Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Related rates of a cone I am in an intro calculus class and I have a problem which I am unsure about. It reads:
Water is being poured into a conical reservoir at a rate of pi cubic feet per second. The reservoir has a radius of 6 fees across the top and a height of 12 feet. At what rate is the depth of the water incre... | Since the dimension of the cone is such that $r = 6$ft and $h=12$ft, we have $\dfrac{r}{h} = \dfrac{6}{12}$, which means $r = \dfrac{h}{2}$. We must substitute this into the equation of the volume of a cone before differentiating. It follows that
$V=\pi r^2 \dfrac{h}{3}$
$V=\pi \bigg(\dfrac{h}{2}\bigg)^2 \dfrac{h}{3}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Extremely ugly integral $\int_{-\pi}^{\pi} \frac{\operatorname{sign}(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)|}\,dx$
Evaluate: $\DeclareMathOperator{\sign}{sign}$ $$\int_{-\pi}^{\pi} \dfrac{\sign(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)|}\,dx$$
My idea:
\begin{align*}I=\int_{-\pi}^{\pi} \frac{\sign(x)\arctan(x^2)-|x|}{\... | Noting
$$ \int_0^{\pi/2}\frac{1}{\sin^2x+\cos x}dx=\int_0^{\pi/2}\frac{1}{\cos^2x+\sin x}dx=\int_0^{\pi/2}\frac{1}{1-\sin^2x+\sin x}dx $$
and using $u=\tan(\frac{x}{2})$, one has
\begin{eqnarray}
\int_0^{\pi/2}\frac{1}{1-\sin^2x+\sin x}dx&=&\int_0^{\pi/2}\frac{1}{1-(\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})})^2+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
How can I solve $\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$ How can I solve the following integral?
$$\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$$
$$\begin{align}
\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx&=\int \frac{1}{\sqrt{\left(x+\frac72\right)^2-\frac{49}{4}} + 3}\, dx
\end{align}$$
$$u = x+\frac{7}{2}, \quad a = \frac{... | Setting $$\sqrt{x^2+7x}=t+x$$ then we have $$x=\frac{t^2}{7-2t}$$ and $$dx=-\frac{2 (t-7) t}{(2 t-7)^2}$$ and $$t+x=\frac{(t-7) t}{2 t-7}$$
and our integral will be $$\int -\frac{2 (t-7) t(2t-7)}{9\left(21-t+t^2\right)}dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding minimum of trigonometric function Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Then find difference between maximum and minimum of $v^2$.
I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?
I tried guessing, and got max... | Let $v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$
$v^2=a^2+b^2+2\sqrt{\left(a^2\cos^2(x)+b^2\sin^2(x)\right)\left(b^2\cos^2(x)+a^2\sin^2(x)\right)}$
$v^2=a^2+b^2+2\sqrt{a^2b^2\left(\cos^4(x)+\sin^4(x)\right)+\left(a^4+b^4\right)\cos^2(x)\sin^2(x)}$
$v^2=a^2+b^2+\sqrt{4a^2b^2\left(1-2\sin^2(x)\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 5
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How to get the limit of this series I need to solve this series:
$$\sum _{ k=2 }^{ \infty } (k-1)k \left( \frac{ 1 }{ 3 } \right) ^{ k+1 }$$
I converted it into $$\sum _{ k=0 }^{ \infty } \frac { { k }^{ 2 }-k }{ 3 } \left(\frac { 1 }{ 3 } \right)^{ k } -\frac { 4 }{ 3 } $$ with the idea, that $$\sum _{ k=0 }^{ \... | By stars and bars the coefficient of $x^n$ in $\frac{1}{(1-x)^3}$, that is the number of ways of writing $n$ as the sum of three natural numbers (or the number of ways for writing $n+3$ as the sum of three positive natural numbers), equals $\binom{n+2}{2}$. It follows that
$$ \frac{1}{(1-x)^3} = \sum_{n\geq 0}\binom{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the exact value of $\frac{\cos( \beta)}{\cos( \beta) -1}$ with $\sin(\beta - \pi) = \frac{1}{3}$
Let $h$ be $$\frac{\cos( \beta)}{\cos( \beta) -1}$$
With $\sin(\beta-\pi)=\frac{1}{3}$ and $\beta \in
{]}\pi,\frac{3\pi}{2}{[}$ determine the exact value of $h(\beta)$.
I tried:
$$\sin(\beta-\pi) = \sin(\beta)\cos(... | 3 things
$\dfrac{\frac{2\sqrt{2}}{3}}{\frac{2\sqrt{2}}{3}-1} =
\dfrac{2\sqrt{2}}{2\sqrt{2}-3}$
you dropped a 2.
$\dfrac{2\sqrt{2}}{2\sqrt{2}-3} = -6\sqrt 2 - 8$
With a little bit of simplification you would see that you were "only" off by a minus sign.
and here is where you lost it.
$\cos \beta = -\frac {\sqrt {2}}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the signal and zeros of $\cos x - \sin x$
Let $f$ be a function of domain $\mathbb{R}$ defined by:
$$f(x) = \cos x - \sin x$$
Find the signal and zeros of the function.
First I tried to find the zeros of $f(x)$:
$$0 = \cos x - \sin x \Leftrightarrow \\
0 = \sqrt{1-\sin^2x}-\sin x \Leftrightarrow \\
\sin x = \sqr... | There is a trigonometric identity $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$. Put $y = 3\pi/4$ to show $f(x) = \sqrt2\sin(x+3\pi/4)$. This quickly gives the answer you want.
You asked about some spurious extra solutions. I think these come from using $\cos(x) = \sqrt{1-\sin^2(x)}$. This is true only if the square ro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\cos\gamma+\sin\gamma$ in a 3-5-7 triangle ($\gamma$ is the greatest angle) Clearly $\gamma$ has to be the angle oposite to the side of lenght 7. The law of cosines gives me $$7^2 = 5^2+3^2 - 2\cdot 5\cdot3\cos{\gamma}\Longleftrightarrow\cos{\gamma}=-\frac{7^2-5^2-3^2}{2\cdot5\cdot 3}=-\frac{1}{2}.$$
This value of $\c... | As $0<\gamma<\pi$ (actually $\frac{\pi}{2}<\gamma<\pi$), we have $\sin\gamma>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What are the values of $a$ that satisfy the equation $\frac{ax^2+ax-1}{2(x+\frac{5}{4})^2+\frac{47}{8}} < 0$? I got this question in a test today, I was stumped on how to do it. I am right to say that since $2(x+\frac{5}{4})^2+\frac{47}{8}$ is always positive $ax^2+ax-1$ is always negative. So, $$ax^2+ax-1<0$$$$Discrim... | Hint: You want $a < 0$, and $a^2 - 4a(-1) < 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Show monotonicity Originally, I want to show that
$$
\frac{\sqrt{a \cdot b + \frac{b}{a}x^2}\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right)}{\sqrt{a \cdot b}\arctan \left(\frac{c}{\sqrt{a \cdot b}}\right)} \geq 1 \ \ \text{for} \ \ x, a,b,c > 0 \ .
$$
To do so, I figured it is sufficient to show tha... | There is an integral for arctan(x):
\[
\arctan(x) = \int_{0}^{x} \frac{dt}{t^2 + 1}, \quad x > 0.
\]
Perhaps this can help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Two six-sided dice are rolled. Find the probability distribution of $\max[X_1,X_2]$ I am not sure if I am understanding this problem correctly. It says;
Two six-sided dice are rolled. Let X1 and X2
denote the outcomes of the first and second die roll respectively.
Define a new random variable Y such that:
Y = max... | It may be useful to construct a table, with $X_1$ down the left, and $X_2$ along the top, and each cell gets the maximum of the two values. The table below has been started for you; fill in the rest of the values.
$$
\begin{array}{|c|c|c|c|c|c|c|} \hline
& \textbf{1} & \textbf{2} & \textbf{3}
& \textbf{4} & \textb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluate: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$ Evaluate: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$.
......
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
It takes $\infty - \infty $ form when $x=\infty $.
Now,
$$=\lim_{x\to \infty } (\sqrt {x-a} - \sqrt {bx})\times \dfrac {\sqrt {x-a}+\sqrt {bx}}{\s... | I'll add to Eugen's comment:
If $b=1$ then we have:
$$ \lim_{x\to \infty} \frac{-a}{\sqrt{x-a}+\sqrt{x}} = \frac{-a}{\infty} = 0$$
otherwise,
$$ \lim_{x\to \infty} \frac{(1-b)x-a}{\sqrt{x-a}+\sqrt{x}} =\lim_{x\to \infty} \frac{(1-b)\sqrt{x} - \frac{a}{\sqrt{x}}}{\sqrt{1-\frac{a}{x}}+1} = \lim_{x\to \infty} \frac{(1-b)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A proof of the approximate expression about totient summatory function I'm a high school student in Korea.
I am preparing for a presentation. so I prove an approximate expression about totient summatory function , but I'm not sure that the proof is correct.
If the proof is incorrect , please tell me what is incorrect. ... | Since $\varphi(n) = \sum_{d | n} \mu(d) \frac{n}{d}$ and $\sum_{d < x} \mu(d) \lfloor x/d \rfloor = 1$ $$\sum_{n < x} \varphi(n) = \sum_{n < x}\sum_{d | n} \mu(d) \frac{n}{d}=\sum_{md < x} \mu(d) m$$
$$=\frac{1}{2} \sum_{d<x}\mu(d) (\left\lfloor \frac{x}{d} \right\rfloor^2 + \left\lfloor \frac{x}{d} \right\rfloor )=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int\limits_{x=\alpha}^{2} x e^{-a x^2}\frac{p/x^2+c}{(p/x^2-t)^2}dx$
Evaluate
$$I(\eta)=2 a \int\limits_{x=\alpha}^{2} x e^{-a x^2}\frac{p/x^2+c}{(p/x^2-t)^2}dx$$
where $a,p,c,t>0,N\in{\mathbb{N}}, \alpha>0$.
Any idea how to evaluate this integral? This comes from the expected value of $\frac{p/X^2+c}... | \begin{align}
I &= \frac{2ap}{p^2} \int_\alpha^2 xe^{-ax^2} \frac{\frac{1}{x^2+c}}{\frac{1}{(x^2-t)}}dx = \frac{2a}{p} \int_\alpha^2 x e^{-ax^2} \frac{(x^2-t)^2}{x^2+c}dx\\ &= \frac{2a}{p} \int_\alpha^2 x e^{-ax^2} \frac{x^4-2tx^2+t^2}{x^2+c}dx \\
&= \frac{2a}{p} \int_\alpha^2 x e^{-ax^2} \left(x^2-(2t+c) + \frac{t^2+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Equivalent $\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(x)\over x^3+1}dx=\int_{0}^{\infty}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx?$ Proposed:
Is $(1)$ equivalent to $(2)$ in term of transformation?
$$\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(x)\over x^3+1}\mathrm dx={5\pi^2\over 2\cdot 6^3}\tag1$$
and
$$\i... | Let
$$I=\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(x)\over x^3+1}\mathrm dx$$
and
$$J=\int_{0}^{\infty}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx.$$
Note
\begin{eqnarray}
J&=&\int_{0}^{1}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm dx+\int_{1}^{\infty}{x^2-x\over x^2+1}\cdot{\ln(1+x)\over x^3+1}\mathrm ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How prove $(2+5x)\ln{x}-6(x-1)>0.\forall x>1$ let $x>1$ show that:
$$(2+5x)\ln{x}-6(x-1)>0.\forall x>1$$
Let
$$f(x)=(2+5x)\ln{x}-6(x-1),~~~f'(x)=\dfrac{2}{x}+5\ln{x}-1$$
since
$f(1)=1$.so it must prove
$$f'(x)=\dfrac{2}{x}+5\ln{x}-1>0?$$
| Let $f(x)=\ln{x}-\frac{6(x-1)}{5x+2}$.
Hence,
$$f'(x)=\frac{1}{x}-6\cdot\frac{5x+2-5(x-1)}{(5x+2)^2}=\frac{1}{x}-\frac{42}{(5x+2)^2}=$$
$$=\frac{25x^2-22x+4}{x(5x+2)^2}=\frac{25x^2-25x+3x+4}{x(5x+2)^2}>0$$
for all $x>1$.
Thus, $f(x)>f(1)=0$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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What is the value of this infinite fraction, where each successive row counts until a power of two? Here is the fraction
$$\frac{1}{\cfrac{2}{\cfrac{4}{8\cdots+9\cdots}+\cfrac{5}{10\cdots+11\cdots}} + \cfrac{3}{\cfrac{6}{12\cdots+13\cdots}+\cfrac{7}{14\cdots+15\cdots}}}$$
I have tried iterating row by row, seeing if th... | Let $a_n$ denote your sequence:
$$ a_n = \dfrac{1}{\dfrac{2}{\dfrac{4}{\vdots} + \dfrac{5}{\vdots}} + \dfrac{3}{\dfrac{6}{\vdots} + \dfrac{7}{\vdots}}}. $$
Let us make an observation. The general rule is that each bottom-most integer $m$ is replaced by $\frac{m}{2m + (2m+1)}$ in the next step. By applying this rule twi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving intersection of translated spiric curve The spiric section is the curve obtained by slicing a torus along a plane parallel to its axis. Assume the torus midplane lies in the $z=0$ plane, $a$ is the radius of the torus tube and $c$ is the distance from the origin of the center of the tube.
Now cut the torus with... | The implicit equation of the torus is
$$(x^2+y^2+z^2+R^2-r^2)^2=4r^2(x^2+y^2)$$
and the parametric equation of the shifted one is
$$\begin{cases}x=(R+r\cos\theta)\cos\phi+p\\y=(R+r\cos\theta)\sin\phi\\z=r\sin\theta+q.\end{cases}$$
The sections are given by
$$y=Y$$ and by eliminating $\theta$,
$$R+r\cos\theta=Y\csc\phi,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Put fraction in "arctan-friendly" form I would like to put $\int\frac{1}{(2x^2+x+1)}dx$ into something like $\int\frac{1}{(u^2+1)}dx$. What is the quickest way to proceed? I know that previous fraction can be rewritten as $2t^2+t+1 = \frac{7}{8}\left( \left( \frac{4t+1}{\sqrt{7}} \right)^2 +1 \right)$, but I don't have... | Multiply numerator and denominator by $4\cdot 2=8$ (the $2$ is the coefficient of $x^2$) and “complete the square”:
$$
\frac{1}{2x^2+x+1}=
\frac{8}{16x^2+8x+8}=
\frac{8}{16x^2+8x+1+7}=
\frac{8}{(4x+1)^2+7}
$$
Now you know that you should set $4x+1=u\sqrt{7}$, so you get
$$
\frac{8}{7}\frac{1}{u^2+1}
$$
Moreover, $4\,dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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If $a,b,c \geq 1$ then prove $\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$ If $a,b,c \geq 1$, then prove:
$$\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$$
My try:
$$A=\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} \\ A^2 =(a-1)+(b-1)+(c-1)+2\sqrt{(a-1)(b-1)}+2\sqrt{(a-1)(c-1)}+2\sqrt{(b-1)(c-1)} $$
$$B=\sqrt{abc... | The given inequality is equivalent to
$$ \sqrt{a-1} + \sqrt{b-1} \leq \sqrt{abc + c} - \sqrt{c-1}.$$
Now set $f(c) := \gamma \sqrt{c} - \sqrt{c-1}$ where $ \gamma := \sqrt{ab+1} $. Then we see that $f'(c) = \gamma \frac{1}{2\sqrt{c}} - \frac{1}{2\sqrt{c-1}}$, from which it is easy to check $f$ takes its minimum when ... | {
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"question_score": "2",
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Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$
Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$ where $n$ is a positive integer.
In order for a positive integer $k$ to begin with a $1$, we need $10^m \leq k < 2 \cdot 10^m$ for some positive integer $m$. Thus since $5^... | First, note that the inequalities must be strict, as we cannot have $5^n = 10^m$ for natural numbers $n,m$. Since $10^m = 2^m5^m$, your last equation becomes
$$2^{n+m+1}5^{m-n} < 2^{n+1} < 2^{n+m+2}5^{m-n}$$
$$\implies 2^{1+2n}10^{m-n} < 2^{n+1} < 2\cdot 2^{1+2n}10^{m-n}$$
Dividing each term by $2^{1+2n}$ gives us
$$10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving $|x^2-2x|+|x-4|>|x^2-3x+4|$
How do I solve $|x^2-2x|+|x-4|>|x^2-3x+4|?$
I can see the difference of the two terms on the left gives the term on the right. Now,what should I do? Is there any general method for solving $$|a|+|b|>|a-b|?$$
Thanks for any help!!
| Solving this is just a matter of using the definition of |x|.
This is |x||x- 2|+ |x+ 4|> |x^2- 3x+ 4|.
x^2- 3x+ 4 is positive for all x so we need to consider three cases:
x< -4. Then all of x, x- 2, and x+ 4 are negative. This is (-x)(2- x)- (x+ 4)= x^2- 3x- 4> x^2- 3x+ 4. That reduces to -4> 4 which is never true. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2293534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Calculate $\lim_{x\to a}(2-\frac{x}{a} )^{\tan( \frac{\pi x}{2a})}$
I would like to calculate
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)},\quad a \in\mathbb{R}^* \,\,\text{fixed} $$
we've
$$\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}=e^{\tan\left... | $$A=\left(2-\frac{x}{a} \right)^{\tan( \frac{\pi x}{2a})}\implies \log(A)=\tan( \frac{\pi x}{2a})\log\left(2-\frac{x}{a} \right)$$ Now, using Taylor expansion around $x=a$
$$\log\left(2-\frac{x}{a} \right)=-\frac{x-a}{a}-\frac{(x-a)^2}{2 a^2}+O\left((x-a)^3\right)$$ Using Laurent expansion around $x=a$
$$\tan( \frac{\p... | {
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"url": "https://math.stackexchange.com/questions/2294071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
} |
Simultaneous triangularization of matrices Let $\mathcal{F}=\{A_1,A_2,\ldots,A_r\}$ be a triangulable commuting family of $n\times n$ matrices (that is, each $A_i$ is triangulable and $A_iA_j=A_jA_i$ for every $i,j$). I know that $\mathcal{F}$ can be simultaneous triangularization, but what is the algorithm of finding ... | Here, it is easy. Since $A$ has $3$ distinct eigenvalues and $B,C$ commute with $A$, we can deduce that $B,C$ are polynomials in $A$ and it suffices to triangularize $A$.
In the general case.
Step 1. Find a common eigenvector of $A,B,C$.
Step 2. Proceed by recurrence.
Moreover, you can choose $P$ as an orthogonal matri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $\tan x$ if $x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right)$ Find $\tan x$ if $$x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right) \tag{1}$$
First i converted $$\frac{3 \sin 2x}{5+4 \cos 2x}=\frac{6 \tan x}{9+\tan^2 x}$$
So
$$\arcsin\left( \frac{6 \tan ... | Let $\alpha=\arctan(2\tan^2x)$ and $\beta=\arcsin(\frac{3\sin2x}{5+4\cos2x})$.
\begin{align}
\sin\beta&=\frac{3\sin2x}{5+4\cos2x}\\
\frac{2\tan\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}&=\frac{3(\frac{2\tan x}{1+\tan^2x})}{5+4(\frac{1-\tan^2x}{1+\tan^2x})}\\
\frac{\tan\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}&=\frac{3\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proof for sum of product of four consecutive integers I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ wil... | HINT:
From the Right Hand Side, if $f(n)=\dfrac{n(n+1)(n+2)(n+3)(n+4)}5$
$$f(m+1)-f(m)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Why can't the quadratic formula be simplified to $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$? I am currently taking Algebra 1 (the school year's almost over ), and we just learned the quadratic formula, another method to solve quadratic equations:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
However, this... | \begin{align}
(b-2)\sqrt{ac} & = \sqrt{(b-2)^2} \sqrt{ac} & & \text{ if } b-2\ge 0 \\[10pt]
& = \sqrt{(b-2)^2 ac}.
\end{align}
Is $(b-2)^2ac$ the same as $b^2-4ac\,$?
If one were to think that $(b-2)^2$ is the same as $b^2-4$ (and it is not) then one would have $(b-2)^2ac = b^2ac-4ac,$ so that is still not the same as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 1
} |
Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ Find Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ without using concept of Eigen Values and Eigen Vectors
I assumed its square root as $$B=\begin{bmatrix} a &b \\ c&d \end{bmatrix}$$
hence
$$B^2=A$$ $\implies$
$$\begin{bmat... | Square roots suck, so I'll do this with polynomials instead. Any matrix $B$ with $B^2=A$ must commute with $A$. Since there exist vectors $e$ such that $(e,Ae)$ form a basis of $\Bbb R^2$ (in fact the first standard basis vector will do, as will almost any other vector), any matrix commuting with $A$ must be a polynomi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find an upperbound for the rational function I know
$$\lim_{(x, y)\to (0,0)} \frac{x^3 + y^4}{x^2 + y^2} = 0$$ so
$$\left\lvert \frac{x^3 + y^4}{x^2 + y^2} \right\rvert \le f(x, y)$$ for some simpler $f(x, y)$ whose limit is also $0$.
How do I find the function $f(x, y)$? In other words, how do I get the upper bound... | Using the inequalities $x^2+y^2\ge x^2$ and $x^2+y^2\ge y^2$, we can write
$$\begin{align}
\left|\frac{x^3+y^4}{x^2+y^2}\right|&\le \frac{|x|^3}{x^2+y^2}+\frac{y^4}{x^2+y^2}\\\\
&\le |x|+y^2
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integration $\int \frac{\sec(x)}{\tan^2(x) + \tan(x)}$ How to evaluate the following integral $$\int \frac{\sec(x)}{\tan^2(x) + \tan(x)} \ dx$$
Thank you.
| Note that:
\begin{align}
I = \frac{\sec x}{\tan^2x + \tan x} &= \frac{\frac{1}{\cos x}}{\frac{\sin^2x}{\cos^2x}+\frac{\sin x}{\cos x}}\\
&= \frac{\cos x}{\sin^2x+\sin x\cos x}\\
&=\frac{(\cos x + \sin x) - \sin x}{\sin x (\sin x + \cos x)}\\
&=\frac{1}{\sin x} - \frac{1}{\sin x + \cos x} \\
&= \csc x - \frac{1}{\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is there any way to simplify the product of cosines? I recently saw a problem: Estimate the following: $cos(\frac{\pi}{15})cos(\frac{2\pi}{15})\ldots cos(\frac{7\pi}{15})$, and the options were between different consecutive power of ten.
How would I do this, no calculator of course, and is there's any way to shorten an... | We have $\cos(\frac{\pi}{15})\cos(\frac{2\pi}{15})\ldots \cos(\frac{7\pi}{15})$. Let's write this as $p=\cos(\frac{\pi}{15})\cos(\frac{2\pi}{15})\ldots \cos(\frac{7\pi}{15})$
We have to multiply both sides of the equation with
$$q= \sin(\frac{\pi}{15})\sin(\frac{2\pi}{15})\ldots \sin(\frac{7\pi}{15})$$
Now, $$p.q = \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What are the different methods to determine the derivative of $f(x)=\sqrt{1+x}$? The methods that I can think of are:
1) Chain rule
2) Binomial series of $f(x)$
3) Through the formula $f\prime(g(x))=\frac{1}{g\prime (x)}$ at a particular point
What are the other methods to determine the derivative of $f(x)$?
| From first principles.
\begin{align}
f'(x)&=\lim_{h\to0}\frac{\sqrt{1+x+h}-\sqrt{1+x}}{h}\\
&=\lim_{h\to0}\frac{(\sqrt{1+x+h}-\sqrt{1+x})(\sqrt{1+x+h}+\sqrt{1+x})}{h(\sqrt{1+x+h}+\sqrt{1+x})}\\
&=\lim_{h\to0}\frac{1+x+h-1-x}{h(\sqrt{1+x+h}+\sqrt{1+x})}\\
&=\lim_{h\to0}\frac{1}{\sqrt{1+x+h}+\sqrt{1+x}}\\
&=\frac{1}{2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Basis Transformation Matrix I have a problem related to transformation matrix.
The problem is like that;
Given basis $B$: $b_1 = \begin{pmatrix}1&2\end{pmatrix}^T$ and $b_2 =\begin{pmatrix}2&1\end{pmatrix}^T$ and $A$: $a_1=\begin{pmatrix}1&2\end{pmatrix}^T$, $a_2=\begin{pmatrix}2&7\end{pmatrix}^T$.
Find the transforma... | The solution by provided by @NDewolf. Let's fill out the details.
Vectors will be colored according to the basis membership, and named based upon color:
$$
\color{blue}{\mathbf{B}\ (standard)}, \qquad
\color{red}{\mathbf{R}\ (a)}, \qquad
\color{green}{\mathbf{G}\ (b)}.
$$
$$
\mathbf{T}_{\color{red}{R}\to \color{bl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7$, find $k$ I need help solving this question:
$$
\text{If }\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7\text{, find k.}
$$
I simplified this down to:
$$
\frac{4\cos^2\frac\pi7-1}{2\cos^2\frac\pi7-1}
$$
But am unable to proceed further. The val... | We need to prove that
$$3-\tan^2\frac{\pi}{7}=4(1-\tan^2\frac{\pi}{7})\cos\frac{\pi}{7}$$ or
$$\frac{3}{2}+\frac{3}{2}\cos\frac{2\pi}{7}-\frac{1}{2}+\frac{1}{2}\cos\frac{2\pi}{7}=4\cos\frac{2\pi}{7}\cos\frac{\pi}{7}$$ or
$$1+\frac{3}{2}\cos\frac{2\pi}{7}+\frac{1}{2}\cos\frac{2\pi}{7}=2\cos\frac{3\pi}{7}+2\cos\frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int_0^1 \sqrt{1 + x ^4 } \, d x $ $$
\int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,\mathrm{d}x
$$
I used substitution of tanx=z but it was not fruitful. Then i used $ (x-1/x)= z$ and
$(x)^2-1/(x)^2=z $ but no helpful expression was derived.
I also used property $\int_0^a f(a-x)=\int_0^a f(x) $
Please help me out
| We can do better than hypergeometric function and elliptic integral:
$$\color{blue}{\int_0^1 {\sqrt {1 + {x^4}} dx} = \frac{{\sqrt 2 }}{3} + \frac{{{\Gamma ^2}(\frac{1}{4})}}{{12\sqrt \pi }}}$$
Firstly, integration by part gives
$$\int_0^1 {\sqrt {1 + {x^4}} dx} = \sqrt 2 - 2\int_0^1 {\frac{{{x^4}}}{{\sqrt {1 + {x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
Solving the limit: $\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]$ Find the value of the limit $$\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]$$ where $[\cdot]$ denotes the greatest integer function or the box function.
My attempt: ... | When $x\to 0 $
$$\sin x\sim x-\frac{x^3}{6} \space , \space \space\space\sin^{-1} x\sim x+\frac{x^3}{6}\\\tan x\sim x+\frac{x^3}{3}\space ,\space\space \space\tan^{-1} x\sim x-\frac{x^3}{3} $$so
$$\lim_{x\to0}\left[100\frac{\sin^{-1}(x)}{x}\right]+\left[100\frac{\tan^{-1}(x)}{x}\right]=\\
\lim_{x\to0}\left[100\frac{x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find a point $P$ on a curve given the gradient of the normal at $P$. The point $P$ lies on curve $y=(x-5)^2$. It is given that the gradient of the normal at $P$ is $-\frac 14$
Find the coordinates of $P$.
| First find the gradient function of the curve:
$$\frac{dy}{dx}=2(x-5)$$
We want to find the point where the gradient is $-\frac{1}{-\frac14}=4$
So we solve \begin{align}2(x-5)&=4\\
x-5&=2\\
x&=7\end{align}
And therefore the point $P$ has an $x$ coordinate of $7$, and we can find its $y$ coordinate:
\begin{align}y&=(x-5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
find the limit of $ (1/\sin\ (x)-1/x)^x$ I am trying to find the limit of $\lim_{x \downarrow 0} (\frac{1}{sinx}- \frac{1}{x})^x$.
My current progress:
$\lim_{x \downarrow 0} (\frac{1}{\sin x}- \frac{1}{x})^x = \lim_{x \downarrow 0} (\frac{x-\sin x}{x\sin x})^x = \lim_{x \downarrow 0} e^{xln(\frac{x-\sin x}{x \sin x})}... | If $ \lim\ (\frac{1}{\sin\ x} - \frac{1}{x} )^x=L$, then $$ \lim\ x\cdot
\ln\ (\frac{1}{\sin\ x} - \frac{1}{x} )=\ln\ L$$
Here by Taylor series, $$ x
\ln\ ( \frac{1}{\sin\ x} - \frac{1}{x} )= x \ln\ \frac{x-\sin\
x}{x\sin\ x} =x\ln\ \frac{x/6 + O(x^2)}{1 + O(x^2)} =\frac{\ln\
(\frac{x}{6} + O(x^2) ) }{1/x}$$
L'Hospital... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Sum $\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$ How can we find the sum
$$\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$$
WolframAlpha says it's $13/4$ but how to compute it? In general how to approach this kind of sums of rational functions where denominator has distinct roots?
| Using partial fractions we have
\begin{eqnarray*}
\frac{3n+7}{n(n+1)(n+2)}= \frac{\frac{7}{2}}{n}+\frac{-4}{n+1}+\frac{\frac{1}{2}}{n+2}
\end{eqnarray*}
It is a telescoping sum ...
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{3n+7}{n(n+1)(n+2)} = \frac{\frac{7}{2}}{1}&+&\color{green}{\frac{-4}{2}} &+&\color{red}{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Find $\delta$ such that if $0 < \vert x - 2 \vert < \delta$, then $\vert f(x)-3 \vert < .2$ I'm looking for assistance with the following problem:
Find $\delta$ such that if $0 < \vert x - 2 \vert < \delta$, then $\vert f(x)-3 \vert < .2$ where $f(x)=x^2-1$.
My attempt:
We need $\delta$ such that $\vert f(x)-2 \vert <... | Since
\begin{eqnarray}
|f(x)-3|&=&|x^2-4|\\
&=&|x-2||(x-2)+4|\\
&\le& |x-2|(|x-2|+4)\\
&=&|x-2|^2+4|x-2|
\end{eqnarray}
if we solve the inequality
$$
y^2+4y\le 0.2
$$
we get
$$
a\le y\le b,
$$
with
\begin{eqnarray}
a&=&-2-\sqrt{4.2}\approx-4.0494\\
b&=&-2+\sqrt{4.2}\approx0.0494\\
\end{eqnarray}
Hence, choosing $\de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$ Find $x\in R$ satisfy $$\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$$
The root of equation very bad
My try 1:
Let $\sqrt[3]{x+2}=a;\sqrt[3]{2x-1}=b\Rightarrow a^3-b^3=3-x$
Have: $a+b=4-x$
=>Root of a system of equation bad, too
My try 2:
Use quality $\sqrt[3]{a}\pm \sqrt[3]... | to solve this equation use $$(a+b)^3=a^3+b^3+3ab(a+b)$$
$$(\sqrt[3]{x+2}+\sqrt[3]{2x-1}=4-x)^3\\x+2+2x-1+3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(\sqrt[3]{x+2}+\sqrt[3]{2x-1})=(4-x)^3$$ put $\sqrt[3]{x+2}+\sqrt[3]{2x-1}=4-x \\$ so
$$3x+1+3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(\sqrt[3]{x+2}+\sqrt[3]{2x-1})=(4-x)^3\\
3x+1+3\sqrt[3]{x+2}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evaluate the following Determinant Find $$ \Delta=\begin{vmatrix}
\frac{1}{a+x} &\frac{1}{b+x} &\frac{1}{c+x} \\
\frac{1}{a+y} &\frac{1}{b+y} &\frac{1}{c+y} \\
\frac{1}{a+z} &\frac{1}{b+z} &\frac{1}{c+z}
\end{vmatrix}$$
I applied $C_1 \to C_1-C_2$ and $C_2 \to C_2-C_3$ we get
$$ \Delta=\begin{vmatrix}
\frac{b-a}{... | \begin{align}
\Delta&=\begin{vmatrix} \frac{1}{a+x} & \frac{1}{b+x} & \frac{1}{c+x} \\ \frac{1}{a+y} & \frac{1}{b+y} & \frac{1}{c+y} \\ \frac{1}{a+z} & \frac{1}{b+z} & \frac{1}{c+z}\end{vmatrix}\\
\left[\prod_{q\in\{x,y,z\}}(a+q)\right]\Delta&=\begin{vmatrix} 1 & \frac{a+x}{b+x} & \frac{a+x}{c+x} \\ 1 & \frac{a+y}{b+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $a \equiv b \pmod{1008}$
Suppose that the positive integers $a$ and $b$ satisfy the equation $$a^b-b^a = 1008.$$ Prove that $a \equiv b \pmod{1008}$.
Taking the equation modulo $1008$ we get $a^b \equiv b^a \pmod{1008}$, but I didn't see how to use this to get that $a \equiv b \pmod{1008}$.
| We prove that $a = 1009, b = 1$ is the only solution.
First we restrict attention to the case in which $a, b \geq 3$.
We can rewrite the equation as
$$\frac{\log a}{a} - \frac{\log b}{b} = \frac{\log\left(1 + \frac{1008}{b^a}\right)}{ab}.$$
Let $f(x) = \frac{\log x}{x}$. We have $f'(x) = \frac{1 - \log x}{x^2} < 0$, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f (x)$
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f(x)$.
$f'(x)$ is the first derivative of $f (x)$.
I have no idea about this question, please help me.
| Assume a polynomial function of the form $f(x)=ax^3+bx^2+cx+d$ then $f^{'}(x)=3ax^2+2bx+c$
Thus the equations becomes $ax^3+(3a+b)x^2+(c+2b)x+d+c=x^3+5x^2+x+2$.
Now you can compare the coefficients to get the solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 4
} |
Trigonometric Equation : $\sin 96^\circ \sin 12^\circ \sin x = \sin 18^\circ \sin 42^\circ \sin (12^\circ -x)$ Please help solving this equation:
$\sin 96^\circ \sin 12^\circ \sin x = \sin 18^\circ \sin 42^\circ \sin (12^\circ -x)$
I used numerical method to solve it and got $x=6^\circ$ but I am not able to solve it ... | Observation
$$
\sin 96^{\circ} \sin 12^{\circ} = \sin 18^{\circ} \sin 42^{\circ}
$$
Implication
$$
\sin x = \sin \left(12^{\circ} - x \right) \qquad \Rightarrow \qquad x = 12^{\circ} - x
$$
Conclusion
$$x=6^{\circ}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find area of curve $(x^2+y^2)^2=2a^2xy$ How can I find the area of this curve?
$(x^2+y^2)^2=2a^2xy$
Should I first assume that $x=a\cos t$, or $y=\sin t$?
And then use the formula with $x$ and $y$ derivative?
| Let's convert to polar coordinates.
We get $\displaystyle (r^2)^2 = 2(a^2)(r\cos(\theta))(r\sin(\theta))$.
That boils to $r^4 = a^2r^2\sin(2\theta)$.
We then get $r^2 = a^2\sin(2\theta)$.
So that means $r = \pm a\sqrt{\sin(2\theta)}$.
Graph analysis.
Notice that it actually doesn't matter which root we take, because $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that each divisor of $5^{(4n+1)}+5^{(3n+1)}+1 =$ $1 \pmod {10}$ Prove that if $d$ divides $5^{(4n+1)}+5^{(3n+1)}+1$ for any $n$, then $d =$ $1 \pmod {10}$.
A similar statement can be proven, where $d$ divides $(3^{(2n+1)}+1)/4$, for any $n$, then $d =$ $1 \pmod {6}$ by first showing that if $-3$ is a quadratic r... | We can restrict to prime divisors $p$. These must be odd, so
all we need to prove is $p\equiv1\pmod5$. We can't have $p=5$ either.
Now
$$5a^4+5a^3+1\equiv0\pmod p$$
where $a=5^n$ and so
$$b^4+5b+5\equiv0\pmod p$$
where $b$ is the inverse of $a$ modulo $p$.
Consider $f(x)=x^4+5x+5$ and let $\zeta$ be a primitive fifth r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Determine if $\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2}$ exist What I tried:
Let $$\ f(x,y) = \frac{x^3y^4}{(x^4 + y^2)^2}$$
For points of the form$\ (x,0)$ then $\ f(x,0)=0$, similarly, for$\ (0,y)$ then $\ f(0,y)=0$, so lets suppose that:
$$\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2} =0$$
So, for$\ ε>0$... | Since
$$
0\leq \frac{y^2}{x^4+y^2} \leq 1 \qquad \forall (x,y)\neq (0,0),
$$
you get
$$
|f(x,y)| = |x|^3 \left[\frac{y^2}{x^4+y^2}\right]^2 \leq |x|^3,
\qquad \forall (x,y)\neq (0,0),
$$
hence $f(x,y) \to 0$ as $(x,y)\to (0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Complex Integration - $\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx$
Exercise :
Show that :
$$\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \frac{4\pi}{5}\sin\bigg(\frac{2\pi}{5}\bigg)$$
Attempt :
$$\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \int_{-\infty}^{\infty} \frac{1}{x^4 + x^3 + x^2 + x + 1}dx $$
$$x^4 + ... | Using the upper part of the complex plane we get:
$\displaystyle \int_{-\infty}^{+\infty}\frac{x-1}{x^5-1}dx=i2\pi Res(\frac{x-1}{x^5-1},e^{i2\pi/5})+ i2\pi Res(\frac{x-1}{x^5-1},e^{i4\pi/5})$
$\displaystyle = i2\pi\frac{x-1}{x^5-1}(x-e^{i2\pi/5})|_{x\to e^{i2\pi/5}}+ i2\pi\frac{x-1}{x^5-1}(x-e^{i4\pi/5})|_{x\to e^{i4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
A Conditional Probability problem with Biased Coins I need explanation for solving the following problem - exactly how one goes about solving it ? Any pointers or hints will certainly be appreciated:
A box contains 5 fair coins and 5 biased coins. Each biased coin has probability of a head as $\frac{4}{5}$. A coin is d... | $$P(\text{second is fair}\mid \text{first is head}) = \frac{P(\text{second is fair and first is head})}{P(\text{1st is head})}.$$
$$P(\text{first is head}) = P(\text{first is biased}) \times P(\text{head}\mid\text{biased}) + P(\text{first is unbiased}) \times P(\text{head}\mid\text{unbiased}) $$ $$= \left(\frac{4}{5} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Is there $n>1$ such that $n^{n+1} \equiv 1 \mod (n+1)^n$? First of all, note that $\frac{n^{n+1}}{(n+1)^n} \sim \frac{n}{e}$.
Question: Is there $n>1$ such that $n^{n+1} \equiv 1 \mod (n+1)^n$?
There is an OEIS sequence for $n^{n+1}\mod (n+1)^n$: https://oeis.org/A176823.
$0, 1, 8, 17, 399, 73, 44638, 1570497, 5077... | We first prove that it is impossible when $n\not\equiv 1\pmod 4$.
Notice how:
$$n^{n+1}=1+(n-1)\sum_{k=0}^{n}n^k$$
So, we would have $n^{n+1}\equiv 1\pmod{(n+1)^n}$ if and only if:
$$(n+1)^n\mid (n-1)\sum_{k=0}^{n}n^k$$
And since the RHS won't be equal to $0$, we'd have:
$$(n-1)\sum_{k=0}^{n}n^k\ge(n+1)^n$$
but, since ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$\sin(x) = \sin(x − \frac\pi3)$, solve for $x$ on interval $[-2\pi, 2\pi]$ According to the answer sheet:
$\sin(x) = \sin(x -\frac\pi3)$ gives:
$x = x-\frac\pi3 + k \cdot 2\pi$ or $x = \pi-(x-\frac\pi3) + k \cdot 2\pi$
^ How did they go from $\sin(x) = \sin(x-\frac\pi3)$ to the equations above?
Thanks in advance!
| \begin{eqnarray}
\sin x&=&\sin\left(x-\frac{\pi}{3}\right)\\
&=&\sin x\cos\frac{\pi}{3}-\cos x\sin\frac{\pi}{3}\\
&=&\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x\\
\frac{1}{2}\sin x&=&-\frac{\sqrt{3}}{2}\cos x\\
\tan x&=&-\sqrt{3}\tag{1}\\
x&=&-\frac{\pi}{3},-\frac{4\pi}{3},\frac{2\pi}{3},\frac{5\pi}{3}
\end{eqnarray}
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Unexpected outcome of integral A given method for calculating $\int^1_{-1} \frac 1 {1+x^2} \, dx$ is
\begin{align}
& \int^1_{-1} \frac 1 {1+x^2} \, dx=\int^1_{-1} \frac 1 {x^2(1+\frac 1 {x^2})} \, dx = -\int^1_{-1} \frac 1 {1+(\frac 1 x)^2} \,d(1/x) = \left.-\arctan\left(\frac 1 x\right)\right|_{x=-1}^1 \\[10pt]
= {} &... | Upon enforcing the substitution $x=1/t$, the domain transforms from $[-1,1]$ to $(-\infty,-1)\cup [1,\infty)$. Hence we have
$$\begin{align}
\int_{-1}^1\frac{1}{1+x^2}\,dx&= \int_{-\infty}^{-1}\frac{1}{1+t^2}\,dt+\int_1^\infty \frac{1}{1+t^2}\,dt\\\\
&=\left(\arctan(-1)+\frac\pi2\right)+\left(\frac\pi2 -\arctan(1)\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Chinese Remainder theorem polynomials Is there a program, small script, or easy method I can use to find the polynomial P(x) such that:
$P(x) = x+1$ $\pmod {x^2+1}$
$P(x) = x+2$ $\pmod {x^2+2}$
$P(x) = x+3$ $\pmod {x^2+3}$
$P(x) = x+4$ $\pmod {x^2+4}$
.........
This is easy to do with integers, but I can't get across t... | Hint Since $k=-X^2 \pmod{X^2+k}$, you have
$$P(X) \equiv X-X^2 \pmod{X^2+1} \\
P(X) \equiv X-X^2 \pmod{X^2+2} \\
P(X) \equiv X-X^2 \pmod{X^2+3} \\
P(X) \equiv X-X^2 \pmod{X^2+4} \\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
show this inequality $f(\sqrt{x_{1}x_{2}})\ge \sqrt{f(x_{1})\cdot f(x_{2})}$ let $$f(x)=\dfrac{1}{3^x}-\dfrac{1}{4^x},x\ge 1$$
For any $x_{1},x_{2}\ge 1$,show that
$$f(\sqrt{x_{1}x_{2}})\ge \sqrt{f(x_{1})\cdot f(x_{2})}$$
or
$$\left(\dfrac{1}{3^{\sqrt{x_{1}x_{2}}}}-\dfrac{1}{4^{\sqrt{x_{1}x_{2}}}}\right)^2 \ge \left(\d... | $f$ is nonincreasing since
\begin{align}
f^\prime(x)&=-3^{-x}\log 3+4^{-x}\log 4<0,\\
f^{\prime\prime}(x)&=3^{-x}(\log 3)^2-4^{-x}(\log 4)^2.
\end{align}
Similarly, denoting with $g(x)=\log(f(x))$, you have
\begin{align}
g^\prime(x)&=f^\prime(x)/f(x)<0,\\
g^{\prime\prime}(x)&=\frac{f^{\prime\prime}(x)f(x)-(f^\prime(x))... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2338165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve this Differential Equation in y and x.
Solve the following DE:
$$y'=\frac{y+y^2}{x+x^2}$$
in particular
$$y(2)=1$$$$\frac{y'}{y+y^2}=\frac{1}{x+x^2}$$$$\int \frac{dy}{y+y^2}=\int \frac{dx}{x+x^2}$$$$y'dx=dy$$$$ln(y)-ln(y+1)=ln(x)-ln(x+1)+C$$
but where to from here?
| Well, when we solve:
$$\text{y}\space'\left(x\right)=\frac{\text{y}\left(x\right)+\text{y}\left(x\right)^2}{x+x^2}\space\Longleftrightarrow\space\int\frac{\text{y}\space'\left(x\right)}{\text{y}\left(x\right)+\text{y}\left(x\right)^2}\space\text{d}x=\int\frac{1}{x+x^2}\space\text{d}x\tag1$$
So, for the intergals we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2338935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I find the term of a recursive sequence? I have $\{a_n\}$ the following sequence:
$a_1 =-1$
$a_k a_{k+1} = - a_k - \frac{1}{4}$
How can I find $a_1 a_2 \cdots a_n$?
| $$a_{k+1} = \frac{-4a_k - 1}{4a_k + 0}$$
Let $p_k / q_k = a_k$ :
$$\frac{p_{k+1}}{q_{k+1}} = \frac{-4 \frac{p_k}{q_k} - 1}{4\frac{p_k}{q_k} + 0} = \frac{-4 p_k - 1q_k}{4p_k + 0q_k}$$
$$\begin{align}
%
\begin{bmatrix} p_{k+1} \\ q_{k+1} \end{bmatrix}
%
& = \begin{bmatrix} -4 & -1 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the limit of the following function? $\lim_{n\rightarrow \infty }\left ( n-1-2\left (\frac{\Gamma(n/2)}{\Gamma((n-1)/2)} \right )^2 \right )$
| $$
\begin{aligned}
\lim_{n\rightarrow \infty }\left (n-1-2\left (\frac{\Gamma(n/2)}{\Gamma((n-1)/2)} \right )^2 \right)
& = \lim _{n\rightarrow \:\infty \:}\:\left(n-1-2\left(\frac{\frac{2\left(\frac{n}{2}\right)!}{n}}{\frac{2\left(\frac{n-1}{2}\right)!}{n-1}}\right)^2\right)
\\& = \lim _{n\rightarrow \:\infty \:}\:\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove that there do not exist positive integers $a,b$ such that $p \mid a+b+1,a^2+b^2+1,a^3+b^3+1$ where $p > 3$ is a prime
Prove that if $p > 3$ is a prime then there do not exist positive integers $a,b$ such that \begin{align*}a+b+1 &\equiv 0 \pmod{p}\\a^2+b^2+1 &\equiv 0 \pmod{p}\\a^3+b^3+1 &\equiv 0 \pmod{p}.\end{... | Let $p\gt 3$ be a prime. In the field $\mathbb F_p$ we have $$\begin{align*}a+b=-1\\a^2+b^2=-1\\a^3+b^3=-1\end{align*}$$ It follows from the first equation $$a^2+b^2+2ab=1 \Rightarrow ab=1$$
$$a^3+b^3+3ab(a+b)=-1 \Rightarrow 3ab(a+b)=0 \Rightarrow ab=0$$ Thus $$ 0=-1\text{ contradiction}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
how to enumerate and index partial permutations with repeats I know how to count permutations of, say, p red balls and q white balls and r blue balls. I also know how to count permutations of part of a set of distinct objects. But I don't know an efficient way to count partial permutations with repetition.
To illustr... | I assume your $45$ cases result from solving the equation
$$a + b + c = 12$$
in the nonnegative integers subject to the restrictions that $a \leq 8$, $b \leq 7$, and $c \leq 6$.
We can reduce the number of cases by considering the number of A's in the sequence.
*
*$8$ A's: There are $\binom{12}{8}$ ways to choose... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How do I solve $\int_0^\infty \frac{x^2}{x^4+1} \, dx$?
$$\int_0^\infty \frac{x^2}{x^4+1} \; dx $$
All I know this integral must be solved with beta function, but how do I come to the form $$\beta (x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\;dt \text{ ?}$$
| note that $$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove $ \left|x-1\right|<1$ implies $ \left|x^2-4x+3\right|<3$. Question:
Let $ x$ be a real number. Prove that if $ \left|x-1\right|<1$ then $ \left|x^2-4x+3\right|<3$.
We can write $ \left|x^2-4x+3\right|<3 $ as $ |x-3||x-1| < 3$. If I start with $ \left|x-1\right|<1$, how can I show that $ |x-3| < 3$ ?
| This is true for
complex numbers as well as
real numbers.
The reason is that,
if $a$, $b$, and $c$
are positive reals,
then
$b \lt c$
implies
$ab \lt ac$.
(Do you see why this is true?)
Since
$|x^2-4x+3|
= |x-1||x-3|
$,
if
$|x-1| < 1$,
then
$|x^2-4x+3|
\lt |x-3|
$.
So,
we are done
if we can show that
$|x-3| < 3$.
But,
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
If $a,b$ are roots of $3x^2+2x+1$ then find the value of an expression It is given that $a,b$ are roots of $3x^2+2x+1$ then find the value of:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3$$
I thought to proceed in this manner:
We know $a+b=\frac{-2}{3}$ and $ab=\frac{1}{3}$. Using this I tried to c... | $$3a^2+2a+1=0 \to 3a^2+3a+1=a\\3b^2+2b+1=0\to 3b^2+3b+1=b\\a-1=3a(a+1)\\b-1=3b(b+1)$$
so $$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=\\
\left(\dfrac{-3a(a+1)}{1+a}\right)^3+\left(\dfrac{-3b(b+1)}{1+b}\right)^3\\=-27(a^3+b^3)=-27(s^3-3ps)\\=-27\left(\left(\frac{-2}{3}\right)^3-3\left(\frac{1}{3}\ti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Flux through sphere I wish to find the flux of $\mathbf{F}=(x^2,y^2,z^2)$ through $S: (x-1)^2+(y-3)^2+(z+1)^2$
Here is what I tried:
I "moved" the sphere to $(0,0)$ by changing the variables to:
$u=x-1$ , $v=y-3$ , $w=z+1$
so now we have $F=((u+1)^2,(v+3)^2,(w-1)^2)$ and $S$ is the unit sphere.
So my calculation is (af... | thanks to zack I found my miscalculation:
from where i stopped:
$\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}divFrdzdrd\theta=\int_0^{2\pi}\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}(2r\cos\theta +2r\sin\theta +2z+6)rdzdrd\theta=
\int_0^1\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}[r(2r\sin\theta-2r\cos\theta+(2z+6)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Are there infinitely many positive integer solutions to the equation $\overline{x^2y^2} = z^2\,$?
Do there exist infinitely many solutions to the equation $$\overline{x^2y^2} = z^2$$ where $x,y,z$ are positive integers? Note: the notation $\overline{xy}$ means a concatenation of the numbers $x$ and $y$ to form one num... | $$\overline{2^2(3\cdot10^k)^2}=(7\cdot10^k)^2$$
and in general, if $(a,b,c)$ is a solution, then so is $(a,b\cdot10^k,c\cdot10^k)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Issue with Baby Rudin Example 7.3 Example 7.3 of Baby Rudin states that the sum
\begin{align}
\sum_{n=0}^{\infty}\frac{x^2}{(1 + x^2)^n}
\end{align}
is, for $x \neq 0$, a convergent geometric series with sum $1 + x^2$. This confuses me. As far as I know, a geometric series is a series of the form
\begin{align}
\sum_{... | $$\begin{align}
\sum_{n=0}^\infty \frac{x^2}{(1+x^2)^n}&=x^2\sum_{n=0}^\infty \left((1+x^2)^{-1}\right)^n\\\\
&=\frac{x^2}{1-(1+x^2)^{-1}}\\\\
&=\frac{x^2(1+x^2)}{1+x^2-1}\\\\
&=1+x^2
\end{align}$$
for $x\ne0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Maximum integer $n$ satisfying Taylor approximation
What is the maximum $n \in \mathbb N$ satisfying the formula
$$\sinh(x^4) = x^4 + \frac 1 6 x^{12} + \mathcal o(x^n) \quad \mathrm{for}\ x\to0 $$
The Taylor's theorem assures me that
$$\sinh x = x + \frac{x^3}{3!} + o(x^3)\ \mathrm{or}\ o(x^4) \quad \mathrm{for}\ ... | As you know the taylor series of the $\sinh(x)$ is:
$$\sinh(x) = \sum_{k=0}^{\infty}\frac{x^{1+2k}}{(1+2k)!} = x + \frac{x^3}{3!} + \frac{x^5}{5!}+\cdots$$
Therfore,
$$\sinh(x^4) = x^4 + \frac{x^{12}}{3!} + \frac{x^{20}}{5!}+\cdots$$
Hence for integer powers of $x$:
$$\sinh(x^4) = x^4 + \frac{x^{12}}{3!} + O(x^{20}) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2350334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to apply Parseval's equation and checking to see if we have convergence in $L^2$
On a previous assignment, you found the sine series on the interval $[0,l]$:
$$x(l-x) = \frac{8l^2}{\pi^3}\sum_{n=1}^{\infty}\frac{1}{n^3}\sin\left(\frac{n \pi x}{l}\right)$$
Note that the sum includes odd terms only.
a.) Show tha... | Parseval's equation,$$\sum_{n=1}^{\infty}|\hat{f}_n|^2 = \|f\|^2$$ We have
$$\sum_{n=1}^{\infty}\left(\frac{8l^2}{\pi^3}\frac{1}{n^3}\right)^2 = \sum_{k=0}^{\infty}\left(\frac{8l^2}{\pi^3}\frac{1}{(2k+1)^3}\right)^2 = \frac{2}{l}\int_{0}^{l}(x(l-x))^2dx = \frac{l^4}{15}$$
Therefore, $$\sum_{n=0}^{\infty}\frac{1}{(2n+1... | {
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"url": "https://math.stackexchange.com/questions/2354034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find a cubic polynomial. If $f(x)$ is a polynomial of degree three with leading coefficient $1$ such that $f(1)=1$, $f(2)=4$, $f(3)=9$, then $f(4)=?,\ f(6/5)=(6/5)^3?$
I attempt:
I managed to solve this by assuming polynomial to be of the form $f(x)=x^3+ax^2+bx+c$, then getting the value of $a,b,c$, back substituting i... | If
*
*$f(x) = x^3 + ax^2 + bx + c$
*$f(1) = 1$
*$f(2) = 4$
*$f(3) = 9$
then
\begin{align}
f(x+1) - f(x) &= (3x^2+3x+1) + a(2x+1) + b \\
\hline
4-1 &= f(2)-f(1) \\
3 &= 7 + 3a + b \\
\hline
9-4 &= f(3) - f(2) \\
5 &= 19 + 5a + b \\
\hline
3a + b &= -4 \\
5a + b &= -14 \\
\hline
a &= -5 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Integration by trig substitution - why is my answer wrong? Attempting integral:
$$-\int \frac{dx}{\sqrt{x^2-9}}$$
Let $x = 3\ sec\ \theta$ so that under the square root we have:
$$\sqrt{9\ sec^2\ \theta - 9}$$
$$\sqrt{9(sec^2\ \theta - 1)}$$
$$\frac{dx}{d\theta} =3\ sec\ \theta\ tan\ \theta$$
The $1/3$ and the $3$ canc... | Your mistake:
$$\sec\theta=\frac x3$$ yields
$$\cos\theta=\frac 3x,\\\sin\theta=\sqrt{1-\frac9{x^2}},\\\tan\theta=\sqrt{\frac{x^2}9-1}.$$
Not $\sqrt{x^2-9}$.
This said, you can rescale the variable with $x=3t$ to get
$$\int\frac{dt}{\sqrt{t^2-1}}.$$
You may recognize another familiar integral,
$$\int\frac{dt}{\sqrt{1-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Heaviside Step Function In studying properties of 1/x, its derivatives and its integral, I came across the following "apparent" identity. Plot it, and found that it appears to be the same as the Heaviside function. Any post with reference to the RHS, or verifying/disproving would be appreciated.
$$\frac{\arctan(1/x)+... | $$\frac{d}{dx}\arctan(X)=\frac{1}{X^2+1}\quad\to\quad \begin{cases}
\frac{d}{dx}\arctan(\frac{1}{x} )=\frac{1}{\frac{1}{x^2}+1}\left(-\frac{1}{x^2} \right) \\
\frac{d}{dx}\arctan(x+\frac{1}{2} )=\frac{1}{\left(x+\frac{1}{2} \right)^2+1} \\
\frac{d}{dx}\arctan(2x^2+x+2)=\frac{1}{(2x^2+x+2)^2+1}(4x+1)
\end{cases}$$
$$f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to find the value of this expression Suppose I have a general quadratic equation $ ax^2 + bx + c = 0 $
And I have $ \alpha , \beta $ as roots, then what is the simplest way to find $ (a\alpha + b)^{-2} + (a\beta + b)^{-2} $ ?
I know the formula for finding the sum and product of roots, but that doesn't helped me i... | Some assumptions:
$c, \alpha, \beta \neq 0$ (the answer will be a bit modified without these assumptions)
Since $\alpha$ is a root therefore
\begin{align*}
a\alpha^2+b\alpha+c &=0\\
a\alpha+b & = \frac{-c}{\alpha}\\
\frac{1}{(a\alpha+b)^2} & = \frac{\alpha^2}{c^2}
\end{align*}
Thus
\begin{align*}
\frac{1}{(a\alpha+b)^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Divisibility of $n\cdot2^n+1$ by 3. I want to examine and hopefully deduce a formula that generates all $n\geq0$ for which $n\cdot2^n+1$ is divisible by $3$. Let's assume that it is true for all $n$ and that there exist a natural number $k\geq0$ such that $$n\cdot2^n+1=3k,$$
Now I want to proceed with induction, but cl... | This is an approach which does not use (directly) modular arithmetic.
Set $a_n = n 2^n + 1$.
Let's define $S:=\{ n\in \mathbb{Z} \mid n \geq 0 \text{ and $3$ divides $a_n$} \}$. Observe that for integers $n,k \geq 0$:
$$ a_{n+k} - a_n = (n+k) 2^{n+k} - n 2^n = 2^n((n+k)2^k - n) = 2^n((2^k-1)n +k2^k) $$
For $k=6$ we get... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Canonical form of a quadratic form
Find the canonical form of $2x^2+y^2+3y=0$
What are the way to approach this?
I know that the eigenvalues are positive and that
$$2x^2+y^2+3y=0\iff2x^2+(y+\frac{3}{2})^2-\frac{9}{4}=0$$
For: $x'=2x$, $y'=y+\frac{3}{2}$
We get: $$x'^2+y'^2-\frac{9}{4}=0$$
How do I categorize this qu... | As you noticed the equation
$$2x^2+y^2+3y=0\tag{1}$$
is equivalent to
\begin{equation*}
2x^{2}+\left[ y-\left( -\frac{3}{2}\right) \right] ^{2}-\left( \frac{3}{2}
\right) ^{2}=0.\tag{2}
\end{equation*}
Dividing by $(3/2)^2 $ and writting $2$ as $1/(\sqrt{2}/2)^2$, we obtain
\begin{equation*}
\frac{x^{2}}{\left( \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Express the differential equation as power series
My attempt: $y=\sum^{\infty}_{n=0}a_n x^n \rightarrow y'=\sum^{\infty}_{n=0}na_n x^{n-1}\rightarrow y''=\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}\\
(2+x^2)y"+x^2y'+3y=(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+x^2\sum^{\infty}_{n=0}na_n x^{n-1}+3\sum^{\infty}_{n=0}a_n x^n\... | You are asked to put the series on the form :
$$(2+x^2)y"+x^2y'+3y=\sum^{\infty}_{n=0}c_n x^n$$
ant to find the coefficients $c_n$.
$$(2+x^2)y"+x^2y'+3y=(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+x^2\sum^{\infty}_{n=0}na_n x^{n-1}+3\sum^{\infty}_{n=0}a_n x^n$$
So, you have to gather the terms of common power $n$.
$$--... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int (x^2+1)/(x^4+1)\ dx$ I have Divided the numerator and Denominator by $x^2$
to get $\dfrac{1+x^{-2}}{x^2+x^{-2}}$ then changed it into $(1+(x^{-2}))/[(x-x^{-1})^2 +2]$ then took $x-(1/x)$ as $u$ and Differentiated it with respect to $x$ to get $dx=du/(1+x^{-2})$ Finally I got this expression:
$$
\int\frac{x^2+1}{... | Express MichaelRozenberg's denominator, $x^2 + \sqrt{2} x + 1$ in the form $u^2 + 1$ so you can use the $\tan^{-1}$ form.
Specifically:
$x^2 + \sqrt{2} x + 1$
Let $\sqrt{2}x + 1 = u$,
so
$u^2 = 2 x^2 + 2 \sqrt{2} x + 1$
so
$u^2 + 1 = 2 x^2 + 2 \sqrt{2} x + 2$
or
${1 \over u^2 + 1} = {1 \over x^2 + \sqrt{2} x + 1}$.
A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
If $A∆B=A∆C$, then $B=C$ If $A\Delta B=A\Delta C$, then $B=C$.
I use the formula $X∆Y=(X\setminus Y)\cup(Y\setminus X)$ in $A∆B=A∆C$. And further use the formula $X\setminus Y=X\cap Y^c$. I am not able to reach the conclusion. Please help me.
| You can do element chasing.
Or you can manipulate symbols and properties and identities.
But you can also divide and conquer.
Divide the space into $8$ disjoint sets determined by the elements that are are not in $A,B,C$.
They are:
1) $A^c \cap B^c \cap C^c$ (the elements that are in none)
2) $A^c \cap B^c \cap C$
3) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$
Find the value of $$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$$
I did it like this
$$\cos^2 76^{\circ}+\cos^2 16^{\circ} = \cos(76^{\circ}+16^{\circ}) \, \cos(76^{\circ}-16^{\circ}).$$
So the expression is $$\cos 92^... | $$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}=$$
$$=\frac{1}{2}\left(1+\cos152^{\circ}+1+\cos32^{\circ}-\cos60^{\circ}-\cos92^{\circ}\right)=$$
$$=\frac{3}{4}+\frac{1}{2}\left(\cos152^{\circ}+\cos32^{\circ}-\cos92^{\circ}\right)=$$
$$=\frac{3}{4}+\frac{1}{2}\left(2\cos92^{\circ}\cos60^{\circ}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2368243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove this trigonometric inequality about the angles of $\triangle ABC$
In $\Delta ABC$ show that
$$\cos{\frac{A}{2}}+\cos\frac{B}{2}+\cos\frac{C}{2}\ge \frac{\sqrt{3}}{2} \left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\frac{A-B}{2}\right)$$
since
$$\frac{\sqrt{3}}{2}\left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\df... | $$2\sum \sin \frac{A}{2}\sin \frac{B}{2}\leq \frac{2}{3}\left(\sum \sin \frac{A}{2}\right)^2\leq\sum \sin \frac{A}{2}$$
$$=\sum \cos \frac{A}{2}\cos \frac{B}{2}-\sum \sin \frac{A}{2}\sin \frac{B}{2},$$
$$ \frac{\sqrt{3}}{2}\sum \cos \frac{A-B}{2}=\frac{\sqrt{3}}{2} \left(\sum \cos \frac{A}{2}\cos \frac{B}{2}+\sum \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2368630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Cross ratio of 4 points on circle I'm in a bit of trouble I want to calculate the cross ratio of 4 points $ A B C D $ that are on a circle
Sadly "officially" it has to be calculated with A B C D as complex numbers and geometers sketchpad ( the geomerty program I am used to) don't know about complex numbers
Now I am wo... | For $A,B,C,D$ on the unit circle,
$$
\begin{align}
\frac{(A-C)(B-D)}{(A-D)(B-C)}
&=\frac{\left(e^{ia}-e^{ic}\right)\left(e^{ib}-e^{id}\right)}{\left(e^{ia}-e^{id}\right)\left(e^{ib}-e^{ic}\right)}\\
&=\frac{\left(e^{i(a-c)}-1\right)\left(e^{i(b-d)}-1\right)}{\left(e^{i(a-d)}-1\right)\left(e^{i(b-c)}-1\right)}\\
&=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2368813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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What is the probability that Cathy wins?
Alice, Bob and Cathy take turns (in that order) in rolling a six sided
die. If Alice ever rolls a 1, 2 or 3 she wins. If Bob rolls a 4 or a 5
he wins, and Cathy wins if she rolls a 6. They continue playing until
a player wins. What is the probability (as a fraction) that ... | You are wrong because that is not the only way in which Cathy can win! Note that if Cathy rolls anything other than a 6, the game repeats itself until someone wins. Using very similar reasoning to yours, show that the probability of her winning is the infinite sum
$$
\sum_{n=0}^\infty\Big(\frac{3}{6}\cdot\frac{4}{6}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374181",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Difficult Homogeneous Differential Equation Solve the differential equation:
$$\frac{dy}{dx}=\frac{\sqrt{x^2+3xy+4y^2}}{x+2y}$$
I tried to solve it by putting $t=x+2y$ but that lead to a very complicated integral. The hint given is that equation is reducible to homogeneous form.
| Partial answer :
Reformulate it as $$\frac{dy}{dx} = \frac{\sqrt{1+3\frac{y}{x}+4\left(\frac{y}{x}\right)^2}}{1+2\frac{y}{x}}$$
The you have $dy/dx = f(y/x)$.
Let $u(x) = \frac{y(x)}{x}$. Then $xu = y$. Then $dy = xdu + udx$. Then $dy/dx = xdu/dx + u$. Then you have
$$x\frac{du}{dx}+u = f(u)\quad \text{where} \quad f(u... | {
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"source": "stackexchange",
"question_score": "3",
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Find A Polynomial Solution For The Legendre equation.
The Legendre equation
$$(1-x^2)y''-2xy'+\alpha(\alpha+1)y=0$$
has a polynomial solution $P_n$ when $\alpha$ is a non-negative number $n$. Find $P_1$ satisfying $P_1(1)=1$ and then find the general solution of the D.E. with $\alpha=1$
$$y=Ax+B$$$$y'=A,y''=0$$$... | $$\alpha=1\qquad\to\qquad (1-x^2)y''-2xy'+2y=0 \tag 1$$
You found a particular solution $y=x$ which is correct.
Or, more general, a family of solutions : $\quad y=C\:x\quad$ where $C$ is a constant.
In order to find the general solution of the ODE, one can use the method of variation of parameter.
In the present case,... | {
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"source": "stackexchange",
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In showing integer sum $(1+2+3+...+n)$ by l'Hopital rule why they take lim as $r$ approaches to $1$ in one of steps? Why $1$? So why lim as $r \to 1$ (why $1$?) Here's the method:
| (Arithmetic - geometric progression)
If we differentiate both sides with respect to $r$, we have
\begin{equation}
1+2r+3r^2+\dots+nr^{n-1}=\frac{nr^{n+2}-(n+1)r^{n+1}+r}{(1-r)^2}
\end{equation}
Letting $r \to 1$ and applying L'Hospital's rule on the fraction, we end up with
\begin{equation}
\frac{n(n+1)}2=1+2+3+\dots+n... | {
"language": "en",
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Show that $\frac{d^n}{dx^n}(\frac{\log x}{x})=(-1)^n\frac{n!}{x^{n+1}}(\log x-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{n})$ Show that $\frac{d^n}{dx^n}\left(\frac{\log x}{x}\right)=(-1)^n\frac{n!}{x^{n+1}}\left(\log x-1-\frac{1}{2}-\frac{1}{3}-\ldots -\frac{1}{n}\right)$
I am not allowed to use induction. I do not know h... | Approach 1: When we calculate the first few derivatives, we have:
$$y'=-\frac{1}{x^2}\ln{x}+\frac{1}{x^2}=-\frac{1}{x^2}(\ln{x}-1).$$
$$y''=\frac{2}{x^3}\left(\ln{x}-1-\frac12 \right).$$
$$y^{(3)} = -\frac{3!}{x^4}\left(\ln{x}-1-\frac12 -\frac13\right).$$
so we can recognize the desired pattern.
Approach 2: We use the ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Find the value of $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}$
Let $a$, $b$ and $c$ be real numbers such that $a + b + c = 0$ and define:
$$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}.$$
What is the value of $P$?
This question came in the regional maths olympiad. I tried AM-GM a... | Write $c=-a-b$.
Then
$$\frac{a^2}{2a^2+bc}=\frac{a^2}{2a^2-ab-b^2}=\frac{a^2}{(2a+b)(a-b)}$$
and
$$\frac{c^2}{2c^2+ab}=\frac{(a+b)^2}{2a^2+5ab+2b^2}=\frac{(a+b)^2}{(2a+b)(a+2b)}.$$
The sum therefore equals
$$\frac{a^2(a+2b)-b^2(2a+b)+(a+b)^2(a-b)}{(2a+b)(a+2b)(a-b)}=\cdots$$
etc. (as long as the denominator is nonzero)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Floor functions of powers increase What is the smallest real number $x$ such that $\lfloor x^n\rfloor<\lfloor x^{n+1}\rfloor$ for all positive integers $n$?
In particular, we have $\lfloor x\rfloor<\lfloor x^2\rfloor<\lfloor x^3\rfloor<\dots$. For the first inequality to hold it must be that $x\geq\sqrt{2}$. But if $x=... | You have correctly shown that $x \ge \sqrt[3]3.\ $ Now if $x^n \ge 3$ and $x \gt \sqrt [3]3, \lfloor x^{n+1}\rfloor= \lfloor x^n\cdot x\rfloor\ge\lfloor x^n\rfloor+\lfloor x^n(x-1)\rfloor \ge \lfloor x^n \rfloor +1$ so any higher $x$ will work as well, so $\sqrt[3]3$ is the smallest $x$
| {
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"timestamp": "2023-03-29T00:00:00",
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show that $7\mid p^3-p$ if $p$ is a prime divisor of $n^3+n^2-2n-1$
Let $p$ be a prime number, and $n$ a positive integer such
$$p\mid n^3+n^2-2n-1, \quad n\ge 2.$$
Show that $$7\mid p^3-p.$$
It maybe can use Fermat's little theorem?
| A blind approach, although it will give the same solution.
The polynomial $f(x)=x^3+ x^2 - 2 x -1$ has discriminant $\Delta=49$, a square. Therefore, given a root $x_1$ of $f(x)$, the other roots can be expressed in terms of $x_1$ from the equality
$$(x_2-x_3) = \frac{(x_1-x_2)(x_1-x_3)(x_2-x_3)}{(x_1-x_3)(x_2-x_3)}= \... | {
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"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 0
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Finding third row of orthogonal matrix?
Find a $3\times3$ orthogonal matrix whose first two rows are $\Big[\frac{1}{3},\frac{2}{3},\frac{2}{3}\Big]$ and $\left[0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right]$.
I tried two approaches.
One, finding vector cross product of given two rows.
Second, assuming third row as $... | Let $[x~y~z]$ be the third row. Due to the orthogonality, we have $$\frac{x}{3}+\frac{2y}{3}+\frac{2z}{3}=0,$$ and $$\frac{y}{\sqrt{2}}-\frac{z}{\sqrt{2}}=0.$$ Consequently, $y=z$, and $x=-4y$. Using $x^2+y^2+z^2=1$, the values of $x$, $y$, and $z$ can easily be found. The last condition ensures that the vector has uni... | {
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"url": "https://math.stackexchange.com/questions/2385989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Geometry : $\frac{GA}{GA'}+\frac{GB}{GB'}+\frac{GC}{GC'}=3$ Let $G$ be centriod of $\triangle ABC$ and GA, GB, GC cut the circumcircle of $\triangle ABC$ again at A', B', C' respectively. Prove that $\frac{GA}{GA'}+\frac{GB}{GB'}+\frac{GC}{GC'}=3$.
My attempt :
Let $D, E, F$ be the midpoints of $BC, CA, AB$ respective... | Say $AD =2x$, $BG = 2y$, $CG=2z$ and $A'D = x'$, $B'E=y'$, $C'F=z'$.
Finally $AB = 2c$, $BC= 2a$ and $AC=2b$.
By the PoP we have: $x'\cdot 3x= a^2$, $y'\cdot 3y=b^2$, $z'\cdot 3z=c^2$ and $$x(x+x') = y(y+y')= z(z+z') =:k$$
Then $$3x^2+a^2 = 3y^2+b^2= 3y^2+c^2=3k$$
So $3(x^2+y^2+z^2)+a^2+b^2+c^2 =9k$
We are interested ... | {
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"url": "https://math.stackexchange.com/questions/2386643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the distance of the centroid of the triangle ABC from the origin? A sphere of radius $r$ passes through the origin and the other points where it meets the coordinate axes are A, B and C. Find the distance of the centroid of the triangle ABC from the origin (in terms of $r$).
i was taking the coordinate of the cent... | If the sphere intersects the three axes at the points A, B and C, it means that the sphere passes through these three points:
$(A,0,0)$; $(0,B,0)$; $(0,0,C)$
i.e., the sphere intersects x-axis at $(A,0,0)$ and y-axis at $(0,B,0)$ and z-axis at $(0,0,C)$
Now, coordinates of the centroid of triangle formed by these three... | {
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} |
Integrating $\frac{1}{(a+b\cos x)}$
Question: How do you prove this integral$$\int\frac {dx}{a+b\cos x}=\frac 2{\sqrt{a^2-b^2}}\arctan\left\{\tan\frac x2\sqrt{\frac {a-b}{a+b}}\right\}$$or$$\int\frac {dx}{a+b\cos x}=\frac 1{\sqrt{b^2-a^2}}\log\frac {\sqrt{b+a}+\sqrt{b-a}\tan\tfrac x2}{\sqrt{b+a}-\sqrt{b-a}\tan\tfrac x... | Lets compute the integral using the half tangent angle sub where
$$
\cos x = \frac{1-t^2}{1+t^2}
$$
where $t = \tan\left(\frac{x}{2}\right)$
we obtain and integral of the form
$$
\int \frac{1}{a+b\left[\frac{1-t^2}{1+t^2}\right]}\frac{2}{1+t^2}dt
$$
or
$$
\int \frac{2}{a(1+t^2)+b(1-t^2) }dt
$$
or
$$
\int \frac{2}{a+b +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Isosceles triangle inscribed in an ellipse.
Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ with its (unique) vertex lying at one of the ends of the major axis of the ellipse.
The three vertices of the triangle would be $(a,0), (x,y), ... | The given answer is correct. Your mistake is in the formula for the area,
which should be $A/2 = 1/2\ {\rm base} \times {\rm height}$ for each half half triangle or
$A = (a-x) b \sqrt{1 - x^2/a^2}$
for the full triangle.
The derivative is then:
$${d A \over d x} = -\frac{b (a+2 x) \sqrt{1-\frac{x^2}{a^2}}}{a+x} .$$
Se... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the Laurent Series of $\frac{z^2}{(z-1)(z^2+4)}$ I have attempted to split this into partial fractions of the form $\frac{A}{(z-1)} + \frac{B}{(z-2i)} + \frac{C}{(z+2i)}$ as well as by $\frac{A}{(z-1)} + \frac{Bz+C}{(z^2+4)}$, but things get very complex (ha!) with i's and z's showing up in the numerator. How shou... | Since you are told to find the series about $0$ and $1$, the first one does not make much trouble using long division.
For the second one, define first $z=x+1$ which makes
$$f=\frac{z^2}{(z-1)(z^2+4)}=\frac{(x+1)^2}{x \left(x^2+2 x+5\right)}=\frac{x^2+2x+1}{x \left(x^2+2 x+5\right)}$$ Perform the long division for $$x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x+y+z+w=29$ where x, y and z are real numbers greater than 2, then find the maximum possible value of $(x-1)(y+3)(z-1)(w-2)$
If $x+y+z+w=29$ where x, y and z are real numbers greater than 2, then find the maximum possible value of $(x-1)(y+3)(z-1)(w-2)$.
$(x-1)+(y+3)+(z-1)+(w-2)=x+y+z+w-1=28$
Now $x-1=y+3=z-1=w-... | Let $$f(x,y,z,w) = (x-1)(y+3)(z-1)(w-2)$$ and $$g(x,y,z,w) = x + y + w - 29$$
We want to $$\max\{f(x,y,x,w)\}$$
subject to:
$$g(x,y,z,w) = 0, \ \ \ x,y,z > 2$$
Let
\begin{align*}
\mathcal{L}(x,y,z,\lambda) &= f(x,y,z,w) + \lambda g(x,y,z,w)\\ &= (x-1)(y+3)(z-1)(w-2) + \lambda(x + y + w - 29)
\end{align*}
Then
$$\nabl... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $ \lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx$
Find the limit of
$$ \lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx$$
Let $$u= 1 +\frac{x}{n} \implies du =\frac{1}{n} dx \implies n \cdot du = dx$$
at $x=0$ $u=1$ and at $x=1$ $u=1+\frac{1}{n}$ ... | You forgot to change the limits of the integral. The integral limits should be $1$ and $1+\frac{1}{n}$.
In particular, $$n\int_{1}^{1+1/n} u^n du = \left(1+\frac{1}{n}\right)^n-\frac{n}{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Maclaurin Series $ \int \frac{\sin(x)}{5x} $ I am supposed to evaluate the indefinite integral as an infinite series centered at $ x=0 $ and give the first five non-zero terms of the series.
$ \int \frac{\sin(x)}{5x} $
Here is what I have done so far:
$ g(x) = \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)... | So, what happened was when I took the indefinite integral, to cover all potential functions which satisfy the indefinite integral, + C needs to be included.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong? The problem says:
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$
I solved it in the following way: $$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1... | $$\cot x =\frac{-7}{24}=\frac{\cos x}{\sin x}$$
$$\frac{49}{576}=\frac{\cos^2 x}{1-\cos^2 x}$$
It gives $$\cos x=\frac{-7}{25}$$
And
$$\cos \frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}=\frac{3}{5}=0.6 $$
Hence (A) is correct
EDIT:
After concerning with experts , I concluded that $\cos \frac{x}{2} $ is negative and hence (D... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Determine IFS of the von Koch Curve I am (still) reading Falconer's Fractal Geometry book and I have reached the section on Iterated Function Systems (IFSs). He defines the IFS for the middle third Cantor set:
as $S_{1}(x)=\frac{1}{3}x$, $S_{2}(x)=\frac{1}{3}x+\frac{1}{3}$ since, at each iteration of the construction,... | TL;DR version: you are not taking rotation into account. This is basically irrelevant in $\mathbb{R}$, but makes a huge difference in $\mathbb{R}^d$, where $d\ge 2$.
In Detail:
A contracting similitude can be characterized as a map
$$ S : \mathbb{R}^{d} \to \mathbb{R}^d $$
which takes the form
$$ S(\vec{x}) = c U \ve... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the exterma of $x^2+y^2+z^2-yz-zx-xy$ s.t. $x^2+y^2+z^2-2x+2y+6z+9=0$ using Lagrange's multiplier,
Using Lagrange's multiplier method, obtain the maxima and minima of $$x^2+y^2+z^2-yz-zx-xy$$ subject to the condition $$x^2+y^2+z^2-2x+2y+6z+9=0$$
My attempt:
I formed the expression
$$F=x^2+y^2+z^2-yz-zx-xy+\la... | You already got the system of equations. So the solution simply follows from there.
You should get $(x,y,z)=(2,-1,-4)$ and $(x,y,z)=(0,-1,-2)$ which gives you the maximum (27) and the minimum (3), respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.