Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Why is the transpose of an interpolation matrix the "inverse" interpolation matrix? Imagine I have some values at $x=0,x=2,x=4,\cdots,x=10$. I want to interpolate to find values at $x=1,x=3,x=5,\cdots,x=9$. I do this
$$\left[\begin{matrix}v_1 \\ v_3 \\ v_5 \\ v_7 \\ v_9\end{matrix}\right]=\left[\begin{matrix}\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{matrix}\right]\left[\begin{matrix}v_0 \\ v_2 \\ v_4 \\ v_6 \\ v_8 \\ v_{10}\end{matrix}\right]$$
Now suppose I want to interpolate back from the odd-index points to the even-index points. Aside from some uncertainty about what to do for $v_0,v_{10}$, which are extrapolated now, a sensible way to do this is
$$\left[\begin{matrix}v_0 \\ v_2 \\ v_4 \\ v_6 \\ v_8 \\ v_{10}\end{matrix}\right]=\left[\begin{matrix}\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{matrix}\right]^T \left[\begin{matrix}v_1 \\ v_3 \\ v_5 \\ v_7 \\ v_9\end{matrix}\right]$$
This seems to be true in general, even in 2D or 3D (maybe only for regularly spaced grids): if we have an interpolation matrix from one grid to another, the interpolation back is given by the transpose of that matrix.
Why is this so? Is there a proof of this? Is there some sort of intuition behind the matrix transpose that makes this obvious? Does the transpose approximate some sort of pseudoinverse to make this work?
| To me this looks more like an artifact of the bidiagonal structure of the interpolation matrix than anything suggestive of a deep relationship between the matrix and its transpose. The nonzero entries are all equal and lie on the two central diagonals. The transpose has the same bidiagonal structure, so aside from the first and last rows, you have the same averaging operations in each row that you had in the original matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all positive integers x,y satisfying $ \frac{1}{\sqrt{x}} +\frac{1}{\sqrt{y}} =\frac{1}{\sqrt{20}}$ Find all positive integers $x$,$y$ satisfying $ \frac{1}{\sqrt{x}} +\frac{1}{\sqrt{y}} =\frac{1}{\sqrt{20}}$
$$ \frac{1}{\sqrt{x}} +\frac{1}{\sqrt{y}} =\frac{1}{\sqrt{20}}\\
\frac{1}{\sqrt{x}} +\frac{1}{\sqrt{y}} =\frac{1}{2\sqrt{5}}$$
By hit and trial I found one pair value of $x$ & $y$ i.e $(80, 80)$
But is there any other way to solve this tricky question and find all possible value of $x$ and $y$.
| After squaring we get
$$
\frac{2}{\sqrt{xy}}=\frac{1}{20}-\frac{1}{x}-\frac{1}{y}
$$
which implies $xy=n^2$ is a square. Thus we can rewrite the original equation as
$$
\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{n}=\dfrac{1}{\sqrt{20}}
$$
or, as well,
$$
\left(\frac{1}{x}+\frac{1}{n}\right)\sqrt{x}=\frac{1}{10}\sqrt{5}
$$
Since the right-hand side belongs to $\mathbb{Q}(\sqrt{5})$, the left-hand side should as well, so we conclude that $x=5X$, $y=5Y$ and so $n=5m$.
$$
\frac{1}{\sqrt{X}}+\frac{\sqrt{X}}{m}=\frac{1}{2}
\qquad\text{also written as}\qquad
\left(\frac{1}{X}+\frac{1}{m}\right)\sqrt{X}=\frac{1}{2}
$$
Now this implies $X$ is a square.
If $x\ge y$, then $20<x\le 80$, so $4<X\le16$. This yields only $X=9$ or $X=16$, which correspond to $x=45,y=180$ or $x=80,y=80$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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sum of series $\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots$
Sum of $n$ terms of the series
$$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots \cdots$$
Plan
$$\frac{x}{1-x^2}=\frac{1}{2}\frac{2x}{1-x^2}=\frac{1}{2}\bigg[\frac{1}{1-x}-\frac{1}{1+x}\bigg]$$
$$\frac{x^2}{1-x^4}=\frac{1}{2}\frac{2x^2}{1-x^4}=\frac{1}{2}\bigg[\frac{1}{1-x^2}-\frac{1}{1+x^2}\bigg]$$
Did not get any pattern to convert into Telescopic sum
How do i solve it Help me please
| we can say that:
$$a_n=\frac{x^{2^n}}{1-x^{2^{n+1}}}$$
and we want to calculate:
$$S=\sum_{n=0}^\infty a_n$$
1 approach would be to try and use generating functions from the fact that:
$$S_{n+1}=S_n+\frac{x^{2^{n+1}}}{1-x^{2^{n+2}}}$$
Now you just need to evaluate this for whatever value of $n$ you need
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $x+ y+ z= 3,\,x^{\,2}+ y^{\,2}+ z^{\,2}= 9\,\therefore\,y- x\leqq 2\sqrt{3}$ . Prove
$$x+ y+ z= 3,\,x^{\,2}+ y^{\,2}+ z^{\,2}= 9\,\therefore\,y- x\leqq 2\sqrt{3}$$
I have a solution, and I'm looking forward to seeing a nicer one(s), thanks for your interests a lot !
We have
$$(\,x+ y+ z\,)^{\,2}+ (\,-\,x+ y+ z\,)^{\,2}+ (\,x- y+ z\,)^{\,2}+ (\,x+ y- z\,)^{\,2}= 4(\,x^{\,2}+ y^{\,2}+ z^{\,2}\,)= 36$$
Or
$$\left ( z+ (\,y- x\,) \right )^{\,2}+ \left ( z- (\,y- x\,) \right )^{\,2}+ (\,3- 2\,z\,)^{\,2}= 27$$
Or
$$3\,z^{\,2}- 6\,z+ (\,y- x\,)^{\,2}= 9$$
Or
$$(\,y- x\,)^{\,2}= -\,3(\,z- 1\,)^{\,2}+ 12\leqq 12\,\therefore\,y- x\leqq |\,y- x\,|\leqq 2\sqrt{3}$$
Q.E.D. The equality condition occurs when $z= 1\,\therefore\,x+ y= 2\,\therefore\,x= 1- \sqrt{3},\,y= 1+ \sqrt{3}$.
Say it (Added)
The @user10354138's solution is so amazing, I try writing the inequality into the homonogeous form, then find $t\!=\!constant$ such that $3(\!y- z\!)^{\!2}\leqq 2t(\!x+ y+ z\!)^{\!2}+ 2(\!1- t\!)(\!x^{\!2}+ y^{\!2}+ z^{\!2}\!)$. That will lead to:
$${\rm discriminant}= 0\,\therefore\,t= -\,2,\,-\,\frac{1}{2},\,1$$
The coefficients of $y^{2}$ and $z^{2}$ both are negative there, I can't make the form like the solution as follow !
| Since $\frac1{\sqrt3}(1,1,1), \frac1{\sqrt2}(-1,1,0), \frac1{\sqrt6}(1,1,-2)$ forms an orthonormal basis of $\mathbb{R}^3$ with respect to the usual inner product, we have
$$
x^2+y^2+z^2=\frac13(x+y+z)^2+\frac12(y-x)^2+\frac16(x+y-2z)^2
$$
So
$$
\frac12(y-x)^2\leq (x^2+y^2+z^2)-\frac13(x+y+z)^2=9-3=6
$$
and hence $y-x\leq 2\sqrt3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $\cos (\theta)$ and $\sin (\theta)$, find $2\theta$ I am working on my scholarship exam. I worked through almost final step but got my answer wrong. Could you please have a look?
If $\cos (\theta) = \sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}}$ and $\sin
(\theta) = -\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}$ with
$0\leq\theta<2\pi$, it follows that $2\theta = ..... \pi$
What I have got is below:
$\sin(2\theta)=2\sin\theta\cos\theta$
Then, $\sin(2\theta)=-\frac{1}{\sqrt{2}}$
Hence, $2\theta = \frac{5\pi}{4}$ or $\frac{7\pi}{4}$ (quadrant 3 or 4)
$\theta=\frac{5\pi}{8}$ or $\frac{7\pi}{8}$
Since $\cos\theta$ is positive and $\sin\theta$ is negative, $\theta$ should be in quadrant 4 but my $\theta$'s are not. So I cannot use my $2\theta$ as a final answer.
However, the answer key provided is $\theta=\frac{15\pi}{4}$, Why do you think that is the case? How did they get to this answer? Please help.
| Remember that if the range of $\theta$ is $0 \leq \theta \lt 2 \pi$ then the range of $2\theta$ will be $0\leq\theta \lt 4\pi$. So $2\theta = \ldots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\frac {1+\frac {2^2}{2!} +\frac {2^4}{3!}+\frac {2^6}{4!} +\dots}{1+\frac {1}{2!}+\frac {2}{3!}+\frac {2^2}{4!}+\dots}$ Evaluate the given series
$$\dfrac {1+\dfrac {2^2}{2!} +\dfrac {2^4}{3!}+\dfrac {2^6}{4!} +....}{1+\dfrac {1}{2!}+\dfrac {2}{3!}+\dfrac {2^2}{4!}+....}$$
If we factor out $\dfrac {1}{2^2}$ from the numerator we are left with
$$\dfrac {2^2}{1!}+\dfrac {(2^2)^{2}}{2!} + \dfrac {(2^2)^{3}}{3!}+.....$$ which is equal to $(e^2)^{2} -1$. But I couldn't manipulate the denominator.
| Series in numerator is the expansion of $e^x-1$ at $x=4$ and series denomirator is the expansion of $e^x-1$ at $x=2$. Substituting the values we get $$\frac{e^4-1}{e^2-1}$$
which on simplification gives $${e^2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Systems of equations involving linear and quadratic terms Can we solve for $y$ in this system using algebra?
$$\left\{
\begin{aligned}
x^2 - yz &= 3 \\
y^2 - xz &= 4 \\
z^2 - xy &= 5
\end{aligned}
\right.$$
I’ve tried to evaluate it using elimination and it just gives another equation with unknowns.
First I've tried to multiply the first equation by $y$, second by $z$ and third by $x$. I get $x^2 - y^2z = 3y, y^2z - xz^2 = 4z,$ and $z^2x - x^2y=5x$. Simplifying I get $5x + 4z + 3y = 0$. I've tried it again by multiplying the 1st and 3rd equation by $z, x$ and $y$ respectively. I get $5y + 4x + 3z = 0$. I don't know where to get my third equation.
| We have
$$\left\{
\begin{aligned}
x^2y - y^2z &= 3y \\
y^2z - xz^2 &= 4z \\
z^2x - x^2y &= 5x
\end{aligned}
\right.
$$ and after summing we obtain:
$$3y+4z+5x=0.$$
Also,
$$
\left\{
\begin{aligned}
x^2z - yz^2 &= 3z \\
y^2x - x^2z &= 4x \\
z^2y - xy^2 &= 5y
\end{aligned}
\right.
$$
and after summing again we obtain:
$$3z+4x+5y=0.$$
The rest is smooth:
From $$5x+3y+4z=0$$ and
$$4x+5y+3z=0$$ we obtain:
$$y=-\frac{x}{11}$$ and $$z=-\frac{13x}{11},$$ which gives
$$x^2-\frac{13x^2}{121}=3$$ and from here $$x=\pm\frac{11}{6}.$$
I got the following answer:
$$\left\{\left(\frac{11}{6},-\frac{1}{6},-\frac{13}{6}\right),\left(-\frac{11}{6},\frac{1}{6},\frac{13}{6}\right)\right\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
} |
Minimize value of the function $a^2+b^2+c^2+2\sqrt{3abc}$ Let $a,b,c$ be the positive real numbers such that $a+b+c=1$. Find Minimize of $$P=a^2+b^2+c^2+2\sqrt{3abc}$$
WA says that $P$ gets only a local minimum. But i think it must be maximum value of $P$.
Then by AM-GM: $$\text{L.H.S}= a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}$$
$$\le a^2+b^2+c^2+2(ab+bc+ca)$$
$$=(a+b+c)^2=1$$
| Also, we can use Shur here.
Indeed, we need to prove that
$$a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}\geq\frac{1}{2}(a+b+c)^2$$ or
$$a^2+b^2+c^2+4\sqrt{3abc(a+b+c)}\geq2(ab+ac+bc).$$
By AM-GM $$4\sqrt{3abc(a+b+c)}\geq4\sqrt{3abc\cdot3\sqrt[3]{abc}}=12\sqrt[3]{a^2b^2c^2}>3\sqrt[3]{a^2b^2c^2}$$ and it's enough to prove that:
$$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}\geq2(ab+ac+bc).$$
Now, let $a^2=x^3$, $b^2=y^3$ and $c^2=z^3$.
Thus, by Schur we obtain:
$$\sum_{cyc}\left(a^2-2ab+\sqrt[3]{a^2b^2c^2}\right)=\sum_{cyc}(x^3-2\sqrt{x^3y^3}+xyz)=$$
$$=\sum_{cyc}(x^3-x^2y-x^2z+xyz)+\sum_{cyc}xy(\sqrt{x}-\sqrt{y})^2\geq0$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the solutions of the next congruence using Chinese Remainder Theorem Find the solutions of the congruence using Chinese Remainder Theorem:
$2x^2 - 3x -2 \equiv 0\mod21$
By now I've done this:
$$
\left\{
\begin{array}{c}
2x^2-3x-2\equiv 0\mod 7\\2x^2-3x-2\equiv 0\mod3
\end{array}
\right.
$$
$$
2x^2-3x-2\equiv 0\mod 7 \\ \Delta = 25 = 5^2 \\x_1,x_2 = (2a)^{-1}*(-b \pm \sqrt\Delta) \\ x_1=4^{-1}*2 = 2*2 = 4 \mod 7 \\ x_2 = 4^{-1}*(-8)= 6*2 = 5 \mod 7
$$
Then I made $\Delta$ for the (mod 3) equation and got $x_3=1 \mod 3, x_4=2 \mod 3$
So I got:
$$
x \equiv 1 \mod 3 \\ x \equiv 2 \mod 3 \\ x \equiv 4 \mod 7 \\ x \equiv 5 \mod 7 \\
$$
And I don't know what to do further. I should apply Chinese Remainder Theorem on it or it would be wrong?
Edit: Thanks for the help! The correct form was:
$$
x \equiv 1 \mod 3 \\ x \equiv 2 \mod 3 \\ x \equiv 2 \mod 7 \\ x \equiv 3 \mod 7
$$
And I understand how to solve it.
| As suggested by saulspatz in a comment to the question,
we have $(x-2)(2x+1)\equiv0\pmod{21}$.
That means $x\equiv 2\pmod{21}$ or $x\equiv-\dfrac12\equiv10\pmod{21}$
or $x\equiv2\pmod7$ and $x\equiv-\dfrac12\equiv1\pmod3$ [which means $x\equiv16\pmod{21}$]
or $x\equiv-\dfrac12\equiv3\pmod7$ and $x\equiv2\pmod3$ [which means $x\equiv17\pmod{21}].$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Number of ordered pairs of integers $(x,y)$ satisfying the equation $x^2+6x+y^2=4$
Find the number of ordered pairs of integers $(x,y)$ satisfying the equation $x^2+6x+y^2=4.$
My attempt:
$x^2+6x+y^2=4$
$x^2+6x+9-9+y^2-4=0$
$(x+3)^2+y^2-13=0$
$(x+3)^2=13-y^2$
$x$ is required to be an integer. Therefore, let us consider $x$ as an integer. Therefore $(x+3)$ is also an integer. Similarly $y$ is an integer.
Now, the square of any integer is a non-negative integer and more specifically a perfect square.
Therefore, $(x+3)^2$ and is a perfect square
$\implies 13-y^2$ is a perfect square
$\implies y=-3,+3,-2,+2$ (By trial and error method)
For $y=-3, +3$, there are two values of $x$ which are $x=-1,-5$
For $y=-2,+2$, there are two values of $x$ which are $x=0,-6$
Hence there are eight ordered pairs in total: $(-1,-3),(-5,-3),(-1,+3),(-5,+3),(0,-2),(-6,-2),(0,+2),(-6,+2)$.
Therefore the number of ordered pairs of integers satifying the equation $x^2+6x+y^2=4$ is $8$.
My problem:
Is my method correct? Is there any other method to solve this problem?
| This is exactly how I would do it, and (shouting out to uday) I am far beyond middle school. Completing the square is a powerful tool for quadratic equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Maximize $(x-1)2y$ subject to $x^3+y^2=3$ I have to do with lagrange multiplier
f = $(x-1)2y + \lambda (x^3+y^2-3) = 0 $
I get
$f_x= y + \lambda x = 0 $
$f_y = y\lambda + x - 1 = 0 $
$x^3+y^2=3$
and
how to solve this
| Hint.
Solving for $x, y$
$$
\left\{
\begin{array}{rcl}
3 \lambda x^2+2 y&=&0 \\
2 (x-1)+2 \lambda y&=&0 \\
\end{array}
\right.
$$
we get
$$
x = \frac{1\pm\sqrt{1-6 \lambda ^2}}{3 \lambda ^2}\\
y = \frac{3\pm\frac{\sqrt{1-6 \lambda ^2}}{\lambda^2}-\frac{1}{\lambda ^2}}{3 \lambda }
$$
substituting those values into $x^3+y^2 = 3$
$$
\left\{
\begin{array}{c}
\frac{\left(1-\sqrt{1-6 \lambda ^2}\right)^3}{27 \lambda ^6}+\frac{\left(\frac{\sqrt{1-6 \lambda ^2}}{\lambda
^2}-\frac{1}{\lambda ^2}+3\right)^2}{9 \lambda ^2}-3=0 \\
\frac{\left(\sqrt{1-6 \lambda ^2}+1\right)^3}{27 \lambda ^6}+\frac{\left(-\frac{\sqrt{1-6 \lambda ^2}}{\lambda
^2}-\frac{1}{\lambda ^2}+3\right)^2}{9 \lambda ^2}-3=0 \\
\end{array}
\right.
$$
after symplifying and solving we obtain
$$
\lambda^* = \pm 0.34548248
$$
so we get
$$
\left[
\begin{array}{cc}
x^*& y^*\\
1.30482 & -0.882306 \\
1.30482 & 0.882306 \\
4.28061 & -9.49573 \\
4.28061 & 9.49573 \\
\end{array}
\right]
$$
as the stationary points.
| {
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"url": "https://math.stackexchange.com/questions/3273512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Finding constants in partial fraction In an example for partial fractions we want to find $A$, $B$, $C$, $D$ and $E$ in the expression:
$$
\frac{x^4-x^3+2x^2-x+2}{(x-1)(x^2+2)^2} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+2)} + \frac{Dx+E}{(x^2+2)^2}
$$
Multiplying through to clear the fractions I obtained:
$$x^4-x^3+2x^2-x+2 = A(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)$$
I found $A=\frac{1}{3}$ by letting $x=1$.
Now in the book they let me know that $B=\frac{2}{3}$, $C=-\frac{1}{3}$, $D=-1$ and $E=0$. But I would really like to figure out how I can find the values for $B, C, D, E$.
| Put $x=1$,
$3=9A$ $\implies$ $A=\dfrac13$.
\begin{align*}
x^4-x^3+2x^2-x+2 &= \frac13(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)\\
\frac23x^4-x^3+\frac23x^2-x+\frac23&= (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)\\
\frac23x^3-\frac13x^2+\frac13x-\frac23&= (Bx+C)(x^2+2) + Dx+E\\
\end{align*}
By division, $\dfrac23x^3-\dfrac13x^2+\dfrac13x-\dfrac23=(x^2+2)\left(\dfrac23 x-\dfrac13\right)-x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can we conclude $0^\circMy attempt
Based on the sine rule and the graph of $\sin A = k a$ (where $k$ is a constant) in interval $(0,\pi)$,
increasing $a$ up to $1/k$ will either
*
*increase $A$ up to $90^\circ$.
*decrease $A$ up to $90^\circ$.
So I cannot conclude that increasing $a$ will increase $A$.
Now I use the cosine rule (it is promising because the cosine is decreasing in the given interval).
\begin{align}
A &= \cos^{-1}\left(\frac{b^2+c^2-a^2}{2bc}\right)\\
B &= \cos^{-1}\left(\frac{a^2+c^2-b^2}{2ac}\right)\\
C &= \cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)\\
\end{align}
It is hard to show that $0^\circ<A\leq B\leq C<180^\circ$ for any $\triangle ABC$ with $0<a\leq b\leq c$. Could you show it?
It means that I need to show that
$$
-1<\frac{a^2+b^2-c^2}{2ab}\leq \frac{a^2+c^2-b^2}{2ac} \leq \frac{b^2+c^2-a^2}{2bc}<1
$$
for $0<a\leq b\leq c$.
| Consider two cases:
Case 1 -- $C \leq 90^\circ$:
Then also $A$ and $B$ are less than $90^\circ$ so we are in the monotonic increasing range of sine, and
$$
A \leq B \leq C \implies \sin A \leq \sin B \leq \sin C \implies a \leq b \leq c
$$
where the second implication uses the law of sines.
Case 2 -- $C > 90^\circ$:
First we need to show is that $a \leq b$.
It is trivial that $A, B < 90^\circ$ because the sum of the angles is only $180^\circ$. So for just $A$ and $B$ we are again in the monotonic region of sine, thus
$$
A \leq B \implies \sin A \leq \sin B \implies a \leq b
$$
Now we need to show that $b < c$. Here the law of cosines is useful, and $\cos C < 0$.
$$
a^2 + b^2 - 2ab \cos C = c^2 \\
a^2 + b^2 = c^2 + 2ab \cos C < c^2\\
0 < a^2 < c^2 - b^2 \\
(c+b)(c-b) > 0 \\
c > b
$$
Thus $c>b$ and $b\geq a$ so
$$
a \leq b < c
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How does $\int_{0}^{\pi/2} \sin x \ln (\sin x)~ dx $ converge to $\ln(2/e)?$ I can solve this integral,
$$I=\int_{0}^{\pi/2} \sin x \ln (\sin x)~ dx $$
by parts as
$$I=\int_{0}^{\pi/2} \sin x \ln (\sin x) dx=\left (\cos x +\ln \tan \left(\frac{x}{2}\right)- \cos x \ln (\sin x) \right)_{0}^{\pi/2}=-1,$$ as the divergent parts $\pm \infty$ cancel out. But Mathematica gives $I=\ln(2/e).$ How to resolve this?
| I appreciate you may not be seeking this, but you can find the exact value for the integral using the Beta and by extension Gamma Function and it's derivative.
Here we will to address your integral:
\begin{equation}
I = \int_0^{\frac{\pi}{2}} \sin(x)\ln\left|\sin(x) \right|\:dx
\end{equation}
Here we will approach this integral by considering a general form:
\begin{equation}
J_{n,m}(a,b) = \int_0^{\frac{\pi}{2}} \ln^n\left|\sin(x) \right| \ln^m\left|\cos(x)\right|\sin^a(x)\cos^b(x)\:dx
\end{equation}
Where $n,m \in \mathbb{Z}$ with $n,m \geq 0$ and $a,b \in \mathbb{R}$. We observe that $I = J_{1,0}(1,0)$. From here we employ the property:
\begin{align}
\frac{\partial}{\partial c} \left[ \sin^c(x) \right]= \ln\left|\sin(x)\right| \sin^c(x) \Longrightarrow \frac{\partial^n}{\partial c^n} \left[ \sin^c(x) \right]= \ln^n\left|\sin(x)\right| \sin^c(x)
\end{align}
And so:
\begin{equation}
\lim_{c \rightarrow a} \frac{\partial^n}{\partial c^n} \left[ \sin^c(x) \right] = \ln^n\left|\sin(x) \right|\sin^a(x)
\end{equation}
Similarly:
\begin{equation}
\lim_{d \rightarrow b}\frac{\partial^m}{\partial b^m} \left[ \cos^b(x) \right]= \ln^m\left|\cos(x)\right| \cos^b(x)
\end{equation}
Thus,
\begin{align}
J_{n,m}(a,b)&= \int_0^{\frac{\pi}{2}} \ln^n\left|\sin(x) \right| \ln^m\left|\cos(x)\right|\sin^a(x)\cos^b(x)\:dx \nonumber \\
& = \int_0^{\frac{\pi}{2}} \lim_{c \rightarrow a} \frac{\partial^n}{\partial c^n} \left[ \sin^c(x) \right] \lim_{d \rightarrow b}\frac{\partial^m}{\partial b^m} \left[ \cos^b(x) \right]\:dx
\end{align}
Employing Leibniz's Integral Rule and the Dominated Convergence Theorem, we can draw the partial derivatives and limits outside of the integrand:
\begin{align}
J_{n,m}(a,b) &= \int_0^{\frac{\pi}{2}}\lim_{c \rightarrow a} \frac{\partial^n}{\partial c^n} \left[ \sin^c(x) \right] \lim_{d \rightarrow b}\frac{\partial^m}{\partial b^m} \left[ \cos^b(x) \right]\:dx \nonumber \\
& = \lim_{c \rightarrow a} \lim_{d \rightarrow b} \frac{\partial^n}{\partial c^n} \frac{\partial^m}{\partial d^m} \int_0^{\frac{\pi}{2}}\sin^c(x)\cos^d(x)\:dx \nonumber \\
&= \lim_{(c,d)\rightarrow (a,b)}\frac{\partial^{n+m}}{\partial c^n \partial d^m} \left[ \frac{1}{2}B\left(\frac{c + 1}{2}, \frac{d + 1}{2} \right) \right]
\end{align}
Where $B(x,y)$ is the Beta Function. Employing the relationship between the Beta and Gamma Function this becomes:
\begin{align}
J_{n,m}(a,b) &= \frac{1}{2} \lim_{(c,d)\rightarrow (a,b)}\frac{\partial^{n+m}}{\partial c^n \partial d^m} \left[ \frac{\Gamma\left( \frac{c + 1}{2}\right)\Gamma\left( \frac{d + 1}{2}\right)}{\Gamma\left( \frac{c + d}{2} + 1\right)} \right]
\end{align}
Returning to your integral $I$:
\begin{align}
I &= J_{1,0}(1,0) = \frac{1}{2}\lim_{(c,d)\rightarrow (1,0)} \frac{\partial}{\partial c} \left[ \frac{\Gamma\left( \frac{c + 1}{2}\right)\Gamma\left( \frac{d + 1}{2}\right)}{\Gamma\left( \frac{c + d}{2} + 1\right)} \right] \nonumber \\
&= \frac{1}{2}\lim_{(c,d)\rightarrow (1,0)} \Gamma\left(\frac{d + 1}{2}\right)\left[\frac{\Gamma^{'}\left( \frac{c + 1}{2}\right)\cdot \frac{1}{2}}{\Gamma\left( \frac{c + d}{2} + 1\right)} -\frac{\Gamma\left( \frac{c + 1}{2}\right) \Gamma^{'}\left( \frac{c + d}{2} + 1\right) \cdot \frac{1}{2}}{\Gamma^2\left( \frac{c + d}{2} + 1\right)} \right] \nonumber \\
&= \frac{1}{4}\Gamma\left( \frac{1}{2}\right)\left[ \frac{\Gamma^{'}\left(\frac{1}{2} \right)}{\Gamma\left(\frac{3}{2}\right)} - \frac{\Gamma\left(\frac{1}{2}\right)\Gamma^{'}\left(\frac{3}{2}\right)}{\Gamma^2\left(\frac{3}{2}\right)} \right]
\end{align}
Now:
\begin{equation}
\Gamma\left(\frac{1}{2} \right) = \sqrt{\pi}, \quad \Gamma\left(\frac{3}{2} \right) = \frac{\sqrt{\pi}}{2} \quad \Gamma^{'}\left(\frac{1}{2} \right) = \sqrt{\pi}\left( - \gamma - 2\ln(2)\right),\quad \Gamma^{'}\left(\frac{3}{2} \right) = \frac{\sqrt{\pi}}{2}\left(2 - \gamma - 2\ln(2)\right)
\end{equation}
Thus,
\begin{align}
I & \frac{1}{4} \cdot \sqrt{\pi}\left[\frac{\sqrt{\pi}\left( - \gamma - 2\ln(2)\right)}{\frac{\sqrt{\pi}}{2}} - \frac{\sqrt{\pi} \cdot \frac{\sqrt{\pi}}{2}\left(2 - \gamma - 2\ln(2)\right)}{\left(\frac{\sqrt{\pi}}{2} \right)^2}\right] \nonumber \\
& = \ln(2) - 1 = \ln\left(\frac{2}{e}\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Expressing the area of the cross section of a cylinder I was able to solve similar questions where a triangle was involved, but this one has had me looping and pondering on it for the past week.
The Condition:
Let $r$ be a positive constant. Consider the cylinder $x^2+y^2\leq r^2$ and let C be part of the cylinder that satisfies $0\leq z\leq y$.
The questions:
(1) Consider the cross section of $C$ by the plane $x=t(-r\leq t\leq r)$ and express its area in terms of $r,t$
(2) Calculate the volume of $C$, and express it in terms of $r$.
(3) Let $a$ be the length of the arc along the base of the circle $C$ from the point $(r,0,0)$ to the point $(r cos\theta,r sin\theta,0)(0\leq\theta\leq\pi)$. Let $b$ be the length of the line segment from the point $(r cos\theta, r sin\theta, 0)$ to the point $(r cos\theta, r sin\theta, r sin\theta)$. Express $a$ and $b$ interms of $r,\theta$
(4)Calculate the are of the side of $C$ with $x^2+y^2=r^2$, and express it in terms of $r$.
Here are the expected answers(which I keep failing to get to):
(1) $\frac{1}{2}(r^2-t^2)$
(2) $\frac{2}{3}r^3$
(3) $r\theta$ and $r sin\theta$
(4) $2r^2$
|
*
*Required area is of triangular form, whose base length is $|y_0 -0|$, where $y_0$ is the coordinate on y-axis at the intersection of plane $y = z$ and $x^2 + y^2 = r^2$, and height is corresponding $z-$ coordinate, say $z_0$ , since $x = t \Rightarrow y = \sqrt{r^2 - t^2} = z$ $$ \Rightarrow A = \cfrac{y_0 z_0}{2} = \cfrac{r^2 - t^2}{2}$$
2.Since, $$V = \iiint dx dy dz$$ And the limits are following: $-\sqrt{r^2 - y^2} \le x \le \sqrt{r^2 - y^2}, 0 \le z \le y$ and $ 0 \le y \le r$, which on evaluating gives $$ V = \cfrac{2}{3} r^3 $$
*Use simple distance formula,
For first part, Simple arc length would suffice, $$a = 2 \pi r \cfrac{\theta}{2 \pi} = r \theta$$
and for second part use simple distance formula, $$b = \sqrt{ ( r \cos \theta - r \cos \theta)^2 + ( r \sin \theta - r \sin \theta)^2 + ( r \sin \theta - 0)^2 } = r \sin \theta$$
4.Required area is enclosed by two curve and is lying on the cylinder:
$dA$ = Infinitesimal arc length $\times$ height at that point
Here, Infinitesmial arc length $ = r d \theta$ , hieght $ r \sin \theta$
$$\Rightarrow dA = r \sin \theta r d\theta$$
$$ \Rightarrow A = \int^{\pi}_{0} r \sin \theta r d\theta = 2r^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that : $\int_0^{1}\frac{1-\sqrt{1-x^{4}}}{x^{2}\sqrt{1-x^4}}dx=1-\frac{\sqrt{2}\pi^{\frac{3}{2}}}{\Gamma(\frac{1}{4})^2}$ Prove that :
$$\int_0^{1}\frac{1-\sqrt{1-x^{4}}}{x^{2}\sqrt{1-x^4}}dx=1-\frac{\sqrt{2}\pi^{\frac{3}{2}}}{\Gamma(\frac{1}{4})^2}.$$
I know how to use the definition of the Beta and Gamma functions so my problem here is when I divide the integral I find:
$$\int_0^{1}\frac{1}{x^2}dx=\infty!!$$
(divergent)
I tried using $y=\frac{1}{x}$ but didn't get the answer.
I also tried $y=x^{4}$ and got the same problem (a divergent integral).
| I begin by making the following observation:
$$\frac{d}{dx} \left (\frac{1 - \sqrt{1 - x^4}}{x} \right )= \frac{x^4 + 1 - \sqrt{1 - x^4}}{x^2 \sqrt{1 - x^4}}.$$
So, for your integral, one can write
\begin{align}
\int_0^1 \frac{1 - \sqrt{1 - x^4}}{x^2 \sqrt{1 - x^4}} \, dx &= \int_0^1 \frac{(x^4 + 1 - \sqrt{1 - x^4}) - x^4}{x^2 \sqrt{1 - x^4}} \, dx\\
&= \int_0^1 \left (\frac{1 - \sqrt{1 - x^4}}{x} \right )' \, dx -\underbrace{\int_0^1 \frac{x^2}{\sqrt{1 - x^4}} \, dx}_{x \, \mapsto \, \sqrt[4]{x}}\\
&= 1 - \frac{1}{4} \int_0^1 x^{-\frac{1}{4}} (1 - x)^{-\frac{1}{2}} \, dx\\[1ex]
&= 1 - \frac{1}{4} \operatorname{B} \left (\frac{3}{4}, \frac{1}{2} \right )\\[1ex]
&= 1 - \frac{1}{4} \frac{\Gamma (\frac{3}{4}) \Gamma (\frac{1}{2})}{\Gamma (\frac{5}{4})}\\
&= 1 - \sqrt{\pi} \frac{\Gamma (\frac{3}{4})}{\Gamma (\frac{1}{4})}\\
&= 1 - \frac{\pi \sqrt{2 \pi}}{\Gamma^2 (\frac{1}{4})},
\end{align}
as required. Here $\Gamma (\frac{3}{4}) \Gamma (\frac{1}{4}) = \pi \sqrt{2}$, has been used.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $a^{2n} -b^{2n}$ is divisible by $a+b$
Prove that for all $n\in\Bbb{Z}$, $a^{2n}-b^{2n}$ is divisible by $a+b$ using induction.
I know that if $a$ is divisible by $b$, then $a=kb$, where $k\in\Bbb{Z}$. Here we have that $a^{2n}-b^{2n}=(a+b)k$, with $k\in\Bbb{Z}$.
For the base case I set $n=1$, so $a^2-b^2=(a+b)(a-b)=(a+b)k$, where $k=a-b\in\Bbb{Z}$.
Now the inductive step (where I have doubts): $$a^{2n}-b^{2n}=(a+b)k\implies a^{2(n+1)}-b^{2(n+1)}=(a+b)m,\;k,m\in\Bbb{Z}.$$ We start from $a^{2(n+1)}-b^{2(n+1)}$. Then $$a^{2n+2}-b^{2n+2}=(a+b)\color{red}{(a^{2n+1}-a^{2n}b+a^{2n-1}b^2-\dots-a^2b^{2n-1}-ab^{2n}+b^{2n+1})},$$ so $a^{2(n+1)}-b^{2(n+1)}=(a+b)m$, where $m=a^{2n+1}-a^{2n}b+a^{2n-1}b^2-\dots-a^2b^{2n-1}-ab^{2n}+b^{2n+1}\in\Bbb{Z}.\qquad\square$
I have two questions:
*
*Is the math in $\color{red}{\text{red}}$ a correct descomposition of $a^{2(n+1)}-b^{2(n+1)}$?
*We have not used the inductive hypothesis. Could we use it?
| Base $n=1$ $$a^2-b^2 = (a-b)(a+b) $$
By induction hypothetis we have $$a^{2n}\equiv b^{2n}\pmod {a+b}$$
From the base case we have also $a^2\equiv b^2 \pmod {a+b}$, so: $$a^{2n+2}= a^2\cdot a^{2n}\equiv b^2 \cdot b^{2n} = b^{2n+2}\pmod {a+b}$$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Integrating the trigonometric function So the problem goes as follows:-
$$\int{{\cos^2x+\sin2x}\over{(2\cos x-\sin x)^2}}~dx$$
My attempt is as follows:-
\begin{align*}\int{{\cos^2x+\sin2x}\over{(2\cos x-\sin x)^2}}~dx&=\int{{\cos^2x+2\sin x\cos x}\over{(2\cos x-\sin x)^2}}~dx\\
&=
\int{{(\cos x)(\cos x+2\sin x)}\over{(2\cos x-\sin x)^2}}~dx
\end{align*}
Now i could see that $\cos x+2\sin x $ is $-1$ times derivative of the denominator:-
$$\cos x+2\sin x=(-1){{d(2\cos x-\sin x)}\over{dx}}$$
And to handle the $\cos x$ term in the numerator i tried the following:-
$$2\cos x=t+\sin x \implies \cos x={{t+\sin x}\over{2}}$$
But the $\sin x$ term in the numerator is annoying.
| Note that
$$\begin{aligned}
\displaystyle \int \frac{\sin \left ( 2x \right ) + \cos^2 \left ( x \right ) }{ \left ( \sin \left ( x \right ) - 2 \cos \left ( x \right ) \right ) ^2} \; \mathrm{d}x
&= \displaystyle \int \frac{2 \tan \left ( x \right ) + 1}{ \left ( \tan \left ( x \right ) - 2 \right ) ^2} \; \mathrm{d}x\\
\end{aligned}$$
by dividing by $\cos^2 \left ( x \right ) $ on numerator and denominator. If we let $u = \tan \left ( x \right ) $, $\mathrm{d}u = \sec^2 \left ( x \right ) \; \mathrm{d}x \implies \mathrm{d}x = \frac{\mathrm{d}u}{u^2+1}$, we get
$$\begin{aligned}
\displaystyle \int \frac{2 \tan \left ( x \right ) + 1}{ \left ( \tan \left ( x \right ) - 2 \right ) ^2} \; \mathrm{d}x
&= \displaystyle \int \frac{2u+1}{ \left ( u - 2 \right ) ^2 \left ( u^2 + 1 \right ) } \; \mathrm{d}u\\
&= \displaystyle \int \frac{2u-1}{5 \left ( u^2 + 1 \right ) } - \frac{2}{5 \left ( u-2 \right ) } + \frac{1}{ \left ( u-2 \right ) ^2} \; \mathrm{d}u\\
&= \frac{1}{5} \ln \left ( u^2 + 1 \right ) - \frac{1}{5} \tan^{-1} \left ( u \right ) - \frac{2}{5} \ln \left ( u - 2 \right ) - \frac{1}{u - 2} + C\\
&= \frac{1}{5} \ln \left ( \sec^2 \left ( x \right ) \right ) - \frac{x}{5} - \frac{2}{5} \ln \left | \tan \left ( x \right ) - 2 \right | - \frac{1}{\tan \left ( x \right ) - 2} + C
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How many even numbers of $3$ digits, between $400$ and $700$ can be formed if repeating digits are not allowed? I have this statement:
How many even numbers of $3$ digits, between $400$ and $700$ can be formed if repeating digits are not allowed?
My attempt was:
In the case that the number start in $4$, so i have $8 * 4 = 32$ ways
In the case that the number start in $5$, so i have $8 * 5 = 40$ ways
In the case that the number start in $6$, so i have $8 * 4 = 32$ ways
In the case that the numbers start in $7$ only can be $700$, so is $1$ way.
Therefore, the total ways are $105$
But the correct answer is $104$ according to the guide.
What is wrong with my development?
| The digits in $700$ are not distinct, so you should not have counted it.
Since the digits of $700$ are not distinct, the leading digit of a number with distinct digits must be 4, 5, or 6.
If the units digit is 4 or 6, there are two choices for the hundreds digit since it must differ from the units digit and eight choices for the tens digit since it must differ from both the units digits and the hundreds digit. Hence, there are $2 \cdot 2 \cdot 8$ such numbers.
If the units digit is 0, 2, or 8, there are three choices for the hundreds digit since 4, 5, 6 are all available and eight choices for the the tens digit since it must differ from both the units digit and the hundreds digit. Thus, there are $3 \cdot 3 \cdot 8$ such numbers.
In total, there are $2 \cdot 2 \cdot 8 + 3 \cdot 3 \cdot 8 = 32 + 72 = 104$ even numbers between $400$ and $700$ with distinct digits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the area between functions $x^3=2y^2;x=0;y=-2$
Take the element of area parallel to the y axis:
$$x^3=2y^2;x=0;y=-2$$
First, I isolated in terms of $y$,
$$y= \pm \sqrt{x^3\over2}
\\ = \pm{\sqrt2\over 2}x^{3 \over 2}$$
Since bounded by $y=-2$, consider the negative portion:
$$= -{\sqrt2\over 2}x^{3 \over 2}$$
and let
$$f(x)= -{\sqrt2\over 2}x^{3 \over 2}$$
$$g(x) = -2$$
since $f(x)$ and $g(x)$ intersect at $x=2$ and bound by $x=0$
So,
$$ \int_0^2 [-g(x) -f(x)]dx
\\ = \int_0^2 (2 + {\sqrt2\over 2}x^{3 \over 2} ) dx
\\ = 2x + {\sqrt2\over 5}x^{5 \over 2}\Bigg]^2_0
\\ = 4 + {\sqrt2\over 5}2^{5 \over 2}$$
However the answer i supposed to be $12\over 5$
And I do not understand what the question means by "The element of area parallel to y axis"
| $f(x)=-\frac{\sqrt2}{2}x^{\frac{3}{2}}$
When intersected with $g(x)=-2$ you get the point $x=2$ as you showed.
So the area between $x=0$ , $y=-2$ and $f(x)$ is the area of $x=0$, $y=2$ and $-f(x)$.
This area is the same as the area of $x=0$ , $x=2$ and $y=2$ minus the area of $-f(x)$ between $0$ and $2$.
So $A=4-\int{f(x)dx}=4-\frac{8}{5}=\frac{12}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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For $0For $0<a<b$, how does $b^n-a^n=(b-a)(b^{n-1}+ab^{b-2}+...+a^{n-1})\implies b^n-a^n<(b-a)nb^{n-1}$?
Clearly, $$\frac{b^n-a^n}{b-a}=b^{n-1}+ab^{n-2}+...+a^{n-1}$$ so I would need to see that $$b^{n-1}+ab^{n-2}+...+a^{n-1}<nb^{n-1}$$
Dividing through by $b^{n-1}$:$$1+\frac{a}{b}+\frac{a^2}{b^2}+...+\frac{a^{n-1}}{b^{n-1}}<n$$
Why should I believe this last inequality even?
| $0 < a < b$ so $a^kb^{n-k-1} < b^{n-1}$.
So $b^{n-1} + ab^{n-2} + ...+ba^{n-2} + a^{n-1} < b^{n-1} + b^{n-1} + ...... + b^{n-1} + ^{n-1} = nb^{n-1}$.
......
$1+\frac{a}{b}+\frac{a^2}{b^2}+...+\frac{a^{n-1}}{b^{n-1}}<n$
"Why should I believe this last inequality even? "
Because $\frac {a^k}{b^k} < 1$ so
$1+\frac{a}{b}+\frac{a^2}{b^2}+...+\frac{a^{n-1}}{b^{n-1}} <1+1+1+......+1 = n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of even numbers N? To sum first N numbers we can use this formula:
$$1 + 2 + 3 + \ldots + n = \frac{(n (n + 1)}{2}$$
To sum even numbers we multiply this formula by 2:
$$2 + 4 + 6 + \ldots + 2n = n (n + 1)$$
Lets check sum of even numbers until 6:
Sum(odd) = 2 + 4 + 6 = 12
Then let's use the formula: 6 * (6 + 1) = 6 * 6 + 6 = 42.
What is wrong with formula or my calculations?
| You use that
$$2+4 + 6+ \ldots + 2n = 2(1 + 2 + 3 + \ldots + n) = 2\cdot \frac{n(n+1)}{2} = n(n+1)$$
So the $n$ used is the number of terms added, not the final term, but half of it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Symmetric matrix (completing the square/diagonal form) Let $A=\begin{pmatrix} 0 & -2 & 1 \\ -2 & 1 & 2 \\ 1 & 2 & 0 \end{pmatrix} \in M_3(\mathbb{R})$.
I want to find an invertible matrix $C$ such that $C^TAC=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$, since there are three eigenvalues, two of them are positive, one is negative.
I want to use completing the square with the bilinear form $s(v,v)=\langle v,Av \rangle$:
$s(v,v)= \langle v,Av \rangle=(v_1,v_2,v_3)\begin{pmatrix} 0 & -2 & 1 \\ -2 & 1 & 2 \\ 1 & 2 & 0 \end{pmatrix}\begin{pmatrix} v_1\\v_2\\v_3 \end{pmatrix}=-4v_1v_2+2v_1v_3+v_2^2+4v_2v_3$
Now I tried to use the formula $ax^2+bx=a(x+\frac{b}{2a})^2-\frac{b^2}{2a}$, but there is no $v_1^2$.
So I started with $v_2^2$:
$s(v,v)=(v_2+2v_3)^2-4v_1v_2+2v_1v_3-4v_3^2$
Here I don't see how to use completing the square again with $-4v_1v_2+2v_1v_3$, since $v_1^2$ is not available. Is there a method how to continue?
| Perhaps not what you're looking for but we can diagonalize the symmetric matrix $A$ as
$$P^TAP =\begin{pmatrix}
-\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} & 0 \\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\
\end{pmatrix} \begin{pmatrix} 0 & -2 & 1 \\ -2 & 1 & 2 \\ 1 & 2 & 0 \end{pmatrix}\begin{pmatrix}
-\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\
-\frac{1}{\sqrt{6}} & \frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
\end{pmatrix}= \begin{pmatrix} -3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
so for $u = P^Tv$ we have
$$\langle Av,v\rangle = \langle A(Pu),(Pu)\rangle = \langle P^TAPu,u\rangle = -3u_1^2+3u_2^2+u_3^2$$
This turns out to be
\begin{align}
\langle Av,v\rangle &= -3 \left(-\frac{v_1}{\sqrt{3}} - \frac{v_2}{\sqrt{3}} + \frac{v_3}{\sqrt{3}}\right)^2 +
3 \left(-\frac{v_1}{\sqrt{6}} + \sqrt{\frac23}v_2 + \frac{v_3}{\sqrt{6}}\right)^2 +\left(\frac{v_1}{\sqrt{2}} + \frac{v_3}{\sqrt{2}}\right)^2\\
&= \left(-\frac{v_1}{\sqrt{2}} + \sqrt{2}v_2 + \frac{v_3}{\sqrt{2}}\right)^2+ \left(\frac{v_1}{\sqrt{2}} + \frac{v_3}{\sqrt{2}}\right)^2-(-v_1-v_2+v_3)^2\\
\end{align}
| {
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Let $S$ be the subset of $M(\mathbb{R})$ consisting of matrices of the form: The proof I read on slader seemed like it only used either $1$ or $2$ matrices to prove this when the question asks for all matrices of a certain form (when $x+y = 1$).
Let $S$ be the subset of $M(\mathbb{R})$ consisting of matrices of the form: $\begin{pmatrix}a & a\\\ b & b\end{pmatrix}$.
If $x + y = 1$, show that $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ is not a left identity in $S$.
If $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ is a left identity, then we have
$\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$$\begin{pmatrix}0 & 0\\\ 1 & 1\end{pmatrix}$ = $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ = $\begin{pmatrix}0 & 0\\\ 1 & 1\end{pmatrix}$.
But $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$$\begin{pmatrix}1 & 1\\\ 0 & 0\end{pmatrix}$ = $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ = $\begin{pmatrix}1 & 1\\\ 0 & 0\end{pmatrix}$
which implies the contradiction that $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ = $\begin{pmatrix}0 & 0\\\ 1 & 1\end{pmatrix}$ = $\begin{pmatrix}1 & 1\\\ 0 & 0\end{pmatrix}$.
Hence $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ is not a left identity in $S$.
Is this a correct proof without using the fact that $x+y = 1$?
| Let $S = \left\{ A \in \mathcal{M}_{2 \times 2}(\mathbb{R}) :\ A= \begin{bmatrix}a &a\\\ b& b \end{bmatrix}\right\}$.
Then, $\begin{bmatrix}x &x\\\ y& y \end{bmatrix}$ is a left identity if $\begin{bmatrix}x &x\\\ y& y \end{bmatrix} \cdot\begin{bmatrix}a &a\\\ b& b \end{bmatrix} = \begin{bmatrix}ax+bx & ax+bx\\\ ay+by& ay+by \end{bmatrix} = \begin{bmatrix}a &a\\\ b& b \end{bmatrix}$ for all $\begin{bmatrix}a &a\\\ b& b \end{bmatrix} \in S, \begin{bmatrix}a &a\\\ b& b \end{bmatrix}\ne 0_{S}.$
$\Rightarrow ax+bx = a$ and $ay+by = b \Rightarrow x= \frac{a}{a+b}, y = \frac{b}{a+b} \Rightarrow x+y = 1$.
So one may conclude that $\begin{bmatrix}x &x\\\ y& y \end{bmatrix}$ is an identity of $S$ if $x+y =1$. But then even if $x+y =1$, by taking
$\begin{bmatrix}a &a\\\ b& b \end{bmatrix} = \begin{bmatrix}0 &0\\\ 1& 1 \end{bmatrix}$ once and $\begin{bmatrix}1 &1\\\ 0& 0 \end{bmatrix}$ second, we obtain $\begin{bmatrix}x &x\\\ y& y \end{bmatrix} = \begin{bmatrix}0 &0\\\ 1& 1 \end{bmatrix}$ and
$\begin{bmatrix}x &x\\\ y& y \end{bmatrix} = \begin{bmatrix}1 &1\\\ 0& 0 \end{bmatrix}$ respectively, which is not possible.
| {
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How to prove this infinite series identity? Prove that:
$$
1-\frac{1}{2^{n}}+ \frac{1}{3^{n}} - \frac{1}{4^{n}} + \cdots = \left(1-\frac{1}{2^{n-1}}\right)\zeta(n)
$$
where $n>1$ and:
$$
\zeta(n) = 1+\frac{1}{2^{n}}+ \frac{1}{3^{n}} + \frac{1}{4^{n}} + \cdots
$$
I have verified this is true using numerical method, but how to get the exact proof?
| Hint:
Rewrite the sum as
$$1-\frac{1}{2^{n}}+ \frac{1}{3^{n}} - \frac{1}{4^{n}} + \cdots=\zeta(n)-2\Bigl(\frac{1}{2^{n}}+ \frac{1}{4^{n}} + \cdots\Bigr)=\zeta(n)-\frac{2}{2^n}\Bigl(1+ \frac{1}{2^{n}} + \cdots\Bigr)$$
Can you end the computation?
| {
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} |
Simplification of arctan-like infinite series I'm trying to find a simplified expression for
$\sum_{n=0}^{\infty} (2n+1) (-1)^n \left[ \frac{a^2}{2 b^2} + \frac{c^2 \pi^2 (2n+1)^2}{2 d^2} \right]^{-1}$
For the special case $a=0$ I obtained
$\frac{2 d^2}{c^2 \pi^2} \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}
= \frac{2 d^2}{c^2 \pi^2} \text{arctan}(1)
= \frac{2 d^2}{c^2 \pi^2} \frac{\pi}{4}
= \frac{d^2}{2 c^2 \pi}$
Does anybody know if I can get a similar simplification for the general case? I have to evaluate this formula repeatedly so I'm looking for a solution that doesn't involve a series or that involves a series whose terms quickly approach zero.
Any help would be appreciated!
| I think that a closed form is possible.
It can be proven, for instance using Fourier series for $t\mapsto\sin(\alpha t)$ that
$$\frac{\pi}{4}\sec\left(\frac{\alpha\pi}{2}\right)=\sum_{n=0}^\infty\left(\frac{(-1)^n(2n+1)}{(2n+1)^2-\alpha^2}\right)$$
For instance, see this for a proof.
Replacing $\alpha$ with $i\alpha$:
$$\frac{\pi}{4\cosh\left(\frac{\alpha \pi}{2}\right)}=\sum_{n=0}^\infty\frac{(-1)^n(2n+1)}{(2n+1)^2+\alpha^2}$$
The expression you're looking to simplify can be written in the following form:
$$\sum_{n=0}^{\infty} (2n+1) (-1)^n \left[ \frac{a^2}{2 b^2} + \frac{c^2 \pi^2 (2n+1)^2}{2 d^2} \right]^{-1} = \beta\sum_{n=0}^{\infty} \frac{(-1)^n(2n+1) }{ \alpha^2 + (2n+1)^2}$$
with $\beta=\frac{2d^2}{c^2\pi^2}$ and $\alpha=\frac{da}{\pi bc}$.
Therefore
$$\sum_{n=0}^{\infty} (2n+1) (-1)^n \left[ \frac{a^2}{2 b^2} + \frac{c^2 \pi^2 (2n+1)^2}{2 d^2} \right]^{-1} = \frac{d^2}{2\pi c^2\cosh\left(\frac{da}{2bc}\right)}$$
| {
"language": "en",
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Let $f (x$) = $\sqrt{−x^2 + 20x + 400} + \sqrt {x^2 − 20x}$. How many elements in the range of $f$ are integers?
Let $f (x$) = $\sqrt{−x^2 + 20x + 400} + \sqrt {x^2 − 20x}$. How many elements in the range of $f$ are integers?
I first let $y= x^2 -20x +100$.
Then substitute it in the function -------> $f(x) = \sqrt{-y+500}+\sqrt{y-100}$.
Which means, $100\leq{y}\leq{500}$. I then tried to find values of $y$ that would make both radicals disappear. I found four values of $y$ that made $f(x)$ integers; $100, 500, 356,$ and $244$. I also checked their discriminants and all of them were greater than $0$, which means each value of $y$ means two solutions for $x$, meaning, there are $8$ elements in the range of $f$ are integers. But the correct answer is $9$ and I can't seem to find the last one.
| By C-S $$\sqrt{400+2x-x^2}+\sqrt{x^2-20x}\leq\sqrt{(1+1)(400+2x-x^2+x^2-20x)}=20\sqrt2.$$
Also, since $$\sqrt{a}+\sqrt{b}=\sqrt{a+b+2\sqrt{ab}}\geq\sqrt{a+b},$$ we obtain: $$\sqrt{400+2x-x^2}+\sqrt{x^2-20x}\geq\sqrt{400+2x-x^2+x^2-2x}=20.$$
We got the maximal and the minimal value of $f$ and $f$ is a continuous function.
Can you end it now?
| {
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"source": "stackexchange",
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How to get the value of the expression? I have this statement:
If $\frac{a}{b+c+d} + \frac{b}{a+c+d} + \frac{c}{a+b+d} + \frac{d}{a+b+c} = 1$
Find the value of $\frac{a^2}{b+c+d} + \frac{b^2}{a+c+d} + \frac{c^2}{a+b+d} + \frac{d^2}{a+b+c}$
My attempt was:
I will denote $k = a+b+c+d$ and $\triangle_n = k-n$
Then,
$\frac{a}{\triangle_a}+\frac{b}{\triangle_b}+\frac{c}{\triangle_c}+\frac{d}{\triangle_d}=1$
Now, multiply by $\frac{a}{a},\frac{b}{b}$ and so on, thus:
$\frac{a^2}{a\triangle_a}+\frac{b^2}{b\triangle_b}+\frac{c^2}{c\triangle_c}+\frac{d^2}{d\triangle_d}=1$
and I do not know how to get more information.
Any hint for this specific problem is aprecciated, and any hint for this type of problem is appreciated too.
| $$a+b+c+d=(a+b+c+d)\sum_{cyc}\frac{a}{b+c+d}=$$
$$=\sum_{cyc}\frac{a^2+a(b+c+d)}{b+c+d}=a+b+c+d+\sum_{cyc}\frac{a^2}{b+c+d},$$
which says $$\sum_{cyc}\frac{a^2}{b+c+d}=0.$$
In the full writing the last equality it's:
$$\frac{a^2}{b+c+d}+\frac{b^2}{c+d+a}+\frac{c^2}{d+a+b}+\frac{d^2}{a+b+c}=0.$$
| {
"language": "en",
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"source": "stackexchange",
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evaluate the summation : $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(n+2x+3)}$
Find
$$S=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(n+2x+3)}$$
for $x≥0$.
At first, I use a partial fraction
$$S=\displaystyle\sum_{n=0}^{\infty}\left(\frac{(-1)^{n}}{2(x+1)(n+1)}-\frac{(-1)^{n}}{2(x+1)(n+2x+3)}\right)
=\frac{1}{2(x+1)}(I-J),$$
where
$$I=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1}
=\ln 2$$
and
$$J=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+3+2x}$$
I think to use:
$$\ln (1+y)=\sum_{n=0}^{\infty}\frac{(-1)^{n}y^{n+1}}{n+1}$$
Is my work correct? How to complete this work ?
| We can write
$$
\eqalign{
& S = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)^{\,n} } \over {\left( {n + 1} \right)\left( {n + 2x + 3} \right)}}} = \cr
& = \sum\limits_{k = 0}^\infty {\left( {{1 \over {\left( {2k + 1} \right)\left( {2k + 2x + 3} \right)}} - {1 \over {\left( {2k + 2} \right)\left( {2k + 2x + 4} \right)}}} \right)} =
\quad \quad (1) \cr
& = {1 \over 4}\sum\nolimits_{k = 0}^{\,\infty } {{1 \over {\left( {k + 1/2} \right)\left( {k + x + 3/2} \right)}}}
- {1 \over 4}\sum\nolimits_{k = 0}^{\,\infty } {{1 \over {\left( {k + 1} \right)\left( {k + x + 2} \right)}}} = \quad \quad (2) \cr
& = {1 \over 4}\left( \matrix{
\sum\nolimits_{k = 0}^{\,\infty } {{1 \over {\left( {k + 1/2} \right)\left( {k + 1/2 + x + 1} \right)}}} + \hfill \cr
- \sum\nolimits_{k = 0}^{\,\infty } {{1 \over {\left( {\left( {k + 1/2} \right) + 1/2} \right)\left( {\left( {k + 1/2} \right) + 1/2 + x + 1} \right)}}} \hfill \cr} \right) = \cr
& = {1 \over 4}\left( \matrix{
\sum\nolimits_{k = 0}^{\,\infty } {{1 \over {\left( {k + 1/2} \right)\left( {k + 1/2 + x + 1} \right)}}} + \hfill \cr
- \sum\nolimits_{k = 1/2}^{\,\infty } {{1 \over {\left( {k + 1/2} \right)\left( {k + 1/2 + x + 1} \right)}}} \hfill \cr} \right) = \cr
& = {1 \over 4}\sum\nolimits_{k = 0}^{\,1/2} {{1 \over {\left( {k + 1/2} \right)\left( {k + 1/2 + x + 1} \right)}}} = \quad \quad (3) \cr
& = {1 \over 4}\sum\nolimits_{k = 1/2\;}^{\,1} {{1 \over {k\left( {k + x + 1} \right)}}} = \quad \quad (4) \cr
& = {1 \over {4\left( {x + 1} \right)}}\sum\nolimits_{k = 1/2\;}^{\,1} {\left( {{1 \over k} - {1 \over {\left( {k + x + 1} \right)}}} \right)} = \quad \quad (5) \cr
& = {1 \over {4\left( {x + 1} \right)}}\left( {\psi \left( 1 \right) - \psi \left( {1/2} \right) - \left( {\psi \left( {x + 2} \right) - \psi \left( {x + 3/2} \right)} \right)} \right) =
\quad \quad (6) \cr
& = {1 \over {4\left( {x + 1} \right)}}\left( {2\ln 2 - \psi \left( {x + 2} \right) + \psi \left( {x + 3/2} \right)} \right) \quad \quad (7) \cr
}
$$
where:
(1) we separate even and odd $n$;
(2) each part is convergent and we can separate into two sums,
at the same time we change the sum into [antidifference][1];
(3) (4)we change the indices and use the properties of anti-diffence;
(5) being the sum limits finite we can apply partial fractions ans separate the sums;
(6) the anti-difference of $1/k$ is the [digamma function][2];
(7) we use the known values for $\psi (1)$ and $\psi(1/2)$.
The final formula can be converted into many equivalent forms
and series developments, depending on the intended scopes.
| {
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"source": "stackexchange",
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Solving $x^5+px^2y^3+p^2y^5 = 0$ where $p$ is prime and $x,y \in \mathbb{Z}$ How do you solve $x^5+px^2y^3+p^2y^5 = 0$ where $p$ is prime and $x,y \in \mathbb{Z}$?
Working
in modulo $p$ we have $x^5 = 0 \pmod{p}$ and $x = 0$ which is the only solution in modulo $p$ since $a^{p} \equiv a \mod{p}$ for any prime $p$ and integer $a$. Taking this back we get $p^2y^5 = 0$ which implies $y=0$. Hence the only solution is $(x,y) = (0,0)$.
Edit/attempt 2:
Since $x = 0 \mod{p}$ we can let $x = pm$ where $m<x$ then we get $p^2 (m^5 p^3 + m^2 p y^3 + y^5) = 0 \implies m^5 p^3 + m^2 p y^3 + y^5$ since $p$ can't be zero. But now $y^5 = 0 \mod{p}$ ; suppose $y = pn$ where $n < y$. Then $p^3 (m^5 + m^2 n^3 p + n^5 p^2) =0$ and now $m^5 + m^2 n^3 p + n^5 p^2=0.$
| Your second attempt looks good (the moment you note explicitly that $(m,n)$ is also a solution), and leads to a contradiction in very much the same way as the standard proof that $x^2-2y^2=0$ has no non-zero integer solution (leading to $\sqrt2$ being irrational).
| {
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Find the minimum of the function $y=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}.$ Find the minimum of the function $$y=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}.$$
By computer I found $\min_y=3;$ then I will prove $y\ge 3$. After squaring we got $$(x+2)(178-37x)\ge 0\quad \forall -2\le x\le 5.$$
My idea is not beautiful so I need some other solution. Thanks!
| Here is a simple demonstration -- not using calculus or concavity arguments --
that the minimum of $y=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}$ is $3$.
Note that $y=\sqrt{-(x-7)(x+3)}+\sqrt{-(x-5)(x+2)}=\sqrt{5^2-(x-2)^2}+\sqrt{\left(\dfrac72\right)^2-\left(x-\dfrac32\right)^2}.$
So $y$ is defined when $-2\le x\le5$ or $\left(x-\dfrac32\right)^2\le\left(\dfrac72\right)^2. $
When that is the case, $-2\le x\le 6$ or $\left(x- 2\right)^2\le4^2. $
Therefore, $y= \sqrt{5^2-(x-2)^2}+\sqrt{\left(\dfrac72\right)^2-\left(x-\dfrac32\right)^2}\ge\sqrt{5^2-(x-2)^2}\ge\sqrt{5^2-4^2}=3,$
and note that $y=3$ when $x=-2$.
| {
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Evaluate $\int_0^{\infty}\frac{\sqrt{x}}{x^2+1}dx$ without Complex Analysis I found this question on a Complex Analysis Qualifying Exam:
How would you evaluate: $$\int_0^{\infty}\frac{\sqrt{x}}{x^2+1}dx$$
I am interested in all methods including Complex Analysis, I'm just less familiar with it.
For instance, since the integral is from $0$ to $\infty$ do we only consider the $x-i$ factor in $x^2+1=(x-i)(x+i)$ since $i$ would be the only root in the upper half plane?
| Without complex analysis? Okay...
Let $$I=\int_0^\infty\frac{\sqrt{x}}{x^2+1}dx$$
Enforce $x:=y^2\implies dx=2y \space dy$. So $$I=\int_0^\infty\frac{\sqrt{x}}{x^2+1}dx=2\int_0^\infty \frac{y^2}{y^4+1}dy=2\int_0^\infty\frac{y^2}{y^2\left(y^2+\frac{1}{y^2}\right)}dy$$ $$=\int_0^\infty \frac{1-\frac{1}{y^2}+1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy=\int_0^\infty\frac{1-1/y^2}{(y+1/y)^2-2}dy+\int_0^\infty\frac{1+1/y^2}{(y-1/y)^2+2}dy\tag1$$
For the first of the two integrals let $t=y+1/y\implies dt=(1-1/y^2)\space dy$. For the second, let $u=y-1/y\implies du=(1+1/y^2)\space dy$. Hence $(1)$ becomes: $$I=\int_{\infty}^\infty\frac{dt}{t^2-2}+\int_{-\infty}^\infty\frac{du}{u^2+2}=0+\int_{-\infty}^\infty\frac{du}{u^2+2}=2\int_0^\infty\frac{du}{2\left(\left(\frac{u}{\sqrt{2}}\right)^2+1\right)}\tag2$$
Finally with the substitution $z=\frac{u}{\sqrt{2}}\implies dz=\frac{du}{\sqrt{2}}$ $(2)$ becomes $$I=\sqrt{2}\int_0^\infty\frac{dz}{z^2+1}=\sqrt{2}\left(\arctan(z)\big|^\infty_0\right)=\sqrt{2}\left(\frac{\pi}{2}\right)=\frac{\pi}{\sqrt{2}}$$
Thus $$\boxed{I=\frac{\pi}{\sqrt{2}}}$$
| {
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How many ways can you give $n$ distinct toys to $3$ children given certain conditions
How many ways can you give $n$ distinct toys to $3$ children given certain conditions:
Child 1 wants an odd number of toys
Child 2 wants at most 3 toys
Child 3 wants any number of toys
I would like to count the number of ways by first coming up with a generating function then determining the coefficient for $n$.
For Child 1, the generating function is: $\frac{e^x-e^{-x}}{2}$
Child 2's generating function is: $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$
Child 3's is: $e^x$
Now we multiply all generating functions to get the generating function for all 3 children giving us:
$(\frac{e^x-e^{-x}}{2})(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!})(e^x)$
Now I would like to find the Taylor expansion for this function and find the $n^{th}$ coefficient.
Am I on the right track so far?
| You have the correct exponential generating function. It only remains to extract the coefficient of $x^n$. So
$$\begin{align}
f(x) &= \frac{1}{2}(e^x-e^{-x})\left(1+x+ \frac{1}{2!}x^2 + \frac{1}{3!}x^3 \right) e^{x} \\
&= \frac{1}{2}(e^{2x}-1)\left(1+x+ \frac{1}{2!}x^2 + \frac{1}{3!}x^3 \right) \\
&=\frac{1}{2} \left( e^{2x} + x e ^{2x} + \frac{1}{2!}x^2 e^{2x} + \frac{1}{3!} x^3 e^{2x} -1 - x -\frac{1}{2!}x^2 - \frac{1}{3!}x^3 \right) \tag{1}
\end{align}$$
Then using the "coefficient of $x^n$" operator $[x^n]$,
$$\begin{align}
[x^n]f(x) &= \frac{1}{2} \left( [x^n]e^{2x} + [x^{n-1}] e ^{2x} + \frac{1}{2!} [x^{n-2}] e^{2x} + \frac{1}{3!} [x^{n-3}] e^{2x} \right) \tag{2}\\
&= \frac{1}{2} \left(\frac{2^n}{n!} + \frac{2^{n-1}}{(n-1)!} +\frac{1}{2!} \frac{2^{n-2}}{(n-2)!} +\frac{1}{3!} \frac{2^{n-3}}{(n-3)!}\right) \tag{3}
\end{align}$$
but since this is an exponential generating function, what we really want is
$$n! [x^n]f(x) = \frac{1}{2} \left( 2^n + n 2^{n-1} +\frac{1}{2!}n(n-1) 2^{n-2} + \frac{1}{3!} n(n-1)(n-2) 2^{n-3} \right) $$
for $ n \ge 4$.
Do you see why this formula is only valid for $n \ge 4$? Hint: I left some terms out at equation $(2)$. Also, the transition from equation $(2)$ to equation $(3)$ is not valid for $n < 4$.
I leave it to you to fill in the missing values for $n=0,1,2,3$. It shouldn't be too hard, if you start with equation $(1)$.
| {
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Number of nine digits numbers whose sum of the digits is even I am reading Mathematical Circle. Problem $48$ in chapter two says that
How many nine-digit numbers have an even sum of their digits?
I am trying in this way, that we can divide the problem in four cases.
*
*$1$ even digit and $8$ odd digits
*$3$ even digits and $6$ odd digits
*$5$ even digits and $4$ odd digits
*$7$ even digits and $2$ odd digits
For the first case we get $4 \cdot 5^8 +5\cdot 5^7 \cdot 5$ number of solution. Because if the even digit is placed in first place (left to right) then we get $4\cdot 5^8$ ways to write the number and if an odd digit is placed in first place then we get $5\cdot 5^7 \cdot 5$ ways to write the number. Similarly for the second case we get $4\cdot 5^8+5^9$ , for the third case we get $4\cdot 5^8+5^9$ and for the fourth case we get $4\cdot 5^8+5^9$ ways to write the number. So total number is $ 4 \cdot (4\cdot 5^8+5^9)$.
The answer is different. So Where I have made a mistake?
Thanks.
| This is an unnecessarily complex solution but shows some combinatorial techniques.
First, we'll count the number of (ordered) sequences of 9 digits where the sum of the digits is even. Then we will subtract the number of such sequences whose first digit is zero.
If the sum of the digits in the 9-digit sequence is even, then we must have an even number of odd digits (which can be 0, 2, 4, or 8). The number of such sequences is:
$$\sum_{n=0,2,4,6,8} {9 \choose n}\times{5^n}\times{5^{9-n}} =\sum_{n=0,2,4,6,8} {9 \choose n}\times{5^9}$$
Among these sequences, the number of the ones that start with $0$ making them non-9-digit numbers is:
$$\sum_{n=0,2,4,6,8} {8 \choose n}\times{5^8} $$
So the total number of 9-digit numbers whose sum of digits is even is:
$${5^8}\times\sum_{n=0,2,4,6,8} 5{9 \choose n}-{8 \choose n} = 450000000$$
| {
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zeta function-deriving reflection equation from functional equation From analytic continuation of zeta function,
\begin{align}
\zeta(s) = \frac{\pi^{\frac{s}{2}}}{\Gamma\left(\frac{s}{2}\right)} \left[ \frac{1}{s(s-1)} + \int_1^{\infty} \left( x^{\frac{s}{2}-1} + x^{-\frac{s}{2}-1} \right) \left(\frac{\theta(x) - 1}{2}\right) dx \right] \end{align}
From this I obtain a functional equation
\begin{align}
\pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right) \zeta(s) = \pi^{-\frac{1-s}{2}} \Gamma\left(\frac{1-s}{2}\right) \zeta(1-s)
\end{align}
My next step is to prove reflection functional equation
\begin{align}
\zeta(1-s) = 2 (2\pi)^{-s} \cos\left(\frac{s\pi}{2}\right) \Gamma(s)\zeta(s)
\end{align}
Reordering a functional equation for $\zeta(s)$, I have
\begin{align}
\zeta(1-s) = \pi^{\frac{1}{2}} \frac{\Gamma\left(\frac{s}{2}\right)}{\Gamma\left(\frac{1-s}{2}\right)} \zeta(s)
\end{align}
I have problem with showing above two equations are indeed same...
| Use
\begin{align}
&\frac{\pi^{\frac{1}{2}}}{2^{s-1}} \Gamma (s) = \Gamma \left(\frac{s}{2}\right) \Gamma \left(\frac{s+1}{2}\right) \\
& \Gamma\left(\frac{1+s}{2}\right) \Gamma\left(\frac{1-s}{2}\right) = \frac{\pi}{\cos\left(\frac{\pi s}{2}\right)}
\end{align}
which comes from Legndre duplication formula and Euler's Reflection formula of gamma function.
Then I can prove the relation well!
| {
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$\frac{2n+3}{n^2+n+1}$ How many integers are there that make an expression an integer? $$\frac{2n+3}{n^2+n+1}$$
How many integers are there that make an expression an integer?
$-1,0,2$ It is seen that the numbers make the fraction integer immediately. But the answer key says there are three more $~n~$ values.
| If $\dfrac{2n+3}{n^2+n+1}$ is an integer, then it is $0$ or $\ge1$ or $\le-1$.
But $\dfrac{2n+3}{n^2+n+1}=0$ would mean $n=\dfrac{-3}2,$ not an integer.
And $n^2+n+1>0$ for all $n$,
so $\dfrac{2n+3}{n^2+n+1}\le-1\implies 2n+3\le-1(n^2+n+1)\implies n^2+3n+4\le0$,
which is true for no real $n$.
Therefore, $\dfrac{2n+3}{n^2+n+1}\ge1,$ which means $2n+3\ge n^2+n+1$
or $n^2-n-2=(n+1)(n-2)\le0$, which means $-1\le n\le 2$.
Of the integers between $-1$ and $2$, $\dfrac{2n+3}{n^2+n+1}$ is an integer for $n=-1$, $0$, and $2$ but not $1$.
| {
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Prove that $11 | 10^{2n+1}+1$ for all $n\in \mathbb{N}\cup \{0\}$. $$11 | 10^{2n+1}+1\;\;\; \forall n\in \mathbb{N}\cup\{0\} \tag{$\star$}$$
My proof of $(\star)$ is as follows:
\begin{align}
10^{2n+1}+1
&= 10\cdot10^{2n}+1 \\
&= (11-1)\cdot10^{2n}+1 \\
&= 11\cdot10^{2n}-10^{2n}+1 \\
&= 11\cdot10^{2n}-\left(10^{2n}-1\right) \\
&= 11\cdot10^{2n}-\left(100^{n}-1\right) \\
&= 11\cdot10^{2n}-\left((99+1)^{n}-1\right) \\
&= 11\cdot10^{2n}-\left(1+\binom{n}{1}99+\binom{n}{2}99^2+\cdots+\binom{n}{n-1}99^{n-1}+99^n-1\right) \\
&= 11\cdot10^{2n}-\underbrace{99}_{11\cdot9} \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right) \\
&= 11\left(10^{2n}-9 \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right)\right)
\end{align}
Is there an easier way to prove $(\star)$? The expansion of $(99+1)^n$ seems unnecessarily complicated, but I wasn't sure how else to go from there. Easier proofs are welcome!
| $$10^{2n+1}+1 \equiv (-1)^{2n+1} + 1 \equiv -1 + 1 \equiv 0 \pmod{11}.$$
| {
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Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
I saw instantly that $$h(x - \frac{1}{x})= x^2 - \frac{1}{x^2} = \left( x -\frac{1}{x} \right)\left( x + \frac{1}{x} \right),$$ but I don't know how to proceed. I tried something like $$\left(x -\frac{1}{x} \right) = a$$ and $$\left( x + \frac{1}{x} \right) = a + \frac{2}{x},$$ trying to apply $h(a)$ but it doesn't seem to work.
Any hints?
| As you have observed, let:
$$h(a) = a(a+\frac{2}{x})$$
So solving the equation
$$a = x - \frac{1}{x} $$
gives $$x = \frac{a \pm \sqrt{a^2 + 4}}{2}$$
Thus, we have $$h(a) = a(a + \frac{4}{a + \sqrt{a^2 + 4}})$$
(After we eliminate the minus case)
Which gives us the function that you are looking for.
| {
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Evaluate $\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$ The problem is as follows:
Find the value of $\textrm{H}$ which belongs to a certain vibration coming from a magnet.
$$H=\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$$
It was easy to spot that each term was related to multiples of two and three of the first angle. So I rewrote that equation like this:
$$H=\sec \frac{2\pi}{7}+\sec \frac{2\times 2\pi}{7}+\sec \frac{3\times 2\pi}{7}$$
One method which I tried was to transform the multiples of each angle into their equivalents as a single one as shown below:
$$\cos^{2}\omega=\frac{1+\cos 2\omega}{2}$$
$$\cos 2\omega= 2 \cos^{2}\omega - 1$$
$$\cos^{3}\omega=\frac{1}{4}\left(3cos\omega+\cos 3\omega \right)$$
$$\cos 3\omega = 4 \cos^{3}\omega - 3 cos\omega$$
Therefore by plugin these expressions into the above equation would become into (provided that secant function is expressed in terms of secant):
$$H=\frac{1}{\cos \frac{2\pi}{7}}+\frac{1}{2\cos^{2}\frac{2\pi}{7}-1}+\frac{1}{4\cos^{3}\frac{2\pi}{7}-3\cos\omega}$$
But from here on it looks convoluted or too algebraic to continue. My second guess was it could be related to sum to product identity but I couldn't find one for the secant.
Does it exist a shortcut or could it be that am I missing something? Can somebody help me to find the answer?
Can this problem be solved without requiring to use Euler's formulas?
| let
$$
r = \cos \frac{2\pi}7+i\sin \frac{2\pi}7
$$
so $r$ is a primitive seventh root of unity and
$$
2 \cos \frac{2\pi}7 = r + r^6 = a$$
$$
2 \cos \frac{4\pi}7 = r^2 + r^5 = b$$
$$
2 \cos \frac{6\pi}7 = r^3 + r^4 = c
$$
and so if
$$
H=\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}
$$
then
$$
\frac{H}2 = \frac1a +\frac1b + \frac1c = \frac{bc+ca+ab}{abc}
$$
by simple drudgery, using $\sum_{k=0}^6 r^k = 0$ (sum of roots of $x^7 = 1$)
$$
bc+ca+ab = (r^2+r^5)(r^3+r^4) + (r^3+r^4)(r^1+r^6) + (r^1+r^6)(r^2+r^5) = -2
$$
and
$$
abc = (r^1+r^6)(r^2+r^5)(r^3+r^4) = 1
$$
from which $H= -4$
| {
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Prove that if $x$ is odd, then $x^2$ is odd
Prove that if $x$ is odd, then $x^2$ is odd
Suppose $x$ is odd. Dividing $x^2$ by 2, we get:
$$\frac{x^2}{2} = x \cdot \frac{x}{2}$$
$\frac{x}{2}$ can be rewritten as $\frac{x}{2} = a + 0.5$ where $a \in \mathbb Z$. Now, $x\cdot\frac{x}{2}$ can be rewritten as:
$$x\cdot\frac{x}{2} = x(a+0.5) = xa + \frac{x}{2}$$
$xa \in \mathbb Z$ and $\frac{x}{2} \notin \mathbb Z$, hence $xa + \frac{x}{2}$ is not a integer. And since $xa + \frac{x}{2} = \frac{x^2}{2}$, it follows that $x^2$ is not divisible by two, and thus $x^2$ is odd.
Is it correct?
| Yes, $ $ your proof is correct. $ $ You have $\ a = \color{#c00}{(x\!-\!1)/2}\in\Bbb Z\ $ so eliminating $\,a\,$ the proof becomes
$$\begin{align} &\dfrac{x^2}2 - \dfrac{x}2\, =\, \dfrac{x(\color{#c00}{x\!-\!1})}{\color{#c00}2}\in \Bbb Z\\[.2em]
\Rightarrow\ \ \ &\dfrac{x^2}2\in\Bbb Z\iff \dfrac{x}2\in\Bbb Z\\[.2em]
\Rightarrow\ \ \ & \ \ \ \ 2\nmid {x^2}\iff\ \ 2\nmid x\\[.2em]
{\rm i.e.}\ \ \ & \ x^2\ {\rm odd}\ \iff x\ \ {\rm odd}
\end{align}\qquad\qquad\qquad\qquad$$
| {
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How to integrate $\sec^{-1}\sqrt\frac{x}{a-x}$ $\int\sec^{-1}\sqrt\frac{x}{a-x}dx$
How to solve the above integration? Please give me some hint about what substitution I should make.
Thanks in advance.
| $$\int \:arcsec\left(\frac{\sqrt{x}}{\sqrt{a-x}}\right)dx$$
Put $u=\sqrt{x}$ so $\frac{du}{dx}=\frac{1}{2\sqrt{x}}$ so $dx=2\sqrt{x}du$
$$2\int \:u\:arcsec\left(\frac{u}{\sqrt{a-u^2}}\right)du$$
Integrate by parts
$∫fg′= fg−∫f′g$
NOTE-Ignore the factor of $2$ and continue with the integral, we will include in end
$$f=arcsec\left(\frac{u}{\sqrt{a-u^2}}\right)$$ and $g$′$=u$
$$f′=\frac{\sqrt{a-u^2}\left(\frac{1}{\sqrt{a-u^2}}+\frac{u^2}{\left(a-u^2\right)^{\frac{3}{2}}}\right)}{u\sqrt{\frac{u^2}{a-u^2}-1}}$$
$$g=\frac{u^2}{2\:}$$
$$=\frac{u^2\:arcsec\left(\frac{u}{\sqrt{a-u^2}}\right)}{2}-\int \:\frac{u\sqrt{a-u^2}\left(\frac{1}{\sqrt{a-u^2}}+\frac{u^2}{\left(a-u^2\right)^{\frac{3}{2}}}\right)}{2\sqrt{\frac{u^2}{a-u^2}-1}}du$$
Now consider the integral in second term only
Put $v=u^2$ so $\frac{dv}{du}=2u$ and $du=\frac{1}{2u}dv$
$$\frac{a}{2}\int \:\frac{u}{\left(a-u^2\right)\sqrt{\frac{u^2}{a-u^2}-1}}du$$
$$\frac{a}{2}\int \frac{1}{\left(a-v\right)\sqrt{\frac{v}{a-v}-1}}dv$$
Put $w=\sqrt{\frac{v}{a-v}-1}$ so $\frac{dw}{\:dv}=\frac{\frac{a}{\left(a-v\right)^2}}{2\sqrt{\frac{v}{a-v}-1}}$ and $w^2+1=\frac{v}{a-v}$
$$\frac{a}{2}\int \:\frac{\left(a-v\right)}{a}dw$$
Put $w^2+2=\frac{a}{a-v}$
$$=\frac{a}{2}\int \:\frac{1}{w^2+2}dw$$
This is a standard integral
$\int \:\frac{1}{y^2+1}dy=arctan\left(y\right)$
$$\left(\frac{a}{2}\right)\frac{arctan\left(\frac{w}{\sqrt{2}}\right)}{\sqrt{2}}$$
Put back $w=\sqrt{\frac{v}{a-v}-1}$ and $v=u^2$
$$\frac{a\:arctan\left(\frac{\sqrt{\frac{u^2}{a-u^2}-1}}{\sqrt{2}}\right)}{2^{\frac{3}{2}}}$$
So combining this with the previously calculated first term
$$\frac{u^2\:arcsec\left(\frac{u}{\sqrt{a-u^2}}\right)}{2}-\frac{a\:arctan\left(\frac{\sqrt{\frac{u^2}{a-u^2}-1}}{\sqrt{2}}\right)}{2^{\frac{3}{2}}}$$
Put back $u=\sqrt{x}$ and multiply the missing factor of $2$ (See note)
So the final answer is
$$\:x\:arcsec\left(\frac{\sqrt{x}}{\sqrt{a-x}}\right)-\frac{a\:arctan\left(\frac{\sqrt{\frac{x}{a-x}-1}}{\sqrt{2}}\right)}{\sqrt{2}}+ C$$
| {
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Verify that an equation is an ellipse I want to show that the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is an ellipse with foci $(\pm\sqrt{a^2-b^2},0)$.
I started with summing the two distances and try to prove a constant but realise it is very tedious.
$\sqrt{(x-\sqrt{a^2-b^2})^2+y^2}+\sqrt{(x+\sqrt{a^2-b^2})^2+y^2}=...$
I tried considering proving the square of this quantity is a constant but I can't get anywhere either.
But by the geometrical definition of ellipse the quantity above must be a constant right?
| It's important to realize that we can plug in $y^2$ from the equation of the ellipse. Solving for $y^2$ in the ellipse, we get $$y^2 = b^2-\frac{b^2x^2}{a^2}$$ Then we plug this in to get $$\sqrt{(x-\sqrt{a^2-b^2})^2+b^2-\frac{b^2x^2}{a^2}}+\sqrt{(x+\sqrt{a^2-b^2})^2+b^2-\frac{b^2x^2}{a^2}}=$$
$$\sqrt{x^2-2x\sqrt{a^2-b^2}+a^2-\frac{b^2x^2}{a^2}}+\sqrt{x^2+2x\sqrt{a^2-b^2}+a^2-\frac{b^2x^2}{a^2}}$$
Simplifying, we get $$\left|x\sqrt{1-\frac{b^2}{a^2}}-a\right|+\left|x\sqrt{1-\frac{b^2}{a^2}}+a\right|$$ Since we are only worried about the interval of $[-a, a]$, we can simplify this to $$a-x\sqrt{1-\frac{b^2}{a^2}}+x\sqrt{1-\frac{b^2}{a^2}}+a = 2a$$ This is a constant value, and thus, the equation is an ellipse.
| {
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I don't understand functional equations. A complicated question.
We are given the following:
$f(x)=5x+3$
$g(2x−1)=x−3$.
We have to find:
$f^{-1}∘g(x) = ?$
I have the solution to the problem, but I don't understand two things:
*
*In order to solve for $g(x)$, we need to have substitution where $t = 2x-1$, and then plug that $x$ onto $x-3$. Why? Given my understanding of functions, it's more logical to me that we should just plug $2x-1$ into $x-3$, so we get $g(x) = (2x-1)-3$.
*Later on in the exercise, after we find $g(x)$ and $f^{-1}(x)$, we are faced with what, for me, seems like the same problem from above. Yet we solve it differently. Here is what I mean:
$$f^{-1}(g(x)) = f^{-1}(\frac{x-5}{2}) = \frac{ \frac{x-5}{2}-3 }{5} $$
Why do we in case (2) simply plug $\frac{x-5}{2}$ into $x$, while in case (1) we need to substitute with $t$ and then plug into $x$? Apologies for the complexity.
| Let's find the inverse of $f$ first. Solving for $x$, we find that $x = \frac{f(x) - 3}{5}$. "Renaming" variables gives you $f^{-1}(x) = \frac{x - 3}{5}$. Now, to find the expression for $g(x)$, we need to perform a change of variables. Let $x^* = 2x - 1$. Then $x = \frac{x^{*} + 1}{2}$. Substituting this into the expression for $g$, you get $g(x^{*}) = \frac{x^{*} + 1}{2} - 3 = \frac{x^{*} - 5}{2}$. Now, substitute this expression into $f^{-1}(x)$, and you get $f^{-1}(g(x)) = \frac{1}{5} \times (\frac{x - 5}{2} - 3) = \frac{x - 11}{10}$.
| {
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why $x^{4} = 1 : x \in \mathbb{C}$ doesn't have infinite solutions? I know that an equation has a number of roots equal to the degree of the equation, but when I was solving this equation:
$$x^{4} = 1$$
I did that:
$$x^{4} = 1 \iff x = \pm \sqrt[4]{1} = \pm 1 \implies x = 1 \text{ or } x = -1$$
then knowing that this equation must have at most 4 distinct roots I replaced $x$ with $a + bi$:
$$(a + bi) = \pm 1$$
And here I realised that $(n + (n - 1)i^{2}) = 1$ and $(n + (n + 1)i^{2}) = -1$.
This means that there are infinite solutions, so where did I go wrong?
| This is a good approach:
Every non zero complex number can be determined by an angle and a length as follows:
$$z=r(\cos\theta+i\sin\theta)~~~~~~~~~~~(\ast)$$
So, you need to find a number $z=x+iy$ (or a set of numbers) such that
$$z^4=1=1+0\cdot i=\cos(2\pi)+i\cdot\sin(2\pi)$$
But $$cos(2\pi)=\cos(4\pi)=\cos(6\pi)=\cdots \mbox{ and } \sin(2\pi)=\sin(4\pi)=\sin(6\pi)=\dots$$
For this reason, the angle for our complex number will be only from $[0;2\pi\rangle$.
Then, as we said we need a lenght and a angle, in complex plane those properties are called module and argument:
If we need a number $z$ such that:
$$z^4=1$$
it must satisfies $$\arg(z^4)=\arg(1)\mbox{ and } ||z^4||=||1||$$
Finding the argument:
$\arg(z^4)=4\cdot\arg(z)$ but multiplying a number of the range $[0;2\pi\rangle$ four times may lead that it escapes from the interval $[0;2\pi\rangle$, so we need to add a correction factor of $2\pi$ several times ($k$ times) because $\cos,\sin$ has a period $2\pi$:$~~\arg(z^4)=4\cdot\arg(z)-2k\pi,~k\in\mathbb{Z}$
So we have: $$4\times\arg(z)-2k\pi = \arg(1)=0\Rightarrow\arg(z)=\frac{2k\pi}{4}$$
Finding the module:
This is easiest:
$$||z^4||=||1||=1\Rightarrow||z||^4=1\Rightarrow ||z||=1 \mbox{ (normal real numbers root)}$$
therefore, the complex $z$ number we are looking for, has a module $1$ ands its argument has to be of the form $\dfrac{2k\pi}{4}$. Since the argument must to be in the range $[0,2\pi\rangle$ we see that if $k>4\Rightarrow\dfrac{2k\pi}{4}>2\pi$ and since $\frac{2k\pi}{4}$ is increasing function respect $k$, the only values for $k$ are $0,1,2\mbox{ and }3$.
our numbers $z$ using $(\ast)$ are:
*
*$k=0: z_1=\cos(0)+i\sin(0)=1$
*$k=1: z_2=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})=0+1\cdot i=i$
*$k=2: \cos(\pi)+i\sin(\pi)=-1+0i=-1$
*$k=3: \cos(\frac{3\pi}{2})+i\sin(\frac{3\pi}{2})=0-1\cdot i=-i $
We conclude: If we are looking for all solutions for the equation $$z^n=w$$
then, those solutions are:
$$z_k=\sqrt[n]{||w||}\cdot\left( \cos\left(\frac{\arg(w)+2k\pi}{n}\right) + i\cdot\sin\left(\frac{\arg(w)+2k\pi}{n}\right) \right),~k=0,1,\cdots,n-1$$
| {
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Solve the following equation: $\sin x \cos x = \frac{1}{2}$ I am required to solve the following equation:
$$\sin x \cos x = \frac{1}{2}$$
My attempt:
Rewriting $\cos x$
$$\sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}$$
Squaring both sides
$$\bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2$$
$$\sin^2 x (1 - \sin^2 x) = \frac{1}{4}$$
Expanding left side and multiplying both sides by 4
$$\sin^2 x - \sin^4 x = \frac{1}{4}$$
$$4\sin^2 x - 4\sin^4 x = 1$$
$$4\sin^2 x - 4\sin^4 x -1 = 0$$
Reordering left side
$$- 4\sin^4 x + 4\sin^2 x -1 = 0$$
$$4\sin^4 x - 4\sin^2 x + 1 = 0$$
Expression above can be factored as
$$(2\sin^2 x - 1)(2\sin^2 x - 1) = 0$$
$$(2\sin^2 x - 1)^2 = 0$$
It follows that
$$2\sin^2 x - 1 = 0 $$
$$\sin^2 x = \frac{1}{2} $$
$$\sin x = ± \frac{1}{\sqrt{2}} $$
So the resulting angles are: $45^{\circ},135^{\circ},225^{\circ},315^{\circ}$
Is my solution correct?
The reason why I am asking is, the author of the book used different method, and the end result he got was:
$$\sin2x = 1$$
So $2x = \sin^{-1}(1) = 90^{\circ},450$, and thus $x = 45^{\circ},225^{\circ}$
| Your solution is not correct, because it assumes that $\cos x=\sqrt{1-\sin^2x}$. In fact, sometimes you indeed have $\cos x=\sqrt{1-\sin^2x}$, but sometimes you have $\cos x=-\sqrt{1-\sin^2x}$.
Note that\begin{align}\sin(x)\cos(x)=\frac12&\iff2\sin(x)\cos(x)=1\\&\iff\sin(2x)=1\\&\iff2x=\frac\pi2+2k\pi\text{ ($k\in\mathbb Z$)}\\&\iff x=\frac\pi4+k\pi\text{ ($k\in\mathbb Z$)}.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 2
} |
Floor function simplification identities I can't seem to find any identity(if any)for division/multiplication involving floor functions: for example
$$\lfloor{\frac{n-1}{2}}\rfloor\cdot 2$$
I know does not simplify down to $$n-1$$.
| We have that
$$
x = \frac{x}
{2} + \frac{x}
{2} = 2\left\lfloor {\frac{x}
{2}} \right\rfloor + 2\left\{ {\frac{x}
{2}} \right\} = 2\left\lfloor {\frac{x}
{2}} \right\rfloor + x\bmod 2
$$
so
$$
2\left\lfloor {\frac{x}
{2}} \right\rfloor = x - 2\left\{ {\frac{x}
{2}} \right\} = x - x\bmod 2
$$
which means that for any $x$
$$
x - 2 < 2\left\lfloor {\frac{x}
{2}} \right\rfloor \leqslant x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Let $a_1,a_2,a_3....$ be an arithmetic progression. Question is
Let $a_1,a_2,a_3....$ be an arithmetic progression. If $\frac{a_1+a_2+a_3+...+a_p}{a_1+a_2+a_3+...+a_q}=\frac{p^2}{q^2},p\not=q$, then find $\frac{a_6}{a_{21}}$
Answer: $\frac{11}{41}$
All options: $\frac{7}{2}$, $\frac{2}{7}$, $\frac{11}{41}$, $\frac{41}{11}$
My approach to the question is
Let $S_r$ be sum of first $r$ terms of this AP
Method 1
we know
$\frac{S_p}{S_q}=\frac{p^2}{q^2}$
So,
$\frac{\frac{p}{2}(a_1+a_p)}{\frac{q}{2}(a_1+a_q)}=\frac{p^2}{q^2}$
$\frac{a_1+a_p}{a_1+a_q}=\frac{p}{q}$
After that I am stuck.
Method 2
But after taking another look at the question, I observed
$\frac{S_6}{S_{21}}=\frac{6^2}{21^2}=\frac{36}{441}$
Also,
$\frac{S_5}{S_{20}}=\frac{5^2}{20^2}=\frac{25}{400}$
And we know $S_r-S_{r-1}=a_r$
Then subtracting numerators and denominators of both
$\frac{S_6-S_5}{S_{21}-S_{20}}=\frac{a_6}{a_{21}}=\frac{36-25}{441-400}=\frac{11}{41}$
which we can't always do in maths, I cannot understand why Method 2 gave correct answer.
| $$\dfrac{p^2}{q^2}=\dfrac{p(2a_1+(p-1)d}{q(2a_1+(q-1)d}$$
$$\implies \dfrac pq=\dfrac{a_1+d(p-1)/2}{a_1+d(q-1)/2}$$
Set $\dfrac{p-1}2=6-1$
and $\dfrac{q-1}2=21-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Expand $\tan z$ in terms of Bernoulli numbers Want to get this:
$$\tan z = \frac{1}{i}\frac{{{e^{iz}} - {e^{ - iz}}}}{{{e^{iz}} + {e^{ - iz}}}} = \frac{1}{i}\left( {1 - \frac{2}{{{e^{2iz}} - 1}} + \frac{4}{{{e^{4iz}} - 1}}} \right)$$
as a start.
Getting lost in the algebra
$$\frac{{{e^{iz}} - {e^{ - iz}}}}{{{e^{iz}} + {e^{ - iz}}}} = \frac{{{e^{iz}} + {e^{ - iz}} - 2{e^{ - iz}}}}{{{e^{ - iz}} + {e^{iz}}}} = \frac{{{e^{iz}} + {e^{ - iz}} - 2{e^{ - iz}}}}{{{e^{ - iz}} + {e^{iz}}}} = 1 - \frac{{2{e^{ - iz}}}}{{{e^{ - iz}} + {e^{iz}}}}$$
and also
$$\frac{{2{e^{ - iz}}}}{{{e^{ - iz}} + {e^{iz}}}} = \frac{2}{{{e^{2iz}} + 1}} = \frac{{ - 2 + 4}}{{{e^{2iz}} + 1}} = \frac{{ - 2}}{{{e^{2iz}} + 1}} + \frac{4}{{{e^{2iz}} + 1}}$$
Please someone, show me the right way.
| Work backwards! We have that
$$\begin{align} {1 - \frac{2}{{{e^{2iz}} - 1}} + \frac{4}{{{e^{4iz}} - 1}}}&=\frac{e^{4iz} - 1-2({e^{2iz}} + 1)+4}{{{e^{4iz}} - 1}}\\&=\frac{e^{4iz}-2e^{2iz}+1}{{e^{4iz}} - 1}=
\frac{(e^{2iz}-1)^2}{{e^{4iz}} - 1}\\&=\frac{e^{2iz}-1}{{e^{2iz}} + 1}=\frac{{{e^{iz}} - {e^{ - iz}}}}{{{e^{iz}} + {e^{ - iz}}}}.\end{align}$$
Supplement: since Bernoulli numbers are related to the expansion of $x/(e^x-1)$, we try to write the LHS in terms of functions like $1/(w^n-1)$ where $w=e^{iz}$,
$$\frac{{{e^{iz}} - {e^{ - iz}}}}{{{e^{iz}} + {e^{ - iz}}}}=\frac{w^2-1}{w^2 + 1}=1-\frac{2}{w^2 + 1}=1-\frac{2(w^2-1)}{w^4 - 1}\\=1-\frac{2(w^2+1)-4}{w^4 - 1}=1-\frac{2}{w^2 - 1}+\frac{4}{w^4 - 1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3318986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim\limits_{n \to \infty} \cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}\right) $, where $n \in \mathbb{N}$ $$
\lim\limits_{n \to \infty} \cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}\right)
$$
where $n \in \mathbb{N}$.
In this question, what I thought was, since $n \to \infty$ and $\cos ^2x$ is periodic , all I need is actually the fractional part of this. And it easy to say that $n+1> \sqrt[3]{n^3+n^2+2n}> n$.
But evaluation of $\sqrt[3]{n^3+n^2+2n} - n$ , is getting tricky. I'm sure there must be a short way to solve it. Can someone help me with it?
| Use the binomial theorem on $(1+x)^{1/3}$ to get
$$
\sqrt[3]{n^3 + n^2 + 2n} = n\sqrt[3]{1 + n^{-1} + 2n^{-2}}= n\left[1 + \frac{n^{-1}+2n^{-2}}{3} - \frac{(n^{-1}+2n^{-2})^2}{9}+...\right] \\= n + \frac{1}{3} + \frac{5}{9n} + O(n^{-2})
$$
So
$$
\lim_{n\rightarrow \infty}\sqrt[3]{n^3 + n^2 + 2n}-n = \frac{1}{3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Factoring $x^4 + 5x^3 + 4x^2 + 2x - 3$
So I have to factor this polynomial
$$x^4 + 5x^3 + 4x^2 + 2x - 3$$
I got $(x^2 + 2x -3)(x^2 + 3x + 1)$
but when I multiplied it, I got a different equation:
$$x^4 + 5x^3 + 4x^2 - 7x - 3$$
I don’t really understand how to find the factors. I only found numbers that could add up to 5 and multiply to -3 but they’re wrong. What did I do wrong and what can I do to fix this?
In the format of:
($x^2$+ __ x +__ ) ($x^2$ + __ x + __ )
or
$$(x^2 + ax + b) (x^2 + cx + d)$$
Also, how does comparing coefficients help factor this polynomial?
| The following way always works:
$$x^2+5x^3+4x^2-7x-3=\left(x^2+\frac{5}{2}x-1\right)^2-\left(\frac{x}{2}+2\right)^2=(x^2+2x-3)(x^2+3x+1)=...$$
We can get this factoring by the following reasoning.
For any $k$ we obtain:
$$x^2+5x^3+4x^2-7x-3=\left(x^2+\frac{5}{2}x+k\right)^2-2kx^2-5kx-\frac{25}{4}x^2-k^2+4x^2-7x-3=$$
$$=\left(x^2+\frac{5}{2}x+k\right)^2-\left(\left(2k+\frac{9}{4}\right)x^2+(5k+7)x+k^2+3\right).$$
Now, we need to choose a value of $k$ for which $2k+\frac{9}{4}\geq0$ and $\left(2k+\frac{9}{4}\right)x^2+(5k+7)x+k^2+3$ is a perfect square.
We immediately see that $k=-1$ is valid, but if we don't see it we can write a condition for a perfect square:
$$(5k+7)^2-4\left(2k+\frac{9}{4}\right)(k^2+3)=0$$ or
$$4k^3-8k^2-23k-11=0$$ or
$$4k^3+4k^2-12k^2-12k-11k-11=0$$ or
$$(k+1)(4k^2-12k-11)=0$$ and since $2\cdot(-1)+\frac{9}{4}>0$, we obtain the difference of squares and the needed factoring.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How to derive $ \frac ab -x= \frac {c-xd}{b}+\left(\frac{b-d}{b}\right) \left(\frac{a-c}{b-d} - x \right) $ $$ \frac ab -x= \frac {c-xd}{b}+\left(\frac{b-d}{b}\right) \left(\frac{a-c}{b-d} - x \right) $$
It is easy to check it by computing right hand side. It feels unnatural and a little magical. I could't derive it starting from LHS
This identity is used in the proof of Stolz Cesaro theorem (https://ru.wikipedia.org/wiki/Теорема_Штольца). It is in russian I understood with the help of google translate .
| Starting from the LHS
$$\frac{a}{b}-x$$
we can subtract and add $\frac{c-xd}{b}$ to form
$$\Big(\frac{a}{b}-x\Big) - \frac{c-xd}{b} + \frac{c-xd}{b}$$
which rearranges to
$$\frac{c-xd}{b}+\Big(\frac{a}{b}-x-\frac{c-xd}{b}\Big)$$
or
$$\frac{c-xd}{b}+\Big(\frac{a-c+xd}{b}-x\Big)$$
which is equivalent to
$$\frac{c-xd}{b}+\Big(\frac{a-bx-c+dx}{b}\Big)$$
where $\left(\frac{b-d}{b}\right) \left(\frac{a-c}{b-d} - x \right) = \Big(\frac{a-bx-c+dx}{b}\Big)$ and therefore we have the RHS
$$\frac{c-xd}{b}+\left(\frac{b-d}{b}\right) \left(\frac{a-c}{b-d} - x \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How is the matrix identity $\det\begin{pmatrix}A&B\\B&A\end{pmatrix}=\det(A+B)\det(A-B)$ proved? The Wikipedia page about the determinant mentions the following matrix identity
$$\det\begin{pmatrix}A&B\\B&A\end{pmatrix}=\det(A+B)\det(A-B),$$
valid for squared matrices $A$ and $B$ of the same size.
How is this result proved?
| $$
\left(
\begin{array}{cc}
I&I \\
0&I \\
\end{array}
\right)
\left(
\begin{array}{cc}
A&B \\
B&A \\
\end{array}
\right)
\left(
\begin{array}{cc}
I& -I \\
0 &I \\
\end{array}
\right) =
\left(
\begin{array}{cc}
A+B& 0 \\
B& A-B \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Prove inequality $(n-1)\cdot a^n +b^n \geq n\cdot a^{n-1}\cdot b$ by induction let $a,b \in \mathbb{R}; a,b>0$
I have to prove the following inequality by induction:
$(n-1)\cdot a^n +b^n \geq n\cdot a^{n-1}\cdot b$ with $n\in \mathbb{N}\setminus{\{1}\}$
I was not able to use the induction hypothesis for hours and of course did not reach the desired result: $n\cdot a^{n+1} +b^{n+1} \geq (n+1)\cdot a^n\cdot b$.
Than I rearranged the inequality to
$(n-1)+(\frac{b}{a})^n \geq n\cdot \frac{b}{a}$ by dividing both sides by $a^n$
Doing induction for the rearranged inequality
Base case: assumption is true for $n=2$
$(2-1)+(\frac{b}{a})^2 \geq 2\cdot \frac{b}{a} \Leftrightarrow a^2+b^2\geq 2ab \Leftrightarrow (a-b)^2\geq 0$
It remains to prove that the assumptions holds for $n+1$
This means $n\cdot a^{n+1} +b^{n+1} \overset{?}{\geq} (n+1)\cdot a^n\cdot b$
Inductive step:
$(n+1-1)+(\frac{b}{a})^{n+1} \geq (n+1)\cdot \frac{b}{a}$
$\Leftrightarrow n+(\frac{b}{a})^n\cdot(\frac{b}{a}) \geq n\cdot (\frac{b}{a})+(\frac{b}{a})$
$\Leftrightarrow \frac{n\cdot a}{b}+(\frac{b}{a})^n\geq n+1$
$\Leftrightarrow \frac{n\cdot a\cdot a^n+b^n\cdot b}{b\cdot a^n}\geq n+1$
$\Leftrightarrow \frac{n\cdot a^{n+1}+b^{n+1}}{b\cdot a^n}\geq n+1$
$\Leftrightarrow n\cdot a^{n+1}+b^{n+1}\geq (n+1)\cdot a^n\cdot b$
So I reached the desired result, but did not use the induction hypothesis.
Can someone please point out the mistake(s) I have done?
| The problem in the inductive step is that, you started with the equation to be proved, and after solving it in a circular fashion, reached back where you started. The correct proof is as follows:
Since, $a$ and $b$ are positive, their ratio will be positive, let $x=\frac{b}{a}$. Now, what you want to show can equivalently be written as,
$$(n-1)+x^n\geq nx$$
as you also noted.
After rearranging,
$$x^n-nx+n-1\geq 0$$
Basis
For $n=2$, case it becomes $(x-1)^2\geq 0$.
Induction hypothesis
$$x^k-kx+k-1\geq 0$$
or equivalently,
$$x^k\geq kx-k+1$$
Inductive step
Consider $$x^{k+1}-(k+1)x+k$$
$$\geq (kx-k+1)x-(k+1)x+k$$
$$=k(x-1)^2\geq 0$$
Hence proved
Hope it helps:)
| {
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"url": "https://math.stackexchange.com/questions/3322947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Co-ordinate Geometry: Equation of the line Find the equation of the line that passes through the point of intersection of the lines
$3 + 2 − 1 = 0$ and $2 − + 7 = 0$ and is perpendicular to the line $4 + 4 = 7$.
Solve for x:
$3 + 2 − 1 = 0$
$2 − + 7 = 0$ (multiply by $2$)
$3 + \not{2y} − 1 = 0$
$4 − \not{2} + 14 = 0$
$7x + 13 = 0$
$x =\frac{-13}{7}$.
Sub $x =\frac{-13}{7}$ into $3 + 2 − 1 = 0$
To get $3\left(\frac{-13}{7}\right) + 2 − 1 = 0$
Which reduces to $y=\frac{23}{7}$.
Line is perpendicular to $4 + 4 = 7$ so $m$ must be $1$.
Equation of line formula: $y-y_1=m(x-x_1)$.
$y-\frac{23}{7}=1(x-\frac{-13}{7})$
$y-1x-\frac{36}{7}=0$ $($Multiply by $7$$)$
$7y-7x-36=0$.
Been at this for about 2 hours now so I'd appreciate any help or guidance, thank you.
| The point of intersection is indeed $\left(-\dfrac{13}7,\dfrac{23}7\right)$. A line perpendicular to $4x+4y=7$ has its linear coefficients proportional to $(1,-1)$ (because $4\cdot1-4\cdot1=0$), hence
$$\left(x+\dfrac{13}7\right)-\left(y-\dfrac{23}7\right)=0$$
or
$$7x-7y+36=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do you do the multiplication of an infinite sided regular polygon by an infinitesimal side to get the perimeter of it? Using the law of cosines to find each side length and multiplying by the number of sides, I see a pattern, which should approach $~2\pi~$ as the number of sides grows to infinity.
How can $~2\pi~$ be calculated, when it's infinity multiplied by an infintesimal?
Using the law of cosines, the missing side of each perimeter section can be found, where $~x~$ is the number of sides the polygon has. In this, the distance from each vertex on the polygon to the center $~= 1~$ unit.
The formula for each polygon's perimeter based on its number of sides $(x)$ is as follows:
perimeter $= x * \sqrt{2 - 2 ~\cos\left(\frac{360}{x}\right)}~.$
So, what is $~2\pi~$ , using this infinite by infinitesimal product? Calculus anyone?
| Using Taylor series for large $x$
$$\cos \left(\frac{2 \pi }{x}\right)=1-\frac{2 \pi ^2}{x^2}+\frac{2 \pi ^4}{3
x^4}+O\left(\frac{1}{x^6}\right)$$
$$2-2\cos \left(\frac{2 \pi }{x}\right)=\frac{4 \pi ^2}{x^2}-\frac{4 \pi ^4}{3
x^4}+O\left(\frac{1}{x^6}\right)$$
$$\sqrt{2-2\cos \left(\frac{2 \pi }{x}\right)}=\frac{2 \pi }{x}-\frac{\pi ^3}{3 x^3}+O\left(\frac{1}{x^5}\right)$$
$$x\sqrt{2-2\cos \left(\frac{2 \pi }{x}\right)}=2 \pi -\frac{\pi ^3}{3 x^2}+O\left(\frac{1}{x^4}\right)$$
Use it for $x=6$; the exact perimeter is $6$ for sure and the above formula gives
$2 \pi -\frac{\pi ^3}{108}\approx 5.99609$.
If you want a still better approximation, using Padé approximants,
$$x\sqrt{2-2\cos \left(\frac{2 \pi }{x}\right)}=2 \pi -\frac{20 \pi ^3}{3 \left(20 x^2+\pi ^2\right)}$$ which, for $x=6$, would give $5.99997$.
| {
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"url": "https://math.stackexchange.com/questions/3328614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
A square inside an equilateral triangle Given an equilateral triangle and a point $D$ on one of its sides, I need to construct a square $DEFG$ with the vertices $E, F$ lying on the other two sides of the triangle and $G$ somewhere inside it (see picture).
I know if $D$ is the midpoint of the respective side, the problem is easy, but how about the general case? Are there any solutions at all? Actually, my intuition says there should not be if $D$ is not quite close to the middle.
Furthermore, I have tried using analytic geometry but it quickly became messed up....so I wonder also if we can construct such a square with compass and ruler only.
Thanks in advance.
| Let
the vertex of the triangle between $D$ and $E$ be $A$
and
the vertex of the triangle between $E$ and $F$ be $C$
Let the length of the side of the triangle be $a$
length of the side of the square be $u$
$\angle ADE = \theta$ (therefore, $\angle EFC = \frac{5 \pi}{6} - \theta$)
length of $AE$ be $x$, therefore length of $CE$ is $a-x$
length of $AD$ be $y$
Then consider the triangle $ ADE$ and use Sine rule:
$\displaystyle \frac{u}{\sin \frac{\pi}{3}} = \frac{x}{\sin \theta} = \frac{y}{\sin \left(\frac{2 \pi}{3} - \theta \right)}$ ........ (1)
Next consider triangle $CEF$ and use Sine rule:
$\displaystyle \frac{u}{\sin \frac{\pi}{3}} = \frac{a-x}{\sin \left( \frac{5 \pi}{6} - \theta \right)}$ ........ (2)
All the above quantities of (1) and (2) are equal to
$\frac{a}{\sin \theta + \sin \left( \frac{5 \pi}{6} - \theta \right)}$
[Ratio of the sum of the numerators and denominators of $\frac{x}{\sin \theta}$ and $\frac{a-x}{\sin \left( \frac{5 \pi}{6} - \theta \right)}$]
Hence
$\displaystyle u = \frac{a \sin \frac{\pi}{3}}{\sin \theta + \sin \left( \frac{5 \pi}{6} - \theta \right)}$
$\displaystyle x = \frac{a \sin \theta}{\sin \theta + \sin \left( \frac{5 \pi}{6} - \theta \right)}$
$\displaystyle y = \frac{a \sin \left(\frac{2 \pi}{3} - \theta \right)}{\sin \theta + \sin \left( \frac{5 \pi}{6} - \theta \right)}$
Note:
*
*Suppose only $y$ is known. You can easily find $\theta$ and then calculate $u$ and $x$
*Not all the values of $y$ are admissible. For example, if $y > \sqrt{3} a$, then the equation does not have any solution. Practically, $0 \leq y \leq a$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Good asymptotic for this recursion? Consider some initial integer values and let the integer sequence continue like
$$\begin{align}f(n) &= f(n-1) \\
&+ n(n+1)(n+2)\dots(n+5) f(n-2) \\
&- \frac{ n(n+1)(n+2)\dots(n+5)}{2} f(n-3) \\
&- \frac{ n(n+1)(n+2)\dots(n+5)}{3} f(n-4) \\
&- \frac{ n(n+1)(n+2)\dots (n+5)}{6} f(n-5) \\
&+ K.
\end{align}$$
For some positive integer $K$.
Consider the cases where eventually the sequence remains strictly increasing.
What are possible closed forms if any ?
What are the best asymptotics !?
I assume the best(?) asymptotics are something like
$ A ( n ! )^6 $ or $ A ( n!)^5 $ ?
( $ A $ some constant )
Big those estimates seem poor and even kinda ignore the variable $K$.
So i doubt my own guesses.
Not sure how to handle it.
Reminder me of the factorial of hypergeo.
Clearly this is just part of a bigger question ; the similar recursion equations
$$ \sum_i p_i(n) f(n-i) = 0 $$
And perhaps it is useful to think of the related equations
$$ \sum_i p_i(x) f(x/i) = 0 $$
$$ \sum_i p_i(x) f(x/2^i) = 0 $$
And look at their Taylor series solutions ?
Those solutions might be found and understood by systems of equations ...
| Under the assumptions that $f$ is positive and strictly increasing we have:
$$
\begin{align}f(n) &= f(n-1) \\
&+ n(n+1)(n+2)\dots(n+5) f(n-2) \\
&- \frac{ n(n+1)(n+2)\dots(n+5)}{2} f(n-3) \\
&- \frac{ n(n+1)(n+2)\dots(n+5)}{3} f(n-4) \\
&- \frac{ n(n+1)(n+2)\dots (n+5)}{6} f(n-5) \\
&+ K.
\\&\Leftrightarrow\\
f(n)&=f(n-2) \\
&+ (n+1)(n+2)\dots(n+6) f(n-3) \\
&- \frac{ (n+1)(n+2)\dots(n+6)}{2} f(n-4) \\
&- \frac{ (n+1)(n+2)\dots(n+6)}{3} f(n-5) \\
&- \frac{ (n+1)(n+2)\dots (n+6)}{6} f(n-6) \\
&+ K\\
&+ n(n+1)(n+2)\dots(n+5) f(n-2) \\
&- \frac{ n(n+1)(n+2)\dots(n+5)}{2} f(n-3) \\
&- \frac{ n(n+1)(n+2)\dots(n+5)}{3} f(n-4) \\
&- \frac{ n(n+1)(n+2)\dots (n+5)}{6} f(n-5) \\
&+ K\\
&\le f(n-2) \\
&+ (n+1)(n+2)\dots(n+6) f(n-3) \\
&+ n(n+1)(n+2)\dots(n+5) f(n-2) \\
&+ 2K\\
&= \\
&+ (n+1)(n+2)\dots(n+6) f(n-3) \\
&+ (1+n(n+1)(n+2)\dots(n+5)) f(n-2) \\
&+ 2K
\end{align}$$
Further we have then
$$
\begin{align}
f(n) &\le\\
&+ (n+1)(n+2)\dots(n+6) f(n-3) \\
&+ (1+n(n+1)(n+2)\dots(n+5)) f(n-2) \\
&+ 2K\\
&\le \\
&+ 2(1+n(n+1)(n+2)\dots(n+5)) f(n-2) \\
&+ 2K
\end{align}
$$
For all $n\ge N$ for some $N$ we further have:
$$
f(n)\le
c\cdot 2(1+n(n+1)(n+2)\dots(n+5)) f(n-2)
$$
You then can solve this recurrence equation, e.g. per WolframAlpha and obtain some bounds using it (you can get an even better than I initially guessed).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Compute $\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}}{(2n+1)^3}$ and $\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}$ How to prove
$$S_1=\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}}{(2n+1)^3}=1+\frac{35}{128}\pi\zeta(3)+\frac{1}{48}\zeta(4)-\frac1{384}\psi^{(3)}\left(\frac14\right)$$
$$S_2=\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}=1+\frac18G\zeta(2)-\frac{35}{64}\pi\zeta(3)-\frac{15}{16}\zeta(4)+\frac1{768}\psi^{(3)}\left(\frac14\right)$$
where $H_n=\sum_{n=1}^\infty\frac1n$ is the $n$th harmonic number, $G$ denotes the Catalan's constant, $\zeta$ denotes the Riemman Zeta function and $\psi^{(n)}$ designates the Polygamma function.
These two sums were proposed by Cornel and can be found here and here . My solution to $S_1$ can be found in the first link but its long, so can we find a better way to find $S_1$ and $S_2$ ?
Thanks.
Note: Using the generating function of $\ \sum_{n=1}^\infty x^n\frac{H_n}{n^3}$ to evaluate $S_1$ is not allowed.
| Different approach to evaluate $S_1$:
From here we have
$$I=\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\frac{\pi^3}{16}\ln2-\frac{7\pi}{64}\zeta(3)-\frac{\pi^4}{96}+\frac1{768}\psi^{(3)}\left(\frac14\right)\tag1$$
On the other hand
$$I=\int_0^1 \frac{\ln^2x\arctan x}{x}\ dx-\int_0^1 \frac{x\ln^2x\arctan x}{1+x^2}\ dx$$
For the first integral, use $\arctan x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}$ and for the second integral, use the identity $\frac{\arctan x}{1+x^2}=\frac12\sum_{n=0}^\infty(-1)^n\left(H_n-2H_{2n}\right)x^{2n-1}$ we have
$$I=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln^2x\ dx-\frac12\sum_{n=0}^\infty(-1)^n(H_n-2H_{2n})\int_0^1x^{2n}\ln^2x\ dx$$
$$=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}-\sum_{n=0}^\infty(-1)^n\frac{H_n-2H_{2n}}{(2n+1)^3}$$
$$=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}-\sum_{n=0}^\infty\frac{(-1)^nH_n}{(2n+1)^3}+2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^3},\quad H_{2n}=H_{2n+1}-\frac{1}{2n+1}$$
$$=\sum_{n=0}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}+2\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^3}\tag2$$
Combine $(1)$ and $(2)$ and substitute
$$\sum_{n=0}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$
we obtain that
$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac1{384}\psi^{(3)}\left(\frac14\right)-\frac{1}{48}\pi^4-\frac{35}{128}\pi\zeta(3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Formula for rotating vectors Let $u = (u_1, u_2)$ and $x=(x,y)$ a rotation of u by an angle θ. Then $\left\lVert u \right\rVert = \left\lVert x \right\rVert$
We know:
$$cosθ = {{u \cdot x}\over \left\lVert u \right\rVert \cdot \left\lVert x \right\rVert} = {{u \cdot x}\over \left\lVert u \right\rVert \cdot \left\lVert u \right\rVert} = {{u \cdot x}\over \left\lVert u \right\rVert^2} = {{u \cdot x} \over u \cdot u} \Leftrightarrow (u \cdot u) cosθ = u \cdot x \tag 1 $$
Expanding $(1)$:
$$(u_1^2 +u_2^2)cosθ = u_1x +u_2y \tag 2 $$
We also know:
$$\left\lVert u \right\rVert = \left\lVert x \right\rVert \Leftrightarrow \sqrt{u_1^2 +u_2^2} = \sqrt{x^2 +y^2} \Rightarrow u_1^2 +u_2^2 = x^2 +y^2 \tag 3$$
For any set of $(u_1,u_2, θ)$ we now have a system of two equations and two unknowns. We should get two vectors that form an angle $θ$ with $u$ out of that.
But I was trying to combine these two formulas ($(2)$ and $(3)$) into a single generic one and I failed. Is it possible to do that?
Or could we solve $(1)$ for $x$ some way to get a vector equation for the rotation? $(x,y) = x_{something}\ cosθ \ (u_1, u_2) $ would be nice.
I know there is a formula for vector rotations using matrix transformations but I would like to know if it is possible to do the same thing with the aforementioned equations.
| Suppose that $u_1,$ $u_2,$ and $\theta$ are given.
First solve for $y$ in Equation $(2)$:
$$ y = \frac1{u_2}(u_1^2 +u_2^2)\cos\theta - \frac{u_1}{u_2}x. \tag4$$
Now let $h = \frac1{u_2}(u_1^2 +u_2^2)\cos\theta$
and let $m = \frac{u_1}{u_2}.$
Note that for any given value of $u_1,$ $u_2,$ and $\theta,$
we can treat $h$ and $m$ as constants.
Likewise we can set $r = \sqrt{u_1^2 + u_2^2}$ and treat $r$ as a constant.
You don't really need to define these extra symbols, but they are a reminder of some important quantities that do not depend on $x$ or $y,$ and they let you write the equations a little quicker and therefore manipulate them a little easier.
Now we can write Equation $(4)$ as
$$ y = h - mx \tag5$$
and Equation $(3)$ as
$$ x^2 + y^2 = r^2 . \tag6$$
Use Equation $(5)$ to substitute for $y$ in Equation $(6)$:
$$ x^2 + (h - mx)^2 = r^2. $$
Expand $(h - mx)^2$ and collect the terms in $x^2$ and $x$; also make a single constant term:
$$ (m^2 + 1)x^2 - 2hmx + (h^2 - r^2) = 0. $$
This is a quadratic equation, $ax^2 + b + c=0$ with $a=m^2+1,$
$b = -2hm,$ and $c = h^2 - r^2.$ Solving it,
\begin{align}
x &= \frac{2hm \pm \sqrt{4h^2m^2 - 4(m^2 + 1)(h^2 - r^2)}}{2(m^2 + 1)}\\
&= \frac{2hm \pm \sqrt{4h^2m^2 - 4(h^2m^2 + h^2 - (m^2+1)r^2)}}{2(m^2 + 1)}\\
&= \frac{hm \pm \sqrt{(m^2+1)r^2 - h^2}}{m^2 + 1}.
\end{align}
Now let's put back the original symbols, but carefully:
\begin{align}
h^2 &= \frac1{u_2^2}(u_1^2 +u_2^2)^2\cos^2\theta = \frac{r^4}{u_2^2}\cos^2\theta, \\
m^2 + 1 &= \frac{u_1^2}{u_2^2} + 1 = \frac{u_1^2 + u_2^2}{u_2^2} = \frac{r^2}{u_2^2},\\
(m^2+1)r^2 - h^2 &= \frac{r^4}{u_2^2} - \frac{r^4}{u_2^2}\cos^2\theta
= \frac{r^4}{u_2^2} \sin^2\theta, \\
\sqrt{(m^2+1)r^2 - h^2} &= \frac{r^2}{u_2} \sin\theta, \\
hm &= \frac{u_1}{u_2^2}(u_1^2 +u_2^2)\cos\theta = \frac{u_1r^2}{u_2^2}\cos\theta.
\end{align}
Therefore
\begin{align}
x &= \frac{\dfrac{u_1r^2}{u_2^2}\cos\theta \pm \dfrac{r^2}{u_2} \sin\theta}
{\dfrac{r^2}{u_2^2}} \\
&= u_1 \cos\theta \pm u_2 \sin\theta, \\[1ex]
y &= \frac{u_1^2 +u_2^2}{u_2}\cos\theta
- \frac{u_1}{u_2}(u_1 \cos\theta \pm u_2 \sin\theta) \\
&= \frac1{u_2}((u_1^2 +u_2^2)\cos\theta - (u_1^2 \cos\theta \pm u_1u_2 \sin\theta)) \\
&= \frac1{u_2}(u_2^2\cos\theta \mp u_1u_2 \sin\theta) \\
&= u_2\cos\theta \mp u_1 \sin\theta, \\
\end{align}
which is the same as you get from the rotation matrix.
The $\pm$ and $\mp$ occur because your equations are not sufficient to distinguish a rotation by $\theta$ from a rotation by $-\theta,$ so we get solutions for two rotation matrices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Unclear with slope field and initial values. So I am working with a first order ODE: $$\frac{dy}{dx}= y\cos^2(x)$$
We are told to draw by hand three separate solution curves for each initial value points: $y(0)=0, y(0)=\frac{1}{2}, y(0)=-\frac{1}{2}.$
I already have the general solution. Am I supposed to solve for $C$ with each of the initial values? I am confused by the idea of the solution curve. I mean the line $y=0$ passes through the singular point $y(0)=0,$ but is that a solution curve? Couldn't you construct an infinite number of lines to pass through any initial value? I am very lost. $x=0$ passes through $y(0)=\frac{1}{2}$. Please help.
| You want to find function $y$, such that $\frac{dy}{dx} = y(x) \cos^2(x)$. Assume for a while that $y$ does not attain $0$ value, then we have: $\frac{dy}{y(x)} = \cos^2(x)dx$, and by integrating side by side, we get:
$\ln|y(x)| = \int \cos^2(x)dx = \frac{1}{2}\int (\cos(2x) + 1)dx = \frac{1}{4}\sin(2x) + \frac{1}{2}x + C = \frac{1}{2}(\sin(x)\cos(x) + x) + C $
By that we get:
$|y(x)| = \exp(\frac{1}{2}(\sin(x)cos(x) + x) + C) = A\exp(\frac{1}{2}(\sin(x)\cos(x) + x)) $, where $A = e^C > 0 $.
Taking rid of absolute value ( and realise that function $y \equiv 0 $ is a solution, too)$, we have:
$y(x) = Ae^{\frac{1}{2}(\sin(x)\cos(x)+x)}$ - general solution.
Now taking your three cases:
$y(0) = 0$, then $0 = Ae^{\frac{1}{2}(0 + 0)} = A $, so your solution is $y \equiv 0$
$y(0) = \pm\frac{1}{2} $, then $A = \pm \frac{1}{2}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Write the quartic equation $4x^4+12x^3-35x^2-300x+625$ as the product of two quadratic expressions. Write the quartic expression
$4x^4+12x^3-35x^2-300x+625$
as the product of two quadratic expressions with real coefficients.
I would like to know the way of solving it other than solving with the simultaneous equation.
| Let
$$f(x)=4x^4+12x^3-35x^2-300x+625$$
Use Rational Root Theorem to find at least 1 root, using the theorem, we can see that:
$a_0=625$, $a_4=4$
The dividers of $a_0=625$ are: $1, 5, 25, 125 , 625$
The dividers of $a_4$=4 are: $1,2 , 4$
So, we check the following numbers: +/- $\frac{1,5,25,125,625}{1,2,4}$.
Out of the above step, we discover that $f(\frac{5}{2})=0$.
Notice that $f'(x)=16x^3+36x^2-70x-300=0$ has a real root at $\frac{5}{2}$ also.
This hints to the fact that $\frac{5}{2}$ is a double root for the main equation $f(x)$.
Since $\frac{5}{2}$ is a root then $(2x-5)$ is a factor.
and since its a repeated root, then $(2x-5)(2x-5)$ are factors.
You could now divide f(x) by its factors, to get the other a second degree equation with complex roots
$$\frac{4x^4+12x^3-35x^2-300x+625}{\left(2x-5\right)\left(2x-5\right)}=x^2+8x+25$$
Now we can write:
$$4x^4+12x^3-35x^2-300x+625=(2x-5)^{2}(x^2+8x+25)$$
Note that maybe a numerical solution could get you the first root faster.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Limit development for $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$. I'm trying to solve this problem: Find a limit development of order
7 for the function $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$. Where
we use the next definition:
Definition. Let $f\colon I\to\mathbb{R}$ be a function and $x_{0}\in I$. We
say that $f$ has a limit development of order $n$ in $x_{0}$ provided
that there exist $a_{0},a_{1},\ldots,a_{n}\in\mathbb{R}$ such that
for $x\in I$
$$
f(x)=a_{0}+a_{1}(x-x_{0})+\ldots+a_{n}(x-x_{0})^{n}+o((x-x_{0})^{n}) \text{ (small $o$)}.
$$
Using Taylor series for $\arctan x$ at 0, we have that $\arctan x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots$,
and therefore
\begin{align*}
\arctan(x^{2}) & =x^{2}-\frac{(x^{2})^{3}}{3}+\frac{(x^{2})^{5}}{5}-\frac{(x^{2})^{7}}{7}+\ldots\\
& =x^{2}-\frac{x^{6}}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+\ldots
\end{align*}
So, replacing for $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$ I have
that:
\begin{align*}
f(x) & =\frac{x}{x^{2}-\frac{x^{6}}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+\ldots}-\frac{1}{x}\\
& =\frac{x}{x\left(x-\frac{x^{5}}{3}+\frac{x^{9}}{5}-\frac{x^{13}}{7}+\ldots\right)}-\frac{1}{x}\\
& =\frac{1}{x}\cdot\left(1-\frac{x^{4}}{3}+\frac{x^{8}}{5}-\frac{x^{12}}{7}+\ldots\right)^{-1}-\frac{1}{x}
\end{align*}
But I couldn't obtain the form that is necessary for the limit development
because I have that part with the inverse of $1-\frac{x^{4}}{3}+\frac{x^{8}}{5}-\frac{x^{12}}{7}+\ldots$.
Could you help me or give me some suggestion?
Thanks.
| You could use the taylor polynomial of $(1-x)^{-1}$ as well, because:
$$\frac{1}{1-x}=\sum_{n \in \mathbb{N}} x^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337730",
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"source": "stackexchange",
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Multivariable limit $\lim_{(x,y,z) \to (0,0,0)} \frac{2x^2+3y^2+z^2}{x^2+y^2+z^2}$ I am studying multivariable limits from the text Vector Calculus by S.J. Colley.
I practiced computing limiting values of the functions of two-variables $f(x,y)$.
Here's a problem involving a function of three variables.
Compute the limit $$\lim_{(x,y,z) \to (0,0,0)} \frac{2x^2+3y^2+z^2}{x^2+y^2+z^2}$$
I thought to myself, this could be for example, the scalar potential of a field in space. So, it's fun to see if the function has a limit as we get closer and closer to the origin. I would like to ask, if my proof is mathematically correct.
Solution.
From solid analytical geometry, a straight line in $\mathbb{R^3}$ is:
$\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}=r$
The straight line passing through the origin and having direction numbers $l,m,n$ is:
$\frac{x}{l}=\frac{y}{m}=\frac{z}{n}=r$
So, along this straight line with direction numbers $l,m,n$, our function takes on values
$f(x,y,z)=f(r)=\frac{2r^2l^2+3r^2m^2+r^2n^2}{r^2l^2+r^2m^2+r^2n^2}=\frac{2l^2+3m^2+n^2}{l^2+m^2+n^2}$
Along the line $x=y,z=0$, $f(x,y,z)=\frac{5}{2}$
Along the line $y=z,x=0$, $f(x,y,z)=\frac{4}{2}=2$
Along the line $x=z,y=0$, $f(x,y,z)=\frac{3}{2}$
Along the line $x=y=z$, $f(x,y,z)=\frac{6}{3}=2$
Hence,
$\lim_{(x,y,z) \to (0,0,0)\\\text{ along }x=y,z=0}\frac{2x^2+3y^2+z^2}{x^2+y^2+z^2}=\frac{5}{2}$
$\lim_{(x,y,z) \to (0,0,0)\\\text{ along }y=z,x=0}\frac{2x^2+3y^2+z^2}{x^2+y^2+z^2}=2$
$\lim_{(x,y,z) \to (0,0,0)\\\text{ along }x=z,y=0}\frac{2x^2+3y^2+z^2}{x^2+y^2+z^2}=\frac{3}{2}$
$\lim_{(x,y,z) \to (0,0,0)\\\text{ along }x=y=z}\frac{2x^2+3y^2+z^2}{x^2+y^2+z^2}=3$
Thus, the limit of this function does not exist as $(x,y,z) \to (0,0,0)$.
| Your limit was $\lim_{r\to0}\frac{2r^2l^2+3r^2m^2+r^2n^2}{r^2l^2+r^2m^2+r^2n^2}=\frac{2l^2+3m^2+n^2}{l^2+m^2+n^2}$ which depends on $(l,m,n)$ i.e. the direction of approach and hence the limit is path dependent.
You can see this by setting arbitrary values for $(l,m,n)\ne(0,0,0)$. For example, for $l=m=n=1$, your line of approach is $x=y=z$ and the limit is $6/3=2$. Along another line, say $\frac x2=\frac y2=z$, you get $21/9$. Hence the limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Algebraic manipulation solving for $x+y$ There are real solutions $x,y$ to
$x^3+2016x+1=0$
$y^3+2016x-1=0$.
Find $x+y$.
My thinking: adding them we get
$x^3+y^3+2016(x+y)=0$
$(x+y)(x^2-xy+y^2)+2016(x+y)=0$
$(x+y)(x^2-xy+y^2+2016)=0$
Then, either
$x+y=0$
or,
$x^2-xy+y^2+2016=0$.
I’m not sure how to proceed with the second equation and the first seems a little odd.
| Given:
$$x^3+2016x+1=0 \tag1$$
$$y^3+2016x-1=0 \tag2$$.
When you solve (1) for $x$:
You get a repeated root at $x=-0.000496032$.
When you solve equation (2) for $y$, you get $y=0.000496032$
We can use the values obtained to calculate $x+y=0$
So the solutions are:
$y=0.000496032$, $x=-0.000496032$ or $x=0.000496032$, $y=-0.000496032$
Another approach is adding equations (1) and (2) to get:
$$x^3-y^3=(x-y)(x^2+xy+y^2)=0 \tag3$$
The expression $(x^2+xy+y^2)$ can't be zero for real values of x. This suggests that $x=y$.
Substituting the value of $x$ in Equation (2) we get:
$$y^3+2016(-y)-1=0$$
which has a real solution $y=0.000496032$.
So the solutions are (for which $x+y=0$ and both $x,y$ are real):
$y=0.000496032$, $x=-0.000496032$ or $x=0.000496032$, $y=-0.000496032$
Note: I use the equal sign loosely here the accurate values may not be fully precise.
| {
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"url": "https://math.stackexchange.com/questions/3341509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equations of the sides an equilateral triangle with centroid at the origin and one side is $x+y=1$
An equilateral triangle has its centroid at the origin and one side is $x+y=1$. Find the equations of the other sides.
My Attempt
$$
OD=\frac{1}{\sqrt{2}}\implies OC=\sqrt{2}\implies C=(-1,-1)\\
m_{AB}=m_1=-1\implies\tan60=\sqrt{3}=|\frac{m+1}{1-m}|\\
\sqrt{3}-m\sqrt{3}=m+1\quad\text{or}\quad m\sqrt{3}-\sqrt{3}=m+1\\
m(1+\sqrt{3})=\sqrt{3}-1\quad\text{or}\quad m(\sqrt{3}-1)=\sqrt{3}+1\\
m=\frac{\sqrt{3}-1}{\sqrt{3}+1}\quad\text{or}\quad m=\frac{\sqrt{3}+1}{\sqrt{3}-1}\\
y+1=\frac{\sqrt{3}-1}{\sqrt{3}+1}(x+1)\quad\text{or}\quad y+1=\frac{\sqrt{3}+1}{\sqrt{3}-1}(x+1)
$$
But, my reference gives the solutions
$$y+1=\frac{\sqrt{3}-1}{\sqrt{3}+1}(x+1)$$
and
$$y\pm1=3+\sqrt{3}(x-1)$$
So, what are the actual solution to the problem and the easiest way to solve it?
| Rotating the coordinate system by $\pm \frac{2 \pi}{3}$ gives the new coordinates $(x', y')$ defined by
$$(x', y') = \left(\frac{-x \mp \sqrt{3} y}{2} , \frac{\pm \sqrt{3} x - y}{2}\right) .$$
So, rotating the line $x + y = 1$ gives the equation $x' + y' = 1$, and substituting the above transformation equation to write this in terms of $x, y$ gives:
$$1 = \left(\frac{-x \mp \sqrt{3} y}{2}\right) + \left( \frac{\pm \sqrt{3} x - y}{2}\right) = \frac{1}{2}\left[\left(-1 \pm \sqrt 3\right) x + \left(-1 \mp \sqrt 3\right) y\right] .$$
But rearranging shows that this equation is equivalent to your solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $1^2 + 2^2 + {...}+ {(n - 1)}^2 < \frac{n^3}{3} < 1^2 + 2^2 + {...}+ {n}^2 $ Prove that $1^2 + 2^2 + {...}+ {(n - 1)}^2 < \frac{n^3}{3} < 1^2 + 2^2 + {...}+ {n}^2 $
I know I need to use induction for this proof, but it feels like a pretty complicated one.
Basis: For $n = 2$,
$$1^2 < \frac{8}3 < 1^2 + 2^2$$
Induction Hypothesis: Assume $P(n)$ holds for $n=k$, that is,
$$1^2 + 2^2 + {...}+ {(k - 1)}^2 < \frac{k^3}{3} < 1^2 + 2^2 + {...}+ {k}^2$$
We need to show that $P(n)$ also holds for $n=k+1$
Proof:
$$1^2+2^2+{...}+{(k)}^2=1^2+2^2+{...}+{(k-1)}^2+{k}^2$$
After this, I'm not sure how to use the assumed inequality to prove it because it's a less than inequality. If I could get a hint that'd be awesome.
| Since $x^2$ is an increasing function on $[0,1]$ we have that $\frac{1}{3}=\int_{0}^{1}x^2\,dx$ can be bounded by two Riemann sums:
$$ \frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^2 < \int_{0}^{1}x^2\,dx < \frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^2 $$
i.e. the lower/upper Riemann sums associated to the partition of $[0,1]$ into $n$ congruent sub-intervals.
By multiplying each term by $n^3$ the claim is readily proved.
Through induction we just have to prove that $\frac{1}{3}\left[(n+1)^3-n^3\right]$ is bounded between $n^2$ and $(n+1)^2$, i.e.
$ n^2<n^2+n+1<n^2+2n+1 $, which is also pretty straightforward.
Yet another way: $\sum_{m=0}^{M}m^2$ can be computed through the hockey stick identity. We have $m^2=2\binom{m}{2}+\binom{m}{1}$, hence
$$ \sum_{m=0}^{M}m^2=2\sum_{m=0}^{M}\binom{m}{2}+\sum_{m=0}^{M}\binom{m}{1} =2\binom{M+1}{3}+\binom{M+1}{2}=\frac{M(M+1)(2M+1)}{6}$$
and the claim is equivalent to
$$ M(M-1)(2M-1) < 2M^3 < M(M+1)(2M+1). $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Theta function identity Let us consider the Theta function
$$\theta(\tau) = \sum_{n \in \mathbb{Z}} e^{\pi i n^2 \tau} \text{ for } \mathrm{Im}(\tau)>0.$$
Then it is rather easy to see that
$$\theta(\tau+2)=\theta(\tau)$$
and
$$\theta(\tau)+\theta(\tau+1)=2 \cdot \theta(4 \tau).$$
Further, one can prove that
$$\theta(-1/ \tau) = \sqrt{\frac{\tau}{i}} \cdot \theta(\tau).$$
All together, this yields
$$\theta(1- \frac{1}{\tau}) = \theta(- \frac{1}{\tau}) + \theta(1- \frac{1}{\tau}) - \theta(- \frac{1}{\tau}) = 2 \cdot \theta(- \frac{4}{\tau}) - \theta(- \frac{1}{\tau}) = \sqrt{\frac{\tau}{i}} \cdot (\theta(\frac{\tau}{4}) - \theta(\tau)).$$
Here $\sqrt{}$ always denotes the principal branch of the square root.
Next, my textbook claims that
$$f(\tau) := \theta^4(\tau)-\theta^4(\tau+1)+\tau^{-2} \cdot \theta^4(1- \frac{1}{\tau})$$
fulfils both $f(\tau+1) = - f(\tau)$ and $f(- \frac{1}{\tau}) = - \tau^{2} \cdot f(\tau)$ (and uses this to deduce that actually $f = 0$). It should be possible to derive these using the above $\theta$-identities, but I am completely stuck here. Any help is appreciated!
| Let us compute explicitly:
$$
\begin{aligned}
f(\tau)+f(\tau+1)
&=
\theta^4(\tau)-\theta^4(\tau+1)
+\tau^{-2} \cdot \theta^4\left(1- \frac{1}{\tau}\right)
\\
&\qquad\qquad
+
\theta^4(\tau+1)-\theta^4(\tau+2)
+(\tau+1)^{-2} \cdot \theta^4\left(1- \frac{1}{\tau+1}\right)
\\
&=
\tau^{-2} \cdot \theta^4\left(1- \frac{1}{\tau}\right)
+(\tau+1)^{-2} \cdot \theta^4\left(1- \frac{1}{\tau+1}\right)
\\
&=
\tau^{-2} \cdot\left[\ \theta^4\left(\frac{\tau-1}{\tau}\right)
+\left(\frac\tau{\tau+1}\right)^2 \cdot \theta^4\left(\frac{\tau}{\tau+1}\right)
\ \right]
\\
&=
\tau^{-2} \cdot\left[\ \theta^4\left(\frac{\tau-1}{\tau}-2\right)
-\theta^4\left(-\frac 1{\qquad\frac{\tau}{\tau+1}\qquad}\right)
\ \right]
\\
&=0\ ,
\\[3mm]
f\left(-\frac1\tau\right)
&=
\theta^4\left(-\frac1\tau\right)
-
\theta^4\left(-\frac1\tau+1\right)
+
\left(-\frac1\tau\right)^{-2}\theta^4\left(1-\frac1{-1/\tau}\right)
\\
&=
\theta^4\left(-\frac1\tau\right)
-
\theta^4\left(1-\frac1\tau\right)
+
\tau^2\theta^4(\tau+1)
\\
&=
-\tau^2\theta^4(\tau)
+\tau^2\theta^4(\tau+1)
-
\tau^2\cdot\tau^{-2}
\theta^4\left(1-\frac1\tau\right)
\\
&=
-\tau^2\;f(\tau)\ .
\end{aligned}
$$
(The power of $\tau$ differs in the last relation.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find the extremal of $\int_0^d \ddot{x}^2-\alpha x \: dt $ We are asked to find the extremal of
$\displaystyle \int_0^d \ddot{x}^2-\alpha x \: dt$
where $x=x(t),\: \alpha$ constant, $d>0$, $x(0)=0, \: x(d)=0$, $\dot x(0) = 0, \: \dot x(d) = 0$,
by considering:
$\displaystyle 0 = \frac{\partial f}{\partial x}- \frac{d}{dt}\frac{\partial
f}{\partial\dot x} + \frac{d^2}{dt^2}\frac{\partial f}{\partial \ddot x}$
What I have is:
$f(t,x,\dot x, \ddot x) = \ddot{x}^2-\alpha x$
and solving:
\begin{align*}
0
&= \frac{\partial f}{\partial x}- \frac{d}{dt}\frac{\partial f}{\partial \dot x} + \frac{d^2}{dt^2}\frac{\partial f}{\partial\ddot x}\\
&= -\alpha + 0 + \frac{d^2}{dt^2}2\ddot x\\
x^{(4)}&=\frac{\alpha}{2}\\
\Longrightarrow x &= \frac{\alpha}{48}t^4 + \beta t^3 + \gamma t^2 + \delta t + \varepsilon, \quad \beta, \gamma, \delta, \varepsilon \in \mathbb R
\end{align*}
From $x(0)=0 \Longrightarrow \varepsilon=0 $, $\dot x(0)=0 \Longrightarrow \delta = 0$.
$x(d)=0 \Longrightarrow 0 = \frac{\alpha}{48} d^4 + \beta d^3 + \gamma d^2 \Longrightarrow 0 = \frac{\alpha}{48} d^2 + \beta d + \gamma \Longrightarrow \gamma = - \frac{\alpha}{48} d^2- \beta d $
$\dot x(d)=0 \Longrightarrow 0 = \frac{\alpha}{12} d^3 + 3\beta d^2 + 2\gamma d$
$\Longrightarrow 0 = \frac{\alpha}{12} d^2 + 3\beta d + 2\gamma = \frac{\alpha}{12} d^2 + 3\beta d + 2(- \frac{\alpha}{48} d^2- \beta d)= \frac{\alpha}{24} d^2 + \beta d \Longrightarrow 0 =\frac{\alpha}{24} d^2 + \beta d $
$\Longrightarrow \beta = - \frac{\alpha}{24} d $
$\Longrightarrow \gamma = - \frac{\alpha}{48} d^2- \beta d = - \frac{\alpha}{48} d^2 - (-\frac{\alpha}{24} d)d =\frac{\alpha}{48} d^2 $
So, $\displaystyle x(t) = \frac{\alpha}{48}t^4 -\frac{\alpha}{24} d t^3 + \frac{\alpha}{48} d^2t^2$ is the extremal.
I'm just checking if this is correct?
| For a critical value of
$$
\int_0^d\left(\ddot x^2-\alpha x\right)\mathrm{d}t\tag1
$$
we need
$$
\begin{align}
0
&=\delta\int_0^d\left(\ddot x^2-\alpha x\right)\mathrm{d}t\\
&=\int_0^d\left(2\ddot x\,\delta\ddot x-\alpha\,\delta x\right)\mathrm{d}t\\
&=\int_0^d\left(-2\dddot x\,\delta\dot x-\alpha\,\delta x\right)\mathrm{d}t\\
&=\int_0^d\left(2\,\ddddot x\,\delta x-\alpha\,\delta x\right)\mathrm{d}t\tag2
\end{align}
$$
Thus, the critical point is when $\ddddot x=\frac\alpha2$. That would mean
$$
x=\frac\alpha{48}t^4+a_3t^3+a_2t^2+a_1t+a_0\tag3
$$
Computing $a_k$ so that $x(0)=x(d)=\dot x(0)=\dot x(d)=0$ and gives
$$
x=\frac\alpha{48}t^4-\frac{\alpha d}{24}t^3+\frac{\alpha d^2}{48}t^2\tag4
$$
So, your answer looks correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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The polynomial $x^3 + 2ax^2 + (2a^2 + b)x + c$ has three (real and not necessairily distinct) roots and $b$ is one of them. Prove that $(ac)^2 \le 3$.
The polynomial $$\large x^3 + 2ax^2 + (2a^2 + b)x + c$$ has three (real and not necessairily distinct) roots and $b$ is one of them. Prove that $(ac)^2 \le 3$.
I'm uncertain of how to prove this.
If $b$ is a root of $x^3 + 2ax^2 + (2a^2 + b)x + c$ then $b$ is also a root of $$(2a + b)x^2 + 2a^2x + (b^2 + c) = 0$$
which means the above polynomial has at least one root $\implies (a^2)^2 - (2a + b)(b^2 + c) \ge 0$
$\iff a^4 - 2ab^2 - 2ca - b^3 - bc \ge 0$
And $b$ is also a root of $x^3 + 2ax^2 + (2a^2 + b)x + c$ then $b$ is also a root of $$2ax^2 + (2a^2 + b)x + (b^3 + c) = 0$$
which means the above polynomial has at least one root $\implies (2a^2 + b)^2 - 4 \cdot 2a(b^3 + c) \ge 0$
$\iff 4a^4 + 4a^2b - 8ab^3 - 8ca + b^2 \ge 0$
But that's all I got.
| ►In the case of three equal roots, it is proved without difficulty that there are only two possible cases $f(x)=x^3$ and $f(x)=(x+\frac23)^3$.
►For three roots $b,r,r$ we have
$$f(x)=x^3-(b+2r)x^2+(2br+r^2)x-br^2\\f(x)=x^3+2ax^2+(2a^2+b)x+c$$ then$$\begin{cases}b+2r=-2a\\2br+r^2=2a^2+b\\-br^2=c\end{cases}\Rightarrow b^2+2b+2r^2=0\Rightarrow b\lt0$$
All the possible $f(x)$ in this second condition with the roots $b,r,r$ are such that the points $(b,r)$ are in the ellipse centered at $(-1,0)$ and having axes $1$ and $\dfrac{1}{\sqrt2}$.
On the other hand
$$(ac)^2=\left(\frac{2r+b}{-2}\right)^2(-br^2)^2=\frac{b^2r^4(2r+b)^2}{4}$$
so, equivalently, we can prove $$b^2r^4(2r+b)^2\le12\qquad(*)$$
Taking the parametrics of the ellipse above we have $(b,r)=\left(\cos(t)-1,\dfrac{\sin(t)}{\sqrt2}\right)$ and the inequality (*) becomes
$$g(t)=(\cos(t)-1)^2\sin^4(t)(\sqrt2\sin(t)+\cos(t)-1)^2\le{24}$$
It is clear that $g(t)$ has an absolute maximum less than $24$ (exactly this absolute maximum is equal to $10.074$ and is taken at point $t=4.325$).
►For three distinct roots $b,r,s$ we have similarly the relation
$$b^2+r^2+s^2+2b=0$$ and we have to prove equivalently that
$$r^2sb+s^2rb+b^2rs\le 2\sqrt3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Ratio distribution of continous uniform Let $X, Y$ be positive random variables with densities $f_X=\mathbb{I}_{[0, 1]}$ and $f_Y=f_X*f_X$. To find the density of $Y$ we observe that when $0\le y\le 1$ we have $f_Y(y)=y$ and when $1< y\le 2$ then $f_Y(y)=2-y$.
Now if $S=X+Y$ and $R={X\over S}$ I want to find $f_R(r)$.
By the change of variables we get:
$$f_{R,S}(r,s)=sf_X(rs)f_Y(s(1-r))$$
$$f_{R,S}(r,s)=\begin{cases}
s^2(1-r)\mathbb{I}_{0\le s\le {1\over r}}\ \mathbb{I}_{0\le s\le {1\over 1-r}}
\\
(2s-s^2(1-r))\mathbb{I}_{0\le s\le {1\over r}}\ \mathbb{I}_{{1\over 1-r}< s\le {2\over 1-r}}
\end{cases}$$
Now we want to integrate w.r.t $s$.
For the first case when $0\le r\le {1\over 2}$ then $0\le s\le {1\over 1-r}$ and
when ${1\over 2}<r\le1$ then $0\le s\le {1\over r}$
For the second case when $0\le r \le{1\over 3}$ then ${1\over 1-r}\le s\le {2\over 1-r}$ and when ${1\over 3}<r \le{1\over 2}$ then ${1\over 1-r}\le s\le {1\over r}$. which gives:
$$f_R(r)=\begin{cases}
{1\over 3(1-r)^2}\ \mathbb{I}_{0\le r\le {1\over 2}}
\\
{1-r\over 3r^3}\ \mathbb{I}_{{1\over 2}< r \le 1}
\\
{2\over 3(1-r)^2} \mathbb{I}_{0\le r \le {1\over 3}}
\\
{2r^3-9r^2+6r-1\over 3r^3(1-r)^2}\mathbb{I}_{{1\over 3}< r \le {1\over 2}}
\end{cases}$$
But this is not a density function. Where did I go wrong?
| The joint density of $R, S$ is correct.
Obviously the support of $R$ is $[0, 1]$. According to the support of the joint, you should split this support into the following $3$ intervals, to facilitate your calculation:
$$ \left(0, \frac {1} {3}\right), \left(\frac {1} {3}, \frac {1} {2}\right), \left(\frac {1} {2}, 1\right)$$
So when $\displaystyle r \in \left(0, \frac {1} {3}\right)$,
$$ \begin{align}
f_R(r) &= \int_0^{1/(1-r)} s^2(1-r)ds + \int_{1/(1-r)}^{2/(1-r)}2s-s^2(1-r)ds
= \frac {1} {(1 - r)^2}
\end{align}$$
When $\displaystyle r \in \left(\frac {1} {3}, \frac {1} {2}\right)$,
$$ \begin{align}
f_R(r) &= \int_0^{1/(1-r)} s^2(1-r)ds + \int_{1/(1-r)}^{1/r}2s-s^2(1-r)ds
= \frac {3r^3 - 9r^2 + 6r - 1 } {3r^3(1 - r)^2}
\end{align}$$
When $\displaystyle r \in \left(\frac {1} {2}, 1\right)$,
$$ \begin{align}
f_R(r) &= \int_0^{1/r} s^2(1-r)ds + 0
= \frac {1-r} {3r^3}
\end{align}$$
Note that the second integral vanish as $1/r < 1/(1-r)$ in this case.
You may further check that
$$ \int_0^1 f_R(r)dr = \frac {1} {2} + \frac {1} {3} + \frac {1} {6} = 1 $$
and it is positive over its support. So it is a valid pdf candidate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Zeros at the end of sum of factorials Find the number of zeros at the end of $15! + 16! + 17! + 18!$ ?
I know the method find the number of zeros at the end of x! where $x = { 15! , 16! , 17! ...}$ by dividing by number by $5,5^2, 5^3$ and so on .
| First divide $15!$ by $10^3$ (in the obvious way, from its expression) to determine its last nonzero digit. Hopefully, it will be the 4th digit from the right.
You're left with
$$N=3\cdot 4\cdot 6\cdot 7\cdot 8\cdot 9\cdot 11\cdot 12\cdot 13\cdot 7\cdot 3= 2^8\cdot 3^6\cdot 7^2\cdot 11\cdot 13$$
Modulo $100$, we have
$$N\equiv 56\cdot 29\cdot 49\cdot 43=56\cdot 29\cdot (50-1)(40+3)\equiv 56\cdot 29\cdot 7\equiv 56\cdot 3\equiv 68 $$
As a conclusion, $\; 15!\equiv 68\,000\mod 10^5$.
Now we'll compute:
\begin{align}M&=1+16+16\cdot 17+16\cdot 17\cdot 18=17^2+16\cdot 17\cdot 18=17(17+17^2-1)\\
&= 17(17+289-1)=17\cdot 305=5185
\end{align}
Thus, $$15!+16!+17!+18!\equiv68\,000\cdot 5185\equiv 8000\cdot 85=680\,000\equiv \color{red}{ 80000} \mod 10^5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3348238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of all entries In a Matrix of Linear Transformation So,few days before I had a Linear Algebra Test and we had following question :
Let $P_3[R]$ denote the vector space of all polynomials with degree $\le 3$. Define the linear Transformation $P_3[R]$ $\to$ $P_3[R]$ defined by :
$T(p(x))$ = $p(x)$ + $p'(x)$ + $p''(x)$ Then find the sum of all the matrix of T relative to the usual basis.
Now the basis for given vector space is :$\{1,x,x^2,x^3\}$ , Using this:
$T(1)$ = $1$
$T(x)$ = $x$ + $1$ +$0$ = $1.1$ + $1.x$ + $0.x^2$ + $0.x^3$
$T(x^2$) = $x^2$ + $2x$ + $2$ = $2.1$ + $2.x$ + $1.x^2$ + $0.x^3$
$T(x^3)$ = $x^3$ + $3x^2$ + $6x$ = $0.1$ + $6.x$ + $3.x^2$ + $1.x^3$
The Matrix is given by :
$$
\begin{pmatrix}
1 & 1 & 2 & 0\\
0 & 1 & 2 & 6 \\
0 & 0 & 1 & 3\\
0 & 0 & 0 & 1\\
\end{pmatrix}$$
Clearly sum of all entries in the matrix is $18$ but the answer given to me is $24$
Can anyone tell me what is the error in my solution ?
Thank you.
| In $$(1+x+x^2+x^3)+(1+2x+3x^2)+(2+6x),$$
the sum of the coefficients is $18$.
In $$(1+x+x^2+x^3)+(1+2x+3x^2)+(2+6x)+(6),$$
it is $24$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all strings of length 4 over the alphabet {0, 1, 2} with the property that there are no 10, 21, or 20 substrings. What is the formula for doing this type of problems?
| Here we use PIE the inclusion-exclusion principle to count the number of valid $4$-letter words from the alphabet $\{0,1,2\}$ which do not contain the bad words $\{10,20,21\}$.
In order to do the job some kind of bookkeeping is helpful. We consider
\begin{align*}
.\ .\ .\ . &-\left(10\ .\ .|20\ .\ .|21\ .\ .\right)\tag{1}\\
&+\left(10\ 10|10\ 20|10\ 21|20\ 20|20\ 21|210\ .|21\ 21\right)\tag{2}\\
\end{align*}
Comment:
*
*In (1) we count all $4$-letter words indicated by four dots which gives $3^4$. Then we subtract all words which contain at least one bad word.
Since $10$ consumes two characters and two are left for free assignment, we count $\binom{3}{1}3^2$ words of this kind and similarly in the other cases with the bad words $20$ and $21$.
*In (2) we add words containing two bad words as compensation for those which we've subtracted twice in (1), noting that we also have to consider overlaps $21$ with $10$ giving $\color{blue}{210}$.
*No more cases are left to consider, since words containing three or more bad words have length $>4$.
We obtain according to (1) to (2):
\begin{align*}
3^4&-\left(\binom{3}{1}3^2+\binom{3}{1}3^1+\binom{3}{1}3^1\right)\\
&\quad+\left(\binom{2}{2}3^0+2\binom{2}{2}3^0+2\binom{2}{2}3^0+\binom{2}{2}3^0+2\binom{2}{2}3^0+\binom{2}{1}3^1
+\binom{2}{2}3^0\right)\\
&=81-(27+27+27)+(1+2+2+1+2+6+1)\\
&=81-81+15\\
&\,\,\color{blue}{=15}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to check if $ x^2 + x = y^2 - 1$ has positive integer solutions? How to check if $ x^2 + x = y^2 - 1$ has positive integer solutions?
I've tried using different modulos to simplify problem, but nothing has worked yet.
| Let $x,y$ be such that $x$ is a positive integer and $x^2+x+1=y^2$. $$x^2<x^2+x+1<x^2+2x+1\\x^2<y^2<(x+1)^2$$
So clearly $y$ cannot be also be a positive integer since there are no positive integers between $x$ and $x+1$.
| {
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An algebra problem of olympiad Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies the relation:
$$a_{k+1} \geq \frac{ka_k}{a_k^2 + (k-1)}$$
for every positive integer $k$. Prove that:
$$a_1+a_2+\cdots+a_n⩾n \text{ for } n⩾2.$$
This is an indian olympiad problem.Can you guys help me solve this out.
edit:
I have tried it by first taking $k = 1$,
then we get $a_2 \geq \frac{1}{a_1}$. By putting $k=2$ we get
$a_3 \geq \frac{2a_2}{a_2^2 + 1}$ and similarly, $a_4 \geq \frac{3a_3}{a_3^2 +2}$.
But I couldn't find any relation between them.
|
Lemma
Let $\{ a_n \} _ {n=1} ^{\infty} $ be as stated in the problem. For
all $n \geq 2$, the following inequality holds: $$ \sum_{1 \leq k < n}
a_k \geq \frac{n-1}{a_n} $$
proof. Use induction on $n$. First of all, we know $a_2 \geq \frac{a_1}{a_1^2 + (1-1)}=\frac{1}{a_1}$. Thus $a_1 \geq \frac{1}{a_2}$. Now assume the claim holds for $n\geq 2$. Then $$a_1 + \cdots + a_n = (a_1 + \cdots + a_{n-1})+a_n \geq \frac{n-1}{a_n}+a_n = \frac{(n-1) + a_n^2}{a_n} \geq \frac{n}{a_{n+1}} $$ The last equality is because the sequence $\{ a_n \} _ {n=1} ^{\infty} $ satisfies the relation $a_{n+1} \geq \frac{na_n}{a_n^2 + (n-1)}$.
Let's prove $$a_1 + \cdots + a_n \geq n$$ by induction on $n$.
If $n=2$, we have $a_1 + a_2 \geq a_1 + \frac{1}{a_2} \geq 2$ by the AM-GM inequality.
Assume the inequality holds for some $n \geq 2$. If $a_{n+1} \geq 1$, it is immediate that $a_1 + \cdots + a_{n+1} \geq n+1$. Let's suppose $0< a_{n+1} < 1$. Observe that $$ X:=\frac{(n-1)+ a_{n}^2 }{a_{n}} \geq \frac{n}{a_{n+1}} \geq n $$
and that $f(x) = x + \frac{n}{x}$ is an increasing function on $[\sqrt{n}, \infty)$. Now \begin{align*} a_1 + \cdots + a_{n+1} &= (a_1 + \cdots + a_{n-1}) + a_{n} + a_{n+1}
\\ &\geq \frac{n-1}{a_{n}} + a_{n} + \frac{n}{ a_{n} + \frac{n-1}{a_{n}} }
\\ &= f(X)
\\ &\geq f(n) = n + 1. \end{align*} so the induction step is achieved.
About a year after answering, I found that this is a problem from the 2015 IMO shortlist.
| {
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Determine $\left(1 - \frac{x_{1}}{y_{1}} \right)\left(1 - \frac{x_{2}}{y_{2}} \right)\left(1 - \frac{x_{3}}{y_{3}} \right)$ $(x_{i},y_{i}), i=1,2,3$ are solutions for
$$ x^{3} - 3xy^{2} = 2010 $$
$$ y^{3} - 3yx^{2} = 2009 $$
What is $\left(1 - \frac{x_{1}}{y_{1}} \right)\left(1 - \frac{x_{2}}{y_{2}} \right)\left(1 - \frac{x_{3}}{y_{3}} \right)$?
Attempt:
My approach is that we should look for $\frac{y-x}{y}=$something.
Notice that
$$ x^{3} - y^{3} + 3(yx^{2} - xy^{2}) = 1 $$
$$ (x-y)(x^{2} + y^{2} + xy) + 3xy(x-y) = 1 $$
$$ (y-x)(x^{2} + y^{2} + 4xy) = -1$$
$$ \frac{y-x}{y} = -\frac{1}{y(x^{2} + y^{2} + 4xy)} (\text{does not seemt to go anywhere})$$
Anther approach if I sum the equations:
$$ x^{3} + y^{3} - 3(xy^{2} + yx^{2}) = 4019 $$
$$ (x + y)(x^{2} + y^{2} - xy) - 3(xy)(x + y) = 4019 $$
$$ (x+y)(x^{2} + y^{2} - 4xy) = 4019 $$
Multiply both we get
$$ (x^{2}-y^{2})((x^{2}+y^{2})^{2} -16 (xy)^{2}) = 4019 $$
???
| It is easy to see none of the $y_i$ vanishes.
Let $u = \frac{x}{y}$ and $u_i = \frac{x_i}{y_i}$. $u_i$ are roots of the equation:
$$x^3 - 3xy^2 - \frac{2010}{2009}(y^3 - 3yx^2) = 0
\iff u^3 - 3u - \frac{2010}{2009}(1 - 3u^2) = 0
$$
This implies
$$(u-u_1)(u-u_2)(u - u_3) = u^3 - 3u - \frac{2010}{2009}(1 - 3u^2)$$
The value we want is the value of this expression at $u = 1$.
$$\prod_{i=1}^3\left(1 - \frac{x_i}{y_i}\right) = \prod_{i=1}^3 (1 - u_i)
= 1 - 3 - \frac{2010}{2009}(1-3) = \frac{2}{2009}$$
| {
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Inequality : $\sum_{cyc}\Big(\frac{x^4}{8x^3+5y^3}\Big)^3\geq \frac{x^3+y^3+z^3}{13^3}$ It's related to this Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$:
$x,y,z >0$, prove
$$\Big(\frac{x^4}{8x^3+5y^3}\Big)^3+\Big(\frac{y^4}{8y^3+5z^3}\Big)^3+\Big(\frac{z^4}{8z^3+5x^3}\Big)^3 \geqslant \frac{x^3+y^3+z^3}{13^3}$$
The inequality is equivalent to :
$$x^3\Big(\frac{1}{8+5(\frac{y}{x})^3}\Big)^3+y^3\Big(\frac{1}{8+5(\frac{z}{y})^3}\Big)^3+z^3\Big(\frac{1}{8+5(\frac{x}{z})^3}\Big)^3 \geqslant \frac{x^3+y^3+z^3}{13^3}$$
Or :
$$\frac{x^3\Big(\frac{1}{8+5(\frac{y}{x})^3}\Big)^3+y^3\Big(\frac{1}{8+5(\frac{z}{y})^3}\Big)^3+z^3\Big(\frac{1}{8+5(\frac{x}{z})^3}\Big)^3}{x^3+y^3+z^3} \geqslant \frac{1}{13^3}$$
Now we apply the Jensen's inequality to the function $f(x)=\frac{1}{(8+5x)^3}$ wich is clearly convex we get :
$$\frac{x^3 f((\frac{y}{x})^3)+y^3 f((\frac{z}{y})^3)+z^3 f((\frac{x}{z})^3)}{x^3+y^3+z^3} \geqslant f\Big(\frac{x^3 (\frac{y}{x})^3+y^3 (\frac{z}{y})^3+z^3 (\frac{x}{z})^3}{x^3+y^3+z^3}\Big)=\frac{1}{13^3}$$
And we are done .
My question is have you an nice alternative proof ?
Thanks a lot for your time .
| By Holder
$$\sum_{cyc}\left(\frac{x^4}{8x^3+5y^3}\right)^3\left(\sum_{cyc}(8x^3+5y^3)\right)^3\geq(x^3+y^3+z^3)^4.$$
Thus, it's enough to prove that
$$(x^3+y^3+z^3)^4\geq\frac{x^3+y^3+z^3}{13^3}\cdot\left(\sum_{cyc}(8x^3+5y^3)\right)^3,$$ which is an equality.
| {
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Use the method of finite differences to determine a formula Please evaluate my solution and let me know of anywhere I can improve or make clearer. Thank you
QUESTION:Use the method of finite differences to determine a formula for the general term of the sequence whose first seven terms are:
$$5, 19, 49, 101, 181, 295, 449, \dots$$
SOLUTION:
The given sequence is
$$S= 5, 19, 49, 101, 181, 295, 449, \dots$$
We can use the method of finite differences to find the general term. We write the difference of consecutive terms of the sequence till we can get a constant sequence.
\begin{align*}
D_1 &= 14, 30, 52, 80, 84, 153, \dots \\
D_2 &= 16, 22, 28, 34, 40, \dots \\
D_3 &= 6, 6, 6, 6, \dots
\end{align*}
So $D_3$ is a constant sequence. This means its general term is a third degree polynomial.
The sequence general term is $X_n = An^3 + Bn^2 + Cn +D$
Now, we know that
\begin{align*}
x_1 &= 5 \\
x_2 &= 19 \\
x_3 &= 49 \\
x_4 &= 101
\end{align*}
Using $x_n = An^3 + Bn^2 + Cn +D$
We get,
\begin{equation} \label{eq1}
5 = A + B + C + D
\end{equation}
\begin{equation} \label{eq2}
19 = 8A + 4B + 2C + D
\end{equation}
\begin{equation} \label{eq3}
49 = 27A + 9B + 3C + D
\end{equation}
\begin{equation} \label{eq4}
101 = 64A + 16B + 4C + D
\end{equation}
Using equation (2) - equation (1), we get
\begin{equation} \label{eq5}
14 = 7A + 3B + C
\end{equation}
Using equation (3) - equation (2), we get
\begin{equation} \label{eq6}
30 = 19A + 5B + C
\end{equation}
Using equation (4) - equation (3), we get
\begin{equation} \label{eq7}
52 = 37A + 7B + C
\end{equation}
Now subtracting again,
Using equation (6) - equation (5), we get
\begin{equation} \label{eq8}
16 = 12A + 2B
\end{equation}
Using equation (7) - equation (6), we get
\begin{equation} \label{eq9}
22 = 18A + 2B
\end{equation}
Again we subtract,
Using equation (9) - equation (8), we get
\begin{equation} \label{eq10}
6 = 6A \Rightarrow A = 1
\end{equation}
Now solving for $B, C,$ and $D$ we get
\begin{equation} \label{eq11}
22 - 18 = 2B \Rightarrow B = 2
\end{equation}
\begin{equation} \label{eq12}
14 = 7 + 6 + C \Rightarrow C = 1
\end{equation}
\begin{equation} \label{eq13}
5 = 1 + 2 + 1 + D \Rightarrow D = 1
\end{equation}
Hence our general term is
$$ n^3 + 2n^2 + n + 1 \text{ for } n \in \mathbb{N}$$
Solution page 1
Solution page 2
| What you did looks basically correct. Nonetheless, one small thing the question is implicitly stating, and you're implicitly using, is that the $D_3$ row of $6$'s will continue indefinitely. There are formulas (although they're somewhat convoluted) which give sequences where the first $7$ values match what you're given, but where not all of the later values of $D_3$ would be $6$.
Another issue is with with the terminology. As stated in the first sentence of Wikipedia's Finite difference page
A finite difference is a mathematical expression of the form $f(x + b) − f(x + a)$.
This does apply to what you're doing. However, note the Wikipedia page says at the start of the third paragraph that
Today, the term "finite difference" is often taken as synonymous with finite difference approximations of derivatives, especially in the context of numerical methods.
Also, Wolfram's MathWorld Finite difference page starts with
The finite difference is the discrete analog of the derivative.
I believe the term "finite difference" likely comes from the idea that you're using one specific, even if very small, finite difference, as opposed to differentiation which looks at using the limiting case of infinitely many small intervals to determine a difference to calculate the derivative function exactly.
Instead, as explained in Mathonline's Difference Sequences page, the $p$'th row you've determined is called the "$p^{\text{th}}$ order difference sequence". The entire set of these rows, as explained in Mathonline's Difference Tables of Sequences, is called a "difference table". This phrase is also used in other places, such as in the Example section of Wikipedia's "Binomial transform" page.
Regarding the solution method, there's an alternate method, as suggested by the Wikipedia article mentioned above, and explained quite well in the Jean-Claude Arbaut's answer to Baffled by resolving number list. Using this, assuming the index of the polynomial values starts at $0$, we get by using the binomial transform with the left-most values of $S$ and $D_i$ that
$$\begin{equation}\begin{aligned}
p_0(n) & = 5{n \choose 0} + 14{n \choose 1} + 16{n \choose 2} + 6{n \choose 3} \\
& = 5 + 14n + 16\left(\frac{n(n-1)}{2}\right) + 6\left(\frac{n(n-1)(n-2)}{6}\right) \\
& = 5 + 14n + 8(n^2 - n) + (n^3 - 3n^2 + 2n) \\
& = n^3 + 5n^2 + 8n + 5
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Converting this to using a starting index of $1$ instead gives
$$\begin{equation}\begin{aligned}
p_0(n) & = p_1(n-1) \\
& = (n-1)^3 + 5(n-1)^2 + 8(n-1) + 5 \\
& = (n^3 - 3n^2 + 3n - 1) + 5(n^2 - 2n + 1) + (8n - 8) + 5 \\
& = n^3 + 2n^2 + n + 1
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
This matches what you got. Alternatively, we can use the pattern to determine the previous values in the sequences. Starting from the $6$'s in $D_3$, we get that the previous value in $D_2$ would be $16 - 6 - 10$, the previous value in $D_1$ would be $14 - 10 = 4$ and the previous previous value in $S$ would be $5 - 4 = 1$. Thus, using the binomial transform with these values values gives
$$\begin{equation}\begin{aligned}
p_1(n) & = 1{n \choose 0} + 4{n \choose 1} + 10{n \choose 2} + 6{n \choose 3} \\
& = 1 + 4n + 10\left(\frac{n(n-1)}{2}\right) + 6\left(\frac{n(n-1)(n-2)}{6}\right) \\
& = 1 + 4n + 5(n^2 - n) + (n^3 - 3n^2 + 2n) \\
& = n^3 + 2n^2 + n + 1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
If we're solving it using a set of linear equations, as you did, then there's a small short-cut we can use. If the degree of the polynomial is $d$ and the coefficient of the highest-order term is $a_d$, then the $D_d$ terms are all $a_d(d!)$. We can determine this by noting the coefficient of the highest order term in each $D_i$ level is $d - i + 1$ times that of the previous level, so these factors combine to form $d!$ times the $S$'s coefficient, i.e., $a_d$. We can also see this from the binomial transform as the highest order term comes from the $D_d$ values times ${n \choose d}$, so it'll be $a_d = \frac{D_d}{d!} \implies D_d = a_d(d!)$. Using this with your set of equations, we can see that $d = 3$, so $3! = 6$, gives $6 = 6A \implies A = 1$. Thus, we only need to use $3$ equations to get
$$5 = 1 + B + C + D \implies 4 = B + C + D \tag{4}\label{eq4A}$$
$$19 = 8 + 4B + 2C + D \implies 11 = 4B + 2C + D \tag{5}\label{eq5A}$$
$$49 = 27 + 9B + 3C + D \implies 22 = 9B + 3C + D \tag{6}\label{eq6A}$$
We can solve \eqref{eq4A}, \eqref{eq5A} and \eqref{eq6A} to get $B = 2$, $C = 1$ and $D = 1$, as you have done in your question text.
| {
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My target is to write the expression $(1)$ in terms of $P_k(N,x)$ Let $P_k(N,x)=\sum_{n=0}^{N-1} \binom{2n}{n} n^kx^n$. Suppose we have the expression as follows:
$A=\sum_{n=1}^{N-1} \binom{2n+1}{n+1} (n+1)^kx^{n+1}, \ \cdots (1)$
My target is to write the expression $(1)$ in terms of $P_k(N,x)$ or simply I want to embed $P_k(N,x)$ into $(1)$.
I have proceed as follows:
\begin{align} A &=\sum_{n=1}^{N-1}\left[ \binom{2n}{n}+\binom{2n}{n+1} \right] (n+1)^kx^{n+1} \\ &=\sum_{n=1}^{N-1} \binom{2n}{n} (n+1)^kx^{n+1}+\sum_{n=1}^{N-1}\binom{2n}{n+1}(n+1)^kx^{n+1} \\ &= \sum_{n=1}^{N-1} \binom{2n}{n} \sum_{l=0}^{k} \binom{k}{l} n^lx^{n+1}+\sum_{n=1}^{N-1}\binom{2n}{n+1}(n+1)^kx^{n+1} \\ &=x\sum_{l=0}^k \binom{k}{l} \sum_{n=1}^{N-1}\binom{2n}{n} n^lx^n+ \sum_{n=1}^{N-1}\binom{2n}{n+1}(n+1)^kx^{n+1} \\ &=x \sum_{l=0}^{k} \binom{k}{l} P_l(N,x)+\color{blue}{\sum_{n=1}^{N-1}\binom{2n}{n+1}(n+1)^kx^{n+1}}\end{align}
So I have been successfully replaced or input $P_k(N,x)$ into the first expression but unable to insert into the 2nd part( blue color part).
Can someone help me with some trick ?
Can I insert $P_k(N,x)$ in expression $(1)$ without splitting it ?
|
Let $$P_k(N,x)=\sum_{n=0}^{N-1}\binom{2n}{n}n^kx^n$$ We obtain
\begin{align*}
\color{blue}{A}&\color{blue}{=\sum_{n=1}^{N-1}\binom{2n+1}{n+1}(n+1)^kx^{n+1}}\\
&=\sum_{n=2}^{N}\binom{2n-1}{n}n^kx^n\tag{1}\\
&=\sum_{n=2}^{N}\binom{2n-1}{n-1}n^kx^n\tag{2}\\
&=\frac{1}{2}\sum_{n=2}^{N}\binom{2n}{n}n^kx^n\tag{3}\\
&\,\,\color{blue}{=\frac{1}{2}\binom{2N}{N}N^kx^N+\frac{1}{2}P_k(N,x)-x}
\end{align*}
Comment:
*
*In (1) we shift the index by $1$ and start with $n=2$.
*In (2) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.
*In (3) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.
| {
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Simplify the following expression $\sum_{j=0}^n 2^n \binom{1/2}{n-j}\binom{j-1/2}{j}$ Simplify the following expression $\sum_{j=0}^n 2^n \binom{1/2}{n-j}\binom{j-1/2}{j}$ to something without a summutation mark (and probably elementary).
The question originates from the $n$th term of the polynomial expansion of $\sqrt{\frac{1+2x}{1-2x}}$, i.e. the coefficient of $x^n$ in its Maclaurin series.
We define $\binom{r}{n}=\frac{a(a-1)(a-2)...(a-b+1)}{b!}$ where $r$ is any real number and $n$ is a natural number.
| We already know for $n\geq 0$:
\begin{align*}
\sum_{j=0}^n 2^n \binom{1/2}{n-j}\binom{j-1/2}{j}=[x^n]\sqrt{\frac{1+2x}{1-2x}}\tag{1}
\end{align*}
We obtain from (1)
\begin{align*}
\color{blue}{\sum_{j=0}^n}&\color{blue}{ 2^n \binom{1/2}{n-j}\binom{j-1/2}{j}}\\
&=[x^n]\sqrt{\frac{1+2x}{1-2x}}\\
&=[x^n]\frac{1+2x}{\sqrt{1-4x^2}}\tag{2}\\
&=[x^n]\left(1+2x\right)\sum_{j=0}^\infty\binom{-\frac{1}{2}}{j}\left(-4x^2\right)^j\tag{3}\\
&=\left([x^n]+2[x^{n-1}]\right)\sum_{j=0}^\infty\binom{-\frac{1}{2}}{j}(-4)^jx^{2j}\tag{4}\\
&\,\,\color{blue}{=}\begin{cases}
\color{blue}{\binom{-\frac{1}{2}}{m}(-4)^m}&\qquad\color{blue}{ n=2m}\\
\tag{5}\\
\color{blue}{2\binom{-\frac{1}{2}}{m}(-4)^m}&\qquad\color{blue}{ n=2m+1}
\end{cases}
\end{align*}
Comment:
*
*In (2) we expand the fraction with $1+2x$.
*In (3) we apply the binomial series expansion.
*In (4) we use $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
*In (5) we select the coefficient of $x^n$ for even and odd $n$.
| {
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How to find the maximum value of the expression $y=2(a-x)(x+\sqrt{x^2+b^2})$ The maximum value of the expression $y=2(a-x)(x+\sqrt{x^2+b^2})$
If we take derivative, then I am not getting anything, please guide how to proceed in such problems will be of great help. Thanks.
| Let $\sqrt{x^2+b^2}+x=z.$ Then $\displaystyle \sqrt{x^2+b^2}-x=\frac{b^2}{z}.$
So $$y=z \bigg(2a-z+\frac{b^2}{z}\bigg)=2az-z^2+b^2=a^2+b^2-(z-b)^2\leq a^2+b^2$$
| {
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For which value of $a$ is the matrix is isometric?
For which value of $a$ is the matrix $\begin{pmatrix} a & 0 \\1 &1\ \end{pmatrix}$ is isometrics ?
My attempt : For isometric it must be unitary so $a = -1$
Is its true ?
| Suppose $\begin{pmatrix} a & 0 \\1 &1\ \end{pmatrix}\begin{pmatrix} x \\y\ \end{pmatrix}$ and $\begin{pmatrix} x \\y\ \end{pmatrix}$ have the same length .
Then $a^2x^2+(x+y)^2=x^2+y^2$ which is not possible for all $x$ and $y$.
| {
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Recursive sequence problem... Consider the recursive sequence
\begin{equation*}
\begin{split}
a_{n+1} = \frac{5}{6-a_n} \quad \textit{with} \quad a_1 = 4.
\end{split}
\end{equation*}
Prove that the sequence $(a_n)$ converges and find its limit, by working out the following steps.
1. First assume that the limit $L = \lim_{n\to \infty} a_n$ exists and find its possible values.
Let $L = \lim_{n\to\infty}a_n$. Then we get
\begin{equation*}
\begin{split}
L &= \lim_{n\to\infty}a_{n+1} \\
&= \lim_{n\to\infty} \frac{5}{6-a_n} \\
&= \frac{5}{6-L}.
\end{split}
\end{equation*}
So we have $L^2-6L+5 = 0 \Longleftrightarrow (L-1)(L-5) = 0$. So the possible values of $L$ are $L = 1$ or $L = 5$.
2. Starting with the initial value $a_1 = 4$, write down the first five entries in the sequence $(a_n)$. Can you see any pattern?
We have
\begin{equation*}
\begin{split}
a_2 &= \frac{5}{6-a_1} = \frac{5}{6-4} = \frac{5}{2} = 2.5 \\
a_3 &= \frac{5}{6-a_2} = \frac{5}{6-5/2} = \frac{10}{7} \approx 1.42857 \\
a_4 &= \frac{5}{6-a_3} = \frac{5}{6-10/7} = \frac{35}{32} = 1.09375 \\
a_5 &= \frac{5}{6-a_4} = \frac{5}{6-35/32} = \frac{160}{157} \approx 1.01091 \\
a_6 &= \frac{5}{6-a_5} = \frac{5}{6-160/157} = \frac{785}{782} \approx 1.00384.
\end{split}
\end{equation*}
3. Find the real valued function $f(x)$ defining the sequence, i.e. $a_{n+1} = f(a_n)$.
This is the question I'm having trouble with. Please help!
| The recuurence relation is $$u_{n+1}(u_n-6)=-5~~~(1)$$
Let $$u_{n}-6=\frac{v_{n-1}}{v_{n-2}}~ in ~(1).$$
We get $$v_n+6 u_{n-1}+5 v_{n-2}~~~(2)$$
Let $v_n=x^n$ in (2), we get $$x^2+6x+5=0 \Rightarrow
x_1=-5,x_2=-1$$; then $$v_n=C_{1} (-5)^n+ C_2(-1)^n~~~(3)$$
$$u_n=\frac{C_1 (-5)^{n-1}+ C_2(-1)^{n-1}}{C_1 (-5)^{n-2}+ C_2(-1)^{n-2}}+6.$$
The single unknown $D=C_1/ C_2$ can be determined by the initial values of $u_n,~$ namely $u_1=4$. Here $D$ coes out to be $5/3$, then
$$U_n=6+\frac{5(-5)^{n-1}+3(-1)^{n-1}}{5(-5)^{n-2} +3 (-1)^{n-2}}.$$
ehere it follows that $\lim_{n \rightarrow \infty} u_n=1.$
| {
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"url": "https://math.stackexchange.com/questions/3372834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does this trick for solving this equation work? The question is to
solve the equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 1.$$
Now if we solve $ \left | 2x-2 \right | = \left | 4x-5 \right |$ first (that is, setting the denominators equal) we find that $x=\frac{3}{2},\frac{7}{6}.$ If we now go back to substitute these values of $x$ in LHS of the original equation, we get that $$\frac{ 4x\left ( \left ( 4x-5 \right )^{2}+3 \right ) + 12x\left ( \left ( 2x-2 \right )^{2}+3 \right ) }{\left ( \left ( 4x-5 \right )^{2}+3 \right )\left ( \left ( 2x-2 \right )^{2}+3 \right )} = 6.$$
Thus, we see that $x=\frac{3}{2},\frac{7}{6}$ both satisfy the equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 6.$$
However, I don't know how this is relevant to the fact that $x=\frac{1}{2},\frac{7}{2}$ are solutions to the original equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 1.$$
That is, what is the relationship between the respective solutions of $\text{original LHS}=6$ and $\text{original LHS}=1$?
Please tell me the mystery behind this process, and can we use this technique on other equations like this? Thank you.
| Note that$$\frac{4x}{(2x-2)^2+3}+\frac{12x}{(4x-5)^2+3}-1=\frac{-16 x^4+100 x^3-200 x^2+175 x-49}{\left(4 x^2-10 x+7\right) \left(4x^2-8 x+7\right)},$$the solutions of your equations are the roots of the polynomial$$-16 x^4+100 x^3-200 x^2+175 x-49.$$Using the rational root theorem, you can deduce that $\frac12$ and $\frac72$ are roots of this polynomial. On the other hand$$\frac{-16 x^4+100 x^3-200 x^2+175 x-49}{\left(x-\frac12\right)\left(x-\frac72\right)}=-4 \left(4 x^2-9 x+7\right).$$So, there are no more real roots and there are two complex non-real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving $2x^3-24x^2+23x+283=0$ I need to solve this equation:
$$\frac{x^2+1}{x-4}+\frac{x^2-1}{x+3}=23$$
When opened up it comes to this:
$$2x^3-24x^2+23x+283=0$$
Then I am stuck...
It feels like it has something to do with complex numbers perhaps?
Any ideas? Thank you!
| Let me be more puristic.
The first thing to notice is that the discriminant is positive $(\Delta=1582748)$; so three real roots with $p=-\frac{73}{2}$ and $q=\frac{119}{2}$.
Now, instead of Cardano, let us use the trigonometric method for cubic equation which write
$$x_k=-\frac b{3a}+2\sqrt{-\frac p 3} \cos\Big(\frac 13 \cos^{-1}\Big(\frac{3q}{2p}\sqrt{-\frac 3p } \Big)-\frac{2\pi}3k\Big)\qquad \text{for}\qquad k=0,1,2$$
Applied to your case this will give
$$x_k=4+\sqrt{\frac{146}{3}} \cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos
^{-1}\left(-\frac{357}{73}\sqrt{\frac{3}{146}}\right)\right)$$
Their decimal representation would be
$$x_0=8.94695974511457058 \qquad x_1=5.78629617321812462\qquad x_2=-2.73325591833269521$$
Edit
If you use your pocket calculator
$$\frac{1}{3} \cos
^{-1}\left(-\frac{357}{73}\sqrt{\frac{3}{146}}\right)\approx 0.782540 \qquad \text{while} \qquad \frac \pi 4\approx 0.785398$$ and $$\sqrt{\frac{146}{3}}\approx 6.97615$$ So, as an approximation,
$$x_k \sim 4+7 \cos \left(\frac{8 k-3}{12} \pi \right)$$ which would give
$$x_1=4+\frac{7}{\sqrt{2}} \approx 8.94975 \quad x_2=4+\frac{7 \left(\sqrt{3}-1\right)}{2 \sqrt{2}}\approx 5.81173\quad x_3=4-\frac{7 \left(\sqrt{3}+1\right)}{2 \sqrt{2}}\approx -2.76148$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374808",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculate the value of $\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+....+\frac{F_n}{2^n}+....$ The Fibonacci sequence starts with 1, 1, 2, 3, 5, 8, 13, ... .(Start from the 3rd term,
each term is the sum of the two previous terms). Let $F_n$ be the $n$th term of this sequence. $S$ is defined as $S=\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+....+\frac{F_n}{2^n}+....$
Calculate the value of $S$
I have no idea how to solve this, hints aswell as solutions would be appreciated
Taken from the 2013 AITMO
| To expand on the hint of @empy2:
$$\begin{array}{rlll}
S(z) &= 1 + & 1z +& 2z^2 + 3z^3 + 5z^4 + ... \\\
z S(z) &= & 1z +& 1z^2 + 2z^3 + 3z^4 + 5z^5 + ... \\\
S(z)-zS(z)-1 &= && 1z^2 + 1z^3 + 2z^4 + 3z^5 \\\
\end{array} \\\
\begin{array}{rlll} \hline
S(z)-zS(z)-1 &= z^2 S(z) &\qquad & \phantom{sdfsdfsdfsdfs} \\\
S(z)(1-z-z^2)& =1 \\\
S(z) &= 1/(1-z-z^2)
\end{array}
$$
Now insert $1/2$ for $z$ and compute $1/2 S(1/2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Truncation error I wonder if the truncation error that I derived in the following approximation really has order 2? If so I also wonder what happens to the $u_{xxxx}$ term since it does not cancel out?
$$u^n_{j+2}-4u^n_{j+1}+6u^n_j-4u^n_{j-1}+u^n_{j-2}$$
$$=\frac{1}{\Delta x^4}(u+2\Delta x u_x+\frac{4}{2}\Delta x^2 u_{xx}+\frac{8}{3!}\Delta x^3u_{xxx}+\frac{16}{4!}\Delta x^4 u_{xxxx}+\frac{32}{5!}\Delta x^5 u_{xxxxx}+\frac{64}{6!}\Delta x^6 u_{xxxxxx}$$
$$
-4(u+\Delta x u_x+\frac{1}{2}\Delta x^2 u_{xx}+\frac{1}{3!}\Delta x^3 u_{xxx}+\frac{1}{4!} \Delta x^4 u_{xxxx}+\frac{1}{5!}\Delta x^5 u_{xxxxx}+\frac{1}{6!}\Delta x^6 u_{xxxxxx})$$
$$
+6u-4(u-\Delta x u_x + \frac{1}{2} \Delta x^2 u_{xx} - \frac{1}{3!} \Delta x^3 u_{xxx}+ \frac{1}{4!}\Delta x^4 u_{xxxx}-\frac{1}{5!}\Delta x^5 u_{xxxxx}+\frac{1}{6!}\Delta x^6 u_{xxxxxx})$$
$$+u-2\Delta x u_x + \frac{4}{2}\Delta x^2 u_{xx}-\frac{8}{3!} \Delta x^3 u_{xxx}+ \frac{16}{4!} \Delta x^4 u_{xxxx}-\frac{32}{5!}\Delta x^5 u_{xxxxx}+\frac{64}{6!}\Delta x^6 u_{xxxxxx}+O(\Delta x^7))$$
$$=O(\Delta x^2)$$
| By taylor series expansions (setting $\Delta x=h$)
\begin{align}u^n_{j+2}&=u+2h u_x+4h^2u_{xx}+8h^3u_{xxx}+16h^4 u_{xxxx}+32h^5u_{xxxxx}+64h^6u_{xxxxxx}\\&+O(h^7),\\
-4u^n_{j+1}&=-4u-4h u_x-4h^2u_{xx}-4h^3u_{xxx}-4h^4 u_{xxxx}-4h^5u_{xxxxx}-4h^6u_{xxxxxx}\\&+O(h^7),\\
6u^n_{j}&=6u,\\
-4u^n_{j-1}&=-4u+4h u_x-4h^2u_{xx}+4h^3u_{xxx}-4h^4 u_{xxxx}+4h^5u_{xxxxx}-4h^6u_{xxxxxx}\\&+O(h^7),\\
u^n_{j-2}&=u-2h u_x+4h^2u_{xx}-8h^3u_{xxx}+16h^4 u_{xxxx}-32h^5u_{xxxxx}+64h^6u_{xxxxxx}\\&+O(h^7),
\end{align}
we see that everything cancels up to the $u_{xxxx}$ terms as
$$16h^4-4h^4-4h^4+16h^4 \ne 0.$$
The $u_{xxxxxx}$ terms also don't cancel because
$$64h^6-4h^6-4h^6+64h^6 \neq 0,$$
so if we make the above cancellations then
\begin{align}\frac {1}{h^4}\left(u^n_{j+2}-4u^n_{j+1}+6u^n_j-4u^n_{j-1}+u^n_{j-2}\right)&=24u_{xxxx}+120h^2u_{xxxxxx}\\&=
24u_{xxxx}+O(h^2).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3380958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find out if the series $\sum_{k=1}^{\infty}k^2\left(\cos\left(\pi k-\frac{1}{k}\right)-(-1)^k\right)$ is divergent. I need to find out if the folowing series is divergent.
$$
\sum_{k=1}^{\infty}k^2\left(\cos\left(\pi k-\frac{1}{k}\right)-(-1)^k\right)
$$
I did the following:
$$
k^2\left(\cos\left(\pi k-\frac{1}{k}\right)-(-1)^k\right)=
k^2\left(\cos\left(\pi k-\frac{1}{k}\right)-\cos(\pi k)\right)=\\
=-2k^2\sin\left(\pi k-\frac{1}{2k}\right)\sin\left(\frac{-1}{2k}\right)=
2k^2\sin\left(\pi k-\frac{1}{2k}\right)\sin\frac{1}{2k}=\\
=2k^2(-1)^{k+1}\sin^2\frac{1}{2k}=A\\
\lim_{k\rightarrow\infty}A=\lim_{k\rightarrow\infty}(-1)^{k+1}\frac{2k^2}{4k^2}=
\lim_{k\rightarrow\infty}\frac{(-1)^{k+1}}{2}\ne0\Rightarrow\\
\text{The initial series is divergent.}
$$
Is my solution correct?
| $\sum_{k=1}^{\infty}k^2\left(\cos\left(\pi k-\frac{1}{k}\right)-(-1)^k\right)
$
Let
$a_k
=\cos\left(\pi k-\frac{1}{k}\right)-(-1)^k
$.
$\begin{array}\\
a_{2n}
&=\cos\left(\pi 2n-\frac{1}{2n}\right)-(-1)^{2n}\\
&=\cos\left(-\frac{1}{2n}\right)-1\\
&=\cos\left(\frac{1}{2n}\right)-\cos(0)\\
&=-2\sin^2(1/(4n))\\
&=-\dfrac1{8n^2}+\dfrac{c_{2n}}{n^4}
\qquad\text{where } |c_{2n}| < 1\\
\text{so}\\
(2n)^2a_{2n}
&=-\dfrac1{2}+\dfrac{c_{2n}}{n^2}\\
a_{2n+1}
&=\cos\left(\pi (2n+1)-\frac{1}{2n+1}\right)-(-1)^{2n+1}\\
&=\cos\left(\pi-\frac{1}{2n+1}\right)+1\\
&=-\cos\left(\frac{1}{2n+1}\right)+\cos(0)\\
&=2\sin^2(1/(2(2n+1)))\\
&=\dfrac1{2(2n+1)^2}+\dfrac{c_{2n+1}}{n^4}
\qquad\text{where } |c_{2n+1}| < 1\\
\text{so}\\
(2n+1)^2a_{2n+1}
&=\dfrac1{2}+\dfrac{c_{2n}}{n^2}\\
\end{array}
$
Since the terms
do not go to zero,
the sum diverges.
Note that if the first $2n$ terms are considered,
the sum does converge,
so this series could be said
to converge to two sums,
one for the even terms
and one for the odd terms.
Also note that if we consider
$\sum ka_k$,
then
$ka_k
\approx \dfrac{(-1)^{k+1}}{4k}
$,
so this will converge
by the alternating series theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Convergence of series $x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots$ Convergence of series $$S=x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots$$
When $x=\frac{1}{2}$
I used ratio test as:
$$a_{n}=\frac{\left(\frac{1}{2}\right)^{2n-1}}{2n-1}$$
Then
$$a_{n+1}=\frac{\left(\frac{1}{2}\right)^{2n+1}}{2n+1}$$
Then we get:
$$\frac{a_{n+1}}{a_n}=\frac{2n-1}{4(2n+1)}$$
Hence we get:
$$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=0.25 \lt 1$$
Hence $\sum a_n$ Converges.
But how to find its value, i guess $\arctan x$ is not useful here
| $$\log(1+x)=x-x^2/2-x^3/3+x^/4-....ad-inf, \log(1-x)=-x-x^2/2-x^3/3-.ad-inf~, if~ |x|<1.$$
Th given series ie quesion is nothing but $\frac{1}{2}\log[(1+x)/(1-x)]$ which is valid (convergent) for $|x|<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find scalars of vectors that equals to zero $x\cdot\begin{bmatrix}0\\-3\\1\end{bmatrix}$ +
$y\cdot\begin{bmatrix}-5\\2\\-4\end{bmatrix}$ +
$z\cdot\begin{bmatrix}-20\\-1\\-13\end{bmatrix}$
=$\begin{bmatrix}0\\0\\0\end{bmatrix}$
x = ？ y = ? z = ?
RREF = $\begin{bmatrix}1&0&3\\0&1&4\\0&0&0\end{bmatrix}$
I started with looking for the Reduced Row Echelon Form, but don't know what to do next.
| $x\cdot\begin{bmatrix}0\\-3\\1\end{bmatrix}$ +
$y\cdot\begin{bmatrix}-5\\2\\-4\end{bmatrix}$ +
$z\cdot\begin{bmatrix}-20\\-1\\-13\end{bmatrix}$
=$\begin{bmatrix}0\\0\\0\end{bmatrix}$
$\implies \begin{bmatrix}-5y-20z\\-3x+2y-z\\x-4y-13z\end{bmatrix}$=$\begin{bmatrix}0\\0\\0\end{bmatrix}$
$\implies y=-4z, -3x-9z=0, x+3z=0$
$\implies y=-4z, x=-3z$
$\implies\begin{bmatrix}x\\y\\z\end{bmatrix}=c\begin{bmatrix}-3\\-4\\1\end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finding second last digit of term in recursive series A sequence of numbers $a_{1}, a_{2}, a_{3},...$ is generated by the formula $a_{1}=3, a_{n+1}=a_{n} + a_{n}^{2}$ for integers $n \geqslant 1$. What is the second last digit of the number $a_{1993}$?
I am not able to recognise a pattern in the last two digits when trying out the recursive formula for the first five iterations. How should this question be approached?
| Since we have to find the second last digit of $a_{1993}$ , it is enough to calculate the remainder $a_{1993}$ leaves upon division by $100$.
Note that $a_{n+1}\equiv a_n +a_{n}^2\pmod {100}$
Note that , $a_1=3$ and $a_{1}^2=9$ and $a_2=2$. Using this we can write $$a_2\equiv12\pmod{100}$$
$$a_{2}^2\equiv 144\equiv 44\pmod{100}$$
$$a_3\equiv a_2+a_{2}^2\equiv 56\pmod{100}$$
$$a_{3}\equiv 56\pmod{100}\implies a_{3}^2\equiv 3136\equiv36\pmod{100}$$
$$a_4\equiv a_3+a_{3}^2\equiv 92\pmod{100}$$
$$a_4\equiv 92\pmod{100}\implies a_{4}^2\equiv 8464\equiv 64\pmod{100}$$
$$a_5\equiv a_4+a_{4}^2\equiv 92+64\equiv 56\pmod{100}$$
Similarly, $a_6\equiv 92\pmod{100}$ ,$a_7\equiv 56\pmod{100}$ and so on.......
This pattern of $56$ and $92$ repeats indefinitely . So , $a_{1993}\equiv 56$ or $92\pmod {100}$. I leave it upto you to find out whether the second last digit is $5$ or $9$.
| {
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"url": "https://math.stackexchange.com/questions/3382635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the two limits without the use of l'Hospital's rule or series expansion. I was asked to evaluate these two limits:
$$\lim_{x\rightarrow0}\frac{x^3}{x-\sin x}$$
$$\lim_{x\rightarrow0}\frac{e^{-x^2}+x^2-1}{\sin(3x^4)}$$
For the first one I tried to divide the numerator and denominator by $x^3$, but I can't get the answer unless I apply l'Hospital's rule or using a series expansion.
I also tried to use a substitution $u=x^2$ for the second limit, but I can't seem to relate anything between the exponential function and sine function.
| Introduction (can skip). This answer aims at producing a solution which rely on as less knowledge as possible. This does not necessarily mean that this solution is easy. Indeed, if we allow more advanced tools, then much shorter proofs are available. Here are some examples:
*
*Integrating $-1 \leq \sin(x) \leq 1$ four times yields $-\frac{x^4}{4!} \leq \sin(x) - \big( x-\frac{x^3}{3!} \big) \leq \frac{x^4}{4!}$. This immediately yields the limit via squeezing.
*Using $\limsup$ and $\liminf$, we can bypass the question of existence of limit. This allows to directly adopt the arguments in this answer.
*Not to mention, L'Hospital's rule or Taylor's theorem gives a one-liner proof.
Solution. In this answer, we will only use following properties of trigonometric functions.
*
*$\text{(P1)} \ $ $0 \leq \sin x \leq x \leq \tan x$ for any $0 < x < \frac{\pi}{2}$.
*$\text{(P2)} \ $ $\sin(2x) = 2\sin(x)\cos(x)$ and $\cos(2x) = 1 - 2\sin^2(x)$.
*$\text{(P3)} \ $ $0 < \cos(y) \leq \cos(x) \leq 1$ for $0 < x < y < \frac{\pi}{2}$.
*$\text{(P4)} \ $ $\sin(-x) = -\sin(x)$.
Define $f(x)$ by
$$f(x) = \frac{x - \sin(x)}{x^3}$$
for $x \neq 0$. From $\text{(P4)}$, we have $f(-x) = f(x)$ and hence it suffices to examine the right-limit of $f$. We also write $\operatorname{sinc}(x) = \sin(x)/x$ for simplicity. By the double-angle formula $\text{(P2)}$,
\begin{align*}
f(x)
&= \frac{x - 2\sin(x/2)}{x^3} + \frac{2\sin(x/2)\left(1 - \cos(x/2) \right)}{x^3} \\
&\quad= \frac{1}{4}f\left(\frac{x}{2}\right) + \frac{\operatorname{sinc}(x/2)\operatorname{sinc}^2(x/4)}{8} \\
&\quad\quad \ldots \\
&\quad\quad\quad = \frac{1}{4^n}f\left(\frac{x}{2^n}\right) + \frac{1}{2} \sum_{k=1}^{n} \frac{\operatorname{sinc}(x/2^k)\operatorname{sinc}^2(x/2^{k+1})}{4^k}.
\end{align*}
Now assume that $x \in (0, \frac{\pi}{2})$. Then by $\text{(P1)}$ and $\text{(P2)}$,
$$ 0 \leq f(x) \leq \frac{\tan(x)-\sin(x)}{x^3} = \frac{\sin(x) (1 - \cos(x))}{x\cos(x)} = \frac{2\sin(x)\sin^2(\frac{x}{2})}{x^3\cos(x)} \leq \frac{1}{2\cos(x)}. \tag{2} $$
Also, by $\text{(P1)}$ and $\text{(P3)}$,
$$ 1 \geq \operatorname{sinc}(x/2^n) = \frac{\tan(x/2^n)}{x/2^n}\cos(x/2^n) \geq \cos(x). \tag{3} $$
Using $\text{(2)}$, $\text{(P3)}$, and $\text{(3)}$, we can bound each term in $\text{(1)}$ from above and below, for $x \in (0, \frac{\pi}{2})$. Indeed, we get
$$ \frac{1}{2\cos^3(x)} \sum_{k=1}^{n} \frac{1}{4^k} \leq f(x) \leq \frac{1}{4^n} \frac{1}{2\cos(x)} + \frac{1}{2} \sum_{k=1}^{n} \frac{1}{4^k}. $$
Since $f(x)$ is independent of $n$, letting $n\to\infty$ and using the geometric series $\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$ for $|r| < 1$ shows that
$$ \frac{1}{6\cos^3(x)} \leq f(x) \leq \frac{1}{6}. $$
By the squeezing theorem, $f(x) \to \frac{1}{6}$ as $x \to 0^+$. (Here, we used the fact that $\cos(x) \to 1$ as $x \to 0$, which itself can be proved from $\cos(x) = 1-2\sin^2(x/2) \in [1-\frac{x^2}{2}, 1]$ together with the squeezing theorem.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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If $f(2x-3) = 4x-2$, then what is $f(x)$? I have this statement:
If $f(2x-3) = 4x-2,$ the function $f(x)$ is ...?
My attempt was:
Move the function $3$ units to the left
$f(2x) = 4(x+3) -2 = 4x+10$
Divide $x$ by $2$
$f(x) = 2x+10$
Verifiy $f(x) = 2x+10 \to f(2x) = 4x+10 \to f(2x-3) = 4(x-3)+10 = \underbrace{4x-2}_{f(2x-3)}$
But according to the guide the correct answer is $2x+4$ and i don't know why. Thans in advance.
| Put $y=2x-3$. Then $x=(y+3)/2$. In particular
$$
f(y)=4x-2=4\left(\frac{y+3}{2}\right)-2=2(y+3)-2=2(y+2)
$$
as desired.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An inequality for polynomials with positives coefficients I have found in my old paper this theorem :
Let $a_i>0$ be real numbers and $x,y>0$ then we have :
$$(x+y)f\Big(\frac{x^2+y^2}{x+y}\Big)(f(x)+f(y))\geq 2(xf(x)+yf(y))f\Big(\frac{x+y}{2}\Big)$$
Where :$$f(x)=\sum_{i=0}^{n}a_ix^i$$
The problem is I can't find the proof I made before . Furthermore I don't know if it's true but I have checked this inequality a week with Pari\Gp and random polynomials defined as before .
So first I just want a counter-examples if it exists .
If it's true if think it's a little bit hard to prove . I have tried the power series but without success .
Finally it's a refinement of Jensen's inequality for polynomials with positives coefficients .
Thanks a lot if you have a hint or a counter-example .
Ps:I continue to check this and the equality case is to $x=y$
| First we rewrite your inequality as
$$ \frac{\frac{1}{2}f(x) +\frac{1}{2}f(y)}{f\left(\frac{1}{2}x + \frac{1}{2}y\right)} \ge \frac{\frac{x}{x+y}f(x) +\frac{y}{x+y}f(y)}{f\left(\frac{x}{x+y}x + \frac{y}{x+y}y\right)}. $$
For $\lambda \in [0, 1]$, let $$ g(\lambda) = \frac{\lambda f(x) + (1 - \lambda) f(y)}{f(\lambda x + (1 - \lambda)y)}. $$
Since $f(x)$ is convex, we know $g(\lambda) \ge 1$. It suffice to show $g\left(\frac{x}{x+y}\right) \le g\left(\frac{1}{2}\right)$. Without losing generality, let's assume $x > y$. Furthermore, it suffice to show $g(\lambda)$ is a non-increasing function for $\lambda \in [\frac{1}{2}, \frac{x}{x+y}]$.
$$g^\prime(\lambda) = \frac{f(x) - f(y)}{f(\lambda x + (1 - \lambda)y)} - \frac{\lambda f(x) + (1 - \lambda) f(y)}{{\left(f(\lambda x + (1 - \lambda)y)\right)}^2}f^\prime(\lambda x + (1 - \lambda)y)(x-y).$$
For $g^\prime(\lambda)$, we are only concerned about its being positive or not. To simplify notation, we use $\stackrel{s}{=}$ to denote sign equality so that we can drop nonnegative terms.
We first consider the special case of $f(x)=x^n, n \ge 1$ (the case $n=0$ is trivial), then $ \frac{f^\prime(x)}{f(x)}=\frac{n}{x}.$ Hence
$$ g^\prime(\lambda) = \frac{1}{{(\lambda x + (1 - \lambda)y)}^n}\left(
{x^n - y^n} - n(x-y)\frac{\lambda x^n + (1-\lambda) y ^n}{\lambda x + (1 - \lambda)y} \right) \\ \stackrel{s}{=} (x^n - y^n)(\lambda x + (1 - \lambda)y) - n(x-y)\left(\lambda x^n + (1-\lambda) y ^n\right) \\ = (x^n - y^n)\left(\frac{x+y}{2}\right) - n(x-y)\left(\frac{x^n + y ^n}{2}\right) \quad\quad\text{(1)} \\ - (n-1)\left(\lambda - \frac{1}{2}\right)(x-y)(x^n - y ^n) \quad\quad\text{(2)}.$$
For $\lambda \in [\frac{1}{2}, \frac{x}{x+y}]$, it is easy to see (2) is non-positive. Note (1) is non-positive as well since
$$ (x^n - y^n)\left(\frac{x+y}{2}\right) - n(x-y)\left(\frac{x^n + y ^n}{2}\right) \le 0 \\ \Longleftrightarrow (x^n - y^n)\left({x+y}\right) \le n(x-y)\left({x^n + y ^n}\right) \\ \Longleftrightarrow x^{n+1} - y^{n+1} -xy^n +yx^n \le n(x^{n+1}-y^{n+1}+xy^n-yx^n) \\ \Longleftrightarrow (n+1)\left(x^ny-xy^n\right) \le (n-1)\left(x^{n+1}-y^{n+1}\right) \\ \Longleftrightarrow (n+1)\left(z^n-z\right) \le (n-1)\left(z^{n+1}-1\right), z=\frac{x}{y}>1 \\ \Longleftrightarrow \ell(z) = (n-1)\left(z^{n+1}-1\right) - (n+1)\left(z^n-z\right) \ge 0. \text{ for } z>1 $$
The last statement is true since $\ell(1) = 0$ and
$$\ell^\prime(z) = (n^2-1)z^n-(n+1)(nz^{n-1}-1)\\ = (n+1)\left(n(z^n-z^{n-1}) - (z^n - 1)\right)\\ = (n+1)(z-1)\left( nz^{n-1}-\sum_{i=0}^{n-1}z^i \right) \ge 0 \text{ for } z \ge 1.$$
For the case of $f(x)=x^n$, we have proven the inequality. I will try to generalize the conclusion to all polynomials with positive coefficients later when I get time.
Noticed that @p4sch put in a counter example for general polynomial cases, I will hence abandon any generalization efforts, and my proof is only valid for the special case of $f(x)=x^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof by induction: $5^n \geq 5n^3 + 2$ for $n \geq 4$ One of the practice problems I have is to prove by induction that for every $n \geq 4$ the following inequality holds:
$5^n \geq 5n^3 + 2$
My progress so far (inequality holds for base case $n=4$):
$5^{n+1} \geq 5(5n^3 + 2)$
$5^{n+1} \geq 25n^3 + 10$
The next step logical step for me is to prove that $25n^3 + 10 \geq 5(k+1)^3 + 2$ but I have no idea how.
| You have $5^{n+1}=5\cdot 5^n\geq 5(5n^3+2)=25n^3+10$, we need to show that
$5^{n+1}\geq 5(n+1)^3+2=5(n^3+3n^2+3n+1)+2=(5n^3+15n^2+15n+5)+2$
We know that $n\geq 4$, so if we write:
$25n^3=5n^3+20n^3\geq 5n^3+80n^2\geq 5n^3+15n^2+65n^2\geq 5n^3+15n^2+260n$
$\geq 5n^3+15n^2+15n+5=5(n+1)^3$
Where we substitute one $n$ by $4$ in every step and derive the result like that.
So: $25n^3+2\geq 5(n+1)^3+2$, which ends the inductive proof.
| {
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"url": "https://math.stackexchange.com/questions/3392592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove geometric sum with induction I'm not certain how to complete the proof:
Question:
Prove by induction that $1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^n = (\frac{3}{4})[1 − (\frac{−1}{3})^{n+1}]$, for every non negative integer $n$.
Solution
Base Step:
Verify that:
$LHS = 1 = (\frac{3}{4})[1 − (\frac{−1}{3})^{0+1}] = RHS$.
$RHS = (\frac{3}{4})[1 − (\frac{−1}{3})^1]\\
= (\frac{3}{4})[1 − (\frac{−1}{3})]\\
= (\frac{3}{4})(1 + \frac{1}{3})\\
= (\frac{3}{4})(\frac{4}{3}) = 1 = LHS.$
Inductive Step:
Assume that:
$1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^k = (\frac{3}{4})[1 − (\frac{−1}{3})^{k+1}]$, for some integer k.
We try to deduce that:
$1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} = (\frac{3}{4})[1 − (\frac{−1}{3})^{k+2} ]$.
$LHS
= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} \\
= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k} + (\frac{−1}{3})^{k+1}\\
= \frac{3}{4}[1-(\frac{-1}{3})^{k+1}] + (\frac{-1}{3})^{k+1}\\
... $
Lost from this point onwards.
| You just about have it. From your last line, you get
$$\begin{equation}\begin{aligned}
\frac{3}{4}\left[1-\left(\frac{-1}{3}\right)^{k+1}\right] + \left(\frac{-1}{3}\right)^{k+1} & = \frac{3}{4} - \frac{3}{4}\left(\frac{-1}{3}\right)^{k+1} + \left(\frac{-1}{3}\right)^{k+1} \\
& = \frac{3}{4} + \frac{1}{4}\left(\frac{-1}{3}\right)^{k+1} \\
& = \frac{3}{4} + \frac{(-3)}{4}\left(\frac{1}{-3}\right)^{k+2} \\
& = \frac{3}{4}\left[1 - \left(\frac{-1}{3}\right)^{k+2}\right]
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
which is the RHS of what you are trying to deduce, thus completing your induction procedure.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a^2+c^2>ab$ and $b^2>4c^2$ , for real x, show that $\frac{x+a}{x^2+bx+c^2}$ cannot lie between two limits If $a^2+c^2>ab$ and $b^2>4c^2$ , for real x, show that $\frac{x+a}{x^2+bx+c^2}$ cannot lie between two limits
My attempt is as follows:
$$y=\frac{x+a}{x^2+bx+c^2}$$
$$yx^2+byx+yc^2=x+a$$
$$yx^2+x(by-1)+yc^2-a=0$$
As x is real,so
$$D>=0$$
$$(by-1)^2-4y(yc^2-a)>=0$$
$$b^2y^2+1-2by-4y^2c^2+4ay>=0$$
$$(b^2-4c^2)y^2+2(2a-b)y+1>=0$$
As it is given $b^2>4c^2$, it means parabola is upwards, now this parabola will not lie between two limits if it does not cut x-axis at two distinct points.
So if $D<=0$ then parabola $(b^2-4c^2)y^2+2(2a-b)y+1>=0$ will not cut x-axis at two distinct points.
So lets calculate D for the equation $(b^2-4c^2)y^2+2(2a-b)y+1=0$
$$D=4(4a^2+b^2-4ab)-4(b^2-4c^2)$$
$$D=4(4a^2+4c^2-4ab)$$
$$D=16(a^2+c^2-ab)$$
But I am getting $D>0$ as $a^2+c^2>ab$
I am getting totally reversed result. What mistake am I doing here. Please help me.
| Suppose $r$ is the smallest and $s$ is the largest roots of the denominator. (There are 2 of them because of your second inequality). Set $f(x)=\frac{x+a} {x^2+bx+c^2}$ .Now if $r+a>0$ then $\lim_{r^{+}}f=-\infty$... Similarly we can prove the case $r+a<0$ and $r+a=0$...
| {
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"timestamp": "2023-03-29T00:00:00",
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Two values of minima of a function by two methods.
I had a problem of finding minima of a function
$$f(x)=2^{x^2}-1+\frac{2}{2^{x^2}+1}$$
I solved it using AM-GM inequality,
$$2^{x^2}-1+\frac{2}{2^{x^2}+1}$$
$$2^{x^2}+1+\frac{2}{2^{x^2}+1}-2$$
$$2^{x^2}+1+\frac{2}{2^{x^2}+1}\ge\ 2\sqrt2$$
$$2^{x^2}-1+\frac{2}{2^{x^2}+1}\ge\ 2\sqrt2-2$$
But in the solution answer was given as 1 and it was solved using differentiation,
$$f'(x)=\frac{2x.ln2.2^{x^2}(2^{x^2}+1-\sqrt2)(2^{x^2}+1+\sqrt2)}{(2^{x^2}+1)^2}$$
$$2^{x^2}\ge1$$
$$2^{x^2}+1-\sqrt2\ge2-\sqrt2>0$$
At $x=0$ ,$f(x)$ is least.
Least value = $f(0)$ $=1$
I cannot understand how can there be two values by two different methods,please help me in the problem.
| The fact is that the equality by AM-GM
$$2^{x^2}+1+\frac{2}{2^{x^2}+1}=\ 2\sqrt2$$
holds if and only if $\exists x$ such that
$$2^{x^2}+1=\frac{2}{2^{x^2}+1}$$
but
*
*$2^{x^2}+1\ge 2$
*$\frac{2}{2^{x^2}+1}\le 1$
therefore the value $2\sqrt 2$ is never reached and the inequality becomes
$$2^{x^2}+1+\frac{2}{2^{x^2}+1}>\ 2\sqrt2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Intersection of curve $ y = x^4 – 6x^3 + 12x^2 + cx + 1$ The number of integers in the range of 'c' such that there exists a line which intersects the curve $ y = x^4 – 6x^3 + 12x^2 + cx + 1$ at four distinct points.
My approach we need to intersect with line $y=mx+C$
Substituting we get $x^4 – 6x^3 + 12x^2 + cx + 1-mx-C=0$
Now this is an equation of polynomial of degree 4
$x^4 – 6x^3 + 12x^2 + (c-m)x + 1-C=0$
All four roots needs to be real.
I don't have any idea how to check whether all the roots are real or not.
| Note that in order for a line to be able to meet this curve at four points, the curve must "change direction" thrice, which is to say that the derivative must have three roots.
The derivative in question is visibly $$g'(x) = 4x^3 -18x^2+24x+c$$
For it to have three roots, it must "change direction" twice, and on the same side of the $x\text{-axis}$. To check that, see its derivative: $$g''(x)= 12x^2 - 36x + 24 = 12(x^2-3x+2)$$ which visibly has roots at $x = 2, 1$. Now all that is left is to ensure that these occur on both sides of the $x$ axis in $g'(x)$.
So, we have: $$g'(2) = 4\cdot 8 - 18 \cdot 4 + 24 \cdot 2 + c = c+8$$
$$g'(1) = 4-18+24+c = c+10$$
So, $c+10$ and $c+8$ have to be on different sides of the $x$ axis. Obviously $c+10>c+8$ so the former would lie above the axis. So, we have $c+10>0$ and $c+8<0$, so our range is $$\boxed{c\in(-10,-8)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3398566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Three polynomials which have the same value for a variable
Let $P_1(x)= ax^2-bx-c$, $P_2(x)=bx^2-cx-a$, and $P_3(x)= cx^2-ax-b$ be three quadratic polynomials, where $a,b$, and $c$ are non-zero real numbers. Suppose that there exists a real number $k$ such that $$ P_1(k) = P_2(k)= P_3(k) $$ Prove that $a=b=c$.
Hello everybody! The above is a question I got stuck on. This problem is from an Indian Olympiad. Here is what I did:
After manipulating the given information, we derive at $(a-b)k^2-(b-c)k-(c-a)=0$ $(b-c)k^2-(c-a)k-(a-b)=0 $ and $(c-a)k^2 - (a-b)k - (b-c)$. Now the solution involves some addition and subtraction to factor it and get $a=b=c$.
But my question is: Can’t I just compare the coefficients of the three equations because they are all zero from which we get $b-c = c-a$, $c-a = a-b $ and $a-b = b-c$ and get $a=b=c$? If not, why?
Thanks
| Write $P_1(k) = P_2(k)= P_3(k)=v$. Then
$$
\pmatrix{ k^2 & -k & -1 \\ -1 & k^2 & -k \\ -k & -1 & k^2}
\pmatrix{ a \\ b \\ c}
=
\pmatrix{ v \\ v \\ v}
$$
and so
$$
\pmatrix{ a \\ b \\ c}
=
\frac{1}{k^6 - 4 k^3 - 1}
\pmatrix{k^4 - k & k^3 + 1 & 2 k^2 \\
2 k^2 & k^4 - k & k^3 + 1 \\
k^3 + 1 & 2 k^2 & k^4 - k}
\pmatrix{ v \\ v \\ v}
$$
Since the sums of the elements in each row are the same, we get $a=b=c$.
Alternatively, without any computation, just note that the first matrix is a circulant matrix and so has a circulant inverse.
On the other hand, we need $k^6 - 4 k^3 - 1 \ne 0$, that is, $k \ne (1\pm\sqrt5)/2$, the roots of $k^2-k-1$. This case needs to be handled separately but it still works. A necessary condition is $v=0$.
(Actually, $a=b=c\ne0$ implies $k^2-k-1=0$.)
| {
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Fractions in Questions and Answers
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