Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
How to add constraints to vectors Is it possible to add constraints to vectors, more specifically, planes? For instance:
In this example, is it possible to constrain the red plane with Cartesian equation $z=0$ to only occupy within the area within the area of the yellow lines, such that only the green line intersects the plane and not the purple one?
|
So you want to parametrize the plane such that a point on the plane is defined by two parameters $$ \vec{r} = \mathbf{ S}(u,v) $$
and then set limits to $u$ and $v$ to define your bounds. In case of a plane, you need a local coordinate system with two mutually orthogonal directions $\mathbf{e}_1$ and $\mathbf{e}_2$ that are also orthogonal to the plane normal $ \mathbf{n} \cdot \mathbf{e}_i = 0$. The origin center is at $\vec{r}_0$.
$$ \vec{r} = \vec{r}_0+ \mathbf{e}_1 u + \mathbf{e}_2 v $$
Now the question becomes: how to set a local coordinate system on a plane defined by the equation $ax +by + cz+d =0 $
The plane normal is $$\mathbf{n} = \frac{1}{\sqrt{a^2+b^2+c^2}} \pmatrix{a\\b\\c}$$ and the point on the plane closest to the origin is
$$ \vec{r}_0 = \pmatrix{ \frac{-a d}{a^2+b^2+c^2} \\ \frac{-b d}{a^2+b^2+c^2} \\ \frac{-c d}{a^2+b^2+c^2} } $$
The choice of local unit vector directions is somewhat arbitrary, but it makes sense to use something related to the bounds you want to define. The only requirement for $\mathbf{e}_1 = \pmatrix{e_x & e_y & e_z}$ is that it has $e_x a + e_y b + e_z c = 0$
One way to achieve this is by finding the direction closest to the x-axis
$$\mathbf{e}_1 = \pmatrix{ \frac{b^2+c^2}{a^2+b^2+c^2} \\ \frac{-a b}{a^2+b^2+c^2} \\ \frac{-a c}{a^2+b^2+c^2} }$$
or the y-axis
$$\mathbf{e}_1 = \pmatrix{ \frac{-a b}{a^2+b^2+c^2} \\ \frac{a^2+c^2}{a^2+b^2+c^2} \\ \frac{-b c}{a^2+b^2+c^2} }$$
or the z-axis
$$\mathbf{e}_1 = \pmatrix{ \frac{-a c}{a^2+b^2+c^2} \\ \frac{-b c}{a^2+b^2+c^2} \\ \frac{a^2+b^2}{a^2+b^2+c^2} }$$
and finally, define the second direction from the cross product of $\mathbf{n}$ and the first direction
$$ \mathbf{e}_2 = \text{unitvector}( \mathbf{n} \times \mathbf{e}_1 ) $$
Summary
With the above scheme you can go from $u$ and $v$ to a point on the plane $\vec{r}=\mathbf{S}(u,v)$. Your limits are set in terms of $(u,v)$ as planar coordinates. For example $u=1$ will produce a line on the plane which is offset from the plane origin by one unit along the local v direction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3321078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Calculus - indefinite integration The integral in which I am interested in is
$$\int x(x^3+1)^{33}\mathrm{d}x$$
I tried to solve by substituting $x^2 = t$, but it didn't help. I find a solution by expanding it with the help of binomial expansion. Can anyone help me with any other method like substitution, by parts?
|
This is not easier than expanding using the Binomial theorem, but it's a different way to approach it which you may at least find interesting, and even potentially useful (in other situations if not this one).
For any integer $n \ge 0$, let
$$f(n) = \int x(x^3 + 1)^n dx \tag{1}\label{eq1}$$
For $n \ge 1$, using integration by parts, where $u(x) = (x^3 + 1)^n$ so $d(u(x)) = 3nx^2(x^3 + 1)^{n-1}dx$, and $d(v(x)) = xdx$ so $v(x) = \frac{x^2}{2}$, you get
$$\begin{equation}\begin{aligned}
f(n) & = \frac{x^2}{2}(x^3 + 1)^n - \frac{3n}{2} \int x^4(x^3 + 1)^{n-1} dx \\
& = \frac{x^2}{2}(x^3 + 1)^n - \frac{3n}{2} \int x(x^3 + 1 - 1)(x^3 + 1)^{n-1} dx \\
& = \frac{x^2}{2}(x^3 + 1)^n - \frac{3n}{2} \int \left(x(x^3 + 1)^n - x(x^3 + 1)^{n-1}\right) dx \\
& = \frac{x^2}{2}(x^3 + 1)^n - \frac{3n}{2} \left(f(n) - f(n-1)\right)
\end{aligned}\end{equation}\tag{2}\label{eq2}$$
This leads to the recursive equation
$$\begin{equation}\begin{aligned}
\left(1 + \frac{3n}{2}\right)f(n) & = \frac{x^2}{2}(x^3 + 1)^n + \frac{3n}{2}f(n-1) \\
f(n) & = \frac{x^2}{2 + 3n}(x^3 + 1)^n + \frac{3n}{3n + 2}f(n-1)
\end{aligned}\end{equation}\tag{3}\label{eq3}$$
You can determine what $f(0)$ is (I'm leaving that to you) and then use \eqref{eq3} to determine each of the rest of the $f$ values up to $f(33)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3322474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Evaluate $\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}$ Evaluate $\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}$.
My approach is to do $\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}} = \lim_{n\to\infty} \frac{\left(\frac{9}{4}\right)^{n}}{\frac{3}{2}\left(\frac{2}{3}\right)^{n}+\frac{3}{2}\left(\frac{9}{4}\right)^{n}}$ but not sure what to do next.
How could I convert the $\left(\frac{2}{3}\right)^{n}$ term into $\left(\frac{9}{4}\right)^{n}$? Thanks.
|
A good general rule: divide numerator and denominator by the largest term:
$$\frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}
=\frac{\frac{2}{3}}{\left(\frac{2}{3}\right)^{3n}+1}$$
and I think you should now be able to see what happens as $n\to\infty$.
Good luck!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3322677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
How to isolate $x$ when $\cos(x/2)\cos(3x)/2−3\sin(x/2)\sin(3x)= 0$ Asked to find all relative and absolute extrema of $f(x) = \sin\left(\frac{1}{2}x\right)\cos(3x)$ on the interval $[0,\pi]$. I've gotten the derivative, $$f'(x)=\frac{\cos(\frac{x}{2})\cos(3x)}{2} - 3\sin(\frac{x}{2})\sin(3x),$$ just fine, but how would I proceed to isolate x after this? I noticed that the situation is similar to the sine angle sum identity, but I don't know if there's an identity for an equation with the form $$h\cos(\alpha)\cos(\beta) - k\sin(\alpha)\sin(\beta)$$ and can't find anything online. I don't have much experience working with Euler's Formula in this context but know enough that I could probably follow an explanation based on it.
|
Expanding and varying my own comment:
$$\begin{split}
f'(x)&=\tfrac{1}{2}\cos\tfrac{x}{2}\cos 3x - 3\sin\tfrac{x}{2}\sin 3x\\
&=\tfrac{1}{2}\cos\tfrac{x}{2}\left(\cos 3x - 6\frac{\sin\tfrac{x}{2}\sin x}{\cos\tfrac{x}{2}} \frac{\sin 3x}{\sin x}\right)\\
&=\tfrac{1}{2}\cos\tfrac{x}{2}\left(\cos 3x - 6(1-\cos x) \frac{\sin 3x}{\sin x}\right)\\
&=\tfrac{1}{2}\cos\tfrac{x}{2}\left(T_3(\cos x) - 6(1-\cos x) U_2(\cos x)\right)\\
\end{split}$$
where
$$\begin{align}
T_3(y)&=4y^3-3y \\
U_2(y)&=4y^2-1\text{.}
\end{align}$$
Thus,
$$f'(x)=\tfrac{1}{2}\cos\tfrac{x}{2}g(\cos x)$$
where
$$g(y)=28y^3-24y^2-9y+6\text{.}$$
Now, we "cheat" a bit and use a CAS to find an optimized representation for $g(y)$. Let
$$y=\frac{z+2}{2(z+4)}\text{.}$$
Then
$$g(y)=-\frac{h(z)}{(z+4)^3}$$
where
$$h(z)=z^3-66 z -172\text{.}$$
Using the trigonometric solution to the cubic, we find
$$\begin{align}
z&=2\sqrt{22}\cos\theta&\cos 3\theta&=\frac{43}{11\sqrt{22}}\text{.}
\end{align}$$
That is,
$$\begin{align}x&=\arccos\frac{1+\sqrt{22}\cos\theta}{2(2+\sqrt{22}\cos\theta)} \\
\theta&=\tfrac{1}{3}\arccos\frac{43}{11\sqrt{22}}+\frac{2k\pi}{3}\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3323680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Evaluate $ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx $ $$ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx = ? $$
Attempt:
$$ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx = \int \frac{3x^{2}}{\sqrt{x^{3}-6} \sin^{2}\sqrt{x^{3}-6}} dx $$
$$ U = \sqrt{x^{3}-6} $$
then the integral is
$$ \int \frac{2U}{\sin^{2} U} dU $$
then using partial integration we will get the above equals:
$$ -2 U \cot U + \ln(\sin U) + C = -2 \sqrt{x^{3}-6} \cot(\sqrt{x^{3}-6}) + \ln (\sin \sqrt{x^{3}-6}) + C$$
But...
$$ \frac{d (-2 \cot(\sqrt{x^{3}-6}))}{dx} = \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} $$
|
When you substitute $U=\sqrt{x^3-6}$, you made a mistake.
It should be $$\int{\dfrac{2U}{U\sin^2{U}}dU}=\int{2\csc^2{U}dU}$$, so the answer is
$$-2\cot^2{\sqrt{x^3-6}}+C$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3323749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$. If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$.
How to prove it without using calculus?
I know if $a , b,c,d \in \mathbb R$ then $a^2+b^2+c^2+d ^2$ will be minimum when $a =b= c =d = \frac{4m+1}{4}$..So the minimum value would have been $4m^2 +2m +1/4$..But what to do in that case?
|
Wlog. $a\le b\le c\le d$. If $d\ge a+2$, then
$$ (a+1)^2+b^2+c^2+(d-1)^2=a^2+b^2+c^2+d^2+2(1+d-a)<a^2+b^2+c^2+d^2,$$
hence for a minimizer $d\le a+1$. If $k$ of the numbers $a,b,c,d$ are $=a+1$, then $4m+1=a+b+c+d=4a+k$. We conclude $k=1$, and then $a=b=c=m$ and $d=m+1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3324144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Verify the following limit using epsilon-delta definition: $ \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0$ Show that $$ \lim\limits_{(x,y)\to(0,0)}\dfrac{x^2y^2}{x^2+y^2}=0$$
My try:
We know that, $$ x^2\leq x^2+y^2 \implies x^2y^2\leq (x^2+y^2)y^2 \implies x^2y^2\leq (x^2+y^2)^2$$
Then, $$\dfrac{x^2y^2}{x^2+y^2}\leq x^2+y^2 $$
So we chose $\delta=\sqrt{\epsilon}$
|
Tips
$\lim\limits_{\left(x,y\right)\rightarrow\left(0,0\right)} \dfrac{x^2 y^2}{x^2+y^2} = \lim\limits_{\left(x,y\right)\rightarrow\left(0,0\right)} \dfrac{1}{\frac{1}{x^2}+\frac{1}{y^2}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3325340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
}
|
Find the sum of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$ up to $n$ terms
Find the sum of first $n$ terms of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$
*
*When $n$ is even.
*When $n$ is odd.
This sum can be written as
$$\sum_{1}^n (2k-1)^{3} +3 \sum_{1}^n (2k)^{2} $$
I can handle the sum up to n terms when it is not specified that $n$ is even or odd.
In this problem I'm confused, what changes should be done to get sum for even or odd $n$.
In my textbook, $n$ is replaced by $2m$ and then they solved the problem for first $m$ terms and then substituted $m = n/2$ and same is done for odd case, by substituting $n=2m-1$.
I didn't get that solution.
Any suggestion would be helpful.
|
HINT
When $n = 2m$ is even, both sums have the same amount of terms, $n/2 = m$ each.
When $n = 2m-1$ is odd, the left sum has one more term than the right, so there must be $m$ terms in the left and $m-1$ in the right.
Also notice that the even $n$ sum and the odd $n$ sum are different by just one last term in the right sum.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3327581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Discrete Mathematics - Combinatorics proof $\\$ I need to prove $\displaystyle\sum\limits_{k=0}^n\left(\frac{ 1^k+(-1)^k}{2}\right)\cdot\left(\begin{array}{c}n\\ k\end{array}\right)4^k=\frac{5^n+(-3)^n}{2}$
my Attempt below
$\displaystyle\sum\limits_{k=0}^n\left(\frac{ 1^k+(-1)^k}{2}\right)\cdot \left(\begin{array}{c}n\\ k\end{array}\right)4^k=\frac{1^0+(-1)^0}{2}\cdot \frac{n!}{0!(n-0)!}\cdot4^0\\ +\frac{1^1+(-1)^1}{2}\cdot \frac{n!}{1!(n-1)!}\cdot4^1+\frac{1^2+(-1)^2}{2}\cdot \frac{n!}{2!(n-2)!}\cdot4^2+\frac{1^3+(-1)^3}{2}\cdot \frac{n!}{3!(n-3)!}\cdot4^3+...+\frac{1^n+(-1)^n}{2}\cdot \frac{n!}{n!(n-n)!}\cdot4^n=\\ =\frac{1+1}{2}\cdot \frac{n!}{1n!}\cdot1+\frac{1+1}{2}\cdot \frac{n!}{2(n-2)!}\cdot16+\dotso+\frac{1^n+(-1)^n}{2}\cdot \frac{n!}{n!(n-n)!}\cdot4^n=\\= \frac{n!}{n!}+ \frac{4n!}{(n-1)!}+1\cdot \frac{16n!}{2(n-2)!} +1\frac{64n!}{6(n-3)!}+1\dotsm\ \frac{4^nn!}{n!(n-n)!}+1=$
$\frac{1}{2}\left[\sum_{k = 0}^{n} \binom{n}{k}4^k + \sum_{k = 0}^{n} \binom{n}{k}(-4)^k\right]=\\$1/2$(1+x)^n=\frac{1}{2}((1+4)^n+(1-4)^n=\frac{5^n+(-3)^n}{2}$
|
$\sum\limits_{k=0}^n 1^{k} 4^{k}\binom {n} {k}$ is the expansion of $(1+4)^{n}$ and $\sum\limits_{k=0}^n (-1)^{k} 4^{k}\binom {n} {k}$ is the expansion of $(1-4)^{n}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3328751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Is $\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$ the same as $\frac{\frac{2-b}{2b}}{b-2}$? Have I subtracted 2 fractions correctly?
$\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$
Starting with the numerator which is a difference of fractions, the least common denominator is 2b? So:
$\frac{2(1)}{2b}-\frac{b(1)}{2b}$ = $\frac{2}{2b}-\frac{b}{2b}$ = $\frac{2-b}{2b}$
So my newly simplified expression is:
$\frac{\frac{2-b}{2b}}{b-2}$
My question - is this correct? If yes, great - now how would I go about simplifying further by taking the denominator $b-2$ into account? If no, where did I go wrong?
|
$\require{cancel}$Yes, it is right. And now:$$\frac{\frac{2-b}{2b}}{b-2}=-\frac{\frac{\cancel{b-2}}{2b}}{\cancel{b-2}}=-\frac1{2b}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3328994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Why does this work? What is the best exponent to approximate the logarithm? Suppose i want to get a upper bound of $log_53$ without other tools than elementary algebra.
I can do this:
Let $x = log_53$, then $5^x=3$
I raise both sides to $4$, then: $(5^x)^4 = 3^4$ And since $3^4 < 5^3$ it follows that $5^{4x} < 5^3$, then $x < \frac{3}{4}$
However, if i raise both sides to $13$ is equal to: $(5^x)^{13} = 3^{13}$ And since $3^{13} < 5^9$ it follows that $5^{13x} < 5^9$, then $x < \frac{9}{13}$
In this case with exponent $13$ the approximation is better.
So, with different exponent i can get different upper bounds. Similarly, the same can be done with the lower bound.
And my question is,
Given $log_ab = x$, what is the best exponent $m$ such that $(a^x)^m = b^m$ gives the better approximation for both cases?(Lower and upper bounds)
|
There's no "best exponent", because you can always get better and better exponents, but you can use continued fractions to do this systematically. With continued fractions, we will write $x$ in the form
$$
x = a_0 + \cfrac1{a_1 + \cfrac1{a_2 + \cfrac1{a_3 + \cfrac1{a_4 + \dots}}}}.
$$
where $a_0, a_1, a_2, a_3, a_4, \dots$ is an infinite sequence of natural numbers.
We can truncate this sequence at any point, getting upper or lower bounds depending on where we stop. (If you stop at an even $a_k$, you get a lower bound; if you stop at an odd $a_k$, you get an upper bound.)
We compute the sequence iteratively: $a_0$ is $\lfloor x\rfloor$ (the greatest integer less than $x$) and to find $a_1$ onward you just replace $x$ by $\frac{1}{x - a_0}$.
All this is abstract and general, so here is how it will go in your example.
Let $x = \log_5(3)$. Then we can estimate that $0 < x < 1$ (because $5^0 < 3 < 5^1$) so $a_0 = \lfloor \log_5(3)\rfloor = 0$, and we will keep going with $\frac1{\log_5(3)} = \log_3(5)$.
We can estimate that $1 < \log_3(5) < 2$, because $3^1 < 5 < 3^2$, so $a_1 = \lfloor \log_3(5)\rfloor = 1$, and we will keep going with $\frac{1}{\log_3(5)-1} = \frac1{\log_3(5/3)} = \log_{5/3}(3)$.
We can estimate that $2 < \log_{5/3}(3) < 3$, because $(\frac53)^2 < 3 < (\frac53)^3$ (this expands to $25 < 27$ but $81 < 125$), so $a_2 = \lfloor \log_{5/3}(3) \rfloor = 2$, and we will keep going with $\frac1{\log_{5/3}(3) - 2} = \frac1{\log_{5/3}(27/25)} = \log_{27/25}(\frac53)$.
We could estimate that $6 < \log_{27/25}(\frac53) < 7$, because $(\frac{27}{25})^6 < \frac53 < (\frac{27}{25})^7$, but this has gotten tricky to check: it involves checking that $3^{19} < 5^{13}$ but $3^{22} > 5^{15}$. This gives us $a_3 = 6$, but maybe we'd better quit while we're ahead, or get a computer.
At this point, our best lower bound is
$$
a_0 + \cfrac1{a_1 + \cfrac1{a_2}} = 0 + \cfrac1{1 + \cfrac12} = \frac23
$$
and our best upper bound is
$$
a_0 + \cfrac1{a_1 + \cfrac1{a_2 + \cfrac1{a_3}}} = 0 + \cfrac1{1 + \cfrac1{2 + \cfrac16}} = \frac{13}{19}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3330557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Finding matrix representation of a linear transformation Find the matrix representation of the linear transformation $\alpha:\mathbb{R}^3 \rightarrow \mathbb{R}^2$ defined by
$$\alpha : \begin{pmatrix} a\\b\\c \end{pmatrix}\rightarrow \begin{pmatrix} a+b+c\\b+c \end{pmatrix}$$
With respect to bases
$\left\{ \begin{pmatrix} -1\\0\\2 \end{pmatrix},\begin{pmatrix} 0\\1\\1 \end{pmatrix},\begin{pmatrix} 3\\-1\\0 \end{pmatrix} \right\}$ of $\mathbb{R}^3$ and $\left\{ \begin{pmatrix} -1\\1 \end{pmatrix},\begin{pmatrix} 1\\0\end{pmatrix} \right\}$ of $\mathbb{R}^2$
My attemp: The first part is
$$A_{B_1}=[\alpha(b_1) ~~\alpha(b_2)~~ \alpha(b_3)]=\left[\alpha\begin{pmatrix} -1\\0\\2 \end{pmatrix}~~\alpha\begin{pmatrix} 0\\1\\1 \end{pmatrix} ~~\alpha\begin{pmatrix} 3\\-1\\0 \end{pmatrix} \right]=\begin{pmatrix} 1 &2&2\\2 &2&-1 \end{pmatrix}$$
I'm confused about the second part , here $\alpha$ takes values in $\mathbb{R}^3$ so how can i proceed with finding matrix for vectors in $\mathbb{R}^2$ ?
|
What you have computed is the matrix for $\alpha$, with respect to the given basis on $\Bbb{R}^3$, but with respect to the standard basis on $\Bbb{R}^2$ (as opposed to the one given). If $B_2$ is the basis on $\Bbb{R}^2$, then the matrix you want is:
$$\left[\begin{array}{c|c|c}\left[\alpha \begin{pmatrix}-1 \\ 0 \\ 2\end{pmatrix}\right]_{B_2} & \left[\alpha \begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}\right]_{B_2} & \left[\alpha \begin{pmatrix}3 \\ -1 \\ 0\end{pmatrix}\right]_{B_2}\end{array}\right],$$
where $[v]_{B_2}$ refers to the coordinate column vector of $v$ with respect to basis $B_2$. That is, it will be $\begin{bmatrix} a \\ b \end{bmatrix}$, where $a$ and $b$ are the unique scalars such that
$$v = a\begin{pmatrix} -1 \\ 1 \end{pmatrix} + b\begin{pmatrix} 1 \\ 0 \end{pmatrix}.$$
Finding such a vector requires solving a system of linear equations.
As an example, take the first column. We have, as you computed,
$$\alpha \begin{pmatrix}3 \\ -1 \\ 0\end{pmatrix} = \begin{pmatrix}1 \\ 2 \end{pmatrix}.$$
Thus, we need to solve
$$\begin{pmatrix}1 \\ 2 \end{pmatrix} = a\begin{pmatrix} -1 \\ 1 \end{pmatrix} + b\begin{pmatrix} 1 \\ 0 \end{pmatrix}.$$
If we turn this into a system of equations, the corresponding augmented matrix is
$$\left[\begin{array}{cc|c}-1 & 1 & 1 \\ 1 & 0 & 2\end{array}\right] \sim \left[\begin{array}{cc|c}0 & 1 & 3 \\ 1 & 0 & 2\end{array}\right] \sim \left[\begin{array}{cc|c}1 & 0 & 2 \\ 0 & 1 & 3\end{array}\right],$$
hence $a = 2$ and $b = 3$, which is to say,
$$\left[\alpha \begin{pmatrix}3 \\ -1 \\ 0\end{pmatrix}\right]_{B_2} = \left[\begin{pmatrix}1 \\ 2 \end{pmatrix}\right]_{B_2} = \begin{bmatrix} 2 \\ 3\end{bmatrix},$$
which is the first column of the matrix you want. The same procedure should find the other two columns.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3330655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$ Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$.
Let $P(x)=x^{2015}-x^{2014}=Q(x)(x-1)^3+ax^2+bx+c.$ If we put $x=1$ in $P(x)$ and $P'(x)$, we get $a+b+c=0$ and $2a+b=1$. Then: $c=a-1$. The second derivative won't help in finding $b$, so, what should I do? Thank you
|
$\dfrac{x^{2014}(x-1)}{(x-1)^3}= \dfrac{x^{2014}}{(x-1)^2}$;
$x^{2014}= Q(x)(x-1)^2+ ax+b$;
Binomial expansion:
$x^{2014}=(1+(x-1))^{2014}=$
$\sum_{k=0}^{n}\binom{2014}{k}1^{n-k}(x-1)^k$
Remainder $ax+b$:
$\binom{2014}{0}1+\binom{2014}{1}(x-1)=$
$1+2014x-2014=2014x-2013$.
Originally:
$x^{2014}(x-1)$ is divided by $(x-1)^3$:
Hence
$x^{2014}(x-1)=Q(x)(x-1)^3+(x-1)(2014x-2013),$
with quadratic remainder
$(x-1)(2014x-2013)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3331414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
}
|
Finding a formula for $(1^3)\cdot(n)+(2^3)\cdot(n-1)+(3^3)\cdot(n-2)+ \cdots + (n^3)\cdot(1)$
I need a formula for this sum:
$$(1^3)\cdot(n)+(2^3)\cdot(n-1)+(3^3)\cdot(n-2)+ \cdots + (n^3)\cdot(1)$$
I have found this formula :
$$1\cdot n +2\cdot(n-1)+3\cdot(n-2)+ \cdots +(n-1)\cdot 2 +n\cdot1= \frac16n(n+1)(n+2)$$
but it is not exactly what I need.
Thanks in advance.
|
Use generating functions.
Write your sum as:
$\begin{align*}
S_{n + 1}
&= \sum_{0 \le k \le n + 1} k^3 (n + 1 - k)
\end{align*}$
So this is a convolution. It is the coefficients of the series:
$\begin{align*}
S(z)
&= \sum_{n \ge 0} S_n z^n \\
&= \left( \sum_{k \ge 0} k^3 z^k \right)
\cdot \left( \sum_{k \ge 0} k z^k \right) \\
&= \frac{z (1 + 4 z + z^2)}{(1 - z)^4}
\cdot \frac{z}{(1 - z)^2} \\
&= \frac{z^2 (1 + 4 z + z^2)}{(1 - z)^6}
\end{align*}$
Extract coefficients, using the generalized binomial theorem:
$\begin{align*}
(1 + u)^{-m}
&= \sum_{k \ge 0} \binom{-m}{k} u^k \\
&= \sum_{k \ge 0} (-1)^k \binom{k + m - 1}{m - 1} u^k
\end{align*}$
giving:
$\begin{align*}
S_n
&= [z^n] \frac{z^2 (1 + 4 z + z^2)}{(1 - z)^6} \\
&= ([z^{n - 2}] + 4 [z^{n - 3}] + [z^{n - 4}]) (1 - z)^{-6} \\
&= \binom{n - 2 + 6 - 1}{6 - 1}
+ 4 \binom{n - 3 + 6 - 1}{6 - 1}
+ \binom{n - 4 + 6 - 1}{6 - 1} \\
&= \frac{3 n^5 - 5 n^3 + 2 n}{60}
\end{align*}$
Your sum is:
$\begin{align*}
S_{n + 1}
&= \frac{3 n^5 + 15 n^4 + 25 n^3 + 15 n^2 + 2 n}{60} \\
&= \frac{n (n + 1) (n + 2) (3 n^2 + 6 n + 1)}{60}
\end{align*}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3332802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Solve System of equation using elimination? \begin{align}
I:&& ~~ x+\frac12y &= 6
\\[.5em]
II:&& ~~ \frac32x + \frac{3}{2}y &= {17 \over 2}
\end{align}
when $x$ was multiplied by $(-3/2)$ in first equation the $x$ will be canceled and the resulting $y = -2/3$ and $x = 19/3$.
But when fractions were simplified first the resulting equation is
$$-8x-4y= -48$$
$$9x+4y = 51$$
here $y$ get canceled and results will be $x=3 , y=6$.
why these two attempts give two different results..I only need answer in elimination technique.
|
This can solved in an entirely algorithmic way, calculating the RREF of the augmented matrix. First we multiply the second equation by $2$. Then:
\begin{align}
\begin{bmatrix}1&\frac 12& 6 \\ 3&3&17
\end{bmatrix}&\rightsquigarrow
\begin{bmatrix}1&\frac 12&\phantom{-}6 \\ 0&\frac32 &-1
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}1&\frac 12&\phantom{-}6 \\ 0& 1 &-\frac23
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}1& 0 &\color{red}{\frac{19}3} \\ 0& 1 &\color{red}{-\frac{2}3}
\end{bmatrix}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3333674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
How to determine algebraically whether an equation has an infinite solutions or not? I was learning for the first time about partial fraction decomposition. Whoever explains it, emphasises that the fraction should be proper in order to be able to decompose the fraction. I was curious about knowing what happens if I try to decompose an improper fraction, So I tried to do one:
$\frac{x^2 - 4}{(x + 5)(x - 3)}$
I got the equation:
$\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)} = \frac{x^2 - 4}{(x + 5)(x - 3)}$. I have 4 unknowns: A, B, C and D.
$\therefore (Ax + B)(x - 3) + (Cx + D)(x + 5) = x^2 - 4$
After expanding and regrouping the coefficients:
$(A + C) x^2 + (-3A + B + 5C + D)x + (-3B + 5D) = x^2 - 4$
Here the coefficient of the term $x^2$ is 1 therefore:
$(A+C) = 1$
similarly:
$(-3A + B + 5C + D) = 0$
$(-3B + 5D) = -4$
I still have to get one more equation to be able to solve this system so I substituted 1 for x and I got this equation:
$-2A - 2B + 6C + 6D = -3$
After getting four equations I used this site to solve the system of equations. Unfortinetly I got no soultion. Tried another site and also the same result.
I've tried to use different values for x and got another equaitons like:
for x = 2 : $-2A - B + 24C + 7D$
for x = -1 : $4A - 4B - 4C + 4D$
for x = -2 : $10A - 5B - 6C + 3D$
But also that didn't work. Always the system of equations have an infinite solutions.
After tring to figure out why this is happening, I've managed to prove logically that this equation:
$(Ax + B)(x - 3) + (Cx + D)(x + 5) = x^2 - 4$
has an infinite solutions and my approach was as follows:
After doing polynomial long division and decomposing the fraction using the traditional way, the result should be:
$\frac{5}{8(x - 3)} - \frac{21}{8(x - 5)} + 1$
Now I can add the last term (the one) to the first term and get the follows:
$\frac{8x-19}{8(x-3)} - \frac{21}{8(x+5)}$
From that solution I can see that $A = 0, B = \frac{-21}{8}, C = 1, D = \frac{-19}{8}$. After all these are just the coefficients of the terms. and this solution worked fine.
Alternatively I can add the one to the second term instead and get:
$\frac{5}{8(x-3)} + \frac{8x + 19}{8(x+5)}$
Now $A = 1, B = \frac{19}{8}, C = 0, D = \frac{5}{8}$
Generally, after adding the one to any of the terms, I can add any number to one of the terms and add its negative to the other term and the equation will remain the same, But the value of the 4 constants (A, B, C, and D) will change. And from that I got convinced that there are an infinite solutions to this equation.
But Algebraically? I'm not able to prove that it has an infinite solutions algebraically. And my questions is how to prove algebraically that this equation has an infinite solutions? Or generally how to know whether the equation has just one solution or an infinite?
|
$$\text{Given: }\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)} = \frac{x^2 - 4}{(x + 5)(x - 3)}$$
$$(Ax^2-3Ax+BX-3B)+(Cx^2+5Cx+Dx+5D)-(x^2-4)=0$$
$$(A+C-1)x^2+(-3A+B+5C+D)x+(-3B+5D+4)=0$$
$$(\mathbf{A+C}-1)x^2+(\mathbf{-3A-3C}+B+\mathbf{5C+3C} -D)x-(3B-5D-4)=0$$
Noting the coefficient of $x^2$: $A+C=1$, we can eliminate $x^2$
$$(B+8C+D-3)x=3B-5D-4$$
Noting the coefficients of $x: (B+8C+D-3)=0\implies3B-5D-4=0$
$$\implies 3B=5D+4\implies B=\frac{5D+4}{3}\implies 5D+4+24C+3D-9=8D+24C-5=0\\
\implies 3C+D=\frac{5}{8}$$
We now have the values for $A+C$ and $3C+D$. If you can set up $4$ simultaneous equations using these and other combinations from the original equations, you can find a unique solution, if the determinant of a matrix of those equations' coefficients is non-zero. Otherwise, you can experiment with one or two values as you did in your question and some solution(s) should pop out. I do not know how you would go about proving there are infinite solutions if that is the case.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3335420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 8,
"answer_id": 5
}
|
A double integral for $\frac{\pi}{2} \ln 2$.
Show that
\begin{eqnarray*}
I=\int_0^1 \int_0^1 \frac{dx \,dy}{\sqrt{1-x^2y^2}} = \frac{\pi}{2} \ln 2.
\end{eqnarray*}
My try ... from this question here we have
\begin{eqnarray*}
\int_0^1 \frac{ \sin^{-1}(x)}{x} \,dx = \frac{\pi}{2} \ln 2 .
\end{eqnarray*}
And from this question here we have
\begin{eqnarray*}
\int_0^1 \ln \left( \frac{1+ax}{1-ax} \right) \frac{dx}{x\sqrt{1-x^2}}=\pi\sin^{-1} a,\qquad |a|\leq 1.
\end{eqnarray*}
It is easy to show
\begin{eqnarray*}
\int_0^1 \frac{dy}{1+yz}=\frac{1}{z} \ln(1+z).
\end{eqnarray*}
So we have (with a little tad of algebra)
\begin{eqnarray*}
\frac{\pi}{2} \ln 2 &=& \int_0^1 \frac{ \sin^{-1}(x)}{x}\, dx \\
&=& \frac{1}{\pi} \int_0^1 \int_0^1 \ln \left( \frac{1+xt}{1-xt} \right) \frac{dt}{xt\sqrt{1-t^2}} x\, dx \\
&=& \frac{2}{\pi} \int_0^1 \int_0^1 \int_0^1 \frac{dx \,dy\, dt}{(1-x^2y^2t^2)\sqrt{1-t^2} } \\
\end{eqnarray*}
This suggests we should consider the integral (sub $t=\sin(\theta)$)
\begin{eqnarray*}
\frac{2}{\pi} \int_0^1 \frac{ dt}{(1-x^2y^2t^2)\sqrt{1-t^2} } \\
= \int_0^{\pi/2} \frac{d \theta }{1- x^2 y^2 \sin^2(\theta)}.
\end{eqnarray*}
Now it is well known (Geometrically expand, integrate term by term & sum the familiar plum) that
\begin{eqnarray*}
\frac{2}{\pi} \int_0^{\pi/2} \frac{d \theta }{1- \alpha \sin^2(\theta)}=\frac{2}{\pi} \frac{1}{\sqrt{1-\alpha}}
\end{eqnarray*}
and using this we have
\begin{eqnarray*}
\frac{\pi}{2} \ln 2 =\int_0^1 \int_0^1 \frac{dx\, dy}{\sqrt{1-x^2y^2}} .
\end{eqnarray*}
The above double integral reminds of
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\int_0^1 \int_0^1 \frac{dx\, dy}{1-x^2y^2} = \frac{\pi^2}{8}
\end{eqnarray*}
which can be evaluated using the substitution $x= \frac{\sin u}{\cos v}$, $y= \frac{\sin v}{\cos u}$.
My solution above used some pretty heavy machinery to establish the result. So my question is: is there an easier method ?
|
Why not go head-on? It seems to work here. That is, write the integral as $$\int_0^1\int_0^1\frac {\mathrm d x \mathrm d y}{\sqrt{1-x^2y^2}}=\int_0^1\int_0^1\frac {\mathrm d x \mathrm d y}{y \sqrt{\frac{1}{y^2}-x^2}}.$$ Setting $x=\frac1y\sin\phi$ as usual helps us evaluate the first integral, or you can simply note that it has the form of an $\arcsin.$ If you make this substitution, the first integral becomes $$\int_0^{\arcsin y}\frac1y\mathrm d \phi=\frac{\arcsin y}{y}.$$ Thus, the integral reduces to $$\int_0^1\frac{\arcsin y}{y}\mathrm d y,$$ which may be done by parts. Forgetting limits for now, this begins as $$\arcsin y\log y-\int\frac{\log y}{\sqrt{1-y^2}}\mathrm dy,$$ etc.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3336333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Integrating $\int \frac{3}{x^2+12x+45}$ My professor is giving a mastery exam over integrals and one of the sections covered is un-factorable quadratics like this problem that are supposed to be done by completing the square.
$$\int \frac{3}{x^2+12x+45}$$
Completing the square and everything gives $$\int \frac{3}{(x+6)^2+9}$$
I figured the answer would be somewhere in the range of $$\frac{1}{3} \tan^{-1}\Biggl(\frac{x+6}{9}\Biggl)+c$$
like I learned in the textbook aside from not knowing what to do with the three. Anyways the answer is somehow $$\tan^{-1}\Biggl(\frac{x}{3}+2\Biggl) + c$$ and I don't know what I did wrong
|
Since\begin{align}\int\frac3{x^2+12x+45}\,\mathrm dx&=\int\frac3{(x+6)^2+9}\,\mathrm dx\\&=\int\frac{\frac39}{\frac{(x+6)^2}9+1}\,\mathrm dx\\&=\int\frac{\frac13}{\left(\frac{x+6}3\right)^2+1}\,\mathrm dx,\end{align}the substitution $y=\frac{x+6}3$ leads indeed to the answer $\arctan\left(\frac{x+6}3\right)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3336430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Find table values of this $y=(x-4)(x-2)(x+1)(x+4)$ how to find this $$y=(x-4)(x-2)(x+1)(x+4)$$
I know that x - intercept is $$4,2,-1,-4$$
and y- intercept is $$32$$
In table values
$$ x = -5,-3,-1,2,3$$
now What is the values of $y$ ?
Thanks in advance
|
Let the function be $f(x)$ for convenience.
Plug in each value:
$f(-5)=(-5-4)(-5-2)(-5+1)(-5+4)=(-9)(-7)(-4)(-1)=252$
$f(-3)=(-3-4)(-3-2)(-3+1)(-3+4)=(-7)(-5)(-2)(1)=-70$
$f(-1)=(-1-4)(-1-2)(-1+1)(-1+4)=(-5)(-3)(0)(3)=0$
$f(2)=(2-4)(2-2)(2+1)(2+4)=(-2)(0)(3)(6)=0$
$f(3)=(3-4)(3-2)(3+1)(3+4)=(-1)(1)(4)(7)=-28$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3336567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How to show that $ \sum_{k=1}^{2^n}\frac{1}{k}\geq 1+\frac{n}{2}$? So in order to prove that the Harmonic series diverge, I want to show that:
$$ \sum\limits_{k=1}^{2^n}\dfrac{1}{k}\geq 1+\dfrac{n}{2}$$
It is clear that if we expand the sum that this inequality holds true,
$$n=3,\ \sum\limits_{k=1}^{2^3=8}\dfrac{1}{k}=1+\frac{1}{2}+\underbrace{\frac{1}{3}+\frac{1}{4}}\limits_{\geq \frac{1}{2}}+\underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}\limits_{\geq \frac{1}{2}}$$
but how can i show that using induction?
Any tips would be helpful, and thanks.
|
*
*Use the fact that $\frac{1}{3}+\frac{1}{4} \geq \frac{1}{4}+\frac{1}{4}$ and $\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8} \geq \frac{1}{8}+\frac{1}{8} +\frac{1}{8}+\frac{1}{8}$
*There are also various other proofs which you might be interested to see. Here is a link: http://scipp.ucsc.edu/~haber/archives/physics116A10/harmapa.pdf
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3338182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Solving an integral with power of $\exp$ Is it possible to get a "closed" form for the following integral
$$\int_{\mathbb{R}}\int_{\mathbb{R}}e^{-\frac{2}{3}\vert x\vert^{\beta}}e^{-\frac{1}{3}\vert y-x\vert^{\beta}}\frac{1}{\vert y\vert^{\gamma}}dxdy,$$
where $1<\beta\le 2\;,0<\gamma<1?$
The case where $\beta=2$ does not seems difficult because of the identity $(x-y)^2=x^2-2xy-y^2$ but as I am interested for the general case and $\beta=2$ will not lead to any idea so I don't think it is "interesting".
*
*We can't expand the two $\exp$ in power series and switch integrals because the result will not converges.
*I am interested in any nice form i.e. using well known functions as for exemple hypergeometric functions.
*In fact we can replace $\frac{2}{3}$ and $\frac{1}{3}$ by $a-b,b$: so is it possible to get a bound in term of $a,b$ ? It will be sufficient for me as well.
|
We have that $\frac{2}{3} = a-b$ and $\frac{1}{3} = b$ so we will try to deal with it as generally as possible (since those specific numbers are unlikely to have any symmetry associated with them).
We can convert the integral into polar coordinates and get an integral of the form
$$\int_{\alpha_1}^{\alpha_2} \int_0^\infty e^{-r^\beta f_\beta(\theta)} \frac{r^{1-\gamma}}{|\pm \sin \theta|^\gamma} drd\theta$$
where
$$f_\beta(\theta) = \begin{cases}
(a-b)\cos^\beta\theta + b(\cos\theta - \sin\theta)^\beta & -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{4} \\
(a-b)\cos^\beta\theta + b(\sin\theta - \cos\theta)^\beta & \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2} \\
\end{cases} $$
There is a symmetry in the regions. There are six distinct angular regions to integrate on but they come in symmetric pairs, so we will only have three angular integrals to do.
Next we will use the substitution $\chi = r^\beta f_\beta(\theta)$ to get
$$\int_{\alpha_1}^{\alpha_2} \int_0^\infty e^{-\chi} \chi^{\frac{2-\gamma}{\beta}-1} \frac{1}{\beta(f_\beta(\theta))^{\frac{2-\gamma}{\beta}}|\pm \sin \theta|^\gamma} d\chi d\theta$$
Using this form, we can say the original expression evaluates to
$$\frac{2}{\beta}\Gamma\left( \frac{2-\gamma}{\beta} \right) (I_1+I_2+I_3)$$
where
$$\begin{align}
& I_1 = \int_0^{\frac{\pi}{4}} \frac{[(a-b)\cos^\beta\theta + b(\cos\theta - \sin\theta)^\beta]^{\frac{\gamma-2}{\beta}}}{\sin^\gamma \theta} d\theta \\
& I_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{[(a-b)\cos^\beta\theta + b(\sin\theta - \cos\theta)^\beta]^{\frac{\gamma-2}{\beta}}}{\sin^\gamma \theta} d\theta \\
& I_3 = \int_0^{\frac{\pi}{2}} \frac{[(a-b)\sin^\beta\theta + b(\sin\theta + \cos\theta)^\beta]^{\frac{\gamma-2}{\beta}}}{\cos^\gamma \theta} d\theta \\
\end{align}$$
If anyone has any idea what these last three integrals could possibly evaluate to, let me know.
$\mathbf{\text{EDIT}}$: In lieu of a formal solution to these integrals, we can bound them. Take the min of each of the "numerators"(since they are raised to negative powers):
$$\begin{align}
& I_1 = \int_0^{\frac{\pi}{4}} \frac{[(a-b)\cos^\beta\theta + b(\cos\theta - \sin\theta)^\beta]^{\frac{\gamma-2}{\beta}}}{\sin^\gamma \theta} d\theta \leq \frac{2^{1-\frac{\gamma}{2}}}{(a-b)^{\frac{2-\gamma}{\beta}}} \int_0^{\frac{\pi}{4}} \frac{1}{\sin^\gamma \theta} d\theta \\
& I_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{[(a-b)\cos^\beta\theta + b(\sin\theta - \cos\theta)^\beta]^{\frac{\gamma-2}{\beta}}}{\sin^\gamma \theta} d\theta \leq \frac{2^{\frac{\gamma}{2}}}{(\min(a-b,b))^{\frac{2-\gamma}{\beta}}}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} d\theta \\
& I_3 = \int_0^{\frac{\pi}{2}} \frac{[(a-b)\sin^\beta\theta + b(\sin\theta + \cos\theta)^\beta]^{\frac{\gamma-2}{\beta}}}{\cos^\gamma \theta} d\theta \leq b^{\frac{\gamma-2}{\beta}}\int_0^{\frac{\pi}{2}} \frac{1}{\cos^\gamma \theta} d\theta \\
\end{align}$$
For the first two equations, both of the terms in the sum are monotonic, so either of the endpoints has to be the min. For the third integral, this isn't the case, but we can take the min of the two functions separately and them together since everything is positive.
Bounding the remaining integrals with singularities:
$$\int_0^{\frac{\pi}{4}} \frac{1}{\sin^\gamma \theta} d\theta = \int_0^{\frac{1}{\sqrt{2}}} \frac{x^{-\gamma}}{\sqrt{1-x^2}}dx \leq \sqrt{2} \int_0^{\frac{1}{\sqrt{2}}} x^{-\gamma} dx = \frac{2^{\frac{\gamma}{2}}}{1-\gamma} $$
$$\int_0^{\frac{\pi}{2}} \frac{1}{\cos^\gamma \theta} d\theta = \int_0^1 \frac{x^{-\gamma}}{\sqrt{1-x^2}} dx$$
$$ = \int_0^{\frac{1}{\sqrt{2}}} \frac{x^{-\gamma}}{\sqrt{1-x^2}} dx + \int_{\frac{1}{\sqrt{2}}}^1 \frac{x^{-\gamma}}{\sqrt{1-x^2}} dx \leq \frac{2^{\frac{\gamma}{2}}}{1-\gamma} + 2^{\frac{\gamma}{2}-2}\hspace{3 pt}\pi$$
All in all, we have bounded our original integral by
$$\frac{2}{\beta}\Gamma\left( \frac{2-\gamma}{\beta} \right) \left( \frac{2(a-b)^{\frac{\gamma - 2}{\beta}}+2^{\frac{\gamma}{2}}b^{\frac{\gamma - 2}{\beta}}}{1-\gamma} + \frac{2^{\frac{\gamma}{2}-2}\hspace{3 pt}\pi}{[\min(a-b,b)]^{\frac{2-\gamma}{\beta}}} + \frac{2^{\frac{\gamma}{2}-2}\hspace{3 pt}\pi}{b^{\frac{2 - \gamma}{\beta}}} \right)$$
$\mathbf{\text{EDIT}}$: Courtesy of Mathematica, those simplified integrals had closed forms, which we can use to get an even tighter bound:
$$\frac{2}{\beta}\Gamma\left( \frac{2-\gamma}{\beta} \right) \left( \frac{\sqrt{2}(a-b)^{\frac{\gamma - 2}{\beta}}}{1-\gamma}{}_2 F_1\left(\frac{1}{2},\frac{1-\gamma}{2},\frac{3-\gamma}{2},\frac{1}{2} \right) + \frac{2^{\frac{\gamma}{2}-2}\hspace{3 pt}\pi}{[\min(a-b,b)]^{\frac{2-\gamma}{\beta}}} + \frac{\sqrt{\pi}}{2b^{\frac{2 - \gamma}{\beta}}}\frac{\Gamma\left( \frac{1-\gamma}{2} \right)}{\Gamma\left( 1-\frac{\gamma}{2} \right)} \right) $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3339384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Proving that $5^n - 1$ is divisible by $4$ by mathematical induction. I have done it, but I am not sure that the inductive step is right. Can anybody please clear me about it?
Basic steps as:
Taking $n=1$: $p(1)=5-1=4$.
Inductive hypothesis: Assume the statement is true for $p(k)$. $5^k - 1$ is divisible by $4$.
Inductive steps: We must show $p(k+1)$ is true when $p(k)$ is true.
\begin{align*}
& 5^k -1 + 5^{k+1} -1\\
& 5^k -1 + 5.5^{k} -1\\
& (5^k -1) + 4
\end{align*}
|
*
*For n=1 the statement is true.
*Assume that $$5^n-1$$ is divisible by 4.
*Let us consider $$5^{(n+1)}-1$$. We have to prove that it is divisible by4 also.
We have that
$$
\begin{gathered}
5^{n + 1} - 1 = 5 \cdot 5^n - 1 = 5 \cdot 5^n - 5 + 5 - 1 = \hfill \\
\hfill \\
= 5\left( {5^n - 1} \right) + 4 \hfill \\
\end{gathered}
$$
Since by inductive step 2 you have that $$5^n-1=4k$$ whit k integer you have that
$$
5\left( {5^n - 1} \right) + 4 = 5 \cdot 4k + 4 = 4\left( {5k + 1} \right)
$$
and this prove step 3
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3339822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Evaluate $\sum\limits_{n=0}^{\infty} \frac{\cos(nx)}{2^n}$ where $\cos x = \frac15$ Evaluate
$$\sum_{n=0}^{\infty} \dfrac{\cos(nx)}{2^n}$$
where $\cos x = \frac{1}{5}$.
This is a complex number question. But I don’t know where to start. Maybe need to use the DeMoivre’s Theorem?
|
If $S=\sum_{r=0}^\infty y^r\cos r x$
Using scale of relation from this or $\#187$ of this(downloadable)
$$(1-2y\cos x+y^2)S$$
$$=1+y\cos x+y^2\cos2x+y^3\cos3x+\cdots$$
$$-2y\cos x-2y^2\cos^2x-2y^3\cos2x\cos x-2y^4\cos3x\cos x-\cdots$$
$$+y^2+y^3\cos x+y^4\cos2x+y^5\cos3x+\cdots$$
$$=1+y(-\cos x)+y^2(\cos2x-2\cos^2x+1)+y^3(\cos3x-2\cos3x\cos x+\cos x)+\cdots$$
$$\implies S=\dfrac{1-y\cos x}{1-2y\cos x+y^2}$$
Here $2y=1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3340563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Proving the inequality $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$ Prove that $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$. When does the inequality hold?
I really don't know how to prove the inequality and would like to know how.
I mainly tried to factorise the LHS-RHS fully but I could never properly do it:
https://imgur.com/user/Khansis/favorites/folder/7408635/math
|
You can simplify first:
$$4(\underbrace{a^6+b^6}_{(1)}) =4\require{cancel}\cancel{(a^2+b^2)}(a^4-a^2b^2+b^4)\ge (a+b)\cancel{(a^2+b^2)}(\underbrace{a^3+b^3}_{(2)}) \Rightarrow\\
4((a^2-b^2)^2+a^2b^2)\ge (a+b)^2((a-b)^2+ab) \Rightarrow \\
3(a^2-b^2)^2+ab(4ab-(a+b)^2)\ge 0 \Rightarrow \\
3(a-b)^2(a+b)^2-ab(a-b)^2\ge 0 \Rightarrow \\
(a-b)^2(\underbrace{3a^2+3b^2+5ab}_{(3)})\ge 0 \quad \color{green}\checkmark$$
The equality holds when $a=b$.
Note:
1) For $(1),(2)$ it was used: $a^3+b^3=(a+b)(a^2-ab+b^2)$.
2) For $(3)$, it was used AM-GM: $3(a^2+b^2)\ge 6|ab|\ge 5ab$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3342070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Solutions to a quadratic given 1 solution in form a+bi I was just really confused as to how I only ended up with 1 of 2 answers for the following question.
Given that $-2+bi$ is a solution of $x^2+ax+(3+a) $ find constants $a$ and $b$ given that they are real.
As soon as I saw that $-2+bi$ was a solution, I immediately jumped to $ -2-bi$ must also be a solution, by the fundamental theorem of algebra. By doing the sum and product of the solution a quadratic could be obtained
Sum $ (-2+bi)+(-2-bi)=-4$
Product $(-2-bi)(-2+bi)=(-2)^2-(-bi)^2=4+b^2$
Thus the quadratic $x^2+4x+(4+b^2)$ is obtained
Equation both sides of the equation
$\\x^2+4x+(4+b^2)=x^2+ax+(3+a)$
$a=4$
Therefore$ b={\sqrt 3}$ or $ b={-\sqrt 3}$
However, the solutions seem to suggest that an extra solution can be
$ b={0}, a=7$
Did I eliminate a solution by doing the sum and product of the solutions to find the quadratic? Or does it have to do with my working process
|
Let the second root be $c-ib$ (because $2+ib+c-ib=a$ must be real). Then
$$(x+2-ib)(x-c+ib)=x^2+(2-c)x-2c+b^2+i(bc+2b)=x^2+ax+a+3.$$
For the imaginary term to vanish, $$b=0\lor c=-2$$
gives us the three solutions
$$a=7,b=0,c=-5$$ and $$a=4,b=\pm\sqrt3,c=-2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3343042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Calculate $\int_0^{\pi} \frac{ \cos ( kx ) }{ 1 - 2 \tau \cos (x ) } dx$ I need to calculate the following integral:
$$\int_0^{\pi} \frac{ \cos ( kx ) }{ 1 - 2 \tau \cos (x ) } dx$$
for $k \geq 0$ and $| \tau | < \frac{1}{2}$. For $k=0$, I use the reparameterization $t = \tan (x /2)$, but I have no idea how to do it for $k \geq 1$.
Thanks!
|
Using the fact that for all $r\in (-1,1)$ and all real $x$,
\begin{align*}
\sum_{n=-\infty}^\infty r^{|n|}e^{inx} =\frac{1-r^2}{1+r^2 - 2r\cos x}
\end{align*} (for background, this is the series representation of the Poisson kernel) we get
\begin{align*}
\int_0^{\pi} \frac{\cos(kx)}{1+r^2 -2r\cos x} dx =& \frac 1 2 \int_{-\pi}^{\pi} \frac{e^{ikx}}{1+r^2 -2r\cos x} dx\\
=&\frac 1{2}\int_{-\pi}^{\pi} \left(\frac 1 {1-r^2}\sum_{n=-\infty}^\infty r^{|n|}e^{inx}\right)e^{ikx} dx\\
=&\frac 1{2(1-r^2)}\sum_{n=-\infty}^\infty r^{|n|}\underbrace{\int_{-\pi}^{\pi} e^{i(n+k)x} dx}_{=0 \ \text{ if }\ n+k\ne 0}\\
=&\frac { \pi r^k}{1-r^2}.
\end{align*} Now solving for $$\frac r{1+r^2} = \tau \ \ \Longrightarrow \ \ r=\frac {1-\sqrt{1-4\tau^2}}{2\tau} \in (-1,1)
$$ we finally get
\begin{align*}
\int_0^{\pi} \frac{\cos(kx)}{1 -2\tau\cos x} dx = &\frac{\pi r^k(1+r^2)}{1-r^2} =\frac{\pi}{ \sqrt{1-4\tau^2}}\left(\frac{1-\sqrt{1-4\tau^2}}{2\tau}\right)^{k}.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3343190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
}
|
Prove that $\left|\left\{\frac{n}{1}\right\} - \left\{\frac{n}{2}\right\} - \cdots - (-1)^n\left\{\frac{n}{n}\right\}\right| \le \sqrt{2n}$.
For all positive integers $n$, prove that $$\large \left|\left\{\frac{n}{1}\right\} - \left\{\frac{n}{2}\right\} + \left\{\frac{n}{3}\right\} - \cdots - (-1)^n\left\{\frac{n}{n}\right\}\right| \le \sqrt{2n}$$
This is the last problem of a book I (wrongly) stole from the shelves from our class. And there definitely were shortcuts taken.
I also don't really understand the last part $$ [\cdots ]\le \frac{m - 2}{2} + \frac{n}{m} < \frac{\sqrt{2n} - 1}{2} + \sqrt{\frac{n}{2}} < 2n$$. Could someone please explain?
Moreover, the answer is lengthy, at least to me. So if you have any other solution that is shorter, please answer below and I would be appreciated.
|
Let $m = \lfloor\sqrt{2n} + 1\rfloor$. We have that $$\left|\left\{\frac{n}{1}\right\} - \left\{\frac{n}{2}\right\} + \left\{\frac{n}{3}\right\} - \cdots - (-1)^n\left\{\frac{n}{n}\right\}\right|$$
equals the absolute value of
$$\underbrace{\left(\left\{\frac{n}{1}\right\} + \left\{\frac{n}{2}\right\} - \left\{\frac{n}{3}\right\} + \cdots - (-1)^{m - 1}\left\{\frac{n}{m - 1}\right\}\right)}_{A}$$
$$ - (-1)^m\underbrace{\left(\frac{n}{m} + \frac{n}{m + 1} - \frac{n}{m + 2} + \cdots + (-1)^{n - m}\frac{m}{n}\right)}_{B}$$
$$ + (-1)^m\underbrace{\left(\left\lfloor\frac{n}{m}\right\rfloor + \left\lfloor\frac{n}{m + 1}\right\rfloor - \left\lfloor\frac{n}{m + 2}\right\rfloor + \cdots + (-1)^{n - m}\left\lfloor\frac{n}{n}\right\rfloor\right)}_{C}$$
Going through each term, we have that
$$ - \left(\left\{\dfrac{n}{2}\right\} + \left\{\dfrac{n}{4}\right\} + \left\{\dfrac{n}{6}\right\} + \cdots\right) \le A \le \left\{\dfrac{n}{1}\right\} + \left\{\dfrac{n}{3}\right\} + \left\{\dfrac{n}{5}\right\} + \cdots$$
with the expressions on the left- and right-hand side having $\left\lfloor\dfrac{m - 1}{2}\right\rfloor$ and $\left\lfloor\dfrac{m}{2}\right\rfloor$ respectively.
(It should be also noted that $\left\{\dfrac{n}{1}\right\} = 0$.)
For all $\left\{\dfrac{n}{p}\right\}$ $(p \in \mathbb Z^+, 1 \le p \le m)$, we have that $\left\{\dfrac{n}{p}\right\} \le \dfrac{p - 1}{p} \le \dfrac{m - 2}{m - 1}$
$$\implies |A| \le \left\lfloor\frac{m - 1}{2}\right\rfloor \cdot \dfrac{m - 2}{m - 1} \le \dfrac{m - 2}{2}$$
It is also evident that $$0 \le \left(\frac{m}{n} - \frac{n}{m + 1}\right) + \left(\frac{n}{m + 2} - \frac{n}{m + 3}\right) + \cdots $$
$$ = B = \frac{m}{n} - \left(\frac{n}{m + 1} - \frac{n}{m + 2}\right) - \cdots \le \dfrac{n}{m}$$
and $$0 \le \left(\left\lfloor\frac{m}{n}\right\rfloor - \left\lfloor\frac{n}{m + 1}\right\rfloor\right) + \left(\left\lfloor\frac{n}{m + 2}\right\rfloor - \left\lfloor\frac{n}{m + 3}\right\rfloor\right) + \cdots $$
$$ = B = \left\lfloor\frac{m}{n}\right\rfloor - \left(\left\lfloor\frac{n}{m + 1}\right\rfloor - \left\lfloor\frac{n}{m + 2}\right\rfloor\right) - \cdots \le \left\lfloor\dfrac{n}{m}\right\rfloor \le \dfrac{n}{m}$$
$$\implies \left|\left\{\frac{n}{1}\right\} - \left\{\frac{n}{2}\right\} + \left\{\frac{n}{3}\right\} - \cdots - (-1)^n\left\{\frac{n}{n}\right\}\right| = |A - (-1)^mB + (-1)^mC|$$
$$ \le \frac{m - 2}{2} + \frac{n}{m} < \frac{\sqrt{2n} - 1}{2} + \sqrt{\frac{n}{2}} < 2n$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3346484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to solve $\int^1_0 (1+7x)^{1/3}dx$? I worked through $\int^1_0 (1+7x)^{1/3}dx$ and I got $\frac{3}{4}+C$ for the answer. However, I forgot about the exponent when I found the difference of the sides of the integral.
I am retrying it, but I've realized I don't know how to find a number to the power of $\frac{4}{3}$. Also, when I went over this with Symbolab, once u-substitution had been applied, the integral changed to $\int^8_1$ for some reason. I'm sure it's the key to solving this, but I have no idea why that's even allowed.
My textbook and Symbolab both say the answer is $\frac{45}{28}$.
Here are the steps I took. Please let me know what I got wrong.
$\int^1_0 (1+7x)^{1/3}dx$
Let $u=1+7x$
Then $du=7dx$ and $dx=\frac{1}{7}du$
so $\int^1_0u^{1/3}\frac{1}{7}du=\frac{1}{7}\int^1_0u^{1/3}$
$$\frac{1}{7}\int^1_0u^{1/3}$$
$$=\frac{1}{7}[\frac{u^{4/3}}{4/3}|^1_0]$$ $$=\frac{1}{7}[\frac{3u^{4/3}}{4}|^1_0]$$
$$=\frac{1}{7}[\frac{3(1+7(1))^{4/3}}{4}-\frac{3(1+7(0))^{4/3}}{4}]$$
$$=\frac{1}{7}[\frac{3(1+7)^{4/3}}{4}-\frac{3(1+0)^{4/3}}{4}]$$
This is as far as I can get.
|
You got $\dfrac{1}{7}\left[\dfrac{3(1+7)^{4/3}}{4}-\dfrac{3(1+0)^{4/3}}{4}\right].$
Note that $(1+7)^{4/3}=8^{4/3}=(8^{1/3})^4=2^4=16$ and $(1+0)^{4/3}=1^{4/3}=1,$
so you actually got the correct answer: $\dfrac17\left[\dfrac{3\times16}4-\dfrac{3\times1}4\right]=\dfrac17\dfrac{45}4=\dfrac{45}{28}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3349010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Solve Diophantine equation with three variables part two I want to find all solutions of $$x^2+y^2+z^2-xy-yz-zx-x-y-z=0$$
Solutions need not to be primitive. I found several parametric family. For example
$(m^2, m^2+m , (m+1)^2)$
$(m^2, m^2+m+2 , (m+1)^2)$
$(m^2+1, m^2+m, (m+1)^2+1)$
$(m^2+1, m^2+m+4, (m+1)^2+1)$
$(m^2-m, m^2, (m+1)^2-(m+1))$
$(m^2-m, m^2+1, (m+1)^2-(m+1))$
$((m+2)^2-m, (m+2)^2, (m+3)^2-(m+1))$
$((m+2)^2-m, (m+2)^2+5, (m+3)^2-(m+1))$
$((m+2)^2+5, (m+2)^2+m+4, (m+3)^2+5)$
$((m+2)^2+5, (m+2)^2+m+12, (m+3)^2+5)$
$((m+3)^2-m+1, (m+3)^2+1, (m+4)^2-m)$
$((m+3)^2-m+1, (m+3)^2+8, (m+4)^2-m)$
$((m+4)^2+8, (m+4)^2+m+8, (m+5)^2+8)$
$((m+4)^2+8, (m+4)^2+m+18, (m+5)^2+8)$
$((m+3)^2+m+13, (m+5)^2+5, (m+4)^2+m+14)$
$((m+3)^2+m+13, (m+5)^2+16, (m+4)^2+m+14)$
I suspect I am missing something. These type of single-valued parametric solutions may exist in infinite numbers. Any help will be much appreciated.
|
I was wrong about the formula... it looks like this....
$$X^2+Y^2+Z^2=XY+XZ+ZY+X+Y+Z$$
$$X=s(3(k^2-kt+t^2)s-k-t)$$
$$Y=s(3(k^2-kt+t^2)s+2k-t)$$
$$Z=s(3(k^2-kt+t^2)s-k+2t)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3350399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Finding $x^3 + y^3$ when $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $ $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $
$x^3 + y^3 =?$
my answer =
$(3 + \sqrt5)^3 = 47 + 32\sqrt5$
$(3 - \sqrt5)^3 = 47 - 32\sqrt5$
$x^3 + y^3 = \frac{47 + 32\sqrt5}{47 - 32\sqrt5} + \frac{47 - 32\sqrt5}{47 + 32\sqrt5} = 2*7329/-2911$
why my answer is wrong? please help me
|
You have made mistakes in computing $(3+\sqrt 5 )^{3}$ and $(3-\sqrt 5 )^{3}$. These are $72+32\sqrt 5$ and $72-32\sqrt 5$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3350519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
find minimum of maximum of two functions
I need to find the angle $\theta$ so that: $$\max(\cos^2(\theta),1-\cos^2(45-\theta))$$ is minimized.
OK, so I wrote \begin{align*}f(\theta)&=\max(\cos^2(\theta),1-\cos^2(45-\theta))\\
&=\frac{\cos^2(\theta)+1-\cos^2(45-\theta)+|\cos^2(\theta)-1+\cos^2(45-\theta)|}{2}\end{align*}
Then I'm stuck.
|
Is known that
$$2\cos^2a = 1+\cos 2a.$$
So
$$f(\theta) = \dfrac12\max\left(1+\cos2\theta,2-\left(1+\cos\left(\dfrac\pi2-2\theta\right)\right)\right),$$
$$f(\theta) = \dfrac12+\dfrac12\max(\cos2\theta,-\sin2\theta).$$
Since
$$\cos2\theta+\sin2\theta = \sqrt2\left(\sin\dfrac\pi4\cos2\theta+\cos\dfrac\pi4\sin2\theta\right) = \sqrt2\sin\left(2\theta+\dfrac\pi4\right),$$
then
$$f(\theta) =
\begin{cases}
\dfrac12+\dfrac12\cos2\theta,\quad\text{if}\quad
2\theta\in\left[2k\pi-\dfrac\pi4,2k\pi+\dfrac{3\pi}4\right]\\
\dfrac12-\dfrac12\sin2\theta,\quad\text{if}\quad
2\theta\in\left[2k\pi+\dfrac{3\pi}4,2k\pi+\dfrac{7\pi}4\right],
\end{cases}$$
where
$$k\in\mathbb Z.$$
If
$$2\theta\in\left(2k\pi-\dfrac\pi4,2k\pi+\dfrac{3\pi}4\right),$$
then
$$\min f(\theta) = \dfrac12-\dfrac{\sqrt2}4\quad\text{at}\quad 2\theta=2k\pi+\dfrac{3\pi}4.$$
If
$$2\theta\in\left[2k\pi+\dfrac{3\pi}4,2k\pi+\dfrac{7\pi}4\right],$$
then
$$\min f(\theta) = \dfrac12-\dfrac{\sqrt2}4\quad\text{at}\quad 2\theta=2k\pi+\dfrac{3\pi}4.$$
Therefore,
$$\boxed{\min f(\theta) = \dfrac12-\dfrac{\sqrt2}4\quad\text{at}\quad \theta=k\pi+\dfrac{3\pi}8.}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3350686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Prove a trigonometric identity: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ when $A+B+C=\pi$ There is a trigonometric identity:
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\equiv 1\text{ when }A+B+C=\pi$$
It is easy to prove it in an algebraic way, just like that:
$\quad\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\\=\cos^2A+\cos^2B+\cos^2\left(\pi-A-B\right)+2\cos A\cos B\cos \left(\pi-A-B\right)\\=\cos^2A+\cos^2B+\cos^2\left(A+B\right)-2\cos A\cos B\cos \left(A+B\right)\\=\cos^2A+\cos^2B+\left(\cos A\cos B-\sin A\sin B\right)^2-2\cos A\cos B\left(\cos A\cos B-\sin A\sin B\right)\\=\cos^2A+\cos^2B+\cos^2A\cos^2B+\sin^2A\sin^2B-2\sin A\cos A\sin B\cos B-2\cos^2A\cos^2B+2\sin A\cos A\sin B\cos B\\=\cos^2A+\cos^2B-\cos^2A\cos^2B+\left(1-\cos^2A\right)\left(1-\cos^2B\right)\\=\cos^2A+\cos^2B-\cos^2A\cos^2B+1-\cos^2A-\cos^2B+\cos^2A\cos^2B\\=1$
Then, I want to find a geometric way to prove this identity, as $A+B+C=\pi$ and it makes me think of the angle sum of triangle. However, it is quite hard to prove it in a geometric way. Therefore, I hope there is someone who can help. Thank you!
|
A purely geometric way doesn't look likely, because the degrees of the cosine terms (two and three respectively) don't match. For what it's worth, here is an alternative trigonometric derivation.
Writing $2\cos B\cos C$ in the second term as $\cos (B+C)+\cos(B-C)$ transforms our expression to$$\cos^2A+\cos^2B+\cos^2C+[\cos(B+C)+\cos(B-C)]\cos A.$$Notice that $\cos(B+C)=-\cos A$, and use this conversion forwards and backwards to give$$\cos^2B+\cos^2C-\cos(B-C)\cos(B+C).$$Now write the last term as $-\frac12(\cos2B+\cos2C)$ and express the first terms also in double-angle format. Then cancellation yields the required result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3350887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 3
}
|
For positive real numbers $a,b,c$ prove that $ a^4 + b^4 + c^4 \ge abc(a+b+c)$ For positive reals $a,b,c$ prove that $$ a^4+b^4+c^4 \ge abc(a+b+c). $$
I tried to pls around trying to reorganize to get AM-GM but i couldn't
Thanks for the help in advance.
|
Hint
$$\frac{a^4+a^4+b^4+c^4}{4}\geq abca \\
\frac{a^4+b^4+b^4+c^4}{4}\geq abcb \\
\frac{a^4+c^4+b^4+c^4}{4}\geq abcc \\$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3353216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Finding all the complex roots of $x^3 - x^2 + 1 = 0$ I'm trying to calculate the exact values of the roots of $x^3 - x^2 +1 =0$ (When I say exact value, here I mean find the roots in the form of $a + bi$ for some $a,b \in \mathbb{R}$
My first thought is to get it to a form where I can use Cardano's method and so, since 0 isn't a root, I took $x = 1/t$ and then got
$t^3 - t +1 = 0$
So by Cardano's formula, I got that
t = $\sqrt[3]{ \frac{-1}{2} + \sqrt[2]{\frac{1}{4}-\frac{1}{27}}} \sqrt[3]{ \frac{-1}{2} - \sqrt[2]{\frac{1}{4}-\frac{1}{27}}}$
= $\sqrt[3]{ \frac{-1}{2} + \sqrt[2]{\frac{23}{108}}} + \sqrt[3]{ \frac{-1}{2} - \sqrt[2]{\frac{23}{108}}}$
In $\mathbb{C}$, both $\sqrt[3]{ \frac{-1}{2} + \sqrt[2]{\frac{23}{108}}}$ and $\sqrt[3]{ \frac{-1}{2} - \sqrt[2]{\frac{23}{108}}}$ have three cube roots, and so taking $u$ and $v$ to be some particular cube roots, I know that I need to compute $\omega ^k u + \omega^l v$ for $k,l \in \{0,1,2\}$ and $\omega$ being a cube root of unity.
But I'm not sure about 1) how to compute any particular values of $u$ and $v$ in this case, as well as 2) if there is a cleaner way to approach this problem and solving monic cubic polynomials of this form in general.
Any advice would be appreciated!
|
Without change of variables, using Cardano method, the raw results are
$$x_1=\frac{1}{3} \left(1-\sqrt[3]{\frac{2}{25-3 \sqrt{69}}}-\sqrt[3]{\frac{1}{2}
\left(25-3 \sqrt{69}\right)}\right)$$
$$x_2=\frac{1}{3}+\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(25-3
\sqrt{69}\right)}+\frac{1-i \sqrt{3}}{3\ 2^{2/3} \sqrt[3]{25-3 \sqrt{69}}}$$
$$x_3=\frac{1}{3}+\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(25-3
\sqrt{69}\right)}+\frac{1+i \sqrt{3}}{3\ 2^{2/3} \sqrt[3]{25-3 \sqrt{69}}}$$
Now, being patient
$$x_{2,3}=a\pm b i$$ where
$$a=\frac{1}{3}+\frac{1}{6} \sqrt[3]{\frac{1}{2} \left(25-3
\sqrt{69}\right)}+\frac{1}{3\ 2^{2/3} \sqrt[3]{25-3 \sqrt{69}}}$$
$$b=\frac{\sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}}{2
\sqrt{3}}-\frac{1}{2^{2/3} \sqrt{3} \sqrt[3]{25-3 \sqrt{69}}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3354712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
For $x$ and $k$ real numbers, for what values of $k$ will the graphs of $f(x)=-2\sqrt{x+1}$ and $g(x)=\sqrt{x-2}+k$ intersect?
For $x$ and $k$ real numbers, for what values of $k$ will the graphs of $f(x)=-2\sqrt{x+1}$ and $g(x)=\sqrt{x-2}+k$ intersect?
I tried to make an equation of them, but I’m stuck with the two variables and I couldn’t solve it. Much appreciation.
We didn’t do calculus yet..
|
Hint: You must solve the equation
$$\sqrt{x-2}+k=-2\sqrt{x+1}$$ for $$x\geq 2$$ Writing this equation in the form
$$-k-\sqrt{x-2}=2\sqrt{x+1}$$ then it must be
$$k^2+2\geq x$$ and you can square it.
After squaring one times we get
$$2k\sqrt{x-2}=3x+6-k^2$$
Squaring again we get
$$-k^4+10 k^2 x+4 k^2-9 x^2-36 x-36=0$$
Solving this we get
$$k\leq -2 \sqrt{3}\land x=\frac{1}{9} \left(5
k^2-18\right)-\frac{4}{9} \sqrt{k^4-9 k^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3355215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Minimizing the error for a second degree interpolating polynomial
Construct the second degree polynomial $q_2(t)$ that approximates $g(t) = \sin(\pi t)$ on the interval [0,1] by minimizing
$$\int_0^1 [g(t) - q_2(t)]^2dt$$
A useful integral: $\int_0^1 (6t^2-6t+1)^2dt=\frac{1}{5}$
My attempt:
We Know $q_2(t)$ will come in the form of $q_2(t) = zt^2 + yt + x$.
First we can expand $[g(t) - q_2(t)]^2dt$ and get the integral
$$\int_0^1 g^2(t)-2q_2(t)g(t)+q_2^2(t)dt$$
$$\rightarrow \int_0^1 \sin^2(\pi t)-2\sin(\pi t)(zt^2+yt+x)+(zt^2+yt+x)^2dt$$
$$\rightarrow \frac{1}{2}-\frac{4x}{\pi}-\frac{2y}{\pi}-\frac{2z(\pi^2-4)}{\pi^3}+\int_0^1 (zt^2+yt+x)^2dt$$
I am having difficulty utilizing the hint to handle $\int_0^1 (zt^2+yt+x)^2dt$. Is the hint implying that the the second degree polynomial $q_2(t)$ is equal to $6t^2-6t+1$? My big picture goal of this problem was to find $x,y,z$ such that the integral is minimized by expanding and solving the integral and then solving for $x,y,z$. What am I missing, or what am I doing wrong? Thank you!
|
I do not see why the hint was even given since
$$I=\int_0^1 (zt^2+yt+x)^2\,dt=\int_0^1 (x^2+2 x yt+ \left(2 x z+y^2\right)t^2+2 y z t^3+ z^2 t^4)\,dt$$ that is to say
$$I=x^2+x y+\frac{2 x z}{3}+\frac{y^2}{3}+\frac{y z}{2}+\frac{z^2}{5}$$ So, for the total
$$R=\frac{1}{2}-\frac{4x}{\pi}-\frac{2y}{\pi}-\frac{2z(\pi^2-4)}{\pi^3}+x^2+x y+\frac{2 x z}{3}+\frac{y^2}{3}+\frac{y z}{2}+\frac{z^2}{5}$$ Computing the partial derivatives
$$\frac{\partial R}{\partial x}=2 x+y+\frac{2 z}{3}-\frac{4}{\pi }=0$$
$$\frac{\partial R}{\partial y}=x+\frac{2 y}{3}+\frac{z}{2}-\frac{2}{\pi }=0$$
$$\frac{\partial R}{\partial z}=\frac{2 x}{3}+\frac{y}{2}+\frac{2 z}{5}-\frac{2 \left(\pi ^2-4\right)}{\pi ^3}=0$$
Solving for $(x,y,z)$
$$x=\frac{12 \left(\pi ^2-10\right)}{\pi ^3} \qquad y=\frac{60 \left(12-\pi^2\right)}{\pi ^3}\qquad z=\frac{60 \left(\pi ^2-12\right)}{\pi ^3}$$ and, for these values,
$$R=\frac{1}{2}-\frac{24 \left(120-20 \pi ^2+\pi ^4\right)}{\pi ^6}\approx 3 \times 10^{-4}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3356458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
On the equation $\sqrt{x^2-9}=\frac{2(x+3)}{(x-3)^2}-x$ I'm trying to solve the equation
$$\sqrt{x^2-9}=\frac{2(x+3)}{(x-3)^2}-x \tag{1}$$
Attempt: I have rewritten the equation as
\begin{align*}
\sqrt{x^2-9} = \frac{2(x+3)}{\left (x-3 \right )^2}-x &\Leftrightarrow \sqrt{x^2-9} = \frac{2(x+3)-x\left ( x-3 \right )^2}{\left ( x-3 \right )^2} \\
&\Leftrightarrow \sqrt{x^2-9} = - \frac{x^3-6x^2+7x-6}{\left ( x-3 \right )^2} \\ &\Leftrightarrow \sqrt{x^2-9} = \frac{2}{x-3} -x + \frac{12}{\left ( x-3 \right )^2}
\end{align*}
This seems manageable (?) but I do not know how to proceed. The solution is $x=8-\sqrt{13}$ as suggested by Mr. Wolfy. I have no idea how to get it.
In the mean time if we square $(1)$ all 6th powers are simplified and we are left with
$$5x^4-96x^3+526x^2-105x+765=0$$
Trying to factorising it we get with some luck that
$$(x^2-16x+51)(5x^2-16x+15) =0$$
and we have to solve this. This is easy but we have to take into account the restriction
$$ -x^3+6x^2-7x+6\geq 0 $$
|
Difference of two squares identity and its Conjugates can help:
$\sqrt{x^2-9} = \frac{2(x+3)}{(x-3)^2} - x$
$(x-3)^2(\sqrt{x^2-9} + x) = 2(x+3)$
$(x-3)^2(\sqrt{x^2-9} + x)(\sqrt{x^2-9} - x) = 2(x+3)(\sqrt{x^2-9} - x)$
$(x-3)^2(x^2-9 - x^2) = 2(x+3)(\sqrt{x^2-9} - x)$
$-9(x-3)^2 = 2(x+3)(\sqrt{x^2-9} - x)$
$\frac{-9(x-3)^2}{2(x+3)} = \sqrt{x^2-9} - x$
$\frac{-9(x-3)^2}{2(x+3)} + x = \sqrt{x^2-9}$
We achieve new equation for $\sqrt{x^2-9}, \ $ and so we can equal this one to the first equation:
$\frac{-9(x-3)^2}{2(x+3)} + x = \sqrt{x^2-9} = \frac{2(x+3)}{(x-3)^2} - x$
$\frac{-9(x-3)^2}{2(x+3)} + x = \frac{2(x+3)}{(x-3)^2} - x$
Now we get ride of radicals, so we can solve it without any restriction caused by squaring.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3357131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
What is the probability that a four-digit number formed by randomly selecting four-digits without replacement is greater than $4321$? A four-digit number is formed by randomly selecting four digits, without replacement, from the set $D = \{ 1, 2, 3, 4, 5, 6, 7\}$. What is the probability that the resulting number is greater than $4321$?
Attempt: The odds of choosing a $5$, $6$, or $7$ are $1/7 + 1/7 + 1/7 = 3/7$. Then the odds of choosing a $4$ for the first number are $1/7$. If my second number is $5$, $6$, or a $7$, I know the number will be greater than $4321$. This is $3/6$ so $1/7 \cdot 3/6 = 3/42$. Then I'm getting stuck.
|
What you have done thus far is correct. That leaves cases in which the first number is $4$ and the second number is $3$. The probability of first obtaining $4$, then obtaining $3$ is $\frac{1}{7} \cdot \frac{1}{6} = \frac{1}{42}$. We now consider two cases: the third digit is larger than $2$ or the third digit is $2$ and the fourth digit is larger than $1$.
The first digit is $4$, the second digit is $3$, and the third digit is larger than $2$: The remaining digits are $1, 2, 5, 6, 7$. Of these, three are larger than $2$. Hence, the probability that the first digit is $4$, the second digit is $3$, and the third digit is larger than $2$ is
$$\frac{1}{7} \cdot \frac{1}{6} \cdot \frac{3}{5}$$
The first digit is $4$, the second digit is $3$, the third digit is $2$, and the fourth digit is larger than $1$: Once $4$, $3$, and $2$ have been selected, the remaining digits are $1, 5, 6, 7$, of which three are larger than $1$. Hence, the probability that the first digit is $4$, the second digit is $3$, the third digit is $2$, and the fourth digit is larger than $1$ is
$$\frac{1}{7} \cdot \frac{1}{6} \cdot \frac{1}{5} \cdot \frac{3}{4}$$
Hence, the probability of randomly selecting a four-digit number without replacement from the set $D = \{1, 2, 3, 4, 5, 6, 7\}$ that is larger than $4321$ is
$$\frac{3}{7} + \frac{1}{7} \cdot \frac{3}{6} + \frac{1}{7} \cdot \frac{1}{6} \cdot \frac{3}{5} + \frac{1}{7} \cdot \frac{1}{6} \cdot \frac{1}{5} \cdot \frac{3}{4}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3358132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to calculate $\int_0^1 \frac{\ln(1-x)}{x} dx$? I have tried to let$f(a)=\int_0^1\frac{\ln(1-ax)}{x} dx$, which means
$f'(x)=-a\int_0^1 \frac{1}{x(1-ax)} dx=a\int_0^1 \left[\frac{a}{ax-1}-\frac{1}{x}\right] dx$.
However, I am stuck at here.
|
Let us consider
$$ I = \int_{0}^{+\infty}\frac{\arctan(x)}{1+x^2}\,dx =\frac{1}{2}\left[\arctan^2(x)\right]_{0}^{+\infty}=\frac{\pi^2}{8}$$
and the parametric integral
$$ I(a) = \int_{0}^{+\infty}\frac{\arctan(ax)}{1+x^2}\,dx. $$
We have $I(0)=0$ and $I(1)=I=\frac{\pi^2}{8}$. By differentiation under the integral sign
$$\frac{\pi^2}{8} = \int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+a^2 x^2)}\,dx\,da$$
and by partial fraction decomposition $\frac{x}{(1+x^2)(1+a^2 x^2)}=\frac{1}{1-a^2}\cdot\left(\frac{x}{1+a^2 x^2}-\frac{x}{1+x^2}\right)$, so
$$ \frac{\pi^2}{8}=\int_{0}^{1}\frac{\log(a)}{a^2-1}\,da=-\sum_{n\geq 0}\int_{0}^{1}a^{2n}\log(a)\,da=\sum_{n\geq 0}\frac{1}{(2n+1)^2}. $$
By termwise integration of a Maclaurin series we have
$$ \int_{0}^{1}\frac{\log(x)}{1-x}\,dx = -\sum_{n\geq 1}\frac{1}{n^2}, $$
where the absolute convergence of the RHS allows to state
$$ S=\sum_{n\geq 1}\frac{1}{n^2}=\sum_{m\geq 1}\frac{1}{(2m)^2}+\sum_{m\geq 0}\frac{1}{(2m+1)^2}=\frac{1}{4}S+\frac{\pi^2}{8}. $$
Connecting the dots:
$$ \int_{0}^{1}\frac{\log(1-x)}{x}\,dx\stackrel{x\mapsto 1-x}{=}\int_{0}^{1}\frac{\log x}{1-x}\,dx = -\frac{4}{3}\int_{0}^{+\infty}\frac{\arctan(x)}{1+x^2}\,dx = \color{red}{-\frac{\pi^2}{6}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3358436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
Two circles tangent to each other and tangent to a line Two circles with centers $A$ and $B$ are externally tangent at point $C$.
tangent to the two circles. Given that the radii of the two circles are $2cm$ and $3cm$, respectively, find $\frac{DC}{FC}$
|
Adjusting Matthew's figure:
$\hspace{2cm}$
Use Cosine theorem:
$$\frac{FC}{DC}=\frac{\sqrt{3^2+3^2-2\cdot 3^2\cdot \cos \beta}}{\sqrt{2^2+2^2-2\cdot 2^2\cdot \cos (180^\circ-\beta)}}=\sqrt{\frac{18-18\cdot \frac15}{8+8\cdot \frac15}}=\sqrt{\frac32}.$$
Note: $\cos \beta =\frac{BE}{AB}=\frac{FB-FE}{AC+BC}=\frac{FE-AD}{AC+BC}=\frac{3-2}{2+3}=\frac15$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3362914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
The closed-form of $\sum_{n=0}^{\infty}\frac{(-1)^n H^{(2)}_{n}}{(2n+1)^2} $ How to Prove that
$$ \sum_{n=0}^{\infty}\frac{(-1)^nH^{(2)}_{n}}{(2n+1)^2} \;\;=\;\;\frac{7 \pi \; \zeta(3)}{4}-\frac{\zeta(2)G}{2}+\frac{45\zeta(4)}{8}-\frac{\Psi^{(3)}\big(\frac{1}{4}\big)}{128}$$
where $H_n^{(m)}=\sum_{k=1}^n\frac1{k^m}$ is the $n$th generalized harmonic number of order $m$, $\zeta$ is the Riemann zeta function and $G$ is Catalan constant?
This problem proposed by Ahmad Albow
|
Solution by Cornel Valean.
\begin{align}
S&=\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{(2n+1)^2}\\
&=\sum_{n=1}^\infty(-1)^{n-1}H_n^{(2)}\int_0^1-x^{2n}\ln x\ dx\\
&=\int_0^1\ln x\sum_{n=1}^\infty(-x^2)^nH_n^{(2)}\ dx\\
&=\int_0^1\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx\\
&=\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx-\underbrace{\int_1^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx}_{x\mapsto 1/x}\\
&=\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx+\int_0^1\frac{\ln x\operatorname{Li}_2(-1/x^2)}{1+x^2}\ dx\\
2S&=\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx+\int_0^1\frac{\ln x[\color{red}{\operatorname{Li}_2(-x^2)+\operatorname{Li}_2(-1/x^2)}]}{1+x^2}\ dx\\
S&=\frac12\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx+\frac12\int_0^1\frac{\ln x[\color{red}{-2\ln^2x-\zeta(2)}]}{1+x^2}\ dx\\
&=\frac12\underbrace{\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx}_{I}-\underbrace{\int_0^1\frac{\ln^3x}{1+x^2}\ dx}_{-6\beta(4)}-\frac12\zeta(2)\underbrace{\int_0^1\frac{\ln x}{1+x^2}\ dx}_{-G}\\
&=\frac12I+6\beta(4)+\frac12G\zeta(2)\tag1
\end{align}
\begin{align}
I&=\int_0^\infty\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx\\
&=\int_0^\infty\frac{\ln x}{1+x^2}\left(\int_0^1\frac{x^2\ln y}{1+yx^2}\ dy\right)\ dx\\
&=\int_0^1\ln y\left(\int_0^\infty\frac{x^2\ln x}{(1+x^2)(1+yx^2)}\ dx\right)\ dy\\
&=\int_0^1\frac{\ln y}{1-y}\left(\int_0^\infty\frac{\ln x}{1+yx^2}\ dx-\underbrace{\int_0^\infty\frac{\ln x}{1+x^2}\ dx}_{0}\right)\ dy\\
&=\int_0^1\frac{\ln y}{1-y}\left(-\frac{\pi}{4}.\frac{\ln y}{\sqrt{y}}\right)\ dy,\quad \sqrt{y}=x\\
&=-2\pi\int_0^1\frac{\ln^2x}{1-x^2}\ dx=-2\pi\left(-\frac74\zeta(3)\right)=\boxed{-\frac72\pi\zeta(3)}\tag2
\end{align}
Plug (2) and (1) we get
$$S=-\frac74\pi\zeta(3)+6\beta(4)+\frac12G\zeta(2)$$
where $\beta(4)=\frac{\psi_3(1/4)}{768}-\frac{15}{16}\zeta(4)$
Addendum:
Another approach is by applying integration by parts to the integral $S$ then we change the limits from $(0,1)$ to $(0,\infty)$ as we did in our solution above.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3364852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Prime Number Theory (Dividers) When $N = a^\alpha × b^β × c^γ$ is divided by $a$, the number of dividers decreases by $m$ units; when divided by $b$, decreases by $n$ and, dividing by $c$, decreases by $p$. Determine the exponents $α, β$ and $γ$.
Mine ended up giving in this system here:
$$
(α-a)(β+1)(γ+1)-(γ+1)(β+1) = (α+1)(β+1)(γ+1)-m;
(β-b)(α+1)(γ+1)-(γ+1)(α+1) = (α+1)(β+1)(γ+1)-n;
(γ-b)(α+1)(β+1)-(α+1)(β+1) = (α+1)(β+1)(γ+1)-p
$$
|
When $N = a^\alpha \times b^{\beta} \times c^{\gamma}$ is divided by $a$, it becomes $\frac{N}{a} = a^{\alpha - 1} \times b^{\beta} \times c^{\gamma}$. Thus, as metamorphy's question comment indicates that if we can assume $a$, $b$ and $c$ are distinct primes, then since the number of divisors decreases by $m$, this means that
$$\begin{equation}\begin{aligned}
(\alpha - 1 + 1)(\beta + 1)(\gamma + 1) & = (\alpha + 1)(\beta + 1)(\gamma + 1) - m \\
(\alpha + 1)(\beta + 1)(\gamma + 1) - (\beta + 1)(\gamma + 1) & = (\alpha + 1)(\beta + 1)(\gamma + 1) - m \\
- (\beta + 1)(\gamma + 1) & = -m \\
(\beta + 1)(\gamma + 1) & = m
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Similarly, we also have
$$(\alpha + 1)(\gamma + 1) = n \tag{2}\label{eq2A}$$
$$(\alpha + 1)(\beta + 1) = p \tag{3}\label{eq3A}$$
Multiplying \eqref{eq1A}, \eqref{eq2A} and \eqref{eq3A} together gives
$$\begin{equation}\begin{aligned}
\left((\alpha + 1)(\beta + 1)(\gamma + 1)\right)^2 & = mnp \\
(\alpha + 1)(\beta + 1)(\gamma + 1) & = \sqrt{mnp}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Dividing \eqref{eq4A} by \eqref{eq1A} gives
$$\begin{equation}\begin{aligned}
\alpha + 1 & = \frac{\sqrt{mnp}}{m} \\
\alpha & = \sqrt{\frac{np}{m}} - 1
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Similarly, you can also get that
$$\beta = \sqrt{\frac{mp}{n}} - 1 \tag{6}\label{eq6A}$$
$$\gamma = \sqrt{\frac{mn}{p}} - 1 \tag{7}\label{eq7A}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3369823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
how to solve $\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}$? $\displaystyle\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+1}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+n}}\right)^{n}$
left=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+1}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+1}}} }=e^{0}=1$
right=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+n}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+n}}} }=e^{-\frac{1}{2}}$
left $\ne$ right ,what to do next?
$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{\displaystyle n \ln \sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}(1)}$
$(1)=\displaystyle \lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n^{2}}}\right)$$=\lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n \cdot 1/n}}\right)$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{1}{\sqrt{1+x/n}} d x$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{nd(x/n+1)}{\sqrt{1+x/n}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 \left.\sqrt{1+\frac{x}{n} }\right|_{0}^{1}}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 (\sqrt{1+\frac{1}{n}}-1)}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} -\frac{1}{2n^2} +o(\frac{1}{n^2}))}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} +\left(\frac{1}{2!}\cdot \frac{1}{2} \cdot \left(\frac{1}{2}-1\right) \right)\frac{1}{n^2} +o(\frac{1}{n^2}))}}=\lim_{n\to\infty}{n \ln{\left(1-\frac{1}{4n}\right)}}=-\frac{1}{4} $
so that $\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{(1)}=e^{-\frac{1}{4}}$
this solution is right.
|
We have
$$\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim_{n\to\infty} \frac1{n^n}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k/n^2}}}\right)^{n}}$$
and
$$\frac{1}{\sqrt{1+k/n^2}}=1-\frac12\frac{k}{n^2}+O\left(\frac{k^2}{n^4}\right)$$
therefore
$$\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k/n^2}}}=n-\frac12\frac{n(n+1)}{2n^2}+O\left(\frac1n\right)$$
and
$$\frac1{n^n}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k/n^2}}}\right)^{n}}=\left(1-\frac{n+1}{4n^2}+O\left(\frac1{n^2}\right)\right)^n\to\frac1{\sqrt[4] e}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3370438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
}
|
Solving $2\sin(x+30^\circ) = \cos(x+150^\circ)$ for $x$ between $0^\circ$ and $360^\circ$
Solve this equation for $x$ between $0^\circ$ and $360^\circ$:
$$2\sin(x+30^\circ) = \cos(x+150^\circ)$$
Anyway thank you so much. This is my first question.
|
$$2\sin(x+30^{\circ}) = \cos(x+150^{\circ})$$
By the Cofunction Identity,
$$2\sin(x+30^{\circ}) = \sin(90^{\circ}-(x+150^{\circ}))$$
$$2\sin(x+30^{\circ}) = \sin(-x-60^{\circ})$$
$$2\sin(x+30^{\circ}) = \sin(-(x+60^{\circ}))$$
By the Negative Angle Identity,
$$2\sin(x+30^{\circ}) = -\sin(x+60^{\circ})$$
$$2\sin(x+30^{\circ})+\sin(x+60^{\circ}) = 0$$
By the Sum Identities,
$$2(\sin(x)\cos(30^{\circ})+\sin(30^{\circ})\cos(x))+\sin(x)\cos(60)+\sin(60)\cos(x)=0$$
$$\sqrt{3}\sin(x)+\cos(x)+\frac{1}{2}\sin(x)+\frac{\sqrt{3}}{2}\cos(x)=0$$
$$\sin(x)(\sqrt{3}+\frac{1}{2})=\cos(x)(-1-\frac{\sqrt{3}}{2})$$
$$\tan(x)=\frac{-1-\frac{\sqrt{3}}{2}}{\sqrt{3}+\frac{1}{2}}$$
$$\arctan{\frac{-1-\frac{\sqrt{3}}{2}}{\sqrt{3}+\frac{1}{2}}}=x$$
I'll let you use reference angles to calculate all the answers over the interval.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3372946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
factor $x^9 + 243x^3 + 729$ Factor the polynomial $$x^{9} + 243x^{3} + 729$$
it might be helpful to see it like this $$x^{9}+3^{5}x^{3}+3^{6}$$
I would imagine this being done without a calculator, but I don't see how to do it.
|
Observe that $x^9+243x^3+729$ = $(x^3+9)^3 - 27x^6$ = $(x^3+9)^3-(3x^2)^3$.
Now using the formula for difference of cubes we obtain:
$(x^3+9)^3-(3x^2)^3$ = $(x^3+9-3x^2)((x^3+9)^2+(x^3+9)3x^2+(3x^2)^2)$ .
Doing the calculations on the second parenthesis and reordering the terms we get that:
$x^9+243x^3+729 = (x^3 - 3 x^2 + 9) (x^6 + 3 x^5 + 9 x^4 + 18 x^3 + 27 x^2 + 81)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3375007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter $a$. Question : Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter a.
I have solved the problem as follows
$\sqrt{a\left(2^{x}-2\right)+1}=1-2^{x}$
$a\left(2^{x}-2\right)+1=\left(1-2^{x}\right)^{2}=2^{2 x}+1-2 \cdot 2^{x}=2^{2 x}-2^{x+1}+1$
$a 2^{x}-2 a=2^{2 x}-2^{x+1}$
$2^{2 x}-(a-1) 2^{x}+2 a=0$
$y^{2}-(a-1) y+2 a=0$
$y=\frac{(a-1) \pm \sqrt{(a-1)^{2}-8 a}}{2}=\frac{(a-1) \pm \sqrt{a^{2}-10 a+1}}{2}$
$2^{x}=\frac{(a-1) \pm \sqrt{(a-5)^{2}-24}}{2}$
After this there are so many conditions on a. Do i need to check for each and every value ?
|
By letting $y=1-2^x$, you can simplify the problem to:
$$
\sqrt{a(3-y)+1} = y
$$
Since root is non-negative, $y\ge0$. Since $2^x$ is always positive, $y<1$. Thus we want to find the solution of the equation above with the condition $0\le y<1$.
Since both sides are non-negative, we can square them and keep the equality:
$$
a(3-y)+1=y^2\\
y^2+ay - (1+3a)=0
$$
This equation will have the solution when $D=a^2+4(1+3a)\ge 0$.
By solving the corresponding equation, we will get $a\in I_1=(-\infty;-6-4\sqrt2]\cup[-6+4\sqrt2;+\infty)$.
Solution for $y$:
$$
y_{1,2}=\frac12\left(-a\pm\sqrt{a^2+12a+4}\right).
$$
Now let's check when $y\ge0$. For smaller root:
$$
y_1\ge0\\
-a-\sqrt{a^2+12a+4} \ge 0\\
-a \ge \sqrt{a^2+12a+4}\\
a\le0\qquad\text{and}\qquad a^2\ge a^2+12a+4
$$
So, smaller root is always smaller than 0. For larger root:
$$
y_2\ge 0\\
-a+\sqrt{a^2+12a+4} \ge 0 \\
\sqrt{a^2+12a+4}\ge a\\
\text{either}\qquad a<0\qquad \text{or }\qquad a^2+12a+4\ge a^2 \\
\text{either}\qquad a<0\qquad \text{or }\qquad a\ge -1/3
$$
Thus, $y_2\ge 0$ when exists.
Finally, we need to check that $y<1$. We don't need to check the $y_1$, since it already failed previous condition. For larger root:
$$
y_1 < 1\\
-a+\sqrt{a^2+12a+4} < 2\\
\sqrt{a^2+12a+4} < 2+a\\
a\ge-2\qquad\text{and}\qquad a^2+12a+4 < (2+a)^2 \\
a\ge-2\qquad\text{and}\qquad 12a+4 < 4a+4 \\
a\ge-2\qquad\text{and}\qquad a< 0 \\
$$
Thus, $a\in I_1 \cap[-2; 0)=[-6+4\sqrt2; 0)$. Finally, the answer:
When $a\in [-6+4\sqrt2; 0)$, the equation has one root, which is:
$$
x = \log_2\left(1-\frac{\sqrt{a^2+12a+4}-a}2\right).
$$
Note. I didn't include analysis of that expression under the root should be non-negative when was calculating $y\ge0$ and $y<1$, because we already did this analysis, and the corresponding interval would be intersecting the result anyway.
Note 2. We didn't need to check that $a(3-y)+1\ge 0$, because we set it equal to $y^2$ which is always non-negative
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3376613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
What is the principal root of $\sqrt{-i}$? Where is the mistake in this solution?
$$\sqrt{-i}=\sqrt {\cos \frac{3 \pi}2+i\sin \frac{3 \pi}2}=\cos \frac{3 \pi}4+i\sin \frac{3 \pi}4=-\frac{\sqrt 2}2+i \frac{\sqrt 2}2$$
WA gives me different result:
$$\frac{\sqrt 2}2-i \frac{\sqrt 2}2$$
Why the principal root must be equal to $\frac{\sqrt 2}2-i \frac{\sqrt 2}2$?
I used Moivre's formula. But I dont know. How can I must choose value of $k$ in Moivre's formula? I chose $k=0$.
$$z=r\left(\cos x+i\sin x\right)$$
$$z^{\frac 1n}=r^\frac1n \left( \cos \frac{x+2\pi k}{n} + i\sin \frac{x+2\pi k}{n}\right)$$
|
Since you are asked to find the principal root, you need to solve according to the definition of the principal root below.
Given the complex number $z$,
\begin{equation}
z = r \mathrm{e}^{i \theta}, ~ -\pi < \theta \le \pi.
\end{equation}
its principal square root is defined as,
\begin{equation}
\sqrt{z} = \sqrt{r} \mathrm{e}^{i \theta/2}
\end{equation}
Note that $\theta$ is expressed within $(-\pi, \pi]$. In your case, $z=e^{-\frac{\pi}{2}i}$. Thus, its principal root is
$$e^{-\frac{\pi}{4}i}=\frac{\sqrt 2}2-i \frac{\sqrt 2}2$$
as given by WA.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3376738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
}
|
Ways to Arrange 7 things given restrictions Let's say you have three $z$'s, two $x$'s, and two $y$'s. How many ways are there to arrange those $7$ variables given that $x$ and $y$ cannot be together?
Ex: $zxzxzyy$ and $xxzzzyy$ are valid while $zyxzyzx$ and $xyzzyxz$ are not.
I tried approaching this problem by finding the total number of ways of arranging the $7$ variables (without any restrictions) which was $\frac{7!}{{3!}\cdot{2}\cdot{2}}$ which is ${7}\cdot{6}\cdot{5}$, or $210$.
Then I could subtract the number of impossible combinations, and I did that by grouping one $x$ and one $y$ together to make a O. Then I just calculated the number of ways to arrange O, one $x$, one $y$, and three $z$'s. That was $\frac{6!}{3!}$, which is $120$. Then I multiplied the product by $2$ because O could either be $xy$ or $yx$ to get $240$ total combinations. Since $240$ is greater than $210$, I knew at once that I had over counted. To account for the over counting, I tried to use P.I.E. But at this point I had totally confused myself and now I have no idea what's the right path to take. (Maybe I should have divided the product $120$ by $4$ to account for the duplicate $x$ and $y$ in O?)
|
We may select three of the seven positions for the $z$'s, two of the remaining four positions for the $x$'s, and fill the remaining two positions with the two $y$'s in
$$\binom{7}{3}\binom{4}{2}\binom{2}{2} = \binom{7}{3}\binom{4}{2}$$
ways. From these, we must subtract the number of arrangements in which an $x$ and $y$ are adjacent.
A pair in which $x$ and $y$ are adjacent: Place the $x$ and $y$ in a box. We then have six objects to arrange: $x, y, z, z, z$ and the box containing an $x$ and a $y$. Choose three of the six positions for the $z$'s. Arrange the remaining three distinct objects in the remaining three positions. Arrange the $x$ and $y$ within the box. There are
$$\binom{6}{3}3!2!$$
such arrangements.
Two pairs in which an $x$ and $y$ are adjacent: The pairs could be overlapping, meaning that we have $xyx$ or $yxy$, or they could be disjoint.
Overlapping pairs: Consider the case in which two $x$'s are adjacent to a $y$. Then we have five objects to arrange: $xyx, y, z, z, z$. Choose three of the five positions for the $z$'s. Arrange the remaining two distinct objects in the remaining two positions. This can be done in
$$\binom{5}{3}2!$$
ways.
By symmetry, there are also
$$\binom{5}{3}2!$$
arrangements in which two $y$'s are adjacent to an $x$.
Two disjoint pairs: We have five objects to arrange, three $z$'s and two boxes containing an $x$ and a $y$. Choose three of the five positions for the $z$'s. Fill the remaining two positions with the boxes. Within each box, arrange the $x$ and $y$. This can be done in
$$\binom{5}{3}2!2!$$
ways.
Three pairs in which an $x$ and $y$ are adjacent: For this to occur, the string must contain either have $xyxy$ or $yxyx$.
Suppose the string contains $xyxy$. Then we have four objects to arrange: $xyxy, z, z, z$. Choose three of the four positions for the $z$'s. The block $xyxy$ must fill the remaining positions. There are
$$\binom{4}{3}$$
such arrangements.
By symmetry, there are also
$$\binom{4}{3}$$
arrangements containing the string $yxyx$.
By the Inclusion-Exclusion Principle, the number of arrangements of the seven letters $x, x, y, y, z, z, z$ in which no $x$ and $y$ are adjacent is
$$\binom{7}{3}\binom{4}{2} - \binom{6}{3}3!2! + \binom{2}{1}\binom{5}{3}2! + \binom{5}{3}2!2! - \binom{2}{1}\binom{4}{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3377314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Interesting function : $f(x)=\frac{\ln (\frac{x+2}{x+1})}{\ln (1+\frac{1}{x})}$ , $x>1$ Question :
Prove this function :
$$f(x)=\dfrac{\ln \left(\dfrac{x+2}{x+1}\right)}{\ln \left(1+\dfrac{1}{x}\right)},\quad x>1$$
is increasing.
I don't know how I solve by my effort is :
Derivative of $f$ is :
$$f'(x)=\dfrac{(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)}{x(x+2)(x+1)\ln^{2} \left(1+\dfrac{1}{x}\right)}$$
Then we will prove that $f'(x)≥0$ for any $x>1$
mean that But I don't know how to prove :
$$(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)\ge0$$
I need some ideas here if any one have.
Thanks!
|
I didn't check all the detail of your derivation, but
$$(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)=\frac{x+2}{x+1}\ln \left(1+\frac1{x+1}\right)^{x+1}-\ln \left(1+\frac1x\right)^x≥0$$
and $$\left(1+\frac1x\right)^x$$ is an increasing function.
EDIT (aimed to clarify any single step)
We start from your last inquality that is
$$(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)\ge0$$
Now we multiply the LHS by a positive factor $1=\frac{x+1}{x+1}$
$$\frac{x+1}{x+1}(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)\ge0$$
then we take the upper factor $(x+1)$ at the LHS inside the logarithm using that $n\log A=\log A^n$
$$\frac{x+2}{x+1}\ln \left(\frac{x+2}{x+1}\right)^{x+1}-x\ln \left(\frac{x+1}{x}\right)\ge0$$
similarly, we take also the upper factor $x$ at the RHS inside the logarithm to obtain
$$\frac{x+2}{x+1}\ln \left(\frac{x+2}{x+1}\right)^{x+1}-\ln \left(\frac{x+1}{x}\right)^x\ge0$$
then we use
*
*$\frac{x+2}{x+1}=\frac{1+x+1}{x+1}=1+\frac{1}{x+1}$
*$\frac{1+x}{x}=1+\frac{1}{x}$
to obtain finally
$$\frac{x+2}{x+1}\ln \left(1+\frac1{x+1}\right)^{x+1}-\ln \left(1+\frac1x\right)^x≥0$$
Now we use that
*
*the function $g(x)=\left(1+\frac1x\right)^x$ is (strictly) increasing
*the factor $\frac{x+2}{x+1}>1$
then we can write the latter inequality as
$$\frac{x+2}{x+1}g(x+1)-g(x)\ge \iff \frac{x+2}{x+1}g(x+1)\ge g(x)$$
which is trivially true indeed
$$\frac{x+2}{x+1}g(x+1)>g(x+1)\ge g(x)\quad \blacksquare$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3379519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Prove that $2^{n+2}$ is a divisor of the number $k^{2^{n}}-1$ Question :
If n is a natural number and k is an odd number
$k\in \mathbb{Z}$ then show that $2^{n+2}$ is a divisor of
the number $k^{2^{n}}-1$
My try as following :
I'm think use induction :
$n=1$ then $2^{3}=8$ since $k$ is odd so : $k^{2}=1\mod 8$
We find : $2^{3}$ divisor of $k^{2}-1$ (correct )
Now for $n+1$ how I prove $2^{n+3}\mid (k^{2^{n+1}}-1)$
And if can prove it without induction ??
|
The trick is $k^{2*m} - 1 = (k^m -1)(k^m+1)$ and if $2^z\mid k^m -1$ then $k^m$ is odd and $k^m + 1$ is even so $2^z|k^m-1$ and $2\mid k^m+1$ so $2^z*2\mid (k^m-1)(k^m+1)$.
So by induction:
Base case: If $k$ is odd the $k = 2m+1$ for some $m$ and $(2m+1)^2-1= 4m^2 + 4m = 4m(m-1)$ either $m$ or $m-1$ is even so $2|m(m-1)$ and $2^3 = 8 = 4*2\mid 4*m(m-1)$
Induction case: So if ${2^{k-1}}\mid (k^{2^{k+1}}-1)$ then $k$ is odd and $2^{2^k}=2^{2^{k-1}}*2\mid (k^{2^{k+1}}-1)(k^{2^{k+1}} + 1) = (k^{2^{k+2}} - 1)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3381826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Find positive integers such that $2n^3 + 5|n^4 +n+1$ $$ 2n^3 + 5 | 2n^4 +2n +2 - 2n^4 - 5n$$
$$= 2n^3+5 | 2-3n$$
$$3n-2≥2n^3 + 5$$
is this correct? Is there a more efficient way?
|
You can use polynomial long division instead:
$$
\require{enclose}
\begin{array}{rll}
\frac{1}{2}n && \\[-3pt]
2n^3+5 \enclose{longdiv}{n^4+0n^3+0n^2+n+1}\kern-.2ex \\[-3pt]
\underline{- (n^4 \quad \quad \quad \quad \ \ + \frac{5}{2}n) \ \ \ } \\[-3pt]
- \frac{3}{2}n+ 1 \\[-3pt]
\end{array}
$$
Therefore we have:
$$\frac{n^4+n+1}{2n^3+5} = \frac{1}{2}n + \frac{-\frac{3}{2}n+1}{2n^3+5}$$
and if $2n^3+5 | n^4+n+1$, then $2n^3+5 | 2n^4 + 2n + 2$:
$$\frac{2n^4+2n+2}{2n^3+5} = n + \frac{-3n+2}{2n^3+5}$$
At $n=0$, $-3(0)+2 = 2$ and $2(0)^3+5 = 5$. However, since $n^3$ is monotonically increasing (equivalent to $(n+1)^3-n^3 > 0$), and $-3x+2$ is monotonically decreasing, then $\frac{-3n+2}{2n^3+5} = 1$ does not have solutions in the positive reals. You can repeat this argument for any positive integer $k$.
For negative integers $-k$, we have $-3n+2 = -k(2n^3+5) \Rightarrow 3n-2 = (2k)n^3 - 5k$. At $n=0$, $3(0)-2=-2$ while $2k(0)^3 + 5k = 5k$. Then when $n$ is positive, cubic functions grow much faster than linear solutions, so there are no solutions. The proof of this fact relies on limits.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3382310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
}
|
Ramanujan's Nested Radical By noting Ramanujan's Nested Radical, we have
$3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$
On the other hand, we can manipulate the number $4$ by applying the similar principle. Here we have
$\begin{aligned} 4 & = \sqrt{16} \\ & = \sqrt{1+15} \\ & = \sqrt{1+2 \cdot \dfrac{15}{2}} \\ & = \sqrt{1+2\sqrt{\dfrac{225}{4}}} \\ & = \sqrt{1+2\sqrt{1+\dfrac{221}{4}}} \\ & = \sqrt{1+2\sqrt{1+3 \cdot \dfrac{221}{12}}} \\ & = \sqrt{1+2\sqrt{1+3\sqrt{\dfrac{44841}{144}}}} \\ & \vdots \\ & = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}} \end{aligned}$
How can it be? Something contradicts?
|
In Ramanujan's radical, if you stop after $n$ nested radicals the last term inside the radical will be $\sqrt{1}$ but in your case, if you stop after $n$ nested radicals the last term is a term which is increasing with $n$. Hence the difference.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3382458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
}
|
Solution of a limit of a sequence $\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}$ I'm trying to solve the limit of this sequence without the use an upper bound o asymptotic methods:
$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\left(\frac{\infty-\infty}{\infty-\infty}\right)$$
Here there are my differents methods:
*
*assuming $f(n)=\sqrt{4n^2+1}, \,$$\ g(n)=2n$, $h(n)=\sqrt{n^2-1}$, $\ \psi(n)= n$ $$f(n)-g(n)=\frac{\dfrac{1}{g(n)}-\dfrac{1}{f(n)}}{\dfrac{1}{f(n)\cdot g(n)}}, \quad h(n)-\psi(n)=\frac{\dfrac{1}{\psi(n)}-\dfrac{1}{h(n)}}{\dfrac{1}{h(n)\cdot \psi(n)}}$$
I always have an undetermined form.
*I've done some rationalizations:
$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}\cdot \frac{\sqrt{4n^2+1}+2n}{\sqrt{4n^2+1}+2n}$$ where to the numerator I find $1$ and to the denominator an undetermined form. Similar situation considering
$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}\cdot \frac{\sqrt{n^2-1}+n}{\sqrt{n^2-1}+n}$$
*$$\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\frac{n\left(\sqrt{4+\dfrac{1}{n^2}}-2\right)}{n\left(\sqrt{1-\dfrac{1}{n^2}}-1\right)}\rightsquigarrow \left(\frac{0}{0}\right)$$
At the moment I am not able to think about other possible simple solutions.
|
From here
$$\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\frac{\sqrt{4+\dfrac{1}{n^2}}-2}{\sqrt{1-\dfrac{1}{n^2}}-1}$$
we can use that
$$\sqrt{4+\dfrac{1}{n^2}}=2\sqrt{1+\dfrac{1}{4n^2}}\sim 2\left(1+\dfrac{1}{8n^2}\right)=2+\dfrac{1}{4n^2}$$
$$\sqrt{1-\dfrac{1}{n^2}}\sim 1-\dfrac{1}{2n^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3384280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Find the sum of $\binom n0 - \binom n2 +\binom n4 -\binom n6 \cdots$ Find the sum of $\binom n0 - \binom n2 +\binom n4 -\binom n6 \cdots$
Using Binomial expansion of $(1+x)^n$,
$$ \binom n0 +\binom n1 x^1 + \binom n 2 x^2 + \cdots + \binom n nx^n $$
Substituting $x = i$,
$$(1+i)^n = \binom n0 +\binom n1 i^1 + \binom n 2 i^2 + \cdots + \binom n ni^n $$
$$(1+i)^n = \binom n0 +\binom n1 i - \binom n 2 - \binom n 3 i\cdots + \binom n ni^n $$
How to proceed further?
|
So you have $$(1+i)^{n}= \binom{n}{0} + \binom{n}{1} i + \binom{n}{2}i^2 + \cdots = \binom{n}{0}-\binom{n}{2} + \binom{n}{4} + i \cdot\left[\binom{n}{1}-\binom{n}{3} + \cdots\right]$$
*
*Write $(1+i)^{n}= (\sqrt{2})^{n}\cdot \left(\frac{1}{\sqrt{2}}+i \cdot\frac{1}{\sqrt{2}}\right)^{n}$ and see if you can find the real part of this term.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3385119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
How to solve this inequality : Do you know what is this problem or how to solve it ?
Let $x,y,z\in\mathbb{R}^+$ such that $x+y+z=1$. Proove that
$$\frac{(xy+yz+zx+1)(3x^3+3y^3+3z^3+1)}{9(x+y)(y+z)(z+x)}\ge\left(\frac{x\sqrt{x+1}}{\sqrt[4]{9x^2+3}}+\frac{y\sqrt{y+1}}{\sqrt[4]{9y^2+3}}+\frac{z\sqrt{z+1}}{\sqrt[4]{9z^2+3}}\right)^2.$$
My attempts:
Let $a=\sqrt[4]{x^2+3}$, $b=\sqrt[4]{y^2+3}$, $c=\sqrt[4]{z^2+3}$ but it seems useless
Thanks a lot !
|
If I understood right you used homogenization and assumption $x+y+z=3$.
Now use C-S:
$$\sqrt[4]{x^2+3}=\sqrt{\frac{1}{2}\sqrt{(1+3)(x^2+3)}}\geq\sqrt{\frac{1}{2}(x+3)}.$$
Also, use Holder: $$9(x^3+y^3+z^3)\geq(x+y+z)^3.$$
After this you'll get something obvious:
$$\sum_{cyc}(x-y)^2(x+y)\geq0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3385718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Let $\xi$ be the fifth primitive root of $1$ an let $\zeta= \xi + \frac{1}{\xi}$. Prove that $\zeta^2 + \zeta =1$ Let $\xi$ be the fifth primitive root of $1$ an let $\zeta= \xi + \frac{1}{\xi}$. Prove that $\zeta^2 + \zeta =1$
In my attempt I used $\xi = \cos(72k) + i\sin(72k)$ with $k=1,2,3,4$
Where I got $\zeta = \frac{1+\cos(144k)+i\sin(144k)}{\cos(72k)+i\sin(72k)}$
Which yielded $\zeta^2 + \zeta = 2\cos(72k)(2\cos(72k)+1)$
I don’t know what else I could do.
|
Write $\zeta= \xi + \dfrac{1}{\xi} = \xi + \xi^4$, because $\xi^5=1$. Then
$$\zeta^2 + \zeta = \xi^2 + 2\xi \xi^4 + \xi^8 + \xi + \xi^4 = \xi^2 + 2 + \xi^3 + \xi + \xi^4 = 1 + 1 + \xi + \xi^2 + \xi^3 + \xi^4 = 1
$$
because $1 + \xi + \xi^2 + \xi^3 + \xi^4 = 0$, since $0=\xi^5-1=(\xi-1)(1 + \xi + \xi^2 + \xi^3 + \xi^4)$ and $\xi \ne 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3386226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Find the roots of $z^4-3z^2+1=0$ in polar form. Question :
Prove that the solutions of $z^4-3z^2+1=0$ are given by :
$$z=2\cos{36^\circ},2\cos{72^\circ},2\cos{216^\circ},2\cos{252^\circ}$$
My work :
First of all, i want ro find the roots with quadratic formula
$\begin{align}
&(z^2)^2-3z^2+1=0\\
&z^2=\dfrac{3\pm \sqrt{5}}{2}\\
&z_{1,2}=\pm\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\\
&z_{3,4}=\pm\dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}\\
\end{align}$
And i'm stuck. I don't know how to transform this complicated roots into polar form. Bcz, i'm not sure that the modulus for each number is exactly $2$?
Btw, i've found and read some possible duplicates
Here's one of the links :
Roots of $z^4 - 3z^2 + 1 = 0$.
But it seems doesn't answer my question...
Please give me a clear hint or another way to solve this without quadratic formula or something else.
|
Set $y = z^2$, your equation is then equivalent to
$$y^2 -3y +1=0.$$
This is a quadratic equation and can be solve with the quadratic formula that you mentioned:
$$y_{1,2} = \dfrac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1}}{2}=\dfrac{3 \pm \sqrt{5}}{2}.$$
Now observe that
$$y_1 = \dfrac{3+\sqrt{5}}{2}= \dfrac{\frac{1}{2}+\sqrt{5} + \frac{5}{2}}{2} = \dfrac{\frac{1+2\sqrt{5} + 5}{2}}{2} = \dfrac{(1 + \sqrt{5})^2}{2\cdot 2}= \dfrac{(1 + \sqrt{5})^2}{4}.$$
Similarly
$$y_2 = \dfrac{3-\sqrt{5}}{2}= \dfrac{\frac{1}{2}-\sqrt{5} + \frac{5}{2}}{2} = \dfrac{\frac{1-2\sqrt{5} + 5}{2}}{2} = \dfrac{(1 - \sqrt{5})^2}{2\cdot 2}= \dfrac{(1 - \sqrt{5})^2}{4}.$$
Therefore the four solutions $z_{1,2,3,4}$ of the original equations are given by
$$z_{1,2} = \pm\sqrt{y_1} = \pm \dfrac{1 + \sqrt{5}}{2}$$
and
$$z_{3,4} = \pm\sqrt{y_2} = \pm \dfrac{-1+ \sqrt{5}}{2}.$$
Now you can use the well-known fact that $\cos(36°)=\dfrac{1+\sqrt{5}}{4}$ to get your result. To get a value of $\cos(72°)$ you can use the trigonometric relation
$$\cos(2x) = \cos^2(x)-1.$$
While for the others you can use reasoning on the unit circle.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3386733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
Domain of inequation inside squared root How should I go when restricting the roots of the inequation:
$\sqrt {x^2+5x+6} - \sqrt {x^2-x+1} \lt 1$?
By restricting both the squared roots, I know that:
$x \le 3$ and $x \ge -2$
However when simplifying the whole inequation, I get the two roots:
$\frac{-13-\sqrt{73}}{16}$ and $\frac{-13+\sqrt{73}}{16}$.
Both roots are valid when swapping them in the first inequation, so how should I restrict my $x$? Should the final answer be: $x \le 3$ and $x \ge -2$?
|
It's $$\sqrt{x^2+5x+6}<1+\sqrt{x^2-x+1}$$ or
$$x^2+5x+6<1+x^2-x+1+2\sqrt{x^2-x+1}$$ or
$$\sqrt{x^2-x+1}>3x+2.$$
Now, consider two cases:
*
*$x< -\frac{2}{3},$ which is $x\leq-3$ or $-2\leq x<-\frac{2}{3}$;
*$x\geq-\frac{2}{3}.$
Can you end it now?
I got the following answer:
$$(-\infty,-3]\cup\left[-2,\frac{-13+\sqrt{73}}{16}\right]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3388042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Showing that $\frac{2^{\sqrt n}}{1+2^{\sqrt n}/x^n}$ is approximated by $2^{\sqrt n}\left(1- \frac{2^{\sqrt n}}{x^n} \right)$ Sorry for this inconvenient question but really I can't see how to show that $$\frac{2^{\sqrt n}}{1+ \frac{2^{\sqrt n}}{x^n}}\quad\text{is approximated by}\quad 2^{\sqrt n } \left(1- \frac{2^{\sqrt n}}{x^n} \right)$$
I know how to move from right to left but I can't achieve the inverse!
|
$$\frac{2^{\sqrt{n}}}{1+\frac{2^{\sqrt{n}}}{x^n}}=\frac{x^n2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}=2^{\sqrt{n}}\left(\frac{x^n}{x^n+2^{\sqrt{n}}}\right)=2^{\sqrt{n}}\left(\frac{x^n+2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}-\frac{2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}\right)=2^{\sqrt{n}}\left(1-\frac{2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}\right)$$
now you have to show that for your values of $x,n$:
$$x^n+2^{\sqrt{n}}\approx x^n$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3388710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following.
If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, show that $$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)=(1-q+s)^2+(p-r)^2.$$
I have tried putting
\begin{align}P(1)&=(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=1+p+q+r+s\\P(-1)&=(\alpha+1)(\beta+1)(\gamma+1)(\delta+1)=1-p+q-r+s.\end{align}
Then \begin{align}P(1)P(-1)&=(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)\\&=(1+p+q+r+s)(1-p+q-r+s)=(1+q+s)^2-(p+r)^2\end{align}
Somehow it does not match the statement given.
|
Denoting $\alpha=x_1$, $\beta=x_2$, $\gamma=x_3$ and $\delta=x_4$, we have $$\small\prod_{i=1}^4(1+x_i^2)-\prod_{i=1}^4(1-x_i^2)=\small2(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)+2(x_1+x_2+x_3+x_4)$$ so taking out a factor of $x_1x_2x_3x_4$ from the first term in brackets, $$\prod_{i=1}^4(1+x_i^2)=2\left(\prod_{i=1}^4x_i\right)\left(\sum_{i=1}^4\frac1{x_i}\right)+2\sum_{i=1}^4x_i+\prod_{i=1}^4(1-x_i^2)$$ which is evaluable by Vieta's formulas. For the sum of reciprocals, see this post.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3390796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
}
|
$x$ and $y$-intercepts of an absolute value function $f(x)=-3|x-2|-1$ I am to find the $x$ and $y$-intercepts of the function $$f(x)=-3|x-2|-1.$$ The solution is provided in my book as
$(0, -7)$; no $x$ intercepts.
I cannot see how this was arrived at. I attempted to find the x intercepts and arrived at $2-\frac{1}{3}$ and $2+\frac{1}{3}$
My working:
$-3|x-2|-1=0$
$-3|x-2|=1$
$|x-2|=-\frac{1}{3}$
Then solve for both the negative and positive value of $\frac{1}{3}$:
Positive version:
$x-2=\frac{1}{3}$
$x=2+\frac{1}{3}$
Negative version:
$x-2=-\frac{1}{3}$
$x = 2-\frac{1}{3}$
How can I arrive at "$(0, -7)$; no x intercepts"? Where did I go wrong in my understanding?
|
$$|x-2|=-\frac{1}{3}$$
Two cases:
1) If $x\ge2$, the equation becomes $x-2 = -\frac{1}{3}$, which has no solution.
2) If $x<2$, the equation becomes $-(x-2 )=-\frac{1}{3}$, which has no solution.
Thus, there is no x intercepts. (0,-7) is the y intercept as seen from $f(0)=-7$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3391033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Taylor series of $\sinh{(x)}$ at $\ln{(2)}$. Determine the Taylor series of $\sinh{(x)}$ about $x = \ln{(2)}$.
Equating each derivative at $x = \ln{(2)}$ gives:
\begin{equation*}
\begin{split}
f'(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \\
f''(\ln{(2)}) &= \frac{2-\frac{1}{2}}{2} = \frac{3}{4} \\
f^{(3)}(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \\
f^{(4)}(\ln{(2)}) &= \frac{2-\frac{1}{2}}{2} = \frac{3}{4} \\
f^{(5)}(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4}
\end{split}
\end{equation*}
and so on. This means the Taylor series is
\begin{equation*}
\begin{split}
\sum_{n=0}^{\infty} \frac{f^{(n)}(\ln{(2)})}{n!}(x-\ln{(2)})^n &= f(\ln{(2)}) + \frac{f'(\ln{(2)})}{1!}(x-\ln{(2)}) + \frac{f''(\ln{(2)})}{2!}(x-\ln{(2)})^2 \\
&+ \frac{f^{(3)}(\ln{(2)})}{3!}(x-\ln{(2)})^3 + \frac{f^{(4)}(\ln{(2)})}{4!}(x-\ln{(2)})^4 + \frac{f^{(5)}(\ln{(2)})}{5!}(x-\ln{(2)})^5 + \ldots \\
&= \frac{3}{4}+\frac{5}{4\cdot1!}(x-\ln{(2)})^1+\frac{3}{4\cdot 2!}(x-\ln{(2)})^2 \\
&+\frac{5}{4\cdot 3!}(x-\ln{(2)})^3+\frac{3}{4\cdot 4!}(x-\ln{(2)})^4+\frac{5}{4\cdot 5!}(x-\ln{(2)})^5+\ldots
\end{split}
\end{equation*}
How do I find the general sum of this? Has it got something to do with the Maclaurin series which is $\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$.
|
$$\sinh(x)=\dfrac{e^x-e^{-x}}2=\sum_{r=0}^\infty\dfrac{x^{2r+1}}{(2r+1)!}$$
Set $x=\ln2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3395794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
If $A$ is a rotation matrix, then $||Ax||=||x||$. Attempt:
Let $$
A =
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix},
x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
$$
Then,
$$Ax =
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \cos\theta \cdot x_1 -\sin\theta\cdot x_2 \\ \sin\theta \cdot x_1 +\cos\theta\cdot x_2 \end{bmatrix} $$
$$
\begin{align}
||Ax|| &= \sqrt{(\cos\theta \cdot x_1 -\sin\theta\cdot x_2)^2 + (\sin\theta \cdot x_1 +\cos\theta\cdot x_2)^2} \\
&=\sqrt{(\cos^2\theta \cdot x^2_1 -2\cos\theta \cdot x_1\sin\theta\cdot x_2+ \sin^2\theta\cdot x^2_2) + (\sin^2\theta \cdot x^2_1 +2\sin\theta \cdot x_1\cos\theta\cdot x_2+ \cos^2\theta\cdot x^2_2)} \\
&=\sqrt{x^2_1+x^2_2}
\end{align}
$$
However, my text seems to suggest that this is only true for $0\leq\theta\leq\pi$ here:
So where in my attempt did I go wrong?
|
You are right and the restriction is not important.
As an alternative we can use that
$$|Ax|^2=(Ax^T)(Ax)=x^TA^TAx=x^Tx=|x|^2$$
since $A$ is orthogonal and thus $A^TA=I$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3401051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Linear Transformation and Basis $$
\begin{array}{l}{\text { 2.) Consider the linear transformation } T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2},\left[\begin{array}{l}{x} \\ {y}\end{array}\right] \mapsto\left[\begin{array}{c}{-2 x-5 y} \\ {2 x+4 y}\end{array}\right]} \\ {\text { Find a basis } \mathcal{B} \text { of } \mathbb{R}^{2} \text { that represents } T \text { in the form }[T]_{\mathcal{B}}=\left[\begin{array}{cc}{a} & {-b} \\ {b} & {a}\end{array}\right]}\end{array}
$$
What I tried is T$\left[\begin{array}{l}{1} \\ {0}\end{array}\right]$ = $\left[\begin{array}{l}{-2} \\ {2}\end{array}\right]$ and T$\left[\begin{array}{l}{0} \\ {1}\end{array}\right] $ = $\left[\begin{array}{l}{-5} \\ {4}\end{array}\right] $ now how to find such matrix with given form?
|
We have that the matrix from basis $B$ to canonical is
$$u_C=M_Bu_B$$
then
$$T_C(u)=\left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{array}\right]u_C=\left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{array}\right]M_Bu_B$$
and then
$$T_B(u)=M_B^{-1}T_C(u)=M_B^{-1}\left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{array}\right]M_Bu_B$$
and we are looking for
$$M_B^{-1}\left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{array}\right]M_B=\left[\begin{array}{cc}{a} & {-b} \\ {b} & {a}\end{array}\right] \iff \left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{array}\right]M_B=M_B\left[\begin{array}{cc}{a} & {-b} \\ {b} & {a}\end{array}\right]$$
that is by $M_B=\left[\begin{array}{cc}{s} & {t} \\ {u} & {v}\end{array}\right]$
*
*$-2s-5u=as+bt \implies -(a+2)s-bt-5u=0$
*$-2t-5v=-bs+at \implies bs-(a+2)t-5v=0$
*$2s+4u=au+bv \implies 2s+(4-a)u-bv=0$
*$2t+4v=-bu+av \implies 2t+bu+(4-a)v=0$
$$\left[\begin{array}{cccc}{-(a+2)} & {-b}&-5&0 \\ {b} & {-(a+2)}&0&-5\\2&0&(4-a)&-b\\0&2&b&(4-a)\end{array}\right]\left[\begin{array}{c}s \\ t\\u\\v \\\end{array}\right]=\left[\begin{array}{c}0 \\ 0\\0\\0 \\\end{array}\right]$$
which can be solved with the condition $a^2+b^2=2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3403899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Find $x$ so that rational function is an integer Find all rational values of $x$ such that $$\frac{x^2-4x+4}{x^2+x-6}$$ is an integer.
How I attempt to solve this: rewrite as $x^2-4x+4=q(x)(x^2+x-6)+r(x)$. If we require that $r(x)$ be an integer then we can get some values of $r(x)$ by solving $x^2+x-6=0$, so that $x=-3$ or $x=2$. In the former case, $r=25$, and in the latter case $r=0$.
[I should note, however, that I'm not really convinced that this step is actually correct, because when $x$ is a root of $x^2+x-6$, the rational function above can have no remainder due to division by $0$. But I read about it as a possible step here and I can't explain it. Q1 How can this be justified?]
Now we want $(x^2+x-6)$ to divide $0$ (trivial) or $x^2+x-6$ to divide $25$. So we set $x^2+x-6=25k$ for some integer $k$ and solve. So $$x=\frac12 (\pm5\sqrt{4k+1}-1)$$
One of the trial-and-error substitutions for $k$ and then for $x$ gives $x=-8$, but that is just one number. Q2: So I'm wondering, how can we find all such $x$? The condition here is that $4k+1$ must be a perfect square, $k\ge 0$. Aren't there infinitely many perfect squares of this form?
I'd appreciate some clarifications about these two questions, Q1 and Q2.
|
For $x \neq 2$ is $$\frac{x^2-4x+4}{x^2+x-6}=\frac{x-2}{x+3}=1-\frac{5}{x+3}.$$
To obtain an integer with an integer $x,\;$ $x+3$ must be a divisor of $5,$ or equivalently $x+3 \in \{-1,1,-5,5\}.$
This gives the solutions $-4,-2$ and $8,$ because $x=2$ is excluded.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3406495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
Let $x>1,y>1$ and $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$, then find the maximum value of $x^{\log_e y}$ Let $x>1,y>1$ and $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$, then find the maximum value of $x^{\log_e y}$
My attempt is as follows:
As $x>1,y>1$ , so $\log_e x>0, \log_e y>0$, hence we can apply $AM>=GM$
$$\dfrac{log_e x+log_e y}{2}>=\sqrt{\log_e x\log_e y}$$
As both the sides are positive, so we can square both the sides without breaking the inequality.
$$(\log_e x)^2+(\log_e y)^2+2\log_e x\log_e y>=4\log_e x\log_e y$$
Using the given condition $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$
$$(\log_e x^2)+(\log_e y^2)>=2\log_e x\log_e y$$
$$(\log_e x)+(\log_e y)>=\log_e x\log_e y$$
$$\log_e xy>=\log_e x^{\log_e y}$$
As $e>1$, so we can safely write $xy>=x^{\log_e y}$
But actual answer is $e^4$, I am not able to think of any other way. Please help me in this.
|
Here is a way where you do not need any standard inequality like AM-GM or Cauchy-Schwarz.
The fist step is, as already shown in an other solution, to get rid of the logarithms by substituting $x=e^a$, $y=e^b$ with $a,b >0$ which gives:
*
*$a^2+b^2 = 2(a+b)$ and
*maximize $x^{\ln y} =e^{\ln x\cdot \ln y} = e^{ab}$.
Now, observe that by using square completion you get
$$a^2+b^2 = 2(a+b) \Leftrightarrow a^2-2a + b^2+2b = 0 \Leftrightarrow \boxed{(a-1)^2+(b-1)^2=2}$$
This is a circle around $(1,1)$ with radius $\sqrt{2}$. So, you can write
*
*$a-1 =\sqrt{2}\sqrt{\cos t} \Leftrightarrow \boxed{a = 1+ \sqrt{2}\cos t}$
*$b-1 =\sqrt{2}\sqrt{\sin t} \Leftrightarrow \boxed{b = 1+ \sqrt{2}\sin t}$
Now, lets maximize $ab$ and let's check at the end whether the maximum is attained for a $t \in [0,2\pi]$ such that $a,b >0$. Here you may use
*
*$(1)$: $\sin\frac{\pi}{4} =\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$
*$(2)$: $(1)\Rightarrow \frac{\sqrt{2}}{2}(\cos t + \sin t) = \sin \left(t+\frac{\pi}{4}\right)$
*$(3)$: $2\sin t \cos t = \sin 2t$
\begin{eqnarray*}ab
& = & (1+ \sqrt{2}\cos t)(1+ \sqrt{2}\sin t)\\
& = & 1+ \sqrt{2}(\cos t + \sin t) + 2\cos t \sin t \\
& \stackrel{(2),(3)}{=} & 1 + 2\sin \left(t+\frac{\pi}{4}\right) + \sin 2t \\
& \leq & 1+2+1 = \boxed{4}
\end{eqnarray*}
Since the maximum is indeed attained for $\boxed{t=\frac{\pi}{4}}$ which satisfies the condition $a,b>0$, you are done.
So, the maximum value of $x^{\log_e y}$ under the given constraints is $\boxed{e^4}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3410401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Show that $(1 - x)^{-k} \le 1 + 2 k x$ when $0 < k x <1/2$ and $k \ge 1$. I think it is true that
$$(1 - x)^{-k} \le 1 + 2 k x$$
when $0 < k x <1/2$ and $k \ge 1$.
This can be proved by using mean-value theorem to show that
$$
(1-x)^{-k} \le 1 + k x / (1-x)^{k+1}
\le
1+ 2kx
$$
under the given condition.
Is there's a more direct way to show it?
|
$(1 - x)^{-k} \le 1 + 2 k x
$
if
$0 \lt x < \dfrac1{2k}$.
This is
$1
\le (1-x)^k(1+2kx)
$.
If $k=1$ this is
$1
\le (1-x)(1+2x)
=1+x-2x^2
$
or $x \ge 2x^2
$
or
$x \le \frac12$
which is true.
Suppose true for $k$.
Want to show
$1
\le (1-x)^{k+1}(1+2(k+1)x)
$.
Assume
$0 \lt x < \dfrac1{2(k+1)}$.
$\begin{array}\\
(1-x)^{k+1}(1+2(k+1)x)
&=(1-x)^{k+1}(1+2kx+2x)\\
&=(1-x)^{k+1}(1+2kx)+(1-x)^{k+1}(2x)\\
&=(1-x)(1-x)^{k}(1+2kx)+(1-x)^{k+1}(2x)\\
&\ge(1-x)+\dfrac{(1-x)}{1+2kx}(2x)\\
&=\dfrac{(1+2kx)(1-x)+2x(1-x)}{1+2kx}\\
&=\dfrac{1+(2k-1)x-2kx^2+2x-2x^2}{1+2kx}\\
&=\dfrac{1+(2k+1)x-2(k+1)x^2}{1+2kx}\\
&=\dfrac{1+2kx+x-2(k+1)x^2}{1+2kx}\\
&=1+\dfrac{x-2(k+1)x^2}{1+2kx}\\
&=1+x\dfrac{1-2(k+1)x}{1+2kx}\\
&\ge 1
\qquad\text{if } 1-2(k+1)x \ge 0\\
\end{array}
$
or
$x \le \dfrac1{2(k+1)}$,
which is the
induction hypothesis.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3413754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Least value of unit vector $|a+b|^2+|b+c|^2+|c+a|^2$ If $a, b, c$ are unit vectors, then least value of $|a+b|^2+|b+c|^2+|c+a|^2$ will be equal to
(1) 1
(2) 3
(3) 9
(4) 12
If am using the concept $a=c=-b$, I am getting the answer 4, but not matching with options provided
|
Let $a,b,c$ be three unit vectors then $$(a+b+c)^2\ge 0 \implies a^2+b^2+c^2+2(a,b+b.c+c.a) \ge 0$$ $$ \implies 2(a.b+b.c+c.a) \ge -3 ~~~(1)$$
Then $$|a+b|^2+|b+c|^2+|c+a|^2=2(a^2+b^2+c^2)+2(a.b+b.c+c.a) \ge 6-3=3.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3415387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Divisibility by 7 of a number consisting of 0 and 1's A decimal number is of arbitrary length consisting of 0 and 1 only i.e. (10,111,10000,11111,1010, number of digits in the number can be upto 100 )
Can this number ever be divisible by 7 if yes, is there any efficient way to list all those numbers
|
Note that $10 \equiv 3 \bmod 7$ and the powers of $3$ modulo $7$ are:
$$3^0=1,\ 3^1=3,\ 3^2=2,\ 3^3=6,\ 3^4=4,\ 3^5=5$$
It follows that $10^k \equiv 3^r \bmod 7$ where $r$ is the remainder of $k$ modulo $6$, so you can find the remainder of a number consisting of only $0$s and $1$s modulo $7$ by splitting its decimal expansion into groups of $6$, replacing each '$1$' by whichever digit it corresponds to in the above, and adding them all together.
For example consider the number $110010 101010 010011$. We can break it up as
$$110010~101010~010011$$
Using the above, this is congruent to
$$(3^5+3^4+3^1)+(3^5+3^3+3^1)+(3^4+3^1+3^0)$$
which is itself congruent to
$$(5+4+3)+(5+6+3)+(4+2+1)$$
modulo $7$. This gives $110010101010010011 \equiv 33 \equiv 6 \bmod 7$.
This gives an algorithm for determining whether such a number $n$ is divisible by $7$ (and its efficiency is $O(\log n)$, which is pretty good). This also hints at a way of listing all such numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3417330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
}
|
Solving a system of modular equations I am trying to solve this system of equations：
$24x ≡ 12 \hspace{5pt} (\mathrm{mod} \hspace{2pt} 63)$
$x ≡ 2\hspace{5pt} (\mathrm{mod} \hspace{2pt} 27)$
So I know that the second equation is equivalent to $x=27y+2$, but when I plug it into $24x ≡ 12\hspace{5pt} (\mathrm{mod} \hspace{2pt} 63)$, I got $2y ≡ 3\hspace{5pt} (\mathrm{mod} \hspace{2pt} 7)$ and I don’t know what to do next... Did I do something wrong?
|
Do you know about modular inverse ?
Here you have $\quad 2y\equiv 3 \pmod 7$
Notice that when we multiply by $4$ we get $\quad 8y\equiv 12\pmod 7$
And since $8\equiv 1\pmod 7$ and $12\equiv 5\pmod 7$ we get in the end $\quad y\equiv 5\pmod 7$
We say that $4$ is the inverse of $2$, since $2\times 4\equiv 1\pmod 7$
Now replace $y=7k+5$ in $x=27y+2=137+189k$
Here is now a way to get to apply the Chinese remainder theorem.
$\begin{cases} 24x\equiv 12\pmod{63}\\x\equiv 2\pmod{27}\end{cases}$
The first thing to do is to use modular inverse to have equations of the form $x\equiv a_i\pmod {m_i}$
Unfortunately here $\gcd(24,63)=3$ thus $24$ does not have an inverse modulo $63$.
But since $12$ is also a multiple of $3$ we can divide the whole equation by the gcd.
We now have $\begin{cases} 8x\equiv 4\pmod{21}\\x\equiv 2\pmod{27}\end{cases}$
Since $8^{-1}\equiv 8\pmod{21}$ we multiply by $8$ in first equation.
$\begin{cases} x\equiv 11\pmod{21}\\x\equiv 2\pmod{27}\end{cases}$
We still can't apply CRT because the $m_i$ have common factors $21=3\times 7$ and $27=3^3$.
So we will split the first equation into two:
$\begin{cases} x\equiv 2\pmod{3}\\x\equiv 4\pmod{7}\\x\equiv 2\pmod{27}\end{cases}$
Now we have to deal with first and third equations, but notice that every solution of the third one is automatically solution of the first one.
So we can reduce the system to:
$\begin{cases} x\equiv 4\pmod{7}\\x\equiv 2\pmod{27}\end{cases}$
It has a unique solution modulo $7\times 27=189$ and applying the CRT gives $x\equiv 137\pmod{189}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3428639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
$x,y,z > 0$, prove that $ 3xyz + x^{3}+ y^{3} + z^{3} \ge 2 \left[ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \right] $ $x,y,z > 0$, prove that
$$ 3xyz + x^{3}+ y^{3} + z^{3} \ge 2 \left[ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \right] $$
Without using Schur's inequality,
Attempt:
By $C.S$:
$$ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} = x^{3/2}y^{3/2} + y^{3/2}z^{3/2} + z^{3/2}x^{3/2} $$
$$ \le \sqrt{ x^{3} + y^{3} +z^{3}} \sqrt{ y^{3} + z^{3} + x^{3}} = x^{3} + y^{3} + z^{3} $$
then I have to prove
$$ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \le 3xyz$$
this one is difficult. Another thing that we know by AM-GM: $ x^{3} + y^{3}+ z^{3} \ge 3xyz$.
|
Let $x=e^a$, $y=e^b$, $z=e^c$ $\;$
and $a\ge b \ge c$ (WLOG)
Our inequality is equivalent to
$$3e^{a+b+c} + e^{3a}+ e^{3b} + e^{3c} \ge 2 \left( e^{3(a+c)/2} + e^{3(a+c)/2} + e^{3(a+b)/2} \right)$$
If $a+c\le 2b$ ,
This inequality is true by Karamata (Majorization Inequality) with convex function $f(x)=e^x$ for all $x\ge 0$ and $$ \left( \frac{3(a+b)}{2}, \frac{3(a+b)}{2}, \frac{3(a+c)}{2}, \frac{3(a+c)}{2}, \frac{3(b+c)}{2}, \frac{3(b+c)}{2} \right ) \prec \left( 3a,3b,(a+b+c),(a+b+c),(a+b+c),3c \right)$$
The majorization holds because,
$$3a\ge \frac{3(a+b)}{2}$$
$$3a+ 3b \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} $$
$$3a+ 3b + (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}$$
$$3a+ 3b + (a+b+c) + (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}$$
$$3a+ 3b + (a+b+c) + (a+b+c)+ (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}+ \frac{3(b+c)}{2}$$
$$3a+ \dots + 3b+ 3c = \frac{3(a+b)}{2}+ \dots + \frac{3(b+c)}{2} + \frac{3(b+c)}{2} $$
$3^{rd}$ and $4^{th}$ inequalities by $a+ c \le 2b$. $\;$ Other inequalities are obvious.
If $a+c\ge 2b$ ,
The desired inequality is true by Karamata (Majorization Inequality) with convex function $f(x)=e^x$ for all $x\ge 0$ and $$ \left( \frac{3(a+b)}{2}, \frac{3(a+b)}{2}, \frac{3(a+c)}{2}, \frac{3(a+c)}{2}, \frac{3(b+c)}{2}, \frac{3(b+c)}{2} \right ) \prec \left( 3a,(a+b+c),(a+b+c),(a+b+c),3b,3c \right)$$
The majorization holds because,
$$3a\ge \frac{3(a+b)}{2}$$
$$3a+ (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} $$
$$3a+ (a+b+c) + (a+b+c) \ge \frac{3(a+b)}{2}+\frac{3(a+b)}{2} + \frac{3(a+c)}{2}$$
$\dots$
$$3a+ \dots + 3b+ 3c = \frac{3(a+b)}{2}+ \dots + \frac{3(b+c)}{2} + \frac{3(b+c)}{2} $$
$3^{rd}$ inequality comes from $a+ c \ge 2b$. $\;$ Other inequalities are obvious.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3435099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
If $f(x)$ is a polynomial of degree $4$ and $f(n)=n+1$ for $n=1,2,3,4$. Find $f(5)$ Question: If $f(x)$ is a polynomial of degree $4$ and $f(n)=n+1$ for $n=1,2,3,4$. Find $f(5)$
If we construct $g(x)=f(x)-(x-2)(x-3)(x-4)(x-5)$, then is it possible to find f(5)?
|
Note that, $f(x)=\lambda(x-1)(x-2)(x-3)(x-4)+x+1$. Here $\lambda$ is a non-zero scalar. Hence $f(5)=\lambda 4\cdot 3\cdot 2\cdot 1+6=6+24\lambda$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3435200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Evaluating integrals involving products of exponential and Bessel functions over the interval $(0,\infty)$ While trying to solve a challenging system of dual integral equations resulting from a mixed boundary value problem arising in a fluid mechanical problem, the four following non-trivial convergent improper integrals emerge:
\begin{align}
I_1 (r,t) &= \int_0^\infty \lambda^{\frac{1}{2}}
e^{-\lambda} J_1 (\lambda r) J_{\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda \, , \\
I_2 (r,t) &= \int_0^\infty \lambda^{\frac{1}{2}}
e^{-\lambda} J_0 (\lambda r) J_{\frac{3}{2}} (\lambda t) \, \mathrm{d} \lambda\, , \\
I_3 (r,t) &= \int_0^\infty \left( \lambda^{\frac{1}{2}}-\lambda^{-\frac{1}{2}} \right)
e^{-\lambda} J_1 (\lambda r) J_{\frac{3}{2}} (\lambda t) \, \mathrm{d}\lambda \, , \\
I_4 (r,t) &= \int_0^\infty \left( \lambda^{\frac{1}{2}}+\lambda^{-\frac{1}{2}} \right)
e^{-\lambda} J_0 (\lambda r) J_{\frac{1}{2}} (\lambda t) \, \mathrm{d}\lambda \, ,
\end{align}
wherein $t$ and $r$ are positive real numbers.
It can be checked that, thanks to the exponential function, these integrals are convergent.
If the integrands do not contain the exponential function, then the evaluation of these integrals is easy and straightforward.
Is there probably a way to evaluate these integrals analytically even as infinite convergent series functions?
Any help or hint is highly appreciated!
|
I believe these integrals have a simple analytical form. I will demonstrate for $I_1$ and I hope you can see how to do the others similarly.
I write $I_1$ out as originally stated:
$$I_1 = \int_0^{\infty} d\lambda \, \lambda^{1/2} \, e^{-\lambda} \, J_1(\lambda r) \, J_{1/2}(\lambda t)$$
Note that
$$J_{1/2}(\lambda t) = \sqrt{\frac{2}{\pi \lambda t}} \sin{(\lambda t)}$$
$$J_1(\lambda r) = \frac1{i \pi} \int_0^{\pi} d\theta \, \cos{\theta} \, e^{i \lambda r \cos{\theta}} $$
Plugging back into the integral definition of $I_1$ and changing the order of integration, we get
$$I_1 = \frac1{i \pi} \sqrt{\frac{2}{\pi t}} \int_0^{\pi} d\theta \, \cos{\theta} \, \int_0^{\infty} d\lambda \, e^{-\lambda} \, \sin{(\lambda t)} \, e^{i \lambda r \cos{\theta}} $$
Rewriting the sine in exponential form, the integral over $\lambda$ is simple, and we are left with the integral over $\theta$:
$$I_1 = -\frac1{2 \pi} \sqrt{\frac{2}{\pi t}} \int_0^{\pi} d\theta \, \cos{\theta} \, \left [\frac1{1-i t - i r \cos{\theta}} - \frac1{1+i t - i r \cos{\theta}} \right ] $$
Now let's consider the integral
$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} $$
where $a$ and $b$ may be complex; in our case $a=1\pm i t$ and $b=-i r$. While there are at least a couple of ways of evaluating this integral, I will demonstrate how it is done using contour integration.
Consider the contour integral
$$-i \oint_C \frac{dz}{z} \, \frac{z^2+1}{b z^2+2 a z+b} $$
where $C$ is the following contour:
The semicircle has unit radius. Note that, because the real integral is only over a half-cycle rather than a full cycle, the contour $C$ includes a traversal along the real axis. Nevertheless, because of the pole at the origin, there needs to be a small detour of radius $\epsilon$ around the origin as shown.
The contour integral is then equal to
$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} - i \, PV \int_{-1}^1 \frac{dx}{x} \, \frac{x^2+1}{b x^2+2 a x+b} - i (i \epsilon) \int_{\pi}^0 d\phi \, e^{i \phi} \, \frac1{\epsilon \, e^{i \phi}} \frac{\epsilon^2 e^{i 2 \phi}+1}{b \epsilon^2 e^{i 2 \phi}+ 2 a \epsilon \, e^{i \phi} + b} $$
The first integral is what we seek (for now). The third integral is, in the limit as $\epsilon \to 0$, $-\pi/b$. The second integral, the principal value integral, may be evaluated as follows:
$$\begin{align} PV \int_{-1}^1 \frac{dx}{x} \, \frac{x^2+1}{b x^2+2 a x+b} &= \frac1{b} \, PV \int_{-1}^1 \frac{dx}{x}\, \left (1 - \frac{2 a x}{b x^2+2 a x+b} \right ) \\ &= \frac1{b} \, PV \int_{-1}^1 \frac{dx}{x} - \frac{2 a}{b} \int_{-1}^1 \frac{dx}{b x^2+2 a x+b}\end{align}$$
Note that the first principal value integral on the RHS vanishes by symmetry. The second integral on the right needs not be expressed using principal value notation because the pole at the origin is removed. Accordingly,
$$\begin{align} PV \int_{-1}^1 \frac{dx}{x} \, \frac{x^2+1}{b x^2+2 a x+b} &= - \frac{2 a}{b} \int_{-1}^1 \frac{dx}{b x^2+2 a x+b} \\ &= -\frac{2 a}{b^2} \int_{-1}^1 \frac{dx}{\left ( x+\frac{a}{b} \right )^2 + 1-\frac{a^2}{b^2}}\\ &= -\frac{2 a}{b^2} \frac1{\sqrt{1-\frac{a^2}{b^2}}} \left [ \arctan{\left ( \frac{x+\frac{a}{b}}{\sqrt{1-\frac{a^2}{b^2}}} \right )} \right ]_{-1}^1 \\ &= \frac{\pi}{b} \frac{a}{\sqrt{b^2-a^2}} \end{align}$$
For convenience later on, we may write
$$ PV \int_{-1}^1 \frac{dx}{x} \, \frac{x^2+1}{b x^2+2 a x+b} = -i \frac{\pi}{b} \frac{a}{\sqrt{a^2-b^2}} $$
This way, we may write that the contour integral is equal to
$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} - \frac{\pi}{b} \frac{a}{\sqrt{a^2-b^2}} - \frac{\pi}{b} $$
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand of the contour integral inside the contour $C$. In this case, the only pole inside the contour is at $z_+ = -\frac{a}{b} + \sqrt{\frac{a^2}{b^2}-1}$. Computing the residue at this pole, the resulting equation for the integral we seek for now is
$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} - \frac{\pi}{b} \frac{a}{\sqrt{a^2-b^2}} - \frac{\pi}{b} = -\frac{2 \pi}{b} \frac{a}{\sqrt{a^2-b^2}}$$
or
$$\int_0^{\pi} d\theta \, \frac{\cos{\theta}}{a+b \cos{\theta}} = -\frac{\pi}{b} \left ( \frac{a}{\sqrt{a^2-b^2}} - 1 \right ) $$
Now we may use this result to determine $I_1$. Again, subbing $a=1 \pm i t$ and $b=-i r$, we get that
$$\begin{align} I_1 &= -\frac1{2 \pi} \sqrt{\frac{2}{\pi t}} \frac{\pi}{i r} \left ( \frac{1-i t}{\sqrt{(1-i t)^2+r^2}} - \frac{1+i t}{\sqrt{(1+i t)^2+r^2}} \right ) \\ &= \sqrt{\frac{2}{\pi t}} \frac1{r} \operatorname{Im}{\left (\frac{1+i t}{\sqrt{(1+i t)^2+r^2}} \right )}\end{align}$$
And with that, we are technically finished. But as someone who likes explicit results, I will take this a bit further and express the result as follows:
$$I_1 = \int_0^{\infty} d\lambda \, \lambda^{1/2} \, e^{-\lambda} \, J_1(\lambda r) \, J_{1/2}(\lambda t) = \\ \frac1{\sqrt{\pi t r^2}} \frac{t \sqrt{\sqrt{(1+r^2-t^2)^2+4 t^2}+(1+r^2-t^2)} - \sqrt{\sqrt{(1+r^2-t^2)^2+4 t^2}-(1+r^2-t^2)}}{\sqrt{(1+r^2-t^2)^2+4 t^2}}$$
I have verified this in Mathematica numerically.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3436474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
Find the smallest value of the following expression $\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2}+ \sqrt{(y-3)^2 +9}$
Find the smallest value of the following expression:
$$\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2} + \sqrt{(y-3)^2 +9}$$
I tried to derive the expression with respect to $x$ and $y$ and then equal the derivative to zero and find the critical points, but I could not do that. It is complicated. Is there a way?
|
Use the Minkowski inequality:
$$
\begin{aligned}
&\sqrt{(x-9)^{2}+4}+\sqrt{x^{2}+y^{2}}+\sqrt{(y-3)^{2}+9}\\
=&\sqrt{(9-x)^{2}+2^2}+\sqrt{x^{2}+y^{2}}+\sqrt{3^2+(3-y)^{2}}\\
\geqslant&\sqrt{(9-x+x)^2+(2+y)^2}+\sqrt{3^2+(3-y)^{2}}\\
\geqslant&\sqrt{(9-x+x+3)^2+(2+y+3-y)^2}=13
\end{aligned}
$$
"$=$" holds iff $(9-x,2)$ and $(x,y)$ are linearly dependent and $(9,2+y)$ and $(3,3-y)$ are linearly dependent, i.e. $x=\dfrac{21}{5},y=\dfrac{7}{4}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3437202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
}
|
General formula for the power series of $\dfrac{1}{(1+x)^3}$ I wish to find the general formula for the following power series:
$\dfrac{1}{(1+x)^3}=1-3x+6x^2-10x^3+15x^4-21x^5+28x^6...$
The difference between the first and second term is $2$
The difference between the second and third term is $3$
The difference between the third and fourth term is $4$
The difference between the fourth and fifth term is $5$
and so on.
How do you find the general formula for this series? It is not quite an arithmetic series, and definitely not a geometric series, so what should I do?
|
$$\frac {1}{1+x} = 1-x+x^2-x^3+x^4-....$$
Differentiate and you get
$$ \frac {-1}{(1+x)^2} = -1+2x-3x^2+4x^3-....$$
Differentiate again and you get
$$ \frac {2}{(1+x)^3} = 2-6x+12x^2-....$$
$$ \frac {1}{(1+x)^3} = 1-3x+6x^2-....$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3437309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Why is the graph of $8x^2+8xy+2y^2+2x+y+5=0$ empty? For the equation
$$8x^2+8xy+2y^2+2x+y+5=0$$
Comparing it with standard equation of conic section we get that
$h^2-ab=0$ and
$$\Delta= \left|\begin{array}{ccc}
a & h & g \\
h & b & f \\
g & f & c \end{array}\right| = 0$$
So as per I know it should represent a pair of straight line which are real and coincident but after plotting it on desmos( an app to plot graphs) it showed nothing. Please help me.
|
$$ 8 \left(x + \frac{y}{2} + \frac{1}{8} \right)^2 + \frac{39}{8} $$
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- \frac{ 1 }{ 8 } & 0 & 1 \\
- \frac{ 1 }{ 2 } & 1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
16 & 8 & 2 \\
8 & 4 & 1 \\
2 & 1 & 10 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - \frac{ 1 }{ 8 } & - \frac{ 1 }{ 2 } \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
16 & 0 & 0 \\
0 & \frac{ 39 }{ 4 } & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
$$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 1 }{ 2 } & 0 & 1 \\
\frac{ 1 }{ 8 } & 1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
16 & 0 & 0 \\
0 & \frac{ 39 }{ 4 } & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 8 } \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
16 & 8 & 2 \\
8 & 4 & 1 \\
2 & 1 & 10 \\
\end{array}
\right)
$$
I think the (real) solution set is empty. Back in a few minutes
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3445951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Prove $S=\{a+b\sqrt[3]{2} +c\sqrt[3]{4}:a,b,c \in \mathbb{Q}\}$ is a ring. Prove $S=\{a+b\sqrt[3]{2} +c\sqrt[3]{4}:a,b,c \in \mathbb{Q}\}$ is a ring.
My professor has gone over how to prove a subring, but I am not sure how to prove a ring.
This is what I have so far, but I am not sure if it is similar to proving a subring, or what I have left to prove.
To prove this set is a ring, we must prove
(i) $S$ is closed under addition, and
(ii) $S$ is closed under multiplication, and
(iii)$S$ is associative on the operations, and
(iv) $S$ is commutative on the operations, and
(v) $S$ is distributive on the operations, and
(vi) $S$ has an additive identity ($0_R$), and
(vii) All operations in $S$ have an additive inverse.
We will begin by proving part (i). That is, we will prove $S$ is closed under addition. We will take $a+b\sqrt[3]{2}+c\sqrt[3]{4} \in S$ and add it to an arbitrary element $d+e\sqrt[3]{2}+f\sqrt[3]{4} \in S$ to get $(a+b\sqrt[3]{2}+c\sqrt[3]{4})+(d+e\sqrt[3]{2} +f\sqrt[3]{4})$. We will then factor out $\sqrt[3]{2}$ and $\sqrt[3]{4}$ using the distributive property to get $(a+d+(b+e)\sqrt[3]{2}+(c+f)\sqrt[3]{4})$. Since $b,e,c,f \in \mathbb{Q}$, $(b+e),(c+f) \in \mathbb{Q}$, we know $S$ is closed under addition which proves part (i) of the proof.
We will now prove part (ii). That is, we will prove $S$ is closed under multiplication. We will take $a+b\sqrt[3]{2}+c\sqrt[3]{4} \in S$ and multiply it to an arbitrary element $d+e\sqrt[3]{2}+f\sqrt[3]{4} \in S$ to get $(a+b\sqrt[3]{2}+c\sqrt[3]{4})(d+e\sqrt[3]{2} +f\sqrt[3]{4})$. We will then multiply to get $ad+ae\sqrt[3]{2}+af\sqrt[3]{4}+bd\sqrt[3]{2}+be\sqrt[3]{2}\sqrt[3]{2}+bf\sqrt[3]{2}\sqrt[3]{4}+cd\sqrt[3]{4}+ce\sqrt[3]{2}\sqrt[3]{4}+cf\sqrt[3]{4}\sqrt[3]{4}$. We know $\sqrt[3]{4}=\sqrt[3]{2}\sqrt[3]{2}$, so we can make a substitution to get $ad+ae\sqrt[3]{2}+af\sqrt[3]{2}\sqrt[3]{2}+bd\sqrt[3]{2}+be\sqrt[3]{2}\sqrt[3]{2}+bf\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}+cd\sqrt[3]{2}\sqrt[3]{2}+ce\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}+cf\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}$. We can then simplify this to $ad+ae\sqrt[3]{2}+af(2\sqrt[3]{2})+bd\sqrt[3]{2}+be(2\sqrt[3]{2})+bf(3\sqrt[3]{2})+cd(2\sqrt[3]{2})+ce(3\sqrt[3]{2})+cf(4\sqrt[3]{2})$ factor out $\sqrt[3]{2}$ using the distributive property to get $ad+(ae+bd+af(2)+bd+be(2)+bf(3)+cd(2)+ce(3)+cf(4))\sqrt[3]{2}$. Since $a,b,c,d,e,f,2,3,4 \in \mathbb{Q}$, $(ae+bd+af(2)+bd+be(2)+bf(3)+cd(2)+ce(3)+cf(4)) \in \mathbb{Q}$. Thus, we know $S$ is closed under multiplication which proves part (ii) of the proof.
We will now prove part (vi). That is, we will prove $S$ has an additive identity, or in other words has $0_R$. We will take $a+b\sqrt[3]{2}+c\sqrt[3]{4} \in S$. Since $a,b,c \in \mathbb{Q}$ and $0 \in Q$, we will let $a$, $b$, and $c$ be $0$ to get $0+0\sqrt[3]{2}+0\sqrt[3]{4}$. We can simplify this to get $0+0\sqrt[3]{2}+0\sqrt[3]{4}=0+0+0=0_R$. Since the ring contains 0, we know it has an additive identity and proves part (vi) of the proof.
Can anyone help me get on track, please?
|
This is a subset of the real numbers with the same ring structure, you just have to see that it is closed under adition and multiplication as distribution, identity and associativity are free. Clearly, it is closed under addition. To prove that it is closed under adition we only have to check that products of (additive) generators are in the ring, Let $x = 2^{1/3}$, $x^2 = 4^{1/3}$ and $x^3 = 2 \in \mathbb{Q}$ so it is also closed under multiplication. Showing that it is a subring of the real numbers takes care of most of the proof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3446309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
If $n\geq 3$, $9^n \equiv a (\mod 100)$ and $9^{n+1} \equiv b (\mod 100)$, then $a+b=90$. I noticed a pattern in the powers of 9 modulo 100.
$9^1 \equiv 9 \pmod{100}$
$9^2 \equiv 81 \pmod{100}$
$9^3 \equiv 29 \pmod{100}$
$9^4 \equiv 61 \pmod{100}$
.
.
.
and conjectured the following:
If $n\geq 3$ is an odd integer where $9^n \equiv a \pmod{100}$ and $9^{n+1} \equiv b \pmod{100}$, then $a+b=90$.
How do I prove this?
|
From the binomial theorem, it follows that,
for $n$ odd, $9^n=(10-1)^n\equiv10n-1\bmod100$,
and $9^{n+1}=(10-1)^{n+1}\equiv-10(n+1)+1\bmod100$.
Therefore, for $n$ odd, $9^n+9^{n+1}\equiv-10\equiv90\bmod100$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3446410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Is $G = (\Bbb Q^*,a*b=\frac{a+b}{2})$ a group, monoid, semigroup or none of these?
Is $G = (\Bbb Q^*,a*b=\frac{a+b}{2})$ a group, monoid, semigroup or none of these?
I tried to solve it by assuming $ a,b,c \in G $ such that $a*(b*c)=(a*b)*c$. Then$$\frac{a+\frac{b+c}{2}}{2} = \frac{\frac{a+b}{2}+c}{2}$$
and by taking $$ 2\times\left( \frac{a+\frac{b+c}{2}}{2} = \frac{\frac{a+b}{2}+c}{2} \right) $$
we have $$ a+\frac{b+c}{2} =\frac{a+b}{2}+c.$$
In my steps I have $(G,*)$ is not associative. Are my steps right?
|
Let $a=2$ and $b=c=1$. This should be a counterexample to associativity.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3448020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
}
|
Computing the cycle decomposition of the composition of permutations. Let $\sigma,\tau\in S_5$ be given by
$$
\sigma = \begin{pmatrix}1&3&5\end{pmatrix}\begin{pmatrix}2&4\end{pmatrix},\quad \tau = \begin{pmatrix}1&5 \end{pmatrix}\begin{pmatrix}2&3\end{pmatrix}.
$$
I want to find the cycle decomposition of $\tau\sigma$. So I write
$$
\tau\sigma = \begin{pmatrix}1&5 \end{pmatrix}\begin{pmatrix}2&3\end{pmatrix}\begin{pmatrix}1&3&5\end{pmatrix}\begin{pmatrix}2&4\end{pmatrix}.
$$
Reading from right to left and starting with the element $1$, I see that $1$ is taken to $3$ and $3$ is taken to $2$. Now, $2$ is taken to $4$, and $4$ is fixed in the other cycles, so our first cycle is $\begin{pmatrix}1&2&4\end{pmatrix}$. Since $3$ is taken to $5$ and $5$ is taken to $1$, the next cycle would be $\begin{pmatrix}3&1\end{pmatrix}$. But these cycles aren't disjoint, so I must have done something wrong. Where is my error?
|
$4$ is taken to $2$ and $2$ is taken to $3$, so it's $(1243)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3448678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Triangle sides $a,b,c$ are in arithmetic progression. Show $\sin^2(A/2)\csc2A$, $\sin^2(B/2)\csc2B$, $\sin^2(C/2)\csc2C$ are in harmonic progression
If sides $a,b,c$ of the triangle ABC are in arithmetic progression, prove that $\sin^2\frac{A}{2}\mathrm{cosec}(2A),\sin^2\frac{B}{2}\mathrm{cosec}(2B),\sin^2\frac{C}{2}\mathrm{cosec}(2C)$ are in harmonic progression.
My attempt is as follows:-
$$T_1=\dfrac{1-\cos A}{2\sin2A}$$
$$T_1=\dfrac{1}{4}\dfrac{\sec A-1}{\sin A}$$
$$T_1=\dfrac{2R}{4}\left(\dfrac{\dfrac{2bc}{b^2+c^2-a^2}-1}{a}\right)$$
$$T_1=\dfrac{R}{2}\left(\dfrac{\dfrac{2bc-b^2-c^2+a^2}{b^2+c^2-a^2}}{a}\right)$$
$$T_1=\dfrac{R}{2}\left(\dfrac{\dfrac{a^2-(b-c)^2}{b^2+c^2-a^2}}{a}\right)$$
$$T_1=\dfrac{R}{2}\left(\dfrac{(a+c-b)(a+b-c)}{(b^2+c^2-a^2)a}\right)$$
By symmetry, we can say $T_2=\dfrac{R}{2}\left(\dfrac{(b+c-a)(b+a-c)}{(a^2+c^2-b^2)b}\right)$
$T_3=\dfrac{R}{2}\left(\dfrac{(c+a-b)(c+b-a)}{(a^2+b^2-c^2)c}\right)$
For $T_1,T_2,T_3$ to be in HP, $\dfrac{1}{T_1},\dfrac{1}{T_2},\dfrac{1}{T_3}$ should be in A.P
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{(b^2+c^2-a^2)a}{(a+c-b)(a+b-c)}+\dfrac{(a^2+b^2-c^2)c}{(c+a-b)(c+b-a)}-\dfrac{2(a^2+c^2-b^2)b}{(b+c-a)(a+b-c)}\right)$$
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{a(b^2+c^2-a^2)(b+c-a)+c(a^2+b^2-c^2)(a+b-c)-2b(a^2+c^2-b^2)(c+a-b)}{(b+c-a)(a+b-c)(c+a-b)}\right)$$
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{a(b^3+b^2c-ab^2+bc^2+c^3-ac^2-a^2b-a^2c+a^3)+c(a^3+a^2b-a^2c+b^2a+b^3-b^2c-c^2a-c^2b+c^3)-2b(a^2c+a^3-a^2b+c^3+ac^2-bc^2-b^2c-ab^2+b^3)}{(b+c-a)(a+b-c)(c+a-b)}-\right)$$
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{a^4+c^4-2b^4+ab^3-a^3b+ac^3-a^3c+a^3c+b^3c-ac^3-bc^3+ab^2c+abc^2+a^2bc+ab^2c-2a^2bc-2abc^2\cdot\cdot}{(b+c-a)(a+b-c)(c+a-b)}\right)$$
It was getting very difficult to solve from here, is there any other method in which we can solve this question?
|
$$f(x)=\dfrac{2\sin2x}{1-\cos x}=\dfrac{4\sin x(1-(1-\cos x))}{1-\cos x}=\dfrac{4\sin x}{1-\cos x}-4\sin x=4\cot\dfrac x2-4\sin x$$
Using https://en.m.wikibooks.org/wiki/Trigonometry/Solving_triangles_by_half-angle_formulae and
https://en.m.wikipedia.org/wiki/Law_of_sines
$f(A)=\dfrac{4s(s-a)}\triangle -\dfrac{4a}{2R}$ where $2s=a+b+c$ and $R$ is the circumradius
Observe that both $s-y$ hence $\dfrac{4s(s-y)}\triangle$ and $\dfrac{4y}{2R}$ will separately form arithmetic progression for $y=a,b,c$ if $a,b,c$ are so.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3450752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Defining a function at some given point to make it continuous So I have a function
$$ f(x,y) = \frac{xy(x^2-y^2)}{x^2+y^2} $$
I have to define the function at $(0,0)$ so that it is continuous at origin. Is there any general approach we follow in this type of problems? I am not sure but I think Sandwich Theorem has to be applied and some suitable function has to be taken .
|
Please check out the new answer
$$ |xy(x^2-y^2)| <= |xy||(x^2-y^2)| <= |x||y||x^2+y^2|$$
Now,
$$ |x||y||(x^2+y^2)| = \sqrt(x^2)\sqrt(y^2)|x^2+y^2| <= (\sqrt{(x^2+y^2)})(\sqrt{(x^2+y^2)})|x^2+y^2| $$
The last expression is nothing but $(x^2+y^2)^2$ . Therefore
$$ \frac{ |xy(x^2-y^2)| }{(x^2+y^2)}<=\frac{(x^2+y^2)^2}{(x^2+y^2)} $$
Thus the Sandwich theorem can be applied now
Then,
$$-(x^2+y^2) < f(x,y) < (x^2+y^2) $$
And I enforce the limit $(0,0)$ which gives $f(x,y)$ as $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3451854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Smallest value of $x^2+5y^2+8z^2$ given $xy+yz+xz=-1$. The question is:
Find the smallest value of $x^2+5y^2+8z^2$ given $xy+yz+xz=-1$.
Here's what I've tried so far. Dividing by $xyz$, I get $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{-xyz}$. Squaring both sizes:
$$\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)^2 = \frac{1}{(xyz)^2}$$
I tried applying Cauchy Schwartz on this:
$$\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)^2 \leq 3 \left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)$$
but after this I'm stuck. Can someone help me out here?
I have also tried AM-GM on $x^2+5y^2+8z^2$ but that doesn't seem to lead anywhere.
Wolfram alpha says the answer is $4$.
|
Let $k$ be a minimal value.
Thus, $k>0$ and $$x^2+5y^2+8z^2\geq k$$ or
$$x^2+5y^2+8z^2+k(xy+xz+yz)\geq0$$ or
$$x^2+k(y+z)x+5y^2+8z^2+kyz\geq0,$$ for which we need
$$k^2(y+z)^2-4(5y^2+8z^2+kyz)\leq0$$ or
$$(20-k^2)y^2+(4k-2k^2)yz+(32-k^2)z^2\geq0,$$ for which we need
$$20-k^2>0$$ and $$(2k-k^2)^2-(20-k^2)(32-k^2)=0$$ or
$$k^3-14k^2+160=0$$ or
$$k^3-4k^2-10k^2+40k-40k+160=0$$ or
$$(k-4)(k^2-10k-40)=0,$$ which gives $k=4$.
Easy to see that the equality occurs, which says that we got a minimal value.
Now, we see that
$$x^2+5y^2+8z^2\geq4$$ it's
$$x^2+4(y+z)x+5y^2+8z^2+4yz\geq0$$ or
$$(x+2y+2z)^2+(y-2z)^2\geq0.$$
The equality occurs for $y=2z$, $x=-6z$ and $xy+xz+yz=-1.$
Id est, for example, for $$(x,y,z)=\left(-\frac{3}{2}, \frac{1}{2}, \frac{1}{4}\right).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3452821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
If there exists a positive integer $n$ such that any element $x$ of a ring $R$ satisfies $x^{4^n+2} = x$, then every element $x$ in $R$ is idempotent Let $R = (A,+,\cdot)$ be a ring. If there exists a positive integer $n$ such that any element $x$ of a ring $R$ satisfies $x^{4^n+2} = x$, then every element $x$ in $R$ is idempotent.
I have studied some group theory but I don't think it's that helpful here. Basically by multiplying both sides a bunch of times with $x^{4^n + 1}$ you can show that
$$x^{k(4^n + 1) + 1} = x \hspace{5px} \forall k \in \mathbb{N}.$$
My intuition goes along the lines of "if $x^a = x$ and $x^b = x$ then $x^{(a,b)} = x$" but in the absence of multiplicative inverses that wouldn't work (the reason I thought this might have been helpful is because the exponent of $2$ in $4^n + 2$ is always $1$, so by finding an appropriate $k$ we could make the gcd $2$).
How should I proceed?
|
Let $N$ be $4^n+2$ for short. It is an even number. Then from
$$
x = x^N=(-x)^N=-x
$$
we obtain $2x=0$ for all $x\in R$. So we work "in the characteristic two". In particular, $(x+y)^{4^n}=x^{4^n}+y^{4^n}$ for all $x,y$ in the ring. This implies:
$$
\begin{aligned}
x+1&=(x+1)^N
\\
&=(x+1)^{4^n}(x+1)^2
\\
&=\left(x^{4^n}+1\right)(x^2+1)
\\
&=x^N +x^{4^n}+x^2+1\ .
\end{aligned}
$$
This implies $x^{4^n}=x^2$. Then the given equation reads simpler,
$$
x = x^N = x^{4^n}\cdot x^2=x^2\cdot x^2=x^4\ .
$$
From here $x=x^4=(x^4)^4=x^{4^2}$, and inductively we get $x=x^{4^n}$. This implies $x=x^N=x\cdot x^2=x^3$. So $(x+1)=(x+1)^3$ and after cancellations finally $x=x^2$.
$\square$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3454587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
}
|
Minimise $2x+y$ subject to $x^2+y^2=1$ using KKT I'm doing this exercise in preparing for the final exam in optimization:
$$\begin{align*}
\text{min} &\quad 2x+y \\
\text{s.t} & \quad x^2+y^2=1
\end{align*}$$
Could you please verify if I correctly understanding the KKT's theorem? Thank you so much for your help!
My attempt:
Let $f(x,y)=2x+y$ and $h(x,y)=x^2+y^2-1$. The problem becomes $$\begin{align*}
\text{min} &\quad f(x,y) \\
\text{s.t} & \quad h(x,y)=0
\end{align*}$$
Let $\mathcal K = \{(x,y) \in \mathbb R^2 \mid h(x,y)=0\}$. Then $\mathcal K$ is compact and $(0,0) \notin \mathcal K$. Moreover, $f$ is continuous. Hence the problem has at least one solution. All solutions $(\bar x,\bar y)$ are different from $(0,0)$.
We have $\mu \nabla h(\bar x,\bar y) = \begin{pmatrix}0 \\0 \\ \end{pmatrix} \iff \mu \begin{pmatrix}2 \bar x \\2\bar y \\ \end{pmatrix} = 0 \iff \mu =0$. So the constraint qualification is satisfied. Consider $$\begin{cases} \nabla f( x, y) + \mu \nabla h( x, y) = \begin{pmatrix}0 \\0 \\ \end{pmatrix} \\ { x}^2+{ y}^2-1=0 \end{cases} \iff \begin{cases} \begin{pmatrix}2 \\1 \\ \end{pmatrix} + \mu \begin{pmatrix}2 x \\2 y \\ \end{pmatrix} = \begin{pmatrix}0 \\0 \\ \end{pmatrix} \\{ x}^2+{ y}^2=1 \end{cases}$$
Solving this system, we get $$(\mu, x, y) = (\sqrt{5}/2,-2/\sqrt{5} , -1/\sqrt{5})$$ and $$(\mu, x, y) = (-\sqrt{5}/2,2/\sqrt{5} ,1/ \sqrt{5})$$
As such, $f(-2/\sqrt{5} , -1/\sqrt{5}) = -\sqrt{5}$ and $f(2/\sqrt{5} , 1/\sqrt{5}) = \sqrt{5}$.
Hence $(-2/\sqrt{5} , -1/\sqrt{5})$ is the solution to the problem.
|
Your final solution is correct. But this problem doesn't use inequalities in the constraint so KKT is not needed. Just basic Lagrangian multipliers. You should consider
$$\begin{align*}
\text{max} &\quad xy \\
\text{s.t} & \quad x+y^2\leq2\\
&\quad x,y\geq 0
\end{align*}$$
Solution can be found here for your practice: http://www.math.ubc.ca/~israel/m340/kkt2.pdf
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3456783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Split into partial fractions $\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}$ This is from "Calculus Made Easy", Exercises 10, Question 15 (page 147). I've worked this one over and over and still haven't made progress.
This is my initial setup:
$$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2} = \frac{A}{(x+2)} + \frac{Bx+C}{(x^2+x+1)} + \frac{Dx+E}{(x^2+x+1)^2}$$
then:
$$3x^2+2x+1 = A(x^2+x+1)^2 + (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
and I can solve for A by setting $x=-2$, which yields $A=1$.
I know the final answer is this:
$$\frac{1}{x+2} - \frac{x-1}{x^2+x+1} - \frac{1}{(x^2+x+1)^2}$$
But I've worked this problem many ways and cannot make progress on the numerators for the other fractions.
|
Now that you know $A=1$ set $A$ equal to $1$. You get
$$3x^2+2x+1 = (x^2+x+1)^2 + (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
$$3x^2+2x+1 - (x^4+2x^3+3x^2+2x+1) = (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
$$-x^4-2x^3 = (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
$$-x^3(x+2) = (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
$$-x^3 = (Bx+C)(x^2+x+1) + (Dx+E)$$
$$-x^3 = Bx^3 + (B+C)x^2 + (B+C+D)x + (C+E)$$
and proceed from there
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3459213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
How to show $B^2=B$ when $rank(B) + rank(I_d - B) = d$? I have a question on this specific question from the past entrance examination of a university.
https://www.ism.ac.jp/senkou/kakomon/math_20190820.pdf
Related post: How to calculate eigenvalues of a matrix $A = I_d - a_1a_1^T - a_2a_2^T$
Problem
$≥3$, $_$ is an identity matrix, and $B$ is a d-dimensional square matrix.
Here, how to show $B^2=B$ when $rank(B) + rank(I_d - B) = d$?
Tried
I specifically considered the case of $d = 2$.
In this case, when $rank(B) + rank(I_d - B) = d$, I can find $rank(B) = rank(I_d - B) = 1$
I assume that
*
*$B = \left[
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right]
\to \left[
\begin{array}{cc}
a & b \\
0 & d-\frac{c}{a}b \\
\end{array}
\right]$
*$I - B = \left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right] - \left[
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right]
= \left[
\begin{array}{cc}
1-a & b \\
c & 1-d \\
\end{array}
\right]
\to \left[
\begin{array}{cc}
1-a & b \\
0 & (1-d)-\frac{c}{(1-a)}b \\
\end{array}
\right]$.
From $rank(B) = rank(I-B) = 1$,
*
*$d - \frac{c}{a}b = 0 \to ad = bc$
*$(1-d)-\frac{c}{(1-a)}b = 0 \to (a+d) = 1$.
And then,
$
\begin{eqnarray}
B^2 &=& \left[
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right]\left[
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right] \\
&=& \left[
\begin{array}{cc}
a^2 + bc & ab+bd \\
ac+dc & ad+d^2 \\
\end{array}
\right] \\
&=& \left[
\begin{array}{cc}
a^2 + ad & ab+bd \\
ac+dc & ad+d^2 \\
\end{array}
\right] \\
&=& \left[
\begin{array}{cc}
a(a+d) & b(a+d) \\
c(a+d) & d(a+d) \\
\end{array}
\right] \\
&=& \left[
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right] = B
\end{eqnarray}$
But I don't know how to generalize this.
I guess I can use $A^2 = A$ when $A = I_d - a_1a_1^T - a_2a_2^T$ (link), but I have no idea how to apply it.
|
Let $B'=I_d-B$. Observe that if $v\in\ker B\cap \ker B'$ then $v=Bv+B'v=0$, hence $\dim(\ker B\cap \ker B')=0$. Thus
$$
\dim(\ker B+\ker B')=\dim \ker B+\dim \ker B'=(d-\textrm{rank} B)+(d-\textrm{rank} B')=d.
$$
On the other hand since $B$ and $B'$ commute, every $v\in \ker B+\ker B'$ satisfies $BB'v=0$. Thus $BB'=0$, as desired.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3460891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Classic Complex Numbers - Given $z+\frac 1z=2\cos 3^\circ$, find least integer greater than $z^{2000}+\frac 1{z^{2000}}$ Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ,$ find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}.$
Solution: We have $z=e^{i\theta}$, so $e^{i\theta}+\frac{1}{e^{i\theta}}=\frac{\cos \theta}{\cos ^2\theta +\sin ^2\theta}+\cos \theta +i\sin \theta - \frac{i\sin \theta}{\cos ^2\theta +\sin ^2\theta}=2\cos \theta$. Therefore, $\theta =3^\circ=\frac{\pi}{60}$. From there, $z^{2000}+\frac{1}{z^{2000}}=e^{\frac{100\pi (i)}{3}}+e^{\frac{-100\pi (i)}{3}}=2\cos \frac{4\pi}{3}=-1$, so our answer is $\boxed{0}$.
How to solve this by applying Tchebyshev?
|
Note that, given $z+\frac1z = 2\cos a$ for any $a$,
$$z^n+\frac1{z^n}=2\cos (na)$$
holds true, which is shown recursively below,
$$z^n+\frac1{z^n}= \left(z+\frac1z\right)\left(z^{n-1}+\frac1{z^{n-1}}\right)
-\left(z^{n-2}+\frac1{z^{n-2}}\right)$$
$$=2\cos a \cdot 2\cos [(n-1)a] - 2\cos [(n-2)a] $$
$$=2[\cos (na) + \cos [(n-2)a]]- 2\cos [(n-2)a]
=2\cos (na) $$
Thus, for $a=3^\circ=\frac\pi{60}$,
$$z^{2000}+\frac 1{z^{2000}}=2\cos\left(\frac{2000\pi}{60}\right)=-1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3463278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Determining the value of $\sum_{n=2}^\infty \frac{1}{n^n-1}$ $$\displaystyle\sum_{n=1}^\infty\sum_{m=2}^\infty \frac{1}{m^{mn}}=\sum_{n=1}^\infty\left(\frac{1}{2^{2n}}+\frac{1}{3^{3n}}+\frac{1}{4^{4n}}+\cdots\right) \tag{$\star$}$$
I have a strong suspicion that the above summation converges, although I'm not sure how to prove it. But I'm more interested in the precise value of $(\star)$, especially a nice representation (if possible).
Here's what I have so far:
\begin{align}
\sum_{n=1}^\infty\left(\frac{1}{2^{2n}}+\frac{1}{3^{3n}}+\frac{1}{4^{4n}}+\cdots\right) & = \sum_{n=1}^\infty\frac{1}{2^{2n}}+\sum_{n=1}^\infty\frac{1}{3^{3n}}+\sum_{n=1}^\infty\frac{1}{4^{4n}}+\cdots \\ & = \sum_{n=1}^\infty\frac{1}{\left(2^{2}\right)^n}+\sum_{n=1}^\infty\frac{1}{\left(3^{3}\right)^n}+\sum_{n=1}^\infty\frac{1}{\left(4^{4}\right)^n}+\cdots \\ & = \frac{1}{2^2-1}+\frac{1}{3^3-1}+\frac{1}{4^4-1}+\cdots\\ & =\sum_{n=2}^\infty \frac{1}{n^n-1}
\end{align}
So $(\star)=\displaystyle\sum_{n=2}^\infty \frac{1}{n^n-1}$. From here, all I know how to do is get an approximate value of the solution. How would you go about finding its exact value?
|
If you want more figures $$\displaystyle \sum_{n=2}^\infty \frac{1}{n^n-1}=0.37605925334160927467565605197313357724916639675$$ All the digits are obtained summing up to $31$.
An amazing approximation of it is
$$\frac{76 \pi ^2+651 \pi-563 }{308 \pi ^2+481 \pi+1385}$$ which is in a relative error of $2.56\times 10^{-18}\text{ %}$.
Edit
Using exact arithmetic, I computed the value for $1000$ exact decimal places (if you want the number, tell me). The result is obtained summing up to $n=386$.
Whet looked interesting (at least to me) is that, negelcting the $-1$ in denominator
$$a_n=\frac{1}{n^n}\implies \frac{a_{n+1}}{a_n}=\frac{1}{e n}-\frac{1}{2 e n^2}+O\left(\frac{1}{n^3}\right)$$ So, writing
$$\sum_{n=2}^\infty \frac{1}{n^n-1}=\sum_{n=2}^{k-1} \frac{1}{n^n-1}+\sum_{n=k}^\infty \frac{1}{n^n-1}=\sum_{n=2}^{k-1} \frac{1}{n^n-1}+e^{\frac{1}{2 e}}I_0\left(\frac{1}{2 e}\right)\frac{ k }{k^k-1}$$
So, if we want the remainder to be less than $\epsilon$ we need
$$k\sim\frac{\log \left(\frac{b}{\epsilon }\right)}{W\left(\log \left(\frac{b}{\epsilon}\right)\right)}\qquad \text{where} \qquad b=e^{\frac{1}{2 e}}I_0\left(\frac{1}{2 e}\right)$$ Applied to the case where $\epsilon =10^{-1000}$, this gives $k=386.55$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3464100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
How can I prove analytically that the golden ratio is less then $\frac{\pi^2}{6}$. Or stated in other terms, prove that
$$\frac{1+\sqrt{5}}{2} < \sum_{n=1}^{\infty}\frac{1}{n^2}$$
|
\begin{align}
\frac{1+\sqrt{5}}{2}\lt\frac{\pi^2}{6}
\iff&3+3\sqrt{5}\lt\pi^2\\
\iff&3\sqrt{5}\lt\pi^2-3\\
\iff&(3\sqrt{5})^2\lt(\pi^2-3)^2\quad(\text{both sides are positive})\\
\iff&45\lt\pi^4-6\pi^2+9\\
\iff&\pi^4-6\pi^2-36\gt0\\
\end{align}
Using, for example, $3.14\lt\pi\lt3.15$ we have that
$$\pi^4-6\pi^2-36\gt3.14^4-6\cdot3.15^2-36\gt0$$
which implies that
$$\frac{1+\sqrt{5}}{2}\lt\frac{\pi^2}{6}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3464638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
What is the probability of the office being deemed functional? An IT office consists of 10 computers of which exactly 4 are working. To check if the office is functional, officials inspect four of the computers picked at random (without replacement). The office is deemed functional if at least three of the four computers are working. What is the probability of the office being deemed functional?
My Solution:
$E$ be event of office being deemed functional.
$E = 3$ out of $4$ computers working or all four computers working
$P(E) = P(3~\text{out of}~4~\text{computers working}) + P(\text{all four computers working})$
(3 out of 4 computers working)={WWWD,WWDW,WDWW,DWWW} where W,D are Working, Defective computers respectively.
(all four computers working)={WWWW}
$P(3~\text{out of}~4~\text{computers working}) = \frac{4 \cdot 3 \cdot 2 \cdot 6}{10 \cdot 9 \cdot 8 \cdot 7} \cdot 4 = 0.114$
$P(\text{all four computers working}) = \frac{4 \cdot 3 \cdot 2 \cdot 1}{10 \cdot 9 \cdot 8 \cdot 7} = 0.004$
$P(E) = 0.114+0.004 = 0.118$
Solution Provided is:
$P(3~\text{out of}~\text{4}~\text{computers working}) = \frac{{4 \choose 3}{6 \choose 1}}{{10 \choose 4}}$
$P(\text{all four computers working}) = \frac{{4 \choose 4}{6 \choose 0}}{{10 \choose 4}}$
The sum of these approximately equals my solution. Looking at the formula though I feel it should not work. The denominator ${10 \choose 4}$ is not size of the sample space for given problem, since all $10$ elements are not distinct. Also ${4 \choose 3}{6 \choose 1} = 24$ is not the number of possibilities that satisfy the condition ($3$ out of $4$ computers working), only $4$ possibilities satisfy this. Why does this answer still work?
|
The two answers are numerically exactly the same. Yours is $\frac{576}{5040}+\frac{24}{5040}=\frac{600}{5040}$ and the provided solution is $\frac{24}{210}+\frac{1}{210}=\frac{25}{210}$. Both are $\frac{5}{42} \approx 0.1190476$.
The factor of $4!=24$ in the numerator and denominator is simply the number of ways of ordering the four selected computers. You distinguished the orderings while the provided solution did not.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3466047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Using Fermat's Little Theorem to Show Divisibility I was asked to prove, using Fermat's Little Theorem, that $11|5^{10n+8}-4$ for $n\ge0$. I proved it but I was wondering whether there's an easier way (still using Fermat's). Here is my proof:
\begin{alignat}{3}
11|5^{10n+8}-4&\iff5^{10n+8}-4&&\equiv0 &&&\mod11\\
\quad&\iff 25^{5n+4}-4&&\equiv0 &&&\mod 11\\
\quad&\iff \qquad3^{5n+4}&&\equiv 4 &&&\mod 11\\
\quad&\iff \qquad3^{5n+5}&&\equiv 12 &&&\mod 11\\
\quad&\iff \qquad3^{5(n+1)}&&\equiv 1 &&&\mod 11.\\
\end{alignat}
For $n\ge1$, let S(n) be the statement
$$ S(n) :3^{5(n+1)}\equiv 1 \mod 11.$$ We will prove by induction on $n$ that $S(n)$ holds.
Base case ($n=1$). By Fermat's Little Theorem, $S(1)$ is true.
Inductive Step. Fix some $k\ge1$ and suppose $S(k)$ is true. To be shown is that the statement
$$S(k+1):3^{5(k+2)}\equiv 1 \mod 11$$ follows. Beginning with the LHS of $S(k+1)$,
\begin{alignat}2
\quad&3^{5(k+2)}&&=3^{5(k+1)+5}\tag{1}\\
\quad&\ \implies &&=3^{5}3^{5(k+1)}\tag{2}\\
\quad& \overset{\text{IH}}{\implies} &&\equiv3^{5}(1)\mod 11\tag{3}\\
\quad&\ \implies &&\equiv1\mod 11\tag{4},\\
\end{alignat}
arriving to the RHS of $S(k+1)$, concluding the inductive step. It is proved, then, by MI that $S(n)$ holds for all $n\ge1.$ Since $S(0)$ holds by $(4)$, then $S(n)$ is true for all $n\ge0$.
|
Much easier way!
By FLT $5^{10} \equiv 1 \pmod{11}$ so $5^{10n+8}\equiv 5^8$ and $5^{10n +8} -4 \equiv 5^8 -4\pmod {11}$.
So you just have to show that one case the $5^8 \equiv 4 \pmod {11}$. Then every case will be $5^{10n + 8} - 4\equiv 0 \pmod{11}$
Admittedly that requirse calculations but there are 3 ways, each more clever than the other
1) $5^2 = 25\equiv 3 \pmod {11}$. $5^4\equiv 3^2 \equiv 9\equiv -2 \pmod {11}$. $5^8\equiv (-2)^2 \equiv 4 \pmod {11}$.
2) $5^8*5^2 \equiv 5^{10} \equiv 1\pmod {11}$
$5^8*5^2 \equiv 5^8*3 \equiv 1\pmod{11}$ so as $11$ is prime $3^{-1}$ exist as is.... $1 \equiv 12=3*4\pmod{11}$ so $5^8*3*4 \equiv 4\pmod {11}$ and $5^8\equiv 4\pmod {11}$.
3) I'll admit I didn't come up with this.
If $5^8 -4 \equiv A\pmod{11}$ then
$(5^8-4)*25 \equiv A*25\pmod{11}$
$5^{10} - 100 \equiv 3A$
$1 - 1 \equiv 3A$
$3A \equiv 0\pmod {11}$ and as $11$ is primes $A\equiv 0 \pmod{11}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3467319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
}
|
The differential equation$ \frac{d^2 x}{dt^2}=f(x)$ We are supposed to solve the diff. eq.
$$x''=\frac{1}{x^3}$$
for $x(0)=1$ and $x'(0)=1$ using a particular approach. I've gotten so far:
Using the interval $I=(0,\infty)$ and $a \in I$ we define $U:I\to \mathbb{R}$:
$$U(x):=-\int_a^x\frac{1}{t^3}dt = \frac{1}{2x^2}-\frac{1}{2a^2}$$
Now we may write
$$x''=-\frac{dU}{dx}(x)$$
with some simple observations we may see that the function $E$ is constant:
$$E(t):=\frac{1}{2}(x'(t)^2)+U(x(t))=\frac{1}{2}(x'(0)^2)+U(x(0))=1-\frac{1}{2a^2}$$
So now we may write for $x'$:
$$x'(t)=\sqrt{2(E-U(x(t)))}=\sqrt{2-\frac{1}{x^2}}$$
which is supposibly a differential equation of seperated variables. I don't see the structure of $y'=f(x)g(y)$! How can we get from here to a solution of the original differential equation?
|
$$\frac{d^2 x}{dt^2}=\frac{1}{x^3}$$
Multiply by $2 \frac{dx}{dt}$ on both sides to get
$$2 \frac{dx}{dt} \frac{d^2x}{dt^2}=2 \frac{dx}{dt} \frac{1}{x^3} \implies \frac{d}{dt} \left( \frac{dx}{dt}\right)^2=\frac{2}{x^3}\frac{dx}{dt}.$$
Next, integrate w.r.t.t both sides
$$\int \frac{d}{dt} \left( \frac{dx}{dt}\right)^2 dt =\int \frac{2}{x^3}\frac{dx}{dt}dt
\implies \left(\frac{dx}{dt} \right)^2 =-\frac{1}{x^2} +A$$
Using $x(0)=1, x'(0)=1$ we get $A=2$, then
$$\frac{dx}{dt}=+\sqrt{2-\frac{1}{x^2}}\ \implies \int \frac{dx}{\sqrt{2-\frac{1}{x^2}}} =\int t+B$$
$$\implies \int \frac{x}{\sqrt{2x^2-1}}dx=t+B \implies $$
$$\sqrt{2x^2-1}=2t+2B \implies 2B=1$$
Finally we get a hyperbolic trajectory as
$$2x^2-(2t+1)^2=1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3468341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
What is the probability that $THTH$ occurs before $HTHH$ in an infinite sequence of coin flips?
What is the probability that $THTH$ occurs before $HTHH$ in an infinite sequence of coin flips?
The expected number of flips until you first see $THTH$ is $6$, while the expected number until you first see $HTHH$ is $10$. Intuitively, I would guess that the probability that $THTH$ occurs before $HTHH$ is $3/4$. Is there a formal argument to compute this probability?
Probabilistic model: We denote by $(X_{n})_{n \geq 1}$ the random variable of coin flips, taking values in $\{H,T\}^{\mathbb{N}}$. We let $t_{THTH}$ be the first time that $THTH$ occurs in the sequence. We have
$$
\mathbb{E}[t_{H}] = \frac{1}{2} \mathbb{E}[t_{H} | X_{1} = H] +
\frac{1}{2} \mathbb{E}[t_{H} | X_{1} = T] = \frac{1}{2} +
\frac{1}{2} (\mathbb{E}[t_{H}] + 1 ).
$$
$$
\mathbb{E}[t_{TH}] = \frac{1}{2} \mathbb{E}[t_{TH} | X_{1} = H] +
\frac{1}{2} \mathbb{E}[t_{TH} | X_{1} = T]
$$
|
We have states $\{H,T,HT,TH,THT,HTH,THTH,HTHH\}$ which are the prefixes of your strings. We can write down a Markov transition matrix for these states:
$$A = \frac{1}{2} \begin{pmatrix} 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0\\
1 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0 & 0 & 2 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \end{pmatrix}$$
This is an absorbing Markov chain already in canonical form \begin{pmatrix}Q&R\\0&I\end{pmatrix} thus we can find its fundamental matrix $N$ with
$$N = (I - Q)^{-1}$$
$$N = \frac{1}{7} \begin{pmatrix}
24 & 20 & 12 & 10 & 8 & 6\\
16 & 32 & 8 & 16 & 10 & 4\\
10 & 20 & 12 & 10 & 8 & 6\\
16 & 18 & 8 & 16 & 10 & 4\\
8 & 16 & 4 & 8 & 12 & 2\\
4 & 8 & 2 & 4 & 6 & 8
\end{pmatrix}.$$
Finally we can get the probabilities of our absorbing states given an initial probability vector of non-absorbing states $\vec p$with ${\vec p}^TNR$. Assuming we threw one coin already our initial state is $[1/2, 1/2, 0, 0, 0, 0]^T$, and thus our final answer is that we end up with $THTH$ $9/14$th of the time and $HTHH$ $5/14$th of the time.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3469403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Limit of $\lim_{x \to 0} \frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) }$ I tried using symbolab to get the limit of
$$\lim_{x \to 0} \frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) }$$
but it couldn't solve it. With WolframAlpha I got $100$ for the limit.
Can someone show me how it's done "per hand"? I tried, but couldn't figure it out.
|
\begin{align*}
&\dfrac{(1+x^{5})^{10}-1}{(\sqrt{1+x^{3}}-1)(\sqrt[5]{1+x^{2}}-1)}\\
&=\dfrac{(\sqrt{1+x^{3}}+1)((1+x^{5})-1)((1+x^{5})^{9}+\cdots+1)((1+x^{2})^{4/5}+\cdots+1)}{((1+x^{3})-1)((1+x^{2})-1)}\\
&=(\sqrt{1+x^{3}}+1)((1+x^{5})^{9}+\cdots+1)((1+x^{2})^{4/5}+\cdots+1)\\
&\rightarrow 2\cdot10\cdot 5\\
&=100.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3469782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
How to Solve this Simultaneous DE? Using laplace transform How to solve this Simultaneous DE using laplace transform
$x''+ y''+3x = 15e^{-t}$
$y''-4x'+3y = 15\sin(2t)$
with initial values $x(0)=35$, $x'(0)= -48$ and $y(0)=27$?
|
Notice that:
\begin{align}
x''+ y''+3x = 15e^{-t} & \Rightarrow_{\mathcal{L}} s^2 X - sx(0) - x'(0) + s^2 Y - sy(0) + y'(0) + 3X= 15 \frac{1}{s+1}\\
& \Rightarrow s^2 X - 35s + 48 +s^2 Y - 27s + a + 3X = \frac{15}{s+1}.
\end{align}
and
\begin{align}
y''-4x'+3y = 15\sin(2t) & \Rightarrow_{\mathcal{L}} s^2 Y - sy(0) + y'(0) - 4(sX - x(0)) + 3Y = 15 \frac{2}{s^2 + 4}\\
& \Rightarrow s^2 Y - 27s + a - 4sX + 140 + 3Y = \frac{30}{s^2+4}.
\end{align}
Here I posed $y'(0) = a$ since you didn't provided its value.
Now, you must solve the following system with respect to $X$ and $Y$:
$$\begin{cases}
s^2 X - 35s + 48 +s^2 Y - 27s + a + 3X = \frac{15}{s+1}\\
s^2 Y - 27s + a - 4sX + 140 + 3Y = \frac{30}{s^2+4}
\end{cases}$$
Finally, you must apply the inverse Laplace transform on both $X$ and $Y$ in order to obtain $x(t)$ and $y(t)$.
It is rather hard for me to solve this if you don't specify $y'(0)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3470923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
What is the eccentricity of the conic given by $r = \frac{2}{1 + \cos \theta - \sin \theta}$? A very simple question. Usually, Conic section in polar graph given as,
$$r = \frac{ke}{1 - e \cos\theta}$$
or $\sin\theta$, or something along this way.
However, given
$$r = \frac{2}{1 + \cos \theta - \sin \theta}$$
how could I find its eccentricity?
|
You have
\begin{align}
\cos\theta-\sin\theta&=-\sqrt2\,\left(-\frac1 {\sqrt2}\cos\theta+\frac1{\sqrt2}\sin\theta\right)\\ \ \\&=-\sqrt2\,\left(\cos\frac{3\pi}4\cos\theta+\sin\frac{3\pi}4\sin\theta\right)\\ \ \\ &
=-\sqrt2\,\cos\left(\theta-\frac{3\pi}4\right)
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3475764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to solve the recursion: $f(n)=3-\frac{1}{f(n-1)}$? I have some problems with the recursion: $f(n) = 3 - \frac{1}{f(n-1)} $, and the initial is $f(1)= \frac{5}{2}$.
Can anyone give me some hints of this?Thanks!!!
|
The given equation is
$$f(n)f(n-1)-3f(n-1)+1=0 \implies f(n-1)(f(n)-3)+1=0.$$
Let $$f(n)-3=\frac{g(n-1)}{g(n)},$$ then
we get $$g(n)+3g(n-1)+g(n-2)=0.$$
Now we take $g(n)= x^n$, then
$$x^2+3x+1=0 \implies x= a,b =\frac{-3 \pm \sqrt{5}}{2}$$
so $g(n)=p a^n+q b^n$. So $$f(n)=\frac{p a^{n-1}+ q b^{n-1}}{pa^n+qb^n}+3= \frac{1}{b}\frac{rc^{n-1}+1}{r c^n+1}+3, c=\frac{a}{b}=\frac{7-3\sqrt{5}}{2}, r=\frac{p}{q}$$
Now, if $f(1)=\frac{5}{2}$, then can find the value of $r$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3476948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Limit of $\frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$ as $n\to\infty$ I've tried to solve the limit
$$ \lim_{n \to \infty} \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$$
but I'm not sure.
$$ \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1} = \frac { 2^{\sqrt{ (\ln n)^2+ 2\ln n}}}{n^2+1} = \frac { 2^{\ln n \sqrt{ 1+ \frac {2}{\ln n}}}}{n^2+1} \sim \frac { 2^{\ln n }}{n^2+1} \rightarrow 0$$
Is it right?
I have another exercize that ends similarly with
$$ \frac { 10^{\ln n }}{n^2+1} \rightarrow 0$$
But the book says that the result is $+\infty$.
|
Your intuition is correct, but the part where you employ the symbol $\sim$ lacks a rigorous justification. To fully go into detail, you have the following expressions and estimates:
$$\frac{2^{\ln{n}\sqrt{1+\frac{2}{\ln{n}}}}}{n^2+1}=\frac{n^{{\ln{2}} \sqrt{1+\frac{2}{\ln{n}}}}}{n^2+1} \leqslant \frac{n^{\sqrt{3}\ln{2}}}{n^2+1}$$ for any $n \geqslant 3$, easily obtained by majorizing the exponent of $n$ in the numerator.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3477804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
Find value of $a_1$ such that $a_{101}=5075$ Let $\{a_n\}$ be a sequence of real numbers where $$a_{n+1}=n^2-a_n,\, n=\{1,2,3,...\}$$
Find value of $a_1$ such that $a_{101}=5075$.
I have
$$a_2=1^2-a_1$$
$$a_3=2^2-a_2=2^2-1^2+a_1$$
$$a_4=3^2-2^2+1^2-a_1$$
$$a_5=4^2-3^2+2^2-1^2+a_1$$
$$\vdots$$
$$a_{101}=100^2-99^2+98^2-97^2+\ldots+2^2-1^2+a_1.$$
Therefore, $$a_{101}=\sum_{i=1}^{50}(2i)^2-\sum_{i=1}^{50}(2i-1)^2.$$
Thus, $$5075=\sum_{i=1}^{50}(4i^2-4i^2+4i-1)+a_1,$$
and, $$a_1=5075-4\sum_{i=1}^{50}(i)+\sum_{i=1}^{50}(1).$$
Hence, $$a_1=5075-4(\frac{50}{2})(51)+50=25,$$
and $a_1=25$.
Is it correct? Do you have another way? Please check my solution, thank you.
|
Yes, your solution is correct. Another method of solution is to note that if $$a_{n+1} = n^2 - a_n,$$ we want to find some (possibly constant) function of $n$ such that $$a_{n+1} - f(n+1) = -(a_n - f(n)).$$ This of course implies $$f(n+1) + f(n) = n^2.$$ A quadratic polynomial should do the trick: suppose $$f(n) = an^2 + bn + c,$$ so that $$n^2 = f(n+1) + f(n) = 2a n^2 + 2(a+b)n + (a+b+2c).$$ Equating coefficients in $n$ gives $a = 1/2$, $b = -1/2$, $c = 0$, hence $$f(n) = \frac{n^2 - n}{2}.$$ It follows that if $$b_n = a_n - f(n) = a_n - \frac{n^2-n}{2},$$ then $$b_{n+1} = - b_n.$$ This gives us $$b_1 = b_{101}$$ which in terms of $a_n$, is $$a_1 = a_1 - \frac{1^2 - 1}{2} = a_{101} - \frac{(101)^2 - 101}{2} = 5075 - 5050 = 25.$$
This solution seems to come out of nowhere, but it is motivated by the idea that if we can transform the given recurrence into a corresponding recurrence for a sequence that is much simpler to determine, we can use this to recover information about the original sequence.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3482467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.