Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Product of generating functions Let $f(x) = \sum_{i=0}^\infty a_ix^i$ and $g(x) = \sum_{i=0}^\infty b_ix^i$ where $a_n = 1$ and $b_n = 2^n$ for all natural numbers $n$
What are the first three terms of the sequence generated by $f(x)g(x)$
?
So I know that $f(x)$ will generate $\{1,1,1,1,1,.... \}$ and $g(x)$ will generate $\{1,2,4,8,16,32,64,....... \}$
The first term $c_0$ of the sequence $f(x)g(x)$will be $1 \times 1 = 1$ and the second term $c_1$ will be $a_0 \times b_1 + a_1 \times b_0 = 1 \times 2 + 1 \times 1 = 3$ and finally $c_2 = 1 \times 4 + 1 \times 2 + 1 \times 1 = 7$
Now what is the general formula for the sequence generated by
$f(x)g(x)$, assume that it generates $\{c_0,c_1,c_2,c_3,..........\}$
If I calculates $c_3$ it would be easier to see that pattern, $c_3 = 8 + 4 + 2 + 1 = 15$
and so the sequence is $\{1,3,7,15,31,.................... \}$
It is basically $\{1,2^{2}-1,2^3-1,2^4-1,.... \}$ and so it is $\{1,a^2-1,a^3-1,a^4-1,.... \}$ where $a=2$, Is this general formula correct ?
I mean what is the formula , I just know the pattern here. So is the formula just $$c_k = 2^k-1$$ where $k>1$
What function generates the sequence $\{c_0,c_1,c_2,..... \}$?
Well I know that $\frac{1}{1-2x}$ generates the sequence $\{1,2,2^2,2^3,... \}$ and that $\frac{-1}{1-x}$ generates $\{-1,-1,-1,.... \}$ but adding these two functions will gives us the sequence $\{0,1,2^2-1,2^3-1 \}$
And what should I do to eliminate that zero, multiply by $x$ will add more zero, what should do I do eliminate one zero ?
| In general:
$\begin{align}
\left( \sum_{n \ge 0} a_n z^n \right)
\cdot \left( \sum_{n \ge 0} b_n z^n \right)
= \sum_{n \ge 0} \left(\sum_{0 \le k \le n} a_k b_{n - k} \right) z^n
\end{align}$
In your particular case, $a_n = 1$, switching the product around:
$\begin{align}
\left( \sum_{n \ge 0} b_n z^n \right)
\cdot \left( \sum_{n \ge 0} z^n \right)
&= \sum_{n \ge 0} \left(\sum_{0 \le k \le n} b_k \right) z^n \\
\frac{1}{1 - 2 z} \cdot \frac{1}{1 - z}
&= \sum_{n \ge 0} \left(\sum_{0 \le k \le n} 2^k \right) z^n \\
\frac{1}{1 - 3 z + z^2}
&= \sum_{n \ge 0} (2^{n + 1} - 1) z^n
\end{align}$
The observation that:
$\begin{align}
g(z)
&= \sum_{n \ge 0} b_n z^n \\
\frac{g(z)}{1 - z}
&= \sum_{n \ge 0} \left(\sum_{0 \le k \le n} b_k\right) z^n
\end{align}$
is quite useful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
In how many ways can two dozen identical robots be assigned to four assembly lines?
In how many ways can two dozen identical robots be assigned to four
assembly lines with
(a) at least three robots assigned to each line?
(b) at least three, but no more than nine assigned to each line ?
Two dozen = 24,
Now we use generating functions so for (a) we would have $(x^3+x^4+.........)^4$ but we can take $x^{12}$ as a common factor so we have $(x^3+x^4+.........)^4 = x^{12}(1+x + x^2 + ........)^4$ and we already know that $\frac{1}{1-x} = (1+x+x^2+......)$ and so $\frac{1}{(1-x)^4} = (1+x+x^2+.......)^4$
and so we have $x^{12}(1+x + x^2 + ........)^4 = x^{12}(1-x)^{-4}$. Now want to find the coefficient of $x^{12}$ in $(1-x)^{-4}$ which is ${-4 \choose 12} = (-1)^{12} {4 + 12 -1 \choose 12} = {15 \choose 12}$
Now for (b) I get stuck.
First we have $(x^3 + x^4 + ........ +x^9)^4$, now again we take $x^{12}$ as a common factor and so we have $x^{12}(1 + x + x^2 + .... + x^6)^4$ , and now I want to find the coefficient of $x^{12}$ in $(1 + x + x^2 + .... + x^6)^4$, But how do we do this ?
| Hint: We have, by the usual formula for the sum of a finite geometric progression, that if $x\ne 0$ then
$$1+x+x^2+\cdots+x^6=\frac{1-x^7}{1-x}.$$
It follows that
$$(1+x+x^2+\cdots+x^6)^4=(1-x^7)^4(1-x)^{-4}.$$
Note that we only need two easy terms of the expansion of $(1-x^7)^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What is the sum of the series $x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n}$? $x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n}$
I'm not sure if this can even be summed. Any help is appreciated.
| \begin{align}
& x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n} \\[10pt]
& x^{n-1}\left( 1 + \frac 2 x + \frac 3 {x^2} + \frac 4 {x^3} + \cdots + \frac n {x^{n-1}} \right) \\[10pt]
= {} & x^{n-1}\left( 1 + 2y + 3y^2 + 4y^3 + \cdots + ny^{n-1} \right) & (\text{where }y = 1/x) \\[10pt]
= {} & x^{n-1} \frac d {dy} \left( 1 + y + y^2 + y^3 + y^4 + \cdots + y^n \right) \\[10pt]
= {} & x^{n-1} \frac d {dy} \, \frac{1 - y^{n+1}}{1-y}.
\end{align}
Now do the differentiation and then finally put $1/x$ wherever $y$ appears.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Computing $\sum{\frac{1}{m^2+n^2}}$ I want to prove that $\sum_{1\leq m^2+n^2\leq R^2}{\frac{1}{m^2+n^2}}=2\pi\log R+O(1)$ as $R\rightarrow\infty$.
For this, I'm trying to approximate the sum by using the integral $\int_{1\leq r\leq R}{\frac{1}{x^2+y^2}}dxdy=\int_{0}^{2\pi}\int_{1}^{R}{\frac{1}{r}}drd\theta=2\pi\log R$
How can I show that $\sum_{1\leq m^2+n^2\leq R^2}{\frac{1}{m^2+n^2}}=\int_{1\leq r\leq R}{\frac{1}{x^2+y^2}}dxdy$ as $R\rightarrow\infty$ rigorously?
Any help will be appreciated.
| Just an idea:
Let's use
$$x-1<[x]\leq x$$
$$\sum \frac{1}{m^2+n^2}=\int \frac{1}{[x]^2+[y]^2}$$
Then
$$\int \frac{1}{x^2+y^2}\leq\int \frac{1}{[x]^2+[y]^2}<\int \frac{1}{(x-1)^2+(y-1)^2}$$
And hopefully we can show
$$\lim_{R\to\infty}\int \frac{1}{x^2+y^2}=\lim_{R\to\infty}\int \frac{1}{(x-1)^2+(y-1)^2}$$
?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
If $z_1$ and $z_2$ are two complex numbers,and if $z_1^3-3z_1^2z_2=2,3z_1z_2^2-z_2^3=11$, If $z_1$ and $z_2$ are two complex numbers,and if $z_1^3-3z_1^2z_2=2,3z_1z_2^2-z_2^3=11$,then find the value of $|z_1^2+z_2^2|$.
$z_1^3-3z_1^2z_2=2$
$3z_1z_2^2-z_2^3=11$
Adding them,we get
$z_1^3-3z_1^2z_2+3z_1z_2^2-z_2^3=13$
$(z_1-z_2)^3=13$
We need to find $|z_1^2+z_2^2|$,
I could not solve from here,I am stuck.Please help me.Thanks.
| $$z_{1}(z^2_{1}-3z^2_{2}) = 2\Rightarrow z^3_{1}-3z_{1}z^2_{2} = 2\cdots \cdots (1)$$
Similarly $$z_{2}(3z^2_{1}-z_{2}^2) = 11\Rightarrow 3z^2_{1}z_{2}-z^3_{2} = 11\cdots (2)\times i$$
Now Add and Subtract these two equation, we get
$$(z_{1}+iz_{2})^3 = 2+11i$$
$$(z_{1}-iz_{2})^3 = 2-11i$$
So $$(z^2_{1}+z^2_{2})^3 = (2+11i)(2-11i) = 125\Rightarrow |z^2_{1}+z^2_{2}|=5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Determine the general solution of the inhomogenous system (Difference&Differential Equations) Determine the general solution of the system
$y(n+1)=\begin{bmatrix} 5 & 1\\ -1 & 3 \\ \end{bmatrix} y(n) + 4^n \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix},\quad n \in \mathbf{N}$
The homogenous part of finding general solution is easy, however I am kinda lost in finding the general solution of this system with the +B[n] part.
| $\begin{bmatrix} a_{1}\\ b_{1} \\ \end{bmatrix}
=\begin{bmatrix} 5 & 1\\ -1 & 3 \\ \end{bmatrix} \begin{bmatrix} a_0\\ b_0 \\ \end{bmatrix} + \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}=
\begin{bmatrix} 5a_0 + b_0+1\\ -a_0+3b_0 -1 \end{bmatrix}
$
$\begin{bmatrix} a_{2}\\ b_{2} \\ \end{bmatrix}
=\begin{bmatrix} 5 & 1\\ -1 & 3 \\ \end{bmatrix} \begin{bmatrix} 5a_0 + b_0+1\\ 3b_0-a_0 -1 \end{bmatrix} + 4 \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}=
4\begin{bmatrix} 6 a_0 + 2 b_0+2\\ - ( 2a_0 - 2b_0+2) \end{bmatrix}
$
$\begin{bmatrix} a_{3}\\ b_{3} \\ \end{bmatrix}=
4^2\begin{bmatrix} 7 a_0+3 b_0+3\\ - (3 a_0-b_0+3) \\ \end{bmatrix}
$
$\begin{bmatrix} a_{4}\\ b_{4} \\ \end{bmatrix}=
4^3\begin{bmatrix} 8 a_0 + 4b_0+4\\ - ( 4 a_0+4) \\ \end{bmatrix}
$
$\begin{bmatrix} a_{5}\\ b_{5} \\ \end{bmatrix}=
4^4\begin{bmatrix} 9 a_0 + 5 b_0+5\\ - ( 5 a_0 + b_0+5) \\ \end{bmatrix}
$
$\vdots$
$\begin{bmatrix} a_{n}\\ b_{n} \\ \end{bmatrix}
=4^{n-1}\begin{bmatrix} 4+n& n \\ -n&4-n \\ \end{bmatrix}\begin{bmatrix} a_0\\ b_0 \\ \end{bmatrix} +
4^{n-1}n \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of solutions to Pell-type equation $k(x^2-2y^2) = d$ Given equation $$k(x^2-2y^2) = d$$
Where d is a constant. k,x,y are variables. All are positive integers.
Is there some characterization for the values of d for which there are unique solutions, multiple solutions or no solutions? If this is unsolved, then any tips towards studying this topic would be appreciated. Thanks.
EDIT:
So the original equation multiplied by k is:
$$k^2x^2-2k^2y^2=kd$$
and substituting $X=kx$ and $Y=ky$ and $Z=kd$
we have
$X^2-2Y^2=Z$
Does this help?
| If there is a solution then there are infinitely many solutions. If $a$ and $b$ are so that $a^2-2b^2=1$ (of which there are infinitely many), and $x,y,k$ are solutions to $k(x^2-2y^2)=d$, then $n=ax+2by$ and $m=ay+bx$ will also satisfy $k(n^2-2m^2)=d$.
This is because $1=a^2-2b^2=(a+\sqrt{2}b)(a-\sqrt{2}b)$ and $(a+\sqrt{2}b)(x+\sqrt{2}y)=n+\sqrt{2}m$. You can use this to shows show that
$$
(n+\sqrt{2}m)(n-\sqrt{2}m)= (a^2-2b^2)(x^2-2y^2)\\
$$
Which is equal to $\frac{d}{k}$. But this also equals $n^2-2m^2$. So if there is a solutions, there are infinitely many.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$
Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$
I'm not sure how to do this integration. It looks like partial fractions but I'm unsure.
| Generally, we can evaluate integrals of the form
$$\int_0^{\infty} dx \, f(x)$$
using a contour integral of the form
$$\oint_c dz \, f(z) \log{z} $$
As long as $f$ is sufficiently well-behaved at the origin and at infinity and in between, we have
$$-i 2 \pi \int_0^{\infty} dx \, f(x) = i 2 \pi \sum_k \operatorname*{Res}_{z=z_k} f(z) \log{z_k}$$
In this case
$$f(z) = \frac1{(1+z)^3+1}$$
so that, for $k \in \{0,1,2\}$:
$$z_k = -1 + e^{i (2 k+1) \pi/3} = -\left [1-\cos{(2 k+1) \frac{\pi}{3}}\right ] + i \sin{(2 k+1) \frac{\pi}{3}}$$
Thus,
$$\int_0^{\infty} \frac{dx}{(1+x)^3+1} = -\sum_{k=0}^2 \frac{\log{z_k}}{3 e^{i 2 (2 k+1) \pi/3}}$$
where
$$\begin{align}\log{z_k} &= \log{|z_k|} + i \arg{z_k} \\ &= \log{2 \left | \sin{(2 k+1) \frac{\pi}{6}} \right |} - i \arctan{\cot{(2 k+1) \frac{\pi}{6}}} \\ &= \log{2} - i \frac{\pi}{2} + \log{\sin{(2 k+1) \frac{\pi}{6}}} + i (2 k+1) \frac{\pi}{6}\end{align}$$
so that
$$\begin{align}\sum_{k=0}^2 \frac{\log{z_k}}{3 e^{i 2 (2 k+1) \pi/3}} &= -i \frac13 \frac{\pi}{3} e^{-i 2 \pi/3} + \frac13 \log{2} e^{-i 2 \pi} + i \frac13 \frac{\pi}{3} e^{i 2 \pi/3}\\ &= -\frac{2 \pi}{9} \sin{\frac{2 \pi}{3}} + \frac13 \log{2}\end{align}$$
Finally, we may conclude that
$$\int_0^{\infty} \frac{dx}{(1+x)^3+1} = \frac{\pi}{3 \sqrt{3}} - \frac13 \log{2} $$
This agrees with a numerical integral in Mathematica v 9.0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Using $y=\sum_{n = 0}^\infty a_nx^n$ in order to find recurrence relation of $a_n$'s for $y'=x-y^2$ I have to solve the differential equation $y'=x-y^2 => y''' = -2y'y' - 2yy''$.
I plugged in the series $\sum_{n = 0}^\infty a_nx^n$ to get:
$$\sum_{n = 0}^\infty a_{n+3}(n+3)(n+2)(n+1)x^n + \sum_{n = 0}^\infty \sum_{n = 0}^\infty 2(a_{n+1}(n+1)x^n)^2 \\
+ \sum_{n = 0}^\infty \sum_{n = 0}^\infty 2(a_nx^n)(a_{n+2}(n+2)(n+1)x^n) = 0$$
How do I further solve this? The double summation is throwing me off.
Thank you in advance.
| Note that $$ \left ( \sum_{n=0}^\infty a_n x^n \right )^2 = \sum_{n=0}^\infty\left ( \sum_{k=0}^na_ka_{n-k} \right ) x^n$$Thus $$y' = \sum_{n=1}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty(n+1)a_{n+1}x^n$$ And you'll get$$y'+y^2 = \sum_{n=0}^\infty(n+1)a_{n+1}x^n + \sum_{n=0}^\infty\left ( \sum_{k=0}^na_ka_{n-k} \right ) x^n = \sum_{n=0}^\infty\left [ \left ( \sum_{k=0}^n a_ka_{n-k} \right )+(n+1)a_{n+1} \right ]x^n = x$$ Equating the coefficients on both sides, we have $$(n+1)a_{n+1} = - \left ( \sum_{k=0}^na_ka_{n-k} \right )$$ for $n \ne 1$ and $2a_2 + 2a_1a_0=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If an integer $n$ is chosen at random from $1$ to $96$ inclusive ,what is the probability that $n(n+1)(n+2)$ is divisible by 8? In this one If I consider that n is even then probability that the number $n(n+1)(n+2)$ will be divisible by 8 will be 1/2 ,Now if n is odd then for n(n+1)(n+2) to be divisible by 8 ,n+1 should be a multiple of 8 ,Now how to find the probability among 96 numbers such that it is a multiple of 8 ?
| HINT:
*
*$n\equiv0\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv0\cdot(0+1)\cdot(0+2)\equiv\color\green0\pmod8$
*$n\equiv1\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv1\cdot(1+1)\cdot(1+2)\equiv\color\red 6\pmod8$
*$n\equiv2\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv2\cdot(2+1)\cdot(2+2)\equiv\color\green0\pmod8$
*$n\equiv3\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv3\cdot(3+1)\cdot(3+2)\equiv\color\red 4\pmod8$
*$n\equiv4\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv4\cdot(4+1)\cdot(4+2)\equiv\color\green0\pmod8$
*$n\equiv5\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv5\cdot(5+1)\cdot(5+2)\equiv\color\red 2\pmod8$
*$n\equiv6\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv6\cdot(6+1)\cdot(6+2)\equiv\color\green0\pmod8$
*$n\equiv7\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv7\cdot(7+1)\cdot(7+2)\equiv\color\green0\pmod8$
Note that the number of elements in the range $[1,96]$ is a multiple of $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Is my integration of $\sqrt{4z^2 - 4z + 2}$ correct? I'm trying to
$$
\int \sqrt{4z^2 - 4z + 2}\ dz
$$
the integrand I first rewrite to (completing the square)
$$
4\left(z^2 - z +\frac{1}{4} - \frac{1}{4}\right) + 2 = 4\left(z - \frac{1}{2}\right)^2 + 1
$$
I apply $\int \sqrt{a^2 + x^2} \ dx = \frac{x}{2} \sqrt{a^2+x^2} + \frac{a^2}{2} \ln \left(x + \sqrt{a^2+x^2}\right)$.
So, before simplification, I get
$$
(z-1/2)\sqrt{4z^2-4z+2} + 1/2 \ln \left[ 2(z-1/2) + \sqrt{4z^2 - 4z +2} \right]
$$
According to my book, the solution is
$$
(z-1/2)\sqrt{z^2-z+1/2} + 1/4 \ln \left[ z - 1/2 + \sqrt{z^2 - z +1/2} \right]
$$
Is my solution equivalent to the book? I dont see how the book gets the `1/4' before the log.
| You have made some little mistakes. I give you a hint to solve it more easier:
$$\int\sqrt{4z^2-4z+2}\space\text{d}z=$$
$$\int\sqrt{(2z-1)^2+1}\space\text{d}z=$$
Subsitute $u=2z-1$ and $\text{d}u=2\space\text{d}z$:
$$\frac{1}{2}\int\sqrt{u^2+1}\space\text{d}u$$
Subsitute $s=\arctan(u)$ and $\text{d}u=\sec^2(s)\space\text{d}s$:
$$\frac{1}{2}\int\sec^3(s)\space\text{d}s=$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_0^2 x(8-x^3)^{\frac{1}{3}}\ dx$ What substitution would you use to get from
$$\int\limits_0^2 x(8-x^3)^{\frac{1}{3}}\ dx$$
to
$$\int\limits_0^1 (1-t)^{a-1}t^{-a}\ dt, \ a\in(0,1)\ ?$$
I know how to evaluate the second integral and I thought that if I substitute $t={x^3\over8}$ I would reduce this to the form above, but what I get is
$$\frac{8}{3}\int\limits_0^1 (1-y)^{\frac{1}{3}}y^{-\frac{1}{3}}\ dy$$
| $$\frac { 8 }{ 3 } \int _{ 0 }^{ 1 }{ \left( { \left( 1-y \right) }^{ \frac { 1 }{ 3 } }{ y }^{ \frac { -1 }{ 3 } } \right) dx } \\ =B\left( \frac { 4 }{ 3 } ,\frac { 2 }{ 3 } \right) \\ =\frac { \Gamma \left( \frac { 4 }{ 3 } \right) \Gamma \left( \frac { 2 }{ 3 } \right) }{ 2 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction
Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$
Therefore, $\frac{0}{0} = 0$.
Q.E.D.
Update (2015-12-01) after your answers:
Proposition 2: $\frac{0}{0}$ is not a real number
Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:
Suppose that $\frac{0}{0}= x$, where $x$ is a real number.
Then, either $x = 0$ or $x$ is not equal to $0$.
1) Suppose $x = 0$, that is $\frac{0}{0} = 0$
Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $
Contradiction
Therefore, it is not the case that $x = 0$.
2) Suppose that $x$ is not equal to $0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction
Therefore, it is not the case that $x$ is a real number that is not equal to $0$.
Therefore, $\frac{0}{0}$ is not a real number.
Q.E.D.
Update (2015-12-02)
If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.
Proposition 3: $\frac{0}{0}$ is not a real number
Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.
$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$
$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$
Q.E.D.
Update (2015-12-07):
How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?
Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.
Then, $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
Suggested definition of division of real numbers:
If $b \ne 0$, then
$\frac{a}{b}=c$ iff $a=bc$
If $a=0$ and $b=0$, then
$\frac{a}{b}=0$
If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.
A somewhat more minimalistic version:
Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.
Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.
$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
| You can define $\frac{0}{0}$ to be anything you want; it can be $0$, or $1$, or $\pi$. But here's the catch: we want the choice of value for $\frac{0}{0}$ to be compatible with the usual laws of arithmetic.
From this standpoint, what you've really proved is that if $\frac{0}{0}$ is anything except $0$, then the law $2\frac{0}{0} = \frac{2 \cdot 0}{0}$ cannot hold. Therefore, the law: $a\frac{b}{c} = \frac{ab}{c}$ cannot hold, either. This suggests that simply defining $\frac{0}{0}=0$ might actually be a good idea. Unfortunately, this breaks another law of arithmetic, namely $\frac{a}{a} = 1.$
Since we cannot have the best of both worlds, perhaps it is best to simply leave $\frac{0}{0}$ undefined.
Actually, I think the best solution is to define division not of real numbers, but of affine subsets of $\mathbb{R}$. These are: the singleton subsets of $\mathbb{R}$, the empty subset, and $\mathbb{R}$ itself. So by passing to the affine subsets, we've effectively adjoined two new "points", namely $\emptyset$ and $\mathbb{R}$.
Now define that given affine subsets $Y$ and $X$ of $\mathbb{R}$, we have:
$$\frac{Y}{X} = \{r \in \mathbb{R} \mid Y \supseteq rX\}$$
You can check that the result of dividing one affine subset by another will itself always be affine.
Under these conventions, we have:
$$\frac{0}{0} = \mathbb{R}, \qquad \frac{1}{0} = \emptyset$$
This justifies the intuition that trying to divide $0$ by $0$ is somehow different from trying to divide a non-zero number by $0$.
Full disclosure: although passing to the affine subsets works algebraically, I'm not entirely sure how to put a topology or uniform structure or metric on the collection of affine subsets of $\mathbb{R}$. Until we can figure out how to do this, my proposed solution probably isn't that useful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 16,
"answer_id": 6
} |
Definite integral $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\ln(\cos x + \sin x) dx$ Evaluate the following integral:
$$\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\ln(\cos x + \sin x) dx$$
| $$I=\frac{1}{2} \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} log(sinx+cosx)^2dx=\frac{1}{2} \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}log(1+sin2x)dx \tag{1}$$ using $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ we have
$$I=\frac{1}{2} \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}log(1-sin2x)dx \tag{2}$$
Adding $(1)$ and $(2)$ we get
$$2I=\frac{1}{2} \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}log(cos^22x)dx= \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}log(cos2x)dx=2\int_{0}^{\frac{\pi}{4}}log(cos2x)dx=\int_{0}^{\frac{\pi}{2}}log(cosx)dx=\frac{-\pi}{2log2}$$
$$I=\frac{-\pi}{4log2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Which is more likely with perfect six-side dice? Find the probability of the following
$(i)$ to roll a $1$,then a $2$ and then a $3$ on consecutive roll of a single dice,
$(ii)$to roll three identical dice and have the outcome be a $1$,a $2$ and a $3$.
Which is more likely with perfect six-side dice?
My Attempt:
Probability to roll a $1$,then a $2$ and then a $3$ on consecutive roll of a single dice$=\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}+\frac{5}{6}\times(\frac{1}{6})^3+(\frac{5}{6})^2\times(\frac{1}{6})^3+(\frac{5}{6})^3\times(\frac{1}{6})^3+.....$
$=\frac{1}{36}$
Probability to roll three identical dice and have the outcome be a $1$,a $2$ and a $3=\frac{6}{216}=\frac{1}{36}$
There are 216 sample points in the sample space when we throw three dices and record the outcomes.Out of them six are favourable.$(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)$
So the two events are equally likely.
But the book answer says $(ii)$ event is more likely.I dont know how.Is some mistake i have made.Please help me.
| Let's rephrase the question:
Which is more likely with perfect six-sided dice?
*
*Roll $1$ die $3$ times and get 1, then 2, then 3
*Roll $3$ dice $1$ time and get 1, 2, 3
Obviously, without even computing the probabilities, the second option is more likely.
This is because in the first option the order matters and in the second option it doesn't.
That said, let's compute the probabilities:
Roll $1$ die $3$ times and get 1, then 2, then 3:
$$\frac{1}{6^3}=\frac{1}{216}$$
Roll $3$ dice $1$ time and get 1, 2, 3:
$$\frac{3!}{6^3}=\frac{6}{216}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How many solutions to the (general) equation? I've been trying to hash this one out for the last few days.
Please determine the number of solutions to the equation $x_1 + x_2 + x_3 + x_4 = n$, where $x_i \in N$, $x_1$ is even, $ 0 \leq x_2 \leq 2$, 3 divides $x_3$, and $x_4 \in \left\{0,1\right\}$.
My thoughts, so far:
The possible combinations of $x_2$ and $x_4$ are fairly straightforward; I'm more concerned about dealing with $x_1, x_3$.
I initially thought, since the range of numbers covered by these two are, for all intents and purposes, infinite, that there would be an infinite amount of solutions. But, n is obviously some finite number, so any value exceeding n would not be a viable solution.
Am I overthinking this? Could this still be the same stars-and-bars problem from previous SE entries?
Thanks for your input.
| Here is a generating function solution.
First your constraints can be encoded by power series which include all powers of $x$ that satisfy your constraint:
$\bullet$ $x_1$ is even$\rightarrow 1 + x^2 + x^4 + \cdots$
$\bullet$ $0 \leq x_2 \leq2 \rightarrow 1+x+x^2$
$\bullet$ $3|x_3 \rightarrow 1+x^3+x^6+ \cdots$
$\bullet$ $x_4 \in \{0,1\} \rightarrow 1+x$.
Now consider the product of all of these:
$$ (1 + x^2 + x^4 + \cdots)(1+x+x^2)(1+x^3+x^6+ \cdots)(1+x)
=\frac{(1+x+x^2)(1+x)}{(1-x^2)(1-x^3)}
$$
If you expand the right hand side above out in a series and find the coefficient of $x^n$, that will tell you the number of solutions to $x_1+x_2+x_3+x_4=n$. This works since the coefficient is the number of ways to pick one term from each factor of the above product so that the exponents add to $n$. For example if $n=8$ you have $4+1+3+0$ as a solution taking $x^4$ from the first factor, $x^1$ from the second, and so on.
So, you're left to expand the rational function in a series. First notice all the cancellation you get:
$$\frac{(1+x+x^2)(1+x)}{(1-x^2)(1-x^3)}=\frac{(1+x+x^2)(1+x)}{(1-x)(1+x)(1-x)(1+x+x^2)}=\left(\frac{1}{1-x}\right)^2$$
Now since $\frac{1}{1-x}=\sum x^i$ we have $\left(\frac{1}{1-x} \right)^2=_*\sum ix^{i-1}$ , so the coefficient of $x^n$ is $n+1$, and there are $n+1$ tuples $(x_1,x_2,x_3,x_4)$ meeting your constraints so that $x_1+x_2+x_3+x_4=n.$
$*:$ you can see this by noticing $\left(\frac{1}{1-x}\right)^2$ is the derivative of $\frac{1}{1-x}$ or just squaring the series for $\frac{1}{1-x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Equation with different bases (exponential) I seem to be stuck with the following equation right here:
$$2^x + 2^{x+1} = 3^{x+2} + 3^{x+3}$$
| \begin{align}
2^x+2^{x+1}&=3^{x+2}+3^{x+3}\\
2^x+2\cdot2^x&=3^2\cdot3^x+3^3\cdot3^x\\
(1+2)2^x&=(3^2+3^3)3^x\\
3\cdot2^x&=36\cdot3^x\\
2^x&=12\cdot3^x\\
\left(\frac23\right)^x&=12\\
\end{align}
Now use logarithms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding the remainder with Modular Arthmetic I have this math question that I'm kind of stuck on.
What is the remainder when $1^5+2^5+3^5+\cdots +99^5+100^5$ is divided
by $4$?
I'm supposed to use modular arithmetic and equivalences. I know that when dividing by $4$ the only possible remainders are $0, 1, 2, 3$ and they cycle, However I'm not sure where to go from here. Thanks.
| The even numbers are not of interest here. If $a$ and $b$ are odd integers from $1$ to $99$, call $a$ and $b$ friends if $a+b=100$.
If $a$ and $b$ are friends, then $a+b\equiv 0\pmod{4}$, so $b\equiv -a\pmod{4}$. It follows that $b^5\equiv -a^5\pmod{4}$, and therefore $a^5+b^5\equiv 0\pmod{4}$.
The sum of the fifth powers of any two friends is therefore congruent to $0$ modulo $4$. It follows that $1^5+3^5+\cdots+99^5\equiv 0\pmod{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$A$ be a real symmetric matrix of size $n$ ; is $I_n+A$ always non-singular ? Is $I_n - A$ always singular ? Let $A$ be a real symmetric matrix of size $n$ ; is $I_n+A$ always non-singular ? Is $I_n - A$ always singular ?
| Take
$$
I= \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
\quad
A= \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}.
$$ We have
$$
I+A= \begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}
$$
which is clearly singular.
Take
$$
I= \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
\quad
A= \begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}.
$$ We have
$$
I-A= \begin{pmatrix}
0 & -1 \\
-1 & 0
\end{pmatrix}
$$
which is clearly non-singular.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Derivative of $10^x\cdot\log_{10}(x)$
Derive $10^x\cdot\log_{10}(x)$
$$10^x\cdot \ln(10)\cdot \log_{10}(x)+\frac{1}{x\cdot \ln(10)}\cdot 10^x$$
But WolframAlpha gives another solution. Where am I wrong?
| \begin{align}
\text{your answer} & = 10^x\cdot \ln10\cdot \log_{10} x+\frac{1}{x\cdot \ln10}\cdot 10^x \\[10pt]
& = 10^x \ln x + \frac{10^x}{x\ln 10} \\[10pt]
& = 10^x \left( \ln x + \frac 1 {x\ln 10} \right) = \text{Wolfram's answer}.
\end{align}
Note that where Wolfram writes $\log x$ or $\log 10$ with no base specified, it means the base is $e$, so it's the natural logarithm.
Note also that we used the identity $\ln10\cdot\log_{10}x = \ln x$. That is an instance of the change-of-base formula for logarithms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
Residue of $\frac{g(z)}{\cos^{2}z}$ I would like to show that the residue of the function $$\text{Res}\left(\frac{g(z)}{\cos^{2}z}\right) = g'(z_n)$$ at $z_{n}=(n+\frac{1}{2})\pi$, where $g$ is analytic.
I tried the Limit formula for higher order poles and it fails. What other method can I use?
| Using the addition theorem for the cosine and the fact that
$\sin(z_n) = \pm 1$, $\cos(z_n) = 0$, you can compute the initial part of the
Taylor series for $\cos^2(z_n + h)$ at $h = 0$:
$$
\cos^2(z_n + h) = \sin^2(h) = \bigl( h + \frac{h^3}{3!} + O(h^5) \bigr)^2 \\
= h^2 \bigl( 1 + \frac{h^2}{6} + O(h^4) \bigr)^2 \\
= h^2 \bigl( 1 + \frac{h^2}{3} + O(h^4) \bigr)
\quad \text{ for } h \to 0
$$
It follows that
$$
\frac{1}{\cos^2(z_n + h)} = \frac{1}{h^2} \bigl( 1 - \frac{h^2}{3} + O(h^4) \bigr) \\
= \frac{1}{h^2} - \frac 13 + O(h^2)
$$
Now multiply that with
$$
g(z_n + h) = g(z_n) + g'(z_n) h + O(h^2)
$$
to get
$$
\frac{g(z_n+h)}{\cos^2(z_n + h)} = \frac{g(z_n)}{h^2} + \frac{g'(z_n)}{h} + O(1)
\quad \text{ for } h \to 0
$$
Finally, substitute $z = z_n + h$:
$$
\frac{g(z)}{\cos^2(z)} = \frac{g(z_n)}{(z-z_0)^2} + \frac{g'(z_n)}{z-z_n} + O(1)
\quad \text{ for } z \to z_n
$$
and the residuum is the coefficient of $(z-z_n)^{-1}$ in the
Laurent series, which is $g'(z_n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$ $$ \lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$$
How do I solve this without using derivatives or integrals.
| Both the numerator and the denominator are of the form
$$
a-\sqrt{bx+a^2}
$$
(with $a>0$). In the numerator we have $a=1$, $b=3$; in the denominator we have $a=2$, $b=5$.
Let's try
\begin{align}
\lim_{x\to0}\frac{a-\sqrt{bx+a^2}}{x}
&=\lim_{x\to0}\frac{a-\sqrt{bx+a^2}}{x}
\frac{a+\sqrt{bx+a^2}}{a+\sqrt{bx+a^2}}\\
&=\lim_{x\to0}\frac{-bx}{x(a+\sqrt{bx+a^2})}\\
&=\lim_{x\to0}\frac{-b}{a+\sqrt{bx+a^2}}\\
&=-\frac{b}{2a}
\end{align}
Thus your limit is
$$
\lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}=
\lim_{x\to 0} \frac{1-\sqrt{3x+1}}{x}\frac{x}{2-\sqrt{5x+4}}=
\left(-\frac{3}{2}\right)\left(-\frac{4}{5}\right)=\frac{6}{5}
$$
When you'll learn Taylor expansions, you'll see that
$$
\sqrt{1+kx}=1+\frac{1}{2}kx+o(x)
$$
so
$$
\frac{1-\sqrt{1+3x}}{2-\sqrt{4+5x}}=
\frac{1-\sqrt{1+3x}}{2-2\sqrt{1+\frac{5}{4}x}}=
\frac{1-(1+\frac{3}{2}x+o(x))}{2-2(1+\frac{5}{8}x+o(x))}=
\frac{-\frac{3}{2}x+o(x)}{-\frac{5}{4}x+o(x)}=
\frac{-\frac{3}{2}+o(1)}{-\frac{5}{4}+o(1)}
$$
and so the limit is
$$
\frac{-3/2}{-5/4}=\frac{6}{5}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
chinese remainder theorem problem in one of the steps I need to calculate the following:
$$x=8 \pmod{9}$$
$$x=9 \pmod{10}$$
$$x=0 \pmod{11}$$
I am using the chinese remainder theorem as follows:
Step 1:
$$m=9\cdot10\cdot11 = 990$$
Step 2:
$$M_1 = \frac{m}{9} = 110$$
$$M_2 = \frac{m}{10} = 99$$
$$M_3 = \frac{m}{11} = 90$$
Step 3:
$$x=8\cdot110\cdot2 + 9\cdot99\cdot9 + 0\cdot90\cdot2 = 9779 = 869\mod 990$$
I have used online calculators to check this result and I know it is wrong (it should be 539 I think) but cannot find out what am I doing wrong. Can you help me?
Thanks
| $x \equiv -1 \pmod 9$ and $x \equiv -1 \pmod {10}.$ So $x \equiv -1 \pmod {90}$ and $x = 90 n - 1.$ But $90 = 88 + 2,$ so $90 \equiv 2 \pmod {11}.$
$$ x = 90 n - 1 \equiv 2n - 1 \pmod {11}. $$
$$ 2n \equiv 1 \pmod {11}, $$
$$ n \equiv 6 \pmod {11}. $$
Start with $n=6,$ so $x = 540 - 1 = 539.$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
A more official way to combine the $90$ and $11$ parts is this: the continued fraction for $90/11$ has penultimate convergent $41/5,$ and
$$ 41 \cdot 11 - 5 \cdot 90 = 1. $$
So
$$ 451 \equiv 1 \pmod {90}, \; \; 451 \equiv 0 \pmod {11}, $$
$$ -450 \equiv 0 \pmod {90}, \; \; -450 \equiv 1 \pmod {11}. $$
We want something $-1 \pmod {90}$ and $0 \pmod {11},$ so we can ignore the second pair and use
$$ -451 \equiv -1 \pmod {90}, \; \; -451 \equiv 0 \pmod {11}. $$
Also note
$$ 990 - 451 = 539. $$
Let's see: the virtue of the continued fraction thing is that, when i want something $A \pmod {90}$ but $B \pmod {11},$ I just take $451 A -450 B.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$ $$\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$$
I'm trying to understand trigonometric substitution better, because I never could get a good handle on it. All I know is that this integral is supposed to reduce to the integral of some power of cosine. I tried $x^2=\tan\theta$, but I ended up with $\sin\theta\cos^3\theta$ as my integrand. Can someone explain how to compute this?
| Note\begin{align*}
\int_{-\infty}^{\infty} \frac{x^2}{x^4+1} dx
&\overset{x\to \frac1x}=\int_{-\infty}^{\infty} \frac{1}{x^4+1}dx
=\int_{0}^{\infty} \frac{x^2+1}{x^4+1}dx
=\int_{0}^{\infty} \frac{1+\frac1{x^2}}{x^2+\frac1{x^2}}dx
\\
&= \int_{0}^{\infty} \frac{d(x-\frac{1}{x} )}{\left(x-\frac{1}{x}\right)^2+2} =\frac{1}{\sqrt{2}} \tan^{-1}{\frac{x-\frac{1}{x}}{\sqrt{2}}} \bigg \rvert_{0}^{\infty}
= {\frac{\pi}{\sqrt{2}}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 13,
"answer_id": 9
} |
Convergence of $\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$ I've been working on an approximation for a problem in my numerical methods course, and I seemed to have run into the series
$$\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$$
I'm trying to figure out the right test for this example. I haven't touched Calc II in a while, and I'm really not interested in getting a weird explanation from Wolfram. I was thinking that I might want to do some kind of ratio test between the $n$th and $n+1$th term.
| Suppose $\sum_{i=1}^\infty \frac{1}{4i + 1}$ converges. Then as for all $i \geq 1$,
$$\frac{1}{4i + 1} > \frac{1}{4i + 2} > \frac{1}{4i + 3} > \frac{1}{4i + 4} $$
it must be the case each of these series also converge:
$$\sum_{i=1}^\infty \frac{1}{4i + 2}, \quad \sum_{i=1}^\infty \frac{1}{4i + 3}, \quad \sum_{i=1}^\infty \frac{1}{4i + 4}$$
As everything in sight is positive and we can rearrange terms, we now have that
$$\sum_{i=1}^\infty \frac{1}{4i + 1} + \sum_{i=1}^\infty \frac{1}{4i + 2} + \sum_{i=1}^\infty \frac{1}{4i + 3}+ \sum_{i=1}^\infty \frac{1}{4i + 4} = \sum_{n=5}^\infty \frac 1n$$
converges. But the last expression on the right does not. Contradiction.
Hence
$$\sum_{i=1}^\infty \frac{1}{4i + 1}$$
must diverge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Determine the form of set containing the real solutions
From the following inequality:
$\dfrac{\log_{2^{x^2+2x+1}-1}(\log_{2x^2 + 2x + 3}(x^2 - 2x)}{\log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10)} \geq 0 $, the set of all real solutions to this inequality is of the form:
$$(a) \ (a,b) \cup (b,c) \\ (b) \ (-\infty,a) \cup (c,\infty) \\ (c) \ (a,b) $$
for some real numbers $a,b,c $ such that $-\infty<a<b<c<+\infty $.
First I found the points at which the numerator and the denominator equal $0$. I found that from the following:
$$\log_{2^{x^2+2x+1}-1}(\log_{2x^2 + 2x + 3}(x^2 - 2x) = 0 \\ \log_{2x^2 + 2x + 3}(x^2 - 2x) = 1 $$
$$2x^2 + 2x + 3 = x^2 - 2x .$$
From this quadratic equation i find that the zeros of the function are $x_1=-3 $ and $x_2=-1$ and let them be $a=-1 $ and $b=-3 $ for now. I also found from the plot of the given quadratic equation that $(-\infty,-3)\cup(-1,\infty) $ makes function a positive one and the values in between $(-3,-1)$ make it negative. I used that later for the rational number chart.
Similarly $$\log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10) = 0 \\
x^2 + 6x + 10 = 1 \\ x = -3.$$ This function is positive for every other value.
Now using the chart of rational numbers, determining when the whole ratio is positive and when is negative I get the positive values to range from:
$(-\infty,-3) \cup (-1,\infty)$ which then makes my answer of the form $(-\infty,b) \cup (a, +\infty)$.
First of all, it's not given, second, it's wrong. The solution I have is for the $(a)$ answer. I just don't know how to get to it.
| Do you know about the change of base formula? If $a,b,x > 0$, $a \not= 1$, and $x \not= 1$ then
$$ \log_a b = \frac{\log_x b}{\log_x a}.$$ Your original inequality reads
$$ \log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0$$ provided that $2^{x^2 + 2x + 1} - 1 > 0$, $2^{x^2 + 2x + 1} - 1 \not= 1$, $x^2 + 6x + 10 > 0$, $x^2 + 6x + 10 \not= 1$, $\log_{2x^2 + 2x + 3} (x^2 - 2x) > 0$, and $2x^2 + 2x + 3 > 0$. These are all true provided that $x \not= 0$, $x \not= -2$, $x \not= -1$, $x \not= -3$, and $x^2 - 2x > 1$.
Next observe $$ \log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0 \iff \log_{2x^2 + 2x + 3} (x^2 - 2x) \ge 1 \\ \iff x^2 - 2x \ge (2x^2 + 2x + 3)^1.$$
The last inequality reduces to $x^2 + 4x + 3 \le 0$ which has the solution $[-3,-1]$. Note that $x^2 - 2x \ge 3$ on this interval, so excluding $x = -1$, $x = -2$, and $x = -3$ you get $$ (-3,-2) \cup (-2,-1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving For $X$ In A 4th Degree Polynomial The examples they use on my book are basically from basic arithmetic like 1, 2 and 3 to calculus and calculating derivatives, which is really annoying because I can't build the fundamental skill required to do harder questions. Anyhow, I have to solve an non-factorable inequality. Normally, I would use the quadratic formula but it's not quadratic, nor cubic.
The question is $x^4 + 2x^3 - 4x^2 - 6x \leq -3$. And I have to get an $x$ = something. I am actually really lost here because I can't even apply factor theorem in this case since no values work out.
I think this is a problem of my fundamentals though, please tell me how to solve things like these when they cannot be factored. Or we can just have a one on one discussion...
Any help is appreciated, thank you!
| After manipulating a little bit and rearranging terms when the whole thing is $\leq 0$:
$$x^4 + 2x^3 - 4x^2 - 6x + 3 \\
x^4+2x^3-6x-4x^2+3 \\
x^4 + 2x(x^2-3)-4x^2+3 \\
x^4 + 2x(x^2-3)-3x^2-x^2+3 \\
x^4 + 2x(x^2-3)-3x^2-(x^2-3) \\
x^4 -3x^2 + 2x(x^2-3)-(x^2-3) \\
x^4 -3x^2 + (x^2-3)(2x-1) \\
x^2(x^2-3) + (x^2-3)(2x-1) \\
(x^2-3)(x^2+2x-1).$$
So your problem is now $(x^2-3)(x^2+2x-1) \leq 0$. So it appears that $x=\pm \sqrt{3}$ will bring about equality. If you use the quadratic formula, you will also find that $x=\pm \sqrt{2}-1$ are 2 more solutions. That means we've found all 4 roots(!).
After testing sample values within each of the intervals delimited by the 4 roots, you can show that the total valid domain is $x \in [-1-\sqrt{2}, -\sqrt{3}] \cup [\sqrt{2}-1,\sqrt{3}]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1568881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
For $n>1$, estimate the product $(n+1)^{n+1}(n/1)^n\dots (1/n)$ from above If $ n $ be a positive integer $>1$, prove that
$$2^{n(n+1)}\gt(n+1)^{n+1}\biggl(\frac{n}{1}\biggr)^{n}\biggl(\frac{n-1}{2}\biggr)^{n-1}...\biggl(\frac{2}{n-1}\biggr)^{2}\biggl(\frac{1}{n}\biggr)$$
| Consider $(n+1)$ positive numbers $\binom{n}{0}, \binom{n}{1}, ...\binom{n}{n}$. $$ A.M.= \frac{\binom{n}{0}+ \binom{n}{1}+ ...+\binom{n}{n}}{n+1}=\frac{(1+1)^n}{n+1}=\frac{2^n}{n+1}$$
$$G.M.=\sqrt[(n+1)]{\binom{n}{0} \binom{n}{1}. ...\binom{n}{n}}=\sqrt[(n+1)]{\biggl(\frac{n}{1}\biggr)^n \biggl(\frac{n-1}{2}\biggr)^{n-1} \biggl(\frac{n-2}{3}\biggr)^{n-2}...\biggl(\frac{2}{n-1}\biggr)^{2} \biggl(\frac{1}{n}\biggr)^{1}} $$ Applying A.M.> G.M., we get $$ \biggl(\frac{2^n}{n+1}\biggr)^{n+1}> \biggl(\frac{n}{1}\biggr)^n \biggl(\frac{n-1}{2}\biggr)^{n-1} \biggl(\frac{n-2}{3}\biggr)^{n-2}...\biggl(\frac{2}{n-1}\biggr)^{2} \biggl(\frac{1}{n}\biggr)^{1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find formula for recursive sequence sum? I have the following sequence:
$$a(1) = 1$$
$$a(n) = a(n-1) + n$$
For example:
$$a(1) = 1$$
$$a(2) =3$$
$$a(3) =6$$
$$a(4) =10$$
$$a(5) =15$$
$$a(6) = 21$$
Which approach should I use in order to find the formula for the sum of elements $a(1)$ through $a(n)$?
Thanks :)
| The equation you write is equivalent to
$$a(n)-a(n-1)=n$$
Sum both sides from $n=1$ to $x$ to find that the left hand side telescopes.
$$\sum_{n=1}^{x} (a(n)-a(n-1))= \sum_{n=1}^{x} n$$
$$a(x)-a(0)=\frac{x(x+1)}{2}$$
But $a(1)-a(0)=1-a(0)=1$ so $a(0)=0$.
Now that we have:
$$a(x)=\frac{x(x+1)}{2}$$
Then you can go from here,
*
*What if I forgot $\sum_{n=1}^{x} n^2$?
We can note that,
$$s(x)-s(x-1)=\frac{x(x+1)}{2}=\frac{1}{2}x^2+\frac{1}{2}x$$
Where $$s(x)=\sum_{n=1}^{x} a(n)$$
Through integration we guess the solution to this equation is of the form:
$$s(x)=ax^3+bx+cx+d$$
Then we plug $s(x)$ back into the equation and equate coefficients to get our final result. Of course noting the initial condition $s(2)=4$.
$$ax^3+bx^2+cx+d-(a(x-1)^3+b(x-1)^2+c(x-1)+d)=\frac{1}{2}x^2+\frac{1}{2}x$$
The algebra can be greatly simplified with Pascal's triangle..or you may skip this all if you have $\sum_{n=1}^{x} n^2$ memorized.
$$3ax^2+(2b-3a)x+(a-b+c)=\frac{1}{2}x^2+\frac{1}{2}x+0$$
$$a=\frac{1}{6} \implies b=\frac{1}{2} \implies c=\frac{1}{3}$$
Finally utilizing $s(2)=4$ we get $d=0$
$$s(x)= \frac{1}{6} x^3+\frac{1}{2} x^2+\frac{1}{3}x$$
$$=\frac{x(x+1)(x+2)}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
What rule is used for this simplification? $$ \frac{8}{(s+1)^2 + 2^2} \times \frac{1}{s} = \frac{8}{5} - \frac{1}{s} + \frac{16}{10}\times \frac{s+1}{(s+1)^2 + 2^2} + \frac{8}{10}\times \frac{2}{(s+1)^2 + 2^2}
$$
| Do some partial fraction magic?
$$\frac{8}{s(s^2+2s+5)} = \frac{As+B}{s^2+2s+5} + \frac{C}{s}$$
This gives $C=8/5, A=-8/5, B=-16/5$ I think, so
$$\frac{8}{s(s^2+2s+5)} = \left(\frac{8}{5}\right)\left[\frac{1}{s}-\frac{s+2}{s^2+2s+5}\right].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\det$ in terms of $k$ Consider the following matrix:
\begin{bmatrix}
1 & 2 & 3 \\
2 & k-3 & 4 \\
3 & 4 & k-4 \\
\end{bmatrix}
I have the following problems:
*
*How to find $\det(B)$ in terms of $k$?
*For what value(s) of $k$ are the column vectors of $B$ linearly dependent?
| You can find the determinant by applying row transformations:
\begin{align}
B=\begin{bmatrix}
1 & 2 & 3\\
2 & k-3 & 4\\
3 & 4 & k-4
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 2 & 3\\
0 & k-7 & -2\\
0 & -2 & k-13
\end{bmatrix}
&&\begin{aligned}R_2&\gets R_2-2R_1\\R_3&\gets R_3-3R_1\end{aligned}
\\&\to
\begin{bmatrix}
1 & 2 & 3\\
0 & -2 & k-13 \\
0 & k-7 & -2
\end{bmatrix}
&&R_2\leftrightarrow R_3
\\&\to
\begin{bmatrix}
1 & 2 & 3\\
0 & 1 & (13-k)/2 \\
0 & k-7 & -2
\end{bmatrix}
&&R_2\gets (-1/2)R_2
\\&\to
\begin{bmatrix}
1 & 2 & 3\\
0 & 1 & (13-k)/2 \\
0 & 0 & (k^2-20k+87)/2
\end{bmatrix}
&&R_3\gets R_3+(7-k)R_2
\end{align}
Thus the determinant is
$$
(-1)(-2)\frac{k^2-20k+87}{2}=
k^2-20k+87
$$
The first factor is due to the row swap, the second factor is due to the pivot reduction, the third is the product of the elements in the diagonal.
Now recall that the columns are linearly independent if and only if the determinant is nonzero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Generalized "square root" of a real positive definite symmetric matrix with respect to another Related post: Square root of Positive Definite Matrix, which is a special case of this question ($A = I$).
Let $A$ and $B$ be real, symmetric, positive definite matrices of the same size $n \times n$. The question is: Does there exist a symmetric matrix $S$, such that $S A S = B$? If there is, how to construct it? If there is not, what conditions we can impose on $A$ and $B$, so that $S$ exists?
I know that in the case of $AB = BA$, $S$ exists. But how is it in the general case?
| Define $S$ by
$$ S = A^{-\frac{1}{2}} \left( A^{\frac{1}{2}} B A^{\frac{1}{2}}\right)^{\frac{1}{2}} A^{-\frac{1}{2}}. $$
Then $S$ is symmetric and so, by Theorem 2 of Wigner's "On Weakly Positive Matrices", $S$ is positive definite and
$$ SAS = A^{-\frac{1}{2}} \left( A^{\frac{1}{2}} B A^{\frac{1}{2}}\right)^{\frac{1}{2}} A^{\frac{1}{2}} A^{-\frac{1}{2}} \left( A^{\frac{1}{2}} B A^{\frac{1}{2}}\right)^{\frac{1}{2}} A^{-\frac{1}{2}} = A^{-\frac{1}{2}} \left( A^{\frac{1}{2}} B A^{\frac{1}{2}}\right) A^{-\frac{1}{2}} = B. $$
The matrix $S$ is the unique positive definite matrix satisfying $SAS = B$. To see this, let $S$ be a positive definite matrix satisfying $SAS = B$ and define $S' = A^{\frac{1}{2}}SA^{\frac{1}{2}}$. Then $S'$ is also positive definite and we have
$$ S'^2 = A^{\frac{1}{2}}SA^{\frac{1}{2}}A^{\frac{1}{2}}SA^{\frac{1}{2}} = A^{\frac{1}{2}}BA^{\frac{1}{2}}. $$
The right hand side is also positive definite and so, by the uniqueness of the positive definite square root, $S'$ is determined uniquely which implies that $S$ is determined uniquely (as $A$ is invertible).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $7^n-3^n$ is divisible by $n>1$, then $n$ must be even. I tried using factorization of $a^n-b^n$ for odd $n$ in an attempt to work through to a situation where the factors are such that they cannot have n as a factor. But I reached nowhere. Here's how I proceeded -
$$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\dots+a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$$
a-b=4 and can be ignored.
The latter term is essentially odd and not divisible by any odd number till $9$ (easy to prove without getting into calculations).
However, for some arbitrary odd number $x=p^k$ where $p \ge 11$ is prime, I cannot say whether the sum of two terms is divisible by $x$ or not when the two terms are individually not divisible by $x$.
| If $n\mid 7^n-3^n$ and $n>1$, then let $p$ be the least prime divisor of $n$.
Clearly $\gcd(21,p)=1$, so $\left(7\cdot 3^{-1}\right)^n\equiv 1\pmod{p}$, i.e. $\text{ord}_p\left(7\cdot 3^{-1}\right)\mid n$.
By Fermat's Little theorem $\text{ord}_p\left(7\cdot 3^{-1}\right)\mid p-1$. Therefore $\text{ord}_p\left(7\cdot 3^{-1}\right)\mid \gcd(n,p-1)=1$, so $7\cdot 3^{-1}\equiv 1\pmod{p}$, so $7\equiv 3\pmod{p}$, so $p\mid 7-3=4$, so $p=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Finding the logarithm of a matrix? Find $B$ if $A=e^B$ and
$A=\begin{bmatrix}
2&1&0\\
0&2&0\\
0&0&4\\
\end{bmatrix}$.
Besides, I would be very happy if give some general remark(Best approach). I have seen the wiki article on log of a matrix but it was too complicated(for me).
| You can make use of the block structure of $A$:
$$
A =
\begin{pmatrix}
C & 0 \\
0 & 4
\end{pmatrix}
= e^B
= \sum_{k=0}^\infty \frac{1}{k} B^k
\Rightarrow
B =
\begin{bmatrix}
D & 0 \\
0 & x
\end{bmatrix}
$$
so we can assume $4 = e^x \Rightarrow x = \ln(4)$.
For the block matrices we get
$$
C =
\begin{pmatrix}
2 & 1 \\
0 & 2
\end{pmatrix}
=
e^D = \sum_{k=0}^\infty \frac{1}{k!} D^k
$$
and try an upper triangular matrix
$$
D =
\begin{pmatrix}
y & z \\
0 & y
\end{pmatrix}
$$
and get the powers
$$
D^2 =
\begin{pmatrix}
y & z \\
0 & y
\end{pmatrix}
\begin{pmatrix}
y & z \\
0 & y
\end{pmatrix}
=
\begin{pmatrix}
y^2 & 2 y z \\
0 & y^2
\end{pmatrix}
\\
D^3 =
\begin{pmatrix}
y^2 & 2 y z \\
0 & y^2
\end{pmatrix}
\begin{pmatrix}
y & z \\
0 & y
\end{pmatrix}
=
\begin{pmatrix}
y^3 & 3 y^2 z \\
0 & y^3
\end{pmatrix}
\\
\vdots
\\
D^k =
\begin{pmatrix}
y^k & k y^{k-1} z \\
0 & y^k
\end{pmatrix}
\quad (k \ge 1)
$$
which suggest
$$
C =
\begin{pmatrix}
2 & 1 \\
0 & 2
\end{pmatrix}
=
e^D
=
\begin{pmatrix}
e^y & e^y z \\
0 & e^y
\end{pmatrix}
$$
so $y = \ln(2)$ and $z = 1/e^y = 1/2$.
This gives
$$
B =
\begin{pmatrix}
\ln(2) & 1/2 & 0 \\
0 & \ln(2) & 0 \\
0 & 0 & \ln(4)
\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{1}{a(a-b)(a-c)} +\frac{1}{b(b-c)(b-a)} +\frac{1}{c(c-a)(c-b)} =\frac{1}{abc}$ for all sets of distinct nonzero numbers $a,b,c$.
Prove that $$\cfrac{1}{a(a-b)(a-c)} +\cfrac{1}{b(b-c)(b-a)}
+\cfrac{1}{c(c-a)(c-b)} =\cfrac{1}{abc}$$ for all sets of distinct nonzero numbers $a,b,c $.
Now my question is not about how to solve this but rather why the technique which shows my book works.
Technique:
Rather than showing that the left side equals $\cfrac{1}{abc}$,we show
that $$\cfrac{1}{a(a-b)(a-c)} +\cfrac{1}{b(b-c)(b-a)}
+\cfrac{1}{c(c-a)(c-b)} -\cfrac{1}{abc}=0 $$
Writing the left side with the common denominator
$abc(a-b)(a-c)(b-c)$,we have
$$\cfrac{bc(b-c)-ac(a-c)+ab(a-b)-(a-b)(a-c)(b-c)}{abc(a-b)(a-c)(b-c)}=0$$
We can show that this is $0$ by showing that the numerator is $0$.We
can do this by looking at the numerator as a polynomial in $c$,meaning
let $a$ and $b$ be constants and $c$ be a variable,or
$$f(c)=bc(b-c)-ac(a-c)+ab(a-b)-(a-b)(a-c)(b-c)$$
Since $f(c)$ is a quadratic equation ,if we can show that this
quadratic has $3$ different roots,then $f(c)=0$ for all $c$.
The proof ends up with showing that $f(a)=0$,$f(b)=0$ and $f(0)=0$.
Now,while I can understand why a quadratic with $3$ roots is the zero polynomial I can't understand why we can treat the numerator as a polynomial and so treat $a,b$ as constants while $c$ as a variable.
Furthermore when we let it be a polynomial we also let $c=a=b$ but the problem in the beginning states that ${a,b,c}$ is all sets of distinct nonzero numbers,so I thought that we can't let $c=a=b$ by definition.
So can someone explain in depth why this is legit to do ?
| For a proof that might satisfy a thirst for more symmetry, but which uses a very similar technique, consider the equivalent identity $$\cfrac{bc}{(a-b)(a-c)} +\cfrac{ac}{(b-c)(b-a)} +\cfrac{ab}{(c-a)(c-b)} =1.$$ Let $f(x)$ be the polynomial function defined by $$f(x) = \cfrac{(x-b)(x-c)}{(a-b)(a-c)} +\cfrac{(x-c)(x-a)}{(b-c)(b-a)} +\cfrac{(x-a)(x-b)}{(c-a)(c-b)}.$$ Observe that $f$ is a quadratic polynomial with $f(a)=f(b)=f(c)=1$. It follows that $f(x)-1$ is a quadratic with three roots, so $f(x)-1=0$ identically. Now compare constant terms of the identity $f(x)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
explicit solution to heat equation without an integral sign Consider the 1-dimensional heat equation:
$$\left\{ \begin{align}
& {{u}_{t}}\left( x,t \right)={{u}_{xx}}\left( x,t \right),\text{ }x\in R,\text{ }t>0 \\
& u\left( x,0 \right)={{e}^{a{{x}^{2}}}},\text{ }x\in R \\
\end{align} \right.$$
Find an explicit solution without integral signs.
I have tried separation of variables, Green's function, and Fourier transform but just couldn't resolve the integral because of the term ${{e}^{a{{x}^{2}}}}$. Please help.
| Inspired by Heat equation with initial value answered by Mercy King.
To simplify, let $a=1$.
Setting
$$\xi=\sqrt{\frac{1-4t}{4t}}\left(y-\frac{x}{1-4t}\right),$$
we have:
\begin{eqnarray}
\frac{(x-y)^2}{4t}-y^2&=&\frac{(1-4t)y^2-2xy+x^2}{4t}\\
&=&\frac{1-4t}{4t}\left[y^2-\frac{2x}{1-4t}y+\frac{x^2}{1-4t}\right]\\
&=&\frac{1-4t}{4t}\left[\left(y-\frac{x}{1-4t}\right)^2+\frac{x^2}{1-4t}-\frac{x^2}{(1-4t)^2}\right]\\
&=&\frac{1-4t}{4t}\left[\left(y-\frac{x}{1-4t}\right)^2-\frac{4x^2t}{(1-4t)^2}\right]\\
&=&\xi^2-\frac{x^2}{(1-4t)}.
\end{eqnarray}
It follows that
\begin{eqnarray}
u(x,t)&=&\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^\infty\exp\left(-\frac{(x-y)^2}{4t}+y^2\right)\, dy\\
&=&\frac{1}{\sqrt{4\pi t}}\sqrt{\frac{4t}{1-4t}}\exp[\frac{x^2}{(1-4t)}]\int_{-\infty}^\infty\exp(-\xi^2)\, d\xi\\
&=&\frac{1}{\sqrt{\pi(1-4t)}}\exp[\frac{x^2}{(1-4t)}]\int_{-\infty}^\infty\exp(-\xi^2)\, d\xi.
\end{eqnarray}
Using the fact that
$$\int_{-\infty}^\infty\exp(-\xi^2)\, d\xi=\sqrt{\pi},$$
we get
$$u(x,t)=\frac{1}{\sqrt{1-4t}}\exp\left(\frac{x^2}{(1-4t)}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using substitution in complex equation $x^{10}-2x^5+\frac{4}{3}=0$
I substituted
$t=x^5$
Then I got with the quadratic formula the result
$t_1=1+\frac{\sqrt\frac{4}{3}}{2}i$
$t_2=1-\frac{\sqrt\frac{4}{3}}{2}i$
How do I calculate the roots now?
When I now resubstitute with
$t=x^5$
and form the polar coordinates do I get the roots for the whole equation?
| You now need to solve $$x^5 = \left(1 \pm \frac{i}{\sqrt{3}}\right) = \frac{2}{\sqrt{3}}\exp\left({i\left(\frac{\pi}{6} + 2\pi k\right)}\right).$$
This should net you ten solutions, $5$ from each - which is intuitively right since you were originally solving a polynomial of degree $10$.
So $$x = \left(\frac{2}{\sqrt{3}}\right)^{1/5}\exp{\left(i\left(\frac{\pi}{30} + \frac{2\pi k}{5}\right)\right)}$$
Then taking various values of $k$ will get you all your roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integrate: $\int \frac{2x^2-3x+8}{x^3+4x} \, dx$ $$\int \frac{2x^2-3x+8}{x^3+4x}\,dx$$
My main problem is calculating the $B$ and $C$. This is the algebra part. Thus, what is a technique I can use that is in line with what I did to calculate $A$?
| Since you're going to set $x=2t$ anyway, because of the factor $x^2+4$, do it at the start: the integral becomes
$$
\int\frac{8t^2-6t+8}{4t(t^2+1)}\,2dt
=
\int\frac{4t^2-3t+4}{t(t^2+1)}\,dt
=
\int\left(\frac{4}{t}-\frac{3}{t^2+1}\right)\,dt=4\log|t|-3\arctan t+c
$$
So your original integral is
$$
\int\frac{2x^2-3x+8}{x^3+4x}\,dx=4\log|x|-3\arctan\frac{x}{2}+c
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
An infinite nested radical problem From this link, problem 36, I found that
$$\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}=2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right).$$
The signs : + + - + + - + + - ... .
How to prove it?
Furthermore, how to represent $\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-...}}}}}}$ by trigonometric function ?
The signs : + - - + - - + - - ... .
Thanks for helping.
| HINT:
Let's try to show that the RHS satisfies the same equation as the LHS.
Consider
$$z = \cos \frac{2\pi}{19} + i \sin \frac{2\pi}{19}$$
Then $z^{19}-1 = 0$ and $z \ne 1$ so
$$\sum_{k=0}^{18} z^k = 0$$
On the right had side you have
$$ z^2 + z^{-2} + z^3 + z^{-3} + z^5 + z^{-5} = z^2 + z^{17} + z^3 + z^{16} + z^5 + z^{14}$$
Take the polynomial equation $P(x) = 0$ obtained for the LHS and show that
$$P(z^2 + z^{17} + z^3 + z^{16} + z^5 + z^{14}) \vdots \sum_{k=0}^{18} z^k $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Trigonometry identity proof sum I am normally good at solving trigonometry sums. However I stumbled upon this and could not figure out how to do it. I solved a few lines and then I got stuck. If anyone can teach me how to prove the equation, I'll be highly grateful.
If $\cos^2 A- \sin^2 A = \tan^2 B$, prove that $\cos^2 B - \sin^2 B = \tan^2 A$
So here is what I did:
$\cos^2 B = \dfrac{\sin^2 B}{\cos^2 A - \sin^2 A}$
So going by the second R.T.P,
$\cos^2 B - \sin^2 B = \dfrac{\sin^2 B}{\cos^2 A - \sin^2 A} - \sin^2 B$
which unfortunately ends up nowhere close to the original proving required.
Thanks for any possible help!
| From $\cos^2A-\sin^2A=\tan^2B$ you get
$$
\cos^2A=\frac{1+\tan^2B}{2}=\frac{1}{2\cos^2B}
$$
and
$$
\sin^2A=\frac{1-\tan^2B}{2}=\frac{\cos^2B-\sin^2B}{2\cos^2B}
$$
Thus
$$
\tan^2A=\frac{\cos^2B-\sin^2B}{2\cos^2B}\,2\cos^2B
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ac}{1+c} \geq 3.$
Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{1+ab}{1+a}+\dfrac{1+bc}{1+b}+\dfrac{1+ac}{1+c} \geq 3.$$
This looks symmetric, so should I prove it for just $a \leq b \leq c$ and then the other cases follow?
| Another way.
Let $a=\frac{y}{x},$ $b=\frac{z}{y}$, where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and by AM-GM we obtain: $$\sum_{cyc}\frac{1+ab}{1+a}=\sum_{cyc}\frac{1+\frac{z}{x}}{1+\frac{y}{x}}=\sum_{cyc}\frac{x+z}{x+y}\geq3\sqrt[3]{\prod_{cyc}\frac{x+z}{x+y}}=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the value of the integral $\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx$ Find the value of the integral $\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx$
$\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx$
$=\int_{0}^{\frac{\pi}{12}}\frac{1}{3}\frac{\tan^2x-\frac{1}{3}+\frac{1}{3}-3}{\tan^2x-\frac{1}{3}}dx$
$=\frac{1}{3}-\frac{8}{9}\int_{0}^{\frac{\pi}{12}}\frac{dx}{\tan^2x-\frac{1}{3}}dx$
$=\frac{1}{3}-\frac{8}{9}\int_{0}^{\frac{\pi}{12}}\frac{\cot^2x dx}{1-\frac{1}{3}\cot^2x}dx$
$=\frac{1}{3}-\frac{8}{9}\int_{0}^{\frac{\pi}{12}}\frac{\cot^2x dx}{\frac{4}{3}-\frac{1}{3}\csc^2x}dx$
I am stuck here and could not solve further.Please help me.
| HINT: convert your integrand into $\frac{4\cos(x)^2-1}{4\cos(x)^2-3}$ and use the tan-half angle formulas
the result should be $$2/3\,\sqrt {3}{\rm arctanh} \left(\cot \left( {\frac {5\,\pi }{12}}
\right) \sqrt {3}\right)+\pi /12
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Replacing in equation introduces more solutions Let's say I have an equation $y=2-x^2-y^2$.
now, since I know that $y$ is exactly the same as $2-x^2-y^2$ I can create the following, equation by replacing $y$ with $2-x^2-y^2$.
$y=2-x^2-(2-x^2-y^2)^2$
doing this replacement introduces new solutions such as $(-1, 0)$. Replacements in other various equations have similar results, although some do not change the equation at all!
What mechanic introduces these new solutions, and what are they?
Edit: one such example of an equation where no solutions are introduced via replacement is $y=x^2+y^2$. That will give $y=x^2+(x^2+y^2)^2$ which upon graphing is the same graph as the original, $y=x^2+y^2$.
Here is an image of the iteration of this replacement on the same function, just for fun.
| We know that every $(x,y)$ such that
$$y=2-x^2-y^2\quad\text{i.e.}\quad x^2+y^2+y-2=0$$
satisfies
$$y=2-x^2-(2-x^2-y^2)^2\quad \text{i.e.}\quad (2-x^2-y^2)^2+x^2+y-2=0\tag1$$
This implies that $(2-x^2-y^2)^2+x^2+y-2$ is divisible by $x^2+y^2+y-2$ :
$$(2-x^2-y^2)^2+x^2+y-2=(x^2+y^2+y-2)(x^2+y^2-y-1)$$
Hence,
$$(1)\iff x^2+y^2+y-2=0\quad\text{or}\quad x^2+y^2-y-1=0.$$
Similarly, for the second example, we know that $(x^2+y^2)^2+x^2-y$ is divisible by $x^2+y^2-y$ :
$$(x^2+y^2)^2+x^2-y=(x^2+y^2-y)(x^2+y^2+y+1)$$
Note here that there are no $(x,y)\in\mathbb R$ such that
$$x^2+y^2+y+1=0\iff x^2+\left(y+\frac 12\right)^2=-\frac 34$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Product of two series to get a series decomposition of zeta in the critical strip $\def\sfrac#1#2{%
\small#1%
\kern-.05em\lower0.1ex/\kern-.025em%
\lower0.4ex\small#2}$I've been working on gaining an intuitive understanding of the analytic continuation of the zeta function, but I've gotten stuck at this part where I have to multiply two very strange series together.
The approach is quite simple: first start with the Dirichlet eta function, defined by $η(s) = \frac{1}{1^s} + \frac{-1}{2^s} + \frac{1}{3^s} + \frac{-1}{4^s} + ... $. Then the following relationship holds:
$ζ(s) = η(s) · \frac{1}{1-\frac{2}{2^s}}$
If $\Re\{s\} > 1$, then the right factor can be expanded into a Taylor series, yielding
$ζ(s) = \left(\frac{1}{1^s} + \frac{-1}{2^s} + \frac{1}{3^s} + \frac{-1}{4^s} + ... \right) \cdot \left(\frac{1}{1^s} + \frac{2}{2^s} + \frac{4}{4^s} + \frac{8}{8^s} + ... \right) \\
= \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + ...$
where the product can be evaluated as a Dirichlet convolution.
This is because the image of $\frac{2}{2^s}$ lies within the radius of convergence of the $1 + x + x^2 + x^3 + ...$ series expansion of $\frac{1}{1-x}$, so the Taylor expansion is valid.
However, if $\Re\{s\} < 1$, then the image of $\frac{2}{2^s}$ lies outside the ROC, so that expansion doesn't work. Instead, we can take advantage of the functional equation
$f(x) + f(\frac{1}{x}) = 1$
where $f(x) = \frac{1}{1-x}$.
Substituting in $x=\frac{2}{2^s}$, and allowing that $\frac{1}{\left(\frac{2}{2^s}\right)} = \frac{\sfrac{1}{2}}{{\sfrac{1}{2}}^s}$, you get
$\frac{1}{1-\frac{2}{2^s}} = 1 - \frac{1}{1-\frac{\sfrac{1}{2}}{{\sfrac{1}{2}}^s}}$
where $\frac{\sfrac{1}{2}}{{\sfrac{1}{2}}^s}$ now has an image lying within the ROC of the aforementioned series expansion for $\Re\{s\} < 1$. So we can expand the whole right side as
$\frac{1}{1-\frac{2}{2^s}} = \frac{-\sfrac{1}{2}}{{\sfrac{1}{2}}^s} + \frac{-\sfrac{1}{4}}{{\sfrac{1}{4}}^s} + \frac{-\sfrac{1}{8}}{{\sfrac{1}{8}}^s} + \frac{-\sfrac{1}{16}}{{\sfrac{1}{16}}^s} + ...$
Finally, putting the whole thing together, we get
$ζ(s) = \left(\frac{1}{1^s} + \frac{-1}{2^s} + \frac{1}{3^s} + \frac{-1}{4^s} + ... \right) \cdot \left(\frac{-\sfrac{1}{2}}{{\sfrac{1}{2}}^s} + \frac{-\sfrac{1}{4}}{{\sfrac{1}{4}}^s} + \frac{-\sfrac{1}{8}}{{\sfrac{1}{8}}^s} + \frac{-\sfrac{1}{16}}{{\sfrac{1}{16}}^s} + ...\right)$
Which, for $0 < \Re\{s\} < 1$, is the product of a conditionally convergent Dirichlet series, and an absolutely convergent "fractional Dirichlet series."
And here I'm stumped. How can I expand this product? I understand the result should be some kind of "fractional Dirichlet series" where the denominators are dyadic rationals raised to the power of s, and I understand I basically want to perform some kind of Dirichlet convolution type thing here.
But how do I actually do it? What does the resulting expression look like?
| the series for $\eta(s)$ is absolutely convergent for $\Re(s) > 0$ if you group the terms by two :
$\eta(s) = \displaystyle\sum_{n=1}^\infty (2n-1)^{-s} - (2n)^{-s} = \sum_{n=1}^\infty \mathcal{O}(s (2n)^{-s-1})$ ( from the Taylor expansion of order 1 of $(1-x)^{-s}$ when $x \to 0$)
$\eta(s) = (1-2^{1-s}) \ \zeta(s)$ and for $\displaystyle \Re(s) < 1 : \ \ \frac{1}{1-2^{1-s}} = -\frac{2^{s-1}}{1-2^{s-1}} = - \sum_{k=1}^\infty 2^{k(s-1)}$.
you get :
$$\zeta(s) = - \sum_{k=1}^\infty 2^{k(s-1)} \eta(s) = - \sum_{n,k} \left( (2n-1)^{-s} - (2n)^{-s} \right) 2^{k(s-1)}$$
which is an absolutely convergent double sum for $\Re(s) \in ]0;1[$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$ It's required to prove that
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
I managed to go about out it two ways:
*
*Show it is equivalent to $\mathsf{true}$:
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
$$\Longleftrightarrow\sin x(1+\cos x+\sin x)\equiv(1+\cos x)(1-\cos x+\sin x)$$
$$\Longleftrightarrow\sin x+\cos x\sin x+\sin^2 x\equiv1-\cos x+\sin x+\cos x-\cos^2 x+\sin x \cos x$$
$$\Longleftrightarrow\sin^2 x\equiv1-\cos^2 x$$
$$\Longleftrightarrow\cos^2 x +\sin^2 x\equiv1$$
$$\Longleftrightarrow \mathsf{true}$$
*Multiplying through by the "conjugate" of the denominator:
$${\rm\small LHS}\equiv\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} $$
$$\equiv\frac{1+\cos x + \sin x}{1 - (\cos x - \sin x)} ~~\cdot ~~\frac{1+(\cos x - \sin x)}{1 +(\cos x - \sin x)}$$
$$\equiv\frac{(1+\cos x + \sin x)(1+\cos x - \sin x)}{1 - (\cos x - \sin x)^2}$$
$$\equiv\frac{1+\cos x - \sin x+\cos x + \cos^2 x - \sin x \cos x+\sin x + \sin x \cos x - \sin^2 x}{1 - \cos^2 x - \sin^2 x + 2\sin x \cos x}$$
$$\equiv\frac{1+ 2\cos x + \cos^2 x- \sin^2 x}{2\sin x \cos x}$$
$$\equiv\frac{1+ 2\cos x + \cos^2 x- 1 + \cos^2 x}{2\sin x \cos x}$$
$$\equiv\frac{2\cos x (1+\cos x)}{2\cos x(\sin x)}$$
$$\equiv\frac{1+\cos x}{\sin x}$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\equiv {\rm\small RHS}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$$
Both methods of proof feel either inelegant or unnecessarily complicated. Is there a simpler more intuitive way to go about this? Thanks.
| For fun, I created a trigonograph:
$$\frac{1 + \cos\theta + \sin\theta}{1 + \sin\theta - \cos\theta} = \frac{1 + \cos\theta}{\sin\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Integral $\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$ It's a follow-up to my previous question.
Can we find an anti-derivative
$$\int\arcsin x\cdot\ln^3x\,dx$$
or, at least, evaluate the definite integral
$$\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$$
in a closed form (ideally, as a combination of elementary functions and polylogarithms)?
| We will outline of a way forward leaving some of the work to the reader.
Denote the integral of interest by $I$ where
$$I=\int \arcsin(x) \log^3(x)\,dx \tag 1$$
Integrating $(1)$ by parts by letting $u=\arcsin(x)$ and $v=x\left(\log^3(x)-3\log^2(x)+6\log(x)-6\right)$, we find that
$$\begin{align}
I&=x\arcsin(x)\left(\log^3(x)-3\log^2(x)+6\log(x)-6\right)\\\\&-\int \left(\frac{\log^3(x)-3\log^2(x)+6\log(x)-6}{\sqrt{1-x^2}}\right)\,x\,dx \tag 2
\end{align}$$
Next, denote the integral on the right-hand side of $(2)$ by $J$. Enforcing the substitution $x=\sqrt{1-y^2}$ yields
$$\begin{align}
J&=-\int \left(\frac{\log^3(x)-3\log^2(x)+6\log(x)-6}{\sqrt{1-x^2}}\right)\,x\,dx\\\\
&=J_3+J_2+J_1+J_0
\end{align}$$
where
$$\begin{align}
J_3&=\int \log^3(\sqrt{1-y^2})\,dy \tag 3\\\\
J_2&=-3\int \log^2(\sqrt{1-y^2})\,dy \tag 4\\\\
J_1&=6\int \log(\sqrt{1-y^2})\,dy \tag 5\\\\
J_0&=-6\int 1\,dy \tag 6
\end{align}$$
The integrals in $(5)$ and $(6)$ can be evaluated in terms of elementary functions with
$$J_0=-6y$$
and
$$J_1=3y\log(1-y^2)-6y-3\log(1-y)+3\log(1+y)$$
The integrals in $(3)$ and $(4)$ can be expressed in terms of polylogarithm functions. For $J_2$ we can write
$$\begin{align}
J_2&=-3\int \log^2(\sqrt{1-y^2})\,dy\\\\
&=-\frac34 \left(K_1+K_2+K_3\right)
\end{align}$$
where
$$\begin{align}
K_1&=\int \log^2(1-y)\,dy \tag 7\\\\
K_2&=\int \log^2(1+y)\,dy \tag 8\\\\
K_3&=2\int \log(1-y)\log(1+y)\,dy \tag 9
\end{align}$$
The integrals $K_1$ and $K_2$ can be written in closed form with
$$\begin{align}
K_1&=(y-1)\left(\log^2(1-y)-2\log(1-y)+2\right)\\\\
\end{align}$$
and
$$\begin{align}
K_2&=(y+1)\left(\log^2(1+y)-2\log(1+y)+2\right)\\\\
\end{align}$$
For $K_3$ we integrate by parts with $u=\log(1-y)$ and $v=(y+1)\log(y+1)-y$ and obtain
$$\begin{align}
K_3&=2(y+1)\log(1-y^2)-2y\log(1-y)+2\int \frac{(y+1)\log(y+1)-y}{1-y}\,dy\\\\
&=2(y+1)\log(1-y^2)-2y\log(1-y)+2y+2\log(1-y)+2\int \frac{(y+1)\log(y+1)}{1-y}\,dy\\\\
&=2(y+1)\log(1-y)+2y\left(1-\log(1-y)\right)+4\int \frac{\log(1+y)}{1-y}\,dy \tag{10}
\end{align}$$
To evaluate the integral in $(10)$, we make the substitution $y=1-2z$. Then,
$$\begin{align}
\int \frac{\log(1+y)}{1-y}\,dy&=-\log(2)\log(w)-\int \frac{\log(1-w)}{w}\,dw\\\\
&=-\log(2)\log\left(\frac{1-y}{2}\right)+\text{Li}_2\left(\frac{1-y}{2}\right)
\end{align}$$
The integral $J_3$ can be evaluated in terms of the dilogarithm function $\text{Li}_2$ and trilogarithm function $\text{Li}_3$ using a similar approach to the one used herein to evaluate $K_2$. We will leave that very tedious analysis to the reader.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Which quantity is greater, $\frac{x^2}{y+\frac1y}$ or $\frac{y^2}{x+\frac1x}$? $x \gt y$, $ xy \neq 0$
A= $ x^2\over {y+{1\over y}}$
B= $ y^2\over {x+{1\over x}}$
Options:
1) Quantity A is greater.
2) Quantity B is greater.
3) The two quantities are equal.
4) The relationship cannot be determined from the given information.
By taking $x=2, y=1 $, I get $A\gt B$ , Thus options 2 and 3 are eliminated.
By taking$ x=2, y=-1$ , I get $B\gt A$ , Thus option 1 is eliminated.
So answer is option 4.
But I am not satisfied with this solution by taking particular values of $x$ and $y$.
Is there any other method to deal with this question?
What should be proper tag for this?
I think it should be comparision but that is not available in tag list.So please edit it.
| A nice approach is as follows: note first of all that $x + 1/x$ is positive if $x$ is positive and negative if $x$ is negative. Now, if $x$ and $y$ have the same sign, then
$$
A > B \iff\\
\frac{x^2}{y+1/y} > \frac{y^2}{x+1/x} \iff \\
x^2(x + 1/x) > y^2(y + 1/y) \iff \\
x^3 + x > y^3 + y
$$
Note, however, that $f(x) = x^3 + x$ is an increasing function. So, $x^3 + x > y^3 + y \iff x > y$. So, $A > B \iff x > y$
On the other hand, if $x$ and $y$ have opposite signs, then
$$
A > B \iff\\
\frac{x^2}{y+1/y} > \frac{y^2}{x+1/x} \iff \\
x^2(x + 1/x) < y^2(y + 1/y) \iff \\
x^3 + x < y^3 + y \iff\\
x < y
$$
So, it suffices to take any positive values $x<y$, then the values $-x,y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How to compute $\lim_{x \to 0} (\frac{x^5 e^{-1/x^2}+x/2 - \sin(x/2))}{x^3})$? I have a problem with this limit. I have no idea where is the problem.
Can you correct my mistake? Thanks
$$\lim\limits_{x \to 0} \left(\frac{x^5 e^\frac{-1}{x^2}+\frac{x}{2} - \sin(\frac{x}{2})}{x^3}\right)$$
I used the developments of McLaurin $e^x$ and $\sin x$
$$\lim\limits_{x \to 0} \left(\frac{x^5 (1-\frac{1}{x^2}+\frac{1}{2x^4})+\frac{x}{2} - ((\frac{x}{2})-(\frac{x^3}{48}))}{x^3}\right) =
\lim\limits_{x \to 0} \left(\frac{x^5-x^3+ \frac{x}{2} +\frac{x}{2} - \frac{x}{2}+\frac{x^3}{48}}{x^3}\right)=$$
$$\lim\limits_{x \to 0} \left(\frac{-x^3+\frac{x^3}{48}}{x^3}\right)=
\lim\limits_{x \to 0} \left(\frac{-\frac{47x^3}{48}}{x^3}\right)=\lim\limits_{x \to 0} \left(-\frac{47x^3}{48x^3}\right)= -\frac{47}{48}$$
but the result is wrong.
| Setting $y=x^{-1}$, it is equivalent to
$$
y^3 \cdot \left(\frac{1}{e^{y^2}y^5}+\frac{1}{2y}-\sin\left(\frac{1}{2y}\right)\right) \approx y^3 \cdot \left(\frac{1}{2y}-\left(\frac{1}{2y}-\frac{1}{48y^3}\right)\right)
$$
as $y\to \infty$. Conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Limit problem with an exponential function without L'Hopital How can I calculate this limit without L'Hospital rule and Taylor series?
$${\lim_{x \to 1} \big(4^x - 3^x\big)^{\frac{1}{x - 1}}}$$
| In THIS ANSWER, I showed using standard, non-calculus based analysis that
$$\frac{x}{x+1}\le \log(1+x)\le x$$
for $x\ge -1$ and
$$1+x\le e^x \le \frac{1}{1-x}$$
for $x<1$.
Now, using $(1)$ we can write
$$\frac{4^x-3^x-1}{(4^x-3^x)(x-1)}\le \frac{\log(4^x-3^x)}{x-1}\le \frac{4^x-3^x-1}{x-1}$$
Next, using $(2)$ we see that
$$\begin{align}
\frac{4^x-3^x-1}{x-1}&=\frac{4(4^{x-1})-3(3^{x-1})-1}{x-1}\\\\
&=\frac{4e^{\log(4)(x-1)}-3e^{\log(3)(x-1)}-1}{x-1}\\\\
&\le \frac{\frac{4}{1-\log(4)(x-1)}-3(1+\log(3)(x-1))-1}{x-1}\\\\
&=\frac{4\log(4)-3\log(3)+3\log(3)\log(4)(x-1)}{1-\log(4)(x-1)}\\\\
&\to 4\log(4)-3\log(3)\,\,\text{as}\,\,x\to 1 \tag 3
\end{align}$$
We also see using $(2)$ that
$$\begin{align}
\frac{4^x-3^x-1}{(4^x-3^x)(x-1)}&=\frac{4(4^{x-1})-3(3^{x-1})-1}{(4^x-3^x)(x-1)} \\\\
&\ge \frac{4(1+\log(4)(x-1))-3\frac{1}{1-\log(3)(x-1)}-1}{(4^x-3^x)(x-1)}\\\\
&=\frac{4\log(4)-3\log(3)-4\log(3)\log(4)(x-1)}{(4^x-3^x)(1-\log(3)(x-1))}\\\\
&\to 4\log(4)-3\log(3)\,\,\text{as}\,\,x\to 1 \tag 4
\end{align}$$
We use $(3)$ and $(4)$ along with the Squeeze Theorem to reveal
$$\lim_{x\to 1}\frac{\log(4^x-3^x)}{x-1}=\log\left(\frac{4^4}{3^3}\right)$$
Finally, we have
$$\lim_{x\to 1}\left(4^x-3^x\right)^{1/(x-1)}=\frac{256}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\sin(1/x)$ - why there is no limit? $$ \lim_{x\to 0+} \sin\left(\frac{1}{x}\right)$$
I know that there is no limit.
but, why there is no limit?
I tried $x=0.4$, $x=0.3$, $x=0.1$, it looks like the limit is $0$.
And how can I show that there is no limit? I tried to calculate it like all the other functions, and I got wrong result and I don't know why:
$$\lim_{x \to 0+} \sin\left(\frac{1}{x}\right) = \sin\left(\frac{1}{0^+}\right) = \sin\left(\frac{1}{\infty}\right) = \sin(0) = 0.$$
| Why there is no limit?
The graphic can help you understand why and suggest you some approach for the proof:
Remark: You have to be careful with tables of values because they can be misleading:
\begin{array}{ c | c c c c }
x & \frac{1}{2\pi} & \frac{1}{3\pi} & \frac{1}{4\pi} &\frac{1}{5\pi} \\ \hline
\sin\left(\frac{1}{x}\right) & 0 & 0 & 0 & 0 \\
\end{array}
\begin{array}{ c | c c c c }
x & \frac{2}{5\pi} & \frac{2}{9\pi} & \frac{2}{13\pi} &\frac{2}{17\pi} \\ \hline
\sin\left(\frac{1}{x}\right) & 1 & 1 & 1 & 1 \\
\end{array}
(The tables above are a sketch of the proof - see Theorem 2.4 here.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Trigonometry with Quadratic Equations If $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, then prove that
$$\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) = q$$
I tried the question but with $q$ other terms came associated.
| $$\text {We have } \tan A+\tan B=-p\; \text { and }\; \tan A \tan B=q.$$ Let $C=A+B.$ Let $F=\sin^2 C + p\sin C \cos C + q\cos^2 C-q.\quad$ CASE 1. $\cos C=0.$ Then $$F=1-q,\; \text {and}$$ $$0=\cos C=\cos (A+B)\implies \tan B=\cot A\implies q=\tan A \tan B=1.$$ CASE 2. $\cos C\ne 0.$ Then $\;\tan C\;$ is defined : $$\tan C=\tan (A+B)=(\tan A+\tan B)/(1-\tan A \tan B)=\frac {-p}{1-q}.$$ And we have $F/(\cos^2 C)= \tan^2 C +p\tan C +q(1-\sec^2 C)=$ $=\tan^2 C+p\tan C+q(-\tan^2 C)= (\tan C) \;([1-q]\tan C+p)=$$(\tan C) \;([1-q]\frac {[-p]}{[1-q]}+p)=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $\sum^n_{k=1} (\frac{k+2}{k+1})^nC_k = \frac{2^n-1}{6}$ then find n. Problem :
If $\sum^n_{k=1} (\frac{k+2}{k+1})^nC_k = \frac{2^n-1}{6}$ then find n.
My approach :
$\sum^n_{k=1} (\frac{k+2}{k+1})^nC_k = \frac{3}{2}^nC_1 +\frac{4}{3}^nC_2 +\frac{5}{4}^nC_3 +\cdots \frac{n+2}{n+1}$
Now how to proceed further please suggest will be of great help thanks.
| We have
$$\sum_{k=1}^n \left(\dfrac{k+2}{k+1}\right) \dbinom{n}k = \sum_{k=1}^n \dbinom{n}k + \sum_{k=1}^n \dfrac1{k+1} \dbinom{n}k$$
We have
$$\sum_{k=0}^n \dbinom{n}k x^k = (1+x)^n \,\,\,\,\, (\spadesuit)$$
Taking $x=1$, we obtain
$$\sum_{k=1}^n \dbinom{n}k x^k = 2^n-1$$
Integrating $(\spadesuit)$, we obtain
$$\sum_{k=0}^n \dfrac1{k+1} \dbinom{n}k x^{k+1} = \dfrac{(1+x)^{n+1}-1}{n+1} \,\,\,\,\, (\clubsuit)$$
Plugging in $x=1$, we obtain
$$\sum_{k=1}^n \dfrac1{k+1} \dbinom{n}k = \dfrac{2^{n+1}-n-2}{n+1}$$
I trust you can finish it off from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The value of the polynomial at given point. Given that:
$f(x)=x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+1$.
Find the value of $f(x)$ at $x=\sqrt{2}-1$
Answer is an integer. I tried factorization but couldn't proceed towards anything promising.
| You can make use of Horner's method. By further applying the formula $(a-b)(a+b) = a^2-b^2$ while sometimes factoring out common factors (-1 and 3 in this case), calculating the result becomes astonishingly nice:
$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|cc}
& 1 & 2 & -2 & -2 & 1 & 0 & 0 & 0 & 3 & 6 & 1\\
& + & + & + & + & + & + & + & + & + & + & + &\\
& 0 & \sqrt{2}-1 & 2-1 & -\sqrt{2}+1 & -1 & 0 & 0 & 0 & 0 & 3\sqrt{2}-3 & 3\\
\hline
x=\sqrt{2}-1 & 1 & \sqrt{2}+1 & -1 & -(\sqrt{2}+1) & 0 & 0 & 0 & 0 & 3 & 3(\sqrt{2}+1) & \boxed{\color{red}{4}}\\
\end{array}
$$
Short explanation:
*
*$1, 2, -2, \ldots, 1$ are the coefficient of the polynom from high to low, including zero ones!
*The third row (below the plus signs) always consists of the result in the last row of the last column multiplied by x. The value in the first column is 0.
*The fourth row is simply the sum of the first and third rows.
For example the first calculations of the scheme above would be as follows:
*
*$1 + 0 = 1$
*$(\sqrt{2}-1)*1 = \sqrt{2}-1$
*$2 + \sqrt{2}-1 = \sqrt{2}+1$
*$(\sqrt{2}+1)(\sqrt{2}-1) = 2 - 1 = 1$
*$\ldots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How to analyze the time complexity $\Theta$ of this recurrence I am trying to understand how to show that
$$T(n) = T(n/2) + T(n/4) + n^2$$
is $\Theta(n^2)$ by using a recursion tree. I tried substitution at first but it got real messy real fast.
This is self-study. Problem is from this pdf, problem 1-12.
| Suppose we start by solving the following recurrence:
$$T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + n^2$$
where $T(1) = 1$ and $T(0) = 0.$
Now let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
be the binary representation of $n.$
We unroll the recursion to obtain an exact formula for $n\ge 1$
$$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor}
[z^j] \frac{1}{1-z-z^2}
\left(\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}\right)^2.$$
We recognize the generating function of the Fibonacci numbers, so the
formula becomes
$$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor} F_{j+1}
\left( \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}\right)^2.$$
We now compute lower and upper bounds which are actually attained and
cannot be improved upon. For the lower bound consider a one digit
followed by a string of zeroes, to give
$$T(n) \ge \sum_{j=0}^{\lfloor \log_2 n \rfloor}
F_{j+1} 4^{\lfloor \log_2 n \rfloor-j}
= 4^{\lfloor \log_2 n \rfloor}
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
F_{j+1} 4^{-j}.$$
Now since $$\left|\frac{1+\sqrt{5}}{2}\right|<4$$
the sum term converges to a number, we have
$$1 \le
\sum_{j=0}^{\lfloor \log_2 n \rfloor-1} F_{j+1} 4^{-j}
\lt \sum_{j=0}^{\infty} F_{j+1} 4^{-j}
= \frac{16}{11}$$
and we thus get for the asymptotics
$$\frac{16}{11} 4^{\lfloor \log_2 n \rfloor}.$$
For an upper bound consider a string of one digits to get
$$T(n) \le \sum_{j=0}^{\lfloor \log_2 n \rfloor}
F_{j+1}
\left( \sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^{k-j}\right)^2
= \sum_{j=0}^{\lfloor \log_2 n \rfloor}
F_{j+1} (2^{\lfloor \log_2 n \rfloor+1-j} - 1)^2
\\ = 4\times 4^{\lfloor \log_2 n \rfloor}
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
F_{j+1} 4^{-j}
- 4\times 2^{\lfloor \log_2 n \rfloor}
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
F_{j+1} 2^{-j}
+ \sum_{j=0}^{\lfloor \log_2 n \rfloor}
F_{j+1}
\\ = 4\times 4^{\lfloor \log_2 n \rfloor}
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
F_{j+1} 4^{-j}
- 4\times 2^{\lfloor \log_2 n \rfloor}
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
F_{j+1} 2^{-j}
+ F_{\lfloor \log_2 n \rfloor + 3} - 1.$$
The same constant appears as in the lower bound as well as an
additional constant. Computing these we get
$$\frac{64}{11} \times 4^{\lfloor \log_2 n \rfloor}
- 16\times 2^{\lfloor \log_2 n \rfloor}
+ F_{\lfloor \log_2 n \rfloor + 3} - 1.$$
Doing the asymptotics of these three terms we obtain a term in $n^2$,
a term in $n$ and a term in $n^{\log_2 \varphi} \in o(n)$ where
$\varphi$ is the golden ratio.
Joining the upper and the lower bound we get for the asymptotics of
this recurrence that it is
$$T(n)\in\Theta\left(4^{\lfloor \log_2 n \rfloor}\right)
= \Theta\left(2^{2\log_2 n}\right) = \Theta(n^2),$$
which, let it be said, could also have been obtained by inspection.
Remark. The evaluation of the constant is done by noting that
the generating function of
$$F_{j+1} 4^{-j}\quad
\text{is}\quad\frac{1}{1-z/4-z^2/16}$$
which at $z=1$ evaluates to $\frac{1}{1-1/4-1/16} = \frac{16}{11}.$
Similarly we have the generating function of
$$F_{j+1} 2^{-j}\quad
\text{is}\quad\frac{1}{1-z/2-z^2/4}$$
which at $z=1$ evaluates to $\frac{1}{1-1/2-1/4} = 4.$
We have a certain flexibility as to what power of four to use in the
constant but this does not affect the asymptotics.
A closely related recurrence that is somewhat simpler may be found at
this MSE link.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $a$ such that $4\sin^2x+4\sin x+a^2-3<0$ for atleast one negative $x$ Find the value of $a$ such that $4\sin^2x+4\sin x+a^2-3<0$ for atleast one negative $x$.
$4\sin^2x+4\sin x+a^2-3<0$
$4\sin^2x+4\sin x<-a^2+3$
or $-a^2+3>4\sin^2x+4\sin x$
I found the minimum and maximum values of $4\sin^2x+4\sin x$.Minimum value is $-1$ and maximum value is $8$.
So $-a^2+3>8$
So $a^2+5<0$ But this gives me no values of $a$.And the answer given is $-\sqrt3<a<\sqrt3$.
I do not know where i am wrong.
| First of all, previous considerations:
a) For such $a$, if there is at least $x<0$ satisfiyng the inequality, then there is an $x^\prime\in[0,2\pi]$ that satisfies it (cause periodicity of sine function), and the inverse is also true. Thus, we can change the problem to find values of $a$ for which inequality is satisfied for at least one $x$ in $[0,2\pi]$.
b) Let $f:[0,2\pi]\to\mathbb{R}$ given by $f(x)=3-4\sin^2x-\sin x$. Then, it is easy to prove that $f(x)>0$ iff $x\in [0,\pi/6)\cup(5\pi/6,2\pi]$.
Then, if $x\in[0,\pi/6)\cup(5\pi/6,2\pi]$, there is an $a$ such that $f(x)>a^2$, due to (b). But, in $[0,\pi/6)\cup(5\pi/6,2\pi]$, the maximum are attained at $x=0,7\pi/6,11\pi/6$, yielding $f(x)=3,4,4$ respectively.
Putting all together: By means of Intermediate Value Theorem, for each $y\in(0,4]$, there is an $x\in[0,\pi/6)\cup(5\pi/6,2\pi]$ such that $f(x)=y$. Finally, if we take $a^2\in[0,4)$, there are $y\in(a^2,4)$ and $x\in[0,\pi/6)\cup(5\pi/6,2\pi]$ such that $f(x)=y>a^2$. So, the interval for $a$ is when $a^2\in[0,4)$, which is equivalent to $-2<a<2$.
Conversely, by the same idea, it is easy to show that for $a\in(-2,2)$ there is an $x\in[0,\pi/6)\cup(5\pi/6,2\pi]$ such that $f(x)>a$.
Conclusion: the desired interval is (-2,2), not $(-\sqrt{3},\sqrt{3})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve irrational inequality $\sqrt{x^2-8x+15} + \sqrt{x^2+2x-15} > \sqrt{4x^2-8x+18}$ Here's what I tried
$$\sqrt{x^2-8x+15} + \sqrt{x^2+2x-15} > \sqrt{4x^2-8x+18}$$
$$\sqrt{(x-3)(x-5)} + \sqrt{(x-3)(x+5)} > \sqrt{4x^2-8x+18}$$
Now,
$$(x-3)(x-5) > 0 \Leftrightarrow x \in (- \infty, 3) \cup (5, \infty)$$
$$(x-3)(x+5) > 0 \Leftrightarrow x \in (-\infty, -5) \cup (3, \infty)$$
$$4x^2-8x+18 > 0, \forall x \in R$$
So, for $x \in (- \infty, -5) \cup (5, \infty)$, we can take squares of both sides and we win this:
$$|x-3| \sqrt{(x-5)(x+5)} > x^2 - x + 9$$
Are my steps correct? And what should I do now? Should I solve for $x < -5,$ and for $x>5$?
Thanks.
| You are correct so far. for $|x|\geq 5$ both sides of your last equation are positive, so you can square the equation again and solve the remaining equation (it is cubic at most).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$$
| $$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{(2x+1)^2(1+\frac{\ln x}{(2x+1)^2})(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{(2x+1)^2(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{(x+2)!}{(2x+1)^2(x)!} + \lim\limits_{x \to +\infty} \frac{4^x}{(2x+1)^2(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{(x+2)(x+1)}{(2x+1)^2} + \lim\limits_{x \to +\infty} \frac{4^x}{(2x+1)^2(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{x^2+3x+1}{4x^2+4x+1} + 0 =$$
$$\lim\limits_{x \to +\infty} \frac{1+\frac{3}{x}+\frac{1}{x^2}}{4+\frac{4}{x}+\frac{1}{x^2}} =\frac{1}{4}$$
The only missing part is
$$\lim\limits_{x \to +\infty} \frac{4^x}{(2x+1)^2(x)!} = 0$$
For this you need to notice that
$$\frac{4^{x+1}}{(x+1)!} = \frac{4 \cdot 4^{x}}{(x+1)x!}=\frac{4}{x+1}\frac{4^{x}}{x!} < \frac{4}{x}\frac{4^{x}}{x!}$$
Since $\lim\limits_{n \to +\infty} \frac{4^n}{x^n}=0$ we have $\lim\limits_{x \to +\infty} \frac{4^{x+1}}{(x+1)!}=0$ as well
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Does $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ converge? Does $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ converges?
Well, I'm trying the approach using Cauchy's test for convergence. If $a_n >= 0$ and $\lim \sqrt[n]a_n = L $ exists. then, I need to check whether it is bigger or lesser than $1$. (if equal then we have a problem)
So.. $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ becomes $\sum_{n=1}^\infty \frac{\sqrt[n]{2\cdot 4\cdot 6\cdot ...\cdot (2n)}}{n}$ becomes $\frac{({2\cdot 4\cdot 6\cdot ...\cdot (2n)})^{1/n}}{n}$ becomes computing the limit of $1/n$ which is zero. Which means it converges?
| Observe this variant of rational test.
$$\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}=\sum_{n=1}^\infty 2^n \frac{n!}{n^n}=\sum_{n=1}^\infty 2^n \prod\limits_{k=1}^{n} \frac{k}{n}$$
Now $$\sum_{n=1}^\infty 2^n \cdot b_{n}$$ could be understood as sort of a binary expansion and we are sure that it converges as long as $$b_{n} < \frac{k}{d^{n}}, \, \, \, d > 2$$ for all $n>M$ for some fixed $k$ and $M$.
This is the same as requiring
$$\frac{b_{n+1}}{b_{n}} < \frac{1}{d}$$
for some $n>M$ but indeed we have
$$\frac{\prod\limits_{k=1}^{n+1} \frac{k}{n+1}}{\prod\limits_{k=1}^{n} \frac{k}{n}}=(\frac{1}{1+\frac{1}{n}})^{n}$$
Since $\lim\limits_{n \to \infty}(\frac{1}{1+\frac{1}{n}})^{n}=\frac{1}{e}$ we are certain that there are $M$ and $d$ for which $$\frac{b_{n+1}}{b_{n}} < \frac{1}{e} < \frac{1}{d} < \frac{1}{2} $$
For this reason the series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Infinite Series $\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{m!\:n!}{(m+n+2)!}$ Evaluating
$$\sum_{m=0}^\infty \sum_{n=0}^\infty\frac{m!n!}{(m+n+2)!}$$
involving binomial coefficients.
My attempt: $$\frac{1}{(m+1)(n+1)}\sum_{m=0}^\infty \sum_{n=0}^\infty\frac{(m+1)!(n+1)!}{(m+n+2)!}=\frac{1}{(m+1)(n+1)} \sum_{m=0}^\infty \sum_{n=0}^\infty\frac{1}{\binom{m+n+2}{m+1}}=?$$
Is there any closed form of this expression?
| One may observe that,
$$
\frac{m!}{(m+n+2)!}=\frac{m!}{(n+1)(m+n+1)!}-\frac{(m+1)!}{(n+1)(m+n+2)!}
$$ giving, by telescoping terms,
$$
\sum_{m=0}^N\frac{m!}{(m+n+2)!}=\frac1{(n+1)(n+1)!}-\frac{(N+1)!}{(n+1)(N+n+2)!}
$$ thus, as $N \to \infty$,
$$
\sum_{m=0}^\infty\frac{m!}{(m+n+2)!}=\frac1{(n+1)(n+1)!}.
$$ Then the initial series reduces to
$$
\sum_{n=0}^\infty \sum_{m=0}^\infty\frac{n!\:m!}{(m+n+2)!}=\sum_{n=0}^\infty\frac{n!}{(n+1)(n+1)!}=\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Explanation of formula for integer sequence with integers being repeated according to polynom I have two questions concerning OEIS sequence A056556:
$m$ is repeated $\frac{1}{2}(m+1)(m+2)$ times:
$$0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5,
5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,
6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, ...$$
Formula: $a(n)=\lfloor x\rfloor$ where x is the (largest real) solution to
$x^3+3x^2+2x-6n=0$
*
*Why is that formula correct?
*Is it possible to derive such a formula for any integer sequence where $m$ is repeated according to a polynom $p(m) = \sum_{i=0}^k c_i\ m^i$?
| Let's study the cubic $\;P_n(x):=x^3+3x^2+2x-6n\,$
Its discriminant is given by $\;\Delta=4\,(1-243\,n^2)\,$ and will thus be negative for $n>0$ :
$P_n(x)$ will always have two conjugate complex solutions with negative real part and one real positive solution (thus the largest).
Since $\;P_n(x):=x(x+1)(x+2)-6\,n\ \ $ we will get the trivial solution $x=m,\ m\in\mathbb{N}$ when $n$ is a "tetrahedral number" A000292 i.e. when $\,n=T(m):=\dfrac {m(m+1)(m+2)}6$.
When $n$ is between two consecutive tetrahedral numbers $T(m)$ and $T(m+1)$ the largest root $x_0$ of $P_n(x_0)=0\,$ will still be given by $x_0=m_0$ but with $m_0$ the real solution of $n=T(m_0)=\dfrac {m_0(m_0+1)(m_0+2)}6\,$ belonging to $(m,m+1)$ and thus verify $\lfloor x_0\rfloor=m$.
(the function $m\mapsto n=T(m)$ is positive and smoothly increasing and thus the inverse function $n\mapsto m(n)=T^{-1}(n)\;$).
We found that for $n$ in $[T(m),T(m+1))\;$ the largest root of $\;P_n(x):=x(x+1)(x+2)-6\,n\ \ $ verified $\lfloor x\rfloor=m\,$ but $[T(m),T(m+1))\;$ has $T(m+1)-T(m)=\dfrac {(m+1)(m+2)(m+3)}6-\dfrac {m(m+1)(t+2)}6=\dfrac {(m+1)(m+2)}2$ elements as wished with $\lfloor x\rfloor=m$.
This method may clearly be generalized to other polynomials of "ascending factorial type" : $x^{(n)}=x(x+1)(x+2)\cdots(x+n-1)\;$ but may be harder to handle in the general case (the "largest root" part and so on...).
Anyway a good starting point for a polynomial $p$ provided should be to search the polynomial $T$ such that : $\;T(m+1)-T(m)=p(m)$ (or define it recursively this way!).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
When is the limit of a sum equal to the sum of limits? I was trying to solve a problem and got stuck at the following step:
Suppose ${n \to \infty}$ .
$$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = 1$$
Let us rewrite $n^3=n \cdot n^2$ as $n^2 + n^2 + n^2 + n^2 \dots +n^2$,$\space$ n times.
Now we have
$$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = \frac {n^2 + n^2 + n^2 + n^2 + n^2 \dots +n^2}{n^3} $$
As far as I understand, we can always rewrite the limit of a sum as the sum of limits ...
$$\dots = \lim \limits_{n \to \infty} \left(\frac{n^2}{n^3} + \frac{n^2}{n^3} + \dots + \frac{n^2}{n^3}\right)$$
...but we can only let ${n \to \infty}$ and calculate the limit if all of the individual limits are of defined form (is this correct?). That would be the case here, so we have:
$= \dots \lim \limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}\right) =$[ letting ${n \to \infty}]$ $= 0 + 0 + \dots + 0 = 0$
and the results we get are not the same.
Where did I go wrong?
| Because the number of terms goes up exactly as the size of each term goes down.
Specifically $$\lim \limits_{n \to \infty} \Big(\underbrace{\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}}_{n\text{ times}}\Big) = \lim \limits_{n \to \infty} \sum_{i=1}^n \frac 1n$$
Does that help?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose the row that they chose. This bothers me. As I should be able to do it as I see fit. I will post my work and someone point out the problem in my work. The matrix is as follows:
$$A = \left(
\begin{matrix}
5&-7&2&2\\
0&3&0&-4\\
-5&-8&0&3\\
0&5&0&-6\\
\end{matrix} \right)
$$
I decided to expand across row one and cross out columns as I found the minors. For the first minor obtaining:
$$
\begin{pmatrix}
3 & 0 & -4 \\
-8 & 0 & 3 \\
5 & 0 & -6 \\
\end{pmatrix}
$$
M1 being row one column one we attain $-1^2 = 1$. This is to be multiplied by the determinate of the minor. Now finding the determinant I did:
3 times $$
\begin{pmatrix}
0 & 3 \\
0 & -6 \\
\end{pmatrix}
$$
giving $3(0-0)= 0$
then:
0 times $$
\begin{pmatrix}
-8 & 3\\
5 & -6\\
\end{pmatrix}
$$
giving 0(48-15)=0
Then:
4 times
$$
\begin{pmatrix}
-8 & 0 \\
5 & 0 \\
\end{pmatrix}
$$
giving $4(0-0)=0$
adding the determinants we get $0+0+0=0$
So det M1 $= 0(1) = 0$
M2--> M(1,2)---> $-1^1+2= -1^3 = -1$
$$
\begin{pmatrix}
0 & 0 & -4 \\
-5 & 0 & 3 \\
0 & 0 & -6 \\
\end{pmatrix}
$$
o*
$$
\begin{pmatrix}
0 & 3 \\
0 & -6 \\
\end{pmatrix}
$$
giving $0(0-0)=0$
obviously the next matrix will look the same as the top term in column two is a zero so the determinant for that will be $0$. Now finally
4 times
$$
\begin{pmatrix}
-8 & 0 \\
5 & 0 \\
\end{pmatrix}
$$
giving 4(0-0)= 0
So the Determinant of Minor 2 is (0+0+0)(-1)= 0 Now on to Minor number 3
M3 --> $-1^4 = 1$
$$
\begin{pmatrix}
0 & 3 & -4 \\
-5 & -8 & 3 \\
0 & 5 & -6 \\
\end{pmatrix}
$$
for the determinant:
0 times
$$
\begin{pmatrix}
-8 & 3 \\
5 & -6 \\
\end{pmatrix}
$$
which gives $0(48-15)=0$
-3 times
$$
\begin{pmatrix}
-5& 3 \\
0 & -6 \\
\end{pmatrix}
$$
which gives $-3(30-0)= -90$
it is redundant to go on from here because after the final computation for this minor I get -100 and as a result get det M3 = -190 and get determinant of zeros for the following determinant of M4.
which gives: $0(5)+ 0(-7) + (-90)(2) + (0)(2)$ giving
Det Ax $= -380.$ The book says its $20$ and when I did it in a calculator it got 20 but the problem is that both the book and calculator expand across the row with the most zeros but theoretically speaking NO MATTER WHICH row or column you choose to expand across you should get the same answer. So what is it? Is my computation wrong or is my assumption that you can expand across any row or column wrong? Isn't it only important if the determinant doesn't equal zero? or does the exact value matter in more advanced cases?
| Write out the determinant. Alongside write the first 3 columns"Downward diagonal multiplication" gives ( O + O + 2OO + O ). "Upward diagonal multiplication" gives -(O + O + O + 18O) . det = 2O.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 5
} |
Integral $\int\left(\frac{1}{x^4+x^2+1}\right)dx$ Someone can halp me to solve this integral:
$$\int\left(\frac{1}{x^4+x^2+1}\right)$$
solution$$\frac{1}{4}\ln\left(\frac{x^2+x+1}{x^2-x+1}\right)+\frac{1}{2\sqrt3}\arctan\frac {x^2-1}{x\sqrt3}$$
I don't manage using partial fraction because $${x^4+x^2+1}$$ has $\Delta\lt 0$ and substituing $t=x^2,$ $2x$ appears.That's a problem.
| HINT:
$$\dfrac2{x^4+x^2+1}=\dfrac{x^2+1-(x^2-1)}{x^4+x^2+1} =\dfrac{1+\dfrac1{x^2}}{x^2+1+\dfrac1{x^2}}-\dfrac{1-\dfrac1{x^2}}{x^2+1+\dfrac1{x^2}}$$
Now as $\int\left(1+\dfrac1{x^2}\right)dx=x-\dfrac1x,$
write the denominator as $\left(x-\dfrac1x\right)^2+3$
Similarly for the second integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to use parametric equation/trigonometric identity to show an ellipse? I have the equation $16x^2+25y^2=400$, and the parametric equation $(x,y)=(5\cos t, 4\sin t)$.
If I plug in the parametric equation into the first equation, I end up with the trigonometric identity $\cos^2 t+ \sin^2 t= 1$. How does this identity show that my non-parametric equation, when graphed, will result in an ellipse?
| The pair of parametric equations $x = 5\cos t, y = 4\sin t$ in fact corresponds to the ellipse described by the Cartesian equation $16x^2 + 25y^2 = 400$. Substituting the first equation into the second would naturally give you an identity (like you got). However, to properly prove this, it's better to manipulate one form into another.
To prove this, manipulate the parametric equations:
$x^2 = 25\cos^2 t$
$\frac{1}{25}x^2 = \cos^2 t$
and similarly,
$\frac{1}{16}y^2 = \sin^2 t$
Add those up:
$\frac{1}{25}x^2 + \frac{1}{16}y^2 = \cos^2 t + \sin^2 t$
And use the trigonometric identity $\cos^2 t + \sin^2 t = 1$ to get:
$\frac{1}{25}x^2 + \frac{1}{16}y^2 = 1$
Rearranging,
$16x^2 + 25y^2 = 400$, which is exactly the Cartesian equation you expect.
You can see this is an ellipse by recognising the earlier derived canonical form:
$(\frac{x}{a})^2 + (\frac{y}{b})^2 = 1$, where $a = 5$ and $b = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determine convergence when AST does not work Consider this series --
$$\frac{1}{2^3}-\frac{1}{3^2}+\frac{1}{4^3}-\frac{1}{5^2}+\frac{1}{6^3}-\frac{1}{7^2}+\frac{1}{8^3}-\frac{1}{9^2}+\dots$$
Apparently, alternating series test (AST) cannot be applied. What other methods should I consider in order to determine if the series converges absolutely, converges conditionally, or diverges? Thanks.
| Generically, if a series converges absolutely, then it converges (i.e. without absolute values around each term). So you might consider the absolute version,
$$ \frac{1}{2^3} + \frac{1}{3^2} + \frac{1}{4^3} + \frac{1}{5^2} + \cdots$$
This looks very similar to some very common series. It's clear that the $n$th term is bounded above by $\frac{1}{n^2}$. So then by comparison, the absolute version of your series is less than
$$ \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6} - 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all $n$ such that $3^{2n+1}+2^{n+2}$ is divisible by $7$ Find all $n$ such that
$3^{2n+1}+2^{n+2}$ is divisible by $7$
Prove that your answer is correct
So I am not allowed to use mods, as is a calculus question, I have tried by induction but can't get to prove that it works for $k+1$, by multiplying the equation by powers of $2$ and $3$.
Thank you for your help
| *
*Setting $n=1$, we get $$3^{2\cdot 1+1}+2^{1+2}=35$$ above number $35$ is divisible by $7$ hence it holds for $n=1$
*Assume the number $3^{2n+1}+2^{n+2}$ is divisible by $7$ for $n=k$ then $$3^{2k+1}+2^{k+2}=7\lambda \tag 1$$
*Setting $n=k+1$, we get $$3^{2k+2+1}+2^{k+1+2}$$
$$=9\cdot 3^{2k+1}+2\cdot 2^{k+1}$$
$$=9\cdot 3^{2k+1}+9\cdot 2^{k+1}-7\cdot 2^{k+1}$$
$$=9(3^{2k+1}+2^{k+1})-7\cdot 2^{k+1}$$
setting the value from (1),
$$=9(7\lambda)-7\cdot 2^{k+1}$$
$$=7(9\lambda-2^{k+1})$$
since, $(9\lambda-2^{k+1})$ is some integer hence the number $7(9\lambda-2^{k+1})$ is divisible by $7$
hence the number $(3^{2n+1}+2^{n+1})$ is divisible by $7$ for all integers $n\ge 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
} |
A simple inequality, does it always hold? I am looking at the following inequality:
$$\Gamma(-\frac{x+1}{x})\lt x , \forall x\gt \frac{7}{2}$$
It seems that the LHS and RHS eventually diverge for large enough $x$, but I have failed in a proof of that inequality. Any help would be greatly appreciated. Thanks.
| We shall show this using the helpful observation made by Claude Leibovici.
First, we shall show that $0 < z < 2/7$,
$$
f(z) \equiv \log\Gamma(1 - z) - z + \frac{ z^2 } { 2} < 0.
\qquad (1)
$$
From the Bohr-Mollerup theorem, $\log\Gamma(1-z)$ is convex, so is $-z + z^2/2$.
So $f(z)$ is convex, and for $z < a = 2/7$, we have
$$
f(z) < \frac{z}{a} f(a) + \left(1 - \frac{z}{a}\right) f(0) = \frac{z}{a} f(a) < 0,
$$
where we have used the fact $f(a) \approx -0.0012 < 0$.
Second, for $0 < z < 1$
$$
\begin{aligned}
\log(1 + z)
&= z - \frac{z^2}2 + \frac{z^3}{3} - \frac{z^4}{4} + \dots
\\
&> z - \frac{z^2}{2} + \frac{z^3(1-z)}{4} + \frac{z^5(1-z)}{6} \dots
\\
&> z - \frac{z^2}{2}.
\qquad (2)
\end{aligned}
$$
Thus, from (1) and (2) we have for $0 < z < 2/7$,
$$
\log \frac{\Gamma(1-z)}{1+z} < f(z) < 0,
$$
or
$$
\frac{\Gamma(1-z)}{1+z} < 1.
$$
Finally, for $x > 7/2$,
$$
\begin{aligned}
\Gamma\left(-\frac{x+1}{x}\right)
&=
\frac{ \Gamma\left( -\frac 1 x \right) }
{\left(-1 - \frac 1 x \right)}
=
\frac{ \Gamma\left( 1 -\frac 1 x \right) }
{\left(-1 - \frac 1 x \right)\left(-\frac 1 x\right) }
=
x \, \frac{ \Gamma\left( 1 -\frac 1 x \right) } {1 + \frac{1}{x} }
< x.
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Pythagorean triples with the same c value $a^2 + b^2 = c^2$
There are, Primitive Pythagorean Triples, that share the same c value. For example,
$63^2 + 16^2 = 65^2$ and $33 ^2 + 56^2 = 65^2$.
I have been trying to figure out why the following theorem for finding such triples works.
Take any set of primes. Ex: $5,13,17$.
Now take their product, $1052$, this is the new $c^2$ value. You can express that $c^2$ value as a triple by factoring the product of primes as Gaussian Integers:
$(2-i)(2+i)(3+2i)(3-2i)(4+i)(4-i)$
Now if you take three of those Gaussian Integers and solve for their product:
ex, $(2+i)(3+2i)(4+i) = 9+32i$ and you have found an $a$ value $9$ and a $b$ value $32$ that works in the theorem. $9^2 + 32^2 = 1105$
You can continue with this:
$(2-i)(3+2i)(4+i) = a + bi$
$(2+i)(3-2i)(4+i) = a + bi$
$(2+i)(3+2i)(4-i) = a + bi$
In fact, the number of triples with this method is $2^{||p|| - 1}$ where $||p||$ is the number of primes used.
Can someone please explain why this method of finding these "c stuck triples" works the way it does?
EDIT: It appears that you cannot take any prime, $p$ but rather must use primes congruent to $1mod4$ according to Fermat's theorem on sums.
| The system of equations:
$$\left\{\begin{aligned}&x^2+y^2=z^2\\&q^2+t^2=z^2\end{aligned}\right.$$
Formulas you can write a lot, but will be limited to this. Will make a replacement.
$$a=p^2+s^2-k^2$$
$$b=p^2+s^2+k^2-2pk-2ks$$
$$c=p^2+k^2-s^2+2ps-2kp$$
$$r=s^2+k^2-p^2+2ps-2ks$$
The solution then is.
$$x=2ab$$
$$y=a^2-b^2$$
$$q=2cr$$
$$t=c^2-r^2$$
$$z=a^2+b^2$$
$p,s,k$ - integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Closed form of $\sum\frac{1}{k}$ where $k$ has only factors of $2,3$
Consider the set containing $A$ all positive integers with no prime
factor larger than $3$, and define $B$ as
$$
B= \sum_{k\in A} \frac{1}{k}
$$
Thus, the first few terms of the sum are:
$$
\frac{1}{1} +\frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\frac{1}{6} +\frac{1}{8} +\frac{1}{9} +\frac{1}{12} +\frac{1}{16} +\frac{1}{18} + \cdots
$$
a) Write a
closed-form expression for $X$ that makes the equation below true. In
other words what expression should $X$ be so that the following
equation is true, i.e. writing $B$ in terms of $X$.
$$
B = \sum_{i=0}^\infty \sum_{j=0}^\infty X
$$
b) Write a closed-form expression for .
| In general, you use the same logic used for the Euler product. Notice that your form becomes $$(1+\sum\limits_{k=1}^{\infty}\frac{1}{2^k})(1+\sum\limits_{k=1}^{\infty}\frac{1}{3^k})$$ since you have all powers of 2 and all powers of 3 and when you multiply them you have all possible combinations without repetition, and since prime decomposition is unique we are sure we have the same series as you have required. But this is
$$(1+\sum\limits_{k=1}^{\infty}\frac{1}{2^k})(1+\sum\limits_{k=1}^{\infty}\frac{1}{3^k})=\frac{1}{1-\frac{1}{2}}\frac{1}{1-\frac{1}{3}}=3$$
If you ask the same question for other cases and you know the prime numbers involved $p_{1},p_{2},...,p_{m}$ you will have the answer as
$$\prod_{k=1}^{m}\frac{p_{k}}{p_{k}-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum roots of $f'(x)\cdot f'''(x)+(f''(x))^2 =0$ given certain conditions on $f(x)$. Problem:
Let $f(x)$ be a thrice differentiable function satisfying:
$$|f(x) - f(4-x)| + |f(4-x)-f(4+x)| = 0, \forall x \in R$$
If $f'(1)=0$, then find the minimum number of roots of $f'(x)\cdot f'''(x)+(f''(x))^2 =0$, on $x \in [0,6]$
My attempt:
We know: $f(x)=f(4-x)$ and $f(4-x)=f(4+x)$
So, $f(x)=f(x+4)$. That is, the period of the given function is $4$.
It can also be noted that the function is symmetric about $2$ and $4$.
I also know that the second equation is nothing but $\frac{d}{dx}(f'(x) \cdot f''(x))$
I don't know how to proceed from here.
| Since $f(x) = f(4-x) = f(4+x)$ we have $f'(x) = -f'(4-x) = f'(4+x)$ for all $x \in \mathbb{R}$.
Since $f'(1) = 0$, using the above identity for $x = 1$ gives us $f'(1) = f'(3) = f'(5) = 0$.
Also, using the above identity for $x = 2$ yields $f'(2) = -f'(2) = f'(6)$, so $f'(2) = f'(6) = 0$.
Similarly, for $x = 0$, we get $f'(0) = -f'(4) = f'(4)$, so $f'(0) = f'(4) = 0$.
By Rolle's theorem, for $n = 1,2,3,4,5,6$, there exists a $x_n \in (n-1,n)$ such that $f''(x_n) = 0$.
Thus, the function $g(x) = f'(x)f''(x)$ has zeros at $x = 0,x_1,1,x_2,2,x_3,3,x_4,4,x_5,5,x_6,6$.
By using Rolle's theorem again, $g'(x) = f'(x)f'''(x)+f''(x)^2$ has at least $12$ zeros in $[0,6]$.
The function $f(x) = \cos\pi x$ satisfies $f(x) = f(4-x) = f(4+x)$ for all $x \in \mathbb{R}$ and $f'(1) = 0$.
For this function, $f'(x)f'''(x)+f''(x)^2 = (-\pi \sin \pi x)(\pi^3\sin\pi x)+(-\pi^2\cos\pi x)^2$ $= \pi^4(\cos^2\pi x - \sin^2\pi x) = \dfrac{\pi^4}{2}\cos 2\pi x$, which has $12$ zeros on $[0,6]$.
Therefore, the minimum number of zeros of $f'(x)f'''(x)+f''(x)^2$ on $[0,6]$ is $12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4\frac{x}{2}}$ Compute the following limit: $$L=\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4\left(\frac{x}{2}\right)}$$
| $$\begin{align}\sin{\log{(1+x)}} &= \sin{\left (x - \frac12 x^2 + \frac13 x^3 - \cdots \right )} \\ &= \left (x - \frac12 x^2 + \frac13 x^3 - \cdots \right ) - \frac16 \left (x - \frac12 x^2 + \frac13 x^3 - \cdots \right )^3 + \cdots \\ &= x - \frac12 x^2 + \frac16 x^3 + \left ( \frac32 \cdot \frac16 - \frac14 \right ) \frac14 x^4 \cdots\end{align}$$
$$\begin{align}\log{(1+\sin{x})} &= \log{\left (1 + x - \frac16 x^3 + \cdots \right )} \\ &= \left ( x - \frac16 x^3 + \cdots \right ) - \frac12 \left (x - \frac16 x^3 + \cdots \right )^2 + \cdots \\ &= x - \frac12 x^2 + \frac16 x^3 + \left (\frac16-\frac14 \right ) x^4 \cdots\end{align}$$
Thus the limit is equal to
$$\frac{1/12}{1/2^4} = \frac43 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove inequality $ a^4+b^4 \leq 2 a^2 b^2 +1. $ Let $a,b$ be such that $|a|\leq 1$, $|b|\leq 1$. Prove that then
$$
a^4+b^4 \leq 2 a^2 b^2 +1.
$$
| Staring at this, I have an uncontrollable urge to complete the square.
$$a^4 + b^4 \le^{\text{?}} 2a^2 b^2 + 1
$$
$$a^4 - 2a^2 b^2 + b^4 \le^{\text{?}} 1
$$
$$(a^2 - b^2)^2 \le^{\text{?}} 1
$$
Now it's easy. If $|a|, |b| \le 1$, it follows that $0 \le a^2, b^2 \le 1$, and so $|a^2 - b^2| \le 1$ and so $(a^2 - b^2)^2 \le 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1608106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum of $\sum\limits_{cyc} {\sqrt{\frac a{2(b+c)}}}$ with $a,b,c \gt 0$ As said in the title, I have to find the minimum of the following:
$$\sum_{cyc} {\sqrt{\frac a{2(b+c)}}} $$ with $a+b+c>0$
In my very last attempt, I tried to work it out using AM-GM:
Since $$\sqrt{\frac a{2(b+c)}}= \frac 1{\sqrt 2}\frac {a}{\sqrt{a(b+c)}}\ge \frac {\sqrt{2}a}{a+b+c}$$
$$\Rightarrow\sum_{cyc} {\sqrt{\frac a{2(b+c)}}}\ge \frac {\sqrt{2}(a+b+c)}{a+b+c}=\sqrt{2}$$
The problem is, to get the minimum value, $$a=b+c$$
$$b=c+a$$
$$c+a+b$$
or $a=b=c=0$. This is wrong since $a,b,c>0$
And after some calculus I found the minimum value is $\frac 32$ when $a=b=c$.
So what mistake I have made?
| By AM-GM $\sum\limits_{cyc}\sqrt{\frac{a}{2(b+c)}}=\sum\limits_{cyc}\frac{\sqrt2a}{2\sqrt{a(b+c)}}\geq\sum\limits_{cyc}\frac{\sqrt2a}{a+b+c}=\sqrt2$,
which is infimum because we can use $c\rightarrow0^+$ and $b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Second derivative of $x^3+y^3=1$ using implicit differentiation I need to find the $D_x^2y$ of $x^3+y^3=1$ using implicit differentiation
So,
$$
x^3 + y^3 =1 \\
3x^2+3y^2 \cdot D_xy = 0 \\
3y^2 \cdot D_xy= -3x^2 \\
D_xy = - {x^2 \over y^2}
$$
Now I need to find the $D_x^2y$.
I am pretty sure that means the second derivative.
How would I do it to find the second derivative? apparently, it is supposed to be$$-{2x \over y^5}$$
| Throughout this solution, I will use the Leibniz notation $\frac{dy}{dx}$ to represent the derivative of $y$ with respect to $x.$ Notice that the second derivative is notated $\frac{d^{2}y}{dx^{2}}.$ We begin as you did - find the first derivative.
Given: $x^{3} + y^{3} = 1$
$3x^{2} + 3y^{2} \cdot \frac{dy}{dx} = 0$ (*)
$\frac{dy}{dx} = -\frac{x^{2}}{y^{2}}$ (**)
We differentiate the (*) equation with respect to $x.$ This yields the following:
$6x + \frac{dy}{dx} \cdot (6y \cdot \frac{dy}{dx}) + (3y^{2}) \cdot (\frac{d^{2}y}{dx^{2}}) = 0$
$6x + 6y \cdot (\frac{dy}{dx})^{2} + 3y^2 \cdot (\frac{d^{2}y}{dx^{2}}) = 0$
$3y^2 \cdot (\frac{d^{2}y}{dx^{2}}) = -6x - 6y \cdot (\frac{dy}{dx})^{2}$
Substituting the (**) equation for $\frac{dy}{dx}$ here, we find our answer:
$\frac{d^{2}y}{dx^{2}} = \frac{-2x - 2y \cdot (-\frac{x^{2}}{y^{2}})^{2}}{y^{2}}$
$\frac{d^{2}y}{dx^{2}} = \boxed{-\frac{2x}{y^{2}}(1 + \frac{x^{3}}{y^{3}})}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6,\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=8$,then find $\frac{a}{b}+\frac{c}{d}$ $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6,\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=8$,then find $\frac{a}{b}+\frac{c}{d}$
I have tried multiplying the two given equations $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6,\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=8$ but not helpful then i tried adding the two equations $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6,\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=8$ but this is also not helpful to find $\frac{a}{b}+\frac{c}{d}$.
Neither $AM-GM$ inequality method is helpful.
Please suggest me a way to find $\frac{a}{b}+\frac{c}{d}$.
| Hint:
Let $$\alpha=\frac ab+\frac cd,\beta=\frac bc +\frac da$$
Consider
$$\alpha\beta=(\frac ab+\frac cd)(\frac bc +\frac da)=\frac ac+\frac bd+\frac db+\frac ca=8$$
And note that
$$\alpha+\beta=6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the probability of a region inside a pyramid I have a region in a 3-D space with a density of
$$
\
f_{x,y,z}(X,Y,Z) = \begin{cases}
1 & \text{if $ (x,y,z)\in W$}; \\
0 & \text{if $(x,y,z)\notin W$};\\
\end{cases}
\ $$
Being $W$ the set of points inside the pyramid $(0,0,0)$, $(2,0,0)$,$(0,3,0)$ and $(0,0,1)$
Im asked to find $P\{\frac{1}{3}<Z<\frac{2}{3}\}$
For doing so i find first the plane situated on the front part of the pyramid:
$z=1-\frac{y}{3}-\frac{x}{2}$
And i will try to find the marginal density probability of $Z$ to find $P\{\frac{1}{3}<Z<\frac{2}{3}\}$
I know the area of the base of the pyramid by doing $y=0 \rightarrow x=2(1-z) $ and $x=0 \rightarrow y=3(1-z)$. So the area is $\frac{xy}{2}=3(1-z)^2$
How can i manage from here to find $P\{\frac{1}{3}<Z<\frac{2}{3}\}$?
| Really, you've done just about all of the work. The only missing part is the easy one. You know the volume of the pyramid is one, so that the probability is the fraction of the volume between $z \in [1/3,2/3]$, or
$$3 \int_{1/3}^{2/3} dz \, (1-z)^2 = \left (\frac23 \right )^3 - \left (\frac13 \right )^3 = \frac{7}{27}$$
Actually, you don't even need integration. You know that the area of the base at $z=2/3$ is $3 (1/3)^2 = 1/3$ and at $z=1/3$ is $3 (2/3)^2 = 4/3$. The volume is then
$$\frac13 \cdot \frac23 \cdot \frac43 - \frac13 \cdot \frac 13 \cdot \frac 13 = \frac{7}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $x,y,z$ are positive real numbers and $ xy+xz+yz = 1$ then $\sqrt{x}+\sqrt{y}+\sqrt{z} > 2$
Prove that if $x,y,z$ are positive real numbers and $ xy+xz+yz = 1$ then $$\sqrt{x}+\sqrt{y}+\sqrt{z} > 2$$
I am having a hard time relating the square roots in the inequality to the given condition. I was thinking that maybe there is some substitution I could do, but I couldn't think of any.
| Another simple solution.
We show that
$$(\sqrt{x}+\sqrt{y}+\sqrt{z})^4 \ge 16(xy+yz+zx) \quad\forall x,y,z \ge 0$$
or
$$(a+b+c)^4 \ge 16(a^2b^2+b^2c^2+c^2a^2) \quad\forall a,b,c \ge 0.$$
WLOG, assume that $a=\max(a,b,c)$, then we have $$a^2b^2+b^2c^2+c^2a^2 = a^2(b+c)^2 + bc(bc-2a^2) \le a^2(b+c)^2.$$
Thus it reduces to prove $$(a+b+c)^4 \ge 16 a^2(b+c)^2$$
or equivalently
$$(a+b+c)^2 \ge 4 a(b+c),$$
which is true because $$(a+b+c)^2 - 4 a(b+c) = (a-b-c)^2 \ge 0.$$
Equality occurs when $a=b$ and $c=0$ or when $a=c$ and $b=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How can I get a irreducible polynomial of degree 8 over $Z_2[X]$? I have got one of degree 5: $x^5+x^2+1$, but I need one of degree 8.
| The only way I know is trial and error. A polynomial of degree $8$ is irreducible/prime over $\mathbb Z_2$ if it divides:
$$\frac{x^{2^8}-x}{x^{2^4}-x}=\frac{x^{255}-1}{x^{15}-1} = 1+x^{15}+x^{30}+\cdots x^{225}+x^{240}$$
This polynomial is exactly the product of all the prime polynomials of degree $8$, once each.
It's probably easier to use repeated squaring to prove that $p(x)$ divides $x^{256}-x$ and not $x^{16}-x$.
This means there are $30$ such polynomials. Since we want an eighth-degree polynomial, and it is not divisible by $x$ or $x-1=x+1$, this means:
$$p(x)=1+\cdots + x^8$$ where the number of terms in $\cdots$ is odd. So there are $\frac{2^7}{2}=64$ subsets of $\{1,2,\dots,7\}$ that are odd, and therefore, if you chose your $p(x)$ at random under these conditions, you'd get approximately $50\%$ chance of finding a prime each trial.
You can remove the cases where the subset of $\{1,\dots,7\}$ are only even. For example, we know:
$$1+x^4+x^6+x^8 = (1+x^2+x^3+x^4)^2$$
That process removes $4$ of the $64$ bad options, and now we have exactly a $50\%$ chance.
We can also avoid duplicate testing by not testing reflections. Assuming $a_0=a_8=1$, then $\sum_0^8 a_ix^i$ is irreducible if and only if $\sum_{0}^8 a_{8-i}x^i$ is irreducible.
So if you've eliminated $1+x+x^8$, you've eliminated $1+x^7+x^8$, for example.
That doesn't reduce our options quite in half, because some polynomials are self-reflective. (And it doesn't increase our initial odds of finding a prime on the first try.)
This still requires trial and error.
Example trial for $p(x)=1+x+x^8$. Then:
$$x^{8}\equiv 1+x\pmod{p(x)}\\
x^{16}\equiv 1+x^2\pmod{p(x)}$$
So $x^{16}-x$ is not divisible by $p(x)$.
$$x^{32}\equiv 1+x^4\pmod{p(x)}\\
x^{64}\equiv 1+x^8\equiv x\pmod{p(x)}\\
x^{128}\equiv x^2\pmod{p(x)}\\
x^{256}\equiv x^4\pmod{p(x)}$$
We could actually stop at $x^{64}$, since it means that $p(x)$ divides $x^{2^6}-x$, which means it must be the product of primes of degree $1,2,3,$ and $6$. Since it is not divisible by a linear polynomial it must be divisible by the only prime of degree $2$, $q(x)=1+x+x^2$. Since $(1+x+x^2)(1+x)=x^3+1$, we get that:
$$p(x)\equiv 1+x+x^5\equiv 1+x+x^2\pmod{1+x+x^2}$$ Indeed, this gives us another way to ensure that $p(x)$ is not prime: $1+x+x^2$ divides $\sum a_ix^i$ with $\sum a_i=1$ and $a_0=a_8=1$ if $$a_3+a_6=0, a_1+a_4+a_7=1\\a_2+a_5=0$$
We get that $1+x^3+x^8$ is divisible by $1+x+x^3$.
I get that $1+x+x^2+x^4+x^8$ is prime, but I'd want to check this.
Another way to check, by the way, is to ensure that $p(x)$ is relatively prime to $x^{8}-x$ (and thus not divisible by any prime of degree $3$) and relatively prime to $x^{16}-x$, and thus not divisible by any prime of degree $1,2,4$.
I get that $1+x+x^3+x^4+x^8$ is irreducible, but I did it on paper, so I'd want to check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the least value of $4\csc^{2} x+9\sin^{2} x$
Find the least value of $4\csc^{2} x+9\sin^{2} x$
$a.)\ 14 \ \ \ \ \ \ \ \ \ b.)\ 10 \\
c.)\ 11 \ \ \ \ \ \ \ \ \ \color{green}{d.)\ 12} $
$4\csc^{2} x+9\sin^{2} x \\
= \dfrac{4}{\sin^{2} x} +9\sin^{2} x \\
= \dfrac{4+9\sin^{4} x}{\sin^{2} x} \\
= 13 \ \ \ \ \ \ \ \ \ \ \ \ (0\leq \sin^{2} x\leq 1)
$
But that is not in options.
I look for a short and simple way.
I have studied maths upto $12$th grade.
| You can use $\displaystyle4\csc^2 x+9\sin^2x=\big(\frac{2}{\sin x}-3\sin x\big)^2+12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
How to show $\frac{1}{2\sqrt{2} + \sqrt{3}} = \frac{2\sqrt{2} - \sqrt{3}}{5}$? Show that:
$$ \dfrac{1}{2\sqrt2+\sqrt3}=\dfrac{2\sqrt2-\sqrt3}{5}$$
So I multiplied everything by $\sqrt3$
Then I got
$$\frac{\sqrt{3}}{2\sqrt{2}+3}$$
Then multiply it by $\sqrt2$ to obtain
$$\frac{\sqrt{2}\sqrt{3}}{2 \cdot 3+3}$$
Which is $$\frac{\sqrt{2}\sqrt{3}}{9}$$ which isn't equal to $$\frac{2\sqrt{2}-\sqrt{3}}{5}$$
What did I do wrong?
| The conjugate of $\sqrt{a} + \sqrt{b}$ is $\sqrt{a} - \sqrt{b}$. To rationalize an expression of the form $$\frac{1}{\sqrt{a} + \sqrt{b}}$$ we multiply the numerator and denominator by $\sqrt{a} - \sqrt{b}$. Since the conjugate of $2\sqrt{2} + \sqrt{3}$ is $2\sqrt{2} - \sqrt{3}$, we obtain
\begin{align*}
\frac{1}{2\sqrt{2} + \sqrt{3}} & = \frac{1}{2\sqrt{2} + \sqrt{3}} \cdot \frac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} - \sqrt{3}} && \text{multiply by the conjugate}\\
& = \frac{2\sqrt{2} - \sqrt{3}}{(2\sqrt{2})^2 - (\sqrt{3})^2} && \text{since $(a + b)(a - b) = a^2 - b^2$}\\
& = \frac{2\sqrt{2} - \sqrt{3}}{8 - 3}\\
& = \frac{2\sqrt{2} - \sqrt{3}}{5}
\end{align*}
As for your mistakes, you did not apply the distributive law when you multiplied the numerator and denominator by $\sqrt{3}$. You should have obtained
$$\frac{1}{2\sqrt{2} + \sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{2\sqrt{2}\sqrt{3} + \sqrt{3}\sqrt{3}} = \frac{\sqrt{3}}{2\sqrt{6} + 3}$$
You also failed to distribute the factor of $\sqrt{2}$ in the denominator in the next step. Once you obtained $$\frac{\sqrt{3}}{2\sqrt{6} + 3}$$ you could have multiplied the numerator and denominator by the conjugate of the denominator, which is $-2\sqrt{6} + 3$, then simplified.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
What is the coefficient of $x^4$ in the expansion of $\sqrt[3]{1+x}$ Here's what I tried:
$$\sum_{n \ge0} {\frac{1}{3} \choose n} x^n= \sum_{n \ge0} = \frac{\frac{1}{3}!}{n!(n-\frac{1}{3})!}x^n=\sum_{n \ge0} \frac{(\frac{1}{3}-1)(\frac{1}{3}-2)\cdot ...\cdot(\frac{1}{3}-(n-1)) }{n!}x^n$$
What to do more, or is this all wrong?
| The fourth term should be:
$\dfrac{\left(\dfrac{1}{3}\right)\left(-\dfrac{2}{3}\right)\left(-\dfrac{5}{3}\right)\left(-\dfrac{8}{3}\right)}{4!}x^4 $
Which should come out to:
$-\dfrac{10}{243}x^4 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find a set that spans the set of all 3 by 3 semimagic matrices and prove it Call a 3 × 3 matrix a semi-magic square if all its rows and columns (but not necessarily its diagonals) have the same sum.
The questions asks me to find a set that spans the set of all 3 by 3 semimagic matrices and prove it
| In order to answer the question we have to cope with two issues
*
*Semimagic matrices: We need a proper characterisation of these specific matrices.
*Linear span: Once we have a candidate for the set of all semimagic matrices, we have to show this set is a linear span, which means a subspace in the vector space of all $\mathcal{M}_{(3\times 3)}$ matrices with the usual addition and scalar multiplication over a field $\mathbb{F}=\mathbb{R}$ (resp. $\mathbb{C}$).
Semimagic matrices:
Let's assume a semimagic $3\times 3$ matrix $A$ has a column-sum, resp. row-sum equal to $n_A$. Whenever we consider two elements $x,y$ in a column or row of $A$ the third element is determined by
\begin{align*}
n_A-x-y\tag{1}
\end{align*}
It follows the shape of a semimagic matrix $A$ is necessarily
\begin{align*}
A=\begin{pmatrix}
a&b&n_A-a-b\\
c&d&n_A-c-d\\
n_A-a-c&n_A-b-d&a+b+c+d-n_A
\end{pmatrix}
\end{align*}
with $a,b,c,d\in\mathbb{F}$ arbitrarily, fixed. Note, the element $a_{3,3}$ of $A$ is according to (1)
\begin{align*}
a_{3,3}=n_A-(n_A-a-b)-(n_A-c-d)=a+b+c+d-n_A
\end{align*}
We conclude, the set of all semimagic $3\times 3$ matrices $\mathcal{S}$ is
\begin{align*}
\mathcal{S}=\left\{\left.\begin{pmatrix}
a&b&n_A-a-b\\
c&d&n_A-c-d\\
n_A-a-c&n_A-b-d&a+b+c+d-n_A
\end{pmatrix}
\right|a,b,c,d,n_A\in\mathbb{F}
\right\}
\end{align*}
Note that we have five degrees of freedom: $a,b,c,d$ and $n_A$.
Linear span:
The set $\mathcal{S}$ is a perfect candidate for the linear span we are looking for. We see the zero matrix
\begin{align*}
\begin{pmatrix}
0&0&0\\
0&0&0\\
0&0&0\\
\end{pmatrix}\in \mathcal{S}
\end{align*}
which is the zero vector of the vector space $\mathcal{M}_{(3\times 3)}$ is an element of $\mathcal{S}$.
Let $\lambda,\mu\in \mathbb{F}$ and $A,B\in \mathcal{S}$ arbitrarily, fixed. We obtain
\begin{align*}
\lambda A + \mu B &=\lambda
\begin{pmatrix}
a_{1,1}&a_{1,2}&n_A-a_{1,1}-a_{1,2}\\
a_{2,1}&a_{2,2}&n_A-a_{2,1}-a_{2,2}\\
n_A-a_{1,1}-a_{2,1}&n_A-a_{1,2}-a_{2,2}&a_{1,1}+a_{1,2}+a_{2,1}+a_{2,2}-n_A\\
\end{pmatrix}\\
&\qquad\quad+\mu
\begin{pmatrix}
b_{1,1}&b_{1,2}&n_B-b_{1,1}-b_{1,2}\\
b_{2,1}&b_{2,2}&n_B-b_{2,1}-b_{2,2}\\
n_B-b_{1,1}-b_{2,1}&n_B-b_{1,2}-b_{2,2}&b_{1,1}+b_{1,2}+b_{2,1}+b_{2,2}-n_B\\
\end{pmatrix}\\
&=\begin{pmatrix}
c_{1,1}&c_{1,2}&n_C-c_{1,1}-c_{1,2}\\
c_{2,1}&c_{2,2}&n_C-c_{2,1}-c_{2,2}\\
n_C-c_{1,1}-c_{2,1}&n_C-c_{1,2}-c_{2,2}&c_{1,1}+c_{1,2}+c_{2,1}+c_{2,2}-n_C\\
\end{pmatrix}\in\mathcal{S}\\
\end{align*}
with $n_C=\lambda n_A+\mu n_B$ and the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\int_{0}^{1}\frac{x^{4}\arctan x}{\sqrt{1+x^{2}}}\mathrm{d}x$ How to evaluate$$\int_{0}^{1}\frac{x^{4}\arctan x}{\sqrt{1+x^{2}}}\mathrm{d}x$$
I tried to define:$$I\left ( \alpha \right )=\int_{0}^{1}\frac{x^{4}\arctan \left ( \alpha x \right )}{\sqrt{1+x^{2}}}\mathrm{d}x$$
and
$$I'\left ( \alpha \right )=\int_{0}^{1}\frac{x^{5}}{\sqrt{1+x^{2}}\left ( 1+\alpha ^{2}x^{2} \right )}\mathrm{d}x$$
but I have no idea how to do next.
| Starting from $$I'\left ( a \right )=\int_{0}^{1}\frac{x^{5}}{\sqrt{1+x^{2}}\left ( 1+a ^{2}x^{2} \right )}\mathrm{d}x$$ change variable $x=\sqrt t$ and get $$I'\left (a \right )=\int_{0}^{1}\frac{t^2}{2 \sqrt{t+1} \left(a^2 t+1\right)}\,dt$$ Now $$\frac{t^2}{2 \sqrt{t+1} \left(a^2 t+1\right)}=\frac{\sqrt{t+1}}{2 a^2 \left(a^2-1\right) \left(a^2 t+1\right)}+\frac{\sqrt{t+1}}{2
a^2}-\frac{1}{2 \left(a^2-1\right) \sqrt{t+1}}$$ For each piece, the antiderivative can be established (even if the first one is quite tricky but doable). So, $I'(a)$ can be computed but the result is a real monster.
At this point, I give up and, as Leg answered, go to numerical integration.
Edit
For the given initial integral, a CAS found (enjoy !) $$\int_{0}^{1}\frac{x^{4}\arctan x}{\sqrt{1+x^{2}}}\mathrm{d}x=\frac{1}{96} \left(-72 C+4 \left(9 i \text{Li}_2\left(-(-1)^{3/4}\right)-9 i
\text{Li}_2\left((-1)^{3/4}\right)-13+11 \sqrt{2}\right)-3 \pi \left(\sqrt{2}+6
\tanh ^{-1}\left((-1)^{3/4}\right)\right)\right)$$
Edit
This was done after Start wearing purple's elegant answer
If we look at the antiderivative, without change of variable, integrating by parts gives $$I=\int\frac{x^{4}\arctan x}{\sqrt{1+x^{2}}}\, dx=\frac{1}{8} \tan ^{-1}(x) \left(x \sqrt{x^2+1} \left(2 x^2-3\right)+3 \sinh
^{-1}(x)\right)- J$$ using $$J=\int \frac{x \sqrt{x^2+1} \left(2 x^2-3\right)+3 \sinh ^{-1}(x)}{8 \left(x^2+1\right)}\,dx$$ Looking at the integrand, it can rewrite $$\frac{x \sqrt{x^2+1} \left(2 x^2-3\right)+3 \sinh ^{-1}(x)}{8 \left(x^2+1\right)}=\frac{1}{4} x\sqrt{x^2+1} -\frac{5 x}{8 \sqrt{x^2+1}}+\frac{3 \sinh ^{-1}(x)}{8
\left(x^2+1\right)}$$ The first and second antiderivatives are quite simple $$\int\frac{1}{4} x\sqrt{x^2+1}\,dx=\frac{1}{12} \left(x^2+1\right)^{3/2}$$ $$\int \frac{5 x}{8 \sqrt{x^2+1}}\,dx=\frac{5 \sqrt{x^2+1}}{8}$$ and we are left with the problem of $$\int\frac{3 \sinh ^{-1}(x)}{8
\left(x^2+1\right)}\,dx$$ which creates the monster.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How do I simplify and evaluate the limit of $(\sqrt x - 1)/(\sqrt[3] x - 1)$ as $x\to 1$? Consider this limit:
$$ \lim_{x \to 1} \frac{\sqrt x - 1}{ \sqrt[3] x - 1}
$$
The answer is given to be 2 in the textbook.
Our math professor skipped this question telling us it is not in our syllabus, but how can it be solved?
| Note the following identities,
$$ y^2-1 = (y-1)(y+1) $$
$$ y^3-1 = (y-1)(y^2+y+1) $$
we can use these to rewrite the numerator and the denomiantor by substituting $\sqrt{x}$ and $\sqrt[3]{x}$ for $y$ respectively,
$$ x - 1 = (\sqrt{x}-1)(\sqrt{x}+1) \Rightarrow \sqrt{x}-1 = \frac{x-1}{\sqrt{x}+1} $$
$$ x - 1 = (\sqrt[3]{x}-1)(\sqrt[3]{x}^2+\sqrt[3]{x}+1) \Rightarrow \sqrt[3]{x}-1 = \frac{x-1}{\sqrt[3]{x}^2+\sqrt[3]{x}+1} $$
substituting this into the ratio you're taking the limit of we get,
$$ \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{\sqrt[3]{x}-1} = \lim_{x\rightarrow 1} \frac{x-1}{\sqrt{x}+1} \frac{\sqrt[3]{x}^2+\sqrt[3]{x}+1} {x-1} = \frac32 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 0
} |
Calculate this integral $\int _{\frac{1}{4}}^2\:\frac{\ln\left(2x\right)}{x\ln\left(4x\right)}dx$ I have to calculate this integral .
$$\int _{\frac{1}{4}}^2\:\frac{\ln\left(2x\right)}{x\ln\left(4x\right)}\,dx$$
I have no idea how to start , help someone ?
Thanks.
| $$\int _{ \frac { 1 }{ 4 } }^{ 2 } \: \frac { ln\left( 2x \right) }{ xln\left( 4x \right) } dx=\int _{ \frac { 1 }{ 4 } }^{ 2 }{ \frac { \ln { 2 } +\ln { \left( x \right) } }{ \ln { 4 } +\ln { \left( x \right) } } d\left( \ln { \left( x \right) } \right) } =\int _{ \frac { 1 }{ 4 } }^{ 2 }{ \frac { \ln { 2 } +\ln { \left( x \right) } }{ 2\ln { 2 } +\ln { \left( x \right) } } d\left( \ln { \left( x \right) } \right) } \\ \ln { \left( x \right) =t } \\ \int _{ \ln { 1/4 } }^{ \ln { 2 } }{ \frac { \ln { 2 } +t }{ 2\ln { 2 } +t } dt } =\int _{ \ln { 1/4 } }^{ \ln { 2 } }{ \frac { 2\ln { 2 } +t-\ln { 2 } }{ 2\ln { 2 } +t } dt } =\int _{ \ln { 1/4 } }^{ \ln { 2 } }{ \left( 1-\frac { \ln { 2 } }{ 2\ln { 2 } +t } \right) dt } =\\ =\int _{ \ln { 1/4 } }^{ \ln { 2 } }{ dt } -\ln { 2 } \int _{ \ln { 1/4 } }^{ \ln { 2 } }{ \frac { d\left( 2\ln { 2+t } \right) }{ 2\ln { 2 } +t } = } $$
$$=\ln { 2-\ln { \frac { 1 }{ 4 } } - } \ln { 2 } \left[ \ln { \left| 2\ln { 2+\ln { 2 } } \right| } -\ln { \left| 2\ln { 2+\ln { \frac { 1 }{ 4 } } } \right| } \right] \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
How to evaluate $\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+sinx}-3}$? $$\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+\sin x}-3}$$
I have tried with Taylor:
$$\lim _{x\to \:0^+}\frac{\left(1+x+\frac{x^2}{2}+1+2x+2x^2\right)^2-4}{\sqrt{9+x-\frac{x^3}{6}+\frac{x^5}{120}}-3}=\lim _{x\to \:0^+}\frac{\frac{25x^4}{4}+15x^3+19x^2+12x}{\sqrt{9+x-\frac{x^3}{6}+\frac{x^5}{120}}-3}$$
But now I dont know how to move forward.
I tried with Hopital and it was found to be so:
$$\lim _{x\to \:0+}\left(\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+\sin \left(x\right)}-3}\right)=\lim _{x\to \:0+}\left(\frac{2e^{2x}\left(e^x+1\right)\left(2e^x+1\right)}{\frac{\cos \left(x\right)}{2\sqrt{\sin \left(x\right)+9}}}\right)$$
$$=\lim _{x\to \:0+}\left(\frac{4e^{2x}\left(e^x+1\right)\left(2e^x+1\right)\sqrt{\sin \left(x\right)+9}}{\cos \left(x\right)}\right)=\frac{4e^{2\cdot \:0}\left(e^0+1\right)\left(2e^0+1\right)\sqrt{\sin \left(0\right)+9}}{\cos \left(0\right)}=\color{red}{72}$$
But I wanted to know if you could get the same result even with Taylor? Thanks
| To simplify, let us first factor $(e^x+e^{2x})^2-4=(e^x+e^{2x}-2)(e^x+e^{2x}+2)$ and get rid of the square root with $(\sqrt{9+\sin x}-3)(\sqrt{9+\sin x}+3)=\sin x$.
Then
$$\frac{e^x+e^{2x}-2}{\sin x}=\frac{1+x+\frac12x^2\cdots+1+2x+2x^2\cdots-2}{x-\frac16x^3\cdots}=\frac{3x+\frac52x^2\cdots}{x-\frac16x^3\cdots}$$tends to $3$.
The final answer is
$$(1+1+2)(\sqrt9+3)\,3=72.$$
The fully painful way, with development up to second order yields
$$\frac{(1+x+\frac12x^2\cdots+1+2x+2x^2\cdots)^2-4}{\sqrt{9+x-\frac16x^3\cdots}-3}\\
=\frac{4+12x+19x^2\cdots-4}{\sqrt9+\frac1{2\sqrt9}(x-\frac16x^3\cdots)-\frac1{2\cdot2\cdot9\sqrt9\cdot2}(x-\frac16x^3\cdots)^2\cdots-3}=\frac{12x+19x^2\cdots}{\frac16x-\frac1{216}x^2\cdots}$$ which tends to $12\cdot6=72$.
[We used $\sqrt{a+x}=\sqrt a+\frac1{2\sqrt a}x-\frac1{2\cdot2\cdot a\sqrt a\cdot2}x^2\cdots$]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}$ converges. Show that
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
converges.
I recognized that that since the integrand is even then
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}} = 2\int_{0}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
Then, I started with the following statement and manipulated it to match the integrand.
$$
\begin{align*}
x^4 +1 &\gt x^4 \\ \sqrt{x^4+1} &\gt x^2 \\ \frac{1}{\sqrt{x^4 +1}} &\lt\frac{1}{x^2}
\end{align*}
$$
It is a well-known result that
$$
\int_1^{\infty}\frac{dx}{x^2}
$$
converges. So would it be correct to say that, by comparison, that
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
converges as well?
| $$
\begin{align}
\int_{-\infty}^\infty\frac{\mathrm{d}x}{\sqrt{x^4+1}}
&=2\int_0^\infty\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag1\\
&=2\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}+2\int_1^\infty\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag2\\
&=2\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}+2\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag3\\
&=4\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag4
\end{align}
$$
Explanation:
$(1)$: symmetry
$(2)$: split the integral into two parts
$(3)$: substitute $x\mapsto\frac1x$
$(4)$: distribute
Because the integrand is between $\frac1{\sqrt2}$ and $1$, the integral converges and is between $2\sqrt2$ and $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Find $\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}\right)$ I don't know how to find the sum of $\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}$. After rationalization we have
$\left(\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+...+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right){/}\left((\sqrt{2}+\sqrt{4})(\sqrt{4}+\sqrt{6})...(\sqrt{2n}+\sqrt{2n+2})\right)=$
$(\sqrt{4}+\sqrt{6})...(\sqrt{2n}+\sqrt{2n+2})+(\sqrt{2}+\sqrt{4})...(\sqrt{2n}+\sqrt{2n+2})+...+(\sqrt{2}+\sqrt{4})...(\sqrt{2n-2}+\sqrt{2n})$
How to reduce this sum to general form?
| $$\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}\right)$$
$$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k+2}+\sqrt{2k}}\right)$$
$$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{\sqrt{2k+2}-\sqrt{2k}}{(\sqrt{2k+2}+\sqrt{2k})(\sqrt{2k+2}-\sqrt{2k})}\right)$$
$$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{\sqrt{2k+2}-\sqrt{2k}}{2k+2-2k}\right)$$
$$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{\sqrt{2k+2}-\sqrt{2k}}{2}\right)$$
$$=\frac{1}{\sqrt{2}} \lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})\right)$$
$$=\frac{1}{\sqrt{2}} \lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\cdot (\sqrt{n+1}-1)\right)$$
$$=\frac{1}{\sqrt{2}} \lim\limits_{n\to\infty}\left( \sqrt{1+\frac{1}{n}}-\frac{1}{\sqrt{n}}\right)$$
$$=\frac{1}{\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $\sum_{n =1}^\infty \frac 1 {5^{n+1}-5^n+1}$ $$\sum_{n = 1}^\infty \dfrac 1 {5^{n+1}-5^n+1}$$
I can factorize denominator to $4\times5^n+1$ to confirm the series does not diverge,
But how do I calculate its actual sum?
The series is not a telescoping series nor I can partial factorise.
I get confused due to $+1$ in the denominator.
Thanks a lot
| The series doesn't have a closed form (except for a very complicated one involving Q-Polygamma function, as was said in a comment), however, we can transform it to get much better convergence.
$$\frac{1}{4 \cdot 5^n+1}=\frac{1}{4 \cdot 5^n} \left(1-\frac{1}{4 \cdot 5^n}+\frac{1}{4^2 \cdot 5^{2n}}-\frac{1}{4^3 \cdot 5^{3n}}+\cdots \right)$$
$$\sum_{n=1}^{\infty} \frac{1}{4^k \cdot 5^{k n}}=\frac{1}{4^k} \left( \dfrac{1}{1-\dfrac{1}{5^{k}}}-1 \right)=\frac{1}{4^k (5^k-1)}$$
$$\sum_{n=1}^{\infty} \frac{1}{4 \cdot 5^n+1}=\sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{4^k (5^k-1)}=0.06001587909991328$$
Why is this series better? Since it is alternating, it provides upper and lower bounds, unlike the first series, which converges monotonely from below.
It also just gives much better approximations, both in terms of their numerical value and the size of their denominators.
Let's denote:
$$A_N=\sum_{n=1}^{N} \frac{1}{4 \cdot 5^n+1}$$
$$B_N=\sum_{k=1}^{N} (-1)^{k+1} \frac{1}{4^k (5^k-1)}$$
Now compare:
$$A_2=\frac{122}{2121}=0.0575200$$
$$A_3=\frac{7027}{118069}=0.0595161$$
$$B_2=\frac{23}{384}=0.0598958$$
$$B_3=\frac{1429}{23808}=0.0600218$$
$$0.0598958<\sum_{n=1}^{\infty} \frac{1}{4 \cdot 5^n+1}<0.0600218$$
Here is the plot of both $A_N$ and $B_N$ up to $N=10$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\lim\limits_{x\to0} \frac{1}{\sin x} \cdot \ln \left(\frac{e^x -1}{x}\right)$ I was trying to calculate the limit of the following function:
$$ \lim_{x\to0} \frac{1}{\sin x} \cdot \ln \left(\frac{e^x -1}{x}\right) $$
My first thought was using L'Hopital's rule since $\Large \frac{e^x -1}{x}$ goes to 1 so the whole $\ln$ goes to 0.
But then I get another complicated expression, and finally I end up using L'Hopital's rule at least 5 times before getting an actual result.
Is there a wiser way for dealing this limit? (I mean, without using this rule?)
Thanks.
| Use Taylor expansion: $e^x-1=x+x^2/2+o(x^2)$. Thus
\begin{align*}
\frac{1}{\sin x}\log\left(\frac{e^x-1}{x}\right)
&=\frac{x}{\sin x}\frac1x\log\left(1+\frac{x^2/2+o(x^2)}{x}\right)\\
&=\frac{x}{\sin x}\frac1x\frac{\log\left(1+\frac{x^2/2+o(x^2)}{x}\right)}{(x^2/2+o(x^2))/x}\frac{x^2/2+o(x^2)}{x}\\
&=\underbrace{\frac{x}{\sin x}\frac{\log\left(1+\frac{x^2/2+o(x^2)}{x}\right)}{(x^2/2+o(x^2))/x}}_{\to1}\underbrace{\frac{x^2/2+o(x^2)}{x^2}}_{\to1/2}\\
\end{align*}
Thus your limit goes to $1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
$ x \ge 0\text{ and } y \ge 0 \implies \frac{x+y}{2} \ge \sqrt{xy} $ The above applies $\forall x,y \in \mathbb{R}$
I've tried: $x + y \ge 0$
$$x + y \ge x$$
$$ (x + y)^2 \ge 2xy$$
$$\frac{(x + y)^2}{2} \ge xy$$
But the closest I get is $\dfrac{x+y}{\sqrt{2}} \ge \sqrt{xy}$
Any ideas?
| $$(x-y)^2 \ge 0$$
$$x^2 - 2xy + y^2 \ge 0 $$
$$x^2 + y^2 \ge 2xy $$
$$x^2 + 2xy + y^2 \ge 4xy $$
$$(x+y)^2 \ge 4xy $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to simplify $\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}$ The answer is 2. But I want to learn how to simplify this expression without the use of calculator.
| *
*$7-5\sqrt 2 = 1-3\sqrt{2} + 6 - 2 \sqrt{2} = 1^3+\dbinom{3}{1}(-\sqrt{2})+\dbinom{3}{2}*(-\sqrt{2})^2+(-\sqrt{2})^3=(1-\sqrt{2})^3$ (By binomial theorem:https://en.wikipedia.org/wiki/Binomial_theorem)
*$7+5\sqrt 2 = 1+3\sqrt{2} + 6 + 2 \sqrt{2} = (1+\sqrt{2})^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
how to prove this inequality $(ab+bc+ac)^2 ≥ 3abc(a+b+c)$
Prove that if $a,b,c$ are non-negative real numbers, then
$(ab + bc + ca)^2 \geq 3abc(a+b+c)$.
I tried to compute from $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$.
|
$$a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2\ge3a^2bc+3ab^2c+3abc^2$$
$$a^2b^2+a^2c^2+b^2c^2-a^2bc-ab^2c-abc^2\ge0$$
$$2a^2b^2+2a^2c^2+2b^2c^2-2a^2bc-2ab^2c-2abc^2\ge0$$
$$(a^2b^2-2a^2bc+a^2c^2)+(a^2c^2-2abc^2+b^2c^2)+(b^2c^2-2ab^2c+a^2b^2)\ge0$$
$$(ab-ac)^2+(ac-bc)^2+(ab-bc)^2\ge0$$
the upper operation is revertable
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix Show if $A^TA = I$ and $\det A = 1$ where
$ A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$, then $A =\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}$.
attempt:
Suppose $ A^TA =\begin{bmatrix}
a & c \\
b & d
\end{bmatrix}
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$ = $\begin{bmatrix}
a^2 + c^2 & ab + cd \\
ab + cd & b^2 + d^2
\end{bmatrix}$ = $\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}$
. Then $a^2 + c^2 = 1$ implies $a = \cos\theta$, and $c = \sin\theta$ or $c = - \sin\theta$ using the unit circle .
Similarly $ d = \cos\theta$, and $b = \sin\theta$ or $b = -\sin\theta$.
So know I am stuck in showing how $b = -\sin\theta$ has to be chosen and $c = \sin\theta$.
Can someone please help? Thank you!
| You have that there is some $\theta$ for which you can write
$$a = \cos\theta \qquad\text{and}\qquad c = \sin\theta$$
Now, note that the vanishing elements of the matrix product give the equation
$$a b + c d = 0 \tag{1}$$
and that the assumed relation $\det A = 1$ gives
$$a d - b c = 1 \tag{2}$$
Simply solve $(1)$ and $(2)$ for $b$ and $d$ in terms of $a$ and $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding a linear transformation with a given null space The problem statement is,
Find a linear transformation $T: \mathbb R^3 \to \mathbb R^3$ such that the set of all vectors satisfying $4x_1-3x_2+x_3=0$ is the (i) null space of $T$ (ii) range of $T$.
For (i),
I found out the basis of the null space for the system,
$$\begin{pmatrix}4&-3&1\end{pmatrix} \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =0$$
viz, $\mathbf v_1 = \begin{pmatrix}3\\4\\0\end{pmatrix} , \mathbf v_2 = \begin{pmatrix}-1\\0\\4\end{pmatrix}$
So, basically, I have to find linear transformation such that $T\begin{pmatrix}3\\4\\0\end{pmatrix} =0$ and $T\begin{pmatrix}-1\\0\\4\end{pmatrix}=0$ such that vector $\mathbf v \in span\{\mathbf v_1, \mathbf v_2\}$ satisfies $T\left(\mathbf v\right) =0$
Now, I'm stuck at this point. I'm not able to describe the such linear transformations.
Further, for (ii), I don't know how to approach the problem.
| For (i), consider the matrix
$$ A = \begin{pmatrix} 4 & -3 & 1 \\ 4 & -3 & 1 \\ 4 & -3 & 1 \end{pmatrix}. $$
Denoting by $T_A \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ the corresponding linear map $T_A(v) = Av$, we have
$$ T_A \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 4x_1 - 3x_2 + x_3 \\ 4x_1 - 3x_2 + x_3 \\ 4x_1 - 3x_2 + x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$
so $\ker(T_A)$ is precisely the solution subspace of the equation $4x_1 - 3x_2 + x_3$.
For (ii), consider the matrix
$$ B = \begin{pmatrix} 3 & -1 & 0 \\ 4 & 0 & 0 \\ 0 & 4 & 0 \end{pmatrix} $$
whose columns are $\bf{v_1},v_2$ and the zero vector. The image of $T_B$ is spanned by the columns of $B$ and thus again is equal to the solution subspace of your equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I finish this trig integral $\int_0^{\pi/4}\frac{\sin^2 \theta}{\cos \theta}d\theta$? I got up to the part where it's $$\frac{9}{125}\int_0^{\large \frac{\pi}{4}}\frac{\sin^2\theta}{\cos\theta}\,\,d\theta$$
but I can't figure out how to finish it off.
By the way the original problem was:
$$\int_0^{0.6}\frac{x^2}{\sqrt{9-25x^2}}\,\,dx$$
| Your substitution is a bit off. Begin with
$$\begin{align*}
\int_0^{0.6}{\frac{x^2}{\sqrt{9-25x^2}}}\, dx
&= \int_0^{0.6}{\frac{x^2}{\sqrt{25\left(\frac{9}{25}-x^2\right)}}}\, dx \\
&= \frac{1}{5}\int_0^{0.6}{\frac{x^2}{\sqrt{\frac{9}{25}-x^2}}}\, dx.
\end{align*}
$$
Now let
$$\begin{align*}
x&=\frac{3}{5}\sin \theta \\
dx&=\frac{3}{5}\cos\theta \, d \theta. \\
x^2&=\frac{9}{25}\sin^2\theta, \\
\frac{9}{25}-x^2&=\frac{9}{25}-\frac{9}{25}\sin^2\theta\\
&=\frac{9}{25}\cos^2\theta.
\end{align*}
$$
To change limits, note that when $x=0$, $\frac{3}{5}\sin \theta=0\Rightarrow \theta=0$, but when $x=0.6=\frac{3}{5}$, $\frac{3}{5}\sin \theta=\frac{3}{5}\Rightarrow \sin\theta=1 \Rightarrow \theta=\frac{\pi}{2}$.
So your integral is now
$$\frac{1}{5}\int_0^{\frac{\pi}{2}}\frac{\frac{9}{25}\sin^2\theta\frac{3}{5}\cos \theta}{\sqrt{\frac{9}{25}\cos^2\theta}}\, d \theta.$$
I'll let you pick it up from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Can an infinite sum of irrational numbers be rational? Let $S = \sum_ {k=1}^\infty a_k $ where each $a_k$ is positive and irrational.
Is it possible for $S$ to be rational, considering the additional restriction that none of the $a_k$'s is a linear combination of the other ?
By linear combination, we mean there exists some rational numbers $u,v$ such that $a_i = ua_j + v$.
| Consider that $$\frac{1}{1-x} = 1 + x + x^{2} + x^{3} + \dots \text{ (for } -1 < x < 1).$$
Then since $\frac{\pi}{5}$ is in $(-1,1)$, we have:
$$\underbrace{\frac{1}{1 - \frac{\pi}{5}}}_{\dfrac{5}{5-\pi}} = 1 + \underbrace{\frac{\pi}{5} + \frac{\pi^{2}}{5^{2}} + \frac{\pi^{3}}{5^{3}} + \dots}_{\text{infinite sum of irrational numbers}}$$
Subtracting $1$ from both sides of the above equation gives:
$$\underbrace{\dfrac{5}{5-\pi} - 1}_{\dfrac{\pi}{5-\pi}} = \frac{\pi}{5} + \frac{\pi^{2}}{5^{2}} + \frac{\pi^{3}}{5^{3}} + \dots$$
Now the right hand side is an infinite sum of terms with each term an irrational number. Unfortunately, the left hand side is also irrational. Let's multiply both sides of the above equation by the reciprocal of the left hand side to induce rationality of that side, i.e., multiply both sides by $\frac{5-\pi}{\pi}$. This gives:
$$\frac{5-\pi}{\pi}\cdot \dfrac{\pi}{5-\pi}= \frac{5-\pi}{\pi} \cdot \left [\frac{\pi}{5} + \frac{\pi^{2}}{5^{2}} + \frac{\pi^{3}}{5^{3}} + \dots \right ]$$
Simplifying gives:
$$\underbrace{1}_{\text{rational number}} = \underbrace{\frac{5-\pi}{5} + \frac{(5-\pi)\pi}{5^{2}} + \frac{(5-\pi)\pi^{2}}{5^{3}} + \dots}_{\text{infinite sum of irrational terms}} $$
EDIT: Further justification/verification that the sum in the right hand side of the equation above equals $1$: Recognize it as a geometric series! The factor we multiply at each step is $r:=\frac{\pi}{5}$. Then we know the series converges to $\frac{\text{first term}}{1 - r} = \dfrac{\frac{5-\pi}{5}}{1 - \frac{\pi}{5}} = \dfrac{\frac{5-\pi}{5}}{\frac{5-\pi}{5}} = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "113",
"answer_count": 23,
"answer_id": 12
} |
Evaluation of $\lim_{n\rightarrow \infty}\sum_{k=1}^n\sin \left(\frac{n}{n^2+k^2}\right)$
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\sin \left(\frac{n}{n^2+1}\right)+\sin \left(\frac{n}{n^2+2^2}\right)+\cdots+\sin \left(\frac{n}{n^2+n^2}\right)$
$\bf{My Try::}$ We can write the Sum as $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\sin\left(\frac{n}{n^2+r^2}\right)$$
Now how can I convert into Riemann Sum, Help me
Thanks
| For $x\gt0$, repeatedly integrating from $0$ to $x$ gives
$$
\cos(x)\le1\implies\sin(x)\le x\implies1-\cos(x)\le\frac{x^2}2\implies x-\sin(x)\le\frac{x^3}6
$$
Noting that both sides are odd, we get
$$
\left|x-\sin(x)\right|\le\frac{\left|x^3\right|}6
$$
Since $\frac{n}{n^2+k^2}\le\frac1n$,
$$
\begin{align}
\sum_{k=1}^n\sin\left(\frac{n}{n^2+k^2}\!\right)
&=\sum_{k=1}^n\frac{n}{n^2+k^2}-\sum_{k=1}^n\left[\frac{n}{n^2+k^2}-\sin\left(\frac{n}{n^2+k^2}\!\right)\right]\\
&=\sum_{k=1}^n\frac1{1+\left(\frac kn\right)^2}\frac1n-\sum_{k=1}^nO\!\left(\frac1{n^3}\right)\\
&\to\int_0^1\frac1{1+x^2}\,\mathrm{d}x-0\\[6pt]
&=\frac\pi4
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$ If
\begin{equation}
\sin(x) + \cos(x) = \frac{7}{5},
\end{equation}
then what's the value of
\begin{equation}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?}
\end{equation}
Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) without using the identities of trigonometry.
The function $\sin x+\cos x$ could be transformed using some trigonometric identities to a single function. In fact, WolframAlpha says it is equal to $\sqrt2\sin\left(x+\frac\pi4\right)$ and there also are some posts on this site about this equality. So probably in this way we could calculate $x$ from the first equation - and once we know $\sin x$ and $\cos x$, we can calculate $\dfrac{1}{\sin x}+\dfrac{1}{\cos x}$. Is there a simpler solution (perhaps avoiding explicitly finding $x$)?
| $$\sin x+\cos x=\frac{7}{5}$$
Let $\sin x=t$
$$\implies t+\sqrt{1-t^2}=\frac{7}{5}$$
Shifting, squaring and simplifying, we get
$$25t^2-35t+12=0$$
$$\implies t=\frac{35 \pm 5}{50}$$
Hence, $$\sin x= \frac{4}{5},\ \cos x=\frac{3}{5} \ \text{or} \ \cos x= \frac{4}{5}, \ \sin x=\frac{3}{5}$$
But as we need to find $$\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$$ it becomes
$$\frac{5}{4}+\frac{5}{3}=\frac{35}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 10,
"answer_id": 5
} |
Probability of a random Permutation Pick up a random permutation in S5(assuming all elements have the equal chance to be picked). Find the probability that the sum of the first three entries of σ is less than or equal to sum of last two.
My try: I mean there will be 5! different combination possible, do I have to look at each of it?
| Hint: the permutations you want are just :
$ (...,5,4)$, $(...,5,3)$, $(...,4,5)$, $(...,3,5)$, $(...,2,5)$ and $(...,2,5)$ and you don't care about the order of the other elements where I put the dots.
So the probability is:
$$\frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.