Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the maximum value of $A$ Let $a;b;c>0$ such that $a+b+c=6$. Find the maximum value of $A=a^2bc+a^2+2b^2+2c^2$
WLOG $b\ge c$. I see maximum value of $A=36$ at $(a;b;c)=(2;1;3)$
So i need to prove $A\le 36$. Or I will prove
$(a+b+c)^4\ge 36a^2bc+(a^2+2b^2+2c^2)(a+b+c)^2$
Or $(2a-b-c)(b^3+c^3+a^2b+a^2c+2ab^2+2ac^2... | Given that $a+b+c=6$ we can substitute $a=6-b-c$ into $A$ to get
$$A=(6-b-c)^2(bc+1)+2(b^2+c^2),$$
where $b,c>0$ and $b+c<6$. Differentiating with respect to $b$ and $c$
yields
$$\frac{\partial A}{\partial b}=-2(6-b-c)(bc+1)+(6-b-c)^2c+4b,$$
$$\frac{\partial A}{\partial b}=-2(6-b-c)(bc+1)+(6-b-c)^2b+4c.$$
If there is a... | {
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"url": "https://math.stackexchange.com/questions/2894315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the prime factorisation of $6500$ and $1120$, and write down, in factorised form, $\gcd(6500, 1120)$ and $\operatorname{lcm}(6500, 1120)$. (i) Find the prime factorisation of $6500$, and of $1120$.
What is the typical way to go about this? Just using common divisibility rules? That's what I did. I'm not sure if th... | One can do the gcd without factoring first. Oh, $$ \operatorname{lcm}(a,b) = \frac{ab}{\gcd(a,b)} $$
$$ \frac{ 6500 }{ 1120 } = 5 + \frac{ 900 }{ 1120 } $$
$$ \frac{ 1120 }{ 900 } = 1 + \frac{ 220 }{ 900 } $$
$$ \frac{ 900 }{ 220 } = 4 + \frac{ 20 }{ 220 } $$
$$ \frac{ 220 }{ 20 } = 11 + \frac{ 0 }{ 20 } $$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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On Lame's Theorem I was trying to study Lame's theorem but I'm a little bit stuck on one specific step... I write here the theorem and the proof:
Lame's theorem:
using the Euclidean algorithm to find the greatest common divisor of two positive integers has number of divisions less than or equal five times the number... | You can play around with things a bit too.
Given $\alpha=\dfrac{1+\sqrt{5}}{2}$, it follows that
\begin{align}
\alpha^2 &= \alpha + 1 \\
\alpha^3 &= \alpha^2 + \alpha \\
&= 2\alpha + 1 \\
\alpha^4 &= 2\alpha^2 + \alpha \\
&= 3\alpha + 2 \\
\alpha^5 &= 3\alpha^2 + 2\alpha \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate the integral $\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2+x^2+(1-x)^2}}dx$ $$I=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2+x^2+(1-x)^2}}dx$$
My attempt:
$$I=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2-2x(1-x)+(x+1-x)^2}}dx$$
$$=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{(1-x(1-x))^2}}dx=\i... | Notice that
$$ I = \frac{1}{2} \int_{0}^{\infty} \frac{\log^2 x}{x^2-x+1} \, dx.$$
To evaluate this integral, we consider the following contour integral
$$ J = \oint_{C}\frac{\operatorname{Log}^3z}{z^2-z+1} \, dz,$$
where $\operatorname{Log}z$ is the complex logarithm with the branch cut $[0,\infty)$ and $C$ is a keyho... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Distance between incentre and excentre of a triangle I am unable to get anywhere regarding the distance between the incentre and an excentre of $\triangle ABC$. I do know that it is generalised as:
$$IJ_c=c\sec\left(\frac C2\right)$$
For the incenter $I$ and excenter $J_c$, where side $c$ is opposite $\angle C$.
B... | Consider this diagram:
Consider that the distance between $I$ and $E_c$ is simply:
$$d=E_cC-IC\tag{1}$$
And we have:
$$\cos \left(\frac C2\right)=\frac{KC}{E_cC}\implies E_cC=\frac{KC}{\cos(C/2)}\\
\cos\left(\frac C2\right)=\frac{T_bC}{IC}\implies IC=\frac{T_bC}{\cos(C/2)}$$
However, we know that $T_bC=\frac{b+a-c}2$.... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the fastest way to estimate the Arc Length of an Ellipse? To estimate the circumference of an ellipse there are some good approximations.
$a$ is the semi-major radius and $b$ is the semi-minor radius.
$$L \approx \pi(a+b) \frac{(64-3d^4)}{(64-16d^2 )},\quad \text{where}\;d = \frac{(a - b)}{(a+b)}$$
Are there a... | I'll assume $\theta_1$ and $\theta_2$ refer to the parametrization
$$ \eqalign{x &= a \cos(\theta)\cr y &= b \sin(\theta)}$$
of the ellipse.
Thus the arc length in question is
$$ L = \int_{\theta_1}^{\theta_2} \sqrt{a^2 \sin^2(\theta) + b^2 \cos^2(\theta)}\; d\theta $$
We want a good approximation of the integrand tha... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a+b+c\le 0$ $a,b,c$ are integer numbers different from zero and
$$\frac{a}{b+c^2}=\frac{a+c^2}{b}$$
Prove that $a+b+c\le 0$.
I know how to prove that $a+b<0$ but do not know what about $c$.
| Since $c \in \mathbb{Z}$, we have $c \le |c|\le c^2$
$$0=(a+b)c^2+c^4$$
$$a+b+c^2=0$$
$$a+b+c \le a+b+c^2=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the sum of the the first n elements in the series 1,2,4,5,7,8.... as a function of n? I see I can split it up into two arithmetic series (1,4,7... and 2,5,8...) and add the two sums, but what is the compact notation for the sum of these two series as a function of n?
| Let us first find the formula for even $n$. In this case, $m = n/2$ is a whole number, and moreover the sum of the first $n$ elements of your sequence is the sum of the first $m$ elements of the two arithmetic progressions, that is,
$$
m + 3\frac{m(m-1)}{2} + 2m + 3\frac{m(m-1)}{2} = 3 (m + m(m-1)) = \frac{3n^2}{4}.
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $\sin(t) + \cos(t) = a$, derive an expression in '$a$' for $(\cos(t))^4 + (\sin(t))^4$ I received this question from a student's trigonometry review assignment. After spending an embarrassing amount of time on it, I consulted others and learned that nobody has been able to solve this problem for multiple semester... | $$\sin t+\cos t=a\\(\sin t + \cos t)^2=a^2\\2\sin t \cos t=a^2-1
\\(\sin t + \cos t)^4=a^4\\ \sin^4 t + \cos^4 t+4\sin t \cos t (\sin^2 t+\cos^2 t)+6\sin^2 t \cos^2 t=a^4\\ \sin^4 t + \cos^4 t +2(a^2-1)+\frac 32(a^2-1)^2=a^4\\
\sin^4 t+\cos^4 t=a^4-2(a^2-1)-\frac 32(a^2-1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\frac{1}{\sin(2x)} + \frac{1}{\sin(4x)} + \frac{1}{\sin(8x)} + \frac{1}{\sin(16x)}$
Evaluate
$$\dfrac{1}{\sin(2x)} + \dfrac{1}{\sin(4x)} + \dfrac{1}{\sin(8x)} + \dfrac{1}{\sin(16x)}$$
It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.
Re... | $$\frac{1}{\sin2x}=\frac{\sin x}{\sin 2x \sin x}=\frac{\sin (2x-x)}{\sin 2x \sin x}=\frac{\sin 2x \cos x - \cos 2x \sin x}{\sin 2x \sin x}=\cot x - \cot2x$$
$$\frac{1}{\sin4x}=\cot 2x - \cot4x$$
$$\frac{1}{\sin8x}=\cot 4x - \cot8x$$
$$\frac{1}{\sin16x}=\cot 8x - \cot16x$$
$$\frac{1}{\sin2x}+\frac{1}{\sin4x}+\frac{1}{\s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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find the value of $p(-\pi)$ and $p(0)$? Let $p(x)$ be a polynomial of degree $7$ with real coefficient such that $p(\pi)=\sqrt3.$ and
$$\int_{-\pi}^{\pi}x^{k}p(x)dx=0$$
for $0\leq k \leq 6$. Then find the value of $p(-\pi)$ and $p(0)$?
If I take general polynomial then it is difficult to find and also time consuming s... | Consider the space $\mathbb{R}_{\le 7}[x]$ of real polynomials of degree at most $7$, equipped with the inner product $\langle f,g\rangle = \int_{-\pi}^\pi fg$.
Your polynomial $p$ satisfies $p \perp \{1, x, \ldots, x^6\}$. We know that a basis for $\mathbb{R}_{\le 7}[x]$ is given by $\{1, x,\ldots, x^7\}$ so applying ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\beg... | As an alternative by Taylor expansion as $x\to 0$
$$\sin x = x -\frac16x^3 + o(x^3)\implies \sin^2 x = \left(x -\frac16x^3 + o(x^3)\right)^2=x^2-\frac13x^4+o(x^4)$$
we have
$$\frac{1}{x^2} - \frac{1} {\sin^2 x} =\frac{\sin^2 x-x^2}{x^2\sin^2 x}=\frac{x^2-\frac13x^4+o(x^4)-x^2}{x^2\left(x^2-\frac13x^4+o(x^4)\right)}=$$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 6
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If $\sin^8(x)+\cos^8(x)=48/128$, then find the value of $x$? If $$\sin^8(x)+\cos^8(x)=48/128,$$ then find the value of $x$? I tried this by De Moivre's theorem:
$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(\theta)=e^{in\theta}$$
But could not proceed further please help.
| \begin{align}
\sin^8 x+\cos^8 x&=(\sin^4 x+\cos^4)^2-2\sin^4 x\cos^4 x\\
&=\left[(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x\right]^2-\cfrac {\sin^4 2x} 8\\
&=(1-\cfrac {\sin^2 2x}2)^2--\cfrac {\sin^4 2x} 8
\end{align}
Let $t=\sin^2 2x$, we have
$$(1-\cfrac {t}2)^2-\cfrac {t^2}8=\cfrac {48}{128}\Leftrightarrow t^2-8t+5=0,$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910845",
"timestamp": "2023-03-29T00:00:00",
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Using geometry to show that $\int_{0}^{b} \sqrt{1-x^2} dx = \frac{1}{2}b\sqrt{1-b^2}+\frac{1}{2}\sin^{-1}b$
Show that for $0\leq b\leq1$,
$$\int_{0}^{b} \sqrt{1-x^2} dx = \frac{1}{2}b\sqrt{1-b^2}+\frac{1}{2}\sin^{-1}b$$
by using geometry.
This is what I have thus far:
$\int_{0}^{b} \sqrt{1-x^2} dx$
$$=y=\sqrt{1-x^2}$... | The area of the triangle with sides $b$ and $a=\sqrt{1-b^2}$ forming the right angle is
$$
\frac 12b\sqrt{1-b^2}\ .
$$
The angle, marked in black, can be extracted from the triangle with sides $a,b,r=1$, it is $\arcsin (b/1)$. So the sector has area
$$\frac 12\arcsin b\cdot 1^2\ .
$$
Adding the two areas we get the ans... | {
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Find all non-negative integer $n$ that satisfy $f(x+1)+f(x-1)=\sqrt nf(x)$
Find all non-negative integer $n$ that there exists a non-periodic function $f:\Bbb R->\Bbb R\quad f(x+1)+f(x-1)=\sqrt nf(x)\forall x$.
My attempt:
$f(x+2)+f(x)=\sqrt n\cdot f(x+1)$
$\sqrt n \cdot (f(x+1)+f(x+3))=n\cdot f(x+2)$
$f(x+2)+f(x+4... | For a nonnegative integer $n$, the roots of the polynomial $t^2-\sqrt{n}\,t+1$ are primitive $k$-th roots of unity for some integer $k>0$ if and only if $n\leq 3$. (Note that $k=4$ for $n=0$, $k=6$ for $n=1$, $k=8$ for $n=2$, and $k=12$ for $n=3$.) This shows that any solution $f:\mathbb{R}\to\mathbb{R}$ to the funct... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding solution to differential equation I'm stuck on this problem:
Find the solution of the given initial value problem:
$ty′+4y=t^2−t+5$, $y(1)=7$, $t>0$
When I multiply both sides by $\mu(t)$ I find that $\mu(t) = e^{4t}$
So:
$$
\frac{d}{dt} (\mu y) = \mu t^2 - \mu t + 5 \mu\\
e^{4t} y= \int t^2 e^{4t}-te^{4t}+5e^... | For $t y′ + 4y = t^2 − t + 5$ then
\begin{align}
\frac{1}{\mu} \frac{d}{dt} \, ( \mu \, y) &= y' + \frac{\mu'}{\mu} \, y
\end{align}
which leads to
$$ \frac{d}{dt} \ln \mu = \frac{4}{t} = \frac{d}{dt} (4 \ln t) = \frac{d}{dt} \ln(t^4),$$
or $\mu(t) = t^4$. Now,
\begin{align}
y' + \frac{4}{t} \, y &= t - 1 + \frac{5}{t... | {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $\int {1+\sin x\over x^2+2}dx$ I've been looking for a suitable approach to solve the following integral:
$$\int {1+\sin x\over x^2+2}dx$$
So far, I've tried by substitution and by parts to no avail. I've also attempted the following manipulations:
$$\int {\sin^2x+\cos^2x+\sin x\over x^2+2}dx = \int {\sin... | $$I=\int\frac{1+\sin(x)}{x^2+2}dx=\int\frac{1}{x^2+2}dx+\int\frac{\sin(x)}{x^2+2}dx$$
$$I_1=\int\frac{1}{x^2+2}dx=\int\frac{1}{2\left[(\frac{x}{\sqrt{2}})^2+1\right]}dx$$
$a=\frac{x}{\sqrt{2}}$ so $dx=\sqrt{2}da$
$$I_1=\frac{\sqrt{2}}{2}\int\frac{1}{a^2+1}da=\frac{\sqrt{2}}{2}\arctan\left(\frac{x}{\sqrt{2}}\right)+C$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the value of a polynomial at certain points $p(x)$ be a polynomial of degree 7 with real coefficients such that $p(π) = √3$ and $$\int_{-π}^{π} x^{k}p(x) = 0, \text{ for} \; 0\leq k \leq 6. $$
I have to find the value of $p(-π)$ and $p(0).$
My initial thoughts were to suppose that $p(x) = a_0 + a_1x + a_2x^2 + ... | Let's start with $$\int_{-\pi}^\pi p(x)dx = 0$$Note that for all of the terms with $x$ having an odd degree, the integral of the function will have that term with even degree, so substituting $\pi$ for $x$ or $-\pi$ will produce the same value. This gives us our key insight: We solely care about the terms with even deg... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Expressing $\sin(18^{°})$ in algebraic form from $z^5-1$ as starting point I am trying to express $\sin(18°)$ in algebraic form using only complex numbers. I know that when I factor $z^5-1.$ I get an expansion that looks like: $(z-1)(z^4+z^3+z^2+z^1+1)$ The exercise then says substitute for $z+\frac{1}{z}$ in the 'long... | I guess you realize that $18^\circ$ is $1/20$ of a circle, so that $a=\sin 18^\circ+i\cos 18^\circ$ is a $5$ root of unity. So $a$ must satisfy $z^{5}=1$.
The polynomial $z^{5}-1$ factors and a bit of thought gets you to the point where you are, that $a$ must satisfy $z^4+z^3+z^2+z+1=0$. So if you can find its root... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Extend $f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$ in a way it is continuous $\forall x \in \mathbb{R}$ Problem
If it is possible to extend $$f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$$ in a way it is continuous $\forall x \in \mathbb{R}$. Extend all $x \not\in X$ where $X$ is domain of $f$. Also define domain $X$.
Atte... | Just do a polynomial long divison.
$$\frac{6x^3-11x^2-16x-4}{2x^2-5x-2} = 2+3x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $z = x+iy$ be a complex number and denote $z^2 = A + Bi$. Solve for $x,y$ in terms of $A,B$. Here, $x, y, A, B$ are all real numbers.
I am just having some problems with the algebra of putting it all together to get a solution.
What I have currently is:
$z^2 = (x+iy)^2 = x^2 + 2xyi - y^2$
This gives $A = x^2 - y^... | You already have $x^2- y^2= A$ and $x^2+ y^2= \sqrt{A^2+ B^2}$. Adding those two equations eliminates $y^2$ and gives $2x^2= A+ \sqrt{A^2+ B^2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to know that $2n^3+9n^2+13n+6$ factors into $(n+1)(n+2)(2n+3)$? Apologies in advance, my math is very rusty.
I'm slowly working my way through Schaum's Outline of Discrete Math for some self-study, occasionally filling in large knowledge gaps in my grasp on algebra. In one of the supplementary questions I'm asked t... | The Rational Root Theorem tells you that any rational root of the polynomial $2 n^3 + 9 n^2 + 13 n + 6$ has the form $\frac{p}{q}$ for integers $q \mid 2$, $p \mid 6$, leaving 12 possible rational roots. Since all of the coefficients are positive, all of the (real) roots are negative, leaving only 6 possible rational r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Aid in identifying an error made whilst evaluating an integral
Evaluate $$I=\int\frac{1}{x^2-8x+8}~dx$$
First, complete the square using the denominator: $$x^2-8x+8=(x-4)^2-8$$
and therefore, let $x=2\sqrt{2}\sec{\theta}+4$, $\therefore dx=2\sqrt{2}\sec{\theta}\tan{\theta}~d\theta$, and hence, we have
$$I=2\sqrt{2}\i... | Your answer is correct and it's only a matter of rewriting.
Let us consider your integral:
$$I=\int\frac{1}{x^2-8x+8}\ dx.$$
I approached this problem by factoring the denominator instead of completing the square:
$$x^2-8x+8=0.$$
The discriminant is given by $\Delta_x=(-8)^2-4(1)(8)=64-32=32$ such that $\sqrt{\Delta_x}... | {
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"timestamp": "2023-03-29T00:00:00",
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How many 3 -digit numbers can be formed from 0 to 8 whose sum is equal to 20 for repetition and non-repetition? I have a range from $0$ to $8$ . I want to find that how many 3-digit numbers can be formed whose sum is equal to $20$.
Non-Repetition of numbers:
$$
5+7+8=20 \\
$$
Repetition of numbers:
$$
4+8+8=20, \\
6... | The problem can be transformed into a couple of related problems as follows:
Problem 1: Consider the non-repetition problem with no restriction on range. Here we are to find natural numbers $0\leq a<b<c$ such that:
$$a+b+c=20$$
Given appropriate $x,y,z\geq0$ we could reformulate this in the following way:
$$
\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving identity $\frac{1}{x+iy}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$ Below are some of the identities provided early in Needhams "Visual Complex analysis"
I want to verify the identity in my title. From the identities given before it, I could only write
$$\frac{1}{x+iy}=\frac{1}{x^2+y^2}\big(-\arctan\frac{y}{x}\big... | $$\frac{1}{x+iy} \frac{x-iy}{x-iy} = \frac{x-iy}{(x+iy)(x-iy)} = \frac{x-iy}{x^2-i^2y^2} = \frac{x-iy}{x^2+y^2} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}$$ as required
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $x^3 + 3x + 3$ is irrational then $x$ is irrational, by proving the contrapositive I don't understand how to do this considering that the contrapositive of $x^3$ is irrational. For example $2$ to the cube root is irrational, but I am trying to prove that is is rational
| A more pedestrian proof (following @Mike's first step):
Suppose $x$ is rational, and the quotient of two integers $x = \frac{a}{b}.$ Then
\begin{align}
f(x)
&= x^3 + 3x + 3 \\[8pt]
&= \left(\frac{a}{b}\right)^3 + 3\frac{a}{b} + 3\frac{1}{1} \\[8pt]
&= \frac{a^3}{b^3} + 3\frac{a}{b}\frac{b^2}{b^2} + 3\frac{1}{1}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$
Show that $$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}.$$
My attempt: I could find a way to develop the LHS by un-nesting the double radical, figuring out that
$$\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqr... | You need to denest $\sqrt{2+\sqrt{3}}$ at some point, but this can be deferred to the last step. Let $s=\sqrt{2}$ and $r=\sqrt{3}$. Then
\begin{align}
\frac{s+\sqrt{s^2+r}}{s-\sqrt{s^2+r}} +3+2r
&=\frac{1+\sqrt{1+r/2}}{1-\sqrt{1+r/2}} +3+2r\\
&=\frac{1+\sqrt{1+r/2}}{1-\sqrt{1+r/2}}\times\frac{1+\sqrt{1+r/2}}{1+\sqrt{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Determining the minimum value of the function $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ I am curious whether there is an algebraic verification for $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ having its minimum value of $\sqrt{2 + \sqrt{3}}$ at $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}$. I have been told the graph of it is that ... | Scale and shift
$$\begin{align}u&=\sqrt{2}x-1&v&=\tfrac{1}{\sqrt{2}}y-\tfrac{1}{2}\text{.}
\end{align}$$
Then it is just as well to minimize
$$v=\tfrac{u}{2}+\sqrt{u^2+1}\text{.}$$
with respect to $u$.
Use the stereographic projection
$$u=\frac{2t}{t^2-1}\text{.}$$
Then it is just as well to minimize
$$v=1+\frac{t+2}{t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove all derivatives of $f(x)=\frac{1}{1+x}$ by induction Problem
Prove all derivatives of:
$$ f(x)=\frac{1}{1+x} $$
by induction.
Attempt to solve
I compute few derivatives of $f(x)$ so that i can form general expression for induction hypothesis. I compute all derivatives utilizing formula:
$$ \frac{d}{dx}x^n=nx^... | Your hypothesis is
$$\frac{d^n}{dx^n}\frac1{1+x}=\frac{(-1)^nn!}{(1+x)^{n+1}}$$
We want to show that
$$\frac{d^{n+1}}{dx^{n+1}}\frac1{1+x}=\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}}$$
Let's verify:
\begin{align}
\frac{d^{n+1}}{dx^{n+1}}\frac1{1+x} &=\frac{d}{dx}\left(\frac{d^n}{dx^n}\frac1{1+x} \right)\\
&=\frac{d}{dx}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find the equation in spherical coordinates of $x^2 + y^2 – z^2 = 4$. Find the equation in spherical coordinates of $x^2 + y^2 – z^2 = 4$.
$$\begin{align}
x^2 + y^2 &= r^2\sin^2(\theta)\\
z^2 &= r^2 \cos(\theta) \\
x^2 + y^2 - z^2&=r^2(\sin^2(\theta) - \cos^2(\theta)) = 4
\end{align}$$
Thus, $$-r^2(\cos(2\theta)) = 4$$
... | $x^2 +y^2-z^2=4 \implies \rho ^2-2z^2=4$$
$$ \implies \rho ^2(1-2\cos ^2 ( \phi ))=4$$
$$\implies \rho ^2(\cos 2( \phi ))=4 \implies \rho ^2=4\sec (2\phi)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the complex square roots of a complex number without a calculator
The complex number $z$ is given by $z = -1 + (4 \sqrt{3})i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to ... | Observe that $\|z\| = \sqrt{(-1)^2+(4\sqrt3)^2} = \sqrt{49} = 7$. Therefore the root of $z$ will have length $\sqrt 7$, so $a^2+b^2=7$. Combine this with $a^2-b^2=-1$ to get $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 5
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Recurrence $a_j = \frac{n}{n-j} + \frac{j}{n-j} a_{j-1}$ Let $n\in \mathbb{N}$. How to solve the recurrence
$$a_j = \frac{n}{n-j} + \frac{j}{n-j} a_{j-1}$$
for $1\leq j <n$, and $a_0=1$?
I calculated it for some $n$s:
$n=2: [1, 3]$
$n=3: [1, 2, 7]$
$n=4: [1, \frac{5}{3}, \frac{11}{3}, 15]$
$n=5: [1, \frac{3}{2}, \frac... | From the recurrence relation, it is easy to check that
$$ \binom{n-1}{j} a_j - \binom{n-1}{j-1}a_{j-1} = \binom{n}{j}. $$
Therefore
$$ a_j = \frac{\sum_{k=0}^{j} \binom{n}{k}}{\binom{n-1}{j}}. $$
Addendum. By a different method, one can also prove that
$$ a_j = n \int_{0}^{1} (1-u)^{n-1-j} (1+u)^j \, du = \sum_{k=0}^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Formalization/Verification of a beginner combinatorics problem I have the following task:
To build a computer chip, 4 non-distinct components are needed. Of 12 existing components, 2 are defective. Given that 4 components are chosen, what are the probabilities of the following combinations?
*
*No defective component... | The probability on $k\in\{0,1,2\}$ defectives among the $4$ chosen is:$$\frac{\binom4k\binom8{2-k}}{\binom{12}2}$$
Hypergeometric distribution is used.
Actually this approach turns things around: there are $4$ chosen items and $8$ non-chosen items. Selecting $2$ items (the defective ones) what is the probability that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit of $\sin(\pi \sqrt{4n^2+n})$ We can notice that $\sqrt{4n^2 + n} = \sqrt{4n^2(1+ \frac{1}{4n})} = 2n\sqrt{1+\frac{1}{4n}}$. Therefore
$$\lim_{n \to \infty} \sin (2n\pi \sqrt{1+ \frac{1}{4n}}) \text{ will be an even number}$$
Because the square root becomes $1$ and we end up with an even number: $\sin(\text{even n... | Consider that
$$\sin(\pi\sqrt{4n^2+n})=\sin(\pi\sqrt{4n^2+n}-2\pi n)=\sin\left(\frac{\pi n}{\sqrt{4n^2+n}+2n}\right)\to\sin\frac\pi4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solution of $X^2+Y^2=Z^2$ under condition $X+Y+Z=1000$ As we know before the general solution of the equation $x^2+y^2=z^2$ in the number theory is as follows:
$x=\pm(a^2-b^2)c$, $y=\pm 2abc$ and $z=\pm(a^2+b^2)c$.
I want to know how to find a solution for the equation $x^2+y^2=z^2$ under condition $x+y+z=1000$?
A hint... | If the solutions you want have to be integer, you can simply add up the formulas for $x, y, z$ under the condition $x^2+y^2=z^2$. For example, let $x=(a^2-b^2)c, y=2abc, z=(a^2+b^2)c$, then $x+y+z=2ac(a+b)=1000\rightarrow ac(a+b)=500=2^2\cdot5^3 $. Since $a,b,c$ need to be integer, we can let $a=2,b=3,c=50$, then we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $ using Cauchy-Schwarz How do I prove this inequality
$$\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $$
I've tried to prove that $$\ \frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1} \ $$ is less than $$\ \frac{1}{1}+\frac{1}{2}+ ... + \frac{1}{... | By C-S
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}+...+\frac{1}{3n}+\frac{1}{3n+1}=$$
$$=\left(\frac{1}{n+1}+\frac{1}{3n+1}\right)+\left(\frac{1}{n+2}+\frac{1}{3n}\right)+...+\left(\frac{1}{2n}+\frac{1}{2n+2}\right)+\frac{1}{2n+1}>$$
$$>\frac{(1+1)^2}{4n+2}+\frac{(1+1)^2}{4n+2}+...+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proof of Bellard's formula I'm reading Bellard's proof for his eponymous formula computing pi digits, and I can't get past the first line.
Given that:
*
*$\displaystyle-\ln(1-x) = \sum_{n=1}^\infty \frac{x^n}{n}$ for $|x| < 1$,
*$\arctan(y) = \Im[-\ln(1-iy)]$ for any real $y$
(I didn't know the last one but Wolfram... | 1. The $\operatorname{arctan}$ formulas.
Consider what it means for a complex number $u$ to be a logarithm of $z$: essentially, that $e^u = z$.
Since $e^u = e^{\Re u + i \Im u} = e^{\Re u} e^{i\Im u} = r e^{i \theta}$ you see that pretty much by definition, $\Im \ln(z) = \Im u = \theta$ is the angular component of $z$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why isn't Legendre's Conjecture resolved by the work done by Nair and Hanson in relation to Least Common Multiples? I recently discovered the work done by Hanson and Nair. As I worked through Hanson's proof, an argument occurred to me regarding Legendre's Conjecture.
Clearly, my argument is wrong. It is too simple no... |
(4) So it follows:$$\text{lcm}(1,2,\dots,(n^2+n)) > \left(\frac{n^2+n}{2}\right)\frac{4^{n^2+n}}{n^2+n} = \frac{4^{n^2+n}}{2}$$
This is incorrect. It should be
$$\text{lcm}(1,2,\dots,(n^2+n)) > \left(\frac{n^2+n}{2}\right)\frac{4^{(n^2+n)/2}}{(n^2+n)/2}=4^{(n^2+n)/2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Find the smallest positive integer $X$ such that $478^{870} \equiv X \ (\text{mod} \ 273)$ Appreciate if one could advise if my solution is correct. Here is my attempt of the problem:
Since $(273, 478) =1,$ by Euler's theorem, $478^{\phi(273)}=478^{144} \equiv 1 \ \ (\text{mod} \ 273) \implies 478^{864} \equiv 1 \ \ (... | Your idea to use Chinese Remainder Theorem is good, but we can apply it more cleanly. We have $273 = 3\cdot 7\cdot 13,$ so we localize at each prime:
$$X\equiv 478^{870} \equiv 1^{870} \equiv 1 \pmod{3}$$
$$X\equiv 478^{870} \equiv 2^{6\cdot 145} \equiv 1 \pmod{7}$$
$$X\equiv 478^{870} \equiv (-3)^{6+12\cdot 72} \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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solve for $x$: $(x^4-13x^2+36)^4+|x^2+x-6|+\sqrt{x^3-7x+6}=0$ There is an equation that I think it is complicated ,a little!
$$(x^4-13x^2+36)^4+|x^2+x-6|+\sqrt{x^3-7x+6}=0$$
Actually we must solve for $x$ here.
I want you to hint me how can I simplify the equation and solve it.
| Hint:
The first expression is biquadratic and
$$x^4-13x^2+36=(x^2-4)(x^2-25).$$
The second is quadratic,
$$x^2+x-6=(x-2)(x+3).$$
The third is cubic and by inspection $x=1$ is a root. So
$$x^3-7x+6=(x-1)(x^2+x-6)=(x-1)(x-2)(x+3).$$
Now you need to find the common roots.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$ The equation is
$$\sqrt{\frac{4-x}x}+\sqrt{\frac{x-4}{x+1}}=2-\sqrt{x^2-12}$$
I tried squaring both left side and right side then bringing them to same numerator but got lost from there ... any ideas of how should this be solved?
I got to:
$$2\cdot\sqr... | Since there are three square roots you will need to square the whole equation at least three times. This is not the best way since it produces false solutions every single time and also the complexity of the equation will increase every time.
Therefore observe the numerator within the square roots of LHS and the square... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Suppose $(a, b, c)\in\Bbb R^3$, $a,b,c$ are all nonzero, and we have $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$. Find the value of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
Here is my attempt:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}.$$
I am having tr... | Just to do it the hard way.
Let $m = a+b$ and $n = b+c$ and $k = a + c$.
$\sqrt m + \sqrt n = \sqrt k$ so $k = m + n + 2\sqrt mn$ and
$\frac 1a + \frac 1b + \frac 1c = \frac 2{m - n +k} + \frac 2{n-k + m} + \frac 2{k-m + n}$
$=2(\frac {(n-k+m)(k-m + n) + (m-n+k)(k-m+n)+ (m-n+k)(n-k+m)}{(m-n+k)(n-k+m)(k-m+n)})$
$n-k +m... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Seeking methods to solve: $\int_{0}^{1} \frac{1}{1 + \arctan(x)} \:dx$ I've been playing with the following definite integral and was wondering if anyone knew of any methods to solve?
$$I = \int_{0}^{1} \frac{1}{1 + \arctan(x)} \:dx$$
| The approach I took:
First let $u = \tan(x)$ to yield:
$$I = \int_{0}^{\frac{\pi}{4}} \frac{\tan^2(u) + 1}{1 + u} \:du$$
Unfortunately I had no luck with my usual tactics, so I decided to use the Taylor series of $\tan^2(u)$ at $u = 0$ which I was greatly helped with here.
We find that:
$$ \tan^{2}(u) + 1 = \sum^{\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2971586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Can't figure out $O(n \log n)$ divide and conquer algorithm For an $n$ that is a power of $2$, the $n × n$ Weirdo matrix $W_n$ is defined as follows. For
$n = 1, W_1 = [1]$. For $n > 1$, $W_n$ is defined inductively by
$W_n =
\left[
\begin{matrix}
W_\frac{n}{2} & -W_\frac{n}{2} \\
I_\frac{n}{2} & W_\frac{n}{2} \\
\end... | I'm going to change indicies so that $W_n$ is a $2^n \times 2^n$ matrix. Let's assume that $x = (a, b)$ is a vector of length $2^n$, split up into two subvectors $a, b$ each of length $2^{n-1}$. Then we have that
$$ W_n x = \begin{pmatrix} W_{n-1} & -W_{n-1} \\ I_{n-1} & W_{n-1} \end{pmatrix} \begin{pmatrix} a \\ b \en... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how to find a, b that satisfy $\displaystyle \lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ How can I find those $a$ and $b$ in $\displaystyle \lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ ?
[my attempt] Since the denominator is $x^2$, it will be $0$. And $e^{-2x}$ is 1 so I get $1-\frac{1+ax}{1+b... | The given limit condition is equivalent to $$\lim_{x\to 0}\frac{(1+bx)-(1+ax)e^{2x}}{x^2}=0\tag{1}$$ We need to make use of the following limits $$\lim_{x\to 0}\frac{e^x-1}{x}=1,\lim_{x\to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}\tag{2}$$ The first one is standard and the second one is derived from first one via L'Hospital's ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right)$ I want to evaluate $$\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right).$$
I have computed
\begin{align*}
a_{n+1}-a_n & = \frac... | Hint: $$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}<\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)(2n)}}$$
$$<\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n-1}$$ and use $\lim_{n \rightarrow \infty} (\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n})=\ln 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Computational Complexity of Euclidean Algorithm for Polynomials Let us assume that the two polynomials that we have are degree $n$ polynomials. The naive Euclidean Algorithm for univariate polynomial does $O(n)$ divisions and each division takes $O(n^2)$. So shouldn't the naive Euclidean algorithm run for $O(n^3)$ time... | Here is an example, see what you think
$$ \left( 6 x^{5} + 5 x^{4} + 4 x^{3} + 3 x^{2} + 2 x + 1 \right) $$
$$ \left( x^{5} + 2 x^{4} + 3 x^{3} + 4 x^{2} + 5 x + 6 \right) $$
$$ \left( 6 x^{5} + 5 x^{4} + 4 x^{3} + 3 x^{2} + 2 x + 1 \right) = \left( x^{5} + 2 x^{4} + 3 x^{3} + 4 x^{2} ... | {
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"timestamp": "2023-03-29T00:00:00",
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If $ \int_{0}^{1} \frac{\ln x}{1-x^2} dx = -\frac{π^2}{\lambda} $ find $\lambda$ given that $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{π^2}{6} $
If $$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{π^2}{6} $$ then $$ \int_{0}^{1} \frac{\ln x}{1-x^2} dx = -\frac{π^2}{\lambda} $$ then the value of $\lambda$ equals?
My attemp... | Use the expansion
$$\ln\frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$$
then
\begin{align}
-\frac{1}{2} \int_{0}^{1} \frac{\ln\frac{1+x}{1-x}}{x} dx
&= -\frac{1}{2} \int_{0}^{1} \dfrac1x2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right) dx \\
&= -\left(1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots\right) \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Functional equation - Cyclic Substitutions Please help solve the below functional equation for a function $f: \mathbb R \rightarrow \mathbb R$:
\begin{align}
&f(-x) = -f(x) , \text{ and } f(x+1) = f(x) + 1, \text{ and } f\left(\frac 1x\right) = \frac{f(x)}{x^2} \\
&\text{ for all } x \in \mathbb R \text{ and } x \ne 0 ... | For $x\neq0$ and $x\neq1$ we obtain: $$f(x)=x^2f\left(\frac{1}{x}\right)=x^2f\left(\frac{1}{x}-1+1\right)=x^2\left(f\left(\frac{1}{x}-1\right)+1\right)=$$
$$=x^2+x^2f\left(\frac{1-x}{x}\right)=x^2+x^2\cdot\frac{f\left(\frac{x}{1-x}\right)}{\frac{x^2}{(x-1)^2}}=x^2+(x-1)^2f\left(\frac{x}{1-x}\right)=$$
$$=x^2+(x-1)^2f\l... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Arranging 5 pairs of brothers in ten tables Suppose I have $10$ people composed of $5$ pairs of brothers. They are seated randomly around $5$ tables. Each table has only two places.
What is the possibility that exactly $3$ pairs of brothers will be sitting at the same table?
$$\\ \frac{ {5 \choose 3} \cdot 3\cdot 2\cdo... | Let's take as our sample space the number of ways of placing two people at each table, which is
$$\binom{10}{2, 2, 2, 2, 2} = \binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
For the favorable cases, choose which three of the five tables will receive a pair of brothers. Choose a pair of brothers to sit... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $B^{225}$ without many computations I have that $B$ is a 4x4 matrix. $B-5I=\begin{pmatrix} -2 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 3 & 0 & -3 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. The question asks to find $B^{225}$ without performing many computations, leaving the answer as a product of 3 matrices (univerted matrice... | Let $A=B-5I$. Then $A^2=-5A$. Therefore $A^n=(-5)^{n-1}A$ for $n\ge1$.
Then
$$B^n=(5I+A)^n=5^n I+\sum_{k=1}^n {n\choose k}5^{n-k}A^k
=5^n I+\sum_{k=1}^n (-1)^{k-1}{n\choose k}5^{n-1}A
=5^n I+5^{n-1}A.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2978749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the Max, Min,Sup and inf of $A = \left \{ \frac{1}{3^\frac{x}{2}+3^\frac{2}{x}} \right \}$ I need to find the min, max, sup and inf of $$A = \left \{ \frac{1}{3^\frac{x}{2}+3^\frac{2}{x}} , x>0, x \in \mathbb{R} \right \}$$
My attempt:
$3^\frac{x}{2} + 3^\frac{2}{x} \geqslant 2\sqrt{3^{2 \left | x \right |... | Consider the function $f(x)=\frac{1}{3^\frac{x}{2}+3^\frac{2}{x}}$ defined on $(0,\infty).$
*
*Clearly, $f(x)>0$ for any $x>0.$ Since $\lim_{x \to 0^+}f(x)=0\quad$ and $\quad\lim_{x \to \infty}f(x)=0,$ $$Inf A=0 \quad \text{and A doesn't have a minimum}.$$
*If $x \in (0,2),$ then $3^\frac{x}{2}<3^\frac{2}{x}$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\int_{0}^{\pi} 2\pi\frac{R^2(z-R\cos\theta)\sin\theta }{\sqrt{(R^2+z^2-2zR\cos\theta)^3}}\,d\theta$ I am doing some physics problem and in order to solve it I need to solve the following integral:
$$ \int_{0}^{\pi} 2\pi\frac{R^2(z-R\cos\theta)\sin\theta }{\sqrt{(R^2+z^2-2zR\cos\theta)^3}}\,d\theta $$ where $z>... | The change of variables $t = \frac{2 z R}{z^2+R^2} \cos(\theta)$ yields
\begin{align} f(z,R) &\equiv 2\pi R^2 \int \limits_{0}^{\pi} \frac{z-R\cos(\theta)}{(z^2+R^2-2zR\cos\theta)^{3/2}} \, \sin(\theta) \mathrm{d} \theta = \frac{\pi R}{\sqrt{z^2+R^2}} \int \limits_{-\frac{2 z R}{z^2+R^2}}^{\frac{2 z R}{z^2+R^2}} \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2980975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For every natural number $m$, $2^{2m}-1$ is divisible by $3$ Can anyone tell me if I missed any small details in this proof?
(basis) $2^{2(1)}-1 = 3*1$ which is divisible by 3.
(induction) Fix $m>1$ so $p<m$. Hence for every natural number $p,$ $2^{2p}-1=3p.$ then for every natural number $(p+1),$ $2^{2(p+1)}-1=3(p+1)... | $2^{2m} - 1 = (2^m)^2 - 1 = (2^m - 1)(2^m + 1); \tag 1$
of the three consecutive integers
$2^m - 1, \; 2^m, \; 2^m + 1 \tag 2$
precisely one of the must be divisible by $3$ (this is true of any three consecutive integers); clearly
$3 \not \mid 2^m; \tag 3$
therefore either
$3 \mid 2^m - 1 \; \text{or} \; 3 \mid 2^m + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981174",
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"source": "stackexchange",
"question_score": "2",
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Prove that $f$ is additive if $f(x)f(x-y)+f(y)f(x+y)= f(x)^2+f(y)^2$
Say $f:\mathbb{R}\to \mathbb{R}$ is non-constant such that $$f(x)f(x-y)+f(y)f(x+y)= f(x)^2+f(y)^2$$
Prove that $f(x+y)=f(x)+f(y)$.
If we put $a=f(0)$ and $y=0$ we get $$f(x)^2+af(x)= f(x)^2+a^2 \implies af(x) = a^2$$
If $a\ne 0$ then $f(x)=a$ is... | Clearly any constant solution is the zero function. So assume that $f$ is nonzero. Let $P(x,y)$ denote the given assertion. We have $f(0)=0$ by $P(x,0)$. Take $f(x)\neq 0$, then $P(x,x)$ implies $f(2x)=2f(x)$ and following an easy induction $f(kx)=kf(x)$ for all $k\in \mathbb N.$ If $f(x_0)=0$ then $P(x_0,-x_0)$ gives ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Integral of $\int \frac{-x^2+2x-3}{x^3-x^2+x-1}dx$ I have this simple integral: $\int \frac{-x^2+2x-3}{x^3-x^2+x-1}dx$ and I can't come up with the correct answer.
Here's what I did:
I found the roots for $x^3-x^2+x-1$ so I could do partial fractions:
$\frac{-x^2+2x-3}{x^3-x^2+x-1}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}=\fra... | Your computation is almost correct, but you made a mistake here.
Note that$\int\dfrac{2}{x^2+1}=2\arctan(x)$, but not $\int\dfrac{2}{x^2+1}\ne2|x^2+1|$
| {
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"url": "https://math.stackexchange.com/questions/2984961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$
Prove for all $n \in \mathbb{Z}^+$ that $5 \mid (n^5-n)$
My proof
Basis step: Since $5 \mid (1^5-1) \iff 5 \mid 0$ and $5 \mid 0$ is true, the statement is true for $n=1$.
Inductive step: Assume the statement is true for $n = k$; that is, assume that $5 \mid (k^... | An alternative proof:
Note that we can write $n^{5} - n$ as follows:
$$n^{5} - n = 5\left[6{n\choose 2} + 30{n\choose 3} + 48{n\choose 4} + 24{n\choose 5}\right].$$
This is very clearly a multiple of $5$ (the expression is $5$ times another integer), so we are done.
Here, ${n\choose k} = {\frac{n!}{k! (n-k)!}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2985270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate a series with binomial coefficients
If
$$y=\frac{3}{5}+\frac{1\cdot 3}{2!}(\frac{2}{5})^2+\frac{1\cdot 3\cdot 5}{3!}(\frac{2}{5})^3+\dots$$
Then find the value 0f $y^2+2y$.
My approach is as follow
$$y-\frac{1}{5}=\frac{2}{5}+\frac{1\cdot 3}{2!}(\frac{2}{5})^2+\frac{1\cdot 3\cdot 5}{3!}(\frac{2}{5})^3+\... | Hint. Recall the definition of double factorial,
$$1\cdot 3\cdots (2k-1)=\frac{k!}{2^k}\binom{2k}{k}.$$
Hence
$$Y=x+\frac{1\cdot 3}{2!}(x)^2+\frac{1\cdot 3 \cdot 5}{3!}(x)^3+\dots=
\sum_{k=1}^{\infty}\binom{2k}{k}(x/2)^k.$$
Now take a look at How to show that $1 \over \sqrt{1 - 4x} $ generate $\sum_{n=0}^\infty \binom... | {
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"source": "stackexchange",
"question_score": "1",
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Extending the question: Is $S$ a basis for $P^3$?
Let $S = \{4 - t, t^3, 6t^2, 3t + t^3, -1 + 4t\}$. Is $S$ a basis for $P^3$?
Polynomial Equation:
0t3 + 0t2 - t + 4 = 0
t3 + 0t2 + 0t + 0 = 0
0t3 + 6t2 + 0t + 0 = 0
t3 + 0t2 + 3t + 0 = 0
0t3 + 0t2 + 4t - 1 = 0
Transform into an augmented matrix:
\begin{bmatrix}0&0&... | $P^3$ is 4 dimensional with a natural basis, $1, t, t^2, t^3$. In this basis, we can represent $S$ as the set of columns of
\begin{pmatrix}
4 & 0 & 0 & 0 & -1 \\
-1 & 0 & 0 & 3 & 4 \\
0 & 0 & 6 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 \\
\end{pmatrix}
This row reduces to
\begin{pmatrix}
1 & 0 & 0 & 0 & -\frac{1}{4} \\
0 & 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987481",
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Prove that $\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$ is a root for $x^3+\sqrt[3]{6}x^2-1$
Prove that
$$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$$
is a root for
$$P(x)=x^3+\sqrt[3]{6}x^2-1$$
Source: list of problems for math contest preparation.
I have n... | Simplify $c$ before substituting to $P(x)=x^3+\sqrt[3]{6}x^2-1$:
$$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\frac{1-\sqrt[3]{2}+\sqrt[3]{2^2}}{\sqrt[3]{9}}=\frac{1+2}{\sqrt[3]{9}(1+\sqrt[3]{2})}=\frac{\sqrt[3]{3}}{1+\sqrt[3]{2}};\\
\begin{align}P\left(\frac{\sqrt[3]{3}}{1+\sqrt[3]{2}}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987981",
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Solution to differential equation system and solution to its conversion into 2nd order differential equation Following is the differential equation system with IVP
$\vec{x}'=\small\begin{pmatrix}3&-9\\4&-3\end{pmatrix}\vec{x},
\vec{x}(0)=\small\begin{pmatrix}2\\-4\end{pmatrix}$ The particular solution to this differ... | That is not the second order equation I get. The system is:
$$\left\{
\begin{matrix}
\dot x=3x-9y \\
\dot y=4x-3y \\
\end{matrix}
\right.$$
From the second equation we get
$$\ddot y=4\dot x-3\dot y\tag 1$$
and
$$4x=\dot y+3y\tag 2$$
Substituting the value for $\dot x$ from the first equation into $(1)$
... | {
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"source": "stackexchange",
"question_score": "3",
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Expectation of total scores when rolling a die until the score is not 6 An unbiased six-sided die 1, 2, ..., 6 is used during a game, where if one player scores a 6, then he rolls the die again. The player continues playing the game until he scores another value which is NOT 6.
Find the expected value of the total scor... | The expected sum after $k$ rolls is not $k$ but $6(k-1)+3$ indeed if we stop at first roll it means that we get $1,2,3,4,5$ and the average sum will the mean i.e. $3$, if we rolled twice means that the first roll we get $6$ and now one of the values $1,2,3,4,5$ and so on. It means that the exact formula is
\begin{alig... | {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite integral of $\frac{\sqrt{ 25x^2 - 4}}{x}$ Right off the bat i factored a 4 from the radicand to get it into a form such that i can leverage ${\tan^2(\theta)}$ = ${\sec^2(\theta)} - 1$
Then i set $\frac{25}{4}$${x^2}$ = ${\sec^2(\theta)}$
${x}$ = $\frac{2}{5}\sec(\theta)$
${dx}$ = $\frac{2}{5} \sec(\theta)\... | Generally, seeing $\sqrt{x^2 - a^2}$ or something of the like begs for the substitution of $x = a\cos(u)$, since $\cos^2(x) - 1 = \sin^2(x)$. (Or perhaps the hyperbolic version; similar process, I'll leave that calculation up to you if you prefer.) So, notice, by factoring out $\sqrt{25}=5$ from the numerator,
$$\mathc... | {
"language": "en",
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$a,b,c,d,e$ are zeroes of $6x^5+5x^4+4x^3+3x^2+2x+1$ find $(a+1)(b+1)(c+1)(d+1)(e+1)$
If $a,b,c,d,e$ are zeroes of the polynomial
$$6x^5+5x^4+4x^3+3x^2+2x+1$$
find the value of $(a+1)(b+1)(c+1)(d+1)(e+1)$.
According to me in order to solve this problem one should first factorize the given polynomial in the form o... | $6(x-a)(x-b)(x-c)(x-d)(x-e) = 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1$
Plugging x = -1
$6(-1-a)(-1-b)(-1-c)(-1-d)(-1-e) = -6(a+1)(b+1)(c+1)(d+1)(e+1) = -6 + 5 - 4 + 3 - 2 + 1 = -3\\
(a+1)(b+1)(c+1)(d+1)(e+1) = \frac 12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990270",
"timestamp": "2023-03-29T00:00:00",
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Easy way of tackling $\sqrt{x^2}+x=\sqrt{5}$ Solve for x
$$\sqrt{x^2}+x=\sqrt{5}\tag1$$
$$\sin(\sqrt{x^2}+x)=\sin({\sqrt{5}})\tag2$$
$$\sin(\sqrt{x^2})\cos(x)+\cos(\sqrt{x^2})\sin(x)=\sin(\sqrt{5})\tag3$$
$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\tan(x)=\sin(\sqrt{5})\sec(x)\tag4$$
$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\sqrt{\s... | Use cases. Certainly $x \neq 0.$
Case I: $x > 0.$ Then $\sqrt{x^2} + x = 2x.$ Thus $x = \frac{\sqrt{5}}{2}.$
Case II: $x < 0.$. Then $\sqrt{x^2} = -x$, but then this says $ 0 = \sqrt{5},$ contradiction. Hence we must have the only solution $x = \frac{\sqrt{5}}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$
Prove the identity
$$8\cos^4 \theta -4\cos^3 \theta-8\cos^2 \theta+3\cos \theta +1=\cos4\theta-\cos3\theta$$
If $7\theta $ is a multiple of $2\pi,$ Show that $\cos4\theta=\cos3\theta$ and deduce,
$$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{... | $\cos\frac{0\pi}{7}, \cos\frac{2\pi}{7}, \cos\frac{4\pi}{7}, \cos\frac{6\pi}{7}$ are distinct roots of the fourth order polynomial $$P(x)=8x^4-4x^3-8x^2+3x+1$$
So $P(x)$ can be re-written $$P(x)=8\left(x-\cos\frac{0\pi}{7}\right)\left(x-\cos\frac{2\pi}{7}\right)\left(x-\cos\frac{4\pi}{7}\right)\left(x-\cos\frac{6\pi}{7... | {
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"url": "https://math.stackexchange.com/questions/2991542",
"timestamp": "2023-03-29T00:00:00",
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Number of garlands which could be formed by using $3$ flowers of same type and $12$ flowers of other type. Question : Find number of garlands which could be formed by using $3$ flowers of same type and $12$ flowers of other type.
By garland I mean a perfectly circular one .
All $3$ flowers are alike of one type, all ... | The numbers that you are after are called compositions (with 0s allowed, rotated such that the largest number comes first, and are not of the form a + b + a with a and b distinct). The compositions of 12 into 3 parts are:
12 + 0 + 0
11 + 1 + 0
11 + 0 + 1
10 + 2 + 0
10 + 1 + 1
10 + 0 + 2
9 + 3 + 0
9 + 2 + 1
9 + 1 + 2
9 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that the following inequality holds when $x>0$ $\require{cancel}$
Show that the following inequality holds for $x>0$
$$1+\frac{x}{2}-\frac{x^2}{8}<\sqrt{x+1}<1+\frac{x}{2}.$$
I proceeded as follows
$$\sqrt{x+1}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}},\quad \xi\in[0,x]$$
and substituing in the in... | The right inequality:
We need to prove that:
$$1+x<\left(1+\frac{x}{2}\right)^2$$ or
$$1+x<1+x+\frac{x^2}{4}.$$
The left inequality.
Let $x+1=t^2,$ where $t>1$.
Thus, we need to prove that:
$$1+\frac{t^2-1}{2}-\frac{(t^2-1)^2}{8}<t$$ or
$$(t-1)^3(t+3)>0.$$
| {
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Given $3^x = 12$ find $(\sqrt{4/3})^\frac{1}{x-2}$ Given $3^x = 12$ find $$\left(\sqrt{\dfrac{4}{3}}\right)^\dfrac{1}{x-2}$$ in simple form.
I've faced this problem, in a high school book, but failed to solve.
I've tried to calculate
*
*$3^{x-2} = \dfrac{12}{9}$
*$x -2 = \log_3 \dfrac{12}{9}$
*$\dfrac{1}{(x -2)} ... | Let $$y= (\sqrt {4/3})^{\frac {1}{x-2}}$$
We have $$\ln y = \frac {\ln (4/3)}{2(x-2)}$$
Since $$x-2 = \frac {\ln (4/3)}{\ln 3}$$
We get $$\ln y = \ln \sqrt 3 $$ which implies $$ y=\sqrt 3 $$
| {
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"url": "https://math.stackexchange.com/questions/3001578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Fast way to solve $(s-2i)^2 (s+2i)^2$ $(s-2i)^2 (s+2i)^2=$
$(s^2-4si-4) (s^2+4si-4)=$
$s^4+4s^3i-4s^2-4s^3i+16s^2+16si-4s^2-16si+16 =$
$(s^4-8s^2+16) =$
$(s^2+4)^2$
Is there a quicker way to see that $(s-2i)^2 (s+2i)^2= (s^2+4)^2$
| We have
$$(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$$
and recall that $(A-B)(A+B)=A^2-B^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What am I doing wrong finding $\lim_{x\to 0} \left( \frac{1+x\cdot2^x}{1+x\cdot3^x} \right)^{1/x^2}$? It has an answer here, but I'd like to know where my solution went wrong.
$$\lim_{x\to 0} \left( \frac{1+x\cdot2^x}{1+x\cdot3^x} \right)^{\frac{1}{x^2}} $$
$$\lim_{x\to 0} \left( \frac{1+x\cdot2^x +x\cdot 3^x-x\cdot 3^... | You want to evaluate:
$$\lim_{x\to 0}\frac{1}{x}\cdot \frac{2^x-3^x}{1+x\cdot3^x}=\lim_{x\to 0}\frac{1}{x}\cdot \frac{(1+x\color{red}{\ln 2}+o(x^2))-(1+x\color{red}{\ln 3}+o(x^2))}{1+x\cdot3^x}=\\
\lim_{x\to 0}\frac{\ln (2/3)+o(x^2)}{1+x\cdot3^x}=\ln \frac23.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2$ If $a, b, c$ are distinct real numbers then you demonstrate that:
$$ S=\frac{|a|}{|b-c|} + \frac{|b|}{|c-a|} + \frac{|c|}{|b-a|} \geq 2.$$
Using inequality $ |x-y|\leq |x|+|y|$ we showed that $ S >\frac{3}{2}.$
For $b = 2a, c = 3a, S=5,$ that is, the ... | Without loss of generality we can assume that $b$ and $c$ have the same sign. Then $|b|=|b-c|+|c|$. Also, $|c-a|\le |c|+|a|$ and $|a-b| \le |a|+|b-c|+|c|$. Therefore
$$
\frac{|a|}{|b-c|}+\frac{|b|}{|c-a|}+\frac{|c|}{|a-b|} \ge \frac{|a|}{|b-c|} + \frac{|b-c|+|c|}{|c|+|a|} + \frac{|c|}{|a|+|b-c|+|c|} = (*).$$
Denote $x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
An AMM-like integral $\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx$
How can we evaluate $$I=\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx=0?$$
I tried substitution $x=\frac{1-t}{1+t}$ and got
$$I=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} \arctan \frac{t-1}{t+1}}{t^2-1}dt\\
=\int_0^1\frac{2 ... | Through the dilogarithm/trilogarithm machinery it can be shown that
$$ \int_{0}^{1}\frac{\log(1+i x)\log(1+x)}{x}\,dx=\\\frac{\pi K}{2}-\frac{9i\pi^3}{64}+3iK\log(2)-\frac{3\pi i}{16}\log^2(2)+\frac{5\pi^2}{32}\log(2)-\frac{\log^3(2)}{8}-\frac{69}{16}\zeta(3)+6\,\text{Li}_3\left(\tfrac{1+i}{2}\right) $$
$$ \int_{0}^{1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
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Limit $ a_{n+1} = \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}), a_1 > 0 $
Let $ a_{n+1} = \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}), a_1 > 0 $. Show the sequence converge and find it's limit.
I did the following:
We see that $a_n$ is positive for every n obviously (considering $a_1 > 0$). Then:
$$ a_{n+1} - a_n = \frac{2a_n^{3... | First observe that
$$
a_{n+1}=\frac{1}{3}\left(2a_n+\frac{5}{a_n^2}\right)=\frac{1}{3}\left(a_n+a_n+\frac{5}{a_n^2}\right)\ge \left(a_n\cdot a_n\cdot \frac{5}{a_n^2}\right)^{1/3}=5^{1/3}.
$$
Hence, $a_n\ge 5^{1/3}$, for all $n>1$.
Next observe that
$$
f(x)=\frac{1}{3}\left(2x+\frac{5}{x^2}\right)<x,
$$
for all $x$ in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Limit of $\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$ (No L'Hôpital)
$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$
I can't get to the end of this limit. Here is what I worked out:
\begin{align*}
& \lim_{x \to 0} \frac{\cos 2x}{\sin 2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\lim_{x \... | Hint:
$$\lim_{x \to 0} \cot (2x)\cot (\frac{\pi }{2}-x)=\lim_{x \to 0} \cot (2x)\tan x=\lim_{x \to 0} \dfrac{\cos2x}{\sin2x}\dfrac{\sin x}{\cos x}=\lim_{x \to 0} \dfrac{\cos2x}{1}\dfrac{2x}{\sin2x}\dfrac{\sin x}{x}\dfrac{1}{\cos x}\dfrac12$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Product of the Riemann sum Hi I want to prove that for all $x,y\in R$, the following holds
$$(\sum_{n=0}^{\infty}\frac{x^n}{n!})(\sum_{n=0}^{\infty}\frac{y^n}{n!})=\sum_{n=0}^{\infty}\frac{(x+y)^n}{n!}$$
without using $e$ nor $\sum_{n=0}^\infty \frac{x^n}{n!}=\lim_{n\rightarrow\infty}(1+\frac{x}{n})^n$
What I know:
$$... | First, prove that $\sum_{n\ge 0} \frac {x^n}{n!}$ and $\sum_{n\ge 0} \frac{y^n}{n!}$ converge absolutely by using the Ratio Test and compute their Cauchy Product:
$$
\begin{align}
\left(\sum_{n\ge 0} \frac {x^n}{n!}\right)\left(\sum_{n\ge 0} \frac {y^n}{n!}\right)&=\sum_{n\ge 0} c_n = \sum_{n\ge 0} \sum_{i=0}^n \frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Ellipse on a Circular Cylinder in Cylindrical Coordinates Is there an equation in cylindrical coordinates for an ellipse (tilted at some angle) on the surface of a right circular cylinder of radius r? For simplicity, I envision the cylinder to be coincident with the x-axis.
I am aware that the cylinder could be "unwra... | As stated before assume the (normalized) equation of the plane is
$$ \frac{a x + b y + c z}{\sqrt{a^2+b^2+c^2}} = d $$
and the parametric equation of the cylinder
$$ \pmatrix{x & y & z}= f(\varphi,z) = \pmatrix{ R \cos\varphi, R\sin\varphi, z} $$
Where the two intersect you have your ellipse in cartesian coordinates
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An Inconsistency in Numerical Approximation Consider the expression
$$
10^5 - \frac{10^{10}}{1+10^5}.
$$
Using the elementary properties of fractions we can evaluate the expression as
$$
10^5 - \frac{10^{10}}{1+10^5} = \frac{10^5 + 10^{10} - 10^{10}}{1+10^5} = \frac{10^5}{1+10^5}\approx 1.
$$
Note that the approximat... | It is simply an issue of accuracy of approximation. Let me write $x = 10^p$. Then your expression is
$$ x - \frac{x^2}{1+x}$$
Note that $$\frac{x^2}{1+x} = \frac{x}{1/x + 1} = x (1 - 1/x + O(1/x^2)) = x - 1 + O(1/x)$$
so that
$$ x - \frac{x^2}{1+x} = x - (x - 1 + O(1/x)) = 1 + O(1/x)$$
In your second calculation you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $ how can I find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $?
It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...
Let consider sum no to infinity, but to n
$$ \sum_... | We have that
$$\sum_{n=1}^{N} \left(n\cdot \ln \frac{2n+1}{2n-1} - 1\right)=\sum_{n=1}^{N} \left(n\cdot \ln (2n+1)-n\ln (2n-1) - 1\right)=$$
$$=(1\cdot \ln 3-1\cdot \ln1-1)+(2\cdot \ln 5-2\cdot \ln3-1)+(3\cdot \ln 7-3\cdot \ln5-1)+\ldots=$$
$$=-\ln(3\cdot 5\cdot 7\cdot \ldots\cdot (2N-1))+N\cdot\ln(2N+1)-N=$$$$=-\ln\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3019050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Ways for Two Dice to Sum to 8: $x_1 +x_2 =8$ I am trying to determine the number of ways two dice can be rolled to have a sum of 8. This question can be easily brute-forced, but I want to use a technique that can be applied to similar questions that cannot be brute-forced so easily.
My solution was as follows,
$$x_1 +x... | Your sum count is wrong because you are including the cases where some of the $x_i>6.$ For example, $7+3+2=12,$ but you don't want to count that case.
Summary:
The number of ways of getting the sum $S$ from rolling $k$ fair $n$-sided dice labeled $1$ to $n$ is:
$$\sum_{i=0}^{k}(-1)^i\binom{k}{i}\binom{S-in-1}{k-1}$$
Y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3020438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use complete induction to prove that $a_n < 2^n$ for every integer $n \geq 2$ Define the sequence of integers $a_0, a_1, a_2, \cdots$ as follows
$$ a_i =
\begin{cases}
i+1 & 0 \leq i \leq 2 \\
a_{i-1} + a_{i-2} + 2a_{i-3} & i > 2 \\
\end{cases}
$$
Use complete induction to prove that $a_n < 2^n$ for eve... | The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove a geometric sequence a, b, c from the arithmetic progression $1/(b-a)$, $1/2b$, $1/(b-c)$ The given task is:
The following forms an arithmetic sequence: $$\frac{1}{b-a}, \frac{1}{2b}, \frac{1}{b-c}.$$
Show, that $a, b, c$ forms an geometric sequence.
It's easily enough to understand that $$ \frac{1}{2b}-\frac{1}{... | Taking it from where you left off, use cross-products and simplify
$$(a+b)(b-c) = (a-b)(b+c) \iff \color{blue}{ab}-ac+b^2\color{green}{-bc} = \color{blue}{ab}+ac-b^2\color{green}{-bc}$$
$$2b^2 = 2ac \iff b^2 = ac \iff \frac{b}{a} = \frac{c}{b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding $\lim\limits_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)$ What is$$\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)?$$So it is$$\lim_{n→∞}\frac{n^3(\sqrt{n^2+\sqrt{n^4+1}})^2-(n\sqrt{2})^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\lim_{n→∞}\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}.$$
I do not ... | HINT
You only need to apply the trick twice, indeed we have that
$$\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}=(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2})\cdot\frac{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=$$$$=\frac{n^2+\sqrt{n^4+1}-2n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}$$
and
$$\frac{\sqrt{n^4+1}-n^2}{\s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving the Diophantine equation $y^2 = x^4+x+ 2$ for $x,y \in \mathbb{Z}$ I want to solve the Diophantine equation $y^2 = x^4+x+ 2$ for $x,y \in \mathbb{Z}$.
I already found 4 solutions: $(x,y) = (1,\pm2)$ and $(x,y)=(-2,\pm4)$. It can probably be solved using some factorization argument, but I don't know how.
| If $x\geq -1$, then $$x^4<y^2<(x^2+2)^2\,.$$
Therefore, for $x\geq -1$, there exists a solution iff $x^4+x+2=(x^2+1)^2$. The only integer root of the last equation is $x=1$, yielding the solutions $(x,y)=(1,\pm2)$.
If $x\leq -3$, then $$(x^2-2)^2<y^2<x^4\,.$$
Therefore, for $x\leq -3$, there exists a solution iff $x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sum\limits_{n=1}^{\infty} a_{f(n)} = \dfrac{3}{2} \ln(2) $ Where $a_n = \dfrac{(-1)^{n+1}} { n}$ and $f:\mathbb{N}\to \mathbb{N} $ is a bijection which for $k \in \mathbb{N}\setminus \{0\}$ is given by: $$f(3k+1)= 4k+1$$ $$f(3k+2)= 4k+3$$ $$f(3k+3)= 2k+2.$$
Could some please help me get started on this proof... | Euler's constant:
The sequence
$$\gamma_n := 1+\frac 12 + \frac 13 + \cdots + \frac 1n - \ln(n)$$
converges to a finite limit $\gamma$, where $\gamma$ is a constant:
$$\lim_{n \rightarrow \infty} \gamma_n = \gamma$$
The subsequence lemma:
The sequence $b_n$ converges to a limit $L$ if and only if all of the subsequenc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Half of Vandermonde's Identity We know Vandermonde's Identity, which states
$\sum_{k=0}^{r}{m \choose k}{n \choose r-k}={m+n \choose r}$.
Does anyone know what happens if we walk bigger steps with k? Say we skip all the odd ks, is something like
$\sum_{k=0}^{r/2}{m \choose 2k}{n \choose r-2k}=\frac{1}{2} {m+n \choose ... | We derive a binomial identity which shows the deviation of OPs sum from $\frac{1}{2}\binom{m+n}{r}$. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n\tag{1}
\end{align*}
We assume wl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Find the sum of these variables. Five real numbers $a_1, a_2, a_3, a_4\;\text{and}\; a_5\;$ are such that
$$\sqrt{a_1- 1} + 2\sqrt{a_2- 4}+3\sqrt{a_3- 9}
+4\sqrt{a_4- 16} + 5 \sqrt{a_4- 25} =\frac{a_1+a_2+a_3+a_4+a_5}{2}.$$
Find $a_1+a_2+a_3+a_4+a_5.$
Thanks for checking this out!
| Define$$b_1=\sqrt{a_1-1}\\b_2=\sqrt{a_2-4}\\b_3=\sqrt{a_3-9}\\b_4=\sqrt{a_4-16}\\b_5=\sqrt{a_5-25}\\$$therefore by substitution we have $$b_1+2b_2+3b_3+4b_4+5b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2\over 2}+{55\over 2}$$or$$2b_1+4b_2+6b_3+8b_4+10b_5={b_1^2+b_2^2+b_3^2+b_4^2+b_5^2}+{55}$$by rearranging we obtain$$(b_1-1)^2+(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{ax^2+bx+c}=\pi$ similar identities I recently found that
$$\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{ax^2+bx+c}=\pi$$
iff
$$b^2-4ac=-4.$$
I found it by integrating
$$I=\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{ax^2+bx+c}.$$
If the reciprocal of the function is to be integrate... | First if we consider the integral:
\begin{equation}
I(a) = \int_{-\infty}^{\infty}\frac{1}{x^2 + a^2}\:dx = \frac{\pi}{a}
\end{equation}
Then we can see:
\begin{equation}
I(a) + cI(b) = \left[\int_{-\infty}^{\infty}\frac{1}{x^2 + a^2}\:dx + c\int_{-\infty}^{\infty}\frac{1}{x^2 + b^2}\:dx \right] = \frac{\pi}{a} + c\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3040185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
} |
Show that $T: K\to K,$ defined by $Tx=(0,x^2_1,x^2_2,x^2_3,\cdots)$ is not non-expansive Let $X = l_\infty$ (the space of sequences of real numbers which are bounded). Let $K=\{x\in l_\infty:\Vert x \Vert_\infty\leq 1\}.$ Define \begin{align} T:& K\to K \\&x\mapsto Tx=(0,x^2_1,x^2_2,x^2_3,\cdots)\end{align}
I want to s... | The only problem I see is the part where you write
$$
\frac{5}{4}\sup \{ 0,\left|\frac{3}{4}-\frac{1}{2}\right|,\left|\frac{3}{4}-\frac{1}{2}\right|\cdots\} \geq\sup \{ \left|\frac{3}{4}-\frac{1}{2}\right|,\left|\frac{3}{4}-\frac{1}{2}\right|\cdots\},
$$
the sign $\ge$ should be $>$ instead because in order to show th... | {
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"url": "https://math.stackexchange.com/questions/3040375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $\sum_{k=0}^{2n} \binom {2n+k}{k} \binom{2n}{k} \frac{(-1)^k}{2^k} \frac{1}{k+1} = 0. $
Let $n$ be a positive integer. Prove that
$$
\sum_{k=0}^{2n} \binom {2n+k}{k} \binom{2n}{k} \frac{(-1)^k}{2^k} \frac{1}{k+1} = 0.
$$
I am trying to solve this by using induction on $n$. I have proven the sum to be zer... | Let's approach the sum through the Hypergeometric Function.
To this purpose let's rewrite it as
$$
\eqalign{
& S(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,2n} \right)} {\binom{2n+k}{k} \binom{2n}{k}
{{\left( { - 1} \right)^{\,k} } \over {2^{\,k} \left( {k + 1} \right)}}} = \cr
& = \sum\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove the inequality $\ln {(1+\frac{1}{x})}> \frac{2}{2x+1}$ Prove the inequality $$\ln {(1+\frac{1}{x})}> \frac{2}{2x+1}$$
for $x>0$.
My attempt: Let $$f(x)=\ln {(1+\frac{1}{x})}-\frac{2}{2x+1}$$
Then $$f'(x)=-\frac{1}{x(x+1)}+\frac{4}{(2x+1)^2}$$
$$f''(x)=\frac{1}{x^2}-\frac{1}{(x+1)^2}-\frac{8}{(2x+1)^3}>0$$
Then th... | With your approach:
$$
f'(x)=-\frac{1}{x(x+1)}+\frac{4}{(2x+1)^2} = \frac{-1}{x(x+1)(2x+1)^2} < 0
$$
so that $f$ is strictly decreasing, and therefore
$$
f(x) > \lim_{t\to \infty} f(t) = 0 \, .
$$
Or you substitute $y = 1/x$ and consider
$$
g(y) = \ln (1+y) - \frac{2y}{2+y}
$$
with $g(0) = 0$ and
$$
g'(y) = \frac{y^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How would I go about solving for $x$ in $\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b$? The question
This is a homework question. Given the following, I am to solve for $x$ in terms of $a$ and $b$:
$$\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b;a>b.$$
My attempt
Although I see ... | Hint: Write your equation in the form
$$\sqrt{x-a}(x+b-2a)=\sqrt{x-b}(a-x)$$ and square it. We get in the case of $$a>b$$ $$x=a$$ or $$x=\frac{1}{3}(4a-b)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Every positive power of $5$ appears in the last digits of bigger power of $5$ Problem. Show that for every positive integer $n$, there is an integer $N > n$ such that the number $5^n$ appears as the last few digits $5^N$. For example, if $n = 3$, we have $5^3 = 125$ and $5^5 = 3125$, so $N = 5$ would satisfy.
This is a... | Fix $n$.
The cases $n \le 3$ can be handled directly.
We now assume $n > 3$.
Let $m = \lceil n \log_{10} 5 \rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.
You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to factor this quadratic expression? A bit confused on how to factor $2x^2 + 5x − 3 = 0$. Firstly, I multiplied $a \cdot c$, so $2(-3)=-6$, however couldn't find two numbers that will add up to $5$. Then, I thought of the factors could be $6(-1)=-6$, and $6+-1=5$. However, shouldn't the $a$ and $c$ values always be... | So we can actually generalize this. Say we have the polynomial
$$p(x)=ax^2+bx+c$$
Fact:
$$p(x)=\bigg(ax+\frac{b-\sqrt{b^2-4ac}}{2}\bigg)\bigg(ax+\frac{b+\sqrt{b^2-4ac}}{2}\bigg)$$
Proof:
Let's assume the existence of three real numbers $r_1$, $r_2$, and $e_1$ such that
$$ax^2+bx+c=e_1(x-r_1)(x-r_2)$$
If we expand th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
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Given $\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$, prove $\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$.
Let $x$ and $y$ be real numbers such that
$$\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$$
Prove that
$$\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt... | Hint :
By the hypothesis of the problem, do some algebra and show that :
$$\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1 \Leftrightarrow \dots \Leftrightarrow y = -x$$
Now, substitute $y=-x$ on the expression $\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right)$ and see what happens.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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If $p$ is prime, then $x^2 +5y^2 = p \iff p\equiv 1,9 $ mod $(20)$. Let $p\neq 2,5$ be prime. I wish to show that: $x^2 +5y^2 = p \Leftrightarrow p\equiv 1,9 $ mod $(20)$.
I proved to $\Rightarrow$ part, means $x^2 +5y^2=p \Rightarrow p\equiv 1,9 $ mod $(20)$.
For $\Leftarrow$ , $p\equiv 1,9(20) \Rightarrow p\equiv 1(4... | Let $p\equiv1$ or $9\pmod{20}$. Then $-5$ is a quadratic residue modulo $p$,
so there are integers $r$ and $s$ with $$r^2+ps=-5.$$
This means that the integer quadratic form
$$f(X,Y)=pX^2+2rXY+sY^2$$
has discriminant $-20$ and is also positive definite. Then $p$
is represented by $f$. Now $f$ is equivalent to a reduced... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
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a problem on complex numbers Let $w\neq 1$ and $w^{13} = 1$.
If $a = w+ w^3 + w^4 + w^{-4} + w^{-3} + w^{-1}$ and $b = w^2+ w^5 + w^6 + w^{-6} + w^{-5} + w^{-2}$, then the quadratic equation whose roots are $a$ and $b$ is ... ?
I got $w=\cos(\frac{2\pi}{13})+i\sin(\frac{2\pi}{13})$
And then I found $a$ and $b$ in tri... | $$ a^2 + a = \frac{ w^{16} + 2w^{15} + w^{14} + 2w^{13} + 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8 + 3w^7 + 3w^6 + 3w^5 + 3w^4 + 2w^3 + w^2 + 2w + 1}{w^8} $$
This is not impressive without
$$ w^{16} + 2w^{15} + w^{14} + 2w^{13} = w^3+2w^2+w+2. $$ Therefore
$$ a^2 + a = \frac{ 3w^{12} + 3w^{11} + 3w^{10} + 3w^9 + 6w^8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$ Let x and a be real numbers > 0. Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$
My idea is that I'm going to use $a>b \iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$\frac{1}{2}(x+\frac{a}{x}) \ge \sq... | $(\sqrt{x}-\sqrt{a/x})^2\geq0\Rightarrow x+a/x-2\sqrt{a}\geq 0\Rightarrow 1/2(x+a/x)\geq \sqrt{a}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Trying to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$ to be $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ I am asked to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$. The solution is provided as $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Here is my wor... | $$\frac{\sqrt{8}}{1-\sqrt{3x}} = \frac{\sqrt{8}}{1-\sqrt{3x}} \cdot \frac{1+\sqrt{3x}}{1+\sqrt{3x}} = \color{blue}{\frac{\sqrt{8}\cdot\left({1+\sqrt{3x}}\right)}{1-3x}} = \frac{\sqrt{8}+\sqrt{24x}}{1-3x}$$
$$= \frac{\sqrt{2^3}+\sqrt{2^3\cdot3x}}{1-3x} = \frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$$
Notice the step I highlighted ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Non-homogenous differential equation help
*a) Given $x = e^u$ and $$x^2\frac{d^2y}{dx^2} - 4x \frac{dy}{dx} + 6y = 12$$ show that $$\frac{d^2y}{du^2} - 5 \frac{dy}{du} + 6y = 12$$
I was able to do this:
$x = e^u$
$$\frac{dx}{du} = e^u = x $$
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = \frac{1}{x}\frac{dy}{du}$$
$... | You're correct that the general solution is
$$ y(x) = Ax^3 + Bx^2 + 2 $$
Using the given boundary conditions we have
\begin{align}
y(1) &= A + B + 2 = 7 \\
y(2) &= 8A + 4B + 2 = 14
\end{align}
or
\begin{align}
A + B &= 5\\
8A + 4B &= 12
\end{align}
Solving the above system gives $A=-2$, $B=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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