Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
System of five equations I'd really appreciate help with this problem, because I'm stuck with it.
$$
\left\{
\begin{array}{c}
ab+ac+ad+ae=-1 \\
ba+bc+bd+be=-1 \\
ca+cb+cd+ce=-1 \\
da+db+dc+de=-1 \\
ea+eb+ec+ed=-1
\end{array}
\right.
$$
I've tried substituting $a+b+c+d+e$ with t but it didn't get me far. I've also... | Let $t=a+b+c+d+e$. Then your equations can be transformed to $a(t-a)=b(t-b)=\ldots=e(t-e)=-1$.
From this $0=a(t-a)-b(t-b)=(a-b)(t-a-b)$ and similarly for all other pairs in place of $a,b$ as you found out yourself.
We conclude that $a=b$ or $a+b=t$ (and equally for other pairs).
If $a=b=c=d=e$, this leads to $4a^2 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/204439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A problem about multiples.
For any positive integers $a$, $ b$, if $ab+1$ is a multiple of $16$, then $a+b$ must be a multiple of $p$. Find the largest possible value of $p$.
I have no idea how to solve this. Please help. Thank you.
| Observe that $(ab,16)=1$
If $a=16A+1,b=16B-1\implies a+b=16(A+B)$
If $a=16A+3,b=16B+5\implies a+b=16(A+B)+8$
If $a=16A+7,b=16B+9\implies a+b=16(A+B+1)$
So, in all the cases, $8\mid (a+b)$
As to find the pairs, as $ab\equiv -1\pmod {16}$ ,so $a(-b)\equiv 1$ i.e.,$-b$ is inverse of $a\pmod {16}$.
we know by this, $\lam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/206509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Correctness of Fermat Factorization Proof I have asked similar questions regarding this proof. But now I would like to know if my reformulation (after perseverance and different thinking) is correct.
Prove: An odd integer $n \in \mathbb{N}$ is composite $\iff$ $n$ can be written as $n = x^2 - y^2 s.t. y+1 < x$
Proof: $... | You can't. However, you can show that $y+1<x$, which is what you should have been asking. Here it is, with $y=(a-b)/2, x=(a+b)/2$:
$$
\begin{align}
y+1 = \frac{a-b}{2}+1&=\frac{a-b+2}{2}<\frac{a+b}{2}\\
&\Longleftrightarrow a-b+2<a+b\\
&\Longleftrightarrow 2<2b \\
&\Longleftrightarrow 1<b \\
\end{align}
$$
which you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/206844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $\sqrt[3]{60}>2+\sqrt[3]{7}$ Prove $$\sqrt[3]{60}>2+\sqrt[3]{7}$$
I try to both sides of the cubic equation, but it is quite complicated
| When you cube it, you get:
$$
60 > \left(2+\sqrt[3]{7} \right)^3 =15+6\sqrt[3]{7}\left(2+\sqrt[3]{7} \right).
$$
Continue to get
$$
\frac{60-15}{6\sqrt[3]{7}} > \left(2+\sqrt[3]{7} \right).
$$
Now cube again to get:
$$
\left(\frac{60-15}{6\sqrt[3]{7}}\right)^3 > 15+6\sqrt[3]{7}\left(2+\sqrt[3]{7} \right).
$$
The rhs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/207036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding the $n$-th derivative of $f(x) =e^x \sin x$, solving the recurrence relation I am trying to find a closed solution for the nth derivative of the function:
$f(x) = e^x \sin x$
So far I have been able to obtain the derivative as:
$f^{(n)}(x) = e^x S_n \sin x + e^x C_n \cos x$
The sequences S and C are defined as ... | \begin{align}
\left( \begin{array}{r} S_n\\C_n\\ \end{array} \right) &=
\left( \begin{array}{rr}
1 &-1\\
1 &1\\
\end{array}
\right)
\left(\begin{array}{c} S_{n-1}\\C_{n-1}\\ \end{array} \right)\\
&=\left( \begin{array}{rr}
1 &-1\\
1 &1\\
\end{array}
\right)^n
\left(\begin{array}{c} S_{0}\\C_{0}\\ \end{array} \right)
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/208002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
How to solve the cubic equation $x^3-12x+16=0$ Please help me for solving this equation $x^3-12x+16=0$
| HINT:
$$
\begin{align*}
x^3-12x-16&=x\underbrace{(x^2-4)}_{(x-2)(x+2)}-8x-16 \tag{1}\\
&=x(x-2)(x+2)-\underbrace{8x-16}_{-8(x+2)}\\
&=(x+2) \underbrace{(x^2-2x-8)}_{\text{quadratic equation}}=0
\\
\end{align*}$$
Now just solve the quadratic equation and you should get something like:
$$(x+2)(x+a)(x+b)=0\tag{2}$$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/208183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Show $\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$ How to show the following equality?
$$\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$$
| Related problems: (I), (II). This problem is a direct application of Fourier transform and Poisson summation formula. Recalling the definition of Fourier transform and the Poisson summation formula respectively
$$ F(w) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) e^{-ixw} dx \,,$$
$$ \sum_{-\infty}^{\infty} f(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/208317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 5,
"answer_id": 2
} |
Solve $5a^2 - 4ab - b^2 + 9 = 0$, $ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0$
Solve $\left\{\begin{matrix} 5a^2 - 4ab - b^2 + 9 = 0\\ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0. \end{matrix}\right.$
I know that we can use quadratic equation twice, but then we'll get some very complicated steps. Are there any elegant wa... | $21a^2+10ab+b^2-10a-8b+12=0$
$(7a+b)(3a+b)-4(7a+b+3a+b)+12=0$
$(7a+b-4)(3a+b-4)=4--->(1)$
$5a^2-4ab-b^2=(5a+b)(a-b)=-9--->(2)$
If $7a+b-4=\frac 2 k, 3a+b-4=2k$,
Express $a,b$ in terms of $k,$ replace their values in (2).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/210454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
How to find limit of a sequence defined by recurrence formula I have the following problem:
Let $a_{n}$ be the recurrence
$$a_{n+1}=a_{n}+2a_{n-1}$$
with $a_{0}=0$ and $a_{1}=1$. Can you help me find
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}}$$
for $n\geq 1$?
| You can solve this recurrence fairly easily using the characteristic equation, to get $a_n = \frac{1}{3}(2^n - (-1)^n)$. Then, as $n \rightarrow \infty$, $a_{n+1}/a_n \rightarrow 2$.
Edit: Given a linear homogeneous recurrence relation with constant coefficients $a_n = \sum_i c_i a_{n-i}$, assume $a_n = \alpha^n$ satis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/217890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Differentiating $x^2 \sqrt{2x+5}-6$ How do I differentiate this function: f(x)= $x^2 \sqrt{2x+5}-6$
I had: I had $2x\sqrt{2x+5} + x^2 \dfrac{1}{2\sqrt{2x+5}}$ but the correction model said it was I had $2x\sqrt{2x+5} + x^2 \dfrac{2}{2\sqrt{2x+5}}$
| Work from the outside in. Begin by differentiating it term by term: $$f\,'(x)=\frac{d}{dx}\left(x^2\sqrt{2x+5}\right)-\frac{d}{dx}(6)=\frac{d}{dx}\left(x^2\sqrt{2x+5}\right)-0\;.$$
Now you have to calculate the derivative of $x^2\sqrt{2x+5}$. This is a product, so you use the product rule:
$$\left[x^2\sqrt{2x+5}\right]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/220858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Partial Fractions - Calculus Evaluate the integral: $$\int \dfrac{9x^2+13x-6}{(x-1)(x+1)^2} dx$$
For some reason I cannot get the right answer. I split up the equation into three partial fractions but I cannot seem to find A, B, or C from the three subsequent equations. Thanks!
| From $(A+B) x^2 + (2A+C)x +(A-B-C) = 9x^2+13 x -6$, you need to solve for $A,B,C$. Since the coefficients of the polynomials in $x$ must match, this gives three equations $A+B=9$, $2A+C = 13$, and $A-B-C = -6$. If you solve these you will get your answer.
To check, move your mouse over the following:
$$\frac{5}{x+1}+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality:
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers.
Thanks :)
| By Holder
$$\sum_{cyc}\frac{a^3}{a+b}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(a+b)}=\frac{(a+b+c)^2}{6}\geq\frac{ab+ac+bc}{2}.$$
Done!
The Holder for three sequences it's the following.
Let $a_i$,$b_i$, $c_i$, $\alpha$, $\beta$ and $\gamma$ be positive numbers. Prove that:
$$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/222934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Evaluate the sum $\sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}$?
Evaluate the series
$$\sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}=?$$
Can you help me ? This is a past contest problem.
| Let
\begin{align}S_n & =\sum_{k=0}^{n}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}\\
& =\sum_{k=0}^{n}\left(\frac{1}{6}\frac{1}{4k+1}-\frac{1}{2}\frac{1}{4k+2}+\frac{1}{2}\frac{1}{4k+3}-\frac{1}{6}\frac{1}{4k+4}\right).
\end{align}
$$A_n=\sum_{k=0}^{n}\left(\frac{1}{4k+1}-\frac{1}{4k+3}\right),$$
$$B_n=\sum_{k=0}^{n}\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/224985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
How do you solve an equation like this? We have an equation:
$ \dfrac{p-1}{q} = \dfrac{q-1}{2p+1} = \dfrac {3}{5}$
How do you solve these types of equations? For example, if we have:
$\dfrac{1}{x} = \dfrac{3}{2} $, we use:
$1\times 2 = 3\times x$
$x = 1.5$
What is a similair approach to my equation?
| $$ \frac{p-1}{q} = \frac{q-1}{2p+1} = \frac {3}{5}$$ so
$$\frac{p-1}{q}=\frac {3}{5}$$ and $$\frac {3}{5}=\frac{q-1}{2p+1}.$$
\begin{eqnarray}
5p-5&=3q&\\
6p+3&=&5q-5
\end{eqnarray}
$$5p-3q=5$$
$$6p-5q=-8$$
$$30p-18q=30$$
$$30p-25q=-40$$
$$-7q=-70$$
$$q=10$$
$$p=7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/226970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Joint PDF. Is my book wrong? Consider that X and Y have the joint pdf $f(x,y) = (2/3)(x+1)$ for $0 < x < 1$ and $0 < y < 1$ and 0 otherwise. What is $P(X < 2Y < 3X)$? My book say the answer is 73/162
But I keep doing $\int_0^1 \int_{x/2}^{3x/2} f(x,y) \, dy \, dx$ and I am not getting 73/162
| The region you need to integrate is the blue region as shown in the figure below.
Hence, your integral should go as follows.
\begin{align}
\int_0^{2/3} \int_{x/2}^{3x/2} f(x,y) \, dy \, dx + \int_{2/3}^1 \int_{x/2}^{1} f(x,y) \, dy \, dx & = \int_0^{2/3} \int_{x/2}^{3x/2} \dfrac23(x+1) \, dy \, dx\\
& + \int_{2/3}^1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/227287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Closed Formula Expression for Sum of Combinatorics I have recently been interested in the problem of summing Combinatorials. I have been beating my brain for the past days to figure out how to find an explicit closed form of:
$n \choose 0 $+$ n \choose 3 $+$ n \choose 6$ + $\dots$ + $n \choose 3K$, where $3K$ is the la... | Let $\omega$ be a third root of 1.
Then
$$(1+1)^n = \binom{n}{0} +\binom{n}{1}+ \binom{n}{2}+ \binom{n}{3}+ ...+\binom{n}{n} \,.$$
$$ (1+\omega)^n = \binom{n}{0} + \binom{n}{1}\omega+ \binom{n}{2} \omega^2+ \binom{n}{3}+ ...+ \binom{n}{n} \omega^n \,.$$
$$ (1+\omega^2)^n =\binom{n}{0} + \binom{n}{1} \omega^2+ \binom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/228481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Area fractal pentagrams I When I saw this image I was a little curious.
How can I find the area of this fractal?
| First,
$ Area\ of\ Star\ with\ 5\ petals\ = Area\ of\ pentagram\ ->\ step\ 1 $
please refer http://mathworld.wolfram.com/Pentagram.html for area formula of pentagram. because the tricky part is identifying the GP.
Idea is that from each of the 5 triangles (assuming the new triangle has side one third the length of b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/229001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Induction: $\sum_{k=0}^n \binom nk k^2 = n(1+n)2^{n-2}$ I found crazy (for me at least) induction example, in fact it just would be nice to prove. (Even have problems with starting) Any hints are highly valued: $$0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} $$
| With induction we prove:
*
*$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n}=2^n$
*$0\binom{n}{0}+1\binom{n}{1}+2\binom{n}{2}+\cdots+n\binom{n}{n}=n2^{n-1}$
*$0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} $
For the proofs we use Pascal's identity: $\binom{n+1}{k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/231596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 1
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Integral table contradiction? In a table of integrals, I see the following two formulas:
$\int \frac{dx}{(a+x)(b+x)} = \frac{1}{b-a}\ln\frac{a+x}{b+x}$, and
$\int \frac{dx}{ax^2+bx+c} = \frac{2}{\sqrt{4ac-b^2}}\tan^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}}$.
How can these both be true? It seems like if we expand $(a+x)(b+x)$ o... | Take a look at the logarithmic form of the complex arctangent:
$$\tan^{-1}x=\frac12i\Big(\ln(1-ix)-\ln(1+ix)\Big)\;.\tag{1}$$
For instance, consider the integral
$$\int\frac{dx}{(1+x)(2+x)}=\int\frac{dx}{x^2+3x+2}\;.$$ The two integration formulas yield the antiderivatives
$$\ln\frac{1+x}{2+x}\tag{2}$$ and
$$\frac2i\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/233725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Sum of the sum of the sum of the first $n$ natural numbers I have here another problem of mine, which I couldn't manage to solve.
Given that: $$x_n = 1 + 2 + \dots + n \\ y_n = x_1 + x_2 + \dots + x_n
\\ z_n = y_1 + y_2 + \dots + y_n $$
Find $z_{20}$.
I know the answer but I'm having a hard time reaching it. I recog... | Here is a slightly longer method than Brian M. Scott's, relying on knowing closed forms for $n$, $n^{2}$ and $n^{3}$.
You have: $$x_{n}=\frac{n(n+1)}{2}=\frac{1}{2}n^{2}+\frac{1}{2}n$$
Therefore: $$\begin{align}y_{n}&=\sum_{i=1}^{n}{\left(\frac{1}{2}i^{2}+\frac{1}{2}i\right)}=\frac{1}{2}\left(\sum_{i=1}^{n}i^{2}+\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/234304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 5
} |
Forward and Backward Euler. I want to consider this differential system:
$$
\ \frac{dx}{dt} = -y(t)\\
\frac{dy}{dt} = \ x(t)
$$
where $t>0$ with initial condition$ (x(0),y(0))=(1,0).$
First I want to show that this differential equation admits an invariant of $I = x(t)^2 + y(t)^2$ Also, Can someone help me to figure ou... | I would like to explain this, using explicit Euler, because it is the most elementary.
We have:
\begin{align}
\begin{pmatrix}
x \\ y
\end{pmatrix}'
=
\begin{pmatrix}
-y \\ x
\end{pmatrix}
= f(x,y)
\end{align}
Explicit euler looks like this: $z_{k+1} = z_k +h \cdot f(z_k)$, applied to our problem it becomes:
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/235929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is $\sqrt{8}/2$ equal to $\sqrt{2}$? I am trying to help my daughter on her math homework and I am having some trouble on some equation solving steps. My current major concern relies on understanding why $\sqrt{8}/2$ equal to $\sqrt{2}$.
Thanks in advance
| Other techniques to prove that result:
Technique $\mathbf 1$:
We may also prove that by solving this equation for $x$:
$$\begin{align}\frac{\sqrt x}{2}&=\sqrt2\\\sqrt x&=2\sqrt 2\\\left(\sqrt x\right)^2&=\left(2\sqrt 2\right)^2\\x&=4\cdot2=\color{green}{\boxed{\color{black}{8}}}\end{align}$$
Therefore: $$\frac{\sqrt8}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/235998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 8,
"answer_id": 0
} |
If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple. Pythagoras stated that there exist positive natural numbers, $a$, $b$ and $c$ such that $a^2+b^2=c^2$. These three numbers, $a$, $b$ and $c$ are collectively known as a
Pythagorean triple. For example, $(8, 15, 17)$ is one of th... | Since I assume that you're new to some Mathematical topics, I'd use basic language.
First of all, the question itself teaches you that, if $(a,b,c)$ is a Pythagorean triplet, then it means that $a^2 + b^2 = c^2$. So if $\rm (foo_1, foo_2, foo_3)$ is a Pythagorean triplet, then $\rm (foo_1)^2 + (foo_2)^2 = (foo_3)^2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/239312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Solve the recurrence $T(n) = 2T(n-1) + n$ Solve the recurrence $T(n) = 2T(n-1) + n$
where $T(1) = 1$ and $n\ge 2$.
The final answer is $2^{n+1}-n-2$
Can anyone arrive at the solution?
| T(n)=2T(n-1)+n -----(1)
T(n-1)=2T(n-2)+n-1 ------(2)
T(n-2)=2T(n-3)+n-2 ------(3)
Substitute (3) in (2) and then (2) in (1)
T(n-1)=$2^2T$(n-3)+2(n-2)+(n-1)
T(n)= $2^3$T(N-3)+$2^2$(n-2)+2(n-1)+n
=> T(k)=$2^k$T(n-k) + $2^{k-1}$(n-(k-1))+........+$2^0$n
Now, since T(1)=1, let n-k=1 => k=n-1
T(n)=$2^{n-1}$T(1) + $2^{n-2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/239974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 7
} |
How would you solve this recurrence equation: $a_{n+1}-2a_{n}=6\cdot 5^n$ for $n\geq 1$ How would you solve $a_{n+1}-2a_{n}=6\cdot 5^n$ for $n\geq 1$ ?
I don't understand the text in my textbook. I Would like somebody to explain it to me.
| \begin{align}
a_{n+1} & = 6 \cdot 5^n + 2 a_n\\
& = 6 \cdot 5^n + 2 (6 \cdot 5^{n-1} + 2a_{n-1})\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2a_{n-1}\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 (6 \cdot 5^{n-2} + 2a_{n-2})\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 \cdot 6 \cdot 5^{n-2} + 2^3a_{n-2}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/243472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Solve $x^2 + 10 = 15$ How do I solve the following equation?
$$x^2 + 10 = 15$$
Here's how I think this should be solved.
\begin{align*}
x^2 + 10 - 10 & = 15 - 10 \\
x^2 & = 15 - 10 \\
x^2 & = 5 \\
x & = \sqrt{5}
\end{align*}
I was thinking that the square root of 5 is iregular repeating 2.23606797749979 number. 2.236 m... | Your equation should be corrected at the second line to be $$x^2 -4 = 0$$ and similarly $$x^2-5=0$$.
Then $$(x-\sqrt{5})(x+\sqrt{5})=0$$ which implies $$x=\sqrt{5}$$ or $$x=-\sqrt{5}$$
The answer being irrational doesn't matter, as $x$ is in the real number set (or can even be complex number set, a quadratic equation a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/246071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What am I doing wrong in calculating this determinant? I have matrix:
$$
A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
And I want to calculate $\det{A}$, so I have written:
$$
\begin{array}{|cccc|ccc}
1 & 2 & 3 & 4 & 1 & 2 & 3 \\
2 & 3 & 3 & 3 & 2 & 3 & 3 \\
0 & ... | The trick you are applying (Rule of Sarrus) only works for $ 3\times 3$ Matrices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/246606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
How do we prove that two parametric equations are drawing the same thing? For example, if I have
$$\begin {align}
x(t) &= r\sin t\cos t\\
y(t) &= r\sin^2 t\\
\end {align}$$
and
$$\begin {align}
x(t) &= \frac r 2 \cos t\\
y(t) &= \frac r 2 (\sin t + 1)
\end {align}$$
How do we show that the two parametric equations ... | Some efforts to establish that two parametrizations represent the same curve are trickier than others. In your first set, you will need the "double-angle" trig identities, or their relatives (I am also going to write '$r$' as '$R$' , since it is intended to be a constant, rather than radius in polar coordinates):
$$\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/247752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Show $2(x+y+z)-xyz\leq 10$ if $x^2+y^2+z^2=9$ If $x,y,z$ are real and $x^2+y^2+z^2=9$, how can we prove that $2(x+y+z)-xyz\leq 10$?
Please provide a solution without the use of calculus. I know the solution in that way.
| Case 1: At least one of $x,y,z < 0$. Let's say $z < 0$. If we let $f(x,y,z) = 2(x+y+z) - xyz$, it's then clear that $f(x,y,z) \leq f(t,t,z)$, where $t = \sqrt{\frac{x^2+y^2}{2}}$. It then suffices to check $f(t,t,z) \leq 10$. Let $z = -u$, so that $u$ is positive. Then we need to show
$$f(t,t,-u) = 2(2t - u) + t^2 u ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/252178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Evaluating$\int {1\over1-\sin2x}dx$ $$
\int {1\over1-\sin 2x}dx = \int {1\over \sin^2 x-2\sin x\cos x+\cos^2x}dx = \int {1\over (\sin x-\cos x)^2}dx
$$
From here I get two different answers, depending on whether I factor out $\sin x$ or $\cos x$.
Factoring out $\sin x$, this one is correct according to WolframAlpha:
$$... | They are both correct. To see this, you can simply take the derivative of each antiderivative to get your original integral back. This shows that both primitives you found are equal, up to a constant.
$$\frac{d}{dx} \left(\frac{1}{\cot x - 1}\right) = \frac{1}{\left(\cos x - \sin x\right)^2}$$
$$\frac{d}{dx} \left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/253879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Choosing a 5 member team out of 12 girls and 10 boys We must choose a 5-member team from 12 girls and 10 boys. How many ways are there
to make the choice so that there are no more than 3 boys on the team?
The correct answer is $\binom{22}{5} - \binom{12}{1} \binom{10}{4} - \binom{10}{5}$.
I understand the $\binom{22}{5... | That solution proceeds by starting with $\binom{22}5$, the total number of possible $5$-person teams, and subtracting the $\binom{12}1\binom{10}4$ teams that have one girl and four boys and the $\binom{10}5$ teams that have five boys.
The problem could also be solved by noting that there are $\binom{12}2\binom{10}3$ te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/254356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
An application of the Mean Value Theorem I'm recalling this question from memory, so I may be messing it up a bit.
Let $a/3+b/2+c=0$. Show that $ax^2+bx+c=0$ has at least one root in $[0,1]$ using the Mean Value Theorem.
Let $f(x)=ax^2+bc+c$. Then $f(0)=c$ and $f(1)=a+b+c$. Also $f'(x)=2ax+b$. So there exists $f(\x... | First, if $a =0$, then we have $bx + c = 0 \implies x = - \frac{c}{b} = \frac{b/2}{b} = \frac{1}{2}$.
Now, suppose $a \neq 0$.
Note that $c = - \frac{a}{3} - \frac{b}{2}$, so you want to prove that the function $f(x) = ax^2 + bx - \frac{a}{3} - \frac{b}{2}$ has a root in $[0,1]$. We have $f(0) = - \frac{a}{3} - \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/255811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How can I find the points at which two circles intersect? Given the radius and $x,y$ coordinates of the center point of two circles how can I calculate their points of intersection if they have any?
| Example 1: Find the points of intersection of the circles given by their equations as follows:
$(x - 2)^2 + (y - 3)^2 = 9$
$(x - 1)^2 + (y + 1)^2 = 16$
Solution to Example 1:
We first expand the two equations as follows:
$x^2 - 4x + 4 + y^2 - 6y + 9 = 9 $
$x^2 - 2x + 1 + y^2 + 2y + 1 = 16 $
Multiply all terms in the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/256100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "59",
"answer_count": 8,
"answer_id": 2
} |
How to get from $\frac{x}{x+1}\;$ to $\;1 - \frac{1}{x+1}$? Please show me how to manipulate $\dfrac{x}{x+1}\;\;$ to get $\;\;1 - \dfrac{1}{x+1}$
| Simply
$$\frac{x}{x+1}=\frac{x+1-1}{x+1}=\frac{x+1}{x+1}-\frac{1}{x+1}=1-\frac{1}{x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/259498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Rationalize Below Equation How can I rationalize the following equation:
$$\frac{22}{4\sqrt[3]{9}+2\sqrt[3]{6}+\sqrt[3]{4}}$$
| Let $a=2\sqrt[3]{3}, b=\sqrt[3]{2}$. Then this is equal to $\dfrac{22}{a^2+ab+b^2} = \dfrac{22(a-b)}{a^3-b^3}$.
Since $a^3-b^3 = 8(3)-2 =22$, we have that our desired answer is $\dfrac{22(a-b)}{a^3-b^3} = a-b = 2\sqrt[3]{3}-\sqrt[3]{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/260112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Complicated limit calculation Find the limit of $$\left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}\right]x^{(mn-m)/n}$$ when $x\to\infty$ and $m,n$ are natural numbers.
Thanks in advance!
| An even quicker way to the answer comes from observing that
$$\left ( x^m + 1 \right )^{\frac{1}{n}} - \left (x^m - 1 \right)^{\frac{1}{n}} = x^{\frac{m}{n}} \left [ \left ( 1 + \frac{1}{x^m} \right )^{\frac{1}{n}} - \left (1 - \frac{1}{x^m} \right)^{\frac{1}{n}} \right ]$$
$$ \approx x^{\frac{m}{n}} \left [ \left ( 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/263306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
For what values of $x$ does $\sum_{n=1}^{\infty}\frac{1}{(x+n)(x+n-1)}$ converge? Please help me to find the values of $x$ in which the series $$\sum_{n=1}^{\infty}\frac{1}{(x+n)(x+n-1)}$$ converges? I applied some tests for it but...:(
Thank you
| We can use partial fractions to decompose this fraction. We need to find $A$ and $B$ such that
$$\frac{A}{x+n} + \frac{B}{x+n-1} \equiv \frac{1}{(x+n)(x+n-1)} \, . $$
If we cross multiply, we get $A(x+n-1)+B(x+n)\equiv1$. When $x=-n$ we get $-A\equiv1$ and when $x=1-n$ we get $B \equiv 1$. It now follows that:
$$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
What is the value of D here? Number $S$ is obtained by squaring the sum of digits of a two digit number $D$. If the difference between $S$ and $D$ is $27$, then the two digit number $D$ is?
My thoughts:
Let the two digit number $D$ be $AB$.
And so $S=(A+B)^2$
If $\,S-D=27,\,$ then $\,(A+B)^2 -AB=27$
$$A^2 + 2AB + B^2 ... | To try and approach systematically:
We know $|S-D|=27$.
Let $D$ is comprised of the digits $A$ and $B$, with each of $A, B$ each a single digit integer. Let $A$ be the left-most digit (the "ten's digit of $D$), and let $B$ be the right-most digit (the "one's" digit of $D$).
So $S = (A + B)^2$ and $D = 10 \cdot A + B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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A nasty integral of a rational function I'm having a hard time proving the following $$\int_0^{\infty} \frac{x^8 - 4x^6 + 9x^4 - 5x^2 + 1}{x^{12} - 10 x^{10} + 37x^8 - 42x^6 + 26x^4 - 8x^2 + 1} \, dx = \frac{\pi}{2}.$$
Mathematica has no problem evaluating it while I haven't the slightest idea how to approach it. Of co... | Some progess: The integrand actually decomposes as
$$\frac{1}{2} \left( \frac{x^2 + 2x + 1}{x^6 + 4x^5 + 3x^4 - 4x^3 - 2x^2 + 2x + 1} + \frac{x^2 - 2x + 1}{x^6 - 4x^5 + 3x^4 + 4x^3 - 2x^2 - 2x + 1} \right).$$
Note that the second term is the same as the first term, except with $-x$ instead of $x$. Thus, with some subs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "68",
"answer_count": 2,
"answer_id": 1
} |
What is a good bound for $\int^{2\pi}_{0}\cos^k(x)dx$? The integral $$\int^{2\pi}_{0}\cos^k(x)dx$$comes up in one of Rudin's problems. He asked me to do a detailed approximation of it. I am wondering how to do it. This feels really elementary, but I need a lower bound that works for any $k$. Integrating by parts would ... | If $k$ is odd, then the integral is zero. If $k$ is even, note that $$\int_0^{2\pi} \cos^{2m}(x) dx = 4 \int_0^{\pi/2} \cos^{2m}(x) dx$$
Let $I_m = \displaystyle \int_0^{\pi/2} \cos^{2m}(x) dx$
Then
\begin{align}
I & = \int_0^{\pi/2} \cos^{2m-1}(x) d(\sin(x))\\
& = \left. \cos^{2m-1}(x) \sin(x)\right \vert_0^{\pi/2} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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What is the valuation of $\lim_{x\to0}\ x^2\left(1+2+3+...+\left[\frac1{|x|}\right]\right)$? Compute the following limitation:
$\lim_{x\to0}\ x^2\left(1+2+3+...+\left[\frac1{|x|}\right]\right)$.
| Recall that $$1 + 2 + 3 + \cdots + \cdots + n = \dfrac{n(n+1)}2$$
If $\vert x \vert \in \left(\dfrac1{n+1}, \dfrac1n\right]$, we have $\lfloor 1/\vert x \vert \rfloor = n$ and hence we get that
$$1 + 2 + 3 + \cdots + \cdots + \lfloor 1/\vert x \vert \rfloor = \dfrac{\lfloor 1/\vert x \vert \rfloor(\lfloor 1/\vert x \v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to compute the determinant of a tridiagonal matrix with constant diagonals? How to show that the determinant of the following $(n\times n)$ matrix
$$\begin{pmatrix}
5 & 2 & 0 & 0 & 0 & \cdots & 0 \\
2 & 5 & 2 & 0 & 0 & \cdots & 0 \\
0 & 2 & 5 & 2 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots & \vdots... | For the more general case of a tridiagonal Toeplitz matrix mentioned at the end on the question, the direct method (not using induction) I used in this answer can be adopted as follows. First note that the answer will depend only on $a$ and the product $p=bc$, since conjugation by a diagonal matrix with as diagonal en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 3
} |
Why do we choose $3$ to be positive after $\sqrt{9 - x^2}$ in the following substitution? The integral $$\int \frac{\sqrt{9 - x^2}}{x^2}dx$$ is solved in my book by letting $x = 3\sin\theta$ where $-\frac {\pi}{2} \le \theta \le \frac {\pi}{2}$.
Then, $dx = 3\cos\theta\,d\theta$
and, $$\sqrt{9-x^2} = 3|\cos\theta| = ... | Think about it: How would using $\;|\,3\,|\;$ change the result?$\quad$ After all, $\;|\,3\,| \;= \quad?$
Also note: $\,|\cos\theta| \,= \,\cos\theta \,\ge \,0\;$ since the range of $\,\theta\,$ is $\;\;-\dfrac{\pi}{2} \,\le\, \theta \,\le \dfrac{\pi}{2}$.
You seem to be confused about the what "$|\;\cdot\;|$" means.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/268830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Rotation around slant line What is the volume when $f(x) = x^2$ is rotated around the line $y = x$?
For each individual $x$, I was considering the difference between $\begin{pmatrix} x\\x \end{pmatrix}$ and the projection of $\begin{pmatrix} x^2\\x \end{pmatrix}$ onto $\begin{pmatrix} x\\x \end{pmatrix}$ (which would g... | [Sorry, previous answer was wrong]
You are right in saying that "Incidentally, $\frac { d {\begin{matrix}x\\x\\ \end{matrix}} }{dx} = \sqrt{2}$. In changing the coordinates, from integrating on the x-axis, to integrating on the line y=x, you also need to account for the difference in volume, which is precisely this am... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/271262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Why does $(2/p)=\prod_{k=1}^{(p-1)/2}2\cos\left(\frac{2\pi k}{p}\right)$? Browsing one of my favorite video game sites, there was a post asking about the derivation of the formula
$$
\left(\frac{2}{p}\right)=\prod_{k=1}^{(p-1)/2} 2\cos(2\pi k/p).
$$
I know that $\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}$, but I'd never... | For odd prime $p,$
$$\cos px+i\sin px=(\cos x+i\sin x)^p=\sum_{0\le r\le p\cos^{p-r}x(i\sin x)^r}$$
Equating the Real parts, $$\cos px=\cos^px-\binom p2 \cos^{n-2}x\sin^2x+\binom p4\cos^4x\sin^4x+\cdots$$
$$=\cos^px(1+\binom p2+\binom p4+\cdots)+\cdots+(-1)^{\frac{p-1}2}\cos x$$
$$=2^{p-1}\cos^px+\cdots+(-1)^{\frac{p-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/272719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Proving:$(1+x)(1+x^2)(1+x^3)\cdots(1+x^n)\ge(1+x^{\frac{n+1}{2}})^n$ How to prove that :
$$(1+x)(1+x^2)(1+x^3)\cdots (1+x^n)\ge(1+x^{\frac{n+1}{2}})^n$$
| $$(1+x^r)(1+x^{n+1-r})=1+x^r+x^{n+1-r}+x^{n+1}\ge 1+x^{n+1}+2\sqrt{x^r\cdot x^{n+1-r}}$$ applying A.M.$\ge$ G.M. with $x^r,x^{n+1-r}$ assuming $x>0$
So, $$(1+x^r)(1+x^{n+1-r})\ge(1+x^{\frac{n+1}2})^2 $$
Now, $1\le r<n+1-r\implies r<\frac{n+1}2$
If $n$ is odd $=2m+1$ (say), $r<\frac{2m+1+1}2\implies 1\le r\le m=\frac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/273383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the equation of the sphere $ x^2+y^2+z^2-2x-4y+8z=15$ I'm not sure how you get from this:
$x^2+y^2+z^2-2x-4y+8z=15$
To:
$(x^2-2x+1) + (y^2-4y+4) + (z^2+8z+16)-1-4-16=15$
How do you get the $1,4,16$?
| It's grouping like terms and completing the square. For example, if you collect the terms in $x$ you have $x^2 - 2x$, so you want a constant term that will make that a perfect square. Since $(x-1)^2 = x^2 - 2x + 1$, you need a $1$. Similarly for the other two.
In general, if you have $ax^2 + bx$, to complete the square... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
find the sum of the alternating series How to find the sum of the infinite series
$$\frac{1}{12}-\frac{1\cdot 4}{12 \cdot 18 } + \frac{1\cdot 4\cdot 7}{12\cdot 18\cdot 24} - \frac{1 \cdot 4 \cdot 7\cdot 10}{12 \cdot 18 \cdot 24 \cdot 30}+...$$
I understood the answer posted in Yahoo Answer till the last but one step:
T... | Using the fact that
$$
\prod_{k=0}^{n}(12+6k) = 6^{n+1}(n+2)! = 3^{n+1} 2^{n+2} \frac{(n+2)!}{2}
$$
we see that the $n^{\text{th}}$ term is
$$
\begin{align*}
(-1)^n \frac{\prod_{k=0}^{n}(1+3k)}{\prod_{k=0}^{n}(12+6k)} &= \frac{\prod_{k=0}^{n}\left(\frac{1}{3}-k\right)}{\frac{(n+2)!}{2}} \cdot \frac{1}{2^{n+2}} \\
&= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/280849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Rational function partial fraction help? So I have to integrate
$$
\frac{3x+2}{(x)(x+1)^3}
$$ so I have the terms
$$
\frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3}
$$
... and then I have
$$
3x+2=A(x+1)^3 + B(x)(x+1)^2 +C(x)(x+1)^2+ D(x)(x+1)^2
$$
... here I tried to equal the quotients and I have... | From $$
\frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3}
$$ you have: $$\frac{A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx}{x(x+1)^3}=\frac{3x+2}{x(x+1)^3}$$ Now let $x=0$ then $$A(0+1)^3=3\times0+2\to A=2$$ Let $x=-1$ then $$D(-1)=3\times(-1)+2=-1\to D=-1$$ Set $x=1$ then $$8A+4B+2C+D=3\times1+2=5\to 16+4B+2C-1=5\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/286767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to transform $2^{n-2}\frac{(2n-5)(2n-7)...(3)(1)}{(n-1)(n-2)...(3)(2)(1)}$ into $\frac{1}{n-1}\binom{2n-4}{n-2}$? Just an algebraic step within the well known solution for the number of triangulations of a convex polygon!
| \begin{eqnarray}
2^{n-2}\frac{(2n-5)(2n-7)\ldots3\cdot1}{(n-1)(n-2)\ldots3\cdot2\cdot1}&=&
2^{n-2}\frac{(2n-4)(2n-5)\ldots\cdot3\cdot2\cdot1}{[(n-1)(n-2)\ldots2\cdot1][(2n-4)(2n-6)\ldots4\cdot2]}\\
&=&2^{n-2}\frac{(2n-4)!}{[(n-1)(n-2)\ldots2\cdot1][2^{n-2}(n-2)(n-3)\ldots2\cdot1]}\\
&=&\frac{(2n-4)!}{(n-1)!(n-2)!}=\fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\lim_{n \to \infty} \oint_{ |z| = 1/4} \frac{1}{(4 z(1-z))^n} \frac{\mathrm{d}z}{z (1-2 z)} = \frac{1}{2}$ While working on an earlier question involving $\sum_{j=0}^n \binom{n+j-1}{j} \frac{1}{2^{n+j}}$ I rewrote the sum as a contour integral, using generating functions:
$$
\sum_{j=0}^n \binom{n+j-1}{... | As $t$ traces around the origin, so does $u=4 t (1-t)$. Solving for $t$ we get two solutions, one, $2t=1-\sqrt{1-u}$ maps a $u$-path around the origin into the $t$-path around the origin, and another, $2t=1+\sqrt{1-u}$, maps a $u$-path around the origin into the $t$-path around $t=1$. This suggests a change of variable... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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closest point to on $y=1/x$ to a given point I feel like I'm missing something basic - given a point $(a,b)$ how do I find the closest point to it on the curve $y=1/x$? I tried the direct approach of pluggin in $y=1/x$ into the distance formula but it leads to an order-4 polynomial...
| Well. We have by the distance formula that the point on $xy = 1$ closest to $(a,b)$ will be the solution to:
$$\frac{d}{dx}\sqrt{(a-x)^2+(b-\frac{1}{x})^2} = 0$$
Which has the same soutions as
$$\frac{d}{dx}\left( (a-x)^2+(b-\frac{1}{x})^2\right)= 0\\
-2(a-x)+\frac{2(b-\frac{1}{x})}{x^2} = 0\\
x^4-ax^3+bx-1=0$$
Which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/290986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proving $ (1)(2) + (2)(3) + (3)(4) + \cdots+ (n) (n+1) = \frac{(n)(n+1)(n+2)}{3} $ by induction I have the Following Proof By Induction Question:
$$ (1)(2) + (2)(3) + (3)(4) + \cdots+ (n) (n+1) = \frac{(n)(n+1)(n+2)}{3} $$
Can Anybody Tell Me What I'm Missing.
This is where I've Gone So Far.
Show Truth for N = 1
LHS =... | It might also be helpful to go backwards to see that it makes sense.
Specifically,
$$S_{n+1} - S_n = \frac{(n+2)(n+1)(n)}{3} - \frac{(n+1)(n)(n-2)}{3} = \frac{1}{3}\left(n^3 + 3n^2 + 2n - n^3 + n\right) = n(n+1)$$
Going backwards, you can more easily see that $$S_{n+1} = S_n + n(n+1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a counterexample to the claim that if $\mathbf y\cdot\mathbf y=1$ and $\mathbf x\cdot\mathbf y=c$ then $\mathbf x=c\mathbf y$? Let $x,y$ be arbitrary vectors where $\mathbf{y} \cdot \mathbf{y} = 1$ and $c$ be a real valued scalar. If
$\mathbf{x} \cdot \mathbf{y} = c = c (\mathbf{y} \cdot \mathbf{y} ) = (c \ma... | $\mathbf{x} = c \mathbf{y} + $ some vector perpendicular to $\mathbf{y}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/293198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Deriving $e^{ix} = 1 + ix + \frac{(ix)^2}{2!}+...$ I know how to prove equation:
$$e = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right)^x$$
How can I now derive the series:
$$e^{ix} = 1 + ix + \frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+...$$
Those two seem very similar to me...
| If we are allowed to use calculus,
let $$z=1+\frac y{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+\frac{y^4}{4!}+\cdots$$
So, $$\frac{dz}{dy}=0+1+\frac y{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+\frac{y^4}{4!}+\cdots=z$$
$$\implies \frac{dz}z=dy$$
Integrating both sides, $\log z=y+c$
For $y=0,z=1\implies c=\log1-0=0\implies \log z=y\iff... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$a+b+c+d+e=abcde$ What is $\max(a)?$ The question from my nephew is this:
If $a, b, c, d, e \in N^+$ and $a+b+c+d+e=a\times b\times c\times d\times e$, then what's the the maximum possible value of $a$? Thanks ahead:)
| $5$. This is clearly possible with $b=2$, $c=d=e=1$. It remains to show that $b+c+d+e< 6(bcde-1)$ or $6bcde>b+c+d+e+6$. However, we cannot have $b=c=d=e=1$, so $bcde\ge 2$ and, thereby, $3bcde\ge 6$. Thus we are left with $3bcde>b+c+d+e$. Assuming that $b$ is the largest number, we have $b\ge 2$, so $\frac 32bcde\ge 3c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $a, b, c, d \in \mathbb{Z}$ such that $2^a=3^b5^c+7^d$ Solve $2^a=3^b5^c+7^d$ over the positive integer.
I know $a$ is even because: $(-1)^a \equiv2^a = 3^b5^c+7^d \equiv1 \ (mod\ 3)$
| One solution to your equation is $a = 2$, $b = 0$, $c = 0$ and $d = 0$. Another one would be $a = 3$, $b = 0$, $c = 0$, $d = 1$. Yet another one would be $a = 4$, $b =1$, $c = 1$, $d = 0$. And $a = 5$, $b = 0$, $c = 2$, $d =1$.
One is clever to recognize that $a + b + c+d = 2(a - 1)$ for all $a,b,c,d \in \mathbb{N}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/298570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Polynomial inequalities Just wanted to check my maths:
a) $\displaystyle \frac{(x^3-4)}{x-2} \leq x+2$ I got that $x\leq 1$.
b) $\displaystyle \frac{(2x-3)(x+1)}{x^2} \leq -2$
I got that $\displaystyle -\frac{3}{4} \leq x \leq 1$
Right/wrong? Thanks :). I just worked on it as if it were an equals sign... so I'm not sur... | In case of the 2nd one,
$\displaystyle \frac{(2x-3)(x+1)}{x^2} \leq -2$
Multiplying both sides by $x^2$ (Considering $x\ne0)$as it is +ve so no problem of sign changing,
$\Rightarrow(2x-3)(x+1)\leq -2x^2$
$\Rightarrow 2x^2-x-3\leq -2x^2$
$\Rightarrow 4x^2-x-3\leq 0$
$\Rightarrow(2x)^2-2.2x.\frac{1}{4}+1/16-1/16-3\leq0$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Suppose that today is tuesday. what day of the week will it be in $3^{100000} + 3^{10000} + 3^{1000}$ days? Suppose that today is tuesday. what day of the week will it be in $3^{100000} + 3^{10000} + 3^{1000}$ days ?
I figured that, eventually, I will add $3^{100000} + 3^{10000} + 3^{1000}$ and $\mod\ 7$ that answer.... | $3^6 = 1 \mod 7$. Hence $3^k \mod 7 = 3^{k-6n} \mod 7$.
$100000 = 6\cdot 16666 + 4$. $10000 = 6\cdot 1666 + 4$. $1000 = 6\cdot 166 + 4$. Hence $3^{100000} + 3^{10000} + 3^{1000} \mod 7 = 4+4+4 \mod 7 = 5 \mod 7$.
Hence it will be Sunday.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Tangent line to solution set of equations Determine the parametric equation for the tangent line at the point $P = (1,1,1)$ to a curve which is described by the solution set of the following equations:
$x^2 + y^2 + z^2 = 3$,
$3x + 4y + 5z = 12$
I dont have a clue where to start. How can I differentiate functions which ... | Hints: to find the solution set, you can set the equations equal to one another, or add them, to find the solution set:
$x^2 + y^2 + z^2 = 3$,
$+$
$3x + 4y + 5z = 12$
$$x^2 + 3x + y^2 + 4y + z^2 + 5z \color{blue}{\bf{-15}} = 0\tag{1}$$
$$x^2 + 3x \color{blue}{\bf -4} + y^2 + 4y \color{blue}{\bf -5} + z^2 + 5z \color{b... | {
"language": "en",
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Algebraically Solve Limit $$\lim_{x \to 0} \dfrac{2\sqrt{x+1}-x-2}{x^2}$$
I can solve it using l'Hôpital but just cannot find a way to do it algebraically.
| $$\begin{align}
\lim_{x\to 0}\frac{2\sqrt{x+1}-x-2}{x^2}
&= \lim_{x\to 0}\frac{2\sqrt{x+1}-(x+2)}{x^2} \frac{2\sqrt{x+1}+(x+2)}{2\sqrt{x+1}+x+2}\\
&= \lim_{x\to 0}\frac{4(x+1)-(x+2)^2}{x^2} \frac{1}{2\sqrt{x+1}+x+2}\\
&= \lim_{x\to 0}\frac{4(x+1)-(x^2+4x+4)}{x^2} \frac{1}{2\sqrt{x+1}+x+2}\\
&= \lim_{x\to 0}\frac{-x^2... | {
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Hensel Lifting and solving with mods I need to use the Hensel-Newton method (aka Hensel Lifting) to find all solutions of:
$$x^2 + x + 47 \equiv 0\:(\text{mod } 343)$$
Note: $$ 343 = 7^3 $$
I don't really understand Hensel Lifting so I am not sure where to begin. Can someone help give me a better understanding of Hens... | Let's start modulo $7$.
By trial and error, we find $x^2+x+5\equiv 0\pmod 7$ has the solutions $x\equiv 1$, $x\equiv 5$.
Now if $x^2+x+47\equiv 0\pmod{49}$ then $x=x_0+7y$ with $x_0\in\{1,5\}$.
Hence $x^2+x+47=(x_0^2+14x_0y+49y^2)+(x_0+7y)+47\equiv (x_0^2+x_0+5)+7\cdot(2x_0y+y+6)\pmod{49}$, where we already know that t... | {
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"timestamp": "2023-03-29T00:00:00",
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Derivative of $x^2\sqrt{1+x}$
Given that $f(x)=x^2\sqrt{1+x}$, show that $f'(x)=\dfrac{x(ax+b)}{2\sqrt{1+x}}$ where $a$ and $b$ are constants to be found.
I first tried using the product rule: $f'(x)=2x\sqrt{1+x}+\dfrac{x^2}{2\sqrt{1+x}}$, but now I'm stuck.
I don't know if it is my algebra or derivation that stops m... | You can avoid the product rule by putting everything under one square root:
$$x^{2} \sqrt{1 + x} \; = \; \sqrt{x^4(1 + x)} \; = \; \sqrt{x^4 + x^5} \; = \; \left(x^4 + x^5 \right)^{\frac{1}{2}}$$
Now apply the power rule for derivatives:
$$f'(x) \; = \; \frac{1}{2} \left(x^4 + x^5 \right)^{-\frac{1}{2}} \cdot \left(4x^... | {
"language": "en",
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Finding minimum and maximum value of complex number Let a be a positive real number and let $M_a =\{z \in C^*: |z+\frac{1}{z}|=a\}$ Find the minimum and maximum value of $|z|$ when z$\in M_a$
| Let $z=r(\cos \theta+i\sin \theta)$
$z^{-1}=r^{-1}(\cos \theta-i\sin \theta)$
We have ,
$|z+\frac{1}{z}|=|(r+1/r)\cos \theta +(r-1/r)i\sin\theta|$
So we have,
$\displaystyle|(r+1/r)\cos \theta +(r-1/r)i\sin\theta|=\sqrt{(r^2+1/r^2)+2\cos 2\theta}$
Thus we have ,
$(r^2+1/r^2)+2\cos 2\theta=a^2$
Let $r^2=x$,
$\Rightarrow... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{j = 0}^{n} (-\frac{1}{2})^j = \frac{2^{n+1} + (-1)^n}{3 \times 2^n}$ whenever $n$ is a nonnegative integer. I'm having a really hard time with the algebra in this proof. I'm supposed to use mathematical induction (which is simple enough), but I just don't see how to make the algebra work.
$\sum_{j = 0... | We start from $\frac{2^{k+1}+(-1)^{k}}{3\cdot 2^k} +\frac{(-1)^{k+1}}{2^{k+1}}$. Multiply numerator and denominator of the first term by $2$, and numerator and denominator of the second term by $3$. Now we can add safely and get
$$\frac{2^{k+2}+2(-1)^{k}+3(-1)^{k+1}}{3\cdot 2^{k+1}}.$$
We need to verify that $2(-1)^k+3... | {
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"timestamp": "2023-03-29T00:00:00",
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Factorization of cyclic polynomial
Factorize $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
Since this is a cyclic polynomial, factors are also cyclic
$$f(a) = a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$f(b) = b(b^2-c^2)+b(c^2-b^2)+c(b^2-b^2) = 0 \Rightarrow a-b$$
is a factor of the given expression. Therefore, other factors are $... | From your last line $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)=m(a-b)(b-c)(c-a);$$
putting $a=0$, $b=1$, $c=2$ we get
$$
0(1^2-2^2)+1(2^2-0^2)+2(0^2-1^2)=m(0-1)(1-2)(2-0)\\
\implies 0+4-2=2m \implies 2m=2 \implies m=1.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Does the Rational Root Theorem ever guarantee that a polynomial is irreducible? I know that the Eisenstein's criterion can guarantee it (when applicable), but what about this?
Also, is there another test that you can use to test irreducibility besides these two? For example, $x^4+2x^2+49$, over the rationals?
| With respect to your specific question (and quartics in general), the RRT doesn't solve the problem but it can help. Once you verify that your quartic has no roots, the only way it can factor is as a product of quadratic polynomials. Additionally, you can guarantee that the quadratic equations are monic and have intege... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/315344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Ellipse in polar coordinates I think Wikipedia's polar coordinate elliptical equation isn't correct. Here is my explanation: Imagine constants $a$ and $b$ in this format -
Where $2a$ is the total height of the ellipse and $2b$ being the total width.
You can then find the radial length, $r$, at any angle $\theta$ to m... | Polar Equation from the Center of the Ellipse
The equation of an ellipse is
$$
\left(\frac{x}{a}\right)^2+\left(\frac{y}{a\sqrt{1-e^2}}\right)^2=1\tag1
$$
Using $x=r\cos(\theta)$ and $y=r\sin(\theta)$ in $(1)$, we get
$$
r^2\cos^2(\theta)+\frac{r^2\sin^2(\theta)}{1-e^2}=a^2\tag2
$$
and we can solve $(2)$ for $r^2$ to g... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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How many solutions to the diophantine equation: $a + b + c + d = 22$
where
$\{a,b,c,d\}$ are distinct integers,
and
for each $x \in \{a,b,c,d\}, 1 \le x \le 9$.
Is there an elegant solution?
| Count only the cases with $a<b<c<d$ and multiply with $4!=24$ afterwards.
Necessarily, $d\ge 7$, as otherwise $a+b+c+d\le 3+4+5+6=18<22$.
If $d=7$, then necessarily $a=4, b=5, c=6$, which gives us one solution $(4,5,6,7)$.
If $d=8$, we have the solution $(3,5,6,8)$, by increasing $c$ $(2,5,7,8)$ and $(3,4,7,8)$.
If $d... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a step by step checklist to check if a multivariable limit exists and find its value? Do we rely on certain intuition or is there an unofficial general crude checklist I should follow?
I had a friend telling me that if the sum of the powers on the numerator is smaller then the denominator, there is a higher c... | I think there is no common method for all types of limits. You need significantly decrease the range of possible functions to get at least some kind of a road map.
For this two particular limits I suggest you the following two "brand new" approaches:
*
*The first one is usage of equivalences (or more general use of ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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separable equation3 Is this solution correct?
Equation:
$x \sin y~dx+(x^2+1)\cos y~dy=0$
solve:
$$x \sin y\,dx=-(x^2+1)\cos y\,dy
\\ \frac{x}{-(x^2+1)}dx=\frac{\cos y}{\sin y}\,dy
\\ \frac{x}{-(x^2+1)}dx=\cot y\,dy
\\ \implies -\frac{1}{2}\ln(|x^2+1|)=-\cot^{-1} y \,dy$$.
| You almost had it, but not quite.
From your second-to-last line:
$$\frac{-x}{(x^2+1}\,dx = \cot(y)\,dy$$
$$\int\frac{-x}{(x^2+1)}\,dx = \int\cot(y)\,dy$$
$$-\frac{1}{2}\ln(x^2+1) + C= \int\frac{\cos(y)}{\sin(y)}\,dy$$
For the left integral, let $u = \sin(y) \implies du=\cos(y) \,dy$.
$$-\frac{1}{2}\ln(x^2+1) + C= \int\... | {
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Complex Analysis Gauss Mean Value Theorem Have I done this correctly?
Evaluate
$\displaystyle\int_{0}^{2\pi}\sin^3(3e^{i\theta} +\frac{\pi}{4})d\theta$
Gauss MVT:
$$f(z_0)=\frac{1}{2\pi}\displaystyle\int_0^{2\pi}f(z_0+re^{i\theta})d\theta$$
So we have the following:
$$\frac{1}{2\pi}f(z_0)=\frac{1}{2\pi}\displaystyle\i... | Since $\sin^3 z$ is a holomorphic function, its average over a circle centered at $\pi/4$ is equal to $\sin^3(\pi/4)=2^{-3/2}$. The average is $\frac{1}{2\pi}\int_0^{2\pi}\dots$, therefore
$$ \int_{0}^{2\pi}\sin^3(3e^{i\theta} +\frac{\pi}{4})\,d\theta = 2\pi \cdot 2^{-3/2}$$
| {
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Please help me to find the value of $ABCDE$ (step by step) $ABCD\times E = DCBA$ with $A,B,C,D$, and $E$ distinct decimal digits (and $ABCD$ representing the concatenation of those digits). How can I find the value each of them?
| First of all, note that $ABCD$ and $DCBA$ have the same digit-sum; this means in particular that they both have the same value $\mod 9$ (call it $r$). But working $\mod 9$, this means that $r\cdot E = r$, meaning that either $E\equiv 1$ (in which case we must have $E=1$ and then trivially $A=D, B=C$ and the solution i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/324018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Range of $\frac{1}{2\cos x-1}$ How can we find the range of $$f(x) =\frac{1}{2\cos x-1}$$
Since range of $\cos x$ can be given as : $-1 \leq \cos x \leq 1$
therefore we can proceed as :$$\begin{array}{rcl}
-2 \leq & 2\cos x & \leq 2 \\
-2-1 \leq & 2\cos x -1 & \leq 2-1\\
-3 \... | I think your calculation is right
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/328493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How can prove this equation. if $a+b=c+d=e+f=\dfrac{\pi}{3}$,
$\dfrac{\sin{a}}{\sin{b}}\cdot\dfrac{\sin{c}}{\sin{d}}\cdot\dfrac{\sin{e}}{\sin{f}}=1$,
Prove that:
$\dfrac{\sin{(2a+f)}}{\sin{(2f+a)}}\cdot\dfrac{\sin{(2e+d)}}{\sin{(2d+e)}}\cdot\dfrac{\sin{(2c+b)}}{\sin{(2b+c)}}=1$
| Consider an equilateral triangle $ABC$, and let $D$ be on $BC$ so that $\angle{BAD}=a$, so $\angle{DAC}=\frac{\pi}{3}-a=b$. Let $E$ be on $AC$ so that $\angle{CBE}=c$, so $\angle{EBA}=\frac{\pi}{3}-c=d$. Let $F$ be on $AB$ so that $\angle{ACF}=e$, so $\angle{FCB}=\frac{\pi}{3}-e=f$.By the sine version of Ceva's theorem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/329316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
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Solve for $X$ in a simple $2\times2$ equation system. I posted a similar question recently but I still have problem with this problem and would appreciate any help!
$$\left[ \begin{array}{cc} 9 & -3\\ 5 & -5\end{array} \right] - X \left[ \begin{array}{cc} -9 & -2\\ 8 & 5\end{array} \right] = E$$ With $E$ i pressume the... | I will express it as equations, as I think that is easier (at least for beginners).
With
$$X=\begin{pmatrix} x_{11} & x_{12}\\ x_{21} & x_{22} \\ \end{pmatrix}$$
At first we make the multiplication
\begin{align*}
X \cdot \begin{pmatrix} -9 & -2\\ 8 & 3 \end{pmatrix}&=
\begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/331048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Showing two graphs isomorphic using their adjacency matrices I don't know how to solve the following problem:
Show that two simple graphs $G$ and $H$ are isomorphic if and only if there exists a permutation matrix $P$ such that $A_G=PA_HP^t$.
Here $A$ is the adjacency matrix. I have a feeling this shouldn't be very d... | Here’s an example that may get you thinking in the right direction. Consider $PAP^t$, where $P$ is the permutation matrix
$$\begin{bmatrix}
0&0&0&1\\
1&0&0&0\\
0&0&1&0\\
0&1&0&0
\end{bmatrix}$$
and, as an illustrative example,
$$A=\begin{bmatrix}
1&2&3&4\\
2&3&4&5\\
0&1&2&3\\
3&2&1&0
\end{bmatrix}\;.$$
We have
$$PA=\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/331233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 1,
"answer_id": 0
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Number of solutions to $x^3 - c^2x^2 + p^2 \equiv 0 \mod p^3$ I would like to find the number of solutions ($\mod p^3$) to $x^3 - c^2x^2 + p^2 \equiv 0 \mod p^3$, where $p$ is prime and $c$ is an integer not divisible by $p$. I think Hensel's Lemma will be useful, but I'm not entirely sure how to apply it.
Help would ... | If $x$ satisfies $x^3 - c^2x^2 + p^2 \equiv 0 \pmod {p^3}$, $x$ also satisfies
$$x^3 - c^2x^2 + p^2 \equiv 0 \pmod {p^2}$$
So $x^2(x-c^2)\equiv 0 \pmod {p^2}$. So $x$ satisfies $p\mid x$ or $p^2\mid(x-c^2)$ and $x$ satisfies only one formula. (Because $c$ is not disible by $p$.)
If $p\mid x$, then there exists $y\in\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/333306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A finite summation involving $2013$ Can you help me compute the summation below?
$$1+\frac{1}{2}+\cdots+\frac{1}{2013}+\frac{1}{1\cdot2}+\frac{1}{1\cdot3}+\cdots+\frac{1}{2012\cdot 2013}+\cdots+\frac{1}{1\cdot2\cdots2013}$$
| Here's how it works for $n=3$.
Find a common denominator:
$$\begin{align}
1 + S &= 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{1\cdot2} + \frac{1}{1\cdot3} + \frac{1}{2\cdot3} + \frac{1}{1\cdot2\cdot3}\\
&=\frac{1\cdot2\cdot3 + 2\cdot3 + 1\cdot3 + 1\cdot2 + 3 + 2 + 1 + 1}{1\cdot2\cdot3}\\
&= \frac{(1+1)(1+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/334160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
$(1-a^2)(1-b^2)(1-c^2)=8abc\; (a,b,c\in \mathbb{Q}^{+})$ has infinitely many solutions Prove that:
$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}$$
has an infinite number of solutions $(a,b,c)$.
Now we found this infinite
$$(a,b,c)=(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+... | The equation,
$$(1-a^2)(1-b^2)(1-c^2)-8abc=0$$
is just a quadratic in any of the variables. Hence,
$$c=\frac{8ab\pm y}{2(-1+a^2+b^2-a^2b^2)}$$
where,
$$64a^2b^2-4(-1+a^2+b^2-a^2b^2)(1-a^2-b^2+a^2b^2) = y^2$$
This quartic polynomial to be made a square is easily reducible to an elliptic curve. From initial rational poin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/334448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a ... | From the original inequality $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
Let’s set $$x= \sqrt{3a + b^3}$$ $$y= \sqrt{3b + c^3}$$ $$z= \sqrt{3c + a^3}$$
By AM-GM we know that $$x^2-2xy+y^2 \ge 0$$ $$y^2-2yz+z^2 \ge0$$ $$z^2-2xz+x^2 \ge 0$$ So we then can set up the following inequalities
$$x^2+y^2\ge2xy$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 4
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Analytic Geometry (high school): Why is the sum of the distances from any point of the ellipse to the two foci the major axis? I don't understand where that formula came from. Could someone explain? For example any point $(x,y)$ on the ellipse from the two foci $(-c,0)$ and $(c,0)$ is equal to $2a$ where $2a$ is the di... | OK, imagine you have tacks at the points $(-c,0$ and $(c,0)$, which hold each end of a string of length $2 a$. We draw an ellipse by holding a pen taut against the string. The sum of the distances to a point on the ellipse from each of the tack points is
$$\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 a$$
The trick is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| Let $s = 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + \dots$
Then $2s = 1 + 2/2 + 3/4 + 4/8 + 5/16 + \dots$
And then subtracting terms with similar denominators gives: $2s - s = 1 +1/2+1/4+1/8+1/16+\dots = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 12,
"answer_id": 4
} |
How to effectively calculate $(1/\sqrt1 + \sqrt2) + (1/\sqrt2 + \sqrt3) +\cdots + (1/\sqrt{99} + \sqrt{100})$ I have this series:
$$\frac{1}{\sqrt1 + \sqrt2} +\frac{1}{\sqrt2 + \sqrt3} +\frac{1}{\sqrt3 + \sqrt4} +\cdots+\frac{1}{\sqrt{99} + \sqrt{100}} $$
My question is, what approach would you use to calculate this pr... | Hint :$\displaystyle \frac{1}{\sqrt{(n+1)}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}$(By multiplying the numerator and the denominator by multiplying $(\sqrt{n+1}-\sqrt{n})$ to both numerator and denominator.)
So we have,
$(1/(\sqrt1 + \sqrt2)) + (1/(\sqrt2 + \sqrt3)) + .. + (1/(\sqrt{99} + \sqrt{100}))=\displaystyle \sum_{n=1}^{9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluate $\lim\limits_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$
*
*Evaluate $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$.
*Examine whether $x^{1/x}$ possesses a maximum ... | $${\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}}=\frac{\sum_{i=1}^{n}i^3}{n^4}=\frac{(n(n+1))^2}{4n^4}=\frac{1}{4}(1+\frac{1}{n})^2$$
$$\Rightarrow \displaystyle \lim_{n\to\infty}{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}}=\lim_{n\to\infty}\frac{1}{4}(1+\frac{1}{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Let $X$ be the number of aces and $Y$ be the number of spades. Show that $X$, $Y$ are uncorrelated. A deck of 52 cards is shuffled, and we deal a bridge of 13 cards. Let $X$ be the number of aces and $Y$ be the number of spades. Show that $X$, $Y$ are uncorrelated.
Here is what I did:
$Cov(X,Y) = E[XY]-E[X]E[Y]$
uncor... | We show that although $X$ and $Y$ are not independent, the conditional expectation of $Y$, given $X=x$, is equal to the plain expectation of $Y$.
Given that $x=0$ (no Aces), we are choosing $13$ cards from the $48$ non-Aces. The expected number of spades is then $13\cdot \frac{12}{48}=\frac{13}{4}$.
Given that $x=1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Finding the locus of a complex number Find the locus of $\arg\left(\frac{z-3}{z}\right) = \frac{\pi}{4}$ where $z$ represent complex number.
Working: $\arg\left(\frac{z-3}{z}\right) $ can be written as $\arg(z-3)-\arg(z) = \frac{\pi}{4}$, or $\arg\left((x-3)+iy\right) - \arg(x+iy)=\frac{\pi}{4}$.
If we take tangent to ... | So, $$\arctan \left(\frac y{x-3}\right)-\arctan \left( \frac yx\right)=\frac\pi4$$
$$\implies \arctan\left(\frac{ \frac y{(x-3)}-\frac yx}{1+ \frac y{(x-3)}\frac yx}\right)=\frac\pi4$$
$$\implies \frac{3y}{(x-3)x+y^2}=\tan\left(\frac\pi4\right)=1$$ assuming $x(x-3)\ne0$
$$\implies x^2-3x+y^2=3y$$
If $x=3,\arctan \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/340799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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If $x > 1$, prove that $f(x) = \dfrac{1}{\sqrt{x^2+1}}\log(x+\sqrt{x^2-1})$ $$f(x) = 2\int_{0}^{1}\dfrac{du}{u^2(1-x)+1+x}$$
I have used partial fractions but solves nothing.
| Since $x>1$ the denominator better reads as $(1+x)-u^2(x-1)$. Now, let's change variables $u = \sqrt{\frac{x+1}{x-1}} w$, giving:
$$
f(x) = \frac{2}{1+x} \sqrt{\frac{x+1}{x-1}} \int_0^\sqrt{\frac{x-1}{x+1}} \frac{\mathrm{d}w}{1-w^2} = \frac{1}{\sqrt{x^2-1}} \int_0^\sqrt{\frac{x-1}{x+1}} \left(\frac{1}{1+w} + \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/341144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Please help with this boundary value problem I came across the following problem and I am stuck on it :
Let $u$ be a solution of the boundary value problem $u''+\dfrac{1}{t}u'=f(t),t \in (0,1)$ and $u'(0)=a,u(1)=b.$ Define for $x^2+y^2 \leq 1,v(x,y)=u(\sqrt {x^2+y^2})$ and $g(x,y)=f(\sqrt {x^2+y^2})$.Then $v$ is a s... | Attacking the problem backwards from Poisson equation to its radial symmetric equation is more intuitive in my opinion.
$\newcommand{\pp}[2]{\frac{\partial #1}{\partial #2}}$
$\newcommand{\pb}[2]{\frac{\partial }{\partial #2}\left(#1\right)}$
$\newcommand{\ppt}[2]{\frac{\partial^2 #1}{\partial #2^2}}$
Consider the Poi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/342395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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The least possible value How to find the least possible value for :$$(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-5)^2$$
For every real $x$
| $(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-5)^2$
$=(y+2)^2+(y+1)^2+y^2+(y-1)^2+(y-2)^2$.Let $(y=(x-3))$
$=5y^2+2(1^2+2^2)\ge2(1^2+2^2)$
Equality occurs at $y=0\Rightarrow x=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/343841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stu... | No. You should have $2(x+4)^2$ on the right after multiplying. Then $$0\le2(x+4)^2-4(x+4)\\0\le(x+4)^2-2(x+4)\\0\le(x+4)\bigl((x+4)-2\bigr)\\0\le(x+4)(x+2).$$ Now, that last inequality is true whenever $x+4$ and $x+2$ have the same sign or one of them is $0$--that is, whenever $x+4\le0$ ($x\le-4$) or $x+2\ge0$ ($x\ge -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/344268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 1
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Find remainder of $F_n$ when divided by $5$
Let $\{ F_n\}$ be the sequence of numbers defined by $F_1=1=F_2;\, F_{n+1}=F_n+F_{n-1}$ for $n \geq 2$. Let $f_n$ be the remainder left when $F_n$ is divided by $5$. Then $f_{2000}$ equals
(A) $0$ $~~~~~~~~~~~~~~~~$ (B) $1$ $~~~~~~~~~~~~~~~~$ (C) $2$$~~~~~~~~~~~~~~~~$ (D) $3... | $$F_n=F_{n-1}+F_{n-2}=F_{n-2}+F_{n-3}+F_{n-2}=2F_{n-2}+F_{n-1}$$
$$=2(F_{n-3}+F_{n-4})+F_{n-3}=3F_{n-3}+2F_{n-4}$$
$$=3(F_{n-4}+F_{n-5})+2F_{n-4}=5F_{n-4}+3F_{n-5}$$
$$\text{ So,} F_n\equiv 3F_{n-5}\pmod 5 $$
As $F_1=F_2=1\implies F_0=F_2-F_1=0\equiv0\pmod 5$
$\implies 5\mid F_n$ if $5\mid n$
Alternatively, from Lemm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/344716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Solving $x+2y+5z=100$ in nonnegative integers I have not done combinatorics since high school, so this is an embarrassingly simple question.
We can solve the diophantine equation $x+y+z=100$ in nonnegative integers using the "bars and boxes" combinatorial method. We have $100$ dots, and we want place 2 partition marker... | We can also solve it without generating functions as follows.
First, as a lemma, what is the number of nonnegative solutions to $x + 2y = n$? For any choice of $y$ such that $0 \le 2y \le n$, we have a unique solution (take $x = n - 2y$), so the number of solutions is the number of such choices, i.e. $\lfloor \frac{n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/346127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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How to find the value of $\sin{\dfrac{\pi}{14}}+6\sin^2{\dfrac{\pi}{14}}-8\sin^4{\dfrac{\pi}{14}}$
Determine
$$
\sin\left(\pi \over 14\right) + 6\sin^{2}\left(\pi \over 14\right)
-8\sin^{4}\left(\pi \over 14\right)
$$
My idea: Let $\displaystyle{\sin\left(\pi \over 14\right)} = x$.
| Try using the Identities
$$\sin^4\theta = \dfrac{3-4\cos2\theta +\cos 4\theta}{8}$$
$$\sin^2\theta = \dfrac{1-\cos2\theta}{2}$$
$$\sin{\dfrac{\pi}{14}}+6\sin^2{\dfrac{\pi}{14}}-8\sin^4{\dfrac{\pi}{14}}$$
Let $\theta = \dfrac{\pi}{14}$
$$\sin{\theta}+6\dfrac{1-\cos2\theta}{2}-8\dfrac{3-4\cos2\theta +\cos 4\theta}{8}$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/346664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
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Please, I need a more detailed explanation of the particular solution of the problem with vectors Here is the problem and its solution (link to the source if you are interested):
Two different points $A$ and $B$ are given. Find a set of such points $M$, that $\overrightarrow{MA}\cdot\overrightarrow{MB}=k^2$, where $k$ ... | Hints:
$$x^2\pm a=\left(x\pm\frac{a}{2}\right)^2-\frac{a^2}{4}$$
and this is what they do in the first step, with both $\,x\,,\,y\,$ . For example,
$$y^2-(y_a+y_b)y=\left(y-\frac{y_a+y_b}{2}\right)^2-\frac{(y_a+y_b)^2}{4}\;,\;\;etc.$$
$$|AB|^2=(x_a-x_b)^2+(y_a-y_b)^2\;\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/346739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Weird calculus limit How to find the following limit?
$$
\lim_{n \to \infty} \dfrac{ 5^\frac{1}{n!} - 4^\frac{1}{n!} }{ 3^\frac{1}{n!} - 2^\frac{1}{n!} }
$$
Edit done to the question.
Thank you!
| Let $x = \dfrac1{n!}$. We then have $$\lim_{n \to \infty} \dfrac{5^{1/n!} - 4^{1/n!}}{3^{1/n!} - 2^{1/n!}} = \lim_{x \to 0} \dfrac{5^x-4^x}{3^x-2^x}$$
Now $$\lim_{x \to 0}\dfrac{a^x - 1}x = \log(a) \tag{$\star$}$$
\begin{align}
\lim_{x \to 0} \dfrac{5^x-4^x}{3^x-2^x} & = \lim_{x \to 0} \dfrac{\dfrac{5^x-1}x-\dfrac{4^x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Convergence or divergence of $\sum\limits^{\infty}_{n=0} \frac{(2n+1)^{2}} {3^{n}(2n)!}$
Show if the series converges or diverges.
$$\sum^{\infty}_{n=0} \frac{(2n+1)^{2}} {3^{n}(2n)!}$$
Can someone please help with proving this? (I think it converges)
| $$\dfrac{(2n+1)^2}{(2n)!} = \dfrac{4n^2+4n+1}{(2n)!} = \dfrac{(2n)(2n-1)+6n+1}{(2n)!} = \dfrac1{(2n-2)!}+ 3 \cdot \dfrac1{(2n-1)!} + \dfrac1{(2n)!}$$
Now $$\sum_{n=0}^{\infty}\dfrac{(2n+1)^2}{(2n)!}x^n = \sum_{n=0}^{\infty}\dfrac{x^n}{(2n)!} + 3 \cdot\sum_{n=1}^{\infty}\dfrac{x^n}{(2n-1)!} + \sum_{n=1}^{\infty}\dfrac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Increasing and decreasing annuity Going in circles.....
Find an expression for the present value of an annuity-immediate where payments start at 1, increase by 1 each period up to a payment of $n$, and then decrease by 1 each period up to a final payment of 1.
Okay... We know that for the first $n$ payments, we have an... | The present values of the payments can be written as
$$\begin{align}
PV &= a_n + a_nv + a_n v^2 + a_n v^3 +a_n v^4 + \ldots +a_n v^n\\
& = a_n(1+v+v^2+v^3+v^4+\ldots + v^n)\\
& = a_n \cdot{\ddot{a}_n}
\end{align}$$
Alternatively, you could write the present value as $$
Ia_n + Da_{n-1} v^n$$ and exploit the relatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/348175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $\alpha $ given $2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha )$ Given:
$$2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha ),$$
where $$0 <\alpha < 90^\circ, $$
find $α.$
The issue I have with this question is the $-3$ on the right hand side, it really complicat... | You can use the so-called double angle formula. Notice that
$$-3\cos(\theta+\alpha) \equiv -3\cos\theta\cos\alpha + 3 \sin\theta\sin\alpha $$
Now you can compare coefficients. You want $-3\cos(\theta+\alpha) \equiv 2\sin\theta - \sqrt{5}\cos\theta$, and so you need to find an $\alpha$ for which $3\sin\alpha = 2$ and $3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/349185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Bessel functions of the first kind How would I show that
$$J_1(x)+J_3(x)=\frac 4x J_2(x)$$
Using the series definition of the Bessel Function, which is
$$J_p(x)=\sum ^\infty _{n=0} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac x2\right)^{2n+p}$$
I know that it can be achieved using some reindexing, but I am not s... | Yes, there is an index shift. Rewrite
$$J_1(x) = \frac{x}{2} + \sum_{n=1}^{\infty} \frac{(-1)^n}{n!(n+1)!} \left ( \frac{x}{2} \right ) ^{2 n+1}$$
$$J_3(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(n-1)!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+1}$$
Then
$$\begin{align}J_1(x)+J_3(x) &= \frac{x}{2} + \sum_{n=1}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/350578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.