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Exact expression of a trigonometric integral Let $a>2$ be a real number and consider the following integral
$$
I(a)=\int_0^\pi\int_0^\pi \frac{\sin^2(x)\sin^2(y)}{a+\cos(x)+\cos(y)} \mathrm{d}x\,\mathrm{d}y
$$
My question. Does there exist a closed-form expression of $I(a)$?
Some comments. Since $a-2<a+\cos(x)+\cos(y)<a+2$ and $\int_0^\pi \int_0^\pi \sin^2(x)\sin^2(y)\ \mathrm{d}x\, \mathrm{d}y=\frac{\pi^2}{4}$, we have the following bounds
$$
\frac{\pi^2}{4(a+2)} < I(a) < \frac{\pi^2}{4(a-2)},
$$
however I didn't manage to find an exact expression for $I(a)$. Any help is welcome!
|
We use the identities
$$
2\sin \phi \sin \psi=\cos(\phi-\psi)-\cos(\phi+\psi)
$$
and
$$
\cos \phi+\cos \psi=2\cos\left(\frac{\phi+\psi}{2}\right)\cos\left(\frac{\phi-\psi}{2}\right)
$$
to get
$$
I=\int^{\pi}_{0}\int^{\pi}_{0}\frac{\sin^2 x \sin^2 y}{a+\cos x+\cos y}dxdy=
$$
$$
\frac{1}{4}\int^{\pi}_{0}\int^{\pi}_{0}\frac{(\cos(x-y)-\cos(x+y))^2}{a+2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)}dxdy
$$
Now we use $\cos \phi=2\cos^2(\phi/2)-1$ and arrive to
$$
I=\int^{\pi}_{0}\int^{\pi}_{0}\frac{(\cos^2\left(\frac{x-y}{2}\right)-\cos^2\left(\frac{x+y}{2}\right))^2}{a+2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)}dxdy=
$$
$$
=a^{-1}\int^{\pi}_{0}\int^{\pi}_{0}\frac{(\cos^2 A-\cos^2 B)^2}{1+2/a \cos A\cos B}dAdB.
$$
Then set $t=\cos A$, $s=\cos B$ and use
$$
\frac{d\left(\cos^{(-1)}(x)\right)}{dx}=-\frac{1}{\sqrt{1-x^2}}
$$
to get
$$
I=1/a\int^{1}_{-1}\int^{1}_{-1}\frac{(t^2-s^2)^2}{1+(2/a)ts}\frac{1}{\sqrt{1-t^2}}\frac{1}{\sqrt{1-s^2}}dtds
$$
If $|2/a|\leq 1$ and $|t|,|s|<1$, we expand $(1+(2/a)ts)^{-1}$ into $\sum^{\infty}_{k=0}(-1)^k(2/a)^kt^ks^k$ we can split the two integrals (that of $t$ and that of $s$). We set
$$
I_k:=1/a\int^{1}_{-1}\int^{1}_{-1}(t^2-s^2)^2(-1)^k(2/a)^kt^ks^k\frac{1}{\sqrt{1-t^2}}\frac{1}{\sqrt{1-s^2}}dtds.
$$
Then $I_{2k+1}=0$ and
$$
I_{2k}=\frac{2^{2k}(1/a)^{2k+1}\pi\Gamma\left(\frac{2k+1}{2}\right)\Gamma\left(\frac{2k+3}{2}\right)}{\Gamma\left(k+2\right)\Gamma(k+3)}.
$$
Hence summing
$$
I=I(a)=\sum^{\infty}_{k=0}I_{2k}=\frac{a\pi^2}{2}-\frac{a\pi^2}{2}{}_2F_1\left(\frac{-1}{2},\frac{1}{2};2;\frac{4}{a^2}\right)=
$$
$$
=\frac{a\pi^2}{2}-\frac{a(a^2+4)\pi}{6}E\left(4a^{-2}\right)+\frac{a(a^2-4)\pi}{6}K(4a^{-2}),
$$
where $K(x)$,$E(x)$ are the complete elliptic integrals of the first and second kind. Examples of evaluations can be given if we consider the singular modulus $k_r$. For example with $a=2\sqrt{2}$, we get $k_1=1/2$ and we have
$$
I\left(2\sqrt{2}\right)=\sqrt{2}\pi^2-2\sqrt{2\pi}\Gamma\left(\frac{3}{4}\right)^2-\frac{8}{3}\sqrt{2\pi}\Gamma\left(\frac{5}{4}\right)^2
$$
...etc
see elliptic integrals singular valus
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that: $\prod_{n=1}^{\infty}\frac{\sqrt{5}F_{2n+2}+1}{\sqrt{5}F_{2n+2}-1}=\phi$ How to show that:
$$P=\prod_{n=1}^{\infty}\frac{\sqrt{5}F_{2n+2}+1}{\sqrt{5}F_{2n+2}-1}=\phi$$
Where $\phi=\frac{1+\sqrt{5}}{2}$
$$P=\frac{\sqrt{5}F_{4}+1}{\sqrt{5}F_{4}-1}\cdot \frac{\sqrt{5}F_{6}+1}{\sqrt{5}F_{6}-1}\cdot \frac{\sqrt{5}F_{8}+1}{\sqrt{5}F_{8}-1}\cdots$$
$$P=\frac{3\sqrt{5}+1}{3\sqrt{5}-1}\cdot \frac{8\sqrt{5}+1}{8\sqrt{5}-1}\cdot \frac{21\sqrt{5}+1}{21\sqrt{5}-1}\cdots$$
I thought maybe I try and alter its form to makes it easy...but it is not apparently
$$P=\frac{2\phi+(3-1)\sqrt{5}}{2\cdot3\phi-(3+1)}\cdot\frac{2\phi+(8-1)\sqrt{5}}{2\cdot8\phi-(8+1)}\cdot\frac{2\phi+(21-1)\sqrt{5}}{2\cdot21\phi-(21+1)}\cdots$$
|
Let $F_n$ denote the Fibonacci sequence and $L_n$ denote the Lucas sequence. Note the following identities
\begin{eqnarray*}
5F_{2n}^2-1&=&L_{2n-1}L_{2n+1} \\
2F_{2n+2}-F_{2n+1}&=&L_{2n+1} \\
5F_{2n+1}F_{2n+2}-2&=&L_{2n+1}L_{2n+2} \\
5F_{2n+2}-L_{2n+2}&=&2L_{2n+1} \\
L_{2n+2}F_{2n+2}-1&=&L_{2n+1}F_{2n+3}.
\end{eqnarray*}
Start by noting that
\begin{eqnarray*}
\frac{3 \sqrt{5}+1}{3 \sqrt{5}-1}= \left( \frac{1+\sqrt{5}}{2}\right) \left(\frac{-2+5 \sqrt{5}}{11} \right)
\end{eqnarray*}
Now we proceed inductively
\begin{eqnarray*}
P_m=\prod_{n=1}^{m}\frac{\sqrt{5}F_{2n+2}+1}{\sqrt{5}F_{2n+2}-1} = \phi
\left( \frac{-2+F_{2m+3}\sqrt{5}}{L_{2m+3}} \right).
\end{eqnarray*}
This follows by multiplying out and using the above formulea.
It remains only to observe that
\begin{eqnarray*}
\lim_{n \rightarrow \infty} \frac{L_n}{F_{n}} = \sqrt{5}
\end{eqnarray*}
and we are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$...
Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$. Then $\sum_{n=1}^{10} f(n)$ is
From the first equation
$$a(x^2+y^2+2xy)+b(x+y)+c=ax^2+bx+c+ay^2+by+c+xy$$
$$2axy=c+xy$$
Also, the summation will be
$$a(1^2+2^2+3^2...10^2)+b(1+2+3+4...+10)+10c$$
$$=385a+55b+10c$$
$$375a+45b+30$$
I don’t know what to do with the x and y terms. How do I proceed?
|
Hint:
$a+b+c=3 \equiv f(1)=3$. Using recurrence relation, $f(2)=2f(1)+1=7$, $f(3)=12$, $f(4)=18$. Notice the pattern.
$$\begin{aligned}a+b+c&=3\\4a+2b+c&=7\\9a+3b+c&=12\end{aligned}$$
This gives $a=1/2$, $b=5/2$, $c=0$. Can you proceed?
|
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|
Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers
I am trying to prove
$$(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$$
for all positive integers.
My attempts so far have been to Taylor expand the left hand side:
$$(n+1)^{2/3} -n^{2/3}\\
=n^{2/3}\big((1+1/n)^{2/3} -1\big)\\
=n^{2/3}\left(\sum_{\alpha=0}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha-1\right)\\
=n^{2/3}\sum_{\alpha=1}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha$$
I also tried proof by induction. Assume that it's true for n=k, so that
$$(n+1)^{2/3} -n^{2/3} < \frac{2}{3}n^{-1/3}\\
n^{2/3}\big((1+1/n)^{2/3} -1\big)<\frac{2}{3} n^{-1/3}\\
(1+1/n)^{2/3} -1<\frac{2}{3} n^{-1}$$
Then I want to prove that $(n+2)^{2/3} -(n+1)^{2/3} < \frac{2}{3}(n+1)^{-1/3}$. The left hand side is:
$$(n+2)^{2/3} -(n+1)^{2/3}\\
=(n+1)^{2/3}\left[\left(1+\frac{1}{n+1}\right)^{2/3}-1^{2/3}\right]\\
<(n+1)^{2/3}\cdot \frac{2}{3} n^{-1}\\
=\frac{2}{3}\frac{(n+1)^{2/3}}{n^{-1}}$$
But this is bigger than $\frac{2}{3}(n+1)^{-1/3}$, so I am stumped!
|
Try multiplying both sides by $n^{1/3}$, expanding $(n(n+1)^2)^{1/3}$ to $(n^3+2n^2+n)^{1/3}$, and moving $n$ from the left hand side to the right hand side, which rewrites the inequality as
$$(n^3+2n^2+n)^{1/3}\lt n+{2\over3}$$
Cubing both sides (which is OK since $x^3\lt y^3\iff x\lt y$) turns the inequality to prove into
$$n^3+2n^2+n\lt n^3+2n^2+{4\over3}n+{8\over27}$$
which is obviously true.
|
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"timestamp": "2023-03-29T00:00:00",
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|
3 x 3 Grid Square For a question that I need to answer, we have to show that there is only one arrangement for which, in a $3\times 3$ grid in which the numbers from $1$ to $9$ are placed and $5$ is the centre number, the sum of the $4$ numbers in every $2\times 2$ grid found within the $3\times 3$ grid all add up to the same number. I have found the answer, which is a grid with the numbers as follows:
\begin{aligned}
&1&8&&3\\\\
&6&5&&4\\\\
&7&2&&9
\end{aligned}
I don't know why this is the only option (except for rotations and reflections). Does anyone know why this is true?
|
The four sums being equal yields four linear equations; for the grid
$$\begin{matrix}a&b&c\\d&5&e\\f&g&h\end{matrix},$$
with $\{a,b,c,d,e,f,g,h\}=\{1,2,3,4,6,7,8,9\}$ you get the equations
\begin{eqnarray*}
a+b+d+5&=&x,\\
b+c+5+e&=&x,\\
d+5+f+g&=&x,\\
5+e+g+h&=&x.
\end{eqnarray*}
First note that $a+b+c+d+e+f+g+h=40$. Summing the first and fourth equations above, as well as the second and third, then shows that
$$2x=a+b+d+e+g+h+10=(40-c-f)+10=50-c-f,$$
$$2x=b+c+d+e+f+g+10=(40-a-h)+10=50-a-h,$$
so opposite corners are congruent mod $2$, i.e. each opposite pair is either both even or both odd.
Summing all four squares together yields the sum
$$4x=a+c+f+h+2(b+d+e+g)+20=60+b+d+e+g,$$
which in turn shows that
$$b+d+e+g\equiv0\pmod{4},$$
and because $a+b+c+d+e+f+g+h=40$ also
$$a+c+f+h\equiv0\pmod{4}.$$
The latter shows that the corners are either all even or all odd. From here, let's simply see how far this restricts the square.
If $a$, $c$, $f$ and $h$ are all even, then $b$, $d$, $e$ and $g$ are all odd. Rotating and reflecting the square if necessary we get $b<d,e,g$ and $d<e$. Because $\{b,d,e,g\}=\{1,3,7,9\}$ we have
$$\begin{matrix}a&1&c\\3&5&7\\f&9&h\end{matrix}
\qquad\text{ or }\qquad
\begin{matrix}a&1&c\\3&5&9\\f&7&h\end{matrix}
\qquad\text{ or }\qquad
\begin{matrix}a&1&c\\7&5&9\\f&3&h\end{matrix}.$$
In the first two cases, from the top left square and the bottom right square we have
$$a+1+3+5=5+7+9+h,$$
and so $a=h+12$, which is of course impossible. For the latter case we have
$$1+c+5+9=x=7+5+f+3,$$
and so $c=f$, a contradiction.
Hence $a$, $c$, $f$ and $h$ must be all odd. Then $b$, $d$, $e$ and $g$ are all even, and again after rotating and reflecting if necessary, we have one of
$$\begin{matrix}a&2&c\\4&5&6\\f&8&h\end{matrix}
\qquad\text{ or }\qquad
\begin{matrix}a&2&c\\4&5&8\\f&6&h\end{matrix}
\qquad\text{ or }\qquad
\begin{matrix}a&2&c\\6&5&8\\f&4&h\end{matrix}.$$
In the latter two cases, comparing the top right square and the bottom left square shows that
$$2+c+5+8=4+5+f+6,$$
and hence that $c=f$, a contradiction. Then we are left only with the first case, and comparing the top left square and the bottom right square shows that
$$a+2+4+5=5+6+8+h,$$
and so $a=h+8$. Then $a=1$ and $h=9$, and it quickly follows that $c=7$ and $f=3$, yielding
$$\begin{matrix}9&2&7\\4&5&6\\3&8&1\end{matrix}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Roots of the equation $(x – 1)(x – 2)(x – 3) = 24$ The equation $(x – 1)(x – 2)(x – 3) = 24$ has the real root equal to 'a' and the complex roots 'b' and 'c'. Then find the value of $\frac{bc}{a}$
My approach is as follow $y=f(x)=(x – 1)(x – 2)(x – 3) - 24=0$
$y'=3x^2-12x+11=0$
Solving we get $x=2\pm\sqrt{\frac{1}{3}}$
$f(2+\sqrt{\frac{1}{3}})<0$ & $ f(2-\sqrt{\frac{1}{3}})<0$
It is Local Minimum at $2+\sqrt{\frac{1}{3}}$ and Local Maximum at $2-\sqrt{\frac{1}{3}}$
By hit and trial I got $f(5)=0$ viz a=5
Given abc=30, therefore bc=6.
Hence the answer is $\frac{6}{5}$ which is correct.
My only concern is to find the real value without using any HIT and TRIAL.
|
There are actual computation methods for cubics but we can poke some number theory style fun into this.
$(x-1)(x-2)(x-3)=24$ are numbers whose product is 24. If we focused on integers, these numbers would be consecutive.
Oh what luck befalls us today. Watch this:
$(x-1)(x-2)(x-3)=4 \cdot 3 \cdot 2$
Let each factor pick a number and the value of $x$ remains the same.
Hence one answer is $x=5$
We have one answer. Perform synthetic division or long division of the full cubic by $x-5$ in order to represent it as $(x-5)P(x)=0$ where P has degree 2.
Doing this would yield $(x-5)(x^2-x+6)$ and the quadratic has no real solutions.
A better way of playing this guessing game would be to try out all factors of the $-30$ in the cubic and see if they work. This would be the rational root theorem.
|
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|
Setting up the particular solution to $y''+4y'=e^{-4x}+3$ $y''+4y'=e^{-4x}+3$
The characteristic polynomial is $r^2+4r=0 \to r = 0,-4$
the complementary solution is: $y_c = C_1+C_2e^{-4x}$
The part that is throwing me off is the addition of 3. I can't even set that up as a polynomial?
|
In this particular case, and similar cases, there is a second derivative with constant coefficients equaling a constant. This suggests a linear polynomial of the order of the highest derivative. This leads to the form (the exponential term being equal to a known solution so other rules apply):
$$f(x) = a_{0} + a_{1} \, x + a_{2} \, x^2 + b \, x \, e^{-4 x}.$$
This solution is to satisfy
$$f'' + 4 \, f' = e^{-4 x} + 3.$$
Now:
\begin{align}
3 + e^{-4 x} &= \frac{d}{dx}(a_{1} + 2 a_{2} x + b (-4 x + 1) \, e^{-4 x}) + 4 \, (a_{1} + 2 a_{2} x + b (-4 x + 1) \, e^{-4 x}) \\
&= 2 a_{2} + b (16 x - 8) \, e^{-4 x} + 4 a_{1} + 8 a_{2} x + b (-16 x + 4) \, e^{-4 x} \\
&= 4 a_{1} + 2 a_{2} + 8 a_{2} \, x - 4 b e^{-4 x}
\end{align}
leading to $4 a_{1} = 3$, $a_{2} = 0$, $-4 b = 1$, and
$$f(x) = a_{0} + \frac{3 \, x}{4} - \frac{x}{4} \, e^{-4 x}.$$
Since the general solution also contains a constant the $a_{0}$ constant in the particular solution can be removed to present the total solution of
$$y(x) = c_{0} + \frac{3 \, x}{4} + \left( c_{1} - \frac{x}{4} \right) \, e^{-4 x}.$$
A second method
By integration one can show that:
$$y' + 4 y = - \frac{1}{4} \, e^{-4 x} + 3 \, x + b_{1}.$$
Now this has equation has the form
\begin{align}
e^{- 4 x} \, \frac{d}{dx} (e^{4 x} \, y) &= - \frac{1}{4} \, e^{-4 x} + 3 \, x + b_{1} \\
\frac{d}{dx} (e^{4 x} \, y) &= - \frac{1}{4} + 3 \, x \, e^{4 x} + b_{1} \, e^{4 x} \\
e^{4 x} \, y &= - \frac{x}{4} + 3 \, \left( \frac{x}{4} - \frac{1}{16} \right) \, e^{4 x} + \frac{b_{1}}{4} \, e^{4 x} + b_{0} \\
y(x) &= \frac{4 b_{1} - 3}{16} + \frac{3 x}{4} + \left(b_{0} - \frac{x}{4}\right) \, e^{- 4 x}.
\end{align}
This solution is in agreement with the first solution.
|
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|
How to determine $x$ in $x^2+3x+7=4$? I'm helping my child with his homework. One of the problems is this: Determine the solution set of the equation
$x^2+3x+7=4$
Here is my attempt to determine $x_1$ and $x_2$:
$a^2+2ab+b^2=(a+b)^2$
$2ab=3x$
$2xb=3x$
$2b=3$
$b=\frac{3}{2}=1.5$
$b^2=2.25$
$x^2+3x+b^2-b^2+7=4$
$x^2+3x+2.25-2.25+7=4$
$(x^2+3x+2.25)-2.25+7=4$
$(x^2+3x+2.25)+7=4+2.25$
$(x^2+3x+2.25)+7=6.25$
$(x^2+3x+2.25)=6.25-7$
$(x^2+3x+2.25)=-0.75$
$(x+1.5)^2=-0.75$
In other similar equations we would take the square root of both sides. But I cannot take a square root of a negative number ($-0.75$).
How can I solve it, i. e. find out the values of $x$?
|
This is actually much simpler if you subtract the 4 to the left hand side first:
$$x^2 + 3x + 3 = 0$$
Then the solution set must satisfy the quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
where we have the quadratic $ax^2 + bx + c = 0$. In this case,
$$a=1, b=3, c = 3$$
So then $x$ must be $$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(3)}}{2a}$$
However, we notice here that inside the square root, we have a negative number $(-3)$, which indicates that the quadratic has no solution set in the real numbers.
If we are allowed to use imaginary numbers, then the solution would be
$$x = \frac{-3\pm \sqrt{-3}}{2} = \frac{-3\pm i\sqrt{3}}{2}$$
|
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|
How to show that $\sum_{n=0}^{N}\frac{(-1)^n}{{(N+n)!}(N-n)!}=\frac{1}{2N!^2}$?
How to show that $\displaystyle\sum_{n=0}^{N}\frac{(-1)^n}{{(N+n)!}(N-n)!}=\frac{1}{2N!^2}$?
How show that $(1)$ is:
$$\sum_{n=0}^{N}\frac{(-1)^n{2n \choose n}{N+n \choose N-n}}{[(n+1)(n+1)(n+2)(n+3)\cdots(n+N)]^2}=\frac{1}{2N!^2}\tag1$$
using : $\frac{\Gamma(n+N+1)}{\Gamma(n+1)}=(n+1)(n+2)\cdots(n+N)$
simplify to:
$$\frac{{2n \choose n}{N+n \choose N-n}}{[(n+1)(n+1)(n+2)(n+3)\cdots(n+N)]^2}=\frac{1}{(N-n)!(N+n)!}$$
$$\sum_{n=0}^{N}\frac{(-1)^n}{{(N+n)!}(N-n)!}\tag2$$
|
Let $ N $ be a positive integer.
Observe that : $$ \left(\forall n\in\mathbb{N}\right),\ \frac{1}{\binom{N+n}{n}}=\left(n+N+1\right)\int_{0}^{1}{x^{n}\left(1-x\right)^{N}\,\mathrm{d}x} $$
Thus, \begin{aligned} \sum_{n=0}^{N}{\frac{\left(-1\right)^{n}\left(N!\right)^{2}}{\left(N+n\right)!\left(N-n\right)!}}&=\sum_{n=0}^{N}{\frac{\left(-1\right)^{n}\binom{N}{n}}{\binom{N+n}{n}}}\\&=\sum_{n=0}^{N}{\left(-1\right)^{n}\binom{N}{n}\left(n+N+1\right)\int_{0}^{1}{x^{n}\left(1-x\right)^{N}\,\mathrm{d}x}}\\ &=\int_{0}^{1}{\left(1-x\right)^{N}\sum_{n=1}^{N}{\left(-1\right)^{n}n\binom{N}{n}x^{n}}\,\mathrm{d}x}+\left(N+1\right)\int_{0}^{1}{\left(1-x\right)^{N}\sum_{n=0}^{N}{\left(-1\right)^{n}\binom{N}{n}x^{n}}\,\mathrm{d}x}\\ &=N\int_{0}^{1}{\left(1-x\right)^{N}\sum_{n=1}^{N}{\left(-1\right)^{n}\binom{N-1}{n-1}x^{n}}\,\mathrm{d}x}+\left(N+1\right)\int_{0}^{1}{\left(1-x\right)^{2N}\,\mathrm{d}x}\\ &=-N\int_{0}^{1}{x\left(1-x\right)^{N}\sum_{n=0}^{N-1}{\left(-1\right)^{n}\binom{N-1}{n}x^{n}}\,\mathrm{d}x}+\frac{N+1}{2N+1}\\ &=-N\int_{0}^{1}{x\left(1-x\right)^{2N-1}\,\mathrm{d}x}+\frac{N+1}{2N+1}\\ &=-\frac{1}{2\left(2N+1\right)}+\frac{N+1}{2N+1}\\ \sum_{n=0}^{N}{\frac{\left(-1\right)^{n}\left(N!\right)^{2}}{\left(N+n\right)!\left(N-n\right)!}}&=\frac{1}{2}\end{aligned}
Hence, $$ \sum_{n=0}^{N}\frac{(-1)^n}{{(N+n)!}(N-n)!}=\frac{1}{2\left(N!\right)^2} $$
|
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|
Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$.
Question: Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $$\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$$.
My approach: Given that $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx, \forall n\in\mathbb{N}.$ Let us make the substitution $x^n=t$, then $$nI_n=\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}}.$$
Now since $0\le t\le 1\implies \frac{1}{t}\ge 1\implies \left(\frac{1}{t}\right)^{2/n}\ge 1 \implies 1+\left(\frac{1}{t}\right)^{2/n}\ge 2\implies \sqrt{1+\left(\frac{1}{t}\right)^{2/n}}\ge \sqrt 2.$
This implies that $$\frac{1}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le\frac{1}{\sqrt 2}\\ \implies \int_0^1 \frac{dt}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le \int_0^1\frac{dt}{\sqrt 2}=\frac{1}{\sqrt 2}.$$
So, as you can see, I am trying to solve the question using Sandwich theorem.
Can someone help me to proceed after this?
Also, in $$\lim_{n\to\infty}nI_n=\lim_{n\to\infty}\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}},$$ the limit and integral interchangeable?
|
Yet another approach is to use integration by parts to obtain
\begin{equation*}
I_{n} + I_{n-2} = \int_{0}^{1}x^{n-2}\sqrt{1+x^{2}}\,\mathrm{d}x = \frac{\sqrt{2}}{n-1} - \frac{1}{n-1}I_{n}
\end{equation*}
for $n \geq 2$, from which we get
\begin{equation*}
nI_{n} = \sqrt{2} - (n-1)I_{n-2} = \sqrt{2} - (n-2)I_{n-2} - I_{n-2}. \tag*{(1)}
\end{equation*}
Define $S_{n} = nI_{n}$ for each $n\in \mathbb{N}$. You have already shown that the sequence $\{S_{n}\}_{n\geq 2}$ is bounded above by $1/\sqrt{2}$, and note also that
\begin{align}
S_{n+1} - S_{n} &= \int_{0}^{1}\frac{(n+1)x^{n+1}-nx^{n}}{\sqrt{1+x^{2}}}\,\mathrm{d}x \\
&\geq \frac{1}{\sqrt{2}}\int_{0}^{1}\big((n+1)x^{n+1} - nx^{n}\big)\,\mathrm{d}x \\
&= \frac{1}{\sqrt{2}(n+1)(n+2)} > 0
\end{align}
so that the sequence $\{S_{n}\}_{n\geq 2}$ is increasing. Thus by the monotone convergence theorem, $S_{n}$ converges to a limit $L$. It is therefore valid to take the limit as $n \to \infty$ on both sides of $(1)$, noting that $I_{n-2} \to 0$, to find that
\begin{equation*}
L=\sqrt{2} - L
\end{equation*}
or
\begin{equation*}
L = \frac{1}{\sqrt{2}}.
\end{equation*}
|
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|
Find the area under the graph of the function $f(x)=(1-2x)^2$ between $x=-2$ and $x=2$ 5.4
Can somebody verify this solution for me?
Find the area under the graph of the function $f(x)=(1-2x)^2$ between $x=-2$ and $x=2$
The area under the graph of $f(x)$ between $x=-2$ and $x=2$ is exactly equal to:
$\int_{-2}^2 (1-2x)^2dx$
$=\int_{-2}^2 1-4x+4x^2dx$
$=(x-\frac{4}{2}x^2+\frac{4}{3}x^3) |_{-2}^2$
$=(x-2x^2+\frac{4}{3}x^3) |_{-2}^2$
$=(2-2(2)^2+\frac{4}{3}(2)^3)-(-2-2(-2)^2+\frac{4}{3}(-2)^3)$
$=(2-8+\frac{32}{3})-(-2-8+\frac{-32}{3})$
$=2-8+\frac{32}{3}+2+8+\frac{32}{3})$
$=4+\frac{64}{3}$
$\frac{12}{3}+\frac{64}{3}$
$=\frac{76}{3}$
|
You would spare your time if you make a substitution $ t=2x-1$. $$\int_{-2}^2 (1-2x)^2dx = {1\over 2}\int_{-5}^3 t^2dt =...$$
|
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|
Show that there do not exist any distinct natural numbers a,b,c,d such that
Show that there do not exist any distinct natural numbers a,b,c,d such that $a^3+b^3=c^3+d^3$ and $a+b=c+d$.
|
Mark $a+b=x$ then $b=x-a$ and $d=x-c$. Notice that $x\ne 0$. No we have: $$a^3+(x-a)^3 = c^3+(x-c)^3$$ and thus $$-3x^2a+3xa^2 = -3x^2c+3xc^2$$
so \begin{align}xa-a^2= xc-c^2& \implies x(a-c) = (a-c)(a+c)\\ &\implies x=a+c \\& \implies a+c=a+b \\&\implies c=b\end{align}
A contradiction.
|
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|
Is the integral $ \int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} - 1} $ converges or diverges. Determine if the following integral diverges/converges, if it converges, is it absolutely or conditionally converges.
$$
\int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} - 1}
$$
What i tried:
In that interval we know that:
$$
0 < x
$$
Therefore we can write:
$$
\frac{1}{\sqrt{x^2-x+1} - 1} < \frac{1}{\sqrt{x^2-2x+1} - 1} = \frac{1}{\sqrt{(x-1)^2} - 1}
$$
$$
= \frac{1}{x - 2}
$$
Now lets try to calculate the integral, using limits, we get:
$$
\lim_{t \to 2^-}\int_{1}^{t}\frac{1}{x-2} = ln(x-2)|_{1}^{t} = ln(t-2) - ln(1-2)
$$
I cant calculate this integral, namely i dont get any limit that is a number.
I thought that maybe i will get to a form like this:
$$
\int_{a}^{b}\frac{1}{(b-x)^\alpha}, \int_{a}^{b}\frac{1}{(x-a)^\alpha}
$$
But also, if i will, and conclude that $ \int \frac{1}{x - 2}$ diverges in that interval, i couldnt use the comparison test to conclude about the original function.
Can i have a hint?
Thank you.
|
First, do a change of coordinates: $y=x-1$. The integral then becomes
$$\int_1^2\frac{1}{\sqrt{x^2-x+1}-1}dx=\int_0^1\frac{1}{\sqrt{y^2+y+1}-1}dy$$
Now, expand using the Taylor series around $0$ to get
$$\frac{1}{\sqrt{y^2+y+1}-1}=\frac{2}{y}-\frac{3}{2}+\frac{15 y}{8}-\frac{33 y^2}{16}+\cdots>\frac{2}{y}-\frac{3}{2}$$
This implies
$$\int_0^1\frac{1}{\sqrt{y^2+y+1}-1}dy>\int_0^1\left(\frac{2}{y}-\frac{3}{2}\right)dy=-\frac{3}{2}+2\int_0^1\frac{1}{y}dy$$
However, this final integral diverges, implying that the original integral also diverges.
|
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|
Finding number of integral solutions to an equation.
Find the number of integral solutions to:
$$x^2+y^2-6x-8y=0.$$
My attempt:
The equation can be rewritten as:
$$x^2+y^2-6x-8y+9+16=25,$$
basically adding 25 to both sides, or equivalently,
$$(x-3)^2+(y-4)^2=25.$$
This is a Pythagorean triplet. The only triplet of this form would be $3$, $4$, $5$, so the possibilities for $x - 3$ and $y - 4$ are:
1.) $3$, $4$
2.) $-3$, $-4$
3.) $-3$, $4$
4.) $3$, $-4$
For each of these pairs, there are two different pairs $(x,y)$. For instance, in the first pair we could have $x - 3 = 3$ and $y - 4 = 4$, or $x - 3 = 4$ and $y - 4 = 3$. So there are $8$ solutions.
But, shouldn't we also consider the pair $(0,5)$? Because
$$0^2+5^2=25.$$
Then we have two new pairs:
1.) $0$, $5$
2.) $0$, $-5$
This would give $4$ new solutions, so the total number of solutions should be $8+4=12$.
How is the answer just $8$ then? What am I missing?
|
There are $12$ different solutions. The square $25$ can be expressed as the sum:
$$5^2+0^2=4^2+3^2=3^2+4^2=0^2+5^2=25$$
So, the possibilities are:
$$\left\{\begin{matrix}
x-3=\pm4 \rightarrow x=7 \vee x=-1
\\ y-4=\pm3 \rightarrow y=7 \vee y=1
\end{matrix}\right.
\vee
\left\{\begin{matrix}
x-3=\pm3 \rightarrow x=6 \vee x=0
\\ y-4=\pm4 \rightarrow y=8 \vee y=0
\end{matrix}\right.
\vee
\left\{\begin{matrix}
x-3=0 \rightarrow x=3
\\ y-4=\pm5 \rightarrow y=9 \vee y=-1
\end{matrix}\right.
\vee
\left\{\begin{matrix}
x-3=\pm5 \rightarrow x=8 \vee x=-2
\\ y-4=0 \rightarrow y=4
\end{matrix}\right.$$
Doing this process for all values of $x,y$, we have $12$ different solutions:
$$(7,7) \vee (-1,1) \vee (3,9) \vee (3,-1) \vee (6,8) \vee (0,0) \vee (8,0) \vee (-2,4) \vee (7,1) \vee (6,0) \vee (-1,7) \vee (8,4)$$
|
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|
How to prove that the series $\sum\limits_{n=1}^{\infty} \log x_n$ is convergent?
Consider the sequence $\{x_n \}$ defined by $$x_n = e \left (\frac {n} {n+1} \right )^{n + \frac 1 2},\ n \geq 1.$$
Prove that the series $\sum\limits_{n=1}^{\infty} \log x_n$ is convergent.
My attempt: I find that \begin{align*} \sum\limits_{n=1}^{\infty} \log x_n & = \sum\limits_{n=1}^{\infty} \left ( 1 - \left (n + \frac 1 2 \right ) \log \left (1 + \frac 1 n \right ) \right ) \\ & = - \sum\limits_{n = 1}^{\infty} \sum\limits_{m=2}^{\infty} (-1)^m \left ( \frac {1} {m+1} - \frac {1} {2m} \right ) \frac {1} {n^m} \\ & = -\sum\limits_{n = 1}^{\infty} \sum\limits_{m=2}^{\infty} (-1)^m \frac {m-1} {2m(m+1)} \frac {1} {n^m} \end{align*}
From here how do I proceed further? Please help me in this regard.
|
Use Taylor series to second and third order:
$$1-\left(n+\frac12\right)\log\left(1+\frac1n\right)=1-\left(n+\frac12\right)\left(\frac1n-\frac1{2n^2}+\frac1{3n^3}-\dots\right)$$
$$\le1-\left(n+\frac12\right)\left(\frac1n-\frac1{2n^2}\right)=\frac1{4n^2}$$
$$1-\left(n+\frac12\right)\log\left(1+\frac1n\right)\ge1-\left(n+\frac12\right)\left(\frac1n-\frac1{2n^2}+\frac1{3n^3}\right)=-\frac{n+2}{12n^3}$$
Thus $|\log x_n|\le\frac1{4n^2}$. Since $\sum_{n=1}^\infty\frac1{4n^2}$ converges, $\sum_{n=1}^\infty\log x_n$ converges.
|
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|
If a,b,c be roots of $2x^3+x^2+x-1=0$ show that some expression is equal to 16. If $a,b,c$ are roots of $2x^3+x^2+x-1=0,$ show that
$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=16$$
My attempt:
Let $\frac{1}{a}=p,\frac{1}{b}=q,\frac{1}{c}=r$
$p+q+r=1$
$pqr=2$
$$pq+qr+rp=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$$
$$=\frac{ab+bc+b^2}{(abc)^2}$$
$$=4\bigg(\frac{1}{pq}+\frac{1}{qr}+\frac{1}{q^2}\bigg)$$
$$pq+qr+rp=4\bigg(\frac{pq+qr+rp}{pq^2r}\bigg)$$
$$pq^2r=4$$
$$\implies q=2 \implies b=\frac{1}{2}$$
So p, r are roots of $x^2+x+1=0$
$\implies p^3=q^3=1$
But this condition gives a different value of the required expression, so what am I doing wrong? Please tell me the right solution.
|
Although this solution is mentioned in the comments to the question, I think it should be put here as an answer because of its simplicity and naturalness.
Looking at the given equation $2x^3+x^2+x-1=0$ one would quickly check possible rational roots using the Rational Root Test:
$$\text{To check: }\pm 1, \pm\frac 12$$
This leads to the root $a=\frac 12$. Factoring gives now
$$2x^3+x^2+x-1 = 2\left(x-\frac 12 \right)\underbrace{(x^2+x+1)}_{=\frac{x^3-1}{x-1}}$$
Hence, the other two roots are the complex conjugated 3rd roots of $1$:
$$b^3=c^3=1$$
Now, plugging in, we get
$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=-6\cdot 8\cdot 8 =-384$$
|
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|
Why doesnt this trignometric substituition work? $$\frac 1 5\int\frac{x+1+5}{(x+1)^2+5}\,\mathrm dx$$
where i substitute $x+1 = \sqrt{5}\tan(\theta)$
After Substituition :
$$\frac15\int\frac{\sqrt5\tan(\theta)+5}{\tan^2(\theta)+1}\sqrt5\sec^2(\theta)\,\mathrm d\theta$$
but if i do $u = x +1$
and then further substitute and integrate i get :
$\int \frac u {(u^2 +5)} + \int \frac 5 {(u^2 +5)} $
which then gives :
$\frac {ln(x^2 + 2x +6)} 2 + \sqrt5 arctan(\frac {x+1}{\sqrt5}))$
not :
$ln(\frac{(x^2 + 2x + 6)} {\sqrt5}) + \sqrt5arctan(\frac {x + 1} {\sqrt5})$
|
As said by Doug, the substitution works. Before applying it, consider decomposing $$\frac{x+1+5}{(x+1)^2+5}=\frac12\frac{((x+1)^2+5)'}{(x+1)^2+5}+\frac{5}{(x+1)^2+5},$$ that makes life simpler.
|
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|
$n$ is prime if $1+x+x^2+\dots+x^{n-1}$ is prime I came across this question in a number theory textbook:
Let $x$ and $n$ be positive integers such that $1+x+x^2+\dots+x^{n-1}$ is prime. Then show that $n$ is prime.
I reasoned that this proof would require Fermat's Little Theorem, and put $1+x+x^2+\dots+x^{n-1} = \frac{x^n-1}{x-1}$, but was unable to continue.
What perplexes me most, however, is a shoddy argument I found in the 'solutions' section of the book. There the author argued that $1+x+x^2+\dots+x^{n-1} = \frac{x^n-1}{x-1} \Rightarrow x-1=1 \Rightarrow x=2.$ This was given without any explanation and I don't know what to make of it.
If that step is correct, why so?
|
If $x=1$, then $1+x+\ldots+x^{n-1}$ is simply $n$ and we are done. So assume $x\gt 1$.
Suppose that $n$ is composite, $n=ab$ with $a\geq 2,b\geq 2$. Then we have $\frac{x^n-1}{x-1}=AB$ where $A=\frac{x^n-1}{x^a-1}=1+x^a+(x^a)^2+\ldots+(x^a)^{b-1}$ and $B=\frac{x^a-1}{x-1}$. So $\frac{x^n-1}{x-1}$ is composite.
|
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|
Question about $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$ $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$
Let $x=t^6$, then it becomes $\lim\frac{t^2 - (\sqrt[6]{2})^2}{t^3 - (\sqrt[6]{2})^3}$=$\lim\frac{(t+\sqrt[6]{2})(t-\sqrt[6]{2})}{(t-\sqrt[6]{2})(t^2+t\sqrt[6]{2}+(\sqrt[6]{2})^2}$=$\lim\frac{(t+\sqrt[6]{2})}{(t^2+t\sqrt[6]{2}+(\sqrt[6]{2})^2}$
But when $x\to 2$, t can go to $\sqrt[6]{2}$ or -$\sqrt[6]{2}$, which gives two different limits, where I was wrong? Thanks!
|
We can write your idea in the following form.
Let $\sqrt[6]{\frac{x}{2}}=y.$
Thus, $$\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}=\frac{\sqrt[3]2}{\sqrt2}\lim_{x\rightarrow2}\frac{\sqrt[3]{\frac{x}{2}}-1}{\sqrt{\frac{x}{2}}-1}=\frac{\sqrt[3]2}{\sqrt2}\lim_{y\rightarrow1}\frac{y^2-1}{y^3-1}=$$
$$=\frac{\sqrt[3]2}{\sqrt2}\lim_{y\rightarrow1}\frac{y+1}{y^2+y+1}=\frac{\sqrt[3]2}{\sqrt2}\cdot\frac{2}{3}=\frac{\sqrt[6]{32}}{3}.$$
I think, in this form just impossible to make a mistake.
|
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|
Real Solution of $4^x + 6^x = 9^x$. So here is the problem
Solve $$4^x + 6^x = 9^x$$ for $x$.
I am trying to find its real solution.
I was trying in this way!!
$$6^x\left( \frac {4^x}{6^x} + 1\right) = 9^x$$
$$\left({\frac 23}\right)^x +1=\left({\frac 32}\right)^x$$ but I'm stuck . . .
Will anyone help ?
|
$$4^x + 6^x = 9^x \implies \left(\frac{4}{9}\right)^x +\left(\frac{6}{9}\right)^x =1 \implies \left(\frac23\right)^{2x}+\left(\frac23\right)^{x}-1= 0 $$
Let $\left(\frac23\right)^x = y$
So, $$y^2+y-1 = 0 \implies y =\left(\frac23\right)^x = \frac{-1\pm \sqrt{5}}{2}$$
Considering the positive root, $$x = \frac{\ln y}{\ln \frac23} =\frac{\ln\left(\frac{-1+ \sqrt{5}}{2}\right)}{\ln \frac23} $$
|
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|
Solution to a non-linear simultaneous equation I'm trying to solve the non-linear, two-variable system of equations
\begin{align*}
y-x+x^5-\frac{y^4x}{2(1+x^2)^2}-\frac{x^3}{1+y^2} &= 0 \\
-x-y+y^5-\frac{x^4y}{2(1+y^2)^2}-\frac{y^3}{1+x^2} &= 0
\end{align*}
with $x,y \in \mathbb{R}$. You might notice that the second equation is almost obtained from the first by swapping $x$ and $y$, except for the first term whose signs are opposed. I think the only solution is the origin $x=y=0$, and this is seems to be confirmed numerically on WolframAlpha, but I'm not sure, and I don't know how to prove it. You can also write the system as
\begin{align*}
y-x+x^5+\frac{1}{4}\partial_x\left(\frac{y^4}{1+x^2}-\frac{x^4}{1+y^2}\right) &= 0 \\
-x-y+y^5-\frac{1}{4}\partial_y\left(\frac{y^4}{1+x^2}-\frac{x^4}{1+y^2}\right) &= 0
\end{align*}
Any hints/ideas?
|
I converted both equations into polynomials and put them into Singular to find all solutions.
LIB "solve.lib";
ring r=0,(x,y),dp;
poly p1=2x9y2+2x9+4x7y2+2x7-4x5+2x4y3+2x4y-4x3y2-6x3+4x2y3+4x2y-xy6-xy4-2xy2-2x+2y3+2y;
poly p2=-x6y-x4y-2x3y4-4x3y2-2x3+2x2y9+4x2y7-4x2y3-2x2y-2xy4-4xy2-2x+2y9+2y7-4y5-6y3-2y;
ideal i=p1,p2;
def T=solve(i,30);
There are $89$ distinct solutions, but $88$ of them involve a complex number.
This shows that the last solution, $x=y=0$, is the only real solution.
|
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"url": "https://math.stackexchange.com/questions/3630174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find $f^{(80)}(27)$ where $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$
Suppose that $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$. Use a Taylor series expansion to find $f^{(80)}(27)$.
I tried the following:
\begin{align}
f'(x) &= (x+3)^{\frac{1}{3}}\cdot 1+(x-27)\cdot \frac{1}{3}(x+3)^{\frac{-2}{3}}\\
%
f''(x)
&= \frac{1}{3}(x+3)^{-\frac{2}{3}}+\frac{1}{3}(x-27)\cdot -\frac{2}{3}(x+3)^{-\frac{5}{3}}+\frac{1}{3}(x+3)^{-\frac{2}{3}} \\
&= \frac{2}{3}(x+3)^{-\frac{2}{3}}-\frac{1\cdot 2}{3^2}(x+3)^{-\frac{5}{3}}(x-27)\\
%
f'''(x) &= -\frac{2^2}{3^3}(x+3)^{-\frac{5}{3}}-\frac{1\cdot 2}{3^2}(x+3)^{-\frac{5}{3}}-\frac{1\cdot 2\cdot 5}{3^3}(x+3)^{-\frac{8}{3}}
\end{align}
This is very nasty, please help me to solve in some easy way.
|
Attention: I suddenly found the OP requires the use of Taylor series expansion. My solution does not meet the requirement.
We have
$$f(x) = (x+3)^{1/3}(x-27) = (x+3)^{4/3} - 30(x+3)^{1/3}.$$
Thus, we have, for $k\ge 1$,
\begin{align}
f^{(k)}(x) &= \tfrac{4}{3}(\tfrac{4}{3}-1)(\tfrac{4}{3}-2) \cdots (\tfrac{4}{3}-(k-1))(x+3)^{4/3-k}\\
&\quad - 30\cdot \tfrac{1}{3}(\tfrac{1}{3}-1)(\tfrac{1}{3}-2) \cdots (\tfrac{1}{3}-(k-1))(x+3)^{1/3-k}.
\end{align}
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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|
Evaluating $\sum_{r=1}^{3n-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{3n}r}$
$$\sum_{r=1}^{3n-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{3n}r}, n \in 2k, k\in \mathbb{Z^+}$$
Answer given (much simpler than expected)
$\dfrac{3n}{3n+2}$
I tried adding and subtracting 1 to $r$ so could use $\dfrac{\binom{n}r}{r+1}=\dfrac{\binom{n+1}{r+1}}{n+1}$, but didn't prove to be useful. I know summation and double summation of binomial coefficients to quite to good extent. If you could help...
|
Let's start by the following : \begin{aligned}\frac{1}{\binom{3n}{r}}=\left(3n+1\right)\int_{0}^{1}{x^{r}\left(1-x\right)^{3n-r}\,\mathrm{d}x}\end{aligned}
Since $\left(\forall x\in\left[0,1\right]\right) $, we have : $$ \sum_{r=0}^{3n-1}{r\left(-1\right)^{r-1}x^{r}\left(1-x\right)^{3n-r}}=\left(-1\right)^{n+1}x^{3n+1}\left(1-x\right)+3\left(-1\right)^{n}n x^{3n}\left(1-x\right)+x\left(1-x\right)^{3n+1} $$
It can be proven by differentiating the formula giving the sum of consecutive terms of a geometric sequence.
Thus, \begin{aligned}\scriptsize \sum_{r=0}^{3n-1}{\frac{\left(-1\right)^{r-1}r}{\binom{3n}{r}}}&\scriptsize=\left(3n+1\right)\int_{0}^{1}{\sum_{r=0}^{3n-1}{r\left(-1\right)^{r-1}x^{r}\left(1-x\right)^{3n-r}}\,\mathrm{d}x}\\ &\scriptsize=\left(-1\right)^{n+1}\left(3n+1\right)\int_{0}^{1}{x^{3n+1}\left(1-x\right)\mathrm{d}x}+3\left(-1\right)^{n}n\left(3n+1\right)\int_{0}^{1}{x^{3n}\left(1-x\right)\mathrm{d}x}+\left(3n+1\right)\int_{0}^{1}{x\left(1-x\right)^{3n+1}\,\mathrm{d}x}\\ &\scriptsize=\frac{\left(-1\right)^{n+1}\left(3n+1\right)}{\left(3n+2\right)\left(3n+3\right)}+\frac{3\left(-1\right)^{n}n}{3n+2}+\frac{\left(3n+1\right)}{\left(3n+2\right)\left(3n+3\right)} \end{aligned}
And hence, if your $ n $ is even, we can get rid of the $ \left(-1\right)^{n} $ and get some cancellations to end with : $$ \sum_{r=0}^{3n-1}{\frac{\left(-1\right)^{r-1}r}{\binom{3n}{r}}}=\frac{3n}{3n+2}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3630624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Calculating determinant of a symmetric matrix where the $k$th row is given by $[a_{k-1},a_k,...,a_0,a_1,...,a_{n-(k-1)}]$
For $j = 0,...,n$ set $a_{j} = a_{0} + jd$, where $a_{0}, d$ are fixed real numbers. Calculate the determinant of the $(n+1)\times (n+1)$ matrix
$$A = \begin{pmatrix}
a_{0} & a_{1} & a_{2} & \dots & a_{n}\\
a_{1} & a_{0} & a_{1} & \dots & a_{n-1}\\
a_{2} & a_{1} & a_{0} & \dots & a_{n-2}\\
\ldots & \ldots & \ldots & \ldots & \ldots\\
a_{n} & a_{n-1} & a_{n-2} & \dots & a_{0}
\end{pmatrix}.$$
How to calculate that? I haven't found any property of determinant of symmetric matrix which could help.
I've tried to use Gaussian elimination (subtracting each row from the row above it), but it didn't work
Gaussian elimination(subtracting each row from the row above it) brings to the matrix:
$$\begin{pmatrix}
-d & d & d & ... & d\\
-d & -d & d & ... & d\\
-d & -d & -d & .... & d\\
\ldots & \ldots & \ldots & \ldots & \ldots\\
a_{n} & a_{n-1} & a_{n-2} & ... & a_{0}
\end{pmatrix} = d^{n-1} \cdot \begin{pmatrix}
-1 & 1 & 1 & ... & 1\\
-1 & -1 & 1 & ... & 1\\
-1 & -1 & -1 & .... & 1\\
\ldots & \ldots & \ldots & \ldots & \ldots\\
a_{n} & a_{n-1} & a_{n-2} & ... & a_{0}
\end{pmatrix}$$
|
Partial Answer: If we use the matrix determinant lemma, then it suffices to consider the case of $a_0 = 0$ and $d = 1$.
With that said, let $A_n$ denote the matrix defined with your formula, with $a_0 = 0$ and $d = 1$. Let $R$ denote an $n \times n$ identity matrix, where the first column has been replaced by $(1,0,\dots,0,n,1-n)$. Note that
$$
RA_nR^T = \pmatrix{-2n(n-1) & 0 & 2n\\
0 & A_{n-1} &-\\
2n & |}.
$$
Conclude that $\det(A_n) = -2n(n-1)\det(A_{n-1}) + 4n^2 \det(A_{n-2})$.
|
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"url": "https://math.stackexchange.com/questions/3633401",
"timestamp": "2023-03-29T00:00:00",
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|
Using that $1 + z + z^{2} + ... + z^{n} = \frac{1-z^{n+1}}{1-z}$ and taking the real parts, prove that: $$ 1 + \cos \theta + \cos2\theta + ... + \cos n\theta = \frac12+\frac{\sin[(n + \frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})} $$
for $0 < \theta < 2\pi$.
Alright. What I have done is this, using the De Moivre's Formula:
$$ 1 + \cos \theta + \cos2\theta + ... + \cos n\theta = \operatorname{Re}(1 + (\cos\theta + i\sin\theta) + (\cos2\theta + i\sin2\theta) + ... + (\cos n\theta + i \sin n \theta))$$
That is equivalent to
$$ \operatorname{Re}(1 + e^{i\theta} + e^{2i\theta} + ... e^{ni\theta}) = \operatorname{Re} \biggl(\frac{1 - e^{(n+1)i\theta}}{1 - e^{i\theta}}\biggr)$$
I've reached to this point, but now I don't know what to do. Any hint or idea?
|
Continue with
$$1 + e^{i\theta} + e^{2i\theta} + ... e^{ni\theta}=\frac{1 - e^{(n+1)i\theta}}{1 - e^{i\theta}}
=\frac{e^{\frac12(n+1)i\theta}}{e^{\frac12i\theta}}\cdot \frac{e^{-\frac12(n+1)i\theta} - e^{\frac12(n+1)i\theta}}{e^{-\frac12i\theta} - e^{\frac12i\theta}}
= e^{\frac12ni\theta} \frac{\sin\left(\frac{n + 1}2\theta\right)}{\sin(\frac{\theta}{2})}$$
Thus,
$$ 1 + \cos \theta + \cos2\theta + ... + \cos n\theta = Re\left( e^{\frac12ni\theta}\frac{\sin\left(\frac{n + 1}2\theta\right)}{2\sin(\frac{\theta}{2})} \right) \\
\frac{\cos\left(\frac12n\theta\right)\sin[(n + \frac{1}{2})\theta]}{\sin(\frac{\theta}{2})}=\frac{\sin[(n + \frac{1}{2})\theta]+\sin(\frac{\theta}{2})}{2\sin(\frac{\theta}{2})}
=\frac{\sin[(n + \frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})}+\frac12$$
Note that the term $\frac12$ is missing in the original expression.
|
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"timestamp": "2023-03-29T00:00:00",
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|
The sequence 1 + (1+1) + (1+1+2) + (1+1+2+4) + (1+1+2+4+8) + .. I am working on a brute-force algorithm for a hard problem where the number of operations $S[n]$ seem to grow as a function of $n$ as follows:
$$
\begin{align}
S[1]& = 1\\
S[2]& = 1 + (1+1)\\
S[3]& = 1 + (1+1) + (1+1+2)\\
S[4]& = 1 + (1+1) + (1+1+2) + (1+1+2+4)\\
S[5]& = 1 + (1+1) + (1+1+2) + (1+1+2+4) + (1+1+2+4+8)
\end{align}
$$
Meaning, the algorithm runs $S[i]$ constant-time operations for an input containing $i$ elements.
Given the above, how would one go about deriving a formula for $S[n]$?
|
$1 + 2 + 4 + ...... + 2^k = 2^{k+1} - 1$.
So $(1+1+2+4+....... + 2^k) = 2^{k+1}$.
So $S[n] = 1 + (1+1) + (1+1+2) + (1+1+2+4) + ..... = 1 + 2^1 + 2^2 + ..... + 2^n = 2^{n+1} -1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3635719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
(Different) derivatives of $f(x) = \arcsin\left(\left(5 x + 12 \sqrt{1-x^2}\right)/13\right)$ via two different substitutions? We have been given a function to differentiate:
$$f(x) = \arcsin \left(\frac{5x + 12\sqrt{1-x^2}}{13}\right)$$
My teacher told me the method to substitute $ x= \sin\vartheta$ which would simplify the argument of $\arcsin$ to $$\frac{5}{13}\sin \vartheta + \frac{12}{13}\cos \vartheta $$ and further into $\sin(\vartheta + \alpha)$ where $ \alpha = \arctan\left(\frac{12}{5}\right)$ therefore the function gets reduced into $f(x) = \arctan\left(\frac{12}{5}\right) + \arcsin(x)$ giving $$f'(x) = \frac{1}{\sqrt{1-x^2}}$$. However when I substitute $x = \cos\vartheta $, the argument reduces to : $$\frac{5}{13}\cos\vartheta + \frac{12}{13}\sin\vartheta$$ and further into $\sin(\alpha + \vartheta)$ where $\alpha = \arctan\left(\frac{5}{12}\right)$ but this time the function gets reduced to $$f(x) = \arctan\left(\frac{5}{12}\right) + \arccos(x)$$
thus $$f'(x) = \frac{-1}{\sqrt{1-x^2}}$$ Hence we obtain two different derivatives for the same function and I cannot understand why. I tried plotting the argument and different simplification and got that they are not always equal but cannot figure out the reason.
|
I assume that the reason for the "paradox" was clarified already in comments, but possibly there is still a need to go a little bit deeper into details:
Your error has its root in the assumption:
$$
\arcsin(\sin x)=x.
$$
However the equality is valid only in the range $-\frac\pi2\le x\le \frac\pi2$ whereas the correct expression for all real $x$ is:
$$
\arcsin(\sin x)=(-1)^m\left(x-m\pi\right),\quad\text{with}\quad m=\left\lfloor\frac{x}\pi+\frac12\right\rfloor.\tag1
$$
In view of this the expression resulting from the substitution $x=\sin\vartheta$ reads:
$$\arcsin\left(\sin\left(\vartheta+\arcsin\frac{12}{13}\right)\right)=\begin{cases}
\frac\pi2-\arcsin\frac5{13}+\vartheta;& \vartheta\le \arcsin\frac{5}{13}\\
\frac\pi2+\arcsin\frac{5}{13}-\vartheta;& \vartheta\ge \arcsin\frac{5}{13}
\end{cases}
$$
or
$$
f(x)=\begin{cases}
\frac\pi2-\arcsin\frac5{13}+\arcsin x;& x\le\frac{5}{13}\\
\frac\pi2+\arcsin\frac{5}{13}-\arcsin x;& x\ge\frac{5}{13}.
\end{cases}
$$
Now it should be obvious that the derivative is discontinuous at $x=\frac5{13}$ taking the values
$$
f'(x)=\pm\frac1{\sqrt{1-x^2}},
$$
to the left and to the right from the point $x=\frac5{13}$.
The same result will be obtained with the substitution $x=\cos\vartheta$ as well.
|
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|
How to solve $x^{-2} + x^{-3} = \dfrac{x^3 + x^2}{4x^6}$? For online school I was given the equation $$x^{-2} + x^{-3} = \frac{x^3 + x^2}{4x^6}.$$ The problem lies in the addition of the bases, as I do not know how to solve that. Could you help me?
|
We have that
$$x^{-2} + x^{-3} = \dfrac{x^3 + x^2}{4x^6}$$
$$x^4+x^3=\frac{1}{4}(x^3+x^2)$$
$$x^4+\frac34 x^3-\frac14 x^2=0$$
$$x^2(4x^2+3x-1)=0$$
However, since $x=0$ is not a solution (it is in the denominator of the original equation), we can divide out by $x$ to get
$$4x^2+3x-1=0$$
This is a simple quadratic equation, can you take it from here?
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is $\frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$ for $x \rightarrow 0$? I am trying to find the limit
$\lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$
for $a,b \in \rm{I\!R}_{+}$. Applying L'Hospital's rule leads to
$\lim_{x \rightarrow0}\frac{\cos x \cdot \sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}{-2 \sin x \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x}) + \frac{\sin x \cdot (a^2 + b^2 + 2 ab \cos x )}{\sqrt{a^2+b^2+ 2 ab \cos x}} }$.
However, this remains with both, cosine and sine. Maybe one could use a trigonometric identity, which I cannot find.
|
You can do it brutally with a limited developement at order 3:
Let's study the parts independantly and start with the denominator:
*
*$a^2+b^2+2ab\cos(x) = (a+b)^2-{abx^2}+O(x^4)$
*From that you have $\sqrt{a^2+b^2+2ab\cos(x)} = a+b - \frac{x^2 a b}{2 (a + b)}
+ O(x^4)$
*Hence $a+b-\sqrt{a^2+b^2+2ab\cos(x)} = \frac{x^2 a b}{2 (a + b)}
+ O(x^4)$
*Then: $(a^2+b^2+2ab\cos(x))(a+b-\sqrt{a^2+b^2+2ab\cos(x)}) = \frac{(a+b)ab}{2}x^2-\frac{a^2b^2}{2(a+b)}x^3+O(x)^4$
*Taking the square root gives a denominator in $x\sqrt{2(a+b)ab}+ o(x^2)$
But the numerator is a $abx+O(x^3)$ so that we have:
$f(x) = \frac{ab}{\sqrt{2(a+b)ab}} +o(1) = \sqrt{\frac{ab}{2(a+b)}} +o(1)$
All in all we therefore have a limit at $0$, which is $\sqrt{\frac{ab}{2(a+b)}}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that if $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ then $ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $ Question -
Let $a, b, c$ be positive real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ . Prove that
$$
\frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1
$$
My try
$$
\frac{a^{2}}{a+2 b^{2}}=a-\frac{2 a b^{2}}{a+2 b^{2}} \geq a-\frac{2 a b^{2}}{3 \sqrt[3]{a b^{4}}}=a-\frac{2(a b)^{2 / 3}}{3}
$$
which implies that
$$
\sum_{c y c} \frac{a^{2}}{a+2 b^{2}} \geq \sum_{c y c} a-\frac{2}{3} \sum_{c y c}(a b)^{\frac{2}{3}}
$$
It suffices to prove that
$$
(a b)^{2 / 3}+( b c)^{2 / 3}+\left (c a)^{2 / 3} \leq 3\right.
$$
beacuse we can easily get that $\sum a \ge 3$
but i am not able to prove it..
note that we have to prove this using only am-gm or weighted am-gm or power mean or any kind of means inequality because author did not introduce any advance inequality yet...
any hints ???
thankyou
|
Another way.
By C-S and P-M we obtain:
$$\sum_{cyc}\frac{a^2}{a+2b^2}-1=\sum_{cyc}\frac{a^4}{a^3+2a^2b^2}-1\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^3+2a^2b^2)}-1=$$
$$=\frac{\sum\limits_{cyc}(a^4-a^3)}{\sum\limits_{cyc}(a^3+2a^2b^2)}=\frac{9(a^4+b^4+c^4)-(a^3+b^3+c^3)(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{9\sum\limits_{cyc}(a^3+2a^2b^2)}\geq0.$$
Because by P-M $$\sqrt[4]{\frac{a^4+b^4+c^4}{3}}\geq\sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$ or $$\sqrt[4]{\left(\frac{a^4+b^4+c^4}{3}\right)^3}\geq\frac{a^3+b^3+c^3}{3}$$ or $$\sqrt[4]{3(a^4+b^4+c^4)^3}\geq a^3+b^3+c^3$$ and
$$\sqrt[4]{\frac{a^4+b^4+c^4}{3}}\geq\left(\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{3}\right)^2$$ or
$$\sqrt[4]{2187(a^4+b^4+c^4)}\geq(\sqrt a+\sqrt b+\sqrt c)^2.$$
Id est, $$9(a^4+b^4+c^4)=\sqrt[4]{3(a^4+b^4+c^4)^3}\cdot\sqrt[4]{2187(a^4+b^4+c^4)}\geq$$
$$\geq(a^3+b^3+c^3)(\sqrt{a}+\sqrt{b}+\sqrt{c})^2.$$
|
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"url": "https://math.stackexchange.com/questions/3642895",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Orthogonally Diagonalize the matricies The Question given was "Orthogonally diagonalize the matrices, giving an orthogonal matrix P and diagonal matrix D."
I was given eigenvalues -3 and 15
and Matrix A=
$$
\begin{bmatrix}
5 & 8 & -4 \\
8 & 5 & -4 \\
-4 & -4 & -1
\end{bmatrix}
$$
So far I've done the identity M-I(eigenvalue) with eigenvalue -3 and have gotten
$$
\begin{bmatrix}
8 & 8 & -4 \\
8 & 8 & -4 \\
-4 & -4 & 2
\end{bmatrix}
$$
Reduced too
$$
\begin{bmatrix}
1 & 1 & -.5 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
$$
I'm struggling with what to do after this. I looked at Slader but whoever did this question did it way differently and extremely confused about how he produced his numbers.
|
The given matrix
$$ \begin{bmatrix}
5 & 8 & -4 \\
5 & 5 & -4 \\ -4 & -4 & -1
\end{bmatrix}
,$$
has eigenvalues $-3$ and two other complex eigenvalues , so I suspect you must have miss-wrote one of the entries of the matrix. Since the matrix is $3\times 3$, by theorem it must have $3$ eigenvalues. So one of the eigenvalues $-3$ or $15$ is repeated twice. You must go over and look which one is repeated. If $-3$ is the one that is repeated , then again by theorem, your diagonal matrix will be
$$
\begin{bmatrix}
-3 & 0 & 0\\
0& -3 & 0\\
0 & 0 & 15
\end{bmatrix}
.$$
Similarly, if your repeated eigenvalue is $15$ the diagonal matrix will be
$$
\begin{bmatrix}
-3 & 0 & 0\\
0& 15 & 0\\
0 & 0 & 15
\end{bmatrix}.$$
For the eigenvalues part , you correctly reduced the matrix. From there
\begin{gather}
\begin{bmatrix}
1& 1 & -5 \\ 0 & 0 & 0 \\ 0 & 0 & 0
\end{bmatrix} \implies a+ b -5c =0
\end{gather}
The way you want to approach this is to set two of the variables as free variables ,i.e.,
$$ b=t \ \ , c = s \ \ \implies a = -t + 5s ,$$
then you want to express $a$ as a linear combination of $t$ and $s$ ;
$$ a = t \begin{pmatrix}
-1 \\ 1 \\ 0
\end{pmatrix} + s
\begin{pmatrix}
5 \\ 0 \\ 1
\end{pmatrix}.$$
Therefore your two eigenvectors corresponding to $\lambda = -3$ are $v_{1} = (-1 \ \ 1 \ \ 0)^{T}$ and $v_{2} = (5 \ \ 0 \ \ 1)^{T}$. Now since you want to have an orthogonal matrix, you must normalize the eigenvectors. To do this, we divide each entry of the eigenvectors by their respective eigenvector norm giving
$$ v_{1} = \begin{pmatrix}
\frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0
\end{pmatrix} \quad \text{and} \quad v_{2} = \begin{pmatrix}
\frac{5}{\sqrt{26}} \\ 0 \\ \frac{1}{\sqrt{26}}
\end{pmatrix}.$$
Your final orthogonal matrix $P$ will have its first and second column the normalized eigenvectors $v_{1}$ and $v_{2}$ respectively. I'll let you do the third eigenvector $v_{3}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Generating function of $(2,2,2,4,4,4,8,8,8,\ldots)$ I know that $F(x)=\dfrac{1}{1-x}$ is a generating function of $(1,1,1,1,\ldots)$ and $F(2x)$ is a generating function of $(1,2,4,8,16,\ldots).$
Then $G(x)=\dfrac{F(2x)-1}{x}=\dfrac{2}{1-2x}$ is a generating function of $(2,4,8,16,\ldots).$
A generating function of $(2,0,0,4,0,0,8,0,0,\ldots)$ is $G(x^3).$
A generating function of $(0,2,0,0,4,0,0,8,0,\ldots)$ is $xG(x^3).$
A generating function of $(0,0,2,0,0,4,0,0,8,\ldots)$ is $x^2G(x^3).$
A generating function of $(2,2,2,4,4,4,8,8,8,\ldots)$ is
$$
H(x)=\left(1+x+x^2\right)G(x^3)=\left(1+x+x^2\right)\cdot \frac{2}{1-2x^3}.
$$
Is this correct?
|
If the book gave the answer as $\frac{1+x+x^2}{1-x}$, that's wrong; your $H$ is correct. Indeed,
$$\begin{align*}
\frac{1+x+x^2}{1-x}&=(1+x+x^2)(1+x+x^2+x^3+\cdots)\\
&=1+2x+3x^2+3x^3+\cdots
\end{align*}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof $\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt2} + \frac{\sqrt{2 + \sqrt2}}{\sqrt2 \sqrt{2 - \sqrt{2 - \sqrt2}}} =\sqrt{2-\sqrt{2+\sqrt2}}$ I want to prove the following equation
$$\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt{2}} + \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2} \sqrt{2 - \sqrt{2 - \sqrt{2}}}}$$
I know this must be equal to $\sqrt{2-\sqrt{2+\sqrt{2}}}$, but I find it pretty hard to prove it manually, without requiring a calculator nor estimates (since the expression should be equal to one of the roots of the polinomyal $f(x)=x^8-8x^6+20x^4-16x^2+2$, and all of them are real and distinct).
Thanks in advance for your help.
|
$a=\sqrt{2-\sqrt{2-\sqrt{2}}}+\sqrt{2-\sqrt{2+\sqrt{2}}}$
$a^2=4+\sqrt{2}$
$a=\sqrt{2(2+\sqrt{2})}$
$b=\sqrt{2-\sqrt{2-\sqrt{2}}}*\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{4+\sqrt{2}-2(\sqrt{2-\sqrt{2-\sqrt{2}}}+\sqrt{2-\sqrt{2+\sqrt{2}}})}=\sqrt{4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})} $
$-\sqrt{2-\sqrt{2-\sqrt{2}}}/\sqrt{2}+\sqrt{2+\sqrt{2}}/\left(\sqrt{2}\sqrt{2-\sqrt{2-\sqrt{2}}}\right)=\sqrt{2-\sqrt{2+\sqrt{2}}}?$
$-\sqrt{2-\sqrt{2-\sqrt{2}}}+\sqrt{2+\sqrt{2}}/\left(\sqrt{2-\sqrt{2-\sqrt{2}}}\right)=\sqrt{2(2-\sqrt{2+\sqrt{2}}})$
$\sqrt{2+\sqrt{2}}-(2-\sqrt{2-\sqrt{2}})=\sqrt{2(2-\sqrt{2+\sqrt{2}})(2-\sqrt{2-\sqrt{2}})}$
$\sqrt{2-\sqrt{2-\sqrt{2}}}+\sqrt{2-\sqrt{2+\sqrt{2}}}-2=\sqrt{4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})}$
$\sqrt{2(2+\sqrt{2})}-2=\sqrt{2(4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})}$
$\sqrt{2+\sqrt{2}}-\sqrt{2}=\sqrt{4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})}$
Squared:
$4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})=4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})$
The proof is done.
|
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|
In how many different ways can you prove that $\sin^2x + \cos^2x = 1$ The standard proof of the identity $\sin^2x + \cos^2x = 1$ (the one that is taught in schools) is as follows: from pythagoras theorem, we have (where $h$ is hypotenuse, $b$ is base and $p$ is perpendicular)
$$h^2 = p^2 + b^2$$
dividing by $h^2$ on both sides:
$$1 = \frac{p^2}{h^2}+\frac{b^2}{h^2}$$
since $\sin x = \frac ph$ and $\cos x = \frac bh$,
$$1 = \sin^2x+\cos^2x$$
Are there any more innovative ways of proving this common identity?
|
1) With identity $a^2-b^2= (a-b)(a+b)$,
$$\cos^2x+ \sin^2x
= (\cos x-i\sin x)(\cos x+i\sin x)=e^{-i x}e^{i x}=1$$
2) With double angle identities,
$$\cos^2x+ \sin^2x
=\frac12 (1+\cos 2 x)+\frac12(1-\cos 2x)=1$$
3) with half-angle subs $t= \tan\frac x2$,
$$\cos^2x+ \sin^2x
=\left(\frac{1-t^2}{1+t^2}\right)^2 +\left(\frac{2t}{1+t^2}\right)^2
=\left(\frac{1+t^2}{1+t^2}\right)^2 =1$$
4) As an integral,
$$f(x)=\cos^2x+ \sin^2x =f(0) + \int_0^x f’(t)dt = 1+0=1$$
5) With $x=it $, apply $\sin(it)= i\sinh t$ and $\cos(it)=\cosh t$
$$\cos^2x+ \sin^2x = \cosh^2 t - \sinh^2 t
= \left(\frac{e^t+e^{-t}}{2}\right)^2 -\left(\frac{e^t-e^{-t}}{2}\right)^2 =1$$
|
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|
What's the value of $\int \left(1-\sqrt{1-x^2} \right) dx$? $$\int \left(1-\sqrt{1-x^2} \right) dx$$
Also, What is it's value from $0$ to $(1+\sqrt7)/2$ ?
|
I believe there might be missing pieces or a mistake in the inquiry, here's why:
let x = sin(u)
Thus, $\frac{dx}{du}=cos(u)$
$$\int 1-\sqrt{1-x^2}dx = \int (1-cos(u))(cos(u))du $$
Using the Double angle formula: $cos(2\theta)=2cos^2\theta -1$ , we have:
$$ sin(u) - \int\frac{cos(2u)-1}{2}du $$ which thus evaluates to (using double angle formula $sin(2\theta)=2sin(\theta)cos(\theta)$):
$$ sin(u)-\frac{1}{2}sin(u)cos(u)+\frac{1}{2}u + C $$
Substituting x back into the equation:
$$ x - \frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}sin^{-1}x+C $$
Coming back to the original integral:
$$ \int_{0}^{\frac{1+\sqrt{7}}{2}}1-\sqrt{1-x^2}dx = [x - \frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}sin^{-1}x]_{0}^{\frac{1+\sqrt{7}}{2}} $$
Here's where it doesn't make sense, the last part of the expression requires you to evaluate $$ sin^{-1}(\frac{1+\sqrt{7}}{2}) $$
which will not be possible since $\frac{1+\sqrt{7}}{2}$ lies outside the function domain of $sin^{-1}x$.
|
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|
Prove that for all positives $a, b$ and $c$, $(\sum_{cyc}\frac{c + a}{b})^2 \ge 4(\sum_{cyc}ca)(\sum_{cyc}\frac{1}{b^2})$.
Prove that for all positives $a, b$ and $c$, $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \ge 4(bc + ca + ab) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Let $ca + ab = m$, $ab + bc = n$ and $bc + ca = p$, we have that $$\left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right)^2 \ge 2(m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
$$\iff \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2} - 1\right)^2 \ge 2 \cdot \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right] + 1$$
Expanding $\displaystyle \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right]$ gives $$2 \cdot \sum_{cyc}\frac{ca}{b^2} + \left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)$$
Let $\dfrac{b + c}{a} = m'$, $\dfrac{c + a}{b} = n'$ and $\dfrac{a + b}{c} = p'$, we have that $$(m' + n' + p' - 1)^2 \ge 2 \cdot \left[2 \cdot \sum_{cyc}\frac{ca}{b^2} + (m' + n' + p')\right] + 1$$
Moreover, $$(m')^2 + (n')^2 + (p')^2 = \sum_{cyc}\left[\left(\frac{c + a}{b}\right)^2\right] \ge 2 \cdot \sum_{cyc}\frac{ca}{b^2}$$
$$\implies (m' + n' + p' - 1)^2 \ge 2 \cdot \left[(m')^2 + (n')^2 + (p')^2 + m' + n' + p'\right] + 1$$
$$\iff -[(m')^2 + (n')^2 + (p')^2] + 2(m'n' + n'p' + p'm') - 4(m' + n' + p') \ge 0$$, which is definitely not correct.
Another attempt, let $(0 <) \ a \le b \le c \implies ab \le ca \le bc \iff ca + ab \le ab + bc \le bc + ca$
$\iff m \le n \le p$ and $a^2 \le b^2 \le c^2 \iff \dfrac{1}{a^2} \ge \dfrac{1}{b^2} \ge \dfrac{1}{c^2}$.
By the Chebyshev inequality, we have that $$3 \cdot \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right) \le (m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Any help would be appreciated.
|
WLOG assume $c\neq \text{mid}\{a,b,c\}$. We have:
$$a^2 b^2 c^2 (\text{LHS-RHS}) =\left( a-b \right) ^{2} \left( ab+ca+bc-{c}^{2} \right) ^{2}+4\,ab{c}
^{2} \left( a-c \right) \left( b-c \right) \geqq 0$$
Done.
|
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|
Integral $\int_0^1(1-x^3+x^5-x^8+x^{10}-x^{13}+\dots)dx$ $$\int_0^1(1-x^3+x^5-x^8+x^{10}-x^{13}+\dots)dx$$
Here's my attempts but I'm not sure that I'm doing well :
$$\text{The integral gives :} 1-\frac1 4+\frac1 6-\frac1 9+ \frac{1}{11}-\frac{1}{14}+\dots$$
This series is :
$$ S=\sum_{k=0}^\infty \frac{3}{(5k+1)(5k+4)}$$
Using Wolfram alpha I got :
$$S=\frac1 5\sqrt{1+\frac{2}{\sqrt{5}}}\pi$$
So If what I did is true the integral must give us this value.
|
Let :
$$I=\int_0^1(1-x^3+x^5-x^8+x^{10}-x^{13}+\dots)dx$$
Let $$S=\sum_{k=0}^\infty \frac{3}{(5k+1)(5k+4)}$$
Let's compute it :
\begin{align}
S&=3\sum_{k=0}^\infty \frac{1}{(5k+1)(5k+4)} \\\\
&=\frac3 4 +\frac1 5 \sum_{k=1}^\infty\bigg(\frac{1}{k+1/5}-\frac{1}{k+4/5}\bigg) \\\
&=\frac{3}{4}+\frac{1}5\sum_{k=1}^\infty\bigg(\frac{1}{k+1/5}-\frac1 k\bigg)+\frac1 5\sum_{k=1}^\infty\bigg(\frac1 k-\frac1{k+4/5}\bigg)\\\
\text{We are introducing the digamma function $\psi$}: \\\
S&=\frac 3 4+\frac 1 5 \Bigg(\psi\bigg(\frac{9}5\bigg)-\psi\bigg(\frac{6}5\bigg)\Bigg)\\\
&=\frac 3 4+\frac 1 5 \Bigg(\psi\bigg(\frac{4}5\bigg)+\frac{5}4\Bigg)-\frac1 5\Bigg(\psi\bigg(\frac1 5\bigg)+5\Bigg)\\\
&=\frac 3 4+\frac 1 5\Bigg(\psi\bigg(\frac4 5\bigg)-\psi\bigg(\frac1 5\bigg)\Bigg)+\bigg(\frac1 4-1\bigg)\\\
&=\frac{1}5 \pi \cot\bigg(\frac{\pi}{5}\bigg) \\\
&=\frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}}
\end{align}
Therefore :
$$I=\frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\ge \frac{a^2+b^2+c^2}{ab+bc+ca}$ For $a,b,c>0$. Prove that: $$\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\geqq \frac{a^2+b^2+c^2}{ab+bc+ca}$$
NguyenHuyen gave the following expression$:$
$$\sum \frac12\, \left( 8\,{a}^{3}b+{a}^{3}c+8\,{a}^{2}{b}^{2}+11\,{a}^{2}bc+7\,a
{b}^{3}+13\,a{b}^{2}c+3\,ab{c}^{2}+3\,b{c}^{3}+2\,{c}^{4} \right)
\left( a+b \right) ^{2} \left( a-b \right) ^{2} \geqslant 0$$
My work with Titu's Lemma and Maple and by lucky!
By Titu's Lemma$,$ we have$:$
$$\text{LHS} \geqq \frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc} a^2(b+c)^2} +\frac{1}{4}\geqq \text{RHS}$$
Last inequality equivalent to$:$
$$\,{\frac {\sum\limits_{cyc} \left( a-b \right) ^{2}\Big[bc \left( 2\,{a}^{2}+ab+ca+{c}^{
2} \right) +2\,ac \left( a-c \right) ^{2}+2\,ab \left( {a}^{2}+{b}^{2}
\right) +c \left( b-c \right) ^{2} \left( a+2\,b \right)\Big]}{ 8\left( {
a}^{2}{b}^{2}+{a}^{2}bc+{a}^{2}{c}^{2}+a{b}^{2}c+ab{c}^{2}+{b}^{2}{c}^
{2} \right) \left( ab+ca+bc \right) }}\geqq 0$$
However$,$ it's very hard with me to find a nice SOS if not have Maple$,$ who have a simple proof for it?
Without $\it{uvw}$ and Buffalo Way if you can!
Thanks a lot!
$\lceil$You can also see here. $\rfloor$
|
Yes, SOS helps!
Indeed, we need to prove that
$$\sum_{cyc}\left(\frac{a^2}{(b+c)^2}-\frac{1}{4}\right)\geq\frac{a^2+b^2+c^2}{ab+ac+bc}-1$$ or
$$\sum_{cyc}\frac{(2a+b+c)(a-b-(c-a))}{(b+c)^2}\geq\frac{2\sum\limits_{cyc}(a-b)^2}{ab+ac+bc}$$ or
$$\sum_{cyc}(a-b)\left(\frac{2a+b+c}{(b+c)^2}-\frac{2b+a+c}{(a+c)^2}\right)\geq\frac{2\sum\limits_{cyc}(a-b)^2}{ab+ac+bc}$$ or
$$\sum_{cyc}(a-b)^2\left(\frac{2a^2+2b^2+3c^2+3ab+5ac+5bc}{(a+c)^2(b+c)^2}-\frac{2}{ab+ac+bc}\right)\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+b)^2(2a^3b+a^2b^2+2ab^3+2a^3c+6a^2bc+6ab^2c+2b^3c+3a^2c^2+5abc^2+3b^2c^2-ac^3-bc^3-2c^4)\geq0,$$ for which it's enough to prove that:
$$\sum_{cyc}(a-b)^2(a+b)^2(2a^3c+2b^3c-2c^4+a^2c^2+2abc^2+b^2c^2-ac^3-bc^3)\geq0$$ or
$$2\sum_{cyc}(a-b)^2(a+b)^2c(a^3+b^3-c^3)+\sum_{cyc}(a-b)^2(a+b)^3c^2(a+b-c)\geq0.$$
Now, let $a\geq b\geq c$.
Thus, $$b^2\sum_{cyc}(a-b)^2(a+b)^2c(a^3+b^3-c^3)\geq$$
$$b^2(a-c)^2(a+c)^2b(a^3+c^3-b^3)+b^2(b-c)^2(b+c)^2a(b^3+c^3-a^3)\geq$$
$$\geq a^2(b-c)^2(a+c)^2b(a^3-b^3)+b^2(b-c)^2(b+c)^2a(b^3-a^3)=$$
$$=ab(b-c)^2(a^3-b^3)(a(a+c)^2-b(b+c)^2)\geq0.$$
Also, $$b^2\sum_{cyc}(a-b)^2(a+b)^3c^2(a+b-c)\geq$$
$$\geq b^2(a-c)^2(a+c)^3b^2(a+c-b)+b^2(b-c)^2(b+c)^3a^2(b+c-a)\geq$$
$$\geq a^2(b-c)^2(a+c)^3b^2(a-b)+b^2(b-c)^2(b+c)^3a^2(b-a)=$$
$$=a^2b^2(b-c)^2(a-b)((a+c)^3-(b+c)^3)\geq0$$ and we are done!
|
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"timestamp": "2023-03-29T00:00:00",
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Probability problem on umbrellas
Two absent-minded roommates, mathematicians, forget their umbrellas in some way or another. $A$ always takes his umbrella when he goes out, while $B$ forgets to take his umbrella with probability $1/2$. Each of them forgets his umbrella at a shop with probability $1/4$. After visiting three shops, they return home. Find the probability that they have only one umbrella.
My Attempt
$A_i/\bar{A}_i$ : $A$ remembers or forgets umbrella at shop $i$
$B_i/\bar{B}_i$ : $B$ remembers or forgets umbrella at shop $i$
$B_o/\bar{B}_o$ : $B$ remembers or forgets umbrella at shop home
$A_i,,B_i=3/4;\bar{A}_i,\bar{B}_i=1/4$ and $B_o,\bar{B}_o=1/2$
$$
\text{Req. Prob.}=A_1A_2A_3\bar{B}_o+A_1A_2A_3B_o\bar{B}_1+A_1A_2A_3B_oB_1\bar{B}_2+A_1A_2A_3B_oB_1B_2\bar{B}_3+\bar{A}_1B_oB_1B_2B_3+A_1\bar{A}_2B_oB_1B_2B_3+A_1A_2\bar{A}_3B_oB_1B_2B_3\\
=\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{1}{4}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)\Big(\frac{1}{4}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^2\Big(\frac{1}{4}\Big)+\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3\\
+\Big(\frac{3}{4}\Big)\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3+\Big(\frac{3}{4}\Big)^2\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3\\
=\frac{27}{128}+\frac{27}{128*4}+\frac{81}{128*16}+\frac{243}{128*64}+\frac{27}{128*4}+\frac{81}{128*16}+\frac{243}{128*64}\\
=\frac{3726}{128*64}=\frac{3726}{8192}
$$
But my reference gives the solution $\frac{7278}{8192}$, so what am I missing in my attempt ?
Note: Please check $b$ part of link for a similar attempt to solve the problem to obtain the solution as in y reference.
|
I get a different answer from either of those. The probability that a professor who leaves the house with his umbrella returns with it is $$\left(\frac34\right)^3=\frac{27}{64}$$ and the probability that he loses it is $\frac{37}{64}$. There are two cases. Both A and B take their umbrellas with them (probability $\frac12$) and exactly one returns with it (probability $2\cdot\frac{27}{64}\cdot\frac{37}{64}$) or only B takes his umbrella (probability $\frac12$,) and he loses it (probability $\frac{37}{64}$.) This gives a total probability of $$
\begin{align}
&\frac12\cdot2\cdot\frac{27}{64}\cdot\frac{37}{64}+\frac12\cdot\frac{37}{64}\\&=\frac12\cdot\frac{37}{64}\left(2\cdot\frac{27}{64}+1\right)\\&=\frac{37}{128}\cdot\frac{121}{64}\\&=\frac{4477}{8192}
\end{align}$$
EDIT
The case where B doesn't take his umbrella and $A$ accounts for $\frac{2109}{8192}$ of this probability, so if we exclude it, we get a substantially smaller probability than you did. It seems to be that are computing the probability that at least one of them leaves his umbrella at a shop, which is the wrong thing, because if both lose their umbrellas, they have zero umbrellas, not one.
|
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|
$3(a+{1\over a}) = 4(b+{1\over b}) = 5(c+{1\over c})$ and $ab+bc+ca=1$ This was the question :
$3(a+{1\over a}) = 4(b+{1\over b}) = 5(c+{1\over c})$ where $a,b,c$ are positive number and $ab+bc+ca=1$ ,
$5({1-a^2\over 1+a^2}+{1-b^2\over 1+b^2}+{1-c^2\over 1+c^2}) = ?$
I have spent a lot of time trying rearranging the terms and combine the two conditions to get the required expression but I couldn't get the right form.
I wrote it like :
$5({2\over 1+a^2}+{2\over 1+b^2}+{2\over 1+c^2} -3) $
This seemed simpler than the required expression , I could not come up with anything else useful.
Could someone please help me in solving this question ?
Thanks !
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The hint.
Let $a=\tan\frac{\alpha}{2},$ $b=\tan\frac{\beta}{2}$ and $c=\tan\frac{\gamma}{2},$ where $\{\alpha,\beta,\gamma\}\subset(0^{\circ},180^{\circ}).$
Thus, $ab+ac+bc=1$ gives $\alpha+\beta+\gamma=180^{\circ}$ and the rest gives
$$\frac{3}{\sin\alpha}=\frac{4}{\sin\beta}=\frac{5}{\sin\gamma},$$ which gives $\gamma=90^{\circ}$ and $c=1.$
Can you end it now?
Since $c=1$, we obtain: $$b\in\{2,\frac{1}{2}\}$$ and $$a\in\{3,\frac{1}{3}\},$$
but since $$ab+ac+bc=1,$$ we obtain:
$$(a,b,c)=\left(\frac{1}{3},\frac{1}{2},1\right).$$
|
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|
Find all natural numbers $a,b$ such that $ab, 4a+b-3$ are perfect squares and $9a-4b$ is a prime number. Find all natural numbers $a,b$ such that $ab, 4a+b-3$ are perfect squares and $9a-4b$ is a prime number.
I can't find any idea for this problem :(
|
If $ab$ is a perfect square, then $a$ and $b$ must be perfect squares, say $a=m^2$ and $b=n^2$. It follows that
$$
9a-4b = (3m+2n)(3m-2n),
$$
so that for $9a-4b$ to be prime it must be or $3m+2n=1$ or $3m-2n=1$. Now we consider $4a+b-3\pmod{3}$: the only squares in $\mathbb{Z}/3\mathbb{Z}$ are $0$ and $1$, thus the only possibilities are $a\equiv 0\pmod{3}\wedge b\equiv 0\pmod{3}$ (which we can exclude immediately because in this case $9a-4b$ cannot be prime), $a\equiv 1\pmod{3}\wedge b\equiv 0\pmod{3}$ (which we can exclude because otherwise $3\mid 9a-4b$) and $a\equiv 0\pmod{3}\wedge b\equiv 1\pmod{3}$.
We now consider this last case. By splitting on whether $3m+2n=1$ or $3m-2n=1$ we get
$$
\begin{cases}
m\equiv 3\pmod{6}\\ n\equiv 2\pmod{3}
\end{cases}
\quad\vee\quad
\begin{cases}
m\equiv 3\pmod{6}\\ n\equiv 1\pmod{3}
\end{cases}.
$$
The first of these two makes it impossible to have $9a-4b$ prime. As for the second case, by writing $n=1+3k$ and $m=3+6h$ we can find that for $9a-4b$ to be prime we must have $k=3h+1$, from which $m=3+6h$ and $n=4+9h$. Now $4a+b-3=225h^2+216h+49$ is a perfect square if and only if $h=0$, so that the only solution to the problem are $a=9$ and $b=16$.
|
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Olympiad Inequality Question I am new to the Olympiad-style questions and I hope someone could correct my proof for this question as I do not have the answer for it. Please leave some constructive criticism if possible so I could improve. Thanks in advance!
Let $a,b,c$ be positive real numbers. Prove that:
$$a^3 +b^3 +c^3\geq a^2b+b^2c+c^2a$$
My attempt:
By AM-GM Inequality,
$$\frac{a^2+b^2}{2}\ge\sqrt{a^2 b^2}=ab$$
$$\frac{a^2+c^2}{2}\ge\sqrt{a^2 c^2} =ac$$
$$\frac{b^2+c^2}{2}\ge\sqrt{b^2 c^2} =bc$$
Next, multiply $a,b,$ or $c$ to get RHS of the inequality wanted above:
$\dfrac{a(a^2+b^2)}{2} \ge a^2b$, $\dfrac{b(b^2+c^2)}{2} \ge b^2c$, $\dfrac{c(a^2+c^2)}{2} \ge ac^2$
Adding up the inequalities gives us:
$$\dfrac{a^3+ab^2+b^3+bc^2+a^2c+c^3}{2} \ge a^2b+b^2c+ac^2$$
and rearranging the inequality gives us:
$$a^3+b^3+c^3 \ge 2(a^2b+b^2c+ac^2)-ab^2-bc^2-a^2c$$ which is generally true.
|
While your final inequality isn't yet what you want, you can make a small modification to get the the answer.
The idea here is that the cyclic sums (but in opposite directions) $Y$ and $Z$ are highly related to each other, so if we have several expressions involving them, we can try to cancel out a term (E.g. via Gaussian elimination).
Let $ X = \sum a^3, Y = \sum a^2b , Z = \sum ab^2$.
You are asked to show that $ X \geq Y$.
You have shown that $ X \geq 2Y - Z$.
Similarly, we can show that $ X \geq 2Z - Y$ by slightly modifying your step. (Do you see how?)
Next, multiply $b,c,$ or $a$ to get: $\frac{b(a^2+b^2)}{2} \geq ab^2$, $\frac{c(b^2+c^2)}{2} \geq bc^2$, $\frac{a(a^2+c^2)}{2} \geq a^2c$.
Adding up the inequalities gives us: $ X \geq 2Z - Y$.
Then, this gives us $ 3X \geq 2 ( 2Y - Z) + (2Z - Y) = 3 Y$.
Hence $ X \geq Y$ as desired.
Moral of the story: Sometimes you're just a stepping stone away.
|
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Find all ordered pairs $(x,y)$ of positive integers such that the expression $x^2+y^2+xy$ is a perfect square. My approach so far:
Let $x^2+y^2+xy=n^2,$ where $n\in\mathbb Z$.
$\implies (x+y) ^2-xy=n^2$
$\implies (x+y) ^2-n^2=xy$
$\implies (x+y+n) (x+y-n) =xy$
Only one case is possible:
When $x+y+n=xy$ and $x+y-n=1$. On adding, we get: $2(x+y) =xy+1$. Since LHS is an even number so is RHS. That's why, $xy$ must be odd$\implies x$ and $y$ both are odds. So I checked for few odd numbers and found out that for $x=3$ and $y=5,\;x^2+y^2+xy$ becomes $49$ which is a perfect square.. I didn't dare to check for more odds as there could be many...
Recently I found out that for $x=5$ and $y=16$(an even number), $x^2+y^2+xy=361=19^2.$ (Surprising!!)
So now I can say, I am stuck very badly.. All of my observations miserably failed...
Please suggest something!
|
$x^2+yx+y^2-n^2=0 \implies \triangle = 4n^2-3y^2= d^2$. At this point I propose you to look up a article by L.J. Mordell or Kneser in internet and they have a formula for all the finite solutions that are bounded above by the products of the coefficients $(4,-3,-1)$ of the Diophantine equation above.
|
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How to prove the following inequality with two variables How to prove the following inequality, which should hold for all $x \in \left]0; 1\right[$, $n \in \mathbb{N}$, n is even.
\begin{align}
n^2 x^{n-2} (x^{n+1}+1)^3\left[2nx^n - (n-1)(x^n-1)\right] > (n+1)^2 x^{n+1} (x^n-1)^3\left[(2(n+1)x^{n+1} - n(x^{n+1}+1)\right]
\end{align}
My attempt so far:
\begin{align}
n^2 x^{n-2} (x^{n+1}+1)^3\left[2nx^n - (n-1)(x^n-1)\right] &> (n+1)^2 x^{n+1} (x^n-1)^3\left[2(n+1)x^{n+1} - n(x^{n+1}+1)\right] \\
n^2 (x^{n+1}+1)^3\left[2nx^n - (n-1)(x^n-1)\right] &> (n+1)^2 x^3 (x^n-1)^3\left[2(n+1)x^{n+1} - n(x^{n+1}+1)\right] \\
\left(\frac{n}{n+1}\right)^2 \left(\frac{x^{n+1}+1}{x^{n+3}-1}\right)^3\frac{2nx^n - (n-1)(x^n-1)}{2(n+1)x^{n+1} - n(x^{n+1}+1)} &> 1
\end{align}
\begin{align}
\frac{1}{4} &\leq \left(\frac{n}{n+1}\right)^2 &< 1 \\
-1 &< \left(\frac{x^{n+1}+1}{x^{n+3}-1}\right)^3 &< 0
\end{align}
But here I stuck.
|
Indeed you need to prove:
\begin{align}
\left(\frac{n}{n+1}\right)^2 \left(\frac{1+x^{n+1}}{x-x^{n+1}}\right)^3\frac{(n-1)+(n+1)x^n}{n-(n+2)x^{n+1}} &> 1
\end{align}
We have
$$ |n-(n+2)x^{n+1}|\leq n.$$
Also
$$ \left(\frac{1+x^{n+1}}{x-x^{n+1}}\right)^3>x^{-3},$$
so we need to prove that
$$(n-1)x^{-3}+(n+1)x^{n-3}>\frac{(n+1)^2}{n}.$$
For $n\geq 4$ we use Cauchy in the next way:
$$(n-1)x^{-3}+(n+1)x^{n-3}=(n-3)*\frac{n-1}{n-3}x^{-3}+3*\frac{n+1}{3}x^{n-3}\geq$$
$$\geq n \sqrt[n]{\left (\frac{n-1}{n-3} \right )^{n-3} \left (\frac{n+1}{3} \right )^{3}}\geq n*\min \left (\frac{n-1}{n-3},\frac{n+1}{3}\right ).$$
The last expression is more than $(n+1)^2/n$ if $n\geq 4$.
If $n=2$ we need to prove:
$$(1+x^3)^3(1+3x^2)>\frac{9}{2}x^3(1-x^2)^3(1-2x^3),$$
which is true because the right side is always less than $1$ while left side is more than $1$.
|
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|
Packs of gum probability On each gum pack there's a prize. There's $6$ different prizes and each gum pack has the same probability to have each prize. Johnny buys a gum pack each day to collect all the different prizes and only then he will stop.
I need to calculate the standard deviation of the number of days that Johnny buys packs of gum.
SO I see it's Geometric distribution and I think I need to calculate the variance first and from that to take a square root to calculate the std. So, to calculate the variance I need the probability to get a prize. So I need to calculate the variance for each prize? First, $1/6$, then $1/5$ etc, and then to sum the variances up?
|
Let $\mathbf{p}_n=(p_1,p_2,p_3,p_4,p_5,p_6)^T$ and $p_i$ be the probability of having $i$ distinct prizes after opening $n$ gum packs. $\mathbf{p}_1=(1,0,0,0,0,0)^T$ and $\mathbf{p}_n=A\mathbf{p}_{n-1}$ where
$$A=\begin{pmatrix}
\frac{1}{6} && \frac{5}{6} && 0 && 0 && 0 && 0\\
0 && \frac{2}{6} && \frac{4}{6} && 0 && 0 && 0\\
0 && 0 && \frac{3}{6} && \frac{3}{6} && 0 && 0\\
0 && 0 && 0 && \frac{4}{6} && \frac{2}{6} && 0\\
0 && 0 && 0 && 0 && \frac{5}{6} && \frac{1}{6}\\
0 && 0 && 0 && 0 && 0 && 1
\end{pmatrix}^T$$
The probability of getting the full of $6$ distinct prizes after opening exactly $n$ gum packs is $\frac16(0,0,0,0,1,0)\mathbf{p}_{n-1}$ as Johnny have to have exactly $5$ distinct prizes after opening $n-1$ gum packs and the $6$th will be different with the probability of $\frac16$.
By computing diagonalization of $A=SDS^{-1}$ we will be able to find $p_5(n-1)$ explicitly, as $\mathbf{p}_{n-1}=A^{n-2}\mathbf{p}_{1}$ and $A^{n-2}=SD^{n-2}S^{-1}$
Diagonalization of $A^T$:
$$S=\begin{pmatrix}
1 && 5 && 10 && 10 && 5 && 1\\
0 && 1 && 4 && 6 && 4 && 1\\
0 && 0 && 1 && 3 && 3 && 1\\
0 && 0 && 0 && 1 && 2 && 1\\
0 && 0 && 0 && 0 && 1 && 1\\
0 && 0 && 0 && 0 && 0 && 1
\end{pmatrix}
$$
$$D=\operatorname{diag}\left(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6},1\right)$$
So $p_5(n-1)=15\cdot2^{3 - n} - 20\cdot3^{2 - n} - 5\cdot2^n\cdot3^{2 - n} + 5\cdot6^{2 - n} + 5^{-1 + n}\cdot6^{2 - n}$
Performing the explicit summations $M[X]=\frac16\sum\limits_{n=2}^\infty n\cdot p_5(n-1)$$=\frac{147}{10}$ and
$M[X^2]=\frac16\sum\limits_{n=2}^\infty n^2\cdot p_5(n-1)$$=\frac{6377}{25}$ and thus the variance is $\sigma^2=D[X]=M[X^2]-\left(M[X]\right)^2=\frac{3899}{100}$
|
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|
What is the total number of positive integers <300 whose digit sum is a multiple of 5? As stated in the title
For one digit number there is $1$
For two digit numbers there are $27(=4+(4+5)+(4+5+5)) (sum
=5n, n=1,2,3)$ $since (5n<18)$
But I'm struggling to find the ones that are of three digits
$(sum)=5n, n=1,...,4)$$for (5n\leq20)$ (without writing out all the possible combinations).
The possible answers are $18,45,60,61$
Can someone please explain which of the above answers is the most appropriate?
|
If the first two digits are $0,0$ than the last digit must be $0$ or $5$. There are two such numbers.
If the first two digits are $0,1$ then the last digit must be $4$ or $9$. There are two such number.
If the first two digits are $2$ and $7$ that then last digit must be $1$ and $6$. There are two such numbers.
And so on.
The first two digits are $00$ to $29$, that's $30$ option and there are two options for $c$ so that there are $2*30 = 60$ such numbers. But $000$ is not acceptable so there are $59$.
So, $59$. Final answer.
....
But we must prove that if the first two digits are $a$ and $b$ there are exactly $2$ digits that third digit can be.
=====
ANd... if you add up $a$ and $b$ and take the remainder from dividing be $5$ (call it $r$) there are exactly $5$ possible remainders values of $r$. For $a+b +c$ to be a multiple of $5$ then the remainder $a+b+c$ must be $0$ and the remainder of $c$ must be the opposite $r$. That is if $r=1$ then the remainder of $c$ (call it $s$) must be $4$ and if $r=2$ then $s = 3$ and $r=3$ then $s=2$ and if $r=4$ then $s=1$ and if $r=0$ then $s = 0$.
And there are exactly $5$ possible opposite remainders that $s$ can be. $c$ can be $s+0$ and $s+5$.
....
Putting that argument in terms of
Modular arithmetic: If $a+b \equiv r \pmod 5$ then $c \equiv - r\pmod 5$. As $\{0,1,2,3,4\}$ and $\{5,6,7,8,9\}$ are two complete residue systems there are exactly one representative for $-r$ in each set so there are two option for digit $c$.
.....
Remainder theorem:
Let $a+b = 5K + r$. $a+b+c = 5M + 0$. Then $c= (5M + 0)- (5K+r)=5(M-K)+r$ and $0\le c \le 9$ and $0 \le r \le 4$. If $r=0$ then $0\le c= 5(M-K)\le 9$ so $c=0, 5$. If $0 < r < 5$ then $0\le c = 5(M-k)-r \le 9$ then $c = 5-r$ or $c = 10-r$.
|
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|
finding the characteristic polynomial of symmetric matrix So I got this symmetric matrix
$$A= \begin{pmatrix}
0 & 3 & 1 & -2 \\
3 & 0 & -2 & 1 \\
1 & -2 & 0 & 3 \\
-2 & 1 & 3 & 0\\
\end{pmatrix}$$
I've got to $$P(t)=\begin{pmatrix}
-t & 3 & 1 & -2 \\
3 & -t & -2 & 1 \\
1 & -2 & -t & 3 \\
-2 & 1 & 3 & -t\\
\end{pmatrix}.$$
My question is does the simple multiplication method not work for this? When I multiply the diagnols to get the determinant I end up with $t^4-2t^2-24$.
I've also tries $a_{i,j}*det\begin{pmatrix}
a&b&c\\
b&c&d\\
c&d&e\\
\end{pmatrix}$ method(does this have a name?), but still it didn't seem right.
When using the Gaussian elimination I get $t(t-2)(t-4)(t+6)$, which seems to be the correct answer.
I'm guessing the other ones doesn't meet the some condition here, But what is it? And is there any other way than using the gaussian elimination here?
|
The Laplace expansion with cofactors works:
$-t\begin{vmatrix}-t&-2&1\\-2&-t&3\\1&3&-t\end{vmatrix}-3\begin{vmatrix}3&-2&1\\1&-t&3\\-2&3&-t\end{vmatrix}+\begin{vmatrix}3&-t&1\\1&-2&3\\2&1&-t\end{vmatrix}+2\begin{vmatrix}3&-t&-2\\1&-2&-t\\-2&1&3\end{vmatrix}$
$=t^4 - 28 t^2 + 48 t=t(t - 4) (t - 2) (t + 6).$
|
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Ahmed integral revisited $\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} \, dx$ How to prove
$$\small \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} dx=-\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{8}+\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{4}+\frac{\pi }{4}\arctan \left(\frac{1}{\sqrt{5}}\right)\;?$$
I came across this Ahmed integral on the site "Art of problem solving", and have found no proof so far. (These two problems seems to be related though). Any help will be appreciated!
|
\begin{align}
J&=\int_0^1 \frac{\arctan\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} \, dx\\
K&=\int_0^1 \int_0^1 \frac{1}{(x^2+2)(y^2+2)}dxdy \\
&=\int_0^1 \int_0^1 \frac{1}{4+x^2+y^2}\left(\frac{1}{2+x^2}+\frac{1}{2+y^2}\right)dxdy\\
&=2\int_0^1 \int_0^1 \frac{1}{(4+x^2+y^2)(2+x^2)}dxdy\\
&=2 \int_0^1 \left[\frac{\arctan\left(\frac{y}{\sqrt{4+x^2}}\right)}{(2+x^2)\sqrt{4+x^2}}\right]_{y=0}^{y=1} dx\\
&=2\int_0^1 \frac{\arctan\left(\frac{1}{\sqrt{4+x^2}}\right)}{(2+x^2)\sqrt{4+x^2}}dx\\
&=\pi \int_0^1 \frac{1}{(2+x^2)\sqrt{4+x^2}}dx-2J\\
&=\frac{\pi}{2} \left[\arctan\left(\frac{x}{\sqrt{4+x^2}}\right)\right]_0^1-2J\\
&=\frac{\pi}{2}\arctan\left(\frac{1}{\sqrt{5}}\right)-2J\\
\end{align}
On the other hand,
\begin{align}K&=\left(\int_0^1 \frac{1}{2+x^2}dx\right)^2\\
&=\left(\frac{1}{\sqrt{2}}\left[\arctan\left(\frac{x}{\sqrt{2}}\right)\right]_0^1\right)^2\\
&=\frac{1}{2}\arctan^2\left(\frac{1}{\sqrt{2}}\right)
\end{align}
Therefore,
$\displaystyle \boxed{J=\frac{\pi}{4}\arctan\left(\frac{1}{\sqrt{5}}\right)-\frac{1}{4}\arctan^2\left(\frac{1}{\sqrt{2}}\right)}$
|
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|
Proof of equation relating to vector triple product and BAC-CAB rule I am currently studying Introduction to Electrodynamics, fourth edition, by David J. Griffiths. Chapter 1.1.3 Triple Products introduces the vector triple product as follows:
(ii) Vector triple product: $\mathbf{A} \times (\mathbf{B} \times \mathbf{C})$. The vector triple product can be simplified by the so-called BAC-CAB rule:
$$\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B}(\mathbf{A} \cdot \mathbf{C}) - \mathbf{C}(\mathbf{A} \cdot \mathbf{B}). \tag{1.17}$$
Notice that
$$(\mathbf{A} \times \mathbf{B}) \times \mathbf{C} = - \mathbf{C} \times (\mathbf{A} \times \mathbf{B}) = - \mathbf{A}(\mathbf{B} \cdot \mathbf{C}) + \mathbf{B}(\mathbf{A} \cdot \mathbf{C})$$
is an entirely different vector (cross-products are not associative). All higher vector products can be similarly reduced, often by repeated application of Eq. 1.17, so it is never necessary for an expression to contain more than one cross product in any term. For instance,
$$(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D}) = (\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot \mathbf{D}) - (\mathbf{A} \cdot \mathbf{D})(\mathbf{B} \cdot \mathbf{C});$$
$$\mathbf{A} \times [ \mathbf{B} \times (\mathbf{C} \times \mathbf{D})] = \mathbf{B}[\mathbf{A} \cdot (\mathbf{C} \times \mathbf{D})] - (\mathbf{A} \cdot \mathbf{B})(\mathbf{C} \times \mathbf{D}). \tag{1.18}$$
I am trying to prove that
$$(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D}) = (\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot \mathbf{D}) - (\mathbf{A} \cdot \mathbf{D})(\mathbf{B} \cdot \mathbf{C})$$
I attempted to prove this in the comments to this answer of a related question that I had asked. However, as you can see, it is still not clear to me how this is done.
I would greatly appreciate it if people would please take the time to clarify how this is done.
|
$(A\times B)\cdot(C\times D)=(A\times B)\cdot E=A\cdot(B\times E)$
where $E=C\times D$, and by the scalar triple product identity.
Then $B\times E=B\times(C\times D)=(B\cdot D)C-(B\cdot C)D$ by
the vector triple product identity. Now dot this with $A$.
|
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How would you find the number of solutions to $m_1+2m_2+3m_3=n$? Assuming $m_i$ are nonnegative integers
I understand that we need to use $C(n+k-1,n-1)$ here but I am not sure how the coefficients of $m_i$ affect the equation?
For example to find the number of solutions to $m_1+m_2+m_2=n$, we could use the equation above to find that it is $C(n+2,2)=\frac{1}{2}(n+2)(n+1)$
|
I'm presuming $n$ is given. The answer (call it $f(n)$) is the coefficient of $x^n$ in the Taylor expansion of the generating function
$$ g(x) = (1+x+x^2 + \ldots)(1+x^2 + x^4 +\ldots)(1 + x^3 + x^6 + \ldots) = \frac{1}{(1-x)(1-x^2)(1-x^3)}$$
Using partial fractions,
$$ g(x) = \frac{1}{6(1-x)^3} + \frac{1}{4(1-x)^2} + \frac{1}{8(1+x)} + \frac{17}{72(1-x)} + \frac{2+x}{9(1+x+x^2)}$$
and we can write
$$ f(n) = \frac{47}{72} + \frac{n}{2} + \frac{n^2}{12} + \frac{(-1)^n}{8} + \frac{\omega^n + \overline{\omega}^n}{9} $$
where $\omega = (-1 + i \sqrt{3})/2$.
Hmm, seems to be OEIS sequence A001399
|
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Solving a nonlinear periodic ODE Is it possible to solve the below ODE for arbitrary real values of $c$?
$$2 (\frac{d \theta}{dx})^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{d^2}{dx^2}\theta=0,$$
where $\theta=\theta(x)$. Also, $\theta$ itself does not have meaning, but ($\cos(2\theta),\sin(2\theta)$) are important and need to be continuous. There is no boundary condition since $x$ is not constrained by a boundary but goes to infinity. It means that the solution must cover the range $(-\infty,\infty)$.
|
$$2 (\frac{d \theta}{dx})^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{d^2}{dx^2}\theta=0$$
Let $y=\frac{d \theta}{dx}=\frac{d \theta}{dy}\frac{dy}{dx}\quad\implies\quad \frac{dy}{dx}=y\frac{dy}{d \theta}$
$$2 y^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{dy}{dx}=0$$
$$2 y^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]y\frac{dy}{d\theta}=0$$
$$\frac{dy}{y}=2\frac{\sin(2\theta) -c \cos(2 \theta)}{\cos(2\theta) +c \sin(2 \theta)}d\theta$$
$$\ln|y|=2\int \frac{\sin(2\theta) -c \cos(2 \theta)}{\cos(2\theta) +c \sin(2 \theta)}d\theta=-\ln|\cos(2\theta) +c \sin(2 \theta)|+\text{constant}$$
$$\frac{d \theta}{dx}=y=\frac{C_1}{\cos(2\theta) +c \sin(2 \theta)}$$
$$C_1x=\int \big(\cos(2\theta) +c \sin(2 \theta)\big)d\theta$$
$$C_1x+C_2=\frac12\big(\sin(2\theta) -c\: \cos(2 \theta)\big)$$
Then there is no difficulty to solve it for $\theta(x)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Second order ODE solution - help me spot a mistake Solve following ODE:
$$
(1-x)x''+2(x')^2=0; x(0)=2, x'(0)=-1
$$
$$
x''=\frac{-2(x')^2}{1-x}
$$
substitute $x'=u(x)$ and assume $u \neq 0$
$$
uu'=\frac{-2(u)^2}{1-x} \\
\frac{dx}{1-x}=-\frac{du}{2u} \\
-\ln{|1-x|}=-\frac{1}{2}\ln{|u|}+c \\
|1-x|=\sqrt(|u|)e^c
$$
undo substitution $x'=u(x)$ and use $x(0)=2, x'(0)=-1$:
$$
|1-2|=\sqrt(|-1|)e^c \\
1 = e^c \\
c = 0
$$
so
$$
|1-x|=\sqrt{|x'|}\\
x' = (1-x)^2 \\
\frac{dx}{dt}=(1-x)^2 \\
dt = \frac{dx}{(1-x)^2} \\
t + c = \frac{1}{1-x}
$$
use $x(0)=2$
$$
0 + c = \frac{1}{1-2} \\
c = -1
$$
finally:
$$
t - 1 = \frac{1}{1-x} \\
\frac{1}{t-1} = 1 - x \\
x = 1 - \frac{1}{t-1} \\
x(t) = \frac{t-2}{t-1}
$$
Wolfram Alpha however, says the ODE is $x(t)=\frac{t+2}{t+1}$. Where is my mistake?
|
The first stage of the solution should be kept with an undetermined constant up to the point
$$
u(x)=C(x-1)^2.
$$
Then using $u(2)=-1$ gives $C=-1$ and thus
$$
x'(t)=-(x-1)^2,
$$
which has a different sign than the equation you got. The sign difference in the solution follows from here.
|
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|
A curious infinite product Let $R$ be the ring $\mathbb{Z[q]}/(q^2)$ whose elements $a+bq$ satisfy $q^2=0.$
Define $g_n\in R$ by $g_1=1$, $g_{2^n}=1-2^{n-1}q$ for $n>0$ and $g_{2n}=0,$ $g_{2n+1}=-q$ else.
It seems that in $R$ the following identity holds in the sense of formal power series with coefficients in $R$:
$$1+x+(1-q)x^2+(1-2q)x^3+(1-3q)x^4+\dots =(1+g_{1}x)(1+g_{2}x^2)(1+g_{3}x^3)\dots.$$
For $q=0$ this reduces to the well-known identity
$$1+x+x^2+x^3+x^4+\dots =(1+x)(1+x^2)(1+x^4)(1+x^8)\dots.$$
Any idea how to prove this?
|
Write $\epsilon = -q$, which will save us some minus signs. Start by substituting $x \mapsto \left( 1 + \frac{\epsilon}{2} \right) x$ into the well-known identity, which gives
$$\prod_{k \ge 0} (1 + (1 + 2^{k-1} \epsilon)) x^{2^k} = \frac{1}{1 - \left( 1 + \frac{\epsilon}{2} \right) x} = \sum_{n \ge 0} \left( 1 + \frac{n\epsilon}{2} \right) x^n.$$
This incorporates all the power-of-$2$ factors except that the factor for $k = 0$ is slightly off; above it's $(1 + \left( 1 + \frac{\epsilon}{2} \right) x)$ when you ask for it to be $1 + x$. We can correct this by multiplying by $\frac{1 + x}{1 + \left( 1 + \frac{\epsilon}{2} \right) x}$, which gives
$$\prod_{k \ge 0} (1 + g_{2^k} x^{2^k}) = \frac{1 + x}{1 - (1 + \epsilon) x^2} = \sum_{n \ge 0} \left( 1 + \left\lfloor \frac{n}{2} \right\rfloor \epsilon \right) x^{n}.$$
Now we need to incorporate the odd factors. The product of the odd factors only is, by direct expansion,
$$\prod_{k \ge 1} (1 + \epsilon x^{2k+1}) = 1 + \epsilon \sum_{k \ge 1} x^{2k+1} = 1 + \epsilon \frac{x^3}{1 - x^2} = \frac{1 - x^2 + \epsilon x^3}{1 - x^2}$$
and taking the product with the power-of-$2$ factors above gives
$$\prod_{n \ge 0} (1 + g_n x^n) = \frac{1 - x^2 + \epsilon x^3}{(1 - x^2 - \epsilon x^2)(1 - x^2)}.$$
This is not yet in the form we want it; we need to move the $\epsilon$ in the denominator to the numerator by multiplying by the "conjugate" $1 - x^2 + \epsilon x^2$, which gives
$$\begin{eqnarray*} \frac{(1 + x)(1 - x^2 + \epsilon x^3)(1 - x^2 + \epsilon x^2)}{(1 - x^2)^3} &=& \frac{(1 - x^2)^2 + \epsilon (x^2 + x^3 - x^4 - x^5)}{(1 - x)(1 - x^2)^2} \\
&=& \frac{1}{1 - x} + \epsilon \frac{x^2}{1 - x} \end{eqnarray*}$$
which is the desired result. I don't have a good explanation for why there is so much cancellation at the end.
|
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|
How do you find appropriate trig substitution for $\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$? I want to solve the below:
$$\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$$
I know for trig substitution, if I have something in the form of $\sqrt{x^2-a^2}$, I can use $x = a\sec{u}$; it just so happens my integral has a numerator in this form: $\sqrt{16x^2 - 3^2}$ so I know to use $x = 3\sec u$:
$$
\begin{align}
& \int \frac{\sqrt{16x^2 - 9}}{x} \, dx \\
= {} & \int \frac{\sqrt{16x^2 - 3^2}}{x} \, dx \\
= {} & \int \frac{\sqrt{16(3\sec u)^2 - 3^2}}{3\sec u} 3\sec u\tan u \, du \\
= {} & \int \frac{(\sqrt{16(3\sec u)^2 - 3^2)}(3\sec u\tan u)}{3\sec u} \, du \\
= {} & \int \sqrt{(16(3\sec u)^2 - 3^2)}(\tan u) \, du
\end{align}
$$
This doesn't seem to make it easy. However, using a calculator online, it suggests I instead use $x = \dfrac{3}{4}\sec{u}$ which simplifies the integral to a crisp $\int 3\tan^2 u \, du$.
My question is, how did the calculator get $a = \dfrac{3}{4}$ and is there a way to determine an ideal trig substitution for a given function?
|
First get rid of the annoying factors,
$$\int\frac{\sqrt{16x^2-9}}xdx=\int\frac{\sqrt{16\left(\dfrac{3y}4\right)^2-9}}{\dfrac{3y}4}d\dfrac{3y}4=3\int\frac{\sqrt{y^2-1}}{y}dy.$$
Then observe the identity
$$\left(\frac 12\left(t+\dfrac1t\right)\right)^2-1=\left(\frac 12\left(t-\dfrac1t\right)\right)^2.$$
Then with $y=\dfrac 12\left(t+\dfrac1t\right)$ and $dy=\dfrac 12\left(1-\dfrac1{t^2}\right)$,
$$\int\frac{\sqrt{y^2-1}}{y}dy=\int\frac{\dfrac 12\left(t-\dfrac1t\right)}{\dfrac 12\left(t+\dfrac1t\right)}\dfrac 12\left(1-\dfrac1{t^2}\right)dt=\frac12\int\left(1+\frac1{t^2}-\frac{4}{t^2+1}\right) dt$$
which is easy.
|
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|
$\int \frac{x^2\,dx}{(a-bx^2)^2}$ How do I integrate $\int \frac{x^2\,dx}{(a-bx^2)^2}$
I've tried substitution and partial fraction decomposition, but I'm not getting anywhere.
|
You haven't placed any restrictions on ${a,b}$, and so the answer depends. We will assume ${a,b >0}$.
In this case we can actually use the hyperbolic trig substitution ${x=\sqrt{\frac{a}{b}}\tanh(t)}$, leading to
$${\left(\frac{a}{b}\right)^{\frac{3}{2}}\int\frac{\tanh^2(t)\text{sech}^2(t)}{(a-a\tanh^2(t))^2}}dt=\frac{1}{a^2}\left(\frac{a}{b}\right)^{\frac{3}{2}}\int\frac{\tanh^2(t)\text{sech}^2(t)}{(1-\tanh^2(t))^2}dt$$
We know that ${\sinh^2(t)-\cosh^2(t)=1}$, and this implies ${\tanh^2(t)-1=\text{sech}^2(t)}$. And so we have
$${\Rightarrow\frac{1}{a^2}\left(\frac{a}{b}\right)^{\frac{3}{2}}\int\frac{\tanh^2(t)\text{sech}^2(t)}{(-\text{sech}^2(t))^2}dt=\frac{1}{a^2}\left(\frac{a}{b}\right)^{\frac{3}{2}}\int\frac{\tanh^2(t)}{\text{sech}^2(t)}dt=\frac{1}{a^2}\left(\frac{a}{b}\right)^{\frac{3}{2}}\int\sinh^2(t)dt}$$
And this can be solved with another identity, ${\sinh^2(t)=\frac{\cosh(2t)-1}{2}}$:
$${\frac{1}{a^2}\left(\frac{a}{b}\right)^{\frac{3}{2}}\left(\int\frac{\cosh(2t)-1}{2}dt\right)=\frac{1}{a^2}\left(\frac{a}{b}\right)^{\frac{3}{2}}\left(\frac{\sinh(2t)}{4}-\frac{t}{2}\right)+c=\frac{1}{4a^2}\left(\frac{a}{b}\right)^{\frac{3}{2}}\left(\sinh(2t)-2t\right)+c}$$
This gives the final answer
$${\frac{1}{4a^2}\left(\frac{a}{b}\right)^{\frac{3}{2}}\left(\sinh\left(2\text{arctanh}\left(\sqrt{\frac{b}{a}}x\right)\right)-2\text{arctanh}\left(\sqrt{\frac{b}{a}}x\right)\right)+C}$$
You can further use the natural log definition for the inverse hyperbolic tangent, and the exponential definition for the hyperbolic sine to simplify the answer.
|
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|
Eigenvectors of $\begin{pmatrix}2&-2\\ -4&-2\end{pmatrix}$ - What am I doing wrong? Eigenvalues are $2\sqrt{3}$ and $-2\sqrt{3}$, I'll calculate the eigenvector for $2\sqrt{3}$ here
We've got:
$\begin{pmatrix}2-2\sqrt{3}&-2\\ -4&-2-2\sqrt{3}\end{pmatrix}\begin{pmatrix}y_1\\ y_2\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}$
I multiply the first row by negative 1, I get:
$\begin{pmatrix}-2+2\sqrt{3}&2\\ -4&-2-2\sqrt{3}\end{pmatrix}\begin{pmatrix}y_1\\ y_2\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}$
I add the first and second row together, I get:
$\left(-6+2\sqrt{3}\right)y_1-2\sqrt{3}y_2=0$
So the solution would be
$\left(-\sqrt{3}+1\right)y_1=y_2$
The eigenvector is $\begin{pmatrix}y\\ \left(1-\sqrt{3}\right)y\end{pmatrix}$
But that's apparently wrong according to this calculator
What's wrong here?
|
It is not clear that your answer disagrees with the source you cite.
\begin{align}
& \left[ \begin{array}{c} 1 \\ 1- \sqrt 3 \end{array} \right] = \frac 1 {1+\sqrt3} \left[ \begin{array}{c} (1+\sqrt3\,)\cdot1 \\[4pt] (1+\sqrt3\,) (1- \sqrt 3\,) \end{array} \right] \\[15pt]
= {} & \frac 1 {1+\sqrt3} \left[\begin{array}{c} 1+ \sqrt 3 \\[4pt] -2 \end{array} \right] = \frac {-2} {1+\sqrt3} \left[\begin{array}{c} (-1- \sqrt 3\,)/2 \\[4pt] 1 \end{array} \right]
\end{align}
|
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|
Prove: $(\forall m, n\in\Bbb N_{>0})(\exists x\in\Bbb R)$ s. t. $2\sin n x \cos m x \ge 1$
Problem 1:
Prove that for any $m,n\in\Bbb N_{>0}$, there exists $x \in\Bbb R$ such that
$2\sin n x \cos m x \ge 1$.
Four months ago, someone asked the above question. However, when I wanted to post my answer,
the question was deleted. I searched by Approach0 without any result.
I think that it is a nice question. I don't know why it was deleted. I post it here. I don't remember who posted it before.
Edit 2021/02/20: We may restrict $x$ to be the form $x = r\pi$ where $r$ is a rational number. We give the following problem:
Problem 2: Prove that for any $m, n \in \mathbb{N}_{>0}$, there exists rational $r$ such that
$2\sin (n r \pi) \cos (m r\pi) \ge 1$.
Any comments and solutions are welcome and appreciated.
Partial results are as follows.
If $n = m$, let $x = \frac{\pi}{4n}$ and we have $2\sin n x \cos m x = \sin 2n x = 1$.
If $n > m$, let $x = \frac{\pi}{2(2n-1)}$. Since $0 < n x < \pi$ and $0 < m x \le (n-1)x < \pi$, we have
\begin{align}
2\sin n x \cos m x &\ge 2\sin n x \cos (n-1)x \\
&= \sin (2n-1)x + \sin x \\
&= 1 + \sin \frac{\pi}{2(2n-1)}\\
& \ge 1.
\end{align}
|
Remarks: Here is my proof for Problem 2. I wrote it for Problem 1 originally, though Problem 1 might be easier. By the way, @mathlove gave a nice proof.
Problem 2: Prove that for any $m, n \in \mathbb{N}_{>0}$, there exists rational $r$ such that
$2\sin (n r \pi) \cos (m r\pi) \ge 1$.
Proof: Let $x = r\pi$. We split into five cases:
*
*$n > m$: Let $x = \frac{\pi}{2(2n-1)}$. Since $0 < n x < \pi$ and $0 < m x \le (n-1)x < \pi$, we have
\begin{align}
2\sin n x \cos m x &\ge 2\sin n x \cos (n-1)x \\
&= \sin (2n-1)x + \sin x \\
&= 1 + \sin \frac{\pi}{2(2n-1)}\\
& \ge 1.
\end{align}
*$n = m$: Let $x = \frac{\pi}{4n}$. We have $2\sin n x \cos m x = \sin 2n x = 1$.
*$\frac{5m}{6} < n < m$: Let $x = -\frac{3\pi}{4m}$.
Since $\frac{5\pi}{8} < \frac{3n\pi}{4m} < \frac{3\pi}{4}$, we have
\begin{align}
2\sin n x \cos mx &= \sqrt{2}\, \sin \frac{3n\pi}{4m}\\
&\ge \sqrt{2}\, \sin \frac{3\pi}{4}\\
& = 1.
\end{align}
*$\frac{m}{6} \le n \le \frac{5m}{6}$: Let $x = -\frac{\pi}{m}$.
Since $\frac{\pi}{6} \le \frac{n\pi}{m} \le \frac{5\pi}{6}$, we have
$$2\sin n x \cos m x = 2\sin \frac{n\pi}{m} \ge 2\sin \frac{\pi}{6} = 1.$$
*$n < \frac{m}{6}$: Let
$$y = \frac{\pi}{m}\Big\lfloor \frac{1}{2}\frac{m}{n} + \frac{1}{2}\Big\rfloor$$
where $\lfloor \cdot \rfloor$ is the floor function.
Clearly, $|\cos my| = 1$. By using $z-1 < \lfloor z \rfloor \le z$ and $\frac{n}{m} < \frac{1}{6}$, we have
$$ny \ge \frac{n\pi}{m}\Big(\frac{1}{2}\frac{m}{n} + \frac{1}{2} - 1\Big)
= \frac{\pi}{2} - \frac{n\pi}{2m} \ge \frac{5\pi}{12}$$
and
$$n y \le \frac{n\pi}{m}\Big(\frac{1}{2}\frac{m}{n} + \frac{1}{2}\Big)
= \frac{\pi}{2} + \frac{n\pi}{2m} < \frac{7\pi}{12}.$$
Thus, we have $\sin ny \ge \sin \frac{5\pi}{12}$ and thus
$$|2\sin ny \cos my| \ge 2 \sin \frac{5\pi}{12} > 1.$$
If $2\sin ny \cos my > 0$, let $x = y$ and we have $2\sin n x \cos mx > 1$.
If $2\sin ny \cos my < 0$, let $x = -y$ and we have $2\sin n x \cos mx > 1$.
We are done.
|
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|
Solving $2\sin(x)= \cot(5x)$ without finding the intersections of the graphs of the two sides I have solved the equation $$2\sin(x)= \cot(5x)$$ by finding the intersections of the graphs of the two sides.
Is there a method to solve this equation without using graphs?
|
We have that $\cos 5x-2\sin x\sin 5x=0,$ or linearising the second term, $$\cos 4x-\cos 5x-\cos 6x=0.$$ Now using the identity $$\cos z=\frac{e^{iz}+e^{-iz}}{2},$$ and with $y=e^{ix}$ we obtain $$y^4+1/y^4+y^5+1/y^5+y^6+1/y^6=0.$$ There are many ways to deal with this. You may explore the substitution $$u=y+1/y.$$ However, clear the fractions to obtain $$y^{12}+y^{11}+y^{10}+y^2+y+1=0,$$ and we can factor the polynomial to get $$y^{10}(y^2+y+1)+y^2+y+1=0,$$ or $$(y^{10}+1)(y^2+y+1)=0,$$ or $$((y^2)^5+1)(y^2+y+1)=0,$$ which becomes $$(y^2+1)((y^2)^4-(y^2)^3+(y^2)^2-y^2+1)(y^2+y+1)=0,$$ and since the quartic in $y^2=z$ gives $$z^4-z^3+z^2-z+1=0,$$ which may be dealt with by dividing by $z^2$ and using the substitution $w=z+1/z,$ the problem is essentially finished. The rest is only a matter of endurance.
|
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|
Finding the equation of a hyperbola, knowing the equations of its asymptotes and a point on it Find the equation of the hyperbola, given that transverse axis parallel to the $x$-axis, equations of asymptotes are $4x + y - 7 = 0$ and $3x - y - 5 = 0$ and the hyperbola passes through point $(4,4)$.
How could I solve this problem?
|
$(4x+y-7)(3x-y-5)=k,$ and it passes through $(4,4)$: $(4\cdot 4+4-7)(3\cdot 4-4-5)=k$ or $k=39,$ so your equation is $$(4x+y-7)(3x-y-5)=39.$$
Edit Expanding to the form $12x^2-xy-y^2-41x+2y-4=0,$ it is a hyperbola since $B^2-4AC=(-1)^2-4\cdot 12\cdot (-1)=37>0.$
To get the canonical form it is convenient to find the center. In maxima CAS: solve([diff(12*x^2-x*y-y^2-41*x+2*y-4,x),diff(12*x^2-x*y-y^2-41*x+2*y-4,y)],[x,y]);. It is $(\frac{12}{7},\frac17).$ (To check expand $12(x-12/7)^2-(x-12/7)(y-1/7)-(y-1/7)^2-39=0$).
Now rotate by $\theta=\frac{\arctan{\frac1{13}}}{2}$ since $\tan{2\theta}=\frac{B}{C-A}$ with the rotation \begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix} eliminates the $xy$ term.
The eigenvalues of the matrix $\begin{pmatrix}A&B/2\\B/2&C\end{pmatrix}=\begin{pmatrix}12&-1/2\\-1/2&-1\end{pmatrix}$ reveal the form $\frac{\sqrt{170}+11}{2} x'^2-\frac{\sqrt{170}-11}{2}y'^2=39.$ And the canonical form is $$(\frac{x'}{\frac{2\cdot 39}{\sqrt{170}+11}})^2-(\frac{y'}{\frac{2\cdot 39}{\sqrt{170}-11}})^2=1.$$
|
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|
Evaluate $\int_0^1 \ln{\left(\Gamma(x)\right)}\cos^2{(\pi x)} \; {\mathrm{d}x}$ I have stumbled across the following integral and have struck a dead end...
$$\int_0^1 \ln{\left(\Gamma(x)\right)}\cos^2{(\pi x)} \; {\mathrm{d}x}$$
Where $\Gamma(x)$ is the Gamma function.
I tried expressing $\Gamma(x)$ as $(x-1)!$ then using log properties to split the integral. Maybe there should be a summation in combination with the integral?? I believe this integral has a closed form but I would like help finding it.
|
The key to evaluating this integral is to utilize Euler's reflection formula, which the proof can be looked up elsewhere, by substituting $u=1-x$ so that the Gamma function "disappears":
$$I=\int_0^1 \ln{\left(\Gamma(1-u)\right)}\cos^2{(\pi u)} \; \mathrm{d}u$$
Now, add the original integral:
\begin{align*}
2I&=\int_0^1 \ln{\left(\Gamma(x)\Gamma(1-x)\right)}\cos^2{(\pi x)} \; \mathrm{d}x \\
I&=\frac{1}{2} \int_0^1 \ln{\left(\frac{\pi}{\sin{(\pi x)}}\right)}\cos^2{(\pi x)} \; \mathrm{d}x \\
I&\overset{\pi x \to x}=\frac{1}{2 \pi} \int_0^{\pi} \ln{\left(\frac{\pi}{\sin{(x)}}\right)}\cos^2{(x)} \; \mathrm{d}x \\
&=\frac{\ln{\pi}}{2 \pi} \int_0^{\pi} \cos^2{(x)} \; \mathrm{d}x-\frac{1}{2 \pi} \int_0^{\pi} \cos^2{(x)} \ln{\left({\sin{(x)}}\right)} \; \mathrm{d}x \\
&= \frac{\ln{\pi}}{4}- \frac{1}{4 \pi}\underbrace{ \int_0^{\pi} \ln{(\sin{x})} \; \mathrm{d}x}_{I_1} - \frac{1}{4 \pi}\underbrace{ \int_0^{\pi} \cos{(2x)} \ln{(\sin{x})} \; \mathrm{d}x}_{I_2}\\
\end{align*}
Now, to calculate $I_1$, use symmetry and let $u=\frac{\pi}{2}-x$, then add the two integrals:
\begin{align*}
I_1&=\int_0^{\frac{\pi}{2}} \ln{(\sin{u})} +\ln{(\cos{u})}\; \mathrm{d}u \\
I_1&=\int_0^{\frac{\pi}{2}} \ln{(\sin{(2u)})}-\ln{2} \; \mathrm{d}u\\
I_1&=\frac{I_1}{2}-\frac{\pi\ln{2}}{2}\\
I_1 &= -\pi\ln{2}\\
\end{align*}
Now, to calculate $I_2$
\begin{align*}
I_2&=2\int_0^{\frac{\pi}{2}} \cos{(2x)} \ln{(\sin{x})} \; \mathrm{d}x\\
&\overset{\sin{x} \to x}=2\int_0^1\frac{\left(1-2x^2\right)\ln{x}}{\sqrt{1-x^2}} \; \mathrm{d}x \\
&=2\int_0^1\frac{\ln{x}}{\sqrt{1-x^2}} \; \mathrm{d}x - 2\int_0^1 \frac{2x^2\ln{x}}{\sqrt{1-x^2}} \; \mathrm{d}x \\
&=-2\int_0^1 \frac{\arcsin{x}}{x} \mathrm{d}x+ 2\int_0^1 \frac{\arcsin{x}-2x\sqrt{1-x^2}}{x} \; \mathrm{d}x \\
&=-2\int_0^1 \frac{\arcsin{x}}{x} \; \mathrm{d}x+2\int_0^1\frac{\arcsin{x}}{x} \; \mathrm{d}x-2\int_0^1 \sqrt{1-x^2} \; \mathrm{d}x \\
&=-\frac{\pi}{2}\\
\end{align*}
Therefore,
\begin{align*}
\int_0^1 \ln{\left(\Gamma(x)\right)}\cos^2{(\pi x)} \; \mathrm{d}x&=\frac{\ln{\pi}}{4}-\frac{1}{4\pi} \left(-\pi \ln{2}-\frac{\pi}{2}\right) \\
&= \boxed{\frac{\ln{(2\pi)}}{4}+\frac{1}{8}}\\
\end{align*}
|
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|
Find the sum of the series with terms given by ${T}_{r}=\frac{r}{(r+1)(r+3)(r+4)}$ The given series has general term as $${T}_{r}=\frac{r}{(r+1)(r+3)(r+4)}$$
I have tried to approach this problem by making a telescopic series as follows, but I end up cancelling $r$ in the numerator, $$\frac{1}{(r+1)(r+3)}-\frac{1}{(r+3)(r+4)}=\frac{3}{(r+1)(r+3)(r+4)}$$ Please provide an alternate approach to form telescopic series.
|
The partial fraction expansion of your summand is
$$-\frac{1}{6}\frac{1}{r+1} + \frac{3}{2}\frac{1}{r+3} - \frac{4}{3}\frac{1}{r+4}.$$
Then notice that $3/2 = 4/3+1/6$ so you have
$$-\frac{1}{6}\frac{1}{r+1} + \frac{1}{6}\frac{1}{r+3}+\frac{4}{3}\frac{1}{r+3} - \frac{4}{3}\frac{1}{r+4}.$$
And now things telescope like crazy.
|
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|
Find integers $1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$ Root numbers Problem (Math Quiz Facebook):
Consider the following equation:
$$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$
Where $a,\,b,\,c,\,d$ are integers. Find $a+b+c+d$
I've tried it like this:
Let $w=\sqrt6,\, x=\sqrt3, \, y=\sqrt2, z=1$
$$\begin{align}
(y+z)^2 &= (y^2 + z^2) + 2yz\\
y+z &= \sqrt{(y^2 + z^2) + 2yz}\\
y+z &= \sqrt{3 + \sqrt{8}}
\end{align}$$
Let $y+z=f$
$$\begin{align}
(x+f)^2 &= (x^2 + f^2) + 2xf\\
x+f &= \sqrt{(x^2 + f^2) + 2xf}\\
x+f &= \sqrt{(9+\sqrt8) + 2\sqrt{9+3\sqrt8}}
\end{align}$$
And I don't think this going to work since there's still a root term on the bracket that is $9+\sqrt8$. I need another way to make it as an integer.
|
Expand out enough to get to
\begin{align*}
(a^2-24a+476-b)+\sqrt{2}(336-16a)+\sqrt{3}(272-12a)+\sqrt{6}(192-8a)&=\sqrt{c+\sqrt{d}}.
\end{align*}
This means, when we square the left side, we need to only have two terms with nonzero coefficient. Note that
$$(w+x\sqrt2+y\sqrt3+z\sqrt6)^2=(w^2+2x^2+3y^2+6z^2)+2\sqrt2(wx+3yz)+2\sqrt3(wy+2xz)+2\sqrt6(wz+xy),$$
so we need two of $\{wx+3yz,wy+2xz,wz+xy\}$ to be $0$. However, if the first two are $0$, then
$$wxy+3y^2z=wxy+2x^2z=0$$
implies that either $z=0$ or $x=y=0$; in the first case, $w=0$. We may get similar conclusions for each of the other selections to be $0$, so we must have that two of the parameters $\{w,x,y,z\}$ are $0$. In particular, since none of our polynomials in $a$ for $x,y,z$ have common roots, we must have that $w=0$. Then, $y\neq 0$ since $y$ has a noninteger root for $a$, so we have $a\in\{21,24\}$ and $a=21\implies b=413$, with $a=24\implies b=476$. If $a=24$, the left side is actually negative (it's $-48\sqrt2-16\sqrt3$), so it can't be the square root of anything. For $a=21$, $b=413$, we may find by direct calculation that
$$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{21+\sqrt{413+\sqrt{4656+\sqrt{16588800}}}}.$$
|
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|
When The Sum Of Squares Of Two Consecutive Integers Is Again A Perfect Square? Find all positive integers $n < 200$, such that $n^2 + (n + 1)^2$ is a perfect square.
Well setting this equal to $k^2$ is important. But before that, since all squares $\equiv 0$ or $1$ (mod $3$ $,4)$ so using this we get that one of the two numbers is divisible by $3$ and same for $4$ (one of the two is divisible by $4$). This gives us cases, like if $4,3 |n$ or $4|n$ and $3|n+1$ and so on. however this seems very tedious, and anyway, for example, the first case, I still don't get how I would solve $(12k)^2+(12k+1)^2=m^2$ for $k,m \in \mathbb N$. Help please; I'm stuck.
|
If we let $m=n+\sqrt{2n^2+(-1)^n}$, starting with $1$, we get a pair of pell numbers that feed directly into Euclid's formula for generating Pythagorean triples which are the ordered pairs (A,B,C) where $A^2+B^2=C^2$.
$$F(m,n):\qquad A=m^2-n^2\qquad B=2mn\qquad C=M^2+n^2$$
Examples:
$$n=1\implies m=1+\sqrt{2+(-1)^1}=1+1=2\quad F(2,1)=(3,4,5)$$
$$n=2\implies m=2+\sqrt{8+(-1)^2}=2+3=5\quad F(5,2)=(21,20,29)$$
$$n=5\implies m=5+\sqrt{50+(-1)^5}=5+7=12\quad F(12,5)=(119,120,169)$$
There an an infinite number of these but if you don't like the alternate $(A<B)$ vs $(A>B)$ you can use this formula, starting with a nonworking $seed$ of $(A_0,B_0,C_0)=(0,0,1)$
$$A_{n+1}=3A_n+2C_n+1\qquad B_{n+1}=3A_n+2C_n+2\qquad C_{n+1}=4A_n+3C_n+2\qquad$$
This produces the triples $\quad(3,4,5)\quad (20,21,29)\quad (119,120,169)\quad ...$
These three are the only triples where $(A,B,C)<200$
|
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|
Each positive integer is painted with some color. we know that $a+b$ and $ab$ have the same colour if $a>1, b>1$. Each positive integer is painted with some color. It is known that for all pairs $a, b$ of integers greater than $1$ the numbers $a+b$ and $ab$ have the same color. Prove that all numbers greater than $4$ are of the same color.
I've found that if $a+b=x$ $ab=k$
are the same colour
then $(a-1)(b+1)=q$ is the same colour too because $a-1+b+1=x$
And like that if some $a+b=x$ there are many $ab$ that have same colour
I think I've to write with rows and come back , go forward again and prove that all numbers are the same colour.
|
Are you familiar with the concept of induction?
Bear with me. Suppose $N=2K$ is a somewhat large even number. Then $N$ is the same color as $2+K$ and $2+K < N=2K$ if $K > 2$ or in other words if $N=2K> 4$
If $N=2K + 1$ is a somewhat large odd number then $N=(2K-2)+3$ so $N$ is the same color as $3(2K-2)=6(K-1)$. so $N$ is the same as $6+(K-1)=K+5$. And $K+5 < N=2K+1$ if $K>4$ or in other words if $N=2K + 1 > 9$.
So if we can show that the colors for $4,5,6,7,8,9$ are all the same color we are done as we can reduce from any larger number down to those.
$2+3=5;2*3=6$
$6=2+4=3+3; 2*4=8;3*3=9$
$12 = 2*6 = 3*4$ and $2+6=8$ and $3+4=7$.
so, yes, $4,5,6,7,8,9$ (and $12$) are the same color.
So that's it, we're done.
|
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|
Find the minimum of $x^3+\frac{1}{x^2}$ for $x>0$ Finding this minimum must be only done using ineaqualities.
$x^3+\frac{1}{x^2}=\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}$
Using inequalities of arithemtic and geometric means:
$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}+1}{6}\geqslant
\sqrt[6]{\frac{1}{2}x^3\frac{1}{2}x^3\frac{1}{3x^2}\frac{1}{3x^2}\frac{1}{3x^2}}=\sqrt[6]{\frac{1}{108}}\Rightarrow x^3+\frac{1}{x^2}\geqslant 6\sqrt[6]{\frac{1}{108}}-1 $
Sadly $\ 6\sqrt[6]{\frac{1}{108}}-1$ is not correct answer, it is not the minimum.
|
Very similar to what you have done:
$$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}}{5}\geq
\sqrt[5]{\frac{1}{2}x^3\frac{1}{2}x^3\frac{1}{3x^2}\frac{1}{3x^2}\frac{1}{3x^2}}=\sqrt[5]{\frac{1}{108}}$$
This gives us
$$x^3+\frac{1}{x^2}\geq \sqrt[5]{\frac{1}{108}}$$
Desmos screenshot:
P.S. Your method fails because equality holds only when
$$\frac{x^3}{2}=\frac{1}{3x^2}=1$$ which is impossible. The extra "one" you added in your AM-GM application screwed your attempt.
|
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|
Solving polynomials using De Moivre's theorem Given $\cos 4\theta=8\cos^4\theta−8\cos^2\theta+1$, solve $16x^4-16x^2+1=0$.
The textbook's answers are $x=\pm\cos\dfrac{\pi}{12},\pm\cos\dfrac{5\pi}{12}$. I managed to get two of the four answers and i can not figure out what i did wrong.
My Attempt
Let $x=\cos\theta$
$16\cos^4\theta-16\cos^2\theta+1=0$
$2(8\cos^4\theta-8\cos^2\theta+1)=1$
$2\cos4\theta=1$
$\cos4\theta=\dfrac{1}{2}$
$\cos4\theta=\cos\dfrac{\pi}{3}$
$\therefore4\theta=\dfrac{\pi}{3}+k\pi$, for any integer k
I chose $0\le k \le 3$ as the other solutions would repeat its values for x.
$\theta=\dfrac{\pi}{12},\dfrac{\pi}{3},\dfrac{7\pi}{12},\dfrac{5\pi}{6}$
$\therefore x=\cos\dfrac{\pi}{12},\cos\dfrac{\pi}{3},\cos\dfrac{7\pi}{12},\cos\dfrac{5\pi}{6}$
$x=\cos\dfrac{\pi}{12},\cos\dfrac{\pi}{3},-\cos\dfrac{5\pi}{12},\cos\dfrac{5\pi}{6}$
|
$$\cos4\theta=\cos\dfrac\pi3$$
$$\implies4\theta=2n\pi\pm\dfrac\pi3=\dfrac\pi3(6n\pm1)$$ where $n$ is any integer
$$\implies\theta=\dfrac\pi{12}(6n+1)$$ where $n=-2,-1,0,1$
$\cos\dfrac\pi{12}(6\cdot-2+1)=\cos\left(\pi-\dfrac\pi{12}\right)=-\cos\dfrac\pi{12}$ etc.
|
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|
For how many natural numbers(<=100) is $1111^n +2222^n+3333^n+4444^n$ divisible by 10? For how many natural numbers (0 not included) $n \leq 100$ is $1111^n +2222^n+3333^n+4444^n$ divisible by 10?
I factored out $1111^n$ and got $1111^n(1+2^n+3^n+4^n)$. So $1+2^n+3^n+4^n$ must be divisible by 10. I figured out that this is divisible by 10 for all odd n, but I don't know how to find the other solutions, if any.
|
$$1^n+2^n+3^n+4^n\equiv0\pmod2\text{ for }n\ge0$$
As $2^3\equiv3,2^2\equiv4\pmod5,$
$$1^n+2^n+3^n+4^n$$
$$\equiv1+2^n+(2^3)^n+(2^2)^n$$
$$\equiv\begin{cases}\dfrac{2^{4n}-1}{2^n-1}\equiv 0 &\mbox{if } 2^n-1\not\equiv0\pmod5\iff4\nmid n \\
4 & \mbox{if } 4\mid n \end{cases}\pmod5$$
$$\implies1^n+2^n+3^n+4^n\equiv0\pmod{[5,2]}\text{ if }4\nmid n$$
|
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|
Integrate $\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$ Evaluate $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$$
I tried $u=\sqrt{2x^4-2x^2+1}$, $u=\dfrac{1}{x}$ and $u=\sqrt{x}$ but none of these worked.
A friend gave this and said it's from IIT JEE and answer is $$\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$$
I would like a hint or suggestion.
|
Substitute $t=\frac1{x^2}$ to get
$$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} {dx}
=\frac12 \int \frac{(t-1)dt}{\sqrt{2-2t+t^2}}
=\frac12 \sqrt{2-2t+t^2}+C
$$
|
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|
$f(xy + x +y) = f(xy) + f(x) + f(y)$ and $f(x)(y - x) + f(y)(x - y) \geq 0$.
Let $f : \mathbb R \to \mathbb R$ that satisfies both 2 conditions ,
$f(xy + x + y) = f(xy) + f(x) + f(y)$ and $f(x)(y - x) + f(y)(x - y) \geq 0$ $\forall x,y \in \mathbb R$.
Determine all such $f$.
My solution.
Let $P(x,y) $ be $f(x)(y - x) + f(y)(x - y) \geq 0$.
Let $Q(x,y)$ be $f(xy + x + y) = f(xy) + f(x) + f(y)$.
First , I show that $f$ is decreasing $\forall x \in \mathbb R$
Proof : Let $x \gt y$ , from $P(x,y)$ we get that $f(x) \leq f(y)$. (Because $(x - y) \leq 0)$
So, $f$ in decreasing.
$Q(0,0)$ $\to$ $f(0) = 3f(0)$ $\to$ $f(0) = 0$.
$Q(x,-x)$ $\to$ $f(-x^2) = f(-x^2) + f(x) + f(-x)$. $\to$ $f(x) = -f(-x)$.
$P(x,2x)$ $\to$ $xf(x) \geq xf(2x)$. $\to$ $f(x) \geq f(2x)$.
Since $f(x) \geq f(2x)$ , take $x \lt 0$ , we get that $f$ is increasing $\forall x \lt 0$. This implies $f(x) = c$ and $f(x) = -c$ ; $c \in \mathbb R$.
$P(x,y)$ when $x,y \gt 0$ implies $-c = -3c$ , $c = 0$.
So, $f(x) = 0$ $\forall x \in \mathbb R$. Q.E.D.
Is my proof correct? (This is from 2016 Thailand POSN Camp 2)
|
You can show that solutions to the functional equation
$$ f ( x y + x + y ) = f ( x y ) + f ( x ) + f ( y ) \tag 0 \label 0 $$
are exactly the additive functions, i.e. those satisfying
$$ f ( x + y ) = f ( x ) + f ( y ) \text . \tag 1 \label 1 $$
It's easy to see that additive functions satisfy \eqref{0}. I try to show the converse, repeating some of your own arguments, for the sake of completeness. First, letting $ x = y = 0 $ in \eqref{0} we get $ f ( 0 ) = 0 $. Then, plugging $ y = - 1 $ in \eqref{0} we find out that
$$ f ( - x ) = - f ( x ) \text . \tag 2 \label 2 $$
Substituting $ - x $ for $ x $ in \eqref{0} and using \eqref{2} we get $ f ( y - x - x y ) = f ( y ) - f ( x ) - f ( x y ) $, which together with \eqref{0} gives
$$ f ( x y + x + y ) + f ( y - x - x y ) = 2 f ( y ) \text . \tag 3 \label 3 $$
For $ y \ne - 1 $, we can substitute $ \frac { x - y } { y + 1 } $ for $ x $ in \eqref{3} and see that $ f ( x ) + f ( 2 y - x ) = 2 f ( y ) $, hence $ f ( 2 y ) = 2 f ( y ) $ by letting $ x = 0 $, which transforms the previous equation to $ f ( x ) + f ( 2 y - x ) = f ( 2 y ) $. Substituting $ \frac { x + y } 2 $ for $ y $ in the last equation, we get \eqref{1} for $ x + y \ne - 2 $. Substituting $ - x $ for $ x $ and $ - y $ for $ y $ and using \eqref{2}, we'll have \eqref{1} for $ x + y \ne 2 $, which shows that \eqref{1} holds for every $ x $ and $ y $.
Finally, it is well-known that monotone additive functions are exactly those of the form $ f ( x ) = c x $, for some constant $ c $. The decreasing ones are exactly those with $ c \le 0 $, which form the class of all solutions of \eqref{0} satisfying
$$ f ( x ) ( y - x ) + f ( y ) ( x - y ) \ge 0 \text . $$
|
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|
Evaluate $\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$ Evaluate $$\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$$
I tried substitutions like $u=\frac{\pi}{4}-x$, and trig identities like $\cos^2{x}=1-\sin^2{x}$ after getting a common denominator.
$$4\int_0^{\frac{\pi}{4}} \frac{\sin^2{(5x)\cos^2{x}-\cos^2{(5x)}\sin^2{x}}}{\sin^2{(2x)}} \mathop{dx}$$
Where should I go from here? Any help is appreciated!
|
Making the problem more general for the antiderivative
$$I_{n,m}=\int\left( \frac{\sin^m{((2n+1)x)}}{\sin^m{(x)}} -\frac{\cos^m{((2+1)x)}}{\cos^m{(x)}} \right)\mathop{dx}$$ using Chebyshev formulae, just as @integrand answered,
$$\frac{\sin{((2n+1)x)}}{\sin{(x)}}$$ is a polynomial of degree $n$ in $\sin^2(x)$ that is to say a polynomial of degree $n$ in $\cos^2(x)$ and the same for
$$\frac{\cos{((2n+1)x)}}{\cos{(x)}}$$ So, the integrand is a polynomial of degree $mn$ in $\cos^2(x)$.
Reversing the problem, the integrand is then a linear combination of cosines of even multiple angles and then the integral is linear combination of sines of the same angles.
If we take the case where $m=2$ as in your case, the result of the integrand would be a linear combination of $\cos(2px)$ with $p=1,2,\cdots,2n-1$.
The table below reports the expression of the integrand
$$\left(
\begin{array}{cc}
n & \text{integrand} \\
1 & 8 \cos (2 x) \\
2 & 8 (2 \cos (2 x)+\cos (6 x)) \\
3 & 8 (3 \cos (2 x)+2 \cos (6 x)+\cos (10 x)) \\
4 & 8 (4 \cos (2 x)+3 \cos (6 x)+2 \cos (10 x)+\cos (14 x)) \\
5 & 8 (5 \cos (2 x)+4 \cos (6 x)+3 \cos (10 x)+2 \cos (14 x)+\cos (18 x)) \\
6 & 8 (6 \cos (2 x)+5 \cos (6 x)+4 \cos (10 x)+3 \cos (14 x)+2 \cos (18 x)+\cos
(22 x))
\end{array}
\right)$$ where you can notice interesting patterns in the coefficients. You could easily generate the general expression
Conderning the value of the integrals from $0$ to $\frac \pi 4$, they generate the sequence
$$\left\{4,\frac{20}{3},\frac{152}{15},\frac{456}{35},\frac{5156}{315},\frac{67028
}{3465},\frac{67952}{3003},\frac{1155184}{45045},\frac{22128676}{765765},\frac
{22128676}{692835}\right\}$$
For sure, we could do similar things for other integer values of $m$.
|
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|
The remainder when $1^1+3^3+5^5+\dots + 1023^{1023} \pmod{1024}$ is $\dots$ The remainder when $1^1+3^3+5^5+\dots + 1023^{1023} \pmod{1024}$ is $\dots$
I've tried for the summation like this.
$\sum_{i=1}^{n} (2i-1)^{2i-1} \equiv 0 \pmod{n}$ when $n$ is even. But, I didn't get the answer when $n$ is odd and didn't yet can proved it.
Any hints or idea? Thanks in advanced.
|
We will prove that for $n\geq 2$ we have
$$
S_n:=\sum_{k=1}^{2^{n-1}}(2k-1)^{2k-1}\equiv 0\pmod{2^n}.
$$
We will induct on $n$ (the case $n=2$ is clear). Consider $S_{n+1}$:
$$
S_{n+1}=\sum_{k=1}^{2^{n}}(2k-1)^{2k-1}=\sum_{k=1}^{2^{n-1}}\left((2k-1)^{2k-1}+(2k-1+2^n)^{2k-1+2^n}\right)\equiv 0\pmod{2^{n+1}}.
$$
Due to Euler's theorem and the fact that $\varphi(2^{n+1})=2^n$ for odd $a$ we have $a^{2^n}\equiv 1\pmod {2^{n+1}}$ (actually $\pmod {2^{n+2}}$ but it's not important now), so
$$
S_{n+1}\equiv\sum_{k=1}^{2^{n-1}}\left((2k-1)^{2k-1}+(2k-1+2^n)^{2k-1}\right)\equiv 0\pmod{2^{n+1}}.
$$
Now, note that because of the Binomial theorem we have ($2^{ns}\equiv 0\pmod {2^{n+1}}$ for $s>1$)
$$
(2k-1+2^n)^{2k-1}\equiv(2k-1)^{2k-1}+\binom{2k-1}{1}(2k-1)^{2k-2}\cdot 2^n\pmod {2^{n+1}},
$$
or
$$
(2k-1+2^n)^{2k-1}\equiv(2^n+1)(2k-1)^{2k-1}\pmod {2^{n+1}}.
$$
Therefore,
$$
S_{n+1}\equiv\sum_{k=1}^{2^{n-1}}\left((2k-1)^{2k-1}+(2^n+1)(2k-1)^{2k-1}\right)=(2^n+2)S_n\equiv 0\pmod{2^{n+1}},
$$
because $2^n+2$ is even and $2^n\mid S_n$ by the inductive assumption.
Thus, $2^n\mid S_n$ for all $n\geq 2$, as desired.
Note. One can extend this argument to prove that map $x\mapsto x^x$ is a bijection on $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$ (just add this to the inductive assumption and prove that $(2k-1)^{2k-1}\not\equiv (2k-1+2^n)^{2k-1+2^n}\pmod{2^{n+1}}$).
|
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|
Is there a quick (hopefully elementary) way to prove that $6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0$ has only one solution? I have the Diophantine equation $$6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0.$$ Numerical calculations suggest this has only one positive integer solution, namely $(b,c)=(2,3)$. Is there a quick way to prove or disprove that?
|
Hint
$$6b^2c^2+3c^2-36bc-4b^4-4b^2+53=(6b^2c^2-36bc+54)-(4b^4+4b^2+1)+3c^2=6(bc-3)^2-(2b^2+1)^2+3c^2=0\\\implies 6(bc-3)^2+3c^2=(2b^2+1)^2$$ Then $3\mid 2b^2+1$. Now proceed like this by checking modulo primes like $2,3$ and so on.
|
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|
$\int \frac{x^3+3x+2}{(x^2+1)^2 (x+1)} \ dx$ without using partial fraction decomposition One way to evaluate the integral $$\int \frac{x^3+3x+2}{(x^2+1)^2(x+1)} \ dx $$ is to rewrite it as $$ \int \frac{x^3+x+2x+2}{(x^2+1)^2(x+1)} dx \\=\int\frac{x(x^2+1) +2(x+1)}{(x^2+1)^2(x+1)}dx\\=\int\frac{x}{(x^2+1)(x+1)}dx+\int\frac{2}{(x^2+1)^2}dx$$ and then proceed by using partial fraction decomposition on the first integral. The second integral could be dealt with by substituting $x=\tan \theta$.
Is there a way to evaluate this integral without using partial fractions, and preferably without splitting it into two integrals as I did here?
|
Let
$$\frac{x^2+1}{(x+1)^2}=t\implies x=\frac{\sqrt{2 t-1}-t}{t-1}\implies dx=\frac{dt}{1-t \left(\sqrt{2 t-1}+2\right)}$$ to make
$$I=\int \frac{x^3+3x+2}{(x^2+1)^2(x+1)} \, dx=\int\left(-\frac{1}{2 t^2}+\frac{1}{4 t}-\frac{3}{4 t \sqrt{2 t-1}}\right)\,dt$$ which does not seem too bad.
$$I=\frac{1}{2 t}+\frac 14 \log(t)-\frac{3}{2} \tan ^{-1}\left(\sqrt{2 t-1}\right)+C$$
|
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|
Simplify $\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$
Simplify:
$$\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$$
After the substitution as $\cos(x)=a$ and $\sin(x)=b$, $(a^2+b^2=1)$, the expression becomes
$$\frac{4(a^2-b^2)^2-4a^2+3b^2}{4b^2-4a^2b^2}=\frac{4a^4-8a^2b^2+4b^4-4a^2+3b^2}{4b^4}=\bigg(\frac{a^2}{b^2}-1\bigg)^2-\frac{4a^2-3b^2}{4b^4}$$But I don't think I got anything useful... Any help is appreciated.
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\begin{align}\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}&=\frac{4(1-2\sin^2x)^2-4(1-\sin^2x)+3\sin^2x}{4\sin^2x-(2\sin x\cos x)^2}\\&=\frac{4-16\sin^2x+16\sin^4x-4+4\sin^2x+3\sin^2x}{4\sin^4x}\\&=\frac{16\sin^4x-9\sin^2x}{4\sin^4x}\\&=4-\frac94\csc^2x\end{align}
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|
prove $\sum\cos^3{A}+64\prod\cos^3{A}\ge\frac{1}{2}$ In every acute-angled triangle $ABC$,show that
$$(\cos{A})^3+(\cos{B})^3+(\cos{C})^3+64(\cos{A})^3(\cos{B})^3(\cos{C})^3\ge\dfrac{1}{2}$$
I want use Schur inequality
$$x^3+y^3+z^3+3xyz\ge xy(y+z)+yz(y+z)+zx(z+x)$$
then we have
$$x^3+y^3+z^3+6xyz\ge (x+y+z)(xy+yz+zx)$$
But I can't use this to prove my question
and I use this post methods links also can't solve my problem,use $AM-GM $ inequality$$\cos^3{A}+\dfrac{\cos{A}}{4}\ge\cos^2{A}$$
so
$$LHS\ge \sum_{cyc}\cos^2{A}-\dfrac{1}{4}\sum_{cyc}\cos{A}+64\prod_{cyc}\cos^3{A}$$use
$$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
it must to prove
$$\frac{1}{2}+64\cos^3{A}\cos^3{B}\cos^3{C}\ge 2\cos{A}\cos{B}\cos{C}+\dfrac{1}{4}(\cos{A}+\cos{B}+\cos{C})$$
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Let $\cos\alpha=\frac{x}{2},$ $\cos\beta=\frac{y}{2}$ and $\cos\gamma=\frac{z}{2}.$
Thus, $x$, $y$ and $z$ are positives, $x^2+y^2+z^2+xyz=4$ and we need to prove that:
$$x^3+y^3+z^3+x^3y^3z^3\geq4.$$
Indeed, let $$f(x,y,z,\lambda)=x^3+y^3+z^3+x^3y^3z^3-4+\lambda(x^2+y^2+z^2+xyz-4).$$
Thus, in the minimum point we need $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=\frac{\partial f}{\partial\lambda}=0,$$
which gives $$3x^2+3x^2y^3z^3+\lambda(2x+yz)=3y^2+3y^2x^3z^3+\lambda(2y+xz)=3z^2+3z^2x^3y^3+\lambda(2z+xy)=0,$$ which gives
$$\frac{x^2+x^2y^3z^3}{2x+yz}=\frac{y^2+y^2x^3z^3}{2y+xz}=\frac{z^2+z^2x^3y^3}{2z+xy}$$ or
$$(x-y)(2x^2y^2z^3(x+y)-z(x^2+xy+y^2)-2xy)=0,$$
$$(x-z)(2x^2z^2y^3(x+z)-y(x^2+xz+z^2)-2xz)=0$$ and
$$(y-z)(2y^2z^2x^3(y+z)-x(y^2+yz+z^2)-2yz)=0.$$
Now, let $(x-y)(x-z)(y-z)\neq0.$
Thus, $$\frac{z(x^2+xy+y^2)+2xy}{z(x+y)}=\frac{y(x^2+xz+z^2)+2xz}{y(x+z)}=\frac{x(y^2+yz+z^2)+2yz}{x(y+z)},$$ which gives $$(x-y)(2z+xy)(xy+xz+yz)=0,$$
$$(x-z)(2y+xz)(xy+xz+yz)=0$$ and $$(y-z)(2x+yz)(xy+xz+yz)=0,$$ which is impossible.
Id est, $(x-y)(x-z)(y-z)=0$.
Let $y=x$.
Thus, the condition gives $$2x^2+z^2+x^2z=4$$ or
$$x^2(2+z)=4-z^2$$ or
$$z=2-x^2,$$ where $0<x<\sqrt2$ and we need to prove that
$$2x^3+(2-x^2)^3+x^6(2-x^2)^3\geq4$$ or
$$(x-1)^2(1+x-x^2)(x^8+3x^7+x^6-4x^5-4x^4-2x^3+4x+4)\geq0,$$ which is true because $$x^8+3x^7+x^6-4x^5-4x^4-2x^3+4x+4>0$$ even for any real $x$.
Also, we need to check, what happens for $xyz\rightarrow0^+$.
Let $z\rightarrow0^+$.
Thus, $x^2+y^2=4$ and
$$x^3+y^3\geq2\left(\sqrt{\frac{x^2+y^2}{2}}\right)^3=4\sqrt2>4$$ and we are done.
|
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|
Point is chosen inside the square, find angle. Inside the square $MNPK$ point $O$ in such a way that $MO:ON:OP =1:2:3$. Find $\angle MON$.
I tried to solve this problem with vectors, but I couldn't find coordinates of point $O$. Can someone explain this to me?
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Apply the Pythagorean theorem to establish the system of three equations
\begin{align}
x^2+y^2=4,\>\>\>\>\> x^2+(a-y)^2=1,\>\>\>\>\> (a-x)^2+y^2=9
\end{align}
The 2nd and 3rd equations lead to $y=\frac{a^2+3}{2a}$ and $x=\frac{a^2-5}{2a}$, respectively. Plug them into the 1st equation to get $a^4 - 10a^2 +17=0$, which yields $a^2 =5+2\sqrt2$. Then, use the cosine rule for the triangle OMN to obtain
$$\cos\angle MON = \frac{OM^2+ON^2-a^2}{2|OM||ON|} = \frac{1+4-(5+2\sqrt2)}{2\cdot 1\cdot2}=-\frac{\sqrt2}2$$
Thus, $\angle MON =135^\circ$.
|
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|
Computation of Definite Integral of Rational Function I am dealing with a definite integral of a rational function that seems quite hard to get a nice closed form/explicit expression for. Let $ -1 < z < 1 $, then my aim is to determine an expression for the integral $ I $ in terms of $ z $:
$$ I(z) = \int_{0}^{\infty} \frac{(1+z)t^4 + (1-z)}{(1+z)^2 t^6 + 3(1+z)(5+z)t^4 + 3(1-z)(5-z)t^2 + (1-z)^2} \ dt $$
Any help would be appreciated.
Current Attempts
Thank you @Claude Leibovici for your answer. Following the process of the comment, one can arrive at the following.
If we let $ (1+z)^2Q(t) := (1+z)^2 t^6 + 3(1+z)(5+z)t^4 + 3(1-z)(5-z)t^2 + (1-z)^2 $ for $ t \in (0,\infty) $, then $ Q $ has 6 roots $ \pm \omega_i \in \mathbb{C} $ for $ i = 1,2,3 $, all of which are dependent on $ z \in (-1,1) $. In particular, it can be shown that
$$ \omega_k(z)^2 = \frac{4\sqrt{4z + 5}\cos\left(\frac{1}{3}\left(\arccos\left(-\frac{2z^2 + 14z + 11}{(4z + 5)^{3/2}}\right) - 2\pi(k-1)\right)\right) - (5+z)}{(1+z)} < 0, $$
for $ z \in (-1,1) $.
Then using a partial fractions approach yields
$$ I(z) = \int_{0}^{\infty} \frac{(1+z)t^4 + (1-z)}{(1+z)^2Q(t)} \ dt \\ = \frac{i\pi}{2(1+z)^2}\frac{\omega_1\omega_2\omega_3(\omega_1\omega_2 + \omega_1\omega_3 + \omega_2\omega_3)(1 + z) + (\omega_1 + \omega_2 + \omega_3)(1 - z)}{\omega_1\omega_2\omega_3(\omega_1 + \omega_2)(\omega_1+\omega_3)(\omega_2 + \omega_3)}, $$
where it is known, by Vieta's formula that
$$ \omega_1^2 + \omega_2^2 + \omega_3^2 = -\frac{3(5+z)}{(1+z)} $$
$$ \omega_1^2\omega_2^2 + \omega_1^2\omega_3^2 + \omega_2^2\omega_3^2 = \frac{3(1-z)(5-z)}{(1+z)^2} $$
$$ \omega_1^2\omega_2^2\omega_3^2 = -\frac{(1-z)^2}{(1+z)^2} $$
A clearer question would be: is there a way to simplify the above evaluation of $ I(z) $ into a nicer expression?
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Consider the integrand $$\frac{(1+z)t^4 + (1-z)}{(1+z)^2 t^6 + 3(1+z)(5+z)t^4 + 3(1-z)(5-z)t^2 + (1-z)^2} $$ and rewrite it as
$$\frac 1 {z+1} \frac{t^4+a}{(t^2-b)(t^2-c)(t^2-d)}$$ where $a=\frac {1-z}{1+z}$ and $(b,c,d)$ are the roots of the cubic equation in $t^2$. Now, using partial fraction decomposition
$$\frac{t^4+a}{(t^2-b)(t^2-c)(t^2-d)}=$$
$$-\frac{a+b^2}{(b-c) (b-d) \left(t^2-b\right)}+\frac{a+c^2}{(b-c) (c-d) \left(t^2-c\right)}-\frac{a+d^2}{(b-d)
(c-d) \left(t^2-d\right)}$$ and the antiderivative will not make any problem.
A numerical analysis for $-1 < z < 1$ shows that $(b,c,d)$ are all negative. So, we face three integrals looking like
$$I_k=\int \frac {dt}{t^2+k}=\frac{\tan ^{-1}\left(\frac{t}{\sqrt{k}}\right)}{\sqrt{k}}\implies J_k=\int_0^\infty \frac {dt}{t^2+k}=\frac{\pi }{2 \sqrt{k}}$$
What is left is just the computation of $(b,c,d)$. Since they are all real, I suggest you use the trigonometric method for cubic equations.
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|
Is there a way to find value for this integral? I am trying to solve this integral by hand and here are my steps for doing that.
The integral is:
$$A=\int_0^{2 \pi} \frac{\cos(\theta)}{\sqrt{1 - a \cos(\theta)}} d\theta$$
Using the trigonometry trick: $\cos(\theta) = 1- 2 \sin\left(\frac{\theta}{2}\right)^{2}$ and the fact that when I plot the graph, the area from $0$ to $\pi$ is the same as the area from $\pi$ to $2\pi$.
Therefore, $$A=2\int_0^{ \pi} \frac{\cos(\theta)}{\sqrt{1 - a \cos(\theta)}} d\theta$$
$$= 4 \int_0^{\frac{\pi}{2}} \frac{\cos(2t)}{\sqrt{1 - a + 2a \sin(t)^{2}}} dt$$
$$= \frac{4}{\sqrt{1-a}} \int_0^{\frac{\pi}{2}} \frac{1 -2 \sin(t)^{2}}{\sqrt{1 + \frac{2a}{1-a} \sin(t)^{2}}} dt$$
Set $x = \sqrt{\frac{2a}{1-a}} \sin(t)$
$\to dx = \sqrt{\frac{2a}{1-a}} \cos(t) dt$
$\leftrightarrow dt = \frac{1}{\sqrt{\frac{2a}{1-a}} \cos(t)} dx$
$\leftrightarrow dt = \frac{1}{\sqrt{\frac{2a}{1-a}} \sqrt{1- \sin(t)^{2}}} dx$
$\leftrightarrow dt = \frac{1}{\sqrt{1-x^2}} dx$
As $x = \sqrt{\frac{2a}{1-a}} \sin(t)$:
$\sin(t) = \frac{x}{ \sqrt{\frac{2a}{1-a}}}$
$$\to A = \frac{4}{\sqrt{1-a}} \int_0^{\sqrt{\frac{2a}{1-a}}} \frac{1 - \frac{1-a}{a} x^2}{\sqrt{1+x^2} \sqrt{1-x^2}} dx$$
$$ \leftrightarrow A = \frac{4}{\sqrt{1-a}} \int_0^{\sqrt{\frac{2a}{1-a}}} \frac{1 - \frac{1-a}{a} x^2}{\sqrt{1-x^4}} dx$$
Set $x = u \sqrt{\frac{2a}{1-a}}$
$\to u = x \sqrt{\frac{1-a}{2a}}$
$\leftrightarrow du = \sqrt{\frac{1-a}{2a}} dx$
$$\to A = \frac{4}{\sqrt{1-a}} \sqrt{\frac{2a}{1-a}}
\int_0^{1} \frac{1 - 2 u^2}{\sqrt{1- \left(\frac{2a}{1-a}\right)^2 u^4}} du$$
I don't really know the way to go from here. Can you help me to get the value for this integral? Thank you so much!
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Let us assume $|a|< 1$ and exploit
$$ \frac{1}{\sqrt{1-z}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}z^n \qquad \text{for }|z|<1. $$
This gives
$$ I(a)=\int_{0}^{2\pi}\frac{\cos\theta}{\sqrt{1-a\cos\theta}}\,d\theta=\sum_{n\geq 0}\frac{a^n}{4^n}\binom{2n}{n}\int_{0}^{2\pi}\left(\cos\theta\right)^{n+1}\,d\theta $$
where the even values of $n$ do not really contribute to the series. We are left with
$$ I(a)=\sum_{m\geq 0}\frac{a^{2m+1}}{4^{2m+1}}\binom{4m+2}{2m+1}\int_{0}^{\pi}(\cos\theta)^{2m+2}\,d\theta=2\pi\sum_{m\geq 0}\frac{a^{2m+1}}{4^{3m+2}}\binom{4m+2}{2m+1}\binom{2m+2}{m+1} $$
where the RHS is a hypergeometric function related to the complete elliptic integrals of the first and second kind:
$$ I(a) = \frac{\pi a}{2}\cdot\phantom{}_2 F_1\left(\frac{3}{4},\frac{5}{4};2;a^2\right)=\frac{4}{a\sqrt{1-a}}\left(K\left(\frac{2a}{a-1}\right)-(1-a)E\left(\frac{2a}{a-1}\right)\right).$$
Here $E$ and $K$ are denoted according to Mathematica's notation, so their argument is the elliptic modulus.
For $a=\pm\frac{1}{3}$ the integral can be expressed in terms of $\pi$ and $\Gamma\left(\frac{1}{4}\right)$ (see Chapter $12$ of my notes):
$$ I\left(\pm\tfrac{1}{3}\right) = \pm\sqrt{\frac{3}{4\pi}}\left(\Gamma\left(\tfrac{1}{4}\right)^2-\frac{(4\pi)^2}{\Gamma\left(\frac{1}{4}\right)^2}\right).$$
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|
Closed form sought for $a_1 = a_2 = 1, a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i $ where $n>2$ I've been working through a problem that I've got as far as getting a recursive answer to. I was hoping to turn this into more of a "closed form" answer, but haven't really gotten anywhere. I'm hoping that someone can help with this, though anything would be greatly appreciated.
The recursive answer I have is a sequence of real numbers given by $$\begin{gather}
a_1 = a_2 = 1 \\
a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i \qquad (n > 2)
\end{gather}$$
The first few non-trivial members of this sequence are
*
*$a_3 = \frac{5}{3}$
*$a_4 = 2$
*$a_5 = \frac{37}{15}$
*$a_6 = \frac{26}{9}$
*$a_7 = \frac{349}{105}$
I've tried to express these in terms of $a_1$ and $a_2$ and and constants and arrived at
*
*$a_3 = 1 + \frac{2}{3} a_1$
*$a_4 = 1 + \frac{2}{4} a_1 + \frac{2}{4} a_2$
*$a_5 = (1 + \frac{2}{5}) + (\frac{2}{5} + \frac{2^2}{3\cdot5} ) a_1 + \frac{2}{5} a_2$
*$a_6 = (1 + \frac{2}{6} + \frac{2}{6}) + (\frac{2}{6} + \frac{2^2}{3 \cdot 6} + \frac{2^2}{4 \cdot 6}) a_1 + (\frac{2}{6} + \frac{2^2}{4 \cdot 6}) a_2$
*$a_7 = (1 + \frac{2}{7} + \frac{2}{7} + \frac{2}{7} + \frac{2^2}{5 \cdot 7}) + (\frac{2}{7} + \frac{2^2}{3\cdot7} + \frac{2^2}{4 \cdot 7} + \frac{2^2}{5 \cdot 7} + \frac{2^3}{3 \cdot 5 \cdot 7}) a_1 + (\frac{2}{7} + \frac{2^2}{4 \cdot 7} + \frac{2^2}{5 \cdot 7}) a_2$
I am not seeing a pattern developing here.
I also rearranged the above noting that $a_1 = a_2 = 1$ and got
*
*$a_3 = 1 + \frac{2}{3} $
*$a_4 = 1 + 2 (\frac{2}{4})$
*$a_5 = 1 + 3 (\frac{2}{5}) + \frac{2^2}{3\cdot5}$
*$a_6 = 1 + 4 (\frac{2}{6}) + \frac{2^2}{3 \cdot 6} + 2 (\frac{2^2}{4 \cdot 6})$
*$a_7 = 1 + 5 (\frac{2}{7}) + \frac{2^2}{3 \cdot 7} + 2 (\frac{2^2}{4 \cdot 7}) + 3 (\frac{2^2}{5 \cdot 7}) + \frac{2^3}{3 \cdot 5 \cdot 7}$
Here I do notice a couple of things
*
*The expression for $a_n$ begins with "$1 + (n-2) \frac{2}{n}$".
*The remaining terms of the expression look like "$k \dfrac{2^{i+1}}{b_1 \cdots b_{i} \cdot n}$" where each $b_j$ is between $3$ and $n-2$ and consecutive numbers cannot appear among them. The $k$ seems to be determined by the smallest number among the $b_j$, but this is more of a guess than anything right now.
These observations are not really helping me much at all.
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After @Gary's answer, using the generating function
$$G(x) = \frac{{1 - e^{ - 2x} }}{{2(x-1)^2}}= \sum_{n = 1}^\infty {a_n x^n }$$ define
$b_n=n! \,a_n$ which gives the sequence
$$\{1,2,10,48,296,2080,16752,151424,1519744,16766208\}$$ which is $A037256$ in $OEIS$ (have a look here).
It does not seem to present any particularity. In year 2002, Vaclav Kotesovec proposed tha approximation
$$b_n \sim \frac{1}{2} \left(1-\frac{1}{e^2}\right) n\, n!$$ which would make
$$a_n\sim \frac{1}{2} \left(1-\frac{1}{e^2}\right) n$$ which shows a relative error smaller than $1$% for $n> 68$.
Edit
As @Gary commented, I missed the constant term, which according to his comment should be
$$a_n\sim \frac{1}{2} \left(1-\frac{1}{e^2}\right) (n+1)$$
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Find a general method to find particular solutions where the sum of the squares of two consecutive integers is equal to the square of another integer Question: If the sum of the squares of two consecutive integers is equal to the square of another integer, then find a general method to find particular solutions. E.g., $27304196^2+27304197^2=38613965^2$.
I tried to figure out other solutions such as $3^2+4^2=5^2$ and $20^2+21^2=29^2$, but I don't know how to find a method for general solutions.
Can anybody help me
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For $A^2+B^2=C^2:\space B=A\pm1\quad$ the $C$-values are the odd Pell numbers $(5,29,169,985\cdots)$ as shown in A000129
and the triples $T_n$ that contain them can be generate sequentially by the formula below with a seed of $\quad T_0=(0,0,1)$
$$A_{n+1}=3A_n+2C_n+1\qquad B_{n+1}=3A_n+2C_n+2\qquad C_{n+1}=4A_n+3C_n+2$$
For example, it generates the following
$$T_1=(3,4,5)\quad T_2=(20,21,29)\quad T_3=(119,120,169)\quad T_4=(697,696,985)$$
They may also be generated directly using Euclid's formula
$$\text{where} \qquad A=m^2-k^2 \qquad B=2mk \qquad C=m^2+k^2\quad \text{and}$$
\begin{equation}
m_n= \frac{(1 + \sqrt{2})^{n+1} - (1 - \sqrt{2})^{n+1}}{2\sqrt{2}}\qquad \qquad\qquad
k_n= \frac{(1 + \sqrt{2})^n - (1 - \sqrt{2})^n}{2\sqrt{2}}
\end{equation}
For example
\begin{align*}
\frac{(1 + \sqrt{2})^{2} - (1 - \sqrt{2})^{2}}{2\sqrt{2}}=2 \qquad &
\frac{(1 + \sqrt{2})^1 - (1 - \sqrt{2})^1}{2\sqrt{2}}=1 \\
&\qquad\qquad F(2,1)=(3,4,5)\\
\frac{(1 + \sqrt{2})^{3} - (1 - \sqrt{2})^{3}}{2\sqrt{2}}=5 \qquad &
\frac{(1 + \sqrt{2})^2 - (1 - \sqrt{2})^2}{2\sqrt{2}}=2 \\
& \qquad\qquad F(5,2)=(21,20,29)\\
\frac{(1 + \sqrt{2})^{4} - (1 - \sqrt{2})^{4}}{2\sqrt{2}}=12 \qquad &
\frac{(1 + \sqrt{2})^3 - (1 - \sqrt{2})^3}{2\sqrt{2}}=5 \\
&\qquad\qquad F(12,5)=(119,120,169)\\
\frac{(1 + \sqrt{2})^{5} - (1 - \sqrt{2})^{5}}{2\sqrt{2}}=29 \qquad &
\frac{(1 + \sqrt{2})^4 - (1 - \sqrt{2})^4}{2\sqrt{2}}=12 \\
&\qquad\qquad F(29,12)=(697,696,985)
\end{align*}
|
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|
Find function $f(x)$ whose expansion is $\sum_{k=0}^{+\infty}k^2x^k$. I know the expansion of $\frac{1}{1-x}$ is $$1+x+x^2+\cdots+x^k+\cdots$$
So taking the derivative of $$\frac{\partial}{\partial{x}} \frac{1}{1-x}=\frac{1}{(1-x)^2}$$
And subsequently the expansion is (after taking derivative): $$1+2x+3x^2+4x^3+...+ kx^{(k-1)}$$
And multiplying by $x$: $$x+2x^2+3x^3+\cdots+kx^k+\cdots$$
I can't seem to figure out how to work with knowing the above expansion and knowing I need to arrive at: $$x+2^2 x^2+3^2 x^3 +\cdots+k^2 x^k +\cdots$$
I thought it would be a matter of $f(x)=\frac{1}{(1-x^2)^2}$ or some form of squaring or adding a constant in there but I'm stumped.
|
Simplify $\displaystyle\sum_{k=0}^{+\infty}k^2x^{k}$.
Differentiate and multiply the below expression by $x$ twice
\begin{align*}
\sum_{k=0}^{+\infty}x^k&=\frac{1}{1-x}\\
\sum_{k=0}^{+\infty}kx^{k-1}&=\frac{1}{(1-x)^2}\\
\sum_{k=0}^{+\infty}kx^{k}&=\frac{x}{(1-x)^2}=-\frac{1}{(1-x)}+\frac{1}{(1-x)^2}\\
\sum_{k=0}^{+\infty}k^2x^{k-1}&=-\frac{1}{(1-x)^2}+\frac{2}{(1-x)^3}=\frac{x+1}{(1-x)^3}\\
\sum_{k=0}^{+\infty}k^2x^{k}&=\frac{x(x+1)}{(1-x)^3}\\
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate: $\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx.$
Prove that:$$\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx=\frac{2\ln c}{\sqrt{4c^2-b^2}}\cot^{-1}\left(\frac{b}{\sqrt{4c^2-b^2}}\right),$$ where $4c^2-b^2>0, c>0.$
We have:
\begin{align}
4(x^2+bx+c^2)&=(2x+b)^2-\left(\underbrace{\sqrt{4c^2-b^2}}_{=k\text{ (say)}}\right)^2\\\\
&=(2x+b+k)(2x+b-k).
\end{align}
Thus,
\begin{align}
I&=4\int_0^{\infty}\frac{\ln x}{(2x+b+k)(2x+b-k)}\,dx\\\\
&=\frac4{2k}\int_0^{\infty}\left(\frac1{2x+b-k}-\frac1{2x+b+k}\right)\ln x\, dx\\\\
&=\frac2k\left[\underbrace{\int_0^{\infty}\frac{\ln x}{2x+b-k}\,dx}_{=I_1}-\underbrace{\int_0^{\infty}\frac{\ln x}{2x+b+k}\,dx}_{=I_2}\right]
\end{align}
For $I_1:$ Letting $2x+b-k=t$ yields
\begin{align}
I_1&=\int_{b-k}^{\infty} \frac{\ln(t-b+k)-\ln 2}t\,dt\\\\
&=\int_{b-k}^{\infty}\frac{\ln(t-b+k)}t\, dt-\ln 2\int_{b-k}^{\infty}\frac{dt}t
\end{align}
Now $\ln(t-b+k)=\ln(k-b)+\ln\left(\frac{t}{k-b}+1\right).$ So,$$I_1=\ln(k-b)\int_{b-k}^{\infty}\frac{dt}t+\int_{b-k}^{\infty} \frac{\ln\left(\frac t{k-b}+1\right)}t\,dt-\ln 2\int_{b-k}^{\infty}\frac{dt}t.$$
If we let: $\frac t{k-b}=-u.$ Then $$I_1=\ln\left(\frac{k-b}2\right)\int_{b-k}^{\infty} \frac{dt}t+\int_1^{\infty}\frac{\ln(1-u)}u\,du.$$
It's getting messier and messier. Also it seems that $I_1$ and hence $I$ diverges. Please tell me, am I heading towards something useful? It has become difficult for me to handle this integral. Please show me a proper way.
|
Hint:
$$ x^2 + bx+ c^2 = (x+b/2)^2 + (c^2 - b^2/4) $$
with $c^2 - b^2/4 > 0$.
This invites change of variables using $\arctan$ as
$$d\arctan x = \frac{1}{x^2+1}dx $$
|
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|
Does $k=1$ follow from $I(5^k)+I(m^2) \leq \frac{43}{15}$, if $p^k m^2$ is an odd perfect number with special prime $p=5$? The topic of odd perfect numbers likely needs no introduction.
Denote the sum of divisors of the positive integer $x$ by $\sigma(x)$, and denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Euler proved that an odd perfect number $n$, if one exists, must have the form
$$n = p^k m^2$$
where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.
Here is my initial question:
Does $k=1$ follow from
$$I(5^k)+I(m^2) \leq \frac{43}{15},$$
if $I(x)$ is the abundancy index of $x$ and $p^k m^2$ is an odd perfect number with special prime $p=5$?
MY ATTEMPT
Since $n = p^k m^2$ is perfect and $I$ is multiplicative, then we have
$$I(m^2) = \frac{2}{I(p^k)}.$$
But $p \mid p^k$. In particular, $I(p) \leq I(p^k)$. This implies that we have the upper bound
$$I(m^2) = \frac{2}{I(p^k)} \leq \frac{2}{I(p)} = \frac{2p}{p+1}.$$
In particular, if $p=5$, then we obtain
$$I(m^2) \leq \frac{5}{3}.$$
Hereinafter, we shall assume that $p=5$.
Now, consider the product
$$\bigg(I(p^k) - \frac{5}{3}\bigg)\bigg(I(m^2) - \frac{5}{3}\bigg).$$
It can be proven that $I(p^k) < I(m^2)$. (For a proof, see [Dris (2012)].) Thus, this product is nonnegative (since $I(m^2) \leq \frac{5}{3}$), whereupon we get
$$\bigg(I(p^k) - \frac{5}{3}\bigg)\bigg(I(m^2) - \frac{5}{3}\bigg) \geq 0 \implies I(p^k)I(m^2) + \bigg(\frac{5}{3}\bigg)^2 \geq \frac{5}{3}\cdot\bigg(I(p^k) + I(m^2)\bigg)$$
$$\implies 2 + \bigg(\frac{5}{3}\bigg)^2 \geq \frac{5}{3}\cdot\bigg(I(p^k) + I(m^2)\bigg) \implies I(p^k) + I(m^2) \leq \frac{6}{5} + \frac{5}{3} = \frac{43}{15}.$$
Now, we compute an exact expression for
$$I(p^k) + I(m^2) = I(p^k) + \frac{2}{I(p^k)}$$
when $p=5$. We obtain
$$I(5^k) + \frac{2}{I(5^k)} = \frac{(5^{k+1} - 1)^2 + {32}\cdot{5^{2k}}}{5^k \cdot {4} (5^{k+1} - 1)} = \frac{57 \cdot {5^{2k}} - 2 \cdot {5^{k+1}} + 1}{4 \cdot {5^{2k+1}} - 4 \cdot {5^k}}.$$
From the same paper cited above, we have the lower bound
$$\frac{57}{20} < I(p^k) + I(m^2)$$
so that collectively we have
$$\frac{57}{20} < I(5^k) + \frac{2}{I(5^k)} = \frac{57 \cdot {5^{2k}} - 2 \cdot {5^{k+1}} + 1}{4 \cdot {5^{2k+1}} - 4 \cdot {5^k}} \leq \frac{43}{15},$$
from which we obtain the (trivial) lower bound
$$k \geq 1,$$
per this WolframAlpha computation.
Here is my follow-up question:
Why does the condition
$$I(5^k) + \frac{2}{I(5^k)} \in \bigg(\frac{57}{20},\frac{43}{15}\bigg]$$
not result in a nontrivial bound for $k$?
|
On OP's request, I am converting my comment into an answer.
I'm not sure if I understand your follow-up question well, but we have
$$\begin{align}\frac{\mathrm d}{\mathrm dk}\bigg(I(5^k)+\frac{2}{I(5^k)}\bigg)&=-\frac{ 7\cdot 5^{2 k} + 2\cdot 5^{k + 1} - 1}{4\cdot 5^k (5^{k + 1} - 1)^2}\log 5\lt 0
\\\\\lim_{k\to\infty}\bigg(I(5^k)+\frac{2}{I(5^k)}\bigg)&=\lim_{k\to\infty} \frac{57 -\frac{ 2}{5^{k-1}} + \frac{1}{5^{2k}}}{20 - \frac{4}{5^k}}=\frac{57}{20}
\\\\I(5)+\frac{2}{I(5)}&=\frac{57 \cdot {5^{2}} - 2 \cdot {5^{2}} + 1}{4 \cdot {5^{3}} - 4 \cdot {5}}=\frac{43}{15}\end{align}$$
|
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|
Use residues to find $\int_0^\pi \frac{d\theta}{5+3\cos\theta}$ I am trying to use residues to find $\int_0^\pi \frac{d\theta}{5+3\cos\theta}$.
My thoughts:
Letting $z=e^{i\theta}$ we get $dz=ie^{i\theta}$. Then, $\int_0^\pi \frac{d\theta}{5+3\cos\theta}=\frac{1}{5}\int_{|z|=1}\frac{dz}{iz(1+\frac{3}{5}(\frac{z+z^{-1}}{2}))}=-2i\int_{|z|=1}\frac{dz}{3z^2+10z+3}$.
Now, using the quadratic formula, we get that the integral becomes $-2i\int_{|z|=1}\frac{dz}{(z+3)(3z+1)}$. So, now we will compute the residue at $z=-\frac{1}{3}$ only since $z=3$ is outside of our circle. The residue is equal to $\frac{3}{8}$, and so the integral is equal to $(2\pi i)(-2i)(\frac{3}{8})=\frac{3\pi}{2}$. But, this integral, I believe should actually be equal to $\frac{\pi}{4}$ based on Wolfram.
I am wondering if I did something wrong, or (hopefully not) is it just some silly algebra mistake somewhere. Any thoughts, suggestions, etc. are always appreciated! Thank you.
|
$$I = \int_0^\pi \frac{d\theta}{5+3\cos\theta}$$
I want the range to be $0$ to $2\pi$ so I will apply the substitution $\tau = \theta / 2$.
$$I = \int_0^{2 \pi} \frac{d\tau}{10+6\cos\tau}$$
Now I can apply my Residue Theorem lemma (from Freitag):
In our case $$f(z) = \frac{1}{i z}\frac{1}{10 + \tfrac{1}{2}6(z + \frac{1}{z})} = \frac{1}{i}\frac{1}{3z^2 + 10z + 3} = \frac{-i}{(z + 3)(3z + 1)} = \frac{-i}{3 (z + 3)(z + \tfrac{1}{3})}$$
There are two simple poles at $z_1=-3$ and $z_2=-\tfrac{1}{3}$ let's calculate the residues:
*
*$\operatorname{Res}(f;z_1) = \frac{-i}{3 \cdot (z_1 + \tfrac{1}{3})} = i/8$
*$\operatorname{Res}(f;z_2) = \frac{-i}{3 \cdot (z_2 + 3)} = -i/8$
we will only use the $z_2$ residue as $z_1$ lies outside $\mathbb E$.
So $$I = 2 \pi i \cdot - i / 8 = \pi / 4$$
|
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"url": "https://math.stackexchange.com/questions/3754861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Estimating $\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$. I'm trying to solve this:
Which of the following is the closest to the value of this integral?
$$\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$$
(A) 1
(B) 1.2
(C) 1.6
(D) 2
(E) The integral doesn't converge.
I've found a lower bound by manually calculating $\int_{0}^{1} \sqrt{1+\frac{1}{3}} \ dx \approx 1.1547$. This eliminates option (A). I also see no reason why the integral shouldn't converge. However, to pick an option out of (B), (C) and (D) I need to find an upper bound too. Ideas? Please note that I'm not supposed to use a calculator to solve this.
From GRE problem sets by UChicago
|
Let's try to use integration by parts to $I = \int\limits_0^1 \sqrt{1 + \frac{1}{3x}}dx$. First, transform integral into $\frac{2}{\sqrt3}\int\limits_0^1\frac{\sqrt{1 + 3x}}{2\sqrt{x}}$. Now $u = \sqrt{3x+1}$ and $dv = \frac{dx}{2\sqrt{x}}$ and what we get after IBP is $$\frac{2}{\sqrt3}\sqrt{x(3x+1)}|_0^1 - \sqrt{3}\int\limits_0^1 \sqrt{\frac{x}{3x+1}}dx = \frac{4}{\sqrt3} - \sqrt{3}\int\limits_0^1 \sqrt{\frac{x}{3x+1}}dx$$. We have $$\frac{5}{2\sqrt3} = \frac{4}{\sqrt3} - \sqrt{3}\int\limits_0^1 \sqrt{\frac{x}{3x + x}}dx < I <\frac{4}{\sqrt3} - \sqrt3\int\limits_0^1 \sqrt{\frac{x}{3 + 1}}dx = \frac{4}{\sqrt3} - \frac{\sqrt3}{2} \frac{2}{3}x\sqrt{x}|_0^1 = \sqrt3$$
$\frac{5}{2\sqrt3} \approx 1.44$ and $\sqrt3 \approx 1.73$, so the answer is (C).
If one doesn't know the value of $\sqrt3$, we can check that $1.7^2 < 3 < 1.8^2$ and then $3 < 1.75^2$. Therefore, $\sqrt3 < 1.75$. From it we have $\frac{5}{2\sqrt3} > \frac{5}{2\cdot1.75} > 1.42$ and, for integral, $1.42 < I < 1.75$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What loops are possible when doing this function to the rationals? What loops are possible when doing this function to the rationals?
Let's define this function on a simplified fraction $\frac{a}{b}$.
$$f\left(\frac{a}{b}\right)=\frac{a+b}{b+1}$$
I started this with $f(\frac{2}{3})=\frac{5}{4}$ then i did the function again and got this sequence of numbers $\frac{2}{3},\frac{5}{4},\frac{9}{5},\frac{7}{3},\frac{5}{2},\frac{7}{3},\dots$ I saw that is starts to loop with $\frac{7}{3},\frac{5}{2}$
Another loop is $\frac{1}{1}$, a one cycle.
Another loop I found was $\frac{2}{1},\frac{3}{2},\frac{5}{3}$.
My first question is: from starting from any rational number does it all ways end in a loop or does it ever go to infinity? And my second question is: what sizes of loops are possible?
If the three loops I stated are the only loops prove it
Dark made a post related What are the possible loops when doing this a type of function to the rationals?
|
Here is a modification of @Steven Stadnicki's proof. The novel contribution of this answer is justifying the reduction step in Steven's solution through the use of a suitable partial order on the set of lattice points.
Step 1. Settings and Useful Observations
Let $\mathbb{N}_1 = \{1, 2, 3, \dots\}$ denote the set of positive integers and define $\mathsf{Red} : \mathbb{N}_1^2 \to \mathbb{N}_1^2$ by
$$ \mathsf{Red}(a, b) = \frac{(a,b)}{\gcd(a,b)}. $$
Also, we equip $\mathbb{N}_1^2$ with the partial order $\leq$ such that1)
$$ (a, b) \leq (c, d) \quad \Leftrightarrow \quad [b < d]\text{ or }[b = d \text{ and } a \leq c]. $$
The following observations are easy to prove but will be useful throughout.
*
*$\text{(P1)} \ $ $a \leq c$ and $b \leq d$ implies $(a, b) \leq (c, d)$.
*$\text{(P2)} \ $ $\mathsf{Red}(\mathrm{p}) \leq \mathrm{p}$ for any $\mathrm{p} \in \mathbb{N}_1^2$.
Step 2. Key Observation
We will identify each pair $(a,b) \in \mathbb{N}_1^2$ satisfying $\gcd(a, b) = 1$ with the simplified fraction $a/b$. Under this identification, we have
$$f(a/b) = \mathsf{Red}(a+b,b+1). $$
Now we will investigate the effect of a suitable number of iterations of $f$. By noting that either $a$ or $b$ must be odd, the following three cases exhaust all the possibilities:
*
*Case 1. Suppose that both $a$ and $b$ are odd. Then both $a+b$ and $b+1$ are even, and so,
\begin{align*}
f(a,b)
= \mathsf{Red}(a+b, b+1)
= \mathsf{Red}(\tfrac{a+b}{2}, \tfrac{b+1}{2})
\stackrel{\text{(P2)}}\leq (\tfrac{a+b}{2}, \tfrac{b+1}{2}). \tag{1}
\end{align*}
*Case 2. Suppose that $a$ is odd and $b$ is even. Then by writing $d=\gcd(a+b,b+1)$,
\begin{align*}
f^{\circ 2}(a,b)
= f(\tfrac{a+b}{d},\tfrac{b+1}{d})
= \mathsf{Red}(\tfrac{a+2b+1}{d},\tfrac{b+d+1}{d}).
\end{align*}
Since $d$ is odd, both $a+2b+1$ and $b+d+1$ are even. This means that both are divisible by $2d$, and so,
\begin{align*}
f^{\circ 2}(a,b)
= \mathsf{Red}(\tfrac{a+2b+1}{2d},\tfrac{b+d+1}{2d})
\stackrel{\text{(P2)}}\leq (\tfrac{a+2b+1}{2d},\tfrac{b+d+1}{2d})
\stackrel{\text{(P1)}}\leq (\tfrac{a+2b+1}{2},\tfrac{b+2}{2}). \tag{2}
\end{align*}
Here, the last inequality follows from the general fact that $\frac{A+Bd}{d}\leq A+B$ for all $A, B \geq 0$ and $d \geq 1$.
*Case 3. Suppose that $a$ is even and $b$ is odd. Since $d = \gcd(a+b, b+1)$ is odd, we find that $\frac{a+b}{d}$ is odd and $\frac{b+1}{d}$ is even. So by applying $\text{(2)}$ and using the inequality in the previous step,
\begin{align*}
f^{\circ 3}(a,b)
= f^{\circ 2}(\tfrac{a+b}{d},\tfrac{b+1}{d})
\stackrel{\text{(2)}}\leq (\tfrac{a+3b+d+2}{2d},\tfrac{b+2d+1}{2d})
\stackrel{\text{(P1)}}\leq (\tfrac{a+3b+3}{2},\tfrac{b+3}{2}). \tag{3}
\end{align*}
Step 3. Proof
Let $(a, b) \in \mathbb{N}_1$ satisfy $\gcd(a, b) = 1$. Then by $\text{(1)}$–$\text{(3)}$, we observe the following:
*
*If $b > 3$, then $\frac{b+3}{2} < b$, and so, a suitable number of iterations by $f$ reduces the second coordinate. This can be repeated finitely many times until the second coordinate becomes $\leq 3$.
*If $b \leq 3$ and $a > 12$, then $\frac{a+3b+3}{2} < a$, and so, a suitable number of iterations by $f$ reduces the first coordinate. Similarly as before, this can be repeated finitely many times until the first coordinate becomes $\leq 12$.
*If $a \leq 12$ and $b \leq 3$, then a suitable number of iterations by $f$ will map $(a, b)$ into another point $(a', b')$ with $a' \leq 12$ and $b' \leq 3$. So by the pigeonhole principle, iteration by $f$ will eventually fall into a cycle.
*By checking all the possible $12+6+8=26$ cases manually, we find that there are only three types of cycles:
$$ (1, 1) \qquad (5, 2), (7, 4) \qquad (2, 1), (3, 2), (5, 3) $$
This completes the proof.
1) Note that this is exactly the colexicographical order induced by the usual order on $\mathbb{N}_1$.
|
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|
How does $e^x\cdot e^X$ equal $e^{x+X}$? I know that they equal each other, but when I'm trying to prove it, something doesn't match. Please mind the difference between the two equations, one is a lowercase $x$ and the other is an uppercase $x.$ I know that the formula to get $e^x$ is $\frac{x^n}{n!}$. So I apply on $e^x$and it becomes $1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots$ and the same for $e^X$. This image shows what I'm thinking, and what happens when I multiply them on each other. But I want to go more deeper and show the full equation of the numbers shown, for example, showing $\frac{1}{6}x^3$ and others as well. If I did, there would be a total of 16 numbers shown that are arranged in order of degrees. So I try to prove it like this picture. But the problem shows when I try to prove it more. I tried proving it more, but $\frac{1}{6}(x+X)^3$ won't work. Than what should I do to make it work? Or what formula should I use?
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Lets look at a couple functions. $Y=e^X$ and $y=e^x$.
Then $\ln Y=X$ and $\ln y = x$. $$X+x=\ln (Y)+\ln (y) = \ln (Yy) \Longrightarrow e^{X+x}=Yy=e^Xe^x$$
Ok, so there is a little informal proof using some properties of the natural log.
$$e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$$ (Note: I am not using z as a complex variable, only a stand in for x because it is used so much in the question.)
So, $$e^x=1+x+\frac{x^2}{2} + \frac{x^3}{6}+\frac{x^4}{24}+...+\frac{x^n}{n!}+...$$
$$e^X=1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...+\frac{X^n}{n!}+...$$
$$\begin{align*} e^xe^X &= \bigg(1+x+\frac{x^2}{2} + \frac{x^3}{6}+\frac{x^4}{24}+...+\frac{x^n}{n!}+... \bigg)\bigg(1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...+\frac{X^n}{n!}...\bigg) \\&= 1\bigg(1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...\bigg) + x\bigg(1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...\bigg)+ \frac{x^2}{2}\bigg(1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...+\frac{X^n}{n!}...\bigg)...\\&= \bigg(1+X+\frac{X^2}{2} + \frac{X^3}{6}...\bigg) + \bigg(x+xX+x\frac{X^2}{2}+x\frac{X^3}{6}...\bigg)+\bigg(\frac{x^2}{2} + \frac{x^2}{2}X + \frac{x^2}{2}\frac{X^2}{2}+\frac{x^2}{2}\frac{X^3}{6}...\bigg)+...\\&= 1+(x+X)+\bigg(\frac{x^2}{2}+xX+\frac{X^2}{2}\bigg)+\bigg(\frac{x^3}{6} + \frac{x^2X}{2}+\frac{X^2x}{2}+\frac{X^3}{6}\bigg)+...\\ &= 1+(x+X)^1 +\frac{(x+X)^2}{2} + \frac{(x+X)^3}{6} + ...\\&= e^{x+X}\end{align*}$$
Another way: $$\begin{align*} e^{x+X}=\sum_{j=0}^{\infty} \frac{1}{j!}(x+X)^j&= \sum_{j=0, 0\le k\le j}^{\infty} \frac{1}{j!} {j\choose k}x^{j-k}X^k \\ &= \frac{1}{0!}\bigg[{0\choose 0}(x^{0-0}X^0)\bigg] + \frac{1}{1!}\bigg[{1\choose 0}(x^{1-0}X^0)+{1\choose 1}(x^{1-1}X^1)\bigg] +\frac{1}{2!}\bigg[{2\choose 0}(x^{2-0}X^0)+{2\choose 1}(x^{2-1}X^1) + {2\choose 2}(x^{2-2}X^2)\bigg] +...\\&= 1+x+X + \frac{x^2}{2} +xX + \frac{X^2}{2} +\frac{x^3}{6} + \frac{x^2X}{2}+\frac{X^2x}{2}+\frac{X^3}{6}+...\\&= \bigg(1+X + \frac{X^2}{2} + \frac{X^3}{6} + ...\bigg)+\bigg(x +xX + x\frac{X^2}{2}+...\bigg) + \bigg(\frac{x^2}{2}+\frac{x^2 X}{2} +\frac{x^2X^2}{2\cdot 2}+...\bigg)+...\\&= \bigg(1+X + \frac{X^2}{2} + \frac{X^3}{6} + ...\bigg)+x\bigg(1+X+\frac{X^2}{2}+...\bigg)+\frac{x^2}{2}\bigg(1+X+\frac{X^2}{2}+...\bigg)+...\end{align*}$$
You can finish it from there.
The exact reason this works is hidden in the way the Taylor series defines a function. I will direct you to this answer, which in my opinion, is a really nice and different way of understanding the intuition behind the Taylor series. The top answer on that page directs to another link of the proof of the Taylor series as well.
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|
What is the residue of $z^2 \cos(\frac{1}{z})$ Can I find it using Laurent series expansion? I tried using the Taylor series expansion of $\cos(1/z)$ but could not reach any conclusion.
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The residue is the coefficient of the term $\frac{1}{z}$ in the Laurent Series.
The Maclaurin Series of $\cos z$ is
$$\cos z = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \cdots $$
It follows that
$$\cos\left(\frac{1}{z}\right) = 1 - \frac{1}{2!}\frac{1}{z^2} + \frac{1}{4!}\frac{1}{z^4} - \frac{1}{6!}\frac{1}{z^6}+\cdots $$
Finally, multiplying both sides by $z^2$, we get
$$z^2\cos\left(\frac{1}{z}\right) = z^2 - \frac{1}{2!} + \frac{1}{4!}\frac{1}{z^2}- \frac{1}{6!}\frac{1}{z^4}+\cdots $$
The coefficient of $\frac{1}{z}$ is $0$, so the residue of $z^2\cos\left(\frac{1}{z}\right)$ is $0$.
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"timestamp": "2023-03-29T00:00:00",
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|
Other way to evaluate $\int \frac{1}{\cos 2x+3}\ dx$? I am evaluating
$$\int \frac{1}{\cos 2x+3} dx \quad (1)$$
Using Weierstrass substitution:
$$ (1)=\int \frac{1}{\frac{1-v^2}{1+v^2}+3}\cdot \frac{2}{1+v^2}dv =\int \frac{1}{v^2+2}dv \quad (2) $$
And then $\:v=\sqrt{2}w$
$$ (2) = \int \frac{1}{\left(\sqrt{2}w\right)^2+2}\sqrt{2} dw$$$$= \frac{1}{2} \int \frac{1}{\sqrt{2}\left(w^2+1\right)}dw$$$$ = \frac{1}{2\sqrt{2}}\arctan \left(w\right) + C$$$$= \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C$$
Therefore,
$$\int \frac{1}{\cos 2x+3} dx = \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C $$
That's a decent solution but I am wondering if there are any other simpler ways to solve this (besides Weierstass). Can you come up with one?
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int{\dd x \over \cos\pars{2x} + 3} & =
\int{\dd x \over \bracks{2\cos^{2}\pars{x} - 1} + 3} =
{1 \over 2}\int{\sec^{2}\pars{x}\,\dd x \over
1 + \sec^{2}\pars{x}}
\\[5mm] & =
{1 \over 2}\int{\sec^{2}\pars{x}\,\dd x \over
\tan^{2}\pars{x} + 2} =
{1 \over 2}\,{1 \over 2}\,\root{2}\int{\bracks{\sec^{2}\pars{x}/\root{2}}
\,\dd x \over
\bracks{\tan\pars{x}/\root{2}}^{2} + 1}
\\[5mm] & =
\bbx{{\root{2} \over 4}\arctan\pars{{\root{2} \over 2}\,\tan\pars{x}} +
\mbox{a constant}}
\end{align}
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|
Is it possible to show that the fifth roots of 1 add up to 0 simply by using trigonometric identities? You can't use geometric sums, minimal polynomials, pentagon, and exact values with radicals.
All the five, fifth-roots of unity are :$1,\left(\cos \left(\frac{2 \pi}{5}\right)+i \sin \left( \frac{2\pi}{5}\right)\right),\left(\cos \left(\frac{4 \pi}{5}\right)+i \sin \left(\frac{4 \pi}{5}\right)\right),\left(\cos \left(\frac{6 \pi}{5}\right)+i \sin \left(\frac{6 \pi}{5}\right)\right), \left(\cos \left(\frac{8 \pi}{3}\right)+i \sin \left(\frac{8 \pi}{5}\right)\right)$
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If you want a simple trigonometric proof, recall
$$\sin a-\sin b=2\sin\frac{a-b}2\cos\frac{a+b}2.$$
Therefore
\begin{align}
0&=\sin(a+2\pi)-\sin a\\
&=(\sin(a+2\pi)-\sin(a+8/\pi/5))+(\sin(a+8/\pi/5)-\sin(a+6/\pi/5))\\
&+(\sin(a+6/\pi/5)-\sin(a+4/\pi/5))+(\sin(a+4/\pi/5)-\sin(a+2/\pi/5))\\
&+(\sin(a+2/\pi/5)-\sin a)\\
&=2(\sin\pi/5)\left[\cos(a+9\pi/5)+\cos(a+7\pi/5)+\cos(a+\pi)+\cos(a+3\pi/5)+\cos(a+\pi/5)\right]
\end{align}
and
$$\cos(a+9\pi/5)+\cos(a+7\pi/5)+\cos(a+\pi)+\cos(a+3\pi/5)+\cos(a+\pi/5)=0.$$
Taking $a=-\pi/5$ gives
$$\cos(8\pi/5)+\cos(6\pi/5)+\cos(4\pi/5)+\cos(2\pi/5)+1=0.$$
and taking $a=-7\pi/10$ gives
$$\sin(8\pi/5)+\sin(6\pi/5)+\sin(4\pi/5)+\sin(2\pi/5)+0=0.$$
These are the real and imaginary parts of the sum of the fifth roots
of zero.
Of course, this method works for other values of $5$.
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Convergence of $\sum \limits_{n=1}^{\infty}\sqrt{n^3+1}-\sqrt{n^3-1}$ Hello I am a high school student from germany and I am starting to study math this october. I am trying to prepare myself for the analysis class which I will attend so I got some analysis problems from my older cousin who also studied maths. But I am stuck on this problem.
Check the following series for convergence/divergence $$\sum \limits_{n=1}^{\infty}\sqrt{n^3+1}-\sqrt{n^3-1}$$ I tried to prove the convergence by comparison test $$\sqrt{n^3+1}-\sqrt{n^3-1}= \frac{2}{\sqrt{n^3+1}+\sqrt{n^3-1}}=\frac{1}{n^2} \cdot \frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}}$$ and then compare it with $$\sum \limits_{n=1}^{\infty}\frac{1}{n^2}$$ But in order to do that, I need to prove that $$\frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} \leq 1$$ But I am having problems to prove that. Does anyone have tip how to solve this problem?
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Note that$$\frac{1}{n^2} \cdot \frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}}=\frac{1}{n^{3/2}} \cdot\underbrace{ \frac{2}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}}}_{<\frac{2}{1+0}=2}<2n^{-3/2}.$$
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Use linearisation of a certain function to approximate $\sqrt[3]{30}$ Background
Find the linearisation of the function
$$f(x)=\sqrt[3]{{{x^2}}}$$
at
$$a = 27.$$
Then, use the linearisation to find
$$\sqrt[3]{30}$$
My work so far
Applying the formula
$${f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) + f^\prime\left( a \right)\left( {x – a} \right),}$$
where
$${f\left( a \right) = f\left( {27} \right) }={ \sqrt[3]{{{{27}^2}}} }={ 9.}$$
Then, the derivative using the power rule:
$${f^\prime\left( x \right) = \left( {\sqrt[3]{{{x^2}}}} \right)^\prime }={ \left( {{x^{\frac{2}{3}}}} \right)^\prime }={ \frac{2}{3}{x^{\frac{2}{3} – 1}} }={ \frac{2}{3}{x^{ – \frac{1}{3}}} }={ \frac{2}{{3\sqrt[3]{x}}}.}$$
then
$${f^\prime\left( a \right) = f^\prime\left( {27} \right) }={ \frac{2}{{3\sqrt[3]{{27}}}} }={ \frac{2}{9}.}$$
Substitute this in the equation for $L(x)$:
$${L\left( x \right) = 9 + \frac{2}{9}\left( {x – 27} \right) }={ 9 + \frac{2}{9}x – 6 }={ \frac{2}{9}x + 3.}$$
Then, to use this linearisation to find
$$\sqrt[3]{30}$$
I perform the following $\Delta x = x – a = 30 – 27 = 3$ as the condition is $x =30$ and the staring point is $a=27$
As the the derivative of this particular function is given by $f\left( x \right) = \sqrt[\large 3\normalsize]{x}$
$${f’\left( x \right) = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^\prime } } = {{\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^\prime } } = {\frac{1}{3}{x^{ – \large\frac{2}{3}\normalsize}} } = {\frac{1}{{3\sqrt[\large 3\normalsize]{{{x^2}}}}},}$$
and its value at point $a$ is equal to
$${f’\left( {a} \right) = \frac{1}{{3\sqrt[\large 3\normalsize]{{{{27}^2}}}}} } = {\frac{1}{{3 \cdot {3^2}}} = \frac{1}{{27}}.}$$
Thus, getting the solution
$${f\left( x \right) \approx f\left( {a} \right) + f’\left( {a} \right)\Delta x,\;\;}\Rightarrow {\sqrt[\large 3\normalsize]{{30}} \approx \sqrt[\large 3\normalsize]{{27}} + \frac{1}{{27}} \cdot 3 } = {3 + \frac{1}{9} } = {\frac{{28}}{9} \approx 3,111.}$$
Is my process correct so far? Or, did I go wrong in the second part? Also, as $a=27$ is from the original linearisation, this would be brought into the linearisation approximation for $\sqrt[3]{30}$?
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The second part of your analysis is correct. The function that you want to approximate is $f(x)=\sqrt[3]{{x}}$ at $a=27$, not $g(x)=\sqrt[3]{{x^2}}$. So
$${f'(x) = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^\prime } } = {{\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^\prime } } = {\frac{1}{3}{x^{ – \large\frac{2}{3}\normalsize}} } = {\frac{1}{{3\sqrt[\large 3\normalsize]{{{x^2}}}}},}$$
and its value at point $a$ equals
$${f'(a) = \frac{1}{{3\sqrt[\large 3\normalsize]{{{{27}^2}}}}} } = {\frac{1}{{3 \cdot {3^2}}} = \frac{1}{{27}}.}$$
Therefore since $f(a)=\sqrt[3]{27}=3$,
$$f(x)\approx L\left( x \right) ={ f\left( a \right) + f^\prime\left( a \right)\left( {x – a} \right)}=3+\frac{1}{27}\left(x-27\right).$$
Hence, the estimate for $\sqrt[3]{30}$ is
$$f(30)\approx3+\frac{1}{27}\left(30-27\right)=3+\frac{1}{9}=3.\overline{1}.$$
The actual value of $\sqrt[3]{30}$ (up to the fifth decimal place) is $3.10723$, so the linear approximation at $a=27$ does quite well.
|
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|
Find the value of $k$ in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$ Find the value of 'k' in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$
My approach is as follow
$\sin {1^\circ} = \sin {179^\circ}$
$T = {\sin ^2}{1^\circ}{\sin ^2}{3^\circ}..{\sin ^2}{89^\circ}$
$\left( {\frac{{1 - \cos {2^\circ}}}{2}} \right) = {\sin ^2}{1^\circ}$
$T = \left( {\frac{{1 - \cos {2^\circ}}}{2}} \right)\left( {\frac{{1 - \cos {6^\circ}}}{2}} \right)\left( {\frac{{1 - \cos {{10}^\circ}}}{2}} \right)....\left( {\frac{{1 - \cos {{178^\circ}}}}{2}} \right)$
$2 + \left( {n - 1} \right)4 = 178 \Rightarrow n = 45$
$T = \frac{1}{{{2^{45}}}}\left( {1 - \cos {2^\circ}} \right)\left( {1 - \cos {6^\circ}} \right)\left( {1 - \cos {{10}^\circ}} \right)....\left( {1 - \cos {{178^\circ}}} \right)$
$\left( {1 - \cos {{178}^\circ}} \right) = \left( {1 + \cos {2^\circ}} \right)$
$T = \frac{1}{{{2^{45}}}}\left( {1 - {{\cos }^2}{2^\circ}} \right)\left( {1 - {{\cos }^2}{6^\circ}} \right)\left( {1 - {{\cos }^2}{10^\circ}} \right)....\left( {1 - {{\cos }^2}{{86}^\circ}} \right)\left( {1 - \cos {{90}^\circ}} \right)$
$T = \frac{{{{\sin }^2}{2^\circ}.{{\sin }^2}{6^\circ}.{{\sin }^2}{{10}^\circ}......{{\sin }^2}{{86}^\circ}}}{{{2^{45}}}}$
Not able to proceed from here
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As $\cos(90^\circ-A)=\sin A,$
$$\prod_{k=0}^{44}\sin(2k+1)^\circ=\prod_{k=0}^{44}\cos(2k+1)^\circ$$
As $\cos180(2k+1)^\circ=-1$
as $\cos180x=2^{180-1}c^{180}x+\cdots+(-1)^{90}$
and for $\cos180x=\cos180^\circ,180x=360^\circ n\pm180^\circ$
$x=(2 n+1)^\circ$ where $0\le n\le179$
If $\cos180x=-1,$ the roots of $$2^{180-1}c^{180}x+\cdots+(-1)^{90}=-1$$ are
$\cos(2 n+1)^\circ$ where $0\le n\le179$
$$\implies\prod_{n=0}^{179}\cos(2 n+1)^\circ=\dfrac2{2^{179}}$$
Now as $\cos(360^\circ-y)=+\cos y$
$$\implies\prod_{n=0}^{179}\cos(2 n+1)^\circ=\prod_{n=0}^{89}\cos^2(2 n+1)^\circ$$
Now $90<2n+1<180\iff 45\le n\le89\implies$ there are $45$(odd) number of multiplicands $<0$
$$\prod_{n=0}^{89}\cos(2 n+1)^\circ<0$$
$$\implies\prod_{n=0}^{89}\cos(2 n+1)^\circ=-\sqrt{\dfrac2{2^{179}}}$$
Finally as $\cos(180^\circ-z)=-\cos z,$
$$\implies\prod_{n=0}^{89}\cos(2 n+1)^\circ=-\prod_{n=0}^{44}\cos^2(2 n+1)^\circ$$
Finally use $$\prod_{n=0}^{44}\cos(2 n+1)^\circ>0$$
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How to evaluate $ \:\sum _{n=3}^{\infty \:}\frac{4n^2-1}{n!}\:\: $? I am trying to evaluate:
$$ \:\sum _{n=3}^{\infty \:}\:\:\frac{4n^2-1}{n!}\:\: \quad (1)$$
My attempt:
$$ \:\sum _{n=3}^{\infty \:}\:\:\frac{4n^2-1}{n!}\:\: \quad = 4\sum _{n=0}^{\infty \:} \frac{(n+3)^2}{(n+3)!} + \sum _{n=0}^{\infty \:} \frac{1}{(n+3)!}$$
The last form is very similar to the expotential series, but I can't get it from here.
Any ideas?
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Note that $$e= \sum _{n=0}^{\infty}\frac{1}{n!}$$
\begin{split}
\sum _{n=3}^{\infty }\frac{4n^2-1}{n!} &= 4\sum _{n=3}^{\infty }\frac{n^2}{n!} -\sum _{n=0}^{\infty}\frac{1}{n!}+\frac{5}{2}\\
&=4\sum _{n=3}^{\infty }\frac{n-1+1}{(n-1)!} -\sum _{n=0}^{\infty}\frac{1}{n!}+\frac{5}{2}\\
&= 4\sum _{n=3}^{\infty }\frac{1}{(n-1)!} + 4\sum _{n=3}^{\infty }\frac{1}{(n-2)!} -\sum _{n=0}^{\infty}\frac{1}{n!}+\frac{5}{2}\\
&= \frac{5}{2}+4\sum _{j=2}^{\infty \:}\:\:\frac{1}{j!} + 4\sum _{k=1}^{\infty }\frac{1}{k!} -\sum _{n=0}^{\infty}\frac{1}{n!} \\
&= \frac{5}{2}+4(e-2)+ 4(e-1)-e=7e-\frac{19}{2}
\end{split}
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How do I evaluate $\sum_{m,n\geq 1}\frac{1}{m^2n+n^2m+2mn}$ I saw a problem here which state to evalute $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2n+n^2m+2mn}$$
My attempt
Let $$f(m,n)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2n+n^2m+2mn}$$ and interchanging $m,n$ as $n,m$ we have $$f'(n,m) = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{mn^2+nm^2+2mn}$$ then I add $f(m,n)+f'(n,m)$ which gives the same which gives me the twice the original series.
I failed to evaluate the series with the process I have applied. How do I evaluate the series? Any help will be appreciated.
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Using integration trick
Note that $$\dfrac{1}{nm^2+n^2m+2mn} =\dfrac{1}{mn(m+n+2)}=\dfrac{1}{mn}\int_0^1 x^{m+n+1}\,dx$$ then we have $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{mn} \int_0^1 x^{m+n+1}\,dx=\int_0^1x\left(\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{x^{m}\cdot x^n}{mn}\right)\,dx \\= \int_0^1 x\ln^2(1-x)\,dx =\int_0^1 (1-x)\ln^2 x\,dx $$ hence integrating (by parts) we have the result $\dfrac{7}{4}$.
Without integration trick
Expanding the inner summation gives us $$\sum_{n\geq 1} \left(\frac{1}{n(n+3)}+\frac{1}{2n(n+4)}+\frac{1}{3n(n+15)}+\cdots\right)$$ making partial fraction of each summand we have $$\sum_{n\geq 1}\left(\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)+\frac{1}{8}\left(\frac{1}{n}-\frac{1}{n+4}\right)+\cdots\right)=\sum_{n\geq 3} \frac{H_n}{n(n-2)}\cdots(1)$$ from here we can carry on our work by using the generation function of harmonic number, ie $$\sum_{n\geq 1} x^nH_n=\frac{\ln(1-x)}{x-1}$$
If we dont wish to make repeated integration and multiplying process then let's continue our work from $(1)$ we have above $$\sum_{n\geq 3}\frac{H_n}{2}\left(\frac{1}{n-2}-\frac{1}{n}\right)=\frac{1}{2}\left(H_3+\frac{H_4}{2}\right)+\frac{1}{2}\sum_{n\geq 3}\left(\frac{H_{n+2}}{n}-\frac{H_n}{n}\right)=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}\right)+\frac{1}{4}\left(1+\frac{1}{2}+\cdots +\frac{1}{4}\right)+\sum_{n\geq 3}\frac{1}{2}\left(\frac{H_n}{n}-\frac{H_n}{n}+\frac{1}{n}\left(\frac{1}{n+1}+\frac{1}{n+2}\right)\right)\\=\frac{1}{2}\left(\frac{11}{6}+\frac{25}{24}\right)+\sum_{n\geq 3}\left(\frac{2}{(n+2)(n+3)}+\frac{3}{n(n+1)(n+2)}\right)=\frac{1}{2}\left(\frac{69}{24}+\frac{1}{2}+\sum_{n\geq 1}\frac{1}{n(n+1)(n+2)}-\frac{3}{6}-\frac{3}{24}\right)=\frac{1}{2}\left(\frac{69}{24}+\frac{1}{2}+\frac{3}{2\cdot 2!}-\frac{5}{8}\right)=\frac{7}{4}$$
Recall that for all $ p>0$ $$\sum_{n\geq 1}\frac{1}{n(n+1)(n+2)\cdots(n+p)}=\frac{1}{p\cdot p!}$$
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Why is the convergence point of $ \sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} $ negative? I am trying to evaluate $\frac1{2^1} - \frac1{2^2} + \frac1{2^3} - \frac1{2^4} + \cdots$
I re-wrote the sum using sigma notation as:
$$ \sum _{n=1}^{\infty } \left( \frac{1}{2^n}-\frac{1}{2^{n+1}} \right) \quad (1) $$
Hence,
$$ (1) = \sum _{n=1}^{\infty } \frac{2^{n+1} - 2^n }{2^n2^{n+1}} = \sum _{n=1}^{\infty } \frac{2^n(2-1) }{2^n2^{n+1}} = - \sum _{n=1}^{\infty } \frac1{2^{n+1}} = - \sum _{n=2}^{\infty } \frac1{2^{n}} = -\left( \sum _{n=0}^{\infty } \left(\frac12\right)^n -1-\frac12\right) = -2+\frac32 =-\frac12 $$
Therefore,
$$ \boxed{\sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} =-\frac12} $$
I can't spot any algebraic mistakes, thus I assume the sum is correct. But I don't understand the result. How can a real valued sum converge to a negative number?
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In the first step
$$(1) = \sum _{n=1}^{\infty } \frac{2^{n+1} - 2^n }{2^n2^{n+1}} = \sum _{n=1}^{\infty } \frac{2^n(2-1) }{2^n2^{n+1}} = \color{red}+\sum _{n=1}^{\infty } \frac1{2^{n+1}} = \ldots$$
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\cos^2(\theta)+\sin^2(\theta) = 1$ $$\cos^2(\theta) + \sin^2(\theta) = 1$$
I solved this by using right triangle,
$$\sin(\theta) = \frac{a}{c}, \quad \cos(\theta) = \frac{b}{c}$$
$$\cos^2(\theta) + \sin^2(\theta) = 1$$
$$\Bigl(\frac{b}{c}\Bigr)^2 + \Bigl(\frac{a}{c}\Bigr)^2 = 1 $$
$$\frac {a^2 + b^2} {c^2} = 1 $$
now using Pythagorean identity: $a^2 + b^2 = c^2$
$$\frac {c^2} {c^2} = 1, \quad 1 = 1 $$
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You solved this by using some right triangle but there isn't a right triangle in your solution? Remember, things like that have to be explicitly written in your proof.
Also, the proof is not very valid because you started off by asserting that:
$$\sin^2(\theta)+\cos^2(\theta) = 1$$
when really, what you should have started with is just the left-hand side of that. In other words, I can see where you're going with the reasoning but it isn't airtight because of your wording.
Really, what you should have done is to say that:
$$\sin(\theta) = \frac{a}{c}$$
$$\cos(\theta) = \frac{b}{c}$$
where $a,b,c$ are the sides of a particular right triangle whose image you're supposed to include. Then, you have:
$$\sin^2(\theta)+\cos^2(\theta) = \frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{a^2+b^2}{c^2}$$
By Pythagoras's Theorem relating the sides of the right triangle, it is the case that $a^2+b^2 = c^2$. Hence, we have:
$$\sin^2(\theta) + \cos^2(\theta) = \frac{a^2+b^2}{c^2} = \frac{c^2}{c^2} = 1$$
which proves the desired result.
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"timestamp": "2023-03-29T00:00:00",
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|
Find the remainder when $\sum_{n=1}^{2015}{n^2\times2^n}$is divided by 23.
Find the remainder when $\sum_{n=1}^{2015}{n^2\times2^n}$is divided by 23.
I am completely stuck at this to even start , here's the samll thing that I have noticed .
When $2^{11}$ is divided by 23 , the remainder is $1$ , so $2^{11k+r}$ is equivalent to $2^r$ (mod 23) , for any natural number $k$.
Apart from this nothing useful thing came to my mind .
Could someone please help me to find the remainder ?
Thanks !
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So what you said and the fact that $n^2 \equiv(n \mod 23)^2 \pmod{23}$ and $12 \equiv -11 \pmod{23}$ and $13 \equiv -10 \pmod {23}$ and so on makes a periodic sum: (let the sum be $S$)
$$S \equiv 1^2\cdot2^1+\dots+11^2\cdot2^{11}$$
$$+(-11)^2\cdot2^{1}+(-10)^2\cdot2^2+\dots+(-1)^2\cdot2^{11}+0+\dots$$
and of course the square makes the negative go out. So we need to find out that $2015=23\cdot87+14$ which makes
$$S \equiv 87(1^2\cdot2^1+\dots+11^2\cdot2^{11}+11^2\cdot2^{1}+10^2\cdot2^2+\dots+1^2\cdot2^{11}+0)$$
$$+1^2\cdot2^1+\dots+11^2\cdot2^{11}+11^2\cdot2^{1}+10^2\cdot2^2+9^2\cdot2^3 \tag{since $14\equiv -9 \pmod{23}$}$$
and that makes the rest easy to compute.
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"timestamp": "2023-03-29T00:00:00",
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|
coordinates of focus of parabola Find the coordinates of focus of parabola $$\left(y-x\right)^{2}=16\left(x+y\right)$$
rewriting:
$(\frac{x-y}{\sqrt{2}})^2=8\sqrt2(\frac{x+y}{\sqrt{2}})$
comparing with $Y^2=4aX$
$4a=8\sqrt2,a=2\sqrt2 $
$\Rightarrow$ coordinates of focus=2,2
Is this the correct approach?
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Rewrite $$(x-y)^2=16(x+y)$$ as
$$(\frac{x-y}{\sqrt{2}})^2=8\sqrt{2}\frac{(x+y)}{\sqrt{2}}\implies Y^2=4AX,~~~ A=2\sqrt{2}$$
The co-0rdinates of focus are given by $(X=A,Y=0) \implies x+y=4,x=y \implies x=2, y=2$
So the focus is at $(2,2)$.
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"timestamp": "2023-03-29T00:00:00",
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|
integral arising in statistical mechanics The following integral arises in statistical mechanics of lattice models:
$
\displaystyle I = \int_{0}^{\pi/2}
\ln\left(\, 1 +
\sqrt{\, 1 - a^{2}\sin^{2}\left(\phi\right)\,}\,\right)\,
\mathrm{d} \phi\quad$ with $\quad a \leq 1
$.
By trial and error, the integral was shown to be nearly equal to
$
\displaystyle I =
{\pi \over 4}\ln\left(2\right) + {\pi \over 4}
\ln\left(1 + \sqrt{\, 1 - Ga^{2}\, }\right)
$.
where $G$ denotes Catalan constant.
Is there a method of obtaining an exact analytical solution or justifying the approximate answer given above ?.
|
Similar to @Maxim's comment
$$I(a) = \int_{0}^{\pi/2}\log\left(\, 1 +\sqrt{\, 1 - a^{2}\sin^{2}\left(\phi\right)\,}\,\right)\,d\phi$$
$$I'(a)=- \int_{0}^{\pi/2} \frac{a \sin ^2(\phi)}{\sqrt{1-a^2 \sin ^2(\phi)} \left(1+\sqrt{1-a^2 \sin^2(\phi)}\right)}\,d\phi=\frac{\pi -2 K\left(a^2\right)}{2 a}$$
$$\int \frac{K\left(a^2\right)}{ a}\,da=-\frac{1}{4} G_{3,3}^{2,2}\left(-a^2|
\begin{array}{c}
\frac{1}{2},\frac{1}{2},1 \\
0,0,0
\end{array}
\right)$$ and $I(0)=\frac{1}{2} \pi \log (2)$.
After simplifications, this led me (laboriously) to @Maxim's result.
$$I(a) = \frac {\pi } 2 \log( 2)- \frac {\pi a^2} {16} {_4 F_3} {\left( 1, 1, \frac 3 2, \frac 3 2; 2, 2, 2; a^2 \right)}$$
The good approximation given in the post
$$ J(a) ={\pi \over 4}\log\left(2\right) + {\pi \over 4}
\log\left(1 + \sqrt{\, 1 - Ca^{2}\, }\right)$$
can easily be justified looking at the series expansions built around $a=0$
$$I(a)=\frac{\pi}{2} \log (2)-\frac{\pi }{16}a^2-\frac{9 \pi }{512}a^4-\frac{25 \pi
}{3072}a^6+O\left(a^8\right)$$
$$J(a)=\frac{\pi}{2} \log (2)-\frac{\pi C}{16} a^2 -\frac{3\pi
C^2}{128} a^4 -\frac{5\pi C^3}{384} a^6 +O\left(a^8\right)$$ Since $C\sim 0.915966 $, this makes the coefficients quite similar (their ratios are $C$, $\frac{4}{3} C^2$, $\frac{8 }{5}C^3$).
If you need approximations, you could probably think about $[2n,2n]$ Padé approximants of the ${_4 F_3}$ hypergeometric function (built around $a=0$).
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.