Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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What is the value of $a$ so that this condition holds? Let $f(x) \colon= x-x^2$, $g(x) \colon= ax$. Determine the value of $a$ so that the region above the graph of $g$ and below the graph of $f$ has area equal to $9/2$.
Here $f(x) - g(x) = (1-a)x - x^2 = x((1-a) - x) = 0$ if and only if $x = 0$ or $x = 1-a$.
So the required area is $$ \int_{0}^{1-a} ( (1-a)x - x^2 ) \ dx = \frac{(1-a)^2}{2} - \frac{(1-a)^3}{3} = \frac{(1-a)^2}{6} ( 3 - 2(1-a) ) = \frac{(1-a)^2}{6} (2a + 1) = \frac{(1-a)^2 (2a+1) }{6} = \frac{(a^2 -2a +1 ) (2a+1)}{6} = \frac{2a^3 - 3a^2 + 1}{6} = \frac{9}{2}, $$
and this last equation simplifies to $$ 2a^3 - 3a^2 - 26 = 0.$$
Is the process so far correct? And if so, then how to solve this last equation for $a$?
| You should have gotten $\displaystyle\int_{0}^{1-a}[(1-a)x-x^2]\,dx = \left[\dfrac{1-a}{2}x^2-\dfrac{1}{3}x^3\right]_{0}^{1-a} = \dfrac{(1-a)^3}{6}$.
Then, you have $\dfrac{(1-a)^3}{6} = \dfrac{9}{2} \leadsto (1-a)^3 = 27$ which should be easy to solve.
| {
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"url": "https://math.stackexchange.com/questions/880825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Decompose a fraction in a sum of two Let's say that I have this fraction:
$$ \frac{2x}{x^2+4x+3}$$
I would like to decompose in two fraction:
$$ \frac{A}{x+3} + \frac{B}{x+1}$$
Which is the procedure for that? :)
| $$x^2+4x+3=0$$
$$\Delta=4^2-4 \cdot 1 \cdot 3=4$$
$$x_{1,2}=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-4 \pm 2}{2}=-3 \text{ or } -1$$
$$x^2+4x+3=(x+3)(x+1)$$
$$\frac{2x}{x^2+4x+3}=\frac{2x}{(x+3)(x+1)}=\frac{A}{x+3}+\frac{B}{x+1}=\frac{A(x+1)+B(x+3)}{(x+3)(x+1)}=\frac{(A+B)x+(A+3B)}{(x+3)(x+1)}$$
Since,$\frac{2x}{(x+3)(x+1)}=\frac{(A+B)x+(A+3B)}{(x+3)(x+1)}$,it must be:
$$A+B=2 \text{ and } A+3B=0$$
$$A+3B=0 \Rightarrow A=-3B$$
$$A+B=2 \Rightarrow -2B=2 \Rightarrow B=-1$$
$$A=3$$
Therefore:
$$\frac{2x}{x^2+4x+3}=\frac{3}{x+3}-\frac{1}{x+1}$$
| {
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Show that the inequality holds $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$ We have to show that:
$\displaystyle\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$
To be honest I don't have idea how to deal with it. I only suspect there will be need to consider two cases for $n=2k$ and $n=2k+1$
| one approach is to pair off the integers in the harmonic sum as $n+k$ and $2n-k$. so if $n$ is even we have:
$$
S_n = \sum_{k=0}^{\frac{n}2-1}\left(\frac1{n+k} + \frac1{2n-k} \right) + \frac2{3n}
$$
now
$$\frac1{n+k} + \frac1{2n-k} = \frac{3n}{(n+k)(2n-k)} \ge \frac{4n}{3n^2} = \frac4{3n}
$$
and there are $\frac{n}2$ such terms, so we have:
$$
S_n \ge \frac23 + \frac2{3n}
$$
when $n$ is odd we arrive (i think) at the numerically similar estimate
$$
S_n \ge \frac{6 \left(1+\frac1{n}\right)}{9-\frac1{n^2}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$a,b,c \geq 0$,prove that $a^2+b^2+c^2+abc+5 \geq3(a+b+c) $
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$a^2+b^2+c^2+abc+5 \geq3(a+b+c).$$
I'm certain that this problem could be solved by using dirchlet's theory.but I do not know how to apply it exactly.
| Since $a^2+b^2+c^2+abc+5 \geq3(a+b+c) \iff (a^2-3a)+(b^2-3b)+(c^2-3c)+abc\ge-5 \iff$
$\;\;\;\;\;(a-\frac{3}{2})^2+(b-\frac{3}{2})^2+(c-\frac{3}{2})^2+abc\ge\frac{7}{4},$
it suffices to show that $f(x,y,z)=(x-\frac{3}{2})^2+(y-\frac{3}{2})^2+(z-\frac{3}{2})^2+xyz$ has
a minimum value of $\frac{7}{4}$
for $x\ge0, y\ge0, z\ge0.$
$\textbf{1)}$ If we consider the values of $f$ on the cube defined by $0\le x\le3, 0\le y\le3, 0\le z\le3$,
$f(x,y,z)\ge\frac{9}{4}$ on the boundary of the cube since $x=0$ or $x=3\implies f(x,y,z)\ge\frac{9}{4}$,
and similarly for the other 4 faces of the cube.
$\textbf{2)}$ At any critical point $(x,y,z)$ in the interior of the cube,
$f_x=2x-3+yz=0$, $f_y=2y-3+xz=0$, and $f_z=2z-3+xy=0$, so
$\;\;\;\;2x+yz=3, \;\;\;2y+xz=3, \;\;\;2z+xy=3$.
Subtracting the 2nd equation from the first gives $(x-y)(2-z)=0$, so either $x=y$ or $z=2$. However, $z=2\implies4+xy=3\implies xy=-1$, which is not possible since $x>0, y>0$; so $x=y$.
Similarly, subtracting the 3rd equation from the 2nd gives $(y-z)(2-x)=0$, so either $y=z$ or $x=2$. As above, $x=2\implies4+yz=3\implies yz=-1$, which is not possible for $y>0,z>0$, so $y=z$.
Substituting into the 1st equation gives $2x+x^2=3$, so $x^2+2x-3=0\implies(x+3)(x-1)=0\implies x=1$
since $x>0$;
so $(1,1,1)$ is the only critical point in the interior of the cube.
Since $f(1,1,1)=\frac{7}{4}$, f has a minimum value of $\frac{7}{4}$ in the cube.
$\textbf{3)}$ Since $f(x,y,z)>\frac{9}{4}$ if $x>3$ or $y>3$ or $z>3$,
it follows that f has a minimum value of $\frac{7}{4}$ for $x\ge0,y\ge0,z\ge0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$
Evaluate $$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$$
I tried by taking $x^2$ out of the root by taking it common.
i.e: $$\lim_{x \to -\infty} \left(\frac{x\sqrt{\frac{1}{x^2}+1}-x}{x} \right)$$
and then cancelling the x in numerator and denominator
$$\lim_{x \to -\infty} \left(\frac{\sqrt{\frac{1}{x^2}+1}-1}{1} \right)$$
then substituting $x= -\infty$ in the equation, we get,
$$\lim_{x \to -\infty} \left(\frac{\sqrt{0+1}-1}{1} \right)$$
which equals to $0$. But it is not the correct answer.
What have I done wrong.
| $$
\displaylines{
\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {1 + x^2 } - x}}{x} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)x^2 } - x}}{x} \cr
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left| x \right|\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)} - x}}{x} \cr
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)} - x}}{x} \cr
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\left( {\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)} + 1} \right)}}{x} = -\mathop {\lim }\limits_{x \to - \infty } \sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)} + 1 =- 2 \cr}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed form of a sum of binomial coefficients? I have the following function:
$T_n(d)=\sum\limits_{k=\frac{n-d}{2}}^{\lceil \frac{n}{2} \rceil}{k\choose \frac{n-d}{2}}$
${n \choose 2k}$, where $n,d\in \mathbb{N}^0$, and $n,d$ have the same parity.
Looking at the sequences for various $d$, it seems that the formula is a polynomial of degree $d$ in $n$. This is speculation, but for example, where defined, it would appear that:
$T_{n}(1)=n$, $T_{n}(2)=\frac{1}{2}n^2$, $T_{n}(3)=\frac{1}{6}n^3-\frac{1}{6}n$, and $T_{n}(4)=\frac{1}{24}n^4-\frac{1}{6}n^2$
In fact, after further numerical testing, it seems that
$T_n(d)=\frac{n}{d!}\prod\limits_{j=1}^{d-1}(n-(2j-d))$
So my question is, is there a way to confirm the above results and show whether or not $T_n(d)$ is a polynomial of degree $d$ in $n$? If so, is there an intuitive reason why it is a polynomial with evenly spaced integer roots? The result seems rather elegant, if it's true.
EDIT: Simplifying the above expression, we have
$$T_n(d)=\frac{n}{d!}\prod\limits_{j=1}^{d-1}(n-(2j-d))=\frac{n}{d!}\frac{(n-1+(d-1))!!}{(n-1-(d-1))!!}=\frac{2^d n}{n+d}\cdot {\frac{n+d}{2} \choose d}$$
The third and fourth expressions in particular, seem like they would be very useful for a combinatorial proof. $n!!$ is the standard double factorial.
| Another useful apparently classic technique uses basic complex
variables.
Suppose we seek to evaluate
$$\sum_{k\ge m} {k\choose m} {2m+d\choose 2k}
= \sum_{k\ge m} {k\choose m} {2m+d\choose 2m+d-2k}$$
where the second binomial coefficient serves to delimit the upper
bound of the range of $k$ which is finite.
Start from
$${2m+d\choose 2m+d-2k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2m+d-2k+1}} (1+z)^{2m+d} \; dz.$$
This gives the following integral for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{k\ge m} {k\choose m}
\frac{1}{z^{2m+d-2k+1}} (1+z)^{2m+d} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2m+d}}{z^{2m+d+1}}
\sum_{k\ge m} {k\choose m} z^{2k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2m+d}}{z^{2m+d+1}}
\sum_{k\ge 0} {k+m\choose m} z^{2k+2m} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2m+d}}{z^{d+1}}
\sum_{k\ge 0} {k+m\choose m} z^{2k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2m+d}}{z^{d+1}}
\frac{1}{(1-z^2)^{m+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+d-1}}{z^{d+1}}
\frac{1}{(1-z)^{m+1}} \; dz.$$
Extracting coefficients we obtain
$$\sum_{q=0}^d {m+d-1\choose q} {d-q+m\choose m}.$$
Use another integral as in
$${d-q+m\choose m}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{d-q+m} \; dz.$$
to obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{q=0}^d {m+d-1\choose q}
\frac{1}{z^{m+1}} (1+z)^{d-q+m} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{d+m}}{z^{m+1}}
\sum_{q=0}^d {m+d-1\choose q} (1+z)^{-q}\; dz.$$
The defining integral is zero for $m+d-1 \ge q>d$ so we may extend the sum to
$m+d-1$, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{d+m}}{z^{m+1}}
\sum_{q=0}^{m+d-1} {m+d-1\choose q} (1+z)^{-q}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{d+m}}{z^{m+1}}
\left(1+\frac{1}{1+z}\right)^{m+d-1} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{d+m}}{z^{m+1}}
\left(\frac{2+z}{1+z}\right)^{m+d-1} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1+z}{z^{m+1}}
(2+z)^{m+d-1} \; dz.$$
We may now extract coefficients to obtain
$$[z^m] (1+z) (2+z)^{m+d-1}
= [z^m] (2+z)^{m+d-1}
+ [z^{m-1} ] (2+z)^{m+d-1}
\\ = 2^{d-1} {m+d-1\choose m}
+ 2^d {m+d-1\choose m-1}.$$
This needs some re-writing to make it match the other answer.
We get
$$2^{d-1} {m+d-1\choose d-1}
+ 2^d {m+d-1\choose d}
\\ = 2^{d-1} \frac{d}{m+d} {m+d\choose d}
+ 2^d \frac{m}{m+d} {m+d\choose d}
\\ = 2^d {m+d\choose d}
\left(\frac{1}{2} \frac{d}{m+d} + \frac{m}{m+d} \right)
\\ = 2^d {m+d\choose d} \frac{2m+d}{2m+2d}.$$
Interesting exercise. There is another one like it at this MSE link.
Observe that these computations were done without intervention by a
computer except to check arithmetic.
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate $\frac{2+5x^3}{2x^3+2}\sqrt{x^3+1}$ How would one find the integral of $\frac{2+5x^3}{2x^3+2}\sqrt{x^3+1}$ with respect to $x$?
I already know the antiderivative (from Wolfram Alpha), but I don't know how to integrate this function with pencil and paper.
Is there maybe a clever variable substitution?
| HINT:
I would say that
$\frac{2+5x^3}{2x^3+2}\sqrt{x^3+1}=\frac{2+5x^3}{2(x^3+1)}\frac{x^3+1}{\sqrt{x^3+1}}=\frac{2+5x^3}{2\sqrt{x^3+1}}=\frac{2(x^3+1)+3x^3}{2\sqrt{x^3+1}}=\frac{x^3+1}{\sqrt{x^3+1}}+\frac{3x^3}{2\sqrt{x^3+1}}=$
$=\sqrt{x^3+1}+\frac{3x^3}{2\sqrt{x^3+1}}=1\cdot\sqrt{x^3+1}+x\cdot\frac{3x^2}{2\sqrt{x^3+1}}=\frac{d}{dx}(x\cdot \sqrt{x^3+1})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/891160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Conditional extreme value of a function Let $x,y,z$ be the positive real numbers, if $x^2+y^2+z^2=1$, then how can we find the minimal value of this function $f(x,y,z)=\dfrac{xz}{y}+\dfrac{yz}{x}+\dfrac{xy}{z}$.
| I will show that,
$$ \frac{xy}{z}+ \frac{yz}{x}+\frac{xz}{y} \ge \sqrt{3}$$
This is equivalent to,
$$ (\frac{xy}{z}+ \frac{yz}{x}+\frac{xz}{y})^2 \ge 3$$
Which is same as,
$$ \frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{x^2z^2}{y^2} + 2(x^2+y^2+z^2) \ge 3 = 3(x^2+y^2+z^2) $$
So it suffice to prove that,
$$ \frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{x^2z^2}{y^2} \ge x^2+y^2+z^2 $$
This is trivial if someone notices inequality, $a^2+b^2+c^2 \ge ab+bc+ac $
Elaborating,
Note that, by AM-GM,Following inequalities are true,
$$\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2} \ge 2y^2 $$
$$\frac{y^2z^2}{x^2}+\frac{x^2z^2}{y^2} \ge 2z^2 $$
$$\frac{x^2y^2}{z^2}+\frac{x^2z^2}{y^2} \ge 2x^2 $$
Adding gives the desired result.
Equality when, $x=y=z \implies x=y=z=\frac{1}{\sqrt{3}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $(x^2-9)+\sqrt{2-x}=0$ Solve the equation $(x^2-9)+\sqrt{2-x}=0$
*
*$(x+3)(x-3)+\sqrt{2-x}=0$
Conditions: $x\neq\pm3 \wedge x\leq2$
*$(x+3)(x-3) = -\sqrt{2-x}$
*$(x+3)^2(x-3)^2 = 2-x$
*$x^4-18x^2+81 = 2-x$
*$x^4-18x^2+x+79 = 0$
I'm positive this isn't going to the right direction.
Please help.
| One way to solve (or approximate) such an equation (or to simply check how many solutions exist) is to graph each side of the equation: $$x^2 - 9 = -\sqrt{2-x}$$
*
*$y_1 = x^2-9$
*$y_2 = -\sqrt{2-x}$
Then look for when $y_1 = y_2$: if there are solutions to this equality, they will appear as points where $y_1$ and $y_2$ intersect.
See the top graph: The blue parabola is $y_1 = x^2 - 9$. The domain of $x$ shown in the purple graph is $x\leq 2$.
There is exactly one real solution: the point of intersection of the two graphs is $$(x, y) \approx (-2.61747, -2.14883)$$
So the one real solution is $x \approx -2.61747$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of elements of order $2$ in $S_n$
How many elements of order $2$ are there in $S_n$?
Using combinatorics I arrived at this:
For $n$ even ($n=2k$) there are ${n\choose2}+{n\choose 2}{n-2\choose 2}\dfrac{1}{2!}+{n\choose 2} {n-2\choose 2}{n-4\choose 2}\dfrac{1}{3!}+\cdots+{n\choose 2}{n-2\choose 2}{n-4\choose 2}\cdots{2\choose 2}\dfrac{1}{k!}$.
For $n$ odd ($n=2k+1$) there are ${n\choose 2}+{n\choose 2}{n-2\choose 2}\dfrac{1}{2!}+{n\choose 2}{n-2\choose 2}{n-4\choose 2}\dfrac{1}{3!}+\cdots+{n\choose 2}{n-2\choose 2}{n-4\choose 2}\cdots{3\choose 2}\dfrac{1}{k!}$
But how do I find the sums?
Seems like I have to use induction. But not quite upto there.
Thanks for the help!!
| An element of order $2$ is a product of $k$, say, disjoint 2-cycles.
*
*For $k=1$, there are $\frac{n(n-1)}{2^1\cdot 1!}$ elements of order two.
*For $k=2$, there are $\frac{n(n-1)(n-2)(n-3)}{2^2\cdot 2!}$ elements of order two.
*For $k=3$, there are $\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{2^3\cdot 3!}$ elements of order two.
In the denominator of each fraction, you have a $2^k k!$, because each 2-cycle could be chosen in the form $(a, b)$ or in the form $(b, a)$ (so you need to divide by $2^k$), while the different permutations of the 2-cycles don't change the element (so you need to divide through by $k!$). Hence, you get in general:
*
*There are $\frac{n!}{(n-2k)!2^k\cdot k!}$ elements of order two which are the product of $k$ disjoint 2-cycles.
Then sum these to get your number!
$$\text{number of elements of order two}=e_2(n)=\sum_{k=1}^{\lfloor n/2\rfloor}\frac{n!}{(n-2k)!2^k\cdot k!}$$
Note that the following recurrence relation holds.
$$e_2(n)=e_2(n-1)+(1+e_2(n-2))(n-1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{a^2+b^2+c^2}{ab+bc+ca} + 8\frac{abc}{(a+b)(b+c)(c+a)} \ge 2$ Let $a,b,c>0$, prove that
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 2.$$
I tried using the equality $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ and the Schur inequality but it's not very helpful.
Thanks.
| WOLG:$a\ge b\ge c$
we have
$$2b(a+c)^2-(a+b)(b+c)(a+c)=(a+c)(a-b)(b-c)\ge 0$$
so
$$\dfrac{8abc}{(a+b)(b+c)(a+c)}\ge\dfrac{4ac}{(a+c)^2}$$
so we only prove
$$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}\ge 2$$
since
$$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}- 2=\dfrac{(a^2+c^2
-ab-bc)^2}{(a+c)^2(ab+bc+ac)}\ge 0$$
| {
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Solution of $\frac{d^2y}{dx^2} - \frac{H(x) y}{b} = H(-x)$ Does the equation
$$\frac{d^2y}{dx^2} - \frac{H(x)}{b} y = c H(x)$$
have a solution where $H(x)$ is the Heaviside step function and $b$ and $c$ are constant?
Update: What about the second step function be $H(-x)$:
$$\frac{d^2y}{dx^2} - \frac{H(x)}{b} y = c H(-x)$$
| I am not sure wether this answer is correct or not, but as far as the step function is only a sign, I brought it out of the integration.
Please note that, for finding the constants, you need the boundary values and compare the value of the function at the point $x=0$.
$$\begin{array}{l}\frac{{{d^2}y}}{{d{x^2}}} - \frac{{H(x)y}}{b} = H(x)\\\frac{{{d^2}y}}{{d{x^2}}} = \frac{{H(x)y}}{b} + H(x)\\\frac{{{d^2}y}}{{d{x^2}}} = \left( {\frac{y}{b} + 1} \right)H(x)\\\frac{{{d^2}y}}{{\left( {\frac{y}{b} + 1} \right)}} = H(x){d^2}x\\\int {\frac{b}{{y + b}}{d^2}y} = H(x)\int {{d^2}x} \\\int {\left\{ {bLn\left( {y + b} \right) + {c_1}} \right\}dy} = H(x)\int {x + {c_3}dx} \\b\left\{ {\left( {y + b} \right)Ln\left( {y + b} \right) - \left( {y + b} \right)} \right\} + {c_1}y + {c_2} = H(x)\left\{ {\frac{1}{2}{x^2} + {c_3}x + {c_4}} \right\}\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/895420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove two of $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6,\frac{2}{b}+\frac{3}{c}+\frac{6}{a}\geq 6,\frac{2}{c}+\frac{3}{a}+\frac{6}{b}\geq 6$ are True
if $a,b,c$ are positive real numbers that $a+b+c\geq abc$, Prove that at least $2$ of following inequalities are true.
$\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6, \space\space\space\space\frac{2}{b}+\frac{3}{c}+\frac{6}{a}\geq 6, \space\space\space\space\frac{2}{c}+\frac{3}{a}+\frac{6}{b}\geq 6$
Additional info: The Proof should be by contradiction.we can use Cauchy , AM-GM and other simple inequalities.
Things I have done so far: I don't have a complete idea for this Problem.I just think
that for starting step I should prove at least one of those inequalities are true.
So, I assume that $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}< 6, \space\space\space\space\frac{2}{b}+\frac{3}{c}+\frac{6}{a}< 6, \space\space\space\space\frac{2}{c}+\frac{3}{a}+\frac{6}{b}<6$
.Summing these
inequalities gives us: $$\frac{11}{a}+\frac{11}{b}+\frac{11}{c}<18$$
using Cauchy and $a+b+c\geq abc$ I can write:$$ab+bc+ac\geq9$$
So I can rewrite Previous inequality as: $$\frac{99}{abc}<18$$
And I stuck here.
UPDATE
Thanks to user169478 help, we proved that at least one of these inequalities is true.So the remaining is to prove that if one of these 3 is true then the another one is true. So any hint for starting this part is appreciated.
As it was Proved that at least one of those 3 inequalities is true, I assume that $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6$ is true. Now we suppose that $\frac{2}{b}+\frac{3}{c}+\frac{6}{a}< 6$ and $\frac{2}{c}+\frac{3}{a}+\frac{6}{b}<6$.So we can say $$\frac{8}{b}+\frac{5}{c}+\frac{9}{a}<12$$
We have $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6$ So we can re write last inequality as $$\frac{7}{a}+\frac{5}{b}-\frac{1}{c}<6$$
and I stuck at proving this inequality is false.
| This is a "brutal force" solution with the help of WolframAlpha. Hopefully someone else will give a more elegant proof later.
Set $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$, we have $xy+yz+xz\geq1$ due to $a+b+c\geq abc$
We can suppose, w.l.o.g, that $x\leq y \leq z$. Thus by the rearrangement inqeuality, $2x + 3y + 6z$ is the biggest one among the three sums. Since you have already proved that at least one of the three is no less than 6, we have $2x + 3y + 6z \geq 6$.
If $y \geq 1$, then we have $2z + 3x + 6y \geq 6$ for free, thus reach the conclusion.
Then we condiser the case when $x\leq y < 1$. Suppose $x= 1-a, y = 1-b$ for $0\leq b\leq a \leq 1$.
If $2y +3z + 6x < 6$ and $2z + 3x +6y < 6$, we have $$2 + 3z < 6a + 2b$$ $$2z + 3 < 3a + 6b$$ i.e.$$z < \frac{6a + 2b-2}{3}$$ $$z <\frac{3a + 6b-3}{2}$$.
Plug $x=1-a,y=1-b$ in $xy+yz+za \geq 1$, we get $z\geq\frac{a+b-ab}{2-a-b}$.
Thus we get $$\frac{6a + 2b-2}{3} > \frac{a+b-ab}{2-a-b}$$ $$\frac{3a + 6b-3}{2} > \frac{a+b-ab}{2-a-b}$$
i.e.$$6a^2 +2b^2+5ab -11a-3b+4 <0$$ $$3a^2+6b^2+7ab -7a-13b+6<0$$.
Draw these two curves in WolframAlpha, we get this.
Therefore we have the conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
How many ways can 5 dice produce a total of 20?
How many ways can $5$ dice produce a total of $20$?
I set up the equation $x_1+x_2+x_3+x_4+x_5 = 20$. The total possible number of combinations is $\binom{19}4$. From there I subtracted the number of possibilities where $1$ of the variables is greater than $6$, which I got as $5\times\binom{13}4$. I also subtracted the possibilities where $2$ variables is greater than $6$, which I used $10 \times \binom74$. I got the $10$ from the number of ways I can choose $2$ of the variables to be greater than $6$ out of the $5$ total variables.
So I have $$\binom{19}4 -5\times\binom{13}4 -10\times\binom74$$
However, I get a negative answer, which can't be right. Can anyone point a flaw in my logic?
| As in this answer, we can approach this question using either generating functions or inclusion-exclusion. Instead of counting the number of ways for $5$ numbers from $1$ to $6$ to sum to $20$, we will count the number of ways for $5$ numbers from $0$ to $5$ to sum to $15$ (then add $1$ to each of the $5$ numbers).
Generating Functions
The generating function for the number of ways for $5$ integers from $0$ to $5$ to sum to a given number is
$$
\begin{align}
\hspace{-1cm}(1+x+x^2+x^3+x^4+x^5)^5
&=\left(\frac{1-x^6}{1-x}\right)^5\\
&=\sum_{k=0}^5\binom{5}{k}(-1)^kx^{6k}\sum_{j=0}^\infty\binom{-5}{j}(-x)^j\\
&=\sum_{k=0}^5\binom{5}{k}(-1)^kx^{6k}\sum_{j=0}^\infty\binom{j+4}{j}x^j\tag{1}
\end{align}
$$
We can get the coefficient of $x^{15}$ in $(1)$ by choosing $6k+j=15$:
$$
\begin{align}
\hspace{-1cm}\sum_{k=0}^5(-1)^k\binom{5}{k}\binom{15-6k+4}{15-6k}
&=\binom{5}{0}\binom{19}{15}-\binom{5}{1}\binom{13}{9}+\binom{5}{2}\binom{7}{3}\\
&=651\tag{2}
\end{align}
$$
Inclusion-Exclusion
Without restriction on the size of the terms, using the standard $\mid$ and $\circ$ argument ($15$ $\circ$s and $4$ $\mid$s), there are $\binom{15+4}{4}$ ways to choose 5 non-negative integers that sum to $15$.
$$
\text{one sum for each arrangement}\\
2+4+6+1+2=\circ\,\circ\mid \circ\circ\circ\,\circ\mid \circ\circ\circ\circ\circ\,\circ\mid\circ\mid \circ\circ
$$
Now let's count how many ways there are to have terms greater than $5$. There are $\binom{5}{1}$ ways to choose which $1$ term should be greater than $5$. To count the number of sums with $1$ term at least $6$, that would be $\binom{15-6+4}{4}$.
$$
\text{consider the red group atomic}\\
2+7+3+2+1=\circ\,\circ\mid\color{#C00000}{\circ\circ\circ\circ\circ\circ}\circ\mid\circ\circ\circ\mid\circ\,\circ\mid\circ
$$
There are $\binom{5}{2}$ ways to choose which $2$ terms should be greater than $5$. To count the number of sums with $2$ terms at least $6$, that would be $\binom{15-12+4}{4}$.
$$
7+0+6+1+1=\color{#C00000}{\circ\circ\circ\circ\circ\circ}\circ\mid\mid\color{#C00000}{\circ\circ\circ\circ\circ\,\circ}\mid\circ\mid\circ
$$
There is no way for $3$ terms to be greater than $5$. Inclusion-Exclusion says there are
$$
\binom{19}{4}-\binom{5}{1}\binom{13}{4}+\binom{5}{2}\binom{7}{4}=651
$$
ways for $5$ terms to sum to $15$ with each term at most $5$.
Problem in the question
With Inclusion-Exclusion, the terms in the sum are alternating. If the last $-$ sign is changed to a $+$, your answer would be correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Simplifying $\sqrt[3]{a\pm\sqrt{b}}$ Let
$$x=\sqrt{a\pm\sqrt{b}}$$
We know that
$$x=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
But, what about cubic root?
Let
$$y=\sqrt[3]{a\pm\sqrt{b}}$$
Is there any formula to find $c$ and $d$ such that $c,d\in\mathbb{Q}$ and $c\pm\sqrt{d}=y$ if $c$ and $d$ exists?
For example, let
$$a=\sqrt[3]{45+\sqrt{1682}}$$
It can be solved factoring terms:
$$a=\sqrt[3]{45+29\sqrt{2}}=\sqrt[3]{27+27\sqrt{2}+18+2\sqrt{2}}=\sqrt[3]{(3+\sqrt{2})^3}=3+\sqrt{2}$$
Is there any formula for cubic root like square root?
| There isn't a formula for $\sqrt[3]{a\pm\sqrt{b}}$ The best method that I know is simplifying $\sqrt{b}$ (if possible) and assuming that it can be denested into $x+y\sqrt{b}$.
More generally, we have $$\sqrt[m]{A+B\sqrt[n]{C}}=a+b\sqrt[n]{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/897149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
When is $f=\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^5-3x+\ln5$ decreasing $\forall\; x$? When is $$f=\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^5-3x+\ln5$$ decreasing $\forall\; x$?
Diffrentiating:
$$f'=5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3$$
If $f$ is decreasing, $f'<0$:
$$5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3<0\implies x^4<\frac3{5\left(\frac{\sqrt{p+4}}{1-p}-1\right)}$$
Since if $\left(\frac{\sqrt{p+4}}{1-p}-1\right)\to0^+\implies x^4<+\infty$, which holds forallx, So:$$p=\frac{3\pm\sqrt{21}}2$$which doesn't ply with textbook-answer, since it mentions an domain for p.
| In order the function be always decreasing, the requirement is that $$f'=5\left(\frac{\sqrt{p+4}}{1-p}-1\right)x^4-3$$ be negative for any value of $x$. Since $x^4 \geq 0$, then $$\frac{\sqrt{p+4}}{1-p}-1 \lt 0$$ which implies either $p \gt 1$ or $-4 \lt p \lt \frac{1}{2} \left(3-\sqrt{21}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/897969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can there ever be infinite number of tuples of $(a,b,c,d)$ such that $ac-bd = k$ and $ad+bc = l$ for fixed $k,l$? Suppose, for now, that all numbers are real numbers. Let us fix numbers $k,l$.
Then can there ever be infinite number of tuples of $(a,b,c,d)$ such that $ac-bd =k$, $ad+bc = l$ for some $k$ and $l$?
What happens if numbers are integers?
Also, what happens if we change the restriction from integers/real numbers to any commutative numbers (field, ring etc.)?
| For real numbers, yes. One infinite family of solutions is for instance
$$
(a,b,c,d) = \left(r\frac{l+k}{2}, r\frac{l-k}{2}, \frac{1}{r}, \frac{1}{r}\right),
$$
for $r \neq 0$.
For integers, if $k = l = 0$, then $(a,b,c,d) = (n,n,0,0)$ gives infinite number of solutions. If $k \neq 0$ or $l \neq 0$,
$$
0 <k^2+l^2 = (ac-bd)^2+(ab+cd)^2= (a^2+b^2)(c^2+d^2),
$$
so $1 \leq a^2+b^2,c^2+d^2$, and
$$
k^2+l^2 = (a^2+b^2)(c^2+d^2) \geq a^2+b^2 \geq a^2,
$$
and similarly $a^2,b^2,c^2,d^2 \leq k^2+l^2$, which implies there are only finitely many solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Intersection of circle and ellipse I'm looking for the points of intersection of a circle
$x^2 + y^2 = r^2$ ($r$ is known, origin is $(0,0)$)
and an ellipse
$(x - x_0)^2 / a^2 + (y-y_0)^2 / b^2 = 1$ ($a,b,x_0,y_0$ are known).
Actually i do only need the Angles $\varphi$ at which the circle with radius $r$ is intersecting the ellipse.
| Hint:$$x^2 + y^2 = r^2\implies x^2 + y^2 - r^2=0$$ and $$(x - x_0)^2 / a^2 + (y-y_0)^2 / b^2 = 1\implies (x - x_0)^2 / a^2 + (y-y_0)^2 / b^2 - 1=0$$
So you're solving $$x^2 + y^2 - r^2=(x - x_0)^2 / a^2 + (y-y_0)^2 / b^2 - 1$$
But the solution will be long, however if $x_0=y_0=0$ then $$y=\pm\frac{\sqrt{a^2r^2+a^2x^2+a^2-x^2}}{\sqrt{\frac{a^2}{b^2}-a^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$
if $a,b,c$ are positive real numbers,Prove:$$a^3+b^3+c^3\geq a^2b+b^2c+c^2a$$
Things I have done so far: I know the fact that $$a^3+b^3+c^3\geq\frac{1}{2}[ab(a+b)+bc(b+c)+ca(c+a)]$$
However i tried to reach the problem inequality,but I was not succesful.
Source of problem: school exam.
| By aM-GM:
$$\frac{2a^3+b^3}3\ge(a^6b^3)^{\frac13}=a^2b$$
$$\frac{2b^3+c^3}3\ge(b^6c^3)^{\frac13}=b^2c$$
$$\frac{2c^3+a^3}3\ge(c^6a^3)^{\frac13}=c^2a$$
Add them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/900973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the sum of $3+4\cdot 3+4^2\cdot 3+\dots +4^{\log n-1} \cdot 3$ I see this:
$$A=3+4\cdot 3+4^2\cdot 3+\dots +4^{\log n-1} \cdot 3=3\cdot ([4^{\log n}-1]/3)=n^2-1$$
The base of logarithm is $2$, and $n$ is $2,4,8,\dots$
Anyone could describe me how this sum was calculated? Some hints or some tutorial for this?
| $$\begin{align}S&=3+3\cdot 4+3\cdot 4^2+3\cdot 4^3+...+3\cdot 4^{\log_2 n-1}\\
&=3[1+4+4^2+4^3+...+4^{\log_2 n-1}]\\
4S&=3[\quad\;\; 4+4^2+4^3+...+4^{\log_2n-1}+4^{\log_2 n}]
\quad \quad \text{as $n=1,2,4,8,...,2^m,...$}
\end{align}$$
Subtracting:
$$\begin{align}
3S&=3[4^{\log_2n}-1]\\
&=3[2^{2\log_2n}-1]\\
&=3[(2^{\log_2 n})^2-1]\\
&=3[n^2-1]\\
S&=n^2-1\end{align}$$
Note: The use of $n$ here is unconventional in that it is not the index count, i.e. $n\neq 1, 2, 3, 4, ....$ but instead $n=2^m=1, 2, 4, 8, 16, ...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/901379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Unique least square solutions There is a theorem in my book that states: If $A$ is $m\times n$, then the equation $Ax = b$ has a unique least square solution for each $b$ in $\mathbb{R}^m$.
But can we find a counter-example to this by providing a matrix $A$ and vector $b$ such that $A^TAx = A^Tb$ produces a general solution with a free variable?
| Your theorem statement is incomplete. Requirements have been omitted.
To amplify the insights of @Troy Woo, given a matrix $\mathbf{A}\in\mathbb{C}^{m \times n}$, a solution vector $x\in\mathbb{C}^{n}$, and a data vector $b\in\mathbb{C}^{m}$ such that $b\notin\mathcal{N}(\mathbf{A}^{*})$, and where $n\in\mathbb{N}$ and $m\in\mathbb{N}$, the linear system
$$
\mathbf{A} x = b
$$
has the least squares solution can be expressed in terms of the Moore-Penrose pseudoinverse $\mathbf{A}^{\dagger}$:
$$
x_{LS} = \mathbf{A}^{\dagger}b + \left(\mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right) y
$$
with the arbitrary vector $y\in\mathbb{C}^{n}$.
If the matrix rank $\rho < m$, the null space $\mathcal{N}\left(\mathbf{A}\right)$ is non-trivial and the projection operator $\left(\mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right)$ is non-zero.
Example
The linear system
$$
\begin{align}
\mathbf{A} x & = b \\
%
\left[
\begin{array}{cc}
1 & 0
\end{array}
\right]
%
\left[
\begin{array}{c}
x_{1} \\
x_{2}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
b_{1}
\end{array}
\right]
\end{align}
$$
has the least squares solution
$$
\begin{align}
x_{LS} & = \mathbf{A}^{\dagger} b + \left( \mathbf{I}_{2} - \mathbf{A}^{\dagger} \mathbf{A}\right) y\\
%
&=
%
\left[
\begin{array}{c}
b_{1} \\
0
\end{array}
\right]
%
+
%
\alpha
\left[
\begin{array}{c}
0 \\
1
\end{array}
\right]
\end{align}
$$
with $\alpha \in \mathbb{C}^{n}$.
The affine space of the solution satisfies
$$
\mathbf{A} \left(
\left[
\begin{array}{c}
b_{1} \\
0
\end{array}
\right]
%
+
%
\alpha
\left[
\begin{array}{c}
0 \\
1
\end{array}
\right]
\right) =
%
\mathbf{A} \left(
\left[
\begin{array}{c}
b_{1} \\
0
\end{array}
\right]
\right)
%
+
%
\alpha
\mathbf{A}
\left(
\left[
\begin{array}{c}
0 \\
1
\end{array}
\right]
\right) =
%
\mathbf{A} \left(
\left[
\begin{array}{c}
b_{1} \\
0
\end{array}
\right]
\right).
$$
The solution vector of least norm,
$$\Bigg\lVert
\left[
\begin{array}{c}
b_{1} \\
0
\end{array}
\right]
%
+
%
\alpha
\left[
\begin{array}{c}
0 \\
1
\end{array}
\right]
\Bigg\rVert_{2}^{2}$$
corresponds to $\alpha=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Two sequences defined by recurrence relations satisfy $x_n/y_n<\sqrt{7}$ for all $n$ Let $(x_n)_{n\geq 1}$ and $(y_n)_{n\geq 1}$ be two sequences such that:
$$x_{n+1}=x_n^2+1 \quad \text{ and } \quad y_{n+1}=x_n y_n$$
with $x_1=2$ and $y_1=1$
Prove that for all $n$
$$\dfrac{x_n}{y_n} < \sqrt{7}.$$
Can I have any help with this situation?
| Given that:
$$\frac{x_{n+1}}{y_{n+1}}=\frac{x_n}{y_n}+\frac{1}{x_n y_n}$$
we have:
$$\frac{x_{n+1}}{y_{n+1}}=\frac{1}{x_n y_n}+\frac{1}{x_{n-1} y_{n-1}}+\ldots+\frac{1}{x_1 y_1}+\frac{x_1}{y_1}$$
or just:
$$\frac{x_{n+1}}{y_{n+1}}=\frac{1}{x_n\cdot\ldots\cdot x_1}+\frac{1}{x_{n-1}\cdot\ldots\cdot x_1}+\ldots+\frac{1}{x_1}+2.$$
Hence we need to prove that:
$$\sum_{m=1}^{+\infty}\frac{1}{\prod_{n=1}^{m}x_n}\leq\sqrt{7}-2.\tag{1}$$
The form of the last inequality strongly suggest a continued fraction approach.
However, we can just notice that $x_1=2,x_2=5$ and for any $n\geq 3$ we have: $$x_n\geq 5^{2^{n-2}},$$
giving that the LHS of $(1)$ is less than
$$\frac{1}{2}+\frac{1}{10}+\frac{1}{10}\sum_{k=1}^{+\infty}\frac{1}{5^{2^k}}\leq \frac{3}{5}+\frac{1}{10}\sum_{k=1}^{+\infty}\frac{1}{5^{k+1}}=\frac{121}{200}.$$
Since $\frac{521}{200}<\sqrt{7}$, we're done.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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The number of positive integers whose digits are all $1$, $3$, or $4$, and add up to $2k$, is a perfect square I have been stuck on this question for a pretty long time. My teacher says that we should find a small pattern, but I can't find one. Can anyone give me a hand?
Let $b_n$ be the number of positive integers whose digits are all $1$, $3$, or $4$, and add up to $n$.
For example, $b_5 = 6$, since there are six integers with the desired property: $41, 14, 311, 131, 113,$ and $11111$.
Prove that $b_n$ is a perfect square if $n$ is even.
| First note that since $F_n=F_{n-1}+F_{n-2}$, we have
$$
\begin{align}
F_n^2
&=F_{n-1}^2+F_{n-2}^2+2F_{n-1}F_{n-2}\\
&=2F_{n-1}^2+2F_{n-2}^2-(F_{n-1}-F_{n-2})^2\\
&=2F_{n-1}^2+2F_{n-2}^2-F_{n-3}^2\tag{1}
\end{align}
$$
a recursion for the squares of the Fibonacci numbers.
The generating function for the count of numbers whose digits are $1$, $3$, or $4$, and whose digits sum to $n$ is
$$
\frac1{1-x-x^3-x^4}=\sum_{k=0}^\infty\left(x+x^3+x^4\right)^k\tag{2}
$$
To examine the coefficients of the even powers of $x$, we compute the even part of $(2)$:
$$
\begin{align}
\frac12\left(\frac1{1-x-x^3-x^4}+\frac1{1+x+x^3-x^4}\right)
&=\frac{1-x^4}{1-x^2-4x^4-x^6+x^8}\\
&=\frac{1-x^2}{1-2x^2-2x^4+x^6}\tag{3}
\end{align}
$$
Since the denominator of $(3)$ is $1-2x^2-2x^4+x^6$, the coefficients of $x^{2n}$ in $(3)$ satisfy the recursion $a_n=2a_{n-1}+2a_{n-2}-a_{n-3}$, which is recursion $(1)$, satisfied by the squares of the Fibonacci numbers.
Computing the beginning of the Taylor series for $(3)$, we get
$$
\frac{1-x^2}{1-2x^2-2x^4+x^6}=1+x^2+4x^4+9x^6+\dots\tag{4}
$$
and since the coefficients satisfy $(1)$, they are the squares of the Fibonacci numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to find that triple integral? How to find the triple integral of $$ \frac{(z-z_0)z}{\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}}$$ over the sphere $ \{(x,y,z):x^2+y^2+z^2 \le 1 \} $ under the assumption $x_0^2+y_0^2+z_0^2 \le 1?$ Its physical interpretation suggests
the integral can be expressed through elementary functions of the parameters.
| Let $\vec{p} = (x,y,z)$ and $\vec{p}_0 = (x_0,y_0,z_0)$. Let $(r,\theta,\phi)$ and $(r_0,\theta_0,\phi_0)$ be their spherical polar coordinates. More precisely,
$$\begin{array}{lll}
\vec{p} &= (x,y,z) &= r(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)\\
\vec{p}_0 &= (x_0,y_0,z_0) &= r_0(\sin\theta_0\cos\phi_0,\sin\theta_0\sin\phi_0,\cos\theta_0)
\end{array}$$
Let $\gamma$ be the angle between $\vec{p}$ and $\vec{p}_0$. i.e
$$\cos\gamma = \cos\theta\cos\theta_0 + \sin\theta\sin\theta_0\cos(\phi-\phi_0)$$
The integral $\mathcal{I}$ at hand can be rewritten as
$$\mathcal{I} = \int_{r\le 1} \frac{r^2\cos\theta^2 -rr_0\cos\theta\cos\theta_0}{\sqrt{r^2 + r_0^2 - 2rr_0\cos\gamma}} r^2 dr d\Omega$$
where $d\Omega = \sin\theta d\theta d\phi$ is the differential element for solid angle.
In terms of Legendre polynomials $P_\ell(x)$, we have
$$r^2\cos\theta^2 -rr_0\cos\theta\cos\theta_0 =
r^2\frac{2P_2(\cos\theta) + P_0(\cos\theta)}{3} - rr_0 P_1(\cos\theta)P_1(\cos\theta_0)$$
and
$$\frac{1}{\sqrt{r^2 + r_0^2 - 2rr_0\cos\gamma}} = \sum_{\ell=0}^\infty \frac{r_<^\ell}{r_>^{\ell+1}} P_\ell(\cos\gamma)
\quad\text{ where }\quad\begin{cases}r_< &= \min(r,r_0)\\ r_> &= \max(r,r_0)\end{cases}
$$
Using following identity for Legendre polynomials
$$\int P_\ell(\cos\gamma)P_{\ell'}(\cos\theta) d\Omega = \frac{4\pi}{2\ell+1}\delta_{\ell\ell'} P_{\ell'}(\cos\theta_0)$$
We find
$$\mathcal{I} =
\frac{8\pi}{15}A_{2,4}(r_0) P_2(\cos\theta)
+ \frac{4\pi}{3}A_{0,4}(r_0) P_0(\cos\theta)
- \frac{4\pi}{3} A_{1,3}(r_0) r_0 P_1(\cos\theta_0)^2$$
where
$$\begin{align}
A_{\ell,s}(r_0)
&= \int_0^1 \frac{r_<^\ell}{r_>^{\ell+1}} r^s dr
= \int_0^{r_0} \frac{r^{\ell+s}}{r_0^{\ell+1}} dr + \int_{r_0}^1 r_0^\ell r^{s-\ell-1} dr\\
&= \frac{r_0^s}{\ell+s+1} + r_0^\ell \left(\frac{1 - r_0^{s-\ell}}{s-\ell}\right)
= \frac{r_0^\ell}{s-\ell} + \frac{r_0^s(2\ell+1)}{\ell(\ell+1) - s(s+1)}
\end{align}
$$
As a result,
$$\begin{align}
\mathcal{I} &=
\frac{8\pi}{15}\left(\frac{r_0^2}{2} - \frac{5r_0^4}{14}\right) \frac{3 \cos\theta_0^2 - 1}{2}
+ \frac{4\pi}{3}\left(\frac14 - \frac{r_0^4}{20}\right)
- \frac{4\pi}{3} \left(\frac{r_0^2}{2} - \frac{3r_0^4}{10}\right) \cos\theta_0^2\\
&= \frac{4\pi}{210}\left(7-5r_0^2\right)\left(3z_0^2 - r_0^2\right)
+ \frac{4\pi}{60}\left(5 - r_0^4\right)
- \frac{4\pi}{30}\left(5-3r_0^2\right)z_0^2
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $a_{1}\;a_{2},a_{3}$ are the Roots of cubic eq. , Then $1000\left(a^2_{1}+a^2_{2}+a^2_{3}\right)$
If $a_{1}\;a_{2},a_{3}$ are three real values of $a$ which satisfy the equation $$\displaystyle \int_{0}^{1}\left(\sin x+a\cdot \cos x\right)^3dx-\frac{4a}{\pi-2}\int_{0}^{1}x\cdot \cos xdx = 2.$$ Then value of $\displaystyle 1000\left(a^2_{1}+a^2_{2}+a^2_{3}\right) = $
$\bf{My\; Trial::}$ Let $\displaystyle I = \int_{0}^{1}\left(\sin x+a\cdot \cos x\right)^3dx=\int_{0}^{1}\left(\sin^3 x+a^3\cos^3 x+3a\sin^2 x\cdot \cos x+3a^2\sin x\cdot \cos^2 x \right)dx$
So $\displaystyle I = -\int_{0}^{1}(1-\cos^2 x)\cdot (\cos x)^{'}dx+a^3\int_{0}^{1}(1-\sin^2 x)\cdot (\sin x)^{'}dx+3a\int_{0}^{1}(\sin x)^2\cdot (\sin x)^{'}dx-3a^2\int_{0}^{1} (\cos x)^2\cdot (\cos x)^{'}dx$
and Let $\displaystyle J = \int_{0}^{1}x\cdot \cos xdx = \left[x\cdot \sin x+\cos x\right]_{0}^{1} = \left(\sin 1+\cos 1-1\right)$
Now How can I solve after that
Help me
Thanbks
| If you can get the cubic into the form $C_3a^3+C_2a^2+C_1a+C_0 = 0$, then the roots satisfy
$a_1+a_2+a_3 = -\dfrac{C_2}{C_3}$
$a_1a_2+a_2a_3+a_3a_1 = \dfrac{C_1}{C_3}$
Thus, $a_1^2+a_2^2+a_3^2 = (a_1+a_2+a_3)^2-2(a_1a_2+a_2a_3+a_3a_1) = \dfrac{C_2^2-2C_1C_3}{C_3^2}$.
So, all you have to do is evaluate the following integrals:
$C_3 = \displaystyle\int_0^1 \cos^3 x\,dx$
$C_2 = \displaystyle\int_0^1 3\sin x \cos^2 x\,dx$
$C_1 = \displaystyle\int_0^1 3\sin^2 x \cos x\,dx - \dfrac{4}{\pi - 2}\int_0^1 x\cos x\,dx$.
Of course, this can get messy.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
A double sum involving the Riemann zeta function Evaluate the sum $S=\sum_{k=2}^{\infty} \frac{\zeta (k)-1}{k+1}$, where $\zeta (s)$ denotes the Riemann zeta function.
The sum is equal to $\sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{(k+1)n^k}$, then switching the order (since the summand converges uniformly) gives $$S=\sum_{n=2}^{\infty} \sum_{k=2}^{\infty} \frac{1}{(k+1)n^k}$$.
For the inner sum, I tried to evaluate it by considering $\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$ for $|x|<1$. Integrating w.r.t $x$ and then divided both sides by $x$ gives
$$\sum_{k=2}^{\infty} \frac{x^k}{k+1}= \frac{-1}{x}\ln |1-x|-1-\frac{x}{2}$$. Then put $x=1/n$ for $n=2,3,\cdots$. But then I dont how to evaluate the outer sum, please helps.
| Remember Stirling's formula
$$\sum_{m=1}^n \log m = \log (n!) = \left(n+\tfrac{1}{2}\right)\log n - n + \tfrac{1}{2}\log (2\pi) + O\left(\tfrac{1}{n}\right).$$
Then we have
$$\begin{align}
\sum_{n=2}^K \left(-n\log\left(1 - \tfrac{1}{n}\right) - 1 - \tfrac{1}{2n}\right)
&= 1-K +\tfrac{1}{2} - \tfrac{1}{2}H_K - \sum_{n=2}^K n\log(n-1) + \sum_{n=2}^K n\log n\\
&= \tfrac{3}{2} - K - \tfrac{1}{2}H_K - \sum_{n=2}^{K-1}\log n + K\log K\\
&= \left(\left(K+\tfrac{1}{2}\right)\log K - K + \tfrac{1}{2}\log (2\pi) - \sum_{n=1}^K\log n\right)\\
&\qquad + \tfrac{3}{2}+ \tfrac{1}{2}\underbrace{\left(\log K - H_K\right)}_{-\gamma + O(K^{-1})} - \tfrac{1}{2}\log (2\pi)\\
&= \tfrac{3}{2} - \tfrac{1}{2}\gamma - \tfrac{1}{2}\log (2\pi) + O\left(\tfrac{1}{K}\right),
\end{align}$$
where $H_K$ is the $K$-th harmonic number.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove logarithmic inequality with greatest integer function. $\left \lfloor n\log_2 n^2 \right \rfloor + \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor \leq \left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1$
How to show this? I tried using $\left \lfloor k \right \rfloor \leq k$, but I'm not sure that that does anything...
LHS $\leq \left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor $ < $\left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1$ because...ugh...
$\left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor \leq \log_2(\left \lfloor n\log_2n^2 \right \rfloor) < 1$?
I don't think the last inequality is true.
Help please?
I think it's supposed to hold for all natural numbers or at least all natural numbers after a certain index.
| hint:
$\left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1 =\left \lfloor (n\log_2 (n+1)^2)+\log_2 (n+1)^2 \right \rfloor +1\ge \left \lfloor (n\log_2 (n+1)^2) \right \rfloor +\left \lfloor \log_2 (n+1)^2\right \rfloor+1$
now prove:
$\left\lfloor (n\log_2 (n+1)^2) \right \rfloor \ge \left\lfloor (n\log_2 n^2) \right \rfloor$
$\left \lfloor \log_2 (n+1)^2\right \rfloor+1 \ge \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/907281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do I integrate $\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}$ How do I evaluate this indefinite integral, for $|k| < 1$:
$$
\int\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}\mathrm{d}x
$$
I tried the change of variable $t=\sin x$, and obtained two integrals, but I can't integrate either.
| The following solution is based on a suggestion of lab bhattacharjee:
$\displaystyle\int \frac{\sqrt{1-k^2\sin^2 x}}{\sin x} dx=\int\frac{\sqrt{1-k^2\sin^2 x}}{\sin^{2} x}\cdot\sin x dx=\int\frac{\sqrt{1-k^2(1-\cos^{2}x)}}{1-\cos^{2}x}\cdot\sin x dx$.
Now let $t=\cos x$, $dt=-\sin x dx$ to get $\displaystyle -\int\frac{\sqrt{1-k^2+k^2t^2}}{1-t^2}dt$; then let $kt=\sqrt{1-k^2}\tan\theta$ to get
$\displaystyle-\int\frac{\sqrt{1-k^2}\sec\theta}{1-\frac{1-k^2}{k^2}\tan^{2}\theta}\cdot\frac{\sqrt{1-k^2}}{k}\sec^{2}\theta d\theta=k(k^2-1)\int\frac{\sec^{3}\theta}{k^2\sec^{2}\theta-\tan^{2}\theta}d\theta$
$=\displaystyle k(k^2-1)\int\frac{\sec\theta}{k^2-\sin^{2}\theta}d\theta=k(k^2-1)\int\frac{\cos\theta}{\cos^{2}\theta(k-\sin\theta)(k+\sin\theta)} d\theta$.
Now let $u=\sin\theta$, $du=\cos\theta d\theta$ and use $\cos^{2}\theta=1-\sin^{2}\theta$ to obtain
$\displaystyle k(k^2-1)\int\frac{1}{(1-u)(1+u)(k-u)(k+u)}du$.
Then $\frac{1}{(1-u)(1+u)(k-u)(k+u)}=\frac{A}{1-u}+\frac{B}{1+u}+\frac{C}{k-u}+\frac{D}{k+u}\Rightarrow A=B=\frac{1}{2(k^2-1)} \text{ and }C=D=\frac{-1}{2k(k^2-1)},$
so $\displaystyle \frac{k}{2}\int\left(\frac{1}{1-u}+\frac{1}{1+u}-\frac{1/k}{k-u}-\frac{1/k}{k+u}\right) du=\frac{k}{2}\left[\ln\left\vert\frac{1+u}{1-u}\right\vert-\frac{1}{k}\ln\left\vert\frac{k+u}{k-u}\right\vert\right]+C$
where $\displaystyle u=\sin\theta=\frac{kt}{\sqrt{ k^{2}t^{2}+1-k^2}}=\frac{k\cos x}{\sqrt{ k^2\cos^{2}x+1-k^2}}=\frac{k\cos x}{\sqrt{1-k^2\sin^{2}x}}$.
| {
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"url": "https://math.stackexchange.com/questions/908576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Proof by induction (exponents)
Use proof by induction and show that the formula holds for all positive integers:$$1+3+3^2+\dots+3^{n-1}=\frac{3^n -1}2$$
The confusing step in my opinion is the first expression: $3^{n-1}$, when I have to show for $k+1$. Any solutions?
| Suppose that $x \neq 1$. We wish to show by inducting on $n$ that $\sum\limits_{k=0}^{n-1}x^{k} = \frac{x^{n} - 1}{x-1}$.
Then $1 = \frac{x-1}{x-1}$ so that the formula holds for $n = 1$.
Suppose that we have $\sum\limits_{k=0}^{n-1}x^{k} = \frac{x^{n} - 1}{x-1}$. We must show that $\sum\limits_{k=0}^{n}x^{k} = \frac{x^{n+1} - 1}{x-1}$.
We have $\sum\limits_{k=0}^{n}x^{k} = \sum\limits_{k=0}^{n- 1}x^{k} + x^{n} = \frac{x^{n} - 1}{x-1} + x^{n}\frac{x -1}{x-1} = \frac{x^{n} - 1 + x^{n+1} - x^{n}}{x - 1} = \frac{x^{n+1} - 1}{x-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/911357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate Integrals $ \int \sqrt{\sec 2x-1}\;dx$ and $ \int \sqrt{\sec 2x+1}\;dx$ Calculation of Integral of
$$\displaystyle \int \sqrt{\sec 2x-1}\;dx,\>\>\>\>\>\displaystyle \int \sqrt{\sec 2x+1}\;dx$$
$\bf{My\; Solution}::$ For $(a)::$
Let $$\displaystyle I = \int \sqrt{\sec 2x-1}\;dx = \int \sqrt{\frac{1-\cos 2x}{\cos 2x}}\;dx $$
$$\displaystyle = \int \sqrt{\frac{1-\cos 2x}{\cos 2x}\times \frac{1+\cos 2x}{1+\cos 2x}}dx = \int \frac{\sin (2x)}{\sqrt{\cos 2x\cdot (1+\cos 2x)}}dx$$
Now Let $\cos (2x) = t\;,$ Then $\displaystyle \sin (2x)dx = -2dt$
So Integral $$\displaystyle I =-2\int\frac{1}{\sqrt{t\cdot (1+t)}}dt=-2\int\frac{1}{\sqrt{(t+\frac{1}{2})^2+(\frac{1}{2})^2}}dt$$
So $\displaystyle I = -2\cdot \ln \left|\left(t+\frac{1}{2}\right)+\sqrt{t^2+t}\right|+\mathcal{C} = -2\cdot \ln \left|\left(\cos 2x+\frac{1}{2}\right)+\sqrt{\cos^2(2x)+\cos (2x)}\right|+\mathcal{C}$
Is There is any Method other then that,
If yes then plz explain here
Thanks
| Integrate as follows
\begin{align}
&\int \sqrt{\sec 2x-1}\;dx \\=&
\int \frac{\sqrt2\sin x}{\sqrt{\cos 2x}}dx=- \int \frac{d(\sqrt2\cos x)}{\sqrt{2\cos^2 x -1}}dx
= -\cosh^{-1}(\sqrt2 \cos x)+C\\
\\
&\int \sqrt{\sec 2x+1}\;dx \\=&
\int \frac{\sqrt2\cos x}{\sqrt{\cos 2x}}dx=\int \frac{d(\sqrt2\sin x)}{\sqrt{1-2\sin^2 x}}dx
= \sin^{-1}(\sqrt2 \sin x)+C\\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the correct way to compare to algebraic quantities? If I have two algebraic quantities, what is the correct way to determine if they are equal, or if not which is greater than the other?
For example, if I have $\frac{2d}{\sqrt{s^2-c^2}}$ and $\frac{2ds}{s^2-c^2}$, and I want to prove $\frac{2d}{\sqrt{s^2-c^2}} < \frac{2ds}{s^2-c^2}$ for all positive $d$, $s$, and $c$ I know that I can set them "equal" and treat them as an equation like so:
$$\frac{2d}{\sqrt{s^2-c^2}}=\frac{2ds}{s^2-c^2}$$
$$\frac{2d}{\sqrt{s^2-c^2}}\cdot\left(s^2-c^2\right) = \frac{2ds}{s^2-c^2}\cdot\left(s^2-c^2\right)$$
$$2d\sqrt{s^2-c^2}=2ds$$
$$\sqrt{s^2-c^2}=s$$
$$s^2-c^2=s^2$$
$$0<c^2$$ because we are given $c>0$.
Then you can replace all the equals signs all the way up with less thans since I didn't multiply by $-1$. This doesn't strike me as particularly formal, so what is the formal way to accomplish this, and in addition, is this even a valid line of reasoning?
| $$\begin{align}\frac{2ds}{s^2-c^2}-\frac{2d}{\sqrt{s^2-c^2}}&=\frac{2d}{\sqrt{s^2-c^2}}\left(\frac{s}{\sqrt{s^2-c^2}}-1\right)\\&=\frac{2d}{\sqrt{s^2-c^2}}\cdot \frac{s-\sqrt{s^2-c^2}}{\sqrt{s^2-c^2}}\\&=\frac{2d}{\sqrt{s^2-c^2}}\cdot \frac{s^2-(s^2-c^2)}{\sqrt{s^2-c^2}(s+\sqrt{s^2-c^2})}\\&=\frac{2dc^2}{(s^2-c^2)(s+\sqrt{s^2-c^2})}\\&\gt 0.\end{align}$$
Or easily, $$\frac{2ds}{s^2-c^2}\gt\frac{2d}{\sqrt{s^2-c^2}}\iff \frac{2ds}{s^2-c^2}\cdot \frac{s^2-c^2}{2d}\gt\frac{2d}{\sqrt{s^2-c^2}}\cdot \frac{s^2-c^2}{2d}\iff s\gt\sqrt{s^2-c^2}$$
$$\iff s^2\gt s^2-c^2\iff c^2\gt 0.$$
Since $c^2\gt 0$ is true, the first inequality is also true.
| {
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Integrate $\int\frac{x^3}{(x^3+1)^2}dx$ any ideas on how I could continue this integral
$$\int\frac{x^3}{(x^3+1)^2}dx$$
I am half way done, by entirely using fraction decomposition
$\int\frac{x^3}{(x^3+1)^2}dx=\int\frac{1}{x^3+1}dx-\int\frac{1}{(x^3+1)^2}dx$
$\frac{1}{x^3+1}=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$ from which it follows that $A=\frac{1}{3}, B=-\frac{1}{3}, C=\frac{2}{3}$
$\frac{1}{x^3+1}=\frac{1}{3(x+1)}-\frac{x-2}{3(x^2-x+1)}$
$\int\frac{1}{3(x+1)}dx=\frac{1}{3}\ln|{x+1}|+C$
$\int\frac{x-2}{3(x^2-x+1)}dx=\frac{1}{3}\left(\int\frac{2x-1}{2(x^2-x+1)}dx-\int\frac{3}{2(x^2-x+1)}dx\right)=\frac{1}{3}\left(\frac{1}{2}\ln|{x^2-1+1}|-\frac{3}{2}\frac{2}{\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})\right)+C$
$\int\frac{dx}{(x^3+1)^2}=?$
This is the part where I am stuck currently, using fraction decomposition seems to take a lot of time and effort, could I be missing anything?
| Hint: Try integrating by parts first, with $u=x$ and $dv={x^2\over(x^3+1)^2}dx$.
| {
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Find the quadratic equation equation of $x_1, x_2$.
Let $x_1 = 1 + \sqrt 3, x_2 = 1-\sqrt 3$. Find the quadratic equation $ax^2+bx+c = 0$ which $x_1, x_2$ are it's solutions.
By Vieta's theorem:
$$x_1\cdot x_2 = \frac{c}{a} \implies c=-2a$$
$$x_1 + x_2 = -\frac{b}{a} \implies b = -2a$$
Therefore, $b=c$
So we have a quadratic equation with the form:
$$-\frac{b}{2}x^2 + bx + b = 0$$
Applying $x_1$ for our equation I get $b=0$. Why?
The final answer is: $x^2-2x-2$.
| From your equation:
$−\frac{b}{2}x^2+bx+b=0$
notice that it's equal to $b*(−\frac{1}{2}x^2+x+1)=0$
Multiplying both sides of the equation by -2 we get:
$b*(x^2-2x-2)=0$
So the solutions to this become either $b=0$
or $x^2-2x-2=0$
Which imply $x=1 \pm \sqrt3$
| {
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Verifying proof of $\lim_{x \to\sqrt{2}}\frac{x^2-2}{x^2+\sqrt{2}x-4} = \frac 2 3$ $$\lim_{x \to\sqrt{2}} \dfrac{x^2-2}{x^2+\sqrt{2}x-4} = \lim_{w \to2} \dfrac{w^2-4}{w^2+2w-8} =\lim_{w \to2} \dfrac{(w-2)(w+2)}{(w+4)(w-2)} = \frac 2 3$$
Change of variable:
$$w=\sqrt{2}x \Rightarrow x=\frac{w}{\sqrt{2}}\Rightarrow x^2=\frac{w^2}{2}\text{ and if }x \to \sqrt{2} \text{ then }w\to 2.$$
Is that correct?
| well, notice that $$\frac{x^2-2}{x^2-\sqrt{2}x-4} = \frac{(x-\sqrt{2})(x+\sqrt{2})}{(x-\sqrt{2})(x+2\sqrt{2})}$$ which gives the same answer as yours.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the last non zero digit of 28!.
Find the last non zero digit of 28!.
It is very hard to multiply and find the last nonzero digit. I just wanna know that, is there any easy technique to solve this type of problem?
| Note that
$$5\times 10\times 15\times 20\times 25=5\times(2\cdot 5)\times (3\cdot 5)\times (2^2\cdot 5)\times 5^2=2^3\cdot 3\cdot 5^6$$
and that
$$10^6=2^6\cdot 5^6=\frac{8\cdot 5\cdot 10\cdot 15\cdot 20\cdot 25}{3}=8\cdot 5\cdot 10\cdot 5\cdot 20\cdot 25.$$
So, in mod $10$, we have
$$\begin{align}\frac{28!}{10^6}&\equiv 1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdot 9\cdot 1\cdot 2\cdot 3\cdot 4\cdot 3\cdot 6\cdot 7\cdot 8\cdot 9\cdot 1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdot 8\\&\equiv (1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7)^3\cdot 3\cdot 8^2\cdot 9^2\\&\equiv \{1\cdot 2\cdot 3\cdot 4\cdot (-4)\cdot (-3)\}^3 \cdot 3\cdot 4\cdot 1\\&\equiv 2\cdot 2\\&\equiv 4.\end{align}$$
| {
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"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Calculate number of integers less than n fitting the form 6n±1 Of course the approximation is n/3, but I am looking for a way to get the number of integers, not an approximation.
| There are several sequences that count the amount of $6k\pm1$ less than or equal to $n$.
Here is a specific sequence:
$$A_n=2\cdot\left\lfloor\frac{n+1}{6}\right\rfloor-\frac{(n\bmod6)^4}{24}+\frac{5(n\bmod6)^3}{12}-\frac{35(n\bmod6)^2}{24}+\frac{25(n\bmod6)}{12}$$
Here is the general solution:
*
*Divide by $3$ and write-down the error next to it:
*
*$\frac{ 0}{3},+\frac{0}{3}$
*$\frac{ 1}{3},+\frac{2}{3}$
*$\frac{ 2}{3},+\frac{1}{3}$
*$\frac{ 3}{3},+\frac{0}{3}$
*$\frac{ 4}{3},-\frac{1}{3}$
*$\frac{ 5}{3},+\frac{1}{3}$
*Rewrite the error as a function of $n\bmod6$:
*
*$f(0)=+\frac{0}{3}$
*$f(1)=+\frac{2}{3}$
*$f(2)=+\frac{1}{3}$
*$f(3)=+\frac{0}{3}$
*$f(4)=-\frac{1}{3}$
*$f(5)=+\frac{1}{3}$
*Compute the interpolating function and add $\frac{n}{3}$ to the result:
$$A_n=\frac{(n\bmod6)^5}{60}-\frac{5(n\bmod6)^4}{24}+(n\bmod6)^3-\frac{55(n\bmod6)^2}{24}+\frac{43(n\bmod6)}{20}+\frac{n}{3}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rational Expression Simplification The function:
$$f(x) = \frac{3x - 4}{x^2 - 2x}$$
is simplified to:
$$f(x) = \frac{2}{x} + \frac{1}{x - 2}$$
How? And in what way?
| We would like to write $\dfrac{3x-4}{x^2-2x} = \dfrac{3x-4}{x(x-2)}$ in the form $\dfrac{A}{x} + \dfrac{B}{x-2}$ for some constants $A,B$.
Now, we just need to solve for $A,B$. First, let's get rid of fractions:
$\dfrac{A}{x} + \dfrac{B}{x-2} = \dfrac{3x-4}{x(x-2)}$
$A(x-2)+Bx = 3x-4$
This equality must be true for all values of $x$, so it must be true for $x = 0$ and $x = 2$:
Plugging in $x = 0$ gives us $A \cdot (0-2) + B \cdot 0 = 3 \cdot 0 - 4$, i.e. $-2A = -4$. Hence, $A = 2$.
Plugging in $x = 2$ gives us $A \cdot (2-2) + B \cdot 2 = 3 \cdot 2 - 4$, i.e. $2B = 2$. Hence, $B = 1$.
Therefore, $\dfrac{3x-4}{x(x-2)} = \dfrac{2}{x} + \dfrac{1}{x-2}$.
You can also solve $A(x-2)+Bx = 3x-4$, by writing it as $(A+B)x+2A = 3x-4$, and equating coefficients. Then, you get the equations $A+B = 3$ and $2A = -4$. Solving this linear system gives $A = 2$ and $B = 1$ as you got with the first method.
Note: This technique is called Partial Fraction Decomposition. I suggest you Google that term for more information on this method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930525",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the circle which passes through two points
Find the equation of a circle which passes through $(4,-3)$ and $(-3,-4)$ with radius $5$.
I tried putting the $x$ and $y$ into the equation $(x-h)^2 + (y-k)^2 = r^2$, but then I don't know how to continue.
| Since we have
$$(4-h)^2+(-3-k)^2=5^2\iff h^2-8h+16+k^2+6k+9=25\tag1$$
$$(-3-h)^2+(-4-k)^2=5^2\iff h^2+6h+9+k^2+8k+16=25$$
subtracting the latter from the former gives you
$$(-8-6)h+7+(6-8)k-7=0\iff k=-7h.$$
Then, use $(1)$ to get $h=0,1$. So, the answer is the followings :
$$x^2+y^2=25,\ \ (x-1)^2+(y+7)^2=25.$$
| {
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"source": "stackexchange",
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} |
Proof by induction that $3^n \geq 2n^2 + 3n$ for $n \ge 4$ Problem:
If $n$ is a natural number and $n\geq4$, then $3^n \geq 2n^2 + 3n$. (Prove by Induction.)
Attempt at solution:
1) Given: $n$ is a natural number, $n \geq 4$.
2) Let $P(n)$ be the statement "$3^n \geq 2n^2 + 3n$."
3) $P(4) = 3^4 > 2(4)^2 + 3(4)$, i.e. $81 > 44$, thus the base case of $P(4)$ is true.
4) Assume $P(k) = 3^k \geq 2k^2 + 3k$ is true, where $k$ is a natural number and $k \geq 4$.
5) $P(k + 1) = 3^k(3) = 3^{k+1} > (2k^2 + 3k)(3) = 6k^2 + 9k$.
And this is where I am stuck. I know that I am trying to get $P(k+1)$ into the form $3^{k+1} > 2(k+1)^2 + 3(k + 1)$, and I have tried all sorts of algebraic manipulation, but I still am nowhere close to arriving at the correct form. How should I proceed?
| $3^n\geq2n^2+3n$
(1) $n=4$
$81\geq44$
(2) Assume true for $n=k$ ,Therfore we have :
$3^k\geq2k^2+3k$
(3) Then for $n=k+1$
$3^{k+1}\geq2(k+1)^2+3k+3$
Simplifying this gives:
$3(3^k)\geq2k^2+7k+5$
Now to prove this ^^^ you can subsititute what we assumed in (2) to show this inequality stands.
$3^k\geq2k^2+3k$
So..
$3(2k^2+3k)\geq2k^2+7k+5$
$6k^2+9k\geq2k^2+7k+5$
$6k^2+7k+2k\geq2k^2+7k+5$
^^^ This is true, therefore its true for n=k+1
So there you have it
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Evaluating $\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}$ I did this:
$$\begin{align}
\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x} &\sim \lim_{x \to 0} \frac{(1 + x + x^2)^{1/x} - (1 + x)^{1/x}}{x} = \\
&= \lim_{x \to 0} \left [ (1+x)^{1/x} \frac{\left ( \frac{1 + x + x^2}{1 + x} \right )^{1/x} - 1}{x} \right ] =\\
&= \lim_{x \to 0} (1+x)^{1/x} \cdot \lim_{x \to 0} \frac{\left ( 1 + \frac{x^2}{1 + x} \right )^{1/x} - 1}{x} =\\
&= e \cdot \lim_{x \to 0} \frac{e^{\frac{1}{x} \cdot \ln \left ( 1 + \frac{x^2}{1+x} \right )} - 1}{x} \sim \\
&\sim e \cdot \lim_{x \to 0} \frac{e^{\frac{x}{1 + x}} - 1}{x} \sim \\
&\sim e \cdot \lim_{x \to 0} \frac{1}{1 + x} = e
\end{align}$$
Is it right? If it is, how to evaluate the limit faster? It was pretty long the way I did it.
| We have $$\lim_{x\to0}(1+\sin x)^{\dfrac1x}\cdot\lim_{x\to0}\left(1+\frac{\sin^2x}{1+\sin x}\right)^{\dfrac1x}$$
$$\lim_{x\to0}(1+\sin x)^{\dfrac1x}=\left(\lim_{x\to0}(1+\sin x)^{\frac1{\sin x}}\right)^{\lim_{x\to0}\dfrac{\sin x}x}=e^1$$
$$\lim_{x\to0}\left(1+\frac{\sin^2x}{1+\sin x}\right)^{\dfrac1x}=\left(\lim_{x\to0}\left(1+\frac{\sin^2x}{1+\sin x}\right)^{\dfrac{1+\sin x}{\sin^2x}}\right)^{\lim_{x\to0}\dfrac{\sin^2x}{x(1+\sin x)}}=e^0$$
As $\displaystyle\lim_{h\to0}\left(1+h\right)^{\frac1h}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 5,
"answer_id": 0
} |
Closed-form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx$ Does the following series or integral have a closed-form
\begin{equation}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx
\end{equation}
where $\Psi_3(x)$ is the polygamma function of order $3$.
Here is my attempt. Using equation (11) from Mathworld Wolfram:
\begin{equation}
\Psi_n(z)=(-1)^{n+1} n!\left(\zeta(n+1)-H_{z-1}^{(n+1)}\right)
\end{equation}
I got
\begin{equation}
\Psi_3(n+1)=6\left(\zeta(4)-H_{n}^{(4)}\right)
\end{equation}
then
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)&=6\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\left(\zeta(4)-H_{n}^{(4)}\right)\\
&=6\zeta(4)\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}-6\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}\\
&=\frac{\pi^4}{15}\ln2-6\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}\\
\end{align}
From the answers of this OP, the integral representation of the latter Euler sum is
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}&=\int_0^1\int_0^1\int_0^1\int_0^1\int_0^1\frac{dx_1\,dx_2\,dx_3\,dx_4\,dx_5}{(1-x_1)(1+x_1x_2x_3x_4x_5)}
\end{align}
or another simpler form
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}&=-\int_0^1\frac{\text{Li}_4(-x)}{x(1+x)}dx\\
&=-\int_0^1\frac{\text{Li}_4(-x)}{x}dx+\int_0^1\frac{\text{Li}_4(-x)}{1+x}dx\\
&=\text{Li}_5(-1)-\int_0^{-1}\frac{\text{Li}_4(x)}{1-x}dx\\
\end{align}
I don't know how to continue it, I am stuck. Could anyone here please help me to find the closed-form of the series preferably with elementary ways? Any help would be greatly appreciated. Thank you.
Edit :
Using the integral representation of polygamma function
\begin{equation}
\Psi_m(z)=(-1)^m\int_0^1\frac{x^{z-1}}{1-x}\ln^m x\,dx
\end{equation}
then we have
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)&=-\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\int_0^1\frac{x^{n}}{1-x}\ln^3 x\,dx\\
&=-\int_0^1\sum_{n=1}^\infty\frac{(-1)^{n+1}x^{n}}{n}\cdot\frac{\ln^3 x}{1-x}\,dx\\
&=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx\\
\end{align}
I am looking for an approach to evaluate the above integral without using residue method or double summation.
| Edited: I have changed the approach as I realised that the use of summation is quite redundant (since the resulting sums have to be converted back to integrals). I feel that this new method is slightly cleaner and more systematic.
We can break up the integral into
\begin{align}
-&\int^1_0\frac{\ln^3{x}\ln(1+x)}{1-x}{\rm d}x\\
=&\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\int^1_0\frac{(1+x)\ln^3{x}\ln(1-x^2)}{(1+x)(1-x)}{\rm d}x\\
=&\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\int^1_0\frac{\ln^3{x}\ln(1-x^2)}{1-x^2}{\rm d}x-\int^1_0\frac{x\ln^3{x}\ln(1-x^2)}{1-x^2}{\rm d}x\\
=&\frac{15}{16}\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\frac{1}{16}\int^1_0\frac{x^{-1/2}\ln^3{x}\ln(1-x)}{1-x}{\rm d}x\\
=&\frac{15}{16}\frac{\partial^4\beta}{\partial a^3 \partial b}(1,0^{+})-\frac{1}{16}\frac{\partial^4\beta}{\partial a^3 \partial b}(0.5,0^{+})
\end{align}
After differentiating and expanding at $b=0$ (with the help of Mathematica),
\begin{align}
&\frac{\partial^4\beta}{\partial a^3 \partial b}(a,0^{+})\\
=&\left[\frac{\Gamma(a)}{\Gamma(a+b)}\left(\frac{1}{b}+\mathcal{O}(1)\right)\left(\left(-\frac{\psi_4(a)}{2}+(\gamma+\psi_0(a))\psi_3(a)+3\psi_1(a)\psi_2(a)\right)b+\mathcal{O}(b^2)\right)\right]_{b=0}\\
=&-\frac{1}{2}\psi_4(a)+(\gamma+\psi_0(a))\psi_3(a)+3\psi_1(a)\psi_2(a)
\end{align}
Therefore,
\begin{align}
-&\int^1_0\frac{\ln^3{x}\ln(1+x)}{1-x}{\rm d}x\\
=&-\frac{15}{32}\psi_4(1)+\frac{45}{16}\psi_1(1)\psi_2(1)+\frac{1}{32}\psi_4(0.5)+\frac{1}{8}\psi_3(0.5)\ln{2}-\frac{3}{16}\psi_1(0.5)\psi_2(0.5)\\
=&-12\zeta(5)+\frac{3\pi^2}{8}\zeta(3)+\frac{\pi^4}{8}\ln{2}
\end{align}
The relation between $\psi_{m}(1)$, $\psi_m(0.5)$ and $\zeta(m+1)$ is established easily using the series representation of the polygamma function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/934981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 5,
"answer_id": 1
} |
finding an indefinite integral of a fraction (a) Show that $\frac{4-3x}{(x+2)(x^2+1)}$ can be written in the form ${\frac{A}{x+2} + \frac{1-Bx}{x^2+1}}$ and find the constants $A$ and $B$.
(b) Hence find $\displaystyle\int\frac{4-3x}{(x+2)(x^2+1)}dx$
For (a) I found that $B=2$ and $A=2$
And I am not quite sure how to integrate. I tried to split them into two $\displaystyle\int\frac{2}{x+2
}dx$ and $\displaystyle\int\frac{1-2x}{x^2+1}dx$ but I don't know how to do after.
| Well,
$$\int\frac{2}{x+2} dx= 2\log(x+2)$$
So you have that part.
But for:
$$\int\frac{1-2x}{x^2 + 1} dx$$
You must further decompose this fraction into partial fractions:
$$\frac{1 - 2x}{x^2 + 1} = \frac{1}{x^2 + 1} - \frac{2x}{x^2 + 1}$$
So this integral becomes:$$\int\frac{1 - 2x}{x^2 + 1} dx= \int\frac{1}{x^2 + 1}dx - \int\frac{2x}{x^2 + 1}dx$$
$$= \tan^{-1}x - \log(x^2 + 1)$$
And overall:
$$\int\frac{4-3x}{(x+2)(x^2 + 1)}= 2\log(x+2) + \tan^{-1}x - \log(x^2 + 1) + c$$
Note that the constant is ommited till the end, just to simplify the answer.
| {
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"url": "https://math.stackexchange.com/questions/936436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to integrate $\frac{y^2-x^2}{(y^2+x^2)^2}$ with respect to $y$? In dealing with the integration,
$$\int\frac{y^2-x^2}{(y^2+x^2)^2}dy$$
I have tried to transform it to polar form, which yields
$$\int\frac{\sin^2\theta-\cos^2\theta}{r^2}d(r\cos\theta)$$
But, what should I do now to continue?
I am sticking on it now.
| Another way to find this integral would be to use
$\displaystyle\int\frac{y^2-x^2}{(y^2+x^2)^2}dy=\int\bigg(\frac{2y^2}{(x^2+y^2)^2}-\frac{1}{y^2+x^2}\bigg)dy$;
Using integration by parts for the first term with $u=y, dv=\frac{2y}{(x^2+y^2)^2}dy$ gives
$\bigg(\displaystyle-\frac{y}{y^2+x^2}+\int\frac{1}{y^2+x^2}dy\bigg)-\int\frac{1}{y^2+x^2}dy=\frac{-y}{y^2+x^2}+g(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/936705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find whether vector w belongs in the span $$v_1=[1,0,1,2]$$
$$v_2 = [0,1,1,3]$$
$$v_3 = [2,1,3,7]$$
$$w = [1,2,3,4]$$
We are supposed to determine if $w$ is in $\operatorname{span}(v_1,v_2,v_3)$.
| The row reduced form:
$$
\begin{align}
\mathbf{A} &\to \mathbf{E}_{\mathbf{A}} \\
%
\left[
\begin{array}{ccc}
1 & 0 & 2 \\
0 & 1 & 1 \\
1 & 1 & 3 \\
2 & 3 & 7 \\
\end{array}
\right]
%
&\to
%
\left[
\begin{array}{ccc}
1 & 0 & 2 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
%
\end{align}
$$
Therefore the matrix rank is $\rho = 2$.
We have two elementary columns, 1 and 2. If $w$ is in the span of the column space then we must have
$$
\left[
\begin{array}{c}
1 \\
2 \\
3 \\
4
\end{array}
\right]
=
\alpha
\left[
\begin{array}{c}
1 \\
0 \\
1 \\
2
\end{array}
\right]
+
\beta
\left[
\begin{array}{c}
0 \\
1 \\
1 \\
3
\end{array}
\right]
$$
The first two elements reveal the answer: $w$ is not in the span.
The vector $w$ can be resolved into $\color{blue}{range}$ and $\color{red}{null}$ space components
$$
\begin{align}
w &=
\color{blue}{w_{\mathcal{R}(\mathbf{A})}} +
\color{red}{w_{\mathcal{N}(\mathbf{A})^{*}}} \\
%
\left[
\begin{array}{c}
1 \\
2 \\
3 \\
4
\end{array}
\right]
&=
\frac{1}{17}
%
\color{blue}{\left[
\begin{array}{c}
21 \\
50 \\
71 \\
56 \\
\end{array}
\right]}
+
\frac{1}{17}
\color{red}{\left[
\begin{array}{r}
-4 \\
-16 \\
-20 \\
12 \\
\end{array}
\right]}
%
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
In $\triangle ABC$, $D$ is a point on side $BC$ that $\angle BAD = \angle CAD =\angle ABC$. If $BD=1$ and $DC=2$, what would be the length of $AB$?
In $\triangle ABC$, $D$ is a point on side $\overline{BC}$ that $\angle BAD = \angle CAD =\angle ABC$. If $\overline{BD}=1$ and $\overline{DC}=2$, what would be the length of $\overline{AB}$ ?
Things I have done: As $\overline{AD}$ is angle bisector this was the first thing came into my mind $$\frac{\overline{BD}}{\overline{DC}}=\frac{1}{2}=\frac{\overline{AB}}{\overline{AC}} \Rightarrow \frac{\overline{AC}}{2}=\overline{AB}$$
Applying Law of sines gives $$\frac{\sin{ \angle C}}{\sin{\angle B}}=\frac{\overline{AB}}{\overline{AC}} \Rightarrow \frac{\sin{ \angle C}}{\sin{\angle B}}\times \overline{AC} = \overline {AB}$$
thus $$\frac{\sin{ \angle C}}{\sin{\angle B}}=\frac{1}{2}$$
taking $\angle B = x$ results $$\frac{\sin{ \angle C}}{\sin{\angle B}}=\frac{\sin{(180-3x)}}{\sin{(x)}}=\frac{\sin{(3x)}}{\sin{(x)}}=\frac{1}{2}$$
And i stuck here. I solved this question previously by adding some elements and using symmetry and ... . but now I want to solve it without adding any element and constructions.Thanks.
Answer According to answer key: $\frac{\sqrt 6}{2}$
| Let $\alpha=\angle ABC$.
Since $AD$ is the bisector of $\angle CAB$, we have
$$AB:AC=BD:CD=1:2\Rightarrow AC=2AB.$$
So, since
$$\angle ACB=180^\circ-3\alpha\Rightarrow \sin(\angle ACB)=\sin(3\alpha)=3\sin \alpha-4\sin^3\alpha,$$we have with $\sin\alpha\not=0$, by the law of sines,
$$\begin{align}\frac{AB}{\sin (\angle ACB)}=\frac{AC}{\sin(\angle ABC)}&\Rightarrow \frac{AB}{3\sin\alpha-4\sin^3\alpha}=\frac{2AB}{\sin\alpha}\\&\Rightarrow \sin\alpha(8\sin^2\alpha-5)=0\\&\Rightarrow \cos^2\alpha=1-\sin^2\alpha=1-\frac 58=\frac 38.\end{align}$$
Hence, we have
$$AB=2\cos \alpha=2\sqrt{\frac 38}=\frac{\sqrt 6}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/938801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this proof by induction for a sum of odd squares correct? Statement: $1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = (n/3)*(2n-1)*(2n+1)$
Proof by induction
-Base case: when $n = 1$
$1^2 = 1/3 * (2 * 1 -1) * (2 * 1 +1) = 1$
$1=1$ hence statement holds for $n = 1$
-Inductive step
assume $n = k$ is true
then $1^2 + 3^2 + 5^2 + .. + (2k-1)^2 = (k/3) * (2k - 1) * (2k + 1)$
let $n = k+1$
then $1^2 + 3^2 + 5^2 + ... (2k - 1)^2 + (2(k+1)-1)^2$
it follows that $1^2 + 3^2 +5^2 .. + (2k-1)^2 = (k/3) * (2k-1) * (2k+1)$
such that $(k/3) * (2k-1)*(2k+1) + (2(k+1) -1)^2 = ((k+1)/3) * (2(k+1) - 1) * (2(k+1) +1)$
hence: $(4k^3+12k^2 + 11k + 3)/3 = (4k^3+12k^2 + 11k + 3)/3$
thus the statement holds for $n+1$ and the statement is therefore true
Is this proof correct? Am I missing anything or detail that I can add?
| After mentioning the case when $n=1$, the way you are writing looks strange to me.
Also, note that you don't need to expand them.
Assuming that $1^2+3^2+\cdots +(2k-1)^2=\frac{k(2k-1)(2k+1)}{3}$ holds, we have
$$\begin{align}1^2+3^2+\cdots +(2k-1)^2+(2(k+1)-1)^2&=\frac{k(2k-1)(2k+1)}{3}+(2k+1)^2\\&=\frac{2k+1}{3}\left(k(2k-1)+3(2k+1)\right)\\&=\frac{2k+1}{3}(2k^2+5k+3)\\&=\frac{2k+1}{3}(k+1)(2k+3)\\&=\frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/939004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to factor $(x^5+1) (x^5-1) $ I have this:
$ (x^5+1) (x^5-1) $
And I don't know how to continue factor.
Geogebra's Factor says:
$(x+1)(x-1)(x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$
| Hint:
$$x^n-1=(x - 1)[x^{n - 1} + x^{n - 2} + ... + x^2 + x + 1] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/939594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$
Integrate $$I=\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$$
Let $$\begin{align}u^2=\frac{\sin(x-a)}{\sin(x+a)}\implies 2udu&=\frac{\sin(x+a)\cos(x-a)-\sin(x-a)\cos(x+a)}{\sin^2(x+a)}dx\\2udu&=\frac{\sin((x+a)-(x-a))}{\sin^2(x+a)}dx\\
2udu&=\frac{\sin(2a)}{\sin^2(x+a)}dx\end{align}$$
Now:
$$\begin{align}u^2&=\frac{\sin(x+a-2a)}{\sin(x+a)}
\\u^2&=\frac{\sin(x+a)\cos(2a)-\cos(x+a)\sin(2a)}{\sin(x+a)}
\\u^2&=\cos(2a)-\sin(2a)\cot(x+a)
\\\cot(x+a)&=(\cos(2a)-u^2)\csc(2a)
\\\csc^2(x+a)=\cot^2(x+a)+1&=(\cos(2a)-u^2)^2\csc^2(2a)+1
\\\csc^2(x+a)&=\frac{\cos^2(2a)+u^4-2u^2\cos2(2a)+\sin^2(2a)}{\csc^2(2a)}
\\\sin^2(x+a)&=\frac{\sin^2(2a)}{u^4-2u^2\cos(2a)+1}\end{align}$$
Now:
$$\begin{align}
I&=\int u.\frac{2udu\sin^2(x+a)}{\sin(2a)}\\
I&=\int\frac2{\sin(2a)}.u^2.\frac{\sin^2(2a)}{u^4-2u^2\cos(2a)+1}du
\\I&=2\sin(2a)\int\frac{u^2}{u^4-2ku^2+1}du\quad k:=\cos 2a
\\\frac If&=\int\frac{2+u^{-2}-u^{-2}}{u^2-2k+u^{-2}}du=\int\frac{1+u^{-2}}{u^2-2k+u^{-2}}du+\int\frac{1-u^{-2}}{u^2-2k+u^{-2}}du\quad \\f:=\sin(2a)
\\&=\int\frac{d(u-u^{-1})}{(u-u^{-1})^2+2-2k}+\int\frac{d(u+u^{-1})}{(u+u^{-1})^2-2-2k}\end{align}$$
Now: $2-2k=2(1-\cos 2a)=4\sin^24a,2+2k=2(1+\cos 2a)=4\cos^24a$
So:
$$I=\sin2a\left(\frac1{2\sin4a}\arctan\left(\frac{u-u^{-1}}{2\sin(4a)}\right)+\frac1{4\cos4a}\ln\left|\frac{u+u^{-1}-2\cos 4a}{u+u^{-1}+2\cos 4a}\right|\right)+C$$
Or:
$$I=\frac1{4\cos2a}\arctan\left(\frac{-\sin a\cos x}{\sin4a\sqrt{\sin(x+a)\sin(x-a)}}\right)+\frac{\sin2a}{4\cos 4a}\ln\left|\frac{\sin x\cos a-\cos4a\sqrt{\sin(x+a)\sin(x-a)}}{\sin x\cos a+\cos4a\sqrt{\sin(x+a)\sin(x-a)}}\right|+C$$
But the textbook answer is:
$$\cos a\arccos\left(\frac{\cos x}{\cos a}\right)-\sin a\ln(\sin x+\sqrt{\sin^2x-\sin^2a})+c$$
| Other answers offer alternative approaches to integrating form the beginning. If you are looking to find where your steps went astray, it starts when you are using: $$2(1-\cos(2a))=4\sin^2(4a)$$ but the correct identity is: $$2(1-\cos(2a))=4\sin^2(a)$$ Just check both sides with $a=\pi/4$ and you'll believe it. And similarly for the next identity that you use: $2(1+\cos(2a))=4\cos^2(4a)$ should be $2(1+\cos(2a))=4\cos^2(a)$.
From there, you would have
$$I=\sin2a\left(\frac1{2\sin a}\arctan\left(\frac{u-u^{-1}}{2\sin(a)}\right)+\frac1{4\cos a}\ln\left|\frac{u+u^{-1}-2\cos a}{u+u^{-1}+2\cos a}\right|\right)+C$$
which gives
$$I=\cos(a)\arctan\left(\frac{u-u^{-1}}{2\sin(a)}\right)+\frac1{2}\sin(a)\ln\left|\frac{u+u^{-1}-2\cos a}{u+u^{-1}+2\cos a}\right|+C$$
And then we start back-substituting. We reach a point where it helps to use $\arctan\left(\frac{A}{B}\right)=\arcsin\left(\frac{A}{\sqrt{A^2+B^2}}\right)=\frac{\pi}{2}-\arccos\left(\frac{A}{\sqrt{A^2+B^2}}\right)$ and the constant can be absorbed into the $C$. Keep in mind that expressions in $a$ are constant for the purposes of $C$. Then we reach a point where by adding a certain logarithmic expression in $a$ (again, absorbed into $C$) we can simplify the appearance of the logarithmic term.
$$
\begin{align}
I&=\cos(a)\arctan\left(\frac{\sqrt\frac{\sin(x-a)}{\sin(x+a)}-\sqrt\frac{\sin(x+a)}{\sin(x-a)}}{2\sin(a)}\right)\\
&\phantom{=}{}+\frac1{2}\sin(a)\ln\left|\frac{\sqrt\frac{\sin(x-a)}{\sin(x+a)}+\sqrt\frac{\sin(x+a)}{\sin(x-a)}-2\cos a}{\sqrt\frac{\sin(x-a)}{\sin(x+a)}+\sqrt\frac{\sin(x+a)}{\sin(x-a)}+2\cos a}\right|+C\\
&=\cos(a)\arctan\left(\frac{\sin(x-a)-\sin(x+a)}{2\sin(a)\sqrt{\sin(x-a)\sin(x+a)}}\right)\\
&\phantom{=}{}+\frac1{2}\sin(a)\ln\left|\frac{\sin(x-a)+\sin(x+a)-2\cos(a)\sqrt{\sin(x-a)\sin(x+a)}}{\sin(x-a)+\sin(x+a)+2\cos(a)\sqrt{\sin(x-a)\sin(x+a)}}\right|+C\\
&=\cos(a)\arctan\left(\frac{-\cos(x)}{\sqrt{\sin(x-a)\sin(x+a)}}\right)\\
&\phantom{=}{}+\frac1{2}\sin(a)\ln\left|\frac{\sin(x)-\sqrt{\sin(x-a)\sin(x+a)}}{\sin(x)+\sqrt{\sin(x-a)\sin(x+a)}}\right|+C\\
&=-\cos(a)\arcsin\left(\frac{\cos(x)}{\sqrt{\cos^2(x)+\sin(x-a)\sin(x+a)}}\right)\\
&\phantom{=}{}+\frac1{2}\sin(a)\ln\left|\frac{\sin(x)-\sqrt{\sin(x-a)\sin(x+a)}}{\sin(x)+\sqrt{\sin(x-a)\sin(x+a)}}\right|+C\\
&=-\cos(a)\left(\frac{\pi}{2}-\arccos\left(\frac{\cos(x)}{\sqrt{\cos^2(x)+\frac12\cos(2a)-\frac12\cos(2x)}}\right)\right)\\
&\phantom{=}{}+\frac1{2}\sin(a)\ln\left|\frac{\sin(x)-\sqrt{\sin(x-a)\sin(x+a)}}{\sin(x)+\sqrt{\sin(x-a)\sin(x+a)}}\right|+C\\
&=\cos(a)\arccos\left(\frac{\cos(x)}{\cos(a)}\right)+\frac1{2}\sin(a)\ln\left|\frac{\sin(x)-\sqrt{\sin(x-a)\sin(x+a)}}{\sin(x)+\sqrt{\sin(x-a)\sin(x+a)}}\right|+C_1
\end{align}$$
Note that $-\cos(a)\frac{\pi}{2}$ has been absorbed into $C$.
It remains to show $$\frac1{2}\ln\left|\frac{\sin(x)-\sqrt{\sin(x-a)\sin(x+a)}}{\sin(x)+\sqrt{\sin(x-a)\sin(x+a)}}\right|=-\ln\left|\sin(x)+\sqrt{\sin^2(x)-\sin^2(a)}\right|+C_2(a)$$
If you add $\ln\left|\sin(x)+\sqrt{\sin^2(x)-\sin^2(a)}\right|$ to the left side, you have $$
\begin{align}
&\ln\left|\sqrt{\sin(x)-\sqrt{\sin(x-a)\sin(x+a)}}\sqrt{\sin(x)+\sqrt{\sin(x-a)\sin(x+a)}}\right|\\
&=\ln\left|\sqrt{\sin^2(x)-\sin(x-a)\sin(x+a)}\right|\\
&=\ln\left|\sqrt{\sin^2(x)-\frac12\cos(2a)+\frac12\cos(2x)}\right|\\
&=\ln\left|\sqrt{\sin^2(a)}\right|=C_2(a)\\
\end{align}$$
which establishes that last relation.
Also this write up has a typo at an earlier line, with $$\csc^2(x+a)=\frac{\cos^2(2a)+u^4-2u^2\cos2(2a)+\sin^2(2a)}{\csc^2(2a)}$$ but you meant $$\csc^2(x+a)=\frac{\cos^2(2a)+u^4-2u^2\cos(2a)+\sin^2(2a)}{\sin^2(2a)}$$ In any case, these things have been corrected at the next line.
| {
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"url": "https://math.stackexchange.com/questions/940037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
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Find side of an equilateral triangle inscribed in a rhombus The lengths of the diagonals of a rhombus are 6 and 8. An equilateral triangle inscribed in this
rhombus has one vertex at an end-point of the shorter diagonal and one side parallel to the longer
diagonal. Determine the length of a side of this triangle.
Express your answer in the form $k\left(4 \sqrt{3} − 3\right)$ where k is a vulgar fraction.
|
Given the picture, let $x$ be the side of the equilateral triangle.
We have:
$$ 6 = \frac{x}{2}\cot\arctan\frac{4}{3}+\frac{x}{2}\cot\frac{\pi}{6},$$
or:
$$ 6 = \frac{3x}{8}+\frac{\sqrt{3}\,x}{2},$$
so:
$$ 48 = x(4\sqrt{3}+3) $$
and:
$$ 48(4\sqrt{3}-3) = 39 x,$$
so $k=\color{red}{\frac{16}{13}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the series of the function 3 / ( 1- x^4) I know that $f(x) = \frac{1}{ 1-x } = \sum_{n=1}^\infty x^n$. We can find that $g(x) = \frac{1}{ 1-x^4 } = \sum_{n=1}^\infty (x^4)^n = \sum_{n=1}^\infty x^{4n}$. Does the sum converge? what is the convergence radius?
| The series $ 1+t+t^2+\cdots$ has radius of convergence $1$, so converges if $|t|\lt 1$ and diverges if $|t|\gt 1$. Thus our series $1+x^4+x^8+\cdots$ converges if $x^4\lt 1$, and diverges if $x^4\gt 1$. So we have convergence if $|x|\lt 1$, divergence if $|x|\gt 1$.
Remark: Your sums should start at $n=0$, not at $n=1$. And $g(x)=3+3x^4+3x^8+ 3x^{12}+\cdots=\sum_{n=0}^\infty (3)(x^{4n})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$
Show this equation holds by squaring both sides and comparing terms up to $x^3$.
I wonder, how can I square the right hand side?
| When you square the right-hand side, you will get another (infinite) polynomial $a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots$ . Now go through the coefficients $a_i$ and ask yourself which terms in the product $\big ( 1+ {1 \over 2} x - {1 \over 8} x^2 + \ldots\big) \big ( 1+ {1 \over 2} x - {1 \over 8} x^2 + \ldots\big)$ will contribute to $a_i$.
For example, there is only one product that contributes to $a_0$, namely $1 \cdot 1$, since the degree of $x$ in every other term is at least 1. Next, to get $a_1$, you can multiply $1$ from the first factor with ${1 \over 2} x$ from the second factor, or ${1 \over 2} x$ from the first and $1$ from the second. Thus $a_1 = 1\cdot {1 \over 2} + {1 \over 2} \cdot 1 = 1$. Continue similarly for the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove $\lim_{n\to\infty}\frac{2n}{n+2}=2$ Prove $$\lim_{n\to\infty} \frac{2n}{n+2} = 2$$
I know for all $\epsilon > 0$ there exists some $k\in\mathbb{N}$ for all $n > K$ and $\left|\frac{2n}{n+2} - 0\right| < \epsilon$.
Proof:
Fix any $\epsilon > 0$ choose $k \in \mathbb{N}$ where there exists $???? < K$
I am getting stuck on what to choose for $K$ in terms of $\epsilon$
I know that $\frac{2n}{n+2} < \frac{2n}{n} = 2$ but from there I get stuck.
| $\textbf{Proof:}$
Let $\epsilon >0$ be given.
$\textbf{RTP:}$ $\exists N_\epsilon \in \mathbb{N} \ni |\frac{2n}{n+2} -2|<\epsilon \ \forall \ n \geq N_\epsilon $.
$\textbf{ROUGH WORK:}$
We know that $|\frac{2n}{n+2}-2| = |\frac{-4}{n+2}| =\frac{4}{n+2} \forall n\in \mathbb{N}$.
Now we need $\frac{4}{n+2} < \epsilon \implies \frac{n+2}{4} > \frac{1}{\epsilon}$ (Since all values are positive)
$\implies n+2 > \frac{4}{\epsilon} \implies n> \frac{4}{\epsilon}-2$.
For the given $\epsilon > 0$, by the Archimedean Property, we may choose $N_\epsilon > \frac{4}{\epsilon}-2$.
We now have, $\forall n \geq N_\epsilon$ :
\begin{align*} n \geq N_\epsilon > \frac{4}{\epsilon} -2\end{align*}
$ \implies n+2 \geq N_\epsilon +2 > \frac{4}{\epsilon} \\ \implies \frac{n+2}{4} \geq \frac{N_\epsilon +2}{4} > \frac{1}{\epsilon} \\ \implies \frac{4}{n+2} \leq \frac{4}{N_\epsilon +2}<\epsilon \\ \implies \frac{4}{n+2}<\epsilon$
And hence, $\lim_{n\to \infty}(\frac{2n}{n+2})=2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $y = a\sin{x} + b\cos{x} +C$ then find maxima and minima for $y$. I was able to solve it till
$$y = \sqrt{(a^2 + b^2)}\sin(\alpha + x) + C.$$
But I don't know how to find maxima and minima from here.
If $C = 0$ then maxima & minima equals the amplitude of the sine curve but when $C$ is non-zero then?
I need help from here onwards.
| The old way:
$$y'=a\cos x-b\sin x=0\iff\tan x=\frac ab\iff x=\arctan\frac ab+k\pi.$$
Then
$$\sin\theta=\pm\frac{a}{\sqrt{a^2+b^2}},\\\cos\theta=\pm\frac{b}{\sqrt{a^2+b^2}},$$ and
$$y=\pm\frac{a^2+b^2}{\sqrt{a^2+b^2}}+c=\pm\sqrt{a^2+b^2}+c.$$
The maximum obviously corresponds to the plus sign.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int\limits^{\infty}_{0} \frac{x^\alpha}{(1+x^2)^2}dx, \ -1 < \alpha<3.$ Evaluate $\begin{align} \int^{\infty}_{0} \dfrac{x^\alpha}{(1+x^2)^2}\end{align}dx, \ -1 < \alpha<3.$
May I verify if my solution is correct? Thank you.
Consider $\gamma_1:=\{x:-\dfrac{1}{r}\leq x \leq-r\}, \ \gamma_2:=\{re^{it}: \pi\leq t \leq 0\}, \gamma_3:\{x:r \leq x \leq\dfrac{1}{r}\}, $ $\gamma_3:= \{\frac{1}{r}e^{it}: 0\leq t\leq \pi\},$ where $0<r<1.$
Let $f(z)=\dfrac{z^\alpha}{(1+z^2)^2}.$
$\begin{align} \int_{\gamma_{1}}f(z)dz=\int^{-r}_{-1/r}f(z)dz= \Biggl[\begin{array}{c} z=-w,w>0 \\\ dz=-dw \end{array}\Biggr]=-\int^{r}_{1/r}\dfrac{(-w)^\alpha}{(1+(-w^2))^2}dw \end{align}$ $= \begin{align} (e^{\pi i})^{\alpha}\int^{1/r}_{r}\dfrac{w^\alpha}{(1+w^2)^2}dw\end{align}.$
$\begin{align}\int_{\gamma_{3}}f(z)dz=\int^{1/r}_{r}\dfrac{x^\alpha}{(1+x^2)^2} dx\end{align}$
$\begin{align}\left|\int_{\gamma_{4}}f(z)dz \right| \to 0, \ r \to 0\end{align},$ since $-1 < \alpha <3.$
Similarly, $\begin{align}\left|\int_{\gamma_{2}}f(z)dz \right| \to 0, \ r \to 0\end{align}$
$i$ is a double pole of $f \implies 2\pi i Res(f,i)=2\pi i\lim_{z \to i}((z-i)^2f(z))^{\prime}= \dfrac{\pi(1-\alpha)i\alpha}{2}$
By Cauchy Residue Thm, $\begin{align}\dfrac{\pi(1-\alpha)i\alpha}{2}= (1+e^{\pi i \alpha})\int^{1/r}_{r}{\dfrac{f(x)}{(1+x^2)^2}dx}+\int_{\gamma_{2} \cup \gamma_{3}}f(z)dz \end{align}.$
Letting $r \to 0,$ we have : $\begin{align} \int^{\infty}_{0} \dfrac{x^\alpha}{(1+x^2)^2}\end{align}dx=\dfrac{\dfrac{\pi(1-\alpha)i\alpha}{2}}{(1+e^{\pi i \alpha})}=\dfrac{\pi(1-\alpha)}{4\text{cos}(\pi \alpha/2)}$
| I am providing an alternative route towards the evaluation of this integral.
Consider a change of variable $x^2=u$ then the integral can be expressed as
$$\int^{\infty}_{0}\frac{x^{\alpha}}{(1+x^2)^2}\,dx=\int^{\infty}_{0}\frac{u^{(\alpha-1)/2}}{2(1+u)^2}\,du=\frac{1}{2}\int^{\infty}_{0}\frac{u^{(\alpha-1)/2}}{(1+u)^2}\,du$$
Now using the Beta function
$$B(x,y)=\int^{\infty}_{0}\frac{u^{x-1}}{(1+u)^{x+y}}\,du$$
defined for $\Re{(x)}>0$ and $\Re{(y)}>0$ one can rewrite the integral as follows
$$\frac{1}{2}\int^{\infty}_{0}\frac{u^{(\alpha-1)/2}}{(1+u)^2}\,du=\frac{1}{2}\cdot B(\frac{\alpha+1}{2},\frac{3-\alpha}{2})$$
Now you can clearly see the restriction why $-1<\alpha<3$. Now lets evaluate the last result
$$\frac{1}{2}\cdot B(\frac{\alpha+1}{2},\frac{3-\alpha}{2})=\frac{1}{2}\cdot B(\frac{\alpha+1}{2},1+\frac{1-\alpha}{2})=\frac{1}{2}\cdot\frac{\frac{1-\alpha}{2}}{\frac{\alpha+1}{2}+\frac{1-\alpha}{2}}\cdot B(\frac{1+\alpha}{2},\frac{1-\alpha}{2})=\frac{1-\alpha}{4}\cdot B(\frac{1+\alpha}{2},\frac{1-\alpha}{2})$$
Using the fact $$B(x,y)=\frac{\Gamma{(x)}\Gamma{(y)}}{\Gamma{(x+y)}}$$
and $$\Gamma{(x)}\Gamma{(1-x)}=\frac{\pi}{\sin{(\pi x)}}$$
then the final result reads as follows
$$\frac{1-\alpha}{4}\cdot B(\frac{1+\alpha}{2},\frac{1-\alpha}{2})=\frac{1-\alpha}{4}\cdot\frac{\Gamma{(\frac{1+\alpha}{2})}\Gamma{(\frac{1-\alpha}{2})}}{\Gamma{(1)}}=\frac{1-\alpha}{4}\cdot\frac{\pi}{\sin{(\frac{\pi(1+\alpha)}{2}})}=\frac{\pi(1-\alpha)}{4\cos{(\frac{\pi\alpha}{2}})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Deriving $\frac{8}{\sqrt{x-2}}$ I'm not sure how to derive this:
$$\frac{8}{\sqrt{x-2}}$$
I tried
$$8 \cdot \frac{1}{\sqrt{x-2}}$$
$$8 \cdot (\sqrt{x-2})^{-1}$$
Differentiating w.r.t. $x$,
$$8 \cdot -1 \cdot (\sqrt{x-2})^{-2}$$
$$8 \cdot -1 \cdot \frac{1}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}{x-2}$$
But the answer is
$$\frac{-4}{(x-2)\sqrt{x-2}}$$
What should I have done?
| $\dfrac{8}{\sqrt{x-2}} = 8\cdot (x-2)^{-1/2}$.
Now you can use $\dfrac{d}{dx}(x^n) = n x^{n-1}$, where $n = -1/2$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Evaluate $\frac{ 1 }{ 1010 \times 2016} + \frac{ 1 }{ 1012 \times 2014} + \frac{ 1 }{ 1014 \times 2012} + \cdots + \frac{ 1 }{ 2016 \times 1010} = ?$ $$\dfrac{ 1 }{ 1010 \times 2016} + \dfrac{ 1 }{ 1012 \times 2014} + \dfrac{ 1 }{ 1014 \times 2012} + \cdots + \dfrac{ 1 }{ 2016 \times 1010} = ?
$$
My attempt so far :
$$\sum\limits_{n=0}^{503}\dfrac{1}{(1010+2n)(2016-2n)} = \dfrac{1}{6052}\sum\limits_{n=0}^{503}\left(\dfrac{1}{n+505} - \dfrac{1}{n-1008}\right)$$
It won't telescope/simplify further. I feel I am in wrong road. Any help ?
| The answers already showed that the expression cannot be more simplified. However, it could be quite accurately approximated since $$S_{m,n}=\sum_{i=m}^n \frac{1}{i}=H_n-H_{m-1}$$ Now, consider that both $m$ and $n$ are large numbers; we can use asymptotic expansions and, at the fourth order, $$S_{m,n}\approx \log \left(\frac{n}{m-1}\right)+\frac{1}{2} \left(\frac{1}{n}-\frac{1}{(m-1)}\right)-\frac{1}{12} \left(\frac{1}{n^2}-\frac{1}{(m-1)^2}\right)+\frac{1}{120} \left(\frac{1}{n^3}-\frac{1}{(m-1)^3}\right)+\cdots$$ Using your numbers, the above approximation gives $0.6926513948612855559$ while the summation leads to $\approx 0.6926513948612855562$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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} |
polynomial with positive integer coefficients divisible by 24? I have to show that $n^4+ 6n^3 + 11n^2+6n$ is divisible by 24 for every natural number, n, so I decided to show that this polynomial is divisible by 8 and 3, but I'm having trouble showing that it is divisible by 8.
Divisible by 3: $$n^4+ 6n^3 + 11n^2+6n \equiv n^4+2n^2 \pmod3 $$$$n^4+2n^2 = n^2(n^2+2)=n^2(n^2-1)=n^2(n+1)(n-1)\equiv 0 \pmod3$$
Divisble by 8: $$n^4+ 6n^3 + 11n^2+6n \equiv n(n^3-2n^2-5n-2) \pmod8 $$$$n(n^3-2n^2-5n-2) = n(n+1)(n^2-3n-2) = n(n+1)(n^2+5n+6) = n(n+1)(n+2)(n+3) = ??? \pmod8$$
So it seems that the polynomial has to be divisible by four, but I'm not sure how to show that it has to be divisible by two one more time to show that it's divisible by 8.
Any other approaches to solving the problem are welcome, as I'm preparing for an exam and would appreciate having multiple ways to tackle the problem. I think this problem was meant to be an exercise in proof by induction, but the other approach seemed more approachable.
| If we write the polynomial as a linear combination of combinatorial polynomials, then the polynomial is divisible by $24$ for all integer arguments if and only if all of the coefficients are divisible by $24$.
$$
n^4+6n^3+11n^2+6n=24\binom{n}{4}+72\binom{n}{3}+72\binom{n}{2}+24\binom{n}{1}
$$
and so the polynomial is divisible by $24$ for all integer arguments.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/958143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Can someone explain how to solve linear inequality?
Can someone explain how to solve this linear inequality?
| Multiply the both sides by $x^2-9=(x-3)(x+3)$.
But note that you need to have two cases as $x^2-9\gt 0$ or $x^2-9\lt 0$.
(Case 1)
If $x^2-9\gt 0\iff x\lt -3\ \text{or}\ x\gt 3$, then
$$\frac{6(x^2-9)}{x^2-9}+\frac{(13-x)(x+3)(x-3)}{x+3}\le\frac{3(x+3)(x-3)}{x+3}-\frac{2(x+3)(x-3)}{-(x-3)}$$
i.e.
$$6+(13-x)(x-3)\le 3(x-3)+2(x+3)\iff x\le 5\ \text{or}\ x\ge 6.$$
Hence, in this case, we have
$$x\lt -3\ \ \text{or}\ \ 3\lt x\le 5\ \ \text{or}\ \ x\ge 6.$$
(Case 2)
If $x^2-9\lt 0\iff -3\lt x\lt 3$, then
$$\frac{6(x^2-9)}{x^2-9}+\frac{(13-x)(x+3)(x-3)}{x+3}\color{red}{\ge}\frac{3(x+3)(x-3)}{x+3}-\frac{2(x+3)(x-3)}{-(x-3)}$$
i.e.
$$6+(13-x)(x-3)\ge 3(x-3)+2(x+3)\iff 5\le x\le 6.$$
Hence, in this case, there is no such $x\in\mathbb R$.
Hence, the answer is
$$x\lt -3\ \ \text{or}\ \ 3\lt x\le 5\ \ \text{or}\ \ x\ge 6.$$
| {
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"source": "stackexchange",
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Evaluating product of exponent and polynomial In a probability theory problem, I need to solve an inequality over $n\in\mathbb{N}$ which can be expressed in a general form like this:
$a^n ( b_1 n^2 + b_2 n + b_3) \leq c$ where $a, b_1, b_2, b_3, c$ are real numbers.
What is the standart way (or a good way) to tackle this? Is there an analytic solution?
For reference, here is the exact inequality:
$0.8^n ( \frac{1}{32} n^2 - \frac{7}{32} n + 1) \leq 0.1$
| Let's clean up a bit
$$ 0.8^n\left(\frac{1}{32}n^2-\frac{7}{32}n+1\right)\leq 0.1 $$
$$ \left(\frac{8}{10}\right)^n\left(\frac{1}{32}n^2-\frac{7}{32}n+1\right)\leq \frac{1}{10} $$
$$ \frac{8^n}{10^n}\left(\frac{1}{32}n^2-\frac{7}{32}n+1\right)\leq \frac{1}{10} $$
$$ \frac{8^n}{8}\left(\frac{1}{4}n^2-\frac{7}{4}n+8\right)\leq \frac{10^n}{10} $$
$$ 8^{n-1}\left(\frac{1}{4}n^2-\frac{7}{4}n+8\right)\leq 10^{n-1} $$
$$ \frac{1}{4}n^2-\frac{7}{4}n+8\leq \frac{10^{n-1}}{8^{n-1}} $$
$$ \frac{1}{4}\left(n^2-7n+32\right)\leq \left(\frac{5}{4}\right)^{n-1} $$
$$ n^2-7n+32\leq 4\left(\frac{5}{4}\right)^{n-1} $$
Now let's find the minimum of the quadratic function
$$ \frac{d}{dn}\left[ n^2-7n+32\right]=2n-7=0 $$
$$ n=\frac{7}{2}=3.5 $$
However since $n\in\mathbb{N}$ then you can start with $n=4$ and work your way up from there. I would choose $n$ in increments of $10$ as it usually doesn't take long before an exponentially increasing function becomes greater than a quadratic function. After a few more than $3$ computations, you'll realize that $n\geq 21$ renders the inequality true.
| {
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How to show that $a,\ b\in {\mathbb Q},\ a^2+b^2=1\Rightarrow a=\frac{s^2-t^2}{s^2+t^2},\ b= \frac{2st}{s^2+t^2} $ I want show the following $$a,\ b\in {\mathbb Q},\ a^2+b^2=1\Rightarrow a=\frac{s^2-t^2}{s^2+t^2},\ b= \frac{2st}{s^2+t^2},\ s,\ t\in{\mathbb Q} $$
How can we prove this ?
[Add] Someone implies that we must use pythagorean triple :
Let $$ a=\frac{n}{m},\ b= \frac{s}{k},\ (m,n)=(s,k)=1$$
Then $$ k^2n^2+s^2m^2=m^2k^2 \Rightarrow n^2|(k^2-s^2),\ k^2|m^2
$$ so that we have $$n^2+ s^2=k^2,\ (n,s)=1,\ k=m$$ We complete the proof by the following
Proof of pythagorean triple : $$a^2+
b^2=c^2,\ (a,b)=1$$ Then which form do $a,\ b,\ c$ have ? We have $$
\frac{c+a}{b}=\frac{A}{B}=\frac{b}{c-a},\ (A,B)=1$$
So $$ (c+a)B^2=bAB=(c-a)A^2 $$
So $$ c-a=B^2t,\ c+a=A^2t$$ That is $$ b=tAB,\
c=\frac{t}{2}(A^2+B^2),\ a= \frac{t}{2}(A^2-B^2)
$$
$(a,b)=1 \Rightarrow t=2$ or $1$ If $t=1$, then $AB$ is odd. So we can derive a contradiction.
| Set $a=\cos\theta$ and $b=\sin\theta$ rational. Then,
Set $s=1$ and $t=\tan(\frac{\theta}{2})$ then $a=\frac{s^2-t^2}{s^2+t^2}$ and $b=\frac{2st}{s^2+t^2}$.
To show that $t$ is rational, remark that $\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}$.
Q.E.D.
And actually it works for all $a,b\in\mathbb R$.
| {
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differentiation of the following equation 3 i already done the differentiation, just wanna confirm either i got it right or wrong. Can someone verify this for me.
1) f(x) = $ -3\over x^{5/2}$
f '(x)
= $ 3({ 5\over 2}x^{3/2})$ . $\frac{1}{x^5}$
= $ { 15\over 2}x^{(3/2)-5}$
= ${ 15\over 2}x^{-7/2}$
2) f(x) = $\frac{2x^2 + 3}{(x^3 - 4)^3}$
f ' (x)
$= \frac{(4x) (x^3 - 4)^3 - (2x^2 + 3)[3(x^3 -4)^2 (3x)}{[(x^3 - 4)^3]^2}$
$= \frac{(4x) (x^3 - 4)^3 - 9x^2 (2x^2 + 3)(x^3 -4)^2}{(x^3 - 4)^6}$
$= \frac{(4x) (x^3 - 4)^3 - 9x^2 (2x^2 + 3)}{(x^3 - 4)^4}$ ---quotient rule
3) f(x) = sin(x cos x)
$f ' (x) = \cos [x \cos(x)] (\cos x - x \sin x)$
4) f(x) = $x^2$ tan 2x
$f ' (x) = 2x \tan (2x)$ + $2x^2$ $\sec^2 2x$
5) f(x) = $ 3 \ln (\cos 3x) $
$f ' (x) = -9 \tan 3x$
| 1) $$\bigg( \frac{-3}{x^{5/2}} \bigg)'=-3x^\frac{-5}{2}=\frac{15}{2}
x^\frac{-7}{2}$$
2) $$ \bigg(\frac{2x^2+3}{(x^3-4)^3}\bigg)' =\bigg( (2x^2+3
)(x^3-4)^{-3}\bigg)' $$ $$= (4x)(x^3-4)^{-3} + (2x^2+3
)(-3)(x^3-4)^{-4}(3x^2) $$$$= (x^3-4)^{-3} [ 4x -9x^2(2x^2+3 )(x^3-4)^{-1}]
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplification of a series so that it converges to a given function I am trying to rearrange the series
$ \frac{1}{1-z} - \frac{(1-a)z}{(1-z)^2} + \frac{(1-a)^2z^2}{(1-z)^3} - \cdots$
In such a way that I can show it converges to
$\frac{1}{1-az} $
What I have so far
Let $ w = \frac{z}{1-z} $, we can then write the series as
$ \frac{w}{z} \left ( 1 - (1-a)w + (1-a)^2 w^2 - \cdots \right ) $
Which is the taylor series expansion about $0$ of
$ \frac{w}{z} \frac{1}{1+(a-1)w} $
Which I can simplify down to
$ \frac{1}{1+(a-2)z}$ as follows
$ \frac{w}{z} \frac{1}{1+(a-1)w} = \frac{1}{1-z} \left ( \frac{1}{1 + (a-1)\frac{z}{1-z}}\right) $
$= \frac{1}{1-z} \left ( {\frac{1-z + az - z}{1-z}}\right)^{-1} $
$ = \frac{1}{1+(a-2)z} $
Obviously this is not the same as $\frac{1}{1-az} $. I would love any guidance people can give as to where I went wrong.
| Even though it has already been answered, I might aswell post my already typed solution:
Your expresion is
$$\frac{1}{1-z} \cdot \sum_{n=0}^\infty \left(-\frac{1-a}{1-z}z\right)^n$$
Using the geometric series, this is
$$\frac{1}{1-z} \cdot \frac{1}{1+\frac{1-a}{1-z}z}$$
which is
$$\frac{1}{1-z} \cdot \frac{1-z}{1-z+(1-a)z} = \frac{1}{1-az}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Conditional expectations of $E(X+Y|z)$ Given: $$f(x,y,z) = \frac23 (x+y+z), \,\,\, 0<x<1,\,\,\, 0<y<1,\,\,0<z<1$$ zero elsewhere.I was instructed to determine the cumulative df of $x,y,z$. Here is my answer $$F (x,y,z) = \frac {xyz (x+xy+z)} {3} $$ Another problem that I can not answer is this one: Find the conditional distribution of $X$ and $Y$, given $Z=z$, and evaluate $E(X+Y|z)$.
| The first thing to notice is that you have mis-interpreted the first question. There is no common notion of the c.d.f. of multiple fvariables; what your professor or book wanted is the c.d.f. of $x$ which is just like the c.d.f. of $y$ or of $z$.
Thus for a value $x = X$,
$$
F(X) = \frac{2}{3}\int_{x=0}^{X} dx\int_{y=0}^{1} dy \int_{z=0}^{1} dz (x+y+z) = \frac{X^2+2X}{3}
$$
The conditional distribution of $(X,Y)$ given $z=Z$ is obtained by plugging $z=Z$ into $f(x,y,z)$ and then normalizing the resulting expression; this gives
$$
f(X,Y |z=Z) = \frac{x+y+Z}{1+Z}
$$
Given that, then
$$
E(X+Y | Z = z) = \frac{1}{1+z}\int_{x=0}^{1} dx\int_{y=0}^{1} dy (x+y) =\frac{1}{1+z}
$$
Finally, to find $E(X|y,z)$ we first find $f(X|y,z)$ by the same business of plugging in and normalizing:
$$
f(X|y,z) = \frac{(X+y+z)}{\int_0^1 (x+y+z) dx} = \frac{2}{1+2y+2z}(X+y+z)
$$
And then
$$
E(X|y,z) = \frac{2}{1+2y+2z} \int_0^1 x(x+y+z)dx
=\frac{2+3(y+z)}{3+6(y+z)}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Log integrals III The integral
\begin{align}
J_{m} = \int_{0}^{1} \frac{t^{m}}{1+t} \, \ln(1+t) \, dt
\end{align}
has the general form
\begin{align}
J_{m} = (-1)^{m} \left[ A_{m} - B_{m} \, \ln(2) + C_{m} \, \ln^{2}(2) \right]
\end{align}.
Is there a general form for the coefficients $A_{m}$, $B_{m}$, and $C_{m}$ ?
| The generating function is, according to Maple,
$$ \sum_{m=0}^\infty J_m x^m = \int_0^1 \dfrac{\ln(1+t)}{(1+t)(1-xt)}\; dt
= (1+x)^{-1} \left(\dfrac{1}{2} \ln(2)^2 + \ln \left(\dfrac{x}{1+x}\right) \ln(1-x) - \text{dilog} \left(\dfrac{x}{1+x}\right) + \text{dilog} \left(\dfrac{2x}{1+x}\right) \right)$$
but I don't know if you can get closed forms from that.
It appears (using the gfun package) that $(-1)^n A_n$ satisfies the recurrence
$$ \left( -{n}^{2}-5\,n-6 \right) a \left( n+1 \right) + \left( -{n}^{2}
-4\,n-3 \right) a \left( n+2 \right) + \left( {n}^{2}+9\,n+19 \right)
a \left( n+3 \right) + \left( {n}^{2}+8\,n+16 \right) a \left( n+4
\right) =0$$
while $(-1)^{n+1} B_n$ satisfies
$$ \left( -n-2 \right) b \left( n+1 \right) + \left( -n-2 \right) b
\left( n+2 \right) + \left( n+4 \right) b \left( n+3 \right) +
\left( n+4 \right) b \left( n+4 \right) =0$$
and $C_n = 1/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Hard Mathematical Induction I have a mathematical induction question and I know what I need to do just not how to do it.
The question is:
Prove the equality of:
$$(1 + 2 + . . . + n)^2 = 1^3 + 2^3 . . . + n^3$$
Base case:
$$(1 + 2)^2 = 1^3 + 2^3\\
(3)^2 = 1 + 8\\
9 = 9$$
and I understand I have to get the sides to equal each other though I'm not sure how to do that:
I use this:
$$(1 + . . . + n + (n + 1))^2 = 1^3 + . . . + n^3 + (n + 1)^3$$
but i can't seem to factor anything in anyway to figure it out . . .
I've tried putting the $S(n)$ in the $S(n + 1)$:
$$(1 + . . . + n + (n + 1))^2 = (1 + . . . + n)^2 + (n + 1)^3$$
but its just getting the $-(n + 1)^3$ on the first side I can't figure out...
Any help would be amazing!!!!
| Let the statement hold for $n \in \mathbb{N}$. For $n+1$, by applying the binomial formula $(a+b)^2=a^2+2ab+b^2$ you have that $$\begin{align*}\left(\underbrace{1+2+\ldots+n}_{=a}+\underbrace{(n+1)}_{=b}\right)^2&=(1+2+\ldots+n)^2+2(1+2+\ldots+n)(n+1)+(n+1)^2=\\&=1^3+2^3+\ldots+n^3+2\frac{(n+1)n}{2}(n+1)+(n+1)^2=\\&=1^3+2^3+\ldots+n^3+\underbrace{n(n+1)^2+(n+1)^2}_{=(n+1)(n+1)^2=(n+1)^3}=\\&=1^3+2^3+\ldots+n^3+(n+1)^3 \end{align*}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to integrate by parts, without changing variable How can I solve:
$$\int \frac{x^2}{\sqrt{1-x^2}}\;dx$$
without changing variables, by parts?
| $$\int\frac{x^2}{\sqrt{1-x^2}}\,dx=-\frac{1}{2}\int\frac{x}{\sqrt{1-x^2}}\,d(1-x^2)=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\int\sqrt{1-x^2}\,dx\Big)=..$$\begin{align}&...=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\frac{1}{2}\int(\sqrt{1-x^2}+\frac{1-x^2}{\sqrt{1-x^2}})\,dx\Big)\\&=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\frac{1}{2}\int(\sqrt{1-x^2}+\frac{1}{\sqrt{1-x^2}}+\frac{-x^2}{\sqrt{1-x^2}})\,dx\Big)\\&=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\frac{1}{2}\int(\sqrt{1-x^2}+\frac{-x^2}{\sqrt{1-x^2}})\,dx-\frac{1}{2}\int\frac{1}{\sqrt{1-x^2}}\,dx\Big)\\&=
-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\frac{1}{2}\int\,d(x\sqrt{1-x^2})-\frac{1}{2}\int\frac{1}{\sqrt{1-x^2}}\,dx\Big)\\&=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\frac{1}{2}x\sqrt{1-x^2}-\frac{1}{2}\arcsin{(x)}+c\Big)\\&=
-\frac{3}{4}x\sqrt{1-x^2}+\frac{1}{4}\arcsin{(x)}+c
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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When $a\to \infty$, $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? What does it mean that $\,\,f(a)=\sqrt{a^2+4}\,\,$ behaves as $\,a+\dfrac{2}{a},\,$
as $a\to \infty$?
How can this be justified?
Thanks.
| For positive $a$:
$$
\sqrt{a^2 + 4} = a\sqrt{1+\frac{4}{a^2}}
$$
and
$$\lim_{a\to\infty}\sqrt{1+\frac{4}{a^2}} = 1.
$$
So, the Taylor series of $\sqrt{1+x}$ around $0$ (when $x\to 0$) is:
$$
\sqrt{1+x} = 1 + \left.\frac{1}{2\sqrt{1+x}}\right\rvert_{x=0} x + O(x^2) = 1+ \frac{x}{2} + O(x^2)
$$
So, when $1<<a$ (it's important for $a$ to be big!) and putting $x = \frac{4}{a^2}$
$$
\sqrt{a^2 + 4} = a\left(1 + \frac{4}{2a^2} + O\left(\frac{1}{a^4}\right)\right)=a + \frac{2}{a} + O\left(\frac{1}{a^3}\right)
$$
EDIT: Corrected my answer.
| {
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how to solve $a\sin x+b\cos x$ Let's solve:
$\sqrt{3}\sin x - \cos x=2$
The left hand side may be expressed as $R\sin(x+ \phi)$
We know that $R=\sqrt{3+1}=2$
We also know that $\tan \phi= \frac{-1}{\sqrt{3}}$
The solution to $\tan \phi=\frac{-1}{\sqrt{3}}$ has many solutions, for example, -30, 150, 330 degrees etc.
Which of these solutions do we accept? Or is it irrelevant which we will accept? Which of these solutions are acceptable?
Thanks!
| $$\frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x=1$$
$$\sin\left(x -\frac{\pi}{6}\right) = 1$$
$$\sin\left(x -\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{2}\right)$$
$$ x = n\pi + (-1)^n \frac{\pi}{2} +\frac{\pi}{6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for all positive integers $x$, $\left\lfloor \frac{x^2 +2x + 2}{4}\right\rfloor =\left\lfloor \frac{x^2 + 2x + 1}{4}\right\rfloor$. Title says it all, basically. I believe it to be true that
$$\left\lfloor \dfrac{x^2 + 2x + 2}{4} \right\rfloor=\left\lfloor \dfrac{x^2 + 2x + 1}{4} \right\rfloor$$ for all positive integers $x$. I am, however, having a difficult time proving this. My current proof reads along the lines of the fact that, when adding $2$ or $1$, it is impossible to cause a large enough difference in the two numbers that, when divided by four and floored, they evaluate to different numbers. This basically comes down to proving that $x^2 + 2x \neq 4k + 2$ for some integer $k$, but I'm not sure if this is a good way of proving it.
Could anyone shine some light on this? Thanks.
| If $x=2k$, then
$$\left\lfloor\frac{x^2+2x+2}{4}\right\rfloor=\left\lfloor\frac{(2k)^2+2\cdot 2k+2}{4}\right\rfloor=\left\lfloor k^2+k+\frac 12\right\rfloor=k^2+k.$$
$$\left\lfloor\frac{x^2+2x+1}{4}\right\rfloor=\left\lfloor\frac{(2k)^2+2\cdot 2k+1}{4}\right\rfloor=\left\lfloor k^2+k+\frac 14\right\rfloor=k^2+k.$$
If $x=2k-1$, then
$$\left\lfloor\frac{x^2+2x+2}{4}\right\rfloor=\left\lfloor\frac{(2k-1)^2+2(2k-1)+2}{4}\right\rfloor=\left\lfloor k^2+\frac 14\right\rfloor=k^2.$$
$$\left\lfloor\frac{x^2+2x+1}{4}\right\rfloor=\left\lfloor\frac{(2k-1)^2+2(2k-1)+1}{4}\right\rfloor=\left\lfloor k^2\right\rfloor=k^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/977003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How do you solve this fraction?
What is the value of:
$$ \frac{2}{3} - \frac{1}{16} $$
My answer: $ \dfrac{1}{48} $ .
I believe that it's incorrect.
Three does not divide into $16$, so I cross multiplied. What am I doing wrong?
| \begin{align*}
\frac{2}{3}+\frac{-1}{16}&=\frac{2}{3}-\frac{1}{16} \\
&=\left(\frac{16}{16}\right)\times\frac{2}{3}-\frac{1}{16}\times\left(\frac{3}{3}\right) \\
&=\frac{32}{48}-\frac{3}{48} \\
&=\frac{29}{48}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/977715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Stuck with modular arithmetic problem using multiplication property I have the following problem:
Given $k\geq 1$, find $h$ such that
$$2^h \frac{4^k-1}{3}-1 \equiv 0 ~(\text{mod}~3).$$
This is my attempt using the invariance of multiplication:
$$2^h \frac{4^k-1}{3} \equiv 1 ~(\text{mod}~3) \Rightarrow 2^h \frac{4^k-1}{3}\cdot3 \equiv 1\cdot3 ~(\text{mod}~3) \Rightarrow 2^h(4^k-1) \equiv 0 ~(\text{mod}~3)$$
Since I know that
$$4^k - 1 \equiv 0 ~(\text{mod}~3) $$
then I can conclude that
$$\forall k, \forall h, 2^h(4^k-1) \equiv 0 ~(\text{mod}~3)$$
Anyway, this seems really wrong, since if I choose $h=3$ and $k=1$, then I get:
$$2^3 \frac{4^1-1}{3}-1 = 7 \not\equiv 0 ~(\text{mod}~3).$$
What's wrong with this proof?
| $$
\begin{align}
\frac{(1+3)^k-1}{3}
&=\frac{\left[\binom{k}{0}3^0+\binom{k}{1}3^1+3^2\sum\limits_{j=2}^k\binom{k}{j}3^{j-2}\right]-1}{3}\\
&=\binom{k}{1}+3\sum\limits_{j=2}^k\binom{k}{j}3^{j-2}\\
&\equiv\binom{k}{1}\pmod3
\end{align}
$$
Therefore,
$$
2^h\frac{(1+3)^k-1}{3}-1\equiv0\pmod3
$$
is the same as
$$
2^hk\equiv1\pmod3
$$
which is solvable when $k\not\equiv0\pmod3$ and then
$$
h=\left\{\begin{array}{}
0&\text{if }k\equiv1\pmod3\\
1&\text{if }k\equiv2\pmod3\\
\end{array}\right.
$$
satisfies the equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$ I am trying to evaluate:
$$\int_0^{\pi/12} \ln(\tan x)\,dx$$
I think the integral is quite simple but I am having a hard time evaluating it. I started with the result:
$$\int_0^{\pi/4} \ln(\tan x)\,dx= -G$$
where $G$ is the Catalan's constant. With the change of variables $x\rightarrow 3x$ and using the fact that $\tan(3x)=\tan x\tan\left(\frac{\pi}{3}+x\right)\tan\left(\frac{\pi}{3}-x\right)$, the integral is:
$$\int_0^{\pi/12}\ln(\tan x)\,dx+\int_0^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx+\int_0^{\pi/12}\ln \tan\left(\frac{\pi}{3}-x\right)\,dx=-\frac{G}{3}$$
$$\Rightarrow \int_0^{\pi/12}\ln(\tan x)\,dx+\int_{-\pi/12}^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx=-\frac{G}{3}$$
But I do not see how to proceed.
Help is appreciated. Thanks!
| An alternative "elementary" method.
Consider,
\begin{align*}
K&=\int_0^1 \frac{\arctan\left(\frac{x}{1-x^2}\right)}{x}\,dx\\
\end{align*}
Perform the change of variable $x=\tan\left(\frac{t}{2}\right) $,
\begin{align*}
K&=\int_0^{\frac{\pi}{2}} \frac{\arctan\left(\frac{1}{2}\tan t\right)}{\sin t}\,dt
\end{align*}
Défine the function $H$ on $\left[\frac{1}{2};1\right]$ to be,
\begin{align*}H(a)&=\int_0^{\frac{\pi}{2}} \frac{\arctan\left(a\tan t\right)}{\sin t}\,dt\end{align*}
Observe that $K=H\left(\dfrac{1}{2}\right)$ and,
\begin{align*}H(1)&=\int_0^{\frac{\pi}{2}} \frac{t}{\sin t}\,dt\\
&=\Big[t\ln\left(\tan\left(\frac{t}{2} \right)\right)\Big]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\ln\left(\tan\left(\frac{t}{2} \right)\right)\,dt\\
&=-\int_0^{\frac{\pi}{2}}\ln\left(\tan\left(\frac{t}{2} \right)\right)\,dt\\
\end{align*}
Perform the change of variable $x=\dfrac{t}{2}$,
\begin{align*}H(1)&=-2\int_0^{\frac{\pi}{4}}\ln\left(\tan\left(t \right)\right)\,dt\\
&=2\text{G}
\end{align*}
\begin{align*}H^\prime (a)&=\int_0^{\frac{\pi}{2}} \frac{\cos x}{1-(1-a^2)\sin^2 x}\,dt\\
&=\left[\frac{1}{2\sqrt{1-a^2}}\ln\left(\frac{1+\sin(x)\sqrt{1-a^2}}{1-\sin(x)\sqrt{1-a^2}}\right)\right]_0^{\frac{\pi}{2}}\\
&=\frac{1}{2\sqrt{1-a^2}}\ln\left(\frac{1+\sqrt{1-a^2}}{1-\sqrt{1-a^2}}\right)
\end{align*}
Therefore,
\begin{align*}H(1)-H\left(\frac{1}{2}\right)&=\int_{\frac{1}{2}}^1 \frac{1}{2\sqrt{1-a^2}}\ln\left(\frac{1+\sqrt{1-a^2}}{1-\sqrt{1-a^2}}\right)\,da\end{align*}
Perform the change of variable $y=\arctan\left(\sqrt{\dfrac{1+\sqrt{1-a^2}}{1-\sqrt{1-a^2}}}\right)$
\begin{align*}H(1)-H\left(\frac{1}{2}\right)&=-2\int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \ln\left(\tan y\right)\,dy\\
&=-2\int_0^{\frac{\pi}{4}} \ln\left(\tan y\right)\,dy+2\int_0^{\frac{\pi}{12}} \ln\left(\tan y\right)\,dy
\end{align*}
But, it is well known that,
\begin{align*}\int_0^{\frac{\pi}{4}} \ln\left(\tan y\right)\,dy=-\text{G}\\\end{align*}
Thus,
\begin{align*}\int_0^{\frac{\pi}{12}} \ln\left(\tan y\right)\,dy=-\frac{1}{2}K\\\end{align*}
On the other hand,
\begin{align}\int_0^1 \frac{\arctan\left( \frac{x}{1-x^2}\right)}{x}\,dx-\int_0^1 \frac{\arctan x}{x}\,dx=\int_0^1 \frac{\arctan \left(x^3\right)}{x}\,dx\end{align}
In the latter integral perform the change of variable $\displaystyle y=x^3$,
\begin{align}\int_0^1 \frac{\arctan\left( \frac{x}{1-x^2}\right)}{x}\,dx-\int_0^1 \frac{\arctan x}{x}\,dx=\frac{1}{3}\int_0^1 \frac{\arctan x}{x}\,dx\end{align}
Therefore,
\begin{align}\int_0^1 \frac{\arctan\left( \frac{x}{1-x^2}\right)}{x}\,dx&=\frac{1}{3}\int_0^1 \frac{\arctan x}{x}\,dx+\int_0^1 \frac{\arctan x}{x}\,dx\\
&=\frac{4}{3}\int_0^1 \frac{\arctan x}{x}\,dx\\
&=\frac{4}{3}\text{G}
\end{align}
Thus,
\begin{align*}\int_0^{\frac{\pi}{12}} \ln\left(\tan y\right)\,dy&=-\frac{1}{2}\times \frac{4}{3}\text{G} \\
&=\boxed{-\frac{2}{3}\text{G}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/983044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 8,
"answer_id": 3
} |
How to find the derivative of $f(x)=(x^3-4x+6)\ln(x^4-6x^2+9)$? Find the derivative of the following: $$f(x)=(x^3-4x+6)\ln(x^4-6x^2+9)$$ Would I use the chain rule and product rule?
So far I have:
$$\begin{align}g(x)=x^3-4x+6
\\g'(x)=2x^2-4\end{align}$$
would $h(x)$ be $\ln(x^4-6x^2+9)$?
If so, how would I find $h'(x)$?
| given $f(x)=(x^3-4*x+6)\ln(x^4-6x^2+9)$ we find by the product and chaine rule
$f'(x)=(3x^2-4)\ln(x^4-6x^2+9)+(x^3-4x+6)\frac{4x^3-12x}{x^4-6x^2+9}$
simplifying this in a few minutes
simplifying this i got
$f'(x)=\frac{4 x^4-16 x^2+3 x^4 \ln \left(x^4-6 x^2+9\right)-13 x^2 \ln \left(x^4-6
x^2+9\right)+12 \ln \left(x^4-6 x^2+9\right)+24 x}{x^2-3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/983200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplifying radicals I am stuck in the following puzzle and couldn't find a way to approach this.
$\sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{14 + \sqrt{180}}}}$
Please help.
| $$
\sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{14 + \sqrt{180}}}} = \sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{(\sqrt{5}+3)^2}}} = \sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{5} + 3}} = \sqrt{5 + \sqrt{5} + \sqrt{6+2\sqrt{5}}} = \sqrt{5 + \sqrt{5} + \sqrt{(\sqrt{5}+1)^2}} = \sqrt{5 + \sqrt{5} + \sqrt{5}+1}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/984040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Recognising that $\sum_{n=0}^\infty \frac{a^2-b^2(2n+1)^2}{(a^2+b^2(2n+1)^2)^2}=-\frac{\pi^2\mathrm{sech}^2\left(\frac{a\pi}{2b}\right)}{8b^2}$ So I know from Mathematica that:
$$\sum_{n=0}^\infty \frac{a^2-b^2(2n+1)^2}{(a^2+b^2(2n+1)^2)^2}=-\frac{\pi^2\mathrm{sech}^2\left(\frac{a\pi}{2b}\right)}{8b^2}$$
I am wondering how someone could come up with this answer logically without having to plug it into a computer. What kind of thought processes could you follow, upon seeing the sum on the left, to try and get a closed form expression?
| As Semiclassical noted, you look at
$$ \sum_{n=0}^\infty \dfrac{r^2 - (2n+1)^2}{(r^2 + (2n+1)^2)^2} $$
Expand the summand in partial fractions:
$$ \dfrac{r^2 - (2n+1)^2}{(r^2 + (2n+1)^2)^2} ={\frac {2 {r}^{2}}{ \left( r^2 + (2n+1)^2 \right) ^{2}}} - \dfrac{1}{ r^2 + (2n+1)^2 }$$
First deal with the term on the right:
$$ F(r) = \sum_{n=0}^\infty \dfrac{1}{r^2 + (2n+1)^2}$$
Let $$G(r) = \sum_{k=1}^\infty \dfrac{1}{r^2 + k^2} $$
so that $F(r)$ consists of the terms of $G(r)$ for odd $k$. But since
$$\dfrac{1}{r^2 + (2n)^2} = \dfrac{1}{4} \dfrac{1}{(r/2)^2 + n^2}$$
we have
$$G(r) = F(r) + \dfrac{1}{4} G(r/2)$$
Now it is a ``well-known'' identity that
$$ \pi \cot(\pi z) = \dfrac{1}{z} + \sum_{n=1}^\infty \dfrac{2z}{z^2 - n^2}
= \dfrac{1}{z} - 2 z G(iz) $$
and this leads to
$$ F(r) = \dfrac{\pi \coth(\pi r)}{2r} - \dfrac{\pi \coth(\pi r/2)}{4r} $$
which can be simplified to
$$ F(r) = \dfrac{\pi \tanh(\pi r/2)}{4r}$$
Now note that $$\dfrac{d}{dr} \dfrac{1}{r^2 + (2n+1)^2} = \dfrac{-2r}{\left(r^2+(2n+1)^2\right)^2}$$
so your sum is
$$ - \dfrac{\pi \tanh(\pi r/2)}{4r} - r \dfrac{d}{dr} \dfrac{\pi \tanh(\pi r/2)}{4r}$$
which should simplify to
$$ - \dfrac{\pi^2 \text{sech}^2(\pi r/2)}{8} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/985162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\int_{0}^{\infty} x \cdot \cos(x^3) dx$ convergence $$\int_{0}^{\infty} x \cdot \cos(x^3) dx$$
I only want to prove, that this integral converges, I don't need to calculate the exact value.
I don't know what to do with the cosinus, I can't get rid of it.
I know that the integral is equal to $$\frac{1}{3} \cdot \int_{0}^{\infty} \frac{\sin(x^3)}{x^2} dx$$
but here is also the problem, that I can't get rid of the sinus...
Any hints?
| Thanks, to everyone!
$$\int_{0}^{\infty} x \cdot \cos(x^3) dx = \frac{1}{3} \cdot \int_{0}^{\infty} \frac{1}{x} \cdot 3 \cdot x^2 \cdot \cos(x^3) dx$$
and because of $\int 3 \cdot x^2 \cdot \cos(x^3) dx = \sin(x^3)$ we get
$$\int_{0}^{\infty} x \cdot \cos(x^3) dx=\frac{1}{3} \cdot \int_{0}^{\infty} \frac{\sin(x^3)}{x^2} dx$$
We can split the integral into 2 parts:
$$\frac{1}{3} \cdot \int_{0}^{\infty} \frac{\sin(x^3)}{x^2} dx = \frac{1}{3} \cdot \left(\int_{0}^{1} \frac{\sin(x^3)}{x^2} dx+\int_{1}^{\infty} \frac{\sin(x^3)}{x^2} dx\right)$$
$\int_{0}^{1} \frac{\sin(x^3)}{x^2} dx$ is finite, so we only need to prove, that $\int_{1}^{\infty} \frac{\sin(x^3)}{x^2} dx$ is finite, too.
Since $\frac{\sin(x^2)}{x^2} \le \frac{1}{x^2}$ for $x > 0$, hence
$$\int_{1}^{\infty} \frac{\sin(x^3)}{x^2} dx \le \int_{1}^{\infty} \frac{1}{x^2} dx$$
$\int_{1}^{\infty} \frac{1}{x^2} dx$ is bounded, hence $\int_{1}^{\infty} \frac{\sin(x^3)}{x^2} dx$ is bounded, too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/986690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Solve complex equation $z^3 = i$ I have this $z^3 = i$ complex equation to solve.
I begin with rewriting the complex equation to $a+bi$ format.
1 $z^3 = i = 0 + i$
2 Calculate the distance $r = \sqrt{0^2 + 1^2} = 1$
3 The angle is $\cos \frac{0}{1}$ and $\sin \frac{1}{1}$, that equals to $\frac {\pi}{2}$.
4 The complex equation can now be rewriten $w^3=r^3(cos3v+i\sin3v)$, $w^3 = 1^3(\cos \frac {\pi}{2} 3 +i \sin \frac {\pi}{2} 3)$ or $w^3 = e^{i \frac {\pi}{2} 3}$.
5 Calculate the angle $3 \theta = \frac {\pi}{2} + 2 \pi k$ where $k = 0, 1, 2$
6 $k = 0$, $3 \theta = \frac {\pi}{2} + 2 \pi 0 = \frac {\pi}{6}$
7 $k = 1$, $3 \theta = \frac {\pi}{2} + 2 \pi 1 = \frac {\pi}{6} + \frac {2 \pi}{3} = \frac {5 \pi}{6}$
8 $k = 2$, $3 \theta = \frac {\pi}{2} + 2 \pi 2 = \frac {\pi}{6} + \frac {4 \pi}{3} = \frac {9 \pi}{6}$
So the angles are $\frac {\pi}{6}, \frac {3 \pi}{6}, \frac {9 \pi}{6}$ but that is no the correct answer. The angle of the complex equation should be $-\frac {\pi}{2}$ where I calculated it to $\frac {\pi}{2}$. I'm I wrong or is there a mistake in the book I'm using?
Thanks!
| Step $4$ is where your mistake happens. Your original equation is
$$z^3=i$$
Then you rewrite $i=1\cdot(\cos\frac\pi2 + i\sin\frac\pi2)$ and rewrite $z = r(\cos v + i\sin v)$, meaning that $$z^3=i$$
will change into $$r^3(\cos3v + i\sin 3v) = 1\cdot(\cos\frac\pi2 + i\sin\frac\pi2)$$
What you made was you also took the third power of $i$, which was wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/987222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Solving $y^2 - yx - y + x = 0$ for $y$? I solved this equation for $y$ by inspection and confirmed it with Wolfram Alpha -
$y^2 - yx - y + x = 0$
I got the values $y = 1$ and $y = x$
However I was wondering is there a formal method for solving it? I expressed it as a polynomial -
$y^2 + (-1 - x)y + x = 0$
and used the quadratic formula but it just left me with an awkward expression involving powers of $x$ and a square root...no $y = 1$ and $y = x$ which is what I was looking for...
So how would I go about solving this formally?
| Well, check again.
For $y^2+(−1−x)y+x=0$ with $y$ as the quadratic variable.
The quadratic formula with $b$ as $(-1-x)$, $a$ as $1$, and $c$ as $x$ tells that the roots are
$$y_{1,2}=\displaystyle \frac{(1+x) \pm \sqrt{1+x^2+2x - 4x}}{2}=\frac{(1+x)\pm\sqrt{1+x^2-2x}}{2}=\frac{(1+x) \pm{1-x}}{2}$$
Hence the solutions are $$y_1=\frac{1+x+1-x}{2}=1 \ , \ y_2=\frac{1+x-1+x}{2}=x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/987558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Beautiful evaluation inequality a,b,c>0 are such that $a^2+b^2+c^2=4$ and $4(a^2+2)=(a^2+b+c)^2$. What is the biggest possible value for a+b+c?
I tried a lot of stuff like $a^2=4-b^2-c^2$. And i think it's somehow connected to QM$\geq$AM where AM is (a+b+c)/3.
| Using the first relation, the left hand side of second relation can be written as $$4(a^2+2)=(a^2+b^2+c^2)(a^2+1^2+1^2)$$ On the right hand side we can apply the Cauchy-Schwarz inequality, so that $$4(a^2+2)=(a^2+b^2+c^2)(a^2+1^2+1^2)\overset{C.S.}\ge(a^2+b+c)^2$$ But due to the second relation we know that it holds actually with equality. Hence $(a,b,c)$ and $(a,1,1)$ must be linearly dependent. Since $a>0$ the only possible solution is $$(a,b,c)=1\cdot(a,1,1)$$ Thus $b=1$, $c=1$ and by substitution $a=\sqrt{2}$, which gives that $$a+b+c=\sqrt{2}+2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/989354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
sum of infinite series with telescopes I need help finding the sum of the infinite series
$$\sum_{k=1}^\infty \frac{1}{n(n+1)(n+2)}$$
I have used the partial fraction decomposition to get this as the sum of
$$\frac{-1}{k+1}+\frac{1}{2(k+2)}+\frac{1}{2k}$$
but don't know where to go from here.
Thanks!
| Hint. First observe that $$
\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)
$$ and conclude by telescoping terms.
You end up with
$$
\sum_{n=1}^{N}\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(1-\frac{1}{N+1}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{N+2}\right)
$$ and then let tend $N$ to $+\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/989562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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what is the limit of f(n)/g(n)? I am trying to solve this problem imagine that
\begin{array}{lr}
f(n) = 2^{\dfrac{1}{\sqrt{5}}\left[\left(\dfrac{1 + \sqrt{5}}{2}\right)^{n} - \left(\dfrac{1 - \sqrt{5}}{2}\right)^{n}\right]} \\
\\
g(n) = 2^{\left(\dfrac{1 + \sqrt{5}}{2}\right)^{n+1}}
\end{array}
and I want to find
\begin{equation*}
\lim_{n\to \infty} \dfrac{f(n)}{g(n)}
\end{equation*}
now in the middle of proof I stock on this limit, how can I solve it? ($\varphi$ is the golden number)
\begin{equation*}
\lim_{n\to \infty} -\frac{3+\sqrt{5}}{2} \varphi^n - \frac{1}{\sqrt{5}}(-\varphi)^{-n}
\end{equation*}
| For simplicity, I will let $\varphi$ denote the golden ratio, $\overline \varphi$ its conjugate, and $F(n)$ the $n^{th}$ Fibonacci number. That is,
$$\varphi = \frac{1 + \sqrt 5}{2} \;\;\;\;\; \overline \varphi = \frac{1 - \sqrt 5}{2} \;\;\;\;\; F(n) = \frac{\varphi^n - \overline \varphi^n}{\sqrt 5}$$
Then we have
$$f(n) = 2^{F(n)} \;\;\;\;\; g(n) = 2^{\varphi^{n+1}}$$
and seek the limit as $n \to \infty$ of $f(n)/g(n)$.
We note that, owing to properties of exponents,
$$\frac{f(n)}{g(n)} = 2^{F(n) - \varphi^{n+1}}$$
Then, by continuity,
$$L := \lim_{n \to \infty} \frac{f(n)}{g(n)} = 2^{\lim \limits_{n \to \infty} F(n) - \varphi^{n+1}}$$
Thus, it will be best for us to find the limit in the exponent. Going forward we will focus entirely on that exponent as well, and use the Binet formula for $F(n)$ as established above.
We see that
$$F(n) - \varphi^{n+1} = \frac{\varphi^n}{\sqrt 5} - \frac{\overline \varphi^n}{\sqrt 5} - \varphi^{n+1}$$
Swap the last two terms and then factor out $\varphi^n$, simplifying its coefficient, and we get the below.
$$F(n) - \varphi^{n+1} = \varphi^n \left( -\frac 1 2 - \frac{3}{2 \sqrt 5} \right) - \frac{\overline \varphi^n}{\sqrt 5}$$
From here, it's pretty easy to conclude our answer already. Why? We note that $\overline \varphi \approx -0.618 \in (-1,1)$. Thus, as $n \to \infty$, that term will go to zero. So, effectively,
$$\lim_{n \to \infty} F(n) - \varphi^{n+1} = \lim_{n \to \infty} \varphi^n \left( -\frac 1 2 - \frac{3}{2 \sqrt 5} \right)$$
The coefficient of the golden ratio here is clearly less than zero - about $-1.1708$ to be precise. Meanwhile, $\varphi \approx 1.618 > 1$, so if $n \to \infty$, then $\varphi^n \to \infty$. Thus,
$$\lim_{n \to \infty} F(n) - \varphi^{n+1} = -\infty$$
Thus, we've found that
$$L = 2^{\lim \limits_{n \to \infty} F(n) - \varphi^{n+1}} = 0$$
since the limit in the exponent is $-\infty$. This intuitively makes sense. Even at $n=10$, $f(n)/g(n) = 2^{F(n) - \varphi^{n+1}} \approx 10^{-44}$ per Wolfram.
| {
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Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$ for positive $a,b,c$
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, if $a,b,c$ are positive.
Well, I got that $bc(a+b+c)+ac(a+b+c)+ab(a+b+c)\geq9abc$.
| it is $\frac{a+b+c}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ $AM-HM$
| {
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If $a_1=1; a_{n}=a_{n-1}+\frac{1}{a_{n-1}}$ if $n>1$ then $a_{100}$ = ? Problem :
If $a_1=1; a_{n}=a_{n-1}+\frac{1}{a_{n-1}}$ if $n>1$ then $a_{100}$ = ?
Putting the value of n=2,3,4, we get :
when n =2 ; $a_2=1+\frac{1}{1} = 2$
when n=3; $a_3 =2+\frac{1}{2}$
when n=4; $a_4 =\frac{5}{2} +\frac{2}{5}$
Now how to get the $a_{100}$ term, please suggest thanks.
| To get a good approximation, note that $a_n^2 = \left(a_{n-1}+\dfrac{1}{a_{n-1}}\right)^2 = a_{n-1}^2+2+\dfrac{1}{a_{n-1}^2}$
Hence, $a_{100}^2 = a_1^2 + \displaystyle\sum_{n = 2}^{100}\left(a_n^2-a_{n-1}^2\right) = 1 + \sum_{n = 2}^{100}\left(2+\dfrac{1}{a_{n-1}^2}\right) = 199 + \sum_{n = 2}^{100}\dfrac{1}{a_{n-1}^2}$.
Since $a_n$ is strictly increasing, $a_n \ge a_3 = \frac{5}{2}$ for $n \ge 3$.
Thus, $\displaystyle\sum_{n = 2}^{100}\dfrac{1}{a_{n-1}^2} = \dfrac{1}{a_1^2}+\dfrac{1}{a_2^2}+\sum_{n = 4}^{100}\dfrac{1}{a_{n-1}^2} \le 1 + \dfrac{1}{4} + 97 \cdot \dfrac{1}{\left(\frac{5}{2}\right)^2} = 16.77$.
Also, $\displaystyle\sum_{n = 2}^{100}\dfrac{1}{a_{n-1}^2} = \dfrac{1}{a_1^2}+\dfrac{1}{a_2^2}+\sum_{n = 4}^{100}\dfrac{1}{a_{n-1}^2} \ge 1 + \dfrac{1}{4} + 97 \cdot 0 = 1.25$.
Therefore, $200.25 \le a_{100}^2 \le 215.77$, and thus, $14.15 \le a_{100} \le 14.69$.
If you compute $a_n$ for a few more small values of $n$, you can get tighter bounds on $\displaystyle\sum_{n = 2}^{100}\dfrac{1}{a_{n-1}^2}$, and hence, a tighter bounds on $a_{100}$.
| {
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Proving the determinant of a tridiagonal matrix with $-1, 2, -1$ on diagonal. Let $A_n$ denote an $n \times n$ tridiagonal matrix.
$$A_n=\begin{pmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{pmatrix} \quad\text{for }n \ge 2$$
Set $D_n = \det(A_n)$
Prove that $D_n = 2D_{n-1} - D_{n-2}$ for $n \ge 4$.
| Here is a way to compute the determinant without using induction. Multiply the matrix $A_n$ to the left by the $n\times n$ upper triangular matrix $U_n$ with all entries on and above the diagonal equal to$~1$:
$$
\begin{pmatrix}1&1&1&\ldots&1\\ 0&1&1&\ddots&\vdots\\0&0&1&\ddots&1\\
\vdots & \ddots & \ddots & \ddots & 1 \\0&0& \ldots &0 & 1
\end{pmatrix}
\begin{pmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\
& \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\
0 & & & -1 & 2 \end{pmatrix} =
\begin{pmatrix}1&0&0&\ldots&0&1\\ -1&1&0&\ddots&0&1\\0&-1&1&\ddots&0&1\\
\vdots & \ddots & \ddots & \ddots & 0&1 \\
\vdots & \ddots & \ddots & -1 & 1&1 \\0&0& \ldots &0&-1 & 2
\end{pmatrix}.
$$
Since $\det(U_n)=1$, we shall get $\det(A_n)$ as the determinant of our resulting matrix$~R$.
Now recognise that $R=I-C_P$ where $C_P$ is the companion matrix of the polynomial $P=X^n+X^{n-1}+\cdots+X^2+X^1+X^0$. Since it is well known that $P$ is the characteristic polynomial $\det(IX-C_p)$ of $C_P$ we get
$$
\det(A_n)=\det(R)=\det(I-C_P)=\det(IX-C_P)[X:=1]=P[X:=1]=n+1.
$$
| {
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Integrating a function from negative infinity to infinity i got this weird question about integration from infinity to infinity.
$$\int_{-\infty}^{\infty} \frac{1}{z^2+25}$$
First idea was to take the factor out to get $(z+5)(z-5)$ but that really did not achieve anything.
Then i tried let $x= z^2$ and the $\frac{dx}{dz}= 2z$ then dz = 1/2z but that also went nowhere
Then i tried..... $x=z$, $\frac{dx}{dz}= 1$ hence $dz = dx$
$$\int_{-\infty}^{\infty} \frac{dx}{x^2+25}$$ which has not achieved anything. Please help!
Wait i just remember how this applies to partial fractions. But still confused how to grom infinity to infinity.
So start wit
$$\frac{1}{z^2+25}=\frac{1}{(z+5)(z-5)}$$
$$\frac{1}{(z+5)(z-5)}=\frac{A}{x+5}+\frac{B}{x-5}$$
$$1=A(x-5)+B(x+5)$$
At x = 5 $1=10B$ hence $B=0.1
At x=-5 $A=1/-10$
But i'm not sure where to go from there
| I think i finally got it
let $z=5tan(x)$ hence $dz=5sec^2(x)dx$ sub this in to the integrand
$$\int_{-\infty}^{\infty} \frac{dz}{z^2+25} =\int_{-\infty}^{\infty} \frac{5sec^2(x)dx}{25tan^2(x)+25}$$
as sec^2(x)=1+tan^2(x)
$$\int_{-\infty}^{\infty} \frac{5(tan^2x+1)}{25(tan^2(x)+1)}= \int_{-\infty}^{\infty} \frac{1dx}{5}$$
$$\int_{-\infty}^{\infty} \frac{dz}{z^2+25} = \frac{x}{5}$$ for z =-$\infty,\infty$
as $$z = 5tan(x)$$ hence $$x=arctan(z/5)$$
observing its graph we find that as x approach positive and negative infinity artctan(z) equal $\frac{\pi}{2},-\frac{\pi}{2}$
this is the same for $x=arctan(z/5)$
hence, $$\int_{-\infty}^{\infty} \frac{dz}{z^2+25} =\frac{1}{5}(\frac{\pi}{2}-(-\frac{\pi}{2}) = \frac{\pi}{5}$$
| {
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Evaluating by real methods $\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx$ $\def\Li{{\rm{Li}}}$I'm sure you guys can briefly get the result by some methods of complex analysis, but now
I'm only interested in real analysis methods of proving the result. What would you propose
for that?
\begin{align*}
\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx=&\,\frac{\pi^6 \sqrt{2}}{768}+\frac{5 \sqrt{2}\pi^4}{64}\Li_2\left(2\sqrt2-3\right)-\frac{15\sqrt{2}\pi^2}{16}\Li_4\left(2\sqrt2-3\right)\\
&+\,\frac{15\sqrt{2}}{8}\bigg[\Li_6\left(2\sqrt2-3\right)-\Li_6\left(3-2\sqrt2\right)\bigg]
\end{align*}
And a supplementary question for another version, that is
$$\int_0^{\pi/2} \frac{x^5}{1+\cos^2(x)}\ dx$$
again, by real analysis methods only.
| Using double angle formula, the integrand can be rewritten as
\begin{equation}
I=\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx=\int_0^{\pi/2} \frac{2x^5}{3-\cos(2x)}\ dx
\end{equation}
Mapping the variable $2x\mapsto x$, we have
\begin{equation}
I=\frac{1}{32}\int_0^{\pi} \frac{x^5}{3-\cos x}\ dx
\end{equation}
Using identity (proof can be seen here)
\begin{equation}
1+2\sum_{n=1}^\infty \left(\frac{b}{a}\right)^n\cos(n x)=\frac{a^2-b^2}{a^2+b^2-2ab\cos x}\qquad,\qquad\mbox{for}\, |b|<a
\end{equation}
and the correspondence values $a=\dfrac{2+\sqrt{2}}{2}$ and $b=\dfrac{2-\sqrt{2}}{2}$, one may find
\begin{equation}
1+2\sum_{n=1}^\infty \left(3-2\sqrt{2}\right)^n\cos(n x)=\frac{2\sqrt{2}}{3-\cos x}
\end{equation}
Therefore
\begin{align}
I&=\frac{1}{64\sqrt{2}}\int_0^{\pi} \left[x^5+2\sum_{n=1}^\infty \left(3-2\sqrt{2}\right)^n x^5\cos(n x)\right]\ dx\\
&=\frac{1}{64\sqrt{2}} \left[\frac{\pi^6}{6}+2\sum_{n=1}^\infty \left(3-2\sqrt{2}\right)^n \int_0^{\pi} x^5 \cos(n x)\ dx\right]\\
\end{align}
The rest part can be done by multiple times integration by parts and using $\sin(n\pi)=0$ for $n\in\mathbb{Z}$. We will obtain
\begin{equation}
I=\frac{\pi^6 \sqrt{2}}{768}+\frac{\sqrt{2}}{64}\sum_{n=1}^\infty \left(3-2\sqrt{2}\right)^n \left[\frac{5\pi^4\cos(\pi n)}{n^2}-\frac{60\pi^2\cos(\pi n)}{n^4}+\frac{120\cos(\pi n)}{n^6}-\frac{120}{n^6}\right]\
\end{equation}
Using
\begin{equation}
\cos(n\pi)=\begin{cases}\,\,+1&,\,\,\mbox{if}\,\, n\,\,\mbox{is even}\\[12pt]
\,\,-1&,\,\,\mbox{if}\,\, n\,\,\mbox{is odd}\\
\end{cases}
\end{equation}
and the representation of polylogarithm function in term of its infinite series, we finally obtain
\begin{align*}
\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx=&\,\frac{\pi^6 \sqrt{2}}{768}+\frac{5 \sqrt{2}\pi^4}{64}\text{Li}_2\left(2\sqrt2-3\right)-\frac{15\sqrt{2}\pi^2}{16}\text{Li}_4\left(2\sqrt2-3\right)\\
&+\,\frac{15\sqrt{2}}{8}\bigg[\text{Li}_6\left(2\sqrt2-3\right)-\text{Li}_6\left(3-2\sqrt2\right)\bigg]
\end{align*}
The same approach can be applied for evaluating the second integral.
| {
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the inequality $\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{a+b+c}2$ How to show that $$\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{a+b+c}2$$ for $a,b,c>0$?
I tried to prove $$\frac{a^4}{a^3+b^3}\ge \frac {5a}4+\frac{-3b}4$$
but could not continue.
Give me ideas, please.
| Here is a solution from "Secrets in Inequalities" by Pham Kim Hung.
| {
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Evaluate $\int_0^2\frac{x^5}{\sqrt{x^3+6}}\,dx.$ I am stuck on the following integral:
$\displaystyle\int_0^2\dfrac{x^5}{\sqrt{x^3+6}}\,dx.$
I have no idea how one can work it out. Normally I'd try $u=x^3+6$ but this surely does not work here.
| $$\frac{x^5}{\sqrt{x^3+6}}=x^3\frac{x^2}{\sqrt{x^3+6}}$$
Set $\sqrt{x^3+6}=u\implies x^3=u^2-6,\dfrac{3x^2dx}{\sqrt{x^3+6}}=du$
| {
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What is the remainder on division of $z^{400} + z^{303} + 1$ by $z^4-1$? I am asked to determine the remainder when the polynomial $f(z)=z^{400}+z^{303}+1$ is divided by the polynomial $g(z)=z^4-1$.
I expressed f as $f(z) = h(z)(z^4-1) + r(z)$ where $r(z)$ is a polynomial
Realising that $z^4-1 = (z^2-1)(z^2+1) = (z-1)(z+1)(z^2+1)$, I expressed f as
$f(z) = h(z)(z-1)(z+1)(z^2+1) + r(z)$
I also wrote f as $f(z) = q(z)(z-1) + 3$, since $f(1)=3$, and q is some polynomial.
So I had $f(z)= q(z)(z-1) + 3 = h(z)(z-1)(z+1)(z^2+1) + r(z)$
Hence, for some constant k, $q(z)=h(z)(z+1)(z^2+1) + k$
We know $f(-1) = 1$
From $f(z) = q(z)(z-1) + 3$, we have $f(-1)=q(-1)(-2) + 3 = 1$, which implies
$q(-1) = 1$.
Then $q(-1)=h(-1)(0) + k = 1$. Thus, $k=1$
We then have $q(z)=h(z)(z+1)(z^2+1) + 1$, and from this, we have
$f(z)= q(z)(z-1) + 3 = (h(z)(z+1)(z^2+1) + 1)(z-1) + 3$
Which simplifies to $f(z) = h(z)(z^4-1) + z+2$
And gives $r(z) =z+2$.
But I am given the answer is $r(z) = z^3+ 2$. What is going wrong here? Can I use modular arithmetic instead? How?
| Another way:
Since
\begin{align*}
z^{400}+z^{303}+1&=z^{400}-1+(z^{300}-1)z^3 +z^3+2 \\
&=(z^4-1)(z^{396}+z^{392}+\ldots +1)+(z^4-1)(z^{296}+z^{292}+\ldots +1)z^3+z^3+2\\
\end{align*}
It follows that $z^3+2$ is the asked remainder.
| {
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How do I simplify $\frac{\sqrt{1-x} + \frac{1}{\sqrt{1+x}}}{1 + \frac{1}{\sqrt{1-x}}}$? $$\frac{\sqrt{1-x} + \frac{1}{\sqrt{1+x}}}{1 + \frac{1}{\sqrt{1-x}}}$$
I've had a go at putting everything over a common denominator in the form of:
$$\frac{\frac{\sqrt{1-x}\sqrt{1+x}+1}{\sqrt{1+x}}}{\frac{\sqrt{1-x}+1}{\sqrt{1-x}}} = \frac{\sqrt{1-x}(\sqrt{1-x^2}+1)}{(\sqrt{1-x}+1)\sqrt{1+x}}$$
Or multiplying the complex fraction by the conjugate of its base, but I'm getting nowhere.
Extensive searching hasn't made me realize the answer either— it has me baffled.
| From where you left off, to remove all radicals from the denominator, we could multiply the top and bottom by $\sqrt{1+x}(1-\sqrt{1-x})$ -- the former factor to remove the factor of $\sqrt{1+x}$, and the latter being the conjugate of $1+\sqrt{1-x}$. This would result with
\begin{align*}
\frac{\sqrt{1-x}(\sqrt{1-x^2}+1)\sqrt{1+x}(1-\sqrt{1-x})}{(1+x)(1-(1-x))} &= \frac{\sqrt{1-x^2}(\sqrt{1-x^2}+1)(1-\sqrt{1-x})}{x(1+x)}
\\
&= \frac{(1-x^2+\sqrt{1-x^2})(1-\sqrt{1-x})}{x(1+x)}
\end{align*}
You could also expand the numerator, but I don't really think it seems "simpler".
\begin{align*}
\frac{(1-x^2+\sqrt{1-x^2})(1-\sqrt{1-x})}{x(1+x)} &= \frac{1-x^2+\sqrt{1-x^2}-\sqrt{1-x}+x^2\sqrt{1-x}-(1-x)\sqrt{1+x}}{x(1+x)} \\
&= \frac{(x^2-1)\sqrt{1-x}+(x-1)\sqrt{1+x}+\sqrt{1-x}\sqrt{1+x}+1-x^2}{x(1+x)}
\end{align*}
There are many different ways to write this expression.
| {
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Conditional probability dice problem, Two players, Adam and Eve, are throwing a die, and the first one to get a 6, will win.
Eve is throwing first, what is the probability that Adam can win?
$P(A)$ = probability Adam to win
$P(E)$ = probability Eve to win
I know the formula for conditional probability
$ P(A|B) = \frac{P(A\cap B)}{P(B)}$
Also, I know that the conditional probability I need to calculate is $ P(A|\overline{E}) $
What I don't know is, what the intersection between these to events is.
$P(A\cap\overline{E}) = ?$
I know that the result should be $0,4545$
Also, what how do I calculate the probability for Adam to win, if the 6 was in the first 3 throws?
Thank you
| If Eve wins, it happens on either the first throw, the third throw, the fifth throw, etc...
If Eve wins on the first throw: $\frac{1}{6}$
If Eve wins on the third throw (his second throw): She missed, he missed, she won: $\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}$
If Eve wins on the fifth throw (her personal third throw): She miss, he miss, she miss, he miss, she hit: $(\frac{5}{6})^4\cdot\frac{1}{6}$
If Eve wins on the $(2n+1)$'th throw: $(\frac{5}{6})^{2n}\cdot\frac{1}{6}$
So, calculate $\sum\limits_{n=0}^{\infty}(\frac{5}{6})^{2n}\frac{1}{6}$ to add all of the possible outcomes where she wins together.
Alternatively, for Adam you have the same situation but offset a bit. You can either take 1 and subtract the previous result, or do it again using:
$\sum\limits_{n=0}^{\infty}(\frac{5}{6})^{2n+1}\frac{1}{6}$
As for winning within the first three throws, it is probably easiest to draw a tree diagram.
For a completely different approach, consider the absorbing markov chain with four states: He won, His turn (hasn't won yet), Her turn (hasn't won yet), She won. There is probability of moving from His turn to He won with probability $\frac{1}{6}$, else probability $\frac{5}{6}$ to become her turn. If he won, it remains at he won with probability $1$. Similarly so for her probabilities.
We get then an absorbing matrix in standard form with order of entries as He won, She won, His turn, Her turn as the following:
\begin{bmatrix} 1 && 0 && \frac{1}{6} && 0\\
0 && 1 && 0 && \frac{1}{6}\\
0 && 0 && 0 && \frac{5}{6}\\
0 && 0 && \frac{5}{6} && 0\end{bmatrix}
This is in the form: \begin{bmatrix} I & S\\0 & R\end{bmatrix} and the limiting matrix becomes \begin{bmatrix} I & S(I-R)^{-1}\\0 & 0\end{bmatrix}
Solving for $S(I-R)^{-1}$:
$I-R = \begin{bmatrix}1&-\frac{5}{6}\\-\frac{5}{6}&1\end{bmatrix}$
$(I-R)^{-1} = \frac{36}{11}\cdot\begin{bmatrix}1&\frac{5}{6}\\ \frac{5}{6}&1\end{bmatrix}$
$S(I-R)^{-1} = \begin{bmatrix}\frac{6}{11} & \frac{5}{11} \\ \frac{5}{11} & \frac{6}{11}\end{bmatrix}$
So, the limiting matrix is then:
\begin{bmatrix} 1& 0 & \frac{6}{11} & \frac{5}{11} \\
0 & 1 & \frac{5}{11} & \frac{6}{11} \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\end{bmatrix}
To see if it is currently Eve's turn, if Adam wins, take the element corresponding to Adam win's row, and Eve's turn column, which was how I labeled it as the first row and fourth column. Thus, he has a $\frac{5}{11}$ chance to win if it is currently her turn.
| {
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Circle inscribed in triangle What's the radius and area of circle of max area that can be inscribed in a isoceles triangle with $2$ equal sides of length $1$? Radius formula is given, $r = \dfrac{2A}{P}$, where $A$ is area of triangle and $P$ is perimeter of triangle.
I have no idea how to do this.
I'm guessing the triangle might be equilateral because equilateral is also isosceles.
| First assume that $\triangle ABC$ has $AB = AC = 1$, and denote $S$ = area of $\triangle ABC$, and $P$ = perimeter of the triangle, then:
$r = \dfrac{2S}{P} = \dfrac{2\cdot \dfrac{1}{2}\cdot 1\cdot 1\cdot \sin A}{1+1+2\sin A} = \dfrac{1}{2}\cdot \dfrac{\sin A}{1+\sin A} = f(A)$.
Taking derivative of $f$ w.r.t $A$ we have: $f'(A) = \dfrac{\cos A}{2(1+\sin A)^2} = 0 \iff \cos A = 0 \iff A = \dfrac{\pi}{2}$. This shows that $A = \dfrac{\pi}{2}$ is a maxima, and thus $r_{max} = f(\dfrac{\pi}{2}) = \dfrac{1}{4}$, and the circle with maximum area equal to: $\pi\cdot \left(\dfrac{1}{4}\right)^2 = \dfrac{\pi}{16}$.
| {
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Large round brackets in equation I cannot recall what the large () brackets mean - Google seems to be full of links on how to create them but not what they actually are or how to resolve them.
$
\left(
\begin{array}{l}
n\\
2
\end{array}
\right)r^2
$
Can someone please clarify? In this instance n = 6 and r = 0.05 if that helps.
| The binomial coefficient
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
is the number of ways of selecting $k$ elements from $n$ elements when order does not matter (the number of subsets with $k$ elements in an $n$ element set).
The number $n!$, read "$n$ factorial," is defined recursively as follows:
*
*$1! = 1$
*$n! = n(n - 1)!$ for $n \geq 1$
If you substitute $1$ in the definition for $n!$, you obtain
\begin{align*}
1! & = 1(1 - 1)!\\
1 & = 1 \cdot 0!\\
1 & = 0!
\end{align*}
For positive integers, $n!$ is the product of the first $n$ positive integers. For instance,
\begin{align*}
6! & = 6 \cdot 5!\\
& = 6 \cdot 5 \cdot 4!\\
& = 6 \cdot 5 \cdot 4 \cdot 3!\\
& = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2!\\
& = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1!\\
& = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1
\end{align*}
Hence,
$$\binom{6}{2} \cdot 0.05^2 = \frac{6!}{2!4!} \cdot 0.0025 = \frac{6 \cdot 5 \cdot 4!}{2 \cdot 1 \cdot 4!} \cdot 0.0025 = 15 \cdot \frac{1}{400} = \frac{3}{80}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simple first order differential equation I've got another first order differential that has me stumped. Please see my work and let me know where I'm going wrong:
$$y' = \frac{6x^2}{y(1+x^3)}$$
$$y'y = \frac{6x^2}{1+x^3}$$
$$ y.\frac{dy}{ dx} = \frac{6x^2}{1+x^3} $$
$$\int y dy = \int 6x^2\frac{1}{1+x^3} dx$$
$$\frac{1}{2}y^2 = 2x^3 \ln|1 + x^3| + c$$
$$y^2 = 4x^3 \ln|1 + x^3| + c$$
$$y = \pm \sqrt{4x^3 \ln |1 + x^3| + c}$$
Unfortunately, my answer is wrong and I'm not quite sure why. Please help.
| Your error is that $\displaystyle\int\dfrac{6x^2}{1+x^3}\,dx = 2\ln|1+x^3|+C$ and not $2x^3\ln|1+x^3|+C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to get the derivative How do I find the derivative of $\displaystyle \frac{1+4\cos x}{2 \sqrt{x+4\sin x}}?$
So I got this as my answer:
$$\frac{(4 \cos x+1)^2}{4\sqrt{4 \sin x+x}}-2 \sin x \sqrt{4\sin x+x}$$
does this look correct?
| Looks like you will need to use the quotient rule and the chain rule to solve this. The quotient rule tells us that $$\left[\frac{1+4\cos x}{2 \sqrt{x+4\sin x}}\right]' = \frac{[1+4\cos(x)]'2 \sqrt{x+4\sin x}-\left(1+4\cos(x)\right)\left[2 \sqrt{x+4\sin(x)}\right]'}{\left(2 \sqrt{x+4\sin(x)}\right)^2}$$ and to evaluate $\left[2 \sqrt{x+4\sin(x)}\right]'$ you need the chain rule. You should get $$\left[2 \sqrt{x+4\sin(x)}\right]' = \frac{2}{2\sqrt{x+4\sin(x)}} \cdot \left(1+4\cos(x)\right) \\ = \frac{1+4\cos(x)}{\sqrt{x+4\sin(x)}}$$ Now you can plug this expression into the one above to get $$ \frac{[1+4\cos(x)]'2 \sqrt{x+4\sin x}-\left(1+4\cos(x)\right) \frac{1+4\cos(x)}{\sqrt{x+4\sin(x)}}}{\left(2 \sqrt{x+4\sin(x)}\right)^2}$$ There is still a tremendous amount of simplification to be done, but it's mostly algebra.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How find this limits with hardly form?
show that:
$$\lim_{n\to\infty}n\left[\left(\dfrac{1}{\pi}\left(\sin{\left(\dfrac{\pi}{\sqrt{n^2+1}}\right)}+
\sin{\left(\dfrac{\pi}{\sqrt{n^2+2}}\right)}+\cdots+\sin{\left(\dfrac{\pi}{\sqrt{n^2+n}}\right)}
\right)\right)^n-\dfrac{1}{\sqrt[4]{e}}\right]=-\dfrac{1}{\sqrt[4]{e}}\left(\dfrac{31}{96}+\dfrac{\pi^2}{6}\right)$$
I only solve this :$$\lim_{n\to\infty}\left(\dfrac{1}{\pi}\sum_{i=1}^{n}\sin{\left(\dfrac{\pi}{\sqrt{n^2+i}}\right)}\right)^n=\dfrac{1}{\sqrt[4]{e}}\tag{1}$$
use
$$\sin{x}\approx x,\Longrightarrow \dfrac{1}{\pi}\sin{(\dfrac{\pi}{\sqrt{n^2+i}})}=\dfrac{1}{\sqrt{n^2+i}}+o(\dfrac{1}{\sqrt{n^2+i}})\to \dfrac{1}{n}\left(1+\dfrac{i}{n^2}\right)^{-\frac{1}{2}},n\to\infty$$
and note
$$(1+x)^{-1/2}=1-\dfrac{1}{2}x+o(x)$$
so
$$\lim_{n\to\infty}\left(\dfrac{1}{\pi}\sum_{i=1}^{n}\sin{\left(\dfrac{\pi}{\sqrt{n^2+i}}\right)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1}{n}\sum_{i=1}^{n}\left(1-\dfrac{i}{2n^2}+o(i/n^2)\right)\right)^n=e^{-\frac{1}{4}}$$
But I can't solve my problem,Thank you
| You seem to be on the right track and I guess we can try to get more reasonable estimates using further terms of the expansion. First we need to use $\sin x \approx x - \dfrac{x^{3}}{6}$ to get $$\begin{aligned}\frac{1}{\pi}\sin\left(\frac{\pi}{\sqrt{n^{2} + i}}\right) &\approx \frac{1}{n}\left(1 + \frac{i}{n^{2}}\right)^{-1/2} - \frac{\pi^{2}}{6n^{3}}\left(1 + \frac{i}{n^{2}}\right)^{-3/2}\\
&\approx \frac{1}{n}\left(1 - \frac{i}{2n^{2}} + \frac{3i^{2}}{8n^{4}}\right) - \frac{\pi^{2}}{6n^{3}}\left(1 - \frac{3i}{2n^{2}} + \frac{15i^{2}}{8n^{4}}\right)\\
&\approx \frac{1}{n}\left(1 - \frac{3i + \pi^{2}}{6n^{2}} + \frac{9i^{2} + 6i\pi^{2}}{24n^{4}}\right)\end{aligned}$$ and then summing from $i = 1$ to $i = n$ we get $$\begin{aligned}\sum_{i = 1}^{n}\frac{1}{\pi}\sin\left(\frac{\pi}{\sqrt{n^{2} + i}}\right) &\approx \frac{1}{n}\left(n - \dfrac{\dfrac{3n(n + 1)}{2} + n\pi^{2}}{6n^{2}} + \dfrac{\dfrac{3n(n + 1)(2n + 1)}{2} + 3n(n + 1)\pi^{2}}{24n^{4}}\right)\\
&= \left(1 - \frac{3(n + 1) + 2\pi^{2}}{12n^{2}} + \frac{(n + 1)(2n + 1) + 2(n + 1)\pi^{2}}{16n^{4}}\right)\\
&\approx \left(1 - \frac{1}{4n} - \frac{2\pi^{2} + 3}{12n^{2}} + \frac{1}{8n^{2}}\right)\\
&= \left(1 - \frac{1}{4n} - \frac{4\pi^{2} + 3}{24n^{2}}\right)\\\end{aligned}$$ and therefore using logs we get $$\begin{aligned}\left(\sum_{i = 1}^{n}\frac{1}{\pi}\sin\left(\frac{\pi}{\sqrt{n^{2} + i}}\right)\right)^{n} &\approx \exp\left(n\log\left(1 - \frac{1}{4n} - \frac{4\pi^{2} + 3}{24n^{2}}\right)\right)\\
&\approx\exp\left(-n\left(\frac{1}{4n} + \frac{\pi^{2}}{6n^{2}} + \frac{5}{32n^{2}}\right)\right)\\
&= \exp\left(-\frac{1}{4} - \frac{\pi^{2}}{6n} - \frac{5}{32n}\right)\\ \end{aligned}$$ We can now see that the desired limit is equal $$\begin{aligned}L &= \lim_{n \to\infty}n\left(\exp\left(-\frac{1}{4} - \frac{\pi^{2}}{6n} - \frac{5}{32n}\right) - \exp(-1/4)\right)\\
&= e^{-1/4}\lim_{n \to\infty}n\left(\exp\left(- \frac{\pi^{2}}{6n} - \frac{5}{32n}\right) - 1\right) = -\frac{1}{\sqrt[4]{e}}\left(\frac{\pi^{2}}{6} + \frac{5}{32}\right)\end{aligned}$$ There might be some calculation mistake in above or there is some typo in the question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
how do you solve $a^2+b^2+c^2=d^3$ let $ a,b,c,d$ be 4 integers such that $\gcd(a,b,c,d)=1$. How do you find the integral solutions of the equation: $$a^2+b^2+c^2=d^3$$
| For the equation:
$$x^2+y^2+z^2=r^3$$
Will make a replacement that formula was compact.
$$c=2(q-p-s)t$$
$$d=s^2+t^2-q^2-p^2+2p(q-s)$$
$$k=p^2+t^2-q^2-s^2+2s(q-p)$$
$$n=p^2+t^2+s^2-q^2$$
$$j=p^2+s^2+t^2+q^2-2q(p+s)$$
$p,s,t,q$ - integers asked us. Then decisions can be recorded.
$$x=dn^2+2cnj-dj^2$$
$$y=cj^2+2dnj-cn^2$$
$$z=k(n^2+j^2)$$
$$r=n^2+j^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Computing $\lim\limits_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum\limits_{k=1}^{n} \binom{2n-1}{n-k}\frac{ 1}{(2k-1)^2+\pi^2}$ What tools would you recommend me for computing the limit below?
$$\lim_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum_{k=1}^{n}\frac{\displaystyle \binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}$$
As soon as any useful idea comes to mind I'll make the proper update with the new findings.
| After I posted my earlier answer, I realized that this can be handled in a much simpler way.
Pick an $\epsilon\gt0$.
Most of the series is contained in a finite sum
This identity is proven in my earlier answer:
$$
\sum_{k=1}^\infty\frac1{(2k-1)^2+\pi^2}=\frac14\tanh(\pi^2/2)\tag{1}
$$
Since the series in $(1)$ converges, we know that there is a $K$ so that
$$
0\le\sum_{k=K+1}^\infty\frac1{(2k-1)^2+\pi^2}=\frac14\tanh(\pi^2/2)-\sum_{k=1}^K\frac1{(2k-1)^2+\pi^2}\le\epsilon\tag{2}
$$
Where most of the series is contained, the coefficients are almost equal
Now, using $(9)$ from this answer, we have
$$
\frac{\sqrt{n}}{4^n}\binom{2n-1}{n}=\frac{\sqrt{n}}{4^n}\frac12\binom{2n}{n}\sim\frac{\sqrt{n}}{4^n}\frac12\frac{4^n}{\sqrt{\pi n}}=\frac1{2\sqrt{\pi}}\tag{3}
$$
Therefore, there is an $M$ so that for $n\ge M$,
$$
\frac1{2\sqrt{\pi}}(1-\epsilon)\le\frac{\sqrt{n}}{4^n}\binom{2n-1}{n}\le\frac1{2\sqrt{\pi}}(1+\epsilon)\tag{4}
$$
Consider that
$$
\binom{2n-1}{n-k}=\binom{2n-1}{n}\frac{n-1}{n+1}\frac{n-2}{n+2}\cdots\frac{n-k+1}{n+k-1}\tag{5}
$$
Since we can choose an $N$ so that for all $n\ge N$ and $k\le K$,
$$
1-\epsilon\le\frac{n-1}{n+1}\frac{n-2}{n+2}\cdots\frac{n-k+1}{n+k-1}\le1\tag{6}
$$
we can combine $(4)$, $(5)$, and $(6)$ to get that for all $n\ge\max(M,N)$ and $k\le K$,
$$
\frac1{2\sqrt{\pi}}(1-\epsilon)^2\le\frac{\sqrt{n}}{4^n}\binom{2n-1}{n-k}\le\frac1{2\sqrt{\pi}}(1+\epsilon)\tag{7}
$$
The coefficients are bounded everywhere
Notice that $(4)$ and $(5)$ also tell us that for all $n\ge M$,
$$
\frac{\sqrt{n}}{4^n}\binom{2n-1}{n-k}\le\frac1{2\sqrt{\pi}}(1+\epsilon)\tag{8}
$$
Put it all together
For $n\ge M$, we have from $(2)$ and $(8)$,
$$
\frac{\sqrt{n}}{4^n}\sum_{k=1}^n\frac{\binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}
\le\frac14\tanh(\pi^2/2)\frac1{2\sqrt{\pi}}(1+\epsilon)\tag{9}
$$
For $n\ge\max(M,N)$, we have from $(2)$ and $(7)$,
$$
\frac{\sqrt{n}}{4^n}\sum_{k=1}^K\frac{\binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}
\ge\left(\frac14\tanh(\pi^2/2)-\epsilon\right)\frac1{2\sqrt{\pi}}(1-\epsilon)^2\tag{10}
$$
Since $\epsilon\gt0$ was arbitrary, we have
$$
\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\sum_{k=1}^n\frac{\binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}=\frac1{8\sqrt\pi}\tanh(\pi^2/2)\tag{11}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
} |
Show that $\ln(1+x)=\ln x+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-\frac{1}{4x^4}+\cdots$ when $x>1$
If $x>1$ show that $\ln(1+x)=\ln x+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-\frac{1}{4x^4}+\cdots$
I know from binomial expansion that $(1+x)$ will produce a divergent series in the form of $1-x+x^2-x^3+\cdots$ but I don't know how to apply that in this situation.
Do I just need to integrate $\ln(1-x+x^2+\cdots$)? If so, why integrate? What role does integration have here?
| Let
$$\begin{align}
f(x)&=\ln(1+x)-\ln x\\
&=\ln\left(1+\frac{1}{x}\right)
\end{align}
$$
By differenting w.r.t. $x$ and writing as sum of infinite GP
$$\begin{align}f'(x)&=\frac{1}{1+\frac1x}\cdot\left(\frac{-1}{x^2}\right)\\
&=-\frac{1}{x^2}\left(1-\frac1x+\frac{1}{x^2}-\cdots\right)\\
f'(x)&=-\frac{1}{x^2}+\frac{1}{x^3}-\frac{1}{x^4}+\cdots\\
\end{align}$$
By integrating
$$\begin{align}
f(x)&=\frac{1}{x}-\frac{1}{2x^2}+\cdots\\
\ln(1+x)-\ln x&=\frac{1}{x}-\frac{1}{2x^2}+\cdots\\
\end{align}$$
$$\ln(1+x)=\ln x+\frac{1}{x}-\frac{1}{2x^2}+\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subtracting Quarters of Squares Equals Multiply?! Can anyone explain to me how/why this works (hopefully in mostly layman's terms)?
It seems pretty magical to me at the moment.
$${{(a+b)^2\over4} - {(a-b)^2\over4}} = a b.$$
| Expanding the squared terms gives
\begin{equation}
\frac{(a + b)^2}{4} - \frac{(a - b)^2}{4} = \frac{a^2 + 2ab + b^2}{4} - \frac{a^2 - 2ab + b^2}{4} = \frac{4ab}{4} = ab.
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 3
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.