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Proof that $y^2=x^3+x$ has a unique integer solution Prove that the equation $y^2=x^3+x$ has only one integer solution, namely $x=y=0$.
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We see that $x$ divides $y^2$. But $x^2+1$ is prime with $x$. Therefore any prime dividing $x$ must appear with an even power. Therefore $x=a^2$ and $a$ divides $y$.
Putting this into the equation we get $y=ay_1$ and
$$a^2y_1^2=a^2(a^4+1).$$
If $a\neq 0$, then $$y_1^2=a^4+1$$
or $$(y+a^2)(y-a^2)=1.$$
Therefore $a=0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What converges the series? $$\sum _{n=4}^{+\infty \:}\frac{4^n\sqrt{n+1}-4^n\sqrt{n}+2^{n-1}\sqrt{n}\sqrt{n+1}}{4^n\sqrt{n}\sqrt{n+1}}$$
I tried different ways to figure this out and at first i did integrals
$$\int _4^{\infty \:}\frac{4^x\sqrt{x+1}-4^x\sqrt{x}+2^{x-1}\sqrt{x}\sqrt{x+1}}{4^x\sqrt{x}\sqrt{x+1}}dx$$
THEN
$=0-\left(4-2\cdot \sqrt{5}-\frac{1}{32\ln \left(2\right)}\right)$
But then i realized it wasn't with integrals and that I can't do convergence D:
I just had to figure out what exactly converges the series without using integrals so I just exploded my brain with everything i did before.
|
$$\sum _{n=4}^{+\infty \:}\frac{4^n\sqrt{n+1}-4^n\sqrt{n}+2^{n-1}\sqrt{n}\sqrt{n+1}}{4^n\sqrt{n}\sqrt{n+1}} =\sum_{n=4}^{\infty} \left(\frac{1}{\sqrt{n}} -\frac{1}{\sqrt{n+1}}\right) +\sum_{n=4}^{\infty}\frac{1}{2^{n+1}} =\frac{1}{\sqrt{4}} +\frac{1}{16}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Can we solve this using stars and bars? The number of ways of distributing 12 identical oranges among 4 children so that every child gets at least one and no child more than 4 is 31.
My try: First of all give each child 1 orange and we are left with 8 oranges.
Then
3 3 2 0
3 3 1 1
3 2 2 1
2 2 2 2
and permuting each, we get total 31. But can we solve this using stars and bars?
|
A simple solution by generating functions, ellipses cover terms that don't influence the result:
\begin{align}
[z^{12}] (z + z^2 + z^3 + z^4)^4
&= [z^{12}] \left( \frac{z (1 - z^4)}{1 - z} \right)^4 \\
&= [z^8] (1 - z^4)^4 (1 - z)^{-4} \\
&= [z^8] (1 - 4 z^4 + 6 z^8 - \cdots)
\sum_{n \ge 0} \binom{-4}{n} (-1)^n z^n \\
&= [z^8] (1 - 4 z^4 + 6 z^8 - \cdots)
\sum_{n \ge 0} \binom{n + 4 - 1}{4 - 1} z^n \\
&= \binom{11}{3} - 4 \binom{7}{3} + 6 \binom{3}{3} \\
&= 31
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve $x$ in $(x+1)^4+(x-1)^4=16$? I'm trying my hand on these types of expressions. How to solve $x$ in $(x+1)^4+(x-1)^4=16$? Please write any idea you have, and try to keep it simple. Thanks.
|
$(x+1)^4+(x-1)^4=16\\
\implies (x^4+4x^3+6x^2+4x+1)+(x^4-4x^3+6x^2-4x+1)-16=0\\
\implies (2x^4+12x^2+2)-16\\
\implies 2x^4+12x^2-14=0\\
\implies 2(x^4+6x^2-7)=0\\
\implies 2((x^2)^2+6(x^2)^1-7(x^2)^0)=0$
Now let $x^2=a$:
$2((a)^2+6(a)^1-7(a)^0)=0\\
\implies 2(a^2+6a-7)=0\\
\implies a^2+6a-7=0\\
\implies (a+7)(a-1)=0\\
\implies a=-7 \text{ or } a=1$.
Letting $a=x^2$ again, we get
$x^2=-7\text{ or } x^2=1$.
Hence the real solutions of $x$ are $\pm1$.
|
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|
Show that $\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$ As the title states, trying to solve $$\int_0^1 \! \frac{1+x^2}{1+x^4} \, \operatorname d \!x=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\cdots$$
|
$$\begin{align}
\int_0^1 \frac{1+x^2}{1+x^4} \operatorname d \!x &=
\int_0^1 \frac{1+x^4-x^4+x^2}{1+x^4} \operatorname d \!x\\
&= 1 + \int_0^1 x^2\frac{1-x^2}{1+x^4} \operatorname d \!x\\
&= 1 + \int_0^1 x^2\frac{1 + x^4 - x^4-x^2}{1+x^4} \operatorname d \!x\\
&= 1 + \frac{1}{3} -\int_0^1 x^4\frac{1+x^2}{1+x^4} \operatorname d \!x\ \ldots
\end{align}
$$
and so forth.
|
{
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|
let a,b,c be the sides of a triangle and $a+b+c=2$ then prove that $ a^2 + b^2 + c^2 +2abc < 2$ let a,b,c be the sides of a triangle and $a+b+c=2$ then prove that
$ a^2 + b^2 + c^2 +2abc < 2$
I used the fact that $ a+b-c>0 $ and multiplied the 3 cyclic equations but coudn't reach the final equation.
|
Just continue your approach, focusing on reaching the goal:
$$(a+b-c)(b+c-a)(c+a-b) > 0 \iff (1-c)(1-a)(1-b) > 0 $$
$$\iff 1-(a+b+c)+(ab+bc+ca)-abc > 0$$
$$\iff 1-(2)+\tfrac12 \left((a+b+c)^2-a^2-b^2-c^2 \right)-abc > 0$$
$$\iff -2+ \left(4-a^2-b^2-c^2 \right)-2abc > 0$$
$$\iff a^2+b^2+c^2 + 2abc < 2$$
|
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|
Inductions and proofs Let $(h_n)$ be a sequence defined by $h_0 =1 , h_1 = 2, h_2 =3$ and $h_n = h_{n-1} + h_{n-2} + h_{n-3}$, for all $n\ge 3$. Prove that $h_n\le 2^n$ , for all $n\ge 0$
Not sure how to go with this problem?
|
Let's prove it by strong induction on $n$.
Base case: $n = 0$. $h_0 = 1 = 2^0$. So assume it is true for $n < k$. We prove it is true for $n = k+1$.
$h_{k+1} = h_k + h_{k-1} + h_{k-2} \leq 2^k + 2^{k-1} + 2^{k-2} = \dfrac{2^{k+1}}{2} + \dfrac{2^{k+1}}{4} + \dfrac{2^{k+2}}{8} = \dfrac{7}{8} \cdot 2^{k+1} < 2^{k+1}$.
So by strong induction, the statement is proven.
|
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|
Coefficients of this generating function For the first part of a problem, I solved the generating function to be $F(x) = \frac{x^3}{(1-x)^2}$
Now it's the easy part that has me a little confused. What would the coefficients be in this case?
I have computed so far:
$$\frac{x^3}{(1-x)^2} = x^3 (1+2x+3x^2 + \dots) =x^3 + 2x^4 + 3x^5 + \dots$$
Which leads me to believe that $f(n)=(n-2)$. But I know this only holds true for $ \ge 2$. What am i missing?
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There is the binomial theorem, which says that
$$(1+x)^n = \sum_{k \ge 0} \binom{n}{k} x^k$$
This is true even when $n$ is not necessarily a positive integer; we just have to define the binomial coefficient properly: for nonnegative integer $k$ and any $n$, we define
$$\binom{n}{k} = \frac{n(n-1)\cdots(n-k+1)}{k!}$$
With that definition, we have
$$\binom{-2}{k} = \frac{(-2)(-3)\cdots(-2-k+1)}{k!} = (-1)^k \frac{2\cdot 3 \cdots (k+1)}{k!} = (-1)^k (k+1).$$
Thus,
$$\frac{x^3}{(1-x)^2} = x^3(1-x)^{-2} = x^3\sum_{k \ge 0} \binom{-2}{k} x^k = x^3\sum_{k \ge 0} (k+1) x^k = \sum_{k \ge 0} (k+1) x^{k+3}$$
As the exponent is $k+3$, we can make the substitution $m = k+3$ (and so $k +1 = m - 2$) to get the above to be equal to
$$\sum_{k \ge 0} (k+1)x^{k+3} = \sum_{m \ge 3} (m-2) x^m$$
This means that the coefficient of any $x^m$ is
$$[x^m]\frac{x^3}{(1-x)^2} = \begin{cases}
(m-2) & \text{if $m \ge 3$} \\
0 & \text{otherwise}
\end{cases}$$
|
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|
Integral of $\frac{-2}{\sqrt{16-x^2}}$ So I am asked to find the anti-derivative of
$$\frac{-2}{\sqrt{16-x^2}}$$
First step I took was to make it easier for me to visualise
$$\int{-2(16-x^2)^{-\frac{1}{2}}}dx$$
I let $u = 16-x^2$
So $\frac{du}{dx} = -2x du$
So substituting in $u$ we have
$$=\int{-2(u)^{-\frac{1}{2}}}dx$$
Substituting in $\frac{du}{dx}$ we have
$$=\int{\frac{u^{-\frac{1}{2}}}{x}}du$$
Since $x$ is a constant I move it to the front
$$=\frac{1}{x}\int{{u^{-\frac{1}{2}}}}du$$
The integral of $u^{-1/2}du$ is $2u^{1/2}$ +c
So I have
$$=\frac{2\sqrt{u}}{x} + c$$
Substituting $u$ back in I have
$$=\frac{2\sqrt{16-x^2}}{x}+c$$
I know I can do this by trigonometric substitution however that wouldn't be obvious to me at this stage and therefore would only be used as a last resort. The first thing I would try is simple integration by substitution but I can't see where I have gone wrong.
|
trigonometric substitution isn't that difficult. It just requires practice.
try expressing
$$\frac{-2}{\sqrt{16-x^2}}$$
as
$$\frac{-2}{\sqrt{16(1-(\frac{x}{4})^2)}} = \frac{-2}{4\sqrt{(1-(\frac{x}{4})^2)}}=\frac{-1}{2\sqrt{(1-(\frac{x}{4})^2)}}=-\frac{1}{2\sqrt{(1-\sin^2(\sin^{-1}(\frac{x}{4})))}}$$
so for the integral
$$\int\frac{-2}{\sqrt{16-x^2}}dx = -\frac{1}{2}\int\frac{1}{\sqrt{1-\sin^2(\sin^{-1}(\frac{x}{4}))}}dx$$
If we try to express $\frac{x}{4}$ as $\sin(\sin^{-1}(\frac{x}{4}))$ the substitution becomes obvious.
$$u = \sin^{-1}(\frac{x}{4})\quad\text{or}\quad 4\sin u=x$$
$$4\cos u\,du/dx=1$$
$$4\sqrt{1-\sin^2 u}\,du/dx=1$$
$$dx=4\sqrt{1-\sin^2 u}\,du$$
and our integral becomes
$$-\frac{1}{2}\int\frac{1}{\sqrt{1-\sin^2u}}4\sqrt{1-\sin^2 u}\,du=-2\int du=-2u+C=-2\sin^{-1}(\frac{x}{4})+C$$
Just one final note on integrating this way...suppose that we integrated a different problem with $u=\sin^{-1}x$, and our final answer was $\tan u+C$, we would want to express the $\tan$ in terms of $\sin$ as follows.
$$\tan u + C = \frac{\sin u}{\cos u}+C=\frac{\sin u}{\sqrt{1-\sin^2 u}}+C$$
this allows for the $\sin$s to cancel out with the $\sin^{-1}$s.
$$\frac{\sin u}{\sqrt{1-\sin^2 u}}+C=\frac{\sin(\sin^{-1}x)}{\sqrt{1-\sin^2(\sin^{-1}x)}}+C=\frac{x}{\sqrt{1-x^2}}+C$$
In general, before deciding on the substitution, we express
$$1-x^2 = 1-\sin^2(\sin^{-1}x)=\cos^2(\sin^{-1}x);$$
$$x^2-1 = \sec^2(\sec^{-1}x)-1=\tan^2(\sec^{-1}x);$$
$$x^2+1 = \tan^2(\tan^{-1}x)+1=\sec^2(\tan^{-1}x).$$
Like I said before, it takes practice, but it does make trigonometric more intuitive and less of just an algorithm.
|
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|
How do I evaluate this limit with an integral function? Could anybody give me some pointers on how do I evaluate this limit:
$$\lim_{t \to +\infty} \frac{\int_0^t x^9e^{-x^2}\,dx}{\int_0^t x^7e^{-x^2}\,dx}$$
I'm not asking for the complete solution, just a hint to point me in the right direction. I appreciate your help. Thanks.
|
$\frac{\int_0^tx^9e^{-x^2}dx}{\int_0^tx^7e^{-x^2}}=\frac{-\frac{1}{2}(e^{-t^2}(t^8+4t^6+12t^4+24t^2+24)+24)}{-\frac{1}{2}(e^{-t^2}(t^6+3t^4+6t^2+6)+6)}=\frac{\frac{1}{2}e^{-t^2}(t^8+4t^6+12t^4+24t^2+24+24e^{t^2})}{\frac{1}{2}(e^{-t^2}(t^6+3t^4+6t^2+6+6e^{t^2}))}=\frac{t^8+4t^6+12t^4+24t^2+24+24e^{t^2}}{t^6+3t^4+6t^2+6+6e^{t^2}}=\frac{e^{-t^2}(t^8+4t^6+12t^4+24t^2+24)+24}{e^{-t^2}(t^6+3t^4+6t^2+6)+6}\to 4\,\,\,,\,as\,t\to\infty$
|
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Area of triangle inscribed in a parabola How can u prove that the area of the triangle inscribed in a parabola is twice the area of the triangle formed by the tangents at the vertices?
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Claim: Given any 2 points on the parabola, the area that is bounded by the line between these 2 points and the parabola, is twice the area that is bounded by the 2 tangents and the parabola.
Proof. Standard calculus techniques work.
WLOG, we normalize the quadratic to be of the form $ y = x^2$. Let the points be $ (a, a^2), (b, b^2)$ with $a<b$. The equation of the tangents are $ y = 2ax - a^2$ and $ y = 2bx -b^2$. They intersect at $x = \frac{ b+a} { 2}, y = ab$.
The area between the two points and the parabola is equal to the trapezium minus the area under the curve. The area of the trapezium is $\frac{ a^2 + b^2}{2} \times (b-a)$. The area under the curve is $\int_a^b x^2 \, dx = [\frac{1}{3} x^3]_a^b = \frac{ b^3 - a^3 } { 3}$. Hence, the area is $ \frac{b^3-ab^2+a^2b-a^3}{2} - \frac{b^3-a^2}{3} = \frac{ b^3-3ab^2+3a^2b-a^3 } { 6}$.
The area between the two points and the intersection of their tangents are given by
$ 2ab^2 + \frac{b^3}{2} + \frac{ab^2}{2} -2ab^2 - \frac{ a^3}{2} - \frac{ a^2b}{2} = \frac{ b^3-3ab^2+3a^2b-a^3 } { 2}$. This is thrice of the area that we calculated above. Hence, the area that is between the parabola and the tangents, is twice the area between the parabola and the line.
Corollary: Apply the above claim to each pair of your 3 points. We get that the area of the triangle within the parabola is twice the area of the triangle formed by tangents.
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How can I find integers $n \gt 1$ such that the average of $1^2,2^2,3^2...n^2$ is itself a perfect square.
$\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$
so we would like to solve
$6k^2=(n+1)(2n+1)$
here we see that $6|(n+1)(2n+1)\implies 2|n+1$
hence we can set $n=2j-1$
$2j(4j-1)=6k^2\implies j(4j-1)=3k^2$
how can i find $j$ ,beside trial-and-error
$n=1$ is trivial
|
We are looking for $n,k\in\mathbb{N}$ such that $k^2=\frac{1}{n}\sum^n_{i=1}i^2=\frac{(n+1)(2n+1)}{6}$. Set $n_0=n+1$.
The equation becomes $n_0(2n_0-1)=6k^2$. So $n_0$ is even.
Set $n_0=2n_1$ to get $n_1(4n_1-1)=3k^2$. Now suppose $3|n_1$ and set $n_1=3n_2$. Then $n_2(12n_2-1)=k^2$. Since $n_2$ and $12n_2-1$ are coprime, both have to be squares. But $12n_2-1$ is never a square because $-1$ is not a quadratic residue modulo $4$.
Hence we must have $3\nmid n_1$. But then $3|4n_1-1$, so $n_1\equiv 1\mod{3}$. Set $n_1=3m+1$.
The equation is now $(3m+1)(4m+1)=k^2$. Since $3m+1$ and $4m+1$ are coprime, $3m+1=k_1^2$ and $4m+1=k_2^2$.
Now just go through the first few squares to find some solutions. The smallest solution is $k_1=13$, $k_2=15$, $m=56$ which gives $n=337$.
|
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|
Find sequence function and general rule the function $$a_{n+2}=3a_{n+1}-2a_n+2$$
is given, and $$a_0=a_1=1, (a_n)_{n\ge0}$$
multiplying everything by $$/\sum_{n=0}^\infty x^{n+2}$$
also adding $$\sum_{n=0}^\infty (a_{n+2}x^{n+2}+a_1x+a_0)-a_1x-a_0=\sum_{n=0}^\infty (3a_{n+1}x^{n+2}+a_0)-a_0-\sum_{n=0}^\infty 2a_nx^{n+2}+\sum_{n=0}^\infty 2x^{n+2}$$
we get $$R(X)-2x-1=3x(R(X)-1)-2x^2R(X)+\frac{2x^2}{1-x}$$
$$R(X)(1-3x+2x^2)+x-1=\frac{2x^2}{1-x}$$
$$R(X)(1-3x+2x^2)=\frac{3x^2-2x+1}{1-x}$$
$$R(X)=\frac{3x^2-2x+1}{(1-x)^2(1-2x)}=\frac{A}{(1-x)^2}+\frac{B}{1-2x}$$
so $$3x^2-2x+1=A(1-2x)+B(1-x)^2$$
$$3x^2-2x+1=x^2(B)+x(-2A-2B)+A+B$$
B=3
-2A-2B=-2
A+B=1$$\quad\Longrightarrow\quad A=-2, B=3$$
$$R(X)=\frac{-2}{(1-x)^2}+\frac{3}{1-2x}=-2\sum_{n=0}^\infty nx^{n-1}+3\sum_{n=0}^\infty (2x)^2=$$
how to proceed?
|
$a_{n+2}=3a_{n+1}-2a_n+2$, $a_0=a_1=1$
Doing it the more pedestrian way, you would first find the basis solutions of the homogeneous equation that are geometric progressions, $a_n=q^n$. This is the case if
$$
q^2=3q-2\iff 0=(q-1)(q-2),
$$
so that the general homogeneous solution is
$$
a_n=C+D\,2^n
$$
Since the inhomogeneity is also a constant, the ansatz of similar type is $a_n=A\,n$, resulting in the equation
$$
A(n+2)=3A(n+1)-2An+2\iff -A=2\iff A=-2
$$
so that the general solution of the inhomogeneous recursion has the form
$$
a_n=-2n+C+D\,2^n
$$
Checking the initial conditions gives
$$
1=C+D=-2+C+2D\iff 0=-2+D\land C=-D+1\iff D=2\land C=-1
$$
so finally
$$
a_n=-(2n+1)+2^{n+1}.
$$
|
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|
find sum of first 2002 terms if $\left \{ a_n \right \}$ is sequence of Real Numbers for $n \ge 1$ such that
\begin{equation}
a_{n+2}=a_{n+1}-a_n \tag{1}
\end{equation}
\begin{equation}
\sum_{n=1}^{999} a_n=1003 \tag{2}
\end{equation}
\begin{equation}
\sum_{n=1}^{1003}a_n=-999 \tag{3}
\end{equation}
Then Find the value of $$\sum_{n=1}^{2002} a_n $$
|
Anaother way to solve the problem is to use the characteristic equation. The characteristic equation is
$$ r^2=r-1 $$
whose solution is $r_1=e^{\frac{\pi}{3}i}, r_2=e^{-\frac{\pi}{3}i}$. So $a_n$ can be expressed as
$$ a_n=C_1e^{\frac{n\pi}{3}i}+C_2e^{-\frac{n\pi}{3}i} $$
where $C_1, C_2$ are constants which will be determined later. Hence
\begin{eqnarray*}
S_n&=&\sum_{k=1}^na_n=C_1\sum_{k=1}^ne^{\frac{k\pi}{3}i}+C_2\sum_{k=1}^ne^{-\frac{k\pi}{3}i}\\
&=&C_1\left(\frac{1-e^{\frac{(n+1)\pi}{3}i}}{1-e^{\frac{\pi}{3}i}}-1\right)+C_2\left(\frac{1-e^{-\frac{(n+1)\pi}{3}i}}{1-e^{-\frac{\pi}{3}i}}-1\right).
\end{eqnarray*}
Easy calculation shows that
$$ a_{999}=(-1+\sqrt{3}i)C_1+(-1-\sqrt{3}i)C_2, a_{1003}=\frac{1+\sqrt{3}i}{2}C_1+\frac{1-\sqrt{3}i}{2}C_1. $$
Solving
$$ a_{999}=1003, a_{1003}=-999 $$
gives
$$ C_1=-\frac{3001}{4}+\frac{995i}{4\sqrt{3}}, C_2=-\frac{3001}{4}-\frac{995i}{4\sqrt{3}} $$
and hence
\begin{eqnarray*}
a_{2002}&=&C_1e^{\frac{2002\pi}{3}i}+C_2e^{-\frac{2002\pi}{3}i}\\
&=&-C_1e^{\frac{\pi}{3}i}-C_2e^{-\frac{\pi}{3}i} \\
&=&Re[-(-\frac{3001}{4}+\frac{995i}{4\sqrt{3}})\frac{1+\sqrt{3}i}{2}]\\
&=&2002.
\end{eqnarray*}
|
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|
Finding $a$ and $b$ from $a^3+b^3$ and $a^2+b^2$ Question 1
Two numbers are such that the sum of their cubes is 14 and the sum of their squares is 6. Find the sum of the two numbers.
I did
$a^2+b^2=6$ and $a^3+b^3=14$ Find $a$ and $b$, two numbers. but got lost when trying to algebraicly solve it.
Thank you, Any help is appreciated
|
There's another way to get to answer if we take $a=c+d$ and $b=c-d$, then using binomial expansions we have
$$
a^2+b^2=2(c^2+d^2)=6\\
a^3+b^3=2c^3+6cd^2=14
$$
We have $d^2=3-c^2$ from first equation and using this in second equation
$$
c(9-2c^2)=7
$$
We can easily see that one of the solutions is c=1.
|
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|
Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$
I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from solve this question...
|
Let $\delta x = y-x$ and $x\ge y$. Then
$$4^x(1+4^{\delta x}) = 2^{x+1}(1+2^{\delta x})$$
then from factorization we can claim that:
$$2x = x+1 ~\text{and}~~ 4^{\delta x} = 2^{\delta x}$$
So, we get $x = y = 1$.
|
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|
Probability of number formed from dice rolls being multiple of 8
A fair 6-sided die is tossed 8 times. The sequence of 8 results is recorded to form an 8-digit number. For example if the tosses give {3, 5, 4, 2, 1, 1, 6, 5}, the resultant number is $35421165$. What is the probability that the number formed is a multiple of 8.
I solved this by listing all possibilities for the last 3 digits that give multiples of 8, and found this to be $\frac{1}{8}$.
The solution key agrees with my answer, but also says that "There are quicker ways to solve the problem using a more advanced knowledge of number theory"
What would be a faster way to solve this using number theory?
|
Let's use Zhuli's notation.
$c$ has a probability of $\frac{1}{2}$ to be even.
Now if $\frac{c}{2}$ is even (i.e., $4|c$), then $4|2b+c \iff b$ is even. If $\frac{c}{2}$ is odd, then $4|2b+c \iff b$ is odd. Both conditional probabilities are $\frac{1}{2}$ therefore the marginal probability $4|2b+c$ is $\frac{1}{4}$.
Next we apply the same reasoning on $2a + (b+\frac{c}{2})$ we get the marginal probability that $4 | 2a + (b+\frac{c}{2})$, or equivalently, $8 | 4a + 2b +c$, is $\frac{1}{8}$. Q.E.D.
============================
Alternatively we could avoid conditional probabilities and just use independence and marginal probabilities.
First note that $\Pr(2|a)=\Pr(2|b)=\Pr(2|c)=\frac 12$ and $a,b,c$ are independent.
Next,
$$
\Pr(4|2b+c)=\Pr(2|c,4|c,2|b) + \Pr(2|c, 4\nmid c, 2\nmid b)
$$
$$
=\Pr(2|c, 4|c) \Pr(2|b)+\Pr(2|c, 4\nmid c)\Pr(2\nmid b) \tag{independence}\\
= \frac 12 \left( \Pr(2|c, 4|c) + \Pr (2|c, 4\nmid c)\right) = \frac 12 \Pr(2|c)=\frac 14.
$$
Finally,
$$
\Pr(8|4a+2b+c)=\Pr(4|2b+c,8|2b+c,2|a) + \Pr(4|2b+c, 8\nmid 2b+c, 2\nmid a)\\
=\Pr(4|2b+c,8|2b+c)\Pr(2|a) + \Pr(4|2b+c, 8\nmid 2b+c)\Pr(2\nmid a)\\
=\frac 12 \left(\Pr(4|2b+c,8|2b+c) + \Pr(4|2b+c, 8\nmid 2b+c)\right)\\
=\frac 12 \Pr(4|2b+c)=\frac 12 \cdot \frac 14 = \frac 18.\blacksquare
$$
Remark: We can prove, via induction, the generalization proposed by Anant using the technique above.
|
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|
Normalizer of a subgroup of $GL_2(\mathbb{R})$ I have following subgroup of $GL_2(\mathbb{R}):$ $$A=\Bigg\{\left( \begin{array}{cc}
1 & 0 \\
0 & 1 \end{array} \right),\left( \begin{array}{cc}
0 & -1 \\
1 & 0 \end{array} \right),\left( \begin{array}{cc}
0 & 1 \\
-1 & 0 \end{array} \right),\left( \begin{array}{cc}
-1 & 0 \\
0 & -1 \end{array} \right)\Bigg\}.$$
It's basically the rotation group of a square centered at the origin in $\mathbb{R}^2.$
I need to compute the normalizer of $A$.
Let's denote the normalizer by $A_N.$ Then, by definition, $$A_N:=\{g\in GL_2(\mathbb{R}): gA=Ag\}.$$ $$\equiv \{g\in GL_2(\mathbb{R}): gag^{-1}\in A,\ \forall \ a\in A\}.$$
I am using the second definition to compute all such $g$ by hand, but the computation is leading me nowhere. I have a feeling that all matrices that have $1$ or $-1$ on one diagonal and $0$ in the other diagonal satisfy the normalization condition. For example: all elements of $A$ are clearly in $A_N$. When I check for other matrices with $1$ or $-1$ on one diagonal and $0$ in the other, they all satisfy the condition. How to make this idea more precise? I am sure there must be a nice way to arrive at such $g$ through computation but I am yet to find one.
|
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\in N_{GL_2(\Bbb R)}(A)\implies $$
$$\begin{align*}
\bullet&\begin{pmatrix}b&\!\!-a\\d&\!\!-c\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&\!\!-1\\1&0\end{pmatrix}=\begin{pmatrix}0&\!\!-1\\1&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}\!\!-c&\!\!-d\\a&b\end{pmatrix}\iff\begin{cases}a=d\\{}\\b=-c\end{cases}\\{}\\\bullet&\begin{pmatrix}\!\!-b&a\\\!\!-d&c\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&1\\\!\!-1&0\end{pmatrix}=\begin{pmatrix}0&1\\\!\!-1&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}c&d\\\!\!-a&\!\!-b\end{pmatrix}\iff\begin{cases}a=d\\{}\\b=-c\end{cases}\end{align*}$$
The other two cases are useless as those are scalar matrices and thus they commute with everything...
We thus get that the general form of an element of the normalizer is
$$\begin{pmatrix}a&b\\\!\!-b&a\end{pmatrix}\;,\;\;\text{and of course}\;\;a^2+b^2\neq0$$
|
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|
Calculating absolute value and argument of a complex number I want to calculate the absolute value and argument of the complex number $a = \left(\sqrt{3} - i\right)^{-2}$.
In order to calculate these two values I tried to reform the number into the form $z = x + y \cdot i$:
$$a =\left(\sqrt{3} - i\right)^{-2} = \frac{1}{\left(\sqrt{3}-i\right)^2} = \frac{1}{2 +\left(-2\sqrt{3}\right)}$$
$$ = -\frac{2}{8} - i \frac{-2\sqrt{3}i}{4-12} = -\frac{1}{4} - i \frac{\sqrt{3}i}{4}$$
I now want to calculate the polar form of a:
$$r = |z| = \sqrt{x^2 + y^2} = \sqrt{\frac{1}{8} + \left( -1 -2i \left( \frac{\sqrt{3}i}{4}\right) + \left(\frac{\sqrt{3}i}{4}\right)^2 \right)}$$
$$ = \sqrt{\frac{1}{8} + \left( -1 -2i \left( \frac{\sqrt{3}i}{4}\right) + \frac{3 + 2 \sqrt{3}i - 1}{16} \right)} =
\sqrt{\frac{1}{8} + \left( -1 -i \left( \frac{\sqrt{3}i}{2}\right) + \frac{3 + 2 \sqrt{3}i - 1}{16} \right)} $$
$$=\sqrt{\frac{1}{8} + -1 \frac{\sqrt{3}-1}{2} + \frac{2 + 2 \sqrt{3}i}{16}}$$
but finally am stuck here. Can you please help me to go on? Is there a better way to get to the absolute value and argument?
|
I would start by finding the polar form of $z = \sqrt{3} - i$. This corresponds to a nice triangle, the so called 30-60-90 triangle. Here one leg is of length $\sqrt{3}$ and the other is of length 1 (in the negative imaginary direction). The hypotenuse of this triangle is the magnitude of $z$, so $|z|=2$.
The argument of this $z$ is $\theta = -\pi/6$.
Now the number you want is $$a = 1/z^2 = \frac{1}{|z|^2e^{i2\theta}} = \frac{1}{|z|^2} e^{-2i\theta}.$$
|
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|
Polar Equation to Rectangular? The equation is:
$$r = \frac{4}{1+2sin(\theta)}$$
I'm confused about how to convert it into rectangular form.
This is what I have so far, although I'm not sure it's correct:
$$r = \frac{ 4(1-2sin(\theta)) }{ 1-2sin^2(\theta) } $$
$$r = \frac{ 4-8sin(\theta) } {( cos^2(\theta) - sin^2(\theta) )} $$
$$r = \frac{( 4-8(\frac{y}{r}) )}{ ( (\frac{x}{r})^2 - (\frac{y}{r})^2 )} $$
$$r ( (\frac{x}{r})^2 - (\frac{y}{r})^2 ) = 4 - 8(\frac{y}{r}) $$
$$\frac{(x^2)}{r} - \frac{(y^2)}{r} - \frac{8y}{r} = 4$$
What do I do after that? Do I complete the square? That leaves me with an r on the right-hand side, though.
|
$$r = \dfrac{4}{1+2 \sin(\theta)} \rightarrow r(1 + 2 \sin(\theta)) = 4$$
A polar plot shows:
We have:
*
*$r = \sqrt{x^2 + y^2}$
*$x = r \cos(\theta)$
*$y = r \sin(\theta)$
So,
$$r(1 + 2 \sin(\theta)) = r + 2 r \sin(\theta) = \sqrt{x^2 + y^2} + 2 y = 4$$
We can re-write this as:
$$3y^2 - 16y - x^2 + 16 = 0$$
|
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|
How to integrate $\int_0^\pi\frac{dx}{\sqrt{3-\cos(x)}}$?
How to integrate $\displaystyle\int_o^\pi\frac{dx}{\sqrt{3-\cos(x)}}$ ?
If I take $y=\sin\left(\frac{x}{2}\right)$ then,
$\displaystyle dx=\frac{2dy}{\cos\left(\frac{x}{2}\right)}=\frac{2dy}{\sqrt{1-y^2}}$
and with $\displaystyle\cos(x)=\cos^2\left(\frac{x}{2}\right)-\sin^2\left(\frac{x}{2}\right)=1-y^2-y^2=1-2y^2$, I obtain:
$\displaystyle\int_o^\pi\frac{dx}{\sqrt{3-\cos(x)}}=\int_0^1\frac{1}{\sqrt{3-(1-2y^2)}}\frac{2dy}{\sqrt{1-y^2}}=\int_0^1\frac{2}{\sqrt{2}\sqrt{1+y^2}\sqrt{1-y^2}}$
$\displaystyle=\sqrt{2}\int_0^1\frac{dy}{\sqrt{1-y^4}}=?$
How can I proceed? Thanks in advance.
|
The integral can be written as an elliptic integral of the first kind with complex elliptic modulus:
$$I=\int_{0}^{\pi}\frac{d\phi}{\sqrt{3-\cos{\phi}}}\\
=\int_{0}^{\pi}\frac{d\phi}{\sqrt{2+2\sin^2{\left(\frac{\phi}{2}\right)}}}\\
=\frac{1}{\sqrt2}\int_{0}^{\pi}\frac{d\phi}{\sqrt{1+\sin^2{\left(\frac{\phi}{2}\right)}}}\\
=\sqrt2\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1+\sin^2{\theta}}}\\
=\sqrt2\,K(-1)$$
The integral could also be written as a beta function by first substituting $x=\sin\theta$, $\theta=\arcsin{x}$, $d\theta=\frac{dx}{\sqrt{1-x^2}}$, and then substituting $t=x^4$, $x=t^{\frac14}$, $dx=\frac14t^{-\frac34}dt$:
$$I=\sqrt2\int_0^1\frac{1}{\sqrt{1+x^2}}\frac{dx}{\sqrt{1-x^2}}\\
=\sqrt2\int_0^1\frac{dx}{\sqrt{1-x^4}}\\
=\sqrt2\int_0^1\frac{dt}{4t^{\frac34}\sqrt{1-t}}\\
=\frac{\sqrt2}{4}\int_{0}^{1}t^{-\frac34}(1-t)^{-\frac12}dt\\
=\frac{\sqrt2}{4}\text{B}(\frac14,\frac12).$$
Using the identity $\text{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ and Euler's reflection formula $\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin{(\pi z)}}$, this becomes
$$I=\frac{\sqrt2}{4}\text{B}(\frac14,\frac12)=\frac{\sqrt2}{4}\frac{\Gamma(\frac14)\Gamma(\frac12)}{\Gamma(\frac34)}=\frac{\Gamma(\frac14)^2}{4\sqrt{2\pi}}$$
|
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|
Proving the area of an equilateral triangle How do you prove that How do you prove that for any equilateral triangle with side length s, area is $\frac{s^2 √3}{4}$ ? I tried using an equilateral triangle in a square, but I keep coming up with a $2x^2√3$ , as shown below. What am I doing wrong?
I started with the following:
The area of the full square is:
$ 2x * 2x = 4x^2$
To find the area of the triangle, I will subtract the non-triangle parts from the square.
The part shaded green is:
$ (2x - x√3) * 2x = 4 x^2-2x^2√3$
The parts shaded blue are:
$ \frac{x * ( x√3)}{2} + \frac{x * ( x√3)}{2} = x^2√3$
Adding blue and green:
$(x^2√3) + (4 x^2-2x^2√3) = 4 x^2-x^2√3 $
Subtract blue and green from whole square:
$(4x^2) -(4 x^2-x^2√3) = x^2√3$
Multiply by 2 because I am referring to the $2x$ side, not half of it ($x$):
Final answer: $2 * (x^2√3) = 2x^2√3$
And of course, $ 2x^2√3 \neq \frac{x^2 √3}{4}$
|
Another approach would be using calculus. For instance, consider the line $f(x) = \sqrt{3}x $. Notice the line from $(0,0)$ to $(\frac{s}{2}, f( \frac{s}{2}) ) $ is the hypothenuse of half the equilateral triangle. The other sides are $\frac{s}{3}$ and $\sqrt{3} \frac{s}{2} $. So, the area of half of the equilateral triangle is
$$ \int\limits_0^{\frac{s}{2}} \sqrt{3} x = \frac{ \sqrt{3} s^2}{8}$$
And hence the area of the equilateral triangle is twice this area which is
$$ \frac{ \sqrt{3} s^2}{4}$$
|
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|
Last two digit of number raised to exponents.
Find the last two digits of $3^{3^{2014}}$.
Attempt:
First I try to work with $3^{2014}$. So we can work on $\text{mod 10}$.
Then, $$\begin{align}&3^1 \equiv 3\pmod{10}\\ &3^2 \equiv 9 \pmod{10}\\ &3^3 \equiv 27 \equiv 7\pmod{10}\\ &3^4 \equiv 81 \equiv 1\pmod{10}\end{align}$$
So $3^{2014} = 3^{(4\cdot503) + 2} \equiv 3^0 \cdot 3^2 \equiv 9 \pmod{10}$.
So the last digit is 9. However I am ask to find the last two digit. Can anyone please help me? I would really appreciate all the help.
|
Hint: by Euler's totient theorem, $3^{40} \equiv 1 \pmod{100}$. Alternatively, note (by playing with a calculator) that $3^{20} \equiv 1 \pmod{100}$.
|
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Formula for $\sum_{k=0}^n k^d {n \choose 2k}$ If $d \geq 1$ is an integer, is there a general formula for $$\sum_{k=0}^n k^d {n \choose 2k}\,?$$
We know that $\sum_{k=0}^n k {n \choose 2k} = \frac{n2^n}{8}$ and $\sum_{k=0}^n k^2 {n \choose 2k} = \frac{n(n+1)2^n}{32}$.
Note that ${n \choose 2k} = 0$ when $2k > n$.
|
This is an half-answer. We have
$$(1+x)^{n}=x^{0}\binom{n}{0}+x^{1}\binom{n}{1}....+x^{n}\binom{n}{n}$$
If we take derivative of this with respect to x.
$$n(1+x)^{n-1}=1x^{0}\binom{n}{1}+2x^{1}\binom{n}{1}....+nx^{n-1}\binom{n}{n}$$
For $x=1$ we have $$n2^{n-1}=1\binom{n}{1}+2\binom{n}{2}....+n\binom{n}{n}$$
Now if we multiply the equation before this one with x and take derivative again.
$$n(1+x)^{n-1}+n(n-1)(1+x)^{n-2}=1^2x^{0}\binom{n}{1}+2^2x^{1}\binom{n}{1}....+n^2x^{n-1}\binom{n}{n}$$
For $x=1$ $$n2^{n-1}+n(n-1)2^{n-2}=1^2\binom{n}{1}+2^2\binom{n}{2}....+n^2\binom{n}{n}$$
We conclude by induction that
$$n2^{n-1}+n(n-1)2^{n-2}..+n(n-1)(n-2)...(n-d+1)2^{n-d}=1^d\binom{n}{1}+2^d\binom{n}{2}....+n^d\binom{n}{n}$$
I don't know what to do next.
|
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Problem of Number Theory Can you give a tip for this question?
For all $ n $ odd positive integer, let be
$ F(n) = \# \{ a($mod $ n): a^{n-1} \equiv 1($mod $n) \} $
that is the number of elements of the set.
*
*Show that
$$ F(n) = \prod_{p|n} gcd(p-1,n-1) $$
*Let be $ F_0(n) $ the numbers $ a($mod $n) $ such that $ a^n \equiv a($mod $n) $. Find a formula like the previous one,
*Show that if $ F_0(n) $ < n, then $ F_0(n) \leq \frac{2}{3} n $. Show that if $ n \neq 6 $ and $ F_0(n) < n $ then $ F_0(n) \leq \frac{3}{5} n $
|
I have skimmed over some steps; you can try filling in the details.
1) Write $n=\prod_{i=1}^{m}{p_i^{a_i}}$, where $p_i$ are odd primes.
2) There exists a primitive root $\pmod{p_i^{a_i}}$.
3) Using 2), conclude that the number of solutions $\pmod{p_i^{a_i}}$ of $a^k \equiv 1 \pmod{p_i^{a_i}}$ is $\gcd(\varphi(p_i^{a_i}), k)$.
4) Apply 4) for $k=n-1$ and Chinese Remainder Theorem to get $F(n)=\prod_{i=1}^{m}{\gcd(\varphi(p_i^{a_i}), n-1)}=\prod_{p\mid n}{\gcd(p-1, n-1)}$.
5) If we want to look at even $n$, then for $2^l \| n$, we get $F(n)=\alpha\prod_{p\mid n, 2 \nmid p}{\left(\gcd(p-1, n-1) \right)}$. where $\alpha$ is the number of solutions to $a^{n-1} \equiv 1\pmod {2^l}$. We get $a \equiv 1 \pmod{2^l}$. Thus $$F(n)=\prod_{p\mid n}{\left(\gcd(p-1, n-1) \right)}$$
For $F_0(n)$, a similar approach applies and we get $\gcd(\varphi(p_i^{a_i}), k)+1$ solutions in step $3$, so $$F_0(n)=\prod_{p\mid n}{\left(\gcd(p-1, n-1)+1 \right)}$$ The analysis of even $n$ is also similar, and we get the same formula.
Note $\gcd(p-1, n-1)+1 \leq p$ so $F_0(n) \leq \prod_{p \mid n}{p} \leq n$. If there is strict inequality, either $n$ is not squarefree or $\gcd(p-1, n-1)<p-1$ for some $p$.
In the former, if $q^2 \mid n$, $F_0(n) \leq \prod_{p \mid n}{p} \leq \frac{n}{q} \leq \frac{n}{2}$.
In the latter case clearly $p>2$, and we have $\gcd(p-1, n-1) \leq \frac{p-1}{2}$ so $\gcd(p-1, n-1) \leq \frac{p+1}{2} \leq \frac{2}{3}p$. Thus $F_0(n) \leq \frac{2n}{3}$.
Note that if $n$ not squarefree or $p \geq 5$ in case 2, we get $F_0(n) \leq \frac{3n}{5}$. Thus if $\frac{3n}{5}<F_0(n)<n$ then we must have $n$ squarefree and $p=3$ in case $2$, $\gcd(3-1, n-1)=1$, $\gcd(p-1, n-1)=p-1$ for $p \mid n, p \not =3$.
We have $\gcd(2, n-1)=1$ so $2 \mid n$. Thus $6 \mid n$
If $p \mid n$ for some $p>3$ then $\gcd(p-1, n-1)=p-1$ implies $n$ odd, a contradiction.
Thus $n$ is squarefree and a multiple of $6$, i.e. $n=6$.
|
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How do I evaluate this integral by hand? TL;DR how do I evaluate $\int_0^{2 \pi } \frac{1}{\cos ^2(\theta )+1} \, d\theta$ by hand?
I'm trying to solve this problem:
Find the volume of the region defined by $x^2+xy+y^2+yz+z^2\le1$.
(The answer is $\frac{4 \sqrt{2} \pi }{3}$)
I have tried several tactics. First of all, I rotated it $45^\circ$ in the $xz$ plane (converting all $x$ to $\frac{(x-z)}{\sqrt{2}}$ and all $z$ to $\frac{(x+z)}{\sqrt{2}}$ to obtain this:
$$x^2+y^2+z^2+xy\sqrt{2}\le1$$
Now if I solve for y at the boundary I get
$$y=\frac{1}{2} \left(-\sqrt{2} x\pm \sqrt{2} \sqrt{-x^2-2 z^2+2}\right)$$
And then I can integrate out the $y$ to get
$$\int _{-1}^1\int _{-\sqrt{2} \sqrt{1-z^2}}^{\sqrt{2} \sqrt{1-z^2}}\sqrt{2}
\sqrt{-x^2-2 z^2+2}\ dxdz$$
Which I can evaluate with Mathematica but not by hand [it does get the right answer in Mathematica, so I know I'm on the right track.]
I can then switch to polar coordinates in $x$ and $z$, yielding
$$\sqrt{2} \int _0^{2 \pi }\int _0^{\sqrt{\frac{2}{2 \sin ^2(\theta )+\cos ^2(\theta
)}}}r \sqrt{-2 r^2 \sin ^2(\theta )-r^2 \cos ^2(\theta )+2}drd\theta$$
This is relatively easy to integrate (the inner integral), so I get
$$\frac{4}{3} \int_0^{2 \pi } \frac{1}{\cos ^2(\theta )+1} \, d\theta$$
which I can't figure out how to evaluate by hand. Where do I go from here?
|
Here is an approach. First evaluate the indefinte integral using the substitution $\theta=\arctan(u)$ which gives
$$ I=\int \frac{d\theta}{1+\cos^2(\theta)} = \int \frac{du}{u^2+2} =\frac{1}{\sqrt{2}}\arctan\left(\frac{u}{\sqrt{2}}\right).$$
Then, split the interval of integration as $[0,\pi/2]\cup[\pi/2,\pi]\cup [\pi,3\pi/2]\cup[3\pi/2,2\pi]$. In fact, you can write the integral as
$$ I = 4\int_{0}^{\infty} \frac{du}{u^2+2}= 4 \frac{\sqrt{2}\pi}{4}=\sqrt{2}\pi. $$
Note:
$$ \cos^2(\tan^{-1}(u)) = \frac{1}{1+u^2}. $$
|
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|
Solve $(2x^3+3x^2y+y^2-y^3)dx+(2y^3+3xy^2+x^2-x^3)dy=0$
Solve the differential equation
$$(2x^3+3x^2y+y^2-y^3)dx+(2y^3+3xy^2+x^2-x^3)dy=0$$
I don't know how to solve it, and I have tried many methods,but they didn't work!
|
Hint: Let:
$$y = v x \rightarrow \dfrac{dy}{dx} = v + x \dfrac{dv}{dx}$$
Substitute this into
$$(2x^3+3x^2y+y^2-y^3)+(2y^3+3xy^2+x^2-x^3)~\dfrac{dy}{dx} = 0$$
It looks nasty!
Anyway, you will get a solvable equation after you use an integrating factor to make it an exact equation and end up with:
$v(x) = \dfrac{\sqrt[3]{2} \left(c_1-x\right)}{\sqrt[3]{27 c_1 x^4+\sqrt{108 x^6 \left(x-c_1\right){}^3+729 \left(x^6-c_1 x^4\right){}^2}-27 x^6}}+\dfrac{\sqrt[3]{27 c_1 x^4+\sqrt{108 x^6 \left(x-c_1\right){}^3+729 \left(x^6-c_1 x^4\right){}^2}-27 x^6}}{3 \sqrt[3]{2} x^2}$
From the original substitution of $y = v x$, eliminate $v$ in the above and solve for $y$.
Final solution:
$y(x)=\dfrac{\sqrt[3]{2} \left(3 c_1+3 x\right)}{3 \sqrt[3]{\sqrt{\left(27 c_1 x+27 x^3\right){}^2+4 \left(3 c_1+3 x\right){}^3}+27 c_1 x+27 x^3}}-\dfrac{\sqrt[3]{\sqrt{\left(27 c_1 x+27 x^3\right){}^2+4 \left(3 c_1+3 x\right){}^3}+27 c_1 x+27 x^3}}{3 \sqrt[3]{2}}$
|
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|
Prove $n^2(n^4-1)$ is divisible by 60 using Mathematical Induction. Base step:
p(2)=4 * 15= 60
Inductive Hypothesis:
Assuming p(k) = $k^2(k^4-1)$ = 60q
Induction:
p(k+1)= $(k+1)^2[(k+1)^4-1]$
= $(k+1)^2[(k+1)^2 + 1][(k+1)^2 - 1]$
= $(k+1)^2(k^2+2k+2)(k^2+2k+1- 1)$
= $(k+1)^2(k^2+2k+2) k (k+2)$
= $k(k+1)(k+2)(k^2+2k+2)(k+1)$
Now $k(k+1)(k+2)$ is the product of 3 consecutive natural numbers. Their product HAS to be a multiple of 2 and 3 and hence also a multiple of 6. So all that I have to do is prove that $(k^2+2k+2)(k+1)$ is a multiple of 10. But I can't seem to do so. Also through my induction, I have not at all made use of my hypothesis! How do I go about solving this question through the Principle of Mathematical Induction?
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If $f(n)=n^2(n^4-1)=n^6-n^2$ is divisible by $60$ for $n=k$
$$f(k+1)-f(k)=(k+1)^6-(k+1)^2-\{k^6-k^2\}$$
$$=6k^5+15k^4+20k^3+15k^2+6k-2k$$
which is $\displaystyle6(k^5-k)+15k^4+20k^3+15k^2+10k$ divisible by $5$ as $k^5-k$
Again, $\displaystyle6k^5+15k^4+20k^3+15k^2+4k\equiv2(k^3-k)\pmod3\equiv0$
finally, $\displaystyle6k^5+15k^4+20k^3+15k^2+4k=2k^5-k^4-k^2\pmod4\equiv2k^4(k-1)+k^4-k^2$
$\displaystyle\equiv2k^3 \underbrace{k\cdot(k-1)}_{\text{ even}}+k^2(k-1)(k+1)\pmod4$
If $k$ is even, $\displaystyle4|k^2$ else $2|(k\pm1)\implies 2^2|(k^2-1)$
So, $\displaystyle f(k+1)-f(k)$ will be divisible by lcm$(4,3,5)=60$
|
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|
Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$ Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$.
I have no clue how to proceed and tried to prove that the whole equation becomes $0$ when $\sin\frac{\pi}{14}$ is placed in place of $x$ but couldn't do anything further. I think the symbols might be different but can be the same. If it is correct, please help me to solve this; if the equation is wrong, then please modify it and solve it.
|
$$\sin\frac\pi{14}=\cos\left(\frac\pi2-\frac\pi{14}\right)=\cos\frac{3\pi}7$$
Let $\displaystyle\cos4x=-\cos3x=\cos(\pi+3x)\implies4x=2n\pi\pm(\pi+3x)$ where $n$ is any integer
'+' $\displaystyle\implies x=(2n+1)\pi\equiv\pi\pmod{2\pi}\implies\cos x=-1$
'-' $\displaystyle\implies x=\frac{(2n-1)\pi}7$
Now $\displaystyle\cos4x=-\cos3x$
Using Multiple angle formula,
$\iff8c^4-8c^2+1=-(4c^3-3c)\iff8c^4+4c^3-8c^2-3c+1=0\ \ \ \ (1)$ where $c=\cos x$
Clearly, the roots of $(1)$ are $\cos x,$ where $\displaystyle x=\pi, \frac{(2n-1)\pi}7 $ where $n\equiv0,1,2\pmod3$
So, the equation whose roots are $\displaystyle\cos\frac{(2n-1)\pi}7 $ where $n\equiv0,1,2\pmod3$ is $$\frac{8c^4+4c^3-8c^2-3c+1}{c+1}=0$$
Here $n=2$
|
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|
If $\sin A + \cos A + \tan A + \cot A + \sec A + \csc A = 7$ then $x^2 - 44x - 36 = 0$ holds for $x=\sin 2A$ If
$$\sin A + \cos A + \tan A + \cot A + \sec A + \csc A = 7$$
then prove that
$$\sin 2A \quad\text{ is a root of }\quad x^2 - 44x - 36 = 0$$
I have no idea how to solve it. Plz help.
|
I think you have the problem wrong, you can show $\sin 2A$ is a root of $x^2-44x+36$ using Awesome's hint.
EDIT:
If $\sin 2A$ is a root of $x^2-44x+36$, then
$$\sin 2A = \frac{44 \pm \sqrt{1936 - 144}}{2}=22\pm 8\sqrt{7}$$
Since $|\sin 2A|\leq 1$, we have that $\sin 2A = 22-8\sqrt{7}$.
Using the hint by Awesome,
\begin{align}
7 &=\sin A + \cos A + \frac{1}{\sin A\cos A} + \frac{\sin A + \cos A}{\sin A\cos A} \\
&= t + \frac{1+t}{\frac{1}{2}(t^2-1)} \\
&= t + \frac{2(1+t)}{t^2-1} \\
&= t + \frac{2}{t-1}. \end{align}
This gives the polynomial $t^2-8t+9=0$, which has roots $t=4 \pm \sqrt{7}$. Since $$(\sin A + \cos A)^2 = 1+\sin 2A \leq 1 + 1= 2,$$ we have $$|\sin A + \cos A| \leq \sqrt{2}.$$
This forces $t=4-\sqrt{7}$. You can now compute $\sin 2A = 22-8\sqrt{7}$ as desired.
|
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|
Calculate $\int_{0}^{2\pi} \frac{\cos((2n+1)t)}{\cos(t)}dt$
Calculate the following integral for $n \in \mathbb{Z}$ with the
residue theorem $$\int_{0}^{2\pi} \frac{\cos((2n+1)t)}{\cos(t)}dt$$
So far I have tried two approaches. Firsty, for $n\geq 0$:
$$\begin{align*}\int_{0}^{2\pi} \frac{\cos((2n+1)t)}{\cos(t)}dt &= \int_{C(0,1)^{+}} \frac{z^{2n+1}+z^{-(2n+1)}}{z+z^{-1}}\cdot \frac{1}{iz}dz \\
& = -i \int_{C(0,1)^{+}} \frac{z^{2n+1}+z^{-(2n+1)}}{z^2+1}dz \\
& = -i \int_{C(0,1)^{+}} \frac{z^{2(2n+1)}+1}{z^{2n+1}(z^2+1)}dz \\
& = -i \cdot 2\pi i \cdot Res_{z=0}\left(\frac{z^{2(2n+1)}+1}{z^{2n+1}(z^2+1)}\right) \\
& = \left.\frac{2\pi}{(2n)!} \left[ \frac{z^{2(2n+1)}+1}{z^2+1} \right]^{(2n)} \right\rvert_{z=0}
\end{align*}$$
For the first equality I used the reversed parametrization $z(t) = e^{it}$, with $0 \leq t \leq 2\pi$. The last equality follows from $Res_{z=a} \left( \frac{f(z)}{(z-a)^{n+1}}\right) = \frac{f^{(n)}(a)}{n!}$. However, I'm not sure how to calculate that derivative.
Seconldy, I tried to integrate the function $f(z) = \frac{e^{i(2n+1)t}}{cos(t)}$. Using a similar technique this yields:
$$\begin{align*}
\int_{0}^{2\pi} \frac{e^{i(2n+1)t)}}{\cos(t)}dt &= \int_{C(0,1)^{+}} \frac{z^{2n+1}}{(z+z^{-1})/2}\cdot \frac{1}{iz}dz \\
& = -2i\int_{C(0,1)^+}\frac{z^{2n+1}}{z^2+1}dz
\end{align*}$$
However, for $n \geq 1$, the singularities lie on my contour over which I integrate. How do I fix this? Can I just ignore that?
Is this the right approach, or should I try differently? Thanks in advance.
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Notice that
$$z^{2(2n+1)} + 1 = (z^2)^{2n+1} + 1^{2n+1} = (z^2 + 1) \sum _{k=0} ^{2n} (z^2)^k (-1)^{2n-k} \ ,$$
and the terms with $k < n$ are killed by the derivative of order $2n$, while the ones with $k > n$ survive it but are killed by the evaluation at $0$, leaving $z^{2n}$ (corresponding to $k=n$) as the sole survivor, so your derivative becomes
$$\frac {2\pi} {(2n)!} \left[ \frac {z^{2(2n+1)}+1} {z^2+1} \right]^{(2n)} (0) = \frac {2\pi} {(2n)!} \left[ \sum _{k=0} ^{2n} (z^2)^k (-1)^{2n-k} \right]^{(2n)} (0) = \frac {2\pi} {(2n)!} \left[ z^{2n} (-1)^n \right]^{(2n)} (0) = 2\pi (-1)^n \ .$$
|
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|
Show $ \frac {a^3}{a^2+b^2+c^2-bc}+\frac {b^3}{a^2+b^2+c^2-ca} +\frac {c^3}{a^2+b^2+c^2-ab} \le \frac {(a+b+c)^2 (a^4+b^4+c^4)}{6abc(ab+bc+ca)} $ If $a,b,c$ are positive real numbers then how do we prove that $ \dfrac {a^3}{a^2+b^2+c^2-bc}+\dfrac {b^3}{a^2+b^2+c^2-ca} +\dfrac {c^3}{a^2+b^2+c^2-ab} \le \dfrac {(a+b+c)^2 (a^4+b^4+c^4)}{6abc(ab+bc+ca)} $
Please help. Thanks
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Noting $ab+bc+ca \le a^2+b^2+c^2$ and $a^4+b^4+c^4 \ge \frac13(a^2+b^2+c^2)^2$, it is sufficient to show:
$$\sum_{cyc} \frac{a^3}{a^2+b^2+c^2-bc} \le \frac{(a+b+c)^2(a^2+b^2+c^2)}{18 abc}$$
Now $a^2+(b^2+c^2-bc) \ge a^2+bc$. So it is enough to show
$$ \frac{(a+b+c)^2(a^2+b^2+c^2)}{18 abc} \ge \sum_{cyc} \frac{a^3}{a^2+bc} = (a+b+c)-abc\sum_{cyc}\frac1{a^2+bc}$$
where we have used $\dfrac{a^3}{a^2+bc} = a - \dfrac{abc}{a^2+bc}$.
This can be written as a quadratic in $\frac{a+b+c}{abc}$:
$$ (a^2+b^2+c^2)\left(\frac{a+b+c}{abc}\right)^2-18\left(\frac{a+b+c}{abc}\right)+18\sum_{cyc}\frac1{a^2+bc} \ge 0$$
and the discriminant condition is then:
$$(a^2+b^2+c^2)\sum_{cyc}\frac1{a^2+bc} \ge \tfrac92$$
which follows readily from Cauchy-Schwarz inequality.
|
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|
I have the pattern: 1 + 2 + 3 + 4 + 5 + 6, but I need the formula for it I'm writing some software that takes a group of users and compares each user with every other user in the group. I need to display the amount of comparisons needed for a countdown type feature.
For example, this group [1,2,3,4,5] would be analysed like this:
1-2, 1-3, 1-4, 1-5
2-3, 2-4, 2-5
3-4, 3-5
4-5
By creating little diagrams like this I've figured out the pattern which is as follows:
Users - Comparisons
2 - 1
3 - 3 (+2)
4 - 6 (+3)
5 - 10 (+4)
6 - 15 (+5)
7 - 21 (+6)
8 - 28 (+7)
9 - 36 (+8)
I need to be able to take any number of users, and calculate how many comparisons it will take to compare every user with every other user.
Can someone please tell me what the formula for this is?
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$$N=2:\ 1 + 2 = (1 + 2) = 1\times3$$
$$N=4:\ 1 + 2 + 3 + 4 = (1 + 4) + (2 + 3) = 2\times5$$
$$N=6:\ 1 + 2 + 3 + 4 + 5 + 6 = (1 + 6) + (2 + 5) + (3 + 4) = 3\times7$$
$$N=8:\ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8= (1 + 8) + (2 + 7) + (3 + 6)+ (4 + 5) = 4\times9$$
More generally, $N/2\times(N + 1)$.
For odd $N$, sum the $N-1$ first terms (using the even formula) together with $N$, giving $(N-1)/2\times N+N=N\times(N+1)/2$.
|
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|
Integral $\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx$ $$I:=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx=\frac{(1+\alpha)\sqrt{2\alpha^3 \pi}}{2\sqrt[\alpha]e},\qquad \alpha>0.$$
This one looks very nice. It has stumped me.
Differentiation with respect to parameter does not seem to work either if I try $I(\alpha)$ and $I'(\alpha)$. at x=0 there seems to be a problem with the integrand also however I am not sure how to go about using this. Perhaps we could try and use a series expansion for $e^x=\sum_{n=0}^\infty x^n /n!$, however the function $e^{-1/x^2}$ is well known that its taylor series is zero despite the function not being.
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Assuming the $3\alpha x$ term is in fact $3\alpha x^2$ (otherwise the numerical results do not match).
$$\begin{align*}
I&=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+3\alpha x^2- 1}{x^6}dx\\
&=\int_0^\infty \ln x\, d\left(-\alpha x^{-3}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right)\\
&=-\alpha\left(\left.\frac{\ln x}{x^3}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right|_0^\infty-\int_0^\infty \frac{1}{x^3}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) d\,\ln x\right)\\
&=\alpha\int_0^\infty \frac{1}{x^4}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) dx\\
&=\alpha\left(\int_0^1 \frac{1}{x^4}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\underbrace{\int_1^\infty \frac{1}{x^4}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx}_{x\to1/x}\right) \\
&=\alpha\left(\int_0^1 \frac{1}{x^4}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\int_1^0 -x^2\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\right) \\
&=\alpha\int_0^1 (x^2+x^{-4})\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\\
&=\alpha\int_0^1 (x^2-1+x^{-2})\exp\left(-\frac{1}{\alpha}-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\
&=\alpha e^{-1/\alpha}\int_0^1 (1+(x-1/x)^2)\exp\left(-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\
&=\alpha e^{-1/\alpha}\int_{-\infty}^0 (1+y^2)\exp\left(-\frac{y^2}{2\alpha }\right)dy\\
&=\alpha e^{-1/\alpha}(\alpha+1)\sqrt{\frac{\alpha\pi}{2}}.
\end{align*}$$
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|
Show $a^2+b^2+c^2=1$ $\implies$ $ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $ If $a$, $b$ and $c$ are real numbers such that $a^2+b^2+c^2=1$ , then how to prove that
$ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $
( don't apply Schur's inequality )? Thanks.
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Let $x=a^2$, $y=b^2$, $z=c^2$ and $f(x,y,z)=\frac{yz}{1+x}+\frac{xz}{1+y}+\frac{xy}{1+z}$ and g(x,y,z)=x+y+z. Consider the restriction $\{(x,y,z) \in \mathbb{R}^3;g(x,y,z)=1\}$. Applying the method of Lagrange multipliers, we have
$$\frac{y}{1+z}+\frac{z}{1+y}-\frac{yz}{(1+x)^2} = \lambda$$
$$\frac{z}{1+x}+\frac{x}{1+z}-\frac{xz}{(1+y)^2} = \lambda$$
$$\frac{x}{1+y}+\frac{y}{1+x}-\frac{xy}{(1+z)^2} = \lambda$$
$$x+y+z=1$$
Subtracting the first two equations, we obtain
$$\frac{y-x}{1+z}+\frac{z}{1+y}(1-\frac{x}{1+y})-\frac{z}{1+x}(1+\frac{y}{1+x})=0 \Rightarrow$$
$$(y-x) \left( \frac{1}{1+z}+\frac{z(x+y+1)(x+y+2)}{(1+x)^2(1+y)^2} \right )=0 $$
The second factor is $\ge 0$, so $x=y$. Analogously, $y=z$. Thus, we conclude that $x=y=z=1/3$. How $f(1/3,1/3,1/3)=1/4$ and $f$ have a maximum in this restriction, we have the desired inequality.
|
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How is this automorphism an inversion? And how is this group abelian? Let $G$ be a finite group, and let $T$ be an automorphism of $G$ which sends more than three-quarters of the elements of $G$ onto their inverses. Then how to demonstrate that $T(x) = x^{-1}$ for all $x$ in $G$? And, how to show that $G$ is abelian?
Can we find an example of a finite, non-abelian group with an automorphism which maps exactly three-quarters of the elements of the group onto their inverses?
|
I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
These problems are Problem 12 and Problem 13 on p.71 in Herstein's book.
I solved Problem 13 as follows:
I used GAP and I found the following example.
Let $H:=\{(), (1\text{ }2), (3\text{ }4), (1\text{ }2)(3\text{ }4), (1\text{ }3)(2\text{ }4), (1\text{ }4)(2\text{ }3), (1\text{ }3\text{ }2\text{ }4), (1\text{ }4\text{ }2\text{ }3)\}$.
Then, $H$ is a non-abelian subgroup of $S_4$. (I believe GAP.)
The identity mapping $\operatorname{id}$ is obviously an element of $\operatorname{Aut}(H)$.
And,
$()=()^{-1}$.
$(1\text{ }2)=(1\text{ }2)^{-1}$.
$(3\text{ }4)=(3\text{ }4)^{-1}$.
$(1\text{ }2)(3\text{ }4)=[(1\text{ }2)(3\text{ }4)]^{-1}$.
$(1\text{ }3)(2\text{ }4)=[(1\text{ }3)(2\text{ }4)]^{-1}$.
$(1\text{ }4)(2\text{ }3)=[(1\text{ }4)(2\text{ }3)]^{-1}$.
$(1\text{ }3\text{ }2\text{ }4)\neq(1\text{ }4\text{ }2\text{ }3)=(1\text{ }3\text{ }2\text{ }4)^{-1}$.
$(1\text{ }4\text{ }2\text{ }3)\neq(1\text{ }3\text{ }2\text{ }4)=(1\text{ }4\text{ }2\text{ }3)^{-1}$.
So, the identity mapping maps exactly three-quarters of the elements of $H$ onto their inverses.
|
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|
How to prove that the following function is convex? I want to prove convexity of the following function:
$$f(x) = log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)$$
for any fixed $a, b \in (0, 1)$ and:
*
*$x\in(0,1)$
*$x\in(1, \infty)$
I'm trying to solve it for a very long time, tried to investigate the sign of second derivative, but it leads to very ugly equation which I am unable to handle. Any hints are appreciated. Thanks!
Edit:
$f''(x)$ equals to:
$$-\frac{{\left(\frac{x^{x - 1} {\left(p^{y} - 1\right)} x}{p - 1} + \frac{{\left(p^{x} - 1\right)}
x^{y - 1} y}{p - 1} - \frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{{\left(p -
1\right)}^{2}}\right)}^{2}}{{\left(\frac{{\left(x^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} +
1\right)}^{2} \log\left(x\right)} +\\
+ \frac{\frac{x^{x - 2} {\left(p^{y} - 1\right)} {\left(x - 1\right)} x}{p - 1} + \frac{2 \, p^{x -
1} x^{y - 1} x y}{p - 1} + \frac{{\left(p^{x} - 1\right)} p^{y - 2} {\left(y - 1\right)} y}{p - 1} -
\frac{2 \, x^{x - 1} {\left(p^{y} - 1\right)} x}{{\left(p - 1\right)}^{2}} - \frac{2 \, {\left(p^{x}
- 1\right)} x^{y - 1} y}{{\left(p - 1\right)}^{2}} + \frac{2 \, {\left(p^{x} - 1\right)}
{\left(x^{y} - 1\right)}}{{\left(p - 1\right)}^{3}}}{{\left(\frac{{\left(p^{x} - 1\right)}
{\left(x^{y} - 1\right)}}{p - 1} + 1\right)} \log\left(p\right)} -\\
- \frac{2 \, {\left(\frac{x^{x - 1} {\left(p^{y} - 1\right)} x}{p - 1} + \frac{{\left(p^{x} -
1\right)} x^{y - 1} y}{p - 1} - \frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{{\left(p
- 1\right)}^{2}}\right)}}{{\left(\frac{{\left(x^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} +
1\right)} x \log\left(p\right)^{2}} + \frac{\log\left(\frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} + 1\right)}{p^{2}
\log\left(x\right)^{2}} + \frac{2 \, \log\left(\frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} + 1\right)}{p^{2} \log\left(p\right)^{3}}
$$
Some elements of that equation are positive, some of them are negative - I do not see any way of how to prove that $f''(x) > 0$.
I have also tried to find counter example by testing the Jensen's inequality on randomly generated data, but everything seems OK - the function seems to be convex, but I am unable to prove it.
|
We have $$f(x) = \log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)
= \frac{\ln \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)}{\ln x}
=\ln \left(\left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)^{1/\ln x}\right)$$
Then we may show Log-Convexity:
For $f(x)=\log g(x)$ (on convex domains $x\in(0,\infty)/1$):
$$ f(x) \text{ is convex.}\Leftrightarrow g(x)\text{ is log-convex.}$$
1) $g(x)$ log-convex iff:
$$g(\theta x + (1 - \theta) y) \leq g(x)^{\theta} g(y)^{1 - \theta}\qquad\theta\in(0,1),x,y\in{(0,\infty)/1}$$
2) $g(x)$ log-convex iff $h(x)=-g(x)$ is log-concave:
$$h(x)h''(x) \leq h'(x)^2$$
Set
$$g(x)=\left(1 + \dfrac{(x^a-1)(x^b - 1)}{x-1}\right)^{1/\ln x} \qquad a, b \in (0, 1).$$
|
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|
Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$
$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$
$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$
$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
|
Here is another solution. Remark that
$$
\sqrt{n^2+n}-n=\sqrt{n^2+n}-\sqrt{n^2}=\frac{(\sqrt{n^2+n}-\sqrt{n^2})(\sqrt{n^2+n}+\sqrt{n^2})}{\sqrt{n^2+n}+\sqrt{n^2}}
$$
Then
$$
\sqrt{n^2+n}-n=\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1}.
$$
Since $\lim\limits_{n\to \infty} \sqrt{1+\frac{1}{n}}=1$, for $\delta>0$, exists $N$ ($N>\frac{1}{2\delta+\delta^2}$) such that $1<\sqrt{1+\frac{1}{n}}<1+\delta$, for all $n>N$. Then
$$
2<\sqrt{1+\frac{1}{n}}+1<2+\delta
$$
So
$$
\frac{1}{2}>\frac{1}{\sqrt{1+\frac{1}{n}}+1}>\frac{1}{2+\delta}.
$$
Then
$$
\frac{\delta}{4}>\frac{1}{2}-\frac{1}{2+\delta}>\frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1}>0
$$
Let $\varepsilon>0$, define $\delta:=4\varepsilon$. So, for $N>\frac{1}{2\delta+\delta^2}=\frac{1}{8\varepsilon+16\varepsilon^2}$,
$$
\varepsilon>\frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1}>0,
$$
i.e.,
$$
\lim\limits_{n\to \infty}\sqrt{n^2+n}-n=\lim\limits_{n\to \infty}\frac{1}{\sqrt{1+\frac{1}{n}}+1}=\frac{1}{1+1}=\frac{1}{2}
$$
|
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|
Rational polynomials I'm not very familiar with algebra and was wondering if there are any results regarding the effective order of rational polynomials (i.e. rational functions).
Specifically: given $P(z)$ and $Q(z)$ as polynomials in $z$ with real coefficients of order $p$ and $q$ respectively, is there a way to know the effective order of the rational function $P(z)Q^{-1}(z)$ - by which I mean to know if there are $r$ roots shared by the numerator and denominator that cancel each other out resulting in a smaller rational polynomial of order $p-r, q-r$?
More importantly, can this be done without explicit factorization ? I just want to know the value of $r$ without knowing the roots or the reduced polynomials.
|
Euclid's GCD algorithm can be used to find common factors between two polynomials. For instance:
$$
\begin{align}
\gcd(x^3 + 1, x^4 + x^2 + 1) & = \gcd(x^3 + 1, x^4 + x^2 + 1 - x(x^3 + 1))
\\ & = \gcd(x^3 + 1, x^2 - x + 1)
\\ & = \gcd(x^3 + 1 - x(x^2 - x + 1), x^2 - x + 1)
\\ & = \gcd(x^2 - x + 1, x^2 - x + 1)
\\ &= \gcd(x^2 - x + 1, 0)
\\ &= x^2 - x + 1
\end{align}
$$
So $x^2 - x + 1$ is the greatest common factor between $x^3 + 1$ and $x^4 + x^2 + 1$.
|
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|
$ ab-1|a^2+ab+b^2 $ I hava a number theory problem. I think on it yestarday night and today, afternoon.
The problem :
$ a,b $ are two natural numbers such that : $ ab>1 $
how many pairs $ (a,b) $ is there such that : $ ab-1|a^2+ab+b^2 $
we have $ ab-1|(a^2+ab+b^2)+(ab-1) = (a+b)^2 - 1^2 = (a+b+1)(a+b-1) $
so $ ab-1|(a+b+1)(a+b-1) $
can you help me to complete this?!
i need a hint!
please don't write a compelete solution : )
|
$a + b + 1$ and $a + b - 1$ are either relatively prime or have $2$ as a common divisor. In the first case, we must have $ab - 1$ divides $a + b + 1$ or $ab - 1$ divides $a + b - 1$. In the second case, there are a few more options to consider.
|
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|
How to prove Fibonacci sequence with matrices? How do you prove that:
$$
\begin{pmatrix}
1 & 1\\
1 & 0
\end{pmatrix}^n
=
\begin{pmatrix}
F_{n+1} & F_n\\
F_{n} & F_{n-1}
\end{pmatrix}$$
|
Let
$$A=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix}$$
And the Fibonacci numbers, defined by
$$\begin{eqnarray}
F_0&=&0\\
F_1&=&1\\
F_{n+1}&=&F_n+F_{n-1}
\end{eqnarray}$$
Then, by induction,
$$A^1=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix} =
\begin{pmatrix}
F_2 & F_1 \\ F_1 & F_0
\end{pmatrix}$$
And if for $n$ the formula is true, then
$$A^{n+1}=A\,A^n=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix}\begin{pmatrix}
F_{n+1} & F_{n} \\ F_{n} & F_{n-1}
\end{pmatrix}=\begin{pmatrix}
F_{n+1}+F_n & F_{n}+F_{n-1} \\ F_{n+1} & F_{n}
\end{pmatrix}=\begin{pmatrix}
F_{n+2} & F_{n+1} \\ F_{n+1} & F_{n}
\end{pmatrix}$$
So, the induction step is true, and by induction, the formula is true for all $n>0$.
|
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|
To solve $ \dfrac1m+\dfrac1n-\dfrac1{mn^2}=\dfrac34$ How do we find all positive integers $(m,n)$ such that $ \dfrac1m+\dfrac1n-\dfrac1{mn^2}=\dfrac34$ ?
|
Since you need $\frac 1m+\frac 1n\gt \frac 34$ you need either $\frac 1m\ge \frac 38\gt \frac 13$ or alternatively $\frac 1n \gt \frac 13$
The number of cases is small and they can be checked quickly.
To be more concrete about the cases involved, we must have either $m\leq 2$ or $n\leq 2$.
$m=1$ gives $\frac 1n-\frac 1{n^2}=-\frac 14$, or $4(n-1)=-n^2$which is impossible for positive integer $n$. Note that $n^2+4n-4=0$ has roots $n=-2\pm 2\sqrt 2$.
$m=2$ gives $\frac 1n-\frac 1{2n^2}=\frac 14$, which becomes $4n-2=n^2$ and this is impossible in integers since an even square is divisible by $4$. Also $n^2-4n+2=0$ has roots $n=2\pm \sqrt 2$
$n=1$ gives $\frac 1m+1-\frac 1m=\frac 34$ which is false.
$n=2$ gives $\frac 1m-\frac 1{4m}=\frac 14$ or $m=3$
|
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|
Need a generalized way to solve $127=1+2+2^2+...$ How can I solve this equation generally? I can solve it by checking in my calculator. But I don't know any generalized way.
$$127=1+2+2^2+2^3+....+ 2^{x-1}$$
|
The most intuitive way:
\begin{align*}
&\;\underbrace{1 + 1} +2+2^2+2^3+ \cdots + 2^{x-1} \\
&= \underbrace{2 + 2} + 2^2 + 2^3 + \cdots + 2^{x-1} \\
&\;\;\;= \underbrace{4 + 2^2} + 2^3 + \cdots + 2^{x-1} \\
& \; \\ & \; \\ %vertical space
&\quad \quad\quad \quad\quad \; \; \; \; \; \cdots \\
& \; \\ & \; \\ %vertical space
&\quad \quad \quad \quad = \underbrace{2^{x-1} + 2^{x-1}} \\
&\quad \quad \quad \quad \quad \quad = 2^x \\
\end{align*}
Hence,
$$
1 + 2 + 2^2 + 2^3 + \cdots + 2^{x-1} = 2^x - 1.
$$
|
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|
Indefinite Integral $\int \frac{dx}{\sqrt {ax^4-bx^2}}$ I am trying to Integrate
$$
I=\int \frac{dx}{\sqrt {ax^4-bx^2}}, \qquad a,b\in \mathbb{R}.
$$
Thanks.
I tried to do $x=\sin \phi$
$$
\int \frac{\cos \phi\, d\phi}{\sqrt{a\sin^4 \phi-b\sin^2 \phi}}=\int \frac{\cot \phi \, b\phi}{\sqrt{a\sin^2\phi-b}}
$$
but get stuck here.
Mathematica gives a closed form result
$$
I=-\frac{x\sqrt{ax^2-2b}}{\sqrt b \sqrt{ax^4-bx^2}}\tan^{-1}\bigg(\frac{\sqrt{2b}}{\sqrt{ax^2-2b}}\bigg).
$$
|
$\displaystyle\int\dfrac{dx}{\sqrt{ax^4-bx^2}}=\displaystyle\int\dfrac{dx}{x\sqrt{ax^2-b}}=\dfrac{1}{\sqrt{a}}\displaystyle\int\dfrac{dx}{x\sqrt{x^2-\dfrac{b}{a}}}$
$\therefore x^2-\dfrac{b}{a}=t^2 \implies x\ dx=t\ dt$
$\therefore \displaystyle\int\dfrac{dx}{x\sqrt{x^2-\dfrac{b}{a}}}=\displaystyle\int\dfrac{dt}{\left(t^2+\dfrac{b}{a}\right)}$
|
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|
$\int_0^\infty x^{-\frac{3}{2}}e^{-\frac{(x-1)^2}{x}}dx=\int_0^\infty x^{-\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx$ This is my third question following the previous post.
Prove that \begin{equation} \int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\frac{2}{3}\int_0^\infty x^{\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx \end{equation}
I am really having trouble to prove it. Please help me to compute the integrals. Every answer would be greatly appreciated. Thank you.
|
Examining the original integral, use the substitution $u=x^{-1}$ so that
$$\frac{du}{dx}=-x^{-2}\Rightarrow dx=-(u^{-2})du$$ and
$u\rightarrow 0$ when $x\rightarrow\infty$, $u\rightarrow \infty$ when $x\rightarrow0$
Therefore
$$\int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=-\int_{\infty}^0u^{\frac{3}{2}}e^{-\frac{(1-u)^2}{u}}(u^{-2})du\\=\int_{0}^{\infty}u^{-\frac{1}{2}}e^{-\frac{(1-u)^2}{u}}du\\=\int_{0}^{\infty}u^{-\frac{1}{2}}e^{-\frac{(u-1)^2}{u}}du$$
If we replace the variable $u$ by $x$ (which is equivalent to the trivial substitution $x=u$) we will end up with the desired result:-
$$\int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx$$
Next we apply integration by parts to the integral in the RHS of the above equation, by setting $dv=x^{-\frac{1}{2}}$ and $u=e^{-\frac{(x-1)^2}{x}}$ so that $v=2x^{\frac{1}{2}}$ and $du=e^{-\frac{(x-1)^2}{x}}\left(\frac{1}{x^2}-1\right)$ and recalling that
$$\int_0^{\infty}(u)(dv)dx=[uv]_0^\infty-\int_0^{\infty}v(du)dx$$
leading to
$$\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\left[2x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}\right]_0^{\infty}-2\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}\left(\frac{1}{x^2}-1\right)dx\\=0-2\int_0^{\infty}x^{-\frac{3}{2}}e^{-\frac{(x-1)^2}{x}}dx+2\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx\\=-2\int_0^{\infty}x^{-\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx+2\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx \\\Rightarrow 3\int_0^{\infty}x^{-\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx=2\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx\\\Rightarrow \int_0^{\infty}x^{-\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx=\frac{2}{3}\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx$$
Let $I_n$ denote the integral we wish to evaluate. Making use of the equality of the integrals in the first part of the question, we have
$$2I_n=\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx+\int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx$$
Using the substitution $u=x^{\frac{1}{2}}$ we have $\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}$ , resulting in
$$2I_n=2\int_0^\infty e^{-\large\left(u-\frac{1}{u}\right)^2}du+2\int_0^\infty\frac{1}{u^2} e^{-\large\left(u-\frac{1}{u}\right)^2}du \\\Rightarrow I_n=\int_0^\infty \left(1+\frac{1}{u^2}\right) e^{-\large\left(u-\frac{1}{u}\right)^2}du$$
If we use the substitution $s=u-\frac{1}{u}$ , we have $ds=du\left(1+\frac{1}{u^2}\right)$ so that the integral reduces to a Gaussian integral (note the change in limits, as $s\rightarrow-\infty$ when $u\rightarrow 0$ and $s\rightarrow\infty$ when $u\rightarrow \infty$):-
$$I_n=\int_{-\infty}^\infty e^{-s^2}ds=\sqrt{\pi}$$
|
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|
Evaluate $a+b+c+d$ If $a$, $b$, $c$, and $d$ are distinct integers such that
$$(x-a)(x-b)(x-c)(x-d)=4$$
has an integral root $r$, what is the value of $a+b+c+d$ in terms of $r$?
I tried to analyze graphically by shifting the graph of
$f(x)=(x-a)(x-b)(x-c)(x-d)$ four units downward but couldn't infer anything due to the random nature of $a$, $b$, $c$ and $d$.
|
Plugging in $x=4$ shows that $4$ can be written as product of four distinct integers. The only divisors of $4$ are $\pm1, \pm2,\pm4$.
The only way to obtain $4$ is as product $(-2)\cdot(-1)\cdot 1\cdot 2$. Then $(r-a)+(r-b)+(r-c)+(r-d)=-2+(-1)+1+2=0$, i.e. $a+b+c+d=4r$.
|
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|
To evaluate $\int_0^{\frac{\pi}2} \frac{\cos x}{x+2} dx $ and $\int_0^4{\frac{\sin x \cos x}{(x+1)^2}} dx$ How to evaluate the following integrals?
$$\int_0^{\frac{\pi}2} \frac{\cos x}{x+2} dx $$
$$\int_0^4{\frac{\sin x \cos x}{(x+1)^2}} dx$$
|
Using what Lucian answered $$\int_0^{\frac{\pi}2} \frac{\cos x}{x+2} dx=\left(\text{Ci}\left(2+\frac{\pi }{2}\right)-\text{Ci}(2)\right) \cos
(2)+\left(\text{Si}\left(2+\frac{\pi }{2}\right)-\text{Si}(2)\right) \sin (2)$$
$$\int_0^4{\frac{\sin x \cos x}{(x+1)^2}} dx=\Big(\text{Ci}(10)-\text{Ci}(2)\Big) \cos (2)-\frac{1}{5} \sin (2) \Big(5 \text{Si}(2)-5
\text{Si}(10)+\cos (2)+\cos (6)\Big)$$ The numerical values are respectively $0.3971615867$ and $0.1441361714$.
|
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|
To solve $1+2^mp^2=q^5$ How do we find all posible solutions of $1+2^mp^2=q^5$ for positive integer $m$ and primes $p,q$ ? $m=1,q=3,p=11$ is a solution , is there any other solution ?
|
Following Ivan Loh, $2^mp^2=(q-1)(q^4+q^3+q^2+q+1)$ implies $(q^4+q^3+q^2+q+1)$ equals either $p$ or $p^2$ (because it's odd). In either case $p\gt q^2$, which implies $p$ cannot divide $q-1$, hence we must have $q^4+q^3+q^2+q+1=p^2$ and thus $q-1=2^m$. Writing now $q=2^m+1$, we wind up with
$$2^{4m}+5\cdot2^{3m}+5\cdot2^{2m+1}+5\cdot2^{m+1}+5=p^2$$
Because $p^2\equiv1$ mod $8$ for any odd number, we must have $m\lt2$. This leaves only $m=1$, $q=3$, $p=11$.
|
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|
Proof for inequality with $a,b,c,d$ with $d =\max(a,b,c,d)$
Let $a,b,c,d$ positive real numbers with $d= \max(a,b,c,d)$. Proof
that
$$a(d-c)+b(d-a)+c(d-b)\leq d^2$$
*
*I believe that the GM-AM inequality with $n=4$ variables might be helpful.
$$\sqrt[n]{x_1 x_2 \dots x_n} \le \frac{x_1+ \dots + x_n}{n}$$
We also know that the Geometric mean is bounded as follows :
$$ \min \{x_1, x_2, \dots x_n\} \le \frac{x_1+ \dots + x_n}{n}
\le \max \{x_1, x_2, \dots x_n\}$$
** I also tried to draw an square and some rectangles, but nothing worked out.
|
Divide both sides by $d^2$ to get an equivalent inequality:
$\dfrac{a}{d}\cdot \left(1 - \dfrac{c}{d}\right) + \dfrac{b}{d}\cdot \left(1 - \dfrac{a}{d}\right) + \dfrac{c}{d}\cdot \left(1 - \dfrac{b}{d}\right) \leq 1$.
Now let $x = \dfrac{a}{d}$, $y = \dfrac{b}{d}$, and $z = \dfrac{c}{d}$, then : $0 \leq x, y, z \leq1$, and we are to prove:
$x(1 - z) + y(1 - x) + z(1 - y) \leq 1$.
Consider $f(x,y,z) = x(1 - z) + y(1 - x) + z(1 - y) - 1 = x + y + z - xy - yz - zx - 1$. We find the critical points of $f$. So take partial derivatives:
$f_x = 1 - y - z = 0 \iff y + z = 1$
$f_y = 1 - x - z = 0 \iff x + z = 1$
$f_z = 1 - x - y = 0 \iff x + y = 1$.
Thus $\nabla{f} = 0 \iff x + y = y + z = z + x = 1 \iff x = y = z = \dfrac{1}{2}$. Thus the maximum of $f$ occurs at either the critical values or the boundary points which are: $(x,y,z) = (0,0,0), (0,1,1), ..., (1,1,1)$. Of these values, the max is $0$. So $f(x,y,z) \leq 0$ which is what we are to prove.
|
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|
Volume of the intersection two elliptical cylinders How can I find the volume of the intersection of the following elliptical cylinders?
$$\dfrac{x^2}{9} + \dfrac{y^2}{4} \leq 1$$
$$\dfrac{y^2}{4}+\dfrac{z^2}{9} \leq 1$$
In the first octant.
This is what I have done:
Let $A$ be the projection of the intersection in the XY plane. Then:
$$A = \{ (x, y, 0) \in \mathbb{R}^2 : x, y \geq 0, \dfrac{x^2}{9} + \dfrac{y^2}{4} \leq 1\}$$
And from the second cylinder I get:
$$z = 3\sqrt{1-\dfrac{y^2}{4}}$$
So the volume $V$ will be:
$$\displaystyle \int_A 3\sqrt{1-\dfrac{y^2}{4}} dxdy $$
Is that right?
|
That is right, but most of the work is actually to split $A$ into the proper bounds for each variable, and then do the actual integration. So write $A$ as \begin{equation} A = \{ (x, y) \in \mathbb{R}^2 : 0\leq x \leq 3\sqrt{1-y^2/4}\;,\; 0\leq y \leq 2\}, \end{equation} and then we have \begin{eqnarray} V&=&\int_0^2 \int_0^{3\sqrt{1-y^2/4}} 3\sqrt{1-\dfrac{y^2}{4}} dxdy \\ &=& \int_0^2 \left (3\sqrt{1-\dfrac{y^2}{4}}\right)^2 dy \\ &=& \int_0^2 9-\dfrac{9y^2}{4} dy \\ &=& (9y-\dfrac{3y^3}{4}) |_0^2 \\ &=& 12 \end{eqnarray}
This problem happens to work out nicely because the ellipse equations for $x$ in terms of $y$ matches with the equation of $z$ in terms of $y$ - so you can get rid of the square root. In general when ellipses or circle are involved I would go for a coordinate transformation to cylindrical coordinates.
|
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|
For which $n$ is $ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx}= \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$? I have been trying to figure out for which $n$ is $$ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx} = \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$$
Using maple I got the following values below 100
$$
n = 1
,\,2
,\,12
,\,13
,\,14
,\,24
,\,25
,\,26
,\,36
,\,37
,\,38
,\,48
,\,49
,\,50
,\,60
,\,61
,\,62
,\,73
,\,74
,\,85
,\,86
,\,98
,\,110
$$
But I am having a hard time seeing any pattern. I tried calculating the integral directly.
The integral seems to converge for all $n$, but a closed form seems hard. Any help would be appreciated =)
EDIT: I ran a few more tests and can not find any more integer values that work. Are the list above exhaustive?
|
Something is amiss. We have
$$\int_0^{\pi/2} \frac{dx}{2+\sin (nx)} = \frac{1}{n}\int_0^{n\pi/2} \frac{dy}{2+\sin y}.$$
Using
$$\int_0^{2\pi} \frac{dy}{2+\sin y} = \frac{2\pi}{\sqrt{3}},$$
writing $n = 4k + r$ with $0 \leqslant r \leqslant 3$, we find
$$\int_0^{\pi/2}\frac{dx}{2+\sin (nx)} = \frac{k}{4k+r}\cdot \frac{2\pi}{\sqrt{3}} + \frac{1}{4k+r}\int_0^{r\pi/2}\frac{dy}{2+\sin y}.$$
In particular for $r = 0$, we have
$$\int_0^{\pi/2}\frac{dx}{2+\sin (4kx)} = \frac{1}{4}\frac{2\pi}{\sqrt{3}} = \frac{\pi}{2\sqrt{3}} \neq \frac{\pi}{3\sqrt{3}}.$$
No $n$ divisible by $4$ satisfies the equation. For $n = 4k+1$, we have the value
$$\frac{k}{4k+1} \frac{2\pi}{\sqrt{3}} + \frac{1}{4k+1}\frac{\pi}{3\sqrt{3}} = \frac{6k+1}{4k+1} \frac{\pi}{3\sqrt{3}},$$
which only equals $\frac{\pi}{3\sqrt{3}}$ for $k = 0$.
For $n = 4k+2$, we have
$$\frac{k}{4k+2}\frac{2\pi}{\sqrt{3}} + \frac{2}{4k+2} \frac{\pi}{3\sqrt{3}} = \frac{6k+2}{4k+2}\frac{\pi}{3\sqrt{3}},$$
which also only equals $\frac{\pi}{3\sqrt{3}}$ for $k = 0$.
For $n = 4k+3$, we have
$$\frac{k}{4k+3}\frac{2\pi}{\sqrt{3}} + \frac{1}{4k+3}\frac{4\pi}{3\sqrt{3}} = \frac{6k+4}{4k+3}\frac{\pi}{3\sqrt{3}},$$
which never equals $\frac{\pi}{3\sqrt{3}}$. Thus $n = 1$ and $n = 2$ are the only solutions.
|
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|
Proof for $\sin(x) > x - \frac{x^3}{3!}$ They are asking me to prove $$\sin(x) > x - \frac{x^3}{3!},\; \text{for} \, x \, \in \, \mathbb{R}_{+}^{*}.$$ I didn't understand how to approach this kind of problem so here is how I tried:
$\sin(x) + x -\frac{x^3}{6} > 0 \\$
then I computed the derivative of that function to determine the critical points. So:
$\left(\sin(x) + x -\frac{x^3}{6}\right)' = \cos(x) -1 + \frac{x^2}{2} \\ $
The critical points:
$\cos(x) -1 + \frac{x^2}{2} = 0 \\ $
It seems that x = 0 is a critical point.
Since $\left(\cos(x) -1 + \frac{x^2}{2}\right)' = -\sin(x) + x \\ $
and $-\sin(0) + 0 = 0 \\$
The function has no local minima and maxima. Since the derivative of the function is positive, the function is strictly increasing so the lowest value is f(0).
Since f(0) = 0 and 0 > 0 I proved that $ \sin(x) + x -\frac{x^3}{6} > 0$. I'm not sure if this solution is right. And, in general, how do you tackle this kind of problems?
|
Recursive integration We know that $$0\le\cos a\le 1\implies \sin t = \int_0^t\cos s ds < t$$ for $0\lt t\lt z\lt x$. Integrating over $\color{blue}{(0,z)}$ we get
$$1-\cos z=\int_0^z\sin tdt < \int_0^ztdt= \frac{z^2}{2}$$
that is for all $0<z<x$ we have,
$$\color{blue}{1-\frac{z^2}{2}< \cos z\le 1}$$
integrating again over $\color{blue}{(0,x)}$ we get
$$\color{red}{x-\frac{x^3}{6} = \int_0^x 1-\frac{z^2}{2} dz< \int_0^x\cos z dz=\sin x}$$
that is
$$\color{blue}{x-\frac{x^3}{6} <\sin x< x}$$
continuing with this process you get,
$$\color{blue}{1-\frac{x^2}{2}< \cos x< 1-\frac{x^2}{2}+\frac{x^4}{24} }$$
$$\color{blue}{x-\frac{x^3}{6} <\sin x< x-\frac{x^3}{6} +\frac{x^5}{5!}}$$
More generally for $n\geq1$, by induction we get
$$\color{blue}{\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k}}{(2k)!}<\cos x<\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k}}{(2k)!}+\frac{x^{4n}}{(4n)!} }$$
$$\color{blue}{\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k+1}}{(2k+1)!} <\sin x<\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k+1}}{(2k+1)!}+\frac{x^{4n+1}}{(4n+1)!}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
For which real numbers $c$ is $\frac{e^x+e^{-x}}{2} \le e^{cx^2}$ for all real numbers $x$? This question comes from the 1980 Putnam exam. My work is shown below.
For all integers $n \ge 1$, \begin{align}
(2n)!&=n!\cdot\underbrace{(n+1)(n+2)(n+3)\cdots(2n-2)(2n-1)(2n)}_{n \text{ terms}} \\
&\ge n! \cdot \underbrace{2 \cdot 2 \cdot 2\cdots 2\cdot2 \cdot 2}_{n \text{ terms}} \\ &=2^nn! \\ \implies \frac{1}{(2n)!} &\le \frac{1}{2^nn!}.\end{align} So we have \begin{align}
\frac{e^x+e^{-x}}{2} = \cosh(x) &= \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \\ &\le \sum_{n=0}^{\infty} \frac{x^{2n}}{2^nn!} \\ &= \sum_{n=0}^{\infty} \frac{\left(\frac{x^2}{2} \right)^n}{n!} \\ &= e^{\frac{1}{2}x^2}
\end{align}
Therefore, $\boxed{c \ge \frac{1}{2}}$.
How do I know that it is $\ge$, not $\le$? Other than that, how did I do on my work?
|
In fact we have the following:
*
*The method presented in the proposal of the problem proves that any $c\geq \frac{1}{2}$ works.
*The solution of Leucippus proves that no $c< \frac{1}{2}$ works. So, these are not alternative solutions, but complementary ones. and the conclusion is that
$$
\left\{c:\forall\, x,~\cosh(x)\leq \exp(c x^2)\right\}=\left[\frac{1}{2},+\infty\right).
$$
|
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"url": "https://math.stackexchange.com/questions/803722",
"timestamp": "2023-03-29T00:00:00",
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|
Can you find a ellipse so that its image is a circle? This is a "fun" question and I have already a solution. I asked this question so that I may see a different approach or an elegant solution.
Let $P$ be a plane with equation $x+y+z=1$. Find an ellipse on $x,y$ plane so that its "shadow" on $P$ is a circle.
Edit: By "shadow", I mean $(x,y)\to(x,y,z)$ where $(x,y,z)\in P$
|
The circle with radius 1 and center $(0,0,1)$ in the plane $x+y+z=1$ can be parametrized by
\begin{align}
\vec{p}(t) &= \langle 0,0,1 \rangle + \cos(t)\langle 1,-1,0 \rangle/\sqrt{2} + \sin(t) \langle 1,1,-2 \rangle/\sqrt{6} \\
&= \left\langle \frac{\sin (t)}{\sqrt{6}}+\frac{\cos
(t)}{\sqrt{2}},\frac{\sin (t)}{\sqrt{6}}-\frac{\cos
(t)}{\sqrt{2}},1-\sqrt{\frac{2}{3}} \sin (t) \right\rangle.
\end{align}
It's shadow can be parametrized by just dropping the $z$ coordinate.
$$\vec{s}(t) = \left\langle \frac{\sin (t)}{\sqrt{6}}+\frac{\cos
(t)}{\sqrt{2}},\frac{\sin (t)}{\sqrt{6}}-\frac{\cos
(t)}{\sqrt{2}} \right\rangle.$$
This is enough to visualize.
Once we have the parametrization, we can generate points on the ellipse by simply plugging in $t$ values. Here are five points on the ellipse in the plane, for example.
$$
\begin{array}{l|l}
t & (x,y) \\
\hline
0 & \left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right) \\
\hline
\frac{\pi }{3} & \left(\frac{1}{\sqrt{2}},0\right) \\
\hline
\frac{\pi }{2} & \left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}\right) \
\\
\hline
\pi & \left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) \\
\hline
\frac{3 \pi }{2} & \left(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}}\right)
\end{array}
$$
Since five points determine an ellipse, we can find the Cartesian formula easily enough. My favorite technique is to set the following determinant equal to zero.
$$
\left|
\begin{array}{cccccc}
x^2 & x y & y^2 & x & y & 1 \\
\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} &
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 1 \\
\frac{1}{2} & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 1 \\
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} &
\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & 1 \\
\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} &
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 1 \\
\frac{1}{6} & \frac{1}{6} & \frac{1}{6} &
-\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{6}} & 1 \\
\end{array}
\right|
$$
After some simplification, this yields
$$
2 x^2 + 2 xy + 2 y^2 = 1.
$$
|
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|
How do you evaluate this limit? $\ln({x^3+2x^2+x})+ \frac{2}{x}$ How do you evaluate the following?
$$\lim_{x \to 0^+} \left [ \ln({x^3+2x^2+x})+ \frac{2}{x} \right ]$$
If I plug in $x$, I get $\infty-\infty$, which is undetermined, and I haven't been able to get the limit at a more manageable form. Can you please help me?
|
$\ln(x^3 + 2x^2 + x) + \dfrac{2}{x} = -ln\left(\dfrac{1}{x}\right) + 2\ln(x+1) + \dfrac{2}{x} = \dfrac{1}{x} + 2\ln(x+1) + y\left(1 - \dfrac{lny}{y}\right) \to +\infty$ as $y = \dfrac{1}{x} \to +\infty$ when $x \to 0^+$, and $\dfrac{lny}{y} \to 0$.
|
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|
Probability Bertsekas Question Question:
We have two jars each containing an equal number of balls. We perform four successive ball exchanges. In each exchange, we pick simultaneously and at random a ball from each jar and move it to the other jar. What is the probability that at the end of the four exchanges all the balls will be in the jar where they started?
Resource (Bertsekas)
Correct Answer:
$$
\frac{1}{n^2}\left( \frac{1}{n^2} + \frac{8(n-1)^2}{n^4} \right)
$$
My Approach:
*
*There are $4$ exchanges in total
*$1^{st}$ swap definitely results in disturbing configuration
*$2^{nd}$ swap can disturb configuration or can set things right
*$3^{rd}$ swap can set things right or can disturb configuration based on $2^{nd}$ swap
*$4^{th}$ swap has to set things right
*Ground rule is one can set things right only if there is a disturbance created before
*$1$ swap you do to create disturbance, $1$ swap you need to do revert that
disturbance. Hence maximum $2$ swaps causing a disturbance allowed. But just making $1$
swap which causes a disturbance cannot just happen by ground rule
I will represent swap causing disturbance by $1$ and swap reverting back balls by $-1$
\begin{align}
P(\text{same configuration after $4$ exchanges})
&= P(1, -1, 1, -1) + P(1, 1, -1, -1) \\
&= 1 \cdot \frac{1}{n^2} \cdot 1 \cdot \frac{1}{n^2} + 1 \cdot \frac{(n-1)^2}{n^2}
\cdot \frac{4}{n^2} \cdot \frac{1}{n^2} \\
&= {\bf \frac{1}{n^2}\left(\frac{1}{n^2} + \frac{4(n-1)^2}{n^4} \right) }
\end{align}
Why I am getting the wrong answer?
|
I can tell you that probability, $\frac{1 + 8(n-1)^2}{n^4}$, is wrong for sure as I wrote a quick simulation to check it and it does not agree (by orders of magnitude). Are you sure you copied it down correctly?
There are essentially two ways in which the jars can have the same configuration after $4$ steps:
*
*The exchange of step $1$ is reversed on step $2$ (e.g. $3 \leftrightarrow 8$ is followed by $8 \leftrightarrow 3$) with the same thing happening on steps $3$ and $4$, with possibly different balls;
*The exchanges of step $1$ and $2$ are not inverses, say $n_1 \leftrightarrow m_1$ followed by $n_2 \leftrightarrow m_2$ with either $n_2 \neq m_1$ or $m_2 \neq n_1$. The sub-possibilities here are
*
*a): $n_2 \neq m_1$, $m_2 \neq n_1$
*b): $n_2 = m_1$, $m_2 \neq n_1$
*c): $n_2 \neq m_1$, $m_2 = n_1$
Now, there are $n^2$ possible exchanges on any step, all equally likely; let $\cal{P}_i$ be the set of all possible exchanges on step $i$. Thus possibility $1$ has probability
$$
\sum_{p_1 \in \cal{P}_1}\left(\frac{1}{n^2}\cdot \frac{1}{n^2}\right)
\sum_{p_3 \in \cal{P}_3}\left(\frac{1}{n^2}\cdot \frac{1}{n^2}\right)
= \frac{1}{n^4}.
$$
For case $2$ a), the possible (ordered) exchanges on steps $3$ and $4$ are
$$
\begin{array}{cc}
(m_1 \leftrightarrow n_1, m_2 \leftrightarrow n_2), &
(m_1 \leftrightarrow n_2, m_2 \leftrightarrow n_1) \\
(m_2 \leftrightarrow n_2, m_1 \leftrightarrow n_1), &
(m_2 \leftrightarrow n_1, m_1 \leftrightarrow n_2) \\
\end{array};
$$
i.e. there are only $4$ exchanges out of a possible $n^4$, thus possibility $2$ a) has probability
$$
\sum_{p_1 \in \cal{P}_1}\left(\frac{1}{n^2}\cdot\left(1- \frac{1}{n}\right)^2\right)
\frac{4}{n^4}
= \frac{4(n-1)^2}{n^6}.
$$
For case $2$ b), the configuration after step $2$ is as follows (ellipses denote unexchanged balls):
$$
\begin{array}{ccc}
m_2, \ldots & \left.\frac{}{}\right| & m_1, n_1 \ldots \\
\end{array}
$$
note that $m_2 \leftrightarrow n_1$ is not a valid exchange since on step $3$ it would restore the original configuration (leaving a perturbed configuration for step $4$), and by symmetry it cannot work on step $4$. The possible (ordered) exchanges on steps $3$ and $4$ are therefore
$$
\begin{array}{cc}
(m_2 \leftrightarrow m_3, m_3 \leftrightarrow n_1), &
(n_3 \leftrightarrow n_1, m_2 \leftrightarrow n_3) \\
\end{array}
$$
for some $m_3 \neq n_1$ and some $n_3 \neq m_2$; so there are $2(n-1)$ possible exchanges. Since the probability of exchange $2$ not being the inverse of exchange $1$ because $n_2=m_1,m_2\neq n_1$ is $\frac{n-1}{n^2}$ (there are $n-1$ choices for $m_2$ and $1$ for $n_2$) the probability of possibility $2$ b) is
$$
\sum_{p_1 \in \cal{P}_1} \frac{1}{n^2}
\frac{n-1}{n^2} \frac{2(n-1)}{n^4}
= \frac{2(n-1)^2}{n^6}
$$
and by symmetry this is also the probability of possibility $2$ c). Therefore, the total probability is (since these possibilities form a partition of the desired event)
\begin{align}
\frac{1}{n^4} + \frac{4(n-1)^2}{n^6} + \frac{4(n-1)^2}{n^6}
&= \frac{1}{n^2}\left(\frac{1}{n^2} + \frac{8(n-1)^2}{n^4}\right),
\end{align}
and this agrees with simulation.
|
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|
Proof that existsts $\delta>0$ such that $2-\frac{1}{3} < x^2 +x < 2+\frac{1}{3}$ I need to prove that exists:
$\delta>0$ such that $1-\delta < x < 1+ \delta\implies 2-\frac{1}{3} < x^2 +x < 2+\frac{1}{3}$
What I did:
$1-\delta < x < 1+ \delta\implies |x-1|<\delta$ and then I suppose that:
$$|x+2|<\delta$$
So:
$$|x-1||x+2|<\delta^2$$
And then I have:
$$|x^2 +x -2|<\delta^2$$
By setting: $\delta^2 = \frac{1}{3}$ we have $\delta=\frac{1}{\sqrt{3}}$ then:
$$|x^2 +x -2|<\frac{1}{3} \tag{for $\delta = \frac{1}{\sqrt{3}}$}$$
But then I graphed this and it didn't work. How, then, it could be done? Where's my error? Is it in supposing $|x+2|<\delta$ too?
|
We continue on the path you started. We want $|x-1||x+2|$ to be "small." The idea is to make $|x-1|$ small, that is, to choose $x$ close to $1$.
Unfortunately, the $x+2$ term could spoil things. But for example if we make $|x-1|\lt \frac{1}{2}$, then $x+2$ will be between $2+\frac{1}{2}$ and $2+\frac{3}{2}$, and in particular will be positive and less than $4$. So if we make $|x-1|\lt \frac{1}{12}$, then $|x-1||x+2|$ will be less than $\frac{1}{3}$.
|
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|
Find a formula for $0 · 1 · 2 + 1 · 2 · 3 + 2 · 3 · 4 + \dots +n(n + 1)(n + 2)$, for $n \in \mathbb N$ $$\sum\limits_{i=1}^n i(i + 1)(i + 2)$$
$$\sum\limits_{i=1}^n i^3 + 3i^2 + 2i$$
$$\sum\limits_{i=1}^n i^3 + 3\sum\limits_{i=1}^ni^2 + 2\sum\limits_{i=1}^ni$$
$$= (\frac14)n^4 + (\frac12)n^3 + (\frac14)n^2 + n^3 + 3(\frac{n^2}2) + (\frac{n}2) + n^2 + n$$
$$ =n^4 + 3n^3 + 8n^2 + 3n$$
Why does what I did above not work?
|
It is very much easier to note that $$n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)=4n(n+1)(n+2)$$ from which you get a telescoping series.
This is related to the binomial identity $\binom nr+\binom n{r+1}=\binom {n+1}{r+1}$ with $r=3$, and the pattern clearly generalises.
|
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How to solve $\int \frac{x^4 + 1 }{x^6 + 1}$?
How to solve $\int \frac{x^4 + 1 }{x^6 + 1}$ ?
The numerator is a irreducible polynomial so I can't use partial fractions.
I tried the substitutions $t = x^2, t=x^4$ and for the formula $\int u\,dv = uv - \int v\,du$ I tried using: $u=\frac{x^4 + 1 }{x^6 + 1} , \,dv=\,dx \\ u=\frac{1}{x^6 + 1} , \,dv= (x^4 + 1) \,dx \\u=x^4 + 1 , \,dv=\frac{\,dx}{x^6 + 1}$
But I always get more complicated integrals.
Any hints are appreciated!
|
$$\int \frac{x^4 + 1 }{x^6 + 1}=\int \frac{x^4 -x^2+ 1 }{x^6 + 1} + \int \frac{x^2 }{x^6 + 1}=\int \frac{1}{x^2 + 1} + \int \frac{x^2 }{x^6 + 1}$$
The first integral is known while the second is easy by $u=x^3$.
|
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|
Integrals of a particular form Recently, I was given a question sheet with a lot of integrals, and I could solve all of them except for a particular type of them:
$$I_1=\int \frac{dx}{(x+1)^2\sqrt{x^2+2x+2}}=\int \frac{dx}{x^2\sqrt{x^2+1}}=\frac{-\sqrt{x^2+1}}{x}+C$$
$$I_2=\int \frac{dx}{(x-1)^2\sqrt{x^2-x+1}}$$
$$I_3=\int \frac{dx}{(2x+1)\sqrt{x^2-8}}$$
$$I_4=\int \frac{dx}{x\sqrt{x^2+x+1}}=\log x + \log\left(x+2+2\sqrt{x^2+x+1}\right)+C$$
And the like($I_2,I_3$ also are expressible in elementary functions, but they are not that cute). The method I see that wolfram is using(trial pro subscription) is to complete the square inside the roots to transform them to $\sqrt{ax^2+b}$ and then substitute $x=\sqrt{\frac{-b}{a}}\sec x$ if $b$ is negative, and $x=\sqrt{\frac{b}{a}}\tan x$ if positive. Finally, it uses the tangent half-angle subsitution. I really don't think this is the best way.
In particular, the trigonometric substitutions in $I_1,I_4$ seem to be unnecessary, since at the end we transform back the resulting composition of trigonometric functions into a function that doesn't need trigonometry at all (e.g. $-\csc(\tan^{-1}x)=-\frac{\sqrt{x^2+1}}{x}$)
So my question is: Is there some easy way to deal with the integrals:
$$\int\frac{dx}{(ax+b)\sqrt{x^2+c}}, \int\frac{dx}{(ax+b)^2\sqrt{x^2+c}}$$
Without having to do the secant-Weiertrass substitution?
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for the type of $\displaystyle \int \frac{1}{(ax+b)^2 \sqrt{cx^2 + dx + e}}dx$, $\displaystyle \frac{1}{ax+b} = t$ transforms into $\displaystyle \frac{at}{\sqrt{bt^2 + ct + d}}$ which can be reduced into standard integrals.
The same seems to work in linear case too.
$$\frac{1}{(ax+b) \sqrt{cx^2+ dx + e}} = \frac{at}{t \sqrt{bt^2+ct+d}} = \frac{a}{\sqrt{bt^2+ct+d}}$$
|
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|
Solving exponential equations using logarithms This is the equation that I am having troubles with:
$$\large x^{\large\log_{10}5}+5^{\large\log_{10}x}=50$$
So the first thing I do, I logarithm the whole expression with $\log_{10}$.
So I get:
$ {\log_{10} 5} \times {\log_{10} x} + {\log_{10} 5} \times {\log_{10} x} = {\log_{10} 50}$
When I solve this one for $x$, I get that $x = 16$, which is totally incorrect because it is supposed to be $100$. Can anyone tell me what am I doing wrong or show me how to solve this equation?
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Let $y=x^{\large\log_{10}5}$, then
$$
\log_{10}y=(\log_{10}5)(\log_{10}x)=\log_{10}5^{\large\log_{10}x}\color{red}{\quad\Rightarrow\quad} y=5^{\large\log_{10}x}.
$$
Hence
\begin{align}
x^{\large\log_{10}5}+5^{\large\log_{10}x}&=50\\
5^{\large\log_{10}x}+5^{\large\log_{10}x}&=50\\
2\times5^{\large\log_{10}x}&=50\\
5^{\large\log_{10}x}&=25\\
5^{\large\log_{10}x}&=5^2\color{red}{\quad\Rightarrow\quad}\log_{10}x=2\color{red}{\quad\Rightarrow\quad}\large\color{blue}{ x=10^2=100}.
\end{align}
|
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|
A little Problem in Trigonometry (Multiple Angle) If $\tan^2 \theta = 1 + 2\tan^2 \phi$, show that $\cos 2\phi = 1 + 2\cos2\theta$.
What I have done..
$$\implies \tan^2 \theta = 1 + 2\tan^2 \phi\\
\implies 1 + \tan^2 \theta = 2 + 2\tan^2 \phi\\
\implies 1 + \tan^2 \theta = 2(1 + \tan^2 \phi)\\
\implies \sec^2 \theta = 2(\sec^2 \phi)$$
|
Apply Componendo & Dividendo on $$\frac{\tan^2\theta}1=\frac{1+2\tan^2\phi}1$$
$$\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-(1+2\tan^2\phi)}{1+(1+2\tan^2\phi)}$$
$$\implies\cos2\theta=-\frac{\tan^2\phi}{1+\tan^2\phi}=-\sin^2\phi$$
$$\implies\cos2\theta=-\frac{1-\cos2\phi}2$$
|
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|
Solve $x^{3}-3x=\sqrt{x+2}$ Solve for real $x$
$$x^{3}-3x=\sqrt{x+2}$$
By inspection, $x=2$ is a root of this equation. So, I squared both sides and divided the six degree polynomial obtained by $x-2$. Then I got a quintic which I couldn't solve despite applying rational root theorem and substitutions. I believe that there must be some nice method to solve this which I can't think about. Please help. Thanks!
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Two preliminary remarks.
If $x$ is solution, then $x(x+\sqrt{3})(x-\sqrt{3}) = x^3-3x \ge 0$, so $-\sqrt{3} \le x \le 0$ or $x \ge \sqrt{3}$.
Observe also that $2$ is solution and no real number larger than $2$ can be solution since for $\ge 2$,
$$\frac{d}{dx}(x^3-3x) = 3x^2-3 \ge 9 > \frac{1}{4} \ge \frac{1}{2\sqrt{x+2}} = \frac{d}{dx}(\sqrt{x+2}).$$
So let $x \in [-2,2]$. Introduce the unique $t \in [0,\pi/2]$ such that $x = 2\cos{2t}$ with $t \in [0,\pi/2]$.
\begin{eqnarray*}
x^3-3x = \sqrt{x+2}
&\iff& 8\cos^3{2t}-6\cos2t = \sqrt{(2\cos2t+2)} \\
&\iff& 2\cos(6t) = 2\cos(t) \textrm{ since } \cos(t) \ge 0 \\
&\iff& 6t \pm t \in 2\pi\mathbb{Z} \\
&\iff& t \in \{0,2\pi/7,22\pi/5\} \textrm{ since } t \in [0,\pi/2] \\
&\iff& X \in \{2,2\cos(2\pi/7),2\cos(2\pi/5)\}.
\end{eqnarray*}
|
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|
Skanavi 2.003, Difference in answers, which is right? In Skanavi book i have a exercise to simplify an equation
$$\begin{align}
((\sqrt[4]{p} - \sqrt[4]{q})^{-2} + (\sqrt[4]{p} + \sqrt[4]{q})^{-2}) : \frac{\sqrt p + \sqrt q}{p-q}
\end{align}$$
Solving:
Replacing:
$$ a =\sqrt[4]{p} $$
$$ b=\sqrt[4]{q} $$
Equation is changed:
$$
((a - b)^{-2} + (a + b)^{-2}): \frac{a^2+b^2}{a^4-b^4}
$$
$$ A = ((a - b)^{-2} + (a + b)^{-2})$$
$$ B = \frac{a^2+b^2}{a^4-b^4}$$
$$ B = \frac{a^2+b^2}{a^4-b^4} => B = \frac{a^2+b^2}{(a^2-b^2)(a^2+b^2)} => B = \frac{1}{a^2 - b^2}$$
$$ A = ((a - b)^{-2} + (a + b)^{-2}) => A = \frac{1}{(a - b)^{2}} + \frac{1}{(a + b)^{2}} => A = \frac{(a + b)^2 + (a-b)^2}{(a^2 - 2ab + b^2)(a^2 + 2ab + b^2)} => A = \frac{a^2 + 2ab + b^2 + a^2 - 2ab + b^2}{a^4 + 2a^2b^2 + b^4 - 2a^2b - 4a^2b^2 - 2ab^2 + a^2b^2+2ab^2+b^4} => A =\frac{2a^2 + 2b^2}{a^4 - 2a^2b^2 + b^4} => A = \frac{2(a^2+b^2)}{(a^2-b^2)^2}$$
$$ A : B $$
$$ \frac{2(a^2+b^2)}{(a^2-b^2)^2} : \frac{1}{a^2 - b^2} = \frac{2(a^2+b^2)(a^2 - b^2)}{(a^2-b^2)^2} = \frac{2(a^2+b^2)}{(a^2-b^2)}$$
My Answer is:
$$
\frac{2(\sqrt p +\sqrt q)}{(\sqrt p-\sqrt q)}
$$
Books answer is:
$$
\frac{2(\sqrt p +\sqrt q)^2}{(p- q)}
$$
Where is my mistake?
|
What Mark said is exactly right.
$$
\frac{2(\sqrt p +\sqrt q)}{(\sqrt p-\sqrt q)} \\ =
\frac{2(\sqrt p +\sqrt q)(\sqrt p + \sqrt q)}{(\sqrt p-\sqrt q)(\sqrt p + \sqrt q)} \\ =
\frac{2(\sqrt p +\sqrt q)^2}{(\sqrt p)^2 - (\sqrt q)^2} \\ =
\frac{2(\sqrt p +\sqrt q)^2}{p-q}
$$
Remember $(a+b)(a-b)= a^2 - b^2$. It's very useful for problems like this.
|
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|
Find volume of revolution Find the volume:
$$x=y^2; x=1-y^2; \text{ rotated around }x=3$$
Use the disk/washer method.
My inner radius is $3+y^2$ and outer radius is $3-y^2$ and the limits of integration are $\frac1{\sqrt2}$ and $\frac{-1}{\sqrt2}$
My final answer is $14\pi\frac{\sqrt2}3$
The answer is supposed to be $10\pi\frac{\sqrt2}3$
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The inner radius is $3-(1-y^2)$, that is, $2+y^2$. By symmetry the volume is
$$2\int_{y=0}^{1/\sqrt{2}}\pi\left((3-y^2)^2-(2+y^2)^2\right)\,dy.$$
Expand and integrate.
Remark: The inner radius of $3+y^2$ used in the OP does not yield volume $14\pi\frac{\sqrt{2}}{3}$.
|
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|
Simple algebra formula for which I can't find the right answer I have the formula $y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$, and I should work to $y = \frac{1}{2}\cdot z \cdot (z + 1)$.
Somebody showed me how it's done:
$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cdot ((z + 1) \cdot (z + 2))$
$y + (z + 1) = \frac{1}{2} \cdot (z^2 + 3z + 2)$
$y + (z + 1) = \frac{1}{2}(z^2) + \frac{1}{2}(3z) + \frac{1}{2}(2)$
$y + (z + 1) = \frac{1}{2}(z^2) + 1\frac{1}{2}z + 1$
$y = \frac{1}{2}(z^2) + 1\frac{1}{2}z + 1$ - z - 1
$y = \frac{1}{2}(z^2) + \frac{1}{2}z$
$y = \frac{1}{2}z(z + 1)$
Great! But, my try went completely wrong, and I don't understand what I'm doing wrong:
$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cdot ((z + 1) \cdot (z + 2))$
$y + (z + 1) = \frac{1}{2} \cdot (z^2 + 3z + 2)$
$y = \frac{1}{2} \cdot z^2 + 3z + 2 - z - 1$
$y = \frac{1}{2} \cdot z^2 + 2z + 1$
$y = \frac{1}{2} \cdot (z^2 + 2z + 1)$
$y = \frac{1}{2}(z^2) + \frac{1}{2}(2z) + \frac{1}{2}1$
$y = \frac{1}{2}(z^2) + z + \frac{1}{2}$
But from this last step, I can't get anywhere near $y = \frac{1}{2}z(z + 1)$, and I do not understand what I did wrong.
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$$y + (z + 1) = \frac{1}{2} (z + 1) (z + 2)$$
$$2y+2(z+1)=(z+1)(z+2)$$
$$2y=(z+1)(z+2)-2(z+1)$$
$$2y=(z+1)(z+2-2)$$
$$2y=z(z+1)$$
$$y=\frac{1}{2}z(z+1)$$
Where is your error? From
$$y + (z + 1) = \frac{1}{2} (z^2 + 3z + 2)$$
you can say
$$y+(z+1)=\frac{1}{2}z^2+\frac{3}{2}z+1$$
where instead you only multiplied $z^2$ by $1/2$.
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|
Primes as a difference of powers Find the smallest prime that cannot be written as
$$|3^a - 2^b|$$
EDIT: I forgot to mention that $a$ and $b$ are whole numbers.
I tried to expand $3^a$ as $(2+1)^a$ using binomial theorem but I couldn't infer much. Please help. Thanks in advance!
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$2$. The second subtrahend is always $0~ (\text{mod}~ 2)$ and the first never is, so their difference can never be $\pm2$.
Conveniently, there are no smaller primes to check.
If on the other hand, $a$ and $b$ can be zero, then you break your problem into eight Pell equations and solve the set sequentially over primes until you find a prime that can't be solved in your desired form. The Pell equations are: \begin{align}
x^2 - y^2 &= p \\
-(x^2 - y^2) &= p \\
3x^2 - y^2 &= p \\
-(3x^2 - y^2) &= p \\
x^2 - 2y^2 &= p \\
-(x^2 - 2y^2) &= p \\
3x^2 - 2y^2 &= p \\
-(3x^2 - 2y^2) &= p
\end{align} This set of Pell equations comes from considering the exponents to be even or odd. If an exponent is odd, extract one power of $2$ or $3$ as a coefficient and the remaining exponent is even. Things with even exponents are squares. For each prime, $p$, see if any of these Pell equations are solvable with $x$ a power of $3$ and $y$ a power of $2$. If not, you have found your solution.
\begin{align}
2 &: &3 \cdot 1^2 - 1^2 = &|3^1 - 2^0| = 2 \\
3 &: &-(1^2 - 2^2) = &|3^0 - 2^2| = 3 \\
5 &: & &|3^2 - 2^2| = 5 \\
& & &|3^1 - 2^3| = 5 \\
7 &: & -(3^2 - 4^2) = &|3^2 - 2^4| = 7 \\
& & 3^2 - 2 \cdot 1^2 = &|3^2 - 2^1| = 7 \\
& & -(1^2-2 \cdot 2^2) = &|3^0 - 2^3| = 7 \\
\vdots \\
37&: & -(3\cdot 3^2 - 8^2) = &|3^3-2^6| = 37 \\
41&: & \text{no solutions}
\end{align}
and we find 41 is the first prime with no solution to any of these Pell equations of the desired form. (It has solutions to the Pell equations where not both $x$ is a power of 3 and $y$ is a power of 2, so not of the desired form.)
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|
$\sum^{\infty}_{n=1} \log(\frac{n+a+b}{n+a} \times \frac{n+b}{n+b+1})$ Prove $$\sum^{\infty}_{n=1} \log(\frac{n+a+1}{n+a} \times \frac{n+b}{n+b+1})=\log\frac{1+b}{1+a}$$
Hints/Answers are appreciated
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I think you mean $n+a+1$ not $n+a+b$.
If we put aside the issue of convergence, then using the property $\log(A)+\log(B) = \log(AB)$, we have
$$\sum_{n = 1}^{\infty} \log \left( \frac{n+a+1}{n+a} \cdot \frac{n+b}{n+b+1} \right) = \log \left( \prod_{n = 1}^{\infty} \frac{(a+1+n)(b+n)}{(n+a)(n+b+1)} \right).$$
Note that the $n+b+1$ cancels everything from the $b+n$ term except $b+1$ and same with $n+a+1$. The conclusion follows.
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|
Finding the explicit notation of $f(n)$, based on it's recursive description. I came across this problem on a HackerRank challenge.
The function $f(n)$ is
*
*$1$ if $n = 0$
*$2f(n - 1)$, if $n$ is odd
*$f(n -1) + 1$, if $n$ is even
I solved the problem using a recursive function and it worked just well. However, I am assuming that a program would work faster if instead of recursion we use an explicit function.
The site gave this solution:
$f(n) = Pow(2, (n+1)/2 + 1) - 1 - (2 \% n)$
Can someone help me arrive at this explicit formation?
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First I would start by writing out the sequence: $$f(0)=1, f(1) = 2, f(2) = 3, f(3) = 6, f(4) = 7, f(5) = 14, f(6) = 15, ...$$
Now notice that $f(0) = 2 - 1 = 2^{0+1} - 1$, $f(2) = 4-1 = 2^{1+1} - 1$, $f(4) = 8-1 = 2^{2+1} - 1$, and $f(6) = 16 - 1 = 2^{3+1}-1$. So we conjecture that $$f(2n) = 2^{n+1} - 1$$
Now it is also clear that $$f(2n-1) = 2f(2n-2) = 2^{n+1}-2$$
From here we just need to verify the conjectures with induction. We see that it certainly holds for $f(k)$ up to $k=6$. Now suppose it holds for $f(k)$ where $k=2n$.
Consider $$f(k+1) = f(2n+1) = 2f(2n) = 2(2^{n+1} - 1) = 2^{n+2} - 2$$ so we have verified this for the odd case.
Now if the conjecture holds for $f(k)$ where $k=2n+1$ then we have $$f(k+1) = f(2n+2) = f(2n+1) + 1 = 2^{n+2}-2+1 = 2^{n+2} - 1$$
Thus we see by induction that $$f(2n) = 2^{n+1}-1$$ and $$f(2n+1) = 2^{n+2} - 2$$ as we conjectured.
Now if you want this in one equation, we write $$f(k) = 2^{[(k+1)/2]+1} - 1.5 + (0.5)(-1)^{k}$$
|
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|
How to calculate the Nth member of this Generating function? The series 012012012... matches the following generating function:
$$T=\frac{x+2x^2}{1-x^3}$$
How could i find a closed expression of the nth member of this series?
|
For a bloated answer:
\begin{align*}
T &= \frac{x+2x^2}{1-x^3} \\
&= \frac{1}{1-x}+\frac{\frac{1}{2}-\frac{\sqrt{3}}{6}\, i}{-\frac{1}{2}+\frac{\sqrt3}{2}\, i - x}-\frac{\frac{1}{2}+\frac{\sqrt{3}}{6}\, i}{\frac{1}{2}+\frac{\sqrt3}{2}\, i + x}
\end{align*}
Extracting $[x^n]$ now gives:
\begin{align*}
a_n &= 1+\frac{\displaystyle \frac{1}{2}-\frac{\sqrt{3}}{6}\, i}{\displaystyle \left(-\frac{1}{2}+\frac{\sqrt3}{2}\, i\right)^{n+1}}- (-1)^n \frac{\displaystyle \frac{1}{2}+\frac{\sqrt{3}}{6}\, i}{\displaystyle \left(\frac{1}{2}+\frac{\sqrt3}{2}\, i\right)^{n+1}}
\end{align*}
Update
Thanks everyone for the discussion and answers!
Using Qiaochu Yuan's answer and notation, we can get the partial fraction,
and in turn, the formula.
For any period $0,1,\ldots, k$, the generating function is:
$$G(x) = \frac{P(x)}{Q(x)}$$
where
\begin{align*}
P(x) &= \sum_{u=1}^k u\, x^u \\
Q(x) &= 1-x^{k+1}
\end{align*}
and the general term is:
\begin{align*}
a_n &= \sum_{v=0}^k \frac{P\left(\zeta^v\right)}{Q'\left(\zeta^v\right)\, \zeta^{v\, (n+1)}}
\end{align*}
where $\zeta=e^{2\pi\, i\, /(k+1)},\;\; i=\sqrt{-1}$
Simplification is looking difficult now!
|
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|
Matrix Power Formula
Prove that for a fixed $a \in \mathbb{R}$ we have the matrix power formula for all $n \in \mathbb{Z}_+$:
$$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^n = \begin{pmatrix}a^n & na^{n-1}\\0 & a^n\end{pmatrix}$$
How would we prove this? Right now we are doing work with proofs by induction... but how would I prove this?
|
By induction:
for $n = 1 $ it's true.
For $n +1 $ $$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^{n+1} = \begin{pmatrix}a & 1\\0 & a\end{pmatrix} \begin{pmatrix}a^n & na^{n-1}\\0 & a^n\end{pmatrix} = \begin{pmatrix}a^{n+1} & a^n + na^n\\0 & a^{n+1} \end{pmatrix} = \begin{pmatrix}a^{n+1} & (n+1)a^n\\0 & a^{n+1} \end{pmatrix}$$
|
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|
Solving combinatorical problem using characteristic polynomial
How many $6$ length strings above $\left\{1,2,3,4\right\}$ are there such that $24$ and $42$ aren't allowed.
The suitable recurrence relation for this problem is: $a_{n+2} = 2a_{n-1} + 4a{n-2}$. Hence, the characteristic polynomial is: $x^2 -2x -4 = 0$.It's roots are: $1+\sqrt5,1-\sqrt5$.
So, the genral function is: $\alpha (1+\sqrt5)^n + \beta (1-\sqrt5)^n$
We know that: $a_0=1$ and $a_1=4$. Hence,
$\beta = \frac{1}{2} - \frac{3}{2\sqrt5}$,
$\alpha = \frac{1}{2} + \frac{3}{2\sqrt5}$
All in all,
$$(\frac{1}{2} + \frac{3}{2\sqrt5})(1+\sqrt5)^n + (\frac{1}{2} - \frac{3}{2\sqrt5})(1-\sqrt5)^n$$
Now, for $n=6$ the result is $1344$, but the book says $1216$.
Who's right?
|
Let $a_n$ be the number of strings of length n,
$b_n$ be the number of strings of length n starting with 1 or 3,
$c_n$ be the number of strings of length n starting with 2, and
$d_n$ be the number of strings of length n starting with 4.
Then $a_n=b_n+c_n+d_n$, $b_n=2a_{n-1}$, $c_n=b_{n-1}+c_{n-1}$, and $d_n=b_{n-1}+d_{n-1}$.
Substituting the last 3 equations into the first equation gives
$a_n=2a_{n-1}+2b_{n-1}+c_{n-1}+d_{n-1}=2a_{n-1}+b_{n-1}+(b_{n-1}+c_{n-1}+d_{n-1})=2a_{n-1}+2a_{n-2}+a_{n-1}$,
so $a_n=3a_{n-1}+2a_{n-2}$.
Starting with $a_0=1$ and $a_1=4$, we can substitute into this recurrence relation to get
$a_2=14$, $a_3=50$, $a_4=178$, $a_5=634$, and $a_6=2258$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Help Evaluating $\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$ Does anyone know how to evaluate the following limit?
$$
\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}
$$
The answer is $\frac{1}{2}$, but I want to see a step by step solution if possible.
|
$$
\begin{align}
\ \lim_{x\rightarrow\infty}\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} &= \lim_{x\rightarrow\infty} \left( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} \right) \cdot \frac{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\
\ &= \lim_{x\rightarrow\infty} \frac{x + \sqrt{x+\sqrt{x}}- x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\
\ &= \lim_{x\rightarrow\infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}} \\
\ &= \lim_{x\rightarrow\infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}} \cdot \frac{1/\sqrt{x}}{1/\sqrt{x}} \\
\ &= \lim_{x\rightarrow\infty} \frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\frac{1}{x}+\frac{1}{x\sqrt{x}}}+1} \\
\ &= \frac{\sqrt{1}}{\sqrt{1}+1} \\
\ &= \frac{1}{2} \\
\end{align}
$$
|
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|
A problem of sum floors let $n$ be a positive integer, prove that $$\sum_{i=0}^{\left\lfloor\frac{n}{3}\right\rfloor}\left\lfloor\frac{n-3i}{2}\right\rfloor=\left\lfloor\frac{n^2+2n+4}{12}\right\rfloor.$$
It looks like we have to consider lots of cases to solve the problem. Are there any simple solutions?
|
I think this is an easy way to handle it. We induct on $n$. See we use a step argument. Call
$$S_n=\sum_{i=0}^{\lfloor{n/3}\rfloor}\left\lfloor{\frac{n-3i}{2}}\right\rfloor$$
We have to prove $S_n=\left\lfloor{\dfrac{(n+1)^2+3}{12}}\right\rfloor \tag{1}$
Now $S_{n+3}-S_{n}=\left\lfloor{\dfrac{n+3}{2}}\right\rfloor$ To prove $(1)$ we need to prove:
$$ \left\lfloor{\dfrac{(n+1)^2+3}{12}}\right\rfloor+\left\lfloor{\dfrac{n+3}{2}}\right\rfloor=\left\lfloor{\dfrac{(n+4)^2+3}{12}}\right\rfloor\tag{2}$$
To prove $(2)$ we replace $n$ by $n-1$ and
$$\left\lfloor{\dfrac{n^2+3}{12}}\right\rfloor+\left\lfloor{\dfrac{n+2}{2}}\right\rfloor-\left\lfloor{\dfrac{(n+3)^2+3}{12}}\right\rfloor=Q_n$$
Now we show
$$Q_{n+6}=Q_n\tag{3}$$ which is a consequence of trivial algebra. Hence to prove $(2)$ for all $n$ we only need prove it for $n=0,1,2,3,4,5$ due to $(3)$. Which is trivial. Now to prove $(1)$ for all $n$ we need to check $n=0,1,2$ and we are again done. Note I haven't done all the calculations, they are left for the OP, but I have checked they are true indeed. I suppose this answers your questions.
|
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|
Please someone help with this nearly impossible integral $$ \int \frac{4x^5 -1} {(x^5 + x +1)^2} dx $$
So this is the integral and I have been stuck on it for ages without getting anywhere at all. Nothing I tried has gotten me anywhere so I'm basically stuck on nowhere with this.
I would really love for someone to go through this step by step, that's how I learn the best way. If it is even solvable by a human being....
|
$$\int \frac{4x^5-1}{(x^5+x+1)^2}\,dx=\int \frac{4x^5+5x^4-5x^4-1}{(x^5+x+1)^2}\,dx$$
$$=\int \frac{4x^5+5x^4}{(x^5+x+1)^2}\,dx-\int \frac{5x^4+1}{(x^5+x+1)^2}\,dx=J_1-J_2$$
$J_2$ can be handled by using the substitution $x^5+x+1=t$ and comes out to be $\dfrac{-1}{x^5+x+1}$. For $J_1$, factor out $x^{5}$ from the expression in parentheses in denominator, i.e:
$$J_1=\int \frac{4x^5+5x^4}{(x^5+x+1)^2}\,dx= \int \frac{\dfrac{4}{x^5}+\dfrac{5}{x^6}}{\left(1+\dfrac{1}{x^4}+\dfrac{1}{x^5}\right)^2}\,dx$$
Now use the substitution $1+\dfrac{1}{x^4}+\dfrac{1}{x^5}=t\Rightarrow -\left(\dfrac{4}{x^5}+\dfrac{5}{x^6}\right)\,dx=dt$, hence
$$J_1=-\int \frac{dt}{t^2}=\frac{1}{t}+C=\frac{x^5}{x^5+x+1}+C$$
Hence,
$$J_1-J_2=\frac{x^5}{x^5+x+1}+\frac{1}{x^5+x+1}+C=\boxed{\dfrac{x^5+1}{x^5+x+1}+C}$$
|
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|
What's the integral of $\int_0^\infty \frac{dx}{(x^4+1)^5}$ $$\int_0^\infty \frac{dx}{(x^4+1)^5}$$
My answer would be : $\dfrac{\Gamma(\tfrac{1}{4})\Gamma(\tfrac{19}{4})}{4\Gamma(5)}$
Solution: You can use this technique. – Mhenni Benghorbal
|
\begin{align}
x^4+1 & = (x^4 +2x^2 + 1)-2x^2 \\[8pt]
& = (x^2+1)^2 - (\sqrt{2}\, x)^2 \\[8pt]
& = (x^2 - \sqrt{2}\, x +1)(x^2 + \sqrt{2}\, x + 1)
\end{align}
Then go to partial fractions.
|
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|
Finding closed form for $1^3+3^3+5^3+...+n^3$ I'd like to find a closed form for $1^3+3^3+5^3+...+n^3$ where $n$ is an odd number.
How would I go about doing this?
I am aware that $1^3+2^3+3^3+4^3+...+n^3=\frac{n^2(n+1)^2}{4}$ but I'm not too sure how to proceed from here.
My gut feeling is telling me to multiply the above series by 8, then subtract it from the original, but it doesn't quite get me there because I'm going to have a whole lot of extra terms that I do not want.
|
Note that
$$\sum 1=n=C_n^1$$
$$\sum\sum1=\sum i=\frac{n(n+1)}{2}=C_{n+1}^2$$
$$\sum\sum\sum1=\sum\frac{i(i+1)}2=\frac{n(n+1)(n+2)}{2.3}=C_{n+2}^3$$$$\sum\sum\sum\sum1=\sum\frac{i(i+1)(i+2)}{2.3}=\frac{n(n+1)(n+2)(n+3)}{2.3.4}=C_{n+3}^4.$$
These are diagonals in Pascal's triangle.
From there, you derive
$$\sum i^2=2\sum\frac{i(i+1)}2-\sum i=2C_{n+2}^3-C_{n+1}^2,$$
and
$$\sum i^3=6\sum\frac{i(i+1)(i+2)}{6}-3\sum i^2-2\sum i=6C_{n+3}^4-6C_{n+2}^3+C_{n+1}^2.$$
|
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|
Values of $6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$ Let $f$ and $a$ such that $6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$.
I need to find the values of $f$ and $a$ that satisfies this condition.
For this i tried:
$F(x) = 6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$ then $F´(x) = \frac{f(x)}{x^2} = \frac{1}{\sqrt{x}}$ since $ \frac{d}{dx} \int_a^x \frac{f(t)}{t^2} dt = \frac{f(x)}{x^2}$ and here i stuck , how can i conclude in get de values of $f$ and $a$ please some healp, thanxs for your time.
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Put $x=a$ in $6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$. This gives you that
$2\sqrt a=6$ and so $a=9$. Differentiating $6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$ we get $\frac{f(x)}{x^2}=\frac{1}{\sqrt x}$ which implies that $f(x)=x^{3/2}$. Note that $\int_a^a \frac{f(t)}{t^2} dt=0$.
|
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|
Series $\sum \frac{1}{n^2\sin^3n}$ Question : Show that series $\sum \cfrac{1}{n^{2}\sin^{3}n}$ is divergent.
Hint:
Show that $$\sum \frac{1}{n|\sin(n)|}$$ is divergent.
I am interested in other possible proofs for this question.
|
Since $\pi$ is irrational, by Hurwitz's theorem, there are infinitely many pairs of relative prime integers $(n,m)$ such that
$$\left|\pi - \frac{n}{m}\right| < \frac{1}{\sqrt{5}{m^2}}
\quad\implies\quad|n - m\pi| < \frac{1}{\sqrt{5}m}
$$
We can assume both $n, m > 0$ and for any such pair, we have
$$
|\sin n | = |\sin(n-m\pi)| < \sin\frac{1}{\sqrt{5}m} < \frac{1}{\sqrt{5}m}
=
\frac{1}{\sqrt{5}n}\left( \pi + ( \frac{n}{m} - \pi )\right)
< \frac{1}{\sqrt{5}n}\left( \pi + 1\right)
\\ \implies \frac{1}{n^2 |\sin n|^3} > \left(\frac{\sqrt{5}}{\pi+1}\right)^3 n$$
This means the sequence $\displaystyle\;\frac{1}{n^2\sin^{3}n}\;$ doesn't converge to $0$
and hence the series diverges.
|
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|
Producing a CDF from a given PDF So I have this PDF:
$$
f(x)=
\begin{cases}
x + 3 & \text{ for } -3 \leq x < -2\\
3 - x & \text{ for } 2 \leq x < 3\\
0 & \text{ otherwise}
\end{cases}
$$
To make this a CDF, I have integrated the PDF from $-\infty$ to some value, $x$.
$$
F(x)=
\begin{cases}
\frac{x^2}{2} + 3x + \frac{9}{2} & \text{ for } -3 \leq x <-2\\
\frac{1}{2} & \text{ for } -2 \leq x<2\\
\frac{-x^2}{2} + 3x + \frac{7}{2} & \text{ for } 2 \leq x<3
\end{cases}
$$
My friend argues that the first term in this CDF which is $(x^2/2 + 3x + 9/2)$ should actually be $(x^2/2 + 3x)$. But isn't this impossible? At $x = -3$, the CDF must be $0$, am I correct?. This is only true in the case where the first term is $(x^2/2 + 3x + 9/2)$.
If someone could shed light on this topic, that would be much appreciated.
|
$ f(x) = \begin{cases} 0 &, \phantom{-3 \leq {}}x< -3 \\ x+3 &, -3 \leq x \leq -2 \\ 0 &, -2 \leq x \leq \phantom{-{}}2 \\ 3-x &, \phantom{-{}}2 \leq x \leq \phantom{-{}}3 \\ 0 &, \phantom{-{}}3 \leq x\end{cases}$
\begin{align}
F(x) = \int_{-\infty}^x f(t) \,\mathrm{d}t &= \begin{cases} 0 &, \phantom{-3 \leq {}}x< -3 \\ 0+\int_{-3}^x t+3 \,\mathrm{d}t &, -3 \leq x \leq -2 \\ 0+\int_{-3}^{-2} t+3 \,\mathrm{d}t + 0&, -2 \leq x \leq \phantom{-{}}2 \\ 0+\int_{-3}^{-2} t+3 \,\mathrm{d}t + \int_{2}^x 3-t \,\mathrm{d}t&, \phantom{-{}}2 \leq x \leq \phantom{-{}}3 \\ 0+\int_{-3}^{-2} t+3 \,\mathrm{d}t + \int_{2}^3 3-t \,\mathrm{d}t + 0 &, \phantom{-{}}3 \leq x\end{cases} \\
&= \begin{cases} 0 &, \phantom{-3 \leq {}}x< -3 \\ 0+\frac{x^2+6x+9}{2} &, -3 \leq x \leq -2 \\ 0+\frac{1}{2} &, -2 \leq x \leq \phantom{-{}}2 \\ 0+\frac{1}{2} + \frac{-x^2+6x-8}{2}&, \phantom{-{}}2 \leq x \leq \phantom{-{}}3 \\ 0+\frac{1}{2} + \frac{1}{2} + 0 &, \phantom{-{}}3 \leq x\end{cases}
\end{align}
As you can see $F(-3) = 0$ because $\frac{9-18+9}{2} = 0$. Your argument is correct that the CDF must increase from zero (starting) at $-3$.
|
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|
If $ x^2+y^2+z^2 =1$ for $x,y,z \in \mathbb{R}$, then find maximum value of $ x^3+y^3+z^3-3xyz $. If $ x^2+y^2+z^2 =1$, for $x,y,z \in \mathbb{R}$, what is the maximum of
$ x^3+y^3+z^3-3xyz $ ?
I factorize it... Then put the maximum values of $x+y+z$ and min value of $xy+yz+zx$...
But it is wrong as they don't hold simultaneously!
Also can it be solved using partial differentiation ?
And plz provide a solution without it.!
|
Yet another way, let $p = x+y+z, q = xy+yz + zx$. Then $x^2+y^2+z^2 = p^2-2q=1$ and $x^3+y^3+z^3-3xyz = p^3-3pq = \frac12p(3-p^2)$ . Also $x^2+y^2+z^2 = 1 \implies p^2\le 3$.
So we want the maximum of $\frac12p(3-p^2)$, subject to $p^2 \le 3$, which easily$^\dagger$ gives $p=1$.
$^\dagger$ Equivalently this is when $\frac14p^2(3-p^2)^2$ gets maximised, or $(p^2)\dfrac{3-p^2}2 \dfrac{3-p^2}2$. As the last is a product of three positive terms with constant sum $3$, it gets maximised when all terms are equal, i.e. $p=1$.
|
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|
Evaluation of $ \int\frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}dx$ Evaluation of $\displaystyle \int\frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}dx$
$\bf{My \; Try::}$ Let $\displaystyle I = \int \frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}dx$
So $\displaystyle I = \int\frac{\sin (x+\alpha)}{\cos x}\cdot \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}dx\cdot \sec^2 x$
$\displaystyle I = \int \frac{\sin x\cdot \cos \alpha+\cos x\cdot \sin \alpha}{\cos x}\cdot \sqrt{\frac{\cos +\sin x}{\cos x-\sin x}}\cdot \sec^2 x dx$
$\displaystyle I = \int\left(\cos \alpha\cdot \tan x+\sin \alpha\right)\cdot \sqrt{\frac{1+\tan x}{1-\tan x}}\cdot \sec^2 xdx$
Now Let $\tan x= t\;,$ Then $\sec^2 xdx = dt$
$\displaystyle I = \int (\cos \alpha\cdot t + \sin \alpha)\sqrt{\frac{1+t}{1-t}}dt$
Now How can I solve after that
Help me
Thanks
|
Let
\begin{align}
I = \int \frac{\sin(x+a)}{\cos^3(x)}\cdot \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}dx
\end{align}
then
\begin{align}
I &= \int (\sin(x) \cos(a) + \cos(x) \sin(a))(\cos(x) + \sin(x)) \ \frac{dx}{\cos^{3}(x) \sqrt{\cos(2x)}} \\
&= \cos(a) \ \int \frac{\sin^{2}(x) \ dx}{\cos^{3}(x) \sqrt{\cos(2x)}} +
\sin(a) \ \int \frac{ dx}{\cos(x) \sqrt{\cos(2x)}} + (\cos(a) + \sin(a)) \ \int \frac{\sin(x) \ dx}{\cos^{2}(x) \sqrt{\cos(2x)}} \\
&= \frac{1}{2} (\cos(a) + 2 \sin(a)) \ \tan^{-1}\left( \frac{\sin(x)}{\sqrt{\cos(2x)}}\right) - \frac{1}{2} \cos(a) \sqrt{\cos(2x)} \ \frac{\sin(x)}{\cos^{2}(x)} - (\cos(a) + \sin(a)) \ \frac{\sqrt{\cos(2x)}}{\cos(x)}
\end{align}
where there integrals
\begin{align}
\int \frac{\sin^{2}(x) \ dx}{\cos^{3}(x) \sqrt{\cos(2x)}} &= \frac{1}{2} \left[
\tan^{-1}\left( \frac{\sin(x)}{\sqrt{\cos(2x)}}\right) - \frac{\sin(x) \ \sqrt{\cos(2x)}}{\cos^{2}(x)} \right] \\
\int \frac{\sin(x) \ dx}{\cos^{2}(x) \sqrt{\cos(2x)}} &= - \frac{ \sqrt{\cos(2x)}}{\cos^{2}(x)} \\
\int \frac{ dx}{\cos(x) \sqrt{\cos(2x)}} &= \tan^{-1}\left( \frac{\sin(x)}{\sqrt{\cos(2x)}}\right)
\end{align}
were used.
|
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|
Proving a sharp inequality After experimenting, I've come to the conclusion that if $x\geq y\geq z\geq 0:$
$$\sum_{x,y,z}\frac{x}{\sqrt{x+y}}\geq \sum_{x,y,z}\frac{y}{\sqrt{x+y}}$$
(the sums are cyclic)
Does anyone know how to prove this? I've tried variants of Cauchy Schwarz, and the rearrangement inequality, but these are not powerful enough.
|
The condition $x^2+y^2+z^2=1$ is redundant.
Let $a=\sqrt{y+z},b=\sqrt{z+x},c=\sqrt{x+y}$ we have $c\ge b \ge a \ge 0$ and $$x=\frac{b^2 + c^2 - a^2}{2},y=\frac{c^2 + a^2 - b^2}{2},z=\frac{a^2 + b^2 - c^2}{2}.$$
Replacing into the original inequality, it becomes
$$\sum_{a,b,c} \frac{b^2 + c^2 - a^2}{2c} \ge \sum_{a,b,c} \frac{c^2 + a^2 - b^2}{2c},$$
which is true because
\begin{align}
LHS-RHS &= \sum_{a,b,c} \frac{b^2 - a^2}{c} \\
&= \frac{b^2 - a^2}{c} + \frac{c^2 - b^2}{a} +\frac{(a^2-b^2) +(b^2 - c^2)}{b}\\
&= (b^2 - a^2)\left(\frac{1}{c}-\frac{1}{b}\right) + (c^2 - b^2)\left(\frac{1}{a}-\frac{1}{b}\right) \\
&= \frac{(b-a)(c-b)}{b} \left(-\frac{b+a}{c}+\frac{b+c}{a}\right) \\
&= \frac{(b-a)(c-b)(c-a)(a+b+c)}{abc}\\
&\ge 0.
\end{align}
|
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|
Show that function is continuous at $x_0$ I want to show that $f: \mathbb R \rightarrow \mathbb R$,
$f(x) = \frac{x-1}{x^2+1}$ is continuous at $x_0=-1$. So if $|x+1|<\delta$, then $|f(x)+1| < \epsilon$.
I rearranged $|f(x)+1|=|\frac{(x-1) +(x^2+1)}{x^2+1}|=|\frac{x+x^2}{x^2+1}|=|\frac{1+x}{x+ \frac{1}{x}}|$, but now I'm stuck. Can anybody help me?
Thanks in advance!
|
You can use that $\displaystyle\left|\frac{x+x^2}{x^2+1}\right|=\frac{|x||x+1|}{x^2+1}<|x+1|$ for all x, since $|x|<x^2+1$
because $x^2<(x^2+1)^2=x^4+2x^2+1$ for all x.
|
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|
Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ I am currently attempting to prove the following inequality
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc.
Using that I get this
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}}$
From here, I used the fact that $(a+b)(b+c)(a+c)\geq 8abc$, which can be easily proven by considering that $a+b\geq 2\sqrt{ab}$
But by using this, I get the following...
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}} \leq 3 \times \sqrt[3]{\dfrac{abc}{8abc}} = \dfrac{3}{2}$
Everything seems so perfect because I get the value $\dfrac{3}{2}$ as required, but this method isn't valid due to the change in direction! What is going on?
Is there a way of proving this inequality otherwise then?
|
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
&& \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}-\frac{3}{2}\\
&=& \frac{2(x^3+y^3+z^3)-(x^2y+xy^2+y^2z+yz^2+z^2x+zx^2)}{2 (x + y) (x + z) (y + z)}\\
&=& \frac{(x+y)(x-y)^2+(y+z)(y-z)^2+(z+x)(z-x)^2}{2 (x + y) (x + z) (y + z)}\geq0
\end{eqnarray*}
with equality holds if and only if $x=y=z$.
|
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|
The limit of a sequence $\lim_{n\rightarrow \infty}\prod_{k=0}^{n-1}( 2+\cos \frac{k\pi }{n})^{\pi/n}$. $$
\lim_{n \to \infty}\prod_{k = 0}^{n - 1}
\left[\,2 + \cos\left(k\pi \over n\right)\right]^{\pi/n}\ =\ ?
$$
Please give some hints.
|
For any $a \in (1,\infty)$, consider the equation
$$\frac{x+x^{-1}}{2} = a\quad\iff\quad x^2 - 2ax + 1 = 0$$
It has two roots and one of it, $x = a + \sqrt{a^2-1}$, is outside the unit circle.
Notice
$$z^{2n}-1
= (z^2 - 1)\prod_{k=1}^{n-1}\left(z-e^{i\frac{k\pi}{n}}\right)\left(z-e^{-i\frac{k\pi}{n}}\right)
= (z^2 - 1)z^{n-1}\prod_{k=1}^{n-1}\left(z + \frac{1}{z} - 2\cos\frac{k\pi}{n}\right)
$$
Substitute $z$ by $-x = -\Big(a + \sqrt{a^2-1}\Big)$, this leads to
$$
\prod_{k=1}^{n-1}\left(a + \cos\frac{k\pi}{n}\right)
= \prod_{k=1}^{n-1}\left(\frac{x+x^{-1}}{2} + \cos\frac{k\pi}{n}\right)
= \frac{x^{2n}-1}{(x^2 - 1)(2x)^{n-1}}
$$
For large $n$, the numerator is dominated by the $x^{2n}$ term. As a result,
$$
\lim_{n\to\infty}\prod_{k=0}^{n-1}\left(a + \cos\frac{k\pi}{n}\right)^{\pi/n}
= \lim_{n\to\infty}\left(\frac{(x+1)(x^{2n}-1)}{(x-1)(2x)^n}\right)^{\pi/n}
= \left(\frac{x}{2}\right)^{\pi}
= \left(\frac{a + \sqrt{a^2-1}}{2}\right)^{\pi}
$$
Substitute $a$ by $2$, the desired limit is
$\displaystyle\;\left(\frac{2+\sqrt{3}}{2}\right)^\pi\approx
7.09761749580862646182\ldots
.$
|
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|
Modular Arithmatic - Solving congruences I'm sure this is pretty basic but I'm struggling to understand how to go about solving this problem for my homework. The question states "Solve the following congruences for x". The first problem is $2x+1\equiv 4\pmod 5$.
|
There are many ways to solve the problem. The conceptually simplest, but most tedious, is to test one by one the possibilities $x\equiv 0\pmod{5}$, $x\equiv 1\pmod{5}$, and so on up to $x\equiv 4\pmod{5}$. Quickly we find that $x\equiv 4\pmod{5}$. (This approach would become quite unpleasant if $5$ were replaced by $97$.)
It is simpler to use some algebra. So rewrite as $2x\equiv 3\pmod{5}$. Since $3\equiv 8\pmod{5}$, it is convenient to rewrite the congruence as $2x\equiv 8\pmod{5}$. Then since $2$ and $5$ are relatively prime, we can divide by $2$, getting $x\equiv 4\pmod{5}$.
A fancier version is to start from $2x\equiv 3\pmod{5}$. Now multiply both sides by $3$ (the modular inverse of $2$). We get $6x\equiv 9\pmod{5}$. But $6\equiv 1\pmod{5}$ and $9\equiv 4\pmod{5}$, so we conclude that $x\equiv 4\pmod{5}$.
Remark: We used congruence notation throughout, since it is very important to get accustomed to it. But $2x\equiv 3\pmod{5}$ means that $5$ divides $2x-3$. So we want to solve $2x-3=5k$, that is, $2x=3+5k$. So we want to find a $k$ such that $3+5k$ is divisible by $2$. It is clear that $k=1$ works, giving $2x=8$ so $x=4$. Any number congruent to $4$ modulo $5$ will also work, giving answer $x\equiv 4\pmod{5}$.
|
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|
If $aI am trying to prove $a^2<\frac{1}{3}(a^2+ab+b^2)<b^2$ if $a<b$. I am having trouble with the left inequality:
$a^2<\frac{1}{3}(a^2+ab+b^2) \Rightarrow 2a^2<ab+b^2$.
If $a>0$, then all is well and adding $a^2<b^2$ to $a^2<ab$ will yield the result I want, but I do not know how to deal with the case $a<0$, or if there is any simpler ways of proving the inequality.
|
Is true only if $0< a<b$, then for the left side:
$$a^2<b^2
$$
$$a^2+a^2+ab<b^2+ab+a^2
$$
$$a^2+a^2+ab<b^2+ab+a^2
$$
If $a<b$ then how $a>0$ then $a^2<ab$
$$a^2+a^2+a^2<a^2+a^2+ab<b^2+ab+a^2
$$
$$3a^2 <b^2+ab+a^2
$$
For the rigth side
$$a^2<b^2
$$
$$a^2+b^2+ab<b^2+b^2+ab
$$
If $a<b$ then how $b>0$ then $ab<b^2$
$$a^2+ab+b^2<b^2+b^2+ab<b^2+b^2+b^2
$$
$$a^2+ab+b^2<3b^2
$$
|
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|
inequality method of solution Im looking for an efficent method of solving the following inequality: $$\left(\frac{x-3}{x+1}\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 <0$$
I've tried first determining when the absolute value will be positive or negative etc, and than giving it the signing in accordance to range it is in, bur it turned out to be quite complex and apperently also wrong. Are there any other ways?
|
When $t \equiv \frac{x-3}{x+1} > 0$, the solution to $ t^2 - 7t + 10 <0$ is $ 2<t<5$. (Factor the quadratic.)
When $t < 0$, the solution to $ t^2 + 7t + 10 <0$ is $ -2<t<-5$.
On the $t < 0$ branch, we need to solve $-2 < \frac{x-3}{x+1} < 5$. We break this up into two possibilities, $x < -1$ and $x > -1$, because when we multiply through by $x+1$ in the $x < -1$case we have to flip the sense of the the inequality.
When $x < -1$ we then get on one side $-2(x+1) < x-3$, which gives $x > +\frac{1}{3}$ and this does not work. But for $x > -1$ we get
$$\begin{array}{c}
-5(x+1) < x-3 < 2(x-1) \\
6x > -2 \rightarrow x > -\frac{1}{3} \\
3x < +1 \rightarrow x < +\frac{1}{3}
\end{array}
$$
which has solution
$$
-\frac{1}{3} < x < \frac{1}{3}
$$
On the $t > 0$ branch, we need to solve $2 < \frac{x-3}{x+1} < 5$. We again break this up into two possibilities, $x < -1$ and $x > -1$.
When $x < -1$ we then get on one side $2(x+1) > x-3$, which gives $x > -2$, and $5(x+1) < x-3$, which gives $x > -5$. For $x > -1$ we get $2(x+1) < x-3$, which holds if $x < -5$, which contradicts $x > -1$ so that case does not give any solutions.
Thus the answer combines the two solution regions:
$$
\left( -5 < x < 2 \right) \bigcup \left( -\frac{1}{3} < x < \frac{1}{3} \right)
$$
|
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|
How to prove that the number $1+4a_{n}a_{n+1}$ is a perfect square.
A sequence of integer $\{a_{n}\}$ is given by the conditions $a_{1}=1,
a_{2}=12,a_{3}=20$,and $$a_{n+3}=2a_{n+2}+2a_{n+1}-a_{n}$$
show that
for every postive integer $n$, the number $1+4a_{n}a_{n+1}$ is a perfect square.
since the
$$r^3=2r^2+2r-1$$
then
$$r^3-2r^2-2r+1=$$
$$(r+1)(r^2-3r+1)=0$$
so
$$r_{1}=-1,r_{2,3}=\dfrac{3\pm \sqrt{5}}{2}$$
so $$a_{n}=A(r_{1})^n+B(r_{2})^n+C(r_{3})^n$$
where $A,B,C$ is constant.
Then I fell follow is very ugly,maybe someone have simple methods.Thank you
|
Given the sequence $a_{n+3} = 2 a_{n+2} + 2 a_{n+1} - a_{n}$ it is evident that a solution of the form $a_{n} \approx r^{n}$ leads to $r^{3} - 2 r^{2} - 2r + 1 = 0$, or $(r+1)(r^{2} - 3r + 1) = 0$ has roots $r_{1} = -1$, $r_{2,3} = (3 \pm \sqrt{5})/2$. Now it is seen that $r_{2,3} = (1 \pm \sqrt{5})^{2}/4$ and
\begin{align}
a_{n} = A (-1)^{n} + B \left( \frac{1+\sqrt{5}}{2} \right)^{2n} + C \left( \frac{1 - \sqrt{5}}{2} \right)^{2n}.
\end{align}
Since the coefficients associated with B and C are elements of the Fibonacci and Lucas numbers the expression for $a_{n}$ can be seen in the form
\begin{align}
a_{n} = A(-1)^{n} + B \ F_{2n} + C \ L_{2n}.
\end{align}
It is easier now to find the coefficients by using $a_{1} = 1$, $a_{2} = 12$, and $a_{3} = 20$ it is seen that
\begin{align}
a_{n} = 3 (-1)^{n} - \frac{F_{2n}}{2} + \frac{3 L_{2n}}{2} = \frac{1}{2} \left( 3 L_{n}^{2} - F_{2n} \right) = \frac{1}{2} \ L_{n}(3 L_{n}-F_{n}) = L_{n} (L_{n} +F_{n-1}).
\end{align}
Now let $S_{n} = 1 + 4 a_{n} a_{n+1}$ for which:
\begin{align}
S_{n} = \left[ 2 L_{n} (L_{n+1} + F_{n}) + (-1)^{n} \right]^{2}
\end{align}
is an integer squared.
|
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|
$\lim_{n \rightarrow \infty} f_n(x) = n^2 \left( 1- \cos \frac{x^3 - 1}{n} \right)$
Let
$$f_n(x) = n^2 \left( 1- \cos \frac{x^3 - 1}{n} \right)$$
Let M be the set of x s.t. $\lim_{n \rightarrow \infty} f_n(x)$
exists. For each $x \in M$ let $f(x) = \lim_{n \rightarrow \infty} f_n(x)$. Then
*
*M is bounded above
*f(x) is even function
*f(x) has slant asymptote
*$f'(1) = 6$
*1-4 are false.
The only thing I see now is that $1 \in M$. Please, give me hints how to deal with this problem.
Update: I try to use L'Hopital's rule
$$\lim \left(x^{3}-1\right)\sin\frac{x^{3}-1}{n}/\frac{2}{n^{2}}\rightarrow \lim \left(x^{3}-1\right)^{2}\cos\frac{x^{3}-1}{n}/\frac{4}{n^{2}}=\infty$$
|
it is easy to see that
$\cos \frac{x^{3}-1}{n} = 1 - \frac{(\frac{x^{3}-1}{n})^{2}}{2} + ... = 1 - \frac{x^{6} - 2x^{3} + 1}{2n^{2}} + ...$ so that $f(x) = n^{2} (1-\cos \frac{x^{3}-1}{n}) \rightarrow \frac{x^{6}}{2} - x^{3} + \frac{1}{2}$
so 1 - false, 2 - false, 3 - false, 4 - false
|
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|
How find the minimum of the $|w^3+z^3|$,if $|z+w|=1,|z^2+w^2|=14$ let complex $z,w$ such
$$|z+w|=1,|z^2+w^2|=14$$
find the minimum of the value
$$|w^3+z^3|$$
My idea: let
$$z=a+bi,w=c+di\Longrightarrow z+w=(a+c)+(b+d)i,z^2+w^2=(a^2+b^2+c^2+d^2)+2(ab+cd)i$$
then we have
$$(a+c)^2+(b+d)^2=1,(a^2+b^2+c^2+d^2)+4(ab+cd)^2=14^2$$
and $$w^3+z^3=(a^3+c^3-3ab^2-3cd^2)+(3a^2b+3c^2d-b^3-d^3)i$$
|
Let $z+w = a$ and $z^2+w^2 = b$. Then, $zw = \dfrac{(z+w)^2-(z^2+w^2)}{2} = \dfrac{a^2-b}{2}$.
Thus, $z^3+w^3 = (z+w)^3-3zw(z+w) = a^3 - 3 \cdot \dfrac{a^2-b}{2} \cdot a = \dfrac{3ab-a^3}{2}$.
If $|a| = |z+w| = 1$ and $|b| = |z^2+w^2| = 14$, then using the reverse triangle inequality, we have
$|z^3+w^3| = \left|\dfrac{3ab-a^3}{2}\right| = \dfrac{|a| \cdot |3b-a^2|}{2} = \dfrac{|3b-a^2|}{2} \ge \dfrac{3|b|-|a|^2}{2} = \dfrac{3\cdot 14 - 1^2}{2} = \dfrac{41}{2}$.
Equality occurs iff $3b$ and $a^2$ are $180^{\circ}$ apart when plotted in the complex plane.
For instance, if $a = i$ and $b = 14$, then we get $z,w = \dfrac{\pm \sqrt{29} + i}{2}$ and $z^3+w^3 = \dfrac{41}{2}i$.
Therefore, the minimum value of $|z^3+w^3|$ is $\dfrac{41}{2}$.
|
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|
An improper integral : $\int_{0}^\infty {\ln(a^2+x^2)\over{b^2+x^2}}dx$ How to evaluate the following improper integral:$$\int_{0}^\infty {\ln(a^2+x^2)\over{b^2+x^2}}dx,$$ where $a,b>0$.
I tried to suppose $$f(a)=\int_0^\infty {\ln(a^2+x^2)\over{b^2+x^2}}dx,$$ based on the convergence theorem, and then I tried $${df(a)\over da}=\int_0^\infty {2a\over {(a^2+x^2)(b^2+x^2)}}dx = {\pi\over b(b+a)},$$and then $$f(a)={\pi\over b}\ln(b+a)+C,$$where $C$ is a constant, but I don't know how to find the constant $C$. Could anyone tell me that, and explain why? Or could anyone find other methods to evaluate the integral? If you could, please explain. Thanks.
|
Letting $y=\frac{x}{b} $ yields
$$
\begin{aligned}
I &=b \int_0^{\infty} \frac{\ln \left(a^2+b^2 y^2\right)}{b^2+b^2 y^2} d y \\
&=\frac{1}{b} \int_0^{\infty} \frac{\ln \left(b^2 y^2+a^2\right)}{1+y^2} d y
\end{aligned}
$$
By my post,
$$
\begin{aligned}
\boxed{I =\frac{1}{b} \pi \ln \left(\sqrt{a^2}+\sqrt{b^2}\right)=\frac{\pi}{b} \ln (a+b)}
\end{aligned}
$$
|
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|
Prove that if $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$ Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.
|
Product of 5 consecutive integer numbers is of course divisible by $5$, so $5|(n-2)(n-1)n(n+1)(n+2)$. If $5\not|n,$ then from primality of $5$ we have
\begin{align*}
5|(n-2)(n-1)(n+1)(n+2) & = (n^2-1)(n^2-4)\\
& = n^4 - 5n^2 +4\\
& = n^4 - 1 - 5(n^2 - 1)
\end{align*}
so $5|n^4-1$.
|
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|
If $a+b+c=1$, then $\frac{2\sqrt{abc}}{a+bc}+\frac{2\sqrt{abc}}{b+ca}+\frac{ab-c}{ab+c} \leq \frac{3}{2}$ Someone can to help me with a hint in the following problem:
Let $a$, $b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove that:
$$\frac{2\sqrt{abc}}{a+bc}+\frac{2\sqrt{abc}}{b+ca}+\frac{ab-c}{ab+c} \leq \frac{3}{2}$$
|
Homogenising, we have to show:
$$\frac{2\sqrt{abc(a+b+c)}}{a(a+b+c)+bc} + \frac{2\sqrt{abc(a+b+c)}}{b(a+b+c)+ca}+\frac{ab-c(a+b+c)}{ab+c(a+b+c)} \le \frac32$$
Now it motivates the triangle substitution $x=a+b, y=b+c, z = c+a$, so we have a triangle with area say $\Delta$, to get the equivalent:
$$\frac{\Delta}{2xz}+\frac{\Delta}{2xy}+\frac{x^2-y^2-z^2}{2yz} \le \frac32$$
So with $X, Y, Z$ angles of some triangle, this is equivalent to the inequality:
$$\sin Y + \sin Z - \cos X =\sin Y + \sin Z + \cos (Y+Z) \le \frac32$$
It is sufficient to show that the unconstrained maximum of $f(A, B) = \sin A + \sin B + \cos(A+B)$ is $\frac32$. This is easy to check as we have maximum when
$$\frac{\partial f}{\partial A} = \frac{\partial f}{\partial B} =0 \implies A=B = \sin^{-1} \tfrac12 = \frac{\pi}6$$
|
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|
If $a^2=b^2+c^2$ and $0If $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, prove
(a) if $n>2$ then $a^n>b^n+c^n$,
(b) if $0<n<2$ then $a^n<b^n+c^n$.
Part (a) was easy to prove: $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, so $a>b$ and $a>c$. Then O can show: $b^n+c^n=b^2(b^{n-2})+c^2(c^{n-2})<b^2(a^{n-2})+c^2(a^{n-2})=a^{n-2}(b^2+c^2)=a^{n-2}\times a^2=a^n \implies b^n+c^n<a^n$
But I stuck in part (b). Can I prove it in a way I proved part (a)?
|
$$a^2+b^2=c^2\iff \left(\frac ac\right)^2=1-\left(\frac bc\right)^2\le1$$
$$\implies\left(\frac ac\right)^n<\left(\frac ac\right)^2$$ for $n>2$
|
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|
Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$. Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$. Then write an explicit expression for (a) $J_\frac{1}{2}(x)-J_\frac{5}{2}(x)$, and (b) an expression for $J_1+2J_3$ in terms of $J_0$.
Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$.
Remember that
$J_k(z)=c_k z^k \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz$ where $c_k=\frac{1}{2^k \Gamma(k+\frac{1}{2})\Gamma(\frac{1}{2})}$
$J_{k-1}(z)=c_{k-1} z^{k-1} \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{3}{2}}dz$ where $c_{k-1}=\frac{1}{2^{k-1} \Gamma(k-\frac{1}{2})\Gamma(\frac{1}{2})}$
$$\frac{d}{dz}(z^kJ_k(z)) =\frac{d}{dz}(z^k c_k z^k \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}\,dz) = c_k \frac{d}{dz}(z^{2k} \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}\,dz) = c_k [2kz^{2k-1}\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz)+ z^{2k} (e^{izs}(1-z^2)^{k-\frac{1}{2}}dz)]_{-1}^1] $$
But $e^{izs}(1-z^2)^{k-\frac{1}{2}}dz)]_{-1}^1=0$. So
$$\frac{d}{dz}(z^kJ_k(z)) =c_k 2kz^{2k-1}\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz) =\frac{1}{2^k \Gamma(k+\frac{1}{2})\Gamma(\frac{1}{2})}2kz^{2k-1}\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz =\frac{1}{2^k-1 (k-\frac{1}{2})\Gamma(k-\frac{1}{2})\Gamma(\frac{1}{2})}kz^{k-1}z^k\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz =z^k c_{k-1}kz^{k-1}\int_{-1}^1 \frac{e^{izs}(1-z^2)^{k-\frac{1}{2}}}{k-\frac{1}{2}}dz$$
How do I contunie from here?
Write an explicit expression for $J_\frac{1}{2}(x)-J_\frac{5}{2}(x)$
We know that $J_\frac{1}{2}(x)=\sqrt{\frac{2}{\pi}}\frac{\sin(x)}{\sqrt{x}}$ when $c_\frac{1}{2}=\frac{1}{\sqrt{2}\Gamma(1)\Gamma(\frac{1}{2})}$. By the formula in the previous part, we now that
$$\frac{d}{dx}(x^kJ_k(x))=x^kJ_{k-1}(x) \iff x^kJ_k(x)=\int_\textbf{R} x^kJ_{k-1}(x)dx \iff J_k(x)=\int_\textbf{R} J_{k-1}(x)dz$$
$$J_\frac{3}{2}(z) =\int_\textbf{R}J_\frac{1}{2}(x) \,dx= \sqrt{\frac{2}{\pi}}\int_\textbf{R}\frac{\sin(x)}{\sqrt{x}} dx =\sqrt{\frac{2}{\pi}}\int_\textbf{R}\frac{\sin(x)}{\sqrt{x}} dx=(1+i)\frac{\pi}{2}$$
$$J_\frac{5}{2}(x) =\int_\textbf{R}J_\frac{3}{2}(x) \,dx =\int_\textbf{R}(1+i){\frac{\pi}{2}} \,dx $$
Does $J_\frac{5}{2}(x)=\infty$?
Write an explicit expression for $J_1+2J_3$ in terms of $J_0$
By the formula in the previous part,
$$J_1(x) =\int_\textbf{R} J_0(x)\,dx$$
$$J_3(x) =\int_\textbf{R}J_2(x)=\int_{\textbf{R}^2} J_1(x)\,dx^2=\int_{\textbf{R}^3} J_0(x)\,dx^3 $$
So $J_1+2J_3=\int_\textbf{R} J_0(x)\,dx+2\int_{\textbf{R}^3} J_0(x)\,dx^3$
Please check
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You have a very bad habit of repeating variables when you should not. This is a grave mistake and causes errors like you have made. I'm going to get a series expansion for the Bessel function because the integrand is fairly difficult to work with, in my opinion. Series are quite nice though-especially with respect to differentiation! By your definition,
$$J_{\nu}(z) = c_{\nu}z^{\nu}\int_{-1}^1 e^{izt}(1-t^2)^{\nu-\frac{1}{2}}\,dt,$$
where $c_{\nu} = \dfrac{1}{\pi^{\frac{1}{2}}2^{\nu}\, \Gamma\left(\nu+\frac{1}{2}\right)}.$ Concentrating on the case where $z$ is real, this becomes
$$J_{\nu}(x) = c_{\nu}x^{\nu}\int_{-1}^1 e^{ixt}(1-t^2)^{\nu-\frac{1}{2}}\,dt.\tag{1}$$
The nonlinearity of this integrand makes it difficult to work with but we can get around this by noting two things: firstly, we needn't consider the imaginary part due to symmetry and by expanding $\cos(xt)$ in a power series. We can interchange integral and summation since the convergence of the power series is uniform.
$$J_{\nu}(x) = c_{\nu}x^{\nu}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\int_{-1}^1 t^{2n}(1-t^2)^{\nu-\frac{1}{2}}\,dt.$$
Making use of evenness,
$$J_{\nu}(x) = 2c_{\nu}x^{\nu}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\int_0^1 t^{2n}(1-t^2)^{\nu-\frac{1}{2}}\,dt.$$
Let's make the change of variable $y = t^2$ then $dy = 2t\,dt$ which gives
$$J_{\nu}(x) = c_{\nu}x^{\nu}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\int_0^1 y^{n-\frac{1}{2}}(1-y)^{\nu-\frac{1}{2}}\,dy.$$
Noting the relationship this integral has to the Gamma function, we get that
$$J_{\nu}(x) = c_{\nu}x^{\nu}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\frac{\Gamma\left(n+\frac{1}{2}\right)\Gamma\left(\nu+\frac{1}{2}\right)}{\Gamma(n+\nu+1)}.$$
We almost have the right expression at our disposal now so that we can easily analyze these problems. We need only to simplify $\Gamma\left(n+\frac{1}{2}\right)$. This can be rewritten as
$$\Gamma\left(n+\frac{1}{2}\right) = \frac{(2n)!\pi^{\frac{1}{2}}}{4^nn!}.$$
(See here for more information.) Thus we get a nice expression for $J_{\nu}$:
$$J_{\nu}(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n+\nu}\Gamma(n+\nu+1)n!}x^{2n+\nu}.$$
This series converges for all $x$ and so we can differentiate it term-by-term. If $\nu=m$, multiplying by $x^m$ gives
$$(x^mJ_m(x))' = \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n+m}\Gamma(n+m+1)n!}(x^{2n+2m})'.$$
$(x^{2n+2m})' = (2n+2m)x^{2n+2m-1} = 2(n+m)x^m(x^{2n+(m-1)})$. This gives
$$(x^mJ_m(x))' = x^m\sum_{n=0}^{\infty}\frac{(-1)^n2(n+m)}{2^{2n+m}\Gamma(n+m+1)n!}x^{2n+(m-1)}.$$
Making use of the fact that $\Gamma(x+1) = x\Gamma(x)$, we get that
$$(x^mJ_m(x))' = x^m\sum_{n=0}^{\infty}\frac{(-1)^n(n+m)}{2^{2n+m-1}(n+m)\Gamma(n+m)n!}x^{2n+(m-1)} = x^m J_{m-1}(x).$$
You don't have to restrict to the real case if you are not so inclined. The work above is quite general but easiest to frame in the real-line case.
With this in mind, we have $J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}}\sin x.$
Therefore by the above,
$$\left(x^{\frac{3}{2}}J_{\frac{3}{2}}(x)\right)' = x^{\frac{3}{2}}J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi}}x\sin x.$$
Integrating both sides gives
$$x^{\frac{3}{2}} J_{\frac{3}{2}}(x) = \sqrt{\frac{2}{\pi}} \int x\sin x\,dx = \sqrt{\frac{2}{\pi}}(\sin x-x\cos x)+C.$$
Setting $x=0$ gives $C=0$ and so
$$J_{\frac{3}{2}}(x) = \sqrt{\frac{2}{\pi x^3}}(\sin x-x\cos x).$$
We employ a very similar procedure for computing $J_{\frac{5}{2}}$:
$$\left(x^{\frac{5}{2}} J_{\frac{5}{2}}(x)\right)' = x^{\frac{5}{2}} J_{\frac{3}{2}}(x).$$
Integrating both sides gives
$$x^{\frac{5}{2}} J_{\frac{5}{2}}(x) = \sqrt{\frac{2}{\pi}}\int x(\sin x-x\cos x) = \sqrt{\frac{2}{\pi}} (-x^2\sin x -3x\cos x+3\sin x)+C.$$
Again, setting $x=0$ gives $C=0$ and so
$$J_{\frac{5}{2}}(x) = \sqrt{\frac{2}{\pi x^5}}(-x^2\sin x - 3x\cos x+3\sin x).$$
I trust that you can take it from here as the remaining case (part (b)) is quite similar in nature.
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"url": "https://math.stackexchange.com/questions/876133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there also an other way to show the equality: $\left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$? I want to show that:
$$ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$$
That's what I have tried:
*
*$ \left\lfloor \frac{n}{2}\right\rfloor=\max \{ m \in \mathbb{Z}: m \leq \frac{n}{2}\}$
*$\left\lceil \frac{n}{2} \right\rceil= \min \{ m \in \mathbb{Z}: m \geq \frac{n}{2}\}$
If $n=2k,k \in \mathbb{Z}$,then: $\frac{n}{2} \mathbb{Z}$,so
$$\left\lfloor \frac{n}{2}\right\rfloor=\max \left\{ \frac{n}{2}, \frac{n-2}{2}, \dots \right\}=\frac{n}{2} \\ \left\lceil \frac{n}{2} \right\rceil=\min \left\{ \frac{n}{2}, \frac{n+2}{2}, \dots \right\}=\frac{n}{2}$$
Therefore, $ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$.
If $n=2k+1, k \in \mathbb{Z}$,then $\frac{n}{2} \notin \mathbb{Z}$.So:
$$\left\lfloor \frac{n}{2}\right\rfloor=\max \left\{ \frac{n-1}{2}, \frac{n-3}{2}, \dots \right\}=\frac{n-1}{2}\\ \left\lceil \frac{n}{2} \right\rceil=\min \left\{ \frac{n+1}{2}, \frac{n+3}{2}, \dots \right\}=\frac{n+1}{2}$$
Therefore, $ \lfloor \frac{n}{2}\rfloor + \lceil \frac{n}{2} \rceil=\frac{n-1}{2}+\frac{n+1}{2}=n$
Is there also an other way to show the equality or is it the only one?
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Let $x,y$ be any two real numbers. Then $\lfloor x\rfloor+\lceil y\rceil>x+y-1$, since $\lfloor x\rfloor>x-1$ and $\lceil y\rceil\geq y$. Similarly $\lfloor x\rfloor+\lceil y\rceil<x+y+1$. So $\lfloor x\rfloor+\lceil y\rceil$ differs from $x+y$ by strictly less than $1$. Since $\lfloor x\rfloor+\lceil y\rceil$ is always an integer, if $x+y$ is also an integer the two must be equal. This holds in particular for $x=y=n/2$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/878854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Polar form of a complex number Question:
Write the polar form of $$\frac{(1+i)^{13}}{(1-i)^7}$$
Well its obviously impractical to expand it and try and solve it.
Multiplying the denominator by $(1+i)^7$ will simplify the denominator, and a single term in the numerator.
Answer I got:
$$(\frac{1}{\sqrt2}(cos(\frac{\pi}{4}) + sin(\frac{\pi}{4})i)^{20}$$
Is this correct?
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You can also convert numerator and denominator into polar form immediately to write
$$ \frac{ [ \ \sqrt{2} \ cis(\frac{\pi}{4}) \ ]^{13} \ }{[ \ \sqrt{2} \ cis(-\frac{\pi}{4}) \ ]^7} \ \ . $$
DeMoivre's Theorem for powers gives us
$$ = \ \frac{ (\sqrt{2})^{13} \ cis(\frac{13\pi}{4}) }{(\sqrt{2})^7 \ cis(-\frac{7\pi}{4})} \ \ . $$
Division of complex numbers in polar form then produces
$$ = \ \frac{ (\sqrt{2})^{13} \ }{(\sqrt{2})^7} \ cis( \ \left[\frac{13\pi}{4} \right] \ - \ \left[-\frac{7\pi}{4} \right] \ ) \ \ . $$
You would simplify things from there. (Since the answer's already been posted, I'll finish this off:
$$ = \ 2^{6/2} \ cis \left( \frac{20 \pi}{4} \right) \ = \ 2^3 \ cis(5 \pi) \ = \ 8 \ cis \ \pi \ \ \text{or} \ \ -8 \ \ . ) $$
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"language": "en",
"url": "https://math.stackexchange.com/questions/878973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Indefinite integral of trignometric function What is the trick to integrate the following
$$\int \frac{1-\cos x}{(1+\cos x)\cos x}\ dx$$
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Hint :
\begin{align}
\int\frac{1-\cos x}{\cos(1+\cos x)}\ dx&=\int\frac{\color{red}{1+\cos x}-2\cos x}{\cos(\color{red}{1+\cos x})}\ dx\\
&=\int\frac1{\cos x}\ dx-\int\frac{2}{1+\cos x}\ dx\\
&=\int\frac{\cos x}{\cos^2 x}\ dx-\int\frac{2}{1+\cos x}\cdot\frac{1-\cos x}{1-\cos x}\ dx\\
&=\int\frac{d(\sin x)}{1-\sin^2 x}-\int\frac{2-2\cos x}{1-\cos^2 x}\ dx\\
&=\int\frac{d(\sin x)}{2(1-\sin x)}-\int\frac{d(\sin x)}{2(1+\sin x)}-\int\frac{2-2\cos x}{\sin^2 x}\ dx\\
&=\int\frac{d(\sin x)}{2(1-\sin x)}-\int\frac{d(\sin x)}{2(1+\sin x)}-\int\frac{2}{\sin^2 x}\ dx+2\int\frac{\cos x}{\sin^2 x}\ dx\\
&=\int\frac{d(\sin x)}{2(1-\sin x)}-\int\frac{d(\sin x)}{2(1+\sin x)}-2\int\csc^2x\ dx+2\int\frac{d(\sin x)}{\sin^2 x}\\
&=\int\frac{d(\sin x)}{2(1-\sin x)}-\int\frac{d(\sin x)}{2(1+\sin x)}+2\int\ d(\cot x)+2\int\frac{d(\sin x)}{\sin^2 x}.
\end{align}
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/880315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.