Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculating limits with Strange Rearrangement How did they get to this third step?
$\lim\limits_{x \to 3} \frac{\sqrt{x+1}-2}{\sqrt{x-2}-1}$ =
$\lim\limits_{x \to 3} \frac{(\sqrt{x+1}-2)(\sqrt{x-2}+1)}{x-3}$ =
$\lim\limits_{x \to 3} \frac{(x-3)(\sqrt{x+1}-2)}{(x-3)(\sqrt{x+1}-2)}$
| $\frac{\sqrt{x+1}-2}{\sqrt{x-2}-1} \cdot \frac{\sqrt{x+1}+2}{\sqrt{x-2}+1} \cdot \frac{\sqrt{x-2}+1}{\sqrt{x+1}+2} \\ \text{ multiply the first two fractions } \\ \text{ Also I multiplied the initial fraction you have by one } $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof by induction that $\frac1{n+1} + \frac1{n+2} + \ldots + \frac1{2n} \geq \frac7{12}$ The question is prove that for every integer greater than or equal to 2
$$\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} \geq \frac{7}{12}$$
So far I have
Base case let $p(2)$
\begin{align*}
\frac{1}{2}+ \frac{1}{2+2} + \... | The three dots $\ldots$ are an ellipsis, indicating continuation of the preceding trend of terms up to the final term.
There are only two terms in the $n=2$ case. The ellipsis form of the formula really only definitely indicates the start point, the end point and the general nature of interim points; it doesn't set a m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability Distribution sum The sum regarding probability distribution is:
There is a group of 50 people who are patriotic out of which 20 believe in non-violence. Two persons are selected at random out of them. Write the probability distribution for the selected persons who are non-violent. Also find the mean of the... | Among the $\binom{50}{2}$ possible couples we can select, $\binom{20}{2}$ of them are made of two non-violent people, $\binom{30}{2}$ of them are made of violent people and $\binom{50}{2}-\binom{30}{2}-\binom{20}{2}=20\cdot 30$ of them are mixed, so:
$$\mathbb{P}[X=0]=\frac{\binom{30}{2}}{\binom{50}{2}}=\frac{87}{245},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1160275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sum $\sum _{k=0}^{\infty } \frac{(-1)^k \psi (k+1)}{\left(k-\frac{3}{2}\right)^2}$ Is it possible to get a closed form for:
$$\sum _{k=0}^{\infty } \frac{(-1)^k \psi (k+1)}{\left(k-\frac{3}{2}\right)^2}$$ where $\psi$ is the polygamma function?
| Since $\psi(k+1)=H_k-\gamma$ our series is given by:
$$ -4+\frac{32}{9}\gamma+4\sum_{k\geq 2}\frac{(-1)^k\left(H_k-\gamma\right)}{(2k-3)^2}=-4+\frac{32}{9}\gamma-4\gamma K+4\sum_{n\geq 0}\frac{(-1)^n H_{n+2}}{(2n+1)^2}$$
where $K$ is the Catalan constant. The last series can be computed by evaluating:
$$S_1 = \sum_{n\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all integers $x$, $y$, and $z$ such that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$ Characterize all positive integers $x$, $y$, and $z$ such that:
$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$$
For example, $\dfrac{1}{x+1} + \dfrac{1}{x(x+1)} = \dfrac{1}{x}$.
| $\frac{1}{x} + \frac{1}{y}=\frac{x+y}{xy}$ then we have to found $(x,y)\in \mathbb N^{2}$ such that $x+y|xy$ then $k(x+y)=xy$ if and only if $x=\frac{ky}{y-k}$ then $k<y$ and $y-k|ky$ , then fixed y you can find a solution choosing $y-k$ as a divisor of y.(Note that if y is prime we not have solution)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
the total differential equation $(y^2z - y^3 +x^2y)dx - (x^2z +x^3 -xy^2)dy +(x^2y - xy^2)dz=0$ In solving the total differential equation $$(y^2z - y^3 +x^2y)dx - (x^2z +x^3 -xy^2)dy +(x^2y - xy^2)dz=0$$
Substituting $x =uz$ and $y=vz$ and reached at a position where, i face the following
$$(v^2 - v^3 +u^2v)du - (u^2+... | It's all equation devided by x^2 y^2.
Now we get
(z-y/x^2 +1/y)dx-(z+x/y^2 - 1/x)dy+(1/y-1/x)dz
d(x/y) +d(y/x) +z(dx-dy) +(1/y-1/x)dz
d(x/y) +d(y/x) +d(z/y) - d(z/x) = 0
x+y + log y - log x = c
x+y+ log y/ x = c
x + y + log y / x = c.
Is a required solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the maximum and minimum of $x^2+2y^2$ if $x^2-xy+2y^2=1$.
Find the maximum and minimum of $x^2+2y^2$ if $x,y\in\mathbb R$ and $$x^2-xy+2y^2=1$$
My attempt:
Clearly, since $x^2-x(y)+(2y^2-1)=0$ and $2y^2-y(x)+(x^2-1)=0$, we have that
$$\Delta_1=y^2-8y^2+4=4-7y^2\ge 0$$
and
$$\Delta_2=x^2-8x^2+8=8-7x^2\ge 0$$
so... | Since your answer is tagged precalculus, I'll stick to precalculus methods. Note that one way to solve this would be using Lagrange multipliers from calculus. That calculation is straight-forward, but the numbers get messy. (In some sense, the solution below is a Lagrange multiplier solution, but without the calculu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Maximum of a trigonometric expression Maximize $f(x)=4\sin x+48\sin x\cos x+3\cos x+14\sin^2x$ I broke it in 2 parts but then realized they don't have their maxima at the same points. So I am stuck. This is what I did. I wrote it as $$3\cos x+4\sin x+2\sin x(24\cos x+7\sin x)$$ Now we can maximize the first two terms a... | $f(x)=4\sin x+48\sin x\cos x+3\cos x+14\sin^2x$
Step 1: Take derivative and equate it to zero
$f'(x)=4\cos x+48(\cos^x-\sin^2 x)-3\sin x+14(2\sin x\cos x)$
$\implies 4\cos x + 48\cos 2x-3\sin x+14\sin 2x=0$
Step 2: Find the solution to above equation.
Step 3: It will attain a maxima if $f''(x)<0$, that means second de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Algebra Splitting fields Take the irreducible polynomial $x^3 + x^2 + 1$ over $F_2$ (field of order $2$). Find the splitting field and its roots in that field.
Where I am:
I understand what splitting fields are, and I also know that the splitting field is $F_8$ because $8=2^3$ and we can work in the field $F_2[\theta]... | in $F_2[Y]/(Y^3+Y^2+1)$, an obvious root of $P(X)=X^3+X^2+1$ is $Y$.
Another one will be $Y^2$ because in $F_2[Y]$, for any polynomial $Q$, $Q(Y^2)=Q(Y)^2$
Hence $$X^3+X^2+1=(X-Y)(X-Y^2)(X-z)$$
But looking at the coefficient of $X^2$, you get $z=Y^2+Y+1$
Or as $Y^2$ is a root $(Y^2)^2=Y^4$ is too. But $Y^4=Y^2+Y+1$.
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
AIME 2013 Solutions (divisiblity) Problem 2
Find the number of five-digit positive integers, $n$, that satisfy the following conditions:
(a) the number $n$ is divisible by $5,$
(b) the first and last digits of $n$ are equal, and
(c) the sum of the digits of $n$ is divisible by $5.$
Obviously, if $n$ satisfies divisibli... | Big hint: Note that the digits $b$ and $c$ can be chosen freely, ($100$ choices total); and then, whatever the choices for $b$ and $c$, there are $2$ choices for $d$. For instance if $b$ and $c$ are chosen to be $7$ and $6$ respectively, then $d$ could be $2$ or $7$ to make $b+c+d$ congruent to $0 \pmod{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Pell Fermat equation If $3n + 1$ and $4n + 1$ are perfect squares then $56$ divides $n$.
I somehow proved $n$ is even so proceeding further proved that $8$ divides $n$ but don't know how to manage for $7$.
| Let $3k+1=x^2,4k+1=y^2$ As you proved $8|k$.
Notice that $z^2-3y^2=1\ \ (*)$ when $z=2a$. All solutions of this Pell equation are obtained by substitution of $(a_n,b_n)$ in the equation: $(2+\sqrt{3})^n=a_n+\sqrt{3}b_n$. Because $z=2a$, w're looking for solutions with even $a_n$.
Let's prove that $a_n$ is even if and o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral: $\int_0^{1/2}\frac{\ln(1+2x)}{1+4x^2}{\rm d}x$ [Other method type question.]
Integral: $$I=\int_0^{1/2}\frac{\ln(1+2x)}{1+4x^2}{\rm d}x$$
One thing to quickly do it take $y=2x$:
$$I=\frac12\int_0^1\frac{\ln(1+y)}{1+y^2}{\rm d}y$$
I took $y=\tan z$:
$$I=\frac12\int_0^{\pi/4}\ln(1+\tan z){\rm d}z$$
Now substi... | We have
\begin{align}I &= \frac{1}{2} \int_0^1 \frac{\ln(1 + y)}{1 + y^2}\, dy\\
& = \frac{1}{2}\int_0^1\int_0^1 \frac{y}{(1 + ry)(1 + y^2)}\, dr\, dy\\
& = \frac{1}{2}\int_0^1 \int_0^1 \frac{y}{(1 + ry)(1 + y^2)}\, dy\, dr\\
&= \frac{1}{2}\int_0^1 \int_0^1 \left(\frac{r}{(1 + r^2)(1 + y^2)} + \frac{y}{(1 + r^2)(1 + y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integer solution to $19x^3-84y^2=1984$ Show that there exist no integer values $x,y$ such that $19x^3-84y^2=1984$.
Please help me in understanding no solution problems.
I tried to check the modulo $7$ of both sides but couldn't reject some cases.
| Working modulo 7,
$$
19x^3 − 84y^2 \equiv 1984 \\
5x^3 \equiv 3 \\
15x^3 \equiv 9 \\
x^3 \equiv 2 \\
$$
By inspection, $x^3 \equiv 2 \mod 7 \,$ has no solutions:
$$
\begin{array}{c|r|c}
x & x^3 & x^3\mod 7 \\
\hline
0 & 0 & 0 \\
1 & 1 & 1 \\
2 & 8 & 1 \\
3 & 27 & 6 \\
4 & 64 & 1 \\
5 & 125 & 6 \\
6 & 216 & 6 \\
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to integrate $\frac{x^n}{1+x}$ How to integrate the function
$ \frac{x^n}{1+x} $
I tried to do it by parts but it was of no use.
I did it by taking x as $-\theta$ and did the series expansion.
I need a better method.
| Write $x^n$ as $(x^n\pm 1)\mp 1$ depending on $n$ being odd or even. By integrating you will get $\pm \log(x+1)$ plus the integral of a polynomial in $x$. For instance:
$$\int\frac{x^3}{1+x}\,dx = -\log(1+x)+\int\frac{x^3+1}{x+1}\,dx = \log(1+x)+\int(x^2-x+1)\,dx\\ =-\log(1+x)+\frac{x^3}{3}-\frac{x^2}{2}+x, $$
$$\int\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Taking the cube root of a sum of radicals I am wondering how to derive the following simplification without knowing it beforehand:
$$^3\sqrt{10 + 6\sqrt{3}} = 1 + \sqrt{3}$$
After the fact, it is easy to verify algebraically. The problem arose when applying Cardano's method to solve
$$y^3 + 6y = 20$$
I was able to deri... | $N(10+6\sqrt{3})=10^2-3\cdot6^2=-8$. So in $\mathbb Z[\sqrt3]$, it is possible that $10+6\sqrt{3}$ is a perfect cube. The only possible solution to $\sqrt[3]{10+6\sqrt{3}} $ will be one with norm $-2$. That yields possible cube roots $(1\pm \sqrt{3})(2\pm \sqrt{3})^k$ for some $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Algebra of trigonometry and some limits I am singlehandedly trying to get through a book on divergent series, however one sequence is really strange and I cant seem to understand what the authors use to come to these conclusions!
Imagine the following is our sum. ($0 < x < 2{\pi}$)
As the authors attempt to determine ... | Rewrite
$$
S_n+\frac{1}{2}=\frac{\sin\left(n+\dfrac{1}{2}\right)x}{2\sin\dfrac{x}{2}}
$$
Then
$$
\sum_{m=1}^n \left(S_m+\frac{1}{2}\right)=
\frac{1}{2\sin(x/2)}\sum_{m=1}^n\sin\left(m+\frac{1}{2}\right)x
$$
or
$$
\frac{n}{2}+\sum_{m=1}^n S_m=
\frac{1}{2\sin(x/2)}\sum_{m=1}^n\sin\left(m+\frac{1}{2}\right)x
$$
that can a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the difference between $\ln(\sin\frac{x}{2}+\cos\frac{x}{2} )- \ln(\sin\frac{x}{2}-\cos\frac{x}{2})$ and $\ln(\tan x+\sec x)$ I've noticed if you ask wolfram what is the integral of $ \sec x$ they will answer
$$\ln\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right) - \ln\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)$$
... | there is no difference. here is a reason
$$\begin{align}\tan x + \sec x &= \frac{\sin x}{\cos x} + \frac 1 {\cos x}\\
& = \frac{1+\sin x}{\cos x} \\
&= \frac{\sin^2x/2 + \cos^2x/2 + 2\sin x/2 \cos x/2}{\cos^2 x/2 - \sin^2 x/2} \\
&= \frac{(\sin x/2 + \cos x/2)^2}{(\cos x/2 - \sin x/2)(\cos x/2 + \sin x/2)}\\
&=\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How do I calculate the number of permutations of the list $(6, 6 ,5, 4)$? I have the list $l = (6, 6, 5, 4)$ and want to how to calculate the possible number of permutations.
By using brute force I know that there are 12 possible permutations:
$$\{(6, 5, 6, 4),
(6, 6, 5, 4),
(5, 6, 6, 4),
(6, 4, 5, 6),
(6, 5, 4, 6)... | Imagine there are two different $6$'s, say $6_a$ and $6_b$. Then there would be $4!=24$ permutations. Now let the two $6$'s be the same, so $(6_a,7,5,6_b)=(6_b,7,5,6_a).$ This halves the number of permutations, giving the answer of $24/2=12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding the roots of $(1 + i)^{\frac{1}{4}}$ The professor says that the $n = 4$ roots of this are in the form: $\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n})$, where $k = 0, 1, 2, 3$.
So to find $\theta$, we find the $r = \sqrt{(1)^2 + (1)^2} = \sqrt{2}$ since $Re(1+i) = 1$ and $Im(1+i) = 1$. So $\s... | $1+i = \sqrt{2}\cdot {e^{i\pi/4}}= \sqrt{2}\cdot e^{i\pi/4+i2k\pi}\to \left(1+i\right)^{1/4}= 2^{1/8}\cdot e^{i(\pi/16+k\pi/2)}, k=0,1,2,3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Problem with definite integral $\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$ $$\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$$
The question says 'evaluate the integral using the suggested substitution. It gives $u=\cos x$. But I think Let $u=1-2\sin x$ is better.
$$\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin... | Well, @Nilan has the better way to go. But, here is another "Brute Force," double substitution method that works.
Write, $\sin x = \sqrt{1-\cos^2 x}$ (the positive square root is appropriate here since $0\le x\le \frac{\pi}{6}$. Then, with $u = \cos x$, $du = -\sin x dx=-\sqrt{1-u^2}du$, and the limits of integration... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\int\frac{1}{5\cos x+\sin x+7}~dx$ Evaluating $$\int\frac{1}{5\cos x+\sin x+7}~dx.$$
This can be done by substituting $$\sin x = \frac{2t}{1 + t^2}$$ and $$\cos x = \frac{1 - t^2}{1 + t^2}.$$
However after I substitute it I cannot simplify it to get anything easier to integrate.
After substituting I got: i... | Generally, consider the real function
$$f(x)=\frac{1}{a+b\cos x+c\sin x}$$
With $a^2>b^2+c^2$ so that $f$ is defined on $\mathbb{R}$.
It is not hard to check that the derivative of
$$F(x)=\frac{x}{d}+\frac{2}{d}\arctan\left(\frac{c\cos x-b\sin x}{d+a+b\cos x+c\sin x}\right)$$
with $d=\sqrt{a^2-b^2-c^2}$, is $f(x)$. So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find a segment value of Isosceles Triangle $M$ and $N$ are points from equal sides $CA$ and $CB$ from isosceles triangle $ABC$ as what $CM = CN$. Determine the segment $CM$, knowing that $AB = 2k$ and that $2P$ and $2p$ are the perimeters respectively from triangle $ABC$ and the trapeze AMNB.
Answer: $\Large CM = \frac... |
Given isosceles $\triangle ABC$,
\begin{align}
AB+BC+AC&=2P, \tag{1}\label{1}\\
AB+BN+MN+AM&=2p. \tag{2}\label{2}
\end{align}
Subtraction $\eqref{1}-\eqref{2}$ gives
\begin{align}
2x-y&=2P-2p. \tag{3}\label{3}
\end{align}
$\triangle MNC\propto\triangle ABC$, hence
\begin{align}
\frac{y}{2k}&=
\frac{x}{AC}=\frac{x}{BC}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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An identity satisfying the divisors of a positive integer I saw a hard competition problem with long and ugly proof in http://solmu.math.helsinki.fi/olympia/valmennus/2013/vt2013_12var.pdf ? The question is from Australian mathematical olympiad 1985. Is there a nice way to solve the following:
A positive integer $n$ ha... | From the equality $n = d_6^2 + d_7^2 - 1$, we see
*
*$d_6$ and $d_7$ are mutually prime,
*$d_7 \mid d_6^2 - 1 = (d_6 - 1)(d_6 + 1)$, and
*$d_6 \mid d_7^2 - 1 = (d_7 - 1)(d_7 + 1)$.
Suppose $d_6 = ab$ and $d_7 = cd$ with $1 < a < b$ and $1 < c < d$.
Then $n$ has $6$ divisors smaller than $d_6$, namely $1$, $a$, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Finding x in matrix algebra The question asked me to calculate $|A|$ for the given matrix below and state whether or not it is invertible.
$$
\left(
\begin{matrix}
1 & 0 & 2 & 1\\
0 & x & x & x\\
x & x & 3x & 1\\
0 & x & x & 0\\
\end{matrix}
\right)
$$
I g... | Okay, here's every step:
$$\left|\begin{matrix} x & x & x \\ x & 3x & 1 \\ x & x & 0\end{matrix}\right|+2\left|\begin{matrix} 0 & x & x \\ x & x & 1 \\ 0 & x & 0\end{matrix}\right| - \left|\begin{matrix} 0 & x & x \\ x & x & 3x \\ 0 & x & x\end{matrix}\right| \\ = x^2\left|\begin{matrix} 1 & 1 & x \\ 1 & 3 & 1 \\ 1 & 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving an equation $\pmod {13}$ Suppose:
$$1 + \frac12 +\frac13 + \dots + \frac1{23} = \frac{a}{23!}$$
I would like to find $a \pmod {13}$.
My attempt:
I'm attempting to use Wilson's theorem which states:
$$(n-1)!= -1 \pmod n$$
consider:
$$\begin{align}
\frac{23!a}{13} & = 23\times22\times21\dots\times14\times a\time... | We have $\frac{23!}{k}\equiv 0\,\text{(mod $13$)}$ for $k=1,2,\dots,23$ except when $k=13$.
Sou you are left with $$
\begin{align}
a&\equiv \frac{23!}{13}=12!\cdot 14\cdot 15\cdots 23\\
&\equiv (-1)\cdot 1\cdot 2\cdots 10\equiv (-1)\cdot 12! \cdot 11^{-1}\cdot 12^{-1}\\
&\equiv (-1)\cdot (-1)\cdot (-2)^{-1}\cdot (-1)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Bernoulli Number analog using Cosine I know that Bernoulli Numbers can be found with the generating function
$$\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}x^n$$
I was wondering if any work has been done using a similar equation
$$\frac{x^2}{\cos{x}-1}=\sum_{n=0}^{\infty}\frac{C_n}{n!}x^{2n}$$
I'm particularly inte... | Your sequence (negated) starts
$$
2, \frac{1}{6}, \frac{1}{60}, \frac{1}{504}, \frac{1}{3600}, \frac{1}{22176}, \frac{691}{82555200}, \frac{1}{570240}, \frac{3617}{8821612800}, \frac{43867}{414096883200}, \frac{174611}{5822264448000}, \ldots
$$
We know that
$$
\frac{1-\cos x}{x^2} = \frac{1}{2!} - \frac{1}{4!}x^2 + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Bound on the Beta function For positive integers x and y, we have that
$$
B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} = \frac{1}{x} \left( \begin{array}{c} x+y-1 \\ x \end{array} \right)^{-1} .
$$
However,
$$
\left( \begin{array}{c} a+b \\ a \end{array} \right) = \left(\frac{b+1}{1}\right)\left(\frac{b+2}{2}\right... | OK, I finally figured it out. This is for integer $x\geq 1$ and real valued $y \geq 1$, which is actually all that I needed. Note that you can swap x and y due to the symmetry of the beta function. I believe that the inequality should hold when neither $x$ nor $y$ are integers (but still at least 1), but I did not actu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
3 urns, each with 4 balls. select one ball from each Three urns are labeled $1,2,3$. Each urn contains $4$ balls labeled $1,2,3,4$. A ball is drawn from each urn such that any ball is equally likely to be drawn. The number on the ball is compared to the number of the urn.
Let $X$ be the number of balls with label highe... | Examine each urn individually, and for urn $i$ for $i\in\{1,2,3\}$ let $X_i=1$ if the the number of the drawn ball is greater than the urn label, or else $X_i=0$.
For urn $1$, the we have $P(X_1=0)=\frac{1}{4}$ and $P(X_1=1)=\frac{3}{4}$
For urn $2$, the we have $P(X_2=0)=\frac{1}{2}$ and $P(X_2=1)=\frac{1}{2}$
For ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Taylor Series for $\frac{1}{1+e^z}$ and radius of convergence I have done some manipulation and got that $$\frac{1}{1+e^z} = \sum_{n=0}^\infty \frac{n!}{n!+z^n}$$ by the fact that:
$$\frac{1}{1+e^z}= \frac{1}{1+\sum_{n=0}^\infty\frac{z^n}{n!}}=\frac{1}{2}+\frac{1}{1+z}+\frac{1}{1+\frac{z^2}{2}}+\ldots = \frac{1}{2}+\fr... | One option is: Write
$$\frac{1}{1 + e^z} \cdot (1 + e^z) = 1$$
and expand both of the factors on the l.h.s. in Taylor series" Since $e^z \sim \sum_{k = 0}^{\infty} \frac{1}{k!} z^k$, we have
$$1 + e^z = 1 + \left(1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + O(z^5)\right) = 2 + z + \frac{z^2}{2} + \frac{z^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $ \sum\limits_{n=1}^\infty \frac{1}{n^2 2^n} $
Evaluate
$$ \sum_{n=1}^\infty \dfrac{1}{n^2 2^n}. $$
I have tried using the Maclaurin series of $2^{-n}$ but it further complicated the question. Moreover, I have also tried taking help from another question when the $n^2$ is in the numerator, but no significa... | Given: $$ \text{S}= \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2.2^n} $$
Consider the following Lemma,
Lemma:
$$ \displaystyle \int_{0}^1 \dfrac{ \ln x }{x+1} \mathrm{d}x = -\dfrac{\pi^2}{12}$$
Proof:
$ \displaystyle \int_{0}^1 \dfrac{ \ln x }{x+1} \mathrm{d}x = \displaystyle \int_{0}^1 \dfrac{ -\ln x }{1-x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
How does this line came? :)
Today I started to learn algebra with my own. And the first chapter which I'm learning is Ratio. I'm kind of confuse in an example, in all the lines starting with (*). Until first three lines of the solution I understand but I didn't get how
$$\frac{ny + mz}a = \frac{lz + nx}b = \frac{mx ... | If we have:$$\frac{a}{b}=\frac{c}{d}$$then this implies:$$\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\tag{1}$$You have:$$\frac{\frac{x}{l}}{mb+nc-la}=\frac{\frac{y}{m}}{nc+la-mb}=\frac{\frac{z}{n}}{la+mb-nc}$$Since these 3 terms are all equal, we can apply (1) to any two of them. So, lets start by applying (1) to the first... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1211782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Interpretation of a combinatorial identity involving iterated binomial coefficients I am trying to find an combinatorial interpretation for the following combinatorial identity involving iterated binomial coefficients, which appeared in the November 1980 edition of The American Math Monthly:
$\dbinom{\binom{n}{b}}{2}=\... | Here is an answer using formal power series that is somewhat simpler
than the first one. We start as before:
$$\frac{1}{2}
\sum_{j=0}^b \left({b\choose j} + e_j\right)
\left({b\choose j} + e_j-1\right)
{n+b-j\choose 2b}$$
where we have $e_j = [[j\;\text{is odd}]].$
Expanding we get three pieces, the first is
$$A = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Prove that $n(r) < 2\pi \sqrt[3]{r^{2}}$
Suppose that $n(r)$ denotes the numbers of points with integer coordinates on a circle of radius $r > 1$. Prove that
$$
n(r) < 2\pi \sqrt[3]{r^{2}}
$$
What process would you use to resolve the last one?
| Label the $n(r)$ points $P_1, P_2, \ldots, P_n$ counterclockwise. A general strategy with problems regarding lattice points is to somehow use the area (using the shoelace formula with integer coordinates gives you good bounds). In this case, looking at any triangle the area will be at least $1/2.$ Now we can in particu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Using factorisation, $A=PJP^{-1}$ to compute $A^k$ Using factorisation, $A=PJP^{-1}$ to compute $A^k$, where $k$ represents an arbitrary positive integer.
$$ \begin{bmatrix} \mathbf{0} & \mathbf{1} \\ \mathbf{-1} & \mathbf{2} \end{bmatrix} = \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{1} \end{bmatri... | You can easily prove (for example with induction) that
$${\begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix}}^k=\begin{pmatrix} 1 & k \\ 0 & 1 \\ \end{pmatrix}$$
Now $A^k$ can be computed as follows:
$A^k=(S^{-1}BS)^{k}=\underbrace{S^{-1}BS\cdot S^{-1}BS\dots S^{-1}BS}_{k \text{ times }}=S^{-1}B^{k}S=S^{-1}\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to solve this limit: $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$ Compute the limit $\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$
my attempt:
I tried to multiply top and bottom by the conjugate
$$\begin{align}
\lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sq... | We have $$\frac{1}{2}=\lim_{x\rightarrow\infty}\frac{\sqrt{x}}{2\sqrt{x}\left(\sqrt{\left(1+\frac{1}{\sqrt{x}}\right)}\right)}\leq\lim_{x\rightarrow\infty}\frac{\sqrt{x}}{\sqrt{x\left(1+\frac{1}{\sqrt{x}}\right)}+\sqrt{x}}\leq\lim_{x\rightarrow\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\leq\lim_{x\rightarrow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
Indefinite Integral with "sin" and "cos": $\int\frac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $ Indefinite Integral with sin/cos
I can't find a good way to integrate: $$\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $$
| We can split this into two integrals:
$$3 \int \frac{\sin(x)}{2\sin(x)+3\cos(x)}dx + 2 \int \frac{\cos(x)}{2\sin(x)+3\cos(x)}dx.$$
Focusing on the second integral we find:
$$\int \frac{\cos(x)}{2\sin(x)+3\cos(x)}dx = \int \frac{1}{2\tan(x)+3}dx$$
Make the substitution $u=2\tan(x)+3$ which makes $$du = 2\sec^2(x)dx = 2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 1
} |
What is a closed form for $\sum_{r=1}^k \binom nr$, where $k\leqslant n$. What will be the closed form of the following equation
$$\sum_{r=1}^k C(n,r), $$ where $n$ and $k$ are positive integers with $k\leqslant n$?
| $$\begin{align}
S=\sum_{\begin{array}{c}0\le k\le n\\1\le r\le k\end{array}}\binom{n}{r}&=\sum_{k=0}^n\sum_{r=1}^k\binom{n}{r}\\
&=\left[\binom{n}{1}\right]+\left[\binom{n}{1}+\binom{n}{2}\right]\dotsm+\left[\binom{n}{1}+\dotsm+\binom{n}{n}\right]\\
&=n\binom{n}{1}+(n-1)\binom{n}{2}+(n-2)\binom{n}{3}+\dotsm+\binom{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find ways to pick 3 numbers such that their product is a multiple of 8 $Let$ $A=\{1,2,3,4,5,6,7,8,9\}$
How many ways are there to pick 3 non-repeated numbers from the set $A$ such that their product is a multiple of 8?
My attempt:
I need to fit three $2$s in the multiplication and after that I can pick whatever to get ... | To obtain a multiple of $8$, we must select numbers containing at least three factors of $2$. We can do this by selecting an $8$, by selecting a $2$ and a $4$, or by selecting a $4$ and a $6$.
Let $E$ be the event that you select $8$ and two other numbers from set $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. Let $F$ be the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Sum of square patterns Can anyone give the name of this pattern
$$136^2+137^2+138^2+139^2+140^2+141^2+142^2+143^2+144^2 =\\
145^2+146^2+147^2+148^2+149^2+150^2+151^2+152^2$$
| If we consider,
$3^2+4^2=5^2$ as triangle,
$10^2+11^2+12^2=13^2+14^2$ as pentagon,
$21^2+22^2+23^2+24^2=25^2+26^2+27^2$ as heptagon.
Then, interestingly, every structure has a right angled triangle as below.
Triangle: $3^2+4^2=5^2,$
Pentagon: $5^2+12^2=13^2,$
Heptagon: $7^2+24^2=25^2.$
The least length leg gives the co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Prove this form $2^{n+2}-7a^2_{n}$ always is square numbers. Let sequence $\{a_{n}\}$ such $a_{1}=1,a_{2}=-1$,and
$$a_{n}+a_{n-1}+2a_{n-2}=0,n\ge 3$$
show that
$2^{n+2}-7a^2_{n}$ is square numbers.
such as
$$2^3-7=1,n=1$$
$$2^4-7=9=3^2,n=2$$
$a_{3}=-1$
$$2^{5}-7=25=5^2,n=3$$
$a_{4}=3$
$$2^6-7\cdot 9=1^2,n=4$$
| One finds
$$a_n=\frac{1}{\sqrt 7i}\left(A^n-B^n\right)\ \ (n\ge 1)$$ where $A=-\frac 12-\frac{\sqrt 7}{2i},B=-\frac 12+\frac{\sqrt 7}{2i}$.
Noting that $AB=2$, one finds
$$\begin{align}2^{n+2}-7{a_n}^2&=2^{n+2}-7\cdot\frac{1}{-7}\left(A^{2n}+B^{2n}-2\cdot 2^n\right)\\&=2^{n+2}+A^{2n}+B^{2n}-2^{n+1}\\&=2^{n+1}(2-1)+A^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating limit at infinity (algebric issue) Q: evaluate $\lim_{x \to \infty}$ $ (x-1)\over \sqrt {2x^2-1}$
What I did:
when $\lim_ {x \to \infty}$ you must put the argument in the form of $1/x$ so in that way you know that is equal to $0$
but in this ex. the farest that I went was
$\lim_{x \to \infty}$ $x \over x \sq... | $$\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-2}}$$
$$\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2}x}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2(x^2-1)}}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1224180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\int^2_0 x(8-x^3)^\frac{1}{3}dx=\frac{16\pi}{9\sqrt{3}}$ I have to prove that:
$$\int^2_0 x(8-x^3)^\frac{1}{3}dx=\frac{16\,\pi}{9\sqrt{3}}.$$
I tried substituting $x^3=8u$ but I just got stuck.
Any help would be appreciated.
| By setting for first $x=2z$ we have:
$$I=\int_{0}^{2}x(8-x^3)^{1/3}\,dx = 8\int_{0}^{1}z(1-z^3)^{1/3}\,dz \tag{1}$$
and by setting $z=t^{1/3}$ we have:
$$I=\frac{8}{3}\int_{0}^{1}t^{-1/3}(1-t)^{1/3}\,dt = \frac{8}{3}\cdot\frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma(2)}\tag{2}$$
through Eul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How to compute $\frac{t^2}{t+1}$ to the form $\frac{1}{t+1} +t -1$ One of my attempts would look like below.
$\frac{t^2}{t+1}$ = $\frac{t \times t+1-1}{t+1}$ = 1+ $\frac{t-1}{t+1}$ = $\frac{t-1+1-1}{t+1} + 1$ = $\frac{-2}{t+1} +2$
Also, I put into t an arbitrary number in both equations and have gotten the same answer.... | we have $$\frac{t^2}{t+1}=\frac{t^2-1+1}{t+1}=\frac{(t-1)(t+1)+1}{t+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Possible solutions of a diophantine equation: $p^2+pq+275p+10q=2008$ What are couples of prime integers that verify this diophantine equation: $$p^2+pq+275p+10q=2008?$$
I tried to solve this equation trough the rules of modular-arithmetic. I rewrite the equation as: $$p^2+pq\equiv 2008 \pmod5.$$ The equation can be rew... | This isn't the most clever approach, but it is a universal approach to quadratic diophantine equations to start by removing linear terms (completing the square).
$$
\begin{align}
p^2+pq+275p+10q
&=2008
\end{align}
$$
Shift to remove linear terms: $p=u+a,q=v+b$
$$
\begin{align}
(u+a)^2+(u+a)(v+b)+275(u+a)+10(v+b)
&=200... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How to determine if an equation represents a cubic spline? Given the equation
$$
f(x) = \left\{
\begin{array}{lr}
2x^3+x^2+4x+5 & : 0 \le x \le 1\\
(x-1)^3 + 7(x-1)^2 + 12(x-1)+12 & : 1 \le x \le 2
\end{array}
\right.
$$
What is the process used to determine if this represents a cubic spl... | Both expression are obviously cubic. If they both have the same value and the same first and second derivatives at $x=1$, your set of expressions fits the definition of a cubic spline. A pretty simple spline, but there it is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve these equations simultaneously (trig) Solve for $ x,y: $
\begin{equation}\cos x -\cos(x+y) = 0
\end{equation}
\begin{equation}\cos y -\cos(x+y) = 0
\end{equation}
The answers are $(0, 0), (\frac{2\pi}{3}, \frac{2\pi}{3})$.
I get $(0, 0)$, but how do you get the latter?
Sorry if this is a bit basic - I don't rem... | Given
$$\cos x-\cos(x+y)=0\\
\cos y-\cos(x+y)=0$$
Subtracting the second equation from the first yields
$$\cos x=\cos y$$
Recall that $\cos x=\pm\sqrt{1-\sin^2x}$ and $\cos y=\pm\sqrt{1-\sin^2x}$.
This now becomes $\sqrt{1-\sin^2x}=\pm\sqrt{1-\sin^2y}$ yielding
$$\sin x=\pm\sin y$$
Recall that $\cos(x+y)=\cos x\cos y-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Working out $\tan x$ using sin and cos expansion
Using only the series expansions $\sin x = x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...$ and $\cos x = 1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ...$
Find the series expansions of the $\tan x$ function up to the $x^5$ term.
So it is:
$$
\frac {x- \dfrac{x^3} {3!} + \d... | You get that the coefficient for $x^5$ is $\dfrac {1}{5!} - \dfrac {1}{4!} + \dfrac {1}{2!2!} - \dfrac {1}{3!2!}$, which is indeed $\dfrac2{15}$. So, no need to change method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Area of triangle formed by angle bisector, altitude and median Question:-
Given a triangle ABC with side length a, b and c. Calculate the area of a triangle in terms of a, b and c formed by angle bisector from vertex A, altitude from vertex B and median from vertex C.
looking for solution using complex number, vectors ... |
Given $\triangle ABC$, $BC=a,\ AC=b,\ AB=c$;
$\angle BAD=\angle CAD$,
$BE\perp AC$, $AF=BF$,
$\angle BAC=\alpha$,
$\angle BCA=\gamma$.
First, define coordinates of the points $A,B,C,E$ and $F$:
\begin{align}
A &=(0,0),\\
B &=(c\cos\alpha,c\sin\alpha), \\
C &=(b,0),\\
E &=(c\cos(\alpha),0),\\
F &=(c/2\cos\alpha,c/2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Complex Numbers Question, IIT JEE [2006]. Please tell me whether I solved it properly? $Q.$The value of $\sum\limits_{k=1}^{10}(\sin{\frac{2k\pi}{11}-i\cos\frac{2k\pi}{11}})$ is-?
I solved it like this-
$\frac{\sum\limits_{k=1}^{10}(\cos{\frac{2k\pi}{11}+i\sin\frac{2k\pi}{11}})}{i}$
If we observe these are the roots o... | $\begin{array}{l}
\\
\sum \limits^{10}_{k=1}(\sin \ \frac{2k\pi }{11} -i\cos \frac{2k\pi }{11} )\\
=(-i)\sum \limits^{10}_{k=1}(\cos \frac{2k\pi }{11} +i\sin \frac{2k\pi }{11} )\\
=(-i)(\sum \limits^{10}_{k=0}(\cos \frac{2k\pi }{11} +i\sin \frac{2k\pi }{11} )-1)\\
=(-i)(-1)\\
=i\\
Note:
\sum \limits^{10}_{k=0}(\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
this sum inequality $\sum_{i=1}^{n}\frac{1}{4i(i+1)-1}<\frac{2}{7}$ show that
$$\sum_{i=1}^{n}\dfrac{1}{4i(i+1)-1}<\dfrac{2}{7}\tag{1}$$
we have
$$4i(i+1)-1>4i^2$$
But
$$\sum_{i=1}^{n}\dfrac{1}{i^2}<\dfrac{8}{7}$$ it is clear not hold, because
$$\sum_{i=1}^{n}\dfrac{1}{i^2}>1+\dfrac{1}{4}>1+\dfrac{1}{7}=\dfrac{8}{7}$$
... | Your estimation of denominator wasn't enough to bound the sum. I have noticed that using $4k^2 + 4k- 8$ your method will work.
\begin{align*}
\sum_{k=1}^n \frac{1}{4k^2+4k-1} & \le \frac 19+\sum_{k=2}^n \frac{1}{4k^2+4k-8} = \frac 19+\sum_{k=2}^n \left(\frac{1}{12(k-1)} - \frac{1}{12(k+2)}\right) \\
& = \frac 19 + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Unilateral Z-Transform of sen($\frac{\pi}{4}n$) I was calculating the Unilateral Z-Transform of $sen(\frac{\pi}{4}n)$
I calculated it this way, but then the calculation become very difficult and i think i have made some mistakes.
It is : $\sum_{n=0}^{+\infty} sen(\frac{\pi}{4}n)z^{-n}$
$sen(\frac{\pi}{4}n) $ is periodi... | For the series
\begin{align}
f(t) = \sum_{n=0}^{\infty} \sin\left(\frac{n \pi}{4} \right) \, t^{n}
\end{align}
then it is seen that
\begin{align}
f(t) &= \sin\left(\frac{\pi}{4} \right) \, t + \sin\left(\frac{2 \pi}{4} \right) \, t^{2} + \sin\left(\frac{3 \pi}{4} \right) \, t^{3} + \cdots \\
&= \sum_{n=0}^{\infty} \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral by using substitution (How to proceed?) Using the substitution $x=a\sin\theta$, or otherwise, find $\int\frac{1}{x^2\sqrt{a^2-x^2}}dx$.
My attempt,
$x=a\sin\theta$
$dx=a\cos (\theta)d\theta$. Then $\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin ^2(\theta)}$
The given answer is $-\frac{\sqrt{a^2-x^2}}{a^2x}+c$
How to proce... | Other possibility:
Use $x=1/y$, the integral becomes:
$$
I(a)=\int -\frac{1}{y^2}\frac{1}{\frac{1}{y^2} \sqrt{a^2-\frac{1}{y^2}}}dy=-\int \frac{y}{\sqrt{a^2y^2-1}}dy=-\frac{1}{a^2}\int \partial_y\left(\sqrt{a^2y^2-1}\right)dy=-\frac{\sqrt{a^2y^2-1}}{a^2}+C
$$
Resubstitute:
$$
I(a)=-\frac{\sqrt{a^2-x^2}}{ a^2 x}+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1244632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove this inequality $25ab+25a+10b\le38$ let $a,b>0$,and such $a^2+b^2=1$,show that
$$25ab+25a+10b\le38$$
Now I have found this inequality $"="$,if and only if $a=\dfrac{4}{5},b=\dfrac{3}{5}$
then How to prove this inequality by AM-GM or other ?
| Given the solution $a=4/5,b=3/5$ gives equality, try $$a=\frac45\cos\theta-\frac35\sin\theta,b=\frac35\cos\theta+\frac45\sin\theta$$
(sorry, my comment was wrong at first) I think it becomes
$$12(\cos^2\theta-\sin^2\theta)+26\cos\theta-7\sin\theta+7\cos\theta\sin\theta)\leq38\\
24(\cos^2\theta-1)+26(\cos\theta-1)+7\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Derivative of $\frac{1}{2} x + x^2 sin{\frac{1}{x}}$ How do you evaluate the derivative of:
$h(x) = \frac{1}{2} x + x^2 \sin{\frac{1}{x}}$ at $x=0$? It seems that the answer is $\frac{1}{2}$ but I don't know how to take care of the $x^2 \sin{\frac{1}{x}}$.
When you differentiate $h$, you get $h'(x) = \frac{1}{2} + 2x\s... | You're right that $x^2 \sin\left(\frac 1 x\right)$ is not defined at $x=0$. But we can get over this discontinuity.
Think about it like we're approaching $x=0$ from the left and the right. That's like taking:
$$\lim_{x\to 0} \left|x^2 \sin\left(\frac 1 x\right)\right|$$
Here's the big trick: we know that $\sin$ can on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1246146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Simplifying Square Roots Frustration Okay, I'm really frustrated with this.
So, when you have $3 \sqrt 5 + 5 \sqrt 5$,
you get $8\sqrt5$, right?
Okay, so what do I do for here:
$\sqrt{11} - 3 \sqrt{11}$
Is it just $-3 \sqrt{11}$ ?
What about for $6 \sqrt 2 + 4 \sqrt{50}$ ?
Do I multiply the $6 \sqrt{2}$ so that I can... | Compare:
\begin{align}
x - 3 x
&= 1x - 3 x \\
&= (1 - 3)x \\
&= -2 x
\end{align}
to
\begin{align}
\sqrt{11} - 3 \sqrt{11}
&= 1\sqrt{11} - 3 \sqrt{11} \\
&= (1 - 3)\sqrt{11} \\
&= -2 \sqrt{11}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How prove that $\frac{1}{\sin^2\frac{\pi}{2n}}+\frac{1}{\sin^2\frac{2\pi}{2n}}+\cdots+\frac{1}{\sin^2\frac{(n-1)\pi}{2n}} =\frac{2}{3}(n-1)(n+1)$ How prove that sum
$$\frac{1}{\sin^2\frac{\pi}{2n}}+\frac{1}{\sin^2\frac{2\pi}{2n}}+\cdots+\frac{1}{\sin^2\frac{(n-1)\pi}{2n}} =\frac{2}{3}(n-1)(n+1)$$
| To develop further on Jean-Claude Arbaut's comment, you'll find here the proof that:
$$\sum_{k=1}^{n-1}{\frac{1}{\sin^2\left(\frac{k\pi}{n}\right)}}=\dfrac{(n-1)(n+1)}{3}$$
All you have to do now, is show that:
$$\sum_{k=1}^{n-1}{\frac{1}{\sin^2\left(\frac{k\pi}{\color{red}{2n}}\right)}}=\color{red}{2}\sum_{k=1}^{n-1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Compute the values of two infinite products whose factors are the same I have the following question:
How to prove that
$(1-\frac{1}{2})\cdot (1+\frac{1}{3})\cdot (1-\frac{1}{4})\cdot (1+\frac{1}{5})\cdot (1-\frac{1}{6})\cdot (1+\frac{1}{7})\cdot (1-\frac{1}{8})\cdot\ ...\ = \frac{1}{2}\ $, but
$(1-\frac{1}{2})\cdot (... | For the first product note that, taking $k\geq3
$ $$\left(1+\frac{1}{k}\right)\left(1-\frac{1}{k+1}\right)=1
$$ then $$\prod_{k\geq2}\left(1+\frac{\left(-1\right)^{k+1}}{k}\right)=1-\frac{1}{2}=\frac{1}{2}.
$$ For the second product, which can be written as kennytm suggest, we have $$\prod_{k\geq1}\left(1-\frac{1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $a^2+b^2+c^2\geq [2(a-b)^2(b-c)^2(a-c)^2]^{1/3}$ Mathematica seems to know that this statement is true, yet I am struggling to prove it. Possible useful inequalities are Minkowski and the geometric mean. Using the geometric mean inequality I can prove a less strict bound:
$$(a-b)^2+(b-c)^2+(a-c)^2 \geq 0 \im... | Hint:
WOLOG, suppose $a>b>c$. Let $b=c+y$ and $a=c+y+x$. One needs to show
$$
(c+x+y)^2+(c+y)^2+c^2 \ge (2x^2y^2(x+y)^2)^{\frac{1}{3}}.
$$
Notice that $(c+x+y)^2+(c+y)^2+c^2$ attains minimal when $c=-\frac{x+2y}{3}$. It remains to show
$$
(x+2y)^2+(x-y)^2+(2x+y)^2 \ge 9 (2x^2y^2(x+y)^2)^{\frac{1}{3}},
$$
or
$$
x^2+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $2^{12^{7}+3}\equiv x \pmod{36}$, then what is the value of $x$?
If $2^{12^{7} + 3} \equiv x \pmod{36}$, then what is the value of $x$?
We have:
$$
\begin{align}
2^5 & \equiv - 4 \pmod{36} \\
2^{10} & \equiv 16 \pmod{36} \\
2^{12} & \equiv - 8 \pmod{36} \\
(2^{12})^{12} & \equiv 8^{12} \pmod{36} \\
2^{12^2} & \equ... | We need $x \in \{0,1,2,\ldots,35\}$ such that $2^{12^7+3} = 36M + x$. Note that $4$ divides $2^{12^7+3}$ and $36M$. Hence, $x$ must be a multiple of $4$, say $4y$. We now need $y$ such that
$$2^{12^7+1} = 9M + y$$
where $y \in \{0,1,2,\ldots,8\}$. Since $\gcd(2,9) = 1$, we have $$2^{\phi(9)} \equiv 1 \pmod9 \implies 2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $x$ be an integer and $n$ be a positive integer. Find the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$. I need help proving the following: Let $x$ be an integer and $n$ be a positive integer. Find the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$. I know that the smallest $n$ is 8 by te... | It is well-known that $x^4+4=\Big(x^2\Big)^2+2^2+\underbrace{2\cdot2x^2-2\cdot2x^2}_0=\Big(x^2+2\Big)^2-4x^2=$
$=\big(x^2-2x+2\big)\big(x^2+2x+2\big)$. However, since $x^2\pm2x+2=\pm1$ has as viable integer
solutions $x=\pm1$, we must extend the search to polynomials of the form $x^2+4a^4$, which
has no such solution... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Domain of the function $f(x) = \sqrt{\frac{3^x-4^x}{x^2-4x-4}}$ will be? I tried solving this question by
$1.$ $-1$ and $4$ will not be in domain because denominator can not be zero .
$2.$ Either both denominator and numerator will be positive or negative so that whole term in root becomes positive.
But I am not able... | To obtain the domain of $f(x) = \sqrt{\frac{3^x-4^x}{x^2-4x-4}}$, the expression inside the radical must be positive and the denominator of the expression must be nonzero.
The numerator $3^ x - 4^x$ is positive when $x < 0$. The denominator $x^2-4x-4$ is positive when $x > 2(1+ \sqrt{2})$ or when $x < 2 (1 - \sqrt{2})$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find a projective transformation that sends the locus $x^2-y^2=z^2 $ to $yz=x^2$ So I'm trying to find a function that sends $x^2-y^2=z^2$ to $yz=x^2$. I know it can be done because all conics are projectively equivalent. I think I have to use a matrix of some kind but I don't know how.
Thanks for any help.
| In general you have three degrees of freedom when mapping one conic to another. One approach is the following: choose three distinct points $A,B,C$ resp. $A',B',C'$ on each conic. Then construct a point $D$ as follows: take the tangents to the first conic in $A$ and $B$, intersecting these in a point $P$ and join that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that $\dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$ is an algebraic integer. Let $m$ be an integer such that $m \equiv 2 \pmod 3$. Show that the number $$\dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$$ is an algebraic integer.
The usual technique, doing $x = \dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$ and trying to find an algebraic exp... | Let $x=\frac {m-\sqrt [3] 2}{\sqrt [3] 3}$
Let $m=3k+2$
Then $x=\frac {3k+2-\sqrt [3] 2}{\sqrt [3] 3}$
$x\sqrt [3] 3=3k+2-\sqrt [3] 2$
$x\sqrt [3] 3+\sqrt [3] 2 =3k+2$
$\left( x\sqrt [3] 3+\sqrt [3] 2 \right)^3 =\left(3k+2 \right)^3$
$3x^3+3\sqrt [3] {18}x^2+3\sqrt [3] {12}x+2=27k^3+54k^2+36k+8$
... no not getting anyw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove that: $2(ab+bc+ca)-a^2-b^2-c^2\le6$. Let $a,b,c>0$ such that: $\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}=1$.
Prove that: $2(ab+bc+ca)-a^2-b^2-c^2\le6$.
I have no idea for solve this problem.
| $2=\dfrac{a^2}{1+a^2}+\dfrac{b^2}{1+b^2}+\dfrac{c^2}{1+c^2}\ge \dfrac{(a+b+c)^2}{a^2+b^2+c^2+3} \iff 2(a^2+b^2+c^2+3)\ge (a+b+c)^2 \iff 2(ab+bc+ac)-a^2-b^2-c^2\le6 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Question regarding proving by induction I am struggling with a math problem I have been assigned. The problem is as follows:
Let $X_1 = -3$ and $X_2 = 0$. Given that for every natural number $n \geq 2, X_{n+1} = 7X_n - 10X_{n-1}$, prove by induction that for every $n$ belonging to $\mathbb{N}$, $X_n = 2 \cdot 5^{n-1} -... | Start with your recurrence relation and substitute your inductive assumption:
$$X_{n+1} = 7(2 \cdot 5^{n-1} - 5 \cdot 2^{n-1}) - 10(2 \cdot 5^{n-2} - 5 \cdot 2^{n-2}) \\
= 14 \cdot 5^{n-1} - 35 \cdot 2^{n-1} - 20 \cdot 5^{n-2} + 50 \cdot 2^{n-2} \\
= 10 \cdot 5^{n-1} - 10 \cdot 2^{n-1} \\
= 2 \cdot 5^n - 5 \cdot 2^n.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solvability of the equation $2a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}$ when $n \geq 5$ is prime? As a natural extension of the question titled
Solvability of $a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}$ when $n \geq 5$ is prime?,
I wonder if the equation
$$2a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}$$
is so... | I do not know why co-prime solutions are more interesting and must these be mutual co-prime or co-prime altogether, but there are few mutual co-prime ones $a_1,a_2,a_3<1000$ (not sure all of these found due to precision trunks):
$$\begin{align}4^5+239^5+659^5=2\cdot 7907819^2\\
40^5+ 617^5+ 633^5= 2\cdot 9773625^2\\
7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$17!=3556xy428096000,$then $(x+y)$ equals?(without using a calculator) $17!=3556xy428096000,$then $(x+y)$ equals?
a)$15$ b)$6$ c)$12$ d)$13$
With help of calculator $(x+y)$ can be easily calculated as $15$.But without a calculator,I can only conclude that the sum of digits $(3+5+5+6+x+y+4+2+8+9+6)$ is a multiple of $3$... | Actually, the number was copied incorrectly (somehow) and it was driving me crazy trying to figure out what I was doing wrong.
Since $17 > 13$, $17!$ must be divisible by $7 \cdot 11 \cdot 13 = 1001$. That means if we take the digits preceding the trailing zeros, and place them in groups of three, like so—$355, 6\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
prove an identity involving beta function and gamma function We know that $B(p,q)=\Gamma(p)\Gamma(q)/\Gamma(p+q)$ where $p, q>0$,
and $B(p,q)$ is related to binomial coefficients if one of $p,q$ is an integer. I want to prove the following identity.
$$\frac{\Gamma(b)}{\Gamma(b+2n+3)}\sum_{k=0}^n \frac{(-1)^{k+1}\Gamma(... | Suppose we are interested in the value of
$$S_b(n) =
\sum_{k=0}^n
\frac{(-1)^{k+1}}{(n-k+1)!(k+1)!}
\frac{\Gamma(b+n+k+2)}{\Gamma(b+k+1)}.$$
This is
$$\sum_{k=0}^n
\frac{(-1)^{k+1}}{(n-k+1)!(k+1)!}
(n+1)! {b+n+k+1\choose n+1}
\\ = \frac{1}{n+2} \sum_{k=0}^n
\frac{(-1)^{k+1} (n+2)!}{(n-k+1)!(k+1)!}
{b+n+k+1\choose n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do I find $\int_{0}^{\infty} \frac{\sin^4 x}{x^2}\,dx$? I need help evaluating the following integral
$$\int_{0}^{\infty} \frac{\sin^4 x}{x^2}\,dx$$
which should probably be equal to $\frac{\pi}{4}$
Using some trigonometric manipulations I got $\frac{3}{8} - \frac{\cos{2x}}{2} + \frac{\cos{4x}}{8}$ which using inte... | Another way to solve this problem is to use parametrized integral: Let
$$ I(s)=\int_0^\infty e^{-sx}\frac{\sin^4x}{x^2}dx. $$
Then
\begin{eqnarray}
I''(s)&=&\int_0^\infty e^{-sx}\sin^4xdx\\
&=&\int_0^\infty e^{-sx}(\frac{3}{8} - \frac{\cos{2x}}{2} + \frac{\cos{4x}}{8})dx\\
&=&\frac{3}{8s}-\frac{s}{2(s^2+4)}+\frac{s}{8(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1276712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
System of equations involving sin and cos I'm trying to solve the following system:
$$
\sin(x) + \cos(y) = 0.6\\
\cos(x) - \sin(y) = 0.2\\
$$
Solving for y in terms of x:
$$
y=\arccos(0.6-\sin(x))=\arcsin(\cos(x) -0.2)
$$
Therefore:
$$
\frac{\pi}{2} - \arcsin(0.6-\sin(x)) = \arcsin(\cos(x) - 0.2)
$$
Forgot to add:
$$
\... | here is one way to do this.
$$\begin{align}\sin x + \cos y &= 0.6\\
\sin (x) + \sin(\pi/2- y) &= 0.6 \\
\sin(\pi/4+x/2-y/2)\cos(x/2+y/2-\pi/4) &= 0.3\tag 1\end{align}$$
in the same way
$$\begin{align}\cos x - \sin y &= 0.2\\
\cos(x) - \cos(\pi/2- y) &= 0.2 \\
\sin(\pi/4+x/2-y/2)\sin(x/2+y/2-\pi/4) &= -0.1\tag 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Solve the inequality $2\left|x+\frac{1}{4}\right| < 9$
I keep trying to figure this out and I can't.
I tried to split up the absolute value first.
$2 (x+\frac{1}{4}) < 9$
$2+x+\frac{1}{4} < 9$
subtract $2$ from both sides? $x+\frac{1}{4} < 7$
and
$2 -(x+\frac{1}{4}) < 9$
$2-x-\frac{1}{4} < 9$
subtract $2$ from both s... | Write $2|\frac{x+1}{4}|<9$ as $\frac{-9}{2}<\frac{x+1}{4}<\frac{9}{2}.$ This gives $-18<x+1<18$ which is $-19<x<17$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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What's wrong?: Find the infinite sum $S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$ The answer I got by hand is not the same to the one I found using a spreadsheet.
$\displaystyle S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$
$\displaystyle \frac{1}{4}S =... | At first, note that $S=\dfrac{2^1-1}{4^{1-1}} + \dfrac{2^2-1}{4^{2-1}}+ \dfrac{2^3-1}{4^{3-1}}+\dfrac{2^4-1}{4^{4-1}} + \dfrac{2^5-1}{4^{5-1}}+\ldots $
So: $$S=\displaystyle\sum_{k=1}^{\infty}\dfrac{2^k-1}{4^{k-1}}?$$
$$S=\displaystyle\sum_{k=1}^{\infty}\dfrac{2^k}{4^{k-1}}-\dfrac{1}{4^{k-1}}$$
$$S=\displaystyle\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Integrate $\int_{-\infty}^\infty \frac{x}{\sinh x}~dx$ From a Problem Set on residues:
Evaluate $$\int_0^\infty \frac{\log x}{(x-1) \sqrt{x}}~dx.$$
After the substitution $x = e^u$ and easy computations, the integral becomes $$\int_{-\infty}^\infty \frac{x}{\sinh x}~dx.$$
| We have
$$\int_1^{\infty} \dfrac{\log(x)}{(x-1)\sqrt{x}}dx = \int_1^0 \dfrac{\log(1/x)}{(1/x-1)1/\sqrt{x}} \left(-\dfrac{dx}{x^2}\right) = \int_0^1 \dfrac{-\log(x)}{(1-x)\sqrt{x}}dx = \int_0^1 \dfrac{\log(x)}{(x-1)\sqrt{x}}dx$$
Hence, the integral is
\begin{align}
I & = -2\int_0^1 \dfrac{\log(x)dx}{(1-x)\sqrt{x}} = -2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
} |
Show that $ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $ The Question reads -
$$ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $$
I tried using the obvious approach by converting $5x , 4x $ and $ 3x$ to either $2x$ or $x$ but all that seemed to do was to further complicate the fraction. Any hints wou... | Main idea: Stick to the $\cos x$ function and use brute force.
LHS:
The numerator
$=\cos5x+\cos4x$
$=\cos3x\cos2x-\sin3x\sin2x+2\cos^22x-1$
$=(\cos2x\cos x-\sin2x\sin x)(2\cos^2x-1)-(\sin2x\cos x+\sin x\cos2x)(2\sin x\cos x)+2(2\cos^2x-1)^2-1$
$=((2\cos^2x-1)\cos x-2\sin^2 x\cos x)(2\cos^2x-1)-(2\sin x\cos^2x+\sin x(2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Integration of $\frac{1}{\sin x+\cos x}$ I'm given this $\int\frac{1}{\sin x+\cos x}dx$.
My attempt,
$\sin x+\cos x=R\cos (x-\alpha)$
$R\cos \alpha=1$ and $R\sin \alpha=1$
$R=\sqrt{1^2+1^2}=\sqrt{2}$,
$\tan\alpha=1$
$\alpha=\frac{\pi}{4}$
So, $\sin x+\cos x=\sqrt{2}\cos (x-\frac{\pi}{4})$
$\int\frac{1}{\sin x+\cos x}d... | Your computations are correct. Now, if you make the substitution $t = \tan\left(\frac{x}{2}\right)$, you get
\begin{align}
\cos x &= \frac{1-t^{2}}{1+t^{2}} \\
\sin x &= \frac{2t}{1+t^{2}} \\
dx &= \frac{2 \, dt}{1+t^{2}}
\end{align} then
\begin{align}
\int \frac{dx}{\cos x + \sin x }
&= -\int \frac{dt}{t^{2}-2t-1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Is it possible to find the sum of all integer values that $x$ can take? Is it possible to find the sum of all integer values that $x$ can take? In:
$$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$$
| Make the substitution $x=t^2+1$
$$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$$
$$\Rightarrow\sqrt{(t^2+1)+3-4\sqrt{(t^2+1)-1}}+\sqrt{(t^2+1)+8-6\sqrt{(t^2+1)-1}}=1$$
$$\Rightarrow\sqrt{t^2+1+3-4|t|}+\sqrt{t^2+1+8-6|t|}=1$$
$$\Rightarrow\sqrt{|t|^2-4|t|+4}+\sqrt{|t|^2-6|t|+9}=1$$
Make the substituion $|t|=a$
$$\Rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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A supposed to be easy calculus problem Find the values of $m$ if the line $y=mx+2$ is a tangent to the curve $x^2-2y^2=1$.
My working:
First we differentiate $x^2-2y^2=1$ with respect to $y$ to get the gradient. We get $y^2=\frac{1}{2}x^2-\frac{1}{2}\implies y=\pm\sqrt{\frac{1}{2}x^2-\frac{1}{2}}$.
We take the positive... | Let the given line be tangent to the curve at the point $(x_0, y_0).$ Then we have $$y_0 = m x_0 + 2, \\ x_0^2 = 1 + 2y_0^2.$$ Using the fist equation in the second, we have $$x_0^2 = 1 + 2 (m x_0^2 + 4 m x_0 + 4),$$ which is $$m = \frac {x_0^2 - 9} {2 x_0^2 + 8 x_0}.$$ We also know from the second equation that $$y' =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
prove using Lagrange multipliers that for $x,y>0,\space n\in \mathbb N,\space (\frac{x+y}2)^n \leq \frac{x^n+y^n}2 $ I have been asked to prove using Lagrange multipliers that for
\begin{equation*}
\space (\frac{x+y}2)^n \leq \frac{x^n+y^n}2,~x,y>0,~n\in \mathbb {N}
\end{equation*}
I am familiar with the proof by in... | Let $a=\frac{x}{x+y},b=\frac{y}{x+y}$. The inequality
$$(\frac{x+y}2)^n \leq \frac{x^n+y^n}2$$
is equivalent to
$$\frac{a^n+b^n}2\ge\frac{1}{2^n} $$
for $a,b>0$ and $a+b=1$. Let
$$ f(a,b,\lambda)=\frac{a^n+b^n}2-\lambda[(\frac{a+b}2)^n-1]. $$
Then solving
$$ \frac{\partial f}{\partial a}=\frac n2a^{n-1}-\frac{\lambda ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding value of a quadratic The polynomial
\begin{equation*}
p(x)= ax^2+bx+c
\end{equation*}
has $1+\sqrt{3}$ as one of it's roots and also $p(2)=-2$. Is there any way to know the value of $a$, $b$ and $c$? I tried but I can form only $2$ equations how can I figure out value of $3$ unknown with $2$ equations so some... | Let $x = 1+\sqrt{3} \Rightarrow (x-1)^2 = 3 \Rightarrow x^2-2x-2 = 0 \Rightarrow a = k, b = -2k,c = -2k$, and $p(2) = -2 \Rightarrow 4a+2b+c=-2\Rightarrow 4k - 4k - 2k = -2 \Rightarrow k = 1 \Rightarrow (a,b,c) = (1,-2,-2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve the following equation: $\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x= 1$ A past examination paper had the following question that I found interesting. I tried having a go at it but haven't come around with any solutions. How would one go about tackling it?
$$\sqrt {x + \sqrt {4x + \sqrt {16x ... | \begin{align*}
\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x & = 1\\
\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} & = 1+\sqrt{x}\\
x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} & = 1+x+2\sqrt{x}\\
\sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} & = 1+2\sqrt{x}\\
4x + \sqrt {16x + \sqrt {64x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Taylor series $\ln(1+e^x)$ about $x=0$ What's the best way to determine this up $x^3$ terms?
I thought it would be to take the series for $\ln(1+x)$ and the series for $e^x$ up to $x^3$ and sub the second series into the first.
$$\ln(1+x) = x - \frac {x^2}{2} +\frac {x^3}{3} + \cdots$$
$$e^x = 1+ x + \frac {x^2}{2} +\... | A standard way to obtain the Taylor series about $0$ is
$$\sum_{k=0}^{\infty} \dfrac{f^{(k)}(0)}{k!}\cdot x^k$$
Since $f(x) = \log(1+e^x)$, we have
$$f(0) = \log(1+e^0) = \log(2)$$
$$f'(0) = \left.\dfrac{e^x}{1+e^x} \right \vert_{x=0} = \dfrac12$$
$$f''(0) = \left.\dfrac{e^x}{(1+e^x)^2} \right \vert_{x=0} = \dfrac14$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the largest triangle inscribed in the unit circle Among all triangles inscribed in the unit circle, how can the one with the largest area be found?
| For the maximum area the normal height of the triangle must be maximum for a given base Hence, the triangle ABC inscribed in unit circle should be an isosceles triangle having, $\angle B=\angle C$, base BC & angle between equal sides AB & AC be $\theta$ ($=\angle A$). Now, from right triangle, we get $$\sin\theta=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 8,
"answer_id": 5
} |
Integral does not 'converge' despite describing a well-defined area.... I have almost evaluated (where all variables are real including the variable $i$)
$$
C_1\int_{a + bt^2}^{i} \frac{r \left(i-r\right)^{\frac{3}{2}\left(i-1\right)}}{\left(j+i-R-r\right)^{\frac{3}{2}\left(j+i-1\right)}}\mathrm{d} r \\ = \left[ \frac... | To answer my own question, after looking at Convergence of the improper integral $\int_{0}^{\pi/2}\tan^{p}(x) \; dx$ I tried the substitution $u = i-r$ and got
$$
\int_{a + bt^2}^{i} \frac{r \left(i-r\right)^{\frac{3}{2}\left(i-1\right)}}{\left(j+i-R-r\right)^{\frac{3}{2}\left(j+i-1\right)}}\mathrm{d} r =\\\frac{2\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
"Rationalizing the denominator" of $1/(a + b\sqrt[3]{2} + c\sqrt[3]{4})$? If $(a, b, c) \in \mathbb{Q}^3 \setminus \{(0, 0, 0)\}$, so that $a + b\sqrt[3]{2} + c\sqrt[3]{4}$ is a nonzero element of $\mathbb{Q}(\sqrt[3]{2})$, is there a formula for $${1\over{a + b\sqrt[3]{2} + c\sqrt[3]{4}}}$$ as a rational linear combin... | Hint:
The product of conjugates elements is an integer (the norm of that element), which is equal to $2$ (see Viète's formulae). The norm of this element, setting $\omega=\mathrm e^{\tfrac{2\mathrm i\pi}3}$, is:
$$(a + b\sqrt[3]{2} + c\sqrt[3]{4})(a + b\omega\sqrt[3]{2} + c\omega^2\sqrt[3]{4})(a + b\omega^2\sqrt[3]{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$ Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$
This is my try
Let $t=\sqrt{2x+3}-2\sqrt{x+1}$ or $t^2=6x+7-4\sqrt{2x^2+5x+3}=1$
The equation is equivalent to: $t^2-4(x+2)t-6x-3=0\qquad(*)$
I have no idea how to solve t... | $$(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$$
$$(x+2)(\sqrt{2x+3}-2\sqrt{x+1})=1-\sqrt{2x^2+5x+3}$$
$$(x+2)^2(6x+7-4\sqrt{(2x+3)(x+1)})=2x^2+5x+4-2\sqrt{2x^2+5x+3}$$
$$(x+2)^2(6x+7)-(2x^2+5x+4)=2(2(x+2)^2-1)\sqrt{(2x+3)(x+1)})$$
$$(6 x^3+29 x^2+47 x+24)^2=4(2x^2+8x+7)^2(2x+3)(x+1)$$
Latter can be factored as ro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Coordinates of the center of the circle I am stuck on this problem:
If the lines $y=x+\sqrt{2}$ and $y=x-2\sqrt{2}$ are two tangents of a
circle and $(0,\sqrt{2})$ lies on this circle then what is the equation of the circle?
I found out the distance between the two tangents $y=x+\sqrt{2}$ and $y=x-2\sqrt{2}$ is ... | Hint: A sketch of the situation should be helpful:
What can we say about the position of the center of a circle with those two tangents?
What is its radius?
The equation of a circle with radious $r$ and center $(x_0,y_0)$ is:
$$
(x - x_0)^2 + (y - y_0)^2 = r^2
$$
Solution:
a) Center line: The center points lie on a li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Given $\csc\theta=-\frac53$ and $\pi<\theta<\frac32\pi$, evaluate sine ,cosine, and tangent of $2\theta$ If $\csc\theta=\frac{-5}{3}$, what is the exact value of $\tan(2\theta)$, $\sin(2\theta)$, and $\cos(2\theta)$ on the interval of $\left(\pi, \frac{3\pi}{2}\right)$?
I think I'm getting the fraction negatives wrong.... | Notice that $$\csc{x} = \frac{1}{\sin{x}}$$
and so $$\csc {x} = \frac{-5}{3} = \frac{1}{\sin{x}}$$ and so $\sin{x} = \frac{-3}{5}$
now since $\color{blue}{\sin^2(x) + \cos^2(x) = 1}$ then we have that $$\big(\frac{-3}{5}\big)^2 + \cos^2(x) = 1$$ and so we get that $$\frac{9}{25} + \cos^2(x) = 1$$ and so $\cos^2{x} = 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Prove that $\sin(a)$ + $\cos(a)\leq\sqrt{2}$ $$\begin{align*}
\sin (a) + \cos(a) &\leq \sqrt{2}\\
(\sin(a)+ \cos(a))^2 &\leq (\sqrt{2})^2\\
\sin^2(a) + 2\sin(a)\cos(a) + \cos^2(a) &\leq \text{2}
\end{align*}$$
Am I doing it right? I need help.
| If @Zev Chonoles guessed right, you can prove this by computing
$$\max_{x\in [0,2\pi]} \sin x + \cos x$$
The critical points are precisely those where $\cos x - \sin x = 0$ (derivative). These are $x = \frac\pi4$ and $x = \frac{5\pi}4$. Evaluating at those and the boundary points gives
$$\sin \frac\pi4 + \cos \frac\pi4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove: $\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+b^2}}\geq 9\frac{\sqrt[3]{abc}}{a+b+c}$ Let $a,b,c\in \mathbb{R^+}$.
Prove: $\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+b^2}}\geq 9\frac{\sqrt[3]{abc}}{a+b+... | $\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}\ge 3 \sqrt[3]{\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}} \\ \iff \sqrt[3]{\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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Zeroes of polynomials and their sum Let $a, b$ are zeroes of the polynomial $x^2-10cx-11d$ and $c,d$ are the zeroes of the polynomial $x^2-10a x-11b $ where $a,b,c, d$ are distinct reals then $a+b+c+d=?$
| K.Dutta gave you the good track to follow.
From the first equation, we have $$a+b=10c\ \ \ \ (1)$$ $$a b=-11d\ \ \ \ (2)$$ From the second equation, we have $$c+d=10a\ \ \ \ (3)$$ $$c d=-11b\ \ \ \ (4)$$ Use the first and the third of these to eliminate $c$ and $d$. This gives $$c=\frac{a+b}{10}$$ $$d=\frac{99 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
I think I can complete the square of any quadratic, is it true? (Any reason to ever use Quad. Formula?) I was taught that you could only complete the square of a quadratic if the coefficient on the $x^2$ term is 1.
However, playing a little bit with other quadratics, I've found that it's just not true. Based on the CT... | The proposed way of completing the square is correct. The identity is:
$$
ax^2 + bx + c = \left(\sqrt{a}x + \frac{b}{2 \sqrt{a}}\right)^2 + \left(c-\frac{b^2}{2a}\right).
$$
One can of course also do what is more usually done and write
$$
ax^2+bx+c = a\left( x + \frac{b}{2a} \right)^2 + \left(c-\frac{b^2}{2a}\right).
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 6,
"answer_id": 0
} |
Prove factorial problem $\forall n\in\mathbb N$ $C_n=\frac{1}{n+1}\left(\begin{matrix}2n\\n\end{matrix}\right)$.
Prove that $C_{n+1}=\left(\begin{matrix}2n\\n\end{matrix}\right)-\left(\begin{matrix}2n\\n-1\end{matrix}\right)$
I tried many ways but got nothing. Appreciate any tips.
My attempt: By defining, $C_{n+1}= \... | You’ve misstated the result: you want to show that
$$C_n=\binom{2n}n-\binom{2n}{n-1}\;,$$
not that
$$C_{n+1}=\binom{2n}n-\binom{2n}{n-1}\;.$$
Note that $\binom{2n-1}{n-1}=\binom{2n-1}n$, so
$$\binom{2n}{n-1}=\frac{2n}{n+1}\binom{2n-1}{n-1}=\frac{n}{n+1}\left(\binom{2n-1}{n-1}+\binom{2n-1}n\right)\;.$$
Now use Pascal’s ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Easier ways to prove $\int_0^1 \frac{\log^2 x-2}{x^x}dx<0$ Prove that
$$\int_0^1 \frac{\log^2 x-2}{x^x}dx<0$$
One way to do this is use the idea in the proof of Sophomore's dream. We have
$$x^{-x}=\exp(-x\log x)=\sum_{n=0}^\infty\frac{(-1)^nx^n\log^n x}{n!}$$
Therefore, using the change of variable $x=\exp(-t/(n+1))$ w... | Using the inequality $\log(x)<x-1$ for all $x>0$ we have
\begin{align}
\int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx} < \int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{x^x }}dx}
\end{align}
Also we use the inequality $e^{x} \ge 1+x$ for all $x>0$ to express the denumeantor
$$x^x=\exp(x\ln(x))\ge 1+x\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$. Hint: $x^2-y^2 = (x+y)(x-y)$. This is the exercise verbatim:
An integer n is a square modulo p if there exists another integer x
such that $n \equiv x^2 \pmod p$. Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equi... | Well, use your hint:
$$x^2 - y^2 \equiv 0 \pmod p$$
$$(x+y)(x-y) \equiv 0 \pmod p$$
Clearly, ths is satisfied if $x=y$ or $x=-y$. Because $\mathbb{Z}_p$ has no zero divisors, these are the only solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Part of an induction question I might have done or not realize something stupid, but I can't seem to prove the following...
Inductive hypothesis
Assume $\exists$k$\in$N such that P(k) is true.
P(k): $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot (2k - 1)}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot (2k)}$ $\leq$ $\frac{1}{\sqrt{k + 1}... | We first (directly) prove:
$$\frac{1}{\sqrt{k+1}}\cdot\frac{2k+1}{2k+2} < \frac{1}{\sqrt{k+2}}$$
for $k>1$. Starting with:
$$9k+2<12k+4$$
which is obviously true for $k>1$. Then, we add $4k^3+12k^2$ to both sides:
$$4k^3+12k^2+9k+2<4k^3+12k^2+12k+4$$
Splitting and factoring:
$$4k^3+4k^2+k+8k^2+8k+2<4k^3+8k^2+4k+4k^2+8k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
Question: Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
I can prove that it is not possible for $n=1,2$, but I am not sure for the general case.
Case $n=1$: $a^2v=-v$ $... | suppose $\lambda,$ necessarily real, is an eigenvalue of the symmetric matrix $A^\top A$ and $x \neq 0$ an eigenvector. then $$A^\top A x = \lambda x \tag 1$$ multiplying by $x^\top$ on the left, we get $$\lambda x^\top x =x^\top A^\top Ax = (Ax)^\top Ax \implies\lambda = \frac{|Ax|^2}{|x|^2} \ge 0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
The maximum of $\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}$
Let $a,b,c$ be positive reals, such that $\frac1a+\frac1b+\frac1c=a+b+c\ (\star)$, find the maximum of $\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}$
This should be an application of Jensen's Inequality, so I have to fin... | I would rather show the maximum is $\frac3{16}$ by getting rid of the cumbersome constraint by homogenizing to:
$$\frac{abc(a+b+c)}{ab+bc+ca}\sum_{cyc} \frac1{(2a+b+c)^2} \le \frac3{16}$$
So if $a+b+c=3$, this is equivalent to
$$16\sum_{cyc} \frac1{(3+a)^2} \le \frac1a+\frac1b+\frac1c$$
which follows from $f(a)+f(b)+f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Combining probability of dependent events?
Two dice are thrown simultaneously. What is the probability of getting a multiple of $2$ on one die and a multiple of $3$ on the other?
According to me, the answer should be $\frac16$, as the probability of getting a multiple of $2$ is $\frac12$, and the probability of getti... | The error you've made is that you haven't considered the possibility that either dice might have the multiple of 2, or the multiple of 3.
For a problem with these kinds of small numbers we can use fairly weak combinatorial methods.
The total number of outcomes is 36, we count the pairs of numbers such that one is a mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ To find the above minimal polynomial, let
$$x=\sqrt{2}+\sqrt{3}+\sqrt{5}$$
$$x^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}$$
Subtracting 10 and squaring gives
$$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$
$$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$
$... | I thought I'd get rid of the surds one-by-one: first get rid of $\sqrt2$, then $\sqrt3$, and finally $\sqrt5$. To get rid of a surd, I would get everything with that surd in it to one side, and then square. (Note that $\sqrt{15}$ has a $\sqrt3$ in it.)
\begin{align}
x&=\sqrt2+\sqrt3+\sqrt5\\
x-\sqrt3-\sqrt5&=\sqrt2\\
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
Expansion of Generating Functions If you roll $10$ dice, how many ways can you get a total sum of top faces of $25$?
I understand how to write the generating function of $(x+x^2+ \dots +x^6)^{10}$ and the fact that you need to find the coefficient of $x^{25}$, but how do you do this? The conventional binomial theorem o... | Write $x+x^2+\cdots+x^6$ as $(1-x^7)(1-x)^{-1}$. Then
$$\begin{align}
[x^{25}]&(x+x^2+\cdots+x^6)^{10}\\
&=[x^{25}](x-x^7)^{10}(1-x)^{-10}\\
&=[x^{15}](1-x^6)^{10}(1-x)^{-10}\\
&=[x^{15}]\sum_{r\ge0}\binom{10}r(-x^6)^r
\sum_{s\ge0}\binom{-10}s(-x)^s\\
&=\sum_{r\ge0}(-1)^r\binom{10}r\binom{10+s-1}{10-1}\quad\textrm{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.