Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proof of $\sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}$ The title pretty much summarizes my question. I am trying to prove the following:
$$\displaystyle \forall N \in \mathbb{N}: \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}.$$
I tried proving this using induction. Startin... | Instead of induction, you could start with
$\displaystyle \sum_{n=1}^{2N}\frac{(-1)^{n-1}}{n}=\sum_{n=1}^{2N}\frac{1}{n}-2\sum_{k=1}^N\frac{1}{2k}=\sum_{n=1}^{2N}\frac{1}{n}-\sum_{n=1}^{N}\frac{1}{n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Find eigenvalues and eigenvectors of this matrix Problem: Let \begin{align*} A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}. \end{align*} Compute all the eigenvalues and eigenvectors of $A$.
Attempt at solution: I found the eigenvalues by computing the characteristic ... | You are right, and to show how this applies more generally, look at the eigenspace corresponding to the other eigenvalue, $\lambda = 0$. This has multiplicity 3, so we expect exactly 3 linearly independent vectors in this space. The main equation looks like $A \vec{x} = 0 \vec{x} = \vec{0}$, in other words,
$$
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find this limit : $x\sin{f(x)}$ How to find the limit
$$\lim_{x\to\infty}x\sin f(x)$$
where $$f(x)=\left(\sqrt[3]{x^3+4x^2}-\sqrt[3]{x^3+x^2}\right)\pi\ ?$$
Is it possible to solve without L'Hospital's rule ?
| Setting $x=u^{-1}$, we have
\begin{eqnarray}
x\sin f(x)&=&\frac{1}{u}\sin\left[\pi\left(\sqrt[3]{u^{-3}+4u^{-2}}-\sqrt[3]{u^{-3}+u^{-2}}\right)\right]=\frac1u\sin\left[\pi\frac{\sqrt[3]{1+4u}-\sqrt[3]{1+u}}{u}\right]\\
&=&\frac1u\sin\left[\pi\frac{(1+\frac{4}{3}u-\frac{16}{9}u^2)-(1+\frac{u}{3}-\frac{u^2}{9})+o(u^2)}{u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\iint_{R}(x+y)^2dxdy$ in $0\leq r\leq 1 \,\, ,\frac{\pi}{3}\leq \theta\leq\frac{2\pi}{3}$
$$\iint_{R}(x+y)^2dxdy$$
$$0\leq r\leq 1 \,\, ,\frac{\pi}{3}\leq \theta\leq\frac{2\pi}{3}$$
My attempt number 1:
$$=\iint_{R}(x^2+2xy+y^2)dxdy$$
$$x:=r\cos \theta \,\,\,,y:=r\cos \theta$$
$$\sqrt{x^2+y^2}=r$$
$$\int_{\... | So you are trying to determine the surface area of the function $(x+y)^2$ in a part of a circle with radius 1. Let's define $R=\left( (r,\theta)\in \mathbb{R}^2 : 0 \leq r \leq 1 , \frac{\pi}{3} \leq \theta \leq \frac{2 \pi}{3}\right)$.
We have the co-ordinate transformation $X(x,y)=(r \cos \theta , r \sin \theta)$.
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find numbers $\overline{abcd}$ so that $\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}$ Find the numbers $\overline{abcd}$, with digits not null that satisfy the equality
\begin{equation}\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}\end{equation}
where
\begin{equation}\overline{abc... | \begin{array}{r}
abcd\\
bcd\\
cd\\
d\\
1\\
\hline
dcba
\end{array}
The leftmost $a$ in the top row cannot recieve more than $1$ as a carry. The worst cases are $(b,c)=(9,8)$ and $(b,c) = (8,9)$ as $(b,c) = (9,9)$ leads quickly to no solution. Hence $d = a + 1$ or $d = a$.
Case: $d = a + 1$
Summing the rightmost column,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve the trigonometric equation: $\sin {3x} = 4 \sin^2 x$ Solve the equation $\sin{3x} = 4 \sin^2 x$.
I tried to change the $\sin{3x}$ to $3\sin x\cos x$ then solve it, but I could not find the correct answer.
| Hint: We have $$\sin 3x = 3\sin x - 4\sin^3 x\neq 3\sin x\cos x$$
So you get (let $\sin x = \alpha$) $$3\alpha - 4\alpha^3 = 4\alpha^2 \iff 4\alpha^3 + 4\alpha^2 - 3\alpha = 0$$
Which is a simple cubic in $\alpha$. Find the roots of the cubic, back-substitute and find the corresponding values of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to show that his series converges or diverges using LCT or CT? $$\sum_{n=1}^{\infty}\left (\sqrt{n^4+1}-n^2\right)$$
The question states that either the limit comparison or comparison test can be used to determine whether the series converge or diverge. I tried finding a $B_n$ in order to test $\frac{A_n}{B_n}$ for... | \begin{align}
\sqrt{n^4+1}-n^2 &=\dfrac{\sqrt{n^4+1}-n^2}{1} \\
&=\dfrac{\left(\sqrt{n^4+1}-n^2\right)\left(\sqrt{n^4+1}+n^2\right)}{\left(\sqrt{n^4+1}+n^2\right)} \\
&=\dfrac{\left(\sqrt{n^4+1}\right)^2-n^4}{\sqrt{n^4\left(1-\dfrac{1}{n^4}\right)}+n^2} \\
&=\dfrac{n^4+1-n^4}{n^2\,\sqrt{1-\dfrac{1}{n^4}}+n^2} \\
&=\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that $\frac{n^3}{3}+\frac{2n}{3}$ is an integer. The question that I am working on is:
Prove that $\dfrac{n^3}{3}+\dfrac{2n}{3} \in \mathbb Z \ \forall \ n \in \mathbb N$
The method that I think would be will work for this question is that I say that $3|(n^3+2n)$ and prove that.Would this be a go... | $$( k + 1 )^3 = k^3 + 3 k^2 + 3k + 1$$
For $n = k + 1$ $$(k + 1)^3 + 2(k + 1) = k^3 + 3k^2 + 3k + 2k + 3$$
If $k^3 + 2k$ is divisible by $3,$ it is ovious that $$k^3 + 3k^2 + 3k + 2k +3=(k^2+2k)+3(k^2+k+1)$$ is dividable by $3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 4
} |
Find the principal solutions of the trigonometric equation $\cos x-\sin x+\sin 2x+3\cos2x+1=0$ I am unable to simplify the expression. If I simplify the double angles, it leaves me with a nasty expression,
$\cos x-\sin x+2\sin x\cos x+6\cos^2 x-2=0$. What do I do next. Some hints, please. Also, is there some elegant s... | You have $\cos x-\sin x+\sin 2x+3\cos 2x+1=0$. Or $\sin x-\sin 2x=\cos x+3\cos 2x+1$. Or $\sin x(1-2\cos x)=\cos x+6\cos^2x-2$.
Either $\cos x =1/2$ or $\sin x=((6\cos^2x+\cos x-2)/(1-2\cos x))$. Or $\sin x=-3\cos x-2$. Squaring we get $10\cos^2x-12\cos x+3=0$. Thus $\cos x=((6\pm \sqrt{6})/(10))$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving $6 \cos x - 5 \sin x = 8$ My attempt:
Using the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi)$$
$$
\sqrt{51} \left(\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x\right) = 8
$$
$$
\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x = \frac{8}{\sqrt{51}}
$$
And... | By the Cauchy-Schwarz inequality:
$$ \left( 6\cos x-5\sin x\right)^2 \leq 36+25 = 61<64 $$
so it is not possible that $6\cos x-5\sin x$ equals $8$ for some real $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Baby Rudin claim: $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}...$ converges This sequence is a rearrangement of the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...$. Note that at this point in the text we do not have any theorem about the convergence ... | By the Riemann-Dini theorem, we may take any series that is conditionally convergent but not absolutely convergent and rearrange it in order to get a series that converges to $\alpha$, for any $\alpha\in\mathbb{R}$.
In our case:
$$\begin{eqnarray*} \sum_{k\geq 0}\left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+2}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
$\lim_{n \to \infty} (\sqrt{9n^2 + 2n + 1} - 3n) = \frac{1}{3}$ Prove
$$\lim_{n \to \infty} (\sqrt{9n^2 + 2n + 1} - 3n) = \frac{1}{3}$$
I got this problem in Harro Heuser's "Lehrbuch der Analysis Teil 1". It is surely smaller than 1 because $\sqrt{9n^2 + 2n + 1} < \sqrt{9n^2 + 6n + 1} = 3n + 1$, but I cannot get closer... | $$\lim_{n\to\infty} (\sqrt{9n^2 + 2n+1} - 3n) = \lim_{n\to\infty} 3n\left(\sqrt{1 + \frac{2}{9n} + \frac{1}{9n^2}} - 1\right)$$
Now, using Taylor expansion for $\sqrt{1 + x} = 1 + \frac{x}{2} + o(x)$
$$\lim_{n\to\infty}3n\left(1 + \frac{1}{9n} + o\left(\frac{1}{n}\right) - 1\right) = \lim_{n\to\infty}\frac{3n}{9n} = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Differentiate the Function: $y=2x \log_{10}\sqrt{x}$ $y=2x\log_{10}\sqrt{x}$
Solve using: Product Rule $\left(f(x)\cdot g(x)\right)'= f(x)\cdot\frac{d}{dx}g(x)+g(x)\cdot \frac{d}{dx}f(x)$
and $\frac{d}{dx}(\log_ax)= \frac{1}{x\ \ln\ a}$
$(2x)\cdot [\log_{10}\sqrt{x}]'+(\log_{10}\sqrt{x})\cdot [2x]'$
$y'=2x\frac{1}{\sqr... | Hint: You'll need to use the product rule with $2x$ and the logarithmic function.
Then, whilst applying the product rule, you use the chain rule on the logarithmic function.
Notice: differentiating $\log_a f(x)$ gives you (this is where you made your mistake) $$(\log_a f(x))' = \frac{f'(x)}{\ln a \cdot f(x)}$$
Use tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
Show that $\frac{1}{b-a_{1}} + \frac{1}{b-a_2}+ ...+ \frac{1}{b-a_n} \geq \frac{n}{b-\frac{1}{n}(a_1+a_2+...+a_n)}$ Let, $b> \max\{a_1,a_2,...,a_n\}.$ Show that $\frac{1}{b-a_{1}} + \frac{1}{b-a_2}+ ...+ \frac{1}{b-a_n} \geq \frac{n}{b-\frac{1}{n}(a_1+a_2+...+a_n)}$
f is convex if $f(t_1x_1+t_2x_2+...+t_nx_n)\leq t_1f(... | Observe the inequality can be proven using the following inequality:
$(x_1+x_2+\cdots + x_n)\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\cdots + \dfrac{1}{x_n}\right)\geq n^2$ with $x_i = b-a_i > 0$. The answer follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding $[T_{|W_i}]_{C_i}$ Let $B=\{v_1,v_2,v_3\}$, a basis of $V$ above $\mathbb{R}$. Let $$ [T]_B = \left(\begin{array}{cccc} 6&-3&-2\\4&-1&-2\\10&-5&-3 \end{array}\right)$$
The characteristic polynomial is $f_T(x) = (x-2)(x^2+1)$. Hence, $m(x) = (x-2)(x^2+1)$
We have $W_1 = \ker (T-2I) = \text{span}\{v_1,2v_3\}$ and... | $ \{v_1 + 2v_3\}=\left(
\begin{array}{c}
2\\
0\\
4\\
\end{array}
\right)\\
\{v_1+v_2,v_3\} =\left(\begin{array}{c} 3\\
3\\
5\\
\end{array}\right),\left(\begin{array}{c} -2\\
-2\\
-3\\
\end{array}\right)$
and $T\left(
\begin{array}{c}
2\\
0\\
4\\
\end{array}
\right)\ = 2\left(
\begin{array}{c}
2\\
0\\
4\\
\end{array}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Point of intersection of $f(x)=\sin(2x)+\cos(2x)$ and the $x$-axis How can I algebraically (without looking at the graph) find the point of intersection of $f(x)=\sin(2x)+\cos(2x)$ and $x$-axis, in the interval $[0, \pi]$?
| Notice, $$f(x)=\sin 2x+\cos 2x$$ $$\implies f(x)=\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos 2x+\frac{1}{\sqrt{2}}\sin 2x\right)$$ $$=\sqrt{2}\left(\cos 2x\cos\frac{\pi}{4}+\sin 2x\sin\frac{\pi}{4}\right)$$ $$=\sqrt{2}\cos \left(2x-\frac{\pi}{4}\right)$$ Now, for the intersection of $f(x)$ with the x-axis, we have $f(x)=0$ $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 2
} |
Matrix Exponential and Logarithm Consider the following matrix $A$:
$A = \begin{bmatrix}
\cos^2(1) & -\sin(2) & \sin^2(1) \\
\cos(1)\sin(1) & \cos(2) & -\cos(1)\sin(1) \\
\sin^2(1) & \sin(2) & \cos^2(1)\\
\end{bmatrix}$
I want to find a matrix $B$ such that $\exp(B)=A$ (or essentially finding $\log(A))$. Is... | First check that one can diagonalize $A$, which should give something like
$$\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & \frac{\cos ^2(1)-2 i \sin (1) \cos (1)-\sin ^2(1)}{\cos ^2(1)+\sin ^2(1)} & 0 \\
0 & 0 & \frac{\cos ^2(1)+2 i \sin (1) \cos (1)-\sin ^2(1)}{\cos ^2(1)+\sin ^2(1)} \\
\end{array}
\right)$$
Calculate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s... | The effect of multiplication by $1/(1-x)$ to the sequence of coefficients is to calculate partial sums: if the original sequence is $c_0,c_1,\ldots$ then the new one is
$$ d_i = c_0 + \cdots + c_i. $$
The starting point is the sequence $1,0,0,\ldots$. Applying this operator twice, we get
$$
1,0,0,0,0,\ldots \\
1,1,1,1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 15,
"answer_id": 0
} |
Raising a number in Rectangular Form What is the value of $(-2 + 3i\sqrt3)^6$?
Answer is $4096$
Convert $(-2 + 3i\sqrt3)^6$ to Polar Form.
$${ (\sqrt{31} \angle 111.05)^6 }$$
I use something called De Moivre's Theorem
$${z^n = r^n( \cos(n\theta) + i\sin(n\theta) ) }$$
$${z^n = (\sqrt{31})^6( \cos(6\cdot 111.05) + i\sin... | First of all we know the following things:
$$a+bi=\left|a+bi\right|e^{\arg\left(a+bi\right)i}=\left|a+bi\right|\left(\cos\left(\arg\left(a+bi\right)\right)+\sin\left(\arg\left(a+bi\right)\right)i\right)$$
$$\left(-2+3i\sqrt{3}\right)^6=$$
$$\left(\left|-2+3i\sqrt{3}\right|e^{\arg\left(-2+3i\sqrt{3}\right)i}\right)^6=$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find $x$ if $\frac {1} {x} + \frac {1} {y+z} = \frac {1} {2}$ I found this question from past year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Find $x$ if \begin{align}\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac... | Choosing
$p = \frac{1}{x} $
$q = \frac{1}{y} $
$r = \frac{1}{z} $
will certainly help writing the equations more neatly, I guess.
Then your equations can be written as:
$$2p + 2qr = 1 \\ 3q + 3pr = 1 \\ 4r + 4pq = 1$$
using first two equations you get:
$$ 2p + 2qr = 3q + 3pr \\ \implies r = \frac{3q - 2p}{3p - 2q}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$\int\dfrac{dx}{x^2(x^4+1)^{3/4}}$
Evaluate $$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$
I thought of rewriting this as $$\large{\int\dfrac{dx}{x^5(1+\frac{1}{x^4})^{3/4}}}$$ and substituting $$u^4=\left(1+\frac{1}{x^4}\right)\Rightarrow u=\left(1+\frac{1}{x^4}\right)^{1/4}$$ and subsequently I got $$du=\dfrac{1}{4}\... | $$\int\frac{1}{x^5}\frac{1}{(1+\frac{1}{x^4})^{3/4}}dx$$
$$u=1+\frac{1}{x^4}$$
$$-\frac{1}{4}du=\frac{1}{x^5}$$
The integral in the variable $u$ is then
$$-\frac{1}{4}\int u^{-3/4}du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Simultaneous equations, $\frac{1}{x}+\frac{1}{y}=1$,$x+y=a$,$\frac{y}{x}=m$ By eliminating $x$ and $y$ from the following equations, I need to find the relation between $m$ and $a$.
\begin{align*}
\frac{1}{x}+\frac{1}{y}=1 \\
x+y=a \\
\frac{y}{x}=m
\end{align*}
I tried different ways, but cannot arrive at the answer. ... | Your system has three equations and only two unknowns, so it is probably incompatible. Nevertheless, it still can have solution for certain values of $a$ and $m$.
From the last equation, we get $y=mx$. With the second eq.:
$$(m+1)x=a$$
We see that the system is incompatible if $m=-1$ and $a\ne 0$. Also, if $m=-1$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h $, find $f(7)$. Problem :
Given a polynomial $f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h$
such that $f(1)= 1, f(2) =2 , f(3) = 3, f(4) =4, f(5)=5, f(6) =6$.
Find $f(7)$ in terms of $h$.
My approach:
We can put the values of $f(1) = 1$ in the given equation and $f(2) = 2$, etc. But ... | Since $f(x) - x$ is a polynomial of degree $6$ and has $6$ roots $1, 2, 3, 4, 5, 6$ by condition, we can factorize $f(x) - x$ as:
$$f(x) - x = C(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6).$$
Plug in $x = 0$ in the above expression, we have
$3 - 0 = C\times 6!$, hence $C = \dfrac{3}{6!}$.
Therefore,
$$f(7) = 7 + (f(7) -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 5
} |
How to rotate a whole rectangle by an arbitrary angle around the origin using a transformation matrix? Suppose, I have a 2D rectangle ABCD like the following:
$A(0,0)$, $B(140,0)$, $C(140,100)$, $D(0,100)$.
I want to rotate the whole rectangle by $\theta = 50°$.
I want to rotate it around the Z-axis by an arbitrary an... | I have solved my problem.
$$Rotation =
\begin{bmatrix}
\
0 & 0 & 0 \\
140 & 0 & 0 \\
140 & 100 & 0 \\
0 & 100 & 0 \\
\end{bmatrix}.\begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
unresolved partial fraction decomposition I'm having some trouble doing this partial fraction decomposition:
$$\frac{1}{t^3-2t+1}$$
using Ruffini rule i get:
$$\frac{1}{t^3-2t+1}= \frac{1}{(t-1)(t^2+t-1)}$$
i would like to decompose the previous result into partial fraction.
I did in this way:
$$\frac{1}{(t-1)(t^2+t-1... | Notice, In general $$\frac{1}{(ax+b)(px^2+qx+r)}=\frac{A}{ax+b}+\frac{Bx+C}{px^2+qx+r}$$ Now, factorizing the expression, we have $$\frac{1}{(t-1)(t^2+t-1)}=\frac{A}{t-1}+\frac{Bt+C}{t^2+t-1}$$ $$\implies \frac{1}{(t-1)(t^2+t-1)}=\frac{A(t^2+t-1)+(Bt+C)(t-1)}{(t-1)(t^2+t-1)}$$ $$\implies (A+B)t^2+(A-B+C)t-(A+C)=1$$ Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
coefficient of $x^{17}$ in the expansion of $(1+x^5+x^7)^{20}$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
find the coefficient of $x^{17}$ in the expansion of $(1+x^5+x^7)^{20}$
(A)3400 (B)3410 (C)3420 (D)3430 (E)3440
so ... | $(1+x^5+x^7)^{20}=\{(1+x^5)+x^7\}^{20}$
$=(1+x^5)^{20}+\binom{20}1(1+x^5)^{20-1}(x^7)^1+\binom{20}2(1+x^5)^{20-2}(x^7)^2+\cdots+(x^7)^{20}$
So the required sum will be
the coefficient of $x^{17}$ in $(1+x^5)^{20}$
$+\binom{20}1\cdot$ the coefficient of $x^{17-7}$ in $(1+x^5)^{20-1}$
$+\binom{20}2\cdot$ the coefficient... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Simplification of rational expressions I have the following expression:
$${2\over x-2} + {2 \over{x^2} -5x +6}$$
So I can simplify this as:
$${2 \over x -2} + {2 \over (x -3) (x-2)}$$
I make the common denominator to be ${(x-3)(x-2)}$
So I then apply ${(x-3)}$ to the left hand side which gives me:
$${2(x-3) + 2 \over (... | Notice, the following steps $$\frac{2}{x-2}+\frac{2}{x^2-5x+6}$$ $$=2\left(\frac{1}{x-2}+\frac{1}{(x-2)(x-3)}\right)$$ $$=\frac{2}{x-2}\left(1+\frac{1}{x-3}\right)$$ $$=\frac{2}{x-2}\left(\frac{x-3+1}{x-3}\right)$$ $$=\frac{2}{x-2}\left(\frac{x-2}{x-3}\right)$$ $$=\frac{2}{x-3}$$
Your book has the correct expression. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction that $\sum_{k=1}^{n} k^3 = \bigg( \sum_{k=1}^{n}k\bigg)^2$ Show the following for all positive integers using proof by induction:
$$\sum_{k=1}^{n} k^3 = \bigg( \sum_{k=1}^{n}k\bigg)^2$$
Base case (n = 1) passes: $1^3 = 1^2$
We assume the following: $$\sum_{k=1}^{p} k^3 = \bigg( \sum_{k=1}^{p}k\bigg)^... | We have $$\sum_{k=1}^{n} k^3 = \bigg( \sum_{k=1}^{n}k\bigg)^2$$
1. Putting $n=1$ in the above equality, we get $$\sum_{k=1}^{1} k^3 = \bigg( \sum_{k=1}^{1}k\bigg)^2$$ $$ (1)^3=(1)^2\iff 1=1$$
Hence, the equality holds for $n=1$
*Assuming that it holds for $n=p$ then we have
$$\sum_{k=1}^{p} k^3 = \bigg( \sum_{k=1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Show that $B$ is invertible if $B=A^2-2A+2I$ and $A^3=2I$ If $A$ is $40\times 40$ matrix such that $A^3=2I$ show that $B$ is invertible where $B=A^2-2A+2I$.
I tried to evaluate $B(A-I)$ , $B(A+I)$ , $B(A-2I)$ ... but I couldn't find anything.
| Here's something I call the "miracle method" for this type of problem. Suspend your disbelief for a moment and suppose $A$ and $B$ were scalars, not matrices. Then, we would simply be looking for
$$ \frac{1}{B} = \frac{1}{A^2 - 2A - 2} = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots$$
where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 5
} |
Proving the inequality $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$
Show that for any two positive real numbers $a$ and $b$, $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$
My attempt: $(\sqrt a-\sqrt b)^2\geq0\\\frac{a+b}{2}\geq \sqrt{ab}$
$(a-b)^2\geq0\\a^2+b^... | By Bernoulli's Inequality,
$$\begin{align}
\sqrt{\frac{a^2+b^2}{2}}
&=\sqrt{\frac{(a+b)^2+(a-b)^2}{4}}=\frac{a+b}{2}\left(1+\left(\frac{a-b}{a+b}\right)^2\right)^{\frac12}
\\
&\leq \frac{a+b}{2}\left(1+\frac{1}{2}\left(\frac{a-b}{a+b}\right)^2\right)\,.
\end{align}$$
Thus,
$$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}\leq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means t... | Use the lagrange multiplier !
minimize the function:
$g(x,y,\lambda) = f(x,y)+\lambda(x^2+y^2-1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
If $\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)+1=0$,show that $\alpha-\beta$ or $\beta-\gamma$ or $\gamma-\alpha$ is multiple of $\pi$. This question is from SL Loney.
If $\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)+1=0$,
then show that $\alpha-\beta$ or $\beta-\gamma$ or $\gamma-\alpha$ is... | using your abbreviations, since
$$
\cos A + \cos B + \cos (A+B) + 1 = 0
$$
we have
$$
(1+\cos A)(1+\cos B) = \sin A \sin B
$$
this leads directly to
$$
\cos \frac{A}2 \cos \frac{B}2 (\cos \frac{A}2 \cos \frac{B}2 -\sin \frac{A}2 \sin \frac{B}2)=0
$$
hence either $A=n\pi$ or $B=n\pi$ or
$$
\cos \frac{A}2 \cos \frac{B}2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Compare $e^2$ and $7$ without using calculator Which is bigger? $e^2$ or $7$? Any tricks?
Don't know quite how to approach those kind of things.
| Here's an alternative. Try to compare $2$ and $\ln(7)$. $\ln(x)$ has a more natural geometric interpretation as area under $y=\frac{1}{x}$, compared to $e^x$, so some geometry might avail us.
Since $\ln(7)=\int_1^7\frac{1}{x}\,dx$, we consider approximating the integral with a trapezoid rule style Riemann sum. The tra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
$\frac{1}{x+1}+\frac{1}{y+1} +\frac{1}{z+1}$ minimum value if $xyz$ =k.$x,y,z$ are positive reals. $\frac{1}{x+1}+\frac{1}{y+1} +\frac{1}{z+1}$ minimum value if $xyz$ =k.$x,y,z$ are positive reals.I think the minimum should be when $x=y=z=k^{1/3}$. How do I show it? I tried to use AM-GM inequality but it doesnt seem to... | First, note that it is not true that always the function takes minimum when $x=y=z$. For e.g., let $r = \sqrt[3]k, \; x = y = rt,\, z= \dfrac{r}{t^2}$ for $r, t>0$. Then our objective function is
$$f(t) = \frac1{rt+1}+\frac1{rt+1}+\frac{t^2}{r+t^2}$$
Now consider $t \to \infty$. Clearly $f(t) \to 1$. Hence for $\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how to solve $3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$ $$A = 3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$$
My answer is:
$$\begin{align}
&A = 3 - \frac {2}{A}\\
\implies &\frac {A^2-3A+2}{A}=0\\
\implies &A^2-3A+2=0\\
\implies &(A-1)\cdot(A-2)=0\\
\implies &A=1\;\text{ ... | Note that as we add more terms to the continued fraction, it oscillates between $1$ and slightly higher than $2$.
$$
\begin{align}
n&=1&
3&=3&
3-2&=1\\\\
n&=2&
3-\cfrac23&=\frac73&
3-\cfrac{2}{3-2}&=1\\\\
n&=3&
3-\cfrac{2}{3-\cfrac23}&=\frac{15}7&
3-\cfrac2{3-\cfrac2{3-2}}&=1\\\\
n&=4&
3-\cfrac2{3-\cfrac2{3-\cfrac23}}&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Quadratic formula does not work If I put the equation: $5x^2-x-4 =0$ in the quadratic formula, than I get
$x = 1$ or $x = \frac{-4}{5}$
but the real zeros are: $x = -1$ or $x = \frac{4}{5}$
Can somebody explain me if the quadratic formula fails or me?
| The quadratic formula comes from "completing the square". Let's do that to your examples:
\begin{eqnarray*}
5x^2-x-4 &=& 0 \\ \\
x^2 - \tfrac{1}{5}x - \tfrac{4}{5} &=& 0 \\ \\
x^2 - \tfrac{1}{5}x &=& \tfrac{4}{5} \\ \\
\left(x - \tfrac{1}{10}\right)^{2} - \tfrac{1}{100} &=& \tfrac{4}{5} \\ \\
\left(x - \tfrac{1}{10}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
evaluate the integral Evaluate the integral:
$$\int_0^{\pi} \frac{\cos 2\theta}{1 -2a\cos \theta +a^2}d\theta$$
The way I approach this problem is:
since, $\cos \theta = \frac{e^{it} + e^{-it}}{2}$; and $cos 2\theta = Re(z^2)$. Then, the integral will be written as follow:
$$\frac{1}{2}\int_0^{2 \pi} \frac{Re(z^2)}{... | Another way is to use the following Fourier expansion: for $|a| < 1$,
$$ \frac{1-a^2}{1 - 2a\cos\theta + a^2} = \sum_{n=-\infty}^{\infty} a^{|n|} e^{in\theta} = 1 + 2 \sum_{n=1}^{\infty} a^n \cos(n\theta). $$
This series immediately gives us
$$ \int_{0}^{\pi} \frac{\cos n\theta}{1 - 2a\cos\theta + a^2} \, dx = \frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$ $\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$
$(A)0\hspace{1cm}(B)\frac{-\pi}{2}\hspace{1cm}(C)\frac{\pi}{2}\hspace{1cm}(D)\frac{7\pi}{2}$
I tried and got the answer but my answer is not matching the options given.Is my method n... | Hint:
$$\arcsin{a} - \arccos{(-a)} = \arcsin{a}+\arcsin{(-a)} - \frac{\pi}{2} = - \frac{\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understa... | First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}\frac{1}{k(k+2)}=\frac34-\frac{2+3}{2(1+1)(1+2)}$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}\frac{1}{k(k+2)}=\frac34-\frac{2n+3}{2(n+1)(n+2)}$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}\frac{1}{k(k+2)}=$
$\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Find $\cos{A}+\cos{B}$
In $\Delta ABC$,if
$$\cos{C}\cdot(\sin{A}+\sin{B})=\sin{C}\cdot\cos{(A-B)}$$
Find $\cos{A}+\cos{B}$
Thus
$$\sin{A}+\sin{B}=\tan{C}\cdot\cos{(A-B)}=\dfrac{\sin{(A+B)}}{\cos{(A+B)}}\cos{(A-B)}$$
$$(\sin{A}+\sin{B})\cos{(A+B)}=\sin{(A+B)}\cos{(A-B)}$$
follow I can't figure it out.
| Let me try. $\cos C (\sin A +\sin B)=\sin C \cos(A-B)$
$\implies2\left(1-2\sin^2\dfrac C2\right)\sin\dfrac{A+B}2\cos\dfrac{A-B}2 = 2\sin \dfrac C2\cos\dfrac C2\left(2\cos^2\dfrac{A-B}2-1\right)\ \ \ \ (1)$
Now as $\sin\dfrac{A+B}2=\sin\left(\dfrac\pi2-\dfrac C2\right)=\cos\dfrac C2$
and $\cos\dfrac C2=0\implies\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $x^3+y^3=72$ and $xy=8$ then find the value of $x-y$. I recently came across a question,
If $x^3+y^3=72$ and $xy=8$ then find the value of $(x-y)$.
By trial and error I found that $x=4$ and $y=2$ satisfies both the conditions. But in general how can I solve it analytically? I tried using $a^3+b^3=(a+b)(a^2+b^2-ab)... | Hint:
$$x^3+y^3=72\text{ and }x^3y^3=512.$$
You know the sum, and you know the product of two cubes.
The cubes are $\dfrac{72\pm\sqrt{72^2-4\cdot 512}}2$, $8$ and $64$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Completing the square of $(x+a)(x+b)$ The problem is simple, to complete the square of $(x+a)(x+b)$. My calculations yield
$$\left(x+\frac{a+b}{2}\right)^2-\frac{(a+b)^2}{4}+ab,$$
But the textbook's answer is different ("problem 361", at the bottom of the page):
$$\left(x+\frac{a+b}{2}\right)^2-\frac{(a-b)^2}{4}$$
Did ... | You did nothing wrong.
Note that
$$-\frac{(a+b)^2}{4}+ab=\frac{-(a+b)^2+4ab}{4}=\frac{-a^2+2ab-b^2}{4}=-\frac{(a-b)^2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find $\int \frac{5x^4+4x^5}{(x^5+x+1)^2}$ $\displaystyle \int \frac{5x^4+4x^5}{(x^5+x+1)^2}$
Since in the denominator of the integrand,$(x^5+x+1)^2$ is there.So the answer must be in the form $\displaystyle \frac{f(x)}{(x^5+x+1)}$, but i could not figure out what $f(x)$ i should suppose.I could not imagine any other wa... | $\displaystyle I = \int \frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \int\frac{5x^4+4x^5}{x^{10}\cdot \left(1+x^{-4}+x^{-5}\right)}dx = \int\frac{(5x^{-6}+4x^{-5})}{(1+x^{-4}+x^{-5})^2}dx$
Now Put $(1+x^{-4}+x^{-5}) = t\;,$ Then $\displaystyle \left(4x^{-5}+5x^{-6}\right)dx = -dt$
So $\displaystyle I = -\int\frac{1}{t^2} = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Family of Lines Problem:
Consider a family of straight lines $(x+y)+\lambda(2x-y+1)=0$. Find the equation of the straight line belonging to this family which is farthest from $(1,-3)$.
$$$$
Any help with this problem would be really appreciated!
| Notice, the equation of straight lines $$x+y+\lambda(2x-y+1)=0$$ $$(2\lambda +1)x+(1-\lambda)y+\lambda=0$$
Now, the distance say $D$ of the above line from the given point $(1, -3)$ is given as follows $$D=\left|\frac{(2\lambda +1)(1)+(1-\lambda)(-3)+\lambda}{\sqrt{(2\lambda +1)^2+(1-\lambda )^2}}\right|$$
$$=\left|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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For what $p$ is $\frac{1}{(x(1+\ln(x)^2))^p}$ Lebesgue integrable? I'm trying to use the fact that given $f:[a,\infty)\to\mathbb{R}$ Riemann integrable for every closed interval $[c,d]\subset [a,\infty)$, then $f$ is Lebesgue integrable if, and only if, $\int_a^\infty|f(x)| \, dx$ exists.
In particular, $f(x)=\frac{1}{... | Using the inequality $\frac{x}{x+1} < \log (1+x) < x$ for ll $x>-1$ and $x \ne 0$. Let $x=u-1$ then the inequality becomes $\frac{u-1}{u} < \log (u) < u-1$ for all $u>0$. Thus, $\frac{(x-1)^2}{x^2}< (\ln x)^2<(x-1)^2$ so that $1+\frac{(x-1)^2}{x^2}<1+ (\ln x)^2<1+(x-1)^2$ which implies that $\frac{1}{1+\frac{(x-1)^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How to solve the equation $x^3+y^3=0$ for real numbers $x$ and $y$? I'm finding stationary points of the function $f(x,y)=2(x-y)^2-x^4-y^4$, but stuck in the equation $x^3+y^3=0$ while solving the equations $f_x=0$ and $f_y=0$.
Please help me. Thanks in advance.
| HINT:
Notice, $$x^3+y^3=(x+y)(x^2+y^2-xy)$$ Then, we have $$(x+y)(x^2+y^2-xy)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find the rightmost 25 digits in $100!$? The question is
Find the rightmost 25 digits in decimal expansion of $100!=1\times 2\times \dotsb \times100$
By counting the number of fives in the prime factorisation of $100!$, I know there are $\lfloor {100 \over 5} \rfloor + \lfloor {100 \over {5^2}} \rfloor =24$ tr... | Just figured out this myself.
It is easy to determine that there are 24 trailing zeros, so the only thing left is to determine $\frac{100!}{10^{24}}\pmod {10}$
Writing the fraction as $$\frac{(1 \cdot 3 \cdot 7 \cdot 9 \ \dotsc \cdot 99)(5 \cdot 10 \cdot 15 \cdot \dotsc \cdot 100)(2 \cdot 4 \cdot 6 \cdot 8 \cdot \dotsc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find inverse of 15 modulo 88. Here the question: Find an inverse $a$ for $15$ modulo $88$ so that $0 \le a \le 87$; that is, find an integer $a \in \{0, 1, ..., 87\}$ so that $15a \equiv1$ (mod 88).
Here is my attempt to answer:
Find using the Euclidean Algorithm, we need to find $\gcd(88, 15)$, that must equal to $1$ ... | $15n=88k+1$. But $15n$ ends either in $0$ or in $5$. So we are looking for a multiple of $88$ which ends either in $9$ or in $4$. The former case is impossible, since $88$ is an even number. Now, what is the first multiple of $8$ to end in a $4$ ? So $k_0=3$ is our first suspect. Unfortunately, in this case, $n_0\not\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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$\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+.....+\cos(\alpha+(n-1)\beta)=0 $ If each side of a regular polygon of $n$ sides subtend an angle $\alpha$ at the center of the polygon and each exterior angle of the polygon is $\beta$,then prove that $\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+.....+\cos(\al... | Notice, in a regular polygon with $n$ number of side,
the angle subtended by each side at the center of the polygon $$\alpha=\frac{2\pi}{n}$$
each exterior angle of the polygon $$\beta=\pi-\frac{(n-2)\pi}{n}=\frac{2\pi}{n}$$
$$\implies \alpha=\beta=\frac{2\pi}{n}$$ Now, we have
$$\cos \alpha+\cos(\alpha+\beta)+\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluation of $\int\frac{x^7+2}{\left(x^2+x+1\right)^2}dx$
Evaluation of $$\int\frac{x^7+2}{\left(x^2+x+1\right)^2}dx$$
$\bf{My\; Try::}$ Let $$\displaystyle \mathop{I = \int\frac{x^7+2}{(x^2+x+1)^2}}dx = \int\frac{(x^7-1)+3}{(x^2+x+1)^2}dx$$
$$\mathop{\displaystyle = \int\frac{x^7-1}{(x^2+x+1)^2}}+\displaystyle \in... | Here is a sketch - not a full solution:
Say we know $\int\cos^2x\tan^jx dx$ for $j=0,1,...,7$. Then using $x+\frac12=\frac{\sqrt3}{2}\tan u$ we have \begin{align}
\int\frac{x^7+2}{(x^2+x+1)^2}dx&=\int\frac{x^7+2}{\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right)^2}dx\\
&=\frac{\sqrt3}{2}\int\frac{\Big(\frac{\sqrt3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$9 \mid 4n^2 + 15n - 1$ for $n \in \mathbb N$ How to prove by induction that $9 \mid 4n^2 + 15n - 1$ for every $n \in \mathbb N$?
For $n = 1$
$4 \cdot 1^2 + 15 \cdot 1 - 1 = 18$
For $n \ge 2$
If $4n^2 + 15n - 1 = 9k$ then $4(n+1)^2 + 15(n+1) - 1 = 4n^2 + 23n + 18 = 9k + 8n + 19$
| If $4n^2+15n-1$ is divisible by $9$, then it's also divisible by $3$; however
$$
4n^2+15n-1\equiv n^2-1\pmod{3}
$$
but $n^2\equiv 1\pmod{3}$ if and only if $3\nmid n$. So, for $n=3k$, the number $4n^2+15n-1$ is not divisible by $3$ and, of course, not divisible by $9$ either.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Linear transformation using non-standard basis Find a linear transformation $T:P_{2}({R})\longrightarrow P_{4}(R)$ so that
$T(1) = x^4$
$T(x+x^2) = 1$
$T(x-x^2) = x+x^3$
I have only solved problems using standard basis, and now I have no idea on how to deal with this.
| You have that
$$x = \frac{(x+x^2) + (x-x^2)}{2}$$
So
$$T(x) = T\left( \frac{(x+x^2) + (x-x^2)}{2} \right) = \frac{T(x+x^2) + T(x-x^2)}{2}$$
$$ = \frac{1+x+x^3}{2}$$
And
$$T(x^2) = T(x+x^2-x) = T(x+x^2)-T(x)$$
$$=1 - \frac{1+x+x^3}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving a rational equation with multiple and nested fractions
This is the equation to solve: $\dfrac{\dfrac{x+\dfrac{1}{2}} {\dfrac{1}{2}+\dfrac{x}{3}}}{\dfrac{1}{4}+\dfrac{x}{5}}=3$
What I did:
$x+\dfrac{1}{2}=\dfrac{2x+1}{2}$
$\dfrac{x}{3}+\dfrac{1}{2}=\dfrac{2x+3}{6}$
$\dfrac{x}{5}+\dfrac{1}{4}=\dfrac{4x+5}{20}$
... | Notice, $$\frac{\frac{x+\frac{1}{2}}{\frac{1}{2}+\frac{x}{3}}}{\frac{1}{4}+\frac{x}{5}}=3$$
$$\frac{\frac{3(2x+1)}{(2x+3)}}{\frac{4x+5}{20}}=3$$
$$\frac{60(2x+1)}{(2x+3)(4x+5)}=3$$
$$8x^2-18x-5=0$$
$$x=\frac{18\pm\sqrt{(-18)^2-4(8)(-5)}}{2(8)}$$
$$x=\frac{18\pm22}{16}$$ $$x=\frac{5}{2}$$ or $$ x=-\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the value of function at $\,x=5$ If $\,f(x)\,$ is a non-constant polynomial of $\,x\,$ such that $\,f\left(x^3\right)-f\left(x^3-2\right)=f^2\left(x\right)+12\,$ is true for all $\,x\,$ then find the value of $\,f\left(5\right).\,$
| HINT: Substitute $\,f(x) = ax^3 + bx^2 + cx + d$ into your equation and collect coefficients of the different powers of $x$.
Solve obtained system of linear equations, thus getting an explicit expression for $f$.
Substitute $\,x = 5\,$ into obtained equation and get the answer.
How do we know that $f$ is the third-o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Probability that the eventually a six on a dice will appear.
Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $ m$ and $ n$ be relatively prime positive integers such that $ \frac{m}{n}$ is the probabilit... | Let $D, L$ be the obvious random variables. Then the desired probability is
\begin{align}
P & = P(D = 1, L = 1) + P(D = 1, L = 2) + P(D = 2, L = 1) \\
& + P(D = 2, L = 2) + P(D = 2, L = 3) + P(D = 3, L = 2) \\
& + P(D = 3, L = 3) + P(D = 3, L = 4) + P(D = 4, L = 3) \\
& + \cdots
\end{align}
Each row constitutes ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving uniqueness of solutions to $\sin^2A + \sin^2B = \sin (A+B)$ without using multivariable calculus In the course of solving a trigonometric problem (see $a^2+b^2=2Rc$,where $R$ is the circumradius of the triangle.Then prove that $ABC$ is a right triangle), in one approach the following equation needed to be solve... | Well, fix $B$ and show that there's a unique solution for $A$. In other words, we want to show that for $0 < B < \pi/2$ the function
$$f_B(A) = \sin^2(A) + \sin^2(B) - \sin(A+B)$$
has a unique zero on the interval $(0, \pi/2)$.
First, note that
$$f_B(0) = \sin^2(B) - \sin(B) = \sin(B)[\sin(B) - 1] < 0$$
since $0 < \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Polynomial with real roots Consider the polynomial: $$f=X^4+4X^3+6X^2+aX+b$$
We know that $f$ has four real roots. Let $x_1,x_2,x_3,x_4$ be the roots of this polynomial. How can one compute $$x_1^{2015}+x_2^{2015}+x_3^{2015}+x_4^{2015}?$$
If $a=4$ and $b=1$, we obtain a self-reciprocal (palindromic) polynomial. We can ... | We know the following to be true:
$$
\tag1x_1+x_2+x_3+x_4 = -4
$$
$$
\tag2x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1x_3+x_2x_4 = 6
$$
$$
\tag3x_1 x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2 = -a
$$
$$
\tag4x_1x_2x_3x_4 = b
$$
Now we can square $(1)$ to get
$$
\begin{align}
(x_1+x_2+x_3+x_4)^2 = &x^2_1+x^2_2+x^2_3+x^2_4\\
&+2[x_1x_2+x_2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Trying to solve the trig equation $\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$ The equation is
$$\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$$
My solution goes like this
$$
\begin{cases}
3+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x) \\
\frac{\sin(x)}{\sqrt 3}+3\cos(x) \g... | $$4t^2-3\sqrt3t-3=0$$
$$\implies t=\dfrac{3\sqrt3\pm\sqrt{(3\sqrt3)^2-4\cdot4(-3)}}{2\cdot4}=\dfrac{3\sqrt3\pm5\sqrt3}8=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Integrating linear/trigonometric I have the following question-
$$\int \frac{x}{1+\cos x}\,\text{d}x$$
Do I do integration by parts or is there some other method?
Thanks for the help.
| Notice, we have $$\int\frac{x}{1+\cos x}dx$$ $$=\int\frac{x}{1+\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}dx$$
$$=\int\frac{x\left(1+\tan^2\frac{x}{2}\right)}{1+\tan^2\frac{x}{2}+1-\tan^2\frac{x}{2}}dx$$
$$=\int\frac{x\sec^2\frac{x}{2}}{2}dx$$
$$=\frac{1}{2}\int x\sec^2\frac{x}{2}dx$$
$$=\frac{1}{2}x \int \sec^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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dice probability - same 2 dice in 6 dice rolls I have this simple probability problem that I am not sure I solved correctly. I am not interested in formulas, but rather the thought process of how to solve it.
Suppose we roll six 6-sided dice that are equal. I want to find the probability that at least two dice have the... | When you say "two with the same dice face" do you mean "at least two"? If so then it is simply 1 minus the probability they are all different. The first one will come up with something. The probability the second is different is $\frac{5}{6}$. The probability the third is different from either of the first two is $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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A geometry question on finding the area of cyclic quadrilaterals The circumcircle of a cyclic quadrilateral $ABCD$ has radius $2$.
$AC, BD$ meet at $E$ such that $AE = EC$. If $AB^2 = 2\cdot AE^2$ and $BD^2 = 12$, what is the area of the quadrilateral?
| Let $O$ be the circumcentre of $ABCD$ and $R$ the circumradius. Since $BD=R\sqrt{3}$, $\widehat{DAB}$ and $\widehat{DCB}$ are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, in some order. Let $AE=x$. We have $AC=2x$ and $AB=x\sqrt{2}$, so:
$$ \arcsin\frac{2x}{4}=\arcsin\frac{x\sqrt{2}}{4}+\arcsin\frac{BC}{4} $$
and:
$$ BC = 2x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum value of $\cos x+\cos y+\cos(x-y)$ What is the minimum value of $$ \cos x+\cos y+\cos(x-y). $$ Here $x,y$ are arbitrary real numbers. Mathematica gives (with NMinimize) $-3/2$. But I don't know if this is correct and if so, how to prove it.
| Since cosine is even, the problem is equivalent to
Minimize $\cos(-x)+\cos(y)+\cos(x-y)$.
which is equivalent to
Minimize $\cos a + \cos b + \cos c$ subject to $a+b+c=0$.
Since cosine is periodic with period $2\pi$, this is equivalent to
Minimize $\cos a + \cos b + \cos c$ subject to $a+b+c\equiv 0\pmod{2\pi}$.
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Integrate $\tan^2(\frac{\pi}{12} \cdot y)$ Integrate $\tan^2(\frac{\pi}{12} \cdot y)$
Wolfram gives the answer:
$$\frac{12 \tan(\frac{\pi y}{12})}{\pi} -y + \text{constant}$$
But why is the value of $\tan^2$ not getting differentiated?
According to this rule, the answer should be:
$$\frac{\tan(\frac{\pi}{12} \cdot y)}{... | Recall that $\int \tan^2 x\ dx = \int {\sin^2 x \over 1 - \sin^2 x} dx = -x + \tan x$, we find:
$\int \tan^2 \left( {\pi y \over 12} \right) d y = \frac{12 \tan \left(\frac{\pi y}{12}\right)}{\pi }-y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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There are 30 tokens numbers from 0 to 30. Find the number of ways of choosing 3 tickets such that the sum of the numbers on the tokens is 30 Then the number of solutions is divisible by
(A) 2 (B) 3 (C) 5 (D) 7
One way to solve its by finding the number of unequal integral solutions of $x+y+z=30$.
The possible cases a... | To determine three such tickets, we can pick two numbers $a<b$ in $\{1,\ldots,29\}$, which is possible in $29\choose 2$ ways; then let the tickets be $a,b-a,30-b$. Then do some inclusion and exclusion: Subtract the $14$ cases where $a=b-a$ ($1\le a\le 14$, $b=2a$); subtract the $14$ cases where $a=30-b$ ($1\le a\le 14$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Which answer is correct for this product rule based probability problem The question reads like this:
A bag contains 5 black and 3 red balls. A ball is taken out of the bag and is not returned to it. If this process is repeated three times, then what is the probability of drawing a black ball in the next draw?
Soluti... | You are overcomplicating things. There are eight balls, and any one of which could be the fourth ball drawn with equal probability. Five of these balls are black. The probability that a black ball is drawn on the fourth draw is thus: $$5/8$$
This agrees with your second solution, which correctly considers that t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Trapped in Induction, how to get out? Example 1:
Prove by induction that $1+3+5+...+(2n-1)=n^2 \text{ for all } n \in \mathbb{N}....(*)$
Proof:
Step 1: For $n=1$, left-side we have $(2(1)-1) = 1$. Right-side we have $(1)^2 = 1$.
Step 2: Suppose (*) is true for some $n=k \in \mathbb{N}$ that is
$$1+3+5+...+(2k-1)=k^2$$... | In Step 3 of the second example, you write "Prove that (*) holds true for $n = k \in N$", but that should be $n = k + 1$.
I find it useful in proofs like this to write down something I call $P(n)$, the proposition that I want to prove, which is typically meant to be true for every integer $n$. In your example 3, $P(n)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Fourth degree polynomial in trajectory problem I have a problem where I'm trying to find the distance needed to lead a moving target with a projectile given variable parameters. I've had limited success so far and I think I may be close to a solution but I'm not sure how to go about the final steps involving a fourth d... | I managed to find the equation for the solution of a depressed fourth degree polynomial on WolframAlpha and it yielded the expected results:
Let $$a = \frac{g^2}{4\,v_1^4}$$ $$c = 1-\frac{v_2^2}{v_1^2}$$ $$d = -2\,d_1\,cos(a_1)$$ $$e = d_1^2$$
$$y = \sqrt[3]{\sqrt{(-72\,e\,a\,c + 27\,a\,d^2 + 2\,c^3)^2 - 4(12\,e\,a + c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find 2 sums with the binomial newton Find the sum of:
i)$\displaystyle\sum_{k=0}^{n} k^2$ $\left(\begin{array}{c} n\\k\end{array}\right)$
ii) $\displaystyle\sum_{k=1}^{n} \frac{2k+5}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$
Thoughts:
i)(After the Edit)$\displaystyle\sum_{k=0}^{n} k^2$ $\left(\begin{array}... | For the first one:
We know that:
$$
\sum_{k=0}^{n}{{n\choose k}x^k}={(1+x)}^{n}
$$
So let's differentiate it to get:
$$
\sum_{k=0}^{n}{k{n\choose k}{x}^{k-1}}=n{(1+x)}^{n-1}
$$
Multiply by $x$ and differntiate again:
$$
\sum_{k=0}^{n}{k^2{n\choose k}{x}^{k-1}}=(nx{(1+x)}^{n-1})'\\
\sum_{k=0}^{n}{k^2{n\choose k}{x}^{k-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Solving $\sqrt{i}$ The results I get not the same as in the book, basically I need to get only this and the $x$'s of course:
$$y = \frac{\sqrt{2}}{2}$$
So this is what I did:
$$\sqrt{i} = x+yi$$
$$i = x^2 +2xyi -y^2$$
$$\begin{cases} x^2-y^2 = 0 \\
2xy = 1 \end{cases}$$
$$xy=0.5$$
$$x=\frac{0.5}{y}$$
$$(1)\space\spac... | You have $y = \pm \dfrac{\sqrt 2} 2$.
If $y= \dfrac{\sqrt 2} 2$ then $x=\dfrac{0.5} y = \dfrac{0.5}{\sqrt 2/2}$, and this simplifies to $x=\dfrac{\sqrt 2} 2$.
So $\dfrac{\sqrt 2} 2 + i \dfrac{\sqrt 2} 2$ is a complex number whose square is $i$.
If you multiply that by $-1$ you get the other complex number whose square ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Integration of an unusual trig function I am trying to solve the following definite integral:
$$ \int_{0}^{\frac{\pi}{4}} \sqrt{\tan^2t + \frac{1}{2}} \, dt $$
I have tried the usual trig substitutions with no avail.
| Let $$\displaystyle I = \int \sqrt{\tan^2 t +\frac{1}{2}}dt\;,$$ Now Put $\displaystyle \tan t = \frac{1}{\sqrt{2}}\tan \phi\;,$
Then $$\displaystyle dt = \frac{1}{\sqrt{2}}\frac{\sec^2 \phi}{\sec^2 t} d\phi =\frac{1}{\sqrt{2}}\frac{\sec^2 \phi}{1+\tan^2 t}d\phi = \sqrt{2}\left(\frac{\sec^2 \phi}{2+\tan^2 \phi}\right)d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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$2(n-2)+1$ does not divide $(n-2)(n-3)/2$ for $n \ge 8$
For $n \ge 8$ the number $2(n-2)+1$ never divides $(n-2)(n-3)/2$.
Any ideas how to prove this? I see that $(n-2)(n-3)/2 = 1 + 2 + \ldots + (n-3)$. If I suppose that $2(n-2)+1$ divides $(n-2)(n-3)/2$ then it should also divide their difference
\begin{align*}
(n... | Note that $2n-3$ is odd, so it is enough to show it does not divide $4(n-2)
(n-3)$. which is $4n^2-20n+24$. Do a polynomial division. We get $2n-7+\frac{3}{2n-3}$.
If $2n-3\gt 3$, then $\frac{3}{2n-3}$ is not an integer.
Another way: Let $n\gt 3$. Note that $2(n-2)+1$ and $n-2$ are relatively prime. So it is enough t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What are better approximations to $\pi$ as algebraic though irrational number? I know that $\pi \approx \sqrt{10}$, but that only gives one decimal place correct. I also found an algebraic number approximation that gives ten places but it's so cumbersome it's just much easier to just memorize those ten places.
What's a... | Some nice approximations can be produced by exploiting the ideas of Archimedes. The difference between a unit circle and an inscribed regular $n$-agon is made by $n$ circle segments. If we approximate them with parabolic segments and call
$$ A_n = \frac{n}{2}\sin\frac{2\pi}{n}=n\sin\frac{\pi}{n}\cos\frac{\pi}{n} $$
the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 9,
"answer_id": 5
} |
Finding roots of a complex number I am asked to find $2^{1/5} $. Using the formula, and noticing $2 = 2e^{0 i} $, we have
$$ 2^{1/5} = 2^{1/5} \bigg( \cos( \frac{ 0}{5} + \frac{ 2 \pi k}{5} ) + i \sin( \frac{ 0}{n} + \frac{ 2 \pi k}{n} ) \bigg) $$
My confusing. I notice we can also write $2 = 2e^{2 \pi k} = 2e^{4 \pi ... | One way to attack this baby is to remember that the number of $n$-th complex roots of a number is always $n$, and that those $n$ roots form the vertices of an $n$-sided regular polygon on the Complex plane. So there are $3$ cube roots, forming a triangle, $4$ cube roots, forming a square, $5$ fifth roots, forming a pe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find $\lim_\limits{x\to 0}{(\sqrt{f^2(x)+2f(x)+3}-f(x))}$ Let $f$ be a function such that $x^2f(x)\geq x^2+x+1,\forall x\in\mathbb{R^*}$. Find the value of: $$\lim_\limits{x\to 0}{\left(\sqrt{f^2(x)+2f(x)+3}-f(x)\right)}$$
I think that we may need to apply a sandwich theorem for the limit, so let $g(x)=\sqrt{f^2(x)+2f(... | If you want to use the sandwich theorem, observe that whenever $x$ is positive,
$f(x)$ also is positive and
\begin{align}
\left(f(x) + 1 + \frac{1}{f(x)}\right)^2
& = (f(x))^2 + 2f(x) + 3 + \frac{2}{f(x)} + \frac{1}{(f(x))^2} \\
& > (f(x))^2 + 2f(x) + 3 \\
& > \left(f(x) + 1\right)^2 \\
& > 0.
\end{align}
Therefore... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$\int \frac{\sqrt{1+x^2}}{1-x^2}dx$ Problem :
$\int \frac{\sqrt{1+x^2}}{1-x^2}dx$
My approach :
Put $x = \tan\theta$
we get $$\int \frac{\sqrt{1+x^2}}{1-x^2}dx = \frac{\frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta \cos\theta}}{\frac{\cos^2\theta -\sin^2\theta}{\cos^2\theta}} d\theta $$
$$= \frac{1}{(\cos^2\theta ... | Let $$\displaystyle I = \int\frac{\sqrt{1+x^2}}{1-x^2}dx = \int\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^2}}dx = -\int\frac{(x^2+1)}{(x^2-1)\sqrt{1+x^2}}dx$$
So $$\displaystyle I =-\int\frac{(x^2-1)+2}{(x^2-1)\sqrt{x^2+1}}dx = -\underbrace{\int\frac{1}{\sqrt{x^2+1}}dx}_{J}+2\underbrace{\int\frac{1}{(1-x^2)\sqrt{1+x^2}}dx}_{K}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
In tetrahedron ABCD, prove that $r<\frac{AB\cdot CD}{2AB+2CD}$
Given tetrahedron $ABCD$, let $r$ be the radius of the sphere inscribed tetrahedron. Prove that $$r<\dfrac{AB\cdot CD}{2\,AB+2\,CD}$$
This is difficult question, my friends and I could not find any hint to solve it.
In the case that $ABCD$ is a regular te... | Let $MN$ be the common perpendicular line of $AB$ and $CD$ with $M$ a point in $AB$ and $N$ in $CD$. The volume $V$ of the tetrahedron $ABCD$ can be calculated in two ways:
$$\begin{align}
V &= \frac16 \times AB \times CD \times MN \times \sin(\alpha) \\
&= \frac13 \times r \times (S_{\triangle ABC} + S_{\triangle ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find $n$, where its factorial is a product of factorials I need to solve $3! \cdot 5! \cdot 7! = n!$ for $n$.
I have tried simplifying as follows:
$$\begin{array}{}
3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\
(3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^3 \cdot 5^2 \cdot 4^2 \... | Notice, $\color{red}{3}$, $\color{red}{5}$, $\color{red}{7}$ are prime numbers (can't be factorized), then the factorials can be successively reduced as follows $$3!5!7!=(3\cdot 2!)\cdot (5\cdot 4!)\cdot (7\cdot 6!)$$
$$=\color{red}{3}\cdot \color{red}{5}\cdot \color{red}{7}\cdot (2!)\cdot (4!)\cdot (6!)$$
$$=\color{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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"answer_id": 3
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Evaluate the limit of ratio of sums of sines (without L'Hopital): $\lim_{x\to0} \frac{\sin x+\sin3x+\sin5x}{\sin2x+\sin4x+\sin6x}$ Limit to evaluate:
$$\lim_{x \rightarrow 0} \cfrac{\sin{(x)}+\sin{(3x)}+\sin{(5x)}}{\sin{(2x)}+\sin{(4x)}+\sin{(6x)}}$$
Proposed solution:
$$
\cfrac{\sin(x)+\sin(3x)+\sin(5x)}{\sin(2x)+\sin... | Use
$$
2·\cos x·\sin(kx) = \sin((k+1)x)+\sin((k-1)x)
$$
or
$$
2\sin(x/2)\sin(kx)=\cos((k-1/2)x)-\cos((k+1/2)x)
$$
or something similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Integral involving cube root and seventh root Find the value of $$\int_{0}^{1} (1-x^7)^{\frac{1}{3}}-(1-x^3)^{\frac{1}{7}}\:dx$$
My Approach:
Let $$I_1=\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dx$$ and
$$I_2=\int_{0}^{1} (1-x^3)^{\frac{1}{7}}dx$$
For $I_1$ substitute $x^7=1-t^3$ so $dx=\frac{-3t^2}{7}(1-t^3)^{\frac{-6}{7}}\:... | Hint. A straightforward approach is to use the Euler beta function
$$
B(a,b) := \int _0^1 x^{a-1}(1-x)^{b-1}dx =\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},\quad a>0, b>0.
$$ Then you get
$$
\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dt-\int_{0}^{1} (1-x^3)^{\frac{1}{7}}dt=0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Find the roots of the equation $x^2-[x]=4$ where $[x]$ is the greatest integer less than or equal to $x$. Show that there is no rational number $x$ satisfying the equation $x^2-[x]=4$. Find the roots of the equation $x^2-[x]=4$ where $[x]$ is the greatest integer less than or equal to $x$.
I tried to solve this equatio... | Write $x = [x] + \varepsilon$. Then you are trying to solve
$$n^2 + 2\varepsilon n + \varepsilon^2 - n - 4= 0 $$
For an integer value of $n$ and some $\varepsilon \in [0,1)$. By the quadratic formula we can express $\varepsilon$ in terms of $n$,
$$\varepsilon^2 + (2n)\varepsilon +[n^2 - n - 4]= 0 $$
$$\varepsilon = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Proving this trig identity:$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$
$$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$$
What I've tried,
$$\frac{((1+\cos\theta)+(\sin\theta))((1+\cos\theta)+(\sin\theta))}{(1+\cos\theta-\sin\theta... | from Second last line, $$\displaystyle \frac{(1+\sin \theta+\cos \theta)^2}{(1+\cos \theta)^2-\sin^2 \theta}=\frac{2\left[1+\sin \theta +\sin \theta\cdot \cos \theta+\sin \theta\right]}{2\cos \theta \cdot (1+\cos \theta)} = \frac{2(1+\cos \theta)(1+\sin \theta)}{2\cos \theta\cdot (1+\cos \theta)}$$
So we get $$\displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Real part of $\sqrt{ai-1}$ Is there a way to find the real and imaginary parts of
$$
z=\sqrt{ai-1},\qquad a>0
$$
where $i=\sqrt{-1}$. Thanks. I do not know what to to do.
Note
$$
i=e^{i\pi/2}=\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}.
$$
| Convert $ai-1$ into polar form. Let $ai-1 = r(\cos\theta + i\sin\theta)$. Then
$$\begin{align*}
a &= r\sin\theta\\-1 &= r\cos\theta\\
r^2 &= (r\sin\theta)^2 + (r\cos\theta)^2\\
&= a^2 + 1\\
r &= \sqrt{a^2 + 1}\\
\tan \theta &= \frac{r\sin\theta}{r\cos\theta}\\
&= -a\\
\theta &= \pi - \tan^{-1} a
\end{align*}$$
Let $r(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Inequality involving an exponent I wish to prove the following inequality
$$x^{\frac{3}{x-1}} > 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}, \quad x > 1.$$
Graphically the above inequality appears to be true since if one plots
$$g(x) = x^{\frac{3}{x-1}} - \left (1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} \rig... | Hint :
Consider the Taylor expansion at infinity of your exponent :
$$x^{\frac{3}{x-1}} = x ^{3\ \big(\frac{1}{x} +\frac{1}{x^2} + \frac{1}{x^3} +...\big)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1452300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Are there $a,b,c,d \in \mathbb N$ such that $\frac{a + b}{a + b + c + d} < \frac{a}{a + c} < \frac{b}{b + d}$? Consider the following $2 \times 2$ contingency table:
\begin{array}{c|cc|c}
& C & \overline C & \Sigma \\ \hline
V & 4000 & 3500 & 7500 \\
\overline V & 2000 & 500 & 2500 \\ \hline
\Sigma & 6000 & 4000 & 100... | No. In general, for $a,b,c,d\in \Bbb N$ with $\frac{a}{b} < \frac{c}{d}$, $$\frac{a}{b} <\frac{a+c}{b+d}<\frac{c}{d}.$$
Proof: The left inequality is equivalent to $$a(b+d) < b(a+c) \iff ad < bc \iff \frac{a}{b}<\frac{c}{d}.$$ Likewise for the right inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating the magnitude. Is this less than 1? Where $|x|<1$, I'm looking to determine if
$$\left|\frac{1}{x}(-1+\sqrt{1-x^2})\right|<1$$
I believe it is, since we can use a Taylor series to approximate
$$\sqrt{1-x^2} = 1 - \frac{1}{2}x^2-\frac{1}{4}(x^2)^2 + O(3) \approx 1 - \frac{1}{2}x^2$$
This then gives
$$\left|\f... | HINT:
Let $2y=\arcsin x,0\le2y\le\dfrac\pi2$
$\implies x=\sin2y,\cos2y=+\sqrt{1-x^2}$
$$\dfrac{\sqrt{1-x^2}-1}x=\dfrac{\cos2y-1}{\sin2y}=\cdots=-\tan y$$
Now $$0\le y\le\dfrac\pi4\implies0\le\tan y\le1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Is this a lower bound for this integral? Let $T$ be the triangular region in the $xy$ plane with vertices $(-2,0) (2,0) $ and $(0.2)$. Then is this inequality true? $$\displaystyle\int\int_T e^{-(x^2+y^2)}dA \geq \pi\left(1-\dfrac{1}{\sqrt{e}}\right)$$
I can inscribe the half disk $D=\{(x,y)\in\mathbb{R}^2$ | $x^2+y^2=... | Twice your integral is the integral over the square with vertices at $(\pm 2,0)$ and $(0,\pm 2)$, hence:
$$2I = \int_{-\sqrt{2}}^{+\sqrt{2}}\int_{-\sqrt{2}}^{+\sqrt{2}} e^{-(x^2+y^2)}\,dx\,dy =\pi\cdot\text{Erf}(\sqrt{2})^2\tag{1}$$
but due to the continued fraction representation for the error function,
$$ \text{Erf}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Li Shanlan's combinatorial identities I am struggling to prove the following combinatorial identities:
$$(1)\quad\sum_{r=0}^m \binom{m}{r}\binom{n}{r}\binom{p+r}{m+n} = \binom{p}{m}\binom{p}{n},\quad \forall n\in\mathbb N,p\ge m,n$$
$$(2)\quad\sum_{r=0}^m \binom{m}{r}\binom{n}{r}\binom{p+m+n-r}{m+n} = \binom{p+m}{m}\bi... | Suppose we seek to verify that
$$\sum_{r=0}^{\min\{m,n,p\}}
{m\choose r} {n\choose r}
{p+m+n-r\choose m+n}
= {p+m\choose m} {p+n\choose n}.$$
Introduce
$${n\choose r} = {n\choose n-r} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-r+1}}
(1+z)^n \; dz$$
and
$${p+m+n-r\choose m+n} = {p+m+n-r\choose p-r} \\ =
\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Prove using factor theorem. Using factor theorem, show that $a+b$,$b+c$ and $c+a$ are factors of
$(a + b + c)^3$ - $(a^3 + b^3 + c^3)$
How do we go about solving this ?
Thanks in advance !
| If a+b is a Factor then -b is a root and vice versa. so just set a=-b and you get $((-b)+b+c)^3 - ((-b)^3+b^3+c^3) =c^3-c^3 = 0$. Therefor a+b is a Factor. Same for a+c and b+c.....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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A cone with diameter $12$cm and height $8$cm. Find the volume of the inscribed Sphere. A cone with diameter $12\ cm$ and height $8\ cm$. Find the volume of the inscribed Sphere. Can someone help me solve this maths problem?
| Notice, let $r$ be the radius of the sphere inscribed in the cone having diameter of base $12\ cm$ (radius, $R=6\ cm$) & the vertical height $H=8\ cm$.
If $\alpha$ is the semi-apex angle of cone then using geometry in a right triangle we get
$$\tan \alpha=\frac{R}{H}=\frac{6}{8}=\frac{3}{4}\ \ (\forall\ \ 0<\alpha<\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A calculation problem about differential equation
Let $n>2$ and $n\in \mathbb{N}_+$, $\lambda \ge - \frac{{{{\left( {n - 1} \right)}^2}}}{4}$.
For $x>0$, we have $$f\left( x \right) = {\left( {\sinh x} \right)^{2 - n}}\int_0^\pi {{{\left( {\cosh x + \cos t} \right)}^\eta }{{\left( {\sin t} \right)}^{2\sqrt {\lamb... | This is not a complete answer, but rather another way of attacking the problem, that might be more fruitful. It is too long to put in a comment.
First, we write the differential equation as
$$
\bigl(f'(x)\sinh^{n-1}x\bigr)'=\lambda f(x)\sinh^{n-1}x.\tag{1}
$$
Differentiating the integral that should equal $f$, and mult... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluate the improper integral: $\int _0^{+\infty }\:\frac{dx}{\left(x^2+9\right)\cdot \sqrt[3]{x^2+9}}$ $$\int _0^{+\infty }\:\frac{dx}{\left(x^2+9\right)\cdot \sqrt[3]{x^2+9}}$$
Wolfram says strange things. And you need to solve analytically. Help.
| $$\begin{align*}I=\int_{0}^{+\infty}(x^2+9)^{-4/3}\,dx = 3^{-5/3}\int_{0}^{+\infty}(1+z^2)^{-4/3}\,dz=\frac{\sqrt{\pi}\cdot\Gamma\left(\frac{5}{6}\right)}{2\cdot 3^{5/3}\cdot\Gamma\left(\frac{4}{3}\right)}\end{align*}$$
by using the substitution $z=\tan\theta$ and Euler's beta function. That simplifies to:
$$ I =\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1464319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Extreme Values of $f(P)$ When $P$ Lies Inside a Triangle with Vectices $A$,$B$, $C$ I would appreciate if somebody could help me with the following problem
Q:$P(x,y)$ given point lie inside or on the boundary of triangle $\triangle ABC$.
Find maximum and minimum $f(a,b,c)$
$$f(P)=\vec{PA}\cdot \vec{PB}+\vec{PB}\cdot \... | I rewrite my friend H. R's answer and reach to solution by simpler math. First you should notice that f is a function of $\overrightarrow P$ not a,b,c ! you want to find appropriate vector $\overrightarrow P$ for fixed triangle.
$$\eqalign{
& f\left( {\overrightarrow P } \right) = \overrightarrow {PA} .\overrightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proof that $n+k+3$ divides $n(n+1)(n+2)(n+3) - k(k+1)(k+2)(k+3)$. I'm looking for proof that
$$
(n+k+3) \mid n(n+1)(n+2)(n+3) - k(k+1)(k+2)(k+3)\\
n,k \in \mathbb N^*, n>k
$$
I tried using induction, but i'm not sure how it would work with 2 parameters.
| Since
$$n+3=n+k+3-k\qquad \text{and}\qquad k+3=n+k+3-n$$
we have
\begin{align}
n(n+1)(n+2)(n+3)&=n(n+1)(n+2)(n+k+3)-n(n+1)(n+2)k\\
k(k+1)(k+2)(k+3)&=k(k+1)(k+2)(n+k+3)-k(k+1)(k+2)n
\end{align}
Then it will be sufficient to show that $n+k+3$ divides $k(k+1)(k+2)n-n(n+1)(n+2)k$
But
\begin{align}
k(k+1)(k+2)n-n(n+1)(n+2)k... | {
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"url": "https://math.stackexchange.com/questions/1469148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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How do i show this equality without using reccurence method:$\prod_{k=0}^{k=n}{\cos\frac{\theta}{2^k}}={\frac {\sin\theta}{2^n\sin(2^{-n}\theta)}}$?? I would like to show this without using reccurence method for all $n$ $\in $ $\mathbb{N}$ and $\theta \in \mathbb{R}$ :
$${\cos\frac{\theta}{2}}\cos\frac{\theta}{2^2}\cos... | First, notice that:
\begin{align}
2\cos x\sin x&=\sin 2x\\
\implies \cos x&=\frac{1}{2}\frac{\sin 2x}{\sin x}
\end{align}
Then, given an integer number $n>0$ we have
$$\cos \left(\frac{\theta}{2}\right)\cdot \cos \left(\frac{\theta}{2^2}\right)\cdot \ldots \cdot \cos \left(\frac{\theta}{2^n}\right)=\frac{1}{2^n}\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How can we show polar coordinates r(theta) is an ellipse? $r(θ) = a(1 − β^2)/(1 + β \cos θ)$
and I want to show this $r(θ)$ is an ellipse described by
$\dfrac{(x+\sqrt{a^2 − b^2})^2}{a^2}+\dfrac{y^2}{b^2}= 1$, when $0<β<1$.
How can we show this?
| With the relations
$$
r=\sqrt{x^2+y^2}, r\cos\theta=x
$$
we can rewrite the equation as
$$
r=\frac{ar(1-\beta^2)}{r+\beta x}
$$
or, equivalently (disregarding $r=0$ that's not a solution),
$$
r+\beta x=a(1-\beta^2)
$$
that becomes $r=a(1-\beta^2)-\beta x$; now square and get
$$
x^2+y^2=a^2(1-\beta^2)^2-2a(1-\beta^2)\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If there exist real numbers $a,b,c,d$ for which $f(a),f(b),f(c),f(d)$ form a square on the complex plane.Find the area of the square. For all real numbers $x$,let the mapping $f(x)=\frac{1}{x-i},\text{where} i=\sqrt{-1}$.If there exist real numbers $a,b,c,d$ for which $f(a),f(b),f(c),f(d)$ form a square on the complex ... | Let $\Phi(z)=\dfrac{1}{\bar{z}}$, this is the inversion with respect to the unit circle. Now, for a real $x$, we have $f(x)=\Phi(x+i)$.
The image of the line $d=\{z:\Im z=1\}$ under the inversion $\Phi$ is the circle
of diameter $[0,i]$, So if $f(a)$, $f(b)$, $f(c)$ and $f(d)$ form a square, it must be inscribed in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$Z^4 = -1$ How do I solve this without a calculator? Basically, the question is to solve $z^8= 1$. I have factored this down to $$(z+1)(z-1)(z^2+1)(z^4+1)=0$$
I have simplified $(z^4+1)$ to $z =\pm \sqrt{i}$ and online I know that this can be simplified to the following four solutions:
z= 0.7071 + 0.7071 i
z= -0.7... | use the Euler's Identity
$$e^{\pi i}=-1$$
and
$$e^{ix}=\cos x+i\sin x$$
$$z^4=-1$$
$$z^4=e^{\pi i}$$
$$z=e^{\frac{\pi }{4}+\frac{n\pi}{2}}$$
$$z_1=\cos(\frac{\pi }{4}+\frac{\pi}{2})+i\sin(\frac{\pi }{4}+\frac{\pi}{2})$$
$$z_2=\cos(\frac{\pi }{4}+\frac{2\pi}{2})+i\sin(\frac{\pi }{4}+\frac{2\pi}{2})$$
$$z_3=\cos(\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Simplify the fraction with radicals I want to simplify this fraction
$$ \frac{\sqrt{6} + \sqrt{10} + \sqrt{15} + 2}{\sqrt{6} - \sqrt{10} + \sqrt{15} - 2} $$
I've tried to group up the denominator members like $ (\sqrt{6} + \sqrt{15}) - (\sqrt{10} + 2) $ and then amplify with $ (\sqrt{6} + \sqrt{15}) + (\sqrt{10} + 2) $... | For this system we implemented finding a split form for the denomiator:
$$a + \sqrt{p}\,b$$
Such that $\sqrt{p}$ is a new radical. For a quotient we then have:
$$\frac{c}{a + \sqrt{p}\,b} = \frac{c\,(a - \sqrt{p}\,b)}{a^2 - p\,b^2}$$
Lets give it a try:
$$\frac{2+\sqrt{6}+\sqrt{10}+\sqrt{15}}{-2+\sqrt{6}-\sqrt{10}+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1476313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the real values of a which satisfies both the equality $|z+\sqrt2|=a^2-3a+2$ and the inequality $|z+i\sqrt 2|
What are the real values of parameter 'a' for which at least one
complex number z satisfies both the equality $|z+\sqrt {2}|=a^2-3a+2$ and
the inequality $|z+i\sqrt 2|<a^2$ ?
Ok I tried to consider the... | The equation
$$|z+\sqrt{2}|=a^2-3a+2 \tag{1}$$
defines a circle of radius $r_1=a^2-3a+2$ centered at $-\sqrt{2}$.
The inequality
$$|z+i\sqrt 2|<a^2 \tag{2}$$
defines the interior of a circle of radius $r_2=a^2$ centered at $-i\sqrt{2}$.
As others have pointed out, we must have
$$\begin{align}
a^2-3a+2>0 &\implies \\[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A trigonometry equation: $3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$
$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$
What are the steps to solve this equation for $ \theta $?
Because, I am always unable to deal with the product $\sin \theta \cos \theta$.
| Following excellent advice by @raul, you write immediately
$$\tan(\theta)=\frac{-5\pm\sqrt{5^2+4\cdot3\cdot2}}{2\cdot3}=\frac13,-2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
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.