Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Uncertain how the following step was accomplished. I'm working through a book example that aims to find the first two nonzero terms of the Laurent expansion of $f(z)=\tan(z)$, about $z=\frac{\pi}{2}$.
The substitution $z=\frac{\pi}{2}+u$ is made
$$f(z)=\frac{\sin(\frac{\pi}{2}+u)}{\cos(\frac{\pi}{2}+u)}=-\frac{\cos(u)}... | Here we do a geometric series expansion
\begin{align*}
\frac{1}{1-u}&=\sum_{n=0}^{\infty}u^n\\
&=1+u+u^2+u^3+\cdots \qquad\qquad\qquad\qquad\qquad |u|<1
\end{align*}
If we substitute $h(u)$ for $u$ we obtain
\begin{align*}
\frac{1}{1-h(u)}&=\sum_{n=0}^{\infty}\left(h(u)\right)^n\\
&=1+h(u)+\left(h(u)\right)^2+\left(h(u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is
$(A)$equilateral
$(B)$isosceles
$(C)$right angled
$(D)$none of these
The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$.
Using sine rule,
$a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get
$\sin^2A+... | If $\triangle ABC$ is equilateral, then $3x^2=x^2(\sqrt{3}+1)$, absurd.
Now, note that the condition is symmetric respect to $b$ and $c$, ie, if $(a,b,c)$ satisfies, $(a,c,b)$ too, and then $\triangle ABC$ is isosceles with $b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
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Proof $\sum \frac{1}{n^a}$ is convergent for a > 1 I get to the fact that $\sum_{k=1}^n \frac{1}{k^a}$ < $\frac{1}{a-1} - \frac{1}{(a-1)(n+1)^{a-1}} - \frac{1}{(n+1)^a} + 1$ and hence $\sum_{k=1}^n \frac{1}{k^a}$ is bounded. How to deduce $\sum \frac{1}{n^a}$ is convergent ?
Note that a > 1
| Recall that a series of positive terms converges if the sequence of partial sums is bounded. Consider the $n$th partial sum of the series,
$$s_n = \sum_{k = 1}^n \frac{1}{n^a}$$
For each $m \ge 0$, $\frac{1}{k^a} \le \frac{1}{2^{ma}}$ for $k = 2^m, 2^m+1,\ldots, 2^{m+1}-1$. Choose a natural number $M$ such that $2^{M+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit a matrix raised to $n$ when $n$ goes to infinity Let $ A $ be a $ 3\times3 $ matrix such that
$$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... | Hint:
Notice that with the given vectors, assuming no $-$ sign,
$$Au=u,Av=2v,Aw=\frac13w,$$ so that
$$A^nu=u,A^nv=2^nv,A^nw=\frac1{3^n}w.$$
Then if you decompose the fourth vectors as a linear combination of $u,v,w$, you should easily see how it is transformed by $A^n$.
If we keep the $-$ sign,
The third equation will... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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'Destruction' of $x$ in second derivative? Can you tell whether I've taken this second derivative and determined its inflection points correctly?
$$f'(x) = (1-x)e^{ x-\frac{1}{2}x^2}$$
Now for the second derivative:
$$f''(x)=(1-x)[e^{ x-\frac{1}{2}x^2}]' + (e^{ x-\frac{1}{2}x^2} [1-x]')$$
$$(1-x)(e^{x-\frac{1}{2}x^2} -... | So, you arrived at the step $e^{x-\frac{1}{2}x^{2}}x=0$ and you want to solve for $x$. Dividing both sides of this equation by $e^{x-\frac{1}{2}x^{2}}$ to conclude that $x=0$ is indeed a legal operation. This is because for all real numbers $r$, $e^{r}>0$. Therefore, for any $x$, $e^{x-\frac{1}{2}x^{2}}>0$. Since you a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Induction proof of the identity $\cos x+\cos(2x)+\cdots+\cos (nx) = \frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}$
Prove that:$$\cos x+\cos(2x)+\cdots+\cos (nx)=\frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}.\ (1)$$
My attempt:$$\sin\left(\frac{x}{2}\right)\sum_{k=1}^{n}\cos{(kx)}... | Here is the induction step: it comes down to proving
\begin{gather*}\frac{\sin \dfrac{nx}{2} \cos\dfrac{(n+1)x}{2}}{\sin\dfrac{x}{2}}+\cos(n+1)x=\frac{\sin\dfrac{(n+1)x}{2}\cos\dfrac{(n+2)x}{2}}{\sin(\dfrac{x}{2})}\\
\text{or}\qquad\sin\dfrac{nx}{2}\cos\dfrac{(n+1)x}{2}+\sin\dfrac{x}{2}\cos(n+1)x=\sin\dfrac{(n+1)x}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the cubic polynomial given linear reminders after division by quadratic polynomials?
A cubic polynomial gives remainders $(13x-2)$ and $(-1-7x)$ when divide by
$x^2-x-3$ and $x^2-2x+5$ respectively. Find the polynomial.
I have written this as:
$P(x)=(x^2-x-3)Q(x)+(13x-2)$
$P(x)=(x^2-2x+5)G(x)+(-1-7x)$
and
$P(x... | Using the Extended Euclidean Algorithm as implemented in this answer, we get
$$
\begin{array}{r}
&&1&x+6&(x-8)/53\\\hline
1&0&1&-x-6&(x^2-2x+5)/53\\
0&1&-1&x+7&(-x^2+x+3)/53\\
x^2-x-3&x^2-2x+5&x-8&53&0\\
\end{array}
$$
That is,
$$
(x+7)(x^2-2x+5)-(x+6)(x^2-x-3)=53\tag{1}
$$
We can now use the Chinese Remainder Theorem.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Positive Definite Proof Method I am interested in an effecient way of showing that the $n \times n$ matrix,
\begin{pmatrix}
2&1&1& & 1\\
1&3&1& \cdots & 1\\
1&1&4& & 1\\
&\vdots& &\ddots& \vdots\\
1&1&1& \cdots & n+1
\end{pmatrix}
is positive definite.
This is an old qualifier question, so it should not require exten... | Observe that
\begin{align}
\begin{pmatrix}
2&1&1& \cdots & 1\\
1&3&1& \cdots & 1\\
1&1&4& \cdots & 1\\
\vdots&\vdots&\vdots &\ddots& \vdots\\
1&1&1& \cdots & n+1
\end{pmatrix}
=
\begin{pmatrix}
1&1&1& \cdots & 1\\
1&1&1& \cdots & 1\\
1&1&1& \cdots & 1\\
\vdots&\vdots&\vdots &\ddots& \vdots\\
1&1&1& \cdots & 1
\end{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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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}) =... | Try a different method.
Take $\sqrt{x^2 +1}$ on RHS Then rationalise $\sqrt{x^2+9}$ - $\sqrt{x^2 +1}$ by it's conjugate.
So your next step would be :
$\dfrac{8}{\sqrt{x^2+1}}$[$\sqrt{x^2+9}$ + $\sqrt{x^2 +1}$] = 8
So 8 gets cancelled. and next step is as follows:
$\dfrac{\sqrt{x^2+9} + \sqrt{x^2 +1}}{\sqrt{x^2+1}}$ = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Any hint to solve given integral $\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}$? Show that for $ab>0$ $$\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}={{2\pi}\over ab}$$
I'm not sure how to go about this. Any solutions or hints are greatly appreciated.
| $\cos^2\theta,\sin^2\theta$ have period $\pi$, so the integral is $2I$ where $I$ is the integral from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. wlog we may take $a,b$ to be positive.
We have $I=\frac{1}{a^2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sec^2\theta}{1+(\frac{b}{a})^2\tan^2\theta}\ d\theta$. Putting $x=\tan\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Show that $\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx={5\pi\over 12}$ Show that $$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx={5\pi\over 12}.$$
Any solutions or hints are greatly appreciated.
I know I can rewrite the integral as $$\int_{-\infty}^\infty {(x-1)(x-2)\over {(x^2+1)(x^2+9)}}dx.$$ ... | Split the integral into two integrals, such that you have only one "bad" point. Now:
$$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx= \int_{-\infty}^{0} {{x^2-3x+2}\over {x^4+10x^2+9}}dx + \int_{0}^{\infty}{{x^2-3x+2}\over {x^4+10x^2+9}}dx$$
Now deal with the integrals separately:
$$\int_{-\infty}^{0} {{x^2-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Extract coefficients for a formal power series using Lagrange Inversion Formula Given $f(x)$ is a formal power series that satisfies $f(0) = 0$
$(f(x))^{3} + 2(f(x))^{2} + f(x) - x = 0$
I know that the Lagrange inversion formula states given f(u) & $\varphi(u)$ are formal power series with respect to u, and $\varphi(0)... | Here we use a somewhat simpler but equivalent variant of the Lagrange Inversion Formula. (See Theorem A.2 in Analytic Combinatorics by P. Flajolet and R. Sedgewick for the equivalence of the variants).
Lagrange Inversion Formula:
Let $g(x), f(x)\in x\mathbb{C}[x]$ be inverses: $g(f(x))=x$. If $g(x)=\frac{x}{\phi(x)}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Inequality, only one solution from algebra I recently came along the following problem:
$$f(x) = {4-x^2 \over 4-\sqrt{x}}$$
Solve for:$$f(x) ≥ 1$$
My Attempt
Now I know that one of the restrictions on the domain is $x≥0$, thus one of the solutions is $0≤x≤1$. There is still one more solution according to the graph, whi... | As already mentioned in the comments, you must also account for the case where $4- \sqrt{x} < 0$, which results in the inequality sign being flipped.
The way to do this in general is to first get $0$ on one side and simplify/factor on the other:
\begin{align}
\frac{4-x^2}{4-\sqrt x} &\ge 1\\[0.3cm]
\frac{4-x^2}{4-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find continuities with square root? I don't understand how to find $$\frac{4-x^2}{3-\sqrt{x^2+5}}$$
The book says to multiply the equation by $\frac{3 + \sqrt {x^2+5}}{3 + \sqrt {x^2+5}}$. I don't understand where that comes from. It says the multiplication simplifies to "$3 + \sqrt {x^2+5}$" - I don't see how t... | $$\frac{4-x^2}{3-\sqrt{x^2+5}}=\frac{4-x^2}{3-\sqrt{x^2+5}}\cdot\frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}}=\frac{(4-x^2)(3+\sqrt{x^2+5})}{3^2-(x^2+5)}$$ $$=\frac{(4-x^2)(3+\sqrt{x^2+5})}{4-x^2}=3+\sqrt{x^2+5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove: $\frac{a+c}{b+d}$ lies between $\frac{a}{b}$ and $\frac{c}{d}$ (for positive $a$, $b$, $c$, $d$) I am looking for proof that, if you take any two different fractions and add the numerators together then the denominators together, the answer will always be a fraction that lies between the two original fractions.
... | Suppose we have positive $a,b,c,d$ with $\frac{a}{b}\ge \frac{c}{d}$. Then multiplying through by $bd$ we get $ad\ge bc$. Adding $ab$ to both sides we get $a(b+d)\ge b(a+c)$. Dividing by $b(b+d)$, we get $\frac{a}{b}\ge\frac{a+c}{b+d}$.
Similarly, add $cd$ to both sides of $ad\ge bc$ to get $d(a+c)\ge c(b+d)$. Dividing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the minimum $k$
Find the minimum $k$, which $\exists a,b,c>0$, satisfies
$$ \frac{kabc}{a+b+c}\geq (a+b)^2+(a+b+4c)^2$$
My Progress
With the help of Mathematica, I found that when $k=100$, we can take $a=1,b=1,c=1/2$. And I'm pretty sure that $k=100$ is the answer, but I couldn't prove it.
| Let $a=b=2$ and $c=1$.
Hence, $k\geq100$.
We'll prove that $100$ it's an answer.
Indeed, let there are positives $a$, $b$ and $c$ for which $ \frac{kabc}{a+b+c}\geq (a+b)^2+(a+b+4c)^2$ and $k<100$.
But it's impossible because we'll prove now that $ \frac{100abc}{a+b+c}\leq (a+b)^2+(a+b+4c)^2$.
Let $c=(a+b)x$.
Hence,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solving $\arcsin(\sqrt{1-x^2}) +\arccos(x) = \text{arccot} \left(\frac{\sqrt{1-x^2}}{x}\right) - \arcsin( x)$ If we have to find the solutions of equation
$$\arcsin(\sqrt{1-x^2}) +\arccos(x) = \text{arccot} \left(\frac{\sqrt{1-x^2}}{x}\right) - \arcsin( x)$$
Using a triangle I rewrite it as
$$2 \arctan \left(\frac{\s... | Straightaway the problem reduces to $$\text{arccot}\dfrac{\sqrt{1-x^2}}x=\dfrac\pi2+\arcsin\sqrt{1-x^2}$$
As $\sqrt{1-x^2}\ge0,$ using the definition of Principal Values
$0\le\arcsin\sqrt{1-x^2}\le\dfrac\pi2$
and consequently, $\dfrac\pi2\le\text{arccot}\dfrac{\sqrt{1-x^2}}x\le\pi$
$\implies x\not>0$ but $x\ne0,$
l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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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... | After the first line of your attempt, multiply both sides by $\sqrt{x+938^2}-\sqrt{x+140^2}$
$$x+938^2-(x+140^2)=1116(\sqrt{x+938^2}-\sqrt{x+140^2})$$
Divide both sides by $1116$.
$$\dfrac{938^2-140^2}{1116}=\sqrt{x+938^2}-\sqrt{x+140^2}$$
Taken with the previous equation
$$\sqrt{x+938^2}+\sqrt{x+140^2}=1116$$
we can a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find $\frac{\mathrm{d}y}{\mathrm{d}x}$ when both number in front and exponent have fractions? I'm not sure how to solve this: $\frac{5}{9}x^\frac{2}{3}$. I applied the product rule and have $\frac{2}{3}\frac{5}{9}x^{-\frac{1}{3}}$.
$\frac{30}{9}x^{-\frac{1}{3}}$, then $\frac{9}{30}x^{\frac{1}{3}}$.
This isn't t... | The comments have already pointed out where you have misunderstood this question.
However, as an answer to your question:
\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x}\Bigg(\frac{5}{9}x^\frac{2}{3} \Bigg)&= \frac{5}{9}\frac{\mathrm{d}}{\mathrm{d}x}(x^\frac{2}{3})\\
&= \frac{5}{9}\Bigg(\frac{2}{3}\Bigg)x^{\frac{2}{3}-1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the simple continued fractions for both $\pm \frac{39}{25}$... Find the simple continued fractions for both $\pm \frac{39}{25}$?
*
*So far for $\frac{39}{25}$ I have:
$39 = 1 \times 25 + 14 $
$ 25 = 1\times 14 + 11 $
$14 = 1 \times 11 + 3$
$11 = 3 \times 3 + 2$
$3 = 1 \times 2 + 1$
$2 = 2 \times 1 + 0$
Giv... | I believe that for $-\frac{39}{25}$ you should first rewrite it as $-2+\frac{11}{25}$. Then
$11=0\times25+11$
$25=2\times11+3$
$11=3\times3+2$
$3=1\times2+1$
$2=2\times1+0$
This gives $-\frac{39}{25}=[-2+0;2,3,1,2]=[-2;2,3,1,2]$
Note that recently there has developed a way of representing negative continued fractions i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Die that never rolls the same number consecutively Suppose we have a "magic" die $[1-6]$ that never rolls the same number consecutively.
That means you will never find the same number repeated in a row.
Now let's suppose that we roll this die $1000$ times.
How can I find the PDF, expected number of times and variance o... | let $P_{k,n}$ be the probability to get $k$ times the result "1" in $n$ throws, and let $$P_{k,n} = p_{k,n} + q_{k,n}$$ where $p,q$ are the corresponding probabilities where you add the constraint that you finish with a 1 (for $p$) or anything but 1 (for $q$).
It is easy to get the equations (for $n \ge 1$) :
$$5p_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Find $\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$ Find $$\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$$
The general term is $\frac{1}{r}\sin\frac{r\pi}{3}$
Let $z=e^{i\frac{\pi}{3}}$
Then, $$\frac{1}{r}z^r=\frac{1}{r}e^{i\frac{r\pi}{3}}$$... | This can be done with Fourier series. Let $f(x)=x$ for $x\in(-\pi,\pi)$, and its periodic extension with period $2\pi$. Then, since $f(x)=-f(-x)$ we can represent it as
$$f(x)=\sum_{n=1}^{\infty}b_n\sin\left(nx\right)$$
Then we can compute
$$\begin{align}\int_{-\pi}^{\pi}f(x)\sin(nx)dx&=\int_{-\pi}^{\pi}x\sin(nx)dx=\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Show that $\sin(x) > \ln(x+1)$ for any $x \in (0,1)$ Show that $\sin(x) > \ln(x+1)$ when $x \in (0,1)$.
I'm expected to use the maclaurin series (taylor series when a=0)
So if i understand it correctly I need to show that:
$$\sin(x) = \lim\limits_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{(-1)^{k-1}}{(2k-1)!} \cdot ... | Let $f(x)=\sin x-\ln(x+1)$. we try to show that $f(x)$ is increasing. In fact
$$ f'(x)=\cos x-\frac{1}{x+1}=\frac{(x+1)\cos x-1}{x+1}. $$
Now we show $(x+1)\cos x>1$ or $\sec x-1<x$ for $0<x<1$.
Note
\begin{eqnarray}
\sec x-1=\frac{1-\cos x}{\cos x}=\frac{2\sin^2\frac{x}{2}}{1-2\sin^2\frac{x}{2}}.
\end{eqnarray}
Since ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb Q$ Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb{Q}$ (by finding a nonzero polynomial $p(x)$ with coefficients in $\mathbb{Q}$ which has $\sqrt[3] 2+\sqrt 5$ as a root).
I first tried letting $a=\sqrt[3]{2} +\sqrt{5}$ and then square both si... | As a vector space over $\mathbb Q$, $\mathbb Q(\sqrt[3]{2},\sqrt 5)$ has a basis $\{1,\sqrt 5,\sqrt[3]{2},\sqrt[3]{4},\sqrt[3]{2}\cdot\sqrt 5,\sqrt[3]{4}\cdot\sqrt 5\}$. Now consider the linear map $\phi:\mathbb Q(\sqrt[3]{2},\sqrt 5)\to\mathbb Q(\sqrt[3]{2},\sqrt 5)$ given by $\phi(y)=(\sqrt[3]{2}+\sqrt 5)y$. With res... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding all real roots of the equation $(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$
Find all real roots of the equation
$$(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$$
I tried squaring the equation, but the degree of the equation became too high and unmanageable. I also tried substitutions, but it didn't work ... | We have
$$x^2+7x+12-(x+1) \sqrt{x+2} - (x+6)\sqrt{x+7}=0$$
Multiplying the both sides by $3$ gives
$$3x^2+21x+36-3(x+1) \sqrt{x+2} - 3(x+6)\sqrt{x+7}=0\tag1$$
Now since
$$\begin{align}&3x^2+21x+36\\&=x^2+5x+4+x^2+13x+42+x^2+3x-10\\&=(x+1)(x+4)+(x+6)(x+7)+(x-2)(x+5)\end{align}$$
we have, from $(1)$,
$$(x+1)(x+4)+(x+6)(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Show $3\cos 2x + 1 = 4\cos^2 x - 2\sin^2 x$ Show $3\cos 2x + 1 = 4\cos^2 x - 2\sin^2 x$
Using the formula $\cos 2x = \cos x - \sin^2 x$
I can say $3\cos 2x + 1 = 3(\cos^2 x - \sin^2 x) + 1$
$\Rightarrow 3\cos x^2 - 3\sin^2 x + 1$
But from there I don't see how I can get the answer.
| I continue from where you left off.
$$ 3\cos^2 x - 3\sin^2 x + 1 = 4\cos^2 x -\cos^2 x - \sin^2 x - 2\sin^2 x + 1 $$
$$ = 4\cos^2 x - 2\sin^2 x \boxed{-\cos^2 x - \sin^2 x + 1 } $$
$$ = 4\cos^2 x - 2\sin^2 x. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Maximum of function containing two variables $x$ and $y$
If $x+y+\sqrt{2x^2+2xy+3y^2} = k(\bf{Const.})\;,$ Then $\max(x^2y)\;,$ Where $x,y\geq 0$
$\bf{My\; Try::}$ Let $x^2y=z\;$ Then we get $$x+\frac{x^2}{z}+\sqrt{2x^2+\frac{2z}{x}+\frac{3z^2}{x^4}} = k$$
Now How can I solve it, Help me
Thanks
| Assuming $y=a x$ ($a\geq0$), the constraint leads to $$x=\frac{k}{\sqrt{3 a^2+2 a+2}+a+1}$$ and $$x^2y=\frac{a k^3}{\left(\sqrt{3 a^2+2 a+2}+a+1\right)^3}$$ Now, computing $$\frac{d}{da}x^2y=-\frac{(2 a-1) \left(3 a+\sqrt{a (3 a+2)+2}+2\right)}{\sqrt{a (3 a+2)+2}
\left(a+\sqrt{a (3 a+2)+2}+1\right)^4}k^3$$ which can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{\cos\theta\cos\delta}{\cos^2\alpha}+\frac{\sin\theta\sin\delta}{\sin^2\alpha}+1=0$
If $$\frac{\cos\theta}{\cos\alpha}+\frac{\sin\theta}{\sin\alpha}=\frac{\cos\delta}{\cos\alpha}+\frac{\sin\delta}{\sin\alpha}=1,$$ where $\theta$ and $\delta$ do not differ by an even multiple of $\pi$, then prove that ... | For future readers, I here combine the comments by @H.Potter and @lab bhattacharjee into an answer.
Let $\frac{\cos \theta}{\cos\alpha} $ be denoted as $(1)$,
$\frac{\sin \theta}{\sin \alpha} $ be denoted as $(2)$,
$\frac{\cos \delta}{\cos\alpha} $ be denoted as $(3)$, and
$\frac{\sin \delta}{\sin\alpha} $ be denoted ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding lim${_{n \rightarrow \infty}}\left( \frac{n^3}{2^n} \right)$ For a class of mine we were given a midterm review; however, I just cannot figure out how to finish this one:
Find the limit $$\lim_{n \rightarrow \infty}\left( \dfrac{n^3}{2^n} \right)$$
My attempt so far:
Let $s_n=\dfrac{n^3}{2^n}$.
Note that $2^... | Thats a long answer.Quick trick:$0 < \dfrac{n^3}{2^n} < \dfrac{1}{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the double integral.
To find:
$$\iint_R(x^2-xy)dA$$
enclosed by
$$y=x, y=3x-x^2,$$
$$x=3x-x^2,$$
$$x=0, x=2,$$
$$y=0, y=2,$$
$$R=((x,y): 0\le x\le 2), (x \le y \le 3x-x^2)$$
$$\int_0^2\int_x^{3x-x^2}(x^2-xy)\,dy\,dx$$
$$\int_0^2 \left. \left(x^2y-x\frac{y^2}{2}\right) \right|_x^{3x-x^2}\,dx$$
where to fro... | With the change of variables $u=x$ and $v=y-x$, that is $x=u$ and $y=v-u$, we have the Jacobian $\left|\frac{\partial(x,y)}{\partial (u,v)}\right|=1$; from $0\le x\le 2$ we have $0\le u\le 2$ and from $x\le y\le 3x-x^2$, we have $0\le y-x\le 2x-x^2$ we have $0\le v\le 2u-u^2$. The integral becomes
\begin{align}
I&=\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Maximum value of the sum of absolute values of cubic polynomial coefficients $a,b,c,d$
If $p(x) = ax^3+bx^2+cx+d$ and $|p(x)|\leq 1\forall |x|\leq 1$, what is the $\max$ value of $|a|+|b|+|c|+|d|$?
My try:
*
*Put $x=0$, we get $p(0)=d$,
*Similarly put $x=1$, we get $p(1)=a+b+c+d$,
*similarly put $x=-1$, we get ... | Let $p(1)=u,$ $p(-1)=v$, $p\left(\frac{1}{2}\right)=w$ and $p\left(-\frac{1}{2}\right)=t$.
Thus, we have the following system:
$$a+b+c+d=u,$$
$$-a+b-c+d=v,$$
$$\frac{a}{8}+\frac{b}{4}+\frac{c}{2}+d=w$$ and
$$-\frac{a}{8}+\frac{b}{4}-\frac{c}{2}+d=t,$$ which gives
$$a=\frac{2u-2v-4w+4t}{3},$$
$$b=\frac{2u+2v-2w-2t}{3},$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Proof of $1^3+1^3+2^3+3^3+5^3+\cdots +F_n^3=\frac{F_nF_{n+1}^2+(-1)^{n+1}[F_{n-1}+(-1)^{n+1}]}{2}$ Fibonacci series
$F_0=0$, $F_1=1$; $F_{n+1}=F_n+F_{n-1}$
This is a well known identity
$1^2+1^2+2^2+3^2+5^2+\cdots +F_n^2=F_nF_{n+1}$
I was curious and look every websites for a closed form of
$1^3+1^3+2^3+3^3+5^3+\cdots ... | This can be proved by induction. It holds for $n=1$. Denoting the right-hand side by $\sigma_n$, we have
\begin{align}
\sigma_n+F_{n+1}^3-\sigma_{n+1}&=\frac{F_nF_{n+1}^2+(-1)^{n+1}F_{n-1}+1}2+F_{n+1}^3-\frac{F_{n+1}F_{n+2}^2+(-1)^{n+2}F_n+1}2
\\
&=
\frac{2F_{n+1}^3+F_nF_{n+1}^2-F_{n+1}F_{n+2}^2+(-1)^{n+1}F_{n+1}}2
\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
LU decomposition using the LU factorization Algorithm with $l_{ii}=1$
For this matrix, I got $U_{31}=1$, but the answer says $U_{31}=1/2$. Since the first two elements of the third row are 0, I think this two answers actually are equivalent. I am just wondering if both of them are correct? Since it is not in the Row ... | We have:
$$A = \begin{bmatrix}1&-1&0\\2 & 2 & 3 \\ -1 & 3 & 2\end{bmatrix} = LU=\begin{bmatrix}1 & 0 & 0\\l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1\end{bmatrix} \begin{bmatrix}u_{11} & u_{12} & u_{13}\\0 & u_{22} & u_{23} \\ 0 & 0 & u_{33}\end{bmatrix}$$
The individual calculations in order they are performed are:
*
*$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the area of the square using co-ordinates Given a square $ABCD$ such that the vertex $A$ is on the $x$-axis and the vertex $B$ is on the $y$-axis. The coordinates of vertex $C$ are $(u,v)$. Find the area of square in terms of $u$ and $v$ only.
What I have done
Let the coordinate of $A$ be $(x,0)$ and $B$ be $(0,y)... | Let $A(x;0), B(0,y), C(u,v)$, and $AB=a$. Then $$S=a^2$$.
$$\begin{cases}
AB=a=\sqrt{x^2+y^2}
\\
BC=a=\sqrt{u^2+(v-y)^2}{}
\\
AC=\sqrt2a=\sqrt{(u-x)^2+v^2}
\end{cases}$$
$$\begin{cases}
a^2=x^2+y^2 (1)
\\
a^2=u^2+(v-y)^2 (2)
\\
2a^2=(u-x)^2+v^2 (3)
\end{cases}$$
(3)+(2) $3a^2=u^2-2ux+x^2+v^2+u^2+v^2-2vy+y^2=2u^2-2ux+2v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove: $\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$
Prove the trigonometric identity
$$\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$$
I've searched high and low on the net and cannot find identities where there is $+$ or $- 1$'s in the equation. Any help is appreciated.
Edit af... | Hint:
$$c(\frac1c-1)c(\frac1c+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integral of the function $\frac{\cos ^2 x}{1+\tan x}$ Evaluate $$\int \frac{\cos ^2 x}{1+\tan x}dx$$ I tried converting in double angle and making the derivative of the denominator in the numerator. But, it didn't work out. Some help please. Thanks.
| Let $$I = \int\frac{\cos^2 x}{1+\tan x}dx = \int\frac{\cos^3 x}{\sin x+\cos x}dx = \frac{1}{2}\int\frac{2\cos^3 x}{\sin x+\cos x}dx$$
So we get $$I = \frac{1}{2}\left[\int\frac{\left(\cos^3 x+\sin^3 x\right)+(\cos^3 x-\sin^3 x)}{\sin x+\cos x}\right]dx$$
So we get $$I = \frac{1}{4}\int (2-\sin 2x)dx+\frac{1}{4}\int \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$ Problem: If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$.
Solution: $3^x +3^y +3^z=9^{13}$
$3^x +3^y +3^z=3^{26}$
I am unable to continue from here.
Any assistance is appreciated.
Edited
$9^{13} =3^{26}$
$=3^{25} (3)$
$=3^{25} (1+1+1)$
$=3^{25} + 3^{25} + 3^{25}$
So $x+y... | Assume $x\geq y \geq z$. Then,
$$
3^{26}=9^{13}=3^x+3^y+3^z\leq 3(3^x)=3^{x+1}
$$
and so $x\geq 25$. Obviously, $x<26$ and so we must have $x=25$. But then
$$
2\times 3^{25}=3^{26}-3^x=3^y+3^z\leq 3^x+3^x=2\times 3^{25}.
$$
It must be the case that $y=x$ and $z=x$. We conclude that $x=y=z=25$ and their sum is $75$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
I have a problem when I go to calculate $\lim_{x\to\infty}\left( \frac {2x+a}{2x+a-1}\right)^{x}.$ The limit:
$\lim_{x\rightarrow\infty}
\left( {\frac {2\,x+a}{2\,x+a-1}} \right) ^{x}$
I make this:
$\left( {\frac {2\,x+a}{2\,x+a-1}} \right) ^{x}$=${{\rm e}^{{\it x\ln} \left( {\frac {2\,x+a}{2\,x+a-1}} \right) }}$
Then... | One may recall that, as $u \to 0$, by using the Taylor series expansion,
$$
\log(1+u)=u+O(u^2)
$$ one gets, as $x \to \infty$,
$$
\begin{align}
\log\left(\frac{2x+a}{2x+a-1}\right)&=\log\left(1+\frac1{2x+a-1}\right)
\\\\&=\frac1{2x+a-1}+O\left(\frac1{(2x+a-1)^2}\right)
\\\\&=\frac1{2x+a-1}+O\left(\frac1{x^2}\right)
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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The elements and operations of the field $C = \Bbb R[x] / \langle x^2 + 1 \rangle$
$$C = \Bbb R[x] / \langle x^2 + 1 \rangle = \{[a + b x_{x^2 + 1}]\}$$
I know $C$ is a field since it has complex roots $(x+i)(x-i)$ and is irreducible over the reals, also since deg is $2$.
How would I find elements of this field? To ... | Any element of $\Bbb R[x]$ has the form $$p(x) + \langle x^2 + 1 \rangle,$$
and by polynomial long division we can find a unique representative of this element of the form
$$a + b x$$
Since any two representatives of a given element differ by a polynomial of the form $(x^2 + 1) q(x)$, which if $q \neq 0$ has degree $\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the range of $y = \sqrt{x} + \sqrt{3 -x}$ I have the function $y = \sqrt{x} + \sqrt{3 -x}$. The range in wolfram is $y \in\mathbb R: \sqrt{3} \leq y \leq \sqrt{6}$
(solution after correction of @mathlove)
$\sqrt{x} + \sqrt{3 -x} = y$
$$
\begin{cases}
x \geq 0\\
x \leq 3
\end{cases}
$$
then
$(\sqrt{x} + \sqrt{3 ... |
$4x^2-12x+y^4-6y^2+9 = 0$
the quadratic equation is verified when the discriminant is $>= 0$, then
$b^2 - 4ac = (-12)^2-4(y^4-6y^2+9) >= 0$
This is the part where you have an error :
$$b^2-4ac=(-12)^2-4\cdot 4(y^4-6y^2+9)\ge 0$$
This is equivalent to $-\sqrt 6\le y\le \sqrt 6$.
With $y^2-3\ge 0$ and $y\ge 0$, the ran... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to find the integral of a quotient of rational functions? How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$
My attempt:
We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$
It is easy to find that
$A=1$,
$B=2$, and
$C=-1$.
Therefore
$$\frac{x^4+1}{x^3+... | You need to combine polynomial division and partial fractions.
$$
\begin{align}
\frac{x^4+1}{x^3+x^2}
&=x-1+\frac{x^2+1}{x^2(x+1)}\tag{1}\\
&=x-1+\frac2{x+1}-\frac1x+\frac1{x^2}\tag{2}
\end{align}
$$
Explanation:
$(1)$: polynomial division
$(2)$: partial fractions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Consider the curve with implicit equation... I have the last question on my assignment which I cannot answer and was wondering if any of you could help me.
The two part question reads
Consider the curve with implicit equation $x^\frac{2}{3} + y^\frac{2}{3} = 4$. By using implicit differentiation show that:
a) $1 + \lef... | You are almost done. To complete (a),
\begin{align}
1+\left(\frac{dy}{dx}\right)^2 &= 1 + \frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}\\
&= 1 + \frac{4-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}\\
&=\frac{4}{x^{\frac{2}{3}}}.
\end{align}
The arc length of $y=f(x)$ from $(a,f(a))$ to $(b,f(b))$ is given by
$$
L=\int_a^b \sqrt{1+(f'(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to calculate $\lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)^{-1}$? $$f (x) = \frac{1}{x^2} - \frac{1}{\sin^2 x}$$
Find limit of $\dfrac1{f(x)}$ as $x\to0$.
| As already noted:
$$\frac1{f(x)}=\frac{x^2\sin^2x}{\sin^2x-x^2}=\color{blue}{\frac {x\sin^2x}{\sin x-x}}\cdot\color{purple}{\frac x{\sin x+x}}\;\color{red}{(**)}$$
Now, using l'Hospital's rule:
$$\begin{align*}&\lim_{x\to0}\color{blue}{\frac{x\sin^2x}{\sin x-x}}\stackrel{\text{l'H}}=\lim_{x\to0}\frac{\sin^2x+x\sin2x}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Integrate $ \int \frac{1}{1 + x^3}dx $ $$ \int \frac{1}{1 + x^3}dx $$
Attempt:
I added and subtracted $x^3$ in the numerator but after a little solving I can't get through.
| Does this help?
$\dfrac{1}{x^3+1} = \dfrac{1}{(x+1)(x^2-x+1)} = \dfrac13\left(\dfrac{1}{x+1} - \dfrac{x-2}{x^2-x+1}\right)$
Can you integrate it from there?
The integral of the second part is a little tricky. Note that:
$\dfrac13\left(\dfrac{x-2}{x^2-x+1}\right) = \dfrac16\left(\dfrac{2x-4}{x^2-x+1}\right)= \dfrac16\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Prove $\sin^2(10 ^\circ)-\sin^2(20^\circ)-\sin^2(40^\circ)=-\frac{1}{2}$ identity 10 degrees
$$\sin^2(10^\circ)-\sin^2(20^\circ)-\sin^2(40^\circ)=-\frac{1}{2}$$
$$\cos^2(10^\circ)-\cos^2(20^\circ)-\cos^2(40^\circ)=-\frac{1}{2}$$
Why are they both have same answer?
The only time they have same answer is at 45 degrees r... | Let me show the first identity. By the double-angle identity:
\begin{align}
& \sin^2(10^\circ) - \sin^2(20^\circ) - \sin^2(40^\circ) \\
= & \frac{1 - \cos(20^\circ)}{2} - \frac{1 - \cos(40^\circ)}{2} - \frac{1 - \cos(80^\circ)}{2} \\
= & -\frac{1}{2} - \frac{1}{2}(\cos(20^\circ) - \cos(40^\circ) - \cos(80^\circ)).\\
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1791258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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First Order PDE, How to Deal With This Boundary Condition? 1. The problem statement
Find a solution of $$\frac{1}{x^2}\frac{\partial u(x,y)}{\partial
x}+\frac{1}{y^3}\frac{\partial u(x,y)}{\partial y}=0$$ Which satisfies
the condition $\frac{\partial u(x,y)}{\partial x}\big |_{y=0}=x^3$ for
all $x$.
2.The attemp... | If $u (x,y) = f \left(\frac{x^3}{3} - \frac{y^4}{4}\right)$, then the boundary condition gives us $f ' \left(\frac{x^3}{3}\right) = x$. Hence, we have
$$f ' (z) = \sqrt[3]{3z}$$
Integrating, we obtain
$$f (z) = \frac{3\sqrt[3]{3}}{4} \, z^{\frac{4}{3}}$$
and, thus,
$$u (x,y) = \frac{3\sqrt[3]{3}}{4} \left(\frac{x^3}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve $\begin{cases}x\equiv-4\pmod {17}\\ x\equiv 3\pmod{23} \end{cases}$
Solve $$\begin{cases}x\equiv-4\pmod {17}\\
x\equiv 3\pmod{23}
\end{cases}$$
My attempt:
$$\gcd (17,23)=1$$
so using the Chinese remainder theorem
there is a solution modulo $17\times 23=391$
$$x=-4+17t\\
\Longrightarrow-4+17t\equiv 3\pmod{23}\... | Once you obtain the equivalence $-4 + 17t \equiv 3 \pmod{23}$, you can add $4$ to each side of the equivalence to obtain
$$17t \equiv 7 \pmod{23}$$
Since $\gcd(17, 23) = 1$, this equivalence has a solution. To solve it, we can apply the extended Euclidean algorithm to find the multiplicative inverse of $17$ modulo $23... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $P$ is an integer polynomial with $P(1)=P(2)=0$, then some coefficient is less than $-1$
Let $P (x)$ be a polynomial with integer coefficients. It is known that the numbers $1$ and $2$ are its roots.
Prove that there exists a coefficient that is less than $-1$.
My work so far:
Let $P(x)=a_nx^n+...+ax+a_0$. $P(1)... | We assume that $deg P(x) \geq 2$ and that $a_n \neq 0$. Since $P(1) = P(2) = 0$, we have $a_n2^n + a_{n - 1}2^{n-1} + .. + a_12 + a_0 = 0$ and $a_n + a_{n - 1} + .. + a_1 + a_0 = 0$. Together, we have $(a_{n - 1} + .. + a_1 + a_0)2^n = a_{n - 1}2^{n - 1} + .. + a_12 + a_0$. If we assume that $a_k \geq -1$ for $k \leq n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Maximum of $(ab+cd)(ac+bd)(ad+bc)$ Let $a,b,c,d\ge 0$ satisfy $a+b+c+d=4$. Find the maximum value of $(ab+cd)(ac+bd)(ad+bc)$.
When all of the variables are $1$, the value is $8$. Using the AM-GM inequality gives $$(ab+cd)(ac+bd)(ad+bc)\leq\left(\frac{ab+cd+ac+bd+ad+bc}{3}\right)^3.$$
Can we upper bound the right-hand s... | We have
$$ab+ac+ad+bc+bd+cd=\frac12\bigl((a+b+c+d)^2-(a^2+b^2+c^2+d^2)\bigr) $$
and for $a+b+c+d=4$ we have $a^2+b^2+c^2+d^2\ge4$ (why?) so that
$$ab+ac+ad+bc+bd+cd\le\frac{16-4}{2}= 6 $$
and you indeed obtain $\le 8$ for the original inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$. Problem :
Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$.
My approach :
Since :
$(a^3+b^3)=(a+b)^3-3ab(a+b)$
$\Rightarrow z^3+\frac{1}{z^3}=(z+\frac{1}{z})^3-3(... | You have $(z+\frac{1}{z})^3-3(z+\frac{1}{z})=z^3+\frac{1}{z^3}$, and $\left|(z+\frac{1}{z})^3-3(z+\frac{1}{z})\right|=\left|z+\frac{1}{z}\right|\cdot\left|(z+\frac{1}{z})^2-3\right|$.
Hence if $\left|z+\frac{1}{z}\right|>2$, then $\left|z^3+\frac{1}{z^3}\right|>2\cdot(4-3)=2$. Contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find the maximum and minimum values of $x-\sin2x+\frac{1}{3}\sin 3x$ in $[-\pi,\pi]$ Find the maximum and minimum values of $x-\sin2x+\frac{1}{3}\sin 3x$ in $[-\pi,\pi]$.
Let $f(x)=x-\sin2x+\frac{1}{3}\sin 3x$
$f'(x)=1-2\cos2x+\cos3x$
Put $f'(x)=0$
$1-2\cos2x+\cos3x=0$ gives $2\sin^2x-\cos2x+\cos3x=0$
I am stuck here.... | Continuing from $1-2\cos 2x+\cos 3x=0$:
$1-2\cos 2x+\cos 3x=0$
$1-2\left(2\cos ^2x-1\right)+\cos \left(x+2x\right)=0$
$1-2\left(2\cos ^2x-1\right)+\cos x\cos 2x-\sin x\sin 2x=0$
$1-2\left(2\cos ^2x-1\right)+\cos x\left(2\cos ^2x-1\right)-\sin x\left(2\sin x\cos x\right)=0$
$1-2\left(2\cos ^2x-1\right)+\cos x\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Maximize system of linear equations Suppose you have the system
$$
\begin{bmatrix}
4 & 3\\
1 & 7\\
5 & 9\\
2 & 4\\
\end{bmatrix}
\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix}b_1\\b_2\\b_3\\b_4\end{bmatrix}
$$
How could one find scalars $x$ and $y$ such that $b_1+b_2+b_3+b_4$ is maximized? I can see the math behind ... | First note
$$
b_1 + b_2 + b_3 + b_4
= (4 + 1 + 5 + 2) x + (3 + 7 + 9 + 4)
= 12 x + 23 y
$$
then we can formulate it as linear program
$$
\begin{array}{rr}
\max & c^\top u \\
\text{w.r.t.} & A u = b \\
& B u \le d \\
& u \ge 0
\end{array}
$$
for cost vector $c = (12, 23)^\top$, vector of unknowns $u = (x, y)^\top$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Recurrence relation $a_r+6a_{r-1}+9a_{r-2}=3$, then find $a_{20}$
Consider the recurrence relation $a_r+6a_{r-1}+9a_{r-2}=3$, given that $a_0=0, a_1=1$. Let $a_{20}=x\times10^9$, then the value of $x$ is______ .
My attempt:
$a_r=3-6a_{r-1}-9a_{r-2}$
I calculated manually, I get $a_{20}=-6465079410=-6.46\times10^9$, ... | Let $f(z)=\sum_{n=0}^\infty a_n z^n$. Multiplying both sides of the recurrence by $z^n$ and summing over $n$ yields
$$\sum_{n=2}^\infty a_n z^n +6\sum_{n=2}^\infty a_{n-1}z^n +9\sum_{n=2}^\infty a_{n-2}z^n = 3\sum_{n=2}^\infty z^n, $$
or
$$f(z)-(a_0+a_1z)+6z(f(z)-a_0) + 9z^2f(z)=\frac{3z^2}{1-z}. $$ Substituting $a_0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to solve this equation using logarithms? I have to solve for all real values of $x$.
$(5+2\sqrt6)^{x^2-3}+(5-2\sqrt6)^{x^2-3}=10$
I tried to take $\log_{10}$ on both sides but could not do this.
How do I do this?Thanks for any hint or answer!!
| Hint.
note that $$(5-2\sqrt{6})=\frac{1}{5+2\sqrt{6}}$$
and use the substitution
$$
\left(5+2\sqrt{6}\right)^{x^2-3x}=y
$$
so the equation becomes:
$$
y+\frac{1}{y}=10
$$
that becomes a second degree equation (multiply by $y \ne 0$). Solve this equation and you can find the final solution (without logarithms).
$$
y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of ($\sqrt{x^2+8x}-\sqrt{x^2+7x}$) as $x$ approaches infinity I've been stuck on this one problem for 3 days now, I don't know how to proceed. Any help would be appreciated.
The problem is asking for the $$\lim_{x\to\infty} (\sqrt{x^2+8x}-\sqrt{x^2+7x}) $$
Every time I attempt this problem, I can never get rid o... | Another way to do it.
Considering that $x\to\infty$ $$\sqrt{x^2+8x}-\sqrt{x^2+7x}=x\left(\sqrt{1+\frac 8x}-\sqrt{1+\frac 7x} \right)$$ Now, using Taylor for small $y$ $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ make $y=\frac 8x$ in the first radical and $y=\frac 7x$ in the second radical to get $$\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
Question: If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
My solution: Consider
$$x^2 + y^2 + z^2 < 2(xy + yz + xz)$$
Notice that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$
Hence... | Since $x,y,z$ are sides of a triangle, we have
$|x-y| < z $
Squaring both sides we get
$(x-y)^2=x^2 +y^2 -2xy < z^2$
Setting up similar inequalities and adding, we get the desired result.
Your proof doesn't work because
$(x+y+z)^2 \geq 0$ doesn't imply that
$(x+y+z)^2 -4(xy+yz+zx) <0$, even when $x,y,z>0$
A simple c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find the value of $a^2-b^2+c^2$ Let $a$, $b$, and $c$ be real numbers such that $a − 7b + 8c = 4$ and $8a + 4b − c = 7$. What is the
value of $a^2-b^2+c^2$ ?
| Given $a + 8c= 7b + 4$ and $8a - c = -4b + 7$
Now Squaring both equation, We get
\begin{align}a^2 + 64c^2 + 16ac&=49b^2 + 16 + 56b\tag{1} \\
64a^2 + c^2 - 16ac&=16b^2 + 49 - 56b\tag{2}\end{align}
Summing these two equations gives
$$ 65a^2 + 65c^2 = 65b^2 + 65 \Rightarrow a^2 - b^2 + c^2 = 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How should I try to evaluate the integral $\int_a^b \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx$ I've tried to evaluate $\displaystyle\int_{-r}^r \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx$ on my own, but I have encountered a problem I cannot get around.
The indefinite integral $\sqrt{\frac{r^2}{r^2-x^2}} \sqrt{r^2-x^2} \tan ^{... | hint...The integrand is equivalent to $$\frac {r}{\sqrt{r^2-x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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A Sine and Inverse Sine integral A demonstration of methods
While reviewing an old text book an integral containing sines and sine inverse was encountered, namely,
$$\int_{0}^{\pi/2} \int_{0}^{\pi/2} \sin(x) \, \sin^{-1}(\sin(x) \, \sin(y)) \, dx \, dy = \frac{\pi^{2}}{4} - \frac{\pi}{2}.$$
One can express $\sin^{-1}(z... | As mickep has observed
\begin{equation*}
I_{1} = \int_0^{\pi/2}\int_0^{\pi/2}\sin(x)\arcsin(\sin(x)\sin(y))\,dxdy = \dfrac{1}{2} \int_{0}^{\pi/2}\ln\left(\dfrac{1+\sin x}{1-\sin x}\right)\dfrac{\cos^{2}x}{\sin x}\, dx.
\end{equation*}
We intend to finish the solution without using series expansion. Instead we will u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
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If $a_{n+1} \leq \left(1-\frac{2}{n+1}\right)a_n + b\left(\frac{2}{n+1}\right)^2$ then $a_n \leq \frac{4b}{n+1}$ A sequence $(a_n)$ satisfy $a_{n+1} \leq (1-\gamma_n)a_n + \frac{\beta R^2}{2}\gamma_n^2$ where $\gamma_n = \frac{2}{n+1}$, $\beta$ and $R$ are constant.
How to verify that
$$a_n \leq \frac{2\beta R^2}{n+1},... | It is easy to show that $a_2 \leq \frac{\beta R^2}{2}$
We use induction to prove the inequality. Suppose that for $i = 2\ldots n$, we have $a_i \leq \frac{2\beta R^2}{i+1}$, then
$$a_{n+1}\leq (1-\gamma_n)a_n + \frac{\beta R^2}{2}\gamma_n^2 = \frac{n-1}{n+1}a_n+\frac{2\beta R^2}{(n+1)^2} \leq \frac{2(n-1)\beta R^2}{(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve $ord_x(2) = 20$ Given that the (multiplicative) order of $2$ mod $x$ is $20$, how can I work out what $x$ is?
| You need $$2^k\cdot2^{20-k}=1$$ for $k=1,2,...,20$. In other words $2^k$ has as inverse $2^{20-k}$ Since $2$ is a primitive root modulo $11$ we examine if this is verified for $x=11$. We have in the field $\mathbb F_{11}$ $$\begin{cases}2^1=2\\2^2=4\\2^3=8\\2^4=5\\2^5=10\\2^6=9\\2^7=7\\2^8=3\\2^9=6\\2^{10}=1\end{cases}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How to find the summation of this infinite series: $\sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!}(1 - \frac{2}{k})$ I've been trying to figure out the following sum for a while now:
$$\sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!}\left(1 - \frac{2}{k}\right)$$
I'm pretty sure that this doesn't evaluate to $0$.
As $k$ increas... | Let's try splitting up the series (since if it converges to a finite number, it will be absolutely convergent, since all terms are positive).
$$\sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!}(1 - \frac{2}{k}) = \sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!} - 2 \sum_{k=1}^{\infty} \frac{1}{(k+1)(k)(k-1)!} \\ = \sum_{k=1}^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluation of $\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Prove that $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \frac{\pi}{2}\frac{1}{4^{n}}\binom{2n}{n}$ and also find value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
$\bf{My\; Try::}$ Let $$\displaystyle I_{n} = \int_{0}^{\fr... | Just completing Olivier's answer:
$$ I=\int_{0}^{\pi/2}\frac{4\,dx}{4-\sin^2 x}=\int_{0}^{\pi/2}\frac{4\,dx}{4-\cos^2 x}=\int_{0}^{+\infty}\frac{dt}{1+t^2}\cdot\frac{4}{4-\frac{1}{1+t^2}}$$
gives:
$$ S = \sum_{n\geq 0}\frac{1}{16^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{+\infty}\frac{4}{4t^2+3} = \color{red}{\frac{2}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
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How does $\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}$ relate to $\sqrt{x^2+y^2+z^2}$? Another possible 'mean' for three positive real numbers $x,y,z$ is made of pairwise quadratic means:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}$$
By QM-AM inequality it is greater than or equal to arithmetic m... | If $\sqrt{x^2+y^2}=c$, $\sqrt{x^2+z^2}=b$ and $\sqrt{y^2+z^2}=a$ so we can use
$\sqrt{3(a^2+b^2+c^2)}\geq a+b+c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Extending the ordered sequence of 'three-number means' beyond AM, GM and HM I want to create an ordered sequence of various 'three-number means' with as many different elements in it as possible.
So far I've got $12$ ($8$ unusual ones are highlighted):
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \color{blue}{ \frac{\sqrt{x^2+... | Suppose $f(x,y,z)$ is a function that is:
*
*symmetric, i.e. $f(x,y,z)=f(y,x,z)=f(x,z,y)$ and so on,
*positive, i.e. $f(x,y,z)\ge0$ for all $x,y,z\ge0$, and
*has power-law scaling with exponent $p$, that is, $f(ax,ay,az)=a^pf(x,y,z)$ for all $a\ge0$.
Suppose $g(x,y,z)$ is a similar function but with exponent $q\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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$\int\frac{\sin x}{\sqrt{1-\sin x}}dx=?$ Calculate this integral $\displaystyle\int\dfrac{\sin x}{\sqrt{1-\sin x}}dx=?$
Effort;
$1-\sin x=t^2\Rightarrow \sin x=1-t^2\Rightarrow \cos x=\sqrt{2t^2-t^4}$
$1-\sin x=t^2\Rightarrow-\cos x dx=2tdt\Rightarrow dx=\frac{2t}{\sqrt{t^4-2t^2}}dt$
$\displaystyle\int\frac{1-t^2}{t}... | HINT:
$$\dfrac{\sin x}{\sqrt{1-\sin x}}=-\sqrt{1-\sin x}+\dfrac1{\sqrt{1-\sin x}}$$
$$1-\sin x=1-\cos\left(\dfrac\pi2-x\right)=2\sin^2\left(\dfrac\pi4-\dfrac x2\right)$$
Now for real $a,$ $$\sqrt{a^2}=|a|=\begin{cases}+a &\mbox{if } a\ge0 \\-a & \mbox{if } a<0 \end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Showing that $\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$ Integrate
$$I=\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$$
Substitution $x=\sqrt{\tan(u)}\rightarrow dx={\sec^2(u)\over 2\sqrt{\tan(u)}}du$
$x=1\rightarrow u={\pi\over 4}$
$x=0\rightarrow u=0$
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{1-... | On the path of Achille Hui,
$\displaystyle J=\int_0^{\infty} \dfrac{x^2}{x^4+4}dx$
Perform the change of variable $y=\dfrac{x}{\sqrt{2}}$,
$\displaystyle J=\dfrac{\sqrt{2}}{2}\int_0^{\infty} \dfrac{x^2}{x^4+1}dx$
1)
Perform the change of variable $y=x^4$,
$\begin{align} J&=\dfrac{\sqrt{2}}{8}\int_0^{\infty} \dfrac{x^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 3
} |
If $x$ and $y$ are positive numbers less than $20$ for which $x+y+xy=76$, what is $x+y$? What is a simple way to solve this problem? I can do it by trying $x$ and $y$, starting from $1$. That does not look like the best way.
If $x$ and $y$ are positive numbers less than $20$ for which $x+y+xy=76$, what is $x+y$?
| \begin{align}xy+x+y&=76\\
x(y+1)+y&=76\\
x(y+1)+y+1&=77\\
(x+1)(y+1)&=77
\end{align}
As $1,7,11,77$ are the only positive divisors of $77$, $(x,y)$ should be $(6,10)$ or $(10,6)$. Thus, $x+y=16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Proving by induction that $\sum\limits_{i=1}^n\frac{1}{n+i}=\sum\limits_{i=1}^n\left(\frac1{2i-1}-\frac1{2i}\right)$ I have a homework problem to prove the following via induction:
$$\sum_{i=1}^n \frac{1}{n+i} = \sum_{i=1}^n \left(\frac{1}{2i-1} - \frac{1}{2i}\right) $$
The base case is true.
So far I've done the follo... | Write the left hand side as $f(n)$.
Note that $$f(n+1)-f(n)=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}$$
That's the key step - that $f(n+1)$ adds two terms to the right of the sum, but subtracts one term from the left.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What are the values of $a_0,a_1,...,a_{10}$ if $\cos^{10} {\theta}=\sum_{k=0}^{10}a_k\cos {k\theta }$? I was thinking of doing the following:
Let $A=a_0+a_1\cos {\theta }+a_2\cos {2\theta }+...+a_{10}\cos {10\theta }$
and
$B=a_1\sin {\theta }+a_2\sin {2\theta }+...+a_{10}\sin {10\theta }$
Then, $S=A+iB=a_0+a_1(\co... | Let $x=\cos {\theta}+i\sin {\theta}$ whence, $x+\frac{1}{x}=2\cos {\theta}$ and $x^n+\frac{1}{x^n}=2\cos {n\theta}$
$\therefore (2\cos {\theta})^{10}=(x+\frac{1}{x})^{10}$
$=(x^{10}+\frac{1}{x^{10}})+10(x^8+\frac{1}{x^8})+45(x^6+\frac{1}{x^6})+120(x^4+\frac{1}{x^4})+210(x^2+\frac{1}{x^2})+252$
$=2\cos {10\theta}+10\tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that: $97|2^{48}-1$ Show that: $97|2^{48}-1$
My work:
$$\begin{align}
2^{96}&\equiv{1}\pmod{97}\\
\implies (2^{48}-1)(2^{48}+1)&=97k\\
\implies (2^{24}-1)(2^{24}+1)(2^{48}+1) &=97k\\
\implies (2^{12}-1)(2^{12}+1)(2^{24}+1)(2^{48}+1)&=97k\\
\implies (2^6-1)(2^6+1)(2^{12}+1)(2^{24}+1)(2^{48}+1) &=97k
\end{align}$... | $2^{48} - 1 = ( (2^{24})^2 - 1) = (2^{24} - 1)(2^{24}+1) = ( (2^{12})^2 - 1)(2^{24}+1) = (2^{12} - 1)(2^{12} + 1)(2^{24} + 1). $
Now $2^6 = 64 $ thus $2^{12} = 64^2 = 4096 = 22 \mod 97$.
Therefore $2^{24} = (2^{12})^2 = 22^2 = 484 = 96 \mod 97$.
Hence $2^{24} + 1 = 96 +1 = 97 = 0 \mod 97$.
Therefore $97$ divides $2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Integration by Parts on $\sqrt{1+x^2}dx$ I have been asked to show that $$\int\sqrt{1+x^2}dx=\frac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\int\frac{1}{{\sqrt{1+x^2}}}dx$$
I'm aware of how to do this with trig substitution, but the question is specifically regarding integration by parts, and the only examples I can find keep usi... | Alternative method, but not really so different from the other answer:
\begin{align}
\int\sqrt{1+x^2}\,dx
&=
\int\frac{1+x^2}{\sqrt{1+x^2}}\,dx
\\[6px]
&=\int\frac{1}{\sqrt{1+x^2}}\,dx+
\int x\frac{x}{\sqrt{1+x^2}}\,dx
\qquad \biggl(u=x, v'=\frac{x}{\sqrt{1+x^2}}\biggr)
\\[6px]
&=\int\frac{1}{\sqrt{1+x^2}}\,dx+x\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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3D Perspective projection I have this following question to answer, however I am not sure how I should combine my calculation into one final answer.
Suppose the Centre of Projection in a viewing space is at an offset
$(0, 0, -5)$ from $(0,0,0)$, and the view plane is the $UV$ plane containing
$c$. Find the transfo... | Let me try based on the theories described in the arXiv.org article Unified Framework of Elementary Geometric Transformation Representation.
A so-called perspective projection in this specific problem can be realized by a central projection defined in definition 3.2 (pages 8-9) and formulated in equation (3.1) or (see ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$ Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n(n$ arbitrary positive integer)
First of all I checked a few cases for small $n$'s and in all cases the rightmost digit was $5$, so maybe this is the case for all values of $n$.
Then I tho... | Our sum is odd, so all we need to do is to compute it modulo $5$.
Note that the congruence class of $k^n$ modulo $5$ is the same as the congruence class of $k^{n+4}$ modulo $5$. This is obvious if $k$ is divisible by $5$. And if $k$ is not divisible by $5$ then $k^4\equiv 1\pmod{5}$.
So to find the last digit for any $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Asymptotes of $(x(t),y(t)) = \bigg(\frac{1+t^2}{2+t^3}, \frac{t}{2+t^3}\bigg)$, collinear points, ... Consider the curve:
\begin{equation}
(x(t),y(t)) = \bigg(\frac{1+t^2}{2+t^3}, \frac{t}{2+t^3}\bigg)
\end{equation}
Question 1: What are his asymptotes?
Answer: In projective space: $[(2+t^3,1+t^2,t)]$. The intersection... | To the third question: Entering into M2
R=QQ[x,y]
S=QQ[t]
KS=frac S
f=map(KS,R,{(1+t^2)/(2+t^3),t/(2+t^3)})
ker f
we see $2x^3-6x y^2+5y^3-x^2-2x y+2y^2+y=0$ is the implicit equation. The partial derivatives
$6x^2-6y^2-2x-2y,-12xy+15y^2-2x+4y+1$
have a common zero at $(x,y)=(2/3,1/3)$ which is on the curve. This is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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given $f(x)$ is continuous and $\int_0^{x^2}f(t)dt = x+sin\frac{\pi}{2}x$,solve $f(1)=?$ given $f(x)$ is continuous and $\int_0^{x^2}f(t)dt = x+\sin\frac{\pi}{2}x$,solve $f(1)=?$
we know
$$[\int_0^{x^2}f(t)dt]^\prime= f(x^2) \cdot2x$$
and
$$[x+\sin\frac{\pi}{2}x]^\prime = 1+ \frac{\pi}{2}\cos\frac{\pi}{2}x$$
therefore
... | The issue is that the identity can only be valid for $x\geq 0$ or $x\leq 0$ but not both. The reason is simple: The LHS of the desired relation is even in $x$, but the RHS is odd! So the formula can't hold for all $x$, and we should limit ourselves to one particular sign.
Regarding the behavior at zero, the conclusion... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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expansion of a generating function I found this formula in a book
$$\sqrt{1-4x}=1-2\sum_{n=1}^{\infty}\frac{1}{n} {{2n-2}\choose {n-1}} x^n$$
How can I prove that?
| Hint. One may apply the Taylor series expansion to $f(x):=\sqrt{1-4x}$,
$$
f(x)=f(0)+\sum_{n=1}^\infty\frac{f^{(n)}(0)}{n!}x^n,\quad |x|<\frac14. \tag1
$$
Then one may prove by induction that
$$
\begin{align}
(\sqrt{u})'=& \frac12\cdot u^{1/2-1}
\\(\sqrt{u})^{(2)}=& \frac12\left(\frac12-1 \right)\cdot u^{1/2-2}
\\(\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find $k$ if Sum of all Trigonometric ratios is $7$ Given $0< A< 90^{\circ}$ and $$\sin A+\cos A+\tan A+\sec A+\operatorname{cosec} A+\cot A=7$$
and if $\sin A$ and $\cos A$ are roots of $4x^2+3x+k=0$
Find the value of $k$
sum of the roots is $$\sin A+\cos A=\frac{-3}{4}$$
Squaring both sides we get
$$1+2\sin A\cos A=... | You're right. This equation is not right.
From this post: If $\sin \theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$, then $\sin 2\theta$ is a root of $x^2 -44x +36=0$ My own bonafide attempt.
it shows that $\sin x \cos x$ is a root of $x^2 -44x +36=0$
=> $\sin x \cos x$ = $22-8\sqrt{7}$ (Another value g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Given that $\tan 2x+\tan x=0$, show that $\tan x=0$ Given that $\tan 2x+\tan x=0$, show that $\tan x=0$
Using the Trigonometric Addition Formulae,
\begin{align}
\tan 2x & = \frac{2\tan x}{1-\tan ^2 x} \\
\Rightarrow \frac{2\tan x}{1-\tan ^2 x}+\tan x & = 0 \\
\ 2\tan x+\tan x(1-\tan ^2 x) & = 0 \\
2+1-\tan ^2 x & = 0 ... | Hint. As written, the assertion is not correct since
$$
\tan\left(\frac{2\pi}3\right)+\tan\left(\frac{\pi}3\right)=0
$$ but $$\tan\left(\frac{\pi}3\right)=\sqrt{3}\color{red}{\neq}0.$$
There is a mistake in your reasoning, starting as you did you obtain
$$
\left(\frac{2\color{blue}{\tan x}}{1-\tan ^2 x}+\color{blue}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove by induction the particular inequality $\left(1.3\right)^n \ge 1 + \left(0.3\right)n$ for every $n \in \mathbb N$ $\left(1.3\right)^n \ge 1 + \left(0.3\right)n$ for every $n \in \mathbb N$
Not sure where I'm going wrong in my Algebra, but I assume it's because I'm adding an extra term. Is the extra term unnecess... | Let's prove Bernoulli's inequality by induction. Assume $x\ge -1$.
Base step $n=1$:
$$(1+x)^1 = 1+1\cdot x.$$
Now induction step. Assume $(1+x)^n \ge 1+nx$. Then
\begin{align*} (1+x)^{n+1} &= (1+x)^n (1+x) \\
& \ge (1+nx) (1+x)\\
& = 1+(n+1)x +nx^2\\
&\ge 1+(n+1)x.
\end{align*}
Note that the first inequality hol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Bound for the sum of the divisors of a number
Let us denote by $s(n) = \sum_{d|n} d$ the sum of divisors of a natural number $n$ ($1$ and $n$ included). If $n$ has at most $5$ distinct prime divisors, prove that $s(n) < \dfrac{77}{16}n$. Also prove that there exists a natural number $n$ for which $s(n) > \dfrac{76}{16... | Let $n = \prod_i^6 p_i^k$.
You've pointed out that $s(n)/n = \prod_i^6 \frac{p_i - \frac 1{p_i^k}}{p_i -1} < \prod_i^6 \frac{p_i }{p_i -1}=77*13/16*12$.
But we can make $k$ arbitrarily large so for any $\epsilon > 0$ in particular any epsilon $77/16 < 77*13/16*12 - \epsilon$ we can find $k$ so that $77*13/16*12 >s(n)/n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integral of $\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx$
Find the integral of the following:
$$\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx$$
Do set $u=x^3+3x^2+1$?
So, $du=(3x^2+6x)dx$?
And, $x^2+2x={u-1-x^2\over x}$?
So then,
$$\int{({u-1-x^2\over x})\over \sqrt u} du$$
This seems very complicated, is there any easier w... | You have $du=3(x^2+2x)dx$, so your integral become :
$$\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx=\int{1\over 3\sqrt{u}} du=\frac{1}{3}\int {1\over \sqrt{u}} du=\frac{2}{3}\sqrt u$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$
Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n}{\left(n+1\right)!} = 1-\frac{1}{\left(n+1\right)!}$
So I proved the base case where $n=1$ and got $\frac{1}{2}$
Then since $n=k$ implies $... | You can simply write it like this also this solution doesn't need induction:
$=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n+1-1}{(n+1)!}$
$=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{n!}-\frac{1}{(n+1)!}$
$=1-\frac{1}{(n+1)!}$
$$solved!$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Use integration by parts $\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$ $$I=\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$$
Clearly $$-2I=\int^{\infty}_{0} \ln x \cdot \frac{-2x }{(1+x^2)^2} dx$$
My attempt :
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \left(\f... | HINT:
Let $x=\tan y$
$$\implies2J=\int_0^{\pi/2}\sin2y\cdot\ln(\tan y)dy$$
Now use $$I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
$$\implies2I=\int_a^bf(x)dx+\int_a^bf(a+b-x)dx=\int_a^b\left(f(x)+f(a+b-x)\right)dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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With $y_n$ a sequence of real numbers, prove that if $y_n=x_{n-1}+2x_{n}$ converges then $x_n$ also converges Let $y_n$ be a sequence of real numbers. Prove that if $y_n=x_{n-1}+2x_{n}$ converges then $x_n$ also converges.
Let us suppose that $y_{n}$ goes to a limit $L$. Then for all $\varepsilon >0$, for sufficiently ... | We have
$$
\begin{aligned}
x_2 &= \frac 12 y_2 - \frac 12 x_1 \\
x_3 &= \frac 12 y_3 - \frac 12 x_2 = \frac 12 y_3 + \frac 14 y_2 - \frac 14 x_1 \\
x_4 &= \frac 12 y_4 - \frac 12 x_3 = \frac 12 y_4 + \frac 14 y_3 - \frac 18 y_2 + \frac 18 x_1 \\
\ldots
\end{aligned}
$$
and generally for $n\ge 2$
$$ \beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove ${2\over \pi}\int_{0}^{\infty}\prod_{k=1}^{n\ge1}{k^2\over k^2+x^2}dx={n\over 2n-1}$ Prove
$$I={2\over \pi}\int_{0}^{\infty}\prod_{k=1}^{n\ge1}{k^2\over k^2+x^2}dx={n\over 2n-1}\tag1$$
Expand out $(1)$
$$I={2n!\over \pi}\int_{0}^{\infty}{1\over (1+x^2)(2^2+x^2)(3^2+x^2)\cdots(n^2+x^2)}dx\tag2$$
Noticing
$${1\o... | I dont know if the follow help you. put $$I_n={2\over \pi}\int_{0}^{\infty}\prod_{k=1}^{n}{k^2\over k^2+x^2}dx$$ By induction we have:
$$I_1={2\over \pi}\int_{0}^{\infty}{1\over 1+x^2}dx={2\over \pi}{ \pi\over2}=1$$
Suppose that it true for $I_n$. Now
\begin{align}I_{n+1}&={2\over \pi}\int_{0}^{\infty}\left(\prod_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Converting from base $2$ to base 3
Transform the binary expansion $y = 0.110110110\ldots$ into a ternary expansion.
We are given that $y = 0.110110110\ldots_2$ and thus $1000_2y = 110_2+y \implies y = \frac{6}{7}$. Then we see that $\frac{6}{7}=0.857142\ldots$. How do I convert this to base $3$?
| Notice that this is $0.\overline{110}$, so it corresponds to $\frac{110_2}{1000_2-1}=\frac 6 7$.
This is how repeating decimals in different bases work: The repeating part can be written as the part that repeats over the difference between the power of the base and $1$.
We need to find where we can do this for $\frac 6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $A$ be a $2 \times 2$ real matrix such that $A^2 - A + (1/2)I = 0$. Prove that $A^n \to 0$ as $n \to \infty$. Question: Let $A$ be a $2 \times 2$ matrix with real entries such that $A^2 - A + (1/2)I = 0$, where $I$ is the $2 \times 2$ identity matrix and $0$ is the $2 \times 2$ zero matrix. Prove that $A^n \to 0$ a... | You can get $A^n=A^{n-1}-1/2A^{n-2}=(A^{n-2}-1/2A^{n-3})-1/2A^{n-2}=1/2A^{n-2}-1/2A^{n-3}=-1/4A^{n-4}$. Then the conclusion is obvious by induction on k, while let n=4k+i.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
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Integrating $\int\frac{x^3}{\sqrt{9-x^2}}dx$ via trig substitution What I have done so far:
Substituting $$x=3\sin(t)\Rightarrow dx=3\cos(t)dt$$ converting our integral to
$$I=\int\frac{x^3}{\sqrt{9-x^2}}dx=\int \frac{27\sin^3(t) dt}{3\sqrt{\cos^2(t)}}3\cos(t)dt\\
\Rightarrow \frac{I}{27}=\int \sin^3(t)dt=-\frac{\sin^... | Substitution $x=3\sin{y} \rightarrow dx=3\cos{y}$ $dy$
Then
$27 \int \frac{\sin^3{y}}{3\cos{y}}.3\cos{y}$ $dy$
$=\frac{27}{4}\int 4\sin^3{y}$ $dy$
Use Identity
$4\sin^3{y}=3\sin{y}-\sin{3y}$
$\frac{27}{4}\int 4\sin^3{y}$ $dy$ $=\frac{27}{4}\int 3\sin{y}-\sin{3y}$ $dy$
$=\frac{27}{4}(-3\cos{y}-\frac{\cos{3y}}{3})$
Fro... | {
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"url": "https://math.stackexchange.com/questions/1826494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$ Can someone point me in the right direction how to solve this?
$3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$
I guess I have to get to logarithms of the same base. But how? What principle should I use here?
Thx
| $$3^{ x+2 }\cdot 4^{ -(x+3) }+3^{ x+4 }\cdot 4^{ -(x+3) }=\frac { 40 }{ 9 } \\ { 4 }^{ -\left( x+3 \right) }{ 3 }^{ x+2 }\left( 1+9 \right) =\frac { 40 }{ 9 } \\ { 4 }^{ -\left( x+3 \right) }{ 3 }^{ x+2 }=\frac { 4 }{ 9 } \\ { \left( \frac { 3 }{ 4 } \right) }^{ x }\frac { 9 }{ 64 } =\frac { 4 }{ 9 } \\ { \left( \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the vertex lies on the surface $z^2(\frac{x}{a}+\frac{y}{b})=4(x^2+y^2)$ Two cones with a common vertex pass through the curves $z^2=4ax,y=0$ and $z^2=4by,x=0.$ The plane $z=0$ meets them in two conics which intersect in four concyclic points.Show that the vertex lies on the surface $z^2(\frac{x}{a}+\frac{y}{... |
The generator line of the first cone is $\frac{x-\alpha}{l_1}=\frac{y-\beta}{m_1}=\frac{z-\gamma}{n_1}.$
Setting $y=0$ gives
$$\frac{x-\alpha}{l_1}=\frac{0-\beta}{m_1}=\frac{z-\gamma}{n_1}\implies x=-\beta\frac{l_1}{m_1}+\alpha,\quad z=-\beta\frac{n_1}{m_1}+\gamma\tag1$$
Also,
$$\frac{l_1}{m_1}=\frac{x-\alpha}{y-\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove this Taylor expansion of $\frac{1}{(1+x)^2}=-1\times\displaystyle\sum_{n=1}^{\infty}(-1)^nnx^{n-1}$? I came across this series of the Taylor Expansion-
$$\frac{1}{(1+x)^2}=1 - 2x + 3x^2 -4x^3 + \dots.=-1\times\sum_{n=1}^{\infty}(-1)^nnx^{n-1}$$
But I have no idea how to prove this...
Thanks for any help!... | Observe that
$$\frac x{(1+x)^2}=x - 2x^2 + 3x^3 -4x^4 + \dots$$ and adding the original
$$\frac1{(1+x)^2}=1-2x+3x^2-4x^3+5x^4\cdots$$
you verify
$$\frac x{(1+x)^2}+\frac1{(1+x)^2}=1-x+x^2-x^3+x^4-\cdots=\frac1{x+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Prove that $\frac{2^a+1}{2^b-1}$ is not an integer
Let $a$ and $b$ be positive integers with $a>b>2$. Prove that $\frac{2^a+1}{2^b-1}$ is not an integer.
This is equivalent to showing there always exists some power of a prime $p$ such that $2^a+1 \not \equiv 0 \pmod{p^a}$ but $2^b-1 \equiv 0 \pmod{p^a}$. How do we pr... | Note for this type of question is always useful to ask what how many times does the denominator easily go into the numerator, here we have $2^a/2^b = 2^{a-b}$, thus $$\frac{2^a + 1}{2^b -1} = \frac{2^a + 1 - 2^{a-b} (2^b - 1)}{2^b - 1} + 2^{a-b} = \frac{2^{a-b} + 1}{2^b - 1} + 2^{a-b}$$
Thus if $a > b$, then
$$\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve definite integral: $\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$ I need to solve:
$$\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$$
Here is my steps, first of all consider just the indefinite integral:
$$\int \arctan(\sqrt{x+2})dx = \int \arctan(\sqrt{x+2}) \cdot 1\ dx$$
$$f(x) = \arctan(\sqrt{x+2})$$
$$f'(x) = \frac{1}{1+x+2} \... | $$\int \arctan(\sqrt{x+2})\, dx$$
$$=\int \arctan (\sqrt{x+2})\, d(x+2)$$
$$=2\int (\arctan(\sqrt{x+2}))(\sqrt{x+2})\, d(\sqrt{x+2})$$
$$=2\int (\arctan u)(u)\, du,$$
where $u=\sqrt{x+2}$. Now use integration by parts:
$$=2\left((\arctan u)\left(\frac{u^2}{2}\right)-\int \frac{u}{u^2+1}\, du\right)$$
$$=2\left((\arctan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + 1/β$ and $β + 1/α$. Quadratic equation question, as specified in the title.
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + \frac{1}{β}$ and $β + \frac{1}{α}$.
I gather that ... | If $\alpha,\beta$ are the roots of $x^2-6x+7=0$, then all quadratic equations with roots $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$ are $$a\left(x-\left(\alpha+\frac{1}{\beta}\right)\right)\left(x-\left(\beta+\frac{1}{\alpha}\right)\right)=0,$$
where $a\in\mathbb R$, $a\neq 0$ (saying "the equation" is wrong... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Prove that $\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$
Prove that $\displaystyle \sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$.
I tried using the partial fraction decomposition $a_j = \frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}$, but I don't see... | More generally,
if
$x_{(n)}
=\prod_{k=0}^{n-1} (x+k)
$,
$\begin{array}\\
\dfrac1{x_{(n)}}-\dfrac1{(x+1)_{(n)}}
&=\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=0}^{n-1} (x+1+k)}\\
&=\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=1}^{n} (x+k)}\\
&=\dfrac1{\prod_{k=1}^{n-1} (x+k)}\left(\dfrac1{x}-\dfrac1{x+n}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$ My question is:
Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer.
I'm stuck at the basis step.
If I started with $1$. I get the right hand side is $1... | For $n=1$, since $3 + 3 \times 5^1= 3+15 = 18$, there is no problem with the righthand side being $18$ too.
Note though that the base case is $n=0$, the condition being $n$ nonnegative, and then the lefthand side is $3$ as is the righthand side, which is $3 \times (5^{0+1}-1)/4 = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
A convergent series: $\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$ I would like to find the value of:
$$\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$$
I could only see that the ratio of two consecutive terms is $\dfrac{1}{27\cos(2\theta)}$.
| Hint. One may observe that
$$
\sin^3 (a)=\frac34\sin (a)-\frac14 \sin(3a)
$$ giving
$$
3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)=\frac{3^{n}}4\sin\left(\frac{\pi}{3^{n+1}}\right)-\frac{3^{n-1}}4\sin\left(\frac{\pi}{3^{n}}\right)
$$
then one gets a telescoping sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find last 5 significant digits of 2017! Since there are less powers of $5$ than of $2$ and since $10 = 2 \cdot 5$, I counted the number of zeros in $2017!$:
$\left \lfloor{ \frac{2017}{5^1}}\right \rfloor + \left \lfloor{ \frac{2017}{5^2}}\right \rfloor +\left \lfloor{ \frac{2017}{5^3}}\right \rfloor +\left \lfloor{ \... | I know how to do the last 3 significant digits, if that helps.
$4! = 24\\
\frac {9!}{5!} = 3024 \equiv 24\pmod {1000}$
$\frac {5n!}{5^n n!} = 24^n \pmod {1000}$
$2017! = \frac {2017!}{2015!}\frac {2015!}{5^{403}403!}\frac {5^{403}403!}{5^{80}80!}\frac {5^{80}80!}{5^{16}16!}\frac {5^{16}16!}{5^{3}3!} 3!\\$
$\frac {2017... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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Solution of Diophantine equation Find all integral solutions of $x^2+1= y^2+z^2$. Actually I have to find all integral solution of $a(a+1)=b(b+1)+c(c+1)$. I reduced this in the above form I.e., $ (2a+1)^2+1= (2b+1)^2+(2c+1)^2$ .
| The case where $z^2-x^2=1-y^2=0$ is self-evident.We have $z=\pm x$ and $y=\pm 1$
Now consider the case where $(1-y^2)(z^2-x^2)\neq 0$
$$z^2-x^2=1-y^2$$
$$(z+x)(z-x)=(1+y)(1-y)$$
$$\dfrac{z+x}{1+y}=\dfrac{1-y}{z-x}
=\dfrac{p}{q}$$ with $pq\neq 0$ and $\gcd(p,q)=1$.
It follows:
$$z+x=\dfrac{p}{q}(1+y)$$
$$z-x=\dfrac{q}{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.