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Integrating $\int_{-\infty}^0e^x\sin(x)dx$ I ask if anyone could solve the following:
$$\int_{-\infty}^0e^x\sin(x)dx=?$$
I can visually see that it will converge and that it should be less than $1$:
$$\int_{-\infty}^0e^x\sin(x)dx<\int_{-\infty}^0e^xdx=1$$
But I am unsure what its exact value is.
Trying to find the definite integral by integrating by parts 4 times only results in $e^x\sin(x)$, which got me nowhere.
How should I evaluate this?
| Integrating by parts two times you get
$$\begin{align}
\int e^x \cdot \sin x \quad dx & = e^x \cdot \sin x - \int e^x \cdot \cos x \quad dx\\
& = e^x \cdot \sin x - \left( \int e^x \cdot \cos x \quad dx \right)\\
& = e^x \cdot \sin x - \left( e^x\cdot \cos x + \int e^x \cdot \sin x\quad dx\right)\\
& = e^x \cdot \sin x - e^x\cdot \cos x - \int e^x \cdot \sin x\quad dx\\
& = e^x \cdot \left( \sin x - \cos x\right)- \int e^x \cdot \sin x\quad dx\\
\end{align}$$
hence
$$\int e^x \cdot \sin x \quad dx = e^x \cdot \left( \sin x - \cos x\right)- \int e^x \cdot \sin x\quad dx\\
\Rightarrow \int e^x \cdot \sin x \quad dx = \frac{e^x \cdot \left( \sin x - \cos x\right)}{2}$$
Finally finding $\int_{-\infty}^0e^x\sin(x)dx$ it's just a matter of substitutions
$$
\begin{align}
\int_{-\infty}^0e^x\sin(x)dx &= \frac{e^x \cdot \left( \sin x - \cos x\right)}{2} \Big|_{-\infty} ^0\\
&= \frac{1 \cdot \left( 0 - 1\right)}{2} - \lim_{x \to -\infty} \frac{e^x \cdot \left( \sin x - \cos x\right)}{2}\\
&=-\frac{1}{2} - 0\\
&=-\frac{1}{2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$? Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$?
The justification is that $\lim_\limits{x\to 0}\frac{\sin(x)}{x} = 1$
But, I am not seeing the connection.
L'Hospital's rule? Is there a double angle substitution happening?
| The key is to use Hopital rule:
$$\begin{align}
\lim_\limits{x\to 0}\frac{\sin(x)}{x} &= \lim_\limits{x\to 0}\frac{\sin'(x)}{1}\\
&= \lim_\limits{x\to 0}\frac{\cos(x)}{1}\\
&= \lim_\limits{x\to 0}\frac{1}{1}\\
&=1
\end{align}
$$
hence
$$
\begin{align}
\lim_{x\to 0} \frac{\sin(6x)}{\sin(2x)} &= \lim_{x\to 0} \frac{\sin(6x)}{\sin(2x)} \cdot \frac{2x}{6x} \cdot \frac{6}{2}\\
&= \lim_{x\to 0} \frac{\sin(6x)}{6x} \cdot \frac{\sin(2x)}{6x} \cdot 3\\
&= 1 \cdot 1 \cdot 3\\
&=3
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 6
} |
Find a basis and dimension of $W_1\cap W_2$ if the span is given Find a basis and dimension of $W_1\cap W_2$ where $W_1$ is a subspace generated by vectors $\left\{ \begin{bmatrix}
1 & 1 \\
0 & 0 \\
\end{bmatrix},\begin{bmatrix}
3 & 1 \\
-1 & 0 \\
\end{bmatrix}\right\}$ and $W_2$ is generated by $\left\{ \begin{bmatrix}
1 & 1 \\
1 & 0 \\
\end{bmatrix},\begin{bmatrix}
2 & -1 \\
1 & -1 \\
\end{bmatrix}\right\}$.
I don't understand why the rank of a matrix $ \begin{bmatrix}
1 & 1 & 3 & 1\\
0 & 0 & -1 & 0\\
\end{bmatrix}$ is $2$, and $rref$ is $\begin{bmatrix}
1 & 0 & 1 & 1\\
0 & 0 & 1 & 0\\
\end{bmatrix}$.
From there, we have that a basis for $W_1$ is $\{\begin{bmatrix}
1 & 0 & 1 & 1\\
\end{bmatrix},\begin{bmatrix}
0 & 0 & 1 & 0\\
\end{bmatrix}\}$ and $\dim(W_1)=2$.
For a matrix $\begin{bmatrix}
1 & 1 & 2 & -1\\
1 & 0 & 1 & -1\\
\end{bmatrix}$ $rref$ is $\begin{bmatrix}
1 & 0 & 1 & 1\\
0 & 1 & 1 & -2\\
\end{bmatrix}$. A basis for $W_2$ is $\{\begin{bmatrix}
1 & 0 & 1 & 1\\
\end{bmatrix},\begin{bmatrix}
0 & 1 & 1 & -2\\
\end{bmatrix}\}$ and $\dim(W_2)=2$.
What is the method for finding a basis of $W_1\cap W_2$ and $\dim(W_1\cap W_2)$?
| To find the matrices in $W_1\cap W_2$, you need to solve the equation
$$\begin{align}
a\begin{bmatrix} 1 & 1\\0 & 0\end{bmatrix}+b\begin{bmatrix}3 & 1\\-1 & 0\end{bmatrix} &=c\begin{bmatrix}1 & 1\\1 & 0\end{bmatrix}+d\begin{bmatrix}2 & -1 \\1 & -1\end{bmatrix}\\
\begin{bmatrix}a+3b & a+b\\-b & 0\end{bmatrix} & =\begin{bmatrix}c+2d & c-d\\c+d & -d\end{bmatrix}
\end{align}$$
Immediately we see that $d=0$. Hence,
$$\begin{bmatrix}a+3b & a+b\\-b & 0\end{bmatrix}=\begin{bmatrix}c & c\\c & 0\end{bmatrix}.$$
So, $b=-c$,and we get
$$\begin{bmatrix}a+3b & a+b\\-b & 0\end{bmatrix}=\begin{bmatrix}-b & -b\\-b & 0\end{bmatrix}$$
which implies
$$\begin{bmatrix}a+4b & a+2b\\0 & 0\end{bmatrix}=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$$
It's pretty easy to see from here that $a=b=c=d=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying nested square roots ($\sqrt{6-4\sqrt{2}} + \sqrt{2}$) I guess I learned it many years ago at school, but I must have forgotten it. From a geometry puzzle I got to the solution
$\sqrt{6-4\sqrt{2}} + \sqrt{2}$
My calculator tells me that (within its precision) the result equals exactly 2, but I have no idea how to transform the calculation to symbolically get to that result.
(I can factor out one $\sqrt{2}$ from both terms, but that does not lead me anywhere, either)
| Since
\begin{align}
6-4\sqrt{2}&=4-2\cdot 2\cdot\sqrt{2}+2\\
&=(2-\sqrt{2})^2
\end{align}
Where $2-\sqrt{2}>0$, it follows
$\sqrt{6-4\sqrt{2}}=2-\sqrt{2}$.
Then $$\sqrt{6-4\sqrt{2}}+\sqrt{2}=\color{blue}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b}{a+c}+\frac{b+c}{b+a}+\frac{c+a}{c+b}.$
Prove that for all positive real numbers $a,b,$ and $c$, we have $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}.$$
What I tried is saying $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} = \dfrac{a^2c+b^2a+c^2b}{abc} \geq \dfrac{3abc}{abc} = 3$. Then how can I use this to prove that $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}$?
| our inequality is equivalent to $$a^4c^2+a^2b^4+b^2c^4+a^3b^3+a^3c^3+b^3c^3\geq a^3bc^2+a^2b^3c+ab^2c^3+3a^2b^2c^2 $$we have $$a^3b^3+a^3c^3+b^3c^3\geq 3a^2b^2c^2$$ by AM-GM, and the rest is equivalent to
$$(a-b)c^2(a^3-b^2c)+(b-c)b^2(a^2b-c^3)\geq 0$$ if we assume that $$a\geq b\geq c$$
and if $$a\geq c\geq b$$ on obtain
$$(a-c)c^2(a^3-b^2c)+(c-b)a^2(ac^2-b^3)\geq 0$$
thank you Michael Rozenberg!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How to find a basis for $W = \{A \in \mathbb{M}^{\mathbb{R}}_{3x3} \mid AB = 0\}$ $B = \begin{bmatrix}
1 & 2 & 1 \\
1 & 3 & 1 \\
1 & 4 & 1
\end{bmatrix}$ and I need to find a basis for $W = \{A \in \mathbb{M}^{\mathbb{R}}_{3x3} \mid AB = 0\}$ .
I know that $AB = A\cdot\begin{bmatrix} 1\\1\\1 \end{bmatrix} \mid A\cdot\begin{bmatrix} 2\\3\\4 \end{bmatrix} \mid A\cdot\begin{bmatrix} 1\\1\\1 \end{bmatrix} = 0 \mid 0 \mid 0$
Then I can conclude that (assume $A_1,...,A_n$ are columns of $A$):
1) $A_1 + A_2 + A_3 = 0$
2) $2A_1 + 3A_2 +4 A_3 = 0$
Meaning: $A_1 + 2A_2 +3A_3 = 0$
But now I got stuck... How should I continue from here?
| Just to simplify notation, instead of $AB=0$, let us have a look at the transpose $B^TA^T=0$.
So we are looking at the subspace
$$W'=\{A\in M_{3\times3}; CA=0\}$$
where $C=B^T=
\begin{pmatrix}
1 & 1 & 1 \\
2 & 3 & 4 \\
1 & 1 & 1 \\
\end{pmatrix}
$.
Let us denote by $\vec a_1$, $\vec a_2$, $\vec a_3$ the columns of matrix $A$.
We are looking at matrices such that
$$CA=C(\vec a_1,\vec a_2,\vec a_3)=(C\vec a_1,C\vec a_2,C\vec a_3)=0,$$
i.e.
\begin{align*}
C\vec a_1&=\vec 0\\
C\vec a_2&=\vec 0\\
C\vec a_3&=\vec 0
\end{align*}
(We are using how multiplication on the left operates on columns, see also: Intuition behind Matrix Multiplication.)
So we found out that the columns of the matrix $A\in W'$ have to fulfill the linear system $C\vec a=\vec 0$
$$\begin{pmatrix}
1 & 1 & 1 \\
2 & 3 & 4 \\
1 & 1 & 1 \\
\end{pmatrix}
\begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.$$
By solving this system of equations we get that the solution are precisely multiples of $(1,-2,-1)^T$. This means that
$$W'=\{
\begin{pmatrix}
a & b & c \\
-2a &-2b&-2c\\
a & b & c
\end{pmatrix}; a,b,c\in\mathbb R\}.
$$
(The space $W'$ consists of those matrices where each column is a multiple of this vector.)
Now to get back to the original question, we have to do the transpose. The space $W$ is equal to
$$W=\{
\begin{pmatrix}
a &-2a & a \\
b &-2b & b \\
c &-2c & c
\end{pmatrix}
\}; a,b,c\in\mathbb R\}.$$
Dimension of this space is $3$ and as a basis we can choose the matrices
$$\begin{pmatrix}
1 &-2 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 \\
1 &-2 & 1 \\
0 & 0 & 0 \\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
1 &-2 & 1
\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Closed form or approximation of $\sum\limits_{i=0}^{n-1}\sum\limits_{j=i + 1}^{n-1} \frac{i + j + 2}{(i + 1)(j+1)} (i + 2x)(j +2x)$ During the solution of my programming problem I ended up with the following double sum:
$$\sum_{i=0}^{n-1}\sum_{j=i + 1}^{n-1} \frac{i + j + 2}{(i + 1)(j+1)}\cdot (i + 2x)(j +2x)$$
where $x$ is some number. Because of the double sum the complexity of the problem will be quadratic (and I have $n$ at the scale of a million), but if I can find a closed form solution, it will reduce dramatically (log or even something close to constant).
After trying to simplify the sum using the fact that $ \frac{i + j + 2}{(i + 1)(j+1)} = \frac{1}{i+1} + \frac{1}{j+1}$ I think that my knowledge is not enough.
Can anyone help me to simplify this problem (or may be find a reasonable approximation)?
| $$
\begin{align}
&\sum_{i=0}^{n-1}\sum_{j=i+1}^{n-1}\frac{i+j+2}{(i+1)(j+1)}\cdot(i+2x)(j+2x)\tag{1}\\
&=\sum_{j=0}^{n-1}\sum_{i=j+1}^{n-1}\frac{i+j+2}{(i+1)(j+1)}\cdot(i+2x)(j+2x)\tag{2}\\
&=\sum_{i=0}^{n-1}\sum_{j=0}^{i-1}\frac{i+j+2}{(i+1)(j+1)}\cdot(i+2x)(j+2x)\tag{3}\\
&=\small\frac12\left(\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\frac{i+j+2}{(i+1)(j+1)}\cdot(i+2x)(j+2x)-\sum_{i=0}^{n-1}\frac{2i+2}{(i+1)^2}\cdot(i+2x)^2\right)\tag{4}\\
&=\small\frac12\left(\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\left(\frac1{i+1}+\frac1{j+1}\right)(i+2x)(j+2x)-2\sum_{i=0}^{n-1}\frac1{i+1}\cdot(i+2x)^2\right)\tag{5}\\
&=\small\frac12\left(2\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\frac1{i+1}\cdot(i+2x)(j+2x)-2\sum_{i=0}^{n-1}\frac1{i+1}\cdot(i+2x)^2\right)\tag{6}\\
&=\small\sum_{i=0}^{n-1}\frac1{i+1}\cdot(i+2x)\left(\frac{n(n-1)}2+2nx\right)-\sum_{i=0}^{n-1}\frac1{i+1}\cdot(i+2x)^2\tag{7}\\
&=\small\left(\frac{n(n-1)}2+2nx\right)(n+(2x-1)H_n)-\left(\frac{n(n+1)}2+2n(2x-1)+(2x-1)^2H_n\right)\tag{8}\\[6pt]
&=\small4(n-1)H_nx^2+\left(2n^2-4n+\left(n^2-3n+4\right)H_n\right)x+\tfrac12\left(n^3-2n^2+3n-\left(n^2-n+2\right)H_n\right)\tag{9}
\end{align}
$$
Explanation:
$(1)$: original formula
$(2)$: symmetry
$(3)$: change order of summation
$(4)$: average $(1)$ and $(3)$
$(5)$: simplify
$(6)$: symmetry
$(7)$: sum in $j$
$(8)$: write $i+2x=(i+1)+(2x-1)$ and sum in $i$
$(9)$: write as a polynomial in $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
To find the sum of the series $\,1+ \frac{1}{3\cdot4}+\frac{1}{5\cdot4^2}+\frac{1}{7\cdot4^3}+\ldots$ The answer given is $\log 3$.
Now looking at the series
\begin{align}
1+ \dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot4^2}+\dfrac{1}{7\cdot4^3}+\ldots &=
\sum\limits_{i=0}^\infty \dfrac{1}{\left(2n-1\right)\cdot4^n}
\\
\log 3 &=\sum\limits_{i=1}^\infty \dfrac{\left(-1\right)^{n+1}\,2^n}{n}
\end{align}
How do I relate these two series?
| hint: You want to know $S(\frac{1}{2})$ whereas $S(x) = 2\displaystyle \sum_{n=1}^\infty \dfrac{x^{2n-1}}{2n-1}, |x| < 1$, $S'(x) = ...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Real values of $x$ satisfying the equation $x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$
Real values of $x$ satisfying the equation $$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$$
We can write it as $$512x^9+576x^6+216x^3-512x+219=0$$
I did not understand how can i factorise it.
Help me
|
Using factor theorem:
If $x-a$ divides $f(x)$, then $f(a)=0$.
As $(\frac{1}{2})^{9}=\frac{1}{512}$, it's worth to try $x=\pm \frac{1}{2}$.
Substituting $x=\frac{1}{2}$, LHS vanishes, so $\frac{1}{2}$ is one possible value.
Further factorize gets,
$$\left( x-\frac{1}{2} \right)
\left( x^{2}-\frac{x}{2}-\frac{3}{4} \right)
\left( x^{6}+x^{4}+\frac{3x^{3}}{4}+x^{2}+\frac{3x}{8}+\frac{7}{3} \right)=0$$ and the second factor can be solved by quadratic formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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To evaluate the given determinant Question: Evaluate the determinant
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|$
My answer:
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|= \left|
\begin{array}{cc} b^2c^2 & bc & c \\
c^2a^2 & ca & a \\
a^2b^2 & ab & b \\
\end{array}
\right| + \left|
\begin{array}{cc} b^2c^2 & bc & b \\
c^2a^2 & ca & c \\
a^2b^2 & ab & a \\
\end{array}
\right|= abc \left|
\begin{array}{cc} bc^2 & c & 1 \\
ca^2 & a & 1 \\
ab^2 & b & 1 \\
\end{array}
\right| +abc \left|
\begin{array}{cc} b^2c & b & 1 \\
c^2a & c & 1 \\
a^2b & a & 1 \\
\end{array}
\right|$
how do I proceed from here?
| This is probably the shortest way:-
$$
\begin{align}
\begin{vmatrix} b^2c^2 & bc & b+c \\\
c^2a^2 & ca & c+a \\\
a^2b^2 & ab & a+b
\end{vmatrix}
&=a^3b^3c^3\begin{vmatrix}a^{-1} &1 & b^{-1}+c^{-1}\\\ b^{-1} &1&a^{-1}+c^{-1}\\\ c^{-1} &1& a^{-1}+b^{-1}\end{vmatrix}\\
&=(abc)^3\left(\frac 1a+\frac 1b+\frac 1c\right)\begin{vmatrix}
a^{-1}&1&1\\\ b^{-1}&1&1\\\ c^{-1} &1 &1
\end{vmatrix}
\\
&=0
\end{align}
$$
What I did was to first take out $bc$ , $ca$ and $ab$ common from the rows individually. Then I took out $abc$ from the first column.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $2\sin^{-1}\sqrt x - \sin^{-1}(2x-1) = \frac{\pi}{2}$. Prove that $2\sin^{-1}\sqrt x - \sin^{-1}(2x-1) = \dfrac\pi2$.
Do you integrate or differentiate to prove this equality? If so, why?
| I thought it might be instructive to present a way forward without the use of calculus, but rather on the use of standard trigonometric identities.
Proceeding, we let $f(x,y) =2\arcsin(x)-\arcsin(y)$ and note that
$$\begin{align}
\sin(f(x,y))&=\sin(2\arcsin(x))\cos(\arcsin(y))-\sin(\arcsin(y))\cos(2\arcsin(x))\\\\
&=2x\sqrt{1-x^2}\sqrt{1-y^2}-y(1-2x^2) \tag 1
\end{align}$$
Substituting $x\to \sqrt{x}$ and $y\to 2x-1$ into $(1)$ reveals
$$\begin{align}
\sin\left(2\arcsin(\sqrt{x})+\arcsin(2x+1)\right)&=2\sqrt{x}\sqrt{1-x}2\sqrt{x}\sqrt{1-x}-(2x-1)(1-2x)\\\\
&=1
\end{align}$$
Therefore, $2\arcsin(\sqrt{x})-\arcsin(2x-1)=\pi/2 +2n\pi$. Inasmuch as the arcsine is bounded in absolute value by $\pi/2$, then we conclude immediately that $n=0$ and
$$\bbox[5px,border:2px solid #C0A000]{2\arcsin(\sqrt{x})-\arcsin(2x-1)=\pi/2 }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate $\int_a^bx^2 dx$ using summation? So for this case, we divide it to $n$ partitions and so the width of each partition is $\frac{b-a}{n}$ and the height is $f(x)$.
\begin{align}
x_0&=a\\
x_1&=a+\frac{b-a}{n}\\
&\ldots\\
x_{i-1}&=a+(i-1)\frac{b-a}{n}\\
x_i&=a+i\frac{b-a}{n}
\end{align}
So I pick left point, which is $x_{i-1}$
I start with
\begin{align}
\sum\limits_{i=1}^{n} \frac{b-a}{n}f\left(a+\frac{(i-1)(b-a)}{n}\right)
&=\frac{b-a}{n}\sum\limits_{i=1}^{n} \left(a+\frac{(i-1)(b-a)}{n}\right)^2\\
&=\frac{b-a}{n}\left(na^2+ \frac{2a(b-a)}{n}\sum\limits_{i=1}^{n}(i-1)+\frac{(b-a)^2}{n^2} \sum\limits_{i=1}^{n}(i-1)^2\right)
\end{align}
Here I am stuck because I don't know what$ \sum\limits_{i=1}^{n}(i-1)^2$is (feel like it diverges). Could someone help?
| The sum that is of concern is multiplied by $\frac1{n^3}$ and the limit therefore converges. Then, proceeding, we have
$$\sum_{i=1}^n(i-1)^2=\sum_{i=0}^{n-1} i^2$$
We can evaluate the sum on the right-hand side of $(1)$ using a number of approaches. The approach we present here exploits telescoping series. We note that
$$\sum_{n=0}^{n-1}\left((i+1)^3-i^3\right)=n^3 \tag 2$$
If we expand the terms in the sum in $(2)$, we find that
$$3\sum_{n=0}^{n-1}i^2+3\sum_{n=0}^{n-1}i+\sum_{n=0}^{n-1} 1=n^3 \tag 3$$
whereupon solving for the sum $\sum_{n=0}^{n-1}i^2$ and using $\sum_{n=0}^{n-1}i=\frac{n(n-1)}{2}$ reveals
$$\begin{align}
\sum_{n=0}^{n-1}i^2&=\frac{n^3-n-3\frac{n(n-1)}{2}}{3}\\\\
&\bbox[5px,border:2px solid #C0A000]{=\frac{n(n-1)(2n-1)}{6}}
\end{align}$$
Dividing by $\frac1{n^3}$ and lettgin $n\to \infty$ yields $\frac13$ as expected!
Adding this to the other terms and letting $n \to \infty$ yields
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(\frac{n\,a^2(b-a)}{n}+\frac{2a(b-a)^2\,\frac{n(n-1)}{2}}{n^2}+\frac{(b-a)^3\,\frac{n(n-1)(2n-1)}{6}}{n^3}\right)=\frac{b^3-a^3}{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $3(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}) \ge 10 + 8\cdot \frac{a^2+b^2+c^2}{ab+bc+ca}$
For the positive real numbers $a, b, c$ prove that $$3\bigg(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\bigg) \ge 10 + 8\cdot \dfrac{a^2+b^2+c^2}{ab+bc+ca}$$
I did the following:
$$\begin{split}\dfrac{a^2+b^2+c^2}{ab+bc+ca} & = \dfrac{(a+b+c)^2-2(ab+bc+ca)}{ab+bc+ca} \\ & = \dfrac{(a+b+c)^2}{ab+bc+ca} - 2 \\ & \le \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} -2\end{split}$$
I got the last one by applying Titu's Lemma
Thus I think it suffices to prove that $$3\bigg(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\bigg) \ge 10 + 8\bigg(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-2\bigg) \\ \Longrightarrow 3a^2b+3b^2c+3c^2a+6abc \ge 5a^2c+5b^2a+5c^2b$$
But by rearrangement inequality this is not always true I think. The direction of the inequality should have been flipped.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, your inequality is a linear inequality of $w^3$.
Hence, it remains to prove it for an extremal value of $w^3$, which happens in the following cases.
*
*$w^3\rightarrow0^+$. In this case our inequality is obvious;
*$b=c=1$, which gives $(a-1)^2(2a+3)\geq0$. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
proving 1/x is convex (without differentiating) I know that $\frac{1}{x}$ is convex when $x \in (0,\infty)$, this can be proven easily by showing that the second derivative is positive. However, I am finding difficulty showing it using the definition of convexity, in other words, for $\alpha \in [0,1]$ and $x_1, x_2 \in \Bbb R^+$, show that:
$$\frac{1}{\alpha x_1 + (1-\alpha) x_2 } \leq \frac{\alpha}{x_1}+\frac{1-\alpha}{x_2}$$
Note that the relation between the harmonic mean and arithmetic mean is just a special case, (take $\alpha = 0.5$).
| We want to show that
$\frac{1}{a x + (1-a) y }
\leq \frac{a}{x}+\frac{1-a}{y}
$
The right side is
$ \frac{a}{x}+\frac{1-a}{y}
=\frac{ay+(1-a)x}{xy}
=\frac{a(y-x)+x}{xy}
$
and the left side is
$\frac{1}{a x + (1-a) y }
=\frac{1}{a (x-y) + y }
$
so we want
$\frac{1}{a (x-y) + y}
\le \frac{a(y-x)+x}{xy}
$.
Cross multiplying,
this is
$\begin{array}\\
xy
&\le (a (x-y) + y)(a(y-x)+x)\\
&=-a^2(x-y)^2+ay(y-x)+ax(x-y)+xy\\
&=-a^2(x-y)^2+a(x-y)(x-y)+xy\\
&=-a^2(x-y)^2+a(x-y)^2+xy\\
&=(a-a^2)(x-y)^2+xy\\
&=a(1-a)(x-y)^2+xy\\
\end{array}
$
And since both
$a(1-a) \ge 0$
and
$(x-y)^2 \ge 0$
the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1622113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
$\int \frac{2x}{9x^2+3}dx=?$ So this one seems very easy. And it really is (I guess). But I have an issue with this one. I solved it this way:
\begin{align}
& \int \frac{2x}{9x^2+3}dx=\frac{1}{6} \int \frac{x}{3x^2+1}dx=\frac{1}{6} \left( x\frac{\arctan(\sqrt3x)}{\sqrt{3}}-\int 1\cdot\frac{\arctan(\sqrt3x)}{\sqrt3} dx\right) \\[10pt]
= {} & \frac{1}{6} \left( x\frac{\arctan(\sqrt3x)}{\sqrt{3}}-x\frac{\arctan(\sqrt3x)}{\sqrt{3}}+\frac{\arctan(\sqrt3x)}{\sqrt3}\right)=\frac{\arctan(\sqrt3x)}{\sqrt3}+C
\end{align}
But at the same time, the textbook says, that it is equal to this:
$$
\frac{1}{9}\log(3x^2+1)+C
$$
Did I do something wrong? It would be quite hard for me to believ that these to functions are the same.
Thanks for the responses!
| $$\int \frac{2x}{9x^2+3}dx$$
$$2\int \frac{x}{9x^2+3}dx$$
Apply Integral Substitution $u=9x^2+3\quad \:du=18xdx$
$$2\int \frac{1}{18u}du = 2\frac{1}{18}\int \frac{1}{u}du=2\frac{1}{18}\ln \left|u\right|=2\frac{1}{18}\ln \left|9x^2+3\right|$$
So
$$\int \frac{2x}{9x^2+3}dx=\color{red}{\frac{1}{9}\ln \left|9x^2+3\right|+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show that the following limit is $AB$. I am interested in showing the following:
$$
\lim\limits_{p\rightarrow 1/2} \frac{B}{1-2p}-\frac{A+B}{1-2p}\frac{r^B-1}{r^{A+B}-1}=AB.
$$
Here $A,B\in(0,\infty)$ and $r=\frac{1}{p}-1$.
This is for a probability theory problem involving a biased random walk, leaving an interval $(-B,A)$
I'm not really sure how to proceed in this problem and could use some help.
Thank you for any assistance.
| Considering the expression $$Z=\frac{B}{1-2p}-\frac{A+B}{1-2p}\frac{r^B-1}{r^{A+B}-1}$$ start changing variable $p=\frac{1-x}{2}$; this makes $$Z=\frac{B}{x}-\frac{(A+B) \left(\left(\frac{2}{1-x}-1\right)^B-1\right)}{x
\left(\left(\frac{2}{1-x}-1\right)^{A+B}-1\right)}$$ and use the generalized binomial theorem or, better, use Taylor expansion around $x=0$ $$\frac{2}{1-x}=2+2 x+2 x^2+2 x^3+O\left(x^4\right)$$ $$\frac{2}{1-x}-1=1+2 x+2 x^2+2 x^3+O\left(x^4\right)$$ $$\left(\frac{2}{1-x}-1\right)^k=1+2 k x+2 k^2 x^2+\frac{2}{3} \left(2 k^3+k\right) x^3+O\left(x^4\right)$$ Using all the above, you should end with $$Z=A B+\frac{1}{3} A B (B-A)x-\frac{1}{3} A B (A B-1)x^2+O\left(x^3\right)$$ and $x\to 0$. This shows the limit and how it is approached.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$? How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$ ?
$A = \left.\frac{s^2-s}{s^2+1} \right\vert_{s=-1} = \frac{1-(-1)}{1+1}=1$
| Cross-multiplying on the right hand side gives
$As^{2} + A + Bs^{2} + Bs + Cs + C$ on the numerator.
Plug A = 1
$(1 + B)s^{2} + (B + C)s + (1 + C) = s^{2} - s$. Equate the coefficients
$1 + B = 1 \rightarrow B = 0$
$B + C = -1 \rightarrow C = -1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
What does $\bigotimes$ and $X^*$ mean? Can someone explain / link me to a linear algebra worked problem where I can see how these work. I've searched and given their statistics and matrix specialty uses, can't find any ready examples.
| $\bigotimes$ is the Kronecker product of two matrices. In this post I used the Kronecker product to understand the model matrices in mixed effects models.
$A^* = A^H = \bar A^T$ is the Hermitian or conjugate transpose of a matrix $A$.
A good example of the use of Hermitian matrices is in obtaining the inverse of a Fourier matrix in the Fast Fourier Transform (FFT). For an $n \times n$ Fourier matrix, $F_n$, the inverse is going to be the Hermitian conjugate, such that $\frac{1}{n}\,F_n^H\,F_n = I.$
For example, a $4\times4$ Fourier matrix ($n=4$) would be:
$$\large \begin{bmatrix}
1 & 1 & 1 & 1\\
1 & W & W^2 & W^3\\
1 & W^2 & W^4 & W^6 \\
1 & W^3 & W^6 & W^9
\end{bmatrix} =
\begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & e^{i\,\frac{2\pi}{4}} & e^{i\,2\frac{2\pi}{4}} & e^{i\,3\frac{2\pi}{4}}\\
1 & e^{i\,2\frac{2\pi}{4}} & e^{i\,4\frac{2\pi}{4}} & e^{i\,6\frac{2\pi}{4}} \\
1 & e^{i\,3\frac{2\pi}{4}} & e^{i\,6\frac{2\pi}{4}} & e^{i\,9\frac{2\pi}{4}}
\end{bmatrix}=
\begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & i & -1 & -i\\
1 & -1 & 1 & -1 \\
1 & -i & -1 & i
\end{bmatrix}
$$
And the Hermitian of this matrix:
$$F_4=\begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & i & -1 & -i\\
1 & -1 & 1 & -1 \\
1 & -i & -1 & i
\end{bmatrix}\implies F_4^H =
\begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & -i & -1 & i\\
1 & -1 & 1 & -1 \\
1 & i & -1 & -i
\end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\frac{1}{\sqrt{1-x}} \le e^x$ on $[0,1/2]$. Is there a simple way to prove $$\frac{1}{\sqrt{1-x}} \le e^x$$ on $x \in [0,1/2]$?
Some of my observations from plots, etc.:
*
*Equality is attained at $x=0$ and near $x=0.8$.
*The derivative is positive at $x=0$, and zero just after $x=0.5$. [I don't know how to find this zero analytically.]
*I tried to work with Taylor series. I verified with plots that the following is true on $[0,1/2]$:
$$\frac{1}{\sqrt{1-x}} = 1 + \frac{x}{2} + \frac{3x^2}{8} + \frac{3/4}{(1-\xi)^{5/2}} x^3 \le 1 + \frac{x}{2} + \frac{3}{8} x^2 + \frac{5 \sqrt{2} x^3}{6} \le 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \le e^x,$$
but proving the last inequality is a bit messy.
| We have
$$-2 \ln \sqrt{1-x}=-\ln(1-x)= \int_{1-x}^1\frac{dt}{t} \leqslant \frac{x}{1-x}.$$
For $0 \leqslant x \leqslant 1/2$, we have $2(1-x) \geqslant 1$ and
$$-\ln \sqrt{1-x} < \frac{x}{2(1-x)} \leqslant x.$$
Hence,
$$\frac{1}{\sqrt{1-x}} = \exp[-\ln(\sqrt{1-x})]\leqslant e^x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$ I saw some resolutions here like $\sqrt{x+\sqrt{x+\sqrt{x}}}- \sqrt{x}$, but I couldn't get the point to find
$\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.
I tried $\frac{1}{x}.(\sqrt{x+\sqrt{x+\sqrt{x}}})=\frac{\sqrt{x}}{x}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)=\frac{1}{\sqrt{x}}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)$ but now I got stuck. Could anyone help?
| For an alternative answer, consider the infinitely nested radical expression $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}}$.
Evidently $y > \sqrt{x+\sqrt{x+\sqrt{x}}}$ for $x > 1$ and therefore $\frac{y}{x}$ is an upper bound for $\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.
Also $y > \sqrt{x}$ and therefore $\lim_{x\to\infty} y = \infty$
But we can express $x$ in terms of $y$ by using the substitution $y = \sqrt{x+y}$ in the infinitely nested expression, leading to $x = y(y-1)$.
Then $\lim_{x\to\infty}\frac{y}{x} = \lim_{y\to\infty} \frac{y}{y(y-1)} = \lim_{y\to\infty}\frac{1}{y-1} = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$ If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$
Using the condition for common root, $$(3c-5b)(b-3a)=(c-5a)^2$$
$$3bc-9ac-5b^2+15ab=c^2+25a^2-10ac$$
$$25a^2+5b^2+c^2=15ab+3bc+ac$$
There is pattern in the equation. The coefficients of the middle terms and first terms of both sides are $(5\times 5,5)$ and $(5\times 3,3)$.
I tried to use Lagrange multipliers. But isnt there any simpler way to minimize $a+b+c$?
| Hint:
The roots of $x^2+3x+5$ are not real, and if $z\in \Bbb C$ is a root of a polynomial with real coefficients but $z$ is not real, then $\bar z$ is also a root of this polynimial.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
induction to prove the equation $3 + 9 + 15 + ... + (6n - 3) = 3n^2$ I have a series that I need to prove with induction. So far I have 2 approaches, though I'm not sure either are correct.
$$3 + 9 + 15 + ... + (6n - 3) = 3n^2$$
1st attempt:
\begin{align*}
& = (6n - 3) + 3n^2\\
& = 3n^2 + 6n - 3\\
& = (3n^2 + 5n - 4) + (n + 1)
\end{align*}
That seems way wrong lol ^^^
2nd attempt:
\begin{align*}
f(n) & = 3 + 9 + 15 + ... + (6n - 3)\\
f(n + 1) & = 6(n + 1) - 3\\
f(n + 1) & = 6(n - 3) + 6(n + 1) - 3\\
& = ?
\end{align*}
I don't know I feel like I'm headed in the wrong direction
I guess another attempt I have would be:
\begin{align*}
f(n) & = 3n^2\\
f(n+1) & = 3(n + 1)^2\\
& = 3(n^2 + 2n + 1)\\
& = 3n^2 + 6n + 3\\
& = f(n) + (6n + 3)
\end{align*}
| If you insist on induction :
Base case $n=1\ :\ 6\times 1-3=3=3\times 1^2$
If you assume $$3+9+15+...+(6n-3)=3n^2$$
You can conclude
$$3+9+15+...+(6n-3)+(6n+3)=3n^2+6n+3=3(n+1)^2$$
completing the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Upper bound on integral: $\int_1^\infty \frac{dx}{\sqrt{x^3-1}} < 4$ I'm going through Nahin's book Inside Interesting Integrals, and I'm stuck at an early problem, Challenge Problem 1.2: to show that
$$\int_1^\infty \frac{dx}{\sqrt{x^3-1}}$$
exists because there is a finite upper-bound on its value. In particular, show that the integral is less than 4.
I've tried various substitutions and also comparisons with similar integrals, but the problem is that any other integral that I can easily compare the above integral to is just as hard to integrate, which doesn't help solve the problem.
I also tried just looking at the graph and hoping for insight, but that didn't work either.
So how doesone one place an upper bound on the integral?
| $$\begin{eqnarray*}\color{red}{I}=\int_{1}^{+\infty}\frac{dx}{\sqrt{x^3-1}}=\int_{0}^{+\infty}\frac{dx}{\sqrt{x^3+3x^2+3x}}&=&\int_{0}^{+\infty}\frac{2\, dz}{\sqrt{z^4+3z^2+3}}\\&\color{red}{\leq}&\int_{0}^{+\infty}\frac{2\,dz}{\sqrt{z^4+3z^2+\frac{9}{4}}}=\color{red}{\pi\sqrt{\frac{2}{3}}.}\end{eqnarray*}$$
A tighter bound follows from Cauchy-Schwarz:
$$\begin{eqnarray*} \color{red}{I} &=&2\int_{0}^{+\infty}\frac{\sqrt{z^2+\sqrt{3}}}{\sqrt{z^4+3z^2+3}}\cdot\frac{dz}{\sqrt{z^2+\sqrt{3}}}\\&\color{red}{\leq}& 2\sqrt{\left(\int_{0}^{+\infty}\frac{z^2+\sqrt{3}}{z^4+3z^2+3}\,dz\right)\cdot\int_{0}^{+\infty}\frac{dz}{z^2+\sqrt{3}}}\\&=&2\pi\cdot\left(\frac{1}{6}-\frac{1}{4\sqrt{3}}\right)^{1/4}\leq\color{red}{\frac{2\pi}{42^{1/4}}}.\end{eqnarray*} $$
The manipulations in the first line show that $I$ is just twice a complete elliptic integral of the first kind, whose value can be computed through the arithmetic-geometric mean.
On the other hand, through the substitution $x=\frac{1}{t}$ and Euler's beta function we have:
$$ I \color{red}{=} \frac{\Gamma\left(\frac{1}{3}\right)\cdot \Gamma\left(\frac{1}{6}\right)}{2\sqrt{3\pi}}\color{red}{\leq}\frac{3\cdot 6}{2\sqrt{3\pi}}=\sqrt{\frac{27}{\pi}}$$
since in a right neighbourhood of the origin we have $\Gamma(x)\leq\frac{1}{x}$.
As a by-product we get:
$$ \frac{\Gamma\left(\frac{1}{3}\right)\cdot \Gamma\left(\frac{1}{6}\right)}{2\sqrt{3\pi}} = \frac{\pi}{\text{AGM}(\frac{1}{2} \sqrt{3+2 \sqrt{3}},3^{1/4})}$$
that allows us to compute $\Gamma\left(\frac{1}{6}\right)$ through an AGM-mean:
$$ \Gamma\left(\frac{1}{6}\right) = \color{red}{\frac{2^{\frac{14}{9}}\cdot 3^{\frac{1}{3}}\cdot \pi^{\frac{5}{6}} }{\text{AGM}\left(1+\sqrt{3},\sqrt{8}\right)^{\frac{2}{3}}}}.$$
The last identity was missing in the Wikipedia page about particular values of the $\Gamma$ function, so I took the liberty to add it and add this answer as a reference.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Find $\alpha^3 + \beta^3$ which are roots of a quadratic equation. I have a question.
Given a quadratic polynomial, $ax^2 +bx+c$, and having roots $\alpha$ and $\beta$. Find $\alpha^3+\beta^3$. Also find $\frac1\alpha^3+\frac1\beta^3$
I don't know how to proceed. Any help would be appreciated.
| First note that $\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)$ and also note that $-\frac{b}{a}=\alpha+\beta$ and $\frac{c}{a}=\alpha\beta$ (do you see why?) We can make $$\alpha^2+2\alpha\beta+\beta^2=(\alpha+\beta)^2=\frac{b^2}{a^2}$$ so our final outcome will be \begin{align}
\alpha^3+\beta^3&=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)\\
&= -\frac{b}{a}(\alpha^2+2\alpha\beta+\beta^2-3\alpha\beta)\\
&= -\frac{b}{a}(\frac{b^2}{a^2}-3\frac{c}{a})\\
&= -\frac{b^3-3abc}{a^3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $z^4+2z^3+3z^2+2z+1 =0$ Solve $z^4+2z^3+3z^2+2z+1 =0$ with $z$: a complex variable.
Attempt at solving the problem:
We divide the polynom by $z^2$ and we get:
$z^2+2z+3+\dfrac{2}{z}+ \dfrac{1}{z^2}=0 $ $ $ We set $w=z+ \dfrac{1}{z}$
We now have $w^2+2w+5=0$
$\bigtriangleup = -16$
Let's find $\omega$ such that $\omega^2=-16$
We have $\omega=4i$
Therefore we have the 2 roots:
$w_ {1}=-1-2i$ and $ w_ {2}=-1+2i $
The issue is: I don't know how to find z
| $$z^4+2z^2+3z^2+2z+1=0\Longleftrightarrow$$
$$\left(z^2+z+1\right)^2=0\Longleftrightarrow$$
$$z^2+z+1=0\Longleftrightarrow$$
$$z^2+z=-1\Longleftrightarrow$$
$$z^2+z+\frac{1}{4}=-\frac{3}{4}\Longleftrightarrow$$
$$\left(z+\frac{1}{2}\right)^2=-\frac{3}{4}\Longleftrightarrow$$
$$z+\frac{1}{2}=\pm\frac{i\sqrt{3}}{2}\Longleftrightarrow$$
$$z=\pm\frac{i\sqrt{3}}{2}-\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632346",
"timestamp": "2023-03-29T00:00:00",
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How to minimize $ab + bc + ca$ given $a^2 + b^2 + c^2 = 1$? The question is to prove that $ab + bc + ca$ lies in between $-1$ and $1$, given that $a^2 + b^2 + c^2 = 1$.
I could prove the maxima by the following approach. I changed the coordinates to spherical coordinates:
$a = \cos A \\
b = \sin A \cos B \\
c = \sin A \sin B$
Using that $\cos X + \sin X$ always lies between $- \sqrt 2$ and $\sqrt 2$ I proved that $a + b + c$ lies between $- \sqrt 3$ and $\sqrt 3$.
Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$ I could prove that $ab + bc + ca$ is maximum at $1$ but I can't prove the minima.
| Hint: $a^2+b^2+c^2+ab+bc+ca=\frac{1}{2}((a+b)^2+(b+c)^2+(c+a)^2)\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Can you find the maximum or minimum of an equation without calculus? Without using calculus is it possible to find provably and exactly the maximum value
or the minimum value of a quadratic equation
$$ y:=ax^2+bx+c $$
(and also without completing the square)?
I'd love to know the answer.
| We assume (for the sake of discovery; for this purpose it is good enough
if this is just an inspired guess)
that the curve $y = ax^2 + bx + c$ is symmetric around a vertical axis.
It's obvious this is true when $b = 0$, and if we have plotted
$y = ax^2 + bx + c$ for various other values of $a$, $b$, and $c$,
we may observe enough appearance of symmetry to suppose that it might be true in general.
So it's reasonable to say: supposing it were true, what would that tell
us about the minimum/maximum value of the polynomial?
We find the points on this curve of the form $(x,c)$ as follows:
\begin{align}
y &= c. \\
c &= ax^2 + bx + c. \\
0 &= ax^2 + bx = (ax + b)x.
\end{align}
Hence if $(x,c)$ is on the curve, then either $ax + b = 0$ or $x = 0$.
Setting $x_1 = -\dfrac ba$ and $x_2 = 0$, we can plug in these two values
for $x$ and confirm that indeed the two points
$\left(-\frac ba, c\right)$ and $(0, c)$ are on the curve.
Using the assumption that the curve is symmetric around a vertical axis,
the vertical axis would have to be halfway between
$\left(-\frac ba, c\right)$ and $(0, c)$, that is, it is
the line $x = -\dfrac b{2a}$.
If there is a global maximum or minimum, it is a reasonable guess that
it would be on this line, so let's see what we have at
$x_0 = -\dfrac b{2a}$. Plugging this into the equation and doing the
algebra to find the point $(x_0, y_0)$ on the curve,
\begin{align}
y_0 &= a\left(-\frac b{2a}\right)^2 + b\left(-\frac b{2a}\right) + c \\
&= c - \frac{b^2}{4a}.
\end{align}
So that's our candidate for the maximum or minimum value.
To prove this is correct, consider any value of $x$ other than
$-\dfrac b{2a}$. Any such value can be expressed by its difference
from $-\dfrac b{2a}$, that is, we let
$$ x = -\frac b{2a} + t$$
where $t \neq 0$. Now plug this value into the equation
and do the algebra:
\begin{align}
y &= a\left(-\frac b{2a} + t\right)^2 + b\left(-\frac b{2a} + t\right) + c
\\[.5ex]
&= at^2 + c - \frac{b^2}{4a}. \tag 1
\end{align}
If $a$ is positive, $at^2$ is positive, hence $y > c - \dfrac{b^2}{4a} = y_0$
for every point $(x,y)$ on the curve such that $x \neq x_0$,
and therefore $y_0 = c - \dfrac{b^2}{4a}$ is a minimum.
But if $a$ is negative, $at^2$ is negative, and similar reasoning
says that $y_0 = c - \dfrac{b^2}{4a}$ is a maximum.
Note that the proof made no assumption about the symmetry of the curve.
On the contrary, the equation $y = at^2 + c - \dfrac{b^2}{4a}$
can be used to prove that the curve is symmetric.
If we take this a little further, we can even derive the standard
quadratic formula from it. The roots of the equation
$y = ax^2 + bx + c$ are the values of $x$ such that $y = 0$.
But as we know from Equation $(1)$, above,
if we make the substitution $x = -\dfrac b{2a} + t$, that means
\begin{align}
0 = y &= ax^2 + bx + c \\ &= at^2 + c - \frac{b^2}{4a}.
\end{align}
A little algebra (isolate the $at^2$ term on one side and divide by $a$)
gives us
$$
t^2 = \frac{b^2}{4a^2} - \frac ca. \tag 2
$$
In general, if $p^2 = q$ then $p = \pm \sqrt q$, so Equation $(2)$
tells us that
\begin{align}
t &= \pm \sqrt{\frac{b^2}{4a^2} - \frac ca} \\
&= \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}\\
&= \pm \frac{\sqrt{b^2 - 4ac}}{\lvert 2a \rvert}\\
&= \pm \frac{\sqrt{b^2 - 4ac}}{2a},
\end{align}
and recalling that we set $x = -\dfrac b{2a} + t$,
\begin{align}
x &= -\frac b{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\
&= \frac{- b \pm \sqrt{b^2 - 4ac}}{2a},
\end{align}
which is precisely the usual quadratic formula.
Is the reasoning above actually just an example of "completing the square,"
or is it sufficiently different from the usual method of "completing the square" that it can be considered a different method?
The equation $x = -\dfrac b{2a} + t$ is equivalent to
$t = x + \dfrac b{2a}$; the method of completing the square involves
expanding $\left(x + \dfrac b{2a}\right)^2$;
and in fact we do see $t^2$ figuring prominently in the equations above.
Certainly we could be inspired to try completing the square after
noticing how neatly the equation
$ax^2 + bx + c = at^2 + c - \dfrac{b^2}{4a}$
simplified the problem; but we never actually expanded the
binomial $\left(x + \dfrac b{2a}\right)^2$, and we never subtracted
the original polynomial from it to find the amount we needed to
"complete" the square.
Instead, the quantity $c - \dfrac{b^2}{4a}$ just "appeared" in the
original equation as the result of a direct substitution.
I think that may be about as different from "completing the square"
as a purely algebraic method can get.
| {
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"url": "https://math.stackexchange.com/questions/1633619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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} |
Generalisation of an already generalised integral Inspired by these two questions:
Closed form for $\int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$
Interesting integral formula
I ask whether the following integral has a closed form:
$$\int_0^\infty \frac{x^n}{(1+x^m)^\alpha} \text{d}x$$
For $\alpha \in \mathbb{N}$ and $0<n<\alpha m+1$
If that is too difficult, then I will also accept the standard restriction from the other questions
$0<n<m+1$
Other generalised forms of the integral are also acceptable, such as $$\int_0^\infty \frac{x^n}{(a+bx^m)^\alpha} \text{d}x$$
And an extension of this question for those who want something more:
Does a closed form exist for $\alpha \in \mathbb{R}^+$?
| $\text {Let } y=\dfrac{1}{1+x^{m}}, \textrm{then } x=\left(\dfrac{1}{y}-1\right)^{\frac{1}{m}} \textrm{ and } d x=\dfrac{1}{m} \left(\dfrac{1}{y}-1\right)^{\frac{1}{m-1}}\left(\dfrac{d y}{-y^{2}}\right).$
Simplifying gives
$$
\begin{aligned}
\int_{0}^{\infty} \frac{x^{n}}{\left(1+x^{m}\right)^{\alpha}} &=\frac{1}{m} \int_{0}^{1} y^{\left(\alpha-\frac{n+1}{m}\right)-1}(1-y)^{\frac{n+1}{m}-1} d y \\
&=\frac{1}{m} B\left(\alpha-\frac{n+1}{m}, \frac{n+1}{m}\right) \\
&=\frac{1}{m} \frac{\Gamma\left(\alpha-\frac{n+1}{m}\right) \Gamma\left(\frac{n+1}{m}\right)}{\Gamma(\alpha)}
\end{aligned}
$$
Using the property of Gamma Function: $\Gamma(z+1)=z \Gamma(z)$ repeatedly yields
$$
\begin{aligned}
\Gamma\left(\alpha-\frac{n+1}{m}\right) &=\left(\alpha-1-\frac{n+1}{m}\right) \Gamma\left(\alpha-1-\frac{n+1}{m}\right) \\
& \qquad\qquad \vdots \\
&=\prod_{k=1}^{\alpha -1}\left(\alpha-k-\frac{n+1}{m}\right) \Gamma\left(1-\frac{n+1}{m}\right)
\end{aligned}
$$
Using Euler’s Reflection Theorem: $
\quad \Gamma(z) \Gamma(1-z)=\pi \csc (\pi z) \textrm{ for } z\notin Z
$ gives
Case 1: When $n=1$,
$$\int_{0}^{\infty} \frac{x^{n}}{1+x^{m}}= \frac{\pi}{m} \csc \frac{(n+1) \pi}{m} $$
Case 2: When $n\geq 2$,
$$
\begin{aligned} \int_{0}^{\infty} \frac{x^{n}}{\left(1+x^{m}\right)^{\alpha}}&=\frac{1}{m(\alpha-1) !} \left[\prod_{k=1}^{\alpha-1}\left(\alpha-k-\frac{n+1}{m}\right) \right]\Gamma\left(1-\frac{n+1}{m}\right) \Gamma\left(\frac{n+1}{m}\right)\\&=\boxed{ \frac{\pi}{m(\alpha-1) !} \csc\frac{(n+1) \pi}{m}\prod_{k=1}^{\alpha-1}\left(\alpha-k-\frac{n+1}{m}\right)} \end{aligned}
$$
For the more general integral $\displaystyle \int_{0}^{\infty} \dfrac{x^{n}}{\left(a+b x^{m}\right)^{\alpha}} d x$, we let $\displaystyle u=\sqrt[m]{\frac{b}{a}} x$ and get $$
\begin{aligned}
\int_{0}^{\infty} \dfrac{x^{n}}{\left(a+b x^{m}\right)^{\alpha}} d x &=\int_{0}^{\infty} \frac{\left(\frac{a}{b}\right)^{\frac{n}{m}} u^{n}}{\left(a+a u^{m}\right)^{\alpha}} \left(\frac{a}{b}\right)^{\frac{1}{m}} d u\\
&=\left(\frac{a}{b}\right)^{\frac{n+1}{m}} \cdot a^{-\alpha} \int_{0}^{\infty} \frac{u^{n}}{\left(1+u^{m}\right)^{\alpha}} d u \\
&=\boxed{\frac{a^{\frac{n+1}{m}-\alpha}\pi}{b^{\frac{n+1}{m}} m(\alpha-1) !} \csc\frac{(n+1) \pi}{m}\prod_{k=1}^{\alpha-1}\left(\alpha-k-\frac{n+1}{m}\right)}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a thousand natural numbers such that their sum equals their product The question is to find a thousand natural numbers such that their sum equals their product. Here's my approach :
I worked on this question for lesser cases :
\begin{align}
&2 \times 2 = 2 + 2\\
&2 \times 3 \times 1 = 2 + 3 + 1\\
&3 \times 3 \times 1 \times 1 \times 1 = 3 + 3 + 1 + 1 + 1\\
&7 \times 7 \times 1 \times 1 \times \dots\times 1 \text{ (35 times) } = 7 + 7 + 1 + 1 .... \text{ (35 times) }
\end{align}
Using this logic, I seemed to have reduced the problem in the following way.
$a \times b \times 1 \times 1 \times 1 \times\dots\times 1 = a + b + 1 + 1 +...$
This equality is satisfied whenever $ ab = a + b + (1000-n)$
Or $ abc\cdots n = a + b + \dots + n + ... + (1000 - n)$
In other words, I need to search for n numbers such that their product is greater by $1000-n$ than their sum. This allows the remaining spots to be filled by $1$'s. I feel like I'm close to the answer.
Note : I have got the answer thanks to Henning's help. It's
$112 \times 10 \times 1 \times 1 \times 1 \times ...$ ($998$ times)$ = 10 + 112 + 1 + 1 + 1 + ...$ ($998$ times)
This is for the two variable case. Have any of you found answers to more than two variables ?
$abc...n = a + b + c + ... + n + (1000 - n) $
| A solution with four numbers different from 1 is:
$$16 \times 4 \times 4 \times 4 \times 1^{996} = 16 + 4 + 4 + 4 + (996 \times 1) = 1024$$
How was this found? $1024 = 2^{10}$ appeared to be a promising candidate for the sum and product because it's slightly larger than 1000 and has many factors. The problem then was to find a, b, c, d such that:
$$a+b+c+d=10$$
$$2^a+2^b+2^c+2^d=1024 -(1000-4)=28$$
None of a-d can be more than 4 since $2^5=32 > 28$. But on trying $a=4$, reducing the problem to finding b,c,d such that $b+c+d=6$ and $2^b+2^c+2^d = 12$, the solution was apparent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim_{(x,y)\in S} \frac{x^2+ay^2}{x^2+by^2}$ as $(x,y) \to (0,0)$ for specified $S$. Find $$\lim_{\stackrel{(x,y)\to(0,0)}{(x,y)\in S}} \frac{x^2+ay^2}{x^2+by^2}$$ where $a,b\in \mathbb{R}$ and
*
*$S=S_1=\{(x,y)\in\mathbb{R}^2 \mid |y| < c|x|^p\}$ with $c > 0, p > 1$
*$S=S_2=\{(x,y)\in\mathbb{R}^2 \mid |y| < |x|\}$.
Here's my attempt by changing to polar coordinate.
$$x = r\cos \theta,\quad y = r\sin \theta$$.
So $$
\begin{align*}
f(x,y) = \frac{x^2+ay^2}{x^2+by^2} &= \frac{r^2(\cos^2\theta+a\sin^2\theta)}{r^2(\cos^2\theta +b\sin^2\theta)} \\
&=\frac{1+a + (1-a)\cos 2\theta}{1+b+(1-b)\cos 2\theta}.
\end{align*}
$$
For $S_1$, I think $|y| < c|x|^p$ "forces" $\theta \to 0$ as $(x,y)\to 0$. So $\cos 2\theta \to 1$ and $f(x,y) \to 1$.
For $S_2$, $\theta$ could be anything in $[-\pi, -\pi+\pi/4) \cup (-\pi/4,\pi/4) \cup (\pi -\pi/4, \pi)$. So the limit does not exist.
How can I make these rigorous?
Thank you!
| Set $$f(x,y)=\frac{x^2+ay^2}{x^2+by^2}$$ I don't think polar coordinates help.
For starters, the limit of $f(x,0)$ as $x\to 0$ is then $1$, so if the limit of $f$ exists for $(x,y)\to (0,0)$, it must be $1$.
For case 1, $|y|<c|x|^p=o(x)$ (i.e., $|y|$ goes to zero faster than $x$) in $S_1$. We have also $ay^2=o(x^2)$ and $by^2=o(x^2)$ as $(x,y)\to (0,0)$, so the limit of $f$ as $(x,y)\to(0,0)$ is $$\lim_{x\to 0} {x^2+o(x^2)\over x^2+o(x^2)} = 1$$ as desired.
For case 2, let's suppose $y=x/c$, $c>1$, so that the line $y=x/c$ lies within $S_2$. We then have $x^2+ay^2=x^2(1+a/c^2)$ and likewise $x^2+by^2=x^2(1+b/c^2)$ so the value of $f$ on the line $y=x/c$ is $(1+a/c^2)/(1+b/c^2)$. Since $c>1$ is arbitrary, the value of $f$ on different lines in $S_2$ passing through the origin is different, and so the limit of $f$ as $(x,y)\to (0,0)$ doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Decompose $\frac{x^4 + 5}{x^5 + 6x^3}$ (partial fraction decomposition)
Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficients.
$$\frac{x^4 + 5}{x^5 + 6x^3}$$
So I factored the denominator to be $x^3(x^2+6)$ and here is the answer i got:
$\frac{Ax+B}{x^3}+\frac{Cx+D}{x^2+6}$
But this isn't the correct answer. Any help is appreciated!
Also have another one:
$\frac{5}{(x^2 − 16)^2}$
Here is what I got: $\frac{A}{x+4}+\frac{Bx+C}{\left(x+4\right)^2}+\frac{D}{x-4}+\frac{Ex+F}{\left(x-4\right)^2}$ which is also wrong... if anyone could help me that would be great
| Question 1:
$$
x^{4}+5=f(x)\left(x^{2}+6\right)+x^{3} g(x)
$$
Let $ \quad \displaystyle \frac{x^{4}+5}{x^{3}\left(x^{2}+6\right)} \equiv \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}+\frac{f(x)}{x^{2}+6}$, then
$$x^{4}+5 \equiv A x^{2}\left(x^{2}+6\right)+B x\left(x^{2}+6\right)+C\left(x^{2}+6\right)+x^{3} f(x) \cdots (1)$$
Putting $x=0$ yields $5=6 C\Rightarrow C=\dfrac{5}{6}.$
Rearranging (1) yields
$$
\begin{aligned}
x^{4}+5-\frac{5}{6}\left(x^{2}+6\right) & \equiv A x^{2}\left(x^{2}+6\right)+B x\left(x^{2}+6\right)+x^{3} f(x) \\
\frac{1}{6} x\left(6 x^{2}-5\right) & \equiv A x\left(x^{2}+6\right)+B\left(x^{2}+6\right)+x^{2} f(x)\cdots (2)
\end{aligned}
$$
Putting $x=0$ in (2) yields $B=0$.
Rearranging (2) yields
$$
\frac{1}{6}\left(6 x^{2}-5\right)=A\left(x^{2}+6\right)+x f(x) \cdots (3)
$$
Putting $x=0$ in (3) yields $ \displaystyle -\frac{5}{6}=6 A \Rightarrow A=-\frac{5}{36}$.
$\begin{aligned}\text{ Then rearranging (3) yields } & \\ x^{2} f(x) &=\frac{1}{6}\left(6 x^{2}-5\right)+\frac{5}{36}\left(x^{2}+6\right) =\frac{41}{6} x^{2} \\ \therefore \quad f(x) &=\frac{41}{6} \\ \therefore \frac{x^{4}+5}{x^{3}\left(x^{2}+6\right)} &=-\frac{5}{36 x}+\frac{5}{6 x^{3}}+\frac{41}{6\left(x^{2}+6\right)} .\end{aligned}$
Question 2:
For this particular fraction, we can do it in a simpler way. $$
\begin{aligned}
\frac{5}{\left(x^{2}-16\right)^{2}} &=5\left[\frac{1}{(x+4)(x-4)}\right]^{2} \\
&=5\left[\frac{1}{8}\left(\frac{1}{x-4}-\frac{1}{x+4}\right)\right]^{2} \\
&=\frac{5}{64}\left[\frac{1}{(x-4)^{2}}-\frac{2}{(x-4)(x+4)}+\frac{1}{(x+4)^{2}}\right] \\
&=\frac{5}{64}\left[\frac{1}{(x-4)^{2}}-\frac{1}{4}\left(\frac{1}{x-4}-\frac{1}{x+4}\right)+\frac{1}{(x+4)^{2}}\right] \\
&=\frac{5}{256}\left[\frac{4}{(x-4)^{2}}-\frac{1}{x-4}+\frac{1}{x+4}+\frac{4}{(x+4)^{2}}\right]
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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About Factorization I have some issues understanding factorization.
If I have the expression $x^{2}-x-7$ then (I was told like this) I can put this expression equal to zero and then find the solutions with the quadratic formula, so it gives me $x_{0,1}= 1 \pm 2\sqrt{2}$ then $$x^{2}-x-7 = (x-1-2\sqrt{2})(x-1+2\sqrt{2}).$$
That is correct I have checked it.
Now for the expression $3x^{2}-x-2$ if I do the same I have $x_{0} = 1$ and $x_1=\frac{-2}{3}$ so I would have $$3x^{2}-x-2 = (x-1)(x+\frac{2}{3})$$
but this is not correct since $(x-1)(x+\frac{2}{3}) = \frac{1}{3}(3x^{2}-x-2)$,
the correct factorization is $3x^{2}-x-2 = (3x+2)(x-1)$.
So I guess finding the roots of a quadratic expression is not sufficient for factorizing.
| For equations of the form $ax^2 + bx + c$ you can use the quadratic equation $x = \frac{-b \pm \sqrt{b^2-4ac }}{2a}$ to help you factor, like you were doing. The adjustment you need to make is to multiply your factorization by a.
In the example you had, $3x^2 -x - 2$, you found the roots were x = 1 and x $= \frac{-2}{3}$. If you do $3*(x+\frac{2}{3})(x−1)$, then you have (3x+2)(x−1) when you multiply. This works in general. I hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find bases for kernel and image of T where $T: P_2 \to M_2$ T is defined as $$T:P_2(\mathbb R) \to M_2 (\mathbb R) \ \text{where} \ T(ax^2 +bx+c)=\begin{pmatrix}-2a +c & b+c\\-3b-3c&6a-3c\\ \end{pmatrix}$$ and I need t find bases for $ Im(T) $ and $ker(T)$
I started with $im(T)$ and got
$$Im(T) = \begin{pmatrix}-2a +c & b+c\\-3b-3c&6a-3c\\ \end{pmatrix} \\
= a\begin{pmatrix}-2 & 0\\0&6\\ \end{pmatrix} +b \begin{pmatrix}0 &1\\-3 &0 \\ \end{pmatrix} +c \begin{pmatrix}1 & 1\\-3 & -3\\ \end{pmatrix} $$
using $$\begin{pmatrix}-2 & 0& 0&6\\
0 & 1& -3&0\\
1 &1& -3&-3\\
\end{pmatrix} $$ reduced to
$$\begin{pmatrix}1 & 0& 0&-3\\
0 & 1& -3&0\\
0 &0& 0&1\\
\end{pmatrix} $$
Is the basis for $Im(T)=\{(1, 0 ,0 ,3),(0,1,-3,0),(0,0,0,1)\}$ ??
$$.....................................................................
$$
To find $Ker(T) = \{(a,b,c) where \begin{pmatrix}-2a+c & b+c\\
-3b-3c& 6a-3c\\
\end{pmatrix} =0 \}$
by solving
$$(-2a+c)(6a-3c)-(b+c)(-3b-3c)=0$$
$$(2a-c)^2 -(b+c)^2=0$$
$$2a+b=0 \ \text{or} \ 2a-b-2c=0$$
but now what??
| Let the basis of $P_2$ is $(1,x,x^2)$ and the basis of $M_2$ is $(a_{11}, a_{12}, a_{21}, a_{22})$
Where
$a_{11}= \begin{pmatrix}1&0 \\ 0&0 \end{pmatrix}$
$a_{12}= \begin{pmatrix}0&1 \\ 0&0 \end{pmatrix}$
$a_{21}= \begin{pmatrix}0&0 \\ 1&0 \end{pmatrix}$
$a_{22}= \begin{pmatrix}0&0 \\ 0&1 \end{pmatrix}$
Then
$T(x^2)=-2\times a_{11}+6\times a_{22}$
$T(x^3)=-2\times a_{11}+6\times a_{22}$
$T(x^2)=1\times a_{12}+-3\times a_{21}$
$T(1)=1\times a_{11}+1\times a_{12}+-3\times a_{21}+(-3)\times a_{22}$
Therefore the matrix of Linear transformation is:
$$\begin{pmatrix}-2 & 0& 1\\0&1&1\\ 0&-3&-3\\6&0&-3\end{pmatrix} \\ $$
Now find basis of column space of this matrix.( i.e, basis of image of T)
Using elementry row transformation this matrix will reduce to :
$$\begin{pmatrix}-2 & 0& 1\\0&1&1\\ 0&0&0\\0&0&0\end{pmatrix} \\ $$
So basis of image of $T$ will be $((-2 ,0, 0, 6), (0 ,1, 0 ,0))$
Ans the basis of null space will be $ ((1/2,-1,1)$
| {
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Prove that $\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$ Without using Mathematical Induction, prove that $$\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$$
I am unable to solve this problem and don't know where to start. Please help me to solve this problem using the laws of inequality. It is a problem of Inequality.
Edit: $n$ is a positive integer such that $n>1$.
| Any statement that needs to be proved for all $n\in\mathbb{N}$, will need to make use of induction at some point.
We have
\begin{align}
S_n & = \sum_{k=1}^n \dfrac1{n+2k-1} = \dfrac12 \left(\sum_{k=1}^n \dfrac1{n+2k-1} + \underbrace{\sum_{k=1}^n \dfrac1{3n-2k+1}}_{\text{Reverse the sum}}\right)\\
& = \dfrac12 \sum_{k=1}^n \dfrac{4n}{(n+2k-1)(3n-2k+1)} = \sum_{k=1}^n \dfrac{2n}{(n+2k-1)(3n-2k+1)}
\end{align}
From AM-GM, we have
$$4n = (n+2k-1) + (3n-2k+1) \geq 2 \sqrt{(n+2k-1)(3n-2k+1)}$$ This gives us that
$$\dfrac1{(n+2k-1)(3n-2k+1)} \geq \dfrac1{4n^2}$$
Hence, we obtain that
$$S_n = \dfrac12 \sum_{k=1}^n \dfrac{4n}{(n+2k-1)(3n-2k+1)} \geq \sum_{k=1}^n \dfrac{2n}{4n^2} = \dfrac12$$
Also, just to note, every step in the above solution requires induction.
Also, as @MartinR rightly points out, the inequality is strictly in our case for almost all $k$ except for $k=\dfrac{n+1}2$ (since equality holds only when $n+2k-1 = 3n-2k+1 \implies k = \dfrac{n+1}2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Proving $\tan A=\frac{1-\cos B}{\sin B} \;\implies\; \tan 2A=\tan B$
If $\tan A=\dfrac{1-\cos B}{\sin B}$, prove that $\tan 2A=\tan B$.
My effort:
Here
$$\tan A=\frac{1-\cos B}{\sin B}$$
Now
$$\begin{align}\text{L.H.S.} &=\tan 2A \\[4pt]
&=\frac{2\tan A}{1-\tan ^2A} \\[6pt]
&=\frac{(2-2\cos B)\over\sin B}{1-\frac{(1-\cos B)^2}{\sin^2 B}}
\end{align}$$
On simplification from here, I could not get the required R.H.S.
| Continuing from your last step,
$$
LHS=\frac{\frac{2-2\cos B}{\sin B}}{1-\frac{(1-\cos B)^2}{\sin^2B}}\\
=\frac{\frac{2-2\cos B}{\sin B}}{\frac{\sin^2B-(1-\cos B)^2}{\sin^2B}} \\
=\frac{2-2\cos B}{\frac{\sin^2B-1-\cos^2B+2\cos B}{\sin B}}
$$
and then use that $1=\sin^2+\cos^2B$ to get
$$
LHS= \frac{2(1-\cos B)(\sin B)}{\sin^2B-\sin^2B-\cos^2B-\cos^2B+2\cos B}\\
=\frac{2(1-\cos B)(\sin B)}{2\cos B(1-\cos B)}\\
=\frac{\sin B}{\cos B}=\tan B=RHS
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find $\lim_{n \rightarrow \infty}\frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$ Find:
$$\lim_{n \rightarrow \infty} \frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$$
The sequence $\frac{1}{nx^2 \log{(1+ \frac{x}{n})}}=\frac{1}{x^3 \frac{\log{(1+ \frac{x}{n})}}{\frac{x}{n}}}$ converges pointwise to $\frac{1}{x^3}$. So if we could apply Lebesgue's Dominated Convergence Theorem, we have:
$\lim_{n \rightarrow \infty} \frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}=\lim_{n \rightarrow \infty} \int_{1}^{\infty} \frac{\mathrm dx}{x^3}=\frac{1}{2}$
I have a problem with finding a majorant. Could someone give me a hint?
| I showed in THIS ANSWER, using only Bernoulli's Inequality the sequence $\left(1+\frac xn\right)^n$ is monotonically increasing for $x>-n$.
Then, we can see that for $x\ge 1$ and $n\ge1$, the sequence $f_n(x)$ given by
$$f_n(x)=n\log\left(1+\frac xn\right)$$
is also monotonically increasing. Therefore, a suitable dominating function is provided simply by the inequality
$$\frac{1}{n\log\left(1+\frac xn\right)}\le \frac{1}{\log(1+x)}\le \frac{1}{\log(2)}$$
Therefore, we have
$$\frac{1}{nx^2\log\left(1+\frac xn\right)}\le \frac{1}{x^2\log(2)}$$
Using the dominated convergence theorem, we can assert that
$$\begin{align}
\lim_{n\to \infty}\int_1^\infty \frac{1}{nx^2\log\left(1+\frac xn\right)}\, dx&=\int_1^\infty \lim_{n\to \infty}\left(\frac{1}{nx^2\log\left(1+\frac xn\right)}\right)\,dx\\\\
&=\int_1^\infty\frac{1}{x^3}\,dx\\\\
&=\frac12.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Intergate $\int \frac{x}{(x^2-3x+17)^2}\ dx$
$$\int \frac{x}{(x^2-3x+17)^2}\ dx$$
My attempt:
$$\int \frac{x}{(x^2-3x+17)^2}\ dx=\int \frac{x}{\left((x-\frac{3}{2})^2+\frac{59}{4}\right)^2}\ dx$$
let $u=x-\frac{3}{2}$
$du=dx$
$$\int \frac{u+\frac{3}{2}}{\left((u)^2+\frac{59}{4}\right)^2}\ du$$
How can I continue from here?
| Notice, $$\int \frac{x}{(x^2-3x+17)^2}\ dx$$
$$=\frac{1}{2}\int \frac{(2x-3)+3}{(x^2-3x+17)^2}\ dx$$
$$=\frac{1}{2}\left(\int \frac{(2x-3)}{(x^2-3x+17)^2}\ dx+3\int \frac{1}{(x^2-3x+17)^2}\ dx\right)$$
$$=\frac{1}{2}\left(\int \frac{d(x^2-3x+17)}{(x^2-3x+17)^2}+3\int \frac{1}{\left(\left(x-\frac{3}{2}\right)^2+\frac{59}{4}\right)^2}\ dx\right)$$
$$=\frac{1}{2}\left( -\frac{1}{x^2-3x+17}+3\int \frac{d\left(x-\frac 32\right)}{\left(\left(x-\frac{3}{2}\right)^2+\frac{59}{4}\right)^2}\right)$$
Now, the integral can be solved using reduction formula: $\int\frac{dx}{(a+x^2)^n}\ dx=\frac{1}{2a(n-1)}\left(\frac{x}{(a+x^2)^{n-1}}+(2n-3)\int \frac{1}{(a+x^2)^{n-1}}\ dx\right)$ as follows
$$\int \frac{d\left(x-\frac 32\right)}{\left(\left(x-\frac{3}{2}\right)^2+\frac{59}{4}\right)^2}$$
$$=\frac{1}{2\left(\frac{59}{4}\right)(2-1)}\left(\frac{x-\frac32}{\left(x-\frac 32\right)^2+\frac{59}{4}}+(2\cdot 2-3)\int \frac{d\left(x-\frac 32\right)}{\left(x-\frac 32\right)^2+\frac{59}{4}}\right)$$
$$=\frac{2}{59}\left(\frac{x-\frac32}{x^2-3x+17}+\frac{1}{\frac{\sqrt{59}}{2}}\tan^{-1}\left(\frac{x-\frac 32}{\frac{\sqrt{59}}{2}}\right)\right)$$
$$=\frac{2}{59}\left(\frac{2x-3}{2(x^2-3x+17)}+\frac{2}{\sqrt{59}}\tan^{-1}\left(\frac{2x-3}{\sqrt{59}}\right)\right)$$
Now, substitute the above value of integral & simplify to get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove $1^3+5^3+3^3=153,16^3+50^3+33^3=165033,166^3+500^3+333^3=166500333,\cdots$? I saw on facebook some image on which these identities that I am going to write below are labeled as "amazing math fact" and on the image there are these identities:
$1^3+5^3+3^3=153$
$16^3+50^3+33^3=165033$
$166^3+500^3+333^3=166500333$
$1666^3+5000^3+3333^3=166650003333$
and then it is written under these identities "and so on and on and on and on!" which suggests that for every $k \in \mathbb N$ we shuld have
$(1 \cdot 10^k + \sum_{i=0}^{k-1} 6 \cdot 10^i)^3 + (5 \cdot 10^k)^3 + (\sum_{i=0}^{k} 3 \cdot 10^i)^3=16...650...03...3$
(on the right hand side of the above stated identity the number of times that number $6$ is shown up is $k-1$, the number of times that number $0$ is shown up is $k-1$ and the number of times that number $3$ is shown up is $k$)
This problem seems attackable with mathematical induction but I would like to see how it could be proved without using mathematical induction in any step(s) of the proof.
| This deserves a shorter proof. Call $x=10^{k+1}$. Then we have:
$$\begin{array}{rcl}166\ldots 666&=&\frac{x}{6}-\frac{2}{3}\\500\ldots 000&=&\frac{x}{2}\\333\ldots 333&=&\frac{x}{3}-\frac{1}{3}\end{array}$$
Now: the sum of the cubes on the left-hand side is:
$$\color{red}{166\ldots 666^3+500\ldots 000^3+333\ldots 333^3}=\left(\frac{x}{6}-\frac{2}{3}\right)^3+\left(\frac{x}{2}\right)^3+\left(\frac{x}{3}-\frac{1}{3}\right)^3=\frac{1}{6^3}((x-4)^3+(3x)^3+(2x-2)^3)=\frac{1}{6^3}(36x^3-36x^2+72x-72)=\color{red}{\frac{1}{6}(x^3-x^2+2x-2)}$$
On the other hand, the big number on the right-hand side is:
$$\color{red}{1666\ldots666\cdot x^2+500\ldots000\cdot x+333\ldots333}=\left(\frac{x}{6}-\frac{2}{3}\right)\cdot x^2+\frac{x}{2}\cdot x+\frac{x}{3}-\frac{1}{3}=\frac{1}{6}(x^3-4x^2+3x^2+2x-2)=\color{red}{\frac{1}{6}(x^3-x^2+2x-2)}$$
which is the same value.
(Note how we used multiplication by $x^2$ and $x$ to "shift those big numbers to the left".)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Find the limit of fraction involving logarithms I am looking for a way to prove the following limit for integer $x$s:
$$\lim_{x\to\infty}{\frac{\log(x+2)-\log(x+1)}{\log(x+2)-\log(x)}}=\frac{1}{2}$$
I could find the result by using a computer program but I cannot formally establish the above equality.
| We will use the fundamental limit $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1\tag{1}$$ We have
\begin{align}
L &= \lim_{x \to \infty}\frac{\log(x + 2) - \log(x + 1)}{\log(x + 2) - \log x}\notag\\
&= \lim_{x \to \infty}\dfrac{\log\left(\dfrac{x + 2}{x + 1}\right)}{\log\left(\dfrac{x + 2}{x}\right)}\notag\\
&= \lim_{x \to \infty}\dfrac{\log\left(1 + \dfrac{1}{x + 1}\right)}{\log\left(1 + \dfrac{2}{x}\right)}\notag\\
&= \lim_{x \to \infty}\dfrac{\log\left(1 + \dfrac{1}{x + 1}\right)}{\dfrac{1}{x + 1}}\cdot\dfrac{\dfrac{1}{x + 1}}{\dfrac{2}{x}}\cdot\dfrac{\dfrac{2}{x}}{\log\left(1 + \dfrac{2}{x}\right)}\notag\\
&= \frac{1}{2}\lim_{y \to 0}\frac{\log(1 + y)}{y}\cdot\lim_{x \to \infty}\frac{x}{x + 1}\cdot\lim_{z \to 0}\frac{z}{\log(1 + z)}\text{ (putting }y = 1/(x + 1), z = 2/x)\notag\\
&= \frac{1}{2}\notag
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to derive the equation of tangent to an arbitrarily point on a ellipse?
Show that the equation of a tangent in a point $P\left(x_0, y_0\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, could be written as: $$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$$
I've tried implicit differentiation $\to \frac{2x\frac{d}{dx}}{a^2}+\frac{2y\frac{d}{dy}}{b^2} = 0$, but not sure where to go from here. Substituting $P$ doesn't seem to help me much, and solving for $y$ from original equation seems to cause me more trouble than help. Please don't give me the solution, rather just give a slight hint or two:)
Thanks in advance!
Edit:
Solving for $\frac{dy}{dx}$ and putting it into point slope gives me:
$\frac{d}{dx} \left[\frac{x^2}{a^2} + \frac{y^2}{b^2}\right] = \frac{d}{dx}\left[1\right] \to \frac{2x}{a^2} + \frac{y}{b^2}\frac{dy}{dx}= 0\to \frac{dy}{dx}=-\frac{xb^2}{ya^2} \to \frac{dy}{dx}(P)=-\frac{x_0b^2}{y_0a^2}$
Then we get:
$y-y_0=\frac{dy}{dx}\left(x-x_0\right)\to y = -\frac{x_0b^2}{y_0a^2}(x-x_0)+y_0\to y = \frac{x_0^2b^2}{y_0a^2}-\frac{xx_0b^2}{y_0a^2} + y_0\to /:b^2,*y_0\to \frac{yy_0}{b^2}+\frac{xx_0}{a^2}=\frac{y_0^2}{b^2}$
Which looks close, but not exactly the expression i wanted. Where is the error?
| We first differentiate the ellipse equation w.r.t. $x$ to find the slope of the tangent $m$,
$$\dfrac{2x_0}{a^2}+\dfrac{2y_0m}{b^2}=0$$
$$m=-\dfrac{x_0}{y_0}\cdot\dfrac{b^2}{a^2}$$
Thus, the tangent is given by
$$\color{darkgreen}{y-y_0}=\color{blue}{-}\dfrac{\color{blue}{x_0}}{\color{darkgreen}{y_0}}\cdot\dfrac{\color{darkgreen}{b^2}}{\color{blue}{a^2}}\color{blue}{(x-x_0)}$$
$$\color{darkgreen}{\dfrac{yy_0}{b^2}-\dfrac{y_0^2}{b^2}}=\color{blue}{\dfrac{x_0^2}{a^2}-\dfrac{xx_0}{a^2}}$$
$$\dfrac{xx_0}{a^2}+\dfrac{yy_0}{b^2}=\dfrac{x_0^2}{a^2}+\dfrac{y_0^2}{b^2}=1$$
as desired.
Hope this helps. :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Numerical evaluation of first and second derivative We start with the following function $g: (0,\infty)\rightarrow [0,\infty)$,
$$ g(x)=x+2x^{-\frac{1}{2}}-3.$$ From this function we need a 'smooth' square-root. Thus, we check $g(1)=0$,
$$g'(x)=1-x^{-\frac{3}{2}},$$ $g'(1)=0$ and
$$g''(x)=\frac{3}{2}x^{-\frac{5}{2}}\geq 0.$$
Therefore, we can define the function $f: (0,\infty)\rightarrow \mathbb{R}$,
$$f(x)=\operatorname{sign}(x-1)\sqrt{g(x)}.$$
We can compute the first derivatives as
$$
f'(x)=\frac{1}{2} \frac{g'(x)}{f(x)}=\frac{1}{2}\frac{1-x^{-\frac{3}{2}}}{\operatorname{sign}(x-1)\sqrt{x+2x^{-\frac{1}{2}}-3}}
$$
and second derivatives as
$$
f''(x)=\frac{1}{2} \frac{g''(x)f(x)-g'(x)f'(x)}{\bigl(f(x)\bigr)^2}\\
=\frac{1}{2}\frac{g''(x)-\frac{\bigl(g'(x)\bigr)^2}{2g(x)}}{f(x)}\\
=\frac{1}{2}\frac{\frac{3}{2}x^{-\frac{5}{2}}-\frac{\bigl(1-x^{-\frac{3}{2}}\bigr)^2}{2\bigl(x+2x^{-\frac{1}{2}}-3\bigr)}}{\operatorname{sign}(x-1)\sqrt{x+2x^{-\frac{1}{2}}-3}}
$$
where we used $f'(x)$ from above, reduced by $f(x)$ and used $f^2(x)=g(x)$.
For $x=1$ the first and the second derivatives of $f$ are of the type $\frac{0}{0}$.
I need a numerical stable evaluation of this derivatives.
But we have numerical cancellation especially in the nominators.
A linear Taylor-polynomial at $x=1$ is possible for $f'(x)$ but the computation of $f'''(1)$ by hand is time-consuming.
*
*Is there a better formulation for $f(x)$?
*Is there an easy way to compute the coefficients of the Taylor-polynomials at $x=1$?
*(Edit 2:) How can I evaluate $f'(x)$ and $f''(x)$ numerically stable for $x\in (0,\infty)$?
Edit 1:
The Taylor-polynomial of degree 1 for $f'(x)$ at $x=1$ is
$$
f'(x)=\frac{\sqrt{3}}{2} - \frac{5}{4\sqrt{3}}(x-1)+\mathcal{O}\bigl(\lvert x-1\rvert^2\bigr)
$$
| Setting $x^{1/2} = 1 + u$ (the content of Octania's comment) gives, for $x > 0$,
$$
g(x) = x + 2x^{-1/2} - 3
= \frac{x^{3/2} - 3x^{1/2} + 2}{x^{1/2}}
= \frac{(1 + u)^{3} - 3(1 + u) + 2}{1 + u}
= \frac{u^{2} (3 + u)}{1 + u}.
$$
Your definition of $f$ amounts to
$$
f(x) = u \sqrt{\frac{3 + u}{1 + u}}
= u \sqrt{1 + \frac{2}{1 + u}}.
$$
This is smooth (and numerically stable) near $u = 0$ (i.e., near $x = 1$).
If you need a Taylor expansion, the geometric series and binomial theorem give the first few terms easily, and (in principle) as many terms as you like:
\begin{align*}
1 + \frac{2}{1 + u}
&= 3\bigl(1 - \tfrac{2}{3}(u - u^{2} + u^{3} + \cdots)\bigr), \\
\sqrt{1 + v}
&= 1 + \tfrac{1}{2} v - \tfrac{1}{8} v^{2} + \tfrac{1}{16} v^{3} - \cdots, \\
\sqrt{1 + \frac{2}{1 + u}}
&= \sqrt{3}\left[1 - \tfrac{1}{2} \cdot \tfrac{2}{3}(u - u^{2} + u^{3} + \cdots)
- \tfrac{1}{8} \cdot \tfrac{4}{9}(u - u^{2} + u^{3} + \cdots)^{2} + \cdots\right].
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
An inequality with the sinus in a triangle. I have solved this problem in a way, rather "inspired". I would like to have a solution found an easier way but I was unable so far.
Let $A,B,C$ the angles of a triangle $\triangle {ABC}$; prove that
$$(\sin^2 A+\sin^2 B+\sin^2 C)\le \frac14 \left(\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}\right) (\sin A+\sin B+\sin C)$$
| Wee need to prove that
$$\sum\limits_{cyc}\frac{4S^2}{b^2c^2}\leq\frac{1}{4}\sum\limits_{cyc}\frac{bc}{2S}\sum\limits_{cyc}\frac{2S}{bc}$$ or
$$\frac{(a^2+b^2+c^2)\sum\limits_{cyc}(2a^2b^2-a^4)}{a^2b^2c^2}\leq\frac{(ab+ac+bc)(a+b+c)}{abc}$$ or
$$abc(ab+ac+bc)\geq(a^2+b^2+c^2)(a+b-c)(a+c-b)(b+c-a)$$ or
$$\sum\limits_{cyc}(a^5-a^4b-a^4c+2a^3bc-a^2b^2c)\geq0,$$
which is true even for all non-negatives $a$, $b$ and $c$.
Indeed, $\sum\limits_{cyc}(a^5-a^4b-a^4c+a^3bc)\geq0$ is true by Schur and
$\sum\limits_{cyc}(a^3bc-a^2b^2c)=\frac{1}{2}abc\sum\limits_{cyc}(a-b)^2\geq0$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $a^5 + b^5 + c^5$
Suppose we have numbers $a,b,c$ which satisfy the equations
$$a+b+c=3,$$
$$a^2+b^2+c^2=5,$$
$$a^3+b^3+c^3=7.$$
How can I find $a^5 + b^5 + c^5$?
I assumed we are working in $\Bbb{C}[a,b,c]$. I found a reduced Gröbner basis $G$:
$$G = \langle a+b+c-3,b^2+bc+c^2-3b-3c+2,c^3-3c^2+2c+\frac{2}{3} \rangle$$
I solved the last equation for $c$ and got 3 complex values. When I plug into the 2nd equation $(b^2+bc+c^2-3b-3c+2)$ I get a lot of roots for $b$, and it is laborious to plug in all these values.
Is there a shortcut or trick to doing this? The hint in the book says to use remainders. I computed the remainder of $f = a^5 + b^5 + c^5$ reduced by $G$:
$$\overline{f}^G = \frac{29}{3}$$
How can this remainder be of use to me?
Thanks. (Note: I am using Macaualay2)
| You have to use Newton Identities. See https://en.wikipedia.org/wiki/Newton%27s_identities
In general if you have $n$ variables $x_1\ldots.x_n$, define the polynomials
$$p_k(x_1,\ldots,x_n)=\sum_{i=1}^nx_i^k = x_1^k+\cdots+x_n^k,$$
and
\begin{align}
e_0(x_1, \ldots, x_n) &= 1,\\
e_1(x_1, \ldots, x_n) &= x_1 + x_2 + \cdots + x_n,\\
e_2(x_1, \ldots, x_n) &= \textstyle\sum_{1\leq i<j\leq n}x_ix_j,\\
e_n(x_1, \ldots, x_n) &= x_1 x_2 \cdots x_n,\\
e_k(x_1, \ldots, x_n) &= 0, \quad\text{for}\ k>n.\\
\end{align}
Then
\begin{align}
p_1 &= e_1,\\
p_2 &= e_1p_1-2e_2,\\
p_3 &= e_1p_2 - e_2p_1 + 3e_3 ,\\
p_4 &= e_1p_3 - e_2p_2 + e_3p_1 - 4e_4, \\
& {}\ \ \vdots\\
\\
e_0 &= 1,\\
e_1 &= p_1,\\
e_2 &= \frac{1}{2}(e_1 p_1 - p_2),\\
e_3 &= \frac{1}{3}(e_2 p_1 - e_1 p_2 + p_3),\\
e_4 &= \frac{1}{4}(e_3 p_1 - e_2 p_2 + e_1 p_3 - p_4),\\
& {} \ \ \vdots
\end{align}
In your case you have only 3 variables. Using the formulas above compute
\begin{align}
e_1 &=p_1=3, \\
e_2 &=2,\\
e_3 &=-\frac{2}{3}\\
e_4 &=0\\
e_5 &=0\\
\end{align}
Then compute $p_4=9$ and $$p_5=\frac{29}{3}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Show that $a_n = 1 + \frac{1}{2} + \frac{1}{3} +\dotsb+ \frac{1}{n}$ is not a Cauchy sequence
Let $$
a_n = 1 + \frac{1}{2} + \frac{1}{3} + \dotsb + \frac{1}{n}
\quad (n \in \mathbb{N}).
$$
Show that $a_n$ is not a Cauchy sequence even though
$$
\lim_{n \to \infty} a_{n+1} - a_n = 0
$$
(Therefore $a_n$ does not have a limit).
| We use the definition of Cauchy sequence to show the sequence $(a_n)$ is not Cauchy.
Let $\epsilon=1/2$. We will show that there does not exist an $m$ such that for any $n\gt m$, we have $|a_n-a_m|\lt \epsilon$.
For let $m$ be given, and let $2^k$ be the smallest power of $2$ that is $\gt m$. Let $n=2^{k+1}-1$. Then
$$a_n-a_m\ge \frac{1}{2^k}+\frac{1}{2^{k}+1}+\cdots +\frac{1}{2^{k+1}-1}\gt \frac{1}{2}.$$
The fact that the above sum is $\gt \frac{1}{2}$ follows from the fact that the sum has at least $2^k$ terms, each $\gt \frac{1}{2^{k+1}}$
The fact that $\lim_{n\to \infty}(a_{n+1}-a_n)=0$ is clear, since $a_{n+1}-a_n=\frac{1}{n+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Triangle angles.
For $\vartriangle$ABC it is given that $$\frac1{b+c}+\frac1{a+c}=\frac3{a+b+c}$$ Find the measure of angle $C$.
This is a "challenge problem" in my precalculus book that I was assigned. How do I find an angle from side lengths like this? I have tried everything I can. I think I may need to employ the law of cosines or sines. Thanks.
| Multiply both sides by $a+b+c$ to get $$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}=3$$
We get
$$\frac{a}{b+c}+\frac{b}{a+c}=1$$
$$a(a+c)+b(b+c)=(a+c)(b+c)$$
$$a^2+ac+b^2+bc=ab+ac+bc+c^2$$
$$a^2+b^2=ab+c^2$$
$$a^2-ab+b^2=c^2$$
By the cosine law, we have $a^2-2\cos(C)ab+b^2=c^2$. Hence $\cos(C)=\frac12$, so $C=60^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Find the sum of all $abc$
Let $T$ be the set of all triplets $(a,b,c)$ of integers such that $1 \leq a < b < c \leq 6$. For each triplet $(a,b,c)$ in $T$ , take number $a \cdot b \cdot c$. Add all these numbers corresponding to all the triplets in $T$. Prove that the answer is divisible by $7$.
Attempt
Let $S = \displaystyle \sum_{b < c} bc $. Then with no restriction on $a$ we have $M = (1+2+3+4+5+6)S = 21S$. This is the sum of the cases where $a < b, a = b, $ and $a > b$. Now how can I use this to find the case where $a<b$?
| A triple $(a, b, c)$ of integers such that $1 \leq a < b < c \leq 6$ is one of the $\binom{6}{3} = 20$ three-element subsets of the set $\{1, 2, 3, 4, 5, 6\}$ because once we choose a subset there is only one way of placing the elements in increasing order. They are
\begin{array}{c c}
(a, b, c) & abc\\
(1, 2, 3) & 6\\
(1, 2, 4) & 8\\
(1, 2, 5) & 10\\
(1, 2, 6) & 12\\
(1, 3, 4) & 12\\
(1, 3, 5) & 15\\
(1, 3, 6) & 18\\
(1, 4, 5) & 20\\
(1, 4, 6) & 24\\
(1, 5, 6) & 30\\
(2, 3, 4) & 24\\
(2, 3, 5) & 30\\
(2, 3, 6) & 36\\
(2, 4, 5) & 40\\
(2, 4, 6) & 48\\
(2, 5, 6) & 60\\
(3, 4, 5) & 60\\
(3, 4, 6) & 72\\
(3, 5, 6) & 90\\
(4, 5, 6) & 120
\end{array}
As you can check, the sum of the products is divisible by $7$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all square numbers $n$ such that $f(n)$ is a square number Find all the square numbers $n$ such that $ f(n)=n^3+2n^2+2n+4$ is also a perfect square.
I have tried but I don't know how to proceed after factoring $f(n)$ into $(n+2)(n^2+2)$. Please help me.
Thanks.
| Your factoring seems to appear difficult to solve.
Instead say $n=t^2$.
If $t>1$, note that $f(t^2)=t^6+2t^4+2t^2+4$, and that $(t^3+t+1)^2=t^6+2t^4+t^2+2t^3+2t+1>f(t^2) >t^6+2t^4+t^2=(t^3+t)^2$ (since $2t^3-t^2+2t-3>0$ from here)
If $t<-1$, note that $(t^3+t-1)^2=t^6+2t^4+t^2-2t^3-2t+1>f(t^2) >t^6+2t^4+t^2=(t^3+t)^2$ (since $-2t^3-t^2-2t-3>0$ from here)
So $t=-1,0,1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $x^2+y^2-xy-x-y+1=0$ ($x,y$ real) then calculate $x+y$ If $x^2+y^2-xy-x-y+1=0$ ($x,y$ real) then calculate $x+y$
Ideas for solution include factorizing the expression into a multiple of $x+y$ and expressing the left hand side as a sum of some perfect square expressions.
| We have
$$2(x^2+y^2-xy-x-y+1)=(x-y)^2+(x-1)^2+(y-1)^2.$$ The right-hand side (for real $x$ and $y$) is equal to $0$ if and only if $x=y=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the limit $\lim_{x\to \infty}( \sqrt{4x^2+x}-2x)$
Evaluate :$$\lim_{x\to \infty} (\sqrt{4x^2+x}-2x)$$
$$\lim_{x\to \infty} (\sqrt{4x^2+x}-2x)=\lim_{x\to \infty} \left[(\sqrt{4x^2+x}-2x)\frac{\sqrt{4x^2+x}+2x}{\sqrt{4x^2+x}+2x}\right]=\lim_{x\to \infty}\frac{{4x^2+x}-4x^2}{\sqrt{4x^2+x}+2x}=\lim_{x\to \infty}\frac{x}{\sqrt{4x^2+x}+2x}$$
Using L'Hôpital $$\lim_{x\to \infty}\frac{1}{\frac{8x+1}{\sqrt{4x^2+x}}+2}$$
What should I do next?
| With the substitution $x=1/t$ (under the unrestrictive condition that $x>0$) you get
$$
\lim_{x\to \infty}(\sqrt{4x^2+x}-2x)=
\lim_{t\to0^+}\left(\sqrt{\frac{4}{t^2}+\frac{1}{t}}-\frac{2}{t}\right)=
\lim_{t\to0^+}\frac{\sqrt{4+t}-2}{t}
$$
which is the derivative at $0$ of $f(t)=\sqrt{4+t}$; since
$$
f'(t)=\frac{1}{2\sqrt{4+t}}
$$
you have
$$
f'(0)=\frac{1}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are, of course, brute force techniques for showing this, but I'm hoping for something elegant.
| If we remember the well-known formulas
$(x+y)^3 - (x^3 + y^3) = 3xy(x+y)$
$(x+y)^5 - (x^5 + y^5) = 5xy(x+y)(x^2 + xy + y^2)$
$(x+y)^7 - (x^7 + y^7) = 7xy(x+y)(x^2 + xy + y^2)^2$
then your identity becomes the observation that there are equal powers of the $xy(x+y)$ and $(x^2+xy+y^2)$ factors on both sides. (Under the change of notation $a,b,c \to -(x+y),x,y$)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
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} |
$\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ without l'Hopital's rule; how is my procedure wrong?
please explain why my procedure is wrong i am not able to find out??
I know the property limit of product is product of limits (provided limit exists and i think in this case limit exists for both the functions).
The actual answer for the given question is $\frac{1}{2}\log(2)$.
My course book has shown that don't use this step but has not given the reason.
AND Please TELL why i am WRONG
| @vim suggested Taylor expansions:
\begin{align}\lim _{x \to 0}\frac{2^{\tan x} - 2^{\sin x}}{x^2 \sin x}
&=\lim _{x \to 0}\frac{2^{x+\frac{x^3}{3}+O(x^5)}-2^{x-\frac{x^3}{6}+O(x^5)}}{x^2(x+\frac{x^3}{3}+O(x^5))}\\
&=\lim _{x \to 0}\frac{\left(2^{x+O(x^5)}\right)\left(2^{\frac{x^3}{3}}-2^{\frac{x^3}{6}}\right)}{x^3\left(1+O(x^5)\right)}\\
&=\lim _{x \to 0}\frac{2^{x-\frac{x^3}{6}+O(x^5)}}{1+O(x^5)}\cdot\lim_{x\to0}\frac{2^{\frac{x^3}{2}} - 1}{x^3 }\\
&=1\cdot\lim_{x\to0}\frac12\cdot\frac{2^{\frac{x^3}{2}} - 1}{\frac{x^3}{2} }\\
&=\frac12\cdot\lim_{u\to0}\cdot\frac{2^{u} - 1}{u }\\
&=\frac12\cdot \ln 2
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate this limit? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to 0+}\left(\frac{\left[\ln\left(\frac{5+x^2}{5+4x}\right)\right]^6\ln\left(\frac{5+x^2}{1+4x}\right)}{\sqrt{5x^{10}+x^{11}}-\sqrt{5}x^5}\right)$$
| Expand the denominator
$$
\sqrt{5}x^5 (\sqrt{1+x/5}-1)\sim \sqrt{5}x^5\left(\frac{x}{10}-\frac{x^2}{200}+\ldots\right)
$$
and the first term of the numerator
$$
\ln (5+x^2)=\ln5+\ln(1+x^2/5)=\ln5+x^2/5-x^4/50+\ldots
$$
$$
\ln(5+4x)=\ln 5+\log(1+4x/5)\sim \ln 5+4x/5-8x^2/25+\ldots
$$
Therefore
$$
\ln \left(\frac{5+x^2}{5+4x}\right)\sim -4x/5+(13/25)x^2+...\ ,
$$
which, raised to the power $6$, goes as $(4096/15625)x^6$. So, noting that $\ln((5+x^2)/(1+4x))$ has a finite limit $\ln 5$, the final value of the limit is (erasing the leading power $x^6$ upstairs and downstairs)
$$
\ln 5\times \frac{4096}{15625}\times \frac{10}{\sqrt{5}}=\frac{8192 \ln (5)}{3125 \sqrt{5}}\ .
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $I= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x$
$$I= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x$$
My Endeavour :
\begin{align}I&= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x\\ &= \int \frac{x}{\sqrt{1+ x^4}}\,\mathrm d x - \int \frac{1}{x\sqrt{1+ x^4}}\,\mathrm d x\end{align}
\begin{align}\textrm{Now,}\;\;\int \frac{x}{\sqrt{1+ x^4}}\,\mathrm d x &= \frac{1}{2}\int \frac{2x^3}{x^2\sqrt{1+ x^4}}\,\mathrm dx\\ \textrm{Taking}\,\,(1+ x^4)= z^2\,\,\textrm{and}\,\, 4x^3\,\mathrm dx= 2z\,\mathrm dz\,\, \textrm{we get} \\ &= \frac{1}{2}\int \frac{z\,\mathrm dz}{\sqrt{z^2-1}\, z}\\ &= \frac{1}{2}\int \frac{\mathrm dz}{\sqrt{z^2-1}}\\ &= \frac{1}{2}\ln|z+ \sqrt{z^2 -1}|\\ &= \frac{1}{2}\ln|\sqrt{1+x^4}+ x^2|\\ \textrm{Now, with the same substitution, we get in the second integral}\\ \int \frac{1}{x\sqrt{1+ x^4}}\,\mathrm d x &= \frac{1}{2}\int \frac{2x^3}{x^4\sqrt{1+ x^4}}\,\mathrm dx\\ &= \frac{1}{2}\int \frac{z\,\mathrm dz}{( z^2 -1)\;z} \\ &=\frac{1}{2}\int \frac{\mathrm dz}{ z^2 -1} \\ &=\frac{1}{2^2}\, \ln \left|\frac{ z+1}{z-1}\right| \\ &= \frac{1}{2^2}\, \ln \left|\frac{ \sqrt{1+ x^4}+1}{\sqrt{1+ x^4}-1}\right|\;.\end{align}
So, \begin{align}I&=\int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x \\ &=\frac{1}{2}\, \ln|\sqrt{1+x^4}+ x^2|- \frac{1}{2^2}\, \ln \left|\frac{ \sqrt{1+ x^4}+1}{\sqrt{1+ x^4}-1}\right| + \mathrm C\;.\end{align}
Book's solution:
\begin{align}I&=\int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x \\ &= \ln\left\{\frac{1+x^2 + \sqrt{1+x^4}}{x}\right\} + \mathrm C\;.\end{align}
And my hardwork's result is nowhere to the book's answer :(
Can anyone tell me where I made the blunder?
| $$\int\frac {1}{\sqrt{z^2-1}}dz=\operatorname{arcosh}z+c$$ not $\operatorname{arcsin z}+c$
| {
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"timestamp": "2023-03-29T00:00:00",
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Where does the normal to the graph of $y = \sqrt x$ at the point $(a, \sqrt a)$ intersect the $x$-axis?
Where does the normal to the graph of $y = \sqrt x$ at the point $(a, \sqrt a)$ intersect the $x$-axis?
I know how to find equation with numbers, but got really confused with this one. If anybody could break it down it would be greatly appreciated.
Here is what I've done: Took derivative $= .5x^{-.5}$, then plugged in $x$ value $a$ to get $1/2a^{-1/2}$, so the slope of tangent is $1/2a^{-1/2}$? so the slope of normal is $2a^{-1/2}$? So $y-\sqrt{a} = 2a^{-1/2}(x-a)$?
Thank you in advance
| \begin{align*}
y' &= \frac{1}{2\sqrt{x}} \\
\text{slope of the normal} &=-2\sqrt{a} \\
y-\sqrt{a} &= -2\sqrt{a}(x-a) \\
2\sqrt{a}x+y &= \sqrt{a}(2a+1) \\
\frac{x}{a+\frac{1}{2}}+\frac{y}{\sqrt{a}(2a+1)} &= 1 \\
x\text{-intercept} &= a+\frac{1}{2}
\end{align*}
| {
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Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\rightarrow \infty$
So Using $\bf{A.M\geq G.M}\;,$ We get $$\frac{x+1+x+2+x+3+x+4+x+5}{5}\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$x+3\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\leq 3$$
and equality hold when $x+1=x+2=x+3=x+4=x+5\;,$ Where $x\rightarrow \infty$
So $$\lim_{x\rightarrow 0}\left[\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right]=3$$
Can we solve the above limit in that way, If not then how can we calculate it
and also plz explain me where i have done wrong in above method
Thanks
| Hint:
Your method can be completed like this
$$HM \le GM \le AM \implies$$
$$\frac5{\frac1{x+1}+\frac1{x+2}+\frac1{x+3}+\frac1{x+4}+\frac1{x+5}} \le \sqrt[5]{(x+1)(x+2)(x+3)(x+4)(x+5)} \le x+3$$
Subtracting $x$ throughout :
$$\frac{15x^4+170x^3+675x^2+1096x+600}{5x^4+60x^3+255x^2+450x+274} \le \sqrt[5]{(x+1)(x+2)(x+3)(x+4)(x+5)} -x \le 3$$
Now take the limit as $x \to \infty$.
| {
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"url": "https://math.stackexchange.com/questions/1666688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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The Sum of the series $\sum\limits_{n=0}^{\infty}\frac{1}{n^2+3}$ I know how to get the sum of geometric series, but otherwise.
How do I get the sum of this series? Thank you.
$$\sum\limits_{n=0}^{\infty}\frac{1}{n^2+3}$$
| The Weierstrass product for the sine function gives:
$$ \forall z\in\mathbb{C},\quad\sin(z) = z\prod_{n\geq 1}\left(1-\frac{z^2}{n^2 \pi^2}\right)\tag{1} $$
hence by replacing $z$ with $iz$ we get:
$$ \forall z\in\mathbb{C},\quad\sinh(z) = \frac{e^z-e^{-z}}{2} = z\prod_{n\geq 1}\left(1+\frac{z^2}{n^2 \pi^2}\right)\tag{2} $$
and by considering the logarithmic derivative, given by $\frac{f'(z)}{f(z)}=\frac{d}{dz}\,\log f(z)$, we have:
$$ \forall z\in\mathbb{C},\quad\coth(z)=\frac{e^z+e^{-z}}{e^{z}-e^{-z}}=\frac{1}{z}+\sum_{n\geq 1}\frac{2z}{z^2+n^2\pi^2}\tag{3} $$
and by replacing $z$ with $\pi\sqrt{3}$ it follows that:
$$ \coth(\pi\sqrt{3})=\frac{e^{2\pi\sqrt{3}}+1}{e^{2\pi\sqrt{3}}-1}=\frac{1}{\pi\sqrt{3}}+\sum_{n\geq 1}\frac{2\pi\sqrt{3}}{\pi^2(n^2+3)}\tag{4} $$
so, by rearranging:
$$ \color{red}{\sum_{n\geq 0}\frac{1}{n^2+3}}=\frac{1}{6}\left(1+\pi\sqrt{3}\coth(\pi\sqrt{3})\right)=\color{red}{\frac{1}{6}+\frac{\pi\sqrt{3}}{6}\cdot\frac{e^{2\pi\sqrt{3}}+1}{e^{2\pi\sqrt{3}}-1}}\approx 1.0736.\tag{5} $$
Another approach (a Fourier-analytic one) is shown in this similar question.
| {
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"timestamp": "2023-03-29T00:00:00",
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$(5 + \sqrt{2})(2-\sqrt{2})=(11-7\sqrt{2})(2+\sqrt{2})$ - Unique factorization? Personal question : We know that $5 + \sqrt{2}$, $2-\sqrt{2}$, $11-7\sqrt{2}$ and $2+\sqrt{2}$ are irreductible in $\mathbb{Z}[\sqrt{2}]$ and that $$(5 + \sqrt{2})(2-\sqrt{2})=(11-7\sqrt{2})(2+\sqrt{2}).$$ Why this fact doesn't contradict the unique factorization in $\mathbb{Z}[\sqrt{2}]$?
Is it because $(5 + \sqrt{2})(2-\sqrt{2})=(5 + \sqrt{2})(2+\sqrt{2})(3-2\sqrt{2})=(11 -7\sqrt{2})(2+\sqrt{2})$?
Is anyone could give me a full explication in ''Answer the question''?
| Unique factorization only means unique factorization upto units.
The units in ${\mathbb Z}[\sqrt{2}]$ are the elements of the form $(1 + \sqrt{2})^n$ with $n \in {\mathbb Z}$.
For your two factorizations:
$$5 + \sqrt{2} = (3 + 2\sqrt{2}) (11 - 7 \sqrt{2}) = (1 + \sqrt{2})^2 (11 - 7 \sqrt{2})$$
so they differ by a unit.
| {
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Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+Ct+D=At+At^3+B+Bt^2 +Ct+D$$
for this I have that $A=0, B=2, C=0, D=-2$
so now I have
$I=\int \frac{2t^2}{(1+t^2)^2}dt= \int\frac{2}{1+t^2}dt - \int\frac{2}{(1+t^2)^2}dt$
Now,
$$ \int\frac{2}{1+t^2}dt = 2\arctan t$$
and
$$\int\frac{2}{(1+t^2)^2}dt$$
using partial integration we have:
$$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{1+t^2}$$
and $$dt=dv \Rightarrow t=v$$
so now we have:
$$\int\frac{2}{(1+t^2)^2}dt =\frac{t}{(1+t^2)^2} + 4\int\frac{t^2}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4\int\frac{t^2 + 1 -1}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4t -4\arctan t$$
so, the final solution should be:
$$I=2\arctan t - \frac{t}{(1+t^2)^2} - 4t +4\arctan t$$
since the original variable was $x$ we have
$$I= 6\arctan \sqrt{x} - \frac{\sqrt{x}}{(1+x)^2} - 4\sqrt{x} $$
But, the problem is that the solution to this in my workbook is different, it says that solution to this integral is $$I=\arctan \sqrt{x} - \frac{\sqrt{x}}{x+1}$$
I checked my work and I couldn't find any mistakes, so i am wondering which solution is correct?
| Notice, you can continue from here without using partial fractions $$\int \frac{2t^2}{(1+t^2)^2}\ dt$$
$$=\int \frac{2t^2}{t^4+2t^2+1}\ dt$$
$$=\int \frac{2}{t^2+\frac{1}{t^2}+2}\ dt$$
$$=\int \frac{\left(1+\frac{1}{t^2}\right)+\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}+2}\ dt$$
$$=\int \frac{\left(1+\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+2}+\int \frac{\left(1-\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+2}$$
$$=\int \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^2+4}+\int \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^2}$$
$$=\frac 12\tan^{-1}\left(\frac{t-\frac 1t}{2}\right)-\frac{1}{\left(t+\frac{1}{t}\right)}+c$$
$$=\frac 12\tan^{-1}\left(\frac{x-1}{2\sqrt x}\right)-\frac{\sqrt x}{x+1}+c$$
let $\sqrt x=\tan\theta\implies \theta=\tan^{-1}(\sqrt x)$,
$$=\frac 12\tan^{-1}\left(\frac{\tan^2\theta-1}{2\tan\theta}\right)-\frac{\sqrt x}{x+1}+c$$
$$=\frac 12\tan^{-1}\left(\tan\left(\frac{\pi}{2}+2\theta\right)\right)-\frac{\sqrt x}{x+1}+c$$
$$=\frac{\pi}{4}+\theta-\frac{\sqrt x}{x+1}+c$$
$$=\theta-\frac{\sqrt x}{x+1}+C$$
$$=\color{red}{\tan^{-1}(\sqrt x)-\frac{\sqrt x}{x+1}+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
If $a^{-1}b^2a=b^3$ and $b^{-1}a^2b=a^3$ then $a=b=e$
Let $G$ be a group and $a,b\in G$ such that
$a^{-1}b^2a=b^3$ and $b^{-1}a^2b=a^3$. Then prove that $a=b=e$, where $e$ is the identity of group.
If we are able to prove $ab = ba$ then we are done. But i am unable to show that.
Can we use somewhere that $\gcd(2,3)=1$,
or if we write $b^2a=ab^3$ and $a^2b=ba^3$ then finding some relation between $a$ and $b$ to prove that.
| The first equation $a^{-1}b^2a=b^3 \Rightarrow a^{-1}b^{2k}a=b^{3k}$ for all $k \in {\mathbb Z}$, so
$$a^{-2}b^{-4}a^2 = a^{-1}b^{-6}a = b^{-9}.$$
The second equation gives $a^{-2}b^{-1}a^2=ab^{-1}$, so
$$a^{-2}b^{-4}a^2 = (ab^{-1})^4$$ and hence
$$b^{-9} = (ab^{-1})^4.$$
Since $b^9$ commutes with $b$ and $(ab^{-1})^4$ commutes with $ab^{-1}$, it follows that $b^{-9} = (ab^{-1})^4 \in Z(G)$.
So $N = \langle b^9 \rangle \le Z(G)$. In the quotient $G/N$, the image of $b^4$ is conjugate to $b^9$, which is trivial, so the image of $b$ is trivial, hence the image of $a^4$ is trivial, but $a^4$ is conjugate to $a^9$, so the image of $a$ is trivial.
Hence $G/N$ has order $1$, so $G=N \le Z(G)$ and hence $G$ is abelian and it is easily proved that $G$ is trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
The indefinite integral $\int x^{13}\sqrt{x^7+1}dx$ I used integration by parts technique to solve this problem
so $u = \sqrt{x^7+1}$ ,$du = \frac{1}{2}\frac{7x^6}{\sqrt{x^7+1}}dx$
$v =\frac{x^{14}}{14}$ $dv=x^{13}dx$
then it becomes
$\frac{x^{14}}{14}\sqrt{x^7+1}-\int\frac{x^{14}}{28}\frac{7x^6}{\sqrt{x^7+1}}dx$
and this is where i got stuck at. I tried to substitute $u =x^7$
but then the integral become $\int \frac{u^2}{28\sqrt{u+1}}du$
the final answer that I found using wolffram calculator is $\frac{2}{105}(x^7+1)^{3/2}(3x^7-2)$
| The integrand is a special case of the more general form $$f(x;m,n,a) = x^{2n-1} (x^n + a)^{1/m}.$$ We observe that the choice $$u^m = x^n + a, \quad m u^{m-1} \, du = n x ^{n-1} \, dx$$ along with $x^{2n-1} = \frac{1}{n} x^n (nx^{n-1})$ gives $$\int f(x;m,n,a) \, dx = \frac{1}{n} \int (u^m - a) u \cdot mu^{m-1} \, du = \frac{m}{n} \left(\frac{u^{2m+1}}{2m+1} - \frac{au^{m+1}}{m+1} \right) + C,$$ which after substituting back, gives $$\frac{m}{n}(x^n+a)^{1 + 1/m} \left( \frac{x^n+a}{2m+1} - \frac{a}{m+1} \right) + C.$$ The given integral is the special case $m = 2$, $n = 7$, $a = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Probability of sum to be divisible by 7 6 fair dice are thrown simultaneously. What is the probability that the sum of the numbers appeared on dice is divisible by 7 ?
| After $n$ rolls, let $p_n$ be the probability that the sum is a multiple of $7$ after $n$ rolls.
Then we get the recursion: $$p_{n+1} = \frac{1}{6}(1-p_n)$$
That's because to get a multiple at roll $n+1$, you have to get a non-multiple of $7$ at roll $n$, and then exactly the right value.
Then start with $p_0=1$, and work from there.
$$\begin{align}p_0&=1\\
p_1&=0\\
p_2&=\frac{1}{6}\\
p_3&=\frac{5}{36}\\
p_4&=\frac{31}{216}\\
p_5&=\frac{185}{6^4}\\
p_6&=\frac{1111}{6^5}
\end{align}$$
(This is essentially a Markov process with two states, much akin to the other answer. You essentially need to compute:
$$\begin{pmatrix}0&\frac{1}{6}\\
1&\frac{5}{6}\end{pmatrix}^6\begin{pmatrix}1\\0\end{pmatrix}$$
The general answer is:
$$p_n=\frac{6^{n-1}-(-1)^{n-1}}{7\cdot 6^{n-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
2,3,7 is the only triple $\geq 2$ that satisfies this property The property being that "taking the product of two of the numbers and adding one yields a number that is divisible by the third".
Clearly, this holds for 2,3,7 since
*
*$2\cdot 3 + 1 = 1\cdot 7$
*$2\cdot7 + 1 = 5\cdot 3$
*$3\cdot 7 + 1 = 11\cdot 2$
but why is this the only such triple?
This is from a (German) textbook on algebraic number theory, reviewing elementary number theory in chapter 1. Up to this point, only divisibility, the gcd and the euclidean algorithm have been introduced (no prime numbers yet). I took an elementary number theory course last semester, but I'm completely puzzled by this problem and don't know where to start. Trying to set up some equations and to manipulate them doesn't lead me anywhere.
My solution attempt
We can, without loss of generality, assume $2 \leq a \leq b \leq c$. From MXYMXY's answer below we have $abc|ab+bc+ca+1$ (which I'll denote (*)) and also that $a < 4$.
Let's then consider the two possibilities for $a$ in turn.
*
*$a =3$: We can show that $b$ and $c$ can't both be $\geq 4$, for if they were we would have $$\frac{3}{c}+1+\frac{3}{b}+\frac{1}{bc} \leq \frac{3}{4}+1+\frac{3}{4}+\frac{1}{16} < 3 \Leftrightarrow 3b + bc + 3c + 1 < 3bc,$$
but from (*) we have $3bc|3b+bc+3c+1$, which is a contradiction.
It follows that $b = 3$ and then we have $ab+1=10$, so $c|10$. Hence $c = 5$ or $c = 10$, but $a = 3$ divides neither $3\cdot5 + 1 = 16$ nor $3\cdot10 + 1 = 31$. Thus, this case is impossible.
*$a = 2$. We will show that $b$ and $c$ can't be both $\geq 5$. Otherwise we'd have $$\frac{2}{c}+1+\frac{2}{b}+\frac{1}{bc} \leq \frac{2}{5}+1+\frac{2}{5}+\frac{1}{25} < 2 \Leftrightarrow 2b + bc + 2c + 1 < 2bc,$$ but from (*) we have that $2bc|2b+bc+2c+1$, which is a contradiction. This implies that $2 \leq b < 5$. Let's consider the different cases for $b$:
*
*$b=4$: Then we have $ab+1=9$, so $c|9$, i.e. $c = 9$. But $a = 2$ does not divide $4\cdot 9 + 1 = 37$, so this is a contradiction.
*$b=2$: Then $ab+1=5$, so $c|5$, i.e. $c = 5$. But $a = 2$ does not divide $2\cdot 5 + 1 = 11$, so we also have a contradiction.
The only case left is where $a = 2, b = 3$. Then, we have $ab+1=7$, so $c|7$. Hence, $c=7$.
Hence, either $2,3,7$ is the only solution or there is none. That it is indeed a solution was verified above.
| Equivalently $a$ divides $bc+1$ and cyclic ones. In particular, they are pairwise coprime. Then $abc$ divides $\prod_{\mathrm{cyc}}(ab+1)$, implying that $abc$ divides $1+\sum_{\mathrm{cyc}}ab$. Suppose wlog $a>b>c\ge 2$. If $c\ge 3$ then
$$
3ab \le abc \le 1+\sum_{\mathrm{cyc}}ab\le 1+(ab-1)+(ab-1)+ab<3ab,
$$
which is impossible. Then $c$ has to be $2$. Then conditions are (i) $2\mid ab+1$ (i.e., $a$ and $b$ are odd), (ii) $a\mid 2b+1$ and (iii) $b\mid 2a+1$. Hence
$$
b\le 2a+1 \le 2(2b+1)+1 \implies \frac{2a+1}{b} \in \{1,2,3,4,5\}.
$$
The ratio between odd numbers has to be odd, hence [$2a+1=3b$ or $5b$] and [$2b+1=a$ or $3a$]. Therefore we find the above unique solution $(2,3,7)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Establish the identity $\frac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta\$ Establish the identity:
$$\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta$$
The first step I got was:
$$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\theta\big)}{\cos\theta + \tan\theta}$$
Then it tells me to rewrite the factor $$\cos\theta + \tan\theta$$
in the numerator using reciprocal identities.
How would I do that?
Here is what the assignment looked like:
| On the one hand
\begin{align}
\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} &=
\dfrac{\dfrac{\cos\theta}{\sin\theta} + \dfrac{1}{\cos\theta}}{\cos\theta + \dfrac{\sin\theta}{\cos\theta}} =
\dfrac{\cos^2\theta+\sin\theta}{\sin\theta\cos\theta} \div\dfrac{\cos^2\theta+\sin\theta}{\cos\theta}
\\ &=
\dfrac{\cos^2\theta+\sin\theta}{\sin\theta\cos\theta} \cdot \dfrac{\cos\theta}{\cos^2\theta+\sin\theta}\\ &= \dfrac{1}{\sin\theta}.
\end{align}
On the other,
\begin{align}
\sec\theta\cot\theta &= \dfrac{1}{\cos\theta}\cdot\dfrac{\cos\theta}{\sin\theta} = \dfrac{1}{\sin\theta} .
\end{align}
Thus we have shown that
$$\bbox[1ex, border:2pt solid #e10000]{\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to think about negative infinity in this limit $\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$
Question:
calculate:
$$\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$$
Attempt at a solution:
This can be written as:
$$\lim_{x \to -\infty} \frac{3 + \frac{1}{x}}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}}}$$
Here we can clearly see that if x would go to $+\infty$ the limit would converge towards $\frac{3}{2}$. But what happens when x goes to $-\infty$.
From the expression above it would seem that the answer would still be $\frac{3}{2}$. My textbook says it would be $- \frac{3}{2}$ and I can't understand why.
I am not supposed to use l'Hospital's rule for this exercise.
| In fact:
$\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$
$ = \lim_{x \to -\infty} \frac{x*(3 + \frac{1}{x})}{(-x)(\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}})}$
$ = \lim_{x \to -\infty} \frac{-3 - \frac{1}{x}}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}}} = -\frac{3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Integral $\int \sqrt{\frac{x}{2-x}}dx$ $$\int \sqrt{\frac{x}{2-x}}dx$$
can be written as:
$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx.$$
there is a formula that says that if we have the integral of the following type:
$$\int x^m(a+bx^n)^p dx,$$
then:
*
*If $p \in \mathbb{Z}$ we simply use binomial expansion, otherwise:
*If $\frac{m+1}{n} \in \mathbb{Z}$ we use substitution $(a+bx^n)^p=t^s$
where $s$ is denominator of $p$;
*Finally, if $\frac{m+1}{n}+p \in \mathbb{Z}$ then we use substitution
$(a+bx^{-n})^p=t^s$ where $s$ is denominator of $p$.
If we look at this example:
$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx,$$
we can see that $m=\frac{1}{2}$, $n=1$, and $p=\frac{-1}{2}$ which means that we have to use third substitution since $\frac{m+1}{n}+p = \frac{3}{2}-\frac{1}{2}=1$ but when I use that substitution I get even more complicated integral with square root. But, when I tried second substitution I have this:
$$2-x=t^2 \Rightarrow 2-t^2=x \Rightarrow dx=-2tdt,$$
so when I implement this substitution I have:
$$\int \sqrt{2-t^2}\frac{1}{t}(-2tdt)=-2\int \sqrt{2-t^2}dt.$$
This means that we should do substitution once more, this time:
$$t=\sqrt{2}\sin y \Rightarrow y=\arcsin\frac{t}{\sqrt{2}} \Rightarrow dt=\sqrt{2}\cos ydy.$$
So now we have:
\begin{align*}
-2\int \sqrt{2-2\sin^2y}\sqrt{2}\cos ydy={}&-4\int\cos^2ydy = -4\int \frac{1+\cos2y}{2}dy={} \\
{}={}& -2\int dy -2\int \cos2ydy = -2y -\sin2y.
\end{align*}
Now, we have to return to variable $x$:
\begin{align*}
-2\arcsin\frac{t}{2} -2\sin y\cos y ={}& -2\arcsin\frac{t}{2} -2\frac{t}{\sqrt{2}}\sqrt\frac{2-t^2}{2}={} \\
{}={}& -2\arcsin\frac{t}{2} -\sqrt{t^2(2-t^2)}.
\end{align*}
Now to $x$:
$$-2\arcsin\sqrt{\frac{2-x}{2}} - \sqrt{2x-x^2},$$
which would be just fine if I haven't checked the solution to this in workbook where the right answer is:
$$2\arcsin\sqrt\frac{x}{2} - \sqrt{2x-x^2},$$
and when I found the derivative of this, it turns out that the solution in workbook is correct, so I made a mistake and I don't know where, so I would appreciate some help, and I have a question, why the second substitution works better in this example despite the theorem i mentioned above which says that I should use third substitution for this example?
|
$$\int \sqrt{\frac{x}{2-x}}dx$$
Set $t=\frac {x} {2-x}$ and $dt=\left(\frac{x}{(2-x)^2}+\frac{1}{2-x}\right)dx$
$$=2\int\frac{\sqrt t}{(t+1)^2}dt$$
Set $\nu=\sqrt t$ and $d\nu=\frac{dt}{2\sqrt t}$
$$=4\int\frac{\nu^2}{(\nu^2+1)^2}d\nu\overset{\text{ partial fractions}}{=}4\int\frac{d\nu}{\nu^2+1}-4\int\frac{d\nu}{(\nu^1+1)^2+\mathcal C}$$
$$=4\arctan \nu-4\int\frac{d\nu}{(\nu^2+1)^2}$$
Set $\nu=\tan p$ and $d\nu=\sec^2 p dp.$ Then $(\nu^2+1)^2=(\tan^2 p+1)^2=\sec^4 p$ and $p=\arctan \nu$
$$=4\arctan \nu-4\int \cos^2 p dp$$
$$=4\arctan \nu-2\int \cos(2p)dp-2\int 1dp$$
$$=4\arctan \nu-\sin(2p)-2p+\mathcal C$$
Set back $p$ and $\nu$:
$$=\color{red}{\sqrt{-\frac{x}{x-2}}(x-2)+2\arctan\left(\sqrt{-\frac{x}{x-2}}\right)+\mathcal C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
What's the formula for this series for $\pi$? These continued fractions for $\pi$ were given here,
$$\small
\pi = \cfrac{4} {1+\cfrac{1^2} {2+\cfrac{3^2} {2+\cfrac{5^2} {2+\ddots}}}}
= \sum_{n=0}^\infty \frac{4(-1)^n}{2n+1}
= \frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \cdots\tag1
$$
$$\small
\pi = 3 + \cfrac{1^2} {6+\cfrac{3^2} {6+\cfrac{5^2} {6+\ddots}}}
= 3 - \sum_{n=1}^\infty \frac{(-1)^n} {n (n+1) (2n+1)}
= 3 + \frac{1}{1\cdot 2\cdot 3} - \frac{1}{2\cdot 3\cdot 5} + \frac{1}{3\cdot 4\cdot 7} - \cdots\tag2
$$
$$\small
\pi = \cfrac{4} {1+\cfrac{1^2} {3+\cfrac{2^2} {5+\cfrac{3^2} {7+\ddots}}}}
= 4 - 1 + \frac{1}{6} - \frac{1}{34} + \frac {16}{3145} - \frac{4}{4551} + \frac{1}{6601} - \frac{1}{38341} + \cdots\tag3$$
Unfortunately, the third one didn't include a closed-form for the series. (I tried the OEIS using the denominators, but no hits.)
Q. What's the series formula for $(3)$?
| Given the symmetric continued fraction found in this post
$$\frac{\displaystyle\Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)}{\displaystyle\Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)}=\cfrac{4(a+b)}{a+b+\cfrac{(2a)(2b)} {3(a+b)+\cfrac{(3a+b)(a+3b)}{5(a+b)+\cfrac{(4a+2b)(2a+4b)}{7(a+b)+\ddots}}}}$$
Your continued fraction $(3)$ is a special case when $a=b=1$.
Moreover ,it has a beautiful q-analogue
$$\begin{aligned}\Big(\sum_{n=0}^\infty q^{n(n+1)}\Big)^2 =\cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q^2(1-q^2)^2}{1-q^5+\cfrac{q^3(1-q^3)^2}{1-q^7+\ddots}}}}\end{aligned}$$ found here and here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1689040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
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Proof of $Δv = c \cdot \tanh \left ( \frac{u}{c}\ln \left( \frac{m_0}{m_1}\right) \right) $
Question: The relativistic rocket equation is given below
$$ \frac{m_0}{m_1} = \left (\frac{1+\frac{Δv}{c}}{1-\frac{Δv}{c}} \right)^{\frac{c}{2u}} $$
Show that the equation rearranges to
$$ Δv = c \cdot \tanh \left ( \frac{u}{c}\ln \left( \frac{m_0}{m_1}\right) \right) $$
where the hyperbolic tangent function is $ \tanh(x) = \frac{e^{2x}-1}{e^{2x}+1} $
My attempt:
$$ \frac{m_0}{m_1} = \left (\frac{1+\frac{Δv}{c}}{1-\frac{Δv}{c}} \right)^{\frac{c}{2u}} $$
$$ \left( \frac{m_0}{m_1} \right)^{\frac{2u}{c}} = \left (\frac{1+\frac{Δv}{c}}{1-\frac{Δv}{c}} \right)$$
$$ \left( \frac{m_0}{m_1} \right)^{\frac{2u}{c}} = \left (\frac{\frac{c+Δv}{c}}{\frac{c-{Δv}}{c}} \right)$$
$$ \left( \frac{m_0}{m_1} \right)^{\frac{2u}{c}} = \frac{c+Δv}{c} \cdot \frac{c}{c-Δv} $$
$$ \left( \frac{m_0}{m_1} \right)^{\frac{2u}{c}} = \frac{c+Δv}{c-Δv} $$
$$ \frac{m_0^{\frac{2u}{c}}}{m_1^{\frac{2u}{c}}} = \frac{c+Δv}{c-Δv} $$
$$ \left(c-Δv\right)m_0^{\frac{2u}{c}} = \left(c+Δv\right)m_1^{\frac{2u}{c}} $$
$$ cm_0^{\frac{2u}{c}} - Δvm_0^{\frac{2u}{c}} = cm_1^{\frac{2u}{c}} + Δvm_1^{\frac{2u}{c}} $$
$$ cm_0^{\frac{2u}{c}} - cm_1^{\frac{2u}{c}} = Δvm_0^{\frac{2u}{c}} + Δvm_1^{\frac{2u}{c}} $$
$$ c(m_0^{\frac{2u}{c}} - m_1^{\frac{2u}{c}}) = Δv(m_0^{\frac{2u}{c}} + m_1^{\frac{2u}{c}})$$
$$ Δv = \frac{ c(m_0^{\frac{2u}{c}} - m_1^{\frac{2u}{c}})}{(m_0^{\frac{2u}{c}} + m_1^{\frac{2u}{c}})} $$
Now I am stuck
| We are given the expression
$$\frac{m_0}{m_1}=\left(\frac{1+\frac{\Delta v}{c}}{1-\frac{\Delta v}{c}}\right)^{c/2u} \tag 1$$
Raising each side of $(1)$ to the $2u/c$ power yields
$$\left(\frac{m_0}{m_1}\right)^{2u/c}=\frac{1+\frac{\Delta v}{c}}{1-\frac{\Delta v}{c}}$$
whereupon solving for $\Delta v/c$ we find that
$$\frac{\Delta v}{c}=\frac{\left(\frac{m_0}{m_1}\right)^{2u/c}-1}{\left(\frac{m_0}{m_1}\right)^{2u/c}+1} \tag 3$$
Then, multiply the numerator and denominator of $(3)$ by $\left(\frac{m_0}{m_1}\right)^{-u/c}$ to obtain
$$\begin{align}
\frac{\Delta v}{c}&=\frac{\left(\frac{m_0}{m_1}\right)^{u/c}-\left(\frac{m_0}{m_1}\right)^{-u/c}}{\left(\frac{m_0}{m_1}\right)^{u/c}+\left(\frac{m_0}{m_1}\right)^{-u/c}} \tag 4\\\\
&=\frac{e^{(u/c)\log(m_0/m_1)}-e^{(u/c)\log(m_0/m_1)}}{e^{(u/c)\log(m_0/m_1)}+e^{(u/c)\log(m_0/m_1)}} \tag 5\\\\
&=\tanh\left(\frac{u}{c}\log\left(\frac{m_0}{m_1}\right)\right) \tag 6\\\\
\Delta v&=c\,\tanh\left(\frac{u}{c}\log\left(\frac{m_0}{m_1}\right)\right)
\end{align}$$
as was to be shown!
NOTES:
In going from $(4)$ to $(5)$, we made use of the identity $x=e^{\log(x)}$.
In going from $(5)$ to $(6)$, we used the definition of the hyperbolic tangent, $\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
ADDRESSING THE POINT AT WHICH THE OP WAS STUCK
Starting with the expression
$$Δv = \frac{ c(m_0^{\frac{2u}{c}} - m_1^{\frac{2u}{c}})}{(m_0^{\frac{2u}{c}} + m_1^{\frac{2u}{c}})} \tag 7
$$
we divide numerator and denominator of $(7)$ by $(m_0\,m_1)^{u/c}$ to obtain
$$\frac{\Delta v}{c}=\frac{\left(\frac{m_0}{m_1}\right)^{u/c}-\left(\frac{m_0}{m_1}\right)^{-u/c}}{\left(\frac{m_0}{m_1}\right)^{u/c}+\left(\frac{m_0}{m_1}\right)^{-u/c}} \tag 8$$
Observe that $(8)$ is identical to $(4)$. Now, finish as in the development that follows $(4)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sin(x) + \cos(x) \geq 1$ $\forall x\in[0,\pi/2]: \sin{x}+\cos{x} \ge 1.$
I am really bad at trigonometric functions, how could I prove it?
| Re-arrange to get $\cos(x) < 1 - \sin(x)$
We know that $\sin^2 (x) + \cos^2(x) = 1$ and therefor that $|\cos(x)| = \sqrt{1-\sin^2(x)}$
Now we want to know if $\sqrt{1-\sin^2(x)} < 1 - \sin(x)$
Now we square both sides: $1-\sin^2(x) < (1 - \sin(x))^2 = 1 - 2\sin(x) + \sin^2(x)$
Re-arranging again we get $0 < 2\sin^2(x) - 2\sin(x)$
In order for $\sin^2(x) > \sin(x)$ we would need a $\sin(x)$ value greater than 1. This is not possible, and therefor neither is the original form of the inequality $\cos(x) < 1 - \sin(x)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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indefinite integration $ \int \frac { x^2 dx} {x^4 + x^2 -2}$ problem : $ \int \frac { x^2 dx} {x^4 + x^2 -2}$
solution : divide numerator and denominator by $x^2$
$ \int \frac { dx} {x^2 + 1 -\frac{1}{x^2}}$
Now whats the next step $?$
Am I doing right $?$
| HINT:
Setting $x^2=y,$
$$A=\frac{x^2}{x^4+x^2-2}=\dfrac y{y^2+y-2}=\dfrac y{(y+2)(y-1)}$$
As $y+2+2(y-1)=3y$
$$3A=\dfrac{y+2+2(y-1)}{(y+2)(y-1)}=\dfrac1{y-1}+\dfrac2{y+2}$$
Replace $y$ with $x^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving for $k$ when $\arg\left(\frac{z_1^kz_2}{2i}\right)=\pi$ Consider $$|z|=|z-3i|$$
We know that if $z=a+bi\Rightarrow b=\frac{3}{2}$
$z_1$ and $z_2$ will represent two possible values of $z$ such that $|z|=3$. We are given $\arg(z_1)=\frac{\pi}{6}$
The value of $k$ must be found assuming $\arg\left(\frac{z_1^kz_2}{2i}\right)=\pi$
My attempt:
We know $z_1=\frac{3\sqrt{3}}{2}+\frac{3}{2}i$ and $z_2=-\frac{3\sqrt{3}}{2}+\frac{3}{2}i$ by solving for $a$.
So let $z_3=\frac{z_1^kz_2}{2i}$ $$z_3=\frac{z_1^kz_2}{2i} = \frac{i\left(\frac{3\sqrt{3}}{2}+\frac{3}{2}i\right)^k\left(-\frac{3\sqrt{3}}{2}+\frac{3}{2}i\right)}{-2}$$
We know $$\arg(z_3)=\arctan\left(\frac{\operatorname{Im}(z_3)}{\operatorname{Re}(z_3)}\right)=\pi \Rightarrow \frac{\operatorname{Im}(z_3)}{Re(z_3)}=\tan(\pi)=0$$
This is the part where I get stuck; I assume that $\operatorname{Re}(z_3)\neq0$ and then make the equation $$\operatorname{Im}(z_3)=0$$
However, I am not sure on how to get the value of $k$ from this, or if I am in the right direction.
What should I do in order to get the value of $k$?
| $$z_1=\frac{3\sqrt{3}}{2}+\frac{3}{2}i$$
and
$$z_2=-\frac{3\sqrt{3}}{2}+\frac{3}{2}i$$
Thus,
$$z_1=3e^{\frac{i\pi}{6}}$$ and $$z_2=3e^{\frac{5i\pi}{6}}$$
$$i=e^{\frac{i\pi}{2}}$$
$$\arg\left(\frac{z_1^kz_2}{2i}\right)=\arg\left(\frac{3^{k+1}}{2}\frac{e^{\frac{ki\pi}{6}}e^{\frac{5i\pi}{6}}}{e^{\frac{i\pi}{2}}}\right)=(k+2)\frac{\pi}{6}$$
$$(k+2)\frac{\pi}{6}=\pi$$
$$k=4$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the expected value of the largest of the three dice rolls?
You toss a fair die three times. What is the expected value of the largest of the three outcomes?
My approach is the following:
calculate the probability of outcome when $\max=6$, which is
$$P(\text{at least one $6$ of the three rolls}) = 1 - P(\text{no }6) = 1 - (5/6)^3$$
and then calculate the probability of outcome when $\max=5$, which is
$$P(\text{at least one $5$ of the three rolls & $5$ is max}) = 1 - P(\text{no $5$ & $5$ is max}) = 1 - (4/6)^3.$$
I wonder if this approach is right.
| Picture the cube of possible outcomes. The cells that represent a maximum of $6$ lie in a greedy half of the outer layer, which has $6^3-5^3=216-125=91$ cells in it.
The next layer represents max $5$, and has $5^3-4^3=125-64=61$ cells in it.
We can proceed in a similar manner and arrive at the sum of the whole cube:
$$6\cdot(6^3-5^3)+5\cdot(5^3-4^3)+4\cdot(4^3-3^3)+3\cdot(3^3-2^3)+2\cdot(2^3-1^3)+1\cdot(1^3-0^3)$$
$$=6^3(6)-5^3(6-5)-4^3(5-4)-3^3(4-3)-2^3(3-2)-1^3(2-1)-0^3(1)$$
$$=6\cdot6^3-5^3-4^3-3^3-2^3-1^3-0^3$$
$$=1296-225$$
$$=1071$$
Divide by the number of cells in the cube $6^3=216$, and the answer is:
$$\frac{1071}{216}=\frac{119}{24}\approx4.96$$
| {
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"url": "https://math.stackexchange.com/questions/1696623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometry conversion rules, why this way? We have domain $\,[0, 2\pi]\,$ and the following functions are given:
$$f(x)=\cos(2x) \text{ and } g(x)=\sin(x-\pi/3)$$Solve exactly: $\,f(x)=g(x)$
Why does one solve:
(right way)
$$\cos(2x)=\sin(x-\pi/3)\\\cos(2x)=\cos(\pi/2-(x-\pi/3))\\ \text{etc.}\ldots
$$
and not (which gives the wrong answer $\rightarrow a - k\,\, \times\,\, 2\pi \cdots$):
(wrong way)
$$\cos(2x)=\sin(x-\pi/3)\sin(\pi/2-(2x))=\sin(x-\pi/3)\\ \text{etc.}\ldots$$
Right way fully worked out:
$$\cos(2x) = \sin(x -\pi/3)\\
\cos(2x) = \cos(\pi/2 - (x -\pi/3))\\
\cos(2x) = \cos(\pi/2 - x + \pi/3)\\
\cos(2x) = \cos(5\pi/6 - x)\\
2x = (5\pi/6 - x) + k\,\, \times \,\, 2\pi\\
3x = 5\pi / 6 + k \,\, \times \,\, 2\pi\\
x = 5\pi /18 + k \,\, \times \,\, 2\pi\\
or:\\
2x = - (5\pi/6 - x) + k\,\, \times \,\, 2\pi\\
x = -5\pi / 6 + k \,\, \times \,\, 2\pi\\
$$
Wrong way fully worked out:
$$\cos(2x) = \sin(x-\pi/3)\\
sin(\pi/2 - x)=\sin(x-\pi/3)\\
\pi/2-2x=(x-\pi/3)+k\,\, \times \,\, 2\pi\\
-3x = (-\pi/3 - \pi/2) + k\,\, \times \,\, 2\pi\\
-3x = -5\pi/6+ k\,\, \times \,\, 2\pi\\
x = 5\pi/18 -k \,\, \times \,\, 2\pi\\
or:\\
\pi/2-2x=\pi-(x-\pi/3) + k \,\, \times \,\, 2\pi\\
\pi/2 - x = \pi - x + \pi/3 + k \,\, \times \,\, 2\pi\\
x= 4\pi/3-\pi/2 + k \,\, \times \,\, 2\pi\,\,\,\\\ \,\,\, =5\pi/6 - k \,\, \times \,\,2\pi
$$
On closer inspection of the worked out examples above you can clearly see that final answers of the wrong way are.. wrong. This because the answers we get aren't within the set domain.
Help is highly appreciated.
-Bowser
| You can choose to transform the sine into cosine or conversely.
First method
\begin{gather}
\cos2x=\cos\left(\frac{\pi}{2}-\left(x-\frac{\pi}{3}\right)\right)
\\[6px]
\cos2x=\cos\left(\frac{5\pi}{6}-x\right)
\\[6px]
2x=\frac{5\pi}{6}-x+2k\pi
\qquad\text{or}\qquad
2x=-\frac{5\pi}{6}+x+2k\pi
\\[6px]
x=\frac{5\pi}{18}+k\frac{2\pi}{3}
\qquad\text{or}\qquad
x=-\frac{5\pi}{6}+2k\pi
\end{gather}
Second method
\begin{gather}
\sin\left(\frac{\pi}{2}-2x\right)=\sin\left(x-\frac{\pi}{3}\right)
\\[6px]
\frac{\pi}{2}-2x=x-\frac{\pi}{3}+2k\pi
\qquad\text{or}\qquad
\frac{\pi}{2}-2x=\pi-x+\frac{\pi}{3}+2k\pi
\\[6px]
x=\frac{5\pi}{18}-k\frac{2\pi}{3}
\qquad\text{or}\qquad
x=-\frac{5\pi}{6}-2k\pi
\end{gather}
What's happening?
Nothing strange. In the solutions above, $k$ denotes an arbitrary integer: for any choice of the integer $k$, you get a solution. So writing $-k$ or $+k$ is irrelevant: the solution for $k=1$ in the first set (on the left side) appears in the solutions of the second set (left side) for $k=-1$.
The two forms give the same solutions.
If you want to find the solutions in the interval $[0,2\pi)$ you can do as follows, for the first method:
$$
0\le\frac{5\pi}{18}+k\frac{2\pi}{3}<2\pi
\iff
0\le5+12k<36
\iff
-5\le12k<31
$$
that gives $k=0,1,2$. Also
$$
0\le-\frac{5\pi}{6}+2k\pi<2\pi
\iff
0\le-5+12k<12
\iff
5\le12k<17
$$
that gives $k=1$. So you have four solutions.
The same procedure in the second method would give $k=0,-1,-2$ in one case and $k=-1$ in the other.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to rearrange this? Can you please help me to derive the second equation from the first one? I'd really appreciate your help!
$$
-\frac{1}{w-s-d} + \frac{1+r}{1+z} \cdot \frac{1}{(1+r)s+(1+n)d} = 0 \tag{1}
$$
$$
(2 + z)(1 + r)s = (1 + r)(w - d)-(1 + z)(1 + n)d \tag{2}
$$
Many thanks!
| \begin{align}
-\frac{1}{w-s-d}+\frac{1+r}{1+z}\cdot \frac{1}{(1+r)s + (1+n)d} &= 0\\
\frac{1+r}{1+z}\cdot \frac{1}{(1+r)s + (1+n)d}&= \frac{1}{w-s-d}\\
\frac{1}{1+z}\cdot \frac{1}{(1+r)s + (1+n)d}&=\frac{1}{(w-s-d)(1+r)}\\
(1+z)[(1+r)s+(1+n)d] &= (w-s-d)(1+r)\\
(1+z)[(1+r)s+(1+n)d]+s(1+r) &= (w-d)(1+r)\\
(1+z)(1+r)s+s(1+r)&= (w-d)(1+r)-(1+z)(1+n)d\\
(2+z)(1+r)s &= (1+r)(w-d)-(1+z)(1+n)d.
\end{align}
Explanation:
First add $\frac{1}{w-s-d}.$ Secondly divide by $(1+r)$. Then take the inverse. Then add $s(1+r)$. Then subtract $(1+z)(1+n)d$. Lastly, do the final small simplifications.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \sin(x)\cos(x) - \frac{1}{2} - \frac{\cos(2x)}{2} = 5$$
$$ \frac{7\cos(2x)}{2} - \sin(x)\cos(x) + \frac{5}{2} = 0 $$
$$ 7\cos(2x) - 2\sin(x)\cos(x) + 5 = 0 $$
$$ 7\cos(2x) - \sin(2x) + 5 = 0 $$
So at this point I am stuck what to do, I have attempted a Weierstrass sub of $\tan(\frac{x}{2}) = y$ and $\cos(x) = \frac{1-y^2}{1+y^2}$ and $\sin(x)=\frac{2y}{1+y^2} $ but I got a quartic and I was not able to solve it.
| It was a good work arriving to $$7\cos(2x) - \sin(2x) + 5 = 0$$ But the substitution to be used was just $t=\tan(\frac{2x}2)=\tan(x)$ which leads to the quadratic $$t^2+t-6=0$$ leading to $$(t-2)(t+3)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701363",
"timestamp": "2023-03-29T00:00:00",
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If a, b, c are three natural numbers with $\gcd(a,b,c) = 1$ such that $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ then show that $a+b$ is a square.
If a, b, c are three natural numbers with $\gcd(a,b,c) = 1$ such that $$\frac{1}{a} + \frac{1}{b}= \frac{1}{c}$$ then show that $a+b$ is a perfect square.
This can be simplified to:
$$a+b = \frac{ab}{c}$$
Also, first few such examples of $(a,b,c)$ are $(12, 4, 3)$ and $(20, 5, 4)$. So, I have a feeling that $b$ and $c$ are consecutive.
I don't think I have made much progress.
Any help would be appreciated.
| Rewrite as $(a-c)(b-c)=c^2$. First we show that $a-c$ and $b-c$ are relatively prime. Suppose to the contrary that the prime $p$ divides $a-c$ and $b-c$. Then $p$ divides $c$ and therefore $a$ and $b$, contradicting the fact that $\gcd(a,b,c)=1$.
Since $a-c$ and $b-c$ are relatively prime, it follows that $a-c=s^2$ and $b-c=t^2$, where $st=c$.
We conclude that $a=s^2+st$ and $b=t^2+st$, so $a+b=(s+t)^2$.
| {
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"url": "https://math.stackexchange.com/questions/1702758",
"timestamp": "2023-03-29T00:00:00",
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Expansion and factorisation I have a little problems with a few questions here and I need help.. Thanks ...
*
*Factorise completely
$$9x^4 - 4x^2 - 9x^2y^2 + 4y^2 $$
My workings ..
$$ (3x^2+2x)(3x^2-2x) - y^2 (9x^2-4) = (3x^2 + 2x)(3x^2 -2x) - y^2 (3x+2)(3x-2) $$
*Factorise $3x^2 + 11x - 20$ and , hence Factorise completely
$$11a - 11b - 20 + 3a^2 + 3b^2 - 6ab$$
My workings ...
$$ 11(a-b) - 20 + (3a-3b)(a-b)$$
*Evaluate the following by algebraic expansion of factorisation
(A) $78^2 + 312 + 4$
(B) $501^2 - 1002 + 1$
Thanks a lot !
| HINTS :
For 1.
The coefficients are $\pm 9,\pm 4$, so
$$9x^4-4x^2-9x^2y^2+4y^2=(9x^4-9x^2y^2)-(4x^2-4y^2)$$
is worth trying.
For 2.
Compare $3x^2+11x-20$ with $$11a-11b-20+3a^2+3b^2-6ab=11(a-b)-20+3(?)=3(?)+11(a-b)-20$$
For 3.
Note that $312=78\times 4$ and that $1002=2\times 501$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
Question: Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
My attempt:
$$\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$$
$$ \frac{\frac {1}{\cos(x)} - \frac{1}{\sin(x)}}{\frac{\sin(x)}{\cos(x)} - \frac{\cos(x)}{\sin(x)}} $$
$$ \frac{\sin(x)-\cos(x)}{\sin^2(x)-\cos^2(x)}$$
$$ \frac{(\sin(x)-\cos(x))}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))} $$
$$ \frac{1}{\sin(x)+\cos(x)} $$
Now this is where I am stuck , I thought of multiplying the numerator and denominator by $$ \frac{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}}{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}} $$ but that did not work out well..
| Take the right hand side:
$$\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}=\frac{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}}{\frac{1}{\cos(x)} + \frac{1}{\sin(x)}} = \frac{\sin^2x+\cos^2x}{\sin x+\cos x}=\frac{1}{\sin x+\cos x} $$
and this is equal to the expression you found for the left hand side.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Unknown Inequality $$ \left( \sqrt{3}\sqrt{y(x+y+z)}+\sqrt{xz}\right)\left( \sqrt{3}\sqrt{x(x+y+z)}+\sqrt{yz}\right)\left( \sqrt{3}\sqrt{z(x+y+z)}+\sqrt{xy}\right) \leq 8(y+x)(x+z)(y+z)$$
I can prove this inequality, but i need know if this inequaliy is known...
| Here is a nice and simple proof (for $x,y,z\ge 0$ of course):
By Cauchy-Schwarz inequality:
$$\left( \sqrt{3}\sqrt{y(x+y+z)}+\sqrt{xz}\right)^2 \le (3+1)\big[(y(x+y+z) + xz\big] = 4(x+y)(y+z).$$
Similarly:
$$\left( \sqrt{3}\sqrt{x(x+y+z)}+\sqrt{yz}\right)^2 \le 4(x+z)(x+y)$$
$$\left( \sqrt{3}\sqrt{z(x+y+z)}+\sqrt{xy}\right)^2 \le 4(x+z)(y+z).$$
Taking the product of the above three inequalities we get the desired inequality.
| {
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Let $x,y,z>0$ and $x+y+z=1$, then find the least value of ${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$ Let $x,y,z>0$ and $x+y+z=1$, then find the least value of
$${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$$
I tried various ways of rearranging and using AM > GM inequality. But I couldn't get it. I am not good at inequalities. Please help me.
I wrote $x$ as $1-(y+z)$ and I took $x+y$ as $a$ and the others as $b$ and $c$. And I am trying.
| Using AM-GM inequality (for $n=6$):
Let $f(x,y,z)=\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}$. Take $x_1=\alpha(2-x), x_2=\alpha(2-y), x_3=\alpha(2-z), x_4=\frac{2}{(2-x)}, x_6=\frac{2}{(2-y)}+\frac{2}{(2-z)}$, where $\alpha=\frac{18}{25}$. Then by AM-GM inequality
$\frac{1}{6}\left(5\alpha+\frac{2}{(2-x)}+\frac{2}{(2-y)}+\frac{2}{(2-y)}\right)=\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}\geq \sqrt[6]{x_1 x_2 x_3 x_4 x_5 x_6}=\sqrt{2\alpha}=\frac{6}{5}$, which gives
$\frac{2}{(2-x)}+\frac{2}{(2-y)}+\frac{2}{(2-y)}\geq \frac{36}{5}-5\frac{18}{25}=\frac{18}{5}$. So,
$f(x,y,z)=-3+\frac{2}{(2-x)}+\frac{2}{(2-y)}+\frac{2}{(2-y)}\geq -3+\frac{18}{5}=\frac{3}{5}=-3+6\sqrt[6]{x_1 x_2 x_3 x_4 x_5 x_6}$, where $x_1=x_2=\ldots=x_6$, which happens when $x=y=z$, since $\alpha\neq 0$, and $\frac{2}{2-x}=\alpha(2-x)$ or $x=\frac{1}{3}$, since $x>0$.
Thus the least value of $f$ is $\frac{3}{5}$.
| {
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Determine whether $p_n$ is decreasing or increasing, if $p_{n+1} = \frac{p_n}{2} + \frac{1}{p_n}$ If $p_1 = 2$ and $p_{n+1} = \frac{p_n}{2}+ \frac{1}{p_n}$, determine $p_n$ is decreasing or increasing.
Here are the first few terms:
$$p_2 = \frac{3}{2}, p_3 = \frac{3}{4} + \frac{2}{3} = \frac{17}{12}, p_4 = \frac{17}{24} + \frac{12}{17} = \frac{577}{408}$$
The sequence seems decreasing to me so I tried to prove it by induction. Need to prove $p_k - p_{k+1} \gt 0$ for all n.
For n = 1, $p_1 - p_2 = 2- \frac{3}{2} = \frac{1}{2} \gt 0$ (true)
However when I tried to prove it for $k+1$, I ran into problems.
Assume $p_k - p_{k+1} \gt 0$ is true for n =k, then it must be also true for $n =k+1$.
$$p_{k+1} - p_{k+2} = \frac{p_k}{2} + \frac{1}{p_k} - \frac{p_{k+1}}{2} - \frac{1}{p_{k+1}} = \frac{p_k - p_{k+1}}{2} + (\frac{1}{p_k}-\frac{1}{p_{k+1}})$$
but $\frac{1}{p_k}-\frac{1}{p_{k+1}} \lt 0$
I don't know what to do from there.
| The sequence appears to be decreasing to $\sqrt2$. (It's the Newton-Raphson iteration for computing $\sqrt 2$ as the root of the function $f(x):=x^2-2$.) So show this in two steps:
(1) First prove by induction that $p_n\ge \sqrt2$ for every $n$. This follows from
$$p_{n+1}-\sqrt 2=\left({p_n\over2} +\frac1{p_n}\right)-\sqrt2={(p_n-\sqrt2)^2\over2p_n}.$$
(2) Then use (1) to prove by induction that $p_{n+1}\le p_n$. You should write your expression $p_n-p_{n+1}$ this way:
$$
p_n-p_{n+1}=p_n-\left({p_n\over2}+\frac1{p_n}\right)={p_n^2-2\over2p_n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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sum of the series $\frac{2^n+3^n}{6^n}$ from $n=1$ to $\infty$ Find the sum of the series $\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}=?$
My thoughts: find $\sum_{n=1}^{\infty} 2^n$, $\sum_{n=1}^{\infty} 3^n$ and $\sum_{n=1}^{\infty} 6^n$ (although I don't know how yet...)
Then, $\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}= \frac{\sum_{n=1}^{\infty} 2^n}{\sum_{n=1}^{\infty} 3^n} + \frac{\sum_{n=1}^{\infty} 3^n}{\sum_{n=1}^{\infty} 6^n}$, am I on the right track? I really don't know how else to solve it.
Thanks.
| Hint:
$$
\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}
=\sum_{n=1}^{\infty} \frac{2^n}{6^n}
+\sum_{n=1}^{\infty} \frac{3^n}{6^n}
=\sum_{n=1}^{\infty} \left(\frac{2}{6}\right)^n
+\sum_{n=1}^{\infty} \left(\frac{3}{6}\right)^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $\csc a +\cot a = \cot\frac{a}{2}$
Prove: $$\csc a + \cot a = \cot\frac{a}{2}$$
All I have right now, from trig identities, is
$$\frac{1}{\sin a} + \frac{1}{\tan a} = \frac{1}{\tan(a/2)}$$
Where do I go from there?
| $\csc a+\cot a=\cot\frac{a}2$
L.H.S.
$\implies\large \frac{1}{\sin a}+\frac{\cos a}{\sin a} \\
\implies \large\frac{1+\cos a}{\sin a}\\
\implies \large\frac{1+2\cos^2\frac{a}{2}-1}{\sin a}\\
\implies \large\frac{2\cos^2\frac{a}{2}}{2\sin\frac{a}2\cos\frac{a}{2}}\\
\implies \cot \frac{a}{2}=\text{R.H.S}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability Problem - Finding a pdf Below is a problem I did. However, I did not come up with the answer in book.
I am thinking that I might have the wrong limits for the integral. I am hoping
somebody can point out what I did wrong.
Bob
Problem
Let $Y = \sin X$, where $X$ is uniformly distributed over $(0, 2 \pi)$. Find
the pdf of $Y$.
Answer:
\begin{eqnarray*}
P(Y<=y_0) &=& P( \sin x <= y_0 ) = P( x <= \arcsin( y_0) ) \\
P(Y<=y_0) &=& \int_0^{\arcsin(y_0)} \frac{1}{2 \pi} dx =
\frac{x}{2 \pi} \Big|_0^{\arcsin(y_0)} =
\frac{\arcsin(y_0)}{2 \pi} - \frac{0}{2 \pi} \\
P(Y<=y_0) &=& \frac{\arcsin(y_0)}{2 \pi} \\
F(Y) &=& \frac{\arcsin(Y)}{2 \pi} \\
f(y) &=& \frac{1}{2 \pi \sqrt{1 - y^2}} \\
\end{eqnarray*}
However, the book gets:
\begin{eqnarray*}
f(y) &=& \frac{1}{\pi \sqrt{1 - y^2}} \\
\end{eqnarray*}
I believe that book is right because
\begin{eqnarray*}
\int_{-1}^{1} \frac{1}{\pi \sqrt{1 - y^2} } dy &=&
\frac{\arcsin{y}}{\pi} \Big|_{-1}^{1} =
\frac{\frac{\pi}{2}}{\pi} - \frac{-\frac{\pi}{2}}{\pi} \\
\int_{-1}^{1} \frac{1}{\pi \sqrt{1 - y^2} } dy &=& 1 \\
\end{eqnarray*}
| If $X$ is uniform in $(0,2\pi)$, then has cdf
$$
F_X(x)=\begin{cases}
0 & x\le 0\\
\frac{x}{2\pi} & x\in (0,2\pi)\\
1 & x\ge 2\pi
\end{cases}
$$
The random variable $Y=\sin (X)$ takes values on $(-1,1)$. Hence, $\Bbb P(Y\le y) = 0$ for $y \le -1$ and $\Bbb P(Y\le y) = 1$ for $y \ge 1$. Let now $y \in (-1, 1)$. We have
$$
F_Y(y)=\Bbb P(Y\le y)=\Bbb P(\sin (X)\le y)
$$
The equation $\sin(x) = y$ has two solutions in the interval $(0, 2\pi)$:
*
*$x = \arcsin(y)$ and $\pi-\arcsin(y)$ for $y > 0$
*$x = \pi-\arcsin(y)$ and $2\pi + \arcsin(y)$ for $y < 0$.
Hence, $Y$ has positive values if $X$ takes values in $A_1=(0,\arcsin y)$ or $A_2=(\pi-\arcsin y,2\pi)$; $Y$ has negative values if $X$ takes values in $B=(\pi-\arcsin y,2\pi + \arcsin y)$.
So we have for $-1<y<1$,
$$
\Bbb P(Y\le y)=\begin{cases}
\Bbb P(X\in B)& -1<y< 0\\
\Bbb P(X\in A_1\cup A_2)=\Bbb P(X\in A_1)+\Bbb P(X\in A_2)& 0<y< 1\\
\end{cases}
$$
that is
$$
F_Y(y)=\begin{cases}
F_X(2\pi + \arcsin y) -F_X(\pi-\arcsin y)& -1<y< 0\\
F_X(\arcsin y) +1-F_X(\pi-\arcsin y)& 0<y< 1\\
\end{cases}
$$
and then
$$
F_Y(y)=\frac{\pi + 2 \arcsin(y)}{2\pi},\qquad y\in (-1,1)
$$
The distribution function of $Y$ is
$$
F_Y(y)=\begin{cases}
0 & y\le 0\\
\frac{\pi + 2 \arcsin(y)}{2\pi} & y\in (-1,1)\\
1 & y\ge 1
\end{cases}
$$
We differentiate the above expression to obtain the probability density:
$$
f_Y(y)=\begin{cases}
\frac{1}{\pi\sqrt{1-y^2}} & y\in (-1,1)\\
0 & y\notin (-1,1)
\end{cases}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For each bounded subset in $\mathbb{C}$, $\lim_{n\rightarrow \infty}(1+x/n)^n=e^x$ uniformly For this question, it is defined that $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$, $\forall x\in\mathbb{C}$.
Though I know why $\lim_{n\rightarrow \infty}(1+x/n)^n=e^x$ pointwise for any real $x$, I feel difficult to show that the sequence of functions converges uniformly on each bounded subset in $\mathbb{C}$. This question comes up in an undergraduate analysis class, but I have little experience working with the complex number. Could you help me on that? Thank you!
| Consider the power series for $\log$ which converges in the complex plane if $|z| < 1:$
$$\log(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \ldots.$$
Hence, if $|z| < n$ we have
$$\log(1+z/n) = \frac{z}{n} - \frac{z^2}{2n^2} + \frac{z^3}{3n^3} - \frac{z^4}{4n^4} + \ldots,$$
and, using the triangle inequality
$$\left|n\log(1+z/n) - z\right| \leqslant \frac{|z|^2}{2n} + \frac{|z|^3}{3n^2} + \frac{|z|^4}{4n^3} + \ldots.$$
In a bounded subset $S \subset \mathbb{C}$, we have a positive real number $R$ such that $|z| \leqslant R$. For all sufficiently large $n$, we have $n > 2R \geqslant 2|z| \implies |z|/n \leqslant 1/2$ for all $z \in S.$
Hence,
$$\left|n\log(1+z/n) - z\right| \leqslant \frac{|z|^2}{n}\left[ \frac{1}{2} + \frac{|z|}{3n} + \frac{|z|^2}{4n^2} + \ldots\right] \\ \leqslant \frac{|z|^2}{n}\left[ \frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 2^2} + \ldots\right] \\ \leqslant \frac{|z|^2}{n}\left[ \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots\right] \\ = \frac{|z|^2}{n} \\ \leqslant \frac{R^2}{n}.$$
Thus,
$$\lim_{n \to \infty} n\log(1+z/n) = \lim_{n \to \infty} \log(1+z/n)^n = z,$$
uniformly for $z \in S$.
The function $z \mapsto \exp(z)$ is uniformly continuous on the compact set $\{z : |z| \leqslant R \},$ and it follows that
$$\lim_{n \to \infty} (1+z/n)^n = e^z, $$
uniformly for $z \in S$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Confusion about nth root of a number Let's say we want to find the 6th roots of 64. Then according to the method from my textbook:
$z^6=64$
$z^6=64+0i=64[\cos(0)+i\sin(0)]=64cis(0)=64cis(0+2k\pi)$
Then by De Moivre's theorem:
$z=\color{red}{64^{\frac{1}{6}}}cis(\frac{2k\pi}{6})$ $,k=0, 1, 2, 3, 4, 5$
$z=\color{red}{2}cis(\frac{k\pi}{3})$
So that $z=cis(0), 2cis(\frac{\pi}{3}), 2cis(\frac{2\pi}{3}), 2cis(\frac{3\pi}{3}), 2cis(\frac{4\pi}{3}), 2cis(\frac{5\pi}{3})$
So the way I understand that is that we want to prove that the equation $z^6=64$ has six different solutions. In other words that there are six different values of $z$ that satisfy $z=64^{\frac{1}{6}}$. If so, then why do we simplify $64^{1/6}$ to just $2$ in the process, what about the other five values?
| Notice:
$$z^6=64\Longleftrightarrow z^6=|64|e^{\arg(64)i}\Longleftrightarrow z^6=64e^{0i}\Longleftrightarrow$$
$$z=\left(64e^{2\pi ki}\right)^{\frac{1}{6}}\Longleftrightarrow z=2e^{\frac{\pi ki}{3}}$$
With $k\in\mathbb{Z}$ and $k:0-5$
So, the solutions are:
$$z_0=2e^{\frac{\pi\cdot0i}{3}}=2$$
$$z_1=2e^{\frac{\pi\cdot1i}{3}}=2e^{\frac{\pi i}{3}}$$
$$z_2=2e^{\frac{\pi\cdot2i}{3}}=2e^{\frac{2\pi i}{3}}$$
$$z_3=2e^{\frac{\pi\cdot3i}{3}}=-2$$
$$z_4=2e^{\frac{\pi\cdot4i}{3}}=2e^{-\frac{2\pi i}{3}}$$
$$z_5=2e^{\frac{\pi\cdot5i}{3}}=2e^{-\frac{\pi i}{3}}$$
| {
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"url": "https://math.stackexchange.com/questions/1715831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On Ramanujan's curious equality for $\sqrt{2\,(1-3^{-2})(1-7^{-2})(1-11^{-2})\cdots} $ In Ramanujan's Notebooks, Vol IV, p.20, there is the rather curious relation for primes of form $4n-1$,
$$\sqrt{2\,\Big(1-\frac{1}{3^2}\Big) \Big(1-\frac{1}{7^2}\Big)\Big(1-\frac{1}{11^2}\Big)\Big(1-\frac{1}{19^2}\Big)} = \Big(1+\frac{1}{7}\Big)\Big(1+\frac{1}{11}\Big)\Big(1+\frac{1}{19}\Big)$$
Berndt asks: if this is an isolated result, or are there others? After some poking with Mathematica, it turns out that, together with $p= 2$, we can use the primes of form $4n+1$,
$$\sqrt{2\,\Big(1-\frac{1}{2^6}\Big) \Big(1-\frac{1}{5^2}\Big)\Big(1-\frac{1}{13^2}\Big)\Big(1-\frac{1}{17^2}\Big)} = \Big(1+\frac{1}{5}\Big)\Big(1+\frac{1}{13}\Big)\Big(1+\frac{1}{17}\Big)$$
(Now why did Ramanujan miss this $4n+1$ counterpart?) More generally, given,
$$\sqrt{m\,\Big(1-\frac{1}{n^2}\Big) \Big(1-\frac{1}{a^2}\Big)\Big(1-\frac{1}{b^2}\Big)\Big(1-\frac{1}{c^2}\Big)} = \Big(1+\frac{1}{a}\Big)\Big(1+\frac{1}{b}\Big)\Big(1+\frac{1}{c}\Big)$$
Q: Let $p =a+b+c,\;q = a b + a c + b c,\;r =abc$. For the special case $m = 2$, are there infinitely many integers $1<a<b<c$ such that,
$$n =\sqrt{\frac{2(p-q+r-1)}{p-3q+r-3}}$$
and $n$ is an integer? (For general $m$, see T. Andrew's comment below.)
Note: A search with Mathematica reveals numerous solutions, even for prime $a,b,c$. It is highly suggestive there may be in fact parametric solutions.
| It can be transformed to next equation.
$$\displaystyle m=\frac{n^2}{n^2-1}\frac{a+1}{a-1}\frac{b+1}{b-1}\frac{c+1}{c-1}$$
so, function m decrease monotonously ,when all values are bigger than 2. If whatever m is, this equation has only finite solutions(when all are integer), $2\leq m\leq12(n=2,a=2,b=3,c=5)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the remainder of $9^2\cdot 13\cdot 21^2$ when divided by $4$
Find the remainder of $9^2\cdot 13\cdot 21^2$ when divided by $4$
How should I approach this type of questions?
Without calculator of course
I did this:
$9^2\cdot 13\cdot 21^2=81\cdot 13\cdot 441=81\cdot 5733=464,373=33\bmod 4=1 \bmod 4$
| $9^2*13*21^2$ is not a multiple of 4 or two.
Hence it cannot be 0 or 2 (mod 4)
$9^2*13*21^2$ can be written as
$9^2*21^2*(3^2+2^2) =$
$(9*21*3)^2 + (9*21*2)^2$
A sum of squares cannot give 3 (mod 4)
Hence it is 1 (mod 4)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I factorize this numerator? $$\lim_{x\to -3} \frac 1{x+3} + \frac 4{x^2+2x-3}$$
I have the solution I just need to know how i turn that into:
$$\frac {(x-1)+4}{(x+3)(x-1)}$$
I know this might be really simple but I'm not sure how to factorise the numerator.
Thanks in advance!
| You have
$$ x^2+2x-3 = (x-1)(x+3),$$
so
$$\frac{1}{x+3}+\frac{4}{x^2+2x-3}=\frac{1}{x+3}+\frac{4}{(x-1)(x+3)}=\frac{1(x-1) +4}{(x-1)(x+3)} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac{1}{3}-\cos x$$
$$ 1 - \cos^2 x = \dfrac{1}{9} + \cos^2 x - \dfrac{2}{3}\cos x$$
$$ -2\cos^2 x + 2/3\cos x +\dfrac{8}{9}=0$$
$$ \boxed{\cos^2 x -\dfrac{1}{3}\cos x -\dfrac{4}{9} = 0}$$
Can I just substitute $\cos x$ by $z$ and solve as if it was a simple second degree equation and then obtain $x$ by taking the inverse cosine? I have tried to do this but I cannot get the right result. If I do this, I obtain the following results:
$$ z_1 = -0.520517 \longrightarrow x_1 = 121.4º\\
z_2= 0.8538509 \longrightarrow x_2 = 31.37º$$
I obtain $x$ from $z$ by taking the inverse cosine.
The correct result should be around 329º which corresponds to 4.165 rad. My question is if what I am doing is wrong because I have tried multiple times and I obtain the same result (or in the worst case, I have done the same mistake multiple times).
| Another way to solving this by the tangent half angle substitution $t = \tan \frac{x}{2}$ (or $x = 2 \arctan t$)
$$ \begin{align} \cos x & \rightarrow \frac{1-t^2}{1+t^2} \\
\sin x & \rightarrow \frac{2 t}{1+t^2} \end{align} $$
Your equation becomes
$$ \frac{2 t}{1+t^2} + \frac{1-t^2}{1+t^2} = \frac{1}{3} $$ which has two solutions
$$ \left. \begin{aligned} t & = \frac{3}{4} + \frac{\sqrt{17}}{4} \\
t & = \frac{3}{4} - \frac{17}{4} \end{aligned} \right\} \begin{aligned} \cos x & = \frac{1}{6} - \frac{17}{6} & \sin x & = \frac{1}{6} + \frac{17}{6} \\ \cos x & = \frac{1}{6} + \frac{17}{6} & \sin x & = \frac{1}{6} - \frac{17}{6} \end{aligned} $$
The two solutions are then
$$ \begin{aligned} x & = \arctan \left( \frac{\sqrt{17}-9}{8} \right) + 2 \pi n \\ x & = \frac{\pi}{2} - \arctan \left( \frac{\sqrt{17}-9}{8} \right) + 2 \pi n \end{aligned} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$ Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$
Answer:
This should be easy enough...
$f'(x) = 2x$
The tangent line in the point $(a, a^2)$ is $y - a^2 = 2 (x - a) \rightarrow y = 2x + a^2 - 2a$
The equation for the normal line is: $y - a^2 = -\frac{1}{2}(x - a) \rightarrow y = -\frac{1}{2}x + a^2 + \frac{1}{2}a$
Now to determine the point of intersection we just see when $f(x)$ and the normal line is equal, i.e.
$x^2 = -\frac{1}{2}x + a^2 + \frac{1}{2}a$
But this seems like a nonsense equation...
| The tangent slope of $f(x)$ at $x = a$ is $f'(a) = 2a$, so you will need to amend the tangent and normal line accordingly.
Normal line is $y - a^2 = -\dfrac{1}{2a}(x - a)$ or $y = -\dfrac{x}{2a}+\dfrac{1}{2}+a^2$.
Set $f(x) = y$, so $x^2 = -\dfrac{x}{2a}+\dfrac{1}{2}+a^2$ or $2ax^2 + x - (a + 2a^3) = 0$. Finish with the quadratic formula
$x = \dfrac{-1 \pm \sqrt{1+4(2a)(a+2a^3)}}{4a} = \dfrac{-1 \pm \sqrt{16a^4 + 8a^2 + 1}}{4a}=\dfrac{-1 \pm \sqrt{(4a^2 + 1)^2}}{4a}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can the system of equations be extracted from its solution? While I was solving the secondary school exam of 2014 I came across a question that states:
After solving those equations: $a_{1}x + b_{1}y = c_{1}$ and $a_{2}x + b_{2}y = c_{2}$, we found that x = $\frac{-7}{\begin{vmatrix}
3 & 1 \\
1 & -2
\end{vmatrix}}$ and y = $\frac{-21}{\begin{vmatrix}
3 & 1 \\
1 & -2
\end{vmatrix}}$, then $c_{1} = .....$ and $c_{2} = .....$
The answer is $c_{1} = 6$ and $c_{2} = -5$ in the ministry's model answers on its website with no steps available ,but it was probably solved by Cramer's rule.
Is this answer right and does $c_{1}$ and $c_{2}$ have only one value?
| Presumably the denominator $\left|\matrix{3 & 1\cr 1 & -2}\right|$ indicates that the coefficient matrix $\pmatrix{a_1 & b_1\cr a_2 & b_2} = \pmatrix{3 & 1\cr 1 & -2}$. Then we have
$$ \pmatrix{c_1 \cr c_2} = \pmatrix{3 & 1\cr 1 & -2} \pmatrix{x\cr y\cr} = \pmatrix{3 & 1\cr 1 & -2} \pmatrix{1\cr 3\cr} = \pmatrix{6\cr -5\cr}$$
agreeing with the ministry's answer.
| {
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What is $\arctan(x) + \arctan(y)$ I know $$g(x) = \arctan(x)+\arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$$
which follows from the formula for $\tan(x+y)$. But my question is that my book defines it to be domain specific, by which I mean, has different definitions for different domains:
$$g(x) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\[1.5ex]
\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\[1.5ex]
-\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases}$$
Furthermore, When I plot the function $2\arctan(x)$, it turns out that the book definition is correct. I don't understand how such peculier definition emerges. Thank you.
| The fact, that $\tan x = \theta$ only implies $x = \arctan \theta + k \pi$, makes the fomular of $g(x)$ a little complicated.
First proof.
There are many good answers pointed by KonKan. The following one is based on the Lagrange mean value theorem. It is not the shortest proof, but the idea is easy.
Denote
$$F(y) = \arctan x + \arctan y - \arctan \frac{x+y}{1-xy}.$$
We find that $F'(y) = 0$.
It is easily to see that
$$
\arctan x + \arctan \frac{1}{x} = \begin{cases} \frac{\pi}{2}, \quad x>0, \\
-\frac{\pi}{2}, \quad x<0, \end{cases}
$$
and
$$
\arctan z = -\arctan (-z),\quad \lim_{z\to -\infty}\arctan z = -\frac{\pi}{2}, \quad \lim_{z\to +\infty}\arctan z = \frac{\pi}{2}.
$$
We divide the problem to three cases.
Case 1. $x = 0$. We have $F(y)\equiv 0$.
Case 2. $x > 0$. The domain of $F(y)$ is $(-\infty, \frac 1 x) \cup (\frac 1 x, +\infty)$. Since $F'(y) = 0$, we have
$$
F(y) \equiv
\begin{cases} \lim\limits_{z\to -\infty}F(z) = \arctan x - \frac \pi 2 + \arctan \frac 1 x = 0, \quad xy<1,\\
\lim\limits_{z\to +\infty}F(z) = \arctan x + \frac \pi 2 + \arctan \frac 1 x = \pi, \quad xy > 1.
\end{cases}
$$
Case 3. $x < 0$. The domain of $F(y)$ is $(-\infty, \frac 1 x) \cup (\frac 1 x, +\infty)$. Since $F'(y) = 0$, we have
$$
F(y) \equiv
\begin{cases} \lim\limits_{z\to -\infty}F(z) = \arctan x - \frac \pi 2 + \arctan \frac 1 x = -\pi, \quad xy>1,\\
\lim\limits_{z\to +\infty}F(z) = \arctan x +\frac \pi 2 + \arctan \frac 1 x = 0, \quad xy < 1.
\end{cases}
$$
Summarize above cases, we get the formular $g(x)$ in the problem.
Second proof.
Denote
$$F(x,y) = \arctan x + \arctan y - \arctan \frac{x+y}{1-xy}.$$
Noticing that the function $xy=1$ separates the $x-y$ plane into three connected domains
$$E_1 =\{(x,y)\mid x,y >0, xy > 1\},$$
$$E_2 = \{(x,y)\mid x,y <0, xy > 1\},$$
$$E_3= \{(x,y)\mid xy < 1\},$$ and $F_x(x,y) = F_y(x,y) =0$, we find that $F$ is constant in each $E_i (i=1,2,3)$.
Thus
$$
F(x,y)=\begin{cases}
F(\sqrt 3,\sqrt 3) = \pi, \quad &(x,y)\in E_1,\\
F(-\sqrt 3,-\sqrt 3) = -\pi, &(x,y)\in E_2,\\
F(0,0)=0, & (x,y) \in E_3.
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 2
} |
Find the sum $\sum _{n=1}^{\infty }\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$ $$\sum _{n=1}^{\infty }\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$$
On their own, all three are divergent, so I thought the best way would be to rewrite it as:
$$\frac{\sqrt{n+2}-2\sqrt{n+1}-\sqrt{n}}{\sqrt{n+2}-2\sqrt{n+1}-\sqrt{n}}\cdot \left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)=\frac{\left(\left(\sqrt{n+2}-2\sqrt{n+1}\right)^2-n\right)}{\left(\sqrt{n+2}-2\sqrt{n+1}-\sqrt{n}\right)}$$
But that doesn't really make anything simpler.
| $$\sqrt{n+2}-2\sqrt{n+1}+\sqrt n=\sqrt{n+2}-\sqrt{n+1}-(\sqrt{n+1}-\sqrt n)$$ $$=\dfrac1{\sqrt{n+2}+\sqrt{n+1}}-\dfrac1{\sqrt{n+1}+\sqrt n}=F(n+1)-F(n)$$
where $F(m)=\dfrac1{\sqrt{m+1}+\sqrt m}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Real solutions of $x^5+ax^4+bx^3+cx^2+dx+e=0$ If $2a^2＜5b$,prove that the equation $x^5+ax^4+bx^3+cx^2+dx+e = 0$
has at least one complex root。Thanks.
| I am assuming all coefficients are real.
If
$f(x)
= x^5+ax^4+bx^3+cx^2+dx+e
$,
then
$f'(x)
= 5x^4+4ax^3+3bx^2+2cx+d
$,
$f''(x)
= 20x^3+12ax^2+6bx+2c
$,
$f'''(x)
= 60x^2+24ax+6b
$.
If $f$ has 5 real roots,
then $f'$ has 4 real roots,
$f''$ has 3 real roots,
and
$f'''$ has 2 real roots.
The roots of $f'''$
are the roots of
$0
= 10x^2+4ax+b
$
and the discriminant of this
is
$(4a)^2-4\cdot10b
=16a^2-40b
=8(2a^2-5b)
$.
If this has real roots,
then
$2a^2-5b \ge 0
$.
Therefore,
if $2a^2< 5b$,
$f'''$ has complex roots
so
$f$ can not have all real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is $2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$ equal to? I came across this question while doing my homework:
$$\Large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty=?$$
$$\small\text{OR}$$
$$\large\prod\limits_{x=1}^{\infty} (2x)^{\frac{1}{4x}} = ?$$
My Attempt:
$\large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$
$\large \Rightarrow 2^{\frac{1}{4}}\cdot2^{\frac{2}{8}}\cdot2^{\frac{3}{16}}\cdot2^{\frac{4}{32}}\cdots\infty$
$\large \Rightarrow 2^{(\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \cdots \infty)}$
OR, $\large 2^{\space (\sum\limits_{x=1}^{\infty} \frac{x}{2^{x+1}})}$
$\cdots$ That's it ... I am stuck here ... It does not resemble any series I know...
How do you do it? Thanks!
| The only thing left that we have to do is to evaluate the following infinite sum:
$$\sum\limits_{x=1}^{\infty} \frac{x}{2^{x+1}}$$
Dividing by $2$ would gives us $$\sum\limits_{x=1}^{\infty} \frac{x}{2^{x+2}}=\sum\limits_{x=2}^{\infty} \frac{x-1}{2^{x+1}}$$
Now, subtract this from the original equation. The limit should now not be too hard to find.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$.
If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$.
I feel that the problem basically uses algebraic manipulation even though it's in a Number Theory textbook. I don't realize how to show $(a^2+b^2+c^2)^2$ as the sum of three squares. I have tried algebraic manipulation but this is the stage I have reached.
$$(b^2 + c^2)^2 + a^2(a^2 + b^2 + c^2 + b^2 + c^2)$$
Could you give me some hints on how to proceed with this question?
Thanks.
| If:
$$x=2ac$$
$$y=2bc$$
$$z=a^2+b^2-c^2$$
$$n=a^2+b^2+c^2$$
Then:
$$x^2+y^2+z^2=n^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to compute this double integral I'm trying to show that $\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dxdy \neq \int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dydx$ by computing these integrals directly.
I tried using polar coordinates with no success as the bounds of integration caused problems.
I also tried the substitution $x=ytan\theta$ but ended up getting something of the form $\infty-\infty$.
Can anyone offer a hint as to how I can compute these directly please???
Thanks in advance!
| Put $$f(x,y) = \arctan\left( \tfrac x y \right), \,\,\,\,\,\, (x,y) \in (0,1).$$ We see $$\frac{\partial f}{\partial y} = \frac{1}{1+\left(\tfrac x y \right)^2}\left( - \frac x {y^2} \right) = -\frac{x}{x^2 + y^2}.$$ Then $$\frac{\partial }{\partial x}\frac{\partial f}{\partial y} = - \frac{1}{x^2 + y^2} + \frac{x}{(x^2 + y^2)^2}(2x) = \frac{x^2 - y^2}{(x^2+y^2)^2}.$$ Next put $$g(x,y) = -\arctan\left( \tfrac y x \right), \,\,\,\,\,\, (x,y) \in (0,1).$$ Then $$\frac{\partial g}{\partial x} = \frac{1}{1+ \left(-\tfrac{y}{x}\right)^2} \left(\frac{y}{x^2} \right) = \frac{y}{x^2 + y^2}$$ so $$\frac{\partial}{\partial y}\frac{\partial g}{\partial x} = \frac{1}{x^2+y^2} - \frac{y}{(x^2+y^2)^2}(2y) = \frac{x^2 - y^2}{(x^2+y^2)^2}.$$ Then \begin{align*} \int^1_0 \int^1_0 \frac{x^2 - y^2}{(x^2+y^2)^2} dx dy &= \int^1_0 \int^1_0 \frac{\partial }{\partial x}\frac{\partial f}{\partial y} dx dy \\
&= \int^1_0 \frac{\partial f}{ \partial y}(1,y) - \frac{\partial f}{ \partial y}(0,y) dy \\
&=- \int^1_0 \frac{1}{1+y^2} dy = - \arctan(1) = - \frac \pi 4
\end{align*} and \begin{align*} \int^1_0 \int^1_0 \frac{x^2 - y^2}{(x^2+y^2)^2} dy dx &= \int^1_0 \int^1_0 \frac{\partial }{\partial y}\frac{\partial g}{\partial x} dy dx \\
&= \int^1_0 \frac{\partial g}{ \partial x}(x,1) - \frac{\partial g}{ \partial x}(x,0) dy \\
&= \int^1_0 \frac{1}{1+x^2} dx = \arctan(1) = \frac \pi 4.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find limit $\lim_{n \to \infty} n \left(n^2\log\left(1+\frac{1}{n^2}\right) + n \log\left(1-\frac{1}{n}\right)\right) $ Define $a_n$ as:
$$
a_n = \left(\left(1+\frac{1}{n^2}\right)^{n^2}\left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)\right)^n
$$
Now I want to calculate $\lim_{n \to \infty} a_n$. So, real question is about other limit.
$$
\lim_{n\to\infty}\left(\left(1+\frac{1}{n^2}\right)^{n^2} \left(1-\frac{1}{n}\right)^n\right)^n
$$
As $\lim_{n\to\infty}\left(1+\frac{1}{n^2}\right)^{n^2} \left(1-\frac{1}{n}\right)^n = 1$, we cannot conclude limit in easy way. ($1^{\infty}$ case.) I don't know, what I should do now.
I tried:
$$
\ln \left(\left(1+\frac{1}{n^2}\right)^{n^2} \left(1-\frac{1}{n}\right)^n\right) =
n \left(n^2\ln(1+\frac{1}{n^2}) + n \ln(1-\frac{1}{n})\right)
$$
So, now problematic is limit:
$$
\lim_{n \to \infty} n \left(n^2\ln(1+\frac{1}{n^2}) + n \ln(1-\frac{1}{n})\right)
$$
It's not better. How should I calculate this limit?
| Let $f_n$ be the sequence given by $f_n=n \left(n^2\log\left(1+\frac{1}{n^2}\right) + n \log\left(1-\frac{1}{n}\right)\right)$. We can evaluate the limit of $f_n$ by expanding the logarithm functions embedded in the limit of interest as
$$\begin{align}
\lim_{n\to \infty}f_n&=\lim_{n \to \infty} n \left(n^2\log\left(1+\frac{1}{n^2}\right) + n \log\left(1-\frac{1}{n}\right)\right)\\\\
&=\lim_{n \to \infty} n\left(n^2\left(\frac{1}{n^2}+O\left(\frac{1}{n^4}\right)\right)+n\left(-\frac{1}{n}-\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right)\right)\right)\\\\
&=-\frac12
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
An inequality involving $\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$
$$\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$$
Let $(x, y, z)$ be non-negative real numbers such that $x^2+y^2+z^2=2(xy+yz+zx)$.
Question: Find the maximum value of the expression above.
My attempt:
Since $(x,y,z)$ can be non-negative, we can take $x=0$, then equation becomes
$$y^2 + z^2=2xy$$
This implies that $(y-z)^2=0$.
So this implies that the required value is $$\frac{y^3 + z^3}{(y+z)(y^2 + z^2)}=\frac{1}{2}$$
But this wrong as the correct answer is $\frac{11}{18}$.
What is wrong with my method?
| Let $P$ be the given expression we want to maximise.
From the hypothesis, we get $xy+yz+zx=\dfrac{1}{4}(x+y+z)^2$
So, $P=\dfrac{x^3+y^3+z^3}{2(x+y+z)(xy+yz+zx)}=\dfrac{2(x^3+y^3+z^3)}{(x+y+z)^3}$
$\qquad \;\;=2\left[\left(\dfrac{x}{x+y+z}\right)^3+\left(\dfrac{y}{x+y+z}\right)^3+\left(\dfrac{z}{x+y+z}\right)^3\right]$
Let $a=\dfrac{x}{x+y+z}; b=\dfrac{y}{x+y+z}; c=\dfrac{z}{x+y+z}$, we have $\left\{\begin{array}{l}a+b+c=1\\ab+bc+ca=\dfrac{1}{4}\end{array}\right]$
Or $\left\{\begin{array}{l}b+c=1-a\\bc=a^2-a+\dfrac{1}{4}\end{array}\right]$
From the inequality $(b+c)^2\ge 4bc$, we get $0\le a\le\dfrac{2}{3}$.
We have:
$P=2(a^3+b^3+c^3)$
$\quad =2(a^3+(b+c)^3-3bc(b+c)$
$\quad =2\left[a^3+(1-a)^3-3\left(a^2-a+\dfrac{1}{4}\right)(1-a)\right]$
$\quad =6a^3-6a^2+\dfrac{3}{2}a+\dfrac{1}{2}$
$\quad =\dfrac{11}{18}+\dfrac{1}{18}(3a-2)(6a-1)^2\le\dfrac{11}{18}$
So, $\max P=\dfrac{11}{18}\approx {\boxed {0.611}}$.
The equality holds when $(x,y,z)=(4k,k,k)$ or any permutations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solve $x^2 = 2^n + 3^n + 6^n$ over positive integers.
Solve $x^2 = 2^n + 3^n + 6^n$ over positive integers.
I have found the solution $(x, n) = (7, 2)$. I have tried all $n$'s till $6$ and no other seem to be there.
Taking $\pmod{10}$, I have been able to prove that if $4|n$ that this proposition does not hold. Can you give me some hints on how to proceed with this problem?
Thanks.
| If $n=2k+1\ge 3$, then $x^2\equiv 3\mod 4$.
If $n=2k\ge 4$, then $(x+2^k)(x-2^k)=x^2-2^{2k}=6^{2k}+3^{2k}=3^{2k}(1+2^{2k})$.
We have $\gcd (x-2^k, x+2^k)=1$, then
$x+2^k\ge 3^{2k} \Rightarrow x-2^k\ge 3^{2k}-2^{2k+1}>2^{2k}+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
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