Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Integrate $\int \frac{x}{x^4+4}dx$ I'm having trouble solving the integral
$\int \frac{x}{x^4+4}dx$
I already tried the following but, I'm getting stuck after step 3.
$\int \frac{x}{(x^2)^2+4}dx $
$u = x^2 , du = 2x dx, \frac{du}{2}=xdx$
$\frac{1}{2}\int \frac{1}{u^2+4}du$
| Use $$\frac{x}{x^4+4}=\frac{x}{x^4+4x^2+4-4x^2}=\frac{x}{(x^2-2x+2)(x^2+2x+2)}=$$
$$=\frac{1}{4}\left(\frac{1}{x^2-2x+2}-\frac{1}{x^2+2x+2}\right)=\frac{1}{4}\left(\frac{1}{(x-1)^2+1}-\frac{1}{(x+1)^2+1}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How many positive integer values of n exist such that $\frac{4^n+2^n+1}{n^2+n+1}$ also a positive integer? My question is as follows: find the number of values of the positive integers $n$ such that the fraction below is also a positive integer:
$$\frac{4^n+2^n+1}{n^2+n+1}$$
I have only been able to establish the bases cases for $n=2$ and $n=4$ which this statement is true for and have been unable to progress further.
| Let's prove first that, if $k = 2^s$, then $x^k + x^{k/2} + 1$ is divisible by $x^2+x+1$ over the integers. This is because we can do this single trick, knowing that $(a+b)(a-b) = a^2-b^2$:
$$(x^2+x+1)(x^2-x+1) = (x^2+1)^2 - x^2 = x^4+x^2+1$$
We can apply it again:
$$(x^4+x^2+1)(x^4-x^2+1) = (x^4+1)^2-x^4 = x^8+x^4+1$$
It's easy to make an induction proof of the previous statement replicating this process to obtain a factorization of $x^k + x^{k/2}+1$ that contains $x^2+x+1$.
Now, in your fraction $\frac{4^n+2^n+1}{n^2+n+1}$ we can make an hypothesis (seeing the first solutions) that $n = 2^{2^m}$ is always a solution, for each $m\in \mathbb{N}$.
Let's study $4^n$:
$$4^n = 4^{2^{2^m}} = 2^{2\cdot 2^{2^m}} = 2^{2^{2^m+1}} = 2^{2^m2^{2^m+1-m}} = \big(2^{2^m}\big)^{2^{2^m+1-m}}$$
This concludes the proof taking $s=2^m+1-m$, $k=2^s$ and applying the previous lemma, because we have:
$$\frac{4^n+2^n+1}{n^2+n+1} = \frac{n^{k}+n^{k/2}+1}{n^2+n+1}$$
This expression can be written as a polynomial with integer coefficients on $n$, and because $n$ is an integer, the result is always an integer. So there is an infinite countable number of solutions.
Side note: These are not the only solutions, $n=215$ also works and it's not one of the mentioned above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the limit $\lim_{x\to 0} x\left(\left[\frac{1}{x}\right] +\left[\frac{2}{x}\right] +\cdots \left[\frac{10}{x}\right] \right)$ Can someone help me finding the following limit
$$ \lim_{x\to 0} x\left(\left\lfloor\frac{1}{x}\right\rfloor +\left\lfloor\frac{2}{x}\right\rfloor +\cdots \left\lfloor\frac{10}{x}\right\rfloor\right)$$
I can somehow guess the limit will be $55$, as $\lim_{x\to 0}x\left\lfloor\frac{1}{x}\right\rfloor=1$. But, I am not able to prove it.
Note: $\left\lfloor x\right\rfloor$ denotes the greatest integer less than or equal to $x$.
| Hint Since $u-1 < \lfloor u \rfloor \leq u$, you have
$$ \left(\frac{1}{x} +\frac{2}{x} +\cdots \frac{10}{x}\right)-10 \leq \left(\left[\frac{1}{x}\right] +\left[\frac{2}{x}\right] +\cdots \left[\frac{10}{x}\right] \right) \leq \left(\frac{1}{x} +\frac{2}{x} +\cdots \frac{10}{x}\right)$$
Therefore,
$$ \frac{55}{x} -10 \leq \left(\left[\frac{1}{x}\right] +\left[\frac{2}{x}\right] +\cdots \left[\frac{10}{x}\right] \right) \leq \frac{55}{x}$$
Now multiply both sides by $x$, splitting the problem into $x >0$ and $x <0$ (since in the second case the inequality flips when you multiply).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Find the $n$-th power of a $3{\times}3$ matrix using the Cayley-Hamilton theorem. I need to find $A^n$ of the matrix $A=\begin{pmatrix}
2&0 & 2\\
0& 2 & 1\\
0& 0 & 3
\end{pmatrix}$ using Cayley-Hamilton theorem.
I found the characteristic polynomial $P(A)=(2-A)^2(3-A)$ from which I got $A^3=7A^2-16A+12$. How to continue?
| Since $p(A)=0$ where $p(x)=(2-x)^2(3-x)$, if we divide $x^n$ by $p(x)$ to get $x^n=p(x)q(x)+r(x)$, then $A^n=p(A)q(A)+r(A)=0q(A)+r(A)=r(A)$, so it suffices to figure out $r$. However, if $n$ is large, actually doing division will not be effective.
Since $p(2)=p(3)=0$, we have that $r(2)=2^n$ and $r(3)=3^n$. However, since all we know about the degree of $r$ is that it is less than $3$, we need another value to specify it. However, if we differentiate
$$x^n=p(x)q(x)+r(x)$$
to get
$$nx^{n-1}=p'(x)q(x)+p(x)q'(x)+r'(x)$$
and use the fact that, because $2$ is a double root of $p(x)$, $p(2)=p'(2)=0$, then we get $n2^{n-1}=r'(2)$.
If we write $r(x)=a_nx^2+b_nx+c_n$, then we get the system of equations:
$$\begin{align} 2^n &=4a_n+2b_n+c_n \\ 3^n&=9a_n+3b_n+c_n \\ n2^{n-1}&=4a_n+b_n \end{align}.$$
Then we can solve
$$\pmatrix{a_n \\ b_n \\ c_n}=\pmatrix{4&2&1\\9&3&1\\4&1&0}^{-1}\pmatrix{2^n\\3^n\\n2^{n-1}}=\pmatrix{3^n-2^n-n2^{n-1}\\4\cdot 2^n+5n\cdot 2^n-3^n\\4\cdot 3^n-3\cdot 2^n-6n\cdot 2^n}.$$
However, we can simplify things if we note that $A$ actually satisfies the quadratic equation $(A-2I)(A-3I)=0$ (the minimal polynomial will always divide the characteristic polynomial and has the same roots but with potentially smaller multiplicities, so there aren't many combos to check here), so we can use this quadratic polynomial instead of the characteristic polynomial to obtain that $A^n=a_nA+b_nI$ for some $a_n,b_n$. If $r_n(x)=a_nx+b_n$ is the remainder polynomial, we have (using the exact same proceedure as before)
$$2^n=2a_n+b_n, \quad 3^n=3a_n+b_n,$$
so $$a_n=3^n-2^n, \quad b_n=-2\cdot 3^n-3\cdot 2^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Asymmetric inequality in three variables $\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$ Consider three real positive variables $a,\ b$ and $c$. Prove that the following inequality holds:
$$\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$$
My progress: We can prove that both sides are greater than $12$ using AM-GM:
$$LHS \geq \frac{3 \cdot 4ab \cdot 4bc}{4a^2bc} = 12$$
and
$$RHS \geq 7+\frac{5[(a+b)^2+(b+c)^2]}{2(a+b)(b+c)} \geq 7+5 = 12$$
So, substract $12$ from both sides and write the inequality into:
$$3\cdot \frac{(a+b)^2(b+c)^2-16ab^2c}{4ab^2c} \geq 5 \cdot \frac{a^2+b^2+c^2-ab-bc-ca}{(a+b)(b+c)}$$
or
$$3\cdot \frac{(b+c)^2(a-b)^2+4ab(b-c)^2}{4ab^2c} \geq \frac{5}{2}\cdot \frac{(a-b)^2+(b-c)^2+(c-a)^2}{(a+b)(b+c)}$$
My next idea was to use $(c-a)^2\leq 2[(a-b)^2+(b-c)^2]$ and write it into a sum of square form with only $(a-b)^2$ and $(b-c)^2$. However, I couldn't reach significant progress.
| Making the coordinates transformation
$$
\cases{
\frac{a+b}{a}=x\\
\frac{b+c}{c}=y\\
a b^2 c=z
}
$$
solving for $a,b,c$ we have
$$
\left\{
\begin{array}{rcl}
a & = & \frac{z}{(y-1)^2 \left(\frac{(x-1) z}{(y-1)^3}\right)^{3/4}} \\
b & = & (y-1) \sqrt[4]{\frac{(x-1) z}{(y-1)^3}} \\
c & = & \sqrt[4]{\frac{(x-1) z}{(y-1)^3}} \\
\end{array}
\right.
$$
conditioned to $x > 1,\ y > 1,\ z>0$. Substituting into
$$
3\frac{(a+b)^2(b+c)^2}{4ab^2c}- 7 - 5\frac{a^2+2b^2+c^2}{(a+b)(b+c)}\ge 0
$$
we have
$$
f(x,y) = \frac{3 x^3 y^3-4 x^2 (y (17 y-27)+15)+4 x (y (27 y-47)+30)-20 (3 (y-2) y+4)}{4 (x-1) x (y-1) y}\ge 0
$$
Now $f(x,y)$ has a minimum for $x=y=2$ such that $f(2,2) = 0$
Follows a plot of $f(x,y)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving a limit by two methods with different results I'm considering this limit
$$\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}.$$
My first attempt was using the following equivalent infinitesimals
$$1-\cos x^2 \sim \frac{x^4}{2},\quad \arctan x \sim x, \quad \sin 2x \sim 2x, \quad e^{x^2} - 1 \sim x^2 \,\,\,\,\text{when}\,\,\,\,x\rightarrow 0,$$
and then
\begin{align*}
\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}&=\lim_{x\to 0}\frac{x^5/2}{2xe^{x^2}-2x+\frac{2}{3}x^3}=\lim_{x\to 0}\frac{x^4}{4(e^{x^2}-1)+\frac{4}{3}x^2}\\
&=\lim_{x\to 0}\frac{x^4}{4x^2+\frac{4}{3}x^2}=\lim_{x\to 0}\frac{3}{16}x^2=0.
\end{align*}
Later, inspecting this other approach combining infinitesimals and the Taylor expansions
\begin{align*}
e^{x^2} &= 1+x^2+\frac{x^4}{2}+o(x^4),\\
\sin 2x &= 2x - \frac{4x^3}{3} + \frac{4x^5}{15} + o(x^5),\\
e^{x^2}\sin 2x &= 2x - \frac{2x^3}{3} - \frac{x^5}{15} + o(x^5),
\end{align*}
I get this other result
\begin{align*}
\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}&=\lim_{x\to 0}\frac{x^5/2}{2x - \frac{2x^3}{3} - \frac{x^5}{15} + o(x^5)-2x+\frac{2}{3}x^3}\\
&=\lim_{x\to 0}\frac{x^5/2}{-x^5/15 + o(x^5)}=-\frac{15}{2}.
\end{align*}
Can someone help me to identify what am I getting wrong?
Thanks.
| Keep in mind that you cannot add equivalents inconsiderately. So you second approach is correct. A simpler counter example is
$$f(x)={\sin{x}-x\over x^2}$$
If I just take equivalents I find $f(x)\to \infty$ obviously incorrect because the limit is $0$ as the Taylor expansions show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Maximize $x^2+2y^2$ subject to $y-x^2+1=0$
Maximize $x^2+2y^2$ subject to $y-x^2+1=0$
I tried using Lagrange multiplier method. We have:
$$L(x,y)=x^2+2y^2+\lambda(y-x^2+1)$$
So we have:
$$L_x=2x(1-\lambda)=0$$
$$L_y=4y+\lambda=0$$
One possible solution with $y-x^2+1=0$ is
$x=0$, $y=-1$, $\lambda=4$
But when we calculate Hessian determinant at $(0,-1)$ for this we have:
$$L_{xx}=2-2\lambda=-6 \lt 0$$
$$L_{yy}=4$$
$$L_{xy}=0$$
So $$L_{xx}L_{yy}-L^2_{xy} \lt 0$$
Does it not mean saddle point at $(0,-1)$
But it actually maximizes $x^2+2y^2$
What am I missing here?
| As in the answer by amd, you need to construct what is called the Bordered Hessian, which is nothing more that the Hessian when $\lambda$ is deliberately included as a new variable. Maybe give it a new name,
$$ h(\lambda, x,y) = x^2 + 2 y^2 + \lambda (y-x^2 +1) $$
The gradient is the triple
$$ h_\lambda = y-x^2 + 1, \; \; h_x = 2x - 2 \lambda x, \; \; h_y = 4y + \lambda $$
while the "bordered Hessian" is
$$
W =
\left(
\begin{array}{rrr}
0 & -2x & 1 \\
-2x & 2-2\lambda & 0 \\
1 & 0 & 4 \\
\end{array}
\right)
$$
First, at one of the side critical points where $x \neq 0,$ we have $x = \frac{\sqrt 3}{2}, \; \; y = \frac{-1}{4}, \; \; \lambda = 1.$ Here
$$
W =
\left(
\begin{array}{rrr}
0 & -\sqrt 3 & 1 \\
- \sqrt 3 & 0 & 0 \\
1 & 0 & 4 \\
\end{array}
\right)
$$
This determinant comes out $-12$ and we have a local min of $x^2 + 2 y^2$
You were worried about
the critical point where $x = 0, \; \; y = -1, \; \; \lambda = 4.$ Here
$$
W =
\left(
\begin{array}{rrr}
0 & 0 & 1 \\
0 & -6 & 0 \\
1 & 0 & 4 \\
\end{array}
\right)
$$
This determinant comes out $6$ and we have a local maximum of $x^2 + 2 y^2$
NOTE: as long as there is just one constraint function, this description of the "second derivative test" is good enough, even if we have more than two original variables. That is, we could ask for critical points of $x^2 + 2 y^2 + 3 z^2$ constrained to $z - xy - 7 = 0$ and do the same steps, in the end calculating just one determinant for each critical point. It does get messier if we have more than one constraint.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find:$\lim_{x\to 0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}$
Evalute: $$\lim_{x\to
0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}$$
My attempt:
I used the standard limits from the table:$$\lim_{x\to 0}\frac{\sin x}{x}=1,\;\;\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2},\;\;\lim_{x\to 0}\frac{\tan x}{x}=1,\;\;\lim_{x\to 0}\frac{e^x-1}{x}=1$$
$$$$
$L=\displaystyle\lim_{x\to 0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}=$
$$$$
$\displaystyle\lim_{x\to 0}\left[\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{e^{1-\cos^3x}-e^{1-\cos^4x}}\cdot\left(\frac{e^{1-\cos^3x}-1}{1-\cos^3x}\cdot\frac{1-\cos^3x}{x^2}-\frac{e^{1-\cos^4x}-1}{1-\cos^4x}\cdot\frac{1-\cos^4x}{x^2}\right)\cdot\frac{x}{\arctan x}\right]$
Substitution: $$[t=\arctan x\implies x=\tan t\;\;\&\;\; x\to 0\implies t\to 0]$$
$$\lim_{x\to 0}\frac{x}{\arctan x}\iff\lim_{t\to 0}\frac{\tan t}{t}=1$$
The next step:$$1-\cos^3x=(1-\cos x)(1+\cos x+\cos^2x)$$
$$1-\cos^4x=(1-\cos x)(1+\cos x)(1+\cos^2x)$$
Now I obtained:
$$L=1\cdot\left(1\cdot\frac{3}{2}-1\cdot 2\right)\cdot 1=-\frac{1}{2}$$
Is this correct?
| Alternatively $$\lim_{x\to0}\dfrac{e^{1-\cos^3x}-e^{1-\cos^4x}}{x\arctan x}$$
$$=\lim_{x\to0} e^{1-\cos^4x}\cdot\lim_{x\to0}\dfrac{e^{\cos^4x-\cos^3x}-1}{x\arctan x}$$
$$=-\lim_{x\to0}\cos^3x\cdot\lim_{x\to0} e^{1-\cos^4x}\cdot\lim_{x\to0}\dfrac{e^{(\cos^4x-\cos^3x)}-1}{\cos^4x-\cos^3x}\cdot\lim_{x\to0}\dfrac{1-\cos x}{x\arctan x}$$
$$=-\lim_{x\to0}\dfrac{1-\cos x}{x\arctan x}$$
$$=-\lim_{x\to0}\left(\dfrac{\sin x}x\right)^2\cdot\lim_{x\to0}\dfrac x{\arctan x(1+\cos x)}$$
$$=-\dfrac1{1+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to prove this symmetric polynomial identity According to this (equation 21 and 26) how to show that
\begin{gather}
\ln(1+\Pi_1t+\Pi_2t^2+\Pi_3t^3+...) =\Pi_1t+\frac{1}{2}(-\Pi^2_1+2\Pi_2)t^2+\frac{1}{3}(\Pi^3_1-3\Pi_1\Pi_2+3\Pi_3)t^3+.. .=\sum_{k=1}^{\infty}\frac{s_k}{k}t^k
\end{gather}
$s_1=\Pi_1 \\ s_2=-\Pi^2_1+2\Pi_2 \\ s_3=\Pi^3_1-3\Pi_1\Pi_2+3\Pi_3 $
And, I've been curious about how did they derive this equation
$(-1)^{p-1}s_p = $
\begin{vmatrix}
\Pi_1 & 1 & 0 & 0 &\dots & 0 \\
2\Pi_2 & \Pi_1 & 1& 0 &\ddots &0 \\
3\Pi_3 & \Pi_2 & \Pi_1 & 1 &\ddots & 0 \\
4\Pi_4 & \Pi_3 &\Pi_2 &\Pi_1 &\ddots&0 \\
\vdots & \vdots & \vdots & \vdots & \ddots &1 \\
p\Pi_{p} & p\Pi_{p-1} & p\Pi_{p-2} & p\Pi_{p-3} & \dots & \Pi_1
\end{vmatrix}
Is there any elementary way to show? Thanks in advance
| Given a generating function
$$ f(x) := 1+a_1t+a_2t^2+a_3t^3+\dots \tag{1} $$
whose logarithm is $$ \ln(f(x)) =
\sum_{k=1}^{\infty}\frac{s_k}{k}t^k, \tag{2} $$
differentiate both sides to get
$$ \frac{d}{dx} \ln(f(x)) = f'(x)/f(x) =
\sum_{k=1}^{\infty}s_k t^{k-1}. \tag{3} $$
Multiply both sides by $\,f(x)\,$ to get
$$ f'(x) = \sum_{k=1}^\infty k\,a_k\,t^{k-1} =
\left(1 + \sum_{k=1}^\infty a_k\,t^k\right)
\left(\sum_{k=1}^{\infty}s_k t^{k-1}\right). \tag{4} $$
For each $\,n>0\,$ this gives a system of equations linear in $\,s_1,s_2,\dots,s_n\,$ with coefficients using $\,a_1,a_2,\dots,a_n.\,$
Use Cramer's rule
to solve for $\,s_n\,$ which leads to the MathWorld determinant
$$(-1)^{n-1}s_n \!=\!
\begin{vmatrix}
1\,a_1 & 1 & 0 & 0 &\dots & 0 \\
2\,a_2 & a_1 & 1& 0 &\ddots & 0 \\
3\,a_3 & a_2 & a_1 & 1 &\ddots & 0 \\
4\,a_4 & a_3 & a_2 & a_1 &\ddots& 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots &1 \\
n\,a_n & a_{n-1} & a_{n-2} & a_{n-3} & \dots & a_1
\end{vmatrix}. \tag{5} $$
Here is an explicit example. Let $\,n=3.\,$ The linear equations are
$$ 1s_1 \!=\! 1a_1,\;\; a_1s_1 \!+\! 1s_2 \!=\! 2a_2,
\;\; a_2s_1 \!+\! a_1s_2 \!+\! 1s_3\!=\! 3a_3. \tag{6} $$
Setting this up as a matrix equation we get
$$ \begin{bmatrix}1 & 0 & 0 \\a_1 & 1 & 0\\
a_2 & a_1 & 1 \end{bmatrix} \begin{bmatrix}
s_1 \\ s_2 \\ s_3 \end{bmatrix} =
\begin{bmatrix}
1\,a_1 \\ 2\,a_2 \\ 3\,a_3 \end{bmatrix}\!. \tag{7} $$
Cramer's rule solving for $\,s_3\,$ and using the
denominator $3\times 3$ determinant $=1$ gives
the solution in equation $(5)$ for $\,n=3\,$ and that is
$$ s_3 = a_1^3 -3a_1a_2 + 3a_3 = (-1)^{3-1}s_3 = \begin{vmatrix}
1\,a_1 & 1 & 0 \\ 2\,a_2 & a_1 & 1 \\
3\,a_3 & a_2 & a_1 \end{vmatrix}. \tag{8} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving the trigonometric equation $483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$
I'm trying the value of $\alpha \in [0,\pi]$ that is solution to this trigonometric equation: $$483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$$
I've tried to write down $\sin\left(\alpha+\frac\pi{3}\right)$ and $\sin\left(2\alpha+\frac{\pi}{3}\right)$with the formula $\sin(\alpha+\beta)=\sin\alpha \cos\beta+\sin\beta\cos\alpha$, but after that I'am stuck and I don't have completely any idea of how to proceed.
| Notice that:
$$\sin\left(\alpha+\frac\pi{3}\right) = \sin\left(\alpha\right)\cos\left(\frac\pi{3}\right) +\cos\left(\alpha\right)\sin\left(\frac\pi{3}\right) = \\
= \frac{1}{2}\sin(\alpha) + \frac{\sqrt{3}}{2}\cos(\alpha).$$
Moreover:
$$\sin\left(2\alpha+\frac\pi{3}\right) = \sin\left(\alpha+\left(\alpha+\frac\pi{3}\right)\right) = \sin\left(\alpha\right)\cos\left(\alpha+\frac\pi{3}\right) +\cos\left(\alpha\right)\sin\left(\alpha+\frac\pi{3}\right).$$
Recall that $\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) -\sin(\alpha)\sin(\beta).$ Then, the previous become:
$$\sin(\alpha)\left(\cos(\alpha)\cos\left(\frac{\pi}{3}\right) - \sin(\alpha)\sin\left(\frac{\pi}{3}\right)\right) + \cos(\alpha)\left(\frac{1}{2}\sin(\alpha) + \frac{\sqrt{3}}{2}\cos(\alpha)\right) = \\
=\sin(\alpha)\left(\frac{1}{2}\cos(\alpha) - \frac{\sqrt{3}}{2}\sin(\alpha)\right) + \cos(\alpha)\left(\frac{1}{2}\sin(\alpha) + \frac{\sqrt{3}}{2}\cos(\alpha)\right) = \\
=\sin(\alpha)\cos(\alpha)-\frac{\sqrt{3}}{2}\sin^2(\alpha) + \frac{\sqrt{3}}{2}\cos^2(\alpha).$$
Let's join together these results!
$$483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0 \Rightarrow \\
\frac{483}{2}\sin(\alpha) + \frac{483\sqrt{3}}{2}\cos(\alpha) + 16\sqrt{3}\sin(\alpha)\cos(\alpha) + 24(\cos^2(\alpha) - \sin^2(\alpha)) + 20 = 0 \Rightarrow \\
\frac{483}{2}\sin(\alpha) + \frac{483\sqrt{3}}{2}\cos(\alpha) + 16\sqrt{3}\sin(\alpha)\cos(\alpha) + 48\cos^2(\alpha) - 4 = 0.
$$
The last equation can be solved by setting $X = \cos(\alpha)$ and $Y = \sin(\alpha)$ with the equation $X^2 + Y^2 = 1.$:
$$\begin{cases}
\frac{483}{2}Y + \frac{483\sqrt{3}}{2}X + 16\sqrt{3}XY + 48X^2 - 4 = 0 \\
X^2 + Y^2 = 1
\end{cases}.$$
Anyway, the last system of equations is a cute
beast, very hard to be solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
value of expression having variables $p,q,r,x,y,z$
If $p,q,r,x,y,z$ are non zero real number such that
$px+qy+rz+\sqrt{(p^2+q^2+r^2)(x^2+y^2+z^2)}=0$
Then $\displaystyle \frac{py}{qx}+\frac{qz}{ry}+\frac{rx}{pz}$ is
what try
$(px+qy+rz)^2=(p^2+q^2+r^2)(x^2+y^2+z^2)$
$p^2x^2+q^2y^2+r^2z^2+2pqxy+2qryz+2prxz=p^2x^2+p^2y^2+p^2z^2+q^2x^2+q^2y^2+q^2z^2+r^2x^2+r^2y^2+r^2z^2$
$2pqxy+2qryz+2prxz=p^2y^2+p^2z^2+q^2x^2+q^2z^2+r^2x^2+r^2y^2$
How do i solve it Help me please
| Write $v=(p,q,r)$ and $w=(x,y,z)$. Then the given relation states
$$v\cdot w+|v||w|=0$$
But $v\cdot w=|v||w|\cos\theta$ where $\theta$ is the angle between them, so
$$|v||w|(\cos\theta+1)=0$$
Since none of the scalars are zero, we get $\cos\theta=-1$, so $v=kw$ for some nonzero $k\in\mathbb R$ and
$$\frac{py}{qx}+\frac{qz}{ry}+\frac{rx}{pz}=\frac{kxy}{kyx}+\frac{kyz}{kzy}+\frac{kzx}{kxz}=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given that $x^2 + ax + b > 0$ and $x^2 + (a + np)x + (b + nq) > 0$, prove that $x^2 + (a + mp)x + (b + mq) > 0, m = \overline{1, n - 1}$.
Given that $$\large \left\{ \begin{align} x^2 + ax + b > 0\\ x^2 + (a + np)x + (b + nq) > 0 \end{align} \right., \forall x \in \mathbb R \ (p, q \in \mathbb R \setminus \{0\}, n \in \mathbb N, n > 1)$$
Prove that $x^2 + (a + mp)x + (b + mq) > 0, m = \overline{1, n - 1}$.
We have that $$\left\{ \begin{align} x^2 + ax + b > 0\\ x^2 + (a + np)x + (b + nq) > 0 \end{align} \right., \forall x \in \mathbb R$$
$$ \iff \left\{ \begin{align} a^2 - 4b < 0 \ (1)\\ (a + np)^2 - 4(b + nq) < 0 \ (2) \end{align} \right.$$
For $(2)$, we obtain that $p^2n^2 + 2(ap - 2q)n + (a^2 - 4b) < 0$
$$\implies (ap - 2q)^2 - p^2(a^2 - 4b) > 0 \iff q^2 - apq + bp^2 > 0 \implies \left\{ \begin{align} (ap)^2 - 4bp^2 < 0\\ (aq)^2 - 4q^2 < 0 \end{align} \right.$$
$\implies \left\{ \begin{align} a^2 - 4b < 0\\ a^2 - 4 < 0 \end{align} \right.$. And that's a dead-end.
| Multiply the first inequality by $(n-m)$ and the second by $m$ and add these together
\begin{eqnarray*}
(n-m) (x^2 +ax+b) >0 \\
m(x^2+(a+np)x+(b+np))>0 \\
n(x^2+(a+mp)x+(b+mp))>0.
\end{eqnarray*}
Now divide the final inequality by $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Given that $2x^2-4xy+6y^2=9$, find the biggest and lowest value of $2x-y$ So, that's the problem. I tried factoring the quadratic equation to get something like $2x-y$ but it doesn't work. Dividing by $y^2$ won't work because the right side is $9$, not $0$. The last idea I have is to say that $2x-y=t => y=2x-t$ and replace $y$ in the quadratic equation with $2x-t$ to get a function but then, I get confused.
| $$2x^2-4xy+6y^2=9$$ and let $$k=2x-y\Rightarrow y=2x-k$$
$2x^2-4x(2x-k)+6(2x-k)^2=9$
$2x^2-8x^2+4xk+6(4x^2+k^2-4xk)=9$
$-6x^2+4xk+24x^2+6k^2-24xk-9=0$
$$18x^2-20kx+6k^2-9=0$$
For real roots, Discriminant $\geq 0$
$$400k^2-4\cdot 18(6k^2-9)\geq 0$$
$$100k^2-108k^2+162\geq 0$$
$$-8k^2+162\geq 0\Rightarrow k^2-\frac{81}{4}\leq 0$$
$$-\frac{9}{2}\leq k\leq \frac{9}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Show $1+a+b+c\leq 2\lfloor\sqrt{at^2+bt+c}\rfloor$ for all $a,b,c,t\in\Bbb N\cup\{0\}$ with $a\neq 0$, $t\ge 2$ and $a,b,c\le t-1$. Related to a project I'm writing, I came across the problem of showing that
$$1+a+b+c \leq 2\lfloor \sqrt{ at^2+bt+c}\rfloor$$
holds for all nonnegative integers $a,b,c,t$ satisfying $a \neq 0$, $t \geq 2$ and $a,b,c \in \{0,1,\ldots,t-1\}$.
I have checked numerically that it holds true for all valid choices of $a,b,c$ for $t\leq 300$, but I'm not able to prove that the statement is always true. I'm suspecting that there is some simple argument to why, but unfortunately I don't see it. Can anyone help ? :)
| You're asking to confirm
$$1+a+b+c \leq 2\lfloor \sqrt{ at^2+bt+c}\rfloor \tag{1}\label{eq1A}$$
Since $a,b,c \in \{0,1,\ldots,t-1\}$, each value is at most $t - 1$, so you get
$$1 + a + b + c \le 1 + (t - 1) + (t - 1) + (t - 1) = 3t - 2 \tag{2}\label{eq2A}$$
With $a \ge 4$, you have
$$2\lfloor \sqrt{ at^2+bt+c}\rfloor \ge 2\lfloor \sqrt{ at^2}\rfloor \ge 2(2t) = 4t \gt 3t - 2 \tag{3}\label{eq3A}$$
This shows \eqref{eq1A} always holds for $a \ge 4$, so you only need to check $a = 1$, $2$ and $3$. Consider first $a = 1$. Then
$$1 + a + b + c \le 1 + 1 + (t - 1) + (t - 1) = 2t \tag{4}\label{eq4A}$$
Since for $a = 1$, you have $t^2 + bt + c \ge t^2$, you thus have $2\lfloor \sqrt{ at^2+bt+c}\rfloor \ge 2(t)$, so \eqref{eq1A} holds.
For $a = 2$, you have $1 + a + b + c \le 2t + 1$. However, if $b \lt t - 1$ or $c \lt t - 1$, then $1 + a + b + c \le 2t$, so the condition will hold since $2t^2 \gt t^2$ which we've shown above works. Thus, only need to check $2t^2 + (t - 1)t + (t - 1) = 3t^2 - 1 \ge (t + 1)^2 = t^2 + 2t + 1$. This is true since with $t \ge 2$, then $t^2 \ge 2t$, so $2t^2 \ge 4t$. Thus, $3t^2 - 1 = t^2 + 2t^2 - 1 \ge t^2 + 4t - 1 \gt t^2 + 2t + 1$.
I'll leave it to you to confirm \eqref{eq1A} also holds for $a = 3$ (hint: $\sqrt{3}t \gt t + 1$ for $t \ge 2$).
Update: The $a = 2$ and $a = 3$ cases can be combined by using that $t \ge a + 1$. With $a \le 3$, you have $1 + a + b + c \le 1 + (t - 1) + (t - 1) + 3 = 2t + 2 = 2(t + 1)$. Also, $(\sqrt{a})t \gt t + 1$, so you can use an argument similar to the one in this answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Splitting field of $\alpha=(2+\sqrt{2})^{1/3}$ Given $\alpha=(2+\sqrt{2})^{1/3}$ The problem asks to calculate the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and find it's splitting field. I have the solution and I understood until the point where it shows that the minimal polynomial is $$x^6-4x^3+2=0$$
Now, since we got this polynomial developing the equation $\alpha=(2+\sqrt{2})^{1/3}$ it is clear that $(2+\sqrt{2})^{1/3}$ is a root of the minimal polynomial.
I don't understand the following: Why are $\left( 2-\sqrt{2} \right) ^{1/3},\left( 2-\sqrt{2} \right) ^{1/3}w,\left( 2-\sqrt{2} \right) ^{1/3}w^2,\left( 2+\sqrt{2} \right) ^{1/3}w,\left( 2+\sqrt{2} \right) ^{1/3}w^2
$ the other roots of the polynomial? $w$ is the cubic root of unity ($w^3=1$).
| $$x^6-4x^3+2=0$$ $$\Leftrightarrow (x^3-2)^2=2$$
$$\Leftrightarrow x^3-2=\pm \sqrt{2}$$
$$\Leftrightarrow x^3=2\pm \sqrt{2}$$
$$\Leftrightarrow x=\omega^i\sqrt[3]{2+\sqrt{2}},\omega^i\sqrt[3]{2-\sqrt{2}},i=0,1,2,$$ where $\omega$ is a root of $x^2+x+1$, so $\omega^3=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proof by induction:$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$ In the very beginning I'm going to refer to similar posts with provided answers:
Induction Inequality Proof with Product Operator $\prod_{k=1}^{n} \frac{(2k-1)}{2k} \leq \frac{1}{\sqrt{3k+1}}$ (answered by Özgür Cem Birler)
Prove that $\prod\limits_{i=1}^n \frac{2i-1}{2i} \leq \frac{1}{\sqrt{3n+1}}$ for all $n \in \Bbb Z_+$
I examined the solutions and tried to apply the methods used there to make sure whether I understand it or not. I'm concerned about the step of induction.
A task from an earlier exam at my university:
Prove by induction:
$$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$$
Attempt:
rewritten:
$$\prod_{i=1}^n\frac{4i-1}{4i+1}<\sqrt{\frac{3}{4n+3}}$$
$(1)$ base case: $\tau(1)$
$$\frac{3}{5}=\sqrt{\frac{4}{7}}<\sqrt{\frac{3}{7}}$$
$(2)$ assumption:
Let:$$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$$
hold for some $n\in\mathbb N$
$(3)$ step: $\tau(n+1)$
$$\frac{4n+3}{4n+5}\cdot\prod_{i=1}^n\frac{4i-1}{4i+1}<\frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}}=\frac{\sqrt{3(4n+3)}}{4n+5}<\sqrt{\frac{3}{4n+7}}$$
$$\frac{12n+9}{16n^2+40n+25}<\frac{3}{4n+7}\iff\frac{48n^2+120n+63-48n^2-120n-75}{\underbrace{(4n+5)^2(4n+7)}_{>0}}<0\iff-\frac{12}{(4n+5)^2(4n+7)}<0$$
Is this combined legitimate?
| It is correct.
Anyways you already ended the proof when you stated that
$\frac{4n+3}{4n+5}\prod_{i=1}^n\frac{4i-1}{4i+1}$ $<$ $\sqrt{\frac{3}{4n+7}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find the area of the largest rectangle that can be inscribed in the ellipse $ \frac {x^2}{2} + \frac {y^2}{6} = 1$ Find the area of the largest rectangle that can be inscribed in the ellipse $$ \frac {x^2}{2} + \frac {y^2}{6} = 1$$
I know the answer is $x=1$ yields $Area = 4\sqrt{3}$.
I'm able to get $x=1$ by taking the derivative and setting equal to zero, but I don't understand where the $4$ in $4\sqrt{3}$ comes from?
| Since your $x=1$ is correct, so I will continue from there. First, solve for y.
$$\frac{1^2}{2}+\frac{y^2}{6}=1$$
$$\frac{y^2}{6}=\frac{1}{2} $$
$$y^2=3$$
$$y=\sqrt{3}$$
Then, we double both of those values to find the side lengths. This is because if we didn’t do that, we would be calculating the area of the rectangle spanning from $(0,0)$ to $(1,\sqrt{3})$. By doubling them we get the area from $(-1,-\sqrt{3})$ to $(1,\sqrt{3})$
You get one side having length $2$, and the other with length $2\sqrt{3}$. Multiplying those gives the final answer of $4\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$ A^3 + B^2 = I_n $ and $A^5=A^2$, then $\det(A^2 + B^2 + I_n) \geq 0 $ and $\operatorname{rank}(I_n + AB^2) = \mathrm{rank}(I_n - AB^2) $ Let $A, B$ be square matrices of size $n$, $n \geq 2$, containing real entries.
$\DeclareMathOperator\rank{rank}$
If the following properties take place: $ A^3 + B^2 = I_n $ and $A^5=A^2$,
then $\det(A^2 + B^2 + I_n) \geq 0 $ and $\rank(I_n + AB^2) = \rank(I_n - AB^2) $.
So far, I've managed to solve the case when $ A $ is invertible.
When $A$ is not invertible, I've obtained $A^2B^2=O_n$, $\rank(A^2) + \rank(B^2) = n $, $\rank(A^2)=\rank(A^m),$ for every $m\geq 2$. All of the aforementioned these also hold when $A$ is invertible.
I would also appreciate it if anyone can recommend me a textbook/book with problems like these.
| From the first condition, $B^2=1-A^3$, so that by the second,
$$ 0=A^2-A^5=A^2B^2=AB^2A=B^2A^2.$$
Now
$$(1-AB^2)(1+AB^2)=1-AB^2AB^2=1 $$
so that
$$\operatorname{rank}(1\pm AB^2)=n. $$
Note that
$$ B^4=(1-A^3)^2=1-2A^3+A^6=1-A^3=B^2.$$
Let $U=\operatorname{im} B^2$ and $W=\operatorname{im} (1-B^2)$. Then for $u\in U$, we have $u=B^2v$ for some $v\in \Bbb R^n$ and so $B^2u=B^4v=B^2v=u$, i.e., $B^2$ acts as identity on $U$. Likewise, $B^2$ acts as zero map on $W$. We conclude $U\cap W=\{0\}$, and as each $v\in \Bbb R^n$ can be written as $v=B^2v+(1-B^2)v$, we conclude obtain the $B^2$-invariant direct sum decomposition $$\Bbb R^n=U\oplus W,$$ where also $U=\ker(1-B^2)$, $W=\ker B^2$.
This decomposition is also $A^2$-invariant:
As $A^2B^2=B^2A^2=0$, we see that $A^2$ maps $\Bbb R^n$ to $\ker B^2=W$ while acting as zero map on $U$.
Then $U,W$ are also $(1+A^2+B^2)$-invariant and we can compute the determinant blockwise.
As $A^2$ acts as zero and $B^2$ as identity on $U$, we see that $(A^2+B^2+1)u$ acts as twice the identity on $U$.
On $W$, $1+A^2+B^2$ coincides with $A^2$. Therefore
$$\begin{align}\det(A^2+B^2+1)&=\det( (A^2+B^2+1)|_U)\det( (A^2+B^2+1)|_W)\\&=\det(2|_U)\det(A^2|_W)\\&=2^{\dim U}\det( A|_W)^2\\&\ge 0.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$
What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$
We have to make this function defined, by change the cosine into sine or tan. So,
$$\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}\right) = \lim_{t \to 0} \left(\frac{a}{t^2} - \frac{6}{t^2}\right) = -18$$
But I am confused what should I do next.
| Your condition implies that $$\lim_{t\to 0}\frac{at\cos^23t-\sin 6t}{t^3}=-18$$ Adding and subtracting $at$ in numerator we get $$\lim_{t\to 0} \frac{at-\sin 6t}{t^3}-9a\cdot\frac{\sin^23t}{(3t)^2}=-18$$ or $$\lim_{t\to 0}\frac {a-6}{t^2}+\frac{6t-\sin 6t}{t^3}=9(a-2)$$ Using substitution $x=6t$ the second term on left side can be written as $$6^3\lim_{x\to 0}\frac {x-\sin x} {x^3}=36$$ (via L'Hospital's Rule or Taylor series) and therefore we have $$\lim_{t\to 0}\frac {a-6}{t^2}=9(a-6)$$ which implies $a=6$ and the above equation also holds true for this value of $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Smarter way to solve $ \int_0^1\arctan(x^2)\,dx$ I'm trying to solve the following definite integral:
\begin{equation}
\int_0^1 \arctan(x^2)dx
\end{equation}
I tryed first by parial integration, finding:
\begin{equation}
x\arctan(x^2)\Bigl|_0^1-\int_0^1 \dfrac{2x^2}{1+x^4}dx
\end{equation}
Then:
\begin{equation}
\int_0^1 \dfrac{2x^2}{1+x^4}dx=\int_0^1 \dfrac{(x^2+1)+(x^2-1)}{1+x^4}dx=\int_0^1 \dfrac{x^2+1}{1+x^4}dx+\int_0^1 \dfrac{x^2-1}{1+x^4}dx
\end{equation}
and i introduced the following substitution:
\begin{equation}
t=x-\dfrac{1}{x}\qquad s=x+\dfrac{1}{x}
\end{equation}
Than I used this weird substitution:
\begin{equation}
\dfrac{\sqrt{2}}{2}\arctan\left( \dfrac{t}{\sqrt{2}} \right)\Bigl|_0^1+\int_0^1\dfrac{ds}{(s-2)(s+2)}
\end{equation}
In the end simple fraction:
\begin{equation}
\left[\dfrac{\sqrt{2}}{2}\arctan\left( \dfrac{t}{\sqrt 2} \right)+\dfrac{1}{4}\log(2-s)-\dfrac{1}{4}\log(s+2)\right]_0^1
\end{equation}
and that is the solution.
Now: does it exist a simpler path to solve this integral?
| First you need to write $x^4+1$ as a product of factors that are irreducible in the ring of real polynomials.$$x^4+1=(x^2+1)^2-(\sqrt 2 x)^2 $$ $$(x^2+1-\sqrt2 x)(x^2+1+\sqrt 2 x)$$ $$=(x^2-\sqrt 2 x+1)(x^2+\sqrt 2 x+1)$$.Then decompose $$\frac{2x^2}{(x^2-\sqrt 2 x+1)(x^2+\sqrt 2 x+1)}$$ into partial fractions as $$\frac{Ax+B}{x^2-\sqrt 2 x+1}+\frac{Cx+D}{x^2+\sqrt 2 x+1} $$. Find $A,B,C,D$. Then you should have no troble integrating to obtain terms involving ln and arctan functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Sum of the series: $\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}$ is? This was a question I confronted in JAM 2016. I tried the following steps:
$$\displaystyle\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}
\implies\sum_{n=2}^\infty\frac{(-1)^n}{(n-1)(n+2)}\implies\sum_{n=2}^\infty(-1)^n\frac{1}{3}\cdot[\frac{1}{(n-1)}-\frac{1}{(n-2)}]$$$$\implies\lim_
{n\to\infty}(-1)^n\cdot\frac{1}{3}[1-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+...] $$
From the options it was evident that the sum had a relation with $\log_{e}2$, however I couldn't manage to solve the problem. What was I missing?
| $$S=\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}$$
Change the indice of the sum
$$m=n-1$$
$$
\begin {align}
S=&-\sum_{m=1}^\infty\frac{(-1)^m}{m(m+3)} \\
S=&-\frac 1 3\sum_{m=1}^\infty{(-1)^m}\left ({\frac 1 m -\frac 1 {m+3}} \right ) \\
S=&\frac 1 3 \ln 2 +\frac 1 3 \left ( \sum_{m=1}^\infty{\frac {(-1)^m} {m+3}} \right ) \\
S=&\frac 1 3 \ln 2 -\frac 1 3 \left ( \sum_{m=4}^\infty{\frac {(-1)^m} {m}} \right ) \\
S=&\frac 2 3 \ln 2 +\frac 1 3 \left ( \sum_{m=1}^3{\frac {(-1)^m} {m}} \right ) \\
S=&\frac 2 3 \ln 2 +\frac 1 3 \left ( -1+\frac 12-\frac 13\right )
\end{align}
$$
Finally
$$\boxed {S=\frac 2 3 \ln 2 -\frac 5{18}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $a^3+b^3+3abc>c^3$ where a,b,c are triangle sides If $a,b,c$ are triangle sides prove
$a^3+b^3+3abc>c^3$
|
Triangle inequality: If $~a,~ b, ~$and$~ c~$ are the lengths of the sides of the triangle, with no side being greater than $~c~$, then the triangle inequality states that
$~c\leq a+b~$.
Now to prove the given inequality $~a^3+b^3+3abc>c^3~$, we have to use the above property. So we have
$$a^3 + b^3 + 3abc = (a + b)(a^2 - ab + b^2) + 3abc $$
$$~~~~~~~~~~~~~~~> c(a^2 - ab + b^2) + 3abc $$
$$~~~~~~~~~~~~~~~~~~~~~~= c[a^2 -ab+ b^2+3ab]$$
$$~~= c(a + b)^2 $$
$$\implies a^3 + b^3 + 3abc > c^3~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3545017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\log_2{\log_x{(x-3y)}}=-1, x\cdot y^{\log_x{y}}=y^{\frac{5}{2}}$ Problem is to solve this system of equation.
$$\log_2{\log_x{(x-3y)}}=-1$$
$$x\cdot y^{\log_x{y}}=y^{\frac{5}{2}}$$
\My attempt:
First equation can be written as $$\log_x{(x-3y)}=\frac{1}{2}$$.
If on the second equation I put $\log_x()$ I get this
$$\log_x{x\cdot y^{\log_x{y}}}=\log_x{y^\frac{5}{2}}$$
$$\log_x{y} + \log_x{y^{\log_x{y}}}=\frac{5}{2}\log_x{y}$$
$$\log_x{y} + \log_x{y}\cdot\log_x{y}=\frac{5}{2}\log_x{y}$$
$$t+t^2-\frac{5}{2}t=0$$
$$t^2 - \frac{3}{2}t=0$$
There are two solutions $t=0$ and $t=\frac{3}{2}$.
$$\log_x{y}=0$$
From that I can easily see that $y=1$.
But this information didn't help me to solve $\log_x{x-3}=\frac{1}{2}$ because I didn't know how to solve this.
| You have some mistakes when you solve the second one. A $x$ is changed to a $y$. But the approach is correct. First of all $y$ is positive, which means $x$ has to be as well. Then
$$x\cdot y^{\log_x{y}}=y^{\frac{5}{2}}\Rightarrow \log_x \color{red}{x}+\log_x y\cdot \log_x y=\frac{5}{2}\log_x y$$
or
$$1+t^2=\frac{5}{2}t$$
with $t = \log_x y$. Solving gives $t=2$ or $t=\frac{1}{2}$.
If $t=2$, we have $y=x^2$, so substituting in the first equation:
$$\log_2{\log_x{(x-3x^2)}}=-1$$
or
$$\log_x(x-3x^2)=\frac{1}{2}$$
or $x-3x^2=\sqrt{x}$. However this is impossible, because from AM-GM:
$$3x^2+\sqrt{x}=3x^2+\frac{1}{2}\sqrt{x}+\frac{1}{2}\sqrt{x}\geq 3\sqrt[3]{\frac{3}{4}}x>x$$
which means no solution in this case. If $t=\frac{1}{2}$, we have $y=\sqrt{x}$, so:
$$\log_x(x-3\sqrt{x})=\frac{1}{2}$$
or $x-3\sqrt{x}=\sqrt{x}$, which gives the solution $x=16$ and $y=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Summation of finite series: Let $f(r)$ be what? Find the sum of the first $n$ terms of
$\displaystyle \frac{1}{1\times4\times7}+\frac{1}{4\times7\times10}+\frac{1}{7\times10\times13}+...$
My working:
Let $\begin{align}\displaystyle\frac{1}{(3x-2)(3x+1)(3x+4)} &\equiv \frac{A}{3x-2}+\frac{B}{3x+1}+\frac{C}{3x+4}\\1&\equiv A(3x+1)(3x+4)+B(3x-2)(3x+4)+C(3x-2)(3x+1)\end{align}$
Let $\displaystyle x=\frac{-1}{3}, B=\frac{-1}{9}$
Let $\displaystyle x=\frac{-4}{3}, A=\frac{1}{18}$
Let $\displaystyle x=\frac{2}{3}, C=\frac{1}{18}$
Thus, $\displaystyle\frac{1}{(3x-2)(3x+1)(3x+4)}\equiv \frac{1}{18(3x-2)}-\frac{1}{9(3x+1)} +\frac{1}{18(3x+4)}$
So, I thought of using the method $$\sum_{r=1}^n u_r= \sum_{r=1}^n [f(r+1)-f(r)]=f(n+1)-f(1)$$ but I’m not sure if finding such a function is possible in this case.
Carrying on using the above method,
$$\sum_{r=1}^n \frac{1}{(3r-2)(3r+1)(3r+4)}$$
$$=\sum_{r=1}^n \Bigl(\frac{1}{18(3r-2)} -\frac{1}{9(3r+1)} + \frac{1}{18(3r+4)}\Bigr)$$
$$=\sum_{r=1}^n \Bigl(\frac{1}{18(3r-2)} -\frac{1}{18(3r+1)}\Bigr) -\sum_{r=1}^n\Bigl(\frac{1}{18(3r+1)} -\frac{1}{18(3r+4)}\Bigr)$$
But now I don’t know what to let $f(r)=$?
How to proceed?
| Do telescopic summation after doing the partial fractions:
$$T_k=\frac{1}{(3k-2)(3k+1)(3k+4)}$$ $$\implies 18 T_k=\left(\frac{1}{3k-2}-\frac{1}{3k+1}\right)-\left( \frac{1}{3k+1}-\frac{1}{3k+4)}\right)$$
Let $F_k=\frac{1}{3k-2}$
$$S_n=\sum_{k=1}^{n} T_k= \frac{1}{18}(\sum_{k=1}^n [F_k-F_{k+1}]-\sum_{k=1}^n [F(k+1)-F(k+2)]).$$
$$\implies S_n=\frac{1}{18}([F_1-F_{n+1}]-[F_2-F_{n+2}])$$
You may complete it now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $n^4 \mod 8$ is identically equal to either 0 or 1, $\forall \ n\in \mathbb{N}$. I feel pretty confident about the first half of my solution for this, however I don't like how I used the induction hypothesis on the case for even integers, it feels like it isn't doing anything useful since it is really easy to show that $P(2x+2) \mod 8 \equiv 0$ without using the inductive hypothesis.
This is also the first semester I've been doing proofs, so am I going about this right?
Proof. Since $1^4 \mod 8 = 1$, $P(1)$ holds. Also, $2^4 \mod 8 = 16 \mod 8 = 0$ so, $P(2)$ holds.
We claim that $P(2x-1)$ holds, then $(2x-1)^4 \mod 8\equiv 1$ for any odd integer $n=2x-1$. So,
$$(2x-1)^4 = 16x^4-32x^3+24x^2-8x+1 \mod 8\equiv 1,$$
now consider
$$(2x+1)^4 = 16x^4+32x^3+24x^2+8x+1 = (16x^4-32x^3+24x^2-8x+1)+64x^3+16x,$$
then since
$$64x^3+16x \mod 8 = 8(8x^3+2x) \mod 8 \equiv 0,$$
we conclude that
$$(16x^4-32x^3+24x^2-8x+1)+64x^3+16x \mod 8 \equiv 1+0 = 1.$$
Thus, $P(2x+1)$ holds, and all odd integers $n$ hold.
Next, we claim that $P(2x)$ holds, then $(2x)^4 \mod 8\equiv 0$ for any even integer $n=2x$. So,
$$(2x)^4 \mod 8 = 16x^4 \mod 8 \equiv 0,$$
now consider
$$(2x+2)^4 = 16x^4+64x^3+96x^2+64x+16,$$
then
$$64x^3+96x^2+64x+16 \mod 8 = 8(8x^3+12x^2+8x+2) \mod 8 \equiv 0,$$
we conclude that
$$16x^4+64x^3+96x^2+64x+16 \mod 8 \equiv 0+0 = 0.$$
Thus, $P(2x)$ holds, and all even integers $n$ hold.
By mathematical induction, if $n=2x-1$ and $n=2x$ hold for the given statement, $n=2x+1$ and $n=2x+2$ also hold. Therefore, the statement holds for all $n \in \mathbb{N}$.
$\mathbb{QED}$
| Your idea is okay, but $(2x+2)^4=16 + 64 x + 96 x^2 + 64 x^3 + 16 x^4$,
unlike what you initially put in the question.
Alternatively, you could argue that if $2|n$ then $8|n^3|n^4$, so $n^4\equiv0\pmod8$,
and if $2\nmid n$ then $2|n-1, n+1, $ and $n^2+1$, so $8|n^4-1$; i.e., $n^4\equiv1\pmod8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given a sequence $a_n =\frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$, prove that $\lim_{n \to \infty} a_{n} = g$. The given Sequence is $a_n = \frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$.
I showed that the Sequence $a_n$ converges towards a Value $g = \frac{1}{3}$.
How do I determine for each $\epsilon > 0$ an $n_0$ so that: $|a_n - g| < \epsilon \ \forall n > n_0$?
This is what I tried so far:
$\begin{array}{rcl}
\left|a_n - g\right| & < & \epsilon \\
\left|\frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5} - \frac{1}{3}\right| & < & \epsilon \\
\left|\frac{3n^2 - 9n + 3 \cdot (-1)^n}{9n^2 - 21n + 15} - \frac{3n^2 - 7n + 5}{9n^2 - 21n + 15}\right| & < & \epsilon \\
\left|\frac{-2n - 5 + 3 \cdot (-1)^n}{9n^2 - 21n + 15}\right| & < & \epsilon \ \text{ Denominator > 0, Numerator < 0 } \forall n \in \mathbf{N} \\
\frac{2n + 5 - 3 \cdot (-1)^n}{9n^2 - 21n + 15} \leq \frac{2n + 5 + 3}{9n^2 - 21n + 15} = \frac{2n + 8}{9n^2 - 21n + 15} & < & \epsilon \\
\end{array}$
Are my Steps correct so far?
How do I proceed from here?
| This follows directly from the definition of a Limit of a sequence. if $a_n$ converges to $g$, then $lim_{n\to \infty} a_n =g$, such that $\forall \epsilon >0 \exists n_0 \in \mathbb{N} \text{ such that } \forall n\ge n_0 : |a_n -g|<\epsilon$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find a limit with sqrt $\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$ $$\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$$
I don't know how to rewrite or rationalize in order to find the limit.
| To write the complete solution, as per the hint in the comments I gave:
$$
\begin{aligned}
\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right) &= \lim_{x \to \infty}x^2\left(\sqrt{(x^2+3)^2} - \sqrt{x^4 + 6x^2}\right)\\
&=\lim_{x \to \infty}x^2\left(\frac{(x^2+3)^2-x^4-6x^2}{\sqrt{(x^2+3)^2} + \sqrt{x^4 + 6x^2}}\right)\\
&=\lim_{x \to \infty}\frac{9x^2}{x^2+3 + \sqrt{x^4 + 6x^2}}\\
&=\lim_{x \to \infty}\frac{9}{1+\frac{3}{x^2} + \sqrt{1 + \frac{6}{x^2}}}\\
&=\frac{9}{2}\\
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
Find $\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )$ I have the following limit to find:
$$\lim\limits_{x \to \infty} x^3 \bigg ( \sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} \bigg )$$
What approah should I use? Since it's an $\infty \cdot 0$ type indeterminate I thought about writing $x^3$ as $\dfrac{1}{\frac{1}{x^3}}$ so I would have the indeterminate form $\dfrac{0}{0}$, but after applying L'Hospital I didn't really get anywhere.
| Let $t=\frac1x$. Then,
$$\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )
=\lim_{t \to 0} \frac1{t^3} \left ( \sin\frac{t}{1 + 2t} - 2 \sin\frac{t}{1+t } + \sin t \right )$$
Use $\frac 1{1+a} = 1-a+a^2+O(a^3)$ to expand,
$$\sin\frac{t}{1 + 2t} - 2 \sin\frac{t}{1+t } + \sin t$$
$$=\sin(t-2t^2+4t^3)+\sin t - 2 \sin(t-t^2+t^3)+O(t^4)$$
$$=2\sin(t-t^2+2t^3)\cos t^2 - 2 \sin(t-t^2+t^3)+O(t^4)$$
$$=2[\sin(t-t^2+2t^3) - \sin(t-t^2+t^3)]+O(t^4)$$
$$=4\cos t\sin\frac{t^3}2+O(t^4)= 4\cdot 1\cdot \frac{t^3}2+O(t^4)=2t^2+O(t^4)$$
where $\cos t^2 = 1 + O(t^4)$ is applied. Thus,
$$\lim_{t \to 0} \frac1{t^3} \left ( \sin\frac{t}{1 + 2t} - 2 \sin\frac{t}{1+t} + \sin t \right )=\lim_{t \to 0} \frac{2t^3+O(t^4)} {t^3}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Closed form of integral $\int_0^1 \frac{1-\frac{1}{\sqrt[5]{x}}}{1-x} \, dx$ Originally this integral was interesting for me as it represents the specific harmonic number $H_{-\frac{1}{5}}$ and also because Mathematica returned the wrong value $0$ for the integral. I posted the probem here https://mathematica.stackexchange.com/questions/215089/possible-bug-integrate1-x-1-5-1-x-x-0-1-0
Later I found a way to get a reasonable result from Mathematica which, finally I could simplify to this form.
$$\int_0^1 \frac{1-\frac{1}{\sqrt[5]{x}}}{1-x} \, dx=\frac{\pi}{2} \sqrt{1+\frac{2 \sqrt{5}}{5}}-\frac{5}{4} \log (5)-\frac{\sqrt{5}}{2} \log \left(\frac{1}{2} \left(\sqrt{5}+1\right)\right)$$
So a presumed bug by has turned finally into a nice statement. Can you prove it?
| Note
$$I=\int_0^1 \frac{1-\frac{1}{\sqrt[5]{x}}}{1-x} dx
\overset{ x=t^5}=\int_0^1 \frac{-5t^3}{t^4+t^3+t^2+t+1}dt$$
$$=\sqrt5\int_0^1\left( \frac{2\phi_-t+1}{t^2+2\phi_-t+1}
- \frac{2\phi_+t+1}{t^2+2\phi_+t+1}\right)dt = \sqrt5 J(\phi_-) - \sqrt5 J(\phi_+)$$
where $\phi_{+}=\frac{1+\sqrt5}4= \cos \frac\pi5$, $\phi_{-}=\frac{1-\sqrt5}4= \cos \frac{3\pi}5$ and
$$J(\phi) = \int_0^1 \frac{2\phi t+1}{t^2+2\phi t+1}dt
= \phi \ln( t^2+2\phi t+1 )|_0^1
+ \frac{1-2\phi^2}{\sqrt{1-\phi^2}} \tan^{-1}\frac{t+\phi}{\sqrt{1-\phi^2}}\bigg|_0^1$$
Then
\begin{align}
I= &\sqrt5 [\phi_-\ln( 2+2\phi_-) - \phi_+\ln( 2+2\phi_+)]
+ \frac{\sqrt5\pi}{10} \left( 3\frac{1-2\phi_-^2}{\sqrt{1-\phi_-^2}} - \frac{1-2\phi_+^2}{\sqrt{1-\phi_+^2}} \right)\\
= &\frac{\sqrt5-5}4\ln\frac{\sqrt5-1}{2\sqrt5} - \frac{\sqrt5+5}4\ln\frac{\sqrt5+1}{2\sqrt5}
+ \frac{\sqrt5\pi}{10} \left( \frac{3(1+\sqrt5)}{\sqrt{10+2\sqrt5}} - \frac{1-\sqrt5}{\sqrt{10-2\sqrt5}} \right) \\= &-\frac{\sqrt5}2\ln\frac{\sqrt5+1}2 -\frac{5}4\ln5
+ \frac{\sqrt5\pi}{10}\sqrt{5+2\sqrt5}\\
=&\frac{\pi}{2} \sqrt{1+\frac{2 \sqrt{5}}{5}}-\frac{5}{4} \ln5-\frac{\sqrt{5}}{2} \ln \frac{\sqrt{5}+1}2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the biggest value of $|a-b|+|b-c|+|c-a|$ when $a^2+b^2+c^2=1$ Given that $a^2 + b^2 + c^2 = 1$
How can I prove that the maximum value of |a - b| + |b - c| + |c - a| is $2\sqrt{2}$ ?
In extends, given that $a_1^2 + a_2^2 + a_3^2 + ... + a_{2020}^2 = 1$ find the maximum value of $|a_1 - a_2| + |a_2 - a_3| +|a_3 - a_4| + ... + |a_{2020} - a_1|$
| Without loss of generality, assume that $a\geq b\geq c$. Then:
$$|a-b|+|b-c|+|c-a| = 2(a-c)$$
And we have:
$$
\begin{aligned}
(a-c)^2&=2(a^2+c^2)-(a+c)^2\\
&\leq 2(a^2+c^2) \\
&\leq 2(a^2+b^2+c^2)\\
&=2
\end{aligned}$$
Therefore $2(a-c)\leq 2\sqrt{2}$. Equality occurs when $(a,b,c) = \left(\frac{1}{\sqrt{2}}, 0,-\frac{1}{\sqrt{2}}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many complex numbers $z$ are there such that $|z|=1$ and $z^{5040} - z^{720}$ is a real number? My attempt:
If $z^{5040} - z^{720}$ is real, that means that their imaginary parts are equal.
$e^{5040i\theta} - e^{720i\theta} = k$ , where $k$ is a real number
$\sin{5040\theta} = \sin{720\theta} $
Let $u=720\theta$
$\sin 7u = \sin u$
By graphing $y=\sin 7u$ and $y=\sin u$, there are $16$ intersections between $0$ and $2\pi$ (including $0$ but not including $2\pi$)
Because the substitution $u=720\theta$ scaled the functions by a factor of $720$ parallel to the x-axis,
there are $11520$ intersections for $\sin{5040\theta} = \sin{720\theta}$ from $\theta=0$ to $\theta= 2\pi$ (not including $2\pi$ because $0+2k\pi=2\pi$ when $k=1$.
Is $11520$ complex numbers correct? There was no answer provided.
| Solve $\sin{7u} = \sin{u} $ as follows,
$$ \sin{7u} - \sin{u} = 2\cos 4u \sin 3u=0$$
$\sin4u = 0$ yields $ u = \frac {\pi n}4$ and $\cos3u=0$ yields $u=\frac\pi6+ \frac {\pi n}3$. Then,
$$z= e^{ \frac { i2\pi n}{8\cdot 720}}, \>\>\>\>\>
z= e^{ \frac{i\pi}{6\cdot 720 }+i\frac { i2\pi n}{6\cdot 720}}$$
Thus, number of distinct solutions are $8\cdot 720 + 6\cdot 720= 10080$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the relative maximum and relative minimum of the function $f(x) = 49x + \frac{4}{x}$ Find the relative maximum and relative minimum of the function $f(x) = 49x + \frac{4}{x}=49x+4x^{-1}$
Solution:
Step 1: Find the values of $x$ where $f'(x)=0$ and $f'(x)$ DNE.
$f'(x)=49-4x^{-2}=49-\frac{4}{x^2}$
We can see that $f'(x)$ DNE when $x=0$. Now lets find the values of $x$ that make $f'(x)=0$.
$f'(x)=49-4x^{-2}=49-\frac{4}{x^2}=0$
$\rightarrow x^2=\frac{4}{49}$
$\rightarrow x= \pm \sqrt{\frac{4}{49}}$
$\rightarrow x= \pm \frac{2}{7}$
Thus our critical points are $x=0, x= \frac{2}{7}, x= -\frac{2}{7}$
Step 2: Make a number line and plot the sign of $f'(x)$ to find where $f(x)$ is increasing and decreasing.
$f'(-3)= 49-\frac{4}{9}>0$
$f'(\frac{-1}{7})=49-\frac{4}{(\frac{-1}{7})^2}=49-4(49)=-3(49)<0$
$f'(\frac{1}{7})=49-\frac{4}{(\frac{1}{7})^2}=49-4(49)=-3(49)<0$
$f'(3)=49-\frac{4}{3^2} >0$
Thus $f(x)$ is increasing as $x$ increases towards $\frac{-2}{7}$ and then $f(x)$ is decreasing as $x$ increases towards $0$. Therefore $x=\frac{-2}{7}$ corresponds to a relative maximum.
Thus $f(x)$ is decreasing $x$ increases towards $\frac{2}{7}$ and then $f(x)$ is increases as $x$ increases towards $\infty$. Therefore $x=\frac{2}{7}$ corresponds to a relative minimum.
Step 3: find the corresponding $y$ values.
$f(\frac{2}{7})=28$
$f(\frac{-2}{7})=-28$
Therefore $(\frac{2}{7},28)$ is a relative minimum and $(\frac{-2}{7},-28)$ is a relative maximum.
This may be weird that our relative maximum is smaller than our relative minimum, but since these are "relative" maximum and minimum, they are only being compared to points on the graph very near to it, so this is okay.
| Rewrite as
\begin{eqnarray*}
49x+\frac{4}{x} = \left( 7 \sqrt{x} - \frac{2}{\sqrt{x}} \right)^2 +28.
\end{eqnarray*}
From this it easy to see that a minimum of $28$ occurs at $x=2/7$.
Also
\begin{eqnarray*}
49x+\frac{4}{x} = \left( 7 \sqrt{x} + \frac{2}{\sqrt{x}} \right)^2 -28.
\end{eqnarray*}
similarly a maximum of $-28$ occurs at $x=-2/7$.
$0$ is not a turning point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}} dx$ I have to find this integral:
$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}} dx$$
My attempt was to split the integral:
$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}}=\int_{\frac{1}{4}}^4\frac{x\arctan x+1}{x\sqrt{x^2+1}}+\int_{\frac{1}{4}}^4\frac{\arctan x-1}{x\sqrt{x^2+1}}$$
The first integral is:
$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}}=\int_{\frac{1}{4}}^4\frac{1}{x}(x\arctan x)'dx=\arctan x|_{\frac{1}{4}}^4+\int_{\frac{1}{4}}^4\frac{1}{x}\arctan x dx$$
but I am stuck here.
| Hint. Let $t=1/x$ then
$$\begin{align}I&:=\int_{\frac{1}{4}}^4\frac{(x+1)\arctan(x)}{x\sqrt{x^2+1}} dx=
\int^{\frac{1}{4}}_4\frac{(1/t+1)\arctan(1/t)}{(1/t)\sqrt{1/t^2+1}} \frac{-dt}{t^2}\\
&=\int_{\frac{1}{4}}^4\frac{(t+1)\arctan(1/t)}{t\sqrt{t^2+1}} \,dt
=\frac{\pi}{2}\int_{\frac{1}{4}}^4\frac{t+1}{t\sqrt{t^2+1}} \,dt-I.
\end{align}$$
where at the last step we used the identity $\arctan(t)+\arctan(1/t)=\pi/2$ for $t>0$.
Now it should be easy to find $I$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that the restriction of $f$ to any straight line through $B$ has a local minimum in $B$ I have the following problem:
We have the function: $f(x,y)=(x^2+y^2-6x)\cdot (\frac{1}{8}y^2-x)$
We also have the point $B=(0,0)$.
Show that the restriction of $f$ to any straight line through $B$ has a local minimum in $B$. Is this enough to prove that $f$ has a local minimum in $B$? Does $f$ actually have local minimum in $B$?
Okay, so I am having some trouble with this. I have found the partiel derivatives for $f$.
$f'_x=(2x-6) \cdot (\frac{y^2}{8}-x)-x^2-y^2+6x$
$f'_y=2y(\frac{y^2}{8}-x)+\frac{(x^2+y^2-6x)y}{4}$
Inserting $(0,0$ into both $f'_x$ and $f'_y$ yields:
$f'_x(0,0)=0$
$f'_y(0,0)=0$
Here, I just showed that $B$ is a stationary point, but I don't really know how to continue from here, and show that any straight line has a local minimum in $B$.
I hope someone can help me.
| Straight lines through $(0, 0)$ take the form
$$y = mx$$
for some $m \in \Bbb{R}$, when not vertical. When vertical, the equation becomes
$$x = 0.$$
This means two cases. For the first case, substitute $(x, y) = (x, mx)$ into $f(x, y)$:
\begin{align*}
f(x, mx) &= (x^2 + m^2 x^2 - 6x)\cdot\left(\frac{1}{8}m^2x^2 - x\right) \\
&= \frac{1}{8}m^2(m^2 + 1)x^4 - \left(1 + \frac{7}{4}m^2\right)x^3 + 6x^2
\end{align*}
Let the above function of $x$ be $g(x)$. Now, it's a one variable function, and we can perform the usual tests to check for local minima. For example, we can differentiate:
\begin{align*}
g'(x) &= \frac{1}{2}m^2(m^2 + 1)x^3 - 3\left(1 + \frac{7}{4}m^2\right)x^2 + 12x \\
g''(x) &= \frac{3}{2}m^2(m^2 + 1)x^2 - 6\left(1 + \frac{7}{4}m^2\right)x + 12.
\end{align*}
Note that $g'(0) = 0$, so we have a stationary point, and $g''(0) = 12 > 0$, hence we have a local minimum.
For the second case, we can also consider $f(x, y)$ along the line $x = 0$, i.e. $f(0, y)$, but I'll leave that to you.
However, as mathcounterexamples.net points out, this is not a local minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Rolling at Least 2 2's on 3 fair dice Say we roll three 6-sided dice.
What is the probability that at least two of the faces are a 2?
The sample space here is 216.
Rolling a 2 on one dice is 1/6.
Rolling two 2s is 1/6 * 1/6 = 1/36
The third dice is at least two 2s so the probability it isn't a 2 is 5/6 and the probability it is a 2 is 1/6.
So do we multiply 1/36 * 1/6 * 5/6 to get the answer?
5/1296 is the probability of rolling at least two 2s on 3 fair dice?
| The following rolls are all possible:
$$2,2,1 \\ 2,1,2 \\ 1,2,2 \\ \vdots \\ 2,2,6 \\ 2,6,2 \\ 6,2,2$$
The probability of getting exactly two $2$'s followed by a non-$2$: $$\dfrac{1}{6}\cdot \dfrac{1}{6}\cdot \dfrac{5}{6}$$
The probability of getting a $2$, a non-$2$, then a $2$:
$$\dfrac{1}{6}\cdot \dfrac{5}{6}\cdot \dfrac{1}{6}$$
The probability of getting a non-$2$, then two $2$'s:
$$\dfrac{5}{6}\cdot \dfrac{1}{6}\cdot \dfrac{1}{6}$$
The probability of getting three $2$'s:
$$\dfrac{1}{6}\cdot \dfrac{1}{6}\cdot \dfrac{1}{6}$$
Adding this up, the probability of at least two $2$'s is:
$$\dfrac{3\cdot 5}{6^3}+\dfrac{1}{6^3} = \dfrac{2}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$\sum_{n=1}^\infty (n!)^{-\frac{1}{n}}$ I have to study the character of this series
$$\sum_{n=1}^\infty (n!)^{-\frac{1}{n}}$$
and I tried with the ratio test:
$ \frac{a_{n+1}}{a_n}= \frac{((n+1)!)^{-\frac{1}{n+1}}}{(n!)^{-\frac{1}{n}}}=
\frac{(n!)^{\frac{1}{n}}}{((n+1)!)^{\frac{1}{n+1}}}=
\frac{(n!)^{\frac{1}{n}}}{((n+1) \star n!)^{\frac{1}{n+1}}}=
\frac{(n!)^{\frac{1}{n}-\frac{1}{n+1}}}{((n+1) )^{\frac{1}{n+1}}} \sim (n!)^{\frac{1}{n}-\frac{1}{n+1}}=
(n!)^{\frac{1}{n(n+1)}}>(n(n+1))^{\frac{1}{n(n+1)}} \sim 1$
So the limit is larger than 1 and the given series diverges.
Is it right?
| $\frac{1}{n}=\frac{1}{n^{n*\frac{1}{n}}}<\frac{1}{n!^{\frac{1}{n}}}$
Applying the comparison test and knowing that the series $\sum_{n=1}^\infty \frac{1}{n}$ diverges, also our series diverges
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the general solution to $\csc \theta + \sec \theta = 1$
Find the general solution to $$\csc\theta + \sec\theta =1$$
This is how I solved.
We have,
\begin{align}
\csc\theta + \sec\theta &=1\\
\frac1{\sin\theta} + \frac1{\cos\theta}& =1\\
\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta} &=1\\
(\sin\theta + \cos\theta)^2 &= (\sin\theta\cos\theta)^2 \\
1 + 2\sin\theta\cos\theta &= \frac{4\sin^2\theta\cos^2\theta}4\\
1 + \sin2\theta &= \frac{(2\sin\theta\cos\theta)^2 }4\\
4 + 4\sin2\theta &= \sin^2 2\theta\\
\sin^2 2\theta - 4\sin2\theta - 4 &= 0\\
\sin2\theta &= 2 - 2\sqrt2\end{align}
Now here I am stuck. Can someone please help me proceed further?
| Your squaring of the equation $$\cos x+\sin x=\cos x\>\sin x\tag{1}$$ has introduced spurious solutions. In fact the value ${1\over2}\arcsin\bigl(2-2\sqrt{2}\bigr)\approx-0.488147$ does not solve the given problem.
Drawing the graphs of $x\mapsto \cos x+\sin x$ and $x\mapsto\cos x\>\sin x$ shows a symmetry with respect to $x={\pi\over4}$. We therefore put $x:={\pi\over4}+t$ and then have
$$\cos x+\sin x=\sqrt{2}\>\cos t,\qquad\cos x\>\sin x={1\over2}\cos(2t)\ .$$
Plugging this into $(1)$ we obtain
$$\sqrt{2}\cos t={1\over2}(2\cos^2 t-1)\ ,$$
so that $\cos t={\sqrt{2}\over2}-1$, or
$$ t=\pm \alpha,\quad{\rm with}\quad \alpha:=\arccos{\sqrt{2}-2\over2}=1.86805\ .$$
This leads to the $x$-values
$$x_1={\pi\over4}-\alpha=-1.08265,\qquad x_2={\pi\over4}+\alpha=2.65345\ .$$
Looking at the graphs we see that these solutions repeat with periodicity $2\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Maximum value of $\frac{a}{1+bc} + \frac b{1+ac} + \frac{c}{1+ab}$ given $a^2 + b^2 + c^2 = 1$ Given that the constraint of $a, b, c$, for which $a, b, c$ are non-negative real numbers, is $a^2+b^2+c^2=1,$ find the maximum value of $$\frac{a}{(1+bc)}+\frac b{(1+ac)}+\frac{c}{(1+ab)}.$$
For this question I have tried using this geometric method, hopefully it can be logically correct
Do you all have actually better method(s) to solve this problem?
| For non-negative variables we obtain: $$\sum_{cyc}\frac{a}{1+bc}\leq\sum_{cyc}\frac{a\sqrt2}{a+b+c}=\sqrt2$$
because $$2(1+bc)^2\geq(a+b+c)^2$$ it's
$$2+4bc+2b^2c^2\geq1+2(ab+ac+bc)$$ or
$$2b^2c^2+(b+c-a)^2\geq0.$$
The equality occurs fot $c=0$ and $a=b=\frac{1}{\sqrt2},$ which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate Variance with Binomial Expansion IQ is normally distributed with mean 100 and standard deviation 15. An IQ of 130 or above is considered gifted, and 150 and above is considered genius. There are 350 million people living in the United States.
Let $X$ be the IQ of a person randomly selected from the population. Compute $\operatorname{Var}(X^2).$
So for this question binomial expansion must be used but I am not sure how to incorporate that into my solution.
| You have
\begin{align}
15^2 & = \operatorname{Var}(X) = \operatorname E\big((X-100)^2\big) \\
& = \operatorname E(X^2) - 200 \operatorname E(X) +100^2 \\
& = \operatorname E(X^2) - 100^2
\end{align}
Therefore
$$
\operatorname E(X^2 ) = 100^2 + 15^2 = 32\,500.
$$
\begin{align}
\operatorname{Var}(X^2) & = \operatorname E\big( (X^2 - 32\,500)^2\big) \\
& = \operatorname E(X^4) - 2\times32\,500\operatorname E(X^2) + 32\,500^2 \\
& = \operatorname E(X^4) - 2\times32\,500\times32\,500 + 32\,500^2 \\
& = \operatorname E(X^4) - 32\,500^2.
\end{align}
Now write $X = 100 + 15Z,$ so that $Z\sim\operatorname N(0,1).$ Then
$$
X^4 = 100^4 + (4\times 100^3 (15Z)) + (6\times 100^2 (15Z)^2) + (4\times100(15Z)^3) + (15Z)^4.
$$
The expected value of $Z$ and of $Z^3$ is $0,$ (that last because the distribution is symmetric about the origin). And that of $Z^2$ is $1.$ So you have
$$
\operatorname E(X^4) = 100^4 + (6\times100^2\times15^2) + 15^4\operatorname E(Z^4).
$$
That last expected value is $3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integral of $\int \sin^2x\cos^4xdx$ $$\int \sin^2x\cos^4xdx$$
I tried
$$I = \int (1-\cos^2x)\cos^4xdx = \int \frac{\sec^2x-1}{\sec^6x}dx = \int \frac{\tan^2x}{\sec^6x}dx$$
Take $\tan x = t \implies \sec^2xdx = dt$
$$I = \int \frac{t^2}{(t^2+1)^4}dt$$
And I could not proceed further from here.
| You can calculate it directly without substitution as follows:
$$\sin^2x\cos^4x=(\sin x\cos x)^2\cos^2x =\frac 14\sin^22x\cdot \frac 12(1+\cos 2x)$$
$$= \frac 18 \cdot\frac 12(1-\cos 4x) +\frac 18\sin^2 2x\cos 2x$$
Hence,
$$\int \sin^2x\cos^4x\; dx = \frac x{16}-\frac{\sin 4x}{64} + \frac{\sin^3 2x}{48}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If $A, B, C$ are angles of $\Delta ABC$ and $\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}$...
If $A, B, C$ are angles of $\Delta ABC$ and $\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}$, then prove that $\sum \tan A \tan B=\sum \tan A$
Solving the given equation, we get$$(\sin A-\cos A)(\sin B -\cos B)(\sin C -\cos C)=1$$
$$(\tan A-1)(\tan B-1)(\tan C-1)=\sec A \sec B \sec C$$
$$\sum \tan A -\sum \tan A \tan B +\sum \tan A-1=\sec A \sec B \sec C$$
How should I proceed?
| Hint:
Both
\begin{align}
\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}
\tag{1}\label{1}
\end{align}
and
\begin{align}
\tan A\tan B+\tan B\tan C+\tan C\tan A
&=
\tan A+\tan B+\tan C
,
\tag{2}\label{2}
\end{align}
expressed in terms of semiperimeter $\rho$,
inradius $r$ and circumradius $R$ of the triangle,
are equivalent to
\begin{align}
\rho^2&=r^2+4\,r\,R+2\,\rho\,r
\tag{3}\label{3}
,
\end{align}
which implies
\begin{align}
\rho&=r+\sqrt{2\,r^2+4\,r\,R}
\tag{4}\label{4}
.
\end{align}
But \eqref{4} is true only for the degenerate triangle
when one angle is $180^\circ$ and the other two are zero.
For the conversion,
use known identities for the angles of triangle
\begin{align}
\cos A\cos B\cos C&=\frac{r}{R}+1
\tag{5}\label{5}
,\\
\sin A\sin B\sin C
&=
\frac{\rho\,r}{2R^2}
\tag{6}\label{6}
,
\end{align}
\begin{align}
\tan A\tan B\tan C
=
\tan A+\tan B+\tan C
&=\frac{2\rho\,r}{\rho^2-(r+2\,R)^2}
\tag{7}\label{7}
,\\
\tan A\tan B+\tan B\tan C+\tan C\tan A
&=1+\frac{4\,R^2}{\rho^2-(r+2\,R)^2}
\tag{8}\label{8}
,\\
\cot A+\cot B+\cot C&=
\frac12\,\left(\frac{\rho}r
-\frac r\rho \right)
-2\,\frac R\rho
\tag{9}\label{9}
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators. My question is: Do we have an splitting equation where we can produce fractions with odd denominators?
To split an Egyptian fraction to Egyptian fractions, we can use the splitting equation below:
$\frac{1}{n}= \frac{1}{n+1}+\frac{1}{n(n+1)}$
The key limitation of the above equation is the following:
If $n$ is even, then $n+1$ is odd and $n(n+1)$ is even, otherwise $n+1$ is even and $n(n+1)$ is even.
Either way, the splitting equation produces with at least one even Egyptian fraction.
An example of a splitting to Odd Egyptian fraction is given below:
$\frac{1}{3}= \frac{1}{5}+\frac{1}{9}+\frac{1}{45}$
$\frac{1}{5}= \frac{1}{9}+\frac{1}{15}+\frac{1}{45}$
$\frac{1}{7}= \frac{1}{15}+\frac{1}{21}+\frac{1}{35}$
$\frac{1}{7}= \frac{1}{9}+\frac{1}{45}+\frac{1}{105}$
$\frac{1}{9}= \frac{1}{15}+\frac{1}{35}+\frac{1}{63}$
$\frac{1}{11}= \frac{1}{21}+\frac{1}{33}+\frac{1}{77}$
The link below is useful for further details about egyptian fraction: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html#section9.5
| A general solution is for every positive positive integer $\ n\ $ :
*
*If $n$ is odd , then $$\frac{1}{3n+2}+\frac{1}{6n+3}+\frac{1}{18n^2+21n+6}=\frac{1}{2n+1}$$ is a solution with odd denominators
*If $n$ is even , then $$\frac{1}{3n+3}+\frac{1}{6n+3}+\frac{1}{6n^2+9n+3}=\frac{1}{2n+1}$$ is a solution with odd denominators
So, for every odd $\ k\ge 3\ $ we can write $\ \frac 1k\ $ with $\ 3\ $ distinct fractions with odd denominators.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Trig substitution for $\sqrt{9-x^2}$ I have an integral that trig substitution could be used to simplify.
$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$
The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\cos \theta)^2}$?
Setting then $x = 3\cos \theta; dx = -3\sin \theta d\theta$
$$-\int \frac{(3\cos\theta)^3}{3\sin\theta}3\sin\theta d\theta$$
$$-27\int\cos^3\theta d\theta$$
$$-27\int(1-\sin ^2\theta)\cos \theta d\theta$$
Substituting again, $u=\sin \theta; du=\cos \theta d\theta$
$$-27\int(1-u^2)du $$
$$-27u + 9u^3 + C$$
$$-27\sin \theta + 9 \sin^3 \theta + C$$
$$-9\sqrt{9-x^2} + 3\sin\theta\cdot 3\sin\theta\cdot \sin \theta + C$$
$$-9\sqrt{9-x^2} + (\sqrt{9-x^2})^2 \cdot \frac{\sqrt{9-x^2}}{3} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)(9-x^2)^{\frac{1}{2}} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)^\frac{3}{2} + C $$
I guess I have more doubts that I've done the algebra correctly than the substitution, but in any case I'm not getting the correct answer. Have I calculated correctly? Is the answer simplified completely?
EDIT
Answer needed to be simplified further:
$$-9\sqrt{9-x^2} + \frac{1}{3}(\sqrt{9-x^2}^2 \sqrt{9-x^2}) + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}((9-x^2)\sqrt{9-x^2}) + C$$
$$\sqrt{9-x^2} \left (-9 + \frac{1}{3}(9-x^2) \right ) + C$$
$$\sqrt{9-x^2} \left (-6 - \frac{x^2}{3} \right ) + C$$
$$ \bbox[5px,border:2px solid red]
{
- \left ( 6+ \frac{x^2}{3} \right ) \sqrt{9-x^2}
}
$$
This is the answer the assignment was looking for.
| You can solve without the trigonometric substitution, decomposing as
$$\frac{x^3}{\sqrt{9-x^2}}=9\frac{x}{\sqrt{9-x^2}}-x\sqrt{9-x^2}$$
and the antiderivatives are immediate (by $u=x^2$):
$$-9\sqrt{9-x^2}-\frac13(9-x^2)^{3/2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Proving $a^2 + b^2 + c^2 \geqslant ab + bc + ca$
Let $a, b, c \in \mathbb{R}$. Show that $a^2 + b^2 + c^2 \geqslant ab + bc + ca$.
My reasoning went as follows and I would like to know if it's correct.
$a^2 + b^2 + c^2 \geqslant ab + bc + ca$
$\Leftrightarrow a^2 + b^2 + c^2 -ab - bc - ca \geqslant 0$
$\Leftrightarrow 2a^2 + 2b^2 + 2c^2 -2ab - 2bc - 2ca \geqslant 0$
$\Leftrightarrow a^2 + a^2 + b^2 + b^2 + c^2 + c^2 -2ab - 2bc - 2ca \geqslant 0$
$\Leftrightarrow (a^2 -2ab + b^2) + (b^2 - 2bc + c^2)+ (c^2 - 2ca + a^2) \geqslant 0$
$\Leftrightarrow (a -b)^2 + (b-c)^2 + (c-a)^2 \geqslant 0$.
And the last inequality holds since squares are non-negative.
I've seen couple of different ways of doing this and they seem to divide by $2$ and I didn't really understand it so instead can I just use the fact that $2a^2 = a^2+a^2$?
| Add up the inequalities below
$$a^2 + b^2 \ge 2ab, \>\>\>\>\>b^2 + c^2 \ge 2bc, \>\>\>\>\>c^2 + a^2 \ge 2ca$$
to arrive at,
$$a^2 + b^2 + c^2 \geqslant ab + bc + ca$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int _0^1\frac{\ln ^2\left(x^2+1\right)}{x+1}\:dx$ using real methods I've stumbled upon that interesting integral here, the OP managed to transform the integral into something more approachable using contour integration and proved that
$$\int _0^1\frac{\ln ^2\left(x^2+1\right)}{x+1}\:dx\:=\:-\pi G+\frac{5}{2}\zeta \left(3\right)+\frac{2}{3}\ln ^3\left(2\right)-\frac{\pi ^2}{24}\ln \left(2\right)$$
I can't come up with a way to attack this one with only real methods, i'd appreciate any help possible.
| Integrate by parts
$$\int _0^1\frac{\ln ^2(x^2+1)}{x+1}~dx
=\ln^32 -4I \tag1$$
where $I$ is given below, along with its twin $J$
$$I=\int_0^1\frac{x\ln(1+x)\ln(x^2+1)}{1+x^2}~dx,\>\>\>\>\>
J=\int_0^1\frac{x\ln(1-x)\ln(x^2+1)}{1+x^2}~dx $$
Then, evaluate their sum and difference as follows
\begin{align}
J+I=& \int_0^1\frac{x\ln(1-x^2)\ln(x^2+1)}{1+x^2}~{dx}
\>\>\>\>\>\>\> \left( x^2\to\frac{1-x}{1+x} \right) \\
=& \frac12\int_0^1 \frac{\ln^2\frac{1+x}2}{1+x}dx
+\frac{\ln2}2 \int_0^1 \frac{\ln x}{1+x}dx+\frac14\int_0^1 \frac{\ln^2(1+x)}{x}dx\\
=&\frac16\ln^32-\frac{\pi^2}{24}\ln2+\frac1{16}\zeta(3)\\
\\
J-I=& \int_0^1\frac{x\ln\frac{1-x}{1+x}\ln(x^2+1)}{1+x^2}~{dx} \>\>\>\>\>\>\> \left( \frac{1-x}{1+x} \to x\right) \\
=& \>\ln2\int_0^1 \frac{\ln x}{1+x}dx
+\frac{\ln2}2 \int_0^1 \frac{\ln(1+x^2)}{x} \overset{x^2\to x}{dx}
+\frac14 \int_0^1 \frac{\ln^2(1+x^2)}{x}
\overset{x^2\to x}{dx}\\
& -2\int_0^1 \frac{\ln x\ln(1+x)}{1+x}\overset{ibp}{dx} - \int_0^1 \frac{\ln (1+x)\ln(1+x^2)}{x}dx\\
= & \>\frac{3\ln2}4\int_0^1 \frac{\ln x}{1+x}dx
+\frac98 \int_0^1 \frac{\ln^2(1+x)}{x} dx
- P
=\frac9{32}\zeta(3) - \frac{\pi^2}{16}\ln2-P\\
\end{align}
where $P$ and its twin $Q$ are
$$P= \int_0^1 \frac{\ln (1+x)\ln(1+x^2)}{x}dx,\>\>\>\>\>
Q= \int_0^1 \frac{\ln (1-x)\ln(1+x^2)}{x}dx
$$
Similarly, their sum and difference are (see here for evaluation details)
\begin{align}
&Q+P=\int_0^1 \frac{\ln (1-x^2)\ln(1+x^2)}{x}dx=-\frac5{16}\zeta(3)\\
&Q-P=\int_0^1 \frac{\ln \frac{1-x}{1+x} \ln(1+x^2)}{x}dx=-\pi G- \frac7{4}\zeta(3)\\
\end{align}
which results in $P= \frac{\pi}{2}G-\frac{33}{32}\zeta(3) $. Substitute into $I-J$ and $I+J$ above to get
$$
I= \frac{\pi}4 G-\frac5{8}\zeta(3)
+ \frac{\pi^2}{96}\ln2+ \frac1{12}\ln^32
$$
and then into (1) to obtain
$$\int _0^1\frac{\ln ^2(1+x^2)}{1+x}\>dx=-\pi G+\frac{5}{2}\zeta (3)-\frac{\pi ^2}{24}\ln2 +\frac{2}{3}\ln^32 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac1{a(1+b)}+\frac1{b(1+c)}+\frac1{c(1+a)}\ge\frac3{1+abc}$ I tried doing it with CS-Engel to get $$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{9}{a+b+c+ a b+b c+a c}
$$
I thought that maybe proof that $$
\frac{1}{a+b+c+a b+b c+a c} \geq \frac{1}{3(1+a b c)}
$$ or $$
3+3 a b c \geq a+b+c+a b+b c+a c
$$, but I don't know how
| For positive variables by AM-GM we obtain:
$$\sum_{cyc}\frac{1}{a(1+b)}=\frac{1}{1+abc}\sum_{cyc}\frac{1+abc}{a(1+b)}=\frac{1}{1+abc}\left(\sum_{cyc}\frac{1+abc}{a(1+b)}+1-1\right)=$$
$$=\frac{1}{1+abc}\sum_{cyc}\frac{1+a+ab+abc}{a(1+b)}-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\left(\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}\right)-\frac{3}{1+abc}\geq$$
$$\geq\frac{6}{1+abc}\sqrt[6]{\prod_{cyc}\left(\frac{1+a}{a(1+b)}\cdot\frac{b(1+c)}{1+b}\right)}-\frac{3}{1+abc}=\frac{3}{1+abc}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$2^{\sin(x) + \cos(y)} = 1$ , $16^{\sin^2(x) + \cos^2(y)} = 4$ (system of equations) My progress so far:
$2^{\sin(x) + \cos(y)} = 1$
$16^{\sin^2(x) + \cos^2(y)} = 4$
||||||||||||||||||||
$\sin(x) + \cos(y) = 0$
$\sin^2(x) + \cos^2(y) = \frac12$
I think I'm on the right track, but I'm not sure how to continue. Thanks in advance to anyone who helps.
| Suppose $\sin x + \cos y = 0$. Then
$$\sin ^2x + \cos ^2y = (\sin x + \cos y) ^2 -2\sin x\cos y = -2\sin x\cos y. $$
Thus we have left to consider
$$\begin{cases} \sin x + \cos y = 0 \\ \sin x\cos y = -\frac{1}{4} \end{cases} $$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Unable to prove an assertion with induction I need to prove:
$ \displaystyle \sum_{k=1}^n\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5n + 6)} $ with mathematical induction for all $n \in \mathbb{N}$.
After successfully proving it for n = 1, I try to prove it in the Induction-Step for n + 1:
$ \displaystyle \sum_{k=1}^{n+1}\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5(n+1) + 6)} $
which can be summarised to:
$ \displaystyle \sum_{k=1}^{n}\frac{1}{(5k + 1) (5k + 6)} + \frac{1}{(5(n+1) + 1) (5(n+1) + 6)} = \frac{1}{30} - \frac{1}{5(5(n+1) + 6)} $
Using the Induction Hyptohesesis this turns to:
$ ( \frac{1}{30} - \frac{1}{5(5n + 6)} ) + \frac{1}{(5n + 6 ) (5n+6+5)} $
when we let s = 5n +6 we get:
$ \frac{1}{30} - \frac{1}{5s} + \frac{1}{(s) (s+5)} = \frac{1}{30} - \frac{1}{5(s + 5)} $
substracting both sides by $\frac{1}{30}$ yields:
$- \frac{1}{5s} + \frac{1}{(s) (s+5)} = - \frac{1}{5(s + 5)}$ and this is where I'm stuck. I'm working on this for more than 6h but my mathematical foundation is too weak to get a viable solution. Please help me.
| You started fine. Note that to prove by induction a sequence of equalities of the type$$a_1+a_2+\cdots+a_n=b_n$$is the same thing as proving that $a_1=b_1$ and that, for each $n\in\mathbb N$, $a_{n+1}=b_{n+1}-b_n$. So, you should prove that\begin{align}\frac1{5\bigl(5(n+1)+1\bigr)\bigl(5(n+1)+6\bigr)}&=\frac1{30}-\frac1{5\bigl(5(n+1)+6\bigr)}-\left(\frac1{30}-\frac1{5(5n+6)}\right)\\&=\frac1{5(5n+6)}-\frac1{5(5n+11)}.\end{align}But\begin{align}\frac1{5\bigl(5(n+1)+1\bigr)\bigl(5(n+1)+6\bigr)}&=\frac1{5(5n+6)(5n+11)}\\&=\frac1{5(5n+6)}-\frac1{5(5n+11)}.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Compute $\left[\begin{smallmatrix}1-a & a \\ b & 1-b\end{smallmatrix}\right]^n$ Compute $\begin{bmatrix}1-a & a \\ b & 1-b\end{bmatrix}^n$, where the power of $n\in\mathbb N$ denotes multiplying the matrix by itself $n$ times; $a,b\in[0,1]$.
Edit:
I considered using induction, computed the desired matrix:
$$\begin{bmatrix}(1-a)^2+ab & a(2-a-b) \\ b(2-a-b) & (1-b)^2+ab\end{bmatrix}$$ and $$\begin{bmatrix}(1-a)^3+ab(1-a)+ba(2-a-b) & a(1-a)^2+a^2b+a^2(2-a-b) \\ (1-a)b(2-a-b)+b(1-b)^2+ab^2 & ab(2-a-b)+(1-b)^3+ab(1-b)\end{bmatrix}$$ for $n=2$ and $3$, respectively. However, I fail to see a pattern here that can be used as an induction hypothesis.
| Let $ n $ be a positive integer, $ a,b\in\left[0,1\right] $ such that $ a+b\neq 0 \cdot $
Denoting $ J=\left(\begin{matrix}-a&a\\b&-b\end{matrix}\right) $, observe that : $ \left(\begin{matrix}1-a&a\\b&1-b\end{matrix}\right)=I_{2}+J \cdot $
We have that: \begin{aligned} J^{2}=\left(\begin{matrix}-a&a\\b&-b\end{matrix}\right)\left(\begin{matrix}-a&a\\b&-b\end{matrix}\right)=\left(\begin{matrix}a^{2}+ab&-a^{2}-ab\\-b^{2}-ab&b^{2}+ab\end{matrix}\right)&=-\left(a+b\right)\left(\begin{matrix}-a&a\\ b& -b\end{matrix}\right)\\&=-\left(a+b\right)J \end{aligned}
Meaning, $ \left(\forall k\in\mathbb{N}\right),\ J^{k+1}=\left(-1\right)^{k}\left(a+b\right)^{k}J \cdot $
Thus, $ \left(\begin{matrix}1-a&a\\b&1-b\end{matrix}\right)^{n}=\sum\limits_{k=0}^{n}{\binom{n}{k}J^{k}}=I_{2}+nJ+\sum\limits_{k=2}^{n}{\binom{n}{k}J^{k}} $
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =I_{2}+nJ+\left(\sum\limits_{k=2}^{n}{\left(-1\right)^{k-1}\binom{n}{k}\left(a+b\right)^{k-1}}\right)J $
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =I_{2}+\frac{1-\left(1-a-b\right)^{n}}{a+b}J $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $p$ and $q=2p−1$ are primes, and $N=pq,$ then $N$ is pseudo-prime for all possible bases $b$ that are quadratic residues modulo $2p−1.$ ... Being $n \in \mathbb Z,$ $n$ is said to be pseudo-prime with respect to the base $b$ if it is compound and also verifies the congruence $b^{n−1}\equiv 1 \pmod n$ where $n|b^{n-1}-1$.
Show that if $p$ and $q=2p−1$ are primes, and $N=pq,$ then $N$ is pseudo-prime for all possible bases $b$ that are quadratic residues modulo $2p−1$.
I can solve it when given concrete examples, but I am struggling with being able to generalize this, as asked here.
For example, $91\equiv 13\cdot 7$, then $13\equiv 2\cdot 7-1$, $$\begin{equation*} \begin{cases} 4^2 = 9^2 \equiv 3 \pmod{13}\\ 5^2 = 8^2 \equiv 12 \pmod{13})\\ 6^2 = 7^2 \equiv 10 \pmod{13}\\ \end{cases} \end{equation*}$$
so \begin{equation*} \begin{cases} 4^2 = 9^2 \equiv 3 \pmod{13}\\ 5^2 = 8^2 \equiv 12 \pmod{13}\\ 6^2 = 7^2 \equiv 10 \pmod{13}.\\ \end{cases} \end{equation*}
But I do not know the general case for $p$ and $2p-1$.
| We want to show that if $p$ and $q:=2p-1$ are prime, then $N:=pq$ satisfies the property that if $$a \equiv u^2 \pmod{q},$$
and $p,q\nmid a$,
then
$$a^{N-1}\equiv 1 \pmod{N}.$$
One solution is to use the Chinese remainder theorem. In this context,
it says that it suffices to show both
$$a^{N-1}\equiv 1 \pmod{p}$$
and
$$a^{N-1}\equiv 1 \pmod{q}.$$
For the first statement,
$$
a^{N-1}=a^{p(2p-1)-1}=a^{2p^2-p-1}
=a^{(2p+1)(p-1)} \equiv 1\pmod{p},$$
since $a^{p-1}\equiv 1\pmod{p}$ by Fermat's little theorem.
For the second statement,
$$
a^{N-1}=a^{q(q+1)/2-1}
\equiv
u^{q(q+1)-2}
= u^{q^2+q-2}
= u^{(q-1)(q+2)}
\equiv 1\pmod{q}
$$
again by Fermat's little theorem, since $u^{q-1}\equiv 1\pmod{q}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For non-negative reals $x,y$ such that $x+y\le 4$ prove that $y(x-3)(y-3) \le 3(4-y)$ For non-negative reals $x,y$ such that $x+y\le 4$ prove that $y(x-3)(y-3) \le 3(4-y)$
ATTEMPT
We transform the equation into $xy^2+12y-3xy-3y^2\le 12$
I noticed that for $x=0, y=2$ equality is achieved, but I am lost here.
| You want to show that if $x$ and $y$ are nonnegative real numbers with $x+y\leq4$ then
$$f(x,y)=y(x-3)(y-3)-3(4-y),$$
is negative. Taking the derivative with respect to $x$ we get
$$\frac{\partial f}{\partial x}=y(y-3),$$
which is nonzero if $y\neq0$ and $y\neq3$. The derivative is everywhere positive for $y>3$ and so for these values of $y$ the maximum is at $x=4-y$. This shows that for $y>3$ we have
$$f(x,y)\leq f(4-y,y)=y(1-y)(y-3)-3(4-y)=-y^3+4y^2-12.$$
Similarly, if $0<y<3$ then the derivative is negative, and so for these values of $y$ the maximum is at $x=0$. This shows that for $0<y<3$ we have
$$f(x,y)\leq f(0,y)=-3y(y-3)-3(4-y)=-3y^2+6y-12.$$
For $y=0$ and $y=3$ it is clear that $f(x,y)=-3(4-y)$ is negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1+\sin^4 x}{\rm d}x$ Someone gives a solution as follows:
\begin{align*} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1+\sin^4 x}{\rm d}x&=2\int_0^{\frac{\pi}{2}}\frac{\sin^2 x}{1+\sin^4 x}{\rm d}x\\ &=2\int_0^{\frac{\pi}{2}}\frac{\cos^2 x}{1+\cos^4 x}{\rm d}x\\ &=2\int_0^{\frac{\pi}{2}}\frac{\sec^2 x}{\sec^4 x+1}{\rm d}x\\ &=2\int_0^{\frac{\pi}{2}}\frac{{\rm d}(\tan x)}{(1+\tan^2 x)^2+1}\\ &=2\int_0^{+\infty} \frac{{\rm d}t}{(1+t^2)^2+1}\\ &\color{red}{=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{{\rm d}u}{\left(u-\frac{1}{u\sqrt{2}}\right)^2+(1+\sqrt{2})}}\\ &=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{{\rm d}u}{u^2+(1+\sqrt{2})}\\ &=\frac{\pi}{2} \sqrt{\sqrt{2}-1} \end{align*}
I wonder how he obtains the sixth equality colored red.
| The red-colored line is obtained with the substitution $t = \frac1u$ as follows
\begin{align}
\int_0^{\infty} \frac{2}{(1+t^2)^2+1} dt
& =\int_{0} ^{\infty} \frac{2}{t^4+2t^2+2}dt\\
&=\int_{0}^{\infty} \frac{1}{\frac1{2u^2}+1+{u^2}} du\\
& =\frac{1}{2}\int_{-\infty}^{+\infty}\frac{1}{\left(u-\frac{1}{u\sqrt{2}}\right)^2+(1+\sqrt{2})}du
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $A$ and $B$ have non-negative eigen values then can we conclude that $A+B$ has non-negative eigen values? Let $A$ and $B$ be $n\times n$ matrices whose all eigen values are non-negative real numbers. What can I say about the signs of the eigen values of $A+B$ ? First of all is it possible that the eigen values of $A+B$ may not be real?
Of course if $A$ and $B$ are symmetric and have non-negative eigen values then it is easy to see that $A+B$ will also have non-negative eigen values (this is just equivalent to positive semidefiniteness). So I am only interested in the non-symmetric case.
Thank you.
| Take $A=\begin{pmatrix} 1& 2\\0&1\end{pmatrix}$, $B=\begin{pmatrix} 1& 0\\2&1\end{pmatrix}$. Then both have $1$ as unique eigenvalue but $A+B$ has zero as eigen value.
Take $C=\begin{pmatrix} 1& 1\\0&1\end{pmatrix}$, $D=\begin{pmatrix} 1& 0\\-1&1\end{pmatrix}$. Then both have $1$ as unique eigen value but $C+D$ has no real eigenvalue.
Take $E=\begin{pmatrix} 1& 1\\0&2\end{pmatrix}$, $F=\begin{pmatrix} 2& 0\\-1&1\end{pmatrix}$. Then both have $1,2$ as distinct eigen value but $E+F$ has no real eigenvalue.
Take $G=\begin{pmatrix} 1& 5\\0&2\end{pmatrix}$, $H=\begin{pmatrix} 2& 0\\5&1\end{pmatrix}$. Then both have $1,2$ as distinct eigen value but $G+H$ has $8,-2$ as eigenvalue.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3601299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that: $\left|\int\limits_a^b\frac{\sin x}{x}dx\right|\le \frac{2}{a}$ Prove that: $$\left|\int\limits_a^b\frac{\sin x}{x}dx\right|\le \frac{2}{a}$$
$0<a<b$
I'm not asking for a complete solution to this, just a hint or a push in the right direction. I've been struggling with this and haven't really made any progress.
| \begin{align*}
I &= \int_{a}^b \dfrac{\sin x}{x}dx \\
&= \dfrac{-\cos x}{x}|_{a}^b+ \int_{a}^b\dfrac{\cos x}{x^2}dx \\
&= \dfrac{\cos a}{a} - \dfrac{\cos b}{b}+ J \\
&\le \dfrac{1}{a}+\dfrac{1}{b}+ \int_{a}^b \dfrac{1}{x^2}dx \\
&= \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{b} \\
&= \dfrac{2}{a} \text{.}
\end{align*}
Also: $I \ge -\dfrac{1}{a} - \dfrac{1}{b}- \displaystyle \int_{a}^b \dfrac{1}{x^2}dx = -\dfrac{2}{a}$. Thus $|I| \le \dfrac{2}{a}$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Fermat's Little Theorem: Proving an integer exists which satisfies the following criteria mod $5$ This question has been tantalizing me for a day now:
Given integers $a,b,c,d$ such that $d \neq 0$ (mod $5$) and $m$ an integer for which $am^3 + bm^2 +cm+d \equiv 0$ (mod $5$), prove there exists an integer $n$ for which $dn^3+cn^2+bn+a \equiv 0 $ (mod $5$).
My attempt:
It is easy to see that $\gcd(5,m)=1$. We know from Fermat's little theorem that $5 \mid m^4 -1 \Rightarrow 5 \mid (m+1)(m-1)(m^2+1)$. The proof is trivial that $5$ only divides one of these factors. Hence we have $3$ cases:
Case $1 \rightarrow m \equiv 1$ (mod $5$). If this is true then it follows that $m^2 \equiv 1$ (mod $5$) and $m^3 \equiv 1$ (mod $5$). We can substitute $m$ into the equation and achieve:
$$a+b+c+d \equiv 0 \pmod5$$
$$\Rightarrow d(1)^3+c(1)^2+b(1)+a \equiv 0 \pmod5$$
Where $1$ satisfies the condition.
Case $2 \rightarrow m \equiv -1$ (mod $5$). If this is true then it follows that $m^2 \equiv 1$ (mod $5$) and $m^3 \equiv -1$ (mod $5$). We can substitute $m$ into the equation and achieve:
$$a-b+c-d \equiv 0 \pmod5$$
$$\Rightarrow d(-1)^3+c(-1)^2+b(-1)+a \equiv 0 \pmod5$$
Where $-1$ satisfies the condition.
Case $3 \rightarrow m^2 \equiv -1$ (mod $5$). If this is true then it follows that $m \equiv 2$ (mod $5$) and $m^3 \equiv -2$ (mod $5$). We can substitute $m$ into the equation and achieve:
$$2a+b-2c-d \equiv 0 \pmod5$$
From where I am unable to proceed. Can I get a hint?
| $m\not\equiv0\bmod5$, since otherwise $d\equiv0\bmod5$, contradicting the given facts. Hence, since $5$ is a prime number, $m$ has an inverse modulo $5$ – let $n$ be this inverse. Multiplying $am^3+bm^2+cm+d\equiv0\bmod5$ by $n^3$, we can easily verify that $n$ satisfies the final congruence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to use prove this $p^4\equiv p\pmod {13}$
Let a prime number $p$, and $n$ a positive integer such $$p\mid n^4+n^3+2n^2-4n+3.$$
Show that $$p^4\equiv p\pmod {13}.$$
A friend of mine suggested that I might be able to use the results problem.
| My take on this problem is to solve the quartic equation using the quartic formula. To use this formula, the $x^3$ term needs to be eliminated, which can be done by substituting $y=x-\frac {1}{4}$. The result is, after quite a bit of algebra, is
$y^4+\frac{13}{8} y^2 - \frac{39}{8}y -\frac{1053}{256}=0$
To solve this, use the cubic resolvent. For $y^4+py^2+qy+r$ this is
$z^3+2pz^2+(p^2-4r)z-q^2=0$
This yields
$z^3+\frac{13}{4}z^2-\frac{221}{16}z-\frac{1521}{64}=0$
Solve this using the cubic formula, which involves eliminating the $z^2$ term with $w=z-\frac{13}{16}$. After a lot more algebra, one finds that the roots are $\frac{13}{4}$, $\frac{-13+2\sqrt(13)}{4}$, and $\frac{-13-2\sqrt(13)}{4}$. That means that a root of the reduced quartic is
$\frac{1}{2} (\sqrt(\frac{13}{4})+\sqrt(\frac{-13+2\sqrt(13)}{4}) + \sqrt(\frac{-13-2\sqrt(13)}{4})$
Going back to the original quartic, one get, as one of the roots
$-\frac{1}{4}+\frac{1}{2} (\sqrt(\frac{13}{4})+\sqrt(\frac{-13+2\sqrt(13)}{4}) + \sqrt(\frac{-13-2\sqrt(13)}{4})$
Now if we take this polynomial mod 13, all the square root stuff beyond the $-\frac{1}{4}$ is zero, so a root mod 13 is $-\frac{1}{4}$. By trial and error or solving a diophantine equation, one gets that $-\frac{1}{4}=3$ mod 13. Then note that $3^4 = 3$ mod 13, which gives the intended result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of $3$ Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck.
Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of 3.
| A bit longer:
Assume $\tau(n)=\top,n\equiv 1\pmod{3}$
$$1+2^n+2^{2n}=7k\implies 2^{2n}=7k-2^n-1,\;k\in\mathbb Z$$
$\tau(n+1)=\top\implies\;n+1\equiv 2\pmod{3}$
$$1+2^{n+1}+2^{2(n+1)}=1+2\cdot 2^{n}+4\cdot2^{2n}=1+2\cdot2^n+4(7k-2^n-1)\\=28k-2\cdot2^n-3=28k-(2^{n+1}+3)$$
$$7\mid2^{n+1}+3\implies7\mid 2\cdot2^n-4=2(2^n-2)\iff 7\mid2^n-2=2(2^{n-1}-1)\\\iff 7\mid 2^{2n-1}-1$$
$$2^6\equiv 1\pmod{7}\implies 2^{6m}\equiv 1\pmod{7},\;m\in\mathbb N$$
$2n-1=6m\implies 3\mid 2n-1\implies 3\mid 2n+2\implies n\equiv 1\pmod{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
This inequality maybe is a form of conditional Chebyshev's inequality let $a_{i},b_{i}>0$, show that
$$\sum_{i=1}^{n}a_{i}b_{i}\ge \dfrac{2}{n+\sqrt{\sum_{i=1}^{n}\dfrac{b_{i}}{a_{i}}\sum_{i=1}^{n}\dfrac{a_{i}}{b_{i}}}}\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i}\tag{1}$$
I try:since
$$\sum_{i=1}^{n}\dfrac{b_{i}}{a_{i}}\sum_{i=1}^{n}\dfrac{a_{i}}{b_{i}}\ge n^2$$
We just have to prove the inequality
$$\sum_{i=1}^{n}a_{i}b_{i}\ge\dfrac{1}{n}\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i}$$
This looks like a Chebyshev's inequality, but unfortunately the monotonicity of the two sequences is not clear, so it can not be applied directly，so How to prove inequality $(1)$
| Proof: We have the following identity:
\begin{align}
&\sum a_i b_i - \frac{2}{n + \sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}}} \sum a_i \sum b_i\\
=\ & \sum \frac{b_i}{a_i}
\left(a_i - \frac{\frac{a_i}{b_i}\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}} + \sum \frac{a_i}{b_i}}
{\left(n+\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}}\right)\sum \frac{a_i}{b_i}}\sum a_i\right)^2.
\end{align}
We are done.
$\phantom{2}$
Remark: It is not hard to verify the identity above. Let $A = \sum \frac{a_i}{b_i}$, $B = \sum \frac{b_i}{a_i}$, $C = \sum a_i$ and $D = \sum b_i$.
We have
\begin{align}
&\sum \frac{b_i}{a_i}
\left(a_i - \frac{\frac{a_i}{b_i}\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}} + \sum \frac{a_i}{b_i}}
{\left(n+\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}}\right)\sum \frac{a_i}{b_i}}\sum a_i\right)^2\\
=\ &
\sum \frac{b_i}{a_i}
\left(a_i - \frac{\frac{a_i}{b_i}\sqrt{AB} + A}{(n+\sqrt{AB})A}C\right)^2\\
=\ &
\sum \frac{b_i}{a_i}
\left(a_i^2 - 2a_i \frac{\frac{a_i}{b_i}\sqrt{AB} + A}{(n+\sqrt{AB})A}C
+ \frac{(\frac{a_i}{b_i}\sqrt{AB} + A)^2}{(n+\sqrt{AB})^2A^2}C^2\right)\\
=\ &
\sum \left(a_ib_i - \frac{2a_i\sqrt{AB} + 2b_iA}{(n+\sqrt{AB})A}C
+ \frac{\frac{a_i}{b_i}AB + 2A\sqrt{AB} + \frac{b_i}{a_i}A^2}{(n+\sqrt{AB})^2A^2}C^2
\right)\\
=\ & \sum a_ib_i - \frac{2C\sqrt{AB} + 2DA}{(n+\sqrt{AB})A}C
+ \frac{A\cdot AB + n\cdot 2A\sqrt{AB} + B\cdot A^2}{(n+\sqrt{AB})^2A^2}C^2\\
=\ &\sum a_ib_i - \frac{2}{n+\sqrt{AB}} CD.
\end{align}
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3618210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
On proving a sequence converges by induction. I would like to prove by induction that given a sequence of positive real numbers $\{x_k \}_{k \in \mathbb{N}}$ such that
$$x_{k+1} \le \left(1 - \frac{2c \ell}{k} + \left(\frac{c}{k} \right)^2 M_1 L^2 \right) x_{k} + \frac{c^2}{k^2} M $$
where $0<c < \max\{ 1/ \sqrt{M_1} L, 2 \ell / M_1 L^2 \}$ and $\ell,M_1, L, M > 0$,
then
$$ x_{k+1} \le v/ (k+1) $$
where $v := \max \{x_1, c^2 M / (2\ell c -2) \}$.
My attempt:
the base case $x_{1} \le v$ is obviously verified, for the inductive step we obtain
\begin{align*}
& x_{k+1} \le \left(1 - \frac{2c \ell}{k} + \left(\frac{c}{k} \right)^2 M_1 L^2 \right) \frac{v}{k} + \frac{c^2}{k^2} M = \\ & \left( \frac{k^2 -2k + c^2 M_1 L^2}{k^3} \right)v - \frac{(2 c \ell -2)v}{k^2} + \frac{c^2M}{k^2}
\end{align*}
at this point we use the fact that $k^3 \ge (k^2-2k +1 ) ( k+1) $ for all $k> 1$ in conjunction with the definition of $c$ that gives
\begin{align*}
k^2 -2k + c^2 M_1 L^2 \le k^2 -2k + 1, \quad 0<\left(1 - \frac{2c \ell}{k} + \left(\frac{c}{k} \right)^2 M_1 L^2 \right)< 1
\end{align*}
while from the definition of $v$
\begin{align*}
- \frac{(2 c \ell -2)v}{k^2} + \frac{c^2M}{k^2} \le 0
\end{align*}
and utilizing these four inequalities we obtain
\begin{align*}
x_{k+1} \le \frac{v}{k+1}
\end{align*}
this should conclude the proof even if I am a bit unsure where inequality $0<\left(1 - \frac{2c \ell}{k} + \left(\frac{c}{k} \right)^2 M_1 L^2 \right)< 1 $ is being utilized, I suspect the bound $c < 2 \ell / M_1 L^2$ is not needed. Does this work?
| This part of the answer is preliminary and is intended to show the problems with the proposed approach.
Put $A= M_1L^2$ and $D=\ell^2-A$. Assume that $M\ge 0$ and $D\le 0$. It follows that $P(k)=1 - \frac{2c \ell}{k} + \left(\frac{c}{k} \right)^2 A\ge 0$ for each $k$.
In the appoach from the question in order to cancel the term $\frac {c^2M}{k^2}$ we need $c\ell>1$, which fails for small $c$. And it seems that anyway we need $2c\ell>1$ to cancel the term of order $O(k)$. Let’s see this.
Consider the induction step of a proof of a bound $x_k\le \frac {u}{k-1}$ for $k\ge 2$ (it is essentially the same as the proposed, but leads to more simple calculations). Since $P(k)\ge 0$, $P(k)x_k\le P(k)\frac u{k-1}$. We are going to show that
$\frac uk\ge \left(1 - \frac{2c \ell}{k} + \left(\frac{c}{k} \right)^2 A\right)\frac u{k-1} + \frac{c^2}{k^2} M$
$uk(k-1)\ge \left(k^2 - 2c \ell k + c^2 A\right) u + c^2(k-1) M$
$u(2c\ell-1) k\ge c^2 Au + c^2(k-1) M$.
Comparing terms of order $O(k)$ we see that we need $2c\ell-1>0$ and $u\ge \frac{c^2M}{2c\ell-1}$. Moreover, the latter inequality usually should be strict because if we put $u=\frac{c^2M}{2c\ell-1}$ then the induction step inequality will cancel to
$0 \ge c^2 A\frac{c^2M}{2c\ell-1} - c^2M$.
$ c^2M\ge c^2 A\frac{c^2M}{2c\ell-1}$.
$1\ge c^2 A\frac{1}{2c\ell-1}$.
$0\ge 1-2c\ell+c^2 A$
$0\ge P(1)$
But we have $P(1)\ge 0$.
PS. But a condition $D\le 0$ is not crucial for the convergence of the sequence $\{x_k\}$ to $0$. If $P(k)<0$ then $x_{k+1}\le \frac {c^2}{k^2}M$, which should converge to zero even faster than $\frac v{k}$. But since when $k$ tends to infinity, $P(k)$ tends to $1$,
we have $P(k)>0$ for sufficiently large $k$. Then similarly to the above arguments we can show that if $P(1)\le 0$ then there exists $u$ such that $x_{k}\le\frac u{k-1}$ for all sufficiently large $k$.
PPS. If $c\ell>0$ then the sequence $\{x_k\}$ converges to $0$. To show this pick any number $0<r<\min\{2c\ell,1\}$. Then there exists $K$ such that $P(k)\le Q(k)=1-\frac rk$ for each $k\ge K$. Consider an auxiliary sequence $\{y_k\}$, where $y_k= x_k+\frac {B}{k-1}$ for each $k\ge 2$ and $B=\frac{c^2M}{1-r}$. We claim that $y_{k+1}\le Q(k)y_k$ for each $k\ge K$. Indeed, we have
$y_{k+1}=x_{k+1}+\frac {B}{k}\le Q(k)x_k+\frac{c^2}{k^2}M+\frac {B}{k}= Q(k)\left(y_k-\frac{B}{k-1}\right) +\frac{c^2}{k^2}M+\frac {B}{k}$.
So it suffices to check that
$0\ge -Q(k)\frac{B}{k-1} +\frac{c^2}{k^2}M+\frac {B}{k}$
$0\ge –(k^2-rk)B +c^2(k-1)M+Bk(k-1)$
$0\ge rkB +c^2(k-1)M-Bk$
$k(B(1-r)-c^2M) \ge -c^2M$.
The last inequality holds by the choice of $B$.
Thus for each $k\ge K$ we have $y_{k+1}\le y_K\prod_{i=K}^k Q(i)$.
Pick any natural $R\ge\frac 1r$. Than for each $k=Rm$ we have
$$\prod_{i=1}^k Q(i)=\prod_{i=1}^{Rm} \left(1-\frac{r}{i}\right)\le \prod_{i=1}^{Rm} \left(1-\frac{1}{iR}\right)\le
\prod_{j=1}^{m} \left(1-\frac{1}{jR}\right)^R\le \prod_{j=1}^{m} \frac j{j+1}=\frac 1{m+1}.$$
Thus the sequence $\{y_k\}$ converges to $0$, and so does $\{x_k\}$.
Remark that an inequality $\left(1-\frac{1}{jR}\right)^R\le \frac j{j+1} $ for each $j$ follows from Bernoulli's inequality.
Indeed, by Bernoulli's inequality, $\left(1+\frac{1}{jR}\right)^R\ge 1+\frac 1j=\frac{j+1}j$. On the other hand,
$\left(1+\frac{1}{jR}\right)^R\left(1-\frac{1}{jR}\right)^R=\left(1-\frac{1}{j^2R^2}\right)^R\le 1$. So
$\left(1-\frac{1}{jR}\right)^R\le \frac j{j+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to integrate with a function in the limit $\int ^a _0 (\int_0 ^ \sqrt {a^2 - x^2} (a^2 - x^2 - y^2)dy)dx = 303$
I'm trying to solve the above question where the integral has a function as the limit.
Seeing the similarity between the limit and the variable integration, I substituted the $\sqrt {a^2 - x^2}$ with $u$.
Hence I end up with:
$\int ^a _0 (\int_0 ^ u (u^2 - y^2)dy)dx = 303$
After that, I'm not too sure how to proceed as there isn't much to go on as far as integrating $y$ with respect to $u$ is concerned.
EDIT:
Building onto the advice given:
$\int ^a _0 (\int_0 ^u (u^2 - y^2)dy)dx = 303$
$\int ^a _0 [u^2y - \cfrac{y^3}{3}]^u _0 dx = 303$
$\int ^a _0 [u^3 - \cfrac{u^3}{3}] dx = 303$
$[u^3x - \cfrac{u^3x}{3}]^a_0$ = 303
$((a^2 - a^2)^{1.5}a - \cfrac{(a^2 - a^2)^{1.5}a}{3} = 303$
EDIT1:
$\int ^a _0 (\int_0 ^\sqrt{a^2 - x^2} (a^2 - x^2 - y^2)dy)dx = 303$
$\int ^a _0 [a^2y - x^2y - \cfrac{y^3}{3}]^\sqrt{a^2 - x^2}_0 dx = 303$
$\int ^a _0 [a^2\sqrt{a^2 - x^2} - x^2\sqrt{a^2 - x^2} - \cfrac{((a^2 - x^2)^{1.5})}{3}dx = 303$
$\int ^a _0 [(a^2 - x^2)\sqrt{a^2 - x^2} - \cfrac{((a^2 - x^2)^{1.5})}{3}dx = 303$
$\int ^a _0 [(a^2 - x^2)^{1.5} - \cfrac{((a^2 - x^2)^{1.5})}{3}dx = 303$
$\int ^a _0 [\cfrac{2}{3}(a^2 - x^2)^{1.5}dx = 303$
| Hint. You do the inner integral first. That is, integrate with respect to $y$ alone, treating $x$ as constant. When you put in the limits, you get an expression that depends on $x$ alone. Then integrate this and substitute the constant limits -- this is the outer integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving $\int_{0}^{\frac{\pi}{2}}\text{erf}(\sqrt{a}\cos(x))\text{erf}(\sqrt{a}\sin(x))\sin(2x)dx=\frac{e^{-a}-1+a}{a}$ How to prove that for $a>0$:
*
*$\int_{0}^{\frac{\pi}{2}}\text{erf}(\sqrt{a}\cos(x))\text{erf}(\sqrt{a}\sin(x))\sin(2x)dx=\frac{e^{-a}-1+a}{a}$
*$\int_{0}^{\frac{\pi}{2}}\text{erf}\ ^2(\sqrt{a}\cos(x))\cos^2(x)dx+\int_0^1 \frac{e^{-\frac{a(1+x^2)}{2}}}{1+x^2}\left(I_0\left(\frac{a(1+x^2)}{2}\right)-I_1\left(\frac{a(1+x^2)}{2}\right)\right)dx=\frac{\pi}{4}$
Here erf denotes error function and $I_\nu$ Bessel. These identities arises from J. M. Borwein's Experiments in mathematics: Computational paths to discovery but the literature offers no proof. Any help will be appreciated!
| As noted by @Zacky the first result can be adapted from this answer from @DrZafarAhmedDSc. Using the series expansion
\begin{equation}
\operatorname{erf}(z)= e^{-z^2} \sum_{n=0}^{\infty} \frac{z^{2n+1}}{\Gamma(n+3/2)}.
\end{equation}
we have
\begin{align}
\mathcal{I}_1&=\int_{0}^{\frac{\pi}{2}}\operatorname{erf}(\sqrt{a}\cos(x))\operatorname{erf}(\sqrt{a}\sin(x))\sin(2x)dx\\
&=2e^{-a}\int_{0}^{\frac{\pi}{2}}\sum_{n=0}^\infty\sum_{m=0}^\infty a^{n+m+1}\frac{\cos^{2n+2}x}{\Gamma(n+3/2)}\frac{\sin^{2m+2}x}{\Gamma(m+3/2)}\,dx\\
&=e^{-a}\sum_{n=0}^\infty\sum_{m=0}^\infty a^{n+m+1}\frac{B(n+3/2,m+3/2)}{\Gamma(n+3/2)\Gamma(m+3/2)}\\
&=e^{-a}\sum_{n=0}^\infty\sum_{m=0}^\infty\frac{a^{n+m+1}}{(n+m+2)!}
\end{align}
where the Beta integral was used. Now the double summation can be rearranged as
\begin{align}
\mathcal{I}_1&=a^{-1}e^{-a}\sum_{m=0}^\infty\sum_{p=m}^\infty\frac{a^{p+2}}{(p+2)!}\\
&=a^{-1}e^{-a}\sum_{p=0}^\infty\sum_{m=0}^{p}\frac{a^{p+2}}{(p+2)!}\\
&=a^{-1}e^{-a}\sum_{p=0}^\infty\frac{(p+1)a^{p+1}}{(p+2)!}\\
&=a^{-1}e^{-a}\left( a\sum_{p=0}^\infty\frac{a^{p+2}}{(p+1)!} -\sum_{p=0}^\infty\frac{a^{p+2}}{(p+2)!}\right)\\
&=a^{-1}e^{-a}\left[a\left( e^a-1 \right)-\left( e^a-1-a \right)\right]\\
&=\frac{1}{a}\left( e^{-a}-1+a\right)
\end{align}
as proposed.
For the second formula, we use the integral representation of $\operatorname{erf}^2$ here:
\begin{equation}
\int_{0}^{1}\frac{e^{-\alpha t^{2}}}{t^{2}+1}\mathrm{d}t=\frac{\pi}{4}e^{a}\left(1-(%
\operatorname{erf}\sqrt{\alpha})^{2}\right)
\end{equation}
(This expression can be derived from the integral definition of $\operatorname{erf}$, by interpreting the product of integrals as a double integral in the unit square and by expressing it in polar coordinates). Then,
\begin{equation}
\operatorname{erf}^2(\sqrt{a}\cos(x))=1-\frac{4}{\pi}\int_0^1e^{-a(1+t^2)\cos^2x}\frac{dt}{1+t^2}
\end{equation}
which can be plug into the integral
\begin{align}
\mathcal{I}_2&=\int_{0}^{\frac{\pi}{2}}\operatorname{erf}^2(\sqrt{a}\cos(x))\cos^2(x)\,dx\\
&=\int_{0}^{\frac{\pi}{2}}\left[1-\frac{4}{\pi}\int_0^1e^{-a(1+t^2)\cos^2x}\frac{dt}{1+t^2}\right]\cos^2x\,dx\\
&=\frac{\pi}{4}-\frac{2}{\pi}\int_0^1\frac{e^{-\frac{a}{2}(1+t^2)}dt}{1+t^2}\int_{0}^{\frac{\pi}{2}}e^{-\frac{a}{2}(1+t^2)\cos2x }\left( 1+\cos 2x \right)\,dx\\
&=\frac{\pi}{4}-\frac{1}{\pi}\int_0^1\frac{e^{-\frac{a}{2}(1+t^2)}dt}{1+t^2}\int_{0}^{\pi}e^{\frac{a}{2}(1+t^2)\cos y }\left( 1-\cos y \right)\,dy\\
&=\frac{\pi}{4}-\int_0^1\frac{e^{-\frac{a}{2}(1+t^2)}}{1+t^2}\left[I_0\left(\frac{a}{2}(1+t^2) \right)-I_1\left(\frac{a}{2}(1+t^2) \right)\right]\,dt
\end{align}
where the classical integral representation for the modified Bessel functions (DLMF) was used. This result corresponds to te proposed formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3651510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that sequence $ x_{(n+1)}= \frac {a}{1+x_n}$ is convergent to positive root of $x^2+x-a=0,$where $a >0$ and $x_1 >0$ Prove that sequence $$ x_{(n+1)}= \frac {a}{1+x_n}$$ is convergent to positive root of $x^2+x-a=0,$where $a >0$ and $x_1 >0$
we have $$x_{(n+2)}-x_n=\frac{-a(x_{(n+1)}-x_{(n-1)})}{(1+x_{(n-1)})(1+x_{(n+1)})}$$
Hence from this we see behaviour of even terms subsequence and odd terms subsequence is opposite. how to proceed from there. any hint please
| Yes, the behavior is opposite indeed, but, as a continuation of your work, let's look at
$$x_{n+2}=\frac{a}{1+x_{n+1}}=\frac{a}{1+\frac{a}{1+x_n}}=\frac{a(x_n+1)}{1+x_n+a}$$
or $x_{n+2}=f(x_n)$ where $f(x)=\frac{a(x+1)}{1+x+a}$. $f(x)$ is ascending because $f'(x)=\left(\frac{a}{1+x+a}\right)^2$. It is also worth mentioning that from $n=2$ we definitely have $0<x_n=\frac{a}{1+x_{n-1}}\leq a$ and $0< f(x)\leq a$ as well, for $x>0$. Thus, technically, we can treat $f(x)$ as a mapping $f:[0,a]\to[0,a]$. It is also a contraction map since, from MVT, $\forall x,y$ there $\exists\epsilon$ in between s.t.
$$|f(x)-f(y)|=|f'(\epsilon)|\cdot|x-y|<\left(\frac{a}{1+a}\right)^2\cdot|x-y|$$
with $q=\left(\frac{a}{1+a}\right)^2\in[0,1)$. As a result
*
*odd terms sub-sequence will converge ...
*even terms sub-sequence will converge ...
... to a $x>0$ that satisfies
$$x=\frac{a(x+1)}{1+x+a}\iff
ax+a=x+x^2+ax \iff\\
x^2+x-a=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3651794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find Maclaurin series for $x^2\ln(1+2x)$ We recently started learning about Maclaurin series. I feel confused a little, can anyone help me with this?
$$f(x) =x^2\ln(1+2x),\;\vert x\vert\lt\frac{1}{2}$$
| One way to assess your problem is to it the following way by using the following equation: \begin{equation}f(x)=f(0)+f'(0)\frac{x}{1!}+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+\dots\end{equation}
So when assessing the problems like one has to realize that $x^2$ is a polynomial form, and $\ln(1+2x)$. So the natural log is what one is going to take the Maclaurin these that are Maclaurin series one can do the following by creating a table of derivatives, and values: \begin{array}{|c|c|}\hline f(x)=\ln(1+2x)& f(0)=\ln(1)=0\\ \hline f'(x)=\frac{2}{1+2x} &f'(0)=2 \\ \hline f''(x)=\frac{-4}{(1+2x)^2} & f''(0)= -4\\ \hline f'''(x) =\frac{16}{(1+2x)^3} & f'''(0)=16 \\ \hline f''''(x)=\frac{-96}{(1+2x)^4} & f''''(0)=-96 \end{array}
\begin{equation}\ln(1+2x)=0+2x-\frac{4}{2!}x^2+\frac{16}{3!}x^3-\frac{96}{4!}x^4+\dots \end{equation}
This equation can then be reduced to:
\begin{equation}\ln(1+2x)=2x-\frac{(2x)^2}{2}+\frac{(2x)^3}{3}-\frac{(2x)^4}{4}+\dots \end{equation}
So one can then write this:
\begin{equation}\sum_\limits{n=0}^\infty (-1)^n\frac{(2x)^n}{n}\end{equation}
All you have to now is multiply it by $x^2$, and get the following:
\begin{equation}\sum_\limits{n=0}^\infty (-1)^n\frac{2^nx^{n+2}}{n}\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find a closed form to the solution of $\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-x}}}}}=x$ Hi I try to solve the following nested radical :
$$\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-x}}}}}=x$$
Miraculously the related polynomials is a quintic .More precisely :
$$ x^5 - x^4 - 4 x^3 + 3 x^2 + 3 x - 1=0$$
I know that we can reduce the quintic to a Bring quintic form and use Jacobi theta function .
My question :
Can we hope to see a closed form with radicals ?
Any helps is greatly appreciated
Thanks a lot for all your contributions.
| One of the solution that can be obtained easily is by substituting $x = 2\cos\theta$
steps as follows
$\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-x}}}} = x$ will become $\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2\cos\theta}}}} = 2\cos\theta$
Now by applying Half angle cosine formula we can simplify as follows
$\sqrt{2+\sqrt{2+\sqrt{2-2\sin\frac{\theta}{2}}}} = 2\cos\theta$
$\sqrt{2+\sqrt{2+\sqrt{2-2\cos(\frac{\pi}{2} - \frac{\theta}{2})}}} = 2\cos\theta$
$\sqrt{2+\sqrt{2+2\sin(\frac{\pi}{4} - \frac{\theta}{4})}} = 2\cos\theta$
$\sqrt{2+\sqrt{2+2\cos(\frac{\pi}{2} -\frac{\pi}{4} + \frac{\theta}{4})}} = 2\cos\theta$
$\sqrt{2+2\cos(\frac{\pi}{4} -\frac{\pi}{8} + \frac{\theta}{8})} = 2\cos\theta$
$2\cos(\frac{\pi}{8} -\frac{\pi}{16} + \frac{\theta}{16}) = 2\cos\theta$
Now solving for $\theta$ will give rise to $\frac{\pi}{16} + \frac{\theta}{16} = \theta$
$\frac{16\theta}{16}- \frac{\theta}{16} = \frac{\pi}{16} $
$\theta = \frac{\pi}{15}$. Therefore $x$ becomes $2\cos\frac{\pi}{15}$ or $2\cos12^\circ$
As this is multiple of 3, we can calculate value of $2\cos12^\circ$ which we can represent as a finite nested radical as $\frac{1}{2}\times\sqrt{9+\sqrt5+\sqrt{(30-6\sqrt5)}}$
Refer here
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can I approach this inequality? Let $a, b$ and $c$ be three non-zero positive numbers. Show that:
$$\sqrt{\frac{2a}{a + b}} + \sqrt{\frac{2b}{b + c}} + \sqrt{\frac{2c}{a + c}} \leq 3$$
I know the triangular inequality would help here, but I don't know how to approach it.
I started by $a+b≥a$ then that gives $\frac{1}{a+b}≤\frac{1}{a}$ by muliplyting both sides by $2a$ we get $\frac{2a}{a+b}≤\frac{2a}{a}$ which leads eventually to $\frac{2a}{a+b}≤2$ and by adding the square root to both sides we get $\sqrt{\frac{2a}{a+b}}\leq\sqrt2$ and doing the same thing to the other terms we get $\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\leq3\sqrt2$ beyond that I don't have any idea if that would lead to anything useful or not.
| By C-S $$\sum_{cyc}\sqrt{\frac{2a}{a+b}}\leq\sqrt{2\sum_{cyc}\frac{a}{(a+b)(a+c)}\sum_{cyc}(a+c)}=$$
$$=\sqrt{\frac{8(ab+ac+bc)(a+b+c)}{\prod\limits_{cyc}(a+b)}}\leq3,$$ where the last inequality it's just $$\sum_{cyc}c(a-b)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
$m(\angle BOD)=60^{\circ}$ iff $k=\sqrt{3}$. Let $\triangle {ABC}$ cu $m(\angle C)=90^{\circ}$ and $D\in [BC], E\in [AC]$ s.t. $\frac{BD}{AC}=\frac{AE}{CD}=k$.
If $BE\cap AD=\{O\}$ show that $m(\angle BOD)=60^{\circ}$ iff $k=\sqrt{3}$.
I tried to prove it with trigonometry, with tangent, but there are a lot of computations. Also I constructed perpendiculars but also I have to do a lot of computations and I am stuck.
|
Let us prove that $k=\sqrt 3\implies \angle BOD=\angle AOE=60^\circ.$
First, construct lines perpendicular to lines $AC$ and $BC$ at points $E$ and $D$. Their intersection is point $F$. Also construct line $OG$ perpendicular to $AC$ at point $G$ and line $OH$ perpendicular to line $BC$ at point $H$.
To simplify calculations, introduce the following lengths: $AC=b$, $EF=CD=x$, $OG=p$ and $OH=q$. Our first task is to calculate $p$ and $q$ in terms of $b$ and $x$.
It is given that:
$$BD=AC\sqrt{3}=b\sqrt3\tag{1}$$
$$AE=CD\sqrt{3}=EF\sqrt{3}=x\sqrt{3}\tag{2}$$
It is obvious that $\triangle ACD\sim\triangle AGO$ and therefore:
$$\frac{GO}{CD}=\frac{AG}{AC}=\frac{AC-GC}{AC}$$
$$\frac{p}{x}=\frac{b-q}{b}\tag{3}$$
It is also obvious that $\triangle ECB\sim\triangle OHB$ and therefore:
$$\frac{EC}{OH}=\frac{BC}{BH}$$
$$\frac{AC-AE}{OH}=\frac{BD+CD}{BD+CD-CH}$$
$$\frac{b-x\sqrt 3}{q}=\frac{b \sqrt 3+x}{b \sqrt 3+x-p}\tag{4}$$
Now solve (3) and (4) for $p$ nad $q$ and you get:
$$p=\frac{x^2}{b^2+x^2}(b\sqrt 3 + x) \tag{5}$$
$$q=\frac{b^2}{b^2+x^2}(b-x\sqrt 3)\tag{6}$$
Let us now compare red angles $\alpha_1=\angle EAO$ and $\alpha_2=\angle EFO$:
$$\tan\alpha_1=\frac{CD}{AC}=\frac xb\tag{7}$$
$$\tan\alpha_2=\frac{EG}{HD}=\frac{AC-AE-CG}{CD-CH}=\frac{b-x\sqrt 3-q}{x-p}\tag{8}$$
If you replace (5) and (6) into (8) you get:
$$\tan\alpha_2=\frac{b-x\sqrt 3-\frac{b^2}{b^2+x^2}(b-x\sqrt 3)}{x-\frac{x^2}{b^2+x^2}(b\sqrt 3 + x)}=\frac{x}{b}$$
In other words, red angles $\alpha_1$ and $\alpha_2$ are equal! The rest is simple: because of this, quadrilateral $AEOF$ is cyclic (angles $\alpha_1$ and $\alpha_2$ above the same quadrilateral side are equal). Consequentially, green angles $\angle AOE$ and $\angle AFE$ are also equal:
$$\angle AOE = \angle AFE$$
But $\tan\angle AFE=\sqrt 3\implies \angle AFE=60^\circ$. It means that $\angle AOE=\angle BOD=60^\circ$.
I'll leave the detailed proof of the opposite statement to you.
HINT: You can show in exactly the same way that quadrilateral $AEOF$ is conclyclic, just replace $\sqrt 3$ with $k$ in previous expressions. It means that angles $\angle EFA$ and $\angle EOA$ are equal to $60^\circ$. Unknown factor $k$ is simply equal to $\tan\angle EFA=\tan 60^\circ=\sqrt 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving a first order differential equation by finding an integrating factor : Edit Problem:
Solve the following differential equations by first finding an integrating factor.
$$ (y^2(x+1) + y ) \, dx + ( 2xy + 1 ) \, dy = 0 $$
Answer:
\begin{align*}
M_y &= 2(x+1)y + 1 = 2xy + 2y + 1 \\
N_x &= 2y \\
\frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] &=
\frac{ 2xy + 2y + 1 - 2y } { 2xy + 1 } \\
\frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] &=
\frac{ 2xy + 1 } { 2xy + 1 } = 1 \\
\end{align*}
This means that:
$$ e ^ { \int \frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] \, dx } $$
is the integrating factor we seek. Call this integrating factor $I$.
\begin{align*}
I &= e ^ { \int 1 \, dx } = e^x \\
(y^2(x+1) + y ) e^x \, dx + ( 2xy + 1 ) e^x \, dy &= 0
\end{align*}
Now we have:
\begin{align*}
M &= (y^2(x+1) + y ) e^x \\
M_y &= ( 2(x+1)y + 1 )e^x = ( 2xy + 2y + 1)e^x \\
N &= ( 2xy + 1 ) e^x \\
N_x &= ( 2xy + 1 ) e^x
\end{align*}
As I understand it, I was suppose to get $M_y = N_x$. That is, the de should have been exact. What did I do wrong?
Now, I have an updated answer. However, It is still wrong. I feel I am much closer to the right answer. Here is my updated answer:
\begin{align*}
M_y &= 2(x+1)y + 1 = 2xy + 2y + 1 \\
N_x &= 2y \\
\frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] &=
\frac{ 2xy + 2y + 1 - 2y } { 2xy + 1 } \\
\frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] &=
\frac{ 2xy + 1 } { 2xy + 1 } = 1 \\
\end{align*}
This means that:
$$ e ^ { \int \frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] \, dx } $$
is the integrating factor we seek. Call this integrating factor $I$.
\begin{align*}
I &= e ^ { \int 1 \, dx } = e^x \\
(y^2(x+1) + y ) e^x \, dx + ( 2xy + 1 ) e^x \, dy &= 0
\end{align*}
Now we have:
\begin{align*}
M &= (y^2(x+1) + y ) e^x \\
M_y &= ( 2(x+1)y + 1 )e^x = ( 2xy + 2y + 1)e^x \\
N &= ( 2xy + 1 ) e^x \\
N_x &= ( 2xy + 1 ) e^x + (2y)e^x = (2xy + 2y + 2)e^x1
\end{align*}
Hence the differential equation is exact. We have:
\begin{align*}
F_x &= (y^2(x+1) + y ) e^x \\
F &= \int (y^2(x+1) + y ) e^x \, dx = \int (x y^2 + y^2 + 1 ) e^x \, dx
\end{align*}
Recall that:
$$ \int x e^x \, dx = x e^x - e^x + C $$
\begin{align*}
F &= y^2 \int xe^x \, dx + (y^2+1) \int e^x \, dx \\
F &= y^2 ( xe^x - e^x) + (y^2 + 1)e^x + \phi(y) \\
F &= y^2 xe^x - y^2 e^x + y^2 e^x + e^x + \phi(y) \\
F &= y^2 xe^x + e^x + \phi(y) \\
F_y &= 2xy e^x + \phi'(y) \\
2xy e^x + \phi'(y) &= ( 2xy + 1 ) e^x \\
\phi'(y) &= e^x \\
\phi(y) &= ye^x + c \\
F &= y^2 xe^x + e^x + ye^x + c
\end{align*}
However, the book gets:
$$ x y^2 e^x + y e^x = c $$
Where did I go wrong?
Problem:
Solve the following differential equations first finding an integrating factor.
$$ ( 5xy + 4y^2 + 1 ) \, dx + ( x^2 + 2xy ) \, dy = 0 $$
Answer:
Now, I try $x^3$ as an integrating factor. This gives me:
$$ ( 5x^4 y + 4 x^3 y^2 + x^3 ) \, dx + ( x^5 + 2x^4 y ) \, dy = 0 $$
Now, we see if it is exact.
\begin{align*}
M_y &= 5x^4 + 8 x^3 y \\
N_x &= 5x^4 + 8 x^3 y
\end{align*}
The equation is exact. Let $F$ be the solution we seek:
\begin{align*}
F_x &= 5x^4 y + 4 x^3 y^2 + x^3 \\
F &= x^5 y + x^4 y^2 + \frac{x^4}{4} + \phi(y) \\
F_y &= 5x^4 + 2x^4 y + \phi'(y) = x^5 + 2x^4 y \\
\phi'(y) &= 0 \\
\phi(y &= C
\end{align*}
Hence the solution we seek is:
$$ 4x^5 y + 4x^4 y^2 + x^4 + C = 0 $$
Where did I go wrong?
| $$(y^2(x+1) + y ) \, dx + ( 2xy + 1 ) \, dy = 0$$
Rearrange some terms:
$$y^2xdx+y^2dx + y dx + xdy^2 + dy = 0$$
$$y^2xdx+(y^2dx +xdy^2)+ y dx + dy = 0$$
$$y^2xdx+dxy^2+ y dx + dy = 0$$
Multiply by $e^x$:
$$y^2xde^x+e^xdxy^2+ y de^x + e^x dy = 0$$
$$dxy^2e^x+ de^xy= 0$$
Integrate:
$$xy^2e^x+ e^xy= C$$
Note that you have to use the product rule for $N_x$:
$$N_x = (( 2xy + 1 ) e^x)'$$
$$N_x= ( 2xy + 1 ) e^x+e^x(2y)$$
$$N_x= ( 2xy + 1 +2y) e^x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to improve $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$ I have proved this inequality $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$.
Using $\left|\sin(nx)\right|\leq n\left|\sin(x)\right|$ on $[0,\frac{\pi}{2n}]$ and $\frac{\left|\sin(nx)\right|}{\left|\sin(x)\right|}\leq\frac{\pi}{2x}$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$,we can have
$$\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{8}+\frac{\pi^{2}}{8}\left(n^{2}-1\right)<\frac{\pi^{2}n^{2}}{4}.$$
But using mathematica I found this inequality can still be improved.
And after calculating some terms I found it seems that when $n\geq 2$ we can have
$$\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{8}.$$
But I cannot prove this.So is there any method to improve my result?Any help will be thanked.
| We have the elementary estimate
$$1 \le \frac{z^4}{\sin^4 z} \le 1 + z^2 \varepsilon$$
where
$$\varepsilon= \frac{\pi^2}{4} - \frac{4}{\pi^2}.$$
Let $z = (y/n)$ and multiply both sides by $\sin^4 y/y^4$. Then for $y \in [0,n \pi/2]$,
one has:
$$ \frac{\sin^4 y}{y^4}
\le
\left(\frac{\sin(y)/n}{\sin (y/n)}\right)^4
\le \frac{\sin^4 y}{y^4} + \frac{\sin^4 y}{y^2 n^2} \cdot \varepsilon
$$
Make the substitution $x = y/n$ in the integral, it becomes
$$I_n:=n^2 \int^{n \pi/2}_{0} y \left(\frac{\sin(y)/n}{\sin(y/n)}\right)^4 dy$$
and thus
$$n^2 \int^{n \pi/2}_{0} \frac{\sin^4 y}{y^3} dy \le I_n
\le n^2 \int^{n \pi/2}_{0} \frac{\sin^4 y}{y^3} dy + \varepsilon \cdot \int^{n \pi/2}_{0} \frac{\sin^4 y}{y} dy$$
The lower bound is asymptotic to
$$n^2 \int^{\infty}_{0} \frac{\sin^4 x}{x^3} dx = n^2 \log 2,$$
and in fact since
$$n^2 \int^{\infty}_{n \pi/2} \frac{\sin^4 x}{x^3}
\le n^2 \int^{\infty}_{n \pi/2} \frac{1}{x^3} = \frac{2}{\pi^2} $$
one even has the lower bound
$$I_n \ge n^2 \log 2 - \frac{2}{\pi^2}$$
On the other hand, an upper bound is given by
$$ n^2 \int^{\infty}_{0} \frac{\sin^4 y}{y^3} dy =
\varepsilon \cdot \int^{1}_{0} \frac{\sin^4 y}{y} +
\varepsilon \cdot \int^{n \pi/2}_{1} \frac{1}{y} dy$$
$$ = n^2 \log 2 + \eta + \varepsilon \log(n \pi/2)$$
where $\eta \sim 0.160629\ldots$ and $\varepsilon \sim 2.062116\ldots$.
From this you can obtain your explicit bound for $n \ge 3$ and check $n = 2$ by hand. Of course, it gives a more precise bound for larger $n$, and it's clear that you can push this much further if you want to.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Calculating the limit of $\lim\limits_{n \to \infty} (\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ...+\frac{1}{2n})$ Hello everyone how can I calculate the limit of:
$\lim\limits_{n \to \infty} (\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ...+\frac{1}{2n})$
| Hint:
Factor out $n$ in all denominators:
$$\frac{1}{n+1} + \frac{1}{n+2} + …+\frac{1}{2n}=\frac1n\biggl(\frac1{1+\frac1n}+\frac1{1+\frac2n}\dots+\frac1{1+\frac nn}\biggr)$$
and recognise a Riemann sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3665329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Any $(x, y, z)$ can satisfy the $5x^2+2y^2+6z^2-6xy-2xz+2yz<0$? Please tell me whether there any $(x, y, z)$
which can satisfy the $5x^2+2y^2+6z^2-6xy-2xz+2yz<0$ ?
No process or just solve it by calculator are both fine.
Thank you.
| The expression $$5x^2+2y^2+6z^2-6xy-2xz+2yz$$ may be written as $$3(x-y)^2+(x-z)^2+(y+z)^2+x^2-2y^2+4z^2$$ Notice that the only term with a negative sign is $-2y^2$, all others are positive. Setting $x=z=0$ we have $$3(x-y)^2+(x-z)^2+(y+z)^2+x^2-2y^2+4z^2\geq2y^2\geq 0$$
Hence, the expression is never negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How do we prove this continued fraction for the quotient of gamma functions Given complex numbers $a=x+iy$, $b=m+in$ and a gamma function $\Gamma(z)$ with $x\gt0$ and $m\gt0$, it is conjectured that the following continued fraction holds
$$\frac{\displaystyle4\Gamma\left(\frac{2a+3}{4}\right)\Gamma\left(\frac{2b+3}{4}\right)}{\displaystyle\Gamma\left(\frac{2a+1}{4}\right)\Gamma\left(\frac{2b+1}{4}\right)}=\cfrac{(2a+1)(2b+1)}{a+b+2+\cfrac{(a-b+1)(b-a+1)} {a+b+4+\cfrac{(2a+3)(2b+3)}{a+b+6+\cfrac{(a-b+3)(b-a+3)}{a+b+8+\ddots}}}}\tag{1a}$$
Corollary
$$\frac{4}{\pi}=\cfrac{(2)^2}{3+\cfrac{(1)^2}{5+\cfrac{(4)^2}{7+\cfrac{(3)^2}{9+\ddots}}}}\tag{1b}$$
Q: How do we prove the continued fraction $(1a)$ rigorously?
| This can be deduced from Gauss' continued fraction
$$\frac{_2F_1(a+1,b;c+1;z)}{_2F_1(a,b;c;z)}=\cfrac{c}{c+\cfrac{(a-c)bz}{c+1+\cfrac{(b-c-1)(a+1)z}{c+2+\cfrac{(a-c-1)(b+1)z}{c+3+\cfrac{(b-c-2)(a+2)z}{c+4+\ddots}}}}}$$
which evaluates the continued fraction in $(1a)$ as
$$\frac{(2a+1)(2b+1)}{a+b+2}\frac{_2 F_1\left(b+\frac32,\frac{b-a+1}{2};\frac{a+b}{2}+2;-1\right)}{_2 F_1\left(b+\frac12,\frac{b-a+1}{2};\frac{a+b}{2}+1;-1\right)},$$
and Kummer's formula $_2 F_1(a,b;a-b+1;-1)=\dfrac{\Gamma(a/2+1)\Gamma(a-b+1)}{\Gamma(a+1)\Gamma(a/2-b+1)}$.
After all the necessary substitutions and cancellations, this yields exactly the expected result. The proofs are sketched in the linked articles, but you may want to follow further references for a deeper treatment.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Polynomial transformations and Vieta's formulas
Let $f(x)$ be a monic, cubic polynomial with $f(0)=-2$ and $f(1)=−5$. If the sum of all solutions to $f(x+1)=0$ and to $f\big(\frac1x\big)=0$ are the same, what is $f(2)$?
From $f(0)$ I got that $f(x)=x^3+ax^2+bx-2$ and from $f(1)=-5$ that $a+b = -4$ however I'm not sure how to use the info about the transformations to find $f(2).$ It seems that $(x+1)$ is a root for $f(x+1)$ and the same logic applies for $f\big(\frac1x\big)$?
Should I use Vieta's here or what's the appropriate way to go?
| When $f(x+1)=0$, we have,
$$(x+1)^3+a(x+1)^2+b(x+1)-2=0$$
Whose sum of roots can be obtained by "negative of coefficient of $x^2$ upon coefficient of $x^3$", which comes out to be $-\frac{3+a}{1}$.
Similarly, when $f(\frac 1x)=0$, we have,
$$\frac {1}{x^3}+a\frac{1}{x^2}+b\frac{1}{x}-2=0$$
Since $x\neq 0$, we can rearrange and get
$$-2x^3+bx^2+ax+1=0$$
Whose sum of roots is simply $\frac b2$. So,
$$\frac b2 = -3-a$$
Use that with the condition for $f(1)=5$ and solve for $a,b$. Then substitute $x=2$. Ta-da!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
how is this solved? $\lim_{x→0} \frac{1}{x^2}\left(\left(\tan(x+\frac{\pi}{4})\right)^{\frac{1}{x}}−e^2\right)$ I know how to solve the trigonometric part of the limit, i.e $\left(\tan(x+\frac{\pi}{4})\right)^{1/x}$, which I think is $e^2$, I do not however, know how to carry forward the question with that.
Also, I am sincerely sorry for the upsetting format, i have no clue how to use mathjax.
| We use Taylor series around $0$:
$$
\tan \left( {x + \frac{\pi }{4}} \right) = 1 + 2x + 2x^2 + \frac{8}{3}x^3 + \cdots .
$$
Hence, by using Taylor series for the logarithm and the exponential function,
\begin{align*}
\left( {\tan \left( {x + \frac{\pi }{4}} \right)} \right)^{\frac{1}{x}} & = \exp \left( {\frac{1}{x}\log \tan \left( {x + \frac{\pi }{4}} \right)} \right) \\ & = \exp \left( {\frac{1}{x}\log \left( {1 + 2x + 2x^2 + \frac{8}{3}x^3 + \cdots } \right)} \right)
\\ &
= \exp \left( {\frac{1}{x}\left( {2x + \frac{4}{3}x^3 + \cdots } \right)} \right) = e^2 \exp \left( {\frac{4}{3}x^2 + \cdots } \right) \\ & = e^2 \left( {1 + \frac{4}{3}x^2 + \cdots } \right).
\end{align*}
This yields the limit $\frac{4}{3}e^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3672397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $x^2+y^2-\sqrt{2}xy < 4$ provided $1\le x,y \le \frac{5}{2}$. If possible I would like an elegant solution to the following problem:
Let $x,y\in\mathbb{R}$ such that $1\le x \le \frac{5}{2}$ and $1\le y \le \frac{5}{2}$. Prove that
$x^2+y^2-\sqrt{2}xy < 4$
I'm aware that you can use Lagrange multipliers but I want an elementary solution using elementary inequalities that are available up to the high school level, or inequalities used in olympiads.
Remarks
I've substituted the obvious specific values for $x,y$ (we can assume without loss of generality that $x\le y$) below
\begin{array}{|c|c|c|}
\hline
x & y &x^2+y^2-\sqrt{2}xy \text{ (rounded to 3 d.p.)}\\ \hline
1 & 1 & \approx 0.587\\ \hline
1 & \frac{5}{2} & \approx 3.714\\ \hline
\frac{5}{2} & \frac{5}{2} & \approx 3.661\\ \hline
\end{array}
Some thoughts:
*
*I've tried solving it in different equivalent forms. Using the following identity
$x^4+y^4 = \left(x^2+y^2+\sqrt{2}xy \right) \left(x^2+y^2-\sqrt{2}xy \right)$
I attempted to prove $x^4+y^4 < 4 \left(x^2+y^2 + \sqrt{2}xy \right)$ (though I'm not sure if it really helps) but to no avail.
*Also, crudely approximating via $x^2+y^2 -\sqrt{2}xy < \left(\frac{5}{2}\right)^2 + \left(\frac{5}{2}\right)^2 - \sqrt{2} \cdot 1 \cdot 1 $ doesn't work since the RHS is clearly larger than 4
| Let $f(x,y)=x^2+y^2-\sqrt2xy.$
Thus, $f$ is a convex function of $x$ and $f$ is a convex function of $y$.
Id est, $$\max{f}=\max\{f(1,1),f(1,2,5),f(2.5,2.5)\}=f(1,2.5)=7.25-2.5\sqrt2<4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Formula for number of permutations that have $1$ flanked by an even integer on each side. Suppose we have the sequence $(1,2,3,...,n)$ permuted. If $n$ is an even integer greater than 3, find a formula for the number of permutations that have $1$ flanked by an even integer on each side.
I'm not sure how to start this, so I decided to look at the case when $n=4$. We would permute $(1,2,3,4)$ and these are the options we would have: $$\{2,1,4,3\}, \{4,1,2,3\}, \{3,2,1,4\}, \{3,4,1,2\}. $$ But I started listing out the permutations if $x=6$ and I'm not sure that's the smartest use of my time...
| Since $1$ must be flanked on each side by an even integer, it can't be at either end. Thus, $1$ must be in one of the $n - 2$ spots in between the end points. Then there are $\frac{n}{2}$ possible even integers that can go before it and, once that is chosen, there are $\frac{n}{2} - 1$ remaining even integers that can go after it. Finally, there are $n - 3$ spots available for the remaining $n - 3$ integers to go in any order, for $(n - 3)!$ possibilities. Thus, the total # of permutations would be
$$\begin{equation}\begin{aligned}
P_n & = (n - 2)\left(\frac{n}{2}\right)\left(\frac{n}{2} - 1\right)((n - 3)!) \\
& = \left(\frac{n}{2}\right)\left(\frac{n - 2}{2}\right)(n - 2)((n - 3)!) \\
& = \frac{n(n-2)((n - 2)!)}{4}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This gives $P_4 = \frac{4(2)(2!)}{4} = \frac{16}{4} = 4$, as you've already determined. Also, $P_6 = \frac{6(4)(4!)}{4} = \frac{576}{4} = 144$, which is a large enough value to help explain why trying to list them became quite tedious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3675284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim_{n\to\infty} \left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}$
Find the following limit without using the L'Hopital rule:
$$\lim_{n\to\infty} \left(\dfrac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}$$
Answer: $e^{-1}$
My attempt: Since the limit is of the form $1^{\infty}$, I decided to use the standard formula:
$$\lim_{x\to a} f^g = e^{\lim_\limits{x\to a}(f-1)g}$$
(See link)
Let $l=(f-1)g$.
We have,
$$l=\left(\dfrac{\sqrt{n^2+n}-(1+n)}{n}\right)(2\sqrt{n^2+n}-1)$$
This on solving boils down to
$$l=2n+3-\sqrt{1+\frac 1n}(2n+3)+\frac 1n$$
Now if I tend $n$ towards infinity, then $l\to 0$ and the limit i.e. $e^l$, is equal to $1$, which contradicts the given answer.
Please help. Thanks!
Edit:
A proof for the "standard formula" I have used.
Edit 2: just noticed a typo in the power which I have now fixed.
| $ \lim_{n \to \infty} l = -1 $, not $ 0 $. Indeed,
\begin{align}
2n + 3 - \sqrt{1 + \frac{1}{n}}(2n + 3) + \frac{1}{n}
&= \left(1 - \sqrt{1 + \frac{1}{n}}\right)(2n + 3) + \frac{1}{n} \\
&= \frac{(1 - \sqrt{1 + 1/n})(1 + \sqrt{1 + 1/n})}{1 + \sqrt{1 + 1/n}} \cdot (2n + 3) + \frac{1}{n} \\
&= \frac{-1/n}{1 + \sqrt{1 + 1/n}} \cdot (2n + 3) + \frac{1}{n} \\
&= - \frac{2 + 3/n}{1 + \sqrt{1 + 1/n}} + \frac{1}{n} \\
&\to -1
\end{align}
as $ n \to \infty $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that a power series satifies that $x^2y''-4xy'+(x^2+6)y=0$ Consider the power series
$$
\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}z^{2n+3}
$$
I have shown that f is differentiable and that
$$
f'(x) = \sum_{n=0}^\infty \frac{(-1)^n(2n+3)}{(2n+1)!}x^{2n+2}
$$
and
$$
f''(x) = \sum_{n=0}^\infty \frac{(-1)^n(2n+3)(2n+2)}{(2n+1)!}x^{2n+1}
$$
and are now asked to prove that $f$ is a solution to the differential equation
$$
x^2y''-4xy'+(x^2+6)y=0
$$
on the interval $]-\infty, \infty[$. I just really have a hard time starting these kind of questions. I have probably tried for 45 minuts and still can't make it work. This is some of what I have gotten. Let $S = x^2y''$. Then
\begin{align*}
S & = x^2 \sum_{n=0}^\infty \frac{(-1)^n(2n+3)(2n+2)}{(2n+1)!}x^{2n+1} \\
& = \sum_{n=0}^\infty \frac{(-1)^n(2n+3)(2n+2)}{(2n+1)!}x^{2n+3} \\
& = 6f(x) + \sum_{n=0}^\infty \frac{(-1)^n(4n^2+10n)}{(2n+1)!} x^{2n+3}
\end{align*}
and then I got stucked and tried some other things but still couldn't get any further in regards what I want to be shown. Do you mind helping me?
| It is plain that the power series is convergent for all $z$ because the factorial denominator grows much faster than any power of $z$. Convergent power series may be differentiated term by term.
It is then a question - as you have started - of working it through.
Let $S = z^2 f''(z) - 4zf'(z) + (z^2 + 6) f(z)$, so that
$$\begin{align}
S&=\sum_{n \geqslant 0} \frac{(-1)^n}{(2n+1)!} \Big\{ (2n+3)(2n+2) z^{2n+3} -4 (2n+3) z^{2n+3} + z^{2n+5} + 6z^{2n+3} \Big\} \\
&= \sum_{n\geqslant 0}\frac{(-1)^n}{(2n+1)!}\Big\{
(2n+1)(2n)z^{2n+3} + z^{2n+5}
\Big\}
\end{align}
$$
We can now shift the indexation of the $z^{2n+3}$ terms. First, peel off the $z^3$ term arising when $n=0$, and the re-index the remaining terms so that powers of $z$ are aligned, to get,
$$\begin{align}
S&= 0 \cdot z^{2n+3} + \sum_{n \geqslant 1} \frac{(-1)^n}{(2n+1)!}(2n+1)(2n) z^{2n+3} + \sum_{n\geqslant 0} \frac{(-1)^n}{(2n+1)!} z^{2n+5} \\
&= \sum_{n\geqslant 0}\Big\{ \frac{(-1)^{n+1}(2n+3)(2n+2)}{(2n+3)!} z^{2n+5} + \frac{(-1)^n}{(2n+1)!} z^{2n+5} \Big\} \\
&= \sum_{n\geqslant 0}\Big\{ \frac{(-1)^{n+1}}{(2n+1)!} z^{2n+5} + \frac{(-1)^n}{(2n+1)!} z^{2n+5} \Big\}
\end{align}
$$
Plainly the terms cancel and we find $S = 0$, which is what we required,
In fact, in case you have not noticed, this is the series for $z^2 \sin z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) = a + bi$, prove that $a^2 + b^2 = (4\cos x \cos\frac x2)^2$ Could anyone help me with this? I'm stuck.
If $$(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) = a + bi$$ prove that $$a^2 + b^2 = \left(4\cos x \cos\frac x2\right)^2$$
For reference, $\operatorname{cis}x = \cos x + i\sin x$.
I found that
$$a = 1 + \cos x + \cos2x + \cos3x \quad\text{and}\quad
b = \sin x + \sin2x + \sin3x$$
| Note
\begin{align}
a^2+b^2 &= |a + bi|^2 \\
&= |(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) |^2\\
&= |1 + e^{i x}|^2|1 + e^{i 2x}|^2\\
&= |e^{-\frac x2}+ e^{-\frac x2}|^2 |e^{-i x}+ e^{ ix}|^2\\
&= |2\cos\frac x2|^2 |2\cos x|^2\\
&= (4\cos x \cos\frac x2)^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proof of the binomial theorem by induction clarification This is an exercise from Spivak's "Calculus".
3.d. Prove the "binomial theorem": If $a$ and $b$ are any numbers and $n$ is a natural number, then
\begin{align} (a+b)^n &= a^n + {n \choose 1}a^{n-1}b + {n \choose 2}a^{n-2}b^2 + \dots + {n \choose n-1}ab^{n-1} + b^n \\ &= \sum_{j=0}^n {n \choose j} a ^{n-j}b^j \end{align}
The solution is given as
The binomial theorem is clear for $n=1$. Suppose that
$$(a+b)^n=\sum_{j=0}^n{n\choose j}a^{n-j}b^j$$
Then
\begin{align}
(a+b)^{n+1}&=(a+b)(a+b)^n=(a+b)\sum_{j=0}^n{n\choose j}a^{n-j}b^j\\
&=\sum_{j=0}^n{n\choose j}a^{n+1-j}b^j+\sum_{j=0}^n{n\choose j}a^{n-j}b^{j+1}\\
&=\sum_{j=0}^n{n\choose j}a^{n+1-j}b^j+\sum_{j=1}^{n+1}{n\choose j-1}a^{n+1-j}b^j\ \text{(replacing }j \text{ by } j-1 \text)\\
&=\sum_{j=0}^{n+1}{n+1\choose j}a^{n+1-j}b^j
\end{align}
My question is similar to the one asked here, but it is not about the change of indices. What were the steps to combine the two summations at the end?
| It is Pascal's rule:
$${n\choose j}+{n\choose j-1}={n+1\choose j}$$
So we have
\begin{align}
(a+b)^{n+1}&=(a+b)(a+b)^n\nonumber \\
&=(a+b)\sum_{j=0}^{n}{n\choose j}a^{n-j}b^j\nonumber \\
&=\sum_{j=0}^{n}{n\choose j}a^{n-j+1}b^j +\sum_{j=0}^{n}{n\choose j}a^{n-j}b^{j+1}\nonumber \\
&={n\choose 0}a^{n+1}+\sum_{j=1}^n{n\choose j}a^{n-j+1}b^j+\sum_{j=1}^{n+1}{n\choose j-1}a^{n-j+1}b^j\nonumber \\
&={n\choose 0}a^{n+1}+{n\choose n}b^{n+1}+\sum_{j=1}^{n}\left[{n\choose j}+{n\choose j-1}\right]a^{n-j+1}b^j\nonumber \\
&={n+1\choose 0}a^{n+1}+{n+1\choose n+1}b^{n+1}+\sum_{j=1}^n{n+1\choose j}a^{n-j+1}b^j\nonumber \\
&=\sum_{j=0}^{n+1}{n+1\choose j}a^{n+1-j}b^j\nonumber
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\int \:\frac{1}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}}dx$ using trig substitution Recently I came accross this integral :
$$ \int \:\frac{1}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}}dx $$
I would evaluate it like this, first start by the substitution:
$$ x=\cos(2u) $$
$$ dx=-2\sin(2u)du $$
Our integral now becomes:
$$\int \:\frac{-2\sin \left(2u\right)du}{\sqrt[3]{\left(\cos \left(2u\right)+1\right)^2\left(\cos \:\left(2u\right)-1\right)^4}}$$
$$\cos(2u)=\cos(u)^2-\sin(u)^2$$
Thus:
$$\cos(2u)+1=2\cos(u)^2$$
$$\cos(2u)-1=-2\sin(u)^2$$
Thus our integral now becomes:
$$\int \:\frac{-\sin \left(2u\right)du}{\sqrt[3]{4\cos \left(u\right)^416\sin \left(u\right)^8}}=\frac{1}{2}\int \:\frac{-\sin \left(2u\right)du}{\sqrt[3]{\cos \left(u\right)^4\sin \left(u\right)^8}}$$
we know:
$$\sin \left(u\right)=\cos \left(u\right)\tan \left(u\right)$$
Thus our integral becomes:
$$\int \:\frac{-\tan \left(u\right)\cos \left(u\right)^2du}{\cos \:\left(u\right)^4\sqrt[3]{\tan \left(u\right)^8}}=\int \frac{-\tan \:\left(u\right)\sec \left(u\right)^2du}{\sqrt[3]{\tan \:\left(u\right)^8}}\:$$
By letting $$v=\tan \:\left(u\right)$$
$$dv=\sec \left(u\right)^2du$$
Our integral now becomes:
$$\int -v\:^{1-\frac{8}{3}}dv=-\frac{v^{2-\frac{8}{3}}}{2-\frac{8}{3}}+C=\frac{3}{2\sqrt[3]{v^2}}+C$$
Undoing all our substitutions:
$$\frac{3}{2\sqrt[3]{\tan \left(u\right)^2}}+C$$
$$\tan \:\left(u\right)^2=\frac{1}{\cos \left(u\right)^2}-1=\frac{2}{1+\cos \left(2u\right)}-1=\frac{2}{1+x}-1$$
Our integral therefore:
$$\frac{3}{2\sqrt[3]{\frac{2}{1+x}-1}}+C$$
However, online integral give me anti-derivative of $$\frac{-3\sqrt[3]{\frac{2}{x-1}+1}}{2}+C$$ so I want to know where I went wrong
| Your result is correct, which can be obtained alternatively by substituting $t= \frac{1+x}{1-x}$ to arrive at
$$ \int \:\frac{dx}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}}
=\frac12\int t^{-2/3}dt = \frac32 t^{1/3}+C=\frac32 \sqrt[3]\frac{1+x}{1-x} +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Technique for simplifying, e.g. $\sqrt{ 8 - 4\sqrt{3}}$ to $\sqrt{6} - \sqrt{2}$ How to find the square root of an irrational expression, to simplify that root. e.g.:
$$
\sqrt{ 8 - 4\sqrt{3} } = \sqrt{6} - \sqrt{2}
$$
Easy to verify:
\begin{align}
(\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} +2
= 8 - 4 \sqrt{3}
\end{align}
But how to work it out in the first place? I feel there's a standard technique (Completing-the-square? Quadratic formula?), but don't recall it or what it's called...
BTW: this came up in verifying equivalence of different calculations of $\cos{75°}$ (the above divided by $4$), as $\cos{\frac{90°+60°}{2}}$ vs $\cos{(45°+30°)}$, from 3Blue1Brown's lockdown video on complex numbers and trigonometry.
| I'm assuming you're looking for a formula to de-nest $\sqrt{a+b\sqrt{c}}$.
To derive such a formula, equate this to $x+y\sqrt{c}$.
$a+b\sqrt{c}=(x+y\sqrt{c})^2=x^2+cy^2+2xy\sqrt{c}$
$a+b\sqrt{c}=x^2+cy^2+2xy\sqrt{c}$
On matching up coefficients, we get:
$
\begin{cases}
x^2+cy^2=a\\
2xy=b
\end{cases}
$
Substitute $y=\frac{b}{2x}$:
$\rightarrow x^2+c(\frac{b^2}{4x^2})=a$
$u:=x^2\rightarrow u+\frac{cb^2}{4u}=a$
$\rightarrow u^2-au+\frac{b^2c}{4}=0$
$\rightarrow u=\frac{a\pm\sqrt{a^2-b^2c}}{2}$
$\because u=x^2\rightarrow x=\sqrt{\frac{a\pm\sqrt{a^2-b^2c}}{2}}$
Now solve for $y$:
$y=\frac{b}{2}(\sqrt{\frac{2}{a\pm\sqrt{a^2-b^2c}}})$
And you're done! This tells us that:
$$\sqrt{a+b\sqrt{c}}=\sqrt{\frac{a+\sqrt{a^2-b^2c}}{2}}+(\frac{b}{2}(\sqrt{\frac{2}{a+\sqrt{a^2-b^2c}}}))\sqrt{c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Why does Stolz- Cesaro fail to evaluate the limit of $\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$, I need to find the limit of the sequence
$\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$,
My strategy is to use Stolz's Cesaro theorem for this sequence.
Now, the numerator is given by :
$x_r = n^1+ n^2 +n^3 + \ldots +n^r$, so $x_{n+1} - x_{n} = n^{n+1}$
Similarly for denominator
$y_r = 1^n + 2^n + 3^n +\ldots +r^n$, so $y_{n+1}- y_{n} = (n+1)^n$
Using Stolz Cesaro, this limit is equivalent to
$\displaystyle \lim \dfrac{n^{n+1}}{(n + 1)^n}$, which diverges to $ +\infty$,
However ans given to me is $\dfrac{e-1}{e}$, Can anyone tell where is the error in my solution ?
Thanks.
| If you approximate the sums with the corresponding integrals it converges to 0
$$
L = \frac{\int_{1}^{n}n^x dx}{\int_{1}^{n} x^n dx} = \frac{n^{n+1} + n^n }{n^{n+1}\log n - \log n} = 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
For angles $A$ and $B$ in a triangle, is $\cos\frac B2-\cos \frac A2=\cos B-\cos A$ enough to conclude that $A=B$? Brief enquiry:
$$\cos\frac B2-\cos \frac A2=\cos B-\cos A$$
Optionally
$$\sqrt\frac{1+\cos B}{2}-\cos B=\sqrt\frac{1+\cos A}{2}-\cos A$$
Is above equality sufficient to prove that it implies $A=B$?
Detailed explanation and motivation for this question:
Consider a triangle with bisectors of equal length:
By definition:
$\lvert AE\rvert = \lvert BD\rvert = D\\\frac A2+\frac A2 = A,\space\space\space\frac B2 + \frac B2 = B$
By cosine law:
$$\lvert AE\rvert^2=y^2+a^2-2ya\cos B\\\lvert BD\rvert^2=x^2+a^2-2xa\cos A\\x^2=\lvert BD\rvert^2+a^2-2\lvert BD\rvert a\cos \frac B2\\y^2=\lvert AE\rvert^2+a^2-2\lvert AE\rvert a\cos \frac A2$$
By sine law:
$$\frac{x}{D}=\frac{\sin \frac B2}{\sin A};\space\space\space \frac{y}{D}=\frac{\sin \frac A2}{\sin B}$$
$$\bigl[D=\lvert AE\rvert = \lvert BD\rvert\bigr]$$
Since bisectors are equal:
$$x^2-2xa\cos A=y^2-2ya\cos B\implies \lvert BD\rvert \cos \frac B2 +x\cos A = \lvert AE\rvert \cos \frac A2 +y\cos B $$
$$D (\cos\frac B2-\cos \frac A2)=y\cos B-x\cos A $$
Dividing by D and substituting y and x we obtain:
$$\cos\frac B2-\cos \frac A2=\frac{\sin \frac A2\cos B}{\sin B}-\frac{\sin \frac B2\cos A}{\sin A}$$
Consider triangles $\Delta$ABE and $\Delta$BAD
Area of triangle $\Delta$ABE:
$$A = \frac{aD}{2}\sin\frac A2 = \frac{aD}{2}\sin B \implies \sin\frac A2 = \sin B$$
Similarly for triangle $\Delta$BAD
$$A=\frac{aD}{2}\sin\frac B2 = \frac{aD}{2}\sin A\implies \sin\frac B2 = \sin A$$
Therefore:
$$\cos\frac B2-\cos \frac A2=\cos B-\cos A$$
Optionally
$$\sqrt\frac{1+\cos B}{2}-\cos B=\sqrt\frac{1+\cos A}{2}-\cos A$$
Is above equality sufficient to prove that it implies A = B?
| After more thinking, I realized that the answer is in fact "Yes".
As before, let $a=\cos\frac{A}{2},b=\cos\frac{B}{2}, 0<a,b<1$
The identity becomes:
$2a^2-a=2b^2-b$
which can be rewritten as:
$(a-b)\left(a+b-\frac{1}{2}\right)=0$
So either $a=b$, which gives $A=B$, as $\cos$ is $1:1$ on $\left[0,\frac{\pi}{2}\right]$
Or $a+b=\frac{1}{2}$
But the latter is impossible, because $A+B<\pi$, so at least one of the angles $A$ and $B$ has to be acute, which gives:
$a=\cos\frac{A}{2}>\cos\frac{\pi}{4}>\frac{1}{2}$
The whole point is that we look at the interior bisectors only. But if we consider the exterior bisectors as well, it is possible to find $A\ne B$ satisfying $\cos\frac{A}{2}+\cos\frac{B}{2}=\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Parabola transformation Find the real affine change of coordinates that maps the parabola in the $xy$-plane to the parabola in the $uv$-plane
$$4x^2 + 4xy + y^2 - y + 1 = 0$$
$$4u^2 + v = 0$$
My attempt:
Since there is an $xy$ term, we know that there is a rotation. Thus suppose there is a $x'y'$ coordinate system before the rotation. We know that $x = x' \cos \theta - y' \sin \theta$ and $y = x' \cos \theta - y' \sin \theta$. If we make this substitution and rewrite the expression as $A'x'^2 + B; x'y'+ C'y'^2 + D'x' + E'y' + F'$, we want $x'y' = 0$. If we solve, we find that $\tan(2 \theta) = \frac{B}{A- C}$. In this case $\tan(2 \theta) = \frac{4}{3}$. Now when we solve this equation, we have $|\sin \theta| = \frac{1}{\sqrt{5}}$ and $|\cos \theta| = \frac{2}{\sqrt{5}}$.
In order to finish this solution, I need help with the following questions
*
*How do I determine sign for the angles? Is there a way to do this other than graphing?
*Once I substitute the angles and find an equation in terms of $x'$ and $y'$ which yields a parabola that is not rotated, I can see how to write $x'$ and $y'$ in terms of $u$ and $v$. Then how how exactly do I account for the rotation when I write $x$ and $y$ in terms of $u$ and $v$? I'm having a difficult time geometrically understanding what has to be done to the parabola coordinates.
| To obtain affine change of coordinates simply observe that
$4x^2+4xy+y^2-y+1=(2x+y)^2+(1-y)=4\left(x+\frac{y}{2}\right)^2+(1-y)$. So take \begin{align*}u&=x+\frac{y}{2}\\v&=1-y\end{align*}
This gives us
$$\begin{bmatrix}u\\v\end{bmatrix}=\begin{bmatrix}1&\frac{1}{2}\\0&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}0\\1\end{bmatrix}.$$
From here we can also get
$$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&\frac{1}{2}\\0&-1\end{bmatrix}\begin{bmatrix}u\\v\end{bmatrix}+\begin{bmatrix}\frac{-1}{2}\\1\end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inequality involving AM-GM but its wierd Let a, b, c be positive real numbers. Prove that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+ \frac{3\cdot\sqrt[3]{abc}}{a+b+c} \geq 4$
Ohk now i know using AM-GM that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 3 \cdot \sqrt[3] {\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}} = 3 \cdot 1=3 $
Now if i could have shown that the other term $\geq 1$. I would be done.
But the problem is that (again using AM-GM)
$\sqrt[3]{abc} \leq \frac{a+b+c}{3}
\Rightarrow 3 \cdot \sqrt[3]{abc} \leq a+b+c \Rightarrow \frac{3 \cdot \sqrt[3]{abc}}{a+b+c} \leq 1$
So if first part is $\geq 3$ and second part is $\leq 1$, How will i show that it is greater than $4$?
Is my approach correct? Or is there something wrong with the question?
Thanks.
(Source: https://web.williams.edu/Mathematics/sjmiller/public_html/161/articles/Riasat_BasicsOlympiadInequalities.pdf ,pg-$13$ exercise $1.3.4.a$ )
| From
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant \frac{(a+b+c)^2}{ab+bc+ca},$$
and $ \sqrt[3]{abc} \geqslant \frac{3abc}{ab+bc+ca},$ we need to prove
$$\frac{(a+b+c)^2}{ab+bc+ca} + \frac{9abc}{(ab+bc+ca)(a+b+c)} \geqslant 4,$$
equivalent to
$$ (a+b+c)^2 + \frac{9abc}{a+b+c} \geqslant 4(ab+bc+ca),$$
or
$$a^2+b^2+c^2 + \frac{9abc}{a+b+c} \geqslant 2(ab+bc+ca).$$
Which is Schur is inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What's the sum function of $\sum_{n=1}^{\infty} \frac{x^n(x-1)}{n(x^{2n+1}-1)}$? Notice that
\begin{align*}
\frac{x^n(x-1)}{n(x^{2n+1}-1)}&=\frac{1}{n}\cdot\frac{x^n}{\sum_{k=0}^{2n}x^k}=\frac{1}{n\sum_{k=0}^{2n}x^{k-n}}=\frac{1}{n\sum_{k=-n}^{n}x^{k}}
\end{align*}
This will help?
| What about
\begin{align*}
(1-x)\sum_{n=1}^\infty \frac{1}{n} \frac{x^n}{1-x^{2n+1}} = (1-x) \sum_{m=0}^\infty \sum_{n=1}^\infty \frac{1}{n} x^{n + (2n +1)m} = (x-1) \sum_{m=0}^\infty x^m \ln(1-x^{2m+1}),
\end{align*}
where we used the series representation of the logarithm.
But I am not sure, if it is possible to give a closed formula for the last series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$ For $a,b,c>0; abc=1.$ Prove$:$ $$(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$$
My proof by SOS is ugly and hard if without computer$:$
$$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{3}-9\,abc \left( {a}^{3}+{b} ^{3}+{c}^{3} \right)$$
$$=\frac{1}{8}\, \left( b-c \right) ^{6}+{\frac {117\, \left( b+c \right) ^{4} \left( b+c-2\,a \right) ^{2}}{1024}}+{\frac {3\,{a}^{2} \left( 40\,{a }^{2}+7\,{b}^{2}+14\,bc+7\,{c}^{2} \right) \left( b-c \right) ^{2}}{ 32}}$$
$$+{\frac {3\, \left( b+c \right) ^{2} \left( 3\,a-2\,b-2\,c \right) ^{2} \left( b-c \right) ^{2}}{32}}+\frac{3}{16}\, \left( a+2\,b+2\,c \right) \left( 4\,a+b+c \right) \left( b-c \right) ^{4}$$
$$+{\frac { \left( 16\,{a}^{2}+24\,ab+24\,ac+11\,{b}^{2}+22\,bc+11\,{c}^{ 2} \right) \left( 4\,a-b-c \right) ^{2} \left( b+c-2\,a \right) ^{2} }{1024}} \geqq 0$$
I think$,$ $uvw$ is the best way here but it's not concordant for student in The Secondary School.
Also$,$ BW helps here, but not is nice, I think.
So I wanna nice solution for it! Thanks for a real lot!
| By applying Vacs's ineq we obtain $$(a^2+b^2+c^2)^3\ge 3(ab^3+bc^3+ca^3)(a^2+b^2+c^2)$$
So it's suffice to prove $$(ab^3+bc^3+ca^3)(a^2+b^2+c^2)\ge 3abc(a^3+b^3+c^3)$$
$$\Leftrightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge \frac{3(a^3+b^3+c^3)}{a^2+b^2+c^2}$$
$$\Leftrightarrow (a-b)^2\left(\frac{a^2+c^2}{b}-a\right)+(b-c)^2\left(\frac{a^2+b^2}{c}-b\right)+(c-a)^2\left(\frac{b^2+c^2}{a}-c\right)$$
$$\Leftrightarrow S=S_c(a-b)^2+S_a(b-c)^2+S_b(c-a)^2\ge 0$$
Assume $b=\text{mid} \{a,b,c\}$ we consider 2 cases
Case 1: $a\ge b\ge c\implies S_a,S_c\ge 0$ and $S_a+2S_b, \,\ S_c+2S_b\ge 0\implies S\ge 0$
Case 2: $c\ge b\ge a\implies S_b,\,\ S_c,\,\ S_a+S_b\ge 0\implies S\ge 0$
So we are done
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Limitations of Arithmetic-Geometric Inequality? I have to find the range of function $f(x) = x+(1/x) +1$, where $x$ is positive. Now I did it with two ways which we can see below, in equations $(1)$ and $(2)$, by using the AM-GM inequality.
$$\frac 1 3 \left( x + \frac 1 x + 1 \right) \ge \sqrt[3]{x \cdot \frac 1 x \cdot 1} \implies x + \frac 1 x + 1 \ge 3 \tag 1$$
$$\frac 1 4 \left( x + \frac 1 x + \frac 1 2 + \frac 1 2 \right) \ge \sqrt[4]{x \cdot \frac 1 x \cdot \frac 1 2 \cdot \frac 1 2} \implies x + \frac 1 x + 1 \ge 4 \cdot (1/4)^4 \tag 2$$
So now after seeing these my question is why is my answer different in each case? Are there any limitations of this inequality?
| look, this isn't a "limitation" of the AM-GM inequality. for example:
$$\frac{1+2}{2} \leq \sqrt{2}$$
and
$$\frac{1+1+1}{3} \leq \sqrt[3]{1}$$
as you can see in the first example the geometric mean isn't certain to be the biggest the amount could be, in your example, the minimum is the first one, but the second one still holds, given it's smaller.
$$(x + \frac{1}{x}+1)'=0$$
$$1-\frac{1}{x^2}=0$$
$$1=\frac{1}{x^2}$$
$$x=1, x=-1 \text{(x is always positive)}$$
$$1+\frac{1}{1}+1=3$$
thus, the minimum is 3.
AM-GM is an inequality, but is in not certain that it is the minimal(or maximal) amount.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Tried to apply the ratio test to determine the convergence interval, but get the limit as a constant. So here is the series: $\Sigma_{n=1}^{\infty} \frac{x^{2n}}{1+x^{4n}}$
$$\left| \frac{x^{2(n+1)} (1+x^{4n})}{x^{2n}(1+x^{4(n+1)})}\right| = \left|\frac{x(1+x^{4n})}{1+x^{4n+4}} \right| \overset{\text{ n } \rightarrow \infty }{\rightarrow}\left[\frac{\infty}{\infty} \right]= \frac{x+x^{4n+1}}{1+x^{4n+4}} = \lim_{n \to \infty} \frac{x}{1+x^{4n+4}} +\lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}} = \\ 0 + \lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}} = \frac{\infty}{\infty} =\left[\text{ apply L'Hospital rule (4n+1)th times } \right] = \frac{1}{(4n+4)(4n+3)(4n+2)x^3} = 0 $$
Where is my mistake? Maybe some other technique would be useful for these series?
$$\lim_{n \to \infty} \frac{x}{1+x^{4n+4}} +\lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}} \\ \text{ if } x = 0 \text{ then limit equals } 0 \\ \text{if x=1 then limit equals to 1 } \\ \text{if } 0<x<1 \text{ then limit is equal to x [ the same goes for } -1<x<0 \\ \text{ if } x> 1 \text{ then limit is equal to} \frac{1}{x^3} \text{ the same goes for x<1} $$
| Let's proceed from your work. Corrections are highlighted in red. We have
$$h(x) = \left| \frac{x^{2(n+1)} (1+x^{4n})}{x^{2n}(1+x^{4(n+1)})}\right| = \left|\frac{x^{\color{red}{2}}(1+x^{4n})}{1+x^{4n+4}} \right|.$$ At this point, we need to consider separate cases. The reason is becuase for $n \ge 1$, the behavior of $|x^n|$ depends on whether $|x| < 1$, $|x| = 1$, or $|x| > 1$.
Case $|x| = 1$: Then $x^2 = x^4 = 1$ and we have $$h(x) = \frac{1(1+1)}{(1+1)} = 1.$$ Going back to the original sum, $$f_n(x) = \frac{x^{2n}}{1+x^{4n}}$$ implies $$f_n(-1) = f_n(1) = \frac{1}{2} > 0$$ so the sum diverges.
Case $|x| > 1$: Then $$\lim_{n \to \infty} 1/x^n = 0$$ and we have, after dividing numerator and denominator by $x^{4n+4}$, $$h(x) = \left|\frac{1/x^{4n+2} + 1/x^2}{1/x^{4n+4} + 1}\right|.$$ Hence $$\lim_{n \to \infty} h(x) = \left|\frac{0 + 1/x^2}{0 + 1} \right| = \frac{1}{x^2} < 1$$ and $\sum f_n(x)$ converges.
Case $|x| < 1$: Then $x^n \to 0$ as $n \to \infty$, hence $$\lim_{n \to \infty} h(x) = \left|\frac{x^2(1 + 0)}{1 + 0}\right| = x^2 < 1$$ and again the sum converges.
Of course, this is not the simplest nor most elegant solution, but it is conforming to how one would apply the ratio test to determine the convergence of the sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$p^2+1=q^2+r^2$. Strange phenomenon of primes Problem:
Find prime solutions to the equation
$p^2+1=q^2+r^2$
I welcome you to post your own solutions as well
I have found a strange solution which I can't understand why it works(or what's the math behind it.) Here it is through examples
Put $r=17$(prime)
Now $17^2-1=16\times 18=288=2 \times 144$ (a particular factorization)
$\frac {2+144}{2}=73$
$\frac {144-2}{2}=71$
Solution pair $(p,q,r)=(73,71,17)$
Put $r=23$,
$23^2-1=22\times 24=8\times 66$
$\frac {8+66}{2}=37$
$\frac {66-8}{2}=29$
Solution pair $(37,29,23)$
It works for each prime except for $2,3,5$ Which generate
$(2,2,1),(3,3,1),5,5,1)$
Please explain me how it's working
| Let $r=193$, a prime. Then $$r^2-1=9313^2-9311^2=3107^2-3101^2=323^2-259^2$$ are the only expressions of $r^2-1$ as a difference of two squares, but $9313=67\times139$, and $3101$ and $259$ are both multiples of $7$. Thus, $p^2+1=q^2+r^2$ is impossible in primes $p,q$ for this prime value of $r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find solution set of $\dfrac{8^x+27^x}{12^x+18^x}=\dfrac{14}{12}$ What I've done is factoring it.
$$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}=\dfrac{7}{2\cdot 3}$$
This looks like it can be factored more but it doesn't work from my attempts.
| Well you've reached a good point:
$$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}$$
$$=\dfrac{2^{3x}+3^{3x}}{2^{x}\cdot 3^{x}(2^x+3^{x})}$$
We can let $2^x=a$ and $3^x=b$ and get
$$\frac{a^3+b^3}{ab(a+b)}=\frac{a^2-ab+b^2}{ab}=\frac{a}{b}-1+\frac{b}{a}$$
Now let $z=\frac{a}{b}$, we get
$$z-1+\frac{1}{z}=\frac{7}{6} \iff z^2-\frac{13}{6}z+1=0 \implies z=\frac{2}{3},\frac{3}{2} \implies x=\pm 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Multiplying $P(x) = (x-1)(x-2) \dots (x-50)$ and $Q(x)=(x+1)(x+2) \cdots(x+50)$
Let $P(x) = (x-1)(x-2) \dots (x-50)$ and $Q(x)=(x+1)(x+2) \cdots(x+50).$
If $P(x)Q(x) = a_{100}x^{100} + a_{99}x^{99} + \dots + a_{1}x^{1} + a_0$, compute $a_{100} - a_{99} - a_{98} - a_{97}.$
I've been quite stuck with this one. If I multiply and group the polynomials with the similar terms e.g $(x-1)$ and $(x+1) ...$ I would get
$P(x)Q(x)= (x^2-1)(x^2-2^2) \dots(x^2-50^2)$ right? From here on if I would keep multiplying the terms wouldn't I end up with $50$ times the term $x^2$ as the initial term, hence $a_{100} = 1$ since it wouldn't have any coefficient? For the other terms I don't have a clue how to find them so any hints would be appreciated
| Clearly $a_{99}=a_{97}=0$, since only even degree terms appear. As you've noticed, $a_{100}=1$; so really we just need to find $a_{98}$. But this is just negative $1$ times the sum of squares up to $50$, i.e. $-\sum_{k=1}^{50} k^2 =- \frac{50\cdot 51\cdot 101}{6}=-42925$. So the answer is $42926$.
Edit: it seems I was a bit sparse with my explanation. Claim: the coefficient of $x^{2n-2}$ in $\prod_{i=1}^{n}(x^2-i^2)$ is $\sum_{i=1}^{n}i^2 = \frac{n(n+1)(2n+1)}{6}$. The claim is immediate when $n=1$ or $n=2$:
$$
(n=1)\qquad \prod_{i=1}^{1}(x^2-i^2)= x^2-1;\, 1 = \frac{1\cdot 2 \cdot 3}{6}
$$
$$
(n=2)\qquad \prod_{i=1}^{2}(x^2-i^2)= x^4-5x^2+1;\, 5 = \frac{2\cdot 3 \cdot 5}{6}
$$Suppose the result is true for some $k$. Then
$$
\prod_{i=1}^{k+1 }(x^2-i^2) = \left(\prod_{i=1}^{k }(x^2-i^2)\right)\cdot(x^2-(k+1)^2)
$$
$$
= \left(x^{2k}-\sum_{i=1}^{k}i^2 \cdot x^{2k-2} +\cdots\right)\cdot(x^2-(k+1)^2)
$$
$$
= \left(x^{2k+2}-((k+1)^2+\sum_{i=1}^{k}i^2) \cdot x^{2k} +\cdots\right),
$$as was to be shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Find the number of functions which satisfy given conditions If f :{1,2,3,4} → {1,2,3,4} and y = f(x) be a function defined such that |f(α) – α| ≤ 1, for α ∈ {1,2,3,4} and m be the number of such functions, then m/5 is equal
| Okay.
$|f(a) - a| \le 1$ so $-1 \le f(a)-a < 1$ so $a-1 \le f(a) \le a+1$.
so $f(a)4$ may be one of three values $f(a)$ could be $a-1$ if $a-1$ is in range; that is if $a-1\ge 1$ or $a \ge 2$.
Or $f(a)$ could be $a$.
Or $f(a)$ could be $a+1$ if $a+1$ is in range; that is if $a+1 \le 4$ of $a \le 3$.
So
For $a = 1$ there are $2$ things $f(1)$ could be. $f(1)=1$ is a possibility and $f(1) =2$ are a possiblity.
Of $ a= 2,3$ there are three things $f(a)$ could be. $a-1, a, a+1$ are the three possibilities.
And for $ a=4$ there are $2$ things for $f(4)$ to be.
So there there are $2$ things $f(1)$ can be and $3$ things $f(2)$ can be and $3$ things $f(3)$ can be and $2$ things $f(4)$ can be.
So how many combinations of ways can $(f(1),f(2),f(3),f(4))$ can be?
$2\times 3\times 3 \times 2= 36$ possible combinations of value so $f(1),f(2),f(3),f(4)$ so $36$ possible functions.
Um..... why does anyone care what $\frac m5$ is?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\int_0^\infty \frac{1}{1+x^4}dx$ using the Residue Theorem I'm trying to evaluate the integral
$$\int_0^\infty \frac{1}{1+x^4}dx $$
using the Residue Theorem.
My approach:
Let's consider
$$\oint_\Gamma f$$
with $f(z)=\frac{1}{1+z^4}$ and $\Gamma = \Gamma_1 + \Gamma_2$, where:
*
*$\Gamma_1:[-R,R]\rightarrow \mathbb{C}$, with $\Gamma_1(t)=t$
*$\Gamma_2:[0, \pi] \rightarrow \mathbb{C}$, with $\Gamma_2(t)=Re^{it}$
So basically $\Gamma$ is the semicircle centers in the origin with imaginary part greater or equal to zero.
First we need to find the isolated singularities $\alpha_i$ of the function $f$. This singularities are the solution of the equation $1 + z^4 = 0$:
Let's call:
*
*$\alpha_1 = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$
*$\alpha_2 = -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$
*$\alpha_3 = -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$
*$\alpha_4 = \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$
So now we have:
$$\oint_\Gamma f = 2 \pi i \sum_i \text{Res}(f,\alpha_i) \text{Ind}_\Gamma(\alpha_i)$$
All singularities are poles of order one, with:
$$\text{Res}(f,\alpha_i)=\frac{1}{1+4\alpha_i^3}$$
So we end up with:
$$\oint_\Gamma f = 2 \pi i \sum_i \frac{\text{Ind}_\Gamma(\alpha_i)}{1+4\alpha_i^3} $$
Because of the shape of our curve $\Gamma$ we have that $\text{Ind}_\Gamma(\alpha_3)=\text{Ind}_\Gamma(\alpha_4)=0$
So we end up with:
$$\oint_\Gamma f = 2 \pi i \underbrace{\left( \frac{1}{1+4\alpha_1^3} + \frac{1}{1+4\alpha_2^3} \right)}_{:=\xi} $$
Now we can work on the left side of this expression:
$$\int_{\Gamma_1} f + \int_{\Gamma_2} f = 2 \pi i \xi$$
We have that:
$$\int_{\Gamma_1}f = \int_{-R}^R \frac{1}{1 + t^4} dt$$
And we also know that
$$\begin{align}
\int_{\Gamma_2}f &\leq \int_0^\pi \left|\frac{Rie^{it}}{1 + R^4e^{4it}} \right| dt
\\
\\ &= \int_0^\pi \frac{R}{\left|1 + R^4e^{4it}\right|} dt
\\
\\ &= \int_0^\pi \frac{1}{\left| \frac{1}{R} + R^3e^{4it}\right|} dt
\end{align}$$
If we let $R \to \infty$ we have that $\int_{\Gamma_1} f = \int_{-\infty}^\infty \frac{dt}{1 + t^4}$ and $\int_{\Gamma_2} f = 0$ and because $\int_{-\infty}^\infty \frac{dt}{1 + t^4} = 2 \int_{0}^\infty \frac{dt}{1 + t^4}$, we end up with:
$$\int_{0}^\infty \frac{dt}{1 + t^4} = \pi i \xi$$
The thing is that $\pi i \xi$ is a complex number, so what did I do wrong?
| The poles lying inside the contour are $\alpha_1=i\pi/4, \alpha_2=i3\pi/4$.
$Res[f(z),\alpha_1]=\frac{1}{4z^3}\rvert_{\alpha_1}=\frac{1}{4}e^{-3i\pi/4}$
$Res[f(z),\alpha_2]=\frac{1}{4z^3}\rvert_{\alpha_2}=\frac{1}{4}e^{-9i\pi/4}$
$ \text{After using the estimation principle to squeeze the integral on upper part to zero}$
$\therefore I=2\pi i\times \left [\frac{1}{4}e^{-3i\pi/4}+\frac{1}{4}e^{-9i\pi/4}\right ]$
$=\pi i/2\times \left (\frac{-1}{\sqrt 2} -i\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}-i\frac{1}{\sqrt 2}\right )$
$=\frac{\pi }{\sqrt 2}$
Now you can use the symmetry of integrand function to change the limits to $(0,\infty)$ so that the value of required integral is $\frac{\pi}{2\sqrt 2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to Prove $\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$ Question:- Prove that
$\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$
On factoring the denominator we get,
$\int_{0}^{\infty}\frac {1}{(x^4+x^2+1)(x^4-x^2+1)}dx$
Partial fraction of the integrand contains big terms with their long integral.So i didn't proceed with partial fraction.I'm unable figure out any other method.I think that there might be some other method for evaluation of this definite integral since its value is
$\frac{π}{2\sqrt{3}}$.
Does anyone have nice way to solve it!
| Let $(a,b,c,d)$ to be the complex roots of $x^8+x^4+1=0$. So, after partial fraction decomposition, the integrand write
$$\frac{1}{(a-b) (a-c) (a-d) \left(x^2-a\right)}+\frac{1}{(b-a) (b-c) (b-d) \left(x^2-b\right)}+$$
$$\frac{1}{(c-a) (c-b) (c-d) \left(x^2-c\right)}+\frac{1}{(d-a) (d-b) (d-c) \left(x^2-d\right)}$$
This makes the definite integral to be
$$\frac \pi 2\frac{ \left(\sqrt{a} \left(\sqrt{b}+\sqrt{c}+\sqrt{d}\right)^2+a
\left(\sqrt{b}+\sqrt{c}+\sqrt{d}\right)+\left(\sqrt{b}+\sqrt{c}\right)
\left(\sqrt{b}+\sqrt{d}\right) \left(\sqrt{c}+\sqrt{d}\right)\right)}{ \sqrt{a}
\sqrt{b} \sqrt{c} \sqrt{d} \left(\sqrt{a}+\sqrt{b}\right)
\left(\sqrt{a}+\sqrt{c}\right) \left(\sqrt{a}+\sqrt{d}\right)
\left(\sqrt{b}+\sqrt{c}\right) \left(\sqrt{b}+\sqrt{d}\right)
\left(\sqrt{c}+\sqrt{d}\right)}$$ Now, using
$$a=\frac{1+i \sqrt{3}}{2}\qquad b=\frac{1-i \sqrt{3}}{2}\qquad c=-\frac{1+i \sqrt{3}}{2}\qquad d=-\frac{1-i \sqrt{3}}{2}$$ leads to the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluating the indefinite Harmonic number integral $\int \frac{1-t^n}{1-t} dt$ It is well-known that we can represent a Harmonic number as the following integral:
$$H_n = \int_0^1 \frac{1-t^n}{1-t} dt$$
The derivation of this integral doesn't need you to derive the indefinite integral first, so now I'm wondering what the indefinite integral is and how one can derive it. According to WolframAlpha the indefinite integral is:
$$\int \frac{1-t^n}{1-t} dt = \frac{t^{n+1}{}_2F_1(1,n+1;n+2;t)}{n+1} - \ln(1-t) + C$$
where ${}_2F_1(a,b;c;z)$ is a Hypergeomtric function. I understand why $-\ln(1-t)$ is at the end, that's the result of splitting up the integrand, but I don't understand how a Hypergeometric function ends up there.
| The integrand is (in the complex field) a meromorphic function, with one pole coinciding with one of the zeros, and therefore removable,
leaving a polynomial defined over the whole complex field. So it is its integral.
A) polynomial form
Let's define
$$
\eqalign{
& I_{\,n} (x,a) = \int_{t\, = \,a}^{\;x} {{{1 - t^{\,n} } \over {1 - t}}dt} = - \int_{t\, = \,a}^{\;x} {{{1 - \left( {1 - u} \right)^{\,n} } \over u}du} = \cr
& = \int_{u\, = \,1 - a}^{\;1 - x} {{{\left( {1 - u} \right)^{\,n} - 1} \over u}du} = \int_{u\, = \,0}^{\;1 - x} {{{\left( {1 - u} \right)^{\,n} - 1} \over u}du} - \int_{u\, = \,0}^{\;1 - a} {{{\left( {1 - u} \right)^{\,n} - 1} \over u}du} = \cr
& = J_{\,n} (1 - x) - J_{\,n} (1 - a) \cr}
$$
so that the integration constant is inglobated in $a$.
Concerning $J_{\,n} (x) =$ we have
$$
\eqalign{
& J_{\,n} (x) = \int_{u\, = \,0}^{\;x} {{{\left( {1 - u} \right)^{\,n} - 1} \over u}du} = \cr
& = \int_{u\, = \,0}^{\;x} {\sum\limits_{1\, \le \,k\, \le \,n} {\left( { - 1} \right)^{\,k} \left( \matrix{
n \cr
k \cr} \right)u^{\,k - 1} } du} = \int_{u\, = \,0}^{\;x} {\sum\limits_{0\, \le \,k\,\left( { \le \,n - 1} \right)} {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
n \cr
k + 1 \cr} \right)u^{\,k} } du} = \cr
& = \sum\limits_{0\, \le \,k\,} {{{\left( { - 1} \right)^{\,k + 1} } \over {k + 1}}\left( \matrix{
n \cr
k + 1 \cr} \right)x^{\,k + 1} } = x\sum\limits_{0\, \le \,k\,} {{{\left( { - 1} \right)^{\,k + 1} } \over {k + 1}}\left( \matrix{
n \cr
k + 1 \cr} \right)x^{\,k} } \cr}
$$
Indicating by $c_k$ the coefficient of $u^k$ we have
$$
\eqalign{
& c_{\,k} = {{\left( { - 1} \right)^{\,k + 1} } \over {k + 1}}\left( \matrix{
n \cr
k + 1 \cr} \right)\quad c_{\,0} = - n\quad \cr
& {{c_{\,k + 1} } \over {c_{\,k} }} = - {{n!} \over {\left( {k + 2} \right)\left( {k + 2} \right)!\left( {n - k - 2} \right)!}}{{\left( {k + 1} \right)\left( {k + 1} \right)!\left( {n - k - 1} \right)!} \over {n!}} = \cr
& = {{\left( {k + 1} \right)\left( {k - n + 1} \right)} \over {\left( {k + 2} \right)\left( {k + 2} \right)}} \cr}
$$
and therefore we can put $J_{\,n} (x) $ into a hypergeometric form as
$$
J_{\,n} (x) = - nx\sum\limits_{0\, \le \,k\,} {{{1^{\,\overline {\,k\,} } \left( { - n + 1} \right)^{\,\overline {\,k\,} } 1^{\,\overline {\,k\,} } } \over {2^{\,\overline {\,k\,} }
2^{\,\overline {\,k\,} } }}{{x^{\,k} } \over {k!}}} = - nx\;{}_3F_{\,2} \left( {\left. {\matrix{ {1,\;1,\; - \left( {n - 1} \right)} \cr
{2,\;2} \cr
} \,} \right|\;x} \right)
$$
which, having a negative upper term, is in fact a polynomial.
B) recursion
Splitting the binomial gives
$$
\eqalign{
& J_{\,n} (x) = \sum\limits_{0\, \le \,k\,} {{{\left( { - 1} \right)^{\,k + 1} } \over {k + 1}}\left( \matrix{
n \cr
k + 1 \cr} \right)x^{\,k + 1} } = \cr
& = \sum\limits_{0\, \le \,k\,} {{{\left( { - 1} \right)^{\,k + 1} } \over {k + 1}}\left( \matrix{
n + 1 \cr
k + 1 \cr} \right)x^{\,k + 1} } - \sum\limits_{0\, \le \,k\,} {{{\left( { - 1} \right)^{\,k + 1} } \over {k + 1}}\left( \matrix{
n \cr
k \cr} \right)x^{\,k + 1} } = \cr
& = J_{\,n + 1} (x) - \sum\limits_{0\, \le \,k\,} {{{\left( { - 1} \right)^{\,k + 1} n^{\,\underline {\,k\,} } } \over {\left( {k + 1} \right)k!}}x^{\,k + 1} } = \cr
& = J_{\,n + 1} (x) - {1 \over {n + 1}}\sum\limits_{0\, \le \,k\,} {{{\left( { - 1} \right)^{\,k + 1} \left( {n + 1} \right)^{\,\underline {\,k + 1\,} } } \over {\left( {k + 1} \right)!}}x^{\,k + 1} } = \cr
& = J_{\,n + 1} (x) - {1 \over {n + 1}}\sum\limits_{0\, \le \,k\,} {\left( { - 1} \right)^{\,k + 1} \left( \matrix{
n + 1 \cr
k + 1 \cr} \right)x^{\,k + 1} } = \cr
& = J_{\,n + 1} (x) - {1 \over {n + 1}}\sum\limits_{1\, \le \,k\,} {\left( { - 1} \right)^{\,k} \left( \matrix{
n + 1 \cr
k \cr} \right)x^{\,k} } = \cr
& = J_{\,n + 1} (x) - {1 \over {n + 1}}\left( {\sum\limits_{0\, \le \,k\,} {\left( { - 1} \right)^{\,k} \left( \matrix{
n + 1 \cr
k \cr} \right)x^{\,k} } - 1} \right) = \cr
& = J_{\,n + 1} (x) - {1 \over {n + 1}}\left( {\left( {1 - x} \right)^{\,n + 1} - 1} \right) \cr}
$$
i.e.
$$
\eqalign{
& J_{\,n + 1} (x) - J_{\,n} (x) = \int_{u\, = \,0}^{\;x} {{{\left( {1 - u} \right)^{\,n + 1} - \left( {1 - u} \right)^{\,n} } \over u}du} = \cr
& = \int_{u\, = \,0}^{\;x} {{{ - u\left( {1 - u} \right)^{\,n} } \over u}du} = \int_{u\, = \,0}^{\;x} {\left( {1 - u} \right)^{\,n} d\left( {1 - u} \right)} = \cr
& = {{\left( {1 - x} \right)^{\,n + 1} - 1} \over {n + 1}} \cr}
$$
and thus
$$ \bbox[lightyellow] {
J_{\,n} (x) - J_{\,0} (x) = J_{\,n} (x) = \sum\limits_{k = 0}^{n - 1} {{{\left( {1 - x} \right)^{\,k + 1} - 1} \over {k + 1}}}
}$$
C) truncated logarithm
If we limit the range of $x$ in $(-1,1)$ then $ I_{\,n} (x,0)$ corresponds to the truncated expansion of $\ln{ \left( \frac{1}{1-x} \right)}$ .
So we may express it as
$$
\eqalign{
& I_{\,n} (x,0) = \cr
& = \int_{t\, = \,0}^{\;x} {{{1 - t^{\,n} } \over {1 - t}}dt} = \int_{t\, = \,0}^{\;x} {\sum\limits_{k = 0}^{n - 1} {t^{\,k} } dt} = \sum\limits_{k = 0}^{n - 1} {{{x^{\,k + 1} } \over {k + 1}}} = \cr
& \quad \left| \matrix{
\,\left| x \right| < 1 \hfill \cr
\;0 \le n \in Z \hfill \cr} \right.\quad = \cr
& = - \ln \left( {1 - x} \right) - \int_{t\, = \,0}^{\;x} {{{t^{\,n} } \over {1 - t}}dt} = \cr
& = - \ln \left( {1 - x} \right) - \sum\limits_{k = n}^\infty {{{x^{\,k + 1} } \over {k + 1}}} = \cr
& = - \ln \left( {1 - x} \right) - x^{\,n + 1} \sum\limits_{0\, \le \,k\,} {{{x^{\,k} } \over {\,k + n + 1}}} = \cr
& = - \ln \left( {1 - x} \right) - {{x^{\,n + 1} } \over {n + 1}}{}_2F_{\,1} \left( {\left. {\matrix{
{n + 1,\;1} \cr
{n + 2} \cr
} \;} \right|\;z} \right) = \cr
& = - \ln \left( {1 - x} \right) - x^{\,n + 1} \hat \Phi \left( {x,1,n + 1} \right) = \cr
& = - \ln \left( {1 - x} \right) - x^{\,n + 1} \int_{t = 0}^\infty {{{e^{\, - \left( {n + 1} \right)t} } \over {1 - xe^{\, - t} }}dt} = \; \cdots \cr}
$$
where
$\hat \Phi $ denotes the Lerch transcendent.
Of course, many other manipulations and transformations may be applied.
However it is to underline that the connection with the logarithm may be deceptive, since the combination
with the hypergeometric, Lerch t., etc. finally shall return a polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ How can I evaluate this integral $$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x } \mathop{dx}$$
I tried using the half angle formula
$$\sin x=\dfrac{2\tan\dfrac x2}{1+\tan^2\dfrac x2}, \cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$$
substituted and simplified I got
$$\int\dfrac{2-2\tan^2\dfrac{x}{2}-2\tan\dfrac{x}{2}}{5\tan^2\dfrac{x}{2}+6\tan\dfrac{x}{2}-5}dx$$
substituted $\tan^2\dfrac x2=\sec^2\dfrac x2-1$
$$\int\dfrac{4-2\sec^2\dfrac{x}{2}-2\tan\dfrac{x}{2}}{5\left(\tan\dfrac{x}{2}+\dfrac{3}{5}\right)^2-\dfrac{34}{5}}dx$$
I can't eliminate $\tan\frac x2$ term in numerator. I think I am not in right direction. your help to solve this integral is appreciated. thank in advance
| Rewrite numerator: $2\cos x-\sin x=\frac{7}{34}(3\sin x+5\cos x)+\frac{11}{34}\frac{d}{dx}(3\sin x+5\cos x)$
$$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x }$$
$$=\int\dfrac{\frac{7}{34}\left(3\sin x+5\cos x\right)+\frac{11}{34}\left(3\cos x-5\sin x\right)}{3\sin x+5\cos x }dx$$
$$=\frac{7}{34}\int\ dx+\frac{11}{34}\int \frac{3\cos x-5\sin x}{3\sin x+5\cos x }dx$$
$$=\frac{7}{34}\int\ dx+\frac{11}{34}\int \frac{d(3\sin x+5\cos x)}{3\sin x+5\cos x }$$
$$=\frac{7x}{34}+\frac{11}{34}\ln|3\sin x+5\cos x|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Solve for $y$ in $\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$ I saw a challenge problem on social media by a friend, solve for $y$ in $$\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$$
I think this is an integration factor ODE
$$\frac{1}{{(2x+1)}^{\frac{3}{2}}} \cdot \frac{dy}{dx}-\frac{3y}{{(2x+1)}^{\frac{5}{2}}}=\frac{3x^2}{{(2x+1)}^{\frac{3}{2}}}$$
Is this correct?
$$\left(\frac{y}{{(2x+1)}^{\frac{3}{2}}} \right)'=\frac{3x^2}{{(2x+1)}^{\frac{3}{2}}}$$
$$\left(\frac{y}{{(2x+1)}^{\frac{3}{2}}} \right)=\int \frac{3x^2}{{(2x+1)}^{\frac{3}{2}}} \mathop{dx}$$
| This ODE is linear. Consider the homogeneous
$$
y_h' -3\frac{y_h}{2x+1}=0
$$
This ODE is separable with solution
$$
y_h = C_0(2x+1)^{\frac 32}
$$
now assuming for the particular $y_p = C_0(x)(2x+1)^{\frac 32}$ after substitution in the complete ODE we obtain
$$
C_0'(x) = \frac{3x^2}{(2x+1)^{\frac 32}}
$$
thus
$$
C_0(x) = \frac{x^2-2x-2}{\sqrt{2x+1}}
$$
and finally
$$
y = y_h + y_p = \left(C_0+\frac{x^2-2x-2}{\sqrt{2x+1}}\right)(2x+1)^{\frac 32}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Letting $k= n+5$ we get that $1^3+2^3+3^3 +\dots+ (k-5)^3 \equiv 17 \text{ (mod $k$)}$.
Knowing the sum of cubes formula we get that ($\frac{(k-5)(k-4)}{2})^2\equiv 17 \text{ (mod $k$)}$.
From here I'm not sure how I should continue. What would be my options?
| Let $$S:=\sum_{k=1}^{n} k^{3}=\left[\frac{n}{2}(n+1)\right]^{2}$$
Then $$\begin{aligned} 4S &=[n(n+1)]^{2} \equiv[-5(-5+1)]^{2}\equiv 400 \quad (\operatorname{mod} n+5) \end{aligned}$$
Given that $S \equiv 17 \bmod (n+5)$, we have
$$
\begin{array}{l}\quad 4(17) \equiv 400 \quad (\bmod n+5) \\
\Rightarrow n+5 \mid 400-4(17)=2^{2} \times 83 \\
\Rightarrow n+51=83,2 \times 83 \text { or } 2^{2} \times 83 \text { (rejected) } \\
\Rightarrow n=78 \text { or } 161
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Limit of a fraction involving square and third roots. $$\lim_{n\to\infty} \frac{\sqrt{n^2+1} - \sqrt{n^2+n}}{\sqrt[3]{n^3+1} - \sqrt[3]{n^3+n^2+1}}$$
Is it a good idea to substitute the numerator and denominator using that $a-b=\frac{a^2-b^2}{a+b}$ and $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$, since I’ll end up having to multiplicate a polynomial by square and third roots?
| Yes, you can make that approach work. It's a little tedious and messy looking, but here's what you get:
$${\sqrt{n^2+1}-\sqrt{n^2+n}\over\sqrt[3]{n^3+1}-\sqrt[3]{n^3+n^2+1}}\\={(n^2+1)-(n^2+n)\over(n^3+1)-(n^3+n^2+1)}\cdot{(\sqrt[3]{n^3+1})^2+\sqrt[3]{n^3+1}\sqrt[3]{n^3+n^2+1}+(\sqrt[3]{n^3+n^2+1})^2\over\sqrt{n^2+1}+\sqrt{n^2+n}}\\
={n-1\over n^2}\cdot{n^2\left(\left(\sqrt[3]{1+{1\over n^3}} \right)^2+\sqrt[3]{1+{1\over n^3}}\sqrt[3]{1+{1\over n}+{1\over n^3}}+\left(\sqrt[3]{1+{1\over n}+{1\over n^3}} \right)^2\right)\over n\left(\sqrt{1+{1\over n^2}}+\sqrt{1+{1\over n}} \right)}\\
={n-1\over n}\cdot{\left(\sqrt[3]{1+{1\over n^3}} \right)^2+\sqrt[3]{1+{1\over n^3}}\sqrt[3]{1+{1\over n}+{1\over n^3}}+\left(\sqrt[3]{1+{1\over n}+{1\over n^3}} \right)^2\over \sqrt{1+{1\over n^2}}+\sqrt{1+{1\over n}}}\\
\to1\cdot{1^2+1\cdot1+1^2\over1+1}={3\over2}$$
The key is to factor an $n$ out of all the square and cube roots after converting the ratio of differences to a ratio of sums, after which the messy stuff that's left inside each of the roots simply tends to $1$.
Remark: I'm prone to making minus sign mistakes, so I always try to doublecheck that the sign of my answer makes sense. In this case it does: the numerator $\sqrt{n^2+1}-\sqrt{n^2+n}$ is negative, since $n^2+n$ is bigger than $n^2+1$, but so is the denominator, since $n^3+n^2+1$ is bigger than $n^3+1$, hence their ratio is positive, so the limit is non-negative, which agrees with the sign of the answer I got.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.