Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluating $\int \frac{1}{x^{7} - x} ~ d{x} $. How do I evaluate the following indefinite integral?
$$
\int \frac{1}{x^{7} - x} ~ d{x}.
$$
Could someone give me some advice as to what method I should use or the steps that I should take?
Note: The OP originally requested for help in evaluating $ \displaystyle \int \lef... | $\displaystyle \int\frac{1}{x^7-x}dx = \int\frac{1}{x^7.\left(1-\frac{1}{x^6}\right)}dx$
Put $\displaystyle \left(1-\frac{1}{x^6}\right) = t$ and $\displaystyle \frac{6}{x^7}dx = dt$
$\displaystyle = \frac{1}{6}\int\frac{1}{t}dt = \frac{1}{6}\ln \mid t \mid+C$
$\displaystyle = \frac{1}{6}\ln \left|\frac{x^6-1}{x^6}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/314611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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probability of drawing marbles based on coin flip
An urn contains 4 red and 3 white marbles. A random marble $M_1$ is drawn and a fair coin is flipped. If the flip is heads then $M_1$ is put back into the urn. On the other hand, if the flip is tails, the marble $M_1$ is not put back into the urn. Now another random ma... | (i) Every marble is just as likely to be picked for $M_2$ as any other marble. That is why the probability is $\frac{4}{7}$. If you want to see this in more detail, lets look at all the possible ways of getting $M_2$ = red. Flip heads: white then red, red then red, flip tails: White then red, red then red. Total pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/315678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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jenny farm and the dozen egg ??? Farmer Jenny decides to expand her business interests and starts to package and sell the eggs produced by her chooks to a local shop.
The cost of producing $x$ dozen eggs per day is given by, in dirhams:
$$C=( 5/12 x^2+ 4x-3)$$
and the selling price of one dozen eggs is given by, in di... | Expanding on the answer by icurays1, the goal is indeed to maximize the quadratic, and the easiest way to do this uses calculus.
But in case you haven't learned calculus, you can maximize a quadratic just by completing the square - no calculator necessary!
The profit for selling $x$ dozen eggs is given by
\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/316057",
"timestamp": "2023-03-29T00:00:00",
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Ratio word problem with eggs? The eggs in a certain basket are either white or brown. If the ratio of the number of white eggs to the number of brown eggs is $2/3$ , each of the following could be the number of eggs in the basket except.....
A 10
B 12
C 15
D 30
E 60
| $$\dfrac 23 = \dfrac 46 = \dfrac 69 = \dfrac{8}{12} = \dfrac{10}{15} = \dfrac {12}{18} = \dfrac {14}{21} = \dfrac {16}{24} = \dfrac{18}{27} = \dfrac{20}{30} = \dfrac{22}{33} = \dfrac{24}{36}$$
For each ratio, the total number of eggs is given by the sum of the numerator with the denominator, ordered as follows $$\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/319033",
"timestamp": "2023-03-29T00:00:00",
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Solving for $b$ in $25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$ What are the steps to get from:
$$25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$$
to:
$$b = \frac{\sqrt{5}}{5} + \frac{2\sqrt{5} - \sqrt{10}}{10}$$
Thanks.
| First do the easy simplification on the lefthand side:
$$25\left(\frac{\sqrt{10}-2\sqrt5}{50}\right)=\frac{25}{50}\left(\sqrt{10}-2\sqrt5\right)=\frac12\left(\sqrt{10}-2\sqrt5\right)\;,$$
so the equation can be rearranged to
$$5b=\sqrt5-\frac12\left(\sqrt{10}-2\sqrt5\right)\;.$$
Now divide through by $5$ to get
$$b=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/320864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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:
$$s_{n} = \left(\frac{3}{10}-\frac{3}{2}\right)+\left(\frac{3}{14}-\frac{1}{2}\right)+\left(\frac{1}{6}-\frac{3}{10}\right)+\left(\fr... | The 3 terms you have mentioned above saying you didnot understand are the (n-2),(n-1) and nth terms of that series.
So if you just substitute 'n' with 'n-2' you get : $$\frac{3}{4n-2} - \frac{3}{4n-10}$$
if you substitute 'n' with 'n-1' you get : $$\frac{3}{4n+2} - \frac{3}{4n-6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321319",
"timestamp": "2023-03-29T00:00:00",
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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] $ ... | OK - if you need a really shorter way (which may work depending on the equation you have):
You have $(a+b+c)^3-(b+c)^3-(c+a)^3-(a+b)^3+a^3+b^3+c^3=6abc$
Looking at the LHS as a cubic polynomial in $a$, we immediately can notice $a=0$ is a root and hence $a$ is a factor. By symmetry, $b$ and $c$ are factors. As $ab... | {
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"url": "https://math.stackexchange.com/questions/321402",
"timestamp": "2023-03-29T00:00:00",
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Proving that $ 2 + 3\sqrt{-2} $ is reducible in $ \mathbb{Z}[\sqrt{-2}] $
Prove that $ 2 + 3\sqrt{-2} $ is irreducible in $
\mathbb{Z}[\sqrt{-2}] $
So far, I have let $ 2 + 3\sqrt{-2} = (a + b\sqrt{-2})(c+ d\sqrt{-2}) $
I then took the norm and got $\mathbf{N}(2 + 3\sqrt{-2}) = 22 = (a^2 + 2 b^2)(c^2 + 2 d^2) $
I thi... |
Let $ 2 + 3\sqrt{-2} = (a + b\sqrt{-2})(c+ d\sqrt{-2}) $ then $\mathbf{N}(2 + 3\sqrt{-2}) = 22 = (a^2 + 2 b^2)(c^2 + 2 d^2)$
That is a good way to start!
Now we just need to show $a^2+2b^2 = 2$ and $c^2+2d^2 = 11$ is impossible, but the first part is possible so we need to show $c^2+2d^2 = 11$ is impossible:
This is ... | {
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"url": "https://math.stackexchange.com/questions/321543",
"timestamp": "2023-03-29T00:00:00",
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Completing a Cayley table with few given spaces \begin{array}{ccc}
* & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6} \\
\textbf{1} & 1 & & & & & \\
\textbf{2} & & & & & 1 & \\
\textbf{3} & & & 1 & & 2 \\
\textbf{4} & & & & & & & \\
\textbf{5} & & & & & & \\
\textbf{6} & & & & & & 1
\en... | Hint: 1 is the identity because the identity is the only idempotent element in a group.
Edit: (I had some nonsense here before.)
Don't forget that the table for a group can only have symbols appear once in each column and row. Also don't forget that the only orders possible are 1,2,3,6. (You can use this to eliminate t... | {
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Finding Laurent series
Find the Laurent series for $\frac{(z+1)}{z(z-4)^3}$ in $0<|z-4|<4$.
How can I do this series and what is the most efficent way of doing it? Do I find the partial fractions or since we are given the region, which is defined by $0<|z-4|<4$, then I know its form must correlate accordingly by $\fr... | Let $\frac{z-4}{4}=t\implies z=4t+4$
then $|t|<1$ and $$\frac{(1+z)}{z(z-4)^3}=\frac{4t+5}{64(4t+4)(t^3)}=\frac{1}{64t^3}+\frac{1}{256t^3(1+t)}\big(=\frac{1}{256t^3}(1-t+t^2-t^3+\cdots)\big)$$ $$=\frac{1}{64t^3}+\frac{1}{256t^3}-\frac{1}{256t^2}+\frac{1}{256t}-\frac{1}{256}+\frac{t}{256}+\cdots$$ $$=\frac{5}{256t^3}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/322310",
"timestamp": "2023-03-29T00:00:00",
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Trouble with factorising a polynomial I'm supposed to show that:
$$y=\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$
and the required answers are: $$ P=5, Q=4, R=-4 $$
I tried to solve this with partial fractions like so:
$$5(x-1)(x+2) = A(x+3) + B(x-2)$$
$\implies$ $A$=4, $B$=-4
$\im... | $$\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$ gives
$$5(x-1)(x+2)=P(x-2)(x+3)+Q(x+3)+R(x-2)........(1)$$
Now put $x=2, x=-3,x=0$ respectively on both sides of $(1)$ to get, $Q=4,R=-4,P=5$ respectively. Hence you can reach the desired result.
| {
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Simplifying repeating square roots. How can I solve the equation:
$$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}} = \frac{1+\sqrt{53}}{2}$$
| Set $y = \sqrt{x + \sqrt{x + \dots}}$.
$y^2 = x+\sqrt{x+\dots} = x + y$
So $y^2 - y = x$.
But $\displaystyle y = \frac{1 + \sqrt{53}}{2}$
Now just substitute.
Addendum for people who want to see something a bit more rigorous.
We will take the infinite square root expression to be defined as the limit of the sequence d... | {
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"url": "https://math.stackexchange.com/questions/326007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the sum $\sum\limits_{n=1,n\neq m^2}^{1000}\left[\frac{1}{\{\sqrt{n}\}}\right]$, Find the value
$\displaystyle\sum_{n=2,n\neq m^2}^{1000}\left[\dfrac{1}{\{\sqrt{n}\}}\right]$,
by $\{x\}=x-[x]$,
$[x]$was bracket function,for example:$[5.4]=5, [2.9]=2,[-1.1]=-2 $and so on.
| The answer is 3843 as computed by brute force.
The rest is not the answer but interesting pieces I find related to this sum.
For $n^2 < x < (n+1)^2$, let $x = n^2 + k$, we have:
$$\left\lfloor\frac{1}{\{\sqrt{x}\}}\right\rfloor
= \left\lfloor\frac{1}{\sqrt{n^2+k}-n}\right\rfloor
= \left\lfloor\frac{\sqrt{n^2+k} + n}{... | {
"language": "en",
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Generating Function Example from class Example: Consider the sequence $(h_n)$ where $h_n$ is the number of nonnegative integer solutions to $$a_1+a_2+a_3+a_4+a_5=n.,$$ where $a_1$ is even, $a_2$ is odd, $a_3$ is a multiple of $5$, $a_4$ is 1,3, or 7, and $a_5$ is 0 or 1.
Without clear explaination, my professor writes ... | In this generating function, the things representing the solutions are the exponents. The condition that $a_1$ is even means that it can be $0, 2, 4, 6, \ldots$. Your professor therefore associates it to the even powers $(1 + x^2 + x^4 + \ldots)$. It's similar for the other cases.
Now, if you understand how multiplicat... | {
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Why does the $ p $-series diverge when $ p = 1 $? We’re currently working with $ p $-series in my Calculus class, and I’ve fallen for the apparently common misconception that the infinite sum $ \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{p}} $, where $ p = 1 $, should converge, whereas in reality, it diverges. Can an... | To answer your question, the integral test says that if $a_n$ is a positive decreasing sequence, and there exists a decreasing positive $f$ such that $f(n)=a_n$ for each $n$, then $$\sum_{n=1}^\infty a_n$$ converges if and only if $$\int_1^\infty f(x) dx$$ does. You have correctly found that with $f(x)=\frac 1 x $ and ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $-\frac{\sqrt{x}}{1+x}\log{x} \leq \log{2}$ for $0 < x < 1$ Graphically and numerically it is obvious but I'm looking for an analytical reasoning.
Just maximizing the left hand side does not yield an analytical expression for the maximum.
I also tried some known bounds for $\log$, but all of them had "oversh... | Put $y=\frac{1}{\sqrt{x}}$ to get $(y+\frac{1}{y})\log 2-2\log y \geq 0, y>1$.
Differentiating gives $\frac{1}{y}((y-\frac{1}{y})\log 2 -2)$, so we have critical points when $(y-\frac{1}{y})\log 2=2$. Solving the quadratic in $y$ and taking the positive root gives $y=\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$. It suffices t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/328906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Series involving log $\sum_{n=1}^{\infty} \left( n\log \left(\frac{2n+1}{2n-1}\right)-1\right) = \frac{1-\log 2}{2}$ Does anybody know how to prove this series?
$$\sum_{n=1}^{\infty} \left( n\log \left(\frac{2n+1}{2n-1}\right)-1\right) = \frac{1-\log 2}{2}$$
I arrived at this through Mathematica.
I tried writing $\log ... | $$n \log \left(\dfrac{2n+1}{2n-1}\right) = n \left(\log(1+1/2n) - \log(1-1/2n)\right)$$
\begin{align}
\log(1+x) - \log(1-x) & = \left(x - \dfrac{x^2}2 + \dfrac{x^3}3 \mp \cdots\right) - \left(-x - \dfrac{x^2}2 - \dfrac{x^3}3 - \cdots\right)\\
& = 2\left(x + \dfrac{x^3}3 + \dfrac{x^5}5 + \cdots \right)
\end{align}
Hence... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Partial Fraction Decomposition Problem... half answered... $$\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{... | I'll denote $f(x)$ this rational function. Here is an alternative to the expanding/equating strategy.
Limit trick: multiply both sides by $x$, compare the degrees, and let $x$ tend to $+\infty$:
$$
\lim_{x\rightarrow+\infty}xf(x)=5=A+C.
$$
Substitution trick: pick a small value in the domain of $f$. Here $0$ is perfect... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Partial Fractions - combinatorics - trouble Having a very hard time with this question:
Q: Use partial fractions to find the power series expansion of $$\frac{1+5x}{1-2x-3x^2}$$
| I just want to show how to apply Partial Fraction Decomposition
$$\frac{1+5x}{1-2x-3x^2}=\frac{1+5x}{(1-3x)(1+x)}=\frac A{1-3x}+\frac B{1+x}$$
$$\implies 1+5x=A(1+x)+B(1-3x)=x(A-3B)+A+B$$
Comparing the coefficients of the different powers of $x,A+B=1,A-3B=5$
$\implies B=-1,A=2$
Alternatively,
putting $x+1=0\implies x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/333565",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Legendre Polynomials: proofs $\int_{-1}^1P_n^2(x)dx=\frac{2}{(2n+1)}$ Does any one know, how to compute any of those two things?
*
*The relationship between Legendre polynomials and Shifted Legendre Polynomials.
*$\displaystyle\int_{-1}^1P_n^2(x)dx=\dfrac{2}{(2n+1)}$ for $n\geq0$.
I tried to use Bonnet's equation... | I will answer your question about determining the value of $\int_{-1}^1 P_n(x)^2dx$, using Rodrigues' formula
$$P_n(x) = \frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$
$\newcommand{\partial}[1]{\left[#1\right]}$
$\newcommand{\bracket}[1]{\left(#1\right)}$
\begin{equation}
\begin{split}
I&=\int_{-1}^1 P_n(x)^2dx
\\ &=\frac{1}{2^{2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/334524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$ Find the value of $\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$, if $x\in \left(0,\frac{1}{2}\right)$. I know it is equal to $\sqrt{2}+1$, but I don't know how to prove it?
| a$\sqrt{1-\sqrt{x}}=a$
$\sqrt{1+\sqrt{x}}=b$
$ab=\sqrt{1-x}$
Now your expression will look something like :
$\frac{b+\sqrt{1+ab}}{a
+\sqrt{1-ab}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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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 ?
| Just for "fun" :
$\displaystyle x^8-x^7+x^2-x+15 = \frac{(128x^4-64x^3-16x^2-8x-9)^2+4(16x^2-11x+121)^2+60(x-49)^2+43055}{2^{14}}>0\;\forall x\in \mathbb{R}$
Another Solution::
If $x<0,$ note that $x^8+(-x^7)+x^2+(-x)>0,$ so the polynomial cannot have any negative roots.
If $x\geq 0,$ then note that from AM-GM inequal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/335653",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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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}... | To elaborate on Ross Millikan's answer and André Nicolas's comment, denote the $n$-th block sum (ignoring the sign) by $B_n$. Then
\begin{align}
B_n &= \sum_{i=1}^n \frac1{k_n+n},\\
k_n &= \frac{n(n-1)}{2},\\
k_{n+1} &= k_n+n = \frac{n(n+1)}{2} \le n^2.\tag{1}
\end{align}
Therefore
\begin{align}
B_n - B_{n+1}
&= \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Proving $\pi(\frac1A+\frac1B+\frac1C)\ge(\sin\frac A2+\sin\frac B2+\sin\frac C2)(\frac 1{\sin\frac A2}+\frac 1{\sin\frac B2}+\frac 1{\sin\frac C2})$ Let $\Delta ABC$, prove that
$$\pi\left(\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\right)\ge \left(\sin{\dfrac{A}{2}}+\sin{\dfrac{B}{2}}+\sin{\dfrac{C}{2}} \right) \left(\dfra... | Maybe this will help you,
$$\sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2} \le \dfrac {1}{8}$$(With equality iff $A=B=C$)
Well here's a proof of it:
We have
$$\sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2}$$
$$=\dfrac{1}{2}[\cos \dfrac{A-B}{2}-\cos \dfrac {A+B}{2}]\sin \dfrac{C}{2}$$
$$\le \dfrac{1}{2}[1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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} |
Having trouble showing that these series are the same. $$\frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\tfrac{n(n+1)}{2}+1}\frac{(x-\pi/4)^n}{n!} $$
$$= \frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\tfrac{n(n-1)}{2}}\frac{(x-\pi/4)^{n+1}}{(n+1)!} + 1$$
*(added the +1) sorry didn't see this before, my answ... | Edit: None of these is the Taylor series of $\sin $ at $\pi/4$...There is a problem with the sign of the coeffcients in both. Here is a safe way to get the sign: use the floor function.
$$
\frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\lfloor n/2\rfloor}\frac{(x-\pi/4)^{n}}{n!}.
$$
Previous answer: They are not t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/337906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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 $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| the sequence $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+.....$$
is an $arithmetico geometric series$
the sum of an$AG$ series of the form
$$S_{\infty}=a+(a+d)r+(a+2d)r^2+(a+3d)r^3+.....\infty=\frac{a}{1-r}+\frac{dr}{(1-r)^2}$$converting the above series inti this form
$$\frac{1}{2}( \frac{1}{1} +\frac{2}{2}+ \frac{3}{4}+..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 12,
"answer_id": 11
} |
Find minimum of $\sin^4 x + \cos^4 x + \sec^4 x$ I tried manipulating the terms but I couldn't get anywhere.
The only other thing I can observe is that the minimum must be greater than $1$ since all the terms are non-negative and the range of $\sec^4 x$ is $[1, \infty)$. Also it can't be $1$ since then $\sin^4 x + \cos... | $\sin^{4}{x}+\cos^{4}{x}+\sec^{4}{x}=\sin^{4}{x}+\left(\cos^{2}{x}-\sec^{2}{x}\right)^2+2$
Minimum value of $\sin^4{x}$ is zero and is attained at $x=n\pi$
The minimum value of $\cos^{2}{x}-\sec^{2}{x}$ is zero and attained when $\cos^{2}{x}=\sec^{2}{x}$ or $x=n\pi$. Both minimum's coincide at $x=n\pi$. And the minimum... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
The generating function for the Fibonacci numbers Prove that $$1+z+2z^2+3z^3+5z^4+8z^5+13z^6+...=\frac{1}{1-(z+z^2)}$$
The coefficients are Fibonacci numbers, i.e., the sequence $\left\{1,1,2,3,5,8,13,21,...\right\}$.
| The proof is quite simple. Let's write our sum in a compact format:
$$
1+z+2z^2+3z^3+5z^4+8z^5+... = \sum_{n=0}^\infty F_nz^n
$$
Where $F_n$ is the $n$th Fibonacci number, starting with $F_0=F_1=1$, and $F_{n+2}=F_n+F_{n+1}$. It is from here that we will prove what needs to be proven.
$$\begin{align}
(1-z-z^2)\sum_{n=0... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
"answer_count": 4,
"answer_id": 0
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Prove $\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}}\ge\sum_{i=1}^{n}\frac{1-a_{i+1}}{1-a_{i}}$ if $a_{i}>0$ and $a_{1}+a_{2}+\cdots+a_{n}=1$ Let $a_{i}>0,i=1,2,\cdots,n$, and $a_{1}+a_{2}+\cdots+a_{n}=1$.
How can we prove that
$$\displaystyle\sum_{i=1}^{n}\dfrac{a_{i}}{a_{i+1}}\ge\displaystyle\sum_{i=1}^{n}\dfrac{1-a_{i+1}}{1-a... | Below is a proof for $n=3$ (which, unfortunately, does not seem to
generalize).
When $n=3$, the inequality to be shown can be rewritten as
$$
\frac{a_1}{a_2}+\frac{a_2}{a_3}+\frac{a_3}{a_1} \geq
\frac{a_1+a_3}{a_2+a_3}+\frac{a_1+a_2}{a_1+a_3}+\frac{a_2+a_3}{a_1+a_2}
\tag{1}
$$
or equivalently,
$$
\bigg(\frac{a_1}{a_2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
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compute line integral where line is boundary of $(x-5)^2 + (y-2)^2 \le 1$ Line integral of $$\frac{-y^3}{(x^2+y^2)^2} dx + \frac{xy^2}{(x^2+y^2)^2} dy$$
Using greene's theorem, this integral is equal to zero. Am i right? $P(x,y)$ and $Q(x,y)$ are not continuous at $(0,0)$. Since the circle does not contain $(0,0)$, gre... | Using Green's Theorem we get:
$$\int_C\frac{-y^3}{(x^2+y^2)^2}dx+\frac{xy^2}{(x^2+y^2)^2}dy=\int\int_D\left(\frac{8xy^3}{(x^2+y^2)^3}-\frac{2x^2y}{(x^2+y^2)^2}\right)dA$$
Then switch to polor coordinates, since $D$ is a unit circle, (just shifted) we get:
$$=\int_0^{2\pi}\int_0^1\left(8\cos\theta \sin^2\theta\cdot r-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/341306",
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"source": "stackexchange",
"question_score": "2",
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How to prove ${{a}^{a}}{{b}^{b}}\ge {{\left(\frac{a+b}{2}\right)}^{a+b}}$ ?thanks. How to prove
$${{a}^{a}}{{b}^{b}}\ge {{\left(\frac{a+b}{2}\right)}^{a+b}}$$
$a>0$,$b>0$,
thanks.
| Let $P=(1+x)^{1+x} (1-x)^{1-x}$ , where $0 ≤ x <1$ ; then
$\log P=(1+x)\log(1+x)+(1-x)\log(1-x)$
$\space$$\space$ $\space$$=x${$\log(1+x)-\log(1-x)$ } +$\log(1+x)+\log(1-x)$ $\space$$=2x\bigg(x+\frac{x^3}{3} + \frac{x^5}{5}+...\bigg) - 2\bigg(\frac{x^2}{2}+\frac{x^4}{4} + \frac{x^6}{6}+...\bigg)$ $=2\bigg(\frac{x^2}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/341439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 4
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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?
... | Yes, it is correct because you are able to cancel out two of the binomials leaving $(x+5)$ on the top and $(x+4)$ on the bottom.
| {
"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": 1
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A condition for a triangle to be isosceles in $\Delta ABC$,and
$$\dfrac{\sin{(\dfrac{B}{2}+C)}}{\sin^2{B}}=\dfrac{\sin{(\dfrac{C}{2}+B)}}{\sin^2{C}}$$
prove that $B=C$
I think $\sin{(\dfrac{B}{2}+C)}\sin^2{C}=\sin{(\dfrac{C}{2}+B)}\sin^2{B}$
then
$$\sin{(\dfrac{B}{2})}\cos{C}\sin^2{C}+\cos{\dfrac{B}{2}}\sin^3C=\sin{\d... |
Its just kind of a hint, I hope its of some help.
Using sine-rule on $\triangle ABC$, you get :
$$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}= 2R$$
$$\dfrac{\sin B}{\sin C}= \dfrac{b}{c} \implies \dfrac{\sin^2 B}{\sin^2 C}= \dfrac{b^2}{c^2}$$
Use sine-rule for $\triangle ADB$ and $\triangle AEC$
$$\dfrac{c}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/344725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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To show that a partial dertivative (of a piecewise function) is continuous at $0$
$$f(z)=\cases{\frac{x^4-6x^2y^2+y^4}{x^2+y^2}
+i\frac{4xy(x^2-y^2)}{x^2+y^2},& $z\ne0$\cr 0, &$z=0$}$$
Let $u=\Re(f)$.
I have shown from first principles that $\frac{\partial u}{\partial x} = 0$ at $0$. To go on to show that $\frac{\... | $$\frac{\partial u}{\partial x}=\frac{(x^2+y^2)(4x^3-12xy^2)-2x(x^4-6x^2y^2+y^4)}{(x^2+y^2)^2}=$$
$$=\frac{2x^5-14xy^4+4x^3y^2}{(x^2+y^2)^2}=2x\frac{x^4+2x^2y^2-7y^4}{(x^2+y^2)^2}\implies$$
$$\left|\;\frac{\partial u}{\partial x}\;\right|\le|2x|\frac{x^4+2x^2y^2+7y^4}{(x^2+y^2)^2}=|2x|\left(1+6\frac{y^4}{(x^2+y^2)^2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/345926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving $\sin \theta + \cos \theta=1$ in the interval $0^\circ\leq \theta\leq 360^\circ$
Solve in the interval $0^\circ\leq \theta\leq 360^\circ$ the equation $\sin \theta + \cos \theta=1$.
I've got the two angles in the interval to be $0^\circ$ and $90^\circ$, it's not an answer I'm after, I'd just like to see diffe... | We have $$\sin(\theta) + \cos(\theta) = 1 \text{ and } \sin^2(\theta) + \cos^2(\theta) = 1$$
If we have $$a+b = 1 \text{ and } a^2 + b^2 =1 \tag{$\star$}$$then either $a=0,b=1$ or $a=1, b=0$. This gives us $$\sin(\theta) = 0, \cos(\theta) = 1 \text{ (or) }\sin(\theta) = 1, \cos(\theta) = 0$$Hence, $$\theta = 0^{\circ},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/346081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 1
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Formula for the $1\cdot 2 + 2\cdot 3 + 3\cdot 4+\ldots + n\cdot (n+1)$ sum Is there a formula for the following sum?
$S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots + n\cdot (n+1)$
| We can use discrete calculus! Let $x^{\overline{k}}$ denote the $k$th rising factorial power of $x$. That is: $$x^{\overline k} = \underbrace{x(x+1)(x+2)\cdots(x+k-1)}_{k\text{ factors}}$$
Then, $S_n = \sum_{k=0}^{n} k(k+1) = \sum_0^{n+1}x^{\overline 2}\,\delta x$. (Notice that I started the summation at $k=0$; th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
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Prove that if $6 \mid (2a + 4)$ and $9 \mid (12 + 3b)$ then $3 \mid (a + b)$ Could you help me with the problem below?
Prove that if $6 \mid (2a + 4)$ and $9 \mid (12 + 3b)$ then $3 \mid (a + b)$.
Thank you!
| Since $6 \mid (2a+4)$, let $$(2a+4) = 6p.$$
$$a = (6p-4)/2.$$
$$a = (3p-2) \tag{1}$$
Again since $9 \mid (12+3b)$, let $$(12+3b) = 9q.$$
$$b = (9q-12)/3.$$
$$b = (3q-4)\tag{2}$$
(1) + (2) gives you: $$(a+b) = 3(p+q)-6 \quad \Rightarrow \quad (a+b) = 3(p+q-2).$$
Hence $3$ divides $(a+b)$ or $3 \mid(a+b)$.
Hope the answe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/348452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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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
| Whenever you have an "$x$ divides $y$" problem, start by stating it in modular form.
$$
a^2+ab+b^2 \equiv 0 \mod 9
$$
Now, we can also write this as
$$
a^2-2ab+b^2 \equiv -3ab \mod 9
$$
Therefore, we must have that $(a-b)^2\equiv 0 \pmod3$, and this can only be satisfied if $a\equiv b\pmod 3$.
So let $b=a+3k$ and subst... | {
"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": 3
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Proofs of $\cos(x+y) = \cos x\cos y - \sin x \sin y$ Define $\sin x $ and $\cos x$ via their infinite series:
$$
\sin x = \sum_n (-1)^{n}\frac{x^{2n+1}}{(2n+1)!}, \qquad
\cos x = \sum_n (-1)^n \frac{x^{2n}}{(2n)!}.
$$
Is there a short, clever proof that $\cos(x+y) = \cos x \cos y - \sin x \sin y$ for all real $x,y$? I... | $$\begin{array} {rcl}
\cos(x + y) + i \sin(x + y)& = & e^{i(x + y)} \\
&=& e^{ix}e^{iy} \\
&=& (\cos(x) + i\sin(x))(\cos(y) + i\sin(y)) \\
&=& (\cos(x)\cos(y) - \sin(x)\sin(y)) + i(\sin(x)\cos(y) + \sin(y)\cos(x)) \\
\end{array}$$
Equating real and imaginary parts you get
$$\cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/349435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 7
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How do I prove that $2\cos (2\theta + {\pi \over 3}) \equiv - 2\sin(2\theta - {\pi \over 6})$ Using the identity $\cos (\theta + {\pi \over 2}) \equiv - \sin\theta $
| Using the identity $\quad\cos\left(\theta + \frac{\pi}{2}\right)\equiv−\sin\theta$,
Let $\;\color{green}{\bf x = 2\theta - \dfrac{\pi}{6}} \implies\left(\implies \color{blue}{\bf x + \dfrac{\pi}{2}}\; = \;\left(2\theta - \dfrac{\pi}{6}\right) + \dfrac{\pi}{2} \;= \;\color{blue}{\bf 2\theta + \dfrac{\pi}{3}}\right)$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/349651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the remainder of $(2012^{2013}+2013^{2012}):(2012\times 2013)$? What is the remainder if the sum $$(2012^{2013}+2013^{2012})$$
is divided by $$2012\times 2013$$
| We don't need Euler Totient Theorem
As $a+1\equiv1 \pmod a,$
$$N=a^{a+1} + (a+1)^a\equiv (a+1)^a\pmod a\equiv1^a \equiv a$$
$\implies N=1+a\cdot A$ for some integer $A$
Similarly, as $a\equiv-1 \pmod{a+1},$
$$N=a^{a+1} +(a+1)^a\equiv a^{a+1}\pmod{a+1}\equiv(-1)^{a+1}\equiv
\begin{cases}
1 & \text{ if } a \text{ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/352873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Factoring into Linear Factors How would one factor the following expression:
$(b - a)(c^2 - a^2) - (c-a)(b^2 - a^2)$
into the set of linear factors:
$(b - a)(c - a)(c - b)$
(This is not for homework but rather exam review. I ran into this issue when required to calculate a matrix's determinant in linear factor form).
T... | Recall that
$$x^2-y^2=(x-y)(x+y)$$
then
\begin{array}((b - a)(c^2 - a^2) - (c-a)(b^2 - a^2)&=(b-a)(c-a)(c+a)-(c-a)(b-a)(b+a)\\&=(b-a)(c-a)[(c+a)-(b+a)]\\&=(b-a)(c-a)(c-b)\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/353446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \frac{x^2 + x+3}{x^2+2x+5}\ dx$ How can we evaluate $$\displaystyle\int \frac{x^2 + x+3}{x^2+2x+5} dx$$
To be honest, I'm embarrassed. I decomposed it and know what the answer should be but
I can't get the right answer.
| $$I=\int \frac{x^2 + x+3}{x^2+2x+5} dx=\int \frac{x(x+1)+3}{(x+1)^2+2^2} dx$$
Putting $x+1=2\tan\theta,dx=2\sec^2\theta d\theta,$
$$I= \frac{2\tan\theta(2\tan\theta-1)+3}{4\sec^2\theta d\theta}2\sec^2\theta d\theta$$
$$2\tan\theta(2\tan\theta-1)+3=4\tan^2\theta-2\tan\theta+3=4\sec^2\theta-2\tan\theta-1$$
$$\text{So,} ... | {
"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": 2
} |
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)... |
Here is a cute picture of the question.
| {
"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": 2
} |
Largest element of the set $\{ \sin{1}, \sin{2}, \sin{3}\}$ I have to find the largest element of the following set $\{ \sin{1}, \sin{2}, \sin{3}\}$.
I converted every element to the first quadrant so I can use the monotony of cosine, the set becomes:
$$\Big\{ \cos{\frac{\pi-2}{2}}, \cos{\frac{\pi-4}{2}}, \cos{\frac{\... | (Let's see how rusty my TeX is...)
We could also use trig identities:
$$\sin 2 = \sin(2·1) = 2 \sin 1 \cos 1 > 2 \sin 1 \cos \frac{\pi}{3} = \sin 1 ,$$
$$\sin 3 = \sin (2+1) = \sin 2 \cos 1 + \cos 2 \sin 1 < \sin 2 \cos \frac{\pi}{4} + \cos \frac{\pi}{2} \sin 1 = \frac{\sqrt{2}}{2} \cdot \sin 2 < \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/357692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solving the improper integral $\int_0^{\infty}\frac{dx}{1+x^3}$ $$\int_0^{\infty} \frac{dx}{1+x^3}$$
So far I have found the indefinite integral, which is:
$$-\frac{1}{6} \ln |x^2-x+1|+\frac{1}{\sqrt{3}} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|$$
Now what do I need to do in order to calculate the i... | As this one has been solved using a keyhole contour I thought I would show that it can also be done using a slice contour. Let $$f(z) = \frac{1}{1+z^3}.$$
The slice consists of three parts parameterized by $R$, which is real and goes to infinity. The first part $\Gamma_1$ is a line going from zero to $R$ along the real... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/358262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
$x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$, what is $3x^3 - 9x$?
Suppose $x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$. Then what is $3x^3 - 9x$?
I tried factorizing $3x^3 - 9x = 3x(x^2 - 3)$ then substituting the values which gives me something very lengthy. I eventually got to the answer — $10$ — after working a bit. A... | Another way:
Since $x= 3^\frac{1}{3} + 3^\frac{-1}{3}$,
$$ \Rightarrow x - 3^\frac{-1}{3} = 3^\frac{1}{3}$$
Take cubes on both sides
$$ x^3 - \frac{1}{3} -3x.3^\frac{-1}{3}(x-3^\frac{-1}{3}) = 3$$
$$ x^3 - \frac{1}{3} -3x.3^\frac{-1}{3}3^\frac{1}{3} = 3$$
$$x^3 - \frac{1}{3} -3x = 3$$
Multiply by 3 to get your answer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/360087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding the limit of $\frac{1-\cos(2x)}{1-\cos(3x)}$ for $x \to 0$ As $x$ goes to $0$, what is the limit of
$$\frac{1-\cos(2x)}{1-\cos(3x)}$$
Thanks.
| One way of doing it is expanding in Maclaurin series:
$$
\lim_{x \to 0} \frac{1-\cos 2x}{1- \cos 3x} = \lim_{x \to 0} \frac{1-1 +\frac{4x^2}{2} - O(x^4)}{1- 1 + \frac{9x^2}{2} - O(x^4)} = \frac{\frac{4}{2}}{\frac{9}{2}}=\frac{4}{9}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/360369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
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} \... | Hint: Use the fact that $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$
| {
"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": 1
} |
Very curious sequence of integrals $I_n=\int_0^1 \frac {(x(1-x))^{4n}} {1+x^2}\mathrm dx$ I was studying the behaviour of very curious sequence of integrals
$$I_n=\int_0^1 \frac {(x(1-x))^{4n}} {1+x^2} \,\mathrm dx$$ which gives a very beautiful result for $n=1$; I tried to calculate for different values of $n$ but ev... | Well, I guess I can't compete with this research cited by Norbert, but maybe this will be of any help.
One can try to expand $(1-x)^n = \sum_{k=0}^n {n \choose k}(-x)^k$, so
$$\int_0^1 \frac {(x(1-x))^{4n}} {1+x^2} \,\mathrm dx= \sum_{k=0}^n {n \choose k}(-1)^k\int_0^1 \frac {x^{4n+k}} {1+x^2} \,\mathrm dx$$
Then one ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/363226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Find the shortest distance from the point $P(0,1)$ to a point on the curve $x² - y² = 1$ , and find the point on the curve closest to $P$.
Find the shortest distance from the point $P(0,1)$ to a point on the curve $x² - y² = 1$, and find the point on the curve closest to $P$.
What I did so far is :
plot the y var fro... | Any point$(Q)$ on the hyperbola $x^2-y^2=1$ can be represented as $(\sec t,\tan t)$
The distance of $Q(\sec t,\tan t)$ from $P(0,1)$ is $$\sqrt{(0-\sec t)^2+(1-\tan t)^2}=\sqrt{2-2\tan t+2\tan^2t}=\sqrt{\frac{(2\tan t-1)^2+3}2}\ge \sqrt{\frac32} $$
We can also apply Second derivative test on $\sqrt{2-2\tan t+2\tan^2t}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/364241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
infinite sum limit how to find the following Hi what is the limit of the following sum:
$$\lim \limits_{n\rightarrow\infty}\frac{2}{n^2}\sum\limits_{j=0}^{n-1}\sum\limits_{k=j+1}^{n-1}\frac{k}{n}$$
Thanks a lot!
| Assuming the problem is correctly written, we have
$$\begin{align}
\lim_{n\rightarrow\infty}
\frac{2}{n^2}\sum_{j=0}^{n-1}\sum_{k=j+1}^{n-1}\frac{j}{n}
&=
\lim_{n\rightarrow\infty}
\frac{2}{n^3}\sum_{j=0}^{n-1}j\sum_{k=j+1}^{n-1}1 \\
&=
\lim_{n\rightarrow\infty}
\frac{2}{n^3}\sum_{j=0}^{n-1}j(n-j-1) \\
&=
\lim_{n\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/366201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Constrained optimization max $ f(x,y) = x+y$ subject to $x^2+y^2 \leq 4, x \geq0, y \geq0$ max $ f(x,y) = x+y$ subject to $x^2+y^2 \leq 4, x \geq0, y \geq0$
I need to solve this by the Kuhn Tucker conditions without using concavity of the Lagrangian.
| The constraint $x^2 + y^2 \leq 4$ may be replaced by $x^2 + y^2 = 4$ since if $f$ were maximized when $x^2 + y^2 < 4$ we could simply increase $x$ or $y$ so that $x^2 + y^2 = 4$, contradicting the fact that $x+y$ was maximized.
As such we may write $y = \sqrt{4-x^2}$. Now we need to maximize $f = x + (4-x^2)^{1/2}$. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/366885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Confusion over a limit. Different ways of solving give different answers? Qn: If it is given that
$$
\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 $$
Since the limit is fini... | In method 2 you are losing too much information. In particular, you lose the linear and constant term in the numerator of the fraction.
To be more particular,
you need to write
$\frac{x^2 - x - 2}{x + 1} - ax - b
=\frac{x^2 - x - 2 - (x+1)(ax+b)}{x + 1}
=\frac{x^2 - x - 2 - (ax^2+x(a+b)+b)}{x + 1}
=\frac{x^2(1-a) - x(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/367080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integer solutions of $x^3+y^3=z^2$ Is there any integer solution other than $(x,y,z)=(1,2,3)$ for $x^3+y^3=z^2$?
| If we take $$x=n(n^2-3)$$ $$y=3n^2-1$$ for $n\in\mathbb{N},$ $$x^2+y^2=(n^6-6n^4+9n^2)+(9n^4-6n^2+1)=n^6+3n^4+3n^2+1=(n^2+1)^3$$ then the above equation has infinitely many integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/369846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 12,
"answer_id": 6
} |
What's the Maclaurin series for $\arcsin(x)$? I solved the problem by using a known series: $\frac{1}{\sqrt{1-x^2}}$, but the solution I got is wrong. Also, I'm not sure what to do with the constant of integration $C$. Where is my mistake?
$$ \frac{1}{\sqrt{1-x^2}} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16}... | You're integrating the right hand side incorrectly. The integral of $\dfrac{x^2}{2}$ isn't $\dfrac{2x^3}{3}$, it's $\dfrac{x^3}{2\cdot 3} = \dfrac{x^3}{6}.$ :)))
| {
"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": 1
} |
Binary Division IF I convert the dividend and divisor into decimal, perform the division and convert the remainder and quotient back in to binary will I get correct answer?
I'm doing this:
*
*$630 ÷ 13$
*Quotient=$48= (110000)_2$
*Remainder=$6= (110)_2$
But if I use this online calculator here, I get
*
*Quoti... | The answer is that "it depends".
If your binary numbers are integers, then the first result is correct:
$$
630=48\cdot13+6,
$$
so the integer part of the quotient is, indeed, 48 and the remainder is six.
The link that you gave specifically states that it is doing polynomial division.
What that means is that it interpre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/371747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding $\lim_{x \to +\infty} \left(1+\frac{\cos x}{2\sqrt{x}}\right)$ Let $f(x)=x+\sin(\sqrt{x})$. I want to find $\lim_{x \to +\infty} f'(x)$.
Attempt 1
We have $$f'(x)=1+\frac{\cos x}{2\sqrt{x}} \leq 1 + \left|\frac{\cos x}{2\sqrt{x}}\right| \leq 1 + \frac{1}{2\sqrt{x}}.$$ As $x \rightarrow \infty$, $\sqrt{x} \right... | The first is almost correct, I guess you intend to do the right thing.
You should rather show
$$|f'(x)-1|\le \frac1{2\sqrt x}.$$
(What you write does not exclude that $f'(x)$ might become awfully negative).
In your second attempt you use only one special sequence instead of all sequences (i.e. you should use an arbitr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/372681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why is every answer of $5^k - 2^k$ divisible by 3? We have the formula $$5^k - 2^k$$
I have noticed that every answer you get from this formula is divisible by 3. At least, I think so. Why is this? Does it have to do with $5-2=3$?
| Let's use induction to prove that:
\begin{equation}
5^k - 2^k
\end{equation}
is divisible by 3
Let's check the proposition for k = 0 is divisible by 3
\begin{equation}
5^0 - 2^0 = 1 - 1 = 0
\end{equation}
which is divisible by 3
Let's assume that for k = n the proposition
\begin{equation}
5^n - 2^n
\end{equation}
is di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/372976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 7,
"answer_id": 4
} |
Laurent expansion of $\frac{z}{z^2+1}$ valid for $|z-3|>2$ how to find the Laurent expansion of $\frac{z}{z^2+1}$ valid for $|z-3|>2$.
$$\frac{z}{z^2+1} = \frac{1}{2}\left( \frac{1}{z-3 + (3 + i)} + \frac{1}{z-3 + (3 - i)}\right)$$
Given $|z-3| < \sqrt{10}$ or $|z-3| > \sqrt{10}$, I think we have two different expansio... | A related problem. First convert to the partial fraction form
$$ \frac{z}{z^2+1}=\frac{1}{2}\frac{1}{z-i}+\frac{1}{2}\frac{1}{z+i}. $$
Then, we consider the first term, since the second is the same
$$ \frac{1}{2}\frac{1}{z-i}=\frac{1}{2}\frac{1}{(z-3)+(3-i)}= \frac{1}{2}\frac{1}{z-3}\frac{1}{1+\frac{3-i}{z-3}}$$
$$ = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/376962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
question in number theory Let $p$ is an odd prime and $n$ is an even natural number. It is clear that $2$ divides $p^n+1$. I would like to know Is the following claim true?
$4$ does not divides $p^n+1$.
| If $p=3, 3^n+1=(4-1)^n+1\equiv1+(-1)^n\pmod 4\implies 4|(3^n+1)$ if $n$ is odd
Other primes can be written as $(6a\pm1)$ where $a$ is any integer
Now, $(6a+1)^n+1\equiv 1+6nk+1\pmod 4\equiv2(nk+1)\pmod 4$
$\implies 4|\{(6a+1)^n+1\}$ if $n\cdot k$ is odd
Again, $(6a-1)^n+1\equiv(-1)^n-6nk+1\pmod 4$
If $n$ is odd $=2m+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A problem on range of a trigonometric function: what is the range of $\frac{\sqrt{3}\sin x}{2+\cos x}$? What is the range of the function
$$\frac{\sqrt{3}\sin x}{2+\cos x}$$
| Note that $y=f(x)=\dfrac{\sqrt 3 \sin x}{2+\cos x}$.
Note that $y^2=\dfrac{3 \sin ^2 x}{(2+\cos ^2 x)^2}=\dfrac{3 (1- \cos ^2 x)}{(2+\cos ^2 x)^2}$
Put $\cos x =u$, we have
\begin{align}
y^2(2+u)^2 =3(1-u^2) \\ \implies (3+y^2)u^2+4y^2u+4y^2-3=0
\end{align}
For this to have a solution, we need discriminant of the qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/382136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Integral of $\int\frac{(x^4+1)\,dx}{x^3+4x}$ I followed the steps to solve this integral and want to know if I did it right and if $C=0? $
$$\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... | There are no constants $A,B,C$ such that
$$\frac{x^4+1}{x^3+4x} = \frac{A}{x}+\frac{Bx+C}{x^2+4},$$
because the integrand is a rational function in $x$ and the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator. So prior to expanding it into partial fractions, the ... | {
"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": 1
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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
| $$ LHS =\frac {1 - \cos A} {\sin A} + \frac {\sin A} {1 - \cos A} $$
$$ = \frac {2 \sin^2 \frac A2} {2\sin \frac A2 \cos \frac A2} + \frac {2\sin \frac A2 \cos \frac A2}{2 \sin^2 \frac A2}$$
$$ = \frac {\sin \frac A2} {\cos \frac A2} + \frac {\cos \frac A2} {\sin \frac A2} $$
Now just cross multiply and you get the ans... | {
"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": 2
} |
Determine the Maclaurin series expansion of $\frac{\mathrm{exp}(z)}{(z+1)}$ Determine the Maclaurin series expansion of $\frac{\mathrm{exp}(z)}{(z+1)}$.
This is the composition of the series expansion of the exponential function centered about $z = -1$. We can rectify the expansion about $\mathrm{exp}(z)$ by writing:
... | The function $e^z$ is entire and has Maclaurin series:
$$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!} + \cdots$$
The function $\frac{1}{1+z}$ has a pole at $z=-1$ and has a Maclaurin series which converges for $|z|<1$:
$$\frac{1}{1+z}=1-z+z^2-z^3+ \cdots$$
Their product has a series that converges for $|z|<1$:
$$\frac{e^z}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/386014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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An easier way to find the integral of: $\int {x\sqrt {2 + x} {\rm{ }}dx} $, where ${u^2} = 2 + x$ My attempt at the question:
$\eqalign{
& \int {x\sqrt {2 + x} {\rm{ }}dx} \cr
& {u^2} = 2 + x \cr
& 2u{{du} \over {dx}} = 1 \cr
& {{du} \over {dx}} = {1 \over {2u}} \cr
& u = \sqrt {2 + x} \cr
& x =... | Let's use your substitution, without the unnecessary manipulations.
Let $u^2=x+2$. Then $2u\,du=dx$ and $x=u^2-2$. Substitute, getting rid of all $x$ all at once. We get
$$\int (u^2-2)(u)(2u)\,du=\int (2u^4-4u^2)\,du=\frac{2u^5}{5}-\frac{4u^3}{3}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A sum with binomial coefficients Show that $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^{n+2}=\frac{2^{n}n(n+2)!}{6}.$$
| Using these three identities for $n^{\text{th}}$ differences:
$$
\begin{align}
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^n&=n!\\
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^{n+1}&=n!\binom{n+1}{2}\\
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^{n+2}&=n!\left(3\binom{n+2}{4}+\binom{n+2}{3}\right)
\end{align}
$$
and the fact that the $n^{\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/386899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Inequality concerning the pairwise correlation coefficients of three random variables I was asked to prove:
The correlation coefficients, $\rho_{12}$, $\rho_{23}$, $\rho_{13}$ between three random variables $X_1$, $X_2$, $X_3$ obey
$$(1+\rho_{12})(1+\rho_{13})(1+\rho_{23})\ge\frac{1}{2}(1+\rho_{12}+\rho_{23}+\rho_{13})... | For ease of notation, I'll let $a = \rho_{12}$, $b = \rho_{13}$ and $c = \rho_{23}$. The matrix
$$
\begin{pmatrix}
1 & a & b \\
a & 1 & c \\
b & c & 1
\end{pmatrix}
$$
is positive semi-definite, so its determinant is non-negative:
$$
1 + 2abc \ge a^2 + b^2 + c^2.
$$
Now consider $(1 + a + b + c)^2$:
\begin{align*}
(1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving $\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$ without induction I was looking at: $$\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$$
It's pretty easy proving the above using induction, but I was wondering what is the actual way of getting this equation?
| $\displaystyle\sum_{k=0}^{n} k^{3} + (n+1)^{3} = 0^{3} + \sum_{k=1}^{n+1}k^{3} = \sum_{k=0}^{n} (k+1)^{3} = \sum_{k=0}^{n} (k^{3} +3k^{2}+3k+1)$
$\displaystyle = \sum_{k=0}^{n} k^{3} + \sum_{k=0}^{n} k^{2} + 3 \frac{n(n+1)}{2} + (n+1)$
$ \displaystyle \implies 3 \sum_{k=0}^{n} k^{2} = 3 \sum_{k=1}^{n} k^{2} = (n+1)^{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
A binomial identity from Mathematical Reflections Here is the problem:
Let $m,n$ be positive integers with $n>m$. Prove that
$\displaystyle\sum_{k=0}^{\lfloor\frac{n+m}2\rfloor} (-1)^{k}\binom{n}{k}\binom{m+n-2k}{n-1}=\binom{n}{m+1}$
This problem is O243 of Mathematical Reflections. A solution had been published using ... | First of all, it is worth stating explicitly that the problem assumes that $\binom{n}{k}$ is zero when $n < 0$ or $n>k$, even for $k\geqslant 0$. Indeed, otherwise
In[61]:= Table[
Sum[(-1)^k Binomial[n, k] Binomial[n + m - 2 k, n - 1], {k, 0,
n}], {n, 1, 5}, {m, 0, n - 1}]
Out[61]= {{0}, {0, 0}, {0, 0, 0}, {0, 0,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/390321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Simple algebra simplification question? Hello everyone I have the following question.
I have the following fraction
$f(x)=-\frac{4}{x^2}+\frac{1}{(x-1)^2}$
But how would I reduce it I know I have to use the multiply the opposite numerator by denominator and I got.
$(x-1)^2(4)=4x^2-8x+4$
$1(x^2)=x^2$
So I got got $f(x)=... | You are close.
Your mistake was applying the "-"
to both terms instead of just one.
You wrote
$f(x)=-\frac{4x^2-8x+4+x^2}{x(x-1)^2}$.
What you should have is
$f(x)=\frac{-(4x^2-8x+4)+x^2}{x(x-1)^2}
=\frac{-4x^2+8x-4+x^2}{x(x-1)^2}
=\frac{-3x^2+8x-4}{x(x-1)^2}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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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:
$$\int\frac{\cos x}{\sin x}x\ \mathrm dx=\ x\log(2\sin x)+\frac{1}{2} \I... | Here's one way to go.
First, note that
$$\begin{eqnarray*}
\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx
&=& \int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx
+\int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx.
\end{eqnarray*}$$
For now I'll simply claim that
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/390957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 1
} |
Test of convergence of $\int_{-\infty}^{\infty} \dfrac{x^6+6}{x^8+8}dx$ I am having some trouble with this problem and don't know if I am doing it right:
$$\int_{-\infty}^{\infty} \dfrac{x^6+6}{x^8+8}dx$$
so the steps I have taken so far are, I split it into
$$\int_0^\infty \dfrac{x^6+6}{x^8+8} + \int_{-\infty}^0 \d... | Your approach is fine, as long as you compare with $1/x^2$ when $|x|\gt1$ and then use the continuity of $\frac{x^6+6}{x^8+8}$ for $|x|\le1$. We can illustrate this with a slightly different comparison function.
First, note that
$$
\lim_{|x|\to\infty}\frac{(x^6+6)(x^2+1)}{x^8+8}=1\tag{1}
$$
Thus, $(1)$ says that there ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/393161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Recurrence Relation How do I solve:
$k(k+1)a_{k}=2(\lambda k-1)a_{k-1}+(a-\lambda^2)a_{k-2}$
where $\lambda$ and $a$ are constants, and similar other recurrence relations?
| Alright, let us substitute for $k-2$ . then the recurrence relation reads:
\begin{equation}
(k+2)(k+3) a_{k+2} - 2 \left( \lambda (k+2)-1 \right) a_{k+1} - \left( a+\lambda^2 \right) a_k = 0
\end{equation}
for $k=0,1,2\cdots$. Here for the sake of simplicity we will make some additional assumption that $a_1 = (\lambda-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/393857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Cubing a simple thing I am trying to expand $\quad (x + 2)^3 $
I am actually not to sure what to do from here, the rules are confusing. To square something is simple, you just foil it. It is easy to memorize and execute. Here though I am not sure if I need to do it like multiplication where I take one $(x + 2)$ term an... | Your idea to do this:
$$(x^2 + 4x + 4)(x + 2)$$
was correct, because $(x + 2)^2 = (x^2 + 4x + 4)$. So it follows that $(x + 2)^3 = (x^2 + 4x + 4)(x + 2)$.
Let's try just multiplying it out:
To multiply $(x^2 + 4x + 4)$ by $(x+2)$, we'll multiply every term in $(x^2 + 4x + 4)$ by every term in $(x+2)$.
That's easy to do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/395142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Limit of a recursive sequence with $u_n$ It is given that $u_{n+1} =1+\dfrac{1}{u_n}$ and $u_1 =1$. Find the limit of $u_n$ as $n\to\infty$.
The limit is $\frac{\sqrt{5}+1}{2}$ from a calculator. Is there an algebraic way to determine this? You can also determine that sequence is bounded but not monotonic.
| Writing $u_n$ as $\frac{p_n}{q_n}$ for two sequences $(p_n)$, $(q_n)$ to be determined. We have:
$$u_{n+1} = 1 + \frac{1}{u_n} \iff \frac{p_{n+1}}{q_{n+1}} = 1 + \frac{q_n}{p_n} = \frac{p_n+q_n}{p_n}$$
Normalize $(p_n)$ and $(q_n)$ appropriately, we can turn the non-linear recurrence equation into a linear one:
$$\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/395468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Composition of systems of equations Suppose $$2x + 3y = u$$ $$x - 4y = v$$
and further that
$$3u - 5v = c$$ $$2u + 3v = d$$
Express c and d in terms of $x$ and $y$ by matrix multiplication.
It's quite easy by direct substitution but I can;t work out how to use matrix multiplication. Any ideas? Thanks in advance!
| $$\begin{align*}
2x + 3y& = u\\
x - 4y &= v
\end{align*}\quad \underset{\substack{\text{convert to}\\ \text{matrix language}}}{\leadsto}\quad \begin{bmatrix} 2 & \hphantom{-}3\\ 1 &-4\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix}u\\ v\end{bmatrix}$$
$$\begin{align*}
3u - 5v &= c\\
2u + 3v &= d
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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1st derivative of $\frac{2x}{\sqrt{x^2 + 1}}$ Another simple question that I can't work out today, yet I would work it out two weeks ago!
I need to find the 1st derivative of $$\frac{2x}{\sqrt{x^2 + 1}}$$.
So I use the Quotient rule and I get: $$\frac{(x^2 + 1)^.5 (2) - (2x)(0.5x^2 + 0.5)^-5}{x^2+1}$$
Am I heading in t... | Without messing up with nominator and denominators:
$x=\tan { u } $ then draw right triangle, see
$\\ \frac { 1 }{ \sqrt { x^{ 2 }+1 } } =\cos { u } \\$
$\\ \frac { x }{ \sqrt { x^{ 2 }+1 } } =\sin { u } \\$
$ \frac { du }{ dx } =\frac { 1 }{ x^{ 2 }+1 } \\$
$ f=\frac { 2x }{ \sqrt { x^{ 2 }+1 } } =2\sin { u } \\$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/397649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Simplifying a Product of Summations I have, for a fixed and positive even integer $n$, the following product of summations:
$\left ( \sum_{i = n-1}^{n-1}i \right )\cdot \left ( \sum_{i = n-3}^{n-1} i \right )\cdot \left ( \sum_{i = n-5}^{n-1}i \right )\cdot ... \cdot \left (\sum_{i=5}^{n-1}i \right )\cdot \left (\sum_... | Let $n=2m$. Your expression is then
$$\begin{align*}
&(2m-1)\cdot\frac32(2n-4)\cdot\frac52(2n-6)\cdot\ldots\cdot\frac{n-1}2n\\\\
&\qquad=1\cdot(2m-1)\cdot3(2m-2)\cdot5(2m-3)\cdot\ldots\cdot(2m-1)m\\\\
&\qquad=\prod_{k=1}^m(2k-1)(2m-k)\\\\
&\qquad=(2m-1)!!\frac{(2m-1)!}{(m-1)!}\\\\
&\qquad=\frac{(2m)!}{2^mm!}\cdot\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/400480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluating a summation of inverse squares over odd indices $$ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$
I want to evaluate this sum when $n$ takes only odd values.
| This can be shown in a similar way to Euler's proof of $\zeta(2) = \frac{\pi^2}{6}$, which starts with the function $\frac{\sin(x)}{x}$
(i.e. the sinc function). Here we start with the cosine function which can be expressed as the infinite product
\begin{align}
\cos(x) &= \prod_{n=1}^\infty \left(1-\frac{4x^2}{\pi^2(2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/402451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Indeterminate Limits I have been studying independently through various online courses and I still have trouble understanding what to do with certain limits. I am hoping for some guidance on the following two problems to help me solve them (I do not need to answer as much as help understanding where I am going).
$$ \li... | I would assume that you typed in the expression for the limit correctly.
For the problem you provided, there is a short way to determine the limit. Given that the limit is:
$$\lim_{x\to 0} \frac{\sqrt{16 + 4x} - \sqrt{16 + 4x}}{x}$$
The numerator expression becomes $0$. Then, we have:
$$\lim_{x\to 0} \frac{0}{x} = \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/408852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How prove that:$\varphi(2)+\varphi(3)+\varphi(4)+\cdots+\varphi(n)\ge\frac{n(n-1)}{4}+1$ Prove that for $n\ge 3$,
$$\varphi(2)+\varphi(3)+\varphi(4)+\cdots+\varphi(n)\ge\dfrac{n(n-1)}{4}+1$$
where $\varphi$ is the Euler's totient function
I think we must use this
$$\sum_{k=1}^{n}\varphi(k)=\dfrac{1}{2}\left(1+\sum_{k=1... | This is only a partial answer, just a possibly profitable approach.
On the rhs we have an expression which contains just the half of the sum of consecutive numbers (with a small deviation), so I would try the problem comparing the double of the sum of totients with that of the natural numbers.
The conjecture is
$$ \va... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/410648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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question about partial fractions why $\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}$ can anyone tell me why
$$\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$$
I don't undestand the $\frac{C}{(x-2)^2}$ and also what is wrong according to basic math rul... | An easy way to see why your proposal doesn't work is the following: Multiply both sides of your equation with $(x-1)(x-2)^2$ and extend the formula continuously to $x=1$ and $x=2$. We now have
$$6x^2+19x+15 = A(x-2)^2+B(x-1)(x-2)+C(x-1)(x-2).$$
If we now enter $x=2$, we get $24+38+15 = A\cdot 0^2+B\cdot 1\cdot 0 + C\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/410962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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sign charts and expressed in interval form Solve by using sign charts and express the solutions in interval form.
$$\text{(a)} \frac{(x+2)(x-3)^{2}}{x^{2}+x-2} \geq 0$$
$$\text{(b)} \frac{1}{x-1} < \frac{2}{x}$$
| Here is a full solution for $(b)$
$$ \frac{1}{x-1} < \frac{2}{x} \implies \frac{2}{x} - \frac{1}{x-1}>0 \implies \frac{x-2}{x(x-1)}>0 $$
$$ \implies \left\{x-2>0\quad \cap \quad x(x-1)>0 \right\} \cup \left\{x-2<0\quad \cap \quad x(x-1)<0 \right\} $$
$$ \implies \left\{x-2>0\, \cap\,(x-1)>0 \right\}\cup \left\{x-2<0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/411430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = 0 $ has precisely two real roots Prove that given $ a < b < c $ this equation: $$\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = 0 $$
has precisely 2 real roots.
I understand there are 3 point of discontinuities, but I have no idea how to prove this. Can you give me a... | HINT:
Rearrange the equation to get $3x^2-2(a+b+c)x+ab+bc+ca=0$
$$\text{The discriminant }\{2(a+b+c)\}^2-4\cdot3\cdot(ab+bc+ca)$$
$$=4(a^2+b^2+c^2-ab-bc-ca)$$
$$=4\cdot\frac{\{(a-b)^2+(b-c)^2+(c-a)^2\}}2>0 \text{ for distinct real }a,b,c$$
What can we conclude from this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 1
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Finding the constant term of a degree $3$ polynomial Let $p(x)=x^3+ax^2+bx+c$, where $a,b,c$ are real constants. If $p(-3)=p(2)=0$ and $p'(-3)<0$, which of the following is a possible value of $c$?
A) $-27$
B) $-18$
C) $-6$
D) $-3$
E) $-\dfrac{1}{2}$
$\textbf{My attempt at this problem}$:
I drew a rough sketch... | $p$ has three roots, $-3$, $2$ and $x_0$, say. We have
$$ p(x) = x^3 + ax^2 + bx + c = (x+3)(x-2)(x-x_0). $$
So $c = 6x_0$. As, as you said, $x_0 < -3$, A) is the only possibility.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Limit of double sum: $\lim\limits_{n\to\infty}n^{-2}\sum\limits_{k=1}^n\sum\limits_{m=k+1}^n\left(\frac{n-2k}{n+2k}\right)^2\frac{n-2m}{n+2m}$ Who is so kind to enlighten me about the steps I need to follow?
$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\sum_{m=k+1}^n\left(\frac{n-2k}{n+2k}\right)^2\frac{n-2m}{n+2m}$$
| We have a Riemann sum:
$$\lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^n \left ( \frac{1-\frac{2k}{n}}{1+\frac{2k}{n}}\right)^2 \sum_{m=k+1}^n \frac{1-\frac{2m}{n}}{1+\frac{2m}{n}} = \int_0^1 dx \left (\frac{1-2 x}{1+2 x}\right )^2 \, \int_x^1 dy \frac{1-2 y}{1+2 y}$$
The evaluation of the above integral is straightforw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Fair and Unfair coin Probability I am stuck on this question.
A coin with $P(H) = \frac{1}{2}$ is flipped $4$ times and then a coin with $P(H) = \frac{2}{3}$ is tossed twice. What is the probability that a total of $5$ heads occurs?
I keep getting $\frac{1}{6}$ but the answer is $\frac{5}{36}$.
Attempt: $P($all hea... | The required probability will be
$P($exactly $4 $ heads from the $4$ flips of $1$st coin$)\cdot P($exactly $1 $ head from the $2$ flips of $2$nd coin $)+$
$P($exactly $3 $ heads from the $4$ flips of $1$st coin$)\cdot P($exactly $2 $ heads from the $2$ flips of $2$nd coin $)$
Using Binomial Distribution, the required ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/415179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to solve this integral $\int^{\infty }_{0} {\frac{x \log x}{(1+x^2)^2}} \, dx$? I will be grateful if you would write me a solution procedure for this integral
$$\int^{\infty }_{0} {\frac{x \log x}{(1+x^2)^2}} \, dx. $$
I am sure that an antiderivative is
$$\frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) ... | Putting $x=\tan\theta$ we get,
$$I=\int_0^{\frac\pi2}\frac{\tan\theta\ln \tan \theta}{\sec^2\theta}d\theta$$
$$=\frac12\int_0^{\frac\pi2}\sin2\theta\ln \tan \theta d\theta$$
Now as $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$
$$I=\frac12\int_0^{\frac\pi2}\sin2\left(\frac\pi2+0-\theta\right)\theta\ln \tan \left(\frac\pi2+0-\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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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 le... | Start from $\sqrt{n}-\sqrt{n-1}$. Multiply top and "bottom" by $\sqrt{n}+\sqrt{n-1}$.
We get $\dfrac{1}{\sqrt{n}+\sqrt{n-1}}$, which is $\gt \dfrac{1}{2\sqrt{n}}$. It follows that $\dfrac{1}{\sqrt{n}}\lt 2(\sqrt{n}-\sqrt{n-1})$.
Add up, $n=1$ to $n=100$. We get
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/420733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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"answer_id": 1
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Rationalizing expressions In my precalc book, I have the following problem:
Calculate $a+b+c$ if $a,b,c\in\mathbb{Q}$ and
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$$
I think that the RHS can stay untouched, while operating the LHS, but I can't find a way to factor $\sqrt[3]{2}-1$ as the third power... | Hint.
$(a,b,c) = \Bigl(\dfrac{1}{9},-\dfrac{2}{9},\dfrac{4}{9}\Bigr)$ $-$ one of rational solutions (ignoring permutations).
So, $a+b+c=\dfrac{1}{3}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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$3^{3n+1} < 2^{5n+6} $ for all non-negative integers $n$. Is my induction solution correct? Show using mathematical induction that $3^{3n+1} < 2^{5n+6} $ for all non-negative integers $n$. I'm not sure whether what I did at the last is valid?
Basis step:
for all non-negative integers
$$P(n) = 3^{3n+1} < 2^{5n+6} $$
... | Excellent work:
You can conclude, since you have shown
$$3^{3(k+1)+1}\;\; <\;\; 3^3 \cdot 2^{5n+6} \;\;{\color{blue}{\bf <}}\;\; 2^{5(k+1)+6}$$
or simply, $$3^{3(k+1)+1}\; {\color{blue}{\bf <}}\; 2^{5(k+1)+6}$$ as desired.
The "blue" strict inequality is all you need. You have shown, prior to your conclusion, that ... | {
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"timestamp": "2023-03-29T00:00:00",
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Understand integral from Gradshteyn and Ryzhik book "Table of integrals, series, products" I was checking useful integrals in this book. I have found one (6.298) that is what I need, but I don't understand how every step towards the final result works.
$$\int_0^{+\infty}\,\left[2\cosh(ab)-e^{-ab}\Phi\left(\frac{b-2ax^... | This example is easier than I thought. Define:
\begin{equation}
f^{(a,b)}(x):= 2 \cosh(a b)- \exp(-a b) Erf[\frac{b- 2 a x^2}{2 x}] - \exp(a b) Erf[\frac{b+ 2 a x^2}{2 x}]
\end{equation}
Then
\begin{eqnarray}
&&\int\limits_0^\infty f^{(a,b)}(x) \cdot x \exp\left(-(\mu-a^2) x^2\right) dx=\\
&&
\int\limits_0^\infty \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/425436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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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: ... | Let's check what @amWhy noted by using Maple. It is a fun.
> f:=n-> 2n^2-3:
> g:=n-> 3n+4:
> s:=n->(g@g)(n):
> simplify((f@s)(n));
162n^2+576n+509
| {
"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": 1
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Narcissistic numbers in other bases It is well known that $153$ is a narcissistic number; that is, it is equal to the sum of the cubes of its digits since $153=1^3+5^3+3^3$.
Other bases have similar numbers. For example, in base $3$, seventeen is $122$; and in base $4$, thirty-five is $203$.
Let $B_3$ be the set of ba... | Proof by magic:
$$
\begin{eqnarray}
(k+1)^3 & + & 0^3 & + & (2k+1)^3 & = & (3k+1)^2 (k+1) & + & (3k+1)0 & + & (2k+1) \\
k^3 & + & 0^3 & + & (2k+1)^3 & = & (3k+2)^2k & + & (3k+2)0 & + & (2k+1) \\
(5k+1)^3 & + & (4k+2)^3 & + & (6k+2)^3 & = & (9k+3)^2(5k+1) & + & (9k+3)(4k+2) & + & (6k+2) \\
(7k+5)^3 & + & (2k+1)^3 & + & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/425901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
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Operator Norm of a Matrix composed of Standard Basis and Fourier Basis Let $\mathbf{A}_n$ be an $n\times 2n$ matrix (where $n=2^k$) composed of Fourier basis and standard basis; that is,
$$\mathbf{A}_n = \begin{bmatrix}\mathbf{I}_n & \mathbf{F}_n\end{bmatrix}$$
where $\mathbf{I}_n$ is the identity matrix and $\mathbf{... | The singular values of $(I,F)$ are the square roots of the eigenvalues of $(I,F)(I,F)^\ast=I+FF^\ast$. Since the DFT matrix $F$ is unitary, $I+FF^\ast=2I$. Consequently, all singular values and in turn the operator $2$-norm of $(I,F)$ are equal to $\sqrt{2}$, regardless of the size $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/427011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Roots of $8x^3-4x^2-4x+1$ It is known that the roots of polynomial $8x^3-4x^2-4x+1$ are $\cos\frac{\pi}{7}$, $\cos\frac{3\pi}{7}$ and $\cos\frac{5\pi}{7}$.
However this is what Wolfram Alpha/Wolfram Mathematica gives:
$$x = \frac{1}{6}+\frac{7^{2/3}}{3 2^{2/3} \sqrt[3]{-1+3 i
\sqrt{3}}}+\frac{1}{6} \sqrt[3]{\frac{7}... | One can show for example, that by putting $a= 2 \cos(\pi/7)$, and $b=a^2-1$, that if $ab=a+b$ and $b^2=1+a+b=1+ab$, then you are indeed dealing with the chords of the heptagon, and that these values are correct.
It is relatively easy to show by using coins on a table, that these two relations correspond to the soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/429883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Prove two similiar integral equalities I am looking to show that
$$ \begin{align*}
I_1 & = \int_{ma}^{na} \frac{ \log(x - a) }{x^2 + a^2 }
= \frac{\log (2a^2)}{2a} \bigl[ \arctan n + \arctan m\bigr] \\
I_2 & = \int_{a/m}^{a/n} \frac{\log(x + a) }{ x^2 + a^2 }
= \frac{\log (2a^2)}{2a}\bigl[ \arctan \frac{... | Hint:
Integrate by parts:
$$\dfrac{1}{a}\int{\ln{(x-a)}\arctan{\left(\dfrac{x}{a} \right)}\ dx}$$
and
$$\dfrac{1}{a}\int{\ln{(x+a)}\arctan{\left(\dfrac{x}{a} \right)}\ dx}.$$
Addition
For example, in the second integral we put
$u=\arctan{\left(\dfrac{x}{a} \right)}, \;\;dv=\ln{(x+a)}\ dx.$ Then
$$\dfrac{1}{a}\int{\ln{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/430276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
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 \f... | you should substitute $dx$ by $\sqrt{2}\cos t dt$
| {
"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": 1
} |
Continued fraction for $\tan(nx)$ I found this beautiful continued fraction expansion of $\tan(nx)$, $n$ being a positive integer, online but I don't remember the source now:
$\displaystyle \tan(nx) = \cfrac{n\tan x}{1 -\cfrac{(n^{2} - 1^{2})\tan^{2}x}{3 -\cfrac{(n^{2} - 2^{2})\tan^{2}x}{5 -\cfrac{(n^{2} - 3^{2})\tan^{... | This continued fraction can be obtained from the Gauss' continued fraction for the ratio of two hypergeometric fractions after some preparation. One can express
\begin{equation}
\tan nx = \frac{\sin nx}{\sin x}\frac{\cos x}{\cos nx}\tan x
\end{equation}
Using the Chebyshev polynomials of the first and of the second k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.