Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Determine the limiting behaviour of $\lim_{x \to \infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$ Determine the limiting behaviour of $\lim_{x \to \infty}{\dfrac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$
Used L'Hopitals to get $\;\dfrac{(x^6+1)^{\frac{2}{3}}}{x^2 \sqrt{x^4+1}}$ but not sure what more i can do after that.
| Dividing numerator and denominator by $x^2$ gives
$$\lim_{x \to +\infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}=\lim_{x\to+\infty} \frac{\sqrt{1+x^{-4}}}{\sqrt[3]{1+x^{-6}}}=\frac{\sqrt{1}}{\sqrt[3]{1}}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Find expansion around $x_0=0$ into power series and find a radius of convergence My task is as in the topic, I've given function $$f(x)=\frac{1}{1+x+x^2+x^3}$$
My solution is following (when $|x|<1$):$$\frac{1}{1+x+x^2+x^3}=\frac{1}{(x+1)+(x^2+1)}=\frac{1}{1-(-x)}\cdot\frac{1}{1-(-x^2)}=$$$$=\sum_{k=0}^{\infty}(-x)^k\c... | Let $x\ne 1$. By the usual formula for the sum of a finite geometric series, we have $1+x+x^2+x^3=\frac{1-x^4}{1-x}$. So your expression is equal to $\frac{1-x}{1-x^4}$.
Expand $\frac{1}{1-x^4}$ in a power series, multiply by $1-x$. Unless we are operating purely formaly, we will need to assume that $|x|\lt 1$.
For de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/369435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $1$ and $2$ are zeros of the following polynomial
Show that $1$ and $2$ are zeros of the polynomial $P(x)=x^4-2x^3+5x^2-16x+12$ and
hence that $(x-1)(x-2)$ is a factor of $P(x)$
| The is an application of the Factor Theorem.
If $\operatorname{f}(x)$ is a polynomial and $\operatorname{f}(p)=0$, then $x-p$ is a factor of $\operatorname{f}(x)$. In particular: if $\operatorname{f}(1)=0$ then $x-1$ is a factor of $\operatorname{f}(x)$ and if $\operatorname{f}(2)=0$ then $x-2$ is a factor of $\operat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/369562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find real solutions of polynomial How with you find the solutions to the following:
$\sqrt{3x+10}-\sqrt{x+2}=2$
This is what I tried so far:
$(\sqrt{3x+10}-\sqrt{x+2})^2=2$
$(3x+10)+(x+2)-2\sqrt{(x+2)(3x+10)}=2$
$2x+5-\sqrt{3x^2+16x+20}=0$
No I do not know where to go from here...
| Hint:
$2x+5-\sqrt{3x^2+16x+20}=0\iff2x+5=\sqrt{3x^2+16x+20} \Longrightarrow\\(2x+5)^2=3x^2+16x+20.$
| {
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"url": "https://math.stackexchange.com/questions/373382",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How show that $\dfrac{a^3}{3}\ge\int_{0}^{a}|F(x)-x|^2dx$ Let $F(x)$ be nonnegative and integrable on $[0,a]$ and such that
$$\left(\int_{0}^{t}F(x)dx\right)^2\ge\int_{0}^{t}F^3(x)dx$$
for every $t$ in $[0,a]$,prove or disprove the conjecture:
$$\dfrac{a^3}{3}\ge\int_{0}^{a}|F(x)-x|^2dx$$
This Problem from SIAM problem... | For simplicity let us write $F = F(x)$ and $G_t = \int_0^t F dx$.
Given $\displaystyle {G_t}^2 \ge \int_0^t F^3 dx$, we need to show $\displaystyle \dfrac{a^3}{3} \ge \int_0^a \mid F - x \mid^2 dx \tag{1}$.
Now, RHS $= \displaystyle \int_0^a \left( F - x \right)^2 dx = \int_0^a F^2 dx - \int_0^a 2x F dx + \frac{a^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/374172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding the number of ordered pairs that satisfies an equation How many number of ordered pairs of positive integer $(a,d)$ satisfies:
$$\frac{1}{\frac{1}{a+2d}-\frac{1}{a+3d}}-\frac{1}{\frac{1}{a}-\frac{1}{a+d}}=2012$$
This question is too complicated. Like I tried to multiply by the common denominator, but it's not ... | The equation simplifies as
\begin{align}
&\frac{1}{\frac{1}{a+2d}-\frac{1}{a+3d}}-\frac{1}{\frac{1}{a}-\frac{1}{a+d}}=\frac{1}{\frac{d}{(a+2d)(a+3d)}}-\frac{1}{\frac{d}{a(a+d)}}\\ &=\frac{(a+2d)(a+3d)}{d}-\frac{a(a+d)}{d}\\&=\frac{a^2+5ad+6d^2-a^2-ad}{d}\\&=\frac{4ad+6d^2}{d}\\&=4a+6d\\&=2012
\end{align}
Thus, $2a+3d=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/374240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability of Population A population consists of $50$% men and $50$% women of a population of $50$ people. A simple random sample (draws at random without replacement) of $4$ people is chosen. Find the chance that in the sample
i) the fourth person is a woman?
ii) The third person is a woman, given that the first ... | We can split it into, the probability that the first three were all women, $2$ women, $1$ woman, or no women. So
$$P(4\text{th Woman})=\frac{\binom{25}{3}}{\binom{50}{3}}\frac{22}{47}+\binom{3}{1}\frac{\binom{25}{2}\binom{25}{1}}{\binom{50}{2}\binom{48}{1}}\frac{23}{47}+\binom{3}{2}\frac{\binom{25}{1}\binom{25}{2}}{\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/376990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$
Find the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$.
Please brief about the concept behind this to solve such problems. Thanks.
| 2^100 + 3^100 + 4^100 + 5^100/7
= (2^3)^33 *2 + (3^3)^33 *3 + (4^3)^33 *3 + (5^3)^33 *3 / 7
= 2*8^33 + 3*27^33 + 4*64^33 + 5*125^33 / 7
= 2*1^33 + 3*-1^33 + 4*1^33 + 5*-1^33 / 7
= 2*1 + 3*-1 + 4*1 + 5*-1 / 7
= 2 - 3 + 4 -5 / 7
= -2/7
= 7-2
= 5 remainder
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculus Reduction Formula
For any integer $k > 0$, show the reduction formula
$$\int^{2}_{-2} x^{2k} \sqrt{4-x^2} \, dx = C_k \int^{2}_{-2} x^{2k-2} \sqrt{4-x^2} \, dx$$
for some constant $C_{k}$.
(original image)
I thought this would be fairly straightforward but im a little confused. Do I start out by doing a ... | It took a little time to check through this, but here is an approach using the substitution I'd proposed in the comments. Applying $\ \sin\theta = \frac{x}{2}$ , we have
$$\int_{-2}^{2} x^{2k} \ \sqrt{4-x^2} \ dx \ \rightarrow \ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2^{2k} \cdot \sin^{2k}\theta \cdot (2 \cos\theta) \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/380025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the point of intersection of the line and surface I have an odd problem with no solution. I am completely lost on how to solve this.
Problem:
Find the coordinates of the point(s) of intersection of the line $x = 1+t$, $y = 2+3t$, $z = 1-t$ and the surface $z = x^2 +2y^2$
Attempt:
$(1) \ x = 1+t$
$(2) \ y = 2+3t$
$... | Your method is entirely correct, but as noted in the comment above, you may pick up unwanted solutions. Your point $(0, -1, 2)$ is on the surface, but this is not the case for $(27/19, 26/19, 11/19)$. (This point doesn't satisfy the equation of the surface. So if you double check your computations, and it results in th... | {
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"url": "https://math.stackexchange.com/questions/380576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the number of real roots of the given equation?
The number of real roots of the equation $$2 \cos \left( \frac{x^2+x}{6} \right)=2^x+2^{-x}$$ is
(A) $0$, (B) $1$, (C) $2$, (D) infinitely many.
Trial: $$\begin{align} 2 \cos \left( \frac{x^2+x}{6} \right)&=2^x+2^{-x} \\ \implies \frac{x^2+x}{6}&=\cos ^{-... | Notice that $2^0+2^0=2$ and that $2^x+2^{-x}> 2$ for $x\ne 0$. Then consider that $2\cos(0)=2$ and that $2\cos\left(\frac{x^2+x}{6}\right)\le2$ for all $x$. What can we conclude?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/380896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Solve for $x$ a trigonometric equation I want to solve for $x$
$$ {{2}^{{{\sin }^{4}}x-{{\cos }^{2}}x}}-{{2}^{{{\cos }^{4}}x-{{\sin }^{2}}x}}=\cos 2x $$
but I don't know how to start. Replacing $\sin x$ or $\cos x$ by $y$ led me nowhere because of the right side.
One of the solutions I've found is $x=\pi/4$ but there ... | Putting $\cos^2x=a,$
$\sin^4x-\cos^2x=(1-a)^2-a=a^2-3a+1$ and $\cos^4x-\sin^2x=a^2-(1-a)=a^2+a-1$
So we get, $2^{a^2-3a+1}-2^{a^2+a-1}=2a-1$
or, $2^{a^2-3a+1}(1-2^{4a-2})=2a-1$
If $2a-1>0,$ $4a-2>0\implies 2^{4a-2}>2^0=1\implies $ the left hand side is $<0$
Similarly, if $2a-1<0$ the right hand side is $>0$
If $2a-1=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/381233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 2
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Proving $\frac{x}{y} +\frac{y}{z} + \frac{z}{x} \ge 3$ for positive $x,y,z$
Suppose $x,y,z$ are real positive numbers, prove that:
$$\dfrac{x}{y} +\dfrac{y}{z} + \dfrac{z}{x} \ge 3$$
with equality when $x=y=z$.
Can someone help me find an easier solution?
I started with assume only two are equal, without loss of gene... | Divide through by 3. From AM-GM, we have for any positive $a,b,c$
$$\frac{a+b+c}{3} \ge (abc)^{\frac{1}{3}}$$
With $a=\frac{x}{y}, b=...$, this becomes:
$$\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} \ge \left( \frac{x}{y} \frac{y}{z} \frac{z}{x} \right) ^{\frac{1}{3}}=1.$$
Now multiply by 3 again to get the ans... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/381673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Find all function $f:\mathbb{R}\mapsto\mathbb{R}$ such that $f(x^2+y^2)=f(x+y)f(x-y)$. Find all function $f:\mathbb{R}\mapsto\mathbb{R}$ such that $f(x^2+y^2)=f(x+y)f(x-y)$.
Some solutions I found are $f\equiv0,f\equiv1$, $f(x)=0$ if $x\neq0$ and $f(x)=1$ if $x=0$.
| let $x=y=0 \to f(0)=f^2(0) \to f(0)=0$ or $ f(0)=1 $,
case I:$f(0)=0$
let $x-y=0 \to f(2x^2)=f(2x)*f(0)=0 \to f(x)=0$ when $x \ge 0 $,
let $x+y=u,x-y=v \to f(\dfrac{u^2+v^2}{2})=f(u)f(v), $if $v=c <0 ,f(c) \not=0$, $ f(u)=\dfrac{f(\dfrac{u^2+c^2}{2})}{f(c)}=f(-u) \to f(x)=0 $ when $x<0$, contradict. so $f(x)=0$ for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/381724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Showing that $\mathbb Q(\sqrt{17})$ has class number 1 Let $K=\mathbb Q(\sqrt{d})$ with $d=17$. The Minkowski-Bound is $\frac{1}{2}\sqrt{17}<\frac{1}{2}\frac{9}{2}=2.25<3$.
The ideal $(2)$ splits, since $d\equiv 1$ mod $8$. So we get $(2)=(2,\frac{1+\sqrt{d}}{2})(2,\frac{1-\sqrt{d}}{2})$ and $(2,\frac{1\pm\sqrt{d}}{2})... | The quadratic integer $n=(1\pm\sqrt{17})/2$ has minimal polynomial $n^2-n-4$, so the ideals
$\left(2,\dfrac{1\pm\sqrt{17}}{4}\right)$
will be principal if natural integers $x,y$ exist such that
$|x^2-xy-4y^2|=2.$
As this condition holds when $x=2,y=1$ we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/382188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Polar equation representation of an exponential spiral involving $ \arctan(z) $ An exponential spiral is formulated as:
$$ \arctan(z) = \ln\left(\sqrt{1+z^2}\right)+\ln(x)+C $$ with: $$z=y/x$$
Represent the equation in polar coordinates and show $y' = \tan(\pi/4 + \theta)$.
Substituting in $ x = r\cos(\theta) $ and $ y... | If $x = r\cos\theta, y = r\sin\theta$, assuming we are in the first/fourth quadrant $\cos\theta >0$, otherwise your equation $\ln x$ would not make sense. Then the polar coordinates can be written as:
$$
r = \sqrt{x^2 + y^2}, \text{ and } \theta = \arctan\frac{y}{x}\tag{1}
$$
You have arrived at
$$ \arctan\frac{y}{x} ... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $a, b, c$ are positive odd integers, then $b^2 - 4ac$ cannot be a perfect square. Prove that if $a, b, c$ are positive odd integers, then $b^2 - 4ac$ cannot be a perfect square.
What I have done:
This has to either be done with contradiction or contraposition, I was thinking contradiction more likely.
| Suppose $b^2-4ac$ is a square. Since $a,b,c$ are odd, $b^2-4ac$ must be an odd square. Hence $x=\frac{-b+\sqrt{b^2-4ac}}{2}$ is an integer. Since $x^2+bx+ac=0$, therefore $x^2+bx+ac\equiv 0\,(mod\,2)$. By FLT $x^2\equiv x\,(mod\,2)$. Hence $(1+b)x+ac\equiv 0\,(mod\,2)$. Since $2|1+b$. Thus $0+ac\equiv 0\,(mod\,2)$ (co... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a set of recurrence relation: Solving a set of recurrence relation:
$a_n=3b_{n-1} , b_n=a_{n-1}-2b_{n-1}$
In addition, It's known that: $a_1=2, b_1=1$.
So i started by isolating $b_n \to b_n=3b_{n-2}-2b_{n-1}$
So i get the current equation: $x^2+2x-3=0 \to(x+3)(x-1)=0$, Which means $b(n)=A_1+A_2(-3)^n$
So what... | Define generating functions $A(z) = \sum_{n \ge 0} a_{n + 1} z^n$ and similarly $B(z)$. Write your recurrences as:
$$
\begin{align*}
a_{n + 1} &= 3 b_n \\
b_{n + 1} &= a_n - 2 b_n
\end{align*}
$$
By properties of generating functions, your recurrences translate to:
$$
\begin{align*}
\frac{A(z) - a_1}{z} &= 3 B(z) \\
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/383192",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$ Find all real numbers $x$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac76$$
I have tried to fiddle with it as follows:
$$2^{3x} \cdot 6 +3^{3x} \cdot 6=12^x \cdot 7+18^x \cdot 7$$
$$ 3 \cdot 2^{3x+1}+ 2 \cdot 3^{3x+1}=7 \cdot 6^x(2^x+3^x)$$
Dividi... | $\Rightarrow $ $6(8^x+27^x)=7(12^x+18^x)$
Divide by$12^x$
$\Rightarrow $$ 6((\frac{2}{3}) ^x+(\frac{3}{2})^{2x}) =7(1+(\frac{3}{2})^{x})$
$\Rightarrow $ $(\frac{2}{3}) ^x+(\frac{3}{2})^{2x}-\frac{7}{6}-(\frac{7}{6})(\frac{3}{2}) ^x=0$
We put :$t=(\frac{3}{2}) ^x$
$\Rightarrow $ $\frac{1}{t} +t^2 - \frac{7}{6}t-\frac{7}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/384090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
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"answer_id": 2
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Laplace equation and integral $$ \int_0^{2\pi} \frac{1+3 \sin{\phi}}{a^2-2ar \cos(\theta - \phi) + r^2 } d\phi$$
Help me plz ... I have tried to solve this. but I still don't know.
| This integral may be evaluated using residue theory. In this case, convert the integral over $\phi$ to a contour integral over the unit circle in the complex plane, and evaluate the residues of the poles inside the unit circle. Let $z=e^{i \phi}$, then $d\phi = dz/(i z)$, $\cos{\phi} = (z+z^{-1})/2$, $\sin{\phi} = (z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/385337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Circular determinant problem I'm stuck in this question:
How calculate this determinant ?
$$\Delta=\left|\begin{array}{cccccc}
1&2&3&\cdots&\cdots&n\\
n&1&2&\cdots&\cdots& n-1\\
n-1&n&1&\cdots&\cdots&n-2\\
\vdots &\ddots & \ddots&\ddots&&\vdots\\
\vdots &\ddots & \ddots&\ddots&\ddots&\vdots\\
2&3&4&\cdots&\cdots&1
\end... | I'll do the case $n=4$ and leave the general case for you.
It is useful to recall how elementary row operations effect the value of determinant, see for example
ProofWiki.
We get
$$
\begin{vmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2 \\
4 & 1 & 2 & 3
\end{vmatrix}
\overset{(1)}=
\begin{vmatrix}
1 & 2 & 3 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/386526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Evaluate a sum with binomial coefficients: $\sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$ $$\text{Find} \ \ \sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$$
I expanded the binomial coefficients within the sum and got $$\binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2$$
What does this equal to? I think this c... | First, use $k\binom{n\vphantom{1}}{k}=n\binom{n-1}{k-1}=n\binom{n-1}{n-k}$
$$
\sum_{k=0}^n(-1)^kk\binom{n}{k}^2
=n\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{n-1}{n-k}\tag{1}
$$
Next compute a generating function. The sum we want is the coefficient of $x^n$
$$
\begin{align}
n\sum_{m,k}(-1)^k\binom{n}{k}\binom{n-1}{m-k}x^m
&=n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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How to calculate ellipse sector area *from a focus* How do you calculate the area of a sector of an ellipse when the angle of the sector is drawn from one of the focii? In other words, how to find the area swept out by the true anomaly?
There are some answers on the internet for when the sector is drawn from the cente... | Following the approach I used for
Evaluating $\int_a^b \frac12 r^2\ \mathrm d\theta$ to find the area of an ellipse,
parametrize the ellipse as:
$x(t)=a \cos (t)$
$y(t)=b \sin (t)$
with $a>b$ (the case $a<b$ is symmetric). Note that $t$ is neither the
central angle nor the focal angle. The foci are then on the x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/388134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
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Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.
Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.($a^2+b^2$, is same $b^2+a... | *
*Let $n = 4x^4$, we have:
$$n(n+1) = (4x^4)^2 + (2x^2)^2 = (4x^4-2x^2)^2 + (4x^3)^2$$
*Let $n = (u^2 + v^2)^2$, we have
$$\begin{align}
n(n+1) = & ((u^2 + v^2)^2)^2 + (u^2+v^2)^2\\
= & (u^4 - 2uv - v^4)^2 + (2uv^3-v^2+2u^3v+u^2)^2\\
= & (u^4 + 2uv - v^4)^2 + (2uv^3+v^2+2u^3v-u^2)^2
\end{align}$$
*Le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/388501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Derivative of Trig Functions (Intuition Help?) Looking for some intuition help here.
I have the following exercise and these are the steps I take:
$$
y = \sin\left(\frac{1}{x}\right)
$$
$$
u=\frac{1}{x}
$$
$$
y = \sin u,\;\;\frac{dy}{du} = \cos u= \cos\left(\frac{1}{x}\right)
$$
$$
u=x^{-1};\;\frac{du}{dx} =-x^{-2}=-\f... | First, you are taking short cuts when you are writing your equations that simply confuse things. Also, LaTeX has \sin and \cos. Start over.
$$
\begin{aligned}
\text{Let}\quad f(x) &= \sin\frac{1}{x}.\\
\text{Let}\quad u &= \frac{1}{x}.\\
f(x) &= \sin u\\
\frac{df}{dx} &= \frac{df}{du} \frac{du}{dx} \quad\text{by the C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$x$ as the shortest alternating sum of $1 \ldots n$ If I have an positive integer $x \in \mathbb{N}$ and I have $Z = \sum_{i = 0}^{n}{i}$ such that $Z \geq x$ and $Z - x \equiv 0 \bmod 2$ and $n$ is the smallest such integer it is possible to create and alternating sum from $1 \ldots n$ such that it equals $x$.
For exa... | Let $S(n)=\{x:x=\pm1 \pm2 \cdots \pm n\}$. Let $T(n)=\{x:|x|\le \frac{n(n+1)}{2}\}\cap\{x:x\equiv \frac{n(n+1)}{2}\pmod{2}\}$.
Claim: $S(n)=T(n)$ for all $n\in \mathbb{N}$.
Proof: Induction on $n$. $n=1,2$: clear
$S(n+1)\subseteq T(n+1)$: consider all $+$ or all $-$
$T(n+1)\subseteq S(n+1)$: $$[0,\frac{(n+1)(n+2)}{2}... | {
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"source": "stackexchange",
"question_score": "2",
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What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists? Here is my candidate for
the most elementary proof that $\lim_{n \to \infty}(1+1/n)^n $ exists.
I would be interested in seeing others.
$***$ Added after some comments:
I prove here by very elementary means that the limit exists.
Calling th... | Could this be ?
$$\begin{align} \frac{d}{dx} \ln x &= \lim_{h \to 0}\; \frac{\ln(x + h) − \ln x}{h} \\
&= \lim_{h \to 0} \;\frac{\ln(1 + \frac{h}{x})}{h} \\
&= \lim_{h \to 0} \;\ln\left((1 + \frac{h}{x})^{\frac{1}{h}}\right) \\
\end{align}$$
let $h=\frac xc$
$$\begin{align} \frac{d... | {
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"source": "stackexchange",
"question_score": "13",
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"answer_id": 0
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GgT, (polynomial) division and finite fields... Exercise: Let $f,g \in \mathbb{Z}_2[x]$ be the polynomials $f = x^6 + x^5 + x^4 + 1$ and $g = x^5 + x^4 + x^3 + 1$. Has the diophantic equation $f u + g u = x^4 + 1$ solutions $u,v \in \mathbb{Z}[x]$? If so, determine them all.
My Solution: First I observed that in $\math... | You are taking polynomials, so it is incorrect to say that $x^n = x$ in $\Bbb{Z}_{2}[x]$. Actually you have that
$$
a_{0} + a_{1} x + \dots + a_{n} x^{n}
=
b_{0} + b_{1} x + \dots + b_{n} x^{n}
$$
for two elements of $\Bbb{Z}_{2}[x]$
if and only if $a_i = b_i$ for all $i$.
Most likely you are mixing up polynomials an... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving recurrence relation, $a_n=6a_{n-1} - 5a_{n-2} + 1$ I'm trying to solve this recurrence relation:
$$
a_n = \begin{cases}
0 & \mbox{for } n = 0 \\
5 & \mbox{for } n = 1 \\
6a_{n-1} - 5a_{n-2} + 1 & \mbox{for } n > 1
\end{cases}
$$
I calculated generator function as:
$$
A = \frac{31x - 24x^2}{1 - ... | $$a_0 =0,a_1=5,a_n=6a_{n-1} - 5a_{n-2} + 1, n > 1$$
$$f(x)=\sum_{n=0}^{\infty}a_nx^n=5x+\sum_{n=2}^{\infty}a_nx^n=$$
$$=5x+\sum_{n=2}^{\infty}(6a_{n-1} - 5a_{n-2} + 1)x^n=$$
$$=5x+6\sum_{n=2}^{\infty}a_{n-1}x^n-5\sum_{n=2}^{\infty}a_{n-2}x^n+\sum_{n=2}^{\infty}x^n=$$
$$=5x+6x\sum_{n=2}^{\infty}a_{n-1}x^{n-1}-5x^2\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/390644",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Integrating $\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^n \, d\theta$, for $n=2$ and $n=3$ How do you integrate the following functions:
$$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^2 \, d\theta$$ and $$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^3 \, d\theta
$$
respectively?
Note: Init... | Multiplying both numerator and denominator by $\sec^2 \theta$ yields
$$
\begin{aligned}
I_2=&\int\left(\frac{\cos \theta}{1+\sin ^{2} \theta}\right)^{2} d \theta\\=& \int \frac{\sec ^{2} \theta}{\left(\sec ^{2} \theta+\tan ^{2} \theta\right)^{2}} d \theta \\
=& \int \frac{d t}{\left(1+2 t^{2}\right)^{2}}, \text { where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/391338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Remainders problem What will be the reminder if $23^{23}+ 15^{23}$ is divided by $19$?
Someone did this way:
$15/19 = -4$ remainder and $23/19 = 4$ remainder So $(-4^{23}) + (4^{23}) =0$ but i didn't understand it
| We have $23\equiv 4\pmod{19}$ and $15\equiv -4\pmod{19}$. Then for any odd positive integer $n$, we have $23^n\equiv 4^n \pmod{19}$ and $15^n\equiv (-4)^n \equiv (-1)^n4^n\equiv -(4^n)\pmod{19}$.
Add. We get $4^n -4^n\equiv 0 \pmod{19}$.
Added: We don't really need to go to $4$ and $-4$. For note that $23\equiv -15 \p... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the general equation of a cubic polynomial? I had this question:
"Find the cubic equation whose roots are the the squares of that of $x^3 + 2x + 1 = 0$"
and I kind of solved it. In that my answer was $x^3 - 4x^2 + 4x + 1$, but it was actually $x^3 + 4x^2 + 4x - 1 = 0$.
I took the general equation of a cubic equ... | Let $a,b,c$ be roots of $x^3 + 2x + 1 = 0$
Using Vieta's Formulas, $a+b+c=0,ab+bc+ca=2,abc=-1$
So, we need to find the cubic equation whose roots are $a^2,b^2,c^2$
i.e., $$(y-a^2)(y-b^2)(y-c^2)=0$$
$$\iff y^3-(a^2+b^2+c^2)y^2+(a^2b^2+b^2c^2+c^2a^2)y+a^2b^2c^2=0$$
Now, $a^2b^2c^2=(abc)^2=(-1)^2=1$
$a^2+b^2+c^2=(a+b+c)^2... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Find $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ if $a+b+c=0$ I'm stuck at this algebra problem, it seems to me that's what's provided doesn't even at all.
Provided: $$a+b+c=0$$
Find the value of: $$\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$$
Like I'm not sure where to start, and t... | Note that:
$$\frac{a^2}{2a^2+bc}=\frac{a^2}{a^2+a(-b-c)+bc}=\frac{a^2}{(a-b)(a-c)}$$
This applies in the same way for:
$$\frac{b^2}{2b^2+ac}=\frac{b^2}{(b-c)(b-a)}\ \text{and}\ \frac{c^2}{2c^2+ab}=\frac{c^2}{(c-a)(c-b)}$$Therefore, the original equation is equal to:
\begin{align}
&\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/393786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How to show this inequality?
Show that $$-2 \le \cos \theta(\sin \theta+\sqrt{\sin^2 \theta +3})\le2$$
Trial: I know that $-\dfrac 1 2 \le \cos \theta\cdot\sin \theta \le \dfrac 1 2$ and $\sqrt 3\le\sqrt{\sin^2 \theta +3}\le2$. The problem looks simple to me but I am stuck to solve this. Please help. Thanks in advanc... | Essentially, we want the extremum of
$$f(a) = a\sqrt{1-a^2} + a\sqrt{4-a^2}$$
We have
$$f'(a) = \sqrt{1-a^2} + \sqrt{4-a^2} - \dfrac{a^2}{\sqrt{1-a^2}} - \dfrac{a^2}{\sqrt{4-a^2}} = \dfrac{1-2a^2}{\sqrt{1-a^2}} + \dfrac{4-2a^2}{\sqrt{4-a^2}}$$
Let $a^2 = x$, we then want to solve for
\begin{align}\dfrac{1-2x}{\sqrt{1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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quadratic equation precalculus from Stewart, Precalculus, 5th, p56, Q. 79
Find all real solutions of the equation
$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{x^2-4}$$
my solution
$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{(x+2)(x-2)}$$
$$(x+2)(x+5)=5(x-2)+28$$
$$x^2+2x-8=0$$
$$\dfrac{-2\pm\sqrt{4+32}}{2}$$
$$\dfrac{... | You multiplied both sides by $(x-2)(x+2)$. If this is zero, you may introduce extra solutions, hence you need to check your final answer to see if you have any extraneous solutions. In this case, you do! For $x=2$, two of the three fractions in the original equation are undefined.
| {
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"source": "stackexchange",
"question_score": "3",
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How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$? The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ... | The square of this is $2x+2\sqrt{x^2-y^2}$. One could find this.
Regarding $\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$, the modulus of this number $35+18\sqrt{-3}$ is 2197, which is $13^3$. This means that we are looking for a number whose cube is $35+18\sqrt{-3}$. An example of such a number is $3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 5
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Fisher information of a Binomial distribution The Fisher information is defined as $\mathbb{E}\Bigg( \frac{d \log f(p,x)}{dp} \Bigg)^2$, where $f(p,x)={{n}\choose{x}} p^x (1-p)^{n-x}$ for a Binomial distribution. The derivative of the log-likelihood function is $L'(p,x) = \frac{x}{p} - \frac{n-x}{1-p}$. Now, to get the... | Fisher information: $I_n(p) = nI(p)$, and $I(p)=-\mathbb{E_p}\Bigg( \frac{\partial^2 \log f(p,x)}{\partial p^2} \Bigg)$, where $f(p,x)={{1}\choose{x}} p^x (1-p)^{1-x}$ for a Binomial distribution. We start with $n=1$ as single trial to calculate $I(p)$, then get $I_n(p)$.
$\log f(p,x) = x \log p + (1-x) \log p$
$\frac ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Getting rid of a floor function in the next expression:$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor $, It is known x is odd. I was wondering if there's a way in which you can get rid of a floor function in the next expression:$$\left\lfloor\frac{(x-2)^2}{4}\right\rfloor $$ It is known x is odd.
| Since $x$ is odd, $x=2n+1$ for some integer $n$, and therefore $(x-2)^2=(2n-1)^2=4n^2-4n+1$. Thus,
$$\left\lfloor\frac{(x-2)^2}4\right\rfloor=\left\lfloor n^2-n+\frac14\right\rfloor=n^2-n\;.$$
And $n=\dfrac{x-1}2$, so the expression reduces to
$$\left(\frac{x-1}2\right)^2-\frac{x-1}2=\frac{x^2-4x+3}4\;.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For x < 5 what is the greatest value of x It can't be $5$. And it can't be $4.\overline{9}$ because that equals $5$. It looks like there is no solution... but surely there must be?
| If $x<5$ then $2x<x+5$ so $x<\frac{x+5}{2}$. Similarly, $x<5$ means $x+5<5+5=10$ or $\frac{x+5}{2}<5$. So if $x<5$ we have $x<\frac{x+5}{2}<5$, and therefore there is a larger number, $\frac{x+5}{2}$ less than $5$.
Basically, the average of two different numbers must be strictly between those two numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 2
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Value of series, Partialsum? given is the following series
$$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$$
And I need to find its value.
How can I start finding it?
Thanks for all
does the Telescop-Summing work here as well?:
$\sum_{n=1}^\infty \frac{1}{4n^2-1} $
now:
$\frac{1}{4n^2-1} = \frac{1}{2} * \frac{(2n+1)-(2n-1... | HINT:
$$\frac{2n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}=\frac1{n^2}-\frac1{(n+1)^2}$$
Can you recognize Telescoping Sum?
| {
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"source": "stackexchange",
"question_score": "3",
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Show $60 \mid (a^4+59)$ if $\gcd(a,30)=1$
If $\gcd(a,30)=1$ then $60 \mid (a^4+59)$.
If $\gcd(a,30)=1$ then we would be trying to show $a^4\equiv 1 \mod{60}$ or $(a^2+1)(a+1)(a-1)\equiv 0 \mod{60}$. We know $a$ must be odd and so $(a+1)$ and $(a-1)$ are even so we at least have a factor of $4$ in $a^4-1$. Was thinkin... | We essentially want to show that $60 \vert (a^4-1)$. Since $\gcd(a,30) = 1$, we have
$$a \equiv \pm1, \pm 7, \pm 11, \pm 13, \pm 17, \pm 19, \pm 23, \pm 29 \pmod{60}$$
Hence,
$$a^2 \equiv \begin{cases}1 \pmod{60} & \text{if }a \equiv \pm1,\pm11,\pm19,\pm29\\ -11 \pmod{60} & \text{if }a \equiv \pm7,\pm13,\pm17,\pm23\en... | {
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"source": "stackexchange",
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Show that for all $a,b,c>0$, $\frac 1 {\sqrt[3]{(a+b)(b+c)(c+a)}}\geq\frac 3 {2(a+b+c)}$.
Show that for all $a,b,c>0$, $\displaystyle\frac 1 {\sqrt[3]{(a+b)(b+c)(c+a)}}\geq\frac 3 {2(a+b+c)}$.
I tried to cube the both sides, and expand it, but that'll be too troublesome, is there simpler ways? Thasnk you.
| HINT:
Using AM-GM inequality,
$$\frac{a+b+b+c+c+a}3\ge \sqrt[3]{(a+b)(b+c)(c+a)}$$
| {
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Sum : $\sum \sin \left( \frac{(2\lfloor \sqrt{kn} \rfloor +1)\pi}{2n} \right)$. Calculate : $$ \sum_{k=1}^{n-1} \sin \left( \frac{(2\lfloor \sqrt{kn} \rfloor +1)\pi}{2n} \right).$$
| Lemma Summation by Pasts (1)
$$\sum_{k=a}^b f_k\Delta g_k=f_kg_k\Bigg|_{k=a}^{b+1}-\sum_{k=a}^b g_{k+1}\Delta f_k=f_{b+1}g_{b+1}-f_ag_a-\sum_{k=a}^b g_{k+1}\Delta f_k$$
$\displaystyle \text{with difference operator }\Delta :\qquad \Delta f_k=f_{k+1}-f_k$
My solution
$\text{Get }j=\left\lfloor \sqrt{kn}\right\rfloor\Rig... | {
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"question_score": "5",
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Compute eigenvalues and eigenvectors problem I really don't know how solve this problem:
Let $V$ be the space of real functions spanned by $\cos(x)$, $\cos(2x)$ and $\cos(3x)$. Let $T\in\mathcal{L}(V,V)$ con $T(\cos(x)) = 3\cos(x) + 2\cos(2x) - \cos(3x)$, $\;T(\cos(2x)) = 3\cos(2x) + \cos(3x)$ and $T(\cos(3x)) = \cos(3... | It's clear that
$$T\left(\begin{array}{c}\cos(x)\\\cos(2x)\\\cos(3x)\end{array}\right)\ =\ \underbrace{\left(\begin{array}{ccc}3 & 2 & -1 \\0 & 3 & 1 \\0 & 0 & 1\end{array}\right)}_{A}\left(\begin{array}{c}\cos(x)\\\cos(2x)\\\cos(3x)\end{array}\right)$$
where $A$ have the eigenvalues 1, 3 and 3. Later,
*
*You have t... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sum_{k=1}^m{k^n}$ is divisible by $\sum_{k=1}^m{k}$ for $ n=2013$ I got an interesting new question, it's about number theory and algebra precalculus. Here is the question:
a positive integer $n$ is called valid if $1^n+2^n+3^n+\dots+m^n$ is divisible by $1+2+3+\dots+m$ for every positive integer $m$.
... | If $n$ is odd, then modulo $m+1$ we have $2(1^n + 2^n + \ldots + m^n) = (1^n+m^n) + (2^n+(m-1)^n) + \ldots + (m^n+1^n) \\ \equiv (1^n-1^n) + (2^n-2^n) + \ldots + (m^n-m^n) = 0 \pmod {m+1}$.
Also, since $m^n \equiv 0 \pmod m$, $2(1^n + 2^n + \ldots + m^n) \equiv 2(1^n + 2^n + \ldots + (m-1)^n) \equiv 0 \pmod m$
Since $m... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Help in calculating the following integral $\int_0^{2\pi}\! \frac{(1+2\cos x)^n \cos (nx)}{3+2\cos x} \, \mathrm{d}x. $ I was asked to calculate this:
$$\int_0^{2\pi}\! \frac{(1+2\cos x)^n \cos (nx)}{3+2\cos x} \, \mathrm{d}x. $$
My idea was to change the integration limits to $|z|=1$ in the complex plane and to use ... | $$ \begin{align} \int_{0}^{2 \pi} \frac{(1+2 \cos x)^{n} \cos (nx)}{3 + 2 \cos x} \ dx &= \text{Re} \int_{0}^{2 \pi} \frac{(1+2 \cos x)^{n} e^{inx}}{3 + 2 \cos x} \ dx \\ &= \text{Re} \int_{0}^{2 \pi}\frac{(1+ e^{ix} + e^{-ix})^{n}e^{inx}}{3 + e^{ix} + e^{-ix}} \ dx \\ &= \text{Re} \int_{|z|=1} \frac{(1+z+z^{-1})^{n}z^... | {
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"url": "https://math.stackexchange.com/questions/410329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Compute $\int_{0}^{1}\left[\frac{2}{x}\right]-2\left[\frac{1}{x}\right]dx$ The question is to find
$$\int_{0}^{1}\left(\left[\dfrac{2}{x}\right]-2\left[\dfrac{1}{x}\right]\right)dx,$$
where $[x]$ is the largest integer no greater than $x$, such as $[2.1]=2, \;[2.7]=2,\; [-0.1]=-1.$
Is there any nice method to solve thi... | We have
$$\left\lfloor \dfrac1x \right\rfloor = n \text{ if }x \in \left(\dfrac1{n+1}, \dfrac1n\right]$$
$$\left\lfloor \dfrac2x \right\rfloor = \begin{cases}2n+1 & \text{ if }x \in \left(\dfrac1{n+1}, \dfrac1{n+1/2}\right] \\ 2n & \text{ if }x \in \left(\dfrac1{n+1/2}, \dfrac1{n}\right] \end{cases}$$
We hence have
$$\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 1
} |
probability of who will be selected "In an office there are 3 secretaries, 4 accountants, and 2 receptionists. If a committee of 3 is to be formed, find the probability that one of each will be selected?
Attempted Solution:
First attempt: (3/9)(4/9)(2/9) = 8/243
Second attempt: (3/9)(4/8)(2/7) = 1/21
Don't if either of... | The standard way to do this is to say there are $\binom{9}{3}$ equally likely ways to choose $3$ people from the $9$. And there are $\binom{3}{1}\binom{4}{1}\binom{2}{1}$ ways to choose one of each kind. For the secretary can be chosen in $\binom{3}{1}$ ways. For each such choice, the accompanying accountant can be cho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving the equation $11x^2-6000x-27500 =0$, preferably without the quadratic formula I obtained this form while solving an aptitude question.
$$\frac{3000}{x-50} + \frac{3000}{x+50} = 11$$
I changed it into quadratic equation
$$11x^2 -6000x - 27500 =0$$
but I don't know how to solve it.
I can't find two factor for 303... | $$11x^2 - 6000x - 27500 = 0$$
$a=11$
$b=-6000 = 2^4 \cdot 3 \cdot 5^3$
$c = -27500 = 2^2 \cdot 5^4 \cdot 11$
$ac=-2^2 \cdot 5^4 \cdot 11^2$
Since $ac$ is negative, we will search for integers $u$ and $v$ such that their product is $=2^2 \cdot 5^4 \cdot 11^2$ and their difference is $2^4 \cdot 3 \cdot 5^3 = 6000$. We wi... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$
We can prove using the Beta-Function identity that
$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac... | Letting $x\mapsto \frac{1}{x}$ reduces the power of the denominator and changes the integral into
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(1+x^n\right)}{\left(1+x^2\right)^2} d x & =\int_0^{\infty} \frac{x^2 \ln \left(1+x^n\right)}{\left(1+x^2\right)^2} d x-n \int_0^{\infty} \frac{x^2 \ln x}{\left(1+x^2\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 6,
"answer_id": 5
} |
How to simplify $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$? $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
I have been staring at it for ages and know that it simplifies to $x$, but have been unable to make any significant progress.
I have tried doing $(\frac{1-x}{1-2x})(\frac{1+2x}{1+2x})$ but that doesn't h... | When four storey fractions are involved, it's better to do computations separately. Write
$$
N=1-\frac{1-x}{1-2x},\qquad
D=1-2\frac{1-x}{1-2x}
$$
and work on $N$ and $D$:
\begin{align}
N&=1-\frac{1-x}{1-2x}\\
&=\frac{(1-2x)-(1-x)}{1-2x}\\
&=\frac{1-2x-1+x}{1-2x}\\
&=\frac{-x}{1-2x}
\end{align}
and
\begin{align}
D&=1-2\... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
How to reconcile the identity $\left( e^{i \theta} \right)^{1/2} = e^{i \theta/2}$ with the fact that $a^2 = b$ is solved by $a = \pm\sqrt{b}$ So I worked along the lines of the following:
$$
\left( \cos \left( \theta \right) + i \sin \left( \theta \right) \right)^{\alpha} = \left( e^{i \theta} \right)^{\alpha} = e^{i ... | Tpofofn's hint is huge:
$$\bullet\;\;\;\left(\cos\frac\theta2+i\sin\frac\theta2\right)^2=\cos^2\frac\theta2-\sin^2\frac\theta2+2i\cos\frac\theta2\sin\frac\theta2=\cos\theta+i\sin\theta$$
$$\bullet\;\;\left(\cos\left(\frac\theta2+\pi\right)+i\sin\left(\frac\theta2+\pi\right)\right)^2=\left(-\cos\frac\theta2-i\sin\frac\t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
derive the Maclaurin series by using partial fractions. derive the Maclaurin series for the function
$(x^3+x^2+2x-2)/(x^2-1)$ by using partial fractions and a known Maclaurin series.
question.
how can I use partial fractions in this case?
Is this case the special one?
plz provide me a hint. thanks.
| Perform polynomial long division
to conclude that
$$\frac{x^{3}+x^{2}+2x-2}{x^{2}-1}=x+1+\frac{3x-1}{x^{2}-1}=x+1+\frac{3x-1}{(x-1)(x+1)}.\tag{1}$$
The theory of partial fractions guarantees that there exist constants $A$ and $B$ such that
$$\frac{3x-1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}.\tag{2}$$
You can find t... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluating an integral in physics question $$U_{C} = \frac{1}{C} \int\!\frac{\cos(100\pi t + \pi/4)}{10}\,dt$$
Find $U_{C}$, the answer is $U_{C}=\left(3.2\times 10^{-4}\right)/C\times \cos(100\pi t - \pi/4)$.
Can someone show to to get this answer?
| First, we make the substitution $u = 100\pi t + \pi/4$, so that $du = 100\pi dt$.
Then, substituting, the integral becomes:
$$U_{c} = \frac{1}{C} \int \cos(u) \cdot \frac{1}{10} \cdot \frac{1}{100\pi}du$$
Integrating, we see that this is:
$$U_{c} = \frac{\frac{1}{1000\pi}}{C}\sin(u) + K$$
or, approximately, with subst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/419616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Jacobi Method and Frobenius Norm Question. I have this linear algebra question concerning the jacobi method and the frobenius norm that I am having a lot of trouble on, I have an exam soon and I would appreciate any help. NOTE: I have read the entire article of wikipedia on jacobi method and frobnenius norm but I have ... | Hint: The Frobenius norm of an $m \times n$ matrix $A$ is defined as the square root of the sum of the absolute squares of its elements.
Example: Consider matrix
$$A = \left(
\begin{array}{ccc}
~~~~1 & -2 & ~~~~~~3 \\
-4 & ~~5 & ~-6 \\
~~~7 & -8 & ~~~~~~9 \\
\end{array}
\right)$$
$\lVert A \rVert _{F} ... | {
"language": "en",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
Let
$$\begin{align*}x &= \frac{1}{1} + \fr... | You are manipulating divergent series. When you do the substraction, you can in fact obtain any limit you want by changing the order in which you are adding or subtracting things.
For example,
$(1 - \frac 12 - \frac 14 - \frac 18 - \ldots) + (\frac 13 - \frac 16 - \frac 1{12} - \frac 1{24}- \ldots) + (\frac 15 - \frac... | {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
$\frac{c}{1+c}\leq\frac{a}{1+a}+\frac{b}{1+b}$ for $0\leq c\leq a+b$ When proving that if $d$ is a metric, then $d'(x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ is also a metric, I have to prove the inequality:
$$\dfrac{c}{1+c}\leq\frac{a}{1+a}+\frac{b}{1+b}$$ for $0\leq c\leq a+b$. This is obvious by expanding, but is there a nicer... | $$\frac{a}{1+a}+\frac{b}{1+b}\ge\frac{a}{1+a+b}+\frac{b}{1+b+a}=\frac{a+b}{1+a+b}=\frac{1}{1+\frac{1}{a+b}}\ge\frac{1}{1+\frac1c}=\frac{c}{1+c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/421099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
multiplication in GF(256) (AES algorithm) I'm trying to understand the AES algorithm in order to implement this (on my own) in Java code.
In the algorithm all byte values will be presented as the concatenation of its individual bit values (0 or 1) between braces with the most significant bit first.
So bytes are interpr... | $$(x^6+x^4+x^2+x+1) * (x^4+x+1) = x^{10}+x^8+x^7+x^3$$
should make you suspicious. What happened to the cross-term $1*1$?
I make that product to evaluate to $x^{10}+x^8+x^7+x^3+1$, and the difference of $+1$ should carry through the remaining computation to give the difference you sought.
The modulo operation is equiva... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Indefinite integration : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$ Problem :
Solve : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$
I tried :
$\frac{1-x^2}{\sqrt{(1-x^2)^3}} + \frac{x}{\sqrt{(1-x^2)^3}}$
But it's not working....Please guide how to proceed
| Hint:when $|x|\lt 1$
$$\int(\frac{1-x^2}{\sqrt{(1-x^2)^3}} + \frac{x}{\sqrt{(1-x^2)^3}})dx=\int\frac{dx}{\sqrt{(1-x^2)}} + \frac{-1}{2}\int\frac{-2xdx}{\sqrt{(1-x^2)^3}}$$ take $$u=1-x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/423286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Evaluating the definite integral $\int^{\pi/2}_0 \frac{x+\sin x}{1+\cos x}\,\mathrm dx$ Problem :
$$\int^{\pi/2}_0 \frac{x+\sin x}{1+\cos x}\,\mathrm dx \tag{i}$$
My approach :
$$\frac{x+\sin x}{1+\cos x}\,\mathrm dx= \left ( \frac{x}{2\cos^2(x/2)} + \tan(x/2) \right )\,\mathrm dx$$
Therefore, (i) will become :
$$\i... | Since $\displaystyle\int\dfrac{x+\sin x}{1+\cos x}dx$ has close-form, I don't think this question is really difficult.
Since
$$
\begin{align}
\int\dfrac{x+\sin x}{1+\cos x}\text{d}x & = \int\dfrac{x}{1+\cos x}dx+\int\dfrac{\sin x}{1+\cos x}dx \\
& = \int\dfrac{x}{2\cos^2\dfrac{x}{2}}dx+\int\tan\dfrac{x}{2}dx \\
& = \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/423757",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Given that $ x + \frac 1 x = r $ what is the value of: $ x^3 + \frac 1 {x^2}$ in terms of $r$? Given that $$ x + \cfrac 1 x = r $$
what is the value of: $$ x^3 + \cfrac 1 {x^2}$$ in terms of $r$?
NOTE: it is $\cfrac 1 {x^2}$ and not $ \cfrac 1 {x^3} $
Where I reached so far:
$$ \Big(x^3 + \cfrac 1 {x^2}\Big) + \cfrac 1... | The answer could be found by direct calculation:
$$ x = \frac{r \pm \sqrt{r^2-4}}{2}$$
$$x^3+\frac{1}{x^2} = -\frac{(- r^2 + r + 1)(r + \sqrt{r^2 - 4} + 2)}{2} \tag{$+$}$$
$$x^3+\frac{1}{x^2} = \frac{4}{(r - \sqrt{r^2 - 4})^2 + (r - \sqrt{r^2 - 4})^{\frac{3}{8}}}\tag{$-$}$$
Edit1 Please note that the domain of $r$ is
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Evaluate $\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx$ I am trying to find a closed form for
$$\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx = 0.094561677526995723016 \cdots$$
It seems that the answer is
$$\frac{\pi^2}{12}\left( 1-\sqrt{3}\right)+\log(2) \log \left(1+\sqrt{3}... | I have found an interesting reference related to this integral. It may be of interest to many users:
A restricted Epstein zeta function
and the evaluation of some definite integrals by Habib Muzaffar and Kenneth S. Williams
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/426325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "169",
"answer_count": 3,
"answer_id": 0
} |
How to derive compositions of trigonometric and inverse trigonometric functions? To prove:
$$\begin{align}
\sin({\arccos{x}})&=\sqrt{1-x^2}\\
\cos{\arcsin{x}}&=\sqrt{1-x^2}\\
\sin{\arctan{x}}&=\frac{x}{\sqrt{1+x^2}}\\
\cos{\arctan{x}}&=\frac{1}{\sqrt{1+x^2}}\\
\tan{\arcsin{x}}&=\frac{x}{\sqrt{1-x^2}}\\
\tan{\arccos{x}}... | I'll be ignoring domains and possible roots of negative numbers. (If you let $\mbox{$x\in \textbf{]}0,\pi/2[$}$ everything works fine).
Given $f\circ g$, the trick is too relate $f$ with $g^{-1}$.
I did some. You should be able to handle the remaining ones.
$\bullet \sin (\arccos (x))=\sqrt {1-(\cos (\arccos (x) ))^2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Trigonometric near-identity The parametrized curve
$$
\left( \sec\theta+\csc\theta,\ 2\sqrt{2}\csc(2\theta) \right), \qquad \frac{10}{100} \le\theta\le\frac{142}{100}
$$
looks to the naked eye like a straight line. The $y$-intercept is not $0$ and the slope is a number that I haven't tried to make sense out of (yet).... | Call $K=\csc\left(\theta\right)+\sec\left(\theta\right)$ and $W=a/\sin\left(2\theta\right)$ for some $a\gt0$ and $\theta\in\left]0,\pi/2\right[$ instead. You are interested in why $K$ and $W$ almost have a linear relationship. First, we know that $$K^2=\frac{1}{\sin^2\left(\theta\right)}+\frac{1}{\cos^2\left(\theta\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/427195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Show that $ \mathbf u^2 \mathbf v^2 = (\mathbf u \cdot \mathbf v)^2 - (\mathbf u \wedge \mathbf v)^2 $ where $ \mathbf u $ and $ \mathbf v $ are vectors. From Linear and Geometric Algebra by Alan Macdonald.
| You can use
$$
\begin{align}
(u\cdot v)^2 - (u\wedge v)^2
& = \frac{1}{4} \left( (uv+vu)^2 - (uv-vu)^2\right) \\
& = \frac{1}{4} \left(uvuv + uvvu + vuuv + vuvu - uvuv + uvvu + vuuv - vuvu \right) \\
& = \frac{1}{2} \left( uvvu + vuuv\right) \\
& = \frac{1}{2} \left( u^2v^2+u^2v^2\right) \\
& = u^2v^2
\end{align}
$$
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/427433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Show that $e^x \geq (3/2) x^2$ for all non-negative $x$ I am attempting to solve a two-part problem, posed in Buck's Advanced Calculus on page 153. It asks "Show that $e^x \geq \frac{3}{2}x^2$ $\forall x\geq 0$. Can $3/2$ be replaced by a larger constant?"
This is after the section regarding Taylor polynomials, so I h... | Use AM-GM (i. e. $a+b \ge 2\sqrt{ab})$:
$$ x+x^3/6 \ge 2\sqrt{x^4/6}=x^2\sqrt{2/3}$$
$$1+x^4/24 \ge 2\sqrt{x^4/24}=x^2\sqrt{1/6}$$
Thus for all $x \ge 0$:
$$e^x \ge 1+x+x^2/2+x^3/6+x^4/24\ge x^2(1/2+ \sqrt{2/3}+\sqrt{1/6}) \ge 3/2 x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/427500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
prove that $\cos x,\cos y,\cos z$ don't make strictly decreasing arithmetic progression let $x,y,z\in R$,and such that $$\sin y-\sin x=\sin z-\sin y\ge 0 $$ show that:
$$\cos x,\cos y,\cos z$$ don't make strictly decreasing arithmetic progression
my idea:
we have $$2\sin y=\sin x +\sin z\cdots\cdots\tag 1$$
and assum... | We have $\sin y = \sin x + p$ and $\sin z = \sin x+2p$, where $p$ is positive.
Suppose that $\cos y = \cos x - r$ and $\cos z = \cos x - 2r$ for positive $r$.
Then $\cos^2 y = 1-(\sin x+p)^2 = (\cos x - r)^2$ and $\cos^2 z = 1-(\sin x+2p)^2 = (\cos x - 2r)^2$.
So $\cos^2y = 1-\sin^2 x-p^2-2p\sin x = \cos^2 x - 2r\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/428748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Showing that $\lim\limits_{n\to\infty}x_n$ exists, where $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$ Let $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$
a) Show that $x_{n} < x_{n+1}$
b) Show that $x_{n+1}^{2} \leq 1+ \sqrt{2} x_{n}$
Hint : Square $x_{n+1}$ and factor a 2 out of the square root
c) ... | Hints: Induction on
$$\bullet\;\;x_n<x_{n+1}\iff 1+\sqrt{2+\sqrt{3\ldots+\sqrt n}}<1+\sqrt{2+\sqrt{3+\ldots+\sqrt{n+\sqrt{n+1}}}}\iff$$
$$2+\sqrt{3+\ldots+\sqrt n}<2+\sqrt{3+\ldots\sqrt{n+1}}\iff\ldots$$
$$\bullet\bullet\;x_{n+1}^2=1+\sqrt{2+\sqrt{3+\ldots+\sqrt{n+1}}}\le 1+\sqrt2\left(\sqrt{1+\sqrt{2+\ldots+\sqrt n}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/428841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Seeking a combinatorial proof of the identity $1+2\cdot 2^1+3\cdot 2^2+\cdots+n\cdot 2^{n-1}=(n-1)2^n+1$ I would appreciate if somebody could help me with the following problem
Q: show that combinatoric identity (using by combinatorial proof)
$$1+2\cdot 2^1+3\cdot 2^2+\cdots+n\cdot 2^{n-1}=(n-1)2^n+1$$
| Combinatoric interpretation.
Sub-task:
There is a ordered sequence (array, vector, chain, cortege, ...) of $k$ balls.
1 ball $-$ red, other balls $-$ black or white.
Let $M(k)$ $-$ number of such arrays.
$M(k) = k \cdot 2^{k-1}$.
Example for $k=3$:
$\Huge{\color{red}\bullet} \circ\circ$,
$\Huge{\color{red}\bullet} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/429084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$ Problem statement:
Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $.
, $n\in \mathbb{N}$
My progress
LHS is e... | Suppose that $P(n)$ which is $\frac{1}{2}\frac{3}{4} \dots \frac{2n-1}{2n} \lt \frac{1}{\sqrt{3n+1}}$. Now $P(n + 1)$ would be $\frac{1}{2}\frac{3}{4} \dots \frac{2n-1}{2n}\frac{2n+1}{2n+2} \overset{?}{\lt} \frac{1}{\sqrt{3n+4}} $. Using the induction hypothesis it would be enough to prove that $\frac{1}{\sqrt{3n+1}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/431234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
On the period of the decimal representation of $n$ when $\gcd(n, 10) \neq 1$ Suppose that $n = 2^a5^bm$, where $n > m > 1$ are integers, with $\gcd(m, 10) = 1$, and $a, b$ are non-negative integers.
How does one show that the lengths of the periods of the decimal expansions of $1/n$ and $1/m$ are the same?
I know that... | [Originally, some of the content below was included as a sort of appendix to the original question, but once I managed to complete the proof (I think), I decided to post it as an answer.]
If the length of the period of the decimal expansion of $m$ is $r > 0$, then $10^r\equiv 1\;(\!\!\!\!\mod m)\;$, and furthermore, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the minimum of this expression This is a problem in my exam and I can't find the solution using elementary inequality knowledge. Can anyone here help me solve this. Thanks
$a,b,c $ are positive real numbers which satisfy $(a+c)(b+c) = 4c^{2}$. Find the minimum of this expression:
$$P = \frac{32a^{3}}{(b+3c)... | We begin with the observation that the change $a=xc, b=yc$ reduces the problem under consideration to the two-dimensional optimization problem
$$\min \frac {x^3} {(y+3)^3} + \frac {y^3} {(x+3)^3}-\sqrt{x^2+y^2}$$
under the constraints $$(x+1)(y+1)=4, x \ge 0, y \ge 0 . $$
Solving it with Mathematica by
Minimize[ 32 x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/436066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$p \mid x^2 +n\cdot y^2$ and $\gcd(x,y)=1 \Longleftrightarrow (-n/p) = 1$ Let $n$ be a nonzero integer, let $p$ be an odd prime not dividing $n$. then $ p \mid x^2 + n\cdot y^2$ and $x,y$ co-prime $ \Longleftrightarrow(-n/p) = 1 $
How can i prove this? by $(-n/p)$ i mean the Legendre symbol.
For $\implies$ i have alre... | Hint: We have $x^2\equiv -ny^2\pmod{p}$. Multiply both sides by $z^2$, where $z$ is the multiplicative inverse of $y$.
Detail: We need to be careful about the statement of the theorem. So we break up the statement and proof into two parts. When we do, we will discover that the result is stated somewhat too informally.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/437720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Proving by induction: $2^n > n^3 $ for any natural number $n > 9$ I need to prove that $$ 2^n > n^3\quad \forall n\in \mathbb N, \;n>9.$$
Now that is actually very easy if we prove it for real numbers using calculus. But I need a proof that uses mathematical induction.
I tried the problem for a long time, but got stuc... | For another way just using $n>9$,
note that when $n=10$, $2^n = 1024 > 1000 = n^3$. Now suppose that $2^n>n^3$ for $n>9$. Then,
$\begin{align*}
2^{n+1} &= 2\cdot2^n \\
&>2n^3 \\
&= n^3 +n^3 \\
&> n^3 + 9n^2 \\
&= n^3 + 3n^2 + 6n^2 \\
&>n^3 + 3n^2 +54n \\
&=n^3+3n^2+3n +51n\\
&>n^3+3n^2+3n+1 \\
&= (n+1)^3.
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
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How can I find a number $n$ such that $(c + d \times n)$ is a multiple of $(a + n)$? As an example, let $a=5$, $c=246$, and $d=316$. My goal is to find a positive integer $n$ such that
$$(a + n) | (c + d \times n)$$
$$(5 + n) | (246 + 316 \times n)$$
I know that one solution is $18$, since
$$(5 + 18) | (246 + 316 \tim... | Make the ansatz
$$d\cdot n + c = (d+k)\cdot(n+a).$$
Expanding the right hand side, you get
$$\begin{align}
d\cdot n + c &= d\cdot n + k\cdot n + d\cdot a + k\cdot a\\
c - d\cdot a &= k\cdot (n+a).
\end{align}$$
So you need to factorise $\lvert c - d\cdot a\rvert$ (if that is $\neq 0$, if $c = d\cdot a$, then all $n$ wo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all real numbers such that $\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$ Find all real numbers such that
$$\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$$
My attempt to the solution :
I tried to square both sides and tried to remove the root but the equation became of 6th degree.Is there an easier metho... | Let's multiply
$$
\sqrt{x-1/x} + \sqrt{1 - 1/x} = x\tag{1}
$$
by $(\sqrt{x-1/x} - \sqrt{1 - 1/x})$. Then we get
$$
x-1=x(\sqrt{x-1/x} - \sqrt{1 - 1/x})
$$
$$
\sqrt{x-1/x} - \sqrt{1 - 1/x}=1-1/x\tag{2}
$$
Sum up $(1)$ and $(2)$ to get
$$
2\sqrt{x-1/x}=x-1/x+1
$$
Now make the substitution $y=\sqrt{x-1/x}$. The rest is cl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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Solve the equation $z^3=z+\overline{z}$ I have been trying to solve an equation $z^3=z+\overline{z}$, where $\overline{z}=a-bi$ if $z=a+bi$. But I cant find any clues on how to move forward on that one. Please help.
| Note that if $z=a+ib$ then $z+\overline{z}=2a$ and
$$
z^3=2a
$$
The cube roots of $2a$ are
\begin{align}
\sqrt[3]{2a}\cdot\omega^{0}
=
&
\sqrt[3]{2a}
\\
\sqrt[3]{2a}\cdot\omega^{1}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)
\\
\sqrt[3]{2a}\cdot\omega^{2}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/440000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 1
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How prove this $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge |(a-1)(b-1)(c-1)|+a+b+c$ let $a,b,c$ are positive numbers,and such $abc\le 1$,prove that
$$\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\ge |(a-1)(b-1)(c-1)|+a+b+c$$
| case 1:
For case $a<1,b<1,c<1$, it is trivial the inequality work(Sudeep's comments)
case 2:
now we prove when one of $a,b,c$ equals $1$,the inequality also works.
WLOG,Let $c=1$, the inequality $\iff b^2+a \ge ab^2+ab$, let $ab=k \le 1 \iff b^3-kb^2-kb+k \ge0 \iff b^3(1-k)+k(b-1)^2(b+1)\ge0$
so the "=" will hold whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/441336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Finding $a + b + c$ given that $\;a + \frac{1}{b+\large\frac 1c} = \frac{37}{16}$ Please help me to find the needed sum:
If $a,b,c$ are positive integers such that $\;a + \dfrac{1}{b+\large \frac 1c} = \dfrac{37}{16},\;$
find the value of $\;(a+b+c)$.
Thanks!
| $$\;\color{blue}{\bf a} + \dfrac{1}{\color{red}{\bf b}+1/\color{green}{\bf c}} = \dfrac{37}{16} = 2 + \dfrac 5{16} = 2 + \dfrac 1{\frac{16}{5}} = \color{blue}{\bf 2} + \dfrac 1{\color{red}{\bf 3} + 1/\color{green}{\bf 5}}$$
Now, match up: $\;\color{blue}{\bf a = \;?}\;\quad \color{red}{\bf b = \;?},\quad \color{green}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/441403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x,y,z>0$ are distinct and $x+y+z=1$ what is the minimum of $\left((1+x)(1+y)(1+z)\right)/\left((1-x)(1-y)(1-z)\right)$? If $x,y,z>0$ are not equal and positive and if $x+y+z=1$ the expression
$$\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}$$
is greater than what quantity?
| we prove
$$\dfrac{1+x}{1-x}\cdot\dfrac{1+y}{1-y}\cdot\dfrac{1+z}{1-z}\ge 8,x+y+z=1$$
$$\Longleftrightarrow \sum \ln{\left(\dfrac{1+x}{1-x}\right)}\ge 3\ln{2}$$
and we have
$$\ln{\left(\dfrac{1+x}{1-x}\right)}\ge\dfrac{9}{4}x-\dfrac{3}{4}+\ln{2}$$
poof:let
$$G(x)=\ln{\left(\dfrac{1+x}{1-x}\right)}-\left(\dfrac{9}{4}x-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/442136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $a,b,c > 0$ satisfy $a^2+b^2+c^2=3$ then $\frac{a}{b+c+3}+\frac{b}{a+c+3}+\frac{c}{a+b+3} \geq \frac{3}{5}$
Given $a,b,c > 0$ satisfying the condition $$a^2+b^2+c^2=3,$$
prove that
$$\frac{a}{b+c+3}+\frac{b}{a+c+3}+\frac{c}{a+b+3} \geq \frac{3}{5}.$$
Thank you all
| The inequality is not TRUE. For example. $a=\sqrt{2}$, b=1, c=0, then
$\frac{\sqrt{2}}{4} + \frac{1}{3+\sqrt{2}} + 0 = 0.3536 + 0.2265 < 0.6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/443214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Die Roll Probability If Zachary rolls a fair die five times, what is the probability that the sum of his five rolls is 20?
1st I did:
Patterns of 5 that can give us 20
66611, 66521, 65531, 65522, 64442, 64433, 64415,
Where do we go from here?
| Assuming that you know generating functions.
We are interested in the coefficient of $x^{20}$ in the expansion $(x+x^2+x^3+x^4+x^5+x^6)^5$.
This is equivalent to the coefficient of $x^{15}$ in the expansion $\left( \frac{1- x^6}{1-x} \right)^5 $
The numerator is easily evaluated as
$$ 1- - 5 x^{6} + 10 x^{12} - 10x^{18... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/447723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What rational numbers have rational square roots? All rational numbers have the fraction form $$\frac a b,$$ where a and b are integers($b\neq0$).
My question is: for what $a$ and $b$ does the fraction have rational square root? The simple answer would be when both are perfect squares, but if two perfect squares are mu... | First,
$$
\sqrt{\frac ab}\in\mathbb{Q}\implies\sqrt{ab}=b\,\sqrt{\frac ab}\in\mathbb{Q}\tag{1}
$$
According to this answer, since $\left(\sqrt{ab}\right)^2=ab\in\mathbb{Z}$, if $\sqrt{ab}\in\mathbb{Q}$, then $\sqrt{ab}\in\mathbb{Z}$. Thus, there is a $c\in\mathbb{Z}$ so that
$$
ab=c^2\tag{2}
$$
If $\gcd(a,b)=1$, then, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/448172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 8,
"answer_id": 5
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Show that $\frac {\sin(3x)}{ \sin x} + \frac {\cos(3x)}{ \cos x} = 4\cos(2x)$ Show that
$$\frac{\sin(3x)}{\sin x} + \frac{\cos(3x)}{\cos x} = 4\cos(2x).$$
| $$\dfrac{\sin(3x)}{\sin x} + \frac{\cos(3x)}{\cos x}$$
$$\dfrac{\sin(3x)\cos x + \cos(3x)\sin x}{\cos x{\sin x}}$$
$$\dfrac {\sin (3x+x)}{\sin x\cos x}$$
$$\dfrac {\sin 4x}{\sin x\cos x}$$
$$\dfrac {2\sin 2x\cos 2x}{\sin x\cos x}$$
$$\dfrac {4\sin x\cos x\cos 2x}{\sin x\cos x}$$
$$4\cos 2x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/448673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
How to solve $ (x + 2y - 4)dx - (2x + y - 5)dy = 0$ How to solve the differential equation $(x + 2y - 4)dx - (2x + y - 5)dy = 0$. It's not separable, nor exact nor homogeneous. The solution is $(x - y -1)^3 = C(x + y - 3)$. How can I achieve it?
The other equations similar to this are:
$(1+ x + y)dy - (1- 3x - 3y)dx = ... | $$ (x + 2y - 4)dx - (2x + y - 5)dy = 0 $$
Let $X = x + a$ and $Y= y+ b$ and let's see if we can choose suitable values for $a$ and $b$.
$$ (X - a + 2Y - 2b - 4)dX - (2X-2a + Y - b - 5)dY =0$$
It's easy to find what the "suitable" values should be, since we just want them to cancel out the constants! So we solve the fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/448839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How prove this $\frac{1}{\cos{A}}-\frac{\sin{\frac{A}{2}}}{\sin{\frac{B}{2}}\sin{\frac{C}{2}}}=4$ in $\Delta ABC $ not An equilateral triangle, let
$$\begin{vmatrix}
2\sin{A}\sin{C}-\cos{B}&2\sin{A}\sin{B}-\cos{C}\\
4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}+\cos{B}&4\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\si... | Hint:
Your identity is equivalent to:
$$\frac{\sin(A/2)\sin(B/2)\sin(C/2)}{\cos A} + \sin^2(A/2) = 4\sin(A/2)\sin(B/2)\sin(C/2)$$
$$\frac{1}{2}\left(1-\cos(A)\right)=\left(\cos(A)+\cos(B)+\cos(C)-1\right)\left(1 - \frac{1}{\cos(A)} \right)$$
Notice that this form is quite close to the fraction you get from the determin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that $(a-d)^2+(b-c)^2\geq 1.6$ if $a^2+4b^2=4$ and $cd=4$ Let $a, b, c, d$ be real numbers. I need to prove this innocent inequality
$$
(a-d)^2+(b-c)^2\geq 1.6
$$
if $a^2+4b^2=4$ and $cd=4$.
I was told that there exist nice and sweet elementary solutions.
| it is supplementary to Alex's method which more like high school method:
The idea is same,we use two parallel lines to calculate. let slop is $k=-\tan{ \alpha}, 0<\alpha <\dfrac{\pi}{2}$,
the tangent line to ellipse $x^2+4y^2=4$ is :$y=kx+b_1 \to x^2+4(kx+b_1)^2=4, \Delta=0,\implies b_1^2=4k^2+1$
the tangent line to H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 2
} |
A divisibility question concerning positive integers Suppose $n$ is a positive integer such that $3n+1$ and $4n+1$ are both perfect squares , then how do we prove that $7|n$ ?
| Suppose $x^2=3n+1$ and $y^2=4n+1$. Then,
$$
4x^2-3y^2=1
$$
Using continued fractions, we find the solutions to this equation are $(x_k,y_k)$ where
$$
\begin{align}
(x_0,y_0)&=(1,1)\\
(x_1,y_1)&=(13,15)\\
(x_k,y_k)&=14(x_{k-1},y_{k-1})-(x_{k-2},y_{k-2})
\end{align}
$$
Looking at $(x_k,y_k)\text{ mod }7$, we see
$$
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/450254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Gaussian Elimination with Scaled Row Pivoting for numerical methods I am solving a system first with basic Gaussian Elimination, and then Gaussian Elimination with scaled row pivoting (used in numerical methods)
Basic Gaussian Elimination on the system $Ax=b$:
\begin{equation} ... | You miscomputed $x_2$ in the back substitution of the row-pivoted system, that's the origin of the discrepancy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/454356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve for c , $y = x + c \big( \frac{mx}{c} + s \big)^a$ I get this equation
$y = x + c \big( \frac{mx}{c} + s \big)^a$
how can I get the $c$ or $m$ ?
I try with $\ln$
$\ln\big(\frac{y-x}{c}\big) = a \ln \big( \frac{mx}{c} + s\big)$
and now ?
| Here is how you solve for $m$:
\begin{align*}
\frac{y-x}{c}=\left(\frac{mx}{c}+s\right)^{a}\\
\left(\frac{y-x}{c}\right)^{\frac{1}{a}}=\left(\frac{mx}{c}+s\right)\\
\left(\frac{y-x}{c}\right)^{\frac{1}{a}}-s=\frac{mx}{c}\\
\end{align*}
Hence
\begin{align*}
m=\frac{c}{x}\left(\left(\frac{y-x}{c}\right)^{\frac{1}{a}}-s\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The values of $N$ for which $N(N-101)$ is a perfect square
For how many values of $N$ (integer), $N(N-101)$ is a perfect square number?
I started in this way.
Let $N(N-101)=a^2$
or $N^2-101N-a^2=0.$
Now if the discriminant of this equation is a perfect square then we can solve $N$.
But I can't progress further.... | We have $$N^2-101N = a^2 \implies (N-101/2)^2 - (101/2)^2 = a^2 \implies (N-101/2)^2 - a^2 = (101/2)^2$$
This gives us
$$(2N-101)^2 - (2a)^2 = 101^2 \implies (2N+2a-101)(2N-2a-101)=101^2$$
Note that $101^2$ can be written as
$$101 \times 101 = 1 \times 101^2 = 101^2 \times 1 = (-101) \times (-101) = (-1) \times (-101^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/460022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Initial value problem differential equation $y' = (x-1)(y-2)$ $$y' = (x-1)(y-2)$$
$y(2)= 4$
$$\frac{1}{y-2}dy = (x-1)dx$$
$$\int \frac{1}{y-2}dy =\int (x-1)dx$$
$$\ln(y-2) = \frac{x^2}{2} - x + c$$
$$y - 2 = e^{\frac{x^2}{2} - x + c} $$
$$y = e^{\frac{x^2}{2} - x + c} + 2$$
plug in the inital value
$$y = e^{\frac{2^2}{... | It would be a good remark that in the step you wrote:
$$\ln(y-2)=x^2/2-x+c$$ we inserted this strong point: $$\ln|y-2|=x^2/2-x+c, ~~y\neq 2$$
Now, if we assume to find just one solution and having $y>2$, then $$y=\exp(x^2/2-x+c)+2=C\exp(x^2/2-x)+2$$ and therefore when $x=2$ then $y=4$ makes the latter equality:
$$4=C\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/460608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Scaling points in circle I have a set of points that fall inside of a circle with a radius of $1$ and a center of $(0, 0)$. I want to know how to scale those points so that they all have a radius between $0.5$ and $0.75$.
For example, if I have $x = 1$, $y = 0$ then $x_1 = 0.75$, $y_1 = 0$. But that is an easy case on ... | For any point $(x,y)$, consider the transformation:
$$
f(x,y)=\left(0.25+\dfrac{0.5}{\sqrt{x^2+y^2}}\right)(x,y)
$$
Notice that the distance from this new point to the origin is:
$$
\left(0.25+\dfrac{0.5}{\sqrt{x^2+y^2}}\right)\sqrt{x^2+y^2}=0.25\sqrt{x^2+y^2}+0.5
$$
Hence, since $0 \leq \sqrt{x^2+y^2} \leq 1$, it foll... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/462254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $a$ such that $P(X\le a)=\frac 1 2$
Given probability density function $f(x)=\begin{cases} 1.5(1-x^2),&0<x<1\\0& \mbox{otherwise}\end{cases}$, calculate for which '$a$', $P(X\le a)=\frac 1 2$ (the solution is $2\cos\left(\frac{4\pi}{9}\right))$.
I calculated the integral and got $P(X\le a)=\displaystyle\int_0^af... | We have the equation $a^3-3a+1=0$. Of course it can be solved numerically. But we show how cosines get into the game.
Let $a=2b$. The equation becomes $8b^3-6b+1=0$, or equivalently $4b^3 -3b=-\frac{1}{2}$.
Recall the identity $\cos 3\theta=4\cos^3\theta-3\cos\theta$. So we can rewrite our equation as $\cos3\theta=-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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interesting Integral , alternative solution. Show the following relation:
$$\int_{0}^{\infty} \frac{x^{29}}{(5x^2+49)^{17}} \,\mathrm dx = \frac{14!}{2\cdot 49^2 \cdot 5^{15 }\cdot 16!}.$$
I came across this intgeral on a physics forum and solved by (1) making a substitution (2) finding a recursive formula (with in... | $$I_{(n,m)}=\int_0^\infty \frac{x^n}{(ax^2+b)^m} \,\mathrm dx$$
$$=x^{n-1}\int_0^\infty\frac{xdx}{(ax^2+b)^m}-\int_0^\infty\left(\frac{d(x^{n-1})}{dx}\int\frac{xdx}{(ax^2+b)^m}\right)dx $$
$$=x^{n-1}\frac1{-2a(m-1)(ax^2+b)^{m-1}}\big|_0^\infty-\frac{n-1}{-2a(m-1)}\int_0^\infty \frac{x^{n-2}}{(ax^2+b)^{m-1}}dx$$
$\lim_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
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Simplify : $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7) $ The question is to simplify $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)$ without using a calculator .
My friend has given me ... | As $$(a+b+c)(-a+b+c)=\{(b+c)+a\}\{(b+c)-a\}=(b+c)^2-a^2,$$
$$(\sqrt5 +\sqrt6 +\sqrt7)(−\sqrt5+\sqrt6+\sqrt7)$$
$$=(\sqrt6+\sqrt7)^2-(\sqrt5)^2=6+7+2\sqrt7\cdot\sqrt6-5=8+2\sqrt{42}=2\sqrt{42}+8$$
Again as $$(a-b+c)(a+b-c)=\{a+(b-c)\}\{a-(b-c)\}=a^2-(b-c)^2,$$
$$(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)=\{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/465103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 4
} |
Factorise: $2a^4 + a^2b^2 + ab^3 + b^4$ Factorize : $$2a^4 + a^2b^2 + ab^3 + b^4$$
Here is what I did:
$$a^4+b^4+2a^2b^2+a^4-a^2b^2+ab^3+b^4$$
$$(a^2+b^2)^2+a^2(a^2-b^2)+b^3(a+b)$$
$$(a^2+b^2)^2+a^2(a+b)(a-b)+b^3(a+b)$$
$$(a^2+b^2)^2+(a+b)((a^2(a-b)) +b^3)$$
$$(a^2+b^2)^2+(a+b)(a^3-a^2b+b^3)$$
At this point I don't... | Just Simple Factorisation:
$2a^4+a^2b^2+ab^3+b^4$
$=2a^4+2ab^3+a^2b^2−ab^3+b^4$
$=2a(a^3+b^3)+a^2b^2−ab^3+b^4$
$=2a(a+b)(a^2-ab+b^2)+b^2(a^2-ab+b^2)$
$=(a^2-ab+b^2)(2a^2+2ab+b^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/465136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Prove the Inequality: $\sum\frac{x^3}{2x^2+y^2}\ge\frac{x+y+z}{3}$ Let $x, y, z>0$. Prove that:
$$\frac{x^3}{2x^2+y^2}+\frac{y^3}{2y^2+z^2}+\frac{z^3}{2z^2+x^2}\ge\frac{x+y+z}{3}$$
| See my solution from 2006 here:
https://artofproblemsolving.com/community/c6h22937p427220
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/468925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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If $\sin(a)\sin(b)\sin(c)+\cos(a)\cos(b)=1$ then find the value of $\sin(c)$ If $$\sin(a)\sin(b)\sin(c)+\cos(a)\cos(b)=1,$$;where abc are the angles of the triangle.!
then find the value of $\sin(c)$. By trial and error put this triangle as right angled isosceles and got the answer..!! But i want a complete proof!
| Consider the two unit vectors on $S^2$ with spherical polar coordinates
$(a, 0)$ and $(b,\frac{\pi}{2}-c)$, its components in $\mathbb{R}^3$ are:
$$\begin{align}&(\sin a, 0, \cos a)\\
\text{ and }\quad&(\sin b \cos(\frac{\pi}{2}-c), \sin b\sin(\frac{\pi}{2}-c), \cos b)
=(\sin b\sin c,\sin b\cos c,\cos b)
\end{align}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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A $4$ variable inequality If $a,b,c,d$ are positive numbers such that $c^2+d^2=(a^2+b^2)^3$, prove that
$$\frac{a^3}{c} + \frac{b^3}{d} \ge 1,$$
with equality if and only if $ad=bc$.
Source: Don Sokolowsky, Crux Mathematicorum, Vol. 6, No. 8, October 1980, p.259.
| Cauchy-Schwarz gives
$$\left( \frac{a^3}{c} + \frac{b^3}{d}\right)(ac+bd)\ge (a^2+b^2)^2.$$
So it suffices to show that
$$(a^2+b^2)^2 \ge ac + bd.$$
Squaring, this is equivalent to showing
$$(a^2+b^2)^3(a^2+b^2) \ge(ac+bd).$$
This follows immediately from the hypothesis that $(a^2+b^2)^3=c^2+d^2$ and Cauchy-Schwarz.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/471739",
"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.