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Computing $\int \frac{1}{(1+x^3)^3}dx$ I tried various methods like trying to break it into partial fractions after factorizing, applying substitutions but couldn't think of any. How will we integrate this?
|
HINT:
$$\int\dfrac{dx}{x^p(1+x^m)^n}=\int\dfrac1{mx^{p+m-1}}\cdot\dfrac{mx^{m-1}}{(1+x^m)^n}dx$$
$$=\dfrac1{mx^{p+m-1}}\int\dfrac{mx^{m-1}}{(1+x^m)^n}dx-\int\left(\dfrac{d\left(\dfrac1{mx^{p+m-1}}\right)}{dx}\cdot\int\dfrac{mx^{m-1}}{(1+x^m)^n}dx\right)dx$$
$$\implies\int\dfrac{dx}{x^p(1+x^m)^n}=-\dfrac1{m(n-1)x^{m+p-1}}\cdot\dfrac1{(1+x^m)^{n-1}}-\dfrac{p+m-1}{m(n-1)}\int\dfrac{dx}{x^{p+m}(1+x^m)^{n-1}}$$
Here start with $p=0,m=3,n=3$
Then $p=0+3,m=3,n=3-1=2$
Finally use, $$\dfrac1{x^6(1+x^3)}=\dfrac{1-x^6+x^6}{x^6(1+x^3)}=\dfrac{1-x^3}{x^6}+\dfrac1{1+x^3}$$
|
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"url": "https://math.stackexchange.com/questions/1884093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Why is my solution of $e^{\sin x } - e^{- \sin x} - 4 = 0$ wrong? $$e^{\sin x } - e^{- \sin x} - 4 = 0$$
Substitute $e^{\sin x} = y$:
$$y - \frac{1}{y} - 4 = 0 \implies y^2 - 4y - 1 = 0$$
Solve for $y$:
$$y = 2 \pm \sqrt{5}$$
$e^{\sin x}$ can't be negative:
$$\therefore y = 2 + \sqrt{5} \implies e^{\sin x} = 2 + \sqrt{5}$$
Differentiating both sides with respect to $x$:
$$\frac{d e^{\sin x}}{d x} = \frac{d (2 + \sqrt{5})}{d x} \implies \frac{d e^{\sin x}}{d \sin x} \times \frac{d \sin x}{d x} = 0 \implies e^{\sin x} \times \cos x = 0$$
If $e^{\sin x} = 0$ then $x$ has no solution. However, when $\cos x = 0$ then $x = 90^{\circ}$
The actual answer of the question states that there is no real solution of the equation. Then why am I getting the solution as $90^{\circ}$ too?
|
$e^{\sin x} = 2 + \sqrt{5}$ has no solutions!
In fact it is equivalent to $\sin(x)=\ln(2 + \sqrt{5})>\ln(4)>1$ which is impossible because $\sin(x)\in[-1,1]$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\sum_{k=1}^n\frac{(b_1+b_2+\cdots+b_k)b_k}{a_1+a_2+\cdots+a_k}<2\sum_{i=1}^n\frac{b_i^2}{a_i}$ How do I prove this?
$$\sum_{k=1}^n\frac{(b_1+b_2+\cdots+b_k)b_k}{a_1+a_2+\cdots+a_k}<2\sum_{i=1}^n\frac{b_i^2}{a_i}$$
Here $a_i,b_i\in\Bbb R^+$. I guess the sum transform works, but I can't prove it.
|
Here is An answer by toshihiro shimizu on AoPS:
I considered this problem long time and finally I found the following inductive solution. I'd like to know the direct solution.
We show the following stronger result by induction;
$$\begin{align*}
\sum_{k=1}^{n} & \frac{\left(b_{1}+b_{2}+\cdots+b_{k}\right)b_{k}}{a_{1}+a_{2}+\cdots+a_{k}}\leq\frac{3}{2}\frac{b_{1}^{2}}{a_{1}}+2\sum_{k=2}^{n}\frac{b_{k}^{2}}{a_{k}}.
\end{align*}$$
$n=1$ is obvious. We assume the result of $n=1$ and we show the
result of $n$. Applying the inductive result to $\left(a_{1}+a_{2},a_{3},a_{4}\ldots,a_{n},b_{1}+b_{2},b_{3},b_{4},\ldots,b_{n}\right)$
we have
$$\begin{align*}
\frac{\left(b_{1}+b_{2}\right)^{2}}{a_{1}+a_{2}} & +\sum_{k=3}^{n}\frac{\left(b_{1}+\cdots+b_{k}\right)b_{k}}{a_{1}+\cdots+a_{k}}\leq\frac{3}{2}\frac{\left(b_{1}+b_{2}\right)^{2}}{a_{1}+a_{2}}+2\sum_{k=3}^{n}\frac{b_{k}^{2}}{a_{k}}
\end{align*}$$
It's sufficient to show that
$$\begin{align*}
\frac{b_{1}^{2}}{a_{1}}+\frac{\left(b_{1}+b_{2}\right)b_{2}}{a_{1}+a_{2}}+\frac{1}{2}\frac{\left(b_{1}+b_{2}\right)^{2}}{a_{1}+a_{2}} & \leq\frac{3}{2}\frac{b_{1}^{2}}{a_{1}}+2\frac{b_{2}^{2}}{a_{2}}\\
\frac{b_{1}^{2}+4b_{1}b_{2}+3b_{2}^{2}}{a_{1}+a_{2}} & \leq\frac{b_{1}^{2}}{a_{1}}+4\frac{b_{2}^{2}}{a_{2}}\\
b_{1}^{2}+4b_{1}b_{2}+3b_{2}^{2} & \leq b_{1}^{2}+\frac{a_{2}}{a_{1}}b_{1}^{2}+4\frac{a_{1}}{a_{2}}b_{2}^{2}+4b_{2}^{2}\\
4b_{1}b_{2} & \leq\frac{a_{2}}{a_{1}}b_{1}^{2}+4\frac{a_{1}}{a_{2}}b_{2}^{2}+b_{2}^{2}
\end{align*}$$
The last inequality is true since
$$\begin{align*}
\frac{a_{2}}{a_{1}}b_{1}^{2}+4\frac{a_{1}}{a_{2}}b_{2}^{2} & \geq2\sqrt{\frac{a_{2}}{a_{1}}b_{1}^{2}\cdot4\frac{a_{1}}{a_{2}}b_{2}^{2}}\\
& =4b_{1}b_{2}
\end{align*}$$
, completing the proof.
|
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|
Better way to evaluate $\int \frac{dx}{\left (a +b\cos x \right)^2}$ $$\int \frac{dx}{\left (a +b\cos x \right)^2}$$
$$u=\frac{b +a \cos x}{a +b\cos x }$$
$$du=\frac{\sin x\left(b^2 -a^2\right)}{ \left (a +b.\cos x \right)^2}$$
$$\frac{du}{\sin x\left(b^2 -a^2\right)}=\frac{dx}{\left (a +b\cos x \right)^2}$$
$$\cos x=\frac{au -b}{a - bu} $$
$$\sin x=\sqrt{1-\left(\frac{au -b}{a - bu}\right)^2}$$
It is becoming messy with this .I also tried it using half angle formula but didn't find it good.
|
Hint:
Rationalize with $z=e^{ix}$ (and $dz=iz\,dx$),
$$\int \frac{dx}{\left (r+\cos x \right)^2}
=\int\frac{dz}{iz\left(r+\dfrac{z+z^{-1}}2\right)^2}
=\int\frac{4z\,dz}{i\left(z^2+2rz+1\right)^2}
=\int\frac{4(z+r-r)\,dz}{i\left((z+r)^2+1-r^2\right)^2}.
$$
This yields a term $\log({z^2+2rz+1})=\log(r+\cos x)$ and, depending on the sign of $1-r^2$, an $\arctan$ or $\text{artanh}$.
|
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"url": "https://math.stackexchange.com/questions/1887194",
"timestamp": "2023-03-29T00:00:00",
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|
Find Maclaurin series of $\frac{x^2+3}{x^2-x-6}$
Find the Maclaurin series of $\frac{x^2+3}{x^2-x-6}$.
So far I have:
$$\frac{x^2+3}{x^2-x-6}=1+\frac{x+9}{(x-3)(x+2)}=1+\frac{x+2}{(x-3)(x+2)}+\frac{7}{(x-3)(x+2)}=1-\frac{1}{(3-x))}+\frac{7}{(x-3)(x+2)}$$
How should I continue?
|
As in this answer, suppose
$$
\frac{a+bx}{6+x-x^2}=\sum_{k=0}^\infty c_kx^k
$$
then
$$
\begin{align}
a+bx
&=\left(6+x-x^2\right)\sum_{k=0}^\infty c_kx^k\\
&=\sum_{k=0}^\infty c_k\left(6x^k+x^{k+1}-x^{k+2}\right)\\
&=\underbrace{\ \ \ \ 6c_0\ \ \ \ \vphantom{()}}_{a}+\underbrace{\left(6c_1+c_0\right)}_b\,x+\sum_{k=2}^\infty\left(6c_k+c_{k-1}-c_{k-2}\right)x^k
\end{align}
$$
Since the coefficient of $x^k$ is $0$ for $k\ge2$, we must have $6c_k+c_{k-1}-c_{k-2}=0$; that is,
$$
c_0=\frac a6\quad c_1=\frac{6b-a}{36}\quad c_k=\frac{c_{k-2}-c_{k-1}}6
$$
$$
\begin{align}
\frac{x^2+3}{x^2-x-6}
&=1+\frac{x+9}{x^2-x-6}\\
&=1+\frac{-9-x}{6+x-x^2}\\
&=1+\sum_{k=0}^\infty c_kx^k
\end{align}
$$
where $c_0=-\frac32$, $c_1=\frac1{12}$, and $c_k=\frac{c_{k-2}-c_{k-1}}6$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1887883",
"timestamp": "2023-03-29T00:00:00",
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|
How to solve this determinant equation in a simpler way
Question Statement:-
Solve the following equation
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=0$$
My Solution:-
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=
\begin{vmatrix}
x+5 & 2 & 3 \\
x+5 & x & 1 \\
x+7 & 2 & 5 \\
\end{vmatrix} \tag{$C_1\rightarrow C_1+C_2+C_3$}$$
$$=\begin{vmatrix}
0 & 2 & 3 \\
0 & x & 1 \\
2 & 2 & 5 \\
\end{vmatrix}+
(x+5)\begin{vmatrix}
1 & 2 & 3 \\
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}\tag{1}$$
On opening the first determinant in the last step above we get $2(2-3x)$.
On simplifying the secind determinant we get,
$$(x+5)\begin{vmatrix}
1 & 2 & 3 \\
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}=(x+5)\begin{vmatrix}
1 & 2 & 3 \\
0 & x-2 & -2 \\
0 & 0 & 2 \\
\end{vmatrix}
(R_2\rightarrow R_2-R_1)
(R_3\rightarrow R_3-R_1)$$
$=2(x+5)(x-2)$
Substituting the values obtained above in $(1)$, we get
$$=\begin{vmatrix}
0 & 2 & 3 \\
0 & x & 1 \\
2 & 2 & 5 \\
\end{vmatrix}+
(x+5)\begin{vmatrix}
1 & 2 & 3 \\
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}=2(2-3x)+2(x+5)(x-2)=2(2-3x+x^2+3x-10)=2(x^2-8)$$
Now, as $\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=0$, $\therefore 2(x^2-8)=0\implies x=\pm2\sqrt2$
As you can see there was lot of work in my solution so if anyone can provide me with some techniques to solve it faster, or a technique which includes less amount of pen and more thinking.
|
Subtract the first row from the last and use Laplace's formula for the third row: $$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}= \begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
0 & 0 & 2 \\
\end{vmatrix} = 2(-1)^{3+3}\begin{vmatrix}
x & 2\\
4 & x
\end{vmatrix} = 2(x^2-8) = 2(x-2\sqrt{2})(x+2\sqrt{2})=0.$$ Now the roots are $x = \pm 2\sqrt{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all integral solutions of $ x^4 + y^4 + z^4 -w^4 = 1995 $ Find all integral solutions of $ x^4 + y^4 + z^4 -w^4 = 1995 $.
Attempt:
From FLT it can be concluded that either all of $ x , y , z$ and $w$ are multiples of 5 ( which is not possible since that would lead to $ x^4 + y^4 + z^4 -w^4 $ being a multiple of $5^4$ which it is not since it's equal to 1995) or
$ w^4 $ and exactly one of $x^4, y^4, z^4 $ are of the form $5k+1$ .
So, there can be 3 cases with each having either $x^4, y^4 $ or $ z^4 $, along with $w^4$ , of the form $5k+1$ .
I am unable to find the solutions from there.
|
Note that fourth powers are always congruent to $0$ or $1$ mod $16$. This can be proved using case analysis. I'll list these cases:
*
*If $n$ is even, then $2 \mid n$ hence $16 \mid n^4$.
*If $n\equiv \pm1 \mod 16$, then $n^4 \equiv 1 \mod 16$
*If $n\equiv \pm3 \mod 16$, then $n^4 \equiv 3^4 \equiv 1 \mod 16$
*If $n\equiv \pm5 \mod 16$, then $n^4 \equiv 5^4 \equiv 1 \mod 16$
*If $n\equiv \pm7 \mod 16$, then $n^4 \equiv 7^4 \equiv 1 \mod 16$
Now, the left hand side can be $-1, 0, 1, 2$ or $3$ mod $16$, but $1995$ is congruent to $11$ mod $16$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integration: $\int\frac{dx}{\sqrt{x(1-x)}}$ My teacher wrote this on the blackboard: $$\int_0^z \frac{1}{\pi \sqrt{x(1-x)}}dx= \frac{2}{\pi} \arcsin(\sqrt{z})$$
But when I try to calculate the integral:
\begin{align}
\int \frac{1}{\pi \sqrt{x(1-x)}}dx&=\int \frac{1}{\pi \sqrt{(\frac{1}{2})^2-(x-\frac{1}{2})^2}}dx\\
&= \int \frac{1}{\pi \sqrt{(\frac{1}{2})^2-k^2}}dk=2 \int \frac{1}{\pi \sqrt{1-(2k)^2}}dk\\
&= \int \frac{1}{\pi \sqrt{1-a^2}}da\\
&= \frac{1}{\pi} \arcsin(a)\\
&=\frac{1}{\pi} \arcsin\left(2(x-\frac{1}{2})\right)
\end{align}
So
$$\int_0^z \frac{1}{\pi \sqrt{x(1-x)}}dx= \frac{1}{\pi} \arcsin(2z-1)-\frac{1}{\pi} \arcsin(-1)= \frac{1}{\pi} \arcsin(2z-1) - \frac{1}{2}$$
How I come to the solution of my teacher?
|
Your answer should be $$\frac {1}{\pi}\arcsin(2z-1)\color{red}{+}\frac 12$$
In fact both answers are equivalent. This is because if $$\phi=\frac{2}{\pi}\arcsin(\sqrt{z}),$$ then$$\sin(\pi\phi)=2\sqrt{z}\cos(\arcsin(\sqrt{z}))=2\sqrt{z}\sqrt{1-z}\Rightarrow\phi=\frac{1}{\pi}\arcsin(2\sqrt{z-z^2})$$
Meanwhile, if $$\theta=\frac {1}{\pi}\arcsin(2z-1)\color{red}{+}\frac 12,$$ then$$\sin(\pi\theta)=\sin(\arcsin(2z-1)+\frac{\pi}{2})$$
$$=\cos(\arcsin(2z-1))=\sqrt{1-(2z-1)^2}=2\sqrt{z-z^2}$$
Hence $$\phi=\theta$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let a, b, c be positive real numbers. Prove that Let a,b,c be positive real numbers. Prove that
$$\frac{a^3+b^3+c^3}{3}\geq\frac{a^2+bc}{b+c}\cdot\frac{b^2+ca}{c+a}\cdot\frac{c^2+ab}{a+b}\geq abc$$
I will post what I had solved originally, however it is unfortunately incorrect. Please help solve and/or aid in finding my mistakes :]
|
The left inequality.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Since $\prod\limits_{cyc}(a^2+bc)=2a^2b^2c^2+\sum\limits_{cyc}(a^3b^3+a^4bc)=8w^6+A(u,v^2)w^3+B(u,v^2)$,
$\prod\limits_{cyc}(a+b)=9uv^2-w^3$ and $a^3+b^3+c^3=27u^3-27uv^2+3w^3$, we see that
our inequality is equivalent to $f(w^3)\geq0$, where $f$ is a concave function.
But the concave function gets a minimal value for an extremal value of $w^3$,
which happens for equality case of two variables and we must check $w^3\rightarrow0^+$.
*
*Let $w^3\rightarrow0^+$. Let $c\rightarrow0^+$.
We get $(a^3+b^3)(a+b)ab\geq3a^3b^3$, which is obvious;
*$b=c=1$, which gives $(a^2-1)^2(2a+1)\geq0$.
Done!
Also we can use a full expanding and we'll get something obvious:
$\sum\limits_{sym}\left(a^5b+a^4b^2-\frac{1}{2}a^4bc-\frac{3}{2}a^3b^3+a^3b^2c-a^2b^2c^2\right)\geq0$, which is Muirhead.
|
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|
Show that $ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, [\arcsin{\frac{x}{a}} / \arccos{\frac{x}{a}}]= \frac{\pi^2}{8} $? I found this very interesting result:
$$ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arcsin{\frac{x}{a}} = \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arccos{\frac{x}{a}} = \frac{\pi^2}{8}, $$
for $a>0.$ I wonder how one could prove this?
|
$$ \arcsin{\frac{x}{a}}=t $$
$$ \frac{dx}{\sqrt{a^2 - x^2}}=dt$$
$$ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arcsin{\frac{x}{a}}=\frac{(\arcsin{\frac{x}{a}})^2}{2}$$
|
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|
By using a geometric series and a factorisation, compute the first three terms of this given Taylor expansion By using the geometric series
$$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$
and the factorisation
$$\frac{1}{1-3x+2x^2}= \left(\frac{1}{1-2x}\right) \left(\frac{1}{1-x}\right)$$
compute the first three terms of the Taylor expansion of
$$\frac{1}{1-3x+2x^2} \text{ around } x=0$$
My theory of how to do this question is input the summation to the factorisation equation and solve so the summation equals
$$\sum_{n=0}^\infty x^n = \frac{1-2x}{1-3x+2x^2}$$
and solve by starting with $x=0$ and going to $x=2$.
But I feel this may be too simple to be true. Are my workings correct?
Thanks in advance
|
Alternative approach: for $x$ near zero we have that $|x(3-2x)|<1$ and
$$\frac{1}{1-3x+2x^2}=\frac{1}{1-x(3-2x)}=\sum_{k=0}^{\infty}(x(3-2x))^k\\=1+(x(3-2x))+(x(3-2x))^2+(x(3-2x))^3+o((x(3-2x))^3)\\
=1+(3x-2x^2)+(9x^2-12x^3+o(x^3))+(27x^3+o(x^3))+o(x^3)\\
=1+3x+7x^2+15x^3+o(x^3).$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove property of greatest integer function:
How can we prove that
$$\lfloor x + y \rfloor = \lfloor y + x - \lfloor x \rfloor \rfloor + \lfloor x \rfloor$$
for all real $x$, where $ \lfloor x \rfloor$ denotes greatest integer less than or equal to $x$?
I was able to prove that $\lfloor x + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor $ or $\lfloor x + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor + 1$ by using the property that $\lfloor x \rfloor = m$ means $m \le x \lt m + 1$ but cannot prove above one by any means. I have tried a lot, can any one please help and please give a simple proof?
|
If $n$ is an integer and $t$ any real number, then it is straightforward to show that$\lfloor t + n \rfloor = \lfloor t \rfloor + n.$
Therefore, since $\lfloor x \rfloor$ is an integer,
\begin{align*}
\quad &\lfloor y + x - \lfloor x \rfloor \rfloor + \lfloor x \rfloor \\
= &\lfloor y + x \rfloor - \lfloor x \rfloor + \lfloor x \rfloor \\
= &\lfloor x + y \rfloor.
\end{align*}
|
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|
How to calculate this limit with cube-roots? How to solve this limit?
$$\lim_{x\to \pm\infty}\sqrt[3]{(x-1)^2}-\sqrt[3]{(x+1)^2}$$
I figured this is the same as:
$$\lim_{x\to \pm\infty}(\sqrt[3]{x-1}+\sqrt[3]{x+1})(\sqrt[3]{x-1}-\sqrt[3]{x+1})$$
But that didn't help much I guess...
|
$$A=\sqrt[3]{(x-1)^2}-\sqrt[3]{(x+1)^2}=(x-1)^{2/3}-(x+1)^{2/3}=x^{2/3}\left( \left(1-\frac 1x \right)^{2/3}-\left(1+\frac 1x \right)^{2/3}\right)$$ Use the generalized binomial theorem or Taylor series.
$$\left(1-\frac 1x \right)^{2/3}=1-\frac{2}{3 x}-\frac{1}{9 x^2}-\frac{4}{81
x^3}+\cdots$$ $$\left(1+\frac 1x \right)^{2/3}=1+\frac{2}{3 x}-\frac{1}{9 x^2}+\frac{4}{81
x^3}+\cdots$$ $$A=x^{2/3} \left(-\frac{4}{3 x}-\frac{8}{81 x^3}+\dots \right)=-\frac{4}{3 x^{1/3}}-\frac{8}{81 x^{7/3}}+\cdots$$
|
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|
Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? I think it diverges although:
$$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$
for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might argue:
$$ \sum_{n \geq 1} (-1)^{n+1} \sqrt{n} = \frac{1}{2}\sum_{m \geq 1} \frac{1}{\sqrt{2m}} = \infty $$
Are these Cesaro summable? For an even number of terms:
$$\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots - \sqrt{2n}
\approx - \frac{1}{2\sqrt{2}}\left( \frac{1}{\sqrt{1}} +
\frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} \right)
\approx \sqrt{\frac{n}{2}}$$
so the Cesaro means tend to infinity. Does any more creative summation method work?
The result is from paper called "The Second Theorem of Consistency for Summable Series" in Vol 6 of the Collected Works of GH Hardy
the series $1 - 1 +1 - 1 \dots$ is summable $(1,k)$ for any $k$ but not summable $(e^n, k)$ for any value of $k$.
The series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ is summable $(n,1)$ but not $(e^{\sqrt{n}},1)$ and so on...
Here things like $(1,k), (n,1)$ refer to certain averaging procedures, IDK
|
Let us show that $\sum_{k=1}^{\infty}(-1)^{k-1}\sqrt{k}$ is Cesaro summable. Once we establish this, then this is also Abel summable and the Cesaro sum is equal the Abel sum, which is
$$ \sum_{k=1}^{\infty}(-1)^{k-1}\sqrt{k} = -\operatorname{Li}_{-1/2}(-1) = (1 - 2^{3/2})\zeta(-1/2). $$
To this end, let $s_n = \sum_{k=1}^{n} (-1)^{k-1}\sqrt{k}$ and notice that $s_n = \mathcal{O}(\sqrt{n})$. This can be easily checked by grouping two successive terms and applying the mean value theorem. Thus it suffices to prove that
$$ \frac{s_1 + \cdots + s_{2n+1}}{2n+1} $$
converges. Now the trick is to consider
$$ s_{2n} + s_{2n+1} = 1 + \sum_{k=1}^{n} (\sqrt{2k-1} + \sqrt{2k+1} - 2\sqrt{2k}). $$
Using Taylor series, it is not hard to check that
$$\sqrt{2k-1} + \sqrt{2k+1} - 2\sqrt{2k} = \mathcal{O}(k^{-3/2}). $$
Thus $s_{2n} + s_{2n+1}$ converges as $n \to \infty$, and the claim follows from Cearso-Stolz theorem.
|
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|
$x+y=xy=w \in \mathbb{R}^+$. Is $x^w+y^w$ real? Question: For $x,y \in \mathbb{C}$, suppose $x+y=xy=w \in \mathbb{R}^+$. Is $x^w+y^w$ necessarily real?
For instance, if $x+y=xy=3$, then one solution is $x = \frac{3 \pm i \sqrt{3}}{2}$, $y = \frac{3 \mp i \sqrt{3}}{2}$, but $x^3 + y^3 = 0$, which is real.
I've checked this numerically for many values of $w$ that give complex $x$ and $y$ (namely, $w \in (0,4)$.)
|
Yes. Write $x=a+bi$, $y=c-di$, then clearly $b=d$ because $x+y$ is real.
So $x=a+bi$, $y=c-bi$. Then $xy=ac-abi+cbi+b^2$, so $a=c$ or $b=0$.
*
*If $a=c$, then $y = \overline{x}$, i.e. the complex conjugate of $x$.
So $x^w+y^w = x^w+(\overline{x})^w = x^w+\overline{x^w}$, which is
real.
*If $b=0$, $x$ and $y$ are real so $x^w+y^w$ is real as $w>0$.
|
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|
In how many ways can I form a set of 6 elements made up of 0s, 1s, 2s, 3s, 4s and 5s such that the sum of its elements is a multiple of 6? I have a set $ S $ which is made up of 6 elements, each one being either 0, 1, 2, 3, 4 or 5 (note that repetition is allowed). I want to evaluate the following problems:
a) In how many ways can such a set be constructed
b) How many of these ways have the property that the sum of all the elements of $ S $ is a multiple of 6 (e.g. {1, 1, 1, 1, 1, 1}; {1, 2, 3, 4, 4, 4})
For a), I tried doing something like $ \frac{6^6}{6!} $: counting the total number and dividing by the number of ways to arrange them; however, that is not an integer. I also tried combinations with replacement: $ \left(\binom{6}{6}\right) = \binom{6+6-1}{6} = \binom{11}{6} = 462$ , however I am not certain if this is correct. I have no clue how to approach b).
|
As a set, $\{1,1,1,1,1,1\}=\{1\}$, hence the sum of its elements is $1$, not $6$.
If the actual problem is
In how may ways we may choose $x_1,x_2,x_3,x_4,x_5,x_6$ from
$\{0,1,2,3,4,5\}$ in such a way that $x_1+x_2+x_3+x_4+x_5+x_6$ is a
multiple of $6$?
Then the answer is just $\frac{6^6}{6}=\color{red}{6^5}$, since we way choose $x_1,x_2,x_3,x_4,x_5$ as we like, then there is a unique option for $x_6$ that ensures the sum being a multiple of $6$.
If the actual problem is
In how many ways we may choose a multi-set of $6$ elements from
$\{0,1,2,3,4,5\}$ such that the sum of its elements is a multiple of
$6$?
we have to consider the sum of the coefficients of $x^0,x^6,x^{12},\ldots$ in the polynomial given by the coefficient of $y^6$ in the product
$$(1+y+y^2+y^3+\ldots)(1+yx+y^2 x^2+y^3 x^3+\ldots)(1+yx^2+y^2x^4+\ldots)\ldots(1+yx^5+y^2 x^{10}+\ldots)$$
that is
$$ \frac{1}{(1-y)(1-yx)(1-yx^2)(1-yx^3)(1-yx^4)(1-yx^5)}. $$
So we have to consider the polynomial
$$ 1+x+2 x^2+3 x^3+5 x^4+7 x^5+10 x^6+12 x^7+16 x^8+19 x^9+23 x^{10}+25 x^{11}+29 x^{12}+30 x^{13}+32 x^{14}+32 x^{15}+32 x^{16}+30 x^{17}+29 x^{18}+25 x^{19}+23 x^{20}+19 x^{21}+16 x^{22}+12 x^{23}+10 x^{24}+7 x^{25}+5 x^{26}+3 x^{27}+2 x^{28}+x^{29}+x^{30} $$
and compute
$$ 1+10+29+29+10+1 = \color{red}{80}.$$
|
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|
Exact Differential Equation - possible textbook mistake - $(3x^2+6xy^2) dx + (6x^2y+4y^2)dy = 0$ and $y(0)=2$ When solving the differential equation
$$
(3x^2+6xy^2) dx + (6x^2y+4y^2)dy = 0
$$
I verified that it is an Exact Equation. By saying that
$$
F_x = 3x^2+6xy^2\\
F_y = 6x^2y+4y^2
$$
and integrating both sides I found $$F(x,y) = 3x^2y^2 + \frac{4}{3}y^3+x^3 = C$$ and since $y(0)=2$ we have $$3x^2y^2+ \frac{4}{3}y^3+x^3 = \frac{32}{3}$$
The problem is that the textbook's solution is $$x^3+3x^2y^2+y^3=16$$
Where did I go wrong?
Thank you.
|
Your book solution is not correct $${ x }^{ 3 }+3{ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 3 }=16\\ 3{ x }^{ 2 }+6{ x }{ y }^{ 2 }+6{ x }^{ 2 }y{ y }^{ \prime }+3{ y }^{ 2 }{ y }^{ \prime }=0\\ 3{ x }^{ 2 }+6{ x }{ y }^{ 2 }+\frac { dy }{ dx } \left( 6{ x }^{ 2 }y+3{ y }^{ 2 } \right) =0\\ \left( 3{ x }^{ 2 }+6{ x }{ y }^{ 2 } \right) dx+\left( 6{ x }^{ 2 }y+\color{red}3{ y }^{ 2 } \right) dy=0\\ \\ \\ $$
which is a differ from original equation form($(3x^2+6xy^2) dx + (6x^2y+\color{red}4y^2)dy = 0$)
|
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|
Prove that $\left(\frac{n}{2}\right)^n \gt n! \gt \left(\frac{n}{3}\right)^n \qquad n\ge 6 $
$$\left(\frac{n}{2}\right)^n \gt n! \gt \left(\frac{n}{3}\right)^n \qquad n\ge 6 $$
I tried it prove it by mathematical induction but failed .
For $n=6$
$$(3)^6 \gt 6!$$
Now for $n=k$
$$\left(\frac{k}{2}\right)^k \gt k!$$
Now for $n=k+1$
$$\left(\frac{(k+1)}{2}\right)^{k+1} \gt (k+1)!$$
$$\left(\frac{(k+1)}{2}\right)^{k} \gt 2(k!)$$
|
$$\begin{align}
\left(\frac{k+1}2\right)^{k+1}&=\frac {k+1}2\left(\frac {k+1}2\right)^k\\
&>\frac{k+1}2\left[\left(\frac k2\right)^k+\binom k1\left(\frac k2\right)^{k-1}\frac12\right]\\
&>\frac{k+1}2[k!+k!]\\
&=(k+1)!
\end{align}$$
|
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|
Prove inequality $\arccos \left( \frac{\sin 1-\sin x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$ I was trying to figure out if the following function can serve as a mean (see mean value theorem):
$$\arccos \left( \frac{\sin y-\sin x}{y-x} \right)$$
And turns out that for $x,y \leq \pi$ it does serve as a mean admirably.
But then I've noticed that for $0<x<1$ the following two functions are very close (see the picture):
Now how would you prove:
$$\arccos \left( \frac{\sin 1-\sin x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$$
It's probably easier to consider another equivalent inequality:
$$\frac{\sin 1-\sin x}{1-x} \geq \cos \sqrt{\frac{1+x+x^2}{3}}$$
Or even:
$$ \text{sinc} \left(\frac{1-x}{2} \right) \cos \left(\frac{1+x}{2} \right) \geq \cos \sqrt{\frac{1+x+x^2}{3}}$$
We could use Taylor series, but that's too cumbersome in my opinion.
Another way would be Mean value theorem itself, but I encounter the same problem.
Is there a simple way to prove this inequality?
My calculus is not as sharp as it used to be (just kidding, it was never sharp).
Edit
Just to confirm (numerically) that the inequality holds, here is the plot of the difference between the two functions:
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This is not a full answer, just a possible way to prove the inequality.
We use the following form of the inequality:
$$\text{sinc} \left(\frac{1-x}{2} \right) \cos \left(\frac{1+x}{2} \right) \geq \cos \sqrt{\frac{1+x+x^2}{3}}$$
It makes sense to try infinite products, mainly because we get rid of the square root:
$$\text{sinc}(t)=\prod_{n=1}^\infty \left(1-\frac{t^2}{\pi^2 n^2} \right)$$
$$\cos (t)=\prod_{n=1}^\infty \left(1-\frac{t^2}{\pi^2 (n-1/2)^2} \right)$$
Thus, our inequality becomes:
$$\prod_{n=1}^\infty \left(1-\frac{(1-x)^2}{4\pi^2 n^2} \right) \left(1-\frac{(1+x)^2}{4\pi^2 (n-1/2)^2} \right) \geq \prod_{n=1}^\infty \left(1-\frac{1+x+x^2}{3\pi^2 (n-1/2)^2} \right)$$
Note that for $x<1$ every term in the infinite products is positive.
If, for example, every term of the product on the left is greater than every term of the product on the right, then our inequality is proven.
$$\left(1-\frac{(1-x)^2}{4\pi^2 n^2} \right) \left(1-\frac{(1+x)^2}{4\pi^2 (n-1/2)^2} \right) \geq^? 1-\frac{1+x+x^2}{3\pi^2 (n-1/2)^2}$$
After expanding and simplifying I obtained the following:
$$-\left(4 \pi^2 n(2n-3)+3(\pi^2-(1+x)^2) \right)(1-x)^2 \geq^? 0$$
And this is highly questionable, i.e. not correct for most cases. Remember, we are interested in the case $x < 1$, so:
$$4 \pi^2 n(2n-3)+3(\pi^2-(1+x)^2) \leq^? 0$$
$$4 \pi^2 n(2n-3)+3\pi^2 \leq^? 3 (1+x)^2$$
For $n=1$ we have a trivial inequality:
$$-\pi^2 < 3 (1+x)^2$$
But for $n \geq 2$ the inequality quickly stops working.
So this method probably doesn't prove anything.
On the other hand, we can compare the two term products on each side, i.e. prove that:
$$\prod_{n=k}^{k+1} \left(1-\frac{(1-x)^2}{4\pi^2 n^2} \right) \left(1-\frac{(1+x)^2}{4\pi^2 (n-1/2)^2} \right) \geq \prod_{n=k}^{k+1} \left(1-\frac{1+x+x^2}{3\pi^2 (n-1/2)^2} \right)$$
If that fails, we can try $3$ product terms and so on. This is just some algebra that a CAS can take care of even for a large number of terms.
Update
I decided to rearrange the second product so the terms are of the same order in $x$ and $n$:
$$\prod_{n=1}^\infty \left(1-\frac{1+x+x^2}{3\pi^2 (n-1/2)^2} \right)=\prod_{k=1}^\infty \left(1-\frac{1+x+x^2}{3\pi^2 (2k-3/2)^2} \right) \left(1-\frac{1+x+x^2}{3\pi^2 (2k-1/2)^2} \right)$$
Still, we have the same relation: only the first terms of the products obey the inequality, while all the rest seem to break it.
Below you can see the plot for:
$$ \dfrac{\left(1-\frac{(1-x)^2}{4\pi^2 n^2} \right) \left(1-\frac{(1+x)^2}{4\pi^2 (n-1/2)^2} \right) }{ \left(1-\frac{1+x+x^2}{3\pi^2 (2n-3/2)^2} \right) \left(1-\frac{1+x+x^2}{3\pi^2 (2n-1/2)^2} \right) }$$
For $x \in [0,1]$ and $n=1,2,3, \dots$.
I see no way to prove that the product of all terms for $n \geq 2$ is still closer to $1$ than the first term (despite numerical evidence), so this way doesn't seem to work as a proof of the original inequality.
|
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|
Generating function for cubes of Harmonic numbers By generalizing the approach in Integral involving a dilogarithm versus an Euler sum. meaning by using the integral representation of the harmonic numbers and by computing a three dimensional integral over a unit cube analytically we have found the generating function of cubes of harmonic numbers. We have:
\begin{eqnarray}
&&S^{(3)}(x) := \sum\limits_{n=1}^\infty H_n^3 x^n =
\frac{-18 \text{Li}_3\left(1-\frac{1}{x}\right)+6 \text{Li}_3\left(\frac{1}{x}\right)-18 \text{Li}_3(x)}{6(1-x)}+
\frac{6 \log ^3(1-x)-9 \log (x) \log ^2(1-x)+3 \left(3 \log ^2(x)+\pi ^2\right) \log (1-x)}{6(x-1)}+\frac{-\log (x) \left(2 \log ^2(x)+ 3 i \pi \log (x)+5 \pi ^2\right)}{6 (x-1)}
\end{eqnarray}
Clearly some of the terms on the right hand side are complex even though the whole expression is of course real. The first two terms in the first fraction on the rhs are complex and the middle term in the last fraction is complex. My question is how do I simplify the right hand side to get rid of the complex terms?
|
By using the functional equations for the trilogarithm we simplified the result as follows:
\begin{eqnarray}
&&S^{(3)}(x)= \\
&&\frac{ \text{Li}_3(x)}{(1-x)}+
3\frac{\text{Li}_3(1-x)-\zeta (3)}{(1-x)}+
\log(1-x)\frac{ \left(-2 \log ^2(1-x)+3 \log (x) \log(1-x)-\pi ^2 \right)}{2 (1-x)}
\end{eqnarray}
For a sanity check we expand each of the terms in the formula in a Taylor series about zero we have:
\begin{eqnarray}
&&\frac{Li_3(x)}{1-x} =\\
&& x+\frac{9 x^2}{8}+\frac{251 x^3}{216} + O(x^4) \\
&&3\frac{\text{Li}_3(1-x)-\zeta (3)}{(1-x)} =\\
&&-\frac{\pi ^2 x}{2}+\frac{9 x^2}{4}-\frac{3}{2} x^2 \log (x)-\frac{3 \pi ^2 x^2}{4}-\frac{11 \pi ^2 x^3}{12}+4 x^3-3 x^3 \log (x) + O(x^4) \\
&&\log(1-x)\frac{ \left(-2 \log ^2(1-x)+3 \log (x) \log(1-x)-\pi ^2 \right)}{2 (1-x)} =\\
&& \frac{\pi ^2 x}{2}+\frac{3 \pi ^2 x^2}{4}+\frac{3}{2} x^2 \log (x) +\frac{11 \pi ^2 x^3}{12}+x^3+3 x^3 \log (x)+ O(x^4)
\end{eqnarray}
As we can see the terms proportional to $\log(x)$being present in the second and the third term exactly cancel each other. The formula is correct.
|
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|
If $a$, $b$, and $c$ are sides of a triangle, then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$.
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
$$\sum_{\text{cyc}}\frac{a}{b+c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2\,.$$
Attempt. By clearing the denominators, the required inequality is equivalent to
$$a^2(b+c)+b^2(c+a)+c^2(a+b)>a^3+b^3+c^3\,.$$
Since $b+c>a$, $c+a>b$, and $a+b>c$, the inequality above is true. Is there a better, non-bruteforce way?
|
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\begin{align}\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&<2\tag{1}\label{1}.\end{align}
Let $\rho,\ r,\ R$
be the semiperimeter,
inradius and circumradius
of the triangle $ABC$, then
we can rewrite \eqref{1} as
\begin{align}
\frac a{2\rho-a}
+\frac b{2\rho-b}
+\frac c{2\rho-c}
&=
\frac{
4\,(a+b+c)\rho^2
-4\,(ab+bc+ca)\,\rho
+3\,abc
}{
8\rho^3-4\,(a+b+c)\,\rho^2+2\,(ab+bc+ca)\,\rho
-abc
}
<2
\tag{2}\label{2}
,
\end{align}
then, rearranging,
\begin{align}
16\rho^3
-12(a+b+c)\rho^2
+8(ab+bc+ca)\rho-5abc
&>0
\tag{3}\label{3}
,\\
16\rho^3
-12(2\rho)\rho^2
+8(\rho^2+r^2+4rR)\rho-5\cdot4\rho r R
&>0
\tag{4}\label{4}
,\\
4r\rho(2r+3R)
&>0
\tag{5}\label{5}
,
\end{align}
which always holds.
|
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|
Max profit question, regarding negative exponents with factors $p= 10q^{-0,5}$
$C(q) = 5q$
In a monopoly the equation for maximum profits is: $p'(q)\cdot q + p(q) - C'(q) = 0$
First order condition:
$-0,5 \cdot 10q^{-1,5} \cdot q + 10q^{-0,5} - q = 0$
can't figure out onwards from here. How do I do with these negative exponents?
the answer is $q = 1$, and $p = 10$
|
As stated above in a comment I believe the correct equation for monopoly profit optimization should be $(-.5)\cdot 10q^{-1.5}\cdot q + 10q^{-.5} - 5 = 0$. We will solve this equation for $q$ below:
$(-.5)\cdot 10q^{-1.5}\cdot q + 10q^{-.5} - 5 = 0 \Rightarrow \frac{-5}{\sqrt{q}} + \frac{10}{\sqrt{q}} - 5 = 0$
Here we have simply used algebra of exponents: $x^{-a} = \frac{1}{x^a}$ and $(x^a)(x^b) = x^{a+b}$.
$\frac{-5}{\sqrt{q}} + \frac{10}{\sqrt{q}} - 5 = 0 \Rightarrow \frac{5}{\sqrt{q}} = 5 \Rightarrow \sqrt{q} = 1 \Rightarrow q = 1$
Thus we arrive at $q = 1$. Plugging $q$ into our equation for $p$ we see $p = 10(q)^{-.5} = 10(1)^{-.5} = 10(1) = 10$. Thus $q = 1$ and $p = 10$.
|
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|
prove $abc \ge 8$ for $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$ I am reading this book about inequalities and the chapter about AM-GM inequalities includes this problem:
Let $a,b,c$ be positive numbers for which $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$, prove that
$$abc \ge 8$$
The book does not provide full solutions but only hints, and the one for this question is that it is similar to this problem:
Let $a,b,c$ be positive numbers with $a+b+c=1$, prove that
$$(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1) \ge 8$$
again there's no solution provided but what I came up with is the following (you can correct me if it is wrong):
$$a+b=1-c$$
$$a+c=1-b$$
$$b+c=1-a$$
$$(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)=(\frac{1-a}{a})(\frac{1-b}{b})(\frac{1-c}{c})=(\frac{b+c}{a})(\frac{a+c}{b})(\frac{a+b}{c})=(\frac{b}{a}+\frac{c}{a})(\frac{a}{b}+\frac{c}{b})(\frac{a}{c}+\frac{b}{c})$$
By AM-GM we have for each term:
$$(\frac{b}{a}+\frac{c}{a}) \ge 2\sqrt{\frac{bc}{a^2}}$$
$$(\frac{a}{b}+\frac{c}{b}) \ge 2\sqrt{\frac{ac}{b^2}}$$
$$(\frac{a}{c}+\frac{b}{c}) \ge 2\sqrt{\frac{ab}{c^2}}$$
By multiplying the three inequalities we have
$$(\frac{b}{a}+\frac{c}{a})(\frac{a}{b}+\frac{c}{b})(\frac{a}{c}+\frac{b}{c}) \ge 8 \sqrt{\frac{a^2b^2c^2}{a^2b^2c^2}}=8$$
as desired.
Can someone please provide me a proof for the first problem as I cannot find any way around it, Thanks.
|
From the condition by AM-GM we obtain:
$$\prod\limits_{cyc}\frac{a}{1+a}=\prod\limits_{cyc}\left(\frac{1}{1+b}+\frac{1}{1+c}\right)\geq\prod\limits_{cyc}\frac{2}{\sqrt{(1+b)(1+c)}}=8\prod\limits_{cyc}\frac{1}{1+a}$$
Id est, $abc\geq8$.
|
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|
show that $\frac{c^2}{a}+\frac{a^2}{c}+16\sqrt{ac}\ge 9(a+c)$ Let $a,c>0$ show that
$$\dfrac{c^2}{a}+\dfrac{a^2}{c}+16\sqrt{ac}\ge 9(a+c)$$
It seem AM-GM inequality.How?
|
Write $a=b^2$ and $c=d^2$, and multiply through by $b^2d^2$. The inequality becomes
$$b^6+d^6+16b^3d^3\geq 9b^4d^2+9b^2d^4.$$
Dividing by $d^6$ and writing $x=\frac{b}{d}$:
$$x^6+1+16x^3\geq 9x^4+9x^2.$$
This follows from
$x^6+16x^3+1-9x^4-9x^2=(x-1)^4(x^2+4x+1)\geq 0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
how I can simplify this limit? So I got $f(x) = 1+\sin^2(x)-e^{x^{2}}$
and the limit :
$$\lim_{x\to 0^+} \cfrac{\ln(1+6x)f(x)}{x^a\cos(f(x))}$$
I have to simplify this limit :(in a form like $\lim_{x\to o} x^{n-a}$ , or something similar)
*
*I noticed that $\cos(f(x))$ for x->$0$ = $1$
*I can see => $\ln(1+6x)$ like $\cfrac{6x\ln(1+6x)} {6x}$ that for $x→0$ is equal to $1$ (and I remain with $6x$)
now I have :
$6\lim_{x\to 0^+} {x^{1-a}f(x)}$
Can you help me simplify $f(x)$ that remains in the limit.
|
By using Taylor expansions, we have that as $x\to 0^+$,
$$f(x) = 1+\sin^2(x)-e^{x^{2}}=1+(x-x^3/6+O(x^5))^2-(1+x^2+x^4/2+O(x^6))\\=
1+x^2-x^4/3-1-x^2-x^4/2+O(x^5)=-5x^4/6+O(x^5).$$
Hence
$$\frac{\ln(1+6x)\cos(f(x)}{x^af(x)}=\frac{(6x+O(x^2))\cos(-5x^4/6+O(x^5))}{x^a(-5x^4/6+O(x^5))}\\
=\frac{(6x+O(x^2))\cdot(1+O(x^8))}{-5x^{4+a}/6+O(x^{5+a})}
\sim-\frac{36}{5x^{3+a}}.$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Taylor/Maclaurin series type question I want to express $\cos(z)$ in a Taylor series centred on $z_0=\frac{\pi}{4}$.
Using the formula $\sum_{k=0}^{\infty} \dfrac{f^{(k)}(z_0)z^k}{k!}
$ and the fact that $f^{(k)}(z)=\cos(z+\frac{k\pi}{2})$ in this case I found that the Taylor series is $$\sum_{k=0}^{\infty} \frac{\cos(\frac{\pi}{4}+\frac{k\pi}{2})}{k!} (z-\frac{\pi}{4})^k$$ which is also what Wolfram Alpha claims the Taylor series to be.
I decided I wanted to get this Taylor series another way. Instead of plugging stuff into the formula directly, I wanted to work from the Maclaurin series of $\cos(z)$ to get the Taylor series. This is what I have so far:
$$\cos(z)=\cos(z-\frac{\pi}{4}+\frac{\pi}{4})\\=\cos(z-\frac{\pi}{4})\cos(\frac{\pi}{4})-\sin(z-\frac{\pi}{4})\sin(\frac{\pi}{4})\\=\frac{1}{\sqrt{2}}\cos(z-\frac{\pi}{4})-\frac{1}{\sqrt{2}}\sin(z-\frac{\pi}{4})\\=\frac{1}{\sqrt{2}}(\sum_{k=0}^{\infty} (-1)^k\frac{(z-\frac{\pi}{4})^{2k}}{(2k)!}-\sum_{k=0}^{\infty} (-1)^k\frac{(z-\frac{\pi}{4})^{2k+1}}{(2k+1)!})\\=\frac{1}{\sqrt{2}}\sum_{k=0}^{\infty} [(-1)^k\frac{(z-\frac{\pi}{4})^{2k}}{(2k)!}(1-\frac{z-\frac{\pi}{4}}{2k+1})]$$
But I don't know how to simplify this to get $\sum_{k=0}^{\infty} \frac{\cos(\frac{\pi}{4}+\frac{k\pi}{2})}{k!} (z-\frac{\pi}{4})^k$.
If somebody could help me out I'd really appreciate it!
|
It is convenient to continue your calculation one line before the last line.
Note, that
\begin{align*}
\cos(z)=\frac{1}{\sqrt{2}}
\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\left(z-\frac{\pi}{4}\right)^{2k}
- \frac{1}{\sqrt{2}}
\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\left(z-\frac{\pi}{4}\right)^{2k+1}\tag{1}
\end{align*}
is a representation in even and odd part of a series according to
\begin{align*}
\sum_{k=0}^\infty a_k\frac{z^k}{k!}
=\sum_{k=0}^\infty a_{2k}\frac{z^{2k}}{(2k)!}+\sum_{k=0}^\infty a_{2k+1}\frac{z^{2k+1}}{(2k+1)!}
\end{align*}
We can merge the series in (1) and obtain
\begin{align*}
\cos(z)=\frac{1}{\sqrt{2}}
\sum_{k=0}^\infty\frac{(-1)^{s(k)}}{k!}\left(z-\frac{\pi}{4}\right)^k
\end{align*}
with
\begin{align*}
s(k)=\begin{cases}
1\quad&k\equiv 0,3(4)\\
-1\quad&k\equiv 1,2(4)\\
\end{cases}\qquad\qquad \qquad k\geq 0
\end{align*}
Since $$\frac{1}{\sqrt{2}}(-1)^{s(k)}=\cos\left(\frac{\pi}{4}+\frac{k\pi}{2}\right)\qquad\qquad\quad\ \ k\geq 0$$
the claim follows.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How can it be proven that $\frac{x}{y}+\frac{y}{x}\geq2$, with $x$ and $y$ positive? So I realized that I have to prove it with the fact that $(x-y)^2+2xy=x^2+y^2$
So $\frac{(x+y)^2}{xy}+2=\frac{x}{y}+\frac{y}{x}$ $\Leftrightarrow$ $\frac{(x+y)^2}{xy}=\frac{x}{y}+\frac{y}{x}-2$
Due to the fact that $(x+y)^2$ is a square, it will be positive
$x>0$ and $y>0$ so $xy>0$
So $\frac{(x+y)^2}{xy}>0$
So $\frac{x}{y}+\frac{y}{x}>2$
But the problem is that I have proven that $\frac{x}{y}+\frac{y}{x}>2$, but in the case of $x=y$, it is equal, so it has to be $\geq$...
Could someone help?
|
$$\left( \frac { \sqrt { x } }{ \sqrt { y } } -\frac { \sqrt { y } }{ \sqrt { x } } \right) ^{ 2 }\ge 0$$
$$\Rightarrow\frac{ { x } }{ { y } }+\frac { { y } }{ { x } } \ge 2$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Shank's Baby-Step Giant-Step for $3^x \equiv 2 \pmod{29}$ Problem: Solve $3^x \equiv 2 \pmod{29}$ using Shank's Baby-Step Giant-Step method.
I choose $k=6$ and calculated $3^i \pmod {29}$ for $i=1,2,...,6$.
$$3^1 \equiv 3 \pmod {29}$$
$$3^2 \equiv 9 \pmod {29}$$
$$3^3 \equiv 27 \pmod {29}$$
$$3^4 \equiv 23 \pmod {29}$$
$$3^5 \equiv 11 \pmod {29}$$
$$3^6 \equiv 4 \pmod {29}$$
Then I have calculated $3^{-1} \equiv 10 \pmod {29}$ and started calculating second list:
$$2 \cdot 3^{-6} \equiv 2 \cdot 10^6 \equiv 2 \cdot 22 = 44 \equiv 15 \pmod {29}$$
$$2 \cdot 3^{-12} \equiv 2 \cdot {3^{-6}}^2 \equiv 2 \cdot 15^2 \equiv 2 \cdot 22 = 44 \equiv 15 \pmod {29}$$
$$2 \cdot 3^{-18} \equiv 2 \cdot {3^{-12}}^2 \equiv 2 \cdot 22^2 \equiv 2 \cdot 20 = 40 \equiv 11\pmod {29}$$
And now I can stop. I can see that:
$$ 3^5 \equiv 11 \pmod {29}\ and\ 2 \cdot 3^{-18} \equiv 11 \pmod {29}$$
Therefore $x = 5 + 18 = 23$
But when I plugin $x=23$ above I get that $3^{23} \equiv 8 \pmod {29}$.
So where am I wrong?
|
I found flaw in my calculus for the second expression in second list, it should be
$$ 2 \cdot 3^{-12} = 2 \cdot (3^{-6})^2 \equiv 2 \cdot 22^2 = 2 \cdot 484 \equiv 11 \pmod {29}$$
Hence, $x = 5 + 12 = 17 $ which is correct.
|
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|
Woodbury formula problem Hello guys i have this linear algebra problem and i want some help on how i should proceed with it. i think the aim is to use the Woodbury formula
Suppose $u,v \in \mathbb{R}^n, \quad A \in \mathbb{R}^{n*n}$ is invertible and consider for scalars $\alpha , \beta \in \mathbb{R} : \\$ $B(\alpha,\beta)= A +\alpha(uu^T +vv^T ) +\beta(uv^T + vu^T ) \\$
give conditions on $\alpha$ and $\beta$ in terms of $m_{ij}$ where :
$\begin{bmatrix} m_{11} & m_{12} \\ m_{21} & m_{22}\end{bmatrix}= \begin{bmatrix} u^T \\ v^T \end{bmatrix} A^{-1} \begin{bmatrix} u & v \end{bmatrix} \\$
That describes precisely when $B(\alpha,\beta)x=0$ has a nontrivial solution $x \neq 0$.
Thank you.
|
We start with recalling two standard results:
*
*If $A$ is invertible then $\begin{pmatrix} A & B \\ C & D \end{pmatrix}$ is invertible iff $D-CA^{-1}B$ is invertible. This follows directly from $ \begin{pmatrix} I & 0 \\ -CA^{-1} & I \end{pmatrix} \begin{pmatrix} A & B \\ C & D \end{pmatrix} \begin{pmatrix} I & -A^{-1}B \\ 0 & I \end{pmatrix} = \begin{pmatrix} A & 0 \\ 0 & D - CA^{-1}B \end{pmatrix}.$
*If $A$ is invertible $A+uv^T$ is invertible iff $1+ v^TA^{-1}u \neq 0.$
Now to this problem.
Note $B(\alpha,\beta) = A + \begin{pmatrix} u & v \end{pmatrix} \begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix} \begin{pmatrix} u^T \\ v^T \end{pmatrix}.$
First we find $\alpha,\beta$ such that $B(\alpha,\beta)$ is singular and $\alpha^2 \neq \beta^2.$
If $\alpha^2 \neq \beta^2$ then $\begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix}$ is invertible and its inverse is $\dfrac{1}{\alpha^2 - \beta^2} \begin{pmatrix} \alpha & -\beta \\ -\beta & \alpha \end{pmatrix}.$
Consider the matrix $$ D = \begin{pmatrix} \dfrac{\alpha}{\alpha^2 - \beta^2} & \dfrac{-\beta}{\alpha^2 - \beta^2} & u^T \\ \dfrac{-\beta}{\alpha^2-\beta^2} & \dfrac{\alpha}{\alpha^2 - \beta^2} & v^T \\
-u & -v & A \end{pmatrix}.$$
Since $\dfrac{1}{\alpha^2 - \beta^2} \begin{pmatrix} \alpha & -\beta \\ -\beta & \alpha \end{pmatrix}$ is the inverse of $\begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix}$ and is hence invertible, $D$ is invertible iff $A + \begin{pmatrix} u & v \end{pmatrix} \begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix} \begin{pmatrix} u^T \\ v^T \end{pmatrix}$ i,e., $B(\alpha,\beta)$ is invertible.
Since $\begin{pmatrix} A & -u & -v \\ u^T & \dfrac{\alpha}{\alpha^2 - \beta^2} & \dfrac{-\beta}{\alpha^2-\beta^2} \\ v^T & \dfrac{-\beta}{\alpha^2-\beta^2} & \dfrac{\alpha}{\alpha^2-\beta^2} \end{pmatrix} = \begin{pmatrix} 0 & 0 & I \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} D \begin{pmatrix} 0 & 1 & 0 \\ 0 &0&1\\I & 0 &0\end{pmatrix}$ can be obtained form permuting rows and columns and $A$ is invertible, $D$ is also invertible iff $\dfrac{1}{\alpha^2 - \beta^2} \begin{pmatrix} \alpha & -\beta \\ -\beta & \alpha \end{pmatrix} + \begin{pmatrix} u^T \\ v^T\end{pmatrix}A^{-1}\begin{pmatrix}u & v\end{pmatrix} =\begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix}^{-1} + M$ is invertible.
So if if $\alpha^2 \neq \beta^2$ then a necessary and sufficient condition for $B(\alpha,\beta)$ to be singular is that $\texttt{det}( \begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix}^{-1} + M ) = 0$ or equivalently $\texttt{det}( I + \begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix}M ) = 0.$
If $\alpha^2 = \beta^2$ then we we have $\alpha = \pm \beta.$
First consider the case $\alpha = \beta.$ In this case $B(\alpha,\beta) = A + \alpha (u+v) (u + v)^T$ and so $B(\alpha,\beta)$ is singular iff $\alpha (u+v)^T A^{-1} (u + v) + 1 = 0$ i.e. $\alpha \mathbf{1}^T M \mathbf{1} + 1 = 0$ where $\mathbf{1}$ is a $ 2 \times 1$ vector of 1's.
If $\alpha = -\beta$ then $B(\alpha,\beta) = A + \alpha(u-v)(u-v)^T$ which leads to then condition that $B(\alpha,\beta)$ is singular iff $\alpha \begin{pmatrix}1 -1 \end{pmatrix} M \begin{pmatrix}1 \\ -1 \end{pmatrix} + 1 = 0.$
|
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|
Unable to justify solution for a problem with exponential constant $e$ involved. For the question,
$e^{2x} + e^x - 2 = 0.$
I was asked to solve for $x$.
What I performed,
$e^{2x} + e^x = 2.$
$e^x(e^x + 1) = 2.$
For the 2 solutions involved,
\begin{align}
e^x &= 2\\
\ln2 &= x\\
.69 &= x
\end{align}
OR
\begin{align}
e^x + 1 &= 2\\
e^x &= 1\\
\ln1 &= x\\
0 &= x
\end{align}
My textbook states a single solution which is $0$.
|
$e^x (e^x+1) = 2$ does not imply that $e^x = 2$ or $e^x+1 = 2$. This type of reasoning only works if the right-hand side is zero. (This is why it's called the zero-product property.)
In general, if $ab = 2$ then we could have $a=2$ and $b=1$, or $a=1$ and $b=2$, or $a=4$ and $b=1/2$, or $a=2\pi$ and $b=\frac1\pi$, etc. There are infinitely many possibilities if the right-hand side is not zero, which is why we need to use a different method in such cases.
The proper way to proceed with $e^{2x} + e^x - 2 = 0$ is to first rewrite $e^{2x}$ as $(e^x)^2$. Then the equation becomes
$$ (e^x)^2 + e^x - 2 = 0.$$
Now we can make the substitution $y = e^x$ to get
$$ y^2 + y - 2 = 0.$$
So now it's a plain quadratic equation. Factor to get $(y+2)(y-1) = 0$. Now we can use the zero-product property to say $y = -2$ or $y=1$. Back-substitute to get
$$ e^x = -2 \qquad \text{ or } \qquad e^x = 1.$$
Since $e^x$ can't be negative for any real value of $x$, the only valid case is $e^x = 1$, and this happens when $x = 0$.
|
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|
Prove inequality $ (a+b+c)(a^7+b^7+c^7)\ge(a^5+b^5+c^5)(a^3+b^3+c^3)$ for $a,b,c\ge 0$ Prove that:
$$ (a+b+c)(a^7+b^7+c^7)\ge(a^5+b^5+c^5)(a^3+b^3+c^3)$$
I already know that this can be proven using Cauchy Schwarz, but I don't really see how to apply it here. I'm looking for hints.
|
Expanding the two sides and cancelling like terms gives
$$a^7b+ab^7+b^7c+bc^7+c^7a+ca^7\geq a^5b^3+a^3b^5+b^5c^3+b^3c^5+c^5a^3+c^3a^5.$$
This follows from Muirhead's inequality, where we set $n=2, a_1=5,a_2=3,b_1=7$, and $b_2=1$.
|
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|
Mistake in basic algebra, I think? Problem prove $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2$
Induction proof: base case $n=1$ assume true for all $n$ prove for $n+1$.
The $n$th or last term becomes $(4(n+1)-1)=4n+3$.
We also sub $n+1$ in for all $n$ the $n-1$ term is $(4n-1)$ and the first term is $2(n+1)+1=2n+3$
The right side is $3(n+1)^2 = 3(n^2 + 2n +1 ) $
Next thing is I appear to be missing the first term in the sum $(2n+1)$ on the left.
Adding $(2n+1)$ to both sides and subtracting $4n+3$ from both sides we get the $n$ case that equals $ 3n^2 $ on the left and
$$3n^2 + 6n +3 + 2n+1 -4n -3= 3n^2 + 4n+1$$
Which leaves me with $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2 +4n +1$
Which is approximately $4n+1$ on the right bigger than what I started with and is exactly the same on the left algebraically I must of done something wrong.
Which leads me to my question, what was it?
|
You can try this approach: Calculate sum of first $k$ odd numbers, namely from 1 to $2k-1$.
Denoting it by $S_{k}$. By induction prove this is $k^2$.
Now the desired sum is $S_{2n}-S_n= 4n^2-n^2=3n^2$.
|
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|
Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?
$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$$
|
First, let's make a common denominator of $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta}$.
Camera? Action!
$$\require{cancel}\begin{aligned}\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta}&=\frac{\sin^4{\theta}+\cos^4{\theta}}{\sin^2{\theta}\cos^2{\theta}}\\&=\frac{\left(\sin^2{\theta}+\cos^2{\theta}\right)^2-2\sin^2{\theta}\cos^2{\theta}}{\sin^2{\theta}\cos^2{\theta}}\\&=\frac{1-2\sin^2{\theta}\cos^2{\theta}}{\sin^2{\theta}\cos^2{\theta}}\\&=\frac{1}{\sin^2{\theta}\cos^2{\theta}}-\frac{2\cancel{\sin^2{\theta}\cos^2{\theta}}}{\cancel{\sin^2{\theta}\cos^2{\theta}}}\\&=\frac{1}{\sin^2{\theta}\cos^2{\theta}}-2\\&=\csc^2{\theta}\sec^2{\theta}-2\\&=\sec^2{\theta}\csc^2{\theta}-2\end{aligned}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\binom{k}{p} \equiv \left\lfloor \frac{k}{p} \right\rfloor \pmod p$ for all odd prime number $p$ and $k\ge 3$ I don't really know how to start this exercise. Do I have to use p-adic valuation ?
If it's the case it will give $\nu_p(\binom{k}{p})= \sum \limits_{r=1}^{\infty}\left(\left\lfloor \frac{k}{p^r} \right\rfloor-\left\lfloor \frac{p}{p^r} \right\rfloor -\left\lfloor \frac{k-p}{p^r} \right\rfloor \right)$.
Thanks in advance !
|
I try to explain better my previous comment on induction's argument:
let's suppose $\binom{k}{p}=n$, so that $np\leq k <np+p$;
it's easy to prove $\binom{k+1}{p}=\frac{k+1}{k+1-p}\binom{k}{p}\equiv 1\left\lfloor \frac{k}{p} \right\rfloor\pmod{p}$,
if $k+1\not\equiv 0\pmod p$, and in this case we are done since $k+1<pn+p$ and so $\left\lfloor \frac{k+1}{p} \right\rfloor=\left\lfloor \frac{k}{p} \right\rfloor$.
If otherwise $k+1=p^{r}a,a\not\equiv 0\pmod p$ then $\left\lfloor \frac{k+1}{p} \right\rfloor=p^{r-1}a$ and $\binom{k+1}{p}=\frac{p^{r-1}a}{p^{r-1}a-1}\binom{k}{p}$;
Now if $r\geq 2$ then $\binom{k+1}{p}\equiv 0\pmod p$ and similarly for $\left\lfloor \frac{k+1}{p} \right\rfloor=p^{r-1}a$;
if $r=1$ then $\binom{k+1}{p}=\frac{a}{a-1}\binom{k}{p}$ and its easy to see that $a=n+1$ so that $\binom{k+1}{p}=\frac{n+1}{n}\binom{k}{p}$;
But now if $n\not \equiv 0 \pmod p$ since $n=\left\lfloor \frac{k}{p} \right\rfloor$ and $n+1=\left\lfloor \frac{k+1}{p} \right\rfloor$ we get $\binom{k+1}{p}=\frac{n+1}{n}\binom{k}{p}\equiv \left\lfloor \frac{k+1}{p} \right\rfloor\pmod p$.
Finally if $n=p^{c}m$ with $m\not\equiv 0\pmod p$ then $\left\lfloor \frac{k+1}{p} \right\rfloor\equiv 1\pmod p$ and $\binom{k}{p}=\binom{p^{c+1}m+p-1}{p}=\frac{(p^{c+1}m+p-1)...(p^{c+1}m)}{p(p-1)!}=p^{c}mw$, with $w\equiv 1\pmod p$
and we get the desired equivalence, since $\binom{k+1}{p}=\frac{p^{c}m+1}{p^{c}m}\binom{k}{p}=(p^{c}m+1)w\equiv 1\pmod p$.
|
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|
Is there a general rule for how to write high order polynomials in matrix form?
Is there a general rule for how to write high order polynomials in matrix form?
For example a linear combination of parameters:
$$a_1 x_1+a_2 x_2+a_3 x_3 + \cdots+ a_n x_n$$
Can be written as
$$\sum^n_{i=1} a_i x_i = \vec{a}^T\vec{x} $$
Second order forms are given by
$$ (a_1 x_1+a_2 x_2+a_3 x_3 + \cdots+ a_n x_n)^2 = \vec{x}^T {\mathbf A} \vec{x}$$
Which ensures all combinations of second order terms. What about the higher orders? i.e.
$$(a_1 x_1+a_2 x_2+a_3 x_3 + \dots +a_n x_n)^k$$
What forms ensure all combinations of terms. Is there a general rule to this? Does it have a name?
|
Linear Form
$$
\begin{pmatrix}
a & b \\
\end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}=
ax+by$$
Quadratic Form
$$
\begin{pmatrix} x & y \end{pmatrix}
\begin{pmatrix}
a & b \\
b & c
\end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}=
ax^2+2bxy+cy^2$$
Cubic Form
$$
\begin{pmatrix} x & y \end{pmatrix}
\begin{pmatrix}
ax+by & bx+cy \\
bx+cy & cx+dy
\end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}=
ax^3+3bx^2y+3cxy^2+dy^3$$
Quartic Form
$$
\begin{pmatrix} x & y \end{pmatrix}
\begin{pmatrix}
ax^2+2bxy+cy^2 & bx^2+2cxy+dy^2 \\
bx^2+2cxy+dy^2 & cx^2+2dxy+ey^2
\end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}$$
$m$-tuple Form
$$ \binom{m}{i_{1},i_{2}, \ldots , i_{n}} a_{i_{1} i_{2} \ldots i_{n}} x_{1}^{i_{1}} x_{2}^{i_{2}} \ldots x_{n}^{i_{n}}$$
where $\boldsymbol{x} \in \mathbb{R}^{n}$ and $i_{1}+i_{2}+\ldots+i_{n}=m$
|
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|
Use De Moivre’s formula to show that $\cos^2 20^{o}+\cos^2 40^{o}+\cos^2 60^{o}+\cos^2 80^{o}=\frac{7}{4}$ I would appreciate if somebody could help me with the following problem.
Q: Use De Moivre’s formula to show that
$$
\cos^2 20^{o}+\cos^2 40^{o}+\cos^2 60^{o}+\cos^2 80^{o}=\frac{7}{4}
$$
I tried to let $z=\cos 20^0+i \sin 20^0$ and find $\cos 20^0=\frac{1}{2}\left(z+\frac{1}{z}\right),\sin 20^0=\frac{-i}{2}\left(z-\frac{1}{z}\right)$,
$\cos 40^0=\frac{1}{2}\left(z^2+\frac{1}{z^2}\right),\sin 40^0=\frac{-i}{2}\left(z^2-\frac{1}{z^2}\right),...$ But///
......
|
Set $z=\cos20^{\circ}+i\sin20^{\circ}$. Then,
$$
z^{2} = \cos40^{\circ}+i\sin40^{\circ}, \quad z^{3} = \cos60^{\circ}+i\sin60^{\circ}, \quad z^{4} = \cos80^{\circ}+i\sin80^{\circ}
$$
and
$$
z^{-1} = \cos20^{\circ}-i\sin20^{\circ}, \qquad z^{-2} = \cos40^{\circ}-i\sin40^{\circ}, \quad z^{-3} = \cos60^{\circ}-i\sin60^{\circ}, \quad z^{-4} = \cos80^{\circ}-i\sin80^{\circ}.
$$
Also note that $z^{9} = -1$, so $z^{-n} = -z^{9-n}$.
Setting $S=\cos^{2}20^{\circ} + \cos^{2}40^{\circ} + \cos^{2}60^{\circ} + \cos^{2}80^{\circ}$, we have
\begin{align*}
4S &= (z+z^{-1})^{2} + (z^{2}+z^{-2})^{2} + (z^{3}+z^{-3})^{2} + (z^{4}+z^{-4})^{2}\\
&= 1 + z^{2} + z^{4} + z^{6} + z^{8} + z^{-2} +z^{-4}+ z^{-6} + z^{-8} + 7\\
&= 1 + z^{2} + z^{4} + z^{6} + z^{8} - z^{7} - z^{5} - z^{3} - z + 7\\
&= 1 - z + z^{2} - z^{3} +z^{4} - z^{5} + z^{6} - z^{7} + z^{8} + 7\\
&= \frac{(-z)^{9}-1}{-z-1} + 7 = 7.
\end{align*}
|
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|
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
|
What are the roots of $x^2-2x-3$? They should also be roots in the 3rd degree polynomial.
|
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|
The value of \zeta(4) with fourier series Given a function as follows :
$f(x) = \pi^2 - x ^2$ on $|x|<\pi$ and $f(x+2\pi)=f(x)$
I did find its Fourier expansion
$f(x)=\frac{2}{3}\pi^2$+$\sum_{n \geq 1}\frac{4}{n^2}(-1)^{n+1}cosx$
And by putting $x=\pi$, I got the zeta of 2 , $\zeta(2) = \sum_{n \geq 1}\frac{1}{n^2} = \frac{\pi^2}{6}$
I d like to find the zeta of 4,$\zeta(4) = \frac{\pi^4}{90}$ with fourier expansion. If possible, what is the method ?
If this function $f(x)$ is Not proper, What shlould i use another function? Please, advise me.
I do guess that it may be used Square Error
|
$f(x) = \pi^4-x^4$, $f(x+2 \pi)=f(x)$
$$f(x) = \sum_{n=0}^{\infty} a_n \cos{n x} $$
$$a_0 = \frac1{2 \pi} \int_{-\pi}^{\pi} dx \, (\pi^4-x^4) = \frac{4 \pi^4}{5}$$
$$a_n = \frac1{\pi} \int_{-\pi}^{\pi} dx \, (\pi^4-x^4) \cos{n x} = (-1)^{n+1} \frac{8 (n^2 \pi^2-6)}{n^4}$$
Thus,
$$f(x) = \frac{4 \pi^4}{5} - 8 \sum_{n=1}^{\infty} (-1)^{n} \frac{n^2 \pi^2-6}{n^4} \cos{n x}$$
$$f(\pi) = 0 \implies \frac{4 \pi^4}{5} - 8 \pi^2 \sum_{n=1}^{\infty} \frac1{n^2} + 48 \sum_{n=1}^{\infty} \frac1{n^4} = 0$$
Using
$$\sum_{n=1}^{\infty} \frac1{n^2} = \frac{\pi^2}{6} $$
We may deduce that
$$\sum_{n=1}^{\infty} \frac1{n^4} = \frac{\pi^4}{90} $$
|
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|
Where am I going wrong when attempting to evaluate the integral $\int { 3tan^{ 2 }(x)sec(x)dx } $ Evaluate the integral:
$$\int 3\tan^2(x)\sec(x) \,dx $$
We begin by taking the constant outside of the integral to get:
$$3\int \tan^2(x)\sec(x)\,dx $$
Next, we use the identity $\tan^2 (x)=\sec^2 (x)-1$ in order to get:
$$3\int (\sec^2 (x)-1)\sec(x)\,dx$$
We can now distribute the $\sec(x)$ in order to get:
$$3\int (\sec^3 (x)-\sec(x))\,dx$$
which can then be simplified further into:
$$3\int \sec^3 (x)\,dx -3\int \sec(x)\,dx $$
We know that $\int \sec(x)\,dx $ is equal to $\ln (|\sec { (x) } + \tan { (x) } ) +C$, so we can simplify to:
$$3\int \sec^2 { (x) }\,dx - 3\ln (|\sec(x)+\tan(x)) +C$$
Now we must integrate $3\int \sec^3(x)\,dx$. We can do this by integrating by parts:
We start by letting $f(x)=\sec(x)$, $f'(x)=\tan(x)\sec(x)$, $g(x)=\tan(x)$, $g'(x)=\sec^2(x)$
Then we evaluate the integral:
$$3\int { \sec ^{ 3 }{ (x) } dx } =3\sec { (x) } \tan { (x) } -3\int { \sec { (x) } dx } \\ 3\int { \sec ^{ 3 }{ (x) } dx } =3\sec { (x) } \tan { (x) } -3\int { \sec ^{ 3 }{ (x) } -\sec { (x) } dx } \\ 3\int { \sec ^{ 3 }{ (x) } dx } =3\sec { (x) } \tan { (x) } -3[\int { \sec ^{ 3 }{ (x) } dx } -\int { \sec { (x) } dx } ]\\ 3\int { \sec ^{ 3 }{ (x) } dx } =3\sec { (x) } \tan { (x) } -3\int { \sec ^{ 3 }{ (x) } dx } +3\int { \sec { (x) } dx } \\ 6\int { \sec ^{ 3 }{ (x) } dx } =3\sec { (x) } \tan { (x) } +3\int { \sec { (x) } dx } \\ 6\int { \sec ^{ 3 }{ (x) } dx } =3\sec { (x) } \tan { (x) } +3\ln { (|\sec { (x) } +\tan { (x) } |) } +C\\ \int { \sec ^{ 3 }{ (x) } dx } =\frac { 3\sec { (x) } \tan { (x) } +3\ln { (|\sec { (x) } +\tan { (x) } |) } }{ 6 } +C\\ \int { \sec ^{ 3 }{ (x) } dx } =\frac { \sec { (x) } \tan { (x) } +\ln { (|\sec { (x) } +\tan { (x) } |) } }{ 2 } +C$$
Finally, we can replace the $3\int { \sec ^{ 3 }{ (x) } dx }$ in $3\int { \sec ^{ 3 }{ (x) } dx } -3\ln { (|\sec { (x) } +\tan { (x) } |) } +C$ to get:
$$\frac { \sec { (x) } \tan { (x) } +\ln { (|\sec { (x) } +\tan { (x) } |) } }{ 2 } -3\ln { (|\sec { (x) } +\tan { (x) } |) } +C$$
So, this appears to be the wrong answer, but even with the aid of wolfram alpha, and other online integral calculators,I cannot seem to pinpoint where I went wrong. I would appreciate help/guidance in this matter.
|
It seems that you forgot to multiply $\int\sec^3(x)dx$ by $3$ :
$$\begin{align}&\int 3\tan^2(x)\sec(x) dx\\&=3\int \sec^3(x)dx-3\int \sec (x)dx\\&=\color{red}{3}\cdot \frac {\sec(x)\tan(x)+\ln|\sec(x)+\tan(x)|}{2}-3\ln|\sec(x)+\tan(x)|+C\\&=\frac 32\sec(x)\tan(x)-\frac 32\ln|\sec(x)+\tan(x)|+C\end{align}$$
|
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|
Proof of the inequality $2^{n} < \binom{2n}{n} < 2^{2n}$? As review for a midterm I am asked to prove the inequality:
$2^{n} < \binom{2n}{n} < 2^{2n}, n > 1.$
What I have is a two-part inductive proof. It is not hard to show for $2^{n} < \binom{2n}{n}$:
Base step:
Let $n=2$:
$2^{2} < \frac{(2n)!}{2!2!} < 2^{2*2}$
$4 < 6 < 16$
Inductive Step:
Show $2^{k+1} < \frac{[2(k+1)]!}{(k+1)!(k+1)!}.$
We have
$2^{k+1} = 2^{k}*2,$
so
$2^{k+1} < 2 * \frac{2k!}{k!k!}.$
Since
$\frac{(2k+2)!}{(k+1)*k!*(k+1)*k!} = \frac{2(k+1)*(2k+1)*2k!}{(k+1)k!(k+1)k!} = (2k+1) * \frac{2k!}{k!k!}$
and
$(2k+1) > 2, k \geq 2,$
we can conclude the first part of the inequality. However, I can't make the second part work. By similar algebra I arrive at:
$(2k+1) * \frac{2k}{k!k!} < 2^{2(k+1)} = 2^{2k}*2^{2},$
but
$ (2k+1) \nless 4, k \geq 2.$
What have I done wrong?
|
Another approach. Using the Stirling's approximation we have $$\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}\leq n!\leq n^{n+\frac{1}{2}}e^{1-n}
$$ hence $$\dbinom{2n}{n}=\frac{\left(2n\right)!}{n!^{2}}\leq\frac{e}{\sqrt{2}\pi n^{1/2}}4^{n}<\color{red}{4^{n}}
$$ and $$\frac{\left(2n\right)!}{n!^{2}}\geq\frac{\sqrt{\pi}2^{2n+1}}{n^{1/2}e^{2}}>\color{red}{2^{n}}.$$
|
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|
Find the average rate of change between two points on a contour map Disclaimer: This a homework question for a multivariable calculus course.
The problem:
Find the average rate of change between $A$ and $C$ using the given contour map.
The average rate of change for a contour map is given by $\frac{\Delta altitude}{\Delta horizontal}$.
What I've done:
$\Delta altitude = -9 - (-3) = -6$.
$\Delta horizontal = \sqrt{(6-2)^2 + (5-4)^2} = \sqrt{17}$
Therefore, the average rate of change between $A$ and $C$ should be $\frac{-6}{\sqrt{17}}$. However, according to the answer key, the average rate of change is $\frac{-9-(-3)}{\sqrt{2^2 + 1^2}} = \frac{-6}{\sqrt{5}}$.
Did I do something wrong along the way, or is the answer key wrong?
|
The textbook answer is incorrect. The correct solution is:
$\Delta altitude = -9 - (-3) = -6$.
$\Delta horizontal = \sqrt{(6-2)^2 + (5-4)^2} = \sqrt{17}$
$\frac{\Delta altitude}{\Delta horizontal}=\frac{-6}{\sqrt{17}}=\frac{-6\sqrt{17}}{17}$.
|
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|
integral points on a simple elliptic curve I'm trying to find all integer points on an elliptic curve $y^2=x(x^2-2x-3)$ or $y^2=x(x+1)(x-3) $. I guess there are only 3 integer points: $(-1,0), (0,0), (3,0) $. It looks like quite an elementary number theory problem, but I couldn't think up any proof... How could I prove it, or disprove it? Please give me at least some hints.
|
Here's an elementary argument. Suppose none of $x$, $x+1$, or $x-3$ is $0$, and that their product is a perfect square. Note that for any prime $p\ge5$, $p$ can divide at most one of $x$, $x+1$, and $x-3$. Thus the only primes that could have an odd power in the prime factorization of any of these three numbers are $2$ and $3$, so each of them has the form $\pm a^2$, $\pm 2a^2$, $\pm 3a^2$, or $\pm 6a^2$ (where $a$ is an integer). Let us say a number has a "spare factor" of a prime if that prime appears with an odd exponent in its prime factorization. In order for $x(x+1)(x-3)$ to be a perfect square, only $x$ or $x-3$ can have a spare factor of $3$ (in which case both of them do), and only $x+1$ or $x-3$ can have a spare factor of $2$ (in which case both of them do).
Let's now consider cases based on whether there are spare factors of $2$ or $3$. If there are no spare factors of $2$ or $3$, then all three numbers are of the form $\pm a^2$. This is impossible: $x$ and $x+1$ can't both be of this form unless one of them is $0$.
If there are spare factors of $2$ but not spare factors of $3$, then $x+1$ and $x-3$ both have the form $\pm2a^2$. Dividing by $2$, we get two numbers of the form $\pm a^2$ with a difference of $2$, which is only possible if they are $-1$ and $1$. This means $x+1=2$ and so $x=1$. But if $x=1$ then $x(x+1)(x-3)=-4$, which is not a perfect square.
If there are spare factors of $3$ but not spare factors of $2$, then $x$ and $x-3$ both have the form $\pm 3a^2$. Dividing by $3$, we get numbers of the form $\pm a^2$, which is impossible since it would imply $x$ or $x-3$ is $0$.
Finally, suppose there are both spare factors of $3$ and spare factors of $2$. Note that $x$ must be $1$ mod $4$ (if it were $3$ mod $4$, then one of $x+1$ and $x-3$ would be divisible by $4$ and not $8$, and thus would not have a spare factor of $2$). Also, $x$ must have the form $\pm 3a^2$ since it has a spare factor of $3$. Now no number of the form $3a^2$ can be $1$ mod $4$, so $x$ must actually have the form $-3a^2$. But if $a\neq 0$, this would imply all three of $x$, $x+1$, and $x-3$ are negative, so their product is negative and not a square.
|
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|
If $\cos A=\tan B$, $\cos B=\tan C$ … If $\cos A=\tan B$, $\cos B=\tan C$ and $\cos C=\tan A$, prove that $\sin A=\sin B=\sin C$.
My Attempt.
Let us consider $x$, $y$ and $z$ as:.
$$x = \tan^2A$$
$$y = \tan^2B$$
$$z = \tan^2C$$
$$\cos^2A = \tan^2B$$
$$\frac {1}{\sec^2A}= \tan^2B$$
$$\frac {1}{1 + \tan^2A} = \tan^2B$$
$$\frac {1}{1 + x} = y$$
$$(1 + x)y = 1\tag{1}$$
Similarly,
$$(1 + y)z = 1\tag{2}$$
$$(1 + z)x = 1\tag{3}$$
Please help me to continue from here.
|
Define $a := \cos^2 A$, and $b := \cos^2 B$, and $c := \cos^2 C$. Then
$$\cos A = \tan B \quad\to\quad \cos^2 A = \tan^2 B = \sec^2 B - 1 \quad\to\quad a = \frac{1}{b} - 1 \quad\to\quad b = \frac{1}{1+a}$$
Likewise,
$$c = \frac{1}{1+b} \qquad\text{and}\qquad a = \frac{1}{1+c}$$
so that, adding a slightly-gratuitous $1$,
$$1+a \;=\; 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+a}}} \\ \tag{$\star$}$$
The evident recursion reveals that $1+a$ (and also $1+b$ and $1+c$) is the "$1$s all the way down" continued fraction, which some will recognize as representing the Golden Ratio, $\phi := \frac{1}{2}(1+\sqrt{5}) = 1.618\ldots$. Consequently,
$$a = b = c = \phi - 1 = \phi^{-1}$$
We finally calculate
$$\sin^2 A = \sin^2 B = \sin^2 C = 1 - \phi^{-1} = \phi^{-2}$$
so that
$$|\sin A| = |\sin B| = |\sin C| = \phi^{-1} = 0.618\ldots$$
That the signs of the sines match is left as an exercise to the reader.
Observe that we'd reach the same conclusion no matter the length of the "loop" of equations. Once $1+a$ appears, at any stage, in the right-hand-side of the counterpart of $(\star)$, the implied continued fraction collapses to the simpler form
$$1 + a = 1 + \frac{1}{1+a} \quad\text{, giving}\quad a = \frac{1}{1+a}$$
which represents the one-equation loop, $\cos A = \tan A$.
|
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|
How to prove $\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$? How to prove this identity?
$$\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$$
Is this a particular case of a more general identity? Also, is it possible to give a geometric proof of this equality?
|
If $(4\cos2A+1)^2=9-8\cos A,$
$9-8\cos A=16\cos^22A+8\cos2A+1=8(1+\cos4A)+8\cos2A+1$
$\iff0=\cos A+\cos2A+\cos4A$
$=\cos A+2\cos A\cos3A=\cos A(1+2\cos3A)$
If $\cos A=0,A=(2n+1)90^\circ$ where $n$ is any integer
Otherwise, $$\cos3A=-\dfrac12=\cos120^\circ$$
$\implies3A=360^\circ m\pm120^\circ$ where $m$ is any integer
The set of values of $A$ can be chosen as $\{40^\circ,80^\circ,160^\circ\}$
For $A=40^\circ,160^\circ;\cos2A>0\implies4\cos2A+1>0$
Consequently, $4\cos2A+1=+\sqrt{9-8\cos A}$
For $A=80^\circ,\cos2A=-\cos20^\circ;$
$4\cos2A+1=1-4\cos20^\circ$ which is $<0$ as $\cos20^\circ>\cos60^\circ=\dfrac12>\dfrac14$
Consequently, $4\cos2A+1=-\sqrt{9-8\cos A}$
|
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|
Inequality $\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}\geq1+ \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ for positive $a$, $b$, $c$
If $A=\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}$ and $B = \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ and $a,b,c>0.$ Then prove that $A\geq 1+B$
$\bf{My\; Try::}$We can write $A$ and $B$ as $$=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} = \frac{3}{A}$$ and $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}= \frac{3}{B}$$
Using $\bf{cauchy \; schwarz }$ Inequality
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \geq \frac{3^2}{1+a+1+b+1+c} = \frac{9}{3+a+b+c}$$
Now How can i solve after that , Help Required, Thanks
|
we need to prove that
$$\frac{3(1+a)(1+b)(1+c)}{\sum\limits_{cyc}(ab+2a+1)}\geq1+\frac{3abc}{ab+ac+bc}$$ or
$$\sum\limits_{cyc}(2a^2b^2-2a^2bc+a^2b+a^2c-2abc)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2(c^2+c)\geq0$$
Done!
|
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|
Why is cross product defined in the way that it is?
$\mathbf{a}\times \mathbf{b}$ follows the right hand rule? Why not left hand rule? Why is it $a b \sin (x)$ times the perpendicular vector? Why is $\sin (x)$ used with the vectors but $\cos(x)$ is a scalar product?
So why is cross product defined in the way that it is?
I am mainly interested in the right hand rule defintion too as it is out of reach?
|
As far as $\sin \theta$ and $\cos \theta$ are concerned,
Using the law of cosines,
\begin{align}
\|\overrightarrow{v_2}\|^2 + \|\overrightarrow{v_1}\|^2
-2\|\overrightarrow{v_2}\| \, \|\overrightarrow{v_1}\| \cos \theta
&= \|\overrightarrow{v_2} - \overrightarrow{v_1}\|^2
\\
2x_1 x_2 + 2y_1 y_2 + 2z_1 z_2
&= 2\|\overrightarrow{v_2}\|\, \|\overrightarrow{v_1}\| \cos \theta
\\
\cos \theta &= \dfrac{x_1 x_2 + y_1 y_2 + z_1 z_2}
{\|\overrightarrow{v_2}\| \, \|\overrightarrow{v_1}\|}
\\
\cos \theta &= \dfrac{\overrightarrow{v_2} \circ \overrightarrow{v_1}}
{\|\overrightarrow{v_2}\| \, \|\overrightarrow{v_1}\|}
\end{align}
And so, $\overrightarrow{v_1} \circ \overrightarrow{v_2}
= \|\overrightarrow{v_2}\| \, \|\overrightarrow{v_1}\| \cos \theta$
Note that $\sin \theta$ is non negative for all $0 \le \theta \le \pi$.
So
\begin{align}
\sin^2 \theta &= 1 - \cos^2 \theta \\
&= 1 - \dfrac{(x_1 x_2 + y_1 y_2 + z_1 z_2)^2}
{\|\overrightarrow{v_2}\|^2 \, \|\overrightarrow{v_1}\|^2}
\\
&= \dfrac{\|\overrightarrow{v_2}\|^2 \, \|\overrightarrow{v_1}\|^2
- (x_1 x_2 + y_1 y_2 + z_1 z_2)^2}
{\|\overrightarrow{v_2}\|^2 \, \|\overrightarrow{v_1}\|^2}
\\
&= \dfrac{x_2^2 y_1^2 - 2x_1 x_2 y_2 y_1 +
x_1^2 y_2^2 + x_2^2 z_1^2 +x_1^2 z_2^2 - 2 x_1 x_2 z_1 z_2
+y_1^2 z_2^2-2 y_2 y_1 z_1 z_2+y_2^2 z_1^2}
{\|\overrightarrow{v_2}\|^2 \, \|\overrightarrow{v_1}\|^2}
\\
&= \dfrac{(-y_2 z_1+y_1 z_2)^2+(x_2 z_1-x_1 z_2)^2+(-x_2 y_1+x_1 y_2)^2}
{\|\overrightarrow{v_2}\|^2 \, \|\overrightarrow{v_1}\|^2}
\\
&= \dfrac
{\|\overrightarrow{v_1} \times \overrightarrow{v_2} \|^2}
{\|\overrightarrow{v_2}\|^2 \, \|\overrightarrow{v_1}\|^2}
\end{align}
And we conclude, for purely formal reasons,
$\|\overrightarrow{v_1} \times \overrightarrow{v_2} \|
= \|\overrightarrow{v_2}\| \, \|\overrightarrow{v_1}\| \sin \theta$
|
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"url": "https://math.stackexchange.com/questions/1941044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ How does one evaluate the following limit?
$$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$
The answer is $6$.
How does one justify this answer?
Edit: So it really was just combine the fraction and use L'hopital's rule twice (because function and its first derivative are of indeterminate form at $x=1$). This problem is more straightforward than it seems at first.
|
This answer does not use L'Hopital (personal taste), only a standard identity restated below, the binomial theorem, and a straightforward Taylor expansion to first order at $0$.
Using the identity $1-x^{2n+1} = (1-x)\sum_{k=0}^{2n} x^k$, we can rewrite
$$\begin{align*}
\frac{23}{1-x^{23}} - \frac{11}{1-x^{11}}
&= \frac{1}{1-x}\left(\frac{23}{\sum_{k=0}^{22}x^k} - \frac{11}{\sum_{k=0}^{10}x^k} \right)\\
&= \frac{1}{1-x}\left(\frac{23\sum_{k=0}^{10}x^k}{\sum_{k=0}^{22}x^k\sum_{k=0}^{10}x^k} - \frac{11\sum_{k=0}^{22}x^k}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k} \right)\\
&= \frac{1}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k}\cdot\frac{1}{1-x}\left(23\sum_{k=0}^{10}x^k - 11\sum_{k=0}^{22}x^k \right)\\
\end{align*}$$
Let us focus on the parenthesis (the first factor converges to $\frac{1}{11\cdot 23}$ by continuity, the second is the problematic one that will be "offset" by the parenthesis).
Writing $x=1+h)$ (where we will have $h\to 0$), we get, for any fixed integer $n$,
$$\begin{align*}
\sum_{k=0}^{n}x^k
&= \sum_{k=1}^{n}(1+h)^k
= \sum_{k=1}^{n} \sum_{\ell=0}^k \binom{k}{\ell} h^\ell \\
&= \sum_{k=0}^{n}(1+kh +o(h)) \\
&= (n+1)+\frac{n(n+1)}{2}h +o(h)
\end{align*}$$
when $h\to 0$, as $n$ is a constant. In particular, this implies
$$\begin{align*}
23\sum_{k=0}^{10}x^k - 11\sum_{k=0}^{22}x^k
&= 23\cdot 11+23\cdot \frac{11\cdot10}{2}h - 11\cdot 23-11\cdot \frac{22\cdot 23}{2}h + o(h)\\
&= 23\cdot 11\cdot (-6h) + o(h)\\
&= 23\cdot 11\cdot 6(1-x) + o(1-x)
\end{align*}$$
Overall, we thus have
$$\begin{align*}
\frac{23}{1-x^{23}} - \frac{11}{1-x^{11}}
&= \frac{23\cdot 11}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k}\cdot\frac{6(1-x)+o(1-x)}{1-x} \\
&= \frac{23\cdot 11}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k}\cdot (6+o(1)) \xrightarrow[x\to1]{} \frac{23\cdot 11}{23\cdot 11} \cdot 6 = 6
\end{align*}$$
as claimed.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Trigonometry find area of triangle Triangle ABC is right angeled with right angle at corner C and angle a at corner A. Calculate triangle area if we know that c=|AB|=10 and tan a = 3/2
I get side |AC| = 2
side |CB| = 3
and side |AB| = 10 but this can not be case because it's not pythagoras definition then
Greetings from RUssia
Vladimir Vlostok
|
From the tangent of $a$, you know that $\frac{BC}{AC} = \frac{3}{2}$.
So let $BC = 3x$ and $AC = 2x$
By Pythagoras, $AC^2 + BC^2 = AB^2 \implies 13x^2 = 100 \implies x^2 = \frac{100}{13}$
Area of triangle = $\frac{1}{2} AC \cdot BC = 3x^2 = \frac{300}{13}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1946426",
"timestamp": "2023-03-29T00:00:00",
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|
Fourier series of $f(x)=\pi-x$ $$f(x)=\pi-x \qquad x \in [0,2 \pi[$$
$$a_0=\frac{1}{\pi} \ \int_0^{2 \pi}f(x) \ dx=\frac{1}{\pi} \ \int_0^{2 \pi} (\pi-x) \ dx=0$$
$$a_n=\frac{1}{\pi} \ \int_0^{2 \pi} \cos(nx) \ dx=\frac{1}{\pi} \ \int_0^{2 \pi}(\pi \ \cos(nx)-x \ \cos(nx)) \ dx=$$
$$\frac{1}{\pi} \ \Big( \ \Big[\frac{\pi}{n} \ \sin(nx) \Big]_0^{2 \pi}-\Big[\frac{x}{n} \ \sin(nx)+\frac{1}{n^2} \ \cos(nx) \Big]_0^{2 \pi} \Big)=0 $$
$$b_n=\frac{1}{\pi} \ \Big( \ \Big[\ -\frac{\pi}{n} \ \cos(nx) \Big]_0^{2 \pi}-\Big[-\frac{x}{n} \ \cos(nx)+\frac{1}{n^2} \ \sin(nx) \Big]_0^{2 \pi} \Big)=\frac{2}{n}$$
$$f(x)=2 \sum_{n=1}^{+\infty} \frac{1}{n} \ \sin(nx)$$
Is it correct?
|
Slight errors:
$$
\begin{align}
%
a_{0} &= \frac{1}{\pi} \int_{-\pi }^{\pi } f(x) \, dx = 2\pi \\[5pt]
%
b_{k} &= \frac{1}{\pi } \int_{-\pi }^{\pi } f(x) \sin (k x) \, dx = (-1)^k \frac{2}{k \pi}
%
\end{align}
$$
The decay of the amplitudes is linear:
The series expansion looks like
$$
\pi - x = \pi - 2 \sin (x) + \sin (2x) - \frac{2}{3} \sin (3x) + \frac{1}{2} \sin \left( 4x \right) - \dots
$$
Convergence sequence:
|
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|
Simplify this expression fully How would i simplify fully the following expression?
$\dfrac{{\sqrt 2}({x^3})}{\sqrt{\frac {32}{x^2}}}$
So far i have got this
$\dfrac{{\sqrt 2}{x^3}}{{\frac{\sqrt 32}{\sqrt x^2}}}$ = $\dfrac{{\sqrt 2}{x^3}}{{\frac{4\sqrt 2}{x}}}$
Am not quite sure if this is correct however, could someone help explain how i would simplify this expression?
|
$\frac{\sqrt{2}x^3}{\sqrt{\frac{32}{x^2}}} = \frac{\sqrt{2}x^3}{\frac{\sqrt{32}}{\sqrt{x^2}}} = \frac{\sqrt{2}x^3}{\frac{4\sqrt{2}}{x}} = \frac{x^3}{\frac{4}{x}} = \frac{x^4}{4}$
|
{
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"url": "https://math.stackexchange.com/questions/1949436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Limit $ \lim_{x \to 1} \frac{x^{\frac{1}{3}}-1}{\sqrt{x}-1}$ $ \lim_{x \to 1} \frac{x^{\frac{1}{3}}-1}{\sqrt{x}-1}$
How would I solve this limit question? Thanks in advance.
|
Another possible way to do it.
Define $x=1+y$ which makes $$A=\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\frac{\sqrt[3]{1+y}-1}{\sqrt{1+y}-1}$$ and use the generalized binomial theorem or Taylor series which write $$(1+y)^a=1+a y+\frac{1}{2} a(a-1) y^2+O\left(y^3\right)$$ which then gives $$A=\frac{1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)-1 } {1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)-1 }=\frac{\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right) }{\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right) }=\frac{\frac{1}{3}-\frac{y}{9}+O\left(y^2\right)} {\frac{1}{2}-\frac{y}{8}+O\left(y^2\right) }$$ Now, using long division $$A=\frac{2}{3}-\frac{y}{18}+O\left(y^2\right)$$ which shows the limit and how it is approached.
Making the problem more general, consider $$B=\frac{\sqrt[p]{x}-1}{\sqrt[q]{x}-1}$$ and the same procedure would give $$B=\frac{q}{p}+\frac{ (q-p)}{2 p^2}y+O\left(y^2\right)$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
The system of equations: $\begin{cases} 2x^2=4y^2+3z^2+2; \\ 13x=4y+3z+29 \end{cases}$ Solve in positive integers the system of equations:
$$\begin{cases}
2x^2=4y^2+3z^2+2;
\\
13x=4y+3z+29
\end{cases}$$
My work so far:
I used wolframalpha: $x=3,y=1,z=2$.
|
$$2x^2=4y^2+3z^2+2\tag1$$
$$13x=4y+3z+29\tag2$$
(1)
$$2x^2=4y^2+3z^2+2 \Rightarrow x>\max\{y,z\}$$
(2)
$$13x=4y+3z+29<4x+3x+29 \Rightarrow 6x<29 \Rightarrow x<5$$
(1) $z -$ even. (2) $x -$ odd. Hence $x \in \{1,3\}$. But $x>y,z\ge1$. Hence $$x=3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1953480",
"timestamp": "2023-03-29T00:00:00",
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|
Find all functions $f:\mathbb Z \rightarrow \mathbb Z$ such that $f(0)=2$ and $f\left(x+f(x+2y)\right)=f(2x)+f(2y)$
Find all functions $f:\mathbb Z \rightarrow \mathbb Z$ such that $f(0)=2$ and
$$f\left(x+f(x+2y)\right)=f(2x)+f(2y)$$
for all $x \in \mathbb Z$ and $y \in \mathbb Z$
My work so far:
1) $x=0$ $$f\left(f(2y)\right)=f(2y)+2$$
2) $y=0$ $$f\left(x+f(x)\right)=f(2x)+2$$
3) Let $n\ge 0$. Use induction we have
If $f(2n)=2n+2$ then $$f(f(2n))=f(2n+2)=2n+2+2=2n+4$$
Hence, if $k=2m\ge0$ then $f(k)=k+2$
4) $n<0$
I need help here
|
Using your results, we find that
$$\tag1f(x+f(x+2y))=2x+2y+4\qquad\text{for }x,y\ge0 $$
In particular,
$$\tag2 f(x+f(x))=2x+4\qquad\text{for }x\ge0$$
Let $S=\{\,k\in\Bbb Z\mid f(2k)=2k+2\,\}$. You essentially showed that $k\in S\implies k+1\in S$ and hence from the given $0\in S$, we have $\Bbb N_0\subseteq S$.
With $x=-2y$ we have
$$ \tag3f(-2y+2)=f(-4y)+f(2y)\qquad\text{for }y\in\Bbb Z$$
Thus if two of $1-y,-2y,y$ are in $S$, then tso is the third.
In particular, for $y<0$ we already known $1-x,-2y\in S$; we conclude
$S=\Bbb Z$, i.e.,
$$\tag4f(x)=x+2\qquad \text{for }x\in2\Bbb Z$$
and hence
$$\tag{1'}f(x+f(x+2y))=2x+2y+4\qquad\text{for }x,y\in\Bbb Z $$
and in particular
$$\tag{2'} f(x+f(x))=2x+4\qquad\text{for }x\in\Bbb Z$$
Assume that for some odd $x=2n+1$ the value $f(x)=2m$ is even.
Then for $k\in\Bbb Z$
$$\begin{align}f(1+2k)&=f\bigl(1+2(k-m)+f(2n+1)\bigr)\\
&=f\Bigl((1+2k-2m)+f\bigl((1+2k-2m)+2(n-k+m)\bigr)\Bigr)\\
&= 2(1+2k-2m)+2(n-k+m)+4\\
&=6+2k-2m+2n\end{align}$$
and so with the odd constant $c:=2n-2m+5$
$$f(x)=x+c\qquad\text{for }x\in 2\Bbb Z+1$$
Then $6=f(1+f(1))=f(2+c)=2+2c$ implies $c=2$, contradicting that $c$ is odd.
Therefore $f(x)$ is odd for all odd $x$. But then $x+f(x)$ is even and from $(4)$, we get $2x+4=f(x+f(x))=x+f(x)+2$ and so
$$f(x)=x+2 $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The function $f(n)$ is defined for all integers $n$, such that $f(x) + f(y) = f(x + y) - 2xy - 1$ for all integers $x$ and $y$ and $f(1) = 1$ The function $f(n)$ is defined for all integers $n$, such that $f(x) + f(y) = f(x + y) - 2xy - 1$ for all integers $x$ and $y$ and $f(1) = 1$. Find $f(n)$.
I started plugging small values in and I got:
$$f(1)=1$$
$$f(2)=5$$
$$f(3)=11$$
I don't see any pattern so far, and I don't know another way to solve this question. Solutions are greatly appreciated!
|
We find the first few values of $f(n)$.
Setting $x = y = 0$, we get $2f(0) = f(0) - 1$, so $f(0)= -1$.
Setting $y = 1$, we get
[f(x) + f(1) = f(x + 1) - 2x - 1,]
so $f(x + 1) = f(x) + 2x + f(1) + 1 = f(x) + 2x + 2$ for all integers $x$.
Then
\begin{align*}
f(2) &= f(1) + 2 \cdot 1 + 2, \\
f(3) &= f(2) + 2 \cdot 2 + 2 = 2 \cdot (1 + 2) + 2 \cdot 2 + f(1), \\
f(4) &= f(3) + 2 \cdot 3 + 2 = 2 \cdot (1 + 2 + 3) + 3 \cdot 2 + f(1), \\
f(5) &= f(4) + 2 \cdot 4 + 2 = 2 \cdot (1 + 2 + 3 + 4) + 4 \cdot 2 + f(1),
\end{align*}
so for any integer $n \ge 1$,
\begin{align*}
f(n) &= 2 \cdot [1 + 2 + \dots + (n - 1)] + (n - 1) \cdot 2 + f(1) \\
&= n(n - 1) + 2(n - 1) + 1 \\
&= n^2 + n - 1.
\end{align*}
Now we must find $f(n)$ when $n$ is a negative integer. Let $n$ be a positive integer. Setting $x = n$ and $y = -n$ in the given functional equation, we get
[f(n) + f(-n) = f(0) + 2n^2 - 1 = 2n^2 - 2.]
Then
\begin{align*}
f(-n) &= -f(n) + 2n^2 - 2 \\
&= -(n^2 + n - 1) + 2n^2 - 2 \\
&= n^2 - n - 1 \\
&= (-n)^2 + (-n) - 1.
\end{align*}
Therefore, $f(n) = \boxed{n^2 + n - 1}$ for all integers $n$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the matrix equation $\sin(X)=\begin{pmatrix}1 & a\\0 & 1\end{pmatrix}$ Solve the equation $\sin(X)=\begin{pmatrix}1 & a\\0 & 1\end{pmatrix}$, where $X\in M_2(\Bbb C)$ and $a\in \Bbb C$.
I discussed the case whether $X$ is diagonalisable or not.
If $X$ is diagonalisable, we have $X\sim \begin{pmatrix}x & 0\\0 & y\end{pmatrix}$, thus $\begin{pmatrix}1 & a\\0 & 1\end{pmatrix}\sim \begin{pmatrix}\sin(x) & 0\\0 & \sin(y)\end{pmatrix}$. There is no solution if $a\neq 0$.
If $X$ is not diagonalisable and then at least triangularizable, we have $X\sim \begin{pmatrix}x & y\\0 & x\end{pmatrix}$, thus $\begin{pmatrix}1 & a\\0 & 1\end{pmatrix}\sim \begin{pmatrix}\sin(x) & y\cos(x)\\0 & \sin(x)\end{pmatrix}$. There is no solution if $a\neq 0$.
Any error?
|
As you correctly infer, there is no diagonalizable solution $X$.
Let $A = \left(\begin{smallmatrix}1&a\\0&1 \end{smallmatrix} \right)$.
Notably, we compute
$$
\sin\pmatrix{\lambda&1\\0&\lambda} = \pmatrix{\sin\lambda &\cos\lambda \\0&\sin \lambda}=:M(\lambda)
$$
and so, $\sin(X) = A$ has a solution if and only if there exists a $\lambda$ such that $A \sim M(\lambda)$.
Now, since $A$ has eigenvalue $1$, it must be that $\lambda = (2k + 1)\pi$ for some $k \in \Bbb Z$. However, this would mean that $\cos \lambda = 0$, which means that $M(\lambda) \sim I \nsim A$.
So, there is indeed no solution.
|
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|
Tricky Limits Problem () Problem:
If $\lim_{x \to 0}{\sin2x\over x^3}+a+{b\over x^2}=0$ then find the value of $3a+b$.
My attempt: $\lim_{x \to 0}{\sin2x\over x^3}+a+{b\over x^2}=\lim_{x \to 0}{\sin2x\over 2x}({2\over x^2})+a+{b\over x^2}={2+b+ax^2\over x^2}$.From this we can conclude that $a=0$ and $b=-2$, hence $3a+b=-2$. However the answer is $2$. Where am I going wrong?
|
We have that
\begin{align}
\lim_{x \rightarrow 0} \frac{\sin 2x}{x^3} + a + \frac{b}{x^2}
& = \lim_{x \rightarrow 0} \ a + \frac{\sin 2x + bx}{x^3} \\
& = \lim_{x \rightarrow 0} \ a + \frac{2 \cos 2x + b}{3x^2},
\end{align}
which gives that $b=-2$, otherwise the second term blows up. Continuing,
\begin{align}
\lim_{x \rightarrow 0} \ a + \frac{2 \cos 2x -2 }{3x^2}
& = \lim_{x \rightarrow 0} \ a + \frac{-4 \sin 2x}{6x} \\
& = \lim_{x \rightarrow 0} \ a + \frac{-8 \cos 2x}{6} \\
& = \lim_{x \rightarrow 0} \ a - \frac{8}{6}.
\end{align}
Hence $a = 8/6$ and $3a+b = 2$.
|
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|
How do you find the area of a parallelogram with the vertices? How do you find the area of a parallelogram with the following vertices; $A(4,2)$, $B(8,4)$, $C(9,6)$ and $D(13,8)$.
|
For this, we plan to use the Shoelace formula.
Shoelace Formula: Given the coordinates of vertices of a polygon, its area is found by $$A=\frac 12\left|\sum_{i=1}^{n-1}x_iy_{i+1}+x_ny_1-\sum_{i=1}^{n-1}x_{i+1}y_i-x_1y_n\right|$$
Or, in other words, we have $$A=\frac 12|x_1y_2+x_2y_3+\ldots x_{n-1}y_n+x_ny_1-x_2y_1-x_3y_2-\ldots -x_ny_{n-1}-x_1y_n|$$
Where $A$ is the area of the polygon, and $(x_i,y_i)$ with $i=1,2,3\dots$ are the vertices of the polyon
So with your case, the vertices are $A(4,2), B(8,4), C(9,6)$ and $D(13,8)$. We let $x_1=13,y_1=8,x_2=9,y_2=6,x_3=4,y_3=2,x_4=8,y_4=4$ and the area is given by $$A=\frac 12|13\cdot 6+9\cdot 2+4\cdot 4+8\cdot 8-9\cdot 8-4\cdot 6-8\cdot 2-13\cdot 4|\\=\frac 12\cdot 12=6$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all functions $f(x)$ such that $f\left(x^2+f(y)\right)$=$y+(f(x))^2$ Let $\mathbb R$ denote the set of all real numbers. Find all function $f: R\to \ R$ such that $$f\left(x^2+f(y)\right)=y+(f(x))^2$$ It is the problem. I tried to it by putting many at the place of $x$ and $y$ but I can't proceed. Please somebody help me.
|
Observe
*
*$f(x^2+f(0)) = [f(x)]^2$
*$f(f(x)) = x+[f(0)]^2$.
Plugging $0$ into expression 1 yields
\begin{align}
f_2(0):=f(f(0)) = [f(0)]^2
\end{align}
and plugging $f(0)$ into expression 1 yields
\begin{align}
f(f(0)^2+f(0)) =[f_2(0)]^2 = [f(0)]^4.
\end{align}
Moreover, by expression 2, we have
\begin{align}
f(f(0)^2+f(0))= f([f(0)]^2)=f(f_2(0)) = f_3(0).
\end{align}
Hence we have
\begin{align}
f_3(0) = [f(0)]^4.
\end{align}
Next, using both 1 and 2, we have
\begin{align}
f(f(x^2+f(0)))=x^2+f(0)+[f(0)]^2. \ \ (\ast)
\end{align}
Plugging $0$ into $(\ast)$ yields
\begin{align}
f_3(0) = f(0)+[f(0)]^2.
\end{align}
Combining everything yields
\begin{align}
[f(0)]^4= f_3(0) = f(0)+[f(0)]^2 \ \ \Rightarrow \ \ f(0)[f(0)^3-f(0)-1] =0.
\end{align}
Solving for $f(0)$ yields that either $f(0) = 0$ or $f(0)$ is a root of the polynomial $p(x) = x^3-x-1$, which only has one real root.
Moreover, observe
\begin{align}
f_4(0) = [f(0)]^4
\end{align}
and
\begin{align}
f_4(0) = f(f(f_2(0))) = f_2(0) + [f(0)]^2 = 2[f(0)]^2
\end{align}
which means
\begin{align}
[f(0)]^4-2[f(0)]^2 = 0.
\end{align}
Thus, $f(0)$ is also a root of $x^4-2x^2 = x^2(x^2-2)$.
In conclusion, $f(0)$ has to be $0$. Thus, $f(f(x)) = x$ and $f(x^2) = [f(x)]^2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Verify that $\int_0^{\pi/2} \frac{{\rm d}\theta}{(1-m^2\cos^2{\theta})^2}= \frac{(2-m^2)\pi}{4(1-m^2)^{3/2}}$ for $0I am told that the integral
\begin{align}
I&=\int_0^{\pi/2} \frac{1}{(1-m^2\cos^2{\theta})^2}\,{\rm d}\theta\\
&= \frac{(2-m^2)\pi}{4(1-m^2)^{3/2}}
\end{align}
Where $0<m<1$. I want to verify this however I have no idea on how this is done. Can someone explain?
I was thinking possibly a substitution so I tried $x=m\cos\theta$ however this makes the problem more complicated so obviously it's wrong.
|
Let $t = \tan \theta = \frac {\sin \theta} {\cos \theta}$. Combining this with the identity $\cos^2 \theta + \sin^2 \theta = 1$ and keeping in mind that $\theta \in [0, \frac \pi 2]$ (where $\cos \theta \ge 0$), we obtain $\cos^2 \theta = \frac 1 {1+t^2}$. With this substitution the integral becomes
$$\int \limits _0 ^\infty \frac 1 {(1 - m^2 \frac 1 {1 + t^2})^2} \frac 1 {1+t^2} \ \Bbb d t = \int \limits _0 ^\infty \frac {t^2 + 1} {(t^2 + 1 - m^2)^2} \ \Bbb d t = \int \limits _0 ^\infty \frac 1 {t^2 + 1 - m^2} \ \Bbb d t + \int \limits _0 ^\infty \frac {m^2} {(t^2 + 1 - m^2)^2} \ \Bbb d t .$$
The first one, by the change of variable $t = \sqrt {1-m^2} u$, becomes
$$\frac 1 {\sqrt {1-m^2}} \int \limits _0 ^\infty \frac 1 {1+u^2} \ \Bbb d u = \frac 1 {\sqrt {1-m^2}} \arctan u \big| _0 ^\infty = \frac \pi 2 \frac 1 {\sqrt {1-m^2}} .$$
With the same change of variable, the second one becomes
$$\frac {m^2} {\sqrt {1-m^2} ^3} \int \limits _0 ^\infty \frac 1 {(1+u^2)^2} \ \Bbb d u .$$
For this last integral, perform the change of variables $u = \tan s$, obtaining
$$\frac {m^2} {\sqrt {1-m^2} ^3} \int \limits _0 ^\frac \pi 2 \frac 1 {(1 + \tan^2 s)^2} \frac 1 {\cos^2 s} \Bbb d s = \frac {m^2} {\sqrt {1-m^2} ^3} \int \limits _0 ^\frac \pi 2 \frac 1 {\frac 1 {\cos^4 s}} \frac 1 {\cos^2 s} \Bbb d s = \frac {m^2} {\sqrt {1-m^2} ^3} \int \limits _0 ^\frac \pi 2 \cos^2 s \ \Bbb d s = \\
\frac {m^2} {\sqrt {1-m^2} ^3} \int \limits _0 ^\frac \pi 2 \frac {1 + \cos 2s} 2 \ \Bbb d s = \frac \pi 4 \frac {m^2} {\sqrt {1-m^2} ^3} .$$
Adding the two integrals, we get
$$\frac \pi 4 \frac 1 {\sqrt {1 - m^2}} \left( 2 + \frac {m^2} {1-m^2} \right) = \frac \pi 4 \frac {2-m^2} {(1-m^2) ^\frac 3 2} .$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\int^{\pi}_0 \ln \left(\frac{b-\cos x}{a- \cos x}\right)dx=\pi \ln (\frac{b+\sqrt{b^2-1}}{a+\sqrt{a^2-1}})$ Given for $\alpha>1$
$$\int^{\pi}_0 \frac{dx}{\alpha-\cos x}=\frac{\pi}{\sqrt{\alpha^2-1}}$$
Prove that $$\int^{\pi}_0 \ln \left(\frac{b-\cos x}{a- \cos x}\right)dx=\pi \left(\frac{b+\sqrt{b^2-1}}{a+\sqrt{a^2-1}}\right)$$
if $a, b>1$
|
First, notice the integrand can be written as
$$
\begin{align}
\ln (\frac{b-\cos x}{a- \cos x}) &=\ln(b-\cos x)-\ln(a-\cos x)\\
& =\ln(y-\cos x)|^{y=b}_{y=a}\\
& =\int^b_a\frac{1}{y-\cos x}dy
\end{align}
$$
So the integration equals
$$
\begin{align}
\int^{\pi}_0 \ln (\frac{b-\cos x}{a- \cos x})dx &=\int^{\pi}_0 [ \int^b_a\frac{1}{y-\cos x}dy]dx \\
&=\int^b_a [\int^{\pi}_0 \frac{1}{y-\cos x}dx ] dy\\
& =\int^b_a(\frac{\pi}{\sqrt{y^2-1}})dy
\end{align}
$$
Let $$y=\sec t$$
so
$$dy=\sec t\tan t dt$$
So the integrantion becomes
$$
\begin{align}
\pi \int^{acrsec b}_{arcsec a}\frac{\sec t\tan t}{\tan t}dt &=\pi \int^{acrsec b}_{arcsec a}\sec t dt\\ & = \pi \ln |\sec t+\tan t|^{arcsec b}_{arcsec a}\\ & = \pi [\ln(b+\sqrt{b^2-1})-\ln(a+\sqrt{a^2-1})] \\ & =\pi \ln (\frac{b+\sqrt{b^2-1}}{a+\sqrt{a^2-1}})
\end{align}
$$
|
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|
Formula for $\sqrt{a-\sqrt{a+\sqrt{a+\ldots}}}$ Supposedly, the infinitely nested radical $$\sqrt{a-\sqrt{a+\sqrt{a+\ldots}}}\tag1\label{1}$$
converges to $$\frac {A-1}{6}+\frac 23\sqrt{4a+A}\sin\left(\frac 13\arctan\frac {2A+1}{3\sqrt 3}\right)\tag2$$
where $A=\sqrt{4a-7}$. In fact, that's how Ramanujan arrived at the famous nested radical $$\sqrt{2-\sqrt{2+\sqrt{2+\ldots}}}=2\sin\left(\frac {\pi}{18}\right)\tag3$$
So I'm wondering how you would prove $\ref{1}$. My best try was to set $(1)$ equal to $x$ and substitute to get $$x=\sqrt{a-\sqrt{a+\sqrt{a+x}}}$$to get an octic polynomial. But I'm not too sure how the trigonometry made its way into the generalization.
And I also wonder if there is a way to generalize this even further to possible $$\sqrt{a-\sqrt{a-\sqrt{a+\sqrt{a+\ldots}}}}$$ which has period $4$ instead of $3$.
|
Here's how I duplicated
Ramanujan's result,
with a lot of help from Wolfy.
Start with
$x=\sqrt{a-\sqrt{a+\sqrt{a+x}}}
$.
Squaring,
$x^2-a
=-\sqrt{a+\sqrt{a+x}}
$.
Squaring again,
$(x^2-a)^2
=a+\sqrt{a+x}
$
or
$(x^2-a)^2-a
=\sqrt{a+x}
$.
A final squaring gives
$((x^2-a)^2-a)^2-a
=x
$.
Setting
$a=2$
gives
$x
=((x^2-2)^2-2)^2-2
$.
According to Wolfy,
there are
8 real roots
of this
(including -1 and 2),
though it identifies
3 as complex roots
while giving them
as real numbers.
These are
$-1, 2,
-1.8794, 0.34730,
1.5321, -1.8019,
-0.44504, 1.2470,
$
though, for some reason,
it calls the last 3 complex.
I have no idea
how Wolfy got the roots.
Each one of them
is expressed with
$\sqrt{3}$ and
expressions involving it
and $i$,
some of them raised to the
$\frac13$ and $\frac23$
powers.
It turns out
the the fourth of these,
0.34730,
is the desired root.
This can be gotten
by asking Wolfy for
"root of
$x=\sqrt{2-\sqrt{2+\sqrt{2+x}}}$"
which gives a value
slightly less than
$0.35$.
Here is how I went
from the exact expression
for this root
given by Wolfy
to Ramanujan's expression:
$\begin{array}\\
x
&=\dfrac{-(1-i \sqrt{3})}{2^{2/3} (-1+i \sqrt{3})^{1/3}}-\frac12 (1+i \sqrt{3}) (\frac12 (-1+i \sqrt{3}))^{1/3} \\
&=\dfrac{-1+i \sqrt{3}}{2^{2/3} (-1+i \sqrt{3})^{1/3}}-\frac12 (1+i \sqrt{3}) (\frac12 (-1+i \sqrt{3}))^{1/3} \\
&=2^{-2/3}(-1+i \sqrt{3})^{2/3}-\frac12 (1+i \sqrt{3}) (\frac12 (-1+i \sqrt{3}))^{1/3} \\
&=(\frac12(-1+i \sqrt{3}))^{2/3}-\frac12 (1+i \sqrt{3}) (\frac12 (-1+i \sqrt{3}))^{1/3} \\
&=(e^{2\pi i/3})^{2/3}-e^{\pi i/3} (e^{2\pi i/3})^{1/3} \\
&=e^{4\pi i/9}-e^{(\pi i)(1/3+2/9)} \\
&=e^{4\pi i/9}-e^{(\pi i)(5/9)}
\qquad\text{(I was worried here)}\\
&=e^{4\pi i/9}-e^{(\pi i)(1-4/9)}
\qquad\text{(but then I thought of this)}\\
&=e^{4\pi i/9}-e^{\pi i}e^{-4\pi i/9} \\
&=e^{4\pi i/9}+e^{-4\pi i/9} \\
&=2\cos(4\pi /9)
\qquad\text{since }2\cos(z)=e^{iz}+e^{-iz} \\
&=2\sin(\pi/2-4\pi /9)
\qquad\text{since } \cos(z) = \sin(\pi/2-z)\\
&=2\sin(\pi/18)\\
\end{array}
$
|
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|
Closed form of a sum $ \sum_{i=1}^{n}\frac{1}{((i-1)x)^2+y^2}$ Consider a sum:
$$ \sum_{i=1}^{n}\frac{1}{((i-1)x)^2+y^2}$$
with $x$ and $y$ being (non-zero) constants. Is it possible to obtain a nice closed form of this expression?
|
------ Complementing Jack's answer -----
Actually, since we have the formula for the infinite summation
$$
S\left( {x,y} \right) = \sum\limits_{0\, \leqslant \,k} {\;\frac{1}
{{\left( {x + k} \right)^{\,2} + y^{\,2} }}} = \frac{1}
{{2y^{\,2} }} + \frac{{\pi \coth \left( {\pi y/x} \right)}}
{{2xy}}
$$
then, for a partial one we will have
$$
\begin{gathered}
S\left( {x,y,n} \right) = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\;\frac{1}
{{\left( {x + k} \right)^{\,2} + y^{\,2} }}} = \hfill \\
= \sum\limits_{0\, \leqslant \,k} {\;\frac{1}
{{\left( {x + k} \right)^{\,2} + y^{\,2} }}} - \sum\limits_{n\, \leqslant \,k} {\;\frac{1}
{{\left( {x + k} \right)^{\,2} + y^{\,2} }}} = \hfill \\
= \sum\limits_{0\, \leqslant \,k} {\;\frac{1}
{{\left( {x + k} \right)^{\,2} + y^{\,2} }}} - \sum\limits_{0\, \leqslant \,k} {\;\frac{1}
{{\left( {\left( {x + n} \right) + k} \right)^{\,2} + y^{\,2} }}} = \hfill \\
= S\left( {x,y} \right) - S\left( {x + n,y} \right) \hfill \\
\end{gathered}
$$
------ in another way -----
$$
\begin{gathered}
S\left( {x,y,n} \right) = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\;\frac{1}
{{\left( {x + k} \right)^{\,2} + y^{\,2} }}} = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\;\frac{1}
{{\left( {x + k} \right)^{\,2} - \left( {iy} \right)^{\,2} }}} = \hfill \\
= \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\;\frac{1}
{{\left( {x + iy + k} \right)\left( {x - iy + k} \right)}}} = \frac{1}
{{2iy}}\left( {\sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\;\frac{1}
{{\left( {x - iy + k} \right)}}} - \sum\limits_{0\, \leqslant \,k} {\;\frac{1}
{{\left( {x + iy + k} \right)}}} } \right) = \hfill \\
= \frac{1}
{{2iy}}\left( {\psi ^{\left( 0 \right)} \left( {x - iy + n} \right) - \psi ^{\left( 0 \right)} \left( {x + iy + n} \right) + \psi ^{\left( 0 \right)} \left( {x + iy} \right) - \psi ^{\left( 0 \right)} \left( {x - iy} \right)} \right) = \hfill \\
= \frac{1}
{y}\left( {\operatorname{Im} \left( {\psi ^{\left( 0 \right)} \left( {x + iy} \right)} \right) - \operatorname{Im} \left( {\psi ^{\left( 0 \right)} \left( {x + iy + n} \right)} \right)} \right) \hfill \\
\end{gathered}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1969295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How should I calculate the determinant? $\left|\begin{array}{cccc}1&a&b&c+d\\1&b&c&a+d\\1&c&d&a+b\\1&d&a&b+c\end{array}\right|=
\left|\begin{array}{cccc}1&a&b&c\\1&b&c&a\\1&c&d&a\\1&d&a&b\end{array}\right|+
\left|\begin{array}{cccc}1&a&b&d\\1&b&c&d\\1&c&d&b\\1&d&a&c\end{array}\right|$
I tried to calculate the determinant but I couldn't do it after separating the determinant by the property. How should I calculate it?
|
$${\begin{vmatrix}1&a & b &c+d\\1 &b &c &d+a \\1 &c &d &a+b\\1&d &a &b+c &\end{vmatrix}} \space c_2+c_3+c_4 \to c_4 \\
{\begin{vmatrix}1&a & b &a+b+c+d\\1 &b &c &a+b+c+d \\1 &c &d &a+b+c+d\\1&d &a &a+b+c+d &\end{vmatrix}} \space factor \space (a+b+c+d)=\\(a+b+c+d)
{\begin{vmatrix}1&a & b &1\\1 &b &c &1 \\1 &c &d &1\\1&d &a &1 &\end{vmatrix}}$$now $c_1-c_4 \to c_1$
$${\begin{vmatrix}0&a & b &1\\0 &b &c &1 \\0 &c &d &1\\0&d &a &1 &\end{vmatrix}} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1971668",
"timestamp": "2023-03-29T00:00:00",
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|
y'' + y = -sin(x) $y'' + y = -\sin(x)$
$y(0) = 0 $
$y'(0) = 0$
I first solved for the homogeneous solution to get:
$y(x) = c_1 \sin(x) + c_2 \cos(x)$
then took the derivative of that:
$y'(x) = c_1 \cos (x) - c_2 \sin(x)$
This is where I am not sure where to go...
$y(0) = c_1 \sin(0) + c_2 \cos(0)
= 0 + c_2 $
so would c2 equal zero?
y'(0) = c1 cos(0) - c2 sin(0)
= c1 - 0
then c2 is zero too?
I don't think that's right, but I'm not sure what else to do.
And where to go from there.
|
$y'' + y = -\sin x$
solve the homogeneous equation
$y = c_1 \sin t + c_2 \cos x + y_p$
Undetermined coefficients.
$y = A x\sin x + Bx\cos x$
Why $A x \sin x + B x\cos x,$ and not the more simple-minded $A \sin x + B \cos x$?
$A \sin x + B \cos x$ is already part of the homogeneous solution. So, when we plug it into the diff eq, it is going to "go away." We need something that when differentiated (twice) has a part that equals $-\sin x$ but is not in the homogeneous solution.
$y' = A\sin x + B\cos x - B x\sin x + Ax\cos x\\
y'' = -2B \sin x + 2A\cos x - A x\sin x + Bx\cos x\\
y''+y = -2B \sin x + 2A\cos x = -\sin x\\
B = \frac 12, A = 0$
$y = c_1 \sin t + c_2 \cos x + \frac 12 x \cos x\\
y(0) = C_2 = 0\\
y'(0) = C_1 = 0$
$y = \frac 12 x\cos x$
|
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"timestamp": "2023-03-29T00:00:00",
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|
find partial fraction $(2x^2-1)/((x^2-1)(2 x^2+3))$
Find partial fraction of $\cfrac{(2x^2-1)}{(x^2-1)(2 x^2+3)}$?
My attempt:
I did google and I tried to solved it as :
Let’s first get the general form of the partial fraction decomposition.
$\cfrac{(2x^2-1)}{(x^2-1)(2 x^2+3)}=\cfrac{A}{(x+1)}+\cfrac{B}{(x-1)}+\cfrac{C+Dx}{(2x^2+3)}$
Setting numerators gives,
$(2x^2-1)=A(x-1)(2x^2+3)+B(x+1)(2x^2+3)+C(x^2-1)$
I stuck here, how to proceed next?
Can you explain it, please?
|
Multiply out and compare coefficients. You'll find for example that the coefficient of $x^3$ on the right hand side is $2A+2B$. As in the left hand side there are zero $x^3$ you conclude $2A+2B=0$.
|
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|
Find the Derivative: This problem for my Calculus 1 class has got me stumped. I am not sure on where to start for this problem. Any help would be much appreciated.
$y=x\tanh^{-1}(x) + \ln(\sqrt{1-x^2})$
|
$y=x\tanh^{−1}(x)+\ln(\sqrt{1−x^2})$
differentiation is linear
$\frac {dy}{dx} = \frac d{dx} (x\tanh^{−1}(x)) +\frac d{dx}(\ln(\sqrt{1−x^2})$
next we need to know the product rule and the chain rule
$\frac {dy}{dx} = \tanh^{−1}(x) + x\frac d{dx} \tanh^{−1}(x) +\frac {1}{\sqrt{1−x^2}} \frac d{dx}\sqrt{1−x^2}\\
\frac {dy}{dx} = \tanh^{−1}(x) + x\frac d{dx} \tanh^{−1}(x) +\frac {1}{\sqrt{1−x^2}} \frac {-x}{\sqrt{1−x^2}}\\
\frac {dy}{dx} = \tanh^{−1}(x) + x\frac d{dx} \tanh^{−1}(x) -\frac {x}{1−x^2}$
Which leave one nasty left to tackle
$u = \tanh^{-1} x\\
\tanh u = x\\
\sech^2 u \frac {du}{dx} = 1\\
\frac {du}{dx} = \frac 1{\sech^2 u}\\
\frac {du}{dx} = \frac 1{1-\tanh^2 u}\\
\frac {du}{dx} = \frac 1{1- x^2}\\
$
$\frac {dy}{dx} = \tanh^{−1}(x) + \frac {x}{1-x^2} -\frac {x}{1−x^2}\\
\frac {dy}{dx} = \tanh^{−1}(x)$
|
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|
General solution to the polynomial of form $x^a(1-x)^b=c$ Given $x\in(0,1)$, I am interested in finding the solution of polynomials in the form of $x^a(1-x)^b=c,$ where $a,b$ are both positive integers, and $c$ is carefully chosen so that a root in $(0,1)$ exists.
For $a+b<5$ the roots can be found painstakingly with radicals, however the problem won't be straightforward for larger $a,b$ due to the Abel–Ruffini theorem. Generally, I know a root can be found using Jacobi's $\vartheta$ functions, but their complexity makes them very difficult to apply in practice.
So here is my question: can the root to this polynomial in $(0,1)$ be expressed, or approximated, by a "nicer" function? By "nicer", I mean a combination of elementary functions and "common" special functions (e.g. Beta, Bessel, etc.).
|
The equation $x^a(1-x)^b=c$ can be written as a generalized trinomial equation
$$
Ax^p+x=1\qquad(p=-\frac{a}{b},~A=c^{{1}/{b}}).
$$
Ramanujan showed that the root of the equation $Ax^p+x=1$ is given by the power series
$$
x=1-A+\sum_{n=2}^\infty\frac{(-A)^n}{n!}\prod_{k=1}^{n-1}(1+pn-k).
$$
So, according to these formulas we have for the root of the equation $x^a(1-x)^b=c$
\begin{align}
x&=1-c^{1/b}+\sum_{n=2}^\infty\frac{(-c^{1/b})^n}{n!}\prod_{k=1}^{n-1}(1-an/b-k)\\&=1-c^{1/b}-\sum_{n=2}^\infty\frac{c^{n/b}}{n!}\frac{\Gamma(an/b+n-1)}{\Gamma(an/b)}.
\end{align}
For rational $a/b$ this series can be written in terms of finite sum of hypergeometric functions (see Glasser's paper for details https://arxiv.org/abs/math/9411224). For given $a,b$ Mathematica can easily simplify this expression in terms of generalized hypergeometric functions.
According to dxiv's analysis there are 2 real roots in the interval $(0,1)$. Let's denote them as $x_1$ and $x_2$. The above formula gives one of the roots, say $x_1$. To obtain $1-x_2$ interchange $a$ and $b$ in this formula.
$\it{Example}$. Let $a=2,b=3$, then the real root $x_1$ of the equation $x^2(1-x)^3=c$ is
$$
x_1=\frac{3}{5}+\frac{2}{5} \, _4F_3\left(-\frac{1}{5},\frac{1}{5},\frac{2}{5},\frac{3}{5};\frac{1}{3},\frac{1}{2},\frac{2}{3};\frac{3125 c}{108}\right)-c^{1/3}{}_4F_3\left(\frac{2}{15},\frac{8}{15},\frac{11}{15},\frac{14}{15};\frac{2}{3},\frac{5}{6},\frac{4}{3};\frac{3125 c}{108}\right)-\frac{2}{3}c^{2/3} {}_4F_3\left(\frac{7}{15},\frac{13}{15},\frac{16}{15},\frac{19}{15};\frac{7}{6},\frac{4}{3},\frac{5}{3};\frac{3125 c}{108}\right).
$$
For the other root $x_2$ we have
$$
x_2=\frac{3}{5}-\frac{3}{5}{}_4F_3\left(-\frac{1}{5},\frac{1}{5},\frac{2}{5},\frac{3}{5};\frac{1}{3},\frac{1}{2},\frac{2}{3};\frac{3125 c}{108}\right)+c^{1/2}{}_4F_3\left(\frac{3}{10},\frac{7}{10},\frac{9}{10},\frac{11}{10};\frac{5}{6},\frac{7}{6},\frac{3}{2};\frac{3125 c}{108}\right),
$$
Numerical check confirms that these formulas are true.
|
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|
How do I prove that $[\frac{x}{n}]+[\frac{x+1}{n}]+[\frac{x+2}{n}]....+[\frac{x+n-1}{n}]=[x]$ How do I prove that $[\frac{x}{n}]+[\frac{x+1}{n}]+[\frac{x+2}{n}]....+[\frac{x+n-1}{n}]=[x]$
How to start?
thanks
|
HINT
Use the fact that $[t]$ is the only integer such that $t - 1 \lt [t] \le t $
$\frac{x}{n} -1 +\frac{x+1}{n} - 1+\frac{x+2}{n} - 1…+\frac{x+n-1}{n} -1 \lt [\frac{x}{n}]+[\frac{x+1}{n}]+[\frac{x+2}{n}]…+[\frac{x+n-1}{n}] \le \frac{x}{n} + \frac{x+1}{n}+\frac{x+2}{n}…+\frac{x+n-1}{n}$
UPDATE
The OP question can be proved using Hermite's identity, by taking $x := \frac x n$
My hint was not useful at all, so please remove the acceptance.
|
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|
Find the minimum value of this Expression (Without Calculus!) Problem:
Suppose that $x,y$ and $z$ are positive real numbers verifying $xy+yz+zx=1$ and $k,l$ are two positive real constants. The minimum value of the expression:
$$kx^2+ly^2+z^2$$ is $2t_0$, where $t_0$ is the unique root of the equation $2t^3+(k+l+1)t-kl=0.$
My Attempt: Let $k=a+(k-a)$ and $l=b+(l-b)$, where $a,b\in \mathbb{R^+}$ and $a\leq k$ and $b\leq l.$ Then from the following Inequalities: $$ax^2+by^2\geq2\sqrt{ab}xy$$ $$(l-b)y^2+z^2/2\geq\sqrt{2(l-b)}yz$$ $$z^2/2+(k-a)x^2\geq\sqrt{2(k-a)}zx$$ We can deduce that $$kx^2+ly^2+z^2\geq2\sqrt{ab}xy+\sqrt{2(l-b)}yz+\sqrt{2(k-a)}zx.$$ Since $xy+yz+zx=1\implies 2\sqrt{ab}=\sqrt{2(l-b)}=\sqrt{2(k-a)}.$ After this stage I get two quadratic expression for $a$ and $b$, which seems far of from the main result. Where am I going wrong?
PS.Please do not use Calculus to solve this Problem.
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We need to find a maximal value of $m$, for which the inequality
$$kx^2+ly^2+z^2\geq m(xy+xz+yz)$$
is true for all reals $x$, $y$ and $z$ or
$$z^2-m(x+y)z+kx^2+ly^2-mxy\geq0$$
for which we need $m^2(x+y)^2-4(kx^2+ly^2-mxy)\leq0$ or
$(4k-m^2)x^2-2(2m+m^2)xy+(4l-m^2)y^2\geq0$ for all reals $x$ and $y$, for which we need
$(2m+m^2)^2-(4k-m^2)(4l-m^2)\leq0$ and $4k-m^2>0$, which gives
$m^3+(1+k+l)m^2-4kl\leq0$ and since for $m=2\sqrt{k}$ we obtain
$m^3+(1+k+l)m^2-4kl=8k\sqrt{k}+(1+k+l)4k-4kl>0$,
we see that the minimal value of $kx^2+ly^2+z^2$ is a positive root of the equation
$$m^3+(1+k+l)m^2-4kl=0$$
which not exactly that you wish.
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"url": "https://math.stackexchange.com/questions/1988064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Integration Calculus Problem How to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{\sin^2{x}}{1+\sin^2x}dx$$
My Try:
I substituted $\displaystyle \sin^2x=\frac{(1-\cos2x)}{2}$ but it gave a complicated expression.
Using the identity$\sin^2x+\cos^2x=1$, the integral becomes:
$$ \int_{0}^{\frac{\pi}{2}}\frac{\sin^2{x}}{1+\sin^2x}dx=\frac{\sin^2 x}{2\sin^2x+\cos^2x}dx=\frac{1}{2+\cot^2x}dx$$
I'm stuck beyond this point.
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Note
$$ \int \frac{ \sin^2 x + 1 - 1}{1 + \sin^2 x } = \int dx - \int \frac{ dx }{1 + \sin^2 x} = x - \int \frac{ ( \sin^2 x + \cos^2 x ) dx }{1 + \sin^2 x} = x - \int \frac{ \sin^2 x }{1 + \sin^2 x} - \int \frac{ \cos^2 x dx}{1 + \sin^2 x } $$
Thus, we have (now putting the limits of integration)
$$ 2 \int_0^{\pi/2} \frac{ \sin^2 x dx }{1 + \sin^2 x} = \frac{\pi}{2} - \underbrace{ \int_0^{\pi/2} \frac{ \cos^2 x dx }{1 + \sin^2 x} }_{I} $$
To integrate $I$, put $x = \tan t$. then $\sin x = \frac{ t }{\sqrt{1+t^2}}$, $\cos x = \frac{ 1}{ \sqrt{1+t^2}}$ and $dx = \frac{ dt }{1 + t^2}$. Thus,
$$ I = \int_0^{\infty} \frac{ \frac{1}{1 + t^2} \cdot \frac{ dt }{1 + t^2} }{ 1 + \frac{ t^2 }{1+t^2}} = \int_0^{\infty} \frac{ dt }{1+2t^2} = \frac{\sqrt{2} }{2} \tan^{-1} (\sqrt{2} t) \bigg|_0^{\infty} = \frac{ \sqrt{2} \pi }{4} $$
Therefore,
$$ \boxed{ \int\limits_0^{ \frac{ \pi }{2} } \frac{ \sin^2 x dx }{1 + \sin^2 x} = \frac{1}{2} \left( \frac{\pi}{2} - \frac{\sqrt{2} \pi }{4} \right) } $$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1990140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Singapore math olympiad Trigonometry question: If $\sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ$, then $ab=$?
$$\text{If}\; \sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ\text{, then}\; ab=\text{?}$$
$\bf{My\; Try::}$ We can write above question as $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = a\sin 50^\circ+b$$
Now for Left side, $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = \sqrt{9\sin^250^\circ-8\sin^350^{\circ}}$$
Now How can i solve it after that , Help required, Thanks
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If you use the formula for the triple angle:
$-8\sin^3(50)=2\sin(150)-6\sin(50)=1-6\sin(50)$
so your last square root becomes
$\sqrt{9\sin^2(50)-6\sin(50)+1}=\sqrt{(3\sin(50)-1)^2}=3\sin(50)-1$
as the last expression is positive. So:
$3\sin(50)-1=a\sin(50)+b$
Now if $a$ and $b$ are rational/integer, we have $a=3$, $b=-1$ so $ab=-3$ but in the case of real numbers there is no unique solution. For example if $a=0,b=3\sin(50)-1$ we have $ab=0$
In any case, the last equation is much simpler to work out than the first one :)
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"language": "en",
"url": "https://math.stackexchange.com/questions/1991029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Unique Factorization in Number Ring I am having a difficulty understanding ideals in a number ring; if I have (2), does it mean that it is a ring generated by $2$ (e.g. {$2k, k \in \mathbb{Z}$}).
If so, why is $(2)=(2,1+\sqrt{-5})^2$? It seems like the ideal $(2,1+\sqrt{-5})^2$ generated by $4$, $2+2\sqrt{-5}$, and $(1+\sqrt{-5})^2=-4+2\sqrt{-5}$, but we $(2)$ does not contain $2+2\sqrt{-5}$. I feel like I'm misunderstanding something....
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Given two ideals $A=(a_1,\ldots)$ and $B=(b_1,\ldots)$, $A=B$ if and only if each $a_i$ is in $B$ and each $b_i$ is in $A$.
Here:
$A=(2)$ and
$B=\left(2,1+\sqrt{-5}\right)^2=\left(2,1+\sqrt{-5}\right)\left(2,1+\sqrt{-5}\right)=\left(4, 2+2\sqrt{-5}, -4+2\sqrt{-5}\right)$.
We have:
$4=2\cdot2\in A$,
$2+2\sqrt{-5}=\left[1+\sqrt{-5}\right]\cdot2\in A$, and
$-4+2\sqrt{-5}=\left[-2+\sqrt{-5}\right]\cdot2\in A$.
Likewise: $2=(-1)\cdot4+1\cdot\left[2+2\sqrt{-5}\right]+(-1)\cdot\left(-4+2\sqrt{-5}\right)\in B$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1991259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$n$-derivative of function $f(x)=e^x \sin x$ at $x=0$ I have function $f(x) = e^x \sin{x}$ and must found $f^{(n)}(0)$
$f'(x) = e^x(\sin{x} + \cos{x}) $
$f''(x) = 2 e^x \cos{x}$
$f'''(x) = 2 e^x (\cos{x} - \sin{x})$
$f''''(x) = -4 e^x \sin{x}$
$f'''''(x) = -4 e^x (\sin{x} + \cos{x})$
I think $f^{(n)}(0) = \alpha (-1)^n x^{2n+1}$ but I can't find $\alpha$
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Let us try to discover a general formula. I see two possible approaches.
The first one is to use induction: assume that $f^{(n)} (x) = \Bbb e ^x (a_n \cos x + b_n \sin x)$. We have $a_0 = 0$ and $b_0 = 1$. Then
$$f^{(n+1)} = \Bbb e ^x (a_n \cos x + b_n \sin x) + \Bbb e ^x (- a_n \sin x + b_n \cos x) = \Bbb e ^x [(a_n + b_n) \cos x + (b_n - a_n) \sin x ] ,$$
which gives us a recurrence relation that can be written in matrix form as
$$\begin{pmatrix} a_{n+1} \\ b_{n+1} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} a_n \\ b_n \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} ^2 \begin{pmatrix} a_{n-1} \\ b_{n-1} \end{pmatrix} = \dots = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} ^{n+1} \begin{pmatrix} a_0 \\ b_0 \end{pmatrix} .$$
If $I$ is the identity matrix, prove by induction (it is easy) that
$$\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} ^{4n} = (-4)^n I .$$
It follows that
$$\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} ^{4n+1} = (-4)^n \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} , \\
\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} ^{4n+2} = (-4)^n \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}^2 = (-4)^n \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} , \\
\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} ^{4n+3} = (-4)^n \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}^3 = (-4)^n \begin{pmatrix} -2 & 2 \\ -2 & -2 \end{pmatrix} ,$$
whence it follows that
$$\begin{pmatrix} a_{4n} \\ b_{4n} \end{pmatrix} = (-4)^n \begin{pmatrix} 0 \\ 1 \end{pmatrix} , \\
\begin{pmatrix} a_{4n+1} \\ b_{4n+1} \end{pmatrix} = (-4)^n \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = (-4)^n \begin{pmatrix} 1 \\ 1 \end{pmatrix} , \\
\begin{pmatrix} a_{4n+2} \\ b_{4n+2} \end{pmatrix} = (-4)^n \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = (-4)^n \begin{pmatrix} 2 \\ 0 \end{pmatrix} , \\
\begin{pmatrix} a_{4n+3} \\ b_{4n+3} \end{pmatrix} = (-4)^n \begin{pmatrix} -2 & 2 \\ -2 & -2 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = (-4)^n \begin{pmatrix} 2 \\ -2 \end{pmatrix} , \\$$
which gives the result
$$f^{(4n)} (x) = \Bbb e^x (-4)^n \sin x , \\
f^{(4n+1)} (x) = \Bbb e^x (-4)^n (\cos x + \sin x), \\
f^{(4n+2)} (x) = \Bbb e^x (-4)^n 2 \cos x , \\
f^{(4n+3)} (x) = \Bbb e^x (-4)^n (2 \cos x - 2 \sin x) ,$$
whence one finally gets
$$f^{(4n)} (0) = 0 , \\
f^{(4n+1)} (0) = (-4)^n , \\
f^{(4n+2)} (0) = 2 (-4)^n , \\
f^{(4n+3)} (0) = 2 (-4)^n .$$
An alternative approach would be to use Leibniz's general formula
$$(fg)^{(n)} = \sum _{k=0} ^n \binom n k f^{(n-k)} g^{(k)}$$
whence it follows that
$$\tag{*} (\Bbb e ^x \sin x) ^{(n)} (0) = \sum _{k=0} ^n \binom n k \sin^{(k)} (0) .$$
It is easy now (again, induction) to see that
$$\sin^{(4n)} = \sin, \quad \sin^{(4n+1)} = \sin' = \cos, \quad \sin^{(4n+2)} = \cos' = -\sin, \quad \sin^{(4n+3)} = -\sin' = -\cos ,$$
meaning that
$$\sin^{(4n)} (0) = 0, \quad \sin^{(4n+1)} (0) = 1, \quad \sin^{(4n+2)} (0) = 0, \quad \sin^{(4n+3)} (0) = -1 .$$
Plugging these in (*) will give you a result, but you would have to work on it a little bit in order to reach the same form obtained through the first method.
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"timestamp": "2023-03-29T00:00:00",
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|
length of the shortest path connecting a point and another point on circle.
The length of the shortest path that begins at $(2,5)$ touches the x-axis and then
end at point on the circle $x^2+y^2+12x-20y+120=0$
$\bf{My\; Try::}$ Equation of circle in standard form :: $(x+6)^2+(y-10)^2=4^2$
So any point on the circle is $x+6=4\cos \theta$ and $y-10 = 4\sin \theta$
So point $P(4\cos \theta-6,4\sin \theta+10)$ on the circle
and let $B(x,0)$ be any point on $x-$ axis and Let $A(2,5)$be a fixed point.
SO we have to minimize $AB+BP = \sqrt{(x-2)^2+5^2}+\sqrt{(x+6-4\cos \theta)^2+(4\sin \theta+10)^2}$
Now how can i solve it after that, Help required, Thanks
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Let $C(2, -5)$ be the reflection of point $A$ across the $x$-axis. Notice that $AB = CB$, so we now only need to minimize $CB + BP$. But there are no restrictions on the path from $C$ to $P$ (since any such path must eventually cross the $x$-axis), so the shortest path between the two fixed points is a straight line. So we now only need to minimize $CP$.
Let $D = (-6, 10)$ be the center of the circle. Notice that $PD = 4$, so we now only need to minimize $CD - 4$. But $CD$ is a fixed distance, so we conclude that the optimal path length is:
$$
\sqrt{(2 + 6)^2 + (-5 - 10)^2} - 4 = 13
$$
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"url": "https://math.stackexchange.com/questions/1992630",
"timestamp": "2023-03-29T00:00:00",
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|
Prove: $\log_{2}{3} < \log_{3}{6}$ How should I prove that $\log_{2}{3} < \log_{3}{6}$?
I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$, $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothing.
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$\log_23$ vs $\log_36$
Multiply each term by $5$:
$5\log_23$ vs $5\log_36$
Apply logarithm rules:
$\log_23^5$ vs $\log_36^5$
Simplify:
$\log_2243$ vs $\log_37776$
Conclude:
$\log_2243<\log_2256=8=\log_36561<\log_37776$
Hence $\log_23<\log_36$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1994320",
"timestamp": "2023-03-29T00:00:00",
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Mathematical Induction for Recurrence Relation I have solved the following recurrence relationship:
$T(1) = 1$
$T(n) = T(n-1) + n + 2$
so
$T(n) = \frac{1}{2}n^2+\frac{5}{2}n -2$
I am now trying to perform mathematical induction to prove this.
$Basis:$
$T(1)=1=3-2=\frac{1}{2} + \frac{5}{2} - 2 $
$Induction:$
$T(k+1) = T(k) + k+1 + 2$
$= \frac{1}{2}k^2 + \frac{5}{2}k -2 + k+1 +2$
$= \frac{1}{2}k^2 + \frac{7}{2}k +1$
What can I do next?
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If induction is not a prerequisite, the following is an alternative that simply telescopes the sequence.
$$T(n) - T(n-1) = n + 2$$
$$\sum_{k=2}^{n} \big(T(k) - T(k-1)\big) = \sum_{k=2}^{n}(k + 2)$$
$$T(n) - T(1) = \frac{(n-1)(n+6)}{2}$$
$$T(n) \;=\; 1 + \frac{n^2 + 5 n - 6}{2} \;=\; \frac{1}{2}n^2 + \frac{5}{2}n - 2$$
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{
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"timestamp": "2023-03-29T00:00:00",
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Prove :(P → Q) ∨ (Q → P) using natural deduction Allowed inference rules: ∨-I, ∨-E, ∧-I, ∧-E, →-I, →-E, ¬-I, ¬-E
I tried to prove a contradiction by assuming $¬ ((P → Q) ∨ (Q → P))$ but got stuck, or am I doing it in the wrong way?
Edit:
My proof attempt
$1.\qquad¬ ((P → Q) ∨ (Q → P))\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad Assum$
$2.\qquad P → Q\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ Assum$
$3.\qquad(P → Q) ∨ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\; ∨-I(2)$
$4.\qquad(¬ ((P → Q) ∨ (Q → P))) ∧ ((P → Q) ∨ (Q → P))\qquad\qquad\qquad\; ∧-I(1,3)$
$5.\qquad ¬ (P → Q)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;\; ¬-I(2,4)$
$6.\qquad Q → P\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad\quad\; Assum$
$7.\qquad (P → Q) ∨ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\; ∨-I(6)$
$8.\qquad (¬ ((P → Q) ∨ (Q → P))) ∧ ((P → Q) ∨ (Q → P))\qquad\qquad\qquad\;\; ∧-I(1,7)$
$9.\qquad ¬ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\; ¬-I(6,8)$
$10.\qquad ¬ (P → Q) ∧ ¬ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\;\qquad\qquad\quad\;\;\, ∧-I(5,9)$
Stuck at this line, can't spot any contradiction to line 1
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Your start is good:
$1.\qquad¬ ((P → Q) ∨ (Q → P))\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad Assum$
$2.\qquad \qquad P → Q\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ Assum$
This second assumption is good too, since you basically want to do a DeMorgan on line 1, and to do those in Natural Deduction, you do a Proof by COntradiction on each of the disjuncts .. so yes, assume the disjunct, but then proceed with getting the full disjunct:
$3.\qquad\qquad (P → Q) ∨ (Q → P) \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \lor Intro \: (2)$
and now you can get a contradiction with line 1:
$4.\qquad \qquad\bot \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad \qquad\qquad\qquad \bot \: Intro \: (1,3)$
So you can close the subproof and get:
$5.\qquad \lnot(P → Q) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \lnot Intro \: (2-4)$
OK, same pattern with the other disjunct, i.e. repeat those same 4 lines:
$6.\qquad \qquad Q → P\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ Assum$
$7.\qquad \qquad (P → Q) ∨ (Q → P) \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \lor Intro \: (6)$
$8.\qquad\qquad \bot \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad \qquad\qquad\qquad \bot \: Intro \: (1,7)$
$9.\qquad \lnot(Q → P) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \lnot Intro \: (6-8)$
OK, we know that $\lnot (P \rightarrow Q) \equiv P \land \lnot Q$, so we should be able to get $P$ from 5, and $\lnot P$ from 9, giving us the desired contradiction. How do we prove that? It's not too hard: two more proofs by contradiction! That is, to get $P$, assume the opposite:
$10.\qquad \qquad\lnot P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ Assum$
And since we want this to contradict with $\lnot (P \rightarrow Q)$, try and prove $P \rightarrow Q$, which we do with a conditional proof:
$11.\qquad\qquad\qquad P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ Assum$
$12.\qquad \qquad\qquad\bot \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad \qquad\qquad \bot \: Intro \: (10,11)$
$13.\qquad \qquad\qquad\ Q \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \bot \: Elim \: (12)$
$14.\qquad\qquad P → Q\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ \rightarrow \: Intro (11-13)$
$15.\qquad\qquad \bot \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad \qquad\qquad\qquad \bot \: Intro \: (5,14)$
$16.\qquad \lnot \lnot P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \lnot \: Intro \: (10-15)$
$17.\qquad P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \lnot \: Elim \: (16)$
Almost there! Do a similar pattern to get $\lnot P$ from 9:
$18.\qquad \qquad P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ Assum$
$19.\qquad\qquad\qquad Q \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ Assum$
$20.\qquad \qquad\qquad\ P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad Reit \: (18)$
$21.\qquad\qquad Q → P\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ \rightarrow \: Intro (11-13)$
$22.\qquad\qquad \bot \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad \qquad\qquad\qquad \bot \: Intro \: (9,21)$
$23.\qquad \lnot P \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \lnot \: Intro \: (18-22)$
And now we're done:
$24.\qquad \bot \qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad \qquad\qquad\qquad\qquad \bot \: Intro \: (17,23)$
$25.\lnot \lnot ((P → Q) ∨ (Q → P))\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \lnot Intro (1-24)$
$26. (P → Q) ∨ (Q → P)\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \lnot Elim (25)$
OK, so it turns out to be a somewhat lengthy proof, but that often happens with Natural Deduction, where we don;t have handy dandy equivalence rules. However, notice that between the set-ups for conditional proof, conditional proof, etc, as well as the repeating patterns for doing things like Demorgan, you end up cranking out these proofs on automatic pilot once you have done them for a while. And finally, oftentimes the instructor, book, or software will start to allow you to simply refer to those patterns as Lemma's, so you don't have to write them out every time.
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"timestamp": "2023-03-29T00:00:00",
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Absolute and conditional convergence of a series with $\tan$ $$\sum_{n=1}^{\infty}\tan{\Big(\pi \sqrt[4]{n^4 + (-1)^n n^2 + 4}} \Big)$$
I tried to do something like this but in the end it don't help:
$$\sum_{n=1}^{\infty}\tan{\Big(\pi \sqrt[4]{n^4 + (-1)^n n^2 + 4}} \Big) =
\sum_{n=1}^{\infty}\tan{\Big(\pi n \sqrt[4]{1 + \Big(\frac{(-1)^n}{n^2} + \frac{4}{n^4}}\Big)} = \\
\sum_{n=1}^{\infty}\tan{\Big(\pi n \Big(1 + \frac{(-1)^n n^2 + 4}{4n^4} \Big) \Big)}.$$
Could you give me tips how to continue solution or solve in another way?
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First, don't try to sum the series when you are transforming its terms. Work on the terms, find equivalents or developments, and then, apply a theorem stating the status of the series.
Here, you took the right start :
$$\sqrt[4]{n^4+(-1)^nn^2+4} = n\left(1+\frac{(-1)^n}{n^2}+\frac{4}{n^4}\right)^{\frac{1}{4}} = n\left(1+\frac{(-1)^n}{4n^2}+O\left(\frac{1}{n^4}\right)\right)$$
so
\begin{align}\tan\left(\pi \sqrt[4]{n^4+(-1)^nn^2+4}\right) &= \tan\left(n\pi+\frac{(-1)^n}{4n}+O\left(\frac{1}{n^3}\right)\right) \\ &= \tan\left(\frac{(-1)^n}{4n}+O\left(\frac{1}{n^3}\right)\right) \\ &= \frac{(-1)^n}{4n}+O\left(\frac{1}{n^3}\right)\end{align}
Therefore $u_n$ is the sum of two terms, one, $\frac{(-1)^n}{4n}$, leads to a convergent, but not absolutely convergent, series, the other, $O\left(\frac{1}{n^3}\right)$, is the term of an absolutely convergent series.
Conclusion : $\sum u_n$ is convergent, but not absolutely convergent ($\left|u_n\right|\sim \frac{1}{4n}$).
|
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|
Find all $\alpha$ such that the series converges Find all values of $\alpha$ such that series $$\sum^\infty_{n=1} \left( \frac{1}{n \cdot \sin(1/n)} - \cos\left(\frac{1}{n}\right) \right)^\alpha$$ converges.
I used Maclaurin for $\sin$ and $\cos$ and got:
$$a_n = \left( \frac{1}{1 - \dfrac{1}{3!n^2} + \ldots} - 1 + \frac{1}{2!n^2} - \frac{1}{4!n^4} + \ldots \right) ^ \alpha$$
Put it together in one fraction seems to be a hard thing to do.
|
Hint
$$n\sin(\frac{1}{n})=1-\frac{1}{6n^2}(1+\epsilon_1(n))$$
$$\frac{1}{n\sin(\frac{1}{n})}=1+\frac{1}{6n^2}(1+\epsilon_2(n))$$
$$\cos(\frac{1}{n})=1-\frac{1}{2n^2}(1+\epsilon_3(n))$$
thus, when $n\to +\infty$, the general term of your series $u_n$, satisfies
$$u_n \sim (\frac{2}{3n^2})^\alpha $$
and by the limit comparison test,
$\sum u_n$ converges $\iff \;
\alpha>\frac{1}{2}$.
|
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|
Pythagoras always divisible by 3 Lets look at the pythagoras $a^2+b^2=c^2$ for the integers $a,b,c$. Proof that one of components $a,b,c$ is always divisible by 3. How do i prove that?
|
Assume none of $a,b,c$ is a multiple of $3$.
then
$a \equiv \pm 1$ mod$ (3)$
$b \equiv \pm 1$ mod $(3)$ and
$c\equiv \pm 1$ mod $(3)$
thus
$a^2\equiv 1$ mod $(3),$
$b^2\equiv 1$ mod $(3)$
$a^2+b^2 \equiv 2$ mod$ (3)$
but
$c^2=a^2+b^2 \equiv 1$ mod$ (3)$
and this is a contradiction.
|
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|
Show that a polynomial family is a base to a subspace QUESTION
Taking $P4 = ax^4+bx^3+cx^2+dx+e$ the vector space of polynomials of degree $4$ or lower and $H$ the subspace of $P_4$ such that its elements satisfy the conditions $p(1)= 0$, $p(-1)=0$
Show that $B = \{q_1,q_2,q_3\}$ is a base of $H$ if the polynomials are the following:
$q_1(x) = x^2-1$
$q_2(x)=x^3-x$
$q_3(x) = (x^2-1)^2$
ANSWER ?
So I know that to be a base you need to be linearly independent and you have to generate $P_4$ (in this specific case).
To prove so, I tried to calculate the matrix of the canonical coordinates of $B$, which gave me
$$\begin{pmatrix}
0&0&1\\
0&1&0\\
1&0&-2\\
0&-1&0\\
-1&0&1
\end{pmatrix}$$
Then I used Gaussian elimination and got to the following point (I had to transform $(x^2-1)^2$ into $x^4-2x^2+1$:
$$\begin{pmatrix}
-1&0&1\\
0&1&0\\
0&0&1\\
0&0&0\\
0&0&0
\end{pmatrix}$$
But I am stuck because how can I show that $B$ is a base of $H$ if $\dim(B) = 3$.
It should be $5$? I recall reading that to be a base of something, you need to have the same dimension.
|
Note that given the conditions $p(1)=p(-1)=0$, it follows that $H$ is the subspace of $P_4$ consisting of all fourth degree polynomials divisible by both $x-1$ and $x+1$ and therefore by $x^2-1$. Using the division algorithm we know that for all $p(x)\in P_4$
\begin{eqnarray}
p(x)&=&ax^4+bx^3+cx^2+dx+e\\
&=&[ax^2+bx+(a+c)](x^2-1)+(b+d)x+(a+c+e)
\end{eqnarray}
so we know that for $p(x)\in H$ the remainder must be zero.
Thus $d=-b$ and $e=-(a+c)$ and every element of $H$ can therefore be expressed in the form
\begin{eqnarray}
p(x)&=&ax^4+bx^3+cx^2-bx-(a+c)\\
&=&a(x^4-1)+b(x^3-x)+c(x^2-1) \tag{1}
\end{eqnarray}
So we know that $B^\prime=\{x^4-1,x^3-x,x^2-1\}\,$ forms a basis for $H$. But we want $(x^2-1)^2=x^4-2x^2+1$, not $x^4-1$. But if in equation $(1)$ we let $a=1,\,b=0,\,c=-2\,$ we get
\begin{equation}
1\cdot(x^4-1)+0\cdot(x^3-x)-2\cdot(x^2-1)=(x^2-1)^2
\end{equation}
Since $(x^2-1)^2$ is a linear combination of the elements of $B^\prime\,$ we may substitute it in place of $x^4-1$ in $B^\prime$ to get the desired basis $B$.
|
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|
How do you evaluate the integral $\int\frac{x^2-1}{(x^4+3 x^2+1) \tan^{-1}\left(\frac{x^2+1}{x}\right)}\,dx$? i'm required to evaluate this integral. I've tried factorizing but it doesn't lead me to anywhere.
$$\int\frac{x^2-1}{(x^4+3 x^2+1) \tan^{-1}\left(\frac{x^2+1}{x}\right)}\,dx$$
I've also tried letting $u = \frac{x^2+1}{x}$, $du/dx$ gets me $1-\frac{1}{x^2}$ but it doesn't seem to be working either.
Hope to receive some advise/ solutions on how to start tackling the question
|
An useful identity to remember is
$$\frac{dx}{x} = \frac{d(x+x^{-1})}{x-x^{-1}} = \frac{d(x-x^{-1})}{x + x^{-1}}$$
Using the first part of this identity, you can rewrite the integral as
$$\begin{align}
& \int \frac{x(x^2-1)}{(x^4+3x^2+1)\tan^{-1}\left(x + x^{-1}\right)}\frac{d(x+x^{-1})}{x-x^{-1}}\\
= & \int \frac{x^2 d(x+x^{-1})}{(x^4+3x^2+1)\tan^{-1}\left(x + x^{-1}\right)}\\
= & \int \frac{d(x+x^{-1})}{(x^2+x^{-2}+3)\tan^{-1}\left(x + x^{-1}\right)}\\
= & \int \frac{d(x+x^{-1})}{((x+x^{-1})^2+1)\tan^{-1}\left(x + x^{-1}\right)}\\
= & \int \frac{d\tan^{-1}\left(x + x^{-1}\right)}{\tan^{-1}\left(x + x^{-1}\right)}\\
= &\log\tan^{-1}\left(x + x^{-1}\right) + \text{constant}.
\end{align}
$$
|
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|
The polynomial has a root in the interval $(a_1,a_1+1).$ Let $a_1,a_2,\ldots,a_k$ and $p$ are positive real numbers where $a_1$ is minimum. If we solve $$\frac{1}{x-a_1}+\frac{1}{x-a_2}+\ldots \frac{1}{x-a_k}-1=0 \quad(1)$$ we have a $k$-degree polynomial.
Consider another polynomial, given by: $$\frac{1}{x-a_1}+\frac{1}{x-a_2}+\ldots \frac{1}{x-a_k}-\frac{p}{x}-1=0\quad(2)$$
I have two observation:
$(a)$ Least root of $(1)$ lies in $(a_1,a_1+1).$
$(b)$ Least root of $(2)$ is less than that of $(1).$
Are the two observations true? If so, how to prove them?
|
Here is part of (a):
If $x < a_1 \le a_i$, then $x-a_i <0$ and so $\displaystyle\frac{1}{x-a_1}+\frac{1}{x-a_2}+\cdots \frac{1}{x-a_k} < 0 < 1$.
Therefore, there is no root smaller than $a_1$.
|
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|
Manipulation of conditions of roots of a quadratic equation. How do I write $a^5+b^5$ in terms of $a+b$ and $ab$. Also is there any general way of writing $a^n+b^n$ in terms of $a+b$ and $ab$?
|
$$(a-b)^2=(a+b)^2-4ab$$ so that
$$a,b=\frac{a+b\pm\sqrt{(a+b)^2-4ab}}2.$$
Then
$$a^5+b^5=\left(\frac{a+b+\sqrt{(a+b)^2-4ab}}2\right)^5+\left(\frac{a+b-\sqrt{(a+b)^2-4ab}}2\right)^5$$ that you can evaluate by the binomial theorem (every other term will cancel out).
Let $m:=(a+b)/2,p=ab$, and after simplification,
$$\frac12(a^5+b^5)=m^5+10m^3\left(\sqrt{m^2-p}\right)^2+5m\left(\sqrt{m^2-p}\right)^4=16m^5-5mp^2-20m^3p.$$
For the general case,
$$\frac12(a^n+b^n)=\sum_{k=0}^{n/2}\binom n{n-2k}m^{n-2k}\left(\sqrt{m^2-p}\right)^{2k}=\sum_{k=0}^{n/2}\binom n{n-2k}m^{n-2k}\left(m^2-p\right)^{k}.$$
This is a polynomial in $m,p$.
|
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|
Distinct roots of $ax^2-bx+c=0$ in $(0,1)$, where $a,b,c\in \mathbb{Z}^+$
Question Statement:-
Let $a,b,c$ be positive integers and consider all the quadratic equations of the form $ax^2-bx+c=0$ which have two distinct real roots in $(0,1)$. Find the least positive integers $a$ and $b$ for which such a quadratic equation exist.
Attempt at a solution:-
Let $f(x)=ax^2-bx+c$. Now as the roots of $f(x)=0$ lie in $(0,1)$ hence $f(0)\cdot f(1)\gt0$.
$$D\gt0\implies b^2-4ac\gt0$$
$$f(0)=c \\ f(1)=a-b+c$$
As, $a,b,c\in\mathbb{Z}^+$, so
$$f(0)\ge1 \implies c\ge1\tag{1}$$
$$f(1)\ge1\implies a-b+c\ge1\tag{2}$$
$$f(0)\cdot f(1)\ge1\implies c(a-b+c)\ge1\implies c^2+(a-b)c-1\ge0\tag{3}$$
Now, for the quadratic inequality in $c$ that we obtained in $(3)$, for it to hold we get
$$D\le0\implies (a-b)^2+4\le0$$
The above inequality doesn't hold $\forall a,b \in \mathbb{Z}^+$.
After arriving at this conclusion I am kind of stuck on what to conclude from the above result or rather am I going in the right direction.
A little push in the right direction in the form of a hint or rather what should be the line of thought for attempting the question. Please don't write the solution, just the line of thought to arrive at the solution would be enough.
|
The brute force approach works out fairly quickly in this case.
*
*$f(0) \gt 0$ $\implies$ $c \gt 0$ and $f(1) \gt 0$ $\implies$ $b \lt a + c$
*the product of the roots must be in $(0,1)$ $\implies$ $c < a$
*the discriminant $\Delta \gt 0$ $\implies$ $b^2 \gt 4 ac$
It follows that $c \ge 1$, $a \ge c+1$ and $b \in (\sqrt{4ac}, a+c)$. Trying the first few values starting at the lowest possible $c=1$ and $a=2$ gives:
$$
\begin{align}
& c = 1 & a = 2 \quad \implies \quad & b \in (\sqrt{8},3) & = & \;\emptyset \\
& & a = 3 \quad \implies \quad & b \in (\sqrt{12},4) & = & \;\emptyset \\
& & a = 4 \quad \implies \quad & b \in (\sqrt{16},5) & = & \;\emptyset \\
& & a = 5 \quad \implies \quad & b \in (\sqrt{20},6) & = & \;\{\;5\;\}
\end {align}
$$
The latter gives the solution $5 x^2 - 5x +1$ which can be easily verified to have both roots in $(0,1)$.
[ EDIT ] To complete the proof that $a=5$ is indeed the smallest possible value, the cases $c=2,3$ must also be checked (we can stop at $3$ since $c \ge 4$ implies $a \ge 5$).
$$
\begin{align}
& c = 2 & a = 3 \quad \implies \quad & b \in (\sqrt{24},5) & = & \;\emptyset \\
& & a = 4 \quad \implies \quad & b \in (\sqrt{32},6) & = & \;\emptyset
\end {align}
$$
$$
\begin{align}
& c = 3 & a = 4 \quad \implies \quad & b \in (\sqrt{48},7) & = & \;\emptyset
\end {align}
$$
|
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|
Solving the differential equation $dy/dx=\frac{6}{x+y}$
Let $ \text{ }\dfrac{dy}{dx}=\frac{6}{x+y}$ where $y(0)=0$. Find the value of $y$ when $x+y=6$.
Let $x+y=v$. Thus $$ \text{ }\dfrac{dy}{dx}=\dfrac{dv}{dx}-1$$
Therefore $$ \dfrac{dv}{dx}=\dfrac{6+v}{v}$$
On separating the variables and integrating, I get
$$y=6\ln(x+y+6
)+C$$
Could somebody please show me where I've gone wrong?
|
$u = x + y \implies u' = 1 + y' $. Thus,
$$ u' - 1 = \frac{6}{u} \implies u' = \frac{6+u}{u} \implies \int \frac{ u du }{6 + u} = x +C \implies u - 6 \ln|u+6| = x + C $$
Since $u = x+y$, then
$$ x + y - 6 \ln | x + y + 6 | = x + C $$
Since $y(0) = 0$, then
$$ 0 + 0 - 6 \ln 6 = C \implies C = - 6 \ln 6 $$
|
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|
Prime factor of $A=14^7+14^2+1$ Find a prime factor of $A=14^7+14^2+1$. Obviously without just computing it.
|
Hint: I've seen the 3rd cyclotomic polynomial too many times.
$$
\begin{aligned}
x^7+x^2+1&=(x^7-x^4)+(x^4+x^2+1)\\
&=x^4(x^3-1)+\frac{x^6-1}{x^2-1}\\
&=x^4(x+1)(x^2+x+1)+\frac{(x^3-1)(x^3+1)}{(x-1)(x+1)}\\
&=x^4(x+1)(x^2+x+1)+(x^2+x+1)(x^2-x+1)
\end{aligned}
$$
|
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|
If the matrices $A^3 = 0$, $B^3=0$ and $AB=BA$ then show this: The question: If $A$ and $B$ are square matrices of the same type such that $A^3=0$, $B^3=0$ and $AB=BA$. Show that
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$
This is how I started.
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{A^2+B^2}{2!}+AB+\frac{AB^2}{2!}+\frac{A^2B}{2!}+\frac{A^2B^2}{2!2!}$$
I tried to get
$$\frac{A^2+B^2}{2!}+AB+\frac{AB^2}{2!}+\frac{A^2B}{2!}+\frac{A^2B^2}{2!2!}$$
to be equal to $$\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$ But I'm not getting there.
|
You can simply match the terms of second order, third order and fourth order to the corresponding terms in the answer:
For second order terms: by multiplication $(A+B)^2=A^2+AB+BA+B^2$ and using $AB=BA$ we see that
$$
\frac{(A+B)^2}{2!} = \frac{A^2+B^2}{2} + AB
$$
For third order terms we have by multiplication $(A+B)^3 = A^3 + (A^2B+ABA+B^2A) + (AB^2+BAB+B^2A) + B^3$. By $A^3=B^3=0$ this becomes
$(A+B)^3=(A^2B+ABA+B^2A) + (AB^2+BAB+B^2A)$ and using $AB=BA$ gives
$(A+B)^3=3(A^2B+AB^2)$. The commutativity property simply allows us to reorder the terms in the multiplications. Therefore
$$
\frac{(A+B)^3}{3!}=\frac{A^2B+B^2A}{2!}
$$.
Finally, for the fourth order term we note that by commutativity we have
$$
(A+B)^4 = A^4+4A^3B+6A^2B^2+4AB^3+B^4
$$
Since $A^3=B^3=0$ only the middle term remains. Hence
$$
\frac{(A+B)^4}{4!} = \frac{A^2B^2}{2!2!}.
$$
|
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|
Number of Integer Solutions to $k^2-2016=3^n$ How many pairs of integers (k,n) satisfy the equation
$k^2-2016=3^n$
The only thing I could think to do was factor out 3^2 from the entire equation to get
$(k/3)^2-224=3^{n-2}$
but that doesn't lead to anything useful. I know that (45, 2) is a solution, but that was from trial and error. Can you show me how to solve this or at least point me in the right direction? Thanks.
|
Note that $2016=224\times 9$ and $x^2\equiv 0,1 \pmod 4$. If $x^2\equiv 0 \pmod 4$, then $3^n\equiv 0 \pmod 4$, but this is impossible. So it must be $x^2\equiv 1 \pmod 4$. Since $3^{2r+1}\equiv 3 \pmod 4$ and $3^{2r}\equiv 1 \pmod 4$ we deduce that $n$ is even. So, let's say $n=2m$ with $m\ge 0$. On the other hand, since $x^2=2016+9^m$ we get that $3|x$, so we can write $x=3y$.
Therefore we have the equation $9y^2-2016=9^m$, which is equivalent to $y^2-224=9^{m-1}$. If $m>1$, $9|9^{m-1}$ and thus $9|y^2-224$. Since $224\equiv -1 \pmod 9$, we get that $y^2\equiv -1\pmod 9$, but for $a\in \mathbb{Z}$ we have $a^2\equiv 0,1,4,7 \pmod 9$, so it's impossible to have $y^2\equiv -1 \pmod 9$. Then $m=1$ and so $n=2$. From this we get $k^2=2016+9=2025$, so $k=\pm 45$.
Hence, all the integer solutions are the pairs $(k,n)=(45,2), (-45,2)$.
|
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Prove $e^{-ax} \le (1-x)^a + \frac{1}{2}ax^2$ for $a > 1$ and $0 \le x \le 1$: Prove that for all $a > 1$ and $0 \le x \le 1$:
$$
e^{-ax} \le (1-x)^a + \frac{1}{2}ax^2
$$
My very limited start:
$f(x) = (1-x)^a + \frac{1}{2}ax^2 - e^{-ax} \ge 0$
$f'(x) = ax + a e^{-ax} - a(1-x)^{a-1}$
|
First, note from the well known $e^{t} \geqslant 1+t$ that
$$e^{-ax} \geqslant (1-x)^a = (1-x)^{a-1}- x(1-x)^{a-1} \geqslant (1-x)^{a-1}-x$$
Multiply by $-a$ to get
$$-a e^{-ax} \leqslant -a(1-x)^{a-1} + ax$$
Integrating this in $[0, x]$ gives
$$e^{-ax}\color{red}{-1} \leqslant (1-x)^a\color{red}{-1} + \frac12ax^2$$
|
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|
If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$
If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\displaystyle \beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$
$\bf{My\; Try::}$ Given $$ \alpha = \arctan \bigg(\frac{2(2\sqrt{2}-1)}{1-(2\sqrt{2}-1)^2}\bigg)=\arctan \bigg(\frac{(2\sqrt{2}-1)}{2\sqrt{2}-4}\bigg)$$
and $$\beta = \arcsin\bigg(1-\frac{4}{27}\bigg)+\arcsin\bigg(\frac{3}{5}\bigg) = \arcsin\bigg(\frac{23}{27}\bigg)+\arcsin\bigg(\frac{3}{5}\bigg)$$
So $$\arcsin\bigg[\frac{23}{27}\cdot \frac{4}{5}+\frac{3}{5}\cdot \frac{10\sqrt{2}}{23}\bigg]$$
Now how can i solve it, Help required, Thanks
|
HINT:
As $\arcsin\dfrac13<\arcsin\dfrac12=\dfrac\pi6$
$3\arcsin\dfrac13=\arcsin\dfrac{23}{27}=\arctan\dfrac{23}{10\sqrt2}$
$\arcsin\dfrac35=\arctan\dfrac34$
As $\dfrac{23}{10\sqrt2}\cdot\dfrac34>1,$
using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$\arctan\dfrac{23}{10\sqrt2}+\arctan\dfrac34=\pi+\arctan\dfrac{\dfrac{23}{10\sqrt2}+\dfrac34}{1-\dfrac{23}{10\sqrt2}\cdot\dfrac34}$$
As $2\sqrt2-1>1$
$$2\arctan(2\sqrt{2}-1)=\pi+\arctan ?$$
Now for $\dfrac\pi2>a>b>-\dfrac\pi2,\arctan a>\arctan b $
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$f(z)=\frac{(iz+2)}{(4z+i)}$ maps the real axis in the $\mathbb{C}$-plane into a circle Find the center and radius of the circle. Also find the points on the complex plane which is mapped onto the center of the circle.
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Let $f(z)=w=x+iy$ and show that $(x,y)$ is on a circle, provided $z \in \mathbb{R}$. First, solve for $z$ so we can get a look at its real and imaginary parts: $$w=\frac{(iz+2)}{(4z+i)} \\
(4z+i)w= iz+2 \\
z(4w-i)=2-iw \\
z=\frac{2-iw}{4w-i} = \frac{2-iw}{4w-i}\frac{4\overline{w}+i}{4\overline{w}+i}=\frac{8\overline{w}+2i-4i|w|^2+w}{|4w-i|^2}$$
Since $z$ is a real number, the imaginary part of the numerator of that last expression must be zero. So
$$-8y+2-4(x^2+y^2)+y=0 \\
x^2 + y^2 +\frac{7}{4}y=\frac{1}{2} \\
x^2 + y^2 +\frac{7}{4}y + \left(\frac{7}{8}\right)^2=\frac{1}{2}+ \left(\frac{7}{8}\right)^2 = \frac{81}{64} \\
x^2 +\left(y+ \frac{7}{8}\right)^2=\left(\frac{9}{8}\right)^2 $$
Now claim the point $f(z)=(x,y)$ does lie on a circle, the circle has center $(0,-7/8)$ and the radius of the circle is $\frac{9}{8}$.
Verified by plotting some points.
To find the point $z_c$ that $f$ maps to the center of the circle $w_c$, we can use our simplest expression for $z$ in terms of $w$ from above.
$$z_c=\frac{2-iw_c}{4w_c-i}=\frac{2-i\left( \frac{-7}{8}i \right)}{4\left( \frac{-7}{8}i \right)-i}=\frac{2-\frac{7}{8}}{-\frac{28}{8}i-i}=\frac{i}{4}$$
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|
$\lim_\limits{n\to \infty}\ (\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})$? what is the value given to this limit?
$$\lim_{n\to \infty}\ \bigg(\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + \cdots + \frac{1}{n+1}\bigg)$$
Is it simply 0 because each term tends to 0 and you are just summing up zeros?
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Note that each term tends to $0$ but the number of terms tends to $\infty$.
Hint. On may write
$$
\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1}=\frac1n \cdot \sum_{k=1}^n\frac1{1+\frac{k}{n}}
$$ and recognize a Riemann sum.
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if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) i want to show that
if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) but im not quite sure how to aproach this problem.
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Remember Eulers theorem applied to the prime $5$ : $$ a^4 = 1 \mod 5 \iff a \mod 5 \neq 0$$ so we have $$(a^4+b^4+c^4+d^4) \mod 5 = 0 \iff a=b=c=d=0 \mod 5$$
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Solving inequations so that $x \rightarrow a^x$
The exponential function $x \rightarrow a^x$, with $0 < a < 1$ is
decrescent in $\mathbb{R}$.
Taking this property in consideration, solve the following
inequalities:
a. $(\frac{1}{3})^{2x} \le (\frac{1}{3})^{x+1}$
b. $0.1^{2-x}>0.1^{3x}$
I did:
a.$$2x \le n+1 \Leftrightarrow 2x-x \le 1 \Leftrightarrow x \le 1$$
b.$$2-x > 3x \Leftrightarrow \frac{2-x}{3} > x \Leftrightarrow -\frac{x}{3}-x > \frac{2}{3} \Leftrightarrow \frac{x}{3}+x < \frac{2}{3} \Leftrightarrow \frac{4x}{3} < \frac{2}{3} \Leftrightarrow 4x < 2 \Leftrightarrow x < \frac{1}{2}$$
But my book says the solution is $x \ge 1$ for a. and $x>\frac{1}{2}$ for b.
What did I do wrong? Did I overlook something regarding the given property?
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When the sides of an inequality are multiplied (or divided) by a negative number, the inequality turns to the opposite side.
It appears you took the $\log$ of the sides to write your inequalities. in the first case, for instance,
$$\log\left(\left(\frac{1}{3}\right)^{2x}\right) \le \log\left(\left(\frac{1}{3}\right)^{x+1}\right)$$
$$2x\log\left(\frac{1}{3}\right) \le (x+1)\log\left(\frac{1}{3}\right)$$
Dividing the sides by $\log\left(\frac{1}{3}\right)$ $\color{red}{\text{which is negative}}$ yields:
$$2x\ge x+1\Rightarrow x\ge1$$
Since $\frac{1}{3}<1$ and $0.1<1$, this will flip the inequalities (since the result of $\log$ is negative). So the result is $x\ge1$ and $x>\frac{1}{2}$.
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Compute $\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})\exp(-\frac{u^2}{2\sigma_1^2\sigma_2^2}(\sigma_2v+\frac{\sigma_1}{v})^2)\,dv$
Compute $$\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})e^{-\frac{u^2}{2\sigma_1^2\sigma_2^2}(\sigma_2v+\frac{\sigma_1}{v})^2}\,dv$$
Does this integral has a close form solution? What if $\sigma_1=\sigma_2=1$, $u=\sqrt2$, i.e, $$\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})e^{-(v+\frac{1}{v})^2}dv$$
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Let $I$ denote the integral. Then by the substitution $v \to \frac{1}{v}$ we find that
\begin{align*}
I
&= \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_1^2\sigma_2^2}\left(\sigma_2 v+\frac{\sigma_1}{v} \right)^2 \bigg\} \,dv \\
&\qquad+ \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_1^2\sigma_2^2}\left(\sigma_1 v+\frac{\sigma_2}{v} \right)^2 \bigg\} \,dv.
\end{align*}
Then using the relation
$$ \left( \sigma_i v+\frac{\sigma_j}{v} \right)^2 = \sigma_i^2 \left( v - \frac{\sigma_j}{\sigma_i v} \right)^2 + 4\sigma_i \sigma_j $$
for $\{i,j\} = \{1,2\}$, we can write
\begin{align*}
I
&= e^{-\frac{2u^2}{\sigma_1 \sigma_2}} \bigg( \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_1^2}\left(v - \frac{\sigma_1}{\sigma_2 v} \right)^2 \bigg\} \,dv \\
&\hspace{5em} + \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_2^2}\left(v - \frac{\sigma_2}{\sigma_1 v} \right)^2 \bigg\} \,dv \bigg).
\end{align*}
Applying the Glasser's Master Theorem, we find that
\begin{align*}
I
&= e^{-\frac{2u^2}{\sigma_1 \sigma_2}} \bigg( \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_1^2}v^2 \bigg\} \,dv + \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_2^2}v^2 \bigg\} \,dv \bigg) \\
&= \sqrt{2\pi } \left( \frac{\sigma_1 + \sigma_2}{|u|} \right) \exp\bigg\{ -\frac{2u^2}{\sigma_1 \sigma_2} \bigg\}.
\end{align*}
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Find the $n$ derivative of $y= e^{2x}\sin^2 x$ We have
\begin{align*}
y&= e^{2x}\sin^2 x\\
&= e^{2x}\left(\frac{1-\cos 2x}{2}\right)\\
&= \frac{e^{2x}}{2} - \frac{e^{2x}\cos 2x}{2}
\end{align*}
Then
\begin{align*}
y^{(n)} &= \left(\frac{e^{2x}}{2}\right)^{(n)} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}\\
&= 2^{n-1}e^{2x} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}
\end{align*}
I don't know how to proceed with the rightmost term. So far I've been applying the Leibniz Rule whenever I've had to find the $n$ derivative of a function of the form $f(x)g(x)$, because is clear that either $f(x)$ or $g(x)$ has a derivative a $k$ derivative ($1<k<n$) equal to zero, which simplifies the expression nicely.
But here
$$\frac{e^{2x}\cos 2x}{2}$$
both functions are infinitely differentiable on $\mathbb{R}$ which makes things a bit different.
My only attempt was to write its n derivative in this form
\begin{align*}
&=\frac{1}{2}\sum_{k=0}^n{n \choose k} \big(2^k e^{2x}\big)\bigg(2^{n-k}\cos \left[2x + \frac{\pi(n-k)}{2}\right]\bigg)\\
&=\sum_{k=0}^n {n \choose k} 2^{n-1}e^{2x}\cos \left[2x + \frac{\pi(n-k)}{2}\right]
\end{align*}
So
\begin{align*}
y^{(n)} &= 2^{n-1}e^{2x} -\sum_{k=0}^n{n \choose k} 2^{n-1}e^{2x}\cos \left[2x + \frac{\pi(n-k)}{2}\right]\\
&= 2^{n-1}e^{2x}\left(1 -\sum_{k=0}^n{n \choose k} \cos \left[2x + \frac{\pi(n-k)}{2}\right]\right)
\end{align*}
But the textbook's answer is
$$2^{n-1}e^{2x}\left(1 -2^{n/2}\cos \left[2x + \frac{\pi n}{4}\right]\right)$$
For some reason I have the feeling that a little of modular arithmetic has to be applied on $\frac{\pi(n-k)}{2}$
|
You're correct that Leibniz's Rule is a sound way forward. Note that we have
$$\begin{align}
e^{2x}\sin^2(x)=\frac12 e^{2x}-\frac12 e^{2x}\cos(2x)
\end{align}$$
Then, taking the $n$'th order derivative, we have
$$\begin{align}
\frac{d^n}{dx^n}\left(e^{2x}\sin^2(x)\right)&=\frac{d^n}{dx^n}\left(\frac12 e^{2x}-\frac12 e^{2x}\cos(2x)\right)\\\\
&=2^{n-1}e^{2x}-\frac12 \sum_{k=0}^n\binom{n}{k}\frac{d^{n-k}e^{2x}}{dx^{n-k}}\frac{d^k\cos(2x)}{dx^k}\\\\
&=2^{n-1}e^{2x}\left(1-\sum_{k=0}^n\binom{n}{k}\frac{d^k\cos(2x)}{d(2x)^k}\right)\\\\
\end{align}$$
So, the problem boils down to taking the $k$'th derivative of the cosine function. But we can express that derivative as $\cos(x+k\pi/2)$. Hence, we have
$$\begin{align}
\frac{d^n}{dx^n}\left(e^{2x}\sin^2(x)\right)&=2^{n-1}e^{2x}\left(1-\sum_{k=0}^n\binom{n}{k}\cos(x+k\pi/2)\right)\\\\
&=2^{n-1}e^{2x}\left(1-\text{Re}\left(\sum_{k=0}^n\binom{n}{k}e^{i(x+k\pi/2)}\right)\right)\\\\
&=2^{n-1}e^{2x}\left(1-\text{Re}\left(e^{ix}(1+i)^n\right)\right)\\\\
&=2^{n-1}e^{2x}\left(1-\text{Re}\left(2^{n/2}e^{i(x+n\pi/4)}\right)\right)\\\\
&=2^{n-1}e^{2x}\left(1-2^{n/2}\cos(x+n\pi/4)\right)\\\\
\end{align}$$
as was to be shown!
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|
Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 - \frac{1}{8} x^4 + O \left( x^5 \right) $$
I notice powers of two in the denominator, but I'm not sure of the pattern (and to calculate the next term to confirm would entail another tedious product rule).
Any ideas? Thanks!
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Herein, we present an approach using partial fraction expansion. Proceeding, we write
$$\begin{align}
\frac{1}{x^2+2x+2}&=\frac{1}{i2}\left(\frac{1}{x+1-i}-\frac{1}{x+1+i}\right)\\\\
&=\text{Im}\left(\frac{1}{x+1-i}\right)\\\\
&=\text{Im}\left(\frac{1}{1-i}\frac{1}{1+\frac{x}{1-i}}\right)\\\\
&=\sum_{n=0}^\infty \frac{(-1)^n\sin((n+1)\pi/4)}{2^{(n+1)/2}}\,x^n\\\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}\sin(n\pi/4)}{2^{n/2}}\,x^{n-1}\\\\
&=\frac12-\frac x2 +\frac{x^2}{4}+\cdots
\end{align}$$
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|
Simplify $G(t) = \sum_\limits{n=1}^{\infty} (\cos\frac{n\pi}{2}-1) \cos(\frac{nt\pi}{3}) $ Simplify $G(t) = \sum_\limits{n=1}^{\infty} (\cos\frac{n\pi}{2}-1) \cos(\frac{n\pi t}{3}) $
I am unsure how to simplify this in the best form, anyone have any ideas?
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$$ (\cos\frac{n\pi}{2}-1) \cos(\frac{nt\pi}{3}) = \\ \frac{1}{2} \cos\left( n \left(\frac{\pi t}{3} - \frac{\pi}{2}\right)\right)+\frac{1}{2} \cos\left( n \left(\frac{\pi t}{3} + \frac{\pi}{2}\right)\right) -\cos \left( \frac{n \pi t}{3} \right) $$
So all terms are cosines.
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Use algebra to calculate the perimeter of a right angle triangle Use algebra to calculate the perimeter of a right angle triangle where the hypotenuse is $(x+4)$, the base is $(x+3)$ and the remaining side is $(x-4)$.
Thanks to the earlier comments I know that I need to use Pythagoras.
So far I know that the Theorem says that $a^2 + b^2 = c^2$
I can write my problem as $(x-4)^2 + (x+3)^2 = (x+4)^2$
What should I do next?
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You're almost done.
$$\begin{align*}(x - 4)^2 + (x + 3)^2 &= (x + 4)^2 \\
\implies x^2 - 8x + 16 + x^2 + 6x + 9 &= x^2 + 8x + 16 \\
\implies x^2 - 10x + 9&= 0 \\
\implies (x - 9)(x - 1) &= 0 \end{align*}$$
So either $x = 9$ or $x = 1$. We know $x \neq 1$ because that would make a side of the triangle $x - 4 = -3$ which is negative. Hence $x = 9$.
Now the perimeter is $(x + 4) + (x + 3) + (x - 4) = 13 + 12 + 5 = 30$.
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.