Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solve the recurrence $T(n) = 2T(n-1) + n$ Solve the recurrence $T(n) = 2T(n-1) + n$
where $T(1) = 1$ and $n\ge 2$.
The final answer is $2^{n+1}-n-2$
Can anyone arrive at the solution?
| \begin{align}
T(n) & = 2 T(n-1) + n = 2(2T(n-2) + n-1) + n = 4T(n-2) + 2(n-1) + n\\
& = 8 T(n-3) + 4(n-2) + 2(n-1) + n = 2^k T(n-k) + \sum_{j=0}^{k-1} 2^j (n-j)\\
& = 2^{n-1} T(1) + \sum_{j=0}^{n-2}2^j (n-j) = 2^{n-1} + \sum_{j=0}^{n-2}2^j (n-j)
\end{align}
\begin{align}
\sum_{j=0}^{n-2}2^j (n-j) & = n \sum_{j=0}^{n-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/239974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 5
} |
Show that $\int_0^\infty \sin\left(x^2\right)dx$ converges, but that $\int_0^\infty \sqrt{\sin^2\left(x^2\right)}dx$ does not. Show that $\int_0^\infty \sin\left(x^2\right)dx$ converges, but that $\int_0^\infty \sqrt{\sin^2\left(x^2\right)}dx$ does not.
The first part I think I proved using triangles, but I could not p... | Your second integral is
$$
\int_0^\infty|\sin x^2|\,dx.
$$
Note that $\sin t\geq1/2$ if $t\in[\frac\pi6+2k\pi,\frac{5\pi}6+2k\pi]$. So $\sin x^2\geq1/2$ if $x\in[\sqrt{\frac\pi6+2k\pi},\sqrt{\frac{5\pi}6+2k\pi}]$, $k\in\mathbb N$.
So
$$
\begin{eqnarray}
\int_0^\infty|\sin x^2|\,dx&\geq&\sum_{k=0}^\infty\int_{\sqrt{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/245296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Conditional probability of cows the question is that if you are a farmer and own six cows: 3 white, 2 black and one that is black on one side and white on the other. Then if you see two black cows (that is 2 black sides of cows) then what is the probability that one of them is the black and white cow?
Here is my attemp... | If the question is asking for the probability that either of the two cows is 2-coloured, we have
$$P(\text {1 cow is 2-coloured | both visible sides are black}) = \frac{P(\text {1 cow is 2-coloured and other is black}) \times P(\text {the black side of the 2-coloured cow is seen})}{P(\text{both visible sides are black}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/246460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
How do we prove that two parametric equations are drawing the same thing? For example, if I have
$$\begin {align}
x(t) &= r\sin t\cos t\\
y(t) &= r\sin^2 t\\
\end {align}$$
and
$$\begin {align}
x(t) &= \frac r 2 \cos t\\
y(t) &= \frac r 2 (\sin t + 1)
\end {align}$$
How do we show that the two parametric equations ... | $\begin{align}
\hline
x(s) &= r\sin s\cos s\\
y(s) &= r\sin^2 s\\
\hline
x(t) &= \frac r2 \cos t\\
y(t) &= \frac r2 (\sin t + 1)\\
\hline
\end {align}$
Let's rewrite the equations to make them more comparable.
$\begin{align}
\hline
x(s) &= \dfrac r2 \sin 2s\\
y(s) &= \dfrac r2(1 - \cos 2s)\\
\hline
x(t) &= \dfrac r2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/247752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Finding the minimum value of a function in an ellipse So I have this problem for my homework:
Consider the ellipse: $\dfrac {x^2}{a^2} + \dfrac{y^2}{b^2}=1$ where $0<b<a$. For every point $(x, y)$ on the ellipse find the the perpendicular line to the ellipse so that the point $(x, y)$ is on that line. This line cuts t... | I think it is not necessary to use Lagrange Multipliers. we can replace $\dfrac{x^2}{a^2}$ with $1- \dfrac{y^2}{b^2}$, then we have:
$D(y)=2\dfrac{(\dfrac{1}{a^2}(1- \dfrac{y^2}{b^2})+ \dfrac{y^2}{b^4})^\frac{3}{2}}{\dfrac{1}{a^4}(1- \dfrac{y^2}{b^2})+ \dfrac{y^2}{b^4}}=2\dfrac{(C_1+C_2y^2)^\frac{3}{2}}{C_3+C_4y^2}$, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/249812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why does $\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}$? As much as it embarasses me to say it, but I always had a hard time understanding the following equality:
$$
\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}
$$
I always thought that the left-hand side of the above equation was equivalent to
$$
\frac{a}{\frac{b}{x}} ... | The problem is that $\frac{a}{\frac{b}{x}}=\frac{\frac{a}{b}}{\frac{1}{x}}=x\frac{a}{b}$ because $\frac{1}{\frac{1}{x}}=x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/251317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 2
} |
How to show two field extensions are equal. How do I go about showing that $\mathbb{Q}(\sqrt{3}+\sqrt{5})=\mathbb{Q}(\sqrt{3},\sqrt{5})$?
Since $\mathbb{Q}(\sqrt{3}+\sqrt{5}) \subset\mathbb{Q}(\sqrt{3},\sqrt{5})$, we need to show the reverse inclusion. Does it suffice to prove that $\sqrt{2} \in \mathbb{Q}(\sqrt{3}+\sq... | We have
$$\frac{(\sqrt{3} + \sqrt{5})^3 - 14(\sqrt{3} + \sqrt{5})}{4} = \frac{18\sqrt{3} + 14\sqrt{5} - 14\sqrt{3} -14 \sqrt{5}}{4} = \frac{4\sqrt{3}}{4} = \sqrt{3}$$
So $\sqrt{3} \in \mathbb{Q}(\sqrt{3} + \sqrt{5})$, and hence so is $\sqrt{5}$.
The general trick in these situations is to write one of the irrational n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/251741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Closed form for $\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$ How can I calculate the following sum involving binomial terms:
$$\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$$
Where the value of n can get very big (thus calculating the binomial coefficients is not feasible).
Is there a closed form for this su... | Suppose we seek to compute
$$S_n = \sum_{k=0}^n {n\choose k} \frac{(-1)^k}{(k+1)^2}.$$
With this in mind we introduce the function
$$f(z) = n! (-1)^n \frac{1}{(z+1)^2} \prod_{q=0}^n \frac{1}{z-q}.$$
We then obtain for $0\le k\le n$
$$\mathrm{Res}_{z=k} f(z) =
(-1)^n \frac{n!}{(k+1)^2} \prod_{q=0}^{k-1} \frac{1}{k-q}
\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/254416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 6
} |
If a,b,c are positive and (a,b)=(b,c)=1 and 1/a + 1/b + 1/c = integer then b=1 and a=c=1 or 2 If you work it out you see that if
$a=b=c=1$, then $(1,1)=(1,1)=1$ and $1/1 + 1/1 + 1/1 = 3 =$ integer
and
$b=1, a=c=2$, then $(2,1)=(1,2)=1$ and $1/2 + 1/1 + 1/2 = 2=$ integer.
But please help me prove this
| So you have proven that the given cases does indeed work, and you need help to show that these are the only cases. So let's try to find another case. Let's see what the fraction sum turns out to be:
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ac + ab}{abc}
$$
Since $b$ have no factors in common with neither... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/254929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Property of Fejer kernel Let
$$
F_n(x) = \frac{1}{n} \left( \frac{ \sin(\frac{1}{2} n x ) } { \sin(\frac{1}{2} x ) } \right)^2
$$
be the n-th Fejer-Kernel. Then
$$
\forall \epsilon > 0, r < \pi : \exists N \in \mathbb{N} : \forall n \ge N : \int_{[-\pi,\pi] \setminus [-r, r]} F_n(t) \, \mathrm{d} t < \epsilon
$$
The... | To see the estimate (*), first note that for $0<r<x<\pi/2$ we have
$$0<\sin r<\sin x<1$$
Hence for $x$ with $0<r<|x|<\pi$
$$0<\sin^2 \frac{r}{2}<\sin^2 \frac{x}{2}<1$$
that is
$$0<\frac{1}{\sin^2 \frac{x}{2}}<\frac{1}{\sin^2 \frac{r}{2}}$$
and we get
$$ \int_{[-\pi,\pi] \setminus [-r, r]} \frac{1}{\sin^2\frac{x}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/255002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to correct this diagonalization argument? I want to diagonalize (with a orthogonal change of coordinates) the quadratic form $F(x,y)=x^2-\frac{n-2}{\sqrt{n-1}}xy-y^2$. I already know that the system $(*)$ bellow
\begin{eqnarray}
x=&\frac{(\sqrt{n-1}+1)u+(\sqrt{n-1}-1)v}{\sqrt{2n}}\\
y=&\frac{(-\sqrt{n-1}+1)u+(\sqr... | It's not far from being orthogonal; you just failed to normalize the columns. Being the eigenvectors of a symmetric matrix, they're already orthogonal; if you normalize them, $P$ will be orthogonal and $P^{-1}AP$ will still be diagonal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/255460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the general solution of $(1+x^2)y''+2xy'-2y=0$. How to express by means of elementary functions?
Find the general solution of $$(1+x^2)y''+2xy'-2y=0$$
in terms of power series in $x$. Can you express this solution by means
of elementary functions?
I know that $y= \displaystyle\sum_{n=0}^{ \infty } a_n... | Let $y=\sum\limits_{n=0}^\infty a_nx^n$ ,
Then $y'=\sum\limits_{n=0}^\infty na_nx^{n-1}$
$y''=\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}$
$\therefore(1+x^2)\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}+2x\sum\limits_{n=0}^\infty na_nx^{n-1}-2\sum\limits_{n=0}^\infty a_nx^n=0$
$\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}+\sum\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/255974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Convergence of Sequence with factorial I want to show that
$$
a_n = \frac{3^n}{n!}
$$
converges to zero. I tried Stirlings formulae, by it the fraction becomes
$$
\frac{3^n}{\sqrt{2\pi n} (n^n/e^n)}
$$
which equals
$$
\frac{1}{\sqrt{2\pi n}} \left( \frac{3e}{n} \right)^n
$$
from this can I conclude that it goes to z... | Alternatively to Stirling:
We have
\begin{align}
a_n =\frac{3^n}{n!}
\end{align}
Now let $n>3$, then
\begin{align}
0\leq a_n &=\frac{3^n}{n!} = \frac{3\cdot3 \cdot 3}{1\cdot 2\cdot 3}\cdot \frac{3^{n-3}}{4\cdot 5 \cdot ...\cdot n}=\frac{3\cdot3 \cdot 3}{1\cdot 2\cdot 3}\cdot \frac{3\cdot 3 \cdot ... \cdot 3}{4\cdot 5 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/260772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integration Question on $\int \frac{1}{x^2+10x+21} \, dx$ How would i do the following indefinite integration $$\int \frac{1}{x^2+10x+21} \, dx$$
so far I've turned the bottom polynomial into $(x+7)(x+3)$ not too sure where to go from here
| Hint: Partial Fractions:
$$\frac{1}{(x+7)(x+3)}=\frac{A}{x+7}+\frac{B}{x+3}$$
Now find $A,B$ and use the linearity of the integral:
$$\int \frac{1}{(x+7)(x+3)} dx=\int \frac{A}{x+7}dx+\int \frac{B}{x+3}dx$$
This should be simple now.
EDIT: Evaluating $A,B$:
$$\frac{1}{(x+7)(x+3)}=\frac{A}{x+7}+\frac{B}{x+3}\iff 1=A(x+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/263680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Synthetic division via the greedy strategy I was looking at the expanded synthetic division within Wikipedia. I was stumped by how to come up with and perform the 'compactified' version of synthetic division.
Does anyone know how to do it?
| I came up with three shorthand methods from noticing certain patterns, but the one I describe below is the simplest:
*
*Write the coefficients of the dividend on a bar
$\begin{array}{cc}
\begin{array}{|rrrrrrrr}
a & b & c & d & e & f & g & h \\
\hline
\end{array}
\end{array}$
*
*Negate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How is this a property of Pascal's triangle? For all non-negative integers $k$ and $n$,
$$
\dbinom{k}{k} + \dbinom{k+1}{k} + \dbinom{k+2}{k} + \ldots + \dbinom{n}{k} = \dbinom{n +1}{k+1}
$$
How is this a property of Pascal's triangle? I do not see at all how it relates. I'm looking for an explanation, as opposed to som... | You're asking, for example, for the sum of all the indicated cells of Pascal's triangle
$$ \begin{matrix} \cdot
\\ \cdot & \cdot
\\ \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet &
\\ \cdot & \cdot & \cdot & \bullet & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$ Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$ .
I am sure this is derived from using roots of unity and Euler's complex number function, but I am very uncomfortable in these areas so some help would be great. It is evident t... | Using this,
the roots of $\displaystyle z^3+z^2-3z-1=0\qquad (1)$ are $\displaystyle 2\cos\frac{2\pi}7, 2\cos\frac{4\pi}7, 2\cos\frac{6\pi}7$
If $\displaystyle\cot^2\frac{r\pi}7=u, \cos\frac{2r\pi}7=\frac{1-\tan^2\frac{r\pi}7}{1+\tan^2\frac{r\pi}7}=\frac{\cot^2\frac{r\pi}7-1}{\cot^2\frac{r\pi}7+1}=\frac{u-1}{u+1}$
$\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Permutation and Combination with divisibility? How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeating?
my explanation:
total number of permutations with 0, 1, 2, 3, 4 and 5
to have 5 digits = 6.5.4.3.2=720 (includes ... | You only substracted the numbers with 0 at the peginning however: the divisibilty rule for three says that for a number to be a multiple of 3 the sum of its digits needs to be a multiple of three. So really it does not matter what order you put the numbers in but the sum of the numbers themselves. There are in total 6 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
simple congruence system problem. This is excercise 1 in George E. Andrews number theory. page 51. How am i supposed to solve this? Thanks very much in advance. I used the cancellation law to get isolate the x. But I don’t know what to do next.
$5x\equiv 4 \pmod 3$
$7x\equiv 6 \pmod 5$
$9x\equiv 8 \pmod 7$
| $$5 x \equiv 1 \pmod 3 \implies x \equiv 2 \pmod 3$$
$$7 x \equiv 1 \pmod 5 \implies x \equiv 3 \pmod 5$$
$$9 x \equiv 1 \pmod 7 \implies x \equiv 4 \pmod 7$$
$$x \equiv 2 \pmod 3 \text{ and }x \equiv 3 \pmod 5 \implies x \equiv a \pmod{15}$$
Since $x \equiv 2 \pmod 3$, we have $a \in \{2,5,8,11,14\}$. Since $x$ is als... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How to compute the determinant of a tridiagonal matrix with constant diagonals? How to show that the determinant of the following $(n\times n)$ matrix
$$\begin{pmatrix}
5 & 2 & 0 & 0 & 0 & \cdots & 0 \\
2 & 5 & 2 & 0 & 0 & \cdots & 0 \\
0 & 2 & 5 & 2 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots & \vdots... | The determinant can also be verified using block determinants.
Define $A^{(k)}$ to be the $k\times k$ matrix of the form$$A^{(k)}:=
\begin{bmatrix}
5 & 2 & 0 &0&0&\cdots & 0\\
2 & 5 & 2 & 0&0&\cdots & 0\\
0 & 2 &5&2&0 & \cdots & 0\\
\vdots & \vdots& \vdots& \vdots& \vdots& \vdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 1
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$\frac{n}{2} < 1+\frac{1}{2} + \frac{1}{3}+ \ldots +\frac{1}{2^n-1}How to show that for $n\geqslant 2$
$$\frac{n}{2} < 1+\frac{1}{2} + \frac{1}{3}+ \ldots +\frac{1}{2^n-1}<n$$
| You can group the sum as follows:
$$\sum_{k=1}^{2^n}{\frac{1}{k}}> 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}\right)...$$
So that:
$$\sum_{k=1}^{2^n-1}{\frac{1}{k}} > 1 + \frac{n}{2}-\frac{1}{2^n}>\frac{n}{2}$$
If instead you group them by the lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/267191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Why is $ab+bc+ac = 0$ in some situation? This is originally a Computer Science question, but I ran a equation that is too hard to solve. Here goes.
So the problem is quite simple, given positive integers $a$, $b$, $c$, and calculate $\sqrt{a^2+b^2+c^2}$
But here's the problem, if I calculate any of the squares, I will ... | In your question you say:
So the problem gives me an additional condition: the difference between the minimum of the three and the bigger other two is perfect squares.
That said, if c is the smallest of the two. $b−c$ and $a−c$ will both be perfect squares. And the answer itself is an integer, which means $a^2+b^2+c^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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$ \frac{1}{3a^2+1}+\frac{1}{3b^2+1}+\frac{1}{3c^2+1}+\frac{1}{3d^2+1} \geq \frac{16}{7}$ Let $a,b,c,d >0$ and $a+b+c+d=2$. Prove this:
$$ \frac{1}{3a^2+1}+\frac{1}{3b^2+1}+\frac{1}{3c^2+1}+\frac{1}{3d^2+1} \geq \frac{16}{7}$$
| Let $a=\frac{x}{2}$, $b=\frac{y}{2}$, $c=\frac{z}{2}$ and $d=\frac{t}{2}$.
Hence, $x+y+z+t=4$ and we need to prove that $\sum\limits_{cyc}\frac{1}{3x^2+4}\geq\frac{4}{7}$.
Let $f(x)=\frac{1}{3x^2+4}$.
Hence, $f''(x)=\frac{6(9x^2-4)}{(3x^2+4)^2}>0$ for all $1\leq x\leq4$.
Thus, by Vasc's RCF Theorem it's enough to prove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Algebra question involving fractions How would I perform the indicated operation.
$$\frac{t+2}{t^2+5t+6}+\frac{t-1}{t^2+7t+12}-\frac{2}{t+4}.$$
I simplified it to
$$
\frac{t+2}{(t+3)(t+2)} + \frac{t-1}{(t+4)(t+3)}-\frac{2}{t+4}.
$$
Then I did the lowest common denominator but I still have problems.
According to my boo... | You did quite well, but note that there's a term $(t+2)$that cancels in your first "reduced"/simplified fraction: $$\frac{t+2}{(t+3)(t+2)} = \frac{1}{t+3}\tag{$t\ne -2$}$$
From the start:
$$\frac{(t+2)}{(t^2+5t+6)}+\frac{(t-1)}{(t^2+7t+12)}-\frac{2}{(t+4)}$$
Factoring denominators gives us:
$$= \frac{(t+2)}{(t+3)(t+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/274912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If x,y,z are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what If $x,y, z$ are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what?
$108$ , $216$ , $405$ , $1048$
| x:y:z=3:2:1 then x=6,y=6/3 and z=6/2 ('.' 1^3=1, 2^3=8, 3^3=27 and 1+2+3=6)
substitute in the equation
6^2+8(3^2)+27(2^2)=216
u can exchange the value of x y z but he has asked for minimum value for expression.so taking these value exactly for x y and z would be appropriate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
How to solve $|x-5|=|2x+6|-1$? $|x-5|=|2x+6|-1$.
The answer is $0$ or $-12$, but how would I solve it by algebraically solving it as opposed to sketching a graph?
$|x-5|=|2x+6|-1\\
(|x-5|)^2=(|2x+6|-1)^2\\
...\\
9x^4+204x^3+1188x^2+720x=0?$
| You have $|x-5|=|2x+6|-1$.
1)If $x\geq5$, $|x-5|=x-5$ and, $|2x+6|=2x+6$, so you have
$x-5=2x+6-1$
$x-5=2x+5$
$x=-10$ (but is not valid)
2)If $-3/2\leq x<5$, $|x-5|=-(x-5)$ and $|2x+6|=2x+6$, then
$-(x-5)=2x+6-1$
$-x+5=2x+5$
$-x=2x$
$x=0$
3)If $x<-3/2$, $|x-5|=-(x-5)$ and $|2x+6|=-(2x+6)$, then
$-(x-5)=-(2x+6)-1$
$-x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
What will be the one's digit of the remainder in: $\left|5555^{2222} + 2222^{5555}\right|\div 7=?$ What will be the ones digit of the remainder in: $$\frac{\left|5555^{2222} + 2222^{5555}\right|} {7}$$
| This is not a solution. I was trying to find out the possible values of $n$ such that $${\underbrace{22\cdots22}_{n\text{ digits }}}^{\underbrace{55\cdots55}_{n\text{ digits }}}+{\underbrace{55\cdots55}_{n\text{ digits }}}^{\underbrace{22\cdots22}_{n\text{ digits }}}$$ is divisible by $7$.
Let $$\underbrace{11\cdots11}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/279333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Radical questions algebra Hello everyone how would I simplify the following radicals.
$$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}}$$
I got $$3 \sqrt{64a^4b^7}$$ I know $64$ square root is $8$ and $a^4$ square root is $a^2$
My second question is how would I simplify the following
$$\frac{3}{4}\sqrt{3t^3}$$
I know $ \sqrt[\large4]... |
$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}}$
$$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}} = 3 \sqrt{64a^4b^7} = 3\sqrt{64a^4}\sqrt{b^7} = 3\cdot 8 a^2 \sqrt{b^7} = 24a^2 \sqrt{b^7} = 24a^2b^{\large\frac72}\quad\quad\quad\tag{1}$$
"$ \;\sqrt[\large4]{3t^3}\;$ is equal to $\;(3t)^{\frac{3}{4}}\;$"
Not quite:
$$\sqrt[\large 4]{3t^3} \;=... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How to solve for $x$ in $x^3+8^x-9=0$ How to find real numbers $x$ that are solutions to
$$x^3+8^x-9=0$$
Please help me.
| Note that
\begin{align}
x^3+8^x-9=0\iff
&
\begin{cases}
x^3-1=2^3-(2^{x})^3\\
x^3-8=1^3-(2^{x})^3\\
\end{cases}
\\
\iff
&
\begin{cases}
(x-1)(x^2+x+1)=[2-(2^{x})][2^2+2(2^{x}) + (2^{x})^2] \\
(x-2)(x^2+2x+4)=[1-(2^{x})][1+(2^{x})+(2^{x})^2] \\
\end{cases}
\\
\end{align}
Then is easy to see that $x=1 \implies x^3+8^x-9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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$\frac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$ Let $a,b,c$ be positive numbers. Prove that $$\dfrac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$$
| A full expanding gives
$$\sum_{cyc}(3a^6+a^5b+a^5c-a^4b^2-a^4c^2+2a^3b^3+3a^4bc-6a^3b^2c-6a^3c^2b+4a^2b^2c^2)\geq0,$$
which is true by Schur and Muirhead:
$$\sum_{cyc}(3a^6+a^5b+a^5c-a^4b^2-a^4c^2+2a^3b^3+3a^4bc-6a^3b^2c-6a^3c^2b+4a^2b^2c^2)=$$
$$=\sum_{cyc}(a^6-a^3b^3)+2\sum_{cyc}(a^6-a^2b^2c^2)+\sum_{cyc}(a^5b+a^5c-a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Converting to polar form
Write each of the given numbers in the polar form $re^{i\theta}$.
a.) $\frac{1-i}{3}$
b.) $-8\pi (1+\sqrt 3 i)$
For a, I got:
r = $\frac{\sqrt 2}{3}$ and $e^{i7\pi /2}$ since $\frac{\frac{1}{3}}{\frac{\sqrt 2}{3}} = \frac{\sqrt 2}{2}$ and $\frac{-\frac{1}{3}}{\frac{\sqrt 2}{3}} = -\frac{\sqr... | Following two points need to be used here:
(1) The principal value of $\arctan \frac yx$ lies in $[-\frac\pi2,\frac\pi2]$
(2)From this, we need to identify the principal value of the argument of a complex number.
Let $$\frac{1-i}3=re^{i\theta}=r(\cos\theta+ i\sin\theta)$$ where $r>0$
Equating the real & the imaginary p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Hyperbola is a pair of straight lines? I'm confused by this question:
If $f(x) = 2x^2 - 6y^2+xy+2x-17y-12=0$ is to represent a pair of
straight lines, one of which has equation $x+2y+3=0$, what must be the
equation of the other line? Verify that $f(x)=0$ does, indeed,
represent a pair of straight lines.
Given t... | Divide the function $f(x, y)=2x^2-6y^2+xy+2x-17y-12$ by given factor $(x+2y+3)$ to get the second linear factor as $(2x-3y-4)$
Hence, the equation of second straight line: $2x-3y-4=0$
For general case, a quadratic equation of two variables $x$ & $y$ is given as
$$ax^2+2hxy+by^2+2gx+2fy+c=0$$
The above equation will... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/283591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Derive the Quadratic Equation Find the Quadratic Equation whose roots are $2+\sqrt3$ and $2-\sqrt3$.
Some basics:
*
*The general form of a Quadratic Equation is $ax^2+bx+c=0$
*In Quadratic Equation, $ax^2+bx+c=0$, if $\alpha$ and $\beta$ are the roots of the given Quadratic Equation, Then,
$$\alpha+\beta=\frac{-b}... | This appears to be a standard question from the R.D. Sharma textbook used in Indian schools. A common trick I use for this type of sums is
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Comparing with the given roots, we get
$$x = \frac{4 \pm \sqrt{4 - 4c}}{2}$$
In this step we find the value of $a$ to be $1$, $b$ to be $-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/284338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
The last digit of $2^{2006}$ My $13$ year old son was asked this question in a maths challenge. He correctly guessed $4$ on the assumption that the answer was likely to be the last digit of $2^6$. However is there a better explanation I can give him?
| The last digit of $2^{2006}$ is $2^{2006} (\text{mod }10)$. So let's look at the behavior of powers of $2$ mod $10$.
$2^2 = 4$,
$2^3 = 8$ (nothing interesting so far),
$2^4 = 16 = 6$ (remember we are working mod $10$, we only keep the last digit).
$2^5 = 32 = 2$, Now THIS is interesting. This suggests we divide $2006$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
"answer_count": 8,
"answer_id": 3
} |
Prove $\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$ I am trying to prove that
$$\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$$
I know how to ... | There have been similar integrals such as this with a factor of $1/\log{t}$ in the integrand. The way I have attacked these is to use the substitution $t=e^{-x}$; here, this produces
$$-\int_0^{\infty} dx \: \frac{e^{-x}}{x} \frac{1-e^{-2 x}}{1+e^{-2 x}} $$
$$ = -\int_0^{\infty} dx \: \frac{1}{x} (e^{-x} - e^{-3 x}) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/285130",
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"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 2
} |
limit of the sum $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} $ Prove that : $\displaystyle \lim_{n\to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=\ln 2$
the only thing I could think of is that it can be written like this :
$$ \lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k+n} =\lim_{n\to \inf... | Set $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}.$$
We can use the following uniform bound
$$a_n = \sum_{k=1}^n \frac{1}{n+k} \leq\sum_{k=1}^n\frac{1}{n+1} = \frac{n}{n+1} < 1$$
Since each $a_n$ is positive, we have that $0\leq a_n < 1$ for all $n$. For monotonicity we use the calculation
$$\be... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 5
} |
How to transform $2^{n-2}\frac{(2n-5)(2n-7)...(3)(1)}{(n-1)(n-2)...(3)(2)(1)}$ into $\frac{1}{n-1}\binom{2n-4}{n-2}$? Just an algebraic step within the well known solution for the number of triangulations of a convex polygon!
| $$\begin{align*}
\frac{1}{n-1}\binom{2n-4}{n-2}&=\frac{(2n-4)!}{(n-1)!(n-2)!}\\\\
&=\frac{\Big((2n-4)(2n-6)\dots(2)\Big)\Big((2n-5)(2n-7)\dots(3)(1)\Big)}{(n-1)!(n-2)!}\\\\
&=\frac{2^{n-2}(n-2)!\Big((2n-5)(2n-7)\dots(3)(1)\Big)}{(n-1)!(n-2)!}\\\\
&=2^{n-2}\frac{(2n-5)(2n-7)\dots(3)(1)}{(n-1)!}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Integration by parts $ \int \sqrt{x-x^2} dx $ I have to integrate $$ \int \sqrt{x-x^2} dx $$
The answer on my textbook is $ \frac 14 \left( \arcsin(\sqrt x)-(1-2x)\sqrt{x-x^2} \right) $ but I want to solve this by myself so can you please give me a clue ? :) Thank you.
| For the real values of $\sqrt{x-x^2},$ we need $x-x^2\ge 0\implies 0\le x\le 1$
If we put $x=\sin^2t$ where $0\le t\le \frac\pi2, 1-x=\cos^2t\ge 0\implies \sqrt{1-x}=+\cos t$ and $dx=2\sin t\cos tdt$
So, $$\int\sqrt{x-x^2}dx=\int\sqrt{x(1-x)}dx$$
$$=\int\sin t\cos t2\sin t\cos tdt$$
$$=\frac12\int (\sin2t)^2dt \text{,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/287238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Show that $\frac{1}{x^2+x+1}=\frac{2}{\sqrt3}\sum_{n=0}^{\infty}\sin\bigg(\frac{2\pi(n+1)}{3}\bigg)x^n$ Need to show that:
$$\frac{1}{x^2+x+1}=\frac{2}{\sqrt3}\sum_{n=0}^{\infty}\sin\bigg(\frac{2\pi(n+1)}{3}\bigg)x^n$$
There is a hint given that $x^3-1=(x-1)(x^2+x+1)$ but I don't seem to get how I could use it. If I tr... | $$let : C_x = \sum_{n=0}^{\infty} \cos \left( \frac{2\pi(n+1)}{3} \right )x^n$$
$$and : S_x = \sum_{n=0}^{\infty} \sin \left( \frac{2\pi(n+1)}{3} \right )x^n$$
$$C_x + iS_x = \large{\sum_{n=0}^{\infty} e^{\frac{2\pi i(n+1)}{3}} x^n = e^{i\frac{2\pi}{3}} \sum_{n=0}^{\infty} (xe^{\frac{2\pi i}{3}} )^n}$$
$$= \frac{e^{i\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/287594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Describe all solutions of $Ax=0$. Describe all solutions of $Ax=0$ in parametric vector form, where $A$ is row equivalent to the given matrix.
$$ \begin{bmatrix}
1 & -2 & 3 & -6 & 5 & 0\\
0 & 0 & 0 & 1 & 4&-6\\
0&0&0&0&0&1\\
0&0&0&0&0&0
\end{bmatrix}$$
I know that I should get this into row reduced echelon ... | Think about what $Ax=0$ means for your $A$:
$$\begin{bmatrix}
1 & -2 & 3 & -6 & 5 & 0\\
0 & 0 & 0 & 1 & 4&-6\\
0&0&0&0&0&1\\
0&0&0&0&0&0
\end{bmatrix}
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\end{bmatrix}=\begin{bmatrix} 0\\0\\0\\0\end{bmatrix},$$
that is,
\begin{align}
x_1-2x_2+3x_3-6x_4+5x_5+0x_6&=0,\\
x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/288199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Bell Numbers: How to put EGF $e^{e^x-1}$ into a series? I'm working on exponential generating functions, especially on the EGF for the Bell numbers $B_n$.
I found on the internet the EGF $f(x)=e^{e^x-1}$ for Bell numbers. Now I tried to use this EGF to compute $B_3$ (should be 15). I know that I have to put the EGF int... | \begin{eqnarray*}
e^{e^x-1}&=&1+(e^x-1)+\frac{(e^x-1)^2}{2!}+\frac{(e^x-1)^3}{3!}\\
&=&1+e^x-1+\frac{e^{2x}-2e^x+1}{2!}+\frac{e^{3x}-3e^{2x}+3e^x-1}{3!}\\
&=&e^x+\frac{e^{2x}}{2!}-e^x+\frac{1}{2!}+\frac{e^{3x}}{3!}-\frac{e^{2x}}{2!}+\frac{e^{x}}{2!}-\frac{1}{3!}\\
&=&\frac{e^{x}}{2!}+\frac{e^{3x}}{3!}-\frac{1}{2!}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/289097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Prove that $1^3 + 2^3 + ... + n^3 = (1+ 2 + ... + n)^2$ This is what I've been able to do:
Base case: $n = 1$
$L.H.S: 1^3 = 1$
$R.H.S: (1)^2 = 1$
Therefore it's true for $n = 1$.
I.H.: Assume that, for some $k \in \Bbb N$, $1^3 + 2^3 + ... + k^3 = (1 + 2 +...+ k)^2$.
Want to show that $1^3 + 2^3 + ... + (k+1)^3 = (1 + ... | IMHO, this fact is a coincidence; a better approach is to prove the closed-form formula for both. As we know
$$ 1 + 2 + \cdots + k = \frac{k(k+1)}{2} $$
the corresponding claim to prove is
$$ 1^3 + 2^3 + \cdots + k^3 = \frac{k^2(k+1)^2}{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/294213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Find the spanning set of the range of the linear transformation $T(x)=Ax$. Let
$$
A=
\begin{bmatrix}
-4 & -4 & 12 & 0 \\
-4 & -4 & 12 & 0 \\
4 & -2 & 0 &-6 \\
1 &-4 &7 &-5 \\
\end{bmatrix}
$$
Find the spanning set of the range of the linear transformation $T(x)=A... | The range of $T$ is the column space of $A$. So the columns of $A$ already form a spanning set. If you want to find a linearly independent spanning set, you should find a column echelon form of $A$ instead of a row echelon form. I found that $\{(-4-4,4,1)^T,\,(0,0,-6,-5)^T\}$ is an answer, but depending on the column o... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Problems regarding $\{x_n \}$ defined by $x_1=1$; $x_n$ is the smallest distinct natural number such that $x_1+...+x_n$ is divisible by $n$. Let me denote a sequence of distinct natural numbers by $x_n$ whose terms are determined as follows:
$x_1$ is $1$ and $x_2$ is the smallest distinct natural number $n$ such that $... | As other posters have noticed, this is closely related to the golden ratio $\varphi = \frac{1+\sqrt{5}}{2}$.
@ Ross Millikan your answer is a great place to start.
$\frac{1}{\varphi}+\frac{1}{\varphi ^2}=1$, so we can partition the natural numbers greater than 1 into numbers of the forms $\lceil {n \varphi} \rceil, \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/295042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 6,
"answer_id": 3
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More on the generalized integral Refer to my previous topic: Definite integral, quotient of logarithm and polynomial: $I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x$
I think we get this :
$$\frac{\sin \theta}{1-2\cos \theta x+x^2}=\sum_{k=1}^{\infty}\sin (k\theta )x^{k-1}$$
Then $$\int_0^1... | Some food for thought that hopefully one of us will pursue in more detail soon.
The function $(1-2 x \cos{\theta} + x^2)^{-1}$ is a generating function for the Chebyshev polynomials of the second kind $U_n(\cos{\theta})$:
$$(1-2 x \cos{\theta} + x^2)^{-1} = \sum_{n=0}^{\infty} U_n(\cos{\theta}) x^n$$
So your integral i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/295377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplifying $\frac{1}{x} + \frac{5+x}{x+1} - \frac{7x^2 + 3}{(x+2)^2}$ I'm having trouble simplifying this expression:
$$\frac{1}{x} + \frac{5+x}{(x+1)} - \frac{7x^2 + 3}{(x+2)^2}$$
Would you first do the addition or subtraction?
What's the steps to solve this?
The final answer is
$$\frac{-6x^4 + 3x^3 + 26x^2 + 25x + 4... | Or, if you set $I=\frac{1}x+\frac{x+5}{x+1}-\frac{7x^2+3}{(x+2)^2}$, then by multiplying $I$ by $x(x+1)(x+2)^2$, you get:
$$x(x+1)(x+2)^2\times I=x(x+1)(x+2)^2\left(\frac{1}x+\frac{x+5}{x+1}-\frac{7x^2+3}{(x+2)^2}\right)\\=(x+1)(x+2)^2+(x+5)x(x+2)^2-(7x^2+3)x(x+1)\\=3x^3+26x^2+25x+4-6x^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/297013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving non-linear congruence $x^2+2x+2\equiv{0}\mod(5)$, $7x\equiv{3}\mod(11)$
My attempt:
$x^2+2x+2\equiv{0}\mod(5)$
$(x+1)^2\equiv-1\mod(5)$, we have $x+1\equiv-1\mod(5)$
since $5$ and $11$ are coprime. We have a solution in $\mathbb{Z}_{11}$
With $[3]$ represent $3$, $[13]$ works for $7x\equiv3\mod11$
so $[3]$ is ... | $\rm mod\ 5\!:\ 0 \equiv x^2+2x+2 \equiv x^2-3x+2 \equiv (x-1)(x-2)\:\Rightarrow\: x\equiv \color{#C00}1,\,\color{#0A0}2.$
$\rm mod\ 11\!:\ 7x\equiv 3\equiv 14\:\Rightarrow\:x\equiv 2.\ $ Solving these congruences yields
*
*$\rm\ \ x\equiv 2\ mod\ 11,\ x\equiv \color{#0A0}2\ mod\ 5\,\Rightarrow \: x\equiv 2\ mod\ 55... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/300940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Permutation question. $qpq^{-1}$, $q,p,r,s \in S_{8}$. Let $p,r,s,q \in S_{8}$ be the permutation given by the following products of cycles:
$$p=(1,4,3,8,2)(1,2)(1,5)$$
$$q=(1,2,3)(4,5,6,8)$$
$$r=(1,2,3,8,7,4,3)(5,6)$$
$$s=(1,3,4)(2,3,5,7)(1,8,4,6)$$
Compute $qpq^{-1}$ and $r^{-2}sr^{2}.$
thanks for your help.
I want ... | If you write all right then
$$
p= \begin{pmatrix}
1 & 2&3&4&5&6&7&8\\4&1&8&3&5&6&7&2
\end{pmatrix}
\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\2&1&3&4&5&6&7&8
\end{pmatrix}
\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\5&2&3&4&1&6&7&8
\end{pmatrix}
$$
and so on (sorry, I write badly the formula). So you have first to calculate $p,q,r,s$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Absolute value inequality - Please guide further
Prove that if the numbers $x$, $y$ are of one sign, then $\left|\frac{x+y}{2}-\sqrt{xy}\right|+\left|\frac{x+y}{2}+\sqrt{xy}\right|=|x|+|y|$.
Expanding the LHS,
$$\left|\frac{x+y}{2}-\sqrt{xy}\right|= \left|\frac{x+y -2\sqrt{xy}}{2}\right| = \frac{(\sqrt{x}-\sqrt{y})^2... | Hint: It may be clearer if you write $x=(-1)^t2a^2$, $y=(-1)^t2b^2$ for some $a,b>0$ and $t\in\{0,1\}$.
Mouse over the gray box for more details if you get stuck.
You now want to show that $$|(-1)^t(a^2+b^2)-2ab|+|(-1)^t(a^2+b^2)+2ab|=2a^2+2b^2.$$ Shuffling the minus signs, we get $$|a^2+b^2+(-1)^{t+1}2ab|+|a^2+b^2+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/307836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Verify that: $2\cot{x}/\tan{2}x = \csc^2x-2$ Verify the following:
$$\frac{2\cot{x}}{\tan{2}x} = \csc^2x-2\;.$$
| Using the same technique as presented here (where it explains why we want to do things this way) we put $z = e^{ix}$ to obtain
$$ \cot x = i \frac{z + \frac{1}{z}}{z - \frac{1}{z}} =
i \frac{z^2+1}{z^2-1}$$
and
$$ \tan 2x = \frac{1}{i} \frac{z^2 - \frac{1}{z^2}}{z^2 + \frac{1}{z^2}} =
\frac{1}{i} \frac{z^4 - 1}{z^4 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/308160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
integrating $\oint_C \frac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ I am doing $\oint_C \dfrac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ and I find that there are poles within the contour at $z = 1$ and at $z = 3i$, both simple poles. I find that the integral $I = 2\pi i \,\text{Res}(1) + 2 \pi i \,\text{Res}(3i)$, or $... | Let $\,C_1,\,C_2,\,C_3\,$ be little circles (say, of radius $\,0.1\,$) around each of the poles $\,1,\,-3i,\,3i\,$ resp., of the function, so by Cauchy's Integral Theorem:
$$\int\limits_{|z|=4}\frac{3z^3+2}{(z-1)(z^2+9)}dz=\int\limits_{C_1}\frac{\frac{3z^3+2}{(z^2+9)}}{z-1}dz+\int\limits_{C_2}\frac{\frac{3z^3+2}{(z-1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/308721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of a sequence with indeterminate form Let $\displaystyle u_n =\frac{n}{2}-\sum_{k=1}^n\frac{n^2}{(n+k)^2}$. The question is: Find the limit of the sequence $(u_n)$.
The problem is if we write $\displaystyle u_n=n\left(\frac{1}{2}-\frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\right)$ and we use the fact that ... | $$\dfrac1{(1+k/n)^2} = \left( 1 + \dfrac{k}n \right)^{-2} = 1 + \dfrac{(-2)}{1!} \dfrac{k}n + \dfrac{(-2)(-3)}{2!} \dfrac{k^2}{n^2} + \dfrac{(-2)(-3)(-4)}{3!} \dfrac{k^3}{n^3} + \cdots $$
Now recall that
$$\sum_{k=1}^n \left(\dfrac{k}n \right)^m = \dfrac{n}{m+1} + \dfrac12 + \mathcal{O}(1/n)$$
Hence,
\begin{align}
\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/309939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then find the largest possible value of $|a-b|$ I came across the following problem that says:
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then the largest possible value of $|a-b|$ is which of the following?
$(1)\sqrt 7$, $(2)\sqrt{7/2}$ , $(3)\sqrt {35}$ $(... | Here's a trick that doesn't involve calculus:
You have seen that
$$(a - b)^2 + 4ab + b^2 = 7$$
Now, adding this equation to S times the original equation, you get:
$$(a - b)^2 + Sa^2 + (4 + 2S)ab + (1 + 2S)b^2 = 7 + 7S$$
You would like to turn the last three terms into a perfect square, ie:
$$Sa^2 + (4 + 2S)ab + (1 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/310348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Find $\delta$ such that $0<|x-3|<\delta \Rightarrow |\frac{1}{x} - \frac{1}{3}| < 10^{-4}$ I was working on a calculus problem:
Find $\delta$ such that $0<|x-3|<\delta \Rightarrow |\frac{1}{x} - \frac{1}{3}| < 10^{-4}$
I did some algebra on the consequent obtaining:
$$
\frac{|x-3|}{|3x|} < 10^{-4}
$$
Then I noted that... | Note that
$0<|x-3|<\delta\iff x\in(3-\delta, 3+\delta)\iff 3-\delta < x < 3+\delta$.
Therefore, you can write $x = 3 + \rho$, such that $|\rho| < \delta$. So you have that
$\bigg\lvert \frac{1}{x}-\frac{1}{3}\bigg\lvert = \bigg\lvert \frac{1}{3 + \rho}-\frac{1}{3}\bigg\lvert = \bigg\lvert \frac{3- 3-\rho}{3(3+\rho)}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum_{j = 0}^{n} (-\frac{1}{2})^j = \frac{2^{n+1} + (-1)^n}{3 \times 2^n}$ whenever $n$ is a nonnegative integer. I'm having a really hard time with the algebra in this proof. I'm supposed to use mathematical induction (which is simple enough), but I just don't see how to make the algebra work.
$\sum_{j = 0... | You have $$\frac{2^{k+1}+(-1)^k}{3\cdot 2^k}+\left(-\frac{1}{2}\right)^{k+1}.$$Getting a common denominator and combining, we have $$\frac{2^{k+2}+2(-1)^k+3\cdot(-1)^{k+1}}{3\cdot 2^{k+1}}.$$Now we can factor out $(-1)^k$ in the part of the numerator which has it and we get $$\frac{2^{k+2}+(2-3)(-1)^k}{3\cdot 2^{k+1}}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Solving Equations with Factorials I am attempting to solve an equation ${{n-2} \choose {2}} + {{n-3} \choose {2}} + {{n-4} \choose {2}} = 136$. With the formula for a combination being $\frac{n!}{r!(n - r)!}$, I simplified the given equation to:
$(n-2)! + (n-3)! + (n-4)! = 272n - 1088$
However, I am not sure how I wou... | Note that
$$\dbinom{n-2}2 = \dfrac{(n-2)(n-3)}{2}$$
$$\dbinom{n-3}2 = \dfrac{(n-3)(n-4)}{2}$$
$$\dbinom{n-4}2 = \dfrac{(n-4)(n-5)}{2}$$
Hence,
$$\dbinom{n-2}2 + \dbinom{n-3}2 + \dbinom{n-4}2 = \dfrac{(n-2)(n-3)}{2} + \dfrac{(n-3)(n-4)}{2} + \dfrac{(n-4)(n-5)}{2}$$
Now solve the quadratic to get the answer.
EDIT
$$\dbin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/312249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Factorization of cyclic polynomial
Factorize $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
Since this is a cyclic polynomial, factors are also cyclic
$$f(a) = a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$f(b) = b(b^2-c^2)+b(c^2-b^2)+c(b^2-b^2) = 0 \Rightarrow a-b$$
is a factor of the given expression. Therefore, other factors are $... | You can put arbitrary values in $f(a,b,c)$ and get the value of $m$
Let's put $a=1, b=2, c=0$ in $$ a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2) = m(a-b)(b-c)(c-a)$$
we get, $m = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/314105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integral $\int\limits_0^\infty \prod\limits_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx$ Does anybody know how to prove this identity?
$$\int_0^\infty \prod_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left(a+\frac{1}{2}\right)\Gamma... | While this is fairly advanced we can at least bring it into a manageable form which allows specific values to be calculated, assuming $a$ and $b$ are positive integers. Start by evaluating the inner product, which has two parts.
First, $$f_1(x) = \prod_{k=0}^\infty \frac{1}{1+ \frac{x^2}{(a+k)^2}} =
\prod_{k=1}^{a-1}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/314856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
} |
Find of $\int e^x \cos(2x) dx$ I did the following.
Using the LIATE rule:
$$\begin{align*}
u &=& \cos(2x)\\
u\prime &=& -2 \sin(2x)\\
v &=& e^x\\
v\prime&=&e^x
\end{align*}$$
We get:
$$\int e^x \cos(2x)dx = e^x \cos(2x) +2 \int e^x \sin(2x)dx $$
Now we do the second part.
$$\begin{align*}
u &=& \sin(2x)\\
u\prime &=& ... | Hint: Take the derivative of both sides with respect to $x$ to see if you did everything right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/316639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Please, help me to find where is a mistake in the solutions of the equation. I have this equation and I will be very thankful to anyone who can provide me any help with the one discrepancy in my solution and the solution from the self-learning website:
$$
\frac{1+\tan(x) + \tan^2(x) + ... + \tan^n(x) + ...}{1-\tan(x) +... | $$
\require{cancel}
\begin{align}
&\frac{\sin(2x)}{\tan(x)} - \frac{\sin(2x)\cancel{\tan(x)}}{\cancel{\tan(x)}} -\frac{2\cancel{\tan(x)}}{\cancel{\tan(x)}} = 0\\ \\
\end{align}
$$
Here you lost the solution $\tan(x)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding a generating-function using partitions Find a generating function for a , the number of partitions of r into
(a.) Even integers
(b.) Distinct odd integers.
I am at a loss of starting this.
| Let $p_E(n)$ be the number of partitions of $n$ into even parts, where we set $p_E(0)=1$. For each $k\in\Bbb Z^+$ let $p_k(n)$ be the number of ways to partition $n$ into parts of size $2k$; clearly
$$p_k(n)=\begin{cases}
1,&\text{if }2k\mid n\\
0,&\text{otherwise}\;,
\end{cases}$$
so $$\sum_{n\ge 0}p_k(n)x^n=1+x^{2k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/317695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$xy=1 \implies $minimum $x+y=$? If $x,y$ are real positive numbers such that $xy=1$, how can I find the minimum for $x+y$?
| Note that $y=1/x$ and $$\left(x+\frac{1}{x}\right)^2=\left( x-\frac{1}{x}\right)^2+4\cdot x\cdot \frac{1}{x}\geq4$$ Hence $x+y\geq 2$. Also for $x=y=1$, you have $x+y=2$, hence $2$ indeed the minimum value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 13,
"answer_id": 0
} |
Evaluating $\frac{d^2y}{dx^2}$ I want to evaluate $\frac{d^2y}{dx^2}$ from the given data: $x=a\cos^3\theta$ and $y=b\sin^3\theta$.
I tried in this way- $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\cos^6\theta+\sin^6\theta=1-3(\frac{xy}{ab})^\frac{2}{3}$. After that I seem to be lost and will appreciate some help in doing this ef... | $\displaystyle x=a\cos^3\theta \Rightarrow \frac {dx}{d\theta}=-3a\cos^2\theta\sin\theta$
$\displaystyle y=b\sin^3\theta \Rightarrow \frac {dy}{d\theta}=3b\sin^2\theta\cos\theta$
$\displaystyle \Rightarrow\frac {dy}{dx}=\frac{\frac {dy}{d\theta}}{\frac {dx}{d\theta}}=\frac{3b\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/319464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Partial Fraction Decomposition $\frac{2x^2+3x+3}{(x+1)(x^2+1)}$ How would one go about decomposing this fraction?
$$
2x^2+3x+3\over
(x+1)(x^2+1)
$$
Here is what I have so far:
$$
{2x^2+3x+3\over
(x+1)(x^2+1)
} =
{ A\over x+1 } + { Bx+C\over x^2+1} $$
$$
{2x^2+3x+3\over
(x+1)(x^2+1)
} =
{ A(x^2+1) } + (Bx+C)(x+1) $$
... | You should have
$2x^2 + 3x + 3 = A(x^2+1) + (Bx+C)(x+1)$
Now, try suitable values of $x$ to find $A, B, C$. For e.g. $x = -1$ will easily give you $A$. Hint, there is another value which easily gives you $C$ after you know $A$.
In case no suitable $x$ values occur to you, just use a few simple values to get as many... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/320260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integration by anti-differentiation I have a question in my assignment, and I highly suspect there is a typo somewhere.
Evaluate the following integral:
$\int \frac{1}{(1-t)^{\frac{3}{2}}} dt = \frac{t}{\sqrt{1-t^2}} + c$
What I did was to perform differentiation and got the following:
$$
\begin{align*}
\frac{d}{dt... | You made an error in differentiating. It should be
\begin{align*}
\frac{d}{dt}\frac{t}{\sqrt{1-t^2}} &= \frac{\sqrt{1-t^2} - (0.5)(1-t^2)^{-\frac 1 2}(-2t)(t)}
{1-t^2}
\\&=\frac{\sqrt{1-t^2} + t^2(1-t^2)^{-\frac 1 2}}{1-t^2}
\\&=\frac{1 - t^2 + t^2}{(1 - t^2)^{3/2}}
\\&=\frac{1}{(1 - t^2)^{3/2}}.
\end{align*}
So the in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/323186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trig Integral (Definite) How do I evaluate this definite integral?
$$\int_{0}^{\frac{\pi}{12}}{\sin^4x \, \cos^4x\, \operatorname{d}\!x}$$
I know this is a trig. function.
| From here, we have
$$\int_0^{\pi/2} \sin^{2n}(x) dx = \dfrac{2n-1}{2n} \cdot \dfrac{2n-3}{2n-2} \cdots \dfrac34 \cdot \dfrac12 \cdot\dfrac{\pi}2$$
Hence,
$$\int_0^{\pi/2} \sin^4(x) \cos^4(x) dx = \dfrac1{16} \int_0^{\pi/2} \sin^4(2x) dx = \dfrac1{32} \int_0^{\pi} \sin^4(t) dt = \dfrac2{32} \int_0^{\pi/2} \sin^4(t) dt$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/323783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Results of the multivariate function $f(x,y) = 3x^2 - 2xy + y^3 $ I was asked to derive 2 experessions for this multivariate function
$$
f(x,y) = 3x^2 -2xy+y^3
$$
The first is $\large{\frac{f(x+h,y) - f(x,y)}{h}}$ and the other is $\large{\frac{f(x,y+k) - f(x,y)}{k}}$
The working is as follows:
$$
\begin{align*}
\frac{... | Yes. The calculations are correct.
Note that when $h\to 0$, the first expression is the partial derivative of $f(x,y)$ by $x$; and when $k\to 0$, the second expression is the partial derivative of $f(x,y)$ by $y$.
That is, we define the partial derivative $$\frac{\partial f}{\partial x}(x,y)=\lim_{h\to0}\frac{f(x+h,y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/324769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show determinant of matrix is non-zero I have $a,b,c\in\mathbb{Q}$ not all zero. ($a^2+b^2+c^2\ne 0$), I want to show that the following determinant is then non-zero. I failed to arrive at an appropriate form of the polynomial. Help please.
$$\left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right|... | Since $a,\ b$ and $c$ are rational, we may clear denominators in $$a^3 + 2b^3 -6abc +4c^3 = 0$$ The above equation is a homogenous equation of degree $3$ so we may cancel common factors. If there exists non-trivial solutions to the equation, we may therefore assume without loss of generality that $a,\ b$ and $c$ are in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/327884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
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$x^2-y^2=n$ has solution in $\Bbb Z$ IFF : $n \not \equiv 2 \pmod 4$ suppose $n\in \Bbb Z $ then how to prove this statement:
$x^2-y^2=n$ has solution in $\Bbb Z$ IFF : $n \not \equiv 2 \pmod 4$
| Observe that $x+y=x-y+2y$
So, $x+y,x-y$ have the parity
If they are odd, $n=x^2-y^2$ will be odd
If they are even, $n=x^2-y^2$ will be divisible by $4$
Conversely, if $n$ is odd $=2m+1$(say)
$(x+y)(x-y)=2m+1$ where $m$ is any integer
So, we can write $x+y=2m+1,x-y=1\implies x=m+1,y=m$
or $x+y=-(2m+1),x-y=-1\implies x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/328195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Range of $\frac{1}{2\cos x-1}$ How can we find the range of $$f(x) =\frac{1}{2\cos x-1}$$
Since range of $\cos x$ can be given as : $-1 \leq \cos x \leq 1$
therefore we can proceed as :$$\begin{array}{rcl}
-2 \leq & 2\cos x & \leq 2 \\
-2-1 \leq & 2\cos x -1 & \leq 2-1\\
-3 \... | As $2\cos x-1$ varies through $[-3,0)$, $\frac{1}{2\cos x-1}$ varies through $(-\infty,-\frac13]$.
As $2\cos x-1$ varies through $(0,1]$, $\frac{1}{2\cos x-1}$ varies through $[1,+\infty)$.
Note that division by $0$ is undefined, so we do not consider the case $2\cos x-1=0$.
Hence the range is $(-\infty,-\frac13]\cup [... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$y^2=x^3+7$ has an integral soln. trivial issues that are imposing me from understanding a proof from a book. I have some trivial issues that are imposing me from understanding a proof from a book.
(i) if $y^2=8k^3+7$ where $k\in \mathbb{Z}$ $\implies$ $y^2\equiv7 \,[8]$ which is not possible, because $(\frac{7}{8})=-1... | Here, I will try to be as explicit as possible to show that $y^2=x^3+7$ has no integer solution.
Case 1: Suppose that $x$ is even, so $x=2k$. It follows that $y^2=8k^3+7$. This implies that $y^2\equiv 7\mod 8$. This is a contradiction because there is no $y$ such that $y^2\equiv 7\mod 8$. You can check this by squa... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Show that $(n - 1)2^{n+1} + 2 + (n+1)2^{n+1} = n(2^{n+2})+2$ I'm having a really hard time showing this equality is true, I've tried several ways of going about it and I just can't seem to make it work. Help!
$(n - 1)2^{n+1} + 2 + (n+1)2^{n+1} = n(2^{n+2})+2$
Thanks!
| Just let $A=2^n$. Then $2^{n+1}=2\cdot 2^n=2A$ and $2^{n+2}=2^2\cdot 2^n = 4A$. Then your equality reads:
$$2(n - 1)A + 2 + 2(n+1)A = 4nA+2.$$
Let us show that this is indeed true:
$$2(n - 1)A + 2 + 2(n+1)A = 2A(n-1+n+1)+2=2A(2n)+2=4nA+2,$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $\int_0^\infty \frac{\ln ^2z} {1+z^2}{d}z$ How to find the value of the integral $$\int_{0}^{\infty} \frac{\ln^2z}{1+z^2}{d}z$$ without using contour integration - using usual special functions, e.g. zeta/gamma/beta/etc.
Thank you.
| \begin{align}J&=\int_0^\infty \frac{\ln^2 x}{1+x^2}dx\\
K&=\int_0^\infty\int_0^\infty \frac{\ln^2(xy)}{(1+x^2)(1+y^2)}dxdy\\
&\overset{u(x)=xy}=\int_0^\infty\int_0^\infty \frac{y\ln^2 u}{(u^2+y^2)(1+y^2)}dudy\\
&=\frac{1}{2}\int_0^\infty \ln^2 u\left[\frac{\ln\left(\frac{1+y^2}{u^2+y^2}\right)}{u^2-1}\right]_{y=0}^{y=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/333672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Trying to find a function whose derivative is $\cfrac{4x^3}{x^4+1}$ any ideas? The closet I can get is
$\arctan(x^4) $
the derivative of $\arctan(x^4)$ is $\dfrac{4x^3}{x^8 + 1} ...$
Any tips?
| Hint: Note that the derivative of $x^4 + 1\;$ is $4x^3$.
If we let $f(x) = x^4 + 1$, then we know $f'(x) = 4x^3$.
Then note that $$\frac{4x^3}{x^4 + 1} = \dfrac{f'(x)}{f(x)}\tag{1}$$
And all integrals of the form $\displaystyle \int \dfrac{f'x}{f(x)}\,dx $ evaluate as $$\int \dfrac{f'x}{f(x)}\,dx = \ln|f(x)| + C \tag... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Solving recursion with generating function I am trying to solve a recursion with generating function, but somehow I ended up with mess.....
$$y_n=y_{n-1}-2y_{n-2}+4^{n-2}, y_0=2,y_1=1 $$
\begin{eqnarray*}
g(x)&=&y_0+y_1x+\sum_2^{\infty}(y_{n-1}-2y_{n-2}+4^{n-2})x^n\\
&=&2+x+\sum^{\infty}_2y_{n-1}x^n-2\sum_2^{\infty}y_{... | I get a slightly different result from you. Let
$$y(x) = \sum_{n=0}^{\infty} y_n x^n$$
Then, summing the recurrence relation from $n=2$ on, I get
$$y(x) - y_0 - y_1 x - x [y(x)-y_0] + 2 x^2 y(x) = \frac{x^2}{1-4 x}$$
Simplifying, using the initial conditions $y_0=2$ and $y_1=1$:
$$(2 x^2-x+1)y(x) = \frac{x^2}{1-4 x} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/335289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a ... | By Holder $$\left(\sum_{cyc}\sqrt{3a+b^3}\right)^2\sum_{cyc}\frac{(a+b)^3}{3a+b^3}\geq8(a+b+c)^3=216.$$
Thus, it's enough to prove that:
$$\sum_{cyc}\frac{(a+b)^3}{3a+b^3}\leq6$$ or $$\sum_{cyc}\frac{(a+b)^3}{a(a+b+c)^2+3b^3}\leq2$$ or
$$\sum_{cyc}(2a^8b+5a^7b^2-3a^6b^3+3a^6c^3-a^5b^4+17a^5c^4)+$$
$$+abc\sum_{cyc}(12a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 2
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Analytic Geometry (high school): Why is the sum of the distances from any point of the ellipse to the two foci the major axis? I don't understand where that formula came from. Could someone explain? For example any point $(x,y)$ on the ellipse from the two foci $(-c,0)$ and $(c,0)$ is equal to $2a$ where $2a$ is the di... | The formula for an ellipse centred at the origin is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \tag{1} $$ where a>b.
Now let us define a constant (note at this point I'm just defining a constant - nothing about foci)
$$f = \sqrt{a^2 - b^2}\tag{2}$$
Let us now pick 2 points on the x axis (f,0) and (-f,0).
The distance from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| Since infinite series with nonnegative terms can be rearranged arbitrarily,
$$\sum_{i=1}^\infty \frac{i}{2^i} =
\sum_{i=1}^\infty \sum_{j=1}^i \frac{1}{2^i}
= \sum_{j=1}^\infty \sum_{i=j}^\infty \frac{1}{2^i}
= \sum_{j=1}^\infty \frac{1}{2^{j-1}}
= 2
$$
More graphically,
1/2 + 2/4 + 3/8 + 4/16 + ...
= 1/2 + 1/4 + 1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 12,
"answer_id": 2
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Recurrence relation for the number of $n$-digit ternary sequences with no consecutive $1$s or $2$s
Find the recurrence relation for the number of $n$-digit ternary sequences with no consecutive $1$'s or $2$'s.
The solution is
$$
a_n = a_{n-1} + 2a_{n-2} + 2a_{n-3} + 2a_{n-4} + \dots. \tag1
$$
I've thought about this... | Call the number of sequences that interest you that end in 0 $a_n$, if they end in 1 $b_n$, $c_n$ if they end in 2, and $d_n$ for ending in 3. Then clearly:
$$
a_0 = b_0 = c_0 = d_0 = 1
$$
Now think how a sequence that ends in 0 can be built: Add a 0 to any of the others, so $a_{n + 1} = a_n + b_n + c_n + d_n$. To make... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Use the $\varepsilon$ - $\delta$ definition to prove $\lim_{x\to\,-1}\frac{x}{2x+1}=1$ Use the $\varepsilon$ - $\delta$ definition of limit to prove that $\displaystyle\lim_{x\to\,-1}\frac{x}{2x+1}=1$.
My working:
$\left|\frac{x}{2x+1}-1\right|=\left|\frac{-x-1}{2x+1}\right|=\frac{1}{\left|2x+1\right|}\cdot \left|x+1\r... | Hint: Note that
\begin{align}
\left| \frac{x}{2\cdot x +1} - 1 \right| < \epsilon
\Longleftrightarrow
&
1-\epsilon< \frac{x}{2\cdot x +1}< 1+\epsilon
\\
\Longleftrightarrow
&
\left\{
\begin{array}
(2x+1)(1-\epsilon)<x \\
\\
x< (2x+1)(1+\epsilon)
\end{array}
\right.
\end{align}
After algebric manipulations of two las... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
generating functions / combinatorics Calculate number of solutions of the following equations:
$$ x_1 + x_2 + x_3 + x_4 = 15 $$
where $ 0 \le x_i < i + 4 $
I try to solve it using generating functions/enumerators :
$$ (1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^4+x^5+x^6)(1+x+x^2+x^3+x^4+x^5+x^6+x^7)$$
and ta... | Use the stars and bars technique to solve for the number of solutions without restrictions, and look at how to count the solutions that violate the restrictions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/339230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How do I write this matrix in Jordan-Normal Form I have the matrix $A=\begin{pmatrix}2&2&1\\-1&0&1\\4&1&-1\end{pmatrix}$, I want to write it in Jordan-Normal Form. I have $x_1=3,x_2=x_3=-1$ and calculated eigenvectors $v_1=\begin{pmatrix}1\\0\\1\end{pmatrix},v_2=\begin{pmatrix}1\\-4\\5\end{pmatrix},v_3=\begin{pmatrix}... | You need a generalized eigenvector for the third eigenvalue and it looks like that went wrong somehow, so lets fix it.
We have $\left(A - \lambda_2 I\right)v_3 = v_2$
From this, we get:
$\begin{pmatrix}3&2&1&1\\-1&1&1&-4\\4&1&0&5\end{pmatrix}$
The RREF yields:
$\displaystyle \begin{pmatrix}1&0&-\frac{1}{5}&\frac{9}{5}\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Sum of $\sum_{n=0}^\infty \frac{(x+2)^{n+2}}{3^n} $ Calculate the sum of the next series and for which values of $x$ it converges:
$$\sum_{n=0}^\infty \frac{(x+2)^{n+2}}{3^n}$$
I used D'Alembert and found that the limit is less than 1, so: $-5 < x < 1$ (because the fraction must be less than 1).
and then I assigned the... | for $|(x+2)/3|<1$ it converges to the limit given by multiplication of geometric series limit and polynomial:
$$
\sum_{n=0}^\infty\frac{(x+2)^{n+2}}{3^n}=(x+2)^2\sum_{n=0}^\infty\left(\frac{x+2}{3}\right)^n=(x+2)^2\frac{1}{1-\frac{x+2}{3}} =\frac{3(x+2)^2}{1-x}
$$
for $|(x+2)/3|\geq 1$, the sum is not convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/348079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Integrate rational function $\frac{x^2}{1+x^4}$ Integrate $$\int\frac{x^2dx}{1+x^4}$$
I've factored the denominator to $(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$ and got stuck.
| The next step is to perform a partial fraction decomposition:
$$
\frac{x^2}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)} = \frac{1}{2 \sqrt{2}} \frac{x}{x^2-\sqrt{2}x+1} - \frac{1}{2 \sqrt{2}} \frac{x}{x^2+\sqrt{2}x+1}
$$
And the than use the table anti-derivative for $\int \frac{x}{x^2 + a x+b} \mathrm{d}x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Help please on complex polynomials I wanted to know if there's any good approaches to these questions
a)By considering $z^9-1$ as a difference of two cubes, write $1+z+z^2+z^3+z^4+z^5+z^6+z^7+z^8$
as a product of two real factors one of which is a quadratic.
b) Solve $z^9-1=0$ and hence write down the 6 solutions of $z... | Hint: $a^3-b^3 = (a-b) (a^2+a b+b^2)$. Think about this with $a=z^3$ and $b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/350879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Cauchy's Residue Theorem I need to evaluate $\int_C \frac{5z-2}{z(z-1)}dz$ where $C$ is the circle $|z|$=2. I used partial fraction decomposition to get $$\frac{5z-2}{z(z-1)}=\frac{2}{z}+\frac{3}{z-1}$$ I have the answer ($10\pi i$) in my book but I don't fully understand how to get it. I think I have to evaluate in th... | I don't think you need the Residue Theorem, but just Cauchy's integral formula? Note that $$\dfrac{5z - 2}{z(z-1)} = \frac{3}{z-1} + \frac{2}{z}$$ and so $$\int_{C} \frac{5z-2}{z(z-1)}dz = \int_{C}\frac{3}{z-1} dz + \int_{C}\frac{2}{z} dz = 2\pi i f_1(1) + 2\pi i f_2(0)$$ where $f_1(z) = 3$ and $f_2(z) = 2$, so $2\pi i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
arithmetic progression of triangle sides Let $gcd(a,b,c)=1$ such that $a^2, b^2, c^2$ are in arithmetic progression. Show they can be written in the form
$a=-p^2+2pq+q^2$
$b=p^2+q^2$
$c=p^2+2pq-q^2$
for relatively prime integers $p,q$ of different parities.
so if $a^2, b^2, c^2$ are in arithmetic progression, then
$a... | $a^2, b^2,$ and $c^2$ are in AP if and only if $a^2+c^2=2b^2$.
Looking at $a,b,$ and $c$ modulo $4$, we see that they must all be odd integers.
Then $u=\frac 12(c-a)$ and $w=\frac 12(c+a)$ are also integers.
In fact $a=w-u$ and $c=w+u$.
So $2b^2=a^2+c^2=(w-u)^2+(w+u)^2=2(u^2+w^2)$.
Hence $u^2+w^2 = b^2$.
So there ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Sketch of the ordinate set of $f$ Let $f$ be defined on $[0,1] \times [0,1]$ as follows:
$f(x,y)= \begin{cases} x+y \mbox{ if } x^2 \leq y \leq 2x^2 \\ 0 \mbox{ otherwise} \end{cases}$
I want to make a sketch of the ordinate set of $f$ over $[0,1] \times [0,1]$ and compute the volume of this ordinate set by double int... | Comment by OP:
If $f$ is nonnegative, the set $S$ of points $(x,y,z)$ in 3-space with
$(x,y)$ in $[0,1]^2$ and $0\le z\le f(x,y)$ is called the ordinate set
of $f$ over $[0,1]$.
My interpretations is as follows. The picture represents a region $R$ bounded by $y=2x^2$, $y=x^2$ (with $0\le x\le 1$) and $y=1$.
$$y... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Really Confused on a surface area integral can't seem to finish the integral off. Basically the question asks to compute $\int \int_{S} ( x^{2}+y^{2}) dA$ where S is the portion of the sphere $x^{2} + y^{2}+ z^{2}= 4$ and $z \in [1,2]$ we start with a chnage of variables
$x=x $
$y=y$
$ z= 2 \cdot(4-(x^{2} + y^{2}... | It seems that spherical coordinates are more appropriate for calculating $\iint\limits_{S} ( x^{2}+y^{2}) dA$.
In the last integral
$$
2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}rdrd\theta=
{4\pi}\int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}r\,dr$$
you can make the substitution $r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/356385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve : $ab(a+b)(a-b)=c^2-1$ As we know that $ab(a+b)(a-b)=c^2$ has no integer solution in $Z^+$.However, it seems that $$ab(a+b)(a-b)=c^2-1$$
has infinite positive integer solutions,could you prove it?
Here are some of them:
$(a,b,c)=(3, 1, 5), (5, 1, 11), (7, 3, 29), (7, 5, 29), (8, 2, 31), (8, 7, 29), (9, 8, 35),... | Let us want
$$a(a-b) = c+1 \text{ and }b(a+b) = c-1$$
For this, we need
$$2 = a(a-b) - b(a+b) = a^2 - 2ab -b^2 = (a-b)^2 - 2b^2$$
Hence, all we want is to solve the Pell's equation
$$x^2-2y^2 =2$$
One trial solution to the above is $(2,1)$ and now generate the rest of $x$ and $y$. From this we get $b=y$ and $a = x+y$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/356748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Computing the integral $\int \sqrt{\frac{x-1}{x+1}}\,\mathrm dx $ How do I compute the next integral:
$$\int \sqrt{\frac{x-1}{x+1}}\,\mathrm dx \;?$$
| Let $\sqrt{\dfrac{x-1}{x+1}} = t$. We then get that
$$\dfrac{x-1}{x+1} = t^2 \implies x-1 = t^2(x+1) \implies x = \dfrac{1+t^2}{1-t^2} \implies dx = \dfrac{4t}{(1-t^2)^2}dt$$
Hence, we get that
$$\int \sqrt{\dfrac{x-1}{x+1}} dx = \int \dfrac{4t^2}{(1-t^2)^2} dt$$ I trust you can take it from here via the method of part... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/357306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Constructing a degree 4 rational polynomial satisfying $f(\sqrt{2}+\sqrt{3}) = 0$ Goal: Find $f \in \mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3}) = 0$.
A direct approach is to look at the following
$$
\begin{align}
(\sqrt{2}+\sqrt{3})^2 &= 5+2\sqrt{6} \\
(\sqrt{2}+\sqrt{3})^4 &= (5+2\sqrt{6})^2 = 49+20\sqrt{6} \\
\end... | You can guess that the conjugates will be $\pm \sqrt 2 \pm \sqrt 3$, and multiply all the corresponding linear factors together.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
$x^4 + 4x^3 - 2x^2 - 12x + k$ has 4 real roots. Find the condition on k. The question is: $f(x) = x^4 + 4x^3 - 2x^2 - 12x + k$ has 4 real roots. What values can k take? Please drop a hint!
| Hint:
Lets call vieta for the rescue.
$f(x) = x^4 + 4x^3 - 2x^2 - 12x + k=0$, let $x_1, x_2,x_3$ and $x_4$are the roots.
$\sum x_i=-4$
$\sum x_ix_j=-2$
$\sum x_ix_jx_k=+12$
$x_1x_2x_3x_4=k$
$\sum (x_i)^2=(\sum x_i)^2-2\sum x_ix_j=16-(-4)=12$
$\dfrac{x_1^2+x_2^2+x_3^2+x_4^2}{4} \ge (x_1x_2x_3x_4)^\frac{1}{2}$
$3 \ge k^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/359513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How do I evaluate $\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots}{n^3}$ How to find this limit : $$\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots+n(n+1)}{n^3}$$ As, if we look this limit problem viz. $\lim_{x \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ then w... | $$\sum_{1\le r\le n}r(r+1)=\sum_{1\le r\le n}r^2+\sum_{1\le r\le n}r=\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2$$
So, $$\lim_{n\to\infty}\frac{\sum_{1\le r\le n}r(r+1)}{n^3}$$
$$=\lim_{n\to\infty}\left(\frac{\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2}{n^3}\right)$$
$$=\lim_{n\to\infty}\left(\frac{(1+\frac1n)(2+\frac1n)}6+\frac{(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/364284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Limits at negative infinity Prove that $\lim_{x\to -\infty}$ $\frac{x}{x+2}$ = 1.
Please help me check if this is correct:
Let $\varepsilon > 0$ be given. Choose $M=2-\frac{2}{\varepsilon}$
$$\begin{align}
x<M \Rightarrow x-2 < M-2\\
\frac{1}{x-2} > \frac{1}{M-2}\\
|f(x)-1|=|\frac{x}{x-2}-1|=|\frac{2}{x-2}|\\
... | Since $x \to -\infty$, we may assume $M < 0$. Then your line $\frac{1}{x-2} > \frac{1}{M-2}$ is a relationship between two negative quantities. Thus $\left|\frac{1}{x-2}\right| < \left|\frac{1}{M-2}\right|$. Can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/364432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing $\int\frac{7x^{13}+5x^{15}}{(x^7+x^2+1)^3}\,dx$
Compute the indefinite integral
$$
\int\frac{7x^{13}+5x^{15}}{(x^7+x^2+1)^3}\,dx
$$
My Attempt:
$$
\int\frac{7x^{13}+5x^{15}}{x^{21}(x^{-7}+x^{-5}+1)^3}\,dx = \int\frac{7x^{-8}+5x^{-6}}{(x^{-7}+x^{-5}+1)^3}\,dx
$$
Let $t=(x^{-7}+x^{-5}+1)$ such that
$$
\begin{... | Your first computation is correct.
If $P$ is a polynomial of degree $n$, then
$$
\frac{d}{dx} \Biggl(\frac{P(x)}{(x^7+x^2+1)^2}\Biggr)=\frac{Q(x)}{(x^7+x^2+1)^3}
$$
where
$$
Q(x)=(x^7+x^2+1)P'(x)-2(7\,x^6+2\,x)P(x)
$$
is a polynomial of degree $\le n+6$. To get $x^{15}$ in the numerator you need $n\ge9$ (in fact, in t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/365313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Determine the limiting behaviour of $\lim_{x \to \infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$ Determine the limiting behaviour of $\lim_{x \to \infty}{\dfrac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$
Used L'Hopitals to get $\;\dfrac{(x^6+1)^{\frac{2}{3}}}{x^2 \sqrt{x^4+1}}$ but not sure what more i can do after that.
| $$ {\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}\sim {\frac{\sqrt{x^4}}{\sqrt[3]{x^6}}}=\frac{x^2}{x^2}=1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
How to find $\cos A \cos B - \sin A \sin B$? Given that:
$\tan A=1$
and
$\tan B = \sqrt{3}$
How would you find $\cos A \cos B - \sin A \sin B$?
EDIT: This is what I've tried after reading bhattacharjee's answer:
$$ \tan(A+B) = \tan A+\tan B−\tan A\tan B$$
so,
$\tan(A+B)= {1+\sqrt{3} \over 1-\sqrt{3}}$
from this I g... | HINT:
$\cos A \cos B - \sin A \sin B=\cos(A+B)$
and $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
Do you know how to find $\cos \theta$ from $\tan\theta?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding the general formula $a_n$ for $a_n = \frac{1}{2a_{n-1}} + 2a_{n-2}$ How to calculate the general formula $a_n$ for the following sequence:
$$a_n = \frac{1}{2a_{n-1}} + 2a_{n-2}$$
where $a_1=\frac{1}{2}, a_2=\frac{1}{4}$
| $$\begin{align}&a_{n} = \frac{1}{2 a_{n-1}} + 2 a_{n-2}\\
\iff & 2 a_{n} a_{n-1} + 1 = 2 ( 2 a_{n-1} a_{n-2} + 1)\\
\implies & 2 a_{n} a_{n-1} + 1 = 2^{n-2} (2 a_2 a_1 + 1 ) = \frac{5}{16} 2^n\\
\implies & a_{n}/a_{n-2} = \frac{\frac{5}{16} 2^n - 1}{\frac{5}{32} 2^n - 1}\\
\implies & a_{n} = \begin{cases}
a_2 \prod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/371233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Expanding complex geometric series I'm having trouble with part $ii.$ of the following question:
$i.$ Express the following in terms of N and z:
$$\sum^N_{n=1}2^{-n}z^n$$
Expanding with geometric series:
$$\sum^N_{n=1}2^{-n}z^n \equiv \sum^N_{n=1}\bigg(\frac{z}2\bigg)^n = \bigg(\frac{z}2\bigg) + \bigg(\frac{z}2\big... | HINT:
$$\frac{\cos t+i\sin t}{2-\cos t-i\sin t}$$
$$=\frac{(\cos t+i\sin t)(2-\cos t+i\sin t)}{(2-\cos t)^2+\sin^2 t}$$
$$=\frac{(2\cos t-1+2i\sin t)}{5-4\cos t}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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.