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Proof by induction with square root in denominator: $\frac1{2\sqrt1}+\frac1{3\sqrt2}+\dots+ \frac1{(n+1)\sqrt n} < 2-\frac2{\sqrt{(n+1)}}$ I need to prove $\frac{1}{2\sqrt1} + \frac{1}{3\sqrt2} + ... + \frac{1}{(n+1)\sqrt n} < 2 - \frac{2}{\sqrt{(n+1)}}$
by induction for every $n \in \mathbb{N} $.
Please help, I am stuck on this from the past 2 days..
Thanks.
|
Great answer by trancelocation, but in case you still want it, here is how to do induction step for an inductive proof. First we note the following general rule of quadratics:
$$(n + \tfrac{a+b}{2})^2 - (n+a)(n+b) = (\tfrac{a-b}{2})^2.$$
Using this rule with $a=1$ and $b=2$, we have:
$$(n+\tfrac{3}{2})^2 > (n+1)(n+2).$$
Rearranging this inequality gives:
$$\frac{(n+\tfrac{3}{2})^2}{n+1} > n+2.$$
Our inductive step can now be accomplished as follows:
$$\begin{equation} \begin{aligned}
\sum_{k=1}^{n+1} \frac{1}{(k+1)\sqrt{k}}
&= \sum_{k=1}^{n} \frac{1}{(k+1)\sqrt{k}} + \frac{1}{(n+2)\sqrt{n+1}} \\[6pt]
&< 2 - \frac{2}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{n+1}} \\[6pt]
&= 2 - \frac{2(n+2)-1}{(n+2)\sqrt{n+1}} \\[6pt]
&= 2 - \frac{2n+3}{(n+2)\sqrt{n+1}} \\[6pt]
&= 2 - \frac{2}{(n+2)} \cdot \frac{n+\tfrac{3}{2}}{\sqrt{n+1}} \\[6pt]
&< 2 - \frac{2}{(n+2)} \cdot \sqrt{n+2} \\[6pt]
&= 2 - \frac{2}{\sqrt{n+2}} . \\[6pt]
\end{aligned} \end{equation}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral $\int_{0}^{1}\frac{\log x}{2 - x} dx$
Calculate: $$\int_{0}^{1}\frac{\log x}{2 - x} dx$$
I've done a lot of research here in the community. I tried varying variants, but I did not get anything. The problem is the $2$ present in the integral.
|
\begin{align}J&=\int_{0}^{1}\frac{\log x}{2 - x} dx\\
\end{align}
Perform the change of variable $x=2y$,
\begin{align}J&=\int_{0}^{\frac{1}{2}}\frac{\log(2y)}{1 - y} dy\\
&=\ln 2\left[-\ln(1-x)\right]_{0}^{\frac{1}{2}}+\int_{0}^{\frac{1}{2}}\frac{\log y}{1 - y} dy\\
&=\ln ^2 2+\int_{0}^{\frac{1}{2}}\frac{\log y}{1 - y} dy\\
&=\ln ^2 2+\left[-\ln(1-y)\ln y\right]_{0}^{\frac{1}{2}}+\int_{0}^{\frac{1}{2}}\frac{\ln(1-y)}{y}\,dy\\
&=\int_{0}^{\frac{1}{2}}\frac{\ln(1-y)}{y}\,dy\\
\end{align}
Perform the change of variable $u=1-y$,
\begin{align}J&=
\int_{\frac{1}{2}}^1\frac{\ln u}{1-u}\,du\\
&=\int_{0}^1\frac{\ln u}{1-u}\,du-\int_0^{\frac{1}{2}}\frac{\ln u}{1-u}\,du\\
&=\int_{0}^1\frac{\ln u}{1-u}\,du+\ln^2 2-J\\
\end{align}
Therefore,
\begin{align}J&=\frac{1}{2}\int_{0}^1\frac{\ln u}{1-u}\,du+\frac{1}{2}\ln^2 2\\
&=\frac{1}{2}\int_0^1\left( \sum_{n=0}^\infty u^n\right)\ln u\,du+\frac{1}{2}\ln^2 2\\
&=\frac{1}{2}\sum_{n=0}^\infty\left(\int_0^1 u^n\ln u\,du\right)+\frac{1}{2}\ln^2 2\\
&=\frac{1}{2}\ln^2 2-\frac{1}{2}\sum_{n=0}^\infty\frac{1}{(n+1)^2}\\
&=\boxed{\frac{1}{2}\ln^2 2-\frac{1}{2}\zeta(2)}\\
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Nice power sum inequality $ab(a-b)\leq a^{ab}-b^{ab}$ for $a+b=2$
Let $a\geq b>0$ such that $a+b=2$ then we have :
$$ab(a-b)\leq a^{ab}-b^{ab}$$
My try :
We study the function $f(x)$ on $[0;1]$ such that :
$$f(x)=(2-x)^{((2-x)x)}-x^{((2-x)(x))}-(2-x)x(2-2x)$$
the derivative is equal to :
$$f'(x)= x^{(x (2 - x))} (-((2 - x) + (2 - 2 x) \log(x))) + (2 - 2 x) x + 2 (2 - x) x - (2 - 2 x) (2 - x) + (2 - x)^{(x (2 - x))} ((2 - 2 x) \log(2 - x) - x)$$
But I can't show that $f'(x)\geq0$
Any hints would be appreciable.
Thanks in advance.
Edit :
If we study the function $f(x)$:
$$f(x)=\frac{a^x-b^x}{a-b}$$
The function is increasing and the minimum is reached for $a=b=1$
So we study the following limit :
$$\lim_{a,b \to 1}\frac{a^x-b^x}{a-b}$$
This limit is equal :
$$\lim_{a,b \to 1}\frac{a^x-b^x}{a-b}=x$$
Put : $x=ab$ we have the result .
|
Let $a=1+x$ and $b=1-x$.
Thus, $x\geq0$ and
$$a^{ab}=(1+x)^{1-x^2}=1+x-x^3-\frac{x^4}{2}+\frac{1}{6}x^5+\frac{5}{12}x^6+O(x^7)$$ and
$$b^{ab}=(1-x)^{1-x^2}=1-x+x^3-\frac{x^4}{2}-\frac{1}{6}x^5+\frac{5}{12}x^6+O(x^7).$$
Can you end it now?
|
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|
Let a,b and c be the side lengths of triangle ABC respectively...find the greatest value of b*c. Let a,b and c be the side lengths of triangle ABC respectively. If the perimeter of $\Delta$ABC is 7, and that $\cos A=-\frac{1}{8}$, find the greatest value of $b*c$.
This is how I start the solution:
$$a+b+c=7 \implies b+c=7-a,\quad \cos A=-\frac{1}{8}\\
a^2=b^2+c^2-2bc\cdot \cos A\\
\implies a^2=b^2+c^2+\frac{bc}{4} \\
\implies a^2=(7-a)^2-2bc+\frac{bc}{4} \\
\implies bc=4(7-2a)$$
|
Apply a.m.-g.m.-inequality on $\sqrt{bc}$.
$$\frac{(7-a)^2}{4}=\left(\frac{b+c}{2}\right)^2 \ge bc = 4(7-2a)$$
$$49-14a+a^2 \ge 16(7 - 2a)$$
$$a^2+18a-63 = (a+21)(a-3)\ge 0$$
$$a \le -21 \text{(rejected) or } a \ge 3$$
$a\mapsto \dfrac{(7-a)^2}{4}$ is the parabola shifted $7$ units to the right multiplied by $1/4$, so it's strictly decreasing on $a\le 7$. Therefore, to maximise $bc$, in the first inequality, we want to set $\dfrac{(7-a)^2}{4}$ as large as possible, and make it an equality, which is equivalent to $b=c$. This gives the solution $a = 3$, $b = c = 2$, so $bc = 4$.
|
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|
recursive relation on derangement of objects Let $a_{n}$ represent the number of derangements of $n$ objects . If $a_{n+2}=p a_{n+1}+q a_{n}\;\forall n\in\mathbb{N}$ then what is $\displaystyle \frac{q}{p}$?
What I have tried:
I have used
$$ a_{n}=n!\bigg[1-\frac{1}{1!}+\frac{1}{2!}+\cdots +(-1)^n\frac{1}{n!}\bigg],$$
$$ a_{n+1}=n!\bigg[1-\frac{1}{1!}+\frac{1}{2!}+\cdots +(-1)^{n+1}\frac{1}{(n+1)!}\bigg],$$
$$ a_{n+2}=n!\bigg[1-\frac{1}{1!}+\frac{1}{2!}+\cdots +(-1)^{n+2}\frac{1}{(n+2)!}\bigg],$$
and comparing coefficients but it is very lengthy.
Please help me to solve it using less complex way
|
The question implicitly seems to assume that there exist $p$ and $q$ that satisfy this relation for all $n$. Then in particular they must satisfy it for $n=1$ and $n=2$. This gives you two linear equations in $p$ and $q$ that are easily solved: For $n=1$ and $n=2$ you get
$$2=1\cdot p+0\cdot q\qquad\text{ and }\qquad 9=2p+q,$$
so $p=2$ and $q=5$. However, plugging in $n=3$ shows that
$$44\neq 2\cdot9+5\cdot2,$$
so the premise of the question is false.
If $p$ and $q$ are allowed to depend on $n$, the question remains ill-posed; for $n=2$ the relation
$$9=2p+q,$$
holds for $(p,q)\in\{(0,9),(1,7),(2,5),(3,3),(4,1)\}$ each yielding different values for $\frac{p}{q}$.
The solution that whomever gave you the question may have had in mind is the following: As
$$a_{n}=n!\left(1-\frac{1}{1!}+\frac{1}{2!}+\cdots +(-1)^n\frac{1}{n!}\right)=n!\sum_{k=0}^n\frac{(-1)^k}{k!},$$
for all $n\in\Bbb{N}$, it follows that
\begin{eqnarray*}
a_{n+1}
&=&(n+1)!\left(1-\frac{1}{1!}+\frac{1}{2!}+\cdots +(-1)^n\frac{1}{(n+1)!}\right)
&=&(n+1)!\sum_{k=0}^{n+1}\frac{(-1)^k}{k!},\\
&=&(n+1)\cdot n!\left(1-\frac{1}{1!}+\frac{1}{2!}+\cdots +(-1)^n\frac{1}{(n+1)!}\right)
&=&(n+1)\cdot n!\sum_{k=0}^{n+1}\frac{(-1)^k}{k!}\\
&=&(n+1)\cdot\left(a_n+n!\frac{(-1)^{n+1}}{(n+1)!}\right)
=(n+1)a_n+(-1)^{n+1}.
\end{eqnarray*}
We can use this identity twice to express $a_{n+2}$ in terms of $a_n$ and $a_{n+1}$ as follows
\begin{eqnarray*}
a_{n+2}
&=&(n+2)a_{n+1}+(-1)^{n+2}\\
&=&(n+1)a_{n+1}+(-1)^{n+2}+a_{n+1}\\
&=&(n+1)a_{n+1}-(-1)^{n+1}+(n+1)a_n+(-1)^{n+1}\\
&=&(n+1)(a_{n+1}+a_n),
\end{eqnarray*}
so in deed $p=q=n+1$ is a valid solution, though just one of many.
|
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|
how to solve this two limit tasks Hello i stumbled across this two limits task and i cant find an answer to them:
*
*Find the limit depending on the parameter $A$
$$\lim \limits_{x\to\infty}\left(\left(\sqrt{x+1} - \sqrt[4]{x^2 + x + 1} \right) \cdot x^A\right)$$
I tried by multipliying with
$$\frac{\sqrt{x+1} + \sqrt[4]{x^2 + x + 1}}{\sqrt{x+1} + \sqrt[4]{x^2 + x + 1}}$$
which equals 1 and then
$$\frac{x+1 + \sqrt{x^2 + x + 1}}{x+1 + \sqrt{x^2 + x + 1}}$$ so i can get rid off roots in denominators but i got tangled up,
*$$\lim\limits_{x\to\infty}\left((x+1) - \sqrt[3]{x^3 + x^2} \right)$$
lim in both tasks goes to +infinity
|
For 2 we have $\lim\limits_{x\to\infty}\left((x+1) - \sqrt[3]{x^3 + x^2} \right)=\lim\limits_{x\to\infty}\frac{2x^2+3x+1}{(x+1)^2+(x+1)\sqrt[3]{x^3 + x^2} +\sqrt[3]{(x^3 + x^2)^2}}$.
Now you can factor $x^2$ both in the numerator and the denominator and see that the limit is $\frac{2}{3}$.The first limit can be done similarly.
Note: What I used was that $a^3-b^3=(a-b)(a^2+ab+b^2)$,$\forall a,b \in \mathbb{R}$.
|
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|
Am I properly simplifying this geometric progression? I'm studying recurrence relations and am given:
$T(n) = 2 \cdot T(n-1) - 1$
with an initial condition that $T(1) = 3$.
I worked through the first few recurrences:
$T(n-1) = 2^2 \cdot T(n-2) - 2 - 1$
$T(n-2) = 2^3 \cdot T(n-3) - 2^2 - 2 - 1$
and so forth, and came the conclusion that the pattern
$2^k \cdot T(n-k) - 2^{k-1} - 2^{k-2} - ... 2 - 1$
represents this relation. Since we have T(1) = 3, there must be some $n$ and $k$ such that $n-k = 1$, so $k= n - 1$. Substituting:
$2^{n-1} \cdot T(1) - 2^{n-2} - 2^{n-3} - ... 2 - 1$
but T(1) = 3, so...
$2^{n-1} \cdot 3 - 2^{n-2} - 2^{n-3} - ... 2 - 1$
Factoring out a $-1$ I have something that looks for all the world like a geometric progression:
$2^{n-1} \cdot 3 - (1 + 2 + ... + 2^{n-3} + 2^{n-2})$
However, I'm convinced my progression is missing a term, namely $2^{n-1}$. It's outside the progression, being multiplied by three. In my notes, my instructor simplified the progression to:
$\frac{2^{n-1} - 1}{2 - 1}$
I'm confused about where $n-1$ in the exponent of the geometric progression simplification is coming from. I eventually solve the recurrence relation as :
$3 \cdot 2^{n-1} - 2^{n-1} + 1 = 2^n + 1$
Is my understanding of the simplification of the geometric progression correct?
|
The left sides of your second and third equations should be $T(n)$, not what you have written. Under "Factoring out a $-1$" it would be better to end with $2^{n-3}+2^{n-2}$ to maintain the pattern of the exponents increasing by $1$. Your progression does end with $2^{n-2}$. That is why the numerator of the sum has $2^{n-1}$. If you look at the formula for the sum of a finite geometric progression, the exponent is $1$ more than the highest term:
$$1+r+r^2+\ldots +r^k=\frac {r^{k+1}-1}{r-1}$$
When you add $1$ to $n-2$ you get $n-1$ so your teacher's sum of the progression is correct. Your final answer is also correct.
|
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|
Find maximum value by using AM-GM inequality I have a problem: Find the maximum value of $P=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$ with $x,y,z>0$.
Is there anyway to solve this problem by using the AM-GM inequality ? Thank for your answer.
The first way, I tried to use AM-GM twice at the two sum in the denominator, but I get the sum of the fractions with square roots of $xy, yz, zx$ at its denominators.
The other way, I tried to factor the denominator and use both AM-GM and the Schwarz inequality but still get the same with the first way, with higher order
|
Another way based on Michael Rozenberg's solution.
If you prove $$\sum_{cyc}\frac{a^3}{a^8+1}\leq\frac{3}{2}$$ you can prove the lemma $$\frac{1}{a^2+a+1}+\frac{1}{b^2+b+1}+\frac{1}{c^2+c+1}\ge 1$$ for $abc=1$ and $a;b;c\in R^+$ $(\text{Prove by C-S and} (a=xy/z^2;b=yz/x^2;c=xz/y^2))$
Note that we have: $$\frac{a^3}{a^8+1}\le \frac{3\left(a^2+1\right)}{4\left(a^4+a^2+1\right)}$$
Or $$-\frac{\left(a-1\right)^2\left(3a^8+6a^7+16a^6+14a^5+16a^4+14a^3+12a^2+6a+3\right)}{4\left(a^2-a+1\right)\left(a^2+a+1\right)\left(a^8+1\right)}\le 0$$
It is simple now. :)
|
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|
$6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}$ = $21\sqrt{6}$ but I get $207\sqrt{6}$ I'm asked to simplify $6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}$
The provided solution is $21\sqrt{6}$ but I arrive at a different amount.
Here is my working, trying to understand where I went wrong:
First expression:
$6\sqrt{24}$ = $6\sqrt{4}$ * $6\sqrt{6}$ = $6*2*6\sqrt{6}$ = $72\sqrt{6}$
Second expression:
$7\sqrt{54}$ = $7\sqrt{9} * 7\sqrt{6}$ = $147\sqrt{6}$
Third expression is already the remaining common expression $12\sqrt{6}$.
So: $147\sqrt{6} + 72\sqrt{6} - 12\sqrt{6}$ = $207\sqrt{6}$
Where did I go wrong?
|
$a\sqrt{bc} = a\sqrt{b}\sqrt c$.
It is FALSE that $a \sqrt{bc} = a\sqrt b\times a\sqrt c$. There is only one $a$; not two.
$6\sqrt{24} = 6\sqrt{4\times 6}= 6\sqrt 4 \times \sqrt 6$.
Your calculation $6\sqrt{4\times 6} = (6\sqrt{4})\times (6\sqrt{6})$ is just plain wrong.
|
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|
Eisenstein and Weierstrass zeta - series identity Let $\zeta_\Lambda$ be the Weierstrass $\zeta$-function for lattice $\Lambda$ and $G_2$ the Eisenstein series of weight $2$.
The quasiperiod is defined by $\eta_\Lambda(\lambda) := \zeta_\Lambda(z + \lambda) - \zeta_\Lambda(z)$.
Then the following identity holds \begin{align*} \eta_{\Lambda_\tau}(1)=G_2(\tau) .\end{align*}
I tried to check the series definition and i am not sure on how to continue with that
\begin{align*}
\zeta_\Lambda(z + 1) - \zeta_\Lambda(z) &= \frac{1}{z+1} + \left( \sum_{(c,d) \neq (0,0)} \frac{1}{(z+1)-(c \tau +d)} +\frac{1}{c \tau +d}+ \frac{z+1}{(c \tau +d)^2}\right) \\ &-\frac{1}{z} + \left( \sum_{(c,d) \neq (0,0)} \frac{1}{z-(c \tau +d)} +\frac{1}{c \tau +d} + \frac{z}{(c \tau +d)^2}\right) \\ &= \left(\frac{1}{z+1} - \frac{1}{z} \right) +...
\end{align*}
Any help is really appreciated
|
$$G_2(\tau)=\sum_c\sum_d\frac{1_{(c,d)\ne (0,0)}}{(c\tau+d)^2}$$
It converges and is holomorphic for $\Im(\tau)> 0$ because $\sum_d \frac{1}{(c\tau+d)^2} = \frac{\pi^2}{\sin^2(\pi c\tau)}= O(e^{-2 \pi |c\Im(\tau)|})$. Then $$\zeta_\tau(z) =\sum_c\sum_d\frac{1}{z-c\tau-d}+1_{(c,d)\ne (0,0)}(\frac{1}{c\tau+d}+\frac{z }{(c\tau+d)^2})$$
$$ =\sum_c\sum_d\frac{1}{z-c\tau-d-1}+1_{(c,d+1)\ne (0,0)}(\frac{1}{c\tau+d+1}+\frac{z }{(c\tau+d+1)^2})$$
From there the result is immediate
$$\zeta_\tau(z+1)-\zeta_\tau(z)-G_2(\tau)\\=\sum_c\sum_d 1_{(c,d)\ne (0,0)}(\frac{1}{c\tau+d}+\frac{z+1}{(c\tau+d)^2})-1_{(c,d+1)\ne (0,0)}(\frac{1}{c\tau+d+1}+\frac{z}{(c\tau+d+1)^2})-\frac{1_{(c,d)\ne (0,0)}}{(c\tau+d)^2}\\=0$$
|
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|
Complete the Square with $4x^2 - 4x + 3$ I am trying to complete the square of $4x^2 - 4x + 3$ and using the shortcut approach outlined here to get $d,e$ faster
Here is my work so far,
$$
\begin{align}
4x^2 - 4x + 3 &= 0 \\
x^2 - x + \frac{3}{4} &= 0 \\
x^2 - 2dx + (d^2 + e) &= 0 \\
d = - \frac{1}{2}, e = \frac{1}{2}
\end{align}
$$
However, this leads me to $(x - \frac{1}{2})^2 + \frac{1}{2}$. The answer in my textbook is $(2x - 1)^2 + 2$. I see how $x - \frac{1}{2} = 2x - 1$, but I don't get how the book gets 2 and why I got $\frac{1}{2}$ instead.
What did I do wrong? I feel like I am not using the first coefficient $a = 4$ correctly, but don't have a good reason as to why this is
edit:
So I guess I needed to add $a=4$ back to my equation and the correct form should have been $a(x+d)^2 + ae$, but I incorrectly learned it as $(x+d)^2 + e$!!!
|
You got a good result, just recall that you actually have
$$ax^2+bx+c=a(x-q)^2+ar$$
instead of $(x-q)^2+r$ for the final result. Since $a=4$, you have
$$4\left(x-\frac 12\right)^2+4\cdot\frac 12\\
=(2x-1)^2+2.$$
|
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In an equilateral $\triangle ABC$ : $ DB^2 + DC^2 + BC^2 = 100 $ I have a question that:
There is a point $D$ inside the equilateral triangle ABC. If
$$ DB^2 + DC^2 + BC^2 = 100 $$
and the area of $DBC$ is $ 5 \sqrt{3} $, find $AD^2$.
This is what I tried: $DB = x ,\: DC = y,\: BC = a$. Then $$ x^2 + y^2 + a^2 = 100 \tag{1}$$
$$ \sqrt{ 2(x^2 y^2 + a^2 x^2 + a^2 y^2 )- (a^4 + x^4 + y^4 ) } = 20 \sqrt{3} \tag{2} $$
I want to find $a$, $x$, and $y$, but I have only two equations. What can I do?
|
Let $|AB|=|BC|=|CA|=a$,
$|DB|=m$, $|DC|=n$, $|AD|=q$,
$|OB|=\tfrac12a$,
$|OH|=x$, $|DH|=y$
We have two constraints
\begin{align}
m^2+n^2+a^2&=100
\tag{1}\label{1}
,\\
S_{DBC}&=5\sqrt3
\tag{2}\label{2}
.
\end{align}
From \eqref{1},
\begin{align}
(\tfrac12a+x)^2+y^2+
(\tfrac12a-x)^2+y^2+a^2=100
,\\
x^2+y^2 = 50-\tfrac34a^2
\tag{3}\label{3}
,
\end{align}
hence one condition is that the point $D$ must be on a circle
with the center in the middle of $BC$.
Next, the area of $\triangle DBC$ is constrained,
\begin{align}
S_{ABC}&=\tfrac12 a y=3\sqrt5
,
\end{align}
and we can express the side of the $a$ inn terms of $y$,
\begin{align}
a&=\tfrac{10\sqrt3}y
.
\end{align}
Given that and assuming $x$ positive,
we have
\begin{align}
x&=\frac1y\cdot\sqrt{50 y^2-225-y^4}=
\frac1y\cdot\sqrt{(45-y^2)(y^2-5)}
\\
&=
\frac1y\cdot\sqrt{(y-\sqrt5)(3\sqrt5-y)(y+\sqrt5)(3\sqrt5+y)}
,\\
\text{thus }\quad
y&\in(\sqrt5,3\sqrt5)
.
\end{align}
Next,
\begin{align}
|AD|^2&=q^2=|AE|^2+|AD|^2
\\
&=
(\tfrac{\sqrt3}2\,a-y)^2+x^2
\tag{4}\label{4}
.
\end{align}
After the substitution of
expressions for $a$ and $x$
in terms of $y$,
everything cancels nicely,
and we have the answer
\begin{align}
|AD|^2&=20
.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives, such that $F(x-1)+f(1-x)=x^3$ Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives $F:\mathbb{R}\to\mathbb{R}$, such that $F(x-1)+f(1-x)=x^3$ for any $x\in\mathbb{R}$. For $1-x$ instead of $x$, $F(-x)+f(x)=(1-x)^3$, $F(x)+f(-x)=(1+x)^3$ (1).
\begin{align*}
\left(\frac{F(x)-F(-x)}{e^x}\right)' &=\frac{(f(x)+f(-x))-(F(x)+F(-x))}{e^x}\\ &=\left(\frac{F(x)}{e^x}\right)'-\left(\frac{F(-x)}{e^x}\right)'.
\end{align*}
From here on however I don't know how to solve it. I feel that the two relations from (1) or similar ones can be used.
|
As you note, the function $f$ with primitive $F$ satisfies $F(x-1)+f(1-x)=x^3$ if and only if
$$F(x)+f(-x)=(x+1)^3.$$
The primitive $F(x)$ uniquely determines $f(x)$, its derivative, so we'll focus solely on $F(x)$. If $F(x)$ is twice differentiable, or equivalently, if $f$ is differentiable, then differentiating the functional equation
$$F(x)+F'(-x)=(x+1)^3
\qquad\text{ yields }\qquad
F(x)-F''(-x)=3(x+1)^2.$$
Plugging $-x$ into the latter and subtracting the two equations yields
$$F(x)+F''(x)=(x+1)^3-3(-x+1)^2=x^3+9x-2.$$
This is a second order linear non-homogeneous ODE, and the standard method yields
$$F(x)=c_1\cos(x)+c_2\sin(x),$$
as the solution to the homogeneous ODE. For the non-homogeneous part, the form suggests that taking $F(x)$ a cubic polynomial will yield a solution. Setting
$F(x)=ax^3+bx^2+cx+d$ yields
$$x^3+9x-2=F(x)+F''(x)=ax^3+bx^2+(6a+c)x+(2b+d),$$
which shows that $a=1$, $b=0$, $c=3$ and $d=-2$. Hence every such $F$ is of the form
$$F(x)=c_1\cos(x)+c_2\sin(x)+x^3+3x-2.$$
Plugging this back into the original equation yields
\begin{eqnarray*}
F(x)+f(-x)
&=&(c_1\cos(x)+c_2\sin(x)+x^3+3x-2)\\
&\ &+(-c_1\sin(-x)+c_2\cos(-x)+3x^2+3)\\
&=&(c_1+c_2)\cos(x)+(c_1+c_2)\sin(x)+x^3+3x^2+3x_1)\\
&=&(c_1+c_2)(\cos(x)+\sin(x))+(x+1)^3.
\end{eqnarray*}
This shows that the function equation is met if and only if $c_1=-c_2$, and so
\begin{eqnarray*}
F(x)&=&c(\cos(x)-\sin(x))+x^3+3x-2\\
f(x)&=&-c(\cos(x)+\sin(x))+3x^2+3.
\end{eqnarray*}
About nondifferentiable solutions $f$ I have no idea at all.
Edit: As Theo Bendit notes in the comments, and shows in his answer, the fact that
$$f(x)=(1-x)^3-F(-x),$$
implies that $f$ is differentiable. So the solutions above are all solutions.
|
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|
Calculate the integral $\int \frac{2-3x}{2+3x} \sqrt{\frac{1+x}{1-x}}dx$ I have to calculate the integral $\int \frac{2-3x}{2+3x} \sqrt{\frac{1+x}{1-x}}dx$. I tried the following substitutions: $x \rightarrow \frac{1+t}{1-t}, x \rightarrow \frac{1-t}{1+t}, x \rightarrow \frac{t^{2}+1}{t^{2}-1}$ but with no good result. Also I observed the symmetry in the integral, by $x\rightarrow\frac{x}{3}$. However I am unable to end it.
|
Hint The standard substitution for integrals like this, where the integrand includes $\sqrt{\frac{1 + x}{1 - x}}$, is simply $u = \sqrt{\frac{1 + x}{1 - x}}$. Rearranging and differentiating gives $$x = \frac{u^2 - 1}{u^2 + 1}, \qquad dx = \frac{4 u \,du}{(u^2 + 1)^2}.$$
So in our case, where the integrand is a product of $\sqrt{\frac{1 + x}{1 - x}}$ and a rational function of $x$, the substitution produces a rational function:
$$\int \frac{2 - 3 x}{2 + 3 x} \sqrt{\frac{1 + x}{1 - x}} \,dx = -\frac{4}{5} \int \frac{(u^2 - 5) u^2\,du}{(u^2 + 1)^2 (u^2 - \frac{1}{5})} .$$
As usual we apply the method of partial fractions, that is write the integrand as
$$\frac{(u^2 - 5) u^2}{(u^2 + 1)^2 (u^2 - \frac{1}{5})} = \frac{A u + B}{(u^2 + 1)^2} + \frac{C u + D}{u^2 + 1} + \frac{E}{u - \frac{1}{\sqrt{5}}} + \frac{F}{u + \frac{1}{\sqrt{5}}} .$$
With six parameters to solve for, this is a priori a computationally involved problem. But the left-hand side is an even function of $u$, so the right-hand side must be, too, which immediately gives $A = C = 0$, $F = -E$, so we only need to solve for three parameters:
$$\boxed{\frac{(u^2 - 5) u^2}{(u^2 + 1)^2 (u^2 - \frac{1}{5})} = \frac{B}{(u^2 + 1)^2} + \frac{D}{u^2 + 1} + \frac{E}{u - \frac{1}{\sqrt{5}}} - \frac{E}{u + \frac{1}{\sqrt{5}}}} .$$
(Instead of decomposing fully, we can also decompose
$$\frac{(u^2 - 5) u^2}{(u^2 + 1)^2 (u^2 - \frac{1}{5})} = \frac{B}{(u^2 + 1)^2} + \frac{D u}{u^2 + 1} + \frac{E'}{u^2 - \frac{1}{5}} ,$$
and then use the elementary integral $\int \frac{du}{u^2 - a^2} = \frac{1}{2 a }\log \left\vert\frac{u - a}{u + a}\right\vert + C$.)
Alternatively, as Lab pointed out in the comments, the form $\sqrt{\frac{1 + x}{1 - x}}$ suggests substitutions using various trigonometric identities, for example, $$\cot \theta = \pm \sqrt\frac{1 + \cos 2 \theta}{1 - \cos 2\theta} .$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
For any positive integer $n$, let $f(n) = 70 + n^2$ and $g(n)$ be the HCF of $f(n)$ and $f(n+1)$ find the highest possible value of $g(n)$. For any positive integer $n$, let $f(n) = 70 + n^2$ and $g(n)$ be the HCF of $f(n)$ and $f(n+1)$ find the highest possible value of $g(n)$.
This question is from HKIMO prelims 2006. I didn't quite understand what was happening when I went through the solution.
So far all I've gotten is:
$g(n) = (70+n^2, 70+(n+1)^2)$
$g(n) = (70+n^2, 2n+1)$
Not sure what to do from here -_-
Thank you!
|
Using your
$$g(n) = (70+n^2, 2n+1) \tag{1}\label{eq1}$$
note $2n + 1$ is odd, so can multiply $70 + n^2$ by $4$ to $280 + 4n^2$, but with it still giving the same HCF. However, note that $2n \equiv -1 \pmod{2n + 1}$, so $4n^2 \equiv 1 \pmod{2n + 1}$. Thus,
$$280 + 4n^2 \equiv 281 \pmod{2n + 1} \tag{2}\label{eq2}$$
I trust you can finish the rest yourself now.
In general, if the $70$ is replaced by a positive integral constant $c$, then you'll get that $2n + 1 | 4c + 1$, so $g(n) | 4c + 1$, which means that $g(n) \le 4c + 1$. To show it reaches this potential maximum, note that at $n = 2c$, you get $f(2c) = c + (2c)^2 = c + 4c^2 = c(4c + 1)$ and $f(2c + 1) = c + (2c + 1)^2 = 4c^2 + 5c + 1 = (4c + 1)(c + 1)$. This shows that $g(2c) \ge 4c + 1$, so it must be $4c + 1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the sum $\sum_{n=1}^{\infty} (-1)^{n-1}\log \left(1- \frac{1}{(n+1)^2}\right)$ based on $a^2-b^2 $? Find the sum of the given series
$$\sum_{n=1}^{\infty} (-1)^{n-1}\log \left(1- \frac{1}{(n+1)^2}\right).$$
My attempt : I used the formula $a^2- b^2 = (a+b)(a-b)$, then I get
$$\sum_{n=1}^{\infty} \left((-1)^{n-1}\log \left(1- \frac{1}{n+1}\right) +(-1)^{n-1}\log \left(1+\frac{1}{n+1}\right)\right).$$
After that I am not able to proceed further
Any hints/solution will be appreciated.
Thank you.
|
Based on the link that Jack provided,
$${\begin{aligned}
\zeta (s)&=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}},\Re (s)>1\\\eta (s)&=(1-2^{1-s})\zeta (s)\\
&=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}},\Re (s)>0
\end{aligned}}$$
where $\zeta (s)$ is the zeta function, and $\eta (s)$ is the eta function. Then we can evaluate your series by using Euler transformation and taking the derivative of the eta function. You just need to know the chain rule and the exponential derivatives.
Using Euler Transform for the eta function, we have
$$\eta (s)={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {1}{n^{s}}}-{\frac {1}{(n+1)^{s}}}\right],\Re (s)>-1$$
Taking the derivative,
$$\begin{aligned}
\eta '(s)&=(1-2^{1-s})\zeta '(s)+2^{1-s}(\ln 2)\zeta (s)\\&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {\ln n}{n^{s}}}-{\frac {\ln(n+1)}{(n+1)^{s}}}\right],\Re (s)>-1
\end{aligned}$$
$$\begin{aligned}
\Rightarrow \eta '(0)&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[\ln n-\ln(n+1)\right]\\&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}\\&=-{\frac {1}{2}}\left(\ln {\frac {1}{2}}-\ln {\frac {2}{3}}+\ln {\frac {3}{4}}-\ln {\frac {4}{5}}+\ln {\frac {5}{6}}-\cdots \right)\\&={\frac {1}{2}}\left(\ln {\frac {2}{1}}+\ln {\frac {2}{3}}+\ln {\frac {4}{3}}+\ln {\frac {4}{5}}+\ln {\frac {6}{5}}+\cdots \right)\\&={\frac {1}{2}}\ln \left({\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot \cdots \right)
\end{aligned}$$
Recall Wallis' product,
$$\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots ={\frac {\pi }{2}}$$
So
$$\eta '(0)=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}=\frac{1}{2}\ln{\frac {\pi }{2}}\\
\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}=-\ln\frac{\pi}{2}$$
Using the same logic, we have
$$\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n+1}{n+2}}=-\ln\frac{\pi}{4}$$
Then,
$$\begin{align}
\sum_{n=1}^{\infty} (-1)^{n-1}\log \left(1- \frac{1}{(n+1)^2}\right)&=\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}+\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n+1}{n+2}}\\
&=-\ln\frac{\pi}{2}-\ln\frac{\pi}{4}\\
&=\color{red}{-\ln\frac{\pi^2}{8}}
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solution verification: finding a Maclaurin series for $f$, interval of convergence, and $f^{(10)}(0)$ I have to find Maclaurin series for function $f(x)$ = $2x^2\over{16+x^4}$, it's interval of convergence and $f^{(10)}(0)$. I managed to calculate Maclaurin series and $10^{th}$ derivative, but I'm not sure if it's done in proper way and if solution is correct.
On determining the series,
$$\begin{align}
f(x) &= \frac{2x^2}{16+x^4} \\
&= \frac{2x^2}{16} \cdot \frac{1}{1-\frac{-x^4}{16}} \\
&= \frac{2x^2}{16} \cdot \sum_{i=0}^\infty \left(-\frac{x^4}{16} \right)^n \\
&= \frac{2x^2}{16} \cdot \sum_{i=0}^\infty \frac{(-1)^n \cdot x^{4n}}{16^n} \\
&= \sum_{i=0}^\infty \frac{(-1)^n \cdot x^{4n+2}}{2^{4n+3}}
\end{align}$$
for $\vert{-x^4\over16}\vert<1 \implies x\in(-2;2)$
On determining $f^{(10)}(0)$,
$$f^{(10)}(0)\cdot \frac{ x^{10}}{ 10!} = \frac{x^{42}}{2^{43}} \implies f^{(10)}(0) = x^{32}\cdot \frac{10!}{2^{43}}$$
Looking for feedback and opinion if the way I solved it is correct.
|
Your computation of the MacLaurin series of $f$ looks right, but remember that $f^{(10)}(0)$ is a number. You know that$$\frac{f^{(10)}(0)}{10!}=\frac{(-1)^2}{2^{11}}.$$Therefore,$$f^{(10)}(0)=\frac{10!}{2^{11}}.$$
Also, $\left\lvert\frac{-x^4}{16}\right\rvert<1$ doesn't just imply that $x\in(-2,2)$; it is actually equivalent to it.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$a,b,c\in\Bbb R^+, x,y,z\in \Bbb R, $ show that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}$ $a,b,c\in\Bbb R^+, x,y,z\in \Bbb R, $ show that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}$ (use Cauchy–Schwarz inequality)
I have trouble finding the two vectors. Is it $(x,y,z)$ and $(a,b,c)$?
I need a hint
|
By C-S $$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=\frac{\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)(a+b+c)}{a+b+c}\geq$$
$$\geq\frac{\left(\frac{x}{\sqrt{a}}\cdot\sqrt{a}+\frac{y}{\sqrt{b}}\cdot\sqrt{b}+\frac{z}{\sqrt{c}}\cdot\sqrt{c}\right)^2}{a+b+c}=\frac{(x+y+z)^2}{a+b+c}.$$
The equality occurs for $$\left(\frac{x}{\sqrt{a}},\frac{y}{\sqrt{b}},\frac{z}{\sqrt{c}}\right)||(\sqrt{a},\sqrt{b},\sqrt{c})$$ or
$$(x,y,z)||(a,b,c).$$
Actually, the last writing is very useful.
|
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"timestamp": "2023-03-29T00:00:00",
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|
closed form of $\prod_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n}$ I am looking for the closed form of this product.
$$\prod_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n}$$
I have sees it somewhere before but I can't remember it closed form. I remember the Glaisher's constant it is invloved alone with $2^{7/6}$ and maybe e (exponential function constant) also.
Does anyone knows it closed form?
|
Let $$a_n=\left(\frac{n}{n+1}\right)^{(-1)^{n-1} n}$$ then
$$a_{2p}= \left(\frac{2p}{2 p+1}\right)^{-2 p}\qquad \text{and}\qquad a_{2p+1}=\left(\frac{2 p+1}{2 p+2}\right)^{2 p+1}$$
Now, using a CAS,
$$\prod_{p=1}^m a_{2p}=\frac{\sqrt[12]{2} \sqrt{\pi } \exp \left(-2 \zeta ^{(1,0)}(-1,m+1)+2 \zeta
^{(1,0)}\left(-1,m+\frac{3}{2}\right)+\frac{1}{4}\right)}{A^3 \,\Gamma
\left(m+\frac{3}{2}\right)}$$
$$\prod_{p=1}^m a_{2p+1}=\frac{2 \sqrt[12]{2} \Gamma (m+2) \exp \left(2 \zeta
^{(1,0)}\left(-1,m+\frac{3}{2}\right)-2 \zeta
^{(1,0)}(-1,m+2)+\frac{1}{4}\right)}{A^3}$$
$$b_m=\frac 12\prod_{p=1}^m a_{2p}\prod_{p=1}^m a_{2p+1}$$ $$b_m=\frac{2^{\frac 16}\sqrt{\pi } \Gamma (m+2) \exp \left(-2 \zeta ^{(1,0)}(-1,m+1)+4
\zeta ^{(1,0)}\left(-1,m+\frac{3}{2}\right)-2 \zeta
^{(1,0)}(-1,m+2)+\frac{1}{2}\right)}{A^6 \,\Gamma \left(m+\frac{3}{2}\right)}$$
$$b_m=\frac{2^{\frac 16} \sqrt{\pi }\, \Gamma (m+2)}{A^4 \,H(m)^2\,\Gamma
\left(m+\frac{3}{2}\right)}\exp \left(4 \zeta ^{(1,0)}\left(-1,m+\frac{3}{2}\right)-2 \zeta
^{(1,0)}(-1,m+2)+\frac{1}{3}\right)$$ where appears the hyperfactorial function.
Taking logarithms and using Stirling like approximations and then continuing with Taylor expansions using $b_m=e^{\log(b_m)}$
$$b_m=\frac{2^{\frac 16} \sqrt \pi}{A^6}\left(1+\frac{1}{8 m}-\frac{49}{384 m^2}+\frac{127}{1024 m^3}+O\left(\frac{1}{m^4}\right) \right)$$
$$\color{blue}{\lim_{m\to \infty } \, b_m=\frac{2^{\frac 16} \sqrt \pi}{A^6}}$$
|
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|
Can I find all solutions of $2^{n-1}\equiv k\mod n$? Suppose$\ k\ge 2\ $ is a positive integer.
Can I find all positive integers $\ n>1\ $ with $$2^{n-1}\equiv k\mod n$$ ?
I only found out yet that there is always a solution if $\ k>2\ $ and $\ k-1\ $ is not a power of $\ 2\ $. In this case, $\ k\ $ has an odd prime factor $\ q\ $, for which we have $\ 2^{q-1}\equiv k\mod q\ $ as desired.
I am particularly interested whether for $\ k=5\ $, there is a solution and whether for $\ k=11\ $, there is a solution besides $\ n=5\ $. Finally, for $\ k=3\ $, is $\ 10669\ $ the only solution?
|
Clearly n can not be prime. I had following experiment:
$2^{4-1}=8=2\times 4 +0$
$2^{6-1}=32=5\times 6 +2$
$2^{8-1}=128=16\times 8+0$
$2^{9-1}=256=28\times 9 +4$
$2^{10-1}=512=51\times 10+2$
$2^{12-1}=2042=170\times 12 +8$
$2^{14-1}=8192=585\times 14 +2$
$2^{15-1}=16384=1092\times 15 +4$
$2^{16-1}=32768=2048\times 16 +0$
$2^{17-1}=65536=3855\times 17 +1$
$2^{18-1}=131072=7281\times 18+14$
$2^{33-1} ≡4 \mod 33$
$2^{27-1} ≡13 \mod 27$
|
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|
Value of $(p,q)$ in indefinite integration Finding value of $(p,q)$ in $\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx=\frac{px^3+qx^8}{x^{10}-2x^5+1}+c$
what i try
$\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx$
put $x=1/t$ and $dx=-1/t^2dt$
$\displaystyle -\frac{2t^5+3x^{10}}{t^{10}-2t^5+1}dt$
How do i solve it Help me please
|
thanks friends got the result
$\displaystyle \int\frac{2x^7+3x^2}{(x^5-1)^2}dx=\int\frac{2x+3x^{-4}}{(x^2-x^{-3})^2}dx$
put $x^2-x^{-3}=t$ and $(2x+3x^{-4})dx=dt$
Integration is $\displaystyle \int t^{-2}dt=-\frac{1}{t}+C=-\frac{x^3}{x^5-1}+C$
$$\int\frac{2x^7+3x^2}{(x^5-1)^2}dx=\frac{x^3-x^8}{(x^5-1)^2}+C$$
|
{
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"url": "https://math.stackexchange.com/questions/3150009",
"timestamp": "2023-03-29T00:00:00",
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|
Generating function with tough restrictions
In how many ways can a coin be flipped $25$ times in a row so that exactly $5$ heads occur and no more than $7$ tails occur consecutively?
For the heads, I think that it is $\binom{25}{5}$, but I do not know what to do with the tails restriction.
Not sure how to approach this problem. It is from the advanced section in the generating functions chapter in my book.
|
This answer is based upon the Goulden-Jackson Cluster Method.
We consider the set words of length $n\geq 0$ built from an alphabet $$\mathcal{V}=\{H,T\}$$ and the set $B=\{TTTTTTTT\}$ of bad words, which are not allowed to be part of the words we are looking for. We derive a generating function $f(s)$ with the coefficient of $s^n$ being the number of searched words of length $n$.
According to the paper (p.7) the generating function $f(s)$ is
\begin{align*}
f(s)=\frac{1}{1-ds-\text{weight}(\mathcal{C})}\tag{1}
\end{align*}
with $d=|\mathcal{V}|=2$, the size of the alphabet and $\mathcal{C}$ the weight-numerator of bad words with
\begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[TTTTTTTT])
\end{align*}
We calculate according to the paper
\begin{align*}
\text{weight}(\mathcal{C}[T^8])&=-s^8-s\cdot \text{weight}(\mathcal{C}[T^8])-\cdots-s^7\cdot\text{weight}(\mathcal{C}[T^8])\tag{1}\\
\end{align*}
and get
\begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[T^8])=-\frac{s^8(1-s)}{1-s^8}
\end{align*}
It follows
\begin{align*}
f(s)&=\frac{1}{1-ds-\text{weight}(\mathcal{C})}\\
&=\frac{1}{1-2s+\frac{s^8(1-s)}{1-s^8}}\tag{2}\\
&=\frac{1+s^8}{1-2s+s^9}\\
&=1 + 2 s + 4 s^2 + 8 s^3 + 16 s^4 +\cdots+\color{blue}{32\,316\,160}s^{25}+\cdots\\
\end{align*}
The last line was calculated with the help of Wolfram Alpha. The coefficient of $s^{25}$ shows that out of $2^{25}=33\,554\,432$ words of length $25$ from the alphabet $\{H,T\}$ there are $\color{blue}{32\,316\,160}$ words which do not contain the word $TTTTTTTT$.
But we also want to keep track of the number of heads and tails. In order to do so we make a refinement of $f(s)$ by marking the heads with $x$ and the tails with $y$. We obtain from (1)
\begin{align*}
\text{weight}(\mathcal{C}[T^8])&=-(ys)^8-(ys)\text{weight}(\mathcal{C}[T^8])-\cdots-(ys)^7\text{weight}(\mathcal{C}[T^8])
\end{align*}
and get
\begin{align*}
\text{weight}(\mathcal{C})=-\frac{(ys)^8(1-ys)}{1-(ys)^8}\tag{2}
\end{align*}
Using this generalized weight we obtain from (2) a generating function $g(s;x,y)$
\begin{align*}
g(s;x,y)&=\frac{1}{1-(x+y)s+\frac{(ys)^8(1-ys)}{1-(ys)^8}}\\
&=1+(x+y)s+(x+y)s^2+(x+y)^3s+\cdots\\
&\qquad+(x^{25} + 25 x^{24} y + 300 x^{23} y^2 + 2\,300 x^{22 }y^3 +\cdots+\color{blue}{ 17\ 892} x^5 y^{20} \\
&\qquad\qquad+ 2\,010 x^4 y^{21} + 84 x^3 y^{22})s^{25}+\cdots\tag{3}
\end{align*}
We finally conclude from (3) there are $\color{blue}{ 17\ 892}$ words of length $25$ which contain exactly $5$ heads and do not contain more than $7$ consecutive tails.
|
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|
Unexpected (incorrect) solution to Lagrange Inversion solution to $x^4 - x^3 - x^2 - x - 1 = 0$ about the solution near $x = 2$ I am developing generalized hypergeometric solutions for a set of such polynomials. With this example we can write $x^4 - x^3 - x^2 - x - 1 = \frac{x^5 - 2 x^4 + 1}{x - 1}$.
Lagrange Inversion setup: Let $f \left({x}\right) = 2\, {x}^{4} - {x}^{5} = 1 = z$. Solve for $x$ about the point ${x}_{0} = 0$. Thus $f \left({{x}_{0}}\right) = 0$. Expanding via Newton's rule we have
$$
\left[{\frac{w - {x}_{0}}
{f \left({w}\right) - f \left({{x}_{0}}\right)}}\right]^{n} =
\left[{\frac{w}{2\, {w}^{4} - {w}^{5}}}\right]^{n} =
\sum_{k = 0}^{\infty} \left({\frac{1}{2}}\right)^{n + k} \binom{n + k - 1}{k} {w}^{k - 3 n}
$$
The $\left({n - 1}\right)$th derivative of ${w}^{k - 3 n}$ is
$$
\frac{{d}^{n - 1}}{d\, {w}^{n - 1}}\, {w}^{k - 3 n} =
\left({n - 1}\right)! \binom{k - 3 n}{n - 1} {w}^{k - 4 n + 1}.
$$
Thus,
$$
\frac{{d}^{n - 1}}{d\, {w}^{n - 1}} \left[{\frac{w - {x}_{0}} {f \left({w}\right) - f \left({{x}_{0}}\right)}}\right]^{n} =
\left({n - 1}\right)! \sum_{k = 0}^{\infty} \left({\frac{1}{2}}\right)^{n + k} \binom{k - 3 n}{n - 1} \binom{n + k - 1}{k} {w}^{k - 4\, n + 1}
$$
then we find that only the terms where $k = - 1 + 4 n$ survive or
$$
\frac{\left({z - {x}_{0}}\right)^{n}}{n!} \lim_{w \rightarrow 0} \frac{{d}^{n - 1}}{d\, {w}^{n - 1}} \left[{\frac{w - {x}_{0}} {f \left({w}\right) - f \left({{x}_{0}}\right)}}\right]^{n} =
\frac{1}{n} \frac{2}{{2}^{5 n}} \binom{5 n - 2}{4 n - 1}.
$$
Together we have
$$
x = 2 \sum_{n = 1}^{\infty} \frac{1}{n} \frac{1}{{2}^{5 n}} \binom{5 n - 2}{4 n - 1}.
$$
Hence, the final solution
$$
x = \frac{8}{5} - \frac{8}{5}\,
{}_4{F}_{3} \left({\left\{{- \frac{1}{5}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}}\right\}, \left\{{\frac{1}{4}, \frac{1}{2}, \frac{3}{4}}\right\}, \frac{3125}{8192}}\right) = 0.724380245...
$$
However the correct solution is
$$
x = \frac{2}{5} + \frac{8}{5}\,
{}_4{F}_{3} \left({\left\{{- \frac{1}{5}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}}\right\}, \left\{{\frac{1}{4}, \frac{1}{2}, \frac{3}{4}}\right\}, \frac{3125}{8192}}\right) = 1.92756197...
$$
Q Why am I getting this incorrect solution? This solution is not any of the roots.
|
You've chosen $w = 2 x^4 - x^5$, so either the inverse function is given by a Puiseux series in powers of $w^{1/4}$ or the center of the expansion is not $(0, 0)$. In the first case,
$$a_k = \lim_{x \to 0^+} \frac 1 {k!} \frac {d^{k - 1}} {d x^{k - 1}}
\left( \frac x {(2 x^4 - x^5)^{1/4}} \right)^{\!k} =
\frac {(-1)^{k - 1} \, 2^{1 - 5 k/4}} {k!}
\left( 2 -\frac {5 k} 4 \right)_{\!k - 1}, \\
x = \sum_{k \geq 1} a_k w^{k/4}.$$
This gives the four roots which behave like the branches of $(w/2)^{1/4}$ at zero. In the second case,
$$a_k = \lim_{x \to 2} \frac 1 {k!} \frac {d^{k - 1}} {d x^{k - 1}}
\left( \frac {x - 2} {2 x^4 - x^5} \right)^{\!k} =
\frac {(-1)^k \, 2^{1 - 5 k}} {k!} (2 - 5 k)_{k - 1}, \\
x = 2 + \sum_{k \geq 1} a_k w^k,$$
which is regular at zero.
|
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|
Integration of $\int^{1}_{-1} \frac {1}{3} \sinh^{-1} \left( \frac {3\sqrt 3}{2} (1-t^2) \right) dt$ Recently I came across with respect to this post of mine hyperbolic solution to the cubic equation for one real root given by
$$
t=-2\sqrt \frac {p}{3} \sinh \left( \frac {1}{3} \sinh^{-1} \left( \frac {3q}{2p} \sqrt \frac {3}{p} \right) \right)
$$
Intuitively I sought to find the related definitely integral,
$$
I=\int^{1}_{-1} \frac {1}{3} \sinh^{-1} \left( \frac {3\sqrt 3}{2} (1-t^2) \right) dt
$$
Unfortunately, there was no closed form solution. However, the Integral is amazingly near $\sqrt 2$.
$$
I=0.8285267994716327, \frac {I}{2} +1=1.4142633998
$$
To investigated more, I tried a heuristic expansion of the integral into Egyptian fractions. Although it gets problematic after the 4th term,
The first four terms are,
$$
\frac {I}{2} +1 = 1+ \frac {1}{2} - \frac {1}{12}-\frac {1}{416}
$$
Here the denominators can be given by,
$$
a_n = \sum_{k=0}^{n} { }^nC_k (2^n - 2^kq)^{n-k}q^k , q=\sqrt 2
$$
(Likewise, the denominators in the expansion for $\sqrt 2$ are related to Pell numbers, which makes me believe that my integral too is somewhat related to the numbers $a_n$.) Therefore, I am finding either a closed form or possibly a fast converging infinite series solution to the integral, just any of these. Thanks for any help.
The indefinite integral
For $t=\sin z$ and applying integration by parts, I get another, somewhat simpler, indefinite integral,
$$
\frac {\sin z}{3} \sinh^{-1} \left( \frac {3\sqrt 3}{2} \cos^2 z \right) + 2\sqrt 3 \int \frac {\sin^2 z \cos z dz}{\sqrt {27\cos^4 z + 4}}
$$
Then again I am stuck. Moreover, this expression ensures that my definite integral is an improper one.
Update
A closed solution in terms of incomplete elliptic integrals with complex arguments is, as given by a user in the comments section,
$$
\frac {4}{9} (9+2\sqrt 3 i) \left[ F \left( \sin^{-1} \sqrt {\frac {3}{31}(9+2\sqrt 3 i)} ; \frac {1}{31} (23-12\sqrt 3 i) \right)-
E \left( \sin^{-1} \sqrt {\frac {3}{31}(9+2\sqrt 3 i)} ; \frac {1}{31} (23-12\sqrt 3 i) \right) \right]
$$
However, I am still wondering how to transform this into a real number, especially the $a_n$ connection of the integral is fascinating my mind.
|
We have from symmetry that $$I=\frac23\int_0^1\sinh^{-1}\left[\frac{3\sqrt3}2(1-x^2)\right]dx$$
So we define
$$f(a)=\int_0^1\sinh^{-1}[a(1-x^2)]dx$$
Then we recall that
$$\sinh^{-1}(x)=x\,_2F_1\left(\frac12,\frac12;\frac32;-x^2\right)=\sum_{n\geq0}(-1)^n\frac{(1/2)_n^2}{(3/2)_n}\frac{x^{2n+1}}{n!}$$
so
$$\sinh^{-1}[a(1-x^2)]=a(1-x^2)\,_2F_1\left(\frac12,\frac12;\frac32;-a^2(1-x^2)^2\right)\\
=\sum_{n\geq0}(-1)^n\frac{a^{2n+1}}{n!}\frac{(1/2)_n^2}{(3/2)_n}(1-x^2)^{2n+1}$$
so
$$f(a)=\sum_{n\geq0}(-1)^n\frac{a^{2n+1}}{n!}\frac{(1/2)_n^2}{(3/2)_n}\int_0^1(1-x^2)^{2n+1}dx$$
For this integral, we use $x=\sin(t)$:
$$j_n=\int_0^1(1-x^2)^{2n+1}dx=\int_0^{\pi/2}\cos(t)^{4n+3}dt$$
I leave it as a challenge to you to show that $$\int_0^{\pi/2}\sin(t)^a\cos(t)^bdt=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$
choosing $b=4n+3$, $a=0$ we have
$$j_n=\frac{\Gamma(1/2)\Gamma(2n+2)}{2\Gamma(2n+5/2)}$$
Then defining
$$t_n=\frac{(1/2)_n^2}{(3/2)_n}j_n$$
we have $$\frac{t_{n+1}}{t_n}=\frac{(n+\frac12)^2(n+1)}{(n+\frac74)(n+\frac54)}$$
Which gives $$f(a)=a\,_3F_2\left(\frac12,\frac12,1;\frac74,\frac54;-a^2\right)$$
And since $I=\frac23f(3\sqrt3/2)$ we have (assuming I've made no mistakes),
$$I=\sqrt3\,_3F_2\left(\frac12,\frac12,1;\frac74,\frac54;-\frac{27}4\right)$$
|
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|
Prove that $(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\dfrac{3}{4}$ where $a+b+c=3$
If $a$, $b$, $c$ are non-negative numbers such that $a + b + c = 3$ then prove
$$(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\frac{3}{4}$$
Here's what I did.
Let $c \ge a \ge b$.
We have that
\begin{align*}
(c - 1)^3 + (a - 1)^3 &= (c + a - 2)(c^2 + a^2 - ca - c - a + 1)\\
&\ge (1 - b)\left[\dfrac{3}{4}(c + a)^2 - b + 4\right]\\
&=(1 - b)\left[\dfrac{3}{4}(3 - b)^2 - b + 4\right]\\
&= \dfrac{1}{4}(1 - b)(3b^2 - 22b + 43)
\end{align*}
That means
\begin{align*}
(a - 1)^3 + (b - 1)^3 + (c - 1)^3 &\ge \dfrac{1}{4}(1 - b)(3b^2 - 22b + 43) + (b - 1)^3\\
&= \dfrac{1}{4}(b - 1)(b^2 + 14b - 39)
\end{align*}
Since $c \ge a \ge b \implies b \le \dfrac{a + b + c}{3} = 1$.
And this is where I am stuck right now.
|
The homogenisation gives
$$\sum_{cyc}(a^3-a^2b-a^2c+2abc)\geq0,$$
which is true by Schur.
Indeed, we need to prove that
$$\sum_{cyc}(a-1)^3+\frac{3}{4}\geq0$$ or
$$\sum_{cyc}\left(a^3-3a^2+3-1+\frac{1}{4}\right)\geq0$$ or
$$\sum_{cyc}(a^3-(a+b+c)a^2)+\frac{27}{4}$$ or
$$4\sum_{cyc}(-a^2b-a^2c)+(a+b+c)^3\geq0$$ or
$$\sum_{cyc}(-4a^2b-4a^2c+a^3+3a^2b+3a^2c+2abc)\geq0$$ or
$$\sum_{cyc}(a^3-a^2b-a^2c+2abc)\geq0,$$ which is obvious.
|
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|
Given $a+b+c+d=4$, find the minimum value of $\Sigma_{cyc}\frac{a}{b^3+4}$ Given $a+b+c+d=4$, find the minimum value of $\Sigma_{cyc}\frac{a}{b^3+4}$
I'm pretty sure this has something to do with Holder's inequality, but I don't know how to solve this. By guessing I found $a=2,b=2,c=d=0$ the smallest solution with $\Sigma_{cyc}\frac{a}{b^3+4}=\frac{2}{3}$
|
Does not exist. Try $a>0$,$c>0$, $d>0$ and $b\rightarrow-\sqrt[3]4^-.$
For positive variables by AM-GM we obtain:
$$\sum_{cyc}\frac{a}{b^3+4}=1+\sum_{cyc}\left(\frac{a}{b^3+4}-\frac{a}{4}\right)=1+\sum_{cyc}\frac{-ab^3}{4(b^3+4)}=$$
$$=1+\sum_{cyc}\frac{-ab^3}{2(b^3+b^3+8)}\geq1+\sum_{cyc}\frac{-ab^3}{2\cdot3\sqrt[3]{b^6\cdot8}}=1-\frac{1}{12}\sum_{cyc}ab=$$
$$=1-\frac{(a+c)(b+d)}{12}\geq1-\frac{\left(\frac{a+c+b+d}{2}\right)^2}{12}=\frac{2}{3}.$$
|
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|
Prove that $\sum_{cyc}\frac{a}{a + b^4 + c^4} \le 1$ where $abc = 1$
If $a$, $b$ anc $c$ are three positives such that $abc = 1$ then prove that $$ \sum_{cyc}\dfrac{a}{a + b^4 + c^4} \le 1$$
Here's what I did.
$$ \sum_{cyc}\frac{a}{a + b^4 + c^4}$$
$$\le \sum_{cyc}\frac{a(b^2 + c^2 + bc)}{(b^2c + c^2b + abc)^2} = \frac{1}{(a + b + c)^2}\sum_{cyc}\frac{ab^2 + ac^2 + 1}{(bc)^2} \le \frac{1}{9}\sum_{cyc}\left(\frac{a}{b^2} + \frac{a}{c^2} + a^2\right)$$
And this is where I gave up.
Thanks for reading this. Even more thanks if you can solve the problem.
(Oh wait, I did try again.)
$$\sum_{cyc}\frac{a}{a + b^4 + c^4}$$
$$= 3 - \sum_{cyc}\frac{b^4 + c^4}{a + b^4 + c^4} \le 3 - 4\sum_{cyc}\frac{a^4}{b + c + c^4 + 2a^4 + b^4} = 2\sum_{cyc}\frac{b + c + b^4 + c^4}{b + c + c^4 + 2a^4 + b^4} - 3$$
...and again...
$$ \sum_{cyc}\frac{a}{a + b^4 + c^4}$$
$$= 3 - \sum_{cyc}\frac{b^4 + c^4}{a + b^4 + c^4} \le 3 - \frac{1}{2}\sum_{cyc}\frac{(b^2 + c^2)^2}{a + b^4 + c^4} = \sum_{cyc}\frac{2(a - bc) + (b^4 + c^4)}{a + b^4 + c^4}$$
$$\le \sum_{cyc}\frac{1}{2(a - bc) + (b^4 + c^4)}\frac{[2(a - bc) + (b^4 + c^4)]^2}{a + 2b^2c^2}$$
$$= \frac{1}{2}\sum_{cyc}\dfrac{(a - bc + b^4)^2 + (a - bc + c^4)^2}{[2(a - bc) + (b^4 + c^4)](a + 2b^2c^2)}$$
And I need help. I would be grateful if you could help with this problem.
|
We can use rearrangement inequality to obtain$$
b^4+c^4\ge b^3c+bc^3=bc(b^2+c^2)=\frac{b^2+c^2}a.
$$ This implies
$$\begin{align*}
\sum_{cyc}\frac{a}{a+b^4+c^4}&\le\sum_{cyc}\frac{a}{a+\frac{b^2+c^2}a}\\&=\sum_{cyc}\frac{a^2}{a^2+b^2+c^2}\\&=\frac{\sum_{cyc}a^2}{a^2+b^2+c^2}\\&=1,
\end{align*}$$ as desired.
|
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|
inequality related to square the sum of any two sides of a triangle with respect to square of other side Consider a,b,c are three side of a triangle. Now we need to find the relation between square the them sum of any two side of triangle with respect to third side.
My approach
As we know that the sum of any two sides of triangle will always be greater then the third side i.e
$(a+b)>c $
$(b+c)>a$
$(a+c)>b$
Take square on both side then
$a^2+b^2+2ab>c^2$
$ c^2+b^2+2bc>a^2$
$a^2+c^2+2ac>b^2$
Add last three equations
$2a^2+2b^2+2c^2-a^2-b^2-c^2>-2(ab+bc+ac)$
$a^2+b^2+c^2>-2(ab+bc+ac)$
Edited :
Corrected equation
But the correct answer is this $a^2 + b^2 + c^2 < 2(ab + bc + ca)$
What am I doing wrong?Please guide.
|
Repeat your reasoning starting with this form of triangular inequalities:
$$
|a-b|<c,\quad |b-c|<a,\quad |c-a|<b,
$$
that is:
$$
(a-b)^2<c^2,\quad (b-c)^2<a^2,\quad (c-a)^2<b^2.
$$
|
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|
series sum $\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots $ Finding value of $\displaystyle \int^{\pi}_{0}\ln(1+k\cos x)dx$ for $0<k<1$
what I try
Let $\displaystyle I =\int^{\pi}_{0}\ln(1+k\cos x)dx$
put $\displaystyle x\rightarrow \frac{\pi}{2}-x$
$\displaystyle I=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\ln(1-k\sin x)dx$
$\displaystyle I =\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\bigg[k\sin x-\frac{k^2\sin^2 x}{2}+\frac{k^3\sin^3 x}{3}-\cdots \bigg]dx$
$\displaystyle I =-2\int^{\frac{\pi}{2}}_{0}\bigg[\frac{k^2\sin^2 x}{2}+\frac{k^4\sin^4 x}{4}+\cdots \bigg]dx$
$\displaystyle I =-\pi\bigg[\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots \bigg]$
How can I find sum of that series. Help me please.
|
Hint: by the binomial theorem, $$\sum_{n\ge 0}\frac{(-\frac{1}{2})(-\frac{3}{2})\cdots(\frac{1}{2}-n)}{n!}(-k^2)^n=(1-k^2)^{-1/2}.$$
|
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|
Proving by induction of $n$ that $\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}} $
$$
\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}}
$$
Base Case:
I did $n = 1$, so..
LHS-
$$\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac3{8}$$
RHS-
$$\frac{1}{2} \ - \frac1{(n+1)2^{n+1}} \ = \frac3{8}$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$\sum_{k=1}^{n+1} \frac {k+2}{k(k+1)2^{k+1}} +\frac {n+3}{(n+1)(n+2)2^{n+2}}$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
\frac{1}{2} \ - \frac1{(n+1)2^{n+1}} +\frac {n+3}{(n+1)(n+2)2^{n+2}}
$$
and then I broke up $\frac {n+3}{(n+1)(n+2)2^{n+2}}$ into
$$\frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$
$$=$$
$$\frac{2}{(n+1)2^{n+2}} - \frac{1}{(n+2)2^{n+2}}$$
$$=$$
$$\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}$$
then put it back in with the rest of the equation, bringing me to
$$\frac{1}2 -\frac {1}{(n+1)2^{n+1}} +\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}$$
then
$$\frac{1}2 -\frac{2}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}$$
and
$$\frac{1}2 -\frac{1}{(n+1)2^{n}} - \frac{1}{(n+2)2^{n+2}}$$
$$\frac{1}2 -\frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$
which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?
$$\frac{1}2 -\frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$
canceling out $2^{n}$
$$\frac{1}2 -\frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$
and I'm stuck, please help!
|
Using a telescoping sum, we get
$$
\begin{align}
\sum_{k=1}^n\frac{k+2}{k(k+1)2^{k+1}}
&=\sum_{k=1}^n\left(\frac1{k2^k}-\frac1{(k+1)2^{k+1}}\right)\\
&=\sum_{k=1}^n\frac1{k2^k}-\sum_{k=2}^{n+1}\frac1{k2^k}\\
&=\frac12-\frac1{(n+1)2^{n+1}}
\end{align}
$$
|
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|
Prove the sequence $\sum_{k=0}^n \frac{1}{(n+k)^2}$ is convergent I'm having a hard time trying to prove that the sequence $\{a_n\}$ whose general term $a_n$ is
$$\sum_{k=0}^n \frac{1}{(n+k)^2}$$
is convergent. I'm trying to prove it by definition, that is to say, by finding a lower/upper bound and by proving that it is decreasing/increasing using induction.
By subtracting $a_n$ from $a_{n+1}$ we obtain the following for $n=m$ (if I'm not mistaken):
$$
\frac{1}{(2m+1)^2} + \frac{1}{(2m+2)^2} - \frac{1}{m^2}
$$
Which is less than $0$ for $n=1$ from which I have assumed the sequence is decreasing and therefore trying to prove that $a_{n+1} - a_n < 0$ using induction. But I am terribly stuck! Thanks in advance for any suggestion.
|
To prove that this sequence converges, we will use the monotone convergence theorem which states
If $(a_n)$ is either
*
*increasing and bounded above
*decreasing and bounded below
then it is a convergent sequence.
In order to use this result, we first look at the difference $a_{n+1}-a_{n}$ to determine if your sequence is increasing or decreasing. We have
\begin{align}
a_{n+1}-a_n &= \sum_{k=0}^{n+1} \frac{1}{(n+1+k)^2} - \sum_{k=0}^n \frac{1}{(n+k)^2}\\
&=\sum_{k=0}^{n+1} \frac{1}{(n+(k+1))^2} - \sum_{k=0}^n \frac{1}{(n+k)^2}\\
&=\sum_{k=1}^{n+2} \frac{1}{(n+k)^2} - \sum_{k=0}^n \frac{1}{(n+k)^2}\\
&=\left(\sum_{k=1}^{n} \frac{1}{(n+k)^2} \frac{1}{(2n+1)^2} + \frac{1}{(2n+2)^2} +\right) - \left(\frac{1}{n^2}+\sum_{k=1}^n \frac{1}{(n+k)^2}\right)\\
&=\frac{1}{(2n+1)^2} + \frac{1}{(2n+2)^2} - \frac{1}{n^2}\\
&\leq \frac{1}{(2n)^2} + \frac{1}{(2n)^2} -\frac{1}{n^2}\\
&= -\frac{1}{2n^2}\leq 0
\end{align}
So your sequence is decreasing. Furthermore, it is clear that $a_n\geq 0$ for any $n\in\mathbb{N}$. That is, your sequence is bounded below. By the monotone convergence theorem, it converges.
|
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|
How to perform polynomial long division on 1/(1 - x)? How do I perform polynomial long division on $\frac{1}{1 - x}$ to obtain the sequence $1 + x + x^2 + x^3 + \cdots$?
In this video, the teacher went about it in the following way...
$$
\require{enclose}
\begin{array}{r}
1 + x + x^2 + x^3 + \cdots \\
1 - x \enclose{longdiv}{\hspace{10pt}1\hspace{85.5pt}} \\
\underline{-\left(1 - x\right)}\hspace{64.5pt} \\
x\hspace{68.5pt} \\
\underline{-\left(x - x^2\right)}\hspace{42pt} \\
x^2\hspace{46.5pt} \\
\vdots\hspace{52.5pt}
\end{array}
$$
I have always performed polynomial long division wrt the term of highest degree, e.g., to find $\frac{-7 + 5x + x^2}{-1 + 2x}$, I would do the following...
$$
\require{enclose}
\begin{array}{r}
x^2 + 5x \hspace{4pt}- 7\hspace{33pt} \\
2x - 1 \enclose{longdiv}{\hspace{10pt}2x^3 + 9x^2 - 19x + 7\hspace{4pt}} \\
\underline{-\left(2x^3 - x^2\right)}\hspace{50.5pt} \\
10x^2 - 19x + 7\hspace{4pt} \\
\underline{-\left(10x^2 - 5x\right)}\hspace{21.5pt} \\
-14x + 7\hspace{4pt} \\
\underline{-\left(-14x + 7\right)}\hspace{0pt} \\
0\hspace{4pt} \\
\end{array}
$$
When should I use the teachers variation of the conventional method?
|
Another way to look at it. Start with the known formulas:
$$\begin{align}1-x^2&=(1-x)(1+x)\\
1-x^3&=(1-x)(1+x+x^2)\\
1-x^4&=(1-x)(1+x+x^2+x^3)\\
&\ \ \vdots\\
1-x^n&=(1-x)(1+x+x^2+x^3+\cdots+x^{n-1}) \Rightarrow \\
\frac{1-x^n}{1-x}&=1+x+x^2+x^3+\cdots +x^{n-1}\end{align}$$
If $|x|<1$ and $n\to \infty$, then $x^n\to 0$ and you get the relation:
$$\frac1{1-x}=1+x+x^3+x^4+\cdots$$
|
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|
Does the equation $a^2+d^2+4=b^2+c^2$ have any solutions? Does the equation have $a^2+d^2+4=b^2+c^2$ where $d<c<b<a$ have any integer solutions? This isn't a homework problem, but I need to know for a separate problem I'm doing. Wolfram Alpha isn't very helpful.
|
Above equation shown below has parameterization:
$a^2+d^2+4=b^2+c^2$
$(2m)^2+(m-2)^2+(2)^2=(m+2)^2+(2m-2)^2$
For, $m=5$, we get:
$(10,3,2)^2=(7,8)^2$
Hence, the integer $4$ can be represented by sum difference of four squares.
|
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|
Help verifying proof to proving that for all natural numbers $\sqrt{1} \leq$ than the sum Prove that for all natural numbers $n$,
$$\sqrt{n} \le 1\,+\,\frac{1}{\sqrt{2}}\,+\,\frac{1}{\sqrt{3}}\,+\,\cdots\,+\,\frac{1}{\sqrt{n}}.$$
Solution: We must prove that $1 + \frac 1 {\sqrt{2}}, + \frac 1 {\sqrt{3}} +\,\cdots\, + \frac 1 {\sqrt{k+1}} \geq {\sqrt{k+1}}$
Add $\sqrt{k+1}-\sqrt{k}$ to both sides of first inequality, we get:
$1+\frac{1}{\sqrt{2}} + \frac 1 {\sqrt{3}} + \cdots + \frac{1}{\sqrt{k}} + \sqrt{k+1}-\sqrt{k} \geq \sqrt{k+1}$
However,
$$\sqrt{k+1}-\sqrt{k}=\frac{(\sqrt{k+1}+\sqrt{k})(\sqrt{k+1}-\sqrt{k})}{(\sqrt{k+1}+\sqrt{k})}=\frac{k+1-k}{(\sqrt{k+1}+\sqrt{k})}=\frac{1}{(\sqrt{k+1}+\sqrt{k})} \leq \frac{1}{\sqrt{k+1}}$$
So:
$\sqrt{k+1} \leq 1 + \frac{1}{\sqrt{2}} + \frac 1 {\sqrt{3}} + \cdots + \frac{1}{\sqrt{k}} + \sqrt{k+1}-\sqrt{k} \leq 1+\frac{1}{\sqrt{2}} + \frac 1 {\sqrt{3}}+\cdots + \frac{1}{\sqrt{k+1}}$
|
Multiply both sides by $\sqrt{n}$. Then the assertion is trivial.
|
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|
Solve $\int \frac{e^{x}(2-x^2)}{(1-x)\sqrt{1-x^2}}\mathrm dx$
Find $$\int \frac{e^{x}(2-x^2)}{(1-x)\sqrt{1-x^2}}\mathrm dx$$
Looking at the numerator, combined with the surd, you can get $$\int\frac{e^x(1+x)\left(\frac{1}{\sqrt{1-x^2}}+\sqrt{1-x^2}\right)}{1-x^2}\mathrm dx$$Then this begins to look like the quotient rule, since the denominator is $\sqrt{1-x^2}^2$, and in the numerator, $1/\sqrt{1-x^2}$ is almost like the derivative of the square root. However, it isn't quite that - there are extra terms. $$\color{red}{e^x\left(\sqrt{1-x^2}+\frac{x}{\sqrt{1-x^2}}\right)}+\frac{e^x}{\sqrt{1-x^2}}+e^x x\sqrt{1-x^2}$$
The red terms are accounted for by quotient rule (giving an integral of $\frac{e^x}{\sqrt{1-x^2}}$). But then what do we do with the remaining terms?
|
Hint
Observe that the exponent of $1-x$ is $-\dfrac32$
So, let us find $$\dfrac{d\left(e^x\dfrac{(1+x)^n}{\sqrt{1-x}}\right)}{dx}$$
Compare with the given expression to find the value of $n$
|
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|
Understanding a proof that, if $xy$ divides $x^2+y^2+1$ for positive integers $x$ and $y$, then $x^2+y^2+1=3xy$ This is a Worked Example from Brilliant.org's entry on Vieta Root Jumping.
Let $x$ and $y$ be positive integers such that $xy$ divides $x^2+y^2+1$. Prove that $$x^2+y^2+1=3xy$$
The solution proposes that $x^2+y^2+1=k(xy)$ where $k$ is an integer.
It claims that there exists a minimum solution $(x,y)$ that has the minimum value of $x+y$.
So, they use $t$ to replace $x$ to show that $t^2-kty+y^2+1=0$
Then $t_1=x$ is one solution. By vieta formula,$t_1+t_2=ky$
Then $t_2=ky-x=\frac{x^2+y^2+1}{x}-x=\frac{y^2+1}{x}$
which implies $t_2\lt y$ then $t_1+t_2\lt x+y$.
So, the minimum condition only exists when $x=y$
I am ok as far, but after that it says, $x^2$ divided by $2x^2+1$, $x^2$divided by $1$.
So $k=3$.
But, why they can get $k=3$? $k=3$ only when $x$ and $y$ be the minimum solution. Why $k$ cannot be multiple of $3$?
|
LEMMA
Given integers $$ m > 0, \; \; M > m+2, $$ there are no integers $x,y$ with
$$ x^2 - Mxy + y^2 = -m. $$
PROOF
Calculus: $m+2 > \sqrt{4m+4},$ since $(m+2)^2 = m^2 + 4m + 4,$ while $\left( \sqrt{4m+4} \right)^2 = 4m + 4.$ Therefore also
$$ M > \sqrt{4m+4} $$
We cannot have $xy < 0,$ as then $x^2 - M xy + y^2 \geq 2 + M > 0. $ It is also impossible to have $x=0$ or $y=0.$ From now on we take integers $x,y > 0.$
With $x^2 - Mxy + y^2 < 0,$ we get $0 < x^2 < Mxy - y^2 = y(Mx - y),$ so that $Mx - y > 0$ and $y < Mx.$ We also get
$x < My.$
The point on the hyperbola $ x^2 - Mxy + y^2 = -m $ has both coordinates $x=y=t$ with $(2-M) t^2 = -m,$ $(M-2)t^2 = m,$ and
$$ t^2 = \frac{m}{M-2}. $$
We demanded $M > m+2$ so $M-2 > m,$ therefore $t < 1.$
More important than first appears, that this point is inside the unit square.
We now begin to use the viewpoint of Hurwitz (1907). All elementary, but probably not familiar. We are going to find integer solutions that minimize $x+y.$ If $2 y > M x,$ then $y > Mx-y.$ Therefore, when Vieta jumping, the new solution given by
$$ (x,y) \mapsto (Mx - y, x) $$
gives a smaller $x+y$ value. Or, if $2x > My,$
$$ (x,y) \mapsto (y, My - x) $$
gives a smaller $x+y$ value. We already established that we are guaranteed $My-x, Mx-y > 0.$
Therefore, if there are any integer solutions, the minimum of $x+y$ occurs under the Hurwitz conditions for a fundamental solution (Grundlösung), namely
$$ 2y \leq Mx \; \; \; \; \mbox{AND} \; \; \; \; 2 x \leq My. $$
We now just fiddle with calculus type stuff, that along the hyperbola arc bounded by the Hurwitz inequalities, either $x < 1$ or $y < 1,$ so that there cannot be any integer lattice points along the arc. We have already shown that the middle point of the arc lies at $(t,t)$ with $t < 1.$ We just need to confirm that the boundary points also have either small $x$ or small $y.$ Given $y = Mx/2,$ with
$$ x^2 - Mxy + y^2 = -m $$ becomes
$$ x^2 - \frac{M^2}{2} x^2 + \frac{M^2}{4} x^2 = -m, $$
$$ x^2 \left( 1 - \frac{M^2}{4} \right) = -m $$
$$ x^2 = \frac{-m}{1 - \frac{M^2}{4}} = \frac{m}{ \frac{M^2}{4} - 1} = \frac{4m}{M^2 - 4}. $$
We already confirmed that $ M > \sqrt{4m+4}, $ so $M^2 > 4m+4$ and $M^2 - 4 > 4m.$ As a result, $ \frac{4m}{M^2 - 4} < 1.$ The intersection of the hyperbola with the Hurwitz boundary line $2y = Mx$ gives a point with $x < 1.$ Between this and the arc middle point, we always have $x < 1,$ so no integer points. Between the arc middle point and the other boundary point, we always have $y < 1.$ All together, there are no integer points in the bounded arc. There are no Hurwitz fundamental solutions. Therefore, there are no integer solutions at all.
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
|
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|
Understanding Mathematical Induction problems I have problem to understand this 3 formulas, I am new in this type of problems.
I have to solve this problems by induction.
\begin{align}
\sum_{j = 1}^{j = n} j^3 &= \left(\frac{n(n + 1)}{2}\right) ^ 2 &\text{where } n \geq 1 \\
\sum_{j = 1}^{j = n} j(j + 1) &= \frac{1}{3}n(n + 1)(n + 2) & \text{where } n \geq 1 \\
\sum_{j = 1}^{j = n} j(j!) &= (n + 1)! - 1
\end{align}
|
Here is an example of using induction that doesn't use sigma notation.
The OP can put it in their #Education Reference Folder.
Show that
$$\tag 1 1 + 2 + \dots + n = \frac{n(n+1)}{2}$$
Base Case:
True when $n = 1$ since $1 = \frac{1(1+1)}{2}$.
Inductive Step:
Assume that $1 + 2 + \dots + k = \frac{k(k+1)}{2}$.
Then $1 + 2 + \dots + k + (k+1) = (1 + 2 + \dots + k) + (k+1) =\frac{k(k+1)}{2} + (k+1) =$
$\quad (k+1) (\frac{k}{2} + 1) =\frac{(k+1)(k+2)}{2}$
So by induction, $\text{(1)}$ is always true.
|
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|
How to find $A^{100}$ where $A$ is a matrix with trigonometric entries?
How to find $A^{100}$ where $A=\left(\begin{matrix}\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta\end{matrix}\right)$. ? (Given$\theta=\frac{2\pi}{7}$)
I am a highschool student and new in linear algebra and seeking for only problem solving strategies. I found somewhere the trick to solve this kind of problem is find characteristic equation and then for some of the cases it is possible to find any power of matrix just by exponentiation of both sides. But here this is not applicable.
The characteristics equation found to be $\lambda^2-2\lambda \cos \theta+1=0$
I
how to proceed now ?
|
From the characteristic equation,
$$\begin{align*}
\lambda^2 - 2\lambda \cos \theta + 1 &= 0\\
\lambda &= \frac{2\cos \theta\pm \sqrt{4\cos^2\theta-4}}{2}\\
&= \cos\theta \pm i\sin\theta
\end{align*}$$
For $\lambda_0 = \cos\theta + i\sin\theta$,
$$\begin{align*}
\pmatrix{\cos \theta-\lambda_0 &-\sin\theta\\\sin\theta &\cos\theta-\lambda_0}\mathbf v &= 0\\
\pmatrix{-i\sin\theta &-\sin\theta\\\sin\theta &-i\sin\theta}\mathbf v &= 0\\
\pmatrix{-i&-1\\1&-i}\mathbf v &= 0\\
\mathbf v &= k\pmatrix{i\\1}\\
A\pmatrix{i\\1} &= (\cos\theta + i\sin\theta)\pmatrix{i\\1} \tag 1
\end{align*}$$
Similarly, for $\lambda_1 = \cos \theta - i\sin\theta$,
$$A\pmatrix{1\\i} = (\cos\theta - i\sin\theta)\pmatrix{1\\i}\tag2$$
$(1)\pmatrix{1 &0} + (2)\pmatrix{0&1}$,
$$\begin{align*}
A\pmatrix{i\\1}\pmatrix{1 &0} + A\pmatrix{1\\i}\pmatrix{0&1} &= \lambda_0\pmatrix{i\\1}\pmatrix{1 &0} + \lambda_1\pmatrix{1\\i}\pmatrix{0&1}\\
&= \pmatrix{i\\1}\pmatrix{\lambda_0 &0} + \pmatrix{1\\i}\pmatrix{0&\lambda_1}\\
A\pmatrix{i&1\\1&i} &= \pmatrix{i&1\\1&i}\pmatrix{\lambda_0 &0\\0&0} + \pmatrix{i&1\\1&i}\pmatrix{0 &0\\0&\lambda_1}\\
&= \pmatrix{i&1\\1&i}\pmatrix{\lambda_0 &0\\0&\lambda_1}\\
A &= \pmatrix{i&1\\1&i}\pmatrix{\lambda_0 &0\\0&\lambda_1}\pmatrix{i&1\\1&i}^{-1}\\
A^{n} &= \pmatrix{i&1\\1&i}\pmatrix{\lambda_0 &0\\0&\lambda_1}^{n}\pmatrix{i&1\\1&i}^{-1}\quad(n\in\mathbb N)\\
&= \pmatrix{i&1\\1&i}\pmatrix{\lambda_0^{n} &0\\0&\lambda_1^{n}}\pmatrix{i&1\\1&i}^{-1}\\
&=\frac1{-2} \pmatrix{i&1\\1&i}\pmatrix{\lambda_0^{n} &0\\0&\lambda_1^{n}}\pmatrix{i&-1\\-1&i}\\
&= \frac1{-2} \pmatrix{i&1\\1&i}\pmatrix{\cos n\theta + i\sin n\theta &0\\0&\cos n \theta - i\sin n\theta}\pmatrix{i&-1\\-1&i}\\
&= \frac1{-2} \pmatrix{i\cos n\theta -\sin n\theta &\cos n\theta - i\sin n\theta\\\cos n\theta + i\sin n\theta &i\cos n\theta +\sin n\theta}\pmatrix{i&-1\\-1&i}\\
&= \frac1{-2} \pmatrix{-2\cos n\theta&2\sin n\theta\\-2\sin n\theta &-2\cos n\theta}\\
&= \pmatrix{\cos n\theta&-\sin n\theta\\\sin n\theta &\cos n\theta}\\
\end{align*}$$
|
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|
Elementary way to evaluate $\lim_{x\to0}\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x}$
Evaluate:
$$
\lim_{x\to0}\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x},\ a>0,\ n\in\Bbb N
$$
I've given it several tries but couldn't find an elementary method to find the limit. Two other ways that worked are L'Hospital's rule and generalized binomial expansion.
First method:
$$
\lim_{x\to0} f(x) = \lim_{x\to0}\frac{g(x)}{h(x)} = \lim_{x\to0}\frac{g'(x)}{h'(x)} \\ = \lim_{x\to0}\left({1\over n}\left(a+x\right)^{{1\over n} -1} + {1\over n}(a-x)^{{1\over n} - 1}\right) = {2\over n}{a^{n-1\over n}} = \frac{2\sqrt[n]{a}}{na}
$$
Second method:
$$
(a+x)^{1\over n} = \sqrt[n]{a}\left(1 + {x\over a}\right)^{1\over n} =\\
\sqrt[n]{a}\left(1 + {1\over n}{x\over a} + \frac{\left({1\over n}\right)\left({1\over n} - 1\right)}{2!}\left({x\over a}\right)^2 + \cdots \right)
$$
Also:
$$
(a-x)^{1\over n} = \sqrt[n]{a}\left(1 - {x\over a}\right)^{1\over n} = \\
\sqrt[n]{a}\left(1 - {1\over n}{x\over a} + \frac{\left({1\over n}\right)\left({1\over n} - 1\right)}{2!}\left({x\over a}\right)^2 + \cdots \right)
$$
Combining those ones may obtain:
$$
\lim_{x\to0}f(x) = \lim_{x\to0}\frac{\sqrt[n]{a}\left({2x\over na} + O(x^2)\right)}{x} = \frac{2\sqrt[n]{a}}{na}
$$
The problem is I'm not supposed to use derivatives for solving that limit. Also, generalized binomial expansion is somewhat too complicated as well.
Are there any elementary methods to evaluate the limit from the problem section?
I've also been trying to cast the expression to the form:
$$
\lim_{x\to a}\frac{x^n - a^n}{x - a} = na^{n-1}
$$
but failed.
|
Just write
*
*$\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x} = 2 \frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{(a+x) - (a-x)}$
Now, use $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$
So, you get
\begin{eqnarray*} \frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x}
& = & 2 \frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{(a+x) - (a-x)}\\
& = & \frac{2}{\sum_{k=0}^{n-1}\sqrt[n]{(a+x)^{n-1-k}(a-x)^k}}\\
& \stackrel{x \to 0}{\longrightarrow} & \frac{2}{\sum_{k=0}^{n-1}\sqrt[n]{a^{n-1-k}a^k}} \\
& = & \frac{2}{n\sqrt[n]{a^{n-1}}} \\
& = & \frac{2\sqrt[n]{a}}{na} \\
\end{eqnarray*}
|
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|
Prove that $\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}<\frac{2}{R'-OO'}$ Let $\Delta ABC$ is acute triangle has $AB>AC$ and $O$ is incircle of $\Delta ABC$ with radius $R$, $O'$ is circumcircle of $\Delta ABC$ with radius $R'$. $OA\cap BC=A';OB\cap AC=B';OC\cap AB=C'$ .Prove that $$\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}<\frac{2}{R'-OO'}$$
I proved that $$\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}=\frac{1}{\frac{2bc}{b+c}\cdot \cos \frac{A}{2}}+\frac{1}{\frac{2ac}{a+c}\cdot \cos \frac{B}{2}}+\frac{1}{\frac{2ab}{b+a}\cdot \cos \frac{C}{2}}$$
And by Euler's formula $OO'=\sqrt{R'^2-2R\cdot R'}$
$$\Rightarrow R'-OO'=\frac{abc}{4S}-\sqrt{\left(\frac{abc}{4S}\right)^2-2\cdot \frac{abc}{4S}\cdot \frac{S}{\frac{a+b+c}{2}}}$$
$$=\frac{abc}{4S}-\frac{a^4bc+abc^4+ab^4c+3a^2b^2c^2-a^3bc^2-a^3b^2c-a^2bc^3-a^2b^3c-ab^2c^3-ab^3c^2}{\left(a+b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)}$$
I tried to change this problem to an inequality with three variables $a,b,c$ but i think it's harder than before.
|
I shall use standard notation in my proof, that is, $R, r, s$ and $\Delta$ are the circumradius, inradius, semi-perimeter and area of the triangle, $AA'$ is the length of $A-\text{internal angle bisector}$ and other similarly.
$$rR(\sin A+\sin B+\sin C)=rs=\Delta=\frac{1}{2}ab\sin C=2R^2 \sin{A}\sin{B}\sin{C}$$$$\implies \frac{\sin A+\sin B+\sin C}{2R \sin{A}\sin{B}\sin{C}}=\frac{1}{r}$$
For completeness, I will include the derivation of $AA'$, which you have already done. Applying sine rule on triangle $AA'C$, we have, $$\frac{AA'}{\sin C}=\frac{b}{\sin{(B+A/2)}}\implies AA'=\frac{2R\sin B\sin C}{\cos\frac{B-C}{2}}\geq2R\sin B\sin C$$
$$\therefore \frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}\leq\frac{\sin A+\sin B+\sin C}{2R \sin{A}\sin{B}\sin{C}}=\frac{1}{r}$$
Now, by Euler's formula, $OI=\sqrt{R^2-2Rr}$.
$$\therefore \frac{2}{R-OI}=\frac{2}{R-\sqrt{R^2-2Rr}}=\frac{R+\sqrt{R^2-2Rr}}{Rr}\geq\frac{1}{r}$$
Therefore, $\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}\leq\frac{2}{R-OI}$, with equality iff the triangle is equilateral.
$\blacksquare$
Here is another proof which does not use trigonometry and is probably more enlightening. Let $h_a, h_b, h_c$ be the respective altitudes of $\triangle ABC$. Then, $AA'\geq h_a$ etc. So, $$\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}\leq\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{1}{\Delta}\bigg(\frac{\Delta}{h_a}+\frac{\Delta}{h_b}+\frac{\Delta}{h_c}\bigg)=\frac{s}{\Delta}=\frac{1}{r}$$ The rest of the proof runs along the same lines as the last one. As can easily be seen from this proof, the result is nothing special about angle bisectors, that is, for any three cevians $l_a$,$l_b$ and $l_c$ from points $A$, $B$ and $C$, $$\frac{1}{l_a}+\frac{1}{l_b}+\frac{1}{l_c}\leq\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{1}{r}\leq\frac{2}{R-OI}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3183855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
If $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$ then $y=2^k$ and $x=1$ Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. Is it necessarily the case that $x=1$ and $y=2^k$?
Equivalently (I prove equivalence at the end): Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two positive divisors of $(2^k-1)(2^km+1)$ which average to $m(2^k-1)+1$. Is it necessarily the case that $m=1$ and that these two divisors are $2^k-1$ and $2^k+1$?
I've tested this up to $y\leq10^{10}$ but I haven't been able to make much progress with standard number theoretic techniques.
If $k=1$ then there are infinitely many solutions of the form $x=y-1$.
Let $(1)$ be the initial version of the problem and let $(2)$ be the supposedly equivalent version of the problem.
$(2)\implies(1)$: Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. We can write $y=m(2^k-1)+1$ for some $m\geq1$. Then
$$2^ky-1=2^k(m(2^k-1)+1)-1=(2^k-1)(2^km+1)$$
so $y-x$ and $y+x$ are two positive divisors of $(2^k-1)(2^km+1)$ which average to $y=m(2^k-1)+1$. By $(2)$, $y-x=2^k-1$ and $y+x=2^k+1$. Then $x=1$ and $y=2^k$.
$(1)\implies(2)$: Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two positive divisors of $(2^k-1)(2^km+1)$ which average to $m(2^k-1)+1$. Let $y=m(2^k-1)+1$. We can write the two divisors as $y-x$ and $y+x$ for some $0<x<y$. Thus,
\begin{align*}
y-x&\bigm|2^ky-1,\\
y+x&\bigm|2^ky-1,
\end{align*}
since $2^ky-1=(2^k-1)(2^km+1)$. Manipulating these divisibility relations shows that
\begin{align*}
y-x&\bigm|2^kx-1,\\
y+x&\bigm|2^kx+1,
\end{align*}
where $\gcd(2^kx-1,2^kx+1)=1$. Then $\gcd(y-x,y+x)=1$ so $y^2-x^2\bigm|2^ky-1$. We clearly have $2^k-1\bigm|y-1$. By $(1)$, $x=1$ and $y=2^k$. Then $m=1$ and the two positive divisors were $2^k-1$ and $2^k+1$.
|
Let $y = 1 + (2^k - 1) i$ and $2^k y = 1 + (y^2 - x^2) j$.
Then
1) $2^k y - 1 = (2^k-1) (2^k i+1) = (y^2 - x^2) j$,
2) $(y-1) (y+i) = i j (y^2 - x^2)$,
3) $(-(2^k-1) + j (y^2 - x^2)) ((2^k i+1) + j (y^2 - x^2)) =
i j (y^2 - x^2) 2^{2k}$,
4) $(2y(i j-1)-(i-1))^2 - (i j-1) i j (2x)^2 = (i - 1)^2 - 4 (i j-1) i$,$\quad$ aka Pell equation,
5) $(j (y - x) - 2^{k - 1}) (j (y + x) - 2^{k - 1}) = 2^{2 (k - 1)} - j$,
6) $(2^k-1) (-(2^k i+1) + (2 + (2^k-1) i) i j) = (x^2 - 1) j$.
gp-code for verifing 5) (actually computable for $2\le k<48)$ :
ijk()=
{
for(k=2,1000, for(i=1,k,
m=2^k-1;
yo=1+m*i;
J=divisors(m*(2^k*i+1));
for(q=2,#J-1,
j=J[q];
z=2^(2*(k-1))-j;
D=divisors(z);
for(l=2,#D-1,
u=D[l]; v=z/u;
s=u+2^(k-1); t=v+2^(k-1);
if(s!=t,
if(s%j==0&&t%j==0,
y=(s+t)/2; x=abs(s-t)/2;
if(y==yo,
print(yo" "k" "i" "j" "s" "t" "x," "y)
)
)
)
)
)
))
};
Code for 4) (evaluate over numbers $d=ij-1$):
ijd()=
{
for(d=3, 10^6,
IJ= divisors(d+1);
for(l=1, #IJ,
i= IJ[l]; j= (d+1)/i;
D= d*i*j;
if(!issquare(D),
C= (i-1)^2-4*d*i;
Q= bnfinit('X^2-D, 1);
if(bnfcertify(Q),
fu= Q.fu[1]; \\print(fu);
N= bnfisintnorm(Q, C);
for(v=1, #N, n= N[v];
for(u=0, 100,
s= lift(n*fu^u);
X= abs(polcoeff(s, 0)); Y= abs(polcoeff(s, 1));
if(Y, if(X^2-D*Y^2==C, if(X==floor(X)&&Y==floor(Y), \\print("(X,Y) = ("X", "Y")");
if(Y%2==0,
x= Y/2;
if((X+i-1)%(2*d)==0,
y= (X+i-1)/(2*d); \\print("(x,y) = ("x", "y")");
if((y-1)%i==0,
k= ispower((y-1)/i+1, , &t),
if(k&&t==2,
if(2^k*y==1+(y^2-x^2)*j,
print(" i= "i" j= "j" k= "k" (x,y)= ("x", "y")")
)
)
)
)
)
)))
)
)
)
)
)
)
};
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3184704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
}
|
Prove with induction $\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)(2k+3)}=\frac{n(n+2)}{3(2n+1)(2n+3)}$ Prove with induction the identity
$\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)(2k+3)}=\frac{n(n+2)}{3(2n+1)(2n+3)}$
How can I solve this problem?
Should i set k= (p+1) and n = (p+1), then try to get the left side equal to the right side?
|
Since $k$ is a dummy variable, you should induct on $n$. In other words, show the claim is right when $n=1$, then show $$\sum_{k=1}^p\tfrac{1}{(2k-1)(2k+1)(2k+3)}=\tfrac{p(p+2)}{3(2p+1)(2p+3)}\implies\sum_{k=1}^{p+1}\tfrac{1}{(2k-1)(2k+1)(2k+3)}=\tfrac{(p+1)(p+3)}{3(2p+3)(2p+5)}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3186411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
If $\tan 9\theta = 3/4$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
If $\tan9\theta=\dfrac{3}{4}$, where $0<\theta<\dfrac{\pi}{18}$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
My approach:-
$$\begin{align*} \tan9\theta &=\frac{3}{4} \\[6pt] \implies \theta & = \frac{37^{\circ}}{3} \end{align*}$$
By using this, we get value of $(3\csc3\theta - 4\sec3\theta) =9.95$ by using calculator.
I want know if there's any way to solve this problem without calculator.
|
We have $$\theta=\frac{1}{9}\arctan\frac{3}{4}.$$
Thus,
$$\frac{3}{\sin3\theta}-\frac{4}{\cos3\theta}=\frac{3}{\sin\frac{1}{3}\arctan\frac{3}{4}}-\frac{4}{\cos\frac{1}{3}\arctan\frac{3}{4}}=$$
$$=\frac{3\cos\frac{1}{3}\arctan\frac{3}{4}-4\sin\frac{1}{3}\arctan\frac{3}{4}}{\sin\frac{1}{3}\arctan\frac{3}{4}\cos\frac{1}{3}\arctan\frac{3}{4}}=$$
$$=\frac{10\left(\frac{3}{5}\cos\frac{1}{3}\arctan\frac{3}{4}-\frac{4}{5}\sin\frac{1}{3}\arctan\frac{3}{4}\right)}{\sin\frac{2}{3}\arctan\frac{3}{4}}=$$
$$=\frac{10\sin\left(\arctan\frac{3}{4}-\frac{1}{3}\arctan\frac{3}{4}\right)}{\sin\frac{2}{3}\arctan\frac{3}{4}}=10.$$
I used $$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta,$$
$$\sin\arctan\frac{3}{4}=\sqrt{1-\cos^2\arctan\frac{3}{4}}=$$
$$=\sqrt{1-\frac{1}{1+\tan^2\arctan\frac{3}{4}}}=\sqrt{1-\frac{1}{1+\left(\frac{3}{4}\right)^2}}=\frac{3}{5}$$ and
$$\cos\arctan\frac{3}{4}=\sqrt{1-\sin^2\arctan\frac{3}{4}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3187142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
Find the minimum and maximum values of $P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$ Let $a,b,c$ be non-negative real numbers such that $c \geq 1$ and that $a+b+c=2$. Find the minimum and maximum values of
$$P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$$
To find the minimum of $P$ I rewrite it as
$$P=2\left( ab+bc+ca+1\right)\left(2-abc\right)$$
Then I show that $P \geq 4$, equivalently show that
$$2\left( ab+bc+ca\right)-abc\left( ab+bc+ca\right) \geq abc$$
Indeed, we have
$$3abc\left( a+b+c\right) \leq \left( ab+bc+ca\right)^2 \Leftrightarrow 6abc \leq \left( ab+bc+ca\right)^2 $$
Thus we need to show
$$ \left( ab+bc+ca\right)^2 \leq 12\left( ab+bc+ca\right) -6abc\left( ab+bc+ca\right)$$
This is equivalent to
$$\left( ab+bc+ca\right)\left( ab+bc+ca+6abc-12\right) \leq 0$$
This implies from the inequalities that $ab+bc+ca \leq \left( a+b+c\right)^2/3=4/3$ and $abc \leq \left( a+b+c\right)^3/27=8/27$. The equality holds if $a=b=0$ and $c=1$
Is this right? And for the maximum value, I have no idea. I only guess it is $8$ attending at $a=0,b=c=1$. Please help me. Thank you.
|
Let $a=b=0$ and $c=2$.
Thus, $P=4$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$(6-a^2-b^2-c^2)(2-abc)\geq4$$ or
$$\left(\frac{3(a+b+c)^2}{2}-a^2-b^2-c^2\right)\left(\frac{(a+b+c)^3}{4}-abc\right)\geq\frac{(a+b+c)^5}{8}$$ or
$$\sum_{cyc}(a^4b+a^4c+3a^3b^2+3a^3c^2+6a^3bc+6a^2b^2c)\geq0,$$ which is obvious.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$ and $M$ is a maximal value.
Thus, the inequality $P\leq M$ is a linear inequality of $w^3$, which says that it's enough to prove this for the extreme value of $w^3$, which happens in the following cases.
*
*$c=1$.
In this case $b=1-a$, where $0\leq a\leq1$ and $$P=8-2(a-a^2)^2\leq8;$$
2. $w^3=0$.
Let $b=0$, $c=2-a,$ where $2-a\geq1,$ which says $0\leq a\leq1.$
Thus, $$P=8-4(a-1)^2\leq8.$$
*Two variables are equal.
Try to end this case by yourself.
Actually, this case we can end by AM-GM.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3188979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Integrate by substitution (Not by parts) $\int(x^2-2x)(x^3-3x^2+9)dx$ I have an integration question and I cannot figure out why my answer is incorrect as online calculators will use integration by parts instead, when technically we haven't (and won't) learn that method so I want to be very comfortable with integration with substitution.
$$\int\big(x^2-2x\big)\big(x^3-3x^2+9\big)dx$$
So what I did was
$$u=x^3-3x^2+9$$
$$=\int(x^2-2x)\;u\;\frac{dx}{3(x^2-2x)}$$
$$=\frac{1}{3} \int u \;du $$
$$=\frac{u^2}{6}$$
$$=\frac{\big(x^3-3x^2+9\big)^2}{6}+C$$
But apparently this is incorrect.
My problem may be misunderstanding what can be substituted when?
Help is appreciated.
|
As per your comment, you can always expand:
$$\int\big(x^2-2x\big)\big(x^3-3x^2+9\big)dx$$
$$= \int x^5-3x^4+9x^2-2x^4+6x^3-18x \ \mathrm{d} x$$
$$= \int x^5-5x^4+6x^3+9x^2-18x \ \mathrm{d} x$$
$$= \frac{1}{6}x^6 - x^5 + \frac{3}{2}x^4 + 3x^3-9x^2 + C$$
$$= \frac{1}{6} \left(x^6 - 6x^5 + 9x^4 + 18x^3 - 54x^2 + C'\right)$$
Which should be equivalent to:
$$\frac{1}{6} \left( \left(x^3-(3x^2-9) \right)\left(x^3-(3x^2-9) \right) \right)$$
$$=\frac{1}{6} \left(x^6-2(x^3)(3x^2-9) + (3x^2-9)^2 \right)$$
$$=\frac{1}{6} \left(x^6-6x^5+18x^3+9x^4-54x^2+81 \right)$$
$$=\frac{1}{6} \left(x^6-6x^5+9x^4+18x^3-54x^2+C' \right) \color{#228b22}{\checkmark}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3189565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
How to compute this improper integral?
Let $n\geq1$ be an integer and let
$$I_n=\int\limits_{0}^{\infty}\dfrac{\arctan x}{(1+x^2)^n} \,\mathrm dx$$
Prove that
$$\sum\limits_{n=1}^{\infty}\dfrac{I_n}{n}=\dfrac{\pi^2}{6} \tag{1}$$
$$\int\limits_{0}^{\infty} \arctan x\cdot\ln\left(1+\frac{1}{x^2}\right) \,\mathrm d x=\dfrac{\pi^2}{6} \tag{2}$$
I have an idea that dominated convergence theorem may help here. But I am not getting the proper way.
|
You can use the monotone convergence theorem to show that the series and the integral are equal:
\begin{align}
\sum \limits_{n=1}^\infty \frac{I_n}{n} &\stackrel{\text{MCT}}{=} \int \limits_0^\infty \arctan(x) \sum \limits_{n=1}^\infty \frac{1}{n (1+x^2)^n} \, \mathrm{d} x = \int \limits_0^\infty \arctan(x) \left[-\ln\left(1 - \frac{1}{1+x^2}\right)\right] \, \mathrm{d} x \\
&\hspace{5pt}= \int \limits_0^\infty \arctan(x) \ln \left(1 + \frac{1}{x^2}\right) \, \mathrm{d} x \, .
\end{align}
The integral can be evaluated by integrating by parts:
\begin{align}
\int \limits_0^\infty \arctan(x) \ln \left(1 + \frac{1}{x^2}\right) \, \mathrm{d} x &= \int \limits_0^\infty \left[\frac{2 \arctan{x}}{1+x^2} - \frac{\ln(1+x^2)}{x (1+x^2)}\right] \, \mathrm{d} x \\
&\hspace{-11pt}\stackrel{x = \sqrt{\mathrm{e}^t - 1}}{=} \frac{\pi^2}{4} - \frac{1}{2} \int \limits_0^\infty \frac{t}{\mathrm{e}^t - 1} \, \mathrm{d} t \\
&= \frac{\pi^2}{4} - \frac{1}{2} \zeta(2) = \frac{\pi^2}{4} - \frac{\pi^2}{12} = \frac{\pi^2}{6} \, ,
\end{align}
The integral over $t$ is found using the geometric series and, once again, the monotone convergence theorem. I have not managed to evaluate the sum directly, however.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3192635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
}
|
Transform $dx/dt$ to $dr/dt$ polar coordinates I've had to screenshot the question and post it as a photo
|
The transformations can be written as
$$\tag 1 r^2 = x^2 + y^2~~ \text{and}~~ \theta = \tan^{-1}\left(\frac{y}{x}\right)$$
Differentiating $(1)$
$$r^2 = x^2 + y^2 \implies 2 r r' = 2x x' + 2y y' \implies r ' = \dfrac{x x' + y y'}{r}$$
Substituting $x = r \cos \theta, y = r \sin \theta$, we have
$$\begin{align} r' &= \dfrac{x x' + y y'}{r} \\&= \dfrac{(r \cos \theta \left(-r \cos \theta \left(r^2 \right)-r \sin \theta +r \cos \theta\right)+r \sin \theta \left(-r \sin \theta \left(r^2 \right)+r \sin \theta+r \cos \theta)\right)}{r}\\ &= r(1-r^2)\end{align}$$
We can also derive
$$\theta = \tan^{-1}\left(\frac{y}{x}\right) \implies \theta' = \dfrac{xy'-yx'}{r^2}$$
Make the appropriate substitutions, do some trig simplifications and cleanup to derive $\theta'$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3194271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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|
Generating function of binomial coefficients We want to evaluate the sum $$\sum_{L=0}^{\infty}\frac{1}{2}L(L+1)x^L$$
From this set of notes (page 2, equation 8) we find the formula $$\sum_{n=0}^{\infty}\binom{n}{k}y^n = \frac{y^n}{(1-y)^{n+1}}$$ which suggests that I can do $$\frac{1}{x}\sum_{L=0}^{\infty}\binom{L+1}{2}x^{L+1} = \frac{x^L}{(1-x)^{L+2}}\tag{1}\label{eqn1}$$ since $$\frac{1}{2}L(L+1) = \frac{(L+1)!}{2!((L+1)-2)!}$$
But if I do
\begin{aligned}
\sum_{L=0}^{\infty}\frac{1}{2}L(L+1)x^L & = \sum_{L=0}^{\infty}\frac{L}{2}\frac{d}{dx}x^{L+1} \\
& = \sum_{L=0}^{\infty} \frac{1}{2} \frac{d}{dx} \left[(L+2)x^{L+1} - 2x^{L+1}\right] \\
& = \frac{1}{2} \frac{d^2}{dx^2} \sum_{L=0}^{\infty} x^{L+2} - \frac{d}{dx} \sum_{L=0}^{\infty} x^{L+1} \\
& = \frac{1}{2} \frac{d^2}{dx^2} \frac{x^2}{1-x} - \frac{d}{dx} \frac{x}{1-x} \\
& = \frac{1}{2} \frac{d}{dx} \left[2x(1-x)^{-1} + x^2(1-x)^{-2}\right] - \left[(1-x)^{-1} + x(1-x)^{-2}\right] \\
& = \frac{1}{2} \frac{d}{dx} \frac{2x-x^2}{(1-x)^2} - \frac{1}{(1-x)^2} \\
& = \frac{1}{2} \frac{d}{dx} \left[\frac{x}{(1-x)^2} + \frac{x}{1-x} \right] - \frac{1}{(1-x)^2} \\
& = \frac{1}{2} \left[(1-x)^{-2} + 2x(1-x)^{-3} \right] \\
& = \frac{1+x}{2(1-x)^3}
\end{aligned}
which is nowhere close to (1). Where did I go wrong?
All help is welcome.
|
Note: The cited formula (8) is not correct. There is a typo since we have
\begin{align*}
\sum_{n=0}^{\infty}\binom{n}{k}y^n = \frac{y^{\color{blue}{k}}}{(1-y)^{\color{blue}{k+1}}}\tag{1}
\end{align*}
We therefore expect
\begin{align*}
\frac{1}{x}\sum_{L=0}^{\infty}\binom{L+1}{2}x^{L+1} =\frac{1}{x}\sum_{L=1}^{\infty}\binom{L}{2}x^L
= \frac{x^{\color{blue}{1}}}{(1-x)^{\color{blue}{3}}}\tag{2}
\end{align*}
Your derivation is fine up to
\begin{align*}
\frac{d}{dx}& \frac{2x-x^2}{(1-x)^2} - \frac{1}{(1-x)^2} \\
&=\frac{1}{2}\left((2-2x)(1-x)^{-2}+(2x-x^2)(+2)(1-x)^{-3}\right)-\frac{1}{(1-x)^2}\\
&=\left(\frac{1}{1-x}+\frac{2x-x^2}{(1-x)^3}\right)-\frac{1}{(1-x)^2}\\
&=\frac{1}{(1-x)^3}-\frac{1}{(1-x)^2}\\
&\,\,\color{blue}{=\frac{x}{(1-x)^3}}
\end{align*}
in accordance with (2).
Hint: With respect to (1) we have
\begin{align*}
\frac{y^k}{(1-y)^{k+1}}&=y^k\sum_{n=0}^\infty \binom{-(k+1)}{n}(-y)^n\tag{3}\\
&=\sum_{n=0}^\infty \binom{k+n}{n}y^{n+k}\tag{4}\\
&=\sum_{n=k}^\infty \binom{n}{n-k}y^n=\sum_{n=k}^\infty\binom{n}{k}y^n\tag{5}\\
&=\sum_{n=0}^\infty\binom{n}{k}y^n\tag{6}
\end{align*}
Comment:
*
*In (3) we use the binomial series expansion.
*In (4) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (5) we shift the index to start with $n=k$.
*In (6) we use $\binom{p}{q}=0$ if $q>p$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3195701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
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Related to angles in tetrahedron Let $OABC$ be a tetrahedron such that $|OA|=|OB|=|OC|$. Denote by $D$ and $E$ the midpoints of segments $AB$ and $AC$ respectively. If $\alpha=\angle(DOE)$ and $\beta=\angle(BOC)$ what is the ratio $\beta/\alpha$?
It is obvious that $|BC|=2|DE|$ since triangles $\Delta(ABC)$ and $\Delta(ADE)$ are similar and $D, E$ are midpoints by assumption. I would expect that $\beta/\alpha\geqslant 2$ but I haven't been able to confirm this. I tried to use the cosine law for the segments $|BC|$ and $|DE|$ but without success.
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This answer attempts to find a formula for the ratio.
I imagine this becomes finding the distance of two mid-points of a spherical triangle.
Construct a sphere centred at $O$ with radius $r=OA$.
Extend $OD$ and $OE$ to meet the sphere at $D'$ and $E'$ respectively; $D'$ and $E'$ will be the mid-points of arcs $\overset{\frown}{AB}$ and $\overset{\frown}{AC}$ respectively.
Following this convention of spherical trigonometry, consider the spherical triangle $ABC$, denote
$$\begin{align*}
a &= \angle BOC = \beta\\
b &= \angle AOC\\
c &= \angle AOB
\end{align*}$$
According to the spherical laws of cosine, if $A$ is the spherical angle made by arcs $\overset{\frown}{AB}$ and $\overset{\frown}{AC}$, consider spherical triangle $ABC$,
$$\begin{align*}
\cos A &= \frac{\cos a -\cos b \cos c}{\sin b \sin c}\\
&= \frac{\cos \beta -\cos b \cos c}{\sqrt{(1+\cos b)(1-\cos b)(1+\cos c)(1-\cos c)}}\tag 1\\
\end{align*}$$
And consider spherical triangle $AD'E'$,
$$\begin{align*}
\cos A &= \frac{\cos \angle D'OE' -\cos \frac b2 \cos \frac c2}{\sin\frac b2 \sin\frac c2}\\
&= \frac{\cos \alpha -\sqrt{\frac{1+\cos b}2} \sqrt{\frac{1+\cos c}2}}{\sqrt{\frac{1-\cos b}2}\sqrt{\frac{1-\cos c}2}}\\
&= \frac{2\cos \alpha - \sqrt{(1+\cos b)(1+\cos c)}}{\sqrt{(1-\cos b)(1-\cos c)}}\tag 2
\end{align*}$$
Equating $\cos A$ in $(1)$ and $(2)$,
$$\begin{align*}
\frac{\cos \beta -\cos b \cos c}{\sqrt{(1+\cos b)(1+\cos c)}} &= 2\cos \alpha - \sqrt{(1+\cos b)(1+\cos c)}\\
\cos \beta -\cos b \cos c &= 2\cos \alpha\sqrt{(1+\cos b)(1+\cos c)} - (1+\cos b)(1+\cos c)\\
\cos \beta &= 2\cos \alpha\sqrt{(1+\cos b)(1+\cos c)} - 1 - \cos b - \cos c\\
\cos \alpha &= \frac{1 + \cos b + \cos c + \cos \beta}{2\sqrt{(1+\cos b)(1+\cos c)}}\\
&= \frac{1 + \cos b + \cos c + \cos \beta}{4\cos\frac b2 \cos \frac c2}\\
\frac \beta\alpha &= \frac{\beta}{\arccos \cfrac{1 + \cos b + \cos c + \cos \beta}{4\cos\frac b2 \cos \frac c2}}
\end{align*}$$
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$11x + 13 \equiv 4$ (mod 37) $11x + 13 \equiv 4$ (mod 37)
My "solution" to the problem
$11x + 13 \equiv 4$ (mod 37) $\rightarrow$
$11x + 13 = 4 + 37y $
$11x - 37y = - 9$
Euclid's algorithm.
$37 = 11*3 + 4$
$11 = 4*2 + 3$
$4 = 3*1 + 1 \rightarrow GCD(37,11) = 1$
$3 = 1*3 + 0$
Write as linear equation
$1 = 4 - 1*3$
$3 = 11 - 2*4$
$4 = 37-3*11$
$1 = 4 -1*3 = 4-1(11-2*4) = 3*4-1*11 = 3(37-3*11)-1*11 = 3*37 -10*11$
$1 = 3*37 -10*11$
$11(10) - 37(3) = -1$
$11(90) - 37(27) = -9$
x = 90
That is my answer. But the correct answer should be:
x = 16
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Mod 11, it reduces to $2\equiv 4y+4$ and $24=2\cdot11+2=4(5)+4$ which then gets $11x+13=189\implies 11x=176\implies x=16$
Note however, this is just one of infinitely many answers. Incrementing x by 35, and y by 11 yields new answers that work. All the x values fall on the line $35z+16$, all the y values fall on the line $11a+5$
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Solving Pell's Equation for $x^2 -7y^2 = 1$ for the first three integral solutions. Like the title states my goal is the find the first three integral solutions of the Diophantine equation. I know $x^2 -7y^2 = 1$ is a Pell's equation where $d = 7$. I found the minimal solution to be $(x,y) = (8,3)$ through brute force. I then found the next pair by using the equation:
$(x + \sqrt{7}y)^i(x-\sqrt{7}y)^i = 1$
$i = 1$ would give me the minimal solution $(8,3)$, so I went to $i = 2$.
$(x + \sqrt{7}y)^2(x-\sqrt{7}y)^2 = 1$
$\left(x^2 + 7y^2 + 2xy\sqrt{7}\right) \left(x^2 + 7y^2 - 2xy\sqrt{7}\right) = 1$
$(x^2 + 7y^2)^2 - 7(2xy)^2 = 1$ (This is of the form $X^2 - 7Y^2 = 1$)
I then let $X = x^2 + 7y^2$ and $Y = 2xy$, so the equation becomes the desired
$X^2 - 7Y^2 = 1$
I then plugged in my minimal solution $(8,3)$ and found $X = (8^3+7(3^3)) = 127$ and $Y = 2(8)(3) = 48$, so my new pair is $(127,48)$. To find the next solution, I let $i = 3$.
$(x + \sqrt{7}y)^3(x-\sqrt{7}y)^3 = 1$
And this is where I got stuck. I tried a few methods to get this new equation into the form of $X^2 - 7Y^2 = 1$, but have been unsuccessful. I tried expanding out the equations completely getting:
$x^6 -21x^4y^2+147x^2y^4-343y^6 = 1$
But I'm fairly sure that's not the right direction. The other method I tried was factoring out a $(x + \sqrt{7}y)^2(x-\sqrt{7}y)^2$, so I got:
$[(x + \sqrt{7}y)(x-\sqrt{7}y)](x + \sqrt{7}y)^2(x-\sqrt{7}y)^2$
which simplifying I got
$(x^2-7y^2) \left( (x^2 + 7y^2)^2 - 7(2xy)^2 \right)$
I'm really not sure how to properly factor this cubic to get to the desired function form. Any help would be greatly appreciated!
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Consider
$$(8+3\sqrt7)^2=127+48\sqrt7,$$
$$(8+3\sqrt7)^3=(8+3\sqrt7)(127+48\sqrt7)=2024+765\sqrt7,$$
$$(8+3\sqrt7)^4=(8+3\sqrt7)(2024+765\sqrt7)=32257+12192\sqrt7$$
etc. Then the first few solutions are $(8,3)$, $(127,48)$, $(2024,765)$,
$(32257,12192)$ etc.
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Laurent series of $f(z)=\frac{4z-z^2}{(z^2-4)(z+1)}$ in different annulus Given $f(z)=\dfrac{4z-z^2}{(z^2-4)(z+1)}$
I need to find the Laurent series in the annulus: $A_{1,2}(0),\;A_{2,\infty}(0),\;A_{0,1}(-1)$
I found the following partial fractions:
$f(z)=\dfrac{-3}{(z+2)}+\dfrac{1}{3(z-2)}+\dfrac{5}{3(z+3)}$,
the power series of these fractions are:
$\dfrac{-3}{(z+2)}=\displaystyle{\frac{-3}{2}\sum_{n=0}^\infty \left( \frac{-z}{2} \right)^n} $
$\dfrac{1}{3(z-2)}=\displaystyle{\frac{-1}{6}\sum_{n=0}^\infty \left( \frac{-z}{2} \right)^n} $
$\dfrac{5}{3(z+1)}=\displaystyle{\frac{5}{3}\sum_{n=0}^\infty \left( -z \right)^n} $
and the principle parts are:
$\dfrac{-3}{(z+2)}=\displaystyle{\frac{-3}{2}\sum_{n=0}^\infty \left( \frac{1}{-2z} \right)^n} $
$\dfrac{1}{3(z-2)}=\displaystyle{\frac{-1}{6}\sum_{n=0}^\infty \left( \frac{1}{-2z} \right)^n} $
$\dfrac{5}{3(z+1)}=\displaystyle{\frac{5}{3}\sum_{n=0}^\infty \left( \frac{1}{-z} \right)^n} $
In the first annuli I take the principle part only of $\dfrac{5}{3(z+1)}$, in the second annuli I take the principle part of all fraction. About the third one, I have $0<\vert z-1\vert<1$, I denoted $w=z-1$ and then I took the power series for all fractions and simply switched the $w$ back to $z-1$ at the end. Is it the right way of doing it?
I received $\displaystyle{\frac{-3}{2}\sum_{n=0}^\infty \left( \frac{1-z}{2} \right)^n - \frac{1}{6}\sum_{n=0}^\infty \left( \frac{1-z}{2} \right)^n + \frac{5}{3}\sum_{n=0}^\infty \left( 1-z \right)^n}$
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There's a typo in your partial fraction decomposition; it should be$$\frac{4z-z^2}{(z^2-4)(z+1)}=-\frac3{z+2}+\frac1{3(z-2)}+\frac5{3(z+1)}.$$If $z\in A_{1,2}(0)$, then:
*
*$\displaystyle-\frac3{z+2}=-\frac32\frac1{1+\frac z2}=-\frac32\sum_{n=0}^\infty\frac{(-1)^n}{2^n}z^n$;
*$\displaystyle\frac1{3(z-2)}=-\frac16\frac1{1-\frac z2}=-\frac16\sum_{n=0}^\infty\frac{z^n}{2^n}$;
*$\displaystyle\frac5{3(z+1)}=\frac53\frac1{1+z}=-\frac53\sum_{n=-\infty}^{-1}(-1)^nz^n$.
In $A_{2,\infty}(0)$, the third expansion remains the same, but now we have:
*
*$\displaystyle-\frac3{z+2}=-\frac32\frac1{1+\frac z2}=\frac32\sum_{n=-\infty}^{-1}\frac{(-1)^n}{2^n}z^n$;
*$\displaystyle\frac1{3(z-2)}=-\frac16\frac1{1-\frac z2}=\frac16\sum_{n=-\infty}^{-1}\frac{z^n}{2^n}$.
In the case of $A_{0,1}(-1)$, you write\begin{align}\frac{4z-z^2}{z^2-4}&=-1+\frac1{z-2}+\frac3{z+2}\\&=1+\frac1{(z+1)-3}+\frac3{(z+1)+1}.\end{align}From this, you get the Taylor series of $\dfrac{4z-z^2}{z^2-4}$ in $D(1,1)$ and then, when you divide everything by $z+1$, you get the Laurent series that you're after.
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Limit of this sum
Evaluate $\lim_{n\to \infty}(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\dots+\frac{1}{n+n})$.
I converted the limit of this sum to the sum of limits. Then in each term, I divided the numerator and denominator by $n$. Each limit came out to be zero. Hence I got the answer as $0$. Have I gone wrong somewhere or is my solution correct? I have mixed feelings. Please help.
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Ok: This is inspired on Riemann Sums
Let $S_n=\sum_{k=0}^{n-1}{\frac{1}{n+k+1}}$ be your sum
We first split $[0;1]$ into $n$ intervals of the form $[\frac{k}{n};\frac{k+1}{n}]$, with $k\in\{0,1,2,...,n-1\}$
Note that if $\frac{k}{n}\le x\le \frac{k+1}{n}$ then
$\frac{1}{\frac{k+1}{n}+1}\le\frac{1}{1+x}\le\frac{1}{1+\frac{k}{n}}$. Now we integrate:
$$\frac{1}{k+n+1}=\int_{\frac{k}n}^{\frac{k+1}{n}}\frac{dx}{\frac{k+1}{n}+1}\le\int_{\frac{k}n}^{\frac{k+1}n}\frac{dx}{1+x}\le\int_{\frac{k}n}^{\frac{k+1}n}\frac{dx}{1+\frac{k}{n}}=\frac{1}{k+n}$$ thus:
$$\sum_{k=0}^{n-1}{\frac{1}{k+n+1}}\le \sum_{k=0}^{n-1}\int_{\frac{k}n}^{\frac{k+1}n}\frac{dx}{1+x}\le\sum_{k=0}^{n-1}{\frac{1}{k+n}}$$
Which means:
$$S_n\le\int_{0}^{1}{\frac{dx}{x+1}}\le S_n-\frac{1}{n}+\frac{1}{n+n}$$
Now take the limit of both side
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$a_{1}=a_{2}=1$ and $ a_{n}=\tfrac{1}{2} (a_{n-1}+\tfrac{2}{a_{n-2}}) $ for $n\geq 3$. Then for all $n\in \mathbb{N}$, $1\leq a_{n} \leq 2$
*
*Let $a_{n}$ be the sequence satisfying $a_{1}=a_{2}=1$ and $ a_{n}=\tfrac{1}{2} (a_{n-1}+\tfrac{2}{a_{n-2}}) $ for $n\geq 3$. Then for all $n\in \mathbb{N}$, $1\leq a_{n} \leq 2$
For $n=3$, then we have $1\leq a_n\leq 2.$
Now assume for $n$ holds. Show for $n+1$ holds.
So
$$ a_{n+1}=\tfrac{1}{2} (a_{n}+\tfrac{2}{a_{n-1}}) $$
Then, what should I do now?
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Assume $1\le a_k\le 2$ for all $k<n.$ Then $\frac12(a_{n-1}+\frac2{a_{n-2}})\leq \frac12(2+2/1)=2$ and similarly $\frac12(a_{n-1}+\frac2{a_{n-2}})\ge \frac12(1+2/2)=1$ because $1\leq a_k\leq 2$ for all $ k\le n-1.$
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What can be said about $(\varepsilon-x)y=y'(-x+y^2-2x^2)$ solutions? There was an unanswered question 4 years ago. OP asked for a solution of ODE $(\varepsilon-x)y=y'(-x+y^2-2x^2)$
The comment to the original question proposes an implicit solution, $2\log y + 2\epsilon\log(x + 2 x\epsilon - y^2) - (1+2\epsilon)\log(\epsilon + 2 x\epsilon - y^2) = C$
Could you explain to me how this solution can be obtained? Are there singular solutions? If there are any orthogonal families that can be described explicitly, I would like to know about them. If it can be reduced to some special function differential equation, that would be also great!
I tried to find an integration factor, but it exists only for $\varepsilon = -\frac{1}{4}$
Here is a graph of solution
$\pm \sqrt{\pm \frac{\sqrt{2cx^2+cx+1}}{c}-\frac{1}{c}}$ for $c = -1$
It's not separable, homogeneous, solvable for $x$ or $y$ or Lagrangian, so I'm stuck.
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$$(\epsilon-x)y=y'(-x+y^2-2x^2)$$
$$(\epsilon-x)y\:dx-(-x+y^2-2x^2)dy=0 \tag 1$$
The integrating factor is
$$\boxed{\mu=\frac{1}{(x+2\epsilon x-y^2)(\epsilon +2\epsilon x-y^2)\:y}}\tag 2$$
Multiplying Eq.$(1)$ by $\mu$ leads to the total differential of the sought function $F(x,y)$ :
$$\frac{ (\epsilon-x)y\:dx-(-x+y^2-2x^2)dy }{(x+2\epsilon x-y^2)(\epsilon +2\epsilon x-y^2)\:y}=0=dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy$$
$$\int \frac{\partial F}{\partial x}dx = \int \frac{ (\epsilon-x)y\:dx }{(x+2\epsilon x-y^2)(\epsilon +2\epsilon x-y^2)\:y}=\frac{1}{1+2\epsilon}\ln\left(x+2\epsilon x-y^2 \right)-\frac{1}{2\epsilon}\ln\left(\epsilon +2\epsilon x-2y^2 \right)+f(y)$$
$$\int \frac{\partial F}{\partial y}dy=\int \frac{ -(-x+y^2-2x^2)dy }{(x+2\epsilon x-y^2)(\epsilon +2\epsilon x-y^2)\:y}=\frac{1}{1+2\epsilon}\ln\left(x+2\epsilon x-y^2 \right)-\frac{1}{2\epsilon}\ln\left(\epsilon +2\epsilon x-2y^2 \right)+\frac{1}{\epsilon(1+2\epsilon)}\ln(y)+g(x)$$
The two integrals are equal to $F(x,y)$, thus :
$$g(x)=0\quad\text{and}\quad f(y)=\frac{1}{\epsilon(1+2\epsilon)}\ln(y)$$
$$F(x,y)=\frac{1}{1+2\epsilon}\ln\left(x+2\epsilon x-y^2 \right)-\frac{1}{2\epsilon}\ln\left(\epsilon +2\epsilon x-2y^2 \right)+\frac{1}{\epsilon(1+2\epsilon)}\ln(y)$$
Since $dF=0$ the function $F$ is constant.
$$\frac{1}{1+2\epsilon}\ln\left(x+2\epsilon x-y^2 \right)-\frac{1}{2\epsilon}\ln\left(\epsilon +2\epsilon x-2y^2 \right)+\frac{1}{\epsilon(1+2\epsilon)}\ln(y)=c$$
Let $c=\frac{C}{2\epsilon(1+2\epsilon)}$
$$\boxed{2\epsilon\ln\left(x+2\epsilon x-y^2 \right)-(1+2\epsilon)\ln\left(\epsilon +2\epsilon x-2y^2 \right)+2\ln(y)=C}$$
I confess that the result already given in the question helped me a lot to find the integrating factor.
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What's the remainder when $x^{7} + x^{27} + x^{47} +x^{67} + x^{87}$ is divided by $x ^ 3 - x$ What's the remainder when $x^{7} + x^{27} + x^{47} +x^{67} + x^{87}$ is divided by $x ^ 3 - x$ in terms of $x$?I tried factoring $x$ from both polynomials but I don't know what to do next since there'd be a $1$ in the second polynomial.
Any help would be appreciated!
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$x^{2n}\equiv1 \pmod{x^2-1}$ so $(x^2-1) | (x^{2n}-1)$ so $(x^3-x) | (x^{2n+1}-x)$
so $x^{2n+1}\equiv x \pmod {x^3-x}$
so $x^7+x^{27}+x^{47}+x^{67}+x^{87}\equiv 5x \pmod {x^3-x}$
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How can I show these series combinations are the same algebraically? We have to show:
$$\frac{1}{2}\sum_{r=1}^{n} {\frac{1}{r}} \ + \ \sum_{r=1}^{n} {\frac{1}{2r+1}} \ = \ \sum_{s=1}^{2n}{\frac{1}{s+1}}$$
I understand how this works intuitively and the most formal way I can think to show it is as follows:
$$\frac{1}{2}\sum_{r=1}^{n} {\frac{1}{r}} \ + \ \sum_{r=1}^{n} {\frac{1}{2r+1}} \ = \ \sum_{r=1}^{n}(\frac{1}{2r}+\frac{1}{2r+1}) \ = \ \frac{1}{2} + \frac{1}{3} + \ ... \ + \frac{1}{2n} + \frac{1}{2n+1} \ = \ \sum_{s=1}^{2n}{\frac{1}{s+1}}$$
This is fine but it feels like there might be a more rigorous method without listing terms. I've tried to combine the two fractions on the left-hand side and change the summation limits but the result looks rather different to the right-hand side.
How can I show this without listing terms?
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You can use the following trick to only count the even terms:
$$\frac{1}{2}\sum_{r=1}^n\frac{1}{r} = \sum_{r=1}^n\frac{1}{2r} = \sum_{r=1}^{2n}\frac{1}{r} \cdot \frac{(-1)^{r}+1}{2}$$
And a similar trick to only count the odd terms. Notice that I've manually added two terms $\left[\frac{1}{2n+1}-\frac{1}{1}\right]$ to make it work out on the given range.
$$\sum_{r=1}^n\frac{1}{2r+1} = \left[\frac{1}{2n+1}-\frac{1}{1}\right] + \sum_{r=1}^{2n}\frac{1}{r} \cdot \frac{(-1)^{r+1}+1}{2}$$
Combing the two sums gives:
$$\left[\frac{1}{2}\sum_{r=1}^n\frac{1}{r}\right] + \left[\sum_{r=1}^n\frac{1}{2r+1}\right]$$
$$=$$
$$\left[\sum_{r=1}^{2n}\frac{1}{r} \cdot \frac{(-1)^{r}+1}{2}\right] + \left[\left[\frac{1}{2n+1}-\frac{1}{1}\right] + \sum_{r=1}^{2n}\frac{1}{r} \cdot \frac{(-1)^{r+1}+1}{2}\right]$$
$$=$$
$$\left[\frac{1}{2n+1}-\frac{1}{1}\right] + \sum_{r=1}^{2n}\frac{1}{r} \left( \frac{(-1)^{r}+1}{2} + \frac{(-1)^{r+1}+1}{2}\right)$$
$$=$$
$$\left[\frac{1}{2n+1}-\frac{1}{1}\right] + \sum_{r=1}^{2n}\frac{1}{r}$$
$$=$$
$$\sum_{r=2}^{2n+1}\frac{1}{r}$$
$$=$$
$$\sum_{r=1}^{2n}\frac{1}{r+1}$$
But honestly, I think your notation (listing the terms) is formal enough and easier to understand :D
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Proving $\tan^{-1}\frac{1}{2\cdot1^{2}}+\tan^{-1}\frac{1}{2\cdot2^{2}}+\cdots+\tan^{-1}\frac{1}{2\cdot n^{2}}=\frac{\pi}{4}-\tan^{-1}\frac{1}{2n+1}$
By using Mathematical Induction, prove the following equation for all positive integers $n$:
$$\tan^{-1}\frac{1}{2 \cdot 1^{2}} + \tan^{-1}\frac{1}{2 \cdot 2^{2}} +\cdots+\tan^{-1}\frac{1}{2 \cdot n^{2}} = \frac{\pi}{4}-\tan^{-1}\frac{1}{2n+1}$$
I am able to prove the statement for n=1 and I got the assumption statement when $n = k$ as follows:
$$\tan^{-1}\frac{1}{2 \times 1^{2}}+\tan^{-1}\frac{1}{2 \times 2^{2}}+\cdots+\tan^{-1}\frac{1}{2 \times k^{2}}=\frac{\pi}{4}-\tan^{-1}\frac{1}{2k+1}$$
The problem is when I have to prove the case for n=k+1. I'm supposed to prove the following:
$$\tan^{-1}\frac{1}{2 \times 1^{2}}+\tan^{-1}\frac{1}{2 \times 2^{2}}+\cdots+\tan^{-1}\frac{1}{2 \times (k+1)^{2}} = \frac{\pi}{4}-\tan^{-1}\frac{1}{2k+3}$$
I simplified the left hand side - with the assumption - into
$$\frac{\pi}{4}-\left(\tan^{-1}\frac{1}{2k+1}-\tan^{-1}\frac{1}{2(k+1)^{2}}\right)$$
I understand that I am supposed to use the identity
$$\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)$$
However, I still am unable to obtain the right-hand side from the expression I have right now.
Is there something I am missing?
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All you need to do is to prove
$$ \tan^{-1}\dfrac{1}{2k+1}-\tan^{-1}\dfrac{1}{2(k+1)^{2}} = \tan^{-1} \dfrac{1}{2k+3}.$$
You can use the formula you mentioned in this way:
$$ \begin{align}\tan^{-1}\dfrac{1}{2k+1} - \tan^{-1} \dfrac{1}{2k+3} &= \tan^{-1} \dfrac{\frac{1}{2k+1}-\frac{1}{2k+3}}{1+\frac{1}{2k+1}\frac{1}{2k+3}}\\
&= \tan^{-1} \dfrac{2}{(2k+1)(2k+3)+1}\\
&= \tan^{-1} \dfrac{1}{2(k+1)^2}.
\end{align}$$
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How to write $[(2+\sqrt{3})^n + (2-\sqrt{3})^n + (2+\sqrt{3})^{n+1} + (2-\sqrt{3})^{n+1}]/6$ to the form $a^2 + 2 b^2$ ($a, b \in \mathbb{N}$). We know that $(2+\sqrt{3})^n + (2-\sqrt{3})^n$ is an integer (See here).
However, we want to write the formula
\begin{align}
&\frac{3+\sqrt{3}}{6} (2+\sqrt{3})^n + \frac{3-\sqrt{3}}{6} (2-\sqrt{3})^n\\
&=\frac{1}{6} \left[(2+\sqrt{3})^{n+1} + (2-\sqrt{3})^{n+1} + (2+\sqrt{3})^n + (2-\sqrt{3})^n\right]
\end{align}
to the form
$$a^2 + 2\,b^2,\ (a, b \in \mathbb{N}).$$
How?
|
Let $f(n)=
\dfrac{3+\sqrt{3}}{6} (2+\sqrt{3})^n + \dfrac{3-\sqrt{3}}{6} (2-\sqrt{3})^n$ and $g(n)=\dfrac{\sqrt3} 6(2+\sqrt3)^n-\dfrac{\sqrt3}6(2-\sqrt3)^n$.
Can you show $f(n)^2+2\times g(n)^2=f(2n)$ and $f(n)^2+2\times g(n+1)^2=f(2n+1)?$
|
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|
What is $\cos\left(\frac{\arctan(x)}{3}\right)$ and $\sin\left(\frac{\arctan(x)}{3}\right)$? I know that $$\cos(\dfrac{\pi}{3} - \arctan(x))= \dfrac{1}{2\sqrt{(1+x^2)}} + \dfrac{\sqrt{3}x}{2\sqrt{(1+x^2)}}$$
$\cos\left(\dfrac{\pi}{3} - \dfrac{\arctan(x)}{3}\right)$ = ?
$\cos\left(\dfrac{\pi}{3} - \dfrac{\arctan(x)}{3}\right) = \cos\left(\dfrac{\pi}{3}\right)\cos\left(\dfrac{\arctan(x)}{3}\right) + \sin\left(\dfrac{\pi}{3}\right)\sin\left(\dfrac{\arctan(x)}{3}\right) = \dfrac{1}{2}\cos\left(\dfrac{\arctan(x)}{3}\right) + \dfrac{\sqrt{3}}{2}\sin\left(\dfrac{\arctan(x)}{3}\right)$
but I can't go further since I don't know how to solve $\sin\left(\dfrac{\arctan(x)}{3}\right)$ and $\cos\left(\dfrac{\arctan(x)}{3}\right)$.
Any suggestion?
|
Triple cosine formula is more well-known: $\cos(3\theta)=4\cos^3\theta-3\cos\theta$. If we let $3\theta=\arctan x$ and $y=\cos\theta$, then
$4y^3-3y=\cos3\theta=\cos\arctan x=\frac{1}{\sqrt{x^2+1}}$. Hence, we have to solve the equation
$$y^3-\frac34y-\frac{1}{4\sqrt{x^2+1}}=0.$$
Its Lagrange resolvent is $z^2--\frac{1}{4\sqrt{x^2+1}}+\frac{1}{4}=0$ with solution $z_{1,2}=\frac{1\pm xi}{8\sqrt{x^2+1}}.$
Hence the real solution of the equation is
$$y=\frac{(1+xi)^{1/3}+(1-xi)^{1/3}}{2(x^2+1)^{1/6}}.$$
For $x=1$ guess what? $y=\cos 15^{\circ}$... We don't need WA for this particular example but when $x=2$ we need.
|
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"timestamp": "2023-03-29T00:00:00",
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|
A geometry problem about triangle angles and perimeter Consider $\Delta ABC$ with three acute angles, we draw its altitudes and make $\Delta MNP$ triangle
if $\frac{PN}{KN}=\frac{3}{2}$ and $\frac{\sin{\alpha}}{\cos{\frac{\alpha}{2}}}+\frac{\sin{\theta}}{\cos{\frac{\theta}{2}}}+\frac{\sin{\gamma}}{\cos{\frac{\gamma}{2}}}=\frac{288}{100}$ then calculate $\frac{MN}{AB+BC+CA}$
Note that $\alpha,\theta,\gamma$ are angles of $\Delta MNP$ and $K$ is the point of concurrency of $MN$ and $CP$
I think it is a famous geomtry problem, I can't remember where I saw this first time but I think it was a famous question...
I thought on this problem a lot but I have no idea to solve that, except that the fraction $\frac{288}{100}$ is $2*\frac{144}{100}$ and I think I should use of this... Maybe I should radical this fraction.
Am I right?
|
I will attempt to solve this problem with as little trigonometry as possible.
The value of $\frac{144}{100}=(1.2)^2$ is actually a red herring. First we note by $a$, $b$, $c$, $\angle{A}$, $\angle{B}$, $\angle{C}$, $S$, $R$ and $r$ the sides, angles, area, circumradius and inradius of $ABC$. Note that: $$\frac{\sin \alpha}{\cos \frac{\alpha}{2}}=\frac{2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}=2\sin\frac{\alpha}{2}=2\sin\frac{\angle{NMP}}{2}=2\sin\angle{AMP}=2\sin\angle{AMP}=2\sin\angle{NBA}=2\sin(90^{\circ}-\angle{AMP})=2\cos\angle{BAC}=2\cos A$$
So we have that $\cos A+\cos B+\cos C=\frac{144}{100}$. Now we will prove that in any triangle we have $\cos A+\cos B+\cos C = 1+\frac{r}{R}$. It can be proven in many ways but one of the nicer ones is this:
Consider the midpoints $D$, $E$, $F$ of $BC$, $CA$, $AB$ respectively which are also the projections of point $O$ - the circumcentre of $ABC$ onto its sides. Denoting by $x$, $y$ and $z$ the lengts of $OD$, $OE$, $OF$ and applying Ptolemy theorem to the cyclic quadrilateral $AEOF$ we obtain:
$$AE \cdot OF + AF \cdot OE = AO \cdot EF$$
$$\frac{b}{2} \cdot z + \frac{c}{2} \cdot y = R \cdot \frac{a}{2}$$
$$bz+cy=aR$$
Writing analogous equations and adding them up we get:
$$x(b+c)+y(c+a)+z(a+b)=R(a+b+c)$$
Since $ax$ is twice the area of $BOC$ and similarly for $by$ and $cz$, $ax+by+cz=2S$ and so:
$$(x+y+z)(a+b+c)=x(b+c)+y(c+a)+z(a+b)+(ax+by+cz)=R(a+b+c)+2S$$
dividing by $(a+b+c)$ and using the fact that $2P=r(a+b+c)$ we get:
$$x+y+z=r+R$$
It's a nice result, but how does it connect to our sum of cosines? Just notice that $\angle{DOB}=\frac{1}{2}\angle{BOC}=A$ so in triangle $BOD$ we have $\cos A=\cos \angle{DOB}=\frac{DO}{OB}=\frac{x}{R}$. Writing analogous equations we obtain:
$$\cos A + \cos B + \cos C = \frac{x}{R}+\frac{y}{R}+\frac{z}{R}=\frac{x+y+z}{R}=\frac{R+r}{R}=1+\frac{r}{R}$$
OK, so far we have $\frac{144}{100}=\cos A+\cos B+\cos C=1+\frac{r}{R}$ so $\frac{r}{R}=0.44$.
Now we will derive the formula for the perimeter of triangle $MNP$. To do this note that reflecting $M$ across $AB$ and $AC$ results in points $Y$ and $Z$ which lie on $PN$. Moreover we have:
$$MN+NP+PM=ZN+NP+PY=YZ$$
So this perimeter is equal to the length of $YZ$. Its half is therefore equal to the length of $Y'Z'$ where $Y'$ and $Z'$ are midpoints of $MY$ and $MZ$ which are also projections of $M$ onto $AB$ and $AC$. Now if we define $A'$ as the antipode of $A$ on the circumcircle of $ABC$ we can say that the quadrilaterals $AY'MZ'$ and $ACA'B$ are (inversely) similar. This in turn yields that the ratios of their diagonals are equal i.e.:
$$\frac{Y'Z'}{AM}=\frac{BC}{AA'}=\frac{a}{2R}$$
Since $a \cdot AM = 2S$ we have:
$$MN+NP+PM=YZ=2Y'Z'=\frac{2AM \cdot a}{2R}=\frac{4S}{2R}=\frac{2S}{R}=\frac{(a+b+c)r}{R}$$
That means that the ratio of the perimeters of $MNP$ and $ABC$ is $\frac{r}{R}=0.44$.
Now let's tackle our main problem - by the angle bisector theorem we have:
$$\frac{3}{2}=\frac{PN}{KN}=\frac{PM}{KM}=\frac{PN+PM}{KN+KM}=\frac{PN+PM}{MN}=\frac{PN+PM+MN}{MN}-1$$
Where in the middle we used the fact that if $\frac{a}{b}=\frac{c}{d}$ then their common value is also equal to $\frac{a+c}{b+d}$. So:
$$\frac{MN}{PN+PM+MN}=\frac{2}{5}=0.4$$
And finally:
$$\frac{MN}{AB+BC+CA}=\frac{MN}{a+b+c}=\frac{MN}{PN+PM+MN} \cdot \frac{PN+PM+MN}{a+b+c}=0.4 \cdot 0.44=0.176$$
|
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|
Power series differential check $(x-1)y'' + y' = 0$ I'm a bit stuck on how to solve this:
$$(x-1)y'' + y' = 0 $$
so assuming y is a solution in this form:
$$\sum_{n=0}^\infty C_nx^n$$
$$\sum_{n=1}^\infty nC_nx^{n-1}$$
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-2}$$
subbing in and distributing:
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-1} - \sum_{n=2}^\infty n(n-1)C_nx^{n-2} + \sum_{n=1}^\infty nC_nx^{n-1}$$
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-1} - \sum_{n=1}^\infty (n+1)(n)C_{n+1}x^{n-1} \sum_{n=1}^\infty nC_nx^{n-1}$$
start at n=2
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-1} - \sum_{n=2}^\infty (n+1)(n)C_{n+1}x^{n-1} + \sum_{n=2}^\infty nC_nx^{n-1} - 2C_2 + C_1$$
so $-2C_2 + C_1 = 0$
and $$(n(n-1)C_n - (n+1)(n)C_{n + 1} + nC_n = 0$$
and $$(n^2C_n - (n+1)(n)C_{n + 1} = 0$$
and $c_{n+1} = \frac{nC_n}{n+1}$ for $n \ge 2$
so writing out the first few terms:
$c_0 = c_0$ and $c_1 = c_1$ and $c_2 = \frac{c_1}{2}$ and $c_3 = \frac{2c_2}{3}$ and $c_4 = \frac{3c_3}{4} = \frac{3 \cdot 2 \cdot c_1}{4!}$ and so $$c_n = \frac{c_1}{n}$$
so can I write that $$y = c_0 + c_1 + \sum_{n=2}^\infty \frac{x^n}{n}$$
|
You initially wrote that,
$$ y = \sum_{n=0}^\infty c_n x^n,$$
and then you determined that for $n \geq 1$ we have $c_n = c_1/n$.
$$ y = c_0 + \sum_{n=1}^\infty c_n x^n,$$
$$ y = c_0 + \sum_{n=1}^\infty \frac{c_1}{n} x^n,$$
$$ y = c_0 + c_1 \sum_{n=1}^\infty \frac{x^n}{n},$$
which is consistent with the answer given by @peter in the comments.
|
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|
find the number of perfect squares of the form : $n^6 + n^4+1$ find the number of perfect squares of the form -
$$n^6 + n^4+1$$
where n is a natural number.
I've reached this equation.
$$n^4(n^2+1)=(k+1)(k-1)$$
I thought this might have something to do with prime factorization but I don't seem to be getting anywhere.
Am I on the right track or should I approach this question differently?
|
We do not have a solution for $n=1$, so $n\geq 2$. Because $(k-1, k+1)\leq 2$ we should have $k+1\geq \frac{n^4}{2}$ and $k-1\leq 2(n^2+1)$. $n^4-2(2(n^2+1)+2)=n^4-4n^2-8=(n^2-2)^2-12>0$ for $n\geq3$. That means we only need to check $n=2: 64+16+1=81=9^2$ - unique solution
Also, there is an alternative way. Considering mod 4 you can show that $n$ should be even. After that consider $(n^3+\frac{n}{2})^2=n^6+n^4+\frac{n^2}{4}> n^6+n^4+1$ for $n>2$ and $(n^3+\frac{n}{2}-1)^2=n^6+n^4-2n^3+\frac{n^2}{4}-n+1=n^6+n^4+1-(n^3-\frac{n^2}{4}-n)<n^6+n^4+1$ for $n>2$
So again you only need to check 2 cases.
|
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|
using trigonometric identities For proving $$\frac {16}{\cos (4x)+7} =\frac{1}{\sin^4x +\cos^2x} +\frac{1}{\sin^2x +\cos^4x} $$
I tried to use that:
\begin{align}
\sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\
&=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\
&=1^2-\frac{1}{2}(2\sin x\cos x)^2\\
&=1-\frac{1}{2}\sin^2 (2x)\\
&=1-\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)\\
&=\frac{3}{4}+\frac{1}{4}\cos 4x
\end{align}
but i can't try more
|
Note that
$$\begin{align}\sin^4x +\cos^2x &=
\sin^2x(1-\cos^2x) +\cos^2x\\
&=\sin^2x+\cos^2x(1-\sin^2x)=\sin^2x +\cos^4x.
\end{align}$$
Hence, according to your work,
$$\begin{align}
2(\sin^4x +\cos^2x)&=(\sin^4x +\cos^2x) +(\sin^2x +\cos^4x)\\
&=\sin^4x +\cos^4x+1=\frac{7+\cos 4x}{4}.\end{align}$$
Can you take it from here?
|
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|
Matrix Normalization From the eigenvectors matrix:
I did normalization
but I think there's an error I could not find.
|
Let$$
v_1=\begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix}
\quad
v_2=
\begin{bmatrix}
1 \\
\frac{-\rho_2 + \sqrt{((\rho_2)^{2} + 8(\rho_1)^2)}}{2\rho_1} \\
1
\end{bmatrix}
\quad
v_3=
\begin{bmatrix}
1 \\
\frac{-\rho_2 - \sqrt{((\rho_2)^{2} + 8(\rho_1)^2)}}{2\rho_1} \\\\
1
\end{bmatrix}
$$
the column vectors of matrix $\Phi^\ast$. Note that $\|v_1\|=\sqrt{2}$ and
\begin{align}
\|v_2\|
=&
\sqrt{
1^2+\left( \frac{-\rho_2 + \sqrt{((\rho_2)^{2} + 8(\rho_1)^2)}}{2\rho_1} \right)^2+1^2}
\\
=&
\sqrt{
2+\frac{\rho_2^2-2\cdot \rho_2\cdot \sqrt{((\rho_2)^{2} + 8(\rho_1)^2)} + ((\rho_2)^{2} + 8(\rho_1)^2)}{4\rho_1^2}}
\\
=&
\sqrt{
\frac{8\rho_1^2+ \rho_2^2-2\cdot \rho_2\cdot \sqrt{((\rho_2)^{2} + 8(\rho_1)^2)} + ((\rho_2)^{2} + 8(\rho_1)^2)}{4\rho_1^2}}
\\
=&
\sqrt{
\frac{\big(16\rho_1^2+ 2\rho_2^2\big)-2\cdot \rho_2\cdot \sqrt{((\rho_2)^{2} + 8(\rho_1)^2)} }{4\rho_1^2}}
\\
=&
\sqrt{
\frac{\big(16\rho_1^2+ 2\rho_2^2\big)-2\cdot \rho_2\cdot \sqrt{((\rho_2)^{2} + 8(\rho_1)^2)}}{4\rho_1^2}}
\\
=&
\frac{\sqrt{16\rho_1^2+ 2\rho_2^2-2\cdot \rho_2\cdot S}}{2\rho_1}
\end{align}
Similarly, we have
$$
\|v_3\|= \frac{\sqrt{16\rho_1^2+ 2\rho_2^2+2\cdot \rho_2\cdot S}}{2\rho_1}
$$
|
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|
Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^3$.
|
Use the following fact: when $x\to 0$,
$$(1+x)^{\alpha}-1\sim\alpha x.$$
So
$$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n}=n\left(\sqrt[3]{1+\frac{3}{n}}-1-\left(\sqrt{1+\frac{2}{n}}-1\right)\right)$$
$$=\frac{\sqrt[3]{1+\frac{3}{n}}-1-\left(\sqrt{1+\frac{2}{n}}-1\right)}{\frac{1}{n}},$$
use the above fact:
$$\sqrt[3]{1+\frac{3}{n}}-1 \sim \frac{1}{n},\sqrt{1+\frac{2}{n}}-1\sim \frac{1}{n},$$
and this implies
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )=0.$$
In fact:
$$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n}\sim -\frac{1}{2n},n\to\infty.$$
|
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|
Show $ f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$ ,$\ g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}$ are bounded on $[0, \infty)$. If $f(x), g(x)$ are defined as following on $[0 , \infty)$,
$$\tag 1 f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$$
$$\tag 2 g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}.$$
. Then how to show that $f,g$ are bounded function on $[0 , \infty)$?
I found this problem on this
Is $f(x)=\sum_{n=1}^\infty\frac{nx^2}{n^3+x^3}$ uniformly continuous on $[0,\infty)$?.
In solution's last steps, i don't know why this is correct answer.
Could you explain for me to elaborate?
Thank you.
|
First note that for $x \ge 0$ and $n \in \Bbb N$
$$
\frac{x^4n}{(n^3 + x^3)^2} =\frac{x^3}{n^3+x^3}\cdot\frac{nx}{n^3 + x^3}
< \frac{nx}{n^3 + x^3}
$$
and therefore $g(x) \le f(x)$, so that it suffices to show that the function $f$ is bounded on $[0, \infty)$.
For $0 \le x \le 1$ we have
$$
\frac{nx}{n^3 + x^3} \le \frac{1}{n^2} \implies f(x) \le \sum_{n=1}^\infty \frac{1}{n^2} .
$$
For fixed $x >1$ and $m =1, 2, 3, \ldots$ consider all $n$ with $x(m-1) \le n < xm$. For each of these $n$,
$$
\frac{nx}{n^3 + x^3} \le \frac{mx^2}{(m-1)^3x^3 + x^3} = \frac 1x \cdot \frac{m}{(m-1)^3+1}
$$
and there are at most $\lfloor x \rfloor +1$ of such $n$. Therefore
$$
f(x) \le \frac{\lfloor x \rfloor +1}{x} \sum_{m=1}^\infty \frac{m}{(m-1)^3+1} \le 2 \sum_{m=1}^\infty \frac{m}{(m-1)^3+1}
$$
for $x > 1$.
Previous solution (more complicated): The idea for estimating $f(x)$ is to replace the infinite sum by a “similar” integral:
$$
\sum_{n=1}^\infty \frac{nx}{n^3+x^3} \lessapprox \int_0^\infty \frac{ux}{u^3+x^3} \, dx = \int_0^\infty \frac{v}{v^3+1} \, .
$$
The two integrals are equal via the substitution $u=xv$, and the last integral is independent of $x$, so that we get a uniform upper bound. Of course the “approximate inequality” must be stated and proved precisely, so here are the gory details:
For fixed $x>0$ we consider the function $\varphi$ defined on $[0, \infty)$ by
$$
\varphi(t) = \frac{tx}{t^3 + x^3} \,.
$$
It is easy to see (by calculating the derivative) that $\varphi $ is increasing on $[0, \frac{x}{2^{1/3}}]$ and decreasing on $[\frac{x}{2^{1/3}}, \infty)$.
If $\frac{x}{2^{1/3}} \le 1$ then $\varphi$ is decreasing on $[1, \infty)$, so that each term in the sum of $f(x)$ (with the exception of the first term) can be estimate above by an integral over $\varphi$:
$$
f(x) = \varphi(1) + \sum_{n=2}^\infty \varphi(n) \\
\le \varphi(1) + \sum_{n=2}^\infty \int_{n-1}^n \varphi(u) \, du
= \varphi(1) + \int_1^\infty \varphi(t) \, dt
\le 1 + \int_0^\infty \frac{ux}{u^3+x^3} \, du \,
$$
and with the substitution $u = xv$ we get
$$ \tag{*}
f(x) \le 1 + \int_1^\infty \frac{v}{v^3+1} \, dv \, .
$$
If $\frac{x}{2^{1/3}} > 1$ then we can proceed similarly. With $m = \lfloor \frac{x}{2^{1/3}} \rfloor$ we estimate
$$
f(x) = \sum_{n=1}^{m-1} \varphi(n) + \varphi(m) + \varphi(m+1) + \sum_{n=m+2}^{\infty} \varphi(n) \\
\le \int_0^\frac{x}{2^{1/3}} \varphi(u) \, du + 2 \varphi(\frac{x}{2^{1/3}}) + \int_\frac{x}{2^{1/3}}^\infty \varphi(u) \, du \\
= \frac{4}{ 2^{1/3} \cdot 3x} + \int_0^\infty \frac{ux}{u^3+x^3} \, du \\
< 1 + \int_0^\infty \frac{ux}{u^3+x^3} \, du = 1 + \int_0^\infty \frac{v}{v^3+1} \, dv
$$
so that $(*)$ holds as well.
|
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|
In diophantine $3b^2=a^2$ where $a$ and $b$ are coprime, does $3|a$? Integers $a$ and $b$ are co-prime and $3\cdot b^2=a^2$.
$3\cdot b^2=a^2$, implies $a^2$ is divisible by 3 since, $3b^2$ is divisible by 3.
Is $a$ divisible by 3?
|
Yes, $a$ must be divisible by $3$. If the integer $a=3k+1$ or $3k+2$ for arbitrary $k$, (which are the cases in which $a$ is not divisible by $3$) then the remainder is $9k^2+6k+1\equiv1\pmod3$ or $9k^2+12k+4\equiv1\pmod3$.
Therefore, $a^2|3$ cannot occur if $a$ is not divisible by $3$, and thus, $a$ must be divisible by $3$.
However, the equation $3a^2=b^2$ cannot be solved for coprime integers, so if you consider that, then it is impossible for $a|3$ since there are no $a$ that fit the equation.
|
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|
Efficient computation of conjugacy classes of a small group. I'm trying to construct a character table for a group of order 54 given by:
$$ \langle a,b : a^9 = b^6 = 1, b^{-1} a b = a^2\rangle $$
To do this first I need to compute conjugacy classes. This feels like a tedious task and I'm looking for some guidance on how to compute those efficiently.
|
$G = \{a^mb^n:0\le m\le 8,0\le n\le 5\}$. For an element $a^mb^n\in G$, $b^{-1}a^mb^nb = a^{2m}b^n$ and thus for a fixed $n$, $a^mb^n$ (with $m = 1,2,4,5,7,8$ or with $m = 3,6$) are in a same conjugacy class. Moreover, $$a^{-1}a^mb^na = a^{m-1}b^nab^{-n}b^n = a^{m-1}a^{5^n}b^n$$since $ba^2b^{-1} = a = (a^2)^5$ and $a^2$ is also a generator of $\langle a\rangle$. Note that a conjugate by $a$ or $b$ does not change $n$. Therefore, we only need to discuss $n$.
(1) $n=0$, then $a^{-1}a^mb^na = a^mb^n$, so the conjugate by $a$ is fixed. So $\{1\},\{a,a^2,a^4,a^5,a^7,a^8\},\{a^3,a^6\}$ are conjugacy classes.
(2) $n=1$, then $a^{-1}a^mb^na = a^{m+4}b^n$. So $\{b,ab,\dots, a^8b\}$ is a conjugacy class, since $m+4$ runs all the $m$'s.
(3) $n=2$, then $a^{-1}a^mb^na = a^{m+24}b^n = a^{m+6}b^n$. Now $a^3b^2$, $a^6b^2$ and $b^2$ are in the same conjugacy class, while other $m$ forms another conjugacy class.
(4) $n=3$, then $a^{-1}a^mb^na = a^{m+124}b^n = a^{m+7}b^n$. The same as (2) since it runs all the $m$'s.
(5) $n=4$, then $a^{-1}a^mb^na = a^{m+3}b^n $. The same as (3).
(6) $n=5$, then $a^{-1}a^mb^na = a^{m+1}b^n$. The same as (2).
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(1+AB)(A^{3} + B^{3})^{2} = (AB)^{2} + 4AB + 4 $$
$$ A^{6} + B^{6} + BA^{7} + A B^{7} = -2 (AB)^{4} - 2(AB)^{3} + (AB)^{2} + 4AB + 4 $$
I also have tried using$A = (1+x), B = (1-x)$ , and some others, but none solves the problem.
I am now trying $A = (1+x)$ and $(1-x) = -(1+x) + 2 = 2 - A$, so:
$$\sqrt{1 + \sqrt{A(2-A)}}\left(\sqrt{(A)^{3}} + \sqrt{(2-A)^{3}} \right) = 2 + \sqrt{A(2-A)} $$
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Let us first simplify the expression. Assume $a^2=1-x$ $ $ & $ $ $b^2=1+x$
Please note that $a^2+b^2=2$
Now our equation becomes $\sqrt{1+ab}(a^3+b^3)=2+ab$
$\sqrt{\frac{a^2+b^2+2ab}{2}}(a+b)(a^2+b^2-ab)=2+ab$ which further simplifies to
$\frac{(a+b)^2}{\sqrt{2}}(2-ab)=2+ab$
$\frac{(2+2ab)(2-ab)}{\sqrt{2}}=2+ab$
Find $ab$ and use the equation $a^2b^2=1-x^2$
Hope this will be helpful!
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Find limit in use of integrals Find limit $\lim_{n \rightarrow \infty} \int_{-1}^{1} x^5 \cdot \arctan{(nx)} dx $
From mean-value-theorem we have
$$ \frac{1}{2} c^5 \cdot \arctan{(nc)} \mbox{ for some c } \in (-1,1) $$
$$ \underbrace{\frac{1}{2} \cdot \arctan{(-n)}}_{\rightarrow - \pi /4} \le \frac{1}{2} c^5 \cdot \arctan{(nc)} \le \underbrace{\frac{1}{2} \cdot \arctan{(n)}}_{\rightarrow \pi /4} $$
so this bounding doesn't help me. Has somebody better idea how to bound that?
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We'll simplify with $y=nx$. Integration by parts gives $$\int y^5\arctan ydy=\frac{y^{6}}{6}\arctan y-\frac{1}{6}\int\frac{y^{6}}{1+y^{2}}dy\\=\frac{y^{6}}{6}\arctan y-\frac{1}{6}\int\left(y^{4}-y^{2}+1-\frac{1}{1+y^{2}}\right)dy\\=\frac{y^6+1}{6}\arctan y-\frac{1}{30}y^5+\frac{1}{18}y^3-\frac16 y+C.$$Hence $$\frac{1}{n^6}\int_{-n}^n y^5\arctan ydy=\frac{\frac{n^6}{3}\arctan n+o(n^6)}{n^6}\stackrel{n\to\infty}{\to}\frac{1}{3}\arctan\infty=\frac{\pi}{6}.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2\lim_{n\to\infty}\int_0^1 x^5\arctan nxdx$ (since the integrand is even), which by dominated convergence is $$\pi\int_0^1 x^5dx=\frac{\pi}{6}.$$
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Find minimize value of $P=(a^2+b^2+c^2)(|a-b|+|b-c|+|c-a|)$
For $a,b,c$ are real number satisfied: $a^3+b^3+c^3=3abc+32$. Find minimize value of $P=(a^2+b^2+c^2)(|a-b|+|b-c|+|c-a|)$
This is my try: WLOG, I suppose that: $a\ge b\ge c$.
Then we have $P=(a^2+b^2+c^2)(a-b+b-c+a-c)=(a^2+b^2+c^2)(2a-2c)$
In the other hand, we also have $(b-a)(b-c)\le 0\iff b^2+ac\le ab+bc\iff b^3+abc\le ab^2+b^2c$.
In addition, we have $(a-b)(b-c)(c-a)\le 0\iff (ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)\le 0$
I have tried exploiting the assumptions that the problem gave, but I can't link these to come final solution !
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Let $a\geq b\geq c$.
Thus, since $(a-c)^2\geq\sum\limits_{cyc}(a^2-ab)$ it's $(a-b)(b-c)\geq0,$ by AM-GM we obtain:
$$P=\sqrt{(a^2+b^2+c^2)^2(a-b+b-c+a-c)^2}=$$
$$=\frac{1}{3}\sqrt{\left((a+b+c)^2+\sum_{cyc}(a-b)^2\right)^2\cdot4(a-c)^2}\geq$$
$$\geq\frac{1}{3}\sqrt{4(a+b+c)^2\sum_{cyc}(a-b)^2\cdot4\sum_{cyc}(a^2-ab)}=$$
$$=\frac{1}{3}\sqrt{32\left((a+b+c)\sum_{cyc}(a^2-ab)\right)^2}=$$
$$=\frac{1}{3}\sqrt{32\left(a^3+b^3+c^3-3abc\right)^2}=\frac{1}{3}\sqrt{32^3}=\frac{128\sqrt2}{3}.$$
The equality occurs for $(a-b)(b-c)=0$, $(a+b+c)^2=\sum\limits_{cyc}(a-b)^2$ and $a^3+b^3+c^3-3abc=32,$ which says that we got a minimal value.
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Exercise about a 2-dimensional random vector I solved the following exercise and want to know, if I did it correct
Consider a 2-dimensional random vector $(X,Y)$ with density
$$f_{X,Y}(x,y)=c\frac{y}{x^3}1_{\{0<x\leq 1\}}1_{\{0<y\leq x^2\}}$$
a) Compute $c$, $f_X$ of $X$, $P[X\leq 1/2]$
b) Compute $E[X/Y]$, $E[Y\mid X=1/2]$
c) Are $X$ and $Y$ independent?
a) use $z=\sqrt{y}$ and $2\sqrt{y}\mathrm{d}z=\mathrm{d}y$ to get
$$2c\int_0^1 \int_0^x \frac{z^3}{x^3}\mathrm{d}z\mathrm{d}x=\frac{c}{2}\int_0^1 x\mathrm{d}x=\frac{c}{4}\implies c=4$$
$$f_X=2x1_{\{0<x\leq 1\}}$$
$$P[X\leq 1/2]=2\int_0^{1/2} x\mathrm{d}x=2[\frac{x^2}{2}]^{1/2}_0=\frac{1}{4}$$
b) $$E[X/Y]=\int \int \frac{x}{y} f_{X,Y}(x,y) \mathrm{d}y\mathrm{d}x=4\int_0^1 \int_0^{x^2} \frac{x}{y}\frac{y}{x^3}\mathrm{d}y\mathrm{d}x$$
$$=4\int_0^1 \int_0^{x^2} \frac{1}{x^2}\mathrm{d}y\mathrm{d}x = 4\int_0^1 \frac{x^2}{x^2}\mathrm{d}x =4$$
$$E[Y\mid X=1/2]=\int y \frac{f_{X,Y}(1/2,y)}{f_{X}(1/2)}\mathrm{d}x \mathrm{d}y=\int y \frac{f_{X,Y}(1/2,y)}{f_{X}(1/2)}\mathrm{d}x \mathrm{d}y
= 4\int_0^{1/4} y\cdot \frac{y}{1/8} \cdot (\frac{2}{2})^{-1}\mathrm{d}y$$
$$= 32 \int_0^{1/4} y^2\mathrm{d}y = \frac{32}{192}$$
c) I would say no because of the indicator function of $y$ depending on $x^2$, but I am not sure how to prove this.
$$f_Y=\frac{f_{X,Y}}{f_{X}}=\frac{4\frac{y}{x^3}1_{\{0<x\leq 1\}}1_{\{0<y\leq x^2\}}}{2x1_{\{0<x\leq 1\}}}=2\frac{y}{x^4}1_{\{0<y\leq x^2\}}$$
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Since the joint pdf factorises as $$f_{X,Y}(x,y)=\frac{2y}{x^4}\,\mathbf1_{0<y<x^2}\underbrace{2x\mathbf1_{0<x<1}}_{f_X(x)}\quad,$$
this automatically shows that the conditional density of $Y\mid X$ is
$$f_{Y\mid X=x}(y)=\frac{2y}{x^4}\mathbf1_{0<y<x^2}$$
This distribution depends on $X$, so that rules out independence of $X$ and $Y$.
And from the pdf you get $$E(Y\mid X=x)=\int_0^{x^2}\frac{2y^2}{x^4}\,dy=\frac{2x^2}{3}$$
This matches with your answer. The rest also looks correct.
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|
Evaluating $\int_0^{\infty} \frac{1}{1-x^2+x^4}\ln\left(\frac{x^2}{1-x^2+x^4}\right)dx$ Can we find the exact value expressed by elementary functions of
$$\int_0^{\infty} \frac{1}{1-x^2+x^4}\ln\left(\frac{x^2}{1-x^2+x^4}\right)dx?$$
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$$\mathcal I=\int_0^{\infty} \frac{1}{1-x^2+x^4}\ln\left(\frac{x^2}{1-x^2+x^4}\right)dx\overset{x\to \frac{1}{x}}=\int_0^{\infty} \frac{x^2}{1-x^2+x^4}\ln\left(\frac{x^2}{1-x^2+x^4}\right)dx$$
Summing up the two integrals from above and rearranging the logarithm gives:
$$2\mathcal I= -\int_0^{\infty} \frac{1+x^2}{1-x^2+x^4}\ln\left(\frac{1-x^2+x^4}{x^2}\right)dx=-\int_0^\infty \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}-1}\ln\left(x^2+\frac{1}{x^2}-1\right)dx$$
$$\Rightarrow \mathcal I=-\frac12 \int_0^\infty \frac{\left(x-\frac{1}{x}\right)'}{1+\left(x-\frac{1}{x}\right)^2}\ln\left(1+\left(x-\frac{1}{x}\right)^2\right)dx\overset{x-\frac{1}{x}=u}=-\frac12\int_{-\infty}^\infty \frac{\ln(1+u^2)}{1+u^2}du$$
$$\overset{u=\tan x}=-\int_0^\frac{\pi}{2} \ln(1+\tan^2 x)dx=2\int_0^\frac{\pi}{2} \ln(\cos x)dx=-\pi\ln 2$$
The integral from above can be found here.
Generalization.
$$\mathcal I(n)=\int_0^{\infty} \frac{1}{1-x^2+x^4}\ln^n\left(\frac{x^2}{1-x^2+x^4}\right)dx=\frac{(-1)^n}{2}\int_{-\infty}^\infty \frac{\ln^n(1+x^2)}{1+x^2}dx$$
$$=2^n \int_0^\frac{\pi}{2}\ln^n(\cos x)dx=\sqrt{\pi}2^{n-1} \lim_{s\to 0} \frac{d^n}{ds^n}\frac{\Gamma\left(\frac{s+1}{2}\right)}{\Gamma\left(\frac{s}2 +1\right)}$$
Which gives for example: $$\int_0^{\infty} \frac{1}{1-x^2+x^4}\ln^2\left(\frac{x^2}{1-x^2+x^4}\right)dx=\frac{\pi^3}{6}+2\pi \ln^2 2$$
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How do I solve this log equation for x? $3\log _5\left(\sqrt{2+x}\right)+\log _5\left(\sqrt{2-x}\right)-3=\log _5\left(\sqrt{4-x^2}\right)$ $3\log _5\left(\sqrt{2+x}\right)+\log _5\left(\sqrt{2-x}\right)-3=\log _5\left(\sqrt{4-x^2}\right)$
I only managed to solve up to this step: $\left(2+x\right)^{\frac{3}{2}}\sqrt{2-x}=\sqrt{4-x^2}\cdot \:125$
However, I'm not too sure about squaring both sides as I would have to handle ridiculously large numbers to find x, so I'm not too sure that I'm doing it correctly.
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$$
(\sqrt{2+x})^3\sqrt{2-x}=125\sqrt{4-x^2}\\
(\sqrt{2+x})^3\sqrt{2-x}=125\sqrt{2-x}\sqrt{2+x}\\
(\sqrt{2+x})^2=125\\
2+x=125\\
x=123.
$$
Note that you have the following constraints on $x$:
$$
\sqrt{2+x}>0\implies 2+x>0\implies x>-2\\
\sqrt{2-x}>0\implies 2-x>0\implies x<2
$$
If the original equation has a solution, it should be a value between the numbers $-2$ and $2$. The answer $123$ that we found does not belong to that interval. So, it looks like the original equation does not have solutions in real numbers.
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Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$ I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$
My Attempt:
I start off by parametizing the integral as $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$
where $I=I(1)$. I then differentiate to get $$I'(a)=\int_0^1 \frac{ax\ln(1+x)}{1-ax}dx=\int_0^1 ax\ln(1+x)\sum_{n=0}^\infty(ax)^ndx=\sum_{n=1}^\infty a^{n+1}\int_0^1 x^{n+1}\ln(1+x)dx$$
Evaluating this integral by integration by parts and geometric series I get
$$\int_0^1 x^{n+1}\ln(1+x)dx=\frac{x^{n+2}}{n+2}\ln(1+x)|_0^1-\frac{1}{n+2}\int_0^1 \frac{x^{n+2}}{1+x}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\int_0^1 x^{n+2}\sum_{k=0}^\infty(-x)^kdx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty(-1)^k\int_0^1 x^{k+n+2}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty\frac{(-1)^k}{k+n+2}=\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)$$ So I arrive at $$I'(a)=\sum_{n=0}^\infty a^{n+1}\left(\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)\right)$$
Re-indexing I get
$$I'(a)=\frac{\ln(2)}{a}\sum_{n=2}^\infty \frac{a^n}{n}+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n}a^{n-1}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n}a^{n-1}$$Integrating both sides from $0$ to $1$ I recover $I(1)$
$$I(1)=\int_0^1 \frac{\ln(2)}{a}\left(-\ln(1-a)-a\right)da+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ Then using the integral equation for the Dilogarithm I arrive at
$$I(1)=\ln(2)\int_0^1 -\frac{\ln(1-a)}{a}da-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$
$$I(1)=\frac{\ln(2)\pi^2}{6}-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$
At this point I could not continue further as I did not know how to simplify the Digamma terms in the sums. I think that by using the Digamma function's relation to the Harmonic Numbers it could be possible to exploit known values of Harmonic sums to arrive at the answer but I could not get the sums in a form where this would work. If anyone could help me continue further or let me know if I am on the right track I would greatly appreciate it. Thank you in advance.
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Some other basic method:
$$\int_0^1 -\log(1+x)\log(1-x) \, {\rm d}x \\
= (1-x)\log(1+x)\log(1-x)\Big|_0^1 - \int_0^1 (1-x) \left[ \frac{\log(1-x)}{1+x} - \frac{\log(1+x)}{1-x}\right] {\rm d}x \\
\stackrel{t=1-x}{=} (1+x)\left[\log(1+x)-1\right] \Big|_0^1 - \int_0^1 \frac{t/2 \cdot \log t}{1-t/2} \, {\rm d}t \\
=2\log 2 - 1 - \sum_{n=1}^\infty \int_0^1 (t/2)^n \log t \, {\rm d}t \\
=2\log 2-1 + \sum_{n=1}^\infty \frac{2^{-n}}{(n+1)^2} \\
=2\log 2 - 2 + 2\,{\rm Li_2}(1/2) \\
=2\log 2 - 2 + \frac{\pi^2}{6} - \log^22
$$
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|
Proof of the ring isomorphism: $ \mathbb{Z}[\sqrt{7}] /(5+2 \sqrt{7}) \cong \mathbb{Z} / (3) $
I am asked to prove that
$$
\mathbb{Z}[\sqrt{7}] /(5+2 \sqrt{7}) \cong \mathbb{Z} / (3)
$$
First, if we define the following homorphism : $\phi:\mathbb{Z} \rightarrow \mathbb{Z}[\sqrt{7}] /(5+2 \sqrt{7})$, we can try to show that this homomorphism has $\ker(\phi)=(3)$ and is surjective to conclude with the first isomorphism theorem.
It is kind of informal but I realised that in $(5+2 \sqrt{7})$, $$5+2 \sqrt{7} \equiv 0$$
$$5 \equiv -2 \sqrt{7}$$
$$25\equiv 28$$
$$3 \equiv 0$$
But then I struggle to show properly that $\phi (3) = 0$.
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Consider the natural homomorphism $$\varphi \colon \mathbb{Z} \hookrightarrow \mathbb{Z}\left[ \sqrt{7} \right] \twoheadrightarrow \frac{\mathbb{Z}\left[ \sqrt{7} \right]}{\left\langle 5 +2 \sqrt{7} \right\rangle}$$ which sends any integer $n \in \mathbb{Z}$ to its class $n +\left\langle 5+ 2 \sqrt{7} \right\rangle \in \frac{\mathbb{Z}\left[ \sqrt{7} \right]}{\left\langle 5 +2 \sqrt{7} \right\rangle}$.
As you said, it is enough to prove that $\text{Ker}(\varphi) = 3 \mathbb{Z}$ and $\varphi$ is surjective.
We have $3 = -\left( 5 -2 \sqrt{7} \right) \left( 5 +2 \sqrt{7} \right) \in \left\langle 5 +2 \sqrt{7} \right\rangle$, and hence $3 \mathbb{Z} \subset \text{Ker}(\varphi)$. Therefore, either $\text{Ker}(\varphi) = \mathbb{Z}$ or $\text{Ker}(\varphi) = 3 \mathbb{Z}$.
Now, if we had $\text{Ker}(\varphi) = \mathbb{Z}$, there would exist $a, b \in \mathbb{Z}$ such that $1 = \left( a +b \sqrt{7} \right) \left( 5 +2 \sqrt{7} \right)$, which would yield $5 a +14 b = 1$ and $2 a +5 b = 0$ since $\sqrt{7}$ is irrational. This is impossible. Thus, $\text{Ker}(\varphi) = 3 \mathbb{Z}$.
Finally, note that $\sqrt{7} = 2 +\left( 8 -3 \sqrt{7} \right) \left( 5 +2 \sqrt{7} \right)$. Therefore, for all $a, b \in \mathbb{Z}$, we have $$a +b \sqrt{7} +\left\langle 5 +2 \sqrt{7} \right\rangle = \varphi(a +2 b) \, \text{,}$$ which proves that $\varphi$ is surjective.
P.S.: In order to avoid possible confusions, I denote by $\left\langle 5 +2 \sqrt{7} \right\rangle$ the ideal of $\mathbb{Z}\left[ \sqrt{7} \right]$ generated by $5 +2 \sqrt{7}$.
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|
Odd & Even Combinations If you have a range of numbers from 1-49 with 6 numbers to choose from, how many combinations are there containing all odd, all even, only 1 odd, and only 1 even number?
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There are $25$ odd and $24$ even numbers in the range $1 - 49$.
The number of selections of six numbers from the set $\{1, 2, 3, \ldots, 49\}$ that contain exactly $k$ odd numbers and $6 - k$ even numbers is
$$\binom{25}{k}\binom{24}{6 - k}$$
Therefore, the number of selections that contain just odd numbers is $$\binom{25}{6}\binom{24}{0} = \binom{25}{6}$$ The number of selections that contain just even numbers is $$\binom{25}{0}\binom{24}{6} = \binom{24}{6}$$ Selections that contain exactly one odd number must contain five even numbers. The number of such selections is $$\binom{25}{1}\binom{24}{5}$$ Selections that contain exactly one even number must contain five odd numbers. The number of such selections is $$\binom{25}{5}\binom{24}{1}$$
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Prove $\lceil z \rceil=z+\frac12-\frac{\tan^{-1}(\tan(\pi(z+0.5)))}{\pi}$ when $z$ is not an integer How can I prove that $\lceil z \rceil=z+\dfrac12-\dfrac{\tan^{-1}(\tan(\pi(z+0.5)))}{\pi}$ for all non-integer real numbers $z$?
Z cannot be an integer because then tan(pi*z + pi/2) would be undefined. I got this equation by messing around with arcsin(sin(x)), arccos(cos(x)), and arctan(tan(x)) and noticed that arctan(tan(x))-x looked liked a weird negative ceiling function, so I made it x-arctan(tan(x)), and saw a disproportional ceiling function:
Then I changed the equation so that it was the arctan(tan(x)) became $\dfrac{\tan^{-1}(\tan(\pi(z+0.5)))}{\pi}$ and I added 0.5 to make it the ceiling function. I am looking for a mathematical proof that is relatively simple and only uses trig, low level calc, and algebra.
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$\tan (x)$ has a period of $\pi$, so $\tan x = \tan (\pi(\frac{x}{\pi}-\lceil \frac{x}{\pi} \rceil + \epsilon))$, where $\epsilon \in \mathbb{Z}$. Consider $x = \pi(z+\frac{1}{2})$, then
$$z + \frac{1}{2} - \frac{\tan^{-1} \tan (\pi(z+\frac{1}{2}))}{\pi} = z + \frac{1}{2} - \frac{\tan^{-1} \tan (\pi(z+\frac{1}{2} - \lceil z+\frac{1}{2}\rceil+\epsilon)))}{\pi}$$
Because we want to invert the tangent/arctangent with its principal value , we want $z+\frac{1}{2}-\lceil z+\frac{1}{2} \rceil + \epsilon \in (-\frac{1}{2},\frac{1}{2})$. This fixes $\epsilon$. So:
$$z + \frac{1}{2} - \frac{\tan^{-1} \tan (\pi(z+\frac{1}{2} - \lceil z+\frac{1}{2}\rceil+\epsilon)))}{\pi} = z + \frac{1}{2} - \frac{\pi(z+\frac{1}{2} - \lceil z+\frac{1}{2}\rceil+\epsilon))}{\pi} = \\ = \lceil z+\frac{1}{2}\rceil-\epsilon$$
Remember when I said "this fixes $\epsilon$"? This trick is such that $$\lceil z+\frac{1}{2}\rceil-\epsilon = \lceil z \rceil$$
Indeed, if $\lceil z+\frac{1}{2}\rceil = \lceil z \rceil +1$, then $\epsilon = 1$: $$z+\frac{1}{2}-\lceil z+\frac{1}{2} \rceil + \epsilon = z - \lceil z \rceil - \frac{1}{2} + \epsilon\in (-\frac{1}{2},\frac{1}{2}) \Rightarrow \epsilon = 1$$
In the same way, if $\lceil z+\frac{1}{2}\rceil = \lceil z \rceil$, then $\epsilon = 0$.
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|
Find Maxima and Minima of $f( \theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta$ Show that, whatever the value of $\theta$, the expression
$$a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta\ $$
Lies between
$$\dfrac{a+c}{2} \pm \dfrac 12\sqrt{ b^2 + (a-c)^2} $$
My try:
The given expression can be reduced as sum of sine functions as:
$$(a-c) \sin^2 \theta + \dfrac b2 \sin 2 \theta + c \tag{*} $$
Now, there is one way to take everything as a function of $\theta$ and get the expression in the form of $ a \sin \theta + b \cos \theta = c$ and dividing it by $ \sqrt{ a^2 + c^2} $ both sides, but square in sine function is a big problem, also both have different arguments.
Other way, I can think of is taking $ \tan \dfrac \theta 2 = t$ and getting sine and cosine function as $ \sin \theta = \dfrac{ 2t}{1+t^2} $ while cosine function as $ \dfrac{ 1-t^2} {1+t^2}$ solving. So getting $(*)$ as a function of $t$, and simplifying we get,
$$ f(t) = \dfrac{2 Rt + 2 R t^3 + R_0 t - R_0 t^3}{1+t^4 + 2t^2} + c\tag{1}$$
For $R_0 = 2b, R = (a-c)$ , but this is where the problem kicks in!, The Range of given fraction seems $ (-\infty,+ \infty)$ and is not bounded!
So what's the problem here? Can it be solved?
Thanks :)
Edit : I'd like to thank @kaviramamurthy for pointing out that as $t \rightarrow \pm \infty, f(t) \rightarrow c$. That's a mistake here.
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By the double angle formulas, the expression is equivalent to
$$\frac12\left(a+c+(a-c)\cos2\theta+b\sin2\theta\right).$$
Now the expression $(a-c)\cos2\theta+b\sin2\theta$ can be seen as the dot product of a vector with a rotating unit vector, which takes its extreme values when the vectors are parallel or antiparallel, giving
$$\pm\|(a-c,b)\|=\pm\sqrt{(a-c)^2+b^2}.$$
The same result can be obtained by differentiation, or by reducing to the sine addition formula.
Yet another way is by finding the extrema of $(a-c)x+by$ under the constraint $x^2+y^2=1$. Using a Lagrange multiplier, the equations are
$$\begin{cases}x^2+y^2&=1,\\a-c&=2\lambda x,\\b&=2\lambda y,\end{cases}$$
easily giving
$$x=\pm\frac{a-c}{\sqrt{(a-c)^2+b^2}},\\y=\pm\frac{b}{\sqrt{(a-c)^2+b^2}}.$$
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|
Rationalizing $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$ Problem
Rationalize $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$
So, I'm training for the Mexican Math Olimpiad. This is one of the algebra problems from my weekly training.
Before this problem, there was other very similar, after proving it, there's an useful property:
If $a+b+c=0$ then $a^3+b^3+c^3=3abc$
It can be easily proved using the following factorization
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
I tried using this one for the rationalization. I got
$$\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}*\frac{(a^2+b^2+c^2-ab-bc-ac)}{(a^2+b^2+c^2-ab-bc-ac)}=\frac{\sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}-\sqrt[3]{ab}-\sqrt[3]{bc}-\sqrt[3]{ac}}{a+b+c-3\sqrt[3]{abc}}$$
But i didn't know how to proceed. I tried looking into it with wolfram alpha and i got there is no racionalization, so i assume $a+b+c=0 $ (not $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$, because the first expression wouldn't have sense) if we want a solution (In fact, if this happens, i've already rationalized it).
So my truly question is not the answer, but how to prove that $a+b+c$ needs to be 0
I would appreciate any help, ideas or suggestions, thanks.
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In M2:
R=QQ[x,y,z,a,b,c]
I=ideal(x^3-a,y^3-b,z^3-c,x+y+z)
gens gb I
toString oo -- note a^3+3*a^2*b+3*a*b^2+b^3+3*a^2*c-21*a*b*c+3*b^2*c+3*a*c^2+3*b*c^2+c^3
-- writing x^3 for a, y^3 for b and z^3 for c:
factor(x^9+3*x^6*y^3+3*x^6*z^3+3*x^3*y^6-21*x^3*y^3*z^3+3*x^3*z^6+y^9+3*y^6*z^3+3*y^3*z^6+z^9)
tex oo
$\left(x+y+z\right)\left(x^{2}-x\,y-x\,z+y^{2}-y\,z+z^{2}\right)\left(x^{2}-x\,y-x\,z+y^{2}+2\,
y\,z+z^{2}\right)\left(x^{2}-x\,y+2\,x\,z+y^{2}-y\,z+z^{2}\right)\left(x^{2}+2\,x\,y-x\,z+y^{2
}-y\,z+z^{2}\right)$
Which means $\frac1{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}=\frac1{x+y+z}=\frac{(x^2-xy-xz+y^2-yz+z^2)(x^2-xy-xz+y^2+2yz+z^2)(x^2-xy+2xz+y^2-yz+z^2)(x^2+
2xy-xz+y^2-yz+z^2)}{x^9+3x^6y^3+3x^6z^3+3x^3y^6-21x^3y^3z^3+3x^3z^6+y^9+3y^6z^3+3y^3z^6+z^9}=\frac{({\sqrt[3]{a}}^2-{\sqrt[3]{a}}{\sqrt[3]{b}}-{\sqrt[3]{a}}{\sqrt[3]{c}}+{\sqrt[3]{b}}^2-{\sqrt[3]{b}}{\sqrt[3]{c}}+{\sqrt[3]{c}}^2)({\sqrt[3]{a}}^2-{\sqrt[3]{a}}{\sqrt[3]{b}}-{\sqrt[3]{a}}{\sqrt[3]{c}}+{\sqrt[3]{b}}^2+2{\sqrt[3]{b}}{\sqrt[3]{c}}+{\sqrt[3]{c}}^2)({\sqrt[3]{a}}^2-{\sqrt[3]{a}}{\sqrt[3]{b}}+2{\sqrt[3]{a}}{\sqrt[3]{c}}+{\sqrt[3]{b}}^2-{\sqrt[3]{b}}{\sqrt[3]{c}}+{\sqrt[3]{c}}^2)({\sqrt[3]{a}}^2+
2{\sqrt[3]{a}}{\sqrt[3]{b}}-{\sqrt[3]{a}}{\sqrt[3]{c}}+{\sqrt[3]{b}}^2-{\sqrt[3]{b}}{\sqrt[3]{c}}+{\sqrt[3]{c}}^2)}{a^3+3a^2b+3a^2c+3ab^2-21abc+3ac^2+b^3+3b^2c+3bc^2+c^3}$
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|
Proving that $\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.$ Various examinations ask students to prove that $$\frac{1}{{n \choose k}}=(n+1) \int_{0}^{1} x^k (1-x)^{n-k} dx ~~~~~(1)$$ by evaluating the integral $\int_{0}^{1}
(tx+1-x)^n~dt$ two ways. When I came across (1), I could prove that $$\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.~~~~(2)$$ as below: $$S_n=\frac{1}{n+1}\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=\int_{0}^1 \sum_{k=0}^{n} (-1)^k x^k (1-x)^{n-k} dx = \int_{0}^{1} (1-x)^n \sum_{k=0}^{n}\left(\frac{-x}{1-x} \right)^k dx =\int_{0}^{1}(1-x)^n \frac{ \left( \frac{-x}{1-x}\right)^{n+1}-1}{\left(\frac{-x}{1-x}\right)-1}dx$$
$$=\int_{0}^{1}[(1-x)^{n+1}-(-x)^{n+1}] dx =\frac{1+(-1)^n}{n+2}.~~~~(3)$$
Therefore, $$S_{2m+1}=0 ~~~~(4) ~~~ \mbox{and}~~~ S_{2m}= \frac{2m+1}{m+1}.~~~~(5)$$
Now the question is: Can one prove (2) in some other way(s)?
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A telescoping approach:
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}}}
&=\sum_{k=0}^n(-1)^k\frac{k!(n-k)!}{n!}\\
&=\sum_{k=0}^n(-1)^k\frac{k!(n-k)!}{n!}\cdot\frac{(n-(k-1))+(k+1)}{n+2}\\
&=\frac{n+1}{n+2}\sum_{k=0}^n(-1)^k\frac{k!(n-(k-1))!+(k+1)!(n-k)!}{(n+1)!}\\
&=\frac{n+1}{n+2}\sum_{k=0}^n\left((-1)^k\frac{1}{\binom{n+1}{k}}-(-1)^{k+1}\frac{1}{\binom{n+1}{k+1}}\right)\\
&\,\,\color{blue}{=\frac{n+1}{n+2}\left(1+(-1)^n\right)}
\end{align*}
and the claim follows.
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|
Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$ Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$
I proved it by induction:
"$n=1$" $ \quad \sum_{k=0}^{1}{\frac{1}{k!}}\leq 3 \quad \checkmark$
"$n \implies n+1$": $$\quad \sum_{k=0}^{n+1}{\frac{1}{k!}}\leq 3 \\ \left(\sum_{k=0}^{n}{\frac{1}{k!}}\right)+\frac{1}{(n+1)!}\leq 3 \\ $$$$3+\overbrace{\frac{1}{(n+1)!}}^{\geq 0}\leq 3 \\$$ That's of course true.
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For any finite $n$,
$$\sum_{k=0}^n \frac{1}{k!} < e < 3.$$
UPDATE
$$1+1+\frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n! }< 1 + 1+ \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}} = 1+2 - \frac{1}{2^{n-1}} <3.$$
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|
Geometric series in a probability question When $A$ and $B$ flip coins, the one coming closest to a given line wins $1$ penny from the other. If $A$ starts with $3$ and $B$ with $7$ pennies, what is the probability that $A$ winds up with all of the money if both players are equally skilled?
What if A were a better player who won $60\%$ percent of the time?
This question I found in the end of the Sheldon Ross introduction (3rd chapter) to probability book. The given answer is $0.3$ but I am not sure for which question exactly, however neither of my answers is equal to that and I am not sure how to exactly apply the basic terms of probility "events" to this type of questions.
My attempt:
first question -
$$ P(A)=P(B)=\frac{1}{2}$$
$$ P(A \,wins \,all) := 0 \,losses : AAAAAAA$$
$$ 1\, loss :BAAAAAAAA, ABAAA..... $$
$$ 2\, loss :....$$
and so on , which can be formulated by:
$$ P(A)= \frac{1}{2^7}*\sum_{0}^{\infty} \frac{1}{2^{2n}}= \frac{1}{2^7}*\frac{4}{3}=0.01$$
Now, I am not sure if I should consider that if A losses 3 matches in a row in the beginning it basically loses I guess?
Second question, I used a similar method:
$$ 0.6^7*\sum_{0}^{\infty} 0.6^n*0.4^n= 0.6^7*\frac{1}{1-\frac{6}{25}} = 0.037 $$
Any hints, or advice greatly appreciated!
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For $1\le n\le 9$, let $p_n$ be the probability that $A$ wins if $A$ has $n$ pennies before starting the next round.
Then we have the following system of $5$ equations in $5$ unknowns . . .
\begin{cases}
p_1=\frac{1}{2}p_2\\[4pt]
p_2=\frac{1}{2}p_3+\frac{1}{2}p_1\\[4pt]
p_3=\frac{1}{2}p_4+\frac{1}{2}p_2\\[4pt]
p_4=\frac{1}{2}p_5+\frac{1}{2}p_3\\[4pt]
p_5=\frac{1}{2}\;\;\;\text{[by symmetry]}\\
\end{cases}
Solving the system yields $p_3=\frac{3}{10}$
If instead of equal skills, we assume that $A$ has probability $\frac{3}{5}$ of winning each round, we get the following system of $9$ equations in $9$ unknowns . . .
\begin{cases}
p_1=\frac{3}{5}p_2
\qquad\qquad\qquad\;\;
\\[4pt]
p_2=\frac{3}{5}p_3+\frac{2}{5}p_1\\[4pt]
p_3=\frac{3}{5}p_4+\frac{2}{5}p_2\\[4pt]
p_4=\frac{3}{5}p_5+\frac{2}{5}p_3\\[4pt]
p_5=\frac{3}{5}p_6+\frac{2}{5}p_4\\[4pt]
p_6=\frac{3}{5}p_7+\frac{2}{5}p_5\\[4pt]
p_7=\frac{3}{5}p_8+\frac{2}{5}p_6\\[4pt]
p_8=\frac{3}{5}p_9+\frac{2}{5}p_7\\[4pt]
p_9=\frac{3}{5}+\frac{2}{5}p_8
\end{cases}
Solving the system yields $p_3={\large{\frac{41553}{58025}}}\approx 0.7161223611$
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How can i get the least $n $ such that $17^n \equiv 1 \mod(100$)? When I solve the problem:
$17^{2018}\equiv r \pmod{100} $
used Euler theorem since $\gcd (17,100)=1$ and so
$\phi(100)=40$
and so
$17^{40}\equiv 1 \pmod{100}$
But i also found that: $17^{20}\equiv 1 \pmod{100}$
How can i get the least n such that $17^{n}=1\pmod{100} $?
Is there any theorm or generalization to this problem?
Thanks for your help
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Euler's theorem says that if $\gcd(a,n)=1$ then $a^{\phi(n)} \equiv 1 \pmod n$ but Euler's theorem does not say $\phi(n)$ is the smallest such number.[1]
But we can easily verify the smallest such power (called the order of $a$) will have to be a factor of $\phi(n)$ [2]
So test $17^k$ where $k|40$.
The powers twos are easy to calculate through repeated squaring.
$17^2 \equiv 89\pmod {100}$ and $17^4 \equiv 89^2\equiv (-11)^2 \equiv 21 \pmod {100}$ and $17^8\equiv 21^2 \equiv 41\pmod {101}$.
The multiples of $5$ are harder but ... well, check this out.
$17^{20} \equiv 21^5 \equiv (20+1)^5 \equiv 20^5 + 5*20^4 + 10*20^3 + 10*20^2 + 5*20 + 1\equiv 0+0+0+0+100 + 1 \equiv 1 \pmod {100}$.
Similarly $17^{10} \equiv (-11)^5 \equiv -(11)^5 \equiv -(10^5 + 5*10^4 + 10*10^3 + 10*10^2 + 5*10 + 1) \equiv -51 \equiv 49\pmod {101}$
Which means the smallest such power is $20$[3]
====
[1] Obviously if $a \equiv b^{k}$ and $k|\phi(n)$ then $a^{\frac {\phi(n)}k} = b^{\phi(n)}\equiv 1 \pmod n$ so this simply can't be true
Taking this to extreme: Obviously $\gcd(1,n) =$ and $1^1 \equiv 1 \pmod n$.
[2] If the order of $a$ is $k < \phi(n)$ and $k\not \mid \phi(n)$ then $\phi(n) = b*k + r$ for some $b$ and $0< r < k$.
So $1\equiv a^{\phi(n)}\equiv (a^{k})^ba^r \equiv 1^ba^r\equiv a^r\pmod n$. But we said $k$ was the least such power so that is a contradiction.
So the order of $a$ must divide $\phi(n)$.
[3]Any factor of $40$ that is less than $20$ is either a power of two (which we've ruled out) or a multiple of $5$. any smaller multiple of $5$ divides $10$. If $17$ to such a smaller power were equivalent to $1$ then $17^{10}$ would also be and it isn't.
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|
Find the equilibria of the following system of ODEs
Find the equilibria of the following system of ODEs
\begin{align*}
\dot{x} = ax - y + kx(x^2 + y^2)\\
\dot{y} = x - ay + ky(x^2 + y^2)
\end{align*}
where $a$ and $k$ are constants, $a > 1$ and $a^2 \geq 1$.
I want to find the equilibria of this system and say something about the behaviour in the phase-plane. Unfortunately, I can't seem to find the equilibria.
What I've tried: I subtracted the second equation from the first one to arrive at
\begin{align}
ax + ay -y -x+kx(x^2 + y^2) - ky(x^2 + y^2) = 0 \\
\Leftrightarrow a(x + y) - (x+y) + (x - y)k(x^2 + y^2) = 0\\
\Leftrightarrow a - 1 + \dfrac{x - y}{x+y}(x^2 + y^2)k = 0
\end{align}
From here I want to find an expression for $x$ or $y$ but I don't know how to do so.
Question: How should I approach this? If I'd get a hint on how to proceed I think I can solve the problem myself.
Thanks!
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Set $\dot x=\dot y=0$. Multiply the first equation with $x$, the second with $y$ and add to get
$$
0=a(x^2-y^2)+k(x^2+y^2)^2.
$$
Now do everything askew, multiply the first equation with $y$, the second with $x$ and subtract to get
$$
0=2axy-(x^2+y^2)
$$
Apart from the origin, there are also solutions for $y=q_\pm x=(a\pm\sqrt{a^2-1})x$, and from the first equation,
$$
y^2-x^2=4kax^2y^2\implies q^2-1=4kaq^2x^2,~~ x^2=\frac{q^2-1}{4kaq^2}
$$
which only gives real solutions for $q=q_+=a+\sqrt{a^2-1}>1$.
Example with $a=2$, $k=3$ using WolframAlpha
streamplot[{2x-y+3x(x^2+y^2), x-2y+3y(x^2+y^2)}, {x,-1.5,1.5}, {y,-1.5,1.5}]
showing a saddle point at the origin and two additional centers at the other two stationary points. At the boundary of the plot all solutions point outwards.
|
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Solve system of congruences $k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$
Solve system of congruences
$$k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$$
Is there any faster way to solve that congurence than looking at table and finding such pairs of $i$ that in both cases it gives $0$? When we have this table it is very fast job but making it is... time-consuming
\begin{array}{|c|c|c|c|}
\hline
i^2 \mod 17 & i^3 \mod 17 \\ \hline
0 & 0 \\ \hline
1 & 1 \\ \hline
4 & 8 \\ \hline
9 & 10 \\ \hline
16 & 13 \\ \hline
8 & 6 \\ \hline
2 & 12 \\ \hline
15 & 3 \\ \hline
13 & 2 \\ \hline
13 & 15 \\ \hline
15 & 14 \\ \hline
2 & 5 \\ \hline
8 & 11 \\ \hline
16 & 4 \\ \hline
9 & 7 \\ \hline
4 & 9 \\ \hline
1 & 16 \\ \hline
0 & 0 \\ \hline
1 & 1 \\ \hline
4 & 8 \\ \hline
\end{array}
|
Clearly if one of them is $0$ in modulo $17$ the other is. We show no other solutions exist.
Assume both $k, l$ are nonzero. Note that since $k^2 + l^2 \equiv 0$, we multiply both sides by $k+l$ and remove the $k^3+l^3$ to get $k^2l + l^2k =0$. Dividing both sides by $lk \not \equiv 0$ gives $k+l \equiv 0$. Squaring gives $k^2+l^2 + 2kl = 0$ which gives $2kl \equiv 0$; dividing by $2$ gives $lk \equiv 0$ which implies one of them is $0$ - a contradiction.
|
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|
Finding unknowns in polynomial with two factors and remainder. $x^2-4x-12$ is a factor of $rx^3-sx^2+36$, find $r$ and $s$.
Long division gives $rx+(-s+4r)$ with remainder $12rx+4(-s+4r)x+36+12(-s+4r)$
Where I have difficulty with the logic is that since $x^2-4x-12=0$ then the remainder is 0?
And from that we can say(how?) that $12r+4(-s+4r)=0$
And $36+12(-s+4r)=0$ also.
The rest is straightforward.
|
Factor Theorem
$x-a$ is a factor of $p(x)$ iff $p(a)=0$
Since $x^2-4x-12=(x-6)(x+2)$ is a factor of $p(x)=rx^3-sx^2+36$,
we know that $x-6$ and $x+2$ are factors of $p(x)$.
Thus $p(6)=0$ and $p(-2)=0$, that is $216r-36s+36=0$ and $-8r-4s+36=0$.
Solving the simultaneous linear equations give you $r=1,s=-7$.
To use long division, you need to divide $p(x)$ by $x-6$ and $x+2$.
By dividing $x-6$, you will get the remainder $216r-36s+36$.
By dividing $x+2$, you will get the remainder $-8r-4s+36$.
Since both are factors of $p(x)$, the remainder should be $0$.
Thus you get $216r-36s+36=0$ and $-8r-4s+36=0$.
If you straight away divide $p(x)$ by $x^2-4x-12$, the remainder you get is $12r+4(-s+4r)x+36+12(-s+4r)$.
Since the remainder must be zero polynomial,
you have $12r+4(-s+4r)=0$ and $36+12(-s+4r)=0$,
However, these two equations are multiple of one another so they cannot give you the solutions for $r$ and $s$.
|
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|
Double integrals involving incomplete beta function
I am trying to solve without success the following double integral
$$I_1^{(p)}(N)\equiv\frac{1}{2^p}\int_0^1\text{d}x\int_0^1\text{d}y(1+y-x)^{N+p}(1+x-y)^{N-2}B\left(\frac{1}{1+y-x};N,p+1\right)\cdot\theta(y-x)\theta(1-x-y),$$
where $N\in\mathbb{N}$, $p>0$, $\theta(x)$ is the Heaviside step function and $B(x;a,b)$ is the incomplete beta function
$$B(x;a,b)=\int_0^xt^{a-1}(1-t)^{b-1}\text{d}t.$$
The product of the two $\theta$ functions can be translated into one of the two following constraints
1) $\quad x\in\left(0,\frac{1}{2}\right)\longrightarrow x<y<1-x$
2) $\quad y\in\left(0,\frac{1}{2}\right)\longrightarrow x<y,\quad y\in\left(\frac{1}{2},1\right)\longrightarrow x<1-y$
so, e.g. in the first case, the integral becomes
$$I_1^{(p)}(N)=\frac{1}{2^p}\int_0^{\frac{1}{2}}\text{d}x\int_x^{1-x}\text{d}y(1+y-x)^{N+p}(1+x-y)^{N-2}B\left(\frac{1}{1+y-x};N,p+1\right).$$
At this point I tried some substitutions, such as $t=\frac{y-x}{1-2x}$ in order to get $\int_x^{1-x}\text{d}y\rightarrow\int_0^1\text{d}t$, but the expression remained not tractable for me.
The same happened when I rewrote the incomplete beta function in terms of a hypergeometric one, e.g. by
$$B(x;a,b)=\frac{x^a(1-x)^{b-1}}{a}{}_2F_1\left(1,1-b;a+1;\frac{x}{x-1}\right),$$
hoping to be able to use one of the relations that I found here.
Any help would be greatly appreciated.
Edit 1. The above integral is part of a larger expression, containing two other similar terms which can be obtained from $I_1^{(p)}(N)$ with the following substitutions respectively
$I_2^{(p)}(N):\quad B\left(\frac{1}{1+y-x};N,p+1\right)\rightarrow -B\left(\frac{1-y-x}{1+y-x};N,p+1\right)$
$I_3^{(p)}(N):\quad (1+y-x)^{N+p}B\left(\frac{1}{1+y-x};N,p+1\right)\rightarrow 2^p(1-x)^{N+p}B\left(\frac{1-y-x}{1-x};N,p+1\right)$
The structure does not change that much, so I thought that the solution procedure could be similar in the three cases. Nonetheless, probably it is better to report every detail, because the hypothesis that a simplication can occur between different terms cannot be discharged, even if I failed in doing that.
Edit 2. Proceeding as illustrated by @G Cab in his answer below, the result I obtained is
\begin{equation}\begin{split}
I_1^{(p)}(N)&=2^{2N-1}\left[B(N+p+1,N)B\left(\frac{1}{2};N,p+1\right)-B(N,p+1)B\left(\frac{1}{2};N+p+1,N\right)\right.\\[6pt]
&\left.\quad+\int_{\frac{1}{2}}^1t^{N-1}(1-t)^pB\left(\frac{1}{2t};N+p+1,N\right)\text{d}t\right].
\end{split}\end{equation}
I am pretty satisfied by the simplication with respect to the starting expression, but now I wonder whether the remaining single integral can be elaborated somehow.
|
a) the Incomplete Beta function is
$$
B \left( {x\;;a,b} \right) = \int_{t\, = \,0}^{\;x} {t^{\,a - 1} \left( {1 - t} \right)^{\,b - 1} dt}
$$
the integration variable is different from the upper bound
b) It might be useful to change the Step function with the Iverson bracket.
So
$$
\begin{split}
I^{\left( p \right)} (N) &= \frac{1}{2^p} \int\limits_{x = 0}^{1} \int\limits_{y = 0}^{1} ( {1 + y - x} )^{\,N + p}( {1 + {x - y} } )^{\,N - 2}
\\
&\qquad\cdot B\left( {{1 \over {1 + y - x}}\;;N,p + 1} \right)[ {0 \le y - x} ][ {y + x \le 1} ]dx\;dy \\
\\
& = \frac{1}{2^p}\int\limits_{x = 0}^{1} \int\limits_{y = 0}^{1} \!\!\int\limits_{t = 0}^{\frac{1}{1 + y - x}} ( {1 + y - x})^{\,N + p}( {1 + {x - y} } )^{\,N - 2}
t^{\,N - 1} \left( {1 - t} \right)^{\,p} \\
&\qquad\cdot[ {0 \le y - x} ][ {y + x \le 1} ]dx\;dy\,dt
\\
\\
& = \frac{1}{2^p}\iiint\limits_{(x,y,t) \in V}
( {1 + y - x})^{N + p} ( {1 + {x - y} } )^{\,N - 2} t^{N - 1} ( {1 - t} )^{\,p} dx dy dt
\end{split}
$$
where
$$
V = \left\{ {(x,y,t)} \right\}:\;\;\left\{ \matrix{
0 \le x \le 1 \hfill \cr
0 \le y \le 1 \hfill \cr
0 \le y - x \hfill \cr
y + x \le 1 \hfill \cr
0 \le t \le {1 \over {1 + y - x}} \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{
0 \le x \le 1/2 \hfill \cr
0 \le x \le y \le 1 - x \hfill \cr
0 \le t \le {1 \over {1 + y - x}} \hfill \cr} \right.
$$
Now it remains to change the variables appropriately, so that we can integrate in $t$ at last, after the others, and to
reformulate accordingly the bounds on $V$.
Proceeding further the change of variables is
$$\begin{cases} v=1+y-x\\ u=1-y-x\end{cases}\quad\Longrightarrow\quad\begin{cases}x=1-\frac{v+u}{2}\\ y=\frac{v-u}{2}\end{cases}$$
and so one obtains $dxdy=\frac{1}{2}dvdu$. It is also easy to check that the domain $V$ splits into two different parts
$$V_1=\{(u,v,t)\}:\begin{cases}0<t<\frac{1}{2}\\ 1<v<2\\ 0<u<2-v\end{cases}\qquad\quad V_2=\{(u,v,t)\}:\begin{cases}\frac{1}{2}<t<1\\ 1<v<\frac{1}{t}\\ 0<u<2-v\end{cases}
$$
This leads us to
$$
\begin{split}
I^{\left( p \right)} (N) &= \frac{1}{2^{p+1}}\iiint\limits_{(u,v,t) \in V_1\cup V_2} v^{N + p} u^{N - 2} t^{N - 1} ( {1 - t} )^{p} du dv dt \\
& = \frac{1}{2^{p+1}}\left[\;\int\limits_{t = 0}^{\frac{1}{2}} t^{N - 1} ( {1 - t} )^{p} dt\int\limits_{v = 1}^{2} v^{N + p} ( {2 - v} )^{N - 2} dv \int\limits_{u = 0}^{2 - v} {du} \right.\\
&\quad\left.+\int\limits_{t = \frac{1}{2}}^{1} t^{N - 1} ( {1 - t} )^{p} dt\int\limits_{v = 1}^{\frac{1}{t}} v^{N + p} ( {2 - v} )^{N - 2} dv \int\limits_{u = 0}^{2 - v} {du} \right]\\
& = \frac{1}{2^{p+1}}\left[\;\int\limits_{t = 0}^{\,\frac{1}{2}} t^{\,N - 1} ( {1 - t} )^{p} dt\int\limits_{v = 1}^{2} {v^{N + p} ( {2 - v} )^{N - 1} dv}\right.\\
&\left.\quad+\int\limits_{t = \frac{1}{2}}^{\,1} t^{\,N - 1} ( {1 - t} )^{p} dt\int\limits_{v = 1}^{\frac{1}{t}} {v^{N + p} ( {2 - v} )^{N - 1} dv} \right]\\
& = 2^{2N -1} \left[\;\int\limits_{t = 0}^{\,\frac{1}{2}} {t^{N - 1} ( {1 - t} )^{p} dt \int\limits_{\frac{v}{2} = \frac{1}{2}}^{1}
{\left( {{v \over 2}} \right)^{N + p} \left( {1 - {v \over 2}} \right)^{N - 1} d\left( {{v \over 2}} \right)} }\right.\\
&\left.\quad+\int\limits_{t = \frac{1}{2}}^{\,1} {t^{N - 1} ( {1 - t} )^{p} dt \int\limits_{\frac{v}{2} = \frac{1}{2}}^{\frac{1}{2t}}
{\left( {{v \over 2}} \right)^{N + p} \left( {1 - {v \over 2}} \right)^{N - 1} d\left( {{v \over 2}} \right)} } \right] = \\
& = 2^{2N -1}\left\{\;\int\limits_{t = 0}^{\frac{1}{2}} {t^{N - 1} \left( {1 - t} \right)^{p} {B\left( {\frac{1}{2};N,N+p+1} \right)
} dt}\right.\\
&\left.\quad+\int\limits_{t = \frac{1}{2}}^{1} {t^{N - 1} \left( {1 - t} \right)^{p} \left[ {B\left( {\frac{1}{2t};N + p + 1,N} \right)
- B\left( {\frac{1}{2};N + p + 1,N} \right)} \right]dt}\right\}
\end{split}
$$
|
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"url": "https://math.stackexchange.com/questions/3257400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Solution of two variables system this might be a very simple question but it has its importance for me. In my multivariable calculus class, I have to find all $(x,y)$ in $R^2$ verifying the system
I will call $(S)$ :
\begin{cases}
3(x-y)^2+8x-3=0 \\
-3(x-y)^2+3=0
\end{cases}
What I did :
This system is equivalent to the following :
\begin{cases}
3(x-y)^2+8x-3=0 \\
3(-(x-y)^2+1)=0
\end{cases}
Equivalent to :
$$\begin{cases}
3(x-y)^2+8x-3=0 \\
3(1-x+y)(1+x-y)=0
\end{cases}$$
By the second last equation, we get
$$(S) \Leftrightarrow \begin{cases}
3(x-y)^2+8x-3=0 \quad (1)\\
(x-y)=1 \text{ or } (x-y)=-1\quad (2)
\end{cases}$$
Then, we substitute in $(1)$.
$$(x-y)=1 \text{ or }(x-y)=-1 \Rightarrow x=0$$
So we found two points $(0,-1)$ and $(0,1)$ verifying $(S)$. My question is : how can we know that there are no other points ? Is it a question of factorisation ?
|
Note that the first equation gives $3(x-y)^2 = 3-8x$, and the second gives $3(x-y)^2 = 3$. So we equate to get $3-8x = 3 \implies x = 0$. Now $3(x-y)^2 = 3$ becomes $(-y)^2 = 1$, so $y$ is $1$ or $-1$, and it is straightforward to check that both $(x,y) = (0,1), (0,-1)$ satisfy both equations.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3257675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Trig identities analogous to $\tan(\pi/5)+4\sin(\pi/5)=\sqrt{5+2\sqrt{5}}$ The following trig identities have shown up in various questions on MSE:
$$-\tan\frac{\pi}{5}+4\sin\frac{2\pi}{5}=\tan\frac{\pi}{5}+4\sin\frac{\pi}{5}=\sqrt{5+2\sqrt{5}}$$
$$-\tan\frac{2\pi}{7}+4\sin\frac{3\pi}{7}=-\tan\frac{\pi}{7}+4\sin\frac{2\pi}{7}=\sqrt{7}$$
$$\tan\frac{\pi}{11}+4\sin\frac{3\pi}{11}=\tan\frac{3\pi}{11}+4\sin\frac{2\pi}{11}=\sqrt{11}$$
$$\tan\frac{6\pi}{13}-4\sin\frac{5\pi}{13}=\tan\frac{2\pi}{13}+4\sin\frac{6\pi}{13}=\sqrt{13+2\sqrt{13}}$$
Does anyone know of any analogous identities for larger primes? I have not been able to find anything similar for $p=17$ or $p=19$.
(I am not asking for proofs of the above equations.)
|
$$\tan\frac{2\pi}{29}+4\left(-\sin\frac{2\pi}{29}+\sin\frac{6\pi}{29}+\sin\frac{8\pi}{29}-\sin\frac{20\pi}{29}+\sin\frac{22\pi}{29}\right)=\sqrt{29-2\sqrt{29}} $$
|
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"url": "https://math.stackexchange.com/questions/3257939",
"timestamp": "2023-03-29T00:00:00",
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|
Solve and asymptotic expansion of $\sum_{a=1}^{H} \sum_{b=a+1}^{H} \left\lfloor{\frac{H}{a\, b}}\right\rfloor$ I am solving constrained polynomial systems resulting in constrained sums. I am looking to see if $$\sum_{a=1}^{H} \sum_{b=a+1}^{H} \left\lfloor{\frac{H}{a\, b}}\right\rfloor$$ is expressible in terms of known functions and very importantly the asymptotic form as $H \rightarrow \infty$. If needed we can make the sum over $b$ symmetric to the sum over $a$.
|
I get
$\frac14 n\ln^2(n)+O(n\ln(n))
$.
If
$\begin{array}\\
s(n)
&=\sum_{a=1}^{n} \sum_{b=a+1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\
\text{then}\\
s(n)
&=\sum_{b=1}^{n} \sum_{a=1}^{b-1} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\
&=\sum_{a=1}^{n} \sum_{b=1}^{a-1} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\
\text{so}\\
2s(n)
&=\sum_{a=1}^{n} (\sum_{b=1}^{a-1} \left\lfloor{\frac{n}{a\, b}}\right\rfloor+ \sum_{b=a+1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor)\\
&=\sum_{a=1}^{n} (\sum_{b=1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor-\left\lfloor{\frac{n}{a^2}}\right\rfloor)\\
&=\sum_{a=1}^{n} \sum_{b=1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor
-\sum_{a=1}^{n}\left\lfloor{\frac{n}{a^2}}\right\rfloor\\
&= u(n)-v(n)\\
v(n)
&= \sum_{a=1}^{n}\left\lfloor{\frac{n}{a^2}}\right\rfloor\\
&= \sum_{a=1}^{\lfloor \sqrt{n} \rfloor}\left\lfloor{\frac{n}{a^2}}\right\rfloor\\
&\le \dfrac{n\pi^2}{6}\\
\text{and}\\
u(n)
&=\sum_{a=1}^{n} \sum_{b=1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\
&=\sum_{a=1}^{n} \sum_{b=1}^{\lfloor n/a \rfloor} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\
&=\sum_{a=1}^{n} \sum_{b=1}^{\lfloor n/a \rfloor} \left\lfloor{\frac{\lfloor n/a \rfloor}{b}}\right\rfloor\\
&\le \sum_{a=1}^{n} \lfloor n/a \rfloor\sum_{b=1}^{\lfloor n/a \rfloor} \left\lfloor{\frac1{b}}\right\rfloor\\
&\le \sum_{a=1}^{n} \lfloor n/a \rfloor\sum_{b=1}^{\lfloor n/a \rfloor} \frac1{b}\\
&\lt \sum_{a=1}^{n} \lfloor n/a \rfloor (\ln(n/a)+1)\\
&\le \sum_{a=1}^{n} (n/a)\ln(n/a)+ \sum_{a=1}^{n} (n/a)\\
&= \sum_{a=1}^{n} (n/a)\ln(n)-\sum_{a=1}^{n} (n/a)\ln(a)+ n\sum_{a=1}^{n} \frac1{a}\\
&= n\ln(n)\sum_{a=1}^{n} \frac1{a}-n\sum_{a=1}^{n} \frac{\ln(a)}{a}+ n\sum_{a=1}^{n} \frac1{a}\\
&= n\ln^2(n)+O(n\ln(n))-n(\frac12 \ln^2(n)+O(\ln(n))+ n(\ln(n)+O(1))\\
&= \frac12 n\ln^2(n)+O(n\ln(n))\\
\text{so}\\
s(n)
&\lt \frac14 n\ln^2(n)+O(n\ln(n))\\
\text{and}\\
u(n)
&=\sum_{a=1}^{n} \sum_{b=1}^{n} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\
&=\sum_{a=1}^{n} \sum_{b=1}^{\lfloor n/a \rfloor} \left\lfloor{\frac{n}{a\, b}}\right\rfloor\\
&\gt\sum_{a=1}^{n} \sum_{b=1}^{\lfloor n/a \rfloor} ({\frac{n}{a\, b}}-1)\\
&=\sum_{a=1}^{n} \sum_{b=1}^{\lfloor n/a \rfloor} {\frac{n}{a\, b}}-\sum_{a=1}^{n} \sum_{b=1}^{\lfloor n/a \rfloor}1\\
&=\sum_{a=1}^{n} \frac{n}{a}\sum_{b=1}^{\lfloor n/a \rfloor} {\frac1{b}}-\sum_{a=1}^{n} \lfloor n/a \rfloor\\
&>\sum_{a=1}^{n} \frac{n}{a}\ln(\lfloor n/a \rfloor)-n\ln(n)\\
&>\sum_{a=1}^{n} \frac{n}{a}(\ln( n/a)-1)-n\ln(n)\\
&= \frac12 n\ln^2(n)+O(n\ln(n))
\quad\text{by the same argument as above}\\
\end{array}
$
|
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|
How is Wolfram Alpha and the reduction formula arriving at a different result for the integral of $\int \sec^4 x\,dx$ than naive $u$-substitution? I calculated the following on paper for the value of $\int \sec^4 x\,dx$.
$$\int \sec^4 x\,dx=\int \sec^2 x \sec^2 x\,dx=\int (\tan^2 x + 1)(\sec^2 x)\,dx.$$
Let $u = \tan x$, $du = \sec^2 x\,dx$ so
\begin{align}\int \sec^4 x\,dx&=\int u^2 + 1\,du\\&=\frac{1}{3} u^3 + u + C\\&=\frac{1}{3} \tan^3 x + \tan x + C\\&=\frac{1}{3} (\tan x)(\tan^2 x + 1) + C\\&=\frac{1}{3} \tan x \sec^2 x + C\end{align}
Wolfram Alpha, however, gives $\int \sec^4(x)\,dx = \frac13(\cos(2 x) + 2) \tan(x) \sec^2(x) + C$. This is notably not equal to my solution.
According to the "step-by-step solution" from the Wolfram Alpha app, the reduction formula was used to produce $$\frac{1}{3}\tan x \sec^2 x + \frac{2}{3} \int \sec^2 x \,dx$$ then $$\frac{2}{3} \tan x + \frac{1}{3} \tan x \sec^2 x + C$$
Why does the reduction formula produce this added term compared to naive $u$-substitution?
|
You made an arithmetic error. It should be
$$
\frac13\tan^3 x+\tan x=\frac13\tan(x)(\tan^2x+3)
$$
not $\frac13\tan(x)(\tan^2x+1)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.
Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.
We have that $xy^2 + y + 1 \mid x^2y + x + y \implies xy^2 + y + 1 \mid y(x^2y + x + y) - x(xy^2 + y + 1)$
$\iff xy^2 + y + 1 \mid y^2 - x \implies xy^2 + y + 1 \mid xy^2 + y + 1 - x(y^2 - x)$
$\iff xy^2 + y + 1 \mid x^2 + y + 1 \implies xy^2 + y + 1 \le x^2 + y + 1$
$\iff xy^2 \le x^2 \iff y^2 \le x$. And I don't know what to do next.
This problem is adapted from a recent competition.
|
OP did most of the work by showing that
*
*$xy^2 + y + 1 \mid x - y^2$.
*$ xy^2 + y+ 1 \mid x^2 + y + 1 $
*Since $x^2 + y + 1 > 0$ hence $x^2 + y + 1 \geq xy^2 + y + 1$ so $x \geq y^2$.
To complete the solution using these ideas, suppose $ x - y^2 > 0 $, then $ x-y^2 \geq xy^2 + y + 1$.
We can verify that this is not possible for any positive integer $y$, since $xy^2 + y + 1 > xy^2 \geq x > x - y^2$.
Hence, we must have $ x - y^2 = 0$.
Plugging this back in, we observe that all $ x = y^2$ are solutions to the divisibility condition.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solving system of linear second order differential equations Solve this system of differential equations:
$$y_1'' + y_2' + 3y_1 = e^{-x}$$
$$y_2''-4y_1' + 3y_2 = \sin(2x)$$
My try: Using $y_1' = p$ and $y_2'= q$ , we get:
$$\begin{pmatrix} p' \\q'\\y_1'\\y_2'\end{pmatrix} =\begin{pmatrix}
0 & -1 & -3 & 0 \\
4 & 0 & 0 & -3 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0
\end{pmatrix}\begin{pmatrix} p \\q\\y_1\\y_2\end{pmatrix} + \begin{pmatrix} e^{-x} \\\sin(2x)\\0\\0\end{pmatrix}$$
I solved the homogeneous part and find :
$$ y_h = c_1e^{-ix}\begin{pmatrix} 1/2 \\-i\\i/2\\1\end{pmatrix} + c_2e^{ix}\begin{pmatrix} 1/2 \\i\\-i/2\\1\end{pmatrix} + c_3e^{3ix}\begin{pmatrix} -3/2 \\3i\\i/2\\1\end{pmatrix} + c_4e^{-3ix}\begin{pmatrix} -3/2 \\-3i\\-i/2\\1\end{pmatrix}$$
I don't know how to solve non-homogeneous part of the equation.
|
I am going to rewrite the system as
$$\tag 1 x'' + y' + 3 x = e^{-t} \\ y'' - 4 x' + 3 y = \sin 2t $$
Using the first equation, we have
$$\tag 2 y' = -x'' - 3 x + e^{-t} \\ y'' = -x''' - 3 x' - e^{-t} \\ y''' = - x'''' - 3 x'' + e^{-t}$$
Taking the derivative of the second equation
$$\tag 3 y''' - 4 x'' + 3 y' = 2 \cos 2 t$$
Substituting $(2)$ into $(3)$
$$(- x'''' - 3 x'' + e^{-t}) - 4 x'' + 3 (-x'' - 3 x + e^{-t}) = 2 \cos 2t $$
This simplifies to
$$-x'''' - 10 x'' - 9x = 2 \cos 2 t - 4 e^{-t} $$
Solving this using your favorite method
$$x(t)= c_1 \cos 3t +c_2 \sin 3t+c_3 \cos t + c_4 \sin t+\dfrac{e^{-t}}{5}+\dfrac{2}{15} \cos 2t$$
Now, using the first equation and $x(t)$, we need to solve
$$\tag 4 y' = -x'' - 3 x + e^{-t} = 6 c_1 \cos 3t+6 c_2 \sin 3t-2 c_3 \cos t -2 c_4 \sin t+\dfrac{e^{-t}}{5}+\dfrac{2}{15} \cos 2t$$
By integrating $(4)$, we get
$$y(t) = 2 c_1 \sin 3t-2 c_2 \cos 3t-2 c_3 \sin t+2 c_4 \cos t-\dfrac{e^{-t}}{5}+\dfrac{1}{15} \sin 2t$$
You can substitute $(1)$ and $(2)$ into the original system and verify they are solutions.
|
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Compute $\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}$ Compute $$\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}.$$
I started by making the substitution $\arccos x = t$. Hence, $-\frac{dx}{\sqrt{1-x^2}}=dt$.
Now I get that $$\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}=-\int\limits_{\pi/3}^{2 \pi/3}\frac{dt}{t^2}=\frac{3}{2\pi}-\frac{3}{\pi}=-\frac{3}{2\pi}.$$
However, the result should be $\dfrac{3}{2\pi}$. What did I do wrong?
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$\arccos(-0.5)=\frac{2\pi}{3},\arccos(0.5)=\frac{\pi}{3}$ thus transformed integral must be $-\int_{\frac{2\pi}{3}}^{\frac{\pi}{3}}(...)$
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the definite integral $\int_0^{2\pi}\arctan\left((a\cos x + b\sin x)^2\right)dx$ Writing:
Integrate[ArcTan[(a Cos[x] + b Sin[x])^2], {x, 0, 2 Pi}, Assumptions -> a^2 + b^2 > 0]
$$\int_0^{2\pi}\arctan\left((a\cos x + b\sin x)^2\right)dx,$$
where $a $ and $b $ are real numbers.
I get:
2 Pi ArcTan[Sqrt[1/2 (-1 + Sqrt[1 + (a^2 + b^2)^2])]]
$$2\pi\arctan\sqrt{\frac{\sqrt{1 + (a^2 + b^2)^2}-1}2} $$
How to derive this result on paper?
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Denote $$C := a^2 + b^2 .$$ Then, we can find an angle $x_0$ such that $a = \sqrt{C} \cos x_0$ and $b = -\sqrt{C} \sin x_0$. The angle sum formula for $\sin$ lets us rewrite the quantity in the inner parentheses of the integrand as $$a \sin x + b \cos x = \sqrt{C} \sin (x - x_0).$$ Then, appealing to the periodicity of the integrand lets us rewrite the integral as
$$I(C) := \int_0^{2 \pi} \arctan (C \sin^2 x) \,dx .$$
Differentiating under the integral sign gives
$$I'(C) = \int_0^{2 \pi} \frac{\sin^2 x\, dx}{1 + C^2 \sin^4 x} .$$
Now, use symmetry to rewrite $I'(C)$ in terms of an integral over $[0, \pi]$, and apply the Euler substitution $x = 2 \arctan t, \,dx = \frac{2\,dt}{1 + t^2}$, giving the rational integral $$I'(C) = 16 \int_0^{\infty} \frac{t^2 (t^2 + 1) \,dt}{(t^2 + 1)^4 + 16 C^4 t^4}.$$
Integrating gives
$$I'(C) = \frac{\pi \sqrt{2}}{\sqrt{1 + \sqrt{1 + C^2}} \sqrt{1 + C^2}} .$$ (This can be done with contour integration, which in this case is tedious but straightforward. Quite possibly there is a better method, and I would be grateful to learn it.)
Since $I(0) = 0$, we have
$$I(C) = \pi \sqrt{2} \int_0^C \frac{dc}{\sqrt{1 + \sqrt{1 + c^2}} \sqrt{1 + c^2}} = 2 \pi \sqrt{2} \int_0^{u_0} \frac{du}{u^2 + 2} ;$$ the latter equality follows from applying the substitution $c^2 + 1 = (u^2 + 1)^2$, and $u_0$ is the $u$-value corresponding to $c = C$. The integral on the right-hand side is elementary, and so one can produce an explicit formula for $I(C)$ in terms of $C$ and hence in terms of $a, b$.
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"timestamp": "2023-03-29T00:00:00",
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Solving $\begin{cases}x' = t \sin^2(\frac 1 t) - x^2 \\ x(0) = 0 \end{cases}$ Is there a way to compute the solution of the ode:
$\begin{cases}x' = t \sin^2(\frac 1 t) - x^2 \\ x(0) = 0 \end{cases}$
where in $t = 0$ we define the field as $-x^2$.
I need for an application of Guiding the solution of ODE with curves. In particular to compute the derivative of the following at $0$:
$\begin{cases}x' = f(t,x) + \epsilon (\gamma(t) - 2x) \\ x(0) = 0\end{cases}$
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Hint:
Let $x=\dfrac{1}{u}\dfrac{du}{dt}$ ,
Then $\dfrac{dx}{dt}=\dfrac{1}{u}\dfrac{d^2u}{dt^2}-\dfrac{1}{u^2}\left(\dfrac{du}{dt}\right)^2$
$\therefore\dfrac{1}{u}\dfrac{d^2u}{dt^2}-\dfrac{1}{u^2}\left(\dfrac{du}{dt}\right)^2=t\sin^2\dfrac{1}{t}-\dfrac{1}{u^2}\left(\dfrac{du}{dt}\right)^2$
$\dfrac{d^2u}{dt^2}-\left(t\sin^2\dfrac{1}{t}\right)u=0$
Let $r=\dfrac{1}{t}$ ,
Then $\dfrac{du}{dt}=\dfrac{du}{dr}\dfrac{dr}{dt}=-\dfrac{1}{t^2}\dfrac{du}{dr}=-r^2\dfrac{du}{dr}$
$\dfrac{d^2u}{dt^2}=\dfrac{d}{dt}\left(-r^2\dfrac{du}{dr}\right)=\dfrac{d}{dr}\left(-r^2\dfrac{du}{dr}\right)\dfrac{dr}{dt}=\left(-r^2\dfrac{d^2u}{dr^2}-2r\dfrac{du}{dr}\right)(-r^2)=r^4\dfrac{d^2u}{dr^2}+2r^3\dfrac{du}{dr}$
$\therefore r^4\dfrac{d^2u}{dr^2}+2r^3\dfrac{du}{dr}-\dfrac{\sin^2r}{r}u=0$
$r\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}-\dfrac{\sin^2r}{r^4}u=0$
Let $v=ru$ ,
Then $\dfrac{dv}{dr}=r\dfrac{du}{dr}+u$
$\dfrac{d^2v}{dr^2}=r\dfrac{d^2u}{dr^2}+\dfrac{du}{dr}+\dfrac{du}{dr}=r\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}$
$\therefore\dfrac{d^2v}{dr^2}-\dfrac{\sin^2r}{r^5}v=0$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3265139",
"timestamp": "2023-03-29T00:00:00",
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Calculating $\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots$
Calculate: $$\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots$$
The solution of this exercise:
Let $$S_1=\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\binom{n}{8}+\cdots$$
$$S_2=\binom{n}{1}-\binom{n}{3}+\binom{n}{5}-\cdots$$
$$S_3=\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots$$
$$S_4=\binom{n}{2}+\binom{n}{6}+\binom{n}{10}+\cdots$$
And we consider $$(1+i)^n=S_1+iS_2=\sqrt2^n\left(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}\right)$$ and $$2^{n-1}+S_1=2S_3$$
The problem is that i didn't get the part with $(1+i)^n$.. from here i got lost.I saw more exercises like this with combinatorial sums whose solution was about complex numbers and i wish that someone explain me that method.Thanks!
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The part with $(1+i)^n$ is explained by De Moivre:
$$(1+i)^n=\sqrt2^ne^{i\frac{n\pi}{4}}=\sqrt2^n(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})^n=\sqrt2^n(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4})$$
Now comparing the $\Re$ parts of the LHS and RHS of the given equation:
$$\Re(\sqrt2^n(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}))=\Re(S_1+iS_2)$$
$$\iff\sqrt2^n\cos\frac{n\pi}{4}=S_1$$
$$\iff \sqrt2^n\cos\frac{n\pi}{4}=2^{n-1}-2S_3.$$ Thereby,
$$S_3=\binom{n}{0}+\binom{n}{4}+\dots=\frac{1}{2}\left(2^{n-1}-\sqrt2^n\cos\frac{n\pi}{4}\right).$$
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"timestamp": "2023-03-29T00:00:00",
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unknown polynomial divided by $x^2(x-1)$, find the remainder. I took an exam today and there's a problem stuck in my head; I still can't figure out yet.
Here's the question (just the concept as I can't remember precisely).
An unknown polynomial divided by $(x-1)^2$ leaves the remainder of $x + 3$ (not sure about the number) and when this polynomial is divided by $x^2 $, it leaves $2x + 4$ (again, not sure about the number). From the given conditions, if this polynomial is divided by $(x-1)x^2$, what would be the remainder?
The solution as far as I figured out is this:
first, from the division of $(x-1)^2$, I got that $f(1) = 3$
in the same way from division of $x^2$, I got $f(0) = 4.$
I can write the polynomial as follows:
$f(x) = (x-1)(x)(x) g(x) + ax^2 +bx +c$
$ax^2 + bx + c$ is the remainder. And to find $a,b,c$, I can use the conditions above, so I got $c = 4$ by substituting $x = 0,$ and I got $a+b+4 = 3$ by substituting $x = 1.$
This leaves $a + b = -1,$ and I can't figure out how to continue; please help.
Edit : I made a mistake $f(1)$ should be equal to $4$ and $a+b+c = 4$
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General method, without the formal derivatives.
Suppose
$$
f(x) = g(x) (x-1)^2 + (x+3)= h(x) x^2 + (2x + 4).
$$
Then
$$
(x-1)^2 f(x) = (x-1)^2 x^2 h(x) + (x-1)^2 (2x+4) \tag 1
$$
and
$$
x^2 f(x) = x^2 (x-1)^2 g(x) + x^2(x+3). \tag 2
$$
Now do the Euclidean algorithm to $x^2, (x-1)^2$:
\begin{align*}
x^2 &= (x-1)^2 + (2x-1), \\
(x-1)^2 &= \frac 14 (2x-1)(2x - 3) + \frac 14,
\end{align*}
therefore
$$
1 = 4(x-1)^2 - (2x - 1) (2x-3) = 4(x-1)^2 - (x^2 - (x-1)^2) (2x-3) = (x-1)^2 (4 + 2x-3) - x^2 (2x - 3) = \color{blue}{(x-1)^2 (2x+1) - x^2 (2x - 3)}.
$$
Thus $(2x+1) \cdot \mathrm {Eq}(1) - (2x - 3) \cdot \mathrm {Eq} (2)$ yields
$$
f(x) = (x-1)^2 x^2 F(x) + (2x+4) (2x+1)(x-1)^2 - (x+3)(2x-3)x^2.
$$
Since
$$
(2x+4) (2x+1)(x-1)^2 - (x+3)(2x-3)x^2 = (4x^2 + 10x + 4)(x-1)^2 - (2x^2 + 3x -9)x^2 = (4x^3 +6x^2 -6x - 4)(x-1) - ((2x+5)(x-1) -4)x^2 = x^2 (x-1) G(x) + ((-6x -4)(x-1) + 4x^2) = x^2(x-1)G(x) + (\color{red}{-2x^2 +2x +4}),
$$
we have
$$
f(x) = x(x-1)^2 (G(x)+ xF(x)) + (\color{red}{-2x^2 +2x +4}),
$$
which means the remainder is
$$
\color{red}{-2x^2 +2x +4}.
$$
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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.