Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculate the maximum possible area of the quadrilateral (functions and derivatives) The total area $A(x)$ in $m^2$ of a quadrilateral is given by:
$$A(x)= 2 \sqrt{ x^2 -16} +\frac 14\sqrt{68x^2-x^4-256},\;\; (4 < x < 8)$$
How do I calculate the maximum possible area of the quadrilateral.
As i have understood I need... | It looks like your derivative you posted in the comment to the other hint is correct.
Hint: Find the common denominator and add the fractions. The stationary points will occur when the derivative is equal to zero, and the derivative will equal zero only when the numerator is equal to zero. What values of $x$ that are w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/682186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rationale for a convention: Why use the semiperimeter in Heron's formula? Heron's formula says that the area of a triangle whose sides have lengths $a, b, c$ is $\sqrt{s(s-a)(s-b)(s-c)}$ where $s=(a+b+c)/2$ is the semiperimeter. It can also be stated by saying that the area is $\frac14\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c... | The reason I think is simplicity:
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
Is much more simple than:
$$A=\frac{1}{4}\sqrt{(a+b+c)(b+c−a)(c+a−b)(a+b−c)}$$
I see in the comments that you say we have to add $\text{where } 2s=a+b+c$, in the standard form, and so the length of the two forms become the same. However, this is a useless ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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An equation, where the solution does not exist, but on solving the equation we got a solution. why this is happening? The solution of the equation
$\sqrt{(x+1)} -\sqrt{(x-1)}= \sqrt{(4x-1)}$
is $\frac{5}{4}$,but when we put $x=\frac{5}{4}$ in the given equation, then it does not satisfy the equation.
Actually, if we ta... | Whenever we square, we immediately introduce extraneous root
Observe that
$\displaystyle\frac54$ is actually a root of $$\sqrt{x+1}=\sqrt{4x-1}-\sqrt{x-1}$$
Also, observe that $\displaystyle2x-1=-2\sqrt{x^2-1}\le0\implies 2x\le1\iff x\le\frac12$ for real $x$
But, $\displaystyle{\sqrt{x-1}}$ is not real unless $x\ge1... | {
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"url": "https://math.stackexchange.com/questions/685687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Let $p$ be prime and $\left(\frac{-3}p\right)=1$. Prove that $p$ is of the form $p=a^2+3b^2$
Let $p$ be prime and $\left(\frac{-3}p\right)=1$, where $\left(\frac{-3}p\right)$ is Legendre symbol. Prove that $p$ is of the form $p=a^2+3b^2$.
My progress:
$\left(\frac{-3}p\right)=1 \Rightarrow$ $\left(\frac{-3}p\right)=\... | Let $\omega\in\Bbb C$ satisfy $\omega^2+\omega+1$. Then $\Bbb Z[\omega]$ is a UFD, since $N(a+b\omega)=a^2-ab+b^2$ defines an Euclidean norm.
Since $\left(\frac{-3}{p}\right)=1$, there is an $s$ such that $p\mid s^2+3$. Working over $\Bbb Z[\omega]$ we have $$p\mid (s+1+2\omega)(s+1+2\omega^2)$$ If $p$ is prime we woul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Approximating $\ln{\frac{x-y}{x+y}}$ For $x>>y$,
$$\ln{\frac{x-y}{x+y}} = \ln{\left[ x\left( 1-y/x \right) \right]} -\ln{\left[ x\left( 1+y/x \right) \right]} \approx -2\frac{y}{x}$$
However, the following does not work:
$$\ln{\frac{x-y}{x+y}} \approx \ln{\frac{x-y}{x}} = \ln{\left( 1-\frac{y}{x} \right)} \approx -\fr... | The correct way to expand $\dfrac{x-y}{x+y}$ is using the binomial theorem.
$$ \frac{1}{x+y} = \frac1x\left(1+\frac{y}{x}\right)^{-1} = \frac1x\left(1-\frac{y}{x} + \ldots\right)$$
Hence
$$ \frac{x-y}{x+y} = \frac{x-y}{x}\left(1-\frac{y}{x} + \ldots\right) \approx 1 -2\frac{y}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/687037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maclaurin series of $\frac{1}{1+\sin x}$ Find the terms through degree four of the Maclaurin series of $f(x)$.
$$f(x) = \frac{1}{1+\sin x}$$
My work:
The Maclaurin series for $\sin x$ up to degree $4$ is $x - \frac{x^3}{6} + \frac{x^5}{120}$
The Maclaurin series for $\frac{1}{1+x}$ up to degree $4$ is $1 - x + x^2 - x^... | You did it right. The answer on WolframAlpha is the same. Try typing "expand" in front on WA.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/692406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I prove the trigonometric Problem? How can I show the following trigonometric problem :
$$\frac{1}{3}\leq \frac{\sec^2\theta-\tan^2\theta}{\sec^2\theta+\tan^2
\theta}\leq 3$$
I have tried in the following way :
$$
\begin{align}
& \phantom{\Rightarrow}
\frac{\sec^2\theta-\tan^2\theta}{\sec^2\theta+\tan^... | You have
$$
\frac 1 3 \le \frac{1-u}{1+u} \le 3
$$
where $u = \sin^2\theta$.
You cannot multiply through by $1+u$ because $1+u$ may be either positive or negative, depending on the value of $u$.
If $\dfrac{1-u}{1+u}\le 3$ then $\dfrac{1-u}{1+u}-3\le0$, so $\dfrac{1-u}{1+u}-\dfrac{3(1+u)}{1+u}\le 0$, and that becomes $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/694637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find this ODE solution $(y-x)\sqrt{x^2+1}\dfrac{dy}{dx}=(1+y^2)^{\frac{3}{2}}$ Find all solutions $y$ to the ODE
$$(y-x)\sqrt{x^2+1}\dfrac{dy}{dx}=(1+y^2)^{\frac{3}{2}}$$
My try:
$$(y-x)\dfrac{d(\sqrt{1+y^2})}{y}=\dfrac{(1+y^2)dx}{\sqrt{x^2+1}}$$
then
$$\dfrac{(y-x)\cdot d(\sqrt{1+y^2})}{y(1+y^2)}=\dfrac{d(\sqrt{1+x^2}... | Another CAS gave this "nice" expression
$$x(y)=\frac{-2 c_1 y+2 c_1 \tan ^{-1}(y)-c_1^2-\tan ^{-1}(y)^2+2 y \tan ^{-1}(y)+1}{c_1^2
y-2 c_1 y \tan ^{-1}(y)-2 c_1-y+y \tan ^{-1}(y)^2+2 \tan ^{-1}(y)}$$ which simplifies if $y(0)=0$ but does not change to be more pleasant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Solving for $\sum_{n = 1}^{\infty} \frac{n^3}{8^n}$? I was trying to solve $ \displaystyle \sum_{n = 1}^{\infty} \frac{n^3}{8^n}$ and I found a way to solve it and I want if there are generalizations for, say, $\displaystyle \sum_{n=1}^{\infty} \frac{n^k}{a^n}$ in terms of $k$ and $a$. I would also like to know if ther... | There turns out to be a standard trick that applies here: let
$$ f(x) = \sum_{i=1}^{\infty} \frac{x^n}{8^n} $$
We can compute this sum because it is a geometric series. The neat idea, now, is that
$$ f'(x) = \sum_{i=1}^{\infty} \frac{n x^{n-1}}{8^n} $$
or alternatively,
$$ x f'(x) = \sum_{i=1}^{\infty} \frac{n x^n}{8^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/698598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Triangle Inequality question with fractions Given $a, b \in \mathbb{R}$, prove that
$$\frac{|a + b|}{1 + |a + b|} \le \frac{|a|}{1 + |a|} + \frac{|b|}{1+|b|}$$
When does equality hold?
The only useful thing I could get (using the triangle inequality) is:
$$\begin{align}1 + \left|\left(a + \frac{ab}{2}\right) + \left(b... | Because the function $f(x)=\frac{1}{1+\frac{1}{x}}=\frac{x}{1+x}$ is increasing, we have
$$\frac{|a+b|}{1+|a+b|} = \frac{1}{1+\frac{1}{|a+b|}} \le \frac{1}{1+\frac{1}{|a|+|b|}}=\frac{|a|}{1+|a|+|b|}+\frac{|b|}{1+|a|+|b|}\le\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$$
In the first inequality we have used the triangle inequalit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/698869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why does the following nonlinear system have 21 solutions? I am curious as to why the following nonlinear system has 21 solutions (according to Wolfram Alpha).
$$y+xy^2-x^3+2xz^4=0 \\ -x-y^3-3x^2y+3yz^4=0 \\ -\frac{5}{2}y^2z^3-2x^2z^3-\frac{z^7}{2}=0$$
I see that there is one real and twenty complex solutions:
$(0,0,0... | When we do the algebra to exactly solve this system, it has $21$ distinct solutions $(x,y,z)$ (disregarding multiplicity). As previously commented, it helps to separate the cases $z=0$ and $z \neq 0$.
Case: $z = 0$
When $z = 0$, the third equation holds automatically. Substituting $z = 0$ in the first two equations... | {
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"url": "https://math.stackexchange.com/questions/704281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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a quadratic equation for two unknown number find values of $p$ such that the equation $4x^2 + 3px - 2p = 0$ has? below are a few choices of the value p:
a) 2 real roots
b) 1 real roots
c) no roots or complex roots
so far i did for a) 2 real roots
$$(3p)^2 - 4(4)(-2p) > 0\\
9p^2 +32p > 0\\
9\left( p^2+\frac{32p}{9} \ri... | The roots are given by the quadratic formula:
$$
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
= \frac{-3C \pm \sqrt{9C^2 + 32p}}{2a}
$$
If the thing under the square root is positive, you'll get two roots. Under what conditions will you get just one? Under what condition will you get none?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/706694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How does $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$? Good afternoon my wonderful friends!
Whenever I do this equation I set it up using the difference of two cubes, which is as follows:
$(a+b)^3 = (a+b)(a^2-ab+b^2) = a^3 + b^3$
Whenever I try to use this formula I always get:
$(x+h) x^2 - xh + h^2$
Simplify:
$x^3 - x^2h... | I don't know where you get "$a^2-ab+b^2$" from. Just use distributivity to carry out the multiplications by hand:
\begin{align}
(a+b)^2 &= (a+b)(a+b) = a(a+b)+b(a+b) \\&= a^2+ab+ba+b^2 = a^2+2ab+b^2
\end{align}
Thus,
\begin{align}
(a+b)^3 &= (a+b)(a+b)^2 = (a+b)(a^2+2ab+b^2) = a(a^2+2ab+b^2)+b(a^2+2ab+b^2) \\&= a^3+2a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/708165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Closed Form for $\int_0^1 \frac{\log(x)}{\sqrt{1-x^2}\sqrt{x^2+2+2\sqrt{2}}}dx$ Is there a closed form for the following integral?
$$\int_0^1 \frac{\log(x)}{\sqrt{1-x^2}\sqrt{x^2+2+2\sqrt{2}}}dx$$
It is approximately equal to $-0.48878092308456029189008$.
Mathematica is unable to find a closed form for this integral. ... | We let $$I(a,b)=\int^1_0\frac{x^{2a}}{\sqrt{1-x^2}\sqrt{x^2+b}}dx$$
so that OP's integral is equal to $\frac12\frac{d}{da}I(0,2+2\sqrt2)$.
According to Mathematica, we have $$I(a,b)=\frac{\sqrt\pi}{2\sqrt b}\frac{\Gamma(a+1/2)}{\Gamma(a+1)}{}_2F_1(\frac12,a+\frac12;a+1\mid-1/b)\\
=\frac{\sqrt\pi}{2\sqrt b}\frac{\Gamma(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/709443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
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Value of this definite integral $\int_{0}^{\infty} \frac{ \ln(x)}{x^2+2x+4} dx $ So I came across this question on brilliant.org and didn't know how to go about it:
$$\int_{0}^{\infty} \frac{ \ln(x)}{x^2+2x+4} dx $$
I tried to complete the squares in the denominator and then use a trigonometric substitution (with tan) ... | This CAN be done with real variables and no, you do not need to differentiate with respect to a parameter. Here's how:
$$I=\int_{0}^{\infty} \frac{\ln(x)}{x^2+2x+4} dx $$
$$I=\int_{0}^{\infty} \frac{\ln(x)}{{(x+1)}^2+3}dx $$
$$I=\frac{1}{3} \int_{0}^{\infty} \frac{\ln(x)}{ ({ \frac{x+1}{\sqrt{3}} })^2 + 1} dx $$
Now, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/709516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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A question on eigenvalues Let $A,B\in M_{2}(\mathbb{R})$ so that $A^2 = B^2 = I$. Which are eigenvalues of $AB$?
1) $1\pm \sqrt 3$
2)$3 \pm 2\sqrt2$
3)$\dfrac {1}{2},2$
4)$2 \pm 2\sqrt 3$
| It's easy to show that possibilities (1) and (4) are impossible. Note that the magnitude of the determinants for both $A$ and $B$ must be $1$, indeed
$$1 = \det(A^2) = \det(B^2) \implies |\det(A)| = |\det(B)| = 1$$
This necessarily implies that $|\det(AB)|=1$, i.e. the product of the eigenvalues must have magnitude $1$... | {
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"url": "https://math.stackexchange.com/questions/713949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Derivatives of Logarithmic function Determine $f'(x)$ for $f(x) = ln(x + \sqrt{x^2 + 1})$
My handbook has the answer as $\displaystyle\frac{1}{\sqrt{x^2 + 1}}$ with no steps on how they got there.
I tried to get there, but somewhere I am getting things wrong. This is what I have
\begin{align}
f'(x) & = \displaystyle\fr... | Using the chain rule: set u = x + (x^2 + 1)^(1/2) ==> du/dx = 1 + x/(x^2 + 1)^(1/2) =
= (x + (x^2 + 1)^1/2)/(x^2 + 1)^1/2, and dy/du = 1/u = 1/(x + (x^2 + 1)^1/2). So:
dy/dx = dy/du*du/dx = 1/(x^2 + 1)^1/2.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Taylor Series of $ \frac{1}{1-x^2} $ about x=2 I am trying to form a taylor series of the following:
$ \frac{1}{1-x^2} $ about $x=2$
I tried factoring the equation such that it becomes the following:
$ \frac{1}{{(1+x)}{(1-x)}} $
I tried to substitute $ x = h + 2 $ into the equation and obtained the following after usi... | Hint: Since $x = (x-2) + 2$, then $x^2 = [(x-2) + 2]^2 = (x-2)^2 + 4(x-2) + 4$ so
\begin{align*}
\frac{1}{1 - x^2} &= \frac{1}{1 - [(x-2)^2 + 4(x-2) + 4]} = \frac{1}{-(x-2)^2 -4(x-2)-3}\\
&= -\frac{1}{[(x-2) + 3][(x-2) + 1]} \, .
\end{align*}
Now use partial fractions to finish it off.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/714997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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I am not sure if the answer for the dividing rational expressions problem should be simplified $$ \frac{x}{x+2} \div \frac{1}{x^2 - 4} $$
-I am not sure if the answer for the problem (which is attached) would be $\frac{x^3-4x}{x+2}$ or if it could be simplified further.
| Remember that:
$$a \div \dfrac{1}{c}=ac$$
Therefore we can convert
$$\frac{x}{x+2} \div \dfrac{1}{x^2-4}$$
into:
$$\left(\frac{x}{x+2}\right)\left(x^2-4\right)$$
We can simplify this by factoring $x^2-4$ using the difference of squares formula, which is:
$$a^2-b^2=(a+b)(a-b)$$
So:
$$\left(\frac{x}{x+2}\right)\left(x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/717238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Find the determinant without row expansion Show that the determinant of the matrix
\begin{bmatrix} 1& a& a^3\\
1& b& b^3\\
1& c& c^3\end{bmatrix}
is $(a-b)(b-c)(c-a)(a+b+c)$ without expanding.
I was able to get out $(a-b)(b-c)(c-a)$ but couldn't complete.
| Use row operations to simplify. In the process, the factorization drops right out.
\begin{align}
\det \begin{bmatrix}
1 & a & a^3 \\
1 & b & b^3 \\
1 & c & c^3
\end{bmatrix}
&= \det \begin{bmatrix}
1 & a & a^3 \\
0 & b-a & b^3-a^3 \\
0 & c-a & c^3-a^3
\end{bmatrix} \\
&= \det \begin{bmatrix}
1 & a & a^3 \\
0 & b-a & (b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/720589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How do I solve this integral with hyperbolic functions? I was studying mechanics when I f ound a problem that lead to an integral that I can't solve. Basically the problem asked to find the period of oscillation function of the energy $E$ of a particle (mass $m$) in a field with a potential defined by:
$\displaystyle U... | When pulling $\sqrt{\lvert E\rvert}$ out of the denominator, you made a sign error. Since $E < 0$, you have
$$E + \frac{U_0}{\cosh^2 (\alpha x)} = (-E)\left(- 1 + \frac{-U_0/E}{\cosh^2 (\alpha x)}\right),$$
so the period would be
$$T = \frac{2\sqrt{2m}}{\alpha\sqrt{\lvert E\rvert}} \int_0^{\alpha x_1} \frac{dx}{\sqrt{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721129",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove that $n^7+7$ can never be a perfect square. Prove that for a positive integer $n$, $n^7+7$ cannot be a perfect square.
I managed to show that $n \equiv 5 \pmod{8}$ or $n \equiv 9 \pmod{16}$. But nothing came from that so I presume another approach is needed???
Thanks for any help.
| Here is an alternative proof that is hopefully slightly simpler.
We will make use of a simple theorem whose proof is skipped here:
For any odd integer number $m$, we have $m^2\equiv 1(mod 4)$.
Assume on the contrary, $n^7+7$ is a perfect square, then there exists an integer $x$ such that $n^7+7=x^2$.
If $x$ is odd, $n$... | {
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"url": "https://math.stackexchange.com/questions/721834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Prove $2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2)$ for $x, y, z \ge 0$
Let $x,y,z\ge 0$. Show that
$$2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2).$$
my idea: let $$x+y+z=p,xy+yz+xz=q,xyz=r$$
since
$$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=p^2-2q$$
and
$$(xy+yz+xz)^2=x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z)$$
$$\Longrightarrow x^2y^2+y^2z^... | Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that $f(w^3)\geq0$, where
$$f(w^3)=2(81u^4-108u^2v^2+18v^4+12uw^3)+2w^3+7-5(9u^2-6v^2).$$
But, $f$ is a linear function, which says that it's enough to prove our inequality
for an extremal value of $w^3$.
We know that $x$, $y$ and $z$ are non-negati... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 5
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How to factor the polynomial $x^4-x^2-2x-1$? By taking advantage of a computer algebra system, I found out that
$$x^4 - x^2 - 2x - 1 = (x^2 + x + 1)(x^2 - x - 1)$$
However, I don't know a straightforward way to solve this by hand. This was in a high-schooler's homework set, so I am assuming it can be solved using eleme... | Upon looking at this expression I immediately noticed that the $-x^2-2x-1$ could be simplified to $-(x+1)^2$. That is how you should be later on. For example, when you see $4x^2-4x+1$, you should right away notice that it equals $(2x-1)^2$
$$x^4-x^2-2x-1=x^4-(x+1)^2$$
$$=\left(x^2\right)^2-(x+1)^2$$
$$=(x^2+x+1)(x^2-(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/723228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Two definite integrals These integrals are closely related since $\frac{\pi^2}{8}=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}$ and $G=\sum_{n\mathop=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2}$. I'm not able to prove them though.
Show that
$$\int_0^1 \frac{1}{\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt... | Note that $\cosh^{-1}{x}=\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)$ and $\frac{d}{dx}\cosh^{-1}{x}=\frac{1}{\sqrt{x^2-1}}$. Substituting $u=\cosh^{-1}{x}$ into the first integral, we get:
$$\int_0^1 \frac{1}{\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\,\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Showing a function $f$ has a zero I have been working on this problem and was wondering if anyone could give me some hints as to how to answer this. So far as $p$ is a polynomial it is continuous and $g$ is given as being continuous, and so by algebra of continuity $f$ is continuous, now does the solution include the u... | Suppose that $g$ is continuous and bounded by $M>0$ and $p(x)= x^n+\ldots+a_0$ where $n$ is odd. Then $f(x)=x^n+ a_{n-1}x^{n-1}\ldots+a_0+g(x)$ which may be written as $x^n\big(1+\ldots\frac{a_0}{x^n}+\frac{g(x)}{x^n}\big)$ when $x\not=0$, the basic idea is that when $|x|$ is big $x^n$ dominate and $f$ contain the sign... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/725079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help with discrete math proof? I am having trouble proving the following:
If $x\in R$ and $x > 0$, then $x^4+1 \geq x^3+x$.
Work: I tried to rearrange the equation as $x^4-x^3-x+1 \geq 1$, but that does not really help. I also tried proof by cases where case 1 would be that x is irrational and case 2 would be that x i... | Hint: try the rearrangement inequality.
details: if $x>1$:
$$
\frac 1x < 1\\
x<x^4\\
x + x^3 = 1\cdot x + \frac 1x \cdot x^4 \le \frac 1x \cdot x + 1\cdot x^4 = 1+x^4
$$
Otherwise:
$$
\frac 1x \ge 1\\
x\ge x^4\\
x + x^3 = 1\cdot x + \frac 1x \cdot x^4 \le \frac 1x \cdot x + 1\cdot x^4 = 1+x^4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/725447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove $\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}$ How to prove:
$\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}$
Please help.
| $$\sin^2\theta=1-\cos^2\theta=(1-\cos\theta)(1+\cos\theta)$$
$$\implies \frac{\sin\theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin\theta}$$
Now add the second term
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/726442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Finding $Ker(T_a)$ and $Im(T_a)$ there will be a matrix $A \in R^{2 x 2}$, $T_A:R^{2 x 2} \rightarrow R^{2 x 2}$
Define by $T_A(X)=$$
\begin{pmatrix}
1 & 2\\
0 & 3 \\
\end{pmatrix}
$$X-X$$
\begin{pmatrix}
1 & 2\\
0 & 3 \\
\end{pmatrix}
$$\forall X \in R^{2... | If we denote
$$X=\begin{pmatrix}
a & b\\
c & d \\
\end{pmatrix}$$
then
$$AX-XA=\begin{pmatrix}
2c & (2d-2b-2a)\\
2c & -2c \\
\end{pmatrix}=\begin{pmatrix}
0 & 0\\
0 & 0 \\
\end{pmatrix}\iff (c=0)\land (d=b+a)$$
hence
$$\operatorname{ker}(T_A)=\op... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/726576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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In Triangle, $\sin\frac A2\!+\!\sin\frac B2\!+\!\sin\frac C2\!-\!1\!=\!4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$ To prove
$$\sin\frac A 2+\sin\frac B 2+\sin\frac C 2-1=4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$$
My approach :
$$
\begin{align}
\text{L.H.S.} & = \sin\frac{A}{2}+\sin \frac{B}{... | Note that
$$\sin\frac{C}{2}=\cos\frac{A+B}{2}$$
and now use following formulas
$$\sin p+\sin q=2\sin\frac{p+q}2\cos\frac{p-q}2$$
$$\cos p-\cos q=-2\sin\frac{p+q}2\sin\frac{p-q}2$$
to obtain
$$\begin{align}\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}&=\sin\frac{A}{2}+\sin\frac{B}{2}+\cos\frac{A+B}{2}\\&=2\sin\frac{A+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/727852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
probability need help on correlation problem A deck of 52 cards is shuffled you are dealt 13 cards. Let $X$ and $Y$ denote, respectively, the number of aces and the number of spades in your hand. Show that $X$ and $Y$ are uncorrelated.
I try to write $X=X_1+X_2+\cdots+X_{13}$, such that $X_1\ldots X_{13}$ is the indicato... | It will suffice to show that $E(XY)=E(X)E(Y)$. The expected number of Aces is $13\cdot \frac{4}{52}$, and the expected number of spades is $13\cdot\frac{13}{52}$. Thus $E(X)E(Y)=\frac{13}{4}$.
We now compute $E(XY)$. Note that $X$ and $Y$ are not independent.
For the calculation, we condition on $X$. The possibilitie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/729894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding max/min of multivariable function The following function
$f(x,y) = 3xy + \frac{6}{1 + x^2 + y^2 }$ within $\frac{1}{3} \leq x^2 +y^2 \leq 4$
I do partial differention
$\frac{\partial z}{\partial x} = 3y - \frac{12x}{1 + x^2 + y^2}$
$\frac{\partial z}{\partial y} = 3x - \frac{12y}{1 + x^2+ y^2 }$
I try to simpl... | This function is begging for the substitution $x=\sqrt r\cos\varphi$, $y=\sqrt r\sin\varphi$, because then you want to find the minimum or the maximum of
$$f(r,\varphi)=3r\cos\varphi\sin\varphi+\frac{6}{1+r}=\frac{3r}2\sin(2\varphi)+\frac6{1+r}$$
where $\dfrac13\leq r\leq4$. Note that
$$\frac6{1+r}-\frac{3r}2\le\frac6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/733127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Harder Trigonometry Identity ($\sec A+\csc A$) How do I prove:
$\sin A (1 + \tan A) + \cos A (1 + \cot A) = \sec A + \csc A$
I've tried expanding the brackets by multiplying sin A and cos A to the left hand side but to no avail. Where should I start from?
| $\tan A=\sin A/\cos A$, and $\cot A=\cos A/\sin A$, so that making this substitution, we have \begin{align}
\sin A\left(1+\frac{\sin A}{\cos A}\right)+\cos A\left(1+\frac{\cos A}{\sin A}\right) & =\sin A + \frac{\sin^2 A}{\cos A}+\cos A +\frac{\cos^2 A}{\sin A} \\
& =\sin A+\cos A+\frac{1-\cos^2 A}{\cos A}+\frac{1-\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/733708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Prove by induction $1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{2^n} \ge 1 + \frac {n}{2}$ Prove by induction $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{2^n} \ge 1 + \frac {n}{2}$
I can't explain in words how the left hand side of the equation is achieved soI shall provide an e... | Hint: Look at your example, appropriately grouped and then with terms made systematically smaller:
$$1+{1\over2}+\left({1\over3}+{1\over4}\right)+\left({1\over5}+{1\over6}+{1\over7}+{1\over8}\right)
\ge1+{1\over2}+\left({1\over4}+{1\over4}\right)+\left({1\over8}+{1\over8}+{1\over8}+{1\over8}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/735911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Demonstrate that the sequence is decreasing $$\left( 1-\frac{1}{n+1}\right)^n$$
How do I demonstrate that the sequence decreases? I tried using the newton binomial but I end up with a terribly complicated expression.
| This is a classic exercise.
Let
$$a_n= \left( 1-\frac{1}{n+1}\right)^n=\left( \frac{n}{n+1}\right)^n$$
$$\frac{a_n}{a_{n+1}} =\left( \frac{n}{n+1}\right)^n\left( \frac{n+2}{n+1}\right)^{n+1}=\left( \frac{n}{n+1}\right)^{n+1}\left( \frac{n+2}{n+1}\right)^{n+1}\frac{n+1}{n}$$
Now write
$$\left( \frac{n}{n+1}\right)^{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/736891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Help on this inequality If a,b,c are positive numbers, prove the inequality
$$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} ≥ \frac{3}{1+abc}
$$
| We need to prove that
$$\sum_{cyc}\frac{1+abc}{a(1+b)}\geq3$$ or
$$\sum_{cyc}\frac{1+abc+a+ab}{a(1+b)}\geq6$$ or
$$\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}\geq6$$ or
$$\sum_{cyc}\frac{1+a}{a(1+b)}+\sum_{cyc}\frac{b(1+c)}{1+b}\geq6,$$
which is AM-GM:
$$\sum_{cyc}\frac{1+a}{a(1+b)}+\sum_{cyc}\frac{b(1+c)}{1+b}\geq3\sqrt[3]{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/739945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate the integral Can some one provide me hint to evaluate the following integral.
$$
\int\csc^{2}\left(x\right)
\ln\left(\cos\left(x\right) - \sqrt{\vphantom{\largeA}\,\cos\left(2x\right)\,}\,\right)
\,{\rm d}x
$$.
| MAPLE returns
$$-\cot(x)\ln(\cos(x)-\sqrt{\cos(2x)})-\cot \left( x \right) +x-\sqrt { \left( -2\,
\left( \cos \left( x \right) \right) ^{2}+1 \right) \left( \left(
\cos \left( x \right) \right) ^{2}-1 \right) } \left( -1/2\,{\frac {-
2\, \left( \cos \left( x \right) \right) ^{3}+2\, \left( \cos \left(
x \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Complex numbers system of equations problem with 5 variables Let $z_0$,$z_1$,$z_2$,$z_3$ and $z_4$ such that $z_i\in C$ that hold:
$$(1)|z_0|=|z_1|=|z_2|=|z_3|=|z_4|=1$$
$$(2)z_0+z_1+z_2+z_3+z_4=0$$
$$(3) z_0z_1+ z_1z_2+z_2z_3+z_3z_4+z_4z_0=0$$
Prove that the solutions $z_i$ of this equation lay on the corners of re... | Let us define $a,b,c,d$ as follows:
$$
a=\frac{z_0}{z_4},~b=\frac{z_1}{z_4},~c=\frac{z_2}{z_4},~d=\frac{z_3}{z_4}.
$$
The proposed equations are equivalent to the equations
$$
|a|=|b|=|c|=|d|=1\tag{1}
$$
$$
a+b+c+d=-1\tag{2}
$$
$$
a b+b c+ c d+ d+a=0\tag{3}
$$
The equations $(2)$ and $(3)$ can be written in the form
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Solving the congruence $7x \equiv 41 \mod{13}$ I have to solve the following linear congruence:
$$7x \equiv 41 \mod{13}$$
The question where I got this from comes in two parts. The first is that it asks to find the set of the inverses of $7 \mod 13$, which turns out to be $[2]_{13}$ but, I found that solution by solvi... | The OP states
... but, I found that solution by solving the linear diophantine equation
From this site I found
Here the equation is $7x\equiv 41 \pmod {13}$ or
$$\tag 1 7x\equiv 2 \pmod {13}$$
Here also, multiplication $7 \times x$ is repeated addition. So you can also get the inverse of ${[7]_{13}}^{-1}$ and/or the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/744390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Several (advanced) L'Hospital problems Problems :
$$
\begin{align}
&\text{A}.\ \lim_{x\rightarrow1}(2-x)^{\tan\left(\frac{\pi x}{2}\right)}\\
&\text{B}.\ \lim_{x\rightarrow 0}\left(\cot x-\frac{1}{x}\right)\\
&\text{C}.\ \lim_{x\rightarrow0}\frac{\sin^{-1}x-x}{\tan^{-1}x-x}\\
&\text{D}.\ \lim_{x\rightarrow 1}(2-x)^{\t... | For example:
$$\lim_{x\to 0}\cot x-\frac1x=\lim_{x\to 0}\frac{x\cos x-\sin x}{x\sin x}\stackrel{\text{l'Hospital}}=\lim_{x\to 0}\frac{\cos x-x\sin x-\cos x}{\sin x+x\cos x}=$$
$$=-\lim_{x\to 0}\frac{\sin x}{\frac{\sin x}x+\cos x}=-\frac0{1+1}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/745427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How to find maximum/minimum of $y=\frac{x(x^2-x+2)}{x^2-9}$? How can I find minimum and maximum of $y=\frac{x(x^2-x+2)}{x^2-9}$?
In other words, points, where $y'=0$.
My current steps are:
$$y=\frac{x(x^2-x+2)}{x^2-9}=x-1+\frac{11x-9}{x^2-9}$$
$$\frac{dy}{dx}=1+\frac{11(x^2-9)-2x(11x-9)}{(x^2-9)^2}$$
\begin{align}
y'=0... | You can use partial fractions here $$\frac{11x-9}{x^2-9}=\frac A{x+3}+\frac B{x-3}=\frac 7{x+3}+\frac 4{x-3}$$ so that $$y=x-1+\frac 7{x+3}+\frac 4{x-3}$$ and $$y'=1-\frac 7{(x+3)^2}-\frac 4{(x-3)^2}$$
Then you should be able to sketch the functions to see what is happening, and to see that any real zeros of the deriva... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/745672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Compute integral $\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$ Compute the integral.
$$\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$$
The answer at the back of the book is
$$\frac{\pi}{4\sin(\frac{3\pi}{8})}$$
| Note
\begin{align}
\int_{-\infty}^\infty \frac{x^4}{1+x^8} dx
& =-\frac1{2i}\int_{-\infty}^\infty \left(\frac{1}{1+i x^4} - \frac{1}{1-i x^4} \right)dx \\
& = -\text{Im} \int_{-\infty}^\infty \frac{1}{1+i x^4}dx
\overset{t=i^{1/4}x} = -\text{Im} \left( e^{-i\frac\pi8}\int_{-\infty}^\infty \frac{1}{1+t^4}dt\right)\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/747136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Solve $2\sqrt{(x-1)(x+2)}\ge|x+1|-2$ Can you please show me how can I solve this inequality. I would like to see how it can be done without the graph of the functions. Thank you!
$$2\sqrt{(x-1)(x+2)}\ge|x+1|-2$$
| If $|x+1|-2$ is negative or zero, the inequality is obvious, so assume $|x+1| >2$, so that $x>1$ or $x < (-3)$.
If $x>1$, then $|x+1|=x+1$ so that your inequality becomes $2\sqrt{(x-1)(x+2)} \geq x-1$, which is equivalent
to $4(x-1)(x+2) \geq (x-1)^2$, or $4(x+2) \geq x-1$, i.e. $3x+9 \geq 0$ which is clear.
If $x<(-3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/748381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the real parameters $a$ and $b$ in limit Can you please help me solve this problem which requires to find the values of real parameters $a$ and $b$ so that the relation below is satisfied:
$$\lim_{x\to -\infty}(\sqrt {x^2+x+1}-ax+b)=0$$
Thank you very much!
| You have
$$
\lim_{x\to -\infty}\left(\sqrt {x^2+x+1}-ax+b\right)=0.
$$
Let $x=-y$, then you have
$$
\begin{align}
\lim_{-y\to -\infty}\left(\sqrt {(-y)^2+(-y)+1}-a(-y)+b\right)&=0\\
\lim_{y\to \infty}\left(\sqrt {y^2-y+1}+ay+b\right)&=0.\tag1
\end{align}
$$
Now, divide both sides of $(1)$ by $y$. It turns out to be
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/749861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the limit of a function?? Determine the following limit:
$$\lim\limits_{x\to0}\frac{(1+x\cdot2^x)^{\frac{1}{x^2}}}{(1+x\cdot3^x)^{\frac{1}{x^2}}}$$
| $$\lim_{x\to0}\frac{(1+x\cdot2^x)^{\frac{1}{x^2}}}{(1+x\cdot3^x)^{\frac{1}{x^2}}}$$
$$=\lim_{x\to0}\left(\dfrac{1+x\cdot2^x}{1+x\cdot3^x}\right)^{\dfrac1{x^2}} =\lim_{x\to0}\left(1+\dfrac{x(2^x-3^x)}{1+x\cdot3^x}\right)^{\dfrac1{x^2}}$$
$$=\left(\lim_{x\to0}\left(1+\dfrac{x(2^x-3^x)}{1+x\cdot3^x}\right)^{\dfrac{1+x\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/750346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Why does $2^{n+1} + 2^{n+1} = 2^{n+2}$? Simple question, why does:
$2^{n+1} + 2^{n+1} = 2^{n+2}$ ?
Furthermore, why does this only work for powers of 2?
Thanks.
| $$2^{n+1} + 2^{n+1} = 2\cdot 2^{n+1} = 2^1 \cdot 2^{n+1} = 2^{n+1 + 1} = 2^{n+2}$$
We use the fact that $a^n\cdot a^m = a^{n+m}$.
Added: For larger bases, say we have an integer base $a$, then
$$\underbrace{a^{n+1}+a^{n+1} + \cdots + a^{n+1}}_{\large a \text{ terms } }= a\cdot a^{n+1} = a^{n+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/750710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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How to closed the sum $\displaystyle \sum_{k=0}^n \dfrac{(-1)^k(2k+1)!!}{(n-k)!k!(k+1)!}$ How to closed the sum $\displaystyle S=\sum_{k=0}^n \dfrac{(-1)^k(2k+1)!!}{(n-k)!k!(k+1)!}$
I'm trying divide two cases $n$ odd and $n$ even. I predict that
$S=\begin{cases}\dfrac{1}{2^n\left[\left(\frac{n}{2}\right)!\right]^2}, &... | Suppose we seek to evaluate
$$\sum_{k=0}^{n} \frac{(-1)^k}{2^k}
{2k+1\choose k} {n\choose n-k}$$
which is
$$\sum_{k=0}^{n} {n\choose k}
\frac{(-1)^k}{2^k} {2k+1\choose k}.$$
Introduce
$${2k+1\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2k+1}}{z^{k+1}} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Volume of a parallelpiped from its sides and diagonals? If a parallelogram has sides of length $a$ and $b$, and diagonals of length $d$ and $e$, then we can find its area in the following way. By the polarization identity, we have $a b \cos\theta = \frac{1}{4}|d^2-e^2|$. Now the area is $a b \sin \theta$, which is eq... | Let the parallelepiped with a vertex $O$ at the origin have vertices $A$, $B$, $C$. Write $D = B+C$; $E= C+A$; $F=A+B$; and $G=A+B+C$. Further, assign these names to the lengths of edges and diagonals:
$$a := |OA|\quad b := |OB| \quad c := |OC| \quad d := |AD| \quad e := |BE| \quad f := |CF| \quad g := |OG|$$
Note that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/752839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum of Binomial Series of form $\binom{2000}{3k-1}$ Find the Value of $$ \binom{2000}{2}+\binom{2000}{5}+\binom{2000}{8}+\cdots+\binom{2000}{1997}+\binom{2000}{2000}$$
| You're off to a good start! In your post you defined quantities $X$ and $Y$ and found that $2X+Y=2^{2000}$. This post builds on that by giving you a second relation between $X$ and $Y$.
Let $\omega=e^{2\pi i/3}$. So $\omega^3=1$. Then by the binomial theorem, $$(1+\omega)^{2000}=\binom{2000}{0}+\binom{2000}{1}\omega+\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/754825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Need help with basic factoring equation I'm just trying to brush up on my factoring of quadratic equations.
$$\frac1{x+3} + \frac1{x^2 + 5x +6}$$
$$\frac1{x+3} + \frac1{(x+2)(x+3)}$$
$$\frac{(x+2)(x+3) + (x+3)}{(x+2)(x+2)(x+3)}$$
Then I'm stuck. I think you can factor out the $(x+2)(x+3)$ from top and bottom?
| Between your second and third step it can be improved: since you have
$
\frac1{x+3} + \frac1{(x+2)(x+3)}
$
in the second step, you want to give all fractions a common denominator - the least common denominator. Since both terms have a $x+3$ in the denom. already, multiply the first term by $\frac{x+2}{x+2}$, and add t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$ If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$
This can be easily done by calculas but is there any way to do do this by algebra
| If $a=b=\frac{1}{\sqrt2}$ then $a+b+\frac{1}{ab}=\sqrt2+2$.
We'll prove that it's a minimal value.
Indeed, let $a+b=2u$ and $ab=v^2$.
Hence, the condition gives $4u^2-2v^2=1$,
which says that $v^2=\frac{4u^2-1}{2}$ and $4u^2=1+2v^2\leq1+2u^2$, which gives $u\leq\frac{1}{\sqrt2}$
and we need to prove that
$$2u+\frac{2}{... | {
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"source": "stackexchange",
"question_score": "1",
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How to find answer to the sum of series $\sum_{n=1}^{\infty}\frac{n}{2^n} $ I have put his on wolfram and obtained answer as follows:
$\sum_{n=1}^{\infty}\frac{n}{2^n} = 2$
And the series is convergent too because $\lim_{n\to\infty} \frac {n}{2^n} = 0$
However I am wondering if there is a convenient way to solve this; ... | In this answer we go for the 'one trick pony' approach - all we know is that for $k \ge 0$,
$\tag 1 \sum_{n=k}^{\infty}\frac{1}{2^n} = 2^{1-k}$
We decompose the summands of $\sum_{n=1}^{\infty}\frac{n}{2^n}$ in a natural/straightforward manner, arranging these numbers into a table:
$$\begin{pmatrix}
\frac{1}{2} & \frac... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Joint distribution probabilities I have a question that is similar to the following(made up here):
The construction of a tower of cards is done is two stages, procrastination and the actual building. The time in minutes needed to complete each stage are independent discrete random variables, X and Y, with probability f... | This explicit computation may help:
\begin{align}
P[X+Y > 6]
&= P[X +Y > 6, X=2] + P[X +Y > 6, X=3] \\
&= P[Y = 5, X = 2] + P[ Y \geq 4, X=3 ] \\
&= P[Y = 5]P[ X = 2] + P[ Y \geq 4]P[ X=3 ] \\
&= \left(\frac{1}{5} \right)\left(\frac{7}{10} \right) +
\left(\frac{3}{5} \right)\left(\frac{3}{10} \right) \\
&= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/757338",
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"source": "stackexchange",
"question_score": "2",
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Express each function in the form $u(x,y) + iv (x,y)$ I was doing some homework with complex numbers and I'm stuck with these two, I hope that someone can solve these and clear it up for me. Thank you.
*
*ln(1+z)
*z/(3+z)
Samples,
| For the first one a function $w=re^{i\theta+2\pi n}$ satisfies $$\ln w=\ln re^{i\theta}=\ln r+i(\theta+2\pi n)$$Now $w=1+z=1+x+iy$. So $$r=\sqrt{(1+x)^2+y^2} \ \ \text{and} \ \ \theta=\arctan\{\frac{y}{1+x}\}$$
For the second one use $$\frac{1}{u+iv}=\frac{1}{u+iv}\cdot \frac{u-iv}{u-iv}=\frac{u-iv}{u^2+v^2}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/758052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A function $f$ has the property that $f(x+y)=f(x)+f(y)+3xy$. If $f(1)=2$, what is $f(8)$? A function $f$ has the property that $f(x+y)=f(x)+f(y)+3xy$. If f(1)=2, what is f(8)?
I would like to try to tackle this problem but I need somewhere to start as I really have no idea at all on how to start.
At first I assumed tha... | Setting $y=x$, we get
$$f(2x) = 2f(x) + 3x^2$$
Hence, if $x=2^n$, we get
$$f(2^{n+1}) = 2f(2^n) + 3 \cdot 2^{2n} = 2(2f(2^{n-1}) + 3 \cdot 2^{2n-2}) + 3 \cdot 2^{2n} = 4 f(2^{n-1}) + 3(2^{2n-1} + 2^{2n})$$
In general, we have
$$f(2^{n+1}) = 2^{k+1} f(2^{n-k}) + 3 \left(2^{2n} + 2^{2n-1} + \cdots +2^{2n-k}\right) = 2^{k... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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If $9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$, then what values can $t$ take?
If $9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$, then what values can $t$ take?
This is what I have done:
Let $y = 3^x$
$$9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$$ $$\implies9y^2 + (t^2 - 4t - 2)y + 1 > 0$$
For the LHS to be greater than zero, $b^2 - 4ac$... | Given $$9^{x+1}+(t^2-4t-2)3^x+1>0$$
We can write it as $$3^x\bigg[9\cdot 3^x+(t^2-4t-2)+\frac{1}{3^x}\bigg]>0$$
So $$\underbrace{3^x}_{>0} \bigg[\bigg(3\cdot 3^{\frac{x}{2}}-\frac{1}{3^{\frac{x}{2}}}\bigg)^2+(t-2)^2\bigg]>0$$
So for Above statement is True $\forall \ x \; \mathbb{R},$ If $t\in \mathbb{R}-\{2\}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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How find the $AP+\frac{1}{2}BP$ minmum value An equilateral triangle $ABC$ such $$AB=BC=AC=2a>0$$
A circle $O$ is inscribed in triangle $ABC$,and the point $P$ on the circle $O$.
Find the minimum $$AP+\dfrac{1}{2}BP$$
My idea: let $$A(-a,0),B(a,0),O(0,\dfrac{\sqrt{3}}{3}a)$$
then the circle equation is $$
x^2+(y-\dfra... | Some ideas:
Taking it from where you left it and correcting the coordinates of $\;O\;$ , we get the circle's equation is
$$x^2+\left(y-\frac a{2\sqrt3}\right)^2=\frac{a^2}{12}\;\;\;\;(I)$$
Observe that since the $\;x$-axis is tangent to the circle, the absolute value of the center's $\;y$-coordinate equals the circle'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/762836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Showing a function has first and second partials everywhere and continuity of the partials Let $f \left( \begin{array}{ccc}
x \\
y \end{array} \right)= \begin{cases}
xy\dfrac{x^2-y^2}{x^2+y^2} & \mbox{if $(x,y) \neq (0,0)$}\\
0 & \mbox{if $(x,y) = (0,0)$} \end{cases}. $
I have this function and I want to show it'... | I'll handle $D_1(f)$.
Let $(x,y)\neq (0,0)$.
Note that
$$\begin{align}
|D_1(f)(x,y)|&=\left|\dfrac {y\left(x^4+4x^2y^2-y^4\right)}{\left(x^2+y^2\right)^2}\right|\\
&\leq \dfrac {y\left(x^4+4x^2y^2+y^4\right)}{\left(x^2+y^2\right)^2}\\
&=\dfrac{y\left(\left(x^2+y^2\right)^2+2x^2y^2\right)}{\left(x^2+y^2\right)^2}\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/763519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Greatest prime factor of $4^{17}-2^{28}$ I have seen the solution to this problem.
What is the greatest prime factor of $ \ 4^{17} - 2^{28} \ $?
Answer: 7
$$ 4^{17}-2^{28} \ = \ 2^{34}-2^{28} \ = \ 2^{28} \ (2^6-1) \ = \ 2^{28} \ \cdot \ 63 \ = \ 2^{28}\ \cdot \ 3^2 \ \cdot \ 7 $$
I understand the problem up to this... | You are asking how you went from $2^{34} - 2^{28} = 2^{28}(2^6-1)$? You are factoring out $2^{28}$ from both of the numbers. $2^{34} - 2^{28}=(2^{34} - 2^{28})*\frac{2^{28}}{2^{28}}=\frac{2^{34} - 2^{28}}{2^{28}}*2^{28}=(2^6-1)*2^{28}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Expression generating $\left( \frac{3}{10}, \, \frac{3}{10} + \frac{33}{100}, \, \frac{3}{10} + \frac{33}{100} + \frac{333}{1000}, \dots \right)$ I'm looking for a closed-form expression (in terms of $n$), that will give the sequence
$$
(s_n) = \left( \frac{3}{10}, \, \frac{3}{10} + \frac{33}{100}, \, \frac{3}{10} + \f... | Built on my answer to the OP's previous post of the similar question, then:
$s_n = \displaystyle \sum_{k=1}^n \left(\dfrac{1}{3} - \dfrac{1}{3\cdot 10^k}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove inequality $(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$ I am trying to prove the following inequality:
$$(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$$
for all $a>0, b>0$.
Does anyone know how to prove it?
Thanks a lot in advance!
| The equality holds for $a=b$, so we can just consider the case for $a>b$. In this case, to show this inequality, it is equivalent to show that
$$4\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\leq(\ln(a)-\ln(b)).$$
As $a>b>0$, we can suppose that $a=e^x$ and $b=e^y$ where $x>y,\ x,y\in\mathbf{R}$. So we have to show tha... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the axis of rotation of a rotation matrix by inspection (NOT by solving $Kv=v$) $$K=\
\begin{pmatrix}
0 & 0 & 1\\
-1 & 0 & 0\\
0 & -1 & 0
\end{pmatrix}$$
Find the axis of rotation for the rotation matrix $K$ by INSPECTION.
This is from my other thread click here to view it
Everything you see below is me finding... | Neglect the sign for the moment and think of
$$
K=\
\begin{pmatrix}
0 & 0 & 1\\
-1 & 0 & 0\\
0 & -1 & 0
\end{pmatrix}
$$
as a permutation matrix: $(1\to 3), (2\to 1), (3\to 2)$ or shorter $(312)$. So you permute your coordinate axes. This points to a rotation axis along one of the vectors $(\pm 1,\pm 1,\pm 1)^T$. Ap... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/766565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Infinite series representation of elliptic integrals $F(k,\phi)$ and $E(k,\phi).$ Here is my working. For a natural number $n,$$$2^{2n-1}\left(-1\right)^n\sin^{2n}\phi=\tfrac{1}{2}\left(-1\right)^n\binom{2n}{n}+\sum\limits_{r=0}^{n-1}\binom{2n}{r}\left(-1\right)^r\cos 2\left(n-r\right)\phi.$$
According to the binomial ... | Assuming that your last line is (you have a typo in the integrand), I just confirm that your development is perfectly correct $$F(k,\pi/2)=\int^{\pi/2}_0\frac{d\phi}{\sqrt{1-k^2\sin^2 \phi}}=\frac{K\left(\frac{k^2}{k^2-1}\right)}{\sqrt{1-k^2}}$$ for which the Taylor expansion built at $k=0$ effectively leads to $$F(k,\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding matching roots If ${4 + \sqrt{2}}$ is one root of a quadratic equation given by ${x^2 - Px + Q =0}$ where P and Q are rational numbers then find the missing root. The answer is ${4 - \sqrt{2}}$.
And I'm a bit confused on how to derive that answer. I remember for complex conjugate roots the answer would be the s... | $$x^2-Px+Q=0\\
\implies x=\dfrac{P\pm\sqrt{P^2-4Q}}{2}=\dfrac{P}{2}\pm\dfrac{\sqrt{P^2-4Q}}{2}$$
One solution, given $x=4+\sqrt{2}$ is $P=8,Q=14$. Then, you have the two solutions $4-\sqrt{2},4+\sqrt{2}$.
In general, however, if $\dfrac{P}{2}+\dfrac{\sqrt{P^2-4Q}}{2}=a+\sqrt{b}$, a solution of $x^2-Px+Q=0$, then $\dfra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Diophantine equation. Three. Diophantine equation. $X^2+Y^2=qZ^3$
I wonder at what values of the coefficient $q$ equation has a solution.
And of course I wonder how she looks like a formula describing their solutions.
For the special case when $X^2+Y^2=Z^3$ You can get a basic formula.
Has the solutions:
$X=2k^6+8tk^... | For the equation: $$X^2+qY^2=Z^3$$
You can write this simple solution:
$$X=(p^2+qs^2)((p^4-q^2s^4)t^3-3(p^2+qs^2)^2kt^2+3(p^4-q^2s^4)tk^2-(p^4-6qp^2s^2+q^2s^4)k^3)$$
$$Y=2ps(p^2+qs^2)((p^2+qs^2)t^3-3(p^2+qs^2)tk^2+2(p^2-qs^2)k^3)$$
$$Z=(p^2+qs^2)((p^2+qs^2)t^2-2(p^2-qs^2)tk+(p^2+qs^2)k^2)$$
$q - $The ratio is given fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/772409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Partial fraction decomposition of a complicated rational function Find the partial fraction decomposition of the rational function $\displaystyle \frac{2x^3+7x+5}{(x^2+x+2)(x^2+1)}$
I have tried dividing first but keep running into problem after problem, please help.
| Step 1: Find $a, b, c, d$ such that: $\dfrac{2x^3 + 7x + 5}{(x^2 + x + 2)(x^2 + 1)} = \dfrac{ax + b}{x^2 + x + 2} + \dfrac{cx + d}{x^2 + 1}$
Step 2: Rewrite:
$\dfrac{ax + b}{x^2 + x + 2} + \dfrac{cx+d}{x^2 + 1} = \dfrac{1}{2}\cdot \dfrac{a(2x+1)}{x^2 + x + 2} + \dfrac{1}{2}\cdot\dfrac{2b - a}{(x + \frac{1}{2})^2 + (\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/773479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proving Identities Can you help me prove this identity:
$$\frac{\sin x}{2\csc x}\left(\tan^2 x + \sin^2 x + \frac{\sin^2x}{\tan^2x}\right)= \frac{1}{2} \tan^2x$$
I have tried working on the left side and changing everything to sine and cosine, but I can't seem to get it down to only $\frac{1}{2} \tan^2 x$.
| Our identity that we need to prove:
$$\frac{\sin x}{2\csc x}\left(\tan^2 x + \sin^2 x +\frac{\sin^2 x}{\tan^2 x}\right)=\frac 12 \tan^2 x$$
Remember that $\dfrac{\sin^2 x}{\tan^2 x}=\cos^2 x$. This follows from the equality $\tan^2 x=\dfrac{\sin^2 x}{\cos^2 x}$.
$$\frac{\sin x}{2\csc x}\left(\tan^2 x+\sin^2 x+\cos^2 x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/773620",
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"source": "stackexchange",
"question_score": "2",
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Can $x^3+x^2+1=0$ be solved using high school methods? I encountered the following problem in a high-school math text, which I wasn't able to solve it: $x^3 + x^2 + 1 = 0$ Am I missing something here, or is indeed a more advanced method necessary to solve this particular cubic?
| Let $x=au+b$. Then $$x^3+x^2+1= a^3u^3+3a^2bu^2+3ab^2u+b^3+a^2u^2+2abu+b^2+1$$ $$=a^3u^3+(3a^2b+a^2)u^2+(3ab^2+2ab)u+(b^3+b^2+1)$$We want $3b+1=0\implies b=-1/3$ and then $$b^3+b^2+1=\frac{-1+3+27}{27}=\frac{29}{27}$$ and $$3ab^2+2ab=a(1/3-2/3)=-\frac{a}{3}$$ So clearly denominators we get $$(3au)^3-3(3au)+29=0$$ Now l... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Question in fraction (not simple ) I have a question and its answer but I don't know how can i solve
$$\frac {37}{13} = 2+ \frac {1}{x+\frac{1}{5+\frac{1}{y}}} $$
the answer $ x =1, y=2$
Could any one explain how to solve this ?? please
| \begin{align*}
\frac {37}{13}
&= 2+ \frac {1}{x+\frac{1}{5+\frac{1}{y}}} \\
\frac {11}{13}
&= \frac {1}{x+\frac{1}{5+\frac{1}{y}}} \\
\frac {13}{11}
&= x+\frac{1}{5+\frac{1}{y}} \\
\frac {13 - 11x}{11}
&= \frac{1}{5+\frac{1}{y}} \\
\frac {11}{13 - 11x}
&= 5+\frac{1}{y} \\
\frac {-54 + 55x}{13 - 11x}
&= \frac{1}{y} \\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/776154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Are these two summations equivalent? Is $\sum_{y=2}^{\infty} (\frac{1}{y})(1-p)^{y-1}$ equivalent to $\sum_{y=1}^{\infty} (\frac{1}{y})(1-p)^{y}$ ?
| We'll go ahead and evaluate both the sums to make things more interesting. Recall that for $|x| < 1$,
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + ...$$
Integrating both sides, we get
$$-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ...$$
Substitute $x = 1-p$ (I'm assuming $| 1- p| < 1$). Then,
$$-\ln p = (... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the full Taylor expansion of the following: I need to find the full Taylor expansion of $$f(x)=\frac{1+x}{1-2x-x^2}$$
Any help would be appreciated. I'd prefer hints/advice before a full answer is given. I have tried to do partial fractions\reductions. I separated the two in hopes of finding a known geometr... | Here is an alternate approach. Suppose that
$$
\frac{1+x}{1-2x-x^2}=\sum_{k=0}^\infty a_kx^k
$$
Multiply by $1-2x-x^2$ to get
$$
\begin{align}
1+x
&=\underbrace{a_0}_{1}+\underbrace{(a_1-2a_0)}_1x+\sum_{k=2}^\infty\underbrace{(a_k-2a_{k-1}-a_{k-2})}_{0}x^k
\end{align}
$$
Equating coefficients of $x^k$ gives
$$
a_0=1,a_... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
proving $\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq 2(\frac{b}{a}+\frac{c}{b}+\frac{a}{c})$ For $a+b+c=1; a,b,c>0$, prove that
$$
\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}
\leq 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right).
$$
| since $$\dfrac{1+a}{1-a}=\dfrac{-(1-a)+2}{1-a}=-1+\dfrac{2}{1-a}=-1+\dfrac{2}{b+c}=-1+\dfrac{2a+2b+2c}{b+c}=1+\dfrac{2a}{b+c}$$
or $$\dfrac{1+a}{1-a}=\dfrac{a+b+c+a}{b+c}=\dfrac{b+c+2a}{b+c}=1+\dfrac{2a}{b+c}$$
so
$$\Longleftrightarrow 2\sum_{cyc}\dfrac{b}{a}-\sum_{cyc}\dfrac{1+a}{1-a}=2\sum_{cyc}\dfrac{b}{a}-\sum_{cyc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/777956",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How many ways can you build a football/soccer team? I'm reading a book on the history of soccer. There are several ways to position a team, for example, four defenders, four midfielders and two forwards is the common 4-4-2. How many ways can you build a team, given that there are 10 players that move because the keeper... | For variety, here's another approach.
Your initial approach looks like you settled on the following way to enumerate all ways to make a list of positive integers that adds to 10:
*
*Write down all lists that start with 1 followed by a list that adds to 9
*Write down all lists that start with 2 followed by a list th... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
integration of $\int \frac{1}{x+ i\:y}\mathrm{d} x$ I can't seem to find where this result comes from
$$
\int \frac{1}{x+ i\:y}\mathrm{d} x = \frac{\ln(x^2
+y^2)}{2} - i \: \arctan \left( \frac{x}{y} \right)
$$
by my calculation the result should be
$$
\int \frac{1}{x+ i\:y}\mathrm{d} x = \frac{\ln(x^2
+y^2)}{2} + i ... | There are different ways to express the same function, and if you compute an unspecified primitive (antiderivative) of some function, the integration constants further enlarge the number of possibilities to express the same family of functions in different ways.
Here, the relations
$$\arctan \frac{1}{\phi} = \int_0^{\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Pie Integral $\int_0^1 \log\frac{(x+\sqrt{1-x^2})^2}{(x-\sqrt{1-x^2})^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$ Hi I am trying to show this astonishing result$$
\int_0^1 \log\frac{\big(x+\sqrt{1-x^2}\big)^2}{\big(x-\sqrt{1-x^2}\big)^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.
$$
Note we can to use $\ln(a/b)=\ln a-\ln b$ b... | I hope it is not too late. Let
$$ I(a)=\int_0^1 \log\frac{\big(x+a\sqrt{1-x^2}\big)^2}{\big(x-\sqrt{1-x^2}\big)^2} \frac{x\, dx}{1-x^2}, -1\le a\le 1. $$
Clearly $I(-1)=0$ and
\begin{eqnarray}
I'(a)&=&2\int_0^1 \frac{x}{(x+a\sqrt{1-x^2})\sqrt{1-x^2}}dx\\
&=&2\int_0^{\frac{\pi}{2}}\frac{\sin t}{\sin t+a\cos t}dt.
\end{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
Why do the even Bernoulli numbers grow so fast? Question is in the title.
We have:
$$B_{2n} \sim (-1)^{n-1} 4 \sqrt {\pi n} \left( \frac n {\pi e} \right)^{2n}$$
| Recall the generating function
$$f(z) = \sum_{q\ge 0} B_q \frac{z^q}{q!} = \frac{z}{e^z-1}.$$
We can treat it as if it were rational as the poles are countable with
no accumulation point, so we may use the techniques of extracting coefficients from rational generating functions, with the asymptotics being given by inve... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof via induction $1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$ (b) Prove that for every integer $n \ge 1$, $$1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$$
This is the second part of a two part question. Part (a) was the following:
Write the sum: $1\cdot3 + 2\cd... | Assume result is true for $n=k$ i.e.
$$S_k=\frac {k(k+1)(2k+7)}6$$
Then, adding the next term gives:
$$\begin{align}
S_{k+1}&=S_k+(k+1)(k+2)\\
&=\frac {k(k+1)(2k+7)}6+(k+1)(k+3)\\
&=\frac{k+1}6 \left( k(2k+7)+6(k+3)\right)\\
&=\frac{k+1}6\left( 2k^2+13k+18\right)\\
&=\frac{(k+1)(k+2)(2k+9)}6\\
&=\frac{(\overline{k+1})(... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Interior point of $\Delta\,ABC$ if $P(\lambda,2)$ is an interior point of $\Delta\,ABC$ formed by the lines
$$x+y=4$$ $$3x-7y=8$$ $$4x-y=31$$ Find $\lambda$
My Idea: The vertices of $\Delta ABC$ are
$A(\frac{18}{5},\frac{2}{5})$ $\:$ $B(7,-3)$ and $C(\frac{209}{25},\frac{61}{25}) $
So using areas of $\Delta ABC$, $\D... | Another way is rough plot all these three lines with $y=2$ With seeing the points of intersection and a rough idea of their positions you will be able to directly tell the range of $\lambda$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate Gauss-like Integral Evaluate Integral
$$\int_0^\infty e^{-ay^{2}-\frac{b}{y^2}}dy $$
Where a and b are real and positive.
This integral is eerily similar to the Gaussian integral
$$\int_0^\infty e^{-\alpha x^2}dx = \frac{1}{2} \sqrt{\frac{\pi}{\alpha}}$$
This is an integral I have come across as a step in a p... | This answer is taken from my answer here.
$$
\begin{align}
\int_{v=0}^\infty \exp\left(-av^2-\frac{b}{v^2}\right)\,dv&=\int_{v=0}^\infty \exp\left(-a\left(v^2+\frac{b}{av^2}\right)\right)\,dv\\
&=\int_{v=0}^\infty \exp\left(-a\left(v^2-2\sqrt{\frac{b}{a}}+\frac{b}{av^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dv\\
&=\int_{v... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Summation of Fibonacci numbers $F_n$ with $n$ odd vs. even Compare the summation below:
$$\begin{align}
\smash[b]{\sum_{i=1}^n F_{2i-1}}&=F_1+F_3+F_5+\cdots+F_{2n-1}\\
&=1+2+5+\cdots+F_{2n-1}\\
&=F_{2n}\\
\end{align}
$$
with this one:
$$\begin{align}
\smash[b]{\sum_{i=1}^n F_{2i}}&=F_2+F_4+F_6+\cdots+F_{2n}\\
&=1+3+8+\... | A clean way to see this is by using generating functions. Define $F(z) = \sum_{n \ge 0} F_n z^n$, take the recurrence:
$$
F_{n + 2} = F_{n + 1} + F_n \qquad F_0 = 0, F_1 = 1
$$
Multiply by $z^n$, sum over all valid values for $n$, i.e., $n \ge 0$, and recognize the resulting sums:
$$
\frac{F(z) - F_0 - F_1 z}{z^2}
=... | {
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"source": "stackexchange",
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"answer_count": 5,
"answer_id": 1
} |
Is this proof by induction correct? Prove by induction that for all $n\in\mathbb N$, $3\mid n^3+3n^2+2n$.
$$P(1) = (1)^3+3(1)^2+2(1) = 6$$
Which is clearly divisble by $3$. Therefore, $P(1)$ is true.
Assume $P(1),\ldots,P(n)$ and show $P(n+1)$ is true.
$$(n+1)^3+ 3(n+1)^2+ 2(n+1) = n^3+6n^2+11n+6$$
This can be rewritte... | Your proof looks solid, but I might suggest an easier proof:
For all $n \in \mathbb{N}$, we have $n \equiv 0, 1, 2 \pmod{3}$. If $n \equiv 0 \pmod{3}$, we are done.
If $n \equiv 1 \pmod{3}$, then we have $n^3 + 3n^2 + 2n \equiv (1+2) \equiv 0 \pmod{3}$.
If $n \equiv 2 \pmod{3}$, then we have $n^3 + 3n^2 + 2n \equiv (2... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why are these two given expressions equivalent? $6\cos^2(\frac {x}{2})-7\cos(\frac {x}{2})+2=0$ is equivalent to the expression $(3\cos(\frac {x}{2})-2)(2\cos(\frac {x}{2})-1) = 0$
Could someone explain the steps involved in evaluating the first expression to equal the second expression?
I noticed the expression in thi... | The answer comes from factoring.
When you foil the two parentheticals from the second expression, their products can be seen to equal the first.
We treat $\cos(\frac {x}{2})$ much like we would the variable $x$ within any given expression.
If given the expression $2x^2-3x+1 = 0$, we are able to factor this to show that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/790358",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
calculating the sum of a series I was wondering if there is a way to use complex-analysis to solve the sum of the following series? (just like you can use it to solve integrals of some kinds (using fourier's transform for integrals).
$$\sum^{\infty}_{n=0}\frac{1}{n^2+25}$$
and
$$\sum^{\infty}_{n=0}\frac{(-1)^n}{n^2+... | We can evaluate the two sums using following expansions of trigonometry functions
which can be proved using Mittag Leffler's theorem:
$$
\frac{\cos x}{\sin x} = \lim_{N\to\infty}\sum_{n=-N}^N \frac{1}{x+n\pi}
\quad\text{ and }\quad
\frac{1}{\sin x} = \lim_{N\to\infty}\sum_{n=-N}^N \frac{(-1)^n}{x+n\pi}
$$
Let's look at... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$ Hi I am trying to solve this integral $$
I:=\int_0^1 \log\left(\frac{1+ax}{1-ax}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}=\pi\arcsin\left(a\right),\qquad
\left\vert a\right\vert \leq 1.
$$
It gives beautiful result for $a = 1$
$$
\int_0^1 \lo... | It is not necessary to use complex analysis or to use power series to compute. From @Random Variable, we have
$$ I'(a) =\int_{-1}^1 \frac{dx}{(1+ax)\sqrt{1-x^2}}=\int_{-1}^1 \frac{1}{1+ax}d\arcsin x. $$
Let $x=\sin t$. Then
$$ I'(a) =\int_{-\pi/2}^{\pi/2} \frac{1}{1+a\sin t}dt. $$
Let $u=\tan\frac{t}{2}$. Then
\begin{e... | {
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"url": "https://math.stackexchange.com/questions/795493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 0
} |
How should I solve integrals of this type? The general form of the integral I want to solve is:
$$ \int e^{bx}\sin(ax) dx$$
Euler's formula has a nice connection, but the i makes it too complicated.
Doing it by parts doesn't seem to get me anywhere.
Do you have any tips for how to begin solving this?
| By using the exponential version of sine:
\begin{align}
\int e^{bx} \ \sin(ax) \ dx &= \frac{1}{2i} \int \left( e^{(b+ai)x} - e^{(b-ai)x}
\right) \ dx \\
&= \frac{1}{2i} \left[ \frac{e^{(b+ai)x}}{b+ai} - \frac{e^{(b-ai)x}}{b-ai} \right] \\
&= \frac{1}{2i (b+ai)(b-ai)} \left[ (b-ai) e^{(b+ai)x} - (b+ai)e^{(b-ai)x}
\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/796730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
How does WolframAlpha simplify sine and cosine? When I feed WolframAlpha an expression like $\sin({\pi\frac{2}{3}})$, it correctly prints that this is equal to $\frac{\sqrt3}{2}$, instead of the decimal expansion $0.866025403\ldots$.
Perhaps it has a lookup table for common fractions of $\pi$. Or is it likely to be mor... | It must be that $\sin\left(\dfrac{p}{q}\pi\right)$ is algebraic. To see why check out this question.
I am almost certain that W|A doesn't use power series unless the value is very small simply because it would be too slow to calculate the value of arbitrary trig functions using power series. It is more likely that ther... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx=-\frac{\pi}{4\alpha}\text{arcsinh}\alpha$ Hi I am trying to prove this
$$
I:=\int_0^{\pi/4}\log\left(\tan\left(x\right)\right)\,
\frac{\cos\left(2x\right)}{1+\alpha^{2}\sin^{2}\left(2x\right)}\,{\rm d}x
=-\,\frac{\pi}{4\alpha}\,\text{arcsinh}\lef... | I use on Ron's results from integrating by parts:
$$\alpha I(\alpha) = - \int_0^{\pi/4} dx \frac{\arctan{(\alpha \sin{2 x})}}{\sin{2 x}},$$
As an alternative way to complete the problem, use the method differentiating under the integral sign:
$$\frac{d}{d\alpha}(\alpha I(\alpha)) = -\int_{0}^{\pi/4}\frac{dx}{\alpha^2... | {
"language": "en",
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"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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Prove that $\lim\limits_{n \rightarrow \infty} \left(1-\frac{1}{n^2}\right)^n= 1.$ Prove that $\lim\limits_{n \rightarrow \infty} \left(1-\frac{1}{n^2}\right)^n= 1.$
I need to show that there exists $N \in \mathbb{N}: \forall n \geq N : \left|(1-\frac{1}{n^2})^n-1\right| \lt \epsilon$ $,\:\:\forall \epsilon \gt 0.$
Sin... | Consider the expansion of $(1-\frac{1}{n^2})^n.$
$$\begin{align}\left(1-\frac{1}{n^2}\right)^n &= 1 - n\dfrac{1}{n^2}+\dfrac{n(n-1)}{2}\dfrac{1}{n^4} - \cdots \\ &= 1-\dfrac{1}{n}+\dfrac{n-1}{2n^3}+\cdots\\&\geq1-\dfrac{1}{n}\end{align}$$
We can construct the inequality $$1-\dfrac{1}{n} \leq \left(1-\frac{1}{n^2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/800219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
The $n$'th derivative of $x^x$ I want to know the $n$'th derivative of $f(x)=x^x$.
Then, I'll calculate $f(0)$ with Taylor expansion of $f(x)$ on $a=1$.
Here is my answer, but it is unfinished.
The derivative of $f(x)=x^x$
$$\begin{align}
f'(x)&=x^x(\log x+1)\\
f''(x)&=x^x(\log x+1)^2+x^{x-1}\\
f'''(x)&=x^x(\log x+1)^3... | For $n\in\{0,1,2,\dotsc\}$, we have
\begin{equation}\label{power-exp-deriv-eq}
(x^x)^{(n)}=n!x^{x-n}\sum_{k=0}^{n} x^{k} \sum_{j=0}^{k}\Biggl[\sum_{q=0}^{n-k} \frac{s(q+j,j)}{(q+j)!} \binom{j}{n-k-q}\Biggr]\frac{(\ln x)^{k-j}}{(k-j)!},
\end{equation}
where $s(n,k)$ for $n\ge k\ge0$ denotes the Stirling numbers of the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/802256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
How prove that there exists $m$ such that $|f(m)|\le \dfrac{\sqrt{b^2-4c}}{2}$ Let the function $$f(x)=x^2+bx+c,\qquad b^2-4c>0$$ Assume that $x_{1},x_{2}$ are the roots of $f(x)$ and $|x_{1}-x_{2}|\ge 1$.
Show that: There exists an integer $m$, such that
$$|f(m)|\le \dfrac{\sqrt{b^2-4c}}{2}$$
My idea: Since $$x_{1... | We consider the following equivalent problem.
For $ d \geq 1 $, define $ g(x) = x^2 + dx$. We want to show that there exists real numbers $ j \leq 0 \leq k$ with $k-j \geq 1$ such that for all $ j \leq x \leq k$, we have
$$ - \frac{d}{2} \leq g(x) \leq \frac{d}{2} . $$
Proof of Equivalent: To see why this is equivalent... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the general term of the recursive relation $a_{n+1}=\frac{1}{n+1}\sum\limits_{k=0}^n a_k 2^{2n-2k+1}$ I need to find the general term of the recursive relation $a_{n+1}=\frac{1}{n+1}\sum\limits_{k=0}^n a_k 2^{2n-2k+1}$
I know it's the central binomial sequence but I can't find a way to show it.
$a_0=1$
| We are given the recurrence relation
$$a_{n+1}=\frac{1}{n+1}\sum_{k=0}^{n}a_{k}2^{2n-2k+1}.$$
Then,
$$a_{n}=\frac{1}{n}\sum_{k=0}^{n-1}a_{k}2^{2n-2k-1}.$$
Multiplying the relation for $a_n$ on both sides by $\frac{2^2n}{n+1}$,
$$\frac{2^2n}{n+1}a_{n}=\frac{1}{n+1}\sum_{k=0}^{n-1}a_{k}2^{2n-2k+1}.$$
Subtracting this fro... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Why this representation of circle is valid? A line passing through two distinct points $P_1(x_1,y_1),P_2(x_2,y_2)$ can be expressed by
$$\det\left| \begin{array}{ccc}
x-x_1&y-y_1 \\
x_2-x_1&y_2-y_1 \\ \end{array} \right|=0$$
Since line is set $A=\{a P_1+bP_2|a+b=1\}$ and it corresponds to determinant properties.
A cir... | Consier the set of equations $$Ax^2+Ay^2+Bx+Cy+D=0 \tag{1} $$
$$Ax_1^2+Ay_1^2+Bx_1+Cy_1+D=0 \tag{2} $$
$$Ax_2^2+Ay_2^2+Bx_2+Cy_2+D=0 \tag{3} $$
$$Ax_3^2+Ay_3^2+Bx_3+Cy_3+D=0 \tag{4} $$
this is a $4\times4$ linear system and it must satisfy,
$$\left( \begin{array}{ccc}
x^2+y^2 & x & y&1 \\
x_1^2+y_1^2 & x_1 & y_1&1 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/808506",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Improper Integral $\int_{1}^{\infty} \sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) dx$ I need to calculate this improper integral.
$$\int_{1}^{\infty} \sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) dx$$
How do I prove that $$ \sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) $$... | Yes, as $x\rightarrow \infty\;$ you have the asmptotic relations
$$\sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right)
\sim \sin \left( \sin\left( \frac {1}{\sqrt{x}} \right) \right)
\sim \sin\left( \frac {1}{\sqrt{x}} \right)
\sim \frac {1}{\sqrt{x}}
$$
because for small $z\rightarrow 0\;$ you have $\sin(z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/809722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Laurent series for $\frac1{z^2+1}$ I have this problem:
Find the Laurent series around $z=0$, for $\dfrac{10}{(z+2)(z^2+1)}$ in the region $1<|z|<2$.
I did partial fractions and found this: $\dfrac{2}{z+2}-\dfrac{2z-4}{z^2+1}$,
then I have to know what's the Laurent series for $\dfrac{1}{z^2+1}$ to solve the problem.
D... | Expanding geometrically,
$$\frac{2}{z+2} = \frac{1}{ 1 + \frac z 2} = \sum_{k=0}^\infty (-1)^k \left( \frac z 2\right)^k $$
Similarly,
$$\frac{2z-4}{z^2 + 1} = \frac{2z-4}{z^2}\cdot \frac 1 {1 + \frac{1}{z^2}} = \frac{2z-4}{z^2} \sum_{k=0}^{\infty}(-1)^k \left( \frac 1 {z^2}\right)^k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Let $a$ and $b$ be relatively prime integers. Show that $\gcd(a+b,a^2-3ab+b^2)= 1$ or $5$ Let $a$ and b be relatively prime integers. Show that $\gcd(a+b,a^2-3ab+b^2)=1$ or $5$
Proof: $s\mid (a+b)$ and $s\mid (a^2-3ab+b^2)$ implies $s\mid(a+b)^2=a^2+b^2+2ab$ and $s\mid (a^2-3ab+b^2)$
implies $s\mid((a+b)^2-(a^2-3ab+b^2... | Note that $$a^2-3ab+b^2=(a+b)^2-5ab$$ So if $p$ is a prime, and such that $p|(a+b)$ and $p|(a+b)^2-5ab$ then $p|5ab$. Now $p\not | ab$ since if $p|a$, say then from $p|a+b$ follows $p|b$ contradiction the fact the $a$ and $b$ are relatively prime therefore $p|5$. The same argument shows that $gcd =5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Maximize the given variable under the following constraint Find the greatest integer $a$ such that $a^5 + 5^5$ is divisible by $a - 5$
| Make a polynomial (euclidean) division of $a^5+5^5$ by $a-5$ : since
$$
a^5+5^5=(a^5-5^5)+2(5^5)=(a-5)(a^4 + 5a^3 + 25a^2 + 125a + 625)+2(5^5)
$$
If $a$ is a solution, then $a-5$ must divide $2(5^5)$. It is then easy to see
that the largest solution is $a=2(5^5)+5=6255$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/811308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Diophantine equation with cubes. Interested in the solution in general Diophantine equations of the form:
$X^3+Y^3+Z^3=3XYZ+q$
$q$ - what some integer.
Solutions similar equations can be written.
Since this equation is easy, as it is quite symmetrical.
$X^3+Y^3+Z^3-3XYZ=R^3$
Such a solution can write.
$X=sp(p+s)$
$Y=s(... | You can use the following identity:
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$$
Fixed integer q has a finite number of factorial expansions. So we get a finite number of systems of two equations:
$$x+y+z = a$$
$$x^2+y^2+z^2-xy-xz-yz = b$$
for all integers $a$ and $b$ such that $ab=q$. Excluding the variable z we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/811456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}} dx$? How to calculate?
$$\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\, \mathrm dx$$
I try to let $x=\cos^2 t$, then
$$\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}=\tan\frac t2,\; dx=-2\sin t\cos t\,\mathrm dt $$
so
$$\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}} \mathrm dx=-2 \int\tan... | Alternate method:
$$ I = \int \sqrt{\frac{1- \sqrt{x}}{1 + \sqrt{x}}}\ dx$$
Putting $x = \cos^{2}{2t}$ so that $dx = -4\cos(2t) \sin(2t)\ dt$ and $t = \frac12 \cos^{-1}({\sqrt x})$
$$\begin{align}\implies I & = \int \frac{\sin(t)}{\cos(t)}\cdot -4\cos(2t) \sin(2t)\ dt\\ & = -4\int \frac{\sin t}{\cos t}[2\cos^2t - 1]\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/813079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
How to find $\lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$ and $\lim_{n\rightarrow 0} \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$ How can I determine these limits:
$$ a) \lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$$
$$ b) \lim_{n\rightarrow 0} \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$$
Note I ca... | \begin{align}
(a)\quad\lim_{n\rightarrow 0} \cos\left(\frac{\pi\sin n \cos n}{n}\right)&=\cos\left(\lim_{n\rightarrow 0}\frac{\pi\sin n \cos n}{n}\right)\\
&=\cos\left(\lim_{n\rightarrow 0}\frac{\pi\sin n \cos n}{n}\right)\\
&=\cos\left(\pi\lim_{n\rightarrow 0}\cos n\frac{\sin n}{n}\right)\\
&=\cos(\pi)\\
&=-1.
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/813769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How does $\frac{t^2}{t+1}$ equal $t-1+\frac{1}{t+1}$? I do the long division:
1: t+1 goes into $t^2$ t times
2: Subtract $t^2$ + 1 from $t^2$ and get -1
3: Answer: t - $\frac{1}{t+1}$
Am I missing something here?
| Polynomial division will also give you the answer quickly:
$$\begin{array}{cccccccc}
& & & t & & -1 & & \\
(t+1)& | & & t^2 & & +0t & &+0 \\
& & -( & t^2 & & +t &) & \\
& & & & & -t & & +0 \\
& & & &-(&-t & -1 & )\\
& & & & & & +1\\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/814410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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Fractions in Questions and Answers
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