Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to simplify $\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}$ The answer is 2. But I want to learn how to simplify this expression without the use of calculator.
| Frequently, numbers of this kind are obtained from Cardano's formula for the roots of a cubic equation $x^3+px+q=0$, that is,
$$
\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}
+
\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}
$$
so you have
$$
-\frac{q}{2}=7,
\qquad
\frac{q^2}{4}+\frac{p^3}{27}=5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Trigonometric Equation Simplification $$3\sin x + 4\cos x = 2$$
To solve an equation like the one above, we were taught to use the double angle identity formula to get two equations in the form of $R\cos\alpha = y$ where $R$ is a coefficient and $\alpha$ is the second angle being added to $x$ when using the double angl... | $$\begin{align}3\sin x + 4\cos x &= \sqrt{3^2 + 4^2 }\sin\left(x+\arctan\frac{4}{3}\right) \\
&= 5\sin(x+\arctan\frac{4}{3})\end{align}$$
In general, if $$a\sin x + b\cos x = c$$
$$\frac{a\sin x}{\sqrt{a^2 + b^2}}+\frac{b \cos x}{\sqrt{a^2 + b^2}} = \frac{c}{\sqrt{a^2 + b^2}}$$
Let $\cos\phi = \dfrac{a}{\sqrt{a^2 + b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\sqrt{x^2+y^2+2x-4y+5} + \sqrt{x^2+y^2-6x+8y+25}$, if $3x+2y-1=0$ As the title says, given $x,y \in \mathbb{R}$ where $3x+2y-1=0$ and $x \in [-1, 3]$, calculate $A = \sqrt{x^2+y^2+2x-4y+5} + \sqrt{x^2+y^2-6x+8y+25}$.
I tried using the given condition to reduce the complexity of the roots, but couldn't get r... | Hint: make a geometrical interpretation and draw a picture:
$x^2+y^2+2x-4y+5=(x+1)^2 +(y-2)^2$ and consider the point $P(-1,2)$.
$x^2+y^2-6x+8y+25=(x-3)^2+(y+4)^2$ and consider the point $Q(3,-4)$.
Observe that $P$ and $Q$ are on the line $3x+2y-1=0$.
Can you now see what $A$ means on the given interval?
The answer is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Area inside polar curve Find the area of the region inside
$r=7\sin\theta$ but outside $r=1$.
I have tried finding the area of both using $A=\frac{1}{2}\int_\alpha^\beta r^2 d\theta$, tried arc length...but I can't find the answer.
| \begin{align*}
A &= \frac{1}{2}
\int_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}}
(r^{2}-1) \, d\theta \\
&= \frac{1}{2} \int_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}}
(49\sin^{2}-1) \, d\theta \\
&= \frac{1}{2} \int_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$x^4 - y^4 = 2z^2$ intermediate step in proof I am ultimately trying to prove, for an Exercise in Burton's Elementary Number Theory, that $x^4 - y^4 = 2z^2$ has no solution in the positive integers.
I can establish that if there is a solution, the solution with the smallest value of x has gcd(x,y)=1
I see that $x^4 - ... | If you want to establish your equalities,
Note that $\gcd(x,y)=1$. This implies that
$$\gcd(x^2+y^2,x+y)=2,1$$$$\gcd(x^2+y^2,x-y)=2,1$$$$ \gcd(x-y,x+y)=2,1$$
This implies that either $x+y=2a^2, x-y=b^2,x^2+y^2=c^2$
Or that $x+y=a^2, x-y=2b^2, x^2+y^2=c^2$
Or that $x+y=a^2, x-y=b^2, x^2+y^2=2c^2$
Or that $x+y=2a^2, x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the integral $\int \frac{1+x}{\sqrt{1-x^2}}\mathrm dx$ The integral can be represented as
$$
\int \frac{1+x}{\sqrt{1-x^2}}\mathrm dx=
\int \left(\frac{1+x}{1-x}\right)^{1/2}\mathrm dx
$$
Substitution $$t=\frac{1+x}{1-x}\Rightarrow x=\frac{t-1}{t+1}\Rightarrow dx=\frac{2}{(t+1)^2}dt\Rightarrow \int\limits \left(\f... | We can also use the substitution $u=\sqrt{1-x}$, then $\mathrm{d}u=-\frac{\mathrm{d}x}{2\sqrt{1-x}}$ and $\sqrt{1+x}=\sqrt{2-u^2}$. We will also use $u=\sqrt2\sin(\theta)$
$$
\begin{align}
\int\frac{\sqrt{1+x}}{\sqrt{1-x}}\,\mathrm{d}x
&=-2\int\sqrt{2-u^2}\,\mathrm{d}u\\
&=-2\int\sqrt2\cos(\theta)\cdot\sqrt2\cos(\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Taylor expansion of $\cos{x}$ I found a pdf file on the internet which gives you known expansions of Taylor's.
There is something I cant understand :
Why is the remainder of $\cos x$ is written like this?
$$\frac{\cos ^{(2n+2)}(c)x^{2n+2}}{(2n+2)!}$$
And not like this:
$$\frac{\cos^{(2n+1)} (c)x^{2n+1}}{(2n+1)!}$$
whe... | I can't guess what a random, unlinked PDF on the Internet says. However, this is not the behaviour of the Taylor series of cosine. For instance, the expansion of $\cos x$ around $x=1$ is $$\cos (1)-(x-1) \sin (1)-\frac{1}{2} (x-1)^2 \cos (1)+\frac{1}{6} (x-1)^3
\sin (1)+\frac{1}{24} (x-1)^4 \cos (1)-\frac{1}{120} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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If $A+B+C=π$, verify the given If $A+B+C=π$, prove that
$$\cos A \sin B \sin C + \cos B \sin C \sin A + \cos C \sin A \sin B=1+\cos A \cos B\cos C$$
ATTEMPT:
Here,
$$A+B+C=π$$
Now,
\begin{align*}
\text{L.H.S} &= \cos A \sin B \sin C + \cos B \sin C \sin A + \cos C \sin A \sin B \\
&=\sin C(\sin A \cos B + \cos A \sin B... | take the trig term fro the RHs to the left and combine to get
$$
\sin C \sin(A+B) - \cos C \cos (A+B) = -\cos (A+B+C)
$$
and now $A+B+C=\pi$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\lim_{n\rightarrow \infty}\sum_{k=1}^n\sin \left(\frac{n}{n^2+k^2}\right)$
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\sin \left(\frac{n}{n^2+1}\right)+\sin \left(\frac{n}{n^2+2^2}\right)+\cdots+\sin \left(\frac{n}{n^2+n^2}\right)$
$\bf{My Try::}$ We can write the Sum as $$\lim_{n\rightarro... | As a consequence of the fact that $\,\,\int_0^1 f(x)\,dx=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f\big(\frac{k}{n}\big)$, we obtain
$$
\sum_{k=1}^n\frac{n}{n^2+k^2}=\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\left(\frac{k}{n}\right)^2}\to\int_0^1\frac{dx}{1+x^2}=\tan^{-1}1=\frac{\pi}{4}
$$
Next, Taylor expansion of $\sin x$, p... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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How many different integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 21$ with restrictions So i was Given this question. How many different integer solutions are there to
the equation
$x_1 + x_2 + x_3 + x_4 = 21$
$0 \leq x_i \leq 9$?
I just assumed it would be
${21+4-4-1 \choose 4-1} = {20 \choos... | All solutions
Split $(x_{1}+x_{2})+(x_{3}+x_{4})=21$. There are 16 different pairs $(x_{1}+x_{2},x_{3}+x_{4})$ (if you count $3+18$ different from $18+3$) simply $(3,18),(4,17),(5,16),...,(10,11),...,(18,3)$
For $(10,11)$ and $(11,10)$ you have $9$ combinations for $10$ and $8$ combinations for $11$. That makes $2\cdot... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sqrt{13a^2+b^2}$ and $\sqrt{a^2+13b^2}$ cannot be simultaneously rational Let $a,b \in \mathbb{N}^{*}$. Prove that $\sqrt{13a^2+b^2}$ and $\sqrt{a^2+13b^2}$ cannot be simultaneously rational.
If $(a,b)=(k,k\cdot6)$, then $\sqrt{13a^2+b^2}$ is rational, but I do not know if those are the only solutions.
| $\sqrt{x}$ (where $x\in\mathbb Z^+$) is rational if and only if $\sqrt{x}$ is a positive integer (see this question).
Therefore, your problem is equivalent to proving the system $13a^2+b^2=k^2$ with $a^2+13b^2=m^2$ has no positive integer solutions.
$$a^2+13\left(k^2-13a^2\right)=m^2\iff m^2+168a^2=13k^2$$
I'll prove t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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What is the procedure to solve for the $\lim\limits_{x\rightarrow 0} \frac{\cos(3x)-1}{x^2}$? What is the procedure to solve for the $\lim\limits_{x\rightarrow 0} \frac{\cos(3x)-1}{x^2}$?
My calculator tells me the answer is -9/2, but I don't know how to solve without substituting values of x.
I suspect there is some t... | From the trigonometric identity $\;\color{blue}{\cos(2\theta)=1-2\sin^2\theta}\;$ we have
\begin{align}
\lim_{x\to 0}\frac{\cos(3x)-1}{x^2}&=\lim_{x\to 0}\frac{1-2\sin^2(3x/2)-1}{x^2}\\
&=\lim_{x\to 0}\frac{-2\sin^2(3x/2)}{x^2}\\
&=-2\cdot\frac{9}{4}\cdot\lim_{x\to 0}\frac{\sin^2(3x/2)}{\frac{9}{4}x^2}\\
&=-2\cdot\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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How to prove $ \left(\sum\limits_{cyc}{xy}\right)^2 \ge3xyz(x+y+z)$ with $x,y,z$ being positive real numbers I have tried to improve the inequality by AM-GM. Here is what I have done:
Since
$$\sum_{cyc}{xy}\ge3\sqrt[3]{x^2y^2z^2}$$
$$\Rightarrow \left(\sum_{cyc}{xy}\right)^2 \ge3xyz\sqrt[3]{xyz}$$ That means we got to... | Simplify the inequality to
$$x^2y^2+y^2z^2+x^2z^2\ge xyz(x+y+z)$$
If some of $x,y,z$ equals $0$, then the inequality is obvious.
So assume $xyz\neq 0$ and divide by $x^2y^2z^2>0$ both sides to get
$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge \frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}$$
This is true by the power of Cauch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Sticks and stones problem Ten squares in a row are labelled 1, 2, 3, . . . 10, in order. A counter starts at square 1. At every step,
the counter can move ahead 1, 2, or 3 squares. However, beginning with the second step, the counter
cannot move the
same
number of squares as it did in the previous step. For example, ... | I get $21$.
$2$ with no $3$'s:
$1,2,1,2,1,2$
$2,1,2,1,2,1$
$12$ with one $3$:
$3,1,2,1,2$
$3,2,1,2,1$
$1,3,2,1,2$
$2,3,1,2,1$
$1,2,3,1,2$
$1,2,3,2,1$
$2,1,3,1,2$
$2,1,3,2,1$
$1,2,1,3,2$
$2,1,2,3,1$
$1,2,1,2,3$
$2,1,2,1,3$
$7$ with two $3$'s:
$3,1,2,3$
$3,2,1,3$
$1,3,2,3$
$3,2,3,1$
$1,3,1,3,1$
$2,3,1,3$
$3,1,3,2$
$2+12+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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A factory makes three types of biased coins with probability of getting head when tossed is $\frac{1}{2}$,$\frac{1}{3}$,$\frac{1}{4}$ respectively
A factory makes three types of biased coins with probability of getting head when tossed is $\frac{1}{2}$,$\frac{1}{3}$,$\frac{1}{4}$ respectively. Each coin is tossed befo... | Your last remark is indeed correct. The probability you calculated is $P(H \cap H_M)$, the probability that a coin from those 3 has its head marked and comes up $H$.
$P(H) = \frac{1}{3}(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = \frac{13}{36}$, by conditioning again.
The asked for probability is (head is marked given ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $\left ( \frac{x}{j}+1 \right )^{-1}\left ( \frac{1}{j}+1 \right )^{x}=1+\frac{x(x-1)}{2j^2}+O(j^{-3})$? On this page, it says
$$ \left ( \frac{x}{j}+1 \right )^{-1}\left ( \frac{1}{j}+1 \right )^{x}=1+\frac{x(x-1)}{2j^2}+O(j^{-3}). $$
Could anyone please enlighten me how it is deduced? I tried to find an approx... | As already said, this is the binomial theorem.
You can also do it with Taylor series. Consider first $$A=\left ( \frac{1}{j}+1 \right )^{x}$$ $$\log(A)=x\, \log\left( \frac{1}{j}+1 \right)=x\left(\frac{1}{j}-\frac{1}{2 j^2}+\frac{1}{3 j^3}+O\left(\frac{1}{j^4}\right)\right)$$ $$A=e^{\log(A)}=1+\frac{x}{j}+\frac{x(x-1)}... | {
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"timestamp": "2023-03-29T00:00:00",
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Does $\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}=\sqrt{1 \sqrt{1 \sqrt{1\dots}}} \implies \cos{\theta}=1$? I was solving this equation:
$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}=1$$
I solved it like this:
The given equation can be written as:
\begin{align*}
\sqrt{\cos{\the... | To answer the question in the title, let $A=\sqrt{a \sqrt{a \sqrt{a\cdots}}}$ and $B=\sqrt{b \sqrt{b \sqrt{b\cdots}}}$.
Then $A=B$ implies $aA=A^2=B^2=bB$ and so $a=b$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Infinity Series to Approximate a fraction I have observed that the following series is a good approximation for
$\frac{1}{10}$.
$\frac{1}{8}- \frac{1}{16} + \frac{1}{32} + \frac{1}{64} - \frac{1}{128} - \frac{1}{256} + \frac{1}{512} + \frac{1}{1024} - \frac{1}{2048} -
\frac{1}{(2\cdot2048)} + \frac{1}{(4\cdot2048)} + ... | The series is similar to $\sum \limits_{n=3}^{\infty} \frac{1}{2^n}=2-1/1-1/2-1/4=1/4$.
Note that $\sum \limits_{n=0}^{\infty} x^n=\frac{1}{1-x}$ for $|x|<1$.
EDIT, since I misread the question:
$$1/8 - 1/16 + 1/32 + 1/64 - 1/128 - 1/256 + 1/512 + 1/1024 - 1/2048 -
1/4096 + 1/8192 + 1/16384 - 1/32768 - 1
/65536 + ...=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666225",
"timestamp": "2023-03-29T00:00:00",
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Prove that $f(x)=8$ for all natural numbers $x\ge{8}$ A function $f$ is such that $$f(a+b)=f(ab)$$ for all natural numbers $a,b\ge{4}$ and $f(8)=8$. Prove that $f(x)=8$ for all natural numbers $x\ge{8}$
| Yes. It is true that $f(x) = 8 \quad \forall\;\;\; x \in N $
Manually, we can prove this for $x \le 20$.
Now, let $x$ be even. $x = 2y$ for some $y$.
$$f(2y)=f((2y-4) +(4))=f(4(2y-4))=f(8(y-2))=f(8+y-2)=f(y+6)$$
Note: This is true only if the $y-2$ factor is greater than $4$, so let $y \ge 6$.
Similarly, if $x$ is odd,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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How can I solve $7^{77}\mod 221$ How is it possible to solve this without calculator etc.: $$7^{77} \mod 221$$
I started with:
\begin{align}
7&\equiv 7 \mod 221 \\
7^2 &\equiv 49 \mod 221 \\
7^4 &\equiv \ ? \mod 221
\end{align}
Sure i can calculate this by my own, but is there a trick to calclulate this with any tools?... | Use the Chinese Remainder Theorem. Note that $221=13\times17$. Modulo $13$ we have
$$7^2\equiv-3\ ,\quad 7^6\equiv(-3)^3\equiv-1$$
and so
$$7^{77}=(7^6)^{12}7^5\equiv7^5=(7^2)(7^2)7\equiv(-3)(-3)7=63\equiv-2\ .$$
Modulo $17$ we have
$$7^2\equiv-2\ ,\quad 7^8\equiv(-2)^4\equiv-1$$
and so
$$7^{77}=(7^8)^97^5\equiv-7^5\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving recurrence relation? Consider the recurrence relation $a_1=8,a_n=6n^2+2n+a_{n−1}$. Let $a_{99}=K\times10^4$ .The value of $K$ is______ .
My attempt :
$a_n=6n^2+2n+a_{n−1}$
$=6n^2+2n+6(n−1)^2+2(n−1)+a_{n−2}$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+a_1$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+... | As $a_n-a_{n-1}$ is a quadratic polynomial, $a_n$ must be a cubic polynomial. By integration, close to $\dfrac63n^3=2n^3$.
Then the factor $10^4$ is very likely to be $(99+1)^2$, and we can gamble on the expression $2n(n+1)^2$. It works for $n=1$ ($a_1=8$) and $n=2$ ($a_2=36$). This is convincing enough.
$$2n=198.$$
B... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How could I find the interval where the solution is valide for the initial value problem : $y'=\frac{1+3x^2}{3y^2-6y}$ with $y(0)=1$? How could I find the interval where the solution is valide for the initial value problem : $y'=\frac{1+3x^2}{3y^2-6y}$ with $y(0)=1$?
I think I have to consider the points where the curv... | See my comment:
The differential equation is only valid on the following continuous intervals: 1) $-\infty < y < 0$, 2) $0 < y < 2$, and 3) $2 < y < \infty$. It might be valid on more but since $1 + 3x^2 = 0$ permits no real solutions, there's no need to worry about possible removeable discontinuities.
Find the "sol... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Looking for a nonrecursive formula for the general derivatives of the quotient of functions I want to prove that the $k$-th derivative $h^{(k)}(x)$ of the function $h(x)=\frac{1}{1+x^2}$ is zero at $x=0$ for all integer values $k>0$.
My only idea was to go the stubborn way applying iteratively the elementary formula fo... | In order to answer OPs second part of the question, we provide an expression of the $n$-th derivative of $\frac{f}{g}$ in terms of derivatives of $f$ and $g$.
Let $D_x$ represent differentiation with respect to $x$. Hence $D^n_x f(x)$ is the $n$-th derivative of $f$ with respect to $x$. The following holds true... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+C... | You have mistake here:
$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{\color{red}(1+t^2\color{red}{)^3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Partition of $\{1,2,3,\cdots,3n\}$ into $n$ subsets, each with $3$ numbers, which have equal sum I want to show, that for every odd $n$ $(n\ge3)$, there exists a partition of $\{1,2,3,\cdots,3n\}$ into $n$ disjoint subsets, where each one has $3$ elements and equal sum.
The first such number is $3$. For $3$ it is obvio... | Let $S$ be the equal sum of each partition, then $nS=1+\ldots+3n$.
i.e. $\: nS=\frac{3n(3n+1)}{2} \implies S=\frac{3(3n+1)}{2}$
Now, $n=1$ is trivial and $n$ should be odd since the sum is an integer.
Note that the median of the sequence is $\frac{3n+1}{2}$.
Observing $1+\frac{3n+1}{2}+3n=S$
Take $a+b+c=0$:
$(1+a)+(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Interval of convergence of $\sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}{n!}$
Given the series
$$ \sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}{n!} \quad \quad k \geq 1 $$
Find the interval of convergence.
I started by applying the Ratio test
$$
\lim_{n\to \infty}... | Your series is
$$
\sum_{n=0}^\infty\binom{-k}{n}(-x)^n=(1-x)^{-k}
$$
This alone should show that there is no convergence at $x=1$ for positive $k$.
For the series at $x=-1$ consider that
$$
\binom{-k}{n}(-1)^n=\binom{n+k-1}{n}=\binom{n+k-1}{k-1}=\frac{(n+1)(n+2)···(n+k-1)}{(k-1)!}
$$
is a polynomial in $n$ of degree $k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
integrate $\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}dxdy, D: (x-a)^2+y^2 \le a^2$ $$I=\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}dxdy, D: (x-a)^2+y^2 \le a^2(a>0)$$
The difficulty is to find a proper and easy way to solve this double integrals.
If do it like this, $ 0\le x \le 2a, -\sqrt{2ax-x^2} \le y \le \sqrt{... | Believe it or not, it's actually easier without polar coordinates.
$\int_{0}^{2a}\int_{-\sqrt{2ax-x^2}}^{\sqrt{2ax-x^2}}\frac{2a}{\sqrt{4a^2-x^2-y^2}}dydx = \int_{0}^{2a}2a\arctan{\frac{y}{\sqrt{4a^2-x^2-y^2}}}\Big\vert_{-\sqrt{2ax-x^2}}^{\sqrt{2ax-x^2}}dx$
$= 4a\int_{0}^{2a}\arctan{\frac{\sqrt{-2x(x-2a)}}{2\sqrt{-a(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integration of rational functions by partial fractions I have the following integral :
$$\int \frac{x^2+2x+3}{x^2-5x+6}~dx$$
Although I tried to solve it by partial fractions, I could not come up with an appropriate answer. My question is more about the technique which I should use here in order to evalue the integral.... | First,
$$
x^2+2x+3=1\cdot (x^2-5x+6)+7x-3.
$$
Thus
$$
\frac{x^2+2x+3}{x^2-5x+6}=1+\frac{7x-3}{x^2-5x+6}.
$$
You know $x^2-5x+6=(x-2)(x-3)$. It is known that there exists $A,B$ such that
$$
\frac{7x-3}{x^2-5x+6}=\frac{A}{x-2}+\frac{B}{x-3}
$$
and so
$$
\frac{7x-3}{x^2-5x+6}=\frac{(A+B)x-(3A+2B)}{x^2-5x+6}.
$$
Thus we ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Nested radical $\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ I am studying the $f(x) = \sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ for $x \in (0,\infty)$ and I am trying to get closed form formula for this, or at least some useful series/expansion. Any ideas how to get there?
So far I've got only trivial values, which... | In the same spirit as Mark Fischler's answer, setting $x=\frac{y^2}2$, for large values of $x$, Taylor expansion is $$y+\frac{1}{4 \sqrt{2}}-\frac{5}{64 }\frac 1 {y}+\frac{85}{256 \sqrt{2} }\frac 1 {y^2}-\frac{1709}{8192
}\frac 1 {y^3}+\frac{6399}{32768 \sqrt{2} }\frac 1 {y^4}+O\left(\frac{1}{y^5}\right)$$ which wou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Finding Maximum Area of a Rectangle in an Ellipse
Question: A rectangle and an ellipse are both centred at $(0,0)$.
The vertices of the rectangle are concurrent with the ellipse as shown
Prove that the maximum possible area of the rectangle occurs when the x coordinate of
point $P$ is $x = \frac{a}{\sqrt{2... | Its easier to solve this question using parametric points.
Let one vertex of the rectangle be $(a\cos\theta,b\sin\theta)$.
The other vertices are $(a\cos\theta,-b\sin\theta)$, $(-a\cos\theta,b\sin\theta)$, $(-a\cos\theta,-b\sin\theta)$
The area of rectangle formed is $$A(\theta)=4ab\cos\theta\sin\theta=2ab\sin2\theta$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
How many tokens would person A have under these conditions? Persons A and B each have a positive integer number of tokens, and the number of tokens B has is a square number less than 100. B says to A, "If you give me all of your tokens, my total number of tokens will still be a square number." A says, "Yes - if on the ... | Let's start by listing all the square numbers between $0$ and $100$ $$1,4,9,16,25,36,49,64,81$$ Note we do not include $100$. We know that $$B \in [1,4,9,16,25,36,49,64,81]\\B+A \in [1,4,9,16,25,36,49,64,81,100,121,144,...]\\ B-A \in [1,4,9,16,25,36,49,64,81,100,121,144,...]$$ So what we need to find in a square equid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Does $L^2 = P^2$ implies that $L = P$? I have encountered a problem when solving this problem:
Assume that $\alpha \in (\pi, \frac{3}{2} \pi)$, then prove $\sqrt{\frac{1 + \sin \alpha}{1 - \sin \alpha}} - \sqrt{\frac{1 - \sin \alpha}{1 + \sin \alpha}} = -2 \tan \alpha$
The most popular way to solve this kind of problem... | (I made it a little more general)
Note that
$\sqrt{x}-\sqrt{\frac1{x}}
=\sqrt{x}-\frac1{\sqrt{x}}
=\frac{x-1}{\sqrt{x}}
$.
If
$x = \frac{1+y}{1-y}
$,
this is
$\begin{array}\\
\frac{x-1}{\sqrt{x}}
&=\frac{\frac{1+y}{1-y}-1}{\sqrt{\frac{1+y}{1-y}}}\\
&=\frac{1+y-(1-y)}{(1-y)\sqrt{\frac{1+y}{1-y}}}\\
&=\frac{2y}{\sqrt{(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$ Calculate:
$$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$$
I don't know how to use L'Hôpital's Rule.
I tried to make $\tan x =\frac{\sin x}{\cos x}$ for the term ${\sqrt{\tan x}}... | $$\frac{\tan x\sqrt{\tan x}-\sin x\sqrt{\sin x}}{x^3\sqrt x}=\left(\frac{\sin x} x\right)^{3/2}\cdot\frac{\frac1{\cos^{3/2}x}-1}{x^2}=$$
$$=\left(\frac{\sin x} x\right)^{3/2}\frac{1-\cos^{3/2}x}{x^2\cos^{3/2}x}=\left(\frac{\sin x} x\right)^{3/2}\frac{1-\cos^2x+\cos^{3/2}x(\cos^{1/2}x-1)}{x^2\cos^{3/2}x}=$$
$$=\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ Solve this equation :
$$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
Such that $a+b+c=\pi$
I don't have any idea. I can't try anything.
| $$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
$$\frac D4= \cos^2 b \cdot \cos^2c-\cos^2b -\cos^2c+1= \left( 1-\cos^2b\right)\left(1-\cos^2 c \right)=\sin^2b \cdot \sin^2c$$
$$x_{1,2}=-\cos b \cdot \cos c \pm \sin b \cdot \sin c=-\cos (b \mp c)$$
$$x_1=-\cos (b+c)=-\cos (\pi - a)=\cos a$$
$$x_2=-\cos(b-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve in positive integers the equation $a^3+b^3=9ab$ Solve in positive integers the equation:
$$a^3+b^3=9ab$$
I try to:
$$\dfrac{a^2}{b}+\dfrac{b^2}{a}=9\Longrightarrow a^2<9b,b^2<9a$$
Of course, I can't solve it. Can anyone help?
| HINT: $x^3+y^3=9xy$ is the Cartesian equation of a Folium of Descartes $x^3+y^3=3axy$ with $a=3$. This is a rational cubic with a double point and its parametric equations are $x=\frac{9t}{1+t^3}$ and $y=tx$ The loop of the curve occurs for the values $t\gt 0$ so the integer points we are looking for belong to this loo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1691052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Criticize my math when I attempt to find the coefficient of $x^2y^6$ in the expansion of $(x+2y^2)^5$ So I look around this site and my textbook (Richmond&Richmond, discrete math) and I know I'm in the right direction but I'm also sure I am doing it wrong.
Original Question: find the coefficients of $x^2y^6$ in the exp... | There are two aspects in your derivation which should be corrected.
When selecting the terms with $x^2$ and $\left(2y^2\right)^3$ you have to multiply them instead of adding them.
\begin{align*}
(x+2 y^2)^5 \implies x^2\left(2y^2\right)^3=2^3x^2y^6
\end{align*}
When identifying the corresponding binomial coef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1691234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x$ How to evaluate
$$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x$$
I completely have no idea how to find the result.Mathematic gave me the following answer part of the integ... | \begin{align}J&=\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x\\
&\overset{y=\frac{1-x}{1+x}}=\frac{\pi}{4}\int_0^1\frac{1}{1+y^2}\ln\left ( \frac{1+y^{2}}{1+y} \right )\mathrm{d}y-J\\
2J&=\frac{\pi}{4}\int_0^1\frac{\ln\left ( 1+y^{2} \right )}{1+y^2}\mathrm{d}y-\frac{\pi}{4}\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Multiple radicals: $(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})$ $(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})$
I don't think multiplying these out will work... | It will be easier if you consider the expression
\begin{align}
(a+b+c)(a+b-c)(a-b+c)(-a+b+c)&=[(a+b)^2-c^2][-(a-b)^2+c^2]\\
&=(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)\\
&=4a^2b^2-(a^2+b^2-c^2)^2\\
&=4a^2b^2-a^4-b^4-c^4-2a^2b^2+2a^2c^2+2b^2c^2\\
&=2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)
\end{align}
And then you put $a=\sqrt{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is Sigma $\Sigma$ a mathematical way of doing a for loop? I've been a programmer for ten years, and once upon a time I was pretty good at math. Those days are long gone. I'm taking some online classes and now I find myself needing to remember the math I learned in college and it has all gone bye-bye.
Specifically, I ha... | In Mathematica I frequently switch between the Sum and the Table commands like in these examples:
Example 1:
$\sum\limits_{n=1}^{n=1} 1 = 1 = 1 $
$\sum\limits_{n=1}^{n=2} 1 = 2 = 1+1 $
$\sum\limits_{n=1}^{n=3} 1 = 3 = 1+1+1 $
$\sum\limits_{n=1}^{n=4} 1 = 4 = 1+1+1+1 $
$\sum\limits_{n=1}^{n=5} 1 = 5 = 1+1+1+1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 1
} |
If $f(x)=\lim_{n\to\infty}[2x+4x^3+\cdots+2nx^{2n-1}]$, $0
If $f(x)=\lim_{n\to\infty}[2x+4x^3+\cdots+2nx^{2n-1}]$, $0<x<1$, then find $\int f(x)\mathrm{d}x$
$$f(x)=\lim_{n\to\infty}2x[1+2x^2+\cdots+nx^{2n}]$$
$$S=\frac{f(x)}{2x}=1+2x^2+\cdots+nx^{2n}$$
$S$ is an AGP. I used the general method for finding $S$.
$$x^2S=x^... | The calculation is fine. Note that
$$\lim_{n\to\infty}x^{2n+1}=0\quad\text{and}\quad \lim_{n\to\infty} nx^{2n+2}=0.$$
Thus
$$f(x)=\frac{2x}{(1-x^2)^2}.$$
Now integrate, using the substitution $u=1-x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\sqrt[4]{\alpha}-\sqrt[4]{\beta}$,where $\sqrt[4]{.}$ denotes the principal value. If $\alpha$ and $\beta$ are the roots of the equation $x^2-34x+1=0$,find the value of $\sqrt[4]{\alpha}-\sqrt[4]{\beta}$,where $\sqrt[4]{.}$ denotes the principal value.
I found out the $\alpha$ and $\beta$.
$\alpha,\... | One may observe that
$$
(\sqrt{2}-1)^2=3-2\sqrt{2}
$$ and that
$$
(3-2\sqrt{2})^2=17-12\sqrt{2}
$$ thus
$$
(\sqrt{2}-1)^4=17-12\sqrt{2}
$$ giving
$$
\sqrt[4]{\alpha}=\sqrt[4]{17-12\sqrt{2}}=\sqrt{2}-1,
$$
similarly
$$
\sqrt[4]{\beta}=\sqrt[4]{17+12\sqrt{2}}=\sqrt{2}+1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Inequality involving four numbers Show that, if $a$, $b$, $c$ and $d$ are four positive numbers with sum $1$, then $$\frac 3 {1-a} + \frac 3 {1-b} + \frac 3 {1-c} + \frac 3 {1-d} \ge \frac 5 {1+a} + \frac 5 {1+b} + \frac 5 {1+c} + \frac 5 {1+d}.$$ I tried to subtract the fractions, but I didn't get to any result.
| Hint: It is enough to show that for $x\in (0,1)$,
$$f(x)=\frac3{1-x}-\frac5{1+x}+\frac{32}{15}(1-4x)=\frac{2(4x-1)^2(4x+1)}{15(1-x^2)}\ge0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Fermat-Torricelli minimum distance The Fermat - Torricelli point minimizes sum of distances $S$ taken from vertices of a triangle of sides $a,b,c. $ Find $S$ in terms of $a,b,c$.
Am trying to set up problem with a Lagrange multiplier or partial derivatives for extremization but it seems tedious even with a CAS.
Althou... | Let
*
*$a, b, c$ be the sides of $\triangle ABC$ whose angles are smaller than $120^\circ$.
*$P$ be the Fermat-Torricelli point for $\triangle ABC$.
*$\alpha = |AP|$, $\beta = |BP|$, $\gamma = |CP|$ and $S = \alpha+\beta+\gamma$.
*$\mathcal{A}$ be the area of $\triangle ABC$.
It is known that for such a trian... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
The right way to cancel out the terms in the following telescoping series So how do I cancel and simplify the terms in the following telescopic series.
Been at it for hours, cant seem to figure it out.
$\sum\limits_{k = 1}^n \frac{1}{2(k+1)} -\frac{1}{k+2}+\frac{1}{2(k+3)} $
Any help would be deeply appreciated.
P.S: I... | Here is another variant
\begin{align*}
\sum_{k=1}^n&\left(\frac{1}{2(k+1)}-\frac{1}{k+2}+\frac{1}{2(k+3)}\right)\\
&=\frac{1}{2}\sum_{k=1}^n\frac{1}{k+1}-\sum_{k=1}^n\frac{1}{k+2}+\frac{1}{2}\sum_{k=1}^n\frac{1}{k+3}\tag{1}\\
&=\frac{1}{2}\sum_{k=1}^{n}\frac{1}{k+1}-\sum_{k=2}^{n+1}\frac{1}{k+1}+\frac{1}{2}\sum_{k=3}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If a chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is a normal to it at $P$,then If a chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is a normal to it at $P$,then show that $\tan \... | The equation of the line $PQ$ is given by
$$y-a\tan\alpha=\frac{a\tan\alpha-a\tan\beta}{a\sec\alpha-a\sec\beta}(x-a\sec\alpha),$$
i.e.
$$\frac{\sin\alpha\cos\beta-\sin\beta\cos\alpha}{\sin\alpha-\sin\beta}x+\frac{\cos\alpha-\cos\beta}{\sin\alpha-\sin\beta}y=a\tag3$$
The $(2)$ you wrote can be written as
$$\frac{\cos\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the line integral of a parabola How can I evaluate :
$$\int_{C} y \;dx + x^2 \; dy$$
where $C$ is the parabola define by
$$y=4x-x^2 \quad \text{from } \; (4,0) \; \text{ to } \; (1,3).$$
Do I need to parameterize the parabola?
| $$\because y = 4x-x^2 , \therefore \frac{dy}{dx} = 4-2x , \therefore dy = (4-2x)dx $$
Hence,
$$\int_{C} y \;dx + x^2 \; dy = \int_{4}^{1} 4x-x^2 dx + \int_{4}^{1} x^2 (4-2x)dx$$
$$\implies \int_4^1 4x-x^2+4x^2-2x^3$$
$$\implies \int_4^1 4x+3x^2-2x^3$$
$$\implies (2x^2+x^3-\frac{x^4}{2})_4^1$$
$$\implies (2+1-\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1700639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{x \to 1}\frac{\sqrt{x^2+35}-6}{x-1}$ I've rationalized the numerator to
$x^2+35-36$
$$= \frac{x^2+35-36}{(x-1)\left(\sqrt{x^2+35}+6\right)} = \frac00$$
(when I substitute $x=1$)
I don't know what to do to the denominator so I can substitute $x=1$ to find the limit.
| You already have been given good answers to the problem.
Let me show you another approach which can be of interest (even just for your curiosity at the present time).
To make life simpler, set $x=1+y$. This makes
$$A=\frac{\sqrt{x^2+35}-6}{x-1}=\frac{\sqrt{y^2+2 y+36}-6}{y}$$ Consider the first portion of the numerator... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$\sqrt{x}-\sqrt{z+y}=\sqrt{y}-\sqrt{z+x}=\sqrt{z}-\sqrt{x+y}$ At a recent maths competition one of the questions was to find for which $x,y,z$ this equation holds true:
$$\sqrt{x}-\sqrt{z+y}=\sqrt{y}-\sqrt{z+x}=\sqrt{z}-\sqrt{x+y}$$
where $x,y,z \in \mathbb{R} \cup \{0\}$. So how am I supposed to approach this problem?... | $x \ge 0; y \ge 0; z \ge 0$.
Suppose $z = 0$
Then $\sqrt{x} - \sqrt{y} = \sqrt{y} - \sqrt{x} = - \sqrt{x+y} \implies \sqrt{x} = \sqrt{y} \implies x = y; x + y = 0 \implies x = y = z = 0$.
Likewise if $x = 0$ or $y = 0$ then $x = y = z = 0$ by the same argument so either they all equal 0 or none do.
So assume no $x, y, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Contraction of an ideal
Let $f: \mathbb{Z}[X] \longrightarrow \mathbb{Z}[\sqrt{2}]$ be a ring homomorphism sending $X$ to $\sqrt{2}$.
I am asked to compute a few contractions, and I am wondering if I could get some help in understanding the reasoning. I also was told that I need to compute the kernel of $f$, which I... | Note that for any ring homomorphism $f \colon R \to S$, and any ideal $J \subseteq S$, we have - due to $0 \in J$ -
$$ J^c = f^{-1}[J] \supseteq f^{-1}[0] = \ker f $$
so the kernel is contained in every contraction.
In the case here, we have $\ker f = (X^2 - 2)$, as you already found. For the three problems, we have
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to calculate this integral without any integration techniques?
Differentiate $f(x) = (5x+2)\ln(2x+1)$ with respect to $x$. Hence, find $\int \ln(2x+1)^3dx$.
Because of the word "Hence" I'm assuming that the question doesn't allow integration techniques such as integration by parts or substitution.
The first part... | The derivative is
$$f'(x) = 5 \log{(2 x+1)} + \frac{10 x+4}{2 x+1} = 5 \log{(2 x+1)} + 5 - \frac1{2 x+1} $$
Integrate both sides wrt $x$:
$$f(x) = (5 x+2) \log{(2 x+1)} = 5 \int \log{(2 x+1)} dx + 5 x - \frac12 \log{(2 x+1)} +C$$
Thus
$$ 5 \int \log{(2 x+1)} dx = \left (5 x + \frac52 \right ) \log{(2 x+1)} - 5 x + C$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Sum of the Powers of $2$ Suppose I have a sequence consisting of the first, say, $8$ consecutive powers of $2$ also including $1$: $1,2,4,8,16,32,64,128$. Why is it that for example, $1 + 2 + 4 = 7$ is $1$ less than the next term in the series, $8$? Even if one was to try, for instance, $524,288 + 1,048,576$ ($2^{19}$ ... | This is a consequence of a well-known formula, but here is an elementary proof for your case, just using distributivity and the fact that $2 - 1 = 1$:
\begin{align}
(1 + 2 + 4 + \dots + 2^n) &= (2-1)(1 + 2 + 4 + \dots + 2^n) \\
&= 2(1 + 2 + 4 + \dots + 2^n) - (1 + 2 + 4 + \dots + 2^n) \\
&= (2 + 4 + 8 + \dots + 2^{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Calculate the limit of: $x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$, $n \rightarrow \infty$ Is it ok to solve the following problem this way? What I have done is to solve parts of the limit first (that converges to $0$), and then solve the remaining expression? Or is this flawed reaso... | Let's try the usual approach of standard limits. We have
\begin{align}
L &= \lim_{n \to \infty}\frac{\log(1 + \sqrt{n} + \sqrt[3]{n})}{\log(1 + \sqrt[3]{n} + \sqrt[4]{n})}\notag\\
&= \lim_{n \to \infty}\dfrac{\log\sqrt{n} + \log\left(1 + \dfrac{1 + \sqrt[3]{n}}{\sqrt{n}}\right)}{\log\sqrt[3]{n} + \log\left(1 + \dfrac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Suppose $X, Y$ are random variables with the equal variance. Show that $X-Y$ and $X+Y$ are uncorrelated. Suppose that $X$ and $Y$ are random variables with the equal variance.
Show that $X-Y$ and $X+Y$ are uncorrelated.
I get I should use the equation $$E[XY] = E[X]E[Y]$$ For the first part I get $$E[(X-Y)(X+Y)] = E[X... | Suppose that
$$
\begin{pmatrix}
X\\
Y
\end{pmatrix}
$$
is a random vector with the covariance matrix
$$
\begin{pmatrix}
\sigma^2&\sigma_{X,Y}\\
\sigma_{X,Y}&\sigma^2
\end{pmatrix}.
$$
The covariance matrix of the random vector
$$
\begin{pmatrix}
1&1\\
1&-1
\end{pmatrix}
\begin{pmatrix}
X\\
Y
\end{pmatrix}
=
\begin{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1708770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Is it true that $x(1-x)=y(1-y)$ if and only if $x=y$? Consider two numbers $0\leq x\leq 1$ and $0\leq y\leq 1$. Is it true that $x(1-x)=y(1-y)$ if and only if $x=y$? If not, can you give a counterexample?
| $$x(1-x) = y(1-y) \implies -x^2+x-y(1-y)=0 \implies x^2-x+y(1-y)=0\text{.}$$
By the quadratic formula,
$$x = \dfrac{1\pm \sqrt{1-4(1)y(1-y)}}{2}=\dfrac{1 \pm \sqrt{1-4y(1-y)}}{2} = \dfrac{1\pm \sqrt{1-4y+4y^2}}{2} = \dfrac{1 \pm |2y-1|}{2}\text{,}$$
so any pair $\left(\dfrac{1 + |2y-1|}{2}, y\right)$ or $\left(\dfrac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Circle inscribed in Equilateral Triangles The circle inscribed in the triangle $ABC$ touches the sides $BC$ , $CA$ , and $AB$ in the points $A_1,B_1,C_1$ respectively. Similarly the circle inscribed in the triangle $A_1B_1C_1$ touches the sides in $A_2,B_2,C_2$ respectively, and so on. If $A_nB_nC_n$ be the $n^{th}$ $... | Quang Hoang has already provided a good hint.
Let us prove that by induction on $n$.
Let $O$ be the incenter of $\triangle{ABC}$.
Then, noting that $OB\perp A_1C_1,OC\perp A_1B_1$, we have
$$\begin{align}\angle{B_1A_1C_1}&=\pi -\angle{B_1A_1C}-\angle{C_1A_1B}\\&=\pi-\left(\pi-\frac{\pi}{2}-\frac C2\right)-\left(\pi-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Derive $\tan(3x)$ in terms of $\tan(x)$ using De Moivre's theorem
Derive the following identity: $$\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$$
The way I approached the questions is that I first derived $\sin(3x)$ and $\cos(3x)$ because $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$. Then substituting:
$$\tan... | Hint:
Notice that
\begin{align*}
\frac{4\sin^2 x-3}{-4\sin^2 x +1}&=\frac{1-4\cos^2 x}{4\cos^2 x-3 }\\[5pt]
&=\frac{\frac{1}{\cos^2 x}-4}{4-\frac{3}{\cos^2x}},\qquad\text{for }\cos x\neq 0\\[5pt]
&=\frac{\sec^2 x-4}{4-3\sec^2 x}\\[3pt]
&=\frac{(\color{blue}{\tan^2 x+1})-4}{4-3(\color{blue}{\tan^2 x+1})}
\end{align*}
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove: $2^k$ is the sum of two perfect squares If $k$ is a nonnegative integer, prove that $2^k$ can be represented as a sum of two perfect squares in exactly one way. (For example, the unique representation of $10$ is $3^2+1^2$; we do not count $1^2+3^2$ as different.)
I understand that $2^{2n}=0+2^{2n}$ and $2^{2n+1}... | I do not see anyone mentioning this simple aspect: if we have integers $u,v$ such that $$ u^2 + v^2 \equiv 0 \pmod 4, $$
then both $u,v$ must be even.
Which means this: take a number that is divisible by $4.$ Suppose we have
$$ x^2 + y^2 = n $$
Keep dividing $n$ by $4$ until the result, $n_0,$ is no longer divisible by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Integer solutions for $n$ for $|{\sqrt{n} - \sqrt{2011}}| < 1$ $$|{\sqrt{n} - \sqrt{2011}}| < 1$$
What is the number of positive integer $n$ values, which satisfy the above inequality.
My effort:
$
({\sqrt{n} - \sqrt{2011}})^2 < 1 \\n + 2011 -2\sqrt{2011n} < 1\\ n+2010<2\sqrt{2011n}\\ n^2+2 \times 2010 \times n +2010^... | $|\sqrt n - \sqrt{2011}| <1 \iff -1 < \sqrt n - \sqrt {2011 }< 1 \iff
\sqrt{2011} - 1 <\sqrt n < \sqrt {2011} + 1 \iff (\sqrt{2011} - 1)^2 < n < (\sqrt{2011}+1)^2.$ You should be able to finish it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Is there a positive integer n such that the fraction $(9n+5)/(10n+3)$ is not in the lowest term? Is there a positive integer n such that the fraction $(9n+5)/(10n+3)$ is not in the lowest term? Please explain.
I found $n=2$ be a solution. Is it correct?
| Your example $n=2$ is correct. We will find all possible examples.
Suppose that $d$ divides $9n+5$ and $10n+3$. Then $d$ divides
$$10(9n+5)-9(10n+3),$$
so $d$ divides $23$.
Since $23$ is prime, the only conceivable (positive) common divisors of $9n+5$ and $10n+3$ are $1$ and $23$. We will find the values of $n$ for w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $abcd=1$,where $a,b,c,d$ are positive reals,then find the minimum value of $a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd$. If $abcd=1$,where $a,b,c,d$ are positive reals,then find the minimum value of $a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd$.
Let $E=a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd=(a+b+c+d)^2-(ab+ac+ad+bc+bd+cd)$
I do not know h... | HINT
Use the AM-GM inequality, which establishes $$\frac{a_1+a_2+\dots+a_n}{n} \ge \sqrt[n]{a_1a_2\dots a_n}$$
When $a_1,a_2, \dots, a_n$ are positive reals.
For example, $a^2+b^2+c^2+d^2 \ge 4\sqrt[4]{a^2b^2c^2d^2}=4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove $1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$ The task is to prove the following non-equality by hand:
$$1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$$
Wolframalpha s... | Here, we will prove the equation by @AndreNicolas and @Blue in a more elementary manner.
$$ 1+4\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) = 0 $$
Note that since $\cos 3x=4\cos^3 x-3\cos x$, and since $\cos 2x=2\cos^2 x-1$, our equation simplifies to $$ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac... | $$\cos(x)+\sin(x)=\frac{1}{3}\Longleftrightarrow$$
Use:
$$\cos(x)+\sin(x)=\sqrt{2}\left[\frac{\cos(x)}{\sqrt{2}}+\frac{\sin(x)}{\sqrt{2}}\right]=$$
$$\sqrt{2}\left(\sin\left(\frac{\pi}{4}\right)\cos(x)+\cos\left(\frac{\pi}{4}\right)\sin(x)\right)=\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)$$
$$\sqrt{2}\sin\left(\frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
} |
integrate $\int \frac{\tan^4x}{4}\cos^3x$
$$\int \frac{\tan^4x}{4}\cos^3x$$
$$\int \frac{\tan^4x}{4}\cos^3x=\frac{1}{4}\int \frac{\sin^4x}{\cos^4x}\cos^3x=\frac{1}{4}\int\frac{\sin^4x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot\sin^2x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot(1-\cos^2x)}{\cos x}=\frac{1}{4}\int \frac... | Omitting the $\frac{1}{4}$, we have that
$\displaystyle\tan^4 x\cos^3 x=\frac{\sin^4 x}{\cos x}=\frac{(1-\cos^2 x)^2}{\cos x}=\frac{1-2\cos^2 x+\cos^4 x}{\cos x}=\sec x-2\cos x+\cos^3 x$
$\hspace{.85 in}\displaystyle=\sec x-2 \cos x+\cos x(1-\sin^2x)=\sec x-\cos x-\sin^2 x\cos x$,
so $\displaystyle\int\tan^4x\cos^3x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How to integrate $\int \frac{ev+f}{av^2 + bv +c} dv$? I want to integrate
\begin{align}
\int \frac{ev+f}{av^2 + bv +c} dv
\end{align}
can you give me some hints or detail procedure for this integral?
From mathematca, i have
\begin{align}
\frac{-\frac{2 (b e-2 a f) \text{ArcTan}\left[\frac{b+2 a v}{\sqrt{-b^2+4 a c}... | $$
\int \frac{ev+f}{av^2 + bv +c} dv = \frac{e}{2a}\int \frac{d(av^2 + bv + c)}{av^2 + bv +c} + \left(f-\frac{be}{2a}\right)\int \frac{1}{av^2 + bv +c} dv
$$
$$
= \frac{e}{2a} \ln(av^2+bv+c) + \left(f-\frac{be}{2a}\right)\int \frac{1}{(\sqrt{a}v+\frac{b}{2\sqrt{a}})^2 - \frac{b^2}{4a} +c} dv
$$
$$
\int \frac{1}{(\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
In $\triangle ABC$, if $\cos A\cos B\cos C=\frac{1}{3}$, then $\tan A\tan B+\tan B \tan C+\tan C\tan A =\text{???}$
In $\triangle ABC$, if
$$\cos A \cos B \cos C=\frac{1}{3}$$
then can we find value of
$$\tan A\tan B+\tan B \tan C+\tan C\tan A\ ?$$
Please give some hint. I am not sure if $\tan A \tan B+\tan ... | The answer to the question is no. The maximum value of $\cos A \cos B \cos C$, where $A$, $B$, and $C$ are the angles of a triangle in the plane, is $\frac{1}{8}$, so there is no plane triangle for which $\cos A \cos B \cos C=\frac{1}{3}$.
The product $\cos A \cos B \cos C$ equals $\frac{1}{8}$ for an equilateral trian... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Use differentiation to find a power series Use differentiation to find a power series of
$f(x) = \frac{1}{(8+x)^2}$
$ f'(x) = \frac{-2}{(8+x)^3} $
how do I find the power series of this? I can not go next step.
| $$
g(x) = \frac{1}{(8+x)} = \sum_{n=0}^{\infty} \frac{(-1)^nx^n}{8^{n+1}}
$$
Taking the derivative in both sides: (n=0 is constant)
$$
g'(x) = \frac{-1}{(8+x)^2} = \sum_{n=1}^{\infty} \frac{n(-1)^nx^{n-1}}{8^{n+1}}
$$
$$
-f(x) = \frac{-1}{(8+x)^2} = \sum_{n=0}^{\infty} \frac{(n+1)(-1)^{n+1}x^{n}}{8^{n+2}}
$$
$$
f(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Last Digit of $x^0 + x^1 + x^2 + \cdots + x^{p-1} + x^p$
Given $x$ and $p$. Find the last digit of $x^0 + x^1 + x^2 + \cdots + x^{p-1} + x^p$
I need a general formula. I can find that the sum is equal to
$\dfrac{x^{p+1}-1}{x-1}$
But how to find the last digit.
P.S: $x\leq 999999$ and $p \leq 10^{15}$
| Assuming that $x>0$.
Let $d$ denote the last digit of the series.
You can split the answer into $10$ different cases.
The following cases are rather simple:
*
*$x\equiv0\pmod{10} \implies d=1$
*$x\equiv1\pmod{10} \implies d=(1+p)\bmod{10}$
*$x\equiv4\pmod{10} \implies d=1+4(p\bmod{2})$
*$x\equiv5\pmod{10} \impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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If the sum to 4 terms of a geometric progression is 15 and the sum to infinity is 16 find the possible values of the common ratio. I can't find a way to get an answer for this. I have tried using the formula for the sum to infinity and dividing it by the sum to 4 terms but i can't get it to work.
| Let the sequence have initial term $a_1$ and common ratio $r$. Then $a_k = a_1r^{k - 1}$. The sum of the first n terms of the geometric series is
$$\sum_{k = 1}^{n} a_1r^{k - 1} = a_1(1 + r + r^2 + \cdots + r^{n - 1}) = a_1 \frac{1 - r^n}{1 - r}$$
provided that $r \neq 1$. If $r = 1$, then the series would not con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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Using squeeze thorem find $ \lim_{n \to \infty}{\frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}}$ It is already solved here at Math.stackexchange, but we haven't learned Stirling's approximation (at our school), so can it be solved using only squeeze theorem?
$$\lim_{n \to \infty}{\fr... | Let
$$
b_n=\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots2n}.
$$
We shall show inductively that
$$
\frac{2}{\sqrt{2n+1}}>b_n.
$$
For $n=1$ it clearly holds.
Assume that
$$
\frac{2}{\sqrt{2k+1}}>\frac{1\cdot 3\cdots (2k-1)}{2\cdot 4\cdots2k}.
$$
Then
$$
\frac{2}{\sqrt{2k+1}}\cdot\frac{2k+1}{2k+2}>\frac{1\cdot 3\cdots (2k-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Compositeness test for repunits Is this proof acceptable ?
Definition
Let $R_p=\frac{10^p-1}{9} $ with $p$ prime be a repunit number .
Theorem
If $R_p$ is prime then $7^{\frac{R_p-1}{2}} \equiv -1 \pmod {R_p}$
Proof
Let $R_p$ be a prime , then by Euler's criterion :
$7^{\frac{R_p-1}{2}} \equiv \left(\frac{7}{R_p}\right... | Looks all good but as Crostul I feel that it is good to add why $R_p \equiv 1,2,4 \pmod 7$. It is easy to prove that $R_p \equiv 1,2 \pmod 7$ for odd $p$:
As $16 \cdot 63 = 1008$, we have $10^3 \equiv -8 \pmod {63}$ and thus $10^6 \equiv 1 \pmod {63}$. This implies $$10^{1+6n} \equiv 10 \pmod {63}$$ and $$10^{5+6n} \eq... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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proof limit of $\frac{1}{(x^{2}-1)}$ = -1 using $\epsilon - \delta$ as $x_{0} \to 0$ Question: proof limit of $\frac{1}{(x^{2}-1)}$ = -1 using $\epsilon - \delta$ as $x_{0} \to 0$
Hi... I am stumped on an intro analysis problem.
so it is stated Show $lim \frac{1}{x^{2}-1}$ = -1 as $x_{0} \to 0$.
here is my work, and I... | Note that $$|f(x)-L|=\left|\frac1{x^2-1}-1 \right|=\left|\frac{x^2}{(x-1)(x+1)} \right|=\frac{x^2}{|x-1||x+1|}, $$ so if $|x|<\frac12$, then $|x-1||x+1|>2|x|$ and $|f(x)-L|<\frac12|x|$. So choosing $\delta=\min\left\{\varepsilon,\frac12\right\}$ we see that $|x|<\delta$ implies $$|f(x)-L|<\frac12|x|<\varepsilon. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Function parameters and cartesian curves Given
$$x = \cos t + \cos 2t \,,\; y = \sin t + \sin 2t ,$$
find the tangent line for the parameter at point $(-1, 1),$ and draw a graph of the curve.
To find the point you could simply do $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ and find $\dfrac{dy}{dx}$ by taking the quotient, ... | Another way by finding implicit equation
\begin{align*}
x^{2} &= \cos^{2} t+2\cos t \cos 2t+ \cos^{2} 2t \\
y^{2} &= \sin^{2} t+2\sin t \sin 2t+ \sin^{2} 2t \\
x^{2}+y^{2} &= 2+2(\cos t \cos 2t+\sin t \sin 2t) \\
&= 2+2\cos t \\
\cos t &= \frac{x^{2}+y^{2}}{2}-1 \\
x &= \cos t+2\cos^{2} t-1 \\
&=(\cos t+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Change in Interval of convergence if center of convergence changes So I have to find a power series that is centered at $-2^{1/2}$
If I choose to use the power series expansion for $e^x$ which converges for all $x$, and change $x$ to $x + 2^{1/2}$ does the interval of convergence remain the same?
| If the radius of convergence of a power series is positive and finite, it is the distance from the center of the series expansion to the nearest singularity or branchpoint. If the center changes, the distance to the nearest singularity or branchpoint may change and so the radius of convergence will also change.
Conside... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find remainder when $1^{5} + 2^{5} \cdots +100^{5}$ divided by 4 I'm studding D.M Burton & want to solve: Find remainder when $1^{5} + 2^{5} \cdots +100^{5}$ divided by $4$. . Please help me by giving your solution to it. I'm new comer to number theory so please don't use theorems above Theory of Congruence.
| You can do it even without knowledge of congruences, we have that $1^5 + 2^5 + ... + n^5= \dfrac {(n(n+1))^2(2n^2+2n-1)}{12}$, now set $n=100$ to get $1^5 + 2^5 + ... + 100^5 = \dfrac {(100 \cdot 101)^2 (2 \cdot 100^2 + 2 \cdot 100 - 1)}{12}=4 \cdot \dfrac {4 \cdot 25^2 \cdot 101^2 \cdot 3 \cdot 6733}{12}$ so your numb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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solve $\sin 2x + \sin x = 0 $ using addition formula
$\sin 2x + \sin x = 0 $
Using the addition formula, I know that
$\sin 2x = 2\sin x \cos x$
=> $2\sin x \cos x + \sin x = 0$
=> $\sin x(2\cos x + 1) = 0$
=> $\sin x = 0$ and $\cos x = -\frac{1}2 $
I know that $\sin x = 0$ in first and second quadrant so $x = 0$ and ... | $$\sin(2x)+\sin(x)=0 $$
First, you have to remember that:
$$\sin(a)+\sin(b)=2\sin(\frac{a+b}{2})\cos(\frac{a-b}{2})$$
Aplying this in the equation:
$$2\sin(\frac{3x}{2})\cos(\frac{x}{2})=0$$
$$\sin(\frac{3x}{2})\cos(\frac{x}{2})=0$$
Finally, you have two equations:
$$\sin(\frac{3x}{2})=0$$
$$\cos(\frac{x}{2})=0$$
F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many strings of length 8 The question - how many strings of length 8, from 4 letter alphabet, using each letter twice. There is to be exactly one pair of same letters next to each other (example of valid string: AABCDBCD).
I tried to check how this develops by drawing a tree, but it seems that this gets really unwi... | If there were no restrictions, we could arrange the letters AABBCCDD in
$$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{8!}{2!6!} \cdot \frac{6!}{4!2!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{0!2!} = \frac{8!}{2!2!2!2!}$$
distinguishable ways since we can place the two A's in two of the eight available positi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Alternative "Fibonacci" sequences and ratio convergence So the well known Fibonacci sequence is
$$
F=\{1,1,2,3,5,8,13,21,\ldots\}
$$
where $f_1=f_2=1$ and $f_k=f_{k-1}+f_{k-2}$ for $k>2$. The ratio of $f_k:f_{k-1}$ approaches the Golden Ratio the further you go:
$$\lim_{k \rightarrow \infty} \frac{f_k}{f_{k-1}} =\p... | From the standard theory of linear recurrence,s $F_n$ is the positive real root of the equation $z^n = z^{n-1} + z^{n-2} + \cdots + z + 1$. Multiplying both sides by $1-z$ and rearranging, you get that $F_n$ is the positive real root of $f_n(z) = z^{n+1} - 2z^n + 1 = 0$ that is not $z = 1$. (By Descartes' rule of sig... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Calculate the sum of the infinite series $ 1 + \frac{1+2}{2!} + \frac{1+2+3}{3!} .... $ Calculate the sum of the infinite series $ 1 + \frac{1+2}{2!} + \frac{1+2+3}{3!} .... $
My attempt : I recognised that this series can be decomposed into the taylor expansion of $ e $ around 0.
So I thought of writing the series as... | Using the well-known formula for $1+2+\dots\;$your sum is
$$\sum_{n=1}^\infty \frac{n(n+1)}{2}\frac{1}{n!}=
\sum_{n=1}^\infty \frac{(n+1)}{2(n-1)!}=
\sum_{n=0}^\infty \frac{(n+2)}{2n!}=
\sum_{n=0}^\infty \left(\frac{n}{2n!} + \frac{2}{2n!}\right)$$
$$=\frac{1}{2}\sum_{n=0}^\infty \frac{n}{n!} + \sum_{n=0}^\infty \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Let $x \ge 0$. Determine a condtion on $|x-4|$ that'll assure $|\sqrt{x} - 2| < 10^{-2}$ I'm trying to understand the logic of this proof.
Let $x \ge 0$. Determine a condition on $|x-4|$ that'll assure $|\sqrt{x} - 2| < 10^{-2}$
Proof
$|\sqrt{x} - 2| = \frac{|(\sqrt{x} - 2)(\sqrt{x} + 2)|}{\sqrt{x} + 2} = \frac{x-4}{\... | What you need to do is find bounds for $\frac{1}{\sqrt x + 2}$
Suppose $|x - 4| < \delta$. Then we can argue that
\begin{align}
|x - 4| < \delta
&\implies4-\delta < x < 4 + \delta \\
&\implies \sqrt{4-\delta} < \sqrt x < \sqrt{4 + \delta}\\
&\implies \sqrt{4-\delta} + 2 < \sqrt x + 2 < \sqrt{4 + \delta... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I show that $\sum_{cyc} \frac {a^6}{b^2 + c^2} \ge \frac {abc(a + b + c)}2?$ Let $a, b, c$ be positive real numbers, show that
$$\frac {a^6}{b^2 + c^2} + \frac {b^6}{c^2 + a^2} + \frac {c^6}{a^2 + b^2} \ge \frac {abc(a + b + c)}2.$$
I think this is likely to turn out to be proved by Hölder, but I can't see how. ... | Note that both sides are homogeneous of degree 4: ie, if we replace $a,b,c$ by $ta,tb,tc$ where $t>0$, both sides get multiplied by $t^4$ which has no effect on the inequality.
If we rescale by $t=1/(a+b+c)$, the effect is the same as requiring that $a+b+c=1$. Often this is formulated as the statement that we can choos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is
$(A)$equilateral
$(B)$isosceles
$(C)$right angled
$(D)$none of these
The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$.
Using sine rule,
$a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get
$\sin^2A+... | Considering cosine law:
\begin{align*}
\sum_{abc} (b^2+c^2-a^2) &= \sum_{abc} 2bc\cos A \\
a^2+b^2+c^2&= 2bc\cos A+2ca\cos B+2ab\cos C
\end{align*}
One possibility is $\cos A=0$, $2\cos B=1$ and $2\cos C=\sqrt{3}$.
Hence $A=90^{\circ}$, $B=60^{\circ}$ and $C=30^{\circ}$.
Proof of Uniqueness of the above solution:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Find the limit a matrix raised to $n$ when $n$ goes to infinity Let $ A $ be a $ 3\times3 $ matrix such that
$$A
\left( \begin{array}{ccc}
1 \\
2 \\
1 \end{array} \right)=\left( \begin{array}{ccc}
1 \\
2 \\
1 \end{array} \right),~~~A
\left( \begin{array}{ccc}
2 \\
2 \\
0 \end{array} \right)=\left( \begin{array}{c... | Use diagonalization. $A=PDP^{-1}$, where $P=\pmatrix{1&2&3\cr 2&2&0\cr 1&0&6\cr}$ and $D=\pmatrix{1&0&0\cr 0&1/2&0\cr 0&0&-1/3\cr}$. Then
$$\lim_{n\to\infty} A^n \pmatrix{6\cr 7\cr 0\cr} = \lim_{n\to\infty} (PD^n P^{-1}) \cdot \pmatrix{6\cr 7\cr 0\cr} = P \lim_{n\to\infty} D^n \cdot P^{-1} \cdot \pmatrix{6\cr 7\cr 0\cr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Limit of tan function This is a question from an old tutorial for a basic mathematical analysis module.
Show that $$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = e^2$$
My tutor has already gone through this in class but I am still confused. Is there anything wrong with the following reasoning?
Since $\... | If the only standard limit available is $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = e$$ then we are out of luck here. It is better to assume the following standard limits $$\lim_{x \to 0}\frac{\tan x}{x} = 1 = \lim_{x \to 0}\frac{\log(1 + x)}{x}$$ and then we can easily evaluate the desired limit. We have
\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Is the following solution correct? Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$
My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$
$=> \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) =... | It would indeed , because then you wont have $x=\pm 3i$ as the root because that would make the denominator zero , which is the greatest offense one can do in algebra!! or more broadly in Mathematics (:P)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Find the limit $\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$ Find the limit if it exists. (this exercise is taken from Calculus - The Classic Edition by Swokowski Chapter 10, section 1, no. 9, p.498)
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$$
since the limit is $0/0$ therefore, we use L'Hopital's ... | You can make it a little simpler by rewriting
$$\frac{\sin(x)-x}{\tan(x)-x}=\cos(x)\frac{\sin(x)-x}{\sin(x)-x\cos(x)}.$$
The first factor tends to $1$ and can be ignored. Then, applying L'Hospital three times,
$$\frac{\sin(x)-x}{\sin(x)-x\cos(x)}\to\frac{\cos(x)-1}{\cos(x)-\cos(x)+x\sin(x)}\to\frac{-\sin(x)}{\sin(x)+x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
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Solve for a in Exponential equation Is it possible to solve for $a$ in the following equation: $a^\alpha=b^\alpha-a$? Currently, I have resorted to using Excel to approximate $a$ (I am given values for $b$ and $\alpha$), but am wondering if it is possible to pinpoint $a$ exactly.
| Although it can not be solved generally, it can be solved if $\alpha \in \{ -1,0, \frac 12,1,2 \}$
Let's consider these in turn:
*
*$\alpha = -1$
$a^{-1} =b^{-1}-a$
$b =a-a^2b$
$a^2b-a+b=0$
$a= \displaystyle \frac {1 \pm \sqrt{(-1)^2-4(b)(b)}}{2b}$
$a= \displaystyle \frac {1 \pm \sqrt{1-4b^2}}{2b}$
*$\alpha = ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given that $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B + \sin C = 0$. If $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B + \sin C = 0$.
The value of $ \sin^3A+\sin^3B+\sin^3C$
What I can see here is that as $\sin A + \sin B + \sin C = 0$ hence $ \sin^3A+\sin^3B+\sin^3C=3\sin A \sin B\sin C$ but I am not ... | Hint: If the centroid of a triangle coincides with it circumcenter, the triangle is equilateral.
Answer: $\sin^3(A)+\sin^3(B)+\sin^3(C)=3\sin^3(A)-\frac{9}{4}\sin(A)=-\frac{3}{4}\sin(3A)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Surface of sphere above/below ellipse I am struggling with the following problem: Find the surface area of $x^2+y^2+z^2=a^2$ enclosed by the cylinder $\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$ $(a>b>0)$. The solution of the problem is supposed to be $4 \pi a^2 - 8a^2 \arcsin\left(\frac{\sqrt{a^2-b^2}}{a}\right)$.
I tried pol... | It turns out we don't really need any calculus to solve this problem.
For completeness, let us first solve this problem the calculus way.
Method 1 - using spherical polar coordinates.
By symmetry, we only need to compute the area on the upper hemisphere and then multiply the result by $2$.
Let $c^2 = a^2-b^2$ and param... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find continuities with square root? I don't understand how to find $$\frac{4-x^2}{3-\sqrt{x^2+5}}$$
The book says to multiply the equation by $\frac{3 + \sqrt {x^2+5}}{3 + \sqrt {x^2+5}}$. I don't understand where that comes from. It says the multiplication simplifies to "$3 + \sqrt {x^2+5}$" - I don't see how t... | Just use the binomial identity
$$(a-b)(a+b)=a^2-b^2,$$
which for $a=3$ and $b=\sqrt{x^2+5}$ yields
$$\frac{(4-x^2)(3+\sqrt{x^2+5})}{(3-\sqrt{x^2+5})(3+\sqrt{x^2+5})}=\frac{(4-x^2)(3+\sqrt{x^2+5})}{9-(x^2+5)}=\frac{(4-x^2)(3+\sqrt{x^2+5})}{(4-x^2)}$$
which is your desired result $(3+\sqrt{x^2+5}).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Expanding a complex function in Taylor series Expand the function
$$ f(z) = \frac {2(z + 2)} {z^2 − 4z + 3} $$
in a Taylor series about the point $ z = 2 $ and find the circle C inside of which the series converges. Find a Laurent series that converges in the region outside of C.
I tried writing the denominator as $ (z... | Split them into partial fraction and upon solving you will get,
$ \frac {2(z+2)}{(z-1)(z-3)} = \frac {3} {(z-(-1))} + \frac {5} {z-3} $
taylor's series expansion is
$ \frac {1}{1-x} = \frac{1}{1-a} + \frac {x-a}{(1-a)^2} + \frac {(x-a)^2}{(1-a)^3} + ...$
so for $\frac {3}{(z-(-1)} $ , value of a=-1 i.e., $ \ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum using The question is as follows:
Find the sum:
$1\cdot2 + 2\cdot3 + ... + (n-1)n$
What I have tried so far:
We can write $(n-1)n$ as $\frac{(n+1)!}{(n-1)!}$ which we can also write as $2\cdot\dbinom{n+1}{2}$
I believe it is possible to use the binomial theorem here, setting $a = b = 1$ in $(a+b)^n$. ... | Hint: Rewrite $k(k+1)$ as $\frac{1}{3}((k+1)^3-k^3)-\frac{1}{3}$. Observe the mass cancellation (telescoping) when we add up.
Another way: As remarked in the post, we want
$$2\left(\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\cdots+\binom{n}{2}\right).$$
We show that the sum of the binomial coefficients is $\binom{n+1}{3}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all function that satisfy $(f(x) + xy) \cdot f(x - 3y) + (f(y) + xy) \cdot f(3x - y) = (f(x + y))^2$ Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all real numbers $x$ and $y$,$$(f(x) + xy) \cdot f(x - 3y) + (f(y) + xy) \cdot f(3x - y) = (f(x + y))^2.$$
| Letting $x=y=0$, we have $ f(0)^2+f(0)^2=f(0)^2$, hence $$\tag1f(0)=0.$$
If $a=x-3y$ and $b=3x-y$ then $x+y=\frac{b-a}2$, whence
$$\tag2f(a)=f(b)=0\implies f\left(\frac{b-a}2\right)=0$$
and in combination with $(1)$
$$ \tag3f(x)=0\implies f\left(\pm\tfrac12x\right)=0.$$
Assume $f(a)=0$ (with $a\ne 0$). Then for $x=\fra... | {
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"timestamp": "2023-03-29T00:00:00",
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$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$ - I keep getting imaginary numbers $$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$$
My attempt
$\sqrt{x+938^2} + \sqrt{x + 140^2} = 1116$
$(\sqrt{x+938^2} + \sqrt{x + 140^2})^2 = (1116)^2$
$x+938^2 + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} + x + 140^2 = 1116^2$
$2x... | Repeated squaring is not necessary. Let $y > -140$ be such that $x = y^2 + 280y$, so that $$\sqrt{x+140^2} - 140 = \sqrt{(y + 140)^2} - 140 = y.$$ Then $$\begin{align*} 38 &= \sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 \\ &= \sqrt{y^2 + 280y + 938^2} - 938 + y, \end{align*}$$ or equivalently, $$976-y = \sqrt{y^2 + ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb Q$ Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb{Q}$ (by finding a nonzero polynomial $p(x)$ with coefficients in $\mathbb{Q}$ which has $\sqrt[3] 2+\sqrt 5$ as a root).
I first tried letting $a=\sqrt[3]{2} +\sqrt{5}$ and then square both si... | $x = \sqrt[3]{2} + \sqrt{5}$
$x - \sqrt{5} = \sqrt[3]{2}$
$x^3 - 3\sqrt{5}x^2 + 15x - 5\sqrt{5} = 2$
$x^3 - 15x - 2 = \sqrt{5}(3x^2 + 5)$
$(x^3 - 15x - 2)^2 = 5(3x^2 + 5)^2$
$x^6 -75x^4 - 4x^3 - 400x^2 + 60x - 121 = 0$
Now, ignoring accuracy and assuming arithmetical errors are both inevitable and irrelevant, there is... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all integer values of $x$ such that $x^2 + 13x + 3$ is a perfect integer square.
Question: Find all integer values of $x$ such that $x^2 + 13x + 3$ is a perfect integer square.
What I have attempted;
For $x^2 + 13x + 3$ to be a perfect integer square let it equal $k^2$ where $k \in \mathbb{Z} $
Hence $$x^2 +... | Observe that $x^2+13x+3=(x+\frac{13}{2})^2-\frac{157}{4}$
As you have assumed, let $(x+\frac{13}{2})^2-\frac{157}{4}=k^2$
Hence, by simplifying, we get that $(2x+13)^2-157=4k^2$
Further simplifying by factorisation yields $(2x+13-2k)(2x+13+2k)=157=157\cdot 1$
So, there are only two possibilities:
*
*$(2x+13-2k)=157$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Derivation of Spherical Law of Cosines I am trying to get a derivation of the spherical law of cosines. The Wikipedia page [https://en.wikipedia.org/wiki/Spherical_law_of_cosines ] contains a proof that I don't understand because there are not enough intermediate steps shown.
The Wikipedia page says that for unit vect... | $$\mathbf{t}_a \cdot \mathbf{t}_b = \frac{\mathbf{v} - \mathbf{u} \cos(a)}{\sin(a)} \cdot \frac{\mathbf{w} - \mathbf{u} \cos(b)}{\sin(b)} \\ = \frac{\mathbf{v} \cdot \mathbf{w} - \mathbf{v} \cdot \mathbf{u} \cos(b) - \mathbf{u} \cdot \mathbf{w} \cos(a) + \mathbf{u} \cdot \mathbf{u}\cos(a)\cos(b)}{\sin(a)\sin(b)} \\ = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Diophantine $4x^3-y^2=3$ I am interested in how to tackle this Diophantine equation:
$$4x^3-y^2=3$$
The solutions I have found so far are $(1,1)$ and $(7,37)$. Are there any more?
I have looked up various material on cubic Diophantines but most of what I’ve found is on equations where the coefficients of $x^3$ and $y^2... | Let $x$ and $y$ be integers such that $4x^3-y^2=3$. Reducing mod $2$ shows that $y$ is odd, so $y=2z+1$ for some integer $z$. Then in the Eisenstein integers $\Bbb{Z}[\omega]$, where $\omega^2+\omega+1=0$, we have
\begin{eqnarray*}
x^3&=&\frac{y^2+3}{4}
=\left(\frac{y+\sqrt{-3}}{2}\right)\left(\frac{y-\sqrt{-3}}{2}\rig... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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How to prove that $(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$, where $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$ $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$, prove
$$(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$$
I try several trig substitutions but feel hopeless with the cyclic term here. The condition $x^2+y^2+z^2... | $\color{brown}{\textbf{Trigonometrical substitution.}}$
From the given conditions should
\begin{cases}
x,y,z \in [0,2]\\[4pt]
(2z+xy)^2 = (4-x^2)(4-y^2).\tag1
\end{cases}
Taking in account $(1),$ can be applied substitution
$$x=2\sin a,\quad y=2\sin b\quad \Rightarrow \quad z = 2\cos(a+b) = 2\sin c,\tag2$$
where
$$a\ge... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Evaluation of $\int_{0}^{1}\frac{x^4(1-x^2)^5}{(1+x^2)^{10}}dx$
Evaluation of $\displaystyle \int_{0}^{1}\frac{x^4(1-x^2)^5}{(1+x^2)^{10}}dx$
$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{x^4(1-x^2)^5}{(1+x^2)^{10}}dx$$
Now Put $x=\tan \theta\;,$ Then $dx = \sec^2 \theta d\theta$ and changing limits, We get
$$I = \in... |
There was a small error in the development in the OP. The integral $I$ should be $$I=\frac{1}{32}\int_0^1 t^4(1-t^2)^2\,dt$$
We can evaluate the integral of interest in terms of the Beta function, $$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt$$and its relationship to the Gamma function, $$B (x,y)=\frac{\Gamma(x)\Gamma(y)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove: $\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$
Prove the trigonometric identity
$$\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$$
I've searched high and low on the net and cannot find identities where there is $+$ or $- 1$'s in the equation. Any help is appreciated.
Edit af... | hint: $(\sec x -1)(\sec x + 1) = \sec^2 x - 1 = \tan^2 x = \dfrac{\sin^2 x}{\cos^2 x}$, and $(1-\cos x)(1+ \cos x) = 1 - \cos^2 x = \sin^2 x$. I hope you can pull it off...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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