the factors of $6^{12}$ that are congruent to $1 \pmod 7$
If the product of the factors of $6^{12}$ that are congruent to $1 \pmod 7$ can be expressed as $2^m \cdot 3^n$, prove that $m=n$ and find $m$.
This may be considered a sub-problem from another problem. Although I have greatly reduced calculation in my solution, if we can give a simple proof that $m=n$ before we compute them, we'd further cut calculation in my answer in half.
My efforts so far (adapted from my solution to the linked problem):
Case 1: $2^x \equiv 3^y \equiv 1$.
*
*There are $3$ cases when $3^y\equiv 1$ for each $x \in \{0,3,6,9,12\}$ so the sum of all these $x$'s is $3 \cdot (3+6+9+12)=3 \cdot 30=90$;
*There are $5$ cases when $2^x\equiv 1$ for each $y\in\{0,6,12\}$ so the sum of all these $y$'s is $5\cdot(0+6+12)=90$;
Case 2: $2^x \not \equiv 1$. Note that we must have $3^y \equiv 2,4$ otherwise we can't have $3^y 2^x \equiv 1$.
*
*There are $2$ cases when $3^y \equiv 2^{-x}$ for each $x\in\{1,2,4,5,7,8,10,11\}$ and the sum of all these $x$'s is $2\cdot(78-30)=96$, where $78=\sum_{i=0}^{12} i$;
*There are $4$ cases when $2^x\equiv 3^{-y}$ for each $y \in \{2,4,8,10\}$ so the sum of all these $y$'s is $4\cdot (2+4+8+10)=96$;
Therefore $m=n=186$.
However I feel that there is a simple argument to just prove $m=n$ so that we can cut our calculation in half. One thing I came up with to prove the sums of $x$'s and $y$'s are equal in Case 1 is as follows:
Denote $o_2, o_3$ as the orders of the elements $2$, $3$, $\pmod 7$, respectively, and $M=12$. Then in Case 1 we need to show $$3\cdot (3+6+9+12) = 5 \cdot (0+6+12)\\ \iff \color{blue}{3} \cdot \color{red}{3} \cdot \color{brown}{\frac{4(4+1)}{2}} = \color{green}{5} \cdot \color{purple}{6} \cdot \color{red}{\frac{2(2+1)}{2}} $$ or equivalently $$\color{blue}{\left(\frac{M}{o_3}+1\right)} \cdot \color{red}{o_2} \cdot \color{brown}{\frac{\frac{M}{o_2} \left(\frac{M}{o_2}+1\right)}{2}} = \color{green}{\left(\frac{M}{o_2} + 1\right)} \cdot \color{purple}{o_3} \cdot \color{red}{\frac{\frac{M}{o_3} \left(\frac{M}{o_3}+1\right)}{2}} \tag 1$$ which is indeed true. And for Case 2 we could do something similar but it would be lengthy and I might as well compute them directly even though there's unnecessary repetition.
My questions:
*
*Is there any better way, e.g., via some simple argument, and better yet combining the two cases together (instead of treating them separately), that $m=n$, $\color{blue}{\it{\text{even before}}}$ we compute them? If it turns out lengthy then it's not worth typing the extra words to replace the few lines of calculation in Case 1 and Case 2 I presented earlier.
*Is there a simpler way to solve the original problem by showing $a=b=c$ before computing them?
If the product of the factors of $30^{12}$ that are congruent to 1 mod 7 can be expressed as $2^{a} \cdot 3^{b} \cdot 5^{c},$ find $a+b+c$.
| If $a$ is a factor of $6^{12}$ and $a \equiv 1 \pmod 7$, then $\frac{6^{12}}{a} \equiv \frac{(-1)^{12}}{1} \equiv 1 \pmod 7$, and these pairs multiply to $6^{12}$. In the special case where $a = 6^6$ and $a = \frac{6^{12}}{a}$, verify that $a \equiv 1 \pmod 7$. Therefore $m = 12k + 6, k \in \mathbb N$.
| {
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$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$
Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$
The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.
The first graph looks pretty weird so please help as to how to proceed with this graphically or an easier algebraic method.
Thanks :)
| WLOG
$$\sqrt{x^2+12y}=33\cos^2t\text{ and }\sqrt{y^2+12x}=33\sin^2t$$
$$(33\cos^2t)^2-(33\sin^2t)^2=x^2-y^2-12(x-y)=(x-y)(x+y-12)$$
$$x+y=23\implies x-y=99(\cos^2t-\sin^2t)=99(2\cos^2t-1)$$
$$\implies x=99\cos^2t-38\implies y=61-99\cos^2t$$
$$\implies(33\cos^2t)^2=x^2+12y=(99\cos^2t-38)^2+12(61-99\cos^2t)$$
Let $33\cos^2t=b$ $$\implies0=b^2-33b+272=(b-16)(b-17)$$
Hope you can take it from here?
| {
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When will matrix multiplication become just "concatenation"? Saw below entertaining matrix multiplication examples.
Obviously they are just coincidences. But I am curious when does below hold?
\begin{equation}
\begin{pmatrix}
a_1 & b_1 \\
c_1 & d_1
\end{pmatrix} \times \begin{pmatrix}
a_2 & b_2 \\
c_2 & d_2
\end{pmatrix} = \begin{pmatrix}
\overline{a_1 a_2} & \overline{b_1 b_2} \\
\overline{c_1 c_2} & \overline{d_1 d_2}
\end{pmatrix}
\end{equation}
here $\overline{a_1 a_2}$ means gluing the integers together, not multiplications.. assuming we are working on integer matrices.
| Let $A$ denote the first matrix in the product, and let $B$ denote the second.
I will consider only the case in which the entries of $B$ have one digit.
In this case, the "gluing" property of this matrix multiplication can be written as
$$
AB = 10A + B.
$$
Note that this equation can be rearranged into
$$
AB - 10 A - B + 10I = 10I \implies\\
(A - I)(B - 10 I)= 10 I,
$$
where $I$ denotes the identity matrix. Now, suppose that we select a matrix $B$ and we want a corresponding matrix $A$. We have
$$
(A - I)(B - 10 I) = 10 I \implies\\
A = 10(B - 10 I)^{-1} + I.
$$
Note that this equation only has a solution if $\det(B - 10 I) \neq 0$. Now, a question remains: how do we ensure that $10 (B - 10 I)^{-1}$ is an integer matrix? As it turns out, this will hold for a given integer matrix $B$ if and only if the determinant of $B - 10 I$ divides $10$.
In fact, we can generate pairs of matrices with non-negative integer entries that have the gluing property via the following steps:
*
*Find a matrix $C$ whose diagonal entries satisfy $-10 \leq c_{ii} \leq -1$ and whose off-diagonal entries are a single digit positive number such that the determinant of $C$ is either $-1$, $-2$, $-5$, or $-10$.
*Take $B = 10 I + C$ and $A = 10C^{-1} + I$.
For example, the matrix
$$
C = \pmatrix{-4 & 3\\7 &-4}
$$
has determinant $16 - 21 = -5$, which divides $10$. The corresponding matrix $B$ is
$$
B = 10 I + \pmatrix{-4 & 3\\7 &-4} = \pmatrix{6&3\\7&6}.
$$
The associated matrix $A$ is
$$
10C^{-1} + I = \frac{10}{-5} \cdot \pmatrix{-4 & -3\\-7 & -4} + \pmatrix{1&0\\0&1}
= \pmatrix{9 & 6\\14 & 9}.
$$
If we compute the product, we indeed find that
$$
\pmatrix{\color{blue}{9} & \color{blue}{6}\\
\color{blue}{14} & \color{blue}{9}} \cdot
\pmatrix{\color{green}{6} & \color{green}{3}\\
\color{green}{7} & \color{green}{6}} =
\pmatrix{\color{blue}{9}\color{green}{6} & \color{blue}{6}\color{green}{3}\\
\color{blue}{14}\color{green}{7} & \color{blue}{9}\color{green}{6}}.
$$
An interesting phenomenon: if $B$ is a single digit matrix for which there exists an $A,B$ pair with this "gluing" property, we will have $A = B$ if and only if $B$ is a "vampire matrix" (cf. my comment on the question), which holds if and only if $B$ has eigenvalues $0,11$, which holds if and only if $C = B - 10 I$ has eigenvalues $-10,1$, which holds if and only if $C$ has determinant $-10$ and trace $-9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 1
} |
Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$ Find a $g\in K$ such that $g^2=x^3+x+1$ Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$. Find a $g\in K$ such that $g^2=x^3+x+1.$
I tried $x^3+x+1$ itself but unfortunately the degree is only $2$. I don't know how I can multiply something and get a polynomial of degree $3$.
| The Quotient ring ${Z_2[x]/(x^4+x^2+x)}$ contains below elements.$${\{1,x,x^2,x^3,x^2+x,x^3+x^2,x^3+x^2+x,x+x^3\}}$$ if we square each of them, then we can see that for ${(x^2+x)^2=x}$, so ${\sqrt{x}=(x^2+x)=p }$(as defined above by Teresa)
*
*now ${(x^3+x+1)^2= x^6+x^2+1= x^3+x^2+x+x^2+1=x^3+x+1}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Is this a Viable/New Proof for Pythagoras Theorem? A triangle with side lengths $a$, $b$, $c$ and a height ($h$) that intercepts the hypotenuse ($c$) such that $c$ is split into two side lengths, $c = m + n$, we can find Pythagorean theorem using simple trigonometry.
We start by finding the sum of tan ratios that are equivalent,
$$\tan^{-1}\frac{a}{b}+\tan^{-1}\frac{b}{a} = 90^\circ$$
$$\tan^{-1}\frac{h}{m}+\tan^{-1}\frac{h}{n} = 90^\circ$$
$$\frac{a}{b}+\frac{b}{a}=\frac{h}{m}+\frac{h}{n}$$
Using $ab = hc$, Substitute $h = \dfrac{ab}{c}$,
$$\frac{a}{b}+\frac{b}{a}=\frac{ab}{cm}+\frac{ab}{cn}$$
After simplifying, we get:
$$\begin{align}
c&=\frac{ma^2b^2+na^2b^2}{mn\left(a^2+b^2\right)} \\[4pt]
&=\frac{(m + n)(a^2b^2)}{mn\left(a^2+b^2\right)} \\[4pt]
&=\frac{c(a^2b^2)}{mn\left(a^2+b^2\right)} \\[4pt]
mn &=\frac{a^2b^2}{\left(a^2+b^2\right)}
\end{align}$$
Note that $\sqrt{mn}=\dfrac{ab}{\sqrt{a^{2}+b^2}}$ and $h = \dfrac{ab}{c}$, this tells us that $\sqrt{mn} = h$.
$$\begin{align}
\frac{ab}{c}&=\frac{ab}{\sqrt{a^2+b^2}} \\[4pt]
c^2 &= a^2 + b^2
\end{align}$$
| This amounts to a well-known similar-triangles argument, splitting the area, that $(a/c)^2+(b/c)^2=1$, which is a bit more direct.
As @DavidK notes, your opening use of trigonometry is invalid, and its conclusion (in fact, something stronger) should instead be deduced as per @WillOrrick's suggestion, giving $\tfrac{a}{b}=\tfrac{h}{m}$ and $\tfrac{b}{a}=\tfrac{h}{n}$, whence $1=\tfrac{h^2}{mn}=\tfrac{(ab)^2}{mnc^2}$, and the Pythagorean theorem is equivalent to $mn=\tfrac{(ab)^2}{a^2+b^2}$. You also use $\tfrac{a^2+b^2}{ab}=\tfrac{hc}{mn}=\tfrac{ab}{mn}$, which completes the proof.
So your strategy amounts to combining two formulae for area with$$u_1=u_2,\,v_1=v_2,\,u_1v_1=1\implies u_1+v_1=\tfrac{u_2+v_2}{u_2v_2}$$(where $u_1=v_1^{-1}=\tfrac{a}{b},\,v_1=\tfrac{h}{m},\,v_2=\tfrac{h}{n}$). But compare $\tfrac{a}{b}=\cot\theta$ with$$\left(\tfrac{a}{c}\right)^2=\tfrac{[\triangle CBH]}{[\triangle ABC]}=\tfrac{m}{m+n}=\tfrac{m/h}{v_1^{-1}+v_2^{-1}}=\tfrac{\cot\theta}{\cot\theta+\tan\theta}.$$Similarly (viz. $\theta\mapsto\tfrac{\pi}{2}-\theta$), $\left(\tfrac{b}{c}\right)^2=\tfrac{\tan\theta}{\cot\theta+\tan\theta}$. The proof technique I pointed to - that of seeing the Pythagorean theorem - notes the sum of these squares is $\tfrac{\color{blue}{blue}+\color{green}{green}}{\text{all}}=1$. Your proof works with fractions $k:=\cot\theta+\tan\theta$ times these, so Einstein's proof is equivalent to$$1=\tfrac{a/b+b/a}{k}=\tfrac{a^2+b^2}{kab}=\tfrac{a^2+b^2}{kch},$$where the last step uses your equation of two formulae for $[\triangle ABC]$, and the proof that $\tfrac{a^2+b^2}{kch}=1$ uses $kh=m+n=c$ by your $AB$-splitting observation that defines $m,\,n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3992151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Prove $(x-1)(y-1)>(e-1)^2$ where $x^y=y^x$, $y>x>0$.
Let $x,y$ be different positive real numbers satisfying $x^y=y^x$.
Prove $(x-1)(y-1)>(e-1)^2$.
We may suppose $a=\frac{y}{x}>1$. Then we obtain $x=a^{\frac{1}{a-1}},y=a^{\frac{a}{a-1}}.$ But how to go on?
| Continuing your idea, let $a = 1+ \dfrac 1t, t > 0.$ Then, you can write your inequality as $xy-x-y > e^2-2e$, which is in terms of the variable $t:$
$$f(t) = t\ln\left(1+\frac 1t\right) + \ln\left(\left(1+\frac 1t\right)^{t+1}-2-\frac 1t\right) > 1+\ln(e-2).$$
Wolfram says that this is indeed a decreasing function in $t$, so the infimum will be:
$$\lim\limits_{t\to\infty} f(t) = 1+ \ln(e-2).$$
Then, the only thing to show now is the fact that $f(t)$ is decreasing, but I suspect this will take an ugly computation.
EDIT: Actually, the derivative did not turn out to be too bad:
\begin{align}f'(t) &= \\
&\dfrac{\dfrac{1}{t^2} + \left(1+\dfrac 1t\right)^{t+1}{\left(2\ln\left(1+\frac 1t\right)-\dfrac 1t-\dfrac{1}{t+1}\right)} + \left(2+\dfrac{1}{t}\right)
\left(\dfrac{1}{1+t} - \ln\left(1+\dfrac{1}{t}\right)\right)}{\left(1+\dfrac 1t\right)^{t+1} - \dfrac 1t-2} &
\\<0\end{align}
Finally, the above is equivalent to:
$$\left(1+\dfrac 1t\right)^{t+1} > \dfrac{\dfrac{1}{t^2} + \left(2+\dfrac 1t\right)\left(\dfrac{1}{1+t} - \ln\left(1+\dfrac{1}{t}\right)\right)}{\dfrac{1}{t}+\dfrac{1}{1+t} - 2\ln\left(1+\dfrac{1}{t}\right)},$$
which I think should be doable by brute-force expanding with a Taylor's series etc.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Integrating an Infinite exponent tower $$ \int_0^1 x^{2^{x^{2^{x^{\ldots}}}}} ~~ dx = ~~?$$
What I've tried so far:
$$ x^{2^{x^{2 \ldots}}} \overbrace{=}^{\text{def}} y \\ x^{2^{y}} = y \\ 2^y \ln (x) = \ln(y)$$
Now applying the chain rule:
$$\begin{align} \\&2^y \cdot \frac{dy}{dx} \cdot \ln(2) \cdot \ln(x) + \frac{1}{x} \cdot 2^y \ln(y) = \frac{dy}{dx} \cdot \frac1y \\ &\implies \frac{dy}{dx} \left (2^y \cdot \ln(x) \cdot \ln(2) - \frac1y \right) = - \frac1x 2^y \cdot \ln(y) \\ &\implies \frac{dy}{dx} = \left ( \frac{- \frac1x 2^y \cdot \ln(y)}{2^y \cdot \ln(x) \cdot \ln(2) - \frac1y} \right) \end{align}$$
The problem that this looks like a pretty harsh equation to solve for $dy$ and then substitute it back to the integral, with the fact that we need to find the value of $x$ in terms of $y$.
Any help would be appreciated!
| Consider
$$I=\int_0^1 y^{2^{-y}}\,dy=\int_0^\frac 12 y^{2^{-y}}\,dy-\int_1^\frac 12 y^{2^{-y}}\,dy=I_1-I_2$$
Each integrand will be developed as a series expansion around the lower bound to order $(n+1)$ and termwise integrated. This does not lead to very complicated expressions for the integrands and the result of integration is just a polynomial in $\log(2)$.
For example, for $n=3$, the integrands are
$$y-y^2 \log (2) \log (y)+\frac{1}{2} y^3 \left(\log ^2(2) \log ^2(y)+\log ^2(2) \log
(y)\right)+O\left(y^4\right)$$
$$1+\frac{y-1}{2}+(y-1)^2 \left(-\frac{1}{8}-\frac{\log (2)}{2}\right)+(y-1)^3
\left(\frac{1}{16}+\frac{\log ^2(2)}{4}\right)+O\left((y-1)^4\right)$$
and the result of integration is
$$\frac{5127+72 \log ^4(2)-36 \log ^3(2)+339 \log ^2(2)-64 \log (2)}{9216}$$
As a function of order $n$, some results
$$\left(
\begin{array}{cc}
n & I_1-I_2 \\
1 & 0.56250000 \\
2 & 0.57249702 \\
3 & 0.56967702 \\
4 & 0.56945295 \\
5 & 0.56962947 \\
6 & 0.56968187 \\
7 & 0.56969293 \\
8 & 0.56969494 \\
9 & 0.56969522 \\
10 & 0.56969526 \\
11 & 0.56969527 \\
12 & 0.56969528
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$.
Evaluate:
$$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$$
The only thing I can think of doing here is long division to simplify the integral down and see if I can work with some easier sections. Here's my attempt:
\begin{align}
\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx &= \int \left(\:x^4+3x^2+4x+3+\frac{18}{x+1} \right )dx \\
&= \int \:x^4dx+\int \:3x^2dx+\int \:4xdx+\int \:3dx+\int \frac{18}{x+1}dx \\
&= \frac{x^5}{5}+x^3+2x^2+3x+18\ln \left|x+1\right|+ c, c \in \mathbb{R}
\end{align}
The only issue I had is that the polynomial long division took quite some time. Is there another way to do this that is less time consuming? The reason I ask this is that, this kind of question can come in an exam where time is of the essence so anything that I can do to speed up the process will benefit me greatly.
| Horner's method for division:
$$(x^2+x+3)(x^3+7)=x^5+x^4+3x^3+7x^2+7x+21$$
\begin{array}{*{7}{r}}
& 1 & 1 & 3 & 7 & 7 & 21 \\
+ & \downarrow & -1 & 0 & -3 & -4 & -3 \\
\hline
\times -1 & \color{red}1 & \color{red}0 & \color{red}3 & \color{red}4 & \color{red}3 & \color{cyan}{18}
\end{array}
so the quotient is $\;x^4+3x^2+4x+3$ and the remainder is $\;18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
The value of $\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$ What is the value of this expression :
$$\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$$
Using calculator and wolfram alpha, the answer is, $-\frac{1}{\sqrt{5}}$
But, by solving it myself the result comes out to be different. My solution is as follow:
$$\begin{aligned} Put,\, &\cot^{-1}\left(\frac{-3}{4}\right) = \theta
\\\implies &\cot(\theta) = \frac{-3}{4} =\frac{b}{p}
\\So,\,&\cos(\theta) = \frac{-3}{5}\end{aligned}$$
$$\begin{aligned}\\Then,
\\\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right) &= \sin\left(\frac{\theta}{2}\right) \\&= \sqrt{\frac{1-\cos{\theta}}{2}}
\\&= \sqrt{\frac{1+\frac{3}{5}}{2}}
\\&= \frac{2}{\sqrt{5}}
\end{aligned}$$
This solution is used by a lot of websites.
So, I got two different values of single expression but I am not sure which one is correct. Can you point out where I have done the mistake?
| Both are correct, but the answer you obtain depends on the definition $\cot^{-1}(x)$, in particular its range, since inverse trigonometric functions are multi-valued.
If you write $\cot^{-1}(x) = \tan^{-1}(1/x)$ [as a calculator seldom has $\cot^{-1}$] and use the range $-\pi/2 < \tan^{-1}x < \pi/2$, we see that $\theta = \cot^{-1}(-3/4)$ must be negative.
In $-\pi/2 < x < 0$, $\cos x > 0$ and $\sin x < 0$. Hence $\cos \theta = \dfrac35$, and $\sin x = -\sqrt{\dfrac {1-3/5}2} = -\dfrac {1}{\sqrt 5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Integrate $\int \prod_{i=0}^n \frac{1}{(x+i)}dx$ I initially tried to find a pattern using partial fraction decomposition:
$$
n= 1 \, : \, \, \frac{1}{x(x+1)} = \frac{1}{x}+\frac{-1}{x+1}
$$
$$
n=2 \, : \, \, \frac{1}{x(x+1)(x+2)} = \frac{1/2}{x}+\frac{-1}{x+1}+\frac{1/2}{x+2}
$$
$$
n=3 \, : \, \, \frac{1}{x(x+1)(x+2)(x+3)} = \frac{1/6}{x}+\frac{-1/2}{x+1}+\frac{1/2}{x+2} + \frac{-1/6}{x+3}
$$
$$
n=4 \, : \, \, \frac{1}{x(x+1)(x+2)(x+3)(x+4)} = \frac{1/24}{x}+\frac{-1/6}{x+1}+\frac{1/4}{x+2} + \frac{-1/6}{x+3} + \frac{1/24}{x+4}
$$
But I don't seem to find a pattern. Is there another method to resolve this? Thank you.
| You may use:
$$\prod_{k=0}^{n} \frac{1}{x+k}=\frac{1}{n!}\sum_{k=1}^{n} (-1)^{k-1} \frac{k{n \choose k}}{x(x+k)}=\frac{1}{n!}\sum_{k=1}^{n}(-1)^{k-1} {n \choose k} [\frac{1}{x}-\frac{1}{x+k}]$$
So $$\int \prod_{k=0}^{n} \frac{1}{x+k} dx=\frac{1}{n!}\sum_{k=1}^{n}(-1)^{k-1} {n \choose k}[\ln x- \ln (x+k)]$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Constructing factor-/quotient groups
Given:$$\Bbb Z_8 =\{0,1,2,3,4,5,6,7\}$$ and $$H=\{0,2,4,6\} = \left<[2]\right>$$, construct the factor group $\frac{\Bbb Z_8}{H}.$
Definition:
For a normal subgroup $H$ of $G$, the group $\frac{G}{H}$ is called the factor-/ or quotient group of $G$ modulo $H$. The set $\frac{G}{H}$ = {$aH | a \in G$} (left-coset) is a group under the operation $(aH)(bH)=abh$.
\begin{align}\\
\frac{\Bbb Z_8}{H}
& =\{gH|g\in \Bbb Z_8\} \\
& =\{g+H|g\in \Bbb Z_8\}
&\text{with }
& g+H=\{g+h|h\in H\}\\
\frac{\Bbb Z_8}{H}
& =\{H, 1+H, 2+H, ..., 7+H\}\\
\end{align}
So we have:
\begin{align}\\
& 0+H=\{0,2,4,6\}&& 4+H=\{4,6,8,10 \}\\
& 1+H=\{1,3,5,7\}&& 5+H=\{5,7,9,11 \}\\
& 2+H=\{2,4,6,8\}&& 6+H=\{6,8,10,12\}\\
& 3+H=\{3,5,7,9\}&& 7+H=\{7,9,11,13\}\\
\end{align}
Taking modulo 8, we get:
\begin{align}\\
&0+H\mod 8=\{0,2,4,6\}&& 4+H\mod 8=\{4,6,0,2\}\\
&1+H\mod 8=\{1,3,5,7\}&& 5+H\mod 8=\{5,7,1,3\}\\
&2+H\mod 8=\{2,4,6,0\}&& 6+H\mod 8=\{6,0,3,4\}\\
&3+H\mod 8=\{3,5,7,1\}&& 7+H\mod 8=\{7,1,3,5\}\\
\end{align}
We see that:
\begin{align}\\
(1+H)\mod 8 =(3+H)\mod 8 =(5+H)\mod 8 =(7+H)\mod 8 &&=\{1,3,5,7\}
\\
(2+H)\mod 8 =(4+H)\mod 8 =(6+H)\mod 8 &&=\{0,2,4,6\}
\end{align}
I calculated the cosets of subgroup $H=\{0,2,4,6\}=\left<[2]\right>$ of $\Bbb Z_8$:
\begin{align}\\
& 0+\left<[2]\right> & =\{0,2,4,6\} \\
& 1+\left<[2]\right> & =\{1,3,5,7\} \\
\end{align}
I don't know how to proceed from here, or even if I'm on the right track. I thougt I would end up with a set of pairs, but I keep ending up calculating the same two sets.
| Those two cosets do indeed form $\Bbb Z_8/H$ (under the operation $(a+H)+(b+H):=(a+b)+H$). Well done! You could improve your answer by noting that $0+H=H$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Solve $\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}$ Solve $$\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}.$$ I was thinking about letting $x^2=u,y^2=v.$ Then we will have $$\begin{cases}u+v^2=20\Rightarrow u=20-v^2\\u^2+v=20\end{cases}.$$ If we substitute $u=20-v^2$ into the second equation, we will get $$v^4-40v^2+v+380=0$$ which I can't solve because we haven't studied any methods for solving equations of fourth degree (except $ax^4+bx^2+c=0$). Any other methods for solving the system?
| By substracting the two equations you get
$$ x^2(1-x^2) + y^2(y^2-1) = 0$$
$$ -(x^2-\frac12)^2 + (y^2-\frac12)^2 = 0 $$
so
$$ x^2 - \frac12 = \pm (y^2-\frac12) $$
that is
$$ x^2 = y^2 $$
or $$ x^2 = 1- y^2 $$
Analysing theses two cases it's easy to solve the equations.
| {
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"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
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Evaluate the integral $\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx$ I try to calculate the integral
\begin{align}
f(y,z):=\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx
\end{align}
on the set $y>0$ and $0<z<1$. I could show by some calculation that in the limit $z\to0$ holds
\begin{align}
f(y,0)=e^{y/2}K_0(y/2),
\end{align}
where $K_0$ denotes the modified Bessel function of the second kind and
\begin{align}
\lim_{z\to1}f(y,z)=f(y,1)=\infty,\\
\lim_{y\to0}f(y,z)=f(0,z)=\infty.
\end{align}
But I can't evaluate the integral in the interior of the defined set. Any help, ideas or hints how to solve this would be appreciated.
Here is a selection of what I've tried:
*
*several substitutions (e.g. $\sinh$,$\dots$),
*integration by parts,
*tried to evaluate the integral $$\partial_zf(y,z)=\int_{\mathbb R} -yx\cdot \exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)\mathrm dx, $$
*tried to evaluate the integrals $$\partial_yf(y,z)=\int_{\mathbb R} -(x^2+zx\sqrt{x^2+1})\cdot \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx $$ and
$$f(y,z)-\partial_yf(y,z)=\int_{\mathbb R} (\sqrt{x^2+1}+zx)\cdot \exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)\mathrm dx. $$
| \begin{align}
f(y,z)=&\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx\\
=&\frac{1}{2}\int_{\mathbb R} \exp\left(-y(\sinh^2\frac{u}{2}+z\cdot \sinh\frac{u}{2}\cosh\frac{u}{2} ) \right)\mathrm du\\
=&\frac{1}{2}\int_{\mathbb R} \exp\left(-\frac{y}{2}(\cosh u -1+z\cdot \sinh u ) \right)\mathrm du\\
=&\frac{1}{2}e^{y/2}\int_{\mathbb R} \exp\left(-\frac{y}{2}(\cosh u+z\cdot \sinh u ) \right)\mathrm du
\end{align}
Until this point I used the substitution $x=\sinh\frac{u}{2}$ and the equation $\sinh^2\frac{u}{2}=\frac{1}{2}(\cosh u -1)$, as well as $\sinh\frac{u}{2}\cosh\frac{u}{2}=\frac{1}{2}\sinh u$. From this I could see for the case $z=0$, that we have $f(y,0)=e^{y/2}K_0(y/2)$ using $$K_0(\phi)=\int_0^\infty\exp\left(-\phi\cosh x \right)\mathrm dx,\qquad \forall \phi>0.$$
To go on from this someone helped me with the hint $$ \cosh u+z\cdot \sinh u =\sqrt{1-z^2}\cosh(u+\tanh^{-1} z).$$
This brings us to
\begin{align}
f(y,z)=&\frac{1}{2}e^{y/2}\int_{\mathbb R} \exp\left(-\frac{y}{2}\sqrt{1-z^2}\cosh(u+\tanh^{-1} z) \right)\mathrm du\\
=&\frac{1}{2}e^{y/2}\int_{\mathbb R} \exp\left(-\frac{y}{2}\sqrt{1-z^2}\cosh u \right)\mathrm du\\
=&e^{y/2}\int_{0}^\infty \exp\left(-\frac{y}{2}\sqrt{1-z^2}\cosh u \right)\mathrm du\\
=&e^{y/2}K_0\left(\frac{y}{2}\sqrt{1-z^2}\right).
\end{align}
The idea behind the hint was to find values $\beta, v$ such that we can use the addition theorem
\begin{align}
\cosh u+z\cdot \sinh u=\frac{1}{\beta}(\beta\cosh u+z\beta\cdot \sinh u)\overset{!}{=}&\frac{1}{\beta}(\cosh v\cosh u+\sinh v\sinh u)\\
=&\frac{1}{\beta}\cosh(u+v),
\end{align}
as a shift $v$ of the argument doesn't affect the integral since we are integrating from $-\infty$ to $\infty$. To find these values we need to have $\beta=\cosh v$ and $z\beta=\sinh v$. This brings us to
$$v=\tanh^{-1}z$$
and $$\beta=\frac{1}{\sqrt{1-z^2}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral involving Airy function I met the following integral when I was reading a paper:
$$\int_0^\infty Ai(y)dy=\frac{1}{3},$$
where
$$Ai(y)=\frac{1}{\pi}\int_0^\infty \cos(\alpha y+\frac{\alpha^3}{3})d\alpha.$$
The paper adopted one asymptotic result of the Airy function,
$$\int_0^x Ai(y)dy \sim \frac{1}{3}-\frac{1}{2}\pi^{-\frac{1}{2}}x^{-\frac{3}{4}}\exp{\left(-\frac{2}{3}x^\frac{3}{2}\right)}, \quad \text{for large x}.$$
You may find this result on Eq.(10.4.82), Handbook of Mathematical Functions, Abramowitz and Stegun. A similar result is given in Eq.(10.4.83),
$$\int_0^x Ai(-y)dy \sim \frac{2}{3}-\frac{1}{2}\pi^{-\frac{1}{2}}x^{-\frac{3}{4}}\cos{\left(\frac{2}{3}x^\frac{3}{2}+\frac{\pi}{4}\right)}, \quad \text{for large x}.$$
However, I was wondering whether we can see the first result by calculating the integrals directly, and
$$\int_0^\infty Ai(-y)dy=\frac{2}{3}$$
as well.
Any advice or references are appreciated.
| You can obtain it using the integral representation
$$
\operatorname{Ai}(y) = \frac{{\sqrt 3 }}{{2\pi }}\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3} - \frac{{y^3 }}{{3t^3 }}} \right)dt} ,\quad |\arg y|<\tfrac{\pi}{6}
$$
(cf. http://dlmf.nist.gov/9.5.E6). Indeed, by the Fubini theorem and the Gauss multiplication theorem for the gamma function, it is found that your integral is
\begin{align*}
& \frac{{\sqrt 3 }}{{2\pi }}\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3}} \right)\int_0^{ + \infty } {\exp \left( { - \frac{{y^3 }}{{3t^3 }}} \right)dy} dt} \\ &\mathop = \limits^{x = y^3 /(3t^3 )} \frac{1}{{2\pi }}\frac{1}{{3^{1/6} }} \int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3}} \right)t \int_0^{ + \infty } e^{ - x} x^{1/3 - 1} dx dt} \\ & = \frac{1}{{2\pi }}\frac{1}{{3^{1/6} }}\Gamma \left( {\frac{1}{3}} \right)\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3}} \right)tdt}
\\ &\mathop = \limits^{s = t^3 /3} \frac{1}{{2\pi }}\frac{1}{{\sqrt 3 }}\Gamma \left( {\frac{1}{3}} \right)\int_0^{ + \infty } {e^{ - s} s^{2/3 - 1} ds} \\ & = \frac{1}{{2\pi }}\frac{1}{{\sqrt 3 }}\Gamma \left( {\frac{1}{3}} \right)\Gamma \left( {\frac{2}{3}} \right) = \frac{1}{3}.
\end{align*}
For a more general result, see http://dlmf.nist.gov/9.10.E17. It can be derived the same way.
| {
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"answer_count": 1,
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} |
General solution of $\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}?$
How to find the general solution of $$\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}$$
| $t=\arctan x\iff \tan t=x$.
Recall that $1+x^2=1+\tan^2 t=\frac{1}{\cos^2 t}$, so $\cos^2 t=\frac{1}{1+x^2}$.
Now, $$\cos(2\arctan x)=\cos 2t=\cos^2t-\sin^2t=2\cos^2t-1=2\frac{1}{1+x^2}-1=\frac{1-x^2}{1+x^2}$$ and it is a tautology.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
} |
maximum and minimum value of $(a+b)(b+c)(c+d)(d+e)(e+a).$
Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Find the maximum and minimum value of $S=(a+b)(b+c)(c+d)(d+e)(e+a).$
I couldn't proceed much, however I think I got the minimum and maximum case.
For minimum, we get $-512$ with equality on $(-1,-1,-1,-1,9).$
For maximum, we get $288$ with equality on $(-1,-1,-1,4,4).$
| For the maximum, you can easily prove that $4$ of the $5$ product terms cannot be negative. Then, the maximum is achieved either all $5$ are positive or only $3$ are positive. In the former case, simple $AM-GM$ gives the maximum of $32.$ For the latter case, assume $a+b\leq 0$ and $b+c\leq 0$ while the other three are non-negative. In that case, $-a-b\leq 2$ and $-b-c\leq 2$ and $d+e = 5-a-b-c\leq 8.$
Then, $$S = (c+d)(d+e)(e+a)(-a-b)(-b-c)\leq 4(c+d)(e+a)(d+e)\leq(c+d+e+a)^2\cdot 8= 8(5-b)^2\leq 8\cdot 36 = 288.$$
Now the other possibility is when two terms involving $4$ different variables are negative - say $a+b\leq 0$ and $c+d\leq 0.$ In this case, $b+c\geq 0$ and so $a+d = (a+b+c+d) - (b+c)\leq 0.$ Finally, $e = 5-a-d -(b+c)\leq 5-a-d\leq 7.$ Now for the kill:
$$S = (-a-b)(-c-d)(d+e)(e+a)(b+c)\leq \dfrac{(e-5)^2}{4}\cdot \dfrac{(e+5)^3}{27}\leq \dfrac{12^3}{27} = 64.$$
You can do similar thing for the minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011875",
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Given Orthocenter of a triangle and length of three segments, need to find area. Let, $O$ be an orthocenter of the triangle $\triangle ABC$, where $AO = 7, AB = 9, CO = 4$. Find the combined area of the triangles $\triangle AOC$ and $\triangle AOB$. Options given are $7\sqrt{3}, 14\sqrt{3}, 21\sqrt{3}, 28\sqrt{3}$.
I started off by assigning $AE =x$. Observing $\Delta AOE$, I can say $OE = \sqrt{49-x^2}$.
Then, $BE = 9-x$ and $OB = \sqrt{130-18x}$. Furthermore, $BC$ would be $\sqrt{146+8\sqrt{49-x^2}}$. But not sure if I'm going in right direction.
| $$\triangle=[AOB]+[AOC]=\frac{1}{2}\cdot BC\cdot AO$$
So we need to find $BC=a$
We use the fact that perpendicular bisector from circumcenter to any side is half the length of segment joining the opposite vertex to the orthocenter.
In the diagram below, not to scale, we see that
$$\left( \frac{c}{2} \right)^2 + \left( \frac{OC}{2} \right)^2 = R^2 = \left( \frac{a}{2} \right)^2 +\left( \frac{AO}{2} \right)^2 $$
$$\Rightarrow a^2 = 4^2+9^2-7^2=48$$
$$\therefore \triangle=14\sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$
If
$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$
I added $x^2$ to both side of the equation:
$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem useful.
| less beautiful
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
- \frac{ 1 }{ 2 } & 1 & 0 & 0 \\
- \frac{ 1 }{ 2 } & - \frac{ 1 }{ 3 } & 1 & 0 \\
- \frac{ 1 }{ 2 } & - \frac{ 1 }{ 3 } & - \frac{ 1 }{ 2 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
0 & \frac{ 3 }{ 2 } & 0 & 0 \\
0 & 0 & \frac{ 4 }{ 3 } & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } \\
0 & 1 & - \frac{ 1 }{ 3 } & - \frac{ 1 }{ 3 } \\
0 & 0 & 1 & - \frac{ 1 }{ 2 } \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrrr}
2 & - 1 & - 1 & - 1 \\
- 1 & 2 & 0 & 0 \\
- 1 & 0 & 2 & 0 \\
- 1 & 0 & 0 & 2 \\
\end{array}
\right)
$$
Back substitution says $t=0$ then $z=0$ then $y=0$ then $x=0.$ All that is going on is that the matrix I am calling $H$ is positive definite. With all real variables, each row in my $Q$ must indicate a zero vector ( when multiplied by $(x,y,z,t)^T \; \; \; \; $ )
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
355/113 and small odd cubes An important approximation to $\pi$ is given by the convergent $\frac{355}{113}$.
The numerator and the denominator of this fraction are at the same distance of small consecutive odd cubes.
$$\frac{355}{113} = \frac{7^3+12}{5^3-12}$$
Is this a consequence of some general formula, such as a series or continued
fraction?
Attempts, drafts
$$\frac{113}{355} = \frac{1}{10}\left(2+\frac{5^3}{3+\frac{7^3}{4}}\right)-\frac{8}{355} = \frac{1}{5}\left(3-\frac{5^3}{3+\frac{7^3}{4}}\right)$$
Similarly, for $\frac{22}{7}$
$$\frac{7}{22} =\frac{1}{2}\left(\frac{3^3}{1+\frac{4^3}{2}}\right) -\frac{1}{11}$$
For the first convergent $3$,
$$\frac{3^3-6}{1^3+6}=\frac{21}{7}=3$$
The tentative expression
$$a(n)=\frac{(4n+3)^3+6(3n-1)}{(4n+1)^3-6(3n-1)}$$
has $a(0)=3$ and $a(1)=\frac{355}{113}$ but $\lim_{n \to \infty} a(n) = 1 \neq \pi$
A similar fraction for the approximation $\frac{223}{71}$ (Archimedes' lower bound) is
$$\frac{98+5^3}{98-3^3}=\frac{2·7^2+5^3}{2·7^2-3^3}=\frac{223}{71}=3+\frac{1}{7+\frac{1}{10}}$$
| For a coprime pair of integers $1 \le p,q$, consider the following diophantine equation:
$$\frac{x^3+z}{y^3-z}= \frac{p}{q}$$
with $x,y,z$ integer unknowns. This is equivalent to
$$z= \frac{y^3p-x^3q}{p+q}$$
Thus a solution exists if and only if there exist $x,y$ such that
$$y^3p-x^3q \equiv 0 \pmod{(p+q)}$$
Now, note that $-q \equiv p$, and that it has an inverse $\mod (p+q)$. This means that we need
$$x^3+y^3 \equiv 0 \pmod{(p+q)}$$
Now, if we pick $p=355$ and $q=113$, we get
$$p+q=468 = 7^3+5^3$$
In other words, the fraction $355/113$ has such a special property not because it is a convergent of $\pi$, but simply because the sum of numerator and denominator is a sum of two small cubes. We can do the same job with the fraction
$$\frac{203}{265} = \frac{5^3+78}{7^3-78}$$
nothing special with $\pi$.
As for $22/7$ we can find $$\frac{22}{7} = \frac{17^3+5^3}{12^3-5^3}$$
which is even more surprising. In my opinion, having three variables gives you a lot of freedom, so that yo can find such nice forms for any rational number.
EDIT: Just for fun, I looked up convergents of a completely different constant, namely $\gamma$: Euler-Mascheroni constant. One of its convergents is
$$\frac{71}{123}= \frac{3^3+470}{11^3-470}$$ Again, there is nothing special about this rational, simply you can find such a form for any number.
| {
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"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Tangent integration $\int \tan (2x)\tan (3x)\mathrm{d}x$ I know how to calculate $\displaystyle\int \tan (x) \tan (2x)\tan (3x)\mathrm{d}x$ with the tangent sum formula
$\tan 3x\tan 2x\tan x =\tan3x -\tan 2x -\tan x$. But I don't know how I can solve
$$\displaystyle\int \tan (2x)\tan (3x)\mathrm{d}x$$
I tried u sub, trig sub, integration by parts. But I couldn't make it.
Thanks.
| Proceed as follows
\begin{align}
\int \tan 2x\tan 3x {d}x
&=\int \frac{\sin 2x\sin 3x }{\cos 2x\cos 3x }dx\\
&=\int \frac{(2\sin x\cos x)[\sin x(4\cos^2x-1)] }{(2\cos^2x-1)[\cos x(4\cos^2x-3) ]}dx\\
&=2\int \frac{(1-\cos^2x)(4\cos^2x-1)}{(2\cos^2x-1)(4\cos^2x-3)}dx\\
&=\int \left( -1 -\frac1{2\cos^2x-1} + \frac2{4\cos^2x-3}\right)dx\\
&=-x -\int \frac{d(\tan x)}{1-\tan^2x} +2\int \frac{d(\tan x)}{1-3\tan^2x}\\
&=-x -\tanh^{-1}(\tan x) +\frac2{\sqrt3}\tanh^{-1}(\sqrt3\tan x)
\end{align}
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to prove that for $n=k+1$ that,
$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$
We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.
Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot 2^{k}}+2^{2\cdot 2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$
I get stuck here, please help me.
| Modulo $7,$ $4^{2^n}+2^{2^n}+1 \equiv 2+4+1\equiv 0$ when $n$ is odd, and
to $4+2+1\equiv 0$ when $n$ is even.
(Start with $4^{2^n}\equiv 2$
and $2^{2^n}\equiv 4$ when $n=1.$)
| {
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"question_score": "13",
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"answer_id": 3
} |
Taylor expansion of $\frac{ \cos( \ln (n+1))}{(n+1)^{1/2}}$ Taylor Series Expansion of$$\frac{ \cos( \ln (n+1))}{(n+1)^{1/2}}$$
My try-
$$\cos(\ln (n+1))= \cos\left (\ln \ n+ \ln\left(1+\frac{1}{n}\right)\right)$$$$=\cos (\ln\ n) \cos\left(\ln\left(1+\frac{1}{n}\right)\right)-\sin(\ln \ n)\sin\left(\ln \left(1+\frac{1}{n}\right)\right) $$
How to proceed?
| Near $x=0$, we have
$$
\log (x + 1) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \cdots ,
$$
$$
\cos (x) = 1 - \frac{1}{2}x^2 + \cdots
$$
and
$$
\frac{1}{{\sqrt {x + 1} }} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 -\frac{5}{16}x^3 + \cdots .
$$
Thus
\begin{align*}
& \frac{{\cos (\log (x + 1))}}{{\sqrt {x + 1} }} = \\ &= \left( {1 - \frac{1}{2}\left( {x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \cdots } \right)^2 + \cdots } \right)\left( {1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{{16}}x^3 + \cdots } \right)
\\ &
= 1 - \frac{1}{2}x - \frac{1}{8}x^2 + \frac{7}{{16}}x^3 + \cdots .
\end{align*}
| {
"language": "en",
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Solve the equation $x=1-5(1-5x^2)^2$ Solve the equation
$$x=1-5(1-5x^2)^2$$
###My work
Let $f(x)=1-5x^2$. Then we have tha equation $f(f(x))=x$. But in this case we don't use the equation $f(x)=x$ because $f(x)$ is not monotonic function
| You are on the right track; a modification of your approach will get you the rest of the way.
Let $y = f(x) = 1 - 5x^2$. Then $f(f(x)) = x$ implies $f(y) = x$, hence the solution to the simultaneous system $$\begin{align} y &= 1 - 5x^2, \\ x &= 1 - 5y^2, \end{align}$$ will yield the desired $x$-values. To this end, we take the difference to obtain $$y - x = 5(y^2 - x^2) = 5(y-x)(x+y),$$ hence $$(y-x)(5(x+y) - 1) = 0.$$ Then $x = y$ amounts to solving $x = 1 - 5x^2$, which are the "obvious" solutions. The other set corresponds to $x + y = \frac{1}{5}$, which are the "nontrivial" solutions, and require solving the quadratic $$5x^2 - x - \frac{4}{5} = 0.$$
| {
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"source": "stackexchange",
"question_score": "3",
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System of equations: $3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y$ This was taken from a local mathematical olympiad in Romania. It is from the year 2003.
Solve for $x, y, z \in \mathbb{R}$:
$$
\left\{
\begin{array}{c}
3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y \\
3^y + 4^y + 5^y = 2^y \cdot 3^{y -1} \cdot z \\
3^z + 4^z + 5^z = 2^z \cdot 3^{z -1} \cdot x
\end{array}
\right.
$$
So, the first thing I notice was $4^x = 2^{2x} = a$ so we get a system of quadratic equations. This proved to be quite hard to deal with, so I gave it up. I also tried dividing by $2^x \cdot 3^{x - 1} \cdot y$ and the others, but it was weird to deal with. Do you have any ideas?
| We have:
$$\left( \frac{3}{6} \right)^x+\left( \frac{4}{6} \right)^x+\left( \frac{5}{6} \right)^x =\frac{y}{3} \tag{1}$$
$$\left( \frac{3}{6} \right)^y+\left( \frac{4}{6} \right)^y+\left( \frac{5}{6} \right)^y =\frac{z}{3}\tag{2}$$
$$\left( \frac{3}{6} \right)^z+\left( \frac{4}{6} \right)^z+\left( \frac{5}{6} \right)^z =\frac{x}{3}\tag{3}$$
Let's denote $f(t) =\left( \frac{3}{6} \right)^t+\left( \frac{4}{6} \right)^t+\left( \frac{5}{6} \right)^t $, this function is decreasing in $t$ (you can prove it easily).
WLOG, suppose $x=\min{\{x,y,z\}}$. This implies that $x \le y$.
From (1) and (3), we have: $x \le y \implies f(x) \ge f(z) \implies x \le z$.
From (2) and (3), we have: $ x \le z \implies f(z) \le f(y) \implies z \ge y$.
From (1) and (2), we have: $ y \le z \implies f(x) \le f(y) \implies x \ge y$.
But we already suppose $x=\min{\{x,y,z\}}$ , hence, $x=y$. And from (1) and (2), we have $f(x) = f(y) \implies y = z$. So, $x=y=z$.
It suffices now to solve the equation
$$f(x)=\left( \frac{3}{6} \right)^x+\left( \frac{4}{6} \right)^x+\left( \frac{5}{6} \right)^x =\frac{x}{3}$$
We notice that $f(x)$ is decreasing function while $=\frac{x}{3}$ is an increasing function. Thus, there is at most 1 solution. And because $x=3$ is the solution, then $x=y=z=3$ is the unique solution of the system equation.
| {
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"source": "stackexchange",
"question_score": "3",
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How to calculate the absolute extrema for this equation Basically, I am trying to find the absolute extrema for this problem, but I am stuck after a few steps.
$$y=\frac{4}{x}+\tan\left(\frac{\pi x}{8}\right),\quad x\in[1,2].$$
I have figured out the derivative, which is
$$y'=-\frac{4}{x^2}+\sec^2\left(\frac{\pi x}{8}\right)\cdot\frac{\pi}{8}.$$
However, after reaching this step, I am stuck. I have tried setting the derivative equal to zero, but I do not get anywhere as I don't know what to do next. How can I solve this problem? Any help is appreciated. Thanks.
| Hint: You can use a difference of squares with the derivative:
$$ \frac{\pi}{8} \sec^2 \left( \frac{\pi x}{8} \right) - \frac{4}{x^2} = 0 $$
$$ \Rightarrow \left(\sqrt{\frac{\pi}{8}} \sec \left( \frac{\pi x}{8} \right) + \frac{2}{x}\right) \left(\sqrt{\frac{\pi}{8}} \sec \left( \frac{\pi x}{8} \right) - \frac{2}{x}\right) = 0 $$
Update:
$$ \Rightarrow \pm \frac{2}{x} = \sqrt{\frac{\pi}{8}} \sec \left( \frac{\pi x}{8} \right)$$
$$ \Rightarrow \pm \frac{x}{4}\sqrt{\frac{\pi}{2}} = \cos \left( \frac{\pi x}{8} \right) $$
Let $f_1(x) = \frac{x}{4}\sqrt{\frac{\pi}{2}}$, $f_2(x) = -\frac{x}{4}\sqrt{\frac{\pi}{2}}$, and $g(x) = \cos \left( \frac{\pi x}{8} \right) $.
We'll start with $f_1(x) = g(x)$. First, note $f_1(0) = 0$, and $g(0) = 1$. The period of $g(x)$ is $16$, so the next time it reaches its maximum value of $1$ is at $x = 16$. At the same time, $f_1(x) = 1$ at $x = \frac{4\sqrt{2}}{\pi} < 16 $. Since $f_1$ strictly increasing, $f_1$ and $g$ intersect exactly once on $x\ge 0$, somewhere on $x \in \left[0,\frac{4\sqrt{2}}{\pi}\right] $.
For $x \le 0$, $f_1(x) = -1$ at $x = -\frac{4\sqrt{2}}{\pi} > -4 $. Again, since $f_1$ is strictly increasing, $f_1$ is below the minimum value of $g$ for all $x < -\frac{4\sqrt{2}}{\pi} $ . Since $g(x) > 0$ for $x \in (-4,0] $, we have $f_1(x) < g(x) $ for all $x \le 0$.
Combining these, we have $f_1$ and $g$ intersect exactly once, let's say at $x = r_1$, where $r_1 \in \left[0,\frac{4\sqrt{2}}{\pi}\right] $.
Then, since $f_2(x) = f_1(-x)$ and $g(x) = g(-x)$, $f_2$ and $g$ intersect exactly once at $x = r_2 = -r_1$.
Further, note that there must be a sign change in $y'$ at $r_1$ because $f_1$ is strictly increasing for all $x$, and $g$ is decreasing on $[0,8]$, implying that $g$ is decreasing at $r_1$. The same logic implies a sign change at $r_2$ as well.
These two solutions are the unique roots of the factored form of the derivative we had above.
Then, we have
$$ y'|_{x=1} = -4 + \frac{\pi}{8}\sec^2 \left( \frac{\pi}{8} \right) $$
$$ = -4 + \frac{\pi}{8}\frac{4}{2+\sqrt{2}} $$
$$ = -4 + \frac{\pi}{4+2\sqrt{2}} $$
$$ < 0 . $$
Similarly,
$$ y'|_{x=2} = -1 + \frac{\pi}{8}\sec^2 \left( \frac{\pi}{4} \right) $$
$$ = -4 + \frac{\pi}{8}*2 $$
$$ = -4 + \frac{\pi}{4} $$
$$ < 0 . $$
Since $y' < 0$ at both $x = 1$ and $x = 2$, we know that $r_1 \notin [1,2]$, since there is no sign change in $y'$ and there is exactly one root (which has a corresponding sign change) for $x \ge 0$.
Then, the extreme values of $y$ must be at $x = 1$ and $x = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
how to determine a transformation matrix for transforming x and y with specific equation? I need to determine a matrix that can be used to:
a) transform $x$ and $y$ using equations
$$
\left\{
\begin{array}{c}
x'=3x + 4y \\
y'=-x + 2y \\
\end{array}
\right.
$$
Then I need to:
b) Transform a triangle $(0,0),(1,0),(1,1)$ using this transformation
c) What are the coordinates of point $(10,10)$ in the basis of transformation?
Could I have some advise? Thanks in advance.
| $\left\{
\begin{array}{c}
x'=3x + 4y \\
y'=-x + 2y \\
\end{array}
\right. \Rightarrow \left\{
\begin{array}{c}
x'=(3,4)(x,y)\\
y'=(-1,2)(x,y) \\
\end{array}
\right.$
$\left\{
\begin{array}{c}
x'=(3,4)(x,y)\\
y'=(-1,2)(x,y) \\
\end{array}
\right. \Rightarrow \begin{pmatrix} 3 & 4 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} x` \\ y` \end{pmatrix}$
$\begin{pmatrix} 3 & 4 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$
$\begin{pmatrix} 3 & 4 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} 3 \\ -1 \end{pmatrix}$
$\begin{pmatrix} 3 & 4 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} 7 \\ 1 \end{pmatrix}$
as itcan be seen, image of triangle is also a triangle (because kernel of matrix is $\{0\}$)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Nested sums - reciprocal of a sum : Find exact value $$\text{Find:} ~~~~~~ \sum_{k=1}^{\infty} \frac{1}{ \left (
\sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}} \right )^2}$$
(Beware of the bounds of $j$, it does not always start from $1$, but starts at $k$ and goes to $k^2$)
At first I thought it has something to do with Riemann's zeta function $\zeta$ and thus the solution is:
$$ \zeta \left( \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}} \right ) $$
However this totally seems incorrect because we cannot find this sum as we have an unknown, $k$.
Which also is being summed to $\infty$.
I thought (however was unsure) of the fact that:
$$ \left ( \sum f \right )^2 \ge \sum f^2 $$
Thus (maybe) it is trivial the sum is converging to some value, which according to a little program I wrote is about: $$ \sum_{k=1}^{\infty} \frac{1}{ \left (
\sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}} \right )^2} \approx 1.596$$
But this is not a rigorous proof, is there a way to solve this so we get a solution by hand - calculating the exact value of this sum if it is finite (if not, why? )
Any mathematical tools (not programs) are acceptable, because this is not a question from a specific course ( I don't have context for this).
| (This is not a solution.)
The question of convergence is raised in the comments, so I show a proof here.
From $k > 0$ and
$$ k \leq j \leq k^2 $$
we have
$$ \frac{1}{\sqrt{k}} \geq \frac{1}{\sqrt{j}} \geq \frac{1}{k} \text{,} $$
using the monotonicity of the square root and standard results about reciprocals and inequalities. So $1/\sqrt{j}$ is lower bounded by $1/k$. Then
\begin{align*}
\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}} &\geq \sum_{j=k}^{k^2} \frac{1}{k} = (k^2 - k + 1)\frac{1}{k} \text{,} \\
\left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2 &\geq \frac{(k^2 - k + 1)^2}{k^2} \text{,} \\
\frac{1}{\left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2} &\leq \frac{k^2}{(k^2 - k + 1)^2} \text{, and} \\
\sum_{k=1}^\infty \frac{1}{\left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2} &\leq \sum_{k=1}^\infty\frac{k^2}{(k^2 - k + 1)^2} \text{.}
\end{align*}
(This last sum can be expressed in terms of polygamma functions, but let's not.)
Factoring out "the big part", we have
$$ \frac{k^2}{(k^2 - k + 1)^2} = \frac{k^2}{k^4} \cdot \frac{1}{1 - 2/k + 3/k^2 - 2/k^3 + 1/k^4} \text{.} $$
This is $\frac{1}{k^2} \cdot \text{[some bounded expression]}$. Let $f(k_) = \frac{1}{1 - 2/k + 3/k^2 - 2/k^3 + 1/k^4}$. For $k \geq 1$, $f'(k) = 0$ at $k = 2$ and $f''(2) > 0$. So $f'(k) < 0$ on $[1,2)$ and $f'(k) > 0$ on $(2,\infty)$. This means the value of $f(k)$ on $[1,\infty)$ is bounded by
$$
\max \{f(1), \lim_{k \rightarrow \infty} f(k)\} = \max \{1,1\} = 1 \text{.} $$
So \begin{align*}
&\sum_{k=1}^\infty \frac{1}{\left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2} \\
&\leq \sum_{k=1}^\infty \frac{1}{k^2} \cdot \frac{1}{1 - 2/k + 3/k^2 - 2/k^3 + 1/k^4} \\
&\leq \sum_{k=1}^\infty \frac{1}{k^2} \cdot 1 \\
&= \frac{\pi^2}{6} \text{.}
\end{align*}
(This last is the Basel problem. Alternatively, one can also use the comparison test with $\int^\infty \frac{\mathrm{d}x}{x^2}$ to show this last sum converges.)
So the sum converges. There's a bit of a gap between this upper bound and the partial sum you cite, but our first approximation (replacing all the $1/\sqrt{j}$ terms with the extremal term in their sum) is a loose step and pretending the bounded constant near the end is $1$ is also loose.
| {
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"url": "https://math.stackexchange.com/questions/4023702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Determine the following sum Let $S(n)$ be the sum of the digits of n in its binary representation. For example the binary notation for 19 is $10011$ and $S(19)=3$. Determine $\sum_{n=1}^{\infty} \frac{S(n)}{n(n+1)}$?
I've made several tables trying to pinpoint any useful pattern, but I've had minimal luck. The best I'd came to was determining that the number of digits is $\left \lceil{\frac{ln(n)}{ln(2)}}\right \rceil$ which gives a bound for $S(n)$. What I'd picked up from those tables is $S(2^k-1)=k$ and $S(2^k)=1$. However, breaking the sum up into those n's that are powers of 2 proved to be a bit of a disaster.
| The key idea is to expand over digits and find a recurrence. Let $$\require{cancel}F(N)=\sum_{n=1}^{2^N}{\frac{S(n)}{n(n+1)}}=\sum_{n=1}^{2^N}{S(n)\left(\frac{1}{n}-\frac{1}{n+1}\right)}$$ and \begin{align*}
G(N)&=\sum_{n=0}^{2^{N-1}-1}{\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)}-\left(\frac{1}{2^N+1}-\frac{1}{2^N+2}\right) \\
&=\sum_{n=1}^{2^N}{\frac{(-1)^{n+1}}{n}}-\frac{1}{(2^N+1)(2^N+2)}\text{;}
\end{align*} we know that $\lim_{N\to\infty}{G(N)}=\ln{(2)}$ and want to find $\lim_{N\to\infty}{F(N)}$.
We can compute \begin{align*}
F(N)&=\sum_{n\in[2^N],2\mid n}{S(n)\left(\frac{1}{n}-\frac{1}{n+1}\right)}+\sum_{n\in[2^N],2\nmid n}{S(n)\left(\frac{1}{n}-\frac{1}{n+1}\right)} \\
&=\sum_{n=1}^{2^{N-1}}{S(2n)\left(\frac{1}{2n}-\frac{1}{2n+1}\right)}+\sum_{n=0}^{2^{N-1}-1}{S(2n+1)\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)} \\
&=\sum_{n=1}^{2^{N-1}}{S(n)\left(\frac{1}{2n}-\frac{1}{2n+1}\right)}+\sum_{n=0}^{2^{N-1}-1}{(1+S(n))\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)} \\
&=\sum_{n=1}^{2^{N-1}}{S(n)\left(\frac{1}{2n}-\cancel{\frac{1}{2n+1}}+\cancel{\frac{1}{2n+1}}-\frac{1}{2n+2}\right)}-\left(\frac{1}{2^N+1}-\frac{1}{2^N+2}\right)+ \\
&\phantom{{}={}}\quad\sum_{n=0}^{2^{N-1}-1}{\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)} \\
&=\frac{F(N-1)}{2}+G(N) \\
&=\frac{F(0)}{2^N}+\frac{G(1)}{2^{N-1}}+\dots+\frac{G(N-1)}{2}+G(N) \\
&=\frac{1}{2^{N+1}}+\sum_{k=0}^{N-1}{\frac{G(N-k)}{2^k}}
\end{align*} Thus $$|F(N)-\ln{(2)}|\leq\frac{1}{2^{N+1}}+\sum_{k=0}^{N-1}{\frac{|G(N-k)-\ln{(2)}|}{2^k}}+\sum_{k=N}^{\infty}{\frac{\ln{(2)}}{2^k}}$$ The outer two terms are clearly tending to $0$ as $N\to\infty$, so it suffices to show the same for the inner.
Fix $\epsilon$. There exists $K$ such that, for $k\geq K$, we have $|G(k)-\ln{(2)}|\leq\epsilon$; let $M=\sup_{k<K}{G(k)}$. Then \begin{align*}
\sum_{k=0}^{N-1}{\frac{|G(N-k)-\ln{(2)}|}{2^k}}&=\sum_{k=1}^N{\frac{|G(k)-\ln{(2)}|}{2^{N-k}}} \\
&\leq\sum_{k=0}^{K-1}{\frac{M}{2^{N-k}}}+\sum_{k=K}^N{\frac{\epsilon}{2^{N-k}}} \\
&\leq\frac{(2^K-1)M}{2^N}+\frac{\epsilon}{2^{K-1}}
\end{align*} Thus $\limsup_{N\to\infty}{|F(N)-\ln{(2)}|}\leq\epsilon$; now take $\epsilon\to0^+$ to get $$\lim_{N\to\infty}{F(N)}=\ln{(2)}$$
Of course, we selected a subsequence of partial sums to get $F(N)$, so theoretically the unselected partial sums could exhibit $\Theta(1)$ variation from $F(N)$. But each term in your original sum is nonnegative, so in fact that can't happen.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$ $\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$
I used the ratio criterion for the calculation and I got to this, can I say now that it is zero or is it still an undefined expression?
$\frac{10+(10/\sqrt{(n+1)})}{\sqrt{(n+2)}+1}$
10/inf < 1 ----> lim an=0
| Using
$$\frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}} \lt \frac{10^n}{2\sqrt{n!}}=\frac{1}{2}\sqrt{\frac{100^n}{n!}}$$
we come to limit $\frac{a^n}{n!}$ for $a \gt 1$. Of course exists natural $k=\lfloor a\rfloor \leqslant a \leqslant \lfloor a\rfloor +1$. So we have
$$0 \lt \frac{a^n}{n!} = \frac{a}{1}\frac{a}{2} \cdots \frac{a}{k} \frac{a}{k+1}\cdots \frac{a}{n} \leqslant \frac{a}{1}\frac{a}{2} \cdots \frac{a}{k} \left(\frac{a}{k+1}\right)^{n-k} \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
sequence limits - correctness of the adjustment I would like to ask why the following modification of the example is not allowed $$\displaystyle\lim_{n\rightarrow \infty} \frac{1+2+ \cdots + 2^n}{1+5+ \cdots + 5^n} = \lim_{n\to \infty}\left(\frac{2^n}{5^n}\right)\cdot \frac{\left(\frac{1}{2^n}+\frac{2}{2^n}+.....+1\right)}{\left(\frac{1}{5^n}+\frac{5}{5^n}+.....+1\right)}$$
and similarly
$$\displaystyle\lim _{n\rightarrow \infty} \left(\frac{3n+1}{4n+5}\right) ^n = \lim_{n \to \infty}
\left(\frac{3}{4}\right)^n\cdot\left(\frac{1 + \frac{1}{3n}}{1 + \frac{5}{4n}}\right)^n$$
For these two examples, I am not concerned with the calculation, but rather with describing why this procedure is wrong.
| To actually evaluate the limits I'd go as follows:
$$\frac{1+2+\ldots+2^n}{1+5+\ldots+5^n}=\frac{\frac{1-2^{n+1}}{1-2}}{\frac{1-5^{n+1}}{1-5}}=4\frac{1-2^{n+1}}{1-5^{n+1}}=4\left(\frac25\right)^{n+1}\;\frac{\frac1{2^{n+1}}-1}{\frac1{5^{n+1}}-1}\xrightarrow[n\to\infty]{}4\cdot0\cdot1=0$$
As for the second one:
$$\left(\frac{3n+1}{4n+5}\right)^n=\frac1{\left(\frac43\right)^n}\left(\frac{3n+1}{3n+\frac{15}4}\right)^n=\frac1{\left(\frac43\right)^n}\left(1-\frac{\frac{11}4}{3n+\frac{15}4}\right)^n=$$
$$=\frac1{\left(\frac43\right)^n}\left[\left(1-\frac{\frac{11}4}{3n+\frac{15}4}\right)^{3n+\frac{15}4}\right]^{-2n-\frac{15}4}\xrightarrow[n\to\infty]{}0\cdot0=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $a^2+b^2+c^2+d^2+ac+bd\ge\sqrt3$? We have $ad-bc=1$ . prove
$$a^2+b^2+c^2+d^2+ac+bd\ge\sqrt3$$
To solve it, I multiplied the inequality by $2$ and added $2(ad-bc)=2$ to it :
$$2a^2+2b^2+2c^2+2d^2+2ac+2bd+2ad-2bc\ge2+2\sqrt3$$
$$(a+c)^2+(b+d)^2+(a+d)^2+(b-c)^2\ge 2+2\sqrt3$$
From here I don't know how to prove the minimum value of sum of these squares is $2+2\sqrt3$ .
| by AM-GM $$a^2+b^2+c^2+d^2\ge 2\sqrt{(a^2+b^2)(c^2+d^2)}=2\sqrt{{(ac+bd)}^2+{(ad-bc)}^2}$$ $$=2\sqrt{{(ac+bd)}^2+1}$$ Let $ac+bd=t$ we have to prove $$2\sqrt{t^2+1}+t\ge \sqrt{3}$$ Can you end it now?
we have to prove $$2\sqrt{t^2+1}\ge \sqrt{3}-t$$ if $\sqrt{3}-t\le 0$ the inequality is obvious other wise squaring both sides of inequality it suffices to prove $$4(t^2+1)\ge t^2-2\sqrt{3}t+3\iff {(\sqrt{3}t+1)}^2\ge 0$$ which is true
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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calculate the limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$ I'm working on finding the limit for this equation, and would kindly welcome your support to my solution:
$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$$
These are my steps in hopefully deriving the correct result:
$$\frac{1-\cos \theta}{\theta \sin \theta} = \frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin\theta}=\frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin \theta}=\frac{\sin\theta}{\theta}\cdot\frac{1}{1+\cos\theta}$$
Given that as $\theta$ approaches $0$ then $\dfrac{\sin\theta}{\theta}$ $\approx$ $0$ and $\cos\theta \approx 1$. I thought the answer would be $0$ given $\sin{\theta}$ tends to $0$ but the answer is $\frac{1}{2}$. Can someone kindly explain and show this to me? Maybe I went wrong in my reasoning or calculation?
EDIT:
Given that:
$\cos \theta < \frac{\theta}{\sin \theta} < \frac{1}{cos \theta} = \sec \theta$
because $\frac{\theta}{\sin \theta}$ is between two variables approaching 1, then it must also approach 1.
hence, $\frac{\sin \theta}{\theta}$ approaches 1.
So the answer is $\frac{1}{2}$
| Here it is a slightly different approach from @amirali.
According to the trigonometric identity $\sin(2x) = 2\sin(x)\cos(x)$, we conclude that
\begin{align*}
\lim_{\theta\to0}\frac{1-\cos(\theta)}{\theta\sin(\theta)} & = \lim_{\theta\to 0}\frac{2\sin^{2}(\theta/2)}{2\theta\sin(\theta/2)\cos(\theta/2)}\\\\
& = \lim_{\theta\to 0}\left[\frac{1}{2}\times\frac{\sin(\theta/2)}{\theta/2}\times\frac{1}{\cos(\theta/2)}\right] = \frac{1}{2}
\end{align*}
where we have applied the continuity of $\cos(x)$ at $x = 0$ as well as the following result:
\begin{align*}
\lim_{x\to0}\frac{\sin(x)}{x} = 1
\end{align*}
Hopefully this helps!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}\,dx$ Someone please help me to solve this integral
$$\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}\,dx$$
I have tried to use the formula
$$ab^2=\frac{(a+b)^3+(a-b)^3+2a^2}{6}$$
to reduce the original integral to three integrals
$$\frac{1}{6}\int_0^1\frac{\ln^3(x+x^2)}{1-x}\,dx+\frac{1}{6}\int_0^1\frac{\ln^3(\frac{x}{1+x})}{1-x}\,dx+\frac{1}{3}\int_0^1\frac{\ln^3x}{1-x}\,dx$$
But I couldn't figure out how to solve these.
Another way is to use series representation of $$\ln(x), \quad \ln(1+x),\quad \frac{1}{1-x}.$$
But I don't know how to solve the multiple sums.
Please help me out.
| $
\displaystyle\int_0^1\frac{\ln^2(x)\ln(1+x)}{1-x}dx=\displaystyle\int_0^1\displaystyle\int_0^1\frac{x\ln^2(x)}{(1+xy)(1-x)}dxdy$
$=\displaystyle\int_0^1\frac{1}{1+y}\left(\displaystyle\int_0^1\frac{\ln^2(x)}{1-x}-\frac{\ln^2(x)}{1+yx}dx\right)dy$
$=\displaystyle\int_0^1\frac{1}{1+y}\left(2\zeta(3)-\displaystyle\sum_{n=1}^{\infty}(-y)^{n-1} \displaystyle\int_0^1x^{n-1}\ln^2(x)dx\right)dy$
$=2\zeta(3)\ln(2)-2\displaystyle\int_0^1\frac{1}{1+y} \displaystyle\sum_{n=1}^{\infty}\frac{(-y)^{n-1}}{n^3}dy=2\zeta(3)\ln(2)+2\displaystyle\int_0^1\frac{\operatorname{Li}_3(-y)\ }{y(1+y)}dy$
$=2\zeta(3)\ln(2)+2\displaystyle\int_0^1\frac{\operatorname{Li}_3(-y)\ }{y}-\frac{\operatorname{Li}_3(-y)\ }{1+y}dy$
$=2\zeta(3)\ln(2)+\frac{7}{4}\zeta(4)-2\left(\ln(1+y)\operatorname{Li}_3(-y)\ \right)_0^1+\displaystyle\int_0^1\frac{\operatorname{Li}_2(-y)\ \ln(1+y)}{y}dy$
$=2\zeta(3)\ln(2)+\frac{7}{4}\zeta(4)+\frac{3\ln(2)\zeta(3)}{2}-2\displaystyle\int_0^1 \operatorname{Li}_2(-y)(-\operatorname{Li}_2^{'}(-y))dy$
$=\frac{7}{4}\zeta(4)+\frac{7\zeta(3)\ln(2)}{2}-2\left(-\frac{(\operatorname{Li}_2(-y))^2}{2}\right)_0^1=\frac{7}{4}\zeta(4)+\frac{7\zeta(3)\ln(2)}{2}+(\operatorname{Li}_2(-1))^2$
$=\frac{7}{4}\zeta(4)+\frac{7\zeta(3)\ln(2)}{2}+\left(-\frac{\pi^2}{12}\right)^2$
| {
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"url": "https://math.stackexchange.com/questions/4035876",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find $\lim_{n \to \infty}\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$ Find the following limit $$\lim_{n \to \infty}\left(\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}\right)$$
I don´t get catch a idea, I notice that
$$\left(\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+\cdots + \frac{2n+1}{2^n} \right)$$ is such that
$$ (\frac{3}{2}-\frac{1}{2})=1, \, (\frac{5}{2^2}-\frac{3}{2^2})=\frac{1}{2},\, \, (\frac{7}{2^3}-\frac{5}{2^3})=\frac{1}{4}\cdots (\frac{2n+1}{2^n}-\frac{2n-1}{2^n})=\frac{1}{2^{n-1}}\text{Which converges to 0 }$$
Too I try use terms of the form $\sum_{n=1}^{\infty}\frac{2n}{2^n}$ and relatione with the orignal sum and consider the factorization and try sum this kind of terms$$\frac{1}{2}\lim_{n \to \infty }\left(1+\frac{3}{2}+\frac{5}{2^2}+ \dots +\frac{2n-1}{2^{n-1}}\right)$$.
Update:
I try use partial sum of the form
$$S_1=\frac{1}{2},S_{2}=\frac{5}{2^2},S_{3}=\frac{15}{2^3},S_{4}=\frac{37}{2^4} $$
and try find $\lim_{n \to \infty }S_{n}$ but I don´t get the term of the numerator.
Unfortunelly I don´t get nice results, I hope someone can give me a idea of how I should start.
| Here is a solution with elementary math.
Let's denote
$$S_n = \frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$$
You have
$$2S_n = 1+\frac{3}{2}+\frac{5}{2^2}+ \dots +\frac{2n-1}{2^{n-1}}$$
Then
\begin{align}
S_n = 2S_n - S_n &= 1+\left(\frac{3}{2}-\frac{1}{2} \right)+\ldots+\left(\frac{2n-1}{2^{n-1}} - \frac{2n-3}{2^{n-1}} \right) - \frac{2n-1}{2^{n}} \\
&= 1+\left(1+\frac{1}{2}+\frac{1}{2^2}\ldots+ \frac{1}{2^{n-2}} \right) - \frac{2n-1}{2^{n}} \\
&= 1+2\left(1- \frac{1}{2^{n-1}} \right) - \frac{2n-1}{2^{n}} \\
&= 3-\frac{1}{2^{n-2}} - \frac{2n-1}{2^{n}} \\
\end{align}
Hence,
$$\lim_{n\to\infty} S_n=3$$
| {
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"url": "https://math.stackexchange.com/questions/4040187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Let $a$ be a non zero real number. Evaluate the integral $\int \frac{-7x}{x^{4}-a^{4}}dx$ I hit a wall on this question. Below are my steps
$$\int \frac{-7x}{x^{4}-a^{4}}dx=-7\int \frac{x}{x^{4}-a^{4}}dx$$
Let $u=\frac{x^2}{2}, dx = \frac{du}{x}, x^{4}=4u^{2}.$
$$-7\int \frac{1}{4u^{2}-a^{4}}du=-7\int \frac{1}{(2u+a^2)(2u-a^2)}du$$
Use partial fraction decomposition,
$$\frac{1}{(2u+a^2)(2u-a^2)}=\frac{A}{2u+a^{2}}+\frac{B}{2u-a^{2}}.$$
Solve for $A$ and $B$:
$$\begin{cases}
A=\frac{1}{-2a^{2}}
\\
B=\frac{1}{2a^{2}}
\end{cases}$$
Now $$\int \frac{1}{(2u+a^2)(2u-a^2)}du=\int \frac{1}{-2a^2(2u+a^{2})}+\int \frac{1}{2a^{2}(2u-a^{2})}$$
Factoring out $a$ yields
$$\frac{7}{2a^{2}}(\int \frac{1}{2u+a^{2}}-\int \frac{1}{2u-a^{2}})$$
Evaluate the integral and substitute $u=\frac{x^{2}}{2}$ back.
My final answer is $$\frac{7}{2a^{2}}(\log(x^2+a^2)-\log(x^2-a^2)).$$
Feedback says my answer is wrong. Where did I mess up?
| You are right up to
$$\frac{7}{2a^{2}}(\int \frac{du}{2u+a^{2}}-\int \frac{du}{2u-a^{2}})
= \frac7{4a^2}\ln\frac{2u+a^2}{|2u-a^2|}= \frac7{4a^2}\ln\frac{x^2+a^2}{|x^2-a^2|}
$$
| {
"language": "en",
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Inequality $\sum_{k=1}^n\frac{k^2}{n^3+n^2}\le\sum_{k=1}^n\frac{k^2}{n^3+k^2}\le\sum_{k=1}^n\frac{k^2}{n^3+1}$ From this question How do you calculate this limit? $\lim_{n\to\infty}( \frac{1^2}{n^3+1^2} + \frac{2^2}{n^3+2^2} ... + \frac{n^2}{n^3+n^2})=?$
I have not understood the reason of the @xpaul's answer for the $(1)$,
$$\sum_{k=1}^n\frac{k^2}{n^3+n^2}\le\sum_{k=1}^n\frac{k^2}{n^3+k^2}\le\sum_{k=1}^n\frac{k^2}{n^3+1} \tag 1$$
| Since the sum ranges from $k=1$ to $k=n$, we have $k\geq 1$ and $k\leq n$; thus $n^2\geq k^2\geq 1$. Translate by $n^3$ to get $n^3+n^2\geq n^3+k^2\geq n^3+1$, then invert (changing the direction of the inequalities) to get $\dfrac1{n^3+n^2}\leq \dfrac1{n^3+k^2}\leq \dfrac1{n^3+1}$. Multiply by $k^2$ to get $\dfrac{k^2}{n^3+n^2}\leq\dfrac{k^2}{n^3+k^2}\leq\dfrac{k^2}{n^3+1}$, and then just sum termwise over $k$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Related rates +implicit differentiation cone problem I'm stuck at the following related rates problem:
As shown in the diagram the apex angle of a cone is $2\theta$ and the slope length is 10cm. The rate of increase of $\theta$ is 0.01 radians per second. The cone starts with $\theta=\frac \pi 6$.
Find the initial rate of change of the volume of the cone.
What I've tried:
The volume $V$ of a cone is given by $V = \frac 1 3 \pi r^2h$. We can replace $r$ and $h$ with $\theta$ using the law of sines if we look at the cross section of the cone:
$\frac {10}{\sin \frac \pi2} = \frac r {\sin \theta}$ => $r = 10\sin \theta$
Similarly
$\frac {10}{\sin \frac \pi2} = \frac h {\sin (\frac \pi 2 -\theta)}$ => $ h = 10 \sin (\frac \pi 2 -\theta) = 10 \cos \theta$
There's probably a better way since this is a right angled triangle but this should work.
From this follows:
$V=\frac 1 3 \pi r^2h = \frac 1 3 \pi (10 \sin \theta)^2 (10 \cos \theta) = \frac {1000} 3 \pi \sin^2 \theta \cos\theta $
So: $\frac {dV} {dt} = \frac {d}{dt} (\frac {1000} 3 \pi \sin^2 \theta \cos\theta)$
$\frac {dV} {dt} = \frac {1000} 3 \pi \frac {d}{dt} (\sin^2 \theta \cos\theta)$
$\frac {dV} {dt} = \frac {1000} 3 \pi ( \cos\theta \frac {d}{dt} \sin^2 \theta + \sin^2 \theta \frac {d}{dt} \cos\theta)$
$\frac {dV} {dt} = \frac {1000} 3 \pi ( 2\cos\theta \sin\theta * \frac {d}{dt} \sin \theta- \sin^2 \theta \sin\theta\frac {d \theta}{dt})$
$\frac {dV} {dt} = \frac {1000} 3 \pi ( 2\cos\theta \sin\theta * \cos\theta\frac {d \theta}{dt}- \sin^2 \theta \sin\theta\frac {d \theta}{dt})$
$\frac {dV} {dt} = \frac {1000} 3 \pi ( 2\cos^2\theta \sin \theta \frac {d \theta}{dt} - \sin^3 \theta \frac {d \theta}{dt})$
Then I plugged in the given values:
$\frac {dV} {dt} = \frac {1000} 3 \pi ( 2 \cos^2\frac \pi 6 \sin \frac \pi 6 (0.01) - \sin^3 \frac \pi 6 (0.01)) = 6.545 \frac {cm^3}{s}$.
However my textbook says that it should be $0.654 \frac {cm^3}s$ which seems pretty close to what I have found. Did I make a blunder somewhere or is my reasoning incorrect?
| Your answer is correct. The book is incorrect by a factor of $10$.
To simplify your calculation, we have $$V(t) = \frac{1000\pi}{3} \sin^2 \theta(t) \cos \theta(t) = \frac{1000\pi}{3} (\cos \theta(t) - \cos^3 \theta(t)).$$ At time $t = 0$, $\theta(0) = \pi/6$ and $\theta'(0) = \frac{1}{100}$, so $$\begin{align}
V'(0) &= \frac{1000\pi}{3} \left(-\theta'(0) \sin \theta(0) + 3 \theta'(0) \cos^2 \theta(0) \sin \theta(0) \right) \\
&= \frac{1000\pi}{3} \theta'(0) \sin \theta(0)(3 \cos^2 \theta(0) - 1) \\
&= \frac{5\pi}{3} \left(3 \cos^2 \frac{\pi}{6} - 1 \right) \\
&= \frac{5\pi}{3} \left(\frac{9}{4} - 1\right) \\
&= \frac{25\pi}{12}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the marginal PDF I am having a hard time trying to find the marginal of the following joint pdf.
$$f_{U,V}(u,v) =2 \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1}$$
where $ u<v<\frac{u+\theta+1}{2}$ and $\theta < u <\theta+1$.
This was my attempt
$\begin{align*}
f_V (v) &= \int_{\theta}^{\theta + 1} \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1} \cdot 2 du\\
&= \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\int_{\theta}^{\theta + 1}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1} \cdot 2 du\\
\end{align*}$
| Your support is: $\theta\lt u< \theta +1$ and $u< v< (u+\theta+1)/2$.
That is equivalently: $\theta\leq v<\theta+1$ and $\max(\theta,2v-\theta-1)\leq u<v$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$ If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$
The solution goes as follows:
$a=\frac{x}{y+z}$, $b=\frac{y}{z+x}$, $c=\frac{z}{x+y}$. We can see that:
$ab+bc+ac+2abc=1$
It's enough if: $(a+b+c)^2\ge 4-14abc$
$(a+b+c)^2\ge 4(ab+bc+ab+2abc)-14abc$
From Schur it is enough if: $6abc\ge\frac{9abc}{a+b+c}$ which is true from Nesbit.
Could you please provide a more intuitive and easier approach?
| Well it's an attempt to delete the square root .
My only idea is to use Bernoulli's inequality and play with the coefficient $\sqrt{2}$.We have :
$$\sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}\leq \sqrt{\alpha}\Big(1+\frac{1}{2}\Big(\frac{4}{\alpha}-1-\frac{14}{\alpha}\frac{xyz}{(x+y)(y+z)(x+z)}\Big)\Big)$$
Remains to show (using the OP's substitution) when:
$$ \sqrt{\alpha}\Big(1+\frac{1}{2}\Big(\frac{4}{\alpha}-1-\frac{14}{\alpha}abc\Big)\Big)\leq a+b+c$$
Wich seems to be easier I think.
The last inequality is equivalent to :
$$\frac{y}{2}+\frac{2}{y}\leq a+b+c+\frac{7}{y}abc$$
With $y^2=\alpha$ so now a bit of algebra to get :
$$y^2+4\leq 2y(a+b+c)+14abc$$
Now putting $y=a+b+c$ we get the inequality of the OP.
The end is the same as the OP.
| {
"language": "en",
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"source": "stackexchange",
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Are there nontrivial rational solutions to $\sqrt{1-x^2} + \sqrt{1-y^2} = \sqrt{1-z^2}$? Obviously $(a,0,a)$,$(-a,0,a)$,$(0,a,a)$,$(0,-a,a)$ are solutions.
I tried finding solutions brute-forcing this problem, but I discovered there are no solutions with numerator and denominator smaller than 100'000.
I can prove that the question is equivalent to finding a bunch of natural numbers that satisfy:
$x'^2+ka^2 = w^2\\y'^2+kb^2 = w^2\\z'^2+kc^2 = w^2\\a+b=c$
(here is $x=x'/w$, $y=y'/w$, $z=z'/w$)
It can be proven they are equivalent as follows:
To go from the title to the 4 equations:
*
*Call $w$ the lcm of the denominators of x, y, z
*Define $x'=xw$, $y'=yw$, $z'=zw$
*The equation is now $\sqrt{w^2-x^2} + \sqrt{w^2-y^2} = \sqrt{w^2-z^2}$
*This equation can only hold if all sqrt are natural numbers, up to a common factor $k$ by this theorem
*Now call $a=\sqrt{w^2-x^2}/\sqrt{k}$, $b=\sqrt{w^2-y^2}/\sqrt{k}$, $c=\sqrt{w^2-z^2}/\sqrt{k}$
*The original equation becomes $a+b=c$, the other 3 can be calculated by the definitions of the previous step.
To go from the 4 equations to the title:
*
*Rewrite the first 3 equations to $a=\sqrt{w^2-x^2}/\sqrt{k}$, $b=\sqrt{w^2-y^2}/\sqrt{k}$, $c=\sqrt{w^2-z^2}/\sqrt{k}$
*Plug in in equation 4.
*Substitude $x=x'/w$, $y=y'/w$, $z=z'/w$
But apparently being able to split a number into two squares in at least three different ways, is quite a rare property, so I would suspect that finding three with the extra property that $a+b=c$ is probably impossible, which makes me believe that there are no solutions, or that solutions are very rare.
But is there a way to find a solution, or prove that non exists except for the trivial case?
Edit: Added proof of equivalence.
| Here is a way to generate solutions for $k=3$.
It is well known that every number whose prime factors are all of the form $6m+1$, is also of the form $n=a^2+3b^2$. Then
$$(2n)^2=4a^4+24a^2b^2+36b^4\\
=(2a^2-6b^2)^2+3(4ab)^2\\
=(a^2+6ab-3b^2)^2+3(a^2-2ab-3b^2)^2\\
=(a^2-6ab-3b^2)^2+3(a^2+2ab-3b^2)^2$$
Note two of the unsquared left-hand terms add up to the third, and two of the unsquared right-hand terms add up to the third.
There are also solutions for $k$ not equal to 3.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x,y,z>0$ with $x+y+z=\sqrt{3}$, prove that: $\sqrt{x^2+1-yz}+\sqrt{y^2+1-zx}+\sqrt{z^2+1-xy}\ge 3$ (Venezuela 1960)
If $x,y,z>0$ with $x+y+z=\sqrt{3}$, prove that: $$\sqrt{x^2+1-yz}+\sqrt{y^2+1-zx}+\sqrt{z^2+1-xy}\ge 3$$
I tried solving it as follows:
The condition we want proved is: $\sqrt{3x^2+3-3yz}+\sqrt{3y^2+3-3zx}+\sqrt{3z^2+3-3xy}\ge 3(x+y+z)$
$\sqrt{3x^2+(x+y+z)^2-3yz}+\sqrt{3y^2+(x+y+z)^2-3xz}+\sqrt{3z^2+(x+y+z)^2-3xy}\ge 3(x+y+z)$
And this is where I got stuck. I couldn't continue it. Could you please explain to me how to solve the question?
| The inequality ia actually very easy with just one hidden point kept in mind
Note that $$\sum \sqrt{x^2+1-yz}\ge \sum \sqrt{x^2+1-\frac{{(y+z)}^2}{4}}$$ Now use the fact $$x^2+1-\frac{{(y+z)}^2}{4}=x^2+\frac{{(x+y+z)}^2}{3}-\frac{{(y+z)}^2}{4}=\frac{1}{12} {(4x+y+z)}^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$ I have to solve the limit
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$$
applying Taylor's series.
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}=\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \frac{\sin 2x}{\cos 2x}}= \lim_{x\to 0^{+}} \frac{\ln (2 \cdot( sin x)^2)}{\ln \sin 2x - \ln \cos 2x}= \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln \sin 2x - \ln \cos 2x}= \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + \ln \sin x + \ln \cos x - 2\ln \cos x + 2 \ln \sin x}= \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + 3\ln \sin x }= \lim_{x\to 0^{+}} \frac{\ln 2( sin x)^2}{\ln 2( sin x)^3}$$
$$\frac{\ln 2( sin x)^2}{\ln 2( sin x)^3} \sim \frac{\ln (2 x^2- \frac{2}{3} x^4+ o(x^4))}{\ln (2 x^3- x^5+ o(x^5))}= \frac{\ln (x^2)+ \ln(2 - \frac{2}{3} x^2+ o(x^2))}{\ln(x^3)+\ln (2 - x^2+ o(x^2))} \sim \frac{2\ln x+ \ln 2}{3\ln x+\ln 2} \sim \frac{2}{3}$$
The suggested solution in my book is $2$. can someone indicate where I made mistakes?
| I think the following step is wrong:$$\lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln \sin x}{\ln \sin 2x - \ln \cos 2x}= \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln \sin x}{\ln 2 + \ln \sin x + \ln \cos x - 2\ln \cos x + 2 \ln \sin x}.$$
I think you have used the false identity $$ \ln\cos2x \equiv 2\ln\cos x - 2\ln\sin x. $$
It is false because $$ \ln\cos 2x = \ln(\cos^2x-\sin^2x) \neq\ln\cos^2x - \ln\sin^2x = 2\ln\cos x - 2\ln\sin^2x. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all real numbers $x$ such that $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number I want to find all values of $x\in \mathbb R$ such that the value of $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number.
I solved it as follows:
$x+2\sqrt{x}-1\ge 0$
$(\sqrt{x}+1)^2-2\ge 0$
$(\sqrt{x}+1)^2\ge 2$
$\sqrt{x}+1\ge 2$ or $\sqrt{x}+1\le -\sqrt{2}$
The second can't hold, from the first $x\ge 3-2\sqrt{2}$.
Similarly $(\sqrt{x}-1)^2\ge 2$, hence $x\ge 3+2\sqrt{2}$.
I find my solution to be very ugly. Is my solution correct and is there a neater approach?
| The requirement $x\ge0$ is obvious. The expression is real iff $(\sqrt{x}\pm1)^2\ge2$ for both choices of $\pm$, i.e. neither $\sqrt{x}\pm1$ is in $(-\sqrt{2},\,\sqrt{2})$, i.e .$\sqrt{x}\notin[0,\,1+\sqrt{2})\cup[0,\,\sqrt{2}-1)=[0,\,1+\sqrt{2})$, since $\sqrt{x}\ge0$. The solution set is $[3+2\sqrt{2},\,\infty)$, as in @alex.jordan's answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Two strange integrals I was asked to solve the two integrals
$$
\int\frac{1}{x^2e^{1/x}-x}dx
$$
$$
\int\frac{x^2+1}{x^3-x^2+x+1}dx
$$
I think the first one is not soluble with the known methods of calculus and the second one is involved with complex roots of cubic equation.
Any improvements appreciated.
Thanks.
| *
*For the integral
$$
\int\frac{x^2+1}{x^3-x^2+x+1}dx
$$
note that the cubic polynomial in the denominator has one real root $r= -0.5437$ and a pair of complex roots, which allows the factorization
$$x^3-x^2+x+1= (x-r)(x^2+(r-1)x-1/r)
$$
Then, decompose the integrand as
\begin{align}
\frac{x^2+1}{x^3-x^2+x+1}=\frac{A}{x-r}+\frac{Bx+C}{x^2+(r-1)x-1/r}
\end{align}
and integrate accordingly.
*For the integral
$$
\int\frac{1}{x^2e^{1/x}-x}dx
\overset{x\to\frac1x}=-\int \frac1{e^x-x}dx
=-\sum_{k\ge0} \int x^ke^{-(k+1)x}dx
$$
where each term can be integrated separately.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Comparison of the largest eigenvalues of two symmetric matrices Let $$A=\left(\begin{array}{ccccc|cccc}0&\cdots&0&0&0&0&0&a\\\vdots&\cdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\0&\cdots&0&0&0&0&0&a\\0&\cdots&0&0&b&0&0&0\\0&\cdots&0&b&0&b&0&0\\\hline 0&\cdots&0&0&b&0&\frac{1}{\sqrt{2}}&0\\0&\cdots&0&0&0&\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\a&\cdots&a&0&0&0&\frac{1}{\sqrt{2}}&0
\end{array}\right),$$ be an $n\times n$ matrix, where $a=\sqrt{\frac{6}{7}}$ and $b=\frac{1}{\sqrt{2}}$.
Let $B$ be same as $A$ except the second diagonal block which is $\left(\begin{array}{ccc}0&\sqrt{\frac{2}{3}}&\sqrt{\frac{8}{21}}\\\sqrt{\frac{2}{3}}&0&0\\\sqrt{\frac{8}{21}}&0&0\end{array}\right).$ With help of matlab I have verified that largest eigenvalue of $A$ is greater than that of $B$. How to prove it? I thought of doing it with calculating determinant of partitioned matrices but couldn't succeed.
| Not an answer, just too long for a comment.
Well, I decided to try and brute-force the computation of the characteristic polynimial for $A$ and got the following:
let $D_n$ be the determinannt of $tI_n-A$ (asumme $n\geq6$ so $A$ will contain the required $a$'s), so:
$$D_n = tD_{n-1} -a^2t^{n-6}(t^4-6b^2t^2+b^4)$$
and:
$$D_6 = t^6+(b-a^2)t^4 -3b^2t^3+(b^4-2b^3)t^2+(3a^2b^2-ab^4)t-a^2b^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4053470",
"timestamp": "2023-03-29T00:00:00",
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Equation with $ \sqrt[5]{}$ How to show that this equation $ \sqrt[5]{20-2x} + \sqrt[5]{7-x}+\sqrt[5]{3x+5}=2$ have 3 solutions 9, -6 and -25
Wolframsays the equation have no solutions !
| If $\sqrt[5]{20-2x}= a$ etc., $a+b+c=2$
$$a^5+b^5+c^5=32=(a+b+c)^5$$
Use How to factor $(a +b+c) ^5 -(a^5+ b^5 + c^5)$? to discover that $(a+b)(b+c)(c+a)=0$ should give three solutions.
| {
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Find the sum of the following geometric series $\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$
Find the sum of the following geometric series
$$\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$$
Attempt:
First I test with the root criterion if its divergent or convergent... $\frac {1}{2} < 1$ so it's convergent...
Now I try to find the sum and I can't get to the solution $\frac{5}{2}$.
Let's try:
$$\sum_{k=2}^{\infty} \frac{5}{2^k}=5\sum_{k=2}^{\infty} \frac{1}{2^k}=5\sum_{k=2}^{\infty} 2^{-^k}=5\cdot2^{-2}\sum_{k=2}^{\infty}2^{-k}$$
I don't know how to work with the $2^{-k}$ in here, hope for your help^^
| You can combine the answers from S.H.W and José Carlos Santos to figure out a solution if you don't know the sum $\sum^\infty_{k=0} 2^{-k}$.
$$
\begin{align}
\sum^\infty_{k=2} 2^{-k} &= \sum^\infty_{k=0}2^{-k} - 1 - \frac{1}{2} \tag{S.H.W} \\
&=2^{-2}\sum^\infty_{k=0}2^{-k} \tag{José Carlos Santos}
\end{align}
$$
If we write $L=\sum^\infty_{k=0}$ then $L-\frac{3}{2} = \frac{L}{4}$. This has as only solution $L=2$. Therefore:
$$\sum^\infty_{k=2} = \frac{L}{4} = \frac{1}{2}$$
| {
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Why is Euler's statement $\exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$ true? Euler, in his paper Variae observationes circa series infinitas [src], makes the following statements in his Theorem 19.
$$ \exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \frac{1}{4}D + \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \dots $$
Where he defines:
$$A = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots $$
$$B = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots $$
$$C = \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots $$
Question: I can't see how that statement is true.
I would appreciate replies that don't assume university level training in mathematics.
| It can be shown that
\begin{align*}
\sum\limits_{n = 1}^\infty {\frac{1}{{n^s }}} & = \prod\limits_p {\frac{1}{{1 - \frac{1}{{p^s }}}}} = \exp \left( { - \sum\limits_p {\log \left( {1 - \frac{1}{{p^s }}} \right)} } \right) \\ &= \exp \left( {\sum\limits_p {\sum\limits_{n = 1}^\infty {\frac{1}{{np^{ns} }}} } } \right)=
\exp \left( {\sum\limits_{n = 1}^\infty \frac{1}{n} {\sum\limits_p {\frac{1}{{p^{ns} }}} } } \right),
\end{align*}
whenever $\Re s>1$. Taking the limit $s\to 1+$ yields Euler's formal identity. I say formal, because the expressions on both sides will diverge. Note that I am not making an attempt to describe this in terms of non-university level mathematics.
| {
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Find maximum natural number k such that for any odd n $n^{12} - n^8 - n^4 + 1$ is divisible by $2^k$ Given $n^{12} - n^8 - n^4 + 1$ it's easy to factorize it: $(n-1)^2(n+1)^2(n^2+1)^2(n^4+1)$. It's stated than this should be divisible by $2^k$ for any odd $n$. So I think that such maximum $k$ can be found by putting the least possible odd $n$ in the expression (because we can easily find such a $k$, that for a bigger $n$ the expression above is divisible by $2^k$, but for a smaller $n$ it isn't). So the least possible $n=3$ (for $n=1$ expression is $0$) and we have $4 \cdot 16 \cdot 100 \cdot 82 = 25 \cdot 41 \cdot 2^9$, so the $k$ we are looking for is $9$.
I am rather weak in number theory so I hope the get some feedback if my solution is correct.
| With $n:=2m+1$, the polynomial is
$$(2m)^2(2m+2)^2(4m^2+4m+2)^2(16m^4+32m^3+24m^2+8m+2)
\\=2^7m^2(m+1)^2(2m^2+2m+1)(8m^4+16m^3+12m^2+4m+1).$$
The last two factors are always odd and the remaining ones form at least a multiple of $2^9$.
More precisely, if the multiplicity of the factor $2$ in the prime factorization of $\dfrac{n-1}2$ is $p>0$, then $k=2p+7$, otherwise $k=2q+7$ where $q$ is the multiplicity of $2$ in $\dfrac{n+1}2$.
E.g., with $n=15$, $p=0,q=3$ and $k=13$. Indeed, $15^{12}-15^8-15^4+1=2^{13}\,15837863153$, and with $n=17$, $p=3$ and $17^{12}-17^8-17^4+1=2^{13}\,71120027025$.
| {
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If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$? If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$?
I tried to solve it as follows:
I state that $p=ab$
$p^2=(2b+15)(2a+15)$
$p^2=4ab+30(a+b)+225$
$p^2=4p+30(a+b)+225$
and this is where I got stuck. I don't know how to get over this hurdle. could you please explain to me how to solve the question?
| $a^2=2b+15$ and $b^2=2a+15$
Subtracting, $a^2-b^2 = -2(a-b)$. As $a \ne b$,
$a+b = - 2$
Also adding both equations, $a^2+b^2 = 2(a+b)+30 = 26$
$(a+b)^2 = a^2+b^2+2ab \implies 4 = 26 + 2ab$
$ab = -11$
| {
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Solving $2^k +k \equiv 0 \pmod {323}$ Find all $k$ such that
$$2^k + k \equiv 0 \pmod{323}.$$
I noticed that $323 = 17\cdot 19$ so I thought about using the Chinese Remainder theorem by considering $2^k+k$ modulo 17 and 19. I got $k \equiv 16 \pmod{17}$ and $k\equiv 18 \pmod{19}$, which gives $k\equiv 322 \pmod{323}$. However, after trying this, the given solution was not valid : $2^{322} + 322 \equiv 156 \not \equiv 0 \mod{323}$.
Can someone explain why this didn't work, and what I can do to solve this problem? Thanks.
| Okay to solve $2^k \equiv \pmod {17}$ we know by FLT than $2^{16}\equiv 1 \pmod {17}$ and as $2^4 =16 \equiv -1$ that $2^8\equiv 1 \pmod {17}$.
So if $k \equiv 0, 1,2,3,4....,7\pmod {16}$ we have $2^k\equiv 1,2,4,8,-1,-2,-4, -8\pmod {17}$ and by CRT will be a unique solution for each pair. $k\equiv a \pmod {16}$ and $k \equiv 2^a \pmod {16}$. So that is eight solutions $\pmod {16*17}$.
(exmaple: If $k\equiv 0\pmod {16}$ and $k \equiv 2^0=1\pmod {17}$ then $k\equiv -16 \equiv 256 \pmod {272}$. Or if $k \equiv 1 \pmod 16$ and $k \equiv 2^1 \equiv 2\pmod {17}$ the solution $\pmod{272}$ is .... whatever)
Then we can do the same thing for $k\equiv b\pmod {18}$ and $k\equiv 2^b\pmod {19}$
So there are $18$ solutions $\pmod{18\cdot 19}$ for those.
However as $\gcd(16,18)=2\ne 1$ we must have $a,b$ either both even or both odd.
Still that is $4$ solutions $\pmod {16*17}$ with $a$ odd, and $4$ \solutions $\pmod{16*17}$ with $a$ even. And $9$ solutions $\pmod{18*19}$ with $b$ even or odd.
So that is $4*9=36$ solutions $\pmod{\frac {16*17*18*19}2}$ for $a,b$ odd and $36$ for $a,b$ even.
So there are $72$ solutions.
And it's way to much work do to.
But to find one solution we can have $k\equiv 256 \pmod {272}$
And $k \equiv 0 \pmod {18}$ and $k\equiv 2^0=1 \pmod{19}$ so $k\equiv -18\equiv 18^2 \mod {18*18}$ or $k \equiv 324 \pmod {342}$.
So if we solve $k\equiv 256\pmod {272}$ and $k \equiv 324{342}$ that solution ought to give us $2^k \equiv k\pmod {323}$ and ... oh, that $2^k - k\equiv 0$.... oh well the same ideaa will hold.
| {
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Where does the Integral went wrong .... $ \displaystyle \int _{0}^{\pi} \frac {\ln ( 1 + \sin \alpha \cos x )}{\cos x} \ \mathrm {d}x $ This is how I solved :
Let $$ \displaystyle y = \int_{0}^{\pi} \frac {\ln ( 1 + \sin \alpha \cos x )}{\cos x} \ \mathrm {d}x
$$
Then I used expansion of $\displaystyle \ln (1+x) = \sum _{n=0}^{\infty} (-1)^n \frac {x^{n+1}}{n+1} $
Now, $\displaystyle y = \sum _{n=0}^{\infty} (-1)^n \frac {\sin ^{n+1} \alpha }{n+1} \int_{0}^{\pi} \cos ^n x \ \mathrm {d}x \implies y = \sum _{n=0}^{\infty} [1+(-1)^n] \frac {\sin ^{n+1} \alpha }{n+1} \int_{0}^{\frac {\pi}{2} } \cos ^n x \ \mathrm {d}x $
Replacing $n$ by $2k$ (as we will end up with $0$ for odd $n$) and using gamma function :
$$ \displaystyle y = 2 \sum _{k=0}^{\infty} \frac {\sin ^{2k+1} \alpha }{2k+1} \cdot \frac { \Gamma \left ( \frac {2k+1}{2} \right ) \sqrt {\pi} }{ \Gamma \left ( \frac {2k+2}{2} \right )} $$
Which can be simplified to $\displaystyle y = 2 \pi \sum _{k=0}^{\infty} \frac {\sin ^{2k+1} \alpha }{2k+1} \cdot \frac {(2k-1)!!}{k! \cdot 2^k}$
Modification :
$$ \displaystyle y = 2 \pi \sum _{k=0}^{\infty} \frac {\sin ^{2k+1} \alpha }{2k+1} \cdot \frac {(2k-1)!!}{k! \cdot 2^k} \cdot \frac {k! 2^k}{k! 2^k} \implies y = 2 \pi \sum _{k=0}^{\infty} \frac {\sin ^{2k+1} \alpha }{2k+1} \binom {2k}{k} \frac {1}{4^k} $$
We get : $\displaystyle \frac {\mathrm {d}y }{\mathrm {d} ( \sin \alpha ) } = 2 \pi \sum _{k=0}^{\infty} \binom {2k}{k} \left ( \frac {\sin ^{2} \alpha}{4} \right ) ^k$
As $\displaystyle (1-4x)^{-1/2} = \sum _{n=0}^{\infty} \binom {2n}{n} x^n$
We will finally get : $\displaystyle y = 2 \pi \int \frac {\mathrm {d} (\sin \alpha )}{ \cos \alpha } \implies y = 2 \pi \alpha + C $
Now as $\alpha = 0 $ the summation is zero $\implies C = 0 $
Hence,
$$ \displaystyle \int_{0}^{\pi} \frac {\ln ( 1 + \sin \alpha \cos x )}{\cos x} \ \mathrm {d}x = 2 \pi \alpha $$
But :
This suggests : $ \displaystyle \int_{0}^{\pi} \frac {\ln ( 1 + \sin \alpha \cos x )}{\cos x} \ \mathrm {d}x = \pi \alpha $
| Your mistake is thinking $\int_0^{\pi/2}\cos^nx\mathrm{d}x$ is a Beta function; it's only half of one, viz.$$\operatorname{B}(a,\,b)=2\int_0^{\pi/2}\sin^{2a-1}x\cos^{2b-1}x\mathrm{d}x.$$
| {
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Find the minimal polynomial of $\sqrt[3]{3} + \sqrt{5}$ over $\mathbb{Q}$. Let $x = \sqrt[3]{3} + \sqrt{5}$
Notice that $(x+y)^2 = x^2 + 2xy + y^2$
Then, $x^2 = (\sqrt[3]{3} + \sqrt{5})^2 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5} + 5$
Then, $x^3 - 5 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5}$
Or notice that $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$
Then, $x^3 = (\sqrt[3]{3} + \sqrt{5})^3 = 3 + 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$
Then, $x^3 - 3 = 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$
Which I think the degree of the minial polynomial of $x$ over $\mathbb{Q}$ should have a degree of $6$.
Which means the minimal polynomial can be written as $x^6 + a_5 x^5 + \cdots + a_0$ for some $a_5, ..., a_0 \in \mathbb{Q}$.
But I just don't how get it from both the ways I provide above.
Or that $\sqrt[3]{3} + \sqrt{5}$ is not algebraic. So, I can not write such minimal polynomial out algebraic ?
| If $x=\sqrt[3]3+\sqrt5$, then $\left(x-\sqrt5\right)^3=3$. In other words $x^3-3 \sqrt{5} x^2+15 x-5 \sqrt{5}-3=0$. But\begin{align}x^3-3 \sqrt{5} x^2+15 x-5 \sqrt{5}-3=0&\iff x^3+15x-3=(3x^2-5)\sqrt5\\&\implies(x^3+15x-3)^2=5(3x^2-5)^2\\&\iff x^6-15 x^4-6 x^3+375 x^2-90 x-116=0.\end{align}
| {
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Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$ Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$.
I find online, one person suggested using Taylor Theorem to expand the right-hand side, and apply Bernoulli's inequality.
So, if $x_0 = 0$, $\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{4(2!)}x^2+R$, where R is larger than $0$, this makes sense to me, but I'm trying to find another way to prove the equality. Like Mean Value Theorem for inequality $\sqrt{1+x} \le 1+\frac{1}{2}x$.
I see the part where $1+\frac{1}{2}x$ is the same, but get trouble to put $\frac{1}{8}x^2$ into the equation.
| We’ll let $a = x + 1$
Transform the inequality by substitute $x$ with $a - 1$.
After some calculations we the get
$$ -a^2 + 6a + 3 \leq 8\sqrt{a} $$
Square both sides and we get
$$ a^4 - 12a^3 + 30a^2 - 28a + 9 \leq 0$$
Which is just $(a-1)^3(a - 9) \leq 0$.
Since $a > 1$ , this statement is true for all $a \in [1,9]$.
Thus, we are left to prove this statement for all $a > 9$.
Since $a = x + 1$ , $x > 8$.
Consider $(x - 2)^2 + 12 > 0$
After some calculations we then get $\frac{1}{2}x < 3 - 1 + \frac{1}{8}x^2$.
But since $x > 8$ we then get $1 + \frac{1}{2}x - \frac{1}{8}x^2 < \sqrt{x + 1}$.
Hence, this inequality is true for all $x > 0$.
| {
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From the given information, how do we use the Remainder Theorem to reach the answer? Why does the sequence for the polynomial alternate when it equals $-30$, and not alternate when it equals $10$? Why is the denominator when the sum of the geometric series is equal to $30$, $1+r$, and not $1-r$? I am unable to correctly relate the Remainder Theorem to its use in the given answer.
The question:
When a polynomial of degree $3$ is written in descending powers of $x$ the coefficients of its four terms form a geometric sequence.
When the polynomial is divided by $x+1$ it s remainder is $-30$. Then it is divided by $x-1$ its remainder is $10$.
The answer:
Use the Remainder Theorem. We have
$-a+ar-ar^2+ar^3=-30$
and
$a+ar+ar^2+ar^3=10$
Solving simultaneously:
$\frac{a(1-r^4)}{1+r}=30$
and
$\frac{a(1-r^4)}{1-r}=10$
Giving $r=-\frac{1}{2}$ and $a=16$
The polynomial is therefore $16x^3-8x^2+4x-2$
| The polynomial is a cubic whose coefficients are a geometric sequence (in the order of descending powers, i.e. starting with $x^3$ and ending with the constant term), meaning that our polynomial must take the form
$$p(x) = ax^3 + arx^2 + ar^2x + ar^3.$$
Remainder theorem states that the remainder of $p(x)$ when divided by $x - \alpha$, is $p(\alpha)$.
So, when dividing $p(x)$ by $x - 1$ (i.e. $\alpha = 1$), the remainder is $p(1)$. Thus,
$$10 = p(1) = a1^3 + ar1^2 + ar^21 + ar^3 = a + ar + ar^2 + ar^3.$$
Similarly, when dividing $p(x)$ by $x + 1$ (i.e. $\alpha = -1$), the remainder is $p(-1)$. Thus,
$$-30 = p(-1) = a(-1)^3 + ar(-1)^2 + ar^2(-1) + ar^3 = -a + ar - ar^2 + ar^3.$$
This should hopefully explain the alternating signs. The next observation you need to make is that both of the above expressions are geometric series. Each have $4$ terms. The first has an initial term of $a$ and a common ratio of $r$, while the second has an initial term of $-a$ and a common ratio of $-r$. The first one is precisely the usual formula for geoemetric series:
$$10 = a + ar + ar^2 + ar^3 = a\frac{1 - r^4}{1 - r}.$$
To get a formula for the second one, we simply replace $r$ with $-r$ and $a$ with $-a$, to get
$$-30 = -a + ar - ar^2 + ar^3 = (-a)\frac{1 - (-r)^4}{1 - (-r)} = -a\frac{1 - r^4}{1 + r}.$$
To get the equation from the solution, simply multiply both sides by $-1$.
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How many five digit numbers can be formed using digits $1,1,2,3,3,4,4$ With digits $1,1,2,3,3,4,4$, how many five digit numbers we can form?
$1)\frac34\times5!\qquad\qquad2)\frac94\times5!\qquad\qquad3)4\times5!\qquad\qquad4)\frac52\times6!$
Ok so the digits $1,3,4$ appears twice and $2$ appears once. I tried to count different cases:
first: having all $1,2,3,4$ digits and choose another digit from $1,3,4$:
$${3\choose1}\times\frac{5!}{2!}$$
second: having two pair of same digit and choose another digit:
$${3\choose2}\times2\times\frac{5!}{2!2!}$$
Summing them we have $3\times\dfrac{5!}{2}+3\times\dfrac{5!}{2}=5!\times3$ but I don't have this in the options.
| Either you don't select $2$ or you select $2$. In the first case you have 3 choices: $(2, 2, 1); (2, 1, 2); (1, 2, 2)$. For each such choice you have $\frac{5!}{2!2!}$ allocations. If you select $2$, you have two paths:$(2, 2, 0), (2, 0, 2), (0,2,2)$, in which case you have $\frac{5!}{2!2!}$ allocations and $(1, 1, 2), (1,2,1),(1,1,2)$, in which case you have $\frac{5!}{2!}$ allocations. Putting it together,
$$
\frac{5!3}{2!2!} + \frac{5!3}{2!2!} +\frac{5!3}{2!} = 5!3
$$
| {
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Congruency application I want to prove whether the following is true using congruencies: If $n$ is odd and $3\not | n$, then $n^2\equiv 1\pmod{24}$.
I tried a direct proof.
Let $n=2k+1$ for an integer $k$ such that $3\not| n$, or $3c=n+r$ such that $r=1$ or $r=2$ for an integer $c$. Then in the first case, $3c=n+1$ implies $n\equiv 1\pmod 3$. Or in the second case, $3c=n+2$ implies $n\equiv 2 \pmod 3$. Then $$n^2\equiv 1 \pmod 3 \text{ or } n^2\equiv 4 \pmod 3$$
iff $$3|{n^2-1} \text{ or } 3|n^2-4$$
or $$8n^2\equiv8, 32 \pmod{24}$$
This was the farthest I got. I'm having a hard time going from mod 3 to mod 24 just given the conjunction that $n$ is odd and that $3\not| n$
| It's the same as proving that $n^2-1 \equiv 0 \pmod 24$.
Since $n \equiv 1 \pmod 2 \implies (n-1)(n+1) \equiv 0 \pmod 8$ (as one of $n-1, n+1$ will be $2 \pmod 4 $ and the other will be $0 \pmod 4$)
and since $n \neq 0 \pmod 3$, one of $n-1,n+1$ must be $0\pmod 3 $ $\implies $(n-1)(n+1) \equiv 0 \pmod 3 $
$\implies n^2-1 \equiv 0 \pmod 24 \implies n^2 \equiv 1 \pmod 24$.
| {
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What is the maximal area between hyperbola and chord $AB$?
Suppose $A$ and $B$ are variable points on upper branch of hyperbola $\mathcal{H}:\;x^2-y^2=-4$ such that $AB = 1$. What is the maximal area between $AB$ and $\mathcal{H}$?
Clearly $y= \sqrt{x^2+4}$. If we set $A\big(t,\sqrt{t^2+4}\big)$ then $B\big(t+h, \sqrt{(t+h)^2+4} \big)$ for some positive $h$, such that $$h^2 +\Big(\sqrt{(t+h)^2+4} -\sqrt{t^2+4} \Big)^2=1$$
and we have to find a minimum value of
$$f(t) = \int_{t}^{t+h}\sqrt{x^2+4}\;dx$$
Notice that if $$F(x):= \int\sqrt{x^2+4}\;dx$$ then $$ F(x)= 2\ln\left(\left|\sqrt{x^2+4}+x\right|\right)+x\sqrt{\dfrac{x^2}{4}+1}$$
and things get complicated very quickly.
| Disclaimer: I am not sure since I am still new to calculus.
Find the minimum value of $f(t) = \displaystyle\int_{t}^{t + h}\sqrt{x^{2} + 4}\,dx$.
The fundamental theorem of calculus can be applied here:
\begin{align*}\dfrac{d}{dt}f(t) &= \dfrac{d}{dt}\int_{t}^{t + h}\sqrt{x^{2} + 4}\,dx \\ f'(t) &=\left(\sqrt{(t + h)^{2} + 4}\right)\left(\dfrac{d}{dt}(t + h)\right) - \left(\sqrt{t^{2} + 4}\right)\left(\dfrac{d}{dt}(t)\right) \\f'(t) &= \left(\sqrt{(t + h)^{2} + 4}\right) - \left(\sqrt{t^{2} + 4}\right)\end{align*}
Then, set $f'(t) = 0$,
\begin{align*}f'(t) &= 0 \\ \sqrt{(t + h)^{2} + 4} - \sqrt{t^{2} + 4} &= 0 \\ \sqrt{(t + h)^{2} + 4} &= \sqrt{t^{2} + 4} \\ (t + h)^{2} + 4 &=t^{2} + 4 \\ t^{2} + 2th + h^{2} + 4 &= t^{2} + 4 \\ 2th + h^{2} &= 0 \\ 2t &= -h \\ t&= -\dfrac{h}{2}\end{align*}
I don't know how to proceed now, but I hope that this can help.
I used the variant of the FTOC which is: $$\dfrac{d}{dx}\int_{u(x)}^{v(x)}f(t)\,dt = f(v(x))v'(x) - f(u(x))u'(x)$$
| {
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"answer_count": 3,
"answer_id": 2
} |
Finding the area of a sector within a triangle.
For part (a) of the question I am getting an answer of $33.6^{\circ}$ of $0.586 radians$ which I am pretty sure is right.
Part (b) is where I am having difficulties because my answer is not matching the mark scheme.
For b (i) I have an answer of $4.516 cm$
for (ii) I have an answer of $$\frac{9*2.49\cdot sin(0.586)}{2}$$
Then I am stuck.
Please could someone give a model answer.
Thanks!
| $|AB|=5, |AC|=|BC|=3=|AX|=|YB|$ and since $|ABC|$ is isosceles, sum of angles in $|ABC|$ is $180°$, then $2\theta_1+\theta_2= 180°$, from cosine rule
$$ \cos(\theta_2) = \frac{ 3^2+3^2-5^2}{2×3^2}$$
$$\theta_2 = \cos^{-1}(\frac{-7}{18})$$
$$\theta_1 = 90° - \frac{1}{2}\theta_2$$
$$\theta_1 = 90° -\frac{1}{2} \cos^{-1}(\frac{-7}{18})°$$
Area of triangle $|ABC|$, the semi-perimeter here is $s = \frac{5+3+3}{2}$, from heron's formula
$$k_1 = \sqrt{\frac{11}{2}(\frac{11}{2}-5)(\frac{11}{2}-3)^2}$$
Perimeter of shape $|CYX| = |YX| +|CY|+|CX|$, $|CY| = |CX|$ now a sector with radius $3$ and angle $[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$ in degree, has curved length as
$$|CY| = |CX| = 3×\frac{\pi}{180}[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$$
The length $|YX| = |AB|-|AY|-|XB|$, $|AY|=|XB|$ meaning that $|YX| = 5-2|AY|$, and also $|AB|-|AX|= |AY|$ meaning that $5-3 = |AY|$, then the length $|YX|$ is
$$|YX| = 5-2(5-3) = 1$$
$$|CYX| = |YX| +2|CY|$$
$$ k_2 = 1+6\frac{\pi}{180}[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$$
The area of sector $|ACX| = \theta_1 × \frac{|AX|^2}{2}$
$$k_3 = \frac{9}{2}\cdot \frac{\pi}{180}[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$$
The area of sector $|CYX| = |ABC| -|ACY| -|BCX|$, $|ACY|=|BCX|$ meaning that $|CYX| = k_1 -2|ACY|$, and also $|ABC| -|ACX| = |BCX|$ meaning that $k_1 - k_3 = |BCX|$, then the sector $|CYX|$ is
$$k_4 = k_1 - 2(k_1 - k_3 )$$
$$k_4 = \sqrt{\frac{11}{2}(\frac{11}{2}-5)(\frac{11}{2}-3)^2}-2\sqrt{\frac{11}{2}(\frac{11}{2}-5)(\frac{11}{2}-3)^2}+2\frac{9}{2}\cdot \frac{\pi}{180}[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$$
$$k_4 = -\sqrt{\frac{11}{2}(\frac{11}{2}-5)(\frac{11}{2}-3)^2}+2\frac{9}{2}\cdot \frac{\pi}{180}[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$$
| {
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Find the minimum value of $P=ab+bc+3ca+\dfrac{3}{a+b+c}$ Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2=3$. Find the minimum value of $$P=ab+bc+3ca+\dfrac{3}{a+b+c}.$$
This is an asymmetric inequality. It is hard for me to find when the equation holds. I guess when it occurs if $a=c=0$ and $b=\sqrt{3}$. Then $\min P=\sqrt{3}$. But I have no idea to solve it. Please help me a hint. Thank you.
| No matter what, by simply calculating one has $P(\sqrt 3, 0, 0) = \sqrt 3$. On the other hand,
We know a minimum must exist by Weierstrass's theorem. This is because $Z := \{(a, b, c) \in \mathbb R ^3\mid a, b, c \geq 0, a^2 + b^2 + c^2 = 3\}$ is a closed bounded set and $P$ is continuous on $Z$.
Write $$P(a,b,c) = S_1 + S_2 \tag{1}$$
where $S_1 :=ab + bc + \dfrac{3}{2} + ac + \dfrac{3}{a + b + c}$ and $S_2 :=2ac - \dfrac{3}{2}$. Note that $S_1$ is symmetric. The choice of $\dfrac{3}{2}$ will be clear should you read on.
If it is possible to simultaneously minimise $S_1$ and $S_2$ (with given constraints), then their sum will be the minimum value of $P$. Let us try to minimise $S_1$.
Plugging $bc = \dfrac{1}{2}(b+c)^2 - \dfrac{3}{2} + \dfrac{a^2}{2}$ into $S_1$ yields
$$ S_1 = \dfrac{1}{2}(b+c)^2 + (b + c) a + \frac{3}{(b + c) + a} + \dfrac{a^2}{2},$$
or with $x := b + c$,
\begin{align}
\begin{split}
S_1(x, a) &= \dfrac{1}{2} x^2 + a\, x + \frac{3}{x + a} + \dfrac{a^2}{2} = \\
&= \dfrac{1}{2} (x + a)^2 + \dfrac{3}{x + a} = \\
&= \dfrac{1}{2} z^2 + \dfrac{3}{z}.
\end{split}
\tag{2}
\end{align}
*
*We are left with the problem of minimising $S_1(z)$ wrt $z$ when $z \geq \sqrt 3$ (because $z^2 \geq a^2 + b^2 + c^2$).
*The derivative with respect to $z$ is $z - \dfrac{3}{z^2}$. This is positive if $z > \sqrt[3]{3}$. So $S_1$ is increasing for $z \geq \sqrt[3]{3}$ and we should pick $z_0 := \sqrt 3$ if we can (choosing $z_0 := \sqrt[3]{3}$ is not possible because $\sqrt[3]{3} < \sqrt 3$).
If one moves up the chain, we have $z_0 = x_0 + a_0 = a_0 + b_0 + c_0 = \sqrt 3$. And we see it is possible to choose $a_0 := \sqrt 3$ and $b_0 := c_0 := 0$ to obtain $z_0 = \sqrt 3$, thus minimising $S_1$. By picking $c_0 = 0$, $S_2$ is minimised as well, so we are done. In conclusion, the minimum value of $P$, subject to given constraints, is $\sqrt 3$. It is obtained at, for example, $(a_0, b_0, c_0) = (\sqrt 3, 0, 0)$.
| {
"language": "en",
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Prove $x^x+y^y\ge x^2+y^2$ for $x,y>0$ and $x+y\le 2$. We may prove the inequality for $x,y\in (0,1]$.
Note that, for $0<x\le 1$, it holds that
\begin{align*}
x^x&=1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3+\cdots\\
&\ge1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3\\
&\ge x^2.
\end{align*}
Similarily, for $y \in (0,1]$, it holds that $$y^y\ge y^2.$$
Thus$$x^x+y^y\ge x^2+y^2.$$
But how to prove under the condition $x+y\in (0,2]$?
| Remarks: @Q. Zhang gave a nice proof. I give an alternative proof here.
Fact 1: $f(v) = v^v - v^2$ is strictly decreasing on $(0, 1)$.
The proof is given at the end.
Fact 2: $u^u \ge \frac{u^2 - u + 2}{3 - u}$ for all $u$ in $(0, 2)$.
The proof is given at the end.
(Note: $\frac{u^2 - u + 2}{3 - u}$ is the Pade $(2,1)$ approximant of $u^u$ at $u = 1$.)
If $x, y \le 1$, we have $x^x \ge x \ge x^2$
and $y^y \ge y \ge y^2$. The desired inequality is true.
In the following, WLOG, assume $x \in (1, 2)$.
By Fact 1, since $0 < y \le 2 - x < 1$, we have $y^y - y^2 \ge (2 - x)^{2 - x} - (2 - x)^2$.
It suffices to prove that, for all $x\in (1, 2)$,
$$x^x + (2 - x)^{2 - x} - x^2 - (2 - x)^2 \ge 0.$$
Using Fact 2, we have
\begin{align*}
&x^x + (2 - x)^{2 - x} - x^2 - (2 - x)^2\\
\ge\ & \frac{x^2 - x + 2}{3 - x} + \frac{(2 - x)^2 - (2 - x) + 2}{3 - (2 - x)} - x^2 - (2 - x)^2\\
=\ & \frac{2(x - 1)^4}{(3 - x)(1 + x)}\\
\ge\ & 0.
\end{align*}
We are done.
Proof of Fact 1: We have $f'(v) = v^v(\ln v + 1) - 2v$.
If $v \in (0, \mathrm{e}^{-1}]$, then clearly $f'(v) < 0$. If $v\in (\mathrm{e}^{-1}, 1)$, then
$f'(v) \le \ln(v) + 1 - 2v \le (v - 1) + 1 - 2v = -v < 0$ where we have used $v^v \le 1$ and $\ln v \le v - 1$. We are done.
Proof of Fact 2: Let
$F(u) = u\ln u - \ln \frac{u^2 - u + 2}{3 - u}$.
We have
\begin{align*}
F'(u) &= \ln u - \frac{(u - 1)(u^2 - 4u + 7)}{(3 - u)(u^2 - u + 2)},\\
F''(u) &= \frac{(u^4 - 5u^3 + 3u^2 - 19u + 36)(u - 1)^2}{u(3 - u)^2(u^2 - u + 2)^2}.
\end{align*}
Let $h(u) = u^4 - 5u^3 + 3u^2 - 19u + 36$.
We have $h'(u) = u^2(4u - 15) + (6u - 19) < 0$.
Thus, $h(u)$ is strictly decreasing on $(0, 2)$.
Also, $h(1) > 0$ and $h(2) < 0$. Thus,
there exists $u_0\in (1, 2)$ such that $h(u_0) = 0$,
$h(u) > 0$ on $(0, u_0)$, and $h(u) < 0$ on $(u_0, 2)$. Thus, $F''(u)\ge 0$ on $(0, u_0]$
and $F''(u) < 0$ on $(u_0, 2)$.
Thus, $F(u)$ is convex on $(0, u_0]$ and concave on $(u_0, 2)$.
Since $F(1) = 0$ and $F'(1) = 0$, we have
$F(u)\ge 0$ on $(0, u_0]$. Also, since $F(u_0)\ge 0$ and $F(2) = 0$,
we have $F(u)\ge 0$ on $(u_0, 2)$.
Thus, $F(u)\ge 0$ on $(0, 2)$. We are done.
| {
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} |
How to proceed solving $(7+4\sqrt3)^m+(7-4\sqrt3)^m=14$? I have$$(7+4\sqrt3)^m+(7-4\sqrt3)^m=14$$
By noticing that $7+4\sqrt3=\frac1{7-4\sqrt3}$ one way to solve the equation is using substitution $(7+4\sqrt3)^m=t$ and solve for $t$ in $t+\frac1t=14$.
But I'm trying to use a little different approach:
We have $7+4\sqrt3+7-4\sqrt3=14$ So by using the substitution $u=7+4\sqrt3$ we have :
$$u^m+\frac1{u^m}=u+\frac1u$$
But from here how can I prove mathematically that the only answers are $m=1$ and $m=-1$ ?
| $m = 1$ is a solution, which is easily seen. Suppose that $m \neq 1$. Clearly, $m = 0$ is not a solution Because $u + \dfrac{1}{u} \neq 2$.
First, suppose that $m > 0$. Then, we would have
$$u^m - u = \dfrac{1}{u} - \dfrac{1}{u^m}.$$
Here, we can use $a^k - b^k = \left( a - b \right) \left( a^{k - 1} + a^{k - 2}b + \cdots + b^{k - 1} \right)$ after taking $u$ (respectively $\dfrac{1}{u}$) common.
Hence, we get
$$u \left( u - 1 \right) \left( 1 + u + u^2 + \cdots u^{m - 2} \right) = \dfrac{1}{u} \left( 1 - \dfrac{1}{u} \right) \left( 1 + \dfrac{1}{u} + \dfrac{1}{u^2} + \cdots + \dfrac{1}{u^{m - 2}} \right).$$
The right-hand side can be simplified as
$$u \left( u - 1 \right) \left( 1 + u + u^2 + \cdots + u^{m - 2} \right) = \dfrac{1}{u^m} \left( u - 1 \right) \left( 1 + u + u^2 + \cdots + u^{m - 2} \right).$$
Now, we know that $u \neq 0$, $u \neq 1$ and $u > 0$. Hence, most of the terms cancel off and we get
$$u = \dfrac{1}{u^m}.$$
This tells us that $m = -1$, which is a contradiction to our assumption.
Now, for $m < 0$, we let $m = -n$ for some $n > 0$. Then, we have
$$u^{-n} - u = \dfrac{1}{u} - \dfrac{1}{u^{-n}}.$$
Then,
$$u^{-n} \left( 1 - u^{n + 1} \right) = \dfrac{1}{u} \left( 1 - u^{n + 1} \right).$$
This gives us $\dfrac{1}{u^n} = \dfrac{1}{u}$, and since $u > 0$, we conclude that $n = 1$. That is, $m = -1$.
| {
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"source": "stackexchange",
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Dirichlet problem on the unit disk using Poisson’s formula I’ve been trying to do the following exercise, of course without success, because I’m struggling with the integral. First things first, here's my exercise:
Writing the unit disk as $D \subset \mathbb{R}^2$, we define $g \in C(\partial D)$ by $g(x,y)=$ $\begin{cases}
y \quad \text{if } y \geq 0\\
0 \quad \text{otherwise}
\end{cases}$.
If $u$ is a solution of the boundary problem
$\begin{cases}
\Delta u = 0 \quad \text{in D}\\
u=g \quad \text{on $\partial D$}
\end{cases}$, compute $u(x,0)$ for $x\in (-1,1)$.
My intuition was to use Poisson's formula $u(x)= \frac{r^2-\lvert x \rvert ^2}{nr \omega_n} \int_{\partial D} \frac{g(y)}{\lvert x-y \rvert ^n} dS(y)$ and try to use the polar coordinates, but I'm instantly stuck because of the denominator. I'd really appreciate a hand with this!
| You can use the Poisson formula for the unit disk on the form
\begin{equation*}
u(r\cos\theta,r\sin\theta) = \frac{1-r^2}{2\pi} \int_0^{2 \pi} \frac{h(\phi)}{1 - 2r \cos(\theta - \phi) + r^2} \, d\phi
\end{equation*}
where $h(\phi)$ is the function giving the boundary data, in this case (since $y=\sin\phi$ on the unit circle)
\begin{equation*}
h(\phi) =
\begin{cases}
\sin \phi, & 0 \le \phi \le \pi
,\\
0, & \pi < \phi < 2\pi
,
\end{cases}
\end{equation*}
so that
\begin{equation*}
u(r\cos\theta,r\sin\theta) = \frac{1-r^2}{2\pi} \int_0^{\pi} \frac{\sin\phi}{1 - 2r \cos(\theta - \phi) + r^2} \, d\phi
.
\end{equation*}
Since the exercise asks for the values of $u$ on the $x$ axis, take $\theta=0$ and $r=x$ (with $0 \le x < 1$ to begin with):
\begin{equation*}
\begin{split}
u(x,0)
&
= \frac{1-x^2}{2\pi} \int_0^{\pi} \frac{\sin\phi}{1 - 2x \cos\phi + x^2} \, d\phi
\\ &
= \frac{1-x^2}{2\pi} \biggl[ \frac{\ln|1 - 2x \cos\phi + x^2|}{2x} \biggr]_0^\pi
\\ &
= \frac{1-x^2}{4\pi x} \Bigl( \ln|1 + 2x + x^2| - \ln|1 - 2x + x^2| \Bigr)
\\ &
= \frac{1-x^2}{4\pi x} \, \ln \frac{|1+x|^2}{|1-x|^2}
\\ &
= \frac{1-x^2}{2\pi x} \, \ln \frac{1+x}{1-x}
.
\end{split}
\end{equation*}
Since this turned out to be an even function, $u(-x,0)=-u(x,0)$, it is correct also for $-1 < x < 0$.
(The solution must be even with respect to $x$, since the boundary values are.)
So that's the answer to your exercise, if you add in the boundary values $u(\pm 1,0)=0$ “by hand”
(the expression above is undefined for $x=\pm 1$, but if you compute the limits as $x \to \pm 1$, you get zero).
As a bonus, note that with a little complex analysis we can compute the full solution for $u$ on the whole unit disk:
\begin{equation*}
u(x,y) = \frac{y}{2} + \operatorname{Re} f(x+iy)
,\qquad
f(z) = \frac{1-z^2}{2\pi z} \, \operatorname{Log} \frac{1+z}{1-z}
,
\end{equation*}
which after some calculations gives the following little expression:
\begin{equation*}
\begin{split}
u(x,y)
& = \frac{y}{2}
+ \frac{x (1-x^2-y^2)}{4\pi (x^2+y^2)} \ln \frac{(1+x)^2 + y^2}{(1-x)^2 + y^2}
\\&
+ \frac{y (1+x^2+y^2)}{2\pi (x^2+y^2)} \arctan \frac{2y}{1-x^2-y^2}
,
\qquad
x^2+y^2 < 1
.
\end{split}
\end{equation*}
| {
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Prove the identity $\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}$ Prove the identity $$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}.$$
We have $$\left(\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}\right)^2=17+2\sqrt{30}-2\sqrt{17+2\sqrt{30}}\cdot\sqrt{17-2\sqrt{30}}+17-2\sqrt{30}=34-2\sqrt{(17)^2-(2\sqrt{30})^2}=34-2\sqrt{289-120}=34-2\sqrt{169}=34-2.13=8=(2\sqrt2)^2, $$
so $$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}.$$
In the hints the authors have written that I should use the fact that the LHS is positive and square it. What would be the problem if it wasn't positive? The identity obviously won't hold because LHS<0, RHS>0...
| Since the roots on the LHS get "annihilated", it makes very much sense to look for natural $m,n$ such that
\begin{eqnarray*}
17+2\sqrt{30} & = & \left(\sqrt n + \sqrt m\right)^2 & = & n + m + 2\sqrt{nm}\\
17-2\sqrt{30} & = & \left(\sqrt n - \sqrt m\right)^2 & = & n + m - 2\sqrt{nm}
\end{eqnarray*}
From this approach you get
\begin{eqnarray*}
n+m & = & 17\\
\sqrt{nm} & = & \sqrt{30}
\end{eqnarray*}
Obviously $n = 15, m=2$ fit, hence
$$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}= \sqrt{15}+\sqrt 2 - (\sqrt{15}-\sqrt 2)=2\sqrt{2}$$
| {
"language": "en",
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"source": "stackexchange",
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If $\tan A, \tan B, \tan C$ are roots of $x^3-ax^2+b=0$, find $(1+\tan^2 A)(1+\tan^2B)(1+\tan^2C)$ Here are all the results I got
$$\tan(A+B+C)=a-b$$
And
$$(1+\tan^2A)(1+\tan^2B)(1+\tan^2C)=(\frac{1}{\cos A\cos B\cos C})^2$$
And
$$\cot A+\cot B + \cot C=0$$
How should I use these results?
| Set $x = 1 + y^2 \Rightarrow y = ±\sqrt{x - 1}$ so that $1 + \tan^2 A, 1 + \tan^2 B, 1 + \tan^2 C$ give the roots of $x^3 - ax^2 + b = 0$.
This gives $±(x-1)^{3/2} - a(x-1) +b=0$. By Vieta, the product of the roots is the constant term, but we do not have a polynomial yet. Squaring both sides of $±(x - 1)^{3/2} = a(x-1) - b$, the constant term is $(-1)^3 - (-a-b)^2 = -1 -(a+b)^2$ if we take the positive branch, and $(-a-b)^2 - (-1) = 1 + (a+b)^2$ if we take the negative branch. However, since $(1 + \tan^2 A)(1 + \tan^2 B)(1 + \tan^2 C)$ is always positive, the answer is $1 + (a+b)^2$.
| {
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Factorize $abx^2-(a^2+b^2)x+ab$ Factorize the quadratic trinomial $$abx^2-(a^2+b^2)x+ab.$$
The discriminant of the trinomial is $$D=(a^2+b^2)^2-(2ab)^2=\\=(a^2+b^2-2ab)(a^2+b^2+2ab)=(a-b)^2(a+b)^2=\\=\left[(a-b)(a+b)\right]^2=(a^2-b^2)^2\ge0 \text{ } \forall a,b.$$
So the roots are $$x_{1,2}=\dfrac{a^2+b^2\pm\sqrt{(a^2-b^2)^2}}{2ab}=\dfrac{a^2+b^2\pm\left|a^2-b^2\right|}{2ab}.$$ How can I expand the modulus here? Have I calculated the discriminant in a reasonable way? Can we talk about "the discriminant of a quadratic trinomial" or only the corresponding quadratic equation (the trinomial=0) has a discriminant? What about "the roots of a trinomial" (or of the corresponding quadratic equation)?
| Everything is OK.
First: Accept
$$a^2≥b^2$$
Then accept
$$a^2≤b^2$$
You will see that the roots have not changed.
We have
$$abx^2-(a^2+b^2)x+ab=(a x - b) (b x - a)$$
Small Supplement:
Note that,
$$\begin{cases} ±|a^2-b^2|=±(a^2-b^2), a^2≥b^2 \\ ±|a^2-b^2|=±(b^2-a^2), a^2≤b^2 \end{cases}$$
$$\implies \pm(a^2-b^2)=\mp(b^2-a^2)$$
Thus,
$$±|a^2-b^2|=±(a^2-b^2).$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Solving an SDE with Ito's Lemma I want to solve the initial value problem
$$ dX_t = \left(\frac{b^2}{4} - X_t\right)dt + b\sqrt{X_t}dw$$
I have the initial condition $X_0 = x > 0$. Note this process stops when $X_t = 0$.
I'm pretty sure I need to apply Ito's formula here but I'm not sure how. I tried making the substitution $Y_t = \sqrt{X_t}$ but have not yet been able to figure the problem out. I'm new to sdes and would really appreciate any help.
| Yes, we need to apply the Ito's formula. We denote $\tau = \inf\{t\geq 0 : X_t = 0\}$ and denote $W$ the Brownian motion.
For $t < \tau$, we can apply Ito's formula to the function $\phi(x) = \sqrt{x}$ using the Ito process $X$. Thus,
\begin{align}
dZ_t = d(\sqrt{X_t}) &= \frac{1}{2\sqrt{X_t}}dX_t - \frac{1}{8(X_t)^{\frac{3}{2}}}d\langle{X}\rangle_t\\
&=\frac{1}{2\sqrt{X_t}}\left(\left(\frac{b^2}{4} - X_t\right)dt + b\sqrt{X_t}dW_t\right) - \frac{1}{8\sqrt{X_t}}b^2dt\\
&=-\frac{1}{2}\sqrt{X_t}dt + \frac{b}{2}dW_t \\
&=-\frac{1}{2}Z_tdt + \frac{b}{2}dW_t \\
\end{align}
This yields to linear SDE which can be easly solved as follows:
Applying IPP to $\{Y_t = e^{\frac{t}{2}}Z_t\}_{t\geq0}$, we have that:
\begin{align}
dY_t &= \frac12Y_tdt + e^{\frac{t}{2}}dZ_t \\
&= \frac12Y_tdt + e^{\frac{t}{2}}\left(-\frac{1}{2}Z_tdt + \frac{b}{2}dW_t\right)\\
&= e^{\frac{t}{2}}\frac{b}{2}dW_t
\end{align}
Thus, $Y_t = Y_0 + \frac{b}{2}\int_0^te^{\frac{s}{2}}dW_s$ and $Y_0 = \sqrt{x}$. Tracing back to the initial quantity, we have
\begin{align}
X_t &= e^{-t}(Y_t)^2 \\
&= e^{-t}\left(\sqrt{x} + \frac{b}{2}\int_0^te^{\frac{s}{2}}dW_s\right)^2
\end{align}
Note that the function $\phi$ is well defined here because the CIR process is always positive for the given parameters.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What are the possible real values of $\frac{1}{x} + \frac{1}{y}$ given $x^3 +y^3 +3x^2y^2 = x^3y^3$? Let $x^3 +y^3 +3x^2y^2 = x^3y^3$ for $x$ and $y$ real numbers different from $0$.
Then determine all possible values of $\frac{1}{x} + \frac{1}{y}$
I tried to factor this polynomial but there's no a clear factors
| HINT Divide by $x^3y^3$ on both sides. You will get a new equation in terms of $1/x$ and $1/y.$ Can you finish from here?
| {
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How to show that $5/(1+2^{2/3}) = 1+2\sqrt[3]{2}-2^{2/3}$? I was trying to simplify the solution, $\frac{1}{5}(1+2\sqrt[3]{2}-2^{2/3})$, from a problem I done and saw that mathematica suggested $\frac{1}{1+2^{2/3}}$. I was wondering if anyone could show how to approach simplification of such expressions manually (it appears to amount to the step in the title of the post)?
| This is the cubic formula from basic algebra:
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$
$$\frac{a^3+b^3}{a+b}=a^2-ab+b^2$$
Observe,
$$\begin{align}(1+2^{2/3})×(1+2\sqrt[3]{2}-2^{2/3})&=1^3+2^2\\
&=5.\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$ Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$
I try $x=0$ We have: $y^3-y^2=0 \Longrightarrow \left\{\begin{array}{l}
y=0 \\
y=1
\end{array}\right.$
I think, this equation only $(x,y)\in ${ $(0,0),(0,1),(1,0)\}$ but I can't prove that.
| Yet an other answer following the same natural lines.
The cases when $x=0$ and/or $y=0$ are clear from the OP. We search for other, "new" solutions.
Let $a=(x,y)$ be the lcm of $x,y$, taken now to be $\ge 1$,
so we can and do write $x=aX$, $y=aY$ with relatively prime integers $X,Y$.
Then the given equation is equivalent to:
$$
a(X+Y)(X^2-XY+Y^2) = X^2+42XY+Y^2\ .
$$
This implies:
*
*$(X+Y)$ divides $40XY=(X^2+42XY+Y^2)-(X+Y)^2$.
So $\color{brown}{(X+Y)}$ divides $\color{brown}{40}$.
*$(X^2-XY+Y^2)$ divides $43XY=(X^2+42XY+Y^2)-(X^2-XY+Y^2)$.
So $\color{brown}{(X^2-XY+Y^2)}$ divides $\color{brown}{43}$.
(If a prime $p$ divides $(X+Y)$ and $XY$, then it divides either $X$ or $Y$, possibly exchanging notations let $p$ divide $X$, the $p$ divides also $(X+Y)-X=Y$, so $p$ divides the lcm $(X,Y)=1$, contradiction. The other case is similar.)
From here, some few cases have to be checked:
*
*$X+Y=d\in\{\pm 1,\pm 2,\pm 4,\pm 5,\pm 8,\pm 10,\pm 20,\pm 40\}$, and
*$X^2-XY+Y^2\in\{1,43\}$.
The first restriction lets us linearly substitute $Y$ in terms of $X$. The substitution
$Y=d-X$, inserted in the second restriction gives $3X^2-3dX+d^2\in\{1,43\}$.
thus getting some few equations of second degree in $X$ of the shape:
$$
\begin{aligned}
3X^2 - 3dX + (d^2-1) &=0\ ,\\
3X^2 - 3dX + (d^2-43) &=0\ .
\end{aligned}
$$
We obtain $(X,Y)$-solutions iff the discriminant is a square. So either
$(4-d^2)$, or
$(172-d^2)$ is a square.
Only the values $d=\pm 1$ and $d=\pm 2$ make the first expression ($1^2$ and $0^2$) a square.
Only the values $d=\pm 5$ and $d=\pm 8$ make the second expression ($7^2$ and $6^2$) a square.
We may and do assume $d>0$ (possibly changing now $a$ into $-a$), since passing from $d$ to $-d$ means replacing $(X,Y)$ by $(-X,-Y)$ (and $a$ by $-a$), which leads finally to the same final solution.
Here are these four cases explicitly displayed, and the four final solutions.
*
*$d=1$ :: The corresponding case is $Y=1-X$ with $3X^2-3X+1=1$. So $X\in\{0,1\}$. No "new" solutions.
*$d=2$ :: The corresponding case is $Y=2-X$ with $3X^2-6X+4=1$. So $X=1=Y$. The new solution is $(X,Y)=(1,1)$. It leads to $a=(1+42+1)/(1+1)=22$, and to the corresponding final solution $\color{blue}{(x,y)=(22,22)}$.
*$d=5$ :: The corresponding case is $Y=5-X$ with $3X^2-15X+25=43$.
The new solutions are $(X,Y)=(6,-1)$, and $(X,Y)=(-1,6)$.
The corresponding value of $a$ is $(36-6\cdot 42+1)/(216-1)=-1$.
So the final solutions are $\color{blue}{(x,y)=(-6,1)}$, and $\color{blue}{(x,y)=(1,-6)}$.
*$d=8$ :: The corresponding case is $Y=8-X$ with $3X^2-24X+64=43$.
The new solutions are $(X,Y)=(7,1)$, and $(X,Y)=(1,7)$.
The corresponding value of $a$ is $(49+7\cdot 42+1)/(343+1)=1$, leading to the copies $\color{blue}{(x,y)=(7,1)}$ and $\color{blue}{(x,y)=(1,7)}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrate $\int\frac{6x^4+4x^2+8x+4}{(1-2x)^3}dx$ Integrate $$\int\frac{6x^4+4x^2+8x+4}{(1-2x)^3}dx$$
I tried $2\cdot\int\frac{3x^4+2x^2+4x+2}{(1-2x)^3}dx\:\:$ now after partial fractions i have:
$$\begin{align}
&\>\>\>\>\>\frac{3x^4+2x^2+4x+2}{(1-2x)^3}\\&=\frac{A}{(1-2x)}+\frac{B}{(1-2x)^2}+\frac{C}{(1-2x)^3}\\[1ex]
&=A(1-2x)^2(1-2x)^3+B(1-2x)(1-2x)^3+C(1-2x)(1-2x)^2\\[1ex]
&=(1-2x)^3\cdot[A(1-2x)^2+B(1-2x)+C]\\[1ex]
&=(1-2x)^3\cdot[A-4Ax+4Ax^2+B-2Bx+C]
\end{align}$$
Need a bit help to find out $A,B,C$ and evaluation of problem.
Thank you in advance:)
| Substitute $t=1-2x$. Then $x=\frac{1-t}2$
\begin{align}
& \int \frac{6x^4+4x^2+8x+4}{(1-2x)^3}dx\\
=& \int \frac{6(\frac{1-t}2)^4+ 4(\frac{1-t}2)^2+ 8(\frac{1-t}2)+4}{t^3}(-\frac{dt}2)\\
=& -\frac1{16} \int \frac{3t^4 -12t^3+26t^2-60t+75}{t^3}dt
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluating $\sum_{cyc}\frac{ab}{c^5}$, where $a$, $b$, $c$ are the roots of $x^3-px+1=0$. How to avoid a large expansion?
Given the polynomial $x^{3}-p{x}+1=0$, evaluate $\frac{bc}{a^{5}}+\frac{ac}{b^{5}}+\frac{ab}{c^{5}}$ in terms of $p$ if $a,b$ and $c$ are the roots of the polynomial.
My attempt involved:
$$\frac{bc}{a^{5}}+\frac{ac}{b^{5}}+\frac{ab}{c^{5}}=\frac{\left(ab+ac+bc\right)^{6}-4...}{\left(-1\right)^{5}}$$
but the power $6$ is just ridiculous. There must be a way to evaluate this without having to expand a trinomial to the power of $6$ by hand. Any ideas?
| As Martin R noted we have to find the value of $$t_6=p^6+q^6+r^6$$ where $p.q.r$ are the roots of $x^3-px^2+1$. Now let $t_k=p^k+q^k+r^k$ for $k\ge 3$ indeed from $x^3=px^2-1$ we get the recurrence $$t_k=pt_{k-1}-3$$ Now we have to find $t_6$ when we know the value of $t_2=p^2$....
The rest should be easy
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the integral: $\frac{1}{2\pi}\int_{-\pi}^{\pi}Q_r^2(\theta) \, d\theta$ Let, $\displaystyle Q_r(\theta) = \frac{r \sin \theta}{1 - 2r\cos \theta + r^2}$ Find the integral: $\frac{1}{2\pi}\int_{-\pi}^{\pi}Q_r^2(\theta) \, d\theta$.
I know one thing that $Q_r(\theta)$ is the imaginary part of the series $\displaystyle\frac{1}{1-re^{i\theta}}$. I am not sure how to proceed next. Squaring the term $Q_r(\theta)$ results in something that is not very suitable to integrate. I am sure there is some easier and shorter method to do it.
| Integrate both sides of the equation
$$\left( \frac{2r\sin \theta}{1 - 2r\cos \theta + r^2}\right)’
=-\frac{1+r^2}{1 - 2r\cos \theta + r^2}+\frac{(1-r^2)^2}{(1 - 2r\cos \theta + r^2)^2}
$$
to establish
$$
\int_0^\pi \frac{(1-r^2)^2}{(1 - 2r\cos \theta + r^2)^2}d\theta
= \int_0^\pi \frac{1+r^2}{1 - 2r\cos \theta + r^2}d\theta
$$
Then
\begin{align}
I(r)=& \int_0^\pi\left( \frac{r \sin \theta}{1 - 2r\cos \theta + r^2}\right)^2d\theta \\
=&\int_0^\pi \left(-\frac14 + \frac{1+r^2}{2(1 - 2r\cos \theta + r^2)}-\frac{(1-r^2)^2}{4(1 - 2r\cos \theta + r^2)^2}\right)d\theta\\
=&-\frac\pi4 +\frac{1+r^2}4 \int_0^\pi \frac{1}{1 - 2r\cos \theta + r^2}d\theta \\
=&-\frac\pi4 +\frac{1+r^2}4 \int_0^\pi \frac{1+r^2}{(1 +r^2)^2- 4r^2\cos^2\theta }d\theta \\
=&-\frac\pi4 +\frac{1}2 \int_0^{\pi/2} \frac{d(\tan \theta)}{\tan^2 \theta + (\frac{1-r^2}{1+r^2})^2}\\
=&-\frac\pi4 +\frac{1+r^2}{1-r^2}\frac\pi4= \frac{\pi r^2}{2(1-r^2)}
\end{align}
Thus
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}Q_r^2(\theta)d\theta
=\frac1\pi I(r)= \frac{r^2}{2(1-r^2)}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$1 - \lim_{x \to \infty} \left( 1 - \frac{2}{x^2} \left( 1 - \exp \left( -\frac{x^2}{2} \right) \right) p \right)^{kx^2}$ I'm currently working on my master thesis and I need to solve this limit.
I forgot almost everything about limits since last time a I saw them was basically in high school.
Anyway, I solved this limit with wolfram and it gives:
$1 - \lim_{x \to \infty} \left( 1 - \frac{2}{x^2} \left( 1 - \exp \left( -\frac{x^2}{2} \right) \right) p \right)^{kx^2} = 1 - \exp(-2 k p)$
$k$ and $p$ are two parameters and, in particular, $k>0$, $p \in (0,1)$.
I'd like to have a step-by-step solution of the limit, or at least I'd like to know which method I could use to solve it.
Thanks!
| We shall use the expansion $\log(1+x) = x + O(x^2)$ as $x\rightarrow 0$.
Write $$
\begin{align*}\left( 1 - \frac{2}{x^2} \left( 1 - \exp \left( -\frac{x^2}{2} \right) \right) p \right)^{kx^2} &= \exp\left\{kx^2\log \left(1-\frac{2p}{x^2}+\frac{2p}{x^2e^{x^2/2}}\right)\right\} \\
&= \exp\left(kx^2\left(-\frac{2p}{x^2}+\frac{2p}{x^2e^{x^2/2}} + O(x^{-4}) \right)\right) \text{ as } x \rightarrow \infty \\
&= \exp\left(-2pk + 2pke^{-x^2/2} + O(x^{-2})\right)\\
&\rightarrow \exp(-2pk) \text{ as } x \rightarrow \infty
\end{align*}
$$
Hence,
$$1-\lim_{x\rightarrow\infty}\left( 1 - \frac{2}{x^2} \left( 1 - \exp \left( -\frac{x^2}{2} \right) \right) p \right)^{kx^2} = 1 - \exp(-2kp)$$
| {
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Convergence of $\sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2}$. My try. Examine the convergence of:
$$\sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2}$$
Pointwise convergence (for every $ x \in \mathbb{R}$ ):
$\displaystyle \lim_{n \to \infty} x\frac{\sin(n^2x)}{n^2} = 0$ , because $\sin(n^2x) \in [-1, 1]$
Uniform convergence (for every $ x \in \mathbb{R}$ and $n \in \mathbb{N}$ ):
form dirichlet criterion:
$$\displaystyle \sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2} = \displaystyle \sum_{n=1}^{\infty} f_n(x) \cdot g_n(x) $$ where:
*
*$f_n(x) = \frac{\sin(n^2x)}{n} \leq 2$ which works for: $ \forall x \in \mathbb{R}, n \in \mathbb{N}$
*$g_n(x) = \frac{x}{n} \implies \displaystyle \lim_{n \to \infty} g_n(x) = 0$ (monotonously)
Therefore the series is uniformly convergent for $x \in \mathbb{R}$.
Is that correct?
| If the series converged uniformly on $\mathbb{R}$, then for any $\epsilon > 0$ there exists $2N+1 \in \mathbb{N}$ such that for all $x \in \mathbb{R}$,
$$\left|\sum_{n= 2N+1}^\infty x \frac{\sin n^2 x}{n^2} \right| < \epsilon$$
In particular, for $x_m = \frac{\pi}{2} + 2m\pi$ we have $\sin n^2x_m = \sin \left( n^2\frac{\pi}{2}+ 2mn^2 \pi\right)= \sin n^2\frac{\pi}{2}$ and
$$\left|\sum_{n= 2N+1}^\infty \frac{\sin n^2 \frac{\pi}{2}}{n^2} \right| < \frac{\epsilon}{\frac{\pi}{2} + 2m\pi}$$
Since the RHS converges to $0$ as $m \to \infty$ and the LHS is independent of $m$, we must have
$$\tag{*}\sum_{n= 2N+1}^\infty \frac{\sin n^2 \frac{\pi}{2}}{n^2}=0$$
However, we can write
$$\sum_{n= 2N+1}^\infty \frac{\sin n^2 \frac{\pi}{2}}{n^2}= \sum_{j=0}^\infty \frac{\sin (2N+1+j)^2 \frac{\pi}{2}}{(2N+1+j)^2}$$
where $\sin (2N+1+j)^2 \frac{\pi}{2} = 1$ if $j$ is even and $\sin (2N+1+j)^2 \frac{\pi}{2} = 0$ if $j$ is odd.
This implies
$$\sum_{n= 2N+1}^\infty \frac{\sin n^2 \frac{\pi}{2}}{n^2}= \sum_{k=0}^\infty \frac{1}{(2N+1+2k)^2} \neq 0,$$
which contradicts (*).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that: $x^x\cdot y^y \geq \left(\frac{x+y}{2} \right)^{x+y} $ Prove that:
$$x^x \cdot y^y \geq \left(\frac{x+y}{2} \right)^{x+y} $$
for every $x, y > 0$
Where I got stuck:
$$x^x \cdot y^y \geq \left(\frac{x+y}{2} \right)^{x+y} \iff \ln(x^x \cdot y^y) \geq \ln \left[\left(\frac{x+y}{2} \right)^{x+y} \right] $$
$$x\ln(x) + y\ln (y) \geq (x+y) \ln \left(\frac{x+y}{2} \right) $$
| First, rewrite your inequality as $$\frac{x\ln(x)+y\ln(y)}2\geq \left(\frac{x+y}2\right)\ln\left(\frac{x+y}2\right)$$
Now if $f(z)=z\ln z$, then $f'(z)=1+\ln z$ and $f’'(z)=\frac1z$ which is larger than $0$ for $z\in(0,\infty)$. Hence, $z\ln z$ is convex in the domain $(0,\infty)$.
As $x,y\in(0,\infty)$, we can conclude by Jensen's inequality that $$\frac{x\ln(x)+y\ln(y)}2\geq \left(\frac{x+y}2\right)\ln\left(\frac{x+y}2\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Area between $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$ and $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ I am trying to find the area between $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$ and $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ using Double integrl in Polar coordinates.
Now as shown in the figure the angle $\theta$ takes values from $0$ to $\frac{\pi}{2}$.
Now $r$ depends on $\theta$.
From $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$,substituting $x=r\cos(\theta)$ and $y=r\sin(\theta)$ we get
$$r=\sin(\theta)$$ and like-wise substituting $x=r\cos(\theta)$ and $y=r\sin(\theta)$ in $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ we get
$$r=\cos(\theta)$$
So the double integral which gives the area is:
$$A=\int_{\theta=0}^{\pi / 2} \int_{r=\sin \theta}^{\cos \theta} r d r d \theta \text { . }$$
But i am getting answer as ZERO.
| Try setting your integral as
$A=2\int_0^\frac{\pi}{4}\int_{\sin\theta}^{\cos\theta}rdrd\theta$
| {
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"timestamp": "2023-03-29T00:00:00",
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$x \mathcal{o}\left(\frac{1}{x^3}\right) =\mathcal{o}\left(\frac{1}{x^2}\right) $? Is this correct:
$$x \mathcal{o}\left(\frac{1}{x^3}\right) = \mathcal{o}\left(\frac{1}{x^2}\right)$$
or the expression $x \mathcal{o}\left(\frac{1}{x^3}\right)$ cannot be simplified?
Here $\mathcal{o}(\cdot)$ stands for little $\mathcal{o}$ of $\ \cdot$.
| $f(x) = o\left(\dfrac{1}{x^2} \right)$ if $\lim_{x \rightarrow \infty} \ x^2f(x) = 0.$
Define $g(x) = \dfrac{1}{x}f(x)$, so $g(x) = \dfrac{1}{x}o\left(\dfrac{1}{x^2}\right)$, but
$$\lim_{x \rightarrow \infty} \ x^3g(x) =\lim_{x \rightarrow \infty} \ x^2f(x) = 0$$ so $g(x) = o\left(\dfrac{1}{x^3}\right)$.
| {
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show this $x_{n+1}-x_{n}\ge 2\pi$ let $f(x)=e^x\cos{x}-\sin{x}-1$,and $n$ be postive integer,such $x_{n}$ be a root of $f(x)=0$ ,and
$\dfrac{\pi}{3}+2n\pi<x_{n}<\dfrac{\pi}{2}+2n\pi$,show that
$$x_{n+1}-x_{n}\ge 2\pi,\forall n\in ^{+}\tag{1}$$
My try: since
$$\dfrac{7\pi}{3}+2n\pi<x_{n+1}<\dfrac{5\pi}{2}+2n\pi$$
and
$$-(\dfrac{\pi}{2}+2n\pi)<-x_{n}<-(\dfrac{\pi}{3}+2n\pi)$$
so
$$x_{n+1}-x_{n}\ge \left(\dfrac{7\pi}{3}+2n\pi\right)-\left(-(\dfrac{\pi}{2}+2n\pi)\right)=\dfrac{11\pi}{6}$$
But $2\pi>\dfrac{11\pi}{6}$,so How to prove this inequality $(1)$?
| We have $f'(x) = \mathrm{e}^x(\cos x - \sin x) - \cos x < 0$ on $(\pi/3 + 2n\pi, \pi/2 + 2n\pi)$. Thus, $f(x)$ is strictly decreasing on $(\pi/3 + 2n\pi, \pi/2 + 2n\pi)$. Also, $f(\pi/3 + 2n\pi) > 0$ and $f(\pi/2 + 2n\pi) < 0$. So, $f(x)=0$ has exactly one real solution (namely, $x_n$)
on $(\pi/3 + 2n\pi, \pi/2 + 2n\pi)$.
Let
\begin{align*}
y_n &= 2n\pi + \frac{\pi}{2} - \frac{1}{3}\mathrm{e}^{-2n\pi},\\
z_n &= 2n\pi + \frac{\pi}{2} - \frac{1}{3}\mathrm{e}^{-2(n- 1)\pi}.
\end{align*}
Clearly, $y_n, z_n \in (\pi/3 + 2n\pi, \pi/2 + 2n\pi)$.
We have
\begin{align*}
f(y_n) &\le \mathrm{e}^{2n\pi + \pi/2}\sin\frac{\mathrm{e}^{-2n\pi}}{3}
- \cos \frac{\mathrm{e}^{-2n\pi}}{3} - 1 \\
&\le \mathrm{e}^{2n\pi + \pi/2} \frac{\mathrm{e}^{-2n\pi}}{3} - \cos \frac{\pi}{6} - 1\\
&= \frac{\mathrm{e}^{\pi/2}}{3} - \frac{\sqrt{3}}{2} - 1 \\
&< 0
\end{align*}
where we have used $\sin u \le u$ for all $u \ge 0$,
and $\frac{\mathrm{e}^{-2n\pi}}{3} < \frac{1}{3} < \frac{\pi}{6}$.
Also, we have
\begin{align*}
f(z_n) &\ge \mathrm{e}^{2n\pi}\sin\frac{\mathrm{e}^{-2(n - 1)\pi}}{3}
- 1 - 1\\
&\ge \mathrm{e}^{2n\pi}\cdot \frac{2}{\pi}\frac{\mathrm{e}^{-2(n - 1)\pi}}{3} - 2\\
&= \frac{2}{3\pi}\mathrm{e}^{2\pi} - 2\\
&> 0
\end{align*}
where we have used $\sin u \ge \frac{2}{\pi}{u} $ for all $u \in [0, \pi/2]$.
Thus, we have $z_n < x_n < y_n$ that is
$$2n\pi + \frac{\pi}{2} - \frac{1}{\pi}\mathrm{e}^{-2(n- 1)\pi} < x_n < 2n\pi + \frac{\pi}{2} - \frac{1}{\pi}\mathrm{e}^{-2n\pi}.$$
So,
$$2(n + 1)\pi + \frac{\pi}{2} - \frac{1}{\pi}\mathrm{e}^{-2n\pi} < x_{n + 1} < 2(n + 1)\pi + \frac{\pi}{2} - \frac{1}{\pi}\mathrm{e}^{-2(n + 1)\pi}$$
and
$$ - 2n\pi - \frac{\pi}{2} + \frac{1}{\pi}\mathrm{e}^{-2n\pi}
< - x_n < - 2n\pi - \frac{\pi}{2} + \frac{1}{\pi}\mathrm{e}^{-2(n- 1)\pi}.$$
Thus, $x_{n + 1} - x_n > 2\pi$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Series convergence by Gauss Test I want to try to prove that the $$\sum_{n=1}^\infty \frac{n!e^n}{n^{n+p}}$$ series convergent for $p>1.5$ with Gauss Test but failed? Gauss Test said if
$\frac{a_n}{a_{n+1}}$ can be represnted as
$\frac{a_n}{a_{n+1}}=\lambda+\frac{\mu}{n}+b_n$
where $\sum_{n=1}^\infty b_n$ absolutely convergent
then
*
*$\lambda>1$ convergent
*$\lambda<1$ divergent
*$\lambda=1, \mu>1 $ convergent
*$\lambda=1, \mu\leq1 $ divergent
My work: I found that
$$\frac{a_n}{a_{n+1}}=\frac{1}{e}{(1+\frac{1}{n})}^{n+p}=\frac{1}{e}(1+1/n)^{n+p}=\frac{1}{e}(1+\frac{n+p}{n}+\frac{(n+p)(n+p-1)}{2}\frac{1}{n^2}+\dots)),$$ and then stopped and didnt see a continuation.
| Proof with Stirling formula :
You have
$$\frac{n!e^n}{n^{n+p}} \sim \sqrt{2\pi n} \frac{n^ne^n}{e^nn^{n+p}} \sim \frac{\sqrt{2\pi}}{n^{p-\frac{1}{2}}}$$
which is the general term of a convergent series, since $p>1,5$. So the given series converges.
Proof with Gauss test :
Let $$a_n = \frac{n!e^n}{n^{n+p}}$$
You have
$$\frac{a_n}{a_{n+1}} = \frac{n!e^n(n+1)^{n+1+p}}{(n+1)!e^{n+1}n^{n+p}} = \frac{1}{e}\left(\frac{n+1}{n} \right)^{n+p} = \exp \left[(n+p)\ln \left( 1+\frac{1}{n}\right)-1 \right]$$
i.e.
$$\frac{a_n}{a_{n+1}} = \exp \left[(n+p)\left(\frac{1}{n} -\frac{1}{2n^2} + O\left( \frac{1}{n^3}\right) \right)-1 \right] = \exp \left[ \frac{p-\frac{1}{2}}{n} + O\left( \frac{1}{n^2} \right) \right]$$ i.e. $$\frac{a_n}{a_{n+1}}= 1+\frac{p-\frac{1}{2}}{n}+O\left(\frac{1}{n^2}\right)$$
Now you can conclude using Gauss test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How does one treat big O when Taylor expanding $\sin(\sin x)$? I am working on finding the Taylor (Maclaurin expansion) of $\sin(\sin(x))$ to the third order.
We have $$\sin{t} = t - \frac{t^3}{6} + \mathcal{O}(t^5)$$
If I then set $t = \sin(x)$, and expand once more, I end up with a big O which is
$$\mathcal{O}((x - \frac{x^3}{6} + \mathcal{O}(x^5))^5)$$
How does one treat this mathematically? How do I know which order the error will be in my final expansion?
| The first line can be written as $\sin t = t - \frac{t^3}{6} + h(t) t^5$ for some function $h$ such that $|h(t)| \le C$ for $t$ near $0$.
Then,
\begin{align}
\sin \sin x
&= \sin x - \frac{(\sin x)^3}{6} + h(\sin x) (\sin x)^5
\\
&= \left(x - \frac{x^3}{6} + h(x) x^5\right)
- \frac{1}{6} \left(x - \frac{x^3}{6} + h(x) x^5\right)^3
+ h\left(x - \frac{x^3}{6} + h(x) x^5\right)
\left(x - \frac{x^3}{6} + h(x) x^5\right)^5
\end{align}
*
*The first term contributes $x - x^3/6$. The $h(x) x^5$ is in $O(x^5)$ by definition, since $h(x)$ is bounded for $x$ near $0$.
*Upon expanding the cube, the second term contributes $-\frac{1}{6} x^3$. All other terms have at least a $x^5$ term, and together are still in $O(x^5)$.
*The factor $h\left(x - \frac{x^3}{6} + h(x) x^5\right)$ is bounded for $x$ sufficiently close to zero since the expression $x - \frac{x^3}{6} + h(x) x^5$ tends to zero as $x \to 0$. Upon expanding $\left(x - \frac{x^3}{6} + h(x) x^5\right)^5$, all terms have at least a $x^5$ term, so the whole expression is $O(x^5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Number of possible paths on a lattice grid given a restriction
On a $6 \cdot 6$ lattice grid, an ant is at point $(0,0).$ There is a teleportation pad at $(2,2)$ and $(3,3).$ When an ant reaches either teleportation pad, it teleports the ant from the pad it is on to the other pad, and then the pad disappears. (Meaning the pad can only be used once) For example, if the ant reaches $(2,2),$ it is automatically brought to $(3,3).$ How many possible routes are there to take from $(0,0)$ to $(5,5),$ such that the ant can only move up or rightwards along the lattice grid.
There is a formula for finding the possible routes from one point to another without having teleportation. I considered the problem if the teleportation condition was voided. Then, the problem would simply become $\binom{10}{5}=252.$ However, I feel like I have overcounted some cases and am not sure how to subtract them. May I have some help? Thanks in advance.
| *
*How many paths are there that avoid both teleportation pads?
*How many paths are there from $(0,0)$ to the first pad at $(2,2)$?
*How many paths are there starting from the second pad at $(3,3)$ to $(5,5)$? (Hint: compare with #2 above.)
*How many paths are there from $(0,0)$ to the second pad at $(3,3)$, avoiding $(2,2)$?
*How many paths are there starting from $(2,2)$ to $(5,5)$, possibly through $(3,3)$?
To get from $(0,0)$ to $(x,y)$, there are $$\binom{x+y}{x} = \frac{(x+y)!}{x! y!}$$ unrestricted possibilities.
There are $\binom{4}{2} = 6$ paths from $(0,0)$ to $(2,2)$, and an equal number of paths from $(3,3)$ to $(5,5)$. And there are $\binom{2}{1} = 2$ paths from $(2,2)$ to $(3,3)$. So there are $\binom{6}{3} - \binom{4}{2}\binom{2}{1} = 20 - 12 = 8$ paths from $(0,0)$ to $(3,3)$ that do not pass through $(2,2)$, and the same number from $(2,2)$ to $(5,5)$ that do not pass through $(3,3)$.
Consequently, ignoring teleportation, there are $8 \binom{4}{2} = 48$ paths from $(0,0)$ to $(5,5)$ that pass through $(3,3)$ do not pass through $(2,2)$, and an equal number of paths that pass through $(2,2)$ but not $(3,3)$. There are $\binom{4}{2}\binom{2}{1}\binom{4}{2} = 72$ paths that pass through both $(2,2)$ and $(3,3)$. All combined, there are $\binom{10}{5} - 2(48) - 72 = \color{red}{84}$ valid paths that do not pass through either $(2,2)$ or $(3,3)$.
Second, count the number of paths to $(5,5)$ through $(2,2)$ and teleporting to $(3,3)$. This is just $\binom{4}{2} \binom{4}{2} = \color{red}{36}$.
Third, we know that the answer to item number 4 above is just $8$. After teleportation back to $(2,2)$, there are another $\binom{6}{3}$ ways to get to $(5,5)$, noting that after teleportation, $(3,3)$ becomes an ordinary point. So this last case has $8(20) = 160$ paths.
All put together, there are $84 + 36 + 160 = \color{red}{280}$ paths.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $b$ and $d$ such that $2x^4+ax^3+bx^2+cx+d $ has roots $x=\sqrt2-3$ and $x=3i+2$
I've got the following function:
$$f(x)=2x^4+ax^3+bx^2+cx+d$$
Given the two roots $x=(\sqrt2-3)$ and $x=(3i+2)$, I need to compute values $b$ and $d$.
Can anyone point me in the right direction of how to start this?
Thanks in advance.
EDIT: Just to clarify, that's the information I've got. No more, no less.
| This is a basic application of Vieta's formulas. Here is one way we can approach the problem:
Assuming rational coefficients, there are two conjugate pairs we are dealing with, namely:
*
*the surd conjugate pair: $\left(\sqrt 2-3\right)$ and $\left(-\sqrt 2 - 3\right)$, whose sum is $-6$ and product is $7$
*the complex conjugate pair: $(3i+2)$ and $(-3i+2)$, whose sum is $4$ and product is $13$
Using the sum and product of roots, a quadratic function that has the surd conjugate pair as zeros must thus have the form $A_1(x^2+6x+7)$; a quadratic function that has the complex conjugate pair as zeros must have the form $A_2(x^2-4x+13)$, where $A_1,A_2\in\mathbb Q$.
We can now think of the original quartic $f(x)$ as a product of two quadratic functions with these respective forms:
$$2x^4+ax^3+bx^2+cx+d=A_1A_2\left(x^4+2x^3-4x^2+50x+91\right)$$
Noticing from the coefficient of $x^4$ that this can only hold if $A_1A_2=2$, we have:
$$2x^4+ax^3+bx^2+cx+d=2x^4+4x^3-8x^2+100x+182$$
From there, we can just compare coefficients to get the values of $a$, $b$, $c$ and $d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral as a limit of sum: $\lim\limits_{n\to\infty}\sum_{k=1}^n\left(\frac k{n^2+n+2k}\right)$
The value of$$\lim_{n\to\infty}\sum_{k=1}^n\left(\frac k{n^2+n+2k}\right)=\,?$$
Can the limit be partially applied to the denominator after converting the numerator into an integral?
I wrote this as $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\left(\frac {k/n}{1+1/n+2k/n^2}\right)$$
From what I know I can write that $1/n$ as $\mathrm{d}x$ and $k/n$ as $x$. Can I apply limit to denominator as $n$ tends to infinity and rewrite the denominator as $1$.
| One the one hand
$$
\mathop {\lim }\limits_{n \to + \infty } \sum\limits_{k = 1}^n {\frac{k}{{n^2 + n + 2k}}} \le \mathop {\lim }\limits_{n \to + \infty } \sum\limits_{k = 1}^n {\frac{k}{{n^2 }}} = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}} = \int_0^1 {x\, dx} = \frac{1}{2}.
$$
On the other hand
\begin{align*}
\sum\limits_{k = 1}^n {\frac{k}{{n^2 + n + 2k}}} \ge \sum\limits_{k = 1}^n {\frac{k}{{n^2 + 3n}}} & = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{k/n}}{{1 + \frac{3}{n}}}} \ge \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}\left( {1 - \frac{3}{n}} \right)} \\ & = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}} - \frac{3}{n}\frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}} ,
\end{align*}
i.e.,
\begin{align*}
\mathop {\lim }\limits_{n \to + \infty } \sum\limits_{k = 1}^n {\frac{k}{{n^2 + n + 2k}}} & \ge \mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}} - \mathop {\lim }\limits_{n \to + \infty } \frac{3}{n}\mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}} \\ & = \int_0^1 {x\, dx} - 0 \cdot \int_0^1 {x\, dx} = \frac{1}{2}.
\end{align*}
Thus, by the squeeze theorem, the limit is $\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4140198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Computing limit based on upper and lower Riemann sums I have a question where it is asking me to compute the $$\lim_{n \to \infty} \frac{(\sqrt{1}+\sqrt{2}+...+\sqrt{n})}{(n\sqrt{n})}$$ based on the integral of sqrt(n) from the interval of $[0,n]$. Based on my previous answer, I was able to use the upper Riemann sum since it overlaps with the question nicely such that the numerator is equaled with the upper Riemann sum, but for the lower riemann sum, it’s kinda awkward since it goes from $\sqrt{1} + \sqrt{2} + ... + \sqrt{n-1}$ so any ideas on how I should proceed? Thanks :)
| $\displaystyle\mathbf{Method}\mathbf{1}:$
$\displaystyle\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\cdots +\sqrt{n}}{n\sqrt{n}}=\lim_{n\rightarrow \infty} \frac{1}{n}\left( \sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots +\sqrt{\frac{n}{n}} \right) =\int_0^1{\sqrt{x}\text{d}x}=\frac{2}{3}x^{\frac{3}{2}}\mid_{0}^{1}=\frac{2}{3}$
$\displaystyle\mathbf{Method}\mathbf{2}:$
$\displaystyle\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\cdots +\sqrt{n}}{n\sqrt{n}}=\lim_{n\rightarrow \infty} \frac{\sqrt{n}}{\sqrt{n^3}-\sqrt{\left( n-1 \right) ^3}}=\lim_{n\rightarrow \infty} \frac{\sqrt{n}\left( \sqrt{n^3}+\sqrt{\left( n-1 \right) ^3} \right)}{3n^2-3n+1}=\frac{2}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4144422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\ell=6m-1$ is prime then $\ell\ne3\frac{j^2+3k^2}{j+3k}$ Let $\ell=6m-1$ for some integer $m\ge1$ be a prime and for any integer $1\le j\le (\ell-1)/2$, $$\dfrac{(j-m)(3j-3m+1)}{2}\neq\dfrac{\ell^2-1}{24}.$$ Then, I need a proof for the inequality that for integers $1\le j,k\le (\ell-1)/2$, $$\dfrac{(j-m)(3j-3m+1)}{2}+\dfrac{3(k-m)(3k-3m+1)}{2}\neq\dfrac{\ell^2-1}{6}.$$
Assuming $$\dfrac{(j-m)(3j-3m+1)}{2}+\dfrac{3(k-m)(3k-3m+1)}{2}=\dfrac{\ell^2-1}{6}$$ and simplification by $\ell=6m-1$ gives that $$\ell=3\dfrac{j^2+3k^2}{j+3k}.$$ If $3\nmid (j+3k)$, $\ell$ is not a prime, which is a contradiction. Now, I have not been able to conclude with a contradiction when $3\mid (j+3k)$. Other proofs are welcome. Thanks in advance.
| For $3|j+3k$, we must have $3|j$. Then, let $j=3t$ for some integer $t$. Then
$$\ell=\frac{3(j^2+3k^2)}{j+3k}=\frac{3(9t^2+3k^2)}{3(t+k)}=\frac{3(k^2+3t^2)}{t+k}.$$
So, $3|t+k$. Let $k=3u-t$; then
$$\ell=\frac{(3u-t)^2+3t^2}u=\frac{4t^2-6tu+9u^2}u.$$
So, $u|4t^2$. Consider any prime $p|u$. Also, since
$$u=\frac j9+\frac k3<\ell-1,$$
$p\neq \ell$. Letting
$$t=p^\alpha r,\ u=p^\beta s$$
with $p\nmid r,s$, we have
$$0=\nu_p(\ell)=\nu_p\left(\frac{4p^{2\alpha}r^2-6p^{\alpha+\beta}rs+9p^{2\beta}s^2}{p^\beta s}\right)\geq 2\min(\alpha,\beta)-\beta,$$
so $2\min(\alpha,\beta)\leq \beta$. Since $\beta>1$, $\min(\alpha,\beta)<\beta$, and so it must be $\alpha$, giving $2\alpha\leq \beta$. However, for $p$ odd, $u|4t^2$ gives $2\alpha\geq\beta$, so $2\alpha=\beta$. For $p=2$, $u|4t^2$ gives $\beta\leq 2\alpha+2$. This means that $u$ is one of $\{t^2,2t^2,4t^2\}$.
If $u=2t^2$, then
$$\ell=2-6t+9u=18t^2-6t+2,$$
which is even. Otherwise,
$$\ell=[1\text{ or }4]-6t+9u\equiv 1\pmod 3,$$
a contradiction since $\ell=6m-1$. So, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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If bisector of angle $C$ of $\triangle ABC$ meet $AB$ in $D$ and circumcircle in $E$ prove that $\frac{CE}{DE}=\frac{(a+b)^2}{c^2}$. If bisector of angle $C$ of $\triangle ABC$ meets $AB$ in point $D$ and the circumcircle in point $E$ then prove that $$\frac{CE}{DE}=\frac{(a+b)^2}{c^2}$$.
My Attempt
Using the fact that $$CD=\frac{2ab}{a+b}\cos\frac{C}{2}$$
and $$AD.DB=CD.DE$$
we get$$\left(\frac{bc}{a+b}\right)\left(\frac{ac}{a+b}\right)=\left(\frac{2ab}{a+b}\cos\frac{C}{2}\right)DE\Rightarrow DE=\frac{c^2}{2(a+b)\cos\frac{C}{2}}$$ and then I did lot of calculation to obtain $$CE=CD+DE=\frac{a+b}{\cos\frac{C}{2}}$$ to get the final ratio.
But I believe there would be a more generic way to do it perhaps a geometrical solution rather than relying upon the formula of angle bisector $CD=\frac{2ab}{a+b}\cos\frac{C}{2}$
| Complete the cyclic quadrilateral $ACBE$ and note,
$$\angle ABE=\angle ACE=\angle BCE=\angle BAE=\frac{\angle C}{2}\implies AE=BE.$$
Using the Angle Bisector Theorem,
$$\frac{a}{b}=\frac{BD}{AD}\implies \frac{a+b}{b}=\frac{c}{AD}\implies AD=\frac{bc}{a+b}$$
Since $\triangle ADC\sim\triangle EDB,$
$$\frac{b}{AD}=\frac{BE}{DE}\implies BE=DE\cdot\frac{a+b}{c} $$
Applying Ptolemy's Theorem,
$$\begin{align*}AE\cdot a+BE\cdot b&=CE\cdot c\\
BE\cdot(a+b)&=CE\cdot c\\
DE\cdot\frac{a+b}{c}\cdot (a+b)&=CE\cdot c \\
\therefore\; \frac{CE}{DE}&=\left(\frac{a+b}{c}\right)^2
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to show that $ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $? By long division, it is easy to show that $$ \frac{1}{1-x} = 1 + x + x^2 +x^3 +... $$
But how to show that
$$ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $$
| $$ \text{Let }S_{n}=1+x+x^2...+x^{n-1}$$
$$\implies xS_{n}=x+x^2....+x^{n}$$
Subtracting both equations,
$$S_{n}(1-x)=1-x^{n}$$
$$\implies S_{n}=\frac{1-x^n}{1-x}$$
Since it is an infinite series, $n\to\infty$ and it converges only when $|x|<1$. When $n\to\infty,x^n\to0$
$$\implies s_{\infty}=\frac{1-0}{1-x}=\frac{1}{1-x} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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How do I evaluate $\int_{0}^{\infty}{\frac{\sqrt{x}\sin(\frac{\cos^3(x)}{x\sqrt x})}{\ln(x+1)}dx}$? I am trying to deduce if the following integral converges or diverges. If it converges I would also like to check for absolut convergence:
$$\int_{0}^{\infty}{\frac{\sqrt{x}\sin(\frac{\cos^3(x)}{x\sqrt x})}{\ln(x+1)}dx}$$
Here is what I have tried:
*
*We know that $|\frac{\sqrt{x}\sin(\frac{\cos^3(x)}{x\sqrt x})}{\ln(x+1)}| \le \frac{\sqrt{x}}{\ln(x+1)}$, therefore I tried to prove that $\int_{0}^{\infty}\frac{\sqrt{x}}{\ln(x+1)}dx$ converges. However, if we notice that $\ln(x+1) \le x+1$, it is easily deducible that it instead diverges.
*The integrals $\int_{0}^{\infty}\frac{1}{\ln(x+1)}dx$ and $\int_{0}^{\infty}\sqrt{x}dx$ both diverge. Therefore dividing the initial function by the former and performing a ratio test will propably not work.
*Since $\int_{0}^{\infty}\frac{1}{\ln(x+1)}dx$ and $\int_{0}^{\infty}\sqrt{x}dx$ both diverge, I tried dividing the initial function by the latter and then performing a ratio test in the hopes that $\lim{\frac{\sin(\frac{\cos^3(x)}{x\sqrt x})}{\ln(x+1)}dx}$ converges. However this limit does not exist, since the $\sin$ component oscillates between negative and positive values indefinitely while $\ln(x+1)$ is always positive.
*I manipulated the initial function by: $\sin(\frac{\cos^3(x)}{x\sqrt x}) = \sin((\frac{\cos(x)}{\sqrt{x}})^3)$ and tried to find some usefull substitution but no luck there either.
| It converges, but not absolutely.
$0<x\le1$:
For $0<x\ll1$ the integrand is approximately $\frac{\sqrt{x}\sin\left(\frac{\cos^3(x)}{x\sqrt{x}}\right)}{\ln(x+1)}\approx x^{-1/2}\sin(x^{-3/2})$, which can be integrated and converges (but in a very oscillating manner):
$$
\int_0^1 x^{-1/2}\sin(x^{-3/2})dx
= E_{\frac{1}{3}}(-i)+E_{\frac{1}{3}}(i)+2 \sin (1) \approx 0.394614
$$
$x\gg1$:
For $x\gg1$ we can approximate $\sin\left(\frac{\cos^3(x)}{x\sqrt{x}}\right)\approx\frac{\cos^3(x)}{x\sqrt{x}}$, which simplifies the limit a bit.
Let's look at positive and negative contributions to the integral:
*
*For $n\in\mathbb{N}$ and $(2n-\frac12)\pi<x<(2n+\frac12)\pi$, the integrand is positive. In the limit $n\to\infty$ it is
$$
\int_{(2n-1/2)\pi}^{(2n+1/2)\pi} dx\frac{\sqrt{x}\sin\left(\frac{\cos^3(x)}{x\sqrt{x}}\right)}{\ln(x+1)}
\approx
\frac{1}{\ln(1+2n\pi)}\int_{(2n-1/2)\pi}^{(2n+1/2)\pi}dx \sqrt{x}\frac{\cos^3(x)}{x\sqrt{x}}
= \frac{1}{\ln(1+2n\pi)} \frac14 \left[
3 \text{Ci}\left(\frac{1}{2} (4n+3) \pi \right)
+\text{Ci}\left(\frac{3}{2} (4n+3) \pi \right)
-3 \text{Ci}\left(\frac{1}{2} (4n+1) \pi\right)
-\text{Ci}\left(\frac{3}{2} (4n+1) \pi \right)
\right]
\approx
\frac{1}{\ln(1+2n\pi)} \frac{2}{3n\pi}
$$
where I've series-expanded the cosine-integral functions for $n\to\infty$.
*For $n\in\mathbb{N}$ and $(2n+\frac12)\pi<x<(2n+\frac32)\pi$, the integrand is negative. In the limit $n\to\infty$ it is
$$
\int_{(2n+1/2)\pi}^{(2n+3/2)\pi}dx \frac{\sqrt{x}\sin\left(\frac{\cos^3(x)}{x\sqrt{x}}\right)}{\ln(x+1)}
\approx
\frac{1}{\ln(1+(2n+1)\pi)}\int_{(2n+1/2)\pi}^{(2n+3/2)\pi}dx \sqrt{x}\frac{\cos^3(x)}{x\sqrt{x}}
\approx
\frac{1}{\ln(1+(2n+1)\pi)} \frac{-2}{3n\pi}
$$
The asymptotic (large-$x$) contribution to your integral is therefore approximated by the sum
$$
\sum_{n=1}^{\infty}\left(\frac{1}{\ln(1+2n\pi)} \frac{2}{3n\pi}-\frac{1}{\ln(1+(2n+1)\pi)} \frac{2}{3n\pi}\right)=
\sum_{n=1}^{\infty}\left(\frac{1}{\ln(1+2n\pi)}-\frac{1}{\ln(1+(2n+1)\pi)}\right) \frac{2}{3n\pi},
$$
which converges, but not absolutely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4159843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving $(x-2)^{x^2-6x+8} >1$ Solving this equation:
$(x-2)^{x^2-6x+8} >1$, by taking log on base $(x-2)$ both the sides, I get the solution as $x>4$. My work: Let $(x-2)>0$
$$(x-2)(x-4)\log_{(x-2)} (x-2) > \log_{(x-2)} 1 =0\implies x<2, or, x>4$$
But this doesn't appear to be the complete solution for instance $x=5/2$ is also a solution. I would like to know how to solve it completely.
| Be careful with $f(x)=\log_a x$, apart from $x>0, a>0$, there are two cases when $0 <a<1$ and $a>1$. In the former $f(x)$ decreases and increases in the latter.
So take
Case 1: $0<(x-2)<1 \implies 2<x<3$, then
$$(x-2)(x-4)\log_{(x-2)} (x-2) < \log_{(x-2)} 1=0 \implies (x-2)(x-4) <0 \implies 2<x<4.$$ Eventually $2<x<3.$
Case 2: $(x-2)>1 \implies x>3$, then
$$(x-2)(x-4)\log_{(x-2)} (x-2) > \log_{(x-2)} 1=0 \implies (x-2)(x-4) >0 \implies x<2, or, x>4.$$
Notice the sign of inequalirt is reversed.So we get $x>4.$
So the total answer is $x\in (2,3) \cup (4,\infty).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4162534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Let $a$ and $b$ be two complex numbers such that $|a+b|=20$ and $|a^2+b^2|=16$ then find maximum and minimum value of $|a^3+b^3|$ Let $a$ and $b$ be two complex numbers such that
$|a+b|=20$ and
$|a^2+b^2|=16$
then find maximum and minimum value of $|a^3+b^3|$
My Attempt:
$|a^3+b^3|=|(a+b)((a+b)^2-3ab|)\leq |a+b|^3+3|ab|$
Couldn't go beyond this
| We have $|a+b|=20\iff a+b=20e^{it}\iff ae^{-it}+be^{-it}=20$
So WLOG we can replace $a,b$ by phase shifted ones and assume $a+b=20$.
We also have $|a^2+b^2|=16\iff a^2+b^2=16e^{i\theta}$
$\begin{align}\Big|a^3+b^3\Big|
&=\Big|(a+b)^3-\tfrac 32(\underbrace{(a+b)^2-(a^2+b^2)}_{2ab})(a+b)\Big|\\
&=\Big|20^3-\frac 32((20^2-16e^{i\theta})\times 20\Big|\\\\
&=\Big|480e^{i\theta}-4000\Big|\\\\
&=160\sqrt{634-150\cos(\theta)}\end{align}$
*
*The minimum is when $\cos(\theta)=1\iff \theta=0$ and the value is $3520$
Initial values would be $\begin{cases}a+b=20\\a^2+b^2=16\end{cases}\iff a,b=10\pm 2i\sqrt{23}$
*
*The maximum is when $\cos(\theta)=-1\iff \theta=\pi$ and the value is $4480$
Initial values would be $\begin{cases}a+b=20\\a^2+b^2=-16\end{cases}\iff a,b=10\pm 6i\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Showing that $(x,y)=(4,5), (5,4)$ are the only positive integer solutions to $x+y=3n, x^2+y^2-xy=7n$
Show that $(x,y)=(4,5), (5,4)$ are the only positive integer solutions to $x+y=3n, x^2+y^2-xy=7n.$
I'm not very certain how to proceed on this problem. I know $x^2+y^2=(x+y)^2-2xy,$ so we essentially have $x+y=3n, (x+y)^2-3xy=7n$ for positive integers $x, y, n.$ However, this doesn't really help. I've also tried writing it as a fraction and doing some algebraic manipulations, but I haven't gotten anywhere either. May I have some help? Thanks in advance.
| We are given
$$x + y = 3n, \; x^2 + y^2 - xy = 7n \tag{1}\label{eq1A}$$
As you showed,
$$\begin{equation}\begin{aligned}
(x + y)^2 - 3xy & = 7n \\
9n^2 - 3xy & = 7n \\
3xy & = 9n^2 - 7n \\
xy & = 3n^2 - \frac{7n}{3}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
From this, plus the first part of \eqref{eq1A}, then using Vieta's formula's gives that $x$ and $y$ are the roots of the quadratic equation
$$z^2 - (3n)z + \left(3n^2 - \frac{7n}{3}\right) = 0 \tag{3}\label{eq3A}$$
The quadratic formula gives
$$\begin{equation}\begin{aligned}
z & = \frac{3n \pm \sqrt{9n^2 - 4\left(3n^2 - \frac{7n}{3}\right)}}{2} \\
& = \frac{3n \pm \sqrt{-3n^2 + \frac{28n}{3}}}{2}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Using that $n$ is positive, then factoring the part in the square root above (i.e., the discriminant), and requiring it to be non-negative, gives
$$-3n^2 + \frac{28n}{3} = n\left(-3n + \frac{28}{3}\right) \implies -3n + \frac{28}{3} \ge 0 \implies n \le \frac{28}{9} = 3 + \frac{1}{9} \tag{5}\label{eq5A}$$
We have \eqref{eq1A} indicating $n$ must be either an integer or a rational value of the form $\frac{m}{3}$ for some positive integer $m$ where $3 \not\mid m$. For the latter case, though, the second part of \eqref{eq1A} would not be an integer. Thus, $n$ is an integer, with \eqref{eq2A} showing it must be a multiple of $3$.
Next, the upper limit in \eqref{eq5A} shows $n = 3$ is the only possible solution, with \eqref{eq4A} giving $z = \frac{9 \pm 1}{2}$, i.e., $z = 4$ or $z = 5$. Thus, we get $(x,y) = (4,5)$ or $(5,4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Symmetric rational functions $x_1,x_2,x_3$ are the roots of the polynomial $x^3+px+q \in \mathbb{R}[x]$ with real coefficients.
Express through $p,q$ the following symmetric rational function
$$
\sum\left(x_1,x_2,x_3\right) = \frac{1}{x_1^2 + x_2^2} + \frac{1}{x_2^2 + x_3^2} + \frac{1}{x_3^2 + x_1^2}
$$
of $x_1,x_2,x_3$.
Any tips? I've tried anything that crossed my mind, but I couldn't cope with that.
| $0=x_1+x_2+x_3$, $p=x_1x_3+x_1x_2+x_2x_3$ and $q=-x_1x_2x_3$
Now $x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)=-2p$.
So your rational expression is
$$\frac{1}{-2p-x_1^2}+\frac{1}{-2p-x_2^2}+\frac{1}{-2p-x_3^2}$$
$$=-\frac{(2p+x_1^2)(2p+x_2^2)+(2p+x_1^2)(2p+x_3^2)+(2p+x_2^2)(2p+x_3^2)}{(2p+x_1^2)(2p+x_2^2)(2p+x_3^2)}$$
Now $x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2=(x_1x_2+x_1x_3+x_2x_3)^2-2(x_1+x_2+x_3)x_1x_2x_3=p^2$
$$=-\frac{12p^2+4p(x_1^2+x_2^2+x_3^2)+(x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2)}{8p^3+4p^2(x_1^2+x_2^2+x_3^2)+2p(x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2)+x_1^3x_2^3x_3^3}$$
$$=-\frac{12p^2+4p(-2p)+p^2}{8p^3+4p^2(-2p)+2p(p^2)+q^2}=-\frac{5p^2}{2p^3+q^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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