Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the area of an equilateral triangle inscribed in a unit square If inside the square $ABCD$ there is an equilateral triangle $CMN$ inscribed, as shown in the figure. If the area of the square is 1, find the area of the triangle.
I attempted to say, I state that $x=MD$
then $MC^2=1+x^2$. $AM=1-x$ and I thought that something might come from Pythagoras, but it didn't. $\triangle AMN$ looks isosceles, but I can't prove that it is, although if it is proved to be isosceles, then the question can easily be solved by use of the pythagorean theorem. We also have that $\angle CNB+\angle CMD=150^o$. I got stuck here, could you please explain to me how to solve the question?
| How about continuing to compute the lengths of the others segments?
You have seen that if $MD = x$, then $AM = 1-x$ and $MC = \sqrt{1+x^2}$. Now, since $MNC$ is equilateral, it follows that $MN = NC = \sqrt{1+x^2}$.
Applying Pythagoras theorem in the triangle $AMN$, we obtain: $AN = \sqrt{2x}$. It follows that $NB = 1- \sqrt{2x}$.
Now, we know all the lengths of the sides of the triangle $NBC$. Using again the Pythagoras theorem we have:
$$NC^2 = NB^2 + BC^2 \implies 1+x^2 = 1+2x-2\sqrt{2x}+1 \\ \implies 2\sqrt{2x} =-x^2+2x+1 \implies 8x = x^4+4x^2+1-4x^3-2x^2+4x \\ \implies x^4 -4x^3+2x^2-4x+1 = 0 \implies (x^2+1)(x^2-4x+1) = 0$$
The roots of this equation are $i, -i, 2-\sqrt{3}$ and $2+\sqrt{3}$. The only convenient value is $x = 2 - \sqrt{3}$.
Later edit: A simpler way of finding $x$ is to observe that the $\triangle MDC \equiv \triangle DBC$. Hence $MD = BN$, so $x = 1 - \sqrt{2x}$, thus $x^2 - 4x+1 = 0$ and from here you get $x = 2-\sqrt{3}$.
| {
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"url": "https://math.stackexchange.com/questions/4072520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rewriting function Given function
$$
f(x) = \frac{-2 + 2 \sqrt{x+1}}{x}
$$
and
$$
g(x) = \frac{2}{1+\sqrt{x+1}}
$$
Prove that $f(x) = g(x)$ where $x\neq 0$.
This question is from a 2018 examen in the Netherlands. I tried rewriting f(x) as follows:
$$
f(x) = \frac{-2 + 2 \sqrt{x+1} * \sqrt{x+1}}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{-2 + 2(x+1)}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{-2 + 2x + 2}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{2x}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{2}{\sqrt{x+1}}
$$
However, I can't find a way to get to the $h(x)$ function where the denominator has a 1 + still.
EDIT:
Just as I post this I notice that I made a mistake in the first step. By not multiplying against the whole numerator. I'll leave the question open anyways.
| Third binomial formula: $(a+b)(a-b)=a^2-b^2$
$$\frac{2}{1+\sqrt{x+1}}=\frac{2(1-\sqrt{x+1})}{(1+\sqrt{x+1})(1-\sqrt{x+1})}=\frac{-2 + 2 \sqrt{x+1}}{1^2-(\sqrt{x+1})^2}=\frac{-2 + 2 \sqrt{x+1}}{x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the trigonometrical solution within the specified region Let $\theta ,\varphi \in \left[ {0,2\pi } \right]$ be such that
$2\cos \theta \left( {1 - \sin \varphi } \right) = {\sin ^2}\theta \left( {\tan \frac{\theta }{2} + \cot \frac{\theta }{2}} \right)\cos \varphi - 1$
$ - 1 < \sin \theta < - \frac{{\sqrt 3 }}{2}$ & $\tan \left( {2\pi - \theta } \right) > 0$, then $\varphi$ cannot satisfy
(A) $0<\varphi<\frac{\pi}{2}$
(B) $\frac{\pi}{2}<\varphi<\frac{4\pi}{3}$
(C) $\frac{4\pi}{3}<\varphi<\frac{3\pi}{2}$
(D) $\frac{3\pi}{2}<\varphi<2\pi$
The official answer is ACD.
My approach is as follow
$ - 1 < \sin \theta < - \frac{{\sqrt 3 }}{2} \Rightarrow \theta \in \left( {\frac{{4\pi }}{3},\frac{{3\pi }}{2}} \right) \cup \left( {\frac{{3\pi }}{2},\frac{{5\pi }}{3}} \right)$
$\tan \left( {2\pi - \theta } \right) > 0 \Rightarrow \tan \theta < 0 \Rightarrow \theta \in \left( {\frac{\pi }{2},\pi } \right) \cup \left( {\frac{{3\pi }}{2},2\pi } \right)$
Hence $\theta \in \left( {\frac{{3\pi }}{2},\frac{{5\pi }}{3}} \right)$
$2\cos \theta \left( {1 - \sin \varphi } \right) = {\sin ^2}\theta \left( {\tan \frac{\theta }{2} + \cot \frac{\theta }{2}} \right)\cos \varphi - 1$
$ \Rightarrow \cos \theta \left( {1 - \sin \varphi } \right) = {\sin ^2}\theta \left( {\frac{{{{\sin }^2}\frac{\theta }{2} + {{\cos }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right)\cos \varphi - \frac{1}{2} \Rightarrow \cos \theta \left( {1 - \sin \varphi } \right) = \sin \theta \cos \varphi - \frac{1}{2}$
$ \Rightarrow \frac{1}{2} + \cos \theta - \sin \varphi \cos \theta = \sin \theta \cos \varphi \Rightarrow \frac{1}{2} + \cos \theta = \sin \theta \cos \varphi + \sin \varphi \cos \theta \Rightarrow \frac{1}{2} + \cos \theta = \sin \left( {\theta + \varphi } \right) = T$
$\theta \in \left( {\frac{{3\pi }}{2},\frac{{5\pi }}{3}} \right) \Rightarrow \frac{{dT}}{{d\theta }} = - \sin \theta $
Therefore T is increasing
Upto to this step I am satisfied with my approach
But from here how do I match $\sin \theta \cos \varphi + \sin \varphi \cos \theta \Rightarrow \frac{1}{2} + \cos \theta = \sin \left( {\theta + \varphi } \right) = T$ with the given answer and arrive at the proper result
| Since $\theta \in (3\pi/2, 5\pi/3)$ is in 4th quadrant. You must have $\frac 12+\cos \theta \gt 0\implies \sin (\theta +\phi)\gt 0\tag 1$
In case $(b),\theta +\phi \in (\pi/2+3\pi/2,4\pi/3+5\pi/3)=(2\pi, 3\pi)$ and that is correct as it satisfies $(1)$.
In case $(a),\theta+\phi \in (3\pi/2, 5\pi/3+\pi/2)=(3\pi/2, 13\pi/6)$ is not correct as $\theta +\phi=2\pi$, for example, will create problem for $(1)$.
Similarly rule out other possibilities.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine point where area between tangent and curve is minimal Given a certain function, how do we determine a point on the function $(a,f(a))$ where the area between its tangent at that point and the function is minimal?
For example, the function given is $f(x)=-x^2 + 4x + 5$ on the interval $(0,3)$.
| Step 1: find the slope of the tangent line:
\begin{align}f(x)=-x^2 + 4x + 5, m=(f(x))'=-2x+4 \end{align}
Step 2: find the intercept and equation of the tangent line at $x=a$:\begin{align} -a^2+4a+5=(-2a+4)a+b, b=a^2+5,y=(-2a+4)x+a^2+5\end{align}
Step 3: find the area bounded by $f(x)$,the tangent line, $x=0,$ and $x=3$ :
\begin{align}S=\int_0^3 (((-2a+4)x+a^2+5)-(-x^2+4x+5)) dx=\int_0^3 (x-a)^2 dx=3(a-\frac{3}{2})^2+\frac{9}{4}\end{align}
Conclusion: when $x=\frac{3}{2}, S$ is minimum.
| {
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"timestamp": "2023-03-29T00:00:00",
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find all the integral solution for $y^2 + 31 = x^3$ Find all the integral solution for $y^2 + 31 = x^3$.
I am reading Ireland's book 'a classical introduction to modern number theory', this is one of the exercises. The hint behind the book is $y^2 + 4 = x^3 - 27$, I don't know how to use it. I use python tried below $1000$ there is no solution. And also there is no prime $p$ below $10000$ such that $y^2 + 31 = x^3 (\text{mod} p)$ not solvable. Thank you
| I found the solution of this problem in Introductory Algebraic Number Theory by Saban Alaca, page 399, Theorem 14.2.5. Here is the solution.
First we prove $x$ is odd impossible. If $x$ is odd, then $y$ is even, by $y^2+4=x^3-27=(x-3)(x^2+3x+9)$, $y^2+4=4(\frac{y}{2}+1)^2$ then any odd prime $p$ divides $y^2+4$ must have $p \equiv 1 \bmod 4$. If $x \equiv 1 \bmod4$, then $y^2 \equiv 2 \bmod4$, contradiction. If $x \equiv 3 \bmod4$, then $x^2 + 3x + 9 \equiv -1 \bmod4$ and greater than 1. Hence there is a prime $q \equiv -1 \bmod4$ divides $x^2+3x+9$, this is a contradiction to we have known that any odd prime $p$ divides $y^2+4$ must have $p \equiv 1 \bmod4$.
Second we prove $x$ is even impossible. Let $K = Q(\sqrt{-31})$. Consider prime 2 in $O_K$. $$2O_K = (2, \frac{3+\sqrt{-31}}{2})(2, \frac{3-\sqrt{-31}}{2}).$$ $$(\frac{y+\sqrt{-31}}{2})(\frac{y-\sqrt{-31}}{2}) = (2, \frac{3+\sqrt{-31}}{2})(2, \frac{3-\sqrt{-31}}{2})(\frac{x}{2})^3.$$ We can prove ideal $(2, \frac{3+\sqrt{-31}}{2})$ is not principal by considering its ideal norm. Also the ideals $(\frac{y+\sqrt{-31}}{2})$ and $(\frac{y-\sqrt{-31}}{2})$ are coprime since the common prime ideal divides them is $(\sqrt{-31})$, which is impossible. Then by $(\frac{y+\sqrt{-31}}{2})(\frac{y-\sqrt{-31}}{2}) = (2, \frac{3+\sqrt{-31}}{2})(2, \frac{3-\sqrt{-31}}{2})(\frac{x}{2})^3$ we know there is an ideal $A$ which has $(\frac{y+\sqrt{-31}}{2})=(2, \frac{3+\sqrt{-31}}{2})A^3$ or $(\frac{y-\sqrt{-31}}{2})=(2, \frac{3+\sqrt{-31}}{2})A^3$. By class number of $K$ equals 3, $A^3$ must be principal. Then $(2, \frac{3+\sqrt{-31}}{2})$ is principal. But we already use ideal norm shows it's not.
| {
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Limit $\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x}$ equals to $\frac{5}{12}$ A friend of mine asked this question to me. It seems it's from Stewart.
Find the values of a and b such that
$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12}$
This is what I tried with better results.
For $b$:
$$\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12} $$
$$\sqrt[3]{a x+b} - 2 = x\frac{5}{12} $$
$$\underset{x\to 0}{\text{lim}} \sqrt[3]{a x+b} - 2 = \underset{x\to 0}{\text{lim}} x\frac{5}{12} $$
$$\sqrt[3]{b} - 2 = 0 $$
$$\sqrt[3]{b} = 2 $$
$$ b = 8$$
For $a$:
$$a x+8 = (x\frac{5}{12} + 2)^3$$
$$a x+8 = \frac{125 x^3}{1728}+\frac{25 x^2}{24}+5 x+8$$
$$a x = \frac{125 x^3}{1728}+\frac{25 x^2}{24}+5 x$$
$$a = \frac{125 x^2}{1728}+\frac{25 x}{24}+5$$
$$\underset{x\to 0}{\text{lim}} a =\underset{x\to 0}{\text{lim}} \frac{125 x^2}{1728}+\frac{25 x}{24}+5$$
$$a = 5 $$
But the limit $$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{5x+8} - 2}{x}$$ doens't go to $\frac {5}{12}$.
May someone help.
| What about multiplying by the conjugate?
\begin{align*}
(ax+b)^{1/3} - 2 = \frac{ax + b - 8}{(ax+b)^{2/3} + 2(ax+b)^{1/3} + 4}
\end{align*}
Then, if we take $b = 8$, the proposed limit equals
\begin{align*}
\lim_{x\to 0}\frac{(ax+b)^{1/3} - 2}{x} & = \lim_{x\to 0}\frac{1}{x}\times\frac{ax + b - 8}{(ax+b)^{2/3} + 2(ax+b)^{1/3} + 4} = \frac{a}{12} = \frac{5}{12}
\end{align*}
thence we conclude that $a = 5$.
Hopefully this helps!
| {
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"timestamp": "2023-03-29T00:00:00",
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The sum of ten distinct natural numbers is $62$. Show that their product is a multiple of $1440$.
The sum of ten distinct natural numbers is $62$. Show that their product is a multiple of $1440$.
I know how to prove that it is a multiple of $60$
Lets write: $a_1+a_2+...+a_{10}=62$. Assuming $a_1<a_2<\cdots <a_{10}$ we search the maximal value of $a_{10}$. This is the difference of $62$ and the minimal value the sum $a_1+a_2+\cdots+a_9$ can take. As this value is given by $1+2+\cdots+9=45$, it follows that $a_{10}=62-45=17$. That means that for any $a_i>17$, there are no $9$ natural numbers to satisfy our initial sum.
So a solution would be $1,2,3,4,5,6,7,8,9,17$.
If $a_{10}=16$, the unique solution is $a_9=10$, the smaller solutions being unchanged as to preserve the distinctness.
If $a_{10}=15$, then $a_9=11$ or $a_8=10$.
If $a_{10}=14$, then $a_9=12$ or $a_8=11$ or $a_7=10$.
If $a_{10}=13$, then $a_8=12$ or $a_7=11$ or $a_6=10$.
If $a_{10}=12$, $a_9=13$ or $a_6=11$ or $a_5=10$. And from here on, the situation repeats.
We observe that the numbers $3,4,5$ or $6,10$ or $5,12$ repeat and as their products are $60$, then $60$ divides the product of $a_1,\cdots,a_{10}$. I don't know how to expand for $1440$. Any help, please?
| Consider potential minimal counter-examples which might not be multiples of the prime factorisation of $1440=2^53^25$ and show that the sums cannot be $62$:
You must have a multiple of $5$ in the product since $1+2+3+4+6+7+8+9+11+12=63$ is too big and any smaller sum must have a multiple of $5$.
You must have a multiple of $9$ in the product since $1+2+3+4+5+7+8+10+11+13=64$ is too big and any smaller sum must have a second multiple of $3$.
You also want to show you must have a multiple of $32=2^5$ in the product, but that is a little more complicated. There must be an even number of odd numbers to get an even sum, and since there are $10$ numbers you must have an even number of even numbers.
*
*If you have six or more even numbers, then you will have at least $2^6=64$ in the product
*If you have four even numbers then at least one of them must be a multiple of $4$ so giving $32$ in the product, since $1+2+3+5+6+7+9+10+11+14=68$ is too big
*If you have two or fewer even numbers then the sum will be too big, since $1+2+3+4+5+7+9+11+13+15=70$ is too big
So you must have $5 \times 9 \times 32=1440$ in the product.
You have given an example of $17 \times 9!= 6168960=1440 \times 4284$ and this is in fact the smallest example.
| {
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Finding Taylor Series from existing Series If the Taylor Series of $\ln(x)$ is known:
$$\ln(x) = (x-1) -\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+\frac{1}{5}(x-1)^5-...$$
Can one find the Taylor series of
$$f(x)= \frac{x}{1-x^2}$$
by manipulating the Taylor series of ln(x)?
| This isn't really an answer on its own, but I'll add this here in case you want to see how to get the series in detail.
$$\ln(x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x-1)^n}{n} \implies \ln(x+1) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x)^n}{n}$$
$$\frac{d}{dx}\ln(x+1) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{n(x)^{n-1}}{n} \implies \frac{1}{1+x} = \sum_{n=1}^{\infty} (-1)^{n+1}x^{n-1}$$
$$\frac{1}{1+(-x)} = \sum_{n=1}^{\infty} (-1)^{n+1}(-x)^{n-1} \implies \frac{1}{1-x} = \sum_{n=1}^{\infty} (-1)^{n+1}(-1)^{n-1}x^{n-1} \implies \frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1}$$
$$ \frac{1}{1-x^2} = \sum_{n=1}^{\infty} x^{2(n-1)} \implies \frac{1}{1-x^2} = \sum_{n=1}^{\infty} x^{2n-2}$$
$$\frac{x}{1-x^2} = \sum_{n=1}^{\infty} x^{2n-1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} dx dy \exp(-5x^2+8xy-5y^2)$ I need to evaluate the following integral
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} dx dy \exp(-5x^2+8xy-5y^2)$$
I know the normal double integrals but this one seem different, moreover, I'm familiar with $$\int_{-\infty}^{\infty} \exp(-x^2)dx = \sqrt\pi$$ but I cannot figure out the polar substitution for $-5x^2+8xy-5y^2$ expression. I tried substituting $x^2+y^2$ at $-5(x^2+y^2)$ with $r^2$ and $8xy$ with $r^2\sin\theta\cos\theta$. And then I got even more lost.
Can someone please help me figure this question out and is there any general solution for such problems? And also can someone provide me a link to similar problems and their solutions in order for me to understand better. Thank you!
| To begin, we complete the square to notice that
$$-5x^{2}+8xy-5y^{2}=-5\left(x-\frac{4}{5}y\right)^{2}-\frac{9}{5}y^{2}$$
so we get that
\begin{align*}
&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp(-5x^{2}+8xy-5y^{2})dxdy\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\left(-5\left(x-\frac{4}{5}y\right)^{2}-\frac{9}{5}y^{2}\right)dxdy.\tag{1}
\end{align*}
For any value of $y$ it holds that
$$\int_{-\infty}^{\infty}\exp\left(-5\left(x-\frac{4}{5}y\right)^{2}\right)dx=\int_{-\infty}^{\infty}\exp\left(-5x^{2}\right)dx$$
because of the substiuttion $u=x-\frac{4}{5}y$, so substituting back into (1) we get that
\begin{align*}
&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp(-5x^{2}+8xy-5y^{2})dxdy\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\left(-5x^{2}-\frac{9}{5}y^{2}\right)dxdy\\
&=\left(\int_{-\infty}^{\infty}\exp\left(-5x^{2}\right)dx\right)\left(\int_{-\infty}^{\infty}\exp\left(-\frac{9}{5}y^{2}\right)dy\right)
\end{align*}
Now making the subs $u=\sqrt{5}x$ and $u=\frac{3}{\sqrt{5}}y$ we get that
\begin{align*}
&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp(-5x^{2}+8xy-5y^{2})dxdy\\
&=\left(\frac{1}{\sqrt{5}}\int_{-\infty}^{\infty}\exp\left(-x^{2}\right)dx\right)\left(\frac{\sqrt{5}}{3}\int_{-\infty}^{\infty}\exp\left(-y^{2}\right)dy\right)\\
&=\left(\frac{\sqrt{\pi}}{\sqrt{5}}\right)\left(\frac{\sqrt{5\pi}}{3}\right)\\
&=\frac{\pi}{3}
\end{align*}
So we are done
| {
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maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$ without Lagrange multiplier What is the maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$? I wonder there is a precalculus method, without using the Lagrange multiplier.
| Let $f(x,y)=x^2-3xy-2y^2$ and $g(x,y)=x^2+xy+y^2-1$. Then,
$$(f_x’, f_y’)= (2x-3y, -3x-4y),\>\>\>\>\>( g_x’, g_y’)= (2x+y, x+2y)$$
At the extrema, the two curve are tangential to each other, i.e. the extreme points satisfying
$$\frac{-3x-4y}{2x-3y}= \frac{x+2y}{2x+y}$$
along with $ g(x,y)=0$.
Solve to obtain $x^2= \frac1{26\pm 7\sqrt{13}}$, $\frac yx = 3\pm \sqrt{13}$.
Then, the extreme values are
$$f_m(x,y)= x^2(1-3\frac yx -2\frac {y^2}{x^2})=
\frac13(1\pm 2 \sqrt{13})$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a closed-form solution to the following summation I am solving a summation that appears in a paper, it claims that
$$\sum_{n=1}^{\infty} \binom{2n}{n+k}z^n=\bigg(\frac{4z}{(1+\sqrt{1-4z})^2}\bigg)^k$$
I found this identity here in equation (66)
$$\sum_{n=0}^{\infty} \binom{2n+k}{n}z^n=1+\sum_{n=1}^{\infty} \binom{2n+k}{n}z^n=1+\sum_{n=1}^{\infty} \binom{2n+k}{n+k}z^n =\bigg(\frac{2}{1+\sqrt{1-4z}}\bigg)^k\frac{1}{\sqrt{1-4z}}$$
I didn't found the paper that has the proof of the identity above, but I notice that
$$\binom{2n}{n+k}+\sum_{j=0}^{k-1} \binom{2n+j}{n+k-1}=\binom{2n+k}{n+k}$$
Therefore we obtain
$$\sum_{n=1}^{\infty} \binom{2n}{n+k}z^n=\bigg(\frac{2}{1+\sqrt{1-4z}}\bigg)^k\frac{1}{\sqrt{1-4z}}-1-\sum_{n=1}^{\infty} \sum_{j=0}^{k-1} \binom{2n+j}{n+k-1}z^n$$
Yet I have no idea for the last summation above, any help would be appreciated.
| We use an approach based upon the Lagrange inversion theorem and prove for $k$ being a non-negative integer
\begin{align*}
\color{blue}{\sum_{n=0}^\infty\binom{2n+k}{n}z^n=\left(\frac{2}{1+\sqrt{1-4z}}\right)^k\frac{1}{\sqrt{1-4z}}\qquad\qquad |z|<\frac{1}{4}}
\end{align*}
We follow thereby the paper Lagrange Inversion: when and how by R. Sprugnoli etal. We consider the generating function
\begin{align*}
F(z)=\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^\infty\binom{2n+k}{n}z^n
\end{align*}
and apply formula (G6) from the paper. The formula (G6) tells us that if there are functions $A(u)$ and $\phi(u)$, so that the coefficient $a_n$ admits a representation
\begin{align*}
a_n=[u^n]A(u)\phi(u)^n
\end{align*}
where $[u^n]$ denotes the coefficient of $u^n$ of a series, then the following is valid:
\begin{align*}
F(z)&=\sum_{n=0}^{\infty}[u^n]A(u)\phi(u)^nz^n\\
&=\left.\frac{A(u)}{1-z\phi^{\prime}(u)}\right|_{u=z\phi(u)}\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{F(z)}&\color{blue}{=\sum_{n=0}^\infty\binom{2n+k}{n}z^n}\\
&=\sum_{n=0}^\infty[u^n](1+u)^{2n+k}z^n\tag{2}\\
&=\left.\frac{(1+u)^k}{1-z\,\frac{d}{du}(1+u)^2}\right|_{u=z(1+u)^2}\tag{3}\\
&=\frac{(1+u)^k}{1-\frac{u}{(1+u)^2}\cdot 2(1+u)}\tag{4}\\
&=\frac{(1+u)^{k+1}}{1-u}\tag{5}\\
&=\frac{\left(\frac{2}{1+\sqrt{1-4z}}\right)^{k+1}}{2-\frac{2}{1+\sqrt{1-4z}}}\tag{6}\\
&\,\,\color{blue}{=\left(\frac{2}{1+\sqrt{1-4z}}\right)^k\frac{1}{\sqrt{1-4z}}}\tag{7}
\end{align*}
and the claim follows.
Comment:
*
*In (2) we use the coefficient of operator $[z^n]$ to denote the coefficient of a series. We observe we can apply the Lagrange Inversion theorem with $A(u)=(1+u)^k$ and $\phi(u)=(1+u)^2$.
*In (3) we use formula (1) with $A$ and $\phi$ as stated in the comment above.
*In (4) we substitute $z=\frac{u}{1+u}$ and do the derivation.
*In (5) we simplify the expression.
*In (6) we recall $u=z(1+u)^2$ which is a quadratic equation in $u=u(z)$. We calculate
\begin{align*}
u(z)&=\frac{1}{2z}\left(1-2z-\sqrt{1-4z}\right)\\
&=\frac{1}{2z}\left(1-\sqrt{1-4z}\right)-1\\
&=\frac{1}{2z}\,\frac{1-(1-4z)}{1+\sqrt{1-4z}}-1\\
&=\frac{2}{1+\sqrt{1-4z}}-1
\end{align*}
and take the solution with the minus sign, since this one can be expanded as generating function.
*In (7) we make again some simplifications to obtain the wanted form.
Note: The relationship with OPs first stated formula is given via
\begin{align*}
\color{blue}{\sum_{n=0}^\infty\binom{2n}{n+k}z^n}
&=\sum_{n=k}^\infty \binom{2n}{n+k}z^n\tag{8}\\
&=\sum_{n=k}^\infty \binom{2n}{n-k}z^n\tag{9}\\
&=\sum_{n=0}^\infty \binom{2n+2k}{n}z^{n+k}\tag{10}\\
&=z^k\left(\frac{2}{1+\sqrt{1-4z}}\right)^{2k}\frac{1}{\sqrt{1-4z}}\tag{11}\\
&\,\,\color{blue}{=\left(\frac{4z}{\left(1+\sqrt{1-4z}\right)^2}\right)^k\frac{1}{\sqrt{1-4z}}}
\end{align*}
Comment:
*
*In (8) we start the right-hand side with index $n=k$, since $\binom{2n}{n+k}=0$ if $n<k$.
*In (9) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.
*In (10) we shift the index to start with $n=0$ again.
*In (11) we use the identity (7) from above with $k\to 2k$ and make finally some simplifications.
| {
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} |
$I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*
*$x \geq 0$
*$n \geq 0$
*$f_n(x)= \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*$I_n= \int_{0}^{ + \infty}f_n(x) $
*What is the limit of $I_n$ ?
*What is an equivalent of $I_n$ ?
$
\begin{align*}
u'&= (-n+1) 4x^3 (1+x^4)^{-n-1}\\
u &=x^{4n-1}\\
v &= x^{4n-1} \\
v' &= (4n-1) x^{4n-2} \\
I_{n+1}&= \int_{0}^{ + \infty} \dfrac{-4n x^3 x^{4n-1} }{ 4 (1+x^4)^{n+1} (-n) }\\
I_{n+1}&= \left[ \dfrac{ x^{4n-1}}{-4n (1+x^4)^n } \right] + \int_{0}^{ + \infty} \dfrac{(4n-1) x^{4n-2}}{4n (1+x^4)^n } \\
I_{n+1} &= \dfrac{4n-1}{4n}I_n \\
I_2 &=\dfrac{4-1}{4 \times 1}I_1 \\
I_{n+1} &= \dfrac{\prod_{k=1}^{n} (4k-1) }{4^n n!} I_1 \\
I_1&= \int_{0}^{ + \infty} \dfrac{ x^{2}}{ (1+x^4) } dx \\
\end{align*}
$
Quanto's answer for $I_1$ :
$$ \int_0^{\infty} \dfrac{ x^{2}}{ 1+x^4 } dx
\overset{x\to \frac1x}=\frac12\int_0^{\infty} \dfrac{ 1+x^2}{ 1+x^4 } dx = \frac12 \int_0^{\infty} \dfrac{ 1+\frac1{x^2}}{ x^2+\frac1{x^2} } dx \\
= \frac12 \int_0^{\infty} \dfrac{ d(x-\frac1{x})}{ (x-\frac1{x})^2+2 } dx =\frac\pi{2\sqrt2}
$$
$I_{n+1} = \dfrac{ \prod_{k=1}^n (4k-1) }{ 4^n n!} I_1 =\prod_{k=1}^n \dfrac{4k-1}{4k} I_1=\prod_{k=1}^n (1- \dfrac{1}{4k}) I_1$
What is the limit ?
$\ln( 1 - u) \sim u$ donc $\ln (1 - \dfrac{1}{4k}) \sim - \dfrac{1}{4k}$
and $- \sum \dfrac{1}{k} = - \infty$ so $I_n \to 0$ ?
| Complex analysis is one possibility. Note that $I_1 = \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^2}{1+x^4} \ dx = J$. Consider the semi-circle contour $C$ in the upper half of the complex plane and note that there are two poles (at $z_1 = e^{\pi i/4}$ and $z_2 = e^{3\pi i/4}$):
$$ 2\pi i \left( \frac{1}{4z_1} + \frac{1}{4z_2} \right) = \oint_C \frac{z^2}{1 + z^4} \ dz = J + \int_{C_R} \frac{z^2}{1+z^4} \ dz $$
where $C_R: z = R e^{i \theta}$ with $\theta \in [0,\pi]$. You can show that $|\int_{C_R}| \to 0$ as $R \to \infty$ and simplifying the above expression you obtain
$$ J = \frac{\pi}{\sqrt 2} \implies I_1 = \frac{\pi}{2\sqrt 2} $$
| {
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An inequality involving three reals greater than or equal to 1 I have been trying to verify the inequality for $x\geq k, y\geq k,$ and $k\geq 1,$
$\left(\frac{\left(x-k\right)}{x+1}\right)\left(\frac{\left(y-k\right)}{y+1}\right)+\left(\frac{\left(x-k\right)}{x+1}\right)+\left(\frac{\left(y-k\right)}{y+1}\right)\geq\left(\frac{\left(xy-k^2\right)}{xy+1}\right) . $ Let me show my attempt.
Let us try to prove $\left(\frac{\left(x-k\right)}{x+1}\right)\left(\frac{\left(y-k\right)}{y+1}\right)+\left(\frac{\left(x-k\right)}{x+1}\right)+\left(\frac{\left(y-k\right)}{y+1}\right)-\left(\frac{\left(xy-k^2\right)}{xy+1}\right)\geq 0 $
But the left hand side of the above inequality is equal to
$$\frac{\left(x-k\right)\left(y-k\right)\left(xy+1\right)+\left(x-k\right)\left(y+1\right)\left(xy+1\right)+\left(y-k\right)\left(x+1\right)\left(xy+1\right)-\left(xy-k^2\right)\left(x+1\right)\left(y+1\right)}{\left(x+1\right)\left(y+1\right)\left(xy+1\right)}$$
$$=\frac{2x^2y^2-2kx^2y+k^2x-2kx-2kxy^2+2xy+2k^2xy-2kxy+x+2k^2-2k-2ky+y+k^2y}{\left(x+1\right)\left(y+1\right)\left(xy+1\right)}. $$
Is there anyway to conclude
$2x^2y^2-2kx^2y+k^2x-2kx-2kxy^2+2xy+2k^2xy-2kxy+x+2k^2-2k-2ky+y+k^2y\geq 0$ using factorization or any other technique?
One can observe that whenever $k=1,$
$$2x^2y^2-2kx^2y+k^2x-2kx-2kxy^2+2xy+2k^2xy-2kxy+x+2k^2-2k-2ky+y+k^2y=2xy(x-1)(y-1)\geq 0$$ and the result is true whenever $k=1.$
| I shall not mention the obvious symmetry.
What I did is to look at the minimum value of function
$$f(x,y)=\frac{x-k}{x+1}\times\frac{y-k}{y+1}+\frac{x-k}{x+1}+\frac{y-k}{y+1}-\frac{x y-k^2}{x y+1}$$
$$f(k,y)=\frac{(k-1) (k-y)}{(y+1) (k y+1)}$$ Its derivative cancels at
$$y_*=k+\frac{\sqrt{k^4+k^3+k^2+k}}{k}$$ and the second derivative test shows that this is a minimum.
Now, computing $$f(k,y_*)=\frac{2 k^2+k+1-2 \sqrt{k (k+1) \left(k^2+1\right)}}{1-k}$$ is never positive. It reaches a minimum value at $k=1+\sqrt 2$ and for such a value we have $-0.0470219$.
Just to be sure, I also computed the partial derivatives and set them equal to $0$. The problem is that this leads to $7$ solutions among which only two are explicit.
Running a few cases, it seems that the minimum effectively occurs at $x=k$ and $y=y_*$. So, to me the inequality does not hold.
I suggest you to do a contour plot of $f(x,y)$ for any $k$ of your choice and ask for the level $-0.01$. You will find here the case for $k=10$; the area on the left of or below the line corresponds to negative values of $f(x,y)$.
Edit
After comments, considering the inequality
$$\frac{x-k}{x+1}\times\frac{y-k}{y+1}+\frac{x-k}{x+1}+\frac{y-k}{y+1}\geq \color{red}{K}\frac{x y-k^2}{x y+1}$$ the problem is to find $K$.
So, as before, I considered the function
$$f(x,y)=\frac{x-k}{x+1}\times\frac{y-k}{y+1}+\frac{x-k}{x+1}+\frac{y-k}{y+1}- K\frac{x y-k^2}{x y+1}$$ then computed $f(x,x)$, assumed $x=k+\epsilon$, developed as a series. The constant term of the expansion is
$$-2 (k+1)^2 \left(k^2+1\right) (k(k+1)K-(1+k^2))$$ making it equal to $0$ gives
$$K=\frac{k^2+1}{k (k+1)}\implies f(x,x)=\frac{(k-x)^2 \,\,A}{k (k+1) (x+1)^2 \left(x^2+1\right)} $$
where
$$A=(2 k^2+3 k-1)x^2+2 \left(k^2-1\right) x+(k-1) (2 k+1)$$ does not show any real root. So, $f(x,x) \geq 0$
| {
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"question_score": "4",
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how to solve $\int\frac{x^2}{x^2+1}dx$ by using series I want to solve $\int\frac{x^2}{x^2+1}dx$ by using series, but I did something wrong.
Correct solution:
\begin{align*}
\int\frac{x^2}{x^2+1}dx &= \int\frac{(x^2+1)-1}{x^2+1}dx \\
&= \int\left(1-\frac{1}{x^2+1}\right)dx\\
&= x-\arctan(x)+C\\
&= x-\left(x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{12}}{12}+\cdots\right)+C\\
&= C+\frac{x^3}{3}-\frac{x^5}{5}+\frac{x^7}{7}-\frac{x^9}{9}+\frac{x^{12}}{12}-\cdots
\end{align*}
My wrong solution:
\begin{align*}
\int\frac{x^2}{x^2+1}dx &= \int\frac{1}{1+(\frac{1}{x^2})}dx\\
&= \int\frac{1}{1+u}dx\text{ ; }u=\frac{1}{x^2}\\
&= \int\left(1-u+u^2-u^3+u^4-\cdots\right)dx\text{ ; }\frac{1}{x+1}=1-x+x^2-x^3+x^4-\cdots\\
&= \int\left(1-\frac{1}{x^2}+\frac{1}{x^4}-\frac{1}{x^6}+\frac{1}{x^8}-\cdots\right)dx\\
&= C+x+\frac{1}{x}-\frac{1}{3x^3}+\frac{1}{5x^5}-\frac{1}{7x^7}+\cdots\\
\end{align*}
What did I do wrong?
Why didn't I get the correct answer?
| Both series are good, but for different domains, Note that when we use this form of binomial expansion:
$$(1+y)^{-1}$$
we require $|y|<1$ and so in the first series it is valid for the domain of the arctan series, whilst the second one is valid for $|x|>1$ which is why there is a difference.
One thing to note is when you integrate make sure it is wrt the correct variable. if you make a substitution $u=f(x)$ you must calculate $du$
| {
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"source": "stackexchange",
"question_score": "4",
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Finding the least 100-digit number, which has no $0$ in its decimal representation, that is divisible by the sum of its digits. An interesting question on number theory:
Find the least 100-digit number, which has no $0$ in its decimal representation, that is divisible by the sum of its digits.
My attempt: let the number be
$$a_1 a_2 a_3.....a_{100}$$
Which can also be written as
$$10^{99}a_1 + 10^{98}a_2+\cdots +10a_{99}+a_{100}=(10^{99}-1)a_1 + (10^{98}-1)a_2+\cdots +(10-1)a_{99}+(a_1 +a_2 + ...a_{99} + a_{100})$$
The last term is divisible by the digit sum. But I don't know how to proceed further.
Thanks for any help.
| A slightly better answer than "by computer" but still requires trial and error: this uses Fermat's Little Theorem.
Obviously ..$11$, ..$13$, ..$15$, ..$17$, ..$19$ are not divisible by their digit sums, which are all even.
..$16$ is not divisible by its digit sum $105$.
Since ..$11$ is divisible by $101$, neither ..$12$ nor ..$21$ are divisible by their digit sum $101$.
The harder part is to check ..$14 \pmod {103}$, ..$18 \pmod {107}$ and $..22 \pmod {102}$.
Note that ..$14 = $..$11 + 3 = (10^{100}-1)/9 + 3 = (10^{100}+26)/9$. Since $103$ is prime, finding $10^{100} \pmod {103}$ would help find ..$14 \pmod {103}$.
Now by FLT, $10^{102} \equiv 1 \pmod {103}$. Also by using Euclidean Algorithm or otherwise, we have:
$103 \times 3 = 309 = 10\times 31-1$, so $10^{-1} \equiv 31 \pmod{103}$.
Finally ..$14\times 9 = 10^{100}+26 \equiv 10^{-2} + 26\equiv 31^2 + 26\equiv 60\pmod {103}$, so ..$14$ is not divisible by $103$.
I leave ..$18 \pmod {107}$ to you.
For $..22 \pmod {102}$, since $102 = 2 \times 3 \times 17$ and divisibility by $2$ and $3$ are very easy to check, we just need to find $(10^{100}-1)/9 + 11 \pmod {17}$, and showing $10^{100}+98\equiv 10^4 + 98$ is divisible by $17$ is trivial.
| {
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"source": "stackexchange",
"question_score": "2",
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Find a generator polynomial of a [15, 10] RS Code Find a generator polynomial of a 16-ary Reed-Solomon code of dimension 10.
My work so far:
First, $$k=q-1-\left(\delta -1\right)$$
$$10=15-1-\left(\delta -1\right)$$
$$\implies \delta =6 $$
and then noting that the generator polynomial $g(x)$ of a Reed-Solomon code will have degree $\delta-1$, which means $g(x)$ will have degree 5.
I then calculated $g(x)$ by noting that any $q$-ary Reed-Solomon code of length $q-1$ will be generated by a $g(x)$ such that
$$g(x) = \prod^{a+\delta -1}_{i=a+1}\left(x-\alpha^{i}\right), a \geq0, 2\leq\delta\leq q-1 $$
Setting $a=0$ and $\delta=6$, and letting $\alpha$ be a root of the the minimal polynomial $x^4+x+1$,
$$\prod^{5}_{i=1}\left(x-\alpha^i\right)=\left(x-\alpha\right)\left(x-\alpha^2\right)\left(x-\alpha^3\right)\left(x-\alpha^4\right)\left(x-\alpha^5\right)$$
expanding this I have
$$x^5-x^4\left(\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha\right)-x^3\left(\alpha^9+\alpha^8+2\alpha^7+2\alpha^6+2\alpha^5+\alpha^4+\alpha^3\right)-x^2\left(\alpha^{12}+\alpha^{11}+2\alpha^{10}+2\alpha^9+2\alpha^8+\alpha^7+\alpha^6\right)=x\left(\alpha^{14}+\alpha^{13}+\alpha^{12}+\alpha^{11}+\alpha^{10}\right)-\alpha^{15}$$
and finally after using a log table for the primitive elements of GF$\left(2^4\right)$, and reducing everything modulo 2, I have $$g(x)=x^5+x^4\left(\alpha^3+\alpha+1\right)+x^3\left(\alpha^2\right)+x^2\left(\alpha^2+\alpha\right)+x\left(\alpha\right)+1$$
So my question is, does this look like the correct generator for this code?
Update: I used the comments below to find $g(x)$ to be $$g(x)=x^5+x^4\left(\alpha^3+\alpha+1\right)+x^3\left(\alpha^2\right)+x^2\left(\alpha^2+\alpha\right)+x\left(\alpha\right)+1$$
Is the final step to find a value $\alpha$ satisfying $\alpha^{3}= \alpha +1$?
| After working through all of the comments a generator polynomial for this [15, 10, 6] Reed-Solomon code, where $\alpha$ is a primitive element of $x^4+x+1$, is $$g(x)=x^5+x^4\left(\alpha^3+\alpha+1\right)+x^3\left(\alpha^2\right)+x^2\left(\alpha^2+\alpha\right)+x\left(\alpha\right)+1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$
Simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$
The problem is from a not so well renowned book for calculus in India - Concepts of Functions & Calculus - Vikas Rahi, ISBN 9780070080805. The answer for this question is given as
$$\lvert \sqrt{x-1} -1 \rvert + \lvert \sqrt{x-1}+1 \rvert$$
My efforts weren't great at all, I tried rationalizing them but the denominator becomes very similar to the question, which is totally not helpful for simplification.
| First, note that
\begin{align}
(1 \pm \sqrt{x-1})^2
&= 1 \pm 2 \sqrt{x-1} + \sqrt{x-1}^2 \\
&= 1 \pm 2 \sqrt{x-1} + x-1 & x \geq 1 \\
&= x \pm 2 \sqrt{x-1}
\end{align}
Thus, since $\sqrt{u^2} = |u|$,
\begin{align}
f(x)
&= \sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}} \\
&= \sqrt{(1-\sqrt{x-1})^2}+\sqrt{(1+\sqrt{x-1})^2} & x \geq 1 \\
&= |1-\sqrt{x-1}|+|1+\sqrt{x-1}| \\
&= \begin{cases}
(1-\sqrt{x-1})+(1+\sqrt{x-1}) & 1 \leq x < 2 \\
-(1-\sqrt{x-1})+(1+\sqrt{x-1}) & 2 \leq x
\end{cases} \\
&= \begin{cases}
2 & 1 \leq x < 2 \\
2 \sqrt{x-1} & 2 \leq x
\end{cases}
\end{align}
| {
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"source": "stackexchange",
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How can I evaluate $\lim_{v \to \frac{\pi}{3}}\frac{1-2\cos{v}}{\sin{\left(v-\frac{\pi}{3}\right)}}$ without using L'Hospital's rule? I found the following limit problem in Differential and Integral Calculus by Piskunov:
$$\lim\limits_{v \to \frac{\pi}{3}}\dfrac{1-2\cos{v}}{\sin{\left(v-\dfrac{\pi}{3}\right)}}$$
I tried searching for a trigonometric identity that could simply this, but couldn't make any work. How can this be solved without the application of L'Hospital's rule?
| just use Taylor expansions:
$
\cos(x) = \frac{1}{2} - \frac{\sqrt{3} \cdot (x - \frac{\pi}{3})}{2} + o\left( x - \frac{\pi}{3}\right), \ x \to \frac{\pi}{3}
$
$
\sin(x) = x - \frac{\pi}{3}, \ o\left( x - \frac{\pi}{3}\right), \ x \to \frac{\pi}{3}
$
$
\lim\limits_{x \to \frac{\pi}{3}} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})} = \lim\limits_{x \to \frac{\pi}{3}} \frac{1 - 1 + \sqrt{3} \cdot (x - \frac{\pi}{3}) + o(x - \frac{\pi}{3})}{x - \frac{\pi}{3} + o(x - \frac{\pi}{3})} = \lim\limits_{x \to \frac{\pi}{3}} \sqrt{3} + o(1) = \boxed{\sqrt{3}}
$
| {
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For every $x,y,z \in[2,7] $ and $x+y+z=12$, prove that $x^3+y^3+z^3\leq 316+x^2+y^2+z^2$ I have tried to use the majorization inequality to approach this question.
I have both $x^3$ and $x^2$ is convex in the region $[2,7]$ since there second derivatives are positive on this interval.
After I rewrite the equation, I have $(x^3-x^2)-(y^3-y^2)-(z^3-z^2)\leq 316$
So, I wonder if I can first say something about $g(x)=x^3-x^2$ by the majorization inequality then generalize to y and z cases.
I would appreciate if someone can help.
| HINT( for another way).
For $(x,y,z)=(2,3,7)$ (and any of the six permutations) we have exactly $x^3+y^3+z^3= 316+x^2+y^2+z^2$.
Fixing $z=7$ we have to prove $x^3+y^3\le 22 + x^2+y^2$ in which the equality is reached at $(x,y)=(2,3)$ so we must take $z\lt7$. An easy calculation more and we are done.
| {
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"source": "stackexchange",
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Solve the equation $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x+4}=0$ Solve the equation $$\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0.$$
For $x\ne -4;-3;-2;-1;0$ we have $$(x+1)(x+2)(x+3)(x+4)+x(x+2)(x+3)(x+4)+x(x+1)(x+3)(x+4)+\text{...}=0$$
Most likely that's not the author's intention. I have tried to substitute $t=x+2$ to get $$\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}=0$$ which actually isn't easier to work with than the original problem.
| You can write it as:
$$\sum_{j=0}^4\prod_{\begin{matrix}i=0\\i\ne j\end{matrix}}^4(x+i)=0$$
which might make it easier to type into an equation expander. Otherwise, note that expanded it gives:
$$5x^4+40x^3+105x^2+100x+24=0$$
(I believe anyway I did not check this by hand). This does in fact have four real roots.
| {
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"question_score": "7",
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how to prove that $1^x+2^x+3^x+4^x+\cdots+N^x$ will never sum to a prime number except $1^x+2^x$? I am a a web developer programming in PhP which is limited to large calculations, but running a quick script shows that $1^x+2^x+3^x+4^x+\cdots+N^x$ can never sum to a prime number unless in the case of $1^x+2^x$, such as in the cases of $x=1$ and $x=2$ where $1^1+2^1=3$ and $1^2+2^2=5$.
As a self learner, I am currently refreshing my learning in Algebra 2 (before moving on), and sometimes my mind wonders into questions that I just can't find the answers to (mostly because I am not familiar with the concerned topics). I tried finding an answer but if this is a duplicate with a relevant answer, please close and refer me to it.
How to prove (if possible) that $1^x+2^x+3^x+4^x+\cdots+N^x$ will never be the sum of a prime number, unless in the case of $1^x+2^x$, such as in the cases of $x=1$ and $x=2$ where $1^1+2^1=3$ and $1^2+2^2=5$?
Edit: $x$ and $N$ are positive integers
I appreciate any answers even if it is just a hint or a reference.
| Faulhaber's formula is frankly not good. It does not deserve to be as popular as it is. A better formula is
$$1^x + 2^x + \cdots + N^x = \sum_{k=1}^x k! \left\{ {x \atop k}\right\} \binom{N+1}{k+1} \qquad (\text{for }x \geq 1).$$
Here $\left\{ {x \atop k}\right\}$ is a Stirling number of the second kind, which is a non-negative integer with simple combinatorial meaning much like the binomial coefficients $\binom{N+1}{k+1}$. This formula is cancellation-free and doesn't involve anything "mysterious" like the Bernoulli numbers.
From this formula, we've got $k!$ in each term, so it's pretty plausible that we'll "usually" get composites. Ok, but when do we actually get a composite?
First off, every term with $k \geq 2$ is divisible by $2$. The $k=1$ term is just $\binom{N+1}{2} = (N+1)N/2$. This will be even when $N=4m$ or $N=4m+3$. So, we can suppose $N=4m+1$ or $N=4m+2$.
Every term with $k \geq 3$ is divisible by $3$. When are the other two terms also divisible by $3$? When
$$\left\{{x \atop 1}\right\} \binom{N+1}{2} + \left\{{x \atop 2}\right\} \binom{N+1}{3} = \frac{(N+1)N}{2} + (2^{x-1} - 1) \frac{(N+1)N(N-1)}{6}$$
is divisible by $3$. This occurs for instance when $3 \mid \frac{(N+1)N}{2}$ and $3 \mid 2^{x-1} - 1$, so when $N=3m$ or $N=3m+2$ and $x$ is odd.
Now reuns has found a counterexample with 5 terms. That fits with this heuristic. A lot of modular congruence conditions have to work out simultaneously to get counterexamples.
| {
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"url": "https://math.stackexchange.com/questions/4122803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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} |
If $\sum_{k=4}^{143} \frac{1}{ \sqrt{k} + \sqrt{k+1}} = a - \sqrt{b}$, then $a$ and $b$ respectively are If $\sum_{k=4}^{143} \frac{1}{ \sqrt{k} + \sqrt{k+1}} = a - \sqrt{b}$, then $a$ and $b$ respectively are
*
*10 and 0
*-10 and 4
*10 and 4
*-10 and 0
This question is from the book, Mathematics, Class 9 (The IIT Foundation Series) , page number 1.25. The answer key present in the book says the first option as the correct answer. I need an explanation to solve this question. Till now I have tried to rationalize $\frac{1}{ \sqrt{k} + \sqrt{k+1}}$ found that $\frac{1}{ \sqrt{k} + \sqrt{k+1}} = \sqrt{k+1}-\sqrt{k}$. I am not sure what to do next.
Thanks!
| The four alternatives result when calculating $a-\sqrt{b}$ in
\begin{align*}
&(10,0)&10-\sqrt{0}&=10\tag{1.}\\
&(-10,4)&-10-\sqrt{4}&=-12\tag{2.}\\
&(10,4)&10-\sqrt{4}&=8\tag{3.}\\
&(-10,0)&-10-\sqrt{0}&=-10\tag{4.}
\end{align*}
We can now use OPs telescoping approach and obtain
\begin{align*}
\color{blue}{\sum_{k=4}^{143} \frac{1}{ \sqrt{k+1} + \sqrt{k}}}
&=\sum_{k=4}^{143}\left(\sqrt{k+1}-\sqrt{k}\right)\\
&=\sqrt{143+1}-\sqrt{4}\\
&=12-2\\
&\,\,\color{blue}{=10}
\end{align*}
showing (1.) is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Line Integral of Sphere Intersected with Cylinder The question:
Compute $\int_{C}(y^2+z^2)dx+(z^2+x^2)dy+(x^2+y^2)dz$ where C is given by:
$
x^2+y^2+z^2=2Rx \\
x^2+y^2=2rx \\
z \geq 0 \\
0 < r \leq R
$
This is a problem I found on the interet, sadly there was no solution provided. I have a hard time understanding how to solve this. I know that the equations given are a sphere and a cylinder, but does that help me in any way? I think usually one has to turn this into a parametric equation to change this to an integral with only 1 variable. However I do not know how to do that. I do no think any of my work is in any way significant for the actual solution, but here it is:
$
x^2 + y^2 + z^2 = 2Rx \implies z^2 = 2Rx - x^2 - y^2 \implies z = \sqrt{2Rx - x^2 - y^2}\\
x^2 + y^2 + z^2 = 2Rx \implies y^2 = 2Rx - x^2 - z^2 \implies y = \sqrt{2Rx - x^2 - z^2}\\
x^2 + y^2 + z^2 = 2Rx \implies x^2-2Rx+y^2+z^2=0 \implies x_{1,2} = R \pm \sqrt{R^2-y^2-z^2}
$
| There are two ways to go about this. Parametrize the intersection curve and do the line integral or applying Stokes' theorem, choose a surface which has the intersection curve as its boundary and do surface integral of the curl of the given vector field. To parametrize the intersection curve,
The given surfaces are $ \ x^2+y^2 = 2 r x, x^2+y^2+z^2 = 2Rx$. At intersection, $x^2+y^2 = 2 rx, z^2 = 2 (R-r)x$. Using polar coordinates,
$x = \rho \cos t, y = \rho \sin t \implies \rho^2 = 2 r \rho \cos t \ $ i.e. $\rho = 2r \cos t$. So we can parametrize the intersection curve as,
$\gamma (t) = (2r\cos^2 t, 2 r \cos t \sin t, 2 \sqrt{R r-r^2} \ \cos t), -\pi/2 \leq t \leq \pi / 2$
Or you can parametrize as,
$\gamma (t) = (r + r \cos t, r \sin t, 2 \sqrt{R r-r^2} \cos \frac{t}{2} \ ), 0 \leq t \leq 2\pi$
The line integral will be $\displaystyle \int_C \vec F(\gamma(t)) \cdot \gamma'(t) \ dt$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Do the graphs of $r=\sec 2\theta$ and $r=\csc 2\theta$ consist of two hyperbolas each? The graphs of the polar equations $r=\sec 2\theta$ and $r=\csc 2\theta$ look like the multiplication and addition signs respectively. They also appear to consist of two hyperbolas each. Is that in fact the case?
If you enter the equations on Desmos, you would get the following:
The red curve is the graph of $r=\sec 2\theta$ and the blue one is the graph of $r=\csc 2\theta$.
| I'll transform $r=\sec{(2\theta)}$ to Cartesian coordinates, and you can do the same for $r=\csc{(2\theta)}$.
$$r=\sec{(2\theta)}$$
$$r=\frac{1}{\cos{(2\theta)}}$$
$$r=\frac{1}{\cos^2(\theta)-\sin^2(\theta)}$$
$$\sqrt{x^2+y^2}=\frac{1}{\cos^2(\arctan(\frac{y}{x}))-\sin^2(\arctan(\frac{y}{x}))}$$
$$\sqrt{x^2+y^2} = \frac{1}{\left(\frac{x}{\sqrt{x^2+y^2}}\right)^2 - \left(\frac{y}{\sqrt{x^2+y^2}}\right)^2}$$
$$\sqrt{x^2+y^2}=\frac{x^2+y^2}{x^2-y^2}$$
$$x^2-y^2=\sqrt{x^2+y^2}$$
So.... not exactly a hyperbola. This last line will give you one of the "hyperbolas", and if you do $-\sqrt{x^2+y^2}$ instead, you'll get the other "hyperbola". So... the actuall equation is $x^2-y^2=\pm \sqrt{x^2+y^2}$. Or, square both sides and get $x^2+y^2=x^4-2x^2y^2+y^4$. Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4125889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$ Theorem. $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$, for all positive integers $n$.
I'm tasked with proving the given theorem by induction. Here's where I've gotten so far...
Proof. Clearly, the theorem is true for $n=1$, establishing the base case. Moreover, it can easily be shown that the theorem works for $n=2,3,4,5,\ldots, 18$.
Suppose, now, that $k$ is an integer for which the given theorem holds. Consider, then, $k+18$.
$$2^{2^{k+18}} + 3^{2^{k+18}} + 5^{2^{k+18}}$$
By Fermat's Little Theorem, if $n \equiv m \pmod{p-1}$, then $a^m \equiv a^n \pmod{p}$. Clearly, $k+18 \equiv k \pmod{18}$. Then,
$$2^{2^{k+18}} \equiv 2^{2^k} \pmod{19}$$
$$3^{2^{k+18}} \equiv 3^{2^k} \pmod{19}$$
$$5^{2^{k+18}} \equiv 5^{2^k} \pmod{19}$$
Hence, $2^{2^{k+18}}$ may be expressed as $\;2^{2^k} + 19n$ for some integer $n$, $\;3^{2^{k+18}}$ as $\;3^{2^k}+ 19m$ for some integer $m$, and $\;5^{2^{k+18}}$ as $\;5^{2^k} + 19q$ for some integer $q$.
Thus, $2^{2^{k+18}} + 3^{2^{k+18}} + 5^{2^{k+18}} = (2^{2^{k}} + 3^{2^{k}} + 5^{2^{k}}) + 19(m+n+q)$, a sum of integers divisible by $19$, and thus clearly divisible by $19$.
By the principle of induction, we've thus shown that $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$, for all positive integers $n$, concluding the proof.
$\blacksquare$
Is this approach valid?
| Note that incrementing $n$ just corresponds to squaring each summand. Just compute the first few cases mod 19 and look for a pattern:
*
*For $n=1$, we have $2^2 \equiv 4$, $3^2\equiv 9$, $5^2 \equiv 6$. We are good since $4+9+6\equiv 19$.
*For $n=2$, we have $2^3=4^2 \equiv -3$, $3^3=9^2 \equiv 5$, $5^3\equiv 6^2 \equiv -2$. Good again since $-3+5-2 \equiv 0$.
Now note that squaring the summands of the $n=2$ case will just give us the $n=1$ case again, as the numbers are the same as what we started with (2, 3, and 5) up to signs. So subsequent increments of $n$ will just cause the summands to alternate between {4, 9, 6} and {-3, 5, -2}. So we're done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How do I evaluate $\int \frac{x+1}{(x^2+1) \sqrt{x^2-6x+1}} dx$? I am trying to evaluate this indefinite integral:
$$\int \frac{x+1}{(x^2+1) \sqrt{x^2-6x+1}} dx$$
What I tried
*
*Substitution $u = \arctan(x)$. However, no luck after this
*Find substitutions to convert $x^2 - 6x + 1$ into $(a + bu)^2$ or $a - (b + cu)^2$, however I could not find any
*Converted $x^2 - 6x + 1$ into $(x-3)^2 - 8$ and used the derivative of $\sec^{-1}{(x)}$ but not any luck after that as well.
*I suspect that it has a real solution, so the solution from wolfram alfa is not what I am looking for.
Any help/solutions would be very much appreciated!
| Dividing numerator and denominator by $x$ and pulling out a factor of $\sqrt{x}$ gives
$$I = \int \frac{1+\frac{1}{(\sqrt{x})^2}}{\left(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+2\right)\sqrt{\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2-4}}\cdot \frac{2dx}{2\sqrt{x}}$$
This suggests using the substitution $$\sqrt{x}-\frac{1}{\sqrt{x}} = 2\cosh t$$
which simplifies the integral to
$$I = \int \frac{dt}{2\cosh^2t+1} = \int \frac{\operatorname{sech}^2 t\:dt}{3-\tanh^2t}=\frac{1}{\sqrt{3}}\tanh^{-1}\left(\frac{\tanh t}{\sqrt{3}}\right)+C$$
Using that
$$\tanh^2t = 1-\operatorname{sech}^2t = 1-\frac{4x}{(x-1)^2} = \frac{x^2-6x+1}{(x-1)^2}$$
we get a final answer of
$$I = \frac{1}{\sqrt{3}}\tanh^{-1}\left(\frac{1}{|x-1|}\sqrt{\frac{x^2-6x+1}{3}}\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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How to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}$ Edit
In this post I computed the following integral
$$\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx=\frac{11}{8}\zeta(3)$$
Now I am trying to compute
$$\boxed{\int_{0}^{1}\frac{\log(1-x)\log(1-x^4)}{x}dx=\frac{67}{32}\zeta(3)-\frac{\pi}{2}G}$$
What I did is
$$\int_{0}^{1}\frac{\log(1-x)\log(1-x^4)}{x}dx=\int_{0}^{1}\frac{\log(1-x)\log[(1-x^2)(1+x^2)]}{x}dx$$
$$=\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx+\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx$$
$$=\frac{11}{8}\zeta(3)+\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx$$
$$=\frac{11}{8}\zeta(3)+I$$
$$I=\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx=\sum_{k=1}^{\infty}\frac{(-1)^{n-1}}{k}\int_{0}^{1}x^{2k-1}\log(1-x)dx$$
Integrating by parts
$$I=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{\frac{\log(1-x)(x^{2n}-1)}{2n}\Big|_{0}^{1}+\frac{1}{2n}\int_{0}^{1}\frac{x^{2n}-1}{1-x}dx \bigg\}$$
$$=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{-\frac{1}{2n}\int_{0}^{1}\frac{1-x^{2n}}{1-x}dx \bigg\}$$
$$=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{-\frac{1}{2n}H_{2n} \bigg\}$$
$$\boxed{I=\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}}$$
So now, the task is to evaluate this Sum
| One idea is to start with the generating function
$$\sum _{n=1}^{\infty } (-1)^n H_{2 n} \,x^{2 n-1}=-\frac{\log \left(x^2+1\right)+2 x \tan ^{-1}(x)}{2x \left(x^2+1\right)}\tag{1}$$
and integrate twice (not forgetting to multiplying the result by $4$). Thus
$$2I=\sum _{n=1}^{\infty } (-1)^n \frac{H_{2n}}{n^2} =-4 \int_0^1 \frac{1}{x} \int \frac{\log \left(x^2+1\right)+2 x \tan ^{-1}(x)}{x \left(2 x^2+2\right)} \, dx\, dx$$
which becomes
$$2I=-2 \int_0^1 \frac{1}{x}\left( \frac{1}{2}\text{Li}_2\left(-x^2\right)+\frac{1}{4} \log ^2\left(x^2+1\right)-\tan ^{-1}(x)^2\right)\, dx\tag{2}$$
alternatively we could of started with
$$\sum _{n=1}^{\infty } (-1)^n H_{2 n} x^{n-1}=-\frac{\log (x+1)+2 \sqrt{x} \tan ^{-1}\left(\sqrt{x}\right)}{x (2 x+2)}\tag{3}$$
to give
$$2I=-\int_0^1 \frac{1}{x} \left( \frac{\text{Li}_2(-x)}{2}+\frac{1}{4} \log ^2(x+1)-\tan ^{-1}\left(\sqrt{x}\right)^2 \right)\, dx\tag{4}$$
So you pick either set of three functions to integrate, or mix and match, since only a factor of 2 is between them.
One of the integrals can be easily integrated:
$$-\frac{1}{2} \int \frac{\text{Li}_2(-x)}{x} \, dx=-\frac{\text{Li}_3(-x)}{2}$$
with
$$-\frac{1}{2} \int_0^1 \frac{\text{Li}_2(-x)}{x} \, dx=\frac{3 \zeta (3)}{8}\tag{5}$$
The next integral also evaluates to $\zeta(3)$
$$-\frac{1}{4} \int_0^1 \frac{\log ^2(x+1)}{x} \, dx=-\frac{\zeta (3)}{16}\tag{6}$$
So interestingly the problem really boils down to integrating: $$2 \int_0^1 \frac{\left(\tan ^{-1}(x)\right)^2}{x} \, dx \;\;\;\;\; \text{or} \;\;\;\;\; \int_0^1 \frac{\left(\tan ^{-1}\left(\sqrt{x}\right)\right)^2}{x} \, dx$$
With an example equivalent integral being
$$4 \int_0^{\frac{\pi }{4}} y^2 \csc (2 y) \, dy=\pi \, G-\frac{7 \,\zeta (3)}{4}\tag{7}$$
$2I$ is the sum of results $(5)$, $(6)$ and $(7)$
$$2I=\frac{3 \,\zeta (3)}{8}-\frac{\zeta (3)}{16}+ \pi \, G -\frac{7\, \zeta (3)}{4}=\pi \, G-\frac{23\, \zeta (3)}{16}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help Solving System of Differential Equations I am working through some practice problems, and I am getting a different answer from what the back of my book says:
Express the general solution of the given system of equations in terms of real valued functions:
$$\begin{equation*}
x' =
\begin{pmatrix}
2 & -5 \\
1 & -2 \\
\end{pmatrix}x
\end{equation*}$$
I get eigenvalues of $$\pm i$$ and eigenvectors
$$\begin{equation*}
v_1 =
\begin{pmatrix}
2+i \\
1 \\
\end{pmatrix}
\end{equation*}$$
$$\begin{equation*}
v_2 =
\begin{pmatrix}
2-i \\
1 \\
\end{pmatrix}
\end{equation*}$$
The general solution with complex values is
$$\begin{equation*}
x = e^{it}
\begin{pmatrix}
2+i \\
1 \\
\end{pmatrix}
\end{equation*}$$
which becomes
$$\begin{equation*}
x = (\cos(t)+i\sin(t))
\begin{pmatrix}
2+i \\
1 \\
\end{pmatrix}
\end{equation*}$$
Expanding this and simplifying, I get
$$\begin{equation*}
x =
\begin{pmatrix}
2\cos(t)-\sin(t) \\
\cos(t) \\
\end{pmatrix}
+ i
\begin{pmatrix}
\cos(t)+2\sin(t) \\
\sin(t) \\
\end{pmatrix}
\end{equation*}$$
and then adding the constants, I get
$$\begin{equation*}
x = c_1
\begin{pmatrix}
2\cos(t)-\sin(t) \\
\cos(t) \\
\end{pmatrix}
+ c_2
\begin{pmatrix}
\cos(t)+2\sin(t) \\
\sin(t) \\
\end{pmatrix}
\end{equation*}$$
which is my final answer.
Looking at the solution at the back of the book, it says the correct answer is
$$\begin{equation*}
x = c_1
\begin{pmatrix}
5\cos(t) \\
2\cos(t)+\sin(t) \\
\end{pmatrix}
+ c_2
\begin{pmatrix}
5\sin(t) \\
-\cos(t)+2\sin(t) \\
\end{pmatrix}
\end{equation*}$$
Where did I go wrong? Thank you.
| $\begin{pmatrix} 2+i \\ 1\end{pmatrix}$ is proportional to $\begin{pmatrix} 5 \\ 2-i\end{pmatrix}$, so your solution is equivalent to the one given. You didn't do anything wrong.
| {
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"url": "https://math.stackexchange.com/questions/4128705",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find the solution to the given differential equation I am trying to find the general solution to the system $x' = Ax + at$ where,
$$ A = \begin{pmatrix}
1 & -1\\
9 & 1
\end{pmatrix} $$
and $$ a = \begin{pmatrix}
-1\\
1
\end{pmatrix} $$
I began by finding the eigenvectors and eigenvalues of $A$ and got the eigevalues $\lambda = 1 \pm 3i$ to correspond to the eigenvectors $ \begin{pmatrix}
i\\
3
\end{pmatrix} $ and $ \begin{pmatrix}
i\\
-3
\end{pmatrix} $
I am confused about how to proceed to finding the final solution here. Any guidance is greatly appreciated!
| Typically, the general method of solving such problems is the Laplace transform which yields
$$
x'=Ax+at\implies sx-x(0)=Ax+L(at)\implies (sI-A)x=x(0)+L(at)\implies x=(sI-A)^{-1}(x(0)+L(at))
.
$$
In our case,
$$x{=
\begin{bmatrix}
s-1&1\\-9&s-1
\end{bmatrix}^{-1}
(x(0)+L(at))
\\
=\begin{bmatrix}
\frac{s-1}{(s-1)^2+9}&-\frac{1}{(s-1)^2+9}\\\frac{9}{(s-1)^2+9}&\frac{s-1}{(s-1)^2+9}
\end{bmatrix}
(x(0)+L(at))
}.
$$
If out experiment starts at $t=0$, we can substitute $t$ with $r(t)$ which is the ramp function with the Laplace transform of $\frac{1}{s^2}$. Hence
$$
x
=\begin{bmatrix}
\frac{s-1}{(s-1)^2+9}&-\frac{1}{(s-1)^2+9}\\\frac{9}{(s-1)^2+9}&\frac{s-1}{(s-1)^2+9}
\end{bmatrix}
\begin{bmatrix}
x_1(0)+\frac{a}{s^2}\\
x_2(0)+\frac{a}{s^2}
\end{bmatrix}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find absolute minimum of $f(x,y,z,w) = x^2 y + y^2 z + z^2 w + w^2 x$ on $S = \{x>0,y>0,z>0,w>0, xyzw=1\} \subset \mathbb{R}^4$ I tried to solve it using the Lagrange multiplier method, but I failed since the system of equations involving partial derivatives was pretty complex.
Any hints will be appreciated.
| The system of Lagrange equations looks complicated, but it possesses an interesting property:
$$ 2xy \ + \ w^2 \ = \ \lambda · yzw \ \ , \ \ x^2 \ + \ 2yz \ = \ \lambda · xzw \ \ , $$
$$ y^2 \ + \ 2zw \ = \ \lambda · xyw \ \ , \ \ z^2 \ + \ 2wx \ = \ \lambda · xyz \ \ . $$
The amount of symmetry among these equations is already suspicious. We can make these easier to work with by applying the constraint equation $ \ xyzw = 1 \ $ to write
$$ 2xy \ + \ w^2 \ = \ \lambda · \frac{1}{x} \ \ , \ \ x^2 \ + \ 2yz \ = \ \lambda · \frac{1}{y} \ \ , $$
$$ y^2 \ + \ 2zw \ = \ \lambda · \frac{1}{z} \ \ , \ \ z^2 \ + \ 2wx \ = \ \lambda · \frac{1}{w} $$
$$ \Rightarrow \ \ \lambda \ \ = \ \ 2x^2y \ + \ xw^2 \ \ = \ \ x^2y \ + \ 2y^2z \ \ = \ \ y^2z \ + \ 2z^2w \ \ = \ \ wz^2 \ + \ 2w^2x \ \ . $$
If we combine these equations pairwise, we produce
$$ x^2y \ = \ 2y^2z - xw^2 \ \ , \ \ y^2z \ = \ 2z^2w - x^2y \ \ , \ \ z^2w \ = \ 2w^2x - y^2z \ \ . \ \ w^2x \ = \ 2x^2y - wz^2 \ \ . $$
It is possible to put pairs of these equations together to find the correspondence between the variables $ \ x \rightarrow y \ , \ y \rightarrow z \ , \ z \rightarrow w \ , \ w \rightarrow \ x \ \ . $ This interchangeability among the variables indicates that $ \ x = y = z = w \ \ , $ and using this result in the constraint equation tells us that all four are equal to $ \ 1 \ $ . Hence, we have an absolute minimum for the function of
$$ f(1,1,1,1) \ = \ 4 \ · \ 1^2 · 1 \ = \ 4 \ \ , $$
since there is no upper limit to the values of the variables.
[As a check, a small perturbation away from the minimum by setting, say, $ \ x = 1 + \epsilon \ , \ y = \frac{1}{1+\epsilon} \ \ , $ produces
$$ f \left(1+\epsilon,\frac{1}{1+\epsilon},1,1 \right) \ = \ \frac{(1+\epsilon)^2}{1+\epsilon} \ + \ \frac{1}{(1+\epsilon)^2} · 1 \ + \ 1 · 1 \ + \ 1^2 · (1+\epsilon) $$ $$ = \ \ (1+\epsilon) \ + \ ( 1 - 2\epsilon + 3\epsilon^2 - 4\epsilon^3 + 5\epsilon^4 + \ldots) \ + \ 1 \ + \ (1+\epsilon) $$ $$ = \ \ 4 + 3\epsilon^2 - 4\epsilon^3 + 5\epsilon^4 + \ldots \ \ . \ ] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129992",
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"source": "stackexchange",
"question_score": "1",
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$25$ people will be divided into $5$ groups ,each group have $5$ individuals . What is the probability that $25$ people will be divided into $5$ groups ,each group have $5$ individuals . What is the probability that
a-) Dennis , John and Jack are in the same group.
b-)Dennis , John and Jack are in different groups.
c-)Dennis and John are in the same group but not Jack
My attempt :
a-) If they are in same group , then there are $5$ ways to select this group. Moreover , we should select $2$ people for the group by $C(22,2)$ . Then $5 \times \frac{C(22,2) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}{C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
b-)If they are in different groups , we can disribute them by $P(5,3)=60$ ways.Then , $60 \times \frac{C(24,4) \times C(19,4) \times C(14,4) \times C(10,5) \times C(5,5)}{C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
c-)We can choose $5$ groups for Dennis and John , 4 groups for Jack ,so $P(5,2)=20$ ways. Then ; $20 \times \frac{C(23,3) \times C(19,4) \times C(15,5) \times C(10,5) \times C(5,5)}{C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
Is my solution way correct ? If not ,can you help..
| I checked the first one, it is correct, but there is a simpler way to compute the probabilities
$\fbox{.}\fbox{.}\fbox{.}\fbox{.}\fbox{.}\quad\fbox{.}\fbox{.}\fbox{.}\fbox{.}\fbox{.}\quad\fbox{.}\fbox{.}\fbox{.}\fbox{.}\fbox{.}\quad\fbox{.}\fbox{.}\fbox{.}\fbox{.}\fbox{.}\quad\fbox{.}\fbox{.}\fbox{.}\fbox{.}\fbox{.}\quad$
A The first friend can be put in any slot, and the remaining two must be put in the same group, thus
$Pr = 1\cdot \frac4{24}\frac3{23} = \frac1{46}$
B Similarly, $Pr = 1\cdot\frac {20}{24}\cdot\frac{15}{23} = \frac{25}{46}$
C John can be anywhere, the other two have to fill two slots in some other group, thus
$Pr = 1\cdot\frac {20}{24}\frac{4}{23} = \frac{10}{69}$
This avoids long expressions and a lot of cancellations.
PS Also avoids errors, which I have found in your working after checking B and C
| {
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"source": "stackexchange",
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show that ${n \choose 1}-\dots+(-1)^{n-1}{n \choose n} \left(1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}\right)=\frac{1}{n}$ Show that
${n \choose 1}$ - $ {n \choose 2}$ (1 + $ \frac{1}{2} $ ) ..............+ $(-1)^{n-1} {n \choose n}$ (1 + $ \frac{1}{2} $ + $ \frac{1}{3} $ +$ \frac{1}{4} $ ............+$ \frac{1}{n} $) = $ \frac{1}{n} $
MY ATTEMPT : I tried by taking the general term
$T_r$ = $(-1)^{r-1}$ ${n \choose r}$ (1 + $ \frac{1}{2} $ + $ \frac{1}{3} $ +$ \frac{1}{4} $ ............+$ \frac{1}{r} $)
It looked some what like ${n \choose r}$ $\int x^{r-1} $ = ${n \choose r}$ $\frac {x^r}{r} $ so I took an expansion say
$T_1(r)$ = $(-1)^{r-1}$ ${n \choose r}$ $(1+ x+x^2+...........+x^{r-1})$ this simplifies to
$T_1(r)$ = $(-1)^{r-1}$ ${n \choose r}$ $(\frac{1-x^r}{1-x})$
I am struck after this I tried to rearrange this and done some simplification but I failed
Please help me
| First, let's summarize what you've already shown: The given sum is equivalent to
$$\sum_{r=1}^n \int_0^1\left((-1)^{r-1}\binom{n}{r}\frac{1-x^r}{1-x}\right)\,dx$$
The first thing we'll do to this is make the sum start at $r=0$ instead of $r=1$ (which doesn't change anything since when $r=0$ the integrand is $0$), move the [finite] sum inside the integral, and then rearrange algebraically:
$$\begin{align*}\sum_{r=1}^n \int_0^1&\left((-1)^{r-1}\binom{n}{r}\frac{1-x^r}{1-x}\right)\,dx = \sum_{r=0}^n \int_0^1\left((-1)^{r-1}\binom{n}{r}\frac{1-x^r}{1-x}\right)\,dx \\ &= \int_0^1 \sum_{r=0}^n \left((-1)^{r-1}\binom{n}{r}\frac{1-x^r}{1-x}\right)\,dx \\ &= \int_0^1 (1-x)^{-1}\left(\sum_{r=0}^n\binom{n}{r}(-x)^r-\sum_{r=0}^n\binom{n}{r}(-1)^r\right)\,dx\end{align*}$$
Now both sums have the form $\sum_{r=0}^n \binom{n}{r} a^r = (1+a)^n$ by the binomial theorem, so by simplifying:
$$\int_0^1(1-x)^{-1}\left((1-x)^n - (1-1)^n\right)\,dx = \int_0^1 (1-x)^{n-1}\,dx = \frac{1}{n}$$
| {
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Solve in positive integers the equation $x^3 + y^3 + z^3 - 3xyz = 1517$ Solve in positive integers the equation
$$x^3 +y^3 +z^3 −3xyz = 1517$$
Initially, I tried to factorize the LHS and my last step was
$$(x+y+z)(x^2+ y^2 + z^2 -xy -zx - yz) = 37×41$$
How do I find the value of $x, y, z$?
Any help would be appreciated
| To continue your argument, we have the following cases:
*
*$x + y + z = 1, x^2 + y^2 + z^2 - xy - yz - zx = 1517$;
*$x + y + z = 37, x^2 + y^2 + z^2 - xy - yz - zx = 41$;
*$x + y + z = 41, x^2 + y^2 + z^2 - xy - yz - zx = 37$;
*$x + y + z = 1517, x^2 + y^2 + z^2 - xy - yz - zx = 1$.
Case 1 is impossible, as $x + y + z \geq 3$.
The three remaining cases can be done in the same way. I will do case 3 as example.
We rewrite the second equality as $(x - y)^2 + (y - z)^2 + (z - x)^2 = 74$. There are only three ways to write $74$ as a sum of three squares:
$$74 = 0^2 + 5^2 + 7^2 = 1^2 + 3^2 + 8^2 = 3^2 + 4^2 + 7^2.$$
Since the sum $(x - y) + (y - z) + (z - x)$ must be zero, we see that the only possibilities, up to permutation of $x, y, z$, are $$x - y = 3, y - z = 4, z - x = -7$$ or $$x - y = 3, y - z = -7, z - x = 4.$$ Together with $x + y + z = 41$, we get $(x, y, z) = (17, 14, 10)$.
With similar calculations, we see that there is no solution in case 2, and $(x, y, z) = (506, 506, 505)$ in case 4.
Therefore all solutions are given by $(x, y, z) = (17, 14, 10), (506, 506, 505)$ up to permutation.
| {
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"source": "stackexchange",
"question_score": "4",
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Prove that $\lim_{n\to\infty} \frac{a_n}{a_{n+1}} = p$
Given a sequence $\{a_n\}_{n\in \mathbb N}$ of real numbers such that
$$\begin{align}
\lim_{n\to\infty}\frac{a_na_{n+1} - a_{n-1}a_{n+2}}{a_{n+1}^2 - a_na_{n+2}} &= p + q && (1)\\[1mm]
\lim_{n\to\infty} \frac{a_n^2 - a_{n-1}a_{n+1}}{a_{n+1}^2 - a_na_{n+2}} &= pq && (2)\end{align}$$
where $|p| < |q|$, prove that
$$\lim_{n\to\infty} \frac{a_n}{a_{n+1}} = p$$
Attempts:
Idea #1: Let us denote $b_n = \dfrac{a_n}{a_{n+1}}$. If we divide both numerators and denominators of $(1)$ and $(2)$ by $a_na_{n+2}$ and $a_{n+1}a_n$, respectively, we have
$$
\begin{align}
\lim_{n\to\infty}\frac{\dfrac{a_{n+1}}{a_{n+2}} - \dfrac{a_{n-1}}{a_{n}}}{\dfrac{a_{n+1}}{a_n}\dfrac{a_{n+1}}{a_{n+2}} - 1} &= p+q && (1') \\[1mm]
\lim_{n\to\infty} \frac{\dfrac{a_n}{a_{a+1}} - \dfrac{a_{n-1}}{a_n}}{\dfrac{a_{n+1}}{a_n} - \dfrac{a_{n+2}}{a_{n+1}}} &= pq && (2')
\end{align}$$
which now can be written as
$$\begin{align}
\lim_{n\to\infty}\frac{b_{n+1}b_n - b_nb_{n-1}}{b_{n+1} - b_n} &= p+q && (1'') \\[1mm]
\lim_{n\to\infty} \frac{b_n - b_{n-1}}{\dfrac{1}{b_{n+1}} - \dfrac{1}{b_{n}}} &= -pq && (2'')
\end{align}$$
Now, every numerator and denominator contains the difference of consecutive terms of some sequence (reminds me of Cesaro-Stolz, but that cannot be applied here).
Idea #2: The given conditions can be written as
$$\begin{align}
\lim_{n\to\infty}\frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+2} & a_{n+1}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}} &= p + q && (1')\\[2mm]
\lim_{n\to\infty} \frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+1} & a_{n}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}} &= pq && (2')\end{align}$$
Now, maybe a bit of Linear Algebra could be incorporated somehow.
If we set (B. Grossman)
$$\begin{align} x_n = \frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+2} & a_{n+1}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}}, \quad y_n = \frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+1} & a_{n}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}} \end{align}$$
Then, we have
$$\pmatrix{a_{n+1} & a_{n+2}\\a_n & a_{n+1}} \pmatrix{x_n\\y_n} = \pmatrix{a_n\\a_{n-1}}$$
This gives
$$\frac{a_{n+1}x_n + a_{n+2}y_n}{a_nx_n + a_{n+1}y_n} = \frac{a_n}{a_{n-1}}$$
Dividing the numerator and the denominator by $a_{n+1}$, we get
$$\frac{x_n + \dfrac{a_{n+2}}{a_{n+1}}y_n}{\dfrac{a_n}{a_{n+1}}x_n + y_n} = \frac{a_n}{a_{n-1}}$$
Noting that $x_n \to p+q$, $y_n \to pq$ and assuming that $\dfrac{a_n}{a_{n-1}} \to A$, and sending $n$ to infinity, we get
$$\frac{p+q + Apq}{\dfrac{1}{A}(p+q) + pq} = A$$
which simplifies to
$$0=0$$
I think I did something wrong somewhere.
Any help is appreciated.
| The following proof is not complete, hoped this is helpful to you.
Denote $b_n=\frac{a_n}{a_{n+1}}$ , then
$$\lim\limits_{n\to\infty}\frac{b_n(b_{n+1}-b_{n-1})}{b_{n+1}-b_n}=p+q,\quad \lim\limits_{n\to\infty}\frac{b_n b_{n+1}(b_{n}-b_{n-1})}{b_{n+1}-b_n}=pq.$$
If you assume that $\lim\limits_{n\to\infty} b_n=x$ exists, since
$$p+q=\lim\limits_{n\to\infty}\frac{b_n(b_{n+1}-b_{n-1})}{b_{n+1}-b_n}=\lim\limits_{n\to\infty}[\frac{b_n(b_{n}-b_{n-1})}{b_{n+1}-b_n}+b_n],$$
hence $\lim\limits_{n\to\infty}\frac{b_n(b_{n}-b_{n-1}\ )}{b_{n+1}-b_n}=y$ also exists. Moreover,
$$x+y=p+q,\quad xy=pq\implies
\left\{\begin{array}{l}
x=p \\ y=q
\end{array}\right.,\ or
\left\{\begin{array}{l}
x=q \\ y=p
\end{array}\right..$$
If you can check that $\lim\limits_{n\to\infty} |b_n|\le \lim\limits_{n\to\infty}|\frac{b_n(b_{n}-b_{n-1}\ )}{b_{n+1}-b_n}|$, then
$$\lim\limits_{n\to\infty} b_n=x=p.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134053",
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"source": "stackexchange",
"question_score": "6",
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Finding the inverse of $(x^2+2)$ in the field $S = \Bbb F_3[x]/(x^4+x^3+x^2+1)$ Let $S = \Bbb F_3[x]/(x^4+x^3+x^2+1)$.
Find the inverse of $(x^2+2)$ in $S$.
I know I'm looking for a polynomial $q(x)$ such that $(x^2+2)q(x) = 1 \mod x^4+x^3+x^2+1$
i.e $(x^2+2)q(x) + k(x)(x^4+x^3+x^2+1) = 1$ for some $k(x)$ in $\Bbb F_3[x]$
I'm not really sure how to find $q(x), k(x)$. I understand the Euclidean algorithm (in polynomials) needs to be applied, but I'm not really sure how to do this working over $\Bbb F_3[x]$.
| In $\mathbb F_3[x]$, you could replace $x^2+2$ with $x^2-1$.
$$\color{purple}{x^4+x^3+x^2+1}=(\color{blue}{x^2-1})(x^2+x-1)+\color{green}x$$
$$\color{blue}{x^2-1}=x(\color{green}x)-\color{orange}1$$
So $$\color{orange}1=x[\color{green}x]-(\color{blue}{x^2-1})=x\left[\color{green}{x^4+x^3+x^2+1-( {x^2-1})(x^2+x-1)}\right]-(\color{blue}{x^2-1}) $$
$$=x(\color{purple}{x^4+x^3+x^2+1})\color{red}{-(x^3+x^2-x+1)}(x^2-1),$$
so an inverse of $x^2-1$ in $\mathbb F_3[x]/(\color{purple}{x^4+x^3+x^2+1})$
is $\color{red}{-(x^3+x^2-x+1)\equiv2x^3+2x^2+x+2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134512",
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"source": "stackexchange",
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$3^{n+2}$ does not divide $2^{3^n}+1$ Prove for all natural numbers $n$ such that $2^{3^n}+1$ is divisible by $3^{n+1}$, but is not divisible by $3^{n+2}$.
I know how to prove why is it divisible, but I need help with why is it not divisible.
I used a similar proof for why is it divisible:
https://math.stackexchange.com/a/1591199/925421
My working for not divisible: Induction proof
Base case: n = 0
$$3^{0+2}=9$$
and
$$2^{3^0}+1=3$$
9 cannot divide 3, so base case is true.
Assume n=k is true. That means $3^{k+2}$ does not divide $2^{3^k}+1$. Now prove $3^{k+3}$ does not divide $2^{3^{k+1}}+1$.
$$(2^{3^k})^3+1=(3^{k+1}*m)^3-3(3^{k+1}*m)^2+3(3^{k+1}*m)$$
$$=3^{3k+3}*m^3-3^{2k+3}*m^2+3^{k+2}*m$$
$$=3^{k+3}(3^{2k}*m^3-3^{k}*m^2+3^{-1}*m)$$
That's where I need help!
I'm so close to finish, I want to show that $$(3^{2k}*m^3-3^{k}*m^2+3^{-1}*m)$$ is not an integer, so that means $3^{n+3}$ does not divide $2^{3^{n+1}}+1$
To show that $(3^{2k}*m^3-3^{k}*m^2+3^{-1}*m)$ is not an integer, I have to show m can never be 3 or multiple of 3. But I don't know how to show that. Please help.
| The linked problem also has an answer involving the Lifting The Exponent (LTE) result. Here is a plain solution using only induction for the proposition:
$$
\color{blue}{
P(n)\text{ : We have $2^{3^n}=-1 + 3^{n+1}(1+3N(n))$ for some integer $N(n)$ .}
}
$$
Let us inductively show that $P(n)$ is true for all integers $n\ge 0$.
$P(0)$ is true, $2^{3^0}=2^1=2=-1+3=-1+3^{0+1}$ and we take $N(0)=0$.
$P(1)$ is true, $2^{3^1}=2^3=9=-1+3^2=-1+3^{1+1}$ and we take $N(1)=0$.
Fix some $n\ge 1$. Assume that (for this $n$) the proposition $P(n)$ is true. Then, working modulo
$3^{(n+1)+2}$ at the places with the $\equiv$ relation symbol:
$$
\begin{aligned}
2^{3^{n+1}}
&=
2^{3^n\cdot 3} = \left(2^{3^n} \right)^3\\
&\overset{P(n)}{=\!=\!=}
\Bigg[\ -1 + 3^{n+1}(1+3N(n))\ \Bigg]^3
\\
&=(-1)^3
+3\cdot 3^{n+1}(1+3N(n))
\\
&\qquad\qquad\qquad
-\underbrace{3\cdot 3^{2(n+1)}}_{\equiv 0}(1+3N(n))^2
+ \underbrace{3^{3(n+1)}}_{\equiv 0}(1+3N(n))^3
\\
&\equiv
-1 + 3^{n+2}(1+3N(n))\qquad\text{ modulo }3^{n+3}
\ .
\end{aligned}
$$
(We have used $n+3\le 1+2(n+1)\le 3(n+1)$, which is valid for $n\ge 1$.)
The claim is now inductively shown.
$\square$
Alternatively, note that $3^n$ is odd, use
$$
1+2^{3^n}=1+(\color{brown}3-1)^{\color{red}{3^n}}\\
=1+(-1)
+ \binom{\color{red}{3^n}}1\cdot \color{brown}3^1
- \binom{\color{red}{3^n}}2\cdot \color{brown}3^2
+ \binom{\color{red}{3^n}}3\cdot \color{brown}3^3
-\dots
\pm\binom{\color{red}{3^n}}k\cdot \color{brown}3^k
\mp\dots
$$
and show (inductively) that the terms corresponding to $k\ge 2$ are each divisible by $3^{n+2}$.
| {
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For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.
For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.
I tried to use Newton's binomial formula as:
$$
(1 + x + x^2)^6 = \sum_{k = 0}^{6}\left( \binom{6}{k}(1 + x)^{n-k} x^{2k}\right)
$$
and that's all I can think of, other then just to compute it.
| This question might be easier to solve by hand.
However, the general way to solve it is to know that:
$$\frac1{(1-x)^k}=\sum_{n=0}^\infty \binom{n+k-1}{k-1}x^k$$
Then use:
$$x^2+x+1=\frac{1-x^3}{1-x}$$
So:
$$(x^2+x+1)^6 =(1-x^3)^6\frac1{(1-x)^3}$$
So the coefficient of $x^n$ is $$\sum_{k=0}^6(-1)^k\binom 6k\binom{n-3k+5}5$$
This gives, for $n=6,$
$$\binom{11}5-\binom61\binom85 +\binom62\binom55$$
because the other terms are zero.
| {
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"url": "https://math.stackexchange.com/questions/4137533",
"timestamp": "2023-03-29T00:00:00",
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How many 3-digit natural numbers have digits that sum to 13 and leave a remainder of 1 when divided by 4? This is how I approached the problem-
Let the digits be of a three digit number be $a, b, c$. We want $a+b+c = 13$ and $10b + c \equiv 1 \;(\bmod\; 4)$ (Since for a number to be divisible by 4, only the last 2 digits need to divide 4
$10b + c \equiv 1 \;(\bmod\; 4)$ would be $2b + c \equiv 1 \;(\bmod\; 4)$
$ b + b + c \equiv 1 \;(\bmod\; 4)$, Since $b + c = 13 - a$,
$b + 13 - a \equiv 1 \;(\bmod\; 4)$
$b - a \equiv -12 \;(\bmod\; 4)$
$b - a \equiv 0 \;(\bmod\; 4)$
Since both $a$ and $b$ are less than 10, for $a - b$ to be a multiple of 4, it must equal either $0$, $4$ or $8$
First possibility: $a - b = 0$
$a = b$
Since $3 < a + b < 14$ (If $a + b < 4$ , $c$, must be bigger than $9$, which can't happen, and if $a + b > 13$, $c$ must have a negative value to satisfy the equation $a + b + c = 13$, which cannot happen)
substituting $a = b$ in $3 < a + b < 14$, we get
$3 < 2a < 14$
$3/2 < a < 7$
This gives us 5 possible values of $a$, to which there are corresponding values of $b$ and $c$. Following the exact same pattern, we solve the other equations-
We get $4$ solutions for $a - b = 4$ and $1$ solution for $a - b = 8$
This gives us a total of $10$ solutions, which must mean there are $0$ numbers that satisfy the conditions. This however, is wrong, since I know the answer is 17. Where did I go wrong here?
| You have not considered the cases $a-b=-4$ and $a-b=-8$; these values are also $0\bmod4$. Here are all possible solutions:
$$193$$
$$157\ 265\ 373\ 481$$
$$229\ 337\ 445\ 553\ 661$$
$$409\ 517\ 625\ 733\ 841$$
$$805\ 913$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Can a symmetric bilinear map $V\times V \to W$ increase subspace dimension? Let $V$ and $W$ be two finite-dimensional real vector spaces, and let $S$ be a subspace of $V$. By the rank-nullity theorem, if $T$ is a linear map $V \to W$, then $\dim T(S) \leq \dim S$.
Now, let $B$ denote a symmetric bilinear map $V \times V \to W$. I would like to know if a similar statement holds for $B$.
More precisely, consider the subspace defined by the span of the image of $B\lvert_{S\times S}$:
$$
B(S)=\mathrm{span}\,\{B(x,y) \mid x,y \in S \}.
$$
Is it true that $\dim B(S) \leq \dim S$?
| Take $V = \mathbb{R}^2_{\textrm{col}}$ and consider the map $B \colon V \times V \rightarrow M_2(\mathbb{R})$ given by $B(x,y) = \frac{1}{2} \left( x \cdot y^T + y \cdot x^T \right)$. Then
$$ B(e_1,e_1) = e_1 \cdot e_1^T = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \\
B(e_1,e_2) = \frac{1}{2} \left( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \end{pmatrix} \right) = \frac{1}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\\
B(e_2,e_2) = e_2 \cdot e_2^T = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}. $$
This means that $B(V)$ is the subspace of $2 \times 2$ symmetric matrices which is three dimensional while $V$ has dimension two.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4142997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there a way to solve this non-linear recurrence relation? While exploring some ideas of my own, I came across the following recurrence relation:
$$a_{n+1}=\frac{1}{4(\frac{1}{2n}-a_n)(2n+1)^2}$$
for $n\ge1$ with $a_1$ arbitrary.
I have no experience with non-linear recurrence relations and I realise that they are generally considered to be intractable. Is this one also?
I managed to guess the solution to a much simpler one that also came up. For this sequence, it turns out that $a_1 = \frac{1}{4a_0}$. In case anyone is better at guessing than me, here are the first few terms of this sequence with $a_0 = 1$. I think it might also help to consider $b_{n+1} = (2n+1)^2a_{n+1}.$
$$\begin{align*} a_2 = \frac{1}{9}, \quad b_2 = 1 \end{align*}$$
$$a_3 = \frac{9}{125}, \quad b_3 = \frac{9}{5}$$
$$a_4 = \frac{375}{6958}, \quad b_2 = \frac{375}{142}$$
At which point, my calculator no longer can write the answer as a fraction.
Also, by construction, I can promise that this sequence tends to zero, if that helps, and it is probably strictly decreasing.
| There is a classical method called fixed point method.
Firstly, let
$$
x_n:=\frac{1+\sqrt{\frac{4n+1}{(2n+1)^2}}}{4n}\quad \text{and}\quad y_n:=\frac{1-\sqrt{\frac{4n+1}{(2n+1)^2}}}{4n}.
$$
It is easy to see that $x_n,y_n$ are the two solutions of the following equation:
\begin{align*}
x=\frac{1}{4(\frac{1}{2n}-x)(2n+1)^2}.
\end{align*}
Subtract $x=x_n,y_n$ from both sides of the recurrence relation to get
\begin{align*}
a_{n+1}-x=\frac{1}{4(\frac{1}{2n}-a_n)(2n+1)^2}-x
\end{align*}
\begin{align}
\Rightarrow 4(a_{n+1}-x)=\frac{(a_n-x)x}{(\frac{1}{2n}-a_n)},\quad x=x_n\text{ or } y_n.\tag{A}
\end{align}
Denote by
\begin{align*}
b_n:=\frac{a_n-x_n}{a_n-y_n}.
\end{align*}
By (A), we have
\begin{align*}
4b_{n+1}=\frac{x_n}{y_n}b_n\Rightarrow b_n=\frac{1}{4^{n-1}}\Pi_{i=2}^n \frac{x_i}{y_i}b_1=\frac{1}{4^{n-1}}\Pi_{i=2}^n \frac{x_i}{y_i}\frac{a_1-x_1}{a_1-y_1}\quad\text{for $n\ge2$}.
\end{align*}
Notice that
\begin{align*}
a_n=\frac{b_ny_n-x_n}{b_n-1},
\end{align*}
which gives the formula of $a_n$.
This method looks a bit complex, and I haven't check if there is any error, like $\frac{1}{0}$, in my formulation. Hope this helps you!
| {
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"url": "https://math.stackexchange.com/questions/4143770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Evaluation of $\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=2-2\ln(2)$ I came across the following statements
$$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2 \qquad \tag{1}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}} \qquad \tag{2}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(4 n+1)}=4-\frac{\pi}{2}-3 \ln 2 \qquad \tag{3}$$
$$\sum_{n=1}^{\infty} \frac{1}{n(6 n+1)}=6-\frac{\sqrt{3} \pi}{2}-\frac{3 \ln 3}{2}-2 \ln 2 \qquad \tag{4}$$
The (1) by partial fractions
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{2}{2n+1}$$
$$=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}$$
Recall the Digamma function
$$\psi(x+1)=\gamma+\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+x}$$
Therefore
$$\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}=\psi(1+\frac{1}{2})-\gamma$$
$$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\psi\left(\frac{3}{2}\right)-\gamma$$
In the same token we can derive the relation for the other three ralations. My Question is: can we calculate the values of the digamma function for those values without resorting in the Gauss´s Digamma formula?
$$\psi\left(\frac{r}{m}\right)=-\gamma-\ln (2 m)-\frac{\pi}{2} \cot \left(\frac{r \pi}{m}\right)+2 \sum_{n=1}^{\left\lfloor\frac{m-1}{2}\right\rfloor} \cos \left(\frac{2 \pi n r}{m}\right) \ln \sin \left(\frac{\pi n}{m}\right)$$
I tried this approach also, but I think the resulting integral is divergent $$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1}x^{2n}dx=\int_{0}^{1}\sum_{n=1}^{\infty} \frac{x^{2n}}{n}=-\int_{0}^{1}\ln(1-x^2)dx $$
| The correct answer should be $2-2\log 2$. Note that
$$
\sum\limits_{n = 1}^\infty {\frac{1}{{n(2n + 1)}}} = \mathop {\lim }\limits_{N \to + \infty } \left( {\sum\limits_{n = 1}^N {\frac{1}{n}} - 2\sum\limits_{n = 1}^N {\frac{1}{{2n + 1}}} } \right).
$$
Now, in terms of the harmonic numbers,
\begin{align*}
\sum\limits_{n = 1}^N {\frac{1}{n}} - 2\sum\limits_{n = 1}^N {\frac{1}{{2n + 1}}} & = \sum\limits_{n = 1}^N {\frac{1}{n}} - 2\left( {-1+\sum\limits_{n = 1}^{2N + 1} {\frac{1}{n}} - \sum\limits_{n = 1}^N {\frac{1}{{2n}}} } \right) \\ & =2+ 2\sum\limits_{n = 1}^N {\frac{1}{n}} - 2\sum\limits_{n = 1}^{2N + 1} {\frac{1}{n}} =2+ 2H_N - 2H_{2N + 1} .
\end{align*}
To finish the proof, use the asymptotics $H_k = \log k + \gamma + o(1)$ as $k\to +\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
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What is the locus of midpoints of the chords of contact of$ x^2+y^2=a^2$ from the points on the $\ell x + my + n = 0$ What is the locus of midpoints of the chords of contact of $ x^2+y^2=a^2$ from the points on the $\ell x + my + n = 0$
My approach is as follow
$\ell x + my + n = 0 \Rightarrow my = - \ell x - n \Rightarrow y = - \frac{{\ell x}}{m} - \frac{n}{m}$
Let us represent the tangent by
$y = px + q \Rightarrow p = - \frac{\ell }{m};q = - \frac{n}{m}$
${x^2} + {y^2} = {a^2} \Rightarrow {x^2} + {\left( {px + q} \right)^2} = {a^2} \Rightarrow {x^2} + {p^2}{x^2} + {q^2} + 2xpq - {a^2} = 0$
$ \Rightarrow \left( {1 + {p^2}} \right){x^2} + 2xpq + {q^2} - {a^2} = 0 \Rightarrow {x^2} + \frac{{2xpq}}{{1 + {p^2}}} + \frac{{{q^2} - {a^2}}}{{1 + {p^2}}} = 0$
$h = - \frac{{pq}}{{1 + {p^2}}} = - \frac{{\frac{{\ell n}}{{{m^2}}}}}{{1 + \frac{{{\ell ^2}}}{{{m^2}}}}} = - \frac{{\ell n}}{{{m^2} + {\ell ^2}}}$, where $h$ represent the abscissa of the mid-point
${x^2} + {y^2} = {a^2} \Rightarrow {\left( {\frac{{y - q}}{p}} \right)^2} + {y^2} = {a^2}$
$ \Rightarrow {\left( {y - q} \right)^2} + {p^2}{y^2} = {p^2}{a^2} \Rightarrow {y^2} + {q^2} - 2qy + {p^2}{y^2} = {p^2}{a^2}$
$ \Rightarrow \left( {1 + {p^2}} \right){y^2} - 2qy + {q^2} - {p^2}{a^2} = 0 \Rightarrow {y^2} - \frac{{2qy}}{{1 + {p^2}}} + \frac{{{q^2} - {p^2}{a^2}}}{{1 + {p^2}}} = 0$
$k = \frac{q}{{1 + {p^2}}} = - \frac{{\frac{n}{m}}}{{1 + \frac{{{\ell ^2}}}{{{m^2}}}}} = - \frac{{mn}}{{{m^2} + {\ell ^2}}}\& h = - \frac{{\ell n}}{{{m^2} + {\ell ^2}}}$ where $k$ represent the ordinate of the mid point
$ \Rightarrow \frac{k}{n} = - \frac{m}{{{m^2} + {\ell ^2}}}\& \frac{h}{n} = - \frac{\ell }{{{m^2} + {\ell ^2}}}$
$ \Rightarrow \frac{k}{n} = - \frac{m}{{{m^2} + {\ell ^2}}}\& \frac{h}{n} = - \frac{\ell }{{{m^2} + {\ell ^2}}}$
$\frac{{{h^2}}}{{{n^2}}} + \frac{{{k^2}}}{{{n^2}}} = \frac{{{\ell ^2} + {m^2}}}{{{{\left( {{m^2} + {\ell ^2}} \right)}^2}}} \Rightarrow {h^2} + {k^2} = \frac{{{n^2}}}{{{m^2} + {\ell ^2}}}$
Not able to approach from here , the term $a^2$ is lost in calculation
| If the tangents are drawn from the point $(\alpha,\beta)$ to the curve $S=0$ then the equation of chord of contact is $T=0\implies\alpha x+\beta y=a^2$
Also, if $(h,k)$ is the mid-point of the chord of contact then the equation of this chord is $T=S_1\implies hx+ky=h^2+k^2$
Comparing the two equations, we get $\dfrac{\alpha}h=\dfrac{\beta}k=\dfrac{a^2}{h^2+k^2}\implies\alpha=\dfrac{ha^2}{h^2+k^2}, \beta=\dfrac{ka^2}{h^2+k^2}$
Putting these values of $(\alpha,\beta)$ in $lx+my+n=0$, we get $\dfrac{hla^2}{h^2+k^2}+\dfrac{kma^2}{h^2+k^2}+n=0$
So, the required locus is: $la^2x+ma^2y+n(x^2+y^2)=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding which complex numbers satisfy $\left|z + 1\right| + \left|z - 1\right| = 4$ This problem is from the book Complex Analysis with Applications by Asmar. Currently I am stuck at, well, everything, as I don't know know how to manipulate the expression $\left|z + 1\right| + \left|z - 1\right| = 4$ into anything meaningful. Squaring and simplifying both sides yields $z\bar{z} + \left|z + 1\right|\left|z - 1\right| + 1 = 8$, and I don't see how that is useful.
The problem is asking us to find those complex numbers whose sum of distances from both $1$ and $-1$ is equal to $4$. I suppose the end result is a circle of some kind, as points $(2, 0)$ and $(-2, 0)$ satisfy the equality.
| If you write $z$ as $a+bi$ with $a,b\in\Bbb R$, then\begin{align}|z+1|+|z-1|=4&\iff\sqrt{(a+1)^2+b^2}+\sqrt{(a-1)^2+b^2}=4\\&\iff2a^2+2b^2+2+2\sqrt{(a+1)^2+b^2}\sqrt{(a-1)^2+b^2}=16\\&\iff\sqrt{(a+1)^2+b^2}\sqrt{(a-1)^2+b^2}=7-a^2-b^2\\&\iff\bigl((a+1)^2+b^2\bigr)\bigl((a-1)^2+b^2\bigr)=(7-a^2-b^2)^2\\&\iff3a^2+4b^2=12.\end{align}So, you get the ellipse $3a^2+4b^2=12$.
However, note that the equivalence\begin{multline}\sqrt{(a+1)^2+b^2}\sqrt{(a-1)^2+b^2}=7-a^2-b^2\iff\\\iff\bigl((a+1)^2+b^2\bigr)\bigl((a-1)^2+b^2\bigr)=(7-a^2-b^2)^2\end{multline}is not obvious, but is not hard to prove that $|z-1|+|z+1|=4\implies|z|^2\leqslant7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Binomial coefficients identity with cases Let $i,j,k$ be non-negative integers such that $i$ is even, $j \leq \frac{i}{2}$, and $k < i$. I would like to show the following identity:
$$\binom{i-j+k}{k} - \sum_{\ell = 0}^{\lfloor\frac{k}{2}\rfloor} \frac{i-2j+k}{\frac{i}{2}-j+k-\ell} \binom{\frac{i}{2}+\ell}{2\ell} \binom{\frac{i}{2}-j+k-\ell}{k-2\ell} =
\begin{cases} 0 & \text{if }j < k \\ (-1)^{k+1}\binom{j}{k} & \text{if } j \geq k \end{cases}.$$
We can change the summation into
$$\sum_{\ell=0}^{\lfloor\frac{k}{2}\rfloor}\binom{\frac{i}{2} + \ell}{2\ell}\left(\binom{\frac{i}{2}-j+k-\ell}{k-2\ell} + \binom{\frac{i}{2}-j+k-\ell-1}{k-2\ell} \right)$$
but I don't know of any identities that could help here. I've also tried induction on $k$, but couldn't see a good way for the induction hypothesis to be used. I would appreciate any ideas!
| In seeking to evaluate
$${q-j+k\choose k}
- \sum_{\ell=0}^{\lfloor k/2 \rfloor}
{q/2+\ell\choose 2\ell}
\left({q/2-j+k-\ell\choose k-2\ell}
+ {q/2-j+k-\ell-1\choose k-2\ell} \right)$$
We get for the first piece of the sum
$$\sum_{\ell=0}^{\lfloor k/2 \rfloor}
{q/2+\ell\choose 2\ell} {q/2-j+k-\ell\choose k-2\ell}
\\ = \frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{1}{z^{k+1}}
(1+z)^{q/2-j+k}
\sum_{\ell=0}^{\lfloor k/2 \rfloor}
{q/2+\ell\choose 2\ell} \frac{z^{2\ell}}{(1+z)^\ell}
\; dz.$$
Now here the residue vanishes when $2\ell \gt k$ so it enforces the
upper limit of the sum and we obtain
$$\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{(1+z)^{q/2-j+k}}{z^{k+1}}
\\ \times \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{q/2}}{w}
\sum_{\ell\ge 0}
\frac{z^{2\ell}}{(1+z)^\ell} \frac{(1+w)^\ell}{w^{2\ell}}
\; dw \; dz
\\ = \frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{(1+z)^{q/2-j+k}}{z^{k+1}}
\\ \times \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{q/2}}{w}
\frac{1}{1-z^2(1+w)/(1+z)/w^2}
\; dw \; dz
\\ = \frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{(1+z)^{q/2-j+k+1}}{z^{k+1}}
\\ \times \frac{1}{2\pi i} \int_{|w|=\gamma} (1+w)^{q/2}
\frac{w}{(w-z)(w(1+z)+z)}
\; dw \; dz.$$
The pole at $w=0$ has been canceled. Now observe that for the geometric
series to converge we must have $$|z^2(1+w)/w^2/(1+z)|\lt 1.$$ We will
choose a contour that includes both simple poles. The first pole is at
$-z/(1+z).$ We thus require $|z/(1+z)| \lt \gamma.$ With $|z/(1+z)| \le
\varepsilon/(1-\varepsilon)$ we get $\varepsilon/(1-\varepsilon) \lt
\gamma$ and we furthermore need $|z^2/(1+z)| \lt |w^2/(1+w)|.$ The latter
holds if $\varepsilon^2 / (1-\varepsilon) \lt \gamma^2/(1+\gamma).$ Both
hold if $\varepsilon \gamma \lt \gamma^2/(1+\gamma)$ or $\varepsilon \lt
\gamma/(1+\gamma).$ So $\varepsilon = \gamma^2/(1+\gamma)$ will work.
Observe that this contour also includes the pole at $w=z.$
First pole. Now to extract the residue at $w=-z/(1+z)$ we write
$$\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{(1+z)^{q/2-j+k}}{z^{k+1}}
\\ \times \frac{1}{2\pi i} \int_{|w|=\gamma} (1+w)^{q/2}
\frac{w}{(w-z)(w+z/(1+z))}
\; dw \; dz$$
and obtain
$$\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{(1+z)^{q/2-j+k}}{z^{k+1}}
(1+z)^{-q/2} \frac{-z/(1+z)}{-z/(1+z)-z} \; dz
\\ = \frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{(1+z)^{k-j}}{z^{k+1}} \frac{1}{z+2} \; dz.$$
Repeating for the second sum we get
$$\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{(1+z)^{k-j-1}}{z^{k+1}} \frac{1}{z+2} \; dz.$$
Adding the two we find
$$\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{(1+z)^{k-j-1} (1+(1+z))}{z^{k+1}} \frac{1}{z+2} \; dz
= {k-j-1\choose k}.$$
Second pole. For the residue at $w=z$ we obtain for the first
sum
$$\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{(1+z)^{q/2-j+k+1}}{z^{k+1}}
(1+z)^{q/2}
\frac{z}{(z(1+z)+z)}
\; dz
\\ = \frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{(1+z)^{q-j+k+1}}{z^{k+1}}
\frac{1}{z+2}
\; dz.$$
Repeating for the second sum we get
$$\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{(1+z)^{q-j+k}}{z^{k+1}}
\frac{1}{z+2}
\; dz.$$
Adding the two we find
$$\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{(1+z)^{q-j+k}(1+(1+z))}{z^{k+1}}
\frac{1}{z+2}
\; dz = {q-j+k\choose k}.$$
Conclusion. Collecting everything we obtain
$${q-j+k\choose k} - {q-j+k\choose k}
- {k-j-1\choose k}.$$
This is $- (k-j-1)^{\underline{k}}/k!.$ Now if $0\le j\lt k$ this is
indeed zero because the falling factorial hits the zero value. If $j\ge
k$ all $k$ terms are negative and we get $-(-j)^{\overline{k}}/k!.$
We have at last
$$\bbox[5px,border:2px solid #00A000]{
(-1)^{k+1} {j\choose k}.}$$
as claimed.
Remark. The potential square roots that appeared in the above all
use the principal branch of the logarithm with branch cut $(-\infty,
-1]$ which means everything is analytic in a neighborhood of zero as
required.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Work and Time Problem : Fill Pipes and Drain Pipes : 2
A fill pipe can fill a tank in $20$ hours, a drain pipe can drain a
tank in $30$ hours. If a system of $n$ pipes (fill pipes and drain
pipes put together) can fill the tank in exactly $5$ hours, which of
the following are possible values of $n$? Options are $32,54,29 \text{ and } 40.$
My $1st$ approach : -
Let the number of fill pipes be $a$ and hence the number of drain pipes will be $n-a$. As per the question ;
$\frac{a}{20} - \frac{n-a}{30}= \frac{1}{5}$
$\Rightarrow n= \frac{5a-12}{2}$ and we can say from the above equation that $n$ should be a multiple of $2$ and hence the possible values of $n$ can be $32,54 \text{ and } 40.$ But to my shock this is the wrong answer.
My $2nd$ approach : -
Let the number of fill pipes be $n-a$ and hence the number of drain pipes will be $a$ (Opposite of what I did in my previous approach). As per the question ;
$\frac{n-a}{20} - \frac{a}{30}= \frac{1}{5}$
$\Rightarrow n= \frac{5a+12}{3}$ and we can say from the above equation that $n$ should be a multiple of $3$ and hence the possible values of $n$ can only be $54.$ But to my shock this is also the wrong answer.
Considering both the cases the common answer should be the answer and hence the option should be 54, right?
Why there is conflict in answers using the two above approach of mine? What I am doing wrong? Please help me with this. How can these types of questions be solved easily?
Thanks in advance !
| Suppose there are $a$ fill pipes and $b$ drain pipes; then $\frac a{20} - \frac b{30} = \frac15$ gives $3a - 2b = 12$.
We now have to be careful. If we get an expression like $a = \frac{12+2b}{3}$, that doesn't tell us $a$ is a multiple of $3$. It tells us that $12+2b$ is a multiple of $3$ (otherwise, $\frac{12+2b}{3}$ will not be an integer).
I prefer to think of it a different way. In the expression $3a-2b=12$, $3a$ and $12$ are both divisible by $3$, so the remaining term $2b$ must also be divisible by $3$, which means $b$ is divisible by $3$. Also, $2b$ and $12$ are both divisible by $2$, so the remaining term $3a$ must also be divisible by $2$, which means $a$ is divisible by $2$.
Anyway, now we can set $a=2x$ and $b=3y$ to get $6x- 6y=12$, or $x-y=2$.
All solutions $(x,y)$ are of the form $(y+2,y)$ where $y$ is a nonnegative integer, and so all solutions $(a,b)$ are of the form $(2y+4, 3y)$ where $y$ is a nonnegative integer. The total number of pipes is $5y+4$, which tells you the restriction on $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Optimization problem Kuhn Tucker i'm stuck on this problem because i don't know what to do next
$$\begin{cases} \text{maximize } &f(x,y) = -(x-2)^2-(y-1)^2 \\ \text{subject to } &g_1(x,y) = -x-y+1 \leq0,\\&g_2(x,y) = x^2+y^2 \leq 1 \\ \end{cases}$$
Since we are in a compact set we can use Kuhn Tucker conditions :
$\lambda_i \geqslant 0$
$L(x,y,\lambda_1, \lambda_2) = -(x-2)^2-(y-1)^2-\lambda_1(-x-y+1 )-\lambda_2(x^2+y^2 -1) $
Let's start with $\lambda_i \neq 0$
\begin{align} \ \frac{dL}{dx} &= -2(x-2)+\lambda_1x-\lambda_22x = x(-2+ \lambda_1-2\lambda_2 ) + 4 = 0\\ \frac{dL}{dy} &= -2(y-1) + \lambda_1y-2\lambda_2y = y(-2+\lambda_1-2\lambda_2) +2 = 0 \\ \frac{dL}{d\lambda_1} &= -x-y+1 = 0 \\ \frac{dL}{d\lambda_2} &= x^2+y^2-1=0 \end{align}
I found this $ x = \frac{-4}{-2+\lambda_1-2\lambda_2}$ $y = \frac{-2}{-2+\lambda_1-2\lambda_2}$
I don't know what to do, any help will be appreciated
| Consider for your Lagrangian instead with $f = -(x-2)^2-(y-1)^2$
$$
L(x,y,\lambda_1, \lambda_2,s_1,s_2) = f+\lambda_1(-x-y+1+s_1^2 )+\lambda_2(x^2+y^2 -1+s_2^2)
$$
Here $s_1, s_2$ are slack variables to transform the inequalities into equations.
The stationary points are the solutions for
$$
\nabla L = \left\{
\begin{array}{l}
\lambda_1=2 (\lambda_2-1) x+4 \\
\lambda_1=2 (\lambda_2-1) y+2 \\
\lambda_1 s_1=0 \\
\lambda_2 s_2=0 \\
s_1^2+1=x+y \\
s_2^2+x^2+y^2=1 \\
\end{array}
\right.
$$
so we obtain
$$
\left(
\begin{array}{ccccccc}
f & x & y & \lambda_1 & \lambda_2& s_1^2& s_2^2\\
-4 & 0 & 1 & 4 & 2 & 0 & 0 \\
-2 & 1 & 0 & 2 & 0 & 0 & 0 \\
-2 \left(3-\sqrt{5}\right) & \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 0 & 1-\sqrt{5} & \frac{3}{\sqrt{5}}-1 & 0 \\
\end{array}
\right)
$$
Here $s_k = 0$ means that the corresponding restriction is actuating. The stationary point with the highest value is represented in the last row. We can verify that the second restriction is actuating. Follows a plot showing in black the level surfaces for $f(x,y)$ in black the $\nabla f$ at each stationary point and in red $\nabla g_1, \nabla g_2$ for each restriction. As we can observe, the stationary points located at the chord extrema, are not feasible maxima because doesn't verify $\nabla f = \mu_1\nabla g_1+\mu_2 \nabla g_2$ with $\mu_1 \ge 0, \mu_2 \ge 0$. The stationary point located over the sector circumference is a feasible maximum because there we have $\nabla f = \mu_2\nabla g_2$ with $\mu_2 > 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Eigenvectors - unknown variables Considering that $v_1$ and $v_2$ are eigenvectors, find the values of $a,b,c,d$
:
$$
A = \begin{bmatrix} 1 & a & b \\1 & c & d\\
1 & 1 & 1\end{bmatrix}\qquad v_1 = \begin{bmatrix} 1 \\1 \\
1 \end{bmatrix}\qquad v_2 = \begin{bmatrix} 1 \\0 \\
-1 \end{bmatrix}
$$
I am honestly quite confused as to how I can solve this. I tried to find det but I found this:
$c-d-a(1-d)+b(1-c)$
which is not very useful
| $Av_1=\lambda_1v_1\Rightarrow \begin{pmatrix}1+a+b\\1+c+d\\3\end{pmatrix}=\lambda_1\begin{pmatrix}1\\1\\1\end{pmatrix}$ and $Av_2=\lambda_2v_2\Rightarrow \begin{pmatrix}1-b\\1-d\\0\end{pmatrix}=\lambda_2\begin{pmatrix}1\\0\\-1\end{pmatrix}$
Then $\lambda_2=0, b,d=1$. Now we can find $a \text{ and }c$ too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use Implicit Differentiation to find $\frac{d^2y}{dx^2}$? Given a system of equation,
\begin{align*}
x &= t^2 + 2t \\
y &= 3t^4 + 4t^3
\end{align*}
I want to find $\frac{d^2 y}{dx^2}$ at $(x,y) = (8, 80)$. Then, $\partial_x(y) = \frac{d y}{dt} \frac{dt}{dx}$. By chain rule,
\begin{align*}
\partial_x^2(y) &= \partial_x \left(\frac{d y}{dt}\right)\frac{dt}{dx} + \frac{dy}{dt} \partial_x \left(\frac{dt}{dx}\right) \\
&= \frac{d^2 y}{dt^2}\left(\frac{dt}{dx}\right)^2 + \frac{dy}{dt}\frac{d^2t}{dx^2}
\end{align*}
Here, how do I find $\frac{d^2 t}{dx^2}$?
|
The solution given by Deepak follows the usual way that the second derivative (parametric) function would be calculated for this curve. The way you followed creates somewhat more work, but will finally achieve the same result.
This approach is somewhat similar to that of Clarinetist, but takes advantage of a particular feature of the parametric curve functions; it also proceeds entirely by implicit differentiation. We may write $$ \ x \ = \ t^2 + 2t \ = \ (t+1)^2 - 1 \ \ , \ \ y \ = \ 3t^4 + 4t^3 \ = \ t^3 · (3t+4) \ = \ t^3 · [ (3t+3) + 1] \ \ . $$ We will then substitute $ \ u = t+1 \ \ ; $ consequently,
$$ x + 1 \ = \ u^2 \ \ \Rightarrow \ \ \frac{d}{dx} \ [x + 1] \ = \ \frac{d}{dx} \ [u^2] \ \ \Rightarrow \ \ 1 \ = \ 2u \ \frac{du}{dx} $$
$$ \Rightarrow \ \ \frac{d}{dx} \ [1] \ = \ \frac{d}{dx} \left[2u \ \frac{du}{dx} \right] \ \ \Rightarrow \ \ 0 \ = \ 2·\left(\frac{du}{dx} \right)^2 \ + \ 2u· \frac{d^2u}{dx^2} \ \ \Rightarrow \ \ \frac{d^2u}{dx^2} \ = \ -\frac{1}{u}·\left(\frac{du}{dx} \right)^2 \ \ . $$
For the other curve equation, we have
$$ y \ = \ t^3 · [ 3(t+1) + 1] \ \rightarrow \ (u-1)^3 · ( 3u + 1) $$
$$ \Rightarrow \ \ \frac{dy}{du} \ = \ 3·(u-1)^2 · ( 3u + 1) \ + \ (u-1)^3 · 3 \ = \ 3·(u-1)^2 · [(3u+1) + (u-1)] $$ $$ = \ 12u·(u-1)^2 \ \ . $$
It follows that $$ \ \ \frac{dy}{dx} \ = \ \frac{dy}{du} · \frac{du}{dx} \ = \ 12u·(u-1)^2 \ · \ \frac{1}{2u} \ = \ 6·(u-1)^2 \ \rightarrow \ 6t^2 \ \ . $$
[We will remark upon this later.]
For the second derivative functions, we obtain
$$ \frac{d^2y}{du^2} \ = \ \frac{d}{du} \ [ 12u·(u-1)^2 ] \ = \ 12 ·(u-1)^2 \ + \ 12u · 2 \ (u-1) $$ $$ = \ 12 ·(u-1) · [(u-1) + 2u] \ = \ 12 ·(u-1) · (3u-1) \ \ ; $$
$$ \frac{d^2y}{dx^2} \ = \ \frac{d}{dx} \ \left[\frac{dy}{dx} \right] \ = \ \frac{d}{dx} \ \left[\frac{dy}{du} · \frac{du}{dx} \right] \ = \ \left(\frac{d}{du} \ \left[\frac{dy}{du} \right] · \frac{du}{dx} \right) · \frac{du}{dx} \ + \ \frac{dy}{du} \ · \frac{d}{dx} \ \left[ \frac{du}{dx} \right] $$
$$ = \ \frac{d^2y}{du^2} · \left(\frac{du}{dx} \right)^2 \ + \ \frac{dy}{du} \ · \frac{d^2u}{dx^2} $$
[similar to your result, since $ \ \frac{du}{dx} = \frac{dt}{dx} ] $
(bringing in the expressions found above)
$$ = \ \frac{d^2y}{du^2} · \left(\frac{du}{dx} \right)^2 \ + \ \frac{dy}{du} \ · \left[-\frac{1}{u}·\left(\frac{du}{dx} \right)^2 \right] \ = \ \left[\frac{d^2y}{du^2} \ - \ \frac{1}{u} · \frac{dy}{du} \right] · \left(\frac{du}{dx} \right)^2 $$
$$ = \ \left[12·(u-1)·(3u-1) \ - \ \frac{1}{u} · 12u·(u-1)^2 \right] · \left(\frac{1}{2u} \right)^2 $$
$$ = \ 12·(u-1) \ · \ [ (3u-1) \ - \ (u-1) ] · \left(\frac{1}{2u} \right)^2 \ = \ 12·(u-1) · 2u · \left(\frac{1}{2u} \right)^2 $$
$$ = \ \frac{6·(u-1)}{u} \ \ \rightarrow \ \ \frac{6t}{t \ + \ 1} \ \ . $$
So this may not be the quickest route to the desired second derivative, but it is effective.
In the graphs at the start of this post, we see that this is a very narrow "hairpin" curve, with two "arms" almost entirely in the first quadrant, with a "turn" close to the origin [shown at lower right]. A curious feature about the curve (which is what interested me in this problem) is that the first derivative parametric function, $ \ 6t^2 \ , $ betrays no hint that there is something peculiar about the turn; the function is defined everywhere and is continuous, and the slope of the curve is always positive, except where it is zero at the origin. However, the second derivative function is not defined at $ \ t = -1 \ $ (the point $ \ (-1,-1) \ ) , $ changes discontinuously from positive for $ \ t < -1 \ $ to negative for $ \ -1 < t < 0 \ , $ then reaches an inflection point at the origin $ \ (t=0) \ $ and returns to "upward" concavity for $ \ t > 0 \ \ . $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$f(f(x))=a^3\left(x^2-(2+b)x+2b-\frac2a\right)\left(x^2-(2+b)x+2b-\frac ba\right)$, $a\ne0$ has exactly one real zeroes $5$.
Let $f(x)=a(x-2)(x-b)$, where $a,b\in R$ and $a\ne0$. Also, $f(f(x))=a^3\left(x^2-(2+b)x+2b-\frac2a\right)\left(x^2-(2+b)x+2b-\frac ba\right)$, $a\ne0$ has exactly one real zeroes $5$. Find the minima and maxima of $f(x)$.
$f(f(x))$ is a quartic equation. So, it would have $4$ roots. Its real root should occur in pair because complex roots occur in pair. So, I failed to understand the meaning of 'exactly one real zero' in the question. Does that mean its real roots are equal and they are equal to $5$?
Since $f(x)$ has a zero at $2$, $f(f(2))=f(0)=2ab$. But even RHS of $f(f(x))$ is coming out to be $2ab$ at $x=2$.
Putting $x=5$ in $f(f(x))$ gives an equation in $a^2$ and $b^2$. Don't know what to do with it.
Also, $f(x)=a(x^2-(2+b)x+2b)$
$f'(x)=a(2x-2-b)=0\implies x=\frac{2+b}2$ is the minima or maxima.
How to proceed next? Do the critical points of $f(x)$ tell us anything about $f(f(x))$?
| From the given form of $f(f(x))$,
$$\begin{align*}
f(f(x)) &= a^3 \left[x^2 - (2+b)x + 2b - \frac 2a\right]\left[x^2 - (2+b)x + 2b - \frac ba\right]\\
&= a^3 \left[(x-2)(x-b) - \frac 2a\right]\left[(x-2)(x-b) - \frac ba\right]
\end{align*}$$
$(x-2)(x-b)$ has roots at $x=2$ and $x=b$. If we ignore the $-\frac2a$ and $-\frac ba$, then $f(f(2)) = f(f(b)) = 0$.
But if $\frac2a$ and $\frac ba$ are just negative enough, then these two factors of $f(f(x))$:
$$\left[(x-2)(x-b) - \frac 2a\right]\left[(x-2)(x-b) - \frac ba\right]$$
One of these factors would be strictly positive, and the other factor would have a double zero at the $x$ that minimises $(x-2)(x-b)$. Then overall, the product would have just one distinct real zero.
At this point, the requirement that $f(f(5))=0$ becomes useful. Consider that if $x=5$ minimises $(x-2)(x-b)$, then $x = \frac{2+b}{2} = 5$ and so $b=8$.
Then choose $a$ (which would be negative) so that within the product
$$\left[(x-2)(x-8) - \frac 2a\right]\left[(x-2)(x-8) - \frac 8a\right],$$
the minimum of the first factor is $0$ at $x=5$. The second factor would be strictly positive.
$$\begin{align*}
(5-2)(5-8) - \frac 2a &= 0\\
a &= -\frac 29\\
\end{align*}$$
To double check:
$$\begin{align*}
f(f(x)) &= a^3\left[(x-2)(x-8) - \frac 2a\right]\left[(x-2)(x-8) - \frac 8a\right]\\
&= a^3 [(x-2)(x-8) + 9][(x-2)(x-8) + 36]\\
&= a^3[(x-5+3)(x-5-3)+9][(x-5+3)(x-5-3)+36]\\
&= a^3 [(x-5)^2 - 3^2+9][(x-5)^2 - 3^2 + 36]\\
&= a^3 (x-5)^2 [(x-5)^2 +27]
\end{align*}$$
And back to the question,
$$\begin{align*}
f(x) &= -\frac 29(x-2)(x-8)\\
f(5) &= -\frac 29(3)(-3) = 2
\end{align*}$$
and the $2$ is a global maximum. This makes sense because
$$f(f(5)) = f(2) = 0,$$
and $f(x)$ is never equal to $8$ for real $x$, otherwise those real $x$ would add zeroes to $f(f(x)) = f(8)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $f(x)=\frac{1}{x}$ is continuous on the interval $(0,1)$
Prove $f(x)=\frac{1}{x}$ is continuous on the interval $(0,1)$.
My attempt:
$$
\left|{\frac{1}{x}-\frac{1}{c}}\right|<\epsilon\\
\left|{\frac{c-x}{xc}}\right|<\epsilon\\
\frac{\left|c-x\right|}{xc}<\epsilon \qquad\text{as $x$ and $c$ are both positive}
$$
Suppose $|x-c|<\frac{c}{2}$ so that $\frac{c}{2}<x<\frac{3c}{2}$. Then we could obtain $\frac{1}{x}<\frac{2}{c}$
$$
\frac{1}{x}<\frac{2}{c}\implies \frac{1}{xc}<\frac{2}{c^2}\implies \frac{\left|c-x\right|}{xc}<\frac{2}{c^2}|c-x|<\epsilon.
$$
By observation, setting $|c-x|<\frac{c^2\epsilon}{2}=\delta$.
Is this proof looks right to you? Any comment and help are much appreciated!
| It is correct for the most part. It would seem that you proved that$$|x-c|<\frac{c^2\varepsilon}2\implies\left|\frac1x-\frac1c\right|<\varepsilon,\tag1$$and that therefore $\delta=\frac{c^2\varepsilon}2$. However, you only proved $(1)$ under the assumption that $|x-c|<\frac c2$ and therefore$$\delta=\min\left\{\frac{c^2\varepsilon}2,\frac c2\right\}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The quadratic equation whose roots are $\sec^2\theta$ and $\csc^2\theta$ can be
The quadratic equation whose roots are $\sec^2\theta$ and $\csc^2\theta$ can be
*
*A) $x^2-2x+2=0$
*B) $x^2-5x+5=0$
*C) $x^2-7x+7=0$
*D) $x^2-9x+9=0$
Method $1$:$$\sec^2\theta+\csc^2\theta=\frac1{\cos^2\theta}+\frac1{\sin^2\theta}\\=\frac1{\sin^2\theta\cos^2\theta}\\=\frac4{\sin^22\theta}\ge4$$
Also, $\sec^2\theta\csc^2\theta=\dfrac1{\sin^2\theta\cos^2\theta}$
So, options $B),C),D)$ are correct.
Method $2$: Let the quadratic equation be $x^2-px+q=0$
So, $\sec^2\theta+\csc^2\theta=p, \sec^2\theta\csc^2\theta=q\implies \csc^2\theta=\dfrac{q}{\sec^2\theta}$
Putting that in the sum of roots, we get $$\sec^2\theta+\frac{q}{\sec^2\theta}=p\\\implies\sec^4\theta-p\sec^2\theta+q=0\\\implies\sec^2\theta=\frac{p\pm\sqrt{p^2-4q}}2\ge1\\\implies p\pm\sqrt{p^2-4q}\ge2\\\implies\pm\sqrt{p^2-4q}\ge2-p\\\implies p^2-4q\ge4+p^2-4p\\\implies p-q\ge1$$
What's wrong in this method?
| As others have pointed out, you’ve assume that if $u\geq v$ then $u^2\geq v^2.$
This is true if $u,v$ are both non-negative, but if both are non-positive, then the inequality reverses, $u^2\leq v^2,$ and there is nothing we can say if $u$ is positive and $v$ negative.
Another way to write the equation is as $$((\cos^2 \theta) x-1)((\sin^2\theta) x-1)=(\cos^2\theta\sin^2 \theta) x^2-x+1$$
So the monic polynomial with these two roots has $$p=q=\frac1{\sin^2\theta\cos^2\theta}=\frac{4}{\sin^22\theta}$$
This is really the same as your first approach, but switches to $\sin$ and $\cos$ immediately.
| {
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"timestamp": "2023-03-29T00:00:00",
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An ellipse with major axis $4$ and minor axis $2$ touches both the coordinate axes. Locus of its Center and Focus is? An ellipse with major axis $4$ and minor axis $2$ touches both the coordinate axes. Locus of its Center and Focus is?
My Approach: For locus of Center.
Since it is touching the coordinate axes, the coordinate axes will act as tangents making angle of $90^{\circ}$, so origin will lies on Director Circle.
Center of director circle will be same as center of Ellipse.
Let center of Ellipse be $(h,k)$, so the equation of director circle will be $$(x-h)^2+(y-k)^2=(\text{semi major axis})^2+ (\text{semi minor axis})^2,$$
$$i.e.\;\;\;(x-h)^2+(y-k)^2=(2)^2+ (1)^2.$$
Because it passes through the origin,
$(0-h)^2+(0-k)^2=(2)^2+ (1)^2$
$\implies $ $h^2+k^2=5$
$\implies $ locus of center of Ellipse is $x^2+y^2=5$.
For Locus of Focus:
I assumed focus as $(x_1,y_1)$ but i cannot proceed further. I know one property, that product of distance of tangents from Foci is constant and equal to square of semi-minor axis and lies on auxiliary circle, but that leads me nowhere.
Note There are many solutions available on internet, but they all did this with the same method of taking axis of ellipse parallel to coordinate axis.
I don't want to do using that method. I want it for slanted ellipse as shown in attached image.
| First, we try to determine the general form of a rotated ellipse in the first quadrant such that it remains tangent to the axes. We suppose it has the parametric equation
$$(x(t),y(t)) = (4 \cos u \cos t - 2 \sin u \sin t + h, 2 \cos u \sin t + 4 \sin u \cos t + k), \quad t \in [0,2\pi)$$ where $(h,k)$ is the center, and $u$ is the counterclockwise rotation angle of the ellipse relative to the coordinate axes. (Note I have modified the direction of rotation compared to the linked answer.)
Such an ellipse has horizontal tangent lines satisfying $$0 = \frac{dy}{dt} = 2 \cos u \cos t - 4 \sin u \sin t,$$ or $$t_{\text{crit}} = \arctan \frac{\cot u}{2}.$$ For these values of $t_{\text{crit}}$, we need to find $k$ such that $y(t_{\text{crit}}) = 0$, placing this ellipse so it is tangent to the $x$-axis; i.e, $$k = 2 \sqrt{\cos^2 u + 4 \sin^2 u}.$$ This gives, as a function of the angle of rotation $u$, the necessary vertical translation to make the ellipse tangent. A similar process using $dx/dt$ gives the necessary horizontal translation, which we show without proof: $$h = 2 \sqrt{4 \cos^2 u + \sin^2 u}.$$ Thus our ellipse is fully parametrized.
The locus of the center is simply $(h,k)$ as a function of $u$:
$$(h(u), k(u)) = \left(2 \sqrt{4 \cos^2 u + \sin^2 u}, 2 \sqrt{\cos^2 u + 4 \sin^2 u}\right).$$
A short computation of $h^2 + k^2$ shows that this locus is an arc of a circle, not the complete circle.
Where are the foci? We can first observe that they are located at some point along the line joining $(x(0), y(0))$ and $(x(\pi), y(\pi))$, so they have coordinates of the form $$(1-\lambda)(x(0), y(0)) + \lambda (x(\pi), y(\pi))$$ for $$\lambda = \frac{4 + 2 \sqrt{3}}{8} = \frac{2 + \sqrt{3}}{4}, \quad \text{and} \quad \lambda = \frac{2 - \sqrt{3}}{4}.$$ We again skip the calculation and show the result: $$(x_f(u), y_f(u)) = 2 \left(\sqrt{3} \cos u + \sqrt{4 \cos^2 u + \sin^2 u}, \sqrt{3} \sin u + \sqrt{\cos^2 u + 4 \sin^2 u} \right). \tag{1}$$ Note that this curve gives the locus of both foci, where $u$ and $u + \pi$ represent the location of each focus for a given rotation angle $u$.
All put together, we can visualize these loci in the following animation:
The conversion of the parametric formula to an implicit curve is tedious but not intractable; one would start with computing the square, then show that the square of the locus satisfies $$(x + y)(16 + xy) = 64 xy.$$
One approach to converting the locus is to note that we can write $$(x_f(u), y_f(u)) = \left(2 \sqrt{3} \cos u + \sqrt{(2 \sqrt{3} \cos u)^2 + 4}, 2 \sqrt{3} \sin u + \sqrt{(2 \sqrt{3} \sin u)^2 + 4} \right),$$ therefore $x_f(u)$ and $y_f(u)$ are roots of the quadratics $$x^2 - (4 \sqrt{3} \cos u) x - 4 = 0, \\ y^2 - (4 \sqrt{3} \sin u) y - 4 = 0,$$ or equivalently, $$48 \cos^2 u = \frac{(4-x^2)^2}{x^2}, \quad 48 \sin^2 u = \frac{(4-y^2)^2}{y^2}.$$ Thus $$48 = \frac{(4-x^2)^2}{x^2} + \frac{(4-y^2)^2}{y^2},$$ and the rest is an exercise in algebra.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality, finding Constant Can anyone help me with this Math Olympiad Task from 2007 from Germany? I want to find the smallest $C$, such that for every $x,y \in \mathbb{R}$ the inequality:
$$
1+(x+y)^2 \leq C(1+x^2)(1+y^2)
$$
holds.
I know that I have to maximize the function
$$a(x, y) = \frac{1+(x+y)^2}{(1+x^2)(1+y^2)}$$
which gives me $\frac{4}{3}.$
P.S.: Thanks for the help y'all :)
| Claim: We will show that $ 1 + (x+y)^2 \leq \frac{4}{3} ( 1+x^2)(1+y^2)$.
Proof: By expanding, WTS
$$ 4x^2y^2 + x^2 + y^2 + 1 \geq 6 xy. $$
This is true by applying AM-GM creatively:
$ 4x^2 y^2 + 1 \geq 4 |xy | \geq 4 xy $.
$x^2 + y^2 \geq 2 | xy | \geq 2xy$.
Equality holds iff $ 2xy = 1$, $ x = y$ and $ xy \geq 0$, which gives the solution set $ x = y = \pm \frac{1}{ \sqrt{2} } $.
This solution sets also what that $\frac{4}{3}$ is the smallest possible value of $C$.
Note:
*
*As to how one can guess the value of $C$, we use the huge wishful thinking simplification that $x=y$, and have the quadratic equation in $t = x^2$ of
$$ C t ^2 + ( 2 C - 4 )t + ( C - 1 ) \geq 0 \quad t \geq 0. $$
To satisfy this, we require
A) If $ t = - \frac{ 2C-4}{C } \geq 0 $, then $f(t) \geq 0$ $\Rightarrow$ if $ 0 < C < 2 $, then $ C \leq \frac{4}{3}$, so the solution set is $ 2 > C \geq \frac{4}{3} $.
B) Else if $ t = - \frac{ 2C-4}{C } \leq 0$, then $ f(0) \geq 0$ $\Rightarrow$ If $ t < 0$ or $ t > 2$, then $ C > 1 $. So the solution set is $ C \geq 2$.
Hence, $ C \geq \frac{4}{3}$, so the minimum value to try is $ C = \frac{4}{3}$.
Note that this might not work because the equality condition might not occur at $ x = y$.
| {
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appropriate bounds involving change of variable Been a long time since I've touched integral so a little help is appreciated!
The integral to be evaluated: $\int_{x = -\sqrt{\frac{2E}{c}}}^{x = +\sqrt{\frac{2E}{c}}} \frac{1}{\sqrt{E-\frac{cx^{2}}{2}}}.dx$
Attempt:
Let $u = E - \frac{cx^{2}}{2}$ so that $\frac{du}{dx} = -cx$ so that $dx = \frac{-1}{cx}du$.
Since the bounds must be $u$ expressed in terms of $x$, $u = E - \frac{cx^{2}}{2} = 0$ for both the lower and upper bound, after substituting $x = -\sqrt{\frac{2E}{c}}$ and $x = +\sqrt{\frac{2E}{c}}$
My bounds appears to be wrong.
| Notice that this is an even function.
Let $v=\sqrt{\frac{c}{2E}}x$, then $\frac{dv}{dx}=\sqrt{\frac{c}{2E}}$.
Hence, we have
\begin{align}
\int_{-\sqrt{\frac{2E}{c}}}^{\sqrt{\frac{2E}{c}}} \frac1{\sqrt{E-\frac{cx^2}2}}\, dx &=2 \int_0^{\sqrt{\frac{2E}c}} \frac1{\sqrt{E-\frac{cx^2}2}}\, dx \\
&=\frac{2}{\sqrt{E}} \int_0^\sqrt{\frac{2E}{c}} \frac1{\sqrt{1-\frac{cx^2}{2E}}} \, dx\\
&= \frac{2}{\sqrt{E}}\cdot \sqrt{\frac{2E}{c}} \int_0^1 \frac1{\sqrt{1-v^2 \, }} \, dv \\
&=2\cdot \sqrt{\frac{2}{c}} (\sin^{-1}(1) - \sin^{-1}(0))\\
&=\pi \sqrt{\frac{2}{c}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I prove this inequality $3(a+b+c+1)\ge 4 \left( \sqrt{\frac{a^2+1}{a+1}}+\sqrt{\frac{b^2+1}{b+1}}+\sqrt{\frac{c^2+1}{c+1}} \right)$? If $a,b,c\gt 0$ and $abc=1$, how do I prove the following inequality $3(a+b+c+1)\ge 4 \left( \sqrt{\frac{a^2+1}{a+1}}+\sqrt{\frac{b^2+1}{b+1}}+\sqrt{\frac{c^2+1}{c+1}} \right)$?
My version:
\begin{gathered}
\sqrt{\frac{a^{2}+1}{a+1}} \leq \sqrt{1+\frac{a^{2}-a}{a+1}} \leq 1+\frac{1}{2} \frac{a^{2}-a}{a+1}=\frac{a}{2}+\frac{1}{a+1} \\
\Leftrightarrow 4\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right) \leq 3+(a+b+c), a b c=1 \\
a+b+c=3 t^{2}=p \geq 3, a b+b c+c a=q \geq \sqrt{3 p r}, a b c=r=1 \\
4\left(1+\frac{1+p}{2+p+q}\right) \leq 4\left(1+\frac{1+p}{2+p+\sqrt{3 p}}\right) \leq 3+p \\
(t-1)\left(3 t^{3}+6 t^{2}+3 t+2\right) \geq 0
\end{gathered}
| Using OP's work, WTS for $ a bc = 1, \sum a + 1 - \frac{4}{1+a } \geq 0 $.
Use the substitution $ a = e^A, b = e^B, c = e^C$,
WTS for $ A + B + C = 0, \sum e^A + 1 - \frac{4}{ 1 + e^A } \geq 0 $.
Let $ f(x) = e^x + 1 - \frac{4}{ 1 + e^x }$, then $f'(x) = e^x + 1 + \frac{ 4 e^ x } { (1 + e^x )^2 } $ and $ f''(x) = e^x + 1 - \frac{ 4e^x ( e^x - 1 )} { ( 1 + e^x ) ^3 } = \frac{ e^{4x} + 4e^{3x} + 2e^{2x} + 8e^x + 1 } { ( 1 + e^x) ^3 } \geq 0.$
Hence, we can apply Jensen's to conclude that
$ f(A) + f(B) + f(C) \geq 3 f( \frac{ A + B + C } { 3 } ) = 3 f(0) = 0 $.
Notes:
*
*This approach of using exponential to convert $abc = 1 $ to $ A+B+C = 0 $ and then apply Jensens is a standard trick.
*It is plausible that Jensen's could have worked directly, but it was too ugly for me to want to try. Have at it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why does the technique to find derivative of $y=\frac{ax^2+bx+c}{a'x^2+b'x+c'}$ quickly, works? I just watched a video about a very good technique to evaluate derivative of $ y=\dfrac{ax^2+bx+c}{a'x^2+b'x+c'} ,\quad a'\neq0$ quickly:
$$\large y'=\large\dfrac{{ \begin{vmatrix}{a} && {b} \\ {a'} && {b'}\end{vmatrix} }x^2+\color{red}2{ \begin{vmatrix}{a} && {c} \\ {a'} && {c'}\end{vmatrix} }x+{ \begin{vmatrix}{b} && {c} \\ {b'} && {c'}\end{vmatrix} }}{(a'x^2+b'x+c')^2}$$
I wonder is it possible to prove this is true? Of course we can calculate derivative of $y$ and expanding the terms and compare it with the formula that I mentioned. But is it possible to check this is true intuitively Or without expanding? ( it is obvious that denominator should be $(a'x^2+b'x+c')^2$ )
| Let $f(x)=ax^2+bx+c$ and $g(x)=a'x^2+b'x+c'$. Then
\begin{aligned}
\frac{d(f/g)}{dx}
&=\frac{f'g-g'f}{g^2}\\
&=\frac{1}{g^2}\left|\begin{matrix}1&0&0\\ 0&f'&f\\ 0&g'&g\end{matrix}\right|\\
&=\frac{1}{g^2}\left|\begin{matrix}1&0&0\\ a&f'&f\\ a'&g'&g\end{matrix}\right|\quad(R_2=R_2+aR_1;\ R_3=R_3+a'R_1)\\
&=\frac{1}{g^2}\left|\begin{matrix}1&-2x&0\\ a&b&f\\ a'&b'&g\end{matrix}\right|\quad(C_2=C_2-2xC_1)\\
&=\frac{1}{g^2}\left|\begin{matrix}1&-2x&x^2\\ a&b&c\\ a'&b'&c'\end{matrix}\right|\quad(C_3=C_3-x^2C_1-xC_2)\\
&=\frac{1}{g^2}\left(\left|\begin{matrix}b&c\\ b'&c'\end{matrix}\right|+2x\left|\begin{matrix}a&c\\ a'&c'\end{matrix}\right|+x^2\left|\begin{matrix}a&b\\ a'&b'\end{matrix}\right|\right).
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4184039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What does$ f(f(x))=0$ mean? I came across a question: If $f(x)=x^3-x+1$ then find the number of real distinct values of $f(f(x))=0$.
Here is what I interpreted the $f(f(x))$ as:
I assumed $a, b$ and $c$ to be the roots of $f(x)$, now if we put $a, b$ or $c$ in the $f(f(x))$ then it becomes $f(0)$ which will be equal to $1$.
I saw a solution where they differentiated the polynomial $f(x)$. They made the graph of $f(x)$ using first order derivative test. Then for $f(f(x))$, they put in $x=a, b, c$ (assumed roots of $f(x)$ ). Then we got three lines for $x=a$, a line below $-1$, for $b$ a line between $0$ and $1$ and for $c$, a line between $1$ and $3$. I am unable to understand why we put $a,b$ and $c$ as $x$ and then how did we get these ranges?
| $f(f(x)) = (x^3 - x + 1)^3 - (x^3 - x + 1) + 1 = (x^3-x)^3+1+3(x^3-x)[x^3-x+1]-(x^3-x) = (x^3-x)^3+3(x^3-x)^2+3(x^3-x)-(x^3-x)+1 = (x^3-x)^3+3(x^3-x)^2+2(x^3-x) +1$
But given $f(f(x) = 0$ so $(x^3-x)^3+3(x^3-x)^2+2(x^3-x)+1 = 0$....eq.(1)
Put $x^3-x = y$ in eq.(1) we get $y^3+3y^2+2y+1 = 0$....eq.(2).
Firstly, solve eq.(2) then solve eq.(1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
Prove that: $(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge9(ab+bc+ca)$ Let $a,b,c>0$ satisfy $a^2+b^2+c^2=3$ . Prove that: $$(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge9(ab+bc+ca)$$
My idea is to use a well-known inequality (We can prove by Schur) $$(a^2+2)(b^2+2)(c^2+2)\ge 9(ab+bc+ca)$$ and the problem is prove $$(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge(a^2+2)(b^2+2)(c^2+2)$$
Anyone have a better idea ? please help me
| Thanks for helping me, I also have a solution after many days of thinking
$$[9(ab+bc+ca)]^2=27(a^2+b^2+c^2)(ab+bc+ca)^2$$
By AM-GM, we have:
$$27(a^2+b^2+c^2)(ab+bc+ca)(ab+bc+ca)\le(a+b+c)^6$$
$$\to 9(ab+bc+ca)\le(a+b+c)^3 $$
The problem is:
$$(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge(a+b+c)^3$$
By AM-GM, we have:
$$a^5-2a+4=\frac{1}{5}(a^5+a^5+1+1+1)-2a+1+\frac{1}{5}(a^5+a^5+a^5+1+1+10)\geq (a-1)^2+\frac{1}{5}(5a^3+10)\geq a^3+2$$
$$\to (a^5-2a+4)(b^5-2b+4)(c^5-2c+4) \ge (a^3+2)(b^3+2)(c^3+2) $$
The problem is:
$$(a^3+2)(b^3+2)(c^3+2) \ge (a+b+c)^3 $$
Which is true by Holder
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to go from $2\cos 2x+1$ to $\frac{\sin 3x}{ \sin x} $ In an exercice I stumbled upon this equation.
$\sin 3x(2\cos 2x+1)=\frac{\sin^23x}{\sin x}$
Or somewhat simplified :
$2\cos 2x+1 = \frac{\sin 3x}{ \sin x} $
I managed to work out the right to the left , using a lot of formula's (starting with rewriting it as $\frac{\sin (2x + x)}{\sin x }$ and although it would be possible to reverse the steps to go from left to right, I have the feeling that I am missing something trivial here (just started out to relearn trigonometry to help my kid out...), it's not obvious for me to see how to startout from left to right...
| There's nothing wrong with what you did:
$$\frac{\sin(2x+x)}{\sin x} = \frac{\sin(2x)\cos x +\cos(2x)\sin x}{\sin x} $$
$$= \frac{2 \sin(x) \cos (x) \cos (x) + (2 \cos^2 x-1)\sin(x)}{\sin x}$$
$$= 2 \cos^2 x + 2 \cos^2 x - 1$$
$$= 2(2 \cos^2 x - 1) + 1 = 2 \cos 2x + 1$$
where $\cos(2x) = \cos^2 x - \sin^2 x = \cos^2 x - (1 - \cos^2 x) = 2 \cos^2 x - 1$, which can be memorised.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that there exists integers such that at least one of three inequalities are satisfied Suppose I am given 4 positive integers $A, B, C, D$, where $A > B$ and $C < \lceil \frac{A}{B} \rceil $, how can I prove that there always exists integers $x,y \geq 0$, where $D = x+y$ such that at least one of these three inequalities must be satisfied for every valid combination of $A, B, C, D \in \mathbb{Z}_{> 0}$.
$$\frac{y}{x+C} \geq \frac{A}{B} \geq \frac{y-1}{x+1}$$
$$\frac{y+C}{x-1} \geq \frac{A}{B} \geq \frac{y+1}{x}$$
$$\frac{y+C}{x} \geq \frac{A}{B} \geq \frac{y}{x+1}$$
I did some algebraic manipulation to arrive at the above three inequalities in that form, but can't seem to reason anything about the existence of such an $x, y$. Simulations seem to suggest that there always exists such an $x, y$ (though not exhaustive!), but I can't find a proof for it.
Does anyone have any insights on how to address this? (Or perhaps a counterexample otherwise).
| Replace $\frac AB$ with $r-1$.
Thus, given $C,D\in\Bbb Z_{>0}$ and a real number $r>C$, the claim is that one of
$$\frac {D-x}{x+C}\ge r-1\ge \frac {D-x-1}{x+1} $$
$$\frac {D-x+C}{x-1}\ge r-1\ge \frac {D-x+1}{x} $$
$$\frac {D-x+C}{x}\ge r-1\ge \frac {D-x}{x+1} $$
has an integer solution $0\le x\le D$ (with $x=0$, $x=1$ excluded for the lower two options).
Equivalently, one of
$$\frac {D+C}{x+C}\ge r\ge \frac {D}{x+1} $$
$$\frac {D+C-1}{x-1}\ge r\ge \frac {D+1}{x} $$
$$\frac {D+C}{x}\ge r\ge \frac {D+1}{x+1} $$
has such a solution. Take reciprocals,
$$\frac {x+C}{D+C}\le \frac1r\le \frac {x+1}D $$
$$\frac {x-1}{D+C-1}\le \frac1r\le \frac {x}{D+1} $$
$$\frac {x}{D+C}\le \frac1r\le \frac {x+1}{D+1} $$
and solve for $x$,
$$ \frac{D+C}r-C \ge x \ge \frac Dr-1 $$
$$ \frac{D+C-1}r+1 \ge x \ge \frac {D+1}r $$
$$ \frac{D+C}r \ge x \ge \frac {D+1}r-1 $$
The differences between upper and lower bound are
$$ \frac Dr+1-C,\qquad \frac{C-2}r+1,\qquad \frac{C-1}r+1$$
As $\frac{C-1}r+1\ge1$, there is always an integer $x$ satisfying $\frac{D+C}r \ge x \ge \frac {D+1}r-1$. As $r>1$, we have $\frac{D+1}r-1<D$ and hence the smallest integer in that range is indeed $\le D$, and as $\frac {D+C}r>0$, the greatest such integer is $\ge 0$. So it seems already the third option always has a solution, but for that, we must require $x\ne 0$. In other words, this works only if $$\tag1\frac{D+C}r\ge 1.$$
On the other hand, if $\frac{D+C}r<1$, then one sees that the second inequality has at most $x=1$ as solution - which is not allowed, and the first inequality cannot have a non-negative solution at all. Thus $(1)$, or equivalently
$$ {B(D+C-1)}\ge A$$
is necessary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Recurrence relation $a_{n} = a_{n-1}+a_{n-2}+n$ How to solve this recurrence relation?
$a_{n} = a_{n-1}+a_{n-2}+n,a_{1}=a_{0}=1$
What I have tried:
$r^2 = r + 1 \rightarrow r = \frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}$
non-homogeneous part:
$$
\begin{split}
a_{n} &= cn+b \implies c(n)+b = c(n-1) + b + c(n-2) + b + n\\
0 &= -3c+b+n(c+1) \implies c+1 = 0 \iff c=-1 \implies b = -4 \\
a_{n} &= c_{0} \left(\frac{1+\sqrt{5}}{2}\right)^n
+ c_{1} \left(\frac{1-\sqrt{5}}{2}\right)^n -n-4
\end{split}
$$
putting $a_{0}$ and $a_{1}$ in recurrence relation gives $c_{0} = (\frac{1}{2})(5+\frac{7}{\sqrt{5}})$ and $c_{1} = (\frac{1}{2})(5-\frac{7}{\sqrt{5}})$.
I am sure my answer is wrong because for n greater than 1 it doesn't work. help me correct my mistake, please.
| Here's another possible approach using generating functions (a standard approach for many linear recurrences):
Let us define $A(x)=\sum_{n=0}^{\infty} a_n x^n$
Then, we have:
$$a_n x^n=a_{n-1} x^n+a_{n-2} x^n+n x^n$$
Summing:
$$A(x)-x-1=x(A(x)-1)+x^2 A(x)+x\frac {d(\sum_{n=1}^{\infty} x^{n})}{dx}$$
Thus, we get:
$$A(x)=\frac { \frac {x}{(1-x)^2}+1}{1-x-x^2}$$
The coefficient of $x^n$ in this polynomial gives $a_n$. That can now be found out using partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the area of the moustache This is more a soft question to calculate :
Let $x\geq 0$ then we have
$$\int_{0}^{\infty}-\left(e^{-x^{2}}-x^{-x}\right)dx-\frac{10}{9}<0$$
My attempt :
We have the following inequalities :
For $1\leq x\leq 2$ :
$$e^{-\left(\frac{x^{0.5}\left(x-1\right)}{2^{0.5}}\right)\cdot2\ln\left(2\right)}\geq x^{-x}$$
For $2\leq x\leq 6$ :
$$e^{\left(-\left(x-1\right)^{\frac{92}{100}}x^{0.5}\cdot2^{0.5}\ln\left(2\right)\right)}\geq x^{-x}$$
And :
$$\int_{0}^{1} x^{-x}\ dx = \sum_{n=1}^{\infty} n^{-n}$$
I let you a picture :
Question :
How to show the first inequality ?
Thanks and have fun!
| We need to prove that
$$\int_0^\infty x^{-x}\, \mathrm{d} x
< \frac{1}{2}\sqrt{\pi} + \frac{10}{9}. \tag{1}$$
It is known that
$$\int_0^1 x^{-x}\, \mathrm{d} x = \sum_{n=1}^\infty \frac{1}{n^n}
\le \sum_{n = 1}^4 \frac{1}{n^n}
+ \sum_{n=5}^\infty \frac{1}{5^n} = \frac{5578603}{4320000}.$$
Also, we have
$$\int_5^\infty x^{-x}\, \mathrm{d} x = \int_5^\infty \mathrm{e}^{-x\ln x}\, \mathrm{d} x \le \int_5^\infty \mathrm{e}^{-x\ln 5}\, \mathrm{d} x = \frac{1}{3125\ln 5}.$$
By Facts 1-4 (given later), we have
\begin{align*}
&\int_1^2 x^{-x}\, \mathrm{d} x
\le \int_0^1 \frac{3 - x}{x^2 - x + 2}\, \mathrm{d} x = \frac{5}{\sqrt{7}}\arctan \frac{\sqrt{7}}{5} - \frac{1}{2}\ln 2,\\
&\int_2^3 x^{-x}\, \mathrm{d} x \le
\frac{21}{2\mathrm{e}} + \frac{1001}{72}
- \frac{1903 + 1518\mathrm{e}}{6\mathrm{e}}\ln 3
+ \frac{2927 + 2286\mathrm{e}}{6\mathrm{e}}\ln 2, \\
&\int_3^4 x^{-x}\, \mathrm{d} x \le \frac{3\mathrm{e} - 1}{81\mathrm{e}\ln 3 + 81 \mathrm{e}}, \\
&\int_4^5 x^{-x}\, \mathrm{d} x \le
\frac{4\mathrm{e} - 1}{2048\mathrm{e}\ln 2 + 1024\mathrm{e}}.
\end{align*}
With these facts, it is easy to verify (1).
We are done.
Fact 1: $x^{-x} \le \frac{3 - x}{x^2 - x + 2}$ for all $x \in [1, 2]$.
Fact 2: For all $x \in [2, 3]$,
$$x^{-x} \le \frac{x}{6\mathrm{e}} - \frac{19 + \mathrm{e}}{6\mathrm{e}} - \frac{293 + 250\mathrm{e}}{2\mathrm{e}x}
+ \frac{213\mathrm{e} + 159}{2\mathrm{e}x^2} - \frac{53}{x^3} + \frac{384\mathrm{e} + 512}{3\mathrm{e}(1 + x)}.$$
Fact 3: For all $x \in [3, 4]$,
$$x^{-x} \le 3^{-x} \mathrm{e}^{-x + 3}.$$
Fact 4: For all $x \in [4, 5]$,
$$x^{-x} \le 4^{-x} \mathrm{e}^{-x + 4}.$$
Remarks: Actually, $x^{-x} \le a^{-x} \mathrm{e}^{-x + a}$ for all $x, a > 0$
which is equivalent to $\ln u \ge 1 - \frac{1}{u}$ for all $u > 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does the sum of the series $\sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt n}{n}$ have an analytic expression? Just out of curiosity, I'd like to know whether or not the sum of the series
$$\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt n}{n}$$
has a known analytic expression.
I stumbled across this series while trying to evaluate
$$\int_1^\infty\frac{1}{\lfloor x^2\rfloor}dx$$
The convergence of this integral can be seen by making use of the inequality $\lfloor x\rfloor > x-1$ and the fact that $\coth^{-1}(t)\to 0$ as $t\to\infty$:
\begin{align*}
\int_1^t\frac{1}{\lfloor x^2\rfloor}dx &= \int_\sqrt{1}^\sqrt{2}\frac{1}{\lfloor x^2\rfloor}dx+\int_\sqrt{2}^t\frac{1}{\lfloor x^2\rfloor}dx\\
&< \int_\sqrt{1}^\sqrt{2}\frac{1}{\lfloor x^2\rfloor}dx+\int_\sqrt{2}^t\frac{1}{x^2-1}dx\\
&= \int_\sqrt{1}^\sqrt{2}\frac{1}{1}dx-\int_\sqrt{2}^t\frac{1}{1-x^2}dx\\
&= \sqrt{2}-1-\left[\coth^{-1}(t)-\coth^{-1}\left(\sqrt 2\right)\right]\\
&= \sqrt{2}-1-\coth^{-1}(t)+\coth^{-1}\left(\sqrt 2\right)\\
&\to \sqrt{2}-1+\coth^{-1}\left(\sqrt 2\right)\text{ as }t\to\infty\\
\end{align*}
Since this implies that $\int_1^t 1/\lfloor x^2\rfloor dx$ is strictly increasing ($1/\lfloor x^2\rfloor >0$ for every $x\geq 1$) and bounded above, the integral necessarily converges. By breaking up the integral
$$\int_1^\sqrt{k+1}\frac{1}{\lfloor x^2\rfloor}dx$$
into integrals indexed by the intervals $\left[\sqrt{i},\sqrt{i+1}\right]$ for $i=1,2,3,...,k$ and simplifying the resulting sum, I was able to show that
$$\int_1^{\sqrt{k+1}}\frac{1}{\lfloor x^2\rfloor}dx=\sum_{n=1}^{k} \frac{\sqrt{n+1}-\sqrt n}{n}$$
is true for every $k\geq 0$, which yields
$$\int_1^\infty \frac{1}{\lfloor x^2\rfloor}dx=\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt n}{n}$$
after letting $k\to\infty$. This equality is the main reason why I'm interested in the sum of the aforementioned series.
After some (unsurprisingly) futile attempts to evaluate the integral, I expect there to be no closed-form expression for the sum, which is why I'm open to an analytic expression (gamma function, Bessel functions, Riemann zeta function, etc.). Any help is appreciated.
Edit: after seeing the bounds provided by Markus Scheuer and Jorge, I thought I'd share some of my own.
From the fact that $x-1<\lfloor x\rfloor<x$ is true for every non-integer $x\geq 1$, we can infer that for every integer $k\geq 1$,
$$\int_\sqrt{k+1}^\infty \frac{1}{x^2}dx<\int_\sqrt{k+1}^\infty \frac{1}{\lfloor x^2\rfloor}dx<\int_\sqrt{k+1}^\infty \frac{1}{x^2-1}dx$$
Using
$$\int_{x}^{\infty}\frac{1}{t^2-1}dt=\coth^{-1}(x)$$
and
$$\int_\sqrt{k+1}^\infty\frac{1}{\lfloor x^2\rfloor}dx=\int_1^\infty\frac{1}{\lfloor x^2\rfloor}dx-\int_1^\sqrt{k+1}\frac{1}{\lfloor x^2\rfloor}dx=\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt n}{n}-\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}$$
we deduce that
$$\frac{1}{\sqrt{k+1}}+\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}<\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt n}{n}<\coth^{-1}\left(\sqrt{k+1}\right)+\sum_{n=1}^{k}\frac{\sqrt{n+1}-\sqrt n}{n}$$
| I'm also doubtful that a closed form exists, but you can get really good approximations using power series and the zeta function.
Note:
$$
\begin{align}
\sum \frac{\sqrt{n+1} - \sqrt{n}}{n}
&= \sum \frac{\sqrt{n} \left ( \sqrt{1 + \frac{1}{n}} - 1 \right )}{n} \\\\
&= \sum \frac{1}{\sqrt{n}} \left ( \frac{1}{2n} - \frac{1}{8n^2} + \ldots \right ) \\\\
&= \frac{1}{2} \sum \frac{1}{n^{3/2}} - \frac{1}{8} \sum \frac{1}{n^{5/2}} + \ldots \\\\
&= \frac{1}{2} \zeta \left ( \frac{3}{2} \right ) - \frac{1}{8} \zeta \left ( \frac{5}{2} \right ) + \ldots
\end{align}
$$
Using the full series for $\sqrt{1+x}$, we find
$$ \sum_{n \geq 1} \frac{\sqrt{n+1} - \sqrt{n}}{n} = \sum_{k \geq 1} \binom{1/2}{k} \zeta \left ( \frac{2k+1}{2} \right ) .$$
Using series expansions for $\zeta \left ( \frac{2k+1}{2} \right )$ (which tends to $1$ as $k \to \infty$) and $\binom{1/2}{k}$ (which decays like $O \left ( k^{-3/2} \right )$) we can find a $k$ which gets us any desired precision.
I hope this helps ^_^
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
How to find the maximum value of $x\cos^{-1}\left(x\right)$? I wanted to find the maximum value of $x\cos^{-1}\left(x\right)$ through differentiation, but upon on differentiation I get $$f'(x) = \arccos\left(x\right) - \frac{x}{\sqrt{1-x^2}}$$ to find the maximum I equated the function to zero: $$\arccos\left(x\right) - \frac{x}{\sqrt{1-x^2}} = 0$$
But I am unable to find the roots of this equation. Could someone show how to find the zeros for this equation?
| As you wrote
$$f(x)=x \cos ^{-1}(x)\qquad \text{and} \qquad f'(x)=\cos ^{-1}(x)-\frac{x}{\sqrt{1-x^2}}$$
The problem would be simple using Newton method since, from the graph of $f'(x)$, you probably notice that the solution is close to $x=0.6$.
Without numerical method, since $x=\frac 12$ is a nice number for the arcosine, perform a series expansion. Thsi would give
$$f'(x)=\left(\frac{\pi }{3}-\frac{1}{\sqrt{3}}\right)-\frac{14
\left(x-\frac{1}{2}\right)}{3 \sqrt{3}}-\frac{10 \left(x-\frac{1}{2}\right)^2}{3
\sqrt{3}}-\frac{152 \left(x-\frac{1}{2}\right)^3}{27 \sqrt{3}}-\frac{724
\left(x-\frac{1}{2}\right)^4}{81
\sqrt{3}}+O\left(\left(x-\frac{1}{2}\right)^5\right)$$ Now, perform a series reversion and obtain
$$x=\frac{1}{2}-t-\frac{5 t^2}{7}+\frac{82 t^3}{441}+\frac{5287
t^4}{9261}+O\left(t^5\right)\quad \text{with}\quad t=\frac{3\sqrt{3}}{14} \left(f'(x)-\frac{\pi }{3}+\frac{1}{\sqrt{3}}\right)$$ and we want $f'x)=0$.
So, this truncated expansion gives as an estimate
$$x=\frac{10117671+3552168 \sqrt{3} \pi -264690 \pi ^2-29184 \sqrt{3} \pi ^3+5287 \pi^4}{39530064}$$ which, numerically, is $0.652206$ quite close to the value @Mariusz Iwaniuk gave in the first comment . For sure, adding a few more terms, the estimate will be closer and closer to the solution.
For example, adding the next term in the initial expansion, the same procedure would give
$$x=\frac{354165816+124246995 \sqrt{3} \pi -9106380 \pi ^2-1074030 \sqrt{3} \pi
^3+211340 \pi ^4-1753 \sqrt{3} \pi ^5}{1383552240}$$ which is $0.652194$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}$?
Evaluate $$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}.$$
I solved the problem with the Taylor series expansion of $\cos x$. Here is my solution:
$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1-\{1-\frac{(7x)^2}{2!}+\frac{(7x)^4}{4!}-\frac{(7x)^6}{6!}+\dots\}}{3x^2}\\ =\lim\limits_{x\to 0}\frac{x^2(\frac{7^2}{2!}-\frac{7^4x^2}{4!}+\frac{7^6x^4}{6!}-\dots)}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1}{3}(\frac{7^2}{2!}-\frac{7^4x^2}{4!}+\frac{7^6x^4}{6!}-\dots)\\ =\frac {49}{6}$
Can this be solved without using the Taylor series?
| An asymptotic approach
As $x \to 0$
$$\frac{1-\cos 7x}{3x^2} \sim \frac{49x^2/2}{3x^2}=\frac{49}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
In a right triangle, the perimeter is equal to 30. How many integer values can the hypotenuse take? (Answer:2)
I did:
$a+b+h = 30\rightarrow a+b = h-30\\
a^2+b^2 = h^2 \rightarrow h = \sqrt{a^2+b^2}\\
a-b<h<a+b \rightarrow a-b < h < \sqrt{a^2+b^2}-30...\\
\text{I didn't find other relationships...}
$
| The sides of a general triangle are the roots of the cubic equation
\begin{align}
x^3-2\rho x^2+(\rho^2+r^2+4rR)x-4\rho r R&=0
\tag{1}\label{1}
\end{align}
where $\rho,r$ and $R$ are the semiperimeter, the inradius and
the circumradius, respectively, of the corresponding triangle.
We are given that $\rho=\tfrac{30}2=15$ and also,
since $c$ is the hypotenuse, $c=2R$. So, \eqref{1} transforms to
\begin{align}
c(r+c-15)^2&=0
\tag{2}\label{2}
\\
\text{or just}\quad
c&=15-r
\tag{3}\label{3}
,
\end{align}
so the integer values of $c$ are defined by integer values of $r$.
Also, since for any valid non-degenerate non-equilateral triangle
\begin{align}
r&\in(0,\tfrac R2)
\tag{4}\label{4}
,
\end{align}
we must have
\begin{align}
15-r&> 4r
\tag{5}\label{5}
\\
\text{or }\quad
r&<3
\tag{6}\label{6}
\end{align}
and we have only two options left:
\begin{align}
r_1&=1
,\quad
c_1=14
\tag{7}\label{7}
\\
\text{and }\quad
r_2&=2
,\quad
c_2=13
\tag{8}\label{8}
.
\end{align}
So, the answer is indeed, $2$, we don't even have to find the other side lengths.
However, it's trivial, since by factoring \eqref{1} they can be found
as a roots of quadratic
\begin{align}
x^2-(30-c) x+450-30c&=0
\tag{9}\label{9}
,
\end{align}
hence for $c_1=14$ two other sides are
$8\pm\sqrt{34}$
and for
$c_2=13$ two other sides are
$5$ and $12$.
Edit
Using geometric approach, when $c=|AB|$ is fixed, the third point $C$
must be located at the intersection of the circle centered at
$O$, the midpoint of $AB$, and the ellipse focused at $A,B$
for which $|AC|+|BC|=30-|AB|$. These are only two possible solutions, that corresponds to $c_1$ and $c_2$.
\begin{align}
\begin{array}{ccc}
R & \text{major semi-axis} & \text{minor semi-axis}
\\ \hline
7 & 8 & \sqrt{15}
\\
6.5 & 8.5 & \sqrt{30}
\\ \hline
\end{array}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$.
Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$
I want to prove the inequality with elementary inequalities ( Mean inequalities, Cauchy-Schwarz etc.). I tried to use Cauchy-Schwarz, but I got $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\frac{9}{2(a^2+b^2+c^2)+ab+bc+ca}.$$
Then we have to show that $$2(a^2+b^2+c^2)+ab+bc+ca\leq (a+b+c)^2 $$
which is not true.
Multiplying the numerators of the fractions in LHS by $a^2,b^2,c^2$ also makes the problem more difficult.
So, how to solve the problem with elementary inequalities?
| The solution by Vo Quoc Ba Can
Multiplying both sides by $a^2+b^2+c^2+ab+bc+ca,$ the inequality is equivalent to
$$\sum {\frac{a^2+b^2+c^2+ab+bc+ca}{a^2+ab+b^2}} \geqslant \frac{9(a^2+b^2+c^2+ab+bc+ca)}{(a+b+c)^2},$$
equivalent to
$$3+(a+b+c)\sum {\frac{c}{a^2+ab+b^2}} \geqslant 9-\frac{9(ab+bc+ca)}{(a+b+c)^2},$$
or
$$(a+b+c)\sum {\frac{c}{a^2+ab+b^2}}+\frac{9(ab+bc+ca)}{(a+b+c)^2}\geq6.$$
By theCauchy-Schwartz inequality, we have
$$\sum {\frac{c}{a^2+ab+b^2}}=\sum {\frac{c^2}{c(a^2+ab+b^2)}}\geq\frac{(a+b+c)^2}{\sum{c(a^2+ab+b^2)}}=\frac{a+b+c}{ab+bc+ca}.$$
Hence, it suffices to prove that
$$\frac{(a+b+c)^2}{ab+bc+ca}+\frac{9(ab+bc+ca)}{(a+b+c)^2}\geq6.$$
Which is just AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
What is the asymptotics of: $\Re\left(\frac{\zeta \left(1+\frac{1}{c}\right) \zeta (s+i t)}{\zeta \left(s+i t+\frac{1}{c}+1-1\right)}\right)$?
What is the asymptotics of the function $f(s,t,c)$:
$$f(s,t,c)=\Re\left(\frac{\zeta \left(1+\frac{1}{c}\right) \zeta (s+i t)}{\zeta \left(s+i t+\frac{1}{c}+1-1\right)}\right)$$
?
For $c=10^4$ the plot of the function in the complex plane is:
(*Mathematica start*)
c = 10^4;
Plot3D[Re[
Zeta[s + I t]*Zeta[1 + 1/c]/Zeta[s + I t + 1 + 1/c - 1]], {s, -60,
60}, {t, -60, 60}]
(*end*)
On the critical line $s=1/2$ the asymptotics appears to be:
$$\Re\left(\frac{\zeta \left(1+\frac{1}{c}\right) \zeta \left(\frac{1}{2}+i t\right)}{\zeta \left(\frac{1}{2}+i t+\frac{1}{c}+1-1\right)}\right) \sim c+\frac{\partial \vartheta (t)}{\partial t}+\gamma$$
where $\vartheta (t)$ is the Riemann-Siegel Theta function.
(*Mathematica start*)
Clear[n, k, t, A, nn];
f[t_] = D[RiemannSiegelTheta[t], t];
nnn = 60
c = 10^1;
g1 = Plot[(f[t] + c + EulerGamma), {t, 0, nnn},
PlotStyle -> {Thickness[0.004], Red}, PlotRange -> {0, c + 5}];
g2 = Plot[
Re[Zeta[1/2 + I*t]*
Zeta[1 + 1/c]/Zeta[1/2 + I*t + 1 + 1/c - 1]], {t, 0, nnn},
PlotStyle -> Thickness[0.02], PlotRange -> {0, c + 5}];
Show[g2, g1, ImageSize -> Large]
(*end*)
| This is for the critical line, the general case may be treated in a similar manner. By Taylor expanding about $c=\infty$, we find
$$
\zeta \!\left( {1 + \frac{1}{c}} \right) = c + \gamma + \mathcal{O}\!\left( {\frac{1}{c}} \right)
$$
and
$$
\zeta \!\left( {\frac{1}{2} + it + \frac{1}{c}} \right) = \zeta\! \left( {\frac{1}{2} + it} \right) + \zeta '\!\left( {\frac{1}{2} + it} \right)\frac{1}{c} + \mathcal{O}\!\left( {\frac{1}{{c^2 }}} \right).
$$
This gives
$$
\frac{{\zeta\! \left( {1 + \frac{1}{c}} \right)\zeta\! \left( {\frac{1}{2} + it} \right)}}{{\zeta\! \left( {\frac{1}{2} + it + \frac{1}{c}} \right)}} = c - \frac{{\zeta '\!\left( {\frac{1}{2} + it} \right)}}{{\zeta \!\left( {\frac{1}{2} + it} \right)}} + \gamma + \mathcal{O}\!\left( {\frac{1}{c}} \right).
$$
Since
$$
- \frac{{\zeta '\!\left( {\frac{1}{2} + it} \right)}}{{\zeta\! \left( {\frac{1}{2} + it} \right)}} = \frac{{d\vartheta (t)}}{{dt}} + i\frac{{Z'(t)}}{{Z(t)}},
$$
we find
$$
\Re \frac{{\zeta\! \left( {1 + \frac{1}{c}} \right)\zeta\! \left( {\frac{1}{2} + it} \right)}}{{\zeta\! \left( {\frac{1}{2} + it + \frac{1}{c}} \right)}} = c + \frac{{d\vartheta (t)}}{{dt}} + \gamma + \mathcal{O}\!\left( {\frac{1}{c}} \right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4199276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solutions of $\sqrt{x^2+5x-14} + |x^2+4x-12|=0$ My attempt:
Given,
$$\sqrt{x^2+5x-14} + |x^2+4x-12|=0 \tag{1}$$
Since $|a|=\sqrt{a^2}$, $$\sqrt{x^2+5x-14}=-\sqrt{(x^2+4x-12)^2}$$
Squaring both sides, $$x^2+5x-14=(x^2+4x-12)^2$$
When I simplify the above, I get two real solutions: $x=2$ and $x=2.138 \text{ (approximately)}$. However, there is only one solution to equation $(1)$ according to Wolfram Alpha, $x=2$.
Is my solution incorrect? If so, where did I go wrong?
| Both $\sqrt{x^2+5x-14}\geq 0$ and $|x^2+4x-12|\geq 0$
So, both of them must equal to zero, since equation is equal to zero.
$$\sqrt{x^2+5x-14}= 0$$
$$x^2+5x-14=0$$
$x=-7$, or $x=2$
$$|x^2+4x-12|= 0$$
$$x^2+4x-12=0$$
$x=-6$, or $x=2$
Finally
$x=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4202536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that if $a,b,c$ are sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$
Prove that, if $a,b,c$ are the sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$.
What I Tried:- Nothing special really came in my mind. I did not find a way to use Triangle Inequality. What I did was, by AM-GM :-
$$a^2 + b^2 + c^2 \geq \frac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg) > \frac{72}{35}(\sqrt{sabc}).$$
But I couldn't proceed from this.
Another Idea I had was :- $a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$.
$\rightarrow a^2 + b^2 + c^2 = 4s^2 - 2(ab + bc + ca).$
But I did not know how to use this here, and would make the calculations a bit messy, especially of the $\dfrac{36}{35}$ part present there.
Can Anyone Help me? Thank You.
| Firstly using $s= \frac{1}{2} (a+b+c)$ the inequality is equivalent to $35(a^2 + b^2 + c^2) \geq 9(a+b+c)^2 + \frac{72abc}{a+b+c}$. Now note that $\frac{3abc}{a+b+c} = \frac{3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}$ and so by the HM-GM-AM inequality it follows that $\frac{3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}} \leq (abc)^\frac{2}{3} \leq \frac{1}{3}(a^2+b^2+c^2)$ hence we have $\frac{72abc}{a+b+c} \leq 8(a^2+b^2+c^2)$. Now using the AM-QM inequality $\frac{a+b+c}{3} \leq \sqrt{\frac{a^2 + b^2 + c^2}{3}} \Rightarrow 9(a+b+c)^2 \leq 27(a^2+b^2+c^2)$. Adding the two inequalities gives $35(a^2 + b^2 + c^2) \geq 9(a+b+c)^2 + \frac{72abc}{a+b+c}$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
What is $S_m(n)$ if $S_0(n) = 1$ and $S_{m+1}(n) = \sum_{k=1}^n S_m(k)$ and $m$ and $n$ are integers? What is $S_m(n)$ if $S_0(n) = 1$ and $S_{m+1}(n) = \sum_{k=1}^n S_m(k)$ and $m$ and $n$ are integers?
I tried to find a pattern:
$S_{m+1}(n) = S_{m}(1) + S_{m}(2) + ... + S_{m}(n)$
$=nS_{m-1}(1) + (n-1)S_{m-1}(2) + ... + S_{m-1}(n)$
$=(n + n-1 + n-2 + ... + 1)S_{m-2}(1) + (n-1 + n-2 + ... 1)S_{m-2}(2) + ... + S_{m-2}(n)$,
but the expression soon became hard to overview. Another attempt was to calculate each summand separately in:
$S_{m+1}(n) = S_{m}(1) + S_{m}(2) + ... + S_{m}(n)$
as follows:
$S_{m}(1) = S_{m-1}(1) = ... = S_1(1) = S_0(1) = 1$,
$S_{m}(2) = S_{m-1}(1) + S_{m-1}(2) = 1 + S_{m-1}(2) = ... = m +1$
...
but soon I got stuck again. It is easily checked that for $m=2$ and $n=5$ we get $S_m(n) = S_2(5) = 15$. Is there a closed general solution? Using a computer is a possibility but for e.g. very large $n$ that seems difficult as well.
| Proposition: $$S_m\left(n\right) = \dfrac{\left(n + m - 1\right)!} {m! \left(n - 1\right)!} = \dfrac{\left(n + m - 1\right) \left(n + m -2\right) \cdots n} {m!}.$$ for $m \ge 0$, $n \ge 1$.
Proof: for $m = 0$, $$S_m\left(n\right) = S_0\left(n\right) = \dfrac{\left(n - 1\right)!} {\left(n - 1\right)!} = 1.$$
For $m > 0$, $$
\begin{array}{rl}
S_{m + 1}\left(n + 1\right) - S_{m + 1}\left( n \right) &= \dfrac{\left(n + m + 1\right)!} {\left(m + 1\right)! n!} - \dfrac{\left(n + m\right)!} {\left(m + 1\right)! \left(n - 1\right)!} \\
& = \dfrac{\left(n + m + 1\right)! - \left( n + m \right)!\cdot n} {(m + 1)! n!} \\
& = \dfrac{\left(n + m\right)!} {m! n!} \\
& = S_m \left(n + 1\right).
\end{array}
$$
Then we have $$
\begin{array}{rl}
S_{m + 1}\left(n + 1\right) & = S_{m}\left(n + 1\right) + S_{m + 1}\left(n\right) \\
& = S_{m}\left(n + 1\right) + S_{m}\left(n\right) + S_{m + 1}\left(n - 1\right) \\
& = S_{m}\left(n + 1\right) + S_{m}\left(n\right) + S_{m}\left(n - 1\right) + S_{m + 1}\left(n - 2\right) \\
& = \cdots \\
& = \sum_{k = 1}^{n + 1} S_m\left( k \right).
\end{array}
$$
Therefore, $$S_m\left(n\right) = \dfrac{\left(n + m - 1\right)!} {m! \left(n - 1\right)!}$$ is the solution. To prove it's the only solution, you can assume there is another solution $S_m^\prime\left( n \right)$, and prove by induction that $S_m^\prime\left( n \right) - S_m\left( n \right) = 0$ for $m \ge 0$ and $n \ge 1$.
You are actually not too far from the answer. You already have $$S_2 \left( n \right) = \dfrac{n \left(n + 1\right)} {2}.$$ Then you easily get $$S_3 \left( n \right) = \dfrac{n \left(n + 1\right) \left(n + 2\right)}{2 \times 3}. $$ Here starts the pattern, and you simply prove it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding modular inverse of $x^4$ in $GF(2^5)\mod (x^5+x^2+1)$ I cannot spot where I am going wrong in this. I am using Extended Euclidean's algorithm here.
$(x^5+x^2+1) = (x^4)(x) + (x^2+1)$
$(x^4) = (x^2+1)(x^2+1) + 1$
let $P(x)=x^4$ and $Q(x)=x^5+x^2+1$
$(x^5+x^2+1) + (x^4)(x) = (x^2+1)$
$(x^4) + (x^2+1)(x^2+1) = 1$
$P(x)+[Q(x)+P(x)(x)][Q(x)+P(x)(x)]=1$
$P(x)(1+xQ(x)+P(x)x^2+xQ(x)) +Q(x)(...)=1$
$P(x)(1+P(x)x^2)+Q(x)(...)=1$
$P(x)(1+x^6) + Q(x)(...)=1$
$(1+x^6) \bmod (x^5+x^2+1)=x^3+x+1$
I am getting the inverse of $x^4 \mod (x^5+x^2+1)$ as $x^3+x+1$ but it seems the right answer is $x^4+x^2+1$. Can anyone tell me where it is wrong?
| We want to solve $$x^4 \cdot (c_0 + c_1 x + c_2 x^2 + c_3 x^4) = 1$$
multiplying this out gives
\begin{align}
& c_0 x^4 + c_1 x^5 + c_2 x^6 + c_3 x^8 \\
=&\, c_0 x^4 + c_1 (x^2 + 1) + c_2 (x^3 + x) + c_3 (x^3 + x^2) \\
=&\, c_1 + (c_1 + c_3) x^2 + (c_2 + c_3) x^3 + c_0 x^4 \\
\end{align}
so we want to solve the linear system:
*
*$c_1 = 1$
*$c_1 + c_3 = 0$
*$c_2 + c_3 = 0$
*$c_0 = 0$
to get our inverse $x + x^2 + x^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the maximum value for $\left\lceil\frac{b-x}{2}\right\rceil + \left\lceil\frac{x-a}{2}\right\rceil$
I have given two numbers $a$ and $b$, we have to choose an optimal $x$ (with $a\le x\le b$) such that
$$\left\lceil\frac{b-x}{2}\right\rceil + \left\lceil\frac{x-a}{2}\right\rceil = \text{maximum}$$
I thought the optimal way to choose $x$ we are such that it got canceled from both sides.
Therefore, maximum sum will be
$$\left\lceil \frac{b}{2}\right\rceil + \left\lceil-\frac{a}{2}\right\rceil$$
It works for some cases, but what am I missing?
| Short answer: choose $x$ which is very slightly larger than $a$. (more precisely: an order of magnitude smaller than the fractional parts of both $-a$ and $b-a$ will suffice.)
Without loss of generality take $a=0$ (since if $(a',b',x')=(0,b-a,x-a)$ then $\left\lceil\frac{b-x}{2}\right\rceil + \left\lceil\frac{x-a}{2}\right\rceil = \left\lceil\frac{b'-x'}{2}\right\rceil + \left\lceil\frac{x'-a'}{2}\right\rceil $), so that $0\leq x\leq b$. It will be convenient to write $B=\lfloor\frac{b}{2}\rfloor$ for the integer part of $\frac{b}{2}$.
If $b=2B$ is an even integer, then $\left\lceil \frac{b-x}{2}\right\rceil + \left\lceil\frac{x}{2}\right\rceil = \frac{b}{2}+\left\lceil\frac{-x}{2}\right\rceil + \left\lceil\frac{x}{2}\right\rceil$, which is $B+1$ for any $x$ which is not an even integer, and this is clearly the maximum (or there is a formal argument similar to the other case).
If $b=2B+\epsilon$ is not even integer, then for all sufficiently small $x$ (explicitly, $x<\epsilon$) we have $\left\lceil\frac{b-x}{2}\right\rceil=\left\lceil\frac{b}{2}\right\rceil=B+1$, and thus $\left\lceil \frac{b-x}{2}\right\rceil + \left\lceil\frac{x}{2}\right\rceil = B+2$.
We claim this is the maximum. Indeed, in general $y\leq \lceil y\rceil<y+1$, so for any $x$ (not necessarily small):
$$ \frac{b-x}{2} + \frac{x}{2} \leq \left\lceil \frac{b-x}{2}\right\rceil + \left\lceil\frac{x}{2}\right\rceil < \frac{b-x}{2} + \frac{x}{2} + 2$$
$$ \frac{b}{2} \leq \left\lceil \frac{b-x}{2}\right\rceil + \left\lceil\frac{x}{2}\right\rceil < \frac{b}{2} + 2$$
But by definition, $B$ is the largest integer such that $B+2<\frac{b}{2}+2$, this proves optimality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f(x):[1,3]\to [-1,1]$ satisfies $\int_1^3 f(x)dx=0$, then find max value of $\int_1^3 \frac{f(x)}{x}dx$ I managed to graph a very rough sketch, which involves $f(2)=0$ and the crests peaking at $-1$ and $1$ on either side of $x=2$
Now for maximization, assume $f(x)=1$ in $[1,2)$ and $f(x) = -1$ in $(2,3]$
So $$\int_1^2 \frac{f(x)}{x}dx + \int_2^3 \frac{f(x)}{x}dx$$
$$=\ln 2$$
However this isn’t in the options. Where am I going wrong
| Let $f(x)$ satisfies the condition. Then using $1-f(x)\ge 0$ and $f(x) +1 \ge 0$,
\begin{align}
\int_1^3 \frac{f(x)}{x} dx &= \int_1^2 \frac 1x dx - \int_1^2 \frac{1-f(x)}{x} dx -\int_2^3 \frac 1x dx + \int_2^3 \frac{f(x) +1}{x}dx\\
&= \ln \frac 43 - \int_1^2 \frac{1-f(x)}{x} dx + \int_2^3 \frac{f(x) +1}{x}dx.\\
& \le \ln \frac 43 -\frac 12 \int_1^2 (1-f(x))dx + \frac 12 \int_2^3 (f(x) + 1) dx\\
&= \ln \frac 43 + \int_1^3 f(x) dx = \ln \frac 43.
\end{align}
Thus the maximum is really $\ln (4/3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find closed form of $a_{n+1} = 4a_n - 2$ Is there a easy way to find the recurrence for $a_{n+1} = 4a_n - 2$, with $a_1 = 2$?
I was learning about finding the closed form of Fibonacci-styled recurrences like $a_{n+2} = 2a_{n+1} - 3a_{n}$ or something, and then this problem appeared. The textbook did a ton of rearranging to cancel out the $-2$ term and get it into the form of a Fibonacci-styled recurrence. I'm wondering if there is a easier way.
| \begin{align*}
a_2 &= 4a_1 - 2\\
a_3 &= 4(4a_1-2)-2 = 4^2a_1 - 4\cdot 2 - 2\\
a_4 &= 4(4^2a_1 - 4\cdot 2 - 2) - 2 = 4^3a_1 - 4^2\cdot 2 - 4\cdot 2 - 2.
\end{align*}
This leads to the conjecture
$$
a_n = 4^{n-1}a_1 - 2\sum_{k=0}^{n-2}4^k = 4^{n-1}a_1 - 2\frac{4^{n-1}-1}{4-1} = 4^{n-1}(a_1-\tfrac 23) + \tfrac 23.
$$
It is easily checked that this sequence satisfies the iteration rule, hence it's the solution.
| {
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Exercise on the following surface $\sqrt[4]{x^3}+\sqrt[4]{y^3}+\sqrt[4]{z^3}=\sqrt[4]{c^3}$ for $c\ge 0$. Consider the surface $\sqrt[4]{x^3}+\sqrt[4]{y^3}+\sqrt[4]{z^3}=\sqrt[4]{c^3}$ for $c\ge 0$. Find the plane tange to the surface at the point $ (x_0, y_0, z_0) $ and show that the sum of the cubes of the intersections of said plane with the axes is equal to one constant.
Try: To find the equation of the tangent plane, let's first find the derivatives. Let $F (x, y, z) = x^{ 3/4} + y ^{3/4} + z^{ 3/4}$. Then,
$$ F_x (x_0, y_0, z_0) = \dfrac {3} {4x_0 ^ {1/4}}, \quad F_y (x_0, y_0, z_0) = \dfrac {3} {4y_0 ^ {1/4} }, \quad F_z (x_0, y_0, z_0) = \dfrac {3} {4z_0 ^ {1/4}}. $$
$$\begin{align*}\dfrac {3} {4x_0 ^ {1/4}}(x-x_0)+\dfrac {3} {4y_0 ^ {1/4} }(y-y_0)+\dfrac {3} {4z_0 ^ {1/4}}(z-z_0)&=0\\ \dfrac{3}{4}\left (\dfrac {1} {x_0 ^ {1/4}}(x-x_0)+\dfrac {1} {y_0 ^ {1/4} }(y-y_0)+\dfrac {1} {z_0 ^ {1/4}}(z-z_0) \right )&=0\\\dfrac {1} {x_0 ^ {1/4}}(x-x_0)+\dfrac {1} {y_0 ^ {1/4} }(y-y_0)+\dfrac {1} {z_0 ^ {1/4}}(z-z_0) &=0 \end{align*}$$
Can someone help me with the other part of the question and tell me if my tangent plane is ok. Thanks a lot.
| You can rewrite the equation of the tangent plane as,
$\dfrac {x} {x_0 ^{1/4}} + \dfrac {y} {y_0 ^ {1/4} } + \dfrac {z} {z_0 ^ {1/4}} = x_0 ^{3/4} + y_0 ^{3/4} + z_0 ^{3/4} = c^{3/4} \tag1$
Now at the intersection of the tangent plane with x-axis, $y = 0, z = 0$.
So plugging into $(1)$,
$x = x_0 ^{1/4} c^{3/4}$
Similarly at intersection with other axes,
with y-axis: $y = y_0 ^{1/4} c^{3/4}$
with z-axis: $z = z_0 ^{1/4} c^{3/4}$
Now using $ \ x_0 ^{3/4} + y_0 ^{3/4} + z_0 ^{3/4} = c^{3/4} , \ $ you can show that the sum of cubes of x, y and z intercepts is a constant $( = c^3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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For $1 \not= \alpha \in \mathbb{C}$ such that $\alpha^7 = 1$, evaluate $\alpha + \alpha^2 + \alpha^4.$ For $1 \not= \alpha \in \mathbb{C}$ such that $\alpha^7 = 1$, evaluate $\alpha + \alpha^2 + \alpha^4.$
My solution :
let $$p = \alpha + \alpha^2 + \alpha^4$$ and $$q = \alpha^3 + \alpha^5 + \alpha^6.$$
We know $$1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 = 0,$$
$$p + q = \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6$$
$$ = -1$$
and
$$pq = (\alpha + \alpha^2 + \alpha^4)(\alpha^3 + \alpha^5 + \alpha^6)$$
$$= \alpha^4 + \alpha^6 + \alpha^7 + \alpha^5 + \alpha^7 + \alpha^8 + \alpha^7 + \alpha^9 + \alpha^{10}$$
$$= 2.$$
Therefore, $p$ and $q$ are the two roots of the following equation :
$$x^2 + x + 2 = 0$$
and
$$p = \alpha + \alpha^2 + \alpha^4$$
$$= \frac{-1 ± \sqrt{7} i}{2}.$$
Would there be other ways of evaluating? I'm thinking of polar forms but not sure how to do this with it.
| Let $p=\alpha+\alpha^2+\alpha^4$. Squaring both sides we obtain
$$p^2= \alpha^2+\alpha^4+\alpha^8+2\alpha^3+2\alpha^6+2\alpha^5.$$ Re arrange the terms we get $$p^2= \alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6+\alpha^3+\alpha^5+\alpha^6.$$
Which is equal to $p^2=-1+ \alpha^3+\alpha^5+\alpha^6$ same as $p^2+1=\alpha^3+\alpha^5+\alpha^6$. Again squaring both sides and re arrange the terms we obtain
$(p^2+1)^2=(\alpha^3+\alpha^5+\alpha^6)^2.$ $(p^2+1)^2=\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6+\alpha+\alpha^2+\alpha^4$. Which is same as $(p^2+1)^2=-1+p \Rightarrow p^4+2p^2-p+2=0.$ The root of this equation gives the desired result.
| {
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"url": "https://math.stackexchange.com/questions/4221162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\frac{P_a}{a^2}+\frac{P_b}{b^2}+\frac{P_c}{c^2}\ge\frac{3}{4R}$. When is the equality reached? Show that $\dfrac{P_a}{a^2}+\dfrac{P_b}{b^2}+\dfrac{P_c}{c^2}\ge\dfrac{3}{4R}$. When is the equality reached?
We're dealing with an acute triangle $ABC$ where $h_a,h_b$ and $h_c$ are altitudes. $P_a,P_b$ and $P_c$ are the distances from the vertices of the triangle to the segments joining the feet of the altitudes. $R$ is the radius of the circumscribed circle of $ABC$.
I discovered a few things in the given configuration. $$\dfrac{P_a}{h_a}=\cos\measuredangle BAC$$ For the proof, we can look at the similar triangles $ABC$ and $AB_1C_1$. In them $\dfrac{AB_1}{AB}=\dfrac{h_{B_1C_1}}{h_{BC}}=\dfrac{P_a}{h_a}=\cos\measuredangle BAC$ (using the definition of cosine of an acute angle in the right triangle $AB_1B$). To show that the mentioned triangles are similar we can use $\triangle AB_1B\sim\triangle AC_1C$ and then SAS.
Similarly, $$\dfrac{P_b}{h_b}=\cos\measuredangle ABC\\ \dfrac{P_c}{h_c}=\cos\measuredangle ACB.$$ Another thing I was able to prove is $$P_a+P_b+P_c=\dfrac{a^2+b^2+c^2}{4R}.$$ We can derive that by simply substituting $$P_a=h_a\cos\alpha,P_b=h_b\cos\beta,P_c=h_c\cos\gamma,h_a=\dfrac{2S}{a},h_b=\dfrac{2S}{b},h_c=\dfrac{2S}{c},$$ where $S$ is the area of triangle $ABC$ and $\alpha,\beta$ and $\gamma$ are the angles of the triangle, and the law of cosines for simplification. Thank you in advance!
| By your work we need to prove that:
$$\sum_{cyc}\frac{S\cos\alpha}{4R^3\sin^3\alpha}\geq\frac{3}{4R}$$ or
$$\sum_{cyc}\frac{\cos\alpha}{\sin^3\alpha}\geq\frac{3R^2}{S}$$ or
$$\sum_{cyc}\frac{b^2+c^2-a^2}{2bc\cdot\frac{8S^3}{b^3c^3}}\geq\frac{3a^2b^2c^2}{16S^3}$$ or
$$\sum_{cyc}b^2c^2(b^2+c^2-a^2)\geq3a^2b^2c^2$$ or
$$\sum_{cyc}c^2(a^2-b^2)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$(A^{-1}+B^{-1})^{-1}=A-A(A+B)^{-1}A$
Let $A, B$ and $A+B$ be invertible matrices. Show that
$(A^{-1}+B^{-1})^{-1}=A-A(A+B)^{-1}A$
Here's my assumption:
\begin{align}
A^{-1} + B^{-1} &= B^{-1} + A^{-1} \\
&= B^{-1}(A + B)A^{-1} \\
\end{align}
Then
\begin{align}
(A^{-1} + B^{-1})^{-1} &= [B^{-1}(A + B)A^{-1}]^{-1} \\
&=A(A+B)^{-1}B
\end{align}
\begin{align}
(A^{-1} + B^{-1})^{-1}B^{-1} &= A(A+B)^{-1}
\end{align}
\begin{align}
-(A^{-1} + B^{-1})^{-1}B^{-1}A &=- A(A+B)^{-1}A
\end{align}
\begin{align}
A-(A^{-1} + B^{-1})^{-1}B^{-1} A&=A- A(A+B)^{-1}A
\end{align}
But I'm stuck. I would be very appreciate if someone could help.
| Hint:
$$
A = A(A+B)^{-1}(A+B)
$$
| {
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Show that order of $2^m t-1$ is same as of order of $2^m t+1$ modulo $2^n$ where $n\geqslant 3$. Given that $n\geqslant 3$ and $n\geqslant m\geqslant 2$. Assume that, for some $t$ within the range $\{0,1,2,\ldots, 2^{n-m}-1\}$, the following holds,
\begin{align*}
&(2^m t+1)^{2^{n-m}}\equiv 1[2^n]\\
&(2^m t+1)^{2^{k}}\not\equiv 1[2^n]
\end{align*}
for any $1\leqslant k\leqslant n-m-1$. How to show that the same is true if we replace $2^m t+1$ with $2^mt-1$ ?
Tried to check through particular value, but general reasoning isn't getting. Any help please ?
| Recall that $5$ has order $2^{n-2}$ modulo $2^n$, that's we have $5^x\equiv 1\pmod{2^n}$ if and only if $x\equiv 0\pmod{2^{n-2}}$.
Let $t\in\Bbb Z$, $h$ such that $2^h\mid t$ and $2^{h+1}\nmid t$ and $a=1+2^mt$.
Then $a\equiv 1\pmod{2^{m+h}}$, hence $a\equiv 5^u\pmod{2^n}$ with $u\equiv 0\pmod{2^{m+h-2}}$.
Since, modulo $2^{n}$, $5$ has order $2^{n-2}$, $5^u$ has order $2^{n-2}/2^{m+h-2}=2^{n-m-h}$.
This proves that $a$ has order $2^{n-m}$ if and only if $h=0$, that's $2\nmid t$.
Consequently, if $(1+2^mt)^{2^k}\not\equiv 1\pmod{2^n}$ for every $0\lt k\lt n-m$ and $(1+2^mt)^{2^{n-m}}\equiv 1\pmod{2^n}$, then $2\nmid t$, hence $2\nmid -t$ and this implies $(1-2^mt)^{2^k}\not\equiv 1\pmod{2^n}$ for every $0\lt k\lt n-m$ and $(1-2^mt)^{2^{n-m}}\equiv 1\pmod{2^n}$ and thus $(2^mt-1)^{2^k}\not\equiv 1\pmod{2^n}$ for every $0\lt k\lt n-m$ and $(2^mt-1)^{2^{n-m}}\equiv 1\pmod{2^n}$.
| {
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A boy named John threw $3$ dice. In how many ways he can get the sum $10$?
A boy named John threw 3 dice. In how many ways can get the sum 10?
My Approach:
So there are 3 dice: $a, b, c$. Since they are dice, we can write: $1 \le a,b,c \le 6$. My equation right now is $$a+b+c=10$$ Now because, each die will roll at least a one, I subtract that from the total sum to remove the bound so my new equation has conditions $a,b,c \le 5$ and is $$a+b+c=7$$ Using stars and bars, I got the answer as $36$ but the correct answer is $36-9=27$. I want to know where the $9$ comes from.
Note: Please answer this using stars and bars
EDIT: The approach to my answer stems from this question
EDIT: stars and bars does not account for the upper bound restriction therefore, I needed to remove the $9$ other cases where either of $a, b, c$ exceeded $5$. These include the $6$ combinations of the case $6 + 1 + 0$ and the $3$ for $7+0+0$.
| Not stars and bars, but an alternative:
You need the coefficient of $x^{10}$ in $(x+x^2 + x^3 + x^4 + x^5+x^6)^3$.
Write the latter expression as $x^3(1+x+ \ldots +x^5)^3$, so we need the coefficient of $x^7$ in $$(1+x+ \ldots +x^5)^3 = \left(\frac{1-x^6}{1-x}\right)^3 = (1-x^6)^3 (1-x)^{-3} = (1-3x^6 + 3x^{12} - x^{18})\sum_{k=0}^\infty \binom{k+2}{k}x^k$$ using the binomial series and standard expansions.
So for the coefficient of $x^7$ we get two terms: $\binom{9}{7} - 3\binom{3}{1} = 36 - 9$ which nicely corresponds with your answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Construct an alternating series such that $\sum (-1)^n a_n$ converge but $\sum (-1)^n a_n^3$ diverge where $a_n>0$ if I write it as the follow:
$a_1-a_2+a_3-a_4...$ converge but $a_1^3-a_2^3+a_3^3-a_4^3...$ diverge
What I have thought is that by making the odd term($a_{2k+1}^3$) a diverge series like $1,\frac{1}{3}, \frac{1}{5}...$,
and the even term($a_{2k}^3$) as something converge with a smaller order such as $\frac{1}{2^3},\frac{1}{4^3}...$, then $\sum (-1)^n a_n^3=1-\frac{1}{2^3}+\frac{1}{3}-\frac{1}{4^3}...$ would be a diverge series, but I can not come up with a satisfactory result to make the original series converge
Hope you could help
| Consider the following sequence for $n \geq 0$:
\begin{align*}
x_{2n} &= \frac{2}{\sqrt[3]{n+1}}\\
x_{4n+1} &= \frac{3}{\sqrt[3]{n+1}}\\
x_{4n+3} &= \frac{1}{\sqrt[3]{n+1}}
\end{align*}
Then $\sum (-1)^n x_n$ converges to $0$ whereas $\sum (-1)^n x_n^3$ diverges since $$x_{4n}^3 - x_{4n+1}^3 + x_{4n+2}^3 - x_{4n+3}^3 = -\frac{20}{n+1}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Roots of $\frac{{2{P^2}}}{x} + \frac{{3{Q^2}}}{{x - 1}} = 6$ Let $\alpha ,\beta $ are the roots of $\frac{{2{P^2}}}{x} + \frac{{3{Q^2}}}{{x - 1}} = 6,P,Q \in {R_0}$ where $R_0$ represent non-zero real number
(A) $\alpha ,\beta \in \left( {0,1} \right)$
(B) $\alpha ,\beta \in \left( {1,\infty } \right)$
(C) One root lies in $(0, 1)$ and other root lies in $(1, \infty)$
(D) $\alpha ,\beta $ are imaginary roots
My approach is as follow $y = \frac{{2{P^2}}}{x} + \frac{{3{Q^2}}}{{x - 1}} - 6$
Putting $x=\beta$ and $y=0$
$ \frac{{2{P^2}}}{\beta} + \frac{{3{Q^2}}}{{\beta - 1}} = 6$
Putting $x=\alpha$ and $y=0$
$ \frac{{2{P^2}}}{\alpha} + \frac{{3{Q^2}}}{{\alpha - 1}} = 6$
I am not be to proceed as this is not a quaratic equatiom
| $2P^2$, $3Q^2$ terms and $6$ in the right hand side are annoying, and the only important thing is that they are positive. So we can let $p = P/\sqrt{3}$ and $q = Q/\sqrt{2}$ to have
$$ \frac{p^2}{x} + \frac{q^2}{x-1} = 1.$$
$x\ne0, 1$, so we have $x(x-1) = p^2(x-1) + q^2 x$, or $$x^2 - (1+p^2+ q^2)x + p^2 =0.$$ Two roots are
$$x_{\pm}=\frac{1+p^2 + q^2 \pm \sqrt{(1+p^2+q^2)^2 - 4p^2 }}{2}. $$
So the situation can be reduced into the problem of quadratic equation.
We can see this in another way. We start with the reduced
$$ f(x) = \frac{p^2}{x} + \frac{q^2}{x-1}.$$
When $x>0$ is small, $q^2/(x-1)$ is just some negative number and $\frac{p^2}{x}$ is very large. (Precisely, we can make this term as big as we want by choose smaller $x>0$.) So at some small $0<x_0<1/2$, $f(x_0)>0$.
When $x<1$ is very close to $1$, $p^2/x$ is just some number and $\frac{q^2}{x-1}$ is negative number with very big absolute value. (like $-10^{100}$.) So at some $1/2<x_1<1$ , $f(x_1)<0$.
Since $f$ is continuous function with $f(x_0)>0>f(x_1)$ with $0<x_0<x_1<1$, there must be a root $y_0\in (x_0, x_1)\subset(0,1)$ with $f(y_0) =0$. This is called the intermediate value theorem. (Google it for further information.)
Similarly by comparing the sign of $f(1+(\text{very small number)})$ and $f(\text{very large number})$, we can assert that there is one other zero $y_1(1, \infty)$.
Since it can be reduced into a quadratic equation $x^2-(1+p^2+q^2)x+p^2$, we can assert that there are 2 or less number of zeros. So $y_0$ and $y_1$ is all of the zeros.
(Thus the answer is option (C).)
| {
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$6n^3+\cos^3n\dots$ Using Comparison Test I have a practice question which I can't figure why the solution comparison is true in the explanation:
$$\frac{6 n^{3}+\cos ^{3} n+5}{5 n^{6}+3 \cos ^{2}(n)+4}$$
Their explaination is:
$-1\leq \cos^3n \leq 1 $ and $ 0 \leq \cos^2n \leq 3$
$\frac{6 n^{3}+\cos ^{3} n+5}{5 n^{6}+3 \cos ^{2}(n)+4} \leq \frac{6 n^{3}+6}{5 n^{6}+4} \leq \frac{12}{5 n^{3}} $
But I don't understand why I can't just make the comparison:
$\frac{6 n^{3}+\cos ^{3} n+5}{5 n^{6}+3 \cos ^{2}(n)+4} \leq \frac{6}{5 n^{3}} $
Thanks.
| Your comparison is not an inequality but an asymptotic approximation for $n$ large that is
$$\frac{6 n^{3}+\cos ^{3} n+5}{5 n^{6}+3 \cos ^{2}(n)+4} \sim \frac{6}{5 n^{3}}$$
as $n \to \infty$ indeed
$$\frac{6 n^{3}+\cos ^{3} n+5}{5 n^{6}+3 \cos ^{2}(n)+4}=\frac{6}{5 n^{3}}\frac{1+\frac{\cos ^{3} n}{6n^3}+\frac{5}{6n^3}}{1+3 \frac{\cos ^{2}(n)}{5n^6}+\frac{4}{5n^6}}$$
with
$$\frac{1+\frac{\cos ^{3} n}{6n^3}+\frac{5}{6n^3}}{1+3 \frac{\cos ^{2}(n)}{5n^6}+\frac{4}{5n^6}}\to 1$$
Edit
For the given series
$$\sum_{n=1}^{\infty} \frac{6 n^{3}+\cos ^{3} n+5}{5 n^{6}+3 \cos ^{2} n+4}$$
we can conclude in both ways by the inequality (direct comparison test) or by the asymptotic approximation (limit comparison test).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
$(1+x+x^2)^n=P_0+P_{1}x + P_{2}x^2+ \cdots +P_{2n}x^{2n}$ Prove that,$ P_0+P_{3}+P_{6}+ \cdots =3^{(n-1)}$ Let's say $$ S_n = (1+x+x^2)^n $$
n=1 $$S_1=1+x+x^2$$
n=2 $$S_2=1+2x+3x^2+2x^3+x^4$$
n=3 $$S_3=1+3x+6x^2+7x^3+6x^4+3x^5+x^6$$
n=4 $$S_4=1+4x+10x^2+16x^3+19x^4+16x^5+10x^6+4x^7+x^8$$
By taking coefficients of the $S_n$ we can form this type of triangle similar to Pascal's Trinagle
$$\begin{matrix}
&&&&&&&&&1\\
&&&&&&&1&&1&&1\\
&&&&&1&&2&&3&&2&&1\\
&&&&1&&3&&6&&7&&6&&3&&1\\
&&1&&4&&10&&16&&19&&16&10&&4&&1
\end{matrix}$$
| You can also prove this based on what you've already observed about the relationship with Pascal's Triangle:
Let $ A(i,n) = \sum P_{3k + i }$ for the coefficients of $ ( 1 + x + x^2 ) ^n$.
Then show by induction that
$$ A(0, n ) = A(1, n) = A(2, n) = A(0, n-1 ) + A(1, n-1) + A(2,n-1) = 3^{n-1}.$$
In fact, the $(1+x)^n$ analogue is the well known fact that the sum of the even terms and odd terms in Pascal's triangle are both $2^{n-1}$.
| {
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"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
solve quadratic inequality for all cases I am in great danger. I am solving a general quadratic inequality $ax^2+bx+c \ge 0$ for $a \neq 0$ as follows:
Let $$ax^2 +bx+c = 0,$$ then the critical points are: $$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}.$$ Then we have three intervals $$\left(- \infty, \frac{-b - \sqrt{b^2-4ac}}{2a}\right], \left[\frac{-b + \sqrt{b^2-4ac}}{2a}, \infty\right), \text{ and } \left[ \frac{-b - \sqrt{b^2-4ac}}{2a}, \frac{-b + \sqrt{b^2-4ac}}{2a}\right].$$
I know that since the inequality is greater than 0, the first two will satisfy the inequality. Hence $$ \left(- \infty, \frac{-b - \sqrt{b^2-4ac}}{2a}\right] \cup \left[\frac{-b + \sqrt{b^2-4ac}}{2a}, \infty\right)$$ is the solution of the given quadratic inequality.
My professor keeps saying "you are wrong, answer me completely and accurately". Anyone, please help me what I am missing and what should be the correct answer?
| This answer is nothing more than a consolidation of all of the responses.
To solve: $ax^2 + bx + c \geq 0.$
Since it is presumed that $a \neq 0$, the first thing to do is to break the analysis into two cases: $a < 0$ or $a > 0$. Then, similar to the manner of deriving the quadratic equation, in each case, you must complete the square.
In both cases, I will use the observation that
$x^2 + \left(\frac{b}{a}\right)x = \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}.$
I will also use the observation that when $r,s \geq 0$, you have that
$(r \geq s) \iff (r^2 \geq s^2).$
$\underline{\text{Case 1:} ~a > 0}$
Then $x^2 + \left(\frac{b}{a}\right)x + \frac{c}{a} \geq 0.$
Therefore $\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} \geq 0.$
Therefore
$$\left(x + \frac{b}{2a}\right)^2 \geq \frac{b^2 - 4ac}{4a^2}.\tag1$$
Within Case 1: there are therefore two possibilities:
Case 1A:
$b^2 - 4ac \geq 0$.
In this event, based on inequality (1) above, then $x$ must satisfy:
$\left|x + \frac{b}{2a}\right| \geq \frac{\sqrt{b^2 - 4ac}}{2a}$.
This is satisfied when
$x \leq - \left(\frac{b}{2a}\right) - \frac{\sqrt{b^2 - 4ac}}{2a}$
or
$x \geq - \left(\frac{b}{2a}\right) + \frac{\sqrt{b^2 - 4ac}}{2a}$.
Case 1B:
$b^2 - 4ac < 0$.
Examining inequality (1) above, since case 1B is presuming that the RHS $< 0$, and since the LHS is always non-negative, inequality (1) will be satisfied for any value of $x$.
$\underline{\text{Case 2:} ~a < 0}$
Then $x^2 + \left(\frac{b}{a}\right)x + \frac{c}{a} \leq 0.$
Therefore $\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} \leq 0.$
Therefore
$$\left(x + \frac{b}{2a}\right)^2 \leq \frac{b^2 - 4ac}{4a^2}.\tag2$$
Case 2A:
$b^2 - 4ac \geq 0$.
In this event, based on inequality (2) above, then $x$ must satisfy:
$\left|x + \frac{b}{2a}\right| \leq \frac{\sqrt{b^2 - 4ac}}{|2a|}$.
This is satisfied when
$x \geq - \left(\frac{b}{2a}\right) - \frac{\sqrt{b^2 - 4ac}}{|2a|}$
and
$x \leq - \left(\frac{b}{2a}\right) + \frac{\sqrt{b^2 - 4ac}}{|2a|}$.
Case 2B:
$b^2 - 4ac < 0$.
Examining inequality (2) above, since case 2B is presuming that the RHS $< 0$, and since the LHS is always non-negative, inequality (2) will never be satisfied for any value of $x$.
| {
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} |
Expressing the $n$–th derivative of $y=\frac{1}{(1+x^2)^2}$ In an attempt to find a way to express the n-th derivative of $y=\frac{1}{(1+x^2)^2}$ using the binomial theorem I got stuck in the last computation.
I'll explain what I did:
$$f(x)=\frac{1}{(1+x^2)^2}=\frac{1}{(x+i)^2(x-i)^2}=\\
\frac{i}{4}\left( \frac{1}{x+i} \right)-\frac{1}{4}\left( \frac{1}{(x+i)^2} \right)-\frac{i}{4}\left( \frac{1}{x-i} \right)-\frac{1}{4}\left( \frac{1}{(x-i)^2} \right)$$
Then I used $y=\frac{1}{ax+b} \Rightarrow y_{n}=\frac{(-1)^nn!a^n}{(ax+b)^{n+1}}$ and $y=\frac{1}{(ax+b)^2} \Rightarrow y_{n}=\frac{(-1)^n(n+1)!a^n}{(ax+b)^{n+2}}$ to get:
$$\frac{(-1)^n}{4}\left( in!\left( \frac{(x-i)^{n+1}-(x+i)^{n+1}}{(x^2+1)^{n+1}} \right)-(n+1)!\left( \frac{(x-i)^{n+2}+(x+i)^{n+2}}{(x^2+1)^{n+2}} \right) \right)$$
Then I stopped for a second and I analized $(x-i)^{n+1}-(x+i)^{n+1}$ and $(x-i)^{n+2}+(x+i)^{n+2}$ using the binomial theorem:
-
$(x-i)^{n+1}-(x+i)^{n+1}$ = $$\sum_{k=0}^{n+1}\binom{n+1}{k}x^{n+1-k}\cdot i^k((-1)^k-1)$$
$i^k((-1)^k-1) =0$ for $k=2n$ ,whereas for $k=2n+1$ $\longrightarrow$$-2i(-1)^{\frac{k-1}{2}}$
-
$(x-i)^{n+2}+(x+i)^{n+2}$ = $$\sum_{k=0}^{n+2}\binom{n+2}{k}x^{n+2-k}\cdot i^k((-1)^k+1)$$
$i^k((-1)^k+1) =0$ for $k=2n+1$ , whereas for $k=2n$ $\longrightarrow$$2(-1)^{\frac{k}{2}}$.
And now I got stuck sorting out the indeces $n,k$ of summations; what I obtain is:
$$\frac{d^n}{dx^n}(f(x))=\\
\frac{(-1)^nn!}{2(x^2+1)^{n+1}}\cdot \sum_{k=1}^{n+1}\binom{n+1}{k}x^{n+1-k}(-1)^{\frac{k-1}{2}} - \frac{(-1)^n(n+1)!}{2(x^2+1)^{n+2}}\cdot \sum_{k=2}^{n+2}\binom{n+2}{k}x^{n+2-k}(-1)^{\frac{k}{2}}$$
But testing it I can see it’s wrong.
Can anyone help me please sorting the $n,k$ indices and perhaps have a look at my computations also?
Thank you
| The analysis and the result stated in OPs answer are sound (+1). Since I also did already the calculation, here it is together with a few hints.
We obtain
\begin{align*}
&\color{blue}{D^n}\color{blue}{f(x)}\\
&=\frac{i}{4}(-1)^nn!\frac{(x-i)^{n+1}-(x+i)^{n+1}}{\left(x^2+1\right)^{n+1}}\\
&\qquad-\frac{1}{4}(-1)^n(n+1)!\frac{(x-i)^{n+2}+(x+i)^{n+2}}{\left(x^2+1\right)^{n+2}}\tag{1}\\
&=\frac{i(-1)^nn!}{4\left(x^2+1\right)^{n+1}}\sum_{k=0}^{n+1}\binom{n+1}{k}x^{n+1-k}i^k\left((-1)^k-1\right)\\
&\qquad-\frac{i(-1)^n(n+1)!}{4\left(x^2+1\right)^{n+2}}\sum_{k=0}^{n+2}\binom{n+2}{k}x^{n+2-k}i^k\left((-1)^k+1\right)\tag{2}\\
&=-\frac{(-1)^{n}n!}{4\left(x^2+1\right)^{n+1}}\sum_{k\geq 0}\binom{n+1}{2k+1}x^{n-2k}(-1)^k(-2)\\
&\qquad-\frac{(-1)^n(n+1)!}{4\left(x^2+1\right)^{n+2}}\sum_{k\geq 0}\binom{n+2}{2k}x^{n+2-2k}(-1)^k\cdot 2\tag{3}\\
&\,\,\color{blue}{=\frac{(-1)^nn!}{2\left(x^2+1\right)^{n+2}}\left(\left(x^2+1\right)\sum_{k\geq 0}\binom{n+1}{2k+1}(-1)^kx^{n-2k}\right.}\\
&\qquad\qquad\qquad\qquad\color{blue}{-\left.(n+1)\sum_{k\geq 0}\binom{n+2}{2k}(-1)^kx^{n+2-2k}\right)}\tag{4}
\end{align*}
Comment:
*
*In (2) we apply the binomial theorem and collect terms with equal power.
*In (3) we observe that in the left-hand sum terms with even powers vanish. We take odd indices $k\to 2k+1$ only. Note we take the index range $k\geq 0$ which is admissible, since $\binom{n}{k}=0$ if $k>n$. In the right-hand sum terms with odd powers vanish and we take even indices $k\to 2k$ only.
*In (4) we simplify further by factoring out the common denominator $\left(x^2+1\right)^{n+2}$.
Note: The answer stated by @Turing is nice and correct. We observe there is just one common denominator $(x-i)^{n+2}(x+i)^{n+2}=\left(x^2+1\right)^{n+2}$. We could think to also transform (1) this way to get another binomial expansion which is somewhat different to (4).
| {
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"source": "stackexchange",
"question_score": "3",
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} |
Limit at (0,0) of $\frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4}$ $$\text{Let}\ f(x,y) = \frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4}$$
WolframAlpha tells me that $\lim_{(x,y) \to (0,0)} f(x,y)$ does not exist. To prove the non-existence of the limit, I tried three different paths ($y=kx, y=kx^2, y=kx^3)$ and they all equal zero. From the graph of the function it looks like the limit is indeed zero. Also I thought I was able to prove the limit does indeed exist at $(0,0)$ with the epsilon-delta definition:
$$y^2 \leq x^2+y^2 \leq x^2 + y^4 \iff \frac{y^2}{x^2+y^4} \leq 1 \overset{|x|<1}{\iff} \frac{|x|y^2}{x^2+y^4} \leq 1 \iff \frac{|x|y^2\sqrt{x^2+y^2}}{x^2+y^4} \leq \sqrt{x^2+y^2} \iff \left|\frac{xy^2\sqrt{x^2+y^2}}{x^2+y^4} - 0\right|\leq \left|\sqrt{(x-0)^2+(y-0)^2}\right| \lt δ := ε $$
So for $δ = ε$ we have that $\|x -(0,0)\| < δ \implies |f(x,y) - 0| < ε$
I'm confused
| One more possible way
$$\frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4}=\sqrt{ \frac{y^4x^4+y^6x^2}{(x^2 +y^4)^2}}\leqslant \sqrt{\frac{y^4x^4+y^6x^2}{2x^2y^4}}=\sqrt{\frac{x^2}{2}+\frac{y^2}{2}}$$
| {
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Proving a function is increasing in $n$. I am trying to prove that the function $\frac{n}{2n+1}$, defined for $n \in \mathbb{N}$, decreases in $\mathbb{N}$. I attempted it by induction, but I'm not convinced that I fully need induction. Why can I not prove that for an arbitrary $n$, $f(n) \leq f(n+1)$ and deduce that, because $n$ was arbitrary, this holds for all $n$? The only thing left out would be the base case, but. I'm not fully sure why I need it here.
Regardless, here is my attempt at the induction:
Let $f: \mathbb{N} \to \mathbb{R}$ be defined by $f(n) = \frac{n}{2n+1}$. We prove by induction on $n$ that $f$ is increasing in $n$. If $n = 1$, we notice that
\begin{align*}
f(1) = \frac{1}{3} \leq \frac{2}{5} = f(2).
\end{align*}
Suppose inductively that we have $f(n) \leq f(n+1)$ for some $n \geq 1$. So we have
$\frac{n}{2n+1} \leq \frac{n+1}{2n+3}$. First, we have
\begin{align*}
\frac{n+1}{2n+3} \leq \frac{n+3}{2n+3}.
\end{align*}
Furthermore, $2n + 5 \geq 2n + 3$, so $\frac{1}{2n + 5} \leq \frac{1}{2n+3}$, so $\frac{n+3}{2n + 3} \leq \frac{n+3}{2n+5}$. Therefore, it follows that
\begin{align*}
\frac{n+1}{2n+3} \leq \frac{n+3}{2n+3} \leq \frac{n+3}{2n + 5} = \frac{(n+2) + 1}{2(n+2) + 1},
\end{align*}
so $f(n+1) \leq f(n+2)$, which closes the induction
| As an alternative, we can directly check $f(n)< f(n+1)$ that is
$$\frac{n}{2n+1}<\frac{n+1}{2n+3} \iff 2n^2+3n<2n^2+3n+1 \iff0<1$$
which is true.
| {
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} |
solving trigonometry Solve $16 \cos^2 x + 6 \sin x = 17$ for $0 < x < 2\pi$
Steps:
$16(1-\sin^2x)+6\sin x = 17$
$16\sin^2x-6\sin x + 1 = 0$
let $y = \sin x$
$16y^2 - 6y + 1 = 0$
I was not able to solve this quadratic equation. It has no real roots. What was done incorrectly?
| As noticed your derivation is fine, the quadratic equation $16y^2 - 6y + 1 = 0$ has not real roots indeed $f(y)=16y^2 - 6y + 1$ represents a parabola that opens upward with vertex at $\left(\frac 3{16},\frac 7{16}\right)$.
As an alternative, we have that
$$16 \sin^2 x - 6 \sin x + 1=0 \iff 16 \sin^2 x +1= 6 \sin x $$
which can't have solutions for $\sin x \le 0$ and for $\sin x >0$ by AM-GM
$$16 \sin^2 x +1\ge 8 \sin x>6 \sin x$$
| {
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"answer_count": 1,
"answer_id": 0
} |
Does the diophantine equation $ax^2+by^2=cz^2+d$ always have solutions? Let $a,b,c,d,x,y,$ and $z \in \mathbb{N}$ where $a,b,c,$ and $d$ are constants but d is allowed to be zero .
$ax^2+by^2=cz^2+d$
First example :
when $a=b=c=1 $ and $d=0$ we have the equation : $x^2+y^2=z^2$ which I know its general solution .
Second example:
$a=1 ,b=4,c=1$ and $d=0$ we have the equation : $x^2+4y^2=z^2$ . it has solutions.
one solution of it is: $x=3,y=2,$ and $z=5$
Third example:
$a=2 ,b=3,c=1$ and $d=0$ we have the equation : $2x^2+3y^2=z^2$ .
this example I tried to find solutions among small numbers but I didn't find any solution.
So does $2x^2+3y^2=z^2$ have solutions but I didn't find any ?
or is there proof that $2x^2+3y^2=z^2$ doesn't have any solution?
| Well $2x^2+3y^2=z^2$ actually has no solution for positive integers $x,y,z$.
Assume the smallest solution of $(x,y,z)$ (with minimal $z$) is $(r,s,t)$. Consider $\mod 8$, as $2x^2+ 3y^2\equiv 0,2,3,4,5,6 \pmod{8}$ and $z^2 \equiv 0,1,4 \pmod{8}$, we can see $z^2 \equiv 0,4 \pmod{8}$, implies that $z$ is even. Let $z=2p$, then $2x^2+3y^2=4p^2$, which also implies $y$ to be even. Let $y=2q$, then $2x^2+12q^2=4p^2$, which becomes $x^2+6q^2=2p^2$, implying $x$ is also even. Let $x=2r$, then the equation becomes $4r^2+6q^2=2p^2$, which is $2r^2+3q^2=p^2$. Therefore, $(r,q,p)$ is also a solution. However, $p<z$, which means $(r,q,p)$ is a smaller solution, which contradicts the first statement. Therefore, there are no solutions for $(x,y,z)$.
| {
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Matrix derivative in multiple linear regression model The basic setup in multiple linear regression model is
\begin{align}
Y &= \begin{bmatrix}
y_{1} \\
y_{2} \\
\vdots \\
y_{n}
\end{bmatrix}
\end{align}
\begin{align}
X &= \begin{bmatrix}
1 & x_{11} & \dots & x_{1k}\\
1 &x_{21} & \dots & x_{2k}\\
\vdots & \dots & \dots\\
1 & x_{n1} & \dots & x_{nk}
\end{bmatrix}
\end{align}
\begin{align}
\beta &= \begin{bmatrix}
\beta_{0} \\
\beta_{1} \\
\vdots \\
\beta_{k}
\end{bmatrix}
\end{align}
\begin{align}
\epsilon &= \begin{bmatrix}
\epsilon_{1} \\
\epsilon_{2} \\
\vdots \\
\epsilon_{n}
\end{bmatrix}
\end{align}
The regression model is $Y=X \beta + \epsilon$.
To find least square estimator of $\beta$ vector, we need to minimize $S(\beta)=\Sigma_{i=1}^n \epsilon_i^2 = \epsilon ' \epsilon = (y-x\beta)'(y-x\beta)=y'y-2\beta 'x'y + \beta 'x'x \beta$
$$\frac{\partial S(\beta)}{\partial \beta}=0$$
My question: how to get $-2x'y+2x'x \beta$?
| The sum of the squared errors can be written as
$$
\left\lVert \epsilon \right\rVert^2 = \left\lVert Y - X\beta \right
\rVert^2
$$
$$
= (Y - X\beta)^T(Y - X\beta) = Y^TY - \beta^TX^TY - Y^TX\beta + \beta^TX^TX\beta
$$
$$
= \left\lVert Y \right\rVert^2 - 2Y^TX\beta + \left\lVert X\beta \right\rVert^2
$$
Then, finding the gradient $\frac{d\left\lVert \epsilon \right\rVert^2}{d\beta}$
$$
\frac{d\left\lVert \epsilon \right\rVert^2}{d\beta} = -2X^TY + 2X^TX\beta
$$
Reviewing term by term in that differentiation (since differentiation is linear operator!)
$\left\lVert Y \right\rVert^2$ does not depend on $\beta$ and becomes $0$.
$2Y^TX\beta$ is a is a sum where the $i^{th}$ term is $2\beta_iY^Tx_i$ where $x_i$ is the $i^{th}$ row of $X$. Therefore, the gradient of this term evaluates to $-2X^TY$.
$\left\lVert X\beta \right\rVert^2$ can be differentiated using the product rule
In general, the Jacobian of $f(x) = Ax $ is $ J_f = A^T$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Solving $\frac{x^{11}+x}{x^7+x^5}=\frac{205}{16}$
Blockquote
Find $x$ satisfy the equation$$\frac{x^{11}+x}{x^7+x^5}=\frac{205}{16}$$
Here is my work
$$\frac{x^{10}+1}{x^6+x^4}=\frac{205}{16},\quad\quad t:=x^2$$
$$\frac{t^5+1}{t^3+t^2}=\frac{205}{16}$$
$$16t^5-205t^3-205t^2+16=0$$
$t=-1$ is a root of the polynomial but not sure how to find the other roots.
Another approach I tried is dividing numerator and denominator of the original fraction by $x^6$,
$$\frac{x^5+\frac1{x^5}}{x+\frac1x}=\frac{205}{16}$$
$$x^4-x^2+1-\frac1{x^2}+\frac1{x^4}=\frac{205}{16}$$
| HINT
You are on the right track!
Notice that
\begin{align*}
x^{2} + \frac{1}{x^{2}} = \left(x + \frac{1}{x}\right)^{2} - 2
\end{align*}
Similarly, we do also have that
\begin{align*}
x^{4} + \frac{1}{x^{4}} & = \left(x^{2} + \frac{1}{x^{2}}\right)^{2} - 2\\\\
& = \left[\left(x + \frac{1}{x}\right)^{2} - 2\right]^{2} - 2
\end{align*}
Now you can make the substitution
\begin{align*}
u = x + \frac{1}{x}
\end{align*}
Then we get that
\begin{align*}
[(u^{2} - 2)^{2} - 2] - (u^{2} - 2) + 1 = \frac{205}{16} \Longleftrightarrow u^{4} - 5u^{2} + 5 = \frac{205}{16}
\end{align*}
which is a biquadratic equation.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4250523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Question about exact ODE. Why in $h'(y)$ contain $x$? Find the solution ODE
$$\left(x^3e^xy+4x^2e^xy+2xe^xy\right)dx+(x^3e^x+x^2e^x)dy=0.$$
Let $M(x,y)=x^3e^xy+4x^2e^xy+2y$ and $N(x,y)=x^3e^x+x^2e^x$.
\begin{align*}
\dfrac{\partial M}{\partial y}&=\dfrac{\partial}{\partial y}\left(x^3e^xy+4x^2e^xy+2xe^xy\right) \\
&=x^3e^x+4x^2e^x+2xe^x.\\
\dfrac{\partial N}{\partial x}&=\dfrac{\partial}{\partial x}\left(x^3e^x+x^2e^x\right) \\
&=3x^2e^x+x^3e^x+2xe^x+x^2e^x\\
&=x^3e^x+4x^2e^x+2xe^x.
\end{align*}
This is exact ODE since $\dfrac{\partial M}{\partial y}= \dfrac{\partial N}{\partial x}$.
Now,
\begin{alignat}{2}
&&\dfrac{\partial F(x,y)}{\partial x}&=M(x,y)\nonumber\\
\Longleftrightarrow\quad
&&\dfrac{\partial F(x,y)}{\partial x}&=x^3e^xy+4x^2e^xy+2y\nonumber\\
\Longleftrightarrow\quad
&&\int\partial F(x,y)&=\int\left(x^3e^xy+4x^2e^xy+2y\right) \partial x\nonumber\\
\Longleftrightarrow\quad
&&\int\partial F(x,y)&=\int x^3e^xy\partial x+\int4x^2e^xy \partial x+\int2y \partial x.\label{ijoet}
\end{alignat}
Consider that
\begin{align}
\int x^3e^xy\partial x&=x^3e^xy-\int e^xy 3x^2 \partial x\nonumber\\
&=x^3e^xy-\int 3x^2 e^x y \partial x\nonumber\\
&=x^3e^xy-\left(3x^2e^x y-\int e^x y 6x \partial x\right)\nonumber\\
&=x^3e^xy-3x^2e^x y+\int 6x e^x y\partial x\nonumber\\
&=x^3e^xy-3x^2e^x y+6x e^x y-\int e^x y 6\partial x\nonumber\\
&=x^3e^xy-3x^2e^x y+6x e^x y- 6 e^x y+h_1(y).\label{meong}
\end{align}
\begin{align}
\int 4x^2e^xy\partial x&=4x^2e^xy-\int e^xy 8x \partial x\nonumber\\
&=4x^2e^xy-\int 8x e^xy \partial x\nonumber\\
&=4x^2e^xy-\left(8x e^xy-\int e^xy 8 \partial x\right)\nonumber\\
&=4x^2e^xy-8x e^xy+\int 8e^xy \partial x\nonumber\\
&=4x^2e^xy-8x e^xy+ 8e^xy +h_2(y).\label{meong1}
\end{align}
\begin{align}
\int2y \partial x&= 2xy+h_3(y).\label{meong2}
\end{align}
So, we have
\begin{alignat}{2}
&&\int\partial F(x,y)&=x^3e^xy-3x^2e^x y+6x e^x y- 6 e^x y+h_1(y)\nonumber\\
&&&\quad +4x^2e^xy-8x e^xy+ 8e^xy +h_2(y)+2xy+h_3(y)\nonumber\\
\Longleftrightarrow\quad
&&F(x,y)&=x^3e^xy+x^2e^x y-2 x e^x y+2e^x y+2xy +h(y)\nonumber
\end{alignat}
which $h(y)=h_1(y)+h_2(y)+h_3(y)$.
Next, consider that
\begin{alignat*}{2}
&&\dfrac{\partial F(x,y)}{\partial y}&=N(x,y)\\
\Longleftrightarrow\quad
&&\dfrac{\partial}{\partial y}\left(x^3e^xy+x^2e^x y-2 x e^x y+2e^x y+2xy +h(y)\right)&=x^3e^x+x^2e^x\\
\Longleftrightarrow\quad
&&x^3e^x+x^2e^x -2 x e^x +2e^x +2x +h'(y)&=x^3e^x+x^2e^x\\
\Longleftrightarrow\quad
&&h'(y)&=2 x e^x -2e^x -2x \\
\end{alignat*}
When I want to find $h(y)$, I have found $h'(y)=2 x e^x -2e^x -2x$.
Why in $h'(y)$ contain $x$? What my mistake?
| You wrote $M(x,y)=x^3e^xy+4x^2e^xy+2y$, but it should be $M(x,y)=x^3e^xy+4x^2e^xy+2\color{red}{xe^x}y.$ Then we have
$$\int2xe^xy\partial x=2xe^xy-2e^xy+h_3(y)$$
So we have
$$F(x,y)=x^3e^xy+x^2e^x y-2 x e^x y+2e^x y+\color{red}{2xe^{x}y-2e^xy }+h(y)$$
$$=x^3e^xy+x^2e^x y+h(y)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve the differential equation. $y^\prime = y^2+\frac{1}{x^4}$, $y=\frac{1}{x^2}\text{ctg}(\frac{1}{x}+c) - \frac{1}{x}$ and $y(+\infty)=0$ Solve the differential equation.
$$y^\prime = y^2+\frac{1}{x^4}.$$
I am given this as a solution.Need to choose one which satisfies given condition.($y(+\infty)=0$)
$$y=\frac{1}{x^2}\cot(\frac{1}{x}+c) - \frac{1}{x}$$ and $$y(+\infty)=0.$$
We need to find $c$ and show that it is solution for differential equation.
So
$$\lim_{x\to +\infty}\frac{1}{x^2}\cot(\frac{1}{x}+c) - \frac{1}{x} =0
\\
\lim_{x\to+\infty}\frac{1}{x^2}\frac{\cos(\frac{1}{x}+c)}{\sin(\frac{1}{x}+c)} =0$$
From here I don't know how to find $c$. Will be glad if you can help me.
| Let's make the variable change
\begin{equation}
t = \frac{1}{x}
\end{equation}
Then
\begin{equation}
\lim_{x\rightarrow +\infty}\left(\frac{1}{x^{2}}\cdot\cot\left(\frac{1}{x} + c\right) - \frac{1}{x}\right) = \lim_{t\rightarrow 0}\left(t^{2}\cot\left(t + c\right) - t\right) = 0
\end{equation}
Therefore any value of $c$ will satisfies this equation.
Next for the left part of equation
\begin{equation}
y’ = - \frac{2\cot\left(\frac{1}{x} + c\right)}{x^{3}} + \frac{1}{x^{4}\sin^{2}\left(\frac{1}{x} + c\right)} + \frac{1}{x^{2}}
\end{equation}
For the right part
\begin{multline}
y^2 + \frac{1}{x^{4}} = \frac{\cot^{2}\left(\frac{1}{x} + c\right)}{x^{4}} - \frac{2\cot\left(\frac{1}{x} + c\right)}{x^{3}} + \frac{1}{x^{4}} + \frac{1}{x^{4}} = \\ = \frac{\cot^{2}\left(\frac{1}{x} + c\right) + 1}{x^{4}} - \frac{2\cot\left(\frac{1}{x} + c\right)}{x^{3}} + \frac{1}{x^{4}} = \\ =\frac{\cos^{2}\left(\frac{1}{x} + c\right) + \sin ^{2}\left(\frac{1}{x} + c\right)} {x^{4}\sin^{2}\left(\frac{1}{x} + c\right)} - \frac{2\cot\left(\frac{1}{x} + c\right)}{x^{3}} + \frac{1}{x^{4}} = \\ = - \frac{2\cot\left(\frac{1}{x} + c\right)}{x^{3}} + \frac{1}{x^{4}\sin^{2}\left(\frac{1}{x} + c\right)} + \frac{1}{x^{2}}
\end{multline}
Therefore
\begin{equation}
y’ = y^{2} + \frac{1}{x^{4}}
\end{equation}
which means that the given $y$ is a solution for this equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trying to prove the Pythagorean theorem using Picks theorem. Picks Theorem Let A be the area of a simply closed lattice square. Let B denote the number of lattice points on the square edges and I the number of points in the interior of the square. Then
$\large A=I + \frac{B}{2}-1$
Define three squares with areas
$\large A_{a} = I_{a} + \frac{B_{a}}{2}-1 = a^2$
$\large A_{b} = I_{b} + \frac{B_{b}}{2}-1 = b^2$
$\large A_{c} = I_{c} + \frac{B_{c}}{2}-1 = c^2$
Theorem
For $B_{c}=4$, $A_{c} = A_{a} + A_{b}$.
Proof
Case $B_{c}=4$,
$\large A_{c} = I_{c} + \frac{4}{2}-1$
$\large A_{c} = I_{c} + 1$
Observe that
$\large I_{c} = A_{a} + A_{b} - 1$
Substituting $I_{c}$
$\large A_{c} = (A_{a} + A_{b} - 1) + 1$
$\large A_{c} = A_{a} + A_{b}$
$\therefore \large c^2 = a^2 + b^2 $
Questions
Is that a valid proof for specific case $B_{c}=4$?
For general cases, can Picks be applied to prove Pythagoras?
Thanks.
| The following describe a possible general proof. You just need to calculate the number of grid points in a triangle and the square as function of the sides of the right angle triangles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Problem while calculating the area of $S^2$ using differential forms I am trying to calculate the area of the semi-sphere $S^2_-$ using differential forms, I have the only local chart $(A,\phi,S^2_-)$ given by $A=\{(x,y)\in\mathbb{R}:x^2+y^2< 1\}$ and $\phi(x,y)={^t(-x, y, -\sqrt{1-x^2-y^2})}$.
Also I have the unitary normal vector field $N(x,y)={^t(-x, y, -\sqrt{1-x^2-y^2})}$, so that $\{N,\frac{\partial{\phi}}{\partial{x}},\frac{\partial{\phi}}{\partial{y}}\}$ form an oriented basis of $\mathbb{R}^3$.
Now if I calculate by hand $\phi^*(\iota(N)dx\wedge dy\wedge dz)$ I get
\begin{align}
&\phi^*(-xdy\wedge dz-y dx\wedge dz-\sqrt{1-x^2-y^2}dx\wedge dy)\\
&=-(-x)dy\wedge\left(\frac{x}{\sqrt{1-x^2-y^2}}dx+\frac{y}{\sqrt{1-x^2-y^2}}dy\right)\\
&\qquad-yd(-x)\wedge\left(\frac{x}{\sqrt{1-x^2-y^2}}dx+\frac{y}{\sqrt{1-x^2-y^2}}dy\right)\\
&\qquad-\sqrt{1-x^2-y^2}d(-x)\wedge dy\\
&=\left(\frac{-x^2}{\sqrt{1-x^2-y^2}}+\frac{y^2}{\sqrt{1-x^2-y^2}}+\sqrt{1-x^2-y^2}\right)dx\wedge dy
\end{align}
so $\int_{S^2_-}\iota(N)dx\wedge dy\wedge dz=\int_{A}\phi^*(\iota(N)dx\wedge dy\wedge dz)=\int_A \frac{1-2x^2}{\sqrt{1-x^2-y^2}}dxdy$.
At this point I know there is a mistake because last integral should be $\int_A \frac{1}{\sqrt{1-x^2-y^2}}dxdy$ but I really can't figure out where this mistake is.
| There's an extra minus sign in the definition of the normal vector field: it should be $$\mathbf{N}=\left(x,y,-\sqrt{1-x^2-y^2}\right).$$ Indeed, consider the point $(1,0,0)$ of the sphere, whose normal vector should be $(1,0,0)$ and not $(-1,0,0)$ since the normal vector by definition points outward and not inward. After fixing that, letting $$\gamma=\iota(\mathbf{N})(dx\wedge dy\wedge dz)=x\,dy\wedge dz - y\,dx\wedge dz -\sqrt{1-x^2-y^2}\,dx\wedge dy$$ be the area form, by definition $$Area(S^2_{-}) = \int_{[S^2_{-}]} \gamma = \int_A \phi^{\star}(\gamma)$$ where \begin{align*} \phi^{\star} (\gamma) &= -x\, dy \wedge \left( \frac{x}{\sqrt{1-x^2-y^2}}dx + \frac{y}{\sqrt{1-x^2-y^2}}dy \right) \\ & \ \ \ \ - y\, d(-x) \wedge \left( \frac{x}{\sqrt{1-x^2-y^2}}dx + \frac{y}{\sqrt{1-x^2-y^2}}dy \right) \\ &\ \ \ \ -\sqrt{1-x^2-y^2}\, d(-x)\wedge dy \\ &= \left(\frac{x^2}{\sqrt{1-x^2-y^2}} +\frac{y^2}{\sqrt{1-x^2-y^2}} +\sqrt{1-x^2-y^2}\right)\,dx\wedge dy \\ &= \frac{1}{\sqrt{1-x^2-y^2}}\,dx\wedge dy\end{align*} and you can then proceed as you intended to.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\frac{3}{n} \sum_{k=1}^n \left[\left(\frac{3k}{n}\right)^2 - \left(\frac{3k}{n}\right)\right]$ using Riemann sums I am asked to find the following limit:
$$\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n \left[\left(\frac{3k}{n}\right)^2 - \left(\frac{3k}{n}\right)\right].$$
I am trying to invoke the following rule
$$\int_a^b f(x) dx = \lim_{n \rightarrow \infty } \sum_{k=1}^n f(x+k \Delta x) \Delta x$$
Where $\Delta x = \frac{b-a}{n}$.
So far what I have is that my function $f$ is $f(x)=x^2$ and that $a=1,b=4$. Then I get $$\int_1^4 x^2 dx$$ which is not the correct answer. Can someone help me with where I went wrong? Thanks in advance. Don't necessarily need answer just a hint or someone point out what obvious mistake I am making. Sorry. Also if this is a repeat question I apologize I will take it down, couldn't find one.
| Alternative approach:
Hello from the rectangle-happy Anti-Calculus League.
$$\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n \left[\left(\frac{3k}{n}\right)^2 - \left(\frac{3k}{n}\right)\right].$$
$\sum_{r=1}^n r^2 = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n.$
$\sum_{r=1}^n r^1 = \frac{1}{2}n^2 + \frac{1}{2}n.$
$\displaystyle \sum_{k=1}^n (3k)^2 = 9\sum_{k=1}^n k^2 =
\frac{9}{3}n^3 + \frac{9}{2}n^2 + \frac{9}{6}n.$
Therefore,
$\displaystyle \sum_{k=1}^{n} \left(\frac{3k}{n}\right)^2 = \frac{1}{n^2} \times \left[\frac{9}{3}n^3 + \frac{9}{2}n^2 + \frac{9}{6}n\right].$
This equals
$\displaystyle \frac{9n}{3} + \frac{9}{2} + \frac{9}{6n}.$
$\displaystyle \sum_{k=1}^n (3k) = 3\sum_{k=1}^n k =
\frac{3}{2}n^2 + \frac{3}{2}n.$
Therefore,
$\displaystyle \sum_{k=1}^{n} \left(\frac{3k}{n}\right) = \frac{1}{n} \times \left[\frac{3}{2}n^2 + \frac{3}{2}n\right].$
This equals
$\displaystyle \frac{3n}{2} + \frac{3}{2}.$
Therefore, ignoring the leading factor of $\displaystyle \frac{3}{n}$ and ignoring that you are taking a limit, you have the intermediate computation of
$\displaystyle \frac{3n}{2} + 3 + \frac{3}{2n}.$
Applying the leading factor $\displaystyle \frac{3}{n}$ gives
$$\frac{9}{2} + \frac{9}{n} + \frac{9}{2n^2}.\tag1$$
Clearly, when applying the limit, as $n \to \infty$, all but the leftmost term in (1) above go to $0$.
Therefore, the limit is $\displaystyle \left(\frac{9}{2}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4261949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to Prove $\lim_{(x, y) \rightarrow (0, 0)}\frac{x^3-y^3}{x^2+y^2}=0$ I wanted to prove how
$$
\lim_{(x, y) \to (0, 0)} \frac{x^3-y^3}{x^2+y^2} = 0.
$$
Specifically, I want to use the Squeeze Theorem for multivariable calculus. Then I know that I should pick an arbitrary function $g(x, y)$ where $|f(x, y)-L| \leq g(x, y)$ and the limit of $g(x, y)$ is $0$. Since the limit of $f(x, y) =0$, I just have to make
$$
\Biggl\lvert \frac{x^3-y^3}{x^2+y^2} \Biggr\rvert \leq g(x, y),
$$
where the limit of $g(x,y)=0$. I was given a hint that $|x^3-y^3| \leq |x|^3+|y|^3$. So I divided both sides by $|x^2+y^2|$,
$$
\frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert}
\leq \frac{\lvert x\rvert^3+ \lvert y\rvert^3}{\bigl\lvert x^2+y^2 \bigr\rvert},
$$
meaning that
$$
\frac{\bigl\lvert x^3-y^3 \bigr\rvert}{\bigl\lvert x^2+y^2 \bigr\rvert}
\leq \lvert x\rvert + \lvert y\rvert.
$$
What do I do next?
| Here it is a similar way to approach it:
\begin{align*}
0\leq x^{2} \leq x^{2} + y^{2} & \Rightarrow 0\leq \frac{x^{2}}{x^{2} + y^{2}}\leq 1\\\\
& \Rightarrow \left|\frac{x^{2}}{x^{2} + y^{2}}\right| \leq 1\\\\
& \Rightarrow \left|\frac{x^{3}}{x^{2} + y^{2}}\right|\leq |x|
\end{align*}
Then you can apply the squeeze theorem as $(x,y)\to(0,0)$.
Similarly, we do also have that
\begin{align*}
\left|\frac{y^{3}}{x^{2} + y^{2}}\right| \leq |y|
\end{align*}
where we can apply the squeeze theorem as well.
Along with the triangle inequality, the desired result holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding inverse function of $y=\sqrt{x+\sqrt{2x-1}}$
If inverse of the function $y=\sqrt{x+\sqrt{2x-1}}$ be equal to
$y=ax^2+bx+c$, then what is the value of $a^2+b^2+c^2$?
$A) 3$
$B) 4$
$C) 5$
$D) 6$
I tried to isolate $x$ from the equation $y=\sqrt{x+\sqrt{2x-1}}$:
$$y^2=x+\sqrt{2x-1}$$
$$(y^2-x)^2=2x-1$$
$$y^4-2xy^2+x^2=2x-1$$
$$y^2(y^2-2x)=-x^2+2x-1$$
But I still have $-2x$ in the parenthesis of the LHS.
| Let $t=\sqrt{2x-1}\geq 0$, then $x= {t^2+1\over 2}$ so $$y = \sqrt{t^2+2t+1\over 2} = {t+1\over \sqrt{2}}$$
Now is easy to finish.
If I didn't make any mistake, then inverse function is
$$y=x^2-x\sqrt{2}+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.