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What is the probability that the sum of 6 four-sided dice is less than or equal to 14? The solution given is shown below.
My question is how did they count the numerator like that?
What is the explanation for it please?
$$\begin{align}
\frac{C_6^{14}-C_1^6\times C_6^{10}+C_2^6 \times C_6^6}{4^6}&=\frac{3003-6\times 210+15\times 1}{4^6}\\
&= \frac{1758}{4^6}\\
&= \frac{879}{2048}
\end{align}$$
I understand the denominator namely because each of the four sided dice has four choices and six of them so, all possible outcomes will be $4^6$.
I believe that the 4-sided dice here has 1, 2, 3, 4 printed on its faces.
Any help is appreciated.
|
The following solution uses a generating function. Readers not familiar with generating functions can find several resources in the answers to this question: How can I learn about generating functions?
There are $4^6$ possible outcomes when rolling six four-sided dice, all of which we assume are equally likely. We want to count the number of arrangements in which the sum is less than or equal to $14$. This is the number of solutions in integers to
$$x_1+x_2+x_3+x_4+x_5+x_6 \le 14$$
sunbject to $1 \le x_i \le 4$ for $i=1,2,3,4,5,6$. Equivalently, we want the number of integer solutions to
$$x_1+x_2+x_3+x_4+x_5+x_6+x_7 = 14 \tag{1}$$
where $0 \le x_7$. The generating function for the number of solutions is
$$\begin{align}
f(x) &= (x+x^2+x^3+x^4)^6 \cdot (1+x+x^2+ \dots) \\
&= x^6 \left( \frac{1-x^4}{1-x} \right)^6 \cdot \frac{1}{1-x} \tag{2} \\
&= x^6 \; (1-x^4)^6 \; (1-x)^{-7} \\
&= x^6 \; (1 -6x^4+15x^8+O(x^{12})) \; \sum_{i=0}^{\infty} \binom{7+i-1}{i} \tag{3} x^i
\end{align}$$
From $(3)$ we can see that the coefficient of $x^{14}$ in $f(x)$ is
$$[x^{14}]f(x) = \binom{7+8-1}{8} - 6 \binom{7+4-1}{4} + 15 = 1758$$
so the number of solutions to $(1)$ is $1758$, and the probability of rolling a sum of $14$ or less is
$$\frac{1758}{4^6}$$
Notes:
$(2)$ Summing geometric series (twice).
$(3)$ Applying the Binomial Theorem, first for a positive exponent and then for a negative exponent.
|
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|
How can I integrate $\int\sqrt{\frac{x^2+bx+c}{x^2+ex+f}}dx?$ How can I integrate
$$\int\sqrt{\dfrac{x^2+bx+c}{x^2+ex+f}}\,dx?$$ I was thinking a substitution $$t=\frac{x^2+bx+c}{x^2+ex+f},$$ which inverts as follows:
$$(x^2+ex+f)t=x^2+bx+c$$
$$(t-1)x^2+(et-b)x+ft-c=0$$
$$x=\dfrac{b-et+\sqrt{(b-et)^2-4(t-1)(ft-c)}}{2(t-1)},$$ but I've never really dealt with integrals of this complexity with the aim of finding a closed form in terms of fundamental integral functions. The best I could do is expand the integrand as a not-so-nice power series.
|
Going off of Travis Willse's answer, I notice that
$$\int \frac{x^2+ax+b}{\sqrt{x^4+\alpha x^2+\beta x+\gamma}}dx$$ can be addressed as follows:
Since $$\frac{1}{\sqrt{x^4+\alpha x^2+\beta x+\gamma}}=\sum_{p,q,r\geqslant 0 }\binom{-\frac{1}{2}}{p,q,r,-\frac{1}{2}-p-q-r}\alpha^{q}\beta^r\gamma^{-\frac{1}{2}-p-q-r}x^{4p+2q+r},$$ we have
$$\int \frac{x^2+ax+b}{\sqrt{x^4+\alpha x^2+\beta x+\gamma}}dx=\int \sum_{p,q,r\geqslant 0 }\binom{-\frac{1}{2}}{p,q,r,-\frac{1}{2}-p-q-r}\alpha^{q}\beta^r\gamma^{-\frac{1}{2}-p-q-r}x^{4p+2q+r}(x^2+ax+b) dx\\=\boxed{\sum_{p,q,r\geqslant 0 }\binom{-\frac{1}{2}}{p,q,r,-\frac{1}{2}-p-q-r}\alpha^{q}\beta^r\gamma^{-\frac{1}{2}-p-q-r}\left(\frac{x^{4p+2q+r+3}}{4p+2q+r+3}+\frac{ax^{4p+2q+r+2}}{4p+2q+r+2}+\frac{bx^{4p+2q+r}}{4p+2q+r}\right) +C}$$ However, this is not the answer I'm looking for, but I put it here just so that it's out there to inspire someone.
|
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Find $a+b +c$, if $\sin{x}+\sin^2{x}=1$ and $\cos^{12}{x}+a\cos^{10}{x}+b\cos^8{x}+c\cos^6{x}=1$ There is my problem :
Find $a+b +c$,
if $$\sin{x}+\sin^2{x}=1$$ and $$\cos^{12}{x}+a\cos^{10}{x}+b\cos^8{x}+c\cos^6{x}=1$$
I'm sorry, I can't solve this problem but I really want to know the solution.
I know that $\cos^2{x}=\sin{x}$, but I can't find $a+b+c$.
Attempt
I used substitute $t=\sin(x)$, and number $1=t^2+t$ put on the left. Then I divided by $t$ as long as I can, then I got polynomial with degree $3$, but I can't conclude what is $a+b+c$.
|
With what you already know, $$\sin^6x+a\sin^5x+b\sin^4x+c\sin^3x-1=0\qquad \sin^2x+\sin x-1=0$$ Now let $X=\sin (x)$. You would like the polynomial $X^2+X-1$ to divide $X^6+aX^5+bX^4+cX^3-1$. How to make that happen?
There would need to be a 4th degree polynomial $v=X^4+v_3X^3+v_2X^2+v_1X+1$ as the quotient.
$$\left(X^2+X-1\right)\left(X^4+v_3X^3+v_2X^2+v_1X+1\right)=X^6+aX^5+bX^4+cX^3-1$$
You can deduce quickly that $v_1=1$ and $v_2=2$ after comparing linear and quadratic coefficients.
$$\left(X^2+X-1\right)\left(X^4+v_3X^3+2X^2+X+1\right)=X^6+aX^5+bX^4+cX^3-1$$
Comparing cubic, quartic, and quintic coefficients:
$$\begin{align}
1+2-v_3&=c\\
2+v_3-1&=b\\
v_3+1&=a
\end{align}$$
You could eliminate $v_3$ by adding the first equation to each of the other two:
$$\begin{align}
4&=b+c\\
4&=a+c
\end{align}$$
This is a system of two equations with three unknowns, and it has an infinitude of solutions, and there is no constant value for $a+b+c$. However, $a+b+2c=8$. If this was a textbook exercise, are you sure it was presented asking for $a+b+c$, not $a+b+2c$?
|
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|
Number of distinct arrangement of $(a,b,c,d,e)$
If $a<b<c<d<e $ be
positive integer such that
$a+b+c+d+e=20$.
Then number of distinct
arrangement of $(a,b,c,d,e)$ is
Here the largest value of $e$ is $10$
like $a\ b\ c\ d\ e$
as $ \ \ 1\ 2\ 3\ 4\ 10$
And least value is $6$
like $ a\ b\ c\ d\ e$
as $\ \ 2\ 3\ 4\ 5\ 6$
Now after that solution given in book as
Total number of ways
$ \displaystyle =\binom{4}{0}+\frac{\binom{4}{1}}{4}+\frac{\binom{4}{2}}{3}+\frac{\binom{4}{3}}{2}+\frac{\binom{4}{4}}{1}$
$\displaystyle = 1+1+2+2+1=7$
I did not understand last $2$ line
i e solution given in book
Please have a look on that part
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This is the same as the number of solutions in positive integers of:
$$5a+4w+3x+2y+z=20 \tag{1}, \\ \text{ where }w=b-a, x=c-b, y=d-c, z=e-d.$$
Equation $(1)$ tells us immediately that $a \in \{1, 2 \}$ and if $a=2$ there is only one possible solution.
Thus, we want to add $1$ to the number of possible solutions in positive integers of:
$$4w+3x+2y+z=15. \tag{2}$$
Again, $w \in \{1, 2 \}$. If $w=2$, there is again only one possible solution. Thus, we want to add $2$ to the number of possible solutions in positive integers of:
$$3x+2y+z=11. \tag{3}$$
Yet again, $x \in \{1, 2 \}$. This time, if $x=2$ there are two possible solutions. If $x=1$ there are three possible solutions. Thus, Equation $(3)$ has $5$ possible solutions in positive integers.
There are, therefore, $7$ possible solutions to the Equation $(1)$.
I really have no idea how the book is doing the counting.
|
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|
Verify my proof: "for all odd integers $a$ and $b$, $b^2-a^2 \neq 4$" I'm learning math without a math professor. I need some feedback from community regarding my proof.
The book is : "Discrete Mathematics with Applications" by Susanna S. Epp, 5th edition.
Exercise 16 from page 226. Please verify my proof.
Prove the following statement: "for all odd integers $a$ and $b$, $b^2-a^2 \neq 4$"
Proof. Let's prove by contradiction. Let say there are two integers $a$ and $b$ and $b^2-a^2 = 4$. Because $a$ and $b$ are odd integers, then $a=2n+1$ and $b=2k+1$, where $n \in Z$ and $k \in Z$. Now let's perform the following computations:
$b^2 - a^2 = (2k+1)^2 - (2n+1)^2 = 4k^2 + 4k + 1 - 4n^2 - 4n - 1 = 4(k^2 + k - n^2 - n) = 4(k^2 - n^2 + k - n) = 4[(k+n)(k-n)+(k-n)] = 4(k-n)(k+n+1) $
So, in order for the expression $b^2 - a^2 = 4(k-n)(k+n+1) $ to be true, $(k-n)(k+n+1)$ must be equal to 1 or -1. Then $k-n = 1$ and $k+n+1 = 1$ or $k-n = -1$ and $k+n+1 = -1$, because both of expressions should be equal to one (or minus one), we have :
$k-n = k+n+1 \implies -n = n + 1 \implies -n-n = 1 \implies -2n = 1 \implies n = -\frac{1}{2} $, so $n$ is not an integer $\blacksquare$
|
I'm merely critiquing your presentation.
Prove the following statement: "for all odd integers $a$ and $b$, $b^2-a^2 \neq 4$"
Proof. Let's prove by contradiction. Let say there are two integers $a$ and $b$ and $b^2-a^2 = 4$. Because $a$ and $b$ are odd integers, then $a=2n+1$ and $b=2k+1$, where $n \in Z$ and $k \in Z$.
I suggest replacing the "where $n\in\mathbb Z$" with "for some $n\in\mathbb Z$" (as opposed to "for all $n\in\mathbb Z$"), so that it is immediately clear that you are not referring to an arbitrary $n$ (consequently, an arbitrary odd integer).
Now let's perform the following computations:
$b^2 - a^2 = (2k+1)^2 - (2n+1)^2 = 4k^2 + 4k + 1 - 4n^2 - 4n - 1 = 4(k^2 + k - n^2 - n) = 4(k^2 - n^2 + k - n) = 4[(k+n)(k-n)+(k-n)] = 4(k-n)(k+n+1) $
So, in order for the expression $b^2 - a^2 = 4(k-n)(k+n+1) $ to be true,
This sentence is confusing: you've already, by assumption, shown it to be true. What you mean is instead something like "Equating both expressions for $b^2-a^2,$ we have that $4(k-n)(k+n+1)=4.$"
$(k-n)(k+n+1)$ must be equal to 1 or -1. Then $k-n = 1$ and $k+n+1 = 1$ or $k-n = -1$ and $k+n+1 = -1$, because both of expressions should be equal to one (or minus one), we have :
$k-n = k+n+1 \implies -n = n + 1 \implies -n-n = 1 \implies -2n = 1 \implies n = -\frac{1}{2} $, so $n$ is not an integer. $\blacksquare$
This is slightly confusing. If I'm merely skimming or if your text is dense, then it is not immediately clear that you are even asserting that $k-n = k+n+1$ (the leftmost antecedent) is true; in other words, it is not immediately clear whether you have actually derived that conclusion $n=-\frac12$ or whether it is provisional on that leftmost antecedent being true. I suggest using words like "so", "thus", "therefore" and "hence" instead, and reserving ⟹ for when you are merely asserting a conditional/implication.
|
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How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$? Background
As I had found the integral
$$I=\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^2} d x =\frac{\pi}{4}, $$
by using $x\mapsto \frac{1}{x}$ yields
$\displaystyle I=\int_0^{\infty} \frac{\frac{1}{x^2}}{\left(\frac{1}{x}+x\right)^2} d x\tag*{} $
Averaging them gives the exact value of the integral
$\displaystyle \begin{aligned}I & =\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2} d x \\& =\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+4} d x \\& =\frac{1}{4}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{2}\right)\right]_0^{\infty} \\& =\frac{1}{4}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right] \\& =\frac{\pi}{4}\end{aligned}\tag*{} $
I guess that we can similarly evaluate the general integral
$$
I_n=\int_0^{\infty} \frac{d x}{\left(x+\frac{1}{x}\right)^{2 n}}
$$
by mapping $x\mapsto \frac{1}{x}$ and then averaging.
$$
I_n=\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^{2 n}} d x=\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left[\left(x-\frac{1}{x}\right)^2+4\right]^n}
$$
Letting $x-\frac{1}{x}=\tan \theta$ yields
$$
\begin{aligned}
I_n & =\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2 \sec ^2 \theta d \theta}{4^n \sec ^{2 n} \theta} =\frac{1}{4^n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d \theta= \boxed{\frac{\pi(2 n-3) ! !}{4^n(2 n-2) ! !}}
\end{aligned}
$$
where the last answer comes from the Wallis cosine formula.
My question: How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$?
|
Letting $x=\tan \theta$ transforms the integral
$$
\begin{aligned}
I_n & = \int_0^{\infty} \frac{x^{2 n}}{\left(1+x^2\right)^{2 n}} d x \\&=\int_0^{\frac{\pi}{2}} \frac{\tan ^{2 n} \theta}{\sec ^{4 n} \theta} \sec ^2 \theta d \theta \\
& =\int_0^{\frac{\pi}{2}} \sin ^{2 n} \theta \cos ^{2n-2} \theta d \theta \\
& =\frac{1}{2} B\left(n+\frac{1}{2}, n-\frac{1}{2}\right)
\end{aligned}
$$
In particular,
$$
I_2=\frac{1}{2} B\left(\frac{5}{2}, \frac{3}{2}\right)=\frac{1}{2} \cdot \frac{\pi}{16}=\frac{\pi}{32}
$$
|
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|
Proving $\sin\frac{\pi}{13}+\sin\frac{3\pi}{13}+\sin\frac{4\pi}{13}=\frac12\sqrt{\frac{13+3\sqrt{13}}2}$
Prove that $$\sin\left(\frac{\pi}{13}\right)+\sin\left(\frac{3\pi}{13}\right)+\sin\left(\frac{4\pi}{13}\right)=\frac{1}{2}\sqrt{\frac{13+3\sqrt{13}}{2}}$$
My Attempt
Let $$x = \frac{1}{2}\sqrt{\frac{13+3\sqrt{13}}{2}} \implies 16x^4-52x^2+13=0$$
And through some donkey work we can calculate the chebyshev polynomial for $\sin\left(\frac{\pi}{13}\right),\sin\left(\frac{3\pi}{13}\right),\sin\left(\frac{4\pi}{13}\right)$ which will all be the same as $\sin(n\pi)=0,\text{ for all } n \in \mathbb{I} $, so
$$P(x) = 4096x^{12}-13312x^{10}+16640x^8-9984x^6+2912x^4-364x^2+13$$
where $x = \sin\left(\frac{2i\pi}{13}\right), \text{ from } 1 \le i \le 12 \text{ where } i \in \mathbb{I}$, are the roots of $P(x)$.
Now I am not getting how to connect these two into a possible solution and even it is possible (probably is), its still a pretty donkey method as you need to find the $13^{th}$ chebyshev polynomial, so if possible maybe give some another method of approach to this question.
|
Let
$$
s=\sin\left(\frac{\pi}{13}\right)+\sin\left(\frac{3\pi}{13}\right)+\sin\left(\frac{4\pi}{13}\right)\tag1
$$
If $\alpha=e^{\pi i/13}$, then
$$
\alpha^{13}+1=0\tag2
$$
and
$$
2is=\underbrace{\alpha+\overbrace{\quad\alpha^{12}\quad}^{-\alpha^{-1}}}_{2i\sin\left(\frac{\pi}{13}\right)}+\underbrace{\alpha^3+\overbrace{\quad\alpha^{10}\quad}^{-\alpha^{-3}}}_{2i\sin\left(\frac{3\pi}{13}\right)}+\underbrace{\alpha^4+\overbrace{\quad\alpha^9\quad}^{-\alpha^{-4}}}_{2i\sin\left(\frac{4\pi}{13}\right)}\tag3
$$
Using the approach in this answer, we get that the minimal polynomial of $x=2is$ is
$$
x^5+13x^4+13x=0\tag4
$$
Since we don't want $s=0$, we have
$$
x^4+13x^2+13=0\tag5
$$
which has roots
$$
x\in\left\{\pm i\sqrt{\frac{13\pm3\sqrt{13}}2}\right\}\tag6
$$
Since $\frac2\pi\theta\le\sin(\theta)\le\theta$ for $0\le\theta\le\frac\pi2$, we have $\frac{16}{13}\lt s\lt\frac{8\pi}{13}$. Therefore, we must have
$$
s=\sqrt{\frac{13+3\sqrt{13}}8}\tag7
$$
|
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Do we have a simpler method for computing $\int_{-\infty}^{\infty} \frac{\ln \left(x^2+ax+b\right)}{1+x^2} d x$, where $b> \frac{a^2}{4} $? Background
After finding the exact value of the integral in my post, I start to investigate a similar integral
$$I(a):
=\int_{-\infty}^{\infty} \frac{\ln \left(x^2+ax+b\right)}{1+x^2} d x=\int_{-\infty}^{\infty} \frac{\ln \left[\left(x+\frac{a}{2}\right)^2+\left(b-\frac{a^2}{4}\right)\right] d x}{1+x^2}$$
where $b> \frac{a^2}{4}.$
By Contour integration along the upper semi-circle
Using the fact that $\ln \left(x^2+y^2\right)=2 \operatorname{Re}(\ln (x+y i))$ to reduce the $x^2$ to $x$ and making the branch point of $\ln$ below the real axis, we change the integral into
$$
$$
\begin{aligned}
I(a) & =2 \operatorname{Re} \int_{-\infty}^{\infty} \frac{\ln \left(x+\frac{a}{2}+i \sqrt{b-\frac{a^2}{4}}\right)}{1+x^2} d x \\
& =2 \operatorname{Re}\left[2 \pi i \lim _{z \rightarrow i} \frac{\ln \left(z+\frac{a}{2}+i \sqrt{b-\frac{a^2}{4}}\right)}{z+i}\right] \\
& =2 \operatorname{Re}\left[2 \pi i \frac{\ln \left(i+\frac{a}{2}+i \sqrt{b-\frac{a^2}{4}}\right)}{2 i}\right] \\
& = \pi \ln \left(1+b+\sqrt{4 b^2-a^2}\right)
\end{aligned}
For example,
$$
\begin{aligned}& \int_{-\infty}^{\infty} \frac{\ln \left(x^2+x+\frac{1}{2} \right)}{1+x^2} d x =\pi \ln \left(\frac{5}{2}\right) \\
& \int_{-\infty}^{\infty} \frac{\ln \left(x^2+x+1\right)}{1+x^2} d x =\pi \ln (2+\sqrt{3})
\end{aligned}
$$
Do we have a simpler method for computing $$\int_{-\infty}^{\infty} \frac{\ln \left(x^2+ax+b\right)}{1+x^2} d x,$$ where $b> \frac{a^2}{4} $?
|
Consider, instead
\begin{align}
&\int_{-\infty}^{\infty} \frac{\ln (x^2+2x\sqrt b \sin \theta+b)}{x^2+1} \ d x\\
=& \int_{-\infty}^{\infty}\bigg(\ln (x^2+b)+ \int_0^{\theta}\frac{2x\sqrt b \cos t}{x^2+2x\sqrt b \sin t+b} dt\bigg) \frac{dx}{x^2+1}\\
=& \ 2\pi\ln(1+\sqrt b) -\int_0^{\theta}\frac{2\pi \sqrt b \sin t}{1+2\sqrt b \cos t+b} dt
= \overset{}{\pi}\ln \left(1+b+2\sqrt b \cos\theta\right)
\end{align}
and then set $a=2\sqrt b \sin \theta$ to obtain
\begin{align}
&\int_{-\infty}^{\infty} \frac{\ln (x^2+ax+b)}{x^2+1} \ d x
= \pi \ln \left(1+b+\sqrt{4 b-a^2}\right)
\end{align}
|
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|
Evaluate $\int_{0}^{\pi/2}x\sin^a (x) dx$, $a>0$
I want to evaluate $$\int_{0}^{\pi/2}x\sin^a (x)\, dx$$ where $a>0$ is a real number.
I tried: $$I(a)= \int_{0}^{\pi/2}x\sin^a(x)\,dx = \int_{0}^{1}\frac{\arcsin x}{\sqrt{1-x^2}}x^a\,dx$$
$$ I(a)=\sum_{m\geq 1}\frac{4^m}{2m\left(2m+a\right)\binom{2m}{m}}$$
$$I(a)=\frac{1}{a+2}\cdot\phantom{}_3 F_2\left(1,1,1+\tfrac{a}{2};\tfrac{3}{2},2+\tfrac{a}{2};1\right)$$
Any other method please.
Any help will be appreciated. Thank you.
edit The series expansion of $ \arcsin(x)^2$ is
$$2\;\arcsin(x)^2=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2\binom{2n}{n}}$$
Differentiating the above series we get the formula used in the question.
|
An elementary (enough) deduction of the (already mentioned) equality $$\boxed{I(a)=\frac{\pi^{3/2}}{4}\frac{\Gamma\left(\frac{a+1}2\right)}{\Gamma\left(\frac{a+2}2\right)}-\sum_{n=1}^\infty\frac{\prod_{k=1}^{n-1}(a+2k)^2}{\prod_{k=1}^{2n}(a+k)}.\quad(\Re a>-1)}$$
We note that $$(a+1)\big(I(a)-I(a+2)\big)=\int_0^{\pi/2}x\cos x\,(\sin^{a+1}x)'\,dx$$ and integrate by parts; this yields the recurrence $$I(a)=\frac{a+2}{a+1}I(a+2)-\frac1{(a+1)(a+2)}.$$
Reusing it with $a+2$ in place of $a$, we get $$I(a)=\frac{(a+2)(a+4)}{(a+1)(a+3)}I(a+4)-\frac1{(a+1)(a+2)}-\frac{a+2}{(a+1)(a+3)(a+4)}$$ and then, by induction (I'm writing it expanded intentionally),
\begin{align}
I(a)&=\frac{(a+2)(a+4)\cdots(a+2n)}{(a+1)(a+3)\cdots(a+2n-1)}I(a+2n)
\\&-\frac{1}{(a+1)(a+2)}-\frac{a+2}{(a+1)(a+3)(a+4)}
\\&-\dots-\frac{(a+2)(a+4)\cdots(a+2n-2)}{(a+1)(a+3)\cdots(a+2n-1)(a+2n)}.
\end{align}
And now we take $n\to\infty$. The fact that $$\lim_{n\to\infty}\frac{(a+2)(a+4)\cdots(a+2n)}{(a+1)(a+3)\cdots(a+2n-1)}\frac1{\sqrt n}=\frac{\Gamma\left(\frac{a+1}2\right)}{\Gamma\left(\frac{a+2}2\right)}$$ may be obtained using infinite product representations of $\Gamma$, or just the known limit $$\lim_{x\to\infty}\frac{\Gamma(x+a)}{x^a\,\Gamma(x)}=1.$$
And Laplace's method gives the remaining piece: $$\lim_{a\to\infty}I(a)\sqrt{a}=(\pi/2)^{3/2}\implies\lim_{n\to\infty}I(a+2n)\sqrt{n}=\frac{\pi^{3/2}}4.$$
|
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|
If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$. I have the following question:
If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$.
Using the first equation, I rearrange to get $x^2=kx-1$. Then, I multiply both sides by x to get get $x^3=(kx-1)^{1.5}$. I can’t think of any other way than to substitute $(kx-1)^{1.5}$ for $x^3$ in the second equation. Ideas?
|
I don't see any way to continue from what you've done, since you're dealing with fractional powers of a polynomial, which are generally hard to work with. Instead, since $x \neq 0$, we can divide both sides by $x$ below to get
$$x^2 + 1 = kx \;\; \to \; \; x+\frac{1}{x}=k$$
Cubing both sides then gives
$$\begin{equation}\begin{aligned}
x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} & = k^3 \\
x^3 + 3\left(x + \frac{1}{x}\right) + \frac{1}{x^3} & = k^3 \\
x^3 + 3k + \frac{1}{x^3} & = k^3 \\
x^3 + \frac{1}{x^3} & = k^3 - 3k
\end{aligned}\end{equation}$$
|
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Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!
Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
$2 \times 0 = 0.$ So it is divisible by $3.$
Induction: Assume that for an arbitrary natural number $n$,
$n^3+ 2n$ is divisible by $3.$
Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use
the induction hypothesis. Got it
$$( n + 1 )^3+ 2( n + 1 ) = ( n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$
$$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying
and regrouping}\}$$
$$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out
the 3}\}$$
which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$
by the induction hypothesis. What?
Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$
|
$$n^3+2n=n(n^2+2)$$
If $n$ is divisible by $3$, then obviously, so is $n^3+2n$ because you can factor out $n$.
If $n$ is not divisible by $3$, it is sufficient to show that $n^2+2$ is divisible by 3. Now, if $n$ is not divisible by $3$, $n=3k+1$ or $n=3k+2$ for some integer $k$. Plug that into $n^2+2$ and you'll get $9k^2+6k+3$ and $9k^2+6k+6$ respectively. Both of which are divisible by 3.
|
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|
Area of a quadrilateral
The perpendicular bisector of the line
joining $A(0,1)$ and $C(-4,7)$ intersects
the $x$-axis at $B$ and the $y$-axis at $D$.
Find the area of the quadrilateral.
Thank you in advance!
|
The perpendicular bisector will pass through the mid-point: $(-2, 4)$ of the line passing through the points $A(0, 1)$ & $C(-4, 7)$ & have a slope: $\frac{2}{3}$ normal to the line $AC$.
The equation of perpendicular bisector is given as $$y-4=\frac{2}{3}(x-(-2))$$ $$\implies y=\frac{2x+4+12}{3}=\frac{2x+16}{3}$$ By substituting $y=0$ & $x=0$ respectively, the points of intersection of perpendicular bisector with the axes are determined as : $B(-8, 0)$ & $D\left(0, \frac{16}{3}\right)$. Now divide the quadrilateral $ABCD$ into two triangles $\Delta ABC$ & $\Delta ACD$ Thus we have $$\text{area of quadrilateral}\space ABCD=\text{area of}\space \Delta ABC \space \text{with vertices}\space (0, 1),(-8, 0)\space \text{&}\space (-4, 7)+\text{area of }\space \Delta ACD \space \text{with vertices}\space (0, 1),(-4, 7)\space \text{&}\space \left(0, \frac{16}{3}\right)$$ $$\implies \text{area of quadrilateral}\space ABCD=\frac{1}{2}\left|0(0-7)-8(7-1)-4(1-0)\right|+\frac{1}{2}\left|0\left(7-\frac{16}{3}\right)-4\left(\frac{16}{3}-1\right)+0(1-7)\right|$$ $$=\frac{1}{2}\left|-60\right|+\frac{1}{2}\left|\frac{-52}{3}\right|=30+\frac{26}{3}$$$$=\frac{116}{3}\space \text{sq.unit}$$
|
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|
Proving $ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $ How to prove this binomial identity :
$$ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$
The left hand side arises while solving a standard binomial problem the right hand side is given in the solution , I checked using induction that this is true but I am inquisitive to prove it in a rather general way.
EDIT: I have gone through all of the answers posted here,I particularly liked Isaac♦
answers after which it was not much difficult for me to figure out something i would rather say an easy and straight algebraic proof, I am posting it here if somebody needs in future :
$$ { 2n \choose n } = \frac{(2n)!}{n! \cdot n!} $$
$$ = \frac{ 1 \cdot 2 \cdot 3 \cdots n \cdot (n+1) \cdots (2n-1)\cdot (2n) }{n! \cdot n!}$$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n! \cdot n!} $$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [(2.1) \cdot (2.2) \cdot (2.3) \cdots (2.n)]}{n! \cdot n!} $$
$$= \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [ 2^n \cdot (n)! ]}{n! \cdot n!} $$
$$ = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$
I do always welcome your comments :-)
|
Your original identity, ${ 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!}$, can be rewritten (by multiplying both sides by $(n!)^2$) as $(2n)!=2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)\cdot n!$. Now, $2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)\cdot n!=$ $1\cdot 3\cdot 5\cdots (2n-1)\cdot 2\cdot 4\cdot 6\cdots 2n=$ $1\cdot 2\cdot 3\cdots (2n-1)(2n)=(2n)!$.
|
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|
The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
|
Define the following series for $ x > 0 $
$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\quad.$$
Now substitute $ x = \sqrt{y}\ $ to arrive at
$$\frac{\sin \sqrt{y}\ }{\sqrt{y}\ } = 1 - \frac{y}{3!}+\frac{y^2}{5!}-\frac{y^3}{7!}+\cdots\quad.$$
if we find the roots of $\frac{\sin \sqrt{y}\ }{\sqrt{y}\ } = 0 $ we find that
$ y = n^2\pi^2\ $ for $ n \neq 0 $ and $ n $ in the integers
With all of this in mind, recall that for a polynomial
$ P(x) = a_{n}x^n + a_{n-1}x^{n-1} +\cdots+a_{1}x + a_{0} $ with roots
$ r_{1}, r_{2}, \cdots , r_{n} $
$$\frac{1}{r_{1}} + \frac{1}{r_{2}} + \cdots + \frac{1}{r_{n}} = -\frac{a_{1}}{a_{0}}$$
Treating the above series for $ \frac{\sin \sqrt{y}\ }{\sqrt{y}\ } $ as polynomial we see that
$$\frac{1}{1^2\pi^2} + \frac{1}{2^2\pi^2} + \frac{1}{3^2\pi^2} + \cdots = -\frac{-\frac{1}{3!}}{1}$$
then multiplying both sides by $ \pi^2 $ gives the desired series.
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6}$$
|
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|
Funny identities Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?
|
$$
\dfrac{1}{2}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\cdots}}}}}}
$$
and more generally we have
$$
\dfrac{1}{n+1}=\frac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\frac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\ddots}}}}}}
$$
|
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|
Find a closed form for $\sum_{k=1}^{x-1} a^{1/k}$ Please find a closed form for partial sum of a function
$$f(x)=a^{1/x}$$
I want it to be expressed in terms of bounded number of elementary functions and/or well known special functions.
No computer algebra systems I have tried so far could find a satisfactory solution. I believe that the expression can exist in the terms of incomplete Gamma function or its generalizations because indefinite integral of this function can be expressed in terms of incomplete Gamma function:
$$\int f(x) dx= x\sqrt[x]{a}-\operatorname{Ei}\left(\frac{\ln a}{x}\right)\ln a$$
|
Just for fun here are some bounds for $\sum_{k=1}^n a^{1/k}$ for $ a \ge 1.$
We have
$$a^{1/k} = 1 + \frac{\log a}{1! k} + \frac{(\log a)^2}{2! k^2}
+ \frac{(\log a)^3}{3! k^3} + \cdots$$
and so
$$\sum_{k=1}^n a^{1/k} = \zeta_n(0) + (\log a)\zeta_n(1) + \frac{(\log a)^2}{2!}\zeta_n(2) +
\frac{(\log a)^3}{3!}\zeta_n(3) + \cdots$$
where $\zeta_n(r) = \sum_{k=1}^n 1/k^r.$
Thus $\zeta_n(0)=n$ and $\zeta_n(1)=H_n = 1 + 1/2 + 1/3 + \cdots + 1/n.$
Now $H_n= \log(n + 1/2) + \gamma + \epsilon(n),$ where $0< \epsilon(n)< 1/24n^2$ and $\gamma$ is the Euler-Mascheroni constant,
and (by comparing the sum with the integral of $1/x^r$)
$$ \frac{1}{(r-1)(n+1)^{r-1}} < \sum_{k=n+1}^\infty \frac{1}{k^r} <
\frac{1}{(r-1)n^{r-1}}.$$
Hence
$$ \frac{1}{(r-1)(n+1)^{r-1}} < \zeta(r) - \zeta_n(r) <
\frac{1}{(r-1)n^{r-1}}$$
and so we have
$$ l(a) \le \sum_{k=1}^n a^{1/k} \le u(a) $$
where
$$l(a) = n + (\log a) \left( \log (n+ 1/2) + \gamma \right)
+\frac{(\log a)^2}{2!} \left( \zeta(2) - \frac{1}{n} \right)$$
$$+\frac{(\log a)^3}{3!} \left( \zeta(3) - \frac{1}{2n^2} \right)
+\frac{(\log a)^4}{4!} \left( \zeta(4) - \frac{1}{3n^3} \right) + \cdots$$
and
$$u(a) = n + (\log a) \left( \log (n+ 1/2) + \gamma + \frac{1}{24n^2} \right)
+\frac{(\log a)^2}{2!} \left( \zeta(2) - \frac{1}{n+1} \right)$$
$$+\frac{(\log a)^3}{3!} \left( \zeta(3) - \frac{1}{2(n+1)^2} \right)
+\frac{(\log a)^4}{4!} \left( \zeta(4) - \frac{1}{3(n+1)^3} \right) + \cdots.$$
Providing $a$ is not enormous and $n$ is sufficiently large, we only need a few terms for a good approximation to our sum.
|
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|
Proving an identity involving terms in arithmetic progression. If $a_1,\ldots,a_n$ are in arithmetic progression and $a_i\gt 0$ for all $i$, then how to prove the following two identities:
$ (1)\large \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$
$(2) \large\frac{1}{a_1 \cdot a_n} + \frac{1}{a_2 \cdot a_{n-1}} + \frac{1}{a_3 \cdot a_{n-2}}+ \cdots + \frac{1}{a_n \cdot a_1} = \frac{2}{a_1 + a_n} \biggl( \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} \biggr)$
|
For the first one, use induction (or) note that $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}$, where $d$ is the common difference between the successive terms. Now use the telescopic summation to cancel out the terms in the numerator and massage it to get the final expression on the right hand side.
For the second one, try to write each term on the Left Hand Side as a difference of two terms and proceed.
|
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|
Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $ How to find the value of
$$\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$$
manually ?
|
If $\cos3\theta=\frac12,$
$$\tan3\theta\cot\theta-4\cos\theta$$
$$=\frac{\sin3\theta\cos\theta}{\cos3\theta\sin\theta}-4\cos\theta$$
$$=\frac{\sin3\theta\cos\theta-4\cos3\theta\sin\theta\cos\theta}{\cos3\theta\sin\theta}$$
$$=\frac{\sin3\theta\cos\theta-\sin2\theta}{\frac12\sin\theta}\text { as }\sin2x=2\sin x\cos x\text{ and } \cos3\theta=\frac12$$
$$=\frac{\sin4\theta+\sin2\theta-2\sin2\theta}{\sin\theta}\text { applying } 2\sin A\cos B=\sin(A+B)+\sin(A-B)$$
$$=\frac{\sin4\theta-\sin2\theta}{\sin\theta}$$
$$=\frac{2\sin\theta\cos3\theta}{\sin\theta}\text { applying } \sin 2C-\sin 2D=2\sin(C-D)\cos(C+D)$$
$$=2\cos3\theta=1$$
Now, $$\cos3\theta=\frac12=\cos60^\circ\implies 3\theta=2n180^\circ\pm60^\circ\text{ where } n \text{ is any integer}$$
So, $\theta=(6n\pm1)20^\circ$ where $n$ is any integer
|
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|
Proving trignometrical identities: If $\sin(A+B) + \sin(B+C) + \cos(C-A) = \frac{3}{2}$ show that, $$1.\sin A + \cos B + \sin C = 0$$ $$2. \cos A + \sin B +\cos C = 0$$
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The problem seems to be missing some assumptions, as noted by Americo.
For instance, If $B = 0$ and $A=C$ are acute angles, such that $\sin A = 1/4$ the we have that
$\sin(A+B) + \sin(B+C) + \cos(A-C) = 3/2$, but none of
$\sin A + \cos B + \sin C$ or $\cos A + \sin B + \cos C$ are $0$.
In any case, this looks like a perfect problem for using complex numbers.
If $B' = \pi/2 - B$ and
$z_1 = \cos A + i \sin A$
$z_2 = \cos B' + i \sin B'$
$z_3 = \cos C + i \sin C$
The given identity is $\cos (A-B') + \cos (C - B') + \cos (A-C) = 3/2$
i.e.
$$\frac{z_1}{z_2} +\frac{z_2}{z_1} +\frac{z_3}{z_2} +\frac{z_2}{z_3} +\frac{z_1}{z_3} +\frac{z_3}{z_1} = 3$$
The two identities
*
*$\sin A + \cos B + \sin C = 0$
*$\cos A + \sin B +\cos C = 0$
are equivalent to showing that $z_1 + z_2 + z_3 = 0$.
Eq 1, says that the imaginary part of $z_1 + z_2 + z_3$ is $0$ and Eq 2 says that the real part is $0$.
Hope that helps.
|
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|
Nice expression for minimum of three variables? As we saw here, the minimum of two quantities can be written using elementary functions and the absolute value function.
$\min(a,b)=\frac{a+b}{2} - \frac{|a-b|}{2}$
There's even a nice intuitive explanation to go along with this: If we go to the point half way between two numbers, then going down by half their difference will take us to the smaller one. So my question is: "Is there a similar formula for three numbers?"
Obviously $\min(a,\min(b,c))$ will work, but this gives us the expression:
$$\frac{a+\left(\frac{b+c}{2} - \frac{|b-c|}{2}\right)}{2} - \frac{\left|a-\left(\frac{b+c}{2} - \frac{|b-c|}{2}\right)\right|}{2},$$
which isn't intuitively the minimum of three numbers, and isn't even symmetrical in the variables, even though its output is. Is there some nicer way of expressing this function?
|
First, define
$$
\Delta=|a-b|+|b-c|+|c-a|\newcommand{\Mu}{\mathrm{M}}\tag{1}
$$
It is somewhat intuitive that
$$
\frac{\Delta}{2}=\max(a,b,c)-\min(a,b,c)\tag{2}
$$
For example, if $a\ge b\ge c$ then $|a-b|+|b-c|+|c-a|=2a-2c$.
Next, define
$$
\Sigma=a\left(1-\frac{|b-c|}{\Delta}\right)+b\left(1-\frac{|c-a|}{\Delta}\right)+c\left(1-\frac{|a-b|}{\Delta}\right)\tag{3}
$$
Again, if $a\ge b\ge c$, then
$$
\begin{align}
\Sigma
&=a\left(\frac{2a-b-c}{2(a-c)}\right)+b\left(\frac{a-c}{2(a-c)}\right)+c\left(\frac{a+b-2c}{2(a-c)}\right)\\
&=a+c
\end{align}
$$
Thus, considering the symmetry of $(3)$, it is evident that
$$
\Sigma=\max(a,b,c)+\min(a,b,c)\tag{4}
$$
Combining $(2)$ and $(4)$ yields
$$
\max(a,b,c)=\frac{\Sigma}{2}+\frac{\Delta}{4}\tag{5}
$$
and
$$
\min(a,b,c)=\frac{\Sigma}{2}-\frac{\Delta}{4}\tag{6}
$$
At least $(5)$ and $(6)$ are symmetric in $a$, $b$, and $c$ since $(1)$ and $(3)$ are. That is,
$$
\begin{align}
\max(a,b,c)
&=\frac{a}{2}\left(\frac{|c-a|+|a-b|}{|a-b|+|b-c|+|c-a|}\right)\\
&+\frac{b}{2}\left(\frac{|a-b|+|b-c|}{|a-b|+|b-c|+|c-a|}\right)\\
&+\frac{c}{2}\left(\frac{|b-c|+|c-a|}{|a-b|+|b-c|+|c-a|}\right)\\
&+\frac{|a-b|+|b-c|+|c-a|}{4}\tag{7}
\end{align}
$$
and
$$
\begin{align}
\min(a,b,c)
&=\frac{a}{2}\left(\frac{|c-a|+|a-b|}{|a-b|+|b-c|+|c-a|}\right)\\
&+\frac{b}{2}\left(\frac{|a-b|+|b-c|}{|a-b|+|b-c|+|c-a|}\right)\\
&+\frac{c}{2}\left(\frac{|b-c|+|c-a|}{|a-b|+|b-c|+|c-a|}\right)\\
&-\frac{|a-b|+|b-c|+|c-a|}{4}\tag{8}
\end{align}
$$
|
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Does closed-form expression exist for integral $\int \frac{dx}{1-x^n}$ Playing with integrals on the form $$\int \frac{1}{1-x^n}\,dx$$ I noticed that for odd values of n > 5, it doesn't appear to be possible to express the integral as a closed-form expression. Is this true? How can I prove it?
|
All of those integrals have closed form expressions, for both even and odd $n$.
The polynomial $x^n-1$ can be factored over the complex numbers as
$$x^n - 1 = (x-1)(x-\zeta)(x-\zeta^2)\cdots(x-\zeta^{n-1}),$$
and when $n$ is odd, the only real solution is $x-1$ and the complex factors pair up in conjugate pairs to give irreducible real quadratic polynomials. So the integral can be written as
$$\int\frac{1}{(1-x)Q_1(x)\cdots Q_{r}(x)}\,dx$$
where $r=\frac{n-1}{2}$ and $Q_i(x)$ is an irreducible quadratic polynomial.
Using the method of partial fractions, you can then rewrite the integral as a sum of integrals
$$\int \frac{1}{1-x^n}\,dx = \int\frac{A}{1-x}\,dx + \int\frac{B_1x+C_1}{Q_1(x)}\,dx + \cdots + \int\frac{B_rx + C_r}{Q_r(x)}\,dx$$
with $A$, $B_i$, and $C_i$ all constants. The problem then reduces to solving these integrals.
The first integral is easy to do with a change of variable, as is any integral of the form $\int \frac{x}{Q_i(x)}\,dx$ with $Q_i$ irreducible quadratic (see below for how to them). So you only need to know how to solve integrals of the form $\int\frac{1}{Q_i(x)}\,dx$; these can be done by completing the square and doing a change of variable, which will lead to an integral of the form $\int\frac{du}{u^2+1}$, which has a closed form. Again, see below for the general method.
In short, they all have closed forms.
When $n$ is even, the only difference is that you have a second linear factor in the factorization, $x+1$, and that the number of irreducible quadratics is then $\frac{n-2}{2}$ instead of $\frac{n-1}{2}$.
To see this in action, let's consider the case of $n=7$. The polynomial $1-x^7$ factors as
\begin{align*}
(1-x^7) &= (1-x)(x-\zeta)(x-\zeta^2)(x-\zeta^3)(x-\zeta^4)(x-\zeta^5)(x-\zeta^6)\\
&= (-x)\Bigl((x-\zeta)(x-\zeta^6)\Bigr)\Bigl((x-\zeta^2)(x-\zeta^5)\Bigr)\Bigl((x-\zeta^3)(x-\zeta^4)\Bigr)
\end{align*}
Now, $\zeta=\cos(\frac{2\pi}{7}) + i\sin(\frac{2\pi}{7})$. It is now easy to check that
\begin{align*}
(x-\zeta)(x-\zeta^6) &= x^2 -2x\cos\frac{2\pi}{7} + 1\\
&= x^2 -c_2 x + 1\\
(x-\zeta^2)(x-\zeta^5) &= x^2 - 2x\cos\frac{4\pi}{7} + 1\\
&= x^2 -c_4x + 1\\
(x-\zeta^3)(x-\zeta^4) &= x^2 - 2x\cos\frac{6\pi}{7} + 1\\
&= x^2 - c_6x + 1
\end{align*}
where for simplicity we have $c_j = 2\cos\frac{j\pi}{7}$.
The discriminant of each of these quadratic polynomials is $4(\cos^2\frac{2k\pi}{7} -1)$, which is always negative, so they are all irreducible quadratics.
Now use the method of partial fractions to obtain a closed form for the integral. It will be in the form of the sum of natural logs and arctangent functions.
In general, the polynomial will have no repeated quadratic factors, which makes it very easy to solve using partial fractions. The irreducible quadratic factors with $n$ odd will be
$$x^2 -2x\cos\frac{2k\pi}{n} + 1,\qquad k=1,\ldots,\frac{n-1}{2}.$$
When $n$ is even, the irreducible quadratic factors will be the same, but $k$ only goes up to $\frac{n-2}{2}$.
Just to give you a guide on solving those integrals in general:
*
*An integral of the form $\int\frac{1}{ax+b}\,dx$ can be solved by doing the change of variable $u=ax+b$.
*An integral of the form $\int\frac{1}{x^2+1}\,dx$ is immediate: you get $\arctan(x)+C$.
*An integral of the form $\int\frac{1}{(x-a)^2+b^2}\,dx$ can be solved by first factoring out $b^2$ from the denominator, then doing the change of variable $u = \frac{x-a}{b}$, which brings it to the form in 2.
*If $Q(x) = x^2+Bx+C$ is irreducible quadratic, it can always be rewritten as $(x-a)^2 + b^2$ for some real numbers $a$ and $b$ (complete the square).
*If $x^2+Bx+C$ is irreducible quadratic, then the integral of the form $\int\frac{x}{x^2+Bx+C}\,dx$ can be solved by rewriting as $\frac{1}{2}\int\frac{2x+B}{x^2+Bx+C}\,dx - \frac{B}{2}\int\frac{1}{x^2+Bx+C}\,dx$. The first integral can be solved by the change of variable $u=x^2+Bx+C$, the second using the method in 3.
Since you can rewrite $\int\frac{1}{1-x^n}\,dx$ into a sum of integrals of the forms described above, there is a closed form for it for all $n$.
Added: In principle, any integral of a rational function (polynomial divided by polynomial) has a closed form, which can be obtained by factoring the denominator into irreducible factors and using a partial fractions decomposition. This is a consequence of the Fundamental Theorem of Algebra; in fact, it was this particular application, the integration of rational functions, which provided the early impetus to prove the Fundamental Theorem of Algebra.
The only practical obstacle to finding the closed form is our ability to factor the denominator into irreducible factors; repeated irreducible quadratic factors introduce some minor complications that are resolved by using integration by parts to obtain reduction formulas. In the case of the polynomials you were looking at, though, factoring the denominator is straightforward, so that there is no practical difficulty in finding a closed form.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find matrices $A$ and $B$ given $AB$ and $BA$ Given that:
$$AB= \left[ {\matrix{
3 & 1 \cr
2 & 1 \cr
} } \right]$$
and
$$BA= \left[ {\matrix{
5 & 3 \cr
-2 & -1 \cr
} } \right]$$
find $A$ and $B$.
|
First note that both the products are of full rank and hence $A$ and $B$ are also of full rank. So we would expect one set of solutions with one degree of freedom since if $A$ and $B$ satisfy our equations, so will $kA$ and $\frac{1}{k}B$.
Let $$A = \left[ {\matrix{
a_1 & a_2 \cr
a_3 & a_4 \cr
} } \right],B = \left[ {\matrix{
b_1 & b_2 \cr
b_3 & b_4 \cr
} } \right]$$
$$AB= \left[ {\matrix{
3 & 1 \cr
2 & 1 \cr
} } \right], BA= \left[ {\matrix{
5 & 3 \cr
-2 & -1 \cr
} } \right]$$
Hence, $$B \left[ {\matrix{
3 & 1 \cr
2 & 1 \cr
} } \right] = B(AB) = (BA)B= \left[ {\matrix{
5 & 3 \cr
-2 & -1 \cr
} } \right] B$$
\begin{align*}
3b_1 + 2b_2 & = 5b_1 + 3b_3\\
b_1 + b_2 & = 5b_2 + 3b_4\\
3b_3 + 2b_4 & = -2b_1 - b_3\\
b_3 + b_4 & = -2b_2 - b_4
\end{align*}
Rearranging, we get,
\begin{align*}
2b_2 & = 2b_1 + 3b_3\\
b_1 & = 4b_2 + 3b_4\\
2b_1 + 4b_3 + 2b_4 & = 0\\
2b_2 + b_3 + 2b_4 & = 0
\end{align*}
Set $b_2 = k$, and $b_4 = m$, we get $b_1 = 4k+3m$ and $b_3 = -2(k+m)$
Hence, $B = \left[ {\matrix{
4k+3m & k \cr
-2(k+m) & m \cr
} } \right]$ where $k,m \neq 0$, $B^{-1} = \frac{1}{3m^2+6mk+2k^2} \left[ {\matrix{
m & -k \cr
2(k+m) & 4k+3m \cr
} } \right]$.
Hence, $A = B^{-1} \times \left[ {\matrix{
5 & 3 \cr
-2 & -1 \cr
} } \right]$
$A = \frac{1}{3m^2+6mk+2k^2} \left[ {\matrix{
5m+2k & 3m+k \cr
2k+4m & 2k+3m \cr
} } \right]$
There are two degrees of freedom.
EDIT
Thanks to Qiang Li for pointing out the other degree of freedom
|
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|
How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$? I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
|
As $42=2.3.7$. Therefore,we need to check that $n^7-n$ is divisible by $2,3$ and 7.
For divisibility by $2$, by fermat's little theorem, $n^2=n\pmod 2 \implies {(n^2)}^3.n=n^4\pmod 2=n^2\pmod 2=n\pmod 2 \implies n^7-n=0\pmod2$.
For divisibility by 3, $n^3=n\pmod 3\implies n^7=n^3\pmod 3=n\pmod 3 \implies n^7-n=0\pmod 3$. For divisibility by 7, $n^7=n\pmod 7 \implies n^7-n=0\pmod 7$. These relations implies that $42|(n^7-n)$.
|
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|
Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four incongruent solutions
Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four incongruent solutions.
My attempt:
We have,
$x^2 - 1 = (x - 1) \times (x + 1)$, then
$(x - 1)(x + 1) \equiv 0 \pmod{2^k}$
which implies,
$2^k|(x - 1)$ or $2^k|(x + 1) \implies x \equiv \pm 1 \pmod{2^k} (1)$
Furthermore, $2^{k-1} \equiv 0 \pmod{2^k} \Leftrightarrow 2^{k-1} + 1 \equiv 1 \pmod{2^k}$.
Multiply both sides by $-1$, we have another congruent namely $-(2^{k-1} + 1) \equiv -1 \pmod{2^k}$
Hence, $x \equiv \pm(1 + 2^{k-1}) \pmod{2^k} (2)$
From $(1)$ and $(2)$, we can conclude that $x^{2} \equiv 1 \pmod{2^k}$ have four incongruent solutions.
Am I in the right track?
Thanks,
|
Existence is easy:
The solutions are $\{1,2^{k-1}-1,2^{k-1}+1,2^k-1\}$.
Squaring them gives $\{1, 2^{2k-2}-2^k+1, 2^{2k-2}+2^k+1, 2^{2k}-2^{k+1}+1\}$.
Reducing $\pmod {2^k}$ gives 1 in each case (given that $2k-2 > k+1$, which forces $k \ge 3$).
For uniqueness you could use the fact that the units group of $\mathbb{Z}/2^k \mathbb{Z}$ is $C_2 \times C_{2^{k-2}}$ on the other hand that also gives existence..
|
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|
If $p - a \equiv -a \pmod{p}$ then what is $\frac{p-1}{2} \equiv ? \pmod{p}$? If $p - a \equiv -a \pmod{p}$ then what is $\frac{p-1}{2} \equiv ? \pmod{p}$? Where $p$ is an odd prime.
I read in the book, they claimed:
$$p - 1 \equiv -1 \pmod{p}$$
$$p - 2 \equiv -2 \pmod{p}$$
$$p - 3 \equiv -3 \pmod{p}$$
$$ ... $$
$$\frac{p - 1}{2} \equiv -(\frac{p - 1}{2}) \pmod{p}$$
However, I realized that the last statement was wrong because if it's true then:
$$\frac{p - 1}{2} + \frac{p - 1}{2}\equiv 0 \pmod{p}$$
$$p - 1 \equiv 0 \pmod{p}$$
Any idea?
Thanks,
|
The sequence as written doesn't make sense, since the last term does not follow the pattern of the previous ones. However, you can argue that
$$p - \left(\frac{p-1}{2}\right) \equiv -\frac{p-1}{2}\pmod{p}.$$
This does follow the same pattern as the rest of the terms.
And since
$$p - \frac{p-1}{2} = \frac{2p-p+1}{2} = \frac{p+1}{2}$$
then you have that
$$\frac{p+1}{2}\equiv -\frac{p-1}{2}\pmod{p},$$
hence $\frac{p-1}{2}\equiv -\frac{p+1}{2}\pmod{p}$.
Or peharps, it went like this:
\begin{align*}
p-1 &\equiv -1\pmod{p}\\
p-2 &\equiv -2 \pmod{p}\\
&\vdots\\
\frac{p-1}{2} = p-\left(\frac{p+1}{2}\right) &\equiv -\frac{p+1}{2}\pmod{p}
\end{align*}
and you have the wrong sign in the numerator on the right hand side?
Added. So, the rest of the proof. According to the scans Sivaram has posted, this was part of the computation of $\left(\frac{p-1}{2}\right)!^2$. The idea then is that we can get $\left(\frac{p-1}{2}\right)!$ by multiplying the terms on the right hand side, and then get it again by multiplying the terms on the left hand side. That is, we look at
\begin{align*}
p-1 &\equiv -1 \pmod{p}\\
p-2 &\equiv -2 \pmod{p}\\
&\vdots\\
p-\left(\frac{p-1}{2}\right) &\equiv -\frac{p-1}{2}\pmod{p}
\end{align*}
and note that if $p\equiv 3\pmod{2}$, then we have an odd number of negative signs on the right hand side. So we have that:
$$(-1)\left(\frac{p-1}{2}\right)! = (-1)\Bigl(1\times 2\times\cdots\times \frac{p-1}{2}\Bigr) \equiv (p-1)(p-2)\cdots\left(p - \frac{p-1}{2}\right)\pmod{p}.$$
Now, $p - \frac{p-1}{2} = \frac{p+1}{2} = \frac{p-1}{2}+1$. So therefore we have that:
\begin{align*}
\left(\frac{p-1}{2}\right)!^2 &= \left(\frac{p-1}{2}\right)!\left(\frac{p-1}{2}\right)!\\
&\equiv \left(\frac{p-1}{2}\right)!\Biggl( -(p-1)(p-2)\cdots\left(\frac{p-1}{2}+1\right)\Biggr)\pmod{p}\\
&\equiv -\left(1\times 2\times 3\times\cdots\times\left(\frac{p-1}{2}\right)\times\left(\frac{p-1}{2}+1\right)\times\cdots\times (p-1)\right)\pmod{p}\\
&\equiv -(p-1)!\pmod{p}.
\end{align*}
But we know that $(p-1)!\equiv -1\pmod{p}$ by Wilson's Theorem, so
$$\left(\frac{p-1}{2}\right)!^2 \equiv -(p-1)!\equiv -(-1) = 1\pmod{p},$$
so if we let $x = \left(\frac{p-1}{2}\right)!$, then $x^2 \equiv 1 \pmod{p}$. This means that $p|x^2-1 = (x-1)(x+1)$, so either $p|x-1$ or $p|x+1$. that is, eithe $x\equiv 1 \pmod{p}$ or $x\equiv-1\pmod{p}$; giving:
$$\text{either }\left(\frac{p-1}{2}\right)!\equiv -1\pmod{p}\quad\text{or}\quad \left(\frac{p-1}{2}\right)!\equiv 1\pmod{p}.$$
|
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|
Constants of integration in integration by parts After finishing a first calculus course, I know how to integrate by parts, for example, $\int x \ln x dx$, letting $u = \ln x$, $dv = x dx$: $$\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2x} dx.$$
However, what I could not figure out is why we assume from $dv = x dx$ that $v = \frac{x^2}{2}$, when it could be $v = \frac{x^2}{2} + C$ for any constant $C$. The second integral would be quite different, and not only by a constant, so I would like to understand why we "forget" this constant of integration.
Thanks.
|
Take your example, $$\int x\ln x\,dx.$$
Note $x\gt 0$ must be assumed (so the integrand makes sense).
If we let $u = \ln x$ and $dv= x\,dx$, then we can take $v$ to be any function with $dv = x\,dx$. So the "generic" $v$ will be, as you note, $v = \frac{1}{2}x^2 + C$. What happens then if we use this "generic" $v$?
\begin{align*}
\int x\ln x\,dx &= \ln x\left(\frac{1}{2}x^2 + C\right) - \int \left(\frac{1}{2}x^2+C\right)\frac{1}{x}\,dx\\
&= \frac{1}{2}x^2\ln x + C\ln x - \int\left(\frac{1}{2}x + \frac{C}{x}\right)\,dx\\
&= \frac{1}{2}x^2\ln x + C\ln x - \frac{1}{4}x^2 - C\ln x + D\\
&= \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + D,
\end{align*}
so in the end, we get the same result no matter what value of $C$ we take for $v$.
This says that we can take any value of $C$ and still get the same answer. Since we can take any value of $C$, why not take the simplest one, the one that does not require us to carry around an extra term that is going to cancel out anyway? Say..., $C=0$?
This works in general. If you replace $v$ with $v+C$ in the integration by parts formula, you have
\begin{align*}
\int u\,dv &= u(v+C) - \int(v+C)\,du = uv + Cu - \int v\,du - \int C\,du\\
&= uv+Cu - \int v\,du - Cu = uv-\int v\,du.
\end{align*}
So the answer is the same regardless of the value of $C$, and so we take $C=0$ because that makes our life simpler.
|
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|
Three-variable system of simultaneous equations
$x + y + z = 4$
$x^2 + y^2 + z^2 = 4$
$x^3 + y^3 + z^3 = 4$
Any ideas on how to solve for $(x,y,z)$ satisfying the three simultaneous equations, provided there can be both real and complex solutions?
|
For a fixed number of variables and a fixed power $n$ the sum of powers $$x^n + y^n + z^n + ... + w^n$$ is a symmetric polynomial.
It is expressible in terms of elementary symmetric polynomials. The elementary symmetric polynomials for three variables are
*
*$e_1 = x + y + z$
*$e_2 = x y + x z + y z$
*$e_3 = x y z$
and your polynomials expressed in terms of them are
*
*$x + y + z = e_1$
*$x^2 + y^2 + z^2 = e_1^2 - 2 e_2$
*$x^3 + y^3 + z^3 = e_1^3 - 3(e_1 e_2 - e_3)$
Now we can find the values of $e_1,e_2,e_3$ evaluated at the given $x,y,z$:
$e_1 = 4$, $e_2 = 6$, $e_3 = 4$.
Now consider the polynomial $(t - x)(t - y)(t - z) = t^3 - e_1 t^2 + e_2 t - e_3 = t^3 - 4 t^2 + 6 t - 4$.
It has the solutions $t = 2, 1 + i$ and $1 - i$.
So now we can check if these are correct:
*
*$(2) + (1+i) + (1-i) = 4$
*$(2)^2 + (1+i)^2 + (1-i)^2 = 4 + 2i - 2i = 4$
*$(2)^3 + (1+i)^3 + (1-i)^3 = 8 + -2 + 2i -2 - 2i = 4$
|
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|
Find all irreducible monic polynomials in $\mathbb{Z}/(2)[x]$ with degree equal or less than 5
Find all irreducible monic polynomials in $\mathbb{Z}/(2)[x]$ with degree equal or less than $5$.
This is what I tried:
It's evident that $x,x+1$ are irreducible. Then, use these to find all reducible polynomials of degree 2. There ones that can't be made are irreducible. Then use these to make polynomials of degree 3, the ones that can't be made are irreducible. Repeat until degree 5.
Doing this way takes way too long and I just gave up during when I was about to reach degree 4 polynomials.
My question is: is there any easier way to find these polynomials?
P.S.: this is not exactly homework, but a question which I came across while studying for an exam.
|
Extrapolated Comments converted to answer:
First, we note that there are $2^n$ polynomials in $\mathbb{Z}_2[x]$ of degree $n$.
A polynomial $p(x)$ of degree $2$ or $3$ is irreducible if and only if it does not have linear factors. Therefore, it suffices to show that $p(0) = p(1) = 1$. This quickly tells us that $x^2 + x + 1$ is the only irreducible polynomial of degree $2$. This also tells us that $x^3 + x^2 + 1$ and $x^3 + x + 1$ are the only irreducible polynomials of degree $3$.
As hardmath points out, for a polynomial $p(x)$ of degree $4$ or $5$ to be irreducible, it suffices to show that $p(x)$ has no linear or quadratic factors. To rule out the linear factors, we can again throw out any polynomial not satisfying $p(0) = p(1) = 1$. That is, we can throw out any polynomial with constant term $0$, and we can throw out any polynomial with an even number of terms. This rules out $3/4$ of the polynomials. For example, the $4^{th}$ degree polynomials which do not have linear factors are:
*
*$ x^4 + x^3 + x^2 + x + 1 $
*$ x^4 + x^3 + 1 $
*$ x^4 + x^2 + 1 $
*$ x^4 + x + 1 $
The $5^{th}$ degree polynomials which do not contain linear factors are:
*
*$x^5 + x^4 + x^3 + x^2 + 1$
*$x^5 + x^4 + x^3 + x + 1$
*$x^5 + x^4 + x^2 + x + 1$
*$x^5 + x^3 + x^2 + x + 1$
*$x^5 + x^4 + 1$
*$x^5 + x^3 + 1$
*$x^5 + x^2 + 1$
*$x^5 + x + 1$
It still remains to check whether $x^2 + x + 1$ (which is the only quadratic irreducible polynomial in $\mathbb{Z}_2[x]$) divides any of these polynomials. This can be done by hand for sufficiently small degrees. Again, as hardmath points out, since $x^2 + x + 1$ is the only irreducible polynomial of degree $2$, it follows that $(x^2 + x + 1)^2 = x^4 + x^2 + 1$ is the only polynomial of degree $4$ which does not have linear factors and yet is not irreducible. Therefore, the other $3$ polynomials listed must be irreducible. Similarly, for degree $5$ polynomials, we can rule out
$$
(x^2 + x + 1)(x^3 + x^2 + 1) = x^5 + x + 1
$$
and
$$
(x^2 + x + 1)(x^3 + x + 1) = x^5 + x^4 + 1.
$$
The other $6$ listed polynomials must therefore be irreducible.
Notice that this trick of throwing out polynomials with linear factors, then quadratic factors, etc. (which hardmath called akin to the Sieve of Eratosthenes) is not efficient for large degree polynomials (even degree $6$ starts to be a problem, as a polynomial of degree $6$ can factor as a product of to polynomials of degree $3$). This method, therefore only works for sufficiently small degree polynomials.
To recap, the irreducible polynomials in $\mathbb{Z}_2[x]$ of degree $\leq 5$ are:
*
*$x$
*$x+1$
*$x^2 + x + 1$
*$x^3 + x^2 + 1$
*$x^3 + x + 1$
*$ x^4 + x^3 + x^2 + x + 1 $
*$ x^4 + x^3 + 1 $
*$ x^4 + x + 1 $
*$x^5 + x^4 + x^3 + x^2 + 1$
*$x^5 + x^4 + x^3 + x + 1$
*$x^5 + x^4 + x^2 + x + 1$
*$x^5 + x^3 + x^2 + x + 1$
*$x^5 + x^3 + 1$
*$x^5 + x^2 + 1$
|
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|
Another magic re-write question, picture! It's the part before and after "Thus".
$$I = \ldots = \int e^{ax} \cos bx \ \mathrm{d}x = \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx - \frac{a^{2}}{b^{2}} I.$$
Thus
$$\left( 1 + \frac{a^{2}}{b^{2}} \right) I = \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx + C_1,$$
and
$$\int e^{ax} \cos bx \ \mathrm{d}x = I = \frac{be^{ax}\sin bx+ae^{ax} \cos bx}{b^2+a^2}+C.$$
Were does the "+1" come from? I thought this was an old "move to other side of equal-sign" until that +1 spawned in my face hehe.
|
The integral calculation is done by integration by parts:
\begin{align}
I = \int e^{ax} \cos bx \ dx = \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx - \frac{a^{2}}{b^{2}} I.
\end{align}
Solving for $I$, we have
\begin{align}
I + \frac{a^{2}}{b^{2}} I = \left( 1 + \frac{a^{2}}{b^{2}} \right) I = \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx + C_1,
\end{align}
(answering your $+1$ question) so
\begin{align}
I = \left( 1 + \frac{a^2}{b^2} \right)^{-1} \left( \frac{1}{b} e^{ax} \sin bx + \frac{a}{b^{2}} \cos bx \right) + C = \frac{b e^{a x} \sin bx + a e^{ax} \cos bx }{a^{2} + b^{2}} + C
\end{align}
by multiplying by $\frac{b^{2}}{b^{2}}$ and clearing the denominator, and where we have added $C_1$ and $C$ as arbitrary integration constants, related by the factor $(1 + \frac{a^2}{b^2})^{-1}$.
|
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|
What will the value of following log expression What will be the value of the expression
$$\log_x \frac{x}{y} + \log_y \frac{y}{x}?$$
I tried:
$$\log_x x - \log_x y + \log_y y - \log_y x = 1 - \log_x y + 1 - \log_y x
= 2 - \log_x y - \log_y x.$$
Now what after this ?
|
If $\log_a b = r$, this means that $a^r = b$, so $b = e^{\ln(a^r)} = e^{r\ln(a)}$. Therefore, $\ln(b) = r\ln (a)$, or
$$\log_a b = \frac{\ln(b)}{\ln a}.$$
Thus, for
$$\log_x y = \frac{\ln y}{\ln x}\quad\text{and}\quad \log_y x = \frac{\ln x}{\ln y},$$
so
$$\log_x y = \frac{1}{\log_y x}.$$
So:
$$\begin{align*}
\log_x\frac{x}{y} +\log_y\frac{y}{x} &= 1-\log_x y + 1 - \log_y x\\
&= 2 - \log_x y - \frac{1}{\log_x y}\\
&= 2 - \left(\log_x y + \frac{1}{\log_x y}\right)\\
&= 2 - \left(\frac{(\log_x y)^2 + 1}{\log_x y}\right)\\
&= -\frac{(\log_x y)^2 - 2\log_x(y) + 1}{\log_x y}\\
&= - \frac{(\log_x y - 1)^2}{\log_x y}.
\end{align*}$$
So for $r=\log_x y$, you get $-\frac{(r-1)^2}{r}$. The value will depend on $r$; if, for example, $y=x$, then you get $0$; if $y=x^2$, then you get $-\frac{1}{2}$; if $y=x^{-1}$, then you get $4$, etc.
|
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|
Where is my (algebra) mistake? Converting parametric to Cartesian equation I'm having a problem with my solution to a textbook exercise:
Find the Cartesian equation of the curve given by this parametric equation:
$$x = \frac{t}{2t-1}, y = \frac{t}{t+1}$$
The textbook's answer is $y = \frac{x}{3x-1}$
My solution is different. I understand how the textbook got to its solution, but I can't find where I made my mistake. Can anyone spot my error below?
$$x = \frac{t}{2t-1} = \frac{t}{2t} - \frac{t}{1}$$
$$\implies x = \frac{1}{2} - t$$
$$\implies x - \frac{1}{2} = -t$$
$$\implies t = -x + \frac{1}{2}$$
Sub this into $y = \frac{t}{t+1} \implies y = \frac{-x + \frac{1}{2}}{-x + \frac{1}{2} + 1}$
$$= \frac{-x + \frac{1}{2}}{-x + \frac{1}{2}} + \frac{-x + \frac{1}{2}}{1}$$
$$= 1 - x + \frac{1}{2}$$
$$= -x + \frac{3}{2}$$
So $y = -x + \frac{3}{2}$
I suspect my error is when I split my fractions up, but if so, why can't I do it like that?
Many thanks!
|
As you said Danny,
$$\frac{a}{b+c} \neq \frac{a}{b} + \frac{a}{c}$$
and Theo pointed out with a simple example why:
$$0.5 = \frac{1}{2} = \frac{1}{1+1} \neq \frac{1}{1} + \frac{1}{1} = 2 \; .$$
It is a common mistake however, so tempting that few people have resisted making it.
P.S: Note you made the mistake twice, once in the formula with $x$ and once in the one with $y$.
|
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|
Solve recursion $a_{n}=ba_{n-1}+cd^{n-1}$ Let $b,c,d\in\mathbb{R}$ be constants with $b\neq d$. Let
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}
\end{eqnarray}$$
be a sequence for $n \geq 1$ with $a_{0}=0$. I want to find a closed formula for this recursion. (I only know the german term geschlossene Formel and translated it that way I felt it could be right. So if I got that wrong, please correct me)
First I wrote down some of the chains and I got
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}\\
&=& b\left(ba_{n-2}+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(ba_{n-3}+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(b\left(ba_{n-4}+cd^{n-4}\right)+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& \dots\\
&=& \sum_{k=0}^{n}b^{k}cd^{n-k-1}\\
&=& \sum_{k=0}^{n}b^{k}cd^{n-\left(k+1\right)}
\end{eqnarray}$$
So I catched the structure in a serie. Now I am asking myself how to proceed. I took the liberty to have a little peek at what WolframAlpha wood say to this serie. I hoped for inspiration and I got
$$\sum_{k=0}^{n-1}b^{k} c d^{n-(k+1)} = (c (b^n-d^n))/(b-d)$$
How did this came to be? And more important: Is my approach useful?
Thank you in advance for any advice!
Edit: My final Solution (recalculated)
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}\\
&=& b\left(ba_{n-2}+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(ba_{n-3}+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(b\left(ba_{n-4}+cd^{n-4}\right)+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& b^{4}a_{n-4}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+bcd^{n-2}+cd^{n-1}\\
&=& b^{5}a_{n-5}+b^{4}cd^{n-5}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+bcd^{n-2}+cd^{n-1}\\
&=& b^{n}a_{0}+b^{n-1}c+\dots+b^{4}cd^{n-5}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+cbd^{n-2}+cd^{n-1}\\
&=& \dots\\
&=& 0+b^{n-1}c+\dots+b^{4}cd^{n-5}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+cbd^{n-2}+cd^{n-1}\\
&=& \sum_{k=0}^{n-1}b^{k}cd^{n-1-k}\\
&=& cd^{n-1}\sum_{k=0}^{n-1}b^{k}d^{-k}\\
&=& cd^{n-1}\sum_{k=0}^{n-1}\left(\frac{b}{d}\right)^{k}\\
&=& cd^{n-1}\frac{1-\left(\frac{b}{d}\right)^{n}}{1-\left(\frac{b}{d}\right)}\\
&=& cd^{n-1}\frac{1-\frac{b^{n}}{d^{n}}}{1-\frac{b}{d}}\\
&=& cd^{n-1}\frac{\frac{d^{n}-b^{n}}{d^{n}}}{\frac{d-b}{d}}\\
&=& cd^{n-1}\frac{d^{n}-b^{n}}{d^{n}}\cdot\frac{d}{d-b}\\
&=& \frac{c\left(d^{n}-b^{n}\right)}{d-b}
\end{eqnarray}$$
|
If $b=0$, then $a_n = cd^{n-1}$.
Now assume $b\neq 0$. You can first divide both sides of the equation by $b^n$, then you get
$$
\frac{a_n}{b^n} = \frac{a_{n-1}}{b^{n-1}} + \frac{cd^{n-1}}{b^n}
$$
Let $x_n = \frac{a_n}{b^n}$ and $q = \frac{d}{b}$, then $x_0 = a_0/b = 0$, $q\neq 1$, and we have
$$
x_n = x_{n-1} + (c/b)q^{n-1} \mbox{ or } x_n - x_{n-1} = (c/b)q^{n-1}
$$
Then we can apply the telescoping sum method,
$$\begin{eqnarray}
x_n & = & x_0 + \sum_{i=1}^n (x_i - x_{i-1})\\
& = & (c/b)\sum_{i=1}^n q^{i-1}\\
& = & (\frac{c}{b})(\frac{1-q^n}{1-q})
\end{eqnarray}
$$
So $$a_n = x_n b^n = (\frac{c}{b})(\frac{1-\frac{d^n}{b^n}}{1-\frac{d}{b}})b^n
= \frac{c(b^n - d^n)}{b - d}$$
|
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|
Interesting problem on "neighbor fractions" This is from I. M. Gelfand's Algebra book.
Fractions $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$ are called
neighbor fractions if their difference $\displaystyle\frac{ad - bc}{bd}$ has numerator of $\pm 1$,
that is, $ad - bc = \pm 1$. Prove that
(a) in this case neither fraction can be simplified (that is, neither has any
common factors in numerator and denominator);
(b) if $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$ are neighbor
fractions, then $\displaystyle\frac{a+c}{b+d}$ is between them and is a neighbor fraction for
both $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$; moreover,
(c) no fraction $\displaystyle\frac{e}{f}$ with positive integer $e$ and $f$ such
that $f < b + d$ is between $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$.
Parts (a) and (b) weren't too difficult, but I'm stuck on part (c). I've included (a) and (b) in case they're related to the solution to (c).
|
Assume $\frac{e}{f}$ is (strictly) between $\frac{a}{b}$ and $\frac{c}{d}$. Then $\left|\frac{a}{b}-\frac{e}{f}\right| + \left|\frac{e}{f}-\frac{c}{d}\right| = \left|\frac{a}{b}-\frac{c}{d}\right| = \frac{1}{bd}$
But $\left|\frac{a}{b}-\frac{e}{f}\right| \geq \frac{1}{bf}$ and $\left|\frac{e}{f}-\frac{c}{d}\right|\geq \frac{1}{df}$. So $\frac{1}{bf} + \frac{1}{df} \leq \frac{1}{bd}$. Multiply both sides by $bdf$ and we get that $b+d\leq f$.
|
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Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
|
Proof 1. (Exercise 2.5.1 in Dias Agudo, Cândido da Silva, Matemáticas Gerais III). Let $S:=\sum_{k=1}^{n}k^{2}$. Consider $(1+a)^{3}=1+3a+3a^{2}+a^{3}$ and sum
$(1+a)^{3}$ for $a=1,2,\ldots ,n$:
$$\begin{eqnarray*}
(1+1)^{3} &=&1+3\cdot 1+3\cdot 1^{2}+1^{3} \\
(1+2)^{3} &=&1+3\cdot 2+3\cdot 2^{2}+2^{3} \\
(1+3)^{3} &=&1+3\cdot 3+3\cdot 3^{2}+3^{3} \\
&&\cdots \\
(1+n)^{3} &=&1+3\cdot n+3\cdot n^{2}+n^{3}
\end{eqnarray*}$$
The term $(1+1)^3$ on the LHs of the 1st sum cancels the term $2^3$ on the RHS of the 2nd, $(1+2)^3$, the $3^3$, $(1+3)^4$, the $4^3$, ..., and $(1+n-1)^3$ cancels $n^3$. Hence
$$(1+n)^{3}=n+3\left( 1+2+\ldots +n\right) +3S+1$$
and
$$S=\frac{n(n+1)(2n+1)}{6},$$
because $1+2+\ldots +n=\dfrac{n\left( n+1\right) }{2}$.
Proof 2. (Exercise 1.42 in Balakrishnan, Combinatorics, Schaum's Outline of Combinatorics). From
$$\binom{k}{1}+2\binom{k}{2}=k+2\frac{k\left( k-1\right) }{2}=k^{2},$$
we get
$$\begin{eqnarray*}
S &:&=\sum_{k=1}^{n}k^{2}=\sum_{k=1}^{n}\binom{k}{1}+2\binom{k}{2}
=\sum_{k=1}^{n}\binom{k}{1}+2\sum_{k=1}^{n}\binom{k}{2} \\
&=&\binom{n+1}{2}+2\binom{n+1}{3} \\
&=&\frac{n\left( n+1\right) \left( 2n+1\right) }{6}.
\end{eqnarray*}$$
|
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|
Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$ Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$.
I know the proof by subtracting LHS by RHS and then doing some arrangement.
But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) >(a+b+c)(a^2+b^2+c^2)$, or an easy proof?
|
By Holder and C-S for non-negatives $a$, $b$ and $c$ we obtain:
$$(3(a^3+b^3+c^3))^2=(1+1+1)^2(a^3+b^3+c^3)(a^3+b^3+a^3)\geq(a+b+c)^3(a^3+b^3+c^3)=$$
$$=(a+b+c)(a^3+b^3+c^3)(a+b+c)^2\geq(a^2+b^2+c^2)^2(a+b+c)^2$$
and we are done!
|
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|
Is this Batman equation for real? HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?
Batman Equation in text form:
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\
&\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\
&\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\
&\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\
&\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0
\end{align}
|
Here's what I got from the equation using Maple...
|
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The number of partitions of $n$ into distinct parts equals the number of partitions of $n$ into odd parts This seems to be a common result. I've been trying to follow the bijective proof of it, which can be found easily online, but the explanations go over my head. It would be wonderful if you could give me an understandable explanation of the proof and let me know how I'd go about finding such a bijection.
|
I'll give a sketch of the bijective proof; ask me if there's some part you don't understand and need fleshed out (or maybe someone else will post a detailed version).
The important idea is that every number can be expressed uniquely as a power of 2 multiplied by an odd number. (Divide the number repeatedly by 2 till you get an odd number.) For instance, $120 = 2^3 \times 15$ and $68 = 2^2 \times 17$ and $81 = 2^0 \times 81$.
Odd parts -> Distinct parts
Suppose you are given a partition of n into odd parts. Count the number of times each odd number occurs: suppose $1$ occurs $a_1$ times, similarly $3$ occurs $a_3$ times, etc. So
$n = a_11 + a_33 + a_55 + \cdots$.
Now write each $a_i$ "in binary", i.e., as a sum of distinct powers of two. So you have
$n = (2^{b_{11}}+2^{b_{12}}+\cdots)1 + (2^{b_{31}}+2^{b_{32}}+\cdots)3 + \cdots$.
Now just get rid of the brackets, and note that all terms are distinct. (Why?)
E.g. for $20 = 5+3+3+3+1+1+1+1+1+1$, which is a partition of $20$ into odd parts, we do
$20 = (1)5 + (3)3 + (6)1$
$20 = (1)5 + (2+1)3 + (4+2)1$, so you get the partition
$20 = 5 + 6 + 3 + 4 + 2$ in which all parts are distinct.
Distinct parts -> Odd parts
Given a partition into distinct parts, we can write each part as a power of 2 multiplied by an odd number, and collect the coefficients of each odd number, and write the odd number those many times, to get a partition into odd parts.
So for example with $20 = 5+6+3+4+2$ which is a partition of $20$ into distinct parts, we write
$20 = 5 + (2)3 + 3 + (4)1 + (2)1$, and then collect coefficients of the odd numbers $5$, $3$ and $1$:
$20 = 5 + (2+1)3 + (4+2)1$
$20 = 5+3+3+3+1+1+1+1+1+1$, which was our original odd partition.
Aside: you asked about the bijective proof, but I cannot resist mentioning that when Euler proved this theorem about partitions, it was with a proof that used generating functions. (When you understand both, maybe you can think about whether this proof is related to the bijective proof.)
Sketch: Let $D(n)$ denote the number of partitions into distinct parts, and $O(n)$ denote the number of partitions into odd parts. Then we have:
$$
\begin{align}
\sum_{n\ge0}D(n)x^n &= (1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)\cdots \\
&= \frac{1-x^2}{1-x}\frac{1-x^4}{1-x^2}\frac{1-x^6}{1-x^3}\frac{1-x^8}{1-x^4}\frac{1-x^{10}}{1-x^5}\cdots \\
&= \frac{1}{(1-x)(1-x^3)(1-x^5)\cdots} \\
&= (1+x+x^{1+1}+\cdots)(1+x^3+x^{3+3}+\cdots)(1+x^5+x^{5+5}+\cdots) \\
&= \sum_{n\ge0}O(n)x^n
\end{align}
$$
which proves the theorem.
|
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How to solve these inequalities? How to solve these inequalities?
*
*If $a,b,c,d \gt 1$, prove that $8(abcd + 1) \gt
(a+1)(b+1)(c+1)(d+1)$.
*Prove that $ \cfrac{(a+b)xy}{ay+bx} \lt \cfrac{ax+by}{a+b}$
*Find the greatest value of $x^3y^5z^7$ when $2x^2+2y^2+2x^2=15$
Any hints/solution are welcome.
|
Let $a=1+x$, $b=1+y$, $c=1+z$ and $d=1+t$. Hence,
$$8(abcd + 1)-(a+1)(b+1)(c+1)(d+1)=$$
$$=8(1+x)(1+y)(1+z)(1+t)+8-(2+x)(2+y)(2+z)(2+t)=$$
$$=4(xy+xz+yz+xt+yt+zt)+6(xyz+xyt+xzt+yzt)+7xyzt>0$$
Done!
|
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Show that $PSL(3,4)$ has no element of order $15$.
$PSL(3,4)$ has no element of order $15$. Thus it is no isomorphic to $A_8$.
Here, $PSL(3,4)$ denotes the $3 \times 3$ projective special linear group on the field with $4$ elements.
As listing all the elements takes too much work, is there any better way to prove there is no element of order $15$ in $PSL(3,4)$?
Thank you very much.
|
I try to write out the proof sketched by user641.
Let $F$ denote the field with 4 elements. Speaking of a polynomial, "irreducible" will mean "irreducible over $F$".
We have to prove that $PSL(3, 4)$ has no element of order 6 and this amounts to prove that if $M$ is a matrix in $SL(3, 4)$ such that $M^{6}$ is scalar, then either $M^{2}$ or $M^{3}$ is scalar.
Assume first that $M^{6} = 1$. Then $M$ annihilates the polynomial $X^{6} - 1 = (X-1)^{2} (X-a)^{2} (X-b)^{2}$, where $a$ and $b$ denote the two nonzero elemets of $F$ other than $1$. Thus the characteristic polynomial of $M$ is of the form $(X-c) (X-d) (X-e)$, with $c$, $d$ and $e$ nonzero in $F$. Since the determinant of $M$ is $1$, the characteristic polynomial of $M$ must be equal to $(X-1) (X-a) (X-b) = X^{3} - 1$ or of the form $(X-f)^{3}$ with $f$ nonzero in $F$. In the first case $M^{3} = 1$, thus $M^{3}$ is scalar. In the second case, $M$ annihilates the polynomial $(X-f)^{3} = X^{3} + fX^{2} + f^{2}X + f^{3}$, thus $M^{4} = f^{4}$. Since $M^{6} = 1$, we have $M^{2} =f^{-4} = f^{2}$, thus $M^{2}$ is scalar.
Thus our thesis is true in the case where $M^{6} = 1$. There remains to prove it when $M^{6} = a$, where $a$ is one of the two nonzero elements of $F$ different from $1$. We have $a^{3} = 1$, thus $a^{4} = a$, thus the polynomial $X^{6} - a$ is equal to $X^{6} - a^{4} = (X^{3} - a^{2})^{2}$. So $M$ annihilates the polynomial $(X^{3} - a^{2})^{2}$. Since $X^{3} - a^{2}$ is irreducible, this implies that the only irreducible factor of the characteristic polynomial of $M$ is equal to $X^{3} - a^{2}$. Since the characteristic polynomial of $M$ is of degree 3, it must be equal to $X^{3} - a^{2}$, so $M^{3} = a^{2}$, so $M^{3}$ is scalar, which completes the proof.
Again, I don't see the usefulness of the rational canonical form.
|
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Elementary central binomial coefficient estimates
*
*How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}<\binom{2n}{n}<\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ > 1 ?
*Does anyone know any better elementary estimates?
Attempt. We have
$$\frac1{2^n}\binom{2n}{n}=\prod_{k=0}^{n-1}\frac{2n-k}{2(n-k)}=\prod_{k=0}^{n-1}\left(1+\frac{k}{2(n-k)}\right).$$
Then we have
$$\left(1+\frac{k}{2(n-k)}\right)>\sqrt{1+\frac{k}{n-k}}=\frac{\sqrt{n}}{\sqrt{n-k}}.$$
So maybe, for the lower bound, we have
$$\frac{n^{\frac{n}{2}}}{\sqrt{n!}}=\prod_{k=0}^{n-1}\frac{\sqrt{n}}{\sqrt{n-k}}>\frac{2^n}{\sqrt{4n}}.$$
By Stirling, $n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, so the lhs becomes
$$\frac{e^{\frac{n}{2}}}{(2\pi n)^{\frac14}},$$
but this isn't $>\frac{2^n}{\sqrt{4n}}$.
|
A way to get explicit bounds via Stirling's approximation is to use the following more precise formulation: $$n! = \sqrt{2\pi n} \left( \frac{n}{e} \right)^n e^{\alpha_n} $$ where $ \frac{1}{12n+1} < \alpha_n < \frac{1}{12n} $.
With this one arrives at $$ \binom{2n}{n} = \frac{4^n}{\sqrt{\pi n}} e^{\lambda_n} $$ where $ \frac{1}{24n+1} - \frac{1}{6n} < \lambda_n < \frac{1}{24n} - \frac{2}{12n+1} $.
|
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|
What is the maximum value of this trigonometric expression What is the maximum value of the expression
$1/(\sin^2 \theta + 3\sin\theta \cos\theta+ 5\cos^2 \theta$).
I tried reducing the expression to $1/(1 + 3\sin\theta$ $\cos\theta + 4\cos^2 \theta)$.
How do I proceed from here?
|
This is the same as minimizing the function $f(\theta)=\sin^2 \theta+3 \sin \theta \cos \theta+5\cos^2 \theta$ (subject to $f(\theta)>0$). We know such a minimum will occur only when $f'(\theta)=0$. We have
\begin{align*}
f'(\theta)&= 2 \sin \theta \cos \theta - 3 \sin^2 \theta +3 \cos^2 \theta-10 \sin \theta \cos \theta=-8 \sin \theta \cos \theta-3\sin^2 \theta + 3 \cos^2 \theta.
\end{align*}
Using trig identities, this is same as solving $3\cos 2\theta -4 \sin 2 \theta=0.$ The critical points occurs when $\tan 2 \theta= \frac{3}{4}$. At such a point $\sin 2 \theta=\frac{\pm 3}{5}$ and $\cos 2 \theta=\frac{\pm 4}{ 5}.$ As a result, $\sin \theta \cos \theta= \frac{\pm 3}{ 10}$ and $\cos^2 \theta -\sin^2 \theta= \frac{\pm 4}{5}$.
Now we have $f(\theta)=2\cos^2 \theta - 2\sin^2 \theta+3 + 3 \sin \theta \cos \theta.$ Substituting the values of $\cos \theta \sin \theta$ and $\cos^2 \theta- \sin^2 \theta$ at the critical points, we can find max/min values for $f$.
|
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|
Number of positive integral solutions for $ab + cd = a + b + c + d $ with $1 \le a \le b \le c \le d$
How many positive integral solutions exist for: $ab + cd = a + b + c +
d $,where $1 \le a \le b \le c \le d$ ?
I need some ideas for how to approach this problem.
|
The equation can be rewritten as
$$(a-1)(b-1)+(c-1)(d-1)=2.$$
Now there are not many possibilities to consider! If the first product is $0$, the second must be $2$, and if the first product is $1$, so is the second.
If $a=1$, then we need to have $(c-1)(d-1)=2$. Since $1\le c\le d$, this forces $c=2$, $d=3$. And $b$ can be $1$ or $2$, giving the solutions $(1,1,2,3)$ and $(1,2,2,3)$.
If $a>1$, we need $a=2$, else the left hand side is too big. That forces $b=c=d=2$, giving the third solution $(2,2,2,2)$.
Comment: Note that in general $ab+pa+qb=(a+q)(b+p) -pq$. This relative of completing the square is occasionally useful.
|
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|
limit of $\frac{f'(x)}{x}$, $f$ convex of class $C^1$ Let $f:\mathbb R\to \mathbb R$ be a convex function of class $C^1$. Let us suppose that there exists the limit
$$L:=\lim_{x\to+\infty}\frac{f(x)}{x^2}$$ and that $0<L<+\infty$.
a) Prove that $$0<\liminf_{x\to+\infty}\frac{f'(x)}{x}\leq\limsup_{x\to+\infty}\frac{f'(x)}{x}<+\infty.$$
b) Prove that there exists the limit $$\lim_{x\to+\infty}\frac{f'(x)}{x},$$ and compute it.
-Mario-
|
Pick any $\epsilon>0$, then there is an $x_0$ big enough so that for $x>x_0$, $\left|\frac{f(x)}{x^2}-L\right|<\epsilon^2$. Then, for $x>y>x_0$,
$$
\begin{align}
\frac{f(x)-f(y)}{x^2-y^2}&\le\frac{(L+\epsilon^2)x^2-(L-\epsilon^2)y^2}{x^2-y^2}\\
&=L+\epsilon^2\frac{x^2+y^2}{x^2-y^2}
\end{align}
$$
and
$$
\begin{align}
\frac{f(x)-f(y)}{x^2-y^2}&\ge\frac{(L-\epsilon^2)x^2-(L+\epsilon^2)y^2}{x^2-y^2}\\
&=L-\epsilon^2\frac{x^2+y^2}{x^2-y^2}
\end{align}
$$
Thus
$$
\left|\frac{f(x)-f(y)}{x^2-y^2}-L\right|\le\epsilon^2\frac{x^2+y^2}{x^2-y^2}\tag{1}
$$
Choose $x>y>z>x_0$ so that $\frac{x^2-y^2}{x^2+y^2}=\epsilon$ and $\frac{y^2-z^2}{y^2+z^2}=\epsilon$. By the Mean Value Theorem, for some $\xi$ and $\eta$ so that $x>\xi>y>\eta>z$, we have
$$
\frac{f'(\xi)}{2\xi}=\frac{f(x)-f(y)}{x^2-y^2}\text{ and }\frac{f'(\eta)}{2\eta}=\frac{f(y)-f(z)}{y^2-z^2}\tag{2}
$$
Using $(1)$ and $(2)$,
$$
\left|\frac{f'(\xi)}{2\xi}-L\right|<\epsilon\text{ and }\left|\frac{f'(\eta)}{2\eta}-L\right|<\epsilon\tag{3}
$$
Note that with our choice of $x$, $y$, and $z$, we have
$$
1\le\frac{\xi}{y}\le\frac{x}{y}=\sqrt{\frac{1+\epsilon}{1-\epsilon}}\text{ and }1\ge\frac{\eta}{y}\ge\frac{z}{y}=\sqrt{\frac{1-\epsilon}{1+\epsilon}}\tag{4}
$$
Since $f$ is convex, $f'$ is non-decreasing. Therefore, $f'(\eta)\le f'(y)\le f'(\xi)$ and so, with $(3)$ and $(4)$, we have
$$
\frac{f'(y)}{y}\le\frac{f'(\xi)}{y}\le\frac{f'(\xi)}{\xi}\sqrt{\frac{1+\epsilon}{1-\epsilon}}\le2(L+\epsilon)\sqrt{\frac{1+\epsilon}{1-\epsilon}}
$$
and
$$
\frac{f'(y)}{y}\ge\frac{f'(\eta)}{y}\ge\frac{f'(\eta)}{\eta}\sqrt{\frac{1-\epsilon}{1+\epsilon}}\ge2(L-\epsilon)\sqrt{\frac{1-\epsilon}{1+\epsilon}}
$$
Thus, for any $y>x_0\sqrt{\frac{1+\epsilon}{1-\epsilon}}$ we have
$$
2(L-\epsilon)\sqrt{\frac{1-\epsilon}{1+\epsilon}}\le\frac{f'(y)}{y}\le2(L+\epsilon)\sqrt{\frac{1+\epsilon}{1-\epsilon}}
$$
Therefore,
$$
\limsup\limits_{x\to+\infty}\frac{f'(x)}{x}\le2(L+\epsilon)\sqrt{\frac{1+\epsilon}{1-\epsilon}}\hspace{.25in}\text{and}\hspace{.25in}\liminf\limits_{x\to+\infty}\frac{f'(x)}{x}\ge2(L-\epsilon)\sqrt{\frac{1-\epsilon}{1+\epsilon}}
$$
Since $\epsilon$ was arbitrary, we get that
$$
\lim\limits_{x\to+\infty}\frac{f'(x)}{x}=2L
$$
|
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|
How many positive integer solutions to $a^x+b^x+c^x=abc$? How many positive integer solutions are there to $a^{x}+b^{x}+c^{x}=abc$? (e.g the solution $x=1$, $a=1$, $b=2$, $c=3$). Are there any solutions with $\gcd(a,b,c)=1$? Any solutions to $a^{x}+b^{x}+c^{x}+d^{x}=abcd$ etc. ?
|
Unfortunately, I don't know how to make comments here, would be better to move it there. Anyways, consider for example, $a^x+b^x+c^x=abc$ with $a\le b\le c.$ Clearly, $c^x<LHS\le c^3,$ therefor $x<3.$ Case $x=1$ is fairly simple as you can bound $ab\le 3$ (following the same lines). So let's consider the case $x=2.$ Our equation is quadratic with respect to $a,$ so it has an integer solution iff $D=(c^2-4)b^2-4c^2=x^2.$ This equation has infinitely many solutions even for some fixed values of $c.$ Take. for example, $c=3.$ Then,
$x^2-5b^2=-36,$ has infinitely many solutions, since it is Pell type equation with one solution being $x=12,$ $b=6.$
By considering equation modulo $3,$ we get $(a,b,c)\ne 1.$
|
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|
Generalisation of the identity $\sum\limits_{k=1}^n {k^3} = \bigg(\sum\limits_{k=1}^n k\bigg)^2$ Are there any generalisations of the identity $\sum\limits_{k=1}^n {k^3} = \bigg(\sum\limits_{k=1}^n k\bigg)^2$ ?
For example can $\sum {k^m} = \left(\sum k\right)^n$ be valid for anything other than $m=3 , n=2$ ?
If not, is there a deeper reason for this identity to be true only for the case $m=3 , n=2$?
|
Here is a curious (and related) identity which might be of interest to you. Let $D_{k} = ${ $d$ } be the set of unitary divisors of a positive integer $k$, and let $\sigma_{0}^{*} \colon \mathbb{N} \to \mathbb{N}$ denote the number-of-unitary-divisors (arithmetic) function. Then it is relatively straightforward to prove
\begin{eqnarray}
\sum_{d \in D_k} \sigma_{0}^{*}(d)^{3} = \left( \sum_{d \in D_k} \sigma_{0}^{*}(d) \right)^{2} \qquad k \in \mathbb{N}.
\end{eqnarray}
Note that $\sigma_{0}^{*}(k) = 2^{\omega(k)}$, where $\omega(k)$ is the number distinct prime divisors of $k$. For example,
\begin{eqnarray}
1^{3} + 2^{3} + 2^{3} + 2^{3} + 4^{3} + 4^{3} + 4^{3} + 8^{3} = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^{2}
\end{eqnarray}
|
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|
Finding asymptotes to $y = \frac{2x^2 + 3x - 6}{2x + 1}$ I need to find the asymptotes of
$y = \frac{2x^2 + 3x - 6}{2x + 1}$.
The asymptote at $x = -1/2$ is clear. If one long divides they can easily see that there is an asymptote of $y = x + 1$ as $x$ goes to infinity.
However, what is wrong with this reasoning? I claim that as $x$ goes to infinity, the $2x^2$ term will dominate, so the graph will be on the order of $y = 2x^2$, which has no asymptote. So $y = x + 1$ is not an asymptote.
|
As $x \to \infty$, $2x^2$ is the dominant term in the numerator, while $2x$ is the dominant term in the denominator. So the leading term in $y$ is $\frac{2x^2}{2x} = x$. You still need the constant term, which you can do this way: $$\frac{2x^2 + 3x - 6}{2x+1} \approx
\frac{2x^2 (1 + 3/(2x))}{2x (1 + 1/(2x))} \approx x \left(1 + \frac{3}{2x}\right) \left(1 - \frac{1}{2x} \right) \approx x \left(1 + \frac{3/2 - 1/2}{x}\right) = x + 1$$
|
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|
System of $\sqrt{7x+y}+\sqrt{x+y}=6$ and $\sqrt{x+y}-y+x=2$ $$\begin{align*}\sqrt{7x+y}+\sqrt{x+y}=6\\\sqrt{x+y}-y+x=2\end{align*}$$
I have tried various things squaring, summing but nothing really helped, got some weird intermediate results which are probably useless such as:
$$(y-x)(y+x+4)+4-x-y=0$$
or
$$x_{1,2}=\frac{2y-3\pm\sqrt{8y-7}}{2}$$
Could you please give me some hints?
|
When we see a square root, it is hard to resist the impulse to square. But resist we will, and such a decision is often useful.
The kind of solution we are looking is not specified, but the expressions $\sqrt{7x+y}$ and $\sqrt{x+y}$ suggest we are looking for real solutions. Let
$$7x+y=p^2 \qquad\text{and}\qquad x+y=q^2,$$
where we can assume that $p$ and $q$ are non-negative.
Note that $(7x+y)-4(x+y)=3x-3y$, so $3(x-y)=p^2-4q^2$. So our equations can be rewritten as
$$p+q=6\qquad \text{and}\qquad q+\frac{1}{3}(p^2-4q^2)=2.$$
Now in principle it's all over! (Substitute $6-q$ for $p$ in the second equation. We get a quadratic in the variable $q$.)
For fun we continue in another way. Observe that the second equation can be written as $p^2-4q^2+3q=6$. Is that a $6$ on the right? Yes, so we have
$p^2-4q^2+3q=p+q$, and therefore
$$p^2-4q^2=p-2q.$$
Since $p^2-4q^2=(p-2q)(p+2q)$, we conclude that (a) $p-2q=0$ or (b) $p+2q=1$.
Case (a): We have $p+q=6$ and $p=2q$, so $p=4$, $q=2$. That gives $7x+y=16$, $x+y=4$, so $x=y=2$. This works.
Case (b): We have $p+q=6$ and $p+2q=1$, and therefore $p=11$, $q=-5$. This contradicts the fact that $q$ is non-negative.
Comment: It is perhaps worth looking at Case (b) more closely. We get $7x+y=121$, $x+y=25$, which yields $x=16$, $y=9$. If we interpret $\sqrt{121}$ as being $11$, and $\sqrt{25}$ as being $-5$, then $x=16$, $y=9$ is a solution. But that is a quite peculiar interpretation of $\sqrt{25}$ if we are working in the reals.
However, the manipulations that we did apply equally well to complex numbers. So we have found all complex solutions. But if we are working in the complex numbers, interpreting $\sqrt{121}$ as $11$ and $\sqrt{25}$ as $-5$ is not a problem. Thus we have the slightly paradoxical fact that if we are working in the reals, there is one real solution, while if we are working in the complex numbers, there are two real solutions!
|
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|
Simplifying a simple fraction with exponent I am trying to simplify this fraction :
$
\dfrac{(3^2)(5^4)}{15^3}
$
The answer is : $ \dfrac{5}{3} $
I am trying to do the following: $ \dfrac{3^2}{15^3} \cdot \dfrac{5^4}{15^3} $ so ... $ \dfrac{1^{-3}}{3} \cdot \dfrac{1^1}{3} $
But that's not giving me the right answer, what am I doing wrong, please ?
Thanks !
|
Here it is explicitly. Notice $15^3=(3\cdot 5)^3=3^3\cdot 5^3$. So
$$
\dfrac{(3^2)(5^4)}{15^3}=\frac{3^2\cdot 5^4}{3^3\cdot 5^3}=\frac{5}{3}.
$$
|
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|
Finding limit of $\lim\limits_{h\to 0} \frac1{h}\left(\frac1{\sqrt{x+h-2}}-\frac1{\sqrt{x-2}}\right)$ As expected, if you plug in 0 into the initial equation, the answer is undefined or indeterminate. I tried multiplying the conjugate $\frac1{\sqrt{x+h-2}}+\frac1{\sqrt{x-2}}$ to the numerator and the denominator, but i couldn't simplify this equation enough to avoid the indeterminate value.
$$\lim_{h\to 0} \dfrac{\frac1{\sqrt{x+h-2}}-\frac1{\sqrt{x-2}}}{h}$$
|
An alternative evaluation. Let $f(x)=\frac{1}{\sqrt{x-2}}$. Then, by definition of $f'(x)$
$$
\lim_{h\rightarrow 0}\frac{1}{h}\left( \frac{1}{\sqrt{x+h-2}}-\frac{1}{\sqrt{
x-2}}\right) =f^{\prime }(x),
$$
which is
$$
\begin{eqnarray*}
f^{\prime }(x) &=&\left( \frac{1}{\sqrt{x-2}}\right) ^{\prime }=\left( \sqrt{
x-2}^{-1}\right) ^{\prime } \\
&=&-1\times \left( \sqrt{x-2}\right) ^{-2}\times \left( \sqrt{x-2}\right)
^{\prime }=-\frac{1}{x-2}\times \frac{1}{2\sqrt{x-2}} \\
&=&-\frac{1}{2\left( x-2\right) ^{\frac{3}{2}}}.
\end{eqnarray*}
$$
|
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|
Solving Radical Equations $x-7= \sqrt{x-5}$ This the Pre-Calculus Problem:
$x-7= \sqrt{x-5}$
So far I did it like this and I'm not understanding If I did it wrong.
$(x-7)^2=\sqrt{x-5}^2$ - The Square root would cancel, leaving:
$(x-7)^2=x-5$ Then I F.O.I.L'ed the problem.
$(x-7)(x-7)=x-5$
$x^2-7x-7x+14=x-5$
$x^2-14x+14=x-5$
$x^2-14x-x+14=x-x-5$
$x^2-15x+14=-5$
$x^2-15x+14+5=-5+5$
$x^2-15x+19=0$
$(x-1)(x-19)=0$
Now this is where I'm stuck because when I tried to see if I got the right numbers in the parentheses I got this....
$x^2-19x-1x+19=0$
$x^2-20x+19=0$
As you may see I'm doing something bad because I don't get $x^2-15x+19$
Could anyone please help me and tell me what I'm doing wrong?
|
$(x-7)=\sqrt{x-5}$
Check the domain first
$x-5 \geq0 \cap x-7 \geq0$
$$x \geq7$$
$(x-7)^2=\sqrt{x-5}^2$
$(x-7)^2=x-5$
$(x-7)(x-7)=x-5$
$x^2-7x-7x+49=x-5$
$x^2 - 15x + 54 = 0$
$(x - 9)(x - 6) = 0$
$x - 9 = 0 $ or $x - 6 = 0$
$x = 9$ or $x = 6$
checking the domain x=6 is an extraneous root.
x = 9
|
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|
Special numbers Our teacher talked about some special numbers.
These numbers total of 2 different numbers' cube. For example :
$x^3+y^3 = z^3+t^3 = \text{A-special-number}$
What is the name of this special numbers ?
|
Here's a sample identity by Ramanujan,
$(3x^2+5xy-5y^2)^3 + (4x^2-4xy+6y^2)^3 = (-5x^2+5xy+3y^2)^3 + (6x^2-4xy+4y^2)^3$
Let {x,y} = {-1,0} and you get the nice $3^3+4^3+5^3 = 6^3$. Or {x,y} = {-1,2} for $1^3+12^3 = 9^3+10^3 = 1729$, the smallest non-trivial "taxicab number" (after transposition and removing common factors). And so on.
|
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|
The $3 = 2$ trick on Google+ I found out this on Google+ yesterday and I was thinking about what's the trick. Can you tell?
How can you prove $3=2$?
This seems to be an anomaly or whatever you call in mathematics. Or maybe I'm just plain dense.
See this illustration:
$$ -6 = -6 $$
$$ 9-15 = 4-10 $$
Adding $\frac{25}{4}$ to both sides:
$$ 9-15+ \frac{25}{4} = 4-10+ \frac{25}{4} $$
Changing the order
$$ 9+\frac{25}{4}-15 = 4+\frac{25}{4}-10 $$
This is just like $a^2 + b^2 - 2a b = (a-b)^2$. Here $a_1 = 3, b_1=\frac{5}{2}$ for L.H.S, and $a_2 =2, b_2=\frac{5}{2}$ for R.H.S. So it can be expressed as follows:
$$ \left(3-\frac{5}{2} \right) \left(3-\frac{5}{2} \right) =
\left(2-\frac{5}{2} \right) \left( 2-\frac{5}{2} \right) $$
Taking positive square root on both sides:
$$ 3 - \frac{5}{2} = 2 - \frac{5}{2} $$
$$ 3 = 2 .$$
I think it's something near the root.
|
It's simply not true in the reals that if $ x^2=y^2 $ then x=y. For example, if $ (-2)^2=2^2 $, but 2 does not equal -2. The scheme of inference used in the last step in general isn't valid.
|
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|
What is the asymptotic bound for this recursively defined sequence? $f(0) = 3$
$f(1) = 3$
$f(n) = f(\lfloor n/2\rfloor)+f(\lfloor n/4\rfloor)+cn$
Intuitively it feels like O(n), meaning somewhat linear with steeper slope than c, but I have forgot enough math to not be able to prove it...
|
Suppose we first study $$g(n) = f(n)-3$$ which has $g(0)=g(1)=0$ and
$$g(n) = g(\lfloor n/2 \rfloor) + g(\lfloor n/4 \rfloor) + cn+3.$$
Then it is not difficult to see that for the binary representation of $n$ being
$$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
we have that for $n\ge 2,$
$$g(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
[z^j] \frac{1}{1-z-z^2}
\left(3 + c\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}\right)$$
which in turn implies
$$f(n) = 3 + \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
[z^j] \frac{1}{1-z-z^2}
\left(3 + c\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}\right).$$
Now note that the polynomial term is the generating function of the Fibonacci numbers shifted down by one, so that this simplifies to an attractive exact formula for all $n$ which is
$$f(n) = 3 + \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1}
\left(3 + c\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}\right)
= 3 F_{\lfloor \log_2 n \rfloor+2} +
c \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1} \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}.$$
Next we compute upper and lower bounds. For an upper bound consider a string of one digits, which gives the following bound which is actually attained and cannot be improved upon:
$$f(n) \le 3 F_{\lfloor \log_2 n \rfloor+2} +
c \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1} \sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^{k-j}
\\ = 3 F_{\lfloor \log_2 n \rfloor+2} +
c \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} (2^{\lfloor \log_2 n \rfloor+1-j}-1)
\\= c + (3-c) F_{\lfloor \log_2 n \rfloor+2} +
c 2^{\lfloor \log_2 n \rfloor+1} \sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} 2^{-j}.$$
Now the sum term converges to a limit. We have
$$ \sum_{j=0}^\infty F_{j+1} 2^{-j}
= \frac{1}{\sqrt{5}}
\left(\varphi \sum_{j=0}^\infty (\varphi/2)^j
- (-1/\varphi) \sum_{j=0}^\infty (-1/\varphi/2)^j\right)
\\ = \frac{1}{\sqrt{5}}
\left(\frac{\varphi}{1-\varphi/2}+\frac{1/\varphi}{1+1/\varphi/2}\right) =4.$$
For a lower bound consider a one followed by a string of zeros, giving
$$f(n) \ge 3 F_{\lfloor \log_2 n \rfloor+2} +
c \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1} 2^{\lfloor \log_2 n \rfloor-j}
=3 F_{\lfloor \log_2 n \rfloor+2} +
c 2^{\lfloor \log_2 n \rfloor} \sum_{j=0}^{\lfloor \log_2 n \rfloor-1}
F_{j+1} 2^{-j}. $$
The sum term converges to four as before. We see that the dominant asymptotics in both bounds come from the two terms
$$F_{\lfloor \log_2 n \rfloor+2} \quad\text{and}\quad 2^{\lfloor \log_2 n \rfloor}.$$
Note that
$$F_n \sim \frac{1}{\sqrt{5}} \left(\frac{1+\sqrt{5}}{2}\right)^n$$
which means that the power of two dominates, giving a final complexity of
$$\Theta\left(2^{\lfloor \log_2 n \rfloor}\right)
= \Theta\left(2^{\log_2 n}\right) = \Theta(n).$$
This MSE link points to a similar calculation.
|
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|
How do I show that $2\ln(1+1/x) > 1/x > \ln(1+1/x)$ for any positive integer $x$? How do I show that $$2\ln\left(1+\frac1x\right) > \frac1x > \ln\left(1+\frac1x\right)$$ for any positive integer $x$?
I know it's true but how do I show it?
|
Let's observe: $\ln(1+\frac{1}{x})<\frac{1}{x}$
We can define $y=\ln(1+\frac{1}{x})-\frac{1}{x}$ ,Now we may find first derivative $y'$:
$y'=\frac{1}{1+\frac{1}{x}}(1+\frac{1}{x})'-(\frac{1}{x})' \Rightarrow y'=\frac{1}{x^2}(\frac{1}{x+1}) \Rightarrow y'>0$ for all positive integers..so function $y$ increases as $x$ increases for all $x$
Next,we will find horizontal asymptote as $\lim_{x \to \infty} y$
$\lim_{x \to \infty} ( \ln(1+\frac{1}{x})-\frac{1}{x})=\ln(1)=0$ so $y=0$ is horizontal asymptote.
Since $y$ increases and has $y=0$ horizontal asymptote it must be that $y<0$ for all $x$ which means
$\ln(1+\frac{1}{x})<\frac{1}{x}$ is true for all positive $x$
For inequality $\frac{1}{x}<2\ln(1+\frac{1}{x})$ we may use next proven inequality which states that
$\frac{n}{n+1}<\ln(n+1)$ ,In your case this means that $2\ln(1+\frac{1}{x})>\frac{\frac{2}{x}}{\frac{1}{x}+1}=\frac{2}{x+1}$ so we have to show :
$\frac{1}{x}<\frac{2}{x+1} \Rightarrow \frac{x-1}{x(x+1)}>0$ which is true for all $x>1$, for $x=1$ we have that $\ln(2)>\frac{1}{2}$ This means that inequality:
$\frac{1}{x}<2\ln(1+\frac{1}{x})$ is true for all positive integers.
|
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|
Finding a Particular Coefficient Using Generating Functions I have a homework question to solve the number of ways to choose 25 ice creams of a selection of 6 types of ice creams and there are only 7 of each ice cream type.
The question requires the use of generating functions and I have gotten to this point:
$$({{x}^{48}}-6{{x}^{40}}+15{{x}^{32}}-20{{x}^{24}}+15{{x}^{16}}-6{{x}^{8}}+1)\cdot {{\left( \frac{1}{1-x} \right)}^{6}}$$
But know I need to find the coefficient of $x^{25}$ and I don't know how to continue.
Can anyone please help?
Thanks a lot,
|
It might be useful (or not) to note that the first factor is $(z^8 - 1)^6$. Try:
$\begin{align*}
[x^{25}] (x^{48}-6 x^{40} + 15 x^{32} - 20 x^{24} + 15 x^{16} -6 x^{8} + 1)
\cdot \left( \frac{1}{1 - x} \right)^6
&= [x^{25}] (1 - 6 x^8 + 15 x^{16} - 20 x^{24})
\cdot \sum_{k \ge 0} (-1)^k \binom{-6}{k} x^k \\
&= ([x^{25}] - 6 [x^{17}] + 15 [x^9] - 20 [z])
\sum_{k \ge 0} \binom{k + 6 - 1}{6 - 1} z^k \\
&= \binom{25 + 5}{5} - 6 \binom{17 + 5}{5} + 15 \binom{9 + 5}{5}
- 20 \binom{1 + 5}{5} \\
&= 14412
\end{align*}$
|
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|
Cauchy-Schwarz inequality and three-letter identities (exercise 1.4 from "The Cauchy-Schwarz Master Class") Exercise 1.4 from a great book The Cauchy-Schwarz Master Class asks to prove the following:
For all positive $x$, $y$ and $z$, one has
$$x+y+z \leq 2 \left(\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y}\right).$$
Introduction to the exercise says:
There are many situations where Cauchy's inequality conspires with symmetry to provide results that are visually stunning.
How to prove that inequality? And how does one benefit from the "symmetry"? What is the general idea behind this "conspiracy"?
|
I think the symmetry here comes from being able to derive a similar set of inequalities incorporating the same splitting method. To build off of @Martin's method, we have
$$ (x+y+z) \leq 2\left(\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y}\right) $$
And using the same splitting method we can derive
$$ (x+y+z) \leq 2\left(\frac{x^2}{x+z} + \frac{y^2}{x+y} + \frac{z^2}{y+z}\right) $$
by splitting $$ (x+y+z) = \frac{x}{\sqrt{x+z}}\sqrt{x+z} + \frac{y}{\sqrt{x+y}}\sqrt{x+y} + \frac{z}{\sqrt{y+z}}\sqrt{y+z}$$
And similarly
$$ (x+y+z) \leq 2\left(\frac{x^2}{x+y} + \frac{y^2}{y+z} + \frac{z^2}{x+z}\right) $$
So we have three different upper bounds just by rotating how we decide to perform the splitting.
|
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|
How to Evaluate $ \int \! \frac{dx}{1+2\cos x} $ ?
Possible Duplicate:
How do you integrate $\int \frac{1}{a + \cos x} dx$?
I have come across this integral and I tried various methods of solving. The thing that gets in the way is the constant $2$ on the $\cos(x)$ term. I tried the conjugate (works without the 2$\cos x$), Weierstrass Substitution (not sure if I was applying it correctly), and others. Is there a way to solve this integral elegantly or some unknown (sneaky) trick when you come across families of similar integrals as this one?:
$$ \int \! \frac{dx}{1+2\cos x} $$
|
Generalization:
Let's consider $\cos x = \frac{1-t^2}{1+t^2}; t = \tan\frac{x}{2}; dx=\frac{2}{1+t^2} dt.$ Then we get that our integral becomes:
$$J = \int \frac{2dt}{(a+b)+(a-b) t^2}$$
I. For the case $a>b$, consider $a+b=u^2$ and $a-b=v^2$, and obtain that:
$$J = 2\int \frac{dt}{u^2+v^2 t^2}=\frac{2}{uv} \arctan\frac{vt}{u} +C.$$
Turning back to our notation we get:
$$I=\frac{2}{\sqrt{a^2-b^2}} \arctan\left(\sqrt{\frac{a-b}{a+b}} \tan\frac{x}{2} \right) + C.$$
II. For the case $a<b$, consider $a+b=u^2$ and $a-b=-v^2$, and obtain that:
$$J = 2\int \frac{dt}{u^2-v^2 t^2}=\frac{1}{uv}\ln\frac{u+vt}{u-vt} \ +C.$$
Turning back again to out initial notation and have that:
$$I=\frac{2}{\sqrt{b^2-a^2}} \ln\frac{b+a \cos x + \sqrt{b^2-a^2} \sin x}{a+b \cos x} + C.$$
Also, note that $x$ must be different from ${+}/{-}\arccos(-\frac{a}{b})+2k\pi$ if $|\frac{a}{b}|\leq1$.
Q.E.D.
|
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|
Find all integers $m$ such that $\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} $ How would you determine all integers $m$ such that the following is true?
$$\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} .$$
Note that $\lfloor \cdot \rfloor$ means the greatest integer function. Also, $x$ must be a positive real number.
|
If $k + j/10 \le x < k + (j+1)/10$ where $k$ and $j$ are nonnegative integers and $j \le 9$, $\lfloor 2x \rfloor = \begin{cases} 2k & 0 \le j \le 4 \\ 2k+1 & 5 \le j \le 9 \end{cases}$ while $\lfloor 5x \rfloor = \begin{cases} 5k + j/2 & j \ \text{even} \\ 5k + (j-1)/2 & j \ \text{odd} \end{cases}$.
Going over the various cases $j = 0$ to $9$, I find two cases where $f(x) = \frac{1}{1/\lfloor 2x \rfloor + 1/\lfloor 5x \rfloor} $ is an integer:
1) if $j = 0$ or $1$ and $k > 0$ is divisible by 7, $f(x) = 10 (k/7)$.
2) if $k=2$ and $j=4$, $f(x) = 3$.
Case (2) comes about as follows: if $j = 4$, $f(x) = \frac{1}{1/(2k) + 1/(5k+2)} = \frac{(2k)(5k+2)}{7k+2}$. Note that $\gcd(k,7k+2) = \gcd(k,2) = 1$ or $2 $ while $\gcd(5k+2,7k+2) = \gcd(5k+2,2k) = \gcd(k+2,2k) =1$, $2$ or $4$. The largest possible value of the denominator that could divide the numerator is thus $2 \times 2 \times 4 = 16$, which occurs for $k=2$.
|
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|
Expanding $(2y-2)^2$ by FOIL Expanding $(2y-2)^2$
Isn't this same as $$(2y-2)(2y-2)\ ?$$
$$4y^2-6y+4$$
This should be FOILd shouldn't it?
|
Yes, $(2y-2)^2=(2y-2)(2y-2)$.
FOILing should work, but will get you $4y^2−8y+4$, rather than $4y^2−6y+4$, as shown:
$(2y-2)(2y-2)=(2y)(2y)+(2y)(-2)+(-2)(2y)+(-2)(-2)$
$=4y^2-4y-4y+4=4y^2−8y+4$
If you want to memorize the formula which will get you the same result, it is
$$(a+b)^2=a^2+2ab+b^2$$
In your example, $a=2y$ and $b=-2$, so you get
$$(2y)^2+2(2y)(-2)+(-2)^2=4y^2-8y+4$$
|
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|
Finding the slope of the tangent line to $\frac{8}{\sqrt{4+3x}}$ at $(4,2)$ In order to find the slope of the tangent line at the point $(4,2)$ belong to the function $\frac{8}{\sqrt{4+3x}}$, I choose the derivative at a given point formula.
$\begin{align*}
\lim_{x \to 4} \frac{f(x)-f(4)}{x-4} &=
\lim_{x \mapsto 4} \frac{1}{x-4} \cdot \left (\frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{4+3 \cdot 4}} \right )
\\ \\ & = \lim_{x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{16}} \right ) \\ \\ & = \lim_{ x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-2\right)
\end{align*}$
But now I can't figure it out, how to end this limit.
I know that the derivative formula for this function is $-\frac{12}{(4+3x)\sqrt{4+3x}}$.
Thanks for the help.
|
Use the L'Hopital method, several equations becomes easy to solve
$$\lim_{x \to 4}\frac{f(x)-f(4)}{x-4}=\lim_{x \to 4}\frac{f'(x)-0}{1-0}=\lim_{x \to 4}f'(x)$$
Where
$f(x)=\frac{8}{\sqrt{3 x+4}}$ and $f'(x)$, the derivative of $f(x)$ is defined by
$f'(x)=-\frac{12}{(3 x+4)^{3/2}}$
The final equation results, just do the final calculus:
$$\lim_{x \to 4}\frac{f(x)-f(4)}{x-4}=\lim_{x \to 4}-\frac{12}{(3 x+4)^{3/2}}$$
|
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|
Group under matrix multiplication I am trying to show that this set P={ $p(\alpha,\beta,\gamma)=\pmatrix{1&\alpha&\beta\\0&1&\gamma\\0&0&1}$ $|$ $\alpha,\beta,\gamma$ $\in R$} is a group under matrix multiplication. I have already proved the closure, identity and associative properties. But with the inverse, Im stuck as to how I should go about proving it. I need to find $pp^{-1} = identity =p^{-1}p$.
So the inverse of $\pmatrix{1&x&y\\0&1&z\\0&0&1}$, I calculated it to be $\pmatrix{1&0&0\\-x&1&0\\xz-y&-z&1}$. But multiplying those 2 matrices it doesnt seem to be the identity matrix?
|
$$\begin{align}
\begin{pmatrix} 1 & x & y & \vdots & 1 & 0 & 0 \\
0 & 1 & z & \vdots & 0 & 1 & 0 \\
0 & 0 & 1 & \vdots & 0 & 0 & 1 \end{pmatrix} &\sim
\begin{pmatrix} 1 & x & 0 & \vdots & 1 & 0 & -y \\
0 & 1 & 0 & \vdots & 0 & 1 & -z \\
0 & 0 & 1 & \vdots & 0 & 0 & 1 \end{pmatrix} \\ &\sim
\begin{pmatrix} 1 & 0 & 0 & \vdots & 1 & -x & -y+xz \\
0 & 1 & 0 & \vdots & 0 & 1 & -z \\
0 & 0 & 1 & \vdots & 0 & 0 & 1 \end{pmatrix}
\end{align}$$
|
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|
If $a^n+n^{a}$ is prime number and $a=3k-1$ then $n\equiv 0\pmod 3$? Is it true that :
If $a^n+n^{a}$ is prime number and $a=3k-1$ then $n\equiv 0\pmod 3$
where $a>1,n>1 ; a,n,k \in \mathbb{Z^+}$
I have checked statement for many pairs $(a,n)$ and it seems to be true.
Small Maple code that prints $(a,n)$ pairs :
Any idea how to prove this statement ?
|
Since it is prime, we know that $a^n+n^a \equiv 1 \text{ or } 5 \mod 6$, and because $a \equiv 2 \mod 3$ we know that $a \equiv 2 \text{ or } 5 \mod 6.$
The case $a \equiv 2 \mod 6:$
Note that $n^2 \equiv n^8 \equiv n^{14} \dots \mod 6.$
When $n \equiv 0, 1, 2, 3, 4, 5 \mod 6, n^2 \equiv 0, 1, 4, 3, 4, 1 \mod 6$ and $2^n \equiv 4, 2, 4, 2, 4, 2 \mod 6.$
Adding these gives $a^n+n^a \equiv 4, 3, 2, 5, 2, 3 \mod 6$. 1 does not appear at all and 5 appears only when $n \equiv 3 \mod 6,$ meaning that $n \equiv 0 \mod 3,$ as required.
The case $a \equiv 5 \mod 6:$
Note that $n^5 \equiv n^{11} \equiv n^{17} \dots \mod 6.$
When $n \equiv 0, 1, 2, 3, 4, 5 \mod 6, n^5 \equiv 0, 1, 2, 3, 4, 5 \mod 6$ and $5^n \equiv 1, 5, 1, 5, 1, 5 \mod 6.$
Adding these gives $a^n+n^a \equiv 1, 0, 3, 2, 5, 4 \mod 6$. 1 appears only when $n \equiv 0\mod 6$, implying that $n \equiv 0 \mod 3,$ as required.
However, 5 appears as well, when $n \equiv 4 \mod 6,$ suggesting the possibility of the result being false for these values of $a$ and $n$.
Edit
I have found that $a=215$ and $n=76$ is a counterexample. $a^n+n^a$ is prime, $a \equiv 2 \mod 3$ but $n\not\equiv 0 \mod 3$.
|
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|
Negative Exponents in Binomial Theorem I'm looking at extensions of the binomial formula to negative powers. I've figured out how to do $n \choose k$ when $n < 0 $ and $k \geq 0$:
$${n \choose k} = (-1)^k {-n + k - 1 \choose k}$$
So now let's look at one case for using the binomial coefficient:
$$(1+x)^n = \sum_{k=0}^n {n \choose k}x^k$$
How do I evaluate $\sum_{k = 0}^{n}$ when $n < 0$? From searching around on the internet I think it's just an infinite series, i.e. $k$ keeps incrementing by 1 forever. But that gets me confused about
$$\begin{align*}
(a + b)^n &= a^n(1 + \frac{b}{a})^n \\ &= a^n \left(\sum_{k = 0}^{n}{n \choose k}\left(\frac{b}{a}\right)^k\right)\\
&= a^n \left(1 + n \left(\frac{b}{a}\right) + \frac{(n)(n-1)}{2}\left(\frac{b}{a}\right)^2 + \cdots\right)
\end{align*}$$
and
$$\begin{align*}
(b + a)^n &= b^n\left(1 + \frac{a}{b}\right)^n\\
&= b^n \left(\sum_{k = 0}^{n}{n \choose k}\left(\frac{a}{b}\right)^k\right)\\
&= b^n \left(1 + n \left(\frac{a}{b}\right) + \frac{(n)(n-1)}{2}\left(\frac{a}{b}\right)^2 + \cdots\right)
\end{align*}$$
Now the two should be equal, but in the first sum I'd never get a $b^n$ and in the second sum I'd never get a $a^n$?
|
For $a=1$, the negative binomial series simplifies to
$(x+1)^{-n}=1-nx+\frac1{2!}n(n+1)x^2-\frac1{3!}n(n+1)(n+2)x^3+....$
|
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|
Distance Between A Point And A Line Any Hint on proving that the distance between the point $(x_{1},y_{1})$ and the line $Ax + By + C = 0$ is,
$$\text{Distance} = \frac{\left | Ax_{1} + By_{1} + C\right |}{\sqrt{A^2 + B^2} }$$
What do I use to get started? All I know is the distance formula $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$.
Kindly Help
|
Nothing to see here, just me preparing for my Calc 3 final by attempting this.
We can Lagrange this. Let $f(x) = \sqrt{(x - x_1)^2 + (y - y_1)^2}$ be the distance of $(x_1, y_1)$ from some $(x, y)$ on the line $Ax + By + C = 0$. Also, let $g(x) = Ax + By + C = 0$ be our constraint function, and $(f(x, y))^2 = (x - x_1)^2 + (y - y_1)^2$ be our target function. We want to minimize the distance of the point to the line, basically.
Observe that $\nabla (f(x, y))^2 = \langle 2(x - x_1),2(y - y_1) \rangle$, and $\nabla g(x, y) = \langle A, B \rangle$. Because of the way the gradient behaves around critical points, $\nabla (f(x, y))^2 = \nabla g(x, y)$, so $\langle 2(x - x_1),2(y - y_1) \rangle = \lambda \langle A, B \rangle$. The solutions to this equation are $x = x_1 + \dfrac{\lambda A}{2}$ and $y = y_1 + \dfrac{\lambda B}{2}$ (this is nice because then $x - x_1 = \dfrac{\lambda}{2}A$ and $y - y_1 = \dfrac{\lambda}{2}B$, both of which are helpful later). Plug this back into $Ax + By + C = 0$, and you obtain $\lambda = \dfrac{2Ax_1 + 2By_1 + 2C}{A^2 + B^2}$.
The distance $f(x) = \sqrt{(x - x_1)^2 + (y - y_1)^2} = \sqrt{\left(\dfrac{\lambda A}{2}\right)^2 + \left(\dfrac{\lambda B}{2}\right)^2} = \dfrac{|\lambda|}{2} \sqrt{A^2 + B^2}$. Putting in the expression for $\lambda$, the closest distance is $$\min f(x) = \dfrac{1}{2}\left|\dfrac{2Ax_1 + 2By_1 + 2C}{A^2 + B^2}\right|\sqrt{A^2 + B^2} = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}},$$as desired.
|
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|
How to compute the characteristic polynomial of $A$ The matrix associated with $f$ is:
$$ \left(\begin{array}{rrr}
3 & -1 & -1 \\
-1 & 3 & -1 \\
-1 & -1 & 3
\end{array}\right) .
$$
First, I am going to find the characteristic equation of $\det(A- \lambda I)$. Please correct me if I'm wrong.
$$= (3-\lambda)(3-\lambda)(3-\lambda)-1-1-(3-\lambda)-(3-\lambda)-(3-\lambda)
=-\lambda^3+9\lambda^2-24\lambda+16 .$$
How to factor this? I know the $\lambda$'s should be $1$, $4$ and $4$. But how am I supposed to find these values?
|
Notice that your characteristic equation is
$$(3-\lambda)^3 - 3(3-\lambda) - 2 = 0.$$
Making the change of variable $x=3-\lambda$, we get the depressed cubic
$$x^3 - 3x - 2 = 0.$$
By the rational root theorem, you can test $1$, $-1$, $2$ and $-2$, which tells you that $x=2$ is a solution. Factoring out $x-2$ we have
$$0 = x^3-3x-2 = (x-2)(x^2+2x+1) = (x-2)(x+1)^2.$$
So the roots are $x=2$ and $x=-1$ (twice), and since $x=3-\lambda$, we get that the roots of the original equation are $\lambda=3-x$, or $\lambda = 1$, and $\lambda=4$ (twice).
|
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|
Right triangle where the perimeter = area*k I was doodling on some piece of paper a problem that sprung into my mind. After a few minutes of resultless tries, I advanced to try to solve the problem using computer based means.
The problem stated is
Does a right angle triangle with integer sides such that
$$P = A\cdot k \quad \quad \text{or} \quad \quad k\cdot P = A$$ where $(P , \ A , \ k )\in \mathbb{N}$ exist?
(Here P is the perimeter of the triangle. Sum of the sides. And A is the area of the triangle. a*b/2 in a triangle with a b c and c is the hypotenuse)
Obviously for the simple cases such triangles exists. For an example
when
$ A=2P \qquad $ the triangle $12,16,20$ works
$ A=P \qquad \; \; $ the triangle $6,8,10$ works
$2A=P \qquad $ the triangle $3,4,5$ works
I tried solving this by hand, first for the special case where $2P=A$. This ended up giving me
$$ \frac{a \cdot b}{2} = A $$
and
$$ P = \frac{A}{2} = \frac{a \cdot b}{4} \qquad \text{also} \qquad P = a + b + c $$
So
$$ \frac{ab}{4} = a + b + c $$
By knowing that this is a right angle this leads to the equation (Using the Pythagorean theorem)
$$ a^2 + b^2 = c^2 $$
Now we have two equations and three unknowns, which also needs to be integers! Sadly I was not able to continue from here. I have only learned how to solve linear Diophantine equations. Not a system of nonlinear Diophantine equations.
Just to restate my question below =)
Is there a right angle triangle with integer sides such that
$$P = A\cdot k \quad \quad \text{or} \quad \quad k\cdot P = A$$ where $(P , \ A , \ k )\in \mathbb{N}$ ?
Regards, Werner
|
We have the following relations:
$$P=a+b+c,\quad 2A=ab,\quad a^2+b^2=c^2$$
In fact, there is one more that is implicit: the triangle inequality $a+b>c$ (which follows from $a^2+b^2=c^2$). We wish to relate the first and second relations, and so after staring at the problem for a while we realize the correct course of action is to multiply $P$ by $a+b-c$ (the latter guaranteed to be positive by the triangle inequality). This gives:
$$\begin{align}P(a+b-c)&=(a+b+c)(a+b-c)=(a+b)^2-c^2\\
&=a^2+b^2+2ab-c^2=2ab=4A
\end{align}$$
Hence in general for a right triangle we have $k=A/P=(a+b-c)/4$, so the question boils down to when the difference between the sum of the lengths of the legs and the length of the hypothenuse divides or is divisible by $4$, for a right triangle with integer side-lengths.
Here we actually need to generate pythagorean triples $(a,b,c)$ such that $a,b,c$ are integers and $a^2+b^2=c^2$; the wikipedia page here gives a good picture of Euclid's formula that I recommend thinking through. Euclid's formula is the following:
$$a=m^2-n^2,\quad b=2mn\quad c=m^2+n^2$$ for integers $m>n$.
Then the quantity $a+b-c$ is $m^2-n^2+2mn-m^2-n^2=2n(m-n)$, so $A/P=k=n(m-n)/2$. The options are now that either $2$ divides $n$ or $2$ divides $m-n$, so we either have $n=2d$ and $m=k/d+2d$ or $n=d$ and $m=2(k/d)+d$ for divisors $d$ of $k$.
(to get $P/A=k$ we need $2n(m-n)$ to divide $4$, so $n(m-n)$ must divide $2$, thus $n=1$, $m=2$, $n=1$, $m=3$, and $n=2$, $m=3$ are the only solutions, giving us $(3,4,5)$, $(8,6,10)$, $(5,12,13)$ with $k=2,1,2$ being the only possibilities).
|
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|
Solving for the smallest $x$ : $1! + 2! + \cdots+ 20! \equiv x\pmod 7$ I know the smallest $x \in \mathbb{N}$, satisfying $1! + 2! + \cdots + 20! \equiv x\pmod7$ is $5$. I would like to know methods to get to the answer.
|
Note that each of $7!$, $8!$, $9!,\ldots, 20!$ is congruent to $0$ modulo $7$, since they are all divisible by $7$.
Note that $6!\equiv -1\pmod{7}$ by Wilson's Theorem, which cancels $1!$.
That leaves $2!+3!+4!+5! = 2! + 3!(1 + 4 + 20)$. But $3!\equiv -1\pmod{7}$, and $20\equiv -1\pmod{7}$, so $2!+3!+4!+5! \equiv 2-(1+4-1) = -2\equiv 5\pmod{7}$.
|
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|
Is there a binomial identity for this expression $\frac{\binom{r}{k}}{\binom{n}{k}}$? What I'm trying to prove is this summation:
$$\sum_{i=0}^{k} \dfrac{\dbinom{r}{i} \cdot \dbinom{n - r}{k - i}}{\dbinom{n}{k}} \cdot i = \dfrac{r}{n} \cdot k$$
I used induction on $k$ as follows:
$$LHS = \sum_{i=0}^{k + 1} \dfrac{\dbinom{r}{i} \cdot \dbinom{n - r}{k + 1 - i}}{\dbinom{n}{k + 1}} \cdot i$$
$$ = \sum_{i=0}^{k} \dfrac{\dbinom{r}{i} \cdot \dbinom{n - r}{k - i}}{\dbinom{n}{k}} \cdot i
+ \dfrac{\dbinom{r}{k + 1} \cdot \dbinom{n - r}{k + 1 - (k + 1)}}{\dbinom{n}{k+1}} \cdot (k + 1)$$
$$ = \dfrac{rk}{n} + \dfrac{\dbinom{r}{k + 1} \cdot \dbinom{n - r}{0}}{\dbinom{n}{k+1}} \cdot (k + 1)$$
$$ = \dfrac{rk}{n} + \dfrac{\dbinom{r}{k + 1}}{\dbinom{n}{k+1}} \cdot (k + 1)$$
Then I was stuck with the expression $\dfrac{\dbinom{r}{k + 1}}{\dbinom{n}{k + 1}}$, where $k + 1 \leq r \leq n$. I guess I need a clever trick here to simplify the left hand side, any idea?
|
We don't need induction:
\begin{align*}
\sum_{i=0}^k\binom ri\binom{n-r}{k-i}i&=\sum_{i=1}^k\frac{r!}{(i-1)!(r-1-(i-1))}\binom{n-r}{k-i}\\
&=r\sum_{l=0}^{k-1}\binom{r-1}{l}\binom{n-r}{k-1-l}\\
&=r\binom{n-r+r-1}{k-1}\\
&=r\binom{n-1}{k-1}\\
&=\frac rn\frac{n!}{(k-1)!(n-k)!}\\
&=\frac {rk}n\binom nk,
\end{align*}
the third equality is Chu-Vandermonde identity.
|
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|
Finding the $n$-th derivatives of $x^n \ln x$ and $\frac{\ln x}{x}$. How can I prove the following identities:
$$ \left( x^n \ln x \right)^{(n)}= n! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{n} \right), \quad x>0, \quad n\ge 1, \tag{a}$$
$$ \left( \frac{\ln x}{x} \right)^{(n)}= (-1)^n \ n! \ x^{-n-1} \left( \ln x - 1 - \frac{1}{2} - \cdots -\frac{1}{n} \right), \quad x>0,\quad n\ge 1, \tag{b}$$
Here $(\cdot )^{(n)}$ indicates the $n$-th derivative with respect to $x$.
I don't know where to start!
|
I prefer using induction to prove the statement. For (a), when $n=1$, by product rule
$$\left( x \ln x \right)'=\ln x+x\cdot\frac{1}{x}=1!(\ln x+1),$$
which shows that (a) is true for $n=1$. Now suppose that (a) is true for $n=k$, i.e.
$$\left( x^k \ln x \right)^{(k)}= k! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{k} \right).$$
Now, for $n=k+1$, we have
$$\left( x^{k+1} \ln x \right)^{(k+1)}=\left[ \big(x^{k+1} \ln x\big)' \right]^{(k)}=
\left[ (k+1)x^k \ln x+x^{k+1}\cdot\frac{1}{x}\right]^{(k)}$$
$$=(k+1)\left( x^k \ln x \right)^{(k)}+(x^k)^{(k)}.$$
Note that $(x^k)^{(k)}=k!$, by induction assumption, we have
$$\left( x^{k+1} \ln x \right)^{(k+1)}=(k+1)k! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{k} \right)+k!$$
$$=(k+1)! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{k}+\frac{1}{k+1} \right),$$
which shows that (a) is true for $n=k+1$.
I will let you try (b) using induction.
|
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|
Can anyone prove this formula? I found formula below$$p_n=6\left \lfloor \frac{p_n}{6}+\frac{1}{2} \right \rfloor+\left ( -1 \right )^\left \lfloor \frac{p_n}{3} \right \rfloor$$ for $n>2$, $p_n$ is prime number sequence.
Can anyone prove this formula?
|
For $n>2$ the prime $p_n$ is greater than $3$, so it must be of the form $6k+1$ or $6k+5$. If $p_n=6k+1$, then $$6\left\lfloor \frac{p_n}6+\frac12\right\rfloor= 6\left\lfloor k+\frac16+\frac12\right\rfloor=6k\;,$$ and $$\left\lfloor\frac{p_n}3\right\rfloor=\left\lfloor\frac{6k+1}3\right\rfloor=2k\;,$$ so $$6\left\lfloor\frac{p_n}6+\frac12\right\rfloor+(-1)^{\left\lfloor\frac{p_n}3\right\rfloor}=6k+(-1)^{2k}=6k+1=p_n\;.$$
If $p_n=6k+5$, then $$6\left\lfloor \frac{p_n}6+\frac12\right\rfloor= 6\left\lfloor k+\frac56+\frac12\right\rfloor=6(k+1)\;,$$ and $$\left\lfloor\frac{p_n}3\right\rfloor=\left\lfloor\frac{6k+5}3\right\rfloor=2k+1\;,$$ so $$6\left\lfloor\frac{p_n}6+\frac12\right\rfloor+(-1)^{\left\lfloor\frac{p_n}3\right\rfloor}=6(k+1)+(-1)^{2k+1}=6(k+1)-1=6k+5=p_n\;.$$
|
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|
The Taylor series of $\int_0^x \operatorname{sinc}(t) dt$ I tried to find what is the Taylor series of the function $$\int_0^x \frac{\sin(t)}{t}dt .$$
Any suggestions?
|
I just want to extend J.M.'s solution to a full solution of the exercise:
$$\int_0^x \frac{\sin(t)}{t} \,dt$$
Is with Maclaurin expansion ($\sin\,t=t\left(1-t^2/3!+t^4/5!-t^6/7!+\cdots\right)$) equals to
$$\int_0^x \frac{t}{t}\left(1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+\cdots\right) \,dt = \int_0^x \left(1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+\cdots\right) \,dt $$
We do now integrate this series to:
$$\int_0^x \left(1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+\cdots\right) \,dt =
x-\frac{x^3}{3 * 3!}+\frac{x^5}{5*5!}-\frac{x^7}{7 * 7!}+\cdots$$
Above we did use the flowing fact:
$$\int_0^x t^k \, dt=\left.\frac{t^{k+1}}{k+1}\right|_0^x = \frac{x^{k+1}}{k+1}-\frac{0^{k+1}}{k+1} = \frac{x^{k+1}}{k+1}$$
We can now write the result as a series:
$$x-\frac{x^3}{3 * 3!}+\frac{x^5}{5*5!}-\frac{x^7}{7 * 7!}+\cdots = \sum_{n=0}^\infty (-1)^n\frac{x^{(2n+1)}}{(2n+1) * (2n+1)!} = Si(x)$$
And see that we get the Sine Integral.
|
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|
Using Dyson's conjecture to give another proof of Dixon's identity. For natural numbers $a_1,\dots,a_n$, Freeman Dyson conjectured (and it was eventually proven) that the Laurent polynomial
$$
\prod_{i,j=1\atop i\neq j}^n\left(1-\frac{x_i}{x_j}\right)^{a_i}
$$
has constant term the multinomial coefficient $\binom{a_1+\cdots+a_n}{a_1,\dots,a_n}$.
A.C. Dixon proved Dixon's Identity:
$$
\sum_{k=-a}^a(-1)^k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}=\frac{(a+b+c)!}{a!b!c!}.
$$
This last quantity is just the multinomial coefficient $\binom{a+b+c}{a,b,c}$. So letting $a_1=a,a_2=b,a_3=c$, it should be the case by Dyson's conjecture that
$$
\prod_{i,j=1\atop i\neq j}^3\left(1-\frac{x_i}{x_j}\right)^{a_i}
$$
that is,
$$
\left(1-\frac{x_1}{x_2}\right)^a\left(1-\frac{x_1}{x_3}\right)^a\left(1-\frac{x_2}{x_1}\right)^b\left(1-\frac{x_2}{x_3}\right)^b\left(1-\frac{x_3}{x_1}\right)^c\left(1-\frac{x_3}{x_2}\right)^c
$$
has constant term $\binom{a+b+c}{a,b,c}$. Does anyone see a clever way to conclude that the constant term is also calculated as $
\sum_{k=-a}^a(-1)^k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}$ to get another proof of the result? Many thanks.
My guess is that one will want to "choose" say the same number of factors say $-\frac{x_1}{x_2}$ from $(1-\frac{x_1}{x_2})^a$ as the number of factors $-\frac{x_2}{x_1}$ from $(1-\frac{x_2}{x_1})^b$, when expanding. So I assume the sum fro $-a$ to $a$ will count how many ways there are to choose a carefully from the six terms in the product to get a constant when it's all multiplied out.
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Setting $a_1=a$, $a_2=b$, $a_3=c$, we can start with
$
\prod_{1\le i,j\le 3, \ i\ne j}\left(1-\frac{x_i}{x_j}\right)^{a_i}
$
and combine pairs of factors involving the same pair of variables to find that
$$
\prod_{1\le i,j\le 3, \ i\ne j}\left(1-\frac{x_i}{x_j}\right)^{a_i}
$$
$$
= (-1)^{a+b+c} x_1^{a-c} x_2^{b-a} x_3^{c-b}
(1-\frac{x_2}{x_1})^{a+b}
(1-\frac{x_3}{x_2})^{b+c}
(1-\frac{x_1}{x_3})^{c+a}. \qquad (1)
$$
Using the binomial theorem, we can expand (1) as
$$
\sum_{i,j,\ell} (-1)^{a+b+c+i+j+\ell} {a+b\choose i} {b+c\choose j} {c+a\choose \ell}
x_1^{\ell-i+a-c}
x_2^{i-j+b-a}
x_3^{j-\ell+c-b}.\qquad (2)
$$
To find the constant term in (2), we need to look at all terms where
$$
\ell-i+a-c=0,\ i-j+b-a=0,\ j-\ell+c-b=0.\qquad (3)
$$
If we set $i:=a+k$ in (3), we immediately get $\ell=c+k$ and $j=b+k$ from the first and second equations.
These choices of $i$, $j$ and $\ell$ also satisfy the third equation. Therefore,
the constant term in (2) is
$$
\sum_k (-1)^{2a+2b+2c+3k} {a+b\choose a+k} {b+c\choose b+k} {c+a\choose c+k},
$$
where the sum is extended over all $k$ where the binomial coefficients are nonzero.
Since $(-1)^{2a+2b+2c+3k}=(-1)^k$, we can now use Dixon's identity and prove that the constant term of (1)
is $\left({a+b+c\atop a\ \ \ b\ \ \ c}\right)$.
|
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|
Trigonometric equality $\cos x + \cos 3x - 1 - \cos 2x = 0$ In my text book I have this equation:
\begin{equation}
\cos x + \cos 3x - 1 - \cos 2x = 0
\end{equation}
I tried to solve it for $x$, but I didn't succeed.
This is what I tried:
\begin{align}
\cos x + \cos 3x - 1 - \cos 2x &= 0 \\
2\cos 2x \cdot \cos x - 1 - \cos 2x &= 0 \\
\cos 2x \cdot (2\cos x - 1) &= 1
\end{align}
So I clearly didn't choose the right path, since this will only be useful if I become something like $a \cdot b = 0$.
All tips will be greatly appreciated.
Solution (Addition to the accepted answer):
The problem was in writing $\cos 3x$ in therms of $cos x$. Anon pointed out that it was equal to $4\cos^3 x - 3\cos x$ but I had to work may way thru it to actually prove that. So I write it down here, maybe it's of use to anybody else.
\begin{align}
\cos 3x &= \cos(2x + x)\\
&= \cos(2x)\cdot \cos x - \sin(2x)\sin x\\
&= (\cos^2 x - \sin^2 x) \cdot \cos x - 2\sin^2 x \cdot \cos^2 x\\
&= \cos^3 x - \sin^2x\cdot \cos x - 2\sin^2x\cdot \cos^2 x\\
&= \cos^3 x - (1 - \cos^2 x)\cos x - 2(1-\cos^2 x)\cos x\\
&= cos^3 x - \cos x + \cos^3 x - 2\cos x + 2\cos^3 x\\
&= 4\cos^3 x - 3\cos x
\end{align}
EDIT: Apparently you can write all of the formulas $\cos(n\cdot x)$ with $n \in \{1, 2, 3, …\}$ in terms of $\cos x$. Why didn't my teacher tell me that! I don't have time to proof it myself now, but I'll definitely adapt my answer tomorrow (or any time soon)!
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Hint: Can you write $\cos 2x$ and $\cos 3x$ in terms of $\cos x$?
|
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|
A symmetric inequality
Possible Duplicate:
Cauchy-Schwarz inequality and three-letter identities (exercise 1.4 from “The Cauchy-Schwarz Master Class”)
Is it true for all $x, y, z > 0$ that
$$ x + y + z \leq 2 \left\{ \frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y} \right\} $$
This is an exercise (1.4) in "The Cauchy-Schwarz Master Class: An Introduction to the Art of mathematical Inequalities"
The solution suggests applying C-S to
$$ x + y + z = \frac{x}{\sqrt{y+z}}\sqrt{y+z} + \frac{y}{\sqrt{x+z}}\sqrt{x+z}
+ \frac{z}{\sqrt{x+y}}\sqrt{x+y} $$
|
It is correct $ x + y + z = \frac{x}{\sqrt{y+z}}\sqrt{y+z} + \frac{y}{\sqrt{x+z}}\sqrt{x+z}
+ \frac{z}{\sqrt{x+y}}\sqrt{x+y}\le \left(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{y+x}\right)^{1/2}(2x+2y+2z)^{1/2}.$ Square both sides, and cancel out $x+y+z$.
|
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|
$(\cos \alpha, \sin \alpha)$ - possible value pairs We introduced the complex numbers as elements of $ \mathrm{Mat}(2\times 2, \mathbb{R})$ with
$$
\mathbb{C} \ni x =
\left(\begin{array}{cc}
a & -b \\
b & a \\
\end{array}\right) =
\frac{1}{\sqrt{a^2+b^2}}
\left(\begin{array}{cc}
\frac{a}{\sqrt{a^2+b^2}} & \frac{-b}{\sqrt{a^2+b^2}} \\
\frac{b}{\sqrt{a^2+b^2}} & \frac{a}{\sqrt{a^2+b^2}} \\
\end{array}\right)
$$
Then we concluded that $0 \leq \frac{a}{\sqrt{a^2+b^2}} \leq 1$ and $0 \leq \frac{b}{\sqrt{a^2+b^2}} \leq 1$ and therefore we could find an $\alpha$ so that $\cos \alpha = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin \alpha = \frac{b}{\sqrt{a^2+b^2}}$. Then we can write the matrix with $\cos$ and $\sin$ and can write it as the Euler form as well. So far so good.
My question ist about the following:
Why is it that we cand find an $\alpha$ so that $\cos \alpha = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin \alpha = \frac{b}{\sqrt{a^2+b^2}}$ for every possible value combination of $a$ and $b$?
Can't be there a combination of $a$ and $b$ where we can't find one and the same angle $\alpha$ so that the identites are true?
|
Find an $\alpha$ so that $\cos(\alpha)= \frac{a}{\sqrt{a^2+b^2}}$. This is possible because the fraction is between $-1$ and $1$.
Then note that
$$\sin^2(\alpha)= 1- \cos^2(\alpha)=1- \frac{a^2}{a^2+b^2}=\frac{b^2}{a^2+b^2} \,.$$
Thus either $\sin(\alpha)=\frac{b}{\sqrt{a^2+b^2}}$ or $\sin(\alpha)=\frac{-b}{\sqrt{a^2+b^2}}$.
In the first situation you are done, while if $\sin(\alpha)=\frac{-b}{\sqrt{a^2+b^2}}$ then
$$\cos(-\alpha)= \frac{a}{\sqrt{a^2+b^2}} \,;\, \sin(-\alpha)=\frac{b}{\sqrt{a^2+b^2}}$$
|
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|
How can I find the area of the region bounded by the hyperbola now I got region question....
The question was "find the area of the region bounded by the hyperbola $9x^{2} - 4y^{2} = 36$ and the line $x = 3$
I drew the graph of $9x^{2} - 4y^{2} = 36$ and $x = 3$ and I got right side is bounded by $x=3$ but since the graph is hyperbola, left side is not bounded by anything. My question is that how I can find the area of the region ?
If you have any idea, could post here ?
Thanks !!
|
Solving for $y$ produces:
$$ y =\frac{1}{2} \sqrt{9x^2-36} $$
What we need to find is the area of $y$ from $x=2$ to $x=3$.
Thus we put:
$$A = \frac{1}{2}\int_2^3 \sqrt{9x^2-36} dx$$
Make an "hiperbolic" substitution:
$$3x = 6 \cosh u$$
$$3dx = 6 \sinh u du$$
We get the new limits are $0$ and $b = \cosh^{-1}\frac{3}{2} = \log({\frac{3}{2} +\sqrt{\frac{5}{4}}})$
$$A = 6\int_0^b {{{\sinh }^2}udu} $$
This integral is very similar to the integral of $\sin^2 x$,
$$A = 6\int_0^b {{{\sinh }^2}udu} =6 \left. {\frac{{\sinh \left( {2u} \right) - 2u}}{4}} \right|_0^b$$
Then you have
$$A = 3\frac{{\sinh 2b - 2b}}{2}$$
After a myriad of algebraic steps I end up with
$$A = \frac{9}{4}\sqrt 5 - 6\log \phi $$
Where $$\phi$$ is the golden ratio number.
Note that
$$\eqalign{
& \log 8 - 3\log \left( {3 + \sqrt 5 } \right) = 3\log 2 - 3\log \left( {3 + \sqrt 5 } \right) = \cr
& - 3\log \left( {\frac{{3 + \sqrt 5 }}{2}} \right) = - 3\log \left( {1 + \frac{{1 + \sqrt 5 }}{2}} \right) = \cr
& - 3\log \left( {1 + \phi } \right) = - 3\log {\phi ^2} = - 6\log \phi \cr} $$
EDIT: Remeber the value is just half the value of the total area.
|
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|
Possibility to simplify $\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{\pi }{{\sin \pi a}}} $ Is there any way to show that
$$\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{1}{a} + \sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^k}\left( {\frac{1}{{a - k}} + \frac{1}{{a + k}}} \right)}=\frac{\pi }{{\sin \pi a}}} $$
Where $0 < a = \dfrac{n+1}{m} < 1$
The infinite series is equal to
$$\int\limits_{ - \infty }^\infty {\frac{{{e^{at}}}}{{{e^t} + 1}}dt} $$
To get to the result, I split the integral at $x=0$ and use the convergent series in $(0,\infty)$ and $(-\infty,0)$ respectively:
$$\frac{1}{{1 + {e^t}}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{e^{ - \left( {k + 1} \right)t}}} $$
$$\frac{1}{{1 + {e^t}}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{e^{kt}}} $$
Since $0 < a < 1$
$$\eqalign{
& \mathop {\lim }\limits_{t \to 0} \frac{{{e^{\left( {k + a} \right)t}}}}{{k + a}} - \mathop {\lim }\limits_{t \to - \infty } \frac{{{e^{\left( {k + a} \right)t}}}}{{k + a}} = \frac{1}{{k + a}} \cr
& \mathop {\lim }\limits_{t \to \infty } \frac{{{e^{\left( {a - k - 1} \right)t}}}}{{k + a}} - \mathop {\lim }\limits_{t \to 0} \frac{{{e^{\left( {a - k - 1} \right)t}}}}{{k + a}} = - \frac{1}{{a - \left( {k + 1} \right)}} \cr} $$
A change in the indices will give the desired series.
Although I don't mind direct solutions from tables and other sources, I prefer an elaborated answer.
Here's the solution in terms of $\psi(x)$. By separating even and odd indices we can get
$$\eqalign{
& \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \sum\limits_{k = 0}^\infty {\frac{1}{{a + 2k}}} - \sum\limits_{k = 0}^\infty {\frac{1}{{a + 2k + 1}}} \cr
& \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} = \sum\limits_{k = 0}^\infty {\frac{1}{{a - 2k}}} - \sum\limits_{k = 0}^\infty {\frac{1}{{a - 2k - 1}}} \cr} $$
which gives
$$\sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right)$$
$$\sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} = \frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) + \frac{1}{a}$$
Then
$$\eqalign{
& \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} + \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} - \frac{1}{a} = \cr
& = \left\{ {\frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right)} \right\} - \left\{ {\frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right)} \right\} \cr} $$
But using the reflection formula one has
$$\eqalign{
& \frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right) = \frac{\pi }{2}\cot \frac{{\pi a}}{2} \cr
& \frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right) = \frac{\pi }{2}\cot \frac{{\pi \left( {a + 1} \right)}}{2} = - \frac{\pi }{2}\tan \frac{{\pi a}}{2} \cr} $$
So the series become
$$\eqalign{
& \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \frac{\pi }{2}\left\{ {\cot \frac{{\pi a}}{2} + \tan \frac{{\pi a}}{2}} \right\} \cr
& \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \pi \csc \pi a \cr} $$
The last being an application of a trigonometric identity.
|
EDIT: The classical demonstration of this is obtained by expanding in Fourier series the function $\cos(zx)$ with $x\in(-\pi,\pi)$.
Let's detail Smirnov's proof (in "Course of Higher Mathematics 2 VI.1 Fourier series") :
$\cos(zx)$ is an even function of $x$ so that the $\sin(kx)$ terms disappear and the Fourier expansion is given by :
$$\cos(zx)=\frac{a_0}2+\sum_{k=1}^{\infty} a_k\cdot \cos(kx),\ \text{with}\ \ a_k=\frac2{\pi} \int_0^{\pi} \cos(zx)\cos(kx) dx$$
Integration is easy and $a_0=\frac2{\pi}\int_0^{\pi} \cos(zx) dx= \frac{2\sin(\pi z)}{\pi z}$ while
$a_k= \frac2{\pi}\int_0^{\pi} \cos(zx) \cos(kx) dx=\frac1{\pi}\left[\frac{\sin((z+k)x)}{z+k}+\frac{\sin((z-k)x)}{z-k}\right]_0^{\pi}=(-1)^k\frac{2z\sin(\pi z)}{\pi(z^2-k^2)}$
so that for $-\pi \le x \le \pi$ :
$$
\cos(zx)=\frac{2z\sin(\pi z)}{\pi}\left[\frac1{2z^2}+\frac{\cos(1x)}{1^2-z^2}-\frac{\cos(2x)}{2^2-z^2}+\frac{\cos(3x)}{3^2-z^2}-\cdots\right]
$$
Setting $x=0$ returns your equality :
$$
\frac1{\sin(\pi z)}=\frac{2z}{\pi}\left[\frac1{2z^2}-\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2-z^2}\right]
$$
while $x=\pi$ returns the $\mathrm{cotg}$ formula :
$$
\cot(\pi z)=\frac1{\pi}\left[\frac1{z}-\sum_{k=1}^{\infty}\frac{2z}{k^2-z^2}\right]
$$
(Euler used this one to find closed forms of $\zeta(2n)$)
The $\cot\ $ formula is linked to $\Psi$ via the Reflection formula :
$$\Psi(1-x)-\Psi(x)=\pi\cot(\pi x)$$
The $\sin$ formula is linked to $\Gamma$ via Euler's reflection formula :
$$\Gamma(1-x)\cdot\Gamma(x)=\frac{\pi}{\sin(\pi x)}$$
|
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|
Find smallest number when divided by $2,3,4,5,6,7,8,9,10$ leaves $1,2,3,4,5,6,7,8,9$ remainder Find smallest number when divided by $2,3,4,5,6,7,8,9,10$ leaves $1,2,3,4,5,6,7,8,9$ remainder.How to go about solving this problem??
|
There is no smallest such number. But there is a smallest positive integer.
Note that $-1$ has the desired property. The integers that have the desired property are all the integers of the form $-1+kM$, where $M$ is the least common multiple of the integers $2,3,4,\dots, 10$ and $k$ ranges over the integers. So the smallest positive integer that works is $M-1$. It is not difficult to find $M$.
Remark: This type of problem is in general much more messy to solve. A specified sequence of remainders may not be achievable, and even when it is, efficiently finding an integer that works involves using the Chinese Remainder Theorem, or some equivalent procedure. In our particular case, finding an integer that works was easy, since $-1$ is clearly a solution. Its only flaw is that it is not positive, but that flaw can be dealt with.
What's going on is easier to see if we use congruence notation. We are told that a certain integer $x$ is congruent to $1$ modulo $2$, congruent to $2$ modulo $3$, congruent to $3$ modulo $4$, and so on up to congruent to $9$ modulo $10$. We can restate these congruences in the form $x$ is congruent to $-1$ modulo $2$, $x$ is congruent to $-1$ modulo $3$, $x$ is congruent to $-1$ modulo $4$, and so on up to $x$ is congruent to $-1$ modulo $10$. So $x=-1$ works for every one of our moduli.
Added details: There is something not very intuitive about the remainder when a negative integer like $-1$ is divided by, say, $6$. The congruence notation is quite helpful here, but we will try to do without. The remainder when the number $a$ is divided by $6$ is the integer $r$ such that $0\le r\le 5$, and there is some integer $q$ such that $a=6q+r$.
Apply this definition to $a=-1$. Note that $-1=6q+r$, where $q=-1$ and $r=5$. So the remainder when $-1$ is divided by $6$ is $5$. The same sort of argument works for all the numbers $2$ to $10$.
To show that all the numbers that satisfy our property have the shape $-1+kM$, we must check (i) All the numbers of this form do satisfy our property, and (ii) Nothing else does.
For (i), suppose that $M$ is the least common multiple of the numbers from $2$ to $10$. Then when we divide $-1+kM$ by any $m$ from $2$ to $10$, we get the same remainder as when we divide $-1$ by $m$, because $m$ divides $M$. Since the remainder when $-1$ is divided by $m$ is $m-1$, the remainder when $-1+kM$ is divided by $m$ is also $m-1$, and therefore $-1+kM$ "works."
To prove (ii), suppose that $x$ has the same remainder as $-1$ when $x$ is divided by any of the numbers from $2$ to $10$. Then $x-(-1)$ has remainder $0$ when it is divided by any one of the numbers from $2$ to $10$. It follows that $x+1$ is divisible by the least common multiple of the numbers from $2$ to $10$. Thus $M$ divides $x+1$, which means that $x$ has the shape $kM-1$.
|
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|
Showing $ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}}$ I would like to show that:
$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}} $$
We have:
$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\sum_{n=0}^{\infty} \frac{1}{3n+1}-\frac{1}{3n+2} $$
I wanted to use the fact that $$\arctan(\sqrt{3})=\frac{\pi}{3} $$ but $\arctan(x)$ can only be written as a power series when $ -1\leq x \leq1$...
|
There is another way to solve the problem. Note
$$ (3n+1)(3n+2)=9\left[(n+\frac{1}{2})^2+\left(\frac{i}{6}\right)^2\right] $$
and
and hence
\begin{eqnarray}
\sum_{n=0}^\infty\frac{1}{(3n+1)(3n+2)}&=&\frac19\sum_{n=0}^\infty\frac1{(n+\frac{1}{2})^2+\left(\frac{i}{6}\right)^2}\\
&=&\frac1{18}\sum_{n=-\infty}^\infty\frac1{(n+\frac{1}{2})^2+\left(\frac{i}{6}\right)^2}\\
&=&\frac{1}{18}\frac{\pi\sinh2\pi b}{b(\cosh2\pi b-\cos2\pi a)}\bigg|_{a=\frac{1}{2},b=\frac{i}{6}}\\
&=&\frac{\pi}{3\sqrt3}
\end{eqnarray}
by using the result from this.
|
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|
Limit of the nested radical $x_{n+1} = \sqrt{c+x_n}$ (Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12)
For $c \gt 0$, consider the quadratic equation
$x^2 - x - c = 0, x > 0$.
Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and then, if $n$ is an index for which $x_n$ has been defined, defining
$$x_{n+1} = \sqrt{c+x_n}$$
Prove that the sequence $\{x_n\}$ converges monotonically to the solution of the above equation.
Note: The answers below might assume $x_1 \gt 0$, but they still work, as we have $x_3 \gt 0$.
|
I am going to do this Ramanujan-style: pick some real positive $a$.
$$
a=\sqrt{a^2}=\sqrt{a^2-a+a}=\sqrt{a^2-a+\sqrt{a^2-a+a}}=\sqrt{a^2-a+\sqrt{a^2-a+\sqrt{a^2-a+a}}}=\dots=\sqrt{a^2-a+\sqrt{a^2-a+\sqrt{a^2-a+\sqrt{a^2-a+\dots}}}}
$$
What you do is keep replacing the last $a$ in the expression by $\sqrt{a^2-a+a}$.
Now let $a^2-a=c$ and we have the given expression. Solve that for $a$ and you have your answer.
|
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|
Finding the minimum value of $\left(\frac{a + 1}{a}\right)^2 + \left(\frac{b + 1}{b}\right)^2 + \left(\frac{c + 1}{c}\right)^2 $ Find the minimum value of
$$\left(\frac{a + 1}{a}\right)^2 + \left(\frac{b + 1}{b}\right)^2 + \left(\frac{c + 1}{c}\right)^2 $$
I tried to expand it and break it into individual terms and use $\text{A.M} \ge \text{G.M}$. inequality. But in this case i stuck where i need to find the value of abc.
|
Assuming $a,b,c$ are positive reals, there is no minimum!
Consider the individual term: $$\left(\frac{a + 1}{a}\right)^2 = \left(1 + \frac{1}{a}\right)^2$$
For positive real $a$ this is strictly greater than $1$, but as $a \to \infty$, this can be made as close to $1$ as we want.
Thus the total sum is always strictly $\gt 3$, and can be made as close to $3$ as we want.
So there is no minimum (but there is an infimum).
If you allow $a,b,c$ to be negative, then $a=b=c=-1$ gives the minimum value of $0$, as pointed out in the comments.
Perhaps you are missing some condition? Like $a,b,c$ positive reals and $abc = 1$ or something like that?
|
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|
Evaluate the integral $\int_{0}^1\frac{x-6}{x^2-6x+8}dx$ I used partial fractions to get:
$\displaystyle\frac{A}{x-4}+\frac{B}{x-2}= \frac{x-6}{(x-4)(x-2)}$
$A = -1$
$B = 2$
$\displaystyle\int_{0}^1\frac{-1}{x-4}+\frac{2}{x-2}dx$
Found the anti-derivative to be:
$(-\ln|x-4| + 2 \ln|x-2|)_0^{1}$
My answer came out to be around -1.1, what am I doing wrong? Thanks for any help.
Edit- Apparently WebAssign thinks that I'm wrong for some reason:
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In the interval $[0,1]$ we have $\frac{-5}{3}\leq\frac{x-6}{x^2-6x+8}\leq\frac{-3}{4}$, hence:
$$ \frac{-5}{3}\leq\int_0^1\frac{x-6}{x^2-6x+8}\leq\frac{-3}{4} $$
|
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|
Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, or both. Here's my (non) proof:
$$\begin{align*}
x^n - y^n &= (x - y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\
&= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\
&\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\
&= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} - x^{n-1}y - y^2 x^{n-2} - \cdots- x y^{n-1} - y^n \\
&= x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n \\
&\neq x^n - y^n
\end{align*}$$
Is there something I can do with $x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n$ that I'm not seeing, or did I make a mistake early on?
EDIT:
I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:
*
*Associate law for addition
*Existence of an additive identity
*Existence of additive inverses
*Commutative law for additions
*Associative law for multiplication
*Existence of a multiplicative identity
*Existence of multiplicative inverses
*Commutative law for multiplication
*Distibutive law
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Using Induction Hypothesis
Let n =k
Then ,
x^k -y^k = (x-y)(x^(k-1) +x^(k-2) y ....+y^k-1)
Let n= (k-1)
Then by Induction Hypothesis,
(x-y){x(x^(k-1) +x^(k-2) y) +y^k }
=» (x-y)(x^k +x^(k-1) y ... +y^k
Hence the formula is true for k+1 whenever true for k .
Therefore by Principle of Induction The formula is true for Positive Integers n≥2 .
Reference = Challenges and Thrills of Pre College Mathematics .
|
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|
For complex $z$, $|z| = 1 \implies \text{Re}\left(\frac{1-z}{1+z}\right) = 0$
If $|z|=1$, show that: $$\mathrm{Re}\left(\frac{1 - z}{1 + z}\right) = 0$$
I reasoned that for $z = x + iy$, $\sqrt{x^2 + y^2} = 1\implies x^2 + y ^2 = 1$ and figured the real part would be:
$$\frac{1 - x}{1 + x}$$
I tried a number of manipulations of the equation but couldn't seem to arrive at any point where I could link the two to show that the real part was = 0.
|
Let $z=a+bi$, where $a^2+b^2=1$ and $z\neq -1$. Then $$\begin{eqnarray}
\Re\left(\frac{1-z}{1+z}\right)
&=&\Re\left(\frac{(1-z)\overline{(1+z)}}{|1+z|^2}\right)\\
&=&\Re\left(\frac{(1-a-bi)(1+a-bi)}{(a+1)^2+b^2}\right)\\
&=&\Re\left(\frac{1-(a^2+b^2)-2bi}{(a+1)^2+b^2}\right)\\
&=&\Re\left(\frac{-2bi}{(a+1)^2+b^2}\right)=0\end{eqnarray}$$
|
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|
How does expanding $(2s+1)^2$ help us prove "$n^2$ even $\implies$ $n$ even"? How can $n^2 = (2s + 1)^2= 4s^2 + 4s + 1$ help in proving "If $n^2$ is even then $n$ is even"?
|
Based on your comment on Lazar Ljubenović answer, here is an attempt to explain how to expand $(2s+1)^2$.
First let us rewrite $(2s+1)^2$ as $(2s+1)(2s+1)$. For the sake of this explanation let us rewrite $(2s+1)(2s+1)$ as $c(2s+1)$ where $c=2s+1$.
Using the distributive law $a(b+c)=ab+ac$ we know $$c(2s+1)=c\cdot 2s+c\cdot 1=c\cdot 2s+c$$
Now if we substitute back $c$ we get
$$2s\cdot (2s+1)+(2s+1)$$
If we use the distributive law once again we get:
$$(2s)\cdot (2s)+2s\cdot 1+2s+1=2\cdot2\cdot s\cdot s+2s+2s+1$$
Which is
$$4s^2+4s+1$$
As for the question: how can this help prove that if $n$ is odd, $n^2$ is also odd?
Since any even number plus 1 is odd, any odd number can be written in the form $2n+1$ where $n$ is an integer. If we square $2n+1$ we get $(2n+1)^2=4n^2+4n+1$. Note that this can be simplified to $4(n^2+n)+1$. Since $4(n^2+n)$ is always even, $4(n^2+n)+1$ must be odd.
|
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|
Show that $2^{3^k}+1$ is divisible by $3^k$ for all positive integers $k$ I attempted this by induction: Here is what I did
For $k=1, 2^{3^{1}}+1 = 2^{3}+1 = 9 = 3^{2}$ is divisible by $3^{1}$ , so the result is true for $k=1$
Now I assume the result to be true for $k=m$,
$2^{3^{m}}+1$ is divisible by $3^m$. To show the result to be true for $k=(m+1)$,
$2^{3^{m+1}}+1 = 2^{3^m} \times 2^3+1$ and I was stuck here.
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We show, as was observed by Sivaram Ambikasaran, that $3^{n+1}$ divides $2^{3^n}+1$. For $\varphi(3^{n+1})=2\cdot 3^n$. By Euler's Theorem, $2^{2\cdot 3^n}\equiv 1\pmod{3^{n+1}}$. Thus $2^{3^n}\equiv \pm 1 \pmod{ 3^{n+1}}$. But $2^{3^n}\not\equiv 1 \pmod{3}$, so $2^{3^n}\equiv -1\pmod{3^{n+1}}$.
Remark: It is not hard to show that $2$ is a primitive root of $3^{e}$ for every positive integer $e$. As a consequence, $2^{3^n}+1$ is not divisible by any power of $3$ greater than $3^{n+1}$.
|
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|
Smith Normal Form Would the Smith Normal Form of the following matrix over $\mathbb Q[x]$
$$\begin{pmatrix}
(x+a)(x+b) & 0 & 0 &0 \\
0 & (x+c)(x+d) & 0 & 0 \\
0 &0 & x^3(x+a) & 0 \\
0 & 0 & 0& x^2(x+b)
\end{pmatrix}$$
simply be
$$\begin{pmatrix}
f(x) & 0 & 0 &0 \\
0 & f(x) & 0 & 0 \\
0 &0 & f(x) & 0 \\
0 & 0 & 0& f(x)
\end{pmatrix}$$
where $f(x)= x^3(x+a)(x+b)(x+c)(x+d)$?
I am not sure because that would make the question quite trivial.
|
No.
Let $s_k$ denote the $k$th entry on the diagonal of Smith form (i.e., the $k$th invariant factor). Then $$s_k = \frac{d_k}{d_{k-1}},$$ where $d_k = \gcd$ of all $k \times k$ minors of the original matrix (aka $k$th determinantal divisor). I'm assuming $a,b,c,d$ are distinct here. So the Smith form is:
$$
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0\\
0 & 0 & x^2(x+a)(x+b) & 0\\
0 & 0 & 0 & x^3(x+a)(x+b)(x+c)(x+d)
\end{pmatrix}
$$
|
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|
What is the largest positive $n$ for which $n^3+100$ is divisible by $n+10$ What is the largest positive $n$ for which $n^3+100$ is divisible by $n+10$
I tried to factorize $n^3+100$, but $100$ is not a perfect cube. I wish it were $1000$.
|
By division we find that $n^3 + 100 = (n + 10)(n^2 − 10n + 100)−900$.
Therefore, if $n +10$ divides $n^3 +100$, then it must also divide $900$.
Since we are looking for largest $n$, $n$ is maximized whenever $n + 10$ is, and since the largest divisor of $900$ is $900$, we must have $n + 10 = 900 \Rightarrow n = 890$
The largest $n$ is therefore $890$
|
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|
Plot $|z - i| + |z + i| = 16$ on the complex plane
Plot $|z - i| + |z + i| = 16$ on the complex plane
Conceptually I can see what is going on. I am going to be drawing the set of points who's combine distance between $i$ and $-i = 16$, which will form an ellipse. I was having trouble getting the equation of the ellipse algebraically.
I get to the point:
$x^2 + (y - 1) ^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} + x^2 + (y+ 1)^2 = 256$
$2x^2 + 2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$
It seems like I'm and doing something the hard way.
|
Sketching the graph gives a centrally symmetric ellipse with the imaginary axis as the major axis. The length of the half the major axis is clearly 8 (square 64), and pythagoras gives the square of half the minor axis as 63. This is reflected in the answer obtained by algebraic manipulation.
$$2x^2+2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$$
$$\sqrt{(x^2 + y^2 + 1) - 2y}\sqrt{(x^2 + y^2 + 1) + 2y} = 127 - x^2 - y^2$$
Square both sides:
$$((x^2 + y^2 + 1) - 2y)((x^2 + y^2 + 1) + 2y) = (127 - x^2 - y^2)^2$$
$$(x^2+y^2+1)^2-4y^2 = 127^2+x^4+y^4-254x^2-254y^2+2x^2y^2$$
Cancelling as we go and gathering terms in the obvious way:
$$256x^2+252y^2=127^2-1=128 \times 126$$
(difference of two squares)
The divisions now become easy and we reach the canonical form:
$$\frac {x^2} {63} + \frac {y^2} {64} = 1$$
|
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|
Proving a complicated inequality involving integers Let $a,b,c,d$ be integers such that $$\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right) \mod 2$$ $$ ad-bc =1$$ $$\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \neq \left( \begin{matrix} \pm 1 & 0 \\ 0 & \pm 1 \end{matrix}\right) .$$
Let $(x,y)\in \mathbf{R}^2$ such that $-1\leq x\leq 1$ and $1/2 \leq y\leq 2$.
I highly suspect that
$$c^2(y^2+2x^2) + a^2+d^2+2cx(d-a) + \frac{1}{y^2}(b-(d-a+3cx)x)^2 \geq 3.$$
The proof is actually very easy and was obtained after Parsa's answer.
|
In the case where $a$ and $d$ have the same sign, rewrite your expression as
$$ c^2(y^2 + x^2) + (cx + (d-a))^2 + 2ad + \frac{1}{y^2}(b-(d-a+3cx)x)^2$$
Since each term is non-negative your inequality follows easily. Perhaps you can adjust this expression so your inequality follows in the case where $a$ and $d$ have opposite signs.
|
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|
Help with basic high school math. What happens to $j$? I know my math is very rusty, actually, its always been that way. but I need help with this. The question below has me stumped. I've tried to show the steps I went through to get the answer. Please tel me where I made the mistake.
If x=a and x=b are two roots of a quadratic equation then (x-a)(x-b) = 0 gives the quadratic equation.
That is $(x - a)(x - b) = x^2 - (a + b)x + ab = 0$.
Here, the two roots are $x= -2 + j\sqrt5$ and $x = -2 - j\sqrt5$ so that
$(x – [-2 + j\sqrt5])(x – [-2 - j\sqrt5]) = 0$
That is $x^2 - x[-2 + j\sqrt5 - 2 - j\sqrt5] + [-2 + j\sqrt5][-2 - j\sqrt5] = 0$
I understand that
$x^2 - x[-2 + j\sqrt5 - 2 - j\sqrt5] + [-2 + j\sqrt5][-2 - j\sqrt5] = 0$
$x^2 - x[-2 - 2] + [-2 + j\sqrt5][-2 - j\sqrt5] = 0$, and
$x^2 - x[-4] + [-2 + j\sqrt5][-2 - j\sqrt5] = 0$, so
$x^2 + 4x + [-2 + j\sqrt5][-2 - j\sqrt5] = 0$
if we separate out the last term for simplicity:
$[-2 + j\sqrt5][-2 - j\sqrt5] = (-2)(-2) + (-2)(-j\sqrt5) + (-2)(+j\sqrt5) + (+j\sqrt5)(-j\sqrt5)$
$= 4 + (2j\sqrt5) - (2j\sqrt5) +(-j\sqrt5)^2$
$ = 4 + (2j\sqrt5) - (2j\sqrt5) +(-j^2)(-\sqrt5)^2$
$ = 4 +j^2 5$
Putting this last term back into the main equation results in:
$x^2 + 4x + (4+j^2 5) = 0$
In the book (Advanced Engineering Mathematics)* this equation works out to
$x^2 + 4x + 9 = 0$
What I don’t understand is what happened to $j^2$ How does it just magically disapear?
*If you use the Amazon "Look inside" feature you can see it on page 4.
|
Here are your steps with mistakes highlighted in red:
$[-2 + j\sqrt5][-2 - j\sqrt5] = (-2)(-2) + (-2)(-j\sqrt5) + (-2)(+j\sqrt5) + (+j\sqrt5)(-j\sqrt5)$
$= 4 + (2j\sqrt5) - (2j\sqrt5) + \color{red}{(-j\sqrt5)^2}$
$ = 4 + (2j\sqrt5) - (2j\sqrt5) + \color{red}{(-j^2)(-\sqrt5)^2}$
$ = 4 + \color{red}{j^2 5}$
Here is fixed, with highlighting in red:
$[-2 + j\sqrt5][-2 - j\sqrt5] = (-2)(-2) + (-2)(-j\sqrt5) + (-2)(+j\sqrt5) + \color{blue}{(+j\sqrt5)(-j\sqrt5)}$
$= 4 + (2j\sqrt5) - (2j\sqrt5) +\color{red}{(-1)\times j^2\times (\sqrt5)^2}$
$ = 4 + (2j\sqrt5) - (2j\sqrt5) +\color{red}{(-1)\times (-1)\times 5}$
$ = 4 +\color{red}{5} = 9$
Note that
$$\color{blue}{(+j\sqrt5)(-j\sqrt5)} =
\color{red}{(-1)\times j^2\times (\sqrt5)^2}$$
and
$$j^2 = -1$$
|
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|
What is the radius of the circle in cm? The rectangle at the corner measures 10 cm * 20 cm.
The right bottom corner of the rectangle is also a point on the circumference of the circle.
What is the radius of the circle in cm?
Is the data sufficent to get the radius of circle?
|
$$r^2=x^2+y^2 \tag{1}$$
$r=y+20$ and $r=x+10$ therefore
$$y+20=x+10 \quad \mbox{then } \quad y=x-10 \tag{2}$$
Substitute $(2)$ into $(1)$
$$(r+10)^2=x^2+(x-10)^2$$
$$x^2+20x+100=x^2+x^2-20x+100$$
$$X^2=40x$$
$$x=40$$
Then substitute $x=40$ into $(2)$
$$y=40-10$$
$$y=30$$
substitute $x=40$ and $Y=30$ into $(1)$
$$r^2=(40)^2+(30)^2$$
$$r=50\rm{cm}.$$
|
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|
Integrate $\sqrt{x^2 - 2x}$ $$\int{\sqrt{x^2 - 2x}}$$
I think I should be doing trig substitution, but which? I completed the square giving
$$\int{\sqrt{(x-1)^2 -1}}$$
But the closest I found is for
$$\frac{1}{\sqrt{a^2 - (x+b)^2}}$$
So I must add a $-$, but how?
|
The standard way to solve these problems is indeed with trigonometric substitutions.
But it is not the only way to solve these.
Another way that was used more before is called Euler substitutions
if we are to integrate $\sqrt{ax^2+bx+c}$ that has real roots $\alpha$ and
$\beta$, then we can use the substitution $\sqrt{ax^2+bx+c}=x \cdot t$
Now, in your problem we have $\alpha=0$ and $\beta=2$, so we may choose to use the substitution
$$ \sqrt{x^2-2x} = (x-0)\cdot t \ \Rightarrow \ x = \frac{2}{1-t^2}$$
Since this is homework and I am lazy, so from here I will only outline the details
$$ \begin{align} I & = \int \sqrt{x^2-2x}\,\mathrm{d}x \\
& = \int \frac{8t^2}{1-3t^2+3t^4-t^6}\,\mathrm{d}t \\
& = \int -\frac{8t^2}{(t+1)^3(t-1)^3}\,\mathrm{d}t \\
& = \int - \frac{1}{2}\frac{1}{t+1} + \frac{1}{(t+1)^3}
- \frac{1}{2}\frac{1}{(t+1)^2} + \frac{1}{2}\frac{1}{(t-1)^2} - \frac{1}{(t-1)^3} - \frac{1}{2}\frac{1}{(t-1)^2} \, \mathrm{d}t \\
\end{align} $$
and from here the rest is obvious =)
|
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|
Extremely hard geometric problem Given a triangle ABC. BL is the bisector of angle ABC, H is the orthocenter and P is the mid-point of AC. PH intersects BL at Q. If $\angle ABC= \beta $, find the ratio $PQ:HQ$.If $QR\perp BC$ and $QS \perp AB$, prove that the orthocenter lies on $RS$.
|
In the figures below, I have added the circumcenter, $U$, and the centroid, $E$. I have also placed $L$ on the circumcircle.
$\hspace{8mm}$
Note that since both are perpendicular to $\overline{AC}$, we have $\overline{BH}\,||\,\overline{UP}$; furthermore, $|\overline{BH}|=2|\overline{UP}|$. The latter is because $\triangle PUE$ is similar to $\triangle BHE$ and
$$
P=\frac{A+C}{2}\text{ and }E=\frac{A+B+C}{3}\tag{1}
$$
so that
$$
P-E=\frac{A-2B+C}{6}\text{ and }E-B=\frac{A-2B+C}{3}\tag{2}
$$
Thus,
$$
|\overline{UP}|=R\cos(B)\text{ and }|\overline{BH}|=2R\cos(B)\tag{3}
$$
where $R$ is the circumradius of $\triangle ABC$.
Since the line containing $\overline{UP}$ is the perpendicular bisector of $\overline{AC}$, the point at which $\overrightarrow{UP}$ intersects the circumcircle of $\triangle ABC$ splits the arc between $A$ and $C$ in half. Of course, the bisector of $\angle ABC$ also splits the arc between $A$ and $C$ in half. Thus, the perpindicular bisector of $\overline{AC}$ and the bisector of $\angle ABC$ meet on the circumcircle at $L$.
$\hspace{8mm}$
Note that $\triangle BHQ$ is similar to $\triangle LPQ$. Equation $(3)$ gives that $|\overline{UP}|=R\cos(B)$ so that
$$
|\overline{PL}|=R(1-\cos(B))\tag{4}
$$
Therefore, $(3)$ and $(4)$ yield
$$
\begin{align}
|\overline{HQ}|/|\overline{PQ}|
&=|\overline{BQ}|/|\overline{LQ}|\\
&=|\overline{HB}|/|\overline{PL}|\\
&=\frac{2\cos(B)}{1-\cos(B)}\tag{5}
\end{align}
$$
which answers the first part.
Because $\triangle BUL$ is isosceles with central angle $2A+B=\pi-(C-A)$, we have
$$
|\overline{BL}|=2R\sin\left(A+\frac{B}{2}\right)=2R\cos\left(\frac{C-A}{2}\right)\tag{6}
$$
Equation $(5)$ yields that $|\overline{BQ}|/|\overline{BL}|=\frac{2\cos(B)}{1+\cos(B)}$. Thus, $(6)$ gives
$$
|\overline{BQ}|=2R\cos\left(\frac{C-A}{2}\right)\frac{2\cos(B)}{1+\cos(B)}\tag{7}
$$
Let $X$ be the intersection of $\overline{BQ}$ and $\overline{RS}$. Since $X$ is on the angle bisector of $\angle ABC$, $\overline{RS}$ is perpendicular to $\overline{BQ}$ and $|\overline{BR}|=|\overline{BS}|$. Thus, $|\overline{BR}|/|\overline{BQ}|=|\overline{BX}|/|\overline{BR}|=\cos(B/2)$. Therefore,
$$
\frac{|\overline{BX}|}{|\overline{BQ}|}=\cos^2(B/2)=\frac{1+\cos(B)}{2}\tag{8}
$$
Equations $(7)$ and $(8)$ yield
$$
|\overline{BX}|=2R\cos\left(\frac{C-A}{2}\right)\cos(B)\tag{9}
$$
Since $\angle HBC=\frac\pi2-C$ and $\angle QBC=\frac{B}{2}$ we get that $\angle HBQ=\frac{C-A}{2}$. Using $(3)$, the orthogonal projection of $\overline{BH}$ onto $\overline{BQ}$ has length is $2R\cos(B)\cos\left(\frac{C-A}{2}\right)$. Thus, the orthogonal projection of $H$ onto $\overline{BQ}$ is $X$. Therefore, $H$ lies on $\overline{RS}$.
|
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|
Partial fraction with a constant as numerator I am trying to express this as partial fraction:
$$\frac{1}{(x+1)(x^2+2x+2)}$$
I have a similar exaple that has $5x$ as numerator, it is easy to understand. I do not know what to do with 1 in the numerator, how to solve it?!
|
You set it up in the usual way as $$\begin{align*}\frac1{(x+1)(x^2+2x+2)}&=\frac{A}{x+1}+\frac{Bx+C}{x^2+2x+2}\\
&=\frac{A(x^2+2x+2)+(Bx+C)(x+1)}{(x+1)(x^2+2x+2)}\;,
\end{align*}$$
so that $$A(x^2+2x+2)+(Bx+C)(x+1)=1\;.$$
Now multiply out the lefthand side to get $$(A+B)x^2+(2A+B+C)x+(2A+C)=1$$
and equate coefficients of $x^2,x$, and $1$ to get (in that order):
$$\left\{\begin{align*}
&A+B=0\\
&2A+B+C=0\\
&2A+C=1
\end{align*}\right.$$
Then solve the system for $A,B$, and $C$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$ Evaluate $$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$$
So ...
$$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011} = (-\sqrt{2}+\sqrt{2}i)^{-2011}$$
$$\theta=\pi - \arctan(\frac{\sqrt{2}}{\sqrt{2}}) = \frac{3\pi}{4}$$
$$-\sqrt{2}+\sqrt{2}i=\cos{\theta} + i \sin{\theta}$$
$$(-\sqrt{2}+\sqrt{2}i)^{-2011}
=\cos{(-2011\theta)} + i \sin{(-2011\theta)} = e^{i(-2011)\theta}$$
Is it correct?
The given answer is
...
$$\arg{z} = \frac{3\pi}{4}, \qquad z=2e^{i \frac{3\pi}{4}}$$
$$LHS
= (\frac{1}{z})^{2011}=2^{-2011}e^{-2011(\frac{3\pi}{4})i}
= 2^{-2011} e^{-(\color{red}{1508\pi i} + \frac{\pi i }{4})}
= 2^{-2011}e^{-\frac {\pi i}{4}}
= 2^{-2011} \color{blue}{\frac{1 - i}{\sqrt{2}}}
= ...
$$
Why is the red $1508\pi i$ removed in the following step?
How do I get the blue $\frac{1 - i}{\sqrt{2}}$ from the prev step?
|
First off, let's get the signs correct:
$$\left(\frac{1}{-\sqrt{2}+\sqrt{2}~i}\right)^{\color{Red}+2011}=(-\sqrt{2}+\sqrt{2}~i)^{-2011}.$$
Second, if $z=-\sqrt{2}+\sqrt{2}i=\sqrt{2}(-1+i)$, then we do have a magnitude of two:
$$|z|=\sqrt{2}\cdot |-1+i|=\sqrt{2}\cdot\sqrt{(-1)^2+1}=2.$$
The angle $\theta=\frac{3}{4}\pi$ of $z$ is correct so finally
$$z^{-2011}=(2e^{\frac{3}{4}\pi i})^{-2011}=2^{-2011}\exp\left(-\frac{6033}{4}\pi i\right)=2^{2011}e^{-\pi i/4}$$
because $6033\equiv1\bmod8$. We could also evaluate $e^{-\pi i/4}$ as $(1-i)/\sqrt{2}$ and get $2^{-(2011+1/2)}(1-i)$.
|
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|
Equivalence of a Real Number to Itself If $\alpha$ and $\beta$ are two real numbers, we say that $\beta$ is equivalent to $\alpha$ if there are integers $a$, $b$, $c$, and $d$ such that $ad-bc=\pm1$ and $\beta=\frac{a\alpha+b}{c\alpha+d}$.
How can I show that a real number $\alpha$ is equivalent to itself?
What I have in mind is to assume that $\alpha=\frac{a\alpha+b}{c\alpha+d}$, and then show that it must be the case that $ad-bc=\pm1$:
$$\begin{align}
c\alpha^2+(d-a)\alpha-b&=0\\
\Longrightarrow\alpha&=\frac{a-d\pm\sqrt{(d-a)^2+4bc}}{2c}\\
&=\frac{a-d\pm\sqrt{d^2-2ad+a^2+4bc}}{2c}\\
&=\frac{a-d\pm\sqrt{-2(ad-2bc)+a^2+d^2}}{2c}\\
&=\frac{a-d\pm\sqrt{a^2+d^2+2bc\pm2}}{2c}
\end{align}$$
I do not know how to proceed. Am I on the correct track?
|
$$ \alpha = \frac{ 1 \cdot \alpha + 0}{0\cdot \alpha + 1} .$$
|
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|
How do you integrate $\int \frac{1}{a + \cos x} dx$? How do you integrate $\int \frac{1}{a + \cos x} dx$? Is it solvable by elementary methods? I was trying to do it while incorrectly solving a homework problem but I couldn't find the answer.
Thanks!
|
Generalization:
Let's consider $\cos x = \frac{1-t^2}{1+t^2}; t = \tan\frac{x}{2}; dx=\frac{2}{1+t^2} dt.$ Then we get that our integral becomes:
$$J = \int \frac{2dt}{(a+b)+(a-b) t^2}$$
I. For the case $a>b$, consider $a+b=u^2$ and $a-b=v^2$, and obtain that:
$$J = 2\int \frac{dt}{u^2+v^2 t^2}=\frac{2}{uv} \arctan\frac{vt}{u} +C.$$
Turning back to our notation we get:
$$I=\frac{2}{\sqrt{a^2-b^2}} \arctan\left(\sqrt{\frac{a-b}{a+b}} \tan\frac{x}{2} \right) + C.$$
II. For the case $a<b$, consider $a+b=u^2$ and $a-b=-v^2$, and obtain that:
$$J = 2\int \frac{dt}{u^2-v^2 t^2}=\frac{1}{uv}\ln\frac{u+vt}{u-vt} \ +C.$$
Turning back again to out initial notation and have that:
$$I=\frac{2}{\sqrt{b^2-a^2}} \ln\frac{b+a \cos x + \sqrt{b^2-a^2} \sin x}{a+b \cos x} + C.$$
Also, note that $x$ must be different from ${+}/{-}\arccos(-\frac{a}{b})+2k\pi$ if $|\frac{a}{b}|\leq1$.
Q.E.D.
|
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|
Solving the recursion $3a_{n+1}=2(n+1)a_n+5(n+1)!$ via generating functions I have been trying to solve the recurrence:
\begin{align*}
a_{n+1}=\frac{2(n+1)a_n+5((n+1)!)}{3},
\end{align*}
where $a_0=5$, via generating functions with little success. My progress until now is this:
Let $A(x)=\sum_{n=0} ^{\infty} a_nx^n$. By multiplying both sides of our recurrence relation by $x^n$ and summing over $n$ from $0$ to $\infty$, we see that
\begin{align}
\sum_{n=0} ^{\infty} a_{n+1} x^n = \frac{2}{3}\sum_{n=0} ^{\infty} (n+1)a_nx^n + \sum_{n=0} ^{\infty} (n+1)!x^n.
\end{align}
Using our definition of $A(x)$ we can rewrite the left hand side as
\begin{align*}
\sum_{n=0} ^{\infty} a_{n+1} x^n=\frac{A(x)-a_0}{x}.
\end{align*}
Such manipulations of the right hand side have been difficult because of the coefficients of the power series.
Is there anyway to proceed from here, or are generating functions not suited to solve such a recurrence?
|
Being lazy and prove it by MI for fun.
Obviously, $a_0=5(0!)$, now assume $a_n=5n!$ for some natural number $n$, then
$$
\begin{aligned}
3 a_{n+1} &=2(n+1) 5 n !+5(n+1)_{0}^{1} \\
&=5 n !(2 n+2+n+1) \\
a_{n+1} &=5(n+1) !
\end{aligned}
$$
proved.
|
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|
Another evaluating limit question: $\lim\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}$ How do I begin to evaluate this limit: $$\lim_{n\to \infty}\ \frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}\;?$$
Thanks a lot.
|
A simple, but famous trick works here: Observe that $(n-1)(n+1) = n^2 - 1 \leq n^2$. Thus we have
$$ \begin{align*}
& \left[ \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \right]^2 \\
&= \frac{1 \cdot 1}{2 \cdot 2}\cdot\frac{3 \cdot 3}{4 \cdot 4}\cdot\frac{5 \cdot 5}{6 \cdot 6}\cdots\frac{(2n-3) \cdot (2n-3)}{(2n-2) \cdot (2n-2)}\cdot\frac{(2n-1) \cdot (2n-1)}{(2n) \cdot (2n)} \\
&= \frac{1 \cdot 3}{2 \cdot 2}\cdot\frac{3 \cdot 5}{4 \cdot 4}\cdot\frac{5 \cdot 7}{6 \cdot 6}\cdots\frac{(2n-3) \cdot (2n-1)}{(2n-2) \cdot (2n-2)}\cdot\left(\frac{2n-1}{(2n)^2}\right) \\
& \leq \frac{2n-1}{(2n)^2} \\
& \leq \frac{1}{2n}.
\end{align*} $$
Thus we have
$$\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \leq \frac{1}{\sqrt{2n}} $$
and the limit is zero. In fact, the sharp estimate
$$ \frac{1}{\sqrt{(\pi + o(1)) n}} \leq \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \leq \frac{1}{\sqrt{\pi n}} $$
holds, so the estimation above is not so far from the truth.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/139494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
}
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.