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Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers I have tried the arithmetic-geometric inequality on $(-a+b+c)(a-b+c)(a+b-c)$ which gives
$$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3$$
and on $abc$ which gives
$$abc \leq \left(\frac{a+b+c}{3}\right)^3.$$
Since both inequalities have the same righthand side, I have tried to deduce something about the lefthand sides, but to no avail. Can somebody help me, please? I am sure it is something simple I have missed.
| Note that no two of $(-a+b+c)$, $(a-b+c)$, and $(a+b-c)$ can be negative. If so, then one of
$$
\begin{align}
(a-b+c)+(a+b-c)&=2a\\
(a+b-c)+(-a+b+c)&=2b\\
(-a+b+c)+(a-b+c)&=2c
\end{align}
$$
would be negative, but each of $a$, $b$, and $c$ is positive. Thus, at most one can be negative. If only one were negative, then the product on the left would be non-positive and the inequality would be trivial. Therefore, we can assume that $a$, $b$, and $c$ are sides of a triangle.
By Heron's Formula, a triangle with sides of length $a$, $b$, and $c$, has area $A$ where
$$
(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=16A^2\tag{1}
$$
Let $r$ be the radius of the inscribed circle and $R$ be the radius of the circumscribed circle. Then
$$
2A=r(a+b+c)\tag{2}
$$
and
$$
4AR=abc\tag{3}
$$
Putting together $(1)$-$(3)$ yields
$$
(-a+b+c)(a-b+c)(a+b-c)R=2rabc\tag{4}
$$
The distance $d$ between the incenter and circumcenter is given by
$$
d^2=R(R-2r)\tag{5}
$$
Combining $(4)$ and $(5)$ gives
$$
\begin{align}
(-a+b+c)(a-b+c)(a+b-c)
&=\left(1-\frac{d^2}{R^2}\right)abc\\[6pt]
&\le abc\tag{6}
\end{align}
$$
Justifications of $(2)$, $(3)$, and $(5)$ can be found in this answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/170813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 1
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Proving inequality on functions $x-\frac{x^2}{2}<\ln(1+x)To prove: $$x-\frac{x^2}{2}<\ln(1+x)<x-\frac{x^2}{2(1+x)},\quad\forall x>0$$
I have used Taylor series expansion at 0 for both the inequalites. The greater than by expanding $\ln(1+x)$ and the less than by expanding $\int \ln(1+x)\,dx$ at 0.
Is there a cleaner / more elegant way of achieving the same?
| Another way to do the lower bound, for example, would be to consider $f(x) = x - \dfrac{x^2}{2} - \ln(1+x)$. $f(0) = 0$.
But also $f'(x) = 1 - x - \dfrac{1}{1+x} = \dfrac{(1-x)(1+x) - 1}{1+x} = \dfrac{1 - x^2 - 1}{1+x} = \dfrac{-x^2}{1+x} < 0$.
So since $f(0) = 0$ and $f' < 0$, $f$ is monotonically and strictly decreasing. Thus $f(x) \leq 0$, and if $x > 0$ we have that $f(x) < 0$. And this says exactly that $x - x^2/2 < \ln(1 + x)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$3\sin^2x=\cos^2x;$ $ 0\leq x\leq 2\pi$ Solve for $x$
$3\sin^2x=\cos^2x;$ $0\leq x\leq 2\pi$ Solve for $x$:
I honestly have no idea how to start this. Considering I'm going to get a number, I am clueless. I have learned about $\sin$ and $\cos$ but I do not know how to approach this problem. If anyone can go step-by-step with hints. That would be greatly appreciated.
EDIT:
$$3\sin^2x=1-\sin^2x$$ $$4\sin^2x=1$$ $$\sin^2x=\frac{1}{4}$$ $$\sqrt{\sin^2x}=\sqrt{\frac{1}{4}}$$ $$\sin x=\pm \left(\frac{1}{2}\right)$$
| Hint: $\cos^2 x=1-\sin^2 x$.${}{}{}{}{}{}{}$
Substitute. We get after some simplification $4\sin^2 x=1$. Can you finish from here?
Added: You just need to find $\sin^2 x$, then $\sin x$. You should get $\sin x=\pm\frac{1}{2}$. Then identify the angles from your knowledge about "special angles." One of the angles will turn out to be $\frac{\pi}{6}$, the good old $30^\circ$ angle. There are $3$ others.
Remark: The way you started is fine too, you got $3(1-\cos^2 x)=\cos^2 x$. Now we need to "isolate" $\cos^2 x$. It is easiest to multiply through by $3$ on the left, getting $3-3\cos^2 x=\cos^2 x$. Bring all the $\cos^2 x$ terms to one side. We get $3=4\cos^2 x$, which I prefer to write as $4\cos^2 x=3$. So we get $\cos^2 x=\frac{3}{4}$.
Take the square roots. We get
$$\cos x=\pm \frac{\sqrt{3}}{2}.$$
Following the hint given at the start happens to be a little easier, same principles, nicer numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of the series : $1 + 2+ 4 + 7 + 11 +\cdots$ I got a question which says
$$ 1 + \frac {2}{7} + \frac{4}{7^2} + \frac{7}{7^3} + \frac{11}{7^4} + \cdots$$
I got the solution by dividing by $7$ and subtracting it from original sum. Repeated for two times.(Suggest me if any other better way of doing this).
However now i am interested in understanding the series 1,2,4,7,11,..... In which the difference of the numbers are consecutive natural numbers.
How to find the sum of
$1+2+4+7+11+\cdots nterms$
This is my first question in MSE. If there are some guidelines i need to follow, which i am not, please let me know.
| Let's first find the general form of the terms of the sum
$1 + 2+ 4 + 7 + 11 +\ldots.$
The terms obey the recurrence
$a_n = a_{n-1} + n$, where $a_0 = 1.$
Using standard techniques we find
$$a_n = \frac{1}{2}n(n+1) + 1.$$
These are basically the triangular numbers, as indicated by @anon in the comments.
The sum of the first $n+1$ terms can be found using
Faulhaber's formula, as indicated by @RossMillikan,
$$\begin{eqnarray*}
\sum_{k=0}^n a_k
&=& \frac{1}{2}\sum_{k=0}^n n^2
+ \frac{1}{2}\sum_{k=0}^n n
+ \sum_{k=0}^n 1 \\
&=& \frac{1}{2}\frac{1}{6}n(n+1)(2n+1) + \frac{1}{2}\frac{1}{2} n(n+1) + n+1 \\
&=& \frac{1}{6}(n+1)(n^2+2n+6).
\end{eqnarray*}$$
The sum
$$\begin{equation*}
\sum_{n=0}^\infty \frac{a_n}{7^n}
= \frac{1}{2}\sum_{n=0}^\infty \frac{n^2}{7^n}
+ \frac{1}{2}\sum_{n=0}^\infty \frac{n}{7^n}
+ \sum_{n=0}^\infty \frac{1}{7^n} \tag{1}
\end{equation*}$$
can be found by a standard trick.
Consider the geometric series
$$\sum_{k=0}^\infty x^n = \frac{1}{1-x}$$
for $|x|<1$.
The last sum on the right hand side of (1) is just $\displaystyle\frac{1}{1-\frac{1}{7}} = 7/6$.
Now notice that for $m\in \mathbb{N}$
$$\begin{eqnarray*}
\sum_{n=0}^\infty n^m x^n
&=& \left(x \frac{d}{dx}\right)^m \sum_{k=0}^\infty x^n \\
&=& \left(x \frac{d}{dx}\right)^m \frac{1}{1-x}.
\end{eqnarray*}$$
Therefore,
$$\begin{eqnarray*}
\sum_{n=0}^\infty \frac{a_n}{7^n}
&=& \left[\frac{1}{2} \left(x \frac{d}{dx}\right)^2 \frac{1}{1-x}
+ \frac{1}{2} \left(x \frac{d}{dx}\right) \frac{1}{1-x}
+ \frac{1}{1-x} \right]_{x=1/7} \\
&=& \frac{1}{2}\frac{7}{27} + \frac{1}{2} \frac{7}{36} + \frac{7}{6} \\
&=& \frac{301}{216}.
\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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Please help me integrate the following: $\int \frac{y^2 - x^2}{(x^2 + y^2)^2}dy$ I'm self-studying a Cramster solution and I came across this integral and I don't know what they've done with it. Help would be appreciated.
$$\int \frac{y^2 - x^2}{(x^2 + y^2)^2} ~dy.$$
| In an integral $dy$, $x$ is a constant. Rewirite this as:
$$\int\frac{y^2-x^2}{\left(x^2+y^2\right)^2}dy=\int\frac{y^2+x^2-2x^2}{\left(x^2+y^2\right)^2}dy=\int\frac{dy}{x^2+y^2}-2x^2\int\frac{dy}{\left(x^2+y^2\right)^2}$$
Can you continue from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating :$\int \frac{1}{x^{10} + x}dx$ $$\int \frac{1}{x^{10} + x}dx$$
My solution :
$$\begin{align*}
\int\frac{1}{x^{10}+x}\,dx&=\int\left(\frac{x^9+1}{x^{10}+x}-\frac{x^9}{x^{10}+x}\right)\,dx\\
&=\int\left(\frac{1}{x}-\frac{x^8}{x^9+1}\right)\,dx\\
&=\ln|x|-\frac{1}{9}\ln|x^9+1|+C
\end{align*}$$
Is there completely different way to solve it ?
| Let we generalise the problem with a slightly different way. Consider
$$\int\frac{\mathrm dx}{x^n+x}$$
Making substitution $x=\frac{1}{y}$ and $z=y^{n-1}$ then reverse it back we get
\begin{align}
\int\frac{\mathrm dx}{x^n+x}&=-\int\frac{y^{n-2}}{1+y^{n-1}}\mathrm dy\\[9pt]
&=-\frac{1}{n-1}\int\frac{\mathrm dz}{1+z}\\[9pt]
&=-\frac{\ln |1+z|}{n-1}+C\\[9pt]
&=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\ln |x|-\frac{\ln \left|1+x^{n-1}\right|}{n-1}+C}}
\end{align}
In your case
$$\int\frac{\mathrm dx}{x^{10}+x}=\ln |x|-\frac{\ln \left|1+x^{9}\right|}{9}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
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Proving:$\tan(20^{\circ})\cdot \tan(30^{\circ}) \cdot \tan(40^{\circ})=\tan(10^{\circ})$ how to prove that : $\tan20^{\circ}.\tan30^{\circ}.\tan40^{\circ}=\tan10^{\circ}$?
I know how to prove
$ \frac{\tan 20^{0}\cdot\tan 30^{0}}{\tan 10^{0}}=\tan 50^{0}, $
in this way:
$ \tan{20^0} = \sqrt{3}.\tan{50^0}.\tan{10^0}$
$\Longleftrightarrow \sin{20^0}.\cos{50^0}.\cos{10^0} = \sqrt{3}.\sin{50^0}.\sin{10^0}.\cos{20^0}$
$\Longleftrightarrow \frac{1}{2}\sin{20^0}(\cos{60^0}+\cos{40^0}) = \frac{\sqrt{3}}{2}(\cos{40^0}-\cos{60^0}).\cos{20^0}$
$\Longleftrightarrow \frac{1}{4}\sin{20^0}+\frac{1}{2}\sin{20^0}.\cos{40^0} = \frac{\sqrt{3}}{2}\cos{40^0}.\cos{20^0}-\frac{\sqrt{3}}{4}.\cos{20^0}$
$\Longleftrightarrow \frac{1}{4}\sin{20^0}-\frac{1}{4}\sin{20^0}+\frac{1}{4}\sin{60^0} = \frac{\sqrt{3}}{4}\cos{60^0}+\frac{\sqrt{3}}{4}\cos{20^0}-\frac{\sqrt{3}}{4}\cos{20^0}$
$\Longleftrightarrow \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{8}$
Could this help to prove the first one and how ?Do i just need to know that $ \frac{1}{\tan\theta}=\tan(90^{\circ}-\theta) $ ?
| Using Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$,
$$\tan20^\circ\tan(60^\circ-20^\circ)\tan(60^\circ+20^\circ)=\tan(3\cdot20^\circ)$$
and $\tan80^\circ=\cot(90^\circ-80^\circ)=\dfrac1{\tan?}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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prove that $x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z)=1$
Prove that $x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z)=1$.
Do I use the terms $x= r^2 - s^2$, $y = 2rs$, and $z = r^2 + s^2$ to prove this problem?
Thanks for any help.
| Actually, a very strong statement is true, and quite easy: given any positive integer $n,$ such that all prime factors $p$ of $n$ satisfy $p \equiv 1 \pmod 4,$ then there is a solution to $$ x^2 + y^2 = n $$ with $$ \gcd(x,y) = 1.$$
For $n$ a prime (so itself $1 \pmod 4$) the existence of a solution was finally proved by Euler. After that, there is a simple mathematical induction, using the two choices in someone's formula (Brahmagupta with $N=-1$), anyway
$$ (a^2 + b^2)(c^2 + d^2) = (ac+bd)^2 + (ad-bc)^2, $$ or
$$ (a^2 + b^2)(c^2 + d^2) = (ac-bd)^2 + (ad+bc)^2 $$
with the induction hypothesis $$ \gcd(a,b) = \gcd(c,d) = 1. $$
Indeed, given such an $n,$ with $r$ distinct prime factors and each $p_i \equiv 1 \pmod 4,$ but we otherwise ignore the exponents, the number of representations $$ x^2 + y^2 = n $$ with
$$ 0 < x < y, \; \; \gcd(x,y) = 1 $$
is exactly $$ 2^{r-1}. $$ For example, about ignoring exponents, we get
$$ 1 +4 = 5, $$
$$ 9+16 = 25 = 5^2, $$
$$ 4 + 121 = 125 = 5^3, $$
$$ 49 + 576 = 625 = 5^4. $$
Alright, nobody is paying attention to this one, but I thought I ought to make sure I knew how the induction proof went. It is easier if we demand the second factor prime, so we are taking
$$ a^2 + b^2 = n, \; \; \gcd(a,b) = 1,$$
$$ c^2 + d^2 = p, \; \; p \equiv 1 \pmod 4, \; \; \mbox{of course} \; \; \gcd(c,d) = 1. $$
We are going to represent $np.$ One way is
$$ np = (ac+bd)^2 + (ad-bc)^2. $$ Suppose these are not relatively prime, there is some positive prime $q$ such that
$$ q | ac+bd \; \; \mbox{and} \; \; q | ad-bc. $$
Multiply the first by $d$ and the second by $c$ and subtract, we get
$$ q | b (c^2 + d^2), \; \; \mbox{or} \; \; q | b p. $$
Back to our pair, multiply the first by $c$ and the second by $d$ and add, we get
$$ q | a (c^2 + d^2), \; \; \mbox{or} \; \; q | a p. $$
Since $ \gcd(a,b) = 1, $ it follows that $q |p$ and then that $q=p.$
So the only possible GCD is a power of $p$ itself. If, in addition,
$$ p | ac-bd \; \; \mbox{and} \; \; p | ad+bc, $$
we get $ p | 2ac $ and $p | 2bc,$ or $p | a$ and $p|b.$ This is false, as $\gcd(a,b) = 1.$
To put it briefly, the only possible obstacle to a proper representation of $np$ is a common factor of $p$ itself, but if $p$ divides all four of $ac+bd, \; ad-bc, \; ac-bd, \; ad+bc, $ then it divides both $a,b,$ a contradiction. So, at least one of the four given expressions is not divisible by $p,$ and the pair using that expression gives a proper/primitive representation of $np$ as the sum of two squares. Naturally, the other expression in the successful pair is also prime to $p.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\left\lfloor \frac{x}{ab} \right\rfloor = \left\lfloor \frac{\left\lfloor \frac{x}{a} \right\rfloor}{b} \right\rfloor$ for $a,b>1$ I'm trying to prove rigorously the following:
$$\left\lfloor \frac{x}{ab} \right\rfloor = \left\lfloor \frac{\left\lfloor \frac{x}{a} \right\rfloor}{b} \right\rfloor$$ for integers $a,b \geq 1$ and real $x$.
So far I haven't gotten far. It's enough to prove this instead:
$$\left\lfloor \frac{z}{c} \right\rfloor = \left\lfloor \frac{\lfloor z \rfloor}{c} \right\rfloor$$ for integers $c \geq 1$ and real $z$
since we can just put $z=\lfloor x/a \rfloor$ and $c=b$.
| This theorem is not true if $b$ is not an integer. Take $x=b=1.5$ and take $a=1$.
If $b$ is an integer, this follows from the rule $$\left\lfloor \frac{y}{b}\right\rfloor = \left\lfloor \frac{\lfloor y \rfloor}b\right\rfloor$$
Setting $y=\frac x a$.
Showing this rule, then, suffices.
Let $y = \lfloor y \rfloor + \{y\}$, where $0\leq \{y\} < 1$.
Use division algorithm to write $\lfloor y \rfloor = qb + r$ with $0\leq r <b$.
The $\frac{\lfloor y \rfloor} b = q + \frac{r}{b}$, and $0\leq \frac{r}{b} <1$, so
$$\left\lfloor \frac{\lfloor y \rfloor}b\right\rfloor = q$$
On the other hand, since $[y]< (q+1)b$, since $0\leq\{y\}<1$, then $y = [y]+\{y\}<(q+1)b$.
So $q \leq \frac y b < q+1$ and again $$\left\lfloor \frac{y}{b}\right\rfloor = q $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find function to make maximum value Let ${f : [0, 1] \rightarrow [-1, 1] }$ is a continuous function such that ${ \int_{0}^{1} x f \left(x\right) dx =0}$
Find $f(x)$ such that ${ \int_{0}^{1} \left(x ^{2 } + \frac{1}{4} \right) f \left(x\right) dx}$ has the maximum value.
| Disregard continuity. Take $f = -1 + g$ so $0 \le g \le 2$ and you want to maximize $F = \int_0^1 (x^2+1/4) (g(x)-1)\ dx$ subject to $C = \int_0^1 x g(x) \ dx = \int_0^1 x\ dx = 1/2$. Since $x^2+1/4$ is strictly convex while $x$ is linear, if you have a bit of "mass" somewhere in $(0,1)$, you can increase $F$ while keeping $C$ the same if you by splitting it up and moving half of it to the right and the other half to the left, by equal distances. So an optimal solution must have $g$ of the form
$$g(x) = \cases{2 & for $x \le a$\cr
0 & for $a < x \le b$\cr
2 & for $b < x \le 1$\cr}$$
where $0 \le a \le b \le 1$, or
$$f(x) = \cases{1 & for $x \le a$\cr
-1 & for $a < x \le b$\cr
1 & for $b < x \le 1$\cr}$$
This makes $C = a^2 + 1 - b^2$ so $a^2 - b^2 = -1/2$, and $F = \dfrac{2}{3} (a^3-b^3) + \dfrac{1}{2}(a-b) + \dfrac{7}{12}$. Using Lagrange multipliers I get
$$ a=\frac{1}{2}\,\sqrt {\sqrt {2}-1},\ b=\frac{ 1+\sqrt {2}
}{2} \sqrt {\sqrt {2}-1}$$
With continuous functions, you can't attain a maximum but you can get arbitrarily close.
| {
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"url": "https://math.stackexchange.com/questions/173132",
"timestamp": "2023-03-29T00:00:00",
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Compute integral $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$ I want to solve $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$ but I get the wrong results:
$$
\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x =
\int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \mathrm{d} x
$$
$$
= \left[ \frac{(4e^{4x} + 8e^{2x} + 4x)2}{e^{2x}} \right]_{-6}^6 =
\left[ \frac{8e^{4x} + 16e^{2x} + 8x}{e^{2x}} \right]_{-6}^6
$$
$$
= (\frac{8e^{24} + 16e^{12} + 48}{e^{12}}) - (\frac{8e^{-24} + 16e^{-12} - 48}{e^{-12}})
$$
$$
= e^{-12}(8e^{24} + 16e^{12} + 48) - e^{12}(8e^{-24} + 16e^{-12} - 48)
$$
$$
= 8e^{12} + 16 + 48e^{-12} - (8e^{-12} + 16 - 48e^{12})
$$
$$
= 8e^{12} + 16 + 48e^{-12} - 8e^{-12} - 16 + 48e^{12})
$$
$$
= 56e^{12} + 56e^{-12}
$$
Where am I going wrong?
| $$ I:=\int_{-6}^6 \frac{(4e^{2x} + 2)^2}{e^{2x}}\ dx$$
Let $u=e^x, du = e^x \ dx$, leaving us with:
$$\int_{e^{-6}}^{e^{6}} \frac{\left( 4u^2 + 2 \right)^2}{u^3} \ du$$
Expand the numerator to get
$$\int_{e^{-6}}^{e^{6}} \frac{16u^4 + 16u^2 + 4}{u^3} \ du$$
Since the highest power in the numerator is greater than the highest power in the denominator, we have to do some long division. Upon dividing, you get:
$$\int_{e^{-6}}^{e^{6}} \frac{4}{u^3} + 16u + \frac{16}{u} \ du$$
Integrate to get:
$$8u^2-\frac{2}{u^2} + 16 \ln |u|$$
Back-substitute $u=e^x$ to get
$$8e^{2x} - 2e^{-2x} + 16 \ln|e^{x}|$$
Since $e^x$ is strictly increasing, we can drop the absolute value. Also, recall that $\ln{e^x} = x$, so you can simplify a bit.
$$8e^{2x} + 16x - 2e^{-2x}$$
Now, simply evaluate at your endpoints to find that
$$I \approx 1.628\times10^6$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine the sum $T=a_0+a_1+a_2+...+a_{2012}$ Let ${a_n}$, $n \ge 0$ be a sequence of positive real numbers, given by
$a_0=1$ and
$a_m<a_n$ for all $m,n \in \mathbb{N}, m<n$ with
$a_n=\sqrt{a_{n+1}a_{n-1}}+1$ and $4\sqrt{a_n}=a_{n+1}-a_{n-1}$ for all $n \in \mathbb{N}, n\neq 0$.
Help me, determining the sum $T=a_0+a_1+a_2+...+a_{2012}$.
| From the conditions given, we have
$$
(a_n-1)^2=a_{n+1}a_{n-1}\tag{1}
$$
and
$$
16a_n=(a_{n+1}-a_{n-1})^2\tag{2}
$$
Adding $4$ times $(1)$ to $(2)$ yields
$$
4(a_n+1)^2=(a_{n+1}+a_{n-2})^2\tag{3}
$$
Taking the square root of each side and subtracting $2a_n$ from both sides, we get
$$
a_{n+1}-2a_n+a_{n-1}=2\tag{4}
$$
Thus, because the second difference of $a_n$ is $2$, $a_n=n^2+bn+c$. Since $a_0=1$, we get that $c=1$. Plugging $n^2+bn+1$ into either $(1)$ or $(2)$ gives $b^2=4$. Since $a_1>a_0$, we must have $b=2$. That is,
$$
a_n=(n+1)^2\tag{5}
$$
To sum consecutive squares, use
$$
\begin{align}
\sum_{k=0}^{n}(k+1)^2
&=\frac{(2n+3)(n+2)(n+1)}{6}\\
&=2721031819\tag{6}
\end{align}
$$
for $n=2012$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
For which angles we know the $\sin$ value algebraically (exact)? For example:
*
*$\sin(15^\circ) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$
*$\sin(18^\circ) = \frac{\sqrt{5}}{4} - \frac{1}{4}$
*$\sin(30^\circ) = \frac{1}{2}$
*$\sin(45^\circ) = \frac{1}{\sqrt{2}}$
*$\sin(67 \frac{1}{2}^\circ) = \sqrt{ \frac{\sqrt{2}}{4} + \frac{1}{2} }$
*$\sin(72^\circ) = \sqrt{ \frac{\sqrt{5}}{8} + \frac{5}{8} }$
*$\sin(75^\circ) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$
*?
Is there is a list of known exact values of $\boldsymbol \sin$ somewhere?
Found a related post here.
| Starting with $\tan(\pi/3)=\sqrt{3}$ and $\tan(\pi/4)=1$ and using
$$
\tan(x/2)=\frac{\sqrt{1+\tan^2(x)}-1}{\tan(x)}\tag{1}
$$
and
$$
\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}\tag{2}
$$
and
$$
(\cos(x),\sin(x))=\frac{(1,\tan(x))}{\sqrt{1+\tan^2(x)}}\tag{3}
$$
we can construct the sine and cosine of all rational multiples of $\pi$ where the denominator is a power of $2$, or $3$ times a power of $2$.
For example, $x=\pi/4$ with $(1)$ gives
$$
\tan(\pi/8)=\sqrt{2}-1\tag{4}
$$
then $(2)$ and $(4)$ yields
$$
\begin{align}
\tan(3\pi/8)
&=\tan(\pi/4+\pi/8)\\
&=\frac{1+\tan(\pi/8)}{1-\tan(\pi/8)}\\
&=\sqrt{2}+1\tag{5}
\end{align}
$$
Then $(3)$ and $(5)$ give
$$
(\cos(3\pi/8),\sin(3\pi/8))=\frac{(1,\sqrt{2}+1)}{\sqrt{4+2\sqrt{2}}}\tag{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/176889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 1
} |
Prove $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots$ converges to $\frac 1 2 $ Show that
$$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots = \frac{1}{2}.$$
I'm not exactly sure what to do here, it seems awfully similar to Zeno's paradox.
If the series continues infinitely then each term is just going to get smaller and smaller.
Is this an example where I should be making a Riemann sum and then taking the limit which would end up being $1/2$?
| consider,$$f(n)=\frac{1}{(2n-1) \cdot (2n-1+2)}$$
where $n$ is a natural number
$$\sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty\frac{1}{(2n-1) \cdot (2n+1)}$$
Let, $$\sum_{n=1}^{\infty} f(n) = S$$
i.e. $$S=\sum_{n=1}^\infty\frac{1}{(2n-1)\cdot(2n+1)}$$
i.e. $$S=\sum_{n=1}^\infty(\frac{1}{2})(\frac{1}{2n-1} - \frac{1}{2n+1})$$
i.e. $$S=(\frac{1}{2})\left(\sum_{n=1}^\infty(\frac{1}{2n-1} - \frac{1}{2n+1})\right)$$
i.e. $$S=\lim_{n \to \infty}\left((\frac{1}{2})(\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2})(\frac{1}{3} - \frac{1}{5}) + (\frac{1}{2})(\frac{1}{5} - \frac{1}{7}) + \cdots + (\frac{1}{2})(\frac{1}{2n-1} - \frac{1}{2n+1})\right)$$
i.e. $$S=\lim_{n \to \infty}(\frac{1}{2})\left((\frac{1}{1} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7}) + \cdots + (\frac{1}{2n-1} - \frac{1}{2n+1})\right)$$
i.e. $$S=\lim_{n \to \infty}(\frac{1}{2})\left(\frac{1}{1} - \frac{1}{2n+1}\right)$$
i.e. $$S=(\frac{1}{2})\left(\frac{1}{1} - \lim_{n \to \infty}(\frac{1}{2n+1})\right)$$
i.e. $$S=(\frac{1}{2})\left(\frac{1}{1} - 0\right)$$
i.e. $$S=(\frac{1}{2})\left(\frac{1}{1}\right)$$
i.e. $$S=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/177373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
What is the units digit of $13^4\cdot17^2\cdot29^3$?
What is the units digit of $13^4\cdot17^2\cdot29^3$?
I saw this on a GMAT practice test and was wondering how to approach it without using a calculator. Thanks.
| $13^1$ will give $3$ at the units place
$13^2$ will give $9$ at the units place, we get $9$ by taking the units digit of the result $3^2=9$
$13^3$ will give $7$ at the units place, we get $7$ by taking the units digit of the result $3^3=27$
$13^4$ gives $1$ at the units place, we get $1$ by taking the units digit of the result $3^4=81$
$17^2$ gives $9$ at the units place, we get $9$ by taking the units digit of the result $7^2=49$
$29^3$ gives $9$ at the units place, we get $9$ by taking the units digit of the result $9^3=729$
units digit of ($13^4$)($17^2$)($29^3$)=(units digit of $13^4$)(units digit of $17^2$)(units digit of $29^3$)
units digit of ($13^4$)($17^2$)($29^3$)=units digit of $(1)(9)(9)$
units digit of ($13^4$)($17^2$)($29^3$)=units digit of $81$
i.e. units digit of ($13^4$)($17^2$)($29^3$)=$1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/178982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$ I'm having trouble computing the integral:
$$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$
I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.
Any suggestions are welcome. Thanks.
| We can write the integrand as
$$\begin{equation*}
\frac{1}{1+\cot x}
\end{equation*}$$
and use the substitution $u=\cot x$. Since $du=-\left( 1+u^{2}\right) dx$ we reduce it to a rational function
$$\begin{equation*}
I:=\int \frac{\sin x}{\sin x+\cos x}dx=-\int \frac{1}{\left( 1+u\right)
\left( u^{2}+1\right) }\,du.
\end{equation*}$$
By expanding into partial fractions and using the identities
$$\begin{eqnarray*}
\cot ^{2}x+1 &=&\csc ^{2}x \\
\arctan \left( \cot x\right) &=&\frac{\pi }{2}-x \\
\frac{\csc x}{1+\cot x} &=&\frac{1}{\sin x+\cos x}
\end{eqnarray*}$$
we get
$$\begin{eqnarray*}
I &=&-\frac{1}{2}\int \frac{1}{1+u}-\frac{u-1}{u^{2}+1}\,du \\
&=&-\frac{1}{2}\ln \left\vert 1+u\right\vert +\frac{1}{4}\ln \left(
u^{2}+1\right) -\frac{1}{2}\arctan u +C\\
&=&-\frac{1}{2}\ln \left\vert 1+\cot x\right\vert +\frac{1}{4}\ln \left(
\cot ^{2}x+1\right) -\frac{1}{2}\arctan \left( \cot x\right) +C \\
&=&-\frac{1}{2}\ln \left\vert 1+\cot x\right\vert +\frac{1}{4}\ln \left(
\csc ^{2}x\right) +\frac{1}{2}x+\text{ Constant} \\
&=&\frac{1}{2}x-\frac{1}{2}\ln \left\vert \sin x+\cos x\right\vert +\text{
Constant.}
\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/180744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "88",
"answer_count": 8,
"answer_id": 2
} |
$f(n)=7^{6n} - 6^{6n}$, where $n$ is a positive integer. $f(n)=7^{6n} - 6^{6n}$, where $n$ is a positive integer. find the divisors of $f(n)$ for odd and even values of $n$. Is there a general solution for the divisors.
$$f(1)=7^6-6^6=(7^3)^{2}-(6^3)^{2}$$
$$f(1)=(7-6)(7^2+(7)(6)+6^2)(7^3+6^3)$$
$$f(1)=(1)(127)(7^3+6^3)$$
| As you have said, you always have $7^{6n}-6^{6n}=(7^{3n})^2-(6^{3n})^2=(7^{3n}-6^{3n})(7^{3n}+6^{3n})$
As Mark Bennet has said $7^{3n}+6^{3n}=(7^n+6^n)(7^{2n}-7^n6^n+6^{2n})$
Also, $7^{3n}-6^{3n}=(7^n-6^n)(7^{2n}+7^n6^n+6^{2n})$
So we have $7^{6n}-6^{6n}=(7^n-6^n)(7^{2n}+7^n6^n+6^{2n})(7^n+6^n)(7^{2n}-7^n6^n+6^{2n})$
Alpha confirms this is the best we can do for $n=1$, but there are more for $2$ through $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/182809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\sum\limits_{n=0}^\infty \left(\operatorname{Si}(n)-\frac{\pi}{2}\right)$? I would like to evaluate the sum
$$
\sum\limits_{n=0}^\infty \left(\operatorname{Si}(n)-\frac{\pi}{2}\right)
$$
Where $\operatorname{Si}$ is the sine integral, defined as:
$$\operatorname{Si}(x) := \int_0^x \frac{\sin t}{t}\, dt$$
I found that the sum could be also written as
$$
-\sum\limits_{n=0}^\infty \int_n^\infty \frac{\sin t}{t}\, dt
$$
Anyone have any ideas?
| A bit late, but here's a more general calculation, which explains where the mysterious $\frac{\pi}{4}$ in Raymond Manzoni's answer comes from:
We will use the Fourier series
$$ \sum \limits_{n=1}^\infty \frac{\cos (ny)}{n} = - \log \left[2 \left \lvert \sin \left(\frac{y}{2}\right) \right \rvert \right] \, , \, y \in \mathbb{R} \setminus 2 \pi \mathbb{Z} \, , \tag{1} $$
and the series
$$ \log[\lvert \operatorname{sinc}(\pi z) \rvert] = \sum \limits_{k=1}^\infty \log \left(1 - \frac{z^2}{k^2}\right) \, , \, z \in \mathbb{R} \setminus \mathbb{Z} \, , \tag{2}$$
connected to Euler's sine product formula.
For $x \in \mathbb{R}^+ \setminus \mathbb{N}$ let
\begin{align}
s(x) &= \sum \limits_{n=1}^\infty \left[\frac{\pi}{2} - \operatorname{Si}(2 \pi x n)\right] = \sum \limits_{n=1}^\infty ~ \int \limits_{2 \pi x n}^\infty \frac{\sin(t)}{t} \, \mathrm{d}t \overset{\text{IBP}}{=} \sum \limits_{n=1}^\infty \left[\frac{\cos(2 \pi x n)}{2 \pi x n} - \int \limits_{2\pi x n}^\infty \frac{\cos(t)}{t^2} \, \mathrm{d} t\right] \\
&= \sum \limits_{n=1}^\infty \left[\frac{\cos(2 \pi x n)}{2 \pi x n} - \int \limits_1^\infty \frac{\cos(2 \pi x n u )}{2 \pi x n u^2} \, \mathrm{d} u\right] = \sum \limits_{n=1}^\infty \int \limits_1^\infty \frac{\cos(2\pi x n)-\cos(2 \pi x n u )}{2 \pi x n u^2} \, \mathrm{d} u \\
&\stackrel{(1)}{=} \frac{1}{2 \pi x} \int \limits_1^\infty \log\left(\left\lvert\frac{\sin(\pi x u)}{\sin(\pi x)}\right\rvert\right) \, \frac{\mathrm{d} u}{u^2} \, .
\end{align}
Interchanging summation and integration is allowed here, as can be shown by integrating by parts one more time. Next we write
$$ s(x) = \frac{1}{2 \pi x} \int \limits_1^\infty \left[\log(u) + \log\left(\left\lvert\frac{\operatorname{sinc}(\pi x u)}{\operatorname{sinc}(\pi x)}\right\rvert\right)\right] \, \frac{\mathrm{d} u}{u^2} = \frac{1}{2 \pi x} \left[1 + \int \limits_1^\infty \log\left(\left\lvert\frac{\operatorname{sinc}(\pi x u)}{\operatorname{sinc}(\pi x)}\right\rvert\right) \, \frac{\mathrm{d} u}{u^2} \right] . $$
At this point we would like to use (2), but here interchanging summation and integration is more problematic. It works on any finite interval of integration by dominated convergence, but not on $(1,\infty)$. We therefore choose $R > 1$ such that $x R \notin \mathbb{N}$ and compute
\begin{align}
& \phantom{={}} \int \limits_1^R \log\left(\left\lvert\frac{\operatorname{sinc}(\pi x u)}{\operatorname{sinc}(\pi x)}\right\rvert\right) \, \frac{\mathrm{d} u}{u^2} \stackrel{(2)}{=} \sum \limits_{k=1}^\infty \int \limits_1^R \log \left(\left \lvert \frac{k^2 - x^2 u^2}{k^2-x^2} \right \rvert \right) \, \frac{\mathrm{d} u}{u^2} \\
&\!\overset{\text{IBP}}{=} \sum \limits_{k=1}^\infty \left[\log \left(\left \lvert \frac{k^2 - x^2 R^2}{k^2-x^2} \right \rvert \right) \left(\frac{x}{k} - \frac{1}{R}\right) + \int \limits_1^R \frac{2 x^2 u}{k^2 - x^2 u^2} \left(\frac{x}{k} - \frac{1}{u}\right) \, \mathrm{d} u\right] \\
&= \sum \limits_{k=1}^\infty \left[\log \left(\left \lvert \frac{k^2 - x^2 R^2}{k^2-x^2} \right \rvert \right) \left(\frac{x}{k} - \frac{1}{R}\right) - \frac{2 x^2}{k} \int \limits_1^R \frac{\mathrm{d} u}{k + x u}\right] \\
&= \sum \limits_{k=1}^\infty \left[\frac{x}{k} \left(\log \left(\left \lvert \frac{k + x}{k-x}\right\rvert\right) - \log \left(\left \lvert \frac{k + xR}{k-xR}\right\rvert\right) \right) - \frac{1}{R} \log \left(\left \lvert \frac{k^2 - x^2 R^2}{k^2-x^2} \right \rvert \right) \right] \\
&\stackrel{(2)}{=} 2 x [f(x) - f(x R)] - \frac{1}{R} \log\left(\left\lvert\frac{\operatorname{sinc}(\pi x R)}{\operatorname{sinc}(\pi x)}\right\rvert\right) .
\end{align}
Here, we have introduced
$$ f \colon \mathbb{R}^+ \setminus \mathbb{N} \to \mathbb{R}^+ \, , \, f(t) = \sum \limits_{k=1}^\infty \frac{1}{2k} \log \left(\left \lvert \frac{k + t}{k-t}\right\rvert\right) = \sum \limits_{k=1}^\infty \frac{1}{k} \operatorname{artanh} \left[\min \left(\frac{k}{t},\frac{t}{k}\right)\right] .$$
Letting $R \to \infty$, we find
$$ s(x) = \frac{1}{2\pi x} + \frac{f(x) - f(\infty)}{\pi} \, .$$
The limit
\begin{align}
f(\infty) &= \lim_{t \to \infty} f(t) = \lim_{\varepsilon \to 0^+} f\left(\frac{1}{\varepsilon}\right) = \lim_{\varepsilon \to 0^+} \varepsilon \sum \limits_{k=1}^\infty \frac{1}{\varepsilon k} \operatorname{artanh} \left[\min\left(\varepsilon k, \frac{1}{\varepsilon k}\right)\right] \\
&= \int \limits_0^\infty \frac{\operatorname{artanh} \left[\min\left(r, \frac{1}{r}\right)\right]}{r} \, \mathrm{d} r = 2 \int \limits_0^1 \frac{\operatorname{artanh} (r)}{r} \, \mathrm{d} r = 2 \frac{\pi^2}{8} = \frac{\pi^2}{4}
\end{align}
can be computed using improper Riemann sums (see this question; the monotonicity condition can be circumvented by splitting the series at $k = \lfloor 1/\varepsilon \rfloor$ and thus the integral at $r=1$). Therefore,
$$ s(x) = \frac{1}{2\pi x} - \frac{\pi}{4} + \frac{1}{\pi} \sum \limits_{k=1}^\infty \frac{1}{k} \operatorname{artanh} \left[\min \left(\frac{k}{x},\frac{x}{k}\right)\right] $$
holds for $x \in \mathbb{R}^+ \setminus \mathbb{N}$. In particular, your series is
$$ - s\left(\frac{1}{2\pi}\right) - \frac{\pi}{2} = -1 - \frac{\pi}{4} - \frac{1}{\pi} \sum \limits_{k=1}^\infty \frac{1}{k} \operatorname{artanh} \left(\frac{1}{2 \pi k}\right) . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/184098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 4
} |
Integrating $\int \frac{dx}{(x+\sqrt{x^2+1})^{99}}$ I am bugged by this problem: how do I evaluate this?
$$\int \frac{dx}{(x+\sqrt{x^2+1})^{99}}.$$
A closed form will be convenient and fine. Thanks (it does not seem particularly inpiring).
| If you sub $u=x+\sqrt{x^2+1}$, that alone will make the integral a simple rational integral. We can solve for $x$ in terms of $u$:
$$\begin{align}u - x & =\sqrt{x^2+1}\\
u^2 - 2ux + x^2& =x^2+1\\
- 2ux& =1-u^2\\
x & = \frac{u^2-1}{2u}\\
dx & = \frac{2u(2u)-2(u^2-1)}{4u^2} du\\
dx & = \frac{u^2+1}{2u^2} du
\end{align}$$
So you have $$\begin{align}\int \frac{u^2+1}{2u^{101}}\,du&=\frac{1}{2}\int u^{-99}+u^{-101}\,du\\&=\frac{1}{2}\left(\frac{u^{-98}}{-98}-\frac{u^{-100}}{100}\right)+C\\&=-\frac{1}{196(x+\sqrt{x^2+1})^{98}}-\frac{1}{200(x+\sqrt{x^2+1})^{100}}+C\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/184217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Sum of Stirling numbers of both kinds Let $a_k$ be the number of ways to partition a set of $n$ elements $orderly$,which means that order of subsets matters, but order of elements in each subset does not.
My task:
Prove, that$$\sum_{k=1}^n \left[n\atop k\right] a_k = n!2^{n-1}$$
The only thing I got so far is:
$$a_k = \sum_{i=1}^{k} \left\{n\atop k\right\} \frac{1}{i!}$$
I tried to change the order of this sum $\sum\limits_{k=1}^n \left[n\atop k\right]\sum\limits_{i=1}^{k} \left\{n\atop k\right\} \frac{1}{i!}$
and got something like that: $\sum\limits_{k=1}^n \frac{1}{k!}\sum\limits_{i=k}^{n} \left[n\atop i\right]\left\{i\atop k\right\}$
My next idea was combinatorial proof, but either I cannot think of one, or I made some crucial mistake above, which makes my efforts pretty useless.
Can anyone guide me how to do this task?
I'd also be most thankful for correcting my mistakes - I'd rather make them here than on my exam sheet...
| Start by observing that the species of orderly partitions has the
specification
$$\mathfrak{S}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z})).$$
This gives the bivariate generating function
$$G(z, u) = \frac{1}{1-u(\exp(z)-1)}.$$
Hence the generating function of the $a_n$ is
$$G(z) = G(z, 1) =
\frac{1}{2-\exp(z)}
= \frac{1}{2} \frac{1}{1-\exp(z)/2}.$$
Substituting this into the sum yields
$$\sum_{k=1}^n
\left[n\atop k\right]
k! [z^k]
\frac{1}{2} \frac{1}{1-\exp(z)/2}.$$
Call this sum $P_n$
and introduce the exponential generating function
$$P(z) = \sum_{n\ge 1} P_n \frac{z^n}{n!}.$$
We have for the sum that
$$\sum_{k=1}^n
\left[n\atop k\right]
k! [z^k]
\frac{1}{2} \frac{1}{1-\exp(z)/2}
= \frac{1}{2} \sum_{k=1}^n
\left[n\atop k\right]
k! [z^k] \sum_{q\ge 0} \frac{\exp(qz)}{2^q}
\\= \frac{1}{2} \sum_{k=1}^n
\left[n\atop k\right]
\sum_{q\ge 1} \frac{q^k}{2^q}.$$
This gives for $P(z)$ that
$$P(z) = \frac{1}{2}\sum_{n\ge 1}
\frac{z^n}{n!} \sum_{k=1}^n
\left[n\atop k\right]
\sum_{q\ge 1} \frac{q^k}{2^q}
= \frac{1}{2} \sum_{q\ge 1} \sum_{k\ge 1}
\frac{q^k}{2^q}
\sum_{n\ge k}
\left[n\atop k\right] \frac{z^n}{n!}.$$
Recall that the species for Stirling numbers of the first kind is
given by
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}_{\ge 1}(\mathcal{Z})).$$
This gives the bivariate generating function
$$H(z, u) = \exp\left(u\log\frac{1}{1-z}\right).$$
Substituting this into $P(z)$ yields
$$\frac{1}{2} \sum_{q\ge 1} \sum_{k\ge 1}
\frac{q^k}{2^q}
\sum_{n\ge k}
\frac{z^n}{n!} n! [z^n] [u^k]
\exp\left(u\log\frac{1}{1-z}\right)
\\ =\frac{1}{2} \sum_{q\ge 1} \sum_{k\ge 1}
\frac{q^k}{2^q}
\sum_{n\ge k}
z^n [z^n] \frac{1}{k!} \left(\log\frac{1}{1-z}\right)^k.$$
Now the innermost sum annihilates the coefficient extractor so we get
$$\frac{1}{2} \sum_{q\ge 1} \sum_{k\ge 1}
\frac{q^k}{2^q}
\frac{1}{k!} \left(\log\frac{1}{1-z}\right)^k
= \frac{1}{2} \sum_{q\ge 1}
\frac{1}{2^q}
\left(-1 + \exp\left(q\log\frac{1}{1-z}\right)\right)
\\ = -\frac{1}{2}
+ \frac{1}{2} \sum_{q\ge 1}
\frac{1}{2^q} \left(\frac{1}{1-z}\right)^q
= -\frac{1}{2}
+ \frac{1}{2} \frac{1/2/(1-z)}{1-1/2/(1-z)}
\\ = -\frac{1}{2}
+ \frac{1}{2} \frac{1}{2(1-z)-1}
= -\frac{1}{2}
+ \frac{1}{2} \frac{1}{1-2z}.$$
The conclusion is that
$$P_n = n! [z^n] P(z) = n! [z^n] \frac{1}{2} \frac{1}{1-2z} =
n! \times \frac{1}{2} \times 2^n
= 2^{n-1} n!$$
as claimed.
Another computation that refererences Wilf's generatingfunctionology.
Remark, somewhat later. The annihilated coefficient extractor is not strictly necessary here, we may also recognize the EGF by inspection.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Evaluating a simple definite integral I'm currently teaching myself calculus and am onto the Mean Value Theorem for Integration.
I am finding the value of $f(c)$ on the function $f(x)=x^3-4x^2+3x+4$ on the interval $[1,4]$.
So, with the equation $(b-a)\cdot f(c)=\int_1^4f(x)dx $, you get
$(4-1)\cdot f(c)=\int_1^4(x^3-4x^2+3x+4)dx$
Now my book says that this equals $3f(c)=\frac{57}{4}$.
I've been racking my brain and can't figure out how $\int_1^4(x^3-4x^2+3x+4)dx=\frac{57}{4}$
So how did the author evaluate that integral to get the answer?
| $\int_1^4(x^3-4x^2+3x+4)dx=\int_1^4x^3dx -4\int_1^4x^2dx +3\int_1^4xdx +4\int_1^4dx=:I$
$$\begin{align*}I &= \left.\left((1/4)x^4 -(4/3)x^3+(3/2)x^2 +4x\right)
\right|_1^4\\ &=
(1/4)(4^4-1^4) -(4/3)(4^3-1^3)+(3/2)(4^2-1^2) +4(4-1)
\\ &= (1/4)(255)-(4/3)(63)+(3/2)(15)+4(3)\\ &= 14.25\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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all integer $(n,m)$, such that $n^3+2^m\cdot n$ is a perfect square What are all integer $(n,m)$, such that $n^3+2^m\cdot n$ is a perfect square
| I’ve left a bit for you to finish, but here’s a pretty detailed guide to a solution.
Notice that $n^3+2^mn=n(n^2+2^m)$, which is a perfect square iff either (1) $n$ and $n^2+2^m$ are both perfect squares, or (2) $n^2+2^m=na^2$ for some positive integer $a$.
In case (1) say $n=a^2$ and $n^2+2^m=b^2$. Then $2^m=b^2-a^4=(b-a^2)(b+a^2)$, and $b-a^2$ and $b+a^2$ must both be powers of $2$; say $b-a^2=2^r$ and $b+a^2=2^s$, where $r+s=m$ and $r<s$. Then $a^2=2^{s-1}-2^{r-1}=2^{r-1}(2^{s-r}-1)$, which is a perfect square iff $r-1$ is even and $2^{s-r}-1$ is a perfect square (why?). Let $t=s-r\ge 1$; for what positive integers $t$ is $2^t-1$ a perfect square? If $t>1$, then $4\mid 2^t$, so $2^t-1\equiv 3\pmod 4$. But $2^t-1$ is odd, and every odd square is congruent to $1$ mod $4$, so we must have $t=1$. Thus, $r$ must be odd, $s=r+1$, and $m=2r+1$.
Conversely, let $r=2k+1$ be any odd positive integer, and let $m=2r+1$. Let $n=2^{r-1}=2^{2k}$. Then $n^3+2^mn=2^{6k}+2^{m+2k}=2^{6k}+2^{6k+3}=2^{6k}(1+2^3)=9\cdot2^{6k}=(3\cdot2^{3k})^2$. This gives us one infinite family of solutions.
In case (2) we must have $n\mid 2^m$, in which case $n=2^k$ for some $k\in\{0,1,\dots,m\}$. Then $na^2=2^{2k}+2^m=2^k(2^k+2^{m-k})$, so $a^2=2^k+2^{m-k}$. If $m=2k$, $a^2=2\cdot2^k=2^{k+1}$, so $k$ must be odd. Conversely, each odd value of $k$ yields a solution of this type; what are these solutions?
If $m\ne 2k$, $a^2=2^k+2^{m-k}$ is the sum of two different powers of $2$, so we need to know when such a sum is a perfect square. Consider the sum $2^r+2^s$, where $r$ and $s$ are non-negative integers, and $r<s$. If $r$ is even, then $2^r+2^s=2^r(1+2^{s-r})$ is a perfect square iff $1+2^{s-r}$ is. If $r$ is odd, then $2^r+2^s=2\cdot2^{r-1}(1+2^{s-r})$ is a perfect square iff $2(1+2^{s-r})$ is. But $2(1+2^{s-r})$ cannot be a perfect square (why?), so we need only consider the case of even $r$ and ask when $1+2^t$ is a perfect square, where $t=s-r\ge 1$. Certainly this is the case when $t=3$. Are there any other solutions?
Suppose that $1+2^t$ is a perfect square, say $1+2^t=c^2$. Then $2^t=c^2-1=(c+1)(c-1)$. What does this tell you about the numbers $c+1$ and $c-1$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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On the number of possible solutions for a quadratic equation. Solving a quadratic equation will yield two roots:
$$\frac{-\sqrt{b^2-4 a c}-b} {2 a}$$
and:
$$\frac{\sqrt{b^2-4a c}-b}{2 a}$$
And I've been taught to answer it like:
$$\frac{\pm\sqrt{b^2-4a c}- b}{2 a}$$
Why does it yields only two solutions? Aren't there infinite solutions for that? Is there a proof on the number of possible solutions for a quadratic equation?
| One can see that there are only two solutions from the way it is solved. If $ax^2+bx+c=0$ with $a\neq 0$, it follows that $ax^2+bx=-c$, so that $x^2+\frac{b}{a}x=-\frac{c}{a}$. Trying to complete the square, one obtains $(x+\frac{b}{2a})^2=x^2+\frac{b}{a}x+(\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}=\frac{b^2-4ac}{4a^2}$. Taking square roots, the result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving the cubic polynomial equation $x^3+3x^2-5x-4=0$
How can I solve the cubic polynomial equation $$x^3+3x^2-5x-4=0$$
I simplified it to:
$$x(x^2+3x-5)=4$$
But I don't know where to go from here.
| The manipulation that you performed does not help. Use the rational root test: since the leading coefficient is $1$ and the constant term is $4$, the only possible rational roots of the cubic are fractions of the form $a/b$, where $a$ is a divisor of $4$ and $b$ is a divisor of $1$. In other words, if the cubic has any rational root at all, it must be one of the numbers $\pm1,\pm2$, and $\pm4$. Actually calculation shows that
$$(-4)^3+3(-4)^2-5(-4)-4=-64+48+20-4=0\;,$$
so $-4$ is a root of the cubic $x^3+3x^2-5x-4$. Now use the fact that $r$ is a root of a polynomial if and only if $x-r$ is a factor of that polynomial to conclude that $x^3+3x^2-5x-4$ is divisible by $x-(-4)=x+4$. When you perform this division $-$ either by polynomial long division or by synthetic division $-$ you’ll get a quadratic as the quotient. Let’s say that you get $x^2+bx+c$ as the quotient; then you know that
$$x^3+3x^2-5x-4=(x+4)(x^2+bx+c)\;.$$
This tells you that $x=-4$ is one solution to your original equation, and the others are the solutions to the quadratic equation $x^2+bx+c=0$, which you can find using the quadratic formula.
| {
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How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ without changing into polar coordinates? How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ other than changing into polar coordinates? It is possible to prove it using infinite series?
| Method 1. Let $I$ denote the integral. Then for $s > 0$, the substitution $t = x\sqrt{s}$ gives
$$ \int_{0}^{\infty} e^{-sx^2} \; dx = \frac{1}{\sqrt{s}} \int_{0}^{\infty} e^{-t^2} \; dt = \frac{I}{\sqrt{s}}. $$
Thus for $u = x^2$,
$$\begin{align*}I^2
&= \int_{0}^{\infty} I e^{-x^2} \; dx
= \int_{0}^{\infty} \frac{I}{2\sqrt{u}}e^{-u} \; du
= \int_{0}^{\infty} \frac{1}{2}e^{-u}\int_{0}^{\infty}e^{-ut^2}\;dtdu\\
&= \int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{2}e^{-u}e^{-ut^2}\;dudt
= \int_{0}^{\infty}\frac{1}{2(t^2+1)}\;dt
= \frac{\pi}{4},
\end{align*}$$
which proves the desired result. Note that this is in fact equivalent to the proof in Wikipedia.
Method 2. We assume that we are aware of the Wallis product
$$ \prod_{n=1}^{\infty} \left(\frac{2n}{2n-1} \cdot \frac{2n}{2n+1}\right) = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2}. $$
Now, it is not hard to find that
$$ f_n(x) = \begin{cases} \left(1 - \tfrac{x^2}{n}\right)^{n} & 0 \leq x \leq \sqrt{n} \\ 0 & \text{otherwise}. \end{cases} $$
increases to $e^{-x^2}$. (Simple application of Bernoulli's inequality will suffice.) Then by monotone convergence theorem,
$$ I := \int_{0}^{\infty} e^{-x^2} \; dx = \lim_{n\to\infty} \int_{0}^{\sqrt{n}} \left(1 - \frac{x^2}{n}\right)^{n} \; dx = \lim_{n\to\infty} \sqrt{n} \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}\theta \; d\theta, $$
where we have used the substitution $x = \sqrt{n}\,\cos\theta$. Now let
$$I_n = \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}\theta \; d\theta. $$
Then for $n \geq 1$,
$$\begin{align*}I_n
&= \left[\sin^{2n}\theta (-\cos\theta)\right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2n \sin^{2n-1}\theta \cos\theta (-\cos\theta) \; d\theta \\
&= \int_{0}^{\frac{\pi}{2}} 2n \sin^{2n-1}\theta (1-\sin^2 \theta) \; d\theta
= 2n(I_{n-1} - I_n)
\end{align*}$$
and we have $I_0 = 1$ and $I_n = \frac{2n}{2n+1} I_{n-1}$. Thus
$$ I_n = \prod_{k=1}^{n} \frac{2k}{2k+1}. $$
Therefore, by Wallis product,
$$ I^2 = \lim_{n\to\infty} n \prod_{k=1}^{n} \left( \frac{2k}{2k+1} \cdot \frac{2k}{2k+1} \right) = \lim_{n\to\infty} \frac{n}{2n+1} \prod_{k=1}^{n} \left( \frac{2k}{2k-1} \cdot \frac{2k}{2k+1} \right) = \frac{1}{2} \cdot \frac{\pi}{2} $$
and we have the desired result. (Conversely, knowing $I$, it can be used to prove the Wallis product formula.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is $a^n - b^n$ divisible by $a-b$? I did some mathematical induction problems on divisibility
*
*$9^n$ $-$ $2^n$ is divisible by 7.
*$4^n$ $-$ $1$ is divisible by 3.
*$9^n$ $-$ $4^n$ is divisible by 5.
Can these be generalized as
$a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer?
But why is $a^n$ $-$ $b^n$$ = (a-b)N$ ?
I also see that $6^n$ $- 5n + 4$ is divisible by $5$ which is $6-5+4$ and $7^n$$+3n + 8$ is divisible by $9$ which is $7+3+8=18=9\cdot2$.
Are they just a coincidence or is there a theory behind?
Is it about modular arithmetic?
| Consider the polynomial $f(x)=x^n-b^n.$ Then $f(b)=b^n-b^n=0.$ So $b$ is a root of $f$ and this implies $(x-b)$ divides $f(x)$. Put $x=a$, then $a-b$ divides $f(a)=a^n-b^n.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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solving for a coefficent term of factored polynomial.
Given: the coefficent of $x^2$ in the expansion of $(1+2x+ax^2)(2-x)^6$ is $48,$ find the value of the constant $a.$
I expanded it and got
$64-64\,x-144\,{x}^{2}+320\,{x}^{3}-260\,{x}^{4}+108\,{x}^{5}-23\,{x}^{
6}+2\,{x}^{7}+64\,a{x}^{2}-192\,a{x}^{3}+240\,a{x}^{4}-160\,a{x}^{5}+
60\,a{x}^{6}-12\,a{x}^{7}+a{x}^{8}
$
because of the given info $48x^2=64x^2-144x^2$ solve for $a,$ $a=3$.
Correct?
P.S. is there an easier method other than expanding the terms?
I have tried using the bionomal expansion; however, one needs still to multiply the terms. Expand $(2-x)^6$ which is not very fast.
| The coefficient of $x^2$ is obtained by adding three contributions :
*
*$a\, 2^6$ : the coefficient of $x^2$ at the left and $2\,x^0$ at the right ($2$ at power $6$)
*$2(6\cdot 2^5(-1))$: the $x$ coefficients on both sides ((constant term $2$)$^5$ $\times\ x$ coef. at the right)
*$1\binom{6}{2}2^4$ : the constant coefficient of the left $\times$ the $x^2$ coefficient at the right ($2^4$)
The sum of all this is : $\quad 2^6\, a-12\cdot 2^5+15\cdot 2^4=48$
Divided by $2^4=16$ this becomes $\ 4a-24+15=3\ $ i.e. $\ \boxed{a=3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How may I prove this inequality? Let $a, b, c$ be positive real, $abc = 1$. Prove that:
$$\frac{1}{1+a+b} + \frac{1}{1+b+c} + \frac{1}{1+c+a} \le \frac{1}{2+a} + \frac{1}{2+b}+\frac{1}{2+c}$$
I thought of Cauchy and AM-GM, but I don't see how to successfully use them to prove the inequality. Any hint, suggestion will be welcome. Thanks.
| $p=a+b+c$ and $q=ab+bc+ca$. Using $AM-GM$ we obtain that : $p,q \geq 3.$
The inequality is equivalent with:
$$\dfrac{3+4p+q+p^2}{2p+q+p^2+pq} \leq \dfrac{12+4p+q}{9+4p+2q} \Leftrightarrow$$
$$3p^2q+pq^2+6pq-5p^2-q^2-24p-3q-27 \geq 0 \Leftrightarrow \\\left(3p^2q-5p^2-12p\right)+\left(pq^2-q^2-3p-3q\right)+\left(6pq-9p-27\right) \geq 0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to solve $x^3=-1$? How to solve $x^3=-1$? I got following:
$x^3=-1$
$x=(-1)^{\frac{1}{3}}$
$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...
| Let $x=a+bi$, where $a$ and $b$ are real. Then $(a+bi)^3 = -1$. Expanding the left-hand side gives $$a^3 +3a^2bi -3ab^2 -b^3i = -1.$$
We can separate the real and imaginary parts of this equation:
$$\begin{align}
a^3 -3ab^2 & = -1 \\
3a^2b - b^3 & = 0
\end{align}$$
Taking the obvious $b=0$ solution gives us $a=-1$ and thus the real solution
$$x =-1.$$
So suppose $b\ne 0$. Then the second equation has $3a^2 = b^2$, so $b = \pm a\sqrt3$. Putting $3a^2$ for $b^2$ in the first equation gives $$ a^3 - 9a^3 = -1$$ so $a = \frac12.$ Since $b = \pm a\sqrt3$, we have $b=\pm\frac{\sqrt3}2$. This gives us the other two solutions, namely
$$x=\frac12 \pm\!\frac{\sqrt3}2i.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$ Please help me for prove this inequality:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$
| Beferore we prove this inequality first prove that the following inequality:
For $a>0$, $b>0$ is true this inequality: $\frac{a}{b}+\frac{b}{a}\geq 2$
$\frac{a}{b}+\frac{b}{a}-2$=$\frac{a^2+b^2-2ab}{ab}$=$\frac{(a-b)^2}{ab}$.
Since $a>0$, $b>0$ $\Rightarrow$ $a-b>0$ $\Rightarrow$ $(a-b)^2>0$, $ab>0$.
Means, $(a-b)^2>0$, $ab>0$ $\Rightarrow$ $\frac{(a-b)^2}{ab}>0$.
Since $a=b$ $\Rightarrow$ $\frac{(a-b)^2}{ab}=0$.
Means, $\frac{(a-b)^2}{ab}\geq 0$ $\Rightarrow$ $\frac{a}{b}+\frac{b}{a}-2$=$\frac{(a-b)^2}{ab}\geq 0$.
$\frac{a}{b}+\frac{b}{a}-2\geq 0$ $\Rightarrow$ $\frac{a}{b}+\frac{b}{a}\geq 2$.
Now turn to prove the first inequality.
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}$
Since $a>0$, $b>0$, $c>0$ $\Rightarrow$ $a+b+c\geq 0$, so
$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(a+b+c)$=$(\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}\frac{c}{a})+(\frac{b}{c}+\frac{c}{b})+3\geq 2+2+2+3=9$
| {
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"timestamp": "2023-03-29T00:00:00",
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More and more limits for sequences So here goes a bit of homework:
$$\lim_{n\to\infty}{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}$$
Well, this would trivially lead to:
$$\lim_{n\to\infty}{\left(\frac{3+\frac{2}{n}+\frac{1}{n^2}}{3+\frac{5}{n^2}}\right)^{\frac{n\left(1+\frac{2}{n^2}\right)}{2+\frac{1}{n}}}}$$
Which is clearly an indetermination of type "$1^\infty$". Now, I can't really get through this step... Any hints?
| Reduce to the limit for the exponential:
$$
\lim_{n \to \infty} \left( 1 + \frac{2}{3 n}\frac{1 - \frac{2}{n}}{1 + \frac{5}{3n}} \right)^{\frac{n}{2} - \frac{n-4}{4n+2}} = \underbrace{\lim_{n \to \infty} \left( 1+\frac{2}{3n} \right)^{n/2}}_{\exp\left(\frac{1}{3}\right)} \cdot \underbrace{\lim_{n \to \infty} \left(1 + \frac{2}{3 n}\frac{1 - \frac{2}{n}}{1 + \frac{5}{3n}}\right)^{-\frac{n-4}{4n+2}} }_{1} \cdot \lim_{n \to \infty} \left(\frac{1 + \frac{2}{3 n}\frac{1 + \frac{2}{n}}{1 + \frac{5}{3n}}}{1-\frac{2}{3n}}\right)^{n/2}
$$
Using
$$
\frac{1 + \frac{2}{3 n}\frac{1 + \frac{2}{n}}{1 + \frac{5}{3n}}}{1-\frac{2}{3n}} = 1 - \frac{22}{(3n+2)(3n+5)}
$$
we conclude that the last limit also equals to 1.
| {
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Integer solutions to $ x^2-y^2=33$ I'm currently trying to solve a programming question that requires me to calculate all the integer solutions of the following equation:
$x^2-y^2 = 33$
I've been looking for a solution on the internet already but I couldn't find anything for this kind of equation. Is there any way to calculate and list the integer solutions to this equation?
Thanks in advance!
| Suppose that $x=y+n$; then $x^2-y^2=y^2+2ny+n^2-y^2=2ny+n^2=n(2y+n)$. Thus, $n$ and $2y+n$ must be complementary factors of $33$: $1$ and $33$, or $3$ and $11$. The first pair gives you $2y+1=33$, so $y=16$ and $x=y+1=17$. The second gives you $2y+3=11$, so $y=4$ and $x=y+3=7$. As a check, $17^2-16^2=289-256=33=49-16=7^2-4^2$.
If you want negative integer solutions as well, you have also the pairs $-1$ and $-33$, and $-3$ and $-11$.
| {
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"timestamp": "2023-03-29T00:00:00",
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how to calculate the exact value of $\tan \frac{\pi}{10}$ I have an extra homework: to calculate the exact value of $ \tan \frac{\pi}{10}$.
From WolframAlpha calculator I know that it's $\sqrt{1-\frac{2}{\sqrt{5}}} $, but i have no idea how to calculate that.
Thank you in advance,
Greg
| $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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\newcommand{\ic}{{\rm i}}%
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$$
\cos\pars{\bracks{n + 1}\theta} + \cos\pars{\bracks{n - 1}\theta}
=
2\cos\pars{n\theta}\cos\pars{\theta}
$$
$$
\cos\pars{\bracks{n + 1}\theta}
=
2\cos\pars{n\theta}\cos\pars{\theta}
-
\cos\pars{\bracks{n - 1}\theta}
$$
Let's $\theta = \pi/10$ and $x = \cos\pars{\theta}$:
\begin{align}
\cos\pars{2\theta}
&=
2x^{2} - 1
\tag{1}
\\
\cos\pars{3\theta}
&=
\bracks{2\cos\pars{2\theta} - 1}x
\tag{2}
\\
\cos\pars{4\theta}
&=
2\cos\pars{3\theta}x - \cos\pars{2\theta}
\tag{3}
\\
0
&=
2\cos\pars{4\theta}x - \cos\pars{3\theta}
\tag{4}
\end{align}
$\pars{3}$ and $\pars{4}$ yield:
$$
0 = \pars{4x^{2} - 1}\cos\pars{3\theta} - 2\cos\pars{2\theta}x
\tag{5}
$$
$\pars{2}$ and $\pars{5}$ yield:
$$
0 = \pars{4x^{2} - 1}\bracks{2\cos\pars{2\theta} - 1}x - 2\cos\pars{2\theta}x
=
\pars{8x^{3} - 4x}\cos\pars{2\theta} - 4x^{3} + x
$$
$$
4\pars{2x^{2} - 1}\cos\pars{2\theta} - 4x^{2} + 1 = 0
\tag{6}
$$
$\pars{1}$ and $\pars{6}$ yield:
$$
0 = 4\pars{2x^{2} - 1}^{2} - 4x^{2} + 1
=
4\pars{2x^{2} - 1}^{2} - 2\pars{2x^{2} - 1} - 1
$$
Then,
$$
\pars{2x^{2} - 1}_{\pm}
=
{2 \pm \sqrt{\pars{-2}^{2} - 4\times 4\times\pars{-1}} \over 2\times 4}
=
{1 \pm \sqrt{5} \over 4}
$$
Obviously, we take the "$+$ sign" as a solution:
$$
2x^{2} - 1
=
{1 + \sqrt{5} \over 4}
\quad\imp\quad
x = \cos\pars{\pi \over 10}
=
\sqrt{{1 \over 2}\bracks{1 + {1 + \sqrt{5} \over 4}}}
=
\sqrt{{5 + \sqrt{5} \over 8}}
$$
Then,
\begin{align}
\tan\pars{\pi \over 10}
&=
\sqrt{\bracks{1 \over \cos\pars{\pi/10}}^{2} - 1}
=
\sqrt{{8 \over 5 + \sqrt{5\,}} - 1}
=
\sqrt{3 - \sqrt{5\,} \over 5 + \sqrt{5\,}}
=
\sqrt{1 - 2\,{1 + \sqrt{5\,} \over 5 + \sqrt{5}}}
\\[3mm]&=
\sqrt{1 - 2\,{\pars{1 + \sqrt{5\,}}\pars{5 - \sqrt{5\,}} \over 20}}
=
\sqrt{1 - {4\sqrt{5\,} \over 10}}
=
\sqrt{1 - {2 \over \sqrt{5\,}}}
\\[5mm]&
\end{align}
$${\large%
\tan\pars{\pi \over 10} = \sqrt{1 - {2 \over \sqrt{5\,}}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/196067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Prove $\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{ (abc)^{\frac{1}{3}}\big( 1+ (abc)^{\frac{1}{3}}\big) }$ using AM-GM I need to proof this inequality by AM-GM method.
Any ideas how to do it?
$$\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{ (abc)^{\frac{1}{3}}\big( 1+ (abc)^{\frac{1}{3}}\big) }$$
| By AM-GM we obtain:
$$(1+abc)\sum_{cyc}\frac{1}{a(1+b)}+3=\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}=$$
$$=\sum_{cyc}\frac{1+a}{a(1+b)}+\sum_{cyc}\frac{b(1+c)}{a(1+b)}\geq\frac{3}{\sqrt[3]{abc}}+3\sqrt[3]{abc},$$
which says that
$$\sum_{cyc}\frac{1}{a(1+b)}\geq\frac{\frac{3}{\sqrt[3]{abc}}+3\sqrt[3]{abc}-3}{1+abc}=\frac{3\left(1-\sqrt[3]{abc}+\sqrt[3]{a^2b^2c^2}\right)}{(1+abc)\sqrt[3]{abc}}=\frac{1}{(1+\sqrt[3]{abc})\sqrt[3]{abc}}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/196176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Generating integral triangles with two equal sides How can I generate all triangles which have integral sides and area, and exactly two of its three sides are equal?
For example, a triangle with sides ${5,5,6}$ satisfies these terms.
| Area of isosceles triangle with sides $a,a,b$ is $\frac{b\sqrt{4a^2-b^2}}{4}$
Now $\sqrt{4a^2-b^2}$ must be integer$=c$(say),
$\implies 4a^2-b^2=c^2\implies b^2+c^2=4a^2$
Clearly, $b,c$ are of same parity.
If $b,c$ are odd $=2C+1,2D+1$ respectively, then $b^2+c^2=(2C+1)^2+(2D+1)^2$
$=4(C^2+D^2+C+D)+2\equiv 2\pmod 4$,not multiple of $4$.
$\implies b,c$ are even
Let $b=2B,c=2C$
So, $B^2+C^2=a^2$
Area of the isosceles triangle becomes $B\sqrt{a^2-B^2}=BC$
The parametric solution of $B^2+C^2=a^2$ is $k(p^2-q^2), 2pqk, k(p^2+q^2)$ where $p,q,k$ are any non-negative integers and $p>q$ .
So, the general solutions are $a=k(p^2+q^2), b=2B=2k(p^2-q^2),c=2C=2(2pqk)=4pqk$
or $a=k(p^2+q^2), b=2B=4pqk,c=2C=2k(p^2-q^2)$, the area being $pq(p^2-q^2)$, in either case.
If $k=1, p=2,q=1,a=2^2+1^2=5,b=2(2^2-1^2)=6$ or $b=4\cdot 2\cdot 1=8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/198034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Showing vectors span a vector space by definition I need to show that the vectors $v_1 = \langle 2, 1\rangle$ and $v_2 = \langle 4, 3\rangle$ span $\mathbb R^2$ by definition. By definition if I can write any vector in $\mathbb R^2$ as a linear combination of $v_1$ and $v_2$ then the vectors span $\mathbb R^2$. How do I show this? Here is what I have been working with:
*
*Let $v_x = \langle c_1, c_2\rangle$ be any vector in $\mathbb R^2$ where $c_1$ and $c_2$ are in $\mathbb R$.
*$v_x = c_1\langle 1, 0\rangle + c_2\langle 0, 1\rangle$
*Set $v_x$ = a linear combination of $v_1$ and $v_2$? How do I proceed from here?
| We want to show that any $\mathbf{v} = (x,y) \in \mathbb{R}^2$ can be written as $v = a\mathbf{v_1}+b\mathbf{v_2}$. This equation can be written more explicitly like this: $\begin{pmatrix}x \\ y\end{pmatrix} = a\begin{pmatrix}2 \\ 1\end{pmatrix} + b\begin{pmatrix}4 \\ 3\end{pmatrix}$, because $\mathbf{v_1} = \begin{pmatrix}2 \\ 1\end{pmatrix}$ and $\mathbf{v_2} = \begin{pmatrix}4 \\ 3\end{pmatrix}$. We'd like to show that no matter what $(x,y)$ is, there always are $a,b$ that satisfy the above equation.
So let's do that. This equation is a vector equation, so it is really two equations. We can separate it:
$$
\begin{align}
2a+4b &= x \\
a+3b &= y \end{align}
$$
Remember that the unknowns here are $a$ and $b$. $x$ and $y$ can be anything; in fact we want to show that no matter what $x$ and $y$ are, there are $a$ and $b$ that satisfy these equations. If we say $A = \begin{pmatrix}2 & 4 \\ 1 & 3\end{pmatrix},\ \mathbf{w} = \begin{pmatrix}a \\ b \end{pmatrix},\ \mathbf{x} = \begin{pmatrix}x \\ y \end{pmatrix}$. This system of equations can be written more succintly as $A\mathbf{w} = \mathbf{x}$. Again, remember that the unknown is $\mathbf{w}$, not $\mathbf{x}$.
We know that this system has a solution $\mathbf{w}$ for every $\mathbf{x}$ if and only if $\det(A) \neq 0$. If we can show that, we have shown that no matter which $(x,y) \in \mathbb{R}^2$ we pick, there are $a,b$ such that $\begin{pmatrix}x \\ y\end{pmatrix} = a\begin{pmatrix}2 \\ 1\end{pmatrix} + b\begin{pmatrix}4 \\ 3\end{pmatrix}$. I leave this calculation to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/199441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
} |
Completing the square with negative x coefficients I know how to complete the square with positive $x$ coefficients but how do you complete the square with negative $x$ coefficients?
For example:
\begin{align*}
f(x) & = x^2 + 6x + 11 \\
& = (x^2 + 6x) + 11 \\
& = (x^2 + 6x + \mathbf{9}) + 11 - \mathbf{9} \\
& = (x+3)^2 + 2.
\end{align*}
For positive you would add 9 inside the parenthesis and subtract 9 outside. However, I get the wrong answer when I do it with a negative coefficient. Do you do the same thing when there is a negative coefficient in front of $x^2$ or is it the other way around? (subtract inside parenthesis and add outside of parenthesis).
Here is an example of a negative coefficient:
$$f(x) = -3x^2 + 5x + 1.$$
I tried to solve this and entered in my answer online but it was wrong.
| $
\begin{split}
f(x) &= -3(x^2-5x/3 - 1/3)\\
&= -3( x^2 - 5x/3 + (5/6)^2 - (5/6)^2 - 1/3)\\
&= -3( (x - 5/6)^2 - 25/36 - 12/36) \\
&= -3(x - 5/6)^2 -3(- 25/36 - 12/36) \\
&= -3(x - 5/6)^2 + 37/12.
\end{split}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/199970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What is the remainder when $4^{100}$ is divided by 6? I am trying to find the remainder when $4^{96}$ is divided by 6.
SO using the cyclicity method,
Dividing $4^1$ by 6 gives remainder 4.
Dividing $4^2$ by 6 gives remainder 4.
Dividing $4^3$ by 6 gives remainder 4.
Dividing $4^4$ by 6 gives remainder 4.
Dividing $4^5$ by 6 gives remainder 4.
..
..
So by this method we get the answer as 4.
But when we reduces this to
$$ 4^{100} /6 $$ to
$$ 2^{200} /6 $$
which is equal to
$$ 2^{199} /3 $$
Then by using the cyclicity method:
Dividing $2^1$ by 3 gives remainder 2.
Dividing $2^2$ by 3 gives remainder 1.
Dividing $2^3$ by 3 gives remainder 2.
Dividing $2^4$ by 3 gives remainder 1.
Dividing $2^5$ by 3 gives remainder 2.
..
..
So accrding to this ,We get the answer as 2.
So which one is correct?
And how we are getting two different answers for the same numbers?
Thanks in advance.
| The first method is correct. Since $4\times 4 \equiv 4 \mod 6$, you can conclude $4^n \equiv 4 \mod 6$ for $n \ge 1$.
The second method is wrong. A simple example is that it suggests the remainder when $8$ is divided by $6$ is the same as the remainder when $4$ is divided by $3$.
If you know that $$2^{2n-1} \equiv 2 \mod 3$$ and $$2^{2n} \equiv 1 \mod 3,$$ then modulo $6$ you can conclude $$2^{2n-1} \equiv 2 \text{ or } 5 \mod 6$$ and $$2^{2n} \equiv 1 \text{ or } 4 \mod 6.$$ You want the latter expression. You can similarly say that since $2^m \equiv 0 \mod 2$ you can conclude $$2^m \equiv 0 \text{ or } 2 \text{ or } 4 \mod 6.$$ Put those two together and you get $$2^{2n} \equiv 4 \mod 6$$ which is the first result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/200618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$ Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!
| Changing into polar form we have $ 1+ i \sqrt{3} = 2 e^{i\pi/3}$ and $1 - i \sqrt{3} = 2e^{-i\pi/3}$ so the left hand side is $$ \sqrt{2} \left( e^{i\pi/6} + e^{-i\pi/6} \right)= 2 \sqrt{2} \cos(\pi/6)= 2\sqrt{2} \cdot \frac{\sqrt{3}}{2}= \sqrt{6}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/203462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 3
} |
How to solve this recurrence relation? $f_n = 3f_{n-1} + 12(-1)^n$ How to solve this particular recurrence relation ?
$$f_n = 3f_{n-1} + 12(-1)^n,\quad f_1 = 0$$
such that $f_2 = 12, f_3 = 24$ and so on.
I tried out a lot but due to $(-1)^n$ I am not able to solve this recurrence?
Any help will be highly appreciated.
Please this is no homework. I came across this while solving a problem on SPOJ
| $f(1)=0$; $f(n)=3f(n-1)+12(-1)^n=3\Big(3f(n-2)+12(-1)^{n-1}\Big)+12(-1)^n$
For $n\ge 3$ you have
$$\begin{align*}
f(n)&=3f(n-1)+12(-1)^n\\
&=3\Big(3f(n-2)+12(-1)^{n-1}\Big)+12(-1)^n\\
&=9f(n-2)+36(-1)^{n-1}+12(-1)^n\\
&=9f(n-2)+12(-1)^{n-1}\Big(3+(-1)\Big)\\
&=9f(n-2)+24(-1)^{n-1}\\
&=9f(n-2)-24(-1)^n\\
&=\begin{cases}9f(n-2)-24,&\text{if }n\text{ is even}\\
9f(n-2)+24,&\text{if }n\text{ is odd}\;.\end{cases}
\end{align*}$$
Now you can solve separately for the even and odd subsequences.
Let $a_0=0$ and $a_{n+1}=9a_n+24$ for $n\ge 0$. Substitute $b_n=a_n-d$ for some as yet unknown $d$, so that $a_n=b_n+d$, and the recurrence becomes $b_{n+1}+d=9(b_n+d)+24=9b_n+9d+24$, or $b_{n+1}=9b_n+8d+24$. If we set $d=-3$, this becomes $b_{n+1}=9b_n$, a simple exponential sequence whose solution has the closed form $b_n=9^nb_0$. And since $b_0=a_0-(-3)=3$, we immediately find the closed form $b_n=3\cdot9^n=3^{2n+1}$, whence $a_n=b_n+d=3^{2n+1}-3$. Now check that $f(2n+1)=a_n$, and you’ll see that $f(2n+1)=3^{2n+1}-3$. In other words, $f(n)=3^n-3$ when $n$ is odd.
You can handle the even subsequence similarly. You find that $d=3$, $b_n=9^{n-1}b_1=9^n$ (since $b_1=a_1-3=9$), and $a_n=9^n+3$. But $a_n=f(2n)$, so $f(2n)=9^n+3=3^{2n}+3$, i.e., $f(n)=3^n+3$ when $n$ is even. Combining the two, we have $f(n)=3^n-3(-1)^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/205372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 1
} |
Real jordan form to complex jordan form then compute P matrix. I have the matrix
$$A =
\begin{bmatrix}
5 & 0 & 1 & 0 & 0 & -6 \\
3 & -1 & 3 & 1 & 0 & -6 \\
6 & -6 & 5 & 0 & 1 & -6 \\
7 & -7 & 4 & -2 & 4 & -7 \\
6 & -6 & 6 & -6 & 5 & -6 \\
2 & 1 & 0 & 0 & 0 & 0
\end{bmatrix}$$
This can be brought in the following Jordan form, i.e. $A = TJT^{-1}$.
$$J =
\begin{bmatrix}
2-3j & 1 & 0 & 0 & 0 & 0 \\
0 & 2-3j & 1 & 0 & 0 & 0 \\
0 & 0 & 2-3j & 0 & 0 & 0 \\
0 & 0 & 0 & 2+3j & 1 & 0 \\
0 & 0 & 0 & 0 & 2+3j & 1 \\
0 & 0 & 0 & 0 & 0 & 2+3j
\end{bmatrix}$$
$$T =
\begin{bmatrix}
2j & 2j & 1+j & -2j & -2j & 1-j \\
1+j & 2j & j & 1-j & -2j & -j \\
0 & 2j & 2j & 0 & -2j & -2j \\
0 & 1+j & 2j & 0 & 1-j & -2j \\
0 & 0 & 2j & 0 & 0 & -2j \\
-1+j & -1+j & j & -1-j & -1-j & -j
\end{bmatrix}
$$
Now I have to bring A into its real Jordan form. This is easy:
$$J^{R} =
\begin{bmatrix}
2 & 3 & 1 & 0 & 0 & 0 \\
-3 & 2 & 0 & 1 & 0 & 0 \\
0 & 0 & 2 & 3 & 1 & 0 \\
0 & 0 & -3 & 2 & 0 & 1 \\
0 & 0 & 0 & 0 & 2 & 3 \\
0 & 0 & 0 & 0 & -3 & 2
\end{bmatrix}
$$
Now I have to compute $V$ such that $VJ^RV^{-1} = A$. My question now is how do I compute this $V$? For real valued jordan forms this is easy. I just have to compute the eigenvectors of A, $\{T_i\}$ and then $T = \begin{bmatrix} T_1 | T_2 | \ldots | T_n\end{bmatrix}$.
| Notice that, with every complex pair of eigenvalues $\lambda = a \pm ib$, there exists a complex pair of eigenvectors $u \pm i v$. If you look at the columns of your matrix $T$, you can observe that you can pair up your eigenvectors according to complex conjugates in this precise way.
In real canonical form, each of your real Jordan blocks $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ comes directly from the eigenspaces generated by the eigenvectors $u \pm iv$. So your $V$ should look like
$$V = [v_1 | u_1 | v_2 | u_2 | v_3 | u_3]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/207397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Unsure about applying series comparison test Does this converge or diverge?
$$
\sum\limits_{n=1}^\infty (a_{n} = \frac{1}{2\sqrt{n} + \sqrt[3]{n}})
$$
The answer is: diverges by limit comparison to $\sum (b_{n} = \frac{1}{\sqrt{n}})$
If I look at $\lim_{n \to \infty}\frac{a_{n}}{b_{n}}$ I get
$$
\frac{\sqrt{n}}{2\sqrt{n} + \sqrt[3]{n}} = \frac{n^\frac{1}{2}} {2n^{\frac{1}{2}} + n^{\frac{1}{2}}n^\frac{2}{3}} = \frac{1}{2 + n^\frac{2}{3}} = \frac{0}{0 + 1} = 0
$$ as $n\to\infty$
I would expect it to be some $c > 0$ or $\infty$
Also, because $a_{n} < b_{n}$ I don't think I can use the comparison test. I'm pretty sure I am missing something but not sure what.
Thanks
Update:
My algebra was wrong, how is this for finding the limit?
$$
\frac{\sqrt{n}}{2\sqrt{n} + \sqrt[3]{n}} = \frac{1} {2 + \frac{1}{n^\frac{1}{6}}} = \frac{1} {\frac{2\sqrt[6]{n} + 1}{\sqrt[6]{n}}} = \frac{\sqrt[6]{n}}{2\sqrt[6]{n} + 1} = \frac{1}{2 + 0} = \frac{1}{2}
$$ as $n \to\infty$
And since $\frac{1}{2} > 0$ and $b_{n}$ diverges, $a_{n}$ diverges
| $$\frac{\sqrt{n}}{2\sqrt{n}+\sqrt[3]{n}}=\frac{\sqrt{n}}{\sqrt{n}\left(2+\frac{\sqrt[3]{n}}{\sqrt{n}}\right)}=\frac{1}{2+\frac{\sqrt[3]{n}}{\sqrt{n}}}\to\frac12,\text{ as $n\to\infty$}$$
because
$$\frac{\sqrt[3]{n}}{\sqrt{n}} = n^{1/3}/n^{1/2} = n^{1/3-1/2}=n^{-1/6}$$
An other argument goes like this $$\sqrt[3]{n}\leq\sqrt{n},$$
so that $$2\sqrt{n}+\sqrt[3]{n}\leq3\sqrt{n} $$
and then $$\frac{\sqrt{n}}{2\sqrt{n}+\sqrt[3]{n}}\geq\frac{\sqrt{n}}{3\sqrt{n}}>1/5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/207663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Use $\epsilon\ -\ \delta$ definition to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$ It's known that $\lim\limits_{x \rightarrow x_0}f(x) = A$, how to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$?
Here's what I've got now:
When $A = 0$, to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = 0$: Since we have $\lim\limits_{x \rightarrow x_0}f(x) = A = 0$, so $|f(x)| < \epsilon$. => $|\sqrt[3]{f(x)}| < \epsilon_0^3 < \epsilon$
When $A \ne 0$, $|\sqrt[3]{f(x)} - \sqrt[3]{A}| = \frac{|f(x) - A|}{|f(x)^{\frac{2}{3}}+(f(x)A)^{\frac{1}{3}} + A^{\frac{2}{3}}|}$...
How can I deal with $(f(x)A)^{\frac{1}{3}}$? Thanks.
| Hint: For $A\neq0$,
$$\left|\sqrt[3]{f(x)}-\sqrt[3]{A}\right|
=\frac{|f(x)-A|}{|f(x)^{\frac{2}{3}}+(f(x)A)^{\frac{1}{3}}+A^{\frac{2}{3}}|}
=\frac{|f(x)-A|}{\bigg(\sqrt[3]{f(x)}+\frac12\sqrt[3]{A}\bigg)^2+A^{\frac{2}{3}}}\leq\frac{|f(x)-A|}{A^{\frac{2}{3}}}.$$
Or just use the inequality:
$$\left|\sqrt[3]{f(x)}-\sqrt[3]{A}\right|\leq\sqrt[3]{|f(x)-A|}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/213386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving trigonometric equations of the form $a\sin x + b\cos x = c$ Suppose that there is a trigonometric equation of the form $a\sin x + b\cos x = c$, where $a,b,c$ are real and $0 < x < 2\pi$. An example equation would go the following: $\sqrt{3}\sin x + \cos x = 2$ where $0<x<2\pi$.
How do you solve this equation without using the method that moves $b\cos x$ to the right side and squaring left and right sides of the equation?
And how does solving $\sqrt{3}\sin x + \cos x = 2$ equal to solving $\sin (x+ \frac{\pi}{6}) = 1$
| I'm assuming $ab\neq 0$, since otherwise trivial. $a\sin x+b\cos x-c=0$
$$\iff a\left(2\sin\frac{x}{2}\cos\frac{x}{2}\right)+b\left(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\right)-c\left(\sin^2\frac{x}{2} +\cos^2\frac{x}{2}\right)=0$$
*
*Assume $\cos\frac{x}{2}\neq 0$. Then
$$\stackrel{:\cos^2 \frac{x}{2}\neq 0}\iff (b+c)\tan^2\frac{x}{2}-2a\tan\frac{x}{2}+(c-b)=0$$
If $b+c\neq 0$, then
$$\iff \tan\frac{x}{2}=\frac{a\pm\sqrt{a^2+b^2-c^2}}{b+c}$$
$$\iff x=2\left(\arctan\left(\frac{a\pm\sqrt{a^2+b^2-c^2}}{b+c}\right)+n\pi\right),\, n\in\Bbb Z$$
Real solutions exist iff $a^2+b^2\ge c^2$.
If $b+c=0$, then
$$\iff \tan\frac{x}{2}=\frac{c-b}{2a}\iff x=2\left(\arctan\frac{c-b}{2a}+n\pi\right),\, n\in\Bbb Z$$
*
*Assume $\cos\frac{x}{2}=0$. Equality holds iff $b+c=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 4
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Proving trigonometric Identity: $\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$ I would like to try and prove
$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$$
using $LHS=RHS$ methods, i.e. pick a side and rewrite it to make it identical to the other side.
I found a quick way by doing this:
$$LHS = \frac{1+\sin x}{\cos x} = \frac{1+\sin x}{\cos x} \cdot \frac{1 - \tan x + \sec x}{1 - \tan x + \sec x}= \frac{1+\sin x+\cos x}{1-\sin x+\cos x} = RHS$$
but I feel that this is not a good way because I am manipulating the denominator of the LHS somewhat artificially, because I know it must be, in the end, $1-\sin x+\cos x$.
Does anyone have a better way of doing this?
| $\frac{1+\sin x}{\cos x}$
$=\frac{(1+\sin x)(1-\sin x+\cos x)}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{1-\sin^2 x+\cos x+\cos x\sin x}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{\cos^2 x+\cos x+\cos x\sin x}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{(\cos x)(1+\sin x+\cos x)}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/213788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Prove that the sequence is convergent How can we show that the sequence $$a_n=\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3-n^2}$$ is convergent?
| $a_n =\sqrt[3]{n^3+n^2} - \sqrt[3]{n^3-n^2}$, so when $n$ is large enough,
$$
\begin{align}
a_n &= n\cdot \sqrt[3]{1+\frac{1}{n}} - n\cdot \sqrt[3]{1 - \frac{1}{n}} \\
&= n\cdot \left(1+\frac{1}{3n} - \frac{1}{9n^2} + \mathcal{O}(n^{-3})\right) - n\cdot \left(1 - \frac{1}{3n} - \frac{1}{9n^2} + \mathcal{O}(n^{-3})\right) \\
&= \frac23 + \mathcal{O}(n^{-2})
\end{align}
$$
It seems that $a_n$ converges to $\frac23$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/220415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Try to solve root in inequality got wrong result I am very confused. So I have to solve this inequality. The result is $13/24$.
But if I try to solve it myself, I get $17/24$. Because:
$$\sqrt{\left(\frac{-5}{24}\right)^2 + \frac{1}{4}} = \frac{5}{24} + \frac{1}{2} = \frac{5}{24} + \frac{12}{24} = \frac{17}{24}.$$
The right solution should be $13/24$.
Where is my failure?
| $\sqrt{(-5/24)^2+1/4} = \sqrt{25/24^2+1/4} = \sqrt{(25+6\cdot 24)/24^2} = \sqrt{169}/\sqrt{24^2} = 13/24$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/220849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Iterated exponentials in modular arithmetic: $a^{b^{c^d}}-a^{b^c}\equiv 0 \mod 2^{2^{2^2}}-2^{2^2}$? How to prove $a^{b^{c^d}}-a^{b^c}\equiv 0 \mod 2^{2^{2^2}}-2^{2^2}$ for all positive integers $a,b,c,d\in \mathbb{N}$? I tried to reduce $d$ to $2$ by induction. But I kind of stuck at the rest of the parameters. Any thoughts?
| Clearly, this is true if at least one of $b,c,d$ is $1$, so we can safely focus
on $b,c,d>1$.
$2^{2^{2^2}}-2^{2^2}=2^{16}-2^4=2^4(2^{12}-1)=2^4(2^6-1)(2^6+1)=2^4\cdot7\cdot 9\cdot 13\cdot 5$
We know,for any prime $p$ either $p\mid a$ or $(p,a)=1$
So, if $p\mid a,p\mid(a^e-a)$ for $e\ge 1$ else $(p,a)=1,$
(i)For $p=13,\phi(13)=12$
if $13\mid a,13\mid(a^{b^{c^d}}-a^{b^c})$
if $(13,a)=1$ it is sufficient to prove $12\mid(b^{c^d}-b^c)$
Now, $b^{c^d}-b^c=b^c(b^{c^d-c}-1)=b^c(b^{c(c^{d-1}-1)}-1)$
As $c,d>1$ so, $b^2(b^2-1)\mid b^c(b^{c(c^{d-1}-1)}-1)$
Now, if $2\mid b,2^2\mid b^2$ else $b$ is odd$=2c+1$(say),$b^2-1=(2c+1)^2-1=8\frac{c(c+1)}2$
and $3\mid (b-1)b(b+1)$.
So, $12\mid b^2(b^2-1)\implies 12\mid b^c(b^{c(c^{d-1}-1)}-1)$
(ii) For $p=7,\phi(7)=6$
if $7\mid a, 7\mid(a^{b^{c^d}}-a^{b^c})$
if $(7,a)=1,$ it is sufficient to prove $6\mid(b^{c^d}-b^c)$
(iii) For $p=5,\phi(5)=4$
if $5\mid a, 5\mid(a^{b^{c^d}}-a^{b^c})$
if $(5,a)=1,$ it is sufficient to prove $4\mid(b^{c^d}-b^c)$
$2^4$ and $9=3^2$ need different approach as they are powers$(>1)$ of primes.
(iv)For $2^4,$
If $2\mid a,$ both $b^c$ and $b^{c^d}\ge 4$ as $b,c,d>1$,
hence, both $a^{b^c}$ and $a^{b^{c^d}}$ are divisible by $2^4$
If $(2,a)=1,\lambda(16)=\frac{\phi(16)}2=4,$ it is sufficient to prove $4\mid(b^{c^d}-b^c)$ where $\lambda$ is the Carmichael Function.
(v)For $9,$
If $3\mid a,$ both $b^c$ and $b^{c^d}\ge 4$ as $b,c,d>1$,
hence, both $a^{b^c}$ and $a^{b^{c^d}}$ are divisible by $3^4,$ hence by $3^2=9$
If $(3,a)=1,\phi(9)=6,$ it is sufficient to prove $6\mid(b^{c^d}-b^c)$
So, in all the cases, $a^{b^{c^d}}-a^{b^c}$ is divisible by $2^4, 7, 9, 13, 5;$ hence by $lcm(2^4, 7, 9, 13, 5)=2^4\cdot7\cdot 9\cdot 13\cdot 5=2^{2^{2^2}}-2^{2^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many strings of length $n$... Let $n$ be a positive integer. If each $a_i$ is chosen from $\left\{ 0,1,2, \dots, 9 \right\},\;$ $1\leq i\leq n$, then how many strings of length $n$ ($a_1a_2\dots a_n$) are there such that the number of occurrences of $0$ is even? Thanks!
| Let us fix the alphabet as $\{0, 1, 2, \dots, M\}$ (we'll later take $M = n$). Let $a_n$ be the number of strings of length $n$ which contain an even number of $0$s, and $b_n$ be the number of strings of length $n$ which contain an odd number of $0$s. Note that there are $(M+1)^n$ strings of length $n$ over that alphabet, so $a_n + b_n = (M+1)^n$. For $n = 0$, we can see that $a_0 = 1$ (the empty string) and $b_0 = 0$.
We can write down recurrence relations for $a_n$ and $b_n$. If we consider how we can get a string of length $n+1$ by appending a symbol (either $0$ or one of $1, \dots, M$) to a string of length $n$, we can see that:
$$\begin{align}
a_{n+1} &= Ma_n + b_n\\
b_{n+1} &= a_n + Mba_n
\end{align}$$
Writing $X_n = \pmatrix{a_n \\ b_n}$, this means
$$\begin{align}
X_0 &= \pmatrix{1 \\ 0} \\
X_{n+1} &= \pmatrix{M & 1 \\ 1 & M}X_n
\end{align}$$
Calling the matrix $T = \pmatrix{M & 1 \\ 1 & M}$, this means $X_n = T^nX_0$. You can verify that
$$T^n = \frac12 \pmatrix{(M-1)^n+(M+1)^n & (M+1)^n-(M-1)^n \\ (M+1)^n-(M-1)^n & (M-1)^n+(M+1)^n}$$
which gives $\pmatrix{a_n \\ b_n} = X_n = T^nX_0 = T^n\pmatrix{1 \\ 0} = \frac12 \pmatrix{(M-1)^n+(M+1)^n \\ (M+1)^n-(M-1)^n}$ and in particular $$a_n = \frac{(M-1)^n+(M+1)^n}{2}$$ or when $M = n$, $$a_n = \frac{(n-1)^n+(n+1)^n}{2}.$$
Alternatively, as $b_n = (M+1)^n - a_n$, the recurrence $a_{n+1} = Ma_n + b_n$ is the same as $a_{n+1} = Ma_n + ((M+1)^n - a_n) = (M-1)a_n + (M+1)^n$. Here, letting $c_n = 2a_n - (M+1)^n$, note that the recurrence becomes:
$$\begin{align}
a_{n+1} &= (M-1)a_n + (M+1)^n \\
2a_{n+1} - (M+1)^{n+1}
&= 2(M-1)a_n + 2(M+1)^n - (M+1)^{n+1} \\
&= 2(M-1)a_n + (M+1)^n(2 - (M+1)) \\
&= 2(M-1)a_n - (M+1)^n(M-1) \\
&= (M-1)(2a_n - (M+1)^n) \\
c_{n+1} &= (M-1)c_n
\end{align}$$
and $c_0 = 2a_0 - 1 = 2 - 1 = 1$, so $c_n = (M-1)^n$, which gives
$a_n = ((M-1)^n + (M+1)^n)/2$ as before.
Of course, there is also probably an easy combinatorial proof showing why $a_n = ((n-1)^n+(n+1)^n)/2$; I can't think of one right now.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{1}{{4n^2 - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$ Could you explain the operation in the third step?
$$\frac{1}{{4n^2 - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$$
It comes from the sumation $$\sum_{n=1}^\infty\frac1{4n^2-1}$$
I could just copy it to my homework, but I'd like to know how this conversion works.
Thanks in advance.
| The motivation is to write $\dfrac1{(2n+1)(2n-1)}$ as $\dfrac{A}{(2n-1)} + \dfrac{B}{(2n+1)}$. Hence, we need $A$ and $B$ such that
$$\dfrac1{(2n+1)(2n-1)} = \dfrac{A}{(2n-1)} + \dfrac{B}{(2n+1)}$$
Simplifying the right hand side, we get that
$$\dfrac1{(2n+1)(2n-1)} = \dfrac{A(2n+1) + B(2n-1)}{(2n+1)(2n-1)} = \dfrac{2(A+B)n + (A-B)}{(2n+1)(2n-1)}$$
Hence, we need $2(A+B)n + (A-B) = 1$ for all $n$. Hence, matching the coefficient of $n$ and the constant term we get that $A-B = 1$ and $2(A+B) = 0$.
This gives us $B = -\dfrac12$ and $A = \dfrac12$. Hence,
$$\dfrac1{(2n+1)(2n-1)}=\dfrac12 \dfrac1{(2n-1)} - \dfrac12 \dfrac1{(2n+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/232600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Simple linear recursion $x_n=\frac{x_{n-1}}{a}+\frac{b}{a}$ with $a>1, b>0$ and $x_0>0$
I tried to solve it using the generating function but it does not work because of $\frac{b}{a}$, so may you have an idea.
| Using generating functions here is easy. Write your recurrence as:
$\begin{equation*}
a x_{n + 1}
= x_n + b
\end{equation*}$
Define the generating function $g(z) = \sum_{n \ge 0} x_n z^n$, multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize some sums:
$\begin{align*}
\sum_{n \ge 0} x_{n + 1} z^n
&= \sum_{n \ge 0} x_n z^n + \sum_{n \ge 0} b z^n \\
\frac{g(z) - x_0}{z}
&= g(z) + \frac{b}{1 - z}
\end{align*}$
Solve for $g(z)$, write as partial fractions if needed:
$\begin{equation*}
g(z)
= \frac{x_0 - (x_0 - n) z}{(1 - z)^2}
\end{equation*}$
Now you want the coefficient of $z^n$ of the above:
$\begin{align*}
[z^n] g(z)
&= [z^n] \frac{x_0 - (x_0 - n) z}{(1 - z)^2} \\
&= x_0 [z^n] (1 - z)^{-2} - (x_0 - b) [z^{n - 1}] (1 - z)^{-2} \\
&= x_0 (-1)^n \binom{-2}{n} - (x_0 - b) (-1)^{n - 1} \binom{-2}{n - 1} \\
&= x_0 \binom{n + 2 - 1}{2 - 1} - (x_0 - b) \binom{n - 1 + 2 - 1}{2 - 1} \\
&= 2 x_0 - b + b n
\end{align*}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/233109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limits with trig, log functions and variable exponents Would someone mind verifying this?
$$ \lim_{x\to \infty} \frac{2 \cdot 3^{5x} + 5}{3^{5x} + 2^{5x}}
= \lim_{x\to \infty} \frac{3^{5x}(2 + \frac{5}{3^{5x}})}{3^{5x}(1 + (\frac{2}{3})^{5x})}
= \lim_{x\to \infty} \frac{2 + \frac{5}{3^{5x}}}{1 + (\frac{2}{3})^{5x}}
= \frac{2 + \frac{5}{3^{5(\infty)}}}{1 + (\frac{2}{3})^{5(\infty)}}
= \frac{2 + 0}{1 + 0}
= 2 $$
$$ \lim_{x\to 0} \frac{e^{2x}-\pi^{x}}{sin(3x)}
= \lim_{x\to 0} \frac{2e^{2x}-\pi^{x} ln(\pi)}{3cos(3x)}
= \frac{2e^{2(0)}-\pi^{(0)} ln(\pi)}{3cos(3(0))}
= \frac{2\cdot 1 - 1\cdot ln(\pi)}{3\cdot1}
= \frac{2-ln(\pi)}{3} $$
| Yes those are both right.
$$\lim_{x\to\infty}\frac{2\cdot 3^{5x}+5}{3^{5x}+2^{5x}}=\lim_{x\to\infty}\frac{2+5\cdot3^{-5x}}{1+\left(3/2\right)^{-5x}}=2$$
and
$$\lim_{x\to0}\frac{e^{2x}-e^{x\ln\pi}}{\sin 3x}=\lim_{x\to0}\frac{(2-\ln\pi)x+O(x^2)}{3x}=\frac{1}{3}(2-\ln\pi).$$
In the second one, you can use that $e^u=1+u+O(u^2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What's the solution to this system of equation? $$
xy^3z^3 = yx^3z^3 = zx^3y^3
$$
Is there a way to solve this system?
I think the answer is 1, but i can't verify my intuition.
| Another way of looking at it, clearly $x=0$, or $y=0$, or $z=0$ is a solution. So, starting with
\begin{alignat*}{3}
xy^3z^3 &= yx^3z^3 &&= zx^3y^3
\end{alignat*}
And considering the case where $x \ne 0$, $y \ne 0$, $z \ne 0$
\begin{alignat*}{3}
y^2z^2 &= x^2z^2 &&= x^2y^2
\end{alignat*}
Considering each pair of equalities, we get:
\begin{align*}
y^2z^2 &= x^2z^2 & x^2z^2 &= x^2y^2 & y^2z^2 &= x^2y^2 \\
\implies y &= \pm x & z &= \pm x & z &= \pm x \\
\implies y &= \pm x & z &= \pm x
\end{align*}
Hence the solutions are points of the form
$(0,a,b)$, $(a,0,b)$, $(a,b,0)$, or $(a, \pm a, \pm a)$
for $a,b \in \mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/237728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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integration by parts from Apostol I working through Apostol's calculus, and I need to prove integrating by parts that :
$\int (a^2 - x^2)^n \,dx = \frac{x (a^2 - x^2)^n}{2n + 1} + \frac{2 a^2 n}{2n+1} \int (a^2 - x^2)^{n-1} \,dx + C $
Now, using the integration by parts formula after first division the integral to parts I arrive at:
$\int (a^2 - x^2)^n \,dx = x(a^2 - x^2)^n + 2n \int x^2 (a^2 - x^2)^{n-1} \,dx $
I could substitute and solve the integral, but I need to do something else. I multiply the first expression on the right by $ \frac{2n + 1}{2n +1}$ which leads to:
$x(a^2 - x^2)^n + 2n \int (a^2 - x^2)^{n-1} \,dx = \frac{x (a^2 - x^2)^n}{2n + 1} + \frac{2nx (a^2 - x^2)^n}{2n + 1} + 2n\int x^2(a^2 - x^2)^{n-1} \,dx$
and I am somewhat close. If I try something else, I end up even closer:
$\int (a^2 - x^2)^n \,dx = \begin{pmatrix}
f(x)= a^2 - x^2 | f'(x)=-2x\\
g'(x)=(a^2-x^2)^{n-1} |g(x)=\int(a^2-x^2)^{n-1} \,dx
\end{pmatrix} = $
$(a^2-x^2)\int(a^2-x^2)^{n-1} \,dx + 2\int x \Big(\int(a^2-x^2)^{n-1} \,dx \Big)\,dx
=$
$(a^2-x^2)\int(a^2-x^2)^{n-1} \,dx +x^2\int(a^2-x^2)^{n-1} \,dx = a^2\int(a^2-x^2)^{n-1} \,dx $
But I think I made a mistake somewhere... could somebody help me out? I'm really stuck!
Thanks!
| NOTE: You can only pull the constant part out of an integral, but not your variable:
*
*This is good: $\int 3x^3 dx = 3 \int x^3 dx$
*This is BAD: $\int 3x^3 dx = \color{red}x \int 3x^2 dx$
So far, so good:
$\int (a^2 - x^2)^n \,dx = x(a^2 - x^2)^n + 2n \int \color{red}{x^2} (a^2 - x^2)^{n-1} \,dx$
The red part $x^2$ is correct, but should be manipulated somehow to make it disappear. Since it does not show up on the RHS of the original equation.
HINT
$\begin{align} \int (a^2 - x^2)^n \,dx &= x(a^2 - x^2)^n + 2n \int \color{red}{x^2} (a^2 - x^2)^{n-1} \,dx \\
&= x(a^2 - x^2)^n + 2n \int \color{red}{(x^2 - a^2 + a^2)} (a^2 - x^2)^{n-1} \,dx\\
&= x(a^2 - x^2)^n + 2na^2 \int (a^2 - x^2)^{n - 1} + 2n \int (x^2 - a^2) (a^2 - x^2)^{n-1} \,dx \\
&= x(a^2 - x^2)^n + 2na^2 \int (a^2 - x^2)^{n - 1} - 2n \int (a^2 - x^2)^{n} \,dx
\end{align}$
It should be straight forward from here. :)
| {
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"source": "stackexchange",
"question_score": "3",
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Solve the recurrence $T(n) = 2T(n-1) + n$ Solve the recurrence $T(n) = 2T(n-1) + n$
where $T(1) = 1$ and $n\ge 2$.
The final answer is $2^{n+1}-n-2$
Can anyone arrive at the solution?
| Let's use the method of annihilators to turn this into a third-order, homogeneous recurrence, and then solve with a characteristic equation. First, we write the recurrence so $n$ is the least index:
$$T(n) - 2T(n-1)= n \implies T(n+1)-2T(n) = n+1$$
Then, we rewrite the recurrence in terms of the shift operator $E$:
$$(E-2)T(n) = n+1$$
Applying the $(E-1)$ operator to both sides of the equation, we have:
$$(E-1)^2(E-2)T(n) = (E-1)^2(n+1)$$
Now, since $(E-1)^2(n+1) = 0$ (it annihilates that term), we have:
$$(E-1)^2(E-2)T(n) = 0$$
The characteristic equation is, then $(r-1)^2(r-2)=0$, so our roots are $r=1$ (with multiplicity $2$) and $r=2$. So, the recurrence has the form:
\begin{align*}
T(n) &= c_1\cdot 2^n + c_2 \cdot n1^n + c_3 \cdot 1^n\\
&= c_1\cdot 2^n + nc_2 + c_3
\end{align*}
Using the recurrence relation, we compute $T(1) = 1, T(2) = 4, T(3) = 11$ and can now solve for $(c_1, c_2, c_3)$, which gives the solution $(2, -1, -2)$.
| {
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Random variable $Z = |X-Y|$ Let's take two independent, random variables $X$ and $Y$ from set $\{1,\ldots,n\}$
probability, that $X =i$ AND $Y = j$ is $\frac{1}{n^2}$.
I want to determine the expected value, so I started:
$$E[Z] = E[|X-Y|] = \sum_i (a_i) \cdot P(|X-Y|=a_i),$$ (sum is to $n$) but I don't now how to count $P(|X-Y|=a_i)$, cause some values values are repeated. Can anybody give a prompt?
| Draw the $n\times n$ grid of points with coordinates $(i,j)$, where $i$ and $j$ range from $1$ to $n$. The points $(x,y)$
such that $x-y=k$ are the points on the line $x-y=k$. For $k=0$ there are $n$ such points, the points on the diagonal. For $x-y=1$, there are $n-1$ points, for $x-y=2$ there are $n-2$ points, and so on. For positive $k$, we are dealing with the points of the grid which are below the diagonal line $y=x$.
Similarly, there are $n-1$ points such that $x-y=-1$, $n-2$ points such that $x-y=-2$, and so on. These are the points of our grid above the diagonal $y=x$.
So for $2(n-1)$ points, the absolute value of the difference is $1$, for $2(n-2)$ points the absolute value of the difference is $2$, and so on. Finally, there are $2$ points where the absolute value of the difference is $n-1$. In all cases, each point has probability $\frac{1}{n^2}$. It follows that
$$E(|X-Y|)=\frac{2}{n^2}\left[(1)(n-1)+(2)(n-2)+(3)(n-3)+\cdots +(n-1)(n-(n-1)) \right].$$
We next obtain a closed form for the above expression. By expanding, we can see that our sum is equal to
$$\frac{2}{n^2}n\left[1+2+3+\cdots+n-1\right] -2\left[1^2+2^2+3^2+\cdots+(n-1)^2\right].$$
By the usual formula for the sum of the consecutive integers, and the sum of consecutive integers, we get, after some algebra,
$$\frac{(n-1)(n+1)}{3n}.$$
Remark: The standard formulas that were used for the closed form are $1+2+\cdots+k=\frac{k(k+1)}{2}$ and $1^2+2^2+\cdots +k^2=\frac{k(k+1)(2k+1)}{6}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find X given Y in a cubic function. Having asked this question on the math overflow boards one of the contributors suggested this may be a more appropriate forum.
I have a cubic function in the form:
$$y = ax^3 + bx^2 + cx + d$$
...(where a, b, c and d are all known constants e.g -0.3, 3.5, 3.83 and 0 respectively) that produces a curve and allows me to calculate a point on that curve given it's x co-ordinate.
What I'd really appreciate help with is: How could I rewrite this equation to calculate the value of x if I knew the value of y?
I understand that cubic curves of this nature can have multiple (upto 3) values for x for any given y though I do not know how to calculate them, and my specific interest is only in the 'rightmost' x (i.e that with highest positive value).
Through research to date I gather my question relates some what to finding the roots of the cubic equation (where the given Y is specifically 0, so the values of x are where the curve crosses the horizontal axis) but I can't expand my admittedly hazy comprehension to calculating for different y values.
The only additional hint I've been able to glean is that because in my circumstances d is always 0 that I may not be dealing with a 'true' cubic function at all and instead might want to "factor out an 'x' and a quadratic".
I have noted there are similar questions posted here but these deal specifically with cubic bezier curves, where the form of the equation deals with t (time) based coefficients and the answers refer to expression of the curve in terms of control points rather then the a, b, c, d form.
Give me anything but mathematics to Google and I'm usually able to work something out for myself but in the sea of coefficients and polynomials I'm afraid I'm lost.
Thank you in adavnce for any pointers, John.
| As your hint says, it is easy to solve your particular equation when $y=0$ as you have $$-0.3 x^3+0.5 x^2 + 3.83x =0$$ so either $x=0$ or $-0.3 x^2+0.5 x + 3.83 =0$ and you can solve the latter as a quadratic.
It is usually not so easy when $y \not = 0$ or in general for a cubic. Wikipedia gives the general (sometimes complex) form of the roots
$$\begin{align}
x_1 =
&-\frac{b}{3 a}\\
&-\frac{1}{3 a} \sqrt[3]{\tfrac12\left[2 b^3-9 a b c+27 a^2 d+\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
&-\frac{1}{3 a} \sqrt[3]{\tfrac12\left[2 b^3-9 a b c+27 a^2 d-\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
x_2 =
&-\frac{b}{3 a}\\
&+\frac{1+i \sqrt{3}}{6 a} \sqrt[3]{\tfrac12\left[2 b^3-9 a b c+27 a^2 d+\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
&+\frac{1-i \sqrt{3}}{6 a} \sqrt[3]{\tfrac12\left[2 b^3-9 a b c+27 a^2 d-\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
x_3 =
&-\frac{b}{3 a}\\
&+\frac{1-i \sqrt{3}}{6 a} \sqrt[3]{\tfrac12\left[2 b^3-9 a b c+27 a^2 d+\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
&+\frac{1+i \sqrt{3}}{6 a} \sqrt[3]{\tfrac12\left[2 b^3-9 a b c+27 a^2 d-\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}
\end{align}$$
| {
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Find three polynomials whose squares sum up to $x^4 + y^4 + x^2 + y^2$ Prove that $$p(x,y) = x^4 + y^4 + x^2 + y^2$$ can be written as a sum of squares of three polynomials over $x,y$ for real numbers.
| $(\sqrt{2\sqrt{2}-2}y^2+x)^2 + (-\sqrt{2\sqrt{2}-2}xy + y)^2 + (x^2 + (1-\sqrt{2})y^2)^2 = x^4+y^4+x^2+y^2$.
In fact, there are many other solutions by using brute force, as suggested by Will Jagy in the comments.
Step 1
Let's suppose that $x^4+y^4+x^2+y^2 = \sum_{i=1}^3 (a_i x^2 + b_i xy + c_i y^2 + d_i x + e_i y)^2$. If we write $\vec{a} = (a_1,a_2,a_3)$ and so forth, we obtain the following identities by comparing coefficients:
Part (a):
$x^2: \| \vec{d} \|^2 = 1$,
$y^2: \| \vec{e} \|^2 = 1$,
$xy: \langle \vec{d}, \vec{e} \rangle = 0$.
Part (b):
$x^3: \langle \vec{a}, \vec{d} \rangle = 0$,
$y^3: \langle \vec{c}, \vec{e} \rangle = 0$,
$x^2y: \langle \vec{b}, \vec{d} \rangle = - \langle \vec{a}, \vec{e} \rangle$,
$xy^2: \langle \vec{b} ,\vec{e} \rangle = - \langle \vec{c}, \vec{d} \rangle$.
Part (c):
$x^4: \| \vec{a} \|^2 = 1$,
$y^4: \| \vec{c} \|^2 = 1$,
$x^2y^2: \| \vec{b} \|^2 + 2 \langle \vec{a}, \vec{c} \rangle = 0$,
$x^3y: \langle \vec{b}, \vec{a} \rangle = 0$,
$xy^3: \langle \vec{b}, \vec{c} \rangle = 0$.
(All inner products are standard Euclidean inner products)
Step 2
We consider an orthonormal basis of $\mathbb{R}^3$: $\vec{d}, \vec{e}, \vec{f}$. This used part (a).
Using part (b), we get
$\vec{a} = x_1 \vec{e} + y_1 \vec{f}$,
$\vec{b} = -x_1 \vec{d} - x_2 \vec{e} + y_3 \vec{f}$,
$\vec{c} = x_2 \vec{d} + y_2 \vec{f}$
for some reals $x_1,x_2,y_1,y_2,y_3$.
Step 3
Part (c) tells us that
$x_1^2 + y_1^2 = 1$, $x_2^2 + y_2^2 = 1$. (From norm of $\vec{a}, \vec{c}$ being 1.)
$y_1y_3 = x_1x_2$, $y_2y_3 = x_1x_2$ (From $\vec{b}$ perpendicular to $\vec{a}, \vec{c}$.)
$x_1^2+x_2^2+y_3^2 + 2y_1y_2 = 0$.
In the second equation, if $y_3$ is not 0, then $y_1 = y_2$, which forces $x_1 = x_2 = y_1 = y_2 = y_3 = 0$ in the third equation, contradicting the first equation.
So $y_3 = 0$, which implies $x_1x_2 = 0$. WLOG assume $x_1 = 0$, then $y_1 = \pm 1$. The third equation becomes
$0 = x_2^2 + 2y_1y_2 = 1 - y_2^2 + 2y_1y_2 = 2 - (y_2 - y_1)^2$.
So $y_2 - y_1 = \pm \sqrt{2}$. The first equation tells us that $y_1,y_2$ has norms less than 1, which implies that $y_2 = \pm (1 - \sqrt{2})$. This gives $x_2 = \pm \sqrt{2\sqrt{2} - 2}$.
To summarize, up to symmetry of $(x_1,y_1) \leftrightarrow (x_2,y_2)$, we have $x_1 = 0$, $y_1 = \pm 1$, $y_2 = \pm (1 - \sqrt{2})$ (same sign as $y_1$), $x_2 = \pm \sqrt{2\sqrt{2} - 2}$ and $y_3 = 0$.
Step 4
By picking arbitrary orthonormal basis $\vec{d}, \vec{e}, \vec{f}$, we can substitute the values for $x_1,x_2,y_1,y_2,y_3$ above to get solutions for $\vec{a}, \vec{b}, \vec{c}$. All these would give the required sum of squares decomposition. The decomposition I took above comes from choosing the standard orthonormal basis of $\mathbb{R}^3$.
| {
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How many Arithmetic Progressions having $3$ terms can be made from integers $1$ to $n$? How many Arithmetic Progressions having $3$ terms can be made from integers $1$ to $n$? The numbers in the AP must be distinct.
For example if $n=6$ then number of AP's possible are $6$
*
*$1,2,3$
*$2,3,4$
*$3,4,5$
*$4,5,6$
*$2,4,6$
| One can select any tow distinct elements of the same parity. Together with their mean, they make such a progression.
There are $\left\lfloor \frac n2\right\rfloor$ even numbers $\le n$ and $\left\lfloor \frac{n+1}2\right\rfloor$ odd numbers.
Since we need to select unordered pairs (or ordered? But you seem not to distinguish between 1,2,3 and 3,2,1), the total number is
$$ f(n) = {\left\lfloor \frac n2\right\rfloor\choose 2}+{\left\lfloor \frac {n+1}2\right\rfloor\choose 2}.$$
If $n=2m$ is even, this amounts to
$$ f(n) = 2{m\choose 2}=m(m-1)$$
and if $n=2m+1$ is odd, to
$$ f(n) = {m\choose 2}+{m+1\choose 2} = m^2.$$
Both may also be summarized as
$$ f(n) = \left\lfloor\frac n2\right\rfloor\cdot\left\lfloor\frac{n-1}2\right\rfloor.$$
Example: Letting $n=6$, we obtain $f(n)=\left\lfloor\frac 62\right\rfloor\cdot\left\lfloor\frac52\right\rfloor=3\cdot 2=6$.
| {
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Generating Pythagorean triples for $a^2+b^2=5c^2$? Just trying to figure out a way to generate triples for $a^2+b^2=5c^2$. The wiki article shows how it is done for $a^2+b^2=c^2$ but I am not sure how to extrapolate.
| Here is a cute way to construct a family of solutions for the given diophantine:
Key Theorem: Let $a, b, c, d \in \mathbb Z$. Then, $$(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$$
Proof: Expand.
Consider a pythagorean triple $(a, b, c)$, which you can easily generate. A simple application of the key theorem yields:
$a^2 + b^2 = c^2$, $\implies$ $5(a^2 + b^2) = 5c^2$, so $(2^2 + 1^2)(a^2 + b^2) = 5c^2$ $\implies$ $(2a + b)^2 + (2b - a)^2 = 5c^2$.
Therefore, if $(a,b,c)$ is a pythagorean triple, then $(2a + b, 2b - a, c)$ is a solution of $x^2 + y^2 = 5z^2$.
$\blacksquare$.
| {
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Is there an easy way of finding the taylor series for $1/(1+x^2)$? I was trying to calculate the fourth derivative and then I just gave up.
| Recall that a geometric series can be represented as a sum by
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots = \sum_{n=0}^{\infty}x^n \quad \quad|x| <1$$
Then we can simply manipulate our equation into that familiar format to get
$$\frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum_{n=0}^{\infty}(-x^2)^{n} = \sum_{n=0}^{\infty}(-1)^n x^{2n}$$
Fun Alternative:
Note that $$\frac{d}{dx} \arctan x = \frac{1}{1+x^2}$$
and that the Taylor Series for $\arctan x$ is
$$\arctan x = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$$
$$\implies \frac{d}{dx} \arctan x = \frac{d}{dx}\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$$
Interchange the sum and $\frac{d}{dx}$ (differentiate term-by-term) on the RHS to get
\begin{eqnarray*}
\frac{1}{1+x^2} &=& \sum_{n=0}^{\infty}\frac{d}{dx}(-1)^n\frac{x^{2n+1}}{2n+1}\\
&=& (-1)^n \frac{1}{2n+1}\sum_{n=0}^{\infty} \frac{d}{dx}x^{2n+1}\\
&=& (-1)^n \frac{1}{2n+1}\sum_{n=0}^{\infty} (2n+1)x^{2n}\\
&=& \sum_{n=0}^{\infty} (-1)^n\frac{(2n+1)x^{2n}}{2n+1}\\
\end{eqnarray*}
Cancel the $(2n+1)$ on the RHS to arrive at
$$\frac{1}{1+x^2} = \sum_{n=0}^{\infty}(-1)^n x^{2n}$$
| {
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Why does $\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}$? As much as it embarasses me to say it, but I always had a hard time understanding the following equality:
$$
\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}
$$
I always thought that the left-hand side of the above equation was equivalent to
$$
\frac{a}{\frac{b}{x}} = \frac{a}{b} \div \frac{x}{1} = \frac{a}{b} \times \frac{1}{x}
$$
What am I doing wrong, here?
| $\frac{a}{b}\times\frac{1}{x}=\frac{a}{b}\div x=\frac{\frac{a}{b}}{x}\neq\frac{a}{\frac{b}{x}}=a\div\frac{b}{x}=a\times \frac{x}{b}$
| {
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Solving a quadratic equation via a tangent half-angle formula (Maybe I'll post my own answer here, but maybe others will make that redundant.)
This is a fun (?) trivia item that fell out of a bit of geometry I was thinking about.
One of the tangent half-angle formulas says
$$
\tan\frac\alpha2 = \frac{\sin\alpha}{1+\cos\alpha}.
$$
Consider the quadratic equation
$$
x^2 + 2bx - 1 = 0.
$$
Solving for $b$, we get
$$
b = \frac{1-x^2}{2x}.
$$
Let $\alpha=\arctan b$. Since $\tan\alpha=(1-x^2)/2x$, we have $\sin\alpha=(1-x^2)/(1+x^2)$ and $\cos\alpha=2x/(1-x^2)$, so
$$
\tan\frac\alpha2=\frac{1-x^2}{(1+x)^2}.
$$
Lo and behold, this fraction is not in lowest terms, and we have
$$
\tan\frac\alpha2= \frac{\sin\alpha}{1+\cos\alpha} = \frac{1-x}{1+x}.
$$
This implies
$$
(1+x)\tan\frac\alpha2= 1-x
$$
and so we seem to have reduced a second-degree equation to a first-degree equation by using the tangent half-angle formula. Solving for $x$, we get
$$
x=\frac{1-\tan\frac\alpha2}{1+\tan\frac\alpha2}.\tag{1}
$$
Since $\tan\alpha=b$, we have $\sin\alpha=b/\sqrt{1+b^2}$ and $\cos\alpha=1/\sqrt{1+b^2}$, so
$$
\tan\frac\alpha2=\frac{\sin\alpha}{1+\cos\alpha}=\frac{b}{1+\sqrt{1+b^2}}.
$$
Sustituting this back into $(1)$, we get
$$
x = \frac{1+\sqrt{b^2-1}-b}{1+\sqrt{b^2+1}+b} = -b+\sqrt{b^2+1}.
$$
This is one of the two solutions of the quadratic equation. What happened to the other one?
| I've up-voted marty cohen's answer, but I would add this:
Maybe the first omission could be said to be the use of $\arctan b$ instead of using both of the points on the circle corresponding to $\tan=b$. For the conventional value of $\arctan b$, the positive square root of $1+b^2$ is the only one that's right.
When you include that other angle whose tangent is $\alpha$, the sine and cosine of the angle are minus what they were before, and then the half-angle, and its tangent, are different from what they otherwise were, and we do get the other solution of the quadratic equation.
But strictly speaking the first actual error is in the sentence "Since $\tan\alpha=(1-x^2)/2x$, we have $\sin\alpha=(1-x^2)/(1+x^2)$ and $\cos\alpha=2x/(1-x^2)$. The fact is there are two points on the circle where $\tan=b$ and we need both of them.
| {
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If $x$,$y$ are odd integers $a$,$b$ and $c$ such that $a^4-b^4=c^4$, argue that... If $x$,$y$ are odd integers $a$,$b$ and $c$ such that $a^4-b^4=c^4$, argue that $\gcd(a,b,c)=1$ implies that $\gcd(a,b)=1$
What I know:
A Pythagorean Triple is a triple of positive integers $a$,$b$,$c$ such that $a^2+b^2=c^2$
A Primitive Pythagorean Triple is a Pythagorean triple $a$,$b$,$c$ with the constraint that $\gcd(a,b)=1$, which implies $\gcd(a,c)=1$ and $\gcd(b,c)=1$.
I'm not sure how to use this, since we are dealing with $a^4-b^4=c^4$.
| If $gcd(a,b)>1$, then there is a prime number $p$ dividing $gcd(a,b)$, i.e., $p|a$ and $p|b$.
Since $a^4-b^4=c^4$, we have $p|c$. Thus $p|gcd(a,b,c)$. But $gcd(a,b,c)=1$, a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a trig equation $1+ \sin (x)=2 \cos(x)$? How would I solve the following equation?
$$
1+ \sin (x)=2 \cos(x)
$$
I am having difficult with it.
| We have $1 + \sin(x) = 2 \cos(x)$. Recall that $1 - \sin^2(x) = \cos^2(x)$. Hence, we get that $$(1 + \sin(x))(1- \sin(x)) = \cos^2(x)$$ i.e.
$$2 \cos(x) (1 - \sin(x)) = \cos^2(x)$$
If $\cos(x) \neq 0$, then we get that $$1 - \sin(x) = \dfrac{\cos(x)}2$$
Hence, have
\begin{align}
1 + \sin(x) & = 2\cos(x)\\
1 - \sin(x) & = \dfrac{\cos(x)}2
\end{align}
This gives us that $2 = \dfrac52 \cos(x) \implies \cos(x) = \dfrac45$. This gives us $\sin(x) = \dfrac35$. Hence, we get one possible solution as $$x = 2n \pi + \theta$$ where $\sin(\theta) = \dfrac35$ with $0 < \theta < \dfrac{\pi}2$.
If $\cos(x) = 0$, then we need $\sin(x) = -1$. Hence, $x = 2n\pi + \dfrac{3 \pi}2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim\limits_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)$ Evaluate
$$
\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)
$$
The answer is $\frac{1}{2}$, have no idea how to arrive at that.
| Method I
By simply applying l'Hôpital's rule, we have
$$\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x \right)=\lim_{x \to \infty}\frac{\displaystyle\sqrt{\frac{x}{x-1}}-1}{\left (\displaystyle \frac{1}{x}\right)}=\lim_{x \to \infty}\frac{x^2}{2(x-1)^2\displaystyle\sqrt{\frac{x}{x-1}}}=\frac{1}{2}.$$
Done.
Method II
Using the same start, we resort to the elementary limit $\lim_{y\to1} \displaystyle \frac{\sqrt{y}-1}{y-1}=\frac{1}{2}.$ Then
$$\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x \right)=\lim_{x \to \infty}\frac{\displaystyle\sqrt{\frac{x}{x-1}}-1}{\left (\displaystyle \frac{1}{x}\right)}=\lim_{x \to \infty}\frac{\displaystyle\sqrt{\frac{x}{x-1}}-1}{\left (\displaystyle \frac{x}{x-1}-1\right) }\cdot \frac{x}{x-1} = \frac{1}{2}.$$
Done.
| {
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Show that $(27!)^6 \equiv 1 \pmod{899}$ Show: $(27!)^6 \equiv 1 \pmod{899}$. Note: $899=30^2 - 1$.
Could anyone show me how to go through this please? I think I have to use Wilson's theorem which is $(p-1)! \equiv -1 \pmod p$ but not exactly sure how else to tackle this. Thanks!
| Using Wilson's Theorem, $28!\equiv-1\pmod{29}\implies 27!(28)\equiv-1$
$\implies 27!(-1)\equiv-1\implies 27!\equiv1\pmod {29}$
$\implies (27!)^6\equiv 1\pmod{29}$
Again $30!\equiv-1\pmod{31} \implies 27!(28)(29)(30)\equiv-1$
$\implies 27!(-3)(-2)(-1)\equiv-1\implies 27!(6)\equiv1 \implies 27!(30)\equiv5$ (multiplying either sides by $5$)
$\implies 27!(-1)\equiv5\implies 27!\equiv-5\pmod{31}$
So,$(27!)^6\equiv 5^6\pmod{31}$
Now, $5^3=125\equiv1\pmod{31} \implies (27!)^3\equiv -1\pmod{31}$
$\implies (27!)^6\equiv (-1)^2\pmod{31}\equiv1$
So, lcm$(31,29)\mid \{(27!)^6-1\}$ but lcm$(31,29)=31\cdot 29=899$
| {
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Limits of three functions Could you help me calculate the following limits:
$$\lim_{x \to 0} x \left[ \frac{1}{x} \right]$$
$$\lim_{x\to 0} \frac{1-\cos x \cdot \sqrt{\cos2x} }{x^2}$$
$$\lim_{x\to 10} \frac{\log _{10}(x) - 1}{x-10}$$
As to the last one I thought I could use $\lim\frac{log _{a}(1+\alpha)}{\alpha} = \log_ae$ but it wouldn't work
| Here is a simpler way to calculate the second limit:
$$\lim_{x\to 0} \frac{1-\cos x \sqrt{\cos2x} }{x^2}=\lim_{x\to 0} \frac{1-\cos x \sqrt{\cos2x} }{x^2} \frac{1+\cos x \sqrt{\cos2x} }{1+\cos x \sqrt{\cos2x}}$$
$$=\lim_{x\to 0} \frac{1-\cos^2 x \cos2x }{x^2} \lim_{x \to 0} \frac{1}{1+\cos x \sqrt{\cos2x}}$$
The second limit is easy, as for the first:
$$\lim_{x\to 0} \frac{1-\cos^2 x \cos2x }{x^2}=\lim_{x\to 0} \frac{1-\cos(2x)+\cos(2x)-\cos(2x)\cos^2 x }{x^2}$$
$$= \lim_{x \to 0} \frac{1-\cos(2x)}{x^2}+\lim_{x \to 0}\cos(2x) \frac{1-\cos^2 x }{x^2}$$
Now, use $1-\cos(2x)=2 \sin^2(x)$ and $1-\cos^2 x =\sin^2(x)$ and you are done.
| {
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Integers that satisfy $a^3= b^2 + 4$ Well, here's my question:
Are there any integers, $a$ and $b$ that satisfy the equation $b^2$$+4$=$a^3$, such that $a$ and $b$ are coprime?
I've already found the case where $b=11$ and $a =5$, but other than that?
And if there do exist other cases, how would I find them? And if not how would I prove so?
Thanks in advance. :)
| Observe that
if $a=3k,b^2=a^3-4=(3k)^3-4\equiv-1\pmod 3$ but $b^2\equiv1,0\pmod 3$
if $a=3k+1,b^2=a^3-4=(3k+1)^3-4=9(3k^3+3k^2+k)-3$ which is divisible by $3,$ but not by $9$
So, $a$ must be of the from $3k+2$
Consequently, $b^2-4=a^3-8=(3k+2)^3-8$
$(b+2)(b-2)=9k(3k^2+6k+4)$
Also, as $(a,b)=1,$ both $a,b$ must be odd $\implies (b+2,b-2)=(b+2,b+2-(b-2))=1$ and $k$ is odd
As $k$ is odd, $(k,3k^2+6k+4)=(k,4)=1$
If $b-2=9k,b+2=9k+4$ and $b+2=3k^2+6k+4\implies 3k^2-3k=0\implies k=0,1$
$k=0\implies b=2,a=2$ (but both $a,b$ are odd)
$k=1\implies b=11,a^3=125,a=5$
If $b+2=9k,b-2=9k-4$ and $b-2=3k^2+6k+4\implies 3k^2-3k+8=0$ whose discriminant is negative.
If $b+2=9,a^3=b^2+4=53$
If $b-2=9\implies b=11,a^3=b^2+4=125\implies a=5$
If $b-2=k\implies b=k+2,b+2=k+4$ and $b+2=9(3k^2+6k+4),27k^2+53k+32=0$ whose discriminant is negative.
If $b+2=k\implies b=k-2,b-2=k-4$ and $b-2=9(3k^2+6k+4),27k^2+53k+40=0$ whose discriminant is negative.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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The volume of the solid from the region bounded by $x=9-y^2$, $y=x-7$, $x=0$ about $y=3$ using cylindrical shells. The volume of the solid from the region bounded by $x=9-y^2$, $y=x-7$, $x=0$ about $y=3$ using cylindrical shells.
I've tried creating two separate regions:
$V_1=2\pi(3-y)(9-y^2)dy$ from 3 to 1
and
$V_2=2\pi(3-y)(y+7)dy$ from 1 to 0
but the answer this gives isn't correct. I can't seem to find any errors in my calculations. Am I devising the solution incorrectly?
| Here again is the graph of the things mentioned in the problem:
$\color{blue}{x=9-y^2}$, $\color{red}{x=y+7}$, $\color{green}{x=0}$, $y=3$ (dashed)
You're correct that the circumference of the shell at $y$ is given by $2\pi(3-y)$, but you should be integrating over all of $-3\leq y\leq 3$, with the height of the shell at $y$ given by
$$\begin{cases} 9-y^2&\text{ if }1\leq y\leq 3,\\ y+7 & \text{ if }-2\leq y\leq 1,\\
9-y^2 & \text{ if }-3\leq y\leq -2.\end{cases}$$
Thus, the volume is
$$V=\int_1^3 (3-y)(9-y^2)\,dy+\int_{-2}^1(3-y)(y+7)\,dy+\int_{-3}^2(3-y)(9-y^2)\,dy$$
which you can check gives the right answer.
| {
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Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Prove the following inequality: for
$a,b,c>0$
$$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$$
What I tried is using substitution:
$p=a+b+c$
$q=ab+bc+ca$
$r=abc$
But I cannot reduce $a^2(b+c)(c+a)+b^2(a+b)(c+a)+c(a+b)(b+c) $ interms of $p,q,r$
| You can write $\frac{a^2}{a+b} = a - \frac{ab}{a+b}$, and similarly with the other terms.
It remains to prove (after rearranging the inequality) that:
$$\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a} \leq \frac{1}{2}\left(a+b+c\right)$$
which follows from AM-HM.
| {
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How is this a property of Pascal's triangle? For all non-negative integers $k$ and $n$,
$$
\dbinom{k}{k} + \dbinom{k+1}{k} + \dbinom{k+2}{k} + \ldots + \dbinom{n}{k} = \dbinom{n +1}{k+1}
$$
How is this a property of Pascal's triangle? I do not see at all how it relates. I'm looking for an explanation, as opposed to some calculation with algebraic manipulations.
| The "usual" property of the Pascal triangle is $$\dbinom{n+1}{k+1} = \underbrace{\dbinom{n}{k+1}}_{\text{Up right term}} + \underbrace{\dbinom{n}k}_{\text{Up left term}}$$
Now proceed along the up right branch i.e. $$\dbinom{n}{k+1} = \underbrace{\dbinom{n-1}{k+1}}_{\text{Up right term}} + \underbrace{\dbinom{n-1}{k}}_{\text{Up right term}}$$
Now again take the up right branch and keep proceeding to finally end with
$$\dbinom{k+1}{k+1} = \dbinom{k}{k}$$
Now put these together to get what you want.
EDIT
$\hskip2.5in$
\begin{align}
\color{red}{\dbinom{n+1}{k+1}} & = \color{blue}{\dbinom{n}{k}} + \color{red}{\dbinom{n}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{red}{\dbinom{n-1}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \color{red}{\dbinom{n-2}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{red}{\dbinom{k+2}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{blue}{\dbinom{k+1}{k}} + \color{red}{\dbinom{k+1}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{blue}{\dbinom{k+1}{k}} + \color{blue}{\dbinom{k}{k}}
\end{align}
| {
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Smart demonstration to the formula $ \sum_{n=1} ^{N} \frac{n}{2^n} = \frac{-N + 2^{N+1}-2}{2^n}$ Someone could give me a smart and simple solution to show the folowing identity?
$$ \sum_{n=1} ^{N} \frac{n}{2^n} = \frac{-N + 2^{N+1}-2}{2^n}$$
| $$\sum_{n=1}^N \dfrac{n}{2^n}$$ can be written as
\begin{matrix}
\dfrac12 & + \dfrac1{2^2} & + \dfrac1{2^3} & + \dfrac1{2^4} & + \cdots & + \dfrac1{2^{N-1}} & + \dfrac1{2^N}\\
& + \dfrac1{2^2} & + \dfrac1{2^3} & + \dfrac1{2^4} & + \cdots & + \dfrac1{2^{N-1}} & + \dfrac1{2^N}\\
& & + \dfrac1{2^3} & + \dfrac1{2^4} & + \cdots & + \dfrac1{2^{N-1}} & + \dfrac1{2^N}\\
& & & + \vdots & + \vdots & + \vdots & + \vdots\\
& & & & & & + \dfrac1{2^N}\\
\end{matrix}
Now sum them row wise to get
$$\left(1 - \dfrac1{2^N} \right) + \dfrac12\left(1 - \dfrac1{2^{N-1}} \right) + \dfrac1{2^2}\left(1 - \dfrac1{2^{N-2}} \right) + \cdots + \dfrac1{2^{N-1}}\left(1 - \dfrac1{2} \right)\\ = \left(1 + \dfrac12 + \dfrac1{2^2} + \cdots + \dfrac1{2^{N-1}} \right) - \left(\dfrac1{2^N} + \dfrac1{2^N} + \dfrac1{2^N} + \cdots + \dfrac1{2^N} \right)\\
= 2 - \dfrac1{2^{N-1}} - \dfrac{N}{2^N}$$
Essentially, we are doing the following
$$\sum_{n=1}^N \dfrac{n}{2^n} = \sum_{n=1}^N \sum_{k=1}^n \dfrac1{2^n} = \sum_{k=1}^N \sum_{n=k}^{N} \dfrac1{2^n} = \sum_{k=1}^N \dfrac1{2^{k-1}}\left(1 - \dfrac1{2^{N+1-k}}\right) = \sum_{k=1}^N \left(\dfrac1{2^{k-1}} - \dfrac1{2^{N}} \right)\\
= 2 - \dfrac1{2^{N-1}} - \dfrac{N}{2^N}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}$?
Let $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}=f(x)$, where $f(u)=c\sin u$.Find $c$.
Trial:$\sum_{k=1}^{\infty}\frac{c \sin (x+k\pi)}{2^k}=c\sin x$.Then $c=0$ is a solution. Is there any other solution of $c$? Mainly I am interested in the sum $\sum_{k=1}^{\infty}\frac{f(x+k\pi)}{2^k}$. please help.
| Using a little trigonometry and the sum of a convergent infinite geometric series:
$$\sin(x+k\pi)=\sin x\cos k\pi+\sin k\pi\cos x=\sin x\cos k\pi=(-1)^k\sin x\Longrightarrow$$
$$c\sin x=\sum_{k=1}^\infty\frac{c\sin(x+k\pi)}{2^k}=c\sin x\sum_{k=1}^\infty\left(-\frac{1}{2}\right)^k=c\sin x\frac{-\frac{1}{2}}{1+\frac{1}{2}}=$$
$$=-\frac{c\sin x}{3}\Longrightarrow c\sin x=-\frac{c\sin x}{3}\Longleftrightarrow c=0\,\,\,\vee\,\,\,\sin x=0\Longleftrightarrow$$
$$\Longleftrightarrow c=0\,\,\,\vee\,\,\,x=n\pi\,\,\,,n\in\Bbb Z$$
| {
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How to calculate $\sum \limits_{x=0}^{n} \frac{n!}{(n-x)!\,n^x}\left(1-\frac{x(x-1)}{n(n-1)}\right)$ What are the asymptotics of the following sum as $n$ goes to infinity?
$$
S =\sum\limits_{x=0}^{n} \frac{n!}{(n-x)!\,n^x}\left(1-\frac{x(x-1)}{n(n-1)}\right)
$$
The sum comes from CDF related to sampling with replacement. Consider a random process where integers are sampled uniformly with replacement from $\{1...n\}$. Let $X$ be a random variable that represents the number of samples until either a duplicate is found or both the values $1$ and $2$ have been found. So if the samples where $1,6,3,5,1$ then $X=5$ and if it was $1,6,3,2$ then $X=4$. This sum is therefore $\mathbb{E}(X)$.
We therefore know that $S = \mathbb{E}(X) \leq \text{mean time to find a duplicate} \sim \sqrt{\frac{\pi}{2} n}$.
Taking the first part of the sum,
$$\sum_{x=0}^{n} \frac{n!}{(n-x)!\,n^x} = \left(\frac{e}{n} \right)^n \Gamma(n+1,n) \sim \left(\frac{e}{n} \right)^n \frac{n!}{2} \sim \sqrt{\frac{\pi}{2} n}. $$
| I get $$S = \sqrt{\frac{\pi n}{2}} + \frac{2}{3} + O(n^{-1/2+4\epsilon}).$$
An asymptotic for the summand.
If $x \leq n^{1/2+\epsilon}$, then Stirling's approximation yields $$\log\left(\frac{n!}{(n-x)!n^x}\right) = - \frac{x^2}{2n} + \frac{x}{2n} - \frac{x^3}{6n^2} + O(n^{-1+4\epsilon}) \tag{1}.$$ To obtain this, we need to rewrite $\log(n-x)$ as $\log(n) \log(1-x/n)$, use the Maclaurin series $\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$ to get the dominant terms of $\log(1-x/n)$, and then crunch through a lot of algebra. Most of the initially dominant terms cancel.
Thus, if $x \leq n^{1/2+\epsilon}$, $$\frac{n!}{(n-x)!n^x} = e^{-x^2/2n} \left(1 + \frac{x}{2n} - \frac{x^3}{6n^2} + O(n^{-1+4\epsilon})\right),$$
where we use the Maclaurin series $e^x = 1 + x + O(x^2)$ on all but the dominant term in Eq. (1).
Bounding the tail sum.
Since $\displaystyle \frac{n!}{(n-x)!n^x}$ is a decreasing function of $x$, if $x > n^{1/2+\epsilon}$, then $\displaystyle \frac{n!}{(n-x)!n^x} = O(e^{-x^2/2n}) = O(e^{-n^{2\epsilon}/2})$. Thus for the tail sum we get $$\sum_{x > n^{1/2} + \epsilon} \frac{n!}{(n-x)!n^x} = O(ne^{-n^{2\epsilon}/2}),$$ which is exponentially small. Therefore we can ignore it from here out for asymptotic purposes.
A useful asymptotic sum.
Problem 9.30 (p. 491) in Concrete Mathematics states that
$$\sum_{k \geq 0} k^r e^{-r^2/n} = \frac{1}{2} n^{(r+1)/2} \Gamma\left(\frac{r+1}{2}\right) - \frac{B_{r+1}}{(r+1)!} + O(n^{-1}) \tag{2},$$ where $B_r$ is the $r$th Bernoulli number. (An outline of the proof is given in the answer section of the book.)
The asymptotic for the first, dominant term.
Equation (2) gives us
\begin{align*}
\sum_{x =0}^n \frac{n!}{(n-x)!n^x} = &\frac{1}{2} \sqrt{2n} \Gamma(1/2) + \frac{1}{2} + O(n^{-1}) + \frac{1}{2n}\left(\frac{1}{2} 2n \Gamma(1) + O(1)\right) \\
&- \frac{1}{6n^2} \left(\frac{1}{2} (2n)^2 \Gamma(2) + O(1)\right) + O(n^{-1/2 + 4 \epsilon}) \\
= & \sqrt{\frac{\pi n}{2}} + \frac{2}{3} + O(n^{-1/2 + 4 \epsilon}).
\end{align*}
An asymptotic for the second term.
For the second term $\displaystyle \sum_{x =0}^n \frac{n!}{(n-x)!n^x} \frac{x(x-1)}{n(n-1)},$ a similar analysis holds. We can ignore the tail sum when $x > n^{1/2+\epsilon}$ because it is even smaller than the tail sum in the first term. If $x \leq n^{1/2+\epsilon}$, we get $$\frac{n!}{(n-x)!n^x} \frac{x(x-1)}{n(n-1)} = e^{-x^2/2n} O(n^{-1+\epsilon}).$$
Therefore, by Equation (2),
$$\sum_{x=0}^n \frac{n!}{(n-x)!n^x} \frac{x(x-1)}{n(n-1)} = O(n^{-1/2+\epsilon}).$$
Finally.
All together, then,
$$S = \sqrt{\frac{\pi n}{2}} + \frac{2}{3} + O(n^{-1/2+4\epsilon}).$$
Additional comments.
Incidentally, the sum $\displaystyle \sum_{x=0}^n \frac{n!}{(n-x)!n^x}$ is one that appears in several places in Knuth's work, and in The Art of Computer Programming, Vol. I, p. 120, he gives the asymptotic (in his notation) $$\sum_{x=0}^n \frac{n!}{(n-x)!n^x} = 1 + Q(n) = \sqrt{\frac{\pi n}{2}} + \frac{2}{3} + \frac{1}{12}\sqrt{\frac{\pi}{2n}} - \frac{4}{135n} + \frac{1}{288} \sqrt{\frac{\pi}{2n^3}} + O(n^{-2}).$$
| {
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Find a plane whose intersection line with a hyperboloid is a circle Find a plane $\pi$ which involves x-axis and its intersection line with $$\frac{x^2}{4}+y^2-z^2=1$$ is a circle.
Because the plane want to be find involves x-axis,so set as $By+Cz=0$,then I must to determine its value such that $$\begin{cases}By+Cz=0\\\frac{x^2}{4}+y^2-z^2=1\end{cases}$$ is a circle in 3-dimensional space,apparently,$B\not=0$,so I replace $y$ with $-\frac{C}{B}z$ in $\frac{x^2}{4}+y^2-z^2=1$,then i get a elliptic cylinder,at last ,i neet to determine $B$ and $C$ such that the intersection line between plane and elliptic cylinder is a circle.$$\begin{cases}By+Cz=0\\\frac{x^2}{4}+\left[(\frac{C}{B})^2-1\right]z^2=1\end{cases}$$
but i can't go any further.
thanks very much
| Assume $\pi$ has an equation like $$by+cz=0$$ for some unknown $c$ and $b$. If $b=0$ then $\pi: z=0$ and then we have an intersection like $x^2/4+y^2=1$ which is an ellipse not an circle. The same story would be for $c=0$. Let $c\neq0,b\neq0$ and so $z=-by/c$ so the intersection would be $$(1/4)x^2+y^2-b^2y^2/c^2 = 1$$ or $$(1/4)x^2+\left(1-b^2/c^2\right)y^2 = 1$$ if we want to have a circle as an intersection we need: $$b^2/c^2=3/4$$ For example set $b=\sqrt{3},c=2$ and look at the following picture:
| {
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An arctangent inequality As in the title, how can I prove
$$
\frac{\arctan(x)}{x}\geq\frac{1}{2}
$$
for $x\in(0,1]$?
I think I can say:
$$\frac{\arctan(x)}{x}$$
is monotonically decreasing in the interval, so its value is greater than the value in $1$, which is $\frac{\pi}{4}$, greater than $\frac{1}{2}$.
There exists some elegant proof of this simple inequality, maybe using series expansions?
| When $-1\le x\le 1$, we have
$$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots,$$
and therefore if $-1\le x\le 1$ and $x\ne 0$,
$$\frac{\arctan x}{x}=1-\frac{x^2}{3}+\frac{x^4}{5}-\frac{x^6}{7}+\cdots. \tag{$1$}$$
The series $(1)$ is an alternating series, and therefore if we truncate just after the term $-\frac{x^2}{3}$ the error term is positive (and less than the first "neglected" term $\frac{x^4}{5}$). It follows that
$$\frac{\arctan x}{x}\gt 1-\frac{x^2}{3}\ge \frac{2}{3}.$$
| {
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Find an expression, in terms of n, for $\sum_{r=2}^n {1\over r-1}-{1\over r+1} $ Working:
$$\sum_{r=2}^n {1\over r-1}-\sum_{r=2}^n {1\over r+1} $$
$$={1\over1} + {1\over2} + {1\over3}+\cdots+{1\over n-1}$$
$$-\left({1\over3}+\cdots+{1\over n-1}+{1\over n+1}\right)$$
This should then make:
$$1+{1\over2}-{1\over n+1}$$
But it is:
$$1+{1\over2}-{1\over n}-{1\over n+1}$$
Why is there the $-{1\over n}$ term?
Also how do you solve $\sum_{r=1000}^\infty {1\over r-1}-{1\over r+1} $ I haven't the faintest.
Thanks
| We want to find a simple expression for
$$\sum_{r=2}^n \left(\frac{1}{r-1}-\frac{1}{r+1}\right).$$
To get an idea about your lost $-\dfrac{1}{n}$, let us add a few terms together, maybe up to $r=7$, to see what's going on. We get
$$\left(1-\frac{1}{3}\right)+ \left(\frac{1}{2}-\frac{1}{4}\right)+ \left(\frac{1}{3}-\frac{1}{5}\right)+ \left(\frac{1}{4}-\frac{1}{6}\right) +\left(\frac{1}{5}-\frac{1}{7}\right)+ \left(\frac{1}{6}-\frac{1}{8}\right) .$$
There is a lot of cancellation. Everything disappears except for the $1$, the $\dfrac{1}{2}$, the $-\dfrac{1}{7}$, and the $-\dfrac{1}{8}$.
The $-\dfrac{1}{7}$ is your missing $-\dfrac{1}{n}$ term.
Added: The OP has been expanded to include the reasoning. Somewhat altered, it says that the sum is equal to
$$\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-2}+\frac{1}{n-1} \right)-\left(\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1} \right).$$
Everything cancels except the first two terms and the last two.
| {
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Find the density function of the random variable $Z=X+Y$ Let $X$ and $Y$ be independent and uniformly distributed random variable on the intervals $[0,3]$ and $[0,1]$, respectively. Find the density function of the random variable $Z=X+Y$.
I find $f(x,y)=1/3$ and the range of $Z$ to be $[0,4]$, but I cannot find the density function of $Z$.
| In general, to find the pdf, you should find the cdf first, and then differentiate the function.
$F[z] = P(Z \leq z) = \begin{cases} \frac {1}{6} z^2 & 0\leq z \leq 1 \\
\frac {1}{6} + \frac {1}{3} (x-1) & 1 \leq z \leq 3 \\
1- \frac {1}{6} (4-z)^2 & 3 \leq z \leq 4 \\ \end{cases}$
This gives that
$f(z) = \begin{cases} \frac {1}{3} z & 0 \leq z \leq 1 \\ \frac {1}{3} & 1 \leq z \leq 3 \\ \frac {1}{3} (4-z) & 3 \leq z \leq 4 \\\ \end{cases}$
| {
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How to solve $|x-5|=|2x+6|-1$? $|x-5|=|2x+6|-1$.
The answer is $0$ or $-12$, but how would I solve it by algebraically solving it as opposed to sketching a graph?
$|x-5|=|2x+6|-1\\
(|x-5|)^2=(|2x+6|-1)^2\\
...\\
9x^4+204x^3+1188x^2+720x=0?$
| Look for where the expressions inside the absolute values change sign: $x-5$ changes sign at $x=5$, and $2x+6$ changes sign at $x=-3$. Thus, when $x<-3$, $x-5$ and $2x+6$ are both negative, and the equation is
$$-(x-5)=-(2x+6)-1\;.$$
When $-3\le x<5$, $x-5$ is negative and $2x+3$ is non-negative, so the equation is
$$-(x-5)=2x+6-1\;.$$
And when $x\ge 5$, both expressions are non-negative, and the equation is
$$x-5=2x+6-1\;.$$
Thus, you need to solve
$$\left\{\begin{align*}
&-x+5=-2x-7&&\text{when }x<-3\\
&-x+5=2x+5&&\text{when }-3\le x<5\\
&x-5=2x+5&&\text{when }x\ge 5\;.
\end{align*}\right.\tag{1}$$
$(1)$ reduces to
$$\left\{\begin{align*}
&x=-12&&\text{and }x<-3\\
&x=0&&\text{and }-3\le x<5\\
&x=-10&&\text{and }x\ge 5\;.
\end{align*}\right.\tag{2}$$
The first two solutions in $(2)$ fall within the intervals on which they are valid; the third does not and therefore is not a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Evaluating the series: $1 +(1/3)(1/4) +(1/5)(1/4^2)+(1/7)(1/4^3)+ \cdots$
Sum the series:
$$1 + \dfrac13 \cdot \dfrac14 + \dfrac15 \cdot \dfrac1{4^2} + \dfrac17 \cdot \dfrac1{4^3} + \cdots$$
How can I solve this? I am totally stuck on this problem.
| Let $$S = \sum_{n=0}^{\infty} \dfrac1{2n+1} \dfrac1{4^n}$$
Note that
$$\dfrac1{2n+1} = \int_0^1 x^{2n} dx$$
Hence, we can write
\begin{align}
S & = \sum_{n=0}^{\infty} \dfrac1{4^n}\int_0^1 x^{2n} dx = \int_0^1 \sum_{n=0}^{\infty} \left(\dfrac{x^2}{4} \right)^n dx\\
& = \int_0^1 \dfrac{dx}{1-\dfrac{x^2}4} = \int_0^1 \dfrac{dx}{2-x} + \int_0^1 \dfrac{dx}{2+x}\\
& = \left. -\log(2-x) \right \vert_0^1 + \left. \log(2+x) \right \vert_0^1 = \log 3
\end{align}
Hence,
$$\sum_{n=0}^{\infty} \dfrac1{2n+1} \dfrac1{4^n} = \log(3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/276753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Showing that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left (\sqrt{a^2+1}-1\right)$. How can I show that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left(\sqrt{a^2+1}-1\right)$?
| Putting $x=a\sin\theta,dx=a\cos\theta d\theta$ and $x=\pm a,\theta=\pm\frac\pi2 $
$$\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx =\int _{-\frac\pi2}^{\frac\pi2}\frac{a^2\cos^2\theta}{1+a^2\sin^2\theta}d\theta$$
$$=\int _{-\frac\pi2}^{\frac\pi2}\frac{a^2\sec^2\theta}{(1+\tan^2\theta)(1+(a^2+1)\tan^2\theta)}d\theta$$ (Diving the numerator & the denominator by $\sec^4\theta$)
$$=\int _{-\infty}^{\infty}\frac{a^2}{(1+t^2)(1+(a^2+1)t^2)}dt$$ (Putting $\tan\theta = t$ as $\tan\theta=\pm\infty, t=\pm\frac\pi2$)
$$=\frac{a^2}{(a^2+1)}\int _{-\infty}^{\infty}\frac{1}{(1+t^2)(\frac1{(a^2+1)}+t^2)}dt$$
$$=\frac{\frac{a^2}{(a^2+1)}}{\left(1-\frac1{1+a^2}\right)}\left(\int _{-\infty}^{\infty}\frac1{(\frac1{(a^2+1)}+t^2)}dt-\int _{-\infty}^{\infty}\frac{1}{(1+t^2)}dt\right)$$ as $\frac1{(c+y)(b+y)}=\frac1{c-b}\frac{(c+y)-(b+y)}{(c+y)(b+y)}=\frac1{c-b}\left(\frac1{y+b}-\frac1{y+c}\right)$
$$=\left(\sqrt{1+a^2}\arctan (t\sqrt{1+a^2} )-\arctan t\right)_{-\infty}^{\infty}$$
So,$$\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx=(\sqrt{1+a^2}-1)\frac\pi2-\left\{-(\sqrt{1+a^2}-1)\frac\pi2\right\}=(\sqrt{1+a^2}-1)\pi$$
Observe that we have put $x=a\sin\theta$ and $\tan\theta = t\implies \left(\frac ax\right)^2- \left(\frac 1t\right)^2=\csc^2\theta-\cot^2\theta=1$
$\implies t^2=\frac{x^2}{a^2-x^2}\iff x^2=\frac{a^2t^2}{1+t^2}, a^2-x^2=\frac{a^2}{1+t^2}$
So, if we straight away take $t=\frac x{\sqrt{a^2-x^2}}$ (assuming $a>0$)
$$\frac{dt}{dx}=\frac1{\sqrt{a^2-x^2}}+x\left(\frac{-1}2\right)\frac1{(a^2-x^2)^{\frac32}}(-2x)=\frac{a^2}{(a^2-x^2)^{\frac32}}$$
$$\sqrt{a^2-x^2} dx=\frac{(a^2-x^2)^2dt}{a^2}=\frac{a^2dt}{(1+t^2)^2}$$
and if $x=\pm a,t=\frac{\pm a}0=\pm\infty$
So, $$\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx$$ becomes $$\int _{-\infty}^{\infty}\frac{a^2}{(1+t^2)(1+(a^2+1)t^2)}dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/281587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Forms $apq +b = r^{n} $ where p,q,r are primes Some small results for $2pq +3 = r^{n} $ p,q,r primes; written in the form (p,q,r,n):
$(3,1093,3,8)
(59,997,7,6)
(73,107,5,6)
(7,223,5,5)
(3,13,3,4)
(11,109,7,4)
(109,131,13,4)
(277,1667,31,4)
(5,491,17,3)
(89,137,29,3)
(11,13,17,2)$
Some small results for $2pq +1 = r^{n} $ p,q,r primes; written in the form (p,q,r,n):
$(13,757,3,9)
(2,19531,5,7)
(11,11,3,5)
(3,2801,7,5)
(29,3541,59,3)
(2,31,5,3)
(2,2,3,2)$ Last one is in fact $2^{3}+ 1=3^{2}$
I do not understand for now, though, why the first form $ 2pq + 3$ produces much more powers than the second. The form $ 3pq + 2$ produces (up to primes < $10^{5}$) releasing the condition for r to be a prime: 1 9-th power; 1 7-th power;
4 5-th powers, 38 cubes, but 0 squares as the form is always 2 mod 3 and squares are 0 or 1 mod 3.
| Primes are $1$ or $3 \mod 4$, else they would be divisible by $2$ or by $4$. Squares are always $0$ or $1 \mod 4$ since $2^2=4 = 0 \mod 4$ and $3^2=9=1 \mod 4$ and so if $a = 2\text{ or }3 \mod 4$, then $a^2= 0\text{ or }1 \mod 4$ and if $a=0\text{ or }1 \mod 4$ then $a^2= 0\text{ or }1 \mod 4$.
1) If $p$ and $q$ are both $1 \mod 4$, then $pq = 1 \mod 4$ and $2pq +1 = 3 \mod 4$ which is never a square.
2) If $p$ and $q$ are both $3 \mod 4$, then $pq = 1 \mod 4$ and $2pq +1 = 3 \mod 4$ which cannot be a square.
3) If one of $p$ and $q$ is $1 \mod 4$ and the other is $3 \mod 4$, then $pq = 3 \mod 4$ and $2pq +1 = 7 \mod 4 = 3 \mod 4$ which cannot be a square.
So no matter what are the primes, the form $2pq +1$ is never a square but when $p=q=2$ and $2pq+1 = 9 \mod 4 = 1 \mod 4 \implies 2 \cdot 2 \cdot 2 +1 = 2^3 +1 = 2^2 + 2^2 +1 = (2 +1)^2 = 3^2$.
I checked first $\mod 3$, but you cannot derive any conclusion. Fortunately $4 = 3+1$ and the intellectual work and effort to do was short.
There is still left the case $p=2$ and $q$ any odd prime. The form is then $4q +1$ and cannot be a square because odd primes are of the form $2k +1 \implies 4q +1 = 8k + 5$ and the squares are $0, 1\text{ or }4 \mod 8 \implies 4q + 1$ is never a square.
But if the form $2pq +1$ does not admit squares but for $p=q=2$, it will not admit any even power which are also squares; thus limiting the number of powers $2pq+1 = a^n$ this form admits.
Forms $9pq +3$ and $9pq +6$ do not admit powers $a^n$ up at least to $n= 11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/282403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$ I am trying to prove that
$$\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$$
I know how to deal with integrals involving cyclotomic polynomials and nested logarithms but I have no idea with this one.
| Let's introduce the parameter $\alpha$, and then differentiate with respect to $\alpha$ that yields
$$I(\alpha)=\int_0^1 \frac{t^\alpha-1}{(t^2+1)\ln t}dt $$
$$I'(\alpha)=\int_0^1 \frac{t^{\alpha}}{(t^2+1)}dt=\frac{1}{4} \left(-\psi_0\left(\frac{1 + \alpha}{4}\right) + \psi_0\left(\frac{3 + \alpha}{4}\right)\right) $$
Then
$$I(\alpha)=\frac{1}{4} \int\left(-\psi_0\left(\frac{1 + \alpha}{4}\right) + \psi_0\left(\frac{3 + \alpha}{4}\right)\right) d\alpha= $$
$$I(\alpha)=\left(\ln \Gamma \left(\frac{3 + \alpha}{4}\right)- \ln \Gamma \left(\frac{1 + \alpha}{4}\right)\right)+C\tag1$$
If letting $\alpha=2$, then
$$I(2)=\ln \left(\frac{\Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}\right)+C$$
On the other hand, by letting $\alpha=0$ in $(1)$ we get
$$C=\ln \left(\frac{\Gamma \left(\frac{1}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}\right)$$
Thus
$$\int_0^1 \frac{t^2-1}{(t^2+1)\ln t}dt=\ln \left(\frac{\Gamma \left(\frac{5}{4}\right)\Gamma \left(\frac{1}{4}\right)}{\Gamma^2 \left(\frac{3}{4}\right)}\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 0
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Solving lyapunov equation, Matlab has different solution, why? I need to solve the lyapunov equation i.e. $A^TP + PA = -Q$. With $A = \begin{bmatrix} -2 & 1 \\ -1 & 0 \end{bmatrix}$ and $Q = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$.
Hence...
$$
\begin{bmatrix}
-2 & -1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
P_{11} & P_{12} \\
P_{12} & P_{22}
\end{bmatrix}
+
\begin{bmatrix}
P_{11} & P_{12} \\
P_{12} & P_{22}
\end{bmatrix}
\begin{bmatrix}
-2 & 1 \\
-1 & 0
\end{bmatrix} +
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} = 0
$$
$$
\begin{bmatrix}
-4P_{11} - 2P_{12} + 1 & P_{11} -2P_{12} - P_{22} \\
P_{11} - 2P_{12} - P_{22} & 2P_{12} + 1
\end{bmatrix} = 0
$$
$$
\begin{bmatrix}
-4 & -2 & 0 \\
1 & -2 & -1 \\
1 & -2 & -1 \\
0 & 2 & 0
\end{bmatrix}
\begin{bmatrix}
P_{11} \\
P_{12} \\
P_{22}
\end{bmatrix} =
\begin{bmatrix}
-1 \\
0 \\
0 \\
-1
\end{bmatrix}
\Rightarrow
\begin{bmatrix}
-4 & -2 & 0 \\
1 & -2 & -1 \\
0 & 2 & 0
\end{bmatrix}
\begin{bmatrix}
P_{11} \\
P_{12} \\
P_{22}
\end{bmatrix} =
\begin{bmatrix}
-1 \\
0 \\
-1
\end{bmatrix}$$
And such we get that $P = \begin{bmatrix} 1/2 & -1/2 \\ -1/2 & 3/2\end{bmatrix}$.
This is the same solution as given by my professor. I wanted to check however if I can also find the solution using Matlab. I entered the following:
A = [-2 1; -1 0];
Q = [1 0; 0 1];
P = lyap(A,Q)
This however tells me that $P = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 3/2\end{bmatrix}$.
What is going on here? Is Matlab correct or wrong? Or is my solution wrong? Or are we both correct?
| X = lyap(A,B,-C) solves the continuous-time Sylvester equation
AX + XB = C
and
X = lyap(A’,Q) solves the continuous-time Lyapunov equation
ATP + PA + Q = 0
so, you can solve the lyapunov function.
A = [-2 1; -1 0];
Q = [1 0; 0 1];
P = lyap(A',Q)
P = [0.5000 -0.5000]
[-0.5000 1.5000]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Simple question - Proof How is $\frac{1}{2}ln(2x+2) = \frac{1}{2}ln(x+1) $ ?
As $\frac{1}{2}ln(2x+2)$ = $\frac{1}{2}ln(2(x+1))$, how does this become$ \frac{1}{2}ln(x+1)$?
Initial question was $ \int \frac{1}{2x+2} $
What I done was $ \int \frac{1}{u}du $ with $u=2x+2$ which lead to $\frac{1}{2}ln(2x+2) $ but WolframAlpha stated it was $\frac{1}{2}ln(x+1)$
| We have $\ln(2x+2)=\ln(2(x+1))=\ln 2+\ln(x+1)$. Thus
$$\frac{1}{2}\ln(2x+2)=\frac{1}{2}\ln 2+\frac{1}{2}\ln(x+1).$$
The two functions thus are definitely not equal. But they differ by a constant.
So the answer to $\int \frac{dx}{2x+2}$ can be equally well put as $\frac{1}{2}\ln(|2x+2|)+C$ and $\frac{1}{2}\ln(|x+1|)+C$. (We can forget about the absolute value part if $x+1$ is positive in our application.)
Remark: A simpler example: It is correct to say $\int 2x\,dx=x^2+C$. It is equally correct (but a little weird) to say $\int 2x\,dx=x^2+47+C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Quintic polynomial with Galois Group $A_5$ A recent question asks what makes degree 5 special when considering the roots of polynomials with integer coefficients etc. One answer is that the Galois Group of $S_5$ is not solvable. What I am looking for is the most straightforward example (with proof) of a polynomial with integer coefficients and Galois Group $A_5$.
Such an object ought to be standard ... if I ever knew one, I have forgotten it.
| Here are some found by computer search
$$x^5 - 55x - 88$$
$$x^5 - 55x + 88$$
$$x^5 + 20x - 16$$
$$x^5 + 20x + 16$$
$$x^5 + 95x - 76$$
$$x^5 + 95x + 76$$
$$x^5 + 3x^3 + 5x - 10$$
$$x^5 + 3x^3 + 5x + 10$$
$$x^5 + 6x^3 - 7x - 8$$
$$x^5 + 6x^3 - 7x + 8$$
$$x^5 + 10x^3 - 10x - 4$$
$$x^5 + 10x^3 - 10x + 4$$
$$x^5 - x^4 + x^3 + 2x^2 + x - 1$$
$$x^5 + x^4 + x^3 - 2x^2 + x + 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Evaluate integral with quadratic expression without root in the denominator $$\int \frac{1}{x(x^2+1)}dx = ? $$
How to solve it? Expanding to $\frac {A}{x}+ \frac{Bx +C}{x^2+1}$ would be wearisome.
| And just to make this problem unnecessarily complicated:
$$I=\int\frac{dx}{x^2\left(x+\frac{1}{x}\right)}=\int\frac{dx}{\left(x+\frac{1}{x}\right)}-\int\frac{\left(1-\frac{1}{x^2}\right)dx}{\left(x+\frac{1}{x}\right)}=\frac{1}{2}\int\frac{d\left(x^2+1\right)}{x^2+1}-\int\frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/287524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Bell Numbers: How to put EGF $e^{e^x-1}$ into a series? I'm working on exponential generating functions, especially on the EGF for the Bell numbers $B_n$.
I found on the internet the EGF $f(x)=e^{e^x-1}$ for Bell numbers. Now I tried to use this EGF to compute $B_3$ (should be 15). I know that I have to put the EGF into a series and have a look at the coefficients. Using $e^{f(x)}=1+f(x)+\frac{f(x)^2}{2!}+\frac{f(x)^3}{3!}+\ldots$ I get
\begin{eqnarray*}
e^{e^x-1}&=&1+(e^x-1)+\frac{(e^x-1)^2}{2!}+\frac{(e^x-1)^3}{3!}\\
&=&1+e^x-1+\frac{e^{x^2}-2e^x+1}{2!}+\frac{e^{x^3}-3e^{x^2}+3e^x-1}{3!}\\
&=&e^x+\frac{e^{x^2}}{2!}-e^x+\frac{1}{2!}+\frac{e^{x^3}}{3!}-\frac{e^{x^2}}{2!}+\frac{e^{x}}{2!}-\frac{1}{3!}\\
&=&\frac{e^{x}}{2!}+\frac{e^{x^3}}{3!}-\frac{1}{2!}+\frac{1}{3!}\\
&=&\frac{1}{2!}e^x+\frac{1}{3!}e^{x^3}-\frac{1}{3}\\
&=&\frac{1}{2!}\left( 1+x^2+\frac{x^4}{2!}+\frac{x^8}{3!} \right)+\frac{1}{3!}\left( 1+x^3+\frac{x^6}{2!}+\frac{x^9}{3!}\right)\\
&=&\frac{1}{2}+\frac{x^2}{2}+\frac{x^4}{4}+\frac{x^8}{24}+\frac{1}{6}+\frac{x^3}{3!}+\frac{x^6}{12}+\frac{x^9}{36}\\
&=&\frac{1}{2}+\frac{x^2}{2}+\frac{x^3}{6}+\ldots
\end{eqnarray*}
I think I can stop here, because the coefficient in front of $\frac{x^3}{3!}$ is not $15$.
Perhaps someone can help me out and give a hint to find my mistake?
| \begin{align}
e^{e^x-1}&= 1+(e^x-1)+\frac{(e^x-1)^2}{2!}+\frac{(e^x-1)^3}{3!}+\cdots+\frac{(e^x-1)^n}{n!}+ \cdots \\[8pt]
&= \cdots\cdots+\frac{e^{nx}+xe^{(n-1)x}+\binom n2 e^{(n-2)x}+\cdots+ne^x + 1}{n!}+ \cdots\cdots
\end{align}
One of the terms in the expansion of $(e^x-1)^n$ is $e^x$. When that is expanded, one of the terms will be $x^4/4!$. No matter how big $n$ is, you don't run out of terms involving $x^4$. So you don't just get finitely many terms involving $x^4$ and add up their coefficients to see if you get $\dfrac{15x^4}{4!}$. Rather, you get an infinite series.
The Wikipedia article titled Exponential formula gives a power series expansion of $$e^{f(x)}= \cdots+\frac{b_n x^n}{n!}+\cdots$$ when the power series expansion of $$f(x) = a_1+\frac{a_2 x^2}{2!}+\cdots+\frac{a_n x^n}{n!}+\cdots $$ is known. Notice this from the article:
$$
b_3 = a_3+3a_2 a_1 + a_1^3
$$
because there is one partition of the set $\{ 1, 2, 3 \}$ that has a single block of size $3$, there are three partitions of $\{ 1, 2, 3 \}$ that split it into a block of size $2$ and a block of size $1$, and there is one partition of $\{ 1, 2, 3 \}$ that splits it into three blocks of size $1$. Apply that to $4$:
$$
b_4 = a_4 + 4a_3a_1 + 3a_2^2+6a_2a_1^2 + a_1^4
$$
since there is one partition of the set $\{1,2,3,4\}$ that has a single block of size 4; there are four partitions into a block of size $3$ and a block of size $1$; three are three partitions into two blocks of size $2$; there are six partitions into a block of size $2$ and two blocks of size $1$; and there is one partition into four blocks of size $1$.
Wikipedia's article titled Dobinski's formula treats the expansion of each individual Bell number as an infinite series.
There is also Faà di Bruno's formula one the derivatives of composite functions.
| {
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"url": "https://math.stackexchange.com/questions/289097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Trigonometric Functions The question is to show that $A\sin(x + B)$ can be written as $a\sin x + b\cos x$ for suitable a and b.
Also, could somebody please show me how $f(x)=A\sin(x+B)$ satisfies $f + f ''=0$?
| If
$$
f(x) = A\sin(x+B)
$$
then
$$
f'(x) = A\cos(x+B)\cdot\frac{d}{dx}(x+B) = A\cos(x+B)\cdot1,
$$
and
$$
f''(x) = -A\sin(x+B)\cdot\frac{d}{dx}(x+B) = -A\sin(x+B).
$$
So
$$
f''(x)+f(x) = -A\sin(x+B)+A\sin(x+B) = 0.
$$
For the initial question, the standard trigonometric identity
$$
\sin(x+B) = \sin x\cos B+ \cos x\sin B
$$
is most of what you need to know. Then you have
$$
A\sin(x+B) = A\Big( \sin x\cos B+ \cos x\sin B \Big)
$$
$$
= \Big(A\cos B\Big) \sin x + \Big( A\sin B\Big) \cos x
$$
$$
=a\sin x+b\cos x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/293026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof of inequality involving surds Determine whether $ \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}$ is greater or less than $\frac{3}{10}$
So what I did was:
Add the fraction with surds to get $\frac{\sqrt{6}-\sqrt{2}}{\sqrt{12}}$
Then squaring both sides results in $\frac{8-4\sqrt{3}}{12} = \frac{2-\sqrt{3}}{3}$
Therefore we have $\frac{2-\sqrt{3}}{3} ? \frac{9}{100}$ (using question mark in place of < or > therefore $200-100\sqrt{3} = 27$, but $1\times1 = 1$ and $2\times2 = 4$ thefore $1<\sqrt{3}<2$ and since $100-100\times1 = 0 <27$, $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}} < \frac{3}{10}$
I checked the answer and they had a method but same result, however would this suffice?
| $$ \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}=\frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) = \frac{\sqrt{2}}{\sqrt{3}(1+\sqrt{3})}=\frac{\sqrt{2}}{3+\sqrt{3}}=\frac{\sqrt{2}(3-\sqrt{3})}{6}<\frac{7\sqrt{2}}{33},$$
where we have used $\sqrt{3}>\frac{19}{11}$. Now $\frac{7\sqrt{2}}{33}<\frac{3}{10}$, since $9800=99^2-1<99^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/293485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limits calculus very short question? Can you help me to solve this limit? $\frac{\cos x}{(1-\sin x)^{2/3}}$... as $x \rightarrow \pi/2$, how can I transform this?
| To evaluate the limit:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{\left(1-\sin x\right)^{\frac{2}{3}}}\right)$$
we can use the fact that as $x$ approaches $\frac{\pi}{2}$, $\sin x$ approaches 1 and $\cos x$ approaches 0. Substituting these values into the expression, we get:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{\left(1-\sin x\right)^{\frac{2}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{0}{\left(1-1\right)^{\frac{2}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{0}{0^{\frac{2}{3}}}\right)$$
The expression is indeterminate in this form ($\frac{0}{0}$), so we can try using L'Hopital's rule to evaluate the limit. Differentiating the numerator and denominator with respect to $x$ gives us:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{-\sin\left(x\right)}{-3\left(1-\sin x\right)^{-\frac{1}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{-\sin\left(x\right)}{-3\left(1-\sin x\right)^{-\frac{1}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{-\cos\left(x\right)}{3\left(-\frac{1}{3}\right)\left(1-\sin x\right)^{-\frac{4}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{-1\left(1-\sin x\right)^{\frac{4}{3}}}\right)$$
Again, the expression is indeterminate in this form ($\frac{0}{0}$), so we can try using L'Hopital's rule again. Differentiating the numerator and denominator with respect to $x$ gives us:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{-\sin\left(x\right)}{-4\left(1-\sin x\right)^{\frac{1}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{-\sin\left(x\right)}{-4\left(1-\sin x\right)^{\frac{1}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{-\cos\left(x\right)}{4\left(\frac{1}{3}\right)\left(1-\sin x\right)^{\frac{2}{3}}}\right)$$
$$= \lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{1\left(1-\sin x\right)^{\frac{2}{3}}}\right)$$
Substituting the values $\sin x = 1$ and $\cos x = 0$, we get:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{0}{\left(1-1\right)^{\frac{2}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{0}{0^{\frac{2}{3}}}\right)$$
The expression is indeterminate in this form ($\frac{0}{0}$), so we can't use L'Hopital's rule any further. However, since $\cos x$ approaches 0 and $\left(1 - \sin x\right)^{\frac{2}{3}}$ approaches 0 as $x$ approaches $\frac{\pi}{2}$, the entire expression approaches $\frac{0}{0}$, which is indeterminate. Therefore, the limit does not exist
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/293560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Factoring $3x^2 - 10x + 5$ How can $3x^2 - 10x + 5$ be factored? FOIL seemingly doesn't work (15 has no factors that sum to -10).
| You can try to Conplete the Square in this kind of problem. What you would want to do, is to manipulate $3x^2-10x+5$ into something like this $a^2 \pm 2ab + b^2 = (a \pm b)^2 \quad \mathbf{(1)}$.
First, factor the 3 out, like this $3x^2-10x+5 = 3 \left( x^2 - \dfrac{10}{3}x + \dfrac{5}{3} \right)$, now, look closely inside the brackets, we'll now complete the square.
$x^2$ will stand for $a^2$ in (1), so basically, we'll have $\color{red}{a = x}$ (remember this); now move onto the second term, we'll find $b$, such that, $\dfrac{10}{3}x$ must stand for $2ab$. Since $a = x$, we will have:
$\begin{align}2xb &= \dfrac{10}{3}x \\
\Leftrightarrow 2b &= \dfrac{10}{3} \quad \mbox{cancel } x \mbox{ from both sides} \\
\Leftrightarrow b &= \dfrac{5}{3}\end{align}$
So, back to the original expression:
The red parts, blue parts, and orange parts, although their forms do change, their values remain the same as we move down.
$\begin{align} 3x^2-10x+5 &= 3 \left( \color{red}{x^2} - \color{blue}{\dfrac{10}{3}x} + \color{orange}{\dfrac{5}{3}} \right)\\
&= 3 \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\dfrac{5}{3}} \right) \quad \mbox{make the }a^2 + 2ab \mbox{ part appear}\\
&= 3 \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\left( \dfrac{5}{3} \right)^2 - \left( \dfrac{5}{3} \right)^2 + \dfrac{5}{3}} \right) \quad \begin{array}{l} \mbox{make the }b^2 \mbox{ part} \\ \mbox{appear, and in other to keep}\\\mbox{the orange part the same,}\\ \mbox{we'll add and subtract simultaneously}\end{array}\\
&= 3 \left[ \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\left( \dfrac{5}{3} \right)^2} \right) \color{orange}{- \left( \dfrac{5}{3} \right)^2 + \dfrac{5}{3}} \right] \quad \mbox{re-grouping}\\
&= 3 \left[ \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\left( \dfrac{5}{3} \right)^2} \right) \color{orange}{- \dfrac{25}{9} + \dfrac{5}{3}} \right]\\
&= 3 \left[ \left( \color{red}{x^2} - \color{blue}{2.x.\dfrac{5}{3}} + \color{orange}{\left( \dfrac{5}{3} \right)^2} \right) \color{orange}{- \dfrac{10}{9}} \right]\\
&= 3 \left[ \left( x - \dfrac{5}{3} \right)^2 \color{orange}{- \dfrac{10}{9}} \right] \quad \mbox{apply } \mathbf{(1)}\\
&= 3 \left[ \left( x - \dfrac{5}{3} \right)^2 \color{orange}{- \left(\dfrac{\sqrt{10}}{3} \right)^2} \right]\\
&= 3 \left[ \left( x - \dfrac{5}{3} - \dfrac{\sqrt{10}}{3} \right) \left( x - \dfrac{5}{3} + \dfrac{\sqrt{10}}{3} \right) \right] \quad \mbox{applying difference of squares formula}\\
&= 3 \left( x - \dfrac{5+\sqrt{10}}{3}\right) \left( x + \dfrac{\sqrt{10} - 5}{3}\right)\quad \mbox{simplify it a bit}
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/297667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How can we show the convergence of $x_n = \sin(2\pi (n^3-n^2+1)^{\frac{1}{3}})$? Show that the sequence $(x_n)_{n\geq 1}$ defined by $$x_n = \sin(2\pi (n^3-n^2+1)^{\frac{1}{3}})$$ converges and compute its limit.
| It smells like an easy high school question
$$\lim_{n\to\infty}\sin(2\pi (\sqrt[3]{n^3-n^2+1}-n)+2\pi n)=$$
$$\lim_{n\to\infty}\sin(2\pi (\sqrt[3]{n^3-n^2+1}-n))=-\frac{\sqrt{3}}{2}$$
because
$$\lim_{x\to0} \frac{\sqrt[3]{x^3-x+1}-1}{x}=\lim_{x\to0} \frac{3x^2-1}{3(x^3-x+1)^{2/3}}=-\frac{1}{3}$$
by the cute l'Hôpital's rule, and then
$$\lim_{n\to\infty} (\sqrt[3]{n^3-n^2+1}-n)=-\frac{1}{3}$$
$$\sin \left(-\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{2}$$
Chris.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/298669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Does $\sin^2 x - \cos^2 x = 1-2\cos^2 x$? I am finishing a proof. It seems like I can use $\cos^2 + \sin^2 = 1$ to figure this out, but I just can't see how it works. So I've got two questions.
Does $\sin^2 x - \cos^2 x = 1-2\cos^2 x$?
And if it does, then how?
| Here is my favorite way to verify trigonometric identities:
First note that the equation of a circle gives us the rational parameterizations
$$\sin\theta=\frac{2t}{1+t^2}\qquad\cos\theta=\frac{1-t^2}{1+t^2}.$$
Substitute these expressions in. Now the equation we want to verify is
$$\left(\frac{2t}{1+t^2}\right)^2-\left(\frac{1-t^2}{1+t^2}\right)^2\overset{?}{=}1-2\left(\frac{1-t^2}{1+t^2}\right)^2.$$
Now just find a common denominator and compare numerators, so we want to know
$$(2t)^2-(1-t^2)^2\overset{?}{=}(1+t^2)^2-2(1-t^2)^2.$$
But this is true because
$$(1+t^2)^2-(1-t^2)^2=(1+2t^2+t^4)-(1-2t^2+t^4)=4t^2=(2t)^2,$$
thus the identity is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/300900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Solve the congruence $x^3+2x-3\equiv{0}\pmod{45}$ Solve (if possible)the congruence involving polynomial
$x^3+2x-3\equiv{0}\pmod{45}$
My work:
Since $45=3^2\cdot5$, we have
$x^3+2x-3\equiv{0}\mod(3)$ and $x^3+2x-3\equiv{0}\pmod 5$
In $\mathbb{Z}_3$,
$x^3+2x-3=(x-1)(x^2+x+3)\equiv{0}\pmod 3$
We have $[0],[1],[2]$
In $\mathbb{Z}_5$,
$x^3+2x-3=(x-1)(x^2+x+3)\equiv{0}\pmod 5$
We have $[0],[1],[2],[3],[4]$
So does it mean I have 3 solutions, $[0],[1],[2]$???
And how should I finish it up?
| Here is one way to find the roots $\rm\,mod\ 9.\:$ Since $\rm\,9\mid(x-1)(x^2+x+3),\:$ by unique factorization, either $\rm\,9\mid x-1,\,\ 9\mid x^2+x+3,\:$ or $\rm\:3\mid x-1,\,x^2+x+3.\:$ The last case is impossible since $\rm\:3\mid x-1\,\Rightarrow\,mod\ 3\!:\ x\equiv 1\,\Rightarrow\, x^2+x+3\equiv 2\not\equiv 0.\:$ In the second case we have $\rm\:mod\ 9\!:\ x^2+x+3\equiv x^2+x-6 = (x-2)(x+3).\:$ Again, either $\rm\,9\mid x-2,\,\ 9\mid x+3,\,$ or $\rm\,3\mid x-2,\,x+3.\:$ The latter is impossible, else $\rm\,2\equiv -3\,\ (mod\ 3).\:$ So this case yields two roots $\rm\:x\equiv 2,\, -3.\:$ Finally, the first case $\rm\,9\mid x-1\,$ yields the root $\rm\:x\equiv 1\,\ (mod\ 9).$
$\rm mod\ 5\!:\ x^2+x+3\equiv x^2+x-2 \equiv (x-1)(x+2),\:$ so there are two roots $\rm\:x\equiv 1,-2\,\ (mod\ 5).$
Now employ the CRT (Chinese Remainder Theorem) to combine the solutions modulo $9$ and $5$ to obtain all of the solutions modulo $\,\rm lcm(9,5) = 45.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/302434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sum\limits_{cyc} \frac 1 {(a-b)^2} \ge \frac{9}{4}$ $a, b, c \in [0;2]$, Prove inequality: $$\frac1{(a-b)^2}+\frac1{(b-c)^2}+\frac1{(c-a)^2}\geq \frac94$$
I tried to:
*
*Use AM-GM: $$LHS \ge \frac{(1+1+1)^2}{(a-b)^2+(b-c)^2+(c-a)^2}=\frac9{2(a^2+b^2+c^2 -ab-bc-ca)}$$ I am proving $a^2 + b^2+c^2 - ab - bc - ca \le 2$ but stuck
*Transform, we have:
$$ (a-b)+(b-c)+(c-a)=0 \implies LHS=\begin{vmatrix}\frac1{a-b}+\frac1{b-c}+\frac1{c-a}\end{vmatrix}$$ stuck here too. One more inequality (may be useful) is $$\frac1{(a-b)^2}+\frac1{(b-c)^2}+\frac1{(c-a)^2}\geq \frac4{ab+bc+ca}$$
| WLOG we can assume $ a < b < c$. Since multiplying $a,b,c$ all by the same constant $> 1$ decreases the left side, we may assume $c = 2$. Since decreasing $a$ and $b$ by the same positive number decreases the left side, we may assume $a=0$. Now there's only one variable left, and I'll leave it to you to minimize $$\frac{1}{b^2} + \frac{1}{(b-2)^2} + \frac{1}{2^2}, \ 0 < b < 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/303826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve equation $ \frac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$? $$ \dfrac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$$
How can I solve this equation in the easiest way?
| Multiply by $2$ to obtain
$$\tag1\sqrt{x^2-16}+\sqrt{x^2-9}=2$$
and multiply by the conjugate $\sqrt{x^2-16}-\sqrt{x^2-9}$ to obtain
$$\tag2 -\frac72=\frac12((x^2-16)-(x^2-9))=\sqrt{x^2-16}-\sqrt{x^2-9}.$$
Add $(1)$ and $(2)$ and divide by $2$ to obtain
$$\sqrt {x^2-16}=-\frac34$$
Which has no real solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
A improper integral with Glaisher-Kinkelin constant Show that :
$$\int_0^\infty \frac{\text{e}^{-x}}{x^2} \left( \frac{1}{1-\text{e}^{-x}} - \frac{1}{x} - \frac{1}{2} \right)^2 \, \text{d}x = \frac{7}{36}-\ln A+\frac{\zeta \left( 3 \right)}{2\pi ^2}$$
Where $\displaystyle A$ is Glaisher-Kinkelin constant
I see Chris's question is a bit related with this Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$
| You may start with the decomposition
$$
\int_0^\infty {\frac{e^{-x}}{x^2} \left( \frac{1}{\left(e^x - 1\right)^2} - \frac{1}{x^2} + \frac{1}{x} - \frac{5}{12} + \frac{x}{12} \right)dx} - 2\int_0^\infty {\frac{e^{-x}}{x^3} \left(\frac{1}{e^x - 1} - \frac{1}{x} + \frac{1}{2}-\frac{x}{12} \right)dx} + \int_0^\infty {\frac{e^{-x}}{x^2} \left(\frac{1}{e^x-1}- \frac{1}{x} + \frac{1}{2}-\frac{x}{12} \right)dx}
$$
and use the theory of the Multiple Gamma functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Solve $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$ for $x$ Is there any smart way to solve the equation: $$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$
Use Maple I can find $x \in \{1;ab+bc+ca\}$
| My solution:
Now $\left( {b - c} \right)\left( {1 + {a^2}} \right) = {a^2}\left( {b - c} \right) + \left( {b - c} \right) + \left( {b - c} \right)x - \left( {b - c} \right)x = \left( {b - c} \right)\left( {x + {a^2}} \right) + \left( {b - c} \right)\left( {1 - x} \right)$.
Then $\dfrac{{\left( {b - c} \right)\left( {1 + {a^2}} \right)}}{{x + {a^2}}} = \dfrac{{\left( {b - c} \right)\left( {x + {a^2}} \right) + \left( {b - c} \right)\left( {1 - x} \right)}}{{x + {a^2}}} = b - c + \dfrac{{\left( {1 - x} \right)\left( {b - c} \right)}}{{x + {a^2}}}$.
Same, $\dfrac{{\left( {c - a} \right)\left( {1 + {b^2}} \right)}}{{x + {b^2}}} = c - a + \dfrac{{\left( {1 - x} \right)\left( {c - a} \right)}}{{x + {b^2}}}\,\,\,\,\& \,\,\,\dfrac{{\left( {a - b} \right)\left( {1 + {c^2}} \right)}}{{x + {c^2}}} = a - b + \dfrac{{\left( {1 - x} \right)\left( {a - b} \right)}}{{x + {c^2}}}$
Because $\left( {a - b} \right) + \left( {b - c} \right) + \left( {c - a} \right) = 0$ so the equation becomes
$\left( {1 - x} \right)\left( {\dfrac{{b - c}}{{x + {a^2}}} + \dfrac{{c - a}}{{x + {b^2}}} + \dfrac{{a - b}}{{x + {c^2}}}} \right) = 0 \Longleftrightarrow \left[ \begin{array}{l}
x = 1\\
\dfrac{{b - c}}{{x + {a^2}}} + \dfrac{{c - a}}{{x + {b^2}}} + \dfrac{{a - b}}{{x + {c^2}}} = 0\left( \bigstar \right)
\end{array} \right.$
$\left( \bigstar \right) \Longleftrightarrow - \dfrac{{\left( {c - a} \right) + \left( {a - b} \right)}}{{x + {a^2}}} + \dfrac{{c - a}}{{x + {b^2}}} + \dfrac{{a - b}}{{x + {c^2}}} = 0 \Longleftrightarrow \dfrac{{c - a}}{{x + {b^2}}} - \dfrac{{c - a}}{{x + {a^2}}} + \dfrac{{a - b}}{{x + {c^2}}} - \dfrac{{a - b}}{{x + {a^2}}} = 0$
$\Longleftrightarrow \left( {c - a} \right)\left( {\dfrac{1}{{x + {b^2}}} - \dfrac{1}{{x + {a^2}}}} \right) + \left( {a - b} \right)\left( {\dfrac{1}{{x + {c^2}}} - \dfrac{1}{{x + {a^2}}}} \right) = 0$
$\Longleftrightarrow \dfrac{{\left( {c - a} \right)\left( {a - b} \right)\left( {a + b} \right)}}{{\left( {x + {a^2}} \right)\left( {x + {b^2}} \right)}} - \dfrac{{\left( {a - b} \right)\left( {c - a} \right)\left( {a + c} \right)}}{{\left( {x + {a^2}} \right)\left( {x + {c^2}} \right)}} = 0 \Longleftrightarrow \dfrac{{\left( {a - b} \right)\left( {c - a} \right)}}{{x + {a^2}}}\left( {\dfrac{{a + b}}{{x + {b^2}}} - \dfrac{{a + c}}{{x + {c^2}}}} \right) = 0$
$\Longleftrightarrow \dfrac{{a + b}}{{x + {b^2}}} = \dfrac{{a + c}}{{x + {c^2}}} \Longleftrightarrow \left( {a + b} \right)x + {c^2}\left( {a + b} \right) = x\left( {a + c} \right) + {b^2}\left( {a + c} \right)$
$\Longleftrightarrow x\left( {b - c} \right) = a\left( {{b^2} - {c^2}} \right) + bc\left( {b - c} \right) \Longleftrightarrow x = a\left( {b + c} \right) + bc \Longleftrightarrow x = ab + bc + ca$.
Hence $S = \left\{ {1;ab + bc + ca} \right\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
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