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How to prove this inequality with the following condition? Let $x$, $y$, $z$ be three positive real numbers satisfying
\begin{equation}
x + y +z + 1 =4xyz.\tag{1}
\end{equation}
Prove that
\begin{equation}
xy + yz + zx \geqslant x + y + z.\tag{2}
\end{equation}
I don't know how to start?
|
Here is a boring, uninspiring but elementary proof ^_^. By $(1)$ and A.M.$\ge$G.M., we get $xyz = (x+y+z+1)/4 \ge (xyz)^{1/4}$. Hence $xyz\ge1$. WLOG, suppose $0<x\le y\le z$. There are three cases. Case 1: $x\ge1$. Clearly $(2)$ holds. Case 2: $0<x<1\le y< z$. Then $(2)$ also holds because
$$1-xyz\le0\le(1-x)(1-y)(1-z) = 1 - (x+y+z) + (xy+yz+zx) - xyz.$$
Case 3: $0<x\le y<1<z$. Then
\begin{align*}
(2)&\Leftrightarrow z(x+y) + xy \ge x+y+z,\\
&\Leftrightarrow z(x+y-1) + xy - x-y\ge0,\\
&\Leftrightarrow z(x+y-1)-1 + (x-1)(y-1) \ge0.
\end{align*}
So it suffices to show that $z(x+y-1)-1\ge0$. Since $x,y,z$ are positive, we must have $4xy>1$, otherwise the LHS of $(1)$ will be strictly greater than the RHS. Now $(1)$ implies that $z=(x+y+1)/(4xy-1)$. Therefore
$$
z(x+y-1)-1
=\frac{(x+y+1)(x+y-1)-(4xy-1)}{4xy-1}
=\frac{(x-y)^2}{4xy-1} \ge0
$$
and we are done. From the details of above three cases, it can be shown that equality in $(2)$ holds iff $x=y=z=1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the solution number of the equation $x^2-x+1\equiv 0 \pmod{p^e}$ What is the solution number of the equation
$$x^2-x+1\equiv 0\pmod{p^e}$$
I know when $e=1$, it is $1+\left(\frac{-3}{p}\right)$, and I guess it is the same for $e>1$, but can anyone provide a proof?
updated:
I know when $e=1$, the number is
$$
1+\left(\frac{-3}{p}\right)
$$
When $e>1$, it is said that the answer is the same, saying that
$$
1+\left(\frac{-3}{p}\right)=1+\left(\frac{-3}{p^e}\right)
$$
That's what puzzling me.
|
If $p>2,$$$p^e\mid (x^2-x+1)\iff p^e\mid (2x-1)^2+3$$
So, $$(2x-1)^2\equiv-3\pmod{p^e}$$
Applying Discrete Logarithm w.r.t some primitive root $g\pmod {p^e}$,
$2ind_g(2x-1)\equiv ind_g(-3)\pmod{p^{e-1}(p-1)}$ as $\phi(p^e)=p^{e-1}(p-1)$
Using Linear congruence theorem, the last equation is solvable iff $(2,p-1)\mid ind_g(-3)\iff 2\mid ind_g(-3)$ and in that case it has exactly $(2,p-1)=2$ solutions.
Now we can prove, $-3$ is a quadratic residue of $p^e,$ iff it is a quadratic residue modulo $p$ (See below)
So, the number of solutions of $$(2x-1)^2\equiv-3\pmod{p^e}$$ is $0=1-1$ or $2=1+1$ i.e. is $1+\left(\frac{-3}p\right)$ for all prime $p>3$
[Proof:
Now, if $-3$ is a quadratic residue modulo $p^s$ so there exists an integer $y$ such that $p^s\mid(y^2+3)\implies y^2+3=a\cdot p^s$ for some positive integer $a$
Now, $(y+b\cdot p^s)^2+3=y^2+3+2y\cdot b\cdot p^s+b^2p^{2s}=a\cdot p^s+2y\cdot b\cdot p^s+b^2p^{2s}$
If $p^{s+1}\mid (a\cdot p^s+2y\cdot b\cdot p^s+b^2p^{2s})\iff p\mid (a+2b)$ if $2s\ge s+1\iff s\ge 1$
$\implies 2b\equiv-a\pmod p$
so if $y^2\equiv-3\pmod{p^s}$ is solvable so will be $y^2\equiv-3\pmod{p^{s+1}}$
Using induction we say $-3$ is a quadratic residue of $p^e$ if it is a quadratic residue of $p$ for $e>1.$
Again if for $e>1,p^e\mid(y^2+3)\implies p\mid (y^2+3),$
so if $-3$ is a quadratic residue of $p^e,$ then it is a quadratic residue modulo $p$
So, $-3$ is a quadratic residue of $p^e,$ iff it is a quadratic residue modulo $p$]
|
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|
Decomposition of polynomial of n-degree into irreducible polynomials. I'm getting ready for the exam in algebra. And I have problem that I can't solve again.
I'm understand what I need to do if I have a polynomial with a given degree. But I very confused by arbitrary degree in this problem.
Given polynomial
$$x^{2n}-2x^n+2$$
The task is find expansion of the polynomial as a product of irreducible polynomials in $\mathbb{R}$.
Thanks in advance.
|
$$x^{2n}-2x^n+2=(x^n-1)^2+1=(x^n-1+i)(x^n-1-i)=(x^n-2^{0.5}e^{\frac{-i\pi}{4}})(x^n-2^{0.5}e^{\frac{i\pi}{4}})$$
Now note that:
$$x^n-2^{0.5}e^{\frac{-i\pi}{4}}=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})$$
$$x^n-2^{0.5}e^{\frac{i\pi}{4}}=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}})$$
Hence:
$$x^{2n}-2x^n+2=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}})(x-2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})$$
$$x^{2n}-2x^n+2=\prod_{k=1}^n(x^2-(2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\frac{2\pi ik}{n}}+2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})x+2^{\frac{1}{n}})$$
$$x^{2n}-2x^n+2=\prod_{k=1}^n(x^2-2(2^{\frac{1}{2n}}\cos({\frac{\pi}{4}+\frac{2\pi k}{n}}))x+2^{\frac{1}{n}})$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$ Some time ago I came across to the following integral:
$$I=\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$$ What are the hints on how to compute this integral?
|
You may like this method. Note
$$ \int_0^1x^ada=-\frac{1-x}{\ln x}, \text{ for }x>0, H_a=\sum_{k=1}^\infty\frac{a}{k(a+k)}, $$
and hence
\begin{eqnarray}
\int_0^1\frac{1-x}{1+x}\frac{dx}{\ln x}&=&-\int_0^1\int_0^1\frac{x^a}{1+x}dadx\\
&=&-\int_0^1\int_0^1\frac{x^a}{1+x}dxda\\
&=&-\frac{1}{2}\int_0^1(H_{\frac{a}{2}}-H_{\frac{a-1}{2}})da\\
&=&-\frac{1}{2}\int_0^1\sum_{k=1}^\infty\left(\frac{a}{k(a+2k)}-\frac{a-1}{k(a+2k-1)}\right)da\\
&=&-\frac{1}{2}\sum_{k=1}^\infty\left(4\ln k-2\ln(k^2-\frac{1}{4})\right)\\
&=&-\frac{1}{2}\ln\prod_{k=1}^\infty\frac{k^4}{(k^2-\frac{1}{4})^2}\\
&=&-\prod_{k=1}^\infty\frac{k^2}{k^2-\frac{1}{4}}\\
&=&-\ln\left(\frac{\pi}{2}\right).
\end{eqnarray}
In the last step, we used the Wallis Formula from here.
|
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|
Finding the integer solutions Find all integer solutions of
$$(a + b^2)(a^2 + b) = (a − b)^3.$$
Obviously $b = 0$ is one. But how to get other solutions?
|
Here is one approach.
Expand both sides and equate terms, so we have:
$$(a + b^2)(a^2 + b) = (a - b)^3$$
$$a^{3} + a^{2}b^{2} + ab + b^{3} = a^{3} - 3a^{2}b + 3a b^2 - b^{3}$$
We can eliminate the $a^{3}$ terms, so we are left with:
$$2b^{3} + a^{2}b^{2} - 3ab^{2} + 3a^{2}b +ab = 0$$
Now, we have a Cubic in the form of:
$f(x) = jx^{3} + kx^{2} +lx + m = 0$, with $j = 2, k = a^{2}-3a, l = 3a^{2}+a$ and $m = 0$.
Using the solution for the cubic formula, we can arrive at the three roots:
$b = 0$
$b = (\frac{1}{4})(-a^{2} + 3a - \sqrt{a - 8}(a+1)\sqrt{a})$
$b = (\frac{1}{4})(-a^{2} + 3a + \sqrt{a - 8}(a+1)\sqrt{a})$
Now, you only want integer solutions, so, we see, by inspection, that at:
$a = -1, a = 0, a = 8, a = 9$, we get integers in $b$.
For $a = -1$, we have $b = -1$
For $a = 8$, we have $b = -10$
For $a = 0$, and $a = 9$, we have $b = 0$
Also note, for $b = 0$, the original equation reduces to, $a^{3} = a^{3}$, so you can also freely choose $a's$ (so you gets lots of integer solutions).
If you do an Implicit Plot using WA, you can see the $b = 0$ cases.
Regards
|
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|
The last digit of $2^{2006}$ My $13$ year old son was asked this question in a maths challenge. He correctly guessed $4$ on the assumption that the answer was likely to be the last digit of $2^6$. However is there a better explanation I can give him?
|
The last digit of the number is the remainder of division by 10, since any number is represented as:
$$ a_n \cdot 10^n + a_{n-1} \cdot 10^{n-1} + \ldots + a_1 \cdot 10 + a_0 = \overline{a_na_{n-1}\ldots a_1a_0} $$ $$\quad \Leftrightarrow \quad $$ $$a_n \cdot 10^n + a_{n-1} \cdot 10^{n-1} + \ldots + a_1 \cdot 10 + a_0 \equiv a_0 \mod 10. $$
Therefore, it is enough to consider the remainder of division by $10$ of the original number.
According to the Chinese remainder theorem
$$ 2^{2006} \equiv x \mod 10 \quad \Leftrightarrow \quad \begin{cases} 2^{2006} \equiv x \mod 5 \\ 2^{2006} \equiv x \mod 2 \end{cases} \quad \Leftrightarrow \quad \begin{cases} 2^{2006} \equiv x \mod 5 \\ 0 \equiv x \mod 2\end{cases}$$
Since the Euler function of 5 is equal to $\phi(5) = 5-1 = 4$ (as well as for any prime number), then by Euler's theorem the system of comparisons can be rewritten as:
$$ \begin{cases} 2^2 \equiv x \equiv 4 \mod 5 \\ 0 \equiv x \mod 2\end{cases}$$
Going through all the numbers from $0$ to $10$, we make sure that only $4$ is suitable, so $x = 4$, which will be the last digit of the number.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to expand $\tan x$ in Taylor order to $o(x^6)$ I try to expand $\tan x$ in Taylor order to $o(x^6)$, but searching of all 6 derivative in zero (ex. $\tan'(0), \tan''(0)$ and e.t.c.) is very difficult and slow method.
Is there another way to solve the problem?
Any help would be greatly appreciated :)
|
This answer is inspired by the answer of coffeemath.
We know the following (from e.g. Wikipedia):
\begin{align}
&\sin(x) = x -\frac{x^3}{6} +\frac{x^5}{120}+ O(x^7) \\
&\cos(x) = 1 -\frac{x^2}{2} +\frac{x^4}{24} +\frac{x^6}{720} +O(x^8) \\
&\tan(x) = a +bx+ cx^2+dx^3+ex^4+fx^5+gx^6+ O(x^7)
\end{align}
Furthermore it holds $\tan(x) = \frac{\sin(x)}{\cos(x)} \Rightarrow \sin(x) =\tan(x)\cos(x)$. So we get
\begin{align}
x -\frac{x^3}{6} +\frac{x^5}{120}+ O(x^7) =& \bigl(1 -\frac{x^2}{2} +\frac{x^4}{24} +\frac{x^6}{720} +O(x^8) \bigr) \\ &\cdot\bigl(a +bx+ cx^2+dx^3+ex^4+fx^5+gx^6+ O(x^7)\bigr)
\end{align}
If we expand the term on the right hand side we obtain
\begin{align}
x -\frac{x^3}{6} +\frac{x^5}{120}+ O(x^7) &=a+bx+(-\frac{a}{2}+c)x^2+(-\frac{b}{2}+d)x^3
\\&+(\frac{a}{24}-\frac{c}{2}+e)x^4+(\frac{b}{24}-\frac{d}{2}+f)x^5
\\&+(\frac{a}{720}+\frac{c}{24}-\frac{e}{2}+g)x^6+ O(x^7)
\end{align}
From this we can deduct the linear system
\begin{align}
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0&0\\
0 & 1 & 0 & 0 & 0 & 0 &0\\
-1/2 &0 &1 & 0& 0& 0&0\\
0 & -1/2 & 0 & 1 & 0 & 0&0 \\
1/24 & 0 & -1/2 & 0 & 1 & 0&0\\
0 &1/24 & 0 & -1/2 & 0 & 1&0\\
1/720 & 0 & 1/24 & 0 & -1/2 & 0 &1
\end{pmatrix}
\begin{pmatrix}
a\\b\\c\\d\\e\\f\\g
\end{pmatrix}
=
\begin{pmatrix}
0\\1 \\ 0 \\ -1/6 \\ 0 \\1/120 \\ 0
\end{pmatrix}
\end{align}
Which has the solution
\begin{align}
\begin{pmatrix}
a\\b\\c\\d\\e\\f\\g
\end{pmatrix}
= \begin{pmatrix}
0 \\ 1 \\ 0 \\ 1/3 \\ 0 \\2/15
\end{pmatrix}
\end{align}
[Note that you can actually extract two linear systems of the one given above. You then only have to solve for the odd powers of $x$ as $\tan(x)$ is odd.]
So you finally can write your Taylor series as:
\begin{align}
&\tan(x) = x+ \frac{1}{3}x^3+\frac{2}{15}x^5+ O(x^7)
\end{align}
Which is (surprisingly) correct, as we can compare our solution with WolframAlpha.
|
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|
Taylor expansion for $\sqrt{x+2}$ I'm enrolled in Coursera's calculus with a single variable and am trying to solve one of the homework problems.
In lecture, it was stated that to expand $\sqrt x$ about $x=a$, you would have:
$$\sqrt{x} = \sqrt{a} + {1 \over 2 \sqrt{a}}(x-a)- {1\over 8 \sqrt{a^3}}(x-a)^2 + H.O.T$$
The homework hint says you can us the Binomial series to find the Taylor series expansion for expressions with non-integer powers.
Wikipedia says the Binomial series expands to
$$(x +1)^{ \alpha }= \sum \limits_{k=0}^{\infty} {\alpha \choose k} x^k$$
$${\alpha\choose{k}} = \frac{\alpha \cdot (\alpha - 1) \cdot (\alpha - 2) \cdot \dots \cdot (\alpha - k + 1)}{k!}$$
My first question is where the term $$a^{1/2 - k}$$ comes from, given the Binomial series formula.
My second question is how to properly evaluate the series about a particular value other than zero.
The homework problem asks me to compute the Taylor series for $$f(x) = \sqrt{x+2}$$ about $x=2$. I also tried to use substitution with $h=x+2$, $x=h-2$ and then compute the Taylor series expansion about h=0 using the definition of Taylor series formula with
$$\sum_{n=0} {{f^{(n)}\over n!}(x-a)^n}$$
$$f(h) = \sqrt{h-2}$$
But with $f(h=0)$, I get imaginary numbers.
|
This is what I have now:
Evaluating derivatives for $\sqrt{x+2}$ at x=2, I get:
$ f(0) = \sqrt{2+ (2)} = 2 $
$ f'(0) = {1\over2} {1 \over \sqrt{2+(2)} }= 1/4$
$ f''(0) = -{1\over 4} { 1 \over (\sqrt{ 2+(2)})^{3} }= -1/32$
The taylor series expansion for $ \sqrt{x+2} \space \space at \space \space x=2 :$
$2 + {1\over 4\cdot 1!} (x-2) - {1 \over 32 \cdot 2! }+...$
Which I think this is correct, but when I try to use the general Binomial theorem, I run into trouble.
$\sqrt{2+(x-2)}=\sqrt{2}\left(1+\frac{x-2}{2}\right)^{1/2}$
$(1+t)^{1/2}$ where $t=\frac{x-2}{2}$
$1 { {1/2} \choose 0} + \sqrt{2}{ 1/ 2 \choose 1}(1+t)+ 2^{3/2}{ 1/ 2 \choose 2}(1+t)^2 = $
$1 + \sqrt{2}\space{ {1\over 2} \over 1!}(1+t)+ 2^{3/2}\space{ {1\over 2} \cdot {-1 \over 2} \over 2!}(1+t)^2 + ...$
$1 + \sqrt{2}\space{ {1\over 2} \over 1!}(1+\frac{x-2}{2})+ 2^{3/2}\space{ {1\over 2} \cdot {-1 \over 2} \over 2!}(1+\frac{x-2}{2})^2 + ...$
I assume that ${ {1/2} \choose 0} = 1$, since there is only one way to choose 0 things, but I'm not sure if that is correct.
|
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|
Working with exponent on series Hi have this sequence:
$$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}$$
I understand that this is a Geometric series so this is what I've made to get the sum.
$$\sum\limits_{n=1}^\infty (-1)^n\frac{3^{n}\cdot 3^{-2}}{4^n}$$
$$\sum\limits_{n=1}^\infty (-1)^n\cdot 3^{-2}{(\frac{3}{4})}^n$$
So $a= (-1)^n\cdot 3^{-2}$ and $r=\frac{3}{4}$ and the sum is given by
$$(-1)^n\cdot 3^{-2}\cdot \frac{1}{1-\frac{3}{4}}$$
Solving this I'm getting the result as $\frac{4}{9}$ witch I know Is incorrect because WolframAlpha is giving me another result.
So were am I making the mistake?
|
$$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}=\sum\limits_{n=1}^\infty (-1)^n\frac{1}{9}\left(\frac{3}{4}\right)^n=\frac{1}{9}\sum\limits_{n=1}^\infty \left(-\frac{3}{4}\right)^n=
\frac{1}{9}\left(\frac{1}{1+3/4}-1\right)$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Determinants of $3\times 3$ Matrix Suppose that $a,b,c,d,e,f$ are numbers such that
$$\det\left(\begin{matrix}
a&1&d\\b&1&e\\ c&1&f
\end{matrix}\right)=7$$ and
$$\det\left(\begin{matrix}
a&1&d\\b&2&e\\ c&3&f
\end{matrix}\right)=11.$$
How do you find the determinant of the Matrix
$$\begin{pmatrix}
a&3&d\\b&5&e\\ c&7&f
\end{pmatrix}?$$
Any suggestions on how to approach the question would be greatly appreciated.
|
Using multilinearity of the determinant (by columns), we get:
$$\begin{vmatrix}
a&3&d\\b&5&e\\ c&7&f
\end{vmatrix}=\begin{vmatrix}
a&1+2&d\\b&1+4&e\\ c&1+6&f
\end{vmatrix}=\begin{vmatrix}
a&1&d\\b&1&e\\ c&1&f
\end{vmatrix}+2\begin{vmatrix}
a&1&d\\b&2&e\\ c&3&f
\end{vmatrix}=7+2\cdot 11=29$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})$ i cannot figure out a way to find this limit. $$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})$$
Its undeterminate form $0\cdot\infty$ so i tried using $$\displaystyle\lim_{n\to \infty}\frac{a^{X_n}-1}{X_n}=\ln{a}$$ this leads to $0\cdot\infty$ again.
I then tried to transform it in $\frac{0}{0}$ $$\displaystyle\lim_{n\to \infty} \frac{\sqrt[n]{2}-\sqrt[n+1]{2}}{\frac{1}{n^2}}$$ here i dont know what i could do to find the limit.
|
We have:
$
\begin{align*}
\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2}) & = \lim_{n\to \infty} n^2\left( 2^{\frac{1}{n}} - 2^{\frac{1}{n+1}} \right) = \lim_{n\to \infty} n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n}-\frac{1}{n+1}} - 1 \right) \\
& = \lim_{n\to \infty} n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n^2+n}} - 1 \right) \\
& = \lim_{n\to \infty} \cfrac{n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n^2+n}} - 1 \right)}{\cfrac{1}{n^2+n}(n^2+n)} \\
& = \lim_{n\to \infty} \dfrac{n^2}{n^2+n} \cdot \lim_{n\to \infty} 2^{\frac{1}{n+1}} \cdot
\lim_{n\to \infty} \cfrac{2^{\frac{1}{n^2+n}} - 1}{\cfrac{1}{n^2+n}} \\
& = 1 \cdot 1 \cdot \ln 2 \\
& = \boxed{\ln 2}.
\end{align*}
$
|
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|
Weird question pertaining to HCF I encountered this question which seems weird/incomplete to me :
Q: H.C.F. of 3240, 3600 and a third number is 36, and their L.C.M. is
$2^4 \cdot 3^5 \cdot 5^2 \cdot 7^2$ . The third number is?
Can anyone please teach me concept wise how to solve it?
|
\begin{align}
3240 & = 2\cdot2\cdot2\cdot3\cdot3\cdot3\cdot3\cdot5 & & = 2^3\cdot3^4\cdot5\\
3600 & = 2\cdot2\cdot2\cdot2\cdot3\cdot3\cdot5\cdot5 & & = 2^4\cdot3^2\cdot5^2 \\
36 & = 2\cdot2\cdot3\cdot3 & & =2^2\cdot3^2
\end{align}
The third number cannot have more than two $2$s in its prime factorization, since then the gcd would not be $36$, but at least $2$ times that. The third cannot have any $5$ in its prime factorization, since then the gcd would have a $5$ in it. The third number must have two $7$s in its prime factorization, since there's nothing else to contribute those two $7$s to the lcm. The third number must have at least two $2$s in its prime factoriation; otherwise it would not be divisible by $36$. The third number must have at least two $3$s in its prime factorization for the same reason.
So the third number could be $2\cdot2\cdot3\cdot3\cdot7\cdot7$.
It could also be $2\cdot2\cdot3\cdot3\cdot3\cdot7\cdot7$ or $2\cdot2\cdot3\cdot3\cdot3\cdot3\cdot7\cdot7$.
|
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|
closest point to on $y=1/x$ to a given point I feel like I'm missing something basic - given a point $(a,b)$ how do I find the closest point to it on the curve $y=1/x$? I tried the direct approach of pluggin in $y=1/x$ into the distance formula but it leads to an order-4 polynomial...
|
The slope of $y = 1/x$ at $x$ is $-1/x^2$,
so the tangent at $(p, 1/p)$ is
$(y-1/p)/(x-p) = -1/p^2$
or $y = 1/p - (x-p)/p^2$.
At $y = 0$, $(x-p)/p^2 = 1/p$
or $x = 2p$;
at $x = 0$, $y = 1/p + p/p^2 = 2/p$.
The intercept points are thus
$(0, 2/p)$ and $(2p, 0)$.
I will minimize the distance by making
the line from $(a, b)$ to
$(p, 1/p)$ perpendicular to this tangent
at $(p, 1/p)$
and do this by making their
dot product zero.
The dot product of the vector
from $(a, b)$ to $(p, 1/p)$
with the tangent is
$\begin{align}
(p-a, 1/p-b)\cdot (2p, -2/p)
&= 2p(p-a) - 2/p^2+2b/p \\
&= (2p^3(p-a) - 2 + 2pb)/p^2 \\
&= (2p^4 - 2ap^3 + 2bp-2)/p^2 \\
&= 2(p^4 - ap^3 + bp-1)/p^2
\end{align}
$
and this is zero when
$f(p; a, b) = p^4 - ap^3 + bp-1 = 0$.
I will consider the special case $a = b$
to check this;
there should be a root at $p = 1$.
If $a = b$,
$\begin{align}
f(p; a, a) &= p^4 - ap^3 + ap-1 \\
&= (p^4-1) - a(p^3-1) \\
&= (p-1)(p^3+p^2+p^1 - a(p^2+p+1)) \\
&= (p-1)(p^3 + (1-a)(p^2+p+1))
\end{align}
$
which has a root at $p = 1$
(as expected) and,
if $a > 1$
(so $(a, a)$ is to the right and above the hyperbola),
another positive real root
since $p^3 + (1-a)(p^2+p+1) < 0$
for small $p$
and $p^3 + (1-a)(p^2+p+1) > 0$
for large $p$.
If $a < 1$,
so $(a, a)$ is between the hyperbola and the origin,
there are no other positive real roots
since $p^3 + (1-a)(p^2+p+1) > 0$
for all $p \ge 0$.
I could argue further about
where the real roots
of $f(p; a, b)$
are using
$f(1/p; a, b) = (1/p)^4 - a(1/p)^3 + b/p-1
= (1 - ap + bp^3 - p^4)/p^4
= -f(p; b, a)/p^4
$
so the roots of
$f(1/p, a, b)$ are the same as the roots of
$f(p; b, a)$,
but that's enough for now for me.
In general, I find it much easier
algebraically to make two vectors
orthogonal by making their dot product zero
than minimizing their distance
using the Pythagorean theorem.
|
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|
Does an integer $9
Does an integer $9<n<100$ exist such that the last 2 digits of $n^2$ is $n$? If yes, how to find them? If no, prove it.
This problem puzzled me for a day, but I'm not making much progress. Please help. Thanks.
|
We are solving $n(n-1)=n^2-n\equiv0\pmod{100}$. Since $\gcd(n,n-1)=1$, one of $n$ or $n-1$ must be a multiple of $4$ while the other must be a multiple of $25$.This leads to the equations
$$
\begin{align}
4x-25y=+1\tag{1}\\
4x-25y=-1\tag{2}
\end{align}
$$
For $(1)$, $n=4x$ and $n-1=25y$. For $(2)$, $n=25y$ and $n-1=4x$.
Using the Euclidean algorithm, $(1)$ has solutions $(x,y)=(-6+25k,-1+4k)$ and $(2)$ has solutions $(6+25k,1+4k)$. The two solutions that give $4x$ and $25y$ between $9$ and $99$ are $(19,3)$ and $(6,1)$.
$(19,3)$ solves $(1)$ so $n=4x=76$ and $76^2=5776\equiv76\pmod{100}$
$(6,1)$ solves $(2)$ so $n=25y=25$ and $25^2=625\equiv25\pmod{100}$
Thus, the two integers that satisfy the given condition are $25$ and $76$.
|
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|
Implicit Differentiation of $x^2y = 1$ I have a slight problem with this: Given the equation
$$ x^2y = 1$$
and asked to find $y''$, I attempted to apply implicit differentiation by differentiation w.r.t. $y$.
$$2xy'y + x^2y' = 0$$
However, it does not seem to be right
UPDATE Solved:
Differentiate w.r.t. $x$
$$2xy + x^2y' = 0$$$$
y' = \frac{-2xy}{x^2} = \frac{-2y}{x}
$$
Subsequently,
$$y'' =\frac{x(-2y') - (-2y)(1)}{x^2} = \frac{-2xy' + 2y}{x^2} = \frac{6y}{x^2}$$
|
We are given that $x^2y = 1 \,\,\,\,\, (\spadesuit)$.
Hence, we have $$\dfrac{d(x^2y)}{dx} = 0 \implies \dfrac{d(x^2)}{dx}y + x^2 \dfrac{dy}{dx} = 0\implies 2xy + x^2 \dfrac{dy}{dx} = 0 \implies 2y + x\dfrac{dy}{dx}=0 \,\,\, (\star)$$
Now differentiate again to get
$$\dfrac{d}{dx} \left(2y + x \dfrac{dy}{dx}\right) = 0 \implies 2 \dfrac{dy}{dx} + \dfrac{d}{dx} \left(x \dfrac{dy}{dx}\right) = 0\\ \implies 2 \dfrac{dy}{dx} + \dfrac{dx}{dx} \dfrac{dy}{dx} + x \dfrac{d^2y}{dx^2} = 0 \implies 3 \dfrac{dy}{dx} + x \dfrac{d^2y}{dx^2} = 0 \,\,\,\, (\dagger)$$
From $(\star)$, we have $\dfrac{dy}{dx} = - \dfrac{2y}x$. Plugging this in $(\dagger)$, we get that
$$x \dfrac{d^2y}{dx^2} + 3 \times \left(- \dfrac{2y}x\right) = 0 \implies \dfrac{d^2y}{dx^2} = \dfrac{6y}{x^2}$$ From $(\spadesuit)$, we have $y = \dfrac1{x^2}$ and hence
$$\dfrac{d^2y}{dx^2} = \dfrac6{x^4}$$
You could do a direct differentiation by noticing that $y = \dfrac1{x^2}$. This implies $$\dfrac{dy}{dx} = -\dfrac2{x^3} \implies \dfrac{d^2y}{dx^2} = \dfrac6{x^4}$$
|
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|
Application for mean value theorem $f(x)$ is three-times differentiable on $[a,b]$, how to show that there is $\varepsilon\in(a,b)$ such that
$$f(b)=f(a)+\cfrac{1}{2}(b-a)[f'(a)+f'(b)]-\cfrac{1}{12}(b-a)^3f'''(\varepsilon)$$
|
Let $$F(x) := f(b)-f(x)- \frac{1}{2} (b-x) \cdot (f'(b)+f'(x)) - K \cdot (b-x)^3$$ where $$K := \frac{f(b)-f(a)-\frac{1}{2} (b-a) \cdot (f'(b)+f'(a))}{(b-a)^3} \tag{1}$$
Then $F$ is differentiable on $[a,b]$ and $F(a)=F(b)=0$. ($K$ is chosen such that $F(a)=0$.) By Rolle's theorem there exists $\varrho \in (a,b)$ such that $F'(\varrho)=0$, i.e.
$$ \begin{align} 0 = F'(\varrho) &= \underbrace{-f'(\varrho)+ \frac{1}{2} (f'(b)+f'(\varrho))}_{\frac{1}{2} (f'(b)-f'(\varrho))}- \frac{1}{2}(b-\varrho) \cdot f''(\varrho) + 3K \cdot (b-\varrho)^2 \\ \Leftrightarrow - \frac{1}{2} f'(b) &= - \frac{1}{2}f'(\varrho) - \frac{1}{2}f''(\varrho) \cdot (b-\varrho) + 3K \cdot (b-\varrho)^2 \tag{2} \\ \Leftrightarrow g(b) &= g(\varrho) + g'(\varrho) \cdot (b-\varrho) +3K \cdot (b-\varrho)^2 \end{align} $$
where $g:=-\frac{1}{2} f'$. This already looks like a Taylor expansion of $g$. In fact, by applying Taylor's theorem to $g$ we obtain
$$\begin{align} g(b) &= g(\varrho) + g'(\varrho) \cdot (b-\varrho) + \frac{g^{(2)}(\varepsilon)}{2!} \cdot (b-\varrho)^2 \\ \Leftrightarrow - \frac{1}{2} f'(b) &= - \frac{1}{2} f'(\varrho) - \frac{1}{2} f''(\varrho) \cdot (b-\varrho) - \frac{1}{2} \cdot \frac{f^{(3)}(\varepsilon)}{2!} \cdot (b-\varrho)^2 \tag{3} \end{align}$$
for some $\varepsilon \in (\varrho,b) \subseteq (a,b)$. Now we subtract $(3)$ from $(2)$:
$$\begin{align} 0 &= 3K \cdot (b-\varrho)^2 + \frac{f^{(3)}(\varepsilon)}{4} \cdot (b-\varrho)^2 \\ \Rightarrow K &= - \frac{1}{12} \cdot f^{(3)}(\varepsilon) \end{align}$$
Hence
$$ \begin{align} - \frac{1}{12} \cdot f^{(3)}(\varepsilon) &= K \stackrel{(1)}{=} \frac{f(b)-f(a)-\frac{1}{2} (b-a) \cdot (f'(b)+f'(a))}{(b-a)^3} \\ \Leftrightarrow f(b) &= f(a) + \frac{1}{2} (b-a) \cdot (f'(b)+f'(a)) - \frac{1}{12} \cdot (b-a)^3 \cdot f^{(3)}(\varepsilon) \end{align}$$
|
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|
Solving complicated fraction for X I realize this rather rudimentary but it has been over a decade since my algebra classes and now I have problem that I can't figure out. I would like someone to walk me through the steps in solving the "X" in this problem:
$$
1.20 \cdot 10^6 = \frac{1}{\left(\dfrac{1}{6 \cdot 10^6}\right) X}
$$
So what I've done so far is this:
$$1,200,000 = \frac{1}{166,666.67X}$$
$$(166,666.67X) * 1,200,000 = \frac{1}{166,666.67X} * (166,666.67X)$$
$$200,000,004,000x = 1$$ But this doesn't make sense.
I supposed to get $x = 5$.
Thanks!
EDIT:
The formula is this:
1
MTTF(system) = ---------------
n 1
E ----------
i=1 MTTFi
This is my best as I can't upload the image and not sure how else to post this. So the "E" is actually the greek letter for sumation. The "n" would be 1 since this only one system.
MTTF(system) = 1,200,000;
MTTFi = 6,000,000
|
$$1.20 \cdot 10^6 = \dfrac{1}{\left(\frac{1}{6\cdot10^6}\right)X} \iff 1.2\cdot 10^6 = \frac{6\cdot 10^6}{X}$$
$$\iff X =\frac{6\cdot 10^6}{1.2\cdot 10^6} = \frac {6}{1.2} = 5$$
The nice thing about the problem is that there is no need to evaluate numerically until the very end, where we need only compute $X =\dfrac{6}{1.2}$ (factors of $10^6$ cancel out, in the process)
Another way to think of the problem is as follows (multiplying both sides of the equation by $\dfrac{1}{6\cdot 10^6}$ to cancel out the fraction in the denominator of the right-hand side: $$1.20 \cdot 10^6 = \dfrac{1}{\left(\frac{1}{6\cdot10^6}\right)X} \iff 1.2\cdot 10^6 \cdot \frac{1}{6\cdot 10^6} = \frac{1}{6\cdot 10^6} \cdot \frac{1}{\frac{1}{6\cdot 10^6}X}$$
$$\iff \frac{1.2}{6} = \frac{1}{X} \iff 1.2x = 6\iff X = \frac{6}{1.2} = 5 $$
|
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|
The maximum of $|x^3 + ax^2 + bx + c|$ on $[-1, 1]$ is at least $1/4$ Let $f(x)=x^{3}+ax^{2}+bx+c$ with a, b, c real.
Show that
$$\frac{1}4 \le \max_{-1 \le x \le 1\hspace{2mm}} |f(x)|=M$$
and find all cases where equality occurs.
|
Note that
$$
\max_{[-1,1]}\,\left|\,x^3-\tfrac34x\,\right|=\tfrac14\tag{1}
$$
Considering the symmetry about $0$ of the domain, we have for any $t\in[0,1]$
$$
\max_{\{t,-t\}}\,\left|\,x^3+ax^2+bx+c\,\right|=\left|\,t^3+bt\,\right|+\left|\,at^2+c\,\right|\tag{2}
$$
Using $(2)$, it is obvious that
$$
M_b=\max_{[-1,1]}\,\left|\,x^3+bx\,\right|\le\max_{[-1,1]}\,\left|\,x^3+ax^2+bx+c\,\right|\tag{3}
$$
It is straightforward to compute
$$
M_b=\left\{\begin{array}{}
2(-b/3)^{3/2}&\text{if }b\in\left[-3,-\tfrac34\right]\\
|\,1+b\,|&\text{otherwise}
\end{array}\right.\tag{4}
$$
and $M_b$ reaches a minimum of $\frac14$ only at $b=-\frac34$. For any other value of $b$, $(3)$ says that
$$
\max_{[-1,1]}\,\left|\,x^3+ax^2+bx+c\,\right|\ge M_b>\tfrac14\tag{5}
$$
Setting $b=-\frac34$ and $t=\frac12$ in $(2)$ yields
$$
\max_{[-1,1]}\,\left|\,x^3+ax^2-\tfrac34x+c\,\right|\ge\tfrac14+\left|\,\tfrac14a+c\,\right|\tag{6}
$$
and this can be $\frac14$ only if $c=-\frac a4$.
At $|x|=\frac12$,
$$
\left|\,x^3+ax^2-\tfrac34x-\tfrac a4\,\right|=\tfrac14\tag{7}
$$
However, at $x=\pm\frac12$, the derivative of $x^3+ax^2-\frac34x-\frac a4$ is $\pm a$. Therefore, the maximum of $\left|\,x^3+ax^2-\tfrac34x-\tfrac a4\,\right|$ will be greater than $\frac14$ unless $a=0$.
Thus,
$$
\max_{[-1,1]}\,\left|\,x^3+ax^2+bx+c\,\right|\ge\tfrac14\tag{8}
$$
where equality holds only for $x^3-\frac34x$.
|
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|
$\det(\exp X)=e^{\mathrm{Tr}\, X}$ for 2 dimensional matrices I want to prove that for $X\in M_2(\mathbb{R})$ the formula $\det(\exp X)=e^{\mathrm{Tr}\, X}$ holds, writing $X$ in normal form gives $X=PJP^{-1}$, where $J$ is the Jordan matrix, now $\exp (PJP^{-1})=P(\exp J)P^{-1}$ and $\det P(\exp J)P^{-1}=\det \exp J$. If $J=\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ where we allow $a=b$, we get that $\det \exp J=e^{a+b}=e^{\mathrm{Tr}\, J}=e^{\mathrm{Tr}\, X}$, since $X$ and $J$ are similar. However, if $J=\begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}$, then $\exp J=\exp\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \exp \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} e^a & e^{a+1} \\ e^a & e^{a} \end{pmatrix}$, so $\det \exp J=e^{2a}(1-e)$, which is not the result I want.
|
$$
\exp\begin{pmatrix}0&1\\0&0\end{pmatrix}=I+\begin{pmatrix}0&1\\0&0\end{pmatrix}+\frac1{2!}\begin{pmatrix}0&1\\0&0\end{pmatrix}^2+\ldots
=I+\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}1&1\\0&1\end{pmatrix}.
$$
|
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|
Rolling three dice...am I doing this correctly? Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?
Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first
For 1) The first die, you have $\frac{6}{6}$. The second die needs to be equal to the first, so you have probability of $\frac{1}{6}$. Then the third die can't be equal to the first and second dice, so it's $\frac{5}{6}$.
All together you get $1 \cdot \frac{1}{6} \cdot \frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3 \cdot \left(1 \cdot\frac{1}{6} \cdot \frac{5}{6}\right)=\frac{5}{12}$$
Did I do this correctly?
Thank you.
|
Yes your answer is correct. $${3\choose 2}\cdot\frac{1}{6}\cdot\frac{5}{6} = \frac{5}{12}$$
Good job!
|
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|
Another integral with Catalan Show that:
$$\int_0^1\frac{\arcsin^3 x}{x^2}\text{d}x=6\pi G-\frac{\pi^3}{8}-\frac{21}{2}\zeta(3)$$
I evaluated this by some Fourier series. Is there any other method?
Start with substitution of $$u=\arcsin x$$
Then we have to integrate $$\int_0^{\frac{\pi}{2}}\frac{u^3\cos u}{\sin^2 u}\text{d}u=-\int_0^{\frac{\pi}{2}}u^3\csc u\text{d}u$$
Since $$\int\csc u\text{d}u=\ln (\csc u-\cot u)=\ln \left(\frac{1-\cos x}{\sin x}\right)=\ln 2+2\ln \left(\sin \frac{x}{2}\right)-\ln \sin x$$
Thus $$\int_0^{\frac{\pi}{2}}u^2\csc u\text{d}u=\int_0^{\frac{\pi}{2}}u^2\text{d}\left(2\ln \frac{\sin u}{2}-\ln \sin u\right)$$
$$=-\frac{\pi^2}{4}\ln 2-2\int_0^{\frac{\pi}{2}}u\left(2\ln \sin \frac{u}{2}-\ln \sin u\right)$$
$$=-\frac{\pi^2}{4}\ln 2-4\int_0^{\frac{\pi}{2}}u\ln \sin \frac{u}{2}\text{d}u+2\int_0^{\frac{\pi}{2}}u\ln \sin u\text{d}u$$
$$=-\frac{\pi^2}{4}\ln 2+4\int_0^{\frac{\pi}{2}}u\left[\ln 2+\sum_{n=1}^{\infty}\frac{\cos nu}{n}\right]\text{d}u-\int_0^{\frac{\pi}{2}}u^2\cot u\text{d}u$$
$$=\frac{\pi^2}{4}\ln 2+4\sum_{n=1}^{\infty}\frac{1}{n}\int_0^{\frac{\pi}{2}}u\cos nu\text{d}u-\frac{\pi^2}{4}\ln 2+\frac78\zeta(3)$$
$$=\frac78\zeta(3)+2\sum_{n=1}^{\infty}\frac{-2+2\cos \frac{n\pi}{2}+n\pi \sin \frac{n\pi}{2}}{n^3}$$
$$=\frac78\zeta(3)-4\zeta(3)-\frac38\zeta(3)+2\pi G=2\pi G-\frac72\zeta(3)$$
Combine these gives $$\int_0^1\frac{\arcsin ^3x}{x^2}\text{d}x=6\pi G-\frac{\pi^3}{8}-\frac{21}{2}\zeta(3)$$
|
Integrating by parts twice, we get
$$
\int_0^{\pi/2}x^2\,e^{ikx}\,\mathrm{d}x
=i^{k-1}\frac{\pi^2}{4k}+i^k\frac\pi{k^2}+\frac2{k^3}\left(i^{k+1}-i\right)
$$
Therefore, using $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$,
$$
\begin{align}
&\int_0^1\frac{\arcsin^3(x)}{x^2}\mathrm{d}x\\
&=\int_0^{\pi/2}\frac{x^3}{\sin^2(x)}\mathrm{d}\sin(x)\\
&=-x^3\csc(x)\Big]_0^{\pi/2}+3\int_0^{\pi/2}\csc(x)\,x^2\,\mathrm{d}x\\
&=-\frac{\pi^3}{8}+3\int_0^{\pi/2}x^2\,\frac{2ie^{-ix}\,\mathrm{d}x}{1-e^{-2ix}}\tag{$\lozenge$}\\
&=-\frac{\pi^3}{8}+6i\sum_{k=0}^\infty\int_0^{\pi/2}x^2\,e^{-(2k+1)ix}\,\mathrm{d}x\\
&=-\frac{\pi^3}{8}+6i\sum_{k=0}^\infty\left((-1)^k\frac{\pi^2}{8k+4}-i(-1)^k\frac\pi{(2k+1)^2}-\left((-1)^k-i\right)\frac2{(2k+1)^3}\right)\\
&=-\frac{\pi^3}{8}+6\sum_{k=0}^\infty\left(\frac{(-1)^k\pi}{(2k+1)^2}-\frac2{(2k+1)^3}\right)\tag{$\ast$}\\
&=-\frac{\pi^3}{8}-\frac{21}{2}\zeta(3)+6\pi\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\\
&=-\frac{\pi^3}{8}-\frac{21}{2}\zeta(3)+6\pi G
\end{align}
$$
where $G$ is Catalan's Constant. In $(\ast)$, we drop the imaginary part (which should be $0$).
There is a question about the convergence in $(\lozenge)$. To handle this, we can consider
$$
x^2\frac{2ie^{-ix}}{1-e^{-2ix}}=\lim_{r\to1^-}x^2\frac{2ire^{-ix}}{1-r^2e^{-2ix}}
$$
and the convergence in the sums is uniform as $r\to1^-$.
|
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|
Irreducible elements of $\mathbb{Z}[i\sqrt{5}]$ Is there any neat way to show that $9$ and $3-3i\sqrt{5}$ are irreducible elements of $\mathbb{Z}[i\sqrt{5}]$, while $1+4i\sqrt{5}$ and $5-2i\sqrt{5}$ are reducible?
|
$9 = 3 \cdot 3$ is definitely reducible, and so is $3-3i\sqrt{5} = 3 \cdot (1-i\sqrt{5})$.
The norm of $a = 1+4i\sqrt{5}$ is $81$. There are no elements of norm $3$, so if it reducible, $a$ could only be the product of two elements of norm $9$, and these are $\pm 3, \pm 2 \pm i \sqrt{5}$. And in fact $a = (-2 + i \sqrt{5}) (2 - i \sqrt{5})$.
Finally, the norm of $b = 5-2i\sqrt{5}$ is $45$, and a similar argument leads to $b = -i \sqrt{5} \cdot (2 + i \sqrt{5})$.
So barring mistakes they all appear to be reducible, it seems.
|
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|
Can $n(n+1)2^{n-2} = \sum_{i=1}^{n} i^2 \binom{n}{i}$ be derived from the binomial theorem? Can this identity be derived from the binomial theorem?
$$n(n+1)2^{n-2} = \sum_{i=1}^{n} i^2 \binom{n}{i}$$
I tried starting from $2^n = \displaystyle\sum_{i=0}^{n} \binom{n}{i}$ and dividing it by $4$
in order to get $2^{n-2}$.
|
Start from the binomial theorem in the form
$$(x+1)^n=\sum_{k=0}^n\binom{n}kx^k$$
and differentiate with respect to $x$:
$$\begin{align*}
n(x+1)^{n-1}&=\sum_{k=0}^n\binom{n}kkx^{k-1}\\\\
&=\sum_{k=1}^n\binom{n}kkx^{k-1}\;.
\end{align*}$$
Differentiate again:
$$\begin{align*}
n(n-1)(x+1)^{n-2}&=\sum_{k=1}^n\binom{n}kk(k-1)x^{k-2}\\\\
&=\sum_{k=1}^n\binom{n}kk^2x^{k-2}-\sum_{k=1}^n\binom{n}kkx^{k-2}\;.
\end{align*}$$
Now let $x=1$ to get
$$\begin{align*}
n(n-1)2^{n-2}&=\sum_{k=1}^n\binom{n}kk^2-\sum_{k=1}^n\binom{n}kk\\\\
&=\sum_{k=1}^n\binom{n}kk^2-\sum_{k=1}^n\binom{n-1}{k-1}n\\\\
&=\sum_{k=1}^n\binom{n}kk^2-n\sum_{k=0}^{n-1}\binom{n-1}k\\\\
&=\sum_{k=1}^n\binom{n}kk^2-n2^{n-1}\\\\
&=\sum_{k=1}^n\binom{n}kk^2-2n2^{n-2}\;,
\end{align*}$$
and solve for $\displaystyle\sum_{k=1}^n\binom{n}kk^2$ to get the desired result.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Identify a formula for each entry of the matrix $\small \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}^n$ Identify a formula for each entry of the matrix $\begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}^n$.
It's easy to find a solution by just looking at the first few results:
\begin{pmatrix} 7^{n-1} & 3\cdot 7^{n-1} \\ 2\cdot 7^{n-1} & 6\cdot 7^{n-1} \end{pmatrix}
But how would I do it if it wasn't so obvious?
|
Fleshing out a little what the comments hint you:
$$\det(tI-A)=\begin{vmatrix}t-1&-3\\-2&t-6\end{vmatrix}=t(t-7)\Longrightarrow$$
the matrix's eigenvalues are $\,0\,,\,7\,$ , and eigenvectors for these values can be found as follows:
$$(1)\,\,\lambda=0:\;\;\;\;\begin{cases}\;\,-x-3y=0\\{}\\-2x-6y=0\end{cases}\;\;\Longrightarrow x=-3y\Longrightarrow\,\,\text{for example}\,\,\binom{3}{\!\!-1}$$
$$(1)\,\,\lambda=t:\;\;\;\;\begin{cases}\;6x-3y=0\\{}\\-2x+y=0\end{cases}\;\;\Longrightarrow y=2y\Longrightarrow\,\,\text{for example}\,\,\binom{1}{2}$$
Form now the matrix
$$P:=\begin{pmatrix}3&1\\\!\!\!-1&2\end{pmatrix}\Longrightarrow P^{-1}=\begin{pmatrix}\frac{2}{7}&\!\!\!-\frac{1}{7}\\{}\\\frac{1}{7}&\;\frac{3}{7}\end{pmatrix}$$
so that
$$P^{-1}AP=\begin{pmatrix}0&0\\0&7\end{pmatrix}=:D$$
And now it is a piece of cake:
$$A^n=\left(PDP^{-1}\right)^n=PD^nP^{-1}\;\;,\;\;\text{and}\;\;\;D^n=\begin{pmatrix}0&0\\0&7^n\end{pmatrix}\ldots$$
|
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|
Finding coefficient of generating function
Find the coefficient of $x^{52}$ in $$(x^{10} + x^{11} + \ldots + x^{25})(x + x^2 + \ldots + x^{15})(x^{20} + x^{21}+ \ldots + x^{45})$$
One thing I tried doing was factoring out $x^{10}, x, x^{20}$ from each of the products, respectively, then using the identity of a product of two polynomials to obtain the coefficients...But I ended up with $1$. I can't figure this out and any help would be much appreciated!
|
The original problem, after factoring out the terms as stated in the post, we have
$$
x^{10}(1+x+\ldots+x^{15})\cdot x(1 + x + \ldots+x^{14})\cdot x^{20}(1+x+\ldots+x^{25}) \tag{1}
$$
Applying some identities, $(1)$ gives
$$ x^{31}\cdot\frac{1-x^{16}}{1-x}\cdot\frac{1-x^{15}}{1-x}\cdot\frac{1-x^{26}}{1-x} \tag{2}$$
Rearranging the terms, we then have
$$
x^{31}\cdot(1-x^{16})\cdot(1-x^{15})\cdot(1-x^{26})\cdot \frac{1}{(1-x)^3} \tag{*}
$$
The last term in the above product is equal to
$$1 + \binom{3}{1}x + \binom{4}{2}x^2 + \ldots \tag3$$
Finally we take the products that can result in $x^{52}$ and obtain the coefficients:
*
*$x^{31}$ (first term in $(*)$) and $x^{21}$ (from $(3)$),
*$x^{31}$, $x^{16}$ (second term in $(*)$), $x^5$ in $(3)$
*$x^{31}$(first term in $(*)$), $x^{15}$ (third term in $(3)$), and $x^6$ in $(3)$
Since all the coefficients of the polynomials equal $1$ or $-1$ except for the polynomial expanded in $(3)$, we have as our coefficient
$$
\binom{21+3-1}{21} - \binom{6+3-1}{6} - \binom{5+3-1}{5} = 204
$$
Note: I hadn't seen Andre's solution prior to typing this.
|
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|
linear algebra, show B is a basis for $\mathbb{R}^3$ Let $B = \{(1,0,1), (1,1,2), (1,2,4)\}$
a) Show that $B$ is a basis for $\mathbb{R}^3$
b) Compute the coordinate vector $[v]_B$ with respect to $B$ of the vector $v = (3, -1, 3).$
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To show it is a basis you have to show it generates $\mathbb{R}^3$, so every $$v=\begin{pmatrix} x \\y \\z \end{pmatrix}$$ can be written as
$$\lambda_1 \begin{pmatrix} 1\\ 0\\1 \end{pmatrix} + \lambda_2 \begin{pmatrix} 1\\1\\2\\ \end{pmatrix} + \lambda_3 \begin{pmatrix} 1\\2\\ 4\\ \end{pmatrix}$$
A basis is minimal, so you have to show that
$$\lambda_1 \begin{pmatrix} 1\\ 0\\1 \end{pmatrix} + \lambda_2 \begin{pmatrix} 1\\1\\2\\ \end{pmatrix} + \lambda_3 \begin{pmatrix} 1\\2\\ 4\\ \end{pmatrix} = 0 \implies \lambda_1=\lambda_2=\lambda_3=0$$
You don't need to calculate this here, because 3 vectors which span $\mathbb{R}^3$ must be linear independent.
If you already know what a determinant is, you can even calculate
$$\begin{vmatrix} 1 & 0 & 1 \\ 1 & 1 & 2 \\ 1 & 2 & 4\\ \end{vmatrix}=1$$
in b) you shall calculate $\lambda_1, \lambda_2, \lambda_3 $ for the given vector.
|
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|
Proof of triangle inequality I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that $|a+b| \leq |a|+|b|$. Any help would be appreciated :)
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A simple proof of the triangle inequality that is complete and easy to understand (there are more cases than strictly necessary; however, my goal is clarity, not conciseness).
Prove the triangle inequality $| x | + | y| ≥ | x + y|$.
Without loss of generality, we need only consider the following cases:
*
*$x = 0$
*$x > 0, y > 0$
*$x > 0, y < 0$
Case $1$. Suppose $x = 0$. Then we have
$| x| = 0$
$| x| + | y| = 0 + | y| = | y|$
Thus $| x| + | y| = | x + y|$.
Case $2$. Suppose $x > 0, y > 0$. Then, since $x + y > 0$, we have
$| x| = x$
$| y| = y$
$| x| + | y| = x + y$
$| x + y| = x + y$
Thus $| x| + | y| = | x + y|$.
Case $3$. Suppose $x < 0, y < 0$. Then, since $x + y < 0$, we have
$| x| = −x$
$| y| = −y$
$| x| + | y| = (−x) + (−y)$
$| x + y| = −(x + y) = (−x) + (−y)$
Thus $| x| + | y| = | x + y|$.
Case $4$. Suppose $x > 0, y < 0$. Then we have
$| x| = x$
$| y| = −y$
$| x| + | y| = x + (−y)$
We must now consider three cases:
a. $x + y = 0$
b. $x + y > 0$
c. $x + y < 0$
Case $4a$. Suppose $x + y = 0$. Then we have
$| x + y | = |0| = 0$
Since $y < 0$, it follows that $−y > 0$ and thus $x + (−y) > 0 + (-y) = -y > 0$.
Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.
Case $4b$. Suppose $x + y > 0$. Then we have
$| x + y| = x + y$
Since $y < 0$, it follows that $−y > 0 > y$ and thus $x + (−y) > x + y$.
Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.
Case $4c$. Suppose $x + y < 0$. Then we have
$| x + y| = −(x + y) = (−x) + (−y)$
Since $x > 0$, it follows that $x > 0 > −x$ and thus $x + (−y) > (−x) + (−y)$.
Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.
This concludes the proof.
|
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|
Computing limits which involve square roots, such as $\sqrt{n^2+n}-n$ Is there a general strategy for this? For example I'm working on the limit
$$\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n $$
I have a simple argument to show that this limit is less than or equal to 1/2, but I can't get much further because it's difficult for me to manipulate the square root symbol. Here is the argument:
$\sqrt{n^2 + n} - n \lt \sqrt{n^2 +n + \frac{1}{4}} - n = n+\frac{1}{2} - n = 1/2$
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$\begin{align}
\sqrt{n^2 + n} - n & = (\sqrt{n^2 + n} - n) \cdot \frac{\sqrt{n^2 + n} + n}{\sqrt{n^2 + n} +n} \\
& = \frac{n^2+n-n^2}{\sqrt{n^2 + n} + n} \\
& = \frac{1}{\frac{\sqrt{n^2 + n} + n}{n}} \\
& = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} \rightarrow \frac{1}{2}\\
\end{align}$
|
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|
How to integrate $\int\frac{x}{(x^2-2x+5)^2} \, dx$? $$\int\frac{x}{(x^2-2x+5)^2} \, dx$$
I tried to complete the square of the bottom like this $\int\frac{x}{((x-1)^2+4)^2} \, dx$ but I'm still not sure what to do.
Any help would be appreciated.
Thanks
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A good strategy, now that you've completed the square is to also "complete the differential":
$$\int\frac{x}{((x-1)^2+4)^2} \, dx=$$
$$\frac 1 2\int\frac{2(x-1)}{((x-1)^2+4)^2} \, dx+\int\frac{1}{((x-1)^2+4)^2} \, dx=$$
$$\frac 1 2\int\frac{du}{u^2} +\int\frac{1}{(u^2+4)^2} \, du=$$
$$-\frac 1 2\frac{1}{u} +\int\frac{1}{(u^2+4)^2} \, du=$$
Now, to solve $$\int\frac{1}{(u^2+4)^2} \, du$$
take $2\tan \rho =u$
It will follow
$$\int\frac{1}{(u^2+4)^2} \, du$$
$$\int\frac{2\sec^2\rho }{16\sec^4 \rho } \,d\rho$$
$$\int\frac{1}{8\sec^2 \rho } \,d\rho=\frac 1 8\int{\cos^2\rho} \,d\rho$$
Now use the fact that $\dfrac{1+\cos(2x)}{2}=\cos^2 x$ to finish it off.
|
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|
Find $\sum\limits_{k\, \text{ odd}} \frac{2(k^2-1)}{k^4+k^2+1}$ How to find $$\sum_{k \text{ odd}} \frac{2(k^2-1)}{k^4+k^2+1}$$ Here we find $\displaystyle\sum_{k=1}^{\infty} \frac{2(k^2-1)}{k^4+k^2+1}=1$ and we know that $\displaystyle\sum_{k \,\text{odd}} + \sum_{k \,\text{even}}=1.$ Can we use this information to find sum? Or, maybe we can find it on other way?
We can say that our sum is equal to $$\color{red}{2}\sum_{k=1}^{\infty}\frac{(2k-1)^2-1}{(2k-1)^4+(2k-1)^2+1}=\sum_{k=1}^{\infty} \left(\frac{1-4k}{4k^2-2k+1}+\frac{4k-3}{4k^2-6k+3}\right),$$ but I don't think we can see something from that.
EDIT: Actually, Wolfram find that $$2\sum_{k=1}^{\infty}\frac{(2k-1)^2-1}{(2k-1)^4+(2k-1)^2+1}=\pi \text{sech}\left(\frac{\sqrt{3} \pi}{2}\right).$$ It's pretty nice closed form.
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I write here so everyone can see. I finally solve it.
First, note that $$\sum_{k \text{ odd}} \frac{2(k^2-1)}{k^4+k^2+1}=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{k^2-k+1},$$ because $$\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{k^2-k+1}=\sum_{k\, \text{odd}}\left(\frac{2k-1}{k^2-k+1}-\frac{2(k+1)-1}{(k+1)^2-(k+1)+1}\right).$$
Now, use formula $$\frac{1}{\cos \pi z}=\frac{4}{\pi}\sum_{k=0}^{\infty} (-1)^k \frac{2k+1}{(2k+1)^2-(2z)^2}=\frac{4}{\pi}\sum_{k=1}^{\infty} (-1)^{k-1} \frac{2k-1}{(2k-1)^2-(2z)^2}$$ and set $z=i\cdot \alpha.$ We find $$\text{sech}(\pi \alpha)=\frac{4}{\pi}\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{(2k-1)^2+4\alpha^2}$$ and then setting $\displaystyle\alpha = \frac{\sqrt{3}}{2}$ we find $$\text{sech}\left(\frac{\sqrt{3}\pi}{2}\right)=\frac{4}{\pi}\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{(2k-1)^2+3}=\frac{1}{\pi}\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{k^2-k+1}$$ or $$\sum_{k \text{ odd}} \frac{2(k^2-1)}{k^4+k^2+1}=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{k^2-k+1}=\pi \text{sech}\left(\frac{\sqrt{3}\pi}{2}\right).$$
Added: Formula $$\frac{1}{\cos \pi z}=\frac{4}{\pi}\sum_{k=0}^{\infty} (-1)^k \frac{2k+1}{(2k+1)^2-(2z)^2}$$ is derived from (same link as above) $$\pi \tan \pi z = \sum_{k=0}^{\infty} \frac{8z}{(2k+1)^2 - 4z^2}, \qquad (2z \neq \pm 1, \pm 3, \dots).$$
From $\dfrac{1}{\sin z}=\cos z + \tan \dfrac{z}{2}$ we find, further, that
$$\frac{\pi}{\sin \pi z}=\frac{1}{z}-\frac{2z}{z^2-1^2}+\frac{2z}{z^2-2^2}\pm\cdots, \qquad (z \neq 0, \pm 1, \pm 2, \dots),$$
and finally, replacing $z$ here by $\dfrac{1}{2}-z,$
$$\frac{\pi}{4 \cos \pi z}=\frac{1}{1^2-(2z)^2}-\frac{3}{3^2-(2z)^2}+\frac{5}{5^2-(2z)^3}\pm \cdots, \qquad (2z \neq \pm 1, \pm 3, \dots).$$
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|
Using Taylor's Theorem to show that $\ln(1 + x^2) \leq x^2$ Can we show that if $\operatorname{abs}(x) \lt 1$, then $$\ln(1+x^2) \leq x^2\;,$$
using Taylor's Theorem?
I am thinking of expanding it about $x=0$ but I got something like
$$f(x) = -x^2 + \frac{x^4}{2} - \dots$$
Is my approach correct? Could you give me some hints/guides here?
Thanks.
|
$$
\begin{array}{rcl}
\ln(1 + x^2) & \le & x^2 \\
e^{\ln(1 + x^2)} & \le & e^{x^2} \\
1 + x^2 & \le & e^{x^2} \\
1 + x^2 & \le & 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots \\ {}
0 & \le & \frac{x^4}{2!} + \frac{x^6}{3!} + \frac{x^8}{4!} + \dots \\ {}
\end{array}
$$
Which is true for all real $x$.
|
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|
Complete induction: $\sum^n_{i=1}\frac{1}{(2i-1)(2i+1)}=\frac{n}{2n+1}$ I am very confused with complete induction. Because in every task there is something different to do, and I never know what to insert (thats my biggest problem).
Here's the example:
Proof with complete induction. Please please help me, because I have exams coming up (I am just becoming a primary school teacher..)
For $n\in\mathbb{N}$:
$$\sum^n_{i=1}\frac{1}{(2i-1)(2i+1)}=\frac{n}{2n+1}$$
|
For a full solution, proceed like this:
$n=1$: $$\sum_{i=1}^1 \frac{1}{(2i-1)(2i+1)} = \frac{1}{(2-1)(2+1)} = \frac{1}{3} = \frac{1}{2 \cdot 1 +1},$$
so it holds for $n=1$.
Assume next that it holds for some generic $n$. You need to show that then it also holds for $n+1$. As it holds for $n$, you can assume that
$$\sum_{i=1}^n \frac{1}{(2i-1)(2i+1)} = \frac{n}{2n+1}. \quad (1),$$
and want to show that
$$\sum_{i=1}^{n+1} \frac{1}{(2i-1)(2i+1)} = \frac{n+1}{2(n+1)+1}. \quad (2)$$
Then:
$$\begin{align}
\sum_{i=1}^{n+1} \frac{1}{(2i-1)(2i+1} &= \sum_{i=1}^n \frac{1}{(2i-1)(2i+1)} + \frac{1}{(2(n+1)-1)(2(n+1)+1)} \\
& = \frac{n}{2n+1} + \frac{1}{(2n+1)(2n+3)} \quad \text{using (1)} \\
& = \frac{n(2n+3)}{(2n+1)(2n+3)} + \frac{1}{(2n+1)(2n+3)} \\
& = \frac{2n^2 +3n +1}{(2n+1)(2n+3)} \\
& = \frac{(n+1)(2n+1)}{(2n+1)(2n+3)} = \frac{n+1}{2(n+1)+1},\\
\end{align}$$
which is (2), and
was to be shown.
|
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|
Trigonometry tangent line question How would I figure this out.
Find all x values between $0$ and $2\pi$ where the line tangent to the graph of
$y=\frac{\cos x}{2+ \sin(x)}$ is horizontal.
I did the deriavative
$\frac{(2+\sin(x)-\sin(x)+\cos(x)\cos(x)}{(2+\sin x)^2}$ but I think I need to find $x$
|
For a horizontal tangent line you need the derivative to be zero. After all, $\operatorname{d}\!y/\!\operatorname{d}\!x$ gives you the gradient of the tangent line to the graph $y=f(x)$. We are told that
$$y = \frac{\cos x}{2+\sin x}$$
Using the quotient rules, and the standard trig identity $\sin^2x+\cos^2x \equiv 1$ we see that
$$\frac{\operatorname{d}\!y}{\operatorname{d}\!x} = \frac{1+2\sin x}{\cos^2x-4\sin x-5} \equiv \frac{1+2\sin x}{(2+\sin x)^2}$$
Since $-1 \le \sin x \le 1$ for all real $x$ it follows that the horizontal tangent lines are gives by $1+2\sin x = 0$. In other words $\sin x = -\frac{1}{2}$. The principal value of which is $x=-\frac{\pi}{6}$. Plotting the graph of $y=\sin x$ and the line $y=-\frac{1}{2}$ can help to find the other solutions. We have:
$$x \in \left\{2\pi n - \frac{\pi}{6} : n \in \mathbb{Z} \right\} \cup \left\{(2n+1)\pi + \frac{\pi}{6} : n \in \mathbb{Z} \right\}$$
Restricting this to the interval $[0,2\pi]$ we have $x=\frac{7}{6}\pi$ and $x=\frac{11}{6}\pi$.
|
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|
Integrating :$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x dx$ How to integrate :
$$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x dx$$
|
$$\begin{aligned}
\int \sqrt{\sin x} \cos^{3/2}x\,\mathrm{d}x &=\int\sqrt{\sin x\cos x}\cos x\,\mathrm{d}x\\
&=\frac{1}{\sqrt{2}}\int\sqrt{\sin 2x}\cos x\,\mathrm{d}x\\
&\overset{(1)}{=}\frac{1}{\sqrt{2}}\left(\sin x\sqrt{\sin 2x} - \int\sin x\frac{\cos 2x}{\sqrt{\sin 2x}}\,\mathrm{d}x\right)\\
&=\frac{\sin x\sqrt{\sin 2x}}{\sqrt{2}}-\frac{1}{2}\int\sqrt{\tan x} \cos 2x\,\mathrm{d}x\\
&\overset{(2)}{=}\frac{\sin x\sqrt{\sin 2x}}{\sqrt{2}}-\frac{1}{2}\left(\frac{\sin 2x \sqrt{\tan x}}{2}-\frac{1}{2}\int\sin 2x \frac{\sec^2 x}{2\sqrt{\tan x}}\,\mathrm{d}x\right)\\
&= \frac{\sin x\sqrt{\sin 2x}}{\sqrt{2}}-\frac{\sin 2x \sqrt{\tan x}}{4} + \frac{1}{4}\int \sqrt{\tan x}\,\mathrm{d}x
\end{aligned}$$
To evaluate $\displaystyle{\mathcal{B}=\int\sqrt{\tan x}\,\mathrm{d}x}$, first set $u = \sqrt{\tan x}$ (or $x = \arctan\left(u^2\right)$) and $\mathrm{d}x=\dfrac{2u}{1 + u^4}\,\mathrm{d}u$ to get
$$
\begin{aligned}
\int\frac{2u^2}{u^4 + 1}\,\mathrm{d}u&=\color{red}{\int\frac{u^2 + 1}{u^4 + 1}\,\mathrm{d}u} + \int\frac{u^2 - 1}{u^4 + 1}\,\mathrm{d}u\\
&=\frac{1}{2}\int\frac{\mathrm{d}u}{\left(u + 1/\sqrt{2}\right)^2 + \left(1/\sqrt{2}\right)^2} + \frac{1}{2}\int\frac{\mathrm{d}u}{\left(u - 1/\sqrt{2}\right)^2 + \left(1/\sqrt{2}\right)^2}+\underbrace{\int\frac{1-\dfrac{1}{u^2}}{\left(u + \dfrac{1}{u}\right)^2 - 2}\,\mathrm{d}u}_{\text{Set }w=u+1/u\text{ and }\mathrm{d}w=\left(1-1/u^2\right)\,\mathrm{d}u}\\
&=\frac{\sqrt{2}}{2}\arctan\left(u\sqrt{2}+1\right)+\frac{\sqrt{2}}{2}\arctan\left(u\sqrt{2}-1\right) + \int\frac{\mathrm{d}w}{w^2 - 2}\\
&=\frac{\sqrt{2}}{2}\arctan\left(u\sqrt{2}+1\right)+\frac{\sqrt{2}}{2}\arctan\left(u\sqrt{2}-1\right)-\dfrac{1}{2\sqrt{2}}\ln\left|\dfrac{w+\sqrt{2}}{w-\sqrt{2}}\right| + C_0\\
&\overset{(3)}{=}\frac{\sqrt{2}}{2}\arctan\left(u\sqrt{2}+1\right)+\frac{\sqrt{2}}{2}\arctan\left(u\sqrt{2}-1\right) - \dfrac{1}{2\sqrt{2}}\ln\!\left(\frac{u^2+u\sqrt{2}+1}{u^2-u\sqrt{2}+1}\right)+C_1\\
&=\frac{\sqrt{2}}{2}\arctan\left(\sqrt{2\tan x}+1\right)+\frac{\sqrt{2}}{2}\arctan\left(\sqrt{2\tan x}-1\right) - \dfrac{1}{2\sqrt{2}}\ln\!\left(\frac{\tan x+\sqrt{2\tan x}+1}{\tan x-\sqrt{2\tan x}+1}\right)+C_1
\end{aligned}
$$
Then,
$$ \int \sqrt{\sin x} \cos^{3/2} x\,\mathrm{d}x = \frac{\sin x\sqrt{\sin 2x}}{\sqrt{2}}-\frac{\sin 2x \sqrt{\tan x}}{4} + \frac{\sqrt{2}}{8}\arctan\left(\sqrt{2\tan x}+1\right)+\frac{\sqrt{2}}{8}\arctan\left(\sqrt{2\tan x}-1\right) - \dfrac{1}{8\sqrt{2}}\ln\!\left(\frac{\tan x+\sqrt{2\tan x}+1}{\tan x-\sqrt{2\tan x}+1}\right) + C $$
Notes:
$(1)$ – Integrate by parts with $u=\sqrt{\sin 2x}$ and $\mathrm{d}v = \cos x\,\mathrm{d}x$
$(2)$ – Integrate by parts with $u=\sqrt{\tan x}$ and $\mathrm{d}v = \cos 2x\,\mathrm{d}x $
$(3)$ – I've removed the absolute value bars because the argument is always positive. It is easier to see if you complete the square.
$\color{red}{(*)}$ – You could use the same technique applied to the integral at right, but, in this case, this trick, although clever, leads to a discontinuity in the primitive (at $u=0$). Look:
$$\begin{aligned} \int \frac{u^2+1}{u^4+1}\,\mathrm{d}u &= \int\frac{1+1/u^2}{\left(u-1/u\right)^2 + 2}\,\mathrm{d}u\\&=\frac{1}{\sqrt{2}}\arctan\!\left(\frac{u - 1/u}{\sqrt{2}}\right)+ C \end{aligned}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving the nonlinear ODE: $ \frac{1}{y^{\prime}}+\left(x-\frac{y}{y^{\prime}}\right)^{2}+1=0$ $$\frac{1}{y^{\prime}}+\left(x-\frac{y}{y^{\prime}}\right)^{2}+1=0$$
I am trying to solve this non-linear ODE and have tried all sorts of substitutions; any hints on how I should progress?
Thanks.
|
Let $y = a x + b$, then $y' = a$, and you have the equation:
$$\begin{align} \dfrac{1}{a} + \left(x - x - \dfrac{b}{a}\right)^2 + 1 & = 0\\ \dfrac{1}{a} + \frac{b^2}{a^2} + 1 & = 0 \\ a + b^2 + a^2 & = 0 \\ \left(a + \frac{1}{2}\right)^2 + b^2 & = \left(\frac{1}{2}\right)^2 \end{align} $$
So the solutions are lines with coefficients lying on a circle
To quickly check if there are any more solutions, we may wish to differentiate the equation and see if $y'' = 0$. Multiplying first by $(y')^2$, we have
$$(1 - 2xy)y' + (x^2 + 1)(y')^2 + y^2 = 0 $$
Differentiating,
$$(-2y)y' - 2x (y')^2 + (1 - 2xy) y'' + 2x(y')^2 + 2(x^2 + 1)y' y'' + 2y y' = 0 $$
Simplifying,
$$(1 - 2xy)y'' + 2(x^2 + 1)y' y'' = 0 $$
Leaving us with
$$y'' = 0$$ or $$1 - 2xy + 2(x^2 + 1)y' = 0 $$
In the second case, we have an ODE which we can solve with an integrating factor
$$y' - \frac{x}{x^2 + 1} y = \frac{-1}{2(x^2 + 1)}$$
This gives us equations of the form $$y = -\frac{x}{2} + C \sqrt{1 + x^2}$$
Substituting this back into the original equation, we see this has a solution when $4 C^2 = 1$, or when $C = \pm \dfrac{1}{2}$.
|
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|
what is the no. of possible parellelograms? A parallelogram having an acute angle of 30 degrees whose area is equal to the perimeter and the sides are whole numbers then number of such parallelograms possible are?
|
We interpret the question to mean that we want to find the number of (unordered) pairs $a$, $b$ of integers that can be the sides. The perimeter is $2a+2b$. The area is twice the area of the triangle we get by joining the two vertices at which there is a $150^\circ$ angle.
The area of each of these triangles is $(1/2)(ab)(\sin 30^\circ)$, for a total of $(1/2)ab$. So we are looking at the equation $2a+2b=(1/2)(ab)$, or equivalently $ab=4a+4b$.
This equation can be rewritten as $(a-4)(b-4)=16$. Thus $a-4$ and $b-4$ must have product $16$. Without loss of generality we may assume that $a\le b$.
It is easy to see that $a-4$ and $b-4$ cannot both be negative. So $a-4=1$, $b-4=16$, or $a-4=2$, $b-4=8$, or $a-4=4$, $b-4=4$.
That gives three parallelograms: one with sides $5$ and $20$; another with sides $6$ and $12$; and finally the rhombus with sides $8$ and $8$.
|
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|
Abstract Algebra: Equivalence Classes Under ~(conjugate) Relation Sorry. Don't know the latex.
Let $g=(1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8)$ in $S_8$
Find a specific permutation $f$ in $S_8$ so that $fgf^{-1} =(1 \ 5 \ 6 \ 3 \ 8 \ 7 \ 2 \ 4)$ (note: $fgf^{-1}$ is short for composition)
|
Use the conjugation rule $fgf^{-1} = (f(1)\; f(2)\; f(3)\; f(4)\; f(5)\; f(6)\; f(7)\; f(8))$.
EDIT: Upon request, I give more details (which in principle are already given by Berci in the comments):
If f and g are any permutations, you can compute $fgf^{-1}$ by simply writing down $g$ as a product of disjoint cycles, and then applying $f$ entry-wise to this representation of $g$.
So in your case, $fgf^{-1} = (f(1)\; f(2)\; f(3)\; f(4)\; f(5)\; f(6)\; f(7)\; f(8))$.
Since your exercise states $fgf^{-1} = (1 \ 5 \ 6 \ 3 \ 8 \ 7 \ 2 \ 4)$, we get $$(f(1)\; f(2)\; f(3)\; f(4)\; f(5)\; f(6)\; f(7)\; f(8)) = (1 \ 5 \ 6 \ 3 \ 8 \ 7 \ 2 \ 4).$$
Now we can simply read off that $f(1) = 1$, $f(2) = 5$, $f(3) = 6$, $f(4) = 3$, $f(5) = 8$, $f(6) = 7$, $f(7) = 2$, $f(8) = 4$.
That is (compare Berci's post) $$f = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\1 & 5 & 6 & 3 & 8 & 7 & 2 & 4\end{pmatrix}$$
or as a product of disjoint cycles
$$f = (2 \ 5\ 8\ 4\ 3\ 6\ 7).$$
|
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|
Prove $\frac{1}{2\sqrt{2}+1}+\frac{1}{3\sqrt{3}+2\sqrt{2}}+\cdots+\frac{1}{100\sqrt{100}+99\sqrt{99}}<\frac{9}{10}$ What would you suggest for the following inequality?
$$\frac{1}{2\sqrt{2}+1}+\frac{1}{3\sqrt{3}+2\sqrt{2}}+\cdots+\frac{1}{100\sqrt{100}+99\sqrt{99}}<\frac{9}{10}$$
Thanks in advance!
Sis.
EDIT: Based upon the nice solution provided by Sasha, I'll try to point out a
possible shortcut.
We might observe and use the fact that
$$a\sqrt{a}+b\sqrt{b}\ge a\sqrt{b}+b\sqrt{a}$$
because
$$(a-b)(\sqrt{a}-\sqrt{b})\ge0$$
|
$$
\sum_{m=1}^{99} \frac{1}{(m+1)^{3/2} + m^{3/2}} < \sum_{m=1}^{99} \left(\frac{1}{\sqrt{m}} - \frac{1}{\sqrt{m+1}} \right) = \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{99+1}} = \frac{9}{10}
$$
The above inequality is true since:
$$ \begin{eqnarray}
\frac{1}{(m+1)^{3/2} + m^{3/2}} &=& \frac{1}{\sqrt{m}\sqrt{m+1}} \left( \frac{m+1}{\sqrt{m}} + \frac{m}{\sqrt{m+1}} \right)^{-1} \\ &=& \frac{1}{\sqrt{m}\sqrt{m+1}} \left( \sqrt{m} + \sqrt{m+1} + \frac{1}{\sqrt{m}} - \frac{1}{\sqrt{m+1}} \right)^{-1} \\ &<& \frac{1}{\sqrt{m}\sqrt{m+1}} \frac{1}{ \sqrt{m} + \sqrt{m+1} } \\ &=& \frac{1}{\sqrt{m}} - \frac{1}{\sqrt{m+1}}
\end{eqnarray}
$$
|
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|
$xy=1 \implies $minimum $x+y=$? If $x,y$ are real positive numbers such that $xy=1$, how can I find the minimum for $x+y$?
|
$x+y=x+y-2\sqrt{xy}+2\sqrt{xy}=(\sqrt{x}-\sqrt{y})^2+2\ge2$
The value will be attained when $(\sqrt{x}-\sqrt{y})^2=0\Rightarrow \sqrt{x}=\sqrt{y}\Rightarrow x=y=1$
Please note that this is the basis for the inequality of A.M.$\ge$ G.M.
If $x,y\ge 0$ then we know that $(\sqrt{x}-\sqrt{y})^2\ge0$
$\Rightarrow x+y-2\sqrt{xy}\ge 0$
$\Rightarrow x+y\ge 2\sqrt{xy}$
|
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|
Finding f intervals How would I find the f intervals for the following two functions.
$f(x)=(x-2)^2(x+1)^2$
using the chain rule I got $(x-2)^2(2)(x+1)(1)+(2)(x-2)(1)(x+1)^2$
then I got f decrease $(-\infty,-1]$
and f increase $[2,\infty)$
but the area between -1 and 2 in confusing me.
My second function is
$f(x)=x+\frac{4}{x^2}$
differentiating I got $\frac{x^3-8}{x^3}$
so I got f increase $(\infty,0)$ and $[2,\infty)$
f decrease $(0,2]$
but did I do this correctly.
|
As you did $f'(x)$ correctly, we have $$f'(x)=2(x-2)(x+1)(2x-1)$$ When:
*
*$x\le -1~~$, $(x-2)\leq0$ and then two terms $(x+1)$ and $(2x-1)$ are negative. So $f'(x)$ is negative.
*$x\le\frac{1}2~~$, $(x-2)\leq0$ and $(x+1)>0$ and $(2x-1)<0$ is negative, so $f'(x)$ is positive.
*$x\le2~~$, $(x-2)\leq0$ and then two terms $(x+1)$ and $(2x-1)$ are positive, so $f'(x)$ is nagative again.
*And if $x>2$ all terms are positive and so $f'(x)$ is positive again.
|
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|
How to determine equation for $\sum_{k=1}^n k^3$ How do you find an algebraic formula for $\sum_{k=1}^n k^3$? I am able to find one for $\sum_{k=1}^n k^2$, but not $k^3$. Any hints would be appreciated.
|
One last Proof :
$$\sum_{k=1}^{n}k^{3}=1+8+27+64+...+n^{3}$$
$$\sum_{k=1}^{n}k^{3}=\underbrace{1}_{1^{3}}+\underbrace{3+5}_{2^{3}}+\underbrace{7+9+11}_{3^{3}}+\cdots+\underbrace{\left(n(n-1)+1\right)+\cdots+\left(n(n+1)-1\right)}_{n^{3}}$$
$$\sum_{k=1}^{n}k^{3}=1+3+5+\cdots+\left(n(n+1)-1\right)$$
Since $n(n+1) -1 $ is odd, then $n(n+1) -1 $ is the form $2k -1$
$$n(n+1) -1 = 2k -1 $$
$$n(n+1) = 2k $$
$$\underbrace{\frac{n}{2}(n+1)}_{\text{Arithmetic progression}} = k $$
$$\sum_{k=1}^{n}k^{3}=\underbrace{\underbrace{\underbrace{\underbrace{1}_{1^{2}}+3}_{2^{2}}+5}_{3^{2}}+\cdots+\left(n(n+1)-1\right)}_{ \left(\frac{n^{2}+n}{2}\right)^{2}}$$
$$\sum_{k=1}^{n}k^{3}=\left(\frac{n^{2}+n}{2}\right)^{2}=\frac{1}{4}n^2(n+1)^2$$
Note:
$\sum_{k=1}^{n}(2k-1)=1+3+5+7+9+...+2n-1=n^{2}$
|
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|
Construct a matrix transform consider
$\frac{dx}{dt} = Ax$ where $A$ is the matrix
$$
\begin{bmatrix}
1 & 0 & 1 \\
0 & 0 & -2 \\
0 & 1 & 0 \\
\end{bmatrix}
$$
construct a real matrix $P$ such that the change of coordinates $x=Py$ transforms our real equation to
$\frac{dy}{dt}=By$ where $B$ is the matrix
$$
\begin{bmatrix}
1 & 0 & 1 \\
0 & 0 & -\sqrt{2} \\
0 & \sqrt{2} & 0 \\
\end{bmatrix}
$$
then solve explicity for $y$ and evaluate the solution in terms of our original $x=Py$
My Solution so far...
I have diagonalized both $A$ and $B$ and got the following matrices respectively
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 &\sqrt{2} & 0) \\
0 & 0 & -\sqrt{2} \\
\end{bmatrix}
$$
and for $B$
$$
\begin{bmatrix}
1 & 0 & 1 \\
0 & \sqrt{2} & 0 \\
0 & 0 & -\sqrt{2} \\
\end{bmatrix}
$$
the eigenvectors for $A$ make the matrix
$$
\begin{bmatrix}
1 & 1+\sqrt{2} & -1-\sqrt{2} \\
0 & 1/\sqrt{2} & -1/\sqrt{2} \\
0 & 1 & 1 \\
\end{bmatrix}
$$
and the eigenvectors for $B$ make the matrix
$$
\begin{bmatrix}
1 & 1/\sqrt{2} & -1/\sqrt{2} \\
0 & 0 & 0 \\
0 & 0 &0 \\
\end{bmatrix}
$$
Thus the determinant of the eigenvector matrix for $B$ is $0$...I am now stuck. Please help!
|
First of all, your eigenvalues (for both $A$ and $B$) should be $1$ and $\pm i\sqrt{2}$. Since you have three distinct eigenvalues, the eigenvectors of either matrix are guaranteed to be linearly independent, and thus constitute a basis.
(By necessity, the eigenvectors corresponding to the complex eigenvalues must be complex.)
If $A=S_1\Lambda S_1^{-1}$ and $B=S_2\Lambda S_2^{-1}$, then $B=S_2S_1^{-1}AS_1S_2^{-1}$. This means that you get the desired transformation $P=S_1S_2^{-1}$, and with some luck this will be a real matrix (both $S_1$ and $S_2$ are complex).
|
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|
$n^2$ for bases to prove that the last digit of base b ends with 0 Consider numbers written in base $b$, where $x\leq q\leq z$. For which bases $q$ $n^2$ ends with if $n$ ends with $0$.
|
Hint: For any base $b$, things will go bad if there is a positive integers $x$ such that $x^2$ is divisible by $b$ but $x$ isn't. For example, $30^2$ is divisible by $50$ but $30$ isn't, so things will go bad for base $50$.
Added: Let us think about the specific bases $5$ to $9$. Let's see what we need to make sure that if $n^2$ ends in $0$, then $n$ ends in $0$. To say that $n^2$ ends in $0$ is just to say that the base $b$ divides $n^2$. To say that $n$ ends in $0$ just says $b$ divides $n$.
So we ask for the various bases, when does $b$ divides $n^2$ force $b$ divides $n$?
Case $b=5$: If $5$ divides $n^2$, does $5$ divide $n$? Sure. After all, $5$ is prime.
Case $b=6$: If $6$ divides $n^2$, does $6$ divide $n$? If $6$ divides $n^2$, then $2$ divides $n^2$ and $3$ divides $n^2$. But then $2$ divides $n$ and $3$ divides $n$, so $6$ divides $n$.
Case $b=7$: same reasoning as for $5$: if $7$ divides $n^2$ then $7$ divides $n$.
Case $b=8$: This one is different. Let $n=4$. Then $8$ divides $n^2$ but $8$ does not divide $n$.
Case $b=9$: This is like $b=8$. Let $n=6$. Then $9$ divides $n^2$ but $9$ does not divide $n$.
Generalization: The "bad" bases $8$ and $9$ are each divisible by a perfect square greater than $1$. Call a positive integer $b$ square-free if no perfect square $\gt 1$ divides $b$. If $b$ is square-free, like $b=10$, then if $n^2$ in base $b$ ends in a $0$, then so does $n$.
If $b\gt 1$ is not square-free, there will be some $n$ such that in base $b$, the number $n^2$ ends in a $0$, but $n$ does not.
|
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|
Gambling puzzle A math friend of mine showed me this strange gambling puzzle. There is a button in a casino and every time you press it you can win either $1$ or $0$ dollars. The probability of winning $1$ dollar depends on how many times you have pressed it so far. If you press it for the $x$-th time you have probability $x/M$ of winning a dollar. If $x$ gets to $M$ or greater then you just win a dollar every time.
How many button presses do you expect to have to press to win $N$ dollars?
|
Caution: This answer is WRONG. But I leave it here since my mistake may help to understand the question correctly. See edit in the last.
Suppose you play $L$ times. If $L > M$ then you will get
$$ \begin{align*}
E(L) &= \frac{1}{M} + \frac{2}{M} + \cdots + \frac{M}{M} + 1 + \cdots + 1 \\
&= \frac{1}{2}(M + 1) + (L - M)
\end{align*}$$
dollars. Otherwise you will get
$$ \begin{align*}
E(L) &= \frac{1}{M} + \frac{2}{M} + \cdots + \frac{L}{M} \\
&= \frac{1}{2M}L(L + 1)
\end{align*}$$
dollars. Hence, all you need to do is solve $E(L) = N$ for $L$: the answer would be
$$L = \begin{cases}
N + \frac{M - 1}{2} \qquad \left(N > \frac{1 + M}{2}\right) \\
\frac{-1 + \sqrt{1 + 8MN}}{2} \qquad \left(N \leq \frac{1 + M}{2}\right).
\end{cases}$$
Edit I: Thanks to Steven Stadnicki, I realize I was completely wrong. Even in case $N > M$, the expected value is
$$ N \frac{M!}{M^M} + \sum_{k = 1}^{M - 1} (N + k) \frac{M!}{M^M} \sum_{0 < n_1 < \cdots < n_k < M} \prod_{i = 1}^k \frac{M - n_i}{n_i} $$
but I cannot get its simpler form.
Edit II: I conduct a computational experiment and get the average values for $10^4$ trials to guess the expected value $E(M, N)$ as follow.
$$\begin{array}{l|lllll}
M \backslash N & 1 & 2 & 3 & 4 & 5 \\
\hline \\
1 & 1 & 2 & 3 & 4 & 5 \\
2 & 1.5 & 2.5 & 3.5 & 4.5 & 5.5 \\
3 & 1.9 & 3.0 & 4.0 & 5.0 & 6.0 \\
4 & 2.2 & 3.5 & 4.5 & 5.5 & 6.5 \\
5 & 2.5 & 3.9 & 5.0 & 6.0 & 7.0
\end{array}$$
|
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|
Generating Functions: Solving a Second-Order Recurrence I'm self-studying generating functions (using GeneratingFunctionology as a text). I came across this programming problem, which I immediately recognized as a modification of the Fibonacci sequence. I wanted to place my newly found generating function techniques to work, so I tried to solve the recurrence.
The problem basically boils down to solving:
$$a_{n+2} = a_{n+1} + ka_n$$
Where $n \ge 0$, and $k$ is some positive integer.
And, $a_0 = 1$, $a_1 = 1$.
However, my solution (below, but not really simplified, since I'm just plugging it into a computer) yields values that too great by a factor of $k$.
My solution:
$$a_n = \frac{r_2^{n+1} - r_1^{n+1}}{(r_2-r_1)r_1^{n+1}r_2^{n+1}}$$
Where:
$$r_1 = \frac{-1 - \sqrt{1+4k}}{2k}$$
$$r_2 = \frac{-1 + \sqrt{1+4k}}{2k}$$
If someone could help me understand where I went wrong, I'd much appreciate it. (I don't see why I'm too great by a factor of $k$.)
My work is as below:
$$a_{n+2} = a_{n+1} + ka_n$$
$$\sum_{n\ge0}a_{n+2}x^n = \sum_{n\ge0}a_{n+1}x^n + k\sum_{n\ge0}a_{n}x^n$$
Let $A(x) = \sum_{n\ge0}a_nx^n$. Then:
$$\frac{A(x) - a_0 - xa_1}{x^2} = \frac{A(x) - a_0}{x} + kA(x)$$
$$A(x)\left(\frac{1}{x^2} -\frac{1}{x} - k\right) = \frac{1+x}{x^2} - \frac{1}{x}$$
$$A(x)\left(\frac{1 - x - kx^2}{x^2}\right) = \frac{1}{x^2}$$
$$A(x)= \frac{1}{1 - x - kx^2}$$
Now, factoring the denominator yields two roots, $r_1$ and $r_2$ as denoted above. This, in turn, yields:
$$A(x) = \frac{1}{(x-r_1)(x-r_2)}$$
Partial fractions:
$$A(x) = \frac{1}{r_2-r_1}\frac{1}{(r_1 - x)} - \frac{1}{r_2-r_1}\frac{1}{(r_2 - x)}$$
$$A(x) = \frac{1}{(r_2-r_1)(r_1)}\frac{1}{(1 - \frac{x}{r_1})} - \frac{1}{(r_2-r_1)(r_2)}\frac{1}{(1 - \frac{x}{r_2})}$$
$$A(x) = \frac{1}{(r_2-r_1)(r_1)}\sum_{n\ge0}{\left(\frac{x}{r_1}\right)^n} - \frac{1}{(r_2-r_1)(r_2)}\sum_{n\ge0}{\left(\frac{x}{r_2}\right)^n}$$
Combining:
$$A(x) = \sum_{n\ge0} x^n \left(\frac{1}{(r_2-r_1)(r_1^{n+1})} - \frac{1}{(r_2-r_1)(r_2^{n+1})}\right)$$
Therefore:
$$a_n = \frac{r_2^{n+1} - r_1^{n+1}}{(r_2-r_1)r_1^{n+1}r_2^{n+1}}$$
|
$$1-x-kx^2=-kx^2-x+1\Rightarrow x_{1,2}=\frac{1\pm\sqrt{1+4k}}{2}$$so
$$1-x-kx^2=-kx^2-x+1=(-k)(x-x_1)(x-x_2)=(-k)(x-\frac{1-\sqrt{1+4k}}{2})(x-\frac{1+\sqrt{1+4k}}{2})$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove this inequality? $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$ let $a,b,c,d\ge 0$,and $a^2+b^2+c^2+d^2=3$,prove that
$ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$
I find this inequality are same as Crux 3059 Problem.
|
Let $a=\sqrt 3\cos x\cos y,b=\sqrt 3\cos x\sin y,c=\sqrt 3\sin x\cos y,d=\sqrt 3\sin x\sin y, \pi/2>x,y>0$.
|
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|
Integral $\int\frac{6^x}{9^x-4^x}dx $ How to solve this integral:
$$\int\frac{6^x}{9^x-4^x}dx $$
(I notice that $\frac{6^x}{9^x-4^x}=\frac{2^x3^x}{(3^x-2^x)(3^x+2^x)}$)
Thank you!
|
Divide top and bottom by $ 4 ^ x $.
We have:
$$\int\dfrac{\left(\frac{3}{2}\right)^x}{\left(\frac{9}{4}\right)^x-1}dx=\int\dfrac{\left(\frac{3}{2}\right)^x}{\left(\left(\frac{3}{2}\right)^x\right)^2-1}dx$$
is replaced:
$$u=\left(\frac{3}{2}\right)^x$$
$$du=\left(\frac{3}{2}\right)^x\cdot\ln\left(\frac{3}{2}\right)dx$$
replacing in the integral:
$$I=\dfrac{1}{\ln\left(\frac{3}{2}\right)}\int\dfrac{1}{u^2-1}du$$
solving:
$$I=\dfrac{1}{\ln\left(\frac{3}{2}\right)}\cdot\dfrac{1}{2}\int\dfrac{2}{u^2-1}du=\dfrac{1}{2\ln\left(\frac{3}{2}\right)}\int\left(\dfrac{1}{u-1}-\dfrac{1}{u+1}\right)du$$
$$I=\dfrac{1}{2\ln\left(\frac{3}{2}\right)}\left(\ln(u-1)-\ln(u+1)\right)=\dfrac{1}{2\ln\left(\frac{3}{2}\right)}\ln\left(\dfrac{u-1}{u+1}\right)$$
Finally:
$$\int\dfrac{6^x}{9^x-4^x}dx=\dfrac{1}{2\ln\left(\frac{3}{2}\right)}\ln\left\vert\dfrac{\left(\frac{3}{2}\right)^x-1}{\left(\frac{3}{2}\right)^x+1}\right\vert+C$$
if you want you can multiply top and bottom within the logarithm, by $ 2 ^ x $.
$$\int\dfrac{6^x}{9^x-4^x}dx=\dfrac{1}{2\ln\left(\frac{3}{2}\right)}\ln\left\vert{\dfrac{3^x-2^x}{3^x+2^x}}\right\vert+C$$
|
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|
Wolframalpha step-by-step of $\int\frac{\cos(x)}{\sqrt{2+\cos(2x)}}dx$ I wonder, where the minus sign goes after the first $u$-substitution of integral $\displaystyle\int\frac{\cos(x)}{\sqrt{2+\cos(2x)}}dx$?
|
Rewrite $2+\cos 2x$ as $3-2\sin^2 x$ and let $u=\sin x$. We end up at
$$\int \frac{du}{\sqrt{3-2u^2}}.$$
Let $\sqrt{2}u=\sqrt{3}{v}$. Then $du=\frac{\sqrt{3}}{\sqrt{2}}\,dv$, and we end up with
$$\int\frac{1}{\sqrt{2}}\frac{dv}{\sqrt{1-v^2}}.$$
Thus our integral is $\frac{1}{\sqrt{2}}\arcsin v+C$. Replace $v$ by $\frac{\sqrt{2}\sin x}{\sqrt{3}}$.
|
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|
Computing $\int \frac{6x}{\sqrt{x^2+4x+8}} dx$ I am trying to compute an indefinite integral. Thanks!
$$\int \frac{6x}{\sqrt{x^2+4x+8}} dx.$$
|
$$
\begin{align}
\int \frac{6x +12 -12}{\sqrt{x^2+4x+8}} dx
&= \int \frac{6x +12}{\sqrt{x^2+4x+8}} dx - \int \frac{12}{\sqrt{x^2+4x+8}} dx \\
&= \int \frac{6(x+2)}{\sqrt{x^2+4x+8}} dx - \int \frac{12}{\sqrt{x^2+4x+8}} dx \\
&= \int \frac{3(2x+4)}{\sqrt{x^2+4x+8}} dx - \int \frac{12}{\sqrt{x^2+4x+8}} dx.
\end{align}$$
Then take
$x^2+4x+8 = t$ substitution in 1st integral and performed perfect polynomial in second integral then simplify to get the answer.
|
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|
$ \sqrt{2-\sqrt{2}} $ simplified So I have a (nested?) square root $ \sqrt{2-\sqrt{2}} $. I know that $ \sqrt{2-\sqrt{3}} = \frac{\sqrt{6}-\sqrt{2}}{2} $. I know how to turn the simplified version into the complex one, but not vice versa. What is $ \sqrt{2-\sqrt{2}} $ simplified in this fashion and what are the steps?
|
A general way to solve such problems:
$$\sqrt{2-\sqrt{2}}=a$$
then
$a^2=2-\sqrt{2}\rightarrow(2-a^2)^2=2\rightarrow 4-4a^2+a^4=2\rightarrow a^4-4a^2+2=0$
Let $b=a^2$
then
Solve $$b^2-4b+2=0$$
As @Martin suggested, this is a way but the conslusion is not right. The right way should be
$$\sqrt{a+b\sqrt{c}}=\sqrt{d}+\sqrt{e}$$ and the following arithmetic similar to above.
|
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|
Express trigonometric expressions in terms of one trigonometric function What is the general process to solve problems such as this:
I'm preparing for this type of exam problem.
For Reference, similar/more advanced problem (with solution): http://prntscr.com/ws6xl
|
(Answered before the edit.) For the first question: If $x=\sin \theta $, then $\cos ^{2}\theta =1-\sin ^{2}\theta =1-x^{2}$ and $\cos \theta =\pm \sqrt{1-x^{2}}$. So $$\sin 2\theta =2\sin \theta \cos \theta =\pm 2x\sqrt{1-x^{2}}.\tag{1}$$
As for the second question divide $2\sin \theta \cos \theta$ by $1=\sin ^{2}\theta +\cos ^{2}\theta$ and express both numerator and denominator in terms of $\tan \theta =x$:
$$\begin{eqnarray*}
\sin 2\theta &=&2\sin \theta \cos \theta =\frac{2\sin \theta \cos \theta }{\sin ^{2}\theta +\cos ^{2}\theta } \\
&=&\frac{\dfrac{2\sin \theta \cos \theta }{\cos ^{2}\theta }}{\dfrac{\sin
^{2}\theta +\cos ^{2}\theta }{\cos ^{2}\theta }}=\frac{2\tan \theta }{1+\tan
^{2}\theta } \\
&=&\frac{2x}{1+x^{2}}\tag{2}.
\end{eqnarray*}$$
Added: apply the same technique to $\cos 2\theta=\cos^2 \theta-\sin^2 \theta$ to obtain
$$\cos 2\theta=\frac{1-x^2}{1+x^2}.\tag{3}$$
ADDED 2: To derive $(2)$ and $(3)$ you can draw a right triangle with horizontal side (cathetus) $1$, vertical side (cathetus) $x$, hypotenuse $\sqrt{1+x^2}$ and angle $\theta$ between the hypotenuse and the horizontal cathetus.
This proves the following
Theorem: all the direct trigonometric functions of the double-angle $2\theta$ can be expressed in terms of the tangent of the angle $\theta$.
As a consequence all the direct trigonometric functions of the angle $\theta$ can be expressed in terms of the tangent of the half angle $\theta/2$.
|
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|
Quadratic Congruence (with Chinese Remainder Thm) How do we solve quadratic congruences such as:
$x^2 \equiv11 \pmod{39}$
I know I must use the chinese remainder theorem with $p = 13, 3$ but I've only done linear examples and am unsure about how to do quadratic ones.
|
$$x^2 \equiv 11 \pmod{39} \implies x^2 \equiv 2 \pmod 3 \implies \text{No solution}$$
EDIT
In general, when you want to solve for $$x^2 \equiv a \pmod n \,\,\, (\spadesuit)$$ and if $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$, the idea is to first solve for $$x^2 \equiv a \pmod {p_l^{a_l}} \,\,\, (\clubsuit)$$ You have a solution for your original problem $(\spadesuit)$ iff you have a solution for each $l \in \{1,2,\ldots,k\}$ in $(\clubsuit)$. Once you find solution for each $l$ in $(\clubsuit)$, put them together using Chinese Remainder theorem.
For instance, if you have $x^2 \equiv 23 \pmod {77}$, then we need to look at $x^2 \equiv 23 \pmod 7$ and $x^2 \equiv 23 \pmod{11}$ i.e. $x^2 \equiv 2 \pmod 7$ and $x^2 \equiv 1 \pmod{11}$.
$$x^2 \equiv 2 \pmod7 \implies x \equiv \pm 3 \pmod 7$$ Similarly, $$x^2 \equiv 1 \pmod{11} \implies x \equiv \pm 1 \pmod{11}$$
Hence,
$$x \equiv 3 \pmod 7 \text{ and } x \equiv 1 \pmod{11} \implies x \equiv 45 \pmod{77}$$
$$x \equiv -3 \pmod 7 \text{ and } x \equiv 1 \pmod{11} \implies x \equiv 67 \pmod{77}$$
$$x \equiv 3 \pmod 7 \text{ and } x \equiv -1 \pmod{11} \implies x \equiv 10 \pmod{77}$$
$$x \equiv -3 \pmod 7 \text{ and } x \equiv -1 \pmod{11} \implies x \equiv 32 \pmod{77}$$
Hence,
$$x \equiv \pm 10, \pm 32 \pmod{77}$$
|
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|
Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a + b^3)(4)} \ge 3\sqrt{a} + \sqrt{b^3}$$
$$\sqrt{(3a + b^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2}$$
Also I get:
$$\sqrt{(3b + c^3)} \ge \frac{3\sqrt{b} + \sqrt{c^3}}{2}$$
$$\sqrt{(3c + a^3)} \ge \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
If I add add 3 inequalities I get:
$$\sqrt{(3a + b^3)} + \sqrt{(3b + c^3)} + \sqrt{(3c + a^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2} + \frac{3\sqrt{b} + \sqrt{c^3}}{2} + \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
Now i need to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} + \frac{3\sqrt{b} + \sqrt{b^3}}{2} + \frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge 6 = 2(a+b+c)$$
It's enough now to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} \ge b+c = 3-a$$
$$\frac{3\sqrt{b} + \sqrt{b^3}}{2} \ge a+c = 3-b$$
$$\frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge b+a = 3-c$$
All three inequalities are of the form:
$$\frac{3\sqrt{x} + \sqrt{x^3}}{2} \ge 3-x$$
$$3\sqrt{x} + \sqrt{x^3} \ge 6-2x$$
$$(3\sqrt{x} + \sqrt{x^3})^2 \ge (6-2x)^2$$
$$9x + x^3 + 6x^2 \ge 36 - 24x + 4x^2$$
$$x^3 + 2x^2 + 33x - 36 \ge 0$$
$$(x-1)(x^2 + 3x + 33) \ge 0$$
Case 1:
$$(x-1) \ge 0 \ \ \ \ \text{ for any }\ x \geq 1$$
$$(x^2 + 3x + 33) \ge 0 \ \ \ \ \text{ for any x in R} $$
Case 2:
$$0 \ge (x-1) \ \ \ \ \text{ for any }\ 1 \geq x$$
$$0 \ge (x^2 + 3x + 33) \ \ \ \ \text{there are no solutions in R} $$
This proves that for $$x \geq 1$$ $$(x-1)(x^2 + 3x + 33) \ge 0$$ is true and so it is
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$$, but a, b, c can be every non-negative number. I proved it's true for $$a,b,c \geq 1$$, but i can't for $$a,b,c \geq 0$$
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I'm not sure if "proof with Mathematica" counts, but here it is:
In order to find the local extrema of function $f(a,b,c) = \sqrt{3a+b^3}+\sqrt{3b+c^3}+\sqrt{3c+a^3}$ under the constraint $a+b+c=3$, we differentiate the corresponding lagrangian and arrive at a system of equations
$$\left\{\begin{aligned}& (a^2c^2-a^2+1)^2(3a+b^3)=(b^2a^2-b^2+1)^2(3b+c^3)=(c^2b^2-c^2+1)^2(3c+a^3) \\& a+b+c=3 \\& c \ge a,b \ge 0 \end{aligned}\right.$$
Obviously, $a=b=c=1$ is one solution; with $f(a,b,c)=6$. In order to find the other solutions I plug this system into Mathematica. After putting the pieces together, Mathematica simplifies this to a 48-degree equation, with only 1 additional solution satisfying the constraints: $a \approx 0.335288, b \approx 0.744425, c \approx 1.920288$. The precision does not matter much, since at this point the value of the function is $f(a,b,c) \approx 6.65093$, which means it's a local maximum.
|
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|
Write in polynomial in factored form in complex number Write the following polynomial in factored form(in complex number):
$$1+z+z^2+z^3+z^4+z^5+z^6$$
Also, is there general solution of factoring for $1+z+z^2...z^n$ types of polynomial?
|
$$1+z+z^2+z^3+z^4+z^5+z^6=\frac{1-z^7}{1-z}$$
$$1+z+z^2+z^3+z^4+z^5+...+z^n=\frac{1-z^{n+1}}{1-z}$$
Denote $$1+z+z^2+...+z^n=S$$ then multiply by z we get
$$z+z^2+...+z^{n+1}=Sz$$
from firs equation subtract the second then solve by $S$
|
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|
Evaluate $\lim\limits_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$
*
*Evaluate $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$.
*Examine whether $x^{1/x}$ possesses a maximum or minimum and determine the same.
|
This begs to be viewed as a Riemann sum:
$$
\frac 1n\left(\frac{1^3}{n^3}+\cdots+\frac{n^3}{n^3}\right) \to \int_0^1 x^3\,dx.
$$
Your second question seems quite different from your first and should be posted separately.
|
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|
Finding determinant (using row reduction) I'm trying to calculate the following
$\det \begin{pmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ -1 & -1 & x-1 & -1 & -1 \\ -1 & -1 &-1 & x-1 & -1 \\ -1 & -1 & -1 & -1 & x-1 \end{pmatrix}$
and I'm sure there must be a way to get this as an upper/lower triangular matrix. However whichever way I try looking at it, I can't see how to make it into one, if you minus the bottom row from each of the other rows you're still left with the bottom row and the same goes for columns. Is there some clever trick to note here? I've also briefly entertained the idea of proving a general form (since I'm sure one exists for a matrix of this form) but it seems a bit overkill...
Many thanks!
|
Hints: make zeros below the entry 2-1 on the first column and develop wrt it and etc.:
$$\begin{vmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ -1 & -1 & x-1 & -1 & -1 \\ -1 & -1 &-1 & x-1 & -1 \\ -1 & -1 & -1 & -1 & x-1 \end{vmatrix}=\begin{vmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ 0 & -x & x & 0 & 0 \\ 0 & -x &0 & x & 0 \\ 0 & -x & 0 & 0 & x \end{vmatrix}=$$
$$=(x-1)\begin{vmatrix} x-1 & -1 & -1 & -1 \\ -x & x & 0 & 0 \\ -x &0 & x & 0 \\ -x & 0 & 0 & x \end{vmatrix}+\begin{vmatrix} -1 & -1 & -1 & -1\\ -x & x & 0 & 0 \\ -x &0 & x & 0 \\ -x & 0 & 0 & x \end{vmatrix}\stackrel{\text{develop by Col. 2}}=$$
$$=(x-1)\left(\,\begin{vmatrix} -x & 0 & 0 \\ -x & x & 0 \\ -x & 0 & x \end{vmatrix}+x\begin{vmatrix} x-1 & -1 & -1 \\ -x & x & 0 \\ -x & 0 & x \end{vmatrix}\,\right)+\begin{vmatrix} -x & 0 & 0 \\ -x & x & 0 \\ -x & 0 & x \end{vmatrix}+x\begin{vmatrix} -1 & -1 & -1 \\ -x & x & 0 \\ -x & 0 & x \end{vmatrix}=$$
$$=(x-1)\left(-x^3+x(x^2(x-1)-2x^2)\right)+(-x^4-3x^3)=$$
$$=(x-1)\left(-x^3+x(x^3-3x^2)\right)-x^3(x+3)=(x-1)(x^4-4x^3)-x^3(x+3)=$$
$$=x^3(x-1)(x-4)-x^3(x+3)=\ldots$$
|
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|
What is the first non zero digit in 50 factorial (50!)? What is the first non zero digit in 50 factorial (50!)?
Any Help or hint will be appreciated.
|
Since $50_{\text{ten}}=200_{\text{five}}$, $\sigma_5(50)=2$. Thus, the number of factors of $5$ in $50!$ is $\frac{50-2}{5-1}=12$.
Since $50_{\text{ten}}=110010_{\text{two}}$, $\sigma_2(50)=3$. Thus, the number of factors of $2$ in $50!$ is $\frac{50-3}{2-1}=47$.
Thus, $\frac{50!}{10^{12}}$ has $35$ factors of $2$.
Little Fermat says that $2^4\equiv1\pmod{5}$, so $2\cdot2^4\equiv2\pmod{10}$, therefore $2^{35}\equiv2\cdot2^{34}\equiv2\cdot2^2\equiv8\pmod{10}$.
Every integer is uniquely representable as $m\cdot2^j\cdot5^k$ where $(m,10)=1$.
Let's compute the product $\bmod{\,10}$ of all of the numbers $m$ so that $(m,10)=1$ and $m\cdot2^j\cdot5^k\le50$, grouped by $2^j\cdot5^k$:
$\overbrace{(1\cdot3\cdot7\cdot9)^5}^{1}\overbrace{(1\cdot3\cdot7\cdot9)^2(1\cdot3)}^{2}\overbrace{(1\cdot3\cdot7\cdot9)^1(1)}^{4}\overbrace{(1\cdot3)^{\vphantom{1}}}^{8}\overbrace{(1\cdot3)^{\vphantom{1}}}^{16}\overbrace{(1)^{\vphantom{1}}}^{32}\\
\overbrace{(1\cdot3\cdot7\cdot9)^1}^{5}\overbrace{(1\cdot3)^{\vphantom{1}}}^{10}\overbrace{(1)^{\vphantom{1}}}^{20}\overbrace{(1)^{\vphantom{1}}}^{40}\\
\overbrace{(1)^{\vphantom{1}}}^{25}\overbrace{(1)^{\vphantom{1}}}^{50}\\
\equiv(1\cdot3\cdot7\cdot9)^9(1\cdot3)^4\equiv9\pmod{10}$
Thus, $\frac{50!}{10^{12}}\equiv8\cdot9\equiv2\pmod{10}$.
|
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|
Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stuck here.
|
An idea:
$$\frac{4}{x+4}\le 2\stackrel{\text{div. by 2}}\iff \frac{2}{x+4}\le 1\iff \frac{2}{x+4}-1\le 0\stackrel{\text{common denom.}}\iff $$
$$\iff \frac{2-x-4}{x+4}\le 0\stackrel{\text{mult. by (-1)}}\iff \frac{x+2}{x+4}\ge0\stackrel{\text{mult. by}\; (x+4)^2}\iff(x+2)(x+4)\ge 0\iff$$
$$\iff x<-4\,\,\vee\,\,x\ge-2\;\ldots$$
|
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|
Show that $\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$ Show that
$$\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$$
|
Hint::
$$\frac{\sin x}{\cos 3x}=$$
$$= \frac{1}{2} \cdot \frac{2 \cdot \sin x \cdot \cos x}{\cos x \cdot \cos 3x}$$
$$= \frac{1}{2} \cdot \frac{\sin 2x}{\cos x \cdot \cos 3x}$$
$$= \frac{1}{2} \cdot \frac{ \sin(3x - x) }{ \cos x \cdot \cos3x }$$
$$= \frac{1}{2} \cdot \frac{ \sin 3x \cdot \cos x - \cos 3x \cdot \sin x}{ \cos x \cdot \cos3x}$$
$$= \frac{1}{2} \cdot \frac {\sin 3x \cdot \cos x}{\cos x \cdot \cos 3x} - \frac{1}{2} \cdot \frac{\cos 3x \cdot \sin x}{\cos x \cdot \cos 3x}$$
$$= \frac{1}{2} \cdot \left (\frac{\sin 3x}{\cos 3x} - \frac{\sin x}{\cos x} \right )$$
$$= \frac{1}{2} \cdot \left (\tan 3x - \tan x \right )$$
Now successively replace $x$ with $3x$ and $9x$
|
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|
$f(x - 1) + f(x − 2) $ and the sum of coeficients If $f(x-1)+f(x-2) = 5x^2 - 2x + 9$
and
$f(x)= ax^2 + bx + c$
what would be the value of $a+b+c$?
I was doing
$f(x-1)+f(x-2)= f(x-3)$
then
$f(x)$
a = 5
b = -2
c = 9
$(5-3)+(-2-3)+(9-3)$
But do not think is is correct
What would be correct approach?
|
$$f(x-1)+f(x-2) = 5x^2 - 2x + 9$$and $$f(x)=ax^2+bx+c$$
for $x=1,2,3$ we get the system
$$f(0)+f(-1)=a-b+2c =12$$
$$f(1)+f(0)=a+b+2c=25$$
$$f(2)+f(1)=5a+3b+2c = 48$$
with solutions
$$a=5/2,b=13/2,c=8$$
so $$f(x)=\frac{5}{2}x^2+\frac{13}{2}x+8$$
|
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|
What goes wrong in this derivative? $$ f(x) = \frac{2}{3} x (x^2-1)^{-2/3} $$
and f'(x) is searched.
So, by applying the product rule $ (uv)' = u'v + uv' $ with $ u=(x^2-1)^{-2/3} $ and $ v=\frac{2}{3} x $, so $ u'=-\frac{4}{3} x (x^2-1)^{-5/3} $ and $ v' = \frac{2}{3} $, I obtain
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3} x^2 + \frac{2}{3} (x^2-1)^{-2/3} $$
whereas according to Wolfram Alpha (see alternate form), the correct result is:
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3} x^2 - \frac{2}{3} (x^2-1)^{-5/3} $$
So apparently, my calculation for $u'v$ is correct, but $uv'$ is wrong. What am I missing here?
|
Doing what you said, you should find
$$f'(x)=-\frac{8}{9}(x^2-1)^{-5/3}x^2+\frac{2}{3}(x^2-1)^{-2/3}.$$
Now
$$
(x^2-1)^{-2/3}=(x^2-1)(x^2-1)^{-5/3}.
$$
So
$$
f'(x)=-\frac{8}{9}(x^2-1)^{-5/3}x^2+\frac{2}{3}(x^2-1)(x^2-1)^{-5/3}
$$
$$
= \left(\frac{2}{3}-\frac{8}{9}\right)(x^2-1)^{-5/3}x^2-\frac{2}{3}(x^2-1)^{-5/3}
$$
$$
=\mbox{Wolfram Alpha's answer}.
$$
|
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|
Weird calculus limit How to find the following limit?
$$
\lim_{n \to \infty} \dfrac{ 5^\frac{1}{n!} - 4^\frac{1}{n!} }{ 3^\frac{1}{n!} - 2^\frac{1}{n!} }
$$
Edit done to the question.
Thank you!
|
$$\frac{5^x-4^x}{3^x-2^x}$$
$$=\frac{4^x\left(\left(\frac54\right)^x-1\right)}{2^x\left(\left(\frac32\right)^x-1\right)}$$
$$=2^x\frac{\frac{ \left(\frac54\right)^x-1 }x}{\frac{\left( \frac32\right)^x-1 }x}$$
So, $$\lim_{x\to0}\frac{5^x-4^x}{3^x-2^x}$$
$$=\lim_{x\to0}2^x\frac{\frac{ \left(\frac54\right)^x-1 }x}{\frac{\left( \frac32\right)^x-1 }x}$$
$$=\lim_{x\to0}2^x\cdot\frac{\lim_{x\to0}\frac{ \left(\frac54\right)^x-1 }x}{\lim_{x\to0}\frac{\left( \frac32\right)^x-1 }x}$$
$$=\frac{\ln\frac54}{\ln\frac32}$$ using $\lim_{h\to0}\left(\frac{a^h-1}h\right)=\ln a$
|
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|
$ \sin^{2000}{x}+\cos^{2000}{x} =1$ equation explanation Solve the equation:
$$ \sin^{2000}{x}+\cos^{2000}{x} =1.$$
What I did:
$\sin^2{x} =1 \land \cos^2{x}=0$ when $x=\frac \pi2 + \pi k $
$\cos^2 {x} =1 \land \sin^2{x}=0$ when $x= \pi k$
I think that these solutions apply for this equation as well but I don't really know how to formally explain it.
Thanks in advance.
|
Since we are concerned with the case when $\theta \neq k\pi/2$, we get
$$0 < \sin^2\theta < 1$$ and $$0 < \cos^2\theta < 1$$
Now observe that for $0 < \theta < \pi/2$
$$1 = (\sin^2\theta + \cos^2\theta)^2 = \sin^4\theta + \cos^4\theta + 2\sin^2\theta\cos^2\theta > \sin^4\theta + \cos^4\theta \quad (1)$$
Now observe that for $0<x<1$, $$x^n + (1-x)^n < 1$$ for $n \geq 2$, and it is monotonically strictly decreasing in n.
To show this, verify it for n=2, then use induction. The inductive step being:
$$1 > x^n + (1-x)^n = (x^n + (1-x)^n)(x + 1-x) = x^{n+1} + (1-x)^{n+1} + x^n(1-x) +(1-x)^nx $$
$$ > x^{n+1} + (1-x)^{n+1}$$
Now put $x=\sin^2 \theta$ and you are done.
$\theta = k\pi/2$ give the only solutions.
|
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|
Prove $\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$ If $a,b$ and $c \ge 0$ and $ab + bc + ca = 1$, prove that the following inequality holds:
$$\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$$
I've tried two aproaches, but it seems like both doesn't work. Here they are:
Cauchy-Scwarz inequality
I've tried using the following formula of Cauchy-Scwarz:
$$\frac{x_1^2}{a_1} + \frac{x_2^2}{a_2} + \frac{x_3^2}{a_3} \ge \frac{(x_1 + x_2 + x_3)^2}{a_1 + a_2 + a_3}$$
And i get:
$$ LHS \ge \frac{(1+1+1)^2}{2a + 2bc + 2b + 2ca + 2c + 2ba + 3}$$
$$ LHS \ge \frac{3^2}{2(a+b+c) + 5}$$
$$ LHS \ge \frac{9}{2(a+b+c) + 5}$$
Now to prove that the RHS is bigger than or equal to 1.
$$ \frac{9}{2(a+b+c) + 5} \ge 1$$
$$ 9 \ge 2(a+b+c) + 5$$
$$ 2 \ge a+b+c$$
And her I'm stuck, what can I do now?
AM - GM inequality
This try doesn't even stand a chance, but I'll still post something, because somebody can recieve an idea.
$$2a + 2bc \ge 2\sqrt{4abc}$$
$$2a + 2bc + 1 \ge 4\sqrt{abc} + 1$$
$$\frac{1}{2a + 2bc + 1} \le \frac{1}{4\sqrt{abc} + 1}$$
And for oher 2 fraction i get the same and if I add them I get:
$$\frac{3}{4\sqrt{abc} + 1} \ge LHS \ge 1$$
Now even if i prove that: $\frac{3}{4\sqrt{abc} + 1} \ge 1$ si true, that doesn't mean the original inequality holds.
|
If none of the terms $a+bc$, $b+ca$, and $c+ab$ is greater than one, then each summand of the LHS is at least $\frac{1}{3}$, and thus the inequality holds. It remains to show the inequality also holds when at least one of $a+bc$, $b+ca$, or $c+ab$ is strictly greater than one. WOLOG, suppose that $c+ab>1$. Since $c=\frac{1-ab}{a+b}$, we must have $a+b<1$. Furthermore, from Cauchy-Schwarz inequality we have $$\frac{1}{2a+2bc+1}+\frac{1}{2b+2ca+1}\geq \frac{4}{2a+2bc+1+2b+2ca+1}\\=\frac{2}{2+a+b-ab}.$$ Therefore, it suffices to show that$$\frac{2}{2+a+b-ab}+\frac{1}{2c+2ab+1} = \frac{2}{2+a+b-ab}+\frac{a+b}{2+a+b-2ab+2ab(a+b)}\\ \geq 1, $$ holds for all $a+b<1$. We have $$A:=\frac{2}{2+a+b-ab}+\frac{a+b}{2+a+b-2ab+2ab(a+b)}\\=\frac{4+4(a+b)-4ab+3ab(a+b)+(a+b)^2}{(2+a+b-ab)(2+a+b-2ab+2ab(a+b))}.$$ For compactness we can let $s=a+b$ and $p=ab$ and obtain $$A=\frac{4+4s-4p+3sp+s^2}{(2+s-p)(2+s-2p+2sp)}\\=\frac{4+4s-4p+3sp+s^2}{4+4s-6p+3sp+s^2+2s^2p-2p^2+2sp^2}.$$ Hence, we need to show that $$2p\geq 2s^2p-2p^2+2sp^2,$$
which is equivalent to $$2p(1-s)(1+p+s)\geq 0.$$ This inequality is valid because by assumption $s<1$.
Equation cannot hold in the last case, since we need to have $p=0$ which implies $a$ or $b$ is zero, but the Cauchy-Schwarz we applied above cannot hold for these values considering $a+b<1$. The only remaining possibility for having equation is in the first case where we should have $a+bc=b+ca=a+bc=1$ which reduces to $(a,b,c)\in\{(0,1,1),(1,0,1),(1,1,0)\}$.
|
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|
Generating function for the solutions of equation $ 2x_1 + 4x_2 = n$ Let's denote $h_n$ as the number of soulutions of the following equation:
$$ 2x_1 + 4x_2 = n$$
where $x_i \in \mathbb N$.
Find generating function of the sequence $h_n$ and calculate $h_{2000}$.
I've found the generating function:
$$\frac{1}{1-x^2}\cdot \frac{1}{1-x^4},$$
but I don't know how to expand it now. Any ideas?
|
As stated, the generating function is:
\begin{align}
f(z) &= \frac{1}{1 - z^2} \cdot \frac{1}{1 - z^4} \\
&= \frac{1}{4 (1 - z^2)} + \frac{1}{4 (1 + z^2)} + \frac{1}{2 (1 - z^2)^2}
\end{align}
(this results from recognizing the generating function as a fraction in $z^2$, and splitting as such into partial fractions). Thus the expansion is:
\begin{align}
f(z) &= \frac{1}{4} \sum_{k \ge 0} \left( 1 + (-1)^k \right) z^{2 k}
+ \frac{1}{2} \sum_{k \ge 0} (-1)^k \binom{-2}{k} z^{2 k} \\
&= \frac{1}{4} \sum_{k \ge 0} \left( 1 + (-1)^k \right) z^{2 k}
+ \frac{1}{2} \sum_{k \ge 0} \binom{n + 1}{1} z^{2 k} \\
&= \frac{1}{4} \sum_{k \ge 0} \left( 1 + (-1)^k \right) z^{2 k}
+ \frac{1}{2} \sum_{k \ge 0} (k + 1) z^{2 k}
\end{align}
This gives the coefficients:
$$
h_n = \begin{cases}
0 & \text{$n$ odd} \\
\left\lfloor \frac{n}{4} \right\rfloor + 1 & \text{$n$ even}
\end{cases}
$$
Thus $h_{2000} = 501$.
|
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|
Prove that for every positive integer $n$, $1/1^2+1/2^2+1/3^2+\cdots+1/n^2\le2-1/n$ Base case: n=1. $1/1\le 2-1/1$. So the base case holds.
Let $n=k\ge1$ and assume
$$1/1^2+1/2^2+1/3^2+\cdots+1/k^2\le 2-1/k$$
We want to prove this for $k+1$, i.e.
$$(1/1^2+1/2^2+1/3^2+\cdots+1/k^2)+1/(k+1)^2\le 2-\frac{1}{k+1}$$
This is where I get stuck. Any help appreciated.
|
For your induction step, all you need is
$$
2-\frac{1}{k}+\frac{1}{(k+1)^2}\leq 2-\frac{1}{k+1}.
$$
That's equivalent to
$$
\frac{1}{k}-\frac{1}{(k+1)^2}-\frac{1}{k+1}\geq 0
$$
i.e.
$$
\frac{(k+1)^2-k-k(k+1)}{k(k+1)^2}=\frac{1}{k(k+1)^2}\geq 0.
$$
So it holds.
|
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|
Sum of all numbers x such that $(3x^2+9x-2012)^{(x^3-2012x^2-10x+1)} = 1$? What is the sum of all $x$ such that $(3x^2 + 9x - 2012)^{(x^3-2012x^2-10x+1)} = 1$?
|
Note that $x^3-2012x^2-10x+1$ has no rational root (it would have to be $\pm1$, which can be checked explicitly).
Also note that yb polynomial division
$$\begin{align}x^3-2012x^2-10x+1 &= (3x^2+9x-2012)\cdot\frac{x-2015}3 + \frac{20117 x -4054177}3, \\
x^3-2012x^2-10x+k &= (3x^2+9x-2013)\cdot\frac{x-2015}3 + 6706 x -1352065+k, \\
x^3-2012x^2-10x+1 &= (3x^2+9x-2011)\cdot\frac{x-2015}3 + \frac{20116 x -4052162}3. \end{align}$$
The first implies that $3x^2+9x-2012$ and $x^3-2012x^2-10x+1$ cannot both be zero as that would lead to a rational root.
Similarly, the third implies that we cannot have base $=1$ and exponent $=0$ at the same time.
Finally, the second equation shows that an $x$ for which the base is $-1$ and the exponent an integer $1-k$, again $x$ must be rational. However, $3x^2+9x-2013$ has no rational solutions.
Therefore, no weird special cases occur and the desired sum is simply the sum of the distinct(!) roots of $x^3-2012x^2-10x+1$ and the distinct(!) roots of $3x^2+9x-2013$. These can be read directly from the coefficients so that we obtain $$2012-\frac 93=2009.$$
|
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|
Residue Integral
Verify the integral with the aid of residues:
$$\int^{\infty}_0\frac{x^2+1}{x^4+1}dx=\frac{\pi}{\sqrt 2}$$
I got:
$f(z)=\frac{z^2+1}{z^4+1}$ and now I must find the residues for $f(z)$ and I got that the poles are: $e^{i\frac{\pi}{4}}$ and $e^{i\frac{3\pi}{4}}$. But I do not know how to finish.
|
An easy way to get the solution is just to use partial fractions in Calculus. Here is the solution:
\begin{eqnarray*}
\int_0^\infty\frac{x^2+1}{x^4+1}dx&=&\int_0^\infty\frac{x^2+1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}dx\\
&=&\frac{1}{2}\int_0^\infty\frac{1}{x^2+\sqrt{2}x+1}dx+\frac{1}{2}\int_0^\infty\frac{1}{x^2-\sqrt{2}x+1}dx\\
&=&\frac{1}{2}\int_0^\infty\frac{1}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dx+\frac{1}{2}\int_0^\infty\frac{1}{(x-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dx\\
&=&\frac{1}{2}\sqrt{2}\left.\left(\arctan\frac{2x+\sqrt{2}}{\sqrt{2}}+ \arctan\frac{2x-\sqrt{2}}{\sqrt{2}}\right)\right|_{0}^{\infty}\\
&=&\frac{\pi}{\sqrt{2}}.
\end{eqnarray*}
|
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|
Show that $\frac{1}{1+x}H(\frac{x}{1+x})=\sum^\infty_{k=0}[\Delta^kh_0]x^k$ For a sequence $\{h_n\}_{\geq 0}$, let $H(x)=\sum_{n\geq0}h_nx^n$. Show that:
$$\frac{1}{1+x}H(\frac{x}{1+x})=\sum^\infty_{k=0}[\Delta^kh_0]x^k$$
What I did was that by proving the $$\Delta^k h_o=\sum^k_{j=0}(-1)^{k-j}{k \choose j}h_j$$
But no clue how to continue.
Help appreciated.
|
What we seek to show here is that
$$\sum_{k=0}^n {n\choose k} (-1)^{n-k} h_k
= [z^n] \frac{1}{1+z} H\left(\frac{z}{1+z}\right)$$
where $$H(z) = \sum_{q\ge 0} h_q z^q.$$
The RHS is given by the integral
$$\frac{1}{2\pi i}\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{1}{1+z} H\left(\frac{z}{1+z}\right) \; dz
\\ = \frac{1}{2\pi i}\int_{|z|=\epsilon} \frac{1}{z^{n+2}}
\frac{z}{1+z} H\left(\frac{z}{1+z}\right) \; dz.$$
Now put $$\frac{z}{1+z} = u \quad\text{so that}\quad
z = \frac{u}{1-u}$$
and $$dz = \frac{1}{1-u} - \frac{u}{(1-u)^2} (-1) \; du
= \frac{1}{(1-u)^2} \; du.$$
We get for the integral
$$\frac{1}{2\pi i}\int_{|u|=\epsilon} \frac{(1-u)^{n+2}}{u^{n+2}}
\times u \times H\left(u\right) \; \frac{1}{(1-u)^2} \; du
\\ = \frac{1}{2\pi i}\int_{|u|=\epsilon} \frac{(1-u)^{n}}{u^{n+1}}
\times H\left(u\right) \; du.$$
This is $$[u^n] H(u) (1-u)^n$$
which evaluates to
$$\sum_{k=0}^n h_k {n\choose n-k} (-1)^{n-k}
= \sum_{k=0}^n {n\choose k} (-1)^{n-k} h_k$$
by inspection.
|
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|
Find the Laurent expansion in powers of $\,z\,$ and $\,1/z\,$ $f(z)=\large\frac{z+2}{z^2-z-2}$ in $1<|z|<2$ and then in $2<|z|<\infty$
I am unsure how the two different regions of $z$ affect the series expansion.
Any help would be appreciated.
|
Hints:
$$1<|z|<2\implies \begin{cases}\frac{1}{2}<\frac{1}{|z|}<1\\{}\\\frac{|z|}{2}<1\end{cases}\;\;\;,\;\;\;\;\text{so}$$
$$\frac{z+2}{(z-2)(z+1)}=\frac{1}{3}\left(\frac{4}{z-2}-\frac{1}{z+1}\right)=$$
$$=-\frac{1}{3}\left(\frac{2}{1-\frac{z}{2}}+\frac{1}{z}\frac{1}{1+\frac{1}{z}}\right)=-\frac{1}{3}\left(4\left(1+\frac{z}{2}+\frac{z^2}{2^2}+\ldots\right)+\frac{1}{z}\left(1-\frac{1}{z}+\frac{1}{z^2}-\frac{1}{z^3}+\ldots\right)\right)=$$
$$=-\frac{1}{3}\left(4+\frac{4z}{2}+\frac{4z^2}{2^2}+\ldots+\frac{4z^n}{2^n}+\ldots+\frac{1}{z}-\frac{1}{z^2}+\frac{1}{z^3}-\ldots\right)=$$
$$=\ldots-\frac{1}{3z^3}+\frac{1}{3z^2}-\frac{1}{3z}+\frac{5}{3}-\frac{4}{3\cdot 2}z-\frac{4}{3\cdot2^2}z^2-\ldots$$
|
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|
If $ a+b+c = \frac{9}{2}$ and $a,b,c>0$, then what is the minimum value of $\frac{a}{b^3+54}+\frac{b}{c^3+54}+\frac{c}{a^3+54}$
If $a+b+c = \dfrac{9}{2}$ and $a,b,c>0$, then what is the minimum value of $$\dfrac{a}{b^3+54}+\dfrac{b}{c^3+54}+\dfrac{c}{a^3+54} \qquad ?$$
My try:
$$\begin{align*}
\frac{a}{b^3+54}+\frac{b}{c^3+54}+\frac{c}{a^3+54} &= \frac{a^2}{ab^3+54a}+\frac{b^2}{bc^3+54b}+\frac{c^2}{ca^3+54c}\\
&\geq \frac{(a+b+c)^2}{ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)}
\end{align*}$$
Now, how can I proceed after that? Thanks.
|
Let $\sum$ denote the cyclic sum in the answer below. Also, let $f$ denote the expression to be minimised. Then,
$$f = \sum \frac{a}{b^3+54} = \frac{1}{54}\left(\sum a - \frac{ab^3}{b^3+54}\right)$$
By AM-GM, $\qquad b^3+54 = b^3 + 27 + 27 \ge 27 b$
So $$54f \ge \frac{9}{2} - \frac{1}{27} \sum ab^2$$
We prove below that $\sum ab^2 \le \frac{27}{2}$. However this gives
$$f \ge \frac{2}{27}$$
Equality is obtained for cyclic permutations of $(\frac{3}{2}, 3, 0)$, so this is the minimum (as surmised by Ivan Loh).
We prove here that $\sum ab^2 \le \dfrac{4}{27}(\sum a)^3 = \dfrac{27}{2}$.
Let $x, y, z$ be a permutation of $a, b, c$; s.t. $x \ge y \ge z \ge 0$.
Then $\sum x = \sum a$ and $xy \ge zx \ge yz$.
Then by rearrangement inequality, we have
$\sum ab^2 \le x(xy) + y(zx) + z(yz) = y(x+z)^2 - xyz$
But $y(x+z)^2 = \dfrac{1}{2} 2y (x+z)^2 \le \dfrac{1}{2} \dfrac{(2y + x + z + x + z)^3}{27} = \dfrac{4}{27}(\sum x)^3$
So $\sum ab^2 \le \dfrac{4}{27}(\sum a)^3 - abc \le \dfrac{4}{27}(\sum a)^3 = \dfrac{27}{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I find this limit: $\lim_{x \to \infty} \sqrt{x^4-3x^2-1}-x^2$ $$
\lim_{x \to \infty} \sqrt{x^4-3x^2-1}-x^2
$$
The answer is
$$
\frac{-3}{2}
$$
according to Wolfram alpha.
|
first, substitute : $t=x^2$, you get :
$$\lim_{t\to +\infty} \sqrt{t^2-3t-1}-t=\lim_{t\to \infty} \frac{-3t-1}{\sqrt{t^2-3t-1}+t}=\lim_{t\to \infty} \frac{-3-\frac{1}{t}}{\sqrt{1-\frac{3}{t} -\frac{1}{t^3}}+1} =\frac{-3}{2}. $$
Explanation : first we use this identity for $a\neq -b$ : $a-b =\frac{a^2-b^2}{a+b}$, then we factor $t$ from the numerator and the denomenator, the limit of the numerator is $-3$ and the limit of the numerator is $2$.
Further more, for any reals $a,b$ : $$\lim_{x\to+ \infty } \sqrt{x^2+ax+b}-x=\frac{a}{2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/359441",
"timestamp": "2023-03-29T00:00:00",
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|
Angle between two 3D vectors is not what I expected. Using the definition of the dot product, you can find the angle between two vectors. I am experiencing an unexpected result, so my question is where did I go wrong.
I have two unit vectors in 3 dimensions. So the angle between these vectors would be just the inverse cosine of the dot product. I picked two unit vectors I know would be $45$ degrees from each other, dotted them, and took the inverse cosine.
$$\arccos\left(\left(\frac{\sqrt3}{3}, \frac{\sqrt3}{3}, \frac{\sqrt3}{3}\right) \cdot \left(\frac{\sqrt2}{2},\frac{\sqrt2}{2},0\right)\right)$$
Look here on Wolfram.
But it does not come out to be $45$ degrees, or $\dfrac{\pi}{4}$ radians. It comes out to be $.61$ which is not $\dfrac{\pi}{4}$. $\dfrac{\pi}{4}$ is $.78$.
So where did I go wrong?
|
Hint: Calculate
$\displaystyle \cos \theta = \frac{a . b}{|a||b|}.$
What do you get?
$\displaystyle a . b = \left(\frac{\sqrt{3} \sqrt{2}}{3 \times 2}\right) + \left(\frac{\sqrt{3} \sqrt{2}}{3 \times 2}\right) + (0) = \frac{\sqrt{6}}{3} = \sqrt{\frac{2}{3}}$
$\displaystyle |a| .|b| = \left|~\sqrt{\left(\frac{\sqrt{3}}{3}\right)^2 + \left(\frac{\sqrt{3}}{3}\right)^2 + \left(\frac{\sqrt{3}}{3}\right)^2 }~\right|. \left|~\sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 +0^2}~\right| = |1|.|1| = 1$
This comes out to: $\displaystyle \cos^{-1} \sqrt{\frac{2}{3}}$
|
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|
Finding this solution to a recurrence relation So, I know that the recurrence relation $a_n = 4a_{n−1} − 3a_{n−2} + 2^n + n + 3$ with $a_0 = 1$ and $a_1 = 4$ has the solution of $a_n = -4(2^n) - n^2 / 4 - 5n / 2 + 1/8 + (39/8)(3^n)$. I just wanted to know how we arrived at this solution. Thank you!
|
Write:
$$
a_{n + 2} = 4 a_{n + 1} - 3 a_n + 4 \cdot 2^n + n + 5 \quad a_0 = 1, a_1 = 4
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$. If you multiply the recurrence by $z^n$ and sum over $n \ge 0$ you get:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 4 \frac{A(z) - a_0}{z} - 3 A(z)
+ \frac{4}{1 - 2 z} + \frac{z}{(1 - z)^2} + 5 \frac{1}{1 - z}
$$
This gives:
$$
\begin{align*}
A(z) &= \frac{1 - 4 z + 14 z^2 - 24 z^3 + 12 z^4}
{1 - 8 z + 24 z^2 - 34 z^3 + 23 z^4 - 6 z^5} \\
&= \frac{39}{8} \cdot \frac{1}{1 - 3 z}
- 4 \cdot \frac{1}{1 - 2 z}
+ \frac{19}{8} \cdot \frac{1}{1 - z}
- \frac{7}{4} \cdot \frac{1}{(1 - z)^2}
- \frac{1}{2} \cdot \frac{1}{(1 - z)^3}
\end{align*}
$$
Expanding the geometric series, and also:
$$
(1 - z)^{-k} = \sum_{n \ge 0} (-1)^n \binom{-k}{n} z^n
= \sum_{n \ge 0} \binom{n + k - 1}{k - 1} z^n
$$
gives:
$$
\begin{align*}
a_n &= \frac{39}{8} \cdot 3^n
- 4 \cdot 2^n
+ \frac{19}{8}
- \frac{7}{4} \cdot \binom{n + 1}{1}
- \frac{1}{2} \cdot \binom{n + 2}{2} \\
&= \frac{39}{8} \cdot 3^n - 4 \cdot 2^n
- \frac{1}{8} (2 n^2 + 20 n - 1)
\end{align*}
$$
|
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|
Can someone check the solution to this recurrence relation? Here's the recurrence relation: $a_n = 4a_{n−1} − 3a_{n−2} + 2^n + n + 3$ with $a_0 = 1$ and $a_1 = 4$
Here's the solution:Write:
$$
a_{n + 2} = 4 a_{n + 1} - 3 a_n + 2^n + n + 3 \quad a_0 = 1, a_1 = 4
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$. If you multiply the recurrence by $z^n$ and sum over $n \ge 0$ you get:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 4 \frac{A(z) - a_0}{z} - 3 A(z)
+ \frac{1}{1 - 2 z} + \frac{z}{(1 - z)^2} + 3 \frac{1}{1 - z}
$$
This gives:
$$
\begin{align*}
A(z) &= \frac{1 - 4 z + 9 z^2 - 12 z^3 + 5 z^4}
{1 - 8 z + 24 z^2 - 34 z^3 + 23 z^4 - 6 z^5} \\
&= \frac{23}{8} \cdot \frac{1}{1 - 3 z}
- \frac{1}{1 - 2 z}
+ \frac{3}{8} \cdot \frac{1}{1 - z}
- \frac{3}{4} \cdot \frac{1}{(1 - z)^2}
- \frac{1}{2} \cdot \frac{1}{(1 - z)^3}
\end{align*}
$$
Expanding the geometric series, and also:
$$
(1 - z)^{-k} = \sum_{n \ge 0} (-1)^n \binom{-k}{n} z^n
= \sum_{n \ge 0} \binom{n + k - 1}{k - 1} z^n
$$
gives:
$$
a_n = \frac{23}{8} \cdot 3^n
- 2^n
+ \frac{3}{8}
- \frac{3}{4} \cdot \binom{n + 1}{1}
- \frac{1}{2} \cdot \binom{n + 2}{2}
= \frac{23}{8} \cdot 3^n - 2^n + \frac{3}{8}
- \frac{1}{6} (n^3 + 6 n^2 + 5 n)
$$
The problem is that when I check this with Wolfram, it has the solution of $a_n = -4(2^n) - n^2 / 4 - 5n / 2 + 1/8 + (39/8)(3^n)$. I just wanted to know if this was an error or what..thanks!
|
But
$a_{n+2}=4a_{n+1}−3a_n+2^{n+2}+(n+2)+3\,$!
|
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|
Iterative recurrence.. Iteration method Here is the Equation and how far I got into solving this problem using the iteration method:
$$T(1) = 8 \\
T(n) = 3T(n-1) - 15$$
Iterations:
$i=1, $
$$T(n) = 3(3T(n-2) - 15) -15$$
$i=2, $
$$ 3(3(3T(n-3) - 15) -15) - 15$$
$i=3,$
$$3(3(3(3T(n-4) - 15) -15) - 15) - 15$$
$i=4,$
$$ 3(3(3(3(3T(n-5) - 15) -15) - 15) - 15) - 15$$
From the iteration pattern I found that
$$T(n) = 3^{i+1} \times T(n-(i+1)) - 15$$
At this point I need to find a summation for this recurrence and make it into closed form... I'm just not sure how.
Can someone explain or help guide me to solving this problem?
|
It is easier to use generating functions. Define $g(z) = \sum_{n \ge 0} T(n + 1) z^n$. Write:
$$
T(n + 2) = 3 T(n + 1) - 15 \quad T(1) = 8
$$
Multiply by $z^n$ and add over $n \ge 0$:
$$
\frac{g(z) - T(1)}{z} = 3 g(z) - 15 \frac{1}{1 - z}
$$
From here:
$$
g(z) = \frac{8 - 23 z}{1 - 4 z + 3 z^2}
= \frac{1}{2} \cdot \frac{1}{1 - 3 z} + \frac{15}{2} \cdot \frac{1}{1 - z}
$$
This gives:
$$
T(n) = \frac{1}{2} \cdot 3^{n + 1} + \frac{15}{2}
= \frac{3^{n + 1} + 15}{2}
$$
|
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|
Find the closest point to the origin On the line,Find the closest point to the origin
$y=\frac{2}{\sqrt{x}}$
What I did so far is :
First point is : $(x,\frac{2}{\sqrt{x}})$ and point two is : (0,0)
$d=\sqrt{x^2+(\frac{2}{\sqrt{x}})^2}$
then - > $d' = 2x-\frac{4}{{x^2}}$
I found the X but what then? I do not think I did right.
Thanks.
|
You are almost there. First note that $y = \dfrac2{\sqrt{x}}$ is not a line. If $r$ is the distance from the origin, we have
$$r^2 = x^2 + \left(\dfrac2{\sqrt{x}} \right)^2 = x^2 + \dfrac4x$$
Now setting $\dfrac{dr^2}{dx} = 0$, we get that
$$2x - \dfrac4{x^2} = 0 \implies x^* =\sqrt[3]{2}$$
Hence, the value of $r^2$ at the point, which is at minimal distance from the origin is $$r^2 = 2^{2/3} + \dfrac4{2^{1/3}} = 2^{2/3} + 2^{5/3} = 3 \cdot 2^{2/3}$$
Hence, the minimum distance from the origin is
$$r = \sqrt{3} \cdot \sqrt[3]2$$
The point on the cuve is given by $$\left(x^*,\dfrac2{\sqrt{x^*}}\right) = \left(2^{1/3},\dfrac2{\sqrt[2]{2^{1/3}}}\right) = \left(2^{1/3}, 2^{5/6}\right)$$
|
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|
Find expansion around $x_0=0$ into power series and find a radius of convergence My task is as in the topic, I've given function $$f(x)=\frac{1}{1+x+x^2+x^3}$$
My solution is following (when $|x|<1$):$$\frac{1}{1+x+x^2+x^3}=\frac{1}{(x+1)+(x^2+1)}=\frac{1}{1-(-x)}\cdot\frac{1}{1-(-x^2)}=$$$$=\sum_{k=0}^{\infty}(-x)^k\cdot \sum_{k=0}^{\infty}(-x^2)^k$$ Now I try to calculate it the following way:
\begin{align}
& {}\qquad \sum_{k=0}^{\infty}(-x)^k\cdot \sum_{k=0}^{\infty}(-x^2)^k \\[8pt]
& =(-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-x^9+\cdots)\cdot(-x^2+x^4-x^6+x^8-x^{10}+\cdots) \\[8pt]
& =x^3-x^4+0 \cdot x^5+0 \cdot x^6 +x^7-x^8+0 \cdot x^9 +0 \cdot x^{10} +x^{11}+\cdots
\end{align}
And now I conclude that it is equal to $\sum_{k=0}^{\infty}(x^{3+4 \cdot k}-x^{4+4 \cdot k})$ ($|x|<1$)
Is it correct? Are there any faster ways to solve that types of tasks? Any hints will be appreciated, thanks in advance.
|
Use the Cauchy product:
$$\sum_{k=0}^\infty a_kx^k\cdot \sum_{k=1}^\infty b_kx^k=\sum_{k=0}^\infty c_kx^k$$
where $$c_k=\sum_{n=0}^k a_n\cdot b_{k-n}$$
In your case: $a_k=(-1)^k$ and $$b_k=\begin{cases}0 & ,k =2l+1 \\(-1)^k&,k=2l\end{cases}$$
|
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|
Prove that $x^{12}-x^9+x^4-x+1>0$ for all $x\in \mathbb{R}$
Prove that the expression $x^{12}-x^9+x^4-x+1>0\; \forall x\in \mathbb{R}$
My try:: Using Interval method::
$\bullet \; $If $x\leq 0$, Then $x^{12}-x^9+x^4-x+1>0$
$\bullet \; $If $0<x\leq 1$, Then $x^{12}+x^4.(1-x^5)+(1-x)>0$
$\bullet \; $If $x>1$, Then $x^9.(x^3-1)+x(x^3-1)+1>0$
So the expression $x^{12}-x^9+x^4-x+1>0\; \forall x\in \mathbb{R}$
My question is How can I solve Using $A.M\geq G.M$ method. or How can I complete the
square so that the expression is $>0$
Thanks
|
$$\begin{aligned}
\frac{1}{4}x^{12}+x^6&\ge |x^9|\ge x^9\\
\frac{3}{4}x^{12}+\frac13&\ge x^6\\
x^4+\frac14&\ge x^2\\
x^2+\frac14&\ge |x|\ge x
\end{aligned}$$
Add them all up, you get that
$$x^{12}-x^9+x^4-x+1\ge \frac16.$$
Another proof:
$$\begin{aligned}
\frac{x^{12}+x^{12}+x^{12}+1}4&\ge \sqrt[4]{x^{36}}\ge x^9\\
\frac{x^4+1+1+1}{4}&\ge \sqrt[4]{x^4}\ge x.
\end{aligned}$$
The inequality follows.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? In this recent answer to this question by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$
I would like to know if this result can be generalized to other triples of
natural numbers.
Question. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right)
\in \mathbb{N} ^{3}.\tag{2} \end{equation} $$
For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$
$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3})
\sqrt{3}
$$
and solve the system
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=26 \\
a^{2}b+b^{3}=5.
\end{array}
\right.
$$
A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.
For $(2)$ the very same idea yields
$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$
and
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=p \\
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$
I tried to solve this system for $a,b$ but since the solution is of the form
$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$
where $x$ satisfies the cubic equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach.
Is this problem solvable, at least partially?
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?
|
Yikes. I am hopind I didn't do any mistake, but that would be a miracle :)
Let $x= \left( p+q\sqrt{3}\right) ^{1/3}\,;\, y= \left(
p-q\sqrt{3}\right) ^{1/3}$.
Then
$$xy= (p^2-3q^2)^\frac{1}{3} $$
Hence
$$2p=x^3+y^3=(x+y)^3-3xy(x+y)=n^3-3n\sqrt[3]{p^2-3q^2}$$
This shows that $\sqrt[3]{p^2-3q^2}$ must be rational, hence integer.
Let $p^2-3q^2=k^3 (*)$. Then $y=\frac{k}{x}$ and thus
$$x+\frac{k}{x} =n \Rightarrow x^2-nx+k=0 \Rightarrow x= \frac{n \pm \sqrt{n^2-4k}}{2} \,.$$
The two roots of this equation must be $x$ and $y$, and thus we get:
$$x=\frac{n + \sqrt{n^2-4k}}{2} \,;\, y= \frac{n - \sqrt{n^2-4k}}{2} \,.$$
Then
$$p+q\sqrt{3}=\left(\frac{n + \sqrt{n^2-4k}}{2} \right)^3 \,.$$
From here, we get that $\sqrt{n^2-4k} \notin \mathbb Q$ and hence
$$p=\frac{n^3+3n^3-12nk}{8}=\frac{n^3-3nk}{2}$$
$$q\sqrt{3}=\frac{3n^2+n^2-4k}{8}\sqrt{n^2-4k}=\frac{n^2-k}{2}\sqrt{n^2-4k}$$
Thus, $n^2-4k=3u^2 $, for some $u$, hence
$$k=\frac{n^2-3u^2}{4} (***)$$
Thus, we get:
$$p=\frac{n^3+9nu^2}{8}$$
$$q=\frac{3n^2+3u^2}{8}u=\frac{3n^2u+3u^3}{8}$$
With this choice we have
$$p+q\sqrt{3}=(\frac{n+u\sqrt{3}}{2})^3$$
$$p-q\sqrt{3}=(\frac{n-u\sqrt{3}}{2})^3$$
Note that $p,q$ integers if and only if $8|n(n^2+u^2)$ and $8|u(n^2+u^2)$. It is easy to check that the first one cannot happen if $n$ is odd. Thus, $n$ must be even and $ 8|u(n^2+u^2) \Rightarrow 2|u^3 \Rightarrow u$ even.
If $n=2s, u=2t$ we get the general solution
$$p=s^3+9st^2$$
$$q=3s^2t+3t^3$$
$$n=2s$$
Note that in this case,
$$\sqrt[3]{p+q\sqrt{3} }=s+t\sqrt{3}$$
$$\sqrt[3]{p-q\sqrt{3} }=s-t\sqrt{3}$$
|
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|
integrating fraction/completing the square Does anyone know how to integrate the following?
$\frac{dx}{9x^2 + 6x + 17}$
I have been trying for ages and cannot get an answer anywhere close to the answer I get on maple?
|
Without many words and since you've already been given some ideas. We're looking for an arctangent (general set up for this particular case of integrals), so try to justify each step:
$$9x^2+6x+17=9\left(x^2+\frac{2}{3}x\right)+17=9\left(x+\frac{1}{3}\right)^2+16=$$
$$16\left[1+\left(\frac{3}{4}x+\frac{1}{4}\right)^2\right]\;\;,\;\;\text{and}\;\;\left(\frac{3}{4}x+\frac{1}{4}\right)'=\frac{3}{4}\implies$$
$$\int\frac{dx}{9x^2+6x+17}=\frac{1}{16}\int\frac{dx}{1+\left(\frac{3}{4}x+\frac{1}{4}\right)^2}=\frac{1}{12}\int\frac{\frac{3}{4}dx}{1+\left(\frac{3}{4}x+\frac{1}{4}\right)^2}=$$
$$=\frac{1}{12}\arctan\left(\frac{3}{4}x+\frac{1}{4}\right)+K$$
|
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|
How to compute the integral $\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx$ I want to compute this integral $$\displaystyle\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx$$
What I did was the following. I substituted $x=t^{6}$, so that my $dx= 6t^{5} \ dt$ and so the integral changes to $$\int_{0}^{\infty} \frac{t^2}{1+t^{12}} \cdot 6 t^{5} \ dt =6 \cdot \int_{0}^{\infty} \frac{t^7}{1+t^{12}} \: dt$$
Now If I substitute $t^{4}=v$ then what I will be having is the following integral $$\frac{6}{4}\cdot \int_{0}^{\infty} \frac{v}{1+v^{3}} \ dv$$
Now I can write $1+v^{3} = (1+v) \cdot (1-v+v^{2})$ and so I have
\begin{align*}
\int_{0}^{\infty} \frac{v}{1+v^{3}}\: dv &= \int_{0}^{\infty}\biggl[\frac{1}{3}\cdot \frac{v+1}{1-v+v^{2}} - \frac{1}{3} \cdot \frac{1}{1+v}\biggr]\: dv
\end{align*}
Now the point is that the integral of $1/(1+v) \to \infty$, so I am not sure if this is the right way to do. Can anyone suggest anything?
|
Substitute $x^{1/3} = t$, i.e., $x = t^3$, i.e., $dx = 3t^2 dt$. Hence,
$$I = \int_0^{\infty} \dfrac{t}{1+t^6} 3t^2 dt = 3 \int_0^{\infty} \dfrac{t^3}{1+t^6}dt$$
$$\int_0^{\infty} \dfrac{t^3}{1+t^6}dt = \int_0^{1} \dfrac{t^3}{1+t^6}dt + \int_1^{\infty} \dfrac{t^3}{1+t^6}dt$$
$$\int_1^{\infty} \dfrac{t^3}{1+t^6}dt = \int_1^0 \dfrac{1/t^3}{1+1/t^6}\left(-\dfrac{dt}{t^2}\right) = \int_0^1 \dfrac{t}{t^6+1} dt$$
Hence,
$$\dfrac{I}3 = \int_0^1 \dfrac{t+t^3}{1+t^6} dt = \int_0^1 \dfrac{t}{1-t^2+t^4} dt = \dfrac1{2 \sqrt{3}} \left(\int_0^1 \dfrac{dt}{t^2 - \sqrt3 t + 1} - \int_0^1 \dfrac{dt}{t^2 + \sqrt3 t + 1} \right)$$
I trust you can finish it off from here.
|
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|
A continued fraction involving Bernoulli numbers Let $B_n$ be the Bernoulli numbers. Then, we can write a function $F(x)$, expressed as a continued fraction involving those $B_n$, such that it gives the form,
$$ \displaystyle
\displaystyle F(x)= B_0+ \cfrac{(1/x)}{B_1+ \cfrac{(2/x)}{B_2+ \cfrac{(3/x)}{B_3+ \cfrac{(4/x)}{B_4+ \cfrac{(5/x)}{B_5+ \cfrac{(6/x)}{B_6+ \cfrac{(7/x)}{B_7+ \cfrac{(8/x)}{\dots}}}}}}}}=\cfrac{x-2}{x}
$$
Can anyone provide a proof for this finding?
Regards
http://tardigrados.wordpress.com/2013/01/29/numeros-de-bernoulli-y-las-fracciones-continuas-espejo/
|
Let's start with your definition:
$$F(x)= B_0+ \cfrac{(1/x)}{B_1+ \cfrac{(2/x)}{B_2+ \cfrac{(3/x)}{B_3+ \cfrac{(4/x)}{B_4+ \cfrac{(5/x)}{B_5+ \cfrac{(6/x)}{B_6+ \cfrac{(7/x)}{B_7+ \cfrac{(8/x)}{\dots}}}}}}}}$$
The only non-zero odd Bernoulli number is $B_1$, so we have
$$F(x)= B_0+ \cfrac{(1/x)}{
B_1+ \cfrac{(2/x)}{
B_2+ \frac{3}{4}\left(
B_4+ \frac{5}{6}\left(
B_6+ \frac{7}{8}\left(B_8 + \ldots\right)\right)\right)}}$$
Define $K = B_2+ \frac{3}{4}\left(
B_4+ \frac{5}{6}\left(
B_6+ \frac{7}{8}\left(B_8 + \ldots\right)\right)\right)$ and we have
$$F(x)= B_0+ \cfrac{(1/x)}{B_1+ \cfrac{(2/x)}{K}} \
= B_0+ \frac{K}{B_1 Kx+ 2} = 1+ \frac{2K}{4 - Kx} = \frac{4 + (2 - x)K}{4 - Kx} $$
So if $F(x)= \frac{x-2}{x}$ then $$\frac{x-2}{x} = \frac{4 + (2 - x)K}{4 - Kx} \\
(x-2)(4 - Kx) = x(4 + (2 - x)K) \\
4x-8-Kx^2+2Kx = 4x + 2Kx-Kx^2 \\
-8 = 0$$
This might be why you have had problems finding a proof, but it might also hint at how your identity can be corrected.
Addendum: actually, $K$ seems to diverge quite fast, so you might be able to take limits and show that as the number of terms increases the partial continued fraction $\frac{4 + (2 - x)k_n}{4 - k_nx}$ tends to $\frac{2-x}{-x}$.
|
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|
Simultaneous solution(s) to $a^2+4b^2+4ab=0$ and $a^2+4b^2+32+16a-8b=0$? Could you tell me just how should I solve this system:
$$
a^2+4b^2+4ab=0\\
a^2+4b^2+32+16a-8b=0
$$
I can't remember the proceeding and it's driving me crazy.
Thanks a lot
|
We have:
$$
a^2+4b^2+4ab=0 \tag{1}$$
$$
a^2+4b^2+32+16a-8b=0\tag{2}
$$
$\bf (I)$ Subtract equation $(2)$ from $(1)$:
$$\begin{align}
& a^2+4b^2+4ab & =0\\
- & a^2+4b^2+32+16a-8b &=0 \\
& \hline \\
= & 4ab -16a + 8b - 32 & = 0 \\
4 & (ab - 4a + 2b - 8) & = 0 \\ \\
= & ab - 4a + 2b -8 = 0 \tag{3}\\
\end{align}
$$
$\bf (II)$ Factor equation $(1)$ $$
a^2+4b^2+4ab=0 \iff (a + 2b)^2 = 0 \iff a = -2b \tag{4}
$$
$\bf (III)$
Substitute $a = -2b$ into equation $(3)$. Then solve for $a$ the solution for $b$ to obtain $a = -2b$.
|
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|
Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$ Inadvertently, I find this interesting inequality. But this problem have nice solution?
prove that
$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$
This problem have nice solution? Thank you.
ago,I find this
$$\ln{2}<\left(\dfrac{1}{2}\right)^{\frac{1}{2}}=\dfrac{\sqrt{2}}{2}$$
following is my some nice methods,
use this inequality
$$\dfrac{x-y}{\ln{x}-\ln{y}}>\sqrt{xy},x>y$$
then we let
$x=2,y=1$
so $$\ln{2}<\dfrac{\sqrt{2}}{2}$$
solution 2:
since
$$\dfrac{1}{n+1}\le\dfrac{1}{2}\cdot\dfrac{3}{4}\cdots\dfrac{2n-1}{2n}$$
then
$$\ln{2}=\sum_{n=0}^{\infty}\dfrac{1}{(n+1)2^{n+1}}<\sum_{n=0}^{\infty}\dfrac{(2n)!}{(n!)^22^{3n+1}}=\dfrac{1}{\sqrt{2}}$$
solution 3
since
$$(1+\sqrt{2})^2(t+1)-(t+1+\sqrt{2})^2=t(1-t)>0$$
so
$$\ln{2}=\int_{0}^{1}\dfrac{1}{t+1}dt<\int_{0}^{1}\left(\dfrac{1+\sqrt{2}}{t+1+\sqrt{2}}\right)^2dt=\dfrac{\sqrt{2}}{2}$$
solution 4:
$$\ln{2}=\dfrac{3}{4}-\dfrac{1}{4}\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(2n+1)}<\dfrac{3}{4}-\dfrac{1}{4}\left(\dfrac{1}{1\times 2\times 3}-\dfrac{1}{2\times 3\times 5}\right)=\dfrac{7}{10}<\dfrac{\sqrt{2}}{2}$$
solution 5
$$\dfrac{1}{\sqrt{2}}-\ln{2}=\sum_{n=1}^{\infty}\dfrac{\sqrt{2}}{(4n^2-1)(17+2\sqrt{2})^n}>0$$
But $$\ln{2}>\left(\dfrac{2}{5}\right)^{\frac{2}{5}}$$ I can't have this nice solution
Thank you everyone can help.
|
$$
\left(2 \over 5\right)^{2/5}
=
0.69314\color{#ff0000}{\Large 4}843155146\ldots\,,
\qquad\qquad
\ln\left(2\right)
=
0.69314\color{#ff0000}{\Large 7}180559945\ldots
$$
So$\ldots$
|
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|
Trig and algebra problem: Finding sides of a triangle Let $ABC$ be a triangle such that $\angle ACB = \pi/6$ and let $a,b,c$ denote the lengths of the sides opposite to $A,B,C$, respectively. What are the value(s) of x for which $a = x^2 + x + 1, b = x^2-1$ and $c = 2x+1$?
I used law of cosines, then saw that I should apply the identity $a^2 + b^2 = (1/2)[(a+b)^2 + (a-b)^2]$, but I didn't know how to handle the remaining terms. (In other words, I got stuck early in the problem).
|
HINT:
Applying Law of Cosines, $$\cos \frac\pi6=\frac{a^2+b^2-c^2}{2ab}$$
Now, $$b^2+a^2-c^2=(x^2-1)^2+(x^2+x+1)^2-(2x+1)^2$$
$$=(x^2-1)^2+(x^2-x)(x^2+3x+2)\text{ applying } (a^2-b^2)=(a+b)(a-b)\text{ for the last two terms}$$
$$=(x^2-1)^2+x(x-1)(x+1)(x+2)$$
$$=(x^2-1)\{x^2-1+x(x+2)\}=(x^2-1)(2x^2+x-1)$$
As $x^2-1\ne0$ (which would make $b=0$)
$$\frac{(x^2-1)(2x^2+x-1)}{2(x^2-1)(x^2+x+1)}=\frac{(2x^2+x-1)}{2(x^2+x+1)}=\cos \frac\pi6=\frac{\sqrt3}2$$
Can you complete it now?
|
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|
Homework: Maclaurin Power Series Help I'm trying to find the Maclaurin Power Series for
$$f(x)=\frac{3x-8}{3x^2+5x-2}$$
but each degree of differentiation gets more complex with no discernible pattern. Any help is appreciated, thanks.
|
$$\begin{align*} f(x)&=\frac{3x-8}{3x^2+5x-2}=\frac{3x-8}{(x+2)(3x-1)}=\{\text{partial fraction decomposition}\}=\\
&=\frac{2}{2+x}-\frac{3}{3x-1}=\frac{1}{1-\left(-\frac{x}{2}\right)}+\frac{3}{1-3x}=\{\text{geometric series for }|x|<\frac{1}{3}\}=\\
&=\sum_{n=0}^\infty\left(-\frac{x}{2}\right)^n+3\sum_{n=0}^\infty(3x)^n=\sum_{n=0}^\infty\left(\left(-\frac{x}{2}\right)^n+3(3x)^n\right)=\\
&=\sum_{n=0}^\infty\left(\frac{1}{(-2)^n}+3^{n+1}\right)x^n \end{align*}$$
|
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|
Integral of $\sin x \cdot \cos x$ I've found 3 different solutions of this integral. Where did I make mistakes? In case there is no errors, could you explain why the results are different?
$ \int \sin x \cos x dx $
1) via subsitution $ u = \sin x $
$ u = \sin x; du = \cos x dx \Rightarrow \int udu = \frac12 u^2 \Rightarrow \int \sin x \cos x dx = \frac12 \sin^2 x $
2) via subsitution $ u = \cos x $
$ u = \cos x; du = -\sin x dx \Rightarrow -\int udu = -\frac12 u^2 \Rightarrow \int \sin x \cos x dx = -\frac12 \cos^2 x = -\frac12 (1 - \sin^2 x) = -\frac12 + \frac12 \sin^2 x $
3) using $ \sin 2x = 2 \sin x \cos x $
$ \int \sin x \cos x dx = \frac12 \int \sin 2x = \frac12 (- \frac12 \cos 2x) = - \frac14 \cos 2x = - \frac14 (1 - 2 \sin^2 x) = - \frac14 + \frac12 \sin^2 x $
So, we have:
$$ \frac12 \sin^2 x \neq -\frac12 + \frac12 \sin^2 x \neq - \frac14 + \frac12 \sin^2 x $$
|
Note: You are calculating indefinite integral and constants can be anything(they may differ). In fact the general solution to that would be just $C+\dfrac{\sin^2 x}{2}$
|
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|
Permutation & Combination - how many numbers smaller than $2.10^8$ and are divisible by $3$ can be written by means of the digits $0$,$1$ and $2$ How many numbers smaller than $2\times10^8$ and divisible by $3$ can be written by means of the digits $0$,$1$ and $2$?
Left zero padding not allowed.
I am getting this as -
$3$ digits - $2\cdot3 = 6$
$4$ digits - $2\cdot3\cdot3 = 18$
$5$ digits - $2\cdot3\cdot3\cdot3 = 54$
$6$ digits - $2\cdot3\cdot3\cdot3\cdot3 = 162$
$7$ digits - $2\cdot3\cdot3\cdot3\cdot3\cdot3\cdot3 = 486$
$8$ digits - $2\cdot3\cdot3\cdot3\cdot3\cdot3\cdot3\cdot3 = 1458$
From $10^8$ to $2\times10^8$ - $2\cdot3\cdot3\cdot3\cdot3\cdot3\cdot3
\cdot3\cdot3/2 = 2187$
I get $4371$ as the answer, but in the book the result is $4351$, am I making an error somewhere?
|
We solve the simpler problem of counting the numbers from $0$ to $10^8-1$ of the right shape that are divisible by $3$. We do allow zero padding to make all the numbers into $8$-digit numbers. This makes no difference to the count.
The same method works for the other half of our interval.
We use a general approach, though the particular case is too simple and quickly collapses. One could then in hindsight give a slicker argument, but we won't do that.
Let $a(n)$ be the number of $n$-digit numbers that use only $0$, $1$, and $2$ and are divisible by $3$. Let $b(n)$ be the number of $n$-digit numbers, of the right form, that give remainder $1$ on division by $3$, and let $c(n)$ be the same thing, except for remainder $2$. We have $a(1)=b(1)=c(1)=1$. Note that
$$a(n+1)=a(n)+b(n)+c(n).$$
This is because a qualifying $n+1$ digit number can be obtained by appending a $0$ to an $n$-digit number divisible by $3$, or a $2$ to an $n$-digit number that has remainder $1$, or a $1$ to an $n$-digit number that has remainder $2$. Similarly, we have $b(n)=a(n)+b(n)+c(n)$ and $c(n)=a(n)+b(n)+c(n)$.
It follows that $a(2)=b(2)=c(2)=3$, and $a(3)=b(3)=c(3)=9$, and so on. Thus $a(8)=3^7$.
For the numbers from $1\times 10^8$ to $2\times 10^8-1$, do the same calculation for numbers of the form $1x$, where $x$ is a (padded) $8$-digit number.
|
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|
A concise distance problem A falsely simple Euclidian geometry problem:
Points $A$, $B$, $C$ are collinear; $\|AB\|=\|BD\|=\|CD\|=1$; $\|AC\|=\|AD\|$.
What is the set of possible $\|AC\|$ ?
I'm after a concise answer, with reasoning, that would get maximum points to an 11th-grader.
A related question asks an appropriate level for the problem (worded in less mathematical terms and thus reduced to distinct points).
To check your answer: the mean of the elements of the set of solutions to the present question is $\approx 1.08$.
|
W.l.o.g. choose coordinates as follows:
\begin{align*}
A &= \begin{pmatrix}0\\0\end{pmatrix} &
B &= \begin{pmatrix}1\\0\end{pmatrix} &
C &= \begin{pmatrix}x\\0\end{pmatrix} &
D &= \begin{pmatrix}\frac{x+1}2\\
\sqrt{x^2-\left(\frac{x+1}{2}\right)^2}\end{pmatrix}
\end{align*}
Point $D$ is choosen on the perpendicular bisector of $BC$. This bisector is only uniquely defined and a strict requirement if $B\neq C$, so the case of $B=C$ will have to be handled separately. It is furthermore chosen at distance $x$ from $A$. So the above coordinates will already ensure the following conditions:
\begin{align*}
\lVert AB\rVert &= 1 &
\lVert BD\rVert &= \lVert CD\rVert &
\lVert AC\rVert &= \lVert AD\rVert
\end{align*}
Now all that remains is the condition $\lVert BD\rVert = 1$ or equivalently $\lVert CD\rVert = 1$. I go for the latter.
\begin{align*}
1 = \lVert CD\rVert^2 &=
\left(x-\frac{x+1}2\right)^2 + \left(\sqrt{x^2-\left(\frac{x+1}{2}\right)^2}\right)^2
\\&= \left(\frac{x-1}2\right)^2 + \left(x^2-\left(\frac{x+1}{2}\right)^2\right) \\
4 &= \left(x^2-2x+1\right) + \left(4x^2-x^2-2x-1\right) \\
0 &= 4x^2 - 4x - 4 = x^2 - x - 1 \\
x_{1,2} &= \frac{1\pm\sqrt{1+4}}{2} \\
x_1 &= \frac{1-\sqrt5}2 \approx -0.618 \\
x_2 &= \frac{1+\sqrt5}2 \approx 1.618
\end{align*}
The special case of $B=C$ gives a third solution $x_3=1$ describing $C$, with an associated point $D_3=\begin{pmatrix}\frac12\\\frac{\sqrt3}2\end{pmatrix}$ (although these coordinates of $D_3$ are not neccessary to answer the question). The requested length is the absolute value of $x$, so the final solution is
$$\lVert AC\rVert = \lvert x\rvert \in \left\{
\frac{\sqrt5-1}2, 1, \frac{\sqrt5+1}2 \right\}$$
The idea of choosing suitable coordinates without loss of generality will probably be the most problematic concept. Whether students will be able to think of this on their own, particularly in an exam situation, depends a lot on the details of their education. After that, chances are good that many will miss the special case, since the coordinates (as perhaps derived from a sketch of a non-degenerate situation) seem to perclude such a third solution. The computation of $x$ on the other hand should be pretty straight forward, once they got this far.
|
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|
If $x^2-xy+2y^2=\frac{a^2}{7}$, find $y'''$.
If $x^2-xy+2y^2=\frac{a^2}{7}$, find $y'''$.
For our 1st derivative we got $$y'=\frac{2x-y}{x-4y}.$$
For the second derivative we got $$y''=\frac{14x^2-14xy+28y^2}{(x-4y)^3}.$$
And for the final answer we got $$y'''=\frac{4(-84x^3+119x^2y-154xy^2-84y^3)}{(x-4y)^5}.$$
Took me 2 hours and 3 white boards to get to that answer, and the teacher said another answer was right, also that there was an $a$ in the final answer, even though the derivative of a constant is zero. any insight or help is greatly appreciated, and will try to post all my work as an answer.
|
As Will Jagy noted, $$y''=\frac{2a^2}{(x-4y)^3}.$$
Then $$y'''=2a^2(-3)(x-4y)^{-4}(1-4y')$$ Now $$1-4y'=\frac{4y-8x-4y+x}{x-4y}=\frac{7x}{x-4y},$$ so your $y'''$ is $$y'''=\frac{42a^2x}{(x-4y)^5}.$$
I used your answer for second derivative to find the third derivative and I got $$y'''=\frac{(x-4y)(28x-14y-14xy'+56yy')-(14x^2-14xy+28y^2)\times 3(1-4y')}{(x-4y)^4}$$ after cancelling $(x-4y)^2$ out from numerator and denominator. Then I substitute $y'=\frac{2x-y}{x-4y}$ and simplify, and I got
$$\frac{294(x^2-xy+2y^2)x}{(x-4y)^5}$$ which agrees with my answer upon substituting $x^2-xy+2y^2=a^2/7$ . This means that if you had calculated correctly, your answer should have been $$y'''=\frac{294(x^2-xy+2y^2)x}{(x-4y)^5}.$$
|
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|
Another trigonometric proof...? ...sigh..another problem how shall I prove the following?
$$ {\cot A\over1- \tan A} + {\tan A \over 1- \cot A} = 1 + \tan A + \cot A$$
so what now? the following's what I've done:
$$\cot A - \cot^2 A + \tan A- \tan^2 A \over 2 - \tan A - \cot A$$
|
$$
\begin{align}
\frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}&=\frac{\cos A\cot A - \sin A \tan A}{\cos A-\sin A}
\\\\&=\frac{\frac{\cos^2 A}{\sin A} - \frac{\sin^2 A}{\cos A}}{\cos A-\sin A}
\\\\&=\frac{\cos^3 A - \sin^3 A}{(\cos A \sin A)(\cos A-\sin A)}
\\\\&=\frac{\cos^2 A + \sin^2 A + \sin A\cos A}{\cos A \sin A}
\\\\&=1+\cot A+\tan A
\end{align}
$$
|
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|
number of solutions to an equation? Given $x$ and $y$ are multiples of $2$ satisfying
$$x^2 - y^2 = 27234702932$$
Find the number of solutions to $x$ and $y$.
|
As $27234702932=2^2\cdot181\cdot37616993$ where the last two factors are primes
If $x=2X,y=2Y$
$$X^2-Y^2=181\cdot37616993$$
If $X^2-Y^2=p\cdot q$ where $p,q$ are primes,
the possible cases for $X+Y,X-Y$ are $\pm pq, \pm 1$ and $\pm p,\pm q $
For example, if $X+Y=1,X-Y=pq$
The number of positive factors of $p\cdot q$ is $(1+1)\cdot(1+1)=4$
So, there should be $4\cdot2 =8$ solutions in integers.
|
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|
Improper integral $\sin(x)/x $ converges absolutely, conditionally or diverges? Improper integral of $\sin(x)/x $ converges absolutely, conditionally or diverges?
We have
$$\int_1^{\infty}\frac{\sin x}{x}\text{d}x$$
Integrating by parts
$$u=\frac{1}{x}$$
$$\text{d}u=-\frac{1}{x^2}\text{d}x$$
$$\text{d}v=\sin x\;\text{d}x$$
$$v=-\cos x$$
$$
\begin{aligned}
\int_1^{\infty} \frac{\sin x}{x} \text{d}x
& = \frac{-\cos x}{x} \Big|_1^{\infty}
- \int_1^{\infty} \frac{\cos x}{x^2} \text{d}x \\
& = \cos 1 - \int_1^{\infty} \frac{\cos x}{x^2} \text{d}x
\end{aligned}
$$
$\int_1^{\infty} \frac{\cos x}{x^2} \text{d}x$ converges absolutely
(using the Comparison Test For Improper Integrals):
$$
\int_1^{\infty} \frac{|\cos x|}{x^2} \text{d}x <
\int_1^{\infty} \frac{1}{x^2} \text{d}x
$$
So $\int_1^{\infty} \frac{\sin x}{x} \text{d}x$
converges.
Now I need to find out if
$\int_1^{\infty} |\frac{\sin x}{x}| \text{d}x$
converges or diverges.
|
Let $N \in \Bbb N, N > 1$, we have:
\begin{align}
\int_0^{2\pi N} \left|\frac{\sin x}{x}\right|\,dx &= \sum_{n=0}^{N-1} \int_{2\pi n}^{2\pi(n+1)} \left|\frac{\sin x}{x}\right|\,dx \\
&\ge \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{2\pi n}^{2\pi(n+1)} \left|\sin x\right|\,dx \\
&= \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{0}^{2\pi} \left|\sin x\right|\,dx \\
&= \sum_{n=0}^{N-1} \frac{2}{\pi (n+1)}
\end{align}
The last sum diverges as $N \to \infty$, and so does the original integral.
Your integral is on $[1, \infty]$, but it also diverges because $\left|\frac{\sin{x}}{x}\right|$ is continuous on $[0, 1]$. My proof is on $[0, \infty]$ because it makes managing the summation slightly easier.
|
{
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"url": "https://math.stackexchange.com/questions/390810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Using Spherical coordinates find the volume: Inside the surfaces $z=x^2+y^2$ and $z=\sqrt{2-x^2-y^2}$
I integrated over the ranges:
$0 \leq \theta \leq 2\pi$
$ 0 \leq \phi \leq \frac{\pi}{2}$
$0 \leq r \leq \sqrt{2}$
I get $\frac{\pi}{2}(4\sqrt{2} -4).$
There answer is the same except a $-\frac{7}{2}$ instead of the 4 at the end. Obviously I'm missing a 1\2 but I seems, I can not find it.
|
This problem is actually better suited for cylindrical coordinates:
$$\begin{align}\int_{0}^{2\pi}\int_0^1\int_{r^2}^{\sqrt{2-r^2}}rdzdrd\theta&=2\pi\int_0^1(r\sqrt{2-r^2}-r^3)dr\\&=2\pi(-\frac{1}{3}(2-r^2)^{3/2}-\frac{r^4}{4})\mid_{0}^1\\&=2\pi(-\frac{1}{3}-\frac{1}{4}+\frac{2\sqrt{2}}{3})\\&=\frac{\pi}{3}(4\sqrt2-\frac{7}{2})\end{align}$$
The problem with spherical coordinates here is that the radius is a (piecewise) function of the azimuthal angle, which makes the integration a bit more difficult.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluating Complex Line Integrals Calculate $\int_{\gamma}\frac{\Re(z)}{z-\frac{1}{2}}dz$ and $\int_{\gamma}\frac{\Im(z)}{z-\frac{1}{2}}dz$ when $\gamma$: $|z|=1$ is positively oriented.
This is what I have tried to do, starting with the first line integral.
Since $\frac{\Re(z)}{z-\frac{1}{2}}$ is not analytical/holomorphic at any point in the plane - it does not satisfy the Cauchy-Riemann equations - we cannot use Cauchy's integral formulae immediately. However, $|z|=1$ implies that the conjugate to $z$, $z*=\frac{1}{z}$, because $1=|z|^{2}=zz*$. Thus, we may write:
\begin{align*} \int_{\gamma}\frac{\Re(z)}{z-\frac{1}{2}}dz &=\frac{1}{2}\int_{\gamma}\frac{z+\frac{1}{z}}{z-\frac{1}{2}}dz \\ &=\frac{1}{2}\int_{\gamma}\frac{z^{2}+1}{z(z-\frac{1}{2})}dz \\
&=\frac{1}{2}\int_{\gamma}(1-\frac{\frac{z}{2}+1}{z(z-\frac{1}{2})}dz \\ &=\frac{1}{2}\int_{\gamma}dz-\frac{1}{2}\int_{\gamma}\frac{\frac{z}{2}+1}{z(z-\frac{1}{2})}dz \\ &=-\frac{1}{2}\int_{\gamma}\frac{-2}{z}+\frac{\frac{5z}{2}}{z-\frac{1}{2}}dz \\ &= \int_{\gamma}\frac{1}{z}dz-\frac{5}{4}\int_{\gamma}\frac{z}{z-\frac{1}{2}}dz \\ &=2\pi{i}(1-\frac{5}{8}) \\ &=\frac{3\pi{i}}{4} \end{align*}
Note that I rewrote the right hand side two times, using partial fractions. Similarly,I have attempted to the second line integral, however, this time a only used partial fractions once.
\begin{align*} \int_{\gamma}\frac{\Im(z)}{z-\frac{1}{2}}dz &= \frac{1}{2i}\int_{\gamma}\frac{z-\frac{1}{z}}{z-\frac{1}{2}}dz \\ &=\frac{1}{2i}\int_{\gamma}\frac{z^{2}-1}{z(z-\frac{1}{2})}dz \\ &=\frac{1}{2i}\int_{\gamma}(1-\frac{2}{z}+\frac{3}{z-\frac{1}{2}})dz \\ &=\frac{1}{2i}\int_{\gamma}dz-\frac{1}{2i}\int_{\gamma}\frac{-2}{z}dz-\frac{1}{2i}\int_{\gamma}\frac{3}{z-\frac{1}{2}}dz \\ &= \frac{1}{i}\int_{\gamma}\frac{1}{z}dz-\frac{3}{4i}\int_{\gamma}\frac{1}{z-\frac{1}{2}}dz \\ &=2\pi{i}(\frac{1}{i}-\frac{3}{4i}) \\ &=\frac{5\pi}{4}\end{align*}
As I have not studied complex analysis, I am unsure whether any of these two answers are correct. I would very much appreciate if someone more experienced person could help me check, most likely correct, my calculations.
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You have some problem on partial fraction $\displaystyle \frac{z^2 + 1}{z(z-1/2)} \neq 1 - \frac{\frac z 2 + 1}{z(z-1/2)}$.
This should be $\displaystyle \frac{z^2 + 1}{z(z-1/2)} = 1 + \frac{\frac z 2 + 1}{z(z-1/2)} = 1+\frac{5/2}{z - 1/2} - \frac 2 z$.
And this should give you $\displaystyle \frac{2 \pi i}{2} ( \frac 52 - 2) = \pi/2$
On the other,
$\displaystyle \frac{z^2 - 1}{z(z-1/2)} = 1 + \frac 2 z - \frac{3/2}{z - 1/2}$ while you have written $\displaystyle \frac{z^2 - 1}{z(z-1/2)} = (1-\frac{2}{z}+\frac{3}{z-\frac{1}{2}})dz $
Also that on here
$$=\frac{1}{2i}\int_{\gamma}(1-\frac{2}{z}+\frac{3}{z-\frac{1}{2}})dz \\
=\frac{1}{2i}\int_{\gamma}dz-\underbrace{\frac{1}{2i}\int_{\gamma}\frac{-2}{z}dz-\frac{1}{2i}\int_{\gamma}\frac{3}{z-\frac{1}{2}}dz}_{\text{why - here ??}} \\
$$
This is wrong step.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluation of a specific determinant.
Evaluate $\det{A}$, where $A$ is the $n \times n$ matrix defined by $a_{ij} = \min\{i, j\}$, for all $i,j\in \{1, \ldots, n\}$.
$$A_2
\begin{pmatrix} 1& 1\\
1& 2
\end{pmatrix};
A_3 = \begin{pmatrix} 1& 1& 1\\
1& 2& 2\\
1& 2& 3
\end{pmatrix};
A_4 = \begin{pmatrix} 1& 1& 1& 1\\
1& 2& 2& 2\\
1& 2& 3& 3\\
1& 2& 3& 4
\end{pmatrix};
A_5 = \begin{pmatrix} 1& 1& 1& 1& 1\\
1& 2& 2& 2& 2\\
1& 2& 3& 3& 3\\
1& 2& 3& 4& 4\\
1& 2& 3& 4& 5
\end{pmatrix}$$
$$A_6 = \begin{pmatrix} 1& 1& 1& 1& 1& 1\\
1& 2& 2& 2& 2& 2\\
1& 2& 3& 3& 3& 3\\
1& 2& 3& 4& 4& 4\\
1& 2& 3& 4& 5& 5\\
1& 2& 3& 4& 5& 6
\end{pmatrix};
A_7 = \begin{pmatrix} 1& 1& 1& 1& 1& 1& 1\\
1& 2& 2& 2& 2& 2& 2\\
1& 2& 3& 3& 3& 3& 3\\
1& 2& 3& 4& 4& 4& 4\\
1& 2& 3& 4& 5& 5& 5\\
1& 2& 3& 4& 5& 6& 6\\
1& 2& 3& 4& 5& 6& 7
\end{pmatrix}
$$
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If you consider the matrix
$$
B_n=\begin{bmatrix}
1&0&\cdots&0\\
1&1&0&\cdots\\
&&\cdots\\
1&1&\cdots &1
\end{bmatrix}
$$
(i.e. the $i,j$ entry of $B_n$ is $1$ is $i\geq j$ and $0$ otherwise), then the $k^{\rm t h}$ column of $A_n$ is obtained by adding the first $k$ columns of $B_n$. So
$$
\det A_n=\det B_n=1.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate by parts: $\int \ln (2x + 1) \, dx$ $$\eqalign{
& \int \ln (2x + 1) \, dx \cr
& u = \ln (2x + 1) \cr
& v = x \cr
& {du \over dx} = {2 \over 2x + 1} \cr
& {dv \over dx} = 1 \cr
& \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr
& = x\ln (2x + 1) - \int 1 - {1 \over {2x + 1}} \cr
& = x\ln (2x + 1) - (x - {1 \over 2}\ln |2x + 1|) \cr
& = x\ln (2x + 1) + \ln |(2x + 1)^{1 \over 2}| - x + C \cr
& = x\ln (2x + 1)^{3 \over 2} - x + C \cr} $$
The answer $ = {1 \over 2}(2x + 1)\ln (2x + 1) - x + C$
Where did I go wrong?
Thanks!
|
Here is a cute variant. Let $u=\ln(2x+1)$ and let $dv=dx$. Then $du=\frac{2}{2x+1}$ and (this is the cute part) we can take $v=x+\frac{1}{2}$. It follows that
$$\int \ln(2x+1)\,dx=\left(x+\frac{1}{2}\right)\ln(2x+1)-\int \left(x+\frac{1}{2}\right)\frac{2}{2x+1}\,dx.$$
But the remaining integrand is just $1$! It follows that our integral is
$$\left(x+\frac{1}{2}\right)\ln(2x+1) -x+C.$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
derive using the chain rule Given the polinomyal $f(x)=\frac{x^3}{(4-x^2)^3}$ find $f'(x)$
So, If I try to derive this, first I must to apply the chain rule in the denominator and then derive of the division (...)
$$f'(x)=\frac{x^3}{3(4-x^2)^2(-2)} = \frac{x^3}{-6(4-x^2)^2}$$
(...)?
Or there is another way to do this?
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You're doing wrong. A good way to do this when still not expert in derivatives, is to write
$$
f(x)=\frac{g(x)}{h(x)}
$$
where
$$
g(x)=x^3,\qquad h(x)=(4-x^2)^3.
$$
Then, first of all, you have to differentiate the quotient:
$$
f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{(h(x))^2}
$$
So you need to compute $g'(x)$ and $h'(x)$. The first one is easy:
$$
g'(x)=3x^2.
$$
The second one needs the chain rule:
$$
h'(x)=3(4-x^2)^2(-2x)=-6x(4-x^2)^2.
$$
Now plug in the formula above:
\begin{align}
f'(x)&=
\frac{3x^2\cdot(4-x^2)^3 - x^3\cdot(-6x(4-x^2)^2)}{(4-x^2)^6}\\[2ex]
&=\frac{3x^2(4-x^2)^3+6x^4(4-x^2)^2}{(4-x^2)^6}\\[2ex]
&=\frac{3x^2(4-x^2)+6x^4}{(4-x^2)^4}\\[2ex]
&=\frac{12x^2+3x^4}{(4-x^2)^4}=\frac{3x^2(4+x^2)}{(4-x^2)^4}
\end{align}
A different strategy may be noting that
$$
f(x)=\left(\frac{x}{4-x^2}\right)^3
$$
so, letting
$$g(x)=\frac{x}{4-x^2}$$
you have, by the chain rule,
$$
f'(x)=3g(x)^2g'(x)
$$
The derivative of $g(x)$ can be computed again with the quotient rule:
$$
g'(x)=\frac{1\cdot(4-x^2)-x\cdot(-2x)}{(4-x^2)^2}=
\frac{4-x^2+2x^2}{(4-x^2)^2}=\frac{4+x^2}{(4-x^2)^2}
$$
and so
$$
f'(x)=3\left(\frac{x}{(4-x^2)}\right)^2\frac{4+x^2}{(4-x^2)^2}
$$
which gives the same result as before.
|
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"timestamp": "2023-03-29T00:00:00",
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|
solution to a root inequality I have the inequality
$$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+ac+bc} \geq \sqrt{a^2+2bc}+\sqrt{b^2+2ac}+\sqrt{c^2+2ab}.$$I tried to do $u=a^2+b^2+c^2$ and $v=ab+ac+bc$ and $x=a^2+2bc$, $y=b^2+2ac$, $z=c^2+2ab$ ...but I did not find any solution. Any help is appreciated.
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We need to prove that
$$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+ac+bc}\geq\sum\limits_{cyc}\sqrt{a^2+2bc}$$ or
$$\sum\limits_{cyc}\left(\sqrt{a^2+b^2+c^2}-\sqrt{c^2+2ab}\right)\geq2\left(\sqrt{a^2+b^2+c^2}-\sqrt{ab+ac+bc}\right)$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2}{\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}}\geq\sum\limits_{cyc}\frac{(a-b)^2}{\sqrt{a^2+b^2+c^2}+\sqrt{ab+ac+bc}}$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2\left(\sqrt{ab+ac+bc}-\sqrt{c^2+2ab}\right)}{\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}}\geq0$$ or
$$\sum\limits_{cyc}\frac{-(a-b)^2(c-a)(c-b)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}\geq0$$ or
$$\sum\limits_{cyc}\left(\tfrac{-(a-b)^2(c-a)(c-b)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}+\tfrac{(a-b)^2(c-a)(c-b)}{2\sqrt{ab+ac+bc}\left(\sqrt{a^2+b^2+c^2}+\sqrt{ab+ac+bc}\right)}\right)\geq0$$ or
$$\sum\limits_{cyc}\tfrac{(a-b)^2(c-a)(c-b)\left(\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)-2\sqrt{ab+ac+bc}\left(\sqrt{a^2+b^2+c^2}+\sqrt{ab+ac+bc}\right)\right)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}\geq0$$ or
$$\sum\limits_{cyc}\tfrac{(a-b)^2(c-a)(c-b)\left(-\sqrt{(a^2+b^2+c^2)(ab+ac+bc)}+\sqrt{c^2+2ab}\left(\sqrt{ab+ac+bc}+\sqrt{a^2+b^2+c^2}\right)+c^2+2ab-2(ab+ac+bc)\right)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}\geq0$$ or
$$\sum\limits_{cyc}\tfrac{(a-b)^2(c-a)(c-b)\left(\left(\sqrt{c^2+2ab}-\sqrt{ab+ac+bc}\right)\left(\sqrt{ab+ac+bc}+\sqrt{a^2+b^2+c^2}\right)+c^2+2ab-(ab+ac+bc)\right)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2(c-a)(c-b)\left(\frac{(c-a)(c-b)\left(\sqrt{ab+ac+bc}+\sqrt{a^2+b^2+c^2}\right)}{\sqrt{c^2+2ab}+\sqrt{ab+ac+bc}}+(c-a)(c-b)\right)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2(c-a)^2(c-b)^2\left(\frac{\sqrt{ab+ac+bc}+\sqrt{a^2+b^2+c^2}}{\sqrt{c^2+2ab}+\sqrt{ab+ac+bc}}+1\right)}{\left(\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}\right)\left(\sqrt{ab+ac+bc}+\sqrt{c^2+2ab}\right)}\geq0$$
Done!
|
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.