Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Euclidean Geometry challenge.
Can someone help me on this one?
I have found that $\frac{1}{(x+1)^2}+1=\frac{1}{x^2}$, but I can't solve the fourth degree equation that comes with it. There must be a easier way!
| here is partial answer:
we need to solve $(1 + \frac{1}{x})^2 = 1 + (1+x)^2$ on clearing you get the quartic $$ 0= f(x) = x^4 + 2x^3 + x^2 -2x - 1 = g(x) + h(x), \text{ where } g(x) = x^4 + x^2 - 1, h(x) = 2x^3 - 2x.$$
we are only interested in the solution $0 < x < 1.$ we can find a lower bound for $x$ this way. $g(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sqrt x$ is Lipschitz on $[1, \infty)$ Prove that $\sqrt x$ is Lipschitz on $[1, \infty)$
I want to show that $|f(x) - f(y)| \leq L |x - y|$
So $|\sqrt x - \sqrt y| = \frac{|x - y|}{\sqrt x + \sqrt y} \leq \frac{1}{2}|x - y|$.
I can see that to get $|\sqrt x - \sqrt y| = \frac{|x - y|}{\sqrt x + \sqrt y}$... | We know $x,y \in [1, \infty)$ thus both $x,y \geq 1$.
Now we can see $$\sqrt{x} + \sqrt{y} \geq 1 + 1 =2$$
and so we know $$\frac{1}{\sqrt{x} + \sqrt{y}} \leq \frac{1}{2}$$
because both values are positive.
Hence all that remains for us is to multiply through by $|x-y|$ and we get $$\frac{|x-y|}{\sqrt{x} + \sqrt{y}} \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Defining the equation of an ellipse in the complex plane
Usually the equation for an ellipse in the complex plane is defined as $\lvert z-a\rvert + \lvert z-b\rvert = c$ where $c>\lvert a-b\rvert$. If we start with a real ellipse, can we define it in the manner below?
For $x,y,h,k,a,b\in\mathbb{R}$ such that $a,b\ne... | Your answer is correct, in the terms you are using. But note that in the standard equation
$$|z-a|+|z-b|=c\ ,$$
it is implicitly assumed that $z=x+iy$. Since your $z$ has a different form from this, you have obtained a different answer.
If you wish to plot your equation in the complex plane, you will have to rescale ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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Prove that a recurrence relation (containing two recurrences) equals a given closed-form formula. Prove that $a_n = 3a_{n-1} - 2a_{n-2} = 2^n + 1$ , for all $n \in \mathbb{N}$ , and $a_1 = 3$ , $a_2 = 5$ , and $n \geq 3$
Basis:
$a_1 = 2^1 + 1 = 2 + 1 = 3$ $\checkmark$
$a_2 = 2^2 + 1 = 4 + 1 = 5$ $\checkmark$
Inductive ... | Inductive Hypothesis:
$a_k = 2^k + 1$ , where $n=k$ , to show:
$$\begin{align} \ a_{k+1} & = 2^{k+1} + 1 \\ \end{align}$$
Inductive Step:
$$\begin{align}
\ a_{k+1} & = 3a_{(k+1)-1} - 2a_{(k+1)-2}\\
& = 3a_k - 2a_{k-1} \\
& = 3(2^k + 1) - 2(2^{k-1} + 1) \\
& = 3(2^k) + 3 - 2(2^{k-1}) - 2 \\
& = (2^k + 2^k + 2^k) - (2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1127212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$f(xy)=\frac{f(x)+f(y)}{x+y}$ Prove that $f$ is identically equal to $0$ For all $x,y\in\mathbb{R}$. also $f : \mathbb{R} → \mathbb{R}$ and $x+y\not=0$.
My attempt:
I restated it as
$a[x^2 y^2 (\frac{x}{y}+\frac{y}{x}-\frac{1}{y^2}-\frac{1}{x^2})] +
b[xy(x+y-\frac{1}{y}-\frac{1}{x})] + c [x+y-2]=0$
because of $f(xy)... | Setting $(x, y) = (1, 0)$ gives $f(1) = f(1) + f(0)$; that is, $f(0) = 0$. For $x\not =0$, setting $y = 0$ in the given equation shows that $(f(x) + f(0))/x = f(x)/x$ vanishes; that is, $f(x) = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$x^2+y^2+9=3(x+y)+xy$ Find all pairs of real $x,y$ that meet this equation $\frac{(x-y)^2}{(y-3)(3-x)} = 1$
That was my attempt, I can't think of anything else here. I'd prefer a hint
| Shift variables to eliminate the linear terms. $x=u+a$ and $y=v+b$, with $a,b$ yet-to-be-determined constants.
$$
\begin{align}
x^2+y^2+9&=3(x+y)+xy\\
(u+a)^2+(v+b)^2+9&=3(u+a+v+b)+(u+a)(v+b)\\
u^2+2au+a^2+v^2+2bv+b^2+9&=3u+3a+3v+3b+uv+av+bu+ab
\end{align}
$$
We would like to be able to cancel the the terms that are li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Solving the inequality $(x^2+3)/x\le 4$ This is the inequality
$$\left(\frac {x^2 + 3}{x}\right) \le 4 $$
This is how I solve it
The $x$ in the left side is canceled and $4x$ is subtracted from both sides.
$$\not{x} \left (\frac {x^2+3} {\not{x}}\right) \le 4x $$
$$ x^2+3 - 4x \le 4x - 4x $$
$$x^2 -4x + 3 \le 0 $$
Then... | To deal away with fractions, we can multiply both sides of $\left(\frac {x^2 + 3}{x}\right) \le 4$ with $x^2>0$ (here, $x\neq 0$):
$$
(x^2+3)x\leq 4x^2\iff x^3+3x-4x^2\leq 0\iff x(x-1)(x-3)\leq 0.
$$
We want the product of 3 factors above to be nonpositive so there is an odd number of nonnegative factors among $x-3<x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
} |
find the length of rectangle based on area of frame A rectangular picture is $3$ cm longer than its width, $(x+3)$. A frame $1$ cm wide is placed around the picture. The area of the frame alone is $42 \text{cm}^2$. Find the length of the picture.
I have tries:
$(x+6)(x+3) = 42 \\
x^2+3x+6x+12 = 42 \\
x^2+9x+12x-42 = 0... | Area $A$ of the $1$-cm-wide frame in cm$^2$ is $2\times $ height $+ 2\times $ width $+ 4$ (for the corners).
$A = 2 (x+3) + 2(x+6)+4 = 4x+22 = 42$
$\implies x=5$
The length is $(x+6) = 11$ cm
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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The inequality. Regional olympiad 2015 Let $a, b, c$ - the positive real numbers, and $ab+bc+ca=1$
Prove that $\sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}+\sqrt{c+\frac{1}{c}} \geqslant 2(\sqrt{a}+\sqrt{b}+\sqrt{c})$
Probably, we should use these facts:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{abc}$
$(a+b+c)^2 = a... | We have
$$
\sum_{cyc} \sqrt{a+\frac{1}{a}}=\sum_{cyc} \sqrt{a+\frac{ab+bc+ca}{a}}=\sum_{cyc} \sqrt{a+\frac{bc}{a}+b+c}\ge\sum_{cyc} \sqrt{2\sqrt{bc}+b+c}=\sum_{cyc} \sqrt{\left(\sqrt b+\sqrt c\right)^2}=\sum_{cyc} \left(\sqrt b+\sqrt c\right)=2\left(\sqrt a+\sqrt b+\sqrt c\right)
$$
as desired.
Note: The inequality is... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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How many three digit numbers are not divisible by 3, 5 or 11? How many three digit numbers are not divisible by 3, 5, or 11?
How can I solve this?
Should I look to the divisibility rule or should I use, for instance,
$$
\frac{999-102}{3}+1
$$
| Assuming that you mean by either $3$ or $5$ or $11$, use inclusion/exclusion principle:
*
*Amount of numbers with at most $3$ digits that are not divisible by $3$ or $5$ or $11$:
$999-\lfloor\frac{999}{3}\rfloor-\lfloor\frac{999}{5}\rfloor-\lfloor\frac{999}{11}\rfloor+\lfloor\frac{999}{3\times5}\rfloor+\lfloor\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
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exercise: ordinary differential equations I am struggling with an exercise. Can you please give me a hint?
Exercise:
Show that the solution curves of the differential equation:
$\frac{dy}{dx}=-\frac{y(2x^3-y^3)}{x(2y^3-x^3)}$, are of the form $x^3+y^3=3Cxy$.
I tried the substitution $u=y/x \rightarrow y=xu, \frac{dy}{d... | Use partial fractions
$$\begin{align}
\frac{2u^{3} - 1}{u^{4} + u} &= \frac{2u^{3} - 1}{u(u^{3} + 1)} \\
&= \frac{A}{u} + \frac{Bu^{2} + Cu + D}{u^{3} + 1} \\
\implies 2u^{3} - 1 &= (A + B)u^{3} + Cu^{2} + Du + A \\
\end{align}$$
Equating coefficients
$$\begin{align}
A &= -1 \\
B &= 3 \\
\end{align}$$
Hence
$$\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Proving that there are at least $n$ primes between $n$ and $n^2$ for $n \ge 6$ I was thinking about Paul Erdos's proof for Bertrand's Postulate and I wondered if the basic argument could be used to show that there are more than $n$ primes between $n$ and $n^2$.
Is this approach valid? Is there a better approach?
Here'... | We have the following result1:
(Chebyshev, 1850) For any $X>30$, we have: $$\pi(X)\cdot\frac{\log
X}{X} \in (A,B),$$ where
$A=\log\left(\frac{2^{\frac{1}{2}}3^{\frac{1}{3}}4^{\frac{1}{4}}}{30^{\frac{1}{30}}}\right)\approx
0.946$ and $B=\frac{6}{5}A\approx 1.135$.
It follows that for any $X>30$ we have:
$$ \pi(X^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Is $4 \times 6$ defined as $4 + 4 + 4 + 4 + 4 + 4$ or $6 + 6 + 6 + 6$? There are long debates among Indonesian netizens about this http://www.globalindonesianvoices.com/15785/is-4x6-the-same-as-6x4-this-primary-school-math-made-controversy-in-social-media/
| Let $A \times B$ mean the addition of $A$ repeated $B$ times.
Since multiplication is commutative in $\mathbb{R}$, that tells us that $A \times B = B \times A$. In other words, the addition of $A$ repeated $B$ times is the same as the addition of $B$ repeated $A$ times.
Thus, $4 \times 6 = 4+4+4+4+4+4 = 6 \times 4 = 6+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 4
} |
finding the minimum value of $\frac{x^4+x^2+1}{x^2+x+1}$ given $f(x)=\frac{x^4+x^2+1}{x^2+x+1}$.
Need to find the min value of $f(x)$.
I know it can be easily done by polynomial division but my question is if there's another way
(more elegant maybe) to find the min?
About my way: $f(x)=\frac{x^4+x^2+1}{x^2+x+1}=x^2-x+... | $$f(x)=\frac{x^4+x^2+1}{x^2+x+1}=x^2-x+1$$
$$f'(x) = 2x-1=0,x=\frac12$$
$$f''(\frac12)=2\gt0\text{ hence this is a local minimum by the second derivative test}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to factorize $n^5+n+1$? How to factorize $n^5+n+1$ ?
I think I should break $a^5$ and use a factorization formula. but how is it done?
| By trial and error, knowing that there must be a simple solution.
*
*Factoring $n^5+1$ leads nowhere.
$$n^5+n+1=(n+1)(n^4-n^3+n^2-n+1)+n.$$
*Factoring $n^5+n$ leads nowhere.
$$n^5+n+1=n(n+1)(n^3-n^2+n-1)+1.$$
*Factoring $n^5+n^2$ with an artifice fails, but shows some hope
$$n^5+n^2-n^2+n+1=n^2(n+1)(n^2-n+1)-n^2+n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ Having the following inequality
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$
To prove it for all natural numbers is it enough to show that:
$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$ or $\frac{1}{(n+1)^2}... | Here is one way...
$$\frac1{k^2} < \frac1{k(k-1)}= \frac1{k-1} - \frac1{k}$$
Now telescope to get
$$1+\sum_{k=2}^n\frac1{k^2} < 1+1-\frac1{n}< 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Argument of $3+4i$ for $z = \sqrt{3 + 4i}$, I am trying to put this in Standard form, where z is complex. I let $w = 3+4i$ and find that the modulus, $|w|=r$, is 5. I am having trouble solving for arg(w).
I find that $\tan^{-1}{\theta} = \frac{4}{3}$. However, this is not an angle well known. How do I find it?
| Note, we have $|w| = 5$. Let $\theta \in Arg(w)$ and then from your corresponding diagram of the triangle form my $w$, $\cos(\theta) = \frac{3}{5}$ and $\sin(\theta) = \frac{4}{5}$. Therefore, from $\sqrt{z} = \sqrt{z}\left( \cos(\frac{\theta}{2}) + i\sin(\frac{\theta}{2})\right )$, we essentially arrive at our answer.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how to evaulate: $\lim \limits_{x \to 0} \frac{\ln(1+x^5)}{(e^{x^3}-1)\sin(x^2)} $ How do I evaluate: $\lim \limits_{x \to 0} \frac{\ln(1+x^5)}{(e^{x^3}-1)\sin(x^2)} $ ?
according to Taylor's series, I did like this:
$$\lim \limits_{x \to 0} \frac{\ln(1+x^5)}{(e^{x^3}-1)\sin(x^2)}=\lim \limits_{x \to 0} \frac{x^5 - \f... | From Taylor's expansions at $x=0$, $\log(1+x^5) \approx x^5$, $e^{x^3}-1 \approx x^3$ and $\sin x^2 \approx x^2$, as $x \rightarrow 0$. Thus,
$$\lim \limits_{x \rightarrow 0} \frac{\log(1+x^5)}{(e^{x^3}-1)(\sin x^2 )} = \lim \limits_{x \rightarrow 0} \frac{x^5}{x^{3} x^{2}} =1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int{\frac{\sqrt[3]{x+8}}{x}}dx$ So I've tried solving the equation below by using $u=x+8$, and I get $\int{\frac{\sqrt[3]{u}}{u-8}}du$ which doesn't seem to lead anywhere, I've also tried taking the $ln$ top and bottom, but I don't know how to proceed. Any hints?
$$\int{\frac{\sqrt[3]{x+8}}{x}}dx$$
Update:... | you can start with
$$\begin{align}
u^3=x+8&\Rightarrow 3u^2du=dx\\
x=u^3-8\\
\int\frac{\sqrt[3]{x+8}}{x}dx&=\int\frac{\sqrt[3]{u^3}}{u^3-8}(3u^2du)\\
&=\int\frac{3u^3}{u^3-8}\\
&=\int\frac{3(u^3-8)+24}{u^3-8}du\\
&=\int3+\frac{24}{u^3-8}du
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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interpreting partitions of integers(generating functions) the question is: In $ f(x) = \left (\frac{1}{1-x} \right )\left (\frac{1}{1-x^2} \right )\left (\frac{1}{1-x^3} \right )$the coefficient of $x^6$ is 7. Interpret this result in terms of partitions of 6.
I checked the answer and it says the number of partition... | Writing each of the terms as power series, we have
$$\eqalign{f(x)
&=(1+x+x^2+\cdots)(1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots)\cr
&=(1+x^1+x^{1+1}+\cdots)(1+x^2+x^{2+2}+\cdots)(1+x^3+x^{3+3}+\cdots)\ .\cr}$$
Now if you look carefully you will see that you can multiply terms, one from each bracket, to get an $x^6$ term i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$
A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\g... | Set $b+c-a=x, a+c-b=y$ and $a+b-c=z$
Due to triangular inequality, $x,y,z$ are positive. Hence we can apply the AM-GM inequality.
We have $x+y=b+c-a+a+c-b=2c \implies c = \frac{x+y}{2}$
Similarly $a = \frac{y+z}{2}$ and $b = \frac{z+x}{2}$
Now we have,
$$\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}=\frac{y+z}{2x}+\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $m+\frac{4}{m^2}\ge3$ How to prove that $m+\frac{4}{m^2}\geq3$ for every $m>0$? I multiplied both sides by $m^2$ and finally got $m^3+4-3m^2\geq0$, yet don't know how to move further. Hint?
| If $m=1$, then
$$m + \frac{4}{m^2}= 1 + \frac{4}{1^2} = 1 + 4 = 5 \geq 3$$
If $m = 2$, then
$$m + \frac{4}{m^2} = 2 + \frac{4}{2^2} = 2 + 1 = 3 \geq 3$$
Consider now $m\geq 3$. Then, since $\frac{4}{m^2}$ is positive for all $m$, it is obvious that also $m + \frac{4}{m^2}$ is greater than $3$.
We can conclude that
$$m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1157949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
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Proof of a limit of a sequence I want to prove that $$\lim_{n\to\infty} \frac{2n^2+1}{n^2+3n} = 2.$$
Is the following proof valid?
Proof
$\left|\frac{2n^2+1}{n^2+3n} - 2\right|=\left|\frac{1-6n}{n^2+3n}\right| =\frac{6n-1}{n(n+3)} $ (because $n \in \mathbb N^+)$. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \... | Your “if and only if” are incorrect in
$$
\frac{6n-1}{n^2+3n}< \varepsilon \iff \frac{n+3}{n(n+3)} < \varepsilon \iff n > \frac{1}{\varepsilon}
$$
What you can do is
$$
n > \frac{6}{\varepsilon}\implies
\frac{6(n+3)}{n(n+3)}<\frac{6}{\varepsilon}\implies
\frac{6n-1}{n(n+3)}<\frac{6}{\varepsilon}
$$
Alternative proof
R... | {
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"url": "https://math.stackexchange.com/questions/1158558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Example of contour integration Could someone help me evaluate the following integral with contour integration ?
$$\int_{0}^{2\pi}\frac{d\theta}{(a+b\cos\theta)^2}.$$
Constraints are: $a>b>0$.
| We have:
$$ I = \int_{0}^{\pi/2}\frac{d\theta}{a+b\cos\theta}=\int_{0}^{\pi/2}\frac{d\theta}{a+b\,\frac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}}=2\int_{0}^{\pi/4}\frac{d\theta}{a+b\,\frac{1-\tan^2(\theta)}{1+\tan^2(\theta)}}$$
or just:
$$ I = 2\int_{0}^{1}\frac{dt}{(a+b)+(a-b)\,t^2}=\frac{2}{\sqrt{a^2-b^2}}\,\arctan\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1160331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that the equation $x^2+y^2=57$ has no integer solution. I am trying to prove that equation $x^2+y^2=57,\{x,y\}\in\mathbb{Z}$ has no integers solution.
In case the equation is of the form $ax^2+by^2=n,\text{where}~~ a\neq 1 ~~\text{or}~~ b\neq1
,\{a,b,n\}\in\mathbb{Z}$ .
Then applying $ax^2+by^2\pmod a=n \pmod a... | $$x^2+y^2\equiv 57\equiv 0\pmod {3}$$
Now, the only quadratic residues modulo $3$ are $0,1$, and since if $a,b\in\{0,1\}$, then $a+b\in\{0,1,2\}$, we know that both $a=0$ and $b=0$ must hold in order for $a+b\equiv 0\pmod{3}$ to be true, and so $$\begin{cases}x^2\equiv 0\pmod{3}\\ y^2\equiv 0\pmod {3}\end{cases}\iff \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1160730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find the inverse function of $y = g(x) = 6 x^3 + 7$: $g^{-1}(y) =?$ The question states,
Find the inverse function of $y = g(x) = 6 x^3 + 7$, $g^{-1}(y) =?$
I have tried setting the equation to $y$ and then solving for $x.$ This resulted in the answer $\dfrac{(x-7)^{1/3}}{6}$. This answer is incorrect. I have also tr... | You are close. Solving for $x$ in $y=6x^3+7$ goes like this:
$$y=6x^3+7 \implies y-7=6x^3 \implies \frac{y-7}{6} = x^3 \implies \left(\frac{y-7}{6}\right)^{1/3} = x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1162111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$ Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$
$$\begin{align}\frac{1}{n^4+n^2+1}&
=\frac{1}{n^4+2n^2+1-n^2}\\
&=\frac{1}{(n^2+1)^2-n^2}\\
&=\frac{1}{(n^2+n+1)(n^2-n+1)}\\
&=\frac{1-n}{2(n^2-n+1)}+\frac{1+n}{2(n^2+n+1)}\\
\end{align}$$
For $n={1,2,3}$ it is not givi... | Just completing Thomas Andrew's answer,
$$\sum_{n=1}^{N}\frac{1}{n^2+n+1}=4\sum_{n=1}^{N}\frac{1}{(2n+1)^2+3}=4\sum_{k=1}^{+\infty}\sum_{k=1}^{N}\frac{3^{k-1}}{(2n+1)^{2k}}\tag{1}$$
can be put in integral form by exploiting:
$$ \int_{0}^{+\infty}\frac{\sin(m x)}{m}\,e^{-\sqrt{3}\,x}\,dx = \frac{1}{3+m^2}.\tag{2}$$
That... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Long limit question If $$\lim_{x \to \infty}\frac{a(2x^3-x^2)+b(x^3+5x^2-1)-c(3x^3+x^2)}{a(5x^4-x)-bx^4+c(4x^4+1)+2x^2+5x}=1$$ then find the value of $a+b+c$. I have done such problems with $x$ tending to some finite amount. Infinity is creating some problems.
| Consider
$$
\lim_{x \to \infty}\frac{a(2x^3-x^2)+b(x^3+5x^2-1)-c(3x^3+x^2)}{a(5x^4-x)-bx^4+c(4x^4+1)+2x^2+5x}=1.
$$
rewrite this as
$$
\lim_{x \to \infty}\frac{x^3(2a+b-3c)+x^2(-a+5b-c)-b}{x^4(5a-b+4c)+2x^2+x(-a+5)+c}=1.
$$
As long as the $x^4$ term is in the denominator the limit cannot be $1$, so the coefficient of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$ How to calculate the following limit?
$$\lim\limits_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$$
| $$\lim_{x\to \infty} \Big(\frac{x^2+1}{x^2-1}\Big)^{x^2}=\lim_{x\to \infty} \Big(1+\frac{2}{x^2-1}\Big)^{\frac{x^2-1}{2}\cdot\frac{2x^2}{x^2-1}}=e^{\lim_{x\to \infty}\frac{2x^2}{x^2-1}}=e^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove... | You can use "Long Division of Polynomials"
$\frac {1-(\frac y x)^3}{1-(\frac y x)}=1+\frac y x+(\frac y x)^2$
Multiply both sides by $x^3$
$\frac {x^3-y^3}{1-(\frac y x)}=x(x^2+xy+y^2)$
Then you have
$x^3-y^3=(x-y)(x^2+xy+y^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 12,
"answer_id": 11
} |
Solving for real x and y If x and y be real numbers such that, $x^3 - 3x^2 + 5x = 1$ and $y^3 - 3x^2 + 5y = 5$; Find $(x + y)$ From an old Russian olympiad.
I tried to make the equations homogenous by substituting for $1 = x^3 - 3x^2 + 5x$ in the second equation for $5 * 1$, which didn't work even by repeated substitut... | I think you have somewrong,I have see a book with Russian olympiad problem
Let $x,y$ are real numbers, and
$$x^3-3x^2+5x=1,y^3-\color{#0a0}{\text{$3y^2$}}+5y=5$$
Find $x+y$
since
$$(x-1)^3+2(x-1)=-2$$
$$(y-1)^3+2(y-1)=2$$
since $f(x)=x^3+2x$ is odd function,and increaing on $R$
since
$$f(x-1)+f(y-1)=0\Longleftrightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Odd numbers expressible as a sum of 3 composite odd integers. How many positive odd integers between $0$ and $999$ inclusive can be
written as sum of $3$ odd composite positive integers?
I am trying to approach this by looking at equivalence classes $\mod 6$ but the calculations are a bit tedious.
| Let's start with integers equivalent to $3\pmod6$. The smallest is obviously $27=9+9+9$. Every following odd multiple of $3$ also works as it can be written as $9+9+3(2n+3)$. $27=6(4)+3$ and $999=6(166)+3$, so we have $163$ odd multiples of $3$.
The first odd number not a multiple of $3$ is $5(5)=25\equiv1\pmod6$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $\frac {1}{2+a} + \frac {1}{2+b} + \frac {1}{2+c} = 1$, prove $\sqrt{ab} + \sqrt{ac} + \sqrt{bc} \leq 3$ Let $a,b,c$ be non-negative numbers such that $$\frac {1}{2+a} + \frac {1}{2+b} + \frac {1}{2+c} = 1.$$
Prove that $ \sqrt{ab} + \sqrt{ac} + \sqrt{bc} \leq 3 $.
| The condition gives:
$$\frac32-1 = \sum_{cyc} \left(\frac12-\frac1{2+a} \right)$$
$$\implies 1 = \sum_{cyc} \frac{a}{2+a} \ge \frac{(\sqrt a+\sqrt b+\sqrt c)^2}{a+b+c+6} \quad \text{by Cauchy-Schwarz inequality}$$
$$\implies 3 \ge \sqrt{ab}+\sqrt{bc}+\sqrt{ca}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
2-adic expansion of (2/3) I have been asked in an assignment to compute the 2-adic expansion of (2/3). It just doesn't seem to work for me though. In our definition of a p-adic expansion we have $x= \sum_{n=0}^{\infty}a_np^n$ with $0 \leq a_0<p$.
So I use the same method as for similar questions which seems to work fin... | I get $2^1+2^2+2^4+2^6+\ldots$, with no $2^5$. In the $2^4$ place, you have $3\times1+1=4=0$ carry $2$, so you need $0\times 2^5$ to give $0$ carry $1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove that $\frac{\sqrt{5(160ab+8b+13)}-5}{20}$ is not an integer
Prove that $\frac{\sqrt{5(160ab+8b+13)}-5}{20}$ is not an integer where $a$ and $b$ are positive integers.
One of the roots of the equation $10x^2+5x-1-(20a+1)b=0$ is $\frac{\sqrt{5(160ab+8b+13)}-5}{20}$. Wolframalpha says that this equation has... | Take $b=14$ and $a=3$. In this case, $\sqrt{5(160ab+8b+13)} = 185$.
This gives $x=\frac{185-5}{20}=9$ as a valid root. So there are integer solutions for $x$ when $a$ and $b$ are positive integers.
If Wolfram told you there were no integer solutions, either Wolfram is wrong, or perhaps you made a typo when entering y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx = \frac{8 \pi ^3}{81 \sqrt{3}}$ I have found myself faced with evaluating the following integral: $$\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx. $$
Mathematica gives a closed form of $8 \pi ^3/(81 \sqrt{3})$, but I have no idea how to arrive at this close... | Note
$$ \int_0^1x^m(\ln x)^2dx=\frac{2}{(m+1)^3} $$
So
\begin{eqnarray}
I&=&\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1}dx\\
&=&\int_0^1\frac{(1-x)(\ln x)^2}{1-x^3}dx\\
&=&\int_0^1\sum_{n=0}^\infty(1-x)x^{3n}(\ln x)^2dx\\
&=&2\sum_{n=0}^\infty\left(\frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3}\right)\\
&=&2\sum_{n=-\infty}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Differentiate $\,y = 9x^2 \sin x \tan x:$ Did I Solve This Correctly? I'm posting my initial work up to this point.
Criticism welcomed!
Using the formula $(fgh)' = f'gh+fg'h + fgh'$, differentiate$$y = 9x^2\sin x \tan x$$
$$\begin{align} y' &= 9\frac d{dx}(x^2)\sin x \tan x + 9x^2 \frac d{dx} (\sin x) \tan x + 9x^2... | You have a parenthesis error
Line 3 should read
$$
9(2x)\sin x \tan x +9x^2(-\cos x)\tan x + 9x^2 \sin x \sec^2 x
$$
You can actually factor out $9x\sin x$ because $\cos x \tan x = \sin x$.
Continuing
$$
9x \sin x (2\tan x - x + x \sec^2 x) \\
9x \sin x (2\tan x +x (\sec^2 x - 1) )\\
$$
Then you can use some trig ident... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve $\int\frac{x}{\sqrt{x^2-6x}}dx$ I need to solve the following integral
$$\int\frac{x}{\sqrt{x^2-6x}}dx$$
I started by completing the square,
$$x^2-6x=(x-3)^2-9$$
Then I defined the substitution variables..
$$(x-3)^2=9\sec^2\theta$$
$$(x-3)=3\sec\theta$$
$$dx=3\sec\theta\tan\theta$$
$$\theta=arcsec(\frac{x-3}{3})$... | I prefer to apply hyperbolic substitution.
More precisely, if we let $x - 3 = 3\cosh(y)$, we arrive at
\begin{align*}
\int\frac{x}{\sqrt{x^{2} - 6x}}\mathrm{d}x & = \int\frac{x}{\sqrt{(x - 3)^{2} - 9}}\mathrm{d}x\\\\
& = 3\int\frac{(1 + \cosh(y))\sinh(y)}{\sqrt{\cosh^{2}(y) - 1}}\mathrm{d}y\\\\
& = 3\int\mathrm{d}y + 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How to solve $\int\sqrt{1+x\sqrt{x^2+2}}dx$ I need to solve
$$\int\sqrt{1+x\sqrt{x^2+2}}dx$$
I've chosen the substitution variables
$$u=\sqrt{x^2+2}$$
$$du=\frac{x}{\sqrt{x^2+2}}$$
However, I am completly stuck at
$$\int\sqrt{1+xu} dx$$
Which let me believe I've chosen wrong substitution variables.
I've then tried lett... | I thought about Wolfram and Maple too, needs a little help. But I was wrong.
$$\int {\sqrt {1 + x\sqrt {{x^2} + 2} } } dx = \int {\sqrt {1 + \frac{1}{2}\sqrt {{x^2} + 2} } } 2xdx$$
\begin{gathered}
y = {x^2} + 2 \\
dy = 2xdx \\
\end{gathered}
\begin{gathered}
\int {\sqrt {1 + \frac{1}{2}\sqrt y } } dy = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
My Attempt: I start with direct proof.
Let $a,b,c$ be consecutive integers and $a< b < c $, there exists a ... | Suppose$$(k-1)^3+k^3=(k+1)^3$$ $$2k^3-3k^2+3k-1=k^3+3k^2+3k+1$$ $$k^3-6k^2-2=0$$ $$k^2(k-6)=2 ,\,\,\,\,\text{where $k$ is an integer. }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$\sup$ and $\inf$ of this set These are exercises from my textbook, and I am not sure if the solutions are correct or not.
Given a set $B = \{\frac{n}{2n+1} : n \in \mathbb{N} \}$
Find the $\sup$ and $\inf$ of $B$, and maxima and minima if they exist.
The solutions give $\sup B = \frac{1}{2}$ and $\inf B = \frac{1}{3}$... | We have
$$
0.333... = \frac{1}{3} < \frac{1}{2} = 0.5
$$
so the $\sup$ is right.
Furthermore, $\frac{1}{3} < \frac{1}{2}$ also implies that the infimum is at most $\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$ How do we compute this integral ?
$$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$$
I have tried partial fraction but it is quite hard to factorize the denominator. Any help is appreciated.
| Let $I$ be the integral given by
$$I= \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$$
Now, make the substitution $x=u^{-1}$. Then $dx=-u^{-2}du$ and the limits of integration over $u$ extend from $\infty$ to $1$. Thus, we may write
$$
\begin{align}
I & = \int_{\infty}^1 \frac{1+u^{-6}}{1-u^{-2}+u^{-4}-u^{-6}+u^{-8}}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me.
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1... | Multiply numerator and denominator by $1 - \sin x$. So that $\displaystyle\int_0^\pi \dfrac{1-\sin x}{1+\sin x} \cdot \dfrac{1-\sin x}{1-\sin x}dx $
$$ = \displaystyle\int_0^\pi \dfrac{1-2\sin x+ \sin^2x}{\cos^2x}dx = \int_0^\pi \sec^2x - 2\dfrac{\sin x}{\cos^2x} + \tan^2x dx$$
We know that $\tan^2x + 1 = \sec^2x$. So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 9,
"answer_id": 8
} |
Evaluate the following limit without L'Hopital I tried to evaluate the following limits but I just couldn't succeed, basically I can't use L'Hopital to solve this...
for the second limit I tried to transform it into $e^{\frac{2n\sqrt{n+3}ln(\frac{3n-1}{2n+3})}{(n+4)\sqrt{n+1}}}$ but still with no success...
$$\lim_{n... | Hints :
$$ \frac{2n^2-3}{-n^2+7} = \frac{2 - \frac{3}{n^2}}{-1+\frac{7}{n^2}},$$
and
$$ \frac{3^n-2^{n-1}}{3^{n+2}+2^n} = \frac{1-\frac{1}{2}\left( \frac{2}{3} \right)^n}{3^2+\left( \frac{2}{3} \right)^n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate $\sum_{n=2}^\infty\frac{(-1)^n}{n^2-n}$ How would you go about evaluating:$$\sum_{n=2}^\infty\frac{(-1)^n}{n^2-n}$$
I split it up to $$\sum_{n=2}^\infty\left[(-1)^n\left(\frac{1}{n-1}-\frac{1}{n}\right)\right]$$
but I'm not sure what to do from here. If the $(-1)^n$ term wasn't there then it would be a... | \begin{align}
& \sum_{n=2}^\infty (-1)^n\left(\frac{1}{n-1}-\frac{1}{n}\right) \\[10pt]
= {} & \underbrace{\left(1-\frac 1 2 \right) - \left( \frac 1 2\right.}_{=\,0} - \left.\frac 1 3 \right) + \left( \frac 1 3 - \frac 1 4 \right) - \left( \frac 1 4 - \frac 1 5\right) + \left( \frac 1 5 - \frac 1 6 \right) - \cdots \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solution of $y''+xy=0$ The differential equation $y''+xy=0$ is given.
Find the solution of the differential equation, using the power series method.
That's what I have tried:
We are looking for a solution of the form $y(x)= \sum_{n=0}^{\infty} a_n x^n$ with radius of convergence of the power series $R>0$.
Then:
$$y'(x)... | Follow-up from my comment, every $a_{n+3}$ can be expressed in terms of $a_n$.
$$a_{n+3} = -\frac{a_{n}}{(n+3)(n+2)}$$
Since $a_2 = 0$, all coefficients with $n = 3k + 2$ will be $0$
The rest will have to based on the initial conditions of $a_0 = y(0)$ and $a_1 = y'(0)$
For $n = 3k$
$$a_3 = -\frac{a_0}{3\cdot2}$$
$$a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
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How to calculate $\int \frac{x^2 }{(x^2+1)^2} dx$? I'm trying to calculate $\int \frac{x^2 }{(x^2+1)^2} dx$ by using the formula:
$$ \int udv = uv -\int vdu $$
I supposed that $u=x$ s.t $du=dx$, and also that $dv=\frac{x}{(x^2+1)^2}dx$, but I couldn't calculate the last integral. what is the tick here?
the answer m... | First, you used parts: with $u = x$ and $dv = \frac{x}{(x^2+1)^2}dx$, you got $$\int\frac{x^2}{(x^2+1)^2}dx = xv - \int v\,dx.\tag{1}$$
From here, I suggested the substitution $t = x^2$ in order to integrate $dv$ and get $v$.
Doing this, we see that $dt = 2x\,dx$, so $x\,dx = \frac{1}{2}dt$.
Thus $$v = \int dv = \int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Parabolic sine approximation Problem Find a parabola ($f(x)=ax^2+bx+c$) that approximate the function sine the best on interval [0,$\pi$].
The distance between two solutions is calculated this way (in relation to scalar product): $\langle u,v \rangle=\int_0^\pi fg$.
My (wrong) solution I thought that I would get the s... | The answer has already been given.
In addition, the numerical solving and the graph are shown below, in order to compare the results.
Note that the result corresponds to the mean least square absolute deviation. A small modification of the method and formulas would give the result corresponding to the mean least squa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How can I express the ration of double factorials $\frac{(2n+1)!!}{(2n)!!}$ as a single factorial? How can I change the double factorial of $$\frac{(2n+1)!!}{(2n)!!}$$ to single factorial?
| $$(2n+1)!!=(2n+1)(2n-1)(2n-3) \cdots (3) (1)$$
$$=\frac{(2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4) \cdots (4)(3)(2) (1)}{(2n)(2n-2)(2n-4) \cdots (4)(2)}=\frac{(2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4) \cdots (4)(3)(2) (1)}{(2)(n)(2)(n-1)(2)(n-2) \cdots (2)(2)(2)(1)}$$
$$=\frac{(2n+1)!}{(2^n)n!}$$
Similarly,
$$(2n)!!=(2n)(2n-2)(2n-4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Derivative Chain Rule $f(x) = (3x^2+2)^2 (x^2 -5x)^3$ I'm learning chain rule in derivative,
and I don't understand this in the example.
$$f(x) = (3x^2+2)^2 (x^2 -5x)^3\\$$
\begin{align}
f'(x)&= (3x^2 +2)^2[3(x^2 -5x)^2(2x-5)] + (x^2-5x)^3[2(3x^2+2)(6x)]
\\&=3(3x^2 +2)(x^2-5x)^2[(3x^2+2)(2x-5)+4x(x^2-5x)]
\\&= 3(3x^2 ... | Hint:
Put $(x^2- 5x)^2$, $(3x^2 + 2)$ and $3$ in evidence. Then
$$\begin{align}f'(x)&= \color{green}{(3x^2 + 2)}^2[\color{red} 3\color {#05f}{(x^2 - 5x)^2} (2x - 5)] + \color {#05f}{(x^2 - 5x)^3}[2\color{green}{(3x^2 + 2)}\color {red} 6 x]\\&=\color{red}3\color{green}{(3x^2 + 2)}\color{#05f}{ (x^2 - 5x)^2} \,[(3x^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Integral of $\log(\sin(x)) \tan(x)$ I would like to see a direct proof of the integral $$\int_0^{\pi/2} \log(\sin(x)) \tan(x) \, \mathrm{d}x = -\frac{\pi^2}{24}.$$ I arrived at this integral while trying different ways to evaluate $\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + ...$ and found this value using the know... | Note that
$$\begin{align}\int_0^{\pi/2} \log(\sin x) \tan x \, \mathrm{d}x &= \frac1{2}\int_0^{\pi/2} \log(\sin^2 x ) \tan x \, \mathrm{d}x \\ &= \frac1{2}\int_0^{\pi/2} \frac{\log(\sin^2 x ) \sin x}{\cos x} \, \mathrm{d}x \\ &= \frac1{2}\int_0^{\pi/2} \frac{\log(1-\cos^2 x ) \cos x \sin x}{\cos^2 x} \, \mathrm{d}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational
How can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?
I know that $\sqrt{3}, \sqrt{5}$ and $\sqrt{7}$ are all irrational and that $\sqrt{3}+\sqrt{5}$, $\sqrt{3}+\sqrt{7}$, $\sqrt{5}+\sqrt{7}$ are all irrational, too. But how can I prove that $\sqrt{... | Suppose $\sqrt{3}+\sqrt{5}+\sqrt{7} = r$ for some rational $r$. Then, $$(\sqrt{3}+\sqrt{5})^2 = (r-\sqrt{7})^2 \implies 8+2\sqrt{15} = 7+r^2-2r\sqrt{7}$$
So, $$1-r^2+2\sqrt{15} =-2r\sqrt{7}$$
Let $1-r^2 = k$, which will be a rational number. So,
$$(k+2\sqrt{15})^2 = k^2+ 60+4k\sqrt{15} = 28r^2$$
The LHS is irrational w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
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What is the name of the following matrix "product" in which elements are not multiplied? Assume that I am given the following matrices:
$A = \begin{bmatrix}
a_{1,1} & \dots & a_{1,n_a} \\
\vdots & \ddots & \vdots \\
a_{m_a,1} & \dots & a_{m_a,n_a}
\end{bmatrix}$ and $B =
\begin{bmatrix}
b_{1,1} & \dots & b_{1,n_b} \\
\... | In terms of operations that already exist, we can write this as the block matrix
$$
\pmatrix{A \otimes X\\ X \otimes B}
$$
Where $X$ is the row-vector $(1,\dots,1)$ and $\otimes$ denotes the Kronecker product.
A nice shorthand for this operation (if the above is not sufficiently compact) could be $\frac{A}{B}$, which h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to find the limit of the following series based on a new operation? Lets define an operation as $$a*b=a+b+ab$$ Now I need to find the following limit $$\lim_{n\to\infty}\frac{1}{1}*\frac{1}{4}*\frac{1}{9}*\frac{1}{16}\cdots*\frac{1}{n^2}$$
Now I observed the series and found that it is nothing but the sum of the m... | If
$$x_n = \frac{1}{1} * \frac{1}{4} * \frac{1}{9} * \cdots * \frac{1}{n^2}$$
then
$$x_n = x_{n-1} * \frac{1}{n^2} = (1 + x_{n-1})\left(1 + \frac{1}{n^2}\right) - 1,$$
so
\begin{align}1 + x_n &= (1 + x_{n-1})\left(1 + \frac{1}{n^2}\right)\\
& = (1 + x_{n-2})\left(1 + \frac{1}{(n-1)^2}\right)\left(1 + \frac{1}{n^2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$ How do I show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$? Someone says I should use the rational root test, but I don't exactly know how that applies. Thanks for any input.
| We have that $p(x)=x^4-10x^2+1$ splits as $(x^2-2)(x^2+2)$ over $\mathbb{F}_5$, hence $p(x)$ has no linear factors over $\mathbb{Q}$ and by assuming it splits over $\mathbb{Q}$, since $p(x)=p(-x)$, it must satisfy:
$$ x^4-10x^2+1 = (x^2+ax+b)(x^2-ax+b) $$
with $b^2=1$ and $a^2-2b=10$, contradiction.
With a little rev... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 2
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Find $p$ for which all solutions of system/equation are real There is system of $5$ equations
$$
a+b+c+d+e = p; \\
a^2+b^2+c^2+d^2+e^2 = p; \\
a^3+b^3+c^3+d^3+e^3 = p; \\
a^4+b^4+c^4+d^4+e^4 = p; \\
a^5+b^5+c^5+d^5+e^5 = p, \\
\tag{1}
$$
where $p\in\mathbb{R}$.
One can prove (like here) that $a,b,c,d,e$ are roots of eq... | In order for the roots to transition from real to complex, they must pass through a double-root first - that is, the gradient must be zero at those points. By examining how many roots are present for one solution between each pair, we can determine the regions for which all solutions are real.
$$
5x^4-4\binom{p}{1}x^3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of series:$ 1+(1+2+4)+(4+6+9)+(9+12+16)+.....+(361+380+400)$ How to find the sum of series
$$ 1+(1+2+4)+(4+6+9)+(9+12+16)+.....+(361+380+400)$$
I have the problem in finding the nth term of the series. Please suggest me the procedure.
| Your terms look something of the form:
$$a_0 = 1, a_1 = (1+2+4) = 1^2 + 1\cdot 2 + 2^2$$
$$a_2 = (4 + 6 + 9) = 2^2 + 2\cdot 3 + 3^2$$
$$a_{19} = (19^2 + 19\cdot 20 + 20^2)$$
$$\sum_{j=0}\big[j^2 + j(j+1) + (j+1)^2\big]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Alternative form of a Trigonometrics Expression Express $8\sin\theta \cos\theta - 6 \sin^2 \theta$ in the form $R \sin(2\theta + \alpha) + k$
Edit: I am sorry, I thought it was a somewhat interesting question. I shall let you know of the progress I have made. I first tried to rewrite it in the form $R \sin (\theta -\al... | $$\begin{align}8\sin(\theta)\cos(\theta)-6\sin^2(\theta)&=4\sin(2\theta)-3(1-\cos(2\theta))\\&=5\left(\frac{4}{5}\sin(2\theta)-\frac{3}{5}\cos(2\theta)\right)-3\end{align}$$
Notice that $\left(\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2=1$, therefore there is an angle $s=\arccos\left(\frac{4}{5}\right)$ such that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Prove that for all naturals $n \ge 6$ there is a set of $n$ positive naturals, $a_1$ to $a_n$ such that $\sum_{i=1}^n \left(\frac{1}{a_i}\right)^2 =1$ I don't know how to prove this. I know that $\{2, 2, 2, 2\}$ is a set for $n = 4$, since $\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\rig... | Extending your observation, you can replace $a$ in any list with $2a, 2a, 2a, 2a$, to get a new valid list. Hence if it is possible for $n$ then it is possible for $n + 3$. Then you only need to prove it for $6, 7, 8$.
Another nice fact:
$$\frac{1}{4} = \frac{1}{9} + \frac{1}{9} + \frac{1}{36}$$
So if $2$ is in your li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solving a 2nd-order linear recurrence with variable coefficients
Question: I am seeking a general closed-form evaluation for the following definite integral.
For nonnegative real parameters $a,b,z\in\mathbb{R}^{+}\land a<z\le b$ and integer index $n\in\mathbb{Z}$, define the auxiliary function $f_{n}{\left(a,b;z\righ... | I think your recurrence may be solved in terms of a generating function
$$G(x) = \sum_{n=0}^{\infty} F_n x^n$$
so that
$$F_n = \frac1{n!}G^{(n)}(0) $$
Because all of the coefficients are no worse than linear, the recurrence relation turns into a first-order differential equation in $G$ by using the relation
$$\sum_{n=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Antiderivative fraction I'm would like to find the antiderivative of
$$\frac{x^2}{(x-1)^5}$$
I tried without using partial fractions but I did not manage, so I started a lesson on partial fractions but I do not succeed in proving $$\frac{x²}{(x-1)^5}=\frac{1}{(x-1)^3}+\frac{2}{(x-1)^4}+\frac{1}{(x-1)^5}$$
Could you tel... | Indeed you are correct that,
$$\frac{x^2}{(x-1)^5} = \frac{1}{(x-1)^3} + \frac{2}{(x-1)^4} + \frac{1}{(x-1)^5}$$
One can integrate each of these terms in turn. I will do the first to help. Let $u=x-1$ and $du=dx$ then,
$$ \int\frac{1}{(x-1)^3} dx = \int u^{-3} du = -\frac{1}{2}u^{-2} = -\frac{1}{2}\frac{1}{(x-1)^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Norm of the sum of two vectors This problem has two parts.
Part a): $x$ and $y$ are vectors. If $||x|| = 7, ||y|| = 11$, what is the smallest value possible for $||x+y||$? (Note: the || || denotes the norm of a vector).
This is what I have tried so far: I put vector $x$ equal to $\begin{pmatrix} a \\ b \end{pmatrix}$... | Your problem is equivalent to finding the smallest value of $\|x-y\|$, which you can obtain by the following triangle inequality
\begin{equation}
|\|x\|-\|y\|| \leq \|x-y\| \leq \|x\|+\|y\|.
\end{equation}
This will give you that the smallest value is $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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how to factorize $x^2+10yz-2xz-2xy-3y^2-3z^2$? How to factorize
$$x^2+10yz-2xz-2xy-3y^2-3z^2$$
It is expanded and we should make them into parts and factorize each part individually.
the last answer is $$(x+y-3z)(x-3y+z)$$
but how to get it ?
| One way to do it is
\begin{align*}
x^2+10yz-2xz-2xy-3y^2-3z^2&=(x^2+y^2+z^2-2xz-2xy+2yz)-(4y^2+4z^2-8yz)\\
&=(x-y-z)^2-(2y-2z)^2\\
&=(x+y-3z)(x-3y+z)
\end{align*}
where going from the second to the third line, we have used $a^2-b^2=(a-b)(a+b)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Find the Laurent Expansion of $f(z)$ Find the Laurent Expansion for
$$f(z)=\frac{1}{z^4+z^2}$$
about $z=0$.
I have found the partial fraction decomposition
$$f(z)=\frac{1}{z^4+z^2}=\frac{1}{z^2}-\frac{1}{2i(z-i)}+\frac{1}{2i(z+i)}.$$
Next I wanted to expand each of the three terms separately. I have
$$\frac{1}{z^2}=\f... | You may just write, as $z \to 0$:
$$\begin{align}f(z)=\frac{1}{z^4+z^2}&=\frac{1}{z^2(1+z^2)}\\\\&=\frac{1}{z^2}(1-z^2+z^4-z^6+z^8...)
\\\\&=\frac{1}{z^2}-1+z^2-z^4+z^6-...
\end{align}
$$ and this gives the Laurent expansion of $f$ near $z=0$, on $0<|z|<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the sum of all odd numbers between two polynomials I was asked this question by someone I tutor and was stumped.
Find the sum of all odd numbers between $n^2 - 5n + 6$ and $n^2 + n$ for $n \ge 4.$
I wrote a few cases out and tried to find a pattern, but was unsuccessful.
Call polynomial 1, $p(n) = n^2 - 5n + 6,$ t... | @Elaqqad gave a very elegant answer, so I'll try to give the "especially pre-calc accessible" answer, exploiting the fact that we have an arithmetic series.
Note $p(n) = n^2 - 5n + 6 = (n-3)(n-2)$, while $q(n) = n^2 + n = n(n+1)$, with $p(n) < q(n)$.
Now, $p(n)$ and $q(n)$ are both even, so that we're adding up odd nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Proving $\lim_{x\to c}x^3=c^3$ for any $c\in\mathbb R$ using $\epsilon$-$\delta$ definition
$\lim_{x\to 3}x^3=c^3$ for any $c\in\mathbb R$
Let $\epsilon>0$. Then
$$|x^3-c^3|=|x-c||x^2+xc+c^2|.$$
Let $\delta=\min\{1,\epsilon/x^2+xc+c^2\}$.
Then if $0<|x-c|<\delta$ and therefore since $|x-c|<\epsilon/(x^2+xc+c^2)$,
$$|... | Suppose $\varepsilon>0$. The inequality
$$
|x^3-c^3|<\varepsilon
$$
becomes
$$
c^3-\varepsilon<x^3<c^3+\varepsilon
$$
or
$$
\sqrt[3]{c^3-\varepsilon}<x<\sqrt[3]{c^3+\varepsilon}
$$
or also
$$
\sqrt[3]{c^3-\varepsilon}-c<x-c<\sqrt[3]{c^3+\varepsilon}-c
$$
Take
$$
\delta=\min\{c-\sqrt[3]{c^3-\varepsilon},\sqrt[3]{c^3+\va... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1227691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Integer solutions to $x^{x-2}=y^{x-1}$
Find all $x,y \in \mathbb{Z}^+ $ such that $$x^{x-2}=y^{x-1}.$$
I can only find the following solutions: $x=1,2$. Are there any other solutions?
| The only solutions are $x=1$ and $y=$ anything, and $x = 2$ and $y = 1$.
Start by rearranging the equation: $(\frac{x}{y})^{x-1} = x$.
If x = 1 then we have: $(\frac{1}{y})^0 = 1$ which is true for any $y$.
If y = 1 then we have: $x^{x-1} = x \implies x^{x-2} = 1$, which means either $x = 1$ which is covered by the cas... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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To find the value of a function at a point where it is continuous Find $f(0)$ so that the function $f(x)=\dfrac{1-\cos(1-\cos x)}{x^4}$ is continuous everywhere.
My attempt:
By applying sandwich theorem $-1 \le \cos(x) \le 1$.
$$1 \ge -\cos(x) \ge -1$$
$$2 \ge 1-\cos(x) \ge 0$$
$\cos(x)$ is a decreasing function from $... | It is enough to note that $\lim_{y \to 0} \frac{1-\cos y}{y^2}=\frac{1}{2}$.
Now you write
$$ \frac{1-\cos(1-\cos x)}{x^4}= \frac{1-\cos(1-\cos x)}{(1-\cos x)^2}\left( \frac{1-\cos x}{x^2}\right)^2 \to \frac{1}{2} \frac{1}{4}=\frac{1}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Sum the infinite series How to solve this:
\begin{equation*}
\sum_{n=1}^{\infty }\left[ \frac{1\cdot 3\cdot 5\cdots \left( 2n-1\right) }{
2\cdot 4\cdot 6\cdots 2n}\right] ^{3}
\end{equation*}
I can make the bracket thing, $\left[ C(2n,n)/4^{n}\right] ^{3}$, but how to proceed now.
| We have:
$$ \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\,x^n = \frac{1}{\sqrt{1-x}},\tag{1} $$
$$\frac{1}{4^n}\binom{2n}{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left(\sin x\right)^{2n}\,dx \tag{2}$$
$$ \sum_{n\geq 0}\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 x^n = \frac{2}{\pi}\,K(x)=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
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$x^3+ (5m+1)x+ 5n+1$ is irreducible over $\Bbb Z$ How to prove that the polynomial: $x^3+ (5m+1)x+ 5n+1$ is irreducible over the set of integers for any integers $m$ and $n$?
I was trying to put $x= y+p$ for some integer $p$ so that I could apply Eisenstein's criterion. But it is very tedious to find some suitable $p$... | If it is not, then it can be written as a product $(x^2 + ax + b)(x + c)$, where $a,b,c$ are integers.
Expanding: $x^3 + (5m + 1)x + 5n + 1 = x^3 + (a + c)x^2 + (ac + b)x + bc$.
Then $a = -c$ and this becomes: $x^3 + (5m+1)x + 5n+1 = x^3 + (b-c^2)x + bc$
Then $c^2 = b - (5m + 1)$, so $c^3 = bc - (5m + 1)c = 5n+1-(5m+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x\over(1-x)$ $y\over(1-y)$ $z\over(1-z)$ >= 8 when $x ,y ,z $ are positive proper fractions and $x+y+z = 2$ Q. Prove that $x\over(1-x)$ $y\over(1-y)$ $z\over(1-z)$ $\geq$ 8 when $x ,y ,z $ are positive proper fractions and $x+y+z = 2$
What I did: From A.M. G.M. inequality,
$(x+y+z)\over3$ $\geq$ $(xyz)^{1/3}$
or, $xyz... | Note that $1 = (x+y+z)/2$. The inequality becomes
$$ \frac{xyz}{(y+z-x)(x+z-y)(x+y-z)} \geq 1$$
which is equivalent to
$$ xyz \geq (y+z-x)(x+z-y)(x+y-z)\ \ \ (1)$$
To see why this is true, apply AM-GM inequality three times:
$$ x = \frac{x+y-z+x+z-y}{2} \geq \sqrt{(x+y-z)(z+x-y)}$$
and the other two for $y,z$. Now ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Maximum value of $ x^2 + y^2 $ given $4 x^4 + 9 y^4 = 64$ It is given that $4 x^4 + 9 y^4 = 64$.
Then what will be the maximum value of $x^2 + y^2$?
I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .
| Without using Lagrange multipliers
It's equivalent to find $x$ that maximizes
$$f(x)=x^2+\frac23\sqrt{16-x^4}$$
You have
$$f'(x)=2x-\frac13\frac{4x^3}{\sqrt{16-x^4}}$$
Hence $f'(x)=0$ iff $x=0$ or
$$\frac{2x^2}{\sqrt{16-x^4}}=3$$
$$2x^2=3\sqrt{16-x^4}$$
$$4x^4=9(16-x^4)$$
$$13x^4=12^2$$
$$x=\pm2\frac{\sqrt{3}}{\sqrt[4]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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How do I integrate this expression? Need to find the following: $$\int \frac{\sqrt{x^2 - 4}}{x^3} dx$$
So far, after using integration by parts, I've arrived at $\frac{-\sqrt{x^2-4}}{2x^2} + \frac12 \int \frac{1}{x\sqrt{x^2-4}} dx$. How to proceed from here?
| $$I=\int\dfrac{\sqrt{x^2-4}}{x^3}dx=\int\dfrac{\sqrt{x^2-4}}{x^4}x\ dx$$
Set $\sqrt{x^2-4}=u\implies x^2-4=u^2\iff x\ dx=u\ du$
$$I=\int\dfrac{u^2}{(u^2+4)^2} du=\int\dfrac1{(u^2+4)} du-4\int\dfrac{du}{(u^2+4)^2}$$
For the second part set $u=2\tan\theta$
Alternatively, set $x=2\sec\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$ Could anybody help me by checking this solution and maybe giving me a cleaner one.
Prove by mathematical induction:
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.
So after I check special cases for $n=2,3$, I have to prove... | In case you want a different approach, here is one:
The red rectangles represent the sum $S_n=\frac 1n+...+\frac 1{n^2}$. Above it is for $n=2$, but the principle is general. From the figure it follows immediately that
$$
\int_n^{n^2}\frac 1x\ dx=\ln(n)<\frac 1n+\frac1{n+1}+...+\frac1{n^2}
$$
and for $n>e$ this implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
} |
Pairs of integers $(a,b)$ such that $\frac{1}{6} =\frac{1}{a} + \frac{1}{b}$ How many pairs of integers are there $(a,b)$ with $a \leq b$ such that $$\frac{1}{6} =\frac{1}{a} + \frac{1}{b}$$
My attempt:
Clearing fractions we get $$ab = 6(a+b)$$
$$ \Longrightarrow \frac{ab}{a+b} = 6$$
Which means that $ab$ is a multiple... | HINT : $$ab=6(a+b)\iff (a-6)(b-6)=36.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Solve the system $ x \lfloor y \rfloor = 7 $ and $ y \lfloor x \rfloor = 8 $.
Solve the following system for $ x,y \in \mathbb{R} $: \begin{align} x \lfloor y \rfloor & = 7, \\ y \lfloor x \rfloor & = 8. \end{align}
It could be reducing to one variable, but it is not that simple.
Another idea is substitution, but it ... | Note: Here we use the following characterization of the floor function
\begin{align*}
\lfloor x\rfloor = p\qquad \text{ with } \qquad p \leq x < p+1, \quad p\in \mathbb Z
\end{align*}
We use this relationship to transform the system of equations into a system of inequalities which can be easily analysed.
Let $x,y \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
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Show the closed form of the sum $\sum_{i=0}^{n-1} i x^i$ Can anybody help me to show that when $x\neq 1$
$$\large \sum_{i=0}^{n-1} i\, x^i = \frac{1-n\, x^{n-1}+(n-1)\,x^n}{(1-x)^2}$$
| The following chain of identities gives the closed form without using calculus.
$$\sum_{i=1}^n i x^i= \sum_{k=0}^{n-1} x^{n-k} \sum_{i=0}^k x^i=$$ $$=\sum_{k=0}^{n-1} x^{n-k} \frac{x^{k+1}-1}{x-1}=\frac{1}{x-1}\sum_{k=0}^{n-1}(x^{n+1}-x^{n-k})=$$ $$=\frac{x^n}{x-1}\sum_{k=0}^{n-1}(x-x^{-k})=\frac{x^n}{x-1}(n x-\frac{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1244297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find basis so matrix is in Jordan Canonical Form $M = \left(\begin{array}{ccc}0 & -3 & -2 \\1 & 3 & 1 \\1 & 2 & 3\end{array}\right)$
I want to find a basis $B$ such that matrix for $M$ w.r.t $B$ has the form:
$\left(\begin{array}{ccc}2 & 1 & 0 \\0 & 2 & 1 \\0 & 0 & 2\end{array}\right)$
The eigenvalues for $M = 2,2,2$.
... | Since $(M-2I)^2=\begin{pmatrix}-1&-1&-1\\0&0&0\\1&1&1\end{pmatrix}$, we can choose any vector $v_3=\begin{pmatrix}a\\b\\c\end{pmatrix}$
which is not in the nullspace of $(M-2I)^2$, so $a+b+c\ne0$, and
then let $v_2=(M-2I)v_3$ and $v_1=(M-2I)v_2$
to get a basis $B=\{ v_1, v_2, v_3\}$ with respect to which M will be in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding maxima of a 3-variable function. Let $x,y,z$ be positive real number satisfy $x+y+z=3$
Find the maximum value of
$P=\frac{2}{3+xy+yz+zx}+(\frac{xyz}{(x+1)(y+1)(z+1)})^\frac{1}{3}$
| We'll prove that $$\frac{2}{3+xy+xz+yz}+\sqrt[3]{\frac{xyz}{(1+x)(1+y)(1+z)}}\leq\frac{5}{6}$$
for all positives $x$, $y$ and $z$ such that $x+y+z=3$.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Since by Holder $\sqrt[3]{(1+x)(1+y)(1+z)}\geq1+\sqrt[3]{xyz}=1+w$, it's enough to prove that
$$\frac{2u^2}{3u^2+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1246298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration: $\int\frac{1}{(x^2+x+1)^{1/2}} dx$ Find the value of $$\int\frac{1}{(x^2+x+1)^{1/2}} dx$$
Anyone can provide hint on how to integrate this, and how you know what method to use? (I mean, is there any general guideline to follow for solving?)
Thank you!
| $\bf{My\; Solution:: }$ Let $\displaystyle I = \int\frac{1}{\sqrt{x^2+x+1}}dx = \int\frac{1}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}dx$
Now Let $\displaystyle \left(x+\frac{1}{2}\right) = t\;,$ Then $dx = dt$ and $\displaystyle \frac{\sqrt{3}}{2} = a>0$
So Integral $$\displaystyle I = \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1246836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Method to integrate $\cos^4(x)$ Here my attempts for integrating $\cos^4(x)$ in few methods.
1st method.
$(\cos^2x)^2=(\frac{1}{2})^2(1+\cos2x)^2$
$=\frac{1}{4}(1+2\cos2x+\cos^22x)=\frac{1}{4}(1+2\cos2x)+\frac{1}{4}(\cos^22x)$
$=\frac{1}{4}(1+2\cos2x)+\frac{1}{4}(\frac{1}{2}(1+\cos4x))$
$=\frac{3}{8}+\frac{1}{2}\cos2... | Let us first establish a general formula, with an obvious notation:
$$(C^cS^s)'=-cC^{c-1}S^{s+1}+sC^{c+1}S^{s-1}=-cC^{c-1}(1-C^2)S^{s-1}+sC^{c+1}S^{s-1}\\
=-cC^{c-1}S^{s-1}+(s+c)C^{c+1}S^{s-1}.$$
Then, rewriting and shifting the indexes,
$$\int C^\color{green}cS^s\,dx=\frac1{s+c}C^{c-1}S^{s+1}+\frac{c-1}{s+c}\int C^{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this a good proof of the binomial identity? Prove that the binomial identity ${n\choose k} = {n-1\choose k-1} + {n-1\choose k}$ is true using the following expression: $(1+x)^n = (1+x)(1+x)^{n−1}$ and the binomial theorem.
What I have:
We know from the binomial theorem that:
$$(x+1)^n= {n\choose 0} x^0 + {n\choose 1... | A much shorter way would be calculating the coefficient of $x^k$ independently from each side of $$(1+x)^n=(1+x)(1+x)^{n-1}$$
Using the binomial theorem, the coefficient of $x^k$ on the LHS is $\binom{n}{k}$.
For the right hand side, $x^k$ can be formed by two ways -
*
*Selecting $1$ from $(1+x)$ and $x^k$ from $(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Is $\sqrt{x^2} = (\sqrt x)^2$? Take $x=4$ for example:
$ \sqrt{(4)^2} = \sqrt{16} = \pm4 $
However:
$ (\sqrt{4})^2 = \sqrt{\pm2}$
Case 1: $ (-2)^2 = 4$
Case 2: $ (2)^2 = 4$
Solution : $+4$
How come the $ \sqrt{(4)^2} = \pm4$; but $ (\sqrt{4})^2 = 4 $ ?
What is missing?
| In general:
$$\sqrt{x^2}=|x|\implies \sqrt{x^2}=\left(\sqrt{|x|}\right)^2=|x|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Number of times $2^k$ appears in factorial
For what $n$ does: $2^n | 19!18!...1!$?
I checked how many times $2^1$ appears:
It appears in, $2!, 3!, 4!... 19!$ meaning, $2^{18}$
I checked how many times $2^2 = 4$ appears:
It appears in, $4!, 5!, 6!, ..., 19!$ meaning, $4^{16} = 2^{32}$
I checked how many times $2^3 = ... | It might be more helpful to do this recursively.
Let $T(n) = \prod_{k=1}^n k!$.
We will use the notation: $2^{r} \| m$ to mean that $2^r$ is the largest power of $2$ that divides $m$.
Then we have $2 \| 2! = T(2)$. We also know that $2 \| 3!$, so $2^2 \| T(3) = 3! T(2)$. Continuing:
$$2^3 \| 4!$$
$$2^3 \| 5!$$
$$2^4 \|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $ \int_{0}^{\pi/4}\left(\cos 2x \right)^{11/2}\cdot \cos x\;dx $
Evaluate the definite integral
$$
I=\int_{0}^{\pi/4}(\cos 2x)^{11/2}\cdot \cos x\;dx
$$
My Attempt:
$$
I = \int \left(1-2\sin^2 x\right)^{11/2}\cdot \cos x\;dx
$$
Now, substitute $\sin x=t$ with $\cos x \,dx = dt$:
$$
I = \int (1-2t^2)^{11/2}\;... | Here's an efficient method using special functions.
Applying the substitution
$$t = 2 \sin^2 x, \qquad ds = 4 \sin x \cos x \,dx $$
gives
$$\int_0^{\pi / 4} (\cos 2x)^{(2 m - 1) / 2} \cos x \,dx = \frac{1}{2 \sqrt{2}} \int_0^1 t^{-1/2} (1 - t)^{(2 m - 1) / 2} dt .$$
By the definition of and then a standard identity for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Proving that $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{13}{24}$ by induction. Where am I going wrong? I have to prove that
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{2n}>\frac{13}{24}$$ for every positive integer $n$.
After I check the special cases $n=1,2$, I have to prove that the given inequality holds fo... | As I see it, here's what might be confusing you:
If you let
$$A=\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2},$$
and
$$B=\frac{13}{24}+\frac{1}{n},$$
then you are trying to prove that $A>B$. Also, if you let
$$C=\frac{13}{24}+\frac{1}{2n+1}+\frac{1}{2n+2},$$
then you have by a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
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Evaluate the integral $\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x$
Evaluate the integral $$\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x.$$
If I transfer $(x^2 + 4)^5$ to the numerator, how do I integrate?
| Put $u = x^2+4 \implies du = 2x\,dx\implies x\,dx = \frac 12 du$.
$$ \begin{align}\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x &= \frac 12\int u^{-5} \,du\\ & = \frac 12 \left(\frac {u^{-4}}{-4}\right) +c \\ &= -\frac{1}{8u^4} + c \\ &= -\frac {1}{8(x^2 + 4)^4} + c\end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
I do not have use derivatives, so I decided to apply the definition of being a strictly increasing function, which should ... | It looks right. Another possible approach is the following one - since:
$$ f(x)=x^2+3x+2 = \left(x+\frac{3}{2}\right)^2-\frac{1}{4} $$
we have that $f(x)$ attains its minimum in $x=-\frac{3}{2}$ (the abscissa of the vertex) and for every $r>-\frac{1}{4}$ the equation
$$ f(x)=r $$
has two solutions, symmetric with respe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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Using implicit differentiation with a fraction How do I solve this? What steps? I have been beating my head into the wall all evening.
$$ x^2 + y^2 = \frac{x}{y} + 4 $$
| We have:
$$ \frac{d(x^2)}{dx} + \frac{d(y^2)}{dy}\cdot \frac{dy}{dx} = \frac{d( \frac{x}{y})}{dx} + \frac{d(4)}{dx}$$
$$2x+2y\frac{dy}{dx} = \frac{y-x \frac{dy}{dx} }{ y^2} +0$$
Can you finish?
Okay, to finish it up:
$$2y\frac{dy}{dx} - \frac{y-x \frac{dy}{dx} }{ y^2} = -2x $$
Multiplying both sides by $y^2$
$$2y^3\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What is wrong with this limit reasoning $$\lim_{x \to 0}\left(\dfrac 1 {\sin x}-\dfrac1 x\right) $$
I solved this problem using the fact that near $x=0$ , $\sin x \cong x$),$(\sin x=x+O(x^2)$ therefore subtracting them results to $0$ which I did verify using other techniques, However, the same reasoning leads to $0$ in... | We will be using the following three Taylor expansions: $\sin{x}=x-\frac{x^3}{6}+O(x^5)$, $\log(1+x)=x-\frac{x^2}{2}+O(x^3)$ and $\frac{1}{1-x}=1+x+O(x^2)$
$$\begin{align}\frac{1}{\sin{x}}&=\frac{1}{x-\frac{x^3}{6}+O(x^5)}\\&=\frac{1}{x(1-x^2/6+O(x^4))}\\&=\frac{1}{x}(1+\frac{x^2}{6}+O(x^4))\\&=\frac{1}{x}+\frac{x}{6}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1257382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.
I was thinking of using induction, but wasn't really sure how to do it.
| Here's a proof based on the fact that among any five consecutive numbers, one of them must be divisible by $5$.
Suppose that $n^2(n^2+1)(n^2-1)$ is not divisible by $5$. Then either $n^2+2$ or $n^2-2$ is divisible by $5$. Let's write this as $n^2+2\sigma=5k$, where $|\sigma|=1$.
But if $n^2(n^2+1)(n^2-1)$ is not divi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1257632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 8
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What is the volume inside $S$, which is the surface given by the level set $\{ (x,y,z): x^2 + xy + y^2 + z^2 =1 \}$? The solution given uses a linear algebraic argument that doesn't seem very instructive -- and may not even be correct, I think.
We notice from the equation, that the surface is a quadratic form, level ... | First look at the elliptical cross-section in xy-plane:
$\{ (x,y): x^2 + xy + y^2 = 1 \}$
It is clear that this is symmetric in $x$ and $y$ so write
$x^2 + xy + y^2 = a(x+y)^2 + b(x-y)^2$ so by equating coefficients $a=\frac{3}{4}, b=\frac{1}{4}$
Therefore $\boxed{\dfrac{3}{4}(x+y)^2 + \dfrac{1}{4}(x-y)^2 = 1}$
Changin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1258463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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On $a^4 + b^4 = c^4 + d^4 = e^5$. Let $a, b, c, d, e$ be distinct positive integers such that $a^4 + b^4 = c^4 + d^4 = e^5$. Show that $ac + bd$ is a composite number.
| Assume that $p=ac+bd$ is prime. Without loss of generality, we may assume that $a>c,d$ from which follows that $b<c,d$. Also, $a,b,c,d<p$, and so must all be relatively prime with $p$.
Computing modulo $p$, we have $ac \equiv -bd$ which leads to $(ac)^4\equiv(bd)^4$. Exploiting $a^4+b^4=c^4+d^4$ we get
$$
(a^4+b^4)c^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1258623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Demonstrating Strassen's method using domain transformation: $T(n)=7T(n/2)+an^2$ I want to solve the recurrence for Strassen's method (for multiplying square matrices) with domain transformation and get a closed form. The equation is given below:
$T(n)=b$, at $n=2$
$T(n)=7T(n/2)+an^2$, at $n>2$
I know that since I have... | At each step, we replace $n$ with $\tfrac{n}{2}$ and scale each term by $7$ as follows:
\begin{align*}
T(n) &= 7T\left(\frac{n}{2}\right) + an^2 \\
7T\left( \frac{n}{2} \right) &= 7^2T\left(\frac{n}{2^2}\right) + \frac{7}{4}an^2 \\
7^2T\left( \frac{n}{2^2} \right) &= 7^3T\left(\frac{n}{2^3}\right) + \left(\frac{7}{4} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1258729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compare $\sum_{k=1}^n \left\lfloor \frac{k}{\varphi}\right\rfloor$ ... Given two integer sequences
\begin{equation*}
\displaystyle A_n=\sum_{k=1}^n \left\lfloor \frac{k}{\varphi}\right\rfloor ,
\end{equation*}
\begin{equation*}
B_n=\left\lfloor\dfrac{n^2}{2\varphi}\right\rfloor-\left\lfloor \dfrac{n}{2\varphi^2}\right\... | Note that $$A_n = \sum_{k=1}^n\lfloor k(\phi - 1) \rfloor = S_n - \frac{1}{2}n(n+1),$$ where $S_n = \sum_{k=1}^n \lfloor k\phi \rfloor$ (appeared as A054347). In The Golden String, Zeckendorf Representations, and the Sum of a Series by Martin Griffiths, he showed that $S_n = \lfloor \frac{n(n+1)\phi}{2} - \frac{n}{2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Find conditions for $\left(\frac{-3}{p}\right)=1$ $\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right)^{\frac{p-1}{2}}$
$=\begin{cases}1,\:p\equiv 1\pmod{4}\text{ or }\begin{cases}p\equiv 3\pmod{4}\\p\equiv 2\pmod {3}\end{cases}\\-1,\:\text{ot... | The case $p=2,3$ is trivial, so suppose that $p\neq3$ is an odd prime with remainder $r$, when divided by three therefore by some observations we have
$$
\left(\frac{-3}p\right)=\left(\frac{-1}p\right)\left(\frac{3}p\right)=(-1)^{(\frac{p-1}2)}(-1)^{(\frac{p-1}2)(\frac{3-1}2)}\left(\frac p3\right)=(-1)^{p-1}\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the ratio of $\frac{XY}{SQ}$ In the picture,$PQRS$ is a prallelogram. $PS$ is parallel to $ZX$ and $\frac{PZ}{ZQ}=\frac{2}{3}$. Then$\frac{XY}{SQ}$ equals:
The answer is $\frac{9}{40}$
Help me with the idea to solve this problem.
Thanks in advance.
| This can be brute-forced using analytical geometry.
We start with
$$
\frac{2}{3}
= \frac{\lVert PZ\rVert}{\lVert ZQ\rVert}
= \frac{\lVert SX\rVert}{\lVert XQ\rVert}
$$
We want the ratio
$$
\frac{\lVert XY\rVert}{\lVert SQ\rVert}
$$
and use the linear interpolation formula
$$
C = (1-t) A + t B \quad (t \in [0,1])
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving an infimum-inequality How does one evaluate $$\inf \{c: c^2 \geq \frac{a^2x^2+b^2y^2}{x^2+y^2} \} $$ for every $a,b,x,y \in \mathbb R$ ?
Hints and tips appreciated
Edit: I tried to write the inequality in polar coordinates, which gives me:
$$ \frac{a^2r^2\cos^2(\phi) + b^2r^2\sin^2(\phi)}{r^2\cos^2(\phi) + r^2... | It's $c = \sqrt{\max(a^2, b^2)}$. Proof:
*
*If $c = \sqrt{\max(a^2, b^2)}$, then $a^2 \leqslant c^2$ and $b^2 \leqslant c^2$, so
$$a^2 \cos^2( \phi ) + b^2 \sin^2( \phi ) \leqslant c^2 \cos^2( \phi ) + c^2 \sin^2( \phi ) = c^2.$$
*If $\gamma^2 \geqslant a^2 \cos^2( \phi ) + b^2 \sin^2( \phi )$ for all $x, y \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to determine if $\sum_{n=1}^{\infty}\left (\frac{n^2-5n+1}{n^2-4n+2}\right)^{n^2}$ converges or diverges $$\sum_{n=1}^{\infty} \left(\frac{n^2-5n+1}{n^2-4n+2}\right)^{n^2}$$
Using root test seems not a efficient way since I got stuck without knowing what to do next
$$\lim_{n\to\infty}\left(\frac{n^2-5n+1}{n^2-4n+2... | The general term decays like $\left(1-\frac{1}{n}\right)^{n^2}\approx e^{-n}$ hence the series is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1263557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is the connection between the discriminant of a quadratic and the distance formula? The $x$-coordinate of the center of a parabola $ax^2 + bx + c$ is $$-\frac{b}{2a}$$
If we look at the quadratic formula
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
we can see that it specifies two points at a certain offset from the cen... | Consider the image below.
The roots of $f(x)=ax^2+bx+c$ will lie on the circle in the complex plane whose center is $\dfrac{-b}{2a}.$ In the case where both roots are real numbers, they will lie on the real line at $R_1$ and $R_2.$ In the case where the roots are purely imaginary, they will lie on the vertical axis ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 8,
"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.