Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Trace constraints and rank-one positive semi-definite matrices. Let $C_1, C_2, \dots, C_N$ be $M \times M$ Hermitian matrices and $c > 0$ be a given constant. Let $W$ be a positive (possibly semi) definite matrix such that
\begin{align}
\text{trace}\{WC_1\} & \geq c \\
\text{trace}\{WC_2\} & \geq c \\
\vdots ... | Here is a counterexample: $$C_1=\begin{pmatrix}2&3\\3&-1\end{pmatrix} \quad C_2=\begin{pmatrix}2&-3\\-3&-1\end{pmatrix} \quad C_3=\begin{pmatrix}2&-3i\\3i&-1\end{pmatrix}\quad C_4=\begin{pmatrix}2&3i\\-3i&-1\end{pmatrix} \\
C_5=\begin{pmatrix}-1&3\\3&2\end{pmatrix} \quad C_6=\begin{pmatrix}-1&-3\\-3&2\end{pmatrix} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/361433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Coefficient of Taylor Series of $\sqrt{1+x}$ The coefficient of $x^3$ in the Taylor series of the function $f(x) = \sqrt{1+x}$ about the point $a = 0$ is $$1\over 16$$
Can someone show me how to get this value?
| Another way to see this is to expand in Binomial series:
$$
(1+x)^{\frac{1}{2}} = \sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}x^k
$$
Although $\binom{\frac{1}{2}}{k}$ looks intimidating, it's nothing when you replace $\frac{1}{2} = \alpha \Rightarrow \binom{\alpha}{k} = \alpha\cdot (\alpha-1)\cdots(\alpha-k+1) \cdot \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/362146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Does $y(y+1) \leq (x+1)^2$ imply $y(y-1) \leq x^2$? Can anyone see how to prove the following?
If $x$ and $y$ are real numbers with $y\geq 0$ and $y(y+1) \leq (x+1)^2$ then $y(y-1) \leq x^2$.
It seems it is true at least according to Mathematica.
| Given $y^2 + y \le x^2 + 2x + 1$, if possible, let $x^2 < y(y-1)$. Clearly $y > 1$.
Then $x^2 + (2x + 1) < y^2 - y + (2x + 1)$
So $y^2 + y < y^2 - y + 2x + 1$, which resolves to $y < x + \frac{1}{2}$.
Hence we also have $y - 1 < x - \frac{1}{2}$.
As $y > 1$, the LHS is positive, and we can multiply the last two to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/363169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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In how many ways i can write 12? In how many ways i can write 12 as an ordered sum of integers where
the smallest of that integers is 2? for example 2+10 ; 10+2 ; 2+5+2+3 ; 5+2+2+3;
2+2+2+2+2+2;2+4+6; and many more
| Let's split the question into a few pieces.
First part: how many ways to sum if all numbers are even? In this case, it's a simple partition problem, and the answer is equally simple. If you imagine the digits like this:
..|..|..|..|..|.. = 2+2+2+2+2+2
Then each of the five dividers can either be there or not, giving ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/364211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Solving a set of 3 Nonlinear Equations In the following 3 equations:
$$
k_1\cos^2(\theta)+k_2\sin^2(\theta) = c_1
$$
$$
2(k_2-k_1)\cos(\theta)\sin(\theta)=c_2
$$
$$
k_1\sin^2(\theta)+k_2\cos^2(\theta) = c_3
$$
$c_1$, $c_2$ and $c_3$ are given, and $k_1$, $k_2$ and $\theta$ are the unknowns. What is the best way to solv... | Re-write in terms of double-angle expressions, and define values $A$, $B$, $C$:
$$\begin{align}
k_1 \cos^2\theta + k_2 \sin^2\theta = c_1 &\quad\implies\quad \cos 2\theta = \frac{2 c_1-k_1-k_2}{k_1-k_2} & =: A \\[5pt]
2 \left(k_1-k_2\right) \sin\theta \cos\theta = c_2 &\quad\implies\quad \sin 2\theta = \frac{c_2}{k_1-k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/364565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Summing $ \sum _{k=1}^{n} k\cos(k\theta) $ and $ \sum _{k=1}^{n} k\sin(k\theta) $ I'm trying to find
$$\sum _{k=1}^{n} k\cos(k\theta)\qquad\text{and}\qquad\sum _{k=1}^{n} k\sin(k\theta)$$
I tried working with complex numbers, defining $z=\cos(\theta)+ i \sin(\theta)$ and using De Movire's, but so far nothing has come ... | HINT:
$$\sum_{1\le k\le n}\left(\cos k\theta+i\sin k\theta\right)=\sum_{1\le k\le n}e^{i k\theta}=e^{i\theta}\cdot\frac{e^{in\theta}-1}
{e^{i\theta}-1}$$
Differentiate wrt $\theta$ and equate the real & the imaginary parts
$$\text{On differentiation, the left hand becomes }\sum_{1\le k\le n}k\left(-\sin k\theta+i\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/364631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Vectors and subspace For which real numbers $w, x, y, z$ do the vectors: $(w, 1, 1, 1)$, $(1, x, 1, 1)$, $(1, 1, y, 1)$, $(1, 1, 1, z)$ NOT form a basis of $\mathbb{R}^4$? For each of the values that you found, what is the dimension of the subspace of $\mathbb{R}^4$ that they span?
| The fact that the vectors does not form a basis of $\mathbb{R}^4$ means that they are not all linearly independent, which in turn is equivalent to:
$$det(A) = \begin{vmatrix} w & 1 & 1 & 1 \\ 1 & x & 1 & 1 \\ 1 & 1 & y & 1 \\ 1 & 1 & 1 & z \end{vmatrix} = 0$$
The determinant can be caluclated by expanding it using co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/366985",
"timestamp": "2023-03-29T00:00:00",
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Maximal Smallest Number Ran into a interesting problem which I have no idea how to solve but have the desire to.
Let a and b be two positive real numbers and let $m$(a,b) be the smallest of the three numbers $a,$ $1/b$ and $1/a + b.$ For which pairs (a,b) is $m$(a,b) maximal? Prove!
| It is not too hard to see that the maximum can only occur for $a=\frac{1}{b}=\frac{1}{a}+b$.
Suppose for instance that $a<\min(\frac{1}{b},b+\frac{1}{a})$, then we can increase $a$ a little and improve our $m(a,b)$.
For $\frac{1}{b}$ and $b+\frac{1}{a}$ similar things happen.
Thus $a=\frac{1}{b} = b+\frac{1}{a}$ implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/367748",
"timestamp": "2023-03-29T00:00:00",
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Symmetry properties of $\sin$ and $\cos$. Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$? For a question such as:
If $\sin(x) = 0.34$, find the value of $\cos\left(\frac{3\pi}{2} - x\right)$.
The solution says that:
\begin{align*}
\cos\left(\frac{3\pi}{2} - x\right) &= \cos\left... | Recall your elementary trigonometric identity $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
So we may begin by expanding $\cos\left(\frac\pi2-x\right)$.$$\begin{align*}\cos\left(\frac\pi2-x\right)&=\cos\frac\pi2\cos(-x)-\sin\frac\pi2\sin(-x)\\&=0(\cos x)-1(-\sin x)\text{ knowing that }\cos x,\sin x\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/367938",
"timestamp": "2023-03-29T00:00:00",
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A question on a proof of $y(y+1) \le (x+1)^2 \implies y(y-1) \le x^2$ I was reading the question posted here:Does $y(y+1) \leq (x+1)^2$ imply $y(y-1) \leq x^2$?. The solution posted was:
"Given $y^2+y≤x^2+2x+1$, if possible, let $x2<y(y−1)$. Clearly $y>1.$
Then $x^2+(2x+1)<y^2−y+(2x+1)$
So $y^2+y<y^2−y+2x+1$, which r... | To reword the proof given there:
We start with real numbers $x,y$ such that
$$\tag1 y\ge0 $$
$$\tag2 y^2+y\le x^2+2x+1$$
$$\tag3 x^2<y(y-1)$$
If we had $y\le1$, then the right hand side in $(3)$ would be the product of a nonnegative (according to $(1)$) and a nonpositve (by assumtion) factor, hence nonnegative, contrad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the values of $x,y$ and $z$ in these equations I'm stuck with these equations. Can somebody help me with solving it?
If $$(a+b) = \frac{x(y+z)}{x+y+z},\quad (b+c) = \frac{y(x+z)}{x+y+z} ,\quad (a+c) = \frac{z(x+y)}{x+y+z}$$ then find $x,y,z$.
| On adding all the 3 we get,
$$a+b+c=\frac{xy+yz+zx}{x+y+z}$$
Subtracting from the given equations we get $$c=\frac{yz}{x+y+z}$$ $$b=\frac{xz}{x+y+z}$$ $$a=\frac{xy}{x+y+z}$$
$$\implies \frac{y}{x}=\frac{c}{b}$$
$$\implies \frac{z}{x}=\frac{c}{a}$$
$$\implies \frac{z}{y}=\frac{b}{a}$$
$$a=\frac{y}{1+\frac{y}{x}+\frac{z}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sums that are pythagorean and normal I noticed that
$3^2+4^2+15^2=9^2+13^2$
and also
$3+4+15=9+13$
Is there an easy way to find all pairs of disjoint sets of positive integers whose sum are the same and whose sum of squares are the same? How common are they?
| set of equations:
$\left\{\begin{aligned}&x+y+z=a+b\\&x^2+y^2+z^2=a^2+b^2\end{aligned}\right.$
Solutions and can be written as follows.
$x=t+qk+qt+k$
$y=t$
$z=qk$
$a=t+qk+k$
$b=t+qk+qt$
And more:
$z=kq$
$y=(k+q+t)^2$
$x=t^2+kq+tk+qt$
$a=k^2+2qk+t^2+qt+2kt$
$b=q^2+2qk+t^2+2qt+kt$
$k,t,q$ - are integers and can be any ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How to estimate the following integral: $\int_0^1 \frac{1-\cos x}{x}\,dx$
How to estimate the following integral?
$$
\int_0^1 \frac{1-\cos x}{x}\,dx
$$
| First note that the integral exists since $$0 \leq \dfrac{1-\cos(x)}x = \dfrac{2 \sin^2(x/2)}x \leq \dfrac{x}2$$ Hence, the integral is between $0$ and $1/4$. To compute the integral, proceed as follows. We have
$$\cos(x) = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} \mp = \sum_{k=0}^{\infty}(-1)^k \dfrac{x^{2k}}{(2k)!}$$
He... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $E[X^2 + Y^2]$ for the given joint density function. Let X and Y have joint probability density function
$f(x,y) = \frac{3}{2}(x^2 +y^2)$, with $0<x<1, 0<y<1$.
Find $E[X^2+Y^2]$.
I came up with $\frac{14}{15}$ by integrating $\int_0^1 \int_0^1 (x^2 +y^2)(x^2+y^2)$, but I've
never had to do anything but E[... | $\mathbf{E}[\varphi(X,Y)] = \int \int_{D} \varphi(x,y) f(x,y) dx dy = \frac{3}{2} \int _{0}^{1} \int _{0}^{1}(x^4 + 2 x^2 y^2 + y^4)dx dy = \frac{3}{2}\int_{0}^{1}[\int_{0}^{1}(x^4 + 2 x^2 y^2 + y^4)dy]dx$
Are you familiar with double integration? Can you handle from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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An application of contraction mapping theorem Suppose $\mathbb{R}$ is equipped with the standard metric and I want to show that the equation
$$
3x^3-3x^2+x-2=0
$$
has a unique root $c\geq1$. The equation is equivalent to
$$
f(x)=x,
$$
where $f(x)=1+\frac{1}{1+3x^2}$, so by showing that $f:[1,+\infty)\to \mathbb{R}$ is ... | Note that $x^2-2x+1=(x-1)^2\ge0$, hence $x^2+1\ge2x$ and similarly $y^2+1\ge2y$, hence indeed (very wastefully)
$$\begin{align}(1+3x^2)(1+3y^2)&=9x^2y^2+3x^2+3y^2+1\\&\ge 9+3x^2+3y^2+1\\&=3(x^2+1)+3(y^2+1)+3\\&\ge 6x+6y+3\\&>6( x+y) \end{align}$$
and you get a contraction factor of $\frac12$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is the polar form of $ z = 1- \sin (\alpha) + i \cos (\alpha) $? How do I change $ z = 1- \sin (\alpha) + i \cos (\alpha) $ to polar? I got $r = (2(1-\sin(\alpha))^{\frac{1}{2}} $. I have problems with the exponential part. What should I do now?
| We have
\begin{eqnarray}
z&=&1-\sin\alpha+i\cos\alpha=1+i(\cos\alpha+i\sin\alpha)=1+e^{i\pi/2}e^{i\alpha}=1+e^{i\frac{\pi+2\alpha}{2}}\\
&=&\left(e^{-i\frac{\pi+2\alpha}{4}}+e^{i\frac{\pi+2\alpha}{4}}\right)e^{i\frac{\pi+2\alpha}{4}}=2\cos\left(\frac{2\alpha+\pi}{4}\right)e^{i\frac{\pi+2\alpha}{4}}.
\end{eqnarray}
For ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove That $x=y=z$ If $x, y,z \in \mathbb{R}$,
and if
$$ \left ( \frac{x}{y} \right )^2+\left ( \frac{y}{z} \right )^2+\left ( \frac{z}{x} \right )^2=\left ( \frac{x}{y} \right )+\left ( \frac{y}{z} \right )+\left ( \frac{z}{x} \right ) $$
Prove that $$x=y=z$$
| Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, and let
$$
G=\sqrt[3]{abc}, A = \frac{a+b+c}{3}, Q=\sqrt{\frac{a^2+b^2+c^2}{3}}
$$
Then the given condition is $Q^2=A$, but by the power mean inequality
$$
Q^2\ge Q \ge A \ge G = 1
$$
with equality in each case only if $a=b=c=1$, i.e. if $x=y=z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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Prove that for any $n \in \mathbb{N}, 2^{n+2} 3^{n}+5n-4$ is divisible by $25$? I have question
Q
Prove that for any $n \in \mathbb{N}, 2^{n+2} 3^{n}+5n-4$ is divisible by $25$?
by using induction
Thanks
| Let $f(n)=2^{n+2}3^n+5n-4=4\cdot6^n+5n-4$
$f(m+1)-f(m)=4\cdot6^{m+1}+5(m+1)-4-(4\cdot6^m+5m-4)$
$=5\{5+4(6^m-1)\}$
Now, we know $(6^m-1)$ is divisible by $6-1=5$
$\implies f(m+1)\equiv f(m)\pmod {25}$
Now, $f(1)=2^3\cdot3+5\cdot1-4=25\implies f(1)$ divisible by $25$
Alternatively,
$2^{n+2}3^n+5n-4=4\cdot6^n+5n-4$
$=4(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/373215",
"timestamp": "2023-03-29T00:00:00",
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sum of squares of the roots of equation The equation is $$x^2-7[x]+5=0.$$
Here $[x]$ the greatest integer less than or equal to $x$. Some other method other than brute forcing. I tried a method of putting $[x]=q$ and $x=q+r$ which gives an equation: $$(q+r)^2-7q+5=0.$$
| HINT:
$0=x^2-7[x]+5\ge x^2-7x+5$
So, $x^2-7x+5\le0$
If $a,b(>a)$ are the roots of $x^2-7x+5=0, a\le x\le b$
Now, $a,b$ are $\approx .8,6.19$
Now, testing for the cases
As $x^2=7[x]-5$
If $0<x<1,[x]=0\implies x^2+5=0$ which does not have any real solution
If $1<x<2,[x]=1\implies x^2=7-5=2$
If $2<x<3,[x]=2\implies x^2=1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Differentiating $\;y = x a^x$ My attempt:
$$\eqalign{
y &= x{a^x} \cr
\ln y &= \ln x + \ln {a^x} \cr
\ln y &= \ln x + x\ln a \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left(x \times {1 \over a} + \ln a \times 1\right) \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({x \over a} + \ln a\... | $$
\frac{d}{dx} \ln a = \frac 1 a \frac{da}{dx}.
$$
If $\dfrac{da}{dx} = 0$, then you need to use that fact.
But here's a quicker way: Since $a$ is constant, $\ln a$ is constant, so its derivative is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/376642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Number of circular placements of $n$ identical letters such that no two letters are adjacent. Suppose I have to place $3$ identical letters on a circular table which has $7$ slots in such a way that no two letters are in consecutive slots. In how many ways can I do this?
Can this be generalized into $n$ identical lette... | Whenever you have a circular table, it's easiest to fix the position of one thing that you're placing, then place the rest relative to that position.
So, let's say your letters are $\color{red}{\operatorname{A}}$, $\color{blue}{\operatorname{B}}$, and $\color{green}{\operatorname{C}}$. Let's fix $\color{red}{\operato... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377890",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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A Permutation problem with sum restrictions In how many permutations of digits 1, 2, 3,...,9 are the following two conditions satisfied:
*
*Sum of digits between 1 and 2 (including both) is 12.
*Sum of digits between 2 and 3 (including both) is 23.
My Attempt
Clearly 3 cannot lie between 1 and 2. So the sequence ... | Properly between $1$ and $2$ we need a total of $9$.
As we cannot use $1,2,3$ to produce this $9$, we find thet the $9$ can only be the digit $9$ itself or $4+5$.
If $1$ is between $2$ and $3$, we need a total of $23-3-12=8$ between $2$ and $3$. This can only be $8$ itself (as $1,2,3$ are already used). This gives us $... | {
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Show that for any odd $n$ it follows that $n^2 \equiv 1 \mod 8$ and for uneven primes $p\neq 3$ we have $p^2 \equiv 1\mod 24$. Show that for any odd $n$ it follows that $n^2 \equiv 1 \mod 8$ and for uneven primes $p\neq 3$ we have $p^2 \equiv 1\mod 24$.
My workings so far: I proceeded by induction. Obviously $1^2 \equi... | If $n$ is odd, $n = 2 k + 1$ and $n^2 = 4 k^2 + 4 k + 1 = 4 k (k + 1) + 1$, and $4 k (k + 1)$ is divisible by 8. Thus $n^2 \equiv 1 \pmod{8}$.
If $p > 3$ is prime, it is odd, and so by the above $p^2 \equiv 1 \pmod{8}$. As it is prime, $p \equiv 1 \pmod{3}$ or $p \equiv 2 \pmod{3}$, so $p^2 \equiv 1 \pmod{3}$. Now, as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/380696",
"timestamp": "2023-03-29T00:00:00",
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Variance for the number of trials before success in an urn problem without replacement This question is asked as an extension of:
Expectation of number of trials before success in an urn problem without replacement
(Note: I am not the author of the original question.)
We have $b$ blue balls and $r$ red balls in an urn.... | The probability of drawing $n$ red balls then a blue ball is
$$
\frac{r}{b+r}\frac{r-1}{b+r-1}\frac{r-2}{b+r-2}\cdots\frac{r-n+1}{b+r-n+1}\frac{b}{b+r-n}
=\frac{\binom{\vphantom{b+}r}{n}}{\binom{b+r-1}{n}}\frac{b}{b+r}\qquad\tag{1}
$$
Since a blue ball must be drawn eventually, $(1)$ indicates that
$$
\sum_{n=0}^r\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/382662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Binomial theorem for prime exponent Could you explain to me why for prime $p$ we have the following?
$$(x+y)^p - (x^p + y^p)= x^p + \binom{p}{1}x^{p-1}y + \binom{p}{2}x^{p-2}y^2 + \binom{p}{p-1}xy^{p-1} + y^p.$$
I found it here: Abstract Algebra - Solutions to Homework (pdf),
solution 5d, page 3.
I would really appr... | As written, what's in the notes is incorrect. We should have
\begin{align*}
(x + y)^p - (x^p + y^p) &= x^p + \begin{pmatrix} p\\ 1\end{pmatrix}x^{p - 1}y + \begin{pmatrix} p\\ 2\end{pmatrix} x^{p - 2}y^2 + \ldots + \begin{pmatrix} p\\ p - 1\end{pmatrix}x y^{p - 1} + y^p - (x^p + y^p)\\
&= \begin{pmatrix} p\\ 1\end{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/385884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the integral of $\int {{2 \over {\sqrt x (x - 4)}}dx} $ given the substitution: $u = \sqrt x $ My stab at it:
$\eqalign{
& \int {{2 \over {\sqrt x (x - 4)}}dx} \cr
& u = \sqrt x \cr
& {{du} \over {dx}} = {1 \over 2}{x^{ - {1 \over 2}}} = {1 \over {2\sqrt x }} \cr
& {{dx} \over {du}} = 2\sqrt x \c... | You did just fine up to $\;\int \frac{4}{u^2-4}\,du\;\;$
Note that $$\int \frac{4}{u^2-4}du = \int \frac{4}{(u - 2)(u+2)}\,du = \int \left(\frac {A}{u - 2} + \frac{B}{u + 2} \right)\,du$$
That is, we can use partial fractions to break our integral into the sum of two nicely integrable functions by simply solving for th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/386691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Are all limits solvable without L'Hôpital Rule or Series Expansion Is it always possible to find the limit of a function without using L'Hôpital Rule or Series Expansion?
For example,
$$\lim_{x\to0}\frac{\tan x-x}{x^3}$$
$$\lim_{x\to0}\frac{\sin x-x}{x^3}$$
$$\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$$
$$\lim_{x\to0}\frac{e^... | Using only trigonometric identities, in this answer, it is shown that
$$
\lim_{x\to0}\frac{x-\sin(x)}{x-\tan(x)}=-\frac12\tag{1}
$$
Therefore, if we subtract from $1$, we get
$$
\lim_{x\to0}\frac{\tan(x)-\sin(x)}{\tan(x)-x}=\frac32\tag{2}
$$
Using the limits proven geometrically in this answer, we can derive
$$
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "121",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int {\sin 2\theta \over 1 + \cos \theta} \, d\theta $, using the substitution $u = 1 + \cos \theta $
Evaluate $$\int_0^{\pi \over 2} {{\sin 2\theta } \over {1 + \cos \theta }} \, d\theta$$ using the substitution $u = 1 + \cos \theta $
Using
$$\begin{align}
u &= 1 + \cos \theta \\
\frac{du}{d\theta} &= -\s... | HINT:
$$\frac{\sin2\theta}{1+\cos\theta}=\frac{2\sin\theta\cos\theta}{1+\cos\theta}=\frac{2\sin\theta(1+\cos\theta-1)}{1+\cos\theta}=2\sin\theta -\frac{2\sin\theta}{1+\cos\theta}$$
Now put $1+\cos\theta=u$ in the second integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What have I done wrong in solving this problem with indices rules? The question asks to simplify:
$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}.$$
So I used $(a^m)^n=a^{mn}$ to get
$$\dfrac{25}{4}x^{-2} = \dfrac{25}{4} \times \dfrac{1}{x^2} = \frac{25}{4x^2} = \frac{25}{4}x^{-2}$$
However, this isn't the answer, and ... | $$(\dfrac{25x^4}{4})^{-\dfrac{1}{2}}$$
$$(\dfrac{25}{4})^{-\dfrac{1}{2}}(x^{-2})$$
$$(\dfrac{4}{25})^{\dfrac{1}{2}}(x^{-2})$$
$$(\dfrac{2}{5})(x^{-2})$$
$$\dfrac {2}{5x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/388748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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If $f(x)=\sqrt{1-\sqrt{1-x^2}}$, then prove that $f(x)$ is continuous on $[-1,1]$ and differentiable on $(-1,0) \cup (0,1)$.
If $f(x)=\sqrt{1-\sqrt{1-x^2}}$, then prove that $f(x)$ is continuous on $[-1,1]$ and differentiable on $(-1,0) \cup (0,1)$.
Please prove using $$\lim_{x\to c^+}f(x)=\lim_{x\to c^-}f(x)=f(c)$$
... | Using $\lim_{x\to c}f(x)=f(c)$ to check continuity, in this case, is just the same as checking that you have a composition of continuous functions, which is continuous wherever it is defined.
Since $1-x^2\ge0$ means $|x|<1$ and implies $\sqrt{1-x^2}\le1$, your function is defined (and continuous) in $[-1,1]$.
The deriv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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I need help to evaluate this integral with a constant $c$ Find the value of the constant $c$ for which the integral converges, and evaluate the integral:
$$\int_0^\infty \left(\frac{9x}{x^2+1}-\frac{9c}{2x+1}\right)dx$$
| We have
\begin{align}
\int_0^a \left(\dfrac{9x}{x^2+1} - \dfrac{9c}{2x+1}\right) dx & = \left. \left(\dfrac92 \log\left(x^2+1 \right) - \dfrac{9c}2 \log(2x+1) \right) \right \vert_0^a\\
& = \dfrac{9}2 \log \left(\dfrac{a^2+1}{(2a+1)^c}\right)
\end{align}
Given this, as $a \to \infty$, what happens if $c<2$ or $c>2$? On... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$ Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + ... | There is a closed quotient form for the polynomial using the geometric series formula:
$$\sum_{k=0}^n x^{2k}=\frac{x^{2n+2}-1}{x^2-1}.$$
The theory of cyclotomic polynomials tells us how to reduce polynomials of the form $x^m-1$. In particular, the irreducible factors of $x^m-1$ are $\Phi_d(x)$ for each divisor $d\mid ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/391086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
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Complex integral help involving $\sin^{2n}(x)$ Show that $$\int_0^\pi \sin^{2n} \theta d\theta=\dfrac{\pi(2n)!}{(2^n n!)^2} $$
So far I have came up with:
$$\sin^{2n} \theta = \left(\dfrac {z-z^{-1}}{2i} \right)^{2n}$$ and I know I should be using:
$$(a+b)^n=\sum_{k=0}^n \dfrac{n!}{(n-k)!k!}a^kb^{n-k}$$
but I'm not sur... | This is my favorite example of integration by parts.
Let $I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^n(x) dx$.
$I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} (n-1) \sin^{n-2}(x) \cos^2(x) dx$
The first expression o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/392083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How prove this $\left(\sqrt{a^2+b^4}-a\right)\left(\sqrt{b^2+a^4}-b\right)\le a^2b^2$ let $a,b\in R$,and such that
$$\left(\sqrt{a^2+b^4}-a\right)\left(\sqrt{b^2+a^4}-b\right)\le a^2b^2$$
prove that $$a+b\ge 0$$
I think this is very beatifull problem, have you nice methods? Thank you,
I have see this problem
$$\left(x+... | Step 1: Multiply by the conjugate, we get that
$$ b^4 a^4 \leq a^2 b^2 ( \sqrt{a^2 + b^4} +a ) ( \sqrt{ b^2 + a^4} + b), $$
$$ a^2 b^2 \leq ( \sqrt{a^2 + b^4} +a ) ( \sqrt{ b^2 + a^4} + b)$$.
This gives us the chain of inequalities
$$ ( \sqrt{a^2 + b^4} -a ) ( \sqrt{ b^2 + a^4} - b) \leq a^2 b^2 \leq ( \sqrt{a^2 + b^4}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$ Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me?
Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$
Isolate one of the square roots: $\sqrt{(2x-5... | $$2x-5 = (1 + \sqrt{x-1})^2$$
to expand RHS use this formula or simple mulipty it with itself(to do square).
formula is:
$(a+b)^2=a^2+b^2+2\times a\times b$
so your expansion will be
$$2x-5 = (1^2 + (\sqrt{x-1})^2+2\times1\times \sqrt{x-1})$$
$$2x-5 = (1 + {x-1}+2\times \sqrt{x-1})$$
$$2x-5 = x+2\sqrt{x-1}$$
$$x-5 = 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/392308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Knowing that $n= 3598057$ is a product of two different prime numbers and that 20779 a square root of $1$ mod $n$, find prime factorization of $n$.
Knowing that $n= 3598057$ is a product of two different prime numbers and that 20779 is a square root of $1$ mod $n$, find prime factorization of $n$.
What I have done so... | From $(x-1)(x+1)\equiv 0 \pmod n$ you cannot conclude that $x-1 = 20779 \lor x + 1=20779$ when $n$ is not prime. You can conclude that $x-1 \equiv 0 \pmod p$ and $x+1 \equiv 0 \pmod q$ (or the other way-we could swap $p,q$) when you know that $n$ has only two prime factors unless $pq$ divides one of $x+1, x-1$. So fa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$ Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?
$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$
Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell with... | And one more... I'm reflecting at the concavity and monotonicty of the curve of the sqrt-function and conclude, that whatever greater or smaller the relation is, it should be the same if I decrease the arguments down to zero. Since we do not yet know whether the relation is $\gt$ or $\lt$ I introduce the "indeterminate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 11,
"answer_id": 5
} |
linear Transformation of polynomial with degrees less than or equal to 2 I would like to determine if the following map $T$ is a linear transformation:
\begin{align*}
T: P_{2} &\to P_{2}\\
A_{0} + A_{1}x + A_{2}x^{2} &\mapsto A_{0} + A_{1}(x+1) + A_{2}(x+1)^{2}
\end{align*}
My attempt at solving:
\begin{align}
T(p + q)... | Why is $T(p+q)=p(x+1)+q(x+1)$? That's basically what you're trying to show.
You could try to just expand $T(p)$ and see that $T$ can be easily expressed as a matrix in base $1,x,x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding invariant factors of finitely generated Abelian group There is this question that I wasn't sure how to do but somehow got the answers partially correct (maybe).
Suppose that the abelian group $M$ is generated by three elements $x,y,z$ subject to the relations $4x+y+2z=0, 5x+2y+z=0,6y-6z=0$. Determine the invari... | When you do the operation $4 R_2 - 5 R_1$, you are altering the determinant by a factor of $4$, since you are multiplying your matrix on the left by
$$
\begin{bmatrix}1 & 0 & 0\\-5&4&0\\0&0&1\end{bmatrix}.
$$
Ditto for $3 R_1 + R_2$, which alters the determinant by a factor of $3$. These two operations have introduced ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Solve equations $\sqrt{t +9} - \sqrt{t} = 1$ Solve equation: $\sqrt{t +9} - \sqrt{t} = 1$
I moved - √t to the left side of the equation $\sqrt{t +9} = 1 -\sqrt{t}$
I squared both sides $(\sqrt{t+9})^2 = (1)^2 (\sqrt{t})^2$
Then I got $t + 9 = 1+ t$
Can't figure it out after that point.
The answer is $16$
| From $\sqrt{t +9} - \sqrt{t} = 1$ you don't get $\sqrt{t +9} = 1 -\sqrt{t}$ but $$\sqrt{t +9} = 1 +\sqrt{t}.$$
You can solve as follows, using the algebraic identity $(a+b)^2=a^2+2ab+b^2$:
$$
\begin{eqnarray*}
\sqrt{t+9}-\sqrt{t} &=&1\Leftrightarrow \sqrt{t+9}=1+\sqrt{t}\tag{1} \\
\text{Square both sides of $(1)$} &\R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/399199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 1
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Finding the sum of a Taylor expansion I want to find the following sum:
$$
\sum\limits_{k=0}^\infty (-1)^k \frac{(\ln{4})^k}{k!}
$$
I decided to substitute $x = \ln{4}$:
$$
\sum\limits_{k=0}^\infty (-1)^k \frac{x^k}{k!}
$$
The first thing I noticed is that this looks an awful lot like the series expansion of $e^x$:
$$
... | Note that $$e^{-x}=\sum_{k=0}^\infty \frac{(-x)^k}{k!}=\sum_{k=0}^\infty (-1)^k\frac{x^k}{k!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$ How to integrate $$\int_0^{\pi/2}\dfrac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$$
What's the approach to it?
Being a high school student , I don't know things like counter integration.(Atleast not taught in India in high school education... | Put $t=\tan x\implies dt=\sec^2x.dx$
$$
\begin{align}
&\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}=\int_0^{\pi/2}\frac{\sec^4x.dx}{(a^2+b^2\tan^2x)^2}\\
&=\int_0^\infty\frac{1+t^2}{(a^2+b^2t^2)^2}dt=\frac{1}{b^2}\int_0^\infty\frac{dt}{a^2+b^2t^2}+\frac{b^2-a^2}{b^2}\int^\infty_0\frac{dt}{(a^2+b^2t^2)^2}=I_1+I_2
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/402223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Simplified form of $\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$. Tried this one a couple of times but can't seem to figure it out.
I am trying to simplify the expression:
$$\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$$
So my attempt at this is:
$$=\bigg(\dfrac{6x}{x}-\dfrac{2}{x}\bigg)\di... | You can factor a $2x^2$ out of the top to get $6x^3-2x^2=2 x^2 (3x-1 )$. After this, you can factor an $x$ out of the bottom to get $(9x^2-1)$, which then factors into $(3x-1) (3x+1)$ as it is a difference of squares. So, putting that all together,
$$\frac{\left(6x^3-2x^2\right)}{\left(9x^3-x\right)}=\frac{2 x^2 (3x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/404280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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$R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$ So the question is:
$R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$.
I don't really know what to do from here. If I substitute
$$R_n = 3(2^n)-4(5^n)$$
into
$$Rn = 7R_{n-1}-10R_{n-2}$$
I end up ge... | We need to eliminate $2^n,3^n$
$$R_n=3\cdot 2^n-4\cdot 5^n \ \ \ \ (1)$$
$$(1)\implies 3\cdot2^n-5\cdot 5^n-R_n=0\ \ \ \ (2)$$
$$(1)\implies R_{n+1}=3\cdot 2^{n+1}-4\cdot 5^{n+1}=6\cdot2^n-20\cdot 5^n$$
$$\implies 6\cdot 2^n-20\cdot 5^n-R_{n+1}=0 \ \ \ \ (3)$$
Solving $(2),(3)$ for $2^n,3^n$ we get $$2^n=\frac{5R_n-R_{... | {
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"url": "https://math.stackexchange.com/questions/405384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$? Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?
I thought it would be 12 this as per pemdas rule:
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$
Wanted to confirm the right answer from you guys. Thanks for your help.
| Your answer of $12$ is numerically correct, but you’ve applied the precedence rules incorrectly. In the absence of parentheses multiplication is done before addition, so
$$1+1+1+1+1+1+1+1+1+1+1+1\cdot 0+1$$
is to be evaluated as
$$1+1+1+1+1+1+1+1+1+1+1+(1\cdot 0)+1\;,$$
not as
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot(0+1)\;.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/405543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Trigonometric identities using $\sin x$ and $\cos x$ definition as infinite series Can someone show the way to proof that $$\cos(x+y) = \cos x\cdot\cos y - \sin x\cdot\sin y$$ and $$\cos^2x+\sin^2 x = 1$$ using the definition of $\sin x$ and $\cos x$ with infinite series.
thanks...
| Let me do a different one. Begin with
$$
\sin x = \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!},
\qquad
\cos x = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k)!}
$$
So compute
$$\begin{align}
\sin(x+y) &= \sum_{k=0}^\infty \frac{(-1)^k(x+y)^{2k+1}}{(2k+1)!}
\\ &=
\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\sum_{j=0}^{2k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/405628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Deriving Reduction formula - Indefinite integration using integration by parts I was working on finding the reduction formula for :
$\int \frac{dx}{(x^2+a^2)^n}$
By using integration by parts formula ( $\int f(x) g(x) dx = f(x) \int g(x)dx -\int (f'(x)\int g(x))$ considering 1.dx as second function :
Let $I_n$ = $\i... | If $$I_n=\int \frac{dx}{(x^2+a^2)^n}$$ then setting $$u=\frac{1}{(x^2+a^2)^n},~~dv=dx$$ leads us to have $$du=\frac{-2nxdx}{(x^2+a^2)^{n+1}},~~v=x$$ and the method of integrating by parts gives us $$I_n=\frac{x}{(x^2+a^2)^n}+2n\int\frac{x^2dx}{(x^2+a^2)^{n+1}}=...=\frac{x}{(x^2+a^2)^n}+2nI_n-2na^2I_{n+1}$$ Now we see ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/405886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What does $\lim\limits_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$ evaluate to? What does $$\lim_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$ evaluate to? This very likely needs substitution.
| $$\frac{1-\sqrt{3}\tan x}{\pi-6x}=\frac{\sqrt3}6\cdot\frac{\tan \frac\pi6-\tan x }{\frac\pi6-x}\left(\text{ as }\tan\frac\pi6=\frac1{\sqrt3}\right)$$
$$=\frac{\sqrt3}6\cdot\frac{\sin\left(\frac\pi6-x\right)}{\left(\frac\pi6-x\right)\cos x\cos \frac\pi6}$$
$$\lim_{x\to\frac\pi6}\frac{1-\sqrt{3}\tan x}{\pi-6x}=\frac{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/408315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why dividing by zero still works Today, I was at a class. There was a question:
If $x = 2 +i$, find the value of $x^3 - 3x^2 + 2x - 1$.
What my teacher did was this:
$x = 2 + i \;\Rightarrow \; x - 2 = i \; \Rightarrow \; (x - 2)^2 = -1 \; \Rightarrow \; x^2 - 4x + 4 = -1 \; \Rightarrow \; x^2 - 4x + 5 = 0 $. Now he... | because he's only dividing by zero when $x$ takes that value (or its conjugate) otherwise he's really just figuring out what the next thing he'll need to multiply is. Yes, it may look a little dodgy, think of this as a sort of back-of-the-envelope calculation. It's only a means to find out the next step and since in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/408527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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If $\frac{\cos^4\theta}{\cos^2\phi}+\frac{\sin^4\theta}{\sin^2\phi}=1$, show $\frac{\cos^4\phi}{\cos^2\theta} +\frac{\sin^4\phi}{\sin^2\theta}=1$
If $\dfrac{\cos^4\theta}{\cos^2\phi}+\dfrac{\sin^4\theta}{\sin^2\phi}=1$, prove that $\dfrac{\cos^4\phi}{\cos^2\theta} +\dfrac{\sin^4\phi}{\sin^2\theta}=1$.
Unable to move ... | We can Write
$$\frac{Cos^4\theta}{Cos^2\phi}+\frac{Sin^4\theta}{Sin^2\phi}=Cos^2\phi+Sin^2\phi$$ $\implies$
$$ \left(Cos^2\phi-\frac{Cos^4\theta}{Cos^2\phi}\right)+\left(Sin^2\phi-\frac{Sin^4\theta}{Sin^2\phi}\right)=0$$ $\implies$
$$ Cos^4\phi+Sin^4\phi=Cos^4\theta+Sin^4\theta$$ $\implies$
$$ Sin\phi Cos\phi=Sin\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/410304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine the base of $\ker\phi$ and fibre of phi of a polynom function Good evening,
given is a base $B$ defined as
$$B:=(x\mapsto1,x\mapsto x,x\mapsto x^2,x\mapsto x^3 ,x\mapsto x^4)$$ a set $V$ defined as $$V:= \{ f: \mathbb{R} \mapsto \mathbb{R}\ |\ \exists\ {a_0},...{a_4} \in \mathbb{R}\ : f(x)=\sum_{i=0}^{4}{a_i... | Imo it'd be easier (or at least slightly less messier) as follows:
$$f(x)=\sum_{k=0}^4a_kx^k\implies $$
$$f(x)\in\ker\phi\iff f''(x)+xf'(x)=f(x+1)\iff$$
$$2a_2+(6a_3+a_1)x+(12a_4+2a_2)x^2+3a_3x^3+4a_4x^4=$$
$$=\sum_{k=0}^4a_k+\left(\sum_{k=1}^4ka_k\right)x+\left(a_2+3a_3+6a_4\right)x^2+\left(a_3+4a_4\right)x^3+a_4x^4$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a contradiction is this exercise? The following exercise was a resolution to this problem
Let $\displaystyle\frac{2x+5}{(x-3)(x-7)}=\frac{A}{(x-7)}+\frac{B}{(x-3)}\space \forall \space x \in \mathbb{R}$. Find the values for $A$ and $B$
The propose resolution was:
In order to isolate $A$ on the right side, m... | If $\frac{2x+5}{x-3}=A+B\frac{x-7}{x-3}$ for all $x\ne 7$ (or $3$), then we can take the limit as $x\to 7$ on both sides to get the given result, and since both sides are continuous, it is equivalent to just plugging in $x=7$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Interestingly restricted compositions of $n$ Let $n$ be a non-negative integer. How many compositions of $n$ are there where the $i$-th part has the same parity as $i$?
The main problem I'm having with this problem is that I can't really formulate a single generating function for the set of all tuples of natural number... | Let $a_n$ be the number of compositions of $n$ whose $i$-the part has the same parity as $i$, and let $b_n$ be the number of compositions of $n$ whose $i$-the part has the opposite parity to $i$. Considering the possible values of the first part of a composition of $n$, we see that
$$a_{2n+1}=\sum_{k=0}^nb_{2k}\quad\te... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that for every $n\in \mathbb{N}^{+}$, there exist a unique $x_{n}\in[\frac{2}{3},1]$ such that $f_{n}(x_{n})=0$ Let $f_{n}(x)=-1+x+\dfrac{x^2}{2^2}+\dfrac{x^3}{3^2}+\cdots+\dfrac{x^n}{n^2}$,
(1) Prove that for every $n\in \mathbb{N}^{+}$, then there exist unique $x_{n}\in[\frac{2}{3},1]$ such that
$f_{n}(x_{n})=0... | $x_{n}-x_{n+p}>0$ trivial, since $f_{n+p}(x_{n})>f_{n}(x_{n})=0$ and these $f$ always increase.
For the other inequality, I couldn't find a neat way to do it, so pardon some of the messy calculation. The idea is to use mean value theorem:
If $n\leq 3$ then trivial. So assume $n\geq 4$:
Now a bit of number crunching go ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is $S=\sum_{r=1}^\infty \tan^{-1}\frac{2r}{2+r^2+r^4}$ finite? Problem:
If $$S=\sum_{r=1}^\infty \tan^{-1}\left(\frac{2r}{2+r^2+r^4}\right)$$ Then find S ??
Solution:
I know that $\tan^{-1} x + \tan^{-1} y= \tan^{-1} \frac {x +y} {1-xy} $
But I have no idea how to such complicated question with it.
| HINT:
As $1+a^2+a^4=(1+a^2)^2-a^2=(1+a^2-a)(1+a^2+a),$
$$\frac{2\cdot a}{2+a^2+a^4}=\frac{(1+a^2+a)-(1+a^2-a)}{1+(1+a^2-a)(1+a^2+a)}$$
$$\implies \arctan \left(\frac{2\cdot a}{2+a^2+a^4}\right)=\arctan(1+a^2+a)-\arctan(1+a^2-a)$$
Can you recognize the Telescoping series?
So, $$\sum_{1\le r\le n}\arctan \left(\frac{2\cd... | {
"language": "en",
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The limit of $\lim\limits_{x \to \infty}\sqrt{x^2+3x-4}-x$ I tried all I know and I always get to $\infty$, Wolfram Alpha says $\frac{3}{2}$. How should I simplify it?
$$\lim\limits_{x \to \infty}\sqrt{(x^2+3x+4)}-x$$
I tried multiplying by its conjugate, taking the squared root out of the limit, dividing everything by... | $$ \lim_{x \rightarrow \infty} \left(x^2 + 3x + 4\right)^{ \frac{1}{2}} - x $$
$$ = \lim_{x \rightarrow \infty} \left(x^2\left(1+\frac{3}{x}+\frac{4}{x^2}\right)\right)^{ \frac{1}{2}} - x $$
$$ = \lim_{x \rightarrow \infty} x\left(1+\frac{3}{x}+\frac{4}{x^2}\right)^{ \frac{1}{2}} - x $$
Then by Taylor expansion, we get... | {
"language": "en",
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How to find $x^{2000}+x^{-2000}$ when $x + x^{-1} = \frac{1}{2}(1 + \sqrt{5})$
Let $x+x^{-1}=\dfrac{1+\sqrt{5}}{2}$. Find $x^{2000}+x^{-2000}$.
How many nice methods do you know for solving this problem? Thank you everyone.
My method: because $x+\dfrac{1}{x}=2\cos{\dfrac{2\pi}{5}}$, so $$x^{2000}+\dfrac{1}{x^{2000}}=... | Here is an algebraic way avoiding trig functions: note that your number $x$ satisfies
$$x^2-(\frac{1+\sqrt{5}}{2})x+1=0 \quad \implies \quad \text{by multiplying by the conjugate} \quad x^4-x^3+x^2-x+1=0$$ and then use the factorization
$$x^{10}-1=(x^6+x^5-x-1)(x^4-x^3+x^2-x+1)$$ to see that $x^{10}=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/416422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Find maximum value of $f(x)=2\cos 2x + 4 \sin x$ where $0 < x <\pi$ Find the maximum value of $f(x)$ where
\begin{equation}
f(x)=2\cos 2x + 4 \sin x \ \
\text{for} \ \ 0<x<\pi
\end{equation}
| We can utilize algebra to avoid calculus as follows
$$f(x)=2\cos2x+4\sin x$$
$$=2(1-2\sin^2x)+4\sin x$$
$$=2-(4\sin^2x-4\sin x)$$
$$=2-(2\sin x-1)^2+1\le 3$$ as for real $x,(2\sin x-1)^2\ge 0$
the equality occurs if $2\sin x-1=0\iff \sin x=\frac12\implies x=n\pi+(-1)^n\frac\pi6$
For $0<x<\pi, x=\frac\pi6$ or $\pi-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/419150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does $x^2>x^3+1 \implies x < -{1\over P}$? How could one prove that:
$$x^2>x^3+1 \implies x < -{1\over P}$$
where $P$ denotes the plastic constant, the unique real root of $x^3-x-1=0$?
| The function $x^2 - x^3 - 1$ is monotone decreasing on $\{x < 0\}$ so we need to look at the (unique) $x$ where $$x^2 = x^3 + 1.$$ But for this $x$, dividing by $x^3$, $$-\frac{1}{x} = -1 - \frac{1}{x^3},$$ so $-\frac{1}{x}$ solves the defining equation for the plastic constant; so $-\frac{1}{x} = P$ and $x = \frac{-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/420396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify a expression with nested radical signs Simplify :
$\sqrt{10+6 \sqrt{2}+5 \sqrt{3}+4 \sqrt{6}}$
I have tried completing square but failed, Can anyone help me please? Thanks.
| $$10+6 \sqrt{2}+5 \sqrt{3}+4 \sqrt{6}=5(2+\sqrt3)+2\sqrt6(2+\sqrt3)=(2+\sqrt3)(5+2\sqrt6)$$
Now, $$5+2\sqrt6=3+2+2\cdot\sqrt2\cdot\sqrt3=(\sqrt3+\sqrt2)^2$$
and $$2+\sqrt3=\frac{4+2\sqrt3}2=\frac{3+1+2\cdot\sqrt3\cdot1}2=\frac{(\sqrt3+1)^2}2$$
Can you take it from here?
| {
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What does an expression $[x^n](1-x)^{-1}(1-x^2)^{-1}(1-x^3)^{-1}(1-x^4)^{-1}...$ mean? I came across the function that describes number of partitions of $n$ (I mean partitions like $5=4+1=3+2=3+1+1$ and so on.
There was defined a Cartesian product: $$\{0,1,1+1,1+1+1,...\}\times\{0,2,2+2,2+2+2,...\}\times\{0,3,3+3,...\}... | Often the notation
$$
[x^n](a_0+a_1x+a_2x^2+a_3x^3+\dots+a_nx^n+\dots)
$$
is used to mean the coefficient of $x^n$ in the given polynomial or power series ($a_n$ in the series I've given).
The formulas presented by the author are not polynomials, but power series. Polynomials are finite versions of power series. That i... | {
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Let $p_n$ be the sequence defined by $p_n=\sum_{k=1}^n\frac{1}{k}$. Show that $p_n$ diverges even though $\lim_{n\to\infty}(p_n-p_{n-1})=0$ I have tried this as :
$$p_n=\sum_{k=1}^n\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n-1}+\frac{1}{n}$$
$$p_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\l... | Recall the inequality $e^x>1+x \forall x>0$
Now Consider $$e^{p_n}=e^{\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)}$$ $$=e\cdot e^{\frac{1}{2}}\cdot e^{\frac{1}{3}}\cdots e^{\frac{1}{n}}$$ $$>\left(1+1\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{n}\right)$$ $$=(2)\left(... | {
"language": "en",
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$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
| So, $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
$\implies ab+bc+ca=-\frac12.$
So $a,b,c$ are the roots of the equation $t^3-\frac12t-u=0$ where $u=abc.$
Using discriminant of the cubic equation fro real roots, $$\triangle= -4\cdot1\cdot\left(-\frac12\right)^3-27(-u)^2=\frac12-27u^2\ge0\iff u^2\le \frac1{54}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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If $10^{20} +20^{10}$ is divided by 4 then what would be its remainder? If $$10^{20} +20^{10}$$ is divided with 4 then what would be its remainder?
| Since
$$
\begin{eqnarray*}
10^{20}+20^{10} &=&\left( 10^{10}\right) ^{2}+2^{10}10^{10} \\
&=&10^{10}\left( 10^{10}+2^{10}\right) \\
&=&2^{10}5^{10}\left( 2^{10}5^{10}+2^{10}\right) \\
&=&2^{10}2^{10}5^{10}\left( 5^{10}+1\right) \\
&=&4^{10}5^{10}\left( 5^{10}+1\right) \\
&=&4\left( 4^{9}5^{10}\left( 5^{10}+1\right)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Using partial fractions to find explicit formulae for coefficients? The set of binary string whose integer representations are multiples of 3 have the generating function
$$\Phi_S(x)={1-x-x^2 \over 1-x-2x^2}$$
Let $a_n=[x^n]\Phi_s(x)$ represent the number of strings in $S$ with length $n$. Using partial fraction expans... | Note that we get very lucky, since $1-x-2x^2=(1-2x)(1+x)$.
The first step, since the numerator has degree $\ge $ the degree of the denominator, is to divide. We get
$$\frac{1-x-x^2}{1-x-2x^2}=\frac{1}{2}+\frac{1}{2}\frac{1-x}{1-x-2x^2}.$$
We now work for a while with the simpler $\frac{1-x}{1-x-2x^2}$, We try to find $... | {
"language": "en",
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Prove $\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A$ by the Pythagorean theorem. How do I use the Pythagorean Theorem to prove that $$\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A?$$
| Let's simplify matters:
In a triangle with hypotenuse equal to $1$ (think of the unit circle, an angle $A$ between the $x$ axis and the hypotenuse, we know that $$\sin A=\frac{\text{opposite}}{\text{hypotenuse}}\quad \text{and}\quad \cos A=\frac{\text{adjacent}}{\text{hypotenuse}}$$ then, since hypotenuse $=1$, we have... | {
"language": "en",
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Show that $7 \mid( 1^{47} +2^{47}+3^{47}+4^{47}+5^{47}+6^{47})$ I am solving this one using the fermat's little theorem but I got stuck up with some manipulations and there is no way I could tell that the residue of the sum of each term is still divisible by $7$. what could be a better approach or am I on the right tra... | One more solution.
By Fermat's Little Theorem,
$a^{p - 1} \equiv 1 \pmod{p}$ for $a \not\equiv 0$. Thus,
$$
a^{48} = \left(a^6\right)^8 \equiv 1^8 = 1 \pmod{7}
$$
for each $a \in \{1, 2, \ldots, 6\}$.
As a consequence, $a^{47} \equiv a^{-1}$, so
$$
\begin{align}
1^{47} + 2^{47} + 3^{47} + 4^{47} + 5^{47} + 6^{47} &\eq... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $a,b$ are roots for $x^2+3x+1=0$.Calculating $(\frac{a}{b+1})^2 +(\frac{b}{a+1})^2$ If $a,b$ are roots for the equation $x^2+3x+1=0$.How to calculate $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$
| HINT:
As $ab=1, \frac a{b+1}=\frac{a}{\frac1a+1}=\frac{a^2}{a+1}=y$(say)
So, $a^2=y(a+1)$
and again $a^2+3a+1=0\implies a^2=-3a-1$
So, $ay+y=-3a-1\implies a=-\frac{y+1}{y+3} $
As $a$ is root of the given eqaution $$\left(-\frac{y+1}{y+3} \right)^2+3\left(-\frac{y+1}{y+3} \right)+1=0$$
Simply to get $y^2+4y-1=0$
Using ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number? $\large S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$
Multiplying by conjugate:
$\large S=\dfrac {-3}{\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}}$
From the original:
$\large S-2\sqrt[3]{5-2 \sqr... | For simplicity, we'll let $S = a+b$ where $a=\sqrt[3]{5+2\sqrt{13}}$ and $b=\sqrt[3]{5-2\sqrt{13}}$.
Note that $S^3=a^3+3a^2b+3ab^2+b^3=5+2\sqrt{13}+5-2\sqrt{13}+3a^2b+3ab^2$.
$S^3+9S-10=3a^2b+3ab^2+9a+9b=3a(ab+3)+3b(ab+3)=3(a+b)(ab+3)=3S(ab+3)$.
Now, $ab=\sqrt[3]{(5+2\sqrt{13})(5-2\sqrt{13})}=\sqrt[3]{25-52}=\sqrt[3]... | {
"language": "en",
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Approximation of DE It depends on my previous question. Closed form solution of DE
I don't want to deal with Airy functions. How can I approximate this DE in continous domain $[0,1]$?
$$y''(x)+(x+1)y(x)=0\quad\text{ with the initial conditions}\quad y(0)=0\quad y'(0)=1$$
What if the conditions change to
$$y''(x)+(x+1)y... | The simplest "brute-force" approach would be to attempt a power series solution. The basic idea is to guess that the solution is of the form $y(x) = \sum_{n=0}^{\infty}a_{n}x^n$ and our task is to find the $a_{n}$. In principle this can be used to solve the DE to arbitrary precision by simply computing more and more te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/431729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Integral $\int_0^\frac{\pi}{2} \sin^7x \cos^5x\, dx $ im asked to find the limited integral here but unfortunately im floundering can someone please point me in the right direction?
$$\int_0^\frac{\pi}{2} \sin^7x \cos^5x\, dx $$
step 1 brake up sin and cos so that i can use substitution
$$\int_0^\frac{\pi}{2} \sin^7(x)... | $$\int_0^\frac\pi2\sin^7x\cos^5xdx=\int_0^\frac\pi2\sin^7x\cos^4x\cos xdx=\int_0^\frac\pi2\sin^7x(1-\sin^2x)^2\cos xdx$$
$$=\int_0^1 u^7(1-u^2)^2 du (\text{ Putting }\sin x=u)$$
$$=\int_0^1 (u^7-2u^9+u^{11}) du$$
$$=\left(\frac{u^8}8-2\frac{u^{10}}{10}+\frac{u^{12}}{12}\right)_0^1$$
$$=\frac18-\frac15+\frac1{12}$$
$$=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
How to integrate these integrals $$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \cos x}$$
$$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \sin x}$$
It seems that substitutions make things worse:
$$\int \frac {dx}{1+ \cos x} ; t = 1 + \cos x; dt = -\sin x dx ; \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (t-1)^2} $$
$$ \Rightarrow
\int \frac ... | You were on the right track in the last line
$$ \frac{1}{2} \int \sec^2 \frac{x}{2} dx = \tan \frac{x}{2} + C $$
That's it. Put the limits in and you're done.
For the second one, substitute $t = \frac{\pi}{2} - x $ and you're back to the first integral
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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How find this value $\frac{a^2+b^2-c^2}{2ab}+\frac{a^2+c^2-b^2}{2ac}+\frac{b^2+c^2-a^2}{2bc}$ let $a,b,c$ such that
$$\left(\dfrac{a^2+b^2-c^2}{2ab}\right)^2+\left(\dfrac{b^2+c^2-a^2}{2bc}\right)^2+\left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2=3,$$
find the value
$$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^... | First, we simplify the initial equation by multiplying out by the denominator. Let
$$f(a,b,c) = c^2(a^2+b^2-c^2)^2 + a^2(b^2+c^2-a^2)^2 + b^2(c^2+a^2-b^2)^2 - 12a^2b^2c^2$$
In Ron Gordon's deleted post, he realized that $f(a,b,c) = 0 $ if $a+b=c$, $b+c=a$, $c+a=b$. This strongly suggests that $a+b-c, b+c-a, c+a-b$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to solve the given initial-value problem? Solve the given problem which the input function $g(x)$ is discontinuous?
$y''+4y = g(x)$, $y(0) = 1$, $y'(0) = 2$, where
$$g(x) = \begin{cases} \sin x, & 0\leq x\leq\frac{\pi}{2}\\
0,& x>\frac{\pi}{2} \end{cases}$$
And the given answer is,
$$y = \begin{cases} \cos 2x+\frac... | First let $g(x)=\sin(x)$.
$$
\begin{align*}
y'' + 4y &= \sin(x)
\end{align*}
$$
Applying the Laplace transform to both sides gives
\begin{align*}
s^2Y(s)-s-2+4Y(s) &= \dfrac{1}{s^2+1}\\
Y(s)(s^2+4) &= \dfrac{1}{s^2+1}+s+2\\
Y(s) &= \dfrac{1}{(s^2+1)(s^2+4)}+\dfrac{s}{s^2+4}+\dfrac{2}{s^2+4}\\
\end{align*}
Via partial f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2}\geq 1$ Let $abcd=1$ and $a,b,c$ and $d$ are all positive.
Prove that
$\dfrac{1}{(1+a)^2}+\dfrac{1}{(1+b)^2}+\dfrac{1}{(1+c)^2}+\dfrac{1}{(1+d)^2}\geq 1$
I am probably able to do this by assuming $a\geq b\geq c\geq d$
and by using deriva... | This inequality I have nice methods:use
$$\dfrac{1}{(1+a)^2}+\dfrac{1}{(1+b)^2}\ge\dfrac{1}{1+ab}$$
pf:use $Cauchy-Schwarz$ inequality
we have
$$(a+1)^2\le\left(\dfrac{a}{b}+1\right)(ab+1)$$
$$(b+1)^2\le\left(\dfrac{b}{a}+1\right)(ab+1)$$
so
$$\dfrac{1}{(1+a)^2}+\dfrac{1}{(1+b)^2}\ge\dfrac{1}{\left(\dfrac{a}{b}+1\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Substitution for $\int \frac {dx} {ax^2 + bx + c}$ I'm looking for the substitution that makes easier to solve integral containing quadratic polynomial in denominator (!) when such polynomial cannot be broken into parts (if it can, then it's possible to use partial fraction decomposition). Example:
$$\int \frac {dx} {5... | HINT:
$$\frac1{ax^2+bx+c}=\frac{4a}{(2ax+b)^2+4ca-b^2}$$
Case $1:$ If $4ca-b^2=0,$ $$\int \frac{dx}{ax^2+bx+c}=4a\int\frac{dx}{(2ax+b)^2} $$ Put $y=2ax+b$
Case $2:$ If $4ca-b^2>0, 4ca-b^2=d^2$(say)
$$\int \frac{dx}{ax^2+bx+c}=4a\int\frac{dx}{(2ax+b)^2+d^2} $$
Put $2ax+b=d\tan\theta$ as $\tan^2\theta+1=\sec^2\theta$
Ca... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factor Equations Please check my answer in factoring this equations:
Question 1. Factor $(x+1)^4+(x+3)^4-272$.
Solution: $$\begin{eqnarray}&=&(x+1)^4+(x+3)^4-272\\&=&(x+1)^4+(x+3)^4-272+16-16\\
&=&(x+1)^4+(x+3)^4-256-16\\
&=&\left[(x+1)^4-16\right]+\left[(x+3)^4-256\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\rig... | For the first, I will put $y=\frac{x+1+x+3}2=x+2$
so that $x +1=y-1, x+3=y+1$ and the odd powers of $y$ vanish in $(y-1)^4+(y+1)^4$
$$\implies (x+1)^4+(x+3)^4-272=(y-1)^4+(y+1)^4-272$$
$$=2\{y^4+6y^2+1\}-272=2(y^4+6y^2-135)$$
$$=2\{y^4+(15-9)y^2-135\}=2(y^2+15)(y^2-9) =2(y^2-15)(y+3)(y-3)$$
$$=2\{(x+2)^2-15\}(x+5)(x-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove that $xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \frac{4}{3}\sqrt{xyz(x+y+z)}$ let $x,y,z>0$,and such that
$x^2+y^2+z^2=1$,prove that
$$xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \dfrac{4}{3}\sqrt{xyz(x+y+z)}$$
Does this have a nice solution? Thank you everyone.
| We need to prove that
$$xyz+\sqrt{(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)}\geq\frac{4}{3}\sqrt{xyz(x+y+z)(x^2+y^2+z^2)}.$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=w^3+3\sqrt{(3u^2-2v^2)(3v^4-2uw^3)}-4\sqrt{u(3u^2-2v^2)w^3}.$$
But by AM-GM
$$f'(w^3)=1-\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/440551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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Induction of inequality involving AP Prove by induction that
$$(a_{1}+a_{2}+\cdots+a_{n})\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)\geq n^{2}$$
where $n$ is a positive integer and $a_1, a_2,\dots, a_n$ are real positive numbers
Hence, show that
$$\csc^{2}\theta +\sec^{2}\theta +\cot^{2}\theta... | For the first part, we need a base case and an inductive step.
Base Case: suppose we have only one number, $a_1$. Then $$(a_1)\left(\frac1{a_1}\right)=1\leq 1$$
Inductive Step: By Isomorphism's work, this amounts to showing that
$$
a_n\left(\frac{1}{a_1} + \frac{1}{a_2} \cdots + \frac{1}{a_{n-1}}\right) + \frac1{a_n}(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/440761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Demonstrate this triangle! Help Give a triangle ABC with
$$\sin{\left(\frac{3A}{2}\right)}+\sin{\left(\frac{3B}{2}\right)}=2\cos{\left(\frac{(A-B)}{2}\right)}$$
Demonstrate that triangle ABC is equilateral triangle!!
Thank all!
P/S: I'm sorry. Because I speak English not good!
| $\sin{\left(\dfrac{3A}{2}\right)}+\sin{\left(\dfrac{3B}{2}\right)}=2\sin{\left(\dfrac{3A+3B}{4}\right)}\cos{\left(\dfrac{3A-3B}{4}\right)}$
let $x=\cos{\dfrac{A-B}{4}},p=\sin{\left(\dfrac{3A+3B}{4}\right)} \implies 1\ge x > \dfrac{\sqrt{2}}{2}, 1 \ge p >0 $,then we have:
$p(4x^3-3x)=2x^2-1$, it is trivial LHS< RHS when... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/441937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find the point where $3x-4y+25=0$ is a tangent to the circle $x^2+y^2=25$ Find the coordinates of the point where the line $3x-4y+25=0$ is a tangent to the circle $x^2+y^2=25$.
Can someone please show me the first few steps of solving the simultaneous equation?
| $$
3x - 4y + 25 = 0\\
4y = 3x + 25\\
m = \frac34
$$
Differentiate circle equation $ x^2 + y^2 =25 $
$ \dfrac{dy}{dx}= -\dfrac xy = m = \dfrac 34 $
Solve for x and y to get, $ x = \mp 3, y =\pm 4 $
Only the first set $ (-3,4)$ satisfies the given equation of straight line tangent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/443136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Simple limit problem: $\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$ While trying to help my sister with her homework she gave me the next limit: $$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$$
I know the conventional way of solving it would be (That's what i showed her):
$$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})=\l... | If $\lim_{x \to 2} \frac{1}{x-2}-\frac 4{x^2-4}=0$, that means $$\lim_{x \to 2} \frac{1}{x-2}- \lim_{x \to 2} \frac 4{x^2-4}=0 \implies \lim_{x \to 2} \frac{1}{x-2}=\lim_{x \to 2} \frac 4{x^2-4}$$.
But $\frac1{x-2} \neq \frac 4{x^2-4}$ or $\frac 1{x-2} \neq \frac1{x-2} \frac4{x+2}$
Thus $\lim_{x \to 2} \frac{1}{x-2}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to prove two trigonometric identities I want to show that
$${\sin}^2 \alpha + 4{\sin}^4\frac{\alpha}{2} = 4{\sin}^2 \frac{\alpha}{2}$$
and
$${\sin}^2 \alpha + 4{\cos}^4\frac{\alpha}{2} = 4{\cos}^2 \frac{\alpha}{2}$$
They should be true, as Wolfram Alpha says so. However, I want to prove them, and I have no idea how... | $${\sin}^2 \alpha + 4{\sin}^4\frac{\alpha}{2}$$
$$\left(2\sin\frac {\alpha} {2}\cos \frac{\alpha}{2}\right)^2 + 4{\sin}^4\frac{\alpha}{2}$$
$$4\sin^2\frac {\alpha} {2}\cos^2 \frac{\alpha}{2} + 4{\sin}^4\frac{\alpha}{2}$$
$$4\sin^2\frac {\alpha} {2}(\cos^2 \frac{\alpha}{2} + {\sin}^2\frac{\alpha}{2})$$
$$4\sin^2\frac {\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/450962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$
Problem: If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$
Solution: $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $
Differenting both sides,... | It is a good start. For simplicity write $y$ for $f(x)$. We can rewrite the result you got as
$$\frac{y'}{y^2}=2(b^2-a^2)\sin x\cos x.$$
Integrate both sides. It may be handy to note that $2\sin x\cos x=\sin(2x)$. Or not, since it is clear that $2\sin x\cos x$ is the derivative of $\sin^2 x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Need help calculating this determinant using induction This is the determinant of a matrix of ($n \times n$) that needs to be calculated:
\begin{pmatrix}
3 &2 &0 &0 &\cdots &0 &0 &0 &0\\
1 &3 &2 &0 &\cdots &0 &0 &0 &0\\
0 &1 &3 &2 &\cdots &0 &0 &0 &0\\
0 &0 &1 &3 &\cdots &0 &0 &0 &0\\
\vdots &\vdots &\vdots&\ddots &\dd... | Let $D_n$ be the determinant of our matrix of size $n$. We can calculate $D_n$ by expansion of the first column: $D_n = 3 D_{n - 1} - 1 \cdot 2 \cdot D_{n - 2}$. For the second term we expanded again by the first row. We can see: $D_1 = 3$, $D_2 = 7$. By our recurence, we can count more terms: 3, 7, 15, 31, 63, …. Now ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/451941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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n is+ve integer, how many solutions $(x,y)$ exist for $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ with $x$, $y$ being positive integers and $(x \neq y)$ I wanted to know, how can i solve this.
For a given positive integer n, how many solutions $(x,y)$ exist for $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ with $x$ and $y$ b... | It is clear that the two positive fractions on the left have to be less than $\frac 1n$, so we have $x,y \gt n$.
Let $x=n+a, y=n+b$ and clear fractions then $$n(n+b)+n(n+a)=(n+a)(n+b)$$ which simplifies to $$ab=n^2$$
Any such $a,b$ give a pair of fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/452137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Differentiability of function involving absolute values I need to check the differentiability at $x=-10$ of the following function.
$f(x)=\cos|x-5|+\sin|x-3|+|x+10|^3-(|x|+4)^2$
Now,
\begin{align*}
\text{LHD} &= \lim_{x\to -10^-}\frac{f(x)-f(-10)}{x+10}\\
&= \lim_{x\to -10^-}\frac{\cos(x-5)-\sin(x-3)-(x+10)^3-(4-x)^2-... | f(x)=cos(x-5)- sin (x-3)-(x-3)^3-(-x+4)^2 when xis less than -10 or = -10
=cos(x-5)-sin(x-3)+(x-3)^3-(-x+4)^2 whenis between-10 and 0
LHD at x=-10
=sin 15- c0s 13 -196
RHD AT x=-10
=sin15-cos 13 -196
icomputed the lrft hand derivative and right hand derivative using the definition and l`hopital rul... | {
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"timestamp": "2023-03-29T00:00:00",
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How to factor $8xy^3+8x^2-8x^3y-8y^2$ How can I factor $8xy^3+8x^2-8x^3y-8y^2$ or the different form $2x(4y^3+4x)-2y(4x^3+4y)$
Is there any general methods that work?
A possible solution should be $8(x^2-y^2)(1-xy)$ But please do not start from here as in the general case I will not know the answer...
Thanks!
Alexander... | $$ 8xy^3+8x^2-8x^3y-8y^2= 8xy^3-8x^3y+8x^2-8y^2$$
$$= 8xy(y^2-x^2)-8(y^2-x^2)= (8xy-8)(y^2-x^2) $$
$$ = 8(1-xy)(x^2-y^2). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/452978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that
$$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$
using different ways
thanks for all
| $$
\begin{aligned}
\int_{0}^{\infty} \frac{\sin ^{3} x}{x^{3}} d x
=& \frac{1}{4} \int_{0}^{\infty} \frac{3 \sin x-\sin 3 x}{x^{3}} d x \\
\stackrel{IBP }{=} & \frac{1}{8} \int_{0}^{\infty} \frac{(3 \sin x-\sin 3 x)^{(2)}}{x} d x\\
=&\frac{1}{8} \int_{0}^{\infty} \frac{-3 \sin x+9 \sin 3 x}{x} d x \\
=&\frac{1}{8}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 3
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For what $n$ is $x^2 + x+ 1\mid x^{2n} + x^n + 1$? For what $n \in \mathbb{N}$ is $x^2 + x+ 1\mid x^{2n} + x^n + 1$?
The only obvious thing that I could see was noticing that $(x^3 - 1) = (x-1)(x^2 + x+1)$. So, if $x^3 - 1\mid x^{2n} + x^n + 1$. I don't even know if this will help.
| HINT:
If $\omega$ is a root of $x^2+x+1=0, \omega^2+\omega+1=0\implies \omega^3=1$
So if $3|n, n=3m$(say),
$$\omega^n=\omega^{3m}=(\omega^3)^m=1 \text{ and } \omega^{2n}=\omega^{6m}=(\omega^3)^{2m}=1$$
$$\implies \omega^{2n}+\omega^n+1=3\ne0$$
$\implies \omega$ is not a root of $x^{2n}+x^n+1=0$
If $n=3m+1,\omega^{2n}+... | {
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"url": "https://math.stackexchange.com/questions/453489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Differentiation, an issue with an exercise I'm currently working on an exercise that involves quite a few fractional exponents. This is it: $$y = \frac {(x^4 + a)^\frac {1}{3}} {(x^3 + a)^ \frac {1}{2}} $$
I take the multiplication route by doing:
$$(x^4 + a)^ \frac{1}{3}(x^3 + a) ^ \frac {-1}{2} $$
I eventually get $$... | Logarithmic differentiation is useful here.
$$\log{y} = \frac13 \log{(x^4+a)} - \frac12 \log{(x^3+a)}$$
Then
$$\frac{y'}{y} = \frac13 \frac{4 x^3}{x^4+a} - \frac12 \frac{3 x^2}{x^3+a}$$
Multiply by the original $y$ and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/454902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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If a,b,c are sides of a triangle, prove: $ \sqrt{a+b-c} + \sqrt{b+c-a} + \sqrt{c+a-b} \le \sqrt{a} + \sqrt{b} + \sqrt{c} $ I did substitute $a=x+y, b=x+z, c=y+z$ and I arrived at $\sqrt{2x} + \sqrt{2y} + \sqrt{2z} \le \sqrt{x+y} + \sqrt{x+z} + \sqrt{y+z}$.
However, after this, I tried various methods like AM-GM and Cau... | Since $\sqrt{x}$ is concave down, Jensen's inequality tells us that
$ \dfrac 12 ( \sqrt{2x} + \sqrt{2y}) \leq \sqrt{ \dfrac{ 2x + 2y } 2 } = \sqrt{x+y}$.
Summing cyclically gives the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$ While I don't doubt that this question is covered somewhere else I can't seem to find it, or anything close enough to which I can springboard. I however am trying to prove $$\frac{1}{1^2} +\frac{1}{2^2} + \cdo... | Hint Prove instead that
$$\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2-\frac{1}{n}$$
Interesting, this stronger exercise is an easy induction problem, while the weaker result you mentioned cannot be proven directly by induction for obvious reasons (LHS increases, RHS is constant < --- in the stronger ver... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Find $a,b\in\mathbb{Z}^{+}$ such that $\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$ find positive intergers $a,b$ such that
$\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$
Here i tried plugging
$x^3=a,y^3=b$
$(x+y-1)^2=x^2+y^2+1+2(xy-x-y)=49+20\sqrt[3]{6} $
the right hand part is a square hence can be... | I did an Excel search and found $(288,48)=(6^2*2^3,6*2^3)$ and the reverse as solutions. This is confirmed by Alpha
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Convergence of the Jacobi iteration method I think I am not quite understanding the Jacobi Method or some related concept for indirectly solving linear systems of equations of the form $Ax=b$. We need the norm
$||I-Q^{-1}A||_\infty < 1$ and the Jacobi Method states to let $Q$ be the matrix with the same diagonal entrie... | Your understanding of the infinity norm for matrices is wrong:
$$ \Vert I - Q^{-1}A \Vert_\infty = \max \bigg \{0+\bigg|-\frac 12\bigg| + \bigg|-\frac14\bigg|, 0 +0+\bigg|-\frac34\bigg|, \bigg|-\frac34\bigg| + \bigg|-\frac18\bigg| + 0 \bigg\} = \frac 7 8 < 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove this result related to progressions I'm stuck with the following problem from Higher Algebra by Hall and Knight:
If $r<1$ and positive, and $m$ is a positive integer, show that $(2m+1)r^m(1-r) < 1-r^{2m+1}$. Hence show that $nr^n$ is indefinitely small when $n$ is indefinitely great.
I have no clue about the se... | You have $0 < r < 1$, so you can divide by $r^m(1-r) > 0$ and get the equivalent inequality
$$\begin{align}2m+1 &< \frac{1}{r^m} \frac{1-r^{2m+1}}{1-r} = \frac{1}{r^m}\left(1 + r + \dotsb + r^{2m-1} + r^{2m}\right)\\
&= \left(\frac{1}{r^m} + \frac{1}{r^{m-1}} + \dotsb + \frac1r + 1 + r + \dotsb + r^{m-1} + r^m \right)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/457976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Closed form for this continued fraction Is there a closed form for this continued fraction?
$$x+\frac{1}{x+\frac{1}{x+\frac{1}{...}}}$$
| Here's a cute, handwavy way to do it:
$$ f(x) = x + \cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\dots}}}$$
Notice then that
$$ \frac{1}{f(x)} = \cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\dots}}} $$
Thus
\begin{align}
x+\frac{1}{f(x)} &= x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\dots}}} \\
&=f(x)
\end{align}
So, we have a functional re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/459938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compare the sum of the squares of the median of a triangle to the sum of the squares of sides You have to compare the sum of the squares of the median of a triangle to the sum of the squares of sides?
| Let $ABC$ be a triangle and let $I,J,K$ be the midpoints of $[BC],[AC],[AB]$ respectively.
$\overrightarrow{AI}^2=\frac{1}{4}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)^2=\frac{1}{4}\left(AB^2+AC^2+2\overrightarrow{AB}\cdot\overrightarrow{AC}\right)$
The same for $BJ^2$ and $CK^2$. Adding the three equations, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/460588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove the inequality $\,\frac{1}{\sqrt{1}+ \sqrt{3}} +\frac{1}{\sqrt{5}+ \sqrt{7} }+\ldots+\frac{1}{\sqrt{9997}+\sqrt{9999}}\gt 24$
Prove the inequality $$\frac{1}{\sqrt{1}+ \sqrt{3}}
+\frac{1}{\sqrt{5}+ \sqrt{7} }+......... +\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$
My work:
Rationalizing the denominator gives
$$\... | Another idea: by considering concavity and a left endpoint approximation, the desired sum is an overestimate of the following integral:
$$\frac{1}{2}\int_0^{2500}\sqrt{4x+3}-\sqrt{4x+1}\approx 24.6528$$
More explicitly, notice that your sum is:
$$\frac{1}{2}\sum_{n=0}^{2499}\sqrt{4n+3}-\sqrt{4n+1}$$
We can think of thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/462118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Factorize $(x+1)(x+2)(x+3)(x+6)- 3x^2$ I'm preparing for an exam and was solving a few sample questions when I got this question -
Factorize : $$(x+1)(x+2)(x+3)(x+6)- 3x^2$$
I don't really know where to start, but I expanded everything to get :
$$x^4 + 12x^3 + 44x^2 + 72x + 36$$
I used rational roots test and Descarte... | $$(x+1)(x+2)(x+3)(x+6)-3x²$$
$$(x+1)(x+6)(x+3)(x+2)-3x²$$
$$(x²+6x+1x+6)(x²+3x+2x+6)-3x²$$
$$(x²+6x+6+1x)(x²+5x+6+x-x)-3x²$$
$$(x²+6x+6+1x)(x²+5x+x+6-x)-3x²$$
$$(x²+6x+6+1x)(x²+6x+6-x)-3x²$$
$$(x²+6x+6)²(+1x)(-x)-3x²$$
$$(x²+6x+6)²-x²-3x²$$
$$(x²+6x+6)²-4x²$$
$$(x²+6x+6)²-(2x)²$$
Apply $a²-b²=(a+b)(a-b)$
$$(x²+6x+6+2x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 4,
"answer_id": 2
} |
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