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Easiest example of rearrangement of infinite leading to different sums I am reading the section on the rearrangement of infinite series in Ok, E. A. (2007). Real Analysis with Economic Applications. Princeton University Press.
As an example, the author shows that
is a rearrangement of the sequence
\begin{align} \frac{(-1)^{m+1}}{m} \end{align}
and that the infinite sum of these two sequences must be different.
My question is : what is to you the easiest and most intuitive example of such infinite series having different values for different arrangements of the terms? Ideally, I hope to find something as intuitive as the illustration that some infinite series do not have limits through $\sum_\infty (-1)^i$.
I found another example in http://www.math.ku.edu/~lerner/m500f09/Rearrangements.pdf but it is still too abstract to feed my intuition...
|
The following variant of the alternating harmonic is computationally easy. Consider the series
$$ 1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{8}-\frac{1}{8}+\cdots.$$
The sum is $0$. For the partial sums are either $0$ or $\frac{1}{2^k}$ for suitable $k$ that go to infinity.
Let us rearrange this series to give sum $1$. Use
$$\begin{align} 1+&\frac{1}{2}+\frac{1}{2}-1+\frac{1}{4}+\frac{1}{4}-\frac{1}{2}+\frac{1}{4}+\frac{1}{4}-\frac{1}{2}+\frac{1}{8}+\frac{1}{8}-\frac{1}{4}+\frac{1}{8}+\frac{1}{8}-\frac{1}{4}+\frac{1}{8}+\frac{1}{8}-\frac{1}{4}\\&+\frac{1}{8}+\frac{1}{8}-\frac{1}{4}+\frac{1}{16}+\frac{1}{16}-\frac{1}{8}+\frac{1}{16}+\frac{1}{16}-\frac{1}{8}+\cdots.\end{align}$$
That the series converges to $1$ follows from the fact that the sum of the first $3n+1$ terms is always $1$, and that the sum of the first $3n+2$ terms, and the sum of the first $3n+3$ terms, differ from the sum of the first $3n+1$ terms by an amount that $\to 0$ as $n\to\infty$.
|
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|
Find the equation of a circle given two points and a line that passes through its center Find the equation of the circle that passes through the points $(0,2)$ and $(6,6)$. Its center is on the line $x-y =1$.
|
First, find the center $(a, b)$. It is equidistant from the two points, so
$$
\sqrt{a^2 + (b - 2)^2} = \sqrt{(a - 6)^2 + (b - 6)^2}
$$
or
$$
a^2 + (b - 2)^2 = (a - 6)^2 + (b - 6)^2.
$$
Now, since the center is on the line $x - y = 1$, we know that $a - b = 1$, or
$$
a = b + 1.
$$
Substitute this into the quadratic equation above and simplify:
$$
\begin{align}
(b + 1)^2 + (b - 2)^2 &= ((b + 1) - 6)^2 + (b - 6)^2 \\
(b^2 + 2b + 1) + (b^2 - 4b + 4) &= (b^2 - 10b + 25) + (b^2 - 12b + 36) \\
2b^2 - 2b + 5 &= 2b^2 - 22b + 61 \\
20b &= 56 \\
b &= \frac{14}{5}.
\end{align}
$$
Therefore,
$$
a = \frac{14}{5} + 1 = \frac{19}{5}.
$$
What is the radius $r$? Well the square of the radius is
$$
\begin{align}
r^2 &= a^2 + (b - 2)^2 \\
&= \left( \frac{19}{5} \right)^2 + \left( \frac{4}{5} \right)^2 \\
&= \frac{377}{25}.
\end{align}
$$
Putting this together, we can write the equation of the circle as
$$
\left( x - \frac{19}{5} \right)^2 + \left( y - \frac{14}{5} \right)^2 = \frac{377}{25}
$$
or, by multiplying by $25$ to clear denominators,
$$
(5x - 19)^2 + (5y - 14)^2 = 377.
$$
Here's a picture.
|
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|
How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks
|
Let $$f(x)=(x-b)^3 + (b-c)^3 + (c-x)^3 -3(x-b)(b-c)(c-x) $$
The $f$ is a polynomial of degree at most $3$. Moreover, $x^3$ appears twice, with coefficients $\pm 1$, thus cancel. It follows that $f$ is at most quadratic and it is obvious that
$$f(c)=f(b)=f(0)=0$$
Since $f$ is at most quadratic and has three roots, $f \equiv 0$.
Note: If you want to avoid the observation that the coefficients of $x^3$ cancel, then use $f(c)=f(b)=f(0)=0$ and
$$f(b+c)=c^3+(b-c)^3-b^3+3c(b-c)(-b)=c^3-b^3+(b-c)^3-3bc(b-c)$$
You can see that this is zero, or if you still want to avoid this computation, repeat the above argument:
Let $g(x)=x^3-b^3+(b-x)^3-3bx(b-x)$. Then $g$ is at most cubic and
$$g(b)=g(0)=g(2b)=g(-b)=0 \Rightarrow g(x)\equiv 0$$
|
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|
Solve these equations simultaneously
Solve these equations simultaneously:
$$\eqalign{
& {8^y} = {4^{2x + 3}} \cr
& {\log _2}y = {\log _2}x + 4 \cr} $$
I simplified them first:
$\eqalign{
& {2^{3y}} = {2^{2\left( {2x + 3} \right)}} \cr
& {\log _2}y = {\log _2}x + {\log _2}{2^4} \cr} $
I then had:
$\eqalign{
& 3y = 4x + 6 \cr
& y = x + 16 \cr} $
Solving:
$\eqalign{
& 3\left( {x + 16} \right) = 4x + 6 \cr
& 3x + 48 = 4x + 6 \cr
& x = 42 \cr
& y = \left( {42} \right) + 16 \cr
& y = 58 \cr} $
This is the wrong answer, I would like to understand where I went wrong so I dont make the same mistake again, your help is greatly appreciated, thanks!
|
The equation wil be $\log_2 y= \log_2 x + \log_2 2^4$.
so $y$ will then be equal to $16x$.
So $4x$ will be $\dfrac y4$.
Substituting,so $3y=\dfrac y4+6$
$\dfrac {11y}4= 6$
So $y=\dfrac {24}{11}$. And $x=\dfrac {3}{22}$.
|
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|
Is my proof correct? $2^n=x^2+23$ has an infinite number of (integer) solutions. This is how I tried to prove it. Is it correct? Thanks!!
$2^n = x^2+23$
$x^2$ must be odd, therefore $x^2 = 4k+1$, where $k \in \mathbb{N}$.
$2^n=4k+24$
$k=2(2^{n-3}-3)$
Since $x^2=4k+1$,$ \ \ \ \ $ $k_1 = \frac{x_1^2-1}{4}$
and $k_2=\frac{(x_1+2)^2-1}{4}=\frac{x_1^2+4x_1+3}{4}$
If we substitute $x_1^2=4k_1 + 1$, we end up with:
$k_2=k_1 + \sqrt{4k_1+1} + 1$
Therefore, finding solutions of $k=2(2^{n-3} - 3)$ is comparable to
finding the solutions of $k+\sqrt{4k+1}+1=2(2^{n-3}-3)$
Let $p=2(2^{n-3}-3)$
Therefore, $(\sqrt{4k+1})^2=(p-k-1)^2$, so
$(p-k)^2 = 2(p+k)$
Since $p$ has an infinite number of solutions $(p-k)^2=2(p+k)$ also has an infinite number of solutions, which implies the the original does also.
|
The substitution $x^2=4k^2+1$ was wrong because $x^2=4k+1$, but nevertheless the next line is correct as if I've substituted the correct thing. The actual problem of the proof comes from when I said "$p$ has infinitely many solutions". I was thinking that $p$ is just an integer, and clearly $2(2^{n-3}-3)$ outputs a positive integer when $n$ is at least $5$. That's why I said that $p$ has an infinite number of solutions. However, I did not realize that since $p=k+\sqrt{4k+a}+1$, I'm already assuming that $2(2^{n-3}-3)=k+\sqrt{4k+a}+1$ has an infinite number of solutions, which would be implied if I was actually correct.Therefore this is a circular argument. But I do have one question about a certain step. Was it correct when I said that finding solutions of $k=2(2^{n-3} - 3)$ is comparable to finding the solutions of $k+\sqrt{4k+1}+1=2(2^{n-3}-3)$?
|
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|
The limit of $\lim\limits_{x \to \infty}\sqrt{x^2+3x-4}-x$ I tried all I know and I always get to $\infty$, Wolfram Alpha says $\frac{3}{2}$. How should I simplify it?
$$\lim\limits_{x \to \infty}\sqrt{(x^2+3x+4)}-x$$
I tried multiplying by its conjugate, taking the squared root out of the limit, dividing everything by $\sqrt{x^2}$, etc.
Obs.: Without using l'Hôpital's.
|
Note that
\begin{align}
\sqrt{x^2+3x-4} - x & = \left(\sqrt{x^2+3x-4} - x \right) \times \dfrac{\sqrt{x^2+3x-4} + x}{\sqrt{x^2+3x-4} + x}\\
& = \dfrac{(\sqrt{x^2+3x-4} - x)(\sqrt{x^2+3x-4} + x)}{\sqrt{x^2+3x-4} + x}\\
& = \dfrac{x^2+3x-4-x^2}{\sqrt{x^2+3x-4} + x} = \dfrac{3x-4}{\sqrt{x^2+3x-4} + x}\\
& = \dfrac{3-4/x}{\sqrt{1+3/x-4/x^2} + 1}
\end{align}
Now we get
\begin{align}
\lim_{x \to \infty}\sqrt{x^2+3x-4} - x & = \lim_{x \to \infty} \dfrac{3-4/x}{\sqrt{1+3/x-4/x^2} + 1}\\
& = \dfrac{3-\lim_{x \to \infty} 4/x}{1 + \lim_{x \to \infty} \sqrt{1+3/x-4/x^2} } = \dfrac{3}{1+1}\\
& = \dfrac32
\end{align}
|
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|
How to prove that $\frac{1}{n+1} = {\sqrt{n\over n+1}}\implies n = \Phi$? Consider:
$$\frac{1}{n+1} = {\sqrt{n\over n+1}}$$
How could one prove that $n$ is of such form that:
$$\frac{1}{n+1} = {\sqrt{n\over n+1}}\implies n = {\sqrt{5\,}-1 \over 2} \implies n = \Phi$$
where $\Phi$ denotes the golden ratio?
Edit:
I noticed that:
$$\frac{1}{n+1} = {\sqrt{n\over n+1}} \implies \left ( \frac{1}{n+1}\right ) ^2 = \left ({\sqrt{n\over n+1}}\right)^2\implies$$
$${1\over(n+1)^2} = {n\over(n+1)}\implies n = {1\over n+1} \implies$$
$${1\over{1+{1\over1+n}}} \approx \Phi$$
as $$\Phi = {1\over{1+{1\over{1+{1\over\dots}}}}}$$
which should be proof enough.
|
HINT:
Just square both sides to get $$\frac1{(n+1)^2}=\frac n{n+1}\implies n^2+n-1=0$$ assuming $n+1\ne0$
Alternatively, $\sqrt{n+1}=(n+1)\sqrt n\implies 1=\sqrt{n(n+1)}$ assuming $n+1\ne0$
Now, square both sides to get $n^2+n=1$
So, $n=\frac{-1\pm\sqrt5}2$
As $\sqrt{\frac n{n+1}}$ is considered to be $\ge 0,\frac1{n+1}\ge 0\implies n\ge -1$
Now, as $-\frac{1+\sqrt5}2<-1,$ it is an extraneous root
|
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|
mathematical induction (sum of squares of first $2n$ numbers) I need to prove with mathematical induction the formula for the sum of the squares of the first $2n$ numbers.
The equation:
$$1^2 + 2^2 + 3^2 + \cdots + (2n)^2 = \dfrac{n(2n+1)(4n+1)}{3}.$$
The equation stands for $n \geqslant 1$.
Here is my solution:
$n=1$:
$1^2 + 2^2 = \dfrac{1 \cdot 3 \cdot 5}{3}$
$5 = 5$
$n = k$:
$1^2+2^2+\cdots+(2k)^2 = \dfrac{k(2k+1)(4k+1)}{3}$
$n = k + 1$:
$1^2+2^2+\cdots+(2k)^2 +(2к + 1)^2 + (2к + 2)^2 = \dfrac{(k+1)(2(k+1)+1)(4(k+1)+1)}{3}$
$\dfrac{k(2k+1)(4k+1)}{3} + \dfrac{3(2k + 1)^2}{3} + \dfrac{3(2k + 2)^2}{3} = \dfrac{(k+1)(2(k+1)+1)(4(k+1)+1)}{3}$
$\dfrac{8k^3 + 2k^2 + 4k^2 + k}{3} + \dfrac{3(4k^2 + 4k + 1)}{3} + \dfrac{3(4k^2 + 8k + 4)}{3} = \dfrac{(k+1)(2k+3)(4k+5)}{3}$
$\dfrac{8k^3 + 6k^2 + k + 12k^2 + 12k + 3 + 12k^2 + 24k + 12}{3} = \dfrac{8k^3 + 10k^2 + 12k^2 + 15k + 8k^2 + 10k + 12k + 15}{3}$
$\dfrac{8k^3 + 30k^2 + 37k + 15}{3} = \dfrac{8k^3 + 30k^2 + 37k + 15}{3}$
In the end, the terms are not equal, and they should be. What am I doing wrong?
|
You are forgetting to add the term $(2k+1)^{2}$ in the LHS,i.e. for $n=k+1$ the LHS is $$1^{2}+2^{2}+\cdots +(2k)^{2}+(2k+1)^{2}+(2k+2)^{2}$$
|
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|
Does $1 + \frac{1}{x} + \frac{1}{x^2}$ have a global minimum, where $x \in \mathbb{R}$? Does the function
$$f(x) = 1 + \frac{1}{x} + \frac{1}{x^2},$$
where $x \in {\mathbb{R} \setminus \{0\}}$, have a global minimum?
I tried asking WolframAlpha, but it appears to give an inconsistent result.
|
$g(x)=1+\frac{1}{x}+\frac{1}{x^2},x\ne 0$
$g'(x)=-\frac{1}{x^2}-2\frac{1}{x^3}=0\Rightarrow \frac{-2}{x^3}=\frac{1}{x^2}\Rightarrow x=-2$
$g''(x)=\frac{2}{x^3}+\frac{6}{x^4}|_{x=-2}> 0$
$\Rightarrow g(-2)$ is the minimum.
Edit:This shows that $g$ has a local minimum at $x=-2$ but this is also the global minimum because
$g(x)-g(-2)=1+\frac{1}{x}+\frac{1}{x^2}-1-(\frac{-1}{2})-\frac{1}{2^2}=\frac{x+2}{2x}+\frac{(2-x)(2+x)}{4x^2}=(x+2)\frac{2x+2-x}{4x^2}=\frac{(x+2)^2}{4x^2}\ge0$
so $g(x)\ge g(-2)$ so $g(-2)$ is a global minimum.
|
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|
Solve $ax - a^2 = bx - b^2$ for $x$ Method 1
Solve for x
$$ax - a^2 = bx - b^2$$
Collect all terms with x on one side of the equation
$$ax - bx = a^2 -b^2$$
Factor both sides of the equation
$$(a -b)x = (a+b)(a - b)$$
Divide both sides of the equation by the coefficient of $x$ (which is $a-b$)
$$x = a + b$$ (where $a \neq b$ since this would mean dividing by $0$)
Method 2
Solve for $x$
$$ax - a^2= bx - b^2$$
Bring all the terms to one side of the equation
$$ax - a^2 -bx + b^2 = 0$$
Rearrange
$$ax - bx -(a^2-b^2)=0$$
Factor
$$(a - b)x - (a + b)(a - b) = 0$$
$$(a - b)( x - (a + b)) = 0$$
which is a true statement if $$a-b=0$$ $$a = b$$ or $$x-(a+b)=0$$ $$x = a + b$$
My question is I don't understand how this second method is consistent with the first in terms of the restriction on $a$ and $b$.
|
In the first method you come to the equation
$$(a-b)x=(a+b)(a-b)$$
In the second method, you come to the equation
$$(a-b)(x-(a+b))=0$$
which is just the first equation rearranged and factored. If we divide the first equation by $(a-b)$ without factoring, we still collect the solution $a=b$, because notice that it does solve the first equation. However, we continue on assuming $a\ne b$ to discover the other solution.
The method is the same with the second equation. We first collect the solution $a=b$, and then we assume $a\ne b$ to see if there are other solutions. By the zero product property, it follows that we must have $x=a+b$.
|
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|
Let $p_n$ be the sequence defined by $p_n=\sum_{k=1}^n\frac{1}{k}$. Show that $p_n$ diverges even though $\lim_{n\to\infty}(p_n-p_{n-1})=0$ I have tried this as :
$$p_n=\sum_{k=1}^n\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n-1}+\frac{1}{n}$$
$$p_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n-1}$$
$$(p_n-p_{n-1})=\frac{1}{n}$$
$$\lim_{n\to\infty}(p_n-p_{n-1})=\lim_{n\to\infty}\frac{1}{n}=0$$
But dont know how to show $p_n$ diverges?
|
$$\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$
$$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$$
Generalising, $$\frac{1}{2^n-2^{n-1}+1}+\dots +\frac{1}{2^n}>\frac{1}{2^n}+\dots +\frac{1}{2^n}\text{ ($2^{n-1}$ terms)}$$
Hence, we can add an infinite number of $\frac{1}{2}$. The sum thus diverges.
|
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|
calculate riemann sum of sin to proof limit proposition $$\lim_{n \to \infty}\frac1n\sum_{k=1}^n\sin(\frac{k\pi}{n})$$
I'm having trouble expressing $\sin(x)$ differently here in order to calculate the riemann sum.
I want to show that this converges to $\frac{2}{\pi}$ so it equals to $\int_0^1 \sin(x\pi)$.
Is there any easy way to express $\sin(x)$ different here?
Added:
$$\frac{1}{2i}(\frac{\cos(\frac{(n+1)\pi}{n})+i\sin(\frac{(n+1)\pi}{n})-\cos(\frac{\pi}{n})+i\sin(\frac{\pi}{n})}{\cos(\frac{\pi}{n})+i\sin(\frac{\pi}{n})-1}-\\\frac{\cos(-\frac{(n+1)\pi}{n})+i\sin(-\frac{(n+1)\pi}{n})-\cos(-\frac{\pi}{n})+i\sin(-\frac{\pi}{n})}{\cos(-\frac{\pi}{n})+i\sin(-\frac{\pi}{n})-1})$$
|
Use that $$\sum_{k=1}^n \sin kx=\frac{\sin\dfrac{kx}2\sin (k+1)\dfrac x2}{\sin\dfrac{x}2}$$
ADD One can deduce the above in several ways. The first is to note it is the imaginary part of $$\sum_{k=0}^n e^{ikx}=\frac{e^{(n+1)ix}-1}{e^{ix}-1}$$
Another choice is to use $$\cos b-\cos a=2\sin\frac{b-a}2\sin\frac{b+a}2$$
Now let $$b=\left(n+\frac 1 2\right)x\\a=\left(n-\frac 1 2\right)x$$
Then you get
$$\cos \left(n+\frac 1 2\right)x-\cos \left(n-\frac 1 2\right)x=2\sin\frac{(n+1)x}2\sin\frac{x}2$$
Then sum and telescope.
|
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|
$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
|
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
$$\implies 0^2=1+2 (ab+bc+ca)\implies ab+bc+ca=-\frac12$$
Again, $ab+bc+ca=a(b+c)+bc=a(-a)+bc\implies bc-a^2$
$$\implies bc-a^2=-\frac12\implies bc=\frac{2a^2-1}2$$
$$\implies abc=a\cdot \frac{2a^2-1}2=\frac{2a^3-a}2$$
Now use Second derivative test for the extreme values of $f(a)=\frac{2a^3-a}2$
As $a^2=1-(b^2+c^2)\le1, -1\le a\le 1$
Alternatively,
we can set $a=\cos A, b=\sin A\cos B,c=\sin A\sin B$
As $a+b+c=0,\cos A+\sin A\cos B+\sin A\sin B=0\implies \cos A=-\sin A(\cos B+\sin B) $
Squaring we get, $\cot^2A=1+2\sin B\cos B$
$$abc=\cos A\cdot\sin A\cos B\cdot\sin A\sin B=\cos A\sin^2A(\sin B\cos B)$$
$$=\cos A\sin^2A\cdot \frac{\cot^2A-1}2=\cos A\cdot\frac{\cos^2A-\sin^2A}2=\frac{2\cos^3A-\cos A}2 $$ which can be handled like the previous method
|
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|
How to use double angle identities to find $\sin x$ and $\cos x$ from $\sin 2x $? If $\sin 2x =\frac{5}{13}$ and $0^\circ < x < 45^\circ$, find $\sin x$ and $\cos x$.
The answers should be $\frac{\sqrt{26}}{26}$ and $\frac{5\sqrt{26}}{26}$
Ideas
The idea is to use double angle identities. One such identity is $\sin 2x=2\sin x\cos x$.
It's easy to use it to find $\sin 2x$ from known $\sin x$ and $\cos x$. But here it's the other way around.
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Suppose we had a right triangle with an angle $2x$, and $\sin2x=\frac{5}{13}$. Further suppose that the hypotenuse of the triangle was 13. We can deduce that the side oppoites $2x$ must be 5. Applying the Pythagorean theorem to find the other side we have
$$13^2=5^2+(\text{adjacent side})^2$$
$$169-25=(\text{adjacent side})^2$$
$$144=(\text{adjacent side})^2$$
implying that the side opposite angle $2x$ is $12$. This allows us to state that
$$\cos 2x=\frac{12}{13}$$
Which is easier to work with because
$$\cos2x=\cos^2x-\sin^2x=2\cos^2x-1$$
Substituting we have
$$\frac{12}{13}=2\cos^2x-1$$
$$\cos^2x=\frac{25}{26}$$
$$\cos x=\frac{5}{\sqrt{26}}=\frac{5\sqrt{26}}{26}$$
Applying the Pythagorean identity we have
$$\cos^2x+\sin^2x=1$$
$$\frac{25}{26}+\sin^2x=1$$
$$\sin^2x=\frac{1}{26}$$
$$\sin x=\frac{1}{\sqrt{26}}=\frac{\sqrt{26}}{26}$$
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Prove $\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A$ by the Pythagorean theorem. How do I use the Pythagorean Theorem to prove that $$\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A?$$
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suppose there is a right angle tri. having b is hypo.,c is base ,a is perpendicular
than $\cos A=\frac cb$ and $\sin A=\frac ab$
using pythgo. theo. $\;b^2=a^2+c^2$
$$\dfrac {\cos^2 A}{1-\sin A}$$
$$\dfrac {\frac {c^2}{b^2}}{1-\frac {a}{b}}$$
$$\dfrac {\frac {c^2}{b^2}}{\frac {b-a}{b}}$$
$$\dfrac {c^2b}{b^2(b-a)}$$
$$\dfrac {c^2}{b(b-a)}$$
$$\dfrac {b^2-a^2}{b(b-a)}$$
$$\dfrac {b+a}{b}$$
$$1+\dfrac {a}{b}$$
$$1+\sin A$$
alternatively
$$\dfrac {\cos^2 A}{1-\sin A}$$
$$\dfrac {1-\sin^2 A}{1-\sin A}$$
$$\dfrac {(1-\sin A)(1+\sin A)}{1-\sin A}$$
$${1+\sin A}$$
|
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|
Question about modular arithmetic and divisibility If $$a^3+b^3+c^3=0\pmod 7$$
Calculate the residue after dividing $abc$ with $7$
My first ideas here was trying the 7 possible remainders and then elevate to the third power
$$a+b+c=x \pmod 7$$
$$a^3+b^3+c^3+3(a+b+c)(ab+bc+ac)-3abc=x^3\pmod 7$$
$$3(a+b+c)(ab+bc+ac)-3abc=x^3 \pmod 7$$
If I replace $x=0$ the result is immediate, $abc=0 \pmod7$. But with $x=1$
$$3(7n+1)(ab+bc+ac)-3abc=x^3 \pmod 7$$
$$3(ab+bc+ac)-3abc=x^3 \pmod 7$$
And there is nothing more to simplify. I know the LHS is a multiple of $3$, but what can i do with that? Is it necessary that $x^3$ or $7-x^3$ is a multiple of $3$? Any help is greatly appreciated
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HINT: What are the possible cubes in modulo 7? Hence what combinations of these cubes allow for $a^3+b^3+c^3=0$ (mod 7). From this $abc$ (mod 7) should be clear.
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|
Given a matrix $A$ and what it maps two vectors to, is $0$ an eigenvalue of it? Studying for my Algebra exam, and this question popped out with no solution in a previous exam:
Given a matrix $A$ such that $A \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix},\ A \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -6 \end{pmatrix}$.
(I) Is $0$ an eigenvalue of the matrix?
(II) Find a matrix like that, where the sum of its' eigenvalues is $0$.
So I (think) solved (I) but have no clue for (II).
Here's my solution for (I):
$A \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} + A \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} = A \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix} + \begin{pmatrix} 2 \\ 4 \\ -6 \end{pmatrix} = 0$, and then, the vector $v = \begin{pmatrix} 2 \\ -1 \\0 \end{pmatrix}$ supplies that $Av = 0v = 0$ meaning that $0$ is an eigenvalue of $A$ with an eigenvector $v$.
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Hint: If a matrix $A$ has column vectors $a_1,..,a_n$, then
$$A\cdot e_i=a_i$$
where $e_1=\pmatrix{1\\0\\0\\ \vdots}$, $e_2=\pmatrix{0\\1\\0\\ \vdots}$, $e_3=\pmatrix{0\\0\\1\\ \vdots}$...
|
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If $a,b$ are roots for $x^2+3x+1=0$.Calculating $(\frac{a}{b+1})^2 +(\frac{b}{a+1})^2$ If $a,b$ are roots for the equation $x^2+3x+1=0$.How to calculate $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$
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Hint: $(a+1)^2 + a = 0$ and so $\frac{1}{(a+1)^2} = -\frac{1}{a}$
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|
Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number? $\large S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$
Multiplying by conjugate:
$\large S=\dfrac {-3}{\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}}$
From the original:
$\large S-2\sqrt[3]{5-2 \sqrt {13}}
=\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}$
Substituting:
$\large S=\dfrac{-3}{S-2\sqrt[3]{5-2 \sqrt {13}}}$
This leads to a quadratic equation in $\large S$ which I checked in wolframalpha and I got imaginary solutions. Why does this happen? I am not looking for an answer telling me how to solve this problem, I just want to know why this is wrong. Thanks.
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Note the identity $$x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)+3xyz$$
So that if $x+y+z=0$ we have $x^3+y^3+z^3=3xyz$
This is often useful with cubic expressions like this. Take $x=-S;\text{ } y=\sqrt[3] {5+2 \sqrt {13}};\text{ }z=\sqrt[3]{5-2 \sqrt {13}}$
Then we have$$-S^3+5+2\sqrt{13}+5-2\sqrt{13}=-3S\sqrt[3]{25-52}$$ whence $$S^3+9S-10=0$$
and you recover the cubic of which $S$ is a root. As it happens this is easy to factor, and you need to identify the root which corresponds to the expression you were given. I shall not repeat what Zarrax has put from this point on. But I think the trick is worth knowing, as it hugely simplifies the arithmetic.
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How to integrate these integrals $$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \cos x}$$
$$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \sin x}$$
It seems that substitutions make things worse:
$$\int \frac {dx}{1+ \cos x} ; t = 1 + \cos x; dt = -\sin x dx ; \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (t-1)^2} $$
$$ \Rightarrow
\int \frac {-\sqrt{1 - (t-1)^2}}{t} = \int \frac {-\sqrt{t^2 + 2t}}{t} = \int \frac {-\sqrt t \cdot \sqrt t \cdot \sqrt{t +
2}}{\sqrt t \cdot t} $$
$$= \int \frac {- \sqrt{t + 2}}{\sqrt t } = \int - \sqrt{1 + \frac 2t} = ? $$
What next? I don’t know. Also, I’ve tried another “substitution”, namely $1 + \cos x = 2 \cos^2 \frac x2) $
$$ \int \frac {dx}{1+ \cos x} = \int \frac {dx}{2 \cos^2 \frac x2} = \int \frac 12 \cdot \sec^2 \frac x2 = ? $$
And failed again. Help me, please.
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HINT:
$$\text{As }\frac{d(\tan mx)}{dx}=m\sec^2mx,$$
$$\int\sec^2mx= \frac{\tan mx}m+C$$
here $m=\frac12$
or using Weierstrass substitution $\tan \frac x2=t,$
$\frac x2=\arctan t\implies dx=2\frac{dt}{1+t^2}$ and $\cos x=\frac{1-t^2}{1+t^2}$
$$\int \frac{dx}{1+\cos x}=\int dt=t+K=\tan\frac x2+K$$
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How find the value $\sum\limits_{k=1}^{\infty}(C_{3k}^k)^{-1}$ find the value
$$\sum_{k=1}^{\infty}\dfrac{1}{C_{3k}^{k}}$$
and long ago,I have see this and is easy
$$\sum_{k=1}^{\infty}\dfrac{1}{C_{2k}^{k}}$$
where
$$C_{n}^{k}=\dfrac{n!}{(n-k)!k!}$$
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For #2:
$$\sum_{k=1}^{\infty}\dfrac{1}{C_{2k}^{k}}=\frac{1}{3}+\frac{2\pi}{9\sqrt{3}}$$
Because,
$$\sum_{n=0}^\infty\frac{x^n}{C^n_{2n}}=\frac{4(\sqrt{4-x}+\sqrt{x}\arcsin(\frac{\sqrt{x}}{2}))}{\sqrt{(4-x)^3}}$$
Which is because
$$\text{If we set, } A_n=\frac{1}{C^{n}_{2n}}$$
Then
$$(4n+2)A_{n+1}=(n+1)A_n$$
And so,
$$\sum_{n=0}^\infty(4n+2)A_{n+1}x^n=\sum_{n=0}^\infty(n+1)A_nx^n$$
So that if $$A(x)=\sum_{n=0}^\infty A_nx^n$$
$$\frac{d}{dx}A(x)-\frac{x+2}{x(4-x)}A(x)=\frac{-2}{4x-x^2}$$
Now with some work you can verify $A(x)$ satisfies the above differential equation.
For #1:
Likewise
If you let $K_n=\frac{1}{C^n_{3n}}$
And,
$$F(x)=\sum_{n=1}^\infty\frac{x^n}{C^n_{3n}}=\sum_{n=1}^\infty K_nx^n$$
We have that, $3(3n+1)(3n+2)K_{n+1}=2(n+1)(2n+1)K_{n}$
So,
$$\sum_{n=1}^\infty 3(3n+1)(3n+2)K_{n+1}x^n =\sum_{n=1}^\infty 2(n+1)(2n+1)K_{n}x^n$$
And so that we get,
$$4x^2\frac{d^2}{dx^2}F(x)-17x\frac{d}{dx}F(x)-\frac{4}{1-x}=0$$
Which is a second order linear ordinary differential equation, and can be solved in terms of elementary functions, namely in terms of logarithms and inverse tangent functions.
Once you have $F$, then just compute $$F(1)=\sum_{n=1}^\infty\frac{1}{C^n_{3n}}$$
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How to solve the given initial-value problem? Solve the given problem which the input function $g(x)$ is discontinuous?
$y''+4y = g(x)$, $y(0) = 1$, $y'(0) = 2$, where
$$g(x) = \begin{cases} \sin x, & 0\leq x\leq\frac{\pi}{2}\\
0,& x>\frac{\pi}{2} \end{cases}$$
And the given answer is,
$$y = \begin{cases} \cos 2x+\frac56\sin2x+\frac13\sin x, & 0\leq x\leq\frac{\pi}{2}\\
\frac23\cos 2x+\frac56\sin2x,& x>\frac{\pi}{2} \end{cases}$$
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Note: It is best to use Laplace Transforms for these sorts of problems and this is where the LT shows it's usefulness, but that is not allowed, so we will use undetermined coefficients instead.
We are given:
$\tag 1 y''+4y = g(x), y(0) = 1, y'(0) = 2$
where
$$g(x) = \begin{cases} \sin x, & 0\leq x\leq\frac{\pi}{2}\\
0,& x>\frac{\pi}{2} \end{cases}$$
The approach will be as mentioned by D.F. in comments, we are going to solve for the lower interval and use that result to solve the upper interval (in essence, we are solving two ODEs).
We start out by finding the homogeneous solution to $(1)$, so we have:
$$m^2 + 4 = 0 \rightarrow m_{1,2} = \pm 2i$$
This gives us:
$$y_h = c_1 \cos 2x + c_2 \sin 2x$$
Next, we will solve for the particular solution, so we guess at a solution of the form:
$$y_p = a \cos x + b \sin x$$
To solve for the constants, $a$ and $b$, we substitute back into $(1)$, yielding:
$-a \cos x - b \sin x + 4a \cos x + 4b \sin x = \sin x$
This yields $a = 0$ and $b = \dfrac{1}{3}$.
So, we can write our solution as: $y = y_h + y_p = c_1 \cos 2x + c_2 \sin 2x + \dfrac{1}{3} \sin x$
Using the IC's, we solve for $c_1$ and $c_2$, yielding $c_1 = 1, c_2 = \dfrac{5}{6}$, so we have:
$$\tag 2 y(x) = \cos 2x + \dfrac{5}{6} \sin 2x + \dfrac{1}{3} \sin x, 0 \le x \le \dfrac{\pi}{2}$$
Next, we want to solve for the upper upper half of the range, so we have, $y'' + 4y = 0$, but we need ICs. We will use $(2)$ to generate an IC at $\dfrac{\pi}{2}$ (do you know what allows us to do this and why it is useful to do so?).
So, using $(2)$, we get $y\left(\dfrac{\pi}{2}\right) = -\dfrac{2}{3}$ and $y'\left(\dfrac{\pi}{2}\right) = -\dfrac{5}{3}$.
We have our solution for $(2)$ from earlier as: $y = c_1 \cos 2x + c_2 \sin 2x$
Using these new ICs, we arrive at, $c_1 = \dfrac{2}{3}, c_2 = \dfrac{5}{6}$, so:
$$\tag 3 y(x) = \dfrac{2}{3} \cos 2x + \dfrac{5}{6} \sin 2x, x \gt \dfrac{\pi}{2}$$
Using $(2)$ and $(3)$, the solution can be written as:
$$y(x) = \begin{cases} \cos 2x + \dfrac{5}{6} \sin 2x + \dfrac{1}{3} \sin x, & 0\leq x\leq\frac{\pi}{2}\\
\dfrac{2}{3} \cos 2x + \dfrac{5}{6} \sin 2x,& x>\frac{\pi}{2} \end{cases}$$
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Finding the value of $\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$ Is it possible to find the value of
$$\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
Does it help if I set it equal to $x$? Or I mean what can I possibly do?
$$x=\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
$$x^2=1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$
$$x^2-1=2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$
$$\frac{x^2-1}{2}=\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$
$$\left(\frac{x^2-1}{2}\right)^2=2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}$$
$$\left(\frac{x^2-1}{2}\right)^2-2=3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}$$
$$\vdots$$
I don't see it's going anywhere. Help appreciated!
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This is meant to follow up on Ethan's comment about using Herschfeld's theorem to prove that the expression converges.
Theorem (Herschfeld, 1935). The sequence
$$
u_n = \sqrt{a_1 + \sqrt{a_2 + \cdots + \sqrt{a_n}}}
$$
converges if and only if
$$
\limsup_{n\to\infty} a_n^{2^{-n}} < \infty.
$$
The American Mathematical Monthly, Vol. 42, No. 7 (Aug-Sep 1935), 419-429.
In our case we have
$$
\begin{align}
u_1 &= \sqrt{1}, \\
u_2 &= \sqrt{1 + 2\sqrt{2}} = \sqrt{1 + \sqrt{2^3}}, \\
u_3 &= \sqrt{1 + 2\sqrt{2 + 3\sqrt{3}}} = \sqrt{1 + \sqrt{2^3 + \sqrt{2^4 3^3}}}, \\
u_4 &= \sqrt{1 + \sqrt{2^3 + \sqrt{2^4 3^3 + \sqrt{2^8 3^4 4^3}}}},
\end{align}
$$
and so on, so that
$$
a_n = n^3 \prod_{k=2}^{n-1} k^{2^{n-k+1}}.
$$
We then have
$$
a_n^{2^{-n}} = n^{3\cdot 2^{-n}} \prod_{k=2}^{n-1} k^{1/2^{k-1}} \sim \prod_{k=2}^{\infty} k^{1/2^{k-1}}
$$
as $n \to \infty$, where the infinite product converges because $k^{1/2^{k-1}} = 1 + O(\log k/2^k)$ as $k \to \infty$. Therefore $u_n$ converges by Herschfeld's theorem.
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|
Find coefficient of $x^{100}$ in the power series expansion of $\frac{1}{(1-x)(1-2x)(1-3x)}$ I'm trying to find to coefficient of $x^{100}$ of
$$\sum_{n=0}^{∞}a_n x^{n}\ =\frac{1}{(1-x)(1-2x)(1-3x)}.$$
I used the sum:
$$\frac{1}{1-x}\ = 1+x+x^2+\ldots.$$
So : $$\frac{1}{(1-x)(1-2x)(1-3x)}= \left(1+x+x^2+\ldots\right)\left(1+2x+(2x)^2+\ldots\right)\left(1+3x+(3x)^2+\ldots\right).
$$
But multiplying out the right hand side to extract the coefficient $x^{100}$ is tedious. Any idea how I can obtain the coefficient in a simpler way ?
Regards
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Use partial fractions: $$\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{1}{2} \cdot \frac{1}{1-x} - 4 \cdot \frac{1}{1-2x} + \frac{9}{2} \cdot \frac{1}{1-3x}.$$
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How to prove that $\left(\sum\limits_{k=1}^{n}a_{k}\right)^2\ge\sum\limits_{k=1}^{n}a_k^3$? Let $$a_{n}\ge a_{n-1}\ge\cdots\ge a_{0}= 0,$$ and for any $i,j\in\{0,1,2\dots,n\},j>i$, there is
$$a_{j}-a_{i}\le j-i.$$
Prove that
$$\left(\sum_{k=1}^n a_k \right)^2\ge\sum_{k=1}^n a_k^3.$$
My idea is by mathematical induction:
Assume that $n$ is true, meaning
$$\left(\sum_{k=1}^n a_k\right)^2\ge\sum_{k=1}^n a_k^3,$$
then for $n+1$,
\begin{align*}
\left(\sum_{k=1}^{n+1} a_k\right)^2&=\left(\sum_{k=1}^n a_k\right)^2+2a_{n+1}\sum_{k=1}^n a_k+a_{k + 1}^2\\
&\ge\sum_{k=1}^n a_k^3 + 2a_{n + 1} \sum_{k=1}^n a_k + a_{n + 1}^2.
\end{align*}
Now it only needs $$2\sum_{k=1}^n a_k+a_{n+1}\ge a_{n + 1}^2.$$
Since for any $i,j\in\{0,1,2\cdots,n\},j>i$, then there is
$$a_j-a_i\le j-i,$$
we have
$$a_1\ge a_{n+1}-n,\ a_2\ge a_{n+1}-(n-1),\ \cdots$$
so
$$2\sum_{k=1}^n a_k+a_{n+1}\ge (2n+1)a_{n+1}-(n+1)n$$
$$\Longleftrightarrow (a_{n+1}-n)(a_{n+1}-n-1)\le 0.$$
From here I can't work out. Thank you everyone.
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We can solve this using induction, like you suggested. The case $n=1$ is quite easy: $(a_1)^2\geq(a_1)^3$ since $0\leq a_1\leq 1$.
Now assume that for some $n > 0$, the property is true for $n-1$. Let $S_n=\sum_{k=0}^n a_k=\sum_{k=1}^n a_k$.
\begin{align*}
(S_n)^2-\sum_{k=1}^n (a_k)^3 &= (a_n)^2 + 2a_nS_{n-1} + (S_{n-1})^2 - \sum_{k=1}^n (a_k)^3 \\
&\geq (a_n)^2 + 2a_nS_{n-1}-(a_n)^3 \\
&\geq a_n \left( -(a_n)^2 + a_n + 2S_{n-1}\right) .
\end{align*}
Since $a_n\geq 0$, we only need to prove the positivity of
$$
-(a_n)^2 + a_n + 2S_{n-1} =: T_n.
$$
Let $x=a_n-a_{n-1} \in [0,1]$.
\begin{align*}
T_n &= -(a_{n-1}+x)^2+a_{n-1}+x+2a_{n-1}+2S_{n-2}\\
&= -(a_{n-1})^2+a_{n-1}+2S_{n-2}-2xa_{n-1}+2a_{n-1}-x^2+x\\
&= T_{n-1} + x(1-x) + 2a_{n-1}(1-x)\\
&\geq T_{n-1}
\end{align*}
Since $T_1=-(a_1)^2+a_1 \geq 0$, another simple induction proves that $T_n\geq 0$, which concludes this proof.
|
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|
Factor Equations Please check my answer in factoring this equations:
Question 1. Factor $(x+1)^4+(x+3)^4-272$.
Solution: $$\begin{eqnarray}&=&(x+1)^4+(x+3)^4-272\\&=&(x+1)^4+(x+3)^4-272+16-16\\
&=&(x+1)^4+(x+3)^4-256-16\\
&=&\left[(x+1)^4-16\right]+\left[(x+3)^4-256\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\right]+\left[(x+3)^2+16\right]\left[(x+3)^2-16\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\right]+\left[(x+3)^2+16\right]\left[(x+3)-4\right]\left[(x+3)+4\right]\end{eqnarray}.$$
Question 2. Factor $x^4+(x+y)^4+y^4$
Solution: $$\begin{eqnarray}&=&(x^4+y^4)+(x+y)^4\\
&=&(x^4+y^4)+(x+y)^4+2x^2y^2-2x^2y^2\\
&=&(x^4+2x^2y^2+y^4)+(x+y)^4-2x^2y^2\\
&=&(x^2+y^2)^2+(x+y)^4-2x^2y^2
\end{eqnarray}$$
I am stuck in question number 2, I dont know what is next after that line.
|
\begin{align*}
(x+y)^4+x^4+y^4&=2(x^4+2x^3y+3x^2y^2+2xy^3+y^4)\\
&=2(x^4+2x^3y+2x^2y^2+x^2 y^2+2 x y^3+y^4)\\
&=2(x^4+2(xy+y^2)x^2+(xy+y^2)^2)\\
&=2(x^2+xy+y^2)^2
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Evaluate $\cos 18^\circ$ without using the calculator I only know $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$, $180^\circ$, $270^\circ$, and $360^\circ$ as standard angles but how can I prove that
$$\cos 18^\circ=\frac{1}{4}\sqrt{10+2\sqrt{5}}$$
|
Let $\theta = 18^\circ$.
Then $2\theta = 36^\circ$ and $3\theta = 54^\circ$.
Note that $90^\circ-3\theta = 2\theta$.
Thus, $\sin(90^\circ - 3\theta) = \sin (2\theta)$ or $\cos(3\theta) = \sin(2\theta)$
So, $4\cos^3\theta-3\cos\theta = 2\sin\theta\cos\theta$.
Since $\cos \theta$ can't be zero, we can divide by it to obtain,
$4\cos^2\theta-3 = 2\sin\theta$
Now use $\cos^2 \theta = 1 - \sin^2 \theta$ to obtain a quadratic in $\sin \theta$.
Solve it and then disregarding the negative root (since $\sin 18^\circ$ can't be negative), solve for $\cos 18^\circ$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Possible to solve this coupled system of vector equations? Let $\gamma, \omega, c$ be positive constants, let $\mathbf{Q}_{a}$ and $\mathbf{Q}_{b}$ be three-dimensional vectors, and let $\mathbf{B}(\mathbf{r})=\mathbf{B}(x,y,z)$ be a vector field. Let $\mathbf{B}_{\circ}= \mathbf{B}(\mathbf{r}_{\circ})$ for some position $\mathbf{r}_{\circ}$ and write $\mathbf{B}=\mathbf{B}_{\circ}+\delta\mathbf{B}$. Consider the following coupled system for the unknown vectors $\mathbf{a}$ and $\mathbf{b}$:
$$
-\frac{\omega^2}{\gamma}\mathbf{a}=c\mathbf{B}_{\circ}\times\mathbf{Q}_{a}+\omega\mathbf{b}\times\mathbf{B}_{\circ} \tag{1}
$$
$$
-\frac{\omega^2}{\gamma}\mathbf{b}=c\mathbf{B}_{\circ}\times\mathbf{Q}_{b}-\omega\mathbf{a}\times\mathbf{B}_{\circ} \tag{2}
$$
If it helps, $\mathbf{Q}_{a}=\int_{0}^{2\pi/\omega}\mathbf{B}\cos(\omega t)\,dt$ and $\mathbf{Q}_{b}=\int_{0}^{2\pi/\omega}\mathbf{B}\sin(\omega t)\,dt$. All the $x, y, z$ are functions of $t$ but $\mathbf{B}$ is (basically) abitrary so those integrals can't be done.
Is it possible to solve $(1)$ and $(2)$ for $\mathbf{a}$ and $\mathbf{b}$ while still keeping everything a vector? I believe that one could simply write out all the components of $(1)$ and $(2)$ to get two linear relationships between $\mathbf{a}$ and $\mathbf{b}$, which is solvable in principle, but I'd really rather not do that. I've tried just subbing one equation into the other, but you end up with the thing to solve for being buried in a double cross product, so I'm not sure.
|
to long for a comment, this is not an answer (yet)
Begin with your equations:
$$
-\frac{\omega^2}{\gamma}\mathbf{a}=c\mathbf{B}_{\circ}\times\mathbf{Q}_{a}+\omega\mathbf{b}\times\mathbf{B}_{\circ} \tag{1}
$$
$$
-\frac{\omega^2}{\gamma}\mathbf{b}=c\mathbf{B}_{\circ}\times\mathbf{Q}_{b}-\omega\mathbf{a}\times\mathbf{B}_{\circ} \tag{2}
$$
take the cross product of the first equation with $\mathbf{a}$ and the second with $\mathbf{b}$ to obtain:
$$
0=c\mathbf{a} \times \left( \mathbf{B}_{\circ}\times\mathbf{Q}_{a}\right)+\omega\mathbf{a} \times \left( \mathbf{b}\times\mathbf{B}_{\circ} \right) \tag{3}
$$
$$
0=c\mathbf{b} \times \left( \mathbf{B}_{\circ}\times\mathbf{Q}_{b}\right)-\omega\mathbf{b} \times \left(\mathbf{a}\times\mathbf{B}_{\circ}\right) \tag{4}
$$
Now, the Jacobi identity is a close as we get to associativity for the cross-product,
$$ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) +\mathbf{b} \times (\mathbf{c} \times \mathbf{a})+\mathbf{c} \times (\mathbf{a} \times \mathbf{b})=0$$
This means we can trade the $-\mathbf{b} \times (\mathbf{a} \times \mathbf{B}_{\circ})$ term for $\mathbf{a} \times (\mathbf{B}_{\circ} \times \mathbf{b})+\mathbf{B}_{\circ} \times (\mathbf{b} \times \mathbf{a})$ to obtain:
$$
0=c\mathbf{b} \times \left( \mathbf{B}_{\circ}\times\mathbf{Q}_{b}\right)+\omega \left(\mathbf{a} \times (\mathbf{B}_{\circ} \times \mathbf{b})+\mathbf{B}_{\circ} \times (\mathbf{b} \times \mathbf{a})\right) \tag{5}
$$
then I could add this to (3.) and use $\mathbf{b} \times \mathbf{B}_{\circ} = -\mathbf{B}_{\circ} \times \mathbf{b} $ to cancel a pair of terms leaving...
Well, you get the idea, maybe I'll finish this later, I just want to post this much now to give you an idea, there may be better ways, certainly you need to look up the triple-product formulas.
|
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"timestamp": "2023-03-29T00:00:00",
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|
"Direct Proof" of the Steiner-Lehmus Theorem The Steiner-Lehmus Theorem is famous for its indirect proof. I wanted to come up with a 'direct' proof for it (of course, it can't be direct because some theorems used, will, of course, be indirect). I started with $\Delta ABC$, with angle bisectors $BX$ and $CY$, and set them as equal. The first obvious step was the Stewart's Theorem:
$a(p^2 + mn) = b^2m + c^2n$
Since an angle bisector divides the third side into the same ratio as the ratio of the other two sides, I set $m = kc$, $n = kb$ and $k = \frac{a}{b + c}$. I plugged them into the equation and got (after some manipulation):
$p^2 = bc\left(1 - \left(\frac{a}{b + c}\right)^2\right)$.
Now, $CY^2 = ab\left(1 - \left(\frac{c}{a + b}\right)^2\right)$ and $BX^2 = ac\left(1 - \left(\frac{b}{a + c}\right)^2\right)$
Since $CY$ = $BX$,the above equations are equal:
$b\left(1 - \left(\frac{c}{a + b}\right)^2\right) = c\left(1 - \left(\frac{b}{a + c}\right)^2\right)$
It is sufficient to prove that $\frac{c}{a+b} = \frac{b}{a+c}$.
Or, $\frac bc = \frac{a + c}{a + b}$
Which can be easily disproved by a counterexample. Did I make a mistake, or is my approach incorrect?
EDIT: Isn't this essentially disproving the theorem?
|
Your method looks fine but I don't know what you mean by "easily disproved by a counterexample."
If $a,b,c>0$ then
$$
\frac{b}{c}=\frac{a+c}{a+b} \iff b^2+ab=c^2+ac \iff b=c
$$
so indeed it is sufficient to prove that. (There are algebraic counterexamples if you let $c=-a-b$, which seems to be why this type of proof is considered "indirect," but which is not normally allowed for the sides of a triangle.)
To be more explicit, you have the statements:
$$
\begin{array}{cl}
\mathbf A & \text{Given a triangle with sides }a,b,c\text{ and two equal angle bisectors.} \\
\mathbf B & 0<a<b+c, 0<b<a+c, 0<c<a+b \\
\mathbf C &b\left(1 - \left(\frac{c}{a + b}\right)^2\right) = c\left(1 - \left(\frac{b}{a + c}\right)^2\right) \\
\mathbf D & \frac{c}{a+b} = \frac{b}{a+c} \\
\mathbf E & b=c \\
\mathbf F & \text{The triangle is isosceles.}
\end{array}
$$
$\mathbf A \implies \mathbf B$ by the definition of "triangle," and $\mathbf E \implies \mathbf F$ by the definition of "isosceles."
The Steiner-Lehmus theorem is $\mathbf A \implies \mathbf F$.
What you have shown above is $\mathbf A \implies \mathbf C$.
The way I interpreted your comment "It is sufficient to prove that $\frac{c}{a+b} = \frac{b}{a+c}$" is that you intended to use that
$$
\mathbf B \wedge \mathbf C \wedge \left(\frac{c}{a+b} = \frac{b}{a+c} = x\right)
\\
\implies (0<x<1) \wedge \left(b(1-x^2)=c(1-x^2)\right) \\
\implies b=c \implies \mathbf F
$$
that is, it is sufficient to somehow prove $\mathbf D$, after which
your proof will be
$$
\mathbf A \implies \mathbf B \wedge \mathbf C \\
\mathbf B \wedge \mathbf C \wedge \mathbf D \implies \mathbf E \implies \mathbf F
$$
It is true that this would be sufficient, but as I showed above $\mathbf B \implies (\mathbf D \iff \mathbf E)$ already, it doesn't simplify the proof because $\mathbf D$ only holds for isosceles triangles from the start.
Your counterexample $b=3,a=4,c=5$ shows only that $\mathbf D$ does not unconditionally follow from $\mathbf B$. But this doesn't disprove the theorem, because this triangle does not have two equal angle bisectors, so $\mathbf A$ is not satisfied.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $(1-\cos x)/\sin x = \tan x/2$ Using double angle and compound angles formulae prove,
$$
\frac{1-\cos x}{\sin x} = \tan\frac{x}{2}
$$
Can someone please help me figure this question, I have no idea how to approach it?
|
$$
\cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2} = 1-2\sin^2\frac{x}{2}
$$
and
$$
\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}
$$
so
$$
\frac{1-\cos x}{\sin x} = \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \tan\frac{x}{2}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/440457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How prove that $xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \frac{4}{3}\sqrt{xyz(x+y+z)}$ let $x,y,z>0$,and such that
$x^2+y^2+z^2=1$,prove that
$$xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \dfrac{4}{3}\sqrt{xyz(x+y+z)}$$
Does this have a nice solution? Thank you everyone.
|
This answer is neither nice nor complete. At the end, I still need to plot a very complicated function in one variable to conclude the inequality is true.
In any event, here is my attempt.
Let $a,b,c$ be the three elementary symmetric polynomials associated with $x,y,z$, i.e:
$$
\begin{cases}
a &= x + y + z\\
b &= xy + yz + zx\\
c &= xyz
\end{cases} \quad\iff\quad (\lambda - x)(\lambda - y)(\lambda - z) = \lambda^3 - a\lambda^2 + b\lambda - c
$$
In terms of $a,b,c$, the inequality we want to prove can be rewritten as:
$$\begin{align}
& c + \sqrt{b^2 - 2ac} -\frac43 \sqrt{ac} \stackrel{?}{\ge} 0\tag{*1}\\
\iff & ( \sqrt{c} - \frac23\sqrt{a} )^2 + \sqrt{b^2 - 2ac} -\frac{4a}{9} \stackrel{?}{\ge} 0\tag{*2}
\end{align}$$
and the condition $x^2 + y^2 + z^2 = 1$ is equivalent to $a^2 = 1 + 2b$.
Over the domain of our problem
$$\mathscr{D} = \{ (x,y,z) : x, y, z \ge 0, x^2 + y^2 + z^2 = 1 \},$$ we have:
$$\begin{align}
& 0 \le b = xy + yz + zx \le x^2 + y^2 + z^2 = 1\\
& 1 \le a = \sqrt{1+2b} \le \sqrt{3}\\
& 0 \le c = xyz = ( (x^2y^2z^2)^{\frac13} )^{\frac32} \le ((x^2+y^2+z^2)/3)^{\frac32} = \frac{1}{\sqrt{27}}.
\end{align}$$
Since both sides of above inequalities are reachable at $(1,0,0)$ and at $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$,
the admissible ranges for $a,b,c$ over $\mathscr{D}$ are $[0,1]$, $[1,\sqrt{3}]$ and $[0, \frac{1}{\sqrt{27}}]$ respectively.
Notice over $\mathscr{D}$,
$$\sqrt{c} \le \frac{1}{\sqrt[4]{27}} < \frac23 \le \frac23 \sqrt{a}$$
In the L.H.S of $(*2)$, if we fix $b$ ( and hence $a = \sqrt{1+2b}$ ) and consider it as
a function of $c$, it is a decreasing function. This implies in order to prove the inequality $(*1 \iff *2)$, we only need to verify $(*1)$ when $c$ attains its maximum admissible value $c_{max}(b)$ subject to given $b$:
$$c_{max}(b) := \sup \{\; xyz : (x,y,z) \in \mathscr{D}, xy + yz + zx = b\;\}$$
Let us rewrite $c$ as a function of $b$ and $t = \min\{x,y,z\}$, we have:
$$\begin{align}
c = & xyz = (b - (a - t)t)t = (b - at + t^2)t \\
= & (t-\frac{a}{3})^3 - \frac{a^2-3b}{3} t + \frac{a^3}{27}\\
= & (t-\frac{a}{3})^3 - \frac{d^2}{3}(t - \frac{a}{3}) + \frac{a^3 - 3ad^2}{27}
\end{align}$$
where $d^2 = a^2 - 3b = 1-b$. Notice $\frac{\partial c}{\partial t} = 3 (t - \frac{a}{3})^2 - \frac{d^2}{3}$. One can verify $c$ achieves its maximum at $t = \frac{a - d}{3}$. As a result, we get:
$$\begin{align}
& c_{max}(b) = -\frac{d^3}{27} + \frac{d^3}{9} + \frac{a^3 - 3ad^2}{27}
= \frac{2 d^3 + a^3 - 3ad^2}{27}
= \frac{(a+2d)(a-d)^2}{27}\\
\implies & a c_{max}(b) = \frac{( (a+d)^2 - d^2 )(a - d)^2}{27}
= \frac{b^2}{3}( 1 - L(b) )
\end{align}$$
where
$$L(b) = \left( \frac{d}{a+d} \right)^2 = \frac{1-b}{(\sqrt{1+2b}+\sqrt{1-b})^2}$$
In terms of $b$ and $L(b)$, the inequality $(*1)$ at $c = c_{max}(b)$ becomes:
$$
\frac{b^2}{3\sqrt{1+2b}}(1 - L(b)) + \sqrt{\frac{b^2}{3}(1 + 2L(b))}
- \frac43 \sqrt{\frac{b^2}{3}( 1 - L(b) )} \stackrel{?}{\ge} 0\tag{*3}
$$
I don't know how to simplify this further but the plot below show that it is true for all $b \in [0,1]$.
So the inequality at hand is indeed true. In fact the inequality is strict except when
$$\begin{cases}
b = 0, &\iff (x,y,z) = (1,0,0) \text{ or } (0,1,0) \text{ or } (0,0,1)\\
b = 1, &\iff (x,y,z) = (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} )
\end{cases}$$
|
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|
Induction of inequality involving AP Prove by induction that
$$(a_{1}+a_{2}+\cdots+a_{n})\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)\geq n^{2}$$
where $n$ is a positive integer and $a_1, a_2,\dots, a_n$ are real positive numbers
Hence, show that
$$\csc^{2}\theta +\sec^{2}\theta +\cot^{2}\theta \geq 9\cos^{2}\theta$$
Please help me.
Thank you!
|
Define $x = a_1 + a_2 \cdots + a_{n-1}$ and $y = \frac{1}{a_1} + \frac{1}{a_2} \cdots + \frac{1}{a_{n-1}}$.
By the induction hypothesis, we have $xy \geq (n-1)^2$.
We need to prove $(x + a_n)(y + \frac1{a_n}) \geq n^2$
But clearly by induction hypothesis,
$(x + a_n)(y + \frac1{a_n}) = xy + ya_{n} + \frac{x}{a_n} + 1 \geq (n-1)^2 + ya_{n} + \frac{x}{a_n} + 1$
So it suffices to prove:
$ya_n + \frac{x}{a_n} \geq 2n - 2 $
Now use the definition of $x$ and $y$, and use $\displaystyle \frac{a_i}{a_n} + \frac{a_n}{a_i} \geq 2$ to obtain $ya_n + \frac{x}{a_n} \geq 2n - 2 $
|
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|
How do I show that $\int_0^\infty \frac{\sin(ax) \sin(bx)}{x^{2}} \, \mathrm dx = \pi \min(a,b)/2$ Recently I found a claim saying that
$$
\int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x= \pi \min(a,b)/2
$$
from what I can see this seems to be true. I already know that
$\int_{0}^\infty \operatorname{sinc}xy\,\mathrm{d}y = \pi/2$, and so independant of $y$. My suspicion is that this is closely related to the integral above.
Can someone give me some suggestions for evaluating the integral above? Also are there any generalizations for the integral? Eg
$$
\int_{0}^{\infty} \left( \prod_{k=1}^N \frac{\sin (a_k \cdot x)}{x} \right) \,\mathrm{d}x
$$
Where $a_k, \cdots, a_N$ are arbitrary positive constants.
It seems related to the Borwein Integral, but there are some subtle differences.
|
First of all, we use compound angle formula and IBP.
$$
\begin{aligned}
& \int_{0}^{\infty} \frac{\sin (a x) \sin (b x)}{x^{2}} d x \\
=& \frac{1}{2} \int_{0}^{\infty}[\cos (a-b) x-(\sin (a+b) x] d\left(-\frac{1}{x}\right) \quad \text{(Via compound angle formula)}\\
\stackrel{IBP}{=} &-\left[\frac{1}{2 x}(\cos (a-b) x-\cos (a+b) x)\right]_{0}^{\infty}+\frac{1}{2} \int_{0}^{\infty} \frac{-(a-b) \sin (a-b) x}{x} d x \\\quad &+\frac{1}{2} \int_{0}^{\infty} \frac{(a+b) \sin (a+b) x}{x} d x
\end{aligned}
$$
Then using the famous formula $$
\int_{0}^{\infty} \frac{\sin k x}{x} d x=(\operatorname{sgn} k) \frac{\pi}{2},
$$
we can now conclude that $$
\begin{aligned}
I &=\left\{\begin{array}{ll}
\frac{1}{2}\left[\frac{-(a-b) \pi}{2}+\frac{(a+b) \pi}{2}\right] & \text { if } a \geqslant b \\
\frac{1}{2}\left[\frac{(a-b) \pi}{2}+\frac{(a+b) \pi}{2}\right] & \text { if } b>a
\end{array}\right.\\
&=\left\{\begin{array}{ll}
\frac{b \pi}{2} & \text { if } a \geqslant b \\
\frac{a \pi}{2} & \text { if } b>a
\end{array}\right.\\
&=\frac{\min (a, b) \pi}{2}
\end{aligned}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simple limit problem: $\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$ While trying to help my sister with her homework she gave me the next limit: $$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$$
I know the conventional way of solving it would be (That's what i showed her):
$$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})=\lim_{x\to2}\left(\frac{x+2-4}{x^2-4}\right)=\lim_{x\to2}\left(\frac{x-2}{(x+2)(x-2)}\right)=\lim_{x\to2}\left(\frac{1}{x+2}\right)=\frac14$$
But she gave me the next answer:
$$\begin{align}
\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})&=\lim_{x\to2}\frac{1}{x-2}-4\lim\frac{1}{x^2-4}\\
&=\lim_{x\to2}\frac{1}{x-2}-4\lim_{x\to2}\frac{1}{x+2}\lim_{x\to2}\frac{1}{x-2}\\
&=\lim_{x\to2}\frac{1}{x-2}-4\frac14\lim_{x\to2}\frac{1}{x-2}\\
&=\lim_{x\to2}\frac{1}{x-2}-\lim_{x\to2}\frac{1}{x-2}\\
&=0
\end{align}$$
I actually couldn't explain her why is she wrong. Cause technically it looks fine. What am i missing?
|
$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})=\lim_{x\to2}\frac{1}{x-2}-4\lim\frac{1}{x^2-4}$ -- it is true only if these limits exist.
|
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|
Use a known Maclaurin series to derive a Maclaurin series for the indicated function. $$f(x)=x\cos(x)$$
I'm not quite sure how to do this. I did two others, which I presume is the right way to do it, as follows:
\begin{align}
e^x&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\\
e^{-x/4}&=1-\frac{x}{4}+\frac{x^2}{4^22!}-\frac{x^3}{4^33!}+\cdots\\
&=\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{4^nn!}
\end{align}
and
\begin{align}
\sin(x)&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\\
\sin(x^6)&=x^6-\frac{(x^6)^3}{3!}+\frac{(x^6)^5}{5!}-\frac{(x^6)^7}{7!}+\cdots\\
&=x^6-\frac{x^{18}}{3!}+\frac{x^{30}}{5!}-\frac{x^{42}}{7!}+\cdots\\
&=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{12n-6}}{(2n-1)!}
\end{align}
But where this one appears to be two separate functions multiplied together, I'm not sure how to approach it.
|
The well known Maclaurin series is $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \dots$
We can use this as follows, (look over @AWertheim's answer before viewing spoiler)
$$ f(x) = x \cos x = x \left( 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \dots \right) = x - \dfrac{x^3}{2!} + \dfrac{x^5}{4!} - \dfrac{x^7}{6!} + \dots $$
|
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|
Finding minimal polynomial of a matrix Let $A\in M\underset{nxn}{(\mathbb{R})}$ so that $\forall\lambda\in\mathbb{R}: A\neq \lambda I$. It also given that $A^4-5A^2+4I=A^3-7A-6I=0$.
Find the minimal polynomial.
I got the polynomial: $x^4-x^3-5x^2+7x+10=0$. After factorization I got:
$(x+2)(x+1)(x^2-4x+5)=0$.
I tried to find the minimal polynomial:
$(x+2)(A)= A+2I \Leftrightarrow A=-2I\rightarrow$ invalid since $A\neq \lambda I$.
$(x+1)(A)=A+I\Leftrightarrow A=-I\rightarrow$ invalid since $A\neq \lambda I$.
$(x^2-4x+5)(A)\rightarrow A^2-4A+5I=0\rightarrow$ irreducible since $F=\mathbb{R}$.
Therefore, the minimal polynomial must be: $(x+2)(x+1)(x^2-4x+5)$.
I am not sure if what I did is correct and how can I check it, can you please give me your way of solving it?
|
Recall that if a polynomial $P$ annihilates a matrix $A$ then the minimal polynomial of $A$ divides $P$ and then the spectrum of $A$ is includ in the set of roots of $P$.
We have
$x^4-5x^2+4=(x-1)(x-2)(x+2)(x+1)$ annihilates $A$ and then $\mathrm{sp}(A)\in\{1,2,-2,-1\}$
and
$x^3-7x-6=(x-3)(x+2)(x+1)$ annihilates $A$ and then $\mathrm{sp}(A)\in\{3,,-2,-1\}$
so
$$\mathrm{sp}(A)\in\{-2,-1\}$$
hence the minimal polynomial $\pi_A(x)=(x+2)(x+1)$ since otherwise $A=-2 I$ or $A=-I$.
|
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|
If $ax^2-bx+c=0$ has two distinct real roots lying in the interval $(0,1)$ $a,b,c$ belongs to natural prove that $\log_5 {abc}\geq2$ If $ax^2-bx+c=0$ has two distinct real roots lying in the interval $(0,1)$ with $a, b, c\in \mathbb N$, prove that $\log_5 {abc}\geq2$.
The equations I could form are:
1) $f(0)>0$ and $f(1)>0$
2) $\frac{b}{2a}$ lies between $0$ and $1$, because: $\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a}<\frac{b}{2a}<\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a} $.
3) $\Delta=b^2-4ac > 0$
|
We need to show that $abc\ge25$. Since both roots are real and distinct, we have that $b^2-4ac>0$ and so $b^2>4ac$. Since both roots are in (0,1), their average $\frac{b}{2a}<1$ and therefore $b<2a$. Since the larger root $\frac{b+\sqrt{b^2-4ac}}{2a}<1$, we have that $b+\sqrt{b^2-4ac}<2a$ and therefore $\sqrt{b^2-4ac}<2a-b$. Squaring both sides gives $b^2-4ac<4a^2-4ab+b^2$, so $4ab<4a^2+4ac$ and therefore $b<a+c$.
Since $2a>b$, we have that $4a^2>b^2>4ac$ and therefore $a>c$. Since $b^2\ge4ac+1$ and $a\ge c+1$, we conclude that $b^2\ge 4c(c+1)+1=(2c+1)^2$ and thus $b\ge 2c+1$. Therefore $abc\ge (c+1)(2c+1)c>25$ if $c\ge 2$.
When $c=1$, $b<a+1$ implies that $b\le a$, so $a^2\ge b^2>4a$ and therefore $a\ge 5$. Then $b^2>4a\ge20$, so $b\ge5$ and $abc=ab\ge25$.
|
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|
Why does $ \frac{2x}{2+x}$ provide a particularly tight lower bound for $\ln(1+x)$ for small positive values of $x$? EDIT:
My question was poorly worded.
I wasn't asking about showing $\ln(1+x) > \frac{2x}{2+x}$ for $x>0$.
What I wanted to know is why the lower bound provided by $ \frac{2x}{2+x}$ was so tight for small positive values of $x$. This is addressed in robjohn's answer.
|
To show that the inequality holds for $x \geq 0$ note that
$$ \log(1+x) = \int_1^{1+x} \frac{1}{t}dt. $$
Approximate the function $t \mapsto \frac{1}{t}$ by its tangent at $t = 1 + \frac{x}{2}$ to get
$$ \frac{1}{t} \geq -\frac{1}{(1+\frac{x}{2})^2}(t - 1 - \frac{x}{2}) + \frac{1}{1 + \frac{x}{2}} = -\frac{4 t}{(x + 2)^2} + \frac{4}{x+2}$$
for all $t \geq 1$. In particular
$$ \log(1+x) \geq \int_1^{1+x} \left(-\frac{4 t}{(x + 2)^2} + \frac{4}{x+2}\right) dt = \frac{2x}{x+2}.$$
This lower bound is optimal among all those that can be obtained by using such a tangent approximation. In general it depends what you mean by "optimal". For example
$$ \log(1+x) - \frac{2x}{a x + 2} = \frac{a-1}{2} x^2 + O(x^3) $$
which implies that $a \geq 1$ to get a possible lower bound. But for $a > 1$ this lower bound is worse than for $a=1$. On the other hand
$$ \log(1+x) \geq \frac{8x}{3(x+4)} $$
for all $x\geq 0$ and this is a better lower bound for $x \geq 4$.
|
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|
Prove that $2^n < \binom{2n}{n} < 2^{2n}$ Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction.
First part: $2^n < \binom{2n}{n}$. The base $n = 1$ is trivial. Assume inductively that some $k$ satisfies our statement. The inductive step can be proved as follows.
$2^k < \binom{2k}{k} \implies 2^{k+1} < 2\binom{2k}{k} = \frac{2(2k)!}{k!k!} = \frac{2(2k)!(k + 1)}{k!k!(k + 1)} = \frac{2(2k)!(k+2)}{(k+1)!k!}<\frac{(2k)!(2k+2)(2k+1)}{(k+1)!k!(k+1)} = \binom{2(k+1)}{k+1}$
Second part: $2^{2n} > \binom{2n}{n}$. Again, the base is trivial. We can assume that for some $k$ our statement is satisfied and prove that inductive step as follows:
$2^{2k} > \binom{2k}{k} \implies 2^{2k + 2} > 2^2\binom{2k}{k} = \frac{2\cdot2(2k)!}{k!k!} = \frac{2\cdot2(2k)!(k+1)(k+1)}{k!k!(k+1)(k+1)} = \frac{(2k)!(2k+2)(2k+2)}{(k+1)!(k+1)!} > \frac{(2k)!(2k+1)(2k+2)}{(k+1)!(k+1)!} = \binom{2(k+1)}{k+1}$
Is there a non-inductive derivation for the inequality?
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To see that $\binom{2n}{n} < 2^{2n}$, apply the binomial theorem $$2^{2n} = (1+1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} > \binom{2n}{n}.$$
To see that $2^n < \binom{2n}{n}$, write it as a product $$\binom{2n}{n} = \frac{2n}{n} \cdot \frac{2n-1}{n-1} \cdot ... \cdot \frac{n+1}{1},$$ where each factor is $\ge 2$, and for $n > 1$, some factors are strictly greater than $2$. The claim is not true if $n=1$.
|
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|
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$.
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
I know how to solve a problem like
"If $\cos\alpha = \frac{\sqrt{3}}{2}$ , find $\sin2\alpha$"
by using the 'double angle' formula: $\sin2\alpha = 2\sin\alpha\cos\alpha$ like this:
Start by computing $\sin\alpha$
$$\sin^2\alpha = 1 -\cos^2\alpha = 1-(\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$$
so
$$\sin\alpha = \pm\frac{1}{2}$$
then it's just a simple matter of plugging $\sin\alpha = \pm\frac{1}{2}$ and $\cos\alpha=\frac{\sqrt{3}}{2}$ into $$\sin2\alpha = 2\sin\alpha\cos\alpha$$ to get $$\sin2\alpha = \pm\frac{\sqrt{3}}{2}$$
Where I can not make progress with the question
"If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$".
Is how do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
What I have tried:
If $\sin\alpha+\cos\alpha = 0.2$
then $\sin\alpha=0.2-\cos\alpha$
and $\cos\alpha=0.2-\sin\alpha$. Should I start by by computing $\sin\alpha$ using
$\sin^2\alpha = 1 -\cos^2\alpha = 1-(0.2-\cos\alpha)^2$?
|
$$\sin\alpha + \cos\alpha = 0.2$$
$$\sin\alpha + \sqrt {1-\sin^2\alpha}= 0.2$$
$$\sqrt{1-\sin^2\alpha}= 0.2-\sin\alpha$$
$$1-\sin^2\alpha=0.04-0.4\sin\alpha+\sin^2\alpha$$
$$2\sin^2\alpha-0.4\sin\alpha-0.96=0$$
$$\sin^2\alpha-0.2\sin\alpha-0.48=0$$
$$\sin\alpha=\frac{0.2\pm1.392...}{2}$$
$$2\sin\alpha=1,592...$$
$$2\sin\alpha=-1,192...$$
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|
Simple looking log problem How would I solve this for $x$?
The original problem is
$$x+x^{\log_{2}3}=x^{\log_{2}5}$$
I have tried to reduce it down to this,
$$x^{\log_{10}3}+x^{\log_{10}2}=x^{\log_{10}5}$$
I have been stuck on this for a while now, can't seem to figure out the trick to this.
Sorry if I tagged this wrong, not sure what tags it should be.
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Oops!
My answer was wrong,
because I wrote the first term of the equation as
$1$
instead of $x$.
I will correct this now,
and hope I do not make any other errors.
The equation is
$x+x^{\log_{2}3}=x^{\log_{2}5}$.
Eliminating the obvious root of
$x=0$,
this becomes
$1+x^{\log_{2}3/2}=x^{\log_{2}5/2}$
or
$1=x^{\log_{2}5/2}-x^{\log_{2}3/2}
$
For any $v$,
$x^{\log_2 v}
= (2^{\log_2 x})^{\log_2 v}
= 2^{(\log_2 x)(\log_2 v)}
= 2^{(\log_2 v)(\log_2 x)}
= (2^{\log_2 v})^{\log_2 x}
= v^{\log_2 x}
$,
so the equation becomes
$1=(5/2)^{\log_2 x}-(3/2)^{\log_2 x}
$
Letting $y = \log_2 x$,
this is
$1 = (5/2)^y-(3/2)^y
$.
Let $f(y) = (5/2)^y-(3/2)^y-1$.
If $y < 0$, letting $z = -y$,
$f(y)=(2/5)^z-(2/3)^z-1
< 0
$
for $z > 0$
since $(2/5)^z<(2/3)^z$.
$f(0) = -1$,
$f(1) = 0$
(aha! a root at $x=2$),
and
$f(2) = (25/4)-(9/4)-1 = 3$.
$\begin{align}
f'(y)
&= (\ln (5/2))(5/2)^y-(\ln (3/2))(3/2)^y\\
&>(\ln (3/2))(5/2)^y-(\ln (3/2))(3/2)^y\\
&=(\ln (3/2))((5/2)^y-(3/2)^y)\\
&=(\ln (3/2))(3/2)^y((5/3)^y-1)\\
&> 0 \text{ for } y>0\\
\end{align}
$
so the only root is at $y=1$,
so the only roots of the original equation are
$x=0$ and $x=2$.
Here is my original answer:
First,
the base does not matter,
since only the ratio of logs is used.
Second,
the basic law of logs is that
if $u = \log_v w$,
then $w = v^u = v^{\log_v w}$.
From this
we can deduce that
$\dfrac{\log_c a}{\log_c b}
= \log_b a$.
To see this,
if $u = \log_c a$ and $v = \log_c b$,
then
$c^u = a$ and $c^v = b$
so
$c = b^{1/v}$
and $a = c^u = (b^{1/v})^u
= b^{u/v}$,
so
$u/v = \log^b a$.
Therefore
$x = 2^{\log_2 x}$,
so
$x^{\log_2 3}
=(2^{\log_2 x})^{\log_2 3}
=2^{(\log_2 x)(\log_2 3)}
=2^{(\log_2 3)(\log_2 x)}
=(2^{\log_2 3})^{\log_2 x}
=3^{\log_2 x}
$
and, similarly
$x^{\log_2 5}
=5^{\log_2 x}
$.
The equation now becomes
$1 + 3^{\log_2 x} = 5^{\log_2 x}$.
Let $y = \log_2 x$.
Then
$1 + 3^y = 5^y$.
As vadim123 stated,
this can only be solved numerically.
Once $y$ has been gotten,
$x = 2^y$.
To show that this is the only root,
let $f(y) = 1+3^y-5^y$.
$f(0) = 1$,
$f(1) = -1$,
and
$f'(y) = (\ln 3)3^y - (\ln 5)5^y
< 0$
for $y \ge 0$,
so the root between $0$ and $1$
is the only one.
|
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|
n is+ve integer, how many solutions $(x,y)$ exist for $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ with $x$, $y$ being positive integers and $(x \neq y)$ I wanted to know, how can i solve this.
For a given positive integer n, how many solutions $(x,y)$ exist for $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ with $x$ and $y$ being positive integers and $(x \neq y)$.
|
If $(x,y)$ is a solution of $\frac{1}{x}+\frac{1}{y}$ = $\frac{1}{n}$ then $x>n$ and $y>n$.
We can write the equation as
$n(x+y)=xy$ = $(x-n)(y-n) = n^2$
and for any $k$ positive $k|n^2$, therefore the solutions are given by $x-n=k$ and $y-n=\frac{n^2}{k}$ or $x=n+k$ and $y=n+\frac{n^2}{k}$.
The only case for $x=y$ is when $n=k$, therefore the number of solutions for $(x,y)$ with $(x\neq y)$ is $d(n^2)-1$ where $d(n^2)$ is the number of divisors of $n^2.$
|
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|
Differentiability of function involving absolute values I need to check the differentiability at $x=-10$ of the following function.
$f(x)=\cos|x-5|+\sin|x-3|+|x+10|^3-(|x|+4)^2$
Now,
\begin{align*}
\text{LHD} &= \lim_{x\to -10^-}\frac{f(x)-f(-10)}{x+10}\\
&= \lim_{x\to -10^-}\frac{\cos(x-5)-\sin(x-3)-(x+10)^3-(4-x)^2-(\cos15-\sin13-14^2)}{x+10}\\
&= \lim_{x\to -10^-}\frac{2\sin(\frac{x+10}{2})\sin(\frac{20-x}{2})+2\cos(\frac{x+10}{2})\sin(\frac{16-x}{2})-(x+10)^3+(10+x)(18-x)}{x+10}\\ \\
&=-\infty
\end{align*}
Similarly,
RHD = $+\infty$. So, not differentiable. But the answer says that $f(x)$ is differentiable at $-10$.
Can anyone please help?
|
Let's concentrate on the $|x+10|^3$ component since near $x=-10$ we have $$|x-5| =5-x $$$$|x-3|=3-x$$ and $$|x|+4=-x+4$$ and the components involving these quantities are readily seen to be differentiable.
Now $|x+10|$ is continuous at $x=-10$ but not differentiable there. We have
$$|x+10|^3= -(10+x)^3 \text{ for }x \le -10$$ and (noting that $x=-10$ is still a point of continuity and doesn't require a special definition) $$|x+10|^3= (10+x)^3 \text{ for }x \ge -10$$
Now if we plug a small change $h$ into the definition of the derivative for $f(x)=|x+10|^3$ we are looking at $\cfrac {f(x+h)-f(x)}h$.
We note that $f(-10)=0$, and that $f(-10+h)=|h|^3$. So $$\frac{f(-10+h)-f(-10)}h=\frac {|h|^3}h=\frac {h\cdot|h|^3}{h^2}=h\cdot|h|$$
(note $h^2=|h|^2$), and you should be able to complete the problem from there.
|
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|
show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that
$$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$
using different ways
thanks for all
|
Related technique. You can use the Laplace transform technique. Recalling the Laplace transform
$$F(s)= \int_{0}^{\infty} f(x) e^{-sx}dx. $$
Taking $ f(x) = \frac{\sin(x)^3}{x^3} $ gives
$$ F(s)= \frac{\pi \,{s}^{2}}{8}+\frac{3\,\pi}{8}- \frac{3( {s}^{2}-1) }{8}\,\arctan \left( s \right) +\frac{( {s}^{2}-9)}{8}\,\arctan \left( \frac{s}{3} \right) $$
$$+\frac{3s}{8}\, \left( -\ln \left( {s}^{2}+9 \right) +\ln \left( {s}
^{2}+1 \right) \right). $$
Taking the limit as $s\to 0$ gives the desired result $\frac{3\pi}{8}$.
Another Laplace transform approach: Referring to the problem, we can use the following relation
$$ \begin{align}
\int_0^\infty F(u)g(u) \, du & = \int_0^\infty f(u)G(u) \, du \\[6pt]
L[f(t)] & = F(s) \\[6pt]
L[g(t)] & = G(s)\end{align} $$
Let
$$ G(u)=\frac{1}{u^3} \implies g(u)=\frac{u^2}{2!}, $$
and
$$ f(u)= \sin(u)^3 \implies F(u) = {\frac {6}{ \left( {u}^{2}+1 \right) \left( {u}^{2}+9 \right) }}. $$
Now,
$$ \int_0^\infty \frac{\sin^3 x}{x^3} \, dx = \frac{6}{2}\int_0^\infty \frac{u^2}{\left( {u}^{2}+1 \right) \left( {u}^{2}+9 \right)} \, du = \frac{3\pi}{8}$$.
|
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|
Providing a closed formula for a linear recursive sequence I am studying for an exam in linear algebra and I have some trouble solving the following:
Let $(a_n)$ be a linear recursive sequence in $\mathbb{Z}_5$ with
\begin{align}
a_0 = 2, a_1 = 1, a_2 = 0 \text{ and } a_{n+3} = 2a_{n+2} + a_{n+1} + 3a_n \text{ for } n \in \mathbb{N}_0
\end{align}
Provide a closed formula for $a_n$.
My findings so far:
$
\begin{pmatrix} a_{n+3} \\ a_{n+2} \\ a_{n+1} \\ \end{pmatrix}
= \begin{pmatrix} 2 & 1 & 3 \\1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix} \cdot \begin{pmatrix} a_{n+2} \\ a_{n+1} \\ a_{n} \\ \end{pmatrix}
$
$
\begin{pmatrix} a_{n} \\ a_{n+1} \\ a_{n+2} \\ \end{pmatrix}
= A^n \cdot \begin{pmatrix} a_{0} \\ a_{1} \\ a_{2} \\ \end{pmatrix} = A^n \cdot \begin{pmatrix} 2 \\ 1 \\ 0 \\ \end{pmatrix}
$
At some point there must hold the following:
\begin{align}
T^{-1}AT = D = \begin{pmatrix} c_1 & 0 & 0 \\0 & c_2 & 0 \\ 0 & 0 & c_3 \\ \end{pmatrix}
\end{align}
such that D is a diagonal matrix containing all Eigenwerte $c_i$, $i \in {\{1,2,3} \}$ of $A$.
How shall I continue to find the matrix $A$?
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Hint: Use elementary row operations on your matrix $A$. You shall get identity matrix as $D$. Represent $T$ by elementary matrixes
|
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|
Euler's phi function $\phi(n)$ is even for all $n \geq 3$; when is it not divisible by $4$? Problem 1: Show that $\phi(n)$ is even for all $n \geq 3$.
Proof: Assume that n is a power of 2, let us say that $n=2^k$, with $k \geq 2$. By the Phi Function Formula, we have $\phi(n) = \phi(2^k)=2^k - 2^{k-1}=2^k(1/2)=2^{k-1}$, which $\phi(n)$ is an even integer.
If n isn't a power of 2, then it is divisible by an odd prime p. Thus, $n=p^km$, where $k \geq 1$ and $gcd(p^k,m)=1$. By the theorem $\phi(mn)=\phi(n)\phi(m)$,
$\phi(n)=\phi(p^km)=\phi(p^k)\phi(m)=p^k(p-1)\phi(m)$,
which is 2|p-1, since p is odd prime.
Problem 2: Describe, with proof, all $n$ for which $\phi(n)$ is divisible by $2$, but not by $4$.
How would you solve this problem? Can you use Problem 1?
|
HINT: If $p$ is prime, $\varphi(p^k)=p^k-p^{k-1}=p^{k-1}(p-1)$. If $m$ and $n$ are relatively prime, then $\varphi(mn)=\varphi(m)\varphi(n)$. And every $n\ge 1$ is a product of powers of distinct primes. These facts make the first question very easy to answer and are sufficient to answer the second question as well.
You might find it helpful to look at some numerical data while you’re thinking about the problem:
$$\begin{array}{c|c|c}
n&\text{prime factorization}&\varphi(n)\\ \hline
2&2^1&1\\
3&\color{blue}{3^1}&2\\
4&2^2&2\\
5&\color{blue}{5^1}&4\\
6&2^1\cdot\color{blue}{3^1}&2\\
7&\color{blue}{7^1}&6\\
8&2^3&4\\
9&\color{blue}{3^2}&6\\
10&2^1\cdot\color{blue}{5^1}&4\\
11&\color{blue}{11^1}&10\\
12&2^2\cdot\color{blue}{3^1}&4\\
13&\color{blue}{13^1}&12\\
14&2^1\cdot\color{blue}{7^1}&6\\
15&\color{blue}{3^1}\cdot\color{blue}{5^1}&8\\
16&2^4&8\\
17&\color{blue}{17^1}&16\\
18&2^1\cdot\color{blue}{3^2}&6\\
19&\color{blue}{19^1}&18\\
20&2^2\cdot\color{blue}{5^1}&8\\ \hline
25&\color{blue}{5^2}&20\\ \hline
27&\color{blue}{3^3}&18\\ \hline
30&2^1\cdot\color{blue}{3^1}\cdot\color{blue}{5^1}&8
\end{array}$$
You might also find this article useful.
|
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|
Geometric Identities involving $π^2$ Are there any known geometric identities that have $π^2$ in the formula?
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The sphere of radius $r$ in $\mathbb R^4$ has volume (4-dimensional volume) $$ \frac{\pi^2 \; r^4}{2} $$ Here, we refer to the set $$ x_1^2 + x_2^2 + x_3^2 + x_4^2 \leq r^2. $$
The sphere of radius $r$ in $\mathbb R^5$ has volume (5-dimensional volume) $$ \frac{8 \pi^2 \; r^5}{15} $$ Here, we refer to the set $$ x_1^2 + x_2^2 + x_3^2 + x_4^2+ x_5^2 \leq r^2. $$
EDIT. My impression is that most students are not aware of the easy way to do volumes of the higher dimensional spheres. Let $V_m$ be the volume of the sphere of radius $1$ in $\mathbb R^m.$ So $$V_1=2, \; V_2 = \pi, \; V_3 = \frac{4 \pi}{3}.$$ In order to find $V_{m+2},$ we integrate $V_m R^m$ over the unit disc in the plane, where this radius $R= \sqrt {1 - x^2 - y^2}.$ Switching to polar coordinates, we get $$ V_{m+2} = \int_0^{2 \pi} \int_0^1 V_m (1-r^2)^{m/2} \, r \, dr \, d \theta = 2 \pi V_m \int_0^1 (1-r^2)^{m/2} \, r \, dr = \frac{2 \, \pi \, V_m}{m+2}. $$ So $$ V_4 = \frac{2 \, \pi \, V_2}{4} = \frac{\pi^2}{2} $$ and $$ V_5 = \frac{2 \, \pi \, V_3}{5} = \frac{2 \pi \cdot 4 \pi}{3 \cdot 5} = \frac{8\pi^2}{15} $$
By induction, $$ V_n = \frac{\pi^{n/2}}{(n/2)!}, $$ where in the case of odd $n$ we need to understand $$ (n/2)! = \Gamma \left( 1 + \frac{n}{2} \right) $$ which is always a rational multiple of $\sqrt \pi$ (for odd $n$ it is).
|
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|
Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$ While I don't doubt that this question is covered somewhere else I can't seem to find it, or anything close enough to which I can springboard. I however am trying to prove $$\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$$ by induction.
I have seen it many times and proved it before but can't remember what it was I did. I see that for the first two terms $n = 1, n=2$ I get:
for $n = 1$, $\frac{1}{1^2} = 1 < 2$
for $n = 2$, $\frac{1}{1^2} + \frac{1}{2^2} = \frac{5}{4} < 2$
Now I am stumped, I know I want to show this works for the $n+1$ term and am thinking, let the series $\sum_{n=1}^\infty \frac{1}{n^2} = A(n)$ Then look to show the series holds for $A(n+1)$ But $A(n+1) = A(n) + \frac{1}{(n+1)^2}$ But now what? If I tried $A(n+1) - A(n) = \frac{1}{(n+1)^2}$ , but would have to show that this is less than $2 - A(n)$. I am stuck.
Thanks for your thoughts,
Brian
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Let $S_n(p) = \displaystyle \sum_{k=1}^n \dfrac1{k^p}$ and $S(p) = \lim_{n \to \infty} S_n(p)$. We then have
$$S_{2n+1}(p) = \sum_{k=1}^{2n+1} \dfrac1{k^p} = 1 + \sum_{k=1}^n \left(\dfrac1{(2k)^p} + \dfrac1{(2k+1)^p}\right) < 1 + \sum_{k=1}^n \dfrac2{(2k)^p} = 1 + \dfrac{S_n(p)}{2^{p-1}}$$
Letting $n \to \infty$, we get that
$$S(p) \leq 1 + \dfrac{S(p)}{2^{p-1}} \implies S(p) \leq \dfrac{2^{p-1}}{2^{p-1}-1}$$
|
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|
Find $a,b\in\mathbb{Z}^{+}$ such that $\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$ find positive intergers $a,b$ such that
$\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$
Here i tried plugging
$x^3=a,y^3=b$
$(x+y-1)^2=x^2+y^2+1+2(xy-x-y)=49+20\sqrt[3]{6} $
the right hand part is a square hence can be written as $(p+q)^2$
|
Firstly, if either $a$ or $b$ is a perfect cube, then we have $$ (\sqrt[3]{a}-N)^2 = 49 + 20 \sqrt[3]{6} $$
Convince yourself that the LHS must involve 2 different surd terms, hence this is not possible. Thus, neither $a$ nor $b$ are a perfect cube.
Like you did, expand the terms and consider what they are. Clearly $a, b, a^2, b^2$ are not perfect cubes. If $ab$ is not a perfect cube, then we must have $1 = 49$, which is a contradiction. Hence, we have $ 2\sqrt[3]{ab} + 1 = 49$, or that $ab = 24^3$. We use the substitution $b = \frac{24^3}{a}$ and $x = \sqrt[3]{a}$, and the equation becomes
$$x^2-2x - 48/x + 24^2/x^2 = 20 \sqrt[3]{6}.$$
Multiplying throughout by $x^2$, this equation has real roots of $x = \sqrt[3]{48}$ and $x= \sqrt[3]{288}$, hence $ (a,b) = (48 288) $ and $(288,48)$ are solutions.
|
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|
$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\
&= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\
&= \sqrt {\left(5-\frac92\right)^2} +\frac92\\
&= 5 + \frac92 - \frac92 \\
&= 5\end{align}$$
Where did I go wrong
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Clearly, $\sqrt{(4-\frac{9}{2})^2}$ is what? Well, $4-\frac{9}{2}$ is $-\frac{1}{2}$. But we all know $(-\frac{1}{2})^2$ is $\frac{1}{4}$ since $(-x)^2=x^2$. That means that $\sqrt{(4-\frac{9}{2})^2}$ is $\frac{1}{2}$. So the assumption that $\sqrt{(4-\frac{9}{2})^2} = 4 - \frac{9}{2}$ leads to $-\frac{1}{2}=\frac{1}{2}$ which is clearly wrong.
So the square root of a positive number squared is a positive number and the square root of a negative positive number squared is also a positive number. So if we assume $\sqrt{(-x)^2} = -x$ to $x = -x$ which is false. Note: x is positive, not negative. This is the correct statement:
$\sqrt{(-x)^2} = \sqrt{x^2} = x \not= -x$
|
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|
What are the coefficients of the polynomial inductively defined as $f_1=(x-2)^2\,\,\,;\,\,\,f_{n+1}=(f_n-2)^2$?
Let $\{f_n(x)\}_{n\in \Bbb N}$ be a sequence of polynomials defined inductively as
$$\begin{matrix} f_1(x) & = & (x-2)^2 &
\\ f_{n+1}(x)& = & (f_n(x)-2)^2, &\text{ for all }n\ge 1.\end{matrix}$$
Let $a_n$ and $b_n$ respectively denote the constant term and the coefficient of $x$ in $f_n(x)$. Then
*
*$(\text A)\,\, a_n=4, b_n=-4^n$
*$(\text B)\,\,a_n=4, b_n=-4n^2$
*$(\text C)\,\, a_n=4^{(n-1)!}, b_n=-4^n$
*$(\text C)\,\, a_n=4^{(n-1)!}, b_n=-4n^2$
Please help me how to solve this problem. I am not able to solve it.
The problem can be found on this link.
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HINT:
$f_1(x)=(x-2)^2=4-4x+x^2$
$f_2(x)=(f_1(x)-2)^2$
$=(x^2-4x+4-2)^2=(2-4x+x^2)^2=2^2+2\cdot2(-4x)+\cdots=4-4^2x+\cdots$
$f_3(x)=(f_2(x)-2)^2$
$=(4-2^4x+\cdots-2)^2=(2-2^4x+\cdots)^2=2^2-2\cdot2\cdot4^2x+\cdots=2^2-4^3x+\cdots$
$f_4(x)=(f_3(x)-2)^2$
$=(2^2-4^3x+\cdots-2)^2=(2-4^3x+\cdots)^2=2^2-2\cdot2\cdot2^6x+\cdots=2^2-4^4x+\cdots$
$f_5(x)=(f_4(x)-2)^2$
$=(4-4^4x+\cdots)^2=(2-4^4x+\cdots)^2=2^2-2\cdot2\cdot2^8x+\cdots=2^2-4^5x+\cdots$
So, $a_n=4$
$b_n=-4^{(n \text{ th term of } 1,2,3,4,5)}=-4^n$
|
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|
How many zeros does $z^{4}+z^{3}+4z^{2}+2z+3$ have in the first quadrant? Let $f(z) = z^{4}+z^{3}+4z^{2}+2z+3$. I know that $f$ has no real roots and no purely imaginary roots.
The number of zeros of $f(z)$ in the first quadrant is $\frac{1}{2\pi i}\int_{C}\frac{f'(z)}{f(z)}\, dz$ by the Argument Principle where $C = C_{1} + C_{2} + C_{3}$ with $C_{1}$ the line from origin to the point $(R, 0)$ on the $x$-axis, $C_{2}$ is the line $Re^{i\theta}$ where $0 \leq \theta \leq \pi/2$, and $C_{3}$ is the line segment from the point $(0, R)$ to the origin.
We have
$$\frac{1}{2\pi i}\int_{C_{2}}\frac{f'(z)}{f(z)}\, dz = \frac{1}{2\pi}\int_{0}^{\pi/2}4 + O(R^{-1})\, d\theta \rightarrow 1$$ as $R \rightarrow \infty$.
I am having an issue computing the integral for $C_{1}$ and $C_{3}$. For $C_{1}$, I want to compute
$$\lim_{R \rightarrow \infty} \frac{1}{2\pi i}\int_{0}^{R}\frac{4x^{3} + 3x^{2} + 8x + 2}{x^{4} + x^{3} + 4x^{2} + 2x + 3}\, dx$$ but doesn't this evaluate to $\infty$ since the inside is $O(1/x)$?
|
Here's an alternative idea, which can be applied to any quartic.
Let $z=x+iy$, expand the powers of $z$, and separate the real and imaginary parts to get
$$x^4+x^3+4x^2+2x+3-(6x^2+3x+4)y^2-y^4=0$$
and
$$(4x^3+3x^2+8x+2)y-(4x+1)y^3=0$$
Since we looking for solutions with $x,y\gt0$, we can rewrite the second equation as
$$y^2={4x^3+3x^2+8x+2\over4x+1}$$
and substitute into the first, obtaining (with a multiplication by $-1$)
$$(4x^3+3x^2+8x+2)^2+(4x+1)(4x^3+3x^2+8x+2)(6x^2+3x+4)-(4x+1)^2(x^4+x^3+4x^2+2x+3)=0$$
Technically one should expand this out, but it seems fairly clear that the coefficients from the first two, positive portions will more than cancel the negative coefficients from the third, leaving a sextic with all positive coefficients and hence no roots with $x\gt0$.
For a general quartic with real coefficients, you can write $y^2$ as a cubic divided by a linear expression in $x$, producing a sextic, which in general may have some positive roots. It may sound as if you could wind up with $6$ real values for $x$, each with two possible values for $y$, giving $12$ roots in all to a quartic. Obviously something's gotta give. Either the sextic has fewer than $4$ real roots, or some of its real roots produce a negative value for $y^2$ (or both).
|
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|
Reasoning that $ \sin2x=2 \sin x \cos x$ In mathcounts teacher told us to use the formula $ \sin2x=2 \sin x \cos x$.
What's the math behind this formula that made it true? Can someone explain?
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A rotation matrix is clearly a linear transformation, so by rotating the elementary basis vectors, a two-dimensional rotation matrix for angle $x$ is
\begin{align}
\begin{bmatrix}
\cos(x) & -\sin(x) \\
\sin(x) & \cos(x)
\end{bmatrix}.
\end{align}
Squaring this matrix is equivalent to applying the rotation twice, so
\begin{align}
\begin{bmatrix}
\cos(2x) & -\sin(2x) \\
\sin(2x) & \cos(2x)
\end{bmatrix}
&=
\begin{bmatrix}
\cos^2(x) - \sin^2(x) & -2\sin(x)\cos(x) \\
2\sin(x)\cos(x) & \cos^2(x) - \sin^2(x)
\end{bmatrix},
\end{align}
which implies that
\begin{align}
\cos(2x) &= \cos^2(x) - \sin^2(x) \\
\sin(2x) &= 2\sin(x)\cos(x).
\end{align}
Using rotation matrices with different angles gives some of the double-angle (or triple-angle, etc.) identities.
|
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|
Calculate the integer part I have to calculate the integer part of this:
$$[(\sqrt{2}+\sqrt{5})^2]$$
I tried to write it like this:
$$[2+5+2\sqrt{10}]=[7+2\sqrt{10}]=7+[2\sqrt{10}]$$
Any ideas?
|
Calvin Lin provided an excellent proof. I am just adding in some details.
Obviously, $\sqrt{ 5} - \sqrt{2} > 0$
Also, $\sqrt{2} > 1$
$2 \sqrt{2} > 2$
$1 + 2 \sqrt{2} + 2 > 5$
$1^2 + 2 \sqrt{2} + (\sqrt{2})^2 > (\sqrt{5})^2$
$(1 + \sqrt{2})^2 > (\sqrt{5})^2$
Taking the positive square root, we have $1 + \sqrt{2} > \sqrt{5}$
Then, $1 > \sqrt{5} - \sqrt{2}$
Combining the results together, we have $0 < \sqrt{ 5} - \sqrt{2} < 1$
Note that squaring a +ve quantity whose value lies between 0 and 1 is again positive but smaller.
Thus, $0 < (\sqrt{ 5} - \sqrt{2})^2 < \sqrt{ 5} - \sqrt{2} < 1$
Since $(\sqrt{5} + \sqrt{2})^2 + (\sqrt{5}- \sqrt{2})^2 = … = 14$
$(\sqrt{5} + \sqrt{2})^2 = 14 - (\sqrt{5}- \sqrt{2})^2$
Therefore, $\lfloor (\sqrt{5} + \sqrt{2})^2 \rfloor = 13$
|
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|
Compare the sum of the squares of the median of a triangle to the sum of the squares of sides You have to compare the sum of the squares of the median of a triangle to the sum of the squares of sides?
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$$ 3( AB^2 + BC^2 + AC^2 ) = 4 ( AD^2 + BE^2 + CF^2)$$
AB , BC , CD are lengths of sides of triangle and AD , BE , CF are lengths of medians of triangle
HINT :
start with appollonius theorem and add all
Apollonius Theorem
so you will get relation as
3 (sum of square of sides) = 4 ( sum of squares of medians )
Proof :
$$ AB^2 + AC^2 = 2 ( AD^2 + \frac{BC^2}{4} )$$
$$ AB^2 + BC^2 = 2 ( BE^2 + \frac{AC^2}{4} )$$
$$ AC^2 + BC^2 = 2 ( CF^2 + \frac{AB^2}{4} )$$
Add all above and bring sides to LHS and medians to RHS to get
$$ 3( AB^2 + BC^2 + AC^2 ) = 4 ( AD^2 + BE^2 + CF^2)$$
|
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|
Prove the inequality $\,\frac{1}{\sqrt{1}+ \sqrt{3}} +\frac{1}{\sqrt{5}+ \sqrt{7} }+\ldots+\frac{1}{\sqrt{9997}+\sqrt{9999}}\gt 24$
Prove the inequality $$\frac{1}{\sqrt{1}+ \sqrt{3}}
+\frac{1}{\sqrt{5}+ \sqrt{7} }+......... +\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$
My work:
Rationalizing the denominator gives
$$\frac{\sqrt{3}-1}{2} +\frac{\sqrt{7}-\sqrt{5}}{2}+......+\frac{\sqrt{9999}-\sqrt{9997}}{2} .$$
Now by taking two as common and separating the positive and negative terms gives
$$\frac{1}{2} [ \{\sqrt{3} +\sqrt{7}+\dots +\sqrt{9999}\} - \{1+\sqrt{5} +\dots+\sqrt{9997}\}].$$
Can we do like this please suggest. Thanks.
|
Hint: Telescope your rationalized sum by adding terms.
|
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|
Show that $\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2$ How to show that $\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2$ for coprime $a$ and $b$?
I know the fact that $\gcd(a,b)=1$ implies $\gcd(a,b^2)=1$ and $\gcd(a^2,b)=1$, but how do I apply this to that?
|
Let positive integer $d$ divides both $a+b,a^2+b^2$
$\implies d$ divides $(a^2+b^2)+(a+b)(a-b)=2a^2$
Similarly, i.e., $d$ divides $(a^2+b^2)-(a+b)(a-b)=2b^2$
$\implies d$ divides $2a^2,2b^2\implies d$ divides $(2a^2,2b^2)=2(a^2,b^2)=2(a,b)^2$
|
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|
How to find the sum $\sum\limits_{k=1}^n (k^2+k+1)k!$? How to find the sum $$\sum^n_{k=1} (k^2+k+1)k!$$
What I tried as follows :
$$\sum^n_{k=1} (k^2+k+1)k!$$
=$$\sum^n_{k=1} (k^2)k!+ \sum^n_{k=1} (k)k! + \sum^n_{k=1} (1)k!$$
Now we can write $\sum^n_{k=1} (k^2) $ as $$\frac{n(n+1)(2n+1)}{6}$$ but what to do with $k!$ please guide thanks.......
|
HINT:
$$(k^2+k+1)k!= (k+1)(k+2) k!+A (k+1)k!+B k!=(k+2)!+A (k+1)!+B k!$$
$$\implies k^2+k+1=(k+1)(k+2)+A(k+1)+B=k^2+k(3+A)+2+A+B $$
Equating the coefficients of $x,A+3=1\implies A=-2$
Equating the constant terms, $A+B+2=1\implies B=-1-A=-1-(-2)=1$
So, $$(k^2+k+1)k!= (k+2)!-2(k+1)!+ k!=(k+2)!-(k+1)!-\{(k+1)!-k!\}$$
Can you recognize the telescopic series?
|
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|
Factorize $(x+1)(x+2)(x+3)(x+6)- 3x^2$ I'm preparing for an exam and was solving a few sample questions when I got this question -
Factorize : $$(x+1)(x+2)(x+3)(x+6)- 3x^2$$
I don't really know where to start, but I expanded everything to get :
$$x^4 + 12x^3 + 44x^2 + 72x + 36$$
I used rational roots test and Descartes rule of signs to get guesses for the roots. I tried them all and it appears that this polynomial has no rational roots.So, what should I do to factorize this polynomial ?
(I used wolfram alpha and got the factorization : $(x^2 + 4x + 6) (x^2 + 8x + 6)$ But can someone explain how to get there ?)
|
I'm assuming that you are looking for a factorization of the polynomial
$$
f = x^4 + 12x^3 + 44x^2 + 72x + 36
$$
in $\mathbb Q[x]$.
By the Lemma of Gauss and the fact that $f$ is monic, this is the same as looking for a factorization in $\mathbb Z[x]$.
Since there are no rational roots, the only remaining possibility is the factorization into two factors of degree 2. Since the polynomial is monic, the factors may be assumed to be monic.
So the putative factors have the form $x^2 + ax + b$ and $x^2 + cx + d$ with $a,b,c,d\in\mathbb Z$.
This leads to
$$
f = (x^2 + ax + b)(x^2 + cx + d) = x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd.
$$
Comparing the coefficients, we get four equations
$$
a+c = 12\\
ac + b + d = 44\\
ad + bc = 72\\
bd = 36
$$
Up to swapping the two factors, the last equation has the solutions
$$
(b,d)\in\{(1,36),(-1,-36),(2,18),(-2,-18),(3,12),(-3,-12),(4,9),(-4,-9),(6,6),(-6,-6)\}.
$$
It's a bit of work, but plugging these values into the remaining three equations, you find that only for $b=d=6$ there is a solution, which is $a = 4$, $b = 8$ or $a = 8$, $b=4$. This gives you the two factors.
|
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|
Why can we use inspection for solving equation with multiple unknowns? In our algebra class, our teacher often does the following:
$a + b\sqrt{2} = 5 + 3\sqrt{2} \implies \;\text{(by inspection)}\; a=5, b = 3
$
I asked her why we can make this statement. She was unable to provide a satisfactory answer. So I tried proving it myself.
$a + b\sqrt{2} = x + y\sqrt{2}$. We are required to prove that $a = x$, and $b = y$. Manipulating the equation, we get $\sqrt{2}(b - y) = x - a$, or $\sqrt{2} = \frac{x-a}{b-y}$. Expanding this, we get $\sqrt{2} = \frac{x}{b-y} + \frac{a}{b-y}$. I tried various other transformations, but nothing seemed to yield a result.
|
HINT:
Assuming $a,b,x,y$ are rationals $\sqrt2(b-y)=x-a$ rational which is only possible
if $b-y=0$
As for $b-y\ne0,\sqrt 2=\frac{x-a}{y-b}$ which is rational
|
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|
Using the definition of the derivative prove that if $f(x)=x^\frac{4}{3}$ then $f'(x)=\frac{4x^\frac{1}{3}}{3}$ So I have that $f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
and know that applying
$f(x)=x^\frac{4}{3}=f\frac{(x+h)^\frac{4}{3} -x^\frac{4}{3}}{h}$
but am at a loss when trying to expand
$(x+h)^\frac{4}{3}$
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Let
$$D = \frac{(x+h)^{4/3}-x^{4/3}}{h}$$
Then
$$\begin{align}D^3 &= \frac{(x+h)^4-3 (x+h)^{8/3} x^{4/3} + 3 (x+h)^{4/3} x^{8/3} - x^4}{h^3}\\ &= \frac{4 x^3 h + 6 x^2 h^2 + 4 x h^3 + h^4 -3 (x+h)^{4/3} x^{4/3} \left [ (x+h)^{4/3}-x^{4/3}\right ]}{h^3}\\ &= \frac{4 x^3 + 6 x^2 h + 4 x h^2 +h^3}{h^2} - \frac{3 (x+h)^{4/3} x^{4/3}}{h^2} D\end{align}$$
Thus,
$$h^2 D^3 + 3 (x+h)^{4/3} x^{4/3} D = 4 x^3 + 6 x^2 h + 4 x h^2 +h^3$$
Taking the limit as $h \to 0$, (and assuming that $\bar{D}=\lim_{h \to 0} D$ remains finite as we do so), we see that
$$3 x^{8/3} \bar{D} = 4 x^3$$
or $\bar{D} = (4/3) x^{1/3}$ as was to be demonstrated.
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|
How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$ How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$
I don't know the solution for this.
Help me!
Thank all!
|
Let $\mathbf v(x) = (\sin x \cdot \sin 2x, \cos x \cdot \cos 2x) \in \mathbb R^2$. Compute:
$$\begin{align} |v|^2 &= \sin^2 x \cdot \sin^2 2x + \cos^2 x \cdot \cos^2 2x \\
&= (\sin^2 x + \cos^2 x)(\sin^2 2x + \cos^2 2x) - \sin^2 x \cdot \cos^2 2x - \sin^2 2x \cdot \cos^2 x \\
&= 1 - \sin^2 x \cdot \cos^2 2x - \sin^2 2x \cdot \cos^2 x \le 1. \end{align}$$
Let $\mathbf u(x) = (\sin 3x, \cos 3x) \in \mathbb R^2$. Then we have $|\mathbf u(x)|=1$. By Cauchy–Schwarz inequality:
$$\begin{align} \sin x \cdot \sin 2x \cdot \sin 3x + \cos x \cdot \cos 2x \cdot \cos 3x &= \mathbf u(x) \cdot \mathbf v(x) \\
&\le |\mathbf v(x)| \\
&\le 1,\end{align}$$
so for equality to hold, it in necessary (but not necessarily sufficient) that $|\mathbf v(x)| =1$, which means that both of these equations must hold:
$$\begin{align} \sin x \cdot \cos 2x &= 0 \\
\sin 2x \cdot \cos x &= 0. \end{align}$$
By $\pi$-periodicity of the expression in $x$ (since $\sin (n(x+\pi)) = (-1)^n \sin (nx)$ and similarly for cosine), it's enough to find the solutions to this in the interval $x \in [0, \pi)$. Obviously, $x=0$ is one such solution. For all other $x$ in this interval, $\sin x \ne 0$ so we need $\cos 2x =0$ to satisfy the first equation. This has two solutions in this interval, namely $x \in \{ \frac{\pi}{4}, \frac{3\pi}{4} \}$, neither of which satisfy the second equation.
Since that was a necessary but not sufficient condition, you need to go back and check that $x=0$ does satisfy the equation (which is trivial). With that out of the way, the solutions are exactly $x = n \pi$ for integer $n$.
Note: This answer is mostly interesting because it proves a more general fact, namely $\sin \theta \sin \phi \sin \psi + \cos \theta \cos \phi \cos \psi \le 1$ for any angles $\theta,\phi,\psi$, and the method can be extended to longer products and gives necessary conditions to be checked for equality which are easy to work with. It's clearly more complicated than strictly required for this problem.
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How to calculate this expression? evaluate the expression [1]:
$$\sum\limits_{n = 1}^\infty {\left( {\frac{1}{n} - \frac{1}{{n + x}}} \right)} $$
where $x$ is a real number, $0\le x\le1$, and $x$ is rounded to 3 digits.
For example, when $x=0.500$, the expression is [2]:
$$\left(\frac11 -\frac1{1.5}\right)+\left(\frac12 -\frac1{2.5}\right)+\left(\frac13 -\frac 1{3.5}\right) + ...$$
For a given $x$, how can I evaluate it?
The answer must be rounded to more than 12 digits.
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For $x=0$, the sum evaluates to $0$, and for $x=1$, it is $1$.
For $0<x<1$, you can derive a power series expression for the given sum as below:
\begin{align}
\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{x+n}\right)=&\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n}\left(1+\frac{x}{n}\right)^{-1}\right)\\
\ =& \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n}\left(1-\frac{x}{n}+\frac{x^2}{n^2}-\frac{x^3}{n^3}+\cdots\right)\right)\\
\ =&\sum_{n=1}^{\infty}\left(\frac{x}{n^2}-\frac{x^2}{n^3}+\frac{x^3}{n^4}-\cdots\right)\\
\ =&\zeta(2)x-\zeta(3)x^2+\zeta(4)x^3-\cdots\\
\ =&\sum_{k=1}^{\infty}\zeta(k+1)x^k
\end{align}
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|
Evalute $\lim_{x\to-\infty} \frac{\sqrt{x^2+4x^4}}{8+x^2}$ Having a hard time with this. So far I have:
$$ \frac{\sqrt{x^2(1+4x^2)}}{8+x^2} = \frac{x\sqrt{1+4x^2}}{8+x^2}$$
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I am assuming that you mean
$\frac{\sqrt{x^2+4x^4}}{8+x^2}$.
Since $x^2$ and $x^4$ are positive,
the limit is the same for
$x \to \infty$ and
$x \to -\infty$.
You can take your last step
further and write
$\begin{align}
\frac{\sqrt{x^2+4x^4}}{8+x^2}
&= \frac{x\sqrt{1+4x^2}}{8+x^2}\\
&= \frac{x^2\sqrt{1/x^2+4}}{x^2(1+8/x^2)}\\
&= \frac{\sqrt{1/x^2+4}}{1+8/x^2}\\
\end{align}
$
As $x \to \pm \infty$,
$\sqrt{1/x^2+4} \to \sqrt{4} = 2$
and
$1+8/x^2 \to 1$,
so I get
$\frac{2}{1} = 1$.
|
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|
Solve the equation for x, y and z: $\sqrt{x-y+z}=\sqrt x - \sqrt y + \sqrt z$ I am having some trouble with this problem,
Solve for $x,y,$ and $z$.
$$\sqrt{x-y+z}=\sqrt x - \sqrt y + \sqrt z$$
Here is my work so far,
$$x - y +z = x+y+z-2\sqrt{xy} + 2\sqrt{xz}- 2\sqrt{zy}$$
$$2y-2\sqrt{xy} + 2\sqrt{xz}- 2\sqrt{zy} = 0 $$
$$2(y-\sqrt{xy} + \sqrt{xz} - \sqrt{zy}) = 0 $$
$$y-\sqrt{xy} + \sqrt{xz} - \sqrt{zy} = 0$$
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Even easier, your last equation can be written as:
$$
-\sqrt{y}\left(\sqrt{x} - \sqrt{y}\right) + \sqrt{z}\left(\sqrt{x} - \sqrt{y}\right) = 0
$$
or
$$
\left(\sqrt{z} - \sqrt{y}\right)\left(\sqrt{x} - \sqrt{y}\right) = 0
$$
Which in turn gives you the same conclusion as Calvin pointed out
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|
Find $x$ in the equation $ax^3+bx^2+cx=d$ $\begin{equation} \tag{A} ax^3+bx^2+cx=d \end{equation}$
We can define Delta for quadratic equation to check whether the equation has answer or not....for $f(x)$ which contains powers higher of $2$ for Is there any method to see how many acceptable roots the polynomial contains? Consider equation $(A)$. Can we say whether the equation has real root? How many roots of the function above are acceptable?
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It happens that equations of the form $ax^3+bx^2+cx+d=0$ and cubic polynomials in general, do have their own so-called discriminant $\Delta$. It is defined as: $$
\Delta=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd.\,
$$
We have $3$ distinct cases:
*
*If $\Delta > 0$, then the equation has three distinct real roots.
*If $\Delta = 0$, then the equation has a multiple root and all its roots are real.
*If $\Delta < 0$, then the equation has one real root and two nonreal complex conjugate roots.
The solutions can then be obtained using the cubic formula: $$\begin{align}x &= \sqrt[\displaystyle3\,]{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2} - \dfrac{d}{2a}\right) + \sqrt{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 + \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}}\\
& + \sqrt[\displaystyle3\,]{\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right) - \sqrt{\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 + \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}} - \dfrac{b}{3a}.\end{align}$$
I hope this helps.
Best wishes, $\mathcal H$akim.
|
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|
question about applying a comparison test to a sequence In lecture, we were asked:
Does $ \sum \limits_{n=0}^\infty \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 }$ converge?
$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 }$
In discussing strategies for applying a comparison test for $a_n$ we rejected several options for choosing a bound for $b_n$. The strategy is to choose an expression larger in the numerator and smaller in the denominator for $b_n$ If I understand correctly, using $2n^5$ in the denominator for $b_n$ doesn't work, because, as n gets large, we have
$$ \frac{2n^3}{n^5} \not < \frac{3n^3}{2n^5}$$
$$ 2 \not < \frac{3}{2} $$
If this is the case, couldn't we choose $b_n = \frac{10n^3}{2n^5}$?
Plugging the first inequality into Wolfram Alpha was not enlightening.
From the lecture:
To apply the comparison test, we need to bound the numerator from above and the denominator from below.
For the numerator of $b_n$, $2n^3$ won't work because we have $3n$ in the numerator of $a_n$. $3n^3$ is large enough.
$2n^5$ in the denominator of $b_n$ won't work.
$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 } \ \substack{ <\\ \\ >} \ \frac{3n^3}{2n^5} =b_n$
Using $n^4$ in the denominator won't work because this gives us a harmonic series for $b_n$, which is divergent.
$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 } \ \substack{ <\\ \\ >} \ \frac{3n^3}{n^4}=b_n $
Using asymptotic analysis, we get $a(x)= \frac{2}{x^2} + \mathcal{O}(\frac{1}{x^4})$, which gives us an integral with p=2, which converges.
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We do an explicit comparison, since that's what the question seems to ask for, but there are simpler ways.
For $n\gt 1$, the top is positive and $\lt 2n^3+3n^3=5n^3$.
The bottom is trickier. Rewrite it as
$$\frac{1}{2}n^5+ \frac{1}{2}n^5-5n^3-n^2+2.$$
Note that if $n\gt 10$, then $\frac{1}{2}n^5-5n^3-n^2+2\gt 0$. This is because then $\frac{1}{2}n^5 \gt 50n^3$. (We could get away with something cheaper than $10$, but why bother?). So the bottom, when $n\gt 10$, is $\gt \frac{1}{2}n^5$.
What we did here was to despatch half of $n^5$ to go crush $5n^3+n^2$. After they are crushed, the bottom is bigger than the remaining $\frac{1}{2}n^5$.
So for $n\gt 10$ we have that our expression is positive and $\lt \frac{5n^3}{x^5/2}=\frac{10}{n^2}$. Good enough to show convergence.
|
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|
Expected value sum of dots We throw $n$-times the die. Let $E_n$ be expected value sum of dots (got in all throws).
Compute
*
*$E_1$
*$E_2$
*$E_3$
*$E_4$
So i know how can I do it, for example in 1. I have:
*
*$E_1 = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} = \frac{7}{2}$.
Similar:
$E_2 = 2 \cdot \frac{1}{36} + 3 \cdot \frac{2}{36} + 4 \cdot \frac{3}{36} + 5 \cdot \frac{4}{36}+ 6 \cdot \frac{5}{36} + 7 \cdot \frac{6}{36} + 8 \cdot \frac{5}{36} + 9 \cdot \frac{4}{36} + 10 \cdot \frac{3}{36} + 11 \cdot \frac{2}{36} + 12 \cdot \frac{1}{36} = 7$
It is easy but for $E_3$ and $E_4$ it will be compute very long. I suppose that exist easier way to do this task. I will grateful for your help.
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If you take $X=X_{1}+,...,+X_{n}$ where $X_{i}, i \in \{1,...,n\}$ is the number of dots on the throw $i$ then you need $E[X]$. So, because expectation of a sum is the sum of the expectations
$$E[X]=E\left[\sum\limits_{i=1}^{n}X_{i}\right]=\sum\limits_{i=1}^{n}E{\left[X_i\right]}$$
but $E[X_i]=E[X_j]$ for all $i,j$ because $X_i$ and $X_j$ are identically distributed and
$$E[X_i]=\sum\limits_{j=1}^{6}j\frac{1}{6}=\frac{1}{6}\frac{42}{2}=\frac{7}{2}$$
then
$$E[X]=\sum\limits_{i=1}^{n}\frac{7}{2}=\frac{7}{2}n.$$
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|
Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$ What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to write $\textrm{cosec}^2\left(\frac{4\pi}7\right)$ as $\textrm{cosec}^2\left(\frac{3\pi}7\right)$. Then converted in $\sin$... But in vain.. Is there any other approach?
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Let $7x=\pi\implies 4x=\pi-3x$
$$\frac1{\sin2x}+\frac1{\sin4x}=\frac{\sin4x+\sin2x}{\sin4x\sin2x}$$
$$=\frac{2\sin3x\cos x}{\sin(\pi-3x)\sin2x}(\text{ using } \sin2A+\sin2B=2\sin(A+B)\cos(A-B))$$
$$=\frac{2\sin3x\cos x}{\sin(3x)2\sin x\cos x}(\text{ using } \sin2C=2\sin C\cos C \text{ and }\sin(\pi-y)=\sin y)$$
$$=\frac1{\sin x}$$
$$\implies \frac1{\sin x}-\frac1{\sin2x}-\frac1{\sin4x}=0$$
Squaring we get,
$$\frac1{\sin^2x}+\frac1{\sin^22x}+\frac1{\sin^24x}=2\left(\frac1{\sin x\sin4x}+\frac1{\sin x\sin2x}-\frac1{\sin2x\sin4x}\right)=2\frac{(\sin2x+\sin4x-\sin x)}{\sin x\sin2x\sin4x}$$
Now, $$\sin2x+\sin4x-\sin x=2\sin3x\cos x-\sin(\pi-6x)\text{ as }x=\pi-6x$$
$$\implies \sin2x+\sin4x-\sin x=2\sin3x\cos x-\sin6x$$
$$=2\sin3x\cos x-2\sin3x\cos3x=2\sin3x(\cos x-\cos3x)=2\sin3x(2\sin2x\sin x)$$ using $\cos2C-\cos2D=\sin(D-C)\sin(C+D)$
$$\implies \frac{\sin2x+\sin4x-\sin x}{\sin x\sin2x\sin4x}=4$$ as $\sin x\sin2x\sin4x\ne0$ if $7x=\pi$
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Prove that: $ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$ How to prove the following trignometric identity?
$$ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$$
Using half angle formulas, I am getting a number for $\cot7\frac12 ^\circ $, but I don't know how to show it to equal the number $\sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$.
I would however like to learn the technique of dealing with surds such as these, especially in trignometric problems as I have a lot of similar problems and I don't have a clue as to how to deal with those.
Hints please!
EDIT:
What I have done using half angles is this: (and please note, for convenience, I am dropping the degree symbols. The angles here are in degrees however).
I know that
$$ \cos 15 = \dfrac{\sqrt3+1}{2\sqrt2}$$
So,
$$\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}$$
$$\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} $$
$$\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 - \sqrt3 + 1}} $$
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Start from $\displaystyle\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{2\tan{15^\circ}}{1-\tan^2{15^\circ}}$.
If $x=\tan{15^\circ}$, then $\displaystyle\tan 15^\circ = x = \frac{2\tan{(\frac{15}{2})^\circ}}{1-\tan^2{(\frac{15}{2})^\circ}}$.
If $y=\tan{(\frac{15}{2})^\circ}$, then $x=\frac{2y}{1-y^2}$. Hence
$\displaystyle\frac{1}{\sqrt{3}} = \frac{2(\frac{2y}{1-y^2})}{1-(\frac{2y}{1-y^2})^2}$.
Simplify the above equation and solve for $y$, then find the reciprocal to find $\cot{(\frac{15}{2})^\circ}$.
EDIT: To simplify your surd, try to multiply both the numerator and denominator by $\sqrt{2\sqrt{2}−\sqrt{3}+1}$.
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|
Use the identity $(r+1)^3-r^3\equiv3r^2+3r+1$ to find $\sum_{r=1}^nr(r+1)$ Use the identity $(r+1)^3-r^3\equiv3r^2+3r+1$ to find $$\sum_{r=1}^nr(r+1)$$
I can obtain $$\sum_{r=1}^n3r^2+3r+1=(n+1)^3-1$$
and I think the next step is
$$3\sum_{r=1}^nr(r^2+1)+\frac13=\left((n+1)^3-1\right)$$
But how do I deal with the constant 3 at the end of the LHS and also tidy up the answer to get $\frac13n(n+1)(n+2)$?
|
We have
$$r(r+1)=\frac{1}{3}((r+1)^3-r^3-1)$$
so
$$\sum_{r=1}^n r(r+1)=\frac{1}{3}\sum_{r=1}^n((r+1)^3-r^3)-\frac{1}{3}\sum_{r=1}^n1=\frac{1}{3}((n+1)^3-1)-\frac{n}{3}$$
|
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|
How many factors does 6N have? Given a number $2N$ having 28 factors another number $3N$ having 30 factors, then find out the number of factors of $6N$.
|
If $2n$ has 28 divisors, then $2n$ is of the form $a^6.b$, where b has 4 divisors. Note that if chose $2^13$ or $2^26$ here, then the third multiple would have more than $30$ divisors, since it would be $3*13* or $2*27$ divisors required.
Since $3n$ does not have a divisor of the form $a^6$, we see that $a=2$, and that n is at least $2^5$. The reminder takes $b$ has $4$ divisors, and 3b has $5$, so $ b=3^3$. So the number is $864$, which has $6*4=24$ divisors, its second and third have $28$ and $30$ divisors.
Likewise $6n$ has 7*5 or $35$ divisors. It is the $5184$.
One can also see that if $6n=2^a 3^b c$, then for $2n$ to have 28 divisors, and $3n$ to have 30 divisors, then $7|a-1$ and $5|b-1$. This, and that $ab+b=28$ and $ab+a=30$, means that $a-b=2$, then $a=6, b=4$ is the only solution here.
|
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|
Is it always possible to factorize $(a+b)^p - a^p - b^p$ this way? I'm looking at the solution of an IMO problem and in the solution the author has written the factorization $(a+b)^7 - a^7 - b^7=7ab(a+b)(a^2+ab+b^2)^2$ to solve the problem. It seems like it's always possible to find a factorization like $(a+b)^p - a^p - b^p=p\cdot(ab)^{\alpha_1}(a+b)^{\alpha_2}(a^2+ab+b^2)^{\alpha_3}P_{\omega}(a,b)$ for any prime number p where $P_{\omega}(a,b)$ is an irreducible homogenous polynomial of degree $\omega$ and $\alpha_1 + \alpha_2 + \alpha_3 + \omega = p-1$. I wonder if it's true in general.
Here are some examples:
$(a+b)^2 - a^2 - b^2 = 2ab$
$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$
$(a+b)^5 - a^5 - b^5 = 5ab(a+b)(a^2+ab+b^2)$
$(a+b)^7 - a^7 - b^7 = 7ab(a+b)(a^2+ab+b^2)^2$
$(a+b)^{11}- a^{11} - b^{11} = 11 ab(a + b)(a^2 + a b + b^2)(a^6 + 3 a^5 b + 7 a^4 b^2 + 9 a^3 b^3 + 7 a^2 b^4 + 3 a b^5 + b^6)$
I've proved the following statements:
For any prime number p:
$p|(a+b)^p - a^p-b^p$
$ab|(a+b)^p - a^p-b^p$
For any odd prime number:
$(a+b) | (a+b)^p - a^p-b^p$ because $(a+b)^p - a^p-b^p$ vanishes when $a=-b$ if p is odd.
For any prime number $p \geq 5$:
$ (a^2 + ab + b^2) | (a+b)^p - a^p - b^p$
Proof: Any prime number greater than 3 is of the form $p=6k+1$ or $p=6k+5$. On the other hand, if we set $\Large \omega= e^\frac{2\pi i}{3}$ we have: $1 + \omega + \omega^2 = 0$, therefore
$(1+\omega)^3 = 1 + 3\omega + 3\omega^2 + \omega^3 = 2 + 3(\omega+\omega^2) = -1$
$(1+\omega^2)^3 = 1 + 3\omega^2 + 3\omega^4 + \omega^3 = 2 + 3(\omega^2+\omega) = -1$
We have $a^2+ab+b^2 = (a-b\omega)(a-b\omega^2)$ in $\mathbb{Z}[\omega]$, I'll show that both $a-b\omega$ and $a-b\omega^2$ divide $(a+b)^p - a^p - b^p$ for every $p>3$:
if $p=6k+1$ and we replace $a=b\omega$ then:
$(a+b)^p=(b\omega+b)^{6k+1} = b^{6k+1} (1+\omega)^{6k} (1+\omega) = b^{6k+1}(1+\omega)$
$(a+b)^p - a^p - b^p = b^{6k+1}(1+\omega) - (b\omega)^{6k+1} - b^{6k+1} = b^{6k+1}(1+\omega) - b^{6k+1}(\omega+1) = 0$
If $p=6k+5$ and we replace $a=b\omega$ then:
$(a+b)^p=(b\omega+b)^{6k+5} = b^{6k+5} (1+\omega)^{6k} (1+\omega)^5 = (-1)b^{6k+1}(1+\omega)^2$
$(a+b)^p - a^p - b^p = (-1)b^{6k+5}(1+\omega)^2 - (b\omega)^{6k+5} - b^{6k+5} = (-1)b^{6k+5}(1+\omega)^2 - b^{6k+5}(\omega^2+1) = p^{6k+5}(-1-2\omega-\omega^2-\omega^2-1)=0$
The same could be shown for $a=b\omega^2$.
Therefore both $(a-b\omega)$ and $(a-b\omega^2)$ are factors of $(a+b)^p-a^p-b^p$ but I'm not sure if that is sufficient to conclude $a^2+ab+b^2=(a-b\omega)(a-b\omega^2) | (a+b)^p - a^p - b^p$
So, I guess I have proved that $p$, $ab$, $a+b$ and $a^2+ab+b^2$ all divide $(a+b)^p-a^p-b^p$, but I have no idea how to show that $P_{\omega}(a,b)$ must be irreducible. I have verified this conjecture up to $p=97$ and it's true I think. Any ideas on how to prove that?
|
The polynomials $P_n(x)=P_n(x,1)$ are called the Cauchy-Mirimanoff polynomials.
They are defined for all $n$ natural $n\geq 2$ by:
$$ (X^n+1)-X^n-1 = X(X+1)^E(X^2+X+1)^eP_n(X) $$
Where $E=0$ if n is even, and 1 otherwise, and, as @KierenMacMillan notice, $e= 0,1,2 $ according to $n= 0,2,1\mod 3$.
We know that $P_{2p}(x)$ and $P_{3p}(x)$ are irreducibles when $p$ is prime. It's still unknown whether all Cauchy-Mirimanoff polynomials are irreducibles...
For example, the proof that $P_{3p}(x)$ is irr, use the fact that those polynomials are self-reciprocals $(a_0=a_n, a_1=a_{n-1}, a_2=a_{n-2}, ...)$, and the degree of $P_n$ is even; this ensures the existence of a unique polynomial $P_n^*$ called the "reciprocal transform" of $P_n$, concretely, if $deg(P_n)=2d$:
$$P_n(x) = x^{d} P_n^*\Big(x+\frac{1}{x}\Big)$$
The irreducibility of $P^*$ and $P$ are directly linked through Dikson's Theorem, which states (in the case of the Cauchy-Mirimanoff polynomials, with even degree) that $P$ is irreducible $\Leftrightarrow$ $P^*$ is irreducible.
Finally, Netwon polygons are used to proof that $P^*$ is irreducible.
The details can be found here:
https://trace.tennessee.edu/utk_graddiss/707/
|
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|
Evaluating $\sum\limits_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$ What is the value of $\displaystyle\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$?
|
Let's observe the relation with $\;\sin(2\pi j/3)$ and rewrite this in an elementary way :
\begin{align}
\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}&= \sum_{k=1}^{\infty}\frac 0{3k+0}+\frac 1{3k+1}-\frac 1{3k+2}\\
&= \frac 2{\sqrt{3}}\Im\left(\sum_{j=1}^\infty\frac {e^{2\pi\,i\,j/3}}{j}\right)-\frac 1{1\cdot 2}\\
&= -\frac 2{\sqrt{3}}\Im\,\ln\left(1-e^{2\pi\,i/3}\right)-\frac 12\\
&= -\frac 2{\sqrt{3}}\Im\,\ln\left(\frac 32-\frac{\sqrt{3}}2i\right)-\frac 12\\
&= -\frac 2{\sqrt{3}}\Im\,\ln\left(\sqrt{3}\;e^{-\pi\,i/6}\right)-\frac 12\\
&= -\frac 2{\sqrt{3}}\left(-\frac{\pi}6\right)-\frac 12\\
&= \frac {\sqrt{3}\,\pi}{9}-\frac 12\\
\end{align}
|
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|
Performing and simplifying equations
The expression is $$\frac {5}{x-2} - \frac 3{x+7} + 2$$
I need to simplify it too.
|
$$\frac {5}{x-2} - \frac 3{x+7} + 2$$
Find the common denominator!
$$\begin{align}\frac {5}{x-2} - \frac 3{x+7} + 2 & = \dfrac{5(x+7)}{(x - 2)(x+ 7)} - \dfrac{3(x - 2)}{(x-2)(x+7)} + \frac{2(x-2)(x+7)}{(x-2)(x+7)} \\ \\& =\dfrac{5(x+7) - 3(x-2) + 2(x-2)(x+7)}{(x - 2)(x+7)} \\ \\ \end{align}$$
Now, expand the numerator, factor it if you can, cancel any like factors, i.e., simplify.
|
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|
I need to calculate $x^{50}$ $x=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}$, I need to calculate $x^{50}$
Could anyone tell me how to proceed?
Thank you.
|
Try to prove by induction that:
$$
A^n=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}^n = \begin{pmatrix}1&0&0\\ \left\lceil\frac{n}{2}\right\rceil&n+1\bmod{2}&n\bmod{2}\\\left\lfloor\frac{n}{2}\right\rfloor&n\bmod{2}&n+1\bmod2\end{pmatrix}
$$
|
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|
Why $ \lim\limits_{n \to \infty} \frac{ (n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{1}{k+1} $? Can anyone explain the following equation?
$$\lim_{n \to \infty} \frac{ (n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{1}{k+1}$$
|
$$\frac{(n+1)^{k}}{(n+1)^{k+1}-n^{k+1}} = \frac{1}{n+1} \frac{1}{1-(1+1/n)^{-(k+1)}}$$
Taylor expand in the denominator in the RHS:
$$\frac{(n+1)^{k}}{(n+1)^{k+1}-n^{k+1}} \approx \frac{1}{n+1} \frac{1}{1-(1-(k+1)/n)}=\frac{n}{n+1} \frac{1}{k+1}$$
The result follows from taking the limit as $n\to\infty$.
|
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|
Can I get a hint on solving this recurrence relation? I am having trouble solving for a closed form of the following recurrence relation.
$$\begin{align*}
a_n &= \frac{n}{4} -\frac{1}{2}\sum_{k=1}^{n-1}a_k\\
a_1 &= \frac{1}{4}
\end{align*}$$
The first few values are $a_1=\frac{1}{4},a_2=\frac{3}{8},a_3=\frac{7}{16},a_4=\frac{15}{32}...$ So it seems the pattern is $a_n = \frac{2^{n}-1}{2^{n+1}}$, but I have been unable to show this algebraically. Here is what I tried:
$$\begin{align*}
2a_n &= \frac{n}{2} - \sum_{k=1}^{n-1}a_k\\
a_n + \sum_{k=1}^n a_k &= \frac{n}{2}\\
a_{n-1} + \sum_{k=1}^{n-1} a_k &= \frac{n-1}{2} \\
2a_n - a_{n-1} & = \frac{1}{2} \\
a_n = \dfrac{2a_{n-1} + 1}{4}
\end{align*}$$
I am so close, I can taste the closed form. Can someone nudge me in the right direction without giving too much away?
|
Subtract the formula for $a_{n-1}$ from that for $a_n$
$$
a_n-a_{n-1}=\frac14-\frac12a_{n-1}
$$
Multiply by $2^n$ and bring $a_{n-1}$ from the left to the right
$$
2^na_n=2^{n-2}+2^{n-1}a_{n-1}
$$
Using the formula for the sum of a geometric series, we get
$$
2^na_n=2^{n-1}+C
$$
Plug in $n=1$ to find that $C=-\frac12$. Thus,
$$
a_n=\frac12-\frac1{2^{n+1}}
$$
|
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|
Limit of $\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot[{(x+\sin{\frac{1}{x}})}^{\frac{1}{3}} -x^{\frac{1}{3}}]}$ I need to find the limit of
$$\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot\left[{\left(x+\sin{\frac{1}{x}}\right)}^{\frac{1}{3}} -x^{\frac{1}{3}}\right]}$$
|
Elementary hint:
Multiply the function by $$1=\frac{\sqrt[3]{(x+\sin(1/x))^2}+\sqrt[3]{x(x+\sin(1/x))}+\sqrt[3]{x^2}}{\sqrt[3]{(x+\sin(1/x))^2}+\sqrt[3]{x(x+\sin(1/x))}+\sqrt[3]{x^2}}$$ and use the basic well known fact that $a^3-b^3=(a-b)(a^2+ab+b^2)$ to get $1/3$.
|
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|
Find maximum and minimum value of function $f(x,y,z)=y+z$ on the circle Find maximum and minimum value of function $$f(x,y,z)=y+z$$ on the circle $$x^2+y^2+z^2 = 1,3x+y=3$$
We have that $$y=3-3x$$ So we would like find minimum and maximum value of function $$g(x,z)=3-3x+z$$ on $$x^2+ (3(1-x))^2 + z^2 = 1$$ And then I was using Lagrange multipliers. I received that $$z=\frac{9-10x}{3}$$ and next I substituted it to circle equal. But the numbers are awful.
This task comes from an exam so I suppose that is the easier way to solve it.
|
Another way. Given that you have $g = 3- 3x + z$ to find extrema, with the constraint
$1 = x^2 + 9(1-x)^2 + z^2 = (10x^2 -18x + 9) + z^2 = 10(x-\frac{9}{10})^2 + z^2 + \frac{9}{10}$
Thus the constraint is $100(x-\frac{9}{10})^2 + 10z^2 = 1$
Now we can let $x - \frac{9}{10} = \frac{\sin(t)}{10}$ and $z = \frac{\cos(t)}{\sqrt{10}}$ so that the constraint is satisfied.
Then our objective becomes $g = 3 - 3(\frac{9}{10}+\frac{\sin(t)}{10})+ \frac{\cos(t)}{\sqrt{10}}$.
$$\implies 10g-3 = - 3 \sin t + \sqrt{10} \cos t$$
Finding the extrema of this RHS should be easy using trigonometry or calculus, extrema of $a \sin t + b \cos t$ is $\pm \sqrt{a^2 + b^2}$. So we have
$$-\sqrt{19} \le 10g - 3 \le \sqrt{19}$$
|
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|
Find the roots of the given equation : $2^{x+2}.3^{\frac{3x}{x-1}} =9$ - Logarithm problem Find the roots of the given equation :
$2^{x+2}.3^{\frac{3x}{x-1}} =9$
My working :
Taking log on both sides we get : $$\log (2^{x+2}.3^{\frac{3x}{x-1}}) =\log 3^2 \Rightarrow (x+2)(\log2) + \frac{3x}{x-1}\log 3 = 2\log 3$$
Now how to proceed further in this problem... please suggest thanks....
|
As $\log_aa=1$ and $\log_a(a^m)=m\log_aa=m$
taking logarithm wrt $3,$
$$(x+2)\log_32+\frac{3x}{x-1}=2$$
$$\implies(x+2)\log_32=2-\frac{3x}{x-1}=-\frac{x+2}{x-1}$$
$$\implies (x+2)\left(\log_32+\frac1{x+1}\right)=0$$
We know if $a\cdot b\cdot c\cdots=0;$ at least one of $a,b,c,\cdots$ is zero
|
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|
Math contest proof equation problem Could someone help me with this?
If $m$ and $n$ are positive integers, then show that $$\frac{m}{
\sqrt n}+ \frac{m}{\sqrt[4]{n}} \neq 1$$.
|
$$\frac{m}{\sqrt n}+ \frac{m}{\sqrt[4]{n}} = 1 \implies \frac{m}{\sqrt n}\left(1+\sqrt[4]{n}\right) = 1 $$
Or $m(1+\sqrt[4]{n}) = \sqrt{n}$. Squaring, $m^2(1+\sqrt[4]{n})^2 = n$.
Suppose $n$ is not a perfect fourth power, so $r = \sqrt[4] n$ is irrational. Then we have $(1+r)^2 = N$, for some factor of $n$. So $r = \sqrt N -1$ and $N$ is not a perfect square.
However $r^4 = N^2+6N+1-4(1+N)\sqrt N= n$ makes $\sqrt N$ rational. Hence this case leads to contradiction.
So we need $n$ to be a perfect fourth power, say $n = k^4$. So $m^2(1+k)^2 = k^4$.
So $m(1+k) = k^2 \implies$ for all primes $p|(1+k)$, we must have $p | k$. This is not possible, hence we cannot have a solution.
|
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|
Effective method to solve $ \frac{x}{3x-5}\leq \frac{2}{x-1}$ I need to solve this inequality. How can I do so effectively?
$$ \frac{x}{3x-5}\leq \frac{2}{x-1}$$
|
Note that $3x - 5 = 0$ when $x = \frac{5}{3}$. When $x < \frac{5}{3}$, $3x - 5 < 0$, and when $x > \frac{5}{3}$, $3x - 5 > 0$.
Also, $x - 1 = 0$ when $x = 1$. When $x < 1$, $x - 1 < 0$, and when $x > 1$, $x - 1 > 0$.
Now you have three cases:
*
*$x < 1$,
*$1 < x < \frac{5}{3}$, and
*$x > \frac{5}{3}$.
In each case, multiply by $(3x - 5)(x - 1)$ which will can be arranged to give you a quadratic inequality. Keep in mind, you need to consider the sign of $(3x - 5)(x - 1)$. If you multiply an inequality by a positive number, the inequality sign remains the same. However, if you multiply by a negative number, the inequality sign reverses.
|
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|
Need help calculating integral $\int\frac{dx}{(x^2+4)^2}$ I need some help calculating the integral
$\int\frac{dx}{(x^2+4)^2}$
I tried integration by parts, but I could not arrive to an answer.
|
As an alternative, note that
$$\int \frac{dx}{(x^2+a^2)^2} = -\frac{1}{2 a} \frac{\partial}{\partial a} \int \frac{dx}{x^2+a^2} = -\frac{1}{2 a} \frac{\partial}{\partial a} \frac{\arctan{(x/a)}}{a}$$
Carry out the differentiation and substitute $a=2$; I get
$$\frac18 \frac{x}{x^2+4} + \frac{1}{16} \arctan{\frac{x}{2}}+C$$
|
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|
If $a,b,c > 0$ and $a+b+c = 1$, then $(\tfrac{1}{a} −1)(\tfrac{1}{b} −1)(\tfrac{1}{c} −1) \geq 8$
If $a, b, c$ are positive real numbers and $a+b+c = 1$, prove that
$$\left(\frac{1}{a} −1\right)\left(\frac{1}{b} −1\right)\left(\frac{1}{c} −1\right) \geq 8.$$
Thank you.
|
Hint: $\dfrac1a-1=\dfrac{b+c}{a}\ge 2\dfrac{\sqrt{bc}}{a}$.
|
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|
Formulas for the polygonal or figurate numbers? Here is an interesting formula for the reciprocal of the heptagonal numbers. Are there any other analogous formulas for the polygonal or figurate numbers?
$$
\sum_{n=1}^\infty \frac{2}{n(5n-3)} =\\
\frac{1}{15}{\pi}{\sqrt{25-10\sqrt{5}}}+\frac{2}{3}\ln(5)+\frac{{1}+\sqrt{5}}{3}\ln\left(\frac{1}{2}\sqrt{10-2\sqrt{5}}\right)+\frac{{1}-\sqrt{5}}{3}\ln\left(\frac{1}{2}\sqrt{10+2\sqrt{5}}\right)
$$
Reference: "Beyond The Basel Problem" L. Downey ,B. Ong ,J. Sellers
|
figurate numbers
$$ \sum_{n=1}^{\infty} \frac{1}{\dbinom{n+r-1}{r}} = \sum_{n=1}^{\infty} \frac{r!(n-1)!}{(n+r-1)!} $$
$$ = r\sum_{n=1}^{\infty} \frac{\Gamma(r) \Gamma(n)}{\Gamma(n +r )} = r\sum_{n=1}^{\infty} \beta(r,n) = r\sum_{n=1}^{\infty} \int_0^1 x^{n-1}(1 - x)^{r-1} \ dx = r\int_0^1 (1-x)^{r-1} \sum_{n=1}^{\infty} x^{n-1} \ dx $$
$$ = r\int_0^1 (1-x)^{r-2} \ dx = -r\left[ \frac{(1-x)^{r-1}}{r-1} \right]_0^1 = \frac{r}{r-1} $$
|
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|
How prove this inequality $\pi<\frac{\sin{(\pi x)}}{x(1-x)}\le 4$ let $x\in (0,1)$ show that
$$\pi<\dfrac{\sin{(\pi x)}}{x(1-x)}\le 4$$
I idea we know
$$\sin{x}<x$$
and
$$x(1-x)\le\dfrac{1}{4}$$
But not usefull for this problem
|
Let
$$y=\frac{\sin{(\pi x)}}{x(1-x)}$$
then
$$\dfrac{dy}{dx}=\dfrac{{\pi {x(1-x)}\cos(\pi x)-(1-2x)\sin(\pi x)}}{x^2(x-x^2)}$$
For the maximum/minimum values we need to know x that make
$${{\pi {x(1-x)}\cos(\pi x)-(1-2x)\sin(\pi x)}}=0$$
If$\quad$ $x=\frac{1}{2}$, we can easily find $\dfrac{dy}{dx}=0$
If$\quad$ $x\not=\frac{1}{2}$,
$$x(1-x)\pi \cos(\pi x)=(1-2x)\sin{\pi x}$$
so
$$x^2(1-x)^2\pi^2 \cos^2(\pi x)=(1-2x)^2\sin^2{\pi x}$$
so
$${\dfrac{\pi^2 x^2(1-x)^2}{(1-2x)^2}}\cos^2(\pi x)=\sin^2{\pi x}=1-\cos^2(\pi x)$$
so
$${\cos^2(\pi x)}{\biggl[1+\dfrac{\pi^2 x^2(1-x)^2}{(1-2x)^2}\biggr]}=1$$
If we put
$$g={\cos^2(\pi x)}{\biggl[1+\dfrac{\pi^2 x^2(1-x)^2}{(1-2x)^2}\biggr]-1}$$
then
$$g \gt0, \quad 0\lt x\lt \frac{1}{2}$$
$$g \lt0, \quad \frac {1}{2}\lt x\lt 1$$
$$g =0, \quad if\ x =0\ or\ 1$$
Because x $\in$ (0,1) only x=$\frac12$ will make $\dfrac{dy}{dx}=0$
and
$$\dfrac{dy}{dx} \gt 0 \quad if \quad 0\lt x \lt \frac12$$
$$\dfrac{dy}{dx} \lt 0 \quad if \quad \frac12 \lt x \lt 1$$
because the denominator of $\dfrac{dy}{dx}$ $$x^2(x-x^2) \gt 0 \quad when \quad x\in (0,1)$$
so, y has its maximum at y($\frac12$)=4
and approaches its minimum value as x approaches to 0 or 1
$$
\lim_{x \to 0}y=\lim_{x \to 1}y = \pi
$$
Therefore $$\pi \lt \frac{\sin{(\pi x)}}{x(1-x)} \le 4$$
|
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|
Proving $\gcd( m,n)$=1 If $a$ and $b$ are co prime and $n$ is a prime, show that:
$\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor unless $a+b$ is a multiple of $n$
Also enlighten me why $n$ has to be prime so that $\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor unless $a$+$b$ is a multiple of $n$?
|
For odd $n$, we have
$$\frac{a^n+b^n}{a+b} = a^{n-1} - a^{n-2}b +-\dotsb -ab^{n-2} + b^{n-1} = \sum_{k=0}^{n-1} (-1)^k a^{n-1-k}b^k.$$
The latter sum can be written as
$$\left(\sum_{k=0}^{n-2} (-1)^k(k+1)a^{n-2-k}b^k\right)(a+b) + (-1)^{n-1}nb^{n-1}$$
(verification by distributing $(a+b)$ over the sum is elementary). Thus we have
$$\gcd\left(\frac{a^n+b^n}{a+b}, a+b\right) = \gcd \left(nb^{n-1},a+b\right) = \gcd(n,a+b),$$
the last because $\gcd(b,a+b) = \gcd(b,a) = 1$.
So if $n$ is an odd prime, we have a common factor (namely $n$) if and only if $a+b$ is a multiple of $n$. For odd composite $n$, we can have common factors also if $a+b$ is not a multiple of $n$, it is sufficient that some prime factor of $n$ divides $a+b$.
For the even prime $n = 2$, in general $\dfrac{a^2+b^2}{a+b}$ is not an integer. For coprime $a,b > 0$, it is only an integer if $a = b = 1$.
|
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|
A necessary and sufficient on the co-efficients of a quadratic to give an integer Le $f(n) := an^2 + bn + c$ for all integers $n$, where $a$, $b$, $c$ are rational. What are the necessary and sufficient conditions on $a$, $b$, and $c$ such that $f(n)$ be an integer for all $n$?
|
We are told that $an^2+bn+c$ is an integer forevery integer value of $n$. Set $n=0$. This tells us that $c$ must be an integer.
Since $c$ is an integer, $an^2+bn+c$ is an integer for all $n$ if and only if $an^2+bn$ is an integer for all $n$.
Suppose that $an^2+bn$ is an integer for all $n$. Then in particular it is an integer for $n=1$ and for $n=-1$. Thus
$a+b$ is an integer, say $k$, and $a-b$ is an integer, say $l$.
From $a+b=k$, $a-b=l$ we get $a=\frac{k+l}{2}$ and $b=\frac{k-l}{2}$. There are two possibilities (i) $k$ and $l$ have the same parity, that is they are both even or both odd and (ii) $k$ and $l$ have opposite parities.
In Case (i), the numbers $a$ and $b$ are integers.
In Case (ii), the numbers $a$ and $b$ are each half of an odd integer.
Finally, we check that (i) if $a$ and $b$ are integers, then $an^2 +bn$ is an integer for all integers $n$. This is obvious.
We also check that (ii) if $a$ and $b$ are each half of an odd integer, then $an^2+bn$ is an integer for all $n$.
So $a=\frac{c}{2}$ and $b=\frac{d}{2}$ where $c$ and $d$ are odd integers.
We have then $an^2+bn=\frac{1}{2}n(cn+d)$. If $n$ is even, then certainly $\frac{1}{2}n(cn+d)$ is an integer. And if $n$ is odd, then $cn+d$ is even, so again $\frac{1}{2}n(cn+d)$ is an integer.
Conclusion: we have $an^2+bn+c$ is an integer for all integers $n$ if and only if $c$ is an integer and $a$ and $b$ are each integers, or each half of an odd integer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
what to do next recurrence relation when solving exponential function? find gernal solution of :$a_n = 5a_{n– 1} – 6a_{n –2} + 7^n$
Homogeneous solution:
$$a_n -5a_{n– 1} + 6a_{n –2} = 7^n$$
put $a_n=b^n$:
$$b^n -5b^{n– 1} + 6b^{n –2} =0
\\b^{n-2} (b^2-5b^{} + 6b) =0
\\b^2-5b^{} + 6b =0
\\(b-2)(b-3)=0\\
b=2,3$$
$$a^h_{(n)} = C_1 3^n+ C_2 2^n$$
Particular solution:
Since RHS is exponent so $a^p_{(n)} = da^n$
put $a^p_{(n)}$ in $a_n -5a_{n– 1} + 6a_{n –2} = 7^n$
$$da^n -5da^{n– 1} + 6da^{n –2} =7^n$$
|
Your computation of the general solution of the homogeneous equation $a_{n}=5a_{n-1}-6a_{n-2}$ is correct. Note that the homogeneous equation does not have $7^n$ on the right.
We want to find a particular solution of the full equation
$$a_{n}=5a_{n-1}-6a_{n-2}+7^n.\tag{1}$$
We look for a solution of the shape $a_n=(k)7^n$, where $k$ is a constant. Substituting in (1), we get
$$(k)7^n =(5k)7^{n-1}-(6k)7^{n-2}+7^n.$$
Dividing both sides by $7^{n-2}$, we get
$$49k=35k-6k+49.$$
Solve for $k$. We get $k=\frac{49}{20}$.
Thus the general solution of (1) is
$$C_1\cdot 3^n+C_2\cdot 2^n +\frac{1}{20}7^{n+2}.$$
|
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|
Difference of number of cycles of even and odd permutations Show that the difference of the number of cycles of even and odd permutations is $(-1)^n (n-2)!$, using a bijective mapping (combinatorial proof).
Suppose to convert a permutation from odd to even we always flip 1 and 2. Then the difference can be easily calculated.
Thus, if I can find the number of cycles of all even permutations of $[n]$ that have $\{1,2\}$ in the same cycle., I can compute the difference. Can somebody show me a way?
|
Here is an answer using generating functions that may interest you. Recall that the sign $\sigma(\pi)$ of a permutation $\pi$ is given by
$$\sigma(\pi) = \prod_{c\in\pi} (-1)^{|c|-1}$$
where the product ranges over the cycles $c$ from the disjoint cycle composition of $\pi$.
It follows that the combinatorial species $\mathcal{Q}$ that reflects the signs and the cycle count of the set of permutations is given by
$$\mathcal{Q} =
\mathfrak{P}(\mathcal{V}\mathfrak{C}_1(\mathcal{Z})
+\mathcal{U}\mathcal{V}\mathfrak{C}_2(\mathcal{Z}))
+\mathcal{U}^2\mathcal{V}\mathfrak{C}_3(\mathcal{Z})
+\mathcal{U}^3\mathcal{V}\mathfrak{C}_4(\mathcal{Z})
+\mathcal{U}^4\mathcal{V}\mathfrak{C}_5(\mathcal{Z})
+\cdots)$$
where we have used $\mathcal{U}$ to mark signs and $\mathcal{V}$ for the cycle count.
Translating to generating functions we have
$$Q(z, u, v) =
\exp\left(v\frac{z}{1}
+ vu\frac{z^2}{2}
+ vu^2\frac{z^3}{3}
+ vu^3\frac{z^4}{4}
+ vu^4\frac{z^5}{5}
+\cdots\right).$$
This simplifies to
$$Q(z,u,v) = \exp\left(\frac{v}{u}
\left(
\frac{zu}{1}
+ \frac{z^2 u^2}{2}
+ \frac{z^3 u^3}{3}
+ \frac{z^4 u^4}{4}
+ \frac{z^5 u^5}{5}
+ \cdots
\right)\right) \\
= \exp\left(\frac{v}{u} \log \frac{1}{1-uz}\right)
= \left(\frac{1}{1-uz}\right)^{\frac{v}{u}}.$$
Now the two generating functions $Q_1(z, v)$ and $Q_2(z, v)$ of even and odd permutations by cycle count are given by
$$Q_1(z,v) = \frac{1}{2} Q(z,+1,v) + \frac{1}{2} Q(z,-1,v) =
\frac{1}{2}\left(\frac{1}{1-z}\right)^v
+\frac{1}{2}\left(\frac{1}{1+z}\right)^{-v}$$ and
$$Q_2(z,v) = \frac{1}{2} Q(z,+1,v) - \frac{1}{2} Q(z,-1,v) =
\frac{1}{2}\left(\frac{1}{1-z}\right)^v
-\frac{1}{2}\left(\frac{1}{1+z}\right)^{-v}.$$
We require the quantity
$$G(z, v) = \left.\frac{d}{dv} (Q_1(z,v)-Q_2(z,v))\right|_{v=1} \\=
\left.\frac{d}{dv} \left(\frac{1}{1+z}\right)^{-v}\right|_{v=1} =
- \left.\log \frac{1}{1+z} \left(\frac{1}{1+z}\right)^{-v}\right|_{v=1}
= -(1+z)\log \frac{1}{1+z}.$$
Finally extracting coeffcients from this generating function we obtain
$$- n! [z^n] (1+z)\log \frac{1}{1+z}
= - n! \left(\frac{(-1)^n}{n} + \frac{(-1)^{n-1}}{n-1}\right)
\\= - n! (-1)^{n-1} \left(-\frac{1}{n} + \frac{1}{n-1}\right)
= n! (-1)^n \frac{n-(n-1)}{n(n-1)}
\\= n! (-1)^n \frac{1}{n(n-1)} = (-1)^n (n-2)!$$
This concludes the proof. I do think this is rather pretty.
|
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|
How to evaluate $\int_{0}^{+\infty}\exp(-ax^2-\frac b{x^2})\,dx$ for $a,b>0$ How can I evaluate
$$I=\int_{0}^{+\infty}\!e^{-ax^2-\frac b{x^2}}\,dx$$
for $a,b>0$?
My methods:
Let $a,b > 0$ and let
$$I(b)=\int_{0}^{+\infty}e^{-ax^2-\frac b{x^2}}\,dx.$$
Then
$$I'(b)=\int_{0}^{\infty}-\frac{1}{x^2}e^{-ax^2-\frac b{x^2}}\,dx.$$
What the other methods that can I use to evaluate it? Thank you.
|
$$\begin{align}
I
= & \int_0^{\infty} e^{-ax^2 - bx^{-2}} dx\\
\stackrel{\color{blue}{[1]}}{=} &
\left(\frac{b}{a}\right)^{1/4}\int_0^{\infty} e^{-\sqrt{ab}(y^2 + y^{-2})} dy\\
= &
\left(\frac{b}{a}\right)^{1/4}\left[ \int_0^{1} + \int_1^{\infty} \right] e^{-\sqrt{ab}(y^2 + y^{-2})} dy\\
\stackrel{\color{blue}{[2]}}{=} &
\left(\frac{b}{a}\right)^{1/4} \int_1^{\infty} e^{-\sqrt{ab}(y^2 + y^{-2})}
\left(\frac{1}{y^2} + 1\right) dy\\
= &
\left(\frac{b}{a}\right)^{1/4} \int_1^{\infty} e^{-\sqrt{ab}((y-y^{-1})^2+2)}
d\left( y - \frac{1}{y}\right)\\
\stackrel{\color{blue}{[3]}}{=} &
\left(\frac{b}{a}\right)^{1/4} e^{-2\sqrt{ab}} \int_0^{\infty} e^{-\sqrt{ab}\,z^2} dz\\
= &
\left(\frac{b}{a}\right)^{1/4} e^{-2\sqrt{ab}} \frac{\sqrt{\pi}}{2(ab)^{1/4}}\\
= &
\sqrt{\frac{\pi}{4a}} e^{-2\sqrt{ab}}
\end{align}
$$
Notes
*
*$\color{blue}{[1]}$ substitute $x$ by $y = \sqrt{\frac{a}{b}} x$.
*$\color{blue}{[2]}$ substitute $y$ by $\frac{1}{y}$ over the interval $[0,1]$.
*$\color{blue}{[3]}$ substitute $y$ by $z = y - \frac{1}{y}$.
|
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|
How prove this $ b^2c^2+abc(b+c)+a(b^3+c^3)+a^3(b+c)\ge 2a^2(b^2+c^2+bc)$ let $a,b,c\ge 0$, show that
$$ b^2c^2+abc(b+c)+a(b^3+c^3)+a^3(b+c)\ge 2a^2(b^2+c^2+bc)$$
my idea use the SOS methods, But I don't work at last. Thank you
|
Since this expression is symmetric in $b$ and $c$, we can make the substitution $x=b+c$, $y=bc$ to obtain the equivalent inequality
$$
y^2+axy+a(x^3-3xy)+a^3x \geq 2a^2(x^2-y) \, ;
$$
the condition $b,c \geq 0$ is equivalent to requiring $x,y \geq 0$.
This inequality is quadratic in $y$. By collecting terms, we see that it suffices to show that
\begin{eqnarray}
f_{x,a}(y):&=&y^2+y(ax-3ax+2a^2)+ax^3+a^3x-2a^2x^2\\
&=&y^2-2ya(x-a)+ax(x-a)^2
\end{eqnarray}
is non-negative whenever $a,x,y$ are.
For fixed $x,a$, the function $f$ is minimized when $y=a(x-a)$. There are two possible cases:
*
*If $a>x$, this minimum value lies outside the domain of consideration. Since $f$ is a quadratic with positive leading coefficient, it is minimized on the domain of consideration at its boundary; that is, when $y=0$. But $a,x \geq 0$, which means $f_{x,a}(0)=ax(x-a)^2 \geq 0$.
*If $a \leq x$, $f$ is minimized on the domain of consideration at $y=a(x-a)$. Moreover,
\begin{eqnarray}
f_{x,a}(a(x-a))&=&a^2(x-a)^2-2a^2(x-a)^2+ax(x-a)^2\\
&=& a(x-a)^3
\end{eqnarray}
which is non-negative as $a \geq 0$ and $x \geq a$.
So, for any choices of $x$ and $a$, $f$'s minimum value in $y$ is non-negative. Thus $f$ is non-negative on the entire domain of consideration.
|
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|
How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$? Well, the original task was to figure out what the following expression evaluates to for any $n$.
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}$$
By trying out different values of $n$, I found the pattern:
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n} = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$$
But I have yet to figure out how to prove it algebraically.
Suggestions?
|
$$
\left(%
\begin{array}{cc}
1 & 1
\\
0 & 1
\end{array}\right)
=
1 + A
\quad\mbox{where}\quad
A
=
\left(%
\begin{array}{cc}
0 & 1
\\
0 & 0
\end{array}\right)
$$
Notice that $A^{2} = 0$. It means that a function ${\rm f}\left(A\right)$ is
${\it linear}$ in $A$. For instance,
$\left(1 + \mu A\right)^{n} = \alpha + \beta A$. Also
$$
nA\left(1 + \mu A\right)^{n -1} = \alpha' + \beta' A
\quad\Longrightarrow\quad
nA\left(\alpha + \beta A\right) = \left(1 + \mu A\right)\left(\alpha' + \beta' A\right)
=
\alpha' + \beta' A + \mu\alpha' A
$$
$$
\mbox{That's means}\quad
0 = \alpha'\,,\quad n\alpha = \beta' + \mu\alpha'= \beta'\,,
\quad\mbox{with}\quad
\left.\alpha\right\vert_{\mu\ =\ 0} = 1\
\mbox{and}\
\left.\beta\right\vert_{\mu\ =\ 0} = 0\
$$
Then, $\alpha = 1$ and $\beta = n\mu$:
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\left(1 + \mu A\right)^{n} \color{#000000}{\ =\ } 1 + n\mu A
\quad\color{#000000}{\Longrightarrow}\quad
\left(1 + A\right)^{n} \color{#000000}{\ =\ } 1 + nA
\quad}
\\ \\ \hline
\end{array}
$$
Or 'easy way':
$$
\left(1 + A\right)^{n}
=
1\ +\ \overbrace{\quad{n \choose 1}\quad}^{=\ n}\ A\
+\
\overbrace{\quad{n \choose 2}\,A^{2} + \cdots\quad}^{=\ 0}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/499646",
"timestamp": "2023-03-29T00:00:00",
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|
Integrals of $\sqrt{x+\sqrt{\phantom|\dots+\sqrt{x+1}}}$ in elementary functions Let $f_n(x)$ be recursively defined as
$$f_0(x)=1,\ \ \ f_{n+1}(x)=\sqrt{x+f_n(x)},\tag1$$
i.e. $f_n(x)$ contains $n$ radicals and $n$ occurences of $x$:
$$f_1(x)=\sqrt{x+1},\ \ \ f_2(x)=\sqrt{x+\sqrt{x+1}},\ \ \ f_3(x)=\sqrt{x+\sqrt{x+\sqrt{x+1}}},\ \dots\tag2$$
The functions $f_0(x)$, $f_1(x)$ and $f_2(x)$ are integrable in elementary functions, e.g.:
$$\int\sqrt{x+\sqrt{x+1}}\,dx=\left(\frac{2\,x}3+\frac{\sqrt{x+1}}6-\frac14\right)\sqrt{x+\sqrt{x+1}}+\frac58\ln\left(2\,\sqrt{x+1}+2\,\sqrt{x+\sqrt{x+1}}+1\right).\tag3$$
Question: Is there an integer $n>2$ such that $f_n(x)$ is integrable in elementary functions?
Update: The question is reposted at MathOverflow as was suggested by moderator.
|
I think there is highly have no chance to have an integer $n>2$ such that $f_n(x)$ is integrable in elementary functions.
The reasons are mainly because of the processes of eliminating the radicals.
For $f_0(x)$ and $f_1(x)$ , they are trivially integrable in elementary functions.
For $\int f_2(x)~dx$ ,
Let $u=\sqrt{x+1}$ ,
Then $x=u^2-1$
$dx=2u~du$
$\therefore\int\sqrt{x+\sqrt{x+1}}~dx$
$=\int2u\sqrt{u^2+u-1}~du$ , which can express in elementary functions.
Starting from $\int f_3(x)~dx$ ,
Let $u=\sqrt{x+1}$ ,
Then $x=u^2-1$
$dx=2u~du$
$\therefore\int\sqrt{x+\sqrt{x+\sqrt{x+1}}}~dx$
$=\int2u\sqrt{u^2-1+\sqrt{u^2+u-1}}~du$
Introduce the Euler substitution:
Let $v=u+\sqrt{u^2+u-1}$ ,
Then $u=\dfrac{v^2+1}{2v+1}$
$du=\dfrac{2v(2v+1)-(v^2+1)2}{(2v+1)^2}dv=\dfrac{2v^2+2v-2}{(2v+1)^2}dv$
$\therefore\int2u\sqrt{u^2-1+\sqrt{u^2+u-1}}~du$
$=\int2\dfrac{v^2+1}{2v+1}\sqrt{\left(\dfrac{v^2+1}{2v+1}\right)^2-1+v-\dfrac{v^2+1}{2v+1}}\dfrac{2v^2+2v-2}{(2v+1)^2}dv$
$=\int4\dfrac{(v^2+1)(v^2+v-1)}{(2v+1)^3}\sqrt{\dfrac{(v^2+1)^2-(v^2+1)(2v+1)+(v-1)(2v+1)^2}{(2v+1)^2}}~dv$ , which is highly have no chance to express in elementary functions.
|
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|
How to solve equation? How to solve this equation in the set of real numbers?
$$(x^{2}+3y^{2}-7)^{2} + \sqrt{3-xy-y^2}=0$$
I tried to solve $x^{2}+3y^{2}-7=0$ and $\sqrt{3-xy-y^2}$=0 for x. But it did not help.
|
As you stated, we must have $x^2 + 3y^2 = 7 $ and $3 = xy + y^2$.
This is an ellipse intersecting a hyperbola, so there are at most 4 points of intersection.
Substtuting $ x = \frac{3-y^2}{y}$ into the first equation, we get a quartic equation
$$ (3-y^2) + 3y^4 = 7y^2. $$
This has solutions $y = -\frac{3}{2}, -1, 1, \frac{3}{2}$.
Hence, this gives us the 4 points of intersection
$$ ( -\frac{1}{2}, - \frac{3}{2} ), ( -2, -1), (2, 1), ( \frac{1}{2}, \frac{3}{2}).$$
|
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|
What is the smallest value of $x^2+y^2$ when $x+y=6$? If $ x+y=6 $ then what is the smallest possible value for $x^2+y^2$?
Please show me the working to show where I am going wrong!
Cheers
|
Reorder $x+y = 6$ to
$y=6-x$
Substituting $y$ in $(x^2+y^2)$ yields
$x^2+(6-x)^2 = 2x^2-12x+36$
The minimum occurs where the derivative equals $0$
$4x -12 = 0$
Therefore at the minimum,
$x=3$
Hence the minimum is $2\cdot 3^2-12\cdot 3+36 = 18$
|
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|
Polynomial long division: different answers when reordering terms When I use polynomial long division to divide $\frac{1}{1-x}$, I get $\;1 + x + x^2 +x^3 + x^4 + \cdots$
But when I just change the order of terms in the divisor: $\frac{1}{-x+1}$, the long division algorithm gives me a very different answer: $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} - \frac{1}{x^4} - \cdots$, which seems somewhat strange to me, because sums are supposed to be commutative, so it looks that these two answers should be equivalent. Am I right?
But I cannot see how could these two answers be equivalent. Could it be true that $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} - \frac{1}{x^4} - \cdots \;=\; 1 + x + x^2 +x^3 + x^4 + \cdots$ ?
If that is the case, how can i "prove" it? Or how can I transform one form into the other?
I already figured out that I can express $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} - \frac{1}{x^4} - \cdots$ as negative exponents: $-x^{-1} - x^{-2} - x^{-3} - x^{-4} - \cdots$ and then factor the $-1$'s inside the exponents: $-x^{-1\cdot1} - x^{-1\cdot2} - x^{-1\cdot3} - x^{-1\cdot4} - \cdots$ which can be seen as a power of a power: $-(x^{-1})^1 - (x^{-1})^2 - (x^{-1})^3 - (x^{-1})^4 - \cdots$ and the minus sign can be factored out too: $-\left[ (x^{-1})^1 + (x^{-1})^2 + (x^{-1})^3 + (x^{-1})^4 + \cdots \right]$ which almost gives me the form of the other series with raising positive exponents. But there's that dangling minus sign in front of it, and the 0th power term is missing :-/ so I don't know how to massage it any further to get the other form of the expansion. Any ideas?
Edit:
I know about Taylor series expansions and I can expand $\frac{1}{1-x}$ this way into $1 + x + x^2 + x^3 + \cdots$ (I don't know how to get the other form this way, though).
Also, I know about the convergence condition for the infinite series, $|x| < 1$ in the case of the first answer, and $|x| > 1$ or $|\frac{1}{x}| < 1$ in the case of the second answer. I just didn't expect that this could matter in my case, when I just reorder the terms in the divisor, which shouldn't produce different answer.
I suspected that this could be somehow related to the fact that raising powers of a fraction is the same as falling powers of its inverse. But I cannot figure out why just one of these or the other is given, depending on the order of terms in the long division, and this is what confuses me the most.
|
Seeing as you tagged this sequences and series, I presume you know that you have to be careful when talking about adding up an infinite sequence of terms.
There is an infinite series hidden in your work when you write
$$
\frac{1}{1-x}=1+x + x^2 +x^3 + \cdots,
$$
but this series only converges when $|x|<1$. This sum is the Taylor series for $\frac{1}{1-x}$.
There is an infinite series hidden in your work when you write
$$
\frac{1}{1-x} = -\frac{1}{x}- \frac{1}{x^2}-\frac{1}{x^3}- \cdots,
$$
and this infinite series only converges when $|x|>1$ (or, equivalently, when $\left | \frac{1}{x} \right|<1$). This sum is the Laurent series for $\frac{1}{1-x}$.
So both expressions are correct (for the right values of $x$) but they are not both correct at the same time. One is valid for $|x|<1$ and the other for $|x|>1$.
Edited to add:
Just to clarify the situation a little further:
You are correct when you say that dividing by $-x+1$ should be the same as dividing by $1-x$. There is no problem with this as long as we stop after finitely many steps:
If we stop after three steps in the polynomial long division algorithm, we get
$$
\frac{1}{1-x}=1+x +x^2 +\frac{x^3}{1-x}.
$$
That is, we get an answer of $1+x + x^2$ with a remainder of $x^3$ that has not yet been divided by $1-x$.
Or if we do it the other way, after three steps, we get
$$
\frac{1}{1-x}= - \frac{1}{x}-\frac{1}{x^2}-\frac{1}{x^3} +\frac{1}{x^3}\times \frac{1}{1-x}.
$$
That is, we get an answer of $- \frac{1}{x}-\frac{1}{x^2}-\frac{1}{x^3}$ with a remainder of $\frac{1}{x^3}$ that has not yet been divided by $1-x$.
Let's assume that $x \neq 0$ and $x \neq 1$ so that everything is well-defined. Then
$$
\frac{1}{1-x}=1+x +x^2 +\frac{x^3}{1-x} =- \frac{1}{x}-\frac{1}{x^2}-\frac{1}{x^3} +\frac{1}{x^3}\times \frac{1}{1-x}.
$$
Both ways of doing it do give us the same answer.
So what's the problem? When we try continuing this to infinitely many terms, we need these infinite series to converge to the correct answer. It is not the order $-x+1$ versus $1-x$ that leads to different answers; it is the change from writing down a sum with finitely many terms to writing down a series with infinitely many terms that we have to be careful about.
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"url": "https://math.stackexchange.com/questions/505817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Integrate $\int x \sqrt{2 - \sqrt{1-x^2}}dx $ it seems that integration by parts with some relation to substitution...
$$
\int x \sqrt{2-\sqrt{1-x^2}}
= \frac{2}{5} \sqrt{2-\sqrt{1-x^2}} \cdot \sqrt{1-x^2}+\frac{8}{15}\sqrt{2-\sqrt{1-x^2}}+c
$$
How can I get that?
|
You don't need any integrations by parts for this. As others have suggested, if you substitute $u^2 = 1 -x^2$ and $2u\,du = -2x\,dx$ the integral becomes
$$-\int u\sqrt{2 - u}\,du$$
Now do another substitution $v = 2 - u$ and the integral becomes
$$\int (2-v)\sqrt{v}\,dv$$
$$= \int 2v^{1 \over 2} - v^{3 \over 2}\,dv$$
$$= {4 \over 3} v^{3 \over 2} - {2 \over 5} v^{5 \over 2} + C$$
Substituting back $v = 2 - u = 2 - \sqrt{1 - x^2}$ this becomes
$${4 \over 3} (2 - \sqrt{1 - x^2})^{3 \over 2} - {2 \over 5} (2 - \sqrt{1 - x^2})^{5 \over 2} + C$$
This is the final answer.
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"timestamp": "2023-03-29T00:00:00",
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|
Prove the Trigonometric Identity: $\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}$ I'm doing some math exercises but I got stuck on this problem.
In the book of Bogoslavov Vene it says to prove that:
$$\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}.$$
It is easy if we do it like this: $\sin^2 x = (1-\cos x)(1+\cos x)=1-\cos^2 x$.
But how to prove it just by going from the left part or from the right
Can anybody help me?
Thank you!
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Note that we must have that $\cos x \neq 1$, so x cannot be any even multiple of $\pi$. Similarly, $\sin x \neq 0$, so x cannot be any multiple of $\pi$.
Why do I note the above?
Because we have an identity, $$\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}$$ which is senseless if $\cos x = 1,$ and/or $\sin x = 0$.
$\require{cancel}$
Since $x$ cannot be any multiple of $\pi$, we can multiply numerator and denominator by $1+ \cos x$, because we already ruled out $x$ = integer multiple of $\pi$.
We also use the Pythagorean Identity $$\sin^2x + \cos^2 x =1 \iff 1 - \cos^2 x = \sin^2 x$$
$$\begin{align}
\frac{\sin(x)}{(1-\cos(x))}&= \frac{\sin(x)}{(1-\cos(x))}\cdot \frac{(1+\cos(x))}{(1+\cos(x))} \\ \\ & = \frac{\sin x(1 + \cos x)}{\underbrace{1- \cos^2 x}_{(a + b)(a - b) = a^2 - b^2}} \\ \\
& = \frac {\sin x(1 + \cos x)}{\sin^2 x} \\ \\
& = \frac{\cancel{\sin x}(1 + \cos x)}{\cancel{\sin x}\cdot \sin x} \\ \\
& = \frac{1 +\cos x}{\sin x}\end{align}$$
for all $x \neq k\pi, k\in \mathbb Z$.
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"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
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