Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$ If
\begin{equation}
\sin(x) + \cos(x) = \frac{7}{5},
\end{equation}
then what's the value of
\begin{equation}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?}
\end{equation}
Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) without using the identities of trigonometry.
The function $\sin x+\cos x$ could be transformed using some trigonometric identities to a single function. In fact, WolframAlpha says it is equal to $\sqrt2\sin\left(x+\frac\pi4\right)$ and there also are some posts on this site about this equality. So probably in this way we could calculate $x$ from the first equation - and once we know $\sin x$ and $\cos x$, we can calculate $\dfrac{1}{\sin x}+\dfrac{1}{\cos x}$. Is there a simpler solution (perhaps avoiding explicitly finding $x$)?
| If
$$\sin x+\cos x=a\qquad\text{and}\qquad{\frac{1}{\sin x}}+{\frac{1}{\cos x}}=b$$ then
$$a=b\sin x\cos x\qquad\text{and}\qquad a^2=1+2\sin x\cos x=1+\frac{2a}{b}$$
from which it follows that
$$b={\frac{2a}{a^2-1}}$$
Letting $a=\frac{7}{5}$, we have
$$b={\frac{\frac{14}{5}}{\frac{49}{25}-1}}={\frac{70}{24}}={\frac{35}{12}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 10,
"answer_id": 2
} |
Solving a determinant using properties of a determinant
$$\begin{vmatrix}
y+z & x & x\\
y & z+x & y\\
z & z & x+y
\end{vmatrix}=k(xyz)$$
Find the value of $k$.
I solved this question by substituting $x=y=z=1$ and then expanding the determinant to get $k=4$ which is the correct answer.
Is there any method other than substitution and expanding the determinant right away?
| Use the property that adding multiples of rows does not affect the determinant. Subtracting row 2 and 3 from row 1, we get
$$
\,\,\,\,\,\,\begin{vmatrix}
y+z & x & x\\
y & z+x & y\\
z & z & x+y
\end{vmatrix}\\=
\begin{vmatrix}
0 & -2z & -2y\\
y & z+x &y\\
z & z & x+y
\end{vmatrix}
$$
Add half row 1 to row 2 and 3,
$$
=\begin{vmatrix}
0 & -2z & -2y\\
y & x &0\\
z & 0 & x
\end{vmatrix}
$$
And now the rule of Sarrus will quickly get you $4xyz$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factor the matrix (scalar $\times A$) into permutations of $A$ Here's an example of $A . B = scalar \times C$, done with magic squares. The last square does not have a consecutive range of digits.
Drop the magic square requirement. In $2\times2$ matrices we have the following, where a matrix times a scalar is factored into permutations of the original matrix. All entries of the matrix are distinct.
$$(\begin{pmatrix} -1 & 0 \\ 2 & 1 \\ \end{pmatrix} + a)\cdot (\begin{pmatrix} 1 & 0 \\ -1 & 2 \\ \end{pmatrix}+a) = (2 a + 1)\times(\begin{pmatrix} -1 & 0 \\ 1 & 2 \\ \end{pmatrix}+a)$$
Here are examples with $3 \times 3$ matrices. Can these be canonicalized in some way and all solutions listed?
$$ \begin{pmatrix} 3 & 2 & 4 \\ 1 & -3 & -1 \\ -4 & 0 & -2 \\ \end{pmatrix} \cdot
\begin{pmatrix} -3 & -1 & 2 \\ -2 & 0 & 3 \\ 4 & 1 & -4 \\ \end{pmatrix} =
\begin{pmatrix} 3 & 1 & -4 \\ -1 & -2 & -3 \\ 4 & 2 & 0 \\ \end{pmatrix}$$
$$ \begin{pmatrix} 4 & -2 & 1 \\ -3 & 2 & -1 \\ -4 & 3 & 0 \\ \end{pmatrix} \cdot
\begin{pmatrix} 1 & -1 & -2 \\ 2 & 0 & -4 \\ 3 & 4 & -3 \\ \end{pmatrix} =
\begin{pmatrix} 3 & 0 & -3 \\ -2 & -1 & 1 \\ 2 & 4 & -4 \\ \end{pmatrix}$$
$$ \begin{pmatrix} -1 & 0 & 1 \\ 3 & 4 & -3 \\ -2 & -4 & 2 \\ \end{pmatrix} \cdot
\begin{pmatrix} 4 & -1 & -3 \\ 0 & -2 & 2 \\ 3 & -4 & 1 \\ \end{pmatrix} =
\begin{pmatrix} -1 & -3 & 4 \\ 3 & 1 & -4 \\ -2 & 2 & 0 \\ \end{pmatrix}$$
$$ \begin{pmatrix} 4 & -3 & -4 \\ -1 & 1 & 0 \\ -2 & 2 & 3 \\ \end{pmatrix} \cdot
\begin{pmatrix} -1 & 4 & -2 \\ 2 & 0 & -4 \\ -3 & 3 & 1 \\ \end{pmatrix} =
\begin{pmatrix} 2 & 4 & 0 \\ 3 & -4 & -2 \\ -3 & 1 & -1 \\ \end{pmatrix}$$
Is there a way to solve "Factor $A =\begin{pmatrix} -4 & 0 & -3 \\ 3 & 4 & -1 \\ 1 & 2 & -2 \\ \end{pmatrix}$ into permutations of $A$." ?
Can examples be made with larger matrices?
| Try
$$ \pmatrix{7 & -3 & 6 & 8 \cr
1 & -5 & 4 & -1 \cr
-7 & 3 & -4 & -6 \cr
0 & -2 & 5 & 2 \cr}
\pmatrix{
-5 & 3 & 7 & -1 \cr
-6 & 4 & 8 & -2 \cr
-4 & 2 & 6 & 0 \cr
5 & -3 & -7 & 1 \cr}
=\pmatrix{
-1 & -3 & 5 & 7 \cr
4 & -6 & -2 & 8 \cr
3 & 1 & -7 & -5 \cr
2 & -4 & 0 & 6 \cr}$$
and
$$
\pmatrix{
4 & -6 & 0 & 8 \cr
-2 & -1 & 3 & 1 \cr
-5 & -4 & -7 & -3 \cr
7 & 6 & 5 & 2 \cr}
\pmatrix{5 & -4 & -6 & -2 \cr
-5 & 4 & 7 & 1 \cr
3 & -1 & -3 & 0 \cr
-7 & 6 & 8 & 2 \cr} =\pmatrix{
-6 & 8 & -2 & 2 \cr
-3 & 7 & 4 & 5 \cr
-5 & -7 & -1 & 0 \cr
6 & 3 & 1 & -4 \cr}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solution Set of System of ODE. I am trying to find the solution of the system
$$\begin{bmatrix}x_1\\x_2\end{bmatrix}'= \begin{bmatrix}1&3\\3&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}$$.
I am given that \begin{bmatrix} e^{4t} \\ e^{4t} \end{bmatrix} and \begin{bmatrix} e^{-2t} \\ -e^{-2t} \end{bmatrix} are solutions, and any linear combination is as well: $$\begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix}\Bigg(A\begin{bmatrix} e^{4t} \\ e^{4t} \end{bmatrix}+B\begin{bmatrix} e^{-2t} \\ -e^{-2t} \end{bmatrix}\Bigg)=\begin{bmatrix} 4Ae^{4t}-2Be^{-2t} \\ 4Ae^{4t}+2Be^{-2t} \end{bmatrix}=\begin{bmatrix} Ae^{4t}+Be{-2t} \\ Ae^{4t}-Be^{-2t} \end{bmatrix}'$$ So the solution set is infinite.
On the other hand, from linear algebra, shouldn't the solution set be just 1 element since the homogenous equation $\begin{bmatrix}1&3\\3&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ has only the trivial solution $x_1=x_2=0$?
| \begin{equation*}
\mathbf{x}^{\prime }=\mathbf{W\cdot x\Rightarrow x}(t)=\exp [\mathbf{W}%
t]\cdot \mathbf{x}(0),\;\mathbf{W}=\left(
\begin{array}{cc}
1 & 3 \\
3 & 1
\end{array}
\right)
\end{equation*}
We can diagonalise $\mathbf{W}$. Thus
\begin{equation*}
\mathbf{W}=\left(
\begin{array}{cc}
1 & 3 \\
3 & 1
\end{array}
\right) =\mathbf{U}^{-1}\mathbf{\cdot }\left(
\begin{array}{cc}
4 & 0 \\
0 & -2
\end{array}
\right) \cdot \mathbf{U}=\mathbf{U}^{-1}\cdot \mathbf{D\cdot U}
\end{equation*}
where $\mathbf{U}$ is a unitary matrix (which can be calculated if so
desired).
\begin{eqnarray*}
\mathbf{x}(t) &=&\exp [\mathbf{W}t]\mathbf{x}(0)=\mathbf{U}^{-1}\mathbf{
\cdot }\exp [\mathbf{D}t]\cdot \mathbf{U\cdot x}(0) \\
\mathbf{U\cdot x}(t) &=&\exp [\mathbf{D}t]\cdot \mathbf{x}(0),\;\exp [
\mathbf{D}t]=\left(
\begin{array}{cc}
e^{4t} & 0 \\
0 & e^{-2t}
\end{array}
\right) \\
\lbrack \mathbf{U\cdot x}(t)]_{1} &=&e^{4t}[\mathbf{U\cdot x}(0)]_{1},\;[
\mathbf{U\cdot x}(t)]_{2}=e^{4t}[\mathbf{U\cdot x}(0)]_{2}
\end{eqnarray*}
This is the most general solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
integrate $\int \frac{dx}{x\sqrt{1-x}}$
$$\int \frac{dx}{x\sqrt{1-x}}$$
$$\int \frac{dx}{x\sqrt{1-x}}$$
$u=1-x$
$du=-dx$
$$-\int \frac{du}{(1-u)\sqrt{u}}$$
$a(1-u)+b\sqrt{u}=1\Rightarrow a-au+b\sqrt{u}=1$
$a=1\Rightarrow b\sqrt{u}-u=0\Rightarrow b=\sqrt{u}$
$$\int \frac{\sqrt{u}}{1-u}du-\int \frac{du}{\sqrt{u}}=\int \frac{\sqrt{u}}{1-u}du-2{\sqrt{u}}$$
$$\int \frac{\sqrt{u}}{1-u}du=\int \frac{1+\sqrt{u}-1}{(1+\sqrt{u})(1-\sqrt{u})}du=\int \frac{1+\sqrt{u}}{(1+\sqrt{u})(1-\sqrt{u})}du-\int \frac{du}{(1+\sqrt{u})(1-\sqrt{u})}du=\int \frac{1}{(1-\sqrt{u})}du-\int \frac{du}{(1+\sqrt{u})(1-\sqrt{u})}du$$
$$\int \frac{1}{(1-\sqrt{u})}du$$
How do I continue from here? it seems that I have made it harder
| HINT:
Let $ x = u^2, \, dx= 2 u du ; \, \int \dfrac{2 du }{u\sqrt{1-u^2} }$
again let $$ u = \sin(t) ; \ du = \cos (t) dt ...$$
leading to
$$ =2 log \dfrac{u}{1+\sqrt{1-u^2}} = 2 log \dfrac{\sqrt x}{1+\sqrt{1-x} }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Clever way of calculating the integral $ \int \frac{dt}{t^2\sqrt{t-2} } $
$$ \int \frac{\text{d}t}{t^2\sqrt{t-2} } $$
I know it can be calculated using somewhat complicated substitutions, but is there possibly some clever way of solving that type of integral? I don't even expect full solution, just ideas.
| \begin{align}
u & = \sqrt{t-2} \\[8pt]
u^2 & = t-2 \\[8pt]
u^2 + 2 & = t \\[8pt]
2u\,du & = dt
\end{align}
\begin{align}
\int \frac{dt}{t^2\sqrt{t-2}} & = \int \frac{2u\,du}{(u^2+2)^2 u} = \int\frac{2\,du}{(u^2+2)^2}
\end{align}
Then let $u = \sqrt 2 \tan \theta$, $du= \sqrt 2 \sec^2\theta\,d\theta$, $u^2+2=2\sec^2\theta$. The integral becomes
$$
\int \frac{2\sqrt 2 \sec^2\theta\,d\theta}{\sec^4\theta} = 2\sqrt2 \int\cos^2\theta\,d\theta, \quad\text{etc.}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
To show the $\epsilon-\delta$ definition for limits holds. Question: Check if the following limit exists, if so show that the $\epsilon$ $\delta$ definition for limits holds.
$$\lim_{(x,y) \to (1,2)} \frac{(x-1)^2(y-2)^2}{x^2+y^2-2xy-4y+5}$$
My answer: So clearly the limit is zero, you can write $\frac{(x-1)^2(y-2)^2}{x^2+y^2-2xy-4y+5}$ as $\frac{(x-1)^2(y-2)^2}{(x-1)^2+(y-2)^2+(1+2x-2xy)}$ and $\lim_{(x,y) \to (1,2)} \frac{1}{(y-2)^2}+\frac{1}{(x-1)^2}+\frac{(1+2x-2xy)}{(x-1)^2(y-2)^2}=0$.
I am having difficulty showing the $\epsilon$-$\delta$ definition holds. The trouble I am having is bounding the denominator in $\frac{(x-1)^2(y-2)^2}{x^2+y^2-2xy-4y+5}$, the numerator can be handled quite easily.
| Hint:
Note we can factorize the denominator $D$ as
$x^2+y^2-2xy-4y+5=(x-\frac{y}{2})^2+\frac{3}{4}(y-2)^2-(y-2)-xy$
If $(x,y)$ closed to $(1,2)$, we know $(x-\frac{y}{2})^2,\frac{3}{4}(y-2)^2,(y-2)$ are very closed to $0$ and $-xy$ is closed to $-2$, hence we know the denominator $D$ satisfies
$$|D|\ge |xy|-|(x-\frac{y}{2})^2+\frac{3}{4}(y-2)^2-(y-2)|\ge 1,$$
when $(x,y)$ closed to $(1,2)$.
Then combine with numerator, which is closed to $0$ when $(x,y)$ closed to $(1,2)$, you can prove the limit is $0$ by translating the idea into $\epsilon-\delta$ language .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $2^{n(n+1)}>(n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}$ If $n$ be a positive integer $>1$, prove that
$$2^{n(n+1)}>(n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\left(\frac{n-2}{3}\right)^{n-2}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}$$
Please help me to prove the above. I have to use laws of inequality like AM-GM. But how to use it for this particular problem.
Edit:
Only use laws of inequality.
Edit 2
I want to solve this by using laws of inequality like weighted AM-GM. My attempt is the following
Consider positive numbers $\left(\frac{n}{1}\right), \left(\frac{n-1}{2}\right), \left(\frac{n-2}{3}\right), \cdots \left(\frac{2}{n-1}\right), \frac{1}{n}$ with corresponding weights $n, n-1, n-2, \cdots 2,1$, respectively and applying weighted AM>GM, we get,
$$\frac{\left(\frac{n^2}{1}\right)+\left(\frac{(n-1)^2}{2}\right)+\left(\frac{(n-2)^3}{3}\right)+\cdots \left(\frac{2^2}{n-1}\right)\frac{1^1}{n}}{n+(n-1)+\cdots +2+1}>\left[\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\left(\frac{n-2}{3}\right)^{n-2}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}\right]^{\frac{n(n+1)}{2}}$$
I am unable to get the result because I am unable to get the sum $\left(\frac{n^2}{1}\right)+\left(\frac{(n-1)^2}{2}\right)+\left(\frac{(n-2)^3}{3}\right)+\cdots \left(\frac{2^2}{n-1}\right)\frac{1^1}{n}$
Please suggest me some possible approach.
| Take $\log$ of both sides and verify that the inequality becomes equality for $n=1$. Take backward difference:
$$\nabla RHS=(n+1)\log (n+1)- n\log n+n\log n-\sum_{i=1}^n \log i\le 2n\log 2 =\nabla LHS \tag 1$$
holds for $n=1$. Take another backward difference $$\nabla^2 RHS=(n+1)\log(1+\frac 1 n)\le 2\log 2 =\nabla^2 LHS\tag 2$$
which holds for $n=1$ as well. Since $((x+1)\log(1+\frac 1 x))'=\log(1+\frac 1 x)-\frac 1 x<0$ on $x>0$, $(2)$ is a strict inequality for $n>1$ hence $(1)$ is a strict inequality for $n>1$ and hence the original (strict) inequality holds for $n>1$. Q.E.D.
ETA: The above also shows that $\frac {LHS}{RHS}\to 2\log 2\approx 1.386$ and possibly monotonically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Power series of $\frac{1+x}{(1-x)^2}$ This question is continuing from the previous question here:
Power Series representation of $\frac{1+x}{(1-x)^2}$
I am trying to calculate the power series representation of the equation:
$$
\begin{align}
f(x) = \frac{1+x}{(1-x)^2}
\end{align}
$$
My workout is as follow:
$$
\begin{align}
\frac{1+x}{(1-x)^2} = \frac{1}{(1-x)^2} + \frac{x}{(1-x)^2}
\end{align}
$$
For $\frac{1}{(1-x)^2}$:
\begin{align}
\frac{1}{(1-x)^2} &= \frac{d}{dx} \frac{1}{1-x}\\
&= \frac{d}{dx} \sum_{n=0}^{\infty} x^n \\
&= \sum_{n=1}^{\infty} nx^{n-1} \\
&= \sum_{n=0}^{\infty} (n+1)x^n
\end{align}
For$\frac{x}{(1-x)^2}$:
$$
\begin{align}
x \frac{1}{(1-x)^2} &= x \sum_{n=0}^{\infty}(n+1)x^n \\
&= \sum_{n=0}^{\infty} (n+1) x^{n+1}
\end{align}
$$
Therefore, $$
\begin{align}\frac{1+x}{(1-x)^2}
= \sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=0}^{\infty} (n+1) x^{n+1} = \sum_{n=0}^{\infty}(n+1)(x^n + x^{n+1}),\end{align}
$$
where range of convergence is $x\in[-1,1)$. When $x=-1$, $(x^n + x^{n+1})$ becomes $0$, and $(\infty)(0) = 0$.
However, the model answer is $\sum_{n=0}^{\infty} (2n+1) x^n$, where range of convergence is $x\in (-1,1)$.
I do not understand what is wrong with my calculation. Any advice will be appreciated!
| Hint: In order to find the power series expansion around $x=0$ you could also use the binomial series representation
\begin{align*}
(1+x)^\alpha=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n\qquad\qquad \alpha\in\mathbb{C}, |x|<1
\end{align*}
We obtain
\begin{align*}
\frac{1+x}{(1-x)^2}&=(1+x)\sum_{n=0}^{\infty}\binom{-2}{n}(-x)^n\tag{1}\\
&=(1+x)\sum_{n=0}^{\infty}\binom{n+1}{n}x^n\\
&=(1+x)\sum_{n=0}^{\infty}(n+1)x^n\\
&=\sum_{n=0}^{\infty}(n+1)x^n+\sum_{n=0}^{\infty}(n+1)x^{n+1}\tag{2}\\
&=\sum_{n=0}^{\infty}(n+1)x^n+\sum_{n=1}^{\infty}nx^{n}\\
&=\sum_{n=0}^{\infty}(2n+1)x^n
\end{align*}
Comment:
*
*In (1) we use the identity
$\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$
*In (2) we shift the index of the right sum by one
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $|\mathbb{N} \times \mathbb{N}| \leq |\mathbb{N}|$ Claim: $|\mathbb{N} \times \mathbb{N}| \leq |\mathbb{N}|$
Proof:
For all $ x \in \mathbb{N}, y \in \mathbb{N}$, we can define a function $f(x,y) = x + \frac{(x+y-1)(x+y-2)}{2}$
We now prove that $f$ is an injection. We will show that if $f(x,y) = f(u,v)$ then $x+y$ = $u+v$
Suppose $f(x,y) = f(u,v)$ but $x+y \neq u+v$. Then either $x+y > u+v$ or $u+v > x+y$.
Case 1: $x+y > u+v$
Then $x+y = u+v + \delta$ for some $\delta > 0$.
$$x + \frac{(x+y -1)(x+y-2)}{2} = u + \frac{(u+v -1)(u+v-2)}{2} + \delta$$
Now, I don't know what to do.
| Here's a way of proving that result that's easier to understand:
$$
\begin{array}{cccccccccc}
1,1 & & 1,2 & \rightarrow & 1,3 & & 1,4 & \rightarrow & 1,5 & & 1,6 & \rightarrow \\
\downarrow & \nearrow & & \swarrow & & \nearrow & & \swarrow & & \nearrow \\
2,1 & & 1,1, & & 2,3 & & 2,4 & & 2,5 & & \cdots \\
& \swarrow & & \nearrow & & \swarrow & & \nearrow \\
3,1 & & 3,2 & & 3,3 & & 3,4 & & 3,5 & & \cdots \\
\downarrow & \nearrow & & \swarrow & & \nearrow \\
4,1 & & 4,2 & & 4,3 & & 4,4, & & 4,5 & & \cdots \\
& \swarrow & & \nearrow \\
5,1 & & 5,2 & & 5,3 & & 5,4 & & 5,5 & & \cdots \\
\downarrow & \nearrow \\
6,1 & & \vdots & & \vdots & & \vdots & & \vdots
\end{array}
$$
There's a first member of $\mathbb N\times\mathbb N$, then a second, then a third, and so on. No matter which member of $\mathbb N\times\mathbb N$ you pick, I know, without knowing which one you picked, that after some finite number of steps, I will reach it by following the arrows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$ Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$
My Try:
$\int\frac{3x^2+1}{(x^2-1)^3}dx$
Put $x=\sec\theta$
$\int\frac{3\sec^2\theta+1}{\tan^6\theta}\sec\theta\tan\theta d\theta$
Converting to $\sin$ and $\cos$ we get
$=\int\frac{(3+\cos^2\theta)\cos^2\theta}{\sin^5\theta}d\theta$
I could not solve it further.
| $\dfrac{d\left(\dfrac{ax+b}{(x^2-1)^n}\right)}{dx}=-\dfrac a{(x^2-1)^n}-\dfrac{2a}{(x^2-1)^{n+1}}-b\cdot\dfrac{2x}{(x^2-1)^{n+1}}$
Integrating both sides wrt $x,$
$$-\dfrac{ax+b}{(x^2-1)^n}=a\int\dfrac{dx}{(x^2-1)^n}+2a\int\dfrac{dx}{(x^2-1)^{n+1}}+b\int\dfrac{2x\ dx}{(x^2-1)^{n+1}}$$
$$\implies2a\int\dfrac{dx}{(x^2-1)^{n+1}}=-\dfrac{ax+b}{(x^2-1)^n}-2a\int\dfrac{dx}{(x^2-1)^{n+1}}+\dfrac b{n(x^2-1)^n}\ \ \ \ (1)$$
Now $\displaystyle\int\dfrac{3x^2+1}{(x^2-1)^3}dx=\int\dfrac{3(x^2-1)+4}{(x^2-1)^3}dx=3\int\dfrac{dx}{(x^2-1)^2}+4\int\dfrac{dx}{(x^2-1)^3}$
Set $n=1,2$ in $(1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Find the value of $\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90}$ Find the value of $p+q$, where $p$ and $q$ are two positive integers such that $p$ and $q$ have no common factor larger than $1$ and
$$\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90} = \frac{p}{q}.$$
By using Wolfram Alpha, I obtain the sum is $\frac{3}{20}$. But I have no idea on how to obtain the sum.
Any hint would be appreaciated.
[The question is taken from SMO 2015 Junior section]
| We can observe that
\begin{align}
20&=4\cdot 5\\
30&=5\cdot6\\
42&=6\cdot7\\
56&=7\cdot8\\
72&=8\cdot9\\
90&=9\cdot10.
\end{align}
Can you use partial fraction?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving for x with matrices $$
\begin{bmatrix}
-5 & -9 \\
-6 & -2 \\
\end{bmatrix} X + \begin{bmatrix}
-3 & 2 \\
-9 & -3 \\
\end{bmatrix} = \begin{bmatrix}
-9 & -7 \\
-2 & -3 \\
\end{bmatrix} X
$$
I am asked to solve for matrix X
I combine the x to the right side and get
$$
\begin{bmatrix}
-14 & -16 \\
-8 & -6 \\
\end{bmatrix} X
$$
Then I move the other matrix to the right and get
$$
\begin{bmatrix}
-14 & -16 \\
-8 & -6 \\
\end{bmatrix} X = \begin{bmatrix}
3 & -2 \\
9 & 3 \\
\end{bmatrix}
$$
I then proceed to find the inverse of the x coefficient matrix which comes out to be \begin{bmatrix}
6/44 & -16/44 \\
-8/44 & 14/44 \\
\end{bmatrix}
I know this is the correct inverse because when I multiply it by the original x coefficient matrix I get the identity matrix.
Now as far as I'm aware, all that's left is multiplying the inverse by the right side, but when I do that, my homework system tells me that I'm wrong. Where is the mistake? This is the answer that I get
\begin{bmatrix}
-63/22 & -15/11 \\
51/22 & 29/22 \\
\end{bmatrix}
| You have formally
$$ AX +B = CX$$
You should proceed as follows
$$ \begin{array}{rll}
AX + B &=& CX \\
B &=& CX - AX \\
B &=& (C- A)X \\
(C-A)^{-1} B &=& (C-A)^{-1} (C- A)X \\
(C-A)^{-1} B &=& X \\
\end{array} $$
Now $$ C-A = -\begin{bmatrix} 9 & 7 \\ 2 &3 \end{bmatrix}
- (-\begin{bmatrix} 5 & 9 \\ 6 &2 \end{bmatrix})
= \begin{bmatrix} -4 & 2 \\ 4 & -1 \end{bmatrix} $$
Then $$ (C-A)^{-1} = -\begin{bmatrix} 1/4 & 1/2 \\ 1 &1 \end{bmatrix} $$
and by $ (C-A)^{-1} B=X$ the result
$$ (C-A)^{-1} B = - \begin{bmatrix} 21/4 & 1 \\ 12 &1 \end{bmatrix} = X $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$
$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac{1-v^2}{(1+v^2)(-v^2+v+1)} \, dv$$
Using partial fractions
$$-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}=\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{2v}{v^2+1}-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}$$
$$=\frac{1}{5}ln|v^2+1|-\frac{4}{5}\arctan(v)+\frac{1}{5}ln|-v^2+v+1|$$ from $\frac{\pi}{8}$ to $0$
$0.02-0+0.299-0+0.04-0=0.359$
But it should come out 0.32
| Universal trigonometric substitution is done in order to get the tangents instead of sines and cosines. In this case it is no longer necessary, so:
$$t= \tan x,\quad x = \arctan t,\quad dx = \dfrac1{t^2+1},$$
$$J =\int\limits_0^{\pi/4}\frac{dx}{2+\tan x} = \int_0^1\dfrac{dt}{(t+2)(t^2+1)}.$$
Let
$$R(t) = \dfrac1{(t+2)(t^2+1)} = \dfrac A{t+2}+\dfrac{Bt+C}{t^2+1},$$
then
$$A = \lim_{t\to-2}(t+2)R(t) = \lim_{t\to-2}\dfrac1{t^2+1} = \dfrac15,$$
$$A+B = \lim_{t\to\infty}tR(t) = 0,\quad B=-\dfrac15,$$
$$\dfrac A2+C = R(0) = \dfrac12,\quad C = \dfrac25.$$
Thus,
$$J=\dfrac15\int_0^1\dfrac{dt}{t+2} - \dfrac15\int_0^1\dfrac{t\,dt}{t^2+1} + \dfrac25\int_0^1\dfrac{dt}{t^2+1},$$
$$J = \dfrac15\log(t+2)\biggr|_0^1 - \dfrac1{10}\log(t^2+1)\biggr|_0^1 +\dfrac25\arctan t\biggr|_0^1,$$
$$J = \dfrac15\log\dfrac32 - \dfrac1{10}\log2 +\dfrac25\dfrac\pi4,$$
$$\boxed{J = \dfrac1{10}\left(\pi +\log\dfrac98\right)}$$
Another way (hinted by Wolfram Alpha):
$$J = \int\limits_0^{\pi/4}\frac{dx}{2+\tan x} = \int\limits_0^{\pi/4}\frac{\cos x\,dx}{\sin x + 2\cos x},$$
$$\cos x = A(\sin x + 2\cos x) + B(\cos x - 2\sin x),$$
$$
\begin{cases}
2A + B = 1\\
A - 2B = 0,
\end{cases}\quad
A = \dfrac25,\quad
B = \dfrac15,
$$
$$J = \dfrac25\int\limits_0^{\pi/4}\,dx + \dfrac15\int\limits_0^{\pi/4}\frac{d(\sin x + 2\cos x)}{\sin x + 2\cos x}\,dx,$$
$$J = \dfrac{2x}5\biggr|_0^{\pi/4} + \dfrac15\log(\sin x + 2\cos x)\biggr|_0^{\pi/4},$$
$$J = \dfrac{2}5\dfrac\pi4 + \dfrac15\log\dfrac3{\sqrt2} - \dfrac15\log2,$$
$$\boxed{J = \dfrac1{10}\left(\pi +\log\dfrac98\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Find the quadratic equation, given three points on the parabola
Given that $f(x)=a(x-b)^2+c$, and that $f(1)=f(5)=-1$ and $f(0)=-11$, find the values of $a,b$,and $c$.
I found the axis of symmetry is $3$, because $f$ has the same value at $1$ and $5$, and $(1+5)/2=3$. Therefore $b=3$. How to find $a$ and $c$?
| You have found $b$ correctly as 3. I.e. $f(x)=a(x-3)^2+c=0$.
From this point you could solve simultaneously to get $a$ and $c$. I.e. solving:
$$f(1)=a(1-3)^2+c=-1$$
$$f(0)=a(0-3)^2+c=-11$$
But there is an even easier way thinking about it graphically:
As you have two points with the same y-value we could also express the quadratic as: $f(x)=a(x-1)(x-5)-1$. Now we only need to solve one equation in one variable:
$$f(0)=a(0-1)(0-5)-1=-11$$
$$5a-1=-11$$
$$5a=-10$$
$$a=-2$$
So the equation is:
$$f(x)=-2(x-1)(x-5)-1$$
$$f(x)=-2(x-3+2)(x-3-2)-1$$
$$f(x)=-2((x-3)^2-4)-1$$
$$f(x)=-2(x-3)^2+7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)$ This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).
If $x, y, z$ are positive real numbers, prove that
$(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$
My best idea was to expand this and simplify. Although that doesn't look very feasible. Another idea is to see that $x^2+y^2+xy \geq x^2+y^2$. Then we just have to show that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(x^2+y^2)(x^2+z^2)(y^2+z^2)$ if that is even true.
| $(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz \\ \le x^2+xy+y^2+y^2+yz+z^2+x^2+xz+z^2$
$3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)-(x^2+xy+y^2+y^2+yz+z^2+x^2+xz+z^2)(yz+zx+xy)^2=(x+y+z)^2(y^2z^2-xyz^2+x^2z^2-xy^2z-x^2yz+x^2y^2) \ge0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Solve $I=\iiint\limits_\Omega z^2 dv$ using spherical coordinate system, $\Omega: x^2 +y^2 + z^2 \le R^2 \cap x^2 +y^2 + z^2 \le 2Rz$ Question: Solve $I=\iiint\limits_\Omega z^2 dv$ using spherical coordinate system. $\Omega$ is the common part of $x^2 +y^2 + z^2 \le R^2 $ and $ x^2 +y^2 + z^2 \le 2Rz$.
My attempt:Because $ r^2 \le R^2 and\ r^2 = 2R r\cos \varphi $, so $ 2R\cos \varphi \le r \le R$.
$$\int_0^{2\pi}d\theta \int_0^{\pi/2}\sin\varphi d\varphi \int_{2R\cos \varphi}^R r^2\cdot (r\cos\varphi)^2 dr = 2\pi\int_0^{\pi/2}\cos^2\varphi \sin \varphi d\varphi \int_{2R\cos \varphi}^R r^4 dr =$$
$$\frac{2\pi}5 \int_0^{\pi/2} cos^2\varphi \sin \varphi [R^5-({2R\cos \varphi})^5 ]d\varphi = \frac{2\pi R^5}5 \int_0^{\pi/2}(cos^2\varphi \sin \varphi-32cos^7\varphi \sin \varphi )d\varphi $$
Can anyone point out where is wrong? Is the lower boundary $ 2R \cos \varphi $ correct？
Update based on the answer:
$$I=\int_0^{2\pi}d\theta \int_0^{\pi/4}\sin\varphi \cos^2\varphi d\varphi \int_0^R r^4 dr - \int_0^{2\pi}d\theta \int_0^{\pi/2}\sin\varphi \cos^2\varphi d\varphi \int_0^{2R\cos \varphi} r^4 dr $$
Is this equation correct?
| Your second equation gives $0\leq r\leq 2R\cos(\varphi)$, so $0\leq r\leq \min{(R,2R\cos{\varphi})}$ is the correct bound.
You now need to break up the integral into the regions where each upper bound applies, integrate, and add them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate the limit $\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
I tried to factorise and to simplify, but I can't find anything good.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$
| Substitute $t=\sqrt{x+2}$; if $x\to2$, then $t\to2$, so, after doing $x=t^2-2$, you get
\begin{align}
\lim_{t\to2}\frac{t(t^2-2)^2-8}{4-(t^2-2)^2}
&=\lim_{t\to2}\frac{t^5-4t^3+4t-8}{4t^2-t^4} \\[4px]
&=\lim_{t\to2}\frac{t^3(t^2-4)+4(t-2)}{t^2(4-t^2)} \\[4px]
&=\lim_{t\to2}\left(-t-\frac{4}{t^2(2+t)}\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Find $\int \frac {x^2}{x^3+1} dx$. What is my mistake? $\int \frac {x^2}{x^3+1} dx$
$ u = 3x+1, du=3x^2 dx$
$\int \frac{3 du}{u} $
Am I wrong something? Why the answer is $\int \frac{du}{3u}$ instead of $\int \frac{3 du}{u} $ ?
Thank you.
| Set $t=x^3+1$ and $dt=3x^2dx$
Therefore $\color{red}{ x^2dx=\frac{du}{3}}$
$$=\frac 1 3\int \frac 1 t dt =\frac{\ln t}{3}+\mathcal C=\boxed{\color{blue}{\frac 1 3 \ln (x^3+1)+\mathcal C}} $$
hope it helps
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $0 \le k \le \frac {n-1}{2}$, then ${n \choose k} \le {n \choose k+1}$, with equality holding if and only if $k = \frac{n - 1}{2}$ Prove that if $0 \le k \le \frac {n-1}{2}$, then ${n \choose k} \le {n \choose k+1}$. Further, prove that equality is met if and only if $k = \frac {n-1}{2}$
I tried to use the contrapositive
$${n \choose k} > {n \choose k+1}$$
$$\frac {n!}{(n-k)! \cdot k!} > \frac {n!}{(n-k+1)! \cdot (k+1)!}$$
$$n! \cdot (n-k+1)! \cdot (k+1)! > n! \cdot (n-k)! \cdot (k)!$$
$$(n-k+1)(k+1) > 1$$
I'm trying to get to $k > \frac{n-1}{2}$ and I can't seem to get there. Can anyone help?
| $$\binom n{k+1}=\frac{n^\underline{k+1}}{(k+1)!}=\frac{n^\underline{k}(n-k)}{(k+1)k!}=\frac{n-k}{k+1}\cdot \frac{n^\underline{k}}{k!}=\frac{n-k}{k+1}\cdot \binom nk\\
\binom n{k+1}\ge\binom nk \;\;\;\text{ if }\frac{n-k}{k+1}\ge 1\;\;\;\Rightarrow n\ge 2k+1 \;\;\;\Rightarrow k\le\frac {n+1}2\qquad\blacksquare$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fraction of length-$n$ binary numbers satisfying a constraint Problem: Derive an expression for the fraction of length-$n$ binary numbers that do not contain the subsequence $010$.
Background: The total number of $n$-bit binary numbers is of course $2^n$. Some of these numbers have the sequence $010$ within it; others do not. In short, the problem is to determine, for a given $n$:
${{\rm number\ of\ sequences\ without\ 010} \over 2^n}$.
Although it is a simple matter to determine this fraction through simulation and curve fitting, it is rather tricky to determine the fraction through combinatorics or other rigorous means.
Clearly for $n=3$, there are $2^3 = 8$ possible binary numbers of length $3$ and only one of them is $010$. Thus for $n=3$, the fraction is:
${7 \over 8}$.
Background reading: Binary strings without zigzags and a related question.
| I upvoted the answer by @Joriki which is exemplary.
This can also be done using generating functions.
We get
$$G(z) = \frac{1}{1-z} + \frac{1}{1-z} \frac{z}{1-z}
\sum_{q\ge 0} \left(\frac{z^2}{1-z}\frac{z}{1-z}\right)^q
\frac{1}{1-z}.$$
Explanation: first term a string of ones. Remaining term: admissible
sequences containing at least one zero. This is
$$1^* 0^+ (11^+ 0^+)^* 1^*.$$
We thus obtain
$$G(z) = \frac{1}{1-z}
+ \frac{z}{(1-z)^3} \frac{1}{1-z^3/(1-z)^2}
= \frac{1}{1-z}
+ \frac{z}{1-z} \frac{1}{1-2z+z^2-z^3}
\\ = \frac{z+1-2z+z^2-z^3}{1-z} \frac{1}{1-2z+z^2-z^3}
= \frac{1-z+z^2-z^3}{1-z} \frac{1}{1-2z+z^2-z^3}
\\ = \frac{1+z^2}{1-2z+z^2-z^3}.$$
Now the dominant contribution (the asymptotical factor being sought)
emerges from the inverses of the poles, i.e. the roots of
$$z^3(1-2/z+1/z^2+1/z^3) = z^3-2z^2+z-1.$$
Using a computer algebra system we find the dominant root to be
$$1.754877667$$
and the rest continues as before.
The dominant contribution emerges from the inverses because the poles
are all simple and hence we have
$$\frac{1+z^2}{1-2z+z^2-z^3}
= \sum_{1-2\rho+\rho^2-\rho^3 = 0}
\frac{1}{z-\rho} \mathrm{Res}_{z=\rho} \frac{1+z^2}{1-2z+z^2-z^3}$$
and
$$\frac{1}{z-\rho} = - \frac{1}{\rho} \frac{1}{1-z/\rho}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Integer solutions for $x^4 + 4xy^3 = z^2$ Find all triplets $(x,y,z)$ of integers so that $$ x^4 + 4xy^3 = z^2. $$
What I've done:
Suppose $x=0$. Then we see $z=0$ and hence $(0,y,0)$ is a solution.
Suppose $y=0$. Then we see $x^4 = z^2$ and hence $(x,0,\pm x^2)$ is a solution.
Suppose $z=0$. Then we see $x^3=-4y^3$, which gives $x=y=0$ because otherwise the number of factors of $2$ on both sides cannot be equal.
We may now assume $x>0$ and $z>0$.
I reasoned, if all three are even, $4$ must divide $z$, hence we may divide both sides by $16$, yielding a contradiction by infinite descent. Rewriting the equation as $4xy^3 = (z+x^2)(z-x^2)$ hasn't resulted in any good. Mathematica has found no solutions for $0<x<1000$ and $0<|y|<500$, so I think there are no more solutions. I haven't come really far yet, so any help is appreciated. Thanks.
| I found an answer based on what MXYMXY posted earlier.
The crucial step is to observe that:
$$
(2x^2)(z-x^2)(z+x^2) = 2x^2(z^2-x^4) = 2x^2(4xy^3) = (2xy)^3.
$$
Note that from my analysis above we may conclude that all factors on the LHS are non-zero.
Also note that $$(2x^2)+(z-x^2) = z+x^2.$$
Let $d = \gcd(2x^2,z-x^2) = \gcd(z-x^2,z+x^2) = \gcd(z+x^2,2x^2)$ and observe that $d^3|(2xy)^3$, thus $d|2xy$. Now set $a = \frac{2x^2}{d}$, $b = \frac{z-x^2}{d}$, $c = \frac{z+x^2}{d}$ and $n =\frac{2xy}{d}$. Note that $a$, $b$ and $c$ are pairwise coprime. It follows that:
$$
abc = n^3 \quad \textrm{and} \quad a+b = c,
$$
from which we conclude that $ab(a+b) = n^3$. Since $a$, $b$ and $a+b = c$ are pairwise coprime, we see that all three must be third powers, which is impossible by Fermat's Last Theorem.
This shows there are no further solutions.
| {
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"url": "https://math.stackexchange.com/questions/1675668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integral using partial fractions I am having trouble using partial fractions to evaluate
$$\int \frac{6x}{x^3-8} dx.$$ I can find the denominator but using equations to find the numerator is difficult.
| First, factor $x^3-8$. $x^3-8=(x^2+2x+4)(x-2)$
$$\frac{6x}{x^3-8}=\frac{ax+b}{x^2+2x+4}+\frac{c}{x-2}$$
Then multiply the lhs denominator out on the right side:
$$6x=(x-2)(ax+b)+c(x^2+2x+4)$$
When $x=2$, $(x-2)(ax+b)$ vanishes and $$6(2)=c(4+4+4)\Rightarrow c=1$$
Now, we can use $c$ to solve for $ax+b$.
$$6x-(1)(x^2+2x+4)=(x-2)(ax+b)$$
which gives us $a=-1$ and $b=2$.
Thus $$\int\frac{6x}{x^3-8}dx=\int\frac{2-x}{x^2+2x+4}+\frac{1}{x-2}dx$$
Which gives you what echzhen just demonstrated.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\int {\frac{1}{(-x^{2}+6x-5)^{1/2}}}\;dx$ such that $(1\lt x\lt5)$ using trigonometric substitutions and pythagorean identities First post so I'll get right to the question;
$$\int {\frac{1}{(-x^{2}+6x-5)^{1/2}}}\;dx,\qquad(1\lt x\lt5)$$
To begin with I completed the square which yields: $$-(x-3)^{2}+4$$
substituting this completed square into the integrand yields:
$$\int {\frac{1}{(-(x-3)^{2}+4)^{1/2}}}\,\,dx$$
solving by substituting the trig substitution:
let $$x = 3+2\tan u$$
hence: $$\frac{dx}{du} = 2\sec^{2}u$$
thus:
$$dx = (2\sec^{2}u) du$$
Rearranging $$x = 3+2\tan u$$
yields $$u=\tan^{-1}(\frac{x-3}{2})$$
Now rewriting the integral using the value of x yields:
$$\int {\frac{1}{(-(3+2\tan u-3)^{2}+4)^{1/2}}}\,\,\,(2\sec^{2}u) du$$
which yields:$$\int {\frac{1}{(-(2\tan u)^{2}+4)^{1/2}}}\,\,(2\sec^{2}u) du$$
which is the same as saying: $$\int {\frac{1}{(-4\tan^{2}u+4)^{1/2}}}\,\,\,(2\sec^{2}u) du$$
which can be rearranged to give: $$\int {\frac{1}{(4-4\tan^{2}u)^{1/2}}}\,\,\,(2\sec^{2}u) du$$
Applying the constant multiple rule to the integral yields:
$$\frac 12\int {\frac{1}{(1-\tan^{2}u)^{1/2}}}\,\,\,(\sec^{2}u) du$$
Now we know that pythagorean identity: $$\tan^{2}u=\sec^{2}u-1$$
so substituting I end up with an indefinite integral that looks like this:
$$\frac 12\int {\frac{1}{(1-\sec^{2}u+1)^{1/2}}}\,\,\,(\sec^{2}u) du$$
which gives:
$$\frac 12\int {\frac{1}{(2-\sec^{2}u)^{1/2}}}\,\,\,(\sec^{2}u) du$$
However this is obviously wrong because I think it is unsolvable. So where did I veer from the right path?
| You are on the right track. You should proceed as follows:
$$\begin{align} &\int {\left(\frac{1}{-x^{2}+6x-5}\right)}^{1/2} dx
\\ &=\int {\left(\frac{1}{4-x^{2}+6x-9}\right)}^{1/2} dx
\\ &=\int \frac{dx}{\sqrt{2^2-(x-3)^2}}
\\ &=\int \frac{d(x-3)}{\sqrt{2^2-(x-3)^2}}
\\ &=\color{blue}{\arcsin \frac{x-3}{2} + c}\end{align}$$ where $c$ is a constant of integration
The last step is according to integration formulae:
$$\color{red}{\int \frac{dx}{\sqrt{x^2+a^2}}=\arcsin \frac{x}{a} + c}$$
$$$$
| {
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"url": "https://math.stackexchange.com/questions/1677308",
"timestamp": "2023-03-29T00:00:00",
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How does one project the gradient at a point on a surface into a plane? I am studying Multivariable Calculus and have come to the following excerpt in my book:
I can see clearly how they get from the given function to
$$ y'(x)\ =\ \frac{3y}{x} $$
And understand that this slope passes through the given point. The following line leaves me totally lost, however:
You can verify that the solution to this differential equation is $$ y\ =\ \frac{4x^3}{27} $$ and the projection of the path of steepest descent in the xy-plane is the curve $$ y\ =\ \frac{4x^3}{27} $$
How did they get from any of the given information--the initial equation, its gradient, the slope of the gradient, etc.--to the above equation for y? Furthermore, how am I to find the projection of the gradient in the xy-plane?
| Thanks to user @Moo for helping out in the comments on the original post.
The solution is simple: you know the first derivative of the line and a point it goes through. If you take the first, indefinite integral of your derivative--you get the original line, plus some constant C. Plug your point into this line, solve for C, and--voila! You have your answer.
\begin{align*}
y'(x) &= 3\frac{y}{x} \\
\frac{dy}{dx}\ &= 3\frac{y}{x} \\
\frac{1}{y} dy\ &= 3\frac{1}{x}\ dx \\
\int\frac{1}{y} dy\ &= 3\int\frac{1}{x}\,dx \\
\ln(y) + C_y\ &= 3\ln(x) +C_x \\
\ln(y) &= 3\ \ln(x) + C \\
\ln(y) - 3 \ln(x)\ &= C \\
\ln(y) - \ln\left(x^3\right)\ &= C \\
\ln \left(\frac{y}{x^3}\right) &= C \\
\ln \left(\frac{4}{3^3}\right) &=\ C \\
\ln \left(\frac{4}{27}\right) &=\ C \\
\ln(y) &= 3 \ln(x) + \ln \left(\frac{4}{27}\right) \\
\ln(y) &= \ln(x^3) + \ln \left(\frac{4}{27}\right) \\
\ln(y) &= \ln\left(\frac{4x^3}{27}\right) \\
y &= \frac{4x^3}{27} \\
\end{align*}
| {
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Factor $16x^4-x^2y^2+y^4$ So basically I have the answer to this problem because it's in the book, but I have no clue how the author solved it. It's from Schaum's Precalculus. Thanks.
Factor $(16x^4)-(x^2y^2)+(y^4) = (4x^2+y^2-3xy)(4x^2+y^2+3xy)$
| Hint
$$16x^4+y^4=(4x^2+y^2)^2-8x^2y^2$$ $$16x^4-x^2y^2+y^4=(4x^2+y^2)^2-9x^2y^2=(4x^2+y^2)^2-(3xy)^2$$ Now, use $a^2-b^2=(a+b)(a-b)$
| {
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Find the solutions of the diophantine equation $(x^2-y^2)(z^2-w^2)=2xyzw$ Let $x,y,z,w$ be postive integers. Find all solutions of:
$$(x^2-y^2)(z^2-w^2)=2xyzw$$
This gives:
$$\left(\dfrac{x}{y}-\dfrac{y}{x}\right)\left(\dfrac{z}{w}-\dfrac{w}{z}\right)=2$$
$$\left(p-\dfrac{1}{p}\right)\left(q-\dfrac{1}{q}\right)=2$$
| Given,
$$\Big(p-\frac{1}{p}\Big)\,r=2\tag1$$
$$q-\frac{1}{q} = r\tag2$$
Solve for $p$ in $(1)$ and $q$ in $(2)$,
$$p=\frac{1\pm\sqrt{r^2+1}}{r}$$
$$q=\frac{r\pm\sqrt{r^2+4}}{2}$$
The discriminant should be a square. We have to find rational number $r=u/v$ such that,
$$u^2+v^2 = w_1^2\\
u^2+4v^2 = w_2^2$$
But $d=4$ is not a concordant form/number.
Thus, there is no solution $u,v$, and your original equation has no integer solutions.
| {
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Positive determinant of a symmetric matrix Let $a,b,c,d,e$ positive reals for which $b<\sqrt{ac}$, $c<\sqrt{bd}$, and $d<\sqrt{ec}$. Then, consider the following (Hankel) matrix
$$
M=\begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a+2b+c \\ a+3b+3c+d & a+2b+c & a+b \\ a+2b+c & a+b & a\\ \end{pmatrix}
$$
Is it possible to prove that the determinant of $M$ is positive by means of "symmetric" arguments?
| The claim is not true. The conditions on the coefficients ensure that certain submatrices of the matrix
$$
\left( \begin{array}{ccc} e & d & c \\ d & c & b \\ c & b & a \end{array} \right)
$$
are positive definite. These conditions are not sufficient to ensure that the whole matrix is positive definite.
Here is a counter-example: choose $a=e=5$, $b=d=2$, $c=1$. Then the conditions on the coefficients are satisfied, however the matrix
$$
\left( \begin{array}{ccc} e & d & c \\ d & c & b \\ c & b & a \end{array} \right)
=
\left( \begin{array}{ccc} 5 & 2 & 1 \\ 2 & 1 & 2 \\ 1 & 2 & 5 \end{array} \right)
$$
has determinant $-8$.
| {
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Indefinite integral of a rational function problem... Edit: Corrected my question.
If we have a rational function in general case:
$$f(x)=\frac{1}{(x^2+a^2)^n}$$
And we denote its integral as:
$$I_n=\int \frac{dx}{(x^2+a^2)^n}$$
For $n=1$ we have the integral in the lists:
$$\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan\frac{x}{a}+C$$
For $n>1$ my professor used integration by parts:
$$u=\frac{1}{(x^2+a^2)^n} \to du=-n\frac{2x}{(x^2+a^2)^{n+1}}dx$$
$$dv=dx \to v=x$$
Then, we have:
$$I_n=uv-\int vdu=\frac{x}{(x^2+a^2)^n}+\int \frac{2nx^2}{(x^2+a^2)^{n+1}}dx=$$
$$=\frac{x}{(x^2+a^2)^n}+2n\int \frac{x^2+a^2-a^2}{(x^2+a^2)^{n+1}}dx=$$
$$\frac{x}{(x^2+a^2)^n}+2n\int \frac{dx}{(x^2+a^2)^n}-2na^2 \int \frac{dx}{(x^2+a^2)^{n+1}}dx=$$
$$\frac{x}{(x^2+a^2)^n}+2nI_n-2na^2I_{n+1}$$
And he got a recursive formula:
$$I_{n+1}=\frac{2n-1}{2na^2}I_n+\frac{1}{2na^2}\cdot\frac{x}{(x^2+a^2)^n}$$
where $n=1,2,3,...$
The procedure seems completely fine to me, I couldn't find a mistake anywhere, but, that recursive formula doesn't give the correct result of the integral of function in that form.
For example, lets say we want to integrate the function:
$$f(x)=\frac{1}{(x^2+4x+5)^2}$$
The correct result would be (using WA):
$$\frac{1}{2}\bigg(\frac{x+2}{x^2+4x+5}+\arctan(x+2)\bigg)+C$$
But if we use the recursive formula for $n=1$ we get:
$$I_2=\frac{3}{4}I_1+\frac{1}{4}\cdot\frac{x+2}{(x^2+4x+5)^2}$$
And since $I_1=\arctan(x+2)+C$ we get:
$$I_2=\frac{3}{4}\arctan(x+2)+\frac{1}{4}\cdot\frac{x+2}{(x^2+4x+5)^2}+C$$
Which is not correct according to WA.
Thank you for your time.
| You obviously made a mistake of putting n=2 in the formula, when you are actually trying to calculate I_2, so n+1=2 and n=1.
Putting n=1, you will get the result you got from Wolfram Alpha.
| {
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Integrate $\int \frac{2x}{x^2 +2x+1}$ Here is my attempt:
$$ \frac{2x}{x^2 +2x+1}= \frac{2x}{(x+1)^2 } = \frac{2}{x+1}-\frac{2}{(x+1)^2 }$$
Then I tried to integrate it,I got $2\ln(x+1)+\frac{2}{x+1}+C$ as my answer. Am I right? please correct me if I'm wrong.
| $$=\int\frac{2x+2}{x^2+2x+1}dx-2\int\frac{dx}{x^2+2x+1}$$
set $t=x^2+2x+1$ and $dt=(2x+2)dx$
$$=\int\frac{1}{t}-2\int\frac{dx}{(x+1)^2}dx$$
Set $\nu=x+1$ and $d\nu=dx$
$$=\int\frac{1}{t}-2\int\frac{1}{\nu ^2}d\nu=\ln|t|+\frac{2}{\nu}+\mathcal C=\color{red}{\ln|x^2+2x+1|+\frac{2}{x+1}+\mathcal C}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solutions of the trigonometric equation $\sin(x+ \frac \pi 4) = \sin(\frac 3 8 \pi-3x)$ The following equation
$$\sin(x+ \frac \pi 4) = \sin(\frac 3 8 \pi-3x)$$
has these solutions:
$$x = \frac {\pi} {32} + k \frac \pi 2 \space \vee x = -\frac 3 {16} \pi + k\pi$$
Is it safe to say the equation has this solution below? Did I group the solutions above correctly?
$$x = \frac \pi {32} + \frac {\pi}{2}k$$
| The equality $\sin\alpha=\sin\beta$ is equivalent to the condition
$$
\alpha=\beta+2k\pi\quad\text{or}\quad \alpha=\pi-\beta+2k\pi
$$
(where $k$ is an integer). So in your case you have
$$
x+\frac{\pi}{4}=\frac{3\pi}{8}-3x+2k\pi
$$
which gives
$$
\boxed{x=\frac{\pi}{32}+k\frac{\pi}{2}}
$$
or the other branch
$$
x+\frac{\pi}{4}=\pi-\frac{3\pi}{8}+3x+2k\pi
$$
that yields
$$
x=-\frac{3\pi}{16}-k\pi
$$
that you can also write as
$$
\boxed{x=-\frac{3\pi}{16}+k\pi}
$$
since $k$ can be an arbitrary integer.
Can you group into just the first solution set? You'd need to write
$$
-\frac{3\pi}{16}=\frac{\pi}{32}+k\frac{\pi}{2}
$$
for some integer $k$, but the equation gives
$$
-6=1+16k
$$
that has no integer solution.
| {
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Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$.
The term inside the bracket is divisible by 6 since we have assumed that the result is true when $n=k$. If we can show that $3k^2+15k+12$ is also divisible by 6, then we are done. But how to proceed?
| One number among $n$, $n+1$, $n+2$ is a multiple of 3.
Since $5 \pmod 3 \equiv 2$, among $n$, $n+1$ and $n+5$, one is a multiple of 3.
Either $n$ or $n+1$ must be an even number (since they are consecutive numbers).
Therefore, n(n+1)(n+5) is a multiple of 2.
As $n(n+1)(n+5)$ is a has both 3 and 2 as factors, it is a multiple of 6.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove this inequality $\sqrt{2x^2-6x+5}+\sqrt{y^2-4y+5}+\sqrt{2x^2-2xy+y^2}\ge \sqrt{10}$ Let $x,y\in R$, show that
$$\sqrt{2x^2-6x+5}+\sqrt{y^2-4y+5}+\sqrt{2x^2-2xy+y^2}\ge \sqrt{10}$$
| Hint:
*
*$\sqrt{2x^2-6x+5}= \sqrt{(x-1)^2+(x-2)^2}$ is the distance from $A(1,2)$ to $B(x,x)$.
*$\sqrt{2x^2-2xy+y^2}=\sqrt{(x-y)^2+x^2}$ is the distance from $B(x,x)$ to $C(y,0)$.
*$\sqrt{y^2-4y+5}=\sqrt{(y-2)^2+(-1)^2 }$ is the distance from $C(y,0)$ to $D(2,-1)$.
The inequality is then equivalent to
$$AB+BC+CD\ge AD.$$
| {
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Recovering a quadratic polynomial from three values using calculus I'm asked to solve this using calculus:
Let $$ f(x) = ax^2 + bx +c .$$ If $ f(1) = 3 $, $f(2) = 7$, $f(3) = 13$, then find $a$, $b$, and $f(0)$.
I know I can solve this using solving three equations simultaneously. And I can also solve this using Gauss Jordan or Gaussian elimination method by writing the augmented matrix. But I'm wondering is there any other method to solve this.
Solving by any method it turns out that $a = b = c = 1$.
| Use calculus to solve this... Well, I can wedge some calculus in here...
The derivative of $f(x) = a x^2 + b x + c$ with respect to $x$ is $2ax+b$. By the Fundamental Theorem of Calculus, we have that \begin{align*}
f(x) &= f(0) + \int_0^x 2 a t + b \,\mathrm{d}t \\
&= f(0) + (a t^2 + bt)_{t=x} - (a t^2 + bt)_{t=0} \\
&= f(0) + a x^2 + bx \text{.}
\end{align*} (This is not a shocking revelation: $c = f(0)$.)
Consequently, \begin{align*}
f(1) = 3 &= f(0) + \int_0^1 2 a t + b \,\mathrm{d}t = f(0) + a + b \text{,} \\
f(2) = 7 &= f(0) + \int_0^2 2 a t + b \,\mathrm{d}t = f(0) + 4a + 2b \text{, and} \\
f(3) = 13 &= f(0) + \int_0^3 2 a t + b \,\mathrm{d}t = f(0) + 9a + 3b \text{.}
\end{align*}
This has the convenient property that it exactly contains the expressions, $a$, $b$, and $f(0)$, that are sought in the Question. Now we look at how the function changes between the given points. \begin{align*}
f(2) - f(1) = 4 &= \int_1^2 2 a t + b \,\mathrm{d}t = 3a+b \text{ and} \\
f(3) - f(2) = 6 &= \int_2^3 2 a t + b \,\mathrm{d}t = 5a + b\text{.}
\end{align*}
These tell us $2a=2$, or $a=1$. Then the first tells us that $b=1$. From $3 = f(1) = f(0) + a + b$, we find $f(0) = 1$.
| {
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Evaluation of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\dotsb$
How can we calculate sum of following infinite series
$\displaystyle \bullet\; 1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$
$\displaystyle \bullet\; 1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\cdots$
$\textbf{My Try:}$ Let $$S = \int_0^1 (1+x^2-x^4-x^6+x^8+x^{10}+\cdots) \, dx$$
So we get $$S=\int_0^1 \left(1-x^4+x^8-\cdots\right)dx+x^2\int_0^1 (1-x^4+x^8-\cdots)\,dx$$
So we get $$S=\int_0^1 \frac{1+x^2}{1+x^4} \, dx= \frac{\pi}{2\sqrt{2}}$$ after that we can solve it
Again for second One, Let $$S=\int_0^1 (1-x^6+x^8-x^{14}+x^{16}+\cdots)$$
So we get $$S=\int_0^1 (1+x^8+x^{16}+\cdots) \, dx-\int_0^1 (x^6+x^{14}+
\cdots)\,dx$$
So we get $$S=\int_{0}^{1}\frac{1-x^6}{1-x^8}dx = \int_{0}^{1}\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$
Now how can i solve after that, Help me
Thanks
| Writing $x^2=y$
$$\dfrac{1+x^2+x^4}{(1+x^2)(1+x^4)}=\dfrac{1+y+y^2}{(1+y)(1+y^2)}$$
Let $$\dfrac{1+y+y^2}{(1+y)(1+y^2)}=\dfrac A{1+y}+\dfrac{By+C}{1+y^2}$$
$$\iff1+y+y^2=A(1+y^2)+(1+y)(By+C)=A+C+y(C+B)+y^2(A+B)$$
Comparing the constants, coefficients $y,y^2$
$$A+B=B+C=C+A=1\iff A=B=C=\dfrac12$$
For $\displaystyle \dfrac{1+x^2}{1+x^4},$ see Evaluating $\int_0^\infty \frac{dx}{1+x^4}$.
Can you take it from here?
| {
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Solving modulus inequality $|x - 1| + |x - 6|\le11$ geometrically Find all possible values of $x$ for which $x$ for which the inequality $$|x - 1| + |x - 6|\le11$$ is true.
I know this can be easily solved by taking $3$ cases for$x$ and then taking the intersection of those $3$ cases. The solution will be $-2\ge x\le9$.
But suppose if I interpret this in this way:
What number $x$ satisfy the condition that the distance between $x$ and $6$ plus the distance between $x$ and $1$ is less than or equal to $11$?
I would be better to get an idea to solve these types of problems by geometrically by intuition using number line.
| $$|1-x| + |x-6| \le 11$$
Let $A, X, B \in \mathbb R^n$ (Euclidean $n$-space). If $X\in \overline{AB}$ ( $X$ is on the line segment $\overline{AB}$ ), we say that $X$ is between $A$ and $B$, and we write this as $A-X-B$.
$$A-X-B \; \text{ if and only if } \; \|A-X\| + \|X-B\| = \|A-B\|$$
On $\mathbb R^1$ (the real number line), every point is on the line through points $1$ and $6$. So there are three possibilities for $1, x$, and $6$ : $x-1-6$, $1-x-6$, or $1-6-x$.
CASE: $x-1-6$
Then $ x \le 1 \le 6$
\begin{align}
|1-x| + |x-6| &\le 11 \\
(1-x) + (6-x) &\le 11\\
-2x &\le 4\\
x &\ge -2\\
x &\in [-2,1]
\end{align}
CASE: $1-x-6$
Then $1 \le x \le 6$ and $|1-x| + |x-6| = |1-6| = 5$
\begin{align}
|1-x| + |x-6| &\le 11 \\
5 &\le 11 \\
x &\in [1,6]
\end{align}
CASE: $1-6-x$
Then $1 \le 6 \le x$
\begin{align}
|1-x| + |x-6| &\le 11 \\
(x-1) + (x-6) &\le 11 \\
2x &\le 18\\
x &\le 9\\
x &\in [6,9]
\end{align}
So $x \in [-2,9]$
| {
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What is the fast way to evaluate the following integral: $\int{\frac{\sqrt{x^2+1}}{x^4}\mathrm{d}x}$? I am trying to evaluate the following integral:
$$\int{\dfrac{\sqrt{x^2+1}}{x^4}\mathrm{d}x}$$
I tried the trigonometric substitution: $u = \tan(x)$. Generally, The whole integral needs two substitutions: $u = \tan(x)$ then $v = \sin(u)$. In order to get rid of trigonometric functions, one needs to know that: $$\sin(\arctan(x))=\dfrac{x}{\sqrt{x^2+1}}$$
My question is: What is the fast substitution that leads to the answer without passing by the above steps?
| We can simplify the expression $\frac{\sqrt{x^2+1}}{x^4}$, assuming that $x>0$, as follows:
$$\frac{\sqrt{x^2+1}}{x^4} = \frac{\sqrt{1+\frac1{x^2}}}{x^3} = \frac1{x^3} \cdot \sqrt{1+\frac1{x^2}}.$$
(We have multiplied both numerator and denominator by $\frac1x$.)
After this simplification $t=\frac1x$ seems as a reasonable substitution. It leads to
$$\newcommand{\dd}{\; \mathrm{d}}\int \frac1{x^3} \cdot \sqrt{1+\frac1{x^2}} \dd x =
\begin{vmatrix} x=\frac1t & \dd x = -\frac{\dd t}{t^2} \\ t=\frac1x & \dd t = -\frac{\dd x}{x^2}\end{vmatrix} =
-\int t\sqrt{1+t^2} \dd t.$$
The last integral is rather simple.
Hint: Try the substitution $u=1+t^2$.
Another reasonable substitution seems to be $t=\frac1{x^2}$, which leads to
$$\int \frac1{x^3} \cdot \sqrt{1+\frac1{x^2}} \dd x =
\begin{vmatrix} t=\frac1{x^2} \\ \dd t=-\frac{\dd x}{x^3}\end{vmatrix} = -\int \sqrt{1+t} \dd t.$$
| {
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Inequality using Cauchy-Schwarz Let $a,b,c\in\mathbb{R}^+$, prove that $$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\leq \sqrt{\frac{3}{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)} $$
Hi everyone, I've been trying to do this exercise but any method that I tried has failed. First I tried to use $v=(\sqrt{\frac{a}{b+c}},\sqrt{\frac{b}{c+a}},\sqrt{\frac{c}{a+b}})$ and $w=(1,1,1)$ (the most obvious) but didn't work. Then I did this $$LHS= \frac{\sqrt{a(c+a)(a+b)}+\sqrt{b(b+c)(a+b)}+\sqrt{c(b+c)(c+a)}}{\sqrt{(b+c)(c+a)(a+b)}}\leq \frac{\sqrt{(a(a+b)+b(b+c)+c(c+a))(c+a+a+b+b+c)}}{\sqrt{(b+c)(c+a)(a+b)}} = \frac{\sqrt{2((a+b+c)^2-(ab+bc+ac))(a+b+c)}}{\sqrt{(b+c)(c+a)(a+b)}}$$
Using the fact that $\sqrt{ax}+\sqrt{by}+\sqrt{cz} \leq \sqrt{(a+b+c)(x+y+z)}$ but I don't know if I'm not looking something. If someone can give me a hint I would really appreciate it
| Just a hint. You can denote:
$x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$
Then we have: $xyz=1$ i.e. $z = \frac{1}{xy}$
Then use that: $\sqrt \frac{a}{b+c} = \sqrt \frac{1}{b/a+c/a} = \sqrt \frac{1}{1/x + z} $ and do analogically for the other terms in the LHS.
Then if you use that $z = \frac{1}{xy}$ it boils down to an inequality containing just $x$ and $y$.
Let me see if something good will come out of this, I think it will.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does a closed form solution exist for $x$? $$ \sqrt{x} + x^2 = \sqrt{2} $$
If so, how would one find it?
| Let $u = \sqrt{x}$: $$u + u^4 = \sqrt{2}$$
There is a general formula for the quartic, though it is rather disgusting.
$$u = \frac{1}{2} \sqrt{-\frac{\sqrt[3]{\frac{1}{2} \left(9+\sqrt{3 \left(27+512
\sqrt{2}\right)}\right)}}{3^{2/3}}+\frac{4\ 2^{5/6}}{\sqrt[3]{3 \left(9+\sqrt{3
\left(27+512 \sqrt{2}\right)}\right)}}+\frac{2}{\sqrt{\frac{\sqrt[3]{\frac{1}{2}
\left(9+\sqrt{3 \left(27+512 \sqrt{2}\right)}\right)}}{3^{2/3}}-\frac{4\
2^{5/6}}{\sqrt[3]{3 \left(9+\sqrt{3 \left(27+512
\sqrt{2}\right)}\right)}}}}}-\frac{1}{2} \sqrt{\frac{\sqrt[3]{\frac{1}{2}
\left(9+\sqrt{3 \left(27+512 \sqrt{2}\right)}\right)}}{3^{2/3}}-\frac{4\
2^{5/6}}{\sqrt[3]{3 \left(9+\sqrt{3 \left(27+512 \sqrt{2}\right)}\right)}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is
A. 0;
B. 3;
C. 5;
D. 1.
I don't know how to solve this.
| Observe by trial and error that $-1$ is a root. Then,
$$x^5+2x^3+x^2+2=(x+1)(x^4-x^3+3x^2-2x+2)=(x+1)(x^4-x^3+x^2+2x^2-2x+2)=(x+1)(x^2(x^2-x+1)+2(x^2-x+1))=(x+1)(x^2+2)(x^2-x+1)$$
Thus, there is only one real root.
| {
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Prove that $1+\frac12+\frac13+\cdots+\frac1{2^n} \ge 1+\frac{n}{2}$ Is it true that $1+\dfrac12+\dfrac13+ \dots +\dfrac{1}{2^n} \geq 1+\dfrac{n}{2}$?
If so, a proof would be great!
Thank you!
| Hint: $$\frac{1}{3}+\frac{1}{4}\ge \frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$
$$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\ge \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$$
| {
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Dice probability with limited reroll What is the probability of getting 2 successes when 3 dice are rolled once, and then one die is optionally rerolled?
$$
n = \text{Number of dice} = 3
$$
$$
p_s = \text{Single die success probability (when the face is 3 or more)} = \frac{1}{2}
$$
$$
k = \text{Number of successful die} \in 0, 1, 2, 3
$$
$$
P(n, k, p) = \binom{n}{k}p^k(1-p)^{n-k}
$$
The probability of getting 2 success $(k=2)$ if each die is rolled only once $(p = p_s = \frac{1}{2})$ is
$$
P\left(3,\ 2,\ \frac{1}{2}\right) = \binom{3}{2}\left(\frac{1}{2}\right)^2\left(1-\frac{1}{2}\right)^{3-2} = \frac{3}{8} = 0.375
$$
If we have the option to reroll any failed die once, then
$$
p = p_s + (1-p_s)p_s = \frac{3}{4}\\
$$
and
$$
P\left(3,\ 2,\ \frac{3}{4}\right) = \binom{3}{2}\left(\frac{3}{4}\right)^2\left(1-\frac{3}{4}\right)^{3-2} = \frac{27}{64} = 0.421875
$$
Following the principles from this answer: Fair and Unfair coin Probability
Out of the 3 dice, 2 $(n_{1,2} = 2)$ are rolled once $(p_{1,2} = p_s = \frac{1}{2})$ and 1 die $(n_3 = 1)$ is rolled twice $(p_3 = p_s + (1 - p_s)p_s = \frac{3}{4})$
The probability of rolling 2 successes when all three dice are rolled and then one is rerolled is:
$$P(\text{Rolling 2 successes on the 1-roll dice})\times P(\text{Rolling 0 successes on the 2-roll die})$$
$$+\ P(\text{Rolling 1 success on the 1-roll dice})\times P(\text{Rolling 1 success on the 2-roll die})$$
$$= P\left(2,\ 2,\ \frac{1}{2}\right)\cdot P\left(1,\ 0,\ \frac{3}{4}\right) + P\left(2,\ 1,\ \frac{1}{2}\right)\cdot P\left(1,\ 1,\ \frac{3}{4}\right)$$
$$= \binom{2}{2}\left(\frac{1}{2}\right)^2\left(1 - \frac{1}{2}\right)^{2 - 2}\cdot\binom{1}{0}\left(\frac{3}{4}\right)^0\left(1 - \frac{3}{4}\right)^{1 - 0}$$
$$ + \binom{2}{1}\left(\frac{1}{2}\right)^1\left(1 - \frac{1}{2}\right)^{2 - 1}\cdot\binom{1}{1}\left(\frac{3}{4}\right)^1\left(1 - \frac{3}{4}\right)^{1 - 1}$$
$$= \frac{7}{16} = 0.4375$$
However this seems wrong because it's greater than if I were to roll 3 dice with rerolling $P\left(3,\ 2,\ \frac{3}{4}\right) = \frac{27}{64} = 0.421875$
What am I missing?
| It seems to me now that there was a misunderstanding regarding your first question. In that question, you didn't state that you stop if you get the desired number of successes on the first attempt ("What is the probability of having exactly $k$ successes from $n$ coins if all $n$ coins are flipped and there are $x$ successes, then $n - x$ coins are re-flipped to give the additional $k - x$ successes?") This (and the fact that you were expecting the success probabilities to add up to $1$) led me to think that you always reflip the failed coins, independent of the result of the first flip. Now I see from the fact that you had $P(2)=f(2)+\cdots$ instead of $P(2)=f(2)f(0)+\cdots$ and $P(1)=f(1)+\cdots$ instead of $P(1)=f(1)f(0)+\cdots$, and from the way you're now trying to apply my answer, that apparently you meant all along that you stop if the first flip gives you the $k$ successes.
In that case, my answer to that other question is wrong, and that also explains what's going wrong here: Your result $\frac{27}{64}$ is the probability to get exactly $2$ successes if all failed coins are reflipped, even if the first flip yielded exactly $2$ successes. This is less than the probability you actually want.
| {
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There are $5$ women, $3$ men. How many ways to form a committee of $3$ with at least $1$ member of the opposite sex? I have looked through several topics for similar solutions and I have attempted an answer to the question. Unfortunately, the sample question itself does not have an answer.
From $5$ women and $3$ men, how many ways are there to form a committee of $3$ where there has to be at least $1$ member of the opposite sex.
My first attempt:
$$\binom{5}{3} \cdot \binom{3}{3} = \frac{5 \cdot 4 \cdot 3}{3 \cdot 2 \cdot 1} = 10 \cdot \frac{3!}{3!}$$
Since there are only two ways for committees with no same gender, I did $10-2 = 8$ as my final answer.
My second attempt:
First case: There has to be at least $1$ male. So that means that there are $\binom{5}{2}$ women available. $\binom{5}{2}$ is $10$. $\binom{3}{1}$ is $3$, so $10 \cdot 3 = 30$.
Second case: There has to be at least $2$ males. So that means $\binom{5}{1}$ women are available. $\binom{5}{1}$ is $5$, and $\binom{3}{2}$ is $3$. So that means there are $15 \cdot 3 = 45$ choices available.
When you add the two choices, you get $30 + 45 = 75$ choices.
Which attempt is right? Or are they both wrong? I would love explanations because the sample question has no answer listed.
| Your first method is conceptually sound, but computationally flawed. With no restrictions on gender we just need to select $3$ people from the available $8$, hence $\binom 83=56$ possible choices. Now, exactly one of these is all male but there are $\binom 53=10$ ways to assemble an all female group. Hence $$56-10-1=\fbox {45}$$
Your second method is also flawed, though again your thinking is fine. To make a committee with $2$ women and $1$ man we have $\binom 52\times \binom 31 = 10\times 3=30$ possibilities. To make a committee with $1$ woman and $2$ men we have $\binom 51\times \binom 32 = 5\times 3=15$ possibilities. And again we get $$30+15=\fbox {45}$$
| {
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Solve the system of equations $\frac{x}{3} + \frac{y}{2} = \frac{2x}{3} - \frac{y}{6} = 7$ for $x$ and $y$. Here is the problem. The answer is suppose to be x = 12, y = 6
$$\frac{ x }{3} + \frac{y}{2} = \frac{ 2x }{3} - \frac{ y }{6} = 7$$
I've tried this. I'm having trouble figuring out what do on the last step I will describe.
$$6 • \frac{ x }{3} + \frac{y}{2} = \frac{ 2x }{3} - \frac{ y }{6} = 7$$
$$2x + 3y = 4x - y = 42$$
$$2x - 4y = 42$$
$$4y = 42 - 2x$$
$$y = \frac{ 42 - 2x}{4}$$
$$2x + 3[ \frac{ 42 - 2x }{4} ] = 7$$
| $6·\left[\frac{x}3+\frac{y}2=7\right]$
$18·\left[\frac{2x}3-\frac{y}6=7\right]$
$====================$
then ...
| {
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How to prove this integral problem? For $0<k<1$ put $$K(k)=\int\limits_{0}^{\pi/2}\frac{\text{d}t}{\sqrt{1-k^2\sin^2 t}} \quad \textbf{and} \quad k'=\sqrt{1-k^2}\in(0,1)$$
*
*Prove that $$K(k)=\frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$$
*Prove that $$K(k)=\frac{2}{1+k'}K\left(\frac{1-k'}{1+k'}\right)$$
I have no idea that how to solve this integral.
| I will show the first result using a substitution of
$$\tan t = \dfrac{\sin 2 \theta}{k + \cos 2 \theta}.$$
Under this substitution it can be shown (it is a bit of a slog) that
$$dt = \frac{2(1 + k \cos 2\theta)}{1 + 2k \cos 2\theta + k^2} \, d\theta,$$
while the term appearing in the denominator of the integral becomes
$$1 - k^2 \sin^2 t = \frac{(1 + k \cos 2\theta)^2}{1 + 2k \cos 2\theta + k^2},$$
The limits of integration remain unchanged.
Thus
\begin{align*}
K(k) &= 2 \int^{\pi/2}_0 \frac{1 + k \cos 2\theta}{1 + 2k \cos 2 \theta + k^2} \cdot \frac{\sqrt{1 + 2k \cos 2 \theta + k^2}}{1 + k \cos 2\theta} \, d\theta\\
&= 2 \int^{\pi/2}_0 \frac{d\theta}{\sqrt{1 + 2k \cos 2\theta + k^2}}\\
&= 2 \int^{\pi/2}_0 \frac{d\theta}{\sqrt{(1 + k)^2 - 4k \sin^2 \theta}}\\
&= \frac{2}{1 + k} \int^{\pi/2}_0 \frac{d\theta}{\sqrt{1 - \left (\frac{2 \sqrt{2}}{1 + k} \right )^2 \sin^2 \theta}}, \quad {\rm since} \,\, 0 < k < 1\\
&= \frac{2}{1 + k} K \left (\frac{2 \sqrt{k}}{1 + k} \right ),
\end{align*}
as required to show.
| {
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"timestamp": "2023-03-29T00:00:00",
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An integral for $2\pi+e-9$ Motivation
Lucian asked about the almost-integer $2\pi+e\approx9$ in a comment to a partially answered why question about $e\approx H_8$. This is more involved than approximations to $\pi$ and logarithms because two transcendental constants are included, as in $e^\pi-\pi\approx20$.
Tried so far
An answer can be crafted from integrals related to $\pi\approx\frac{22}{7}$ and $e\approx\frac{19}{7}$
$$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx =\frac{22}{7}-\pi$$
$$\frac{1}{14}\int_0^1 x^2(1-x)^2e^xdx=e-\frac{19}{7}$$
to obtain
$$\int_0^1 x^2 (1-x)^2 \left(\frac{e^x}{14}-\frac{2 x^2 (1-x)^2}{1+x^2}\right) dx = 2\pi+e-9 $$
The visual representation of this integral provided by WolframAlpha shows that $2\pi+e-9$ is positive and small (the integrand is between $0$ and $0.004$ for $0<x<1$), although this is not immediate from the analytic expression.
Moreover, two maxima appear, instead of the single one that is usual in this type of integrals.
Question
Is there a simpler integral with positive integrand in (0,1) that proves $2\pi+e\approx 9$?
| Approximations $e\approx \frac{163}{60} $ and $ \pi \approx \frac{377}{120}$ are related to the integrals
$$\frac{1}{2}\int_0^1 (1-x)^2\left(e^x-1-x-\frac{x^2}{2}\right)dx = e-\frac{163}{60}$$
and
$$\frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx = \frac{377}{120}-\pi.$$
Combining both, we can build
$$\frac{1}{2} \int_0^1 (1 - x)^2 \left(e^x - 1 - x - \frac{x^2}{2} - \frac{2 x^5 (1 - x)^4}{1 + x^2}\right) dx = 2\pi+e-9,$$
which explains the result with nonnegative small integrand and a single maximum in $(0,1)$.
WolframAlpha link
This sets the lower bound
$$9<2\pi+e$$
For an upper bound, we may take the integral from a failed attempt to match $9$
$$\int_0^1 x^4(1-x)^4\left(-\frac{e^x}{24024}+\frac{2}{1+x^2}\right) dx = 9+\frac{4}{1001}-2\pi-e$$
and write
$$2\pi+e<9+\frac{4}{1001}$$
Finally,
$$9<2\pi+e<9+\frac{4}{1001}$$
In short, $2\pi+e$ is close to $9$ because
$\pi\approx\dfrac{377}{120}$
from above,
$e \approx \dfrac{163}{60}$
from below and
$$2·\frac{377}{120}+\frac{163}{60}=9$$
| {
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"timestamp": "2023-03-29T00:00:00",
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An alternative to integration by trigonometric substitution? When I was a Calculus II student many (about 4800) moons ago, our professor taught us an alternative to trig sub. For example, if we have
$$
\int \frac{dx}{x^2\sqrt{x^2 - 9}},
$$
we would evaluate with trig sub by letting $x = 3\sec\theta$ and we'd get:
\begin{align}
\int \frac{dx}{x^2\sqrt{x^2 - 9}} &= \int\frac{3\sec\theta\tan\theta \, d\theta}{9\sec^2\theta\sqrt{9\sec^2\theta - 9}}\\[0.3cm]
&= \frac{1}{3}\int\frac{\tan\theta\,d\theta}{\sec\theta \cdot 3\tan\theta}\\[0.3cm]
&= \frac{1}{9} \int \cos\theta \, d\theta\\[0.3cm]
&= \frac{1}{9} \sin\theta + C\\[0.3cm]
&= \frac{\sqrt{x^2-9}}{9x} + C
\end{align}
The alternative method he showed us goes like this for this problem:
\begin{align}
\int\frac{dx}{x^2\sqrt{x^2 - 9}} &= \int \frac{dx}{x^2\sqrt{x^2(1 - 9x^{-2})}}\\[0.3cm]
&= \int\frac{dx}{x^3\sqrt{1 - 9x^{-2}}}\\[0.3cm]
&= \int\frac{x^{-3} \, dx}{\sqrt{1 - 9x^{-2}}}
\end{align}
Yes, I know that technically $\sqrt{x^2} = |x|$. But trig sub also comes with domain restrictions.
Now let $u = 1 - 9x^{-2}$. Then $du = 18x^{-3} \, dx$ and we have:
\begin{align}
\int\frac{x^{-3} \, dx}{\sqrt{1 - 9x^{-2}}} &= \frac{1}{18}\int\frac{du}{\sqrt{u}}\\[0.3cm]
&= \frac{1}{18} \cdot 2\sqrt{u} + C\\[0.3cm]
&= \frac{\sqrt{1 - 9x^{-2}}}{9} + C
\end{align}
Same answer, different form. My question has two parts, sort of (since an answer to #1 could point to an answer to #2).
*
*Does anyone know the history of this method?
*Is this method perfectly interchangeable with trig sub? In other words, can an integral be done using this method iff it can also be done using trig sub?
IIRC, which I may not because of all the moons, our professor said there's no "general formula" to get this to work. You just kind of have to eyeball it and try it. But I don't remember him saying whether or not it would work all the time.
Thanks!
| set $$\sqrt{x^2-9}=t+x$$ then you will get $$x=\frac{t^2+9}{2t}$$ then we will have $$dx=\frac{1}{2}\frac{t^2-9}{t^2}dt$$
| {
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Simplifying the integral $\int\frac{dx}{(3 + 2\sin x - \cos x)}$ by an easy approach $$I=\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)}$$
If $$\tan\left(\frac{x}{2}\right)=u$$
or $$x=2\cdot\tan^{-1}(u)$$
Then,
$$\sin{x}=\dfrac{2u}{1+u^2}$$
$$\cos{x}=\dfrac{1-u^2}{1+u^2}$$
$$dx=\dfrac{2}{1+u^2}$$
Substitute $$\tan\left(\dfrac{x}{2}\right)=u$$
Let us simplify the integrand before integrating
$$\dfrac{1}{3+2\sin{x}-\cos{x}}$$
$$=\dfrac{1}{3+2\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}}$$
$$=\dfrac{1}{3+\frac{4u-1+u^2}{1+u^2}}$$
$$=\dfrac{1}{\frac{4u-1+u^2+3+3u^2}{1+u^2}}$$
$$=\dfrac{1+u^2}{4u^2+4u+2}$$
$$=\dfrac{1+u^2}{(2u+1)^2+1}$$
$$I=\displaystyle\int\dfrac{1+u^2}{(2u+1)^2+1}\cdot\dfrac{2}{1+u^2}\ du$$
$$=\displaystyle\int\dfrac{1}{(2u+1)^2+1}\ 2\,du$$
Now,
Take : $$v=2u+1$$
Therefore, $$dv=2\,du$$
$$I=\displaystyle\int\dfrac{1}{v^2+1}\ dv$$
$$I=\tan^{-1}(v)$$
Substitute everything back
$$I=\tan^{-1}(2u+1)$$
$$I=\tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)$$
$$\boxed{\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)} = \tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)+C}$$
I know that my approach is also not so difficult but still I think there must be an relatively easy approach to this integral. I have tried many different things using trigonometric identities but nothing seems to bring to the solution easily. Kindly help me out.
| You could write $2\sin(x)-\cos(x) = \sqrt{5} (\frac{2}{\sqrt{5}} \sin x - \frac{1}{\sqrt{5}} \cos x) = \sqrt{5} \sin (x - \alpha)$, with $\tan \alpha = \frac{1}{2}$, so that the required integral becomes $\int \dfrac{dx}{3 + \sqrt{5} \sin(x-\alpha)}$ and then change the integration variable from $x$ to $y = x - \alpha$. Your integral is then of the form $\int \dfrac{dy}{A + B\sin y}$ which is a bit easier to solve using standard methods.
| {
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Find $\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$$
What I have done is to take the $\ln$
$$e^{\lim_{x \to 0}\ln\left(\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}\right)}$$
$$y={\lim_{x \to 0}{\frac{1}{x^2}}\cdot \ln\frac{\cos x}{\cos(2x)}}$$
Applying l'hopital to get
$$y={\lim_{x \to 0}{\frac{-2}{x^3}}\cdot \frac{\cos(2x)}{\cos x}\cdot \frac{-\sin x\cos(2x)+2\sin(2x)\cos x}{(\cos(2x)^2)}}={\lim_{x \to 0}\frac{-2}{x^3\cdot \cos x}\cdot \frac{-\sin x\cos(2x)+2\sin(2x)\cos x}{(\cos(2x))}}$$
Should I apply l'hopital again?
| If you start from $g(x) = \frac{\cos x}{\cos(2x)}^{\frac{1}{x^2}}$ and first take logs you have $\ln g(x) = {\frac{1}{x^2}} (\ln \frac{\cos x}{\cos(2x)})$
Expanding $\frac{\cos(x)}{\cos(2x)} \approx 1 + 3/2 x^2$ and $\ln (1+x) \approx x$ you have $\ln g(x) \approx \frac{3/2 x^2}{x^2} = 3/2$, so that the limit is $e^{3/2}$.
| {
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Let $a,b,c$ be positive reals. Prove that $ab(a+b)+bc(b+c)+ca(c+a)\geq \sum_{cyc} ab\sqrt {{a\over b}(c+a)(c+b)}$
Let $a,b,c$ be positive reals. Prove that
$$ab(a+b)+bc(b+c)+ca(c+a)\geq \sum_{cyc} ab\sqrt {{a\over b}(c+a)(c+b)}$$
I have no clue and I am not good at inequalities. Please help.
| Denote $A = a^2b+b^2c+c^2a,B=ab^2+bc^2+ca^2$, we have $LHS=A+B$.
By AM-GM inequality $$ab\sqrt{{a\over b}(c+a)(c+b)} = \sqrt{a^2(b+c)\cdot ab(a+c)} \le \frac{a^2(b+c) + ab(a+c)}{2},$$
Thus
$$RHS = \sum ab\sqrt{{a\over b}(c+a)(c+b)} \le \frac{1}{2}\sum [a^2(b+c) + ab(a+c)]=\frac{1}{2}(2A+B+3abc)\quad (1)$$
By AM-GM inequality again: $$B=ab^2+bc^2+ca^2 \ge 3\sqrt[3]{ab^2\cdot bc^2\cdot ca^2}=3abc\quad (2)$$
From $(1)$ and $(2)$ we have $$RHS \le \frac{1}{2}(2A+B+3abc) \le \frac{1}{2}(2A+B+B) = A+B = LHS,$$
QED.
| {
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"url": "https://math.stackexchange.com/questions/1713225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find polynomials $f(x),g(x)$ such that $f(x)p(x) + g(x)q(x) = 2x^2 + 6x +2$ Let $$p(x) = x^4 + 7x^3 + 14x^2 + 7x+1 $$$$q(x) = x^4 +10x^3 + 23x^2 + 10x+1$$
Find polynomials $f(x),g(x)$ with rational coefficients such that $$f(x)p(x) + g(x)q(x) = 2x^2 + 6x +2$$
I totally have no idea to solve this problem... Please help me to do it. Thank you!!
| If you factor both polynomials in $\mathbb Q[x]$ you'll get
$p(x) = (x^2+3x+1) (x^2+4x+1)$
$q(x) = (x^2+3x+1) (x^2+7x+1)$
These factors are irreducible (Why?).
Therefore $gcd(p,q) = x^2+3x+1$. Now with the Euclidean algorithm (or some guessing) you should be able to get two polynomials $r,s$ such that $$r(x^2+4 x+1)+s(x^2+7 x+1) = gcd(x^2+4 x+1,x^2+7 x+1) = 1$$
If you multiply this by $2(x^2+3x+1)$ you get exactly what you want:
$$(2r) p + (2s)q = 2x^2+6x+2.$$
Hint 1:
$r,s$ are of degree 1
Hint 2:
$x+7$,$x+4$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove that $\lim_{x \to 3} \sqrt{x+1} = 2$ Prove that $\displaystyle\lim_{x \to 3} \sqrt{x+1} = 2$
Attempt:
$0 < |x - 3| < \delta \Rightarrow |\sqrt{x+1} - 2| < \epsilon$
Well $|\sqrt{x+1} - 2| = |(\sqrt{x+1} - 2) \cdot \displaystyle\frac{\sqrt{x+1} + 2}{\sqrt{x+1}+2}| = |\frac{x-3}{\sqrt{x+1} + 2}| = |x-3| \cdot \frac{1}{|\sqrt{x+1}+2|}$
Here, the second term is the nuisance so maybe do something like the following:
Suppose we want $|x-3| < 1$. Then, $2 < x < 4$.
So $|x-3| \cdot \displaystyle\frac{1}{\sqrt{x+1}+2} < \frac{1}{\sqrt{3}+2} \cdot |x-3|$
Looks like we want $\delta = \min((\sqrt{3}+2)\epsilon, 1)$
Then $|\sqrt{x+1} - 2| = |x-3| \cdot \displaystyle\frac{1}{\sqrt{x+1}+2} < \frac{1}{\sqrt{3}+2} \cdot |x-3| < \frac{1}{\sqrt{3}+2} \cdot (\sqrt{3} + 2)\epsilon = \epsilon$
| Hint:
If $f(x)$ is continuous,
$$\lim_{x\to c} f(x)=f(c)$$
and $\sqrt{x+1}$ is continuous on $[-1,\infty)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find $(a,b)$ such that $\lim\limits_{x\to 0}\frac{f(x)}{x}=1$ implies $\lim\limits_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$
Let $f$ denote any function such that $\lim\limits_{x\to 0}\frac{f(x)}{x}=1$. Find the value of $a$ and $b$, assuming that $\lim\limits_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$.
My attempt:
$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$
$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3\frac{(f(x))^3}{x^3}}=1$
$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$
$\lim_{x\to 0}\frac{1+a\cos x-b\frac{\sin x}{x}}{x^2}=1$
$\lim_{x\to 0}\frac{1+a\cos x-b}{x^2}=1$
As denominator is tending to zero,so numerato will also tend to zero.
$1+a-b=0............................(1)$
Applying L Hospital rule,
$\lim_{x\to 0}\frac{-a\sin x}{2x}=1$
So $a=-2$ and $b=-1$
But the answer given in my book is $a=\frac{-5}{2}$ and $b=-\frac{3}{2}$.
I do not understand where have i gone wrong?
| Others have pointed to your mistake.
The numerator in your third line can be written as
$$x\bigl(1+a(1-{x^2\over2}+?x^4)\bigr)-b\bigl(x-{x^3\over6}+?x^5\bigr)=(1+a-b)x+\bigl(-{a\over2}+{b\over6}\bigr)x^3+?x^5\ ,$$
where each question mark represents some convergent power series. It follows that $a$ and $b$ have to satisfy
$$1-a-b==0,\qquad-{a\over2}+{b\over6}=1\ ,$$
from which we obtain
$$a=-{5\over2},\qquad b=-{3\over2}\ .$$
| {
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"url": "https://math.stackexchange.com/questions/1713981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Algebraic vector proof of Lagrange's identity $$|v · w|
^2 + |v × w|
^2 = |v|^2 |w|^2 $$
Edit
Despite doing it multiple times it seems I have made a meal of the expansion see Jean-Claude's answer for a great explanation
Using $v = (v_1,v_2,v_3)$ and $w = (w_1, w_2, w_3)$ i have expanded the LHS and gotten
$$(v_2)^2(w_3)^2 + (v_3)^2(w_2)^2 + (v_3)^2(w_1)^2 + (v_1)^2(w_3)^2 + (v_1)^2(w_2)^2
+ (v_2)^2(w_1)^2 +(v_1)^2(w_1)^2 + (v_2)^2(w_2)^2 + (v_3)^2(w_3)^2 -\mathbf{2(v_2 w_3 w_2 v_3 + v_3 w_1 v_1 w_3 + v_1 w_2 v_2 w_1)}$$
Now this is RHS minus the bolded terms and I dont know how to get rid of the bolded terms.
| You have
\begin{align*}
|v\cdot w|^2 &= (v_1w_1+v_2w_2+v_3w_3)^2 \\
&= v_1^2w_1^2 + v_2^2w_2^2 + v_3^2w_3^2 + 2v_1w_1v_2w_2 + 2v_2w_2v_3w_3 + 2v_3w_3v_1w_1 \text{.} \\
v\times w &= \begin{pmatrix}v_2w_3-v_3w_2 \\ v_3w_1-v_1w_3 \\ v_1w_2-v_2w_1\end{pmatrix} \text{.} \\
|v\times w|^2 &= (v_2w_3-v_3w_2)^2+ (v_3w_1-v_1w_3)^2 +(v_1w_2-v_2w_1)^2 \\
&= v_2^2w_3^2+v_3^2w_2^2-2v_2w_2v_3w_3+v_3^2w_1^2+v_1^2w_3^2-2v_1w_1v_3w_3 \\
&\qquad {} + v_1^2w_2^2+v_2^2w_1^2-2v_1w_1v_2w_2 \text{.} \\
|v\cdot w|^2+|v\times w|^2 &= v_1^2w_1^2+v_2^2w_2^2+v_3^2w_3^2+v_2^2w_3^2+v_3^2w_2^2+v_3^2w_1^2+v_1^2w_3^2 \\
&\qquad {} + v_1^2w_2^2+v_2^2w_1^2 \\
&= v_1^2(w_1^2+w_2^2+w_3^2)+v_2^2(w_1^2+w_2^2+w_3^2)+v_3^2(w_1^2+w_2^2+w_3^2) \\
&= (v_1^2+v_2^2+v_3^2)(w_1^2+w_2^2+w_3^2) \\
&= |v|^2|w|^2 \text{.}
\end{align*}
See also Lagrange's identity on Wikipedia, for other approaches.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the roots of quadratic polynomial given one root of another quadratic polynomial? if $a,b,c$ are Real numbers and $1$ is a root of equation $ax^2+bx+c=0$ then curve $y = 4ax^2+3bx+2c$ , (a is not zero) intersects $x$ axis at how many points?
I get a relation $a+b+c = 0$ I tried substituting this in given polynomial and tried finding its discriminant but cannot figure out how to find whether it is positive, negative or zero? can someone help?
| You found that $a+b+c=0$, which means that $c=-a-b$, so we have $y=4ax^2+3bx-(2a+2b)$. The discriminant is as follows:
$$(3b)^2-4(4a)(-(2a+2b))=9b^2+16a(2a+2b)=9b^2+32a^2+32ab$$
Now, if we plug in $a=1$ and $b=1$ into the discriminant, we clearly get a positive number, meaning there are two solutions. We need to prove that we get a positive number for all $b \in \Bbb{R}$ and all $a \in \Bbb{R}-\{0\}$.
In order to do this, let's try to express the discriminant as the sum of two squares. We can do this by completing the square with $32a^2+32ab$. First, factor out a $32$ to get $32(a^2+ab)$. Then, divide the coefficient of $ab$ by $2$ to get $\frac 1 2$. This will be the coefficient of $b$ in the monomial squared. Then, square that to get $\frac 1 4$ to get the coefficient of $b^2$. Finally, add $\frac 1 4 b^2$ inside the parentheses to get the following:
$$32(a^2+ab+\frac 1 4 b^2)=32a^2+32ab+8b^2=32\left(a+\frac 1 2 b\right)^2$$
We now know that $32a^2+32ab+8b^2=32\left(a+\frac 1 2 b\right)^2$. However, the discriminant is $32a^2+32ab+9b^2$ which can be expressed as the sum of two squares as follows:
$$32a^2+32ab+9b^2=32a^2+32ab+8b^2+b^2=32\left(a+\frac 1 2 b\right)^2+b^2$$
Thus, our discriminant is the sum of two squares with one of the squares being multiplied by a positive number, which means that the discriminant
is always non-negative. The only case where it is $0$ is if $b=0$ and $a+\frac 1 2 b=0$. Combining these two equations finds that $a=0$, which you said is not under the constraints. Therefore, the discriminant is always positive.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Steps to solve $\lim_{n \to \infty} (\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} ) = 4 $ $$\lim_{n \to \infty} \left(\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} \right) = 4 $$
I think $\sqrt{n^4+1} - \sqrt{n^4-1}$ is approaching to zero, but it is not correct. What steps can evaluate above limit to 4?
| You could also write $\sqrt{n^4 \pm 1} = n^2 \sqrt{1 \pm \frac{1}{n^4}}$ and expand the square roots out using the Binomial Theorem (for large $n, \, \frac{1}{n^4}$ is small), and see what this leaves you.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sigma(n^2)=\sum_{d\mid n} 2^{\omega(d)}$ Let $\omega(n)$ denote the number of distinct prime divisors of $n>1$, with $\omega(1)=0$.
(a) Show that $2^{\omega(n)}$ is a multiplicative function.
(b) Prove that $$\sigma(n^2)=\sum_{d\mid n} 2^{\omega(d)}.$$
I have done the part (a) and I am stuck by (b).
First, I set $d=p_1^{e_1}\cdots p_k^{e_{k}}$ be a factor of $n$. Then I don't know what's the next step.
| Suppose we seek to show that
$$\tau(n^2) =
\sum_{d|n} 2^{\omega(d)}.$$
This can be done using Dirichlet series and Euler products.
We have for the RHS and
$$\sum_{n\ge 1} \frac{1}{n^s} 2^{\omega(n)}$$
the Euler product
$$\prod_p
\left(1 + \frac{2}{p^s} + \frac{2}{p^{2s}} + \frac{2}{p^{3s}}
+\cdots\right).$$
which is
$$\prod_p \left(-1 + 2\frac{1}{1-1/p^s}\right)
= \prod_p \frac{-1+1/p^s+2}{1-1/p^s}
\\ = \prod_p \frac{1+1/p^s}{1-1/p^s}
= \prod_p \frac{1-1/p^{2s}}{(1-1/p^s)^2}
= \frac{\zeta(s)^2}{\zeta(2s)}.$$
Therefore
$$\sum_{n\ge 1} \frac{1}{n^s} \sum_{d|n} 2^{\omega(d)}
= \frac{\zeta(s)^3}{\zeta(2s)}.$$
On the other hand we have
$$\sum_{n\ge 1} \frac{1}{n^s} \tau(n^2)
\\= \prod_p
\left(1 + (2+1) \frac{1}{p^s}
+ (4+1) \frac{1}{p^{2s}}
+ (6+1) \frac{1}{p^{3s}}
+ (8+1) \frac{1}{p^{4s}}
+ \cdots\right).$$
This is
$$\prod_p \left(1+\frac{1/p^s}{1-1/p^s}
+ \sum_{k\ge 1} \frac{2k}{p^{ks}}
\right)
\\ = \prod_p \left(1+\frac{1/p^s}{1-1/p^s}
+ 2 \frac{1/p^s}{(1-1/p^s)^2}
\right).$$
To aid in simplification we put $z=1/p^s$ to get
for the inner term
$$1 + \frac{z}{1-z}
+ \frac{2z}{(1-z)^2}$$
This simplifies to
$$\frac{1+z}{(1-z)^2}.$$
On the other hand
$$\frac{\zeta(s)^3}{\zeta(2s)}
= \prod_p \frac{1-z^2}{(1-z)^3}
= \prod_p \frac{1+z}{(1-z)^2}.$$
We have equality, QED.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to show pattern in iterated differentiation Generally, how would one go about proving general patterns of $n$th derivative?
The specific problem:
$$f(x)=(4-x)^{-0.5}$$
Show that:
$$f^n(0) = 0.5 \left( \frac{(1)(3)(5)\cdots(2n-1)}{8^n} \right)$$
| We can write this formally (using the odd factorial)
$$ f^{(n)}(0) = \frac{1}{2^{3n+1}} \prod_{k=1}^n (2k-1) = \frac{1}{2^{4n+1}} \frac{(2n)!}{n!}
$$
Using the power rule,
$$ f^{(n)}(x) = (-1)^n(4-x)^{-1/2 - n} \prod_{k=1}^{n} \left(\frac{1}{2} - k\right) = (4-x)^{-1/2 - n} \frac{1}{2^n} \prod_{k=1}^{n} \left(2k - 1\right)
$$
Let $x = 0$, we have
$$ (4-0)^{-1/2 - n} \frac{1}{2^n} \prod_{k=1}^{n} \left(2k - 1\right)= \frac{1}{2^{2n+1}} \frac{1}{2^n} \frac{(2n)!}{2^n n!} = \frac{1}{2^{4n+1}} \frac{(2n)!}{n!}
$$
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does the series $\sum_{n=1}^{\infty}{\frac{\sin^2(\sqrt{n})}{n}}$ converge?
Does the following series converge?
$$\sum_{n=1}^{\infty}{\frac{\sin^2(\sqrt{n})}{n}}$$
It shouldn't, but I have no idea how to prove it. I was wondering about Integral Criterion, but the assumptions are not satisfied. Or perhaps Dirichlet test would help, but then it should be shown that $\sum_{k=1}^n(\sin^2(\sqrt{k}))$ is bounded.
| Observe that $\sin^2(\sqrt{n}) \geq 1/4$ iff $|\sin(\sqrt{n})| \geq 1/2$ iff
$$\frac{\pi}{6} + k\pi \leq \sqrt{n} \leq \frac{5\pi}{6} + k\pi$$
for some nonnegative integer $k$. This chain of inequalities is equivalent to
$$\left(\frac{\pi}{6} + k\pi\right)^2 \leq n \leq \left(\frac{5\pi}{6} + k\pi\right)^2$$
For a fixed $k$, the number of values of $n$ which satisfy the above is approximately
$$\left(\frac{5\pi}{6} + k\pi\right)^2 - \left(\frac{\pi}{6} + k\pi\right)^2
= \frac{2\pi^2}{3} + \frac{4\pi^2}{3}k > 6+13k$$
Therefore,
$$\sum_{n=1}^{\infty}\frac{\sin^2(\sqrt{n})}{n} > \sum_{k=1}^{\infty}\frac{6+13k}{4}\frac{1}{\left(\frac{5\pi}{6} + k\pi\right)^2}$$
which diverges by limit comparison with $\sum\frac{1}{k}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Find the minimal polynomial of $\sqrt{3} +i$ over $Q(i)$ and over $Q(\sqrt{3})$
Find the minimum polynomial of $\sqrt{3} +i$ over $Q(i)$ and $Q(\sqrt{3})$.
My solution:
*
*Over $Q(i)$ : Suppose $a = \sqrt{3} +i$, $a-\sqrt{3} =i$;
so, the minimal polynomial is $x-\sqrt{3} -i=0$.
*Over $Q(\sqrt{3})$: Suppose a=$\sqrt{3} +i$, $a-\sqrt{3} =i$; by squaring both side of this equation, $a^2-2\sqrt{3}a+3=-1$. So, we get $x^2-2\sqrt{3}x+4=0$.
Is this correct?
| Hint Consider the product of the monomial $$x - (\sqrt{3} + i)$$ with its conjugate, in two different senses of the word conjugate. (One of the products recovers the correct partial solution from the question: The minimal polynomial of $\sqrt{3} + i$ over $\Bbb Q (\sqrt{3})$ is $x^2 - 2\sqrt{3} x + 4 = 0$.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Summation $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+....$ I came across a question today...
Q. The sum $\dfrac{3}{1^2}+\dfrac{5}{1^2+2^2}+\dfrac{7}{1^2+2^2+3^2}+....$ upto $11$ terms is?
Okay, I think it can be written as
$$\sum_{r=1}^{11}\dfrac{2r+1}{1^2+2^2+...+r^2}$$
I can't see how to simplify it.
Another way I tried is:
$$\dfrac{2^2-1^2}{1^2}+\dfrac{3^2-2^2}{1^2+2^2}+\dfrac{4^2-3^2}{1^2+2^2+3^2}+....$$
I simplified it to
$$-1+2^2(\dfrac{1}{1^2}-\dfrac{1}{1^2+2^2})+3^2(\dfrac{1}{1^2+2^2}-\dfrac{1}{1^2+2^2+3^2})+....$$
$$\Rightarrow -1+\dfrac{2^4}{1^2(1^2+2^2)}+\dfrac{3^4}{(1^2+2^2)(1^2+2^2+3^2)}+....$$
I can't see any way after it too... :(
Can you please help?
| As $1^2+2^2+\cdots+r^2=\dfrac{r(r+1)(2r+1)}6$
$$\dfrac{2r+1}{1^2+2^2+\cdots+r^2}=6\left(\dfrac{r+1-r}{r(r+1)}\right)=?$$
Can you recognize the Telescoping nature?
| {
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The number of words of length $n$ from specific alphabet with rule of creating. Determination of the number of words of length n formed from the alphabet $\{ a, b , c, d \} $, where the letters $a , b $ are not adjacent.
How to find out a recurrence and explicit formula for it ?
| If $a$ and $b$ are not adjacent in a sequence, we call it a valid sequence. Let the number of valid sequences of length $n$ ending with $a$ or $b$ be $p_n$, and let the number of valid sequences of length $n$ ending with $c$ or $d$ be $q_n$. Let the total number of valid sequences of length $n$ be $k_n$.
The first equation is
$$k_n = p_n + q_n$$
We also know
$$p_n = p_{n-1} + 2q_{n-1}$$
because a sequence ending with $a$ or $b$ can be formed by appending a $b$ or a $a$ to a sequence in $p_{n-1}$ ending with $a$ or $b$ (respectively), or by appending $a$ or $b$ to any sequence in $q_{n-1}$. Also,
$$q_n = 2k_{n-1}$$
as we can append $c$ or $d$ to any valid sequence of length $n-1$.
Using these three, we can write
$$
\begin{align}
k_n &= p_n + q_n \\
&= p_{n-1} + 2q_{n-1} + 2k_{n-1} \\
&= p_{n-2} + 2q_{n-2} + 4k_{n-2} + 2k_{n-1} \\
&= \ldots \\
&= p_1 + 2q_1 + 4k_1 + 4k_2 + \dotsb + 4k_{n-2} + 2k_{n-1} \\
&= p_1 + 2q_1 + 4(k_1 + k_2 + \dotsb + k_{n-2}) + 2k_{n-1}
\end{align}
$$
We see that $p_1 = 2$ and $q_1 = 2$, and $k_1 = 4$. Thus
$$k_n = 4(k_1 + k_2 + \dotsb + k_{n-2}) + 2k_{n-1} + 6$$
Though we cannot solve this recurrence directly, the first few values of $k_n$ are $4, 14, 50, 178, 634$ for $n = 1, 2, 3, 4, 5$. According to OEIS, there is only the closed-form
$$k_n = \frac{5+\sqrt{17}}{2\sqrt{17}} \cdot \bigg(\frac{3+\sqrt{17}}{2}\bigg)^n - \frac{5-\sqrt{17}}{2\sqrt{17}} \cdot \bigg(\frac{3-\sqrt{17}}{2}\bigg)^n$$
as obtained by @ChistianBlatter.
| {
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"timestamp": "2023-03-29T00:00:00",
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Tangent line problems Problem 1
Find common tangent to the curve:
$y+x^2=-4$ and $x^2+y^2=4$.
My idea:
Let $t1... y=ax+b$ is a tangent line to the first curve.
Let $t2... y=cx+d$ is a tangent line to the second curve.
But we are seracing for common tangent so that mean that $a=c$ and $b=d$.
By using implicit differentiation we can find $a$ and $b$.
$$y+x^2=-4$$
$$y'+2x=0$$
$$y'=-2x$$
$$x^2+y^2=4$$
$$2x+2yy'=0$$
$$x+yy'=0$$
$$y'=\frac{-x}{y}$$
$$-2x=\frac{-x}{y} \to y=\frac{1}{2},$$$x$ is any real number
Problem 2
For which value of the coefficients $a$, $b$ and $c$ $\in$ R is the x-coordinate axis tangent to the curve?
$y=ax^2+bx+c$
My idea:
Curve and the tangent line must have just one common point. Our curve is a parabola and the number of dots on x axis is a number of solution of quadratic equation. So we need $a,b,c$ to be equal to $b^2-4ac=0$
Problem 3
Find a line that is tangent to the curve
$y = x^4 - 2x^3 - 3x^2 + 5x + 6$
in at least two points .
My idea:
Let that tangent be $t=ax+b$.
If we that line be a tangent line at two point to the curve equation $y(x)-t(x)=0$ must have at least two (not equal) solution.
$$x^4 - 2x^3 - 3x^2 + 5x + 6-(ax+b)=0$$
$$x^4 - 2x^3 - 3x^2 + 5x + 6=ax+b$$
$$a=5 , b=6$$
$$x^4 - 2x^3 - 3x^2=0$$ $$x_1=0, x_2=-1, x_3=3$$
Problem 4
I am trying to find the number of tangents to a curve that all pass through the origin. The curve's equation is $y=x^3+x^2−22x+20.$ I also need to find the equation of said tangents.
My work:
Let's use formula for tangent line:
$$y-y_0=y'(x_0)(x-x_0)$$
We know that $x=0$ and $y=0$.
Let's find $y'(x_0)$:
$y'(x_0)=3x_0^2+2x_0-22$
Plugin what we know:
$$-y_0=(3x_0^2+2x_0-22)*-x_0$$
We aslo know that $y_0=x_0^3+x_0^2-22x_0+20$
$$-x_0^3-x_0^2+22x_0-20=-3x_0^3-2x_0^2+22x_0$$
$$2x_0^3+x_0^2-20=0$$
Only real solution is 2. $x_0=2$ and $y_0=-12$.
Let tangent line be $y=kx+l$.
$k=y'(2)=-6$
$-12=-6*2+l\to l=0$
But we want l to be zero.
Our solution is $y=-6x$.
| Problem$\#1:$
The equation of tangent of $$x^2+y+4=0$$ at $(t,-(4+t^2)),$
$$x(t)+\dfrac{y+\{-(4+t^2)\}}2+4=0\iff2t x+y+4-t^2=0$$
Now this will be a tangent of $$x^2+y^2=4$$ if the perpendicular distance of tangent to center of circle $=$ radius of circle
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $\sin y+\cos y=\frac{1}{2}$ then find $\frac{\sin^3y}{\cos^2y}+\frac{\cos^3y}{\sin^2y}$
If $$\sin y+\cos y=\frac{1}{2} \tag{1}$$ Then find $$x=\frac{\sin^3y}{\cos^2y}+\frac{\cos^3y}{\sin^2y} \tag{2}$$
Given that $$\sin y=\frac{1}{2}-\cos y$$
Squaring both sides we get
$$8\cos^2y-4\cos y-3=0$$
Hence
$$\cos y=\frac{1-\sqrt{7}}{4}$$ and so $$\sin y=\frac{1+\sqrt{7}}{4}$$
substituting the above values in Eq $(2)$ and using the Binomial Theorem we get
$$x=\frac{\sin^5y+\cos^5y}{\sin^2y \cos^2y}$$ so
$$x=\frac{\dfrac{\left(1-\sqrt{7}\right)^5+\left(1+\sqrt{7}\right)^5}{4^5}}{\dfrac{36}{256}}$$
Now by the Binomial theorem $$\left(1-\sqrt{7}\right)^5+\left(1+\sqrt{7}\right)^5=2\left(1+10 \times 7+5 \times 49\right)=632$$
So $$x=\frac{\dfrac{632}{4^5}}{\dfrac{36}{256}}=\frac{79}{18}$$
I feel this is a very lengthy approach; can I get a better approach?
| Write
$$
x = s (s^2/c^2) + c (c^2/s^2)
$$
then replace $c^2 = 1-s^2$ and vice versa to get
$$
x = s (1-c^2)/c^2 + c (1-s^2)/s^2 = (s/c^2) - s + (c/s^2) - c
$$
so that
$$
x + s + c = \frac{s}{c^2} + \frac{c}{s^2} = \frac{sc^2 + cs^2}{c^2s^2} = \frac{c + s}{cs}
$$
Since $c + s = 1/2$, this gives
$$
x + \frac{1}{2} = \frac{\frac{1}{2}}{cs}
$$
so
$$
2x + 1 = \frac{1}{cs}
$$
Now
$$
(c+s) = 1/2 \\
(c+s)^2 = 1/4 \\
c^2 + 2cs + s^2 = 1/4\\
2cs + 1 = 1/4\\
2cs= -3/4 \\
cs = -3/8
$$
so the formula above becomes
$$
2x + 1 = \frac{1}{-3/8} = -\frac{8}{3}
$$
which you can solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Linear transformation of a subspace How can I calculate the image of a linear transformation of a subspace?
Example:
Given a subspace $A$ defined by $x + y + 2z=0$, and a linear transformation defined by the matrix
$$M=
\left(
\begin{matrix}
1 & 2 & -1\\
0 & 2 & 3\\
1 & -1 & 1\\
\end{matrix}\right)
$$
What is the image of the transformation of A by M?
| x+ y+ 2x is more commonly written as 3x+ y so I am going to assume you mean x+ y+ 2z= 0. From that equation, x= -y- 2z so any vector satisfying that can be written <-y- 2z, y, z>. Applying M to that,
$\begin{pmatrix}1 & 2 & -1 \\ 0 & 2 & 3 \\ 1 & -1 & 1 \end{pmatrix}\begin{pmatrix}-y- 2z \\ y \\ z \end{pmatrix}= \begin{pmatrix}-y- 2z+ 2y-z\\ 2y+ 3z \\ -y- 2z- y+ z\end{pmatrix}= \begin{pmatrix} y- 3z \\ 2y+ 3z \\ -2y- z\end{pmatrix}$ $= y\begin{pmatrix}1 \\2 \\-2 \end{pmatrix}+ z\begin{pmatrix}-3 \\ 3 \\-1 \end{pmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Use Darboux sums to calculate the area of $\sqrt x$ in $[0,1]$ I know that $$\Delta x=\frac{b-a}{n}=\frac{1-0}{n}=\frac{1}{n}$$
So $$\overline{D}=\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^n\sup f(x_i)\Delta x=\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^nf(x_i)\Delta x=\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^n \frac{1}{n}\sqrt {\frac{i}{n}}=\displaystyle\lim_{n\to\infty}\frac{1}{n\sqrt n}\displaystyle\sum_{i=1}^n \sqrt {i} $$
I don't know how to calculate the last part $\displaystyle\sum_{i=1}^n \sqrt {i}$
| It is possible to compute the area (or integral) directly as the limit of the upper Darboux sum.
As you showed
$$\overline{D} = \lim_{n\to\infty}\frac{1}{n\sqrt n} \sum_{i=1}^n \sqrt {i}.$$
We can use the binomial expansion $(1-x)^{\alpha} =1 - \alpha x + O(x^2).$
For $i > 1,$ we have
$$\begin{align}(i-1)^{3/2} &= i^{3/2}(1 - i^{-1})^{3/2} \\ &= i^{3/2}\left[ 1 - \frac{3}{2}i^{-1} + O(i^{-2})\right] \\ &= i^{3/2} - \frac{3}{2}\sqrt{i} + O(1/\sqrt{i}).\end{align}$$
Hence,
$$\sqrt{i} = \frac{2}{3}\left[i^{3/2} - (i-1)^{3/2}\right] + O(1/\sqrt{i}),$$
and,
$$\begin{align}\frac1{n\sqrt{n}}\sum_{i=1}^n\sqrt{i} &=\frac{1}{n \sqrt{n}}\left(1 + \frac{2}{3}\sum_{i=2}^n[i^{3/2}-(i-1)^{3/2} + O(1/\sqrt{i})]\right) \\ &= \frac{1}{n \sqrt{n}}\left(1 + \frac{2}{3}n^{3/2}- \frac{2}{3} + O(n)\right) \\ &= \frac{2}{3} +\frac{1}{3n \sqrt{n}} + O(1/\sqrt{n}). \end{align}$$
Thus,
$$\lim_{n \to \infty}\frac1{n \sqrt{n}}\sum_{i=1}^n\sqrt{i} = \frac{2}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Simple differential equation $\frac{x}{y}=\frac{y'}{x+1}$ $$\frac{x}{y}=\frac{y'}{x+1}$$
The solution to this is very easy, we just multiply both sides by
$$(x+1)y\ \mathrm dx$$
Then we have
$$(x^2+x)\ \mathrm dx=y\ \mathrm dy$$
Then we integrate both sides to get solution
$$\frac{x^3}{3}+ \frac{x^2}{2}=\frac{y^2}{2} + C, C\in \mathbb{R}$$
But I don't understand how can we divide by $y$ and $x+1$. What if $y=0$ or $x=-1$? What happens then?
| $$\frac{x}{y}=\frac{1}{x+1}\frac{\mathrm dy}{\mathrm dx}$$
The above equation does not hold for $x=-1$ and $y=0$ because division by zero is undefined. This implies that $-1$ is excluded from the domain of $y$ and $0$ is excluded from the codomain of $y$. Therefore we can safely divide by $x+1$ and $y$, respectively. We can also proceed as follows
$$x^2+x=y\frac{\mathrm dy}{\mathrm dx}$$
$$\int x^2+x\ \mathrm dx=\int y\frac{\mathrm dy}{\mathrm dx}\ \mathrm dx$$
$$\int x^2\ \mathrm dx+\int x\ \mathrm dx=\int y\ \mathrm dy$$
$$\frac{x^3}{3}+\frac{x^2}{2}+C=\frac{y^2}{2}$$
$$y^2=\frac{2x^3}{3}+x^2+C$$
$$\left|y\right|=\sqrt{\frac{2x^3}{3}+x^2+C}$$
Therefore the solutions are
$$y=\pm\sqrt{\frac{2x^3}{3}+x^2+C}$$
Where $-1$ is excluded from the domain of $y$ and $0$ is excluded from the codomain of $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1734580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show $\frac{\pi}{\cos \pi z}=4\sum\limits_{n=0}^\infty \frac{(-1)^n(2n+1)}{(2n+1)^2-4z^2}$ using partial fraction decomposition of $\cot$ and $\tan$ I am asked to show that $\frac{\pi}{\cos \pi z}=4\sum\limits_{n=0}^\infty \frac{(-1)^n(2n+1)}{(2n+1)^2-4z^2}$. Here is what I have got so far,
$\frac{\pi}{\cos (\pi z)}=\frac{\pi}{\sin(\pi(\frac{1}{2}-z))}=$
$\pi[\cot(\pi(\frac{1}{2}-z))+\tan(\frac{\pi}{2}(\frac{1}{2}-z))]=\frac{1}{(\frac{1}{2}-z)}+\sum\limits_{n=1}^\infty \frac{2(\frac{1}{2}-z)}{(\frac{1}{2}-z)^2-n^2}+\sum\limits_{n=0}^\infty \frac{-2(\frac{\pi}{2}(\frac{1}{2}-z))}{(\frac{\pi}{2}(\frac{1}{2}-z))^2-(n+\frac{1}{2})^2\pi^2}$
I cannot see where to go from here. Any suggestions or hints would be appreciated.
| Since
$
\frac{1}{\sin z}=\cot z+\tan\left(\frac{z}{2}\right)
$, and using the partial fraction decomposition of the tangent and cotangent
$$
\pi \tan(\pi z) = 8z\sum_{n=0}^{+\infty}\frac{1}{(2n + 1)^2 − 4z^2}\quad\text{and}\quad
\pi \cot(\pi z) = \frac{1}{z}+2z\sum_{n=1}^{+\infty}\frac{1}{z^2 − n^2}
$$
we have
\begin{align}
\frac{\pi}{\sin \pi z}&=\frac{1}{z}+2z\left[\sum_{n=1}^{+\infty}\frac{1}{z^2 − n^2}-\sum_{n=0}^{+\infty}\frac{2}{z^2-(2n + 1)^2 }\right]\\
&=\frac{1}{z}+2z\left[\frac{1}{z^2 − 1^2}+\frac{1}{z^2 − 2^2}+\frac{1}{z^2 − 3^2}+\frac{1}{z^2 − 4^2}+\frac{1}{z^2 − 5^2}+\cdots\right.\\
&\left.\qquad\qquad\quad-\frac{2}{z^2-1^2}-\frac{2}{z^2-3^2}-\frac{2}{z^2-5^2}\right]\\
&=\frac{1}{z}+2z\left[-\frac{1}{z^2 − 1^2}+\frac{1}{z^2 − 2^2}-\frac{1}{z^2 − 3^2}+\frac{1}{z^2 − 4^2}-\frac{1}{z^2 − 5^2}+\cdots\right]
\end{align}
that is
$$
\frac{\pi}{\sin \pi z}=\frac{1}{z}+\sum_{n=1}^{+\infty}(-1)^{n-1}\frac{2z}{n^2-z^2}
$$
Observing that $\sin\left(\pi\left(\frac{1}{2}-z\right)\right)=\cos \pi z$, upon writing $\frac{2z}{n^2-z^2}=\frac{1}{n-z}-\frac{1}{n+z}$ and then replacing $z$ with $\frac{1}{2}-z$, we have
$$
\frac{\pi}{\cos \pi z}=\frac{2}{1-2z}+\left(\frac{2}{1+2z}-\frac{2}{3-2z}\right)-\left(\frac{2}{3+2z}-\frac{2}{5-2z}\right)+-\cdots
$$
Here we may omit the parentheses and combine pairs of terms having the same sign; this yields
$$
\frac{\pi}{\cos \pi z}=\frac{4\cdot 1}{1^2-4z^2}-\frac{4\cdot 3}{3^2-4z^2}+\frac{4\cdot 5}{5^2-4z^2}-+\cdots
$$
that is
$$
\frac{\pi}{\cos \pi z}=4\sum\limits_{n=0}^\infty (-1)^n\frac{(2n+1)}{(2n+1)^2-4z^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Define the image of the function $f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$ Question
Define the range of the function:
$$f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$$
Answer attempt
I assume that the domain for $f(x)$ is in ${\rm I\!R}$.
If we draw a triangle with the sides: $1, x$ and $\sqrt{1+x^2}$, with the angle $\theta$ opposite of $x$, we get:
$$\sin{\frac{x}{\sqrt{1+x^2}}} \rightarrow$$ $$\theta = \arcsin{\frac{x}{\sqrt{1+x^2}}}$$
$$\tan{\frac{x}{1}} \rightarrow$$ $$\theta = \arctan{x}$$
This means that:
$$\arctan x = \arcsin{\frac{x}{\sqrt{1+x^2}}}$$
The original expression can thus be written:
$$2 \arctan x + \arcsin \frac {2x}{1+x^2} = 2 \arcsin \frac{x}{\sqrt{1+x^2}} + \arcsin \frac {2x}{1+x^2}$$
We know that $\arcsin$ has the domain $-1 \leq x \leq 1$ and the range $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$
If we start with: $\arcsin \frac {2x}{1+x^2}$
$x=1 \rightarrow \arcsin{\frac{2}{2}} = \arcsin 1 = \frac{\pi}{2}$
$x=-1 \rightarrow \arcsin{\frac{-2}{2}} = \arcsin -1 = -\frac{\pi}{2}$
$-1 \leq x \leq 1 \rightarrow -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
And analyzing: $2\arcsin \frac{x}{\sqrt{1+x^2}}$
$x=1 \rightarrow 2\arcsin{\frac{1}{\sqrt{2}}} \rightarrow \sin{\theta} = \frac{1}{\sqrt{2}} \rightarrow \theta = \frac{\pi}{4}$
$x=-1 \rightarrow 2\arcsin{-\frac{1}{\sqrt{2}}} \rightarrow \sin{\theta} = -\frac{1}{\sqrt{2}} \rightarrow \theta = -\frac{\pi}{4}$
$-1 \leq x \leq 1 \rightarrow -2(\frac{\pi}{4}) \leq y \leq 2(\frac{\pi}{4}) \rightarrow -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
The range of $f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$ is therefore $-\pi \leq y \leq \pi$
I wonder if you feel that my reasoning is correct? Is there some better way to solve this problem?
| This range is correct. While you start with the assumption that the domain for $f(x)$ is $x\in\mathbb{R}$, we know that the domain of $f(x)$ is limited by the domain of $\arcsin(x)$ which is $[-1,1]$.
So, this means that $-1\leq\dfrac{2x}{1+x^2}\leq 1$ which is precisely $x\in[-1,1]$
Since $\arctan(x)$ is an increasing function, we know that the maximum value of $f(x)$ is at $x=1$ and and the minimum at $x=-1$ If we substitute these values, we get the same range that you found.
For a more rigorous verification, we could take the derivative, but I think this is a more intuitive explanation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Fewest steps to show $\sin^2(s) = \frac{1}{2} - \frac{1}{2} \cos(2s)$? Is there a method for getting from this statement
$
\sin^2(s)
$
to this statement
$
\frac{1}{2}\ -\ \frac{1}{2}\cos(2s)
$
in fewer steps than the following?
$\begin{align}
\cos(2s) &= \cos^2(s) - \sin^2(s) \\
\frac{1}{2} \cos(2s) &= \frac{1}{2} \cos^2(s) - \frac{1}{2} \sin^2(s) \\
-\frac{1}{2} \cos(2s) &= -\frac{1}{2} \cos^2(s) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \frac{1}{2} - \frac{1}{2} \cos^2(s) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \frac{1}{2} ( 1 - \cos^2(s) ) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \frac{1}{2} ( \sin^2(s) ) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \sin^2(s) \\
\end{align}$
| $$\frac{1-\cos(2s)}2=\frac{\cos^2(x)+\sin^2(x)-\cos^2(s)+\sin^2(x)}2=\sin^2(s).$$
What else ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$ Prove that $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$
$$\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{\frac{2}{\cos\theta} +\frac{3\sin\theta}{\cos\theta}+5\sin\theta-7\cos\theta+5}{\frac{2\sin\theta}{\cos\theta} +\frac{3}{\cos\theta}+5\cos\theta+7\sin\theta+8}$$
$$=\frac{2+3\sin\theta+5\sin\theta\cos\theta-7\cos^2\theta+5\cos\theta}{2\sin\theta+3+5\cos^2\theta+7\sin\theta\cos\theta+8\cos\theta}$$
I am stuck here.I tried to factorize numerator and denominator but does not succeed.
| Let's simplify our notation by writing $c=\cos\theta$ and $s=\sin\theta$. First note that $$\frac{1-\cos\theta}{\sin\theta}=\frac{\sin\theta}{1+\cos\theta}.$$ So, for arbitrary $a$ and $b$, not both zero, we can write either fraction as $$\frac{a(1-c)+bs}{as+b(1+c)}.$$It's not hard to spot that setting $a=5c$ and $b=3+5c+2s$ gives us the original expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$? How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...
| Using partial fraction decomposition
$$
\frac{1}{n(n+3)} = \frac{1}{3} \cdot \frac{1}{n} - \frac{1}{3} \cdot \frac{1}{n+3}
$$
convince yourself that for any $f$, we have
$$
\sum_{n=1}^{m} \left(f(n)-f(n+3)\right) = f(1) + f(2) + f(3) - f(m+3) - f(m+2) - f(m+1)
$$
and now take the limit of $m \to \infty$ getting
$$
\sum_{n=1}^\infty \frac{1}{n(n+3)} = \frac{1}{3} \left(1 + \frac{1}{2} + \frac{1}{3} \right) = \frac{11}{18}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Solving exponential equation (quadratic type) I fail trying to solve the following equation:
$9^x-6^x-2^{2x+1}=0$
Trying to write it as a quadratic equation makes my constant term exponential
$(3^x)^2-2^x3^x-2^{2x+1}=0$
How can I solve this type of problem most efficiently?
| \begin{align*}
9^x-6^x-2^{2x+1} &=0 \\
(\color{red}{3^x})^2-(\color{red}{3^x})(\color{blue}{2^x})-
2(\color{blue}{2^x})^2 &=0 \\
(\color{red}{3^x}+\color{blue}{2^x})
(\color{red}{3^x}-2\cdot \color{blue}{2^x}) &= 0
\end{align*}
Rejecting $\color{red}{3^x}+\color{blue}{2^x}=0$, we have
\begin{align*}
\left( \frac{3}{2} \right)^{x} &= 2 \\
x(\ln 3-\ln 2) &= \ln 2 \\
x &= \frac{\ln 2}{\ln 3-\ln 2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why is the double angle formula not used in this solution? solve $\sin 2x = 0.5$
I am looking at the answer to this question and it does not use the double angle formula to make this:
$\sin 2x = 2\sin x \cos x$
And instead use
$2x = \arcsin 0.5$
Why is the double angle formula not used in this situation?
| Using the double angle formula here ends you up with two things that must multiply together to become $0.5$, and we would need to solve a nasty quadratic to discover $\sin x$ and $\cos x$. Just using $\arcsin$ directly makes your life easier: $\arcsin 0.5$ is very easy to obtain, and we reduce the rest of the problem to simple manipulation.
To give some sense here, using double angle formula you get
$$2\sin x \cos x = \frac{1}{2}$$
This with the pythagorean identity $\sin^2 x + \cos^2 x = 1$ can combine to make the following two equations:
$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \frac{3}{2}$$
$$\sin^2 x - 2\sin x \cos x + \cos^2 x = \frac{1}{2}$$
Factor the left side:
$$(\sin x + \cos x)^2 = \frac{3}{2}$$
$$(\sin x - \cos x)^2 = \frac{1}{2}$$
Take the square root:
$$\sin x + \cos x = \pm\sqrt{\frac{3}{2}}$$
$$\sin x - \cos x = \pm\sqrt{\frac{1}{2}}$$
Add the two equations:
$$2\sin x = \pm\sqrt{\frac{3}{2}} \pm \sqrt{\frac{1}{2}}$$
And now finally we can discover the sine of the original angle:
$$\sin x = \pm \frac{\sqrt{6}\pm\sqrt{2}}{4}$$
And then we need to apply $\arcsin$ and do some fiddling with "which equivalent angles do I take?". What a mess. Instead, just find $\arcsin \frac{1}{2} = 2x$ and use supplemental and coterminal angles to find the other values of $2x$ that work.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$ Solve the following equation: $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$
Unfortunately I have no idea.
| Put $x=\cos \theta$, then:
$\sqrt{1-\cos \theta} = 2\cos^2 \theta -1 \pm 2\cos \theta\sin\theta = \cos 2\theta \pm \sin 2\theta$.
Squaring ,we get:
$1-\cos \theta = 1 \pm \sin 4\theta$.
Therefore $\cos \theta = \pm\sin 4\theta$. The solutions of this equation are easy to find. Also, one must eliminate extraneous solutions e.g. $126^\circ$.
| {
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Proving $1 \cdot 2 + 2 \cdot 3 + \cdots + n(n + 1) = 2\binom{n + 2}{3}$by math induction? I am working on a problem, but I don't know whether or not to use math induction on it. Here's the problem:
Prove that for all integers $n \geq 1$,
$$1 \cdot 2 + 2 \cdot 3 + \cdots + n(n + 1) = 2\binom{n + 2}{3}$$
If I used math induction, I'd first pick out a basis which would be $1$. Then I'd substitute $n$ for $1$. Which would be:
$$1(1 + 1) = 2\binom{1+2}{3}$$
the left hand side is equal to the right hand side, thus we have verified this part of the proof.
Next we look at $n + 1$:
$$n + 1(n + 2) = 2\binom{n + 3}{3}$$
How do we prove the last part?
| Check carefully and algebraically each step:
$$\overbrace{1\cdot2+2\cdot3+\ldots+n(n+1)}^{=2\binom{n+2}3}+(n+1)(n+2)=2\frac{(n+2)!}{3!(n-1)!}+(n+1)(n+2)=$$
$$=\frac{2(n+2)!n+6(n+2)!}{6n!}=\frac{(n+2)!}{3n!}\left(n+3\right)=2\frac{(n+3)!}{3!n!}=2\binom{n+3}3$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Epsilon delta proof with fractions, given what delta must be Choose $\delta = \min (1, \frac{\epsilon}{10})$
is the following statement true?
$$0 < |x − 1| < δ\text{ implies that }\left|\frac{x^2+3x}{x^2+1} − 2\right| < ε$$
Okay so this is what I have so far
$$\left|\frac{x^2+3x}{x^2+1} − 2\right| = |x-1|\frac{|-x+2|}{|x^2+1|}$$
I then chose $\delta = \frac{1}{2}$ (because if I chose $\delta = 1$, $x$ would come out as $2$ which would give me $|x-1|\cdot 0 < \epsilon$ which doesn't tell me much?)
So if $\delta = \frac{1}{2}$, then
$\frac{1}{2}< x < \frac{3}{2}$ (because it's a fraction I used $x > \frac{1}{2}$)
$$|x-1|\cdot\frac{\left|-\frac{1}{2}+2\right|}{\left|(\frac{1}{2})^2+2\right|}=\frac{2}{3}|x-1|< \epsilon$$
$$|x-1|< \frac{3}{2}\epsilon$$
I'm just so confused because I don't know how to relate the value for $\delta$ (which I found to be $\frac{3}{2}\epsilon$ to the delta they're making me choose of $\min (1, \frac{\epsilon}{9})$. I'm also not sure if I chose the correct value for $x$ as it is a fraction.
| An alternative answer keeping the $\delta$:
We know that $|x-1|<\delta$ and
$$\left|\frac{x^2+3x}{x^2+1}-2\right|= |x-1|\frac{|x+2|}{|x^2+1|}$$
Now, we use the bounds
$$|x-2| = |x-1-1|\leq |x-1|+1 \leq\delta +1$$
and
$$|x^2+1|=x^2+1\geq 1$$
to obtain
$$\left|\frac{x^2+3x}{x^2+1}-2\right|= |x-1|\frac{|x+2|}{|x^2+1|}\leq \delta(1+\delta) \leq \delta2 \leq \frac{\varepsilon}{5}<\varepsilon$$
using $\delta\leq \min\{1,\frac{\varepsilon}{10}\}$.
| {
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How do I show that $\frac {a^2}b + \frac {b^2}c + \frac {c^2}a \ge \frac {(a + b + c)(a^2 + b^2 + c^2)}{ab + bc + ca}?$ For positive real numbers $a, b, c$, show that
$$\frac {a^2}b + \frac {b^2}c + \frac {c^2}a \ge \frac {(a + b + c)(a^2 + b^2 + c^2)}{ab + bc + ca}.$$
I don't know how to solve this at all. Can you provide any hints?
| $$LHS-RHS=\frac{1}{2(ab+bc+ca)}\cdot \left[\frac{b(ab-c^2)^2}{ca}+\frac{c(bc-a^2)^2}{ab}+\frac{a(ca-b^2)^2}{bc}\right]\ge 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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You cast a pair of dice. If you get a 6 and an 8 before 7 comes up twice, you win. What is the probability of winning? You cast a pair of dice. If you get a 6 and an 8 before 7 comes up twice, you win. What is the probability of winning?
What I tried was
\begin{align*}
P(X=6) & = \frac{5}{36}\\
P(X=8) & = \frac{5}{36}\\
P(X=7) & = \frac{6}{36} = \frac{1}{6}
\end{align*}
I try finding the probability of getting $6$ and $8$ and not $7$
$$P(X=6)+P(X=8)+P(X \neq 7) = \frac{5}{26} + \frac{5}{36} + \frac{5}{6} = 1.11$$
I stick since I know that the probability cannot be greater than one. The solution key says the answer is $0.5456$.
| It is easier to compute the probability of first getting two $7's\;\;$ and take the complement
Totals other than $6,7,8$ don't matter, so we can simply take their odds in favor as $5:6:5$
Consider the three mutually exclusive starts $(a)\;\;7-7,\;\;(b)\;\; 7-(6\;\;or\;\; 8),\;\;(c)\;\; (6\;\; or\;\; 8)-7$
In $(a),\;\; 7$ wins with $Pr = \frac{6}{16}\frac6{16}$
For $(b)\;\;and\;\;(c)$, note that for subsequent throws, it becomes a direct race between $7$ and whichever of $6$ and $8$ has not yet appeared, thus the odds in favor of $7$ becomes $6:5$
In $(b),\;\; 7\;\;$ wins with $Pr = \frac6{16}\frac{10}{16}\frac6{11}$
In $(c),\;\; 7\;\;$ wins with $Pr = \frac{10}{16}\frac6{11}\frac6{11}$
Adding up, $7$ wins with $Pr = 0.4544$
Finally, answer = $1 - 0.4544 = 0.5456$
| {
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Let $\textbf{v} = (1, 1, 1, 1)$. Find a basis for... Let $\textbf{v}=(1,1,1,1)$. Find a basis for $\{\textbf{u}\in\Bbb{R}^4\ |\ \textbf{u}\cdot\textbf{v}=0\}$
How can I do this? In particular, I do not understand $\textbf{u}\cdot\textbf{v}=0.$
| Denote $W=\{u\in\mathbb{R}^4\mid u\cdot v=0\}$, we claim that the set
$\beta=\left\{
\begin{pmatrix}-1\\1\\0\\0\end{pmatrix},
\begin{pmatrix}-1\\0\\1\\0\end{pmatrix},\begin{pmatrix}-1\\0\\0\\1\end{pmatrix}
\right\}$ is a basis for $W$. First, given
$u=(u_1,u_2,u_3,u_4)\in W$, then we have
$$0=u\cdot v=
\begin{pmatrix}u_1\\u_2\\u_3\\u_4\end{pmatrix}\cdot
\begin{pmatrix}1\\1\\1\\1\end{pmatrix}=
u_1+u_2+u_3+u_4,$$ that is, $u_1=-u_2-u_3-u_4$. So we can write
$$u=\begin{pmatrix}-u_2-u_3-u_4\\u_2\\u_3\\u_4\end{pmatrix}
=u_2\begin{pmatrix}-1\\1\\0\\0\end{pmatrix}
+u_3\begin{pmatrix}-1\\0\\1\\0\end{pmatrix}
+u_4\begin{pmatrix}-1\\0\\0\\1\end{pmatrix},$$
which follows that $\beta$ spans $W$. Next, we show that $\beta$ is linearly independent, but it is easy to prove so I leave it to you. Hence we conclude that
$\beta$ is a basis for $W$, and the claim is proved.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this inequality: $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$
Let $a$,$b$,$c$ be positive real numbers such that $abc=1$. Prove that
$$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$$
I tried various methods. But, couldn't solve it. It'd be great if anyone can help.
| Id est, we need to prove that $\sum\limits_{cyc}(a^2b+a^2c-a^{\frac{4}{3}}b^{\frac{5}{6}}c^{\frac{5}{6}}-abc)\geq0$
which is AM-GM and Muirhead because $(2,1,0)\succ\left(\frac{4}{3},\frac{5}{6},\frac{5}{6}\right)$
Your another problem can be solved by the same way.
| {
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Find the limit $\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$ Find the limit if it exists. (this exercise is taken from Calculus - The Classic Edition by Swokowski Chapter 10, section 1, no. 9, p.498)
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$$
since the limit is $0/0$ therefore, we use L'Hopital's rule, that is,
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x} = \lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 }$$
since $\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \frac {1 - 1}{ 1- 1} = \frac00$. Thus, we use the L'Hopital's rule again. that is,
$$\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \lim_{x\rightarrow 0}\frac{-\sin x}{2 \sec^2 x \sec x \tan x}$$
since $\lim_{x\rightarrow 0} \frac{-\sin x}{2 \sec^2 x \sec x \tan x}= \frac{0}{2(1)(0)} = \frac00$. Thus, it always goes to zero by zero. But the answer for this question is $\frac{-1}2$. How is that?!! there must be something missing that either I forget or misunderstand.
| $$\frac{\cos(x)-1}{\sec^2(x)-1}=\frac{\cos(x)-1}{\frac{1}{\cos^2(x)}-1}=\frac{\cos^2(x)(\cos(x)-1))}{1-\cos^2(x)}=\frac{\cos^2(x)(\cos(x)-1)}{(1-\cos(x))(1+\cos(x))}=\frac{- \cos^2(x)(1-\cos(x))}{(1-\cos(x))(1+\cos(x))}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I find $[T]_E$ when given this linear transformation $T(p(x)) = (1+2x^2)p''(x)+(1-2x)p'(x)+p(x)+p(0)$? In this question I have been asked to find $[T]_E$.
I was given that $T: R_3[x] \rightarrow R_3[x]$ is a linear transformation
defined as: $T(p(x)) = (1+2x^2)p''(x)+(1-2x)p'(x)+p(x)+p(0)$,
$p(x),p(x)',p(x)'' \in R_3[x]$ and $E = \{ 1,x,x^2 \}$ is the basis of the transformation.
For solving this question I went to the definition of matrix that represents a transformation that says:
$[T]_E = [[Te_1],[Te_2],[Te_3]]$
Now, my problem is how do I map each vector in E with this transformation (first time seeing transformation with derivatives)
Thank you in advance.
| Actually, $[T]_E = [[Te_1]_E,[Te_2]_E,[Te_3]_E].$ Compute
$$\begin{align}
Te_1 & = T(1)\\
& = (1 + 2x^2)1'' + (1 - 2x)1' + 1 + 1\\
& = 2\\
& = \color{#08F}2 \cdot 1 + \color{#08F}0 \cdot x + \color{#08F}0 \cdot x^2\\
& = \color{#08F}2 e_1 + \color{#08F}0e_2 + \color{#08F}0e_3,\\
[Te_1]_E & = {\begin{bmatrix}\color{#08F}{2\\ 0\\ 0}\end{bmatrix}};\\\\
Te_2 & = T(x)\\
& = (1 + 2x^2)x'' + (1 - 2x)x' + x + 0\\
& = 1 - x\\
& = \color{#08F}1 \cdot 1 \color{#08F}{- 1} \cdot x + \color{#08F}0 \cdot x^2\\
& = \color{#08F}1 e_1 \color{#08F}{- 1}e_2 + \color{#08F}0e_3,\\
[Te_2]_E & = {\begin{bmatrix}\color{#08F}{1\\ -1\\ 0}\end{bmatrix}};\\\\
Te_3 & = T(x^2)\\
& = (1 + 2x^2)(x^2)'' + (1 - 2x)(x^2)' + x^2 + 0^2\\
& = 2 - 2x + x^2\\
& = \color{#08F}2 \cdot 1 \color{#08F}{- 2} \cdot x + \color{#08F}1 \cdot x^2\\
& = \color{#08F}2 e_1 \color{#08F}{- 2}e_2 + \color{#08F}1e_3,\\
[Te_3]_E & = {\begin{bmatrix}\color{#08F}{2\\ -2\\ 1}\end{bmatrix}}.\end{align}$$
Thus,
$$[T]_E = {\begin{bmatrix}
2 & 1 & 2\\
0 & -1 & -2\\
0 & 0 & 1\end{bmatrix}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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For the given u(x,y) find a v(x,y) such that f(z) = u(x,y) + iv(x,y) is holomorphic in some region of $\mathbb{C}$ I used the Cauchy Riemann equations and got two different functions v(x,y) by integrating. Should I just add them up?
My u(x,y) is $\frac{x}{x^2 + y^2}$
My partial wrt x is $\frac{y^2 - x^2}{x^2 + y^2}$ and my partial wrt y is $\frac{-2xy}{x^2 + y^2}$.
| $u(x,y)$ partial wrt $x$ is $\frac{y^{2}-x^{2}}{(x^2+y^2)^2}$, partial wrt $y$ is
$\frac{-2xy}{(x^2+y^2)^2}$.
By the rule of Cauchy-Riemann Equations,we have
$v(x,y)$ partial wrt $y$ is $\frac{y^{2}-x^{2}}{(x^2+y^2)^2}$, partial wrt $x$ is
$\frac{-2xy}{(x^2+y^2)^2}$.
Then we can let $v(x,y)=\frac{-y}{x^2+y^2}$.
| {
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Prove that $ζ(4)=π^4/90$ knowing that $\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$ The question
Knowing that:
$$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right) \tag{1}$$
obtain the Taylor series expansion of $\frac{\sin(πz)}{πz}$ to deduce:
$$ \sum_{1 ≤ n_1 < n_2 < … < n_k} \frac{1}{n_1^2n_2^2 … n_k^2} = \frac{π^{2k}}{(2k + 1)!} \tag{2}$$
Also deduce that:
$$ ζ(4) = \sum_{n=1}^∞ \frac{1}{n^4} = \frac{π^4}{90} \tag{3}$$
What I've obtained
For (2) I've done the following (doing what the question suggests):
$$ 1 - \frac{z^2π^2}{3!} + \frac{z^4π^4}{5!} - … = (1 - z^2)\left(1 - \frac{z^2}{2^2}\right)\left(1 - \frac{z^2}{3^2}\right)\left(1 - \frac{z^2}{4^2}\right)…= $$
$$ = 1 - z^2\left( 1 + \frac{1}{2^2} + \frac{1}{3^2} + … \right) + z^4 \left( \frac{1}{1^22^2} + \frac{1}{1^23^2} + … + \frac{1}{2^23^2}+…\right) - z^6\left( \frac{1}{1^22^23^2} +\frac{1}{1^22^24^2} + … + \frac{1}{1^23^24^2} + … \right) + … =$$
$$ = 1 - z^2\sum_{1 ≤ n_1}\frac{1}{n_1^2} + z^4 \sum_{1 ≤ n_1 < n_2}\frac{1}{n_1^2n_2^2} + … $$
So one sees that (2) necessarily holds.
For (3) I've tried taking $\sum_{1 ≤ n_1<n_2}\frac{1}{n_1^2n_2^2}$ in order to have exponent 4 in the numerator, but according to (2) that gives $π^4/120$. That try is obviously not correct since $n_1≠n_2$ and $90≠120$, but I don't know how how to prove it.
Any help?
| Hint: $\frac{\pi^4}{90}=\left(\frac{\pi^2}{6}\right)^2-2\left(\frac{\pi^4}{120}\right)$.
| {
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Finding out a limit using Taylor series. So the limit is the following:
$$\lim_{x \to 0}{\frac{x^2-\frac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}}$$
Expansions for $\sin(x)$ and $\cos(x)$ are given:
$$\sin x = x-\frac{x^3}{3!} + \frac{x^5}{5!}-...+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!} + o(x^{2n})$$
$$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-...+(-1)^n\frac{x^{2n}}{(2n)!}+o(x^{2n+1})$$
Here is what I tried:
$$\lim_{x->0}{\frac{x^2-\frac{x^6}{2}-x^2(1-\frac{x^4}{2}+\frac{x^8}{4!}+o(x^{2*5}))}{x^{10}+o(x^{10*2})}}=\lim_{x->0}{\frac{-\frac{x^{10}}{4!}-o(x^{12})}{x^{10}+o(x^{20})}}$$
This is where I am stuck. I figured that the problem occurs when expending $o()$. What am I missing here?
| How about factoring by $x^{10}$?
| {
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Does an analytical form exist for the following integral I have an integral
$$f(n,a)=\int_0^{2\pi}\mathop{\mathrm{d}x}\frac{\cos(nx)\cos^2x}{1-a\cos^2x},$$
where $n$ is an even integer and $0<a<1$ is a real number. Does an analytical form exist for this integral?
Edit: Here is a partial answer that I discovered:
$$\int_0^{2\pi}\mathop{\mathrm{d}x}\frac{\cos(nx)}{1-a\cos^2x}=-\frac{2\pi}{\sqrt{1-a}}\left(\frac{1-\sqrt{1-a}}{\sqrt{a}}\right)^n,\qquad a<1, n>0.$$
$n$ is even.
| I'm kind of surprised that this problem hasn't been attacked with a geometric series yet. Let $0<b<1$. Then
$$\sum_{k=0}^{\infty}b^ke^{ik\theta}=\sum_{k=0}^{\infty}b^k\cos k\theta+i\sum_{k=0}^{\infty}b^k\sin k\theta=\frac1{1-be^{i\theta}}=\frac{1-b\cos\theta+ib\sin\theta}{1-2b\cos\theta+b^2}$$
Taking real parts and setting $\theta=2x$,
$$\begin{align}1+\sum_{k=1}^{\infty}b^k\cos2kx&=\frac{1-b(2\cos^2x-1)}{1-2b(2\cos^2x-1)+b^2}\\
&=\frac1{1+b}+\frac{\frac{2b(1-b)}{1+b}\cos^2x}{(1+b)^2-4b\cos^2x}\end{align}$$
Then
$$\frac b{1+b}+\sum_{k=1}^{\infty}b^k\cos2kx=\frac{\frac{2b(1-b)}{(1+b)^3}\cos^2x}{1-\frac{4b}{(1+b)^2}\cos^2x}$$
Solving $a=\frac{4b}{(1+b)^2}$ for
$$b=\frac{2-a\pm\sqrt{1-a}}a=\frac{\left(1-\sqrt{1-a}\right)^2}a$$
And inserting this result for $b$ throughout,
$$\frac{1-\sqrt{1-a}}2+\sum_{k=1}^{\infty}\frac{\left(1-\sqrt{1-a}\right)^{2k}}{a^k}\cos2kx=\frac{\frac a2\sqrt{1-a}\cos^2x}{1-a\cos^2x}$$
This shows that $$f(0,a)=2\pi\frac{1-\sqrt{1-a}}{a\sqrt{1-a}}$$
And
$$f(n,a)=\frac{2\pi}{a\sqrt{1-a}}\frac{\left(1-\sqrt{1-a}\right)^n}{a^{n/2}}$$
For positive even n, in agreement with other results, but I think the tools used here are a little simpler.
| {
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Show $x_n := \frac{1-p^{n+1}}{1-p^n} \frac{n}{n+1}$ is increasing in n for $p \in (0,1)$ I need to show that $x_n := \frac{1-p^{n+1}}{1-p^n} \frac{n}{n+1}$ is increasing in $n$ for $p \in (0,1)$. My attempts have involved trying to show
$\begin{eqnarray*} \frac{1-p^{n+2}}{1-p^{n+1}} \frac{n+1}{n+2} &>& \frac{1-p^{n+1}}{1-p^n} \frac{n}{n+1} \\
\left(1 + \frac{p^{n+1}}{\sum_{k=0}^n p^k} \right) \frac{n+1}{n+2} &>& \left(1 + \frac{p^{n}}{\sum_{k=0}^{n-1} p^k} \right) \frac{n}{n+1} \\
\frac{n+1}{n+2} - \frac{n}{n+1} &>& \left(\frac{p^{n}}{\sum_{k=0}^{n-1} p^k} \right)\frac{n}{n+1}-\left(\frac{p^{n+1}}{\sum_{k=0}^n p^k} \right) \frac{n+1}{n+2}\\
\frac{n+1}{n+2} - \frac{n}{n+1} &>& p^n \left[\left(\frac{1}{\sum_{k=0}^{n-1} p^k} \right)\frac{n}{n+1}-\left(\frac{p}{\sum_{k=0}^n p^k} \right) \frac{n+1}{n+2} \right]\end{eqnarray*}$
but this does not seem to simplify.
I have also attmpted to treat $x_n$ as a continuous function in $n$ and differentiated (this leaves a mess so I won't put it up here).
Sadly, I seem to be unable to get my required result from either.
EDIT: It might be helpful to note that $x_n = \frac{\sum_{k=0}^n p^k}{\sum_{k=0}^{n-1} p^k}\frac{n}{n+1}$
EDIT 2: I can show that $x_n < 1$ for all $n \in \mathbb{N}$ and that $x_n \to 1$. But I don't think this suffices...
EDIT 3: If we can show that
$\frac{\sum_{k=0}^{n+1}p^k}{n+2} \cdot \frac{\sum_{k=0}^{n-1}p^k}{n} \geq \left(\frac{\sum_{k=0}^{n+1}p^k}{n+2} \right)^2$, we are done. But I don't know how to show this.
| Writing $x_n = \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}$, we see:
$$x_n>x_{n-1}\iff \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}>\frac{1-p^{n}}{1-p^{n-1}}\frac{n-1}{n}$$
$$\iff(1-p^{n-1})(1-p^{n+1})n^2>(1-p^n)^2(n^2-1)$$
$$\iff (1-p^n)^2 > n^2[(1-p^n)^2-(1-p^{n-1})(1-p^{n+1})]=n^2[p^{n-1}+p^{n+1}-2p^n]$$
$$\iff\frac{(1-p^n)^2}{p^n}>n^2\left(\frac{1}{p}+p-2\right)=n^2\frac{(p-1)^2}{p}$$
$$\iff \frac{1-p^n}{p^{n/2}}>n\frac{1-p}{p^{1/2}}\iff\frac{p^{n/2}}{p^{1/2}}\frac{1-p^{-n}}{1-p^{-1}}>n$$
$$\iff p^{\frac{n-1}{2}}(1+p^{-1}+p^{-2}+\cdots+p^{-(n-1)})>n$$
$$\iff \frac{1+p^{-1}+p^{-2}+\cdots+p^{-(n-1)}}{n}>p^{-\frac{n-1}{2}}$$
which is true by the AM-GM inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\lim_{x\rightarrow 0}\left(\frac{16^x+9^x}{2}\right)^{\frac{1}{x}}$
Evaluation of $\displaystyle \lim_{x\rightarrow 0}\left(\frac{16^x+9^x}{2}\right)^{\frac{1}{x}}$
$\bf{My\; Try::}$ I am Using above question using Sandwich Theorem
So Using $\bf{A.M\geq G.M\;,}$ We get
$$\frac{16^x+9^x}{2}\geq (16^x\cdot 9^x)^{\frac{1}{2}}\Rightarrow \lim_{x\rightarrow 0}\left(\frac{16^x+9^x}{2}\right)^{\frac{1}{x}}\geq \lim_{x\rightarrow 0}(16^x\cdot 9^x)^{\frac{1}{2x}}=12$$
But I did not Understand How can I Calculate it for Upper bond, Help me
Thanks
| Whenever the base and the exponent are variables it makes sense to take logs. If $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{x \to 0}\left(\frac{16^{x} + 9^{x}}{2}\right)^{1/x}\right\}\notag\\
&= \lim_{x \to 0}\log\left(\frac{16^{x} + 9^{x}}{2}\right)^{1/x}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{x}\log\left(\frac{16^{x} + 9^{x}}{2}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x}\cdot\frac{16^{x} + 9^{x} - 2}{2}\cdot\dfrac{\log\left(1 + \dfrac{16^{x} + 9^{x} - 2}{2}\right)}{\dfrac{16^{x} + 9^{x} - 2}{2}}\notag\\
&= \lim_{x \to 0}\frac{16^{x} + 9^{x} - 2}{2x}\cdot 1\notag\\
&= \frac{1}{2}\lim_{x \to 0}\left(\frac{16^{x} - 1}{x} + \frac{9^{x} - 1}{x}\right)\notag\\
&= \frac{1}{2}\cdot(\log 16 + \log 9)\notag\\
&= \log\sqrt{16\cdot 9} = \log 12\notag
\end{align}
Hence $L = 12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762733",
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How to prove that $\sin x = x \, _{0}F_{1}(-;\frac{3}{2};-\frac{x^{2}}{4})$? How to prove that $\sin x = x \, _{0}F_{1}(-;\frac{3}{2};-\frac{x^{2}}{4})$?
where $_{0}F_{1}$ is the hypergeometric series?
| The function $_0F_1(;a;z)$ is a solution to the differential equation
\begin{equation}\tag{1}
z \frac{d^2}{dz^2}y(z)+a\frac{d}{dz}y(z)= y(z).
\end{equation}
Knowing this, consider the function $u(x)=x y(-x^2/4)$ and take its second derivative. We obtain
$$\tag{2} \frac{d^2}{dx^2}u(x)=x\left[\frac{x^2}{4}y''\left(\frac{x^2}{4}\right)-\frac{3}{2}y'\left(\frac{x^2}{4}\right)\right] $$
Now choose $y(z)=$$_0F_1(;3/2;z)$, so that it satisfies eq.(1) with $a=3/2$ and $z=-x^2/4$. Using it we obtain
$$ \frac{d^2}{dx^2}u(x)=-xy\left(\frac{x^2}{4}\right)=-u(x) \tag{3}.$$
The general solution to (3) is $u(x)=\alpha \sin(x)+\beta \cos(x).$
Since $u$ is odd $u(x)=-u(x)$, $\beta=0$. Finally,
$$ \alpha = \lim_{x\to0} \frac{u(x)}{x} =\; _0F_1\left(;\frac{3}{2};0\right)=1$$
so
$$ u(x)= \sin(x) = x \; _0F_1\left(;\frac{3}{2};-\frac{x^2}{4}\right)$$
| {
"language": "en",
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"source": "stackexchange",
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Determine the maximum and the minimum of an expression Let $x,y,z \in \Bbb R, x,y,z \gt 0$ such that $x^2+y^2+z^2=1$. Determine tha maximum and the minimum possible values of the expression
$$\frac {x^3+y^3+z^3} {x+y+z}.$$
| The rearrangement inequality implies that
$$3(x^3+y^3+z^3)\ge (x+y+z)(x^2+y^2+z^2).$$
So the minimum is $\frac13$, attained when $x=y=z$.
Since obviously $x,y,z\in(0,1)$,
$$\frac{x^3+y^3+z^3}{x+y+z}<1\tag1.$$
On the other hand, if we set $y=z=\varepsilon$ and $x=\sqrt{1-2\varepsilon^2}$, then
$$\lim_{\varepsilon\to 0}\frac{x^3+y^3+z^3}{x+y+z}=1.$$
That proves the upper bound $1$ in (1) is sharp and the said function does not have a maximum.
| {
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"source": "stackexchange",
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$\frac {2 \cdot 3^{m + 1}}{k} - \frac {2 \cdot 3^{m}}{k + 1} \in \Bbb N_+$, find all possible values of $k, m$. If $$\frac {2 \cdot 3^{m + 1}}{k} - \frac {2 \cdot 3^{m}}{k + 1} \in \Bbb N_+$$ and $$\frac {2 \cdot 3^{m}}{k} - \frac {2 \cdot 3^{m}}{k + 1} \le 1$$ where $k, m \in \Bbb N_+$ and $k \ge 2$, find all possible values of $k, m$.
This comes from a problem in graph theory, which may be simplified to this (though I don't know if this method works). According to the solution to the original question, $(k, m) = (2, 1)$.
Can I say that $\frac {2 \cdot 3^{m + 1}}{k} \in \Bbb N_+$ and $\frac {2 \cdot 3^{m}}{k + 1} \in \Bbb N_+$? I found that if that's true then $(k, m)$ has only one solution, i.e., $(2, 1)$. But I'm unsure whether I can get that result. If I can't, how can I solve the problem above?
This is the original question:
Consider a graph $G(V, E)$ of order $n$ such that $$\forall x, y \in V ((x \not = y) \implies (|E(G - \{ x, y \})| = 3^m)),$$ where $m \in \Bbb N_+$, find all possible values of $n$.
I observed that (by double counting) $$|E(G)| = \frac {\binom {n}{n - 2}}{\binom {n - 2}{n - 4}} 3^m \le \binom {n}{2},$$ which is equivalent to this question (with $k = n - 3$).
| \begin{align*}
\frac {2 \cdot 3^{m + 1}}{k} - \frac {2 \cdot 3^{m}}{k + 1} \in \Bbb N_+ \\
\implies 2\cdot 3^m \Bigg(\frac {3}{k} - \frac {1}{k + 1}\Bigg) \in \Bbb N_+ \\
\implies 2\cdot 3^m\frac {2k+3}{k(k+1)} \in \Bbb N_+ \\
\implies 2\cdot 3^m\frac {2t+1}{t(t-1)} \in \Bbb N_+ \tag{$t\to k+1$} \\
\end{align*}
Clearly, $t\not = 1$.
Now, $t \nmid 2t+1$ since $\gcd(2t+1,t)=\gcd(2t+1-2t,t)=\gcd(1,t)=1$. Therefore $t$ must divide $2\cdot3^m$ So $t$ has to be of the form $2^\alpha 3^\beta$ where $\alpha \in \{0,1\}$ and $\beta \in \{0,1,\dots m\}$
If $t$ is odd ($\alpha=0$), $t-1$ is even and does not divide $3^m(2k+1)$ which is always odd. But $t-1$ divides $2\cdot3^m(2t+1)$ for the expression to be $\mathbb{N}_+$ Hence, $t-1|2\implies t-1=2\implies t=3$ and $(\alpha, \beta) = (0,1)$. $k=t-1=2$. Then, $m\in\{1,2,\dots \infty\}$. ($m\not =0$ as $t\not =1$)
But then $$\frac{1}{k}-\frac{1}{k+1}=\frac{1}{12}\implies 2\cdot 3^m\Bigg(\frac{1}{k}-\frac{1}{k+1}\Bigg) = \frac{3^m}{6}\le 1 $$ and your inequality holds only when $m\in\{0,1\}$. Taking the intersection of the two solutions for $m$, we find $m=1$. Hence, $(k,m)=(2,1)$
If $t$ is even ($\alpha=1$), $t-1$ is odd and for $t-1$ to divide $3^m(2t+1)$, $$\gcd(3^m(2t+1),t-1)=t-1=\gcd(3^m(2t+1)-2\cdot3^m(t-1),t-1)= \gcd(3^{m+1},t-1)$$
Either $t-1=1\implies t=2$ or $t-1=3^{m+1} \implies t=1+3^{m+1} \implies k=3^m$. But $1+k$ should divide $2\cdot 3^m$ as the other fraction $\Big(\frac {2 \cdot 3^{m + 1}}{k}\Big)$ is an integer. Hence, $1+k=2 \implies k=1$ and $m=0$. But you need $k\ge2$ and $m\in \mathbb{N}_+$
Thus the only acceptable solution is $(k,m)=(2,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767752",
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"source": "stackexchange",
"question_score": "3",
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$A^2+B^2=AB$ and $BA-AB$ is non-singular The question is:
Are there square matrices $A,B$ over $\mathbb{C}$ s.t. $A^2+B^2=AB$ and $BA-AB$ is non-singular?
From $A^2+B^2=AB$ one could obtain $A^3+B^3=0$. Can we get something from this?
Edit:
$$A^2+B^2=AB\implies\\
A(A^2+B^2)=A^2B \implies\\ A^3+AB^2=A^2B \implies \\
A^3+(A^2+B^2)B=A^2B \implies \\
A^3+A^2B+B^3=A^2B \implies \\
A^3+B^3=0
$$
| The answer is yes. We shall construct a concrete example below. First, suppose $A$ and $B$ are invertible and let $C=AB^{-1}$. From $A^2+B^2=AB$ we get $C + A^{-1}C^{-1}A = I$ and $A^{-1}CA=(I-C)^{-1}$. Therefore the spectrum of $C$ is invariant under the transformation $z\mapsto \frac1{1-z}$.
Let $w$ be the cubic root of unity. If $A^2+B^2=AB$, then $(A+wB)(A+w^2B)=w(BA-AB)$. So, if we want $BA-AB$ to be invertible, neither $-w$ nor $-w^2$ can be an eigenvalue of $C$. As $-w$ and $-w^2$ are fixed points of $z\mapsto\frac1{1-z}$, the spectrum of $C$ has to be splittable into three cycles of the form $\{z,\ \frac1{1-z},\ \ 1-\frac1z\}$, where $z\ne-w,-w^2$, and for every feasible pair of invertible matrices $A,B$, their dimensions must be divisible by 3.
Now we can construct an example on $M_3(\mathbb C)$. Take $z=2$ and
$C=\operatorname{diag}(2, -1, \frac12)$. So $(I-C)^{-1}=\operatorname{diag}(-1, \frac12, 2)$. To make $A^{-1}CA=(I-C)^{-1}$ or $A^2+B^2=AB$, we pick $A=\pmatrix{0&0&1\\ 1&0&0\\ 0&1&0}$ and $B=C^{-1}A=\pmatrix{0&0&\frac12\\ -1&0&0\\ 0&2&0}$. One can readily verify that $BA-AB=\pmatrix{0&-\frac32&0\\ 0&0&-\frac32\\ 3&0&0}$ is invertible.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Prove $(1+x)^p+(1-x)^p \ge 2(1+x^p)$ for $0\le x\le1$ and real number $p\ge2$. I don't know how to prove the following questions: If $p\ge2$ is real, then
$$
(1+x)^p+(1-x)^p \ge 2(1+x^p) \quad \text{for } 0\le x\le1;
$$
if $1\le p<2$, then opposite direction of the inequality holds.
I try to find a convex function but I could not. And I take the derivative of function $f(x):=(1+x)^p+(1-x)^p-2x^p$ but it still doesn't work.
EDIT
I tried the "Binomial series", but it could only be workable for the case $p\ge2$ but not for the case $1\le p<2$.
|
Lemma. for $f(x)=x^k$. if $k\geq 1$ then for $a,b>0$,then $a^k+b^k\leq (a+b)^k$. if $0\leq k\leq 1$, then $a^k+b^k\geq (a+b)^k$.
proof.when $k\geq 1$, then we have $(\frac{a}{a+b})^k\leq\frac{a}{a+b}$ and $(\frac{b}{a+b})^k\leq\frac{b}{a+b}$ for $\frac{a}{a+b}\leq 1$, $\frac{b}{a+b}\leq 1. $
Hence:$(\frac{a}{a+b})^k+(\frac{b}{a+b})^k\leq \frac{a}{a+b}+\frac{b}{a+b}=1$,which implies:$a^k+b^k\leq (a+b)^k$. In the similar way, we can prove if $0\geq k\leq 1$, then $a^k+b^k\geq (a+b)^k$.
Back to the problem:Set $f(x)= (1+x)^p+(1-x)^p-2(1+x^p)$,we have:
$$
f'(x)= p\times (1+x)^{p-1}-p\times (1-x^{p-1})-2p\times x^{p-1}
$$
When $p\geq 2$, then $p-1 \geq 1$.
Then for $0\leq x \leq 1$:
$$
(1+\frac{1}{x})^{p-1}\geq 2^{p-1}+(\frac{1}{x}-1)^{p-1}\geq 2+(\frac{1}{x}-1)^{p-1}
$$
which implies $f'(x)\geq 0$
Thus $f(x)\geq f(0)=0$
Applying the same method, when $1\leq p <2$, we have:
$$
f(x)\leq f(1)=2^p-2\times (1+1)\leq 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this integration question correctly answered? So we have:
$$y = (x^3 + 2)^7$$
And we want to find:
$$I = \int y \space dx \\$$
From the chain rule, we know:
$${d\over dx} g \Big( f(x) \Big) = g' \Big( f(x) \Big) \cdot f'(x)$$
Hence:
$$\int g' \Big( f(x) \Big) \cdot f'(x) \space dx = g \Big( f(x) \Big) \\ $$
Let's call this identity $(1)$ so we don't have to rewrite it.
So I looked at $\space (x^3 + 2)^8 $. Which differentiates to give:
$${24x^2} (x^3 + 2)^7$$
So if we integrate this we get:
$$8 \int {3x^2} (x^3 + 2)^7 \space dx= (x^3 + 2)^8 $$
(Note I took $8$ out the integral).
Now we divide by $8$ and get:
$$\int {3x^2} (x^3 + 2)^7 \space dx= {(x^3 + 2)^8 \over 8}$$
The integral on the left is in the same format as the identity $(1)$.
Does this mean that:
$$I = {(x^3 + 2)^8 \over 8} + C$$
Or have I gone wrong somewhere?
| You can apply binomial theorem to integrate $(x^3+2)^7$ directly:
\begin{align}
\int (x^3+2)^7 dx &= \int \left(\sum_{k=0}^7\binom{7}{k} 2^{7-k}x^{3k}\right) dx\\
&=\sum_{k=0}^7 \binom{7}{k}\frac{2^{7-k}}{3k+1}x^{3k+1}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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use little $o$ notation. out from apostol's book.
$f(x)=o(g(x))$ if $\frac{f}{g}\rightarrow 0$ when $x\rightarrow a$ and gives some properties.
1)$o(g(x))+o(g(x))=o(g(x))$
2)$o(cg(x))=o(g(x))$
3)$f(x).o(g(x))=o(f(x).g(x))$
4)$o(o(g(x)))=o(g(x))$
5)$\frac{1}{1+g(x)}=1-g(x)+o(g(x))$ when $g(x)\rightarrow 0$ when $x\rightarrow a$
we can get another results:
6)$o(f(x)).o(g(x))=o(o(f(x)).g(x))=o(o(f(x).g(x)))=o(f(x).g(x))$
7)$x^n=o(x^{n-1})$ more general $x^a=o(x^b)$ if $a>b$
Give a example.
Show that $\tan(x)= x+\frac{x^2}{2}+o(x^3)$ when $x\rightarrow 0$.
we have $\cos(x)=1-\frac{x^2}{2}+o(x^3)$ then using $(5)$ with $g(x)= -\frac{x^2}{2}+o(x^3)$ have $\frac{1}{\cos(x)}=\frac{1}{1-\frac{x^2}{2}+o(x^3)}= 1 +\frac{x^2}{2}+o(x^2)$
we have $\sin(x)= x-\frac{x^3}{6}+o(x^4)$ then $\tan(x)=\frac{\sin(x)}{\cos(x)}=(x-\frac{x^3}{6}+o(x^4)).(1 +\frac{x^2}{2}+o(x^2))= x+\frac{x^2}{2}+o(x^3)$.
my question: using $(5)$ with $g(x)= -\frac{x^2}{2}+o(x^3)$ we have $\frac{1}{1-\frac{x^2}{2}+o(x^3)}= 1 -(-\frac{x^2}{2}+o(x^3))+o(-\frac{x^2}{2}+o(x^3))= 1+\frac{x^2}{2}-o(x^3)+o(-\frac{x^2}{2}+o(x^3))$
how he gets to this $-o(x^3)+o(-\frac{x^2}{2}+o(x^3))=o(x^2)$
| I'm not sure about this but $\text o(x^3)=\text o(x^2)$ from $(7)$ and, by $(3)$, $$\text o\left(-\dfrac{x^2}{2}+\text o(x^3)\right)=\text o\left(x^2\cdot\left(-\dfrac{1}{2}+\text o(x)\right)\right)=\left(-\dfrac{1}{2}+\text o(x)\right)\cdot\text o(x^2)$$
then the above expression becomes $-\dfrac{1}{2}\text o(x^2)+\text o(x^3)$ and again $\text o(x^3)=\text o(x^2)$ with the fact that $-\dfrac{1}{2}\text o(x^2)=\text o(x^2)$, the last one follows by thinking the left-handed side to be some function which tends to $0$ when divided by $x^2$ and $x\to 0$ which is nothing but $\text o(x^2)$ itself and then combine all of these to obtain what we want.
| {
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"source": "stackexchange",
"question_score": "4",
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$\displaystyle\lim _{x\to 0}\frac{\sqrt{1+x+x²}-1}{\sqrt{1+x}-\sqrt{1-x}}$ First I need to prove that this limit; $\displaystyle\lim _{x\to 0}\frac{\sqrt{1+x+x^2}-1}{\sqrt{1+x}-\sqrt{1-x}}$ converges, then I have to find its limit. Now I don't know how to prove that it converges (these epsilon proofs are still something that I'm trying to learn). And to actually find the limit; I tried to rewrite it by multiplying by $\displaystyle{\sqrt{1+x}+\sqrt{1-x}\over \sqrt{1+x}+\sqrt{1-x}}$ but I didn't get any further..
Edit: I know that it converges to $1/2$, but that's what wolfram alpha says :|
| $$\frac{\sqrt{1+x+x^2}-1}{\sqrt{1+x}-\sqrt{1-x}}=\frac{x+x^2}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x+x^2}+1)}=$$ $$=\frac{x+x^2}{\sqrt{1+x}-\sqrt{1-x}}*\frac{1}{\sqrt{1+x+x^2}+1}$$ It follows
$$\lim _{x\to 0}\frac{\sqrt{1+x+x^2}-1}{\sqrt{1+x}-\sqrt{1-x}}=\lim _{x\to 0}\frac{x+x^2}{\sqrt{1+x}-\sqrt{1-x}}*\lim _{x\to 0}\frac{1}{\sqrt{1+x+x^2}+1}$$ The second factor is clearly equal to $\color{red}{\frac 12}$ and $$\lim _{x\to 0}\frac{x+x^2}{\sqrt{1+x}-\sqrt{1-x}}=\lim _{x\to 0}\frac{(x+x^2)(\sqrt{1+x}+\sqrt{1-x})}{2x}=\lim _{x\to 0}\frac{(1+x)(\sqrt{1+x}+\sqrt{1-x})}{2}=\frac 22=1$$ Thus the asked limit is equal to $$1\cdot \frac 12=\color{red}{\frac 12}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$
Evaluation of $$\lim_{n\rightarrow \infty}\left(\tan \frac{\pi}{2n}\cdot \tan \frac{2\pi}{2n}\cdot \tan \frac{\pi}{3n}\cdot ...............\tan \frac{(n-1)\pi}{2n}\right)^{\frac{1}{n}} = $$ without using Limit as a sum.
$\bf{My\; Try::}$ Using the formula $$\displaystyle \sin\left(\frac{\pi}{n}\right)\cdot \sin\left(\frac{2\pi}{n} \right)....\sin\left(\frac{(n-1)\pi}{n}\right) = \frac{n}{2^{n-1}}$$
Replace $n\rightarrow 2n$
$$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(2n-1)\pi}{2n}\right) = \frac{2n}{2^{2n-1}}$$
Now How can I calculate $$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right)$$
and also How can I calculate $$\displaystyle \cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$$
Help required, Thanks
| For fixed $n$, call the expression $A$, and take its $\log[.] $ to get:
$$ \log (A)=\frac{1}{n} \left(\log\tan \frac{\pi}{2n}+ \tan \frac{2\pi}{2n}+ .............. + \tan \frac{(n-1)\pi}{2n}\right) $$
$$ =\frac{1}{n} \sum_{k=1}^{n-1}\log\tan (\frac{\pi k}{2n})$$
Doesn't it look familiar?! Yes, Riemann sum as in the definition of definite integrals. Here, if $\ k/n$ is seen as our discretized parameter $x$, then
$$\lim_{n\rightarrow \infty} \log (A) = \int_0^1\log\tan(\frac{\pi }{2}x)dx$$
Once this value is calculated, $A$ will be the exponential of it. But how do we get $$ \int_0^1\log\tan(x)dx \ ?$$
Despite its nasty look, there seems to be hope to find an antiderivative. But I don't know right away.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Splitting fractions with a linear denominator: $\frac{2x-1}{x+2}$ How can $$\frac{2x-1}{x+2}$$ be split to give $$A-\frac{B}{x+2}$$
where $A$ and $B$ are integers?
The solution is $$2-\frac{5}{x+2}.$$
| Imagine that you divide polynomials as usual numbers.
For example, let's divide $2x-1$ by $x+2$. Firstly, find the number such that if you multiply the divisor by it and substract from a dividend, you vanish the first term. Here $a=2$, because $2x-1 - 2(x+2) = -5$, which doesn't contain a term with $x$. -5 has the degree less than $x+2$, so we must stop. Then just take the result, which is 2, and write the expression: $2x-1=2(x+2) - 5$. Divide each part by $x+2$ and you'll get what you are looking for.
Try on the harder example:
$\frac{4x^2 + 3x + 1}{x-1}$.
Find such monome that $4x^2+3x+1 - a(x-1)$ doesn't contain $x^2$. It's $4x$:
$4x^2+3x+1 - 4x(x-1) = 7x + 1$.
Then find the next term such that $7x+1 - a(x-1)$ doesn't contain $x$. It is $7$:
$7x+1 - 7(x-1) = 8$.
The degree of $8$ is less than the degree of $x-1$, so stop.
Now write the full expression:
$4x^2+3x+1 = (x - 1) (4x+7) + 8$
And divide it by $x-1$:
$\frac{4x^2 + 3x + 1}{x-1} = 4x+7 + \frac{8}{x-1}$.
Everything works just the same way as the usual division of the numbers (https://en.wikipedia.org/wiki/Long_division), because numbers are polynomials with $x=10$.You can write this division the same way as you write numbers division.
This algorithm is extremely useful. For example, it is a standart way to find the integrals from functions like $\frac{f(x)}{g(x)}$, where $f(x)$ and $g(x)$ are polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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$ \sum_{n = 1}^{\infty} \frac{1}{n^4}$? Using Fourier series I have managed to show that
$$ \frac{x^4}{12} = \frac{\pi^2 x^2}{6} + 4 \sum_{n = 1}^{\infty} \frac{(-1)^n}{n^4}(1-\cos(nx)) , x \in [-\pi,\pi]$$
From here apparently one need just one step to find an expression for $ \sum_{n = 1}^{\infty} \frac{1}{n^4}$. I need a hint what this step might be.
| For $\;x=\pi\;$:
$$\frac{\pi^4}{12}=\frac{\pi^4}6+4\sum_{n=1}^\infty\frac{(-1)^n}{n^4}\left(1-(-1)^n\right)\stackrel{\text{Only odd index matters}}\implies\frac{\pi^4}{96}=\sum_{n=1}^\infty\frac1{(2n-1)^4}$$
and from here:
$$I:=\sum_{n=1}^\infty\frac1{n^4}=\frac1{16}\sum_{n=1}^\infty\frac1{n^4}+\sum_{n=1}^\infty\frac1{(2n-1)^4}=\frac1{16}\sum_{n=1}^\infty\frac1{n^4}+\frac{\pi^4}{96}\implies$$
$$\color{red}{I=\sum_{n=1}^\infty\frac1{n^4}=\frac{\pi^4}{96}\cdot\frac{16}{15}=\frac{\pi^4}{90}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Fractions in Questions and Answers
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