Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Use Integration by Parts to prove that $\int x^{n}\ln{x}\ dx=\frac{x^{n+1}}{(n+1)^{2}}\left[-1+(n+1)\ln{x}\right]+c$ I've gotten most of the way, but I can't see how I can transform my answer into the form in the assignment:
Use Integration by Parts to prove that $\displaystyle\int x^{n}\ln{x}\ dx=\frac{x^{n+1}}{(n+1)^{2}}\left[-1+(n+1)\ln{x}\right]+c$
\begin{align}
\int x^n\ln{x}\ dx&=\frac{\ln{x}\cdot x^{n+1}}{n+1}-\int\frac{x^{n+1}}{n+1}\cdot\frac{1}{x}\ dx\\
&=\frac{\ln{x}\cdot x^{n+1}}{n+1}-\frac{1}{n+1}\int x^n\ dx\\
&=\frac{\ln{x}\cdot x^{n+1}}{n+1}-\frac{1}{n+1}\left[\frac{x^{n+1}}{n+1}\right]+C\\
&=\frac{x^{n+1}}{(n+1)^2}+\dots
\end{align}
| Your work is fine. We just need to algebraically "manipulate" the result a bit to get the answer to match the given equality (as stated in your title):
Starting with your second-to-last line, we find a common denominator, $(n+1)^2$, and then factor out the common factor, which is the term you list in your last line, but is a factor (which multiplies over a sum/difference). So we have the common factor $\;\dfrac{x^{n+1}}{(n+1)^2}\times\Big[\cdots\Big]$:
$$
\begin{align}
\int x^n\ln{x}\ dx
& = \frac{\ln{x}\cdot x^{n+1}}{n+1}-\frac{1}{n+1}\left[\frac{x^{n+1}}{n+1}\right]+C \\ \\
& = \frac{(n+1)\ln x\cdot x^{n+1}}{(n+1)^2} - \frac{x^{n+1}}{(n+ 1)^2} + C \\ \\
& = \frac{x^{n+1}}{(n+1)^2}\Big[\left((n+1)\cdot \ln x\right) - 1\Big] + C \\ \\
& = \frac{x^{n+1}}{(n+1)^2}\Big[-1 + (n+1)\cdot \ln x\Big] + C \\ \\
\end{align}
$$
...which is now in the desired form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to find last items of order to get its sum Find the sum of order:
$$\sum_{n=1}^{∞}\left(\frac{\frac{3}{2}}{2n+3}-\frac{\frac{3}{2}}{2n-1}\right)$$
There is how they count it in book:
$$s_{n} = \left(\frac{3}{10}-\frac{3}{2}\right)+\left(\frac{3}{14}-\frac{1}{2}\right)+\left(\frac{1}{6}-\frac{3}{10}\right)+\left(\frac{3}{22}-\frac{3}{14}\right)+...+\left(\frac{3}{4n-2}-\frac{3}{4n-10}\right)+\left(\frac{3}{4n+2}-\frac{3}{4n-6}\right)+\left(\frac{3}{4n+6}-\frac{3}{4n-2}\right)$$
$$s_{n} = \frac{-3}{2}+\frac{-1}{2}+\frac{3}{4n+2}+\frac{3}{4n+6}$$
$$s = \lim_{n->∞}s_{n} = \lim_{n->∞}\left [\frac{-3}2 - \frac12 + \frac3{4n+2} + \frac3{4n+6}\right ] = -2$$
I understand how to solve the lim, I just dont understand, how to get those last items from order, I mean this items:
$$\left(\frac{3}{4n-2}-\frac{3}{4n-10}\right)+\left(\frac{3}{4n+2}-\frac{3}{4n-6}\right)+\left(\frac{3}{4n+6}-\frac{3}{4n-2}\right)$$
UPDATE
Now I understand how to get those "last" items. But I'm confused now, why are they even in the sum inside of lim of $s_n$? If I would keep counting next items, they would get canceled. For example, there is $\frac3{4n+2}$ in $s_n$, if I count n+1 item, this would get canceled. So why do we count them in $s_n$ if only first two fractals {$\frac{-3}2, \frac12 $} couldn't be canceled (if not thinking of negative n).
Could anyone explain please?
| I hope i didn't missunterstood your question, this sum is a so called telescoping sum, the other terms are canceled so you only have those left which are mentioned.
Those are
\begin{align*}
\frac{3}{4n-10}&= \frac{3}{4(n-1)-6}\\
\frac{3}{4n-6}&= \frac{3}{4(n-1)-2} \\
\end{align*}
And $$\frac{3}{4n-2}$$ is canceled from the last paranthesis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Factorization of three variables Prove that : $(a+b+c)^3-(b+c)^3-(c+a)^3-(a+b)^3+a^3+b^3+c^3=6abc$
Since $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a) $
Therefore the equation becomes : $2(a^3+b^3+c^3)+3(a+b)(b+c)(c+a) - [(c+a)^3 +(a+b)^3 +(b+c)^3]$
Putting $A:=(c+a)$ ; $B:=(a+b)$ ; $C:=(b+c)$
$[(c+a)^3 +(a+b)^3 +(b+c)^3] $ becomes $(A^3+B^3+C^3)$ now again using the formulae:$ a^3+b^3+c^3 = (a+b+c)^3-3(a+b)(b+c)(c+a)$
Could you please guide if any other easier way of doing this...
| If you simply expand everything and collect terms, everything will cancel out and reduce down to 6abc. However, if you want to factorize it, think about the other terms you have -
As you said, $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$. So then what does $(a+b)^3$ equal? If you expand it out following the same format as $(a+b+c)^3$ you get $a^3+b^3+3ab(a+b)$
So now your expression becomes
$$a^3+b^3+c^3+3(a+b)(b+c)(c+a)-a^3-b^3-3ab(a+b)-b^3-c^3-3bc(b+c)-a^3-c^3-3ac(a+c)+a^3+b^3+c^3$$
Clean up all the cubes in there and you get
$$3(a+b)(b+c)(c+a)-3ab(a+b)-3bc(b+c)-3ac(a+c)$$
Expand the first part of that to get
$$3(2abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)-3ab(a+b)-3bc(b+c)-3ac(a+c)$$
Which becomes
$$3(2abc+ab(a+b)+bc(b+c)+ac(a+c))-3(ab(a+b)+bc(b+c)+ac(a+c))$$
Now just cancel out all those like terms and you get $6abc$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the critical point and local extreme value? How would I find the critical point of the following function.
$f(x)=x^{2/3}+2x^{-1/3}$
This is what I did.
$\frac{2}{3}x^{-1/3}-\frac{2}{3}x{^\frac{-4}{3}}$
$\frac{2}{3\sqrt[3]{x}}-\frac{2}{3\sqrt[3]{x^4}}$
but how do I simplify it and get the critical point.
| $$
\begin{align}
f'(x) = \frac{2}{3}x^{-1/3}-\frac{2}{3}x^{-4/3}
& = \frac 23\left(x^{-1/3} - x^{-4/3}\right) \\ \\
& = \frac 23\left(x^{3/3}x^{-4/3} - x^{-4/3}\right) \\ \\
& = \frac 23 x^{-4/3}\left(x^{3/3} - 1\right) \\ \\
& = \frac 23 x^{-4/3}(x - 1) \\ \\
& = \frac{2(x-1)}{3x^{4/3}} \\ \\
& = \frac{2(x-1)}{3\sqrt[\large 3]{x^4}} \\
\end{align}
$$
Now, we have critical points where
*
*$f'(x)$ is undefined: at $x = 0$, as is $f(x)$
*$f'(x) = 0:$ This will occur if and only when the numerator is zero. In this case, $f'(x) = 0$ when $(x - 1) = 0 \implies x = 1$.
Let's look at the Wolgram Alpha graph below (the blue curve): We see that $f(x)\to \infty$ as $x \to 0$. That is, there exists a vertical asymptote at $x = 0$. We also see that at $x = 1$ we have a local minimum. You can use the derivative to tell you when $f(x)$ is increasing ($f'(x) > 0), when decreasing ($f'(x) \lt 0$), and you already know where it is neither increasing nor decreasing, given the points we've found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate these $4$ integrals?
\begin{align}
(1) & \int \frac{\sqrt{x^4 + \frac{1}{x^4} + 2}}{x^3}dx\\
(2) & \int \sqrt{1 - \sin(2x)}dx & \forall x \in [-\pi/4, \pi/4]\\
(3) & \int \cos^2(2x) \cos(x)dx\\
(4) & \int \frac{dx}{\sqrt{x +1}(x+5)}
\end{align}
I'd be very grateful for any help and hints!
| $1$. $\sqrt{x^4 + \dfrac1{x^4} + 2} = x^2 + \dfrac1{x^2}$.
$2$. $\sqrt{1-\sin(2x)} = \sqrt{\cos^2(x) + \sin^2(x) -2 \sin(x) \cos(x)} = \cos(x) - \sin(x)\,\,\, \forall x \in [-\pi/4,\pi/4]$.
$3$. $\cos^2(2x) \cos(x) = \dfrac{1+\cos(4x)}2 \cdot \cos(x) = \dfrac{\cos(x)}2 + \dfrac{\cos(5x) - \cos(3x)}4$.
$4$. $\dfrac{dx}{\sqrt{x+1} (x+5)} \underset{\sqrt{x+1} = t}{\to} \dfrac{2tdt}{t(t^2+4)} = \dfrac{2dt}{t^2+4} = d \left(\arctan \left(\dfrac{t}2 \right)\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Closed form for the sum of even fibonacci numbers? I recently took a look a project euler, and I am trying to think of a smart way to do number 2. I looked at the sequence, and I saw that the question is basically asking for
$$
\sum_{i=1}^n F_{3i}
$$
For whatever n is gives me the nth even under a million. So, I did some fiddling around with this, and I got this to be equivalent to
$$
\sum_{i=0}^{n-1} F_{3i-1}(3^{n-i}-1)
$$
I was wondering, is there even a closed form for something like this? Or am I wasting my time? I couldn't find a closed form online anywhere.
| $$F_k=\frac{\left(1+\sqrt{5}\right)^k}{2^k\sqrt{5}}-\frac{\left(1-\sqrt{5}\right)^k}{2^k\sqrt{5}}$$
$$\sum_{k=1}^nF_{3k}=\sum_{k=1}^n\frac{\left(1+\sqrt{5}\right)^{3k}}{2^{3k}\sqrt{5}}-\sum_{k=1}^n\frac{\left(1-\sqrt{5}\right)^{3k}}{2^{3k}\sqrt{5}}$$
$$=\frac{1}{\sqrt5}\sum_{k=1}^n\left(\frac{1+\sqrt{5}}{2}\right)^{3k}-\frac{1}{\sqrt5}\sum_{k=1}^n\left(\frac{1-\sqrt{5}}{2}\right)^{3k}$$
$$\text{ but we have , } x^3+x^6+x^9...x^{3n}=x^3\frac{x^{3n}-1}{x^3-1}$$
$$\text{ so then, }$$
$$=\frac{1}{\sqrt5}\sum_{k=1}^n\left(\frac{1+\sqrt{5}}{2}\right)^{3k}-\frac{1}{\sqrt5}\sum_{k=1}^n\left(\frac{1-\sqrt{5}}{2}\right)^{3k}$$
$$=\frac{1}{\sqrt5}\left(\left(\frac{1+\sqrt{5}}{2}\right)^3\frac{\left(\frac{1+\sqrt{5}}{2}\right)^{3n}-1}{\left(\frac{1+\sqrt{5}}{2}\right)^3-1}-\left(\frac{1-\sqrt{5}}{2}\right)^3\frac{\left(\frac{1-\sqrt{5}}{2}\right)^{3n}-1}{\left(\frac{1-\sqrt{5}}{2}\right)^3-1}\right)=\frac{F_{3n+2}-1}{2}$$
$$=\sum_{k=1}^{n}F_{3k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
Integrating $\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$ Could someone help with the following integration:
$$\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$$
So far I have done the following, but I am stuck:
I denoted $ y=-\cos x $ then:
$$\begin{align*}&\int^{1}_{-1} \frac{\arccos(-y) \sin x}{1+y^2}\frac{\mathrm dy}{\sin x}\\&= \arccos(-1) \arctan 1+\arccos 1 \arctan(-1) - \int^1_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\\&=\frac{\pi^2}{4}-\int^{1}_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\end{align*}$$
Then I am really stuck. Could someone help me?
| Let $y=x-\frac{\pi}{2}$.
\begin{array}{l}\displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x&=\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(y+\frac{\pi}{2}\right) \cos y}{1+\sin ^{2} y} d y\\&=\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{y \cos y}{1+\sin ^{2} y} d y +\frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos y}{1+\sin ^{2} y} d y
\\\displaystyle &=\displaystyle \pi \int_{0}^{\frac{\pi}{2}} \frac{d(\sin y)}{1+\sin ^{2} y}
\\&=\displaystyle 0+\pi\left[\tan ^{-1}(\sin y)\right]_{0}^{\frac{\pi}{2}}\\&=\displaystyle \frac{\pi^{2}}{4} \end{array}
Hence
$$\int_{\pi}^{0} \frac{x \sin x}{1+\cos ^{2} x} d x =-\frac{\pi^2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Using Hensel's Lemma to Factor a Polynomial over $\mathbb{Z}_4[x]$ We recently learned about codes over $\mathbb{Z}_4$, and Hensel's Lemma. The lemma is as follows:
Let $f(x) \in \mathbb{Z}_4[x]$. Suppose $\mu(f(x)) = h_1(x)h_2(x) \cdots h_k(x)$, where $h_1(x), h_2(x), \ldots , h_k(x)$ are pairwise coprime polynomials in $\mathbb{F}_2[x]$. Then, there exist $g_1(x),g_2(x), \ldots , g_k(x)$ in $\mathbb{Z}_4[x]$ such that:
(i) $\mu(g_i(x)) = h_i(x)$ for $1 \leq i \leq k$,
(ii) $g_1(x), g_2(x), \ldots , g_k(x)$ are pairwise coprime, and
(iii) $f(x) = g_1(x)g_2(x) \cdots g_k(x)$.
The map $\mu: \mathbb{Z}_4[x] \rightarrow \mathbb{F}_2[x]$ is defined by $\mu(f(x)) = f(x)(\mbox{mod } 2)$. It is also known at the reduction homomorphism.
I am interested in trying to factor $x^7 + 2x^6 + 2x^4 + 2x + 3$ as a product of basic irreducible polynomials in $\mathbb{Z}_4[x]$. I'm trying to follow the proof of this theorem, which can be found in Fundamentals of Error Correcting Codes by Huffman and Pless, on page 477.
So far, I figured out that $\mu(f(x)) = x^7 + 1$, which can be factored into: $$(x + 1)(x^3 + x^2 + 1)(x^3 + x + 1).$$
Now, I know these are pairwise coprime in $\mathbb{F}_2[x]$, but I am having trouble finding the pairwise coprime polynomials $g_1(x), g_2(x),$ and $g_3(x)$ such that $f(x) = g_1(x)g_2(x)g_3(x)$ and $\mu(g_i(x)) = h_i(x)$ for $i=1,2,3$.
I've been messing around with this, but I can't get anywhere. Any help would be greatly appreciated.
EDIT
After messing around with various combinations of $x+1$, $x^3 + 2x^2 + x + 1$, and $x^3 + x^2 + 2x + 1$ on WolframAlpha, I somehow stumbled across a combination in $Z_4[x]$ that works, but I'm not sure how to figure it out using a more concrete method.
$$g_1(x) = x+1,$$
$$g_2(x) = x^3 + 3 x^2 - 1,$$
$$g_3(x) = x^3 - 2x^2 + x + 1.$$
*
*These are pairwise coprime
*$\mu(g_1(x)) = x+1$, $\mu(g_2(x)) = x^3 + x^2 + 1$, $\mu(g_3(x)) = x^3
+ x + 1$
*$g_1(x)g_2(x)g_3(x) = x^7+2 x^6-4 x^5-2 x^4+8 x^3+4 x^2-2 x-1 = x^7 + 2x^6 + 2x^4 + 2x + 3 = f(x)$.
| We are going to lift the $\mathbb{F}_2$-factorization $$\bar{f} = (\underbrace{x^3 + x + 1}_{=:g})(\underbrace{x^4 + x^2 + x + 1}_{=:h})$$ to $\mathbb{Z}_4$. That is, considering $g$ and $h$ as polynomials in $\mathbb{Z}_4[x]$, we look for lift-polynomials $a,b\in\mathbb{F}_2[x]$, $\deg(a)\leq 2$, $\deg(b)\leq 3$ such that $$f = (g + 2a)(h + 2b).$$
Since $2\cdot 2 = 0$, we get
$$f = gh + 2(ah + gb)$$
With $gh = x^7 + 2x^5 + 2x^4 + 2x^3 + 2x^2 + 2x + 1$ this implies
$$2(ah + gb) = 2x^6 + 2x^5 + 2x^3 + 2x^2 + 2.$$
Equivalently, in $\mathbb{F}_2[x]$
$$ah + gb = x^6 + x^5 + x^3 + x^2 + 1.$$
Writing $a = a_2x^2 + a_1x + a_0$ and $b = b_3x^3 + b_2x^2 + b_1x + b_0$, expanding the left hand side gives $$(a_2 + b_3)x^6 + (a_1 + b_2)x^5 + (a_0 + a_2 + b_3)x^4 + (a_1 + a_2 + b_0 + b_2 + b_3)x^3 + (a_0 + a_1 + a_2 + b_1 + b_2)x^2 + (a_0 + a_1 + b_0 + b_1)x + (a_0 + b_0)\\ = x^6 + x^5 + x^3 + x^2 + 1.$$
Rewriting this as a system of $\mathbb{F}_2$-linear equations:
$$\begin{pmatrix}
0 & 0 & 1 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 & 1 & 0 & 1 \\
0 & 1 & 1 & 1 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 & 1 & 1 & 0 \\
1 & 1 & 0 & 1 & 1 & 0 & 0 \\
1 & 0 & 0 & 1 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
a_0 \\ a_1 \\ a_2 \\ b_0 \\ b_1 \\ b_2 \\ b_3
\end{pmatrix}
=
\begin{pmatrix}
1 \\
1 \\
0 \\
1 \\
1 \\
0 \\
1
\end{pmatrix}.
$$
Note that the equation system matrix is the Sylvester matrix of $g$ and $h$. It is invertible, because its determinant is the resultant of $g$ and $h$ which is nonzero since $g$ and $h$ are coprime.
So there is a unique solution $(a_0,a_1,a_2,b_0,b_1,b_2,b_3) = (0,0,1,1,1,1,0)$ which finally gives the $\mathbb{Z}_4$-factorization $$f = (x^3 + 2x^2 + x + 1)(x^4 + 3x^2 + 3x + 3)$$
You can do the same again to find the $\mathbb{Z}_4[x]$-factorization of $x^4 + 3x^2 + 3x + 3$. The final result is $$f = (x^3 + 2x^2 + x + 1)(x+1)(x^3 + 3x^2 + 3).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Consider the ordinary differential equation, $\frac{dy}{dx}=\frac{4+y^2}{1+x^2}$ Consider the ordinary differential equation, $\frac{dy}{dx}=\frac{4+y^2}{1+x^2}$
If $y(1)=2$, then find the value of $y(2)$
$\int \frac{dy}{4+y^2}=\int \frac{dx}{1+x^2} -- (|)$
$\frac{1}{2}\ tan^{-1}\frac{y}{2}=\tan^{-1}x+c$
$tan^{-1}\frac{y}{2}=2(\tan^{-1}x+c)$
$\frac{y}{2}=tan(2(tan^{-1}x+c))$
$y=2\tan(2(\tan^{-1}x+c))$
$$y(x) = 2 tan(2(\tan^{-1}x+c)) -- (||)$$
$y(1) = 2 \tan(2(tan^{-1}1+c))=2$
$2 tan(2(\tan^{-1}1+c))=2$
$tan(2(\tan^{-1}1+c))=1$
$tan(2(\tan^{-1}1+c))=tan(\frac{\pi}{4})$
$2(tan^{-1}1+c))=\frac{\pi}{4}$
$tan^{-1}1+c=\frac{\pi}{8}$
$\frac{\pi}{4}+c=\frac{\pi}{8}$
$c=\frac{\pi}{8}-\frac{\pi}{4}$
$c=-\frac{\pi}{8} -- (|||)$
From, (||) and (|||) we have,
$y(x) = 2 \tan(2(tan^{-1}x-\frac{\pi}{8}))$
| $$\frac12\arctan \frac y2=\arctan x+c$$
As $y(1)=2, \frac12\arctan 1=\arctan 1+c$
$\implies c=-\frac12\arctan 1=-\frac\pi8$
So, $$\frac12\arctan\frac y2=\arctan x-\frac\pi8\iff \arctan y=2\arctan x-\frac\pi4$$
$$\implies \frac y2=\tan\left(2\arctan x-\frac\pi4\right)$$
$$=\frac{\tan(2\arctan x)-\tan\frac\pi4}{1+\tan(2\arctan x)\tan\frac\pi4}\text{ applying }\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$
As $2\arctan x=\arctan \frac{2x}{1-x^2}, \tan(2\arctan x)=\tan\left(\arctan \frac{2x}{1-x^2}\right)=\frac{2x}{1-x^2}$
$$\implies \frac y2=\frac{\frac{2x}{1-x^2}-1}{1+\frac{2x}{1-x^2}}=\frac{2x-1+x^2}{1-x^2+2x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Variable Exponents Question If $a,b$ and $c$ are different positive integers and $2^a\cdot2^b\cdot2^c =64$ then $2^a+2^b+2^c$=?
This is so far my work: I got $2^a\cdot2^b\cdot2^c=2^6$ then $abc=6$ is this so far in the right track?
| Here's the $log$ solution,
$log 2^{a+b+c}=log 64$
$(a+b+c).(0.3010)=6.(0.3010)$
$a+b+c=6$
$a \neq b \neq c \implies log 2^a\neq log 2^b \neq log 2^c$
$log 2^{a+b+c}=log2^a+log2^b+log2^c=6.log2$
$a,b,c$ are distinct, therefore $(a,b,c)=(1,2,3),(3,2,1),(1,3,2),(2,3,1),(2,1,3),(3,1,2)$
Therefore, $2^a+2^b+2^c= 2+ 4+ 8 = 14$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/325127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
A recurrence relation on two variables How to solve the recurrence relation for $n >=m$:
$$P_{n,m}=\frac{n}{n+m}P_{n-1, m} + \frac{m}{m+n}P_{n,m-1}$$
$$P_{11}=\frac{1}{2}; P_{i,0}=1 \forall i > 0; P_{i,j}=0 \forall i<j$$
| Hint: Have you tried calculating the first few numbers. There is a interesting pattern.
$$\begin{array}{c|cccc}
&0&1&2&3\\
\hline
1&\frac{2}{2} & \frac{3}{3} & \frac{4}{4}& \frac{5}{5}\\
2&\frac{1}{2} & \frac{2}{3} & \frac{3}{4} & \ddots\\
3&0 & \frac{1}{3} & \frac{2}{4} & \ddots\\
4&0 & 0 & \frac{1}{4} & \ddots\\
\end{array}$$
It looks like, that your formular for $P_{n,m}$ is
$$P_{n,m} = \left\{\begin{array}{c,l}0\quad &\text{if }m>n\\
\frac{n+1-m}{n+1}\quad &\text{else}\end{array}\right.$$
Try to proof that by using induction over $n$ and over $m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/325437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Norm $N$ on $\Bbb Z[\sqrt d]$ is multiplicative, $N(x)$ prime $\Rightarrow x$ irreducible For the ring $\Bbb{Z}[\sqrt{d}]=\{a+b\sqrt{d}\mid a,b \in \Bbb{Z}\}$ where $d$ is not divisible by the square of a prime, prove that the norm $N(a+b\sqrt{d})= |a^2-db^2|$ satisfies the assertions: $N(x)=0$ if and only if $x=0$; $N(xy)=N(x)N(y)$ for all $x$ and $y$; $x$ is a unit if and only if $N(x)=1$; and if $N(x)$ is prime, then $x$ is irreducible in $\Bbb{Z}[\sqrt{d}]$.
| We define $\mathbb Z[\sqrt d] = \{a + b \sqrt d : a, b \in \mathbb Z\}$ where $d$ is a square-free integer.
Claim: $N(x) = 0$ if and only if $x = 0$.
Proof: If $N(x) = N(a + b \sqrt d) = 0$, then $N(x) = \vert a^2 - db^2 \vert = 0$ which means $a^2 = \vert d \vert b^2$. If either $a$ or $b$ are not zero, the other one isn't and we get that $d$ is not square-free, a contradiction. Conclude that $a = b = 0$ and hence $x = 0$.
Conversely, if $x = a + b \sqrt d = 0$, then $a = b = 0$ which means $N(x) = \vert a^2 - d b^2 \vert = 0$.
Claim: $N(xy) = N(x)N(y)$ for all $x, y \in \mathbb Z[\sqrt d]$.
Proof: Let $x, y \in \mathbb Z[\sqrt d]$ with $x = a + b \sqrt d$ and $y = m + n \sqrt d$. Observe that
\begin{align*}
N(x)N(y) &= \vert a^2 - db^2 \vert \vert m^2 - dn^2 \vert \\
&= \vert(am)^2 + (dnb)^2 - \big( (an)^2 - (bm)^2 \big) d \vert
\end{align*}
and
\begin{align*}
N(xy) &= N\big((a + b \sqrt d)(m + n \sqrt d) \big) \\
&= N\big( (am + bnd) + (an + bm) \sqrt d\big) \\
&= \vert (am + bnd)^2 - d (an + bm)^2 \vert \\
&= \vert(am)^2 + (dnb)^2 - \big( (an)^2 - (bm)^2 \big) d \vert.
\end{align*}
Conclude that $N(xy) = N(x)N(y)$ for all $x, y \in \mathbb Z[\sqrt d]$.
Claim: $x$ is a unit if and only if $N(x) = 1$.
Proof: Suppose that $x$ is a unit, then there exists $y$ such that $xy = 1$. Observe that $$1 = N(1) = N(xy) = N(x)N(y)$$ which means $N(x) = 1$.
Conversely, suppose that $\vert a^2 - d b^2 \vert = N(x) = 1$. Choosing $y = a - b \sqrt d$, we see that $$xy = (a + b \sqrt d)(a - b \sqrt d) = a^2 - db^2 = \pm 1.$$ Conclude that $x$ is a unit.
Claim: If $N(x)$ is prime, then $x$ is irreducible in $\mathbb Z[\sqrt d]$.
Proof: Suppose $N(x) = N(a + b \sqrt d) = p$ is prime (and hence irreducible since prime elements are always irreducible). Suppose, by way of contradiction, that $x$ is not irreducible, then we can write $x = yz$ where $y, z$ are not units. But then $$p = N(x) = N(y)N(z)$$ where $N(y), N(z) \neq 1$ by previous result. This contradicts that $p$ is irreducible (since the only units in $\mathbb Z$ are $\pm 1$). Conclude that $x$ is irreducible in $\mathbb Z[\sqrt d]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/325562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve using the definition of Big-$\mathcal{O}$ By using the definition of Big-$\mathcal{O}$, show that
\begin{align} 2^{(n+2)} + 3^{(n+1)} & \text{ is } \mathcal{O}(3^n)\\ \sqrt{10n^2+7n+3} & \text{ is } \mathcal{O}(n)\end{align}
I'm not sure where to start with this, especially with the last one. Can someone help please?
| By the definition of the Big-$O$ function:
$$ \left| 2^{n+2} + 3^{n+1} \right| \le A \cdot \left| 3^n \right| $$
$$ \left| 4 \cdot 2^n + 3 \cdot 3^n \right| \le A \left| 3^n \right| $$
By the triangular inequality:
$$ \left| 4 \cdot 2^n + 3 \cdot 3^n \right| \le \left| 4 \cdot 2^n \right| + \left| 3 \cdot 3^n \right| $$
We can show that for all $n \ge 0$ (since $3>2$):
$$ \left| 4 \cdot 2^n \right| + \left| 3 \cdot 3^n \right| \le \left| 4 \cdot 3^n \right| + \left| 3 \cdot 3^n \right| $$
So we have:
$$ 7 \cdot \left| 3^n \right| \le A \cdot \left| 3^n \right| $$
$$ 7 \le A $$
For $n$ sufficiently large (viz. $n\ge\frac{7 + \sqrt{61}}{2}$), we have:
$$ \sqrt{10n^2 + 7n + 3} \le \sqrt{11n^2} \le A \cdot n$$
$$ \sqrt{11} \le A $$
You can show that $11 n^2 > 10n^2 + 7n + 3$ by solving the quadratic appropriately to get that it is true for $n$ not between $\frac{7 \pm \sqrt{61}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/326850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Linear algebra proof, dependent or independent Suppose that $p_1=4-3x+6x^2+2x^3$, $p_2=1+8x+3x^2+x^3$, and $p_3=3-2x-x^2$ are vectors in $P_3$. Determine if $p_1$,$p_2$,and $p_3$ are linearly independent or dependent. Justify your answer.
So far, what I did was say:
Suppose $k_1$, $k_2$ and $k_3$ are constants.
Then I said:
$$k_1
\begin{pmatrix}
4\\
-3\\
2\\
6
\end{pmatrix}
+ k_2
\begin{pmatrix}
1\\
8\\
3\\
1
\end{pmatrix}
+ k_3 \begin{pmatrix}
3\\
-2\\
-1\\
0
\end{pmatrix}$$
Then I made the augmented matrix $$\begin{pmatrix}
4 & 1 & 3 &0 \\
-3 & 8 & -2 & 0\\
6 & 3 & -1 & 0\\
2 & 1 & 0 & 0
\end{pmatrix}$$
and found that the RREF is
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0
\end{pmatrix}
Therefore, $k_1 = k_2 = k_3 = 0$.
Therefore, $p_1$, $p_2$ and $p_3$ are linearly independent.
So far this is what I have. If this is right then how to I go about justifying my answer?
Also, is there a better way of writing vectors on this website?
| That absolutely justifies that $p_1,p_2,p_3$ are linearly independent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/328722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Approximating a sequence with funny recurrence Consider the sequence $a_n$ defined as $a_1=a_2=1, a_{n+1}(1+a_{n})=n+1$.
This sequence describes the average number of fixed points of an involution on an $n$-set, and one can approach the problem of approximating it using complex analysis, but I want to avoid it. Can some approximation be obtained by elementary methods?
EDIT: If it is true that the sequence is monotone increasing, we get the estimates $-\frac{1}{2} + \sqrt{n+\frac{1}{4}} \le a_n \le -\frac{1}{2} + \sqrt{n+\frac{5}{4}}$, i.e. $a_n = -\frac{1}{2} + \sqrt{n} + O(\frac{1}{\sqrt{n}})$, but why is it monotone?
EDIT2: Just to clarify the generating function part, note that $\exp(x+x^2/2)$ is the exponential generating function of the number of involutions and $\frac{\partial(\exp(tu+t^2/2))}{\partial u}|_{u=1} = t\exp(t+t^2/2)$ is the exponential generating function of $a_n \times b_n$ where $a_n$ is the number of involutions and $b_n$ the average of fixed points. This gives $b_n = na_{n-1} / a_{n}$ and the recurrence of $a_n$ gives a recurrence on $b_n$, the one I posted here. This can also be proved combinatorially.
| After you have guessed the bounds, it is now straightforward to prove them using induction.
We shall induct on $n$ to prove the statement $-\frac{1}{2} + \sqrt{n+\frac{1}{4}} \leq a_n \leq -\frac{1}{2} + \sqrt{n+\frac{5}{4}}$.
When $n=1$, we have $-\frac{1}{2}+\sqrt{1+\frac{1}{4}} \leq 1=a_1=-\frac{1}{2}+\sqrt{1+\frac{5}{4}}$.
Suppose it holds for $n=k$. Now $\frac{1}{2} + \sqrt{k+\frac{1}{4}} \leq a_k+1 \leq \frac{1}{2} + \sqrt{k+\frac{5}{4}}$, so using $a_{k+1}=\frac{k+1}{a_k+1}$, we get
$$-\frac{1}{2} + \sqrt{k+\frac{5}{4}}=\frac{(k+1)(-\frac{1}{2} + \sqrt{k+\frac{5}{4}})}{(k+\frac{5}{4})-\frac{1}{4}} =\frac{k+1}{\frac{1}{2} + \sqrt{k+\frac{5}{4}}} \leq a_{k+1} \leq \frac{k+1}{\frac{1}{2} + \sqrt{k+\frac{1}{4}}}$$
We now need to prove that $$k+1 \leq \left(\frac{1}{2}+\sqrt{k+\frac{1}{4}}\right)\left(-\frac{1}{2}+\sqrt{k+\frac{9}{4}}\right)$$.
Let $x=k+\frac{5}{4} \geq \frac{9}{4}>2$, so it suffices to prove that
\begin{align}
&x-\frac{1}{4} \leq -\frac{1}{4}+\frac{1}{2}\sqrt{x+1}-\frac{1}{2}\sqrt{x-1}+\sqrt{x^2-1} \\
\Leftrightarrow &\sqrt{x^2}-\sqrt{x^2-1} \leq \frac{1}{2}(\sqrt{x+1}-\sqrt{x-1}) \\
\Leftrightarrow &\frac{1}{\sqrt{x^2}+\sqrt{x^2-1}} \leq \frac{1}{\sqrt{x+1}+\sqrt{x-1}} \\
\Leftrightarrow & \sqrt{x+1}+\sqrt{x-1} \leq \sqrt{x^2}+\sqrt{x^2-1}
\end{align}
This last inequality is true since $x \geq 2$ implies $x^2 \geq 2x>x+1$.
We are thus done by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/328891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Prove $((n+1)!)^n < 2!\cdot4!\cdots(2n)!$ so I know I need to prove this via induction, but I am somewhat stuck. Here is what I have does so far.
Let $p(n) = (n+1)!^n \le 2!\cdot4!\cdot\ldots\cdot(2n)!$
*
*$p(2) = 3!^2\le 2!\cdot4!$
*Assume $p(n)$ is true.
*Prove $p(n+1)$
*$p(n+1) = (n+2)!^{n+1}\le 2!\cdot4!\cdot\ldots\cdot(2n+2)!$
$$\begin{align*}
&=(n+2)!^n\cdot(n+2)!\le\ldots\\
&=(n+1)!^n\cdot(n+2)^n\cdot(n+2)!\le\ldots
\end{align*}$$
Given that $(n+1)!^n\le 2!\cdot4!\cdot\ldots\cdot(2n)!$ (Using Inductive Hypothesis)
$$\begin{align*}
&=(n+2)^n\cdot(n+2)!\le(2n+2)!\\\\
&=(n+2)^n\le\frac{(2n+2)!}{(n+2)!}
\end{align*}$$
And I am stuck here. Any help anyone can give would be helpful.
| Since you have reached $ (n+2)^n.(n+2)! < (2n+2)! $ , I will lead you from here.
From this we can observe that there are 'n' number of $(n+2)$ terms on the LHS.
we can easily conclude that:
$(n+2) < (n+3)$ ---> $1st (n+2)$
$(n+2) < (n+4)$ ---> $2nd (n+2)$
.......and so on till
$(n+2) < (n+n+2)$ ---> $nth (n+2)$
So, we can clearly say that $(n+2)^n.(n+2)! < (n+3).(n+4)....(2n+2).(n+2)!$
=>$(n+2)^n.(n+2)! < (2n+2)!$
Hope the answer is clear !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/329039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Finding $a_n$ for $a_{n+1}=\frac{1}{2}\left(a_n+\frac{1}{a_n}\right)(n=1,2,3,\cdots),~a_1=2$ I would appreciate if somebody could help me with the following problem:
Q: find $a_n=?$
$$a_{n+1}=\frac{1}{2}\left(a_n+\frac{1}{a_n}\right)(n=1,2,3,\cdots),~a_1=2$$
There are quite a few posts on the main site concerning convergence of this sequence, for example, Calculating square roots using the recurrence $x_{n+1} = \frac12 \left(x_n + \frac2{x_n}\right)$ and more generally Calculating square roots using the recurrence $x_{n+1} = \frac12 \left(x_n + \frac2{x_n}\right)$. So from those posts we know that it converges to $\sqrt2$. But finding the $n$-th term is a slightly different task from finding the limit.
| Rewrite $a_n$ as $\frac{p_n}{q_n}$ for two sequences $p_n, q_n$ that need to be determined.
We have:
$$a_{n+1} = \frac12 ( a_n + \frac{1}{a_n} ) \iff \frac{p_{n+1}}{q_{n+1}} = \frac{p_n^2+q_n^2}{2p_nq_n}$$
We can choose $p_{n+1} = p_n^2 + q_n^2$ and $q_{n+1} = 2p_nq_n$ and get:
$$ p_{n+1} \pm q_{n+1} = ( p_n \pm q_n )^2 \implies p_{n} \pm q_{n} = (p_1 \pm q_1)^{2^{n-1}}$$
Since $a_1 = 2$, we take $p_1 = 2, q_1 = 1$ to get:
$$ p_{n} + q_{n} = 3^{2^{n-1}} \text{ and } p_{n} - q_{n} = 1$$
This leads to
$$ a_{n} = \frac{p_n}{q_n} = \frac{3^{2^{n-1}}+1}{3^{2^{n-1}}-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/329240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
The sum of infinite series $\sum_{k=1}^{\infty}2\sin\left(\frac{1}{k}-\frac{1}{k+1}\right)\cos\left(\frac{1}{k}+\frac{1}{k+1}\right)$
Determine the sum of the infinite series
$$\sum_{k=1}^{\infty}2\sin\left(\frac{1}{k}-\frac{1}{k+1}\right)\cos\left(\frac{1}{k}+\frac{1}{k+1}\right).$$
| This turns out to be a telescoping sum:
$$\sum_{k=1}^{\infty} \left ( \sin{\frac{2}{k}} - \sin{\frac{2}{k+1}}\right ) = \sin{2} - \sin{1} + \sin{1} - \sin{\frac{2}{3}} + \ldots = \sin{2}$$
This sum converges because, as $k \rightarrow \infty$, the summand approaches
$$\frac{2}{k (k+1)} \sim \frac{2}{k^2}$$
so the sum converges by the comparison test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/330639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve for $X$ in a simple equation system. I cannot really understand how to read this question, so please help me out here.
$$\left[ \begin{array}{ccc} 0 & -6 & 4\\ 1 & 2 & 7\end{array} \right] = 4X + 5 \left[ \begin{array}{ccc} 3 & 4 & 3\\ 8 & -4 & 8\end{array} \right]$$
First, how should i read this? Secondly how do I procede and solve for $X$, a full development would be very much appreciated!
Thank you kindly for you help!
| Two matrices are the same when all of their values are the same.
Lets call
$$X=\begin{pmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \end{pmatrix}$$
So your equations are
\begin{align*}
0&= 4 x_{11} +5 \cdot 3\\
-6&= 4 x_{12} + 5 \cdot 4\\
4&=4 x_{13} + 5 \cdot 3\\
1&= 4 x_{21} + 5\cdot 8\\
2&= 4x_{22} + 5 \cdot (-4)\\
7&= 4x_{23} + 5 \cdot 8
\end{align*}
Just solve those and write what $X$ is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/331014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Number of permutations of a specific cycle decomposition
Let $X(n)$ denote number of all permutations of $\left\{1,\ldots,n \right\}$ that have only cycles of even length. Let $Y(n)$ denote number of all permutations of the same set that have only cycles of odd length. Count $X(2n)-Y(2n)$.
I don't know even how to start. I counted this sum for $n=1,2,3$ and it seems that it is always equal to $0$, but it doesn't help with calculating it in general.
| The exponential generating function for permutations containing only even cycles is
$$ G_1(z) =
\exp\left( \sum_{k\ge 1} \frac{z^{2k}}{2k} \right) =
\sqrt{ \frac{1}{1-z^2}} = \frac{\sqrt{1-z^2}}{1-z^2}$$
and the one for odd cycles is
$$ G_2(z) = \exp\left( \sum_{k\ge 0} \frac{z^{2k+1}}{2k+1} \right).$$
We are looking for $$(2n)! [z^{2n}] \left(G_1(z)-G_2(z)\right).$$
Now we have
$$G_2(z) = \exp \log \frac{1}{1-z} \exp\left( - \sum_{k\ge 1} \frac{z^{2k}}{2k} \right)
= \frac{1}{1-z} \frac{1}{G_1(z)} =
\frac{1}{1-z}\sqrt{1-z^2}.$$
It follows that
$$ G_1(z) - G_2(z) = \sqrt{1-z^2} \left(\frac{1}{1-z^2} - \frac{1}{1-z}\right)
= - \sqrt{1-z^2} \frac{z}{1-z^2} = -z \frac{\sqrt{1-z^2}}{1-z^2}.$$
Now clearly this last term consists of a term whose series expansion contains only even powers of $z$, multiplied by $z$, hence there are only odd powers of $z,$ the two counts are equal for $2n$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/331504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Partial Fraction Decomposition Problem... half answered... $$\int \frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}dx$$
I know I will be using partial fraction decomposition on this problem, at least it seems that way. so far, what I have is this:
$$\frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}=\frac{A}{x+3}{}+\frac{B}{(x+3)^2}+\frac{Cx+D}{x^2+3}$$
Multiplying by the LCD : $(x+3)^2(x^2+3)$
I am left with :
$$5x^3+19x^2+27x-3=A(x+3)(x^2+3)+B(x^2+3)+(Cx+D)(x+3)^2$$
By setting $x=-3:B=-4$
Now is where I am running into trouble. Now that I can substitute B into the original decomposition equation, There is no value of x that will leave only one variable to solve for. Please lend me a hand you guys(and girls). Thanks!
| There are many options:
*
*Differentiate and substitute $x=-3$.
*Substitute simple values such as $x=0$ and $x=1$ and $x=-1$ to get three equations in three unknowns.
*The method @Shu mentions in the comments.
*Substitute $x=\sqrt{-3}$ and $x=-\sqrt{-3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Partial Fractions - combinatorics - trouble Having a very hard time with this question:
Q: Use partial fractions to find the power series expansion of $$\frac{1+5x}{1-2x-3x^2}$$
| The partial fraction decomposition is
$$-\frac{1}{1+x} - \frac{2}{-1+3x}$$
As the geometric series is
$$\frac{1}{1-x}=\sum_{k=0}^\infty x^k $$
So we have
$$\frac{1+5x}{1-2x-3x^2}=-\frac{1}{1+x}+\frac{2}{1-3x}=-\sum_{k=0}^\infty
(-1)^k x^k + 2\cdot \sum_{k=0}^\infty 3^k x^k=\sum_{k=0}^\infty (2 \cdot 3^k -(-1)^k) x^k $$
As you see that the partial fraction is really helpful here I will make it more explicit. we can use partial fraction decomposition if we know the zeroes of the denominator, when $x=-1$ we have
$$1 - 2 (-1)-3(-1)^2=1+2-3=0$$ so $x=-1$ is a zero. the other zero is at $x=\frac{1}{3}$.
Now we have
$$\frac{1+5x}{1-2x-3x^2}=\frac{1+5x}{(1+x)(1-3x)}$$
We try to write it as
$$\frac{1+5x}{(1+x)(1-3x)}=\frac{A}{1+x} + \frac{B}{1-3x}$$
The way to achiefe $A$ and $B$ is the following: we know that
$$\frac{A}{1+x} + \frac{B}{1-3x}= \frac{A(1-3x)+ B(1+x)}{1-2x-3x^2}$$
As we have
$$\frac{A(1-3x)+ B(1+x)}{1-2x-3x^2}=\frac{x(B-3A)+1(A+B)}{1-2x-3x^2}$$
Ok now we compare the coeffients of $1+5x$ and $x(B-3A)+1(A+B)$
We have
$$1=A+B$$
and
$$5 = B-3A$$
I hope I made no mistake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding real roots of $ P(x)=x^8 - x^7 +x^2 -x +15$ Let $ P(x)=x^8 - x^7 +x^2 -x +15 $, Descartes' Rule of Signs tells us that the polynomial has 4 positive real roots , but if we group the terms as $$ P(x)= x(x-1)(x^6+1) +15 $$ we find that $ P(x) $ does not have any positive roots.
Where did I err ?
| we have to calculate no. of ral roots of $x^8-x^7+x^2-x+15=0$
$\bf{Solution::}$ Let $f(x) = x^8-x^7+x^2-x+15$, we check real solution in $\bf{x\in \mathbb{R}}$
$\bullet$ If $x\leq 0$, Then $f(x) = x^8-x^7+x^2-x+15>0$
$\bullet$ If $0<x<1$, Then $f(x) = x^8+x^2(1-x^5)+(1-x)+14>0$
$\bullet$ If $x\geq 1$, Then $f(x)= x^7(x-1)+x(x-1)+15>0$
So we observe that $f(x)= x^8-x^7+x^2-x+15>0\;\forall x\in \mathbb{R}$
So $f(x)=x^8-x^7+x^2-x+1=0$ has no real roots for all $x\in \mathbb{R}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/335653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$
The number of signs increases by one in each "block".
I have an idea. Group the series like this: $1-(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\frac{1}{6})-...$
We can show that $1, \frac{1}{2}+\frac{1}{3},\frac{1}{4}+\frac{1}{5}+\frac{1}{6},...$ converges to 0. I'm trying to use Dirichlet's Test. However, I'm not sure wether this sequence is decreasing.
Any idea? Or any other method to establish the convergence?
| We can see that the general term of this series is
$$a_n=(-1)^n\sum_{k=\frac{n(n+1)}{2}+1}^{\frac{n(n+1)}{2}+n+1}\frac{1}{k}$$
and we have
$$H_n=\sum_{k=1}^n\frac{1}{k}=\log n+\gamma+\frac{1}{2n}+O(\frac{1}{n^2})$$
so
\begin{align}|a_n|=H_{\frac{n(n+1)}{2}+n+1}-H_{\frac{n(n+1)}{2}+1}&=\log(1+\frac{2}{n})+\frac{1}{(n+1)(n+2)}-\frac{1}{n(n+1)}+O(\frac{1}{n^2})\\
&=\frac{2}{n}+O(\frac{1}{n^2})\end{align}
hence we have
$$a_n=\frac{2(-1)^n}{n}+O(\frac{1}{n^2})$$
which allows us to conclude the convergence of the series since it's sum of two convergent series, one by alternating series test and the other by comparaison with Riemann series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 3
} |
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| Consider the function
$$
f(z) = \sum_{i=0}^{\infty}\frac{z^i}{2^i} = \frac z{2-z}.
$$
Its radius of convergence is $2$.
Its derivative is
$$
f'(z) = \sum_{i=1}^{\infty}\frac{i}{2^i}z^{i-1} = \frac 2{(2-z)^2}.
$$
Since $1<2$, it is allowed to evaluate the derivative at $z=1$ with the written formula and obtain
$$
\sum_{i=1}^{\infty}\frac{i}{2^i} = f'(1) = \frac 2{(2-1)^2} = 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 12,
"answer_id": 10
} |
Simplify : $\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a}$? Just out of curiosity, is $$\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a},a \gt0\quad?$$
Thanks
| $$\sqrt{\frac{a}{2}} + \sqrt{\frac{a}{2}} = 2\sqrt{\frac{a}{2}} = \frac{2}{\sqrt{2}}\sqrt{a}\geq \sqrt{a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Rank of the elliptic curve $y^2=x^3+px$ I need to prove that the rank of the curve $y^2=x^3+px$ is $0$, if $p\equiv 7 \pmod {16}$ is a prime.
Using the standard technique, we need to show that none of the following two equations admits an integer solution in M, N and e (with M, N and e pairwise coprime; M, e non-zero):
$2M^4-2pe^4=N^2$
$4M^4-pe^4=N^2$
I have got this after going modulo 16. But, $2M^4-2pe^4=N^2$ and $4M^4-pe^4=N^2$ do
admit solutions in $Z/16Z$
| For $2M^4-2pe^4=N^2$, we see that $N=2n$ for some integer $n$. Substituting in the equation and reducing by 2 we get $2n^2=M^4-pe^4$. Now if $p\equiv 7(mod 16)$ then we get
$n^2\equiv -3 (mod 16)$ so $n^2\equiv -3 (mod 8)$ which is a contradiction. Because
$x^2\equiv 0,1,4(mod 8) $ for integer $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
factorisation of algebra homework question that asks to perform the multiplication and division and simplification.
$$\frac{x^2+7x+10}{x^2+5x+4} \times \frac{x^2+3x+2}{x^2+4x+4} =$$
$$\frac{(x+5)(x+2)}{(x+4)(x+1)} \times \frac{(x+1)(x+2)}{(x+2)(x+2)} =$$
$$ = \frac{(x+5)}{(x+4)}$$
is my working out and answer correct?
I am typing this extra sentence in order to meet your quality standards!!!
| Yes, it is correct. The $x+2$'s and $x+1$ terms do cancel. If there was no multiplication, the second part would be $\frac{x+1}{x+2}$, and the left side would be as is. However, you're OK.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/342606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Can't find solutions for $\tan{2x} = \tan{x}$ Solving $\tan{2x}=\tan{x}$
Reducing the left side: $$\frac{\sin{2x}}{\cos{2x}} = \frac{2\sin{x}\cos{x}}{2\cos^{2}(x)-1}.$$
Reducing the right side: $$\tan{x} = \frac{\sin{x}}{\cos{x}}.$$
therefore:
$$\frac{2\sin x\cos x}{2\cos^2x-1} = \frac{\sin x}{\cos x}$$
$$\frac{2\cos x}{2\cos^2x-1} = \frac{1}{\cos x}$$
$$2\cos{^2}(x) = 2\cos{^2}(x) - 1$$
$$0 = -1 ????$$
| We know $\tan A=\tan B\implies A=n\pi+B$ where $n$ is any integer (Proof Below )
So, $\tan2x=\tan x\implies 2x=n\pi+x\implies x=n\pi$ where $n$ is any integer
[
Proof:
$\tan A=\tan B\implies \sin A\cos B=\cos A\sin B$
$\implies \sin(A-B)=0\implies A-B=n\pi$ where $n$ is any integer
]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/344015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Lagrange multipliers - maximum and minimum values given constraint Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer does not exist, enter DNE.)
$ \ \ f(x, y, z) = xyz \ ; \ \ x^2 + 2y^2 + 3z^2 = 96$
What I have gotten to:
$\Delta f = \ <yz, xz, xy>$
and $\Delta g = \ λ<2x, 4y, 6z>$
set them equal and get:
$x^2 = 2y^2$ and
$z^2 = \frac{2}{3} y^2$ .
Then:
$$
\begin{cases}x^2 + 2y^2 +3z^2 = 96 \\
6y^2 = 96 \\
y = \pm16 \end{cases}
$$
Plugging $y$ into $z^2$ and $x^2$ equations above, you get the points:
$(\pm16\sqrt{2}, \ \pm 16, \ \pm16\sqrt{2/3} ) \ .$
Plugging the points back into $f$, I got a maximum of $32\sqrt{3}$ and a minimum of $-32\sqrt{3}$.
Where did I go wrong?
Thank you in advance!
| I do not see where you used Lagrange multipliers. Any critical points will satisfy the Lagrange multipliers equation
$$
\begin{bmatrix}
yz & xz & xy
\end{bmatrix} = \lambda \begin{bmatrix}
2x & 4y & 6z
\end{bmatrix} \ .
$$
This gives you the system of equations
$$
\begin{align}
yz &= 2\lambda x & (1) \\
xz &= 4\lambda y & (2) \\
xy &= 6\lambda z & (3) \\
96 &= x^2+2y^2+3z^2 & (4)
\end{align}
$$
First consider $\lambda = 0$ and then consider other cases. Try this and see how it goes for you. I hope this helps.
EDIT:
If you assume $x,y,z\neq 0$, you can solve for $\lambda$ in each of $(1), (2), (3),$ and $(4)$. You then obtain (by equating these $\lambda$), as you did, $6y^2=96$, which gives $y=\pm4$. You similarly obtain $x^2 = 32$, or $x=\pm4\sqrt{2}$, and $z^2 = \frac{32}{3}$, or $z=\pm4\sqrt{\frac{2}{3}}$. With all of the $\pm$'s, you get a few critical points. Plug them in to see which ones are the largest/smallest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/344214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving $\gcd(n^2(n^2+1),2n+1)=\gcd(2n+1,5)$ We suppose $\forall n \in \mathbb {N}\setminus{0}$.
How can I prove that $\gcd(n^2(n^2+1),2n+1)=\gcd(2n+1,5)$?
| \begin{align}
(2n+1)^4 - 16n^2(n^2+1) & = 32n^3 + 8n^2 + 8n+1\\
4(2n+1)^3 - (32n^3 + 8n^2 + 8n+1) & = 40n^2+16n+3\\
10(2n+1)^2 - (40n^2+16n+3) & = 24n+7\\
12(2n+1) - (24n+7) &= 5
\end{align}
Hence,
\begin{align}
\gcd(n^2(n^2+1),2n+1) & = \gcd(32n^3 + 8n^2 + 8n+1,2n+1)\\
& = \gcd(40n^2+16n+3,2n+1)\\
& = \gcd(24n+7,2n+1)\\
& = \gcd(2n+1,5)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/345154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Sum of two quadratic forms Suppose I have two quadratic forms $Q_i(x)=(x-a_i)^T A_i(x-a_i)+c_i$, $i=1,2$ where $x,a_i \in \Bbb{R}^n$ and $A_i$ are positive-definite $n\times n$ matrices.
Then $Q(x)=Q_1(x)+Q_2(x)$ is also a quadratic form, $Q(x)=(x-a)^T A(x-a)+c$, with $A=A_1+A_2$ (easy to see by considering just the quadratic terms).
How do I find $a$ and, especially, $c$?
| Let's turn to homogenous coordinates. For example, for $n=2$ you get
\begin{align*}
Q(x) &= (x-a)^T\cdot A\cdot (x-a) + c
\\ &=
\begin{pmatrix}x_1\\x_2\\1\end{pmatrix}^T\cdot\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
-a_1 & -a_2 & 1
\end{pmatrix}\cdot\begin{pmatrix}
A_{1,1} & A_{1,2} & 0 \\
A_{2,1} & A_{2,2} & 0 \\
0 & 0 & c
\end{pmatrix}\cdot\begin{pmatrix}
1 & 0 & -a_1 \\
0 & 1 & -a_2 \\
0 & 0 & 1
\end{pmatrix}\cdot\begin{pmatrix}x_1\\x_2\\1\end{pmatrix}
\\&=
\begin{pmatrix}x_1\\x_2\\1\end{pmatrix}^T\cdot B
\cdot\begin{pmatrix}x_1\\x_2\\1\end{pmatrix}
\\
B &=\begin{pmatrix}
A_{1,1} & A_{1,2} & -(A_{1,1}a_1 + A_{1,2}a_2) \\
A_{2,1} & A_{2,2} & -(A_{2,1}a_1 + A_{2,2}a_2) \\
-(A_{1,1}a_1 + A_{2,1}a_2) &
-(A_{1,2}a_1 + A_{2,2}a_2) &
A_{1,1}a_1^2 + (A_{1,2}+A_{2,1})a_1a_2 + A_{2,2}a_2^2 + c
\end{pmatrix}
\end{align*}
Now you can simply compute the sum of two or more $Q_i$ by adding their $B_i$ matrices. From the upper left $n\times n$ block you can directly deduce the $A$ matrix of the result, just as you already wrote in your question. From the first $n$ entries of the last column you can then deduce the offset vector $a$, since they form a linear system of equations in its coordinates once you know $A$. Other answers have explicitely formulated its solution this using matrix inversion. When you have both $A$ and $a$, you can use the bottom right element to obtain $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/345789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Help evaluating $\lim_{\left|x\right| \to \infty}y$, given $\frac{y^2}{x^2}=\frac{b^2}{a^2}-\frac{b^2}{x^2}$ I'm trying to understand this step in a derivation of the standard equation of a hyperbola. We have constants $a$ and $b$, and variables $x$ and $y$. We've gotten to a point where we have
$$\frac{y^2}{x^2}=\frac{b^2}{a^2}-\frac{b^2}{x^2}$$
The text then states:
As x and y attain very large values, the quantity $\frac{b^2}{x^2}\to0$, so that $\frac{y^2}{x^2}\to \frac{b^2}{a^2}$. This shows that $y\to\pm\frac{b}{a}x$ as $\left|x\right|$ grows large.
I'm having a hard time getting from the first sentence to the second. How do we make that logical jump?
| Since $$\left(\frac yx\right)^2=\frac{y^2}{x^2}\to\frac{b^2}{a^2}=\left(\frac ba\right)^2$$ as $|x|$ grows large, then $$\left(\frac yx-\frac ba\right)\left(\frac yx+\frac ba\right)=\left(\frac yx\right)^2-\left(\frac ba\right)^2\to 0$$ as $|x|$ grows large, so $$\frac yx-\frac ba\to 0\quad\text{or}\quad\frac yx+\frac ba\to 0$$ as $|x|$ grows large, so $$\frac yx\to\frac ba\quad\text{or}\quad\frac yx\to-\frac ba$$ as $|x|$ grows large, so $$\frac yx\to\pm\frac ba$$ as $|x|$ grows large. Then $y\sim\pm\frac bax$ as $|x|$ grows large.
We can't actually take the limit of $y$ as $|x|\to\infty$, as $|y|$ necessarily grows without bound.
An alternate way to see this, if you're still having trouble, is to note that $$y^2=\frac{b^2}{a^2}x^2-b^2,$$ so $$\left(y-\frac bax\right)\left(y+\frac bax\right)=y^2-\left(\frac bax\right)^2=y^2-\frac{b^2}{a^2}x^2=-b^2.$$ Thus, $$\lim_{x\to\infty}\left(y-\frac bax\right)\left(y+\frac bax\right)=-b^2\in\Bbb R,\tag{$\heartsuit$}$$ but since $\left|\frac bax\right|\to\infty$ as $x\to\infty$, then $(\heartsuit)$ can only happen if $y-\frac bax\to 0$ or $y+\frac bax\to 0$ as $x\to\infty$, so $y\to\pm\frac bax$ as $x\to\infty$. Similarly, $y\to\pm\frac bax$ as $x\to-\infty$, so $y\to\pm\frac bax$ as $|x|\to\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/345924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then $3$ divides both $a$ and $b$. If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then show that $3$
divides both $a$ and $b$.
can anyone tell me please how to solve these types of problem oe which formula is required
| The divisibility is equivalent to $$3^2\mid 4(x^2+xy+y^2)=(2x+y)^2+3y^2$$. Since $3\mid 3y^2$ then $3\mid (2x+y)^2$. But $(2x+y)^2$ is a square so $3^2\mid (2x+y)^2$, implying that $3^2\mid 3y^2$, i.e. $3\mid y$. Together with $3\mid 2x+y$ we get also $3\mid x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/348964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Find the Laplace transform of $f(t) = \begin{cases} 0, & \text{if $t<5$} \\ t^2−10t+31, & \text{if $t\ge 5$} \\ \end{cases} $ Find the Laplace transform of
$$f(t) =
\begin{cases}
0, & \text{if $t<5$} \\
t^2−10t+31, & \text{if $t\ge 5$} \\
\end{cases}
$$
$F(s)=$ __________?
Here is my work. I went wrong somewhere. Can someone tell me the correct answer
First, rewrite $t^2 - 10t + 31$ in powers of $t - 5$:
$$\begin{split}
t^2 - 10t + 31
&= [(t - 5) + 5]^2 - 10 [(t - 5) + 5] + 31 \\
&= [(t - 5)^2 + 10(t - 5) + 25] - [10(t - 5) + 50] + 31 \\
&= (t - 5)^2 + 1.
\end{split}
$$
Hence,
$$\begin{split}
f(t) &= ((t - 5)^2 + 1) u(t - 5), \text{which implies}\\
F(s) &= \frac{2!}{s^3} + \frac{\exp(-5s)}{s}
\end{split}$$
by the shifting theorem.
| Your simplification is wrong. Note that
$(t-5)^2+1 = t^2 -10t + 26$ and you should have $t^2-10t+31$, so you need to use $(t-5)^2+6$ instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/350490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Trigonometrical limit $\lim\limits_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}?$ Can you help me solve this without using de l'Hôpital's rule (just using Standard rules):
$$ \lim_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}? $$
| Use that
$$\sin(x)=x-\frac{x^3}{3!} + \frac{x^5}{5!}\mp \cdots$$
and that
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}\pm\dots$$
Hence
$$\frac{\sin(x)-x\cos(x)}{x^3}=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}- x+\frac{x^3}{2}- \frac
{x^5}{4!} \pm \dots}{x^3}=\frac{1}{3} +\frac{x^2}{5!}-\frac{x^2}{4!}\pm \dots$$
Hence the Limit of
$$\lim_{x\to 0} \frac{\sin(x)-x\cos(x)}{x^3} =\lim_{x\to 0} \frac{1}{3} +\frac{x^2}{5!}-\frac{x^2}{4!}=\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Finding Determinants Recursively From the MIT OCW Linear Algebra (18.06) final exam, question 9:
For square matrices with 3's on the diagonal, 2s on the diagonal
above, and 1s on the diagonal below:
$$A_1=\begin{pmatrix} 3 \end{pmatrix}, A_2=\begin{pmatrix} 3 & 2 \\ 1
& 3\\ \end{pmatrix}, A_3=\begin{pmatrix} 3&2&0\\ 1&3&2\\ 0&1&3\\
\end{pmatrix},... $$
The determinant of $A_n$, $D_n$, can be defined recursively as:
$$D_n=aD_{n-1}+bD_{n-2}$$
Find $D_5$ using eigenvalues.
I found $D_n=3D_{n-1}-2D_{n-2}$ and used
$$\begin{pmatrix}
D_n\\
D_{n-1}\\
\end{pmatrix}
=
\begin{pmatrix}
a&b\\
1&0\\
\end{pmatrix}
\begin{pmatrix}
D_{n-1}\\
D_{n-2}\\
\end{pmatrix}
$$
and $\begin{pmatrix}
D_2\\
D_1\\
\end{pmatrix}
=\begin{pmatrix}
7\\
3\\
\end{pmatrix}$
to solve:
$$\begin{pmatrix}
D_n\\
D_{n-1}\\
\end{pmatrix}
=(-1)1^n\begin{pmatrix}1\\1\end{pmatrix} + 2^n\begin{pmatrix}2\\1\end{pmatrix}
$$
Plugging in $n=5$ gives $D-5=63$, but the answer key says $D_5=207$.
What am I doing wrong?
| It should
$$\begin{pmatrix} D_2\\ D_1\end{pmatrix} = \begin{pmatrix} 7\\ 3\end{pmatrix}$$
and not
$$\begin{pmatrix} D_2\\ D_1\end{pmatrix} = \begin{pmatrix} 3\\ 7\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/353368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
3 complex-variable equation
Moderator Note: This is a current contest question on Brilliant.org.
$x,y,z$ are complex numbers satisfying
$$
\begin{align}
x+y+z & =1\\
x^2+y^2+z^2 & =2\\
x^3+y^3+z^3 & =3
\end{align}
$$
The value of $x^4+y^4+z^4$ can be expressed as $\dfrac ab$, where $a$ and $b$ are positive co-prime integers. What is the value of $a+b$ ?
| HINT:
$(x+y+z)^2=(x^2+y^2+z^2)+2(xy+yz+zx)$
$\implies 2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)=1^2-2=-1$
Again, $x^3+y^3+z^3-3xyz=(x+y+z)\{(x+y+z)^3-3(xy+yz+zx)\}$
So, we have found $x+y+z,xy+yz+zx, xyz$
So, we can form the cubic equation whose roots are $x,y,z$
Can you take it from here?
Alternatively,
$$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+y^2+z^2+z^2x^2)$$
$$=4-2(x^2y^2+y^2+z^2+z^2x^2)$$
Now, $x^2y^2+y^2+z^2+z^2x^2=(xy+yz+zx)^2-2(xy\cdot yz+yz\cdot zx+zx\cdot xy)$
$=(xy+yz+zx)^2-2xyz(x+y+z)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/353900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{x^2 + x+3}{x^2+2x+5}\ dx$ How can we evaluate $$\displaystyle\int \frac{x^2 + x+3}{x^2+2x+5} dx$$
To be honest, I'm embarrassed. I decomposed it and know what the answer should be but
I can't get the right answer.
| Hint Use the decomposition
$$\frac{x^2 + x+3}{x^2+2x+5}=1-\frac{ x+2}{x^2+2x+5}=1-\frac{1}{2}\frac{ 2x+2}{x^2+2x+5}-\frac{ 1}{x^2+2x+5}$$
and
$$\frac{ 1}{x^2+2x+5}=\frac{ 1}{(x+1)^2+4}=\frac{1}{4}\frac{ 1}{(\frac{x+1}{2})^2+1}$$
the first fraction is on the form $\frac{f'}{f}$ and the second have the form $\frac{1}{u^2+1}$ by change of variable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/355296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the volume of the region contained above $z=1$ and below $x^{2}+y^{2}+z^{2}=4$ Why doesn't this work?
Find the volume of the region contained above $z=1$ and below $x^{2}+y^{2}+z^{2}=4$
going to cylindrical this should be easy. $z=(4-r^{2})^{\frac {1}{2}}$ and $z=1$
Clearly $z=1$ yields $x^{2}+y^{2}=3 \to r=(3)^{\frac {1}{2}}$
$$\int^{2\pi}_{0}\int^{0}_{(3)^{\frac {1}{2}}} [(4-r^{2})^{\frac {1}{2}} -1] \, r\,dr\,d\theta$$
$$\int^{2\pi}_{0}\left[-\frac {1}{3} (4-r^{2})^{\frac {3}{2}}-\frac{r^{2}}{2}\right]\Bigg|^{0}_{(3)^{\frac {1}{2}}} d\theta$$
$$-\frac{8}{3}-\left[-\frac{1}{3}-\frac{3}{2}\right] \to -\frac{16}{6} +\frac{11}{6} \to -\frac{5}{6} \to -\frac{5\pi}{3}$$
I don't get why the bound isn't supposed to be $\int^{0}_{(3)^{\frac {1}{2}}}$ - why is it $\int^{(3)^{\frac {1}{2}}}_{0}?$ Clearly when I start integrating, my radius is root 3 and it reduces to 0?
| $$\displaystyle{\int _0^{2\pi }\int _0^{\sqrt{3}}\int _1^{\sqrt{4-r^2}}rdzdrd\theta} $$
$$\displaystyle{2\pi \int_0^{\sqrt{3}} \left(r\sqrt{4-r^2}-r\right) \, dr} $$
The first part you solve it by changing variables:
$$u=4-r^2 \qquad du=-2rdr \qquad \frac{-du}{2}=rdr$$
After solving that, the volume is $\frac{5\pi}{3}$
There are 2 ways to solve the problem using cylindrical coordenates
$$0 < \theta <2\pi$$
$$0 < r < \sqrt{3} $$
$$ 1 < z < \sqrt{4-r^2} $$
or
$$0 < \theta <2\pi$$
$$1 < z < 2 $$
$$ 0 < r < \sqrt{4-z^2} $$
r will always go from the inside to the outside, z from de first plane it intersects to the other, and $\theta$ will spin totally in this case
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/355531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the point or points on C closest to the origin.
A curve $C$ in space is defined implicitly on the cylinder $x^2 + y^2 = 1$ by the additional equation $x^2 - xy + y^2 - z^2 = 1$. Find the point or points on $C$ closest to the origin.
This is an optimization problem. I tried constraining the distance $d = (x^2 + y^2 + z^2)^{1/2}$ to $x^2 - xy + y^2 - z^2 = 1$ by substituting $z^2$ into the distance equation and then finding the partial derivatives but I get $x=0, y=0$ which seems incorrect. Alternatively, I tried plugging $x^2 + y^2 = 1$ into the additional equation and then tried the same approach. I still got $x=0$ and $y=0$.
| The square of the distance from the origin is $x^2+y^2+z^2=1+z^2$ for any point on the cylinder. Any point on the curve satisfies $xy+z^2=0$, or $z^2=-xy$, so the square of the distance to the origin is $1-xy$, where $x^2+y^2=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/355607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove the following ceiling and floor identities? Could someone help me prove these identities? I'm completely lost:
$$\begin{align*}
&(1)\quad \left\lceil \frac{\left\lceil \frac{x}{a} \right\rceil} {b}\right\rceil = \left\lceil {\frac{x}{ab}}\right\rceil\\\\
&(2)\quad\left\lfloor \frac{\left\lfloor \frac{x}{a} \right\rfloor} {b}\right\rfloor = \left\lfloor {\frac{x}{ab}}\right\rfloor\\\\
&(3)\quad \left\lceil {\frac{a}{b}} \right\rceil \le \frac{a + b - 1}{b}\\\\
&(4)\quad \left\lfloor {\frac{a}{b}} \right\rfloor \ge \frac{a - b + 1}{b}
\end{align*}$$
| You can work directly from the definitions. I’ll do $(2)$ and $(3)$ to illustrate.
$(2)$ First note that you want $a$ and $b$ to be positive. If not, you might have $x=\frac12$ and $a=b=-1$, in which case the lefthand side is $$\left\lfloor-\left\lfloor-\frac12\right\rfloor\right\rfloor=\lfloor-(-1)\rfloor=1\;,$$ and the righthand side is $\left\lfloor\frac12\right\rfloor=0$. You also want them to be integers: if $x=2$ and $a=b=\sqrt2$, the lefthand side is $$\left\lfloor\frac{\lfloor\sqrt2\rfloor}{\sqrt2}\right\rfloor=\left\lfloor\frac1{\sqrt2}\right\rfloor=0\;,$$ and the righthand side is $1$. I will assume, then, that $a$ and $b$ are positive integers.
Let $m=\left\lfloor\dfrac{x}a\right\rfloor$ and $n=\left\lfloor\dfrac{x}{ab}\right\rfloor$. Then $$m\le\frac{x}a<m+1\quad\text{and}\quad n\le\frac{x}{ab}<n+1\;,\tag{1}$$ and you need to show that $\left\lfloor\dfrac{m}b\right\rfloor=n$, i.e., that $$n\le\frac{m}b<n+1\;.\tag{2}$$ Divide the first inequality in $(1)$ by $b$ to get $$\frac{m}b\le\frac{x}{ab}<\frac{m}b+\frac1b\;,$$ where $b\ge 1$. Now what would happen if $(2)$ failed? That would mean that either $\frac{m}b<n$, or $\frac{m}b\ge n+1$. It’s not hard to show that both are impossible. Suppose first that $\frac{m}b<n$. Then $\frac{m}b\le n-\frac1b$. (Why? Use the fact that $m,n$, and $b$ are integers.) Thus, $$\frac{x}{ab}<\frac{m}b+\frac1b\le n\;,$$ contradicting the second inequality in $(1)$. And if $\frac{m}b\ge n+1$, then $$\frac{x}{ab}\ge\frac{m}b\ge n+1\;,$$ again contradicting that inequality. So $(2)$, which is exactly what we set out to prove, must be true.
$(3)$ Let $m=\left\lceil\dfrac{a}b\right\rceil$, so that $m-1<\dfrac{a}b\le m$. Because $a,b$, and $m-1$ are integers, $$\frac{a}b-(m-1)\ge\frac1b\;,$$ and therefore $$m\le\frac{a}b+1-\frac1b=\frac{a+b-1}b\;,$$ which is exactly what we wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/356168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
On comparing fractions , fraction with smaller difference between numerator and denominator is greater than the other A text book proposed that "when comparing fractions ,if the compared fractions's are such that numerator is smaller than denominator ,then fraction with more difference(absolute) between numerator and denominator is the smallest among the fractions compared "
And I found many text books support this. Even on looking at the video http://www.youtube.com/watch?v=rJz-f7uCBns#t=6m35s , here he has used this idea , but for a case where numerator is greater than denominator .
But consider the case
$2/3$ & $ 20/30 $ , but as per the theory proposed 2/3 > 20/30.but actually they are the same.
Even taking a bit complex case
if $2/7 = 0.285714 $ definitely we will be able to find another number with different difference but same answer as $ 3/0.2857514 = 10.5 $ so
$ 3/ 10.5 = 0.285714 $
here difference is 7.5 but value of the ratio is still 0.285714 .
So am I going wrong in understanding this concept preferred in many popular text books . If so please spot the error and help me by giving conditions when this fact holds good .
| I have never seen this claim in any textbook; in any case it's wrong. The claim seems to be that if you have two fractions $\frac{a}{b}$ and $\frac{c}{d}$ with $a < b$ and $c < d$, then $|a - b| < |c - d| \iff \frac{a}{b} < \frac{c}{d}$.
This is false. We can write $\frac{a}{b} = 1 - \frac{b-a}{b}$ and $\frac{c}{d} = 1 - \frac{d-c}{d}$, so you're comparing $\frac{b-a}{b}$ and $\frac{d-c}{d}$ (whichever is greater, the corresponding fraction is smaller). The first numerator may be smaller than the second, but the actual comparison of these fractions can of course go either way.
For example,
*
*Here is one with $b - a < d - c$ but $\frac{a}{b} > \frac{c}{d}$: consider $\frac{2}{3} > \frac{3}{5}$.
*Here is one with $b - a < d - c$ but $\frac{a}{b} = \frac{c}{d}$: consider $\frac{2}{3} = \frac{4}{6}$.
*Here is one with $b - a < d - c$ but $\frac{a}{b} < \frac{c}{d}$: consider $\frac{2}{3} < \frac{5}{7}$.
So all results are possible; the test is nonsense.
Edit: Just for fun/completeness, here is a table showing pairs $(\frac{a}{b}, \frac{c}{d})$ with each possible combination of the two comparisions:
$$\begin{array}{c|c|c|c}
& \frac{a}{b}<\frac{c}{d} & \frac{a}{b}=\frac{c}{d} & \frac{a}{b}>\frac{c}{d}\\
\hline \\
b-a<d-c & \frac23,\frac57 & \frac23,\frac46 & \frac23,\frac35\\
\hline \\
b-a=d-c & \frac23,\frac34 & \frac23,\frac23 & \frac23,\frac12\\
\hline \\
b-a>d-c & \frac57,\frac23 & \frac46,\frac23 & \frac35,\frac23 \\
\end{array}$$
(If you want examples involving fractions greater than $1$, turn each of the fractions upside down. Each of the inequalities between the fractions will reverse direction, so you'll still have a complete set of examples.)
Edit: On looking at that segment of the video, it's possible (not very clear) that what he may have been saying is equivalent to the following claim, which is true: if you have two fractions $\frac{a}{b}$ and $\frac{c}{d}$ with $a > b$ and $c > d$, and if $a - b < c - d$ and $b > d$, then $\frac{a}{b} < \frac{c}{d}$. Proof:
$$\frac{a}{b} = 1 + \frac{a-b}{b} < 1 + \frac{c-d}{b} < 1 + \frac{c-d}{d} = \frac{c}{d}$$
With fractions less than $1$, the corresponding statement would be that if you have two fractions $\frac{a}{b}$ and $\frac{c}{d}$ with $a < b$ and $c < d$, and if $b - a < d - c$ and $b > d$, then $\frac{a}{b} > \frac{c}{d}$:
$$\frac{a}{b} = 1 - \frac{b-a}{b} > 1 - \frac{b-a}{d} > 1 - \frac{d-c}{d} = \frac{c}{d}$$
But these are so many conditions on the hypothesis that I wonder how often it will be useful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/358704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How can I determine the Jordan Form of a matrix? How can I go about proving that the characteristic polynomial, minimal polynomial, and the dim(eigenspace) is enough to determine the Jordan Form of a matrix for n<7?
| I realize this is 5 years old, but the only answer is wrong, given that the two matrices in DonAntonio's answer do not have matching minimal polynomials. So in case anyone comes across it, here is an answer:
If Dimension=7, then you could have two matrices such as
$$
A=\begin{pmatrix}
0 & 1 \\
& 0 & 1 \\
& & 0 \\
& & & 0 & 1 \\
& & & & 0 & 1 \\
& & & & & 0 \\
& & & & & & 0
\end{pmatrix}
\quad,\quad
B=\begin{pmatrix}
0 & 1 \\
& 0 & 1 \\
& & 0 \\
& & & 0 & 1 \\
& & & & 0 & \\
& & & & & 0 & 1 \\
& & & & & & 0
\end{pmatrix}
$$
Notice that the eigenspace is just the kernel, and both matrices are rank 4, so both have kernels of dimension 3. Both have characteristic polynomial $x^7$, and both have minimal polynomial $x^3$. But let's take a moment to examine what failed.
If we examine the number of integer partitions of $7$, we have
7=3+3+1 and also 7=3+2+2. These are both integer partitions of 7 into 3 parts with largest part 3. If we associate each part with a Jordan block, the number of parts is the dimension of the Eigenspace, and the largest part in the decomposition is the degree of the minimal polynomial. However, if we look at all partitions of 6, we have
6=6
6=5+1
6=4+2
6=4+1+1
6=3+3
6=3+2+1
6=3+1+1+1
6=2+2+2
6=2+2+1+1
6=2+1+1+1+1
6=1+1+1+1+1+1
Notice that for $6$, or any number smaller than 6 (as all of those are implicitly in this list as well), there is no integer partition for which the largest part and number of parts is the same simultaneously. This is why knowing the dimension of the eigenspace (tells you the number of parts), minimal polynomial (degree tells you the largest part), and characteristic polynomial (degree tells you the dimension of the ambient space) is sufficient, but it is not sufficient for dimension 7.
Best,
--Kris
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/358777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\sum_{n=2}^\infty ({n^4+2n^3-3n^2-8n-3\over(n+2)!})$ Calculate $\sum_{n=2}^\infty ({n^4+2n^3-3n^2-8n-3\over(n+2)!})$
I thought about maybe breaking the polynomial in two different fractions in order to make the sum more manageable and reduce it to something similar to $\lim_{n\to\infty}(1+{1\over1!}+{1\over2!}+...+{1\over n!})$, but didn't manage
| First step, we find the Taylor series of $x^4+2x^3-3x^2-8x-3$ at the point $x=-2$ and then use it to write
$$ n^4+2n^3-3n^2-8n-3 = 1-4\, \left( n+2 \right) +9\, \left( n+2 \right)^{2}-6\, \left( n+2
\right) ^{3}+ \left( n+2 \right) ^{4}.$$
Using the above expansion and shifting the index of summation ($n \longleftrightarrow n-2$ ), we have
$$ \sum_{n=2}^\infty {n^4+2n^3-3n^2-8n-3\over(n+2)!}= \sum_{n=2}^\infty {1-4\, \left( n+2 \right) +9\, \left( n+2 \right)^{2}-6\, \left( n+2\right) ^{3}+ \left( n+2 \right)^{4}\over(n+2)!} $$
$$ = \sum_{n=4}^\infty {1-4\, n +9\, n^{2}-6\, n^{3}+ n^{4} \over n! }+\sum_{n=0}^3 {1-4\, n +9\, n^{2}-6\, n^{3}+ n^{4} \over n! }$$
$$ -\sum_{n=0}^3 {1-4\, n +9\, n^{2}-6\, n^{3}+ n^{4} \over n! }$$
$$= c+ \sum_{n=0}^\infty {1-4\, n +9\, n^{2}-6\, n^{3}+ n^{4} \over n! } $$
$$ = c+e(1-4B_1 + 9 B_2 -6B_3 +B_4), $$
where $B_n$ are the bell numbers
$$ B_n = \frac{1}{e}\sum_{k=0}^{\infty} \frac{k^n}{k!}, $$
and $c$ is given by
$$ c=-\sum_{n=0}^3 {1-4\, n +9\, n^{2}-6\, n^{3}+ n^{4} \over n! }. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/360293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to compare two ratios of gamma functions? How would I show that for $x \ge 2$:
$$\frac{\Gamma(x-1)}{[\Gamma(\frac{x}{2})]^2} \le \frac{\Gamma(x)}{[\Gamma(\frac{x+1}{2})]^2}$$
Any hints or suggestions are greatly appreciated.
| The answer is to show that $\frac{d}{dx}(\frac{\Gamma(x-1)}{\Gamma([\frac{x}{2}]^2)}) > 0$
This will be established if I show that $\frac{d}{dx}[\ln{\Gamma(x-1)}- 2\ln\Gamma(\frac{x}{2})] > 0$.
I will use this series ψ:
$$\frac{d}{dx}(\ln\Gamma(x)) = \frac{\psi(x)}{dx} = -\gamma + \sum_{k=0}^\infty(\frac{1}{k+1} - \frac{1}{k + x})$$
Applying this gives:
$$\frac{d}{dx}[\ln{\Gamma(x-1)}- 2\ln\Gamma(\frac{x}{2})] = \psi(x-1) - \frac{2\psi(\frac{x}{2})}{2} = \psi(x-1) - \psi(\frac{x}{2})$$
$$\psi(x-1) - \psi(\frac{x}{2}) = \sum_{k=0}^{\infty}(\frac{1}{k+\frac{x}{2}} - \frac{1}{k+x-1}) $$
We can see that for $x \ge 2$, $\frac{x}{2} < x-1$ so it follows that $\frac{1}{k + \frac{x}{2}} > \frac{1}{k+x-1}$ and it is therefore increasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/361898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Minimum value of inverse trigo function (cubic) The minimum value of $(\sin^{-1}x)^3+(\cos^{-1}x)^3$ is equal to ( following options)
a) $\displaystyle \frac{\pi^3}{32}$
b) $\displaystyle\frac{5\pi^3}{32}$
c) $\displaystyle\frac{9\pi^3}{32}$
d) $\displaystyle\frac{11\pi^3}{32}$
Can we go like this :
$ -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2}$ Therefore minimum value of $(\sin^{-1}x)^3 = (\frac{-\pi}{2})^3$= -$\frac{\pi}{8}$
Please guide..
| As $\sin^{-1}x+\cos^{-1}x=\frac\pi2$
$$(\sin^{-1}x)^3+(\cos^{-1}x)^3$$
$$=(\sin^{-1}x)^3+\left(\frac\pi2-\sin^{-1}x\right)^3$$
$$=\left(\frac\pi2\right)^3-3\left(\frac\pi2\right)^2\sin^{-1}x+3\left(\frac\pi2\right)(\sin^{-1}x)^2$$
$$=\left(\frac\pi2\right)^3+3\left(\frac\pi2\right) \{(\sin^{-1}x)^2-\frac\pi2\sin^{-1}x\}$$
$$=\left(\frac\pi2\right)^3+3\left(\frac\pi2\right) \{\left(\sin^{-1}x-\frac\pi4\right)^2-\left(\frac\pi4\right)^2\}$$
$$\ge \left(\frac\pi2\right)^3-3\left(\frac\pi2\right) \left(\frac\pi4\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/362185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
how to prove that $(1 + \frac{1}{n})^{n+1}$ is decreasing? Please, help me to prove that $$x_n=\left(1+\frac{1}{n}\right)^{n+1}$$decreases.
I know I must to prove that that $$\frac{x_n}{x_{n+1}}> 1$$
What to do next?
| Write out
$$
\frac{x_{n+1}}{x_n} = \frac{\left(1+\frac{2}{n}\right)^2}{\left(1+\frac{1}{n}\right)^3} \left( \frac{1+\frac{2}{n}}{\left(1+\frac{1}{n} \right)^2} \right)^n =
\frac{\left(1+\frac{2}{n}\right)^2}{\left(1+\frac{1}{n}\right)^3} \left( 1 - \frac{\frac{1}{n^2}}{\left(1+\frac{1}{n} \right)^2} \right)^n =
\frac{\left(1+\frac{2}{n}\right)^2}{\left(1+\frac{1}{n}\right)^3} \left( 1 - \frac{1}{\left(n+1 \right)^2} \right)^n = \left(1+\frac{n}{(n+1)^2} - \frac{1}{(n+1)^3} \right) \left( 1 - \frac{1}{\left(n+1 \right)^2} \right)^n
$$
we now have to use $(1-x)^n \leqslant \frac{1}{1+n x}$ for $0<x<1$ and $n \geqslant 1$:
$$
\frac{x_{n+1}}{x_n} \leqslant \frac{1+\frac{n}{(n+1)^2} - \frac{1}{(n+1)^3}}{ 1+\frac{n}{(n+1)^2} } < 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/365680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Calculate the number of real roots of $x^8-x^5+x^2-x+1 = 0$ Calculate the number of real roots of $x^8-x^5+x^2-x+1 = 0$
My try: $$\left(x^4-\frac{x}{2}\right)^2+\frac{3}{4}x^2-x+1 = \left(x^4-\frac{x}{2}\right)^2+\frac{3}{4}\left(x^2-\frac{4}{3}x+\frac{4}{3}\right)$$
$$\implies \left(x^4-\frac{x}{2}\right)^2+\frac{3}{4}\left(x-\frac{2}{3}\right)^2+1-\frac{4}{9}>0\quad \forall x\in \mathbb{R}$$
My question is any other method like Using Inequality to solve Given Question
If Yes then please explain here.
| AM-GM Inequality gives us,
$$\large\frac{1}{2} \left(x^{8}+x^{2}\right)\geq x^{5}$$
$$\large\frac{1}{2} \left(x^{2}+1\right)\geq x$$
$$\large\therefore x^{8}+x^{2}+1>\frac{1}{2} \left(x^{8}+x^{2}\right)+\frac{1}{2} \left(x^{2}+1\right)\geq x^{5}+x$$
$$\large\implies x^{8}-x^{5}+x^{2}-x+1> 0$$
Hence there cannot exist any real solution for $\large x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/366093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Confusion over a limit. Different ways of solving give different answers? Qn: If it is given that
$$
\lim_{x\to\infty} \frac{x^2 - x - 2}{x + 1} - ax - b = 1
$$
then a and b must be?
Now, I tried doing this by 2 methods.
Method 1:
$$ \frac{x^2 - x - 2}{x + 1} - ax - b $$
$$ = (x - 2) - ax - b $$
Since the limit is finite, $a$ must be $= 1$ and so, $b = -3$
Method 2:
$$ \frac{x^2 - x - 2}{x + 1} - ax - b $$
$$ = \frac{x^2(1 - \frac 1 x - \frac2 {x^2})}{x(1 + \frac1x)} - ax - b $$
$$ = \frac{x - 1 - \frac2x}{(1 + \frac1x)} - ax - b $$
as $x \to \infty$, we have the above expression
$$ = x - ax - b$$
So, $a = 1$ and $b = -1$
Which of the above is correct?
| Method $1$ is correct
In Method $2$, you cannot apply limit to a part of the fraction. So
$$ = \frac{x^2(1 - \frac 1 x - \frac2 {x^2})}{x(1 + \frac1x)} - ax - b $$
is not equal to
$$ = x - ax - b$$
In the above step you have applied limit to just a part of that fraction leaving behind $\dfrac{x^2}{x}$.
hope the answer is clear !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show $x\sqrt{n} - n \ln\left(1+\frac{x}{\sqrt{n}}\right) \to \frac{x^2}{2}$ With $x > 0$, show
$$
L=\lim_{n \to \infty} x\sqrt{n} - n \ln\left(1+\frac{x}{\sqrt{n}}\right) = \frac{x^2}{2}.
$$
I tried to write
$$
x\sqrt{n}=\ln \left( e^{x\sqrt{n}} \right),
$$
so that
$$
L = \lim_{n\to \infty}
\frac{e^{x\sqrt{n}}}{\left(1+x/\sqrt{n}\right)^n}.
$$
The last expression is of the form $\frac{\infty}{\infty}$.
However, l'hospital rule won't change the denumerator.
I don't see what to do.
| $$\begin{aligned}x\sqrt{n}-n\ln\left(1+\frac{x}{\sqrt{n}}\right)&=x\sqrt{n}-n\left(\frac{x}{\sqrt{n}}-\frac{x^2}{2n}+\frac{x^3}{3n\sqrt{n}}-\cdots\right)\\&=\frac{x^2}{2}-\frac{x^3}{3\sqrt{n}}+\cdots\to\frac{x^2}{2}\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/368630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving two simultaneous recurrence relations If we have the two recurrence relations $$a_n = 3a_{n-1} + 2b_{n-1}$$ $$b_n = a_{n-1} + 2b_{n-1}$$ with $a_0 = 1$ and $b_0 = 2$.
My solution is that we first add two equations and assume that $f_n = a_n + b_n$. The result is $f_n = 4f_{n-1}$. This can be solved easily and the solution is $f_n = a_n + b_n = 4^nf_0=4^n(3)$. We subtract the two equations from each other and we get $a_n - b_n = 2a_{n-1}$. By adding these two results we get $2a_n = 2a_{n-1} +4^n (3) $.
$a_n = a_{n-1} + (3/2) 4^n$ is annihilated by $(E-1)(E-4)$. Thus, the Generic solution is $a_n = \sigma + \beta4^n$. The constants $\sigma, \beta$ satisfy the equations: $a_0 = 1 = \sigma + \beta$ and $a_1 = 7 = \sigma + 4\beta$. Thus, the final solution is $a_n = -1 + (2) 4^n$ and $b_n = 4^n + 1$. Is this right? Thank you!
| Use generating functions, $A(z) = \sum_{n \ge 0} a_n z^n$ and similarly $B(z)$.
Write:
$$
\begin{align*}
a_{n + 1} &= 3 a_n + 2 b_n \\
b_{n + 1} &= a_n + 2 b_n
\end{align*}
$$
By properties of generating functions:
$$
\begin{align*}
\frac{A(z) - a_0}{z} &= 3 A(z) + 2 B(z) \\
\frac{B(z) - b_0}{z} &= A(z) + 2 B(z)
\end{align*}
$$
With the given initial values:
$$
\begin{align*}
A(z) &= \frac{1 + 3 z}{1 - 4 z + z^2} \\
B(z) &= \frac{2 - 5 z}{1 - 4 z + z^2} \\
\end{align*}
$$
Next step is to split $A(z)$ and $B(z)$ into partial fractions,
and expand the results by geometric series to extract the terms. Sadly, in this case the zeros of the denominator are surds, and the expŕessions are unwidely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/369377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Finding eigenvectors with square root eigenvalues I have a matrix
$$\begin{bmatrix}1 &-1 &2\\2 &-2 &4\\0 &1 &1\end{bmatrix}$$
Its eigenvalues are $0$, $\sqrt{5}$ and $-\sqrt{5}$
(These are checked in MATLAB to be correct). I have found its eigenvectors for $0$, and also don't seem to have a problem in similar situations when the eigenvalues are not "root". But when I have root, and are not allowed to turn into descimal-numbers, I have no clue how to proceed.
Does anyone know how to find eigenvectors when the values are $\sqrt{x}$? I'd be happy to watch a youtube video, or an example of how to proceed.
| If you want results to compare against, you should get for $A = \begin{bmatrix}1 &-1 &2\\2 &-2 &4\\0 &1 &1\end{bmatrix}$, the following characteristic polynomial:
$$5 \lambda - \lambda^3 = 0$$
This CP yields the following eigenvectors and eigenvalues:
$\displaystyle \lambda_1 = -\sqrt{5}, v_1 = \left(\frac{1}{2} (-1-\sqrt{5}), -1-\sqrt{5}, 1\right)$
$\displaystyle \lambda_2 = \sqrt{5}, v_2 = \left(\frac{1}{2} (-1+\sqrt{5}), -1+\sqrt{5}, 1\right)$
$\lambda_3 = 0, v_3 = (-3, -1, 1)$
Update
To find the eigenvectors, for each eigenvalue, we solve:
$$|A -\lambda_i I|v_i = 0$$
For $\lambda_3 = 0$, we have:
$$|A -\lambda_3 I|v_3 = \begin{bmatrix} 1 & -1 & 2 \\2 & -2 & 4\\0 & 1 &1\end{bmatrix}v_3 = 0$$
The RREF is:
$$\begin{bmatrix}1 & 0 & 3 \\0 & 1 & 1\\0 & 0 & 0\end{bmatrix}v_3 = 0$$
Here, we have:
$a + 3 c = 0$, and
$ b + c = 0 \rightarrow b = -c, \text{so pick}~~ c = 1 \rightarrow b = -1$
So, $a = -3c \rightarrow a = -3$
Thus for $\lambda_3 = 0$, we have $v_3 = (-3, -1, 1)$
For $\lambda_2 = \sqrt{5}$, we have:
$$|A -\lambda_2 I|v_2 = \begin{bmatrix} 1-\sqrt{5} & -1 & 2 \\2 & -2-\sqrt{5} & 4\\0 & 1 &1-\sqrt{5}\end{bmatrix}v_2 = 0$$
The RREF is:
$$\begin{bmatrix}1 & 0 & \frac{1}{2}(1-\sqrt{5} \\0 & 1 & 1-\sqrt{5}\\0 & 0 & 0\end{bmatrix}v_2 = 0$$
So, we have:
$a + \frac{1}{2}(1-\sqrt{5})c = 0$, and
$b + (1-\sqrt{5})c = 0$
So, we can choose: $c = 1 \rightarrow b = (-1+\sqrt{5})$
This leads to: $a = \frac{1}{2}(-1 + \sqrt{5})$
So, for $\lambda_2 = \sqrt{5}$, we get $v_2 = \left(\frac{1}{2} (-1+\sqrt{5}), -1+\sqrt{5}, 1\right)$.
Can you repeat this process for the last eigenvalue? If not give a yell and I will add details.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/369507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What's the Maclaurin series for $\arcsin(x)$? I solved the problem by using a known series: $\frac{1}{\sqrt{1-x^2}}$, but the solution I got is wrong. Also, I'm not sure what to do with the constant of integration $C$. Where is my mistake?
$$ \frac{1}{\sqrt{1-x^2}} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} +... $$
$$ \int\frac{1}{\sqrt{1-x^2}}dx = \int1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} +... dx$$
$$ \arcsin(x) + C = x + \frac{2x^3}{3} + \frac{3x^5}{24} + \frac{5x^7}{112}+... \tag{what happens to $C$?}$$
The right solution is:
$$ x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} +... $$
| You did a great job:
Just small mistakes.
*
*To find constant of integration, substitute known value of $\arcsin(x)$, $x=0$ is a good choice. You need to have a well defined interval while dealing with inverse trigonometric functions.
*The formula to integrate is $$\int x^n \, dx=\frac{x^{n+1}}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/371611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
What's the value of $ y^{(n)}$when $ y=\frac{x^n}{(x+1)^2(x+2)^2}$ What's the value of $\displaystyle y^{(n)}$when $\displaystyle y=\frac{x^n}{(x+1)^2(x+2)^2}$?
My Try:Let $\displaystyle y_n=\frac{x^n}{(x+1)^2(x+2)^2}$,so $\displaystyle y_n=xy_{n-1}$.According to Leibniz's formula,$$y_n^{(n)}=ny_{n-1}^{(n-1)}+xy_{n-1}^{(n)}$$.But I don't konw how to achieve it.
| Assertion:
$$y_n^{(n)} = \frac{A_nx+B_n}{(x+1)^{n+2}}+\frac{C_nx+D_n}{(x+2)^{n+2}}$$
Now, we are going to need
$$
xy_{n}^{(n+1)}=\frac{-(n+1)A_nx^2+(A_n-B_n(n+2))x}{(x+1)^{n+3}}+\frac{-(n+1)C_nx^2+(2C_n-(n+2)D_n)x}{(x+2)^{n+3}}
$$
and
$$
(n+1)y_{n}^{(n)}=(n+1)\frac{(A_nx+B_n)(x+1)}{(x+1)^{n+3}}+(n+1)\frac{(C_nx+D_n)(x+2)}{(x+2)^{n+3}}
$$
Combining these and applying the Leibniz rule, we get
$$
y_{n+1}^{(n+1)}=\frac{((n+2)A_n-B_n)x+(n+1)B_n}{(x+1)^{n+3}}+\frac{(2(n+2)C_n-D_n)x+2(n+1)D_n}{(x+2)^{n+3}}
$$
Matching with the original assertion, we have
$$
A_{n+1}=(n+2)A_n-B_n\\
B_{n+1}=(n+1)B_n\\
C_{n+1}=2(n+2)C_n-D_n\\
D_{n+1}=2(n+1)D_n
$$
And as $y_0^{(0)} = -\frac{2x+1}{(x+1)^2}+\frac{2x+5}{(x+2)^2}$, we also have that
$$
A_0 = -2\\
B_0 = -1\\
C_0 = 2\\
D_0 = 5
$$
Now, we can solve for $B_n$ and $D_n$ relatively easily. $A_n$ and $C_n$ will require a little more work. But I'll leave the rest to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $y=2^{2/3} + 2^{-1/3}$ I am working on a calculus problem where I have to find the local minimum. The value I got was $$y=2^{2/3} + 2^{-1/3}.$$ I simplified it and got this:
$$ y=2^{2/3} + \frac{1}{2^{1/3}}$$
$$y=\frac{2^{2/3}2^{1/3}}{2^{1/3}} + \frac{1}{2^{1/3}}\frac{2+1}{2^{1/3}}=\frac{3}{2^{1/3}}$$
According to the online homework and Wolfram Alpha, the correct answer is :
$$\frac{3}{2^{2/3}}$$
Did I miss a step while simplifying the answer? I don't understand why it is $2/3$ instead of $1/3$.
| Your simplification is entirely correct. Using your original evaluation of $y$, $$y=e^{2\frac{\ln2}{3}}+e^{-\frac{\ln2}{3}}=2^{2/3} + 2^{-1/3} = \dfrac{3}{2^{1/3}}$$
Are you sure your value for $y$ is a correct mimimum?
EDIT:
Original function: $$y(x)=e^{2x}+e^{-x}$$
Critical point (minimum) at $x = \dfrac{-\ln 2}{3} \implies$
$$y=e^{-2\frac{\ln2}{3}}+e^{\frac{\ln2}{3}}=2^{-2/3} + 2^{1/3} = \dfrac{3}{2^{2/3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/373115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Closed form for $(a_n)$ such that $a_{n+2} = \frac{a_{n+1}a_n}{6a_n - 9a_{n+1}}$ with $a_1=1$, $a_2=9$ $$a_1 = 1; a_2 = 9; a_{n+2} = \frac{a_{n+1}a_n}{6a_n - 9a_{n+1}}$$
I need to find non-recurring formula for $a_n$. Is there any good way to do this? The only one comes to mind is to guess the formula and then prove it using mathematical induction.
Thanks in advance!
I've got the result and it looks like this: $a_n = \frac{-3*2^{n-1} + 2^{2n - 1} + 1}{3}$ but I really don't like this way and would love to know how to solve this properly.
| As in Did's answer, but simpler to handle:
$$
b_{n + 2} = 6 b_{n + 1} - 9 b_n \qquad b_1 = 1, b_2 = 1/9
$$
Define $B(z) = \sum_{n \ge 0} b_{n + 1} z^n$. By the properties of generating functions:
$$
\frac{B(z) - b_1 - b_2 z}{z^2} = 6 \frac{B(z) - b_1}{z} - 9 B(z)
$$
From here:
$$
B(z) = \frac{9 - 53 z}{9 - 54 z + 81 z^2}
= \frac{53}{27} \frac{1}{1 - 3 z} - \frac{26}{27} \frac{1}{(1 - 3 z)^2}
$$
By the expansion (a form of the binomial theorem):
$$
(1 - u)^{-m} = \sum_{n \ge 0} \binom{n + m - 1}{m - 1} z^n
$$
we get:
$$
b_{n + 1} = \frac{53}{27} 3^n - \frac{26}{27} \binom{n + 1}{1} 3^n
= (27 - 26 n) \cdot 3^{n - 3}
$$
so that finally:
$$
a_n = \frac{1}{(53 - 26 n) \cdot 3^{n - 4}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/375215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to count permutations with restrictions on how items are grouped I am trying to solve the following problem:
A town contains $4$ people who repair televisions. If $4$ sets break down, what is the probability that exactly $i$ of the repairers are called? Solve the problem for $i=1,2,3,4$.
For $i=1$, there are ${}_4P_1$ ways to assign $1$ person to $4$ televisions, so the probability is $\frac{{}_4P_1}{4^4}=\frac{1}{64}$.
For $i=4$, there are ${}_4P_4$ ways to assign $4$ people to $4$ televisions, so the probability is $\frac{{}_4P_4}{4^4}=\frac{3}{32}$.
I am having trouble with $i=2,3$. How should I go about these cases?
| *
*Probability that exactly 2 repairers are called.
There are two ways how two repairers are called:
*
*Both gets call for two broken sets each
First repairer $A$ can be anyone. Hence, will have probability $\frac{4}{4}=1$. The same repairer will be called for any one of 2nd, 3rd or 4th broken TV set. [Hence the total possible sequences can be 3 $(AABB,ABAB,ABBA)$.] At that time $A$ can be selected with probability $\frac{1}{4}$, that is, he should be same as first one. The 2nd repairer can be any one apart from $A$, Thus, the first call to 2nd repairer $B$ can happen with probability $\frac{3}{4}$. The 2nd call to $B$ can happen with probability $\frac{1}{4}$. Final probability $\frac{4}{4}\times\frac{3}{4}\times\frac{1}{4}\times\frac{1}{4}\times3$
*One gets call for three broken TV sets, while the other one gets call for
remaining one broken set, then there are four ways:
$(AAAB,AABA,ABAA,BAAA)$.
Each one will have same probability as first bullet point. Hence final probability: $\frac{4}{4}\times\frac{3}{4}\times\frac{1}{4}\times\frac{1}{4}\times 4$.
Final probability of 1st and 2nd bullet point combined: $\frac{4}{4}\times\frac{3}{4}\times\frac{1}{4}\times\frac{1}{4}\times 7 =\frac{21}{64}$
*Probability that exactly 3 repairers are called.
Let the three repairmen called be $A,B,C$.
There can be six sequences in which call are made for 1st, 3nd, 3rd and 4th broken TV sets: $(ABCA,ABAC,AABC,ABBC,ABCB,ABCC)$. First repairman can be called with probability $\frac{4}{4}$. Second new repairman can be called with probability $\frac{3}{4}$. Third new repairman can be called with probability $\frac{2}{4}$. Any repairman can be repeated with probability $\frac{1}{4}$. Final probability $=\frac{4}{4}\times\frac{3}{4}\times\frac{2}{4}\times\frac{1}{4}\times 6=\frac{36}{64}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/377270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Integral Question - $\int\frac{1}{x^2-6x}\,\mathrm dx$ How I can evaluate the indefinite integral? :
$$\int\frac{1}{x^2-6x}\,\mathrm dx$$
Do I need to bring it to this format? : $\displaystyle \int\frac{1}{x^2-a^2}\,\mathrm dx\;$?
Thanks!
| $$\int\frac{1}{x^2-6x}\,\mathrm dx = \int\frac{1}{x(x-6)}\,\mathrm dx$$
Now use partial fraction decomposition to obtain an integral of the form $$\int\frac{1}{x(x-6)}\,\mathrm dx =\int \left(\frac{A}{x} + \frac{B}{x-6}\right)\,\mathrm dx$$
Now all you need to do is to determine the (constant) values of $A$ and of $B$ so that $$ \left(\frac{A}{x} + \frac{B}{x-6}\right) = \frac{1}{x(x-6)}.$$
$$\left(\frac{A}{x} + \frac{B}{x-6}\right) = \frac{A(x-6) + B(x)}{x(x - 6)} \iff A(x - 6) + B(x) = 1\iff (A + B)x - 6A = 1 $$
$$\iff A+B = 0 \;\;\text{and}\;\; -6A = 1 \iff \color{blue}{\bf A = -\frac 16}, \;\;\text{and}\;\; B = - A \iff \color{red}{\bf B = \frac 16}$$
This gives us, then, Now use
$$\int\frac{1}{x(x-6)}\,\mathrm dx =\int \left(\frac{A}{x} + \frac{B}{x-6}\right)\,\mathrm dx = \int \left(\color{blue}{\bf -}\frac{\color{blue}{\bf 1}}{\color{blue}{\bf 6}x} + \frac{\color{red}{\bf 1}}{\color{red}{\bf 6}(x-6)}\right)\,\mathrm dx$$
$$ = -\frac 16 \int \left(\dfrac 1x \right) \,\mathrm dx + \frac 16 \int \left(\frac 1{x-6}\right)\,\mathrm dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/377410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Small inequality on unit open disc For $|u|,|z|<1$, $u,z$ complex numbers, how to show the inequality:
$|\frac{u-z}{1-\bar uz}|<1$?
| Let $u=a+ib, z=c+id$
So, $|u-z|=\sqrt{(a-c)^2+(b-d)^2}=\sqrt{a^2+c^2-2ca+b^2+d^2-2bd}$
$1-\bar uz=1-(a-ib)(c+id)=1-ac-bc+i(bc-ad)$
So, $|1-\bar uz|=\sqrt{(1-ac-bc)^2+(bc-ad)^2}$
$|u-v|<|1-\bar uz|$
$\implies \sqrt{a^2+c^2-2ca+b^2+d^2-2bd}<\sqrt{(1-ac-bc)^2+(bc-ad)^2}$
$\implies a^2+c^2-2ca+b^2+d^2-2bd <(1-ac-bc)^2+(bc-ad)^2$
$\implies \{1-(a^2+b^2)\}\{1-(c^2+d^2)\}<0$
$\implies \{1-|u|^2\}\{1-|z|^2\}<0$
which is satisfied if the modulus of $u,z$ both $>1$ or both $<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/378074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the limit of $ x_n = \prod_{j=2}^{n} \left(1 - \frac{2}{j(j+1)}\right)^2$ I am stuck on the following problem:
Let $x_n=(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2 \ldots...(1-\frac{1}{n(n+1)/2})^2, \text{where} \space n \geq 2$. Then $\lim_{n \to \infty}x_n=?$
I see that $x_n^{\frac{1}{2}}=\frac{2}{3}\frac{5}{6}\frac{9}{10} \ldots..\frac{(n-1)(n+2)}{n(n+1)}$. now not sure what to do next ?Any idea?
| Note you really have a telescoping product. To see this intuitively, write out a few terms. You start with $\frac{1 \cdot 4}{2 \cdot 3}$
$$\frac{1 \cdot 4}{2 \cdot 3} \times \frac{2 \cdot 5}{3 \cdot 4} = \frac{1 \cdot 5}{3 \cdot 3}$$
and adding another term
$$\frac{1 \cdot 5}{3 \cdot 3} \times \frac{3 \cdot 6}{4 \cdot 5} = \frac{1 \cdot 6}{3 \cdot 4}$$
and proceed by induction to prove that $x_n = \frac{1 \cdot (n+2)}{3 \cdot n}$ which in the limit should converge to 1/3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Why is $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$? Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$
I can't seem to wrap my head around that.
| Simple arithmetic:
\begin{equation}
n^2 - \dfrac{n^2}{2} = n^2(1 - \dfrac{1}{2}) = n^2(\dfrac{1}{2}) = \dfrac{1}{2}n^2 = \dfrac{n^2}{2}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 15,
"answer_id": 5
} |
Simplifying $\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}$ How do I simplify $\sqrt{(4+2\sqrt{3})}+\sqrt{(4-2\sqrt{3})}$?
I've tried to make it $x$ and square both sides but I got something extremely complicated and it didn't look right.
I got $2\sqrt{3}$ on wolframalpha, but I'm not sure how is it possible?
Help would be appreciated! Thanks!
| Since $(4+2\sqrt3)(4-2\sqrt3)=16-12=4$, try squaring:
$$
\begin{align}
\left(\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}\right)^2
&=(4+2\sqrt3)+(4-2\sqrt3)+2\sqrt{(4+2\sqrt3)(4-2\sqrt3)}\\
&=8+2\sqrt{16-12}\\[6pt]
&=12
\end{align}
$$
Therefore, $\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}=2\sqrt3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/383975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 0
} |
Can both $x^2 + y+2$ and $y^2+4x$ be squares? Prove that there exist no positive integers $x$ and $y$ such that both $x^2+y+2$ and $y^2+4x$ are perfect squares.
I thought I could perhaps solve this by square bounding but I couldn't get anywhere with it.
Thanks in advance for any help.
| Thanks to Erick Wong for setting me on the right track.
Assume for sake of contradiction that $x^2+y+2$ and $y^2+4x$ are both perfect squares.
Then as $y$ is a positive integer, $x^2+y+2 \geq (x+1)^2=x^2 +2x+1$
$$\implies y+2\geq 2x+1$$
$$y+1 \geq 2x \; \; \; \; \; (1)$$
Following a similar argument, as $x$ is a positive integer, $y^2+4x \geq (y+1)^2=y^2+2y+1$
$$\implies 4x \geq 2y+1$$
But, $4x$ is even and $2y+1$ is odd so equality can never hold and hence
$$y^2+4x \geq (y+2)^2=y^2+4y+4$$
$$\implies x \geq y+1 \; \; \; \; (2)$$
Combining (1) and (2) we have that $x \geq y+1 \geq 2x$
$$\implies x \geq 2x$$
Which is a contradiction as $x \geq 1$.
Hence there are no positive integers $x$ and $y$ satisfying the requirements. QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/384100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Integral of $\int\frac{(x^4+1)\,dx}{x^3+4x}$ I followed the steps to solve this integral and want to know if I did it right and if $C=0? $
$$\int\frac{(x^4+1)\,dx}{x^3+4x} = \int\frac{(x^4+1)\,dx}{x(x^2+4)} = \frac{A}{x}+\frac{Bx+C}{x^2+4}$$
$$(x^2+4)A+x(Bx+C)=x^4+1$$
$$x=0 => 4A=1 => A=\frac{1}{4}$$
$$Ax^2+4A+Bx^2+Cx=x^4+1 = > (A+B)x^2+4A+Cx=x^4+1$$
$$A+B=0 => B=-\frac{1}{4}, C=0$$
Thanks!
| Whatever $A$, $B$, $C$ may be, the numerator of $\frac{A}{x} + \frac{B x+C}{x^2+4}$ is no greater than 3. The proper ansatz is
$$
\frac{x^4+1}{x^3+4x} = x + \frac{1-4x^2}{x(x^2+4)} = x + \frac{A}{x} + \frac{B x+ C}{x^2+4}
$$
and you should get $A = \frac{1}{4}$ and $C=0$, $B = -\frac{17}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/384496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I prove the arithmetic-geometric mean inequality? I am following along with this bare-bones proof of the arithmetic-geometric mean inequality with two real numbers. I'm having difficulty understanding the logic behind this step:
$$
\frac{a}{2}+\frac{a}{2}< \frac{b}{2}+\frac{a}{2}\Rightarrow a< \frac{b+a}{2}
$$
How is the step of adding $\frac{a}{2}$ to both sides valid? Also, if you have a better/easier way of proving this, please let me know. Thanks!
| Assume $0< a < b$.
The first thing we want to show for this proof is that given $$a<b,$$ we want $$a<\frac{a+b}{2} <b.$$ To do this, we show the left and right inequalities separately. Now, it is a rule of elementary algebra that addition preserves ordering, i.e. $$a+k < b+k$$ for all $k\in \mathbb{R}$. If we choose $k=\frac{a}{2}$ and apply this to $$\frac{a}{2} < \frac{b}{2},$$ then we have
$$
\frac{a}{2} + \frac{a}{2} < \frac{b}{2} + \frac{a}{2}
$$
so
$$
a < \frac{a+b}{2}.
$$
This is the left side of the inequality. To get the right side, we add $\frac{b}{2}$ to the same inequality:
$$
\frac{a}{2} + \frac{b}{2} < \frac{b}{2} + \frac{b}{2}
$$
so
$$
\frac{a+b}{2} < b.
$$
Combining these two inequalities we have
$$
a< \frac{a+b}{2} < b
$$
as desired.
For the remainder, we wish to show that
$$
a < \sqrt{ab} < \frac{a+b}{2}.
$$
First, the left side. Using the fact that $0<a<b$, we have $\sqrt{a}<\sqrt{b}$. Multiplying both sides by $\sqrt{a}$, we have
$$
\sqrt{a}\sqrt{a} <\sqrt{ab}
$$
$$
a < \sqrt{ab}
$$
This is the left side. For the right, we use the fact that
$$
0 < (a-b)^2 = a^2 - 2ab + b^2
$$
so
$$
ab < \frac{a^2 + b^2}{2}.
$$
But because $a, b > 0$,
$$
\sqrt{ab} < \frac{a + b}{2}.
$$
This is the right side. Combining all of these, we finally have
$$
0<a<\sqrt{ab}<\frac{a+b}{2}<b.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/385029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove this trigonometric expression? How would you go about proving the following?
$${1- \cos A \over \sin A } + { \sin A \over 1- \cos A} = 2 \operatorname{cosec} A $$
This is what I've done so far:
$$LHS = {1+\cos^2 A -2\cos A + 1 - \cos^2A \over \sin A(1-\cos A)}$$
....no idea how to proceed .... X_X
| Hint:
$1-\cos A=1-(1-2 \sin^2\dfrac{A}{2})=2\sin^2 \dfrac{A}{2}$
$\dfrac{2 \sin^2 \dfrac{A}{2}}{2 \sin \dfrac{A}{2} \cos \dfrac{A}{2}}=\tan \dfrac{A}{2}$
The other expression will be $\cot \dfrac{A}{2}$
$(\tan^2 \dfrac{A}{2}+1) /\tan\dfrac{A}{2}= \dfrac{\sec^2 A \cos \dfrac{A}{2}}{\sin \dfrac{A}{2}}= \dfrac{1}{2 \sin A}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/385537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
An easier way to find the integral of: $\int {x\sqrt {2 + x} {\rm{ }}dx} $, where ${u^2} = 2 + x$ My attempt at the question:
$\eqalign{
& \int {x\sqrt {2 + x} {\rm{ }}dx} \cr
& {u^2} = 2 + x \cr
& 2u{{du} \over {dx}} = 1 \cr
& {{du} \over {dx}} = {1 \over {2u}} \cr
& u = \sqrt {2 + x} \cr
& x = {u^2} - 2 \cr
& so: \cr
& \int {x\sqrt {2 + x} {\rm{ }}dx} = \int {x\sqrt {2 + x} } {\rm{ }}{{dx} \over {du}}du \cr
& = \int {x\sqrt {2 + x} } {\rm{ }} \times 2\sqrt {2 + x} du \cr
& = \int {2x} (2 + x)du \cr
& = \int {4x} + 2{x^2}du \cr
& = \int {4({u^2} - 2)} + 2{({u^2} - 2)^2}du \cr
& = \int {4{u^2} - 8} + 2({u^4} - 4{u^2} + 4)du \cr
& = \int {2{u^4} - 4{u^2}} du \cr
& = {2 \over 5}{u^5} - {4 \over 3}{u^3} + C \cr
& = {2 \over 5}{(\sqrt {2 + x} )^5} - {4 \over 3}{(\sqrt {2 + x} )^3} + C \cr
& = {2 \over 5}{(2 + x)^{{5 \over 2}}} - {4 \over 3}{(2 + x)^{{3 \over 2}}} + C \cr} $
A few questions I have:
Given ${u^2} = 2 + x$, $u = \pm \sqrt {2 + x} $, so why is it that we only take the principal square root and not the negative one for substitution?
The second question I have is a general one; is there an easier way of finding the integral? Have I done things in a manner that isn't overly longwinded? If so please suggest ways that would allow me to reach an answer quicker.
I'm on shakey grounds with integration at the moment so I was wondering if I could integrate this part of my working out without expanding out:
$ = \int {4({u^2} - 2)} + 2{({u^2} - 2)^2}du$
Thank you for all your help!
| If $u=2+x \implies dx=du$ then the integral becomes
$$\int (u-2)\sqrt u\,du$$
which can now be integrated easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to derive $\cos\frac{n\pi}{3}=\frac{1+3(-1)^{[\frac{n+1}{3}]}}{4}$ Consider the following formula to calculate a trigonometric function:
$$\cos\frac{n\pi}{3}=\frac{1+3(-1)^{[\frac{n+1}{3}]}}{4}$$ $[x]$ denotes the integer part of $x$. The formula is valid for $n=0,2,4,6,...$
I'm curious how this formula is derived? Is there any analytical method or is it a pure guessing?
| HINT:
The formula can be written as $$\cos\frac{2m\pi}3=\frac{1+3(-1)^{\left[\frac{2m+1}3\right]}}4$$ for all integer $m\ge0$
Now, the value of $(-1)^{\left[\frac{2m+1}3\right]}$ can be $+1,-1$ based on whether $\left[\frac{2m+1}3\right]$ is even or is odd
Observe that the fractional part of $\frac{2m+1}3$ can be $\frac a3$ where $0\le a\le2$
$(1)$ If $\left[\frac{2m+1}3\right]$ is even, the right hand side will be $\frac{1+3}4=1$
and $\frac{2m+1}3=2r+\frac a3$ where $r$ is a non-negative integer
So, $2m+1=6r+a\implies a=2(m-3r)+1$ i.e. odd $\implies a=1$
$\implies m=3r\implies $ the Left hand side $=\cos \frac{2(3r)\pi}3=\cos2r\pi=1$
$(2)$ If $\left[\frac{2m+1}3\right]$ is odd, the right hand side will be $\frac{1-3}4=-\frac12$
and $\frac{2m+1}3=2r+1+\frac a3$ where $r$ is a non-negative integer
So, $2m+1=6r+3+a\implies a=2(m-3r-1)$ i.e. even $\implies a=0,2$
$a=0\implies m=3r+1\implies $ the Left hand side becomes $=\cos\frac{2(3r+1)\pi}3=\cos\frac{2\pi}3=\cos\left(\pi-\frac\pi3\right)=-\cos\frac\pi3=-\frac12$ as $\cos(\pi-y)=-\cos y$
$a=2\implies m=3r+2\implies $ the Left hand side becomes $\cos\frac{2(3r+2)\pi}3=\cos\frac{4\pi}3=\cos\left(\pi+\frac\pi3\right)=-\cos\frac\pi3=-\frac12$ as $\cos(\pi+y)=-\cos y$
Alternatively,
$m$ can be of the form $3n,3n+1,3n+2$ where $n$ is any integer
Putting $m=3n,\cos\frac{2m\pi}3=\cos\frac{2(3n)\pi}3=\cos2n\pi=1$
and $\left[\frac{2m+1}3\right]=\left[\frac{2(3n)+1}3\right]=\left[2n+\frac13\right]=2n$
$\implies 1+3(-1)^{\left[\frac{2m+1}3\right]}=1+3(-1)^{2n}=1+3=4$
$\implies \frac{1+3(-1)^{\left[\frac{2m+1}3\right]}}4=1$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proof of the identity $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$ I just found this identity but without any proof, could you just give me an hint how I could prove it?
$$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$
I know that $$2^n = \sum\limits_{k=0}^n \binom{n}{k}$$ but that didn't help me
| Suppose we seek to verify that
$$S_n = \sum_{k=0}^n 2^{-k} {n+k\choose k} = 2^n.$$
We introduce the Iverson bracket
$$[[k\le n]] =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} \frac{1}{1-z}
\; dz$$
so we may extend $k$ to infinity, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{1-z}
\sum_{k\ge 0} 2^{-k} {n+k\choose n} z^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{1-z}
\frac{1}{(1-z/2)^{n+1}}
\; dz.$$
We evaluate this using the negative of the residues at $z=1, z=2$ and
$z=\infty.$ Here the contour does not include the other two finite poles
which also ensures that the geometric series converges. We could choose
$\epsilon = 1/2.$ We get for the residue at $z=1$
$$- \frac{1}{(1/2)^{n+1}} = - 2^{n+1}.$$
For the residue at $z=2$ we write
$$(-1)^{n+1} \mathrm{Res}_{z=2}
\frac{1}{z^{n+1}} \frac{1}{1-z}
\frac{1}{(z/2-1)^{n+1}}
\\ = (-1)^{n+1} 2^{n+1} \mathrm{Res}_{z=2}
\frac{1}{z^{n+1}} \frac{1}{1-z}
\frac{1}{(z-2)^{n+1}}
\\ = (-1)^{n} 2^{n+1} \mathrm{Res}_{z=2}
\frac{1}{(2+(z-2))^{n+1}} \frac{1}{1+(z-2)}
\frac{1}{(z-2)^{n+1}}
\\ = (-1)^{n} \mathrm{Res}_{z=2}
\frac{1}{(1+(z-2)/2)^{n+1}} \frac{1}{1+(z-2)}
\frac{1}{(z-2)^{n+1}}.$$
This is
$$(-1)^n \sum_{q=0}^n (-1)^q {n+q\choose q} 2^{-q} (-1)^{n-q}
= \sum_{q=0}^n {n+q\choose q} 2^{-q} = S_n.$$
Finally do the residue at $z=\infty$ getting (this also follows by
inspection having degree zero in the numerator and degree $2n+3$ in the
denominator)
$$\mathrm{Res}_{z=\infty} \frac{1}{z^{n+1}} \frac{1}{1-z}
\frac{1}{(1-z/2)^{n+1}}
\\ = - \mathrm{Res}_{z=0} \frac{1}{z^2}
z^{n+1} \frac{1}{1-1/z} \frac{1}{(1-1/2/z)^{n+1}}
\\ = - \mathrm{Res}_{z=0} \frac{1}{z}
z^{n+1} \frac{1}{z-1} \frac{z^{n+1}}{(z-1/2)^{n+1}}
\\ = - \mathrm{Res}_{z=0} z^{2n+1}
\frac{1}{z-1} \frac{1}{(z-1/2)^{n+1}} = 0.$$
Using the fact that the residues sum to zero we thus obtain
$$S_n - 2^{n+1} + S_n = 0$$
which yields
$$\bbox[5px,border:2px solid #00A000]{
S_n = 2^n.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 8,
"answer_id": 2
} |
Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$ Please help me to solve this integral:
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx.$$
I managed to calculate an indefinite integral of the left part:
$$\int\frac{\cos x}{\sin x}x\ \mathrm dx=\ x\log(2\sin x)+\frac{1}{2} \Im\ \text{Li}_2(e^{2\ x\ i}),$$
where $\Im\ \text{Li}_2(z)$ denotes the imaginary part of the dilogarithm. The corresponding definite integral $$\int_0^\pi\frac{\cos x}{\sin x}x\ \mathrm dx$$ diverges. So, it looks like in the original integral summands compensate each other's singularities to avoid divergence.
I tried a numerical integration and it looks plausible that
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx\stackrel{?}{=}\pi \log 54,$$
but I have no idea how to prove it.
| Begin with $u$-substitution using
$$\begin{align}
u & =x\\
dv & = \frac{3\cos(x)+\sqrt{8+\cos^2(x)}}{\sin(x)}\,dx\end{align}$$
so that $du=dx$, and my CAS tells me (which I suppose could be verified through differentiation and identities) that
$$\begin{align}
v & = \sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))
\end{align}$$
Now we have
$$\begin{align}
\left[x\left(\sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))\right)\right]_0^\pi\\
-\int_0^\pi\left(\sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))\right)dx
\end{align}$$
and most of the integral part can be evaluated by taking advantage of symmetry about $\pi/2$:
$$\begin{align}
\left[x\left(\sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))\right)\right]_0^\pi\\
-3\int_0^\pi\ln(1-\cos(x))dx
\end{align}$$
($\sinh^{-1}$ is odd and $\cos(x)$ has odd symmetry about $\pi/2$. For the logarithmic term, the input to $\ln()$ at $x$ is the reciprocal of the input at $\pi/2-x$.)
Some of the nonintegral-part can be cleanly evaluated:
$$\begin{align}
\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\left[3x\ln(1-\cos(x))\right]_0^\pi\\
-3\int_0^\pi\ln(1-\cos(x))dx
\end{align}$$
and now moving the "unclean" part back into an integral:
$$\begin{align}
\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\int_0^\pi\left(3\ln(1-\cos(x))+\frac{3x\sin(x)}{1-\cos(x)}\right)\,dx\\
-3\int_0^\pi\ln(1-\cos(x))dx\\
=\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\int_0^\pi\frac{3x\sin(x)}{1-\cos(x)}\,dx
\end{align}$$
My CAS says this is
$$\begin{align}
\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\pi\ln(64)
\end{align}$$
which is the only thing the CAS does that I don't quite get. But it's nothing special about endpoints: even WA can give an antiderivative if we can use the dilogarithm. It looks like an integral that might even appear somewhere on this site. A conversion of the arcsinh and logarithm rules yields
$$\begin{align}
\pi\ln(2^{-\frac{1}{2}})+\pi\ln\left(\frac{27}{8^{3/2}}\right)+\pi\ln(64)=\pi\ln(54)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 0
} |
Finding asymptotes to general functions I stumbled upon this problem and didn't quite understand why the solution worked. I have two questions:
*
*Why is it that the taylor series is enough to find the oblique asymptote? Specifically, why do the first two terms suffice? It seems like that may not work for all functions.
*From pre-calculus I recall finding asymptotes of rational functions, but how do I find vertical, horizontal, and oblique asymptotes of a more general function (such as an exponential or square root, etc.)?
| In general, the Taylor Series is the last thing you should be looking at.
In fact, the Taylor Series was not looked at; they calculated the Laurent Series. The point is that:
$$\operatorname{e}^y = 1 + y + \frac{1}{2}y^2 + \frac{1}{3!}y^3 + \cdots $$
Replacing $y$ with $2/x$ gives:
\begin{array}{ccc}
\operatorname{e}^{2/x} &=& 1 + \frac{2}{x} + \frac{1}{2}\left(\frac{2}{x}\right)^2 + \frac{1}{3!}\left(\frac{2}{x}\right)^3 + \cdots \\ \\
&=& 1 + \frac{2}{x} + \frac{2}{x^2} + \frac{4}{3x^3} + \cdots
\end{array}
The actual function under consideration was $x\operatorname{e}^{2/x}+1$ and so we have:
\begin{array}{ccc}
x\operatorname{e}^{2/x}+1 &=& x\left(1 + \frac{2}{x} + \frac{2}{x^2} + \frac{4}{3x^3} + \cdots\right) + 1 \\ \\
&=& \left(x + 2 + \frac{2}{x} + \frac{4}{3x^2} + \cdots\right) + 1 \\ \\
&=& x + 3 + \frac{2}{x} + \frac{4}{3x^2} + \cdots
\end{array}
The point here is that as $x$ gets larger, anything with an $x$ in the denominator gets smaller: $1/x \to 0$ as $x \to \infty$. So as $x \to \infty$ we have $2/x \to 0$, $4/3x^2 \to 0$, etc. All of "the tail" gets smaller and smaller very quickly. The only terms that we are left with are the $x$ and the $3$.
There is a general definition for an asymptote. They don't even need to be lines.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Further explanation needed :Finding all $z$ such that the modulus of $f(z)=e^{(z+1)/(z-1)}$ is equal to/at most $1$ I was solving a previous exam paper and there I got stuck on the following problem:
Let $f(z)=e^{\frac{z+1}{z-1}}$. Then find all $z \in \Bbb C$ for which
*
*$|f(z)|=1$,
*$|f(z)|\le 1$.
Any idea about how to tackle this problem will be highly appreciated. Thanks for your time.
| Let $ z = a + bi, (a, b) \in \mathbb{R}^2 $. Hence $$\begin{align*} \left|e^{\frac{z+1}{z-1}}\right| &= \left|e^{\frac{(a + 1 + bi)(a - 1 - bi)}{(a - 1 + bi)(a - 1 - bi)}} \right| \\&= \left|\ e^{\frac{a^2 - 1 + b^2 -2bi}{(a - 1)^2 + b^2}}\ \ \right| \\ &= \left|\ e^{ \frac{a^2 -1 + b^2}{(a - 1)^2 + b^2}}\ \ \ \ \ \right|\cdot\left|e^{-\frac{2bi}{(a-1)^2 +b^2}}\right|\\&=\ \ e ^{\frac{a^2 - 1 + b^2}{(a - 1)^2 + b^2)}}\end{align*}$$
Hence, the modulus is $ 1 $ when $ a^2 + b^2 = 1 $, which is a unit circle. Note that the point $ (1, 0) $ should be excluded because $ f(1) $ is not defined.
Also, the modulus is less than $1 $ on the entirety of the disk that is contained by that circle, i.e. $ a^2 + b^2 < 1 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving that $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{100}}<20$ How do I prove that:
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{100}}<20$$
Do I use induction?
| Prove the following claim using induction on $n$:
$$\sum_{k=1}^n \dfrac1{\sqrt{k}} < 2 \sqrt{n}$$
In the induction, you will essentially need to show that
$$2\sqrt{n} +\dfrac1{\sqrt{n+1}} < 2 \sqrt{n+1} \tag{$\star$}$$
To prove $(\star)$, note that
$$\sqrt{n} < \sqrt{n+1} \implies \sqrt{n} + \sqrt{n+1} <2 \sqrt{n+1} \implies \dfrac1{\sqrt{n+1}} < \dfrac2{\sqrt{n} + \sqrt{n+1}}$$
Multiplying and divding the right hand side by $(\sqrt{n+1} - \sqrt{n})$, we get
$$\dfrac1{\sqrt{n+1}} < \dfrac2{\sqrt{n} + \sqrt{n+1}}\cdot \dfrac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = 2({\sqrt{n+1} - \sqrt{n}})$$
which gives us $(\star)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/395979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Laurent Series Expansion Problems Expand in a laurent Series :
1- $f_{1} (z) = \frac{z^{2} - 2z +5 }{(z^{2}+1) (z-2)}$
in the ring : $1 < |z| < 2 $
2- $ f_{2} (z) = \frac{1 }{(z-3) (z+2)}$
In :
$i. 2 < |z| < 3
\\ ii. 0 < |z+2| < 5$
I managed to solve the second one but not sure if it is correct
For i. $2 < |z| < 3$ :
$ \frac{-1}{5} * \frac{1}{z(1+ \frac{2}{z}) } + \frac{1}{5} * \frac{1}{-3(1- \frac{z}{3}) }
= \frac{-1}{5} \sum_{n=0}^ \infty (-1)^{n} (\frac{2}{z})^{n} - \frac{1}{15}\sum_{n=0}^ \infty (\frac{z}{3})^{n}$
For ii. $0 < |z+2| < 5$ :
$ \frac{-1}{5} * \frac{1}{z+2} + \frac{1}{5} * \frac{1}{(z+2 -5) }
= \frac{-1}{5} * \frac{1}{z+2} + \frac{1}{25} * \frac{1}{-5 (1- \frac{Z+2}{5} ) } \\ = \frac{-1}{5} * \frac{1}{z+2}- \frac{1}{25} \sum_{n=0}^ \infty (\frac{z+2}{5})^{n} $
| In case 1, note that $1\lt|z|\lt2$ means that $1/|z|\lt1$ and $|z|/2\lt1$, hence the well known expansion
$1/(1-u)=\sum\limits_{n\geqslant0}u^n$, valid for every $|u|\lt1$, applied twice, yields
$$
\frac1{z^2+1}=\frac1{z^2(1+1/z^2)}=\sum\limits_{n\geqslant0}\frac{(-1)^n}{z^{2n+2}},$$
and
$$
\frac1{z-2}=-\frac12\frac1{1-z/2}=-\sum\limits_{n\geqslant0}\frac{z^n}{2^{n+1}}.
$$
The expansion of $f_1(z)$ as a Laurent series follows from the decomposition
$$
f_1(z)=\frac1{z-2}-\frac2{z^2+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Telescoping sum of powers
$$
\begin{array}{rclll}
n^3-(n-1)^3 &= &3n^2 &-3n &+1\\
(n-1)^3-(n-2)^3 &= &3(n-1)^2 &-3(n-1) &+1\\
(n-2)^3-(n-3)^3 &= &3(n-2)^2 &-3(n-2) &+1\\
\vdots &=& &\vdots & \\
3^3-2^3 &= &3(3^2) &-3(3) &+1\\
2^3-1^3 &= &3(2^2) &-3(2) &+1\\
1^3-0^3 &= &3(1^2) &-3(1) &+1\\
\underline{\hphantom{(n-2)^3-(n-3)^3}} & &\underline{\hphantom{3(n-2)^2}} &\underline{\hphantom{-3(n-2)}} &\underline{\hphantom{+1}}\\
n^3-0^3 &= & 3f_2(n) &-3f_1(n) &+n
\end{array}
$$
Can somebody explain me how these results are disposed intuitively? I didn't understand why $$(n-1)^3 -(n-2)^3$$ became equals to $$3(n-1)^2 - 3(n-1) + 1$$
How do this transformation was done?
Thanks!
| An idea: since
$$A^3-B^3=(A-B)(A^2+AB+B^2)$$
we get then
$$(n-1)^3-(n-2)^3=\left[(n-1)-(n-2)\right]\left[(n-1)^2+(n-1)(n-2)+(n-2)^2\right]\ldots\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/402445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
A tricky logarithms problem? $ \log_{4n} 40 \sqrt{3} \ = \ \log_{3n} 45$. Find $n^3$.
Any hints? Thanks!
| By elementary arithmetic operations (after / describing next action):
$$\log_{4n}40\sqrt{3}=\log_{3n}45\ \ \ \mbox{ / definition of logarithm}$$
$$(4n)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / } 4=\frac{4}{3}\cdot 3$$
$$\left({4\over 3}\cdot 3n\right)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / }(ab)^c=a^cb^c$$
$$\left({4\over 3}\right)^{\log_{3n}45}\cdot (3n)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / } a^{\log_ab}=b$$
$$\left({4\over 3}\right)^{\log_{3n}45}\cdot 45=40\sqrt{3}\ \ \ \mbox { / }\cdot\frac{1}{45}$$
$$\left({4\over 3}\right)^{\log_{3n}45}={8\over 9}\sqrt{3}\ \ \ \mbox { / }\left({a\over b}\right)^c=\frac{a^c}{b^c}$$
$$\left({4\over 3}\right)^{\log_{3n}45}=\left({4\over 3}\right)^{3\over 2}\ \ \ \mbox { / }a^b=a^c\Rightarrow b=c\ \ (a\neq 0,1)$$
$$\log_{3n}45=\frac{3}{2}\ \ \ \mbox { / definition of logarithm}$$
$$(3n)^{3\over 2}=45\ \ \ \mbox { / powered by } \frac{2}{3}\mbox{ and divided by }3$$
$$n=\frac{\sqrt[3]{45^2}}{3}\ \ \ \mbox { / powered by 3}$$
$$n^3=\frac{45^2}{27}$$
$$n^3=75$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/404389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$? I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$,
where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$.
Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ where $s_k$ satisfies the recurrence relation
\begin{align}
& s_{1}=-1,\hspace{5mm} s_{2}=\frac18,\hspace{5mm} s_{3}=-\frac{215}{216},\hspace{5mm} s_{4}=\frac{155}{1728},\hspace{5mm} \text{for all} \quad k>4, \\ s_{k} &=\frac1{k^3(2k-3)}\left(\left(-4k^4+18k^3-25k^2+12k-2\right)s_{k-1}+\left(12k^3-39k^2+38k-10\right)s_{k-2} \right.\\
& \hspace{5mm} \left. +\left(4k^4-18k^3+25k^2-10k\right)s_{k-3}\\+\left(2k^4-15k^3+39k^2-40k+12\right)s_{k-4}\right),
\end{align}
but it could not express $S$ or $s_k$ in a closed form.
Can you suggest any ideas how to calculate $S$?
| Here is a solution using simple tools
We have
$$\sum_{n=1}^\infty x^nH_n=-\frac{\ln(1-x)}{1-x}$$
Replace $x$ with $-x$ then multiply both sides by $-\frac{\ln(1-x)}{x}$ and use the fact that $-\int_0^1 x^{n-1}\ln(1-x)\ dx=\frac{H_n}{n}$
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n}=\int_0^1\frac{\ln(1-x)\ln(1+x)}{x(1+x)}\ dx$$
$$=\underbrace{\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\ dx}_{-5/8\zeta(3)}-\underbrace{\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ dx}_{\frac{1}{1+x}=y}$$
$$=-\frac58\zeta(3)-\int_{1/2}^1\frac{\ln\left(\frac{y}{2y-1}\right)\ln y}{y}\ dy=-\frac58\zeta(3)-I$$
$$I=\int_{1/2}^1\frac{\ln^2y}{y}\ dy-\int_{1/2}^1\frac{\ln(2y-1)\ln y}{y}\ dy=\frac13\ln^32-\Re\int_{1/2}^1\frac{\ln(1-2y)\ln y}{y}\ dy$$
$$=\frac13\ln^32+\Re\sum_{n=1}^\infty \frac{2^n}{n}\int_{1/2}^1 y^{n-1}\ln y\ dy=\frac13\ln^32+\Re\sum_{n=1}^\infty\frac{2^n}{n}\left(\frac{\ln2}{n2^n}+\frac{1}{n^22^n}-\frac{1}{n^2}\right)$$
$$=\frac13\ln^32+\ln2\zeta(2)+\zeta(3)-\Re\text{Li}_3(2)=\frac18\zeta(3)-\frac12\ln2\zeta(2)+\frac13\ln^32$$
where we used $\Re\text{Li}_3(2)=\frac78\zeta(3)+\frac32\ln2\zeta(2)$
Plug the result of $I$ we get $$\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n}=\frac12\ln2\zeta(2)-\frac34\zeta(3)-\frac13\ln^32$$
A different way to find $\int\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx$
First, add and subtract $\ln2$ and note that $\int\frac{\ln\left(\frac{1-x}{2}\right)}{1+x}\ dx=-\text{Li}_2\left(\frac{1+x}{2}\right)$
$$\int\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx=\int\frac{\ln\left(\frac{1-x}{2}\right)\ln(1+x)}{1+x} \ dx+\ln2\int\frac{\ln(1+x)}{1+x}\ dx$$
$$\overset{IBP}{=}-\ln(1+x)\text{Li}_2\left(\frac{1+x}{2}\right)+\int\frac{\text{Li}_2\left(\frac{1+x}{2}\right)}{1+x}\ dx+\frac12\ln2\ln^2(1+x)$$
$$=-\ln(1+x)\text{Li}_2\left(\frac{1+x}{2}\right)+\text{Li}_3\left(\frac{1+x}{2}\right)+\frac12\ln2\ln^2(1+x)$$
Therefore
$$\small{\int_0^a\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx=\text{Li}_3\left(\frac{1+a}{2}\right)-\text{Li}_3\left(\frac{1}{2}\right)-\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)+\frac12\ln2\ln^2(1+a)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 5,
"answer_id": 3
} |
Determine the general solution for $2\cos^2x-5\cos x+2=0$ Determine the general solution for $ 2\cos^{2}x-5\cos x+2=0$
My attempt:
$2u^2 - 5u + 2 = 0$
$(2u - 1)(u - 2) = 0$
$u = \frac{1}{2}$ or $u = 2$
$\cos(x) = \frac{1}{2}$ or $\cos(x) = 2$
The answers on the worksheet are as follows but my answer does not match...
A. $2.89$, $3.39$
B. $3.39$, $6.03$
C. $0.25$, $6.03$
D. $4.46$, $6.03$
| [\begin{array}{l}
\\
\\
{\rm{ }}2{\cos ^2}x - 5\cos x + 2 = 0\\
2\cos x(\cos x - 2) - 1(\cos x - 2) = 0\\
{\rm{ }}(2\cos x - 1)(cosx - 2) = 0\\
2\cos x - 1 = 0{\rm{ OR }}\cos x - 2 = 0\\
\\
{\rm{Case 1}}\\
2\cos x - 1 = 0\\
{\rm{ }}\cos x = \frac{1}{2}\\
{\rm{ }}\cos x{\rm{ }} = \cos \frac{\pi }{3}\\
{\rm{ }}\therefore x = 2n\pi \pm \frac{\pi }{3}{\rm{ }}\left( {n \in Z} \right)\\
\\
{\rm{Case2}}\\
\cos x - 2 = 0\\
{\rm{ }}\cos x = 2\\
{\rm{but }}\\
{\rm{ cosx}} \ne 2\\
{\rm{because - 1}} \le {\rm{cosx}} \le {\rm{1 }}\\
\therefore {\rm{ }}In{\rm{ }}this{\rm{ }}case{\rm{ }},there{\rm{ }}is{\rm{ }}no{\rm{ }}solution{\rm{ }}for{\rm{ }}\cos x{\rm{ }}
\end{array}]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/406368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Given that $\;\sin^3x\sin3x = \sum^n_{m=0}C_m\cos mx\,,\; C_n \neq 0\;$ is an identity . Find the value of n. Problem :
Given that $\sin^3 x \sin 3x = \sum^n_{m=0}C_m \cos mx, C_n \neq 0 $ is an identity. Find the value of n.
I tried : $\sin3x = 3\sin x - 4\sin^3 x$ but unable to reach to any point.... Please suggest further ....Thanks..
| HINT:
As you have noticed, $4\sin^3x=3\sin x-\sin3x,$
$$\text{So, }\sin^3x\sin3x=\frac{(3\sin x-\sin3x)\sin3x}4=\frac{3(2\sin x\sin3x)-2\sin^23x}8$$
As $2\sin A\sin B=\cos(A-B)-\cos(A+B),$
$\implies 2\sin x\sin3x=\cos(3x-x)-\cos(3x+x)=\cos2x-\cos4x$
and as $\cos2y=1-2\sin^2y\implies 2\sin^2y=1-\cos2y,$
$\implies 2\sin^23x=1-\cos6x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Equation in polar coordinates to rectangular coordinates The equation in polar coordinates is $r=2\sin(\theta)\tan(\theta)$.
Show that the equation in rectangular coordinates is $y^2=\frac{x^3}{2-x}$ x cannot equal 2.
$r=2\sin(\theta)\tan(\theta)$
1.) I multiplied both sides by r so. $r^2=2r\sin(\theta)\tan(\theta)$
2.) $r\sin(\theta)=y$ and $\tan(\theta)=\frac{y}{x}$ and $r^2=y^2+x^2$ so I plugged these in and got
$y^2+x^2=2y\frac{y}{x}$
3.) Then got $y^2=\frac{2y^2}{x}-x^2$ which equals $y^2=\frac{2y^2-x^3}{x}$
this is close but not the right answer; I don't know how to get it to be $y^2=\frac{x^3}{2-x}$.
| Good start! You want to gather the $y^2$ terms on one side, though, so instead, from $$x^2+y^2=\frac{2y^2}x,$$ subtract $y^2$ from both sides and factor, yielding $$x^2=\frac{2y^2}x-y^2=\left(\frac2x-1\right)y^2=\frac{2-x}xy^2.$$ Can you finish it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
help understanding step in derivation of correlation coefficient I'm looking to understand the starred step in the derivation below (also, if someone could help with the LaTex alignment, I'd appreciate it).
The regression line is $y= b_0 + b_1 x$, where $b_0$ and $b_1$ can be found by:
1) taking the difference between each observed value $y_i$ and the expected point regression line, $b_0 + b_1 x_i$
$$\text{ difference } = y_i - b_0 -b_1 x_i$$
2) summing the square of the differences from 1) to get the sum of squares
$$SS = \sum \limits_{i=1}^n (y_i - b_0 -b_1 x_i)^2$$
3) taking the partial derivative with respect to $b_0$ and $b_1$, then solving for each
$$
\begin{align}
\text{ solving for } b_0 \\
SS &= \sum(y_i - b_0 -b_1 x_i)^2\\
SS &= \sum (Y_i ^2 - 2Y_i b_0 - 2 Y_i b_1+ 2b_0 b_1X_i + b_1^2X_i^2+b_0^2) &\text{expand the square}\\
\frac{ \partial }{\partial_{b_0} }SS &= \sum (-2Y_i + 2b_1 X_i + 2b_0) &\text{partial derivative wrt} b_0\\
0 &= \sum 2(-Y_i + b_1 X_i + b_0) &\text{factor out 2 from the sum}\\
0 &= \sum (-Y_i + b_1 X_i + b_0) &\text{divide both sides by 2}\\
0 &= \sum -Y_i + \sum b_1 X_i + \sum b_0 &\text{split summation into parts}\\
\sum Y_i &= \sum b_1 X_i + \sum b_0 \\
\sum Y_i &= b_1 \sum X_i + n b_0 \\
\frac{1}{n}(\sum Y_i - b_1 \sum X_i ) &= b_0 \\
\bar Y - b_1 \bar X &= b_0 \text { rewrite sums as averages since } \frac{1}{n} \sum Y_i = \bar Y\\
\end{align}
$$
$$
\begin{align}
\\
\text{solving for } b_1\\
\frac{ \partial }{\partial_{b_1} }SS &= \sum (-2Y_iX_i + 2b_0 X_i + 2 b_1 X_i^2) &\text{ partial derivative wrt } b_1\\
0 &= 2\sum (-Y_iX_i + b_0 X_i + b_1 X_i^2) \\
0 &= \sum (-Y_iX_i + b_0 X_i + b_1 X_i^2) \\
0 &= -\sum Y_iX_i + b_0 \sum X_i + b_1 \sum X_i^2 &\text{ split summation}\\
0 &= -\sum Y_iX_i + (\bar Y - b_1 \bar X) \sum X_i + b_1 \sum X_i^2 &\text{ substitue } b_0\\
0 &= -\sum Y_iX_i + (\bar Y \sum X_i - b_1 \bar X \sum X_i) + b_1 \sum X_i^2 &\text{ distribute sum}\\
b_1 \bar X \sum X_i - b_1 \sum X_i^2 &= -\sum Y_iX_i + \bar Y \sum X_i &\text{ collect } b_1 \text{ terms}\\
b_1 (\bar X \sum X_i - \sum X_i^2) &= -\sum Y_iX_i + \bar Y \sum X_i \\
b_1 &= { \bar Y \sum X_i -\sum Y_iX_i \over (\bar X \sum X_i - \sum X_i^2) }\\
b_1 &= { \frac{1}{n} \sum Y_i \sum X_i -\sum Y_iX_i \over (\frac{1}{n} \sum X_i \sum X_i - \sum X_i^2) } \biggr ( \frac{-n}{-n} \biggr )\\
b_1 &= { n \sum Y_iX_i - \sum Y_i \sum X_i \over n \sum X_i^2 -(\sum X_i)^2 } \\
\end{align}
$$
$$
\begin{align}
b_0 &= \frac{1}{n} \sum y_i - b_1 \frac{1}{n} \sum x_i\\\\\\
b_1 &= {n \sum x_i y_i - \sum x_i \sum y_i \over n \sum x_i^2-(\sum x_i)^2}
\end{align}
$$
(derivation shown in http://polisci.msu.edu/jacoby/icpsr/regress3/lectures/week2/5.LeastSquares.pdf)
From this point you can use $b_1$ to get the correlation coefficient as follows:
$$
\begin{align}
b_1 &= {\frac{1}{n} \sum x_i y_i - (\frac{1}{n}\sum x_i) (\frac{1}{n} \sum y_i ) \over (\frac{1}{n} \sum x_i^2) -(\frac{1}{n}\sum x_i)^2} & \text{ divide top and bottom by } n^2 \\\\
b_1 &= {\frac{1}{n} \sum x_i y_i - (\bar x) (\bar y ) \over (\frac{1}{n} \sum x_i^2) -(\bar x)^2} & \text{ rewrite product of sums as averages } \\\\
b_1 &= {\frac{1}{n} \sum (x_i - \bar x)(y_i - \bar y ) \over \frac{1}{n} \sum (x_i - \bar x)^2} & \color{red} *\text{application of inscrutably arcane magic} \\\\
b_1 &= { \sum (x_i - \bar x)(y_i - \bar y ) \over \sqrt{\sum (x_i - \bar x)^2} \sqrt{\sum (x_i - \bar x)^2} } & \text{cancel } \frac{1}{n}\text{, factor denominator }\\\\
b_1 &= { \sum (x_i - \bar x)(y_i - \bar y ) \over \sqrt{\sum (x_i - \bar x)^2} \sqrt{\sum ( x_i - \bar x)^2} } \biggr({\sqrt{\sum(y_i - \bar y)^2} \over \sqrt{\sum(y_i - \bar y)^2}}\biggr) & \text{multiply by 1 } \\\\
b_1 &= { \sum (x_i - \bar x)(y_i - \bar y ) \over \sqrt{\sum (x_i - \bar x)^2} \sqrt{\sum(y_i - \bar y)^2}} \biggr({\sqrt{\sum(y_i - \bar y)^2} \over \sqrt{\sum ( x_i - \bar x)^2} }\biggr) & \text{re-arrange } \\\\
b_1 &= R \frac{S_x}{S_y}
\end {align}
$$
| For an alternative method for the same derivation:
$$
\begin{align}
\text{solving for } b_1\\
\frac{ \partial }{\partial_{b_1} }SS &= \sum (-2Y_iX_i + 2b_0 X_i + 2 b_1 X_i^2) &\text{ partial derivative wrt } b_1\\
0 &= 2\sum (-Y_iX_i + b_0 X_i + b_1 X_i^2) \\
0 &= \sum(X_iY_i - X_i\bar{Y} + b_1X_i\bar{X}-b_1X_i^2) &\text{replace }b_0 \\
0 &= \sum(X_iY_i - X_i\bar{Y}) - b_1\sum(X_i^2-X_i\bar{X}) &\text{separate into two sums} \\
b_1 &= \frac{\sum(X_iY_i - X_i\bar{Y})}{\sum(X_i^2-X_i\bar{X})} \\
\end{align}
$$
Noting that:
$$
\begin{align}
\sum X_i\bar{Y} = \sum X_i\left( \frac{1}{n}\sum Y_i \right) = n\left(\frac{1}{n}\sum X_i\right) \left( \frac{1}{n}\sum Y_i \right) = n\bar{X}\bar{Y} \\
\sum X_i\bar{X} = \sum X_i\left( \frac{1}{n}\sum X_i \right) = n\left(\frac{1}{n}\sum X_i\right) \left( \frac{1}{n}\sum X_i \right) = n\bar{X}^2 \:
\end{align}
$$
$b_1$ can be solved as:
$$
\begin{align}
b_1 &= \frac{\sum(X_iY_i) - n\bar{X}\bar{Y}}{\sum(X_i^2)-n\bar{X}^2}
\end{align}
$$
Finally, to rewrite in a more intuitive form, noting that:
$$
\sum(\bar{X}^2-X_i\bar{X})=0 \: \text{ and } \: \sum(\bar{X}\bar{Y}-Y_i\bar{X})=0
$$
$b_1$ can be written as the ratio of covariance to predictor variance:
$$
b_1 = \frac{\sum(X_iY_i-X_i\bar{Y}) + \sum(\bar{X}\bar{Y}-Y_i\bar{X})}{\sum(X_i^2-X_i\bar{X}) + \sum(\bar{X}^2-X_i\bar{X})} = \frac{\frac{1}{n}\sum(X_i - \bar{X})(Y_i-\bar{Y})}{\frac{1}{n}\sum(X_i-\bar{X})^2} = \frac{Cov(X,Y)}{Var(X)}
$$
Ref (equations 6-9):
http://seismo.berkeley.edu/~kirchner/eps_120/Toolkits/Toolkit_10.pdf
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Dividing polynomial by binomial How can I divide this using long division?
$$\frac{ax^3-a^2x^2-bx^2+b^2}{ax-b}$$
Edit
Sorry guys I wrote it wrong... Fixed it now.
| $$
\begin{array}{rccccccccccccc}
& & x^2 & - & ax & - & b \\[12pt]
ax-b & ) & ax^3 & - & (a^2+b)x^2 & + & b^2 \\
& & ax^3 & - & bx^2 \\[12pt]
& & & & -a^2x^2 & + & b^2 \\
& & & & -a^2x^2 & + & abx \\[12pt]
& & & & & & -abx & + & b^2 \\
& & & & & & -abx & + & b^2 \\[12pt]
& & & & & & & & 0
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Interval of convergence of $\sum\limits_{n\geq0} \binom{2n}{n} x^n$ We consider the power series $\displaystyle{\sum_{n\geq0} {2n \choose n} x^n}$. By Ratio Test, the radius of convergence is easily shown to be $R=\frac{1}{4}$.
For $x=\frac{1}{4}$, Stirling equivalent and Ratio Test imply that the series is divergent. For $x=\frac{-1}{4}$, Stirling equivalent and Alternating Series Test can be used to show the convergence of the series. Thus the interval of convergence is $\left[\frac{-1}{4},\frac{1}{4}\right)$.
My question: is there a (preferably simple ^^) method to determine the interval of convergence without using equivalents? My students don't know about equivalents.
| Let $a_n=\frac{1}{4^n}\binom{2n}{n}$. Then
\begin{eqnarray}
a_n&=&\frac{(2n)\cdot(2n-1)\cdot\dots\cdot 2 \cdot 1}{(2n)^2 \cdot (2n-2)^2 \cdot \dots \cdot 2} \\
&=&\frac{(2n-1) \cdot (2n-3) \cdot \dots \cdot 3 \cdot 1}{(2n) \cdot (2n-2) \cdot \dots \cdot 4 \cdot 2} \\
&=& \prod_{k=1}^n\left(1-\frac{1}{2k}\right)\\
&<& \prod_{k=1}^n \sqrt{\left(1-\frac{1}{2k+1}\right)\left(1-\frac{1}{2k}\right)}\\
&=&\sqrt{\frac{(2n)!}{(2n+1)!}}\\
&=&\sqrt{\frac{1}{2n+1}} \, .
\end{eqnarray}
So $a_n$ tends to $0$ as $n \to \infty$; as $a_n$ is clearly monotone, the alternating series test proves convergence for $x=-1/4$.
You can make a similar comparison to show that $a_n>\sqrt{\frac{1}{4n}}$, which proves divergence for $x=1/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How can I evaluate $\lim_{x\to7}\frac{\sqrt{x}-\sqrt{7}}{\sqrt{x+7}-\sqrt{14}}$? I need to evaluate the following limit, if it exists:
$$\lim_{x\to7}\frac{\sqrt{x}-\sqrt{7}}{\sqrt{x+7}-\sqrt{14}}$$
How can I solve it without using differentiation or L'Hôpital?
| We can get a nice function whose limit will not be indeterminate, by multiplying numerator and denominator each by the conjugate of $\sqrt{x + 7} - \sqrt{14}$:
$$\lim_{x\to7}\frac{\sqrt{x}-\sqrt{7}}{\sqrt{x+7}-\sqrt{14}}\cdot\dfrac{(\sqrt{x+7} + \sqrt{14})}{(\sqrt{x+7} + \sqrt{14})}$$
In the denominator, we obtain a difference of squares:
$(\sqrt{x + 7} - \sqrt{14})(\sqrt{x + 7} + \sqrt{14})\; =\; (x + 7) -14 \; =\; x-7 $
$$\lim_{x\to7}\frac{(\sqrt x-\sqrt 7)(\sqrt{x+7} + \sqrt{14})}{(x - 7)}$$
Hmmm...this doesn't seem to help us any, since both the numerator AND denominator will still evaluate to $0$.
But wait! We can view the term in the denominator as a difference of squares: $$(x - 7) = (\sqrt x)^2 - (\sqrt 7)^2 = (\sqrt{x} - \sqrt 7)(\sqrt x + \sqrt 7)$$
That gives us:
$$\lim_{x\to7}\dfrac{(\sqrt x-\sqrt 7)(\sqrt{x+7} + \sqrt {14})}{(\sqrt{x} - \sqrt 7)(\sqrt x + \sqrt 7)}$$
Canceling the common factor in numerator and denominator, we have (provided $x \neq 7$):
$$\lim_{x\to7}\dfrac{(\sqrt{x+7} + \sqrt {14})}{(\sqrt x + \sqrt 7)} = \frac {2\sqrt{14}}{2 \sqrt 7} = \sqrt{\frac {14}{7}} = \sqrt 2$$
which can now be evaluated without problems. We are taking the limit as $x \to 7$ after all, and so, while the original function is undefined at $x = 7$, the limit as $x \to 7$ indeed exists.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/415903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
$1=2$ | Continued fraction fallacy It's easy to check that for any natural $n$
$$\frac{n+1}{n}=\cfrac{1}{2-\cfrac{n+2}{n+1}}.$$
Now,
$$1=\frac{1}{2-1}=\frac{1}{2-\cfrac{1}{2-1}}=\frac{1}{2-\cfrac{1}{2-\cfrac{1}{2-1}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-1}}}}=\ldots
=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\dots}}}}},$$
$$2=\cfrac{1}{2-\cfrac{3}{2}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{4}{3}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{5}{4}}}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{6}{5}}}}}=\ldots
=\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\ldots}}}}}.$$
Since the right hand sides are the same, hence $1=2$.
| Another example where dots are misleading: $$1= \frac{ 1 \cdot \color{blue}{2} \cdot \color{green}{3} \cdot \color{red}{4} \cdots}{ 2 \cdot \color{blue}{3} \cdot \color{green}{4} \cdot \color{red}{5} \cdots} \leq \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "89",
"answer_count": 7,
"answer_id": 0
} |
Showing that $\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$
Show that $$\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$$
Using the substitution $t=\tan\frac{1}{2}x$
$\frac{\mathrm dt}{\mathrm dx}=\frac{1}{2}\sec^2\frac{1}{2}x$
$\mathrm dx=2\cos^2\frac{1}{2}x\,\mathrm dt$
$=(2-2\sin^2\frac{1}{2}x)\,\mathrm dt$
How do you get this in the form of $t$ instead of $x$ using $\sin A=\dfrac{2t}{1+t^2}$ ?
$$=\int_0^{1/\sqrt3}\frac{2-2\sin^2\frac{1}{2}x}{1-\frac{2t}{1+t^2}}\mathrm dt\,??$$
| The Weierstrass substitution sets $t = \tan \frac{x}{2}$ and it's possible that's what you mean to do. In that case, $$\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{2} \mathrm{sec}^2 \frac{x}{2} = \frac{1}{2} (1 + \tan^2 \frac{x}{2}) = \frac{1 + t^2}{2}$$ which gives you $$\mathrm{d}x = \frac{2}{1+t^2} \mathrm{d}t.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$ How am I suppose to prove that:
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$$
Do I use the way like how we count $1+2+ \cdots+100$ to estimate?
So $1/5050 \lt 20$, implying that it is indeed less than $20$?
| We can estimate the sum by integration: because the inverse square root function is strictly descreasing, hence for $n\in \mathbb{N}$
$$
\frac{1}{\sqrt{n}}< \frac{1}{\sqrt{x}} \quad \text{ for } x\in (n-1,n).
$$
Integrating both sides on this interval:
$$
\frac{1}{\sqrt{n}}= \int^n_{n-1}\frac{1}{\sqrt{n}}\,dx< \int^{n}_{n-1}\frac{1}{\sqrt{x}}\,dx.
$$
Therefore
$$
\frac{1}{\sqrt{1}} < \int_0^1 \frac{1}{\sqrt{x}}\,dx
\\
\frac{1}{\sqrt{2}} < \int_1^2 \frac{1}{\sqrt{x}}\,dx
\\
\cdots
\\
\frac{1}{\sqrt{100}} < \int_{99}^{100} \frac{1}{\sqrt{x}}\,dx
$$
and
$$
\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{100}} < \int_0^{100}\frac{1}{\sqrt{x}}\,dx =20.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/420733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 0
} |
Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here? I have $\sin 2x=\frac 23$ , and I'm supposed to express $\sin^6 x+\cos^6 x$ as $\frac ab$ where $a, b$ are co-prime positive integers. This is what I did:
First, notice that $(\sin x +\cos x)^2=\sin^2 x+\cos^2 x+\sin 2x=1+ \frac 23=\frac53$ .
Now, from what was given we have $\sin x=\frac{1}{3\cos x}$ and $\cos x=\frac{1}{3\sin x}$ .
Next, $(\sin^2 x+\cos^2 x)^3=1=\sin^6 x+\cos^6 x+3\sin^2 x \cos x+3\cos^2 x \sin x$ .
Now we substitute what we found above from the given:
$\sin^6 x+\cos^6+\sin x +\cos x=1$
$\sin^6 x+\cos^6=1-(\sin x +\cos x)$
$\sin^6 x+\cos^6=1-\sqrt {\frac 53}$
Not only is this not positive, but this is not even a rational number. What did I do wrong? Thanks.
| $(\sin^2 x + \cos^2 x)^3=\sin^6 x + \cos^6 x + 3\sin^2 x \cos^2 x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Given: $f(n)=2n^2-3$ and $g(n)=3n+4$, find $k(n)=(fgg)(n)$? Given: $f(n)=2n^2-3$ and $g(n)=3n+4$, find $k(n)=(fgg)(n)$?
solution:
\begin{align*}
fgg(n) &= f(g(g(n)))\\
&= f(g(3n+4)))\\
&= f(3(3n + 4) + 4)\\
&= f(9n + 16)\\
&= 2(9n + 16)^2 - 3\\
&= 2(81n^2 + 288n + 256) - 3 \\
&= 162n^2 + 576n + 509
\end{align*}
Check: Try $n = 1$, $f(g(g(1)) = f(g(7)) = f(25) = 2(25)^2 - 3 = 1247$,
$162(1^2) + 576(1) + 509 = 1247$
The practice test multiple choice answers are...
A. $6n^3+8n^2-9n-12$
B. $n^4+4n^3+4n^2+16n$
C. $-6n^3+8n^2+9n-12$
D. $n^4+4n^3+4n^2+16n$
Which one is it? What am I doing wrong?
| It looks to me as though you need to multiply functions $(f\cdot g \cdot g),$ given the solutions. Here, you are doing function composition, which given your earlier problem, seems you denote as $f\circ g\circ g$. Composing the functions, you did that correctly, no mistakes (I ran through it myself.) So solution to composing the functions is correct, but matches none of the answers you provide.
$$(f\circ g\circ g)(x) = 162n^2 + 576n + 509$$
So, given the notation used in earlier problems you posted to day, try to multiply $f(x)\cdot g(x)\cdot g(x)$
So, you want to expand the following:
$$(f\cdot g\cdot g)(x)=(2n^2 - 3)(3n+ 4)^2 = 18n^4 + 48n^3 + 5n^2 - 72n - 48$$
But again, the correct result is not a match to any of the given solutions you post.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing that $1^k+2^k + \dots + n^k$ is divisible by $n(n+1)\over 2$
For any odd positive integer $k\geq1$, the sum $1^k+2^k + \dots + n^k$ is divisible by $n(n+1)\over 2$.
I used induction principle for the solution but cannot prove it.
I took $P(k) = 1^k+2^k+\dots+n^k$.
For $P(1)$ it is true.
For $P(n)$ let it be true.
But for $P(n+1)$ I cannot solve it.
| For odd $k$ we have that
$$
a^k+b^k=(a+b)(a^{k-1}-a^{k-2}b+a^{k-3}b^2-\dots+b^{k-1})
$$
Thus, each column of
$$
\begin{align}
&0^k+\hphantom{(n-\ )}1^k+\hphantom{(n-\,\,)}2^k+\dots+n^k\\
&n^k+(n-1)^k+(n-2)^k+\cdots+0^k
\end{align}
$$
is divisible by $n$ and each column of
$$
\begin{align}
&1^k+\hphantom{(n-\ )}2^k+\hphantom{(n-\,\,)}3^k+\dots+n^k\\
&n^k+(n-1)^k+(n-2)^k+\cdots+1^k
\end{align}
$$
is divisible by $n+1$. Since $(n,n+1)=1$ we get that
$$
n(n+1)\,|\,2(1^k+2^k+3^k+\dots+n^k)
$$
and therefore, since $n(n+1)$ is even,
$$
\left.\frac{n(n+1)}{2}\middle|1^k+2^k+3^k+\dots+n^k\right.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/427744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 2
} |
Find number of solutions of $|2x^2-5x+3|+x-1=0$
Problem: Find number of solutions of $|2x^2-5x+3|+x-1=0$
Solution:
Case 1: When $2x^2-5x+3 \geq 0$
Then we get, $2x^2-5x+3+x-1=0$
x=1,1
Case 2: When $2x^2-5x+3 < 0$
Then we get, $-2x^2+5x-3+x-1=0$
x=1,2
In both cases, common value of x is 1
Hence solution is x=1
Am I doing right ??
Somebody told me solution is x=2
| $$|2x^2-5x+3|+x-1=
\begin{cases}
2x^2-5x+3+x-1 & x\ge\frac32 \quad \text{and} \quad x\le1 \\
-(2x^2-5x+3)+x-1 & 1\lt x\lt\frac32\\
\end{cases}\Rightarrow\begin{cases}
2x^2-4x+2 & x\ge\frac32 \quad \text{and} \quad x\le1 \\
-2x^2+6x-4 & 1\lt x\lt\frac32\\
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/428303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Determine the determinant and the inverse to the matrix A. I am currently working on determining the determinant and the inverse to the matrix A.
$A =\begin{bmatrix} 1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0\\0 & 0 & 0 & a & b\\0 & 0 & 0 & c & d\\\end{bmatrix}$
The determinant is $ad-bc$. But how do I find out the inverse? I guess there should be some kind of way to see what the inverse would be...
Any help would be much appreciated, as I am a beginner in this area a good explanation is always helpful :)!
| Since the OP asked in several comments whether this can be done by Gauss elimination/row operations, I wil also provide this computation. I will only do this for the right lower block.
$$
\left(\begin{array}{cc|cc}
a & b & 1 & 0\\
c & d & 0 & 1
\end{array}\right)\sim
\left(\begin{array}{cc|cc}
ac & bc & c & 0\\
ac & ad & 0 & a
\end{array}\right)\sim
\left(\begin{array}{cc|cc}
ac & bc & c & 0\\
0 & ad-bc &-c & a
\end{array}\right)\sim
\left(\begin{array}{cc|cc}
ac & bc & c & 0\\
0 & 1 &-\frac{c}{ad-bc} & \frac{a}{ad-bc}
\end{array}\right)\sim
\left(\begin{array}{cc|cc}
ac & 0 & \frac{acd}{ad-bc} & -\frac{abc}{ad-bc}\\
0 & 1 &-\frac{c}{ad-bc} & \frac{a}{ad-bc}
\end{array}\right)\sim
\left(\begin{array}{cc|cc}
1 & 0 & \frac{d}{ad-bc} & -\frac{b}{ad-bc}\\
0 & 1 &-\frac{c}{ad-bc} & \frac{a}{ad-bc}
\end{array}\right)
$$
The above calculation shows that
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$$
Note that the operations we did are only valid if $ad-bc\ne0$ (since we have divided by this expression).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/431398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Evaluating $\int^1_0 \frac2{\sqrt{2-x^2}} dx$ $$\int^1_0 \frac2{\sqrt{2-x^2}} dx$$
using substitution $x=\sqrt 2 \sin \theta$
$$\int^{\pi/4}_0 \frac{2\cos \theta d\theta}{\sqrt{2-2\sin^2 \theta}} = \int^{\pi/4}_0 \frac{2\cos\theta d\theta}{\sqrt2 \cos\theta} = \int^{\pi/4}_0 \frac{2d\theta}{\sqrt2} = \int^{\pi/4}_0 \frac{2\cdot\sqrt2 d\theta}{\sqrt2\cdot\sqrt2} = \int^{\pi/4}_0 \sqrt2d\theta = \sqrt2 \theta = \sqrt2 (\frac\pi4 - 0) = \sqrt2 \frac\pi4 $$
The problem is that the answer is $\frac\pi2$. Where did I make a mistake?
UPDATE:
using substitution $x=\sqrt 2 \sin\theta \rightarrow dx=\sqrt2\cos\theta$
$$\int^{\pi/4}_0 \frac{2\sqrt2\cos\theta}{\sqrt{2-2\sin^2\theta}} = \int^{\pi/4}_0 \frac{2\sqrt2\cos\theta}{\sqrt2 cos\theta} = \int^{\pi/4}_0 \frac{2\sqrt2}{\sqrt2} = \int^{\pi/4}_0 2 = 2 \theta = 2 (\frac\pi4 - 0) = \frac\pi2 $$
| HINT:
Putting $x=\sqrt2\sin\theta, dx=\sqrt2\cos\theta d\theta$ (you missed this $\sqrt2$)
$$\int^1_0 \frac2{\sqrt{2-x^2}}dx=2\int_0^{\frac\pi4}\frac{\sqrt2\cos\theta d\theta}{\sqrt2\cos\theta}=2\int_0^{\frac\pi4}d\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/432533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Which of the following sets is a complete set of representatives modulo 7? Which of the following sets is a complete set of representatives modulo 7?
1) (1, 8, 27, 64, 125, 216, 343)
1 mod 7 = 1
8 mod 7 = 1
27 mod 7 = 6
64 mod 7 = 1
125 mod 7 = 6
216 mod = 6
343 mod = 0
2) (1, -3, 9, -27, 81, -243, 0)
1 mod 7 = 1
-3 mod 7 = 4
9 mod 7 = 2
-27 mod 7 = 1
81 mod 7 = 4
-243 mod 7 = 2
0 mod 7 = 0
3) (0, 1, -2, 4, -8, 16, -32)
0 mod 7 = 0
1 mod 7 = 1
-2 mod 7 = 5
4 mod 7 = 4
-8 mod 7 = 6
16 mod 7 = 2
-32 mod 7 = 3
i think its only 1. but wanted to make sure.
| $3)~~ (0, 1, -2, 4, -8, 16, -32)$ is only complete set of $\pmod 7$.
$$0 \pmod 7 = 0$$
$$1 \pmod 7 = 1$$
$$-2 \pmod 7 = 5$$
$$4 \pmod 7 = 4$$
$$-8 \pmod 7 = 6$$
$$16 \pmod 7 = 2$$
$$-32 \pmod 7 = 3$$
they all have a unique solution from $0$ to $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/432676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Unintentional Negative Sign in Limit Evaluation I've been working on evaluating the following limit:
$$\lim_{x\to 0} \left(\csc(x^2)\cos(x)-\csc(x^2)\cos(3x) \right)$$
According to my calculator, the limit should end up being 4.
Though I've tried using the following process to find the limit, I continue to get -4:
$$\begin{align}
\lim_{x\to 0} \left(\frac{\cos(x)-\cos(3x)}{\sin(x^2)}\right)&=\lim_{x\to 0} \left(\frac{\cos(x) - \cos(2x+x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\cos(x) - \cos(2x)\cos(x) - \sin(2x)\sin(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\cos(x) - \left(1-2\sin^2(x)\right)\cos(x) - 2\sin^2(x)\cos(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\cos(x) - \cos(x)-2\sin^2(x)\cos(x)-2\sin^2(x)\cos(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\cos(x) - \cos(x)-4\sin^2(x)\cos(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{-4\sin^2(x)\cos(x)}{\sin(x^2)}\right)\\
&=\lim_{x\to 0} \left(\frac{\frac{d}{dx}\left(-4\sin^2(x)\cos(x)\right)}{\frac{d}{dx}\left(\sin(x^2)\right)}\right)\\
&=\lim_{x\to 0} \left(\frac{-4\cos^2(x)\cos(x)-\sin^2(x)\sin(x)}{\cos(x^2)}\right)\\
&=\frac{-4\cdot1 - 0}{1} \\
&=-4 \end{align}$$
Is there a specific place in my steps where I am going wrong with respect to negative sign notation or distribution?
(Also, apologies for the extra steps... and possible incorrect ordering of labels, functions, and symbols.)
| HINT:
Using Prosthaphaeresis Formulas,
$$\cos x-\cos3x=2\sin 2x\sin x$$
$$\implies \dfrac{\cos x-\cos3x}{\sin (x^2)}=2\dfrac{\sin 2x\sin x}{\sin (x^2)}=2\cdot2\cdot\dfrac{\dfrac{\sin2x}{2x}\cdot\dfrac{\sin x}x}{\dfrac{\sin (x^2)}{x^2}}$$
Now use $\displaystyle\lim_{h\to0}\frac{\sin h}h=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/432762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\sum_{n=1}^{\infty} {(-1)^n \cdot \frac{2^{2n-1}}{(2n+1)\cdot 3^{2n-1}}}$
Calculate the summation $\sum_{n=1}^{\infty} {(-1)^n \cdot \frac{2^{2n-1}}{(2n+1)\cdot 3^{2n-1}}}$.
So I said:
Mark $x = \frac{2}{3}$. Therefore our summation is $\sum_{n=1}^{\infty} {(-1)^n \cdot \frac{x^{2n-1}}{(2n+1)}}$.
But how do I exactly get rid of the $(-1)^n$? Also I notice it is a summation of the odd powers of $x$, how can I convert it to a full sum? (I know I should subtract from the full sum) but the signs of this summation is different than the signs of the full sum
| One can notice that
$$\frac{1}{x^2}\int x^{2n}\mathrm dx=\frac{x^{2n-1}}{(2n+1)}$$
So:
$$\sum_{n=1}^{\infty} {(-1)^n \frac{x^{2n-1}}{(2n+1)}}=\frac{1}{x^2}\sum_{n=1}^{\infty} {(-1)^n }\int x^{2n}\mathrm dx=\frac{1}{x^2}\int \left(\sum_{n=1}^{\infty} {(-1)^n }x^{2n}\right)\mathrm dx$$
And (keeping in mind that $x\leq 1$)
$$\sum_{n=1}^{\infty} {(-1)^n }x^{2n}=-\frac{x^2}{x^2+1}$$
Then
$$\sum_{n=1}^{\infty} {(-1)^n \frac{x^{2n-1}}{(2n+1)}}=\frac{1}{x^2}\int \left(-\frac{x^2}{x^2+1}\right)\mathrm dx=\frac{\tan^{-1}(x)-x}{x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/435956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Using trig substitution to evaluate $\int \frac{dt}{( t^2 + 9)^2}$ $$\int \frac{\mathrm{d}t}{( t^2 + 9)^2} = \frac {1}{81} \int \frac{\mathrm{d}t}{\left( \frac{t^2}{9} + 1\right)^2}$$
$t = 3\tan\theta\;\implies \; dt = 3 \sec^2 \theta \, \mathrm{d}\theta$
$$\frac {1}{81} \int \frac{3\sec^2\theta \mathrm{ d}\theta}{ \sec^4\theta} = \frac {1}{27} \int \frac{ \mathrm{ d}\theta}{ \sec^2\theta} = \dfrac 1{27}\int \cos^2 \theta\mathrm{ d}\theta $$
$$ =\frac 1{27}\left( \frac{1}{2} \theta + 2(\cos\theta \sin\theta)\right) + C$$
$\arctan \frac{t}{3} = \theta \;\implies$
$$\frac{1}{27}\left(\frac{1}{2} \arctan \frac{t}{3} + 2 \left(\frac{\sqrt{9 - x^2}}{3} \frac{t}{3}\right)\right) + C$$
This is a mess, and it is also the wrong answer.
I have done it four times, where am I going wrong?
| The integral $$ \int \cos^2 \theta \, \mathrm d\theta=\frac{\theta}{2}+\frac{1}{2} \sin \theta \cos \theta+C. $$
(You had the coefficient wrong)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove that: $a^2+b^2+(1-a-b)^2\ge \frac {1}{3}$ Where $a$ and $b$ are any given real number. I have tried solving it using partial derivative.
$$
s=a^2+b^2+(1-a-b)^2$$
$$\frac{\partial s}{\partial a}=2a-2(1-a-b) \tag{1}$$
$$\frac{\partial s}{\partial b}=2b-2(1-a-b) \tag{2}$$
for maxima both (1) and (2) are 0..from here we get two equations from where we get values of $a$ and $b$
$$2a-2(1-a-b)=0 \tag{3}$$
$$2b-2(1-a-b)=0 \tag{4}$$
putting the values of $a$ and $b$ we find from (3) and (4) in the function the maximum value of the function should be $\frac {1}{3}$.which in this case is not
| Taking it from where you left it:
$$a=1-a-b\implies a=\frac{1-b}2\\
b=1-a-b\implies b=\frac{1-a}2$$
So substituting the first eq. into the second one and using the symmetry of both:
$$b=\frac{1-\frac{1-b}2}2=\frac{1+b}4\implies b=\frac13\;\;\text{, and thus also}\;\;a=\frac13$$
Continuing with the partial derivatives
$$\frac{\partial^2s}{\partial a^2}=4=\frac{\partial^2s}{\partial b^2}\;\;,\;\;\frac{\partial^2s}{\partial a\partial b}=\frac{\partial^2s}{\partial b\partial a}=2$$
$$\text{Thus, at}\;\;\left(\frac13\,,\,\frac13\right)\;,\;\;\text{the Hessian is positive definite and we get a minimum there.}$$
From here, we get that
$$\forall\,a,b\in\Bbb R\;,\;\;a^2+b^2+(1-a-b)^2\ge\frac29+\left(1-\frac23\right)^2=\frac29+\frac19=\frac13\;\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Does $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ converge or diverge? I have to show whether $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ is convergent or divergent. I tried using the squeeze theorem to prove it was convergent. So what I did was
bound ${y_n}$ in between $\frac{-1}{n}$ and $\frac{1}{n}$. That is $\frac{-1}{n} \leq \frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n} \leq \frac{1}{n}$ Since we also proved earlier that $lim\frac{1}{n}=0$ and $lim\frac{-1}{n}= -lim\frac{-1}{n}=0$ It follows by the squeeze theorem that $lim(y_n)=0$. Would this be correct?
Edit: This what I have now.
Well this what I get so far that $ \sum_{k=n+1}^{2n}\frac{1}{k}≥ \sum_{k=n+1}^{2n}\frac{1}{k+1}$ Hence its divergent. I think it works since $ \sum_{k=n+1}^{2n}\frac{1}{k+1}=\frac{1}{(n+1)+1}+\frac{1}{(n+2)+1}+...+\frac{1}{2n+1}$ ≤ $ \sum_{k=n+1}^{2n}\frac{1}{k}= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} $.
| Note that
$\frac1{k+1}
\le \int_k^{k+1} \frac{dt}{t}
\le \frac1{k}
$.
Summing from $n+1$ to $n^2$,
$\sum_{k=n}^{n^2-1}\frac1{k+1}
\le \sum_{k=n}^{n^2-1}\int_k^{k+1} \frac{dt}{t}
\le \sum_{k=n}^{n^2-1}\frac1{k}
$,
or
$\sum_{k=n+1}^{n^2}\frac1{k}
\le \int_n^{n^2} \frac{dt}{t}
\le \sum_{k=n}^{n^2-1}\frac1{k}
$.
Using the right-hand inequality,
$
\sum_{k=n+1}^{n^2}\frac1{k}
=-\frac1{n}+\frac1{n^2}+\sum_{k=n}^{n^2-1}\frac1{k}
\ge -\frac1{n}+\frac1{n^2}+\int_n^{n^2} \frac{dt}{t}
>-\frac1{n}+ \ln(n^2)-\ln(n)
= \ln(n)-\frac1{n}
$
which diverges as $n \to \infty$.
If we use the left-hand inequality,
$
\sum_{k=n+1}^{n^2}\frac1{k}
\le \ln(n)
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/437244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
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.