Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Counting Possible Sums in a Group of Integers Say you have 5 buckets.
Each bucket contains a distinct number $[1, 10]$.
So why can it be guaranteed that the total sum of 1 or more numbers from the 5 buckets will equal the total sum of 1 or more OTHER numbers from the 5 buckets?
Example:
6, 7, 8, 9, 10.
We know that $6+... | This is the pigeonhole principle at work, although it takes a little digging to get it to work.
Firstly we can always assume that the numbers are different, or we could just match two that are identical. So although you gave me that the numbers were distinct, that's actually not needed to make this work.
So given the ... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$ without using L'Hôpital's rule $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$
To do this I tried 2 approaches:
1: If $\lim_{x \rightarrow 0} \frac{\ln(x+1)}{x} = 1$, $\lim_{x\to1}\frac{\ln(x)}{x-1}=1$ and $\lim_{x\to0}\frac{\ln(x+1)}x=\lim_{u\to1}\fra... | $$\lim_{x\to3}\frac{\ln \sqrt {x-2}}{x^2-9} = \lim_{x\to3}\frac{\ln (1+\sqrt{x-2}-1)}{\sqrt {x-2}-1}\cdot\frac{\sqrt {x-2}-1}{x^2-9} = \lim_{x\to3}\frac{x-3}{1+\sqrt {x-2}}\cdot\frac{1}{x^2-9} = \frac{1}{12}$$
| {
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Find $a, b\in \mathbb Z$ where $(a^2+b)(a+b^2)=(a-b)^3$ Find all non-zero $a, b\in \mathbb Z$ where
$$(a^2+b)(a+b^2)=(a-b)^3$$
I actually had no clue on what to try. Thanks for your help.
I believe I've already tried but per the 1st comment let me expand both sides and see what I cancel.
$$a^3+a^2b^2+ab+b^3=a^3-3a^2b+... | Let me post it as an answer to mark this answered.
Thanks again to астон вілла олоф мэллбэрг!
$$a^3+a^2b^2+ab+b^3=a^3-3a^2b+3ab^2-b^3$$
$$b(a^2b+2b^2+3a^2-3ab+a)=0$$
$$b=0\quad or \quad 2b^2+(a^2-3a)b+3a^2+a=0$$
Applying quadratic formula on b,
$$b=\frac{a(3-a)\pm\sqrt{a^2(a-3)^2-24a^2-8a}}{4}=\frac{a(3-a)\pm (a+1)\sqr... | {
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Formal way of showing inequality? At the end of my class today, there was a proof that involved using the fact that given $f(x)=3+4x-x^2-x^4$
$f(x)\leqslant 0 $ if $|x|>2 $
this seems obvious but I was wondering if anyone could show me a more formal way of proving it?
| For $x\geq2$ we have $$x^4+x^2-4x-3=x^4-2x^3+2x^3-4x^2+3x^2-6x+2x-4+1=$$
$$=(x-2)(x^3+2x^2+3x+2)+1>0$$
For $x\leq-2$ we obtain:
$$x^4+x^2-4x-3=x^4+3x^3+4x^2-4x^3-16x^2-16x+13x^2+52x+52-40x-80+25=$$
$$=(x+2)^2(x^2-4x+13)-40(x+2)+25>0$$
| {
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How to take the integral $\int\frac{dx}{\sin^3x}$ There's
$$\int\frac{\mathrm dx}{\sin^3x}.$$
I tried to write it like
$$\int\frac{(\sin^2x+\cos^2x)}{\sin^3x}\,\mathrm dx,$$
and then made partial fractions from it, but it didn't help much, the answer is still incorrect.
| $$
\begin{aligned}
\int \frac{1}{\sin^3x}dx
& = \int \frac{\left(t^2+1\right)^2}{4t^3}dt
\\& =\frac{1}{4}\int \:\frac{1}{t^3}+t+\frac{2}{t}dt
\\& = \color{red}{\frac{1}{4}\left(\frac{1}{2}\tan ^2\left(\frac{x}{2}\right)-\frac{1}{2}\cot ^2\left(\frac{x}{2}\right)+2\ln \left|\tan \left(\frac{x}{2}\right)\right|\right)+... | {
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Which of the following numbers is greater? Which of the following numbers is greater?
Without using a calculator and logarithm.
$$7^{55} ,5^{72}$$
My try
$$A=\frac{7^{55} }{5^{55}×5^{17}}=\frac{ 7^{55}}{5^{55}}×\frac{1}{5^{17}}= \left(\frac{7}{5}\right)^{55} \left(\frac{1}{5}\right)^{17}$$
What now?
| Observe that $7^2/5^2 = 49/25 < 2$. Cubing both sides, $(7^2/5^2)^3 < 8$. Since $7/5^3 = 7/125 < 1/8$ we have $7^7/5^9 < 8 \times 1/8 = 1$.
Then $7^{55}/5^{72} = 1/7 \times (7^7/5^9)^8 < 1/7 < 1$ and so $7^{55} < 5^{72}$.
| {
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How to solve $p^2 - p + 1 = q^3$ over primes?
How to solve $p^2 - p + 1 = q^3$ over primes?
I checked that $p$ and $q$ must be odd and I don't know what to do next. I don't have any experience with such equations.
| COMMENT.-Primes $p$ of the form $6x-1$ cannot be solution: in fact
$$(6x-1)^2-(6x-1)+1=36x^2-18x+3=3(12x^2-6x+1)$$ so $q=3$ which is not compatible with $12x^2-6x+1=3^2$.
It follows $p$ must be of the form $6x+1$ so we have
$$(6x+1)^2-(6x+1)+1=36x^2+6x+1=q^3$$ Solving the equation, where $X=6x$,
$$X^2+X+1-q^3=0$$ we ha... | {
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If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle what is the value of ab? If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle, find the value of $ab$
| Pretty lengthy equations (and probably bad method to solve). So here are the equations :
$$a^2+11^2=b^2+37^2=(b-a)^2+(37-11)^2$$
$$\implies b^2+a^2-2ab +676 = a^2+121\implies b^2-2ab+555=0\,\,\,\,(1)$$
$$b^2 = a^2-1278 \implies b=\sqrt{a^2-1278}\,\,\,\,\,\,(2)$$
substitute $b$ in $(1)$$$\implies (a^2-1278)+555 = 2a(\sq... | {
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Determinant of a matrix that is almost lower triangular Calculate the determinant of
$$ \left[
\begin{array}{cccc}
1 & 0 & 0 & 0 & \cdots & 1\\
1 & a_1 & 0 & 0 & \cdots & 0 \\
1 & 1 & a_2 & 0 & \cdots & 0 \\
1 & 0 & 1 & a_3 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cd... | Expanding from the first one will force a product along the leading diagonal; the one at the end of the first row will leave you with another determinant
\begin{eqnarray*}
\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 & \cdots & 1\\
1 & a_1 & 0 & 0 & \cdots & 0 \\
1 & 1 & a_2 & 0 & \cdots & 0 \\
1 & 0 & 1 & a_3 & \cd... | {
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"source": "stackexchange",
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Solving: $(b+c)^2=2011+bc$
Solve $$(b+c)^2=2011+bc$$ for integers $b$ and $c$.
My tiny thoughts:
$(b+c)^2=2011+bc\implies b^2+c^2+bc-2011=0\implies b^2+bc+c^2-2011=0$
Solving in $b$ as Quadratic.$$\implies b=\frac{-c\pm \sqrt{8044-3c^2}} {2}.$$
So $8044-3c^2=k^2$, as $b$ and $c$ are integers. We also have inequalitie... | Decided to see what I could do in my head.
$2011 \equiv 1 \pmod 3$ is a prime. This means it has both expressions as $u^2 + uv + v^2$ and $x^2 + 3 y^2.$ Note that $45^2 = (4 \cdot 5) \cdot 100 + 25 = 2025$ is too large for $x^2.$ Then $44^2 = 45^4 - 2 \cdot 45 + 1 = 2025 - 90 + 1 = 1935 + 1 = 1936$ is small enough. The... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Quadratic equation find all the real values of $x$ Find all real values of $x$ such that
$\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$
I tried sq both sides by taking 1 in RHS but it didn't worked out well...
| Divide by $x$:
$$\sqrt{\frac1x-\frac1{x^3}}+\sqrt{\frac1{x^2}-\frac1{x^3}}=1$$
Change $t=1/x$:
$$\sqrt{t-t^3}+\sqrt{t^2-t^3}=1$$
Square and rearrange:
$$2\sqrt{t^3-t^4-t^5+t^6}=t+t^2-2t^3-1$$
Square and rearrange again:
$$4t^6-4t^5-4t^4+4t^3=4t^6-4t^5-3t^4+6t^3-t^2-2t+1$$
Finally we get
$$t^4+2t^3-t^2-2t+1=0$$
That is,... | {
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Multiple choice question of indefinite integral, $\int \frac{x + 9}{x^3 + 9x} dx$. If $\int \frac{x + 9}{x^3 + 9x} dx = k\arctan(mx) + n\ln (x) + p \ln (x^2 + 9) + c$, then $(m+n)/(k+p) = $
(A) 6
(B) -8
(C) -3
(D) 4
I tried solving it by differentiating the R.H.S. but couldn't arrive at the answer.
| Try to separate the denominator like this
$$\frac{x+9}{x^3+9x}=\frac{x+9}{x(x^2+9)}$$
$$\qquad\qquad\ \,=\frac Ax+\frac{Bx+C}{x^2+9}$$
Then construct an equality for $A,B,C$ and we have
$$Ax^2+9A+Bx^2+Cx=x+9$$
Here we get $A=1,\ B=-1,\ C=1$, and that gives us
$$\int\frac{x+9}{x^3+9x}\,dx\ =\ \int\frac1x+\frac1{x^2+9}-\... | {
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Show that $\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$
Prove the following:
$\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$
How do you go about proving this?
| Previous comments and other answers show you the path to the solution.
I would like to propose another formula of the same kind, thats illustrates the fact that cancellation of nested square roots may sometimes occur ... but not necessarily in any obvious way !
Question Find a simpler form for $x=\sqrt{2+\sqrt{3}}+\sqr... | {
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Regarding complex roots of a polynomial I am having difficulty in finding the roots of the following polynomial when it is given that all the roots are complex:-
$$f(x)= x^4+4x^3+8x^2+8x+4$$
How can I factorize the polynomial to get its roots?
| Using binomial theorem;
$$\begin{align}f(x)&=x^4+4x^3+8x^2+8x+4\\
&=\left(x^4+4x^3+6x^2+4x+1\right)+2\left(x^2+2x+1\right)+1\\
&=(x+1)^4+2(x+1)^2+1\\
&=\left((x+1)^2+1\right)^2\\
&=\left(x^2+2x+2\right)^2\end{align}$$
| {
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If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$ If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$.
Find the value of $2b + \dfrac {c}{a}$.
My Attempt:
$$\sin A+\sin^2 A=1$$
$$\sin A + 1 - \cos^2 A=1$$
$$\sin A=\cos^2 A$$
Now,
$$a(\cos^2 A)^{6}+b(\cos^2 A)^{4}+c(\cos^2 A)^{3}=1$... | @MyGlasses .. In the third line from below you forgot that there is $(-a)$ twice in your attempt to prove the required identity, resulting in a mistake in your answer.
| {
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How many integers between $10000$ and $99999$ ( inclusive) are divisible by $3$ or $5$ or $7$? How many integers between $10000$ and $99999$ (inclusive) are divisible by $3$ or $5$ or $7$ ?
My Try :
Total Integers between $10000$ and $99999$ are $89999$.
$\left\lfloor\frac{89999}{3}\right\rfloor+\left\lfloor\frac{899... | Another way:
$10000=3\times 3333+1\implies$ First integer $>10000$ divisible by $3$ is $10002$.
So if we count a number of integers in sequence $10002,10005,\cdots,99999$ we have our answer. the common difference is $3$ first term is $10002$ hence, $a_n=10002+(n-1)3=99999\implies n=30000$. Similarly for others too.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I determine a basis of the vector space of polynomials degree 3 or less that satisfy $\int_{0}^{1}[xp'(x)-p(x)]=p(1)-2p(0)$? I have proven that the set of polynomials of degree 3 or less that satisfy $\int_{0}^{1}[xp'(x)-p(x)]=p(1)-2p(0)$ is indeed a subspace of $\mathscr{P}_{3}(\mathbb{R})$.
Now I need to find ... | Let $p(x) = a + bx + cx^2 + dx^3$. Then the equation reads
$$
\int_0^1 x(b + 2cx + 3d x^2) - (a + bx + cx^2 + dx^3) \mathrm{d} x = a+b+c+d - 2a
$$
Evaluating the left-hand side gives
$$
\frac b2 + \frac{2c}{3} + \frac{3d}{4} - a - \frac{b}{2} - \frac{c}{3} - \frac{d}{4} = a + b + c + d - 2a,
$$
or
$$
b = - \frac{2}... | {
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Sum of reciprocal of the positive divisors of $1800.$ Sum of reciprocal of the positive divisors of $1800.$
Attempt: divisors of $1800 = 2^3\times 3^2 \times 5^2$
so sum of divisors of $1800$ is $\displaystyle (1+2+2^2+2^3)\times (1+3+3^2)\times (1+5+5^2)$
$ \displaystyle = \bigg(\frac{2^4-1}{2-1}\bigg)\times \bigg(\fr... | HINT:
$$\sum_{d=1,d|n}^n\dfrac1d=\dfrac1n\sum_{d=1,d|n}^n\dfrac nd=\dfrac1n\sum_{d=1,d|n}^nd$$
Now use this
See also: Is there a formula to calculate the sum of all proper divisors of a number?
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine: $S = \frac{2^2}{2}{n \choose 1} + \frac{2^3}{3}{n \choose 2} + \frac{2^4}{4}{n \choose 3} + \cdots + \frac{2^{n+1}}{n+1}{n \choose n}$ We are given two hints: consider $(n+1)S$; and use the Binomial Theorem. But we are not to use calculus.
My consideration of $(n+1)S$ goes like this:
\begin{align*}
\sum\... | The trick is to get rid of the factors $1/(k+1)$.
You do so by absorbing $k+1$ in $k!$, and
$$\frac{\binom nk}{k+1}=\frac{n!}{(k+1)k!(n-k)!}=\frac{n!}{(k+1)!(n+1-k-1)!}=\frac{\binom{n+1}{k+1}}{n+1}.$$
Now,
$$S=\sum\limits_{k=1}^{n}\frac{2^{k+1}}{k+1}{n \choose k}=\frac1{n+1}\sum\limits_{k=1}^{n}2^{k+1}{n+1 \choose k+1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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why is $2^{\log^{2}\frac{1}{2}} = 1/2$? I have no understanding of $\log^2n$, wouldn't $2^{\log^{2}\frac{1}{2}} = 1/4$ since $2^{\log\frac{1}{2}} = 1/2$?
$\log$ here has the base of $2$.
| Using a combination of exponent rules and the one log rule you mentioned,
$$2^{\log^2\frac{1}{2}} = 2^{\log\frac{1}{2} \cdot \log\frac{1}{2}} = (2^{\log\frac{1}{2}})^ {\log\frac{1}{2}} = \left(\frac{1}{2}\right)^{\log\frac{1}{2}} = \frac{1}{2^{\log\frac{1}{2}}} = \frac{1}{\frac{1}{2}} = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Fisher information of exponential distribution using the generic formula. I am trying to calculate Fisher information when $X_1...X_n$ are IID and follow the exponential distribution:
$$f(x)=\frac{1}{\theta}e^{-\frac{x}{\theta}}$$
I use the following formula of Fisher information to confirm that the result is indeed th... | Ok using that parameterisation I agree your likelihood is correct! So method one we differentiate again to get
$$
\ell_{\theta \theta} = -\frac{2 \sum x_i }{\theta^3} + \frac{n}{\theta^2}
$$
now since $\mathbb{E} \sum_i x_i = n \theta$ we get
$$
\begin{align}
\mathcal{I} &= - \mathbb{E}\left[ \ell_{\theta \theta} \rig... | {
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Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$
Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$
My work so far:
1) $$y =\frac{2x+9}{\sqrt{-x^2+4x+12}+\sqrt{-x^2+2x+3}} $$
2) I used a derivative and found the answer ($y=\sqrt3$ at $x=0$). Is there any other way?
| Domain gives $-1\leq x\leq3$.
If $x=0$ so $y=\sqrt3$.
We'll prove that it's a minimal value of $y$.
Id est, we need to prove that
$$\sqrt{-x^2+2x+3}+\sqrt3\leq\sqrt{-x^2+4x+12}$$ or after squaring of the both sides we need to prove that
$$\sqrt{3(-x^2+2x+3)}\leq x+3$$ or since $x+3>0$, we need to prove that
$$3(-x^2+2... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $b_n = b_{n+100}$
Let $b_n$ denote the units digit of $\displaystyle\sum_{a=1}^n a^a$. Prove that $b_n = b_{n+100}$.
I tried rewriting the sum, but didn't see how to prove the equality. For example, if $n = 178$ we have $$\displaystyle\sum_{a=1}^{178} a^a = (1^1+2^2+3^3+\cdots+78^{78})+(79^{79}+80^{80}+81^... | It is sufficient to prove that for each integer $k$
$$\sum_{a=t+1}^{t+100} a^a \equiv 0 \mod10$$
Consider a subsequence with indices $k,k+10,\dots,k+90$. We have
$$
b(k) = k^k + (k+10)^{k+10} +\dots+(k+90)^{k+90}
$$
Obviously, the following holds $$b(k) \equiv k^k + k^{k+10} + \dots+k^{k+90}\mod10$$
The right side coul... | {
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Evaluate the sum of series $\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$ Evaluate the sum of series
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$$
I have tried two methods:
1) using power series
2) using partial sums
but I can't find the sum.
1) Using power series:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{... |
With telescoping we obtain
\begin{align*}
\sum_{k=1}^\infty& (-1)^{k-1}\frac{1}{k(4k^2-1)}\\
&=\lim_{N\to\infty}\sum_{k=1}^N(-1)^{k-1}\frac{1}{k(4k^2-1)}\\
&=\lim_{N\to\infty}\sum_{k=1}^N(-1)^{k-1}\left(\frac{1}{2k-1}+\frac{1}{2k+1}-\frac{1}{k}\right)\\
&=1-\sum_{k=1}^\infty(-1)^{k-1}\frac{1}{k}\\
&=1-\ln 2
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Variation of Coupon Collectors Problem for the case where $k\le n$. I was working on this problem which says,
Suppose one draws balls with replacement from an urn containing $n$ unique balls and records its number. Then what is the expected number of draws required for getting
(a) the set of balls numbered $\{1,2,\cdo... | By way of enrichment here is a generating function approach to the
question of when the first $k$ coupons where $k\le n$ have been seen.
Using the notation from the following MSE
link we get from
first principles for the probability of $m$ draws that
$$P[T=m] = \frac{1}{n^m} \times {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I derive the conditions of Positive semidefinite cone in $2\times2$ matrix. By the definition, in order for $X$ to be positive semidefinite cone in $S^2$, it should satisfy that
\begin{equation}
X=\left[
\begin{array}{cc}
x & y \\
y & z
\end{array}
\right]\in S_+^2 \quad\Longleftrightarrow\quad x\ge0,\quad z\ge... | For the 2x2 case, you can form the characteristic polynomial, obtaining an expression for the eigenvalues using the quadratic formula. Setting the eigenvalues greater than or equal to zero yields the inequalities:
$2 \lambda_{+} = (x + z) + \sqrt{(x+z)^2 + 4(y^2-xz)} \ge 0$
$2 \lambda_{-} = (x + z) - \sqrt{(x+z)^2 + 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many $2 \times 2$ matrices are there with entries from the set ${\{0,1,2,...,i}\}$ in which there are no zeros rows and no zero columns? How many $2 \times 2$ matrices are there with entries from the set ${\{0,1,2,...,i}\}$ in which there are no zeros rows and no zero columns?
attempt:
Suppose we have a matrix $ \... | You may count the number of two tuples that can be formed from the elements of your set. clearly there are $(i+1)^2$ two tuples. Now the first entry should not be the vector $(0,0)$ so you have $(i+1)^2-1$ options for first row of marix. You have number of options for second row but then you have to subtract matrices o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluate: $\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$ Here is what I did:
$\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$
Put $x^2=\tan (\theta )$
Then: $x=\sqrt{\tan (\theta )}$ and $dx=\frac{1}{2} \tan ^{-\frac{1}{2}}(\theta ) \sec ^2(\theta )d\theta$
So the integral ... | These are three successive substitutions that will get you to the answer:
First, use: $ a) \, x^2 = \tanh{\theta}$
Then, use: $ b) s^2 = \sinh{\theta}$
Finally, use: $ c) \, u = 2s^4$
Then use the following definition of the Beta Function:
$\displaystyle \text{B}(p,q) = \int_0^\infty \frac{x^{p-1}}{(1+x)^{p+q}} \text{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Differentiation of a function We need to find out the derivative $\frac{dy}{dx}$ of the following:
$x^m$$y^n$=$(x+y)^{m+n}$
*
*I know how to differentiate the function and on solving we get $\frac{dy}{dx}$=$\frac{y}{x}$. But we notice that $\frac{dy}{dx}$ is independent of values of $m$ and $n$. So what I did was aga... | I'm not quite sure how you got $\frac{dy}{dx}=\frac yx$. You should instead have
$$\frac d{dx}x^my^n=mx^{m-1}y^n+nx^my^{n-1}\frac{dy}{dx}$$
$$\frac d{dx}(x+y)^{m+n}=(m+n)(x+y)^{m+n-1}\left(1+\frac{dy}{dx}\right)$$
Put these two together and we get
$$mx^{m-1}y^n+nx^my^{n-1}\frac{dy}{dx}=(m+n)(x+y)^{m+n-1}\left(1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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The probability that $a$ and $b$ satisfy the inequality $a-2b+10>0$? Nine identical balls are numbered $1,2,3,.........,9$ are put in a bag.$A$ draws a ball and gets the number $a$ and puts back in the bag. Next $B$ draws a ball and gets the number $b$.
The probability that $a$ and $b$ satisfy the inequality $a-2b+10>0... | $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2187290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that there exist integers $p$ and $q$ such that $\det(A^3+B^3) = p^3+q^3$
Let $A$ and $B$ be $3 \times 3$ matrices with integer entries so that $AB = BA$ and $\det(A) = \det(B) = 0$. Prove that there exist integers $p$ and $q$ such that $\det(A^3+B^3) = p^3+q^3$.
I thought about factorizing $A^3+B^3$. We have $... | In general it is not true that $\det(A+xB) = x(a+bx)$, but in this case you can prove it.
Consider the polynomial
$$P(X)=\det(A+XB)$$
This is a polynomial of degree at most $3$ with integer coefficients (this is an easy exercise, just write out the formula for the determinants). Write it as
$$
P(x)=aX^3+bX^2+cX+d
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2189377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Three Distinct Points and Their Normal Lines
Suppose That three points on the graph of $y=x^2$ have the property that their normal lines intersect at a common point. Show that the sum of their $x$-coordinates is $0$.
I have a lot going but can not finish it.
Proof:
Let $(a,a^2)$, $(b,b^2)$, and $(c,c^2)$ be three dis... | Michele Maschio’s answer picks up nicely where you left off, although neither your partial solution nor that answer deals with the case that one of the points is the origin, where the slope of the normal is undefined. This is a slightly different approach that doesn’t need to consider that case separately.
If $y=f(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving Ax=b Using the Basis of the Nullspace Let B = \begin{bmatrix}2&3&1&-1\\1&2&1&2\\3&5&2&1\\4&7&3&3\end{bmatrix}
Find the complete solution to the nonhomogenous system Bx=\begin{bmatrix}6\\-4\\2\\-2\end{bmatrix} by first computing a basis for the nullspace of B.
x = $c_1$ + $sc_2$ + $tc_3$
Where $c_1$, $c_2$ and $... | Start with the reduced row echelon form:
$$
\begin{align}
\mathbf{B} &\mapsto \mathbf{E}_{\mathbf{B}} \\
%
\left[
\begin{array}{rrrr}
2 & 3 & 1 & -1 \\
1 & 2 & 1 & 2 \\
3 & 5 & 2 & 1 \\
4 & 7 & 3 & 3 \\
\end{array}
\right]
%
&\mapsto
%
\left[
\begin{array}{rrrr}
\color{blue}{1} & 0 & -1 & -8 \\
0 & \color{blue}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Deducing a supremum from a given property Consider $$ S = \bigg\{ { \frac{x^2}{1+2x^2} } :x\in\mathbb{R}\bigg\}$$
we may guess that $\sup A=\frac{1}{2}$
But how does one prove this without taking limits?
| $\frac{1}{2}$ is an upper bound for $S$. In fact
$$\frac{x^2}{1+2x^2} =\frac{1}{2} \frac{2x^2}{1+2x^2}\le\frac{1}{2}\frac{1+2x^2}{1+2x^2}= \frac{1}{2} $$
for all $x \in \mathbb R$. Moreover,
$$\lim_{x \to +\infty}\frac{x^2}{1+2x^2} =\frac{1}{2} $$
and so for any $\varepsilon >0$
$$\left|\frac{x^2}{1+2x^2}-\frac{1}{2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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When is $\frac{x^2+xy+y^2}{49}$ an integer?
Find the number of distinct ordered pairs $(x, y)$ of positive integers such that $ 1\leq x, y \leq 49$ and $\frac{x^2+xy+y^2}{49}$ is an integer.
Multiplying the given equation by $(x-y)$ gives $x^3 \equiv y^3 \pmod{49}$. Thus we can't have $7 \mid x$ while $7 \nmid y$ or ... | The square roots of $-3 \pmod 7$ are $2,5.$ We ask when
$$ (2 + 7t)^2 \equiv -3 \pmod {49}, $$
with only solution $t =5$ and the square root being $37 \pmod{49}.$
We ask when
$$ (5 + 7t)^2 \equiv -3 \pmod {49}, $$
with only solution $t =1$ and the square root being $12 \pmod {49}.$
This little procedure is the beginnin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How can I prove this trigonometric equation with squares of sines? Here is the equation:
$$\sin^2(a+b)+\sin^2(a-b)=1-\cos(2a)\cos(2b)$$
Following from comment help,
$${\left(\sin a \cos b + \cos a \sin b\right)}^2 + {\left(\sin a \cos b - \cos a \sin b\right)}^2$$
$$=\sin^2 a \cos^2b + \cos^2 a \sin^2 b + \sin^2 a \c... | We have $$
\begin{align}
\sin^2(a+b)+\sin^2(a-b)&=\frac{1-\cos(2(a+b))}{2}+\frac{1-\cos(2(a-b))}{2}\\
&=1-\frac{\color{red}{\cos(2a+2b)}+\color{blue}{\cos(2a-2b)}}{2}\\
&=1-\frac{\color{red}{\cos 2a\cos 2b-\sin 2a\sin 2b}+ \color{blue}{\cos 2a\cos 2b+\sin 2a\sin 2b}}{2}\\
&=1-\frac{2\cos 2a\cos 2b}{2}\\
&=1-\cos 2a\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 2
} |
Solve $\lim_{x \rightarrow \infty}\frac{\ln(x+1)-\ln(x)}{x}$ without using L'Hopital or Taylor Series
Solve $$\lim_{x \rightarrow \infty}\frac{\ln(x+1)-\ln(x)}{x}$$ without using L'Hopital or Taylor Series
I tried:
\begin{align}\lim_{x \rightarrow \infty}\frac{\ln(x+1)-\ln(x)}{x} &=
\frac{\ln(x+1)}{x}-\frac{\ln(x)}{x... | Recall that
$$ \ln x := \int_1^x \frac{1}{t} ~dt.$$
Thus
\begin{align}
\lim_{x \to \infty} \frac{\ln (x+1) - \ln x}{x} &= \lim_{x \to \infty} \frac{1}{x} \left[ \int_1^{x+1} \frac{1}{t} ~dt - \int_1^x \frac{1}{t} ~dt \right] \\
&= \lim_{x \to \infty} \frac{1}{x} \int_x^{x+1} \frac{1}{t} ~dt
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Show that $\lim_{n \rightarrow \infty} \frac{\Sigma_{i=0}^k \binom{n}{3i}}{2^n} = \frac{1}{3}$ Let $n = 3k$ . Show that $\displaystyle{\lim\limits_{n \rightarrow \infty} \frac{\Sigma_{i=0}^k \binom{n}{3i}}{2^n} = \frac{1}{3}}$ . In other words, the sum of every third element of the nth row of the Pascal triangle is r... | Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. The characteristic function of the integers $n$ of the form $3k$ can be written as $\frac{1+\omega^n+\omega^{2n}}{3}$ (this is the principle behind the discrete Fourier transform) hence
$$ \sum_{i\geq 0}\binom{n}{3i} = \frac{1}{3}\sum_{k=0}^{n}\binom{n}{k}(1+\omega^k+\ome... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculate the following determinant How do I calculate the following determinant?
$$
\begin{vmatrix}
a_n & -1 & 0 & \dots & 0 & 0 \\
a_{n-1} & x & -1 & \dots & 0 & 0 \\
a_{n-2} & 0 & x & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
a_1 & 0 & 0 & \dots & x & -1 \\
a_0 & 0 & 0 & \d... | Let $F(a_0, a_1, \dots, a_n)$ be the determinant in your question. Expanding by minors across the first row, we get that $F(a_0, a_1, \dots, a_n)$ is equal to
$$
a_n
\begin{vmatrix}
x & -1 & \dots & 0 & 0 \\
0 & x & \dots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & x & -1 \\
0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving binomial theorem via induction I'm trying to prove binomial theorem by induction, but I'm a little stuck. I would look at online resources as this problem has been done many times, but the version I am trying to prove the binomial theorem in a different form.
$$(1 + x)^n = \sum_{k = 0}^{n} \binom{n}{k} x^k$$
I'... |
We show by induction the following is valid for $n\geq 0$
\begin{align*}
(1 + x)^n = \sum_{k = 0}^{n} \binom{n}{k} x^k
\end{align*}
Base step: $n=0$
We have to show
\begin{align*}
(1 + x)^0 = \sum_{k = 0}^{0} \binom{0}{k} x^k
\end{align*}
Since the left-hand side is $$(1+x)^0=1$$
and the right-hand side is $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show: $\cos \left( \frac{ 3\pi }{ 8 } \right) = \frac{1}{\sqrt{ 4 + 2 \sqrt{2} }}$ I'm having trouble showing that:
$$\cos\left(\frac{3\pi}{8}\right)=\frac{1}{\sqrt{4+2\sqrt2}}$$
The previous parts of the question required me to find the modulus and argument of $z+i$ where $z=\operatorname{cis{\theta}}$. Hence, I found... | Hint: by the double angle formula:
$$-\,\frac{1}{\sqrt{2}}=\cos\left(\frac{3\pi}{4}\right)=2\,\cos^2\left(\frac{3\pi}{8}\right)-1 \;\;\implies\;\; \cos\left(\frac{3\pi}{8}\right) = \sqrt{\frac{1}{2}\left({1-\frac{1}{\sqrt{2}}}\right)}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$
The above is equivalent to the posted form since $\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to prove this condition if three vectors are colinear? So I was given this problem:
Let $\vec{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\vec{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\vec{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$.
Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_... | We have $|v \times w| = |v||w|\sin \theta$, where $\theta$ is the angle between $u$ and $v$. Therefore three points $a$, $b$ and $c$ are collinear if and only
$$(b - a) \times (c - a) = 0$$
Expanding
$$b \times c - b\times a - a \times c + a \times a = 0$$
Since $a \times a = 0$ and anticommutivity
$$a\times b + b \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a+b+c = 13$; if $b/a=c/b$, find the maximum and minimum values of $a$ and the corresponding $b$ and $c$ Question :
The sum of $3$ integers $a,b$ and $c$ is $13$. If $\dfrac{b}{a}=\dfrac{c}{b}$, find the maximum and minimum values of $a$ and the corresponding $b$ and $c$.
To tackle this problem I let $x=\dfrac{b}{a}=... | ∵ $x=b/a$ is a solution of (1)
∴ (1) has at least one solution.
$△=1-4(1-(13/9))≥0$
$52/a≥3...(2)$
From (2) $a>0$ and $52/3≥a$
$∴0<a≤52/3$
$1<a≤17$
Minimum of $a=1$ and maximum of $a =17$.
Sub these values of $a$ into (1) to find the corresponding maximum and minimum of b and c.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Calculating value of a variable in combination? Using the formula (bionomial theorem):
$${}_nC_r \cdot a^{n-r} \cdot b^{r}$$
i was trying to find out the term with the coefficient $400$ from $(2x + 5y)^5$? Then i came up with this equation which i have no idea to solve beside trying numbers between 1 and 5. Is there an... | $\frac{2^5}{2^r} \cdot 5^r \cdot \binom 5r = 400$
$\frac{32}{2^r} \cdot 5^r \cdot \binom 5r = 400$
$\frac{1}{2^r} \cdot 5^r \cdot \binom 5r = \frac{25}{2}$
$\frac{1}{2^r} \cdot 5^r \cdot \binom 5r = \frac{5 \cdot 5}{2}$
$\frac{1}{2^r} \cdot 5^r \cdot \binom 5r = \frac{5}{2} \cdot \binom 51$
On comparing both sides,
$2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluating the indefinite integral $\int \frac{x^3}{\sqrt{4x^2 -1}}~dx$.
Find $$\int {x^3\over \sqrt{4x^2 -1}}\,dx.$$
Let
$2x = \sec u$, $2 =\sec (u) \tan(u) u^{'}(x).$ Then
$$
\begin{align*}
\int {x^3\over \sqrt{4x^2 -1}}\,dx &= \frac1{16}\int {\sec^3 u\over \tan u}\tan u \sec u \, du\\
&= \frac1{16}\int {\sec^4... | Hint:
$$\tan(\sec^{-1}(x))=\pm\sqrt{x^2-1}$$
(that is, your answer simplifies a bit further)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
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Proving $\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$.
Prove the following identity:
$$\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$$
How can I express $\cos(4\theta) $ in other terms?
| Since $\cos(4\theta)=2\cos^2(2\theta)-1$ and $\cos(2\theta)=1-\sin^2(\theta)$, we have
$$\begin{align}
\cos(4\theta) - 4\cos(2\theta)&=2\cos^2(2\theta)- 4\cos(2\theta)-1 \\
&=2(1-2\sin^2(\theta))^2-4(1-2\sin^2(\theta))-1\\
&=2-8\sin^2(\theta)+8\sin^4(\theta)-4+8\sin^2(\theta)-1\\
&=8\sin^4(\theta)-3.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2218085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Need direction for solving integral
Evaluate: $$\int \frac{\sin^2x}{\cos^2x +4}dx.$$
I tried to this things:
First
$$\tan\left(\frac{x}{2}\right) = t$$
$$dx = \frac{2~dt}{1+t^2}$$
$$\sin x= \frac{2t}{1+t^2}$$
$$\cos x= \frac{1-t^2}{1+t^2}$$
Tried also but this is same thing?
$$\tan\left(\frac{x}{2}\right) = t$$
There... |
$$\int \frac{1}{\cos^2x +4}dx$$
Divide top and bottom of integrand by $\cos^2 x$,
$$=\int \frac{\sec^2 x}{1+4\sec^2 x} dx $$
Now use $\tan^2 x+1=\sec^2 x$ to get,
$$=\int \frac{\sec^2 x}{1+4(\tan^2 x+1)} dx$$
$$=\int \frac{\sec^2 x}{5+4 \tan^2 x} dx$$
Let $u=\tan x$.
$$=\int \frac{1}{5+4u^2} du$$
$$=\frac{1}{5} \int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum of Sequence with Squares of Fibonacci Numbers in Denominator Find the sum of this sequence:
$$\frac{1}{1^2+1}-\frac{1}{2^2-1}+\frac{1}{3^2+1}-\frac{1}{5^2-1}+\frac{1}{8^2+1}-...$$
So, alternating series, but I've got nothing. I tried regrouping by pairs and got $$\frac{1}{6}+\frac{7}{120}+\frac{103}{10920}$$
which,... | Note: most of this is the identity
$$ F_{n+1} F_{n-1} - F_n^2 = (-1)^n $$ which is how I how I saw the two parts telescope.
The $+$ part is
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 13} + \frac{1}{13 \cdot 34} + $$
Do the $\pm$ parts separately. The partial sums for the $+$ part are
$$ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $2^p+p^2$ is prime for $p=3$ only I do know that all prime numbers larger than $3$ can be expressed as $3k + 1$ and or $3k + 2$. Plugging those in I still see no solution.
EDIT: $p$ can only be a prime number.
| From Fermat's little theorem, it follows that $2^p \equiv 2 \pmod p$. Then $2^p + p^2 \equiv 2 \pmod p$... oops, that doesn't help, never mind.
Oh, wait a minute: since $2 = 3 - 1$, then $p^2 \equiv 1 \pmod 3$ by Fermat's little theorem. That means $p^2$ is of the form $3k + 1$.
So as long as $2^p \not \equiv 2 \pmod 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How do I solve this system of equations (with squares and square roots)? Consider the system of equations:
\begin{align*}
x+y+z&=6\\
x^2+y^2+z^2&=18
\\\sqrt{x}+\sqrt{y}+\sqrt{z}&=4.
\end{align*}
How do I solve this? I've tried squaring, adding equations side by side, substituting, etc., but without success, e.g.
$$x^2+... | Let $s = \sqrt{x}$, $t = \sqrt{y}$, $u = \sqrt{z}$. Then your system is
$$ \eqalign{s^2 + t^2 + u^2 &= 6\cr
s^4 + t^4 + u^4 &= 18\cr
s + t + u &= 4\cr}$$
Solving the third equation for $u$ and substituting in the others gives
$$ \eqalign{s^2+t^2+(4-s-t)^2-6 &= 0\cr s^4+t^4+(4-s-t)^4-18 &= 0\cr}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find all possible solutions to $a^2 + b^2 = 2^k$? I'm working on the following problem, this is for an introductory discrete mathematics class.
Find all possible solutions to the equation $a^2 + b^2 = 2^k, k\geq1$ and $a$ and $b$ positive integers.
I've observed that the following are answers:
$2^1 = 1^2 + 1^2$
... | easier than you think. If $a^2 + b^2$ is divisible by $4,$ then $a,b$ are both even. Easy to prove. Induction says if $a^2 + b^2$ is divisible by $16,$ then $a,b$ are both divisible by $4.$ If if $a^2 + b^2$ is divisible by $64,$ then $a,b$ are both divisible by $8.$ And so on.
Which means that there are just two cases... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Will there always be integer solutions for $x$ and $y$ in $3^y = 2^x - 1$? Is there a way to find the solutions of $3^y=2^x-1$ where $(x,y)$ are integer coordinates (maybe within a certain interval)? Are there an infinite number of integer coordinate pairs for this equation? Is it possible to even determine this? I'm n... | There is an entirely elementary way to show that $(x,y)=(1,0)$ and $(2,1)$ are the only integer solutions to $3^y=2^x-1$.
If $y\ge1$, $x$ must be even since $2^{2u+1}-1\equiv1$ mod $3$. Writing $x=2u$, we have
$$3^y=2^{2u}-1=(2^u-1)(2^u+1)$$
which implies $2^u-1$ and $2^u+1$ are each a power of $3$, say $2^u-1=3^r$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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A determinant from analytic geometry? I have a question regarding the following determinant:
$\begin{vmatrix}
+ax - by - cz & bx+ay & cx+az \\
bx+ay& -ax+by-cz & bz+cy \\
cx+az & bz+cy & -ax-by+cz
\end{vmatrix}
=
(a^2 + b^2 + c^2)(x^2 + y^2 +z^2)(ax+by+cz).$
I can prove the above equality by performing row operations ... | Now I can answer the first question. We can write the matrix as a product of two rectangular matrices.
$$\left[ \begin{matrix}
+ax - by - cz & bx+ay & cx+az \\
bx+ay& -ax+by-cz & bz+cy \\
cx+az & bz+cy & -ax-by+cz
\end{matrix}\right] = \left[ \begin{matrix}
a & -b & c & 0\\
b & a & 0 & -c \\
c & 0 & -a & b \\
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Need help in an elementary number theory problem from Higher Algebra by S. BARNARD Can someone help me solve this question:
If a and b are coprime and n is a prime,prove that $$\frac{(a^n+b^n)}{(a+b)}$$ and $$a+b$$ have no common factor,unless $a+b$ is a multiple of $n$.
I've tried it out but not sure about the sol... | We have $$\frac{a^n+b^n}{a+b}=\sum_{j=0}^{n-1}a^j(-b)^{n-1-j}\equiv \sum_{j=0}^{n-1}a^ja^{n-1-j}=na^{n-1}\mod a+b$$ for every odd $n$
Hence if $n$ is an odd prime and $q\ne n$ a prime dividing both $a+b$ and $\frac{a^n+b^n}{a+b}$, we get $$0\equiv \frac{a^n+b^n}{a+b}\equiv na^{n-1}\mod q$$ hence $q$ must didvide $a$ an... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I solve $5^{2x} + 4(5^x) - 5 = 0$? This is a math problem I'm currently working on.
$$5^{2x} + 4(5^x) - 5 = 0$$
I've used logarithm to try solve the problem. Here's what I've done so far:
\begin{align}5^{2x} + 4(5^x) - 5 &= 0\\
5^{2x} + 5^x &= \frac{5}{4}\\
\log_5{2x} + \log_5{x} &= \log_5\left(\frac{5}{4}\rig... | Hint:
A useful trick is to substitute $u=5^x$ to obtain a quadratic equation:
$$u^2+4u-5=0$$
Can you solve for $u$, and then solve for $x$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Answering a question containing $\sqrt{1.5}$ without using calculator. Of the following which is the best approximation of $\sqrt{1.5}(266)^{\frac{3}{2}}$?
(A)1,000
(B)2,700
(C)3,200
(D)4,100
(E)5,300
How can I answer this without using a calculator and in about 2.5 minutes?
| $\sqrt{1.5}(266)^{\frac{3}{2}}
= 266 \times\sqrt{1.5 \times 266}
= 266 \times\sqrt{266 + 133}
= 266 \times\sqrt{399}$
Note that $\sqrt{399}$ is very close to $\sqrt{400} = 20$.
So $266 \times\sqrt{399} \approx 266 \times 20 = 5320.$
By the way, if you want a bit more accurracy, let $f(x) = \sqrt{x}$. Then
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $f(x-3) = x^2$, $f(x) = (x+3)^2$? I was wondering suppose you have $f(x-3) = x^2$. Is it correct to say that $f(x) = f(x-3 +3 ) = (x+3)^2$. If so, why or why not? Why can we say that if $g(x) = x^2 + x$, then $g(x+2) = (x+2)^2 + (x+2)$ but not the other way around?
| I would interpret
$f(x-3) = x^2$
as composition
$f \circ u$
where $u(x) = x - 3$.
In this case $x = u + 3$ and $f(u) = x^2 = (u+3)^2$ or $f(x) = (x + 3)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Computing $\int_0^\pi (4 + \sin^2 \theta)^{-1} d\theta$ with the residue theorem It's asked to find the value of the real integral
$$
I = \int\limits_0^{\pi}\frac{\rm{d}\theta}{4+\sin^2{\theta}}
$$
I don't understand how to apply the theorem, basically because I cannot parametrise $z = e^{i\theta}$ as I'm not on a circ... | Note that $\sin^2(\theta)$ is an even function. Hence, we assert that
$$\int_0^\pi \frac{1}{4+\sin^2(\theta)}\,d\theta=\frac12\int_{-\pi}^\pi \frac{1}{4+\sin^2(\theta)}\,d\theta \tag 1$$
And now one can proceed with the substitution $z=e^{i\theta}$ in $(1)$ such that
$$\int_0^\pi \frac{1}{4+\sin^2(\theta)}\,d\theta=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Ellipse tangent to two circles We're given two circles with radii $p$ and $q$, one centered at the origin, and one centered at the point $(w,0)$.
I want to construct an ellipse of the form
$$
\frac{(x-c)^2}{a^2} + \frac{y^2}{b^2} = 1
$$
that is tangent to the two circles, as shown here:
If $b$ is known, can we obtain... | For ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Equation of normal at $(x_1,y_1)$:
$$a^2y_1(x-x_1)=b^2x_1(y-y_1)$$
Put $y=0$,
$$x=\frac{a^2-b^2}{a^2}x_1=e^2x_1 \tag{$y_1 \ne 0$}$$
which is the $x$-intercept.
Radius $p$:
\begin{align*}
p^2 &= (x_1-e^2x_1)^2+y_1^2 \\
&= \frac{b^4x_1^2}{a^4}+y_1^2 \\
&= \frac{b^4x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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General form of multiplication of a matrix by its transpose When multiplying $A^T*A$, does it usually end up in the form $\begin{pmatrix} a&b\\b&-2b\end{pmatrix}$? If not, are there any special cases where it does? If so, why?
| Per the comment, we generally have
$$
\begin{pmatrix} a&b\\c&d\end{pmatrix} \begin{pmatrix} a&c\\b&d\end{pmatrix} = \begin{pmatrix} a^2+b^2 & ac+bd \\ ac+bd & c^2+d^2\end{pmatrix} \overset{?} = \pmatrix{\alpha & -\frac 12 \beta\\ - \frac 12 \beta & \beta}
$$
Your question amounts to asking when $a,b,c,d$ are such tha... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove inequality $\sum\frac{{{a^3} - {b^3}}}{{{{\left( {a - b} \right)}^3}}}\ge \frac{9}{4}$ Given $a,b,c$ are positive number. Prove that $$\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge \frac{9}{4}$$
$$\Leftrightarrow \sum \frac{3(a+b)^2+(a-b)^2}{(a-b)^2} ... | $$(x+y+z)^2\ge 0\\x^2+y^2+z^2+2xy+2xz+2yz\ge 0\\x^2+y^2+z^2\ge -2(xy+yz+zx)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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solve for $x$ and $y$ in the following equation $x^2 +y^2 -3=3xy$ and $2x^2 +y^2 =6$ Solve for $x$ and $y$ in the following equations: $x^2 +y^2 -3=3xy$ and $2x^2 +y^2 =6$. I made $y^2$ the subject of the formula in eqn 2.
This gives $y^2 = 6 -2x^2$.
I substitute this into the first eqn.
This gives $x^2+3xy -3$. There'... | multiplying the first equation by $-1$ and adding to the second we get
$$x^2+3xy=3$$ thus $$y=\frac{3-x^2}{3x}$$ if $$x\ne 0$$ and you can eliminate $y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the rational canonical form of a matrix from its minimal and characteristic polynomials
What is the rational canonical form of $A$?
$$A=\begin{bmatrix}
1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
2 & 3 & -1 & 4\\
1 & 1 & -1 & 3\\
\end{bmatrix}$$
I found that the minimal polynomial $m_... | The "first" $2\times 2$ principal block is clearly not an $x-1,x-1$ block as it is not the identity. Nor is the "other" $2\times 2$ principal block since it is not the identity. So we have two $(x-1)^2$ blocks.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Finding $\int x \sqrt{1- x^2 \over 1 + x^2}dx$.
$$J =\int x \sqrt{1- x^2 \over 1 + x^2}dx$$
Substituting $u = \sqrt{1 + x^2}$
$$J = \int \sqrt{2 - u^2} du = \sqrt{2}\int \sqrt{1 - \left({u \over \sqrt{2}}\right)^2} du$$
Now substituting $\sin z = u/\sqrt{2}$
$$J = 2\int \cos^2 z dz = z + (\sin 2z)/2 + C = \arcsin\l... | HINT:
Let $x^2=\cos2y$ as $x^2\ge0,0\le2y\le\dfrac\pi2\ \ \ \ (1)$
$\implies x\ dx=-\sin2y\ dy,\sin2y=+\sqrt{1-x^4}$
Now $\sin2y=2\sin y\cos y$
and $\sqrt{\dfrac{1-x^2}{1+x^2}}=+\tan y\text{ by } (1)$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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For a given condition is it true that $a+b+c=3$. Suppose a, b, c are positive real numbers such that
$$(1+a+b+c)\left(1+\frac 1a+\frac 1b+\frac 1c\right)=16$$
Then is it true that we must have $a+b+c=3$ ?
Please help me to solve this. Thanks in advance.
| By Cauchy-Schwarz $$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq\left(1\cdot1+\sqrt{a}\cdot\frac{1}{\sqrt{a}}+\sqrt{b}\cdot\frac{1}{\sqrt{b}}+\sqrt{c}\cdot\frac{1}{\sqrt{c}}\right)^2= 16$$
The equality occurs for $a=b=c=1$, which says that $a+b+c=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Prove by induction that $\forall n\geq 1,\ 7\mid 3^{2n+1} + 2^{n-1}$ Prove by induction that $$7 \mid 3^{2n+1} + 2^{n-1},\ \forall n\geq 1$$
Base case $n=1$:
$$3^{2 × 1+1} + 2^{1-1} = 28.$$
Induction:
$$P(k): 3^{2k+1} + 2^{k-1},\ P(k+1): 3^{2(k+1)+1} + 2^{(k+1)-1}.$$
$$3^{2k+3} + 2^k = 9 \times 3^{2k+1} + 2^{k-1} \tim... | The correct expression is obviously $$3^{2n+1}+ 2^{n-1}$$.
$n=1: 3^3 +2^0 = 27 +1 =28$
$n=2: 3^5 +2^1 = 245 = (35)(7)$
$n=3: 3^7 +2^2 = 2191 = (313)(7)$
So
3^[2(n+1)+1] +2^(n+1-1) = [3^(2n+1)](9) + 2^n
= [7k-2^(n-1)](9) +2^n
= 63k -2^(n-1)(9-2)
= (7)[9k - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
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How $10 +2\sqrt{2} \sqrt{10}+2 > 17$ becomes $2\sqrt{2} \sqrt{10} > 5$ due to theorem $3$? I'm reading Kazarinoff's Analytic inequalities. He gives the theorem 3 and 6:
*
*If $a>b$ and $c>d$ then $a+c> b+d$.
*If $a>b>0$, then $a^{\frac{p}{q}}> b^{\frac{p}{q}}$.
There is an exercise a little bit further: Show that $... | It does not follow directly from Theorem 3, but one can make a corollary:
$a>b$ $\implies$ $a+c>b+c$
Proof: By Definition 1, p. 2, $a>b$ means $a=b+h$, $h>0$. As $h/2>0$ too (by Theorem 4 that says $a>b$, $c>0$ $\implies$ $a/c>b/c$), the same definition gives that $a>b+h/2$. Now take $c$ and $d=c-h/2$. We have $c>d$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some ... | We have an instance of the least-norm problem
$$\begin{array}{ll} \text{minimize} & \| {\rm x} \|_2^2 \\ \text{subject to} & {\Bbb 1}_3^\top {\rm x} = 1\end{array}$$
The minimizer is
$${\rm x}_{\min} := {\Bbb 1}_3 \left( {\Bbb 1}_3^\top {\Bbb 1}_3 \right)^{-1} = \frac13 {\Bbb 1}_3$$
and, thus, the minimum is $\| {\rm x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 15,
"answer_id": 14
} |
Showing $\frac{1}{x^3+x}$ is continuous at $x=1$ I've been attempting to use the definition of continuity and felt a little uncertain about my working on this question.
We define $\epsilon>0$ and choose $\delta $ s.t .....
Let $\lvert x-1\rvert < 1 $:
$\lvert \frac{1}{x^3+x}-\frac{1}{2} \rvert = \lvert \frac{2-x^3-x... | I like to my limits
go to zero,
so,
let $x = 1+y$.
Then
$\begin{array}\\
\lvert \frac{1}{x^3+x}-\frac{1}{2} \rvert
&=\lvert \frac{1}{(1+y)^3+(1+y)}-\frac{1}{2} \rvert\\
&=\lvert \frac{1}{y^3+3y^2+4y+2}-\frac{1}{2} \rvert\\
&=\lvert \frac{2-(y^3+3y^2+4y+2)}{2(y^3+3y^2+4y+2)}\rvert\\
&=\lvert \frac{-(y^3+3y^2+4y)}{2(y^3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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why$2 (-1 + 2^{1 + n})$ is the answer to the recurrence relation $a_{n}=2a_{n-1}+2$? $a_{0}=2$
$a_{1}=2(2)+2$
$a_{2}=2(2(2)+2)+2$
$a_{3}=2(2(2(2)+2)+2)+2$
$a_{4}=2(2(2(2(2)+2)+2)+2)+2$
$a_{5}=2(2(2(2(2(2)+2)+2)+2)+2)+2$
To simplifiy
$a_{6}=2^{6}+2^{5}...2^{1}$
so my answer is
$a_{n}=2^{n+1}+2^{n}+...2^{1}$
The correct... | You can also do this in base 2 fairly easily:
$$
\begin{align*}
a_0 &= 10_2 \\
a_1 &= 2a_0 + 2 = 100_2 + 10_2 = 110_2 \\
a_2 &= 1100_2 + 10_2 = 1110_2.
\end{align*}
$$
We can see that by induction we will have.
$$a_n = \underbrace{11\cdots1}_{n+1}0_2 = 2(\underbrace{11\cdots1_2}_{n+1}).$$
Now note that
$$\underbrace{11... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of trigonometric function $\lim_{x\to\pi/3} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})}$ I want to compute this limit:
$\displaystyle \lim_{x\to\pi/3} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})}$
Using L'Hopital, is easy to get the result, which is $\sqrt{3}$
I tried using linear approximation (making $u = x - ... | HINT:
Using Prosthaphaeresis & Double angle Formula as $\cos\dfrac\pi3=?$
$$2\cdot\dfrac{\cos\dfrac\pi3-\cos x}{\sin\left(x-\dfrac\pi3\right)}=2\cdot\dfrac{2\sin\dfrac{\pi+3x}6\sin\dfrac{3x-\pi}6}{2\sin\dfrac{3x-\pi}6\cos\dfrac{3x-\pi}6}$$
Now as $x\to\dfrac\pi3\iff3x\to\pi,3x\ne\pi\implies\sin\dfrac{3x-\pi}6\ne0$, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
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For an even and $2\pi$ periodic function, why does $\int_{0}^{2\pi}f(x)dx = 2\int_{0}^{\pi}f(x)dx $ How does one derive the fact that $$\int_{0}^{2\pi}f(x)dx = 2\int_{0}^{\pi}f(x)dx $$ whenever $f(x)$ is an even and $2\pi$ periodic function. I do know the result that for even functions, we have $$\int_{-a}^{a}f(x)dx = ... | Hint:
We have,
$$\int_{0}^{2\pi}\frac{1}{a+b\cos x}dx$$
$$=\int_{0}^{\pi} \frac{1}{a+b \cos x} dx+\int_{\pi}^{2\pi} \frac{1}{a+b \cos (x-2\pi)} dx$$
Can you see why? What happens if you let $x-2\pi=u$ on the second part?
\begin{align} \int_{0}^{2\pi} \frac{1}{a+b\cos x}dx \ =\int_{0}^{\pi} \frac{1}{a+b\cos x}dx+\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Geometry Problem Concerning Lengths with a Square and Two Right Triangles Square $ABCD$ has side length 13, and points $E$ and $F$ are exterior to the square such that $BE = DF = 5$ and $AE = CF = 12$. Find $EF^2$.
Click Here for the Attached Diagram
| Solution:
First, by Pythagoras, triangles $ABE$ and $CDF$ are right triangles. We extend lines $\overline{EA}$, $\overline{EB}$, $\overline{FC}$, and $\overline{FD}$ to construct quadrilateral $EGFH$.
Right triangles $ABE$ and $CDF$ are congruent, so $\angle DCF = \angle EAB$, which means $\angle GDA = \angle EAB$. Al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Suppose $ \sum_{n=1}^{\infty} b_n x^n= \frac{x^3}{(x^4-1)^2}$. What could be an expression of $b_n$? A practice problem reads:
Suppose $$\sum_{n=1}^{\infty} b_n x^n = \frac{x^3}{(x^4-1)^2}.$$
What could be an expression of $b_n$?
Some of the possible answers read $ 2^{3n}nx^{3n-1}, nx^{3n-1}, nx^{4n-1}$
There are a fe... | Another variation is based upon the binomial series expansion.
\begin{align*}
(1+x)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}x^n\qquad\qquad |x|<1
\end{align*}
We obtain
\begin{align*}
\frac{x^3}{(x^4-1)^2}&=x^3\sum_{n=0}^\infty\binom{-2}{n}(-x^4)^n\tag{1}\\
&=x^3\sum_{n=0}^\infty\binom{n+1}{1}x^{4n}\\
&=\sum_{n=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find a quadratic integer polynomial that annihilates a given $2\times2$ integer matrix Suppose $A=\begin{bmatrix}3&2\\2&3\end{bmatrix}$. How can we find integers $b$ and $c$ such that $A^2+bA+cI_2=0$?
| $A=\begin{pmatrix}3&2\\ 2&3\end{pmatrix}\implies A^2= \begin{pmatrix}3&2\\ 2&3\end{pmatrix}\begin{pmatrix}3&2\\ 2&3\end{pmatrix} =\begin{pmatrix}13&12\\ 12&13\end{pmatrix}$
Then $$A^2+bA+cI_2=0\implies \begin{pmatrix}13&12\\ 12&13\end{pmatrix}+b\begin{pmatrix}3&2\\ 2&3\end{pmatrix}+c\begin{pmatrix}1&0\\ 0&1\end{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Proving a variable always returns true in an inequality I have the following inequality
$$(1+a)^n\leq 1+(2^n−1)a$$ for $0 ≤ a ≤ 1$
What's the approach to prove that any value of the variable returns true ?
| We will prove this by induction on $n$, assuming $n > 0$. First, the base for the induction, $n=1$:
$$
(1 + a)^1 \leq 1 + (2^1 - 1)a
\iff 1 + a \leq 1 + a
$$
which is true. Now, the induction hypothesis is $(1 + a)^k \leq 1 + (2^k - 1)a$. We can add $2^ka$ to both sides to get
$$\begin{align}
1 + (2^{k+1} - 1)a &\geq (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding minimum of trigonometric function Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Then find difference between maximum and minimum of $v^2$.
I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?
I tried guessing, and got max... | Let $$p=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)},q=\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Now $p^2+q^2=a^2+b^2$
and $$2(p^2+q^2)-(p+q)^2=(p-q)^2\ge0\iff(p+q)^2\le2(p^2+q^2)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
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Evaluate $\displaystyle \lim_{n \to \infty} n \int_0^1 (\cos x - \sin x)^n dx$ Evaluate $$\lim_{n \to \infty} n \int_0^1 (\cos x - \sin x)^n dx$$
The answer should be $1$.
I tried to solve it similar to how one user solved this integral Limit of integral with cos and sin
but it seems like it doesn't work because the up... | If you know how to deal with
$$n \int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx,$$
just split the integral at $\frac{\pi}{4}$ and see whether you can find something useful for the other part.
Since $x \mapsto \cos x - \sin x$ is strictly decreasing on $\bigl[0, \frac{\pi}{2}\bigr]$, and $\cos \frac{\pi}{4} = \sin \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Is there a possible explanation in plain English how to use the Chinese Reminder Theorem? For example, if it is the problem of
Find the smallest integer that leave a remainder of 3 when divided by
5, a remainder of 5 when divided by 7, and a remainder of 7 when
divided by 11
In some explanation such as in this ar... | Find the smallest integer, $n$, that leaves
$\text{A remainder of $3$ when divided by $5$.}\tag{1}$
$\text{A remainder of $5$ when divided by $7$.}\tag{2}$
$\text{A remainder of $7$ when divided by $11$.}\tag{3}$
Without using moduli, the Chinese Remainder theorem says that the integers $n$ that meet the above three re... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Locating a line perpendicular to an ellipse At any given point on an ellipse, the line of the perpendicular is easy to calculate; what about the inverse? Given a line of known orientation, at what point on the circumference of an ellipse would it be perpendicular?
| Given the perpendicular line, you can find the slope of the tangent line by taking the negative reciprocal of the slope of the perpendicular line. Then, once you have the slope of the tangent at a point on the ellipse, find the first derivative of the ellipse and set it equal to that quantity, and then solve. There wil... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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$A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ $A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ and $\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$. Then $A+2B$ is equal to:
$1$. $\dfrac {\pi}{4}$
$2$. $\dfrac {\pi}{3}$
$3$. $... | Hint: Square both sides of the second equation, and and replace $\sin^2 $ by $1 - \cos^2$, and coupled with the first solve a system of equations for $\cos A, \cos B$. Can you manager to proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find the shortest distance between the curve $y^2=x$ to the line $y=x+1$
Find the shortest distance between the curve $y^2=x$
to the line $y=x+1$.
Im not sure how to attack a problem like this.
| Let the parabola and the line be parameterized as follows
$$\mathcal P := \{ (t_1^2, t_1) \mid t_1 \in \mathbb R \} \qquad \qquad \qquad \mathcal L := \{ (t_2, t_2+1) \mid t_2 \in \mathbb R \}$$
The squared distance between $\mathcal P$ and $\mathcal L$ is given by the minimum of the quartic objective function
$$f (t_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2274561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\gcd(x^3, x^3+x+1)$ Suppose $f(x) = x^3$ and $g(x) = x^3+x+1$. I am trying to show that the $\gcd(f,g)= 1$ but am running into some trouble...
My attempt: (Division algorithm)
$x^3+x+1 = (1)(x^3)+(x+1)$
$x^3 = x^2(x+1)-x^2$
But I cannot go further than this, what is going wrong here?
| The gcd of $x^3+x+1$ and $x^3$ is the same as the gcd of $x^3$ and $(x^3+x+1)-x^3=x+1$.
Now $x+1$ is irreducible and doesn't divide $x^3$. End.
By the way, the remainder of the division of $f(x)=x^3$ by $x+1$ is $f(-1)=-1$ and, indeed, $x^3+1=(x+1)(x^2-x+1)$, but it's not needed.
| {
"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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Summation of the Sine Function I was messing around on Wolfram Alpha's summation calculator and when I plugged in the summation
$$\sum_{i=1}^n\sin\frac{i\pi}{180}$$
and it gave me the value
$$\frac12\left(\cot\frac\pi{360}-\csc\frac\pi{360}\cos\frac{(2n+1)\pi}{360}\right)$$
I don't understand... how does it arrive at t... | We have
$$ 2\sin{A}\sin{B} = -\cos{(A+B)}+\cos{(A-B)} $$
by using the angle addition formulae. Putting $A=ak+b$, $B=a/2$ gives
$$ 2\sin{\left(\frac{a}{2}\right)}\sin{(ak+b)} = -\cos{\left(a\left(k+\frac{1}{2}\right)+b\right)} + \cos{\left(a\left(k-\frac{1}{2}\right)+b\right)}. $$
Summing from $k=1$ to $n$ gives
$$ 2\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For $x \ge 1$, is $2^{\sqrt{x}} \ge x$ For $x \ge 1$, is $2^{\sqrt{x}} \ge x$
The answer appears to me to be yes since:
$$\sqrt{x} \ge \log_2 x$$
Here's my reasoning:
(1) For $x > 16$, $5\sqrt{x} > 4\sqrt{x} + 4$
(2) For $x > 25$, $x > 5\sqrt{x}$
(3) So, $x > 25$, $2x > x + 4\sqrt{x} + 4$ and it follows that $\sqrt{2x}... | The inequality doesn't always hold, as can be seen by taking $x=9$.
Notice that your question is about comparing $a^b$ and $b^a$ with $a=2,b=\sqrt{x}$. It is well known that $a^b \leq b^a$ if and only if $$a^{1/a} \leq b^{1/b} $$
which in our case is
$$\sqrt{2} \leq b^{1/b}.$$
This inequality holds iff $$2 \leq b \leq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$. If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.
$$\sec \theta + \tan \theta = x$$
$$\dfrac {1}{\cos \theta }+\dfrac {\sin \theta }{\cos \theta }=x$$
$$\dfrac {1+\sin \theta }{\sqrt {1-\sin^2 \theta }}=x$$
$$1+\sin \theta =... | The equation becomes
$$
1+\sin\theta=x\cos\theta
$$
Set $X=\cos\theta$ and $Y=\sin\theta$, so the equation becomes
$$
\begin{cases}
X^2+Y^2=1 \\[4px]
1+Y=xX
\end{cases}
$$
Note that $x\ne0$ and substitute $X=x^{-1}(1+Y)$ in the first equation getting
$$
(1+Y)^2+x^2Y^2=x^2
$$
that simplifies to
$$
(1+x^2)Y^2+2Y+1-x^2=0
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280453",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Solving $|x-1|^{\log^2(x)-\log(x^2)}=|x-1|^3$
Solve the equation:$$|x-1|^{\log^2(x)-\log(x^2)}=|x-1|^3.$$
There are three solutions of $x$: $10^{-1}$, $10^3$ and $2$. I obtained the first two solutions but I have been unsuccessful in getting $2$ as a solution. Please help.
| First, the presence of $\ln x$ in the equation, i.e. in $\ln^2 x - \ln(x^2)$, dictates that $x > 0$. Next, if $|x - 1| = 0$, then $x = 1$ and $\ln^2 x - \ln(x^2) = 0$, but $0^0$ is undefined and thus $x = 1$ is excluded.
Now the equation reduces to$$
|x - 1|^{\ln^2 x - \ln(x^2) - 3} = 1,
$$
which is equivalent to$$
|x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inverse element of
Given the extension $ \dfrac{\mathbb Q[x]}{\langle x^3-5\rangle}: \mathbb Q$ and $\alpha =1+x^2+\langle x^3-5\rangle \in \dfrac{\mathbb Q[x]}{\langle x^3-5 \rangle}$, find $\alpha^{-1}$.
Can somebody give a hint? or an answer? please?
| Since $x^3-5$ is a polynomial of degree $3$, every element in $k:= \dfrac{\mathbb{Q}[x]}{\langle x^3-5\rangle}$ has a representative of degree at most $2$. Thus we are looking for a polynomial $\beta:=a + bx + cx^2 + \langle x^3 - 5\rangle$ where $a,b,c\in \mathbb{Q}$ and such that $\alpha\beta = 1 + \langle x^3 + 5\ra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove this inequality $(2+x)\ln{x}-e(x-1)>0,x>1$
Let $x>1$,show that
$$(2+x)\ln{x}-e(x-1)>0$$
The simaler problem:[How prove $(2+5x)\ln{x}-6(x-1)>0.\forall x>1$
this problem idea: $f(x)=\ln{x}-\dfrac{e(x-1)}{x+2}$
then
$$f'(x)=\dfrac{x^2+(4-3e)x+4}{x(x+2)^2}=\dfrac{\left(x-\dfrac{3e-4}{2}\right)^2+4-\dfrac{(3e-4)... | It's easier to deal with $f(x)=(2+x)\ln x-e(x-1)$ directly. Note that $f(1)=0$.
Since $f'(x)=\ln x+\frac2{x}+1-e$ and $f''(x)=\frac{x-2}{x^2}$, $f'(x)$ is monotone decreasing in $1<x<2$ and monotone increasing in $x>2$.
From $f'(2)=\ln 2+2-e<0$ and $f'(2.6)=\ln 2.6+\frac2{2.6}+1-e>0$, there is a constant $2<c<2.6$ for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a,b,c,d$ are real numbers such that....
Suppost that $a,b,c,d$ are real numbers such that $$a^2+b^2=1$$ $$c^2+d^2=1$$ $$ac+bd=0$$
I've to show that $$a^2+c^2=1$$ $$b^2+d^2=1$$ $$ab+cd=0$$
Basically,I've no any idea or tactics to tackle this problem. Any methods? Thanks in advance.
EDITED.
The given hint in the book... | Note that
$$0=(ac+bd)^2 = a^2c^2+b^2d^2+2abcd = a^2c^2+(1-a^2)(1-c^2) +2ac(-ac)$$
$$ = a^2c^2+1-a^2-c^2+a^2c^2-2a^2c^2 = 1-a^2-c^2$$
And so $a^2+c^2=1$. A similar argument shows that $b^2+d^2=1$. Further, now we can say
$$(ac+bd)^2-(ab+cd)^2 =a^2c^2+b^2d^2-a^2b^2-c^2d^2$$
$$ = a^2c^2+b^2d^2-a^2(1-d^2)-c^2(1-b^2)$$
$$=a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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How can we prove the equality of the following determinants? Assume that $A$ is an $n\times n$ real matrix whose entries are all $1$. Then how can we show the following determinant equality for any $x$?
$\det(A-xI)$=\begin{vmatrix}
1 -x & 1 & 1 & \cdots & 1 \\
1 & 1 -x & 1 & \cdots & 1 \\
... | Note that
$$
\begin{bmatrix}
n-x & n-x & n-x & \cdots & n -x \\
1 & 1 -x & 1 & \cdots & 1 \\
1 & 1 & 1 -x & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1 & 1 & \cdots & 1 -x
\end{bmatrix} = \\
\begin{bmatrix}
1&1&1&\cdots&1\\
&1&0&\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Generating functions ( recurrence relations ) Find $a_n$ using Generating Functions : $a_n = -a_{n-1} + 2a_{n−2}$, $n\ge2$ and $a_0 = 1$, $a_1 = 2$.
Approach : So I will form a characteristic equation $ r^2 + r - 2 = 0$ whose roots are $r_1 = -2$, $r_2 = 1$.
So my general solution is $a_n = α_1r_1^n + α_2r_2^n$.
$a_n =... | The standard generating function approach yields
$$GF=\frac{(1+3x)}{1+x-2x^2}$$
which has a denominator which can be factorised
$$GF=\frac{1+3x}{(1+2x)(1-x)}$$
Applying the theory of partial fractions gives
$$GF=-\frac{1}{3(1+2x)}+\frac{4}{3(1-x)}$$
which are all recognizable contributions to a formula for the $n$th te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2288360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Conditional Probability with dice
Two dice are rolled and their number is added. If the sum equal 7, the game is lost. If the sum equal 9, the game is won. If the sum is any other number, the dice are rolled again, until the sum equals 7 or 9. What is the probability that someone wins 1 out of 5 games?
My approach:
S... | Let us consider $(a, b)$ a pair of values for the two dice, with $a$ the value for die 1 and $b$ the value for die 2. There are six equiprobable cases in which you lose:
$$(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$$
Similarly, there are four equiprobable cases in which you win:
$$(3, 6), (4, 5), (5, 4), (6, 3)$$
A... | {
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"url": "https://math.stackexchange.com/questions/2294544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Show that $\int_0^\infty \frac{\ln x}{(x^2+1)(x^2-1)}dx=\frac{\pi^2}{8}$ How do I show that: $\int_0^\infty \frac{\ln x}{(x^2+1)(x^2-1)}dx=\frac{\pi^2}{8}$ using contours and residues
My attempt:
I know that the singular points are $i,-i,-1,1,0$
consider $f(z)= \frac{\ln z}{(z^2+1)(z^2-1)}$
and the branch $|z|>0$, $0<\... | First, enforcing the substitution $x\to 1/x$ reveals
$$\int_0^\infty \frac{\log(x)}{(x^2+1)(x^2-1)}\,dx=\int_0^\infty \frac{x^2\log(x)}{(x^2+1)(x^2-1)}\,dx\tag 1$$
From $(1)$ it is evident that
$$\int_0^\infty \frac{\log(x)}{(x^2+1)(x^2-1)}\,dx=\frac12\int_0^\infty \frac{\log(x)}{x^2-1}\,dx\tag 2$$
We evaluate the i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Help with finding the length of one of the sides of a triangle
\begin{align}\frac z{\sin E}&=\frac y{\sin D}\tag{1}\\
z^2 &= x^2 + y^2 - 2xy\cos E\tag{2}\\
y^2 &= z^2 + x^2 - 2zx\cos D\tag{3}\end{align}
Solve for $z$
My solution is as follow.
I stopped at the equation:
$$(2xy)^2 - (x^2 + y^2 - z^2)^2 = (2zx)^2 - (z^... | If you expand your final equation $$(2xy)^2 - (x^2 + y^2 - z^2)^2 = (2zx)^2 - (z^2 + x^2 - y^2)^2$$ you will get an equation which has equal terms in $x^4$, $y^4$, and $z^4$ on both sides. Removing these gets you:
$$2x^2y^2 + 2x^2z^2 + 2y^2z^2 = 2x^2y^2 + 2x^2z^2 + 2y^2z^2$$
which as you can see is always true.
The pro... | {
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"url": "https://math.stackexchange.com/questions/2295867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the following are real numbers
*
*$$\frac{1}{z}+\frac{1}{\overline{z}}$$
*$$z^3\cdot\overline{z}+z\cdot\overline{z}^3$$
1.$$\frac{1}{z}+\frac{1}{\overline{z}}$$
$$\frac{\overline{z}}{z\cdot \overline{z}}+\frac{z}{z\cdot\overline{z}}$$
$$\frac{\overline{z}+z}{z\cdot \overline{z}}$$
$$\frac{2Re(z... | Alternatively:
*
*$\cfrac{1}{z}+\cfrac{1}{\overline{z}}=\left(\cfrac{1}{z}\right)+\overline{\left(\cfrac{1}{z}\right)}= 2 \operatorname{Re}\left(\cfrac{1}{z}\right)$
*$z^3\cdot\overline{z}+z\cdot\overline{z}^3 = z \bar z(z^2 + \bar z ^2) = |z|^2\cdot 2 \operatorname{Re}(z^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to convert the given set of equations into a symmetric form How do I convert the equations
$$\left\{\begin{array}{} 3a +2b+ c & = & 0\\ a+4b+4c & = & 0 \end{array}\right.$$
to the form
$$\frac{a}{4}=\frac{b}{-11}=\frac{c}{10}=k.$$
Thank you
| One way of doing this is to relate $\frac a4$ to $b$ and $c$, separately.
First, multiplying the first equation by $2$ gives
$2\times (3a+2b+c)=2\times0 \Leftrightarrow 6a+4b+2c=0$.
Subtracting the second equation from this result gives
$5a-2c=0$
$5a=2c$
$a=\frac{2}{5}c$
$\frac{a}{4}=\frac{c}{10}$.
Similarly, to relat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove an inequality I didn't know where can I start because it doesn't fit to any theories or formulas.
If $a>0$, $b>0$, $c>0$ and $a+b+c=1$, prove that:
$$\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\geq \frac{1}{2}$$
| We need to prove that
$$\sum_{cyc}\frac{a^3}{a^2+b^2}\geq\frac{1}{2}$$ or
$$\sum_{cyc}\frac{a^3}{a^2+b^2}\geq\frac{a+b+c}{2}$$ or
$$\sum_{cyc}\left(\frac{a^3}{a^2+b^2}-\frac{a}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{a(a^2-b^2)}{a^2+b^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{a(a^2-b^2)}{a^2+b^2}-(a-b)\right)\geq0$$ or
$$\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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General Solution for a PDE I would like to know if someone could provide me the solutions for the next PDE:
\begin{align*}
f_{x}(x,y) - f_{y}(x,y) = \frac{y-x}{x^{2}+y^{2}}f(x,y)
\end{align*}
I empirically found the following particular solution given by $f(x,y) = \displaystyle\frac{x+y}{\sqrt{x^{2}+y^{2}}}$. Any contr... | $$f_{x}(x,y) - f_{y}(x,y) = \frac{y-x}{x^{2}+y^{2}}f(x,y)$$
The set of characteristic ODEs is:
$$\frac{dx}{1}=\frac{dy}{-1}=\frac{df}{\frac{y-x}{x^{2}+y^{2}}f}$$
The equation of a first family of characteristic curves comes from :
$$\frac{dx}{1}=\frac{dy}{-1} \quad\to\quad x+y=c_1$$
The equation of a second family of c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Finding solutions to $2^x+17=y^2$
Find all positive integer solutions $(x,y)$ of the following equation:
$$2^x+17=y^2.$$
If $x = 2k$, then we can rewrite the equation as $(y - 2^k)(y + 2^k) = 17$, so the factors must be $1$ and $17$, and we must have $x = 6, y = 9$.
However, this approach doesn't work when $x$ is ... | $x=3,5,9$ work with $y=5,7,23$ but fail nor $x=1,7$. If $x $ works for any other value then $x>4$.
$2^x+17=y^2$
$2^4 (2^{x-4}+1)=y^2-1=(y+1)(y-1) $
$y\pm 1$ must be even and $\gcd(x+1,x-1)=2$ and one is a multiple of $4$ while the other isn't.
That means::
$y\pm 1=8m;y\mp 1=2n;mn=2^{x-4}+1$
Also means
$2^x+8=y^2-9$
$... | {
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"url": "https://math.stackexchange.com/questions/2298402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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