Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Simplification of ${0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots}$ Simplify
$$0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots,$$ where $n \ge 2.$
I think we can write this as the summation $\displaystyle\sum_{i=0}^{n} 2i\binom{n}{2i},$ which simplifies to $\boxed{n\cdot2^{n-2}}.$ Am I on the right track?
|
Here's an alternative approach, motivated by the appearance of even numbers in the summand.
Because $$\frac{1+(-1)^k}{2}=\begin{cases}1&\text{if $k$ is even}\\0&\text{if $k$ is odd}\end{cases}$$
we have $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2}.$$
Now take $a_k=k\binom{n}{k}$ to obtain
\begin{align}
\sum_{k\ge 0} 2k\binom{n}{2k}
&=\sum_{k\ge 0} k\binom{n}{k}\frac{1+(-1)^k}{2} \\
&=\sum_{k\ge 1} n\binom{n-1}{k-1}\frac{1+(-1)^k}{2} \\
&=\frac{n}{2}\sum_{k\ge 1} \binom{n-1}{k-1} +\frac{n}{2}\sum_{k\ge 1} \binom{n-1}{k-1}(-1)^k \\
&=\frac{n}{2}2^{n-1} +\frac{n}{2}(1-1)^{n-1} \\
&=\frac{n}{2}2^{n-1} +\frac{n}{2}[n=1] \\
&=n2^{n-2}
\end{align}
for $n \ge 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3715757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
}
|
How to prove that $1-\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} \geq \frac{2^{N-1}+1}{2^N}$? The question is in the title. Numerical computation suggests the result is true, but I don't know how to prove it rigorously.
|
I think I got it, but my answer isn't very slick. Improvements are very welcome.
Rearranging, it suffices to show that
$$\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} \leq \frac{1}{2} - \frac{1}{2^N}.$$
Now, I can verify this inequality by hand for $N \in \{1,2\}$. Let $N > 2$. First observe that
$$\frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} < \frac{1}{2^{n+1}}.$$
In particular, for $n=1$, the LHS is $0$ while the RHS is $1/4$. Thus,
$$\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} + \frac{1}{4} < \sum_{n=1}^N\frac{1}{2^{n+1}} < \frac{1}{2},$$
which implies
$$\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} < \frac{1}{4} < \frac{1}{2} - \frac{1}{2^N}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3716642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Using residue theorem to calculate integral $\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}$ - where is my mistake? I am to calculate:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}
$$
We can set $\gamma(t)=e^{it}$ for $t \in [0, 2\pi]$ and then $z = e^{it}$, $\dfrac{dz}{iz}=dt$, $\sin t =\dfrac{1}{2i}(z-\frac{1}{z})$ so that:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = \int\limits_\gamma \frac{dz}{\left(10+\frac{3}{i}(z-\frac{1}{z})\right)iz} = \int\limits_\gamma \frac{dz}{\left(10iz+3z^2-3\right)}
$$
Roots of denominator are $-3i$ and $\frac{-i}{3}$ but since the winding number of $-3i$ is equal to 0 we have:
$$
\int\limits_\gamma \frac{dz}{10iz+3z^2-3} = 2 \pi\, i\, Res\left(f,\frac{-i}{3}\right)\cdot1
$$
Calculating residue:
$$
Res\left(f,\frac{-i}{3}\right) = \lim_{\large z \to \frac{-i}{3}} \frac{1}{(z+3i)}=\frac{3}{8i}
$$
summing up:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = 2\pi i \cdot \frac{3}{8i} = \frac{3\pi}{4}
$$
But wolfram says it is equal to $\dfrac{\pi}{4}$. Could you help me spot my mistake?
|
$$
\operatorname{Res}\left(f,\frac{-i}{3}\right) = \lim_{z \to \frac{-i}{3}} \frac{1}{\color {red}3(z+3i)}=\frac{1}{8i}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3717891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Solve for $a_{n}$ where $a_{n} = 4a_{n-1} + 2^{2n-1}$, $a_0=1$ and $a_1=6$ . $a_{n} = 4a_{n-1} + 2^{2n-1}$, $a_0=1$ and $a_1=6$.
So I am trying to find $a_n$ by using the generating function let's call it $A(x)$.
the equation then is written as (if I doing this correctly) :
$A(x)-ao = 4xA(x) + \sum_{k=1}2^{2n-1}*x^k$
now for $\sum_{k=0}2^{2n-1}*x^k = \sum_{k=1}\frac{1}{2}4^{k}x^k \mapsto \frac{1}{2}\frac{1}{1-4x} - \frac{1}{2}$
Then we have: $(1-4x)A(x)= 1+\frac{1}{2}\frac{1}{1-4x} -\frac{1}{2}\Rightarrow A(x)=\frac{1}{2}\frac{1}{1-4x}+\frac{1}{2}\frac{1}{(1-4x)^2}\\ \Rightarrow a_n= ? $
$ (1-4x)^{-2} \mapsto 1 + \sum_{k=1}\binom{2+k-1}{k}*4^kx^{k}\\ \Rightarrow \frac{1}{(1-4x)^2}= 1+ \sum_{k=1}\binom{1+k}{k}*4^kx^{k} = 1+ \sum_{k=1}(k+1)4^kx^{k}= \sum_{k=0}(k+1)4^kx^{k}$
$And, \frac{1}{1-4x} \mapsto 4^n$
So $an = \frac{1}{2}4^n + \frac{1}{2}(n+1)4^n$ agrees with a1 = 6
|
Note, you didn't provide the boundary value for $a_1$.
Do you specifically want to use a GF? If you set
$$
b_n = \frac{a_n}{2^{2n-1}}
$$
to get (since $2(n-1)-1 = 2n -3$)
$$
b_n = b_{n-1} + 1
$$
and,
$$
a_n = 2^{n-2}a_1 + 2^{n-1}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3719192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Projection of triangle on coordinate axes? A triangle in the $xy$-plane is such that when projected onto the $x$-axis, $y$-axis and the line $y=x$ the results are line segments whose end points are $(1,0)$ and $(5,0)$, $(0,8)$ and $(0,13)$ and $(5,5)$ and $(7.5,7.5) $ respectively. If the area of triangle is $\Delta$,
I don’t know to how to find side length from projections length?
|
Considering the inequalities, which should hold for every triangle point, namely $\begin{cases}1\le x\le 5\\8\le y\le 13\\10\le x+y\le 15\end{cases}$, we have the following figure containing all the triangle:
The border of the figure consists of $6$ straight line segments, each segment should contain a triangle vertex, or otherwise the projections would be shorter.
It leaves only $2$ possibilities:
Knowing that the figure area is $5\cdot 4-\frac{1}{2}\cdot 1\cdot 1-\frac{1}{2}\cdot 3\cdot 3=15$ we compute the area of the triangle,
*
*in the first case it's $15-\frac12\cdot 4\cdot 1-\frac12\cdot 2\cdot 3-\frac12\cdot 3\cdot 1=\frac{17}{2}$
*in the second case it's $15-\frac12\cdot 1\cdot 3-\frac12\cdot 2\cdot 3-\frac12\cdot 4\cdot 1=\frac{17}{2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3719705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove $x^4 + x^2 +1$ is always greater than $x^3 + x$ Let's say P is equal to $x^4 + x^2 +1$ and $Q$ is equal to $x^3 + x$.
For $x <0$, $P$ is positive and $Q$ is negative. Hence, in this region, $P>Q$.
For $x=0$, $P>Q$.
Also, for $x = 1$, $P>Q$.
For $x > 1$, I factored out $P$ as $x^2(x^2+1) + 1$ and $Q$ as $x(x^2+1)$. For $x > 1$, $x^2(x^2+1) > x(x^2+1)$, hence $P>Q$.
The part where I have the problem is I can't prove this for the range $0 < x < 1$ without the help of a graphing calculator. Can anyone help?
What I've done in this region so far is:
*
*Prove that $P$ and $Q$ is always increasing in this region,
*The range for $P$ starts from $1 < P < 3$, and
*The range for $Q$ starts from $0 < Q < 2$.
The only thing I need to prove now is that $P$ and $Q$ will not intersect at $0 < x < 1$, but I can't prove this part.
|
For $x\lt1$ we have $x(x^2+1)\lt(x^2+1)$, since $x^2+1\gt0$, and therefore
$$x^3+x=x(x^2+1)\lt x^2+1\le x^4+x^2+1$$
For $x\ge1$, we have $x^3+x\le x(x^3+x)$, and therefore
$$x^3+x\le x(x^3+x)=x^4+x^2\lt x^4+x^2+1$$
Thus $x^3+x\lt x^4+x^2+1$ for all $x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3721404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 3
}
|
Turning a fraction with repeating decimals into a mixed number: why doesn't this work? Problem:
Turn $\frac{0.\overline{48}}{0.\overline{15}} $ into a mixed number.
My solution:
$0.\overline{15}$ goes into $0.\overline{48}$ 3 times, with a remainder of $0.\overline{48} - 3 x 0.\overline{15} = 0.\overline{48}-0.\overline{45}= 0.\overline{03}$
$100x -x = 3.\overline{03} - 0.\overline{03} = 99 x = 3$, hence $x = \frac{99x}{99} = \frac{3}{99}= \frac{1}{33}$. Since the remainder is $\frac{1}{33}$, the mixed number I'm looking for is $3\frac{1}{33}$ but the book gives $3\frac{1}{5}$ as a result, where am I wrong? Does it have anything to do with the way I multiplied $0.\overline{15}$ by 3?
|
When you want to compute, say,
$$
\frac{25}{7}
$$
you say "$7$ goes into $25$ three times, with a remainder of $4$."
But does that mean that
$$
\frac{25}{7} = 3 + 4?
$$
Not at all! It means that
$$
\frac{25}{7} = 3 + \frac{4}{7}.
$$
By analogy, in your case, you have
$$
\frac{0.\overline{48}}{0.\overline{15}}
$$
is $3$, with a remainder of $0\overline{.03}$, which means that
\begin{align}
\frac{0.\overline{48}}{0.\overline{15}} = 3 + \frac{0.\overline{03}}{0.\overline{15}}
\end{align}
You still have to simplify that last fraction but that's relatively easy: You can write
\begin{align}
\frac{0.\overline{03}}{0.\overline{15}}
&= \frac{1}{10} \frac{0.\overline{30}}{0.\overline{15}} \\
&= \frac{1}{10} 2 \\
&= \frac{2}{10}\\
&= \frac{1}{5},
\end{align}
although your text may have some other way of reducing that to get the same answer --- I just happened to notice that the "3" and the 15" could be made to cancel nicely if it was "30" and "15" instead.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3721887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Construct a Reed Solomon code: find the parity check matrix I am trying to solve the following exercise, but I need a check/opinion on how to solve it.
Construct a Reed-Solomon code with dimensions $[12,7]$ over $\mathbb{F}_{13}$ and find a parity check matrix for the code $C$. Hint: $2$ is a primitive element of $\mathbb{F}_{13}$.
First thing: I have $\delta=12-7+1=6$, so the minimum distance is exactly $6$. Also, I choose to build a narrow-sense code, so the defining set is $T = \mathcal{C}_1 \cup \ldots \cup \mathcal{C}_{5}$.
As $12=n=13-1$, then $\mathcal{C}_i=\{ i \}$, so the generator polynomial is $$g(x)=(x-2)(x-2^2)(x-2^3)(x-2^4)(x-2^5)=(x-2)(x-4)(x-8)(x-3)(x-6)$$
Now, I can work out the computations and find $h(x)$,check polynomial, dividing $x^{12}-1$ by $g(x)$, but it seems a bit heavy to me. Is there any other possibility to compute the check polynomial faster? And so also the parity check matrix.
|
As an alternative to @rschweib's answer and possibly requiring little computation you have good look-up tables, a cyclic Reed-Solomon code whose generator polynomial has $2, 2^2, 2^3, 2^4, 2^5$ as roots has parity check matrix
$$H = \left[\begin{matrix}
1&2&2^2&2^3&\quad \cdots&2^{11}\\
1&2^2&(2^2)^2&(2^2)^3&\quad \cdots&(2^2)^{11}\\
1&2^3&(2^3)^2&(2^3)^3&\quad \cdots&(2^3)^{11}\\
1&2^4&(2^4)^2&(2^4)^3&\quad \cdots&(2^4)^{11}\\
1&2^5&(2^5)^2&(2^5)^3&\quad \cdots&(2^5)^{11}
\end{matrix}\right]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3723530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Solving $(D^2-1)y=e^x(1+x)^2$ I did like this:
$$\text{Let,} y=e^{mx} \text{ be a trial solution of } (D^2-1)y=0$$
$$\therefore \text{The auxiliary equation is } m^2-1=0$$
$$\therefore m=\pm1\\
\text{C.F.} = c_1e^x+c_2e^{-x}$$
$$\begin{align}
\text{P.I.}& =\frac{1}{D^2-1}e^x(1+x)^2\\
& =e^x\frac{1}{(D+1)^2-1}(1+2x+x^2)\\
& =e^x\frac{1}{D^2+2D}(1+2x+x^2)\\
& =\frac{e^x}{2}\left[\frac{1}{D}-\frac{1}{D+2}\right](1+2x+x^2)\\
& =\begin{aligned}
\frac{e^x}{2}\frac{1}{D}(1+2x+x^2)-\frac{e^x}{2}\frac{1}{D+2}(1+2x& +x^2)\\
\end{aligned}\\
& =\frac{e^x}{2}\left(x+x^2+\frac{x^3}{3}\right)-\frac{e^x}{4}\left(x^2+x+\frac{1}{2}\right)\\
& =e^x\left(\frac{x^3}{6}+\frac{x^2}{4}+\frac{x}{4}-\frac{1}{8}\right)\\
\end{align}$$
$$\therefore \text{The solution is}$$
$$y=c_1e^x+c_2e^{-x}+e^x\left(\frac{x^3}{6}+\frac{x^2}{4}+\frac{x}{4}-\frac{1}{8}\right)$$
But, in my book the answer is:
$$y=c_1e^x+c_2e^{-x}+\frac{xe^x}{12}(2x^2+3x+3)$$
$$\text{Please, check if there is any } \color{red}{mistake}.$$
|
It is the same. Your coefficient for $e^x$ is $c_1-\frac 18$, which you can rename it to say $c_3$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3728427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Solving trigonometric equations: $\cos(3x)+ \cos(x)=0$ Recently I worked on a problem where I had to solve $$\cos(3x)+\cos(x)=0$$
When I tried calculating it by evaluating $x$, $x=2nπ+\dfrac{\pi}{4}$, $x=2n\pi-\dfrac{\pi}{4}$, or $x=2n\pi+\dfrac{\pi}{2}$ was the solution I reached.
Unfortunately, it was apparently wrong. Is there another way to solve this problem?
|
So, we have ${\cos(2x + x) + \cos(x) = 0}$. Using the compound trig formulae:
$${\Rightarrow \cos(2x)\cos(x) - \sin(2x)\sin(x) + \cos(x) = 0}$$
Using the double angle identity for ${\sin(2x)}$, and ${\cos(2x)}$ gives
$${(2\cos^2(x) - 1)\cos(x) - 2\cos(x)\sin^2(x) + \cos(x)= 0}$$
$${\Rightarrow 2\cos^3(x) - 2\cos(x)\left(1 - \cos^2(x)\right)=0}$$
$${\Rightarrow 4\cos^3(x) - 2\cos(x) = 0}$$
$${\Rightarrow \cos(x)\left(4\cos^2(x) - 2\right)=0}$$
So either ${\cos(x)=0}$ or ${4\cos^2(x) - 2 = 0}$. Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3729544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
}
|
Prove that $\sum_nF_{2n+1}x^n=\frac{1-x}{1-3x+x^2}$
Prove that $$\sum_nF_{2n+1}x^n=\frac{1-x}{1-3x+x^2}$$
This is a problem of generating functions. I know that $\sum_nF_{2n}x^=\frac{x}{1-3x+x^2}$ and I am guessing this is a good point to start. But I don't really know where to go from here. Could someone help me?
|
$$\sum\limits_{n=0}^\infty F_{2n}x^n=\dfrac{x}{1-3x+x^2}$$
$$=F_0+\sum_{n=0}^\infty F_{2n+2}x^{n+1}=x\sum_{n=0}^\infty F_{2n+2}x^n.$$
$$\therefore \sum_{n=0}^\infty F_{2n+2}x^n=\dfrac{\sum\limits_{n=0}^\infty F_{2n}x^n}x=\dfrac{1}{1-3x+x^2}.$$
$$\therefore\sum_{n=0}^\infty F_{2n+1}x^n=\sum_{n=0}^\infty F_{2n+2}x^n-\sum_{n=0}^\infty F_{2n}x^n=\dfrac{1-x}{1-3x+x^2},$$
since $F_{2n+1}=F_{2n+2}-F_{2n}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3730708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Integrate $\int_0^{\infty} \frac{e^{\frac{2}{1+x^2}} \cos{\left(\frac{2x}{1+x^2}\right)}}{x^2+1} \mathop{dx}$ A challenge problem $$\int_0^{\infty} \frac{e^{\frac{2}{1+x^2}} \cos{\left(\frac{2x}{1+x^2}\right)}}{x^2+1} \mathop{dx}$$
Someone said that differentiation in the integral should be used, I dont know how? I try $u=\frac{1}{x^2+1}$, $u=\frac{x}{1+x^2}$, and others but I am confused and want help. Thanks
|
Expanding on my comment, the integrand is the real part of the complex function $$f(z) = \frac{\exp \left(\frac{2}{1-iz} \right)}{1+z^{2}}. $$
And by letting $z= x+iy$, we find that the real part of $\frac{2}{1-iz}$ is $$\frac{2(y+1)}{x^{2}+(y+1)^{2}}.$$
Therefore, in the upper half of the complex plane, the magnitude of $\exp \left(\frac{2}{1-iz} \right) $ is less than $e^{2}$.
So by integrating $f(z)$ around a contour that consists of the real axis and the infinitely large semicircle above it, it follows that $$ \int_{-\infty}^{\infty} f(x) \, \mathrm dx = 2\pi i \operatorname{Res} \left[f(z), i \right] = 2\pi i \left( \frac{e}{2i} \right)= \pi e.$$
Now just equate the real parts on both sides of the equation.
In general, we have $$\int_{0}^{\infty} \frac{\exp \left(\frac{a}{1+x^{2}} \right) \cos \left(\frac{ax}{1+x^{2}} \right)}{b^{2}+x^{2}} \, \mathrm dx = \frac{\pi}{2b} \exp \left(\frac{a}{1+b} \right), \quad \left(a \in \mathbb{R}, \ b > 0 \right).$$
The same approach using the fact that $$\lim_{R \to \infty} \int_{R \exp \left(i[0, \pi] \right)} \frac{z \exp \left( \frac{az}{1-iz}\right)}{b^{2}+z^{2}} = \pi i $$ shows that $$\int_{0}^{\infty} \frac{x \exp \left(\frac{a}{1+x^{2}} \right) \sin\left(\frac{ax}{1+x^{2}} \right)}{b^{2}+x^{2}} \, \mathrm dx = \frac{\pi}{2} \left( \exp \left(\frac{a}{1+b} \right)-1 \right), \quad \left(a \in \mathbb{R}, \ b \ge 0 \right).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3731649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
For real values with $abc\neq0$, if $\frac{xb+(1-x)c}{a}=\frac{xc+(1-x)a}{b}=\frac{xa+(1-x)b}{c}$, show that $x^3 = -1$ and $a=b=c.$
Let $a,b,c$ and $x$ are real numbers such that $abc \neq 0$ and $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c}.$$ Prove that $x^3=-1$ and $a=b=c.$
My attempt $:$ If $a+b+c \neq 0$ then $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c} = \frac {\left [\{xb+(1-x)c\} + \{xc + (1-x)a\} + \{xa+(1-x)b\} \right]} {a+b+c} =1.$$ Therefore $$x = \frac {a-c}{b-c} = \frac {b-a} {c-a} = \frac {c-b} {a-b}.$$
Comparing the first two expressions of $x$ and simplifying we get \begin{align*} a^2+b^2+c^2 - ab - bc -ca & = 0 \\ \implies (a-b)^2+(b-c)^2 +(c-a)^2 & = 0. \end{align*} Therefore we have $a=b=c.$ But then $x$ would be an indeterminant form. Does it imply that $a+b+c = 0$? How to proceed further? Any help will be highly appreciated. Thank you very much.
|
A Comment and the full scenario:
$a,b,c,x$ are real and $abc \ne 0$
Case (1) : $a+b+c \ne 0$
(i): $a \ne b \ne c:$
We get $x^3=-1\implies x=-1$ (from OP's work if multiply three expressions of $x$) without $x$ becoming indeterminate ($0/0$).
(i): $a=b=c$ : After OP's second-step we can write two equations for $a,b,c$
as $$-a+xb+(1-x)c=0~~~(1),~~~ (1-x)a-bx+xc=0$$ Solving by Cramer's method, we get
$$\frac{a}{x^2-x+1}=\frac{b}{x^2-x+1}=\frac{c}{x^2-x+1} \implies a=b=c, ~as~ x^2-x-1 \in R, \forall ~x \in R-\{0\}.$$ So when $a+b+c \ne 0$ and $a=b=c$, $x$ is an arbitrary real number.
Case (2): $a+b+c=0:$ The value $x=1/2$ satisfies the the main equation of the question.
Finally we comment that if $a+b+c \ne 0$ and $a=b=c$, then $x=-1$ is only one out of infinitely many real numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3733834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
If $ 3a+2b+c=7$ then find minimum value of $ a^2+b^2+c^2$ Question:- If $ 3a+2b+c=7$ then find the minimum value of $ a^2+b^2+c^2 $.
I used vectors to solve this problem.
Let $$α=3\hat{i}+2\hat{j}+\hat{k}$$
$$β=a\hat{i}+b\hat{j}+c\hat{k}$$
Using Cauchy-Schwarz inequality
we have, $|α.β|\le |α| |β|$
$=|3a+2b+c|\le\sqrt{14}\sqrt{a^2+b^2+c^2}$
$= 7\le\sqrt{14}\sqrt{a^2+b^2+c^2}$
So, $a^2+b^2+c^2\ge \frac72$
Therefore, the minimum value of $a^2+b^2+c^2$ is $\frac72$
I want to know are there any other method to find the minimum value of
$a^2+b^2+c^2$ such as using inequalities and calculus by assuming function $f(x,y,z)=x^2+y^2+z^2$.
|
There is a variant of the Cauchy-Schwarz inequality and its name is by a certain community of problem solvers as Titu’s lemma:
$a^2+b^2+c^2 = \dfrac{(3a)^2}{9}+\dfrac{(2b)^2}{4}+\dfrac{c^2}{1} \ge \dfrac{(3a+2b+c)^2}{9+4+1} = \dfrac{49}{14} = \dfrac{7}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3735483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
If $x, y, z\in\mathbb R^+ $ and $x^3+y^3=z^3,$ then prove that $x^2+y^2-z^2>6(z-x) (z-y). $ I made several unsuccessful attempts. Still couldn't think of a proper way to prove the inequality. Please suggest how to approach this problem. Thanks in advance.
EDIT 1. My approach (that I was talking about):
Given: $z^3=x^3+y^3.$
We have to prove:
$x^2+y^2-z^2>6(z-x) (z-y)$ i.e., $\underbrace{(z^2+zx+x^2) (z^2+zy+y^2) (x^2+y^2-z^2)}_{=E\text{ (say)}}>6(z^3-x^3) (z^3-y^3)=6x^3y^3.$
(Here one thing which I noticed is that $(x^2+y^2-z^2)>0,$ since each of the terms on the LHS except this one is positive and $6x^3y^3$ is also positive for $x, y, z>0.$)
Using AM $\ge$ GM, we have:
$E\ge 3zx\cdot3zy\cdot(x^2+y^2-z^2)\ge 9xyz^2(2xy-z^2).$
From here I couldn't think of a proper way to prove $E>6x^3y^3.$ But I'm still working on it. At present I'm trying to manipulate the expression $9xyz^2(2xy-z^2)$ to get the job done. If I find something useful I'll update here.
|
Let $x^2+y^2=2uxy$.
Thus, $u\geq1$ and we need to prove that:
$$x^2+y^2-\sqrt[3]{(x^3+y^3)^2}>6\left(\sqrt[3]{x^3+y^3}-x\right)\left(\sqrt[3]{x^3+y^3}-y\right)$$ or
$$x^2+y^2-6xy+6(x+y)\sqrt[3]{x^3+y^3}-7\sqrt[3]{(x^3+y^3)^2}>0$$ or
$$(x^2+y^2-6xy)^3+216(x+y)^3(x^3+y^3)-343(x^3+y^3)^2+$$
$$+126(x^2+y^2-6xy)(x+y)(x^3+y^3)>0$$ or
$$4(u-3)^3+432(u+1)^2(2u-1)-343(u+1)(2u-1)^2+$$
$$+252(u-3)(u+1)(2u-1)>0$$ or
$$129u-127>0,$$ which is obvious.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3735618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
}
|
Determinant of a Toeplitz matrix How can I calculate the determinant of the following Toeplitz matrix?
\begin{bmatrix}
1&2&3&4&5&6&7&8&9&10\\
2&1&2&3&4&5&6&7&8&9 \\
3&2&1&2&3&4&5&6&7&8 \\
4&3&2&1&2&3&4&5&6&7 \\
5&4&3&2&1&2&3&4&5&6 \\
6&5&4&3&2&1&2&3&4&5 \\
7&6&5&4&3&2&1&2&3&4 \\
8&7&6&5&4&3&2&1&2&3 \\
9&8&7&6&5&4&3&2&1&2 \\
10&9&8&7&6&5&4&3&2&1 \\
\end{bmatrix}
|
We define the following $n \times n$ (symmetric) Toeplitz matrix
$${\rm A}_n := \begin{bmatrix}
1 & 2 & 3 & \dots & n-1 & n \\
2 & 1 & 2 & \dots & n-2 & n-1 \\
3 & 2 & 1 & \dots & n-3 & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
n-1 & n-2 & n-3 & \dots & 1 & 2 \\
n & n-1 & n-2 & \dots & 2 & 1 \\
\end{bmatrix}$$
Hence,
$${\rm A}_{n+1} = \begin{bmatrix} {\rm A}_n & {\rm c}_n\\ {\rm c}_n^\top & 1\end{bmatrix}$$
where ${\rm c}_n = {\rm A}_n {\rm e}_n + {\Bbb 1}_n$. Computing the determinant,
$$\det \left( {\rm A}_{n+1} \right) = \det \begin{bmatrix} {\rm A}_n & {\rm c}_n\\ {\rm c}_n^\top & 1\end{bmatrix} = \left( 1 - {\rm c}_n^\top {\rm A}_n^{-1} {\rm c}_n \right) \det \left( {\rm A}_n \right)$$
where
$$\begin{aligned} {\rm c}_n^\top {\rm A}_n^{-1} {\rm c}_n &= \left( {\rm A}_n {\rm e}_n + {\Bbb 1}_n \right)^\top {\rm A}_n^{-1} \left( {\rm A}_n {\rm e}_n + {\Bbb 1}_n \right)\\ &= \underbrace{{\rm e}_n^\top {\rm A}_n {\rm e}_n}_{= 1} + \underbrace{{\rm e}_n^\top {\Bbb 1}_n}_{= 1} + \underbrace{{\Bbb 1}_n^\top {\rm e}_n}_{= 1} + \underbrace{{\Bbb 1}_n^\top {\rm A}_n^{-1} {\Bbb 1}_n}_{= \frac{2}{n+1}} = 3 + \frac{2}{n+1}\end{aligned}$$
and, thus,
$$\boxed{ \quad \det \left( {\rm A}_{n+1} \right) = -2 \left( \frac{n+2}{n+1} \right) \det \left( {\rm A}_n \right) \quad }$$
and, since $\det \left( {\rm A}_1 \right) = 1$, after some work, we eventually conclude that
$$\color{blue}{\boxed{ \quad \det \left( {\rm A}_n \right) = (-1)^{n-1} \left( n + 1 \right) 2 ^{n-2} \quad }}$$
which is related to integer sequence A001792, as pointed out by Somos.
Addendum
How to show the following?
$${\Bbb 1}_n^\top {\rm A}_n^{-1} {\Bbb 1}_n = \frac{2}{n+1}$$
Note that the $n$-th column of matrix ${\rm A}_n$ is the reversal of its first column. Hence,
$${\rm A}_n \left( {\rm e}_1 + {\rm e}_n \right) = (n+1) {\Bbb 1}_n$$
Left-multiplying both sides by ${\Bbb 1}_n^\top {\rm A}_n^{-1}$,
$$\underbrace{{\Bbb 1}_n^\top {\rm A}_n^{-1} {\rm A}_n \left( {\rm e}_1 + {\rm e}_n \right)}_{= {\Bbb 1}_n^\top \left( {\rm e}_1 + {\rm e}_n \right) = 2} = (n+1) {\Bbb 1}_n^\top {\rm A}_n^{-1} {\Bbb 1}_n$$
and, thus,
$${\Bbb 1}_n^\top {\rm A}_n^{-1} {\Bbb 1}_n = \frac{2}{n+1}$$
SymPy code
>>> from sympy import *
>>> [ Matrix(n, n, lambda i,j: 1 + abs(i-j)).det() for n in range(1,11) ]
[1, -3, 8, -20, 48, -112, 256, -576, 1280, -2816]
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3736861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
There exists $c$ such that $\int_a^{(a+b)/2}f(x)dx=(b-a)/4(f(a)+f((a+b)/2))-(b-a)^3/96f''(c)$ I am struggling to understand a statement from a solution to a problem concerning integrals.
The hypotesis of the problem is that $f:[a,b]\to\mathbb{R}$ is twice differentiable with continuous second derivative and (but I don't think it is needed for this part) $\displaystyle \int_a^b f(x)\mathop{}\!\mathrm{d}x=0$.
The solution begins with the following statement which I can't prove: There exist $c_1\in\left( a,\frac{a+b}{2}\right)$ and $c_2\in\left(\frac{a+b}{2},b\right)$ such that:
$$\int_a^\frac{a+b}{2}f(x)\mathop{}\!\mathrm{d}x=\frac{b-a}{4}\left( f(a)+f\left(\frac{a+b}{2}\right)\right)-\frac{(b-a)^3}{96}f''(c_1)$$
and
$$\int_\frac{a+b}{2}^bf(x)\mathop{}\!\mathrm{d}x=\frac{b-a}{4}\left(f\left(\frac{a+b}{2}\right)+f(b)\right)-\frac{(b-a)^3}{96}f''(c_2).$$
I tried several aproaces including Taylor's formula or integrating by parts: $\displaystyle\int_a^\frac{a+b}{2}f(x)\mathop{}\!\mathrm{d}x=\int_a^\frac{a+b}{2}f(x)(x-c)'\mathop{}\!\mathrm{d}x=\ldots$.
|
We have that
$$\int_a^\frac{a+b}{2}f(x)\mathop{}\!\mathrm{d}x=\int_{a}^\frac{a+b}{2}f(x)\left(x-\frac{3a+b}{4}\right)^\prime \mathop{}\!\mathrm{d}x=$$
$$=f(x)\left(x-\frac{3a+b}{4}\right)\Bigg|_a^\frac{a+b}{2}-\int_a^\frac{a+b}{2}f'(x)\left(x-\frac{3a+b}{4}\right)\mathop{}\!\mathrm{d}x=$$
$$=\frac{b-a}{4}\left(f\left(\frac{a+b}{2}\right)+f(a)\right)-\int_a^\frac{a+b}{2}f'(x)\left(\frac{\left(x-\frac{3a+b}{4}\right)^2}{2}-\frac{(b-a)^2}{32}\right)^\prime \mathop{}\!\mathrm{d}x=$$
$$=\frac{b-a}{4}\left(f(a)+f\left(\frac{a+b}{2}\right)\right)-f'(x)\left(\frac{\left(x-\frac{3a+b}{4}\right)^2}{2}-\frac{(b-a)^2}{32}\right)\Bigg|_a^\frac{a+b}{2}-\int_a^\frac{a+b}{2}f''(x)\left(\frac{\left(x-\frac{3a+b}{4}\right)^2}{2}-\frac{(b-a)^2}{32}\right)\mathop{}\!\mathrm{d}x=$$
$$=\frac{b-a}{4}\left(f(a)+f\left(\frac{a+b}{2}\right)\right)+\int_a^\frac{a+b}{2}f''(x)\left(\frac{(b-a)^2}{32}-\frac{\left(x-\frac{3a+b}{4}\right)^2}{2}\right)\mathop{}\!\mathrm{d}x.$$
Now notice that
$$\frac{(b-a)^2}{32}-\frac{\left(x-\frac{3a+b}{4}\right)^2}{2}\geq 0$$
for any $x\in\left[a,\frac{a+b}{2}\right]$ so, by the Mean Theorem, there exists $c\in\left(a,\frac{a+b}{2}\right)$ such that the last integral is equal to
$$f''(c)\int_a^\frac{a+b}{2}\left(\frac{(b-a)^2}{32}-\frac{\left(x-\frac{3a+b}{4}\right)^2}{2}\right)\mathop{}\!\mathrm{d}x=$$
$$=f''(c)\left(\frac{(b-a)^3}{64}-\frac{(b-a)^3}{192}-\frac{(b-a)^3}{192}\right)=f''(c)\frac{(b-a)^3}{96},$$
which concludes the proof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3736988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Maximize $\boxed{\mathbf{x}+\mathbf{y}}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$ Maximize $\mathbf{x}+\mathbf{y}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$
My approach
$\frac{x^{2}}{1 / 2}+\frac{y^{2}}{1 / 3} \leq 1$
Let $z=x+y$
$\mathrm{Now}, 4 \mathrm{x}+6 \mathrm{y} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{2 x}{3 y}$
$2 x^{2}+3 y^{2}=1$
What to do next? Any suggestion or Hint would be greatly appreciated!
|
Alternatively, the objective function $z=x+y$ will achieve its maximum when the contour line $y=-x+z$ will touch the ellipse $2x^2+3y^2=1$ from above. So, the slope of the tangent line must be equated to the slope of the contour line:
$$4x_0^2+6y_0^2y'=0\Rightarrow y'=-\frac{2x_0}{3y_0}=-1\Rightarrow x_0=\frac32y_0$$
Now plug this into the equation of ellipse:
$$2\cdot \left(\frac32y_0\right)^2+3y_0^2=1\Rightarrow y_0=\sqrt{\frac2{15}}\Rightarrow x_0=\sqrt{\frac3{10}}.$$
Hence, the maximum is $z(x_0,y_0)=\sqrt{\frac3{10}}+\sqrt{\frac2{15}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3740226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
}
|
Number of Real Solutions of $\frac{7^{1+\cos(\pi x)}}{3}+3^{x^2-2}+9^{\frac{1}{2}-|x|}=1$
Find the number of real solutions of the equation $$\frac{7^{1+\cos(\pi x)}}{3}+3^{x^2-2}+9^{\frac{1}{2}-|x|}=1\,.$$
By hit and trial i got the solution at $x=\pm 1$ but i am not able to solve it as it involves power of 7
|
Note that $1+\cos(\pi x)\geq 0$ for all $x\in\mathbb{R}$. Therefore,
$$1-\frac{7^{1+\cos(\pi x)}}{3}\leq 1-\frac{1}{3}=\frac{2}{3}\,.$$
Thus, if $x$ is a real solution of the required equation, then
$$3^{x^2-1}+3^{2\big(1-|x|\big)}=3\,\left(3^{x^2-2}+9^{\frac12-|x|}\right)=3\,\left(1-\frac{7^{1+\cos(\pi x)}}{3}\right)\leq 2\,.$$
By the AM-GM Inequality,
$$3^{x^2-1}+3^{2\big(1-|x|\big)}\geq 2\,\sqrt{3^{x^2-1}\cdot3^{2\big(1-|x|\big)}}=2\cdot3^{\frac{\big(|x|-1\big)^2}{2}}\geq 2\,.$$
Therefore, the inequality above must be an equality, which means $x^2-1=2\big(1-|x|\big)$ and $\big(|x|-1\big)^2=0$. This shows that $|x|=1$, or $x=\pm1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3740954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How many positive integers less than or equal to $500$ have exactly $3$ positive divisors?
How many positive integers less than or equal to $500$ have exactly $3$ positive divisors?
My approach was to use the following inequality $N \leqslant 500$, where $N$ is the number of positive integers that have exactly $3$ positive divisors. Prime factorizing $500$ one would get that $500 = 2^2 \cdot 5^3$. Wouldn't this imply that $N$ would have to be of the form $N = 2^a \cdot 5^b$ and from the given condition we would get that $(a+1)(b+1) = 3$? However I couldn't get this any further. Any tips would be appreciated.
|
For $n$ to have exactly $3$ positive divisors, $1$ must be a factor, $n$ must be a factor and also we need another factor say $p$.
We need $\frac{n}{p}=p$ and $p$ is a prime. Prime is needed or additional divisors exists.
Hence $n$ must be a prime square.
Since $\sqrt{500}\approx 22.3$, the numbers are $2^2, 3^2, 5^2, 7^2, 11^2, 13^2, 17^2, 19^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3742838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
For which $k$ does $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ hold? By generalizing this (1) and this (2) questions and performing some research
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt[3]{abc}},\hbox{ for }a,b,c>0$$
for all $0\le k<k_0\approx 11.108$.
The main goal was to prove the original inequality from (2), however, letting $a=x^3,\,b=y^3,\,c=z^3$ and clearing the denominator, the inequality becomes
$$3 k x^3 y^3 z^3 + 3 \sum\limits_{sym}x^6 y^3 z^0 - \left(3+\frac k2\right)\sum\limits_{sym} x^5 y^2 z^2\ge 0\tag{1}$$
and I'm failing to apply Muirhead's inequality.
The method from this answer works only for $k\le 3$, and even with calculus I don't think that solving system of $3$ equations like $\frac{\partial}{\partial x}$LHS(1)$=0$:
$$5 k x^3 y^2 z^2 - 9 k x y^3 z^3 + 2 k y^5 z^2 + 2 k y^2 z^5 - 18 x^4 y^3 - 18 x^4 z^3 + 30 x^3 y^2 z^2 - 9 x y^6 - 9 x z^6 + 12 y^5 z^2 + 12 y^2 z^5=0$$
may lead to something neat.)
Any help is appreciated. Thanks.
The question: what is $k_0$.
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to find a maximal $k$ for which the following inequality is true for any positives $a$, $b$ and $c$.
$$\frac{9uv^2}{w^3}+k-3\geq \left(2+\frac{k}{3}\right)\frac{3u}{w},$$
which says that it's enough to show it for a minimal value of $v^2$.
Now, $a$, $b$ and $c$ are roots of the equation $$(x-a)(x-b)(x-c)=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$$ or
$$3v^2x=-x^3+3ux^2+w^3.$$
Id est, the line $y=3v^2x$ and the graph of $f(x)=-x^3+3ux^2+w^3$ have three common points
(maybe less of three common points if this line is a tangent line to the graph).
We can draw a graph of $f$: $$f'(x)=-3x(x-2u),$$ which gives that $(0,w^3)$ is a minimum point and $(2u,f(2u))$ is a maximum point.
Now, we see that $v^2$ will get a minimal value, when $y=3v^2x$ would be a tangent to the graph of $f$,
which happens for equality case of two variables.
Since our inequality is homogeneous and symmetric we can assume $b=c=1$ and $a=x^3$, which gives
$$\frac{6(x^4+x^3+x^2+2x+1)}{x^2(x+2)}\geq k,$$ which says $$k_0=\min_{x>0}\frac{6(x^4+x^3+x^2+2x+1)}{x^2(x+2)}\approx11.10864$$
Since $$\left(\frac{6(x^4+x^3+x^2+2x+1)}{x^2(x+2)}\right)'=\frac{6(x^2+x+1)(x^3+3x^2-3x-4)}{x^3(x+2)^2},$$ we see that this minimum occurs, when $x$ is a positive root of the equation: $x^3+3x^2-3x-4=0,$ which gives $$x_{min}=2\sqrt2\cos\left(\frac{1}{3}\arccos\left(-\frac{1}{4\sqrt2}\right)\right)-1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3746078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Intuition for weighted average. Why $\frac{w_1}{w_1 + w_2}x_1 + \frac{w_2}{w_1 + w_2}x_2 = \frac{\sum_i w_ix_i}{\sum_i w_i}$? I know $\dfrac{w_1}{w_1 + w_2}x_1 + \dfrac{w_2}{w_1 + w_2}x_2 = \dfrac{\sum_i w_ix_i}{\sum_i w_i}$, because $\sum_i w_i$ is common denominator. I'm not asking about this algebra.
It's intuitive that $\dfrac{w_i}{w_1 + w_2}$ weighs $x_i$.
Intuitively, why's $\dfrac{\sum w_ix_i}{\sum w_i}$ Weighted Average? You are summing $w_ix_i$ and $w_i$ separately. Thus you lost information, because the weight for $x_i$ doesn't appear. When you sum $\sum w_ix_i$ and and $\sum w_i$, these end as totals. They inform nothing about weights! And you can't recover the weights for just these sums!
Can picture explain?
|
Here is an example from statistics.
The table shows the sales of sugar (in kilograms) during $10$ days:
$$\begin{array}{c|c|c}
\text{Sales of sugar (in kg)}, x & \text{Number of days}, f & \text{Percentage of days}, P(x)\\
\hline
0&1&0.1\\
1&3&0.3\\
2&4&0.4\\
3&2&0.2\\
\hline
&10&1
\end{array}$$
On $3$ days (or during $30\%$ of the $10$-day period) $1$ kilo of sugar was sold each day. Now we need to find the average sales during the $10$-day period.
Method 1. Convert the table data to raw data. Let's assume the following sales took place each day:
$$3,0,3,2,2,1,3,1,1,2$$
So, the average sale is:
$$\frac{\sum x}{n}=\frac{3+0+2+2+2+1+3+1+1+2}{10}=1.7$$
Method 2. Let's simplify the above expression:
$$\frac{\sum x}{n}=\frac{0+1+1+1+2+2+2+2+3+3}{10}=\\
\frac{0\cdot 1+1\cdot 3+2\cdot 4+3\cdot 2}{10}=\\
0\cdot \frac{1}{10}+1\cdot \frac{3}{10}+2\cdot \frac4{10}+3\cdot \frac{2}{10}=\\
0\cdot 0.1+1\cdot 0.3+2\cdot 0.4+3\cdot 0.2=1.7$$
So, the sales figures are elements (x) and the percentage of days (P(x)) are weights. The more percentage a partiular sales figure occurs, the more its impact on the average sales figure.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3747408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
}
|
Solve the inequality $|3x-5| - |2x+3| >0$. In order to solve the inequality $|3x-5| - |2x+3| >0$, I added $|2x+3|$ to both sides of the given inequality to get $$|3x-5| > |2x+3|$$ Then assuming that both $3x-5$ and $2x+3$ are positive for certain values of $x$, $$3x-5 > 2x+3$$ implies $$x>8$$ If $3x-5$ is positive and $2x-3$ is negative for certain values of $x$, then $$3x-5 > -2x-3$$ implies $$5x >2$$ implies $$x > \dfrac{2}{5}$$ I'm supposed to get that $x < \dfrac{2}{5}$ according to the solutions, but I'm not sure how to get that solution.
|
Needless to explicit the absolute values in function of the intervals where $x$ lives:
$$|3x-5| > |2x+3| \iff (3x-5)^2>(2x+3)^2\iff 5x^2-42x+16>0,$$
so it comes down to a quadratic inequality.
The reduced discriminant is $\Delta'=21^2-80=361=19^2$, and the quadratic is positive outside the interval of the roots.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3747907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
}
|
If $a_{n+1}=2a_n −n^2+n$ Define a sequence $a_n$ that satisfy the recurrence relation as described above, with $a_1 = 3$ If
$$a_{n+1}=2a_n −n^2+n$$
Define a sequence $a_n$
that satisfy the recurrence relation as described above, with $a_1 = 3$
Find the value of $$\dfrac{ |a_{20} - a_{15} | }{18133} $$
Attempt
First evaluate $a_{0}$
$$a_{1} = 2a_{0} \Rightarrow a_{0}= \frac{3}{2}$$
Then, use Z-transform: $$a_{n+1} - 2a_{n} + n^2 - n = 0$$
$$z(\mathbf{A}(z)-a_{0}) - 2\mathbf{A}(z) + \dfrac{z(z+1)}{(z-1)^3} - \dfrac{z}{(z-1)^2} = 0$$
$$\Rightarrow \mathbf{A}(z) = \dfrac{z(3z^3 -9z^2 + 9z - 7)}{2(z-2)(z-1)^3}$$
$$\Rightarrow \mathbf{A}(z) = \dfrac{2z}{z-1} + \dfrac{z}{(z-1)^2} + \dfrac{z(z+1)}{(z-1)^3} - \dfrac{z}{2(z-2)}$$
The inverse of the Z-transform will be: $$\boxed{a_{n} = 2 + n + n^2 - 2^{n-1}}$$
Now: $$a_{20} = 422-2^{19}$$ $$a_{15} = 242-2^{14}$$
Is it Correct??
Any other precise solution will be highly appreciated
|
As an alternative, you can use more elementary methods for linear difference equations. In fact, the general solution will be of the form
$$
a_n = a_n^h + a_n^*
$$
where $a_n^h$ is the general solution of the homogeneous equation, i.e. $a_n^h = c 2^n$, and $a_n^*$ is a particular solution of the full equation. If you try a particular solution "similar" to $-n^2+2n$, namely $a_n^* = k_1 n^2+ k_2n + k_3$, you'll see that
$$
a_n = c 2^n +n^2+n+2
$$
The constant $c$ can be computed from the condition $a_1=3$, yielding the solution you have obtained using the Z-Transform.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3749804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
}
|
How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{u}{(u^2+1)^2}-\dfrac{u}{(u^2+1)^3}$$
Substitute $u^2+1=t$, $2u\ du=dt$, $u\ du=dt/2$
$$\int\frac{u^3}{(u^2+1)^3}du=\int \frac{dt/2}{t^2}-\int \frac{dt/2}{t^3}$$
$$=\frac12\dfrac{-1}{t}-\frac{1}{2}\dfrac{-1}{2t^2}$$
$$=-\dfrac{1}{2t}+\dfrac{1}{4t^2}$$
$$=-\dfrac{1}{2(u^2+1)}+\dfrac{1}{4(u^2+1)^2}+c$$
My question: Can I integrate this with suitable substitution? Thank you
|
Letting
$v = u^2+1$,
$dv = 2u\, du, u^2 = v-1$
so
$\begin{array}\\
\int\dfrac{u^3}{(u^2+1)^3}du
&=\int\dfrac{(v-1)}{2v^3}dv\\
&=\frac12\int(v^{-2}-v^{-3}dv\\
&=\frac12\left(\dfrac{v^{-1}}{-1}-\dfrac{v^{-2}}{-2}\right)\\
&=\frac12\left(-v^{-1}+\frac12 v^{-2}\right)\\
&=\frac12\left(-\dfrac1{u^2+1}+\dfrac1{2(u^2+1)^2}\right)\\
&=\frac12\dfrac{-2(u^2+1)+1}{2(u^2+1)^2}\\
&=\dfrac{-2u^2-1}{4(u^2+1)^2}\\
&=-\dfrac{2u^2+1}{4(u^2+1)^2}\\
\end{array}
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 8
}
|
Calculate the Covariance of X and Y for a Uniformly Distributed Quadrilateral
Exercise 8.15: Let $(X,Y)$ be a uniformly distriubted random point on the quadrilateral $D$ with vertices $(0,0), (2,0), (1,1),$ and $(0,1)$. What is the Covariance of $X$ and $Y$?
After numerous attempt I calculate Cov($X,Y) = \frac{83}{324}$ when the answer the book gives is $-\frac{13}{324}$. Intuitively, my answer does not make sense as I suspect there should be a negative correlation based on the shape of the quadrilateral.
Anyways below is my work:
First, after graphing the quadrilateral, we have that the area of the quadrilateral is $\frac{3}{2}$. Therefore, it follows from the uniform distributed nature of the points that
\begin{equation*}
f(x,y):=\begin{cases}\frac{2}{3}, \quad x \in \text{Quadrilateral},\\0, \quad\text{
otherwise}.\end{cases}
\end{equation*}
Given this, recall that Cov($X,Y$) = $E[XY] - E[X]E[Y]$. Now, we calculate $E[X]$:
\begin{equation*}
E[X] = \int_0^1\int_0^1\frac{2}{3}xdxdy + \int_1^2\int_0^{2-x}\frac{2}{3}xdydx = \frac{4}{9}.
\end{equation*}
Similarly, we calculate $E[Y]$:
\begin{equation*}
E[Y] = \int_0^1\int_0^1\frac{2}{3}ydxdy + \int_1^2\int_0^{2-x}\frac{2}{3}ydydx = \frac{1}{9}.
\end{equation*}
Lastly,
\begin{equation*}
E[XY] = \int_0^1\int_0^1 \frac{2}{3}xydxdy + \int_1^2\int_0^{2-x}\frac{2}{3}xydydx = \frac{11}{36}.
\end{equation*}
Thus,
\begin{equation*}
Cov(X,Y) = E[XY] - E[X]E[Y] = \frac{11}{36} - \frac{4}{9}\frac{1}{9} = \frac{83}{324}.
\end{equation*}
Remark: Now, my only intuition regarding what I got wrong would be how I set up the expectations, perhaps since I computed the expectation by decomposing the quadrilateral into two easier computations (a square and triangle) I should have used different density functions for both? But I don't know if that should be the case.
|
Thanks to the comment from YJT, I realize that it was a computational error. The integrals were set up correctly, but I forgot to add the integrals for the squares into my expectation, i.e., while computing by hand I simply dropped a term.
Correction:
\begin{equation*}
E[X] = \int_0^1\int_0^1\frac{2}{3}xdxdy + \int_1^2\int_0^{2-x}\frac{2}{3}xdydx = \frac{1}{3} + \frac{4}{9} = \frac{7}{9},
\end{equation*}
\begin{equation*}
E[Y] = \int_0^1\int_0^1\frac{2}{3}ydxdy + \int_1^2\int_0^{2-x}\frac{2}{3}ydydx = \frac{1}{3} + \frac{1}{9} = \frac{4}{9}.
\end{equation*}
With the already found $E[XY]$, it immediately follows that:
\begin{equation*}
Cov(X,Y) = E[XY] - E[X]E[Y] = \frac{11}{36} - \frac{7}{9}\cdot \frac{4}{9} = -\frac{13}{324}
\end{equation*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3755695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Geometric sequence problem including sum of the numbers Numbers: $a,b,c,d$ generate geometric sequence and $a+b+c+d=-40. $ Find these numbers if $a^2+b^2+c^2+d^2=3280$
I tried this problem and I have system of equations which I can't solve. I think there should be different way to handle this problem.
|
Let $r$ be the common difference of the GP. Then using GP sum formula, we have,
\begin{align}
a+b+c+d=-40&\implies a\left(\dfrac{r^4-1}{r-1}\right)=-40\tag1\\
a^2+b^2+c^2+d^2=3280&\implies a^2\left(\dfrac{r^8-1}{r^2-1}\right)=3280\tag2
\end{align}
Now, divide the second equation from the square of the first equation to get,
\begin{equation}
\dfrac{(r^4+1)(r-1)}{(r^4-1)(r+1)}=\dfrac{41}{20}\tag3
\end{equation}
Solve to get,
\begin{equation}
r\in\left\{-3,-\dfrac13\right\}
\end{equation}
Now make two cases for $r$ and substitute in the first equation to get the desired answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3756260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \theta$. Now I can change the form of the denominator to become easy enough to substitute as follows:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +x^2\bigg)\bigg) ^\frac{3}{2} $$
Which makes it clear that $x$ needs to be substituted as $x = \frac{3}{2} \tan \theta $, and $dx = \frac{3}{2} \sec^2 \theta $
for later use. At this point I can represent $(1)$ in terms of my substituted trignometric function. The only problem comes with the denominator where I get stuck on the power. Here is how I went about solving it:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\left(\frac{3}{2} \tan \theta\right)^2 \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\frac{9}{4} \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \frac{9}{4}\bigg (\ 1 + \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$ 9^\frac{3}{2}\bigg( \ 1 + \tan^2 \theta \bigg) ^\frac{3}{2} $$
$$ 27\ ( \sec^2 \theta ) ^\frac{3}{2} $$
Now I have no idea how to evaluate this power of $sec$. The author says that it changes into $sec^3 \theta$ but I just can't fathom how that would go about. If what I understand is correct, the power it is raised to would be added to it's own making it $ \sec^\frac{7}{2} \theta$. My question is that how exactly is my reasoning wrong here?
|
The OP correctly identified the power rule for exponents
$$(b^n)^m= b^{n\times m}=b^{nm}.$$
Therefore for $b=\sec\theta,n=2,$ and $m=3/2$,
$$27\ ( \sec^2 \theta ) ^\frac{3}{2}=27(\sec\theta)^{2\times\frac32}=27(\sec^3\theta).$$
To finish the problem, the OP can apply the substitution $x = \frac{3}{2} \tan \theta \implies dx = \frac{3}{2} \sec^2 \theta \,d\theta.$
$$\int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx=\int\frac{\frac{27}{8}\tan^3\theta}{27\sec^3\theta}\frac{3}{2} \sec^2 \theta \,d\theta$$
$$=\frac{3}{16}\int\frac{\tan^3\theta}{\sec^3\theta}\sec^2 \theta \,d\theta$$
$$=\frac{3}{16}\int\frac{\tan^3\theta}{\sec\theta} \,d\theta$$
$$=\frac{3}{16}\int\frac{(\sec^2\theta-1)\tan\theta}{\sec\theta} \,d\theta$$
$$=\frac{3}{16}\left(\int\,d(\sec \theta)+\int\,d(\cos\theta)\right)$$
$$=\frac{3}{16}\left(\sec \theta+\cos\theta\right)+C.$$
The substitution $x = \frac{3}{2} \tan \theta \implies \sqrt{4x^2+9}=3\sec\theta$ which forms
$$=\frac{3}{16}\left(\frac{\sqrt{4x^2+9}}{3}+\frac{3}{\sqrt{4x^2+9}}\right)+C$$
$$=\frac{3}{16}\left(\frac{4x^2+18}{3\sqrt{4x^2+9}}\right)+C$$
$$=\boxed{\frac{2x^2+9}{8\sqrt{4x^2+9}}+C}$$
As an alternative approach, I observed that
$$x^3=\frac{x(4x^2+9)-9x}{4},$$
therefore
$$\int \frac{x^3}{(4x^2+9)^{3/2}}\,dx=\frac{1}{4}\int \frac{x(4x^2+9)-9x}{(4x^2+9)^{3/2}}\,dx$$
$$=\frac{1}{4}\int \frac{x}{\sqrt{4x^2+9}}\,dx-\frac{9}{4}\int\frac{x}{(4x^2+9)^{3/2}}\,dx$$
Substituting $u=4x^2+9 \implies du=8x\,dx$ on both integrals forms
$$=\frac{1}{32}\int\frac{1}{\sqrt{u}}\,du-\frac{9}{32}\int\frac{1}{u^{3/2}}\,du$$
$$=\frac{1}{16}\sqrt{u}+\frac{9}{16\sqrt{u}}+C$$
$$=\frac{1}{16}\left(\frac{u+9}{\sqrt{u}}\right)+C$$
$$=\frac{1}{16}\left(\frac{4x^2+18}{\sqrt{4x^2+9}}\right)+C$$
$$=\boxed{\frac{2x^2+9}{8\sqrt{4x^2+9}}+C}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3758050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
}
|
Small-angle approximation of $ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} $ I need to show the following:
$$ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} \approx 1-\frac{x^2}{6} $$ when $ x $ is small.
I think this problem is trickier than most other questions like it because in the original source there is comment saying "if you got $ 1+\frac{x^2}{6} $ [what I got] then think again!". My attempt was:
When $ x $ is small, $ \sin x \approx x $ so
$$ \frac{\sin^2{x}}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} = \frac{x^2}{x^2 \sqrt{1-\frac{x^2}{3}}} = \left ( 1 - \frac{x^2}{3} \right )^{-\frac{1}{2}} $$
Then using the binomial series approximation,
$$ \left ( 1 - \frac{x^2}{3} \right )^{-\frac{1}{2}} \approx 1 - \frac{1}{2}\left ( -\frac{x^2}{3} \right ) + ... = 1 + \frac{x^2}{6} $$
...and so it looks like I've fallen into whatever trap the question set.
Where is my error?
|
Suppose for simplicity you had two polynomials $P(z) = 1+ z + z^2 + 4z^4 + 7z^5$ and $Q(z) = 1 + 2z + 3z^2 + 4z^3$, and I asked you to calculate the product $P(z)Q(z)$... but not the entire thing. Suppose I only want the terms up to quadratic term; i.e if we write
\begin{align}
P(z)Q(z) &= a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4 + a_5z^5 +a_6z^6 + a_7z^7 +a_8z^8
\end{align}
then I'm asking you to find the coefficients $a_0,a_1,a_2$ (but for now, let's just say for some reason I'm interested in what happens when $|z|$ is very small up to an accuracy of quadratic order, so I don't really care
about the rest of the terms ). Well, we just multiply everything out:
\begin{align}
P(z)Q(z) &= (1+ z + z^2 + 4z^4 + 7z^5)(1 + 2z + 3z^2 + 4z^3) \\
&= 1 + (1\cdot 2z + z \cdot 1) + (1\cdot 3z^2 + z\cdot 2z + z^2 \cdot 1) \\
&+ \text{(terms involving $z^3$ or higher, which I don't care about for now)} \\
&= 1 + 3z + 6z^2 + O(z^3)
\end{align}
In other words, because in my final product, I'm only interested in calculating up to the quadratic term, I can simply truncate the polynomials $P$ and $Q$ to quadratic order, and then multiply them out (and then again only keep terms up to quadratic order):
\begin{align}
P(z)Q(z) &= (1 + z + z^2 + \cdots)(1 + 2z + 3z^2 + \cdots) \\
&= 1 + 3z + 6z^2 + O(z^3)
\end{align}
Again, because I'm only interested up to quadratic order, there's no need for me to keep any terms beyond that for $P(z)$ and $Q(z)$, because if I approximate $P(z) \approx 1+ z + z^2 + \color{red}{4z^4}$ (i.e I keep the $4^{th}$ order term) and I multiply with $Q(z) = 1+2z+3z^2 + 4z^3$, then the red term multiplied with anything in $Q(z)$ will yield terms which are $4^{th}$ order or higher (which I don't care about).
But what you should not do is truncate $P(z)$ and $Q(z)$ up to linear order, and say that
\begin{align}
P(z)Q(z) & \approx (1+z)(1+2z) = 1 + 3z + 2z^2
\end{align}
Because in this way, you're missing out other second order contributions (by multiplying constant term of $P$ with quadratic term of $Q$ and vice-versa).
This is how you know how many terms you need to use in your approximation. In your case, you want to approximate
\begin{align}
f(x) &= \dfrac{\sin^2x}{x^2\sqrt{1 - \frac{\sin^2x}{3}}}
\end{align}
up to $2^{nd}$ order. So, what do we do? We write things as a product first:
\begin{align}
f(x) &= \left(\dfrac{\sin x}{x}\right)\cdot\left(\dfrac{\sin x}{x}\right) \cdot \left(\dfrac{1}{\sqrt{1- \frac{\sin^2x}{3}}}\right)\tag{$1$}
\end{align}
Now, we have to expand each bracketed term up to atleast second order in $x$ and then multiply the result together. First:
\begin{align}
\dfrac{\sin x}{x} &= \dfrac{x - \dfrac{x^3}{6} + O(x^4)}{x} = 1 - \dfrac{x^2}{6} + O(x^3) \tag{$2$}
\end{align}
Next, we recall that
\begin{align}
\dfrac{1}{\sqrt{1-z}} &= 1+ \dfrac{z}{2} + \dfrac{3z^2}{8} + O(z^3)
\end{align}
Now, plug in $z= \frac{\sin^2x}{3} = x + O(x^3)$, to get
\begin{align}
\dfrac{1}{\sqrt{1-\frac{\sin^2x}{3}}} &= 1+ \dfrac{1}{2}\left(\dfrac{\sin^2x}{3}\right) + \dfrac{3}{8}\left(\frac{\sin^2x}{3}\right)^2 + O((\sin^2 x)^3) \\
&= 1 + \dfrac{1}{2}\left(\dfrac{x^2 + O(x^4)}{3}\right) + O(x^4) + O(x^6) \\
&= 1 + \dfrac{1}{6}x^2 + O(x^4)\tag{$3$},
\end{align}
where in the second line, hopefully it's clear how I got the various terms: for example $\sin x = x + O(x^3)$, so $\left(\frac{\sin^2x}{3}\right)^2 = \frac{1}{9}\sin^4x = \frac{1}{9} (x + O(x^3))^4 = O(x^4)$. Therefore, the final answer is obtained by plugging $(2)$ and $(3)$ into $(1)$ :
\begin{align}
f(x) &= \left(1 - \dfrac{x^2}{6} + O(x^3)\right)^2 \cdot \left(1 + \dfrac{1}{6}x^2 + O(x^4)\right) \\
&= 1 - \dfrac{x^2}{6} + O(x^4)
\end{align}
Long story short, if your end goal is to calculate up to second order, then at each stage of your algebra make sure you're keeping terms atleast up to $x^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3760642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Let $f(x) = 2x^2 + nx - 6$ and $g(x) = mx^2 + 2x - 4$? The functions are combined to form the new functions $h(x) = f(x) - g(x)$. Points $(2, 4)$ and $(-3, 17)$ are given to satisfy the new function. Determine the values of $m$ and $n$.
I know these formulas but I don't know what is $m$ and $n$ values are.
$$h(x)=(2-m)(x^2)+(n-2)(x)-2$$
$$h(2)=4=(2-m)(2^2)+(n-2)(2)-2$$
$$h(-3)=17=(2-m)(-3^2)+(n-2)(-3)-2$$
Help, please.
|
Just continue... But be careful of arithmetic
$h(x) = f(x) - g(x) = (2-m)x^2 + (n-2)x -2$
$h(2) = 4(2-m) + 2(n-2)-2 = -4m + 2n +2 = 4$.
so simplify $-4m + 2n = 2$ and $-2m + n = 1$
$h(-3) = 9(2-m)-3(n-2) -2 = -9m -3n+22 = 17$
so simplify $-9m -3n = -5$ so $3m +n = \frac 53$
Now just solve for $n$ and $m$.
Simple substitution $n = 2m + 1$ and so $3m + 2m+1 = \frac 53$ so
$5m = \frac 23$ and $m = \frac 2{15}$
ANd $n = 2\cdot\frac 2{15} + 1 = \frac {19}{15}$
Let's verify as I've seen answers all over the map.
$f(x) = 2x^2 +\frac {19}{15}x -6$ so $f(2)= 8+\frac {38}{15}-6 = 4\frac 8{15}$. and $f(-3) = 18-\frac {19}5-6=12-3\frac 45= 8\frac 15$.
Ahd $h(x) = \frac 2{15}x^2 +2x -4$ so $g(2) = \frac 8{15}+4+4=\frac 8{15}$ and $g(-3)=\frac {18}{15}-6-4= -10+1\frac 15= -8\frac 45$.
And $h(2)=4\frac 8{15}-\frac 8{15} = 4$. And $h(-3) = 8\frac 15 - (-8\frac 45)=17$.
Hmmm.... what do you know. My arithmetic was correct and everyone else's was wrong.
That's got to be a first.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3761472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\\
&=\int \dfrac{3\sec^2\theta d\theta}{81\sec^4\theta}\\
&=\dfrac{1}{27}\int \cos^2\theta d\theta\\
&=\dfrac{1}{27}\int \frac{1+\cos2\theta}{2} d\theta\\
&=\dfrac{1}{54}\left(\theta+\frac{\sin2\theta}{2}\right)+C
\end{align*}
This is where I got stuck. How can I get the answer in terms of $x$?
Can I solve it by other methods?
|
$\begin{aligned} I&=\int \frac{d x}{\left[(x-2)^{2}+9\right]^{2}} \\ &=\int \frac{d y}{\left(y^{2}+9\right)^{2}}, \text { where } y=x-2 \\ &=-\frac{1}{2} \int \frac{1}{y} d\left(\frac{1}{y^{2}+9}\right) \quad \text{(By IBP)}\\ &--\frac{1}{2 y\left(y^{2}+9\right)}-\frac{1}{2} \int\left(\frac{1}{y^{2}} \cdot \frac{1}{y^{2}+9}\right) d y \\ &=-\frac{1}{2 y\left(y^{2}+9\right)}-\frac{1}{18} \int\left(\frac{1}{y^{2}}-\frac{1}{y^{2}+9}\right) d y \\ &=-\frac{1}{2 y\left(y^{2}+9\right)}+\frac{1}{18 y}+\frac{1}{54} \tan ^{-1}\left(\frac{y}{3}\right)+C \\ &=\frac{y}{18\left(y^{2}+9\right)}+\frac{1}{54} \tan ^{-1}\left(\frac{y}{3}\right)+C \\ &=\frac{1}{54}\left(\frac{3(x-2)}{x^{2}-4 x+13}+\tan ^{-1}\left(\frac{x-2}{3}\right) \right)+C \end{aligned}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3761986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 8
}
|
How to evaluate $\int \frac{\cos x}{a-\cos x} \mathrm{d}x $ in a more elegant way? I am trying to evaluate
$$\int \frac{\cos x}{a-\cos x} \mathrm{d}x \quad (1)$$
Since there is a ratio of trigonometric functions, I tried to reduce the problem to a polynomial ratio by using Weierstrass substitutions:
$$t = \tan\left(\frac{x}{2}\right) \iff x = 2\arctan(t) \iff \mathrm{d}x = \frac{2}{t^2+1}$$
$$\cos x = \dfrac{1 - t^2}{1 + t^2}$$
$$\sin x = \dfrac{2t}{1 + t^2}$$
Therefore from $(1)$:
$$ \int \frac{1-t^2}{1+t^2} \frac{1}{a-\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}\mathrm{d}t = a \int \frac{1-t^2}{1+t^2} \frac{1}{1+(a+1)t^2} \mathrm{d}t $$
The function is a ratio such as: $\deg(1-t^2) = 2 < 4 = \deg((1+3t^2)(1+t^2))$ therefore we can break the relationship to partial fractions. We notice that the denominator has complex roots, therefore:
$$ \int \frac{1-t^2}{(1+t^2)(1+(a+1)t^2)} \mathrm{d}t = \cdots = \int -\frac{1}{t^2+1}+\frac{2}{(a+1)t^2+1} \mathrm{d}t$$
Therefore,
$$\int \frac{\cos x}{a-\cos x} \mathrm{d}x = -x + \frac{2a}{\sqrt{a^2-1}} \tan^{-1} \left(\sqrt{\frac{a+1}{a-1}} \tan \frac{x}{2} \right) + C$$
I feel there is an easier way to evaluate this integral. Can you come up with one?
|
Let $|a| > 1$. Observe $$f_a(x) = \frac{\cos x}{a - \cos x} = \frac{(\cos x - a) + a}{a - \cos x} = -1 + \frac{a}{a - \cos x} = -1 + \frac{a}{a - (1 - 2 \sin^2 \frac{x}{2})}$$ where we have employed the half-angle identity $$\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}.$$ Consequently
$$\begin{align} f_a(x)
&= -1 + a \frac{1}{(a-1) + 2 \sin^2 \frac{x}{2}} \\
&= -1 + a \frac{\sec^2 \frac{x}{2}}{(a-1) \sec^2 \frac{x}{2} + 2 \tan^2 \frac{x}{2}} \\
&= -1 + a \frac{\sec^2 \frac{x}{2}}{(a-1)(1+\tan^2 \frac{x}{2}) + 2 \tan^2 \frac{x}{2}} \\
&= -1 + \frac{2a}{a-1} \frac{\frac{1}{2}\sec^2 \frac{x}{2}}{1+ \frac{a+1}{a-1} \tan^2 \frac{x}{2}}.
\end{align}$$
With the substitution $$u = \sqrt{\frac{a+1}{a-1}}\tan \frac{x}{2}, \quad du = \frac{1}{2}\sqrt{\frac{a+1}{a-1}} \sec^2 \frac{x}{2} \, dx,$$ we obtain
$$\begin{align}
\int f_a(x) \, dx
&= -x + \frac{2a}{a-1} \sqrt{\frac{a-1}{a+1}} \int \frac{du}{1 + u^2} \\
&= -x + \frac{2a}{\sqrt{a^2-1}} \tan^{-1} u + C \\
&= -x + \frac{2a}{\sqrt{a^2-1}} \tan^{-1} \left(\sqrt{\frac{a+1}{a-1}} \tan \frac{x}{2} \right) + C.
\end{align}$$
Of course, I have taken the trouble to write out each step in detail which makes the solution longer than it really needs to be.
As an exercise for the reader, how is this approach modified if $|a| < 1$?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3762875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find the matrix $A^{15}$. Let $I=
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}$ and $O=\begin{pmatrix}
0 & 0 \\
0 & 0 \\
\end{pmatrix}$.
1.Let $A=\begin{pmatrix}
1 & 3 \\
3 & 5 \\
\end{pmatrix}$ and $ B=\begin{pmatrix}
x & 3 \\
3 & 6 \\
\end{pmatrix}$. Find the value of $x$ which satisfies $AB=BA$.
$AB=\begin{pmatrix}
1 & 3 \\
3 & 5 \\
\end{pmatrix}\cdot\begin{pmatrix}
x & 3 \\
3 & 6 \\
\end{pmatrix}=\begin{pmatrix}
x+9 & 21 \\
3x+15 & 39 \\
\end{pmatrix}$
$BA=\begin{pmatrix}
x & 3 \\
3 & 6 \\
\end{pmatrix}\cdot\begin{pmatrix}
1 & 3 \\
3 & 5 \\
\end{pmatrix}=\begin{pmatrix}
x+9 & 3x+15 \\
21 & 39 \\
\end{pmatrix}$
.So that we get $3x+15=21 \Rightarrow x=2$
2.Let $A=\begin{pmatrix}
1 & 2 \\
2 & 4 \\
\end{pmatrix}$ and $ B=\begin{pmatrix}
-2 & x \\
4 & y \\
\end{pmatrix}$.Find the values of x and y which satify $BA=O$.
$BA=\begin{pmatrix}
-2 & x \\
4 & y \\
\end{pmatrix}\cdot\begin{pmatrix}
1 & 2 \\
2 & 4 \\
\end{pmatrix}=\begin{pmatrix}
-2+2x & 0 \\
4+2y & 8+4y \\
\end{pmatrix}=\begin{pmatrix}
0 & 0 \\
0 & 0 \\
\end{pmatrix}$
S0 that we can get $x=1,y=-2$.
3.Let $A$ satisfying $A^2=A-I$. Find $A^{15}$.
Please help to show me about this. Thank you in advance!
|
If you recognize $x^2-x+1$ as a cyclotomic polynomial, then
$$x^{15} + 1 = (x^3+1)a(x) = (x^2-x+1)b(x)$$ gives $A^{15}+I=0$.
The systematic way, which does not need insights, is to use polynomial division:
$$
x^{15}=(x^2-x+1)q(x)-1
$$
where $q(x)=x^{13} + x^{12} - x^{10} - x^9 + x^7 + x^6 - x^4 - x^3 + x + 1$ is not really relevant. Only the remainder matters here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3763415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Strange Cube Root Offense in an Inequality I don't know how to tackle the unusual cube root present in this inequality-
$1.$For real numbers $a,b,c > 0$ and $n\le3$ prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c}\right)\ge 3+n$$
Here is another question with the same lesser side (and of course I couldn't prove)-
$2.$Let $a, b, c$ be positive real numbers such that $a + b + c = ab + bc + ca$
and $n ≤ 3$. Prove that
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2}\ge 3+n$$
What I attempted was this-
$$\left(a+b+c\right)\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2}\right)\ge \left(a+b+c\right)\left(3+n\right)$$
Avoiding the RHS for some time-
$$\left(a+b+c\right)\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2}\right)\ge (a+b+c)^2+\frac{3n(a+b+c)}{a^2+b^2+c^2}$$
After this step I don't know where to use $a+b+c=ab+bc+ca$.
These are very basic. I need a solution using AM-GM Inequality.
Any help will be appreciated.
|
A proof for $n=3$ of the second one.
By AM-GM $$\sum_{cyc}\frac{a^2}{b}+\frac{9}{a^2+b^2+c^2}\geq2\sqrt{\sum_{cyc}\frac{a^2}{b}\cdot\frac{9}{a^2+b^2+c^2}}.$$
Thus, it's enough to prove that:
$$\sum_{cyc}\frac{a^2}{b}\geq a^2+b^2+c^2$$ or
$$\sum_{cyc}\frac{a^2}{b}\sum_{cyc}ab\geq\sum_{cyc}a^2\sum_{cyc}a$$ or
$$\sum_{cyc}(a^4c^2-a^3c^2b)\geq0,$$ which is true by AM-GM again:
$$\sum_{cyc}a^4c^2=\frac{1}{6}\sum_{cyc}\left(4a^4c^2+b^4a^2+c^4b^2\right)\geq\sum_{cyc}\sqrt[6]{(a^4c^2)^4\cdot b^4a^2\cdot c^4b^2}=\sum_{cyc}a^3c^2b.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3764022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Evaluating an improper integral - issues taking the cubic root of a negative number Problem:
Evaluate the following integral.
$$ \int_{-1}^{-1} \frac{dx}{x^\frac{2}{3}} $$
Answer:
This integral includes the point $x = 0$ which results in a division by $0$. To get around this difficulty, we break the integral into two integrals.
\begin{align*}
\int_{-1}^{-1} \frac{dx}{x^\frac{2}{3}} &= \int_{-1}^{0} \frac{dx}{x^\frac{2}{3}} + \int_{0}^{1} \frac{dx}{x^\frac{2}{3}} \\
\int_{-1}^{0} \frac{dx}{x^\frac{2}{3}} &= \int_{-1}^{0} x^{-\frac{2}{3}} \,\,\, dx \\
\int_{-1}^{0} \frac{dx}{x^\frac{2}{3}} &= 3x^{\frac{1}{3}} \Big|_{-1}^0
= \lim_{x \to 0} 3x^{\frac{1}{3}} - \lim_{x \to -1} 3x^{\frac{1}{3}} \\
\lim_{x \to 0} 3x^{\frac{1}{3}} &= 0 \\
\end{align*}
Am I right so far? I do not know how to evaluate the following limit:
$$ \lim_{x \to -1} 3x^{\frac{1}{3}} $$
The problem is taking the cube root of a negative number.
|
Cube root of a real number $p$ is the unique real number $q$ such that $q^3=p$. Therefore, $x^{\frac{1}{3}} \to -1$ as $x \to -1$ because $(-1)^3=-1$ and so $(-1)^{\frac{1}{3}}=-1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3765835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
What is the value of $1 -\omega^h + \omega^{2h} -...+(-1)^{n-1} \omega^{(n-1)h}$ when $\omega$ is a root of unity? I'm reading Ahlfors' complex analysis book. One of the problems in the book says as follows
What is the value of $1 -\omega^h + \omega^{2h} -...+(-1)^{n-1} \omega^{(n-1)h}$?
where $h$ is some integer and $ \omega = \cos\left(\frac{2\pi}{n}\right) + i \sin \left(\frac{2 \pi}{n}\right)$, for some fixed $n \in \mathbb{N}$, is one of the $n$-th roots of unity.
The first thing I noticed is that I could write the series in terms of $-\omega^h$ as
$$
1 +\left(-\omega^h\right) + \left(-\omega^h\right)^2 +...+\left(-\omega^h\right)^{n-1}
$$
Inspired by this, I separated the problem into 2 cases
*
*If $h$ is an integer of the form $ h = \frac{n(2k+1)}{2}$ for some $k \in \mathbb{Z}$, then I get the following
$$
-\omega^h = -\cos\left(\frac{2\pi}{n}h\right) - i \sin \left(\frac{2 \pi}{n}h\right)= -\cos\left(\pi + 2\pi k\right) - i \sin \left(\pi + 2\pi k\right) = 1
$$
which means the sum evaluates to $\sum_{j=0}^{n-1} 1 = n$.
*If $-\omega^h \neq 1$, then using the fact that the sum in question is a sum of the first $n$ terms in a geometric series, I can write $$
1 -\omega^h + \omega^{2h} -...+(-1)^n \omega^{(n-1)h} = \frac{1 - \left(-\omega^h\right)^n}{1 - \left(-\omega^h\right)} = \frac{1 - (-1)^n\omega^{nh}}{1 +\omega^h}
$$
and since $h$ is an integer, I see that
$$
\omega^{nh} = \cos\left(\frac{2\pi}{n}nh\right) + i \sin \left(\frac{2 \pi}{n}nh\right) = \cos\left(2\pi h\right) + i \sin \left(2\pi h\right) = 1
$$
which means the sum simplifies to $\frac{1 - (-1)^n}{1 +\omega^h}$. From here I see that if $n$ is odd the sum will become $0$ because of the numerator, but for the case of $n$ being an even number, I don't see a way to simplify $\frac{2}{1 +\omega^h}$ more than it already is.
Is my solution correct? And if so, is this as simplified as I can write the solution, or can it still be simplified further? Thank you very much!
|
A bit late of an answer, but I think there is a simpler form for the even case (provided I haven't made any mistakes).
Startig with the form $\frac{2}{1+\omega^h}$, we begin with the usual fare of multiplying on top and bottom by the conjugate, leaving us with
$$\frac{2(1+\bar{\omega}^h)}{1+\omega^h + \bar{\omega}^h + |\omega|^{2h}} = \frac{2(1+\bar{\omega}^h)}{1+\omega^h + \bar{\omega}^h + 1^{2h}} = \frac{2(1+\bar{\omega}^h)}{2 + 2\operatorname{Re}(\omega^h)} = \frac{1+\bar{\omega}^h}{1+\operatorname{Re}(\omega^h)}.$$
Now, reintroducing the trigonometric form, $\omega = \cos(\frac{2\pi}{n}) + i\sin(\frac{2\pi}{n})$, this becomes
$$\frac{1+\cos(\frac{2\pi h}{n}) + i\sin(\frac{2\pi h}{n})}{1+\cos(\frac{2\pi h}{n})} = 1 + i\left(\frac{\sin(\frac{2\pi h}{n})}{1+\cos(\frac{2\pi h}{n})}\right).$$
Using the identities $1+\cos(2\theta) = 2\cos^2(\theta)$ and $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, we get
$$\frac{\sin(\frac{2\pi h}{n})}{1+\cos(\frac{2\pi h}{n})} = \frac{2\sin(\frac{\pi h}{n})\cos(\frac{\pi h}{n})}{2\cos^2(\frac{\pi h}{n})} = \frac{\sin(\frac{\pi h}{n})}{\cos(\frac{\pi h}{n})} = \tan\left(\frac{\pi h}{n}\right).$$
And, so the expression for the whole sum becomes
$$1 + i\tan\left(\frac{\pi h}{n}\right).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3767285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Finding coefficients in expansions Show that the coefficient of $x^{−12}$ in the expansion of
$$\left(x^4−\frac{1}{x^2}\right)^5\left(x−\frac{1}{x}\right)^6$$
is $−15$, and calculate the coefficient of $x^2$. Hence, or otherwise, calculate the coefficients of $x^4$ and $x^{38}$ in the expansion of $$(x^2−1)^{11}(x^4+x^2+1)^5.$$
The first part of this exercise is solved easily considering "partitions", i.e. what can be added to get the exponents and turns out to be $-15$ and $215$ respectively. I was able to solve the second part of this exercise easily using multinomial theorem. However, this exercise is given before multinomial theorem is introduced, and I would really like to know what the "Hence" was gunning for. Once I factor out to get$$x^{21}\left(x-\frac{1}{x}\right)^{11}\left(x^2+\frac{1}{x^2}+1\right)^5,$$I can see similarities with the first question. I can even deduce that since $4-21=-17$, $38-21=17$, and inspecting signs, that the coefficients are equal in magnitude and opposite in sign. Please explain how I can use the first part of the question to proceed, rather than going "otherwise"/multinomial theorem.
|
\begin{align*}
(x^2 - 1)^{11}(x^4 + x^2 + 1)^5
&=(x^2 - 1)^{5}(x^4 + x^2 + 1)^5(x^2 - 1)^{6} \\
&=(x^6 - 1)^5(x^2 - 1)^6 \\
&=x^{16}\left(x^4 - \frac{1}{x^2}\right)^5\left(x - \frac{1}{x}\right)^6
\end{align*}
Thus, the coefficient of $x^4$ in the new equation is the same as the coefficient of $x^{-12}$ in the original equation, which is $-15$.
And like you mentioned, the coefficient of $x^4$ and $x^{42 - 4} = x^{38}$ should be the same but opposite in sign. So the coefficient of $x^{38}$ is $15$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3769728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Evaluate $\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2 $ I'm trying to calculate:
$$T = \lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2$$
Here is my attempt.
Put $x^2=\dfrac{1}{t}$ so when $x\to \infty, t \to 0$ and the limit become
\begin{align*}
T &= \lim\limits_{t \to 0} \sqrt[n]{\left(1+\dfrac{1}{t}\right)\left(2+\dfrac{1}{t}\right)...\left(n+\dfrac{1}{t}\right)}-\dfrac{1}{t}\\
&=\lim\limits_{t \to 0} \sqrt[n]{\left(\dfrac{t+1}{t}\right)\left(\dfrac{2t+1}{t}\right)...\left(\dfrac{nt+1}{t}\right)}-\dfrac{1}{t} \\
&=\lim\limits_{t \to 0} \dfrac{\sqrt[n]{(t+1)(2t+1)...(nt+1)}-1}{t}
\end{align*}
My idea is to use $\lim\limits_{x\to0}\dfrac{(ax+1)^{\beta}-1}{x} =a\beta .$ But after some steps (above), now I'm stuck.
Thanks for any helps.
|
$$\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2=\lim\limits_{x \to \infty}x^2\left[ e^{\frac{1}{n}\ln \left(1+ \frac{1}{x^2} \right)\cdots \left(1+ \frac{n}{x^2} \right)}-1 \right] =\\=\lim\limits_{x \to \infty}\frac{1}{n}x^2\left[ \ln \left(1+ \frac{1}{x^2} \right)+\cdots+\ln \left(1+ \frac{n}{x^2} \right)\right]=\frac{1}{n}\left[1+2+ \cdots+n\right] =\frac{n+1}{2} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3770941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
}
|
Show that for $p \geq 1$, $[\Gamma(\frac{p+1}{2})]^{1/p}=O(\sqrt{p})$ as $p \rightarrow \infty$ Actually it is Exercise 2.5.1 in High Dimensional Probability by Vershynin.
I got stuck at showing that for $p \geq 1$,
$$\left[\Gamma\left(\frac{p+1}{2}\right)\right]^{1/p}=O(\sqrt{p}) \ \ \ \text{as }p\rightarrow\infty$$
I can show it if $p\in \mathbb{N}$ but I have no idea for real $p\geq 1$.
Any kinds of help is appreciated, thank you!
|
$$
\begin{align}
\Gamma\!\left(\frac{p+1}2\right)
&=\Gamma\!\left(1+\left\{\frac{p-1}2\right\}\right)\prod_{k=1}^{\left\lfloor\frac{p-1}2\right\rfloor}\left(k+\left\{\frac{p-1}2\right\}\right)\tag1\\
&\le1\cdot\left[\frac1{\left\lfloor\scriptstyle{\frac{p-1}2}\right\rfloor}\sum_{k=1}^{\left\lfloor\frac{p-1}2\right\rfloor}\left(k+\left\{\frac{p-1}2\right\}\right)\right]^{\left\lfloor\frac{p-1}2\right\rfloor}\tag2\\
&=\left(\frac{p+1}4+\frac12\left\{\frac{p+1}2\right\}\right)^{\left\lfloor\frac{p-1}2\right\rfloor}\tag3
\end{align}
$$
Explanation:
$(1)$: use the recurrence formula $\Gamma(x+1)=x\,\Gamma(x)$
$(2)$: $\Gamma(x)\le1$ for $1\le x\le2$ and the AM-GM inequality
$(3)$: evaluate the sum
Therefore,
$$
\begin{align}
\Gamma\left(\frac{p+1}2\right)^{1/p}
&\le\left(\frac{p+3}4\right)^{\frac{p-1}{2p}}\tag4\\[6pt]
&\le\sqrt{p}\tag5
\end{align}
$$
Explanation:
$(4)$: inequality $(3)$ and $0\le\left\{\frac{p+1}2\right\}\lt1$
$(5)$: for $p\ge1$, $\left(\frac{p+3}{4p}\right)^p\le1\le\frac{p+3}4\implies\left(\frac{p+3}4\right)^{\frac{p-1}{2p}}\le\sqrt{p}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3771462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Find limits of $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor Expansion. Find the limit $\displaystyle \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor expansion.
My Try
$\displaystyle =\lim _{x \to 0} \frac {\cos x - \frac{\sin x}{x}} {x \sin x}$
$=\frac {\displaystyle\lim _{x \to 0}\cos x - \lim _{x \to 0}\frac{\sin x}{x}} {\displaystyle\lim _{x \to 0}x \sin x}$
$=\frac{1-1}{0}$
But still I end up with $\frac00$
Any hint for me to proceed would be highly appreciated.
P.S: I did some background check on this question on mathstack and found they have solved this with l'Hopital's rule and the answer seems to be $\frac{-1}{3}$.
What is $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$?
|
While searching more easy way, let me suggest one possible way to solve main difficult part of suggested limit. I change denumerator to $x^3$, for simplicity, as it's equivalent $x^2\sin x$
Suppose we know existence of limit. Then
$$L=\lim_{x\to0}\frac{x-\sin x}{x^3} = \lim_{x\to0}\frac{x-3\sin \frac{x}{3}+4 \sin^3 \frac{x}{3}}{x^3}=\\ =\lim_{x\to0}\left(3\frac{\frac{x}{3} - \sin \frac{x}{3}}{x^3} + \frac{4 \sin^3 \frac{x}{3}}{x^3}\right) =\lim_{x\to0}\left(\frac{\frac{x}{3} - \sin \frac{x}{3}}{9\left(\frac{x}{3}\right)^3} + \frac{4 \sin^3 \frac{x}{3}}{x^3}\right)=\frac{L}{9}+\frac{4}{27}$$
From obtained equation $L=\frac{1}{6}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3772486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in this equation gives us:
$$a\left( \frac{1 + \sqrt5}{2}\right)^9 + b\left( \frac{1 + \sqrt 5}{2}\right)^8 + 1 = 0$$
I was able to solve this far, but I gave up because the calculation past this point gets too tedious. The textbook has gone ahead and simplified this to
$$2^9 a + 2^8b(\sqrt 5 - 1) + (\sqrt5 - 1)^9 = 0$$
after which it simplifies to (divide by $2^8$ and solve the binomial expression)
$$2a + b(\sqrt 5 -1) = 76 - 34\sqrt5$$
Is there a more elegant way to solve this problem? Preferably one that does not include the magical use of a calculator or the evaluation of that ugly binomial expansion?
|
Put $A = x^2 - x - 1$, $B = ax^9 + bx^8 + 1$ now we want $B/A$ to divide without remainder. We can subtract multiples of $A$ from $B$ to kill off high order terms and see what the remainder would be:
B
- a*x^7*A
= (a + b)*x^8 + a*x^7 + 1
- (a + b)*x^6*A
(2*a + b)*x^7 + (a + b)*x^6 + 1
- (2*a + b)*x^5*A
(3*a + 2*b)*x^6 + (2*a + b)*x^5 + 1
We find
Q = a*x^7 + (a + b)*x^6 + (2*a + b)*x^5
+ (3*a+2*b)*x^4 + (5*a + 3*b)*x^3 + (8*a + 5*b)*x^2
+ (13*a + 8*b)*x + (21*a + 13*b)
R = (34*a + 21*b)*x + 21*a + 13*b + 1
via B - Q A = R
So we need to solve the system
*
*$34a + 21b = 0$
*$21a + 13b + 1 = 0$
To get no remainder.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3773143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
}
|
Evaluate integral $\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}$ How can I evaluate this integral $$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\;\;?$$
My attempt:
I tried using substitution $x=\sec\theta$, $dx=\sec\theta\ \tan\theta d\theta$,
$$\int (\sec^2\theta-1)(\sec^3\theta-3\sec\theta)^{4/3}
\sec\theta\ \tan\theta d\theta $$
$$=\int \tan^2\theta \sec^4\theta(1-3\cos^2\theta)^{4/3} \sec\theta\ \tan\theta d\theta $$
$$=\int \tan^3\theta \sec^5\theta(1-3\cos^2\theta)^{4/3}\ d\theta $$
$$=\int\dfrac{ \sin^3\theta}{ \cos^8\theta}(1-3\cos^2\theta)^{4/3}\ d\theta $$
I can't see if this substitution will work or not. This has become so complicated.
Please help me solve this integral.
|
$$t=x^{3}-3x\\
I=\frac{1}{3}\int t^{\frac{4}{3}}dt=\frac{1}{7}t^{2}\sqrt[3]{t}+c=\frac{1}{7}(x^{3}-3x)^{2}\sqrt[3]{x^{3}-3x}+c$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3773220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
}
|
How to prove $\int_0^\pi\biggl |\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le c(k+1)$ for some constant $c$? Let $|a_i|\le1$, prove that $$\int_0^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr|\,dx \le c(k+1)$$ for some constant $c$.
I tried to solve the question as follows
$$\int_0^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le
\int_0^\pi \sum_{i=0}^k \biggl| \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr|\,dx \\=
\sum_{i=0}^k|a_{i}|\int_0^\pi \biggl| \frac{\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr|\,dx \le
\sum_{i=0}^k\int_0^\pi \biggl| \frac{\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx.$$
Then it looks like a classic equality about Lebesgue constant,
$$\frac{1}{\pi}\int_{-\pi}^\pi \biggl| \frac{\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr|\,dx=\frac{4}{\pi^2}\log i+O(1).$$
However, it didn't work. The above process of estimation doesn't seem to be careful enough.
How to make more careful and effective estimates to solve the question?
Thank you so much!
|
On the other forum, I found the answer to this question.The website link is here, https://www.zhihu.com/question/410940277/answer/1400850373.
I copied the proof roughly here.
$$\int_0^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx = \int_0^\frac{\pi}{k+1} \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx+\int_\frac{\pi}{k+1}^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx$$
It is easy to get that $$\frac{\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x} \le i+\frac12.$$
Then $$\int_0^\frac{\pi}{k+1} \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le \frac{\pi}{k+1}\sum_{i=0}^{k}|a_i|\biggl(i+\frac12\biggr) \le \frac\pi2(k+1).$$
According to Cauchy-Schwarz inequality,we have
$$\int_\frac{\pi}{k+1}^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le \sqrt{\int_\frac{\pi}{k+1}^\pi \biggl(\frac{1}{2\sin \frac{1}{2}x}\biggr)^2 \, dx}\sqrt{\int_\frac{\pi}{k+1}^\pi \biggl(\sum_{i=0}^k a_i\sin(i+\frac{1}{2})x\biggr)^2 \, dx}.$$
After calculation,we get that $$\int_\frac{\pi}{k+1}^\pi \biggl(\frac{1}{2\sin \frac{1}{2}x}\biggr)^2 \, dx \lt 2k+2,$$
and $$\int_\frac{\pi}{k+1}^\pi \biggl(\sum_{i=0}^k a_i\sin(i+\frac{1}{2})x\biggr)^2 \, dx \le \int_{-\pi}^\pi \biggl(\sum_{i=0}^k a_i\sin(i+\frac{1}{2})x\biggr)^2 \, dx=\pi\sum_{i=0}^{k}a_i^2 \le \pi(k+1).$$
Therefore, $$\int_\frac{\pi}{k+1}^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le \sqrt{2\pi}(k+1).$$
In brief, we come to the conclusion $$\int_0^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le (\frac\pi2+\sqrt{2\pi})(k+1).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3774467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $a$, $b$, $c$, $d$ are positive reals so $(a+c)(b+d) = 1$, prove the following inequality would be greater than or equal to $\frac {1}{3}$.
Let $a$, $b$, $c$, $d$ be real positive reals with $(a+c)(b+d) = 1$. Prove that $\frac {a^3}{b + c + d} + \frac {b^3}{a + c + d} + \frac {c^3}{a + b + d} + \frac {d^3}{a + b + c} \geq \frac {1}{3}$.
So I've been trying to solve this problem, and I've been trying to find a way to modify it into using AM-GM. The issue is that the $(a+c)(b+d) = 1$ is really throwing me off, as I haven't dealt with any inequalities that have used that as a condition yet (most other conditions I have seen go along the lines of $abcd = 1$ or something like that), and I'm not sure how exactly to deal with this inequality. Does anyone have any ideas?
|
Also, we can use Holder here:
$$\sum_{cyc}\frac{a^3}{b+c+d}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c+d)}=\frac{(a+b+c+d)^2}{12}\geq\frac{\left(2\sqrt{(a+c)(b+d)}\right)^2}{12}=\frac{1}{3}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3774895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Evaluating $\int \left(\frac{1}{3x}-2\sec^2\left(\frac x2\right)-e^{-2x+3}\right)dx$ Given that $\displaystyle f(x)=\frac{1}{3x}-2\sec^2\left(\frac{x}{2}\right)-e^{-2x+3}$, evaluate $\int f(x) \mathrm{d}x$.
Attempt: $$\begin{aligned} \int f(x) \mathrm{d}x&=\int \left[ \frac{x}{3}-2\sec^2\left(\frac{x}{2}\right)-e^{-2x+3}\right]\mathrm{d}x \\ &=\frac{x^2}{6}-4\tan\left(\frac{x}{2}\right)-\frac{e^{-2x+3}}{-2}+C \\ &= \frac{x^2}{6}-4\tan\left(\frac{x}{2}\right)+\frac{e^{-2x+3}}{2}+C \end{aligned}$$
Did I find the antiderivative correctly? Any feedback would be appreciated.
|
Instead of evaluating
$$\int \left(\frac{1}{3x}-2\sec^2\left(\frac x2\right)-e^{-2x+3}\right)dx$$
You seem to have evaluated
$$\int \left(\frac{x}{3}-2\sec^2\left(\frac x2\right)-e^{-2x+3}\right)dx$$
Your evaluation for the incorrect integrand is correct though, and to get the actual answer, you only need to change the first term from $\frac{x}{3}$ to $\frac{1}{3x}$, whose integration will involve the natural logarithm.
Can you finish?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3775449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
The triangle that has the longest possible smallest side of a triangle inscribed in a unit square is equilateral The points $A$, $B$ and $C$ lie on the sides of a square of side $1$ cm and no two points lie on the same side. Show that the length of at least one side of the triangle $ABC$ must be less than or equal to $(√6−√2)$ cm.
The given result can easily be derived by asserting that (1) the triangle is equilateral and (2) touches one of the corners.
This is a perfectly intuitive result, but how can I prove (1) and (2) rigorously?
|
Here is an entirely analytical attempt. Let $(x_1,0), (1,y),(x_2,1)$ be the vertices of the triangle inscribed in the unit square $[0,1]^2$. Since only one of them can be in a vertex of the square, we will allow only $x_1 \in [0,1\rangle$ and $x_2,y \in \langle 0,1\rangle$. Also we can assume $x_1 \le x_2$. The length of the smallest side is then given by the function $f : \{(x_1,x_2,y) \in [0,1\rangle \times \langle 0,1\rangle^2 : x_1\le x_2\}\to \Bbb{R}$ defined as
$$f(x_1,x_2,y) = \min\left\{\sqrt{(x_2-x_1)^2+1},\sqrt{(1-x_1)^2+y^2},\sqrt{(1-x_2)^2+(1-y)^2}\right\}.$$
Let $(x_1,x_2,y)$ be the triple which maximizes $f$. Assume that this triangle is not equilateral. Then we can offset the vertices a bit so that the same side remains the smallest but it is a bit larger than before. For example if
$$\sqrt{(x_2-x_1)^2+1} < \sqrt{(1-x_1)^2+y^2}, \sqrt{(1-x_2)^2+(1-y)^2}$$
then by continuity there exists $\varepsilon > 0$ such that
$$\sqrt{((x_2+\varepsilon)-x_1)^2+1} < \sqrt{(1-x_1)^2+y^2}, \sqrt{(1-(x_2+\varepsilon))^2+(1-y)^2}$$
and hence $$f(x_1,x_2+\varepsilon,y) =\sqrt{((x_2+\varepsilon)-x_1)^2+1} > \sqrt{(x_2-x_1)^2+1} = f(x_1,x_2,y)$$
which contradicts maximality of $(x_1,x_2,y)$. Similarly in the other cases.
We conclude that the triangle is equilateral, so in particular $$f(x_1,x_2,y) = \sqrt{(x_2-x_1)^2+1}=\sqrt{(1-x_1)^2+y^2}=\sqrt{(1-x_2)^2+(1-y)^2}.$$
The term $\sqrt{(1-x_1)^2+y^2}$ is maximized when $x_1 = 0$ (since $y=1$ is not allowed). Therefore
$$f(x_1,x_2,y) = \sqrt{x_2^2+1}=\sqrt{1+y^2}=\sqrt{(1-x_2)^2+(1-y)^2}.$$
and hence $y=x_2$ and $1+y^2=2(y-1)^2$. This yields $$x_1=0,\quad x_2=y=2-\sqrt{3}$$
and $f(x_1,x_2,y) = \sqrt{1+y^2} = \sqrt{6}-\sqrt{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3777495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Maximum value of $|z|$ given $\lvert z-\frac 4z \rvert = 8$? The question is $$ \left|z-\frac 4z \right| = 8$$ Find the max value of $ |z|$
You know how the triangle inequality is:
$$ \bigg| | z_1 | - | z_2| \bigg| \leqq | z_1 \pm z_2 | \leqq | z_1 | + | z_2 | $$
The solutions used only the left hand side inequality, and also ignoring the absolute values outside of $ | z_1 | - | z_2 | $, i.e. they solved
$ | z | - \left| \frac 4z \right| \leqq 8$ to obtain the answer $ | z |_{max} = 4 + 2 \sqrt{5} $
I am confused about this in two ways, firstly, the way they solved it aren't they assuming here that $| z | \geqq | \frac 4z |$ ? Also can you just ignore the right hand side inequality?
|
Redo:
We have three potential inequalities
$|z| - |\frac z4| \le ||z| - |\frac z4|| \le |z -\frac z 4|=8$ or
*
*$|z| -|\frac z4| \le 8$.
$-|z| + |frac z4| \le ||z| - |\frac z4|| \le |z -\frac z 4|=8$ or
*$|\frac z4| -|z| \le 8$.
$8=|z-\frac z4|\le |z| + |\frac z4|$ or
*$|z| +|\frac z4| \ge 8$.
All three of these equations are true for any possible value of $z$ where $|z +\frac 4z| =8$.
The first yields $|z|\le 4 + 2\sqrt 5$ AND $|z|\ge 2\sqrt 5-4$ so $2\sqrt 5-4\le |z|\le 4+2\sqrt 5$. That is always true.
The second yields $|z|\ge 2\sqrt 5-4$ OR that $|z|\le -4 -\sqrt 5$ but that is impossible. So $|z|\ge 2\sqrt 5 -4$. That is always true.
The third yields $|z| \ge 2\sqrt 3$ OR $|\le 4-2\sqrt 3$.
Putting these results together and noting that $2\sqrt 5-4 \le 4-2\sqrt 3 \iff \sqrt 5 + \sqrt 3 < 4 \iff 5+2\sqrt {15}+3 < 16\iff 8+2\sqrt {15}< 16=8+2\sqrt {16}$ which it is we get.
$2\sqrt 5 - 4\le |z| \le 4-2\sqrt 3$ or $4+2\sqrt 3 \le |z| \le 4+ 2\sqrt 5$
So $|z|_{max} = 4+2\sqrt 5$.
As we were only asked for the maximum value and we get that from inequality 1), inequality 1) is the only one we need to consider.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3778267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Laurent Series of $\frac{1}{(z-1)(z-2)}$ for $|z-1|>1$ I am calculating Laurent series of
$$
f(z)=\frac{1}{(z-1)(z-2)}
$$
which converges when $|z-1|>1$.
I started as
$$
\frac{1}{(z-1)(z-2)}=\frac{1}{z-1}\cdot\frac{1}{(z-1)-1}=\frac{1}{(z-1)^2-(z-1)}.
$$
Is this even the right way to transform function to some kind of geometric series, which can be easily expandet as Taylor series? How to calculate this Laurent seires?
|
$$f(z) = \frac{1}{(z-1)(z-2)}$$
We write $f(z)$ in its partial fraction expansion, and then we expand $f(z)$ in powers of $\frac{1}{z-1}$:
$$\begin{aligned}
f(z) &= \frac{1}{z-2}-\frac{1}{z-1} \\
&= \frac{1}{(z-1)-1} - \frac{1}{z-1} \\
&= \frac{1}{z-1} \frac{1}{1-\frac{1}{z-1}}- \frac{1}{z-1}\\
&= \frac{1}{z-1} \left( \frac{1}{z-1} + \frac{1}{(z-1)^2} + \cdots \right) \\
&= \frac{1}{(z-1)^2} + \frac{1}{(z-1)^3} + \cdots
\end{aligned}$$
This is the Laurent series for $f(z)$ in the region $|z-1|>1.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3780020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Proving $\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$
Proving $\displaystyle\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$
My atempt:
\begin{align*}
\int_0^1 \int_0^1\frac{-x\ln(xy)}{1-x^2y^2} \, dx \, dy &=\int_0^1I_x(y)\,dy\\[6pt]
\text{where }I_x(y)=\int_0^1\frac{-x\ln(xy)}{1-x^2y^2} \, dx
\end{align*}
\begin{align*}
I_x(y)&=\int_0^1\frac{-x\ln(xy)}{2(1-xy)}-\frac{x\ln(xy)}{2(1+xy)} \, dx\\[6pt]
&=\frac{1}{2}\int_0^1\frac{-x\ln(xy)}{1-xy}\, dx+\frac{1}{2} \int_0^1\frac{-x\ln(xy)}{1+xy} \, dx\\[6pt]
&=\frac{1}{2} \sum_{n=0}^\infty-y^n \int_0^1x^{n+1}\ln(xy)\,dx+\frac{1}{2} \sum_{n=0}^\infty(-1)^{n+1} \int_0^1y^nx^{n+1}\ln(xy) \, dx\\[6pt]
&=\frac{1}{2} \sum_{n=0}^\infty-y^n\left(\frac{\ln(y)}{n+2}-\int_0^1 \frac{x^{n+1}}{(n+2)y} \, dx\right)+\frac{1}{2} \sum_{n=0}^\infty (-1)^{n+1} y^n \left(\frac{\ln(y)}{n+2}-\int_0^1\frac{x^{n+1}}{(n+2)y} \, dx\right)\\[6pt]
&=\frac{1}{2} \sum_{n=0}^\infty \frac{y^n\ln(y)}{n+2} ((-1)^{n+1}-1) +\frac{1}{2} \sum_{n=0}^\infty \frac{y^{n-1}}{(n+2)^2}(1-(-1)^{n+1})
\end{align*}
|
\begin{align}
\int _0^1\int _0^1-\frac{x\ln \left(xy\right)}{1-x^2y^2}\:dx\:dy
&=\int _0^1\int _0^1\left(\underbrace{-\frac{x\ln \left(x\right)}{1-x^2y^2}-\frac{x\ln \left(y\right)}{1-x^2y^2}}_{t=x^2}\right)\:dx\:dy
\\[3mm]
&=\int _0^1\int _0^1\left(-\frac{1}{4}\underbrace{\frac{\ln \left(t\right)}{1-ty^2}}_{K}-\frac{1}{2}\frac{\ln \left(y\right)}{1-ty^2}\right)\:dt\:dy
\\[3mm]
&=\frac{1}{4}\int _0^1\frac{\:\operatorname{Li}_{2}\left(y^2\right)}{y^2}\:dy+\frac{1}{2}\int _0^1\frac{\ln \left(y\right)\ln \left(1-y^2\right)}{y^2}\:dy
\\[3mm]
&=\frac{1}{4}\sum _{k=1}^{\infty }\frac{1}{k^2\left(2k-1\right)}+\frac{1}{2}\sum _{k=1}^{\infty }\frac{1}{k\left(2k-1\right)^2}
\\[3mm]
&=\ln \left(2\right)-\frac{1}{4}\zeta \left(2\right)+\frac{3}{4}\zeta \left(2\right)-\ln \left(2\right)
\\[3mm]
&=\frac{1}{2}\zeta \left(2\right)
\end{align}
I used $\displaystyle \int _0^1\frac{a\ln ^n\left(t\right)}{1-at}\:dt=\left(-1\right)^nn!\operatorname{Li}_{n+1}\left(a\right)$ to evaluate $K$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3783459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
How to find the determinant of this $6\times 6$ X-matrix? This question was asked in my quiz and i was unable to solve it, so I am asking it here.
Find the value of determinant of this particular matrix .
$$\begin{pmatrix}1&0&0&0&0&2\\0&1&0&0&2&0\\0&0&1&2&0&0\\0&0&2&1&0&0\\0&2&0&0&1&0\\2&0&0&0&0&1\end{pmatrix}$$
I have no clue on how this kind of matrices can be solved. Can anyone give a general strategy on how to solve matrices whose size are greater that $3\times 3$?
That would be really helpful.
|
Permuting the rows and columns, we obtain a block diagonal matrix.
$$\det \begin{bmatrix} \color{red}{1} & 0 & 0 & 0 & 0 & \color{red}{2}\\ 0 & \color{orange}{1} & 0 & 0 & \color{orange}{2} & 0\\ 0 & 0 & \color{magenta}{1} & \color{magenta}{2} & 0 & 0\\ 0 & 0 & \color{magenta}{2} & \color{magenta}{1} & 0 & 0\\ 0 & \color{orange}{2} & 0 & 0 & \color{orange}{1} & 0\\ \color{red}{2} & 0 & 0 & 0 & 0 & \color{red}{1}\end{bmatrix} = \det \begin{bmatrix} \color{red}{1} & \color{red}{2} & & & & \\ \color{red}{2} & \color{red}{1} & & & & \\ & & \color{orange}{1} & \color{orange}{2} & & \\ & & \color{orange}{2} & \color{orange}{1} & & \\ & & & & \color{magenta}{1} & \color{magenta}{2} \\ & & & & \color{magenta}{2} & \color{magenta}{1} \end{bmatrix} = \left( \det \begin{bmatrix} 1 & 2\\ 2 & 1\end{bmatrix} \right)^3 = (-3)^3 = \color{blue}{-27}$$
matrices block-matrices permutation-matrices determinant
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3784764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
}
|
How can we take the derivative of this function: $y = \frac{x}{x^2+1}$ from first principles (using the limit definition of the derivative)? I was taking the derivative of the function: $y = \frac{x}{x^2+1}$. I know that we can solve it by the quotient rule. But I tried using the limit definition of differentiation. This is how I did it:
$$\lim_{h \to 0} \frac{(x+h)/\left((x+h)^2+1\right) - x/(x^2+1)}{h}$$
$$=\lim_{h \to 0}\frac{(x+h)/(x^2+h^2+2xh+1) - x/(x^2+1)}{h}$$
Then I multiplied both the denominator and numerator by the common factor $(x^2+h^2+2xh+1)(x^2+1)$ and expanded:
$$=\lim_{h \to 0} \ \Biggl(\frac{x^3+x+x^2h+h - x^3-xh^2-2x^2h-x}{x^4+x^2h^2+2x^3h+x^2+x^2+h^2+2xh+1}\Biggr) \cdot \frac{1}{h}$$
I then simplified the expression:
$$=\lim_{h \to 0} \ \frac{h-x^2h-xh^2}{x^4h + x^2h^3+2x^3h^2+2x^2h+h^3+2xh^2+h}$$
Now what is to be done? It seems that I am very far from the actual derivative.
|
Factor out $h$ from the numerator and the denominator. You can then cancel the factored term:
$$\frac{1-x^2-xh}{x^4+x^2h^2+2x^3h+2x^2+2xh+h^2+1}$$
Then take the limit as $h$ tends to $0$ and you get:
$$\frac{1-x^2}{x^4+2x^2+1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3785924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Prove $\int_0^{\infty} \frac{\ln^2(x^3+1)}{x^3+1} dx = \frac{\sqrt{3} \pi}{18} \left(9\ln^2(3)+4\psi ^{\prime} \left(\frac{2}{3}\right)\right)-\ldots$ Prove $$\int_0^{\infty} \frac{\ln^2{(x^3+1)}}{x^3+1} \; \mathrm{d}x = \frac{\sqrt{3} \pi}{18} \left(9\ln^2{(3)}+4\psi ^{\prime} \left(\frac{2}{3}\right)\right)-\frac{\pi^3 \sqrt{3}}{54}-\frac{\pi}{3}\ln{(3)}$$
I tried feynman method with $ I(a)=\int_0^{\infty} \frac{\ln^2\left(ax^3+1\right)}{x^3+1} \; \mathrm{d}x$ but this got ugly because we need to differentiate wrt a twice I think. I also try to factor $x^3+1$ and either partial fraction decomposition denominator or log property for numerator but these did not work. I am not sure what to do now. Im not very good with contour integration so can people who respond try to use real methods?? Any help is appreciated
|
Consider the parameterized integral $I(a)$ where the integral in question is equal to $I''(1)$:
$$I(a)=\int_0^{\infty} \frac{1}{\left(x^3+1\right)^a} \; \mathrm{d}x$$
First, let $x^3 \to x$:
\begin{align}
I(a) &= \frac{1}{3} \int_0^{\infty} \frac{t^{-\frac{2}{3}}}{(t+1)^a} \notag \\
& \; \; = \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)}{\Gamma\left(a\right)} \notag
\end{align}
Where we used the definition of the beta function.
Now we will find $I''(1)$:
\begin{align}
I'(a) &= \frac{d}{da} \left( \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)}{\Gamma\left(a\right)} \right) \notag \\
&= \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)\left(\psi\left(a-\frac{1}{3}\right)-\psi\left(a\right)\right)}{\Gamma\left(a\right)} \notag \\
I''(a) &= \frac{d}{da} \left( \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)\left(\psi\left(a-\frac{1}{3}\right)-\psi\left(a\right)\right)}{\Gamma\left(a\right)} \right) \notag \\
&= \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)\left(\psi\left(a-\frac{1}{3}\right)^2-2\psi\left(a\right)\psi\left(a-\frac{1}{3}\right)+\psi\left(a\right)^2+\psi^{\prime}\left(a-\frac{1}{3}\right)-\psi^{\prime}\left(a\right)\right)}{\Gamma\left(a\right)} \notag \\
I''(1) &= \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(\frac{2}{3}\right)\left(\psi\left(\frac{2}{3}\right)^2-2\psi\left(1\right)\psi\left(\frac{2}{3}\right)+\psi\left(1\right)^2+\psi^{\prime}\left(\frac{2}{3}\right)-\psi^{\prime}\left(1\right)\right)}{\Gamma\left(1\right)} \notag \\
&= \frac{2\pi}{3\sqrt{3}}\left(-\frac{\pi^2}{12}+\frac{9\ln^2{(3)}}{4}-\frac{\pi \sqrt{3} }{2} \ln{(3)} + \psi^{\prime} \left(\frac{2}{3}\right)\right) \notag \\
&= \boxed{\frac{\sqrt{3} \pi}{18} \left(9\ln^2{(3)}+4\psi ^{\prime} \left(\frac{2}{3}\right)\right)-\frac{\pi^3 \sqrt{3}}{54}-\color{red}{\frac{\pi^2}{3}\ln{(3)}}} \notag
\end{align}
It appears that the result of the integral you provided is slightly off (see the red term). Wolfram Alpha agrees with the answer I provided as well.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3788447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Find limit of $f(x)$ as $x$ tends to $0$ I need some help answering this question:
$$f(x) = \frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}$$
find the limit of $f(x)$ as $x$ tends to $0$ by writing $f(x)$ as a quotient of two powers series.
I have so far:
$$\frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{x(x + \frac{x^3}{3!}+\cdots)}= \frac{(x+\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{(x^2 + \frac{x^4}{3!}+\cdots)}$$
but I don't know how to reduce this further.
|
We have that $$ \lim_{x \to 0} \frac{\sinh(x)}{\cosh(x)} - \frac{1}{x} = \lim_{x \to 0} \frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{x(x + \frac{x^3}{3!}+\cdots)}$$ $$= \lim_{x \to 0} \frac{(x+\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{(x^2 + \frac{x^4}{3!}+\cdots)} = \lim_{x \to 0} \frac{(\frac{x^3}{2!} + \frac{x^5}{4!} + \cdots) - (\frac{x^3}{3!} + \cdots) }{ ({x^2} + \frac{x^4}{3!} + \cdots) }$$ $$ = \lim_{x \to 0} \frac{(\frac{x}{2!}+\frac{x^3}{4!}+\cdots) - (\frac{x}{3!} + \cdots) }{(1 + \frac{x^2}{3!} + \cdots)} = 0 $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3788612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
}
|
What is the range of $\vec{z}^{ \mathrm{ T } }A\vec{z} $? Let A be a 3 by 3 matrix
$$\begin{pmatrix}
1 & -2 & -1\\
-2 & 1 & 1 \\
-1 & 1 & 4
\end{pmatrix}$$
Then we have a real-number vector $\vec{ z }= \left(
\begin{array}{c}
z_1 \\
z_2 \\
z_3
\end{array}
\right)$ such that
$$\vec{z}^{ \mathrm{ T } }\vec{z} = 1$$
$$z_1+z_2+z_3=1$$
What is the range of $\vec{z}^{ \mathrm{ T } }A\vec{z} $?
I have found that $A$'s eigenvalues are -1,2, and 5 and eigenvectors are
$\left(
\begin{array}{c}
1 \\
1 \\
0
\end{array}
\right)$$\left(
\begin{array}{c}
1 \\
-1 \\
1
\end{array}
\right)$$\left(
\begin{array}{c}
-1 \\
1 \\
2
\end{array}
\right)$ for each.
Can anyone help me?
|
If we eliminate $z_3$ by replacing it with $1-z_1-z_2$, you want to find the minimum and maximum of
$$\{z^TAz + b^Tz + c : zQz+q^Tz = 0\}$$
with
$$A=\begin{pmatrix}7 & 2 \\ 2 & 3\end{pmatrix}, \; b=\begin{pmatrix}-10\\-6\end{pmatrix}, \; c=4, \; Q=\begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix}, \; q=\begin{pmatrix}-2\\ -2\end{pmatrix}.$$
Via the Lagrangian we find that an extremum must satisfy $2Az+b+\lambda(2 Qz + q)=0$ and $z^TQz+q^Tz = 0$, but I do not see an easy solution. The problem is now in a format that allows for this numerical procedure.
Instead I will go on and eliminate $z_2$ to get an unconstrained problem in $z_1$. The constraint is $2z_2^2+(2z_1-2)z_2+(2z_1^2-2z_1)=0$, so $z_2=\frac{1}{2}(1-z_1) \pm \sqrt{\frac{1}{4}-\frac{3}{4}z_1^2+\frac{1}{2}z_1}$. Plugging this into the objective function no longer gives a nice expression. Numerical analysis shows that the positive branch has a maximum of $41/9$ at $z_1=-1/3$ and a minimum of $\approx-0.53$ at $z_1\approx 0.538$ while the negative branch has a maximum of $\approx 4.92$ at $z_1 \approx -0.29$ and a minimum of $1$ at $z_1=1$.
So the range is approximately $-0.53$ to $4.92$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3790521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Plese help me to express this Jacobian determinant $\frac{ \partial (x,y) }{ \partial (u,v) }$ only using u and v We define u and v as
$$u=\frac{2x}{x+y+1}$$
$$v=\frac{2y}{x+y+1}$$
I'm trying to get Jacobian determinant $\frac{ \partial (x,y) }{ \partial (u,v) }$ and express it only using $u$ and $v$.
What I have tried
$$x=\frac{-uy-u}{u-2}=-\frac{2y}{v^2}$$
$$y=-\frac{2x}{u^2}=\frac{-vx-v}{v-2}$$
$$\frac{ \partial x }{ \partial u }=\frac{2y+2}{(u-2)^2}$$
$$\frac{ \partial y }{ \partial u }=-\frac{2x}{u^2}$$
$$\frac{ \partial x }{ \partial v }=\frac{2y}{v^2}$$
$$\frac{ \partial y }{ \partial v }=-\frac{2x+2}{(v-2)^2}$$
But even if you calculate the determinant with these values, it end up with result below and fail to cancel x and y out.
$$\frac{-16yu^2x+64yux-64yx+4yu^2v^2+16yuxv^2-16yxv^2+16yu^2xv-64yuxv+64yxv+4u^2xv^2+4u^2v^2}{u^2v^2\left(u-2\right)^2\left(v-2\right)^2}$$
Can anyone help me?
|
Calculate $\frac{\partial(u,v)}{\partial(x,y)}$ first
$$\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \frac{2}{x+y+1} - \frac{2x}{(x+y+1)^2} & \frac{-2x}{(x+y+1)^2} \\ \frac{-2y}{(x+y+1)^2} & \frac{2}{x+y+1}-\frac{2y}{(x+y+1)^2} \\ \end{vmatrix} = \frac{4}{(x+y+1)^3}$$
Then notice that
$$u+v = 2-\frac{2}{x+y+1} \implies x+y+1 = \frac{2}{2-u-v}$$
which means
$$\frac{\partial(x,y)}{\partial(u,v)} = \frac{(x+y+1)^3}{4} = \frac{2}{(2-u-v)^3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3791897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
If $V_n(a)$ counts sign changes in the sequence $\cos a, \cos2a,\cos3a,\ldots,\cos na,$ show that $\lim_{n\to\infty}\frac{V_n(a)}n=\frac{a}\pi$
Let $0\leq\alpha\leq \pi $. $V_n (\alpha) $ denote the number of sign changes in the sequence $\cos\alpha,\cos2\alpha,\cos3\alpha,\ldots,\cos n\alpha $. Then prove that $$\lim\limits_{n\to\infty}\dfrac{V_n (\alpha)}{n}=\dfrac{\alpha}{\pi}.$$
I saw a hint where $\dfrac{V_n (\alpha)}{n}$ is considered as the probability. I mean how this expression is a probability of something. If it is, how can I progress further in this way?
Update: I have a solution to this problem
In $n\alpha$ rotation the number of times full circle rotation occures $=\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor$
In one full circle rotation sign change occures 2 times. Hence in $\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor$ full rotation sign change occures $=2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor$
Now the rest angle is $n\alpha-\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor\times2\pi$
If we consider 0 as a change of sign in case of $\cos\left( \dfrac{\pi}{2}\right)$ and $\cos\left(\dfrac{3\pi}{2}\right)$ then:-
(1) If $0\leq n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi<\dfrac{\pi}{2 }$ sign changes 0 times
(2) If $\dfrac{\pi}{2 }\leq n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi<\dfrac{3\pi}{2 }$ sign changes 1 times
(3) If $\dfrac{3\pi}{2 }\leq n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi<2\pi$ sign changes 2 times
Let $f$ be a function such that $$f\left(\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor\right)=\begin{cases}0,\text{ when }\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=0\\ 1,\text{ when }\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=1\\ 1,\text{ when }\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=2\\ 2,\text{ when } \left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=3\end{cases}$$
Therefore $\dfrac{V_n(\alpha)}{n}=\dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor+ f\left(\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor\right)}{n}$
Hence $$\dfrac{V_n(\alpha)}{n}\geq \dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor}{n}$$ and $$\dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor+ 2}{n}\leq \dfrac{V_n(\alpha)}{n}$$
$\lim\limits_{n\to \infty}\dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor}{n}=\dfrac{\alpha}{\pi}$ and $\lim\limits_{n\to\infty} \dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor+ 2}{n}=\dfrac{\alpha}{\pi}$
Hence by Sandwich Theorem We get $\lim\limits_{n\to \infty}\dfrac{V_n(\alpha)}{n}=\dfrac{\alpha}{\pi}$ [Proved]
Is this correct?
|
HINT: let $ b_n\equiv n a \pmod {2\pi}$ indicate the angle formed with the $x$- axis in the $n^{th}$ term of the sequence. Assume that $b$ is uniformly distributed in the range between $0$ and $2\pi$.
Now firstly consider the case in which $0<b_n<\pi/2$ or $3\pi/2<b_n<2\pi$. In the next step, a change of sign will occur only if $b_{n+1}>\pi/2$. What is the probability that this occurs, given that $b_{n+1}=b_n+a$?
Then repeat the same considerations for the case in which $\pi/2<b_n<3\pi/2$. A change of sign will occur only if $b_{n+1}>3\pi/2$. What is the probability that this occurs?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3794745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .
What I Tried :
I used some clever ways to get $x + y + z = 26xyz$ , but I suppose we have some solution as a number .
All all $3$ to get :-
$$2(x + y + z) = 5xy + 6yz + 7zx$$
Or,
$$ 2(x + y + z) = (xy + xy + xy + xy + xy) + (yz + yz + yz + yz + yz + yz) + (zx + zx + zx + zx + zx + zx + zx)$$
That is,
$$ 2(x + y + z) = (xy + zx) + (xy + zx) + (xy + zx) + (yz + zx) + (yz + zx) + (yz + zx) + (yz + zx) + (xy + yz) + (xy + yz)$$
Now see that $(xy + zx) = x(y + z) = 6xyz$ , similarly $(yz + zx) = 5xyz$ and $(xy + yz) = 7xyz$
So
$$2(x + y + z) = 3(6xyz) + 4(5xyz) + 2(7xyz)$$
$$\Rightarrow (x + y + z) = \frac{52xyz}{2} = 26xyz$$
I tried till this , then I have no idea . Can anyone help?
|
Assuming $x,y,z \neq 0$, we have that,
$$x + y = 5xy \iff \frac1x+\frac1y =5$$
$$y + z = 6yz \iff \frac1y+\frac1z =6$$
$$z + x = 7zx \iff \frac1z+\frac1x =7$$
then solve for $1/x$, $1/y$, $1/z$.
The case $x=0 \lor y=0 \lor z=0$ is trivial.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3795093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Finding $a$ such that the complex solutions of $z^4 - 6z^3 + 11az^2 - 3(2a^2 + 3a - 3) z + 1 = 0$ form a parallelogram in the complex plane Find all values of the real number $a$ so that the four complex roots of
$$z^4 - 6z^3 + 11az^2 - 3(2a^2 + 3a - 3) z + 1 = 0$$ form the vertices of a parallelogram in the complex plane.
I set $z = x + yi$ due to the given knowledge that $z$ is complex. I then substituted it in, which gave me $$(x+yi)^4 - 6(x+yi)^3 + 11a(x+yi)^2 - 3(2a^2 + 3a - 3)(x+yi) + 1 = 0.$$ However, I'm afraid it's a bit too bashy for it to be viable, so are there any other ways I can get started on this question?
|
Expanding on @AlexeyBurdin's answer (in case anyone wants to see how to do it by hand), since $x=3/2$ we have $y^2+z^2=27/2-11a$ and$$6a^2+9a-9=27/2-3(y^2+z^2)=33a-27\implies a\in\{1,\,3\}.$$Each such $a$ works viz.$$1=x^4-(y^2+z^2)x^2+y^2z^2=81/16-(27/2-11a)9/4+y^2z^2,$$which we can simplify to obtain $y^2z^2$. We then have the sum and product of $y^2,\,z^2$, and solutions in $\Bbb C$ exist.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3800690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
If $(a-b^2)b>0$, then $\sqrt[3]{a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}+\sqrt[3]{a-\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}$ is rational From Hardy´s "A course of pure mathematics" 10th edition, problem 31 miscellaneous problems of chapter I.
If $(a-b^2)b>0$, then
$$
\sqrt[3]{a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}+\sqrt[3]{a-\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}
$$
is rational.
"Note from the book":
Each of numbers under a cube root is of the form
$$
\left( \alpha + \beta\sqrt{\frac{a-b^3}{3b}} \right)^3
$$
where $\alpha$ and $\beta$ are rational.
|
No proof, but making the start. From the hint we have $$\left( \alpha + \beta\sqrt{\frac{a-b^3}{3b}} \right)^3 = a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}} \\ =\frac{\beta \left(\beta^2(a-b^3)+9\alpha^2b\right)}{3b} \sqrt{\frac{a-b^3}{3b}} + \frac{\alpha\left(\beta^2(a-b^3) + \alpha^2b\right)}{b}$$
so by comparison we have the set of equations $${\alpha\left(\beta^2(a-b^3) + \alpha^2b\right)}=ab \tag{1}$$ $${\beta \left(\beta^2(a-b^3)+9\alpha^2b\right)} = {9b^3+a} \tag{2} \, .$$ Replacing $\beta^2(a-b^3)$ from (1) into (2) gives $$\beta=\frac{9b^3+a}{\frac{ab}{\alpha}+8\alpha^2 b} \, .$$ This shows that if $\alpha$ is rational, then so is $\beta$. Furthermore, if $(\alpha,\beta)$ is a solution to (1)+(2), then $(\alpha,-\beta)$ is a solution to (1)+(2) with $-\frac{9b^3+a}{3b}$ instead of $+$. So the sought after expressions for the cube roots are indeed "conjugate" to each other.
Now, plugging $\beta$ into (1) gives the equation $$\left(\frac{ab}{\alpha}+8\alpha^2b\right)^2\left(\frac{a}{\alpha}-\alpha^2\right)=\left(9b^3+a\right)^2 \left(\frac{a}{b}-b^2\right) \, .$$
Multiplying by $\alpha^3$ leads to a cubic polynomial in $\alpha^3$ for which rational solutions $\alpha=\frac{p}{q}$ are sought (gcd(p,q)=1). Again, inserting this expression into the polynomial and collecting by common denominator leads to the equation $$-64\,{b}^{3}{p}^{9}+48\,a{b}^{3}{p}^{6}{q}^{3}+ \left( 81\,{b}^{9}-63
\,a{b}^{6}-2\,{a}^{2}{b}^{3}-{a}^{3} \right) {q}^{6}{p}^{3}+{a}^{3}{b}
^{3}{q}^{9}=0 \, .$$
We can assume that $a,b$ are integers (otherwise multiply by a power of the common denominator of a and b), so reducing this equation mod $p^3$ gives $p|ab$ and mod $q^3$ gives $q|4b$.
to be continued.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3805796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Maximum value of $abc$ for $a+b+c=5$ and $a^2+b^2+c^2=11$
$a,b,c$ are three real numbers such that $a+b+c=5$ and $a^2+b^2+c^2=11$,
what's the maximum value of $abc$?
I thought of a way, $ab+bc+ca$ is not hard to find, $a,b,c$ satisfy the cubic equation $x^3 - 5 x^2 + 7 x - abc = 0$ , then use the discriminant of the cubic equation non-negative.
The discriminant of $x^3 + A x^2 + B x + C=0$ is
$A^2 B^2 - 4 B^3 - 4 A^3 C + 18 A B C - 27 C^2$
Is there an easier way?
|
Plugging in $c=5-a-b$ into the quadratic and simplifying shows that
$$b^2+(a-5)b+(a^2-5a+7)=0.\tag{1}$$
Because $b+c=5-a$ we see that $b$ and $c$ are precisely the roots of this quadratic, so
$$abc=a(a^2-5a+7)=a^3-5a^2+7a.\tag{2}$$
Then the quadratic $(1)$ has two real roots so its discriminant is positive, i.e.
$$0\leq(a-5)^2-4(a^2-5a+7)=-3a^2+10a-3=-(3a-1)(a-3),$$
which shows that $\tfrac13\leq a\leq3$. Hence checking $(2)$ for extrema on the interval $[\tfrac13,3]$ shows that it is maximal if and only if $a\in\{1,3\}$. By symmetry $a,b,c\in\{1,3\}$ and because $a+b+c=5$ it follows that $(a,b,c)=(1,1,3)$ up to permutation, so the maximum value of $abc$ is $3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3809100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
Solve $x^5 - 1 = y^2$ for integer values of $(x,y)$ .
Solve $x^5 - 1 = y^2$ for integer values of $(x,y)$ .
What I Tried :- I see that $x^5 = y^2 + 1$ , from here I can conclude that $x$ has to be positive , because if $x \leq 0$ , then $x^5 \leq 0$ , but $y^2 + 1 > 0$.
Also I thought that maybe $(x^\frac{5}{2} + 1)(x^\frac{5}{2} - 1) = 1$ would do the trick [which would have implied that both the terms are $(1,1)$ or $(-1,-1)$], but that dosen't necessarily mean that $x^\frac{5}{2}$ is an integer .
Then I see that :-
$x^5 - 1 = y^2$
$\rightarrow (x-1)(x^4 + x^3 + x^2 + x + 1) = y^2$
From here the only idea is that $x \neq 2$ , because if it is then it can't be a perfect square (it's a foolish idea though) .
But I tried playing this problem with many ways , but I am stuck at the same place . Can anyone help ?
Note :- It's given that answer is only $(1,0)$ , but how is it coming?
|
So
$$x^5=y^2+1=(y+i)(y-i).$$
Considering this modulo $4$ gives $y$ even and $x$ odd.
The gcd of $y\pm i$ in the Gaussian integers divides $2$ and $y^2+1$ (which is odd)
so it is $1$. As the Gaussian integers has unique factorisation
and has four units,
both $y\pm i$ are fifth powers, so
$$y+i=(a+bi)^5=(a^5-10a^3b^2+5ab^4)+(5a^4b-10a^2b^3+b^5)i.$$
So $b\mid 1$ and $b=\pm1 $ and
$$5a^4-10a^2+1=\pm1.$$
The only integer solution of this is $a=0$ leading to $y=0$ and $x=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3813107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Proving continuous differentiability over the domain I have the following problem from Tao.
Let $f: \mathbb R^2 \to\mathbb R$ be the function defined by $f(x,y)=\dfrac {xy^3}{x^2+y^2}$ when $(x,y) \neq (0,0)$ and $f(0,0):=0$. Show that $f$ is continuously differerentiable.
I have $\dfrac{\partial{f(x,y)}}{\partial{x}} = \dfrac{y^5 - x^2y^3}{(x^2+y^2)^2}$ when $(x,y) \neq (0,0)$ and $\dfrac{\partial{f(x,y)}}{\partial{x}} = (0,0)$ when $(x,y) \neq (0,0)$.
I am not sure how to proceed with proving continuity, because it seems that the derivative is discontinuous. I am inclined to proceed with $\epsilon-\delta$ definition if possible, but I don't see the point.
|
Let $f(x,y)$ be given by
$$f(x,y)=\begin{cases}\frac{xy^3}{x^2+y^2}&,(x,y)\ne (0,0)\\\\0&,(x,y)=(0,0)\end{cases}$$
For $(x,y)\ne(0,0)$, we have
$$\begin{align}
\frac{\partial f(x,y)}{\partial x}&=\frac{y^3(y^2-x^2)}{(x^2+y^2)^2}\\\\
\frac{\partial f(x,y)}{\partial y}&=\frac{xy^2(3x^2+y^2)}{(x^2+y^2)^2}
\end{align}$$
Now, note that from the limit definition of the partial derivatives at $(0,0)$ we have
$$f_x(0,0)=f_y(0,0)=0$$
We now use the bounds $|x^2-y^2|\le x^2+y^2$, $|x|\le \sqrt{x^2+y^2}$ and $|y|\le \sqrt{x^2+y^2}$ to establish the estimates
$$\begin{align}
\left|\frac{\partial f(x,y)}{\partial x}\right|&=\left|\frac{y^3(y^2-x^2)}{(x^2+y^2)^2}\right|\\\\
&\le \frac{(x^2+y^2)^{5/2}}{(x^2+y^2)^2}\\\\
&=\sqrt{x^2+y^2}
\end{align}$$
Similarly, we have
$$\left|\frac{\partial f(x,y)}{\partial y}\right|=\left|\frac{x(3x^2y^2+y^4)}{(x^2+y^2)^2}\right|\le 3\sqrt{x^2+y^2}$$
Can you wrap things up now?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3813691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Diophantine equation:$x^5+x^4+1=p^y$
find all triplets $(x,y,p)$ satisfying $x^5+x^4+1=p^y$ where x, y are positive integers and p is a prime.
My attempt:I didn't know how to start. So I tried finding some triplets. Interestingly, $(1,1,3)$ satisfies the given equation, but I am not able to find any more.
Next, I tried factorising the equation i.e. $x^5+x^4+1=(x^2+x+1)(x^3-x+1)=p^y$
Now I am stuck. I tried considering the gcd of the common factors but it does not help.
Any ideas??
|
For anybody like me who is unable to view the solution at Diophantus Era begins! that Math Lover's question comment linked to because they're not a member of Brilliant and doesn't want to join, here is a solution.
First, apart from $x = 1$, which leads to the solution of $(1, 1, 3)$ you've already found, then both of the factors on the left are greater than $1$ and, thus, must be positive powers of $p$. This gives
$$x^2 + x + 1 \equiv 0 \pmod{p} \tag{1}\label{eq1A}$$
$$x^3 - x + 1 \equiv 0 \pmod{p} \tag{2}\label{eq2A}$$
Next, \eqref{eq2A} minus \eqref{eq1A} gives
$$x^3 - x^2 - 2x \equiv 0 \pmod{p} \implies x(x + 1)(x - 2) \equiv 0 \pmod{p} \tag{3}\label{eq3A}$$
This means $x \equiv 0 \pmod{p}$, $x \equiv -1 \pmod{p}$ or $x \equiv 2 \pmod{p}$. The first $2$ cases give in \eqref{eq1A} that $1 \equiv 0 \pmod{p} \implies p = 1$, which is not allowed. With the third case, \eqref{eq1A} gives $7 \equiv 0 \pmod{p} \implies p = 7$ is the only possibility.
Checking $x = 2$ itself shows it works to give on the left side $49 = 7^2$, so $(2, 2, 7)$ is another solution. Next, consider $x \gt 2$, so
$$x = 7z + 2 \tag{4}\label{eq4A}$$
for some integer $z \ge 1$. Substituting this into the $x^2 + x + 1$ factor, and letting it be equal to $7^m$ for some integer $m \ge 1$, gives
$$\begin{equation}\begin{aligned}
7^m & = (7z + 2)^2 + (7z + 2) + 1 \\
& = 49z^2 + 28z + 4 + 7z + 3 \\
& = 49z^2 + 35z + 7 \\
& = 7(7z^2 + 5z + 1)
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Thus, since $z \ge 1 \implies 7z^2 + 5z + 1 \gt 1$, then $7 \mid 7z^2 + 5z + 1$, so
$$5z + 1 \equiv 0 \pmod{7} \tag{6}\label{eq6A}$$
Next, letting $x^3 - x + 1 = 7^n$ for some integer $n \ge 1$ and using \eqref{eq4A} gives
$$\begin{equation}\begin{aligned}
7^n & = (7z + 2)^3 - (7z + 2) + 1 \\
& = (7)^3z^3 + 3(7^2)(2)z^2 + 3(7)(4)z + 8 - 7z - 1 \\
& = (7)^3z^3 + 6(7^2)z^2 + (11)(7)z + 7 \\
& = 7((7)^2z^3 + 6(7)z^2 + (11)z + 1)
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Similar to before, this results in
$$11z + 1 \equiv 0 \pmod{7} \implies 4z + 1 \equiv 0 \pmod{7} \tag{8}\label{eq8A}$$
Next, \eqref{eq6A} minus \eqref{eq8A} gives
$$z \equiv 0 \pmod{7} \tag{8}\label{eq9A}$$
However, this contradicts both \eqref{eq6A} and \eqref{eq8A} since it results in $1 \equiv 0 \pmod{7}$, showing there is no such positive integer $z$.
In summary, the only $2$ triplets $(x, y, p)$ that satisfy the equation are $(1, 1, 3)$ and $(2, 2, 7)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3813806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
How to find $\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$? I tried to solve it, but the answer I got was different from the answer given.
Answer given:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)} = \frac{n(2n+1)}{4(2n-1)(2n+3)}$$
My working:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$$
$$= \sum_{r=1}^{n} \Biggl[\frac{1}{16(2r-1)}+\frac{1}{8(2r+1)}-\frac{3}{16(2r+3)}\Biggl]$$
$$= \frac{1}{16} + \frac{1}{24} - \require{cancel} \cancel{\frac{3}{80}}$$
$$+ \frac{1}{48} + \require{cancel} \cancel{\frac{1}{40}} - \require{cancel} \cancel{\frac{3}{112}}$$
$$+ \require{cancel} \cancel{\frac{1}{80}} + \require{cancel} \cancel{\frac{1}{56}} - \require{cancel} \cancel{\frac{3}{144}}$$
. . .
. . .
. . .
$$+ \require{cancel} \cancel{\frac{1}{16(2n-3)}} + \require{cancel} \cancel{\frac{1}{8(2n-1)}} - \frac{3}{16(2n+1)}$$
$$+ \require{cancel} \cancel{\frac{1}{16(2n-1)}} + \frac{1}{8(2n+1)} - \frac{3}{16(2n+3)}$$
$$= \frac{1}{16} + \frac{1}{24} + \frac{1}{48} - \frac{3}{16(2n+1)} + \frac{1}{8(2n+1)} - \frac{3}{16(2n+3)}$$
$$= \frac{2n^{2}+6n+3}{4(2n+1)(2n+3)}$$
|
Substituting $n=1$, the "correct answer" gives a wrong result as mentioned in the comments.
Indeed $$\sum_{r=1}^{1}\frac{r}{(2r-1)(2r+1)(2r+3)}=\frac{1}{1\times3\times5}=\frac{1}{15}$$ but $$\frac{n(2n+1)}{4(2n-1)(2n+3)}\big|_{n=1}=\frac{3}{4\times1\times5}=\frac{3}{20}\neq \frac{1}{15}.$$
Also you should have:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$$
$$=\frac{1}{16}+\frac{1}{24}+\frac{1}{48}-\frac{3}{16(2n+1)}+\frac{1}{8(2n+1)}-\frac{3}{16(2n+3)}$$
$$=\frac{1}{8}-\frac{3(2n+3)}{16(2n+1)(2n+3)}+\frac{2(2n+3)}{16(2n+1)(2n+3)}-\frac{3(2n+1)}{16(2n+1)(2n+3)}$$
$$=\frac{2(2n+1)(2n+3)}{16(2n+1)(2n+3)}-\frac{2(4n+3)}{16(2n+1)(2n+3)}=\frac{(2n+1)(2n+3)-(4n+3)}{8(2n+1)(2n+3)}$$
$$=\frac{4n^{2}+8n+3-4n-3}{8(2n+1)(2n+3)}=\frac{n(n+1)}{2(2n+1)(2n+3)}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3820237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
How does one go about solving $arg(\frac{z-2i}{z-6}) = \frac{1}{2}\pi$ This should give $$\frac{z-2i}{z-6} = bi$$
but solving that gives me $$z = \frac{-2b +6b^2-6bi +2i}{1+b^2}$$
and substituting $z$ for $x + yi$ gives me $x = \frac{-2b +6b^2}{1+b^2}$ and $y=\frac{-6b +2}{1+b^2}$
And I have no clue how to continue now. The answer is suppose to be $x = -by$, but I have no clue how to get there (have I made a calculation error??)
EDIT: Thanks everyone!!
|
As you said solving gets you here:
$$z = \frac{-2b +6b^2-6bi +2i}{1+b^2}$$
and finally, substituting gets u here:
$$x = \frac{-2b +6b^2}{1+b^2}$$ and $$y=\frac{-6b +2}{1+b^2}$$
Just divide above two equations you get
$$x = \frac{-2b + 6b^2}{1 + b^2} = -\frac{2 - 6b}{1 + b^2}\cdot b = -by$$
$$\implies x+by=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3826976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Strange series that converges I'm trying to evaluate this series
$$S= \sum_{n=2}^{\infty} a_n \frac{\ln (n)}{n}$$
But I have some conditions on $a_n$ making the problem hard. Namely, $a_n=3$ for $n = 2 \mod 4$ and $a_n = -1$ otherwise.
Albeit 2 mod 4 =2 is just a number, a friend in the comments suggested that the 2[mod 4] takes values 2,6,10,... which of course makes sense. Meaning that $a_n$ will go like $(3 -1 -1 -1 +3 -1 -1 -1 +3 + ...)$ for $n=2,3,4 ...$.
I don't really know how to do convergence test when I have constraints on a summation constant like this. I have tried to rewrite the sum to a simpler form, tried to separate is at two sums for the two different $a_n$'s but I don't really know how to do it and there is where I'm stuck...
|
As I pointed out in the comments, this is Question B-4 from the 2017 William Lowell Putnam competition. For convenience, I am providing the first of two solutions provided from the link in my comment. I take no credit--only minor changes to phrasing were made.
The key insight is to define an auxiliary telescoping series with terms $$a_k = \frac{\log k}{k} - \frac{\log (k+1)}{k+1},$$ for which we trivially have $$\sum_{k=1}^\infty a_k = 0. \tag{1}$$ Since $a_k > 0$ for $k \ge 3$, we also see that $(1)$ is absolutely convergent.
Having constructed such a series permitting rearrangement of its terms, we next observe $$3a_{4k+2} + 2a_{4k+3} + a_{4k+4} = (a_{4k+2} + a_{4k+4}) + 2(a_{4k+2} + a_{4k+3}),$$ hence
$$\begin{align}
S &= \sum_{k=0}^\infty \left(3 \frac{\log(4k+2)}{4k+2} - \frac{\log(4k+3)}{4k+3} - \frac{\log(4k+4)}{4k+4} - \frac{\log(4k+5)}{4k+5}\right) \\
&= \sum_{k=0}^\infty (3a_{4k+2} + 2a_{4k+3} + a_{4k+4}) \\
&= \sum_{k=1}^\infty a_{2k} + \sum_{k=0}^\infty 2(a_{4k+2} + a_{4k+3}). \tag{2}
\end{align} $$
We next observe $$2(a_{4k+2} + a_{4k+3}) = \frac{\log(4k+2)}{2k+1} - \frac{\log(4k+4)}{2k+2} = a_{2k+1} + \left(\frac{1}{2k+1} - \frac{1}{2k+2}\right)\log 2.$$ Hence
$$\sum_{k=0}^\infty 2(a_{4k+2} + a_{4k+3}) = \sum_{k=0}^\infty a_{2k+1} + \log 2 \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} = \sum_{k=0}^\infty a_{2k+1} + \log^2 2. \tag{3}$$
Finally, putting $(1)$, $(2)$, and $(3)$ together yields the desired result:
$$S = \log^2 2 + \sum_{k=1}^\infty a_{2k} + \sum_{k=0}^\infty a_{2k+1} = \log^2 2 + \sum_{k=1}^\infty a_k = \log^2 2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3827440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
Epsilon Delta Proof Verification for $14 + \frac{12}{x^2}$ Can someone check if my steps are completely justified:
Question: Prove $\lim\limits_{x \to \infty} 14 + \frac{12}{x^2} = 14$
For all $\epsilon > 0 \ \exists \ N>0$ such that $|14+\frac{12}{x^2} -14| < \epsilon$ for all $x > N$
Let $N = \frac{2 \sqrt{3}}{\sqrt{\epsilon}}$, then $x > \frac{2 \sqrt{3}}{\sqrt{\epsilon}} \rightarrow x^2 > \frac{4 \times 3}{\epsilon} = \frac{12}{\epsilon} \rightarrow |\frac{1}{x^2}|<\frac{\epsilon}{12} \rightarrow|\frac{12}{x^2}|< \epsilon \rightarrow |14 +\frac{12}{x^2} -14| = |f(x) - L| < \epsilon $ QED
Thanks
|
Write your proof clearer:
Proof: Let $\varepsilon>0$ and define $N := \sqrt{\frac{12}{\varepsilon}}$. Then, for all $x > N$ we have $x^2 > \frac{12}\varepsilon$, and so
$$\bigg|\Big( 14 + \frac{12}{x^2} \Big) - 14 \bigg| = \frac{12}{x^2} < \varepsilon.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3830549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $a^2+b^2+c^2+d^2=4$ then $(a+2)(b+2)\geq cd$ Let $a,b,c,d$ be real numbers with $a^2+b^2+c^2+d^2=4$. Prove that $(a+2)(b+2)\geq cd$.
My approach: I have considered an expression $$\begin{aligned}(a+2)(b+2)-cd=&4+2(a+b)+(ab-cd)\\=&(a^2+b^2+c^2+d^2)+2a+2b+(ab-cd)\end{aligned}$$ I was trying to write it as the sum of squares but I failed.
Can anyone show how to solve this problem please?
|
Using the AM-GM inequality, we have
$$(a+2)(b+2) = \frac{4-a^2-b^2-c^2-d^2}{2}+\frac{(a+b+2)^2}{2}+\frac{c^2+d^2}{2}$$
$$\geqslant \frac{c^2+d^2}{2} \geqslant cd.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3830865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
How can I prove that $y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}>0$ when $x>0$ and $1I would like to prove that
$$y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}>0$$
for all real numbers $x > 0$ and $1 < y < 1.5$. This seems true when plotted on WolframAlpha, but I don't know how to prove it. I tried replacing some of the terms using the given inequalities to obtain a simpler function, but any perturbation I make seems to render the inequality untrue. How would you approach this problem?
|
Let $f(y)=y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}.$
Thus, $$f''(y)=-\frac{6xy^2}{(1+x^2)^2}+2x^3\left(\frac{y}{(1+y^2)^2}\right)'=$$
$$=-\frac{6xy^2}{(1+x^2)^2}+2x^3\left(\frac{1}{(1+y^2)^2}-\frac{4y^2}{(1+y^2)^3}\right)=$$
$$=-\frac{6xy^2}{(1+x^2)^2}+\frac{2x^3(1-3y^2)}{(1+y^2)^3}<0,$$
which says that $f$ is a concave function.
But the concave function gets a minimal value for extreme value of $y$,
which says that it's enough to prove our inequality for $y\in\{1,1.5\}.$
Can you end it now?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3831191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
What is the solution to $\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$ $$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$$
I've already tried this many times but still not get the answer, this was one of my solution which I can't go further anymore.
$\frac{{x}^{2}}{(x+1)({\sqrt{x+1}}-1)^{2}}< \frac{{x}^{2}+3x+18}{({x+1})^{2}}$
$\frac{{x}^{2}(x+1)}{({\sqrt{x+1}}-1)^{2}}< {x}^{2}+3x+18$
$\frac{{x}^{3}+{x}^{2}}{({\sqrt{x+1}}-1)^{2}}< \frac{({x}^{2}+3x+18)({\sqrt{x+1}-1})^{2}}{({\sqrt{x+1}-1})^{2}}$
$\frac{{x}^{3}+{x}^{2}}{({\sqrt{x+1}}-1)^{2}}< \frac{{x}^{3}+5{x}^{2}+24x+36-(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}-1})^{2}}$
$\frac{{x}^{3}+{x}^{2}-{x}^{3}-5{x}^{2}-24x-36+(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
$\frac{-4{x}^{2}-24x-36+(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
$\frac{-4(x+3)^{2}+2({x}^2+3x+18)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
Does anyone have a hint for the solution?
|
The domain gives $x>-1$ and we need to solve $$\frac{x^2}{(\sqrt{x+1}-1)^2}<\frac{x^2+3x+18}{x+1}$$ or
$$x^2(x+1)<(x+2-2\sqrt{x+1})(x^2+3x+18)$$ or
$$2(x+3)^2>(x^2+3x+18)\sqrt{x+1}$$
or $$4(x+3)^4>(x^2+3x+18)^2(x+1)$$ or after factoring
$$x^2(3-x)(x^2+6x+21)>0,$$ which gives $$(-1,3)\setminus\{0\}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3833035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Can a triangle ABC be made if $\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$
I would like to know the simplest approach to find out whether a triangle ABC will be made if $$\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$$
The counterpart questions for sine and tangent can be handled as follows:
*
*If $\dfrac{\sin A}{2}=\dfrac{\sin B}{3}=\dfrac{\sin C}{7}$, we can rule out triangle because by the Sine Rule $a=2k$, $b=3k$, $c=7k \implies a+b <c.$
*If $\dfrac{\tan A}{2}=\dfrac{\tan B}{3}=\dfrac{\tan C}{7}$, we can see that a triangle will be made as $\tan A=2k, \tan B =3k,\tan C=7k$, when inserted in the identity $\tan A+ \tan B+ \tan C= \tan A \tan B \tan C \implies k=\sqrt{2/7}$.
|
Strating from @Teresa Lisbon's answer, the exact results are
$$k=\frac{31}{126} \left(2 \cos \left(\frac{1}{3} \left(2 \pi n-\cos
^{-1}\left(-\frac{17884}{29791}\right)\right)\right)-1\right)\qquad (n=0,1,2)$$ and this gives angles (in degrees) $a=76.358$, $b=69.281$, $c=34.361$.
Using algebra, the problem is very simple since it reduces to the equation
$$a+\cos ^{-1}\left(\frac{3}{2} \cos (a)\right)+\cos ^{-1}\left(\frac{7 }{2} \cos (a)\right)=\pi$$ which has only one solution.
Using series expansion around $a=\frac \pi 2$ gives
$$0=\frac{\pi }{2}+6 \left(a-\frac{\pi }{2}\right)+\frac{55}{8} \left(a-\frac{\pi
}{2}\right)^3+\frac{4627}{128} \left(a-\frac{\pi
}{2}\right)^5+O\left(\left(a-\frac{\pi }{2}\right)^7\right)$$ and series reversion leads to
$$a \sim\frac{5 \pi }{12}+\frac{55 \pi ^3}{82944}+\frac{89 \pi ^5}{10616832}\approx 1.33212$$ while the "exact" solution is $a=1.33270$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3834491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
}
|
Understanding convergence of $a_n := \frac{2n^3 +n^2 +3}{n^3-4}$ We proved by definition that the sequence $a_n := \frac{2n^3 +n^2 +3}{n^3-4}$ converges.
Let $\epsilon > 0$. Choose $N := \lceil \frac{24}{\epsilon} \rceil + 2 $.
Then for all $n \geq N$ it holds, that
$$|a_n - 2| = \big |\frac{2n^3+n^2+3}{n^3-4} - \frac{2n^3-8}{n^3-4} \big | = \big | \frac{n^2+11}{n^3-4} \big | = \frac{n^2+11}{n^3-4} \leq \frac{n^2+11}{n^3 - \frac{1}{2}n^3} \leq \frac{12n^2}{\frac{1}{2} n^3} = \frac{24}{n} < \epsilon $$
What I don't understand is how we get to $$- \frac{2n^3-8}{n^3-4} $$
Could someone explain? Thanks
|
We have $$-\frac{2n^{3}-8}{n^3-4}=-\frac{2(n^3-4)}{n^3-4}=-2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3834940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
question from Euclid 2011 about proving that $\frac{(a^3+b^3)c}{a^2+b^2}+\frac{(b^3+c^3)a}{b^2+c^2}+\frac{(c^3+a^3)}{c^2+a^2}\ge 1$ I just did the following question:
If $a, b, c$ positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}$ prove that $\frac{(a^3+b^3)c}{a^2+b^2}+\frac{(b^3+c^3)a}{b^2+c^2}+\frac{(c^3+a^3)}{c^2+a^2}\ge 1$
I solved it in the following way:
$ab+bc+ac=1$
From Tchebychev we have that $2(a^3+b^3)\ge (a^2+b^2)(a+b)$
So $\frac{(a^3+b^3)c}{a^2+b^2}+\frac{(b^3+c^3)a}{b^2+c^2}+\frac{(a^3+c^3)b}{c^2+a^2}\ge \frac{(a+b)c+(b+c)a+(a+c)b}{2}$
$=ab+bc+ac=1$
This question really troubled me, as I had difficulties thinking of braking $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}$ into $ab+bc+ac=1$. Can someone please show me either a more intuitive approach to the question, or why I should have thought of braking up the original equation, into the second one, earlier and more intuitively?
|
SOS helps:
$$\sum_{cyc}\frac{(a^3+b^3)c}{a^2+b^2}-1=\sum_{cyc}\frac{(a^3+b^3)c}{a^2+b^2}-ab-ac-bc=$$
$$=\sum_{cyc}\left(\frac{(a^3+b^3)c}{a^2+b^2}-\frac{c(a+b)}{2}\right)=$$
$$=\sum_{cyc}\frac{c(a+b)}{2}\left(\frac{2(a^2-ab+b^2)}{a^2+b^2}-1\right)=\sum_{cyc}\frac{c(a+b)(a-b)^2}{2(a^2+b^2)}\geq0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3836259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Directional derivatives at the origin and conditions for differentiability
Let $f:\mathbb{R^2} \to \mathbb{R}$ $$f(x) = \left\{
\begin{array}{ll}
\frac{xy^2}{x^2+y^4}, & (x,y)\ne0 \\
0, & (x,y) =0 \\
\end{array}
\right.$$ Find the directional derivatives at the origin $D_af(0,0)$ for every direction $a=(a_1,a_2)$, when $||a||=1.$ Show that $f$ is not differentiable at the origin.
For the partials I found that $\frac{\partial}{\partial x} = \frac{y^2(y^4-x^2)}{(x^2+y^4)^2}$ and $\frac{\partial}{\partial y} = \frac{2xy(x^2-y^4)}{(x^2+y^4)^2}$
so $\nabla f=(\frac{y^2(y^4-x^2)}{(x^2+y^4)^2}, \frac{2xy(x^2-y^4)}{(x^2+y^4)^2})$.
The directional derivative is then $D_af=\nabla f\cdot a = (\frac{y^2(y^4-x^2)}{(x^2+y^4)^2}, \frac{2xy(x^2-y^4)}{(x^2+y^4)^2})\cdot(a_1,a_2)$
I'm not sure I understand what they mean by $D_af(0,0) = \nabla f(0,0)\cdot a$ this would lead to division by $0$ right?
Also for the differentiability I tried to use the definition of the partials and see if they're both continuous at the origin, but that lead to a very messy expression for instance $$\frac{\partial}{\partial x} = \frac{f(x+h,y)-f(x,y)}{h} = \frac{\frac{(x+h)y^2}{(x+h)+y^4}-\frac{xy}{x^2+y^4}}{h}$$
and this didn't seem to simplify to anything usable... What should I do here?
|
We are requested to find the directional derivatives at the origin that is for $a\cdot b \neq 0$
$$\lim_{(ah,bh)\to(0,0)} \frac{\frac{ab^2h^3}{a^2h^2+b^4h^4}-0}{h} =\lim_{(ah,bh)\to(0,0)} \frac{ab^2h^3}{a^2h^3+b^4h^5}=\frac{b^2}a$$
with $f_x=f_y=0$.
For differentiability just note that for $x=y^2$
$$\lim_{(x,y)\to(0,0)} \frac{xy^2}{x^2+y^4}=\lim_{(y^2,y)\to(0,0)} \frac{y^4}{y^4+y^4}=\frac12$$
therefore $f(x,y)$ is not continuous at the origin and then it is not differentiable.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3837022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Induction proof of a known harmonic sum I want to prove that $$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n} \leq 1$$ only by induction!
I check for the first one, $\frac12 \leq 1 $ correct.
Then I assume for $n=k$ : $$\frac12 + \frac14 + \dots + \frac{1}{2^k} \leq 1$$
And Try and prove for $n=k+1$
$$\frac12 + \frac14 + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} \leq 1$$
But I know that $$\frac12 + \frac14 + \dots + \frac{1}{2^k} \leq \frac12 + \frac14 + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} \leq 1$$ and so:
But I am stuck, this tells me that the sum for $n=k+1$ is always $1$ , not $S \leq 1$ I am so confused, because I can't use the geometric series sum formula.. any help would be appreciated!
|
$\frac12 + \frac14 + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} \leq 1 \iff \frac14 + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} \leq \frac12 \iff \frac12 + \dots + \frac{1}{2^{k-1}} + \frac{1}{2^{k}} \leq 1$.
The last step is multiplying by $2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3837387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Use mathematical induction to prove that $\sum_{j=1}^n \frac{1}{(j^2-1)}$ = $\frac{3}{4}$ - $\frac{2n+1}{2n(n+1}$ for all n $\geq$ 2 so I know how to do induction and all I'm just struggling with the algebra for proving the case for n = k+1. My proof is below:
For the case n=2 we have
$\sum_{j=1}^2 \frac{1}{j^2-1}$ = $\frac{1}{3}$ and $\frac{3}{4}$ - $\frac{2*2 + 1}{2n(n+1)}$ = $\frac{1}{3}$.
Thus the equality holds for n = 2.
Assume for n = k for k $\in$ $\mathbb{Z}$ that the equality holds.
For the case n = k + 1 we must show that
$\sum_{j=1}^{k+1} \frac{1}{j^2-1}$ = $\frac{3}{4}$ - $\frac{2(k+1)+1}{2(k+1)((k+1)+1)}$
For the LHS we have:
$\sum_{j=1}^{k+1} \frac{1}{j^2-1}$ = $\sum_{j=1}^{k} \frac{1}{j^2-1}$ + $\frac{1}{(k+1)^2-1}$
I'm not gonna type all the algebra out because that's what I'm having trouble with and there's just a lot of it.
For the RHS we have:
$\frac{3}{4}$ - $\frac{2(k+1)+1}{2(k+1)((k+1)+1)}$ = $\frac{3}{4}$ - $\frac{2(k+1)+1}{(2k+2)(k+2)}$ = $\frac{3}{4}$ - $\frac{2k+3}{2k^2+6k+4}$
When you expand both sides I'm not seeing any way to manipulate them to equal one another... I know it's just basic algebra that's messing this up because the instructions are to prove it.
|
\begin{align}
\frac34 - \frac{2k+1}{2k(k+1)} + \frac{1}{(k+1)^2-1} &= \frac34 - \frac{2k+1}{2k(k+1)} + \frac{1}{(k+1)^2-1} \\
\end{align}
Your goal is to simplify $- \frac{2k+1}{2k(k+1)} + \frac{1}{(k+1)^2-1} $ and show that it is equal to $-\frac{2(k+1)+1}{2(k+1)(k+2)}$
\begin{align}
- \frac{2k+1}{2k(k+1)} + \frac{1}{(k+1)^2-1} &= - \frac{2k+1}{2k(k+1)} + \frac{1}{k(k+2)}\\
&= \frac{-(2k+1)(k+2) + 2(k+1)}{2k(k+1)(k+2)}
\end{align}
Task that I left for you, simplify the numerator and you should see the solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3838113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Calculate the value of the following limit So I have given the limit:
$$\lim_{x\to0} \frac{2\sin\left ( e^{-\frac{x^2}{2}} -\cos x \right)}{(\arctan(\sinh(x^2))^2}$$
I have been struggling for hours with it. Since i got the undefined form when i put $x=0$ i tried out with L'Hopital method and I come to this point:
$$\lim_{x\to 0} \frac{2\cos(e^{-\frac{x^2}{2}}-\cos x)(e^{-\frac{x^2}{2}}(-x)+\sin x)}{\frac{4x \arctan(\sinh(x^2))\cosh(x^2)}{\sinh(x^2)+1}}$$
Still here when i substitute x with $0$ i still get $0$. I tried factorising the x, i also tried using the identities $cos(A-B)$ and so on, but nothing.
I think the answer that should come out is $\frac{1}{6}$
I would be very thankful for your help,
Annalisa
|
If you are aware of Taylor expansions then:
$$
e^{\frac{-x^2}{2}} =_{x \rightarrow 0} 1-\frac{x^2}{2} + \frac{x^4}{8} + o(x^4)
$$
$$
\cos(x) =_{x \rightarrow 0} 1 - \frac{x^2}{2} + \frac{x^4}{4!} + o(x^4)
$$
Then
$$
e^{\frac{-x^2}{2}} - \cos(x) =_{x \rightarrow 0} \frac{x^4}{12} + o(x^4)
$$
Now because $sin(x) =_{x \rightarrow 0} x + o(x)$ then
$$
2 \sin( e^{\frac{-x^2}{2}} - \cos(x)) =_{x \rightarrow 0} \frac{x^4}{6} + o(x^4)
$$
And for the denominator:
$$
\sinh(x^2) =_{x \rightarrow 0} x^2 + o(x^2)
$$
$$
\arctan(x) =_{x \rightarrow 0} x + o(x)
$$
So:
$$
(\arctan(\sinh(x^2))^2 =_{x \rightarrow 0} x^4 + o(x^4)
$$
And thus:
$$
\frac{2 \sin( e^{\frac{-x^2}{2}} - \cos(x))}{(\arctan(\sinh(x^2))^2} =_{x \rightarrow 0} \frac{1}{6} + o(1)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3838421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Find the height of a trapezoid
A trapezoid $ABCD$ is given $(AB$ $||$ $CD)$ with side lengths $AB=18,BC=\sqrt{74},CD=5$ and $AD=\sqrt{61}.$ Find the sines of $\measuredangle A$ and $\measuredangle B.$
(I have only studied trig functions of acute angles)
Since we are given the four sides of the trapezoid, it is enough to find the height. I am not sure how to approach the problem. If $CP$ is parallel to $AD,$ we can find the area of triangle $PBC$ by Heron's formula (the area of a triangle when the length of all three sides are known) but we still haven't studied it. Can you give me a hint for another solution? Thank you in advance!
|
First Solution
$PC=AD=\sqrt{61}$, $BC=\sqrt{74}$, $BP=13$
$cos \angle{B}=\frac{74+169-61}{26\sqrt{74}}=\frac{7}{\sqrt{74}}$
$sin \angle{B}=\frac{5}{\sqrt{74}}$
$cos \angle{A}= cos \angle{P}=\frac{61+169-74}{26\sqrt{61}}=\frac{6}{\sqrt{61}}$
$sin\angle{A}=\frac{5}{\sqrt{61}}$
Second Solution
Let $AD_1=x, BC_1=13-x, CC_1=DD_1=h$
$h^2=61-x^2=74-(13-x)^2$
$x=6, h=5$
Third Solution
$BP=13$ since $AP=CD=5$
$CP=AD=\sqrt{61}$
Let $PC_1=x$, $BC_1=13-x$, $CC_1=h$
$h^2=61-x^2=74-(13-x)^2$
$x=6, h=5$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3839126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find the area bounded by the curve $x^4+y^4=x^2+y^2$ I am stuck with this problem which deals with evaluating an Area
The problem reads :
Find the area bounded by the curve $x^4+y^4=x^2+y^2$.
I tried factorizing the expression and expressing $y$ in terms of $x$, not able to proceed with that idea. Someone please help me out.
|
Use polar co-ordinates $x=r \cos t, y=r \sin t$, then this curve is $$r^2=\frac{1}{\sin^4 t+ \cos^4 t}=\ $$, the area the area is
$$A=4 ~ \frac{1}{2}\int_{0}^{\pi/2} r^2 dt=2 \int_{0}^{\pi/2} \frac{1}{\cos^4 t+\sin^4 t}dt =2 \int_{0}^{\pi/2}\frac{\sec^4 t}{1+\tan^4 t} dt $$
Let $z=\tan t \implies dz=\sec^2 t dt$
$$=2\int_{0}^{\infty} \frac{1+z^2}{1+z^4} dz=2 \int_{0}^{\infty}\frac{1+1/z^2}{z^2+1/z^2}=2\int_{0}^{\infty} \frac{1+1/z^2}{(z-1/z)^2+2}$$ $$\implies A=2 \int_{-\infty}^{\infty} \frac{du}{u^2+2}=\sqrt{2} \pi.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3840121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
A probability question about choosing 6 shoes are chosen randomly from the 20 distinct pairs of shoes drawer. $\mathbf{Question:}$
A drawer has 20 distinct pairs of shoes. 6 shoes are chosen randomly from the drawer. The drawer contain only one my favorite pair.
(a) what is the probability that my favorite pair is chosen ?
(b) what is the probability that I get no complete pair ?
(c) what is the probability that I get exactly one complete pair ?
(d) what is the probability that I get at two complete pairs ?
$\mathbf{My~Attempt:}$
(a) The probability that my favorite pair is shosen $= \frac{2}{40} \cdot \frac{1}{39} \approx 0.0013$.
(b) The probability that I get no complete pair $= \frac{40 \cdot 38 \cdot 36 \cdot 34 \cdot 32 \cdot 30}{40 \cdot 39 \cdot 38 \cdot 37 \cdot 36 \cdot 35} = \frac{34 \cdot 32 \cdot 30}{39 \cdot 37 \cdot 35} \approx 0.6463$.
(c) The probability that I get exactly one complete pair $= \frac{\binom{20}{1} \cdot \binom{18}{4} \cdot 2^4}{\binom{40}{6}} \approx 0.2551$.
(d) The probability that I get at least one complete pair $= 1 - \text{The probability I get no complete pair} = 0.3537$.
$~~~~~$ So, The probability that I get at least two complete pairs
$~\hspace{11mm}$ $= \text{The probability I get at least one complete pair}$
$~\hspace{15mm}$ $- \text{The probability I get exactly one complete pair}$
$~\hspace{11mm}$ $= 0.3537 - 0.2551 = 0.0986$
$\textbf{Is that my attempt of (a), (b), (c) and (d) correct ?}$
$\textbf{If they are all correct, then are there any other ways to think about (d)}$
$\textbf{without using the answer of (b) and (c) ?}$
|
Not all of your answers are correct.
(a) If you choose both shoes from your favorite pair, you must also choose four shoes from the remaining $40 - 2 = 38$ shoes. Thus, the number of favorable cases is
$$\binom{2}{2}\binom{38}{4}$$
You failed to account for the fact that you are choosing six shoes, not two. Selecting more shoes increases your chances of choosing both shoes from your favorite pair.
(b) Your answer is correct. Another way to count the favorable cases is to observe we must choose six different pairs, and draw one of the two shoes from each pair. Hence, the number of favorable cases is
$$\binom{20}{6}2^6$$
(c) You made a minor error. There are $20$ ways to choose the pair from which both shoes are taken. This leaves $20 - 1 = 19$ pairs. To ensure that exactly one complete pair is chosen, we must select four of these $19$ pairs from which to extract one shoe each. There are two ways to choose a shoe from each of these four pairs. Hence, the number of favorable cases is
$$\binom{20}{1}\dbinom{19}{4}2^4$$
(d) Your method is sound. However, with the above observations, the answer should be
$$1 - \frac{\dbinom{20}{6}2^6 + \dbinom{20}{1}\dbinom{19}{4}2^4}{\dbinom{40}{6}}$$
You could also add the probabilities of obtaining exactly two complete pairs and exactly three complete pairs.
Exactly two complete pairs: Choose from which two of the $20$ pairs of shoes both shoes will be taken. You must pick two additional shoes. To ensure that exactly two pairs are selected, you must select two of the remaining $20 - 2 = 18$ pairs and choose one of the two shoes from each of those pairs.
$$\binom{20}{2}\binom{18}{2}2^2$$
Exactly three complete pairs: Choose from which three of the $20$ pairs both shoes will be taken.
$$\binom{20}{3}$$
Thus, the desired probability is
$$\frac{\dbinom{20}{2}\dbinom{18}{2}2^2 + \dbinom{20}{3}}{\dbinom{40}{6}}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3840509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}$ for $a,b,c\in\mathbb{R}^+$ with $abc=1$ Suppose that $a,b,c$ are positive reals such that $abc=1$. Prove that $$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}.$$
Hint: Use Titu's lemma.
My approach: I am trying to use Titu's lemma directly but it is not working. I meant that I wrote each term in the following way: $$\frac{a^3}{b+c}=\frac{a^3bc}{bc(b+c)}=\frac{a^2}{bc(b+c)}.$$ Then I applied Titu's lemma to the sum $$\frac{a^2}{bc(b+c)}+\frac{b^2}{ac(a+c)}+\frac{c^2}{ab(a+b)}\geq \frac{(a+b+c)^2}{bc(b+c)+ac(a+c)+ab(a+b)}=\frac{\sigma_1^2}{\sigma_1\sigma_2-3},$$ where $\sigma_1=a+b+c$ and $\sigma_2=ab+ac+bc$ are elementary symmetric functions.
This is what I got so far.
Would thankful if someone shows correct solution.
|
Notice that $$\sum_{cyc}\frac{a^3}{b+c}=\sum_{cyc}\frac{a^4}{a(b+c)}\geqslant \frac{(a^2+b^2+c^2)^2}{2(ab+bc+ca)}\geqslant \frac{a^2+b^2+c^2}{2}\geqslant \frac{3\sqrt[3]{abc}}{2}=\frac32 $$ Where we first used Titu's Lemma, then the well-known $a^2+b^2+c^2\geqslant ab+bc+ca$, and finally AM-GM.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3842654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
What does the geometric series $ \sum\limits_{n=0}^{\infty} n \cdot \left( \frac{1}{4} \right) ^{n}$ converge to? I am trying to find the value to which the geometric series
$$
\sum\limits_{n=0}^{\infty} n \cdot \left( \frac{1}{4} \right) ^{n},
$$
converges. Now I know how to find a sum when we have a common ratio, but since $n$ is increased with each term, I am stuck at how to tackle this.
Any suggestions would be appreciated.
|
Hint
$$S_n=1\cdot\left( \frac{1}{4} \right) ^{1}+2\cdot\left( \frac{1}{4} \right) ^{2}+...+n\cdot\left( \frac{1}{4} \right) ^{n}$$
Now, multiply by $1/4$
$$\frac{1}{4}S_n=1\cdot\left( \frac{1}{4} \right) ^{2}+2\cdot\left( \frac{1}{4} \right) ^{3}+...+n\cdot\left( \frac{1}{4} \right) ^{n+1}$$
Now subtract,
$$S_n-\frac{1}{4}S_n=1\cdot\left( \frac{1}{4} \right) ^{2}+(2-1)\cdot\left( \frac{1}{4} \right) ^{2}+...+(n-(n-1))\left( \frac{1}{4} \right) ^{n}+n\cdot\left( \frac{1}{4} \right) ^{n+1}$$
Did you understand the idea? Can you finish?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3845365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Find $E[B 1_{\{ B\ge \frac{n}{2} \} }]$ where $B$ is Binomial $(n,\frac{1}{2})$. How to find the following expectation
$$E[B 1_{\{ B\ge \frac{n}{2} \} }]$$
where $B$ is Binomial random variable with $(n,\frac{1}{2})$.
Here is what I did. The pmf in this case is given by
\begin{align}
E[B 1_{\{ B\ge \frac{n}{2} \} }]= \sum_{k \ge \lceil n/2 \rceil }^n k {n \choose k} \left( \frac{1}{2}\right)^n
\end{align}
However, I am not sure how to compute this sum. An upper bound would be fine too.
|
Note that we have
$$
E[B 1_{\{ B\ge \frac{n}{2} \} }]=\frac n2 - E[B1_{\{B< \frac n2\}}].
$$
Then we evaluate $E[B1_{\{B< \frac n2\}}]$. We have
$$
E[B1_{\{B< \frac n2\}}]=\sum_{k<\frac n2} k\binom nk \frac 1{2^n}=\sum_{k<\frac n2} n \binom{n-1}{k-1} \frac1{2^n}.
$$
The second equality is due to $k\binom nk = n \binom{n-1}{k-1}$.
Now, the index $k$ of the sum is up to $k\leq \frac n2 -1$ if $n$ is even, and $k\leq \frac{n-1}2$ if $n$ is odd. Then $k-1\leq \frac n2-2$ if $n$ is even, and $k-1\leq \frac{n-3}2$ if $n$ is odd.
We apply the symmetry of binomial coefficients $\binom{n-1}k=\binom{n-1}{n-1-k}$.
Then the last sum is
$$
\begin{cases}
\frac n4 - \frac n{2^n}\binom{n-1}{\frac n2-1} &\mbox{if } n \mbox{ is even}\\
\frac n4- \frac12 \frac n{2^n}\binom{n-1}{(n-1)/2} &\mbox{if } n \mbox{ is odd}\\
\end{cases}.
$$
Therefore,
$$
E[B 1_{\{ B\ge \frac{n}{2} \} }]=\begin{cases}
\frac n4 + \frac n{2^n}\binom{n-1}{\frac n2-1} &\mbox{if } n \mbox{ is even}\\
\frac n4+ \frac12 \frac n{2^n}\binom{n-1}{(n-1)/2} &\mbox{if } n \mbox{ is odd}\\
\end{cases}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3845626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Is there an identity to combine a sum of more than two sines; eg $\sin(a)+\sin(b)+\sin(c)+\sin(d)$? I get these trigonometric product to sum formulas like:
$$\sin(a)+\sin(b)=2\sin\frac12(a+b)\cos\frac12(a-b)$$
And that's useful, but I'm not too sure what to do if I need to turn a product into a sum if there's more than two variables.
What would I do with something like this?
$$\sin(a)+\sin(b)+\sin(c)+\sin(d)$$
|
The following formula works:
$$
\begin{align}
\sin(a)+sin(b)+sin(c)+sin(d) = 4*\sin\left(\frac{a+b+c+d}{4}\right)*\cos\left(\frac{a-b+c-d}{4}\right) \\
*\cos\left(\frac{a+b-c-d}{4}\right)*\cos\left(\frac{a-b-c+d}{4}\right)- \\
4*\cos\left(\frac{a+b+c+d}{4}\right)*\sin\left(\frac{a-b+c-d}{4}\right) \\
*\sin\left(\frac{a+b-c-d}{4}\right)*\sin\left(\frac{a-b-c+d}{4}\right) \\
\end{align}
$$
I wrote this Python script in Google colab to "prove" it:
import numpy as np
def C(x):
return np.cos(180/np.pi*x)
def S(x):
return np.sin(180/np.pi*x)
a = 37
b = 6
c = 88
d = 7
x = S(a) + S(b) + S(c) + S(d)
y1 = 4*S((a+b+c+d)/4.0)*C((a-b+c-d)/4.0)*C((a+b-c-d)/4.0)*C((a-b-c+d)/4.0)
y2 = 4*C((a+b+c+d)/4.0)*S((a-b+c-d)/4.0)*S((a+b-c-d)/4.0)*S((a-b-c+d)/4.0)
print(x)
print(y1-y2)
When the input is $(a,b,c,d) = (37,6,88,7)$, with angles measured in degrees, the outputs for $x$ and $(y1-y2)$ agree to $11$ decimal places at $-1.02707706592$. I also tried $(a,b,c,d) = (10,27,18,68)$, and the two sides of the equation again agreed to $11$ decimal places. That isn't a mathematical proof, but the probability of my proposed identity holding true by accident for randomly selected angles is essentially nil. I tried making one of the angles obtuse, and it still worked.
Steps in the derivation:
*
*Use angle sum formulas to derive an expression for $\sin\left(\frac{A+B+C+D}{4}\right)$ in terms of $\sin\left(\frac{A}{4}\right)$, $\sin\left(\frac{B}{4}\right)$, $\sin\left(\frac{C}{4}\right)$, $\sin\left(\frac{D}{4}\right)$, $\cos\left(\frac{A}{4}\right)$, $\cos\left(\frac{B}{4}\right)$, $\cos\left(\frac{C}{4}\right)$ and $\cos\left(\frac{D}{4}\right)$.
*Use the first expression, and the fact that sine and cosine are odd and even functions, respectively, to derive expressions for $\sin\left(\frac{A-B+C-D}{4}\right)$, $\sin\left(\frac{A+B-C-D}{4}\right)$ and $\sin\left(\frac{A-B-C+D}{4}\right)$.
*Add the four expressions together.
*Make the following substitutions:
$$
A = a+b+c+d \\
B = a-b+c-d \\
C = a+b-c-d \\
D = a-b-c+d
$$
...and you should get the formula.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3848224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Convergence of $ \prod_{n=1}^{\infty}\Bigg\{ \left(1+\frac{1}{n} \right)^x\left(1-\frac{x}{n}\right)\Bigg\}$ I have to discuss the convergence of the product
$ \prod_{n=1}^{\infty}\Bigg\{ \left(1+\frac{1}{n} \right)^x\left(1-\frac{x}{n}\right)\Bigg\}$
Here is my solution:
Based on the binomial formula, we have
$ \left(1+\frac{1}{n} \right)^x\left(1-\frac{x}{n}\right)=1+\frac{x}{n}+\frac{x(x-1)}{2n^2}+\frac{x(x-1)(x-2)}{6n^3}+\dots -\frac{x}{n}-\frac{x^2}{n^2}-\frac{x^2(x-1)}{2n^3}-\frac{x^2(x-1)(x-2)}{6n^4}-\dots$
$ = 1-\frac{x^2}{n^2}-\frac{x(x-1)}{2n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)$
So we have
$e^{ln C}=\exp\left[\sum_{n=0}^{\infty}\left(\log \left(1-\frac{3x^2-x}{2n^2}+\mathcal{O}\left(\frac{1}{n^3}\right) \right)\right)\right] $
$ \Big|e^{ln C}\Big|\leq\Bigg| \exp\left[\sum_{n=0}^{\infty}\left( -\frac{3x^2-x}{2n^2}+\mathcal{O}\left(\frac{1}{n^3} \right)\right)\right] \Bigg|$
The integral test shows that this series converges, since
$ \int \limits_1^\infty \frac{1}{n^2}\, \mathrm{d}x = \lim_{b \to \infty} \left[-\frac{1}{n}\right]_1^b=1$
Is this right?
|
Yes, this is mostly right, except that it should be $\int \limits_1^\infty \frac{1}{n^2}\, \mathrm{d}n$. And I'd use some simpler test for that, $1/n^2\le1/n(n-1)$, or condensation test, but that's a matter of taste, naturally.
Looking at https://en.wikipedia.org/wiki/Gamma_function#Euler's_definition_as_an_infinite_product , we see that the infinite product is $$1/(-x)!=1/\Gamma(1-x).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3848469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$? From the Pre-Regional Mathematics Olympiad, 2019:
If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$?
I have provided one answer below, and would be interested in alternative solutions.
|
Using a bit of Galois theory: first, note that the intermediate fields between $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ are $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, and $\mathbb{Q}(\sqrt{6})$, none of which contains $\sqrt{2} + \sqrt{3} + \sqrt{6}$. Therefore, $\sqrt{2} + \sqrt{3} + \sqrt{6}$ is a primitive generator of the extension $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ of $\mathbb{Q}$, so the degree of its minimal polynomial is $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}] = 4$.
Now, also note that if the minimal polynomial is $p(x) = x^4 + ax^3 + bx^2 + cx + d$, then $a + b + c + d = p(1) - 1$. On the other hand, we also know that $p(x) = \prod_{\sigma \in G} (x - \sigma(\alpha))$ where $G = \operatorname{Gal}(\mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q})$ and $\alpha = \sqrt{2} + \sqrt{3} + \sqrt{6}$. Therefore, $p(1) = N^{\mathbb{Q}(\sqrt{2}, \sqrt{3})}_{\mathbb{Q}}(1 - \alpha)$.
But also,
$$
N^{\mathbb{Q}(\sqrt{2}, \sqrt{3})}_{\mathbb{Q}} (1 - \alpha) = N^{\mathbb{Q}(\sqrt{2})}_{\mathbb{Q}} N^{\mathbb{Q}(\sqrt{2}, \sqrt{3})}_{\mathbb{Q}(\sqrt{2})} [(1 - \sqrt{2}) - (1 + \sqrt{2}) \sqrt{3}] = \\
N^{\mathbb{Q}(\sqrt{2})}_{\mathbb{Q}} [(1 - \sqrt{2})^2 - 3 (1 + \sqrt{2})^2] = N^{\mathbb{Q}(\sqrt{2})}_{\mathbb{Q}} (-6 - 8 \sqrt{2}) = 6^2 - 2 \cdot 8^2 = -92.
$$
Thus, $a + b + c + d = -93$, and $|a + b + c + d| = 93$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3851795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
What is $\tan \alpha$, if $(a+2)\sin\alpha +(2a - 1)\cos\alpha =2a + 1$? I tried the following:
$$\begin{aligned}a\sin\alpha +2\sin\alpha + 2a\cos\alpha - \cos\alpha &= 2a+1\\
a(\sin\alpha +2\cos\alpha)+(2\sin\alpha-\cos\alpha)&=2a+1\end{aligned}$$
Therefore,
$$\sin\alpha +2 \cos\alpha=2$$
$$2\sin\alpha - \cos\alpha=1$$
From these two equations, we get
$$\sin\alpha=\frac{4}{5},\cos\alpha=\frac{3}{5}$$
Therefore,
$$\tan\alpha = \frac{\sin\alpha} {\cos\alpha} = \frac{4} {3}$$
Is this a correct method to solve the question? Since $a$ is a constant, it does not seem necessary to me that its coefficients on the two sides of the equation be equal. Should I find $\sin\alpha$ and $\cos\alpha$ using some other method? Are there specific cases where this method of equating the coefficients will break?
|
The problem is that $a$ is a constant, not a variable. Therefore equating by parts does not work here: if we separate $2a - 1$ into $2(a-1) + 1$, $2(a-2) + 3$ and so on, we end up with a different set of solutions each time. If you try substituting $\sin \alpha$ and $\cos \alpha$ back into the original question, the LHS does not equal $2a-1$, so clearly something must have gone wrong.
Instead, notice that as this is an identity, so it must hold for all $a$. Substituting $a = -2, \frac{1}{2}$ to cancel one of the terms, we get:
$$-5 \cos a = -5 \Rightarrow \cos a = 1$$
$$\frac{5}{2} \sin a = 0 \Rightarrow \sin a = 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3852291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
How to prove this inequality: $\sum_{cyc} \frac {1}{\alpha + \log_a {b}} \le \frac {2}{\alpha}$? This is another Olympiad problem my teacher gave us to train.
$$\sum_{cyc} \frac {1}{\alpha + \log_a{b}} \le \frac {2}{\alpha}$$ with $\alpha \in (0, 2]$ and $a, b, c \in (0, 1)$ or $ a, b, c \in(1, \infty)$. After working a little bit with it, we get $$\sum_{cyc} \log_{a^\alpha b} {a} \le \frac {2}{\alpha}$$ but I don't know how to go on.
|
Starting from @Robert Z's answer, consider that we look for the maximum value of function
$$f(x,y,z)=\frac {1}{\alpha + x}+\frac {1}{\alpha + y}+\frac {1}{\alpha + z}$$ Using the constraint $xyz=1$, we look at the maximum of function
$$g(x,y)=\frac{2 \alpha +x^2 y (2 \alpha +y)+x \left(2 \alpha y^2+3 \alpha ^2
y+1\right)+y}{(\alpha +x) (\alpha +y) (\alpha x y+1)}$$ Computing the partial derivatives
$$\frac {\partial g(x,y)}{\partial x}=\frac{y}{(\alpha x y+1)^2}-\frac{1}{(\alpha +x)^2}=0$$
$$\frac {\partial g(x,y)}{\partial y}=\frac{x}{(\alpha x y+1)^2}-\frac{1}{(\alpha +y)^2}=0$$
The only real solutions are
$$x=1 \quad y=1\quad z=1 \implies f(x,y,z)=\frac{3}{\alpha +1}$$
$$x=\frac 1{\alpha^2}\quad y=\alpha^4\quad z=\frac 1{\alpha^2}\implies f(x,y,z)=\frac{2 \alpha ^3+1}{\alpha(\alpha ^3+1) }$$
$$x={\alpha^2}\quad y=\alpha^2\quad z=\frac 1{\alpha^4}\implies f(x,y,z)=\frac{2 \alpha ^3+1}{\alpha(\alpha ^3+1) }$$
$$x={\alpha^4}\quad y=\frac 1{\alpha^2}\quad z=\frac 1{\alpha^2}\implies f(x,y,z)=\frac{2 \alpha ^3+1}{\alpha(\alpha ^3+1) }$$
So, if $\alpha=2$, the maximum value is $\frac 12$
If $0 <\alpha<2$, the maximum value is $\frac 2 \alpha$
If $\alpha>2$, the maximum value is $\frac 3 {\alpha+1}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3855023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Why does this equality Dirichlet series hold? Following on from my question here, I have hit a second roadblock.
I am working (very slowly!) through a paper here that demonstrates Riemann's analytic continuation of the zeta function $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$ to the complex plane (except for the pole at $s=1$).
At the top of page 6 in equation 14, the paper asserts that
$$\frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} = \ldots = \frac{1}{s-1} \sum_{n=1}^\infty \biggl(\frac{n}{(n+1)^s}-\frac{n-s}{n^s}\biggr)$$
What are the logical steps that give this result? I assume that the expression $\frac{n}{(n+1)^s}-\frac{n-s}{n^s}$ is somehow arrived at by splitting odd and even $n$, but this gives me
$$\begin{aligned}
\frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} &= \frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \biggl( \frac{1}{2^sn^s}- \frac{1}{(2n-1)^s} \biggr)
\\ &= \sum_{n=1}^\infty \biggl( \frac{1}{n^s(2^s-2)}- \frac{1}{\bigl(n- \frac{1}{2}\bigr)^s(2^s-2)} \biggr)
\end{aligned}$$
But I can't see how to extract the factor $\frac{1}{s-1}$ to produce the desired result.
|
For $\Re s >2$ so everything is absolutely convergent and then extending by analytic continuation wherever RHS converges:
$\frac{1}{s-1} \sum_{n=1}^\infty (\frac{n}{(n+1)^s}-\frac{n-s}{n^s})=\frac{1}{s-1} \sum_{n=1}^\infty (\frac{n+1}{(n+1)^s}-\frac{n}{n^s}-\frac{1}{(n+1)^s}+\frac{s}{n^s})=$
(by telescoping the first two terms so remaining with just a term of $-1$ and then inserting a $0=1-(1/1^s)$ to make the third term go from $1$)
(edited - more detail as per comment: note that for $n=1$ the first two terms inside the four term bracket are $2/2^s-1/1^s$ for $n=2$ the first two terms are $3/3^s-2/2^s$, for $n=3$ the first two terms are $4/4^s-3/3^s$; since we chose $\Re s >2$ and we have absolute convergence in the sense that the sum of absolute values of each individual term inside each summand is finite, so$\sum_{n=1}^\infty (|\frac{n+1}{(n+1)^s}|+|\frac{n}{n^s}|+|\frac{1}{(n+1)^s}|+|\frac{s}{n^s}|) < \infty$, we can move terms around at will so clearly all the terms noted above cancel out except the first which is the $-1$ we took out; similarly, the third term in the original 4 term bracket gives $-1/2^s-1/3^s-...$ so adding $1-1=0$ and taking the $1$ out gives $1-\zeta(s)$ there, while the last term is self evidently gives $s\zeta(s)$
$=\frac{1}{s-1}(-1+1+\sum_{n=1}^\infty (-\frac{1}{n^s}+\frac{s}{n^s}))= \sum_{n=1}^\infty \frac{1}{n^s}= \zeta(s)=\frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}$
which is the required equality
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3857805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Evaluating $\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$. My attempt.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$$
$$=\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k}\binom{2k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}$$
The first sum can be evaluated easily if one uses the central binomial coefficient generating function
, the closed form is $2\zeta \left(2\right)$.
For the remaining sum consider the $\arcsin$ series expansion.
$$\sum _{k=0}^{\infty }\frac{x^{2k+1}}{4^k\left(2k+1\right)}\binom{2k}{k}=\arcsin \left(x\right)$$
$$\sum _{k=1}^{\infty }\frac{x^k}{4^k\left(2k+1\right)}\binom{2k}{k}=\frac{\arcsin \left(\sqrt{x}\right)}{\sqrt{x}}-1$$
$$-\sum _{k=1}^{\infty }\frac{1}{4^k\left(2k+1\right)}\binom{2k}{k}\int _0^1x^{k-1}\ln \left(1-x\right)\:dx=-\int _0^1\frac{\arcsin \left(\sqrt{x}\right)\ln \left(1-x\right)}{x\sqrt{x}}\:dx$$
$$+\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx$$
$$\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}=-2\int _0^1\frac{\arcsin \left(x\right)\ln \left(1-x^2\right)}{x^2}\:dx-\zeta \left(2\right)$$
But I got stuck with:
$$\int _0^1\frac{\arcsin \left(x\right)\ln \left(1-x^2\right)}{x^2}\:dx$$
Anything I try yields more complicated stuff, is there a way to calculate the main sum or the second one (or the integral) elegantly\in a simple manner?
|
It seemed something was missing, so with the right tools the proof isn't difficult.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$$
Consider:
$$\sum _{k=1}^{\infty }\frac{x^k}{4^k}H_k\binom{2k}{k}=\frac{2}{\sqrt{1-x}}\ln \left(\frac{1+\sqrt{1-x}}{2\sqrt{1-x}}\right)$$
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k}\binom{2k}{k}\int _0^1x^{2k}\:dx=2\int _0^1\frac{\ln \left(1+\sqrt{1-x^2}\right)}{\sqrt{1-x^2}}\:dx-2\int _0^1\frac{\ln \left(\sqrt{1-x^2}\right)}{\sqrt{1-x^2}}\:dx$$
$$-2\ln \left(2\right)\int _0^1\frac{1}{\sqrt{1-x^2}}\:dx$$
$$=2\int _0^1\frac{\ln \left(1+x\right)}{\sqrt{1-x^2}}\:dx-2\int _0^1\frac{\ln \left(x\right)}{\sqrt{1-x^2}}\:dx-\pi \ln \left(2\right)$$
$$\int _0^1\frac{\ln \left(1+x\right)}{\sqrt{1-x^2}}\:dx=\frac{\pi }{2}\ln \left(2\right)-\int _0^1\frac{\arcsin \left(x\right)}{1+x}\:dx$$
$$=\frac{\pi }{2}\ln \left(2\right)-\int _0^{\frac{\pi }{2}}\frac{x\cos \left(x\right)}{1+\sin \left(x\right)}\:dx=\int _0^{\frac{\pi }{2}}\ln \left(1+\sin \left(x\right)\right)\:dx$$
$$=4\int _0^1\frac{\ln \left(1+t\right)}{1+t^2}\:dt-2\int _0^1\frac{\ln \left(1+t^2\right)}{1+t^2}\:dt$$
This means that:
$$\int _0^1\frac{\ln \left(1+x\right)}{\sqrt{1-x^2}}\:dx=-\frac{\pi }{2}\ln \left(2\right)+2G$$
Thus:
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}=-\pi \ln \left(2\right)+4G$$
Bonus.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}=\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k}\binom{2k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}$$
And so we find that:
$$\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}=2\zeta \left(2\right)+2\pi \ln \left(2\right)-8G$$
And in the body of the question we had:
$$\int _0^1\frac{\ln \left(1-x^2\right)\arcsin \left(x\right)}{x^2}\:dx=-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}-\frac{1}{2}\zeta \left(2\right)$$
Hence:
$$\int _0^1\frac{\ln \left(1-x^2\right)\arcsin \left(x\right)}{x^2}\:dx=-\frac{3}{2}\zeta \left(2\right)-\pi \ln \left(2\right)+4G$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3860138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
}
|
Prove $\forall z\in\mathbb C-\{-1\},\ \left|(z-1)/(z+1)\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$ I'm trying to prove
$$\forall z\in\mathbb C-\{-1\},\ \left|\frac{z-1}{z+1}\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$$
thus showing that the solutions to $\left|(z-1)/(z+1)\right|=\sqrt2$ form the circle of center $-3$ and radius $\sqrt8$. But my memories of algebra in $\mathbb C$ fail me. The simplest I get is writing $z=x+i\,y$ with $(x,y)\in\mathbb R^2-\{(-1,0)\}$ and doing the rather inelegant
$$\begin{align}\left|\frac{z-1}{z+1}\right|=\sqrt2&\iff\left|z-1\right|=\sqrt2\,\left|z+1\right|\\
&\iff\left|z-1\right|^2=2\,\left|z+1\right|^2\\
&\iff(x-1)^2+y^2=2\,((x+1)^2+y^2)\\
&\iff0=x^2+6\,x+y^2+1\\
&\iff(x+3)^2+y^2=8\\
&\iff\left|z+3\right|^2=8\\
&\iff\left|z+3\right|=\sqrt8\\
\end{align}$$
How can I avoid the steps with $x$ and $y$ ?
|
For $r>0, a,b \in \mathbb{C}, a\neq b\;$ the equation $\left|\frac{z-a}{z-b}\right|=r$ defines a hyperbolic pencil of Apollonian circles. Their centers lie on the line $AB,$ where $A(a), B(b).$
To find the Apollonian circle (its center and radius) in the particular case
$$\left|\frac{z-1}{z+1}\right|=\sqrt2,$$
it suffices to consider $z\in \mathbb{R}$ because $AB$ is the real axis.
We find two real values $z=-3\pm 2\sqrt2.$ They are limit points of a diameter of the circle.
The midpoint $C(-3)$ is the center, the radius is $2\sqrt2.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3860623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Evaluate $\int \frac{1-x^2}{(1+x^2)\sqrt {1+x^4}} dx$ $$\int \frac{\frac{1}{x^2}-1}{(x+\frac 1x)\sqrt{\frac{1}{x^2} + x^2}}dx$$
Let $x+\frac 1x = t$
$$-\int \frac{dt}{t\sqrt {t^2-2}}$$
Let $\sqrt{t^2-2} =u$
$$-\int \frac{du}{t^2}$$
$$-\int \frac{du}{u^2+2}$$
$$-\frac{1}{\sqrt 2} \arctan (\frac{u}{\sqrt 2})$$
$$-\frac{1}{\sqrt 2} \arctan (\frac{1}{\sqrt 2} \sqrt { x^2 +\frac{1}{x^2}}$$
But the given answer is$ \frac {1}{\sqrt 2} \arcsin (\frac{\sqrt {2 x}}{x^2+1})$
Where am I going wrong?
|
You need to take care of the sign in your approach. The result you derived is valid only for $x>0$. The full result valid for all domain is, instead
$$\int \frac{1-x^2}{(1+x^2)\sqrt {1+x^4}} dx
=-\frac{\text{sgn}(x)}{\sqrt 2} \arctan \frac{\sqrt { x^2+\frac{1}{x^2}}}{\sqrt 2}
=-\frac{1}{\sqrt 2} \arctan \frac{ \sqrt { x^4 +1}}{{\sqrt 2}x}
$$
which, as expected, differs from the given answer by a constant, i.e.
$$-\frac{1}{\sqrt 2} \arctan \frac{ \sqrt { x^4 +1}}{{\sqrt 2}x}- \frac {1}{\sqrt 2} \arcsin \frac{\sqrt {2}x}{x^2+1}=-\frac{\pi}{2\sqrt2}$$
Note
$$-\int \frac{dt}{t\sqrt {t^2-2}}
= \int \frac{d(\frac1t)}{\sqrt {1-\frac2{t^2}}}= \frac1{\sqrt2}\arcsin\frac{\sqrt2}t= \frac {1}{\sqrt 2} \arcsin \frac{\sqrt {2}x}{x^2+1}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3867040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
From 7 men & 4 women, 4 are to be selected to form a committee so that at least a woman is there on the committee. In how many ways can it be done?
From $7$ men & $4$ women, $4$ people are to be selected to form a committee so that at least a woman is there on the committee. In how many ways can it be done?
I was trying this in the following way, but surely I am missing something:
As there must be $1$ woman, there are $\binom{4}{1}$ (using the other notation this is $C_4^1$) ways to select $1$ woman from $4$
There are $\binom{10}{3}$ ( or$C_{10}^3$) ways to select $3$ people from the remaining $10$.
So $\binom{4}{1}\times \binom{10}{3} = 480$ ways.
|
There are 7 men and 4 women, and we need 4 members on the committee such that at least there is one woman on it. In that regard, we can make the following combinations.
We can have 4 women and 0 men. Or we can have 3 women and 1 man. Or we have 2 women and 2 men. Or we can have 1 woman and 3 men. We have to stop at this point because we need at least one woman on the committee. Start with the first case, where we have 4 women and 0 men. There $\binom{4}{4}\times\binom{7}{0}$ ways to form the committee.
In the second case where we have 3 women and 1 man, there are $\binom{4}{3}\times\binom{7}{1}$ ways to form the committee. In the third case, there are $\binom{4}{2}\times\binom{7}{2}$ ways to form the committee. In the last case, there are $\binom{4}{1}\times\binom{7}{3}$ ways to form the committee. To find the total number of ways to form the 4-member committee, we need to add the ways from the individual cases. Therefore, there are
$$\binom{4}{4}\times\binom{7}{0}+\binom{4}{3}\times\binom{7}{1}+\binom{4}{2}\times\binom{7}{2}+\binom{4}{1}\times\binom{7}{3}\\ =1+28+126+140\\=295$$ ways to form the committtee.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3867924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Solve $\cos\theta-3\cos2\theta+\cos3\theta=\sin\theta-3\sin2\theta+\sin3\theta$ My attempt:
\begin{align*}
\cos\theta-3\cos2\theta+\cos3\theta&=\sin\theta-3\sin2\theta+\sin3\theta\\
\cos\theta-3\cos2\theta+4\cos^3\theta-3\cos\theta&=\sin\theta-3\sin2\theta+3\sin\theta-4\sin^3\theta\\
-2\cos\theta-3\cos2\theta+4\cos^3\theta&=4\sin\theta-3\sin2\theta-4\sin^3\theta
\end{align*}
I have faced several symmetric trigonometry problems most of them need to use product to sum identities, but this one I can't continue.
|
It is a false identity. Click here to see the graphs.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3869446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Find $\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2}$ Please help me find:
$$
\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2}
$$
I cannot use L'Hospital's rule.
I tried to eliminate $x-3$, but I have no idea what to do next.
$$
\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2} =
\lim_{x \to 3} \frac{x^{14} + 3x^{13} + ... + 3^{14} - 15 \cdot 3^{14}}{x-3}
$$
|
Let $x=3+u$ with $u\to 0$ then use binomial formula to expand $(3+u)^{15}$.
You can ignore terms in $u^k$ with $k\ge 3$ because when divided by $u^2$ they converge to $0$.
$\require{cancel}\begin{align}f(x)&=\dfrac{(3+u)^{15}-3^{15}-15\cdot3^{14}u}{u^2}\\\\&=\dfrac{\bigg(\cancel{3^{15}}+\cancel{15\cdot3^{14}u}+\binom{15}{2}3^{13}u^2+o(u^2)\bigg)-\cancel{3^{15}}-\cancel{15\cdot3^{14}u}}{u^2}\\\\&=3^{13}\times\frac{14\times15}{2}+o(1)\to 35\times 3^{14}\end{align}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3869567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Application of the Cauchy-Schwarz Inequality Need to prove the following:
$(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c)^2\leq \frac{1}{2}a^2+\frac{1}{3}b^2+\frac{1}{6}c^2$
using the Cauchy-Schwarz inequality however direct application yields $(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c)^2\leq \frac{7}{18}(a^2+b^2+c^2)$ which is a dead end.
|
Because by C-S:
$$\frac{1}{2}a^2+\frac{1}{3}b^2+\frac{1}{6}c^2=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)\left(\frac{1}{2}a^2+\frac{1}{3}b^2+\frac{1}{6}{c^2}\right)\geq\left(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c\right)^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3869873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Given $\cos x + 3\sin x = \sqrt{10} \cos(x-71.6)$, find the second solution in the interval $0 < \theta < 90$ $$\cos x + 3\sin x = \sqrt{10} \cos(x-71.6)$$
I've proven the above true. I must now solve this:
$$\cos 2\theta + 3\sin 2\theta = 2, 0<\theta<90$$
Here's what I've done so far:
$$\alpha = 2\theta - 71.6$$
$$\cos\alpha = \frac{2}{\sqrt{10}}$$
$$\alpha = 50.77$$
Now I've found $\theta = 61.2$, but there's another solution -- $\theta = 10.4$.
I don't understand how to find the second solution. The cos graph does not become positive again until the $270-360$ range, so the fourth quadrant, and they've restricted me to $0 < \theta < 90$.
Can someone explain (in some detail if possible) why and how to find my second solution? Does that same process work in general for problems like this? I often miss roots for the less simple trig angle business.
|
$\cos 2x + 3\sin 2x = \sqrt 10\cos (2x - \arctan 3)=2\\
\cos (2x - \arctan 3) = \frac {2}{\sqrt 10}\\
2x - \arctan 3 = \arccos\frac {2}{\sqrt 10}\\
x = \frac 12 (\arccos \frac {2}{\sqrt {10}} + \arctan 3)$
Regarding the second solution.
Since $\cos x = \cos -x$
Then $\alpha = -\arccos \frac {2}{\sqrt 10}$ will point to the second solution.
$x = \frac 12 (-\arccos \frac {2}{\sqrt {10}} + \arctan 3)$
Is also a solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3870569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Find the internal angle of the triangle between $\overrightarrow a \& \overrightarrow b $. If $\overrightarrow a = \hat j + \sqrt 3 \hat k;\overrightarrow b = - \hat j + \sqrt 3 \hat k;\overrightarrow c = 2\sqrt 3 \hat k$ form a triangle then find the internal angle of the triangle between $\overrightarrow a \& \overrightarrow b $.
My approach is as follow
Let $\overrightarrow a + \overrightarrow b = \overrightarrow c $
$\overrightarrow a = \hat j + \sqrt 3 \hat k;\overrightarrow b = - \hat j + \sqrt 3 \hat k;\overrightarrow c = 2\sqrt 3 \hat k$
$ \Rightarrow {\left( {\overrightarrow a + \overrightarrow b } \right)^2} = {\left( {\overrightarrow c } \right)^2} \Rightarrow {\overrightarrow a ^2} + {\overrightarrow b ^2} + 2\overrightarrow a .\overrightarrow b = {\overrightarrow c ^2}$
$ \Rightarrow \overrightarrow a .\overrightarrow b = 2 \Rightarrow $
$\cos \theta = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|}} = \frac{1}{2} = \frac{\pi }{3}$
But official answer is $\frac{2\pi }{3}$ where I am making mistake
|
The best way to find the angle is tp find the lengths of the side vectors here they are $a=2,b=2, c=\sqrt{12}$, so the required angle is $C$, then by cosine law
$$\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{2+2-12}{8}=-\frac{1}{2} \implies C=\frac{2\pi}{3}.$$
If you want to it by dot product of $\vec a$ and $\vec b$, the angle between $\vec a$ and $\vec B$ is given by $$\cos \phi =\frac{\vec a. \vec b}{a b}=\frac{1}{2} \implies \phi =\frac{\pi}{3}.$$ But the internal abgke of the trangle is given by $\theta=\pi-\phi= \frac{2\pi}{3}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3873356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Understanding the solution to $4^{3x+6}+3^{2x+3}=4^{3x+4}+2\times3^{2x+4}$ For the last 5 hours I tried to solve the equation below (without luck).
$$4^{3x+6}+3^{2x+3}=4^{3x+4}+2\times3^{2x+4}$$
A friend of mine then told me that the solution is
$$\frac{\ln(16/3)}{\ln(3/8)}$$
Can anyone help me understanding why this is the solution? No matter how I turn it, I never get to this solution.
My last final solution was
$$\frac{\ln(135)-\ln(16)}{\ln(64)-\ln(9)}$$
I'm looking forward to your answers, as I'm quite desperate at the moment.
|
$$4^{3x+6}+3^{2x+3}=4^{3x+4}+2 \times 3^{2x+4}$$
$$\Longleftrightarrow 16(4^{3x+4})+3^{2x+3}=4^{3x+4}+6(3^{2x+3}) $$
$$\Longleftrightarrow 15(4^{3x+4})=5(3^{2x+3}) $$
$$\Longleftrightarrow 3(4^{3x+4})=3^{2x+3} $$
$$\Longleftrightarrow \ln(3)+(3x+4)\ln(4)=(2x+3)\ln(3)$$
$$\Longleftrightarrow x(3\ln(4)-2\ln(3))=2\ln(3)-4\ln(4)$$
$$\Longleftrightarrow x=\frac{2\ln(3)-8\ln(2)}{6\ln(2)-2\ln(3)} = - \frac{\ln\left( \frac{16}{3}\right)}{\ln\left( \frac{8}{3}\right)}$$
so your friend is right !
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3876760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Conditional Expectation of Square of a Gaussian Consider two Gaussian random variables $X\sim N(0,\sigma_X^2)$ and $Y\sim N(0,\sigma_Y^2)$ with known $E[XY] = \sigma_{XY}$. The question is how to find $E[X^2\mid Y]$ assuming $X$ and $Y$ are jointly Gaussian.
My approach: I thought of the following steps:
*
*Find the pdf $f_Y(y)$;
*Find the joint pdf $f_{X,Y}(x,y)$;
*Now, we have the conditional pdf $f_{X\mid Y}(x\mid y) = f_{X,Y}(x,y)/f_Y(y)$;
*$E[X^2\mid Y=y] = \int_{-\infty}^{\infty}x^2f_{X\mid Y}(x\mid y)\ dx$.
Is my approach correct? If yes, are there any other more elegant solutions?
|
Your approach is right if the distribution of $X$ and $Y$ is jointly Gaussian, rather than only each by itself being Gaussian. For example, suppose one had $Y = \begin{cases} +X & \text{if } |X|>c, \\ -X & \text{if } |X|<c. \end{cases}\quad$ Then it can be shown that $X$ and $Y$ are both Gaussian if $X$ is Gaussian, but the pair $(X,Y)$ is not Gaussian, since the probability that it is $0$ is positive.
Also, I would write $f_{X,Y}$ rather than $f_{XY}$ so as not to confuse it with the density of the product $XY.$
You don't need step 1.
The density is
\begin{align}
& c\cdot\exp \left( \frac{-1}{2(1-\rho^2)} \left( \left( \frac x {\sigma_X}\right)^2 + \left( \frac y {\sigma_Y} \right)^2 - 2\rho \left( \frac x {\sigma_X} \right) \left( \frac y {\sigma_Y} \right) \right) \right) \\[6pt]
& \qquad \text{where } \rho = \frac{\sigma_{X,Y}}{\sigma_X \sigma_Y} = \operatorname{cor} (X,Y). \\[4pt]
& \qquad \text{and } c = \tfrac 1 {2\pi\sqrt{\sigma_X^2\sigma_Y^2- \sigma_{X,Y}^2}}
\end{align}
To find the conditional density of $X$ given $Y$ we view the quadratic function of $x$ and $y$ just as a function of $x$ and complete the square:
\begin{align}
& \left( \frac x {\sigma_X}\right)^2 + \left( \frac y {\sigma_Y} \right)^2 - 2\rho \left( \frac x {\sigma_X} \right) \left( \frac y {\sigma_Y} \right) \\[8pt]
= {} & \left[ \left( \frac x {\sigma_X}\right)^2 - 2\rho \left( \frac x {\sigma_X} \right) \left( \frac y {\sigma_Y} \right) \right] + \left( \frac y {\sigma_Y} \right)^2 \\[8pt]
= {} & \left[ \left( \frac x {\sigma_X}\right)^2 - 2\rho \left( \frac x {\sigma_X} \right) \left( \frac y {\sigma_Y} \right) + \rho^2 \left( \frac y {\sigma_Y} \right)^2 \right] + \left( \frac y {\sigma_Y} \right)^2 - \rho^2 \left( \frac y {\sigma_Y} \right)^2 \\[8pt]
= {} & \left[ \frac x {\sigma_X} - \rho\cdot\frac y {\sigma_Y} \right]^2 + {} \underbrace{ (1-\rho^2) \left( \frac y {\sigma_Y} \right)^2}_\text{No “$x$” appears here.}
\end{align}
Things not depending on $x$ are in this context constants, so we have
\begin{align}
f_{X\,\mid\,Y\,=\,y} (x) & = \text{constant} \times\exp\left( -\frac 1 {2(1-\rho^2)} \left[ \frac x {\sigma_X} - \rho\cdot\frac y {\sigma_Y} \right]^2 \right) \\[8pt]
& = \text{constant} \times \exp\left( -\frac 1 {2\sigma_X^2(1-\rho^2)} \left[ x - \frac{\sigma_{X,Y}}{\sigma_Y^2}\cdot y \right]^2 \right)
\end{align}
This is a Gaussian density with variance $\sigma_X^2 (1-\rho)^2$ and expectation $\dfrac{\sigma_{X,Y}}{\sigma_Y^2} \cdot y.$
The expected value of the square of a random variable is the sum of its variance and the square of its expected value, thus it is
$$
\sigma_X^2(1-\rho^2) + \left( \frac{\sigma_{X,Y}}{\sigma_Y^2} \cdot y\right)^2.
$$
This can be viewed as a weighted average:
$$
(1-\rho^2) \sigma_X^2 + \rho^2\left( \frac{\sigma_X}{\sigma_Y}\cdot y \right)^2.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3877089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.