Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Polynomial division in 3 variables I'm trying to compute $$\frac{x^3-2x^2-xyz+2xy+yz}{x-2}$$
I have written this in the form $$\begin{align}x^3-2x^2-xyz+2xy+yz&=(x-2)(x^2+bx+cy+dz)\\
x^3-2x^2-(yz+2y)x+yz&= x^3 + (b-2)x^2 + (cy+dz-2b)x+(-2cy-2dz)
\end{align}$$
This leaves me with the following equations
$$\begin{align}b... | Let $F(x)={x}^{3}-xyz-2\,{x}^{2}+2\,xy+yz$. Then by Bézout's theorem we have
$F(x)=(x-2) Q(x)+F(2).$
Now $F(2)=4y-yz$.
$$
F(x)-F(2)={x}^{3}-xyz-2\,{x}^{2}+2\,xy+2\,yz-4\,y=\left( x-2 \right) \left( {x}^{2}-yz+2\,y \right).
$$
Thus
$$
\frac{{x}^{3}-xyz-2\,{x}^{2}+2\,xy+yz}{x-2}={x}^{2}-yz+2\,y+\frac{4y-yz}{x-2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2167213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Beta function-like integral $$\int\limits_0^1 \frac{x^{1-\alpha} (1-x)^\alpha}{(1+x)^3} \, dx $$
After the substitution $z=\frac{1}{x} - 1$,
I've got this:
$$\int\limits_0^\infty \left(\frac{1}{z}-1\right)^\alpha\left(\frac{1}{z}+1\right)^{-3} dz $$
But it is still far from the Beta function.
|
The key is the linear-fractional transformation,
$$x\mapsto\frac{1-t}{1+t}.$$
Then $1-x=\frac{2t}{1+t}$ and $1+x=\frac{2}{1+t}$, and we find
$$\begin{align}
\int_{0}^{1}\frac{x^{1-a}\left(1-x\right)^{a}}{\left(1+x\right)^{3}}\,\mathrm{d}x
&=\frac12\int_{1}^{0}\left(\frac{1-x}{x}\right)^{a}\frac{x}{1+x}\cdot\frac{\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solving: $3^m-2=n^2$
Solve $3^m-2=n^2$ for positive pairs of integers $(n,m)$.
My try:
$3^m-6=n^2-4\implies 3(3^{m-1}-2)=(n+2)(n-2)\implies 3=n+2\ or\ 3=n-2$ so we got $n=\pm 5 , \pm 1$. Used $\pm$ as we have $n^2$.
Putting $n=\pm 5 , \pm 1$ gives $m=3,1$ respectively. We kick out negative ones as problem ask for pos... | If $m=3k$, then the equation is $n^2 = (3^k)^3 -2$. I put the elliptic curve $y^2=x^3-2$ into SAGE. The only integral point is $(x,y) = (3,5)$ which corresponds to the the solution $m=3, n=5$.
If $m=3k+1$, then multiply through by $9$ to get $(3n)^2 = (3^{k+1})^3 - 18.$ I put the curve $y^2 = x^3-18$ into SAGE. The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Variation of parameters (issue with the constants) I have to solve this ODE using the variation of parameters method:
$$4y''+y=\frac{2}{\cos \left( \frac{x}{2} \right)}$$
Solving the homogeneous problem yields
$$y_h(x)=
c_1 \cos \left( \frac{x}{2} \right)+
c_2 \sin \left( \frac{x}{2} \right)$$
Now, to solve the variati... | Try with the equation writen in this form. The method needs the coefficient to be 1 for the highest derivative (standard form).
$$y''+\frac{y}{4}=\frac{1}{2\cos \left( \frac{x}{2} \right)}$$
With it, we get $c_2'/2=1/2$ and $c_1'=-\tan \left( \frac{x}{2} \right)$. With them, the expected solution follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry - Obtuse angle An obtuse angle $\theta$ has a $\sin \theta = \frac{x+1}{x+2}$ and $\cos \theta= \frac{x}{x+2}$.
Which angle is it?
I've been frustrated with this for the last few hours now. My problem is that I can't seem to find the length of any of the sides. I've tried working it out with the $\sin \th... | $$\sin \alpha = \frac{x+1}{x+2}, \,\cos \alpha=\frac{x}{x+2},\, \alpha\in \left(\frac{\pi}{2}, \pi\right), \, x\neq -2$$
Using $\sin^2\alpha+\cos^2\alpha=1$:
$$1=\left(\frac{x+1}{x+2}\right)^2+\left(\frac{x}{x+2}\right)^2$$
$$1=\frac{2x^2+2x+1}{x^2+4x+4}$$
$${2x^2+2x+1}=x^2+4x+4$$
$${x^2-2x-3}=0$$
$$(x+1)(x-3)=0$$
So $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Co-ordinates of the vertices an icosahedron relative to its centroid.
This is a picture of an icosahedron. I need to know the coordinates of the vertices of the icosahedron relative to it's centroid in order to programme a projection of one on a three dimensional plane. (By the way, it has twelve vertices, so it is go... | With $20$ faces, each face has an area of $\frac\pi5$ steradians. That means that the spherical excess in each face is $\frac\pi5$ radians. Thus, each angle in each spherical triangular face has an angle of $\frac\pi3+\frac\pi{15}=\frac{2\pi}5$.
We can use the Spherical Law of Cosines for Angles to get the angle subten... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
What does this series converge to, if anything? $$\arctan{1} + \arctan{\frac{1}{2}} + \arctan{\frac{1}{3}} + \arctan{\frac{1}{4}} ...= ?$$
The infinite series for arctan is
$$\arctan{x} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ...$$
So I want to sum up the $\arctan{1\over n}$ where $n$ starts at $1$ and goes... | For $0 < x \leq 1$, the series
$$ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ... $$
is an alternating series with decreasing terms. This means the terms alternate between overshooting and undershooting the actual value of $\arctan(x)$. In particular,
$$ 0 < x \leq 1 \implies x - \frac{x^3}{3} < \arctan(x) < x $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2175526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Factor $(x+y)^4+x^4+y^4$ Title says it all, I just want to know how to factor $(x+y)^4+x^4+y^4$. I only know that it's possible to factor, but got no idea how to do it. If it were a single-variable polynomial I could try to find rational roots or something, but I'm lost with this one.
| $$x^4 + y^4 + (x + y)^4
=2 x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + 2 y^4$$
$$=x (x (x (2 x + 4 y) + 6 y^2) + 4 y^3) + 2 y^4
=2 x^4 + y (4 x^3 + y (6 x^2 + y (4 x + 2 y)))$$
$$=4 y^2 (x^2 + x y) + 2 (x^2 + x y)^2 + 2 y^4
=2 (x^4 + 2 x^3 y + 3 x^2 y^2 + 2 x y^3 + y^4)$$
$$=2 (x^2 + x y + y^2)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $x^3+y^3 \le 2$. As in the title. If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $$x^3+y^3 \le 2.$$
This seems to be a very tricky one. I tried applying various inequalities like AM-GM, unfortunately, none of techniques I'm familiar with seem to work here.
I'd greatly ap... | Another way. Let $y=kx$.
Hence, the condition gives $x^4x^2-(k^3-1)x-1\leq0$ or
$$x\leq\frac{k^3-1+\sqrt{k^6+4k^4-2k^3+1}}{2k^4}.$$
Thus, we need to prove that
$$\left(\frac{k^3-1+\sqrt{k^6+4k^4-2k^3+1}}{2k^4}\right)^3(k^3+1)\leq2$$ or
$$3k^{12}-3k^{10}+2k^9+3k^4-2k^3+1\geq(k^9+k^7-k^6+k^4-k^3+1)\sqrt{k^6+4k^4-2k^3+1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2178764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Sketching the graph of $Y=\sin(3x)+\sin(x)$ How do I sketch the graph of
$Y = \sin(3x) + \sin (x)$?.
I realize the amplitude is $1$, but am not sure. I found the period to be $2\pi$. My problem is that I don't know if this graph will be similar to the graph of $Y = \sin (x)$. I want to know how the graph will be like.
| A simple method to sketch periodic functions like this is:
1) find all zeroes ($x$ for which $Y=0$) in the first period
$$ \sin(3x) + \sin(x) = 0 $$
$$ x = n\pi,\, x=n\pi - \frac{\pi}{2}, \quad n\in\mathbb{Z} $$
So $Y$ crosses the x-axis at $x=0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ for $x \in [0,2\pi]$
Edit How ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sixth degree polynomial problem If the graph of $$y = x^6 - 10x^5 + 29x^4 - 4x^3 + ax^2$$ always lies above the line $y = bx + c$, except for $3$ points where the curve intersects the line.
What is the largest value of $x$ for which the line intersects the curve?
*
*A) 4
*B) 5
*C) 6
*D) 7
Through general idea ... | I begin the computation by the same expression as @Ákos Somogyi
$$\tag{1}x^6-10 x^5+29 x^4-4 x^3+ax^2-bx-c=\underbrace{(x-\alpha)^2(x-\beta)^2(x-\gamma)^2}_{p(x)^2}$$
But I consider at once that this polynomial is equal to
$$\tag{2}p(x)^2=(x^3+ux^2+vx+w)^2$$
for certain coefficients $u,v,w.$ Expanding the square in (2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How did they simplify this sum? Im having trouble seeing how this simplification is justified:
$$\frac{2}{\pi}\sum_{k=-\infty}^\infty \frac{e^{i(2k+1)x}}{(2k+1)^{2}} =\frac{4}{\pi}\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}} $$
There are no steps on how they simplified, could anyone help me with it?
| Note that we can write
$$\frac{2}{\pi}\sum_{k=-\infty}^\infty \frac{e^{i(2k+1)x}}{(2k+1)^{2}} =\frac{2}{\pi}\sum_{k=-\infty}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}+i\frac{2}{\pi}\sum_{k=-\infty}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}} \tag1$$
Next, we split the series for the cosine into series over negative and non-nega... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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division of polynomials in $\mathbb{Z_2}$ I'm working on a problem involving division of polynomials in $\mathbb{Z_2}$. The question goes like this...
Find a greatest common divisor of the polynomials $x^6+x^4+x+1$ and $x^7+x^4+x^3+1$.
My textbook gave some examples but skipped a lot of the work and cannot follow what... | I don't know how to typeset long division here, so I'll try to describe the procedure step-by-step instead. Say, we want to divide $x^5+x^2+x+1$ by $x^4+x^3+x+1$ with remainder over $\mathbb{Z}_2$. Dividing the leading terms gives us $x^5/x^4=\color{magenta}{x}$ as the first term of the quotient. Now,
$$\color{magenta}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Constant sequence limit Let us consider the sequence $(a_n)_{n \ge 1}$ of positive real numbers such that
$$\lim _ {n \to \infty} \left( \frac {a_1}{a_2} + \frac {a_2}{a_3} + \dots + \frac {a_{n-1}}{a_n} + \frac {a_n}{a_1} -n\right)=0.$$
Prove that the sequence is constant.
My try: I considered the $AM -GM$ inequality ... | We use induction to prove that $ a_1 = \cdots = a_m$ for all $m = 1, 2, \cdots$. Since the base case is trivial, we focus on the induction step.
Assume that $a_1 = \cdots = a_m$. Then for $n > m$, we have
\begin{align*}
\frac{a_1}{a_2} + \cdots + \frac{a_n}{a_1} - n
&= \overbrace{\left(\frac{a_1}{a_2} + \cdots + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a+b=2$ so $2+\frac{ab(a-b)^2}{4}\leq a^{1+\sqrt{b}}+b^{1+\sqrt{a}}\leq2+\frac{(a-b)^2}{4}$
Let $a$ and $b$ be non-negative numbers such that $a+b=2$. Prove that:
$$2+\frac{ab(a-b)^2}{4}\leq a^{1+\sqrt{b}}+b^{1+\sqrt{a}}\leq2+\frac{(a-b)^2}{4}$$
In the left inequality the equality occurs also for $a=2$ and $b=0$... | (Partial solution) For the right inequality:
WLOG, assume $a \le b$. Then $0 < a \le 1 \le b < 2$.
We have $\sqrt{b} = 1 + \frac{b - 1}{\sqrt{b} + 1} \ge 1 + \frac{b - 1}{\frac{b + 1}{2} + 1} = \frac{3b + 1}{b + 3}$
and $\sqrt{a} = 1 - \frac{1 - a}{\sqrt{a} + 1} \le 1 - \frac{1 - a}{(a + 1)/2 + 1} = \frac{3a + 1}{a + 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find $f'(1)$ given $f(x)$ and $f(1)$ Question:
Let $$f(x)=\int\frac{x-1}{x+1}\frac{dx}{\sqrt{x^3+x^2+x}}$$ such that $f(1)=\frac{2\pi}{3}$
Then $f'(1)$ is equal to
A) $0$
B) $\cfrac \pi3$
C) $\cfrac\pi4$
D) $2$
Attempt:
\begin{align}f(x)&=\int\frac{x-1}{x+1}\frac{1}{\sqrt{x^3+x^2+x}}dx\\
f(1)&=\int0dx\\
f(1)&=0+c\\
... | $f'(x)$ is simply the term under integration.$$f'(x)=\frac{x-1}{x+1}\frac{1}{\sqrt{x + x^2 + x^3}}$$
Take an example.
$$\text{Let} ~~~g(x) = \int xdx ; ~~\text{find g(1), given g(0)=0.}$$.
As you have done , it should be $$g(1) = \int 1dx= x+c$$ using $ Since g(0)=0 \implies c=0$
So, $g(1)$ should $x$, isn't it weird?... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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The remainder when $33333\ldots$ ($33$ times) is divided by $19$ $A= 33333\ldots$ ($33$ times). What is the remainder when $A$ is divided by $19$?
I don't know the divisibility rule of $19.$
What I did was
$32\times(33333\times100000)/19$ and my remainder is not zero and this is completely divisible by $19.$
This is a... | Here's something crude but it takes less than the whole $33$ steps:
$$
\begin{array}{rccccccccccccccccc}
& & & 1 & 7 & 5 & 4 & \\[12pt]
& 19) & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & \ldots \\
& & 1 & 9 \\[12pt]
\text{remainder} = 14 \rightarrow & & 1 & 4 & 3 \\
& & 1 & 3 & 3 \\[12pt]
\text{remainder} = 10 \rightarrow &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
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Find all solutions in positive integers to $a^b -b^a = 3$
Find all solutions in positive integers to $a^b -b^a = 3$
It appears that the only solution is $a=4, b=1$.
Modular arithmetic is not of much help (apart from deriving that $a,b$ must have opposite parity).
We know that $a^x$ dominates the polynomial $x^a$ bu... | Consider $g(t) = a^t - t^a .$
Then $g'(t) = (\ln a) a^t - at^{a-1}$.
Note that
$$ \left(\frac{t}{\ln (t)} \right )' = \frac{(-1 + \ln(t))}{\ln^2(t)}$$
and
$$\left( \frac{t-1}{\ln(t)}\right)' = \frac{t(\ln (t) -1)+1}{t \ln^2(t)} $$
Thus both $t /\ln t$ and $(t-1)/\ln t$ are strictly increasing for $t > e$.
Therefore, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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if such $a^2+b^2+c^2+1=d+\sqrt{a+b+c-d}$ show $2(a+b+c)=3$ Let $a,b,c,d\in R$ and such
$$a^2+b^2+c^2+1=d+\sqrt{a+b+c-d}$$
show that $$2(a+b+c)=3$$
It seem can use inequality to solve it?
| Let $a+b+c-d=x^2$, where $x\geq0$.
Hence, $$a^2+b^2+c^2+1=a+b+c-x^2+x$$ or
$$x^2-x+a^2+b^2+c^2-a-b-c+1=0$$ or
$$\left(x-\frac{1}{2}\right)^2+\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2=0,$$
which gives $a=b=c=\frac{1}{2}$ and $2(a+b+c)=3$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that all odd derivate of $\tan(x)$ at $x=0$ is at least $1$. This was the original excercise:
Prove that $$\frac{\sin{x}+\tan{x}}{2} \geq x$$ where $x \in (0,\frac{\pi}{2})$
This is how I did it:
Knowing that
\begin{align}\sin{0}&=0\\
\cos{0}&=1\\
\frac{\mathrm{d} \sin{x}}{\mathrm{d} x}&=\cos{x}\\
\frac{\mathr... | Nice exercise. We have that $f(x)=\tan(x)$ fulfills the differential equation
$$ f'(x) = 1 + f(x)^2 \tag{1}$$
with $f(0)=0$ and $f'(0)=1$. The even derivatives at the origin are obviously zero, since $\tan(x)$ is an odd function, and the odd derivatives can be computed by applying $\frac{d^{2k}}{dx^{2k}}$ to both sides... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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When, where and **how often** do you find polynomials of higher degrees than two in mathematical, pure/applied, research? A formula for solving a polynomial of degree three, see this link; $ax^3+bx^2+cx+d=0$, is
$$\begin{align}
x\quad&=\quad
\sqrt[3]{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{... | Gaussian quadrature is a class of methods of numerical integration based on polynomial approximation. Typically, polynomials of degree $5$ or $10$ are employed. For many of the functions that arise in applied mathematics, these give very accurate results. Polynomials of arbitrarily high degree can in principle be used,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2197171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 13,
"answer_id": 11
} |
find the value of x at second derivative The quesion is :
\begin{align}
\ y= \sqrt{x} ~\text{ find the value of x for which}~ y+ \frac{d^2y}{dx^2} = 0
\end{align}
\begin{align}
\
\end{align}
The answer given is : x= 1/2
I am not sure how they got this answer so any help would be appreciated.
| $y=\sqrt { x } $ so plug this in the equation $$\sqrt { x } +\frac { d }{ dx } \left( \frac { d\left( \sqrt { x } \right) }{ dx } \right) =0\\$$ now here $\frac { d }{ dx } \left( \frac { d\left( \sqrt { x } \right) }{ dx } \right) $ means you should find derivative twice of $y=\sqrt { x } $
$$ \sqrt { x } +\fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
$$\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{\cos^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{1}{1+... | Replacing $x$ by $\pi/2-x$, the integral becomes
$$ I = \int_0^{\pi/2} \frac{\cos^3{x}}{\cos^3{x}+\sin^3{x}} \, dx. $$
Adding together with the original form of $I$,
$$ 2I = \int_0^{\pi/2} \frac{\sin^3{x}+\cos^3{x}}{\cos^3{x}+\sin^3{x}} \, dx = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2}, $$
so $I = \pi/4$.
This trick works... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Multi-scale approximation from periodic standard form I solved the first order two-scale approximation of $\ddot y + y + \epsilon y^3=0$ ($y(0)=\alpha , \dot y(0)=0$) to be $\alpha \cos(t(1+\frac{3\alpha^2}{8}\epsilon))$. Then I put that equation into periodic standard form using $u=y$, $v=\dot y$, $u=a \cos t + b \sin... | To be a bit more precise: you're using a particular form of multi-scale approximation, namely the Poincaré-Lindstedt method. The leading order term of the expansion is, as you calculated, equal to
\begin{equation}
\alpha\,\cos\left( \left(1+\epsilon \frac{3 \alpha^2}{8}\right)t\right).
\end{equation}
The second method... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inverse of matrix of ones + nI Having a vector $\mathbf{1} \in \mathbb{R}^{n}$ containing only ones, following equality should be true according to a paper I am currently reading:
\begin{equation}
\left( nI+\mathbf{1}\mathbf{1}^T \right)^{-1}= \frac{1}{n}\left( I - \frac{1}{2n} \mathbf{1}\mathbf{1}^T \right)
\end{equ... | Example
Example for $n=5$:
$$
\mathbf{A}(n) =
\left(
\begin{array}{ccccc}
6 & 1 & 1 & 1 & 1 \\
1 & 6 & 1 & 1 & 1 \\
1 & 1 & 6 & 1 & 1 \\
1 & 1 & 1 & 6 & 1 \\
1 & 1 & 1 & 1 & 6 \\
\end{array}
\right), \qquad
\mathbf{A}^{-1}(n) =
\frac{1}{50}
\left(
\begin{array}{rrrrr}
9 & -1 & -1 & -1 & -1 \\
-1 & 9 & -1 & -1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$
Let $n,x,y,z$ be positive integers such that $$(x-y)^2+(y-z)^2+(z-x)^2 = (2n+1)xyz.$$ Prove that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$.
The original equation doesn't have $x^3,y^3,z^3$, so how can we show that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$? Expanding the ... | We're given that
$$(x-y)^2+(y-z)^2+(z-x)^2=(2n+1)xyz$$
Hence
$$\frac{(x-y)^2+(y-z)^2+(z-x)^2}{xyz}=(2n+1)$$
Substitute this into the expression given:
$$
\begin{aligned}
(2n+1)(x+y+z)+6
=
\frac{(x-y)^2+(y-z)^2+(z-x)^2}{xyz}(x+y+z)+\frac{6xyz}{xyz}
\end{aligned}
$$
After expanding and simplifying, you will get to
$$\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $c$ is equal to $\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\binom{99}{6}+\cdots+\binom{99}{96}-\binom{99}{98},$ then find $\log_2{(-c)}$. If $c$ is equal to $$\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\binom{99}{6}+\cdots+\binom{99}{96}-\binom{99}{98},$$then find $\log_2{(-c)}$.
I think the binomial theorem might h... | $$\sum_{99 \geq k \geq 0, \text{even}} i^k {99 \choose k}$$
$$=\sum_{99 \geq k \geq 0} \frac{(-1)^k+1^k}{2}i ^k{99 \choose k}$$
$$=\frac{1}{2} \sum_{k=0}^{99} i^k {99 \choose k}+\frac{1}{2} \sum_{k=0}^{99} (-i)^k {99 \choose k}$$
$$=\frac{1}{2}(1+i)^{99}+\frac{1}{2}(1-i)^{99}$$
$$=\frac{1}{2}(\sqrt{2})^{99}e^{99 \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How should i approximate this? Find the taylor series for the given function about the given point.
$$f(x,y)= e^{x^2+y^2}$$
My solution I thought was $e^{x^2}\cdot e^{y^2}$.
How should I make a summation about this?.
| $e^{x^2} = 1 + x^2 + \frac 12 x^4 + \frac 16 x^6\cdots\\
(e^{x^2})(e^{y^2} = (1 + x^2 + \frac 12 x^4 + \frac 16 x^6\cdots)(1 + y^2 + \frac 12 y^4 + \frac 16 y^6\cdots)\\
(e^{x^2})(e^{y^2} = 1 + x^2 + y^2 + \frac 12 (x^4 + y^4 + 2x^2y^2)+\frac 16 (x^6 + y^6 + 3x^4y^2 + 3x^2y^4) \cdots$
I am just multiplying the terms fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove this $1+2abcxyz \geq a^2x^2 + b^2y^2+c^2z^2$ Let $a,b,c,x,y,z\in\mathbb{R}$ such that \begin{align}\{a,b,c,x,y,z\}&\subset[-1,1]\\
1 + 2abc &\geq a^2+b^2+c^2\\
1+2xyz&\geq x^2+y^2+z^2\end{align}
Prove that:
$$1+2abcxyz \geq a^2x^2 + b^2y^2+c^2z^2$$
| We'll rewrite our conditions in the following forms.
$(1-b^2)(1-c^2)=(a-bc)^2$ and $(1-y^2)(1-z^2)\geq(x-yz)^2$.
Let $b^2=\frac{1}{1+v^2}$, $c^2=\frac{1}{1+w^2}$, $y^2=\frac{1}{1+q^2}$, $z^2=\frac{1}{1+r^2}$, $a-bc=ubc$ and $x-yz=pyz$,
where $v$, $w$, $q$ and $r$ are non-negatives.
Hence, our conditions give $vw\geq|u|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is $\displaystyle\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}$ equal to $-\dfrac 12$ or just $-\dfrac{\sin y}{2y}$ Question:
$$\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}=?$$
Here is my try:
\begin{align}\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}&=\lim_{x\rightarrow y}\dfrac{-2 \sin (\frac 12(x+... | You could also have used Taylor series around $x=y$. This is $$\cos(x)=\cos (y)-(x-y) \sin (y)-\frac{1}{2} (x-y)^2 \cos (y)+O\left((x-y)^3\right)$$ Then $$\cos(x)-\cos (y)=-(x-y) \sin (y)-\frac{1}{2} (x-y)^2 \cos (y)+O\left((x-y)^3\right)$$ $$\frac{\cos(x)-\cos (y) }{x^2-y^2}=-\frac{\sin (y)}{2 y}+(x-y) \left(\frac{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove: Im$\left(\frac{1+re^{i\theta}}{1-e^{i\theta}}\right) = \frac{2r\sin\theta}{1-2r\cos\theta + r^2}$ Prove the identity: $\operatorname{Im}\left(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\right) = \dfrac{2r\sin\theta}{1-2r\cos\theta + r^2}$
My only lead is to assume $r$ and $\theta$ are real.
Then,
$\operatorname{Im}\l... | RTP: $Im\bigg(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\bigg) = \dfrac{2r\sin\theta}{1-2r\cos\theta + r^2}$=$\frac{2r\sin\theta}{(r-\cos\theta)^2+1-\cos^2\theta}$
$$Im\bigg(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\bigg) = Im\bigg(\frac{(1+r\cos \theta+ri\sin\theta )(1-\cos\theta +i\sin\theta)}{(1-\cos\theta-i\sin\theta)(1-\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\frac{(\sqrt{x^2+12}-4)}{2-\sqrt{x^3-4}}$ as $x$ goes to 2
$$\lim_{x\to 2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}$$
I know it must be very trivial, but yet I'm stuck at this problem and other similar problems for quite a long time. I'd be glad if someone showed me a full solution without using L'Hospital and... | You need to get used to the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ which solves these algebraic limit problems instantly.
Thus we have
\begin{align}
L &= \lim_{x \to 2}\frac{\sqrt{x^{2} + 12} - 4}{2 - \sqrt{x^{3} - 4}}\notag\\
&= \lim_{x \to 2}\frac{\sqrt{x^{2} + 12} - \sqrt{16}}{(x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How many different Strings can I create with the characters of the String *SPEISESTAERKE*? How many different Strings can I create with the characters of the String SPEISESTAERKE ?
According to that post, I have tried the following. Note, that I use all the given characters at once to create all possible combinations: ... | alternative answer - with correctly counted letters:
$\binom {13}{3} \cdot \binom {10}{4} \cdot \binom {6}{1} \cdot \binom {5}{1} \cdot \binom {4}{1} \cdot \binom {3}{1} \cdot \binom {2}{1} \cdot \binom {1}{1} = 432 432 00$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that the sum of numbers written on the board is always equal to $\binom{20}{2}$ We write the number $20$ as the sum of two numbers $a,b$ and write $ab$ on the board where $a,b \ge 1$. We do this again for $a,b$ until we get only (multiple instances of) the number $1$. Prove that the sum of numbers written on the ... | This process can be described as follows:
$F(n) = a \cdot (n-a) + F(a) + F(n-a)$ where $F(0) = F(1) = 0$.
(In other words, given some $n$, we're choosing a value for $a$ and letting $b=n-a$, so that $a+b=n$)
Try with $a=1$:
$F(n) = 1 \cdot (n - 1) + F(1) + F(n-1) = n - 1 + F(n-1) = \sum_{k=1}^{n} (k-1) = \frac{n(n-1)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Limit question: $\lim_{x \rightarrow 2} \frac{x+1}{x+2} = \frac{3}{4}$
Show that $$\lim_{x \rightarrow 2} \frac{x+1}{x+2} = \frac{3}{4}$$ by the definition.
This is what I've done so far:
$$\left\vert\frac{x+1}{x+2} - \frac{3}{4} \right| = \left| \frac{x-2}{4(x+2)} \right|$$
and I see that
$$|x-a| = |x-2|$$ which is... | Note that
$$
|\frac{x-2}{4(x+2)}| = \frac{|x-2|}{4|x+2|}
$$
for all $x \neq -2$.
We have to "bound away" the annoying term $|x+2|$ by a preliminary upper bound for $|x-2|$. If $|x-2| < 1$, then $-1 < x-2 < 1$, so $1 < x < 3$, and hence $3 < x+2 < 5$; hence $3 < |x+2| < 5$. Then
$$
\frac{|x-2|}{4|x+2|} < \frac{|x-2|}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Question on calculating $\int_{\partial B_{3/2}(1)} \frac{z^7 + 1}{z^2(z^4 + 1)}dz$ My task is to calculate $$\int_{\partial B_{3/2}(1)} \frac{z^7 + 1}{z^2(z^4 + 1)}dz$$ using the Cauchy integral formula. My question is: Is there a simple trick or do I have to perform a partial fraction decomposition?
| A partial fraction decomposition is not necessary.
The function
\begin{align*}
f(z)=\frac{z^7 + 1}{z^2(z^4 + 1)}
\end{align*}
has a double pole at $z=0$ and simple poles at
$z\in\left\{e^{\frac{\pi i}{4}},e^{-\frac{\pi i}{4}},e^{\frac{3\pi i}{4}},e^{-\frac{3\pi i}{4}}\right\}$. The contour of integration $\partial B_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find shortest distance between lines in 3D Find shortest distance between lines given by
$$\frac{x-2}{3}=\frac{y-6}{4}=\frac{z+9}{-4}$$
and
$$\frac{x+1}{2}=\frac{y+2}{-6}=\frac{z-3}{1}$$
Is there any shortcut method for this problems?
| Hint:
write the equations of the two lines in the form $\vec x=\vec p+t\vec q$:
$$
r_1) \qquad \begin{pmatrix}
x\\y\\z
\end{pmatrix}=\begin{pmatrix}
2\\6\\-9
\end{pmatrix}+t\begin{pmatrix}
3\\4\\-4
\end{pmatrix}
$$
$$
r_2) \qquad \begin{pmatrix}
x\\y\\z
\end{pmatrix}=\begin{pmatrix}
-1\\-2\\3
\end{pmatrix}+t\begin{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 2
} |
Solve the area on quadrant one given by the two curves $x^2+y^2=9$ and $ 4x^2+3y^2=36$
I'm having difficulties solving this, but I was told the answer is $\frac{3}{4}\pi$
However, when I equate the two and get $x = 3$ I try to solve them but can't get the correct answer after integration. I was thinking if these are li... | The required region (shaded portion) is given below:
The upper curve is given by $y=\sqrt{\frac{36-4x^2}{3}}$ and the lower curve is $y=\sqrt{9-x^2}$. Thus, the area is given by
$$A=\int_{0}^3\sqrt{\frac{36-4x^2}{3}}dx-\frac{\pi 3^2}{4}$$
Using Wolfram Alpha,
$$\int_{0}^3\sqrt{\frac{36-4x^2}{3}}dx=\frac{3\sqrt{3}\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$1^2$+$2^2$-$2×3^2$+$4^2$+$5^2$-$2×6^2$+$...$+$9997^2$+$9998^2$-$2×9999^2$ ∵ For any positive integer $n$
$n^2-2(n+1)^2+(n+2)^2$
= $n^2-2n^2-4n-2+n^2+4n+4=2$
∵ $1^2$+$2^2$-$2×3^2$+$4^2$+$5^2$-$2×6^2$+$...$+$9997^2$+$9998^2$-$2×9999^2$
= $1^2+(2^2-2×3^2+4^2)+(5^2-2×6^2+7^2)+...+(9995^2-2×9996^2+9997^2)+9998^2-2×9999^2$
... | Filling a gap for the benefit of other readers:
$$\begin{align}
(n-1)^2 -2n^2 + (n+1)^2 &= (n^2-2n+1) -2n^2+(n^2+2n+1) \\
&=2n^2-2n^2+2n-2n+2 \\
&=2
\end{align}$$
or:
$$\begin{align}
(n-1)^2 -2n^2 + (n+1)^2 &= ((n+1)^2-n^2) - (n^2-(n-1)^2) \\
&=(n+1+n)(n+1-n) - (n+n-1)(n-(n-1)) \\
&=(2n+1) - (2n-1) \\
&=2
\end{align}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Geometric transformation (symmetric point to line) I'm trying my own exercise:
Let $(ε):y=3x-4$ be a line in $\Bbb R^2$ given in cartesian coordinates. Find the geometric transformation that sends every point in $\Bbb R^2$ to it's symmetrical as to the line $(ε)$.
My attempt:
Let $$\begin{array}\\f_1: \Bbb R^2 \righ... | Let us look for a solution of the form:
$$\tag{1}\begin{pmatrix}x'\\ y'\end{pmatrix}=\begin{pmatrix}a & \ \ b \\b & -a\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}+\begin{pmatrix}e\\ f\end{pmatrix}$$
(see Edit 2 below)
(a symmetry matrix is orthogonal with determinant -1)
It suffices to write that points of intersecti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Setting up inequalities is proving difficult Given $$(a+b)^2\leq4$$
$$(a-b)^2\leq4$$
Is this true?
$$2(a+b)^2+2(a-b)^2-b^2\leq16$$
The problem I need to solve is to find the maximum value of $(4a^2+3b^2)$ Please help.
| $(a+b)^2 \le 4$ so $2(a+b)^2 \le 8$.
$(a-b)^2 \le 4$ so $2(a-b)^2 \le 8$.
So $2(a+b)^2 + 2(a+b)^2 \le 8+8 = 16$.
And $b^2 \ge 0$ so
$2(a+b)^2 + 2(a+b)^2-b^2 \le 2(a+b)^2 + 2(a+b)^2 \le 16$.
$16 \ge 2(a+b)^2 + 2(a+b)^2-b^2 = 2(a^2 + 2ab +b^2) + 2(a^2 -2ab + b^2) -b^2 = 4a^2 + 3b^2$. So $16$ is a potential maximum. But... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
sum of two determinants If
\begin{vmatrix}a&c\\b&d\end{vmatrix} and \begin{vmatrix}a&e\\b&f\end{vmatrix}, then sum of these determinant can be written as in terms of another determinant given by
\begin{vmatrix}a&c+e\\b&d+f\end{vmatrix}
is it right?
| The determinant is a multilinear function of its columns and so
$$
\begin{vmatrix}a&c+e\\b&d+f\end{vmatrix}
=\begin{vmatrix}a&c\\b&d\end{vmatrix} + \begin{vmatrix}a&e\\b&f\end{vmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to know if equation adds more info to a set of equations? I'm looking for a formalized way to be able to know whether adding an equation to a set of linear equations adds new information or not, does anyone know of a way?
For a really simple case, let's say I have these equations:
$A = 3\\
B+C = 4\\
D = 8
$
Obvious... | Given a set of equations,
$$
\begin{cases}
A+B = 5\\
2D = 3\\
A + B + C + 3D = 10,
\end{cases}
$$
You can identify it with a matrix like this,
$$
\begin{bmatrix}
1 & 1 & 0 & 0 & 5 \\
0 & 0 & 0 & 2 & 3 \\
1 & 1 & 1 & 3 & 10 \\
\end{bmatrix}
$$
Notice that each column of the matrix corresponds to each v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
$c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$ so prove $e - c_n \le \frac{1}{n! * n}$ $c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$, so $e - c_n \le \frac{1}{n! * n}$
I absolutely have no idea how to solve it, could anyone tell me the approach?
| Hint. Using the infinite summation representation of $e$, we have
$$ e - c_n = \sum_{r=0}^\infty \frac1{r!} - \sum_{r=0}^n \frac1{r!} = \sum_{r=n+1}^\infty \frac1{r!} = \frac1{n!}\left(\frac1{n+1} + \frac1{(n+1)(n+2)} + \cdots \right). $$
All that's left to do now is to show that the sum in brackets $()$ is $\le 1/n$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Area of a right triangle if $a/b$ is $1.05$ and the difference between the radii of the circumscribed and inscribed circles is $17$? What's the area of a right triangle if the quotient of its legs is $1.05$ and the difference between the radii of the inscribed and circumscribed circles is $17$?
I've been trying to solv... | The inradius of a right triangle in terms of its sides is given by the relationship $$r \frac{a+b+c}{2} = rs =|\triangle ABC| = \frac{ab}{2},$$ hence $$r = \frac{ab}{a+b+c}.$$ The circumradius is trivially $$R = \frac{c}{2}.$$ Thus the given conditions may be summarized as $$17 = R - r = \frac{c}{2} - \frac{ab}{a+b+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Calculate $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{2n+1}\right)$. Calculate
$$\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{2n+1}\right)$$
Question: I want to verify that my next attempt is correct, I do it too exhausted and in that state I do not trust my abilities.
My attempt... | For all $n\in\Bbb{Z}^+ $,
$$
\sum_1^n \cos\frac{2\pi k}{2n+1} = -\frac12$$
Proof:
$$
\sum_1^{2n+1} \cos\frac{2\pi k}{2n+1} = 0$$
because it is the real part of
$$
S=\sum_1^{2n+1} e^{i\frac{2\pi k}{2n+1}}$$
and $S$ can be multiplied by any $e^{i\frac{2\pi r}{2n+1}}$ for integer $r$ with the effect of just rotating the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$
Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$
My Approach:
$\int (x^2+a^2)^{-3/2} \cdot dx$
$\int (x^2+a^2)^{-3/2} \cdot d(a^2+x^2)\cdot \frac{dx}{d(x^2+a^2)}$
But this doesn't give the right answer. I showed this to my friend and he said $d(x^2+a^2)$ is not possib... | We use a trick to get a reduction formula:
$$ \frac{1}{(x^2+a^2)^{3/2}} = \frac{1}{a^2} \frac{x^2+a^2-x^2}{(x^2+a^2)^{3/2}} = \frac{1}{a^2} \frac{1}{\sqrt{x^2+a^2}} - \frac{1}{a^2}\frac{x^2}{(x^2+a^2)^{3/2}}. $$
The first term is lower-order, the second can be integrated by parts:
$$ \int \frac{x^2}{(x^2+a^2)^{3/2}} \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Values for $a$ and $b$ in $y=\cos(x)+a\cos(bx)$ such that every real value for $x$ has either a positive or $0$ value for $y$ If there is a function in the form $y=\cos(x)+a\cos(bx)$ do there exists real number values for $a$ and $b$ such that for every real number value for $x$ there is either a positive number value ... | Let $f(x) = \cos(x)+a\cos(bx)$, and suppose $f(x) \ge 0$ for all $x \in \mathbb{R}$.
First suppose $b=0$. Then $f(x) = \cos(x) + a,$ which has range $[a-1,\infty).$
Thus, for $b=0$, we get the solution pairs $(a,b) = (t,0),\,$ for any $t \ge 1.$
Next, suppose $b \ne 0$.
If $|a| > 1$, then when $a\cos(bx)$ realizes ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2223465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
The value of $\prod_2^\infty \left(1-2/(n(n+1))\right)$ I'm trying to find the value of $\prod_2^\infty \left(1-2/(n(n+1))\right)$. So far I have the following.
\begin{align*}
\prod_2^\infty \left(1-2/(n(n+1))\right) &= (1-2/6)*(1-2/12)*(1-2/20)*(1-2/30)... \\&= (1-1/3)*(1-1/6)*(1-1/10)*(1-1/15)... \\&= (2/3)*(5/6)*(9/... | $$\prod _{ 2 }^{ \infty } \left( 1-\frac { 2 }{ n\left( n+1 \right) } \right) =\prod _{ 2 }^{ \infty } \left( \frac { { n }^{ 2 }+n-2 }{ n\left( n+1 \right) } \right) =\prod _{ 2 }^{ \infty } \left( \frac { \left( n+2 \right) \left( n-1 \right) }{ n\left( n+1 \right) } \right) =\\ =\prod _{ 2 }^{ \infty } \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to solve $a_n = 2a_{n-1} + 1, a_0 = 0, a_1 = 1$? $a_n = 2a_{n-1} + 1, a_0 = 0, a_1 = 1$
So to get the closed form of this recurrence relation, I would usually try to get it into the form of $a_n = ra_{n-1}$ and then $a_n = r^na_0$. But what am I supposed to do with the $1$?
Thanks!
| $a_2=2\cdot 1+1=3=2^2-1,\quad a_3=2\cdot 3+1=7=2^3-1,\quad a_4=2\cdot7+1=15=2^4-1,\quad a_5=2\cdot 15+1=31=2^5-1$
So in general we guess that $a_{n}=2^n-1,\quad n\in\mathbb{N}$
So we know it's true for $n=0$
Assume true for $n$, now we must prove true for $n+1$:
$a_{n+1} = 2\cdot a_n+1=2\cdot(2^n-1)+1=2^{n+1}+1$ as req... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the equation of the tangent line to the circle $3x^2+3y^2+3z^2-2x-3y-4z=22;3x+4y+5z=26$ at the point $(1,2,3)$. I try take the centre of the sphere, then equation of the perpendicular line joining the centre of circle and the centre of sphere, then find the centre of circle.Tell me if I am going right way?
| Let $(x,y,z)=(1,2,3)+t(a,b,c)$ is an equation of out tangent line.
Thus, $3a+4b+5c=0$.
Rewrite an equation of our sphere in the following form:
$$\left(x-\frac{1}{3}\right)^2+\left(y-\frac{1}{2}\right)^2+\left(z-\frac{2}{3}\right)^2=\frac{293}{36}$$
Thus, $\left(1-\frac{1}{3}\right)a+\left(2-\frac{1}{2}\right)b+\left(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Generators of $GL_2(\mathbb{Q}_p)$ A well known fact is that the group $GL_2(\mathbb{Q}_p)$ is generated by the following matrices:
$1) \text{ }
w=
\left( {\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array} } \right)
$
$2) \text{ } \mathbb{Q}_p^{\star}
\left( {\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\e... | We first note that multiplication on the left by $(1)$ switches the rows of a matrix and multiplication on the right by $(1)$ switches the columns call these observations $(*)$. Using both parts of $(*)$ we see conjugation by $(1)$ transforms $\begin{pmatrix} 1 & b \\ 0 & 1\end{pmatrix}$ to $\begin{pmatrix} 1 & 0 \\ b ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Limit of $S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$ when $n\to\infty$ $$S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$$
the task is to find $$\lim_{n \to \infty} S_n$$
I've tried different ways, but all I could do is to make an estimation that the limit is somewhere betwee... | The sum can be written as
$$S(n) = \sum_{k=n}^{2n} \frac{1}{2k-1}$$
If we replace the sum by an integral we find
$$I(n) = \int_n^{2 n} \frac{1}{2 k-1} \, dk = \frac{1}{2} \log (2+\frac{1}{2n-1})$$
And the limit $n\to\infty$ is
$$S(n\to\infty) = \frac{1}{2} \log (2)\tag{1}$$
The justification of the replacement of the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Area of equilateral triangle inscribed in right triangle An Isosceles right triangle $ABC$ , $AB=BC= 4 cm$
Point $p$ is a midpoint of $BC$ , points $q$, $s$ lies on $AC$,$AB$ respectively , such that the triangle $pqs$ is an equilateral triangle ;
what is the area of triangle $pqs$
| Based on your description, let's define:
$$ \begin{align}
\mbox{Hypotenuse AC}:\; y &= -x +4 \\
p &= (2,0) \\
q &= (x, y=-x+4) \\
s & = (0,h)
\end{align}$$
The three sides are equal, hence:
$$ \begin{align}
\mbox{sp = pq} \Rightarrow
4+h^2 &= (x-2)^2 + (x-4)^2 \quad (1)\\
\mbox{sp = sq} \Rightarrow
4+h^2 &= x^2 + (h+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find tan(C/2) in The triangle ABC . $ABC$ is a triangle.
$\tan\frac{A}{2} = 0.5$
$\tan\frac{B}{2} =\frac{1}{3}$.
Find $\tan\frac{C}{2}$.
I tried to find it :
when $A+B+C = 180^{\circ}$
So $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^{\circ}$.
How can I complete on these information?
| ∴ $\dfrac C2 = 90^0 –(\dfrac A2 + \dfrac B2)$
Then, $\tan (\dfrac {C}{2}) = \cot (\dfrac A2 + \dfrac B2) = \dfrac {1}{\tan (\dfrac A2 + \dfrac B2)} = \dfrac {1 -\tan (\dfrac A2) \tan (\dfrac B2)}{\tan (\dfrac A2) + \tan (\dfrac B2)} = ...$
Added:- It is the application of the compound angle formula $\tan (x + y) = \dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit proof by definition I have the following limit:
$$\lim_{x \to 6} \frac{x+1}{x-5} = 7$$
How do I prove this equation by the famous definition of the limit? (the one includes delta and epsilon).
I know how to prove simple limits like this one: $$\\\lim_{x \to 6} x-5 = 1
\\ |x - x_0| < \delta \implies |x - 6| < \de... | We have $$\bigg|\frac{x+1}{x-5}-7\bigg|=\bigg|\frac{x+1-7x+35}{x-5}\bigg|=\bigg|\frac{-6x+36}{x-5}\bigg|=6\cdot\bigg|\frac{x-6}{x-5}\bigg|=6\cdot\bigg|\frac{1}{x-5}\bigg|\cdot|x-6|$$
Assume that $|x-6|<\frac{1}{2}$. Then we get
$$-\frac{1}{2}<x-6<\frac{1}{2}$$ and so we get
$$-\frac{1}{2}+1<x-6+1<\frac{1}{2}+1$$ that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
A proof that $\frac{\log^k(1)+\log^k(2)+\dotsb +\log^k(n)}{1^k+2^k+\dotsb +n^k} \to 0$ Let $\left(a_n\right)$ be the following sequence:
$$a_n =\frac{\log^k\left(1\right)+\log^k\left(2\right)+\dotsb +\log^k\left(n\right)}{1^k+2^k+\dotsb +n^k},$$ for a fixed $k \in \mathbb{N}$. Prove that $a_n \to 0$. There are many pr... | I have perhaps a more elegant proof:
$$\begin{align}a_n&=\frac{\log^k(1)+\dots+\log^k(n)}{1^k+\dots+n^k}\\&<\frac{\log^k(n)+\dots+\log^k(n)}{n^k}\\&=\frac{n\log^k(n)}{n^k}\\&=\frac{\log^k(n)}{n^{k-1}}\\&\to0\end{align}$$
The first step follows from the fact that $\frac ab<\frac cd$ if $c>a$ and $d<b$ for positive numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Given a $2\times 2$ matrix $A$, compute $A^7$. Let $A =
\begin{pmatrix}
e^{2x} & -1 \\
0 & e^{2x}-1 \\
\end{pmatrix}
$. Compute $A^7$.
I've tried the obvious way of multiplying $A$ with $A$, then $A^2$ with $A^2$, but I arrived at a messy result in the top right member of the matrix. ... | For the 7th power, it pays already off to diagonalize the matrix. It is easy to see that the eigenvalues are $e^{2x}$ and $e^{2x}-1$ with the eigenvectors $(1,0)^T$ and $(1,1)^T$. Thus, we can write $A$ as
$A = T D T^{-1}$
with $$T= \begin{pmatrix}1 &1\\
0 &1 \end{pmatrix}, \quad D= \begin{pmatrix}e^{2x} &0\\
0 &e^{2x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Can you explain why this series converges? $$1 - \frac{1}{3 \cdot 2!} + \frac{1}{5 \cdot 3!} - \frac{1}{7 \cdot 4!}+\cdots$$
I am new to math over flow, and I do not know how to format the math, sorry! Also, what should this converge to?
| $$1 - \frac{1}{3 \cdot 2!} + \frac{1}{5 \cdot 3!} - \frac{1}{7 \cdot 4!}+\cdots$$
A standard theorem says this converges if the corresponding series of absolute values converges:
$$1 + \frac{1}{3 \cdot 2!} + \frac{1}{5 \cdot 3!} + \frac{1}{7 \cdot 4!}+\cdots$$
The "comparison test" says this converges if the following ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that the sequence $\left\{\frac{4n^2}{2n^3-5}\right\}$ converges to $0$. Prove that the sequence $$\left\{\frac{4n^2}{2n^3-5}\right\}$$ converges to $0$.
The first thing I would like to do is further bound the sequence so I can get rid of the $-5$ in the bottom of the fraction. How can I do this? For negatives it... | For $n\ge 2$, we have for all $\epsilon>0$
$$\frac{4n^2}{2n^3-5}\le \frac{4n^2}{n^3}=\frac4n<\epsilon$$
whenever $n>N=\max\left(2,\frac{4}{\epsilon}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve $\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}=13$ for $x$
Solve $$\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}=13$$ for $x$
Any help is appreciated. I'm entering a challenge and can't reach the solution.
| First, we can rearrange the given equation as follows:
\begin{align}\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}&=13\\\\
\frac{(2^2)^{2x}+(2^2)^x+1}{2^{2x}+2^x+1}&=13\\\\
(2^2)^{2x}+(2^2)^x+1&=13\left(2^{2x}+2^x+1\right)\\
2^{4x}+2^{2x}+1&=13\left(2^{2x}+2^x+1\right)\\
(2^x)^4+(2^x)^2+1&=13\left((2^x)^2+2^x+1\right)\\
y^4+y^2+1&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Random walk with 3 cases. Let the walking start be at $x=0$. With probability $p_1$ new $x=x+1$, with probability $p_2$: $x=x-1$ and with probability $1-p_1-p_2 \geq 0$ walking ends.
The question is what is the probability of ending on point $n$.
I've computed it numerically and getting something close to normal distri... |
We derive a closed formula and show the following is valid for $n\geq 0$
\begin{align*}
\sum_{k=0}^\infty\binom{n+2k}{k}x^k&=
\frac{1}{\sqrt{1-4x}}
\left(\frac{1-\sqrt{1-4x}}{2x}\right)^n\tag{1}\\
\end{align*}
Note that
\begin{align*}
c(x)=\frac{1-\sqrt{1-4x}}{2x}=1+x+2x^2+5x^3+14x^4+\cdots
\end{align*}
is the gen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Test the series $\sum_{k=0}^{\infty}\frac{(-1)^k 1 \cdot3\cdot5 \cdots(2k+1)}{1\cdot4\cdot7\cdots(3k+1)}$ Use the ratio test for absolute convergence to determine
whether the series absolutely or
diverge.
$$\sum_{k=0}^{\infty}\frac{(-1)^k 1 \cdot3\cdot5 \cdots(2k+1)}{1\cdot4\cdot7\cdots(3k+1)}$$
I don't understand how ... | Let us write
$$a_k=\frac{ 1 \cdot3\cdot5 \cdots(2k+1)}{1\cdot4\cdot7\cdots(3k+1)}.$$
Then, we have
$$\begin{align}
\frac{a_{k+1}}{a_k}&=\frac{1\cdot3\cdot5\cdots [2(k+1)+1]}{1\cdot 4\cdot 7\cdots [3(k+1)+1]}\quad\cdot\quad\frac{1\cdot 4\cdot 7\cdots (3k+1)}{1\cdot3\cdot5\cdots(2k+1)}\\
&=\frac{1\cdot3\cdot5\cdots (2k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Ramanujan's Surd : Proof required. I came across a problem on the web, it is as follows:
Question:If $ x $, $ y $ and $ z $ are rational numbers such that
$ \sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}$ then find $ x,y,z $.
From Wikipedia|Nested Radicals|Some identities by Ramanujan, it is clear... | You want a proof of the identity $$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$$ which can be written as $$\sqrt[3]{x-1}=\frac{1-x+x^{2}}{\sqrt[3]{9}}$$ where $x=\sqrt[3]{2}$. Thus we let the RHS be denoted by $y$ and show that $y^{3}=x-1$. Since $x, y$ are real this proves... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solve : $x+y=2010$, $(\sqrt[3]{x}+\sqrt[3]{y})(\sqrt[3]{x}+11)(\sqrt[3]{y}+11)=2942$
Given$$x+y=2010,\\(\sqrt[3]{x}+\sqrt[3]{y})(\sqrt[3]{x}+11)(\sqrt[3]{y}+11)=2942,$$
find the value of $\sqrt[3]{x}+\sqrt[3]{y}$.
My attempts:
Let $u=\sqrt[3]{x} , v=\sqrt[3]{y}$.
$(u+v)(u^2-uv+v^2)=2010 \tag{1}$
$(u+v)(uv+11u+11v+1... | u and v look quite similar, so I will just use a and b instead:
$$3(a+b)(ab+11a+11b+121)+(a+b)(a^2-ab+b^2)=3\times2942+2010=10836$$
$$(a+b)(a^2+2ab+b^2+33a+33b+363)=10836$$
$$(a+b)((a+b)^2+33(a+b)+363)=10836$$
$$(a+b)^3+33(a+b)^2+363(a+b)-10836=0$$
Solving for $a+b$, the real root to this is $a+b=12$, so the answer is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , ... | You can solve it more easily by splitting the modulus and squaring both sides:
Assuming $x\neq2$,
$$|3x+1|<|x-2|\implies(3x+1)^2-(x-2)^2<0$$
$$\implies(3x+1+x-2)(3x+1-x+2)<0$$
So $$(4x-1)(2x+3)<0\implies-\frac 32<x<\frac 14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
Simple inequality $\frac{4x+1}{2x-3}>2$
$$\frac{4x+1}{2x-3}>2$$
I have started with looking at positive/negative situations and there are 2 or that both expressions are positive or both are negative.
If both a positive we can solve for $$\frac{4x+1}{2x-3}>2$$
$${4x+1}>2(2x-3)$$
$${4x+1}>4x-6$$
$${0x}>-7$$
Or both a... | HINT
Notice that
$$
\frac{4x+1}{2x-3}
= \frac{4x-6}{2x-3} + \frac{7}{2x-3}
= 2 + \frac{7}{2x-3}
$$
which is more than $2$ when the fraction is positive. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove expression is not prime $$
(n + 4)^4 + 4
$$
If n is natural number, how to prove that above expression is not prime?
I am stuck here
$$
(n+4)^2 \cdot (n+4)^2 + 2 . 2
$$
$$
\left(\left(n^2+4^2\right) \cdot 2\right)\left(\left(n^2+4^2\right) \cdot 2\right)
$$
| From the context of the question, I assume $n ≥ 0$.
Clearly for even $n$, $(n+4)^4+4$ is even. Hence suppose $n$ is odd, so $n = 2k+1$. Then $(n+4)^4+4 = (2k+5)^4+4 = 16 k^4 + 160 k^3 + 600 k^2 + 1000 k + 629 = (4 k^2 + 16 k + 17) (4 k^2 + 24 k + 37)$.
Clearly this is composite, as for $n≥0$, this is a product of two n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is $1+x+x^2+x^3...$? What is the difference between these two series?
$$
\begin{align}
1+x+x^2+x^3+...+x^n+\mathcal O(x^{n+1})&=\frac{1}{1-x}\\
\\
1+r+r^2+r^3+...+r^{n-1}&=\frac{r^n-1}{r-1}\\
\end{align}
$$
I can't wrap my head around it; they both start with $1+x+x^2+x^3...$? Of course the first is the maclaurin... | The two series actually do produce the same result. You just have to compare them fairly.
To begin with, you've listed $n$ terms in the first series
(not including the "big-O" notation) and only $n-1$ terms in the second series. Let's make both series have $n$ terms, as follows.
(I'm also putting the "closed expressio... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to deal with a Pell's equation type problem with two primes I have some understanding of Pell's equation $$x^2-dy^2=1$$ where $d$ is a prime. I know that you can take the continued fraction of $\sqrt{d}$ and use information about the period and convergents to find things out. Now I need to show that if $p,q$ are pr... | You have a primitive solution to $x^2-pqy^2=1$ ($x$, $y$ positive, $y$ minimal). Then $x$ is odd and $y$ is even (think modulo $4$). Also
$$x^2-1=(x+1)(x-1)=pqy^2$$
so that
$$\frac{x+1}2\frac{x-1}2=pq\left(\frac{y}{2}\right)^2.$$
As $(x\pm 1)/2$ are coprime integers then we have one of the following
*
*$(x+1)/2=u^2$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some ... | Hint:
Expand $(3x-1)^2+(3y-1)^2+(3z-1)^2\geqslant 0$ And simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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In a triangle $ABC$, if its angles are such that $A=2B=3C$ then find $\angle C$? In a triangle $ABC$, if its angles are such that $A=2B=3C$ then find $\angle C$?
$1$. $18°$
$2$. $54°$
$3$. $60°$
$4$. $30°$
My Attempt:
$$A=2B=3C$$
Let $\angle A=x$ then $\angle B=\dfrac {x}{2}$ and $\angle C=\dfrac {x}{3}$
Now,
$$x+\dfr... | Add:
$$180^\circ=A+B+C=2B+B+\frac23B=\frac{11}3B\implies B=\frac{540}{11},\ldots etc.$$
Or also
$$180^\circ=A+B+C=3C+\frac32C+C=\frac{11}2C\implies C=\frac{360}{11}$$
By the way, none of the options given in your question matches the above...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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There exist $3$ positive numbers $x,y,z$ such that $x\cdot y\cdot z=100$ but $x^2+y^2+z^2<65$. I have to show this using a form of calculus.
I know the answer is the cube root of $100$. I did this using intuition and upper and lower bounds but I have no idea how to show it from a calculus standpoint. Maybe showing the ... | Hint: by AM-GM $\,
x^2+y^2+z^2 \ge 3 \sqrt[3]{x^2 y^2 z^2} = 30 \sqrt[3]{10} \simeq 64.63\,$, with equality iff $x=y=z\,$.
[ EDIT ] Equivalent proof using Jensen's inequality for the concave function $\ln(x)$ on $\mathbb{R}^+\,$:
$$
\ln \left(\frac{1}{3}x^2+\frac{1}{3}y^2+\frac{1}{3}z^2\right) \ge \frac{1}{3}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Express the value of $s\left(m\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}$ in terms of $m$. The previous question was: Find the range such that the equation $|x^2 -3x +2|=mx $ has 4 distinct real solutions: $a,b,c,d$, and that turned out to be $0<m<3-2\sqrt{2}$. The book says that the solution is $\... | Hint: $a+b=m+3$ and $ab=2$ by Vieta's relations for $x^2-(m+3)x+2=0\,$, so:
$$\frac{1}{a^2}+\frac{1}{b^2}=\frac{a^2+b^2}{a^2b^2} = \frac{(a+b)^2-2\cdot ab}{(ab)^2}=\frac{(m+3)^2-2 \cdot 2}{2^2}=\frac{m^2+6m+5}{4}$$
Repeat the similar calculation for $\,\cfrac{1}{c^2}+\cfrac{1}{d^2}\,$, then add them up.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve recurrence relation: $a_n=3a_{n−1}+5a_{n−2}−7a_{n−3}$ I hope can you help me, I want to solve this recurrence relation:
$a_n=3a_{n−1}+5a_{n−2}−7a_{n−3}$ with initial values $a_1=4, a_2=8, a_3=2$
Using synthetic division I get the roots: $1, 1-2\sqrt{2}$ and $1-2\sqrt{2}$
What are the characteristic equation for t... | Recurrence $a_n=3a_{n−1}+5a_{n−2}−7a_{n−3}$ has characteristic polynomial $x^3 - 3x^2 - 5x + 7 = 0$ which has roots $x=1$, $x=1 - 2\sqrt{2}$, and $x=1 +2\sqrt{2}$.
That means your recurrence has form $a_n = \alpha \cdot 1^n + \beta \cdot (1-2\sqrt{2})^n + \gamma \cdot (1+2\sqrt{2})^n$.
Substitute in the values you kn... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a problem on limits
$$\lim_{x\to 1}\left(\frac{p}{1-x^p}-\frac{q}{1-x^q}\right) , p,q\in N$$
I tried fitting the formula $\lim_{x\to a}\frac{x^n - a^n}{x-a} = na^{n-1}$ but could not proceed further. Also I don't think that we should directly apply L'Hôpital's rule as it will be too lengthy.
| Using Taylor expansions around $0$, namely:
*
*$(1+u)^\alpha = 1+u+\frac{\alpha(\alpha-1)}{2}u^2 + o(u^2)$
*$\frac{1}{1+u} = 1-u+o(u)$ (a specific case of the first, for $\alpha=-1$)
You have, setting $h\stackrel{\rm def}{=} 1-x \xrightarrow[x\to1]{}0$ (to center things around $0$, with which I am much more familia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value of $\alpha^6+\beta^6$ is
If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value
of $\alpha^6+\beta^6$ is
I know that, here, $\alpha\beta=4$ and $\alpha + \beta = 2$ and use that result to find $\alpha^2 + \beta^2$ using the expansion... | Use this one:
$$[(a+b)^2-2ab)][((a+b)^2-2ab)^2-3a^2b^2]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Solve the equation in integers $a,b$: $20a^3-b^3=1$
Solve the equation in integers $a,b$: $$20a^3-b^3=1.$$
Assume that $a \neq 0$. Then simplifying and rearranging the equation gives $$20a^3 = 2^2 \cdot 5 \cdot a^3 = b^3+1 = (b+1)(b^2-b+1).$$ Note that neither factors can be divisible by $3$, since both of them must ... | With the equation $20a^3 - b^3 = 1$, if $b$ = $55860$, $a \approx 20578.999951$. In fact, the prime factorisation of $55860$ is $2^2\times3\times5\times7^2\times19$.
$\frac{55860}{19} = 2940$ and funnily enough, $\frac{20579}{7} \approx 2939.857136$ which is close to $2940$. The prime factorisation of $20579$ is $13\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Solve $2\cos^2{x}=\sqrt{3}\sin{2x}$.
Problem:
Solve $2\cos^2{x}=\sqrt{3}\sin{2x}$ and give the sum of all the solutions in the interval $0\leq x\leq2\pi.$
Attempt: Using the fact that $\sin{2\theta}= 2\cos{\theta}\sin{\theta}$ on the RHS I get $$2\cos^2{x}=2\sqrt{3}\cos{x}\sin{x}.$$
Dividing by $2\cos{x}$ I get $$\co... | The right way to solve is:
$$2\cos^2{x}=2\sqrt{3}\cos{x}\sin{x} \to \cos x(\cos x-\sqrt{3}\sin x)=0$$
what give you:
$$\cos x=0 \text{ or } \cos x-\sqrt{3}\sin x=0$$
For
$$\cos x=0\to x=\frac{\pi}{2}+k\pi$$
and for,
$$\cos x-\sqrt{3}\sin x=0\to \tan x=\frac{\sqrt{3}}{3}\to x=\frac{\pi}{6}+k\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Three real roots of equation is the length of three sides of a right triangle, find the parameter Three real roots of the equation $x^3-2p(p+1)x^2+(p^4+4p^3-1)x-3p^3=0$ is the length of three sides of a right triangle, what is the value of $p$?
| Hint: let $a,b,c$ be the roots, then by Vieta's relations:
$$\require{cancel}
\begin{align}
a^2+b^2+c^2 &= (a+b+c)^2 - 2(ab+bc+ca) \\
&= \left(2p(p+1)\right)^2 - 2(p^4+4p^3-1) \\
&= 4p^2(p^2+\bcancel{2p}+1) - 2p^4-\bcancel{8p^3}+2 \\
&= 2(p^4 + 2p^2+1) \\
&= 2(p^2+1)^2
\end{align}
$$
For the roots to be the sides o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the inequality $\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}$ Solve for $x$ real the inequality $$\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}.$$ Obviously $x\ge-2$. After that I tried to square the whole inequality, which led me to $x < - \frac { 18} { 4\sqrt { 6} - 5}$. Now, the answer... | First lets multiply the inequality by $-1$ so we can square. $$\sqrt{x+2}-\sqrt{x+3}<\sqrt2-\sqrt3\\\sqrt{x+3}-\sqrt{x+2}>\sqrt{3}-\sqrt{2}$$Now lets square$$2x+5-2\sqrt{(x+3)(x+2)}>5-2\sqrt{6}\\x+\sqrt6>\sqrt{(x+3)(x+2)}$$
Now since $x+\sqrt{6}\geq-2+\sqrt 6>0$ we can square the inequality again
$$x^2+2\sqrt{6}x+6>x^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To prove $\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$ if $a,b,c$ are non zero positive reals prove $$\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$$ I have used A.M G.M inequality as follows:
$$a^3+b^3+c^3 \ge 3abc$$ adding $abc$ both sides we get
$$a^3+b^3+abc \ge 4abc-c^3=c(4ab-c^2)$$ so
$$\frac{1}{a^3+b... | Notice that $$\sum_{cyc}\frac{abc}{a^3+b^3+abc} = \sum_{cyc}\frac{1}{\frac{a^2}{bc}+\frac{b^2}{ac}+1} \leq \sum_{cyc}\frac{1}{\frac{a+b}{c}+1} = \sum_{cyc}\frac{c}{a+b+c} = 1 $$ Then divide both sides by $abc$.
Note: In the inequality part we used Cauchy-Schwartz to get $$\left(\frac{a^2}{b}+\frac{b^2}{a}\right)(b+a)\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the nth element of a sequence Suppose we have sequence given by
$$ (1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,...) $$
I need to prove that $a_n = \lfloor{ \sqrt{2n} + \frac{1}{2} \rfloor} $.
Try:
I notice the following. The last occurrence of 1 occurs at $a_1$, the last occurrence of $2$ occurs at $a_3$, the last occurrenc... | In your sequence the first occurrence of $m\in\{1,2,\ldots\}$ appears at the position $1+2+\ldots+(m-1)+1 = \frac{m^2-m}{2}+1$ and the last one at the position $1+2+\ldots+m=\frac{m^2+m}{2}$. To prove the claim it is enough to show that the function $f:x\mapsto\left\lfloor\sqrt{2x}+\frac{1}{2}\right\rfloor$ maps every ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Matrix determinant as Dickson polynomial $\frac{x^{n+1}-y^{n+1}}{x-y}$
Given matrix $$
A=\begin{bmatrix}
x+y&xy&0& .&.&. &0\\
1&x+y&xy&0& .&.&0 \\
0&1&x+y&xy&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&0 \\
.&.&.&.&.&.&xy \\
0&.&.&.&0&1&x+y
\end{bmatrix}
$$
prove by induction that $$|A|=\frac{x^{n+1}-y^{n+1}}{x-y}$$ $x \neq... | A linear algebra approach to the recurrence relation is to write it in matrix form: $$\mathcal{D}_n\equiv \begin{pmatrix} D_{n} \\ D_{n-1}\end{pmatrix}=\begin{pmatrix} x+y & -xy\\1 & 0 \end{pmatrix}\begin{pmatrix} D_{n-1} \\ D_{n-2}\end{pmatrix}\equiv M \mathcal{D}_{n-1},\hspace{4mm} \mathcal{D}_0 = \binom{1}{0}.$$ (I ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2262488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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How to tell two ideals belong to the same ideal class group Potentially dumb question here: how can you tell if two ideals belong to the same ideal class group?
Let's say we're looking at $\textbf{Z}[\sqrt{10}]$. It has infinitely many ideals, but by the Minkowski bound we need only concern ourselves with ideals with a... | Let's show that the ideal $(3, 1+ \sqrt{10})\cdot (3, 1-\sqrt{10})^{-1}$ is principal. We have
$$\frac{(3, 1+ \sqrt{10})}{(3, 1-\sqrt{10})} = \frac{(3, 1+ \sqrt{10})\cdot(3, 1+ \sqrt{10})}{(3, 1-\sqrt{10})\cdot (3, 1+\sqrt{10})}$$
Now, for fractional ideals of number fields we have
$$(\alpha_1, \alpha_2)\cdot( \beta_1,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 3
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The sides $a$, $b$, $c$ of a triangle with area $D$ are such that $ab+bc+ca\ge 4\sqrt{3}D$
The sides $a$, $b$, $c$ of a triangle with area $D$ are such that $ab+bc+ca\ge 4\sqrt{3}D$.
I think I need to figure out a way to apply the A.M.-G.M. inequality here, but couldn't find a proper way of doing so.
| I realise this isn't answering your question, but it was interesting so I have decided to post it anyway. tl;dr I showed that $ab+ac+bc \geq 6D$, a weaker inequality, using AM-GM.
The area of triangle (in terms of its $3$ sides) is given by:
$D=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$. This is known as Heron's ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Inequality $\frac{b}{ab+b+1} + \frac{c}{bc+c+1} + \frac{a}{ac+a+1} \ge \frac{3m}{m^2+m+1}$ Let $m=(abc)^{\frac{1}{3}}$, where $a,b,c \in \mathbb{R^{+}}$. Then prove that
$\frac{b}{ab+b+1} + \frac{c}{bc+c+1} + \frac{a}{ac+a+1} \ge \frac{3m}{m^2+m+1}$
In this inequality I first applied Titu's lemma ; then Rhs will come 9... | By Holder:
$$\sum_{cyc}\frac{b}{ab+b+1}=1-\frac{(abc-1)^2}{\prod\limits_{cyc}(ab+b+1)}\geq1-\frac{(m^3-1)^2}{(\sqrt[3]{a^2b^2c^2}+\sqrt[3]{abc}+1)^3}=$$
$$=1-\frac{(m-1)^2(m^2+m+1)^2}{(m^2+m+1)^3}=1-\frac{(m-1)^2}{m^2+m+1}=\frac{3m}{m^2+m+1}.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Am I doing it wrong or this derivative is really lengthy? We have:
$$f(x)=\begin{vmatrix} (\alpha x+1)\cos^2 x & x & 1-x\\\beta\sin x& x^2 & 2x \\(\gamma x^2+1)\tan x & x & 1-x^2\end{vmatrix}$$
And we are supposed to calculate $$\lim_{x\rightarrow 0}\frac{1}{x^6}\int_{x^2}^{x^3}f(x).dx=A$$
and tell if $[A]$ (where $[x]... | Let $(u,v,w)$ be the entries in $1^{st}$ column of the determinant $f(x)$, we have
$$f(x)
= \begin{vmatrix} u & x & 1-x\\ v & x^2 & 2x \\ w & x & 1-x^2\end{vmatrix}
= x\begin{vmatrix} u & 1 & 1-x\\ v & x & 2x \\ w & 1 & 1-x^2\end{vmatrix}
= x\begin{vmatrix} u & 1 & -x\\ v & x & x \\ w & 1 & -x^2\end{vmatrix}
= x^2\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Direct/transverse tangents of two circles If equations of two circles in a plane are given as
$$ f(x,y)=0,\, g(x,y)=0,\, $$
*
*Find equations of direct tangents in terms of $f,g$
and
*Find equations of transverse tangents in terms of $f,g$
We could calculate using coincident roots/point etc. of a transverse l... | If $f(x,y)=0$ is the equation of a circle with center $(h,k)$ and radius $r,$ then the most general form of $f$ is
$$f(x,y) = a((x-h)^2 + (y-k)^2 - r^2)$$
where $a\neq0.$ Consider the following values of $f$:
\begin{align}
f(0,0) &= a(h^2 + k^2 + r^2), \\
f(1,0) &= a(h^2 - 2h + 1 + k^2 + r^2), \\
f(-1,0) &= a(h^2 + 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\frac{\tan \alpha + \sqrt{5}\sin \alpha}{2}(\sqrt{5}\cos \alpha -1)$ is equivalent to $\frac{5}{4}\sin(2\alpha)-\frac{1}{2}\tan \alpha$
Show that $\frac{\tan \alpha + \sqrt{5}\sin \alpha}{2}(\sqrt{5}\cos
\alpha -1)$ is equivalent to $\frac{5}{4}\sin(2\alpha)-\frac{1}{2}\tan
\alpha$.
I tried:
$$\frac{\fra... | Now simply $$\frac{5\sin\alpha \cos\alpha-\tan\alpha}{2}$$
$$\frac{5\times2.\sin\alpha \cos\alpha}{4}-\frac{\tan\alpha}{2}$$
$$\frac{5\times\sin 2\alpha}{4}-\frac{\tan\alpha}{2}\tag{sin2x=2 sinx cox}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Using generating functions in combinatorics I know these questions get asked a lot but I cannot figure it out. It requires the use of generating functions to find the number of solutions (coefficient) to the equation:
$u_i+u_2+u_3+u_4 = 20$, where $1 \leq u_i \leq 5, i = 1,...,4$
How do I solve this question step-by-st... |
The constraint $1\leq u_i \leq 5$ can be encoded using the finite geometric series formula as
\begin{align*}
z^1+z^2+z^3+z^4+z^5=\frac{z\left(1-z^5\right)}{1-z}\tag{1}
\end{align*}
Since (1) holds for each $u_i, 1\leq i\leq 4$ all possible configurations can be encoded as
\begin{align*}
\left(\frac{z\left(1-z^5\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$ For real numbers $a, b, c, d >0$ I have to prove that $$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$$ always holds. I tried using the mean inequalities for multiple numbers but I always get in trouble with the d... | Since
$$\frac{a^2+b^2+c^2+d^2}{4}\geq\frac{ab+ac+bc+ad+bd+cd}{6}$$
it's just
$$(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2\geq0,$$ it's enough to prove that
$$\sqrt{\frac{ab+ac+bc+ad+bd+cd}{6}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}},$$
which I proved here:
How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+b... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove the sqrt of 4 is irrational, where did I go wrong? So I know that $\sqrt{4}=2$, and therefore is rational, but, it seems like I proved it to be irrational? Could I have some help as to where I went wrong? Thank you!
Prove $\sqrt4$ is irrational.
Suppose per contradiction $\sqrt4$ is rational, that is, $\sqrt4 = \... | So $4|a^2$ does not imply that $4|a$. But your real error comes from
$$4b^2 = (4n)^2 4b^2 = 16n^2 b^2 = 4n^2$$ and you conclude that $4|b^2$. You should divide by $4$ to get
$$b^2=n^2$$ from which little can be concluded about divisibility by $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help solving the differential equation $y''+y'+y=0$ with initial conditions $y(0)=4,y'(0)=-3$ using Laplace transform. UPDATED WTH ANSWER:
Please help me solve the following differential equation using Laplace transform:
$$y''+y'+y=0;y(0)=4,y'(0)=-3$$
My answer so far:
$$\mathcal{L}\{y''\} + \mathcal{L}\{y'\} + \mathca... | $$\mathcal{L}(s)=\frac{4\left(s+\frac{1}{2}\right)-1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}$$
therefore
$$ f(x)=e^{-\frac{1}{2}x}\left(4\cos \left(\frac{\sqrt{3}}{2}x\right)-\frac{2\sqrt{3}}{3}\sin \left(\frac{\sqrt{3}}{2}x\right)\right)$$
In case there remains some mystery concerning how to know to replace the $4s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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The value of infinite product At a first look I observed that every term of this product is less than 1 so the limit will also be less than 1 but clueless about how to find limit .
$$P=\frac{7}{9}\cdot\frac{26}{28}\cdot\frac{63}{65}\cdots\frac{n^3-1}{n^3+1}\cdots.$$
| To restate your question:
$$
\prod_{n=2}^\infty \frac{n^3-1}{n^3+1}
$$
First we begin by expanding the terms:
$$
n^3-1 = (n-1)(n^2+n+1)
$$
$$
n^3+1 = (n+1)(n^2-n+1)
$$
By treating each term as a separate product,and shifting the index of the product, a telescope-like method can be used to simplify the terms into the de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Coefficient of $x^9$ Find the coefficient of $x^9$ in $(1+x)(1+x^2)(1+x^3)\cdots(1+x^{100})$.
Approach:-
By inspection, coefficient of $x^9$ will not be there in terms after $(1+x^{9})$
Now checking coefficient in $(1+x)(1+x^2)...(1+x^9)$.
Now by hit and trial, I checked what are the possible combinations of getting $x... | This question is equivalent to: How many sums of natural numbers equal $9$, subject to the condition that the terms be increasing? The terms in the sum are just exponents coming from all of the factors you mention.
For example, the sum $2+7=9$ corresponds to the term in your product where we take $x^2$, $x^7$, and $1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as... If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as
a) $\frac{1+t^2}{1-t^2}$
b) $\frac{2t}{1+t^2}$
c) $\frac{1-t^2}{1+t^2}$
d) $\frac{2t}{1-t^2}$
Attempt: I tried using the half angle formula but it just leaves me with an expression in terms of $... | Following your idea you can do:
$$\tan x=\frac{2\tan (x/2)}{1-\tan^2(x/2)}=\frac{2t}{1-t^2}$$
so,
$$\frac{\sin x}{\cos x}=\frac{2t}{1-t^2}\to \frac{\sin^2 x}{\cos^2 x}=\frac{4t^2}{(1-t^2)^2}$$
using proportion properties we get:
$$\frac{\cos^2 x}{\sin^2 x+\cos^2 x}=\frac{(1-t^2)^2}{4t^2+(1-t^2)^2}\to \cos^2x=\frac{(1-t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$x=(a+b\omega)^3$ with $\omega$ a primitive third root of unity. Then $x=(c+d\sqrt{-3})^3$. ($a,b,c,d \in \mathbb{Z}$) Let $x=(a+b\omega)^3$ with $\omega$ a primitive third root of unity. Then $x=(c+d\sqrt{-3})^3$. ($a,b,c,d \in \mathbb{Z}$). Can someone give me a hint?
Approach:
$x= a^3 + 3a^2 b\omega + 3ab^2\omega^2 ... | Hint:
$$ (a + b \omega)^3 = (a \omega + b \omega^2)^3 = (a \omega^2 + b)^3.$$
Now write $$\omega = - \frac 1 2 + \frac{\sqrt{-3}}2, \ \ \ \ \ \ \ \ \omega^2 = - \frac 1 2 - \frac{\sqrt{-3}}2,$$
and separate into cases depending on whether $a$ and $b$ are odd or even...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I find the generating function of this sequence with recurrence relation? I looked up related documents and saw a similar problem solving.
But, I don't understand well and still can not solve the problem.
How can I find the sequence of generating function ???
$${ G }_{ n }(x)=\sum _{ k=0 }^{ n }{ f(n,k){ x }^{ ... | Try writing this as a series in two indeterminates $x$ and $y$ such that
$$H(x,y)=\sum_{n\ge 0}G_n(x)y^n=\sum_{k,n\ge 0}f(n,k)y^nx^k$$
$$\implies H(x,y)=\sum_{n\ge 0}f(n,0)y^n + \sum_{k\ge 1}f(0,k)x^k + \sum_{k,n\ge 1}f(n,k)y^nx^k$$
From the i.c's we get
$$H(x,y)=1+\sum_{k,n\ge 1}f(n,k)y^nx^k$$
and using $f(n,k)=f(n,k-... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Solve for $x$ in $\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$ Solve for $x$ in $$\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$$
i have done in a lengthy way:
By inspection we observe that $x=41$ and $x=-24$ are the solutions
we have
$$\sqrt[4]{57-x}=5-\sqrt[4]{x+40}$$ squaring both sides we get
$$\sqrt{57-x}=25+\sqrt{x+40}-10 \sqrt[4]{x+40}$$... | $$\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$$
Let $y=\sqrt[4]{57-x}$, and $z=\sqrt[4]{x+40}$
Then
$$y\ge 0, z \ge 0, y+z=5$$
$$x=57- y^4=z^4-40$$
Thus
$$y^4+ (5-y)^4-97$$
$$=2(y-3)(y-2)(y^2-5y+44)=0$$
Thus $y=2$ or $y=3$, which means $x=41$ or $x=-24$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Fractions in Questions and Answers
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