Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Consecutive Prime Gap Sum (Amateur) List of the first fifty prime gaps:
1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 14, 4, 6, 2, 10, 2, 6, 6, 4, 6, 6, 2, 10, 2, 4, 2, 12, 12, 4, 2, 4.
My conjecture is that the sum of consecutive prime gaps is always prime whenever a prime gap ... | Let $p_n$ denote the $n^{\text{th}}$ prime. Then $p_{n+1}-p_n$ is the $n^{\text{th}}$ prime gap. The sum of the first $k$ prime gaps is $$\sum_{n=1}^k(p_{n+1}-p_n) = p_{k+1} - p_1 = p_{k+1} - 2.$$ Now, if the $k^{\text{th}}$ prime gap is $2$, that is $p_{k+1} - p_k = 2$, then $p_{k+1} - 2 = (p_k + 2) - 2 = p_k$ which i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Finding the asymptotes of a general hyperbola I'm looking to find the asymptotes of a general hyperbola in $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ form, assuming I know the center of the hyperbola $(h, k)$. I came up with a solution, but it's too long for me to be confident that I didn't make a mistake somewhere, so I wa... | So a fairly simple calculus based solution arises from knowing that the Asymptotes are the points where the slope tends toward being constant
That is given
$$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$
We begin by deriving with respect to $x$
$$2Ax + 2By \frac{dy}{dx} + Cy+Cx\frac{dy}{dx} + D + E\frac{dy}{dx} = 0 $$
And no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Determining if a recursively defined sequence converges and finding its limit Define $\lbrace x_n \rbrace$ by $$x_1=1, x_2=3; x_{n+2}=\frac{x_{n+1}+2x_{n}}{3} \text{if} \, n\ge 1$$ The instructions are to determine if this sequence exists and to find the limit if it exists.
I prove that the sequence is eventually decr... | This is a linear recurrent sequence of order $2$. The characteristic
polynomial $X^2-\frac{X+2}{3}$ has roots $1$ and $-\frac{2}{3}$. So there
are two constants $a$ and $b$ such that $x_n=a(1^n)+b(\frac{-2}{3})^n$.
Using the initial conditions, we obtain $a=\frac{11}{5}$ and $b=\frac{9}{5}$.
The limit is therefore $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Integrate $\int \left(A x^2+B x+c\right) \, dx$ I am asked to find the solution to the initial value problem:
$$y'=\text{Ax}^2+\text{Bx}+c,$$
where $y(1)=1$,
I get:
$$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$
But the answer to this is:
$$y=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1.$$
Co... | Use the condition $y(1)=1$, i.e. substitute $x=1$ and then you have
\begin{equation}
\frac{A}{3} + \frac{B}{2} + c +d = 1
\end{equation}
So, you get
\begin{equation}
d = 1 -\frac{A}{3} - \frac{B}{2} - c
\end{equation}
By substituting in $\frac{A}{3}x^3+\frac{B}{2}x^2+cx+d$, you get that equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Partial fraction decomposition of type $1/(x^2+k)$ I know that partial fraction of this can be written as:
$$\frac{3x}{(1+x)(2+x)}=\frac{-3}{1+x}+\frac{6}{2+x}$$
Which can be done in these ways:
$$\frac{3x}{(1+x)(2+x)}=\frac{A}{1+x}+\frac{B}{2+x}\implies3x=A(2+x)+B(1+x),\forall\;x$$
And now solving it to get A and B.Al... | Decomposing $1/(x+n)(x^2 + k)$ is only difficult (for me) when k is a positive quantity. Otherwise just use $a^2 - b^2$. For example:
$$\frac1{(x-3)(x^2-1)} = \frac1{(x-3)(x-1)(x+1)} = \frac{A}{x-3} + \frac{B}{x-1} + \frac{C}{x+1}$$
Here,
$$A = \frac1{(3-1)(3+1)} = \frac1{8}\\
B = \frac1{(1-3)(1+1)} =\frac{-1}{4} \\
C ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/902874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Simple use of log I am struggling to see how we can go from the first expression to the second:
$$\begin{align}
2\log_3 12 - 4\log_3 6 &= \log_3 \left ( \frac{4^2 \cdot 3^2}{2^4 \cdot 3^4} \right )\\
&= \log_3 (3^{-2}) = -2
\end{align}$$
| Remember, $$\log ab = \log a + \log b$$
and
$$m\log a=\log a^m.$$
HINT:
$$12^2 = (4\cdot 3)^2=4^2\cdot 3^2.$$
Similarly,
$$6^4 = (2\cdot 3)^4 = 2^4\cdot 3^4$$
I think you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/903168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find $\int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$ $$I=\int x.\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$$
Try 1:
Put $z= \ln(x+\sqrt{1+x^2})$, $\mathrm dz=1/\sqrt{1+x^2}\mathrm dx$
$$I=\int \underbrace{x}_{\mathbb u}\underbrace{z}_{\mathbb v}\mathrm dz=x\int zdz-\int (z^2/2)\mathrm dz\tag{... | Hint :
Use IBP by setting $u=x$ and $dv=\dfrac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\ dx$. Note that: $\text{arsinh}\ x=\ln(x+\sqrt{1+x^2})$ and $\dfrac{d}{dx}\text{arsinh}\ x=\dfrac{1}{\sqrt{1+x^2}}$, then
\begin{align}
\int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\ dx&=x\left[\int\text{arsinh}\ x\ d(\text{arsinh}\ x)\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Inequality involving Pochhammer symbols Let $m,S$ be integers satisfying $2\leq m\leq S$. I would like to show that
$$h_1\left(x\right) h_3\left(x\right) \leq h_2^2\left(x\right)$$
for all $x\geq 0$ where
$$h_k\left(x\right) \equiv {}_2F_1\left(k,k-1-m;S-m+k,-x\right).$$
${}_2F_1$ denotes the hypergeometric function.
... | I'll prove here some other easy cases, the first being in the comments.
My numerical experiments indicate that one should require $2 \le k \le m \le S$ to show
\begin{align}
q_k(x) := h_{k-1}(x)h_{k+1}(x) - h_{k}^2(x) \le 0
\end{align}
Step 0, one sees immediately that $h_k(x) = h_{k,S,m}(x)$ is a polynomial of degree ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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How do you solve for x in this equation? I tried expanding, but I still can't get rid of the exponents to isolate x.
$$\frac{(1+x)^4-1}{x}=4.374616$$
Thank you in advance for your help.
| \begin{align}
(1+x)^{4} - 1 &= [(1+x)^{2} - 1] [ (1+x)^{2} + 1] = (1+x-1)(1+x+1)(x^{2} + 2x +1)
\nonumber\\
&= x(x+2)(x^{2}+2x+2)
\end{align}
or
\begin{align}
\frac{(1+x)^{4} -1}{x} = (x+2) (x^{2} + 2x +2).
\end{align}
Now, $4.374616$ can be factored into
\begin{align}
4.374616 = (2.06)(2.1236) = (2 + .06)((.06)^{2} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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To test the convergence of series 2 To test the convergence of series 2
$\displaystyle \frac{a+x}{1!}+\frac{(a+2x)^2}{2!}\frac{(a+3x)^3}{3!}+...\infty$
My Attempt:
$$\begin{align}
\frac{u_n}{u_{n+1}} & = \frac{(a+nx)^n}{n!}\frac{(n+1)!}{(a+(n+1)x)^{n+1}} \\
& = \frac{(a+nx)^n}{1}\frac{n+1}{(a+nx+x)^{n+1}} \\
& = \f... | $$\begin{align}
lim_{n \to \infty}\frac{u_n}{u_{n+1}}
& = lim_{n \to \infty} (1+a/nx)^n\frac{1+1/n}{(1+a/nx+1/n)^{n+1}}\frac{1}{x} \\
& =lim_{n \to \infty} (1+a/nx)^n\frac{1}{ lim_{n \to \infty}(1+(a+x)/nx)^n}\frac{1}{x}\\
&= e^{a/x} \dfrac{1}{xe^{a+x/x}}\\ &=\dfrac{e^{-1}}{x}\end{align}$$
Therefore your series is c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Indefinite integral of $\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}$ How do I find $$\int\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}\mathrm dx$$
I used partial fractions by breaking up $x^2 + 2x - 2$ into $(x+1)^2 - 3$ and split it into $(a+b)(a-b)$ but as u can see it's extreme tedious. I was won... | $$\int\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}\mathrm dx\\
=\int\left(+\frac{2x+2}{2(x^2+2x+2)}+\frac{2x+2}{2(x^2+2x-2)}-\frac1{\frac{(2x+2)^2}4+1}\right)\mathrm dx\\
=\frac12\ln(x^2+2x+2)+\frac12\ln(x^2+2x-2)-\arctan\left(\frac{2x+2}2\right)+\mathbb C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/905084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show that $ax^2+2hxy+by^2$ is positive definite when $h^2The question asks to "show that the condition for $P(x,y)=ax^2+2hxy+by^2$ ($a$,$b$ and $h$ not all zero) to be positive definite is that $h^2<ab$, and that $P(x,y)$ has the same sign as $a$."
Now I've seen questions similar to this before where it's a two variab... | In higher dimensions (than 1), there is a elegant matrix based approach. In this approach, the sign is determined by the diagonal matrix. Specifically, note that for your question $$a{x^2} + 2hxy + b{y^2} = \left[ {\begin{array}{*{20}{c}}x&y\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a&h\\h&b\end{array}} \right]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Evaluation of a dilogarithmic integral
Problem. Prove that the following dilogarithmic integral has the indicated value:
$$\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(x)}\operatorname{Li}_2{(x)}}{1-x}\stackrel{?}{=}-11\zeta{(5)}+6\zeta{(3)}\zeta{(2)}.$$
My attempt:
I began by using the polylogarithmic expansion in terms ... | It is easy to see that
$$2\sum^\infty_{n=1}\frac{H_n^{(2)}}{(n+1)^3}=2\sum^\infty_{n=1}\frac{H_{n+1}^{(2)}}{(n+1)^3}-2\sum^\infty_{n=1}\frac{1}{(n+1)^5}=2\sum^\infty_{n=1}\frac{H_n^{(2)}}{n^3}-2\zeta(5)$$
Consider $\displaystyle f(z)=\frac{\pi\cot{\pi z} \ \Psi^{(1)}(-z)}{z^3}$. We know that
$$\pi\cot{\pi z}=\frac{1}{z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
How to express $\log_2 (\sqrt{9} - \sqrt{5})$ in terms of $k=\log_2 (\sqrt{9} + \sqrt{5})$? If $$k=\log_2 (\sqrt{9} + \sqrt{5})$$
express $\log_2 (\sqrt{9} - \sqrt{5})$ in terms of $k$.
| Add to both sides the term $\log_2 (\sqrt{9} - \sqrt{5})$, then you have
$$
\log_2 (\sqrt{9} - \sqrt{5}) + k = \log_2 (4),
$$
so that
$$
\log_2 (\sqrt{9} - \sqrt{5}) = 2 - k.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/907687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limit of a quotient with a radical in the numerator I have a limit but I'm so confused in how to rationalize the numerator because it has two numbers separated. How should I change the signs, please help me out.
$$\lim \limits_{x \to 6}{\sqrt{x+10}-x+2\over 3x-18}$$
| The numerator is $\sqrt{x+10} - (x-2)$, so let's multiply the numerator and denominator by $\sqrt{x+10} + (x-2)$. This gives us:
$\dfrac{\sqrt{x+10} - (x-2)}{3x-18}$
$= \dfrac{\sqrt{x+10} - (x-2)}{3x-18} \cdot \dfrac{\sqrt{x+10} + (x-2)}{\sqrt{x+10} + (x-2)}$
$= \dfrac{(x+10)-(x-2)^2}{(3x-18)(\sqrt{x+10} + (x-2))}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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help in Laplace and partial fractions Can any one teach me how to solve C2.(a) and (b) step by step?
C2. (a) Resolve $\frac{1}{s^2(s^2+s+1)}$ into partial fractions of the form $\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+s+1}$.
Hence, resolve $\frac{1}{s(s^2+s+1)}$ into its partial fractions.
C2. (b) Find
$$\mathcal{... | (a) The function
$$ f(z)=\frac{1}{z^2(z^2+z+1)}$$
has a double pole in $z=0$ and a simple pole in $z=e^{\pm 2\pi i/3}$, since $z^2+z+1=\frac{z^3-1}{z-1}$. This gives:
$$ f(z)=\frac{A}{z^2}+\frac{B}{z}+\frac{C}{z-\omega}+\frac{D}{z-\omega^2}+g(z)$$
where $g(z)$ is a holomorphic function. By computing:
$$\lim_{z\to 0}z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Check if two vector equations of parametric surfaces are equivalent Give the vector equation of the plane through these lines:
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}\,\,\,$ and $\,\,\,\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\... | Since the first equation describes a plane whereas the second one describes a straight line the equations aren't equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/912048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Equation $3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$ Solve the equation
$$3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$$
Having a complex root of modulus $1$.
To get the solution, I tried to take a complex root $\sqrt{\frac{1}{2}} + i \sqrt{\frac{1}{2}}$ but couldn't get the solution right. Please help me.
| Note that you have an equation with integer coefficients (real coefficients would do), so that any complex root $a$ of modulus $1$ implies a conjugate root $\bar {a }$ with $a\bar a=1$. You will therefore have a real factor of the form $x^2-bx+1$ with $b=a+\bar a$. Making obvious deductions from the leading and constan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/913039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Closed-forms of infinite series with factorial in the denominator How to evaluate the closed-forms of series
\begin{equation}
1)\,\, \sum_{n=0}^\infty\frac{1}{(3n)!}\qquad\left|\qquad2)\,\, \sum_{n=0}^\infty\frac{1}{(3n+1)!}\qquad\right|\qquad3)\,\, \sum_{n=0}^\infty\frac{1}{(3n+2)!}\\
\end{equation}
Of course Wolfra... | This problem can be solved without advanced techniques. We have the Taylor series
$$
e^x = \sum_{n\geq 0} \frac{x^n}{n!}.
$$
As the questioner noted, plugging $x=1$ yields the equation $A+B+C=e$ connecting the three unknown sums. However, plugging in any cube root of unity also sheds light on the question because the n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/914176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
The line is tangent to a parabola The line $y = 4x-7$ is tangent to a parabola that has a $y$-intercept of $-3$ and the line $x=\frac{1}{2}$ as its axis of symmetry. Find the equation of the parabola. I really need help solving this question. THx
| Let the formula for the parabola be $f(x)=ax^2+bx+c$
1. y-intercept: $y$-intercept is @ $(0,-3)$ so we have: $f(0)=-3 \Rightarrow c=-3$.
2. Vertex: The vertex is @ $(x,\frac{1}{2})$ and we now that at the vertex, the derivative is equal to zero, so we have:
$f'(\frac{1}{2})=0 \Leftrightarrow 2a\cdot(\frac{1}{2})+b=0 \R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $\int\frac{1-5\sin^2 x}{\cos^5 x\cdot \sin^2 x}dx = \frac{f(x)}{\cos^5 x}+\mathcal {C}$. Then value of $f(x)$. If $\displaystyle \int\frac{1-5\sin^2 x}{\cos^5 x\cdot \sin^2 x}dx = \frac{f(x)}{\cos^5 x}+\mathcal {C}$. Then value of $f(x)$.
$\bf{My\; Try::}$ Given $$\displaystyle \int\frac{1-5\sin^2 x}{\cos^5 x\cdot ... | Consider the differentiation of both sides of the integral to find $f(x)$. In this view differentiating
\begin{align}
\int \frac{1-5 \sin^{2}(x)}{\cos^{5}(x) \, \sin^{2}(x)} \, dx = \frac{f(x)}{\cos^{5}(x)} + c
\end{align}
leads to
\begin{align}
f'(x) + 5 \tan(x) \, f(x) = \csc^{2}(x) -5.
\end{align}
Now, the right-ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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The limit of $(4-\sqrt{16-7\sin(x)})/(8x)$ at zero without using L'Hôpital I stumbled across this silly limit and I am perplexed at how I can arrive to a solution by only relying on the simplest rules of limits.
$$
\lim_{x \to 0}\frac{4-\sqrt{16-7\sin(x)}}{8x}
$$
Any help is appreciated, thanks in advance.
| $$ \lim_{x\rightarrow 0}\frac{4-\sqrt{16-7\sin x}}{8x}=\\
\lim_{x\rightarrow 0}\frac{4-\sqrt{16-7\sin x}}{8x}\frac{4+\sqrt{16-7\sin x}}{4+\sqrt{16-7\sin x}}=\\
\lim_{x\rightarrow 0}\frac{16-{16+7\sin x}}{8x}\frac{1}{4+\sqrt{16-7\sin x}}=\\
\lim_{x\rightarrow 0}\frac{{7\sin x}}{8x}\frac{1}{4+\sqrt{16-7\sin x}}=\\
\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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General formula of repeated roots. Prove that $$\underbrace{\sqrt{k\sqrt{k\sqrt{k\sqrt{\cdots\sqrt{k}}}}}}_{n\text { times}}=k^{1-1/2^n}$$
How do I derive this formula?
| $$\underbrace{\sqrt{k\sqrt{k\sqrt{k\sqrt{\cdots\sqrt{k}}}}}}_{n\text { times}} = \sqrt{k}\times\sqrt[4]{k}\times \ldots \times \sqrt[2^n]{k} = k^{1/2}\times k^{1/4}\times \ldots \times k^{1/2^n} = \\ = k^{\Large\frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^n}} = k^{\Large1-\frac{1}{2^n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/919313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Proof by Induction - Algebra Problem (Steps included but not understood) I do not quite understand this proof, if anyone could explain the steps for me it would be greatly appreciated. It's probably something glaringly obvious I'm not seeing, thanks in advance.
Prove that for every integer $n \ge 0,$ the number $4^{2n+... | Here is a variant on the same idea. $4^{2n+1}+3^{n+2}=4(13+3)^n+9\times 3^n$. Expand the bracket by the binomial theorem. Every term has a factor 13 except $3^n$, so we have that $4^{2n+1}+3^{n+2}$ is a multiple of 13 plus $3^n(4+9)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/919380",
"timestamp": "2023-03-29T00:00:00",
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"answer_count": 4,
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To find the determinant of a matrix Given $A_{n\times n}$=$(a_{ij}),$ n $\ge$ 3, where $a_{ij}$ = $b_{i}^{2}$-$b_{j}^2$ ,$i,j = 1,2,...,n$ for some distinct real numbers $b_{1},b_{2},...,b_{n}$. I have to find the determinant of A. I can see that A is a skew-symmetric matrix. So determinant of A is $0$ when n is odd. B... | Let $v := \begin{bmatrix}
\dfrac{b_2^2 - b_3^2}{b_2^2 - b_1^2} &
\dfrac{b_1^2 - b_3^2}{b_1^2 - b_2^2} &
-1 &
0 &
\ldots &
0
\end{bmatrix}^T$ if $b_1^2 \ne b_2^2$, or else
$v := \begin{bmatrix}
1 &
-1 &
0 &
\ldots &
0
\end{bmatrix}^T$. Then $Av = 0$, so $\ker A \ne 0 \implies \det A = 0$.
Proof of $Av = 0$: Let $r_j :... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Evaluating $\int \:\sqrt{1+e^x}dx$ , why I got different answers? I've got 2 steps to evaluating $\int \:\sqrt{1+e^x}dx$ which lead to different values
first step :
$\int \:\sqrt{1+e^x}dx$
let $u\:=\:\sqrt{1+e^x}$ , $du\:=\:\frac{e^x}{2\sqrt{1+e^x}}dx$ , but $e^x\:=\:u^2-1$ and substitute $u$ into $du$, i get $dx\:=\... | Suppose $y = \tanh^{-1} x$. Then $$x = \tanh y = \frac{\sinh y}{\cosh y} = \frac{(e^y - e^{-y})/2}{(e^y + e^{-y})/2} = \frac{e^{2y} - 1}{e^{2y}+1}.$$ Hence $(e^{2y}+1)x = e^{2y}-1$, or $e^{2y}(1-x) = x+1$, or $$e^{2y} = \frac{1+x}{1-x}.$$ It follows that $$y = \frac{1}{2} \log \frac{1+x}{1-x} = \tanh^{-1}x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Polynomial factοrisation Problem
Let $P(x)=x^n+64$ be a polynomial. Find the form of the natural number $n$ such that $P(x)=P_1(x)\times P_2(x)$, $\deg P_1(x),\deg P_2(x)\geq1$.
I thought of taking $n \mod 4$. For $n=0\mod 4$ we can use Sophie Germain's identity to show that $P(x)$ is reducible but after this poin... | Here are the the ones I found
\begin{align}
x^{12k+0}+64 =\\ (x^{2k}-2x^k+2)(x^{2k}+2x^k+2)(x^{4k}-2x^{3k}+2x^{2k}-4x^k+4)(x^{4k}+2x^{3k}+2x^{2k}+4x^k+4)\\
x^{12k+3}+64 = (x^{4k+1}+4)(x^{8k+2}-4x^{4k+1}+16)\\
x^{12k+4}+64 = (x^{6k+2}-4x^{3k+1}+8)(x^{6k+2}+4x^{3k+1}+8)\\
x^{12k+6}+64 = (x^{4k+2}+4)(x^{8k+4}-4x^{4k+2}+16... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/926774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $a+b+c=1$ and $abc>0$, then $ab+bc+ac<\frac{\sqrt{abc}}{2}+\frac{1}{4}.$ Question:
For any $a,b,c\in \mathbb{R}$ such that $a+b+c=1$ and $abc>0$, show that
$$ab+bc+ac<\dfrac{\sqrt{abc}}{2}+\dfrac{1}{4}.$$
My idea: let
$$a+b+c=p=1, \quad ab+bc+ac=q,\quad abc=r$$
so that
$$\Longleftrightarrow q<\dfrac{\sqrt{r}}{2... | OK. Let me try to complete the answer.
We want to prove that for $a,b,c\in \mathbb{R},abc>0$,
$$ab+bc+ac<\dfrac{\sqrt{abc}}{2}+\dfrac{1}{4}. \tag{1}$$
In a previous post, it is proved for $a,b,c\ge 0$
Because $abc>0$, the only other possibility is that two of the numbers are negative and one is positive. We can assume ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Calculus limit epsilon delta Prove using only the epsilon , delta - definition
$\displaystyle\lim_{x\to 2}\dfrac{1}{x} = 0.5$
Given $\epsilon > 0 $, there exists a delta such that
$ 0<|x-2|< \delta$ implies $|(1/x) – 0.5| < \epsilon$.
Therefore, $$\begin{align}|(1/x) – 0.5| & = |(2-x)/2x| \\
& =|\frac{-(x-2)}{2(x-2+... | Hint: You want the following to hold for all $x$ such that $2-\delta < x < 2+\delta$:
$\left|\dfrac{1}{x} - \dfrac{1}{2}\right| < \epsilon$
$-\epsilon < \dfrac{1}{x} - \dfrac{1}{2} < \epsilon$
$\dfrac{1}{2} -\epsilon < \dfrac{1}{x}< \dfrac{1}{2} + \epsilon$
For what range of values for $x$ is this inequality true? Tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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The area not covered by six pointed star In a circle with radius $r$, two equi triangles overlapping each other in the form of a six pointed star touching the circumference is inscribed! What is the area that is not covered by the star?
Progress
By subtracting area of the star from area of circle , the area of the surf... | If the six-pointed star is regular, then the answer is $r^2(\pi-\sqrt{3})$. If it is not, then the answer can be larger, up to a limit of $r^2\big(\pi-\frac{3}{4}\sqrt{3}\big)$.
Proof
The required area is the area of the circle ($\pi r^2$) minus the area of the star.
The area of the star is the area of a large equilat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/932172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove that $\frac{1}{1^4}+\frac{1}{2^4}+\cdots+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ I have to show: $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ for natural $n$
I tried to show it by induction (but I think it could be possible to show it using some ineqaulity of means) so for... | You want to show that
$$2-\frac{1}{\sqrt{n}}+\frac1{(n+1)^4}\le 2-\frac1{\sqrt{n+1}}.\tag{1}$$
If (1) is true, then using induction hypothesis one gets
$$1+\frac1{2^4}+\cdots+\frac{1}{n^4}+\frac{1}{(n+1)^4}\le 2-\frac1{\sqrt{n}}+\frac1{(n+1)^4}\le 2-\frac1{\sqrt{n+1}},$$
and we are done by induction.
Showing (1) should... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/932427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Gosper's Identity $\sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k]=1 $ The page on Binomial Sums in Wolfram Mathworld http://mathworld.wolfram.com/BinomialSums.html (Equation 69) gives this neat-looking identity due to Gosper (1972):
$$\sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k]=1 $$
Would anyon... | Consider the expression
\begin{align}
\sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k]
\end{align}
in the following way. First consider the generating function of the series
\begin{align}
S_{n} = \sum_{k=0}^{n} \binom{k+n}{k} \, x^{k}
\end{align}
for which it is seen that
\begin{align}
\phi(x, t) &= \sum_{n=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Why this gamma function reduces to the factorial? $$\Gamma(m+1) = \frac{1\cdot2^m}{1+m}\frac{2^{1-m}\cdot3^m}{2+m}\frac{3^{1-m}\cdot4^m}{3+m}\frac{4^{1-m}\cdot5^m}{4+m}\cdots$$
My books says that in a letter from Euler to Goldbach, this expression reduces to $m!$ when $m$ is a positive integer, but that Euler verified ... | The Product
$$
\begin{align}
\prod_{k=1}^n\frac{\color{#C00000}{k^{1-m}}\color{#00A000}{(k+1)^m}}{\color{#0000FF}{k+m}}
&=\color{#C00000}{n!^{1-m}}\color{#00A000}{(n+1)!^m}\color{#0000FF}{\frac{m!}{(n+m)!}}\\
&=n!\left(\frac{(n+1)!}{n!}\right)^m\frac{m!}{(n+m)!}\\
&=n!(n+1)^m\frac{m!}{(n+m)!}\\
&=m!\color{#00A000}{(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/934767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluating an indefinite integral $\int\sqrt {x^2 + a^2} dx$ indefinite integral $$\int\sqrt {x^2 + a^2} dx$$
After some transformations and different substitution, I got stuck at this
$$a^2\ln|x+(x^2+a^2)| + \int\sec\theta\tan^2\theta d\theta$$
I am not sure I am getting the first step correct. Tried substituting $ x=... | Let $x=a\tan\theta$, so $dx=a\sec^{2}\theta d\theta$ to get
$\int\sqrt{x^2+a^2}\;dx=a^2\int\sec^{3}\theta\;d\theta$. Using integration by parts with $u=\sec\theta$ and $dv=\sec^{2}\theta\;d\theta$ gives
$\hspace{.6 in}\sec^{3}\theta\;d\theta=\frac{1}{2}\left(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\right)+C$,
s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/935519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
How do I solve $\lim_{x\to -\infty}(\sqrt{x^2 + x + 1} + x)$? I'm having trouble finding this limit:
$$\lim_{x\to -\infty}(\sqrt{x^2 + x + 1} + x)$$
I tried multiplying by the conjugate:
$$\lim_{x\to -\infty}(\frac{\sqrt{x^2 + x + 1} + x}{1} \times \frac{\sqrt{x^2 + x + 1} - x}{\sqrt{x^2 + x + 1} - x}) = \lim_{x\to -\i... | As $\;x<0\;$
$$\frac{x+1}{\sqrt{x^2+x+1}-x}\cdot\frac{-\frac1x}{-\frac1x}=\frac{-1-\frac1x}{\sqrt{1+\frac1x+\frac1{x^2}}+1}\xrightarrow[x\to-\infty]{}-\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solving Diophantine equations involving $x, y, x^2, y^2$ My father-in-law, who is 90 years old and emigrated from Russia, likes to
challenge me with logic and math puzzles. He gave me this one:
Find integers $x$ and $y$ that satisfy both $(1)$ and $(2)$
$$x + y^2 = 8 \tag{1} $$
$$x^2 + y = 18 \tag{2}$$
I found one sol... | Here is an alternative way to solve for the original problem.
$$\left\{\begin{array}{ll}x+y^2&=&8\\x^2+y&=&18\end{array}\right. \tag{1}$$
Instead of adding, subtract the two equations and get
$$x^2-y^2+y-x=10\Rightarrow\\(x-y)(x+y-1)=10$$
For two integers $a$ and $b$ to have product $10$, list all possibilities. $$1\ti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/937891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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How find this integral $\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$ let $$D=\{(x,y)|y\ge x^3,y\le 1,x\ge -1\}$$
Find the integral
$$I=\dfrac{1}{2}\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$$
My idea:
$$I=\int_{0}^{1}dx\int_{x^3}^{1}(x^2y+2y^2)dy+\int_{-1}^{0}dx\int_{0}^{-x^3}(xy^2+2x+2y^2)dy$$
so
$$I=\int_{0}^{1}[\dfrac{1}{2}x^2y^2+\dfrac... | The domain $D$ looks roughly like a right triangle $ABC$ with the right angle $B$ at $(-1,1)$, $A$ at $(-1,-1)$, $C$ at $(1,1)$ and a curve $y=x^3$ instead of a straight line from $A$ to $C$. Since the curve does not do anything tricky (one value of $x$ maps to one value of $y$ and vice-versa) you can do this as a sing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/937955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Closed Form of Recursion $a_n = \frac{6}{a_{n-1}-1}$ Given that $a_0=2$ and $a_n = \frac{6}{a_{n-1}-1}$, find a closed form for $a_n$.
I tried listing out the first few values of $a_n: 2, 6, 6/5, 30, 6/29$, but no pattern came out.
| I think that Semiclassical proposed a very nice solution rewriting $$a_n = \frac{6}{a_{n-1}-1}$$ $$\dfrac{1}{a_n+2}=\dfrac{1}{2}-\dfrac{3/2}{a_{n-1}+2}$$ So, let us define $$b_n=\dfrac{1}{a_n+2}$$ (with $b_0=\dfrac{1}{4}$); so the recurrence equation is simply $$b_n=\dfrac{1}{4}-\dfrac{3}{2}b_{n-1}$$ from which $$b_n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/939725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Probably of 2 six in 5 dice rolls What is the probability of obtaining exatcly 2 six when rolling a dice 5 times?
In order to obtain this probability, I will need to devide the number of favorable events by the number of possible events. I believe the denominator is $6^5$. But I am having troubles figuring out the nume... | The generating function for the five throws is
$$ \left(\frac{s}{6}+\frac{5 o}{6}\right)^5 $$ where o are the other faces (1,2,3,4,5) and s represents the six. You need the coefficient of $$ s^2o^3 $$. This you get from Pascals triangle which tells us that the expansion of $$(a+b)^5$$ is
$$(a+b)^5 = a^5 + 5 a^4 b + 10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/945277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Prove $\frac{n^2+2}{(2 \cdot n^2)-1} \to \frac{1}{2}$
Prove $\frac{n^2+2}{(2 \cdot n^2)-1} \to \frac{1}{2}$ for $n \to \infty$.
I've been looking at this for hours! Also, sorry I don't have the proper notation.
This is where I'm at:
$$
\left| \frac{n^2 + 2}{2 \cdot n^2 - 1} - \frac{1}{2}\right| = \left| \frac{5}{4 ... | You might find it easier to write
$$\dfrac{n^2+2}{2n^2-1} = \dfrac{1}{2} \times \dfrac{2n^2+4}{2n^2-1} = \dfrac{1}{2} + \dfrac{5}{4n^2-2}$$ and note you can make $\dfrac{5}{4n^2-2}$ as small as you want by making $n$ large enough.
If you must use epsilon-delta or epsilson-$N_0$, then note that $\dfrac{5}{4n^2-2} \lt \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How do I solve this geometric series I have this geometric series $2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{128}$to solve. So I extract the number two and get $2(\frac{1}{2}^0+ \frac{1}{2}^1+...+ \frac{1}{2}^7)$
I use the following formula $S_n= \frac{x^{n+1}-1}{x-1}$ so I plug in the values in this formula and get ... | $$S=2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{128}$$
Now if we subtract $2$ from both sides of the equation we get
$$S-2=1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{128}$$
$$S-2=\frac{1}{2}\cdot \left(2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{64}\right)$$
Now notice that $2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/948146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Integral of voltage, $\int_{-a}^a \frac{dy}{\sqrt{x^2 + y^2}}$ This is (probably) a very easy integral to solve, but for some reason the answer just isn't coming to me (or at least the one my professor got isn't). He gave us a formula for voltage along the x-axis from a charged wire of length 2a, but none of the detail... | The ideal substitution in this case would be $\sqrt{x^2+y^2}+y=t$. The substitution relation gives $t$ as a function of $y$, but we also need to know $y$ as a function of $t$. We can isolate $y$ as follows:
$$\begin{align}
\sqrt{x^2+y^2}+y&=t\\
\sqrt{x^2+y^2}&=t-y\\
x^2+y^2&=(t-y)^2\\
x^2+y^2&=t^2-2ty+y^2\\
x^2&=t^2-2t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/948502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\int\frac{2x+1}{x^2+2x+5}dx$ by partial fractions $$\int\frac{2x+1}{x^2+2x+5}dx$$
I know I'm supposed to make the bottom a perfect square by making it $(x+1)^2 +4$ but I don't know what to do after that. I've tried to make $x+1= \tan x$ because that's what we did in a class example but I keep getting stuck.
| $$\frac{d}{dx} \ln(x^2 + 2x + 5) = \frac{2x + 2}{x^2 + 2x + 5}dx$$
$$\int\,\left(\dfrac{2x+2}{x^2 + 2x + 5}- \dfrac{1}{x^2 + 2x + 5}\right)\,dx = \ln(x^2 + 2x + 5) + C - \int\,\dfrac{1}{x^2+2x+5}dx$$
=
Let $u=x+1$, so $du = dx$, $x = u - 1$, and $$x^2 + 2x + 5= u^2 - 2u + 1 + 2u - 2 + 5 = u^2 + 4$$
$$\begin{align}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/950354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Given $a+b+c=4$ find $\max(ab+ac+bc)$ $a+b+c = 4$.
What is the maximum value of $ab+ac+bc$?
Could this be solved by a simple application of Jensen's inequality? If so, I am unsure what to choose for $f(x)$. If $ab+ac+bc$ is treated as a function of $a$ there seems no easy way to express $bc$ in terms of $a$.
EDIT: The ... | @James, first we square both sides:
$$(a+b+c)^2=1 \implies 1=a^2+b^2+c^2+2(ab+bc+ac)\tag{1}$$
$$\because a^2+b^2+c^2 \geq ab +ac + bc\tag{2}$$
For a,b,c>0
$$ \\ 1 \geq 3(ab+bc+ac) \tag{3}$$ $$ \frac{1}{3} \geq ab +bc+ca\tag{4} \\ \square$$
$(2)$ can be easily proven by considering: $(a-b)^2 \geq 0 \\ (b-c)^2 \geq 0 \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/951997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $ \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x $ In a related question the following integral was evaluated
$$
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
=\int_0^{\pi} \frac{\mathrm{d}x/2}{1 +... | The integrands are both periodic with period $\pi$, so it suffices to verify the identity over any interval of length $\pi$, including $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Then, since both integrands are even functions, it suffices to check that the identity holds when integrating over $\left[0, \frac{\pi}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/955294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 3
} |
Any shorter way to solve trigonometric problem? If $10 \sin^4\theta + 15 \cos^4 \theta=6$, then find value of $27 \csc^2 \theta + 8\sec^2 \theta$
I know the normal method o solve this problem in which we need to multiply L.H.S. of $10 \sin^4\theta + 15 \cos^4 \theta=6$ by $(\sin^2\theta + \cos \theta^2)^2$ and then sim... | Following your way:
$$10 \sin^4\theta + 15 \cos^4 \theta=6(\sin^2\theta + \cos \theta^2)^2$$
$$\iff4(\sin^2\theta)^2-12\sin^2\theta\cos^2\theta+9(\cos^2\theta)^2=0$$
Dividing either sides by $\cos^4 \theta,$
$$4(\tan^2\theta)^2-12\tan^2\theta+9=0\iff(2\tan^2\theta-3)^2=0$$
We can find $\sec^2\theta=1+\tan^2\theta, \cot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/955698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Does an elementary solution exist to $x^2+1=y^3$? Prove that there are no positive integer solutions to $$x^2+1=y^3$$
This problem is easy if you apply Catalans conjecture and still doable talking about Gaussian integers and UFD's. However, can this problem be solved using pre-university mathematics?
I am talking about... | An alternative proof and a generalisation
As stated in the accepted answer, it is sufficient to prove that $3a^4+3a^2+1=b^2$ has no positive integer solution. With a change of notation we shall prove the following generalisation by using the same method used for Does the equation $y^2=3x^4-3x^2+1$ have an elementary s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/955967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 1,
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Show that $x(\pi - x)= \frac{\pi^2}{6}-\sum_{k=1}^{\infty} \frac{\cos(2kx)}{k^2}$ Show that
$$x(\pi - x)= \frac{\pi^2}{6}-\sum_{k=1}^{\infty} \frac{\cos(2kx)}{k^2}$$
for $ 0<x<\pi$
My idea:
I've defined the periodic function
$$f(x) = 0 \text{ if } x \in [- \pi, 0) \text{ and }$$
$$f(x)=x(\pi-x) \text{ if } x \in [0, \... | Target function
$$
f(x) = x \left(\pi - x \right)
$$
To reproduce the answer, use an even reflection of the function:
$$
\begin{align}
r(x) &= x \left(\pi - x \right), \quad x\ge 0, \\
l(x) &= -x \left(\pi + x \right), \quad x < 0.
\end{align}
$$
Fourier amplitudes
$$
\begin{align}
%
a_{0} &= \frac{1}{\pi} \left(
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum the series (real analysis) $$\sum_{n=1}^\infty {1 \over n(n+1)(n+2)(n+3)(n+4)}$$
I tried to sum the above term as they way I can solve the term $\sum_{n=1}^\infty {1 \over (n+3)}$ by transforming into ${3\over n(n+3)} ={1\over n}-{1\over(n+3)}$ but I got stuck while trying to transform $12\over n(n+1)(n+2)(n+3)(n+4... | Hint
Notice that
$${1 \over n(n+1)(n+2)(n+3)(n+4)}=\frac14{(n+4 )-n\over n(n+1)(n+2)(n+3)(n+4)}\\=\frac14\left({1\over n(n+1)(n+2)(n+3)}-{1\over (n+1)(n+2)(n+3)(n+4)}\right)$$
and then telescope.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How did we find the solution? In my lecture notes, I read that "We know that $$x^2 \equiv 2 \pmod {7^3}$$ has as solution $$x \equiv 108 \pmod {7^3}$$"
How did we find this solution?
Any help would be appreciated!
| For example, $3^2=2\bmod 7$. So $(3+7k)^2=2+7+42k\bmod 7^2$. So $k=1$ and $10^2=2\bmod 7^2$. So $(10+7^2h)^2=2+2\cdot7^2(1+10h)\bmod 7^3$. So $h=2$. Bother! Is the crummy formatting my fault or the iPhone 6+ ?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sums with squares of binomial coefficients multiplied by a polynomial It has long been known that
\begin{align}
\sum_{n=0}^{m} \binom{m}{n}^{2} = \binom{2m}{m}.
\end{align}
What is being asked here are the closed forms for the binomial series
\begin{align}
S_{1} &= \sum_{n=0}^{m} \left( n^{2} - \frac{m \, n}{2} - \frac... | Lemma:
$$
\begin{align}
\sum_{n=0}^m\binom{n}{k}\binom{m}{n}^2
&=\sum_{n=0}^m\binom{n}{k}\binom{m}{n}\binom{m}{m-n}\tag{1}\\
&=\sum_{n=0}^m\binom{m}{k}\binom{m-k}{n-k}\binom{m}{m-n}\tag{2}\\
&=\binom{m}{k}\binom{2m-k}{m-k}\tag{3}\\
&=\binom{m}{k}\binom{2m-k}{m}\tag{4}\\
&=\binom{2m-k}{k}\binom{2m-2k}{m-k}\tag{5}
\end{a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculation of $\int\frac1{\tan \frac{x}{2}+1}dx$ Calculation of $\displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$
$\bf{My\; Try}::$ Let $\displaystyle I = \displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$, Now let $\displaystyle \tan \frac{x}{2}=t\;,$ Then $\displaystyle dx=\frac{2}{1+t^2}dt$
So $\displaystyle I = 2\... | Partial fractions can be avoided, but we need to integrate both $\sec x$ and $\tan x$:
\begin{align}
\frac{1}{\tan(x/2)+1} & = \frac{1}{\tan(x/2)+1} \cdot \frac{\cos(x/2)}{\cos(x/2)} \cdot \frac{\cos(x/2)-\sin(x/2)}{\cos(x/2)-\sin(x/2)} \\[0.1in]
& = \frac{\cos(x/2)}{\sin(x/2)+\cos(x/2)} \cdot \frac{\cos(x/2)-\sin(x/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to derive the closed form of this recurrence? For the recurrence, $T(n) = 3T(n-1)-2$, where $T(0)= 5$, I found the closed form to be $4\cdot 3^n +1$(with help of Wolfram Alpha). Now I am trying to figure it out for myself. So far, I have worked out: $T(n-1) = 3T(n-2)-2, T(n-2) = 3T(n-3)-2, T(n-3) = 3T(n-4)-2$ leadi... | There are many ways to solve such linear recurrences. We describe one of them.
Let the $n$-th term be $T_n$. We are told that
$$T_n=3T_{n-1}-2.$$
It would be nice to get rid of the constant term $-2$. Let $T_n=U_n+d$, where we will choose $d$ soon. Then
$$U_n+d=3(U_{n-1}+d)-2.$$
The constant term disappears if we choo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How find this sum $\frac{1}{1^2}+\frac{2}{2^2}+\frac{2}{3^2}+\frac{3}{4^2}+\frac{2}{5^2}+\frac{4}{6^2}+\cdots+\frac{d(n)}{n^2}+\cdots$ Question:
Find the value
$$\dfrac{1}{1^2}+\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{3}{4^2}+\dfrac{2}{5^2}+\dfrac{4}{6^2}+\cdots+\dfrac{d(n)}{n^2}+\cdots$$
where $d(n)$ is The total numb... | Let $p_1, p_2, \ldots$ be the list of primes and
$\mathcal{E}$ be the collection of non-negative integer sequences
$( e_k )_{k\in \mathbb{Z}_{+}}$ with finitely many non-zero terms.
Evey $n \in \mathbb{Z}_{+}$ can be represented as a product of the form
$\prod\limits_{k=1}^\infty p_k^{e_k}$ for an unique $e \in \mathc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/961034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Sketch the set of points determined by $\lvert z-i \rvert^2 + \lvert z+i \rvert^2 \le 4$ Sketch the set of points determined by the following conditions $\lvert z-i \rvert^2 + \lvert z+i \rvert^2 \le 4$ So I started by $(z-i)(\bar{z} -\bar{i}) + (z+i)(\bar{z} + \bar{i}) \le 4$
$z\bar{z} - z\bar{i} - i\bar{z} + i\bar{i... | In your first derivation, you have a sign error:
$$z\bar{z} - z\bar{i} - i\bar{z} \oplus i\bar{i} + z\bar{z} + z\bar{i} + i\bar{z} + i\bar{i}\le 4$$
The plus sign is circled for clarity. From there on, it works.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\gcd(a, 42) = 1$, then $a^6 \equiv 1 \pmod{168}$. I have just learned Fermat's little theorem.
That is,
If $p$ is a prime and $\gcd(a,p)=1$, then $a^{p-1} \equiv 1 \mod p$
Well, there's nothing more explanation on this theorem in my book.
And there are exercises of this kind
If $\gcd(a,42)=1$, then show that $a^... | If $x \equiv 1 \bmod y$ and $x \equiv 1 \bmod z$, then $x \equiv 1 \bmod yz$ if $y$ and $z$ are coprime.
It also turns out that $a \equiv 1 \bmod 2 \Rightarrow a^2 \equiv 1 \bmod 2^3$. This is because if $a = 2k + 1$, $a^2 = 4(k^2 + k) + 1$. Since $k^2 \equiv k \bmod 2$, $k^2 + k$ is even. Thus $8 \mid 4(k^2+k)$.
Then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
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If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$. and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $ If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $.
Progress
We have $\gcd(a,b)=1\implies \exis... | Let $d=\text{gcd}(a+b,a-b)$ then $d$ divides $2a$ and $2b$ so that $d|\text{gcd}(2a,2b)=2$. The latter equality is because $\text{gcd}(a,b)=1$ so, in particular, $a$ and $b$ cannot be both even. In any case, $d|2$ means $d=1$ or $2$.
Edit: this is an edit to respond to your own edit:
Let $e=\text{gcd}(a^2+b^2,a^2-b^2)$... | {
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Prove the gcd $(4a + b, a + 2b) $ is equal to $1$ or $7$. So in the question it says to let $a$ and $b$ be nonzero integers such that $\gcd(a,b) = 1$.
So based on that I know that $a$ and $b$ are relatively prime and that question is basically asking if the GCD divides $7$.
So I tried to prove through the Euclidean alg... | As you saw, we have $2(4a+b)-(a+2b)=7a$. Also, $4(a+2b)-(4a+b)=7b$.
So any common divisor of $4a+b$ and $a+2b$ is a common divisor of $7a$ and $7b$.
Since $\gcd(a,b)=1$, there are integers $x$ and $y$ such that $ax+by=1$. It follows that $(7a)x+(7b)y=7$, so any common divisor of $7a$ and $7b$ divides $7$.
Remark: We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/964537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find all solutions of $\tan(x) = 2 + \tan(3x)$ without a calculator? Find all solutions of the equation $\tan(x) = 2 + \tan(3x)$ where $0<x<2\pi$.
By replacing $\tan(3x)$ with $\dfrac{\tan(2x) + \tan(x)}{1-\tan(2x)\tan(x)}$
I've gotten to $\tan^3 (x) - 3 \tan^2 (x) + \tan(x) + 1 =0$.
I am not sure how to procee... | Put $t=\tan(x)$, so that your equation become:
$t^3-3t^2+t+1=0$.
Since $t^3-3t^2+t+1=(t-1)(t^2-2t-1)$, it follows that the solutions are $t=1$, $t=1\pm\sqrt 2$, from which
$x\equiv\arctan(1)\equiv\frac\pi 4\pmod \pi$,
$x\equiv\arctan(1+\sqrt 2)\equiv\frac{3\pi}8\pmod\pi$,
$x\equiv\arctan(1-\sqrt 2)\equiv-\frac\pi8\pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/970284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Possible eigenvalues of a matrix $AB$ Let matrices $A$, $B\in{M_2}(\mathbb{R})$, such that $A^2=B^2=I$, where $I$ is identity matrix.
Why can be numbers $3+2\sqrt2$ and $3-2\sqrt2$ eigenvalues for the Matrix $AB$?
Can be numbers $2,1/2$ the eigenvalues of matrix $AB$?
| Set
$$
A=\left(\begin{matrix}0 & 3-2\sqrt{2} \\ 3+2\sqrt{2} & 0\end{matrix}\right),\quad
B=\left(\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right).
$$
Then
$$
AB=\left(\begin{matrix} 3-2\sqrt{2} & 0 \\ 0 & 3+2\sqrt{2}\end{matrix}\right).
$$
The eigenvalues of $A,B$ are $\pm 1$, and hence $A^2=B^2=I$, while the eigenvalue... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/970409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\sum_{n=1}^\infty \frac{1}{a_n}$ converges, must $\sum_{n=1}^\infty \frac{n}{a_1 + \dots + a_n}$ converge? Suppose $\sum_{n=1}^\infty \frac{1}{a_n} = A$ is summable, with $a_n > 0,$ $n = 1,2,3,\cdots.$ How can we prove that $\sum_{n=1}^\infty \frac{n}{a_1 + \dots + a_n}$ is also summable?
This question came from a ... | Allow me to present a short, "magical" solution by thoughtful application of Cauchy-Schwarz.
Define $A_n := a_1 + \cdots + a_n.$ We will first prove the convergence of $\sum_{n=1}^\infty \frac{n^2}{A_n^2}a_n$. Now,
\begin{eqnarray} \sum_{n=1}^N \frac{n^2}{A_n^2}a_n &<& \sum_{n=2}^N \frac{n^2}{A_nA_{n-1}}a_n + \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/975257",
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"source": "stackexchange",
"question_score": "6",
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Sum of roots of an equation $\sqrt{x-1}+\sqrt{2x-1}=x$ Find the sum of the roots of the equation $\sqrt{x-1}+\sqrt{2x-1}=x$
My attempt: Squaring the equation: $(x-1)+(2x-1) +2\sqrt{(x-1)(2x-1)}=x^2$
$\implies x^2-3x+2=2\sqrt{(x-1)(2x-1)} $
$\implies (x-1)(x-2)=2\sqrt{(x-1)(2x-1)} $
$\implies (x-2)=2\sqrt{\displaystyle ... | HINT:
$x=1$ is also a valid root where $x-1$ has been cancelled out assuming $x-1\ne0$
Observe that, though $0$ being additive identity, $x=0$ does not effect the sum is not a valid root
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Integrate $\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$ I would like some guidance regarding the following integral:
$$\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$$
EDIT: The upper problem was derived from the following integral $$\int\frac{\sqrt{x^2+2}}{x^2+1}dx$$
Where I rationalized the numerator which followed into: $$\int\frac{dx}{... | If you aren't aware of the posted formula for calculating an integral of this form (which would completely understandable!), we can start with more familiar approaches, such as using trigonometric substitution.
Try putting $x = \sqrt 2\tan \theta$. Then $$dx = \sqrt 2 \sec^2 \theta\,d\theta$$
$$\text{ and }\,x^2 + 1 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/978177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$
I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
b^2 &= 4 - a^2\\
b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\... | Let $z=a+bi$. Then $z^3=8$ and $8=|z^3|=|z|^3$, and since $|z|$ is a positive real number, $|z|=2$, that is, $a^2+b^2=|z|^2=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 0
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Find the continued fraction of the square root of a given integer How to find the continued fraction of $\sqrt{n}$, for an integer $n$?
I saw a site where they explained it, but it required a calculator. Is it possible to do it without a calculator?
| The easiest way to find a continued fraction for $\sqrt{n}$ is to use a generalized continued fraction (continued fractions where the numerator is something besides 1).
Starting from $\sqrt{n}$, partition ${n}$ as the sum of the perfect square which is less than or equal to ${n}$, ${s^{2}}$, and a remainder, ${r}$, lik... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
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How to prove this inequality relating to trigonometric function? In a triangle, A, B, C are three corners of the triangle, try to prove that :
$$\root 3 \of {1 - \sin A\sin B} + \root 3 \of {1 - \sin B\sin C} + \root 3 \of {1 - \sin C\sin A} \geqslant {3 \over 2}\root 3 \of 2 $$
So complicated that I have no idea...... | \begin{align}
\sin^2C &= \sin^2(A+B) = (\sin A \cos B + \cos A \sin B)^2 \\
&= \sin^2A \cos^2B + \sin^2B \cos^2A + 2\sin A \sin B \cos A \cos B \\
&\leqslant \sin^2A \cos^2B + \sin^2B \cos^2A + \sin A \sin B (cos^2A + cos^2B) \\
&= \sin^2A(1-\sin^2B)+\sin^2B(1-\sin^2A)+\sin A \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Regular Octagon Area Doing some maths homework I came across the area of a regular octagon on Google. This was given by:
$$ A=2(1+\sqrt{2})a^2 $$
I thought this looked rather ugly and slightly complicated and so began to look at regular octagons myself (Yes, I'm a nerd :)!). I managed to re-write the equation to
$$ A=\... | This result is known. More generally, a regular $n$-gon with side-length $x$ has area $\frac{n}{4}x^2\cot(\frac{\pi}{n})$.
To see this: by translating and rotating we can assume that the vertices of the regular $n$-gon are at $(r\cos k\theta, r\sin k\theta)$ for $0 \le k < n$, where $\theta = \frac{2\pi}{n}$ and $r$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/981754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Scaling variables in homogeneous equation of degree two in a,b,c The problem I'm having trouble with is:
Let $a,b,c$ be nonzero real numbers and let $a^2 - b^2 = bc$ and $b^2 - c^2 = ca$. Prove that $a^2 - c^2 = ab$.
The solution strategy given in the course was to scale the two given equations by $s=\frac{1}{c}$, res... | You can think of this scaling as dividing all equalities by $c^2$, which is valid because $c\neq 0$. This division basically reduces a 3-variables problem into a 2-variables problem, which should be more tractable.
Let $A=a/c$, $B=b/c$ and $C=c/c=1$. Then, we have
$$
a^2-b^2=bc\implies A^2-B^2=B;\\
b^2-c^2=ca\implies B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/982136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cardinality of a set of polynomials where the sum of the cubes of the roots is zero Let $C\subseteq \mathbb Z\times \mathbb Z$ be the set of integer pairs $(a,b)$ for which the 3 complex roots $r_1,r_2,r_3$ of the polynomial $p(x)=x^3-2x^2+ax+b$ satisfy $r_1^3+r_2^3+r_3^3=0$ .Then what will be the cardinality of $ C $... | Applying Vieta's formulas, we know that $$\begin{align}
r_1+r_2+r_3 & = 2 \tag{A}\\
r_1r_2 + r_1r_3 + r_2r_3 & = a\tag{B} \\
r_1r_2r_3 & = -b \tag{C}
\end{align}$$
Multiplying $(A)$ by $(B)$ yields:
$$r_1^2r_2 + r_1^2r_3 + r_1r_2^2 + r_1r_3^2 + r_2^2r_3 + r_2r_3^2 +
3r_1r_2r_3 = 2a \tag{D}$$
and then subtracting three... | {
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"url": "https://math.stackexchange.com/questions/984584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find series expansion of 1/cosx Find the series expansion of 1/cosx from basic series expansions.
I tried to find 1/cosx from the expansion of cosx but was unsure how to continue. When I found 1/cosx from the basic formula for finding series expansions I didn't get the same answer.
| $\sec\left(x\right)=\dfrac{1}{\cos \left(x\right)}$ is well-behaved about $x=0$, so assume
$$
\frac{1}{\cos \left(x\right)} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots
$$
Also, $\cos \left(x\right)$ is
$$
\cos \left(x\right) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots
$$
So, on performing multiplication of the ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Using a recursion tree to obtain an algorithm classification with n^2 time I'm having trouble getting the classification of this recurrence relation using a recursion tree.
$$T(n) = 3T(n/2) + n^2$$
I have the tree written out correctly (I hope):
n^2
... | At each level, the number of nodes triples and $n$ is cut in half. Extending the pattern in your table, we see that at level $k$, the time is $3^k(\tfrac{n}{2^k})^2 = (\tfrac{3}{4})^kn^2$. At the second-to-last level, we have $k = \log_2 n - 1$, which corresponds to the term $(\tfrac{3}{4})^{\log_2 n - 1}n^2$. At the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/986239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Exact value of polynomial at trigonometric argument
Given that $$\cos 8\theta= 128\cos^8 \theta −256\cos^6 \theta +160 \cos^4 \theta −32\cos^2 \theta +1$$
Find the exact value of:
$$4x^4 −8x^3 +5x^2 −x$$
where $x=\cos^2 (\frac{\pi}{8})$
My try:
$$\cos 8\theta= 128\cos^8 \theta −256\cos^6 \theta +160 \cos^4 \theta −32... | Your approach is correct, but you should probably say that you let $\theta=\frac{\pi}{8}$ before you introduce $x$. Continuing after your third line, you could write something like this:
Let $\theta = \frac{\pi}{8}$, and denote $x=\cos^2\left(\frac{\pi}{8}\right)$. This gives
$$ \frac{\cos\pi-1}{32}=4x^4-8x^3+5x^2-x$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/987344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Optimization problem: rowing across a lake
A woman at a point A on the shore of a circular lake with radius $r=3$ wants to arrive at the point $C$ diametrically opposite $A$ on the other side of the lake in the shortest possible time. She can walk at the rate of 10 mph and row a boat at 5 mph. What is the shortest amo... | If you call the center of the circle $O$ and let $\alpha=\angle AOB$, then the woman's rowing distance is $2r\sin \frac{\alpha}{2}=6\sin \frac{\alpha}{2}$; and her walking distance is $r\cdot (\pi-\alpha)=3(\pi-\alpha)$.
So her travelling time is $T(\alpha)=\frac{6}{5}\sin\frac{\alpha}{2}+\frac{3}{10}(\pi-\alpha)$.
I t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/988865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested w... | I try to tackle the second integral
$$
\int_{0}^{2\pi} \frac{\cos \theta}{(1-a \cos \theta)^{3}} d \theta
$$
by the result in my post
$$I_n(a)=\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}=\frac{(-1)^{n-1} \pi}{(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{1}{\sqrt{a^{2}-1}}\right) $$
I start with the integral
$$
\int_{0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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How to express $z^8 − 1$ as the product of two linear factors and three quadratic factors
Verify
$$(z-e^{i \theta} ) (z - e^{-i \theta} ) ≡ z^2 - 2\cos \theta + 1$$
Hence express $z^8 − 1$ as the product of two linear factors and three quadratic factors, where all
coefficients are real and expressed in a non-trigonome... | $$z^8-1=(z^2-1)(z^6+z^4+z^2+1)=(z-1)(z+1)(z^2+1)(z^4+1)$$
$$z^4+1=(z^2+1)^2-2z^2=(z^2+\sqrt2 z+1)(z^2-\sqrt2 z+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $3x^2-y^2=2$ for Integers If $x$ and $y$ are integers, then solve (using elementary methods)
$$3x^2-y^2=2$$
I tried the following
If $y$ is even, then $4|y^2$ and hence $2|y^2+2$ (and $4$ doesn't divide it), but $3x^2=y^2+2$ and since for R.H.S. to be even, $2|x^2 \implies 4|x^2$, and we get a contradictio... | For the equation: $$3y^2-x^2=2$$
Knowing the first solution: $(y,x) - (3;5)$
The equation to find the rest:
$$y_2=x+2y$$
$$x_2=2x+3y$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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An equation for the third powers of the roots of a given quadradic polynomial The roots of the equation $3x^2-4x+1=0$ are $\alpha$ and $\beta$. Find the equation with integer coefficients that has roots $\alpha^3$ and $\beta^3$.
GIVEN
SR: $\alpha + \beta = \frac43$
PR: $\alpha\beta = \frac13$
REQUIRED
SR: $\alpha^3 + \... | solving the equation we get
$x_1=1$ and $x_2=\frac{1}{3}$ thus we have $\alpha^3=1$ and $\beta^3=\frac{1}{27}$
and our equation is given by
$(x-1)(x-1/27)=0$ therefore we get
$27x^2-28x+1=0$
the equation $3x^2-4x+1=0$ is equivalent to
$x^2-\frac{4}{3}x+\frac{1}{3}=0$
after the formula for a quadratic equation we obtai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.
Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.
Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ im... | You are nearly finishing the proof.
$$|x - a| (\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)})\le |x - a| (\frac{1}{2(1 + a^2)} + \frac{1}{2(1 + x^2)})\le |x-a|$$
Take $\delta=\epsilon$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Quadratic equation, math olympiad question So this is a 9-10th grade, math olympiad problem I found. Define the parabola $y=ax^2+bx+c$ such that $a,b,c$ are positive integers. Suppose that the roots of the quadratic equation $ax^2+bx+c=0$ are $x_1,x_2$ such that $|x_1|,|x_2|>1$, then compute the minimum value of $abc$ ... | $f(x) = ax^2+bx+c$ cannot have any positive root when $a$, $b$, $c$ are all positive. So we have $x_1, x_2<-1$.
The midpoint between the roots is $-b/2a$. That must be less than $-1$, so we have $$ \tag{1} \frac{-b}{2a}<-1 \implies \frac{b}{2a}>1 \implies b>2a $$
Furthermore the larger root is less than $-1$ too, so
$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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derivative of $y=\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$ $y=\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$
The answer is $\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5} \left(\dfrac{2}{x}+\dfrac{1}{2(x+1)}-\dfrac{1}{x+2}-\dfrac{5}{x-3}\right)$
I know that the quotient rule is used but I don't know how to do this problem. Would you multipl... | $$y=\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}\\ln (y)=ln(\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5})=\\ln(y)=2lnx +\frac{1}{2}ln(x+1)-ln(x+2)-5ln(x-3)\\\frac{y'}{y}=(2\frac{1}{x}+\frac{1}{2(x+1)}-\frac{1}{x+2}-5\frac{1}{x-2})\\so\\y'=y*(2\frac{1}{x}+\frac{1}{2(x+1)}-\frac{1}{x+2}-5\frac{1}{x-2})\\y'= \frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}... | Some observations:
$$A=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \tag{1}$$
$$A=\sqrt{1+a}+\sqrt{1+b}+\sqrt{1+c} +\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} -\left(\frac{2}{\sqrt{1+a}}+\frac{2}{\sqrt{1+b}}+\frac{2}{\sqrt{1+c}} \right)$$
$$A\ge(2+2+2)-\left(\frac{2}{\sqrt{1+a}}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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solve quadratic equation I'm trying to solve the following equation $2t^2 + t - 3 = 0$
I start by dividing by 2, $t^2 + \frac {t}{2} - \frac {3}{2} = 0$
Then I solve for t $t = - \frac{ \frac {1}{2} }{2} \binom{+}{-} \sqrt{(\frac {1}{2})^2 + \frac {3}{2}}$
$t = - \frac{1}{4} \binom{+}{-} \sqrt{(\frac {1}{4}) + \frac {6... | we get $t_{1,2}=-\frac{1}{4}\pm\sqrt{\frac{1}{16}+\frac{24}{16}}$ and we get
$t_1=-\frac{1}{4}+\frac{5}{4}=\frac{4}{4}=1$ and
$t_2=-\frac{1}{4}-\frac{5}{4}=-3/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\int_0^1\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\,dx=\frac{5\pi^2}{48}-\frac{\ln^22}{4}$ How does one prove the following integral
\begin{equation}
\int_0^1\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\,dx=\frac{5\pi^2}{48}-\frac{\ln^22}{4}
\end{equation}
Wolfram Alpha and Mathematica can easily evaluate this integral... | Start with subbing $x=\frac{1-y}{1+y}$
$$\mathcal{I}=\int_0^1\frac{\ln(2)-\ln(1+x^2)}{1-x}dx=\int_0^1\frac{2\ln(1+y)-\ln(1+y^2)}{y(1+y)}dy$$
$$=2\int_0^1\frac{\ln(1+y)}{y}dy-2\int_0^1\frac{\ln(1+y)}{1+y}dy-\int_0^1\frac{\ln(1+y^2)}{y(1+y)}dy$$
$$=-2\operatorname{Li}_2(-y)|_0^1-\ln^2(1+y)|_0^1-\mathcal{J}$$
$$=\frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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If $x,y,z.\geq 0$ and $x+y+z = 10$ , Then Max. value of $xyz+xy+yz +zx$, If $x,y,z.\geq 0$ and $x+y+z = 10$ , Then Maximum value of $xyz+xy+yz +zx$, is
$\bf{My\; Try::}$ First we can write the given expression
$xyz+xy+yz+zx = (x+1)(y+1)(z+1)-(x+y+z)-1 = $
$\displaystyle = (x+1)(y+1)(z+1)-\left\{(x+1)+(y+1)+(z+1)\right\... | From there: you know that $a+b+c = 13$, so it simplifies into $abc-14$. Then, since the max of the product of three numbers is if all the numbers are equal, so the max of $abc$ is $\left(\frac{13}{3}\right)^3 = \frac{2107}{27}$. So, the answer would be $\frac{2107}{27}-14 = \frac{1729}{27}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of positive integral solutions to $x+y+z+w=20$ with $xWhat is the number of positive unequal integral solution of the equation $x+y+z+w=20$, if $\,x<y<z<w\,$ and $\,x,y,z,w\ge1\;?$
How to solve this question?
| There are following posible solution:
1,2,3,14 $\,\,\,\,\,\, $ 1,3,4,12, $\,\,\,\,\,\, $ 1,4,5,10 $\,\,\,\,\,\, $ 1,5,6,8
1, 2,4,13 $\,\,\,\,\,\, $ 1,3,5,11 $\,\,\,\,\,\, $ 1,4,6,9
1,2,5,12 $\,\,\,\,\,\, $ 1,3,6,10 $\,\,\,\,\,\, $ 1,4,7,8
1,2,6,11 $\,\,\,\,\,\, $ 1,3,7,9
1,2,7,10
1,2,8,9
So th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$ Question:
let $x,y,z>0$ and such $xyz=1$, show that
$$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$
My idea: use AM-GM inequality
$$x^3+x^3+1\ge 3x^2$$
$$y^3+y^3+1\ge 3y^2$$
$$z^3+z^3+1\ge 3z^2$$
so
$$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$
But this is not my inequality,so How pro... | This is hardly a pretty answer, but then again this inequality seems rather tight.
Consider the function $f(t)=e^{3t}-2e^{2t}+1+t$. It has the following properties:
*
*$f'(t)=e^{2t}(3e^t-4)+1$, so $f(t)$ is tangent to $y=0$ at $t=0$;
*$f''(t)=e^{2t}(9e^t-8)$, so $f(t)$ is concave down on $(-\infty,\ln\frac89)$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Solve $\sqrt{3}\cos2\theta+\sin2\theta-1=0$ I tried using the identities $\cos2\theta=1-2\sin^2\theta$ and $\sin2\theta=2\sin\theta\cos\theta$. These give
$\sqrt{3}(1-2\sin^2\theta)+2\sin\theta\cos\theta-1=0$
which doesn't seem to lead anywhere. Perhaps I must equate the function to something like $R\sin(2\theta+\alpha... | In general you can write:
\begin{align}
a\cos\theta+b\cos\theta&=
\sqrt{a^2+b^2}\left(
\frac{a}{\sqrt{a^2+b^2}}\cos\theta+\frac{b}{\sqrt{a^2+b^2}}\sin\theta
\right)\\[2ex]
&=\sqrt{a^2+b^2}(\sin\alpha\cos\theta+\cos\alpha\sin\theta)
\end{align}
Hope this is helpful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that: $\lim_{x\to 0}\frac{x}{\sin^2(x) + 1} = 0$ Prove
$$\displaystyle \lim_{x\to 0} \frac{x}{\sin^2(x) + 1} = 0$$
The proof:
Let $$|x| \le 1 \implies -1 \le x \le 1$$
$$\displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \epsilon\text{ for }\displaystyle |x| < \delta$$
$$-1 \le x \le 1
\\\implies \sin(-1) \le \sin(x) \... | Remarks on your new approach
Since in your attempt, $x\in[-1, 1]$, lets try $x=0$
$$ \sin^2(-1)+1\lt \sin^2(0)+1 $$
$$ \sin^2(-1)\lt \sin^2(0) $$
$$ \sin^2(1)\lt 0 $$
Is this correct to you? Also if you wish to start from $|x|$, then why not start with the fact that
$$
|x|\le |x|\left|\sin^2(x)+1\right|
$$
$$
\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving a simple first order differential equation I've got the following first order differential equation that is doing something I can't quite figure out. The question is to solve:
$$y' = 2x^2y^2$$
Here is my solution:
$$y' \frac{1}{y^2} = 2x^2$$
$$\int \frac{1}{y^2} y' dx = \int 2x^2 dx$$
$$\int \frac{1}{y^2}dy = ... | Instead of keeping $y'$ as $y'$, you should change it into $\frac{dy}{dx}$. Then,$$\frac{dy}{dx} = 2x^2y^2$$ $$y^2 dy = 2x^2 dx$$ $$\int y^2\, dy = \int2x^2\, dx$$ $$\frac{y^3}{3} = \frac{2}{3}x^3 + C_1$$ $$y = \sqrt[3]{2x^3+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Taylor series expansion for $e^{-x}$ could anyone show me the Taylor series expansion for $e^{-x}$.I was trying to find out how
$e^{-i\theta}$=$\cos\theta-i\sin\theta$.
More specifically could you show me how $e^{-i\theta}$=$\cos\theta-i\sin\theta$ is obtained from Taylor series.
| $$e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}+\cdots$$
So, taking into account the fact that $i^0=1$, $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, etc., we get
\begin{align}
e^{i\theta} &=
1+(i\theta)+\frac{(i\theta)^2}{2!}+\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a$ and $b$ are positive real numbers, then $a + b \geq 2 \sqrt{ab}$. If $a$ and $b$ are positive real numbers, then $a + b \geq 2 \sqrt{ab}$.
I know how to do the direct proof, but in this case, I want to try proving it by contradiction. I have tried manipulating the inequality $a + b < 2 \sqrt{ab}$ after making t... | \begin{align}
a > 0, b > 0, a + b < 2\sqrt{ab}
&\implies a - 2\sqrt{ab} + b < 0 \\
&\implies (\sqrt a - \sqrt b)^2 < 0 \\
\end{align}
Which is a contradiction.
by the way
$a + b < 2 \sqrt{ab}\;$ implies $a^2-2ab+b^2 > 0\;$, not what you said.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality proof by induction, what to do next in the step I have to prove that for $n = 1, 2...$ it holds: $2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}$
Base: For $n = 1$ holds, because $2\sqrt{2}-2 < 1$
Step: assume holds for $n_0$.
$2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt2} + \f... | You assume it holds for $n$, i.e.
$$ 2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}}. $$
You now want to prove that it holds for $n+1$, i.e.
$$ 2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}}. $$
Subtracting the... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the positive integer numbers to get $\frac{\pi ^2}{9}$ As we know, there are many formulas of $\pi$ , one of them $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}......
$$
and this $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}......$$
Now,find the positive integer numbers $(a_{0}, a_{... | From $$\frac{\pi^2}8=\sum_{2\nmid n}\frac1{n^2}$$
we find that
$$\frac{\pi^2}{72}=\frac1{3^2}\sum_{2\nmid n}\frac1{n^2}=\sum_{2\nmid n, 3\mid n}\frac1{n^2}$$
and hence
$$\begin{align}\frac{\pi^2}9=\frac{\pi^2}8-\frac{\pi^2}{72}&=\sum_{2\nmid n, 3\nmid n}\frac1{n^2}\\&=\frac1{1^2}+\frac1{5^2}+\frac1{7^2}+\frac1{11^2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 3
} |
An annoying Pell-like equation related to a binary quadratic form problem Let $A,B,C,D$ be integers such that $AD-BC= 1 $ and $ A+D = -1 $.
Show by elementary means that the Diophantine equation
$$\bigl[2Bx + (D-A) y\bigr] ^ 2 + 3y^2 = 4|B|$$
has an integer solution (that is, a solution $(x,y)\in\mathbb Z^2$). If po... | Reduction can also be accomplished in a conjugacy class in $SL_2 \mathbb Z,$ in a manner very similar to that for quadratic forms. As matrices, there is a little difference: for quadratic forms, we start with symmetric $G$ and work with $P^T G P, $ for conjugacy classes we work with $P^{-1} \gamma P.$
Lemma: given real... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Linear independence of $\sin^2(x)$ and $\cos^2(x)$ The Wronskian for $\sin^2x, \cos^2x$ is
\begin{align}
& \left| \begin{array}{cc} \sin^2 x & \cos^2 x \\ 2\sin x\cos x & -2\cos x\sin x \end{array} \right| \\[8pt]
= {} & -2\sin^2x \cos x \sin x - 2 \cos^2 x \sin x \cos x,
\end{align}
with $x = \frac{π}{6},$ this is $=$... | No calculus needed --
If $a\sin^2 x + b\cos^2 x=0$, then $a$ must be $0$ because that's the only way to make the sum $0$ at $x=\pi/2$ where $\cos^2 x=0$.
Similarly $b=0$ is the only way to make the sum zero at $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to compute the smallest integer which is sum of cubes in 13 ways? I would like to know the smallest integer which is sum of three positive cubes in 13 ways, such that $1^3+2^3+3^3=2^3+1^3+3^3$ are same ways of representation? What kind of theory there is behind that? I saw the question Sum of squares in at least 64... | I guess it is
119095488
= 24^3+204^3+480^3
= 48^3+ 85^3+491^3
= 72^3+384^3+396^3
= 113^3+264^3+463^3
= 114^3+360^3+414^3
= 149^3+336^3+427^3
= 176^3+204^3+472^3
= 190^3+279^3+449^3
= 207^3+297^3+438^3
= 226^3+332^3+414^3
= 243^3+358^3+389^3
= 246^3+328^3+410^3
= 281^3+322^3+399^3
The solution was given by Matti K. Si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the $a$ to make the sum of series equal to zero,$\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=0$ Find all values of $a$ which make the sum of series $\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=0$
| Here is one solution: $a = 9/4$.
Then
$$\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=-4/9 + \frac1{3}\sum_{n=1}^{\infty }\frac{1}{n^2+n-3/4}=0$$
Note that
$$\frac1{3}\sum_{n=1}^{\infty }\frac{1}{n^2+n-3/4}=\frac1{6}\sum_{n=1}^{\infty }\left[\frac{1}{n-1/2}-\frac{1}{n+1/2}\right]+\frac1{6}\sum_{n=1}^{\infty }\left[\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$ Evaluation of Integral $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$
$\bf{My\; Try::}$ First we will convert $x^4+x^3+x^2+x+1$ into closed form, which is $\displaystyle \left(\frac{x^5-1}{x-1}\right)$
So Integral is $\displaystyle... | $$
I=\int^{\infty}_{0}\frac{dx}{\sqrt{x^4+x^3+x^2+x+1}}
$$
Making the change of variable $x\rightarrow y/(1+y)$ we get
$$
I=\int^{1}_{0}\frac{dy}{\sqrt{y^4-2y^3+4y^2-3y+1}}=
$$
$$
=\int^{1}_{0}\frac{dy}{\sqrt{1/16(2y-1)^4+5/8(2y-1)^2+5/16}}.
$$
Also setting $(2y-1)\rightarrow u$, then
$$
I=\frac{1}{2}\int_{-1}^{1}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Writing $\frac{x^4(1-x)^4}{1+x^2}$ in terms of partial fractions How does one write
$$\frac{x^4(1-x)^4}{1+x^2}$$
in terms of partial fractions?
My Attempt
$$\frac{x^4(1-x)^4}{1+x^2}=\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}$$
$$=Ax^6+Bx^5+Cx^4+Dx^3+Ex^2+Fx+\frac{G}{1+x^2}$$
Multiplying out and comparing coefficient gives:... | To answer your comment about long division: you want to divide $x^8-4x^7+6x^6-4x^5+x^4$ by $x^2+1$. Then,
$$x^8-4x^7+6x^6=x^6(x^2+1)-4x^7+5x^6$$
$$-4x^7+5x^6-4x^5=-4x^5(x^2+1)+5x^6$$
$$5x^6+x^4=5x^4(x^2+1)-4x^4$$
$$-4x^4=-4x^2(x^2+1)+4x^2$$
$$4x^2=4(x^2+1)-4$$
All in all,
$$x^8-4x^7+6x^6-4x^5+x^4=(x^6-4x^5+5x^4-4x^2+4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
When does $x^3+y^3=kz^2$? For which integers $k$ does
$$
x^3+y^3=kz^2
$$
have a solution with $z\ne0$ and $\gcd(x,y)=1$? Is there a technique for counting the number of solutions for a given $k$?
| Consider the equation: $$(a+b)^3+(a-b)^3=2a(a^2+3b^2)$$ Now let $a^2+3b^2=c^2$. We want to find a general equation for this. Note that $$(m-n)^2+2mn=(m+n)^2$$ Therefore, substitute $m\mapsto m^2$ and $n\mapsto n^2k$ then $$(m^2-k\cdot n^2)^2+k(2mn)^2=(m^2+k\cdot n^2)^2$$ Let $k=3$, and it follows that $a=m^2-3n^2$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
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.