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Converse to Jensen's inequality for $1/x$ on a positive bounded interval? Consider the function $f(x) = 1/x$ on the interval $I = [a, b]$, where $0 < a \leq b$.
By Jensen's inequality, we have for any $\{x_j \}_{j=1}^n \subset I$,
$$
f(\overline{x}) \leq \frac{1}{n} \sum_{i=1}^n f(x_i).
$$
Above, $\overline{x} = (1/n)\sum_{j=1}^n x_j$.
Is there a converse inequality of the form
$$
\frac{1}{n} \sum_{i=1}^n f(x_i) \leq C f\Big(\frac{1}{n}\sum_{i=1}^n x_i\Big),
$$
where $C$ may depend on $a, b$?
One thing I thought of is a second-order Taylor expansion.
By Taylor expansion, there are $\xi_i$ in the interval between $x_i, \overline{x}$ such that
$$
\frac{1}{n} \sum_i f(x_i) = f(\overline{x}) + \frac{1}{2n} \sum_{i=1}^n f''(\xi_i)
(x_i - \overline{x})^2.
$$
Additionally, we have that $f''(\xi) = \tfrac{2}{\xi^3}$, so that the remainder is
$$
\frac{1}{n} \sum_{i=1}^n \frac{(x_i - \overline{x})^2}{\xi_i^3}
\leq
\frac{1}{n} \sum_{i=1}^n \max \Big\{
\frac{(x_i - \overline{x})^2}{x_i^3},
\frac{(x_i - \overline{x})^2}{\overline{x}^3} \Big\}.
$$
I struggled to bound this remainder in terms of $C(a, b)~f(\overline{x})$, though.
|
Yes, there is a bound available.
Let $t_i := \max \Big\{
\tfrac{(x_i - \overline{x})^2}{x_i^3},
\tfrac{(x_i - \overline{x})^2}{\overline{x}^3} \Big\}$.
We have that the remainder is the average $\frac{1}{n} \sum_{j=1}^n t_j$.
Consider the two cases separately. First suppose $x_i > \overline{x}$.
Then
$$
t_i = \frac{(x_i - \overline{x})^2}{\overline{x}^3} \leq \frac{1}{\overline{x}} \frac{x_i^2}{\overline{x}^2} \leq (b/a) \frac{1}{\overline{x}} \frac{x_i}{\overline{x}}
$$
On the other hand, if $\overline{x} > x_i$,
$$
t_i = \frac{(x_i - \overline{x})^2}{x_i^3} \leq
\frac{1}{\overline{x}} \frac{\overline{x}^3}{x_i^3} \leq (b/a)^3 \frac{1}{\overline{x}}.
$$
Therefore,
$$t_i \leq (b/a)^3 \frac{1}{\overline{x}} + (b/a) \frac{1}{\overline{x}} \frac{x_i}{\overline{x}}$$.
If we average over $i=1, \dots, n$, we get
$$
\frac{1}{n}\sum_{i=1}^n f(x_i) \leq f(\overline{x}) + (b/a)^3 f(\overline{x}) + (b/a) f(\overline{x}),
$$
so that we may take $C(a, b) = 1 + (b/a)^3 + (b/a)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to differentiate $y=\sqrt{\frac{1-x}{1+x}}$? It is an example question from "Calculus Made Easy" by Silvanus Thompson (page 68-69). He gets to the answer $$\frac{dy}{dx}=-\frac{1}{(1+x)\sqrt{1-x^2}}$$ The differentiation bit of the question is straightforward, but I'm having trouble simplifying it to get the exact answer. My working is the following:
$$y=\sqrt{\frac{1-x}{1+x}}$$
This can be written as $$y=\frac{(1-x)^\frac{1}{2}}{(1+x)^\frac{1}{2}}$$
Using the quotient rule we get $$\frac{dy}{dx}=\frac{(1+x)^\frac12\frac{d(1-x)^\frac12}{dx}-(1-x)^\frac12\frac{d(1+x)^\frac12}{dx}}{1+x}$$
Hence $$\frac{dy}{dx}=-\frac{(1+x)^\frac12}{2(1+x)\sqrt{1-x}}-\frac{(1-x)^\frac12}{2(1+x)\sqrt{1+x}}$$
$$=-\frac{1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}$$
What is the next step?
|
I will directly follow your step.
$$
\begin{aligned}
\frac{\operatorname{d}y}{\operatorname{d}x}&=-\frac{1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}\\
&=-\frac{1}{2}\frac{1}{\sqrt{1+x}}\Big(\frac{1}{\sqrt{1-x}}+\frac{\sqrt{1-x}}{1+x}\Big)\\
&=-\frac{1}{2\sqrt{1+x}}\frac{2}{(1+x)\sqrt{1-x}}\\
&=-\frac{1}{(1+x)\sqrt{1-x^2}}
\end{aligned}
$$
|
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|
Find all bounded infinite sequences of positive integers $(a_n)$ with $a_n = (a_{n-1}+a_{n-2})/(\gcd(a_{n-1},a_{n-2}))$ for $n\ge 2$.
Find all bounded infinite sequences of positive integers $(a_n)$ with $a_n = (a_{n-1}+a_{n-2})/(\gcd(a_{n-1},a_{n-2}))$ for $n\ge 2$.
I think the only possible sequence is the constant sequence with each term equal to $2$. Let $g_n = \gcd(a_{n},a_{n+1})$. One can show that $(g_n)$ is eventually constant by proving that it is nonincreasing, since it is bounded below by zero and a convergent sequence of integers is eventually constant (the sequence $g_n$ converges by the Monotone Convergence theorem). If $d$ divides $a_{n+1}$ and $a_{n+2}$, then it divides $g_n a_{n+2} - a_{n+1} = a_n,$ so $g_{n+1} | g_n$, proving the claim. Clearly one cannot have $g=1$, since then for all $n$ sufficiently large, $a_n = a_{n-1}+a_{n-2}$ so $(a_n)$ has a strictly increasing, unbounded subsequence. One should also be able to eliminate the case $g\ge 3$. Now I think once one proves $g=2$, one can use backwards induction to prove that the sequence $(a_n)$ must be as claimed, though I'm not sure about the details.
Here's some additional justification of the third case of the given answer by acreativename:
For the third case, I think they're assuming $A_{n} = 2$ for $n\ge k-2$ (even if not, the proof should still work if you instead assume this). We have that for all $n\ge k-2, A_n = 2.$ So in particular, $a_n$ and $a_{n+1}$ cannot both be divisible by $4$ for any $n\ge k-2.$ We have $a_{k+1} = \dfrac{a_{k-1}+a_{k}}2.$ So $a_k$ is even. If $a_k\equiv 0\mod 4,$ then $a_{k-1} + a_k\equiv 0\mod 4$ (as $a_{k+1}$ is even) implies $a_{k-1}\equiv 0\mod 4,$ contradicting $\gcd(a_k, a_{k-1}) = 2.$ Hence $a_k\equiv 2\mod 4,$ and so $a_{k-1}\equiv 2\mod 4.$ But from this point on, I don't know why $a_{k-1}=a_{k-2} = 2$ must hold. Once we show that $a_{k-1} = a_{k-2} = 2,$ we can then conclude the result by backwards induction ($a_{k-j-1} =a_{k-j} = 2\Rightarrow a_{k-j-2}=2$).
|
$\textbf{Lemma:}$ when $j \in \mathbb{N}$ we have that $\text{gcd}(a_{j+1},a_{j+2})$ divides $\text{gcd}(a_{j},a_{j+1})$
$\textbf{Proof:} $ Put $\text{gcd}(a_{j},a_{j+1}) = A_{j}$ and write $a_{j} = A_{j}v_{j}, a_{j+1} = A_{j+1}v_{j+1}$ where $\text{gcd}(v_{j}, v_{j+1}) = 1$. We have $a_{j+2} = v_{j}+v_{j+1}$ and $\text{gcd}(a_{j+1}, a_{j+2}) = \text{gcd}(A_{j}v_{j+1}, v_{j}+v_{j+1}) = \text{gcd}(A_{j}, v_{j}+v_{j+1})$ as $\text{gcd}(v_{j}, v_{j+1}) = 1$.
Thus by the lemma the sequence $A_{j} := \text{gcd}(a_{j},a_{j+1})$ is a non-increasing sequence with $A_{j+1}|A_{j}$.
We now have three cases to consider;
$\bullet$ If $\lim_{j \rightarrow \infty} A_{j} = 1$ then there exists an integer $k$ so that $\text{gcd}(a_{k}, a_{k+1}) = 1$ and $a_{k}, a_{k+1} \geq 1$; whence by the above lemma we will have $a_{n} = a_{n-1} + a_{n-2}$ for $n \geq k+2$ and therefore will have $a_{n} \rightarrow \infty$.
$\bullet$ If $\lim_{j \rightarrow \infty} A_{j} \geq 3$ then $a_{n} \leq \frac{a_{n-1}+a_{n-2}}{3}$ for some $n \geq k$ and will therefore have $a_{n} \rightarrow 0$ (trivial exercise) which is impossible.
$\bullet$ If $\lim_{j \rightarrow \infty} A_{j} = 2$ then we have $a_{n} = \frac{a_{n-1}+a_{n-2}}{2}$ for some $n \geq k$. The only way $a_{k-2},a_{k-1},...$ are positive integers under this condition is when $a_{k-1}=a_{k-2} = 2$.
Thus such bounded sequences must eventually collapse to $2$. Now note that if $w \in \mathbb{N}$ and $\frac{w+2}{\text{gcd}(w,2)} = 2$ then $w$ is forced to be $2$. Hence we must have $a_{1} = a_{2} = 2$.
|
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|
How to solve this quadratic system of equations? The equations are:
$$ \begin{aligned} b + d &= c^2 - 6 \\ b - d &= -\frac{1}{c} \\ b d &= 6 \end{aligned} $$
and I want integer solutions for this.
I tried using various methods such as using $(a+b)^2-(a-b)^2=4ab$ or trying to solve for $b+d$ using the last two equations and substituting into the first. None of these methods worked. How do I do this?
|
If we call $ \ b \ $ and $ \ d \ $ the roots of a quadratic equation, the Viete relations tell us that the equation is $ \ x^2 + (6 - c^2)x + 6 \ = 0 \ \ \ $ (from the sum and product of the roots). The difference of the zeroes of a monic quadratic polynomial is equal to the square-root of its discriminant $ \ ( \ b - d \ = \ \sqrt{D} \ ) \ \ , \ $ so we have
$$ (b \ - \ d)^2 \ \ = \ \ (6 - c^2)^2 \ - \ 4·1·6 \ \ = \ \ c^4 \ - \ 12c^2 \ + \ 12 \ \ = \ \ \frac{1}{c^2} \ \ . $$
This leads to the apparently unpleasant sixth-degree equation $ \ c^6 - 12c^4 + 12c^2 - 1 \ = \ 0 \ \ . \ $ But the polynomial is an even function and (anti-)palindromic, so both $ \ c \ = \ 1 \ $ and $ \ c \ = \ -1 \ $ are among its roots, leading to the factorization $ \ (c^2 - 1)·(c^4 \ - \ 11c^2 + 1) \ \ , $
using polynomial or synthetic division.
Since the second factor is a biquadratic polynomial, it is not hard to calculate using the quadratic formula that none of its zeroes are integers (the Rational Zeroes Theorem also tells us this). We thus have only one possibility for obtaining integer roots of our original equation
$$ \mathbf{c \ = \ \pm 1 \ \ : } \quad \quad x^2 + 5x + 6 \ = \ (x + 3)·(x + 2) \ = \ 0 \ \ \Rightarrow \ \ b \ , \ d \ \ = \ \ -3 \ , \ -2 \ \ . $$
$$ \ \ $$
ADDENDUM --
We can also interpret this system geometrically. The equation $ \ xy \ = \ 6 \ $ describes a (rectangular) hyperbola with its major axis on the line $ \ y \ = \ x \ \ . \ $ The line $ \ x - y \ = \ -\frac{1}{c} \ $ runs parallel to this major axis, but we are not permitted $ \ c \ = \ 0 \ \ . \ $ There will be just two intersections of such a line with the hyperbola, one in the first and one in the third quadrant; thus $ \ b \ $ and $ \ d \ $ must have the same sign. Since we wish both of these numbers to be integers, the only possibilities are those laid-out in marty cohen's answer: the "diagonal" symmetry of the hyperbola allow us to write these as
$$ (b \ , \ d) \ \ = \ \ (\pm 1 \ , \pm 6) \ \ , \ \ (\pm 2 \ , \pm 3) \ \ \ \text{[either permutation].} $$
These correspond to the lines $ \ x - y \ = \ \pm 5 \ \ $ or $ \ x - y \ = \ \pm 1 \ \ , \ $ respectively.
But these points must also lie on a line $ \ x + y \ = \ c^2 - 6 \ \ge \ -6 \ \ . \ $ This eliminates the pair $ \ (-1 \ , \ -6) \ \ . \ $ As for the others, $ \ (1 \ , \ 6) \ $ requires $ \ x + y \ = \ 7 \ \Rightarrow \ c \ = \ \pm \sqrt{13} \ \ , \ $ which is inconsistent with the difference equation; similarly, $ \ (2 \ , \ 3) \ \Rightarrow \ x + y \ = \ 5 \ \Rightarrow \ c \ = \ \pm \sqrt{11} \ \ . \ $ The remaining pair $ \ (-2 \ , \ -3) \ $ produces $ \ x + y \ = \ -5 \ \Rightarrow \ c \ = \ \pm \sqrt1 \ \ , \ $ giving us the sole consistent solution to the given system of equations.
$$ \text{rejected solutions} $$
$$ \text{only permissible solutions lie on green line} $$
|
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|
There is a disk of radius $r(x)=2-\frac{2}{5}|x|$ at every point on $[-5, 5]$. Find equation of a union of all disks Modeling shows that the figure look like this:
But I don't know how to get an equation for it analytically
Looks like for some $0<x_0<5$ for $(x_0, 5]$ and $[-5, x_0)$ correspoing disks touch final shape at only one point each. It would help to find at least this $x_0$.
|
Based on the diagram and the comment from @ajotatxe, I will assume that you meant to write $r(x) = 2 - \frac{2}{5}|x|$. Fix a point $a$ on the interval $[0,5]$ and consider the disk
$$(x-a)^2 + y^2 = r(a)^2 = \left(2 - \frac{2}{5}a\right)^2 = 4 - \frac{8}{5}a + \frac{4}{25}a^2$$
centered at $a$. There is a unique coordinate $x_a$ so that the tangents to this circle at $(x_a,y_a)$ passes through the point $(5,0)$, where $y_a = \sqrt{r(a)^2-(x_a-a)^2}$. To find it, we calculate
$$2(x-a) + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x-a}{y}$$
and so $x_a$ satisfies
$$0 - y_a = -\frac{x_a - a}{y_a}(5 - x_a) \implies (x_a-a)(5-x_a) =y_a^2 = r(a)^2 - (x_a-a)^2 $$
$$\implies -x_a^2 - 5a + (a+5)x_a = r(a)^2 - x_a^2 + 2ax_a - a^2 $$
$$\implies (5-a)x_a = r(a)^2 + 5a - a^2 = 4 + \frac{17}{5}a - \frac{21}{25}a^2 = \frac{-1}{25}(a-5)(21a+20) $$
and so $x_a = \frac{21}{25}a + \frac{4}{5}$. Thus $5-x_a = \frac{21}{25}(5 - a)$ and $x_a -a =\frac{4}{25}(5 - a)$, so $y_a = \pm\frac{2\sqrt{21}}{25}(5-a)$. This means the equation of the desired tangent line is $y - y_a = \mp\frac{2}{\sqrt{21}}(x-x_a)$. Notice that the slope of this line, $\frac{\mp 2}{\sqrt{21}}$, does not depend on $a$. Since by construction the tangent line always includes the point $(5,0)$, we see that they are all in fact the exact same line! Therefore, as the disks move from being centered at $(5,0)$ to $(0,0)$, they slide tangentially along the lines $y = \frac{\mp 2}{\sqrt{21}}(x-5)$.
Now since the biggest circle is at $a = 0$, we substitute to find that the last points in the shaded region along these common tangents will be at $(x_0,y_0) = \left(\frac{4}{5},\pm \frac{2\sqrt{21}}{5}\right)$, and on the interval $-\frac{4}{5} \leq x \leq \frac{4}{5}$ the upper and lower boundaries are the same as that of the circle $x^2 + y^2 = 4$. So to summarize,the region is:
$$(x,y) \text{ so that } \begin{cases} |y| \leq \sqrt{4 - x^2} & \text{ if } |x| \leq \frac{4}{5} \\ |y| \leq \frac{2}{\sqrt{21}}(5-x) & \text{ if } \frac{4}{5} \leq |x| \leq 5 \end{cases} $$
I also graphed this region in Desmos with a slider if you want to play around with it.
|
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|
Equation of a line orthogonal to two other lines Consider the two lines $\ell_1:\mathbf{r}=\begin{pmatrix}1\\1\\0\end{pmatrix}+s\begin{pmatrix}-1\\2\\-2\end{pmatrix}$ and $\ell_2:\mathbf{r}=\begin{pmatrix}-2\\3\\-2\end{pmatrix}+t\begin{pmatrix}0\\-1\\2\end{pmatrix}$. I need to determine the equation of the line $\ell_3$ whcich intersects both $\ell_1$ and $\ell_2$ at right angles.
I know that the direction of $\ell_3$ will be given by $\begin{pmatrix}2\\2\\1\end{pmatrix}$, by the cross product of the two direction vectors of the first two lines. I am not sure how to proceed from here.
|
If
\begin{align}
\mathbf{r}_1&=\begin{pmatrix}1\\1\\0\end{pmatrix}+t_1\begin{pmatrix}-1\\2\\-2\end{pmatrix}\ \text{and}\\
\mathbf{r}_2&=\begin{pmatrix}-2\\3\\-2\end{pmatrix}+t_2\begin{pmatrix}0\\-1\\2\end{pmatrix}
\end{align}
are the points of intersection of $\ \ell_1\ $ and $\ \ell_2\ $ with $\ \ell_3\ $, respectively, then $\ \mathbf{r}_2-\mathbf{r}_1\ $ must be in the same direction as $\ \ell_3\ $. That is
\begin{align}
\mathbf{r}_2-\mathbf{r}_1&=\begin{pmatrix}-3\\2\\-2\end{pmatrix}-t_1\begin{pmatrix}-1\\2\\-2\end{pmatrix}+t_2\begin{pmatrix}0\\-1\\2\end{pmatrix}\\
&=t_3\begin{pmatrix}2\\2\\1\end{pmatrix}\ .
\end{align}
for some $\ t_3\ $. Can you see how to determine the values of $\ t_1,t_2\ $and $\ t_3\ $?
|
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|
Everywhere differentiable function whose derivative is not locally integrable My question is all in the title: I can’t come up with an example of a function $u \colon \mathbb R \to \mathbb R$ such that
*
*$u$ is differentiable at every $x \in \mathbb R$
*$u’ \notin L^1_{\text{loc}}(\mathbb R)$
Any ideas?
|
To combine all the ideas from the comments, define $$f(x)=\begin{cases} x^2\sin(\frac{1}{x^2}) & x\not=0 \\ 0 &x=0\end{cases}$$
Then it's easy to see $f(x)$ is differentiable everywhere, and whenever $x\not=0$, $$f'(x)=2x\sin(\frac{1}{x^2})+x^2\cos(\frac{1}{x^2})\frac{-2}{x^3}=2x\sin(\frac{1}{x^2})+\frac{-2\cos(\frac{1}{x^2})}{x}$$
Note that $2x\sin(\frac{1}{x^2})$ is continuous and hence locally integrable, it's sufficient to show $g(x):=\frac{\cos(\frac{1}{x^2})}{x}$ is not around $0$. It's clearly unbounded, but this is insufficient to conclude it's not locally integrable.
Indeed, in the inverval $\frac{1}{x^2}\in [2n\pi, 2n\pi + \frac{1}{3}\pi]$, we have $\frac{1}{x}\ge \sqrt{2n\pi}\gg\sqrt{n}$, and $\cos(\frac{1}{x})\ge \frac{1}{2}\gg 1$. And the length of the interval is $$\frac{1}{\sqrt{2n\pi}}-\frac{1}{\sqrt{2n\pi + \frac{1}{3}\pi}}=\frac{\pi/3}{\sqrt{2n\pi + \frac{1}{3}\pi}\cdot\sqrt{2n\pi}\cdot(\sqrt{2n\pi + \frac{1}{3}\pi}+\sqrt{2n\pi})}\gg \frac{1}{n^{3/2}}$$
Hence $\int_0^\epsilon g^+(x)dx\gg\sum_{n=m(\epsilon)}^\infty \frac{1}{n}=\infty$.
|
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|
Tricky limitting sum Find the sum of the following infinite series, given $|x|<1$
$$2+4x+\frac{9}{2}x^2+\frac{16}{3}x^3+\frac{25}{4}x^4+\frac{36}{5}x^5+\frac{49}{6}x^6+\frac{64}{7}x^7+\frac{81}{8}x^8+ \ldots $$
I have tried turning this into a geometric series, but I just didn't even know where to begin. I also tried relating this to well-known sums but that didn't work out either.
I would love some help with this.
|
So your sum looks like
$$
s(x) = 2 + \sum_{k=1}^\infty \frac{(k+1)^2}{k} x^k.
$$
One approach is to expand the sum to
$$
\begin{split}
\sum_{k=1}^\infty \frac{(k+1)^2}{k} x^k
&= \sum_{k=1}^\infty \frac{k^2+2k+1}{k} x^k \\
&= \sum_{k=1}^\infty kx^k
+ 2 \sum_{k=1}^\infty x^k
+ \sum_{k=1}^\infty \frac{x^k}{k}.
\end{split}
$$
The first sum you can get by differentiating a geometric series:
$$
\begin{split}
g(x) &= \sum_{k=0}^\infty x^k\\
g'(x) &= \sum_{k=1}^\infty k x^{k-1}\\
x g'(x) &= \sum_{k=1}^\infty k x^k
\end{split}
$$
The second sum is a geometric series. Differentiate the third sum to get a geometric series:
$$
\begin{split}
f(x) &= \sum_{k=1}^\infty \frac{x^k}{k}\\
f'(x) &= \sum_{k=1}^\infty x^{k-1}
\end{split}
$$
which you can now find...
Just make sure to be careful about the term where $k=0$ everywhere...
|
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|
At which points does the graph of the $f(x)$ have horizontal tangent line? $(2x^2-2x-1)/(x^4+1)$
I tried to solve with the derivative method. But came across a fifth degree Polynomial involving $5$ terms. After that my math failed to factorize it. This is the derivative $2x^5 - 3x^4-2x^3-2x+1$. Kindly help.
|
With derivative method, we get
$$f'(x) = \frac{(4x-2)(x^4+1)-4x^3(2x^2-2x-1)}{(x^4+1)^2} = 0.$$
We have to find $x$ so that the equality can be fulfilled.
We obtain
$$(4x-2)(x^4+1) - 4x^3(2x^2-2x-1) = 0$$
$$ \iff (4x^5-2x^4 + 4x-2)- 8x^5+8x^4+4x^3 = 0 $$
$$\iff -4x^5 + 6x^4+4x^3+4x-2 = 0 $$
$$\iff 2x^5 - 3x^4 -2x^3-2x+1 = 0.$$
Define $g(x) = 2x^5 - 3x^4 -2x^3-2x+1$.
If you substitute $x = -1$, then $g(-1) = 0$. So $(x+1)$ is the factor of $g(x)$.
We get
$$g(x) = (x+1)(2x^4-5x^3+3x^2-3x+1).$$
From this equation, we obtain at $x=-1$, $f(x)$ has a horizontal tangent line.
You can find 2 more solution (if I'm not mistaken).
Hint: Find $x$ as a solution of $2x^4-5x^3+3x^2-3x+1 = 0.$
|
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show that all the roots of a given equation are in $[-1,1]$
*
*Suppose $c\in \mathbb{R}$ is such that one of the roots of the equation $x^3 - \dfrac{3}4 x+c=0$ belongs to $[-1,1]$.
*Show that all the roots belong to $[-1,1]$.
*The equation is equivalent to $p(x)=0$ where $p(x) = 4x^3 - 3x+4c.$
*
*Observe that $4\cos^3\theta - 3\cos \theta = \cos(3\theta).$
*Let the root $r$ of the equation in $[-1,1]$ be equal to $\cos \theta$ for some $\theta \in [0,\pi]$.
*From the above observation, $c = -\dfrac{\cos \theta_2}4$ for some $\theta_2 \in [0,\pi]$.
*The equation has two more roots, and if one of them is complex, then because the polynomial $p(x)$ has real coefficients, both of the others must be complex. We know that if $r_2,r_3$ are the other roots, then $r r_2 r_3 = -c, r + r_2 + r_3 = 0, r r_2 + r r_3 + r_2r_3 = -3/4.$
I'm not sure how to derive a contradiction from here.
|
Without calculus:
Let $r$ be a root of the equation in $[-1, 1]$. We have $c = -r^3 + \frac34 r$.
The equation becomes
$$x^3 - \frac34x - r^3 + \frac34 r = 0$$
or
$$\frac14(x-r)(4x^2 + 4rx + 4r^2 - 3) = 0$$
which has exactly three real roots
$$x_1 = r,\quad x_2 = -\frac12 r + \frac12\sqrt{3 - 3r^2}, \quad
x_3 = -\frac12 r - \frac12\sqrt{3-3r^2}.$$
It is easy to prove that $x_2, x_3 \in [-1, 1]$.
Indeed, first, we have $x_2 \ge - \frac12 r \ge -\frac12$
and $x_3 \le -\frac12 r \le \frac12$.
Second, we have
$$1 - x_2 = 1 + \frac12 r - \frac12\sqrt{3 - 3r^2}
\ge 0$$
since $1 + r/2 > 0$ and $1 + r + \frac14 r^2 - \frac{3 - 3r^2}{4} = (r + 1/2)^2 \ge 0$.
Third, we have
$$1 + x_3 = 1 - \frac12 r - \frac12\sqrt{3 - 3r^2} \ge 0$$
since $1 - r/2 > 0$ and $1 - r + \frac14r^2 - \frac{3-3r^2}{4} = (r - 1/2)^2 \ge 0$.
We are done.
|
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|
Let $x,y,z\in\mathbb{R}^{+}$ such that $x+y+z = 1.$ T/F: $\frac{x^2 y^2}{(x+y)^2}+\frac{y^2 z^2}{(y+z)^2}+\frac{z^2 x^2}{(z+x)^2}\geq\frac{9}{4}xyz$ In Springer's "Inequalities" book, Exercise $2.5$ is the following:
Let $x,y,z\in\mathbb{R}^{+}\ $ such that $\ x+y+z = 1.\ $ Then:
$$ xy + yz + zx \geq 9xyz.\qquad (1)$$
Proof: Applying AM $\geq$ GM, we get
$$ xy + yz + zx = (xy + yz + zx)(x + y + z) \geq 3 \sqrt[3]{ (xy)(yz)(zx)} \cdot 3 \sqrt[3]{ xyz } = 9xyz. \square $$
Using the square of the HM-GM inequality; that is, for all $a,b\geq 0, $ we have:
$$ \frac{ 4 a^2 b^2 }{ (a+b)^2 } \leq ab\qquad (2), $$
and then comparing $\frac{ 4 a^2 b^2 }{ (a+b)^2 }$ to $ab$ in each term in $(1),$ I propose a stronger inequality than $(1):$
$$ \frac{ x^2 y^2 }{ (x+y)^2 } + \frac{ y^2 z^2 }{ (y+z)^2 } + \frac{ z^2 x^2 }{ (z+x)^2 } \geq \frac{9}{4} xyz\qquad (3) $$
and so if $(3)$ is true, then it can be viewed as an improvement on inequality $(1).$
I have checked the following cases:
*
*$x=y=z=\frac{1}{3}:\ $ This is the only case where I found equality of $(3)$, although since $(3)$ is stronger than $(1),$ it is not clear that this is the only case where we have equality in $(3);$
*$x=y=\frac{1}{100},\ z=\frac{98}{100};$
*$x=y=\frac{1}{1000},\ z=\frac{998}{1000};$
*$x=y=\frac{1}{4},\ z=\frac{1}{2};$
*$x=\frac{1}{2},\ y=\frac{1}{3},\ z=\frac{1}{2};$
*$x=\frac{1}{3},\ y=\frac{1}{5},\ z=\frac{7}{15};$
and $(3)$ holds in all these cases.
I have been recreationally searching for ways in which the HM-GM inequality might be able to make other inequalities stronger and therefore improve upon the original inequality, but in other inequalities I have tried this for, the proposed improvement using HM-GM has turned out to be false.
|
$$\sum_{cyc}\frac{ 4x^2 y^2 }{ (x+y)^2 } \ge 9xyz$$
$$\sum_{cyc}\frac{ 4x y z(x+y+z)}{ (x+y)^2z^2 } \ge 9$$
define $a=yz,b=xz,c=xy$
$$\sum_{cyc}\frac{ 4(ab+bc+ac)}{ (a+b)^2 } \ge 9$$
$$\sum_{cyc}\frac{4}{ (a+b)^2 } \ge \frac{9}{ab+bc+ac}$$
which is well known prove thatt $\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$
|
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|
Write $ z = \frac{(1-i)^3(√3+i)}{4i}$ to polar form Write the complex number in polar form:
$$ z = \frac{(1-i)^3(\sqrt 3+i)}{4i}$$
So my try goes as follows:
\begin{align}
\frac{(1−i)^3(\sqrt 3+i)}{4i} &=
\frac{(1−i)^3(\sqrt 3+i) \times -4i}{16}\\& =
\frac{(1-3i-3+i)(\sqrt 3+i)\times-4i}{16}\\& =
\frac{(-2-2i)(\sqrt 3+i)\times-4i}{16} \\&=
\frac{(-1-i)(\sqrt 3+i)\times-4i}{8}\\& =
\frac{(4i - 4)(\sqrt 3+i)}{8} \\&=
\frac{(i-1)(\sqrt 3+i)}{2} \\&=
\frac{(\sqrt 3 \times i-1-\sqrt 3-i)}{2}\\& =
\frac{-(\sqrt 3+1)}{2} + \frac{(\sqrt 3-1)i} {2}\end{align}
Polar form:
$$r =\sqrt {\left(\frac {-1+\sqrt 3)}{2}\right)^2 + \left(\frac{\sqrt 3-1}{2}\right)^2} = \sqrt{2} $$
$$\tan(v) = \frac{(\sqrt 3-1)} { -(1+\sqrt 3)} \iff v = \arctan(\sqrt 3-2) $$
put it equal to $K$
$$z = \sqrt{2}(\cos K, i \sin K), \\ K = \arctan(\sqrt 3-2)$$
To me, this doesn’t seem like a clean answer, since this question could potentially be on an upcoming exam. Is there something I am missing, like a better approach?
|
Problems
The absolute value $r$ is correct, but the angle is not!
You used the wrong formula for the angle...
If you got a complex number $z = a + b \cdot \mathrm{i} = r \cdot \operatorname{cis}(\theta)$ then your angle is $\theta = \operatorname{arctan2}(b, ~a)$. But you used another formula.
If you use $\theta = \operatorname{arctan2}(b, ~a)$ you will get: $\theta = \operatorname{arctan2}\left(\Im\left(\frac{\left(1 - \mathrm{i}\right)^{3} \cdot \left(\sqrt{3} + \mathrm{i}\right)}{4 \cdot \mathrm{i}}\right), ~\Re\left(\frac{\left(1 - \mathrm{i}\right)^{3} \cdot \left(\sqrt{3} + \mathrm{i}\right)}{4 \cdot \mathrm{i}}\right)\right) = \arctan\left(\frac{-1 + \sqrt{3}}{-1 - \sqrt{3}}\right) + \pi = \frac{11}{12} \cdot \pi + 2 \cdot k \cdot \pi$.
better way to approach this
Is there something I am missing, like a better way to approach this?
Yes there is.
But you could also write the individual terms that are multiplied or divided with each other in polar form and then summarize them using the power laws.
If you don't whant to this 'cause you don't like the polar form.
You just can say/use (algebraic):
$$
\begin{align*}
z &= a + b \cdot \mathrm{i}\\
z &= r \cdot \operatorname{cis}(\theta)\\
\\
r &= |z| = \sqrt{a^{2} + b^{2}}\\
\theta &= \arg(z) = \operatorname{arctan2}\left( b, ~a \right)\\
\\&\text{with}\\\\
\operatorname{arctan2}\left( b, ~a \right) &=
\begin{cases}
\arctan\left({\frac {b}{a}}\right) \qquad\quad~~~ \text{for } a > 0\\
\arctan\left({\frac {b}{a}}\right) + \pi \qquad \text{for } a < 0 \quad y > 0\\
\pi \qquad\qquad\qquad\quad~~~ \text{for } a < 0 \quad y = 0\\
\arctan\left({\frac {b}{a}}\right) - \pi \qquad \text{for } a < 0 \quad y < 0\\
\frac{\pi}{2} \qquad\qquad\qquad\quad~~ \text{for } a = 0 \quad y > 0\\
-\frac{\pi}{2} \qquad\qquad\qquad~~~ \text{for } a = 0 \quad y < 0\\
\end{cases} + 2 \cdot k \cdot \pi
\end{align*}
$$
If you use that you'll get:
$$
\begin{align*}
z &= \frac{\left(1 - \mathrm{i}\right)^{3} \cdot \left(\sqrt{3} + \mathrm{i}\right)}{4 \cdot \mathrm{i}}\\
z &= -\frac{1 + \sqrt{3}}{2} + \left( \frac{-1 + \sqrt{3}}{2} \right) \cdot \mathrm{i}\\
z &= \sqrt{\left( -\frac{1 + \sqrt{3}}{2} \right)^{2} + \left( \frac{-1 + \sqrt{3}}{2} \right)^{2}} \cdot \operatorname{cis}\left(\operatorname{arctan2}\left( -\frac{1 + \sqrt{3}}{2}, ~\frac{-1 + \sqrt{3}}{2} \right)\right)\\
z &= \sqrt{2} \cdot \operatorname{cis}\left(\arctan\left(\frac{-1 + \sqrt{3}}{-1 - \sqrt{3}}\right) + \pi\right)\\
z &= \sqrt{2} \cdot \operatorname{cis}\left(\frac{11}{12} \cdot \pi + 2 \cdot k \cdot \pi\right)\\
z &= \sqrt{2} \cdot \operatorname{cis}\left(\frac{11}{12} \cdot \pi\right)\\
z &= \sqrt{2} \cdot \left(\cos\left(\frac{11}{12} \cdot \pi\right) + \sin\left(\frac{11}{12} \cdot \pi\right) \cdot \mathrm{i} \right)\\
\end{align*}
$$
|
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|
Solve the system $ \left\lbrace \begin{array}{ccc} \sqrt{1-x} + \sqrt{1-y} &=& \frac{y-x}{2} \\ x+y &=& 2xy \end{array}\right. $ I have to solve the following system:
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} + \sqrt{1-y} &=& \frac{y-x}{2} \\
x+y &=& 2xy
\end{array}\right.
$$
I've showed that it is equivalent to
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} - \sqrt{1-y} &=& 2 \\
x+y &=& 2xy
\end{array}\right. \text{ or } x=y= 1
$$
And i can't conclude nothing frome this system:
$$
\left\lbrace \begin{array}{ccc}
\sqrt{1-x} - \sqrt{1-y} &=& 2 \\
x+y &=& 2xy
\end{array}\right.
$$
|
We can still solve the system from its original form. From the first equation you get: $y \le 1$, and $y - x \ge 0$ or $y \ge x$. This leads to: $2x \le x+y = 2xy \le 2x\implies 2x \le 2xy \le 2x\implies 2x = 2xy \implies 2x(1-y) = 0\implies x = 0$ or $y = 1$ or both. If $x = 0$ then $y = 0$ from the second equation, but the first equation is not satisfied. So $y = 1$, and then $x = 1$. So the only solution is $x = y = 1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
if $\sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y $ then $y\leq 1$ or $y \geq 4$ Let $x \in \mathbb{R}^*$ and $y \geq \frac{1}{4}$.
Show that if $\sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y $ then $y\leq 1$ or $y \geq 4$
I tried the following idea:
$$
2\sqrt{y-\frac{1}{4}}\leq \sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y
$$
which leads to:
$$
y^2 - 4 y +1 \geq 0
$$
then $y \leq 2-\sqrt{3} $ or $ y \geq 2+\sqrt{3} $
|
Claim: $y \ge 1$.
Proof: For $y \ge \dfrac{1}{4}$ we have: $y = \sqrt{x^2+x+y} + \sqrt{x^2-x+y}\ge \sqrt{x^2+x+\dfrac{1}{4}}+\sqrt{x^2-x+\dfrac{1}{4}}=\left|x+\dfrac{1}{2}\right|+\left|x-\dfrac{1}{2}\right|$. So if $x \le -\dfrac{1}{2}$, then $y \ge -x-\dfrac{1}{2}-x+\dfrac{1}{2}=-2x \ge 1$. If $-\dfrac{1}{2} \le x \le \dfrac{1}{2}$, then $y \ge x+\dfrac{1}{2} - x+\dfrac{1}{2} = 1$. And if $x \ge \dfrac{1}{2}$, then $y \ge x+\dfrac{1}{2}+x-\dfrac{1}{2} = 2x \ge 1$. Thus we have shown that for any real number $x$, $y \ge 1$. Observe this is the range of $y$ as well, and you can choose an $x$ so that $y \ge 4$. Let $x = \dfrac{7}{2}$, then $y \ge \left|\dfrac{7}{2}+\dfrac{1}{2}\right|+\left|\dfrac{7}{2}-\dfrac{1}{2}\right|=4+3=7 > 4$.
Note: The statement that $y \le 1$ or $y \ge 4$ is changed to $y \ge 1$.
|
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|
How many natural numbers have $ n$ digits such that the sum of its digits is $ m.$ How many n-digit numbers such that the sum of its digits is $ m.$
My attempt :
We have the gf:
$\begin {align*}
f(x)&=x(1-x^9)(1-x^{10})^{n-1}(1-x)^{-n}\\&=(x-x^{10})(1-x^{10})^{n-1}(1-x)^{-n}
\end{align*}$
Extracting the coefficients of term of $x^m$ in expansion :
$[x^m]f(x)=\left ( [x^{m-1}] -[x^{m-10}]\right )\sum_{k=0}^{n-1 }(-1)^k\binom{n-1}{k}x^k\sum_{l=0}^{n}\binom{l+n-1}{n-1}x^l$
...and I am stuck here please help.
|
We obtain
\begin{align*}
\color{blue}{[x^m]}&\color{blue}{\left(x-x^{10}\right)\left(1-x^{10}\right)^{n-1}(1-x)^{-n}}\\
&=\left([x^{m-1}]-[x^{m-10}]\right)\sum_{j=0}^{\infty}\binom{-n}{j}(-x)^j\left(1-x^{10}\right)^{n-1}\tag{1}\\
&=\left([x^{m-1}]-[x^{m-10}]\right)\sum_{j=0}^{\infty}\binom{n+j-1}{j}x^j\left(1-x^{10}\right)^{n-1}\tag{2}\\
&=\sum_{j=0}^{m-1}\binom{n+j-1}{j}[x^{m-1-j}]\left(1-x^{10}\right)^{n-1}\\
&\qquad-\sum_{j=0}^{m-10}\binom{n+j-1}{j}[x^{m-10-j}]\left(1-x^{10}\right)^{n-1}\tag{3}\\
&=\sum_{j=0}^{m-1}\binom{n+m-2-j}{m-1-j}[x^j]\left(1-x^{10}\right)^{n-1}\\
&\qquad-\sum_{j=0}^{m-10}\binom{n+m-11-j}{m-10-j}[x^j]\left(1-x^{10}\right)^{n-1}\tag{4}\\
&=\sum_{j=0}^{\left\lfloor\frac{m-1}{10}\right\rfloor}\binom{n+m-2-10j}{m-1-10j}[x^{10j}]\left(1-x^{10}\right)^{n-1}\\
&\qquad-\sum_{j=0}^{\left\lfloor\frac{m-10}{10}\right\rfloor}\binom{n+m-11-j}{m-10-10j}[x^{10j}]\left(1-x^{10}\right)^{n-1}\tag{5}\\
&\,\,\color{blue}{=\sum_{j=0}^{\left\lfloor\frac{m-1}{10}\right\rfloor}\binom{n+m-2-10j}{m-1-10j}\binom{n-1}{j}(-1)^j}\\
&\,\,\color{blue}{\qquad-\sum_{j=0}^{\left\lfloor\frac{m}{10}\right\rfloor-1}\binom{n+m-11-j}{m-10-10j}\binom{n-1}{j}(-1)^j}\tag{6}\\
\end{align*}
Comment:
*
*In (1) we use the binomial series expansion.
*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (3) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$. We also set the upper limit to $m-1$ resp. $m-10$, since other terms do not contribute.
*In (4) we change the order of summation $j\to m-1-j$ resp. $j\to m-10-j$.
*In (5) we substitute $j$ with $10j$ since the expansion of $\left(1-x^{10}\right)^{n-1}$ contains powers of $x$ which are all multiples of $10$.
*In (6) we select the coefficient of $x^{10j}$.
|
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|
Finding $\lim_{x\to 0}\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}$ How to evaluate the following limit without L'Hôpital's rule
?$$\lim_{x\to 0}\left(\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}\right)$$
My attempt:
$$\begin{align}L &= \lim_{x\to 0}\left(\frac{1}{(\sin^{-1}(x))^2} - \frac1{x^2}\right)\tag1\\& = \lim_{x\to 0}\left(\frac{x^2}{(\sin^{-1}(x))^2x^2} - \frac1{x^2}\right)\tag2\\& = \lim_{x\to 0}\left(\frac{x}{\sin^{-1}(x)}\right)^2\cdot \frac1{x^2} - \frac1{x^2}\tag3\\& = \lim_{x\to 0}\left(\frac{x}{\sin^{-1}(x)}\right)^2\cdot\lim_{x\to 0} \frac1{x^2} - \frac1{x^2}\tag4\\& =\left(1\right)^2\cdot\lim_{x\to 0} \frac1{x^2} - \frac1{x^2}\tag5\\& =\cdot\lim_{x\to 0} \frac1{x^2} - \frac1{x^2} \tag6\\& =0\tag7\end{align}$$
But later I realized, $\lim\limits_{x\to a} f(x)\ g(x) = \lim\limits_{x\to a} f(x)\lim\limits_{x\to a} g(x) $ if and only if both $\lim\limits_{x\to a} f(x)$ and $\lim\limits_{x\to a} g(x)$ are well defined and finite. Thus moving from $(3)$ to $(4)$ is absolutely wrong.
I don't have any more ideas about the problem.
|
Using $$ \sin^{-1}(x) = x + \frac{x^3}{6} + \frac{3 \, x^5}{40} + \mathcal{O}(x^7) $$ then $$\frac{1}{(\sin^{-1}(x))^2} = \frac{1}{x^2} - \frac{1}{3} - \frac{x^2}{15} - \frac{31 \, x^4}{945} + \mathcal{O}(x^6). $$ This gives $$ \frac{1}{(\sin^{-1}(x))^2} - \frac{1}{x^2} = - \frac{1}{3} - \frac{x^2}{15} - \frac{31 \, x^4}{945} + \mathcal{O}(x^6) $$ and the limit $$ \lim_{x \to 0} \left( \frac{1}{(\sin^{-1}(x))^2} - \frac{1}{x^2} \right) = - \frac{1}{3}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int{\sqrt{x^2+4x+13}\, dx}$ I was trying to calculate with $\sinh$ :
$$\begin{align}I&=\int{\sqrt{x^2+4x+13}\, dx}\\&=\int{\sqrt{(x+2)^2+9}\, dx}\end{align}$$
Now with $x+2=3\sinh(u)$, $dx=3\cosh(u)\,du$
$$\begin{align}I&=3\int{\left(\sqrt{9\sinh^2(u)+9}\, \right)\cosh(u)\, du}\\&=9\int{\cosh(u)\sqrt{\sinh^2(u)+1}\, du}\\&=9\int{\cosh^2(u)\,du}\\&=\frac{9}{2}\int{\cosh(2u)+1\, du}\\&=\frac{9}{2}\left(\frac{1}{2}\sinh(2u)+u\right)+C\end{align}$$
How can I rewrite the $\frac{1}{2}\sinh(2u)$ in terms of $x$? I tried with the double angle formula, but this seemingly just complicates things by introducing $\cosh(u)$ also.
|
Letting $x+2=3\tan \theta$ changes the integral into
$$
I=9 \int \sec ^3 \theta d \theta
$$
$$
\begin{aligned}
\int \sec ^3 \theta d \theta &=\int \sec \theta d(\tan \theta) \\
&=\sec \theta \tan \theta-\int \sec \theta \tan ^2 \theta d \theta \\
&=\sec \theta \tan \theta-\int \sec \theta\left(\sec ^2 \theta-1\right) d \theta \\&= \sec \theta \tan \theta-\int \sec ^3 \theta d \theta-\int \sec \theta d \theta\\
&=\frac{1}{2}(\sec \theta \tan \theta+\ln \left|\sec \theta+\tan \theta \right|+C\\
&=\frac{1}{2}\left(\frac{(x+2) \sqrt{x^2+4 x+13}}{9}+\ln \left| \frac{x+2+\sqrt{x^2+4 x+13}}{3}\right|\right)+C
\end{aligned}
$$
Plugging back yields
$$
I=\frac{(x+2) \sqrt{x^2+4 x+13}}{2}+9 \ln \left|x+2+\sqrt{x^2+4 x+13}\right|+C
$$
|
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|
Evaluating $\int_1^{\pi/2}\frac{\sin(x)\ln(x)}{x^2}dx$ The question goes:
Evaluate $\int_1^{\pi/2}\frac{\sin(x)\ln(x)}{x^2}dx$.
I have tried by parts, substitutions and other techniques I have learned from first year...but I didn't get any thing look good...
|
The only viable solution using first-year knowledge is a series solution. We start by replacing the sine function with its respective Taylor series. Then we can integrate term by term
$$I=\int_1^{\frac{\pi}{2}}\frac{\ln{x}\sin{x}}{x^2}dx=\int_1^{\frac{\pi}{2}}\frac{\ln{x}}{x^2}\left(x+\sum_{k=1}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}\right)dx$$$$=\frac{1}{2}\ln^2\frac{\pi}{2}+\sum_{k=1}^\infty\frac{(-1)^k}{(2k+1)!}\int_1^{\frac{\pi}{2}}x^{2k-1}\ln{x}dx$$$$=\frac{1}{2}\ln^2\frac{\pi}{2}+\sum_{k=1}^\infty\frac{(-1)^k}{(2k+1)!}\left(\frac{\ln{\frac{\pi}{2}}}{2}\frac{(\pi^2/4)^k}{k}-\frac{1}{4}\frac{(\pi^2/4)^k}{k^2}+\frac{1}{4}\frac{1}{k^2}\right)$$$$=\frac{1}{2}\ln^2\frac{\pi}{2}+\frac{1}{2}\ln\frac{\pi}{2}\sum_{k=1}^\infty\frac{(-\pi^2/4)^k}{k(2k+1)!}-\frac{1}{4}\sum_{k=1}^\infty\frac{(-\pi^2/4)^k}{k^2(2k+1)!}+\frac{1}{4}\sum_{k=1}^\infty\frac{(-1)^k}{k^2(2k+1)!}$$
Without special functions, this is probably the closest you can get to a solution.
|
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|
Finding $x+y$, given $xy= 1$, $x^2+y^2=5$, $x^3+y^3=8$ This problem is from a math competition, but I think is wrong:
Find the value of $x+y$ if:
$$\begin{align}
xy &= 1 \\
x^2 + y^2 &= 5 \\
x^3+y^3 &= 8
\end{align}$$
Solution (I think is wrong):
$x^3 + y^3 = (x + y)(x^2-xy+y^2) = (x+y)(5-1) = 4(x+y)$
So we have:
$x^3+y^3 = 4(x+y)$
$x^3+y^3 = 8$
Then:
$8 = 4(x+y)$
$x+y = 2$
However if we replace that value in $(x+y)^2$ we have:
$(x+y)^2 = 2^2 =4$
$(x+y)^2 = x^2+y^2+2xy = 5 + 2 = 7$
As you can see $4 \neq 7$, what is happening?
|
The following method has the advantage that it can be generalized
for similar kinds of problems with more variables.
Consider the equation satisfied by $\;z=x\;$ and $\;z=y\;$
$$ z^2 - (x+y)z + (xy) = (z-x)(z-y) = 0. \tag1 $$
Then for all integer $n\ge 0$ it is true that
$$ z^{n+2} - (x+y)z^{n+1} + (xy)z^n = 0. \tag2 $$
Substitute $x$ and $y$ for $z$ in this equation and add the two equations to get
$$ (x^{n+2} + y^{n+2}) - (x+y)(x^{n+1} + y^{n+1}) + (x^n + y^n) = 0. \tag3 $$
Define $\;c := x^1+y^1 = x+y \;$ and note that $\;x^0 + y^0 = 2.\;$
Now assume that
$$ x\,y = 1, \quad x^2 + y^2 = 5, \quad \text{ and } \quad x^3 + y^3 = 8. \tag4 $$
Use equation $(3)$ with $\;n=1,\;n=0\;$ and assumptions $(4)$ to get
$$ 8 - c\,5 + c = 0 \qquad \text{ and } \qquad 5 - c\,c + 2 = 0. \tag5 $$
The equation for $\;n=1\;$ implies that $\;c=2\;$ but this is inconsistent
with the equation for $\;n=0.\;$
Thus the assumptions in $(4)$ are inconsistent and there is no solution.
|
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|
Where is my mistake when I am trying to solve $| \int_{\Gamma} \frac{1} {(z^2 +1)^2} dz | $
Find an upper bound for
$$ \bigg| \int_{\Gamma} \frac{1} {(z^2 +1)^2} dz \bigg|.$$
where Γ is the upper half-circle |z| = a with radius a > 1 traversed once in the counter clockwise direction.
I know the correct answer but I didn't know where I get wrong.
The following is my solution
$$
\bigg|\int_{0}^{\pi} \frac{1} {(a^2 e^{2i \theta} +1)^2} d (a e^{i \theta}) \bigg |
$$
$$
\bigg |\int_{0}^{\pi} \frac{1} {(a^2 e^{2i \theta} +1)^2} (a i e^{i \theta}) d \theta \bigg |
$$
$$
\bigg |\int_{0}^{\pi} \frac{(a i e^{i \theta})} {(a^2 e^{2i \theta} +1)^2} d \theta \bigg |
$$
$$
\bigg |\int_{0}^{\pi} \frac{(a i e^{i \theta})} {(a^2 e^{2i \theta} +1)^2} d \theta \bigg |
$$
$$
\bigg |\int_{0}^{\pi} \frac{(a i e^{i \theta})} {(a^2 e^{2i \theta} +1)^2} d \theta \bigg |
\leq
\int_{0}^{\pi} \frac{\bigg |(a i e^{i \theta}) \bigg |} {|(a^2 e^{2i \theta} +1)^2|} d \theta
$$
$$
\int_{0}^{\pi} \frac{|(a i (\cos \theta + i \sin \theta)|} {|(a^2 e^{2i \theta} +1)^2|} d \theta
$$
$$
\int_{0}^{\pi} \frac{|(a i \cos \theta - a \sin \theta)|} {|(a^2 e^{2i \theta} +1)^2|} d \theta
$$
$$
\int_{0}^{\pi} \frac{a} {|(a^2 e^{2i \theta} +1)^2|} d \theta
$$
$$
\int_{0}^{\pi} \frac{a} {|(a^4 e^{4i \theta} + 2 a^2 e^{2i \theta} + 1)^2|} d \theta
$$
$$
\int_{0}^{\pi} \frac{a} {|(a^4 \cos 4\theta + 2 a^2 \cos 2\theta + 1) +
(a^4 \sin 4\theta + 2 a^2 \sin 2\theta) i|} d \theta
$$
$$
\int_{0}^{\pi} \frac{a} {\sqrt{(a^4 \cos 4\theta + 2 a^2 \cos 2\theta + 1)^2 +
(a^4 \sin 4\theta + 2 a^2 \sin 2\theta)^2}} d \theta
$$
$$
\int_{0}^{\pi} \frac{a} {\sqrt{
{\left(1+2\ \cos\left(2\ \theta \right)\ a^2+a^4\right)}^2
}} d \theta
$$
Because a > 1 and $\theta \in [0, \pi]$, $\Rightarrow 1+2\ \cos\left(2\ \theta \right)\ a^2+a^4 \geq 0$
$$
\int_{0}^{\pi} \frac{a} {
{\left(1+2\ \cos\left(2\ \theta \right)\ a^2+a^4\right)}
} d \theta
$$
Use matlab symbolic function.
$$
\int_{0}^{\pi} \frac{a} {
{\left(1+2\ \cos\left(2\ \theta \right)\ a^2+a^4\right)}
} d \theta =
\frac{a \pi} {a^4 -1}
$$
$$
\bigg| \int_{\Gamma} \frac{1} {(z^2 +1)^2} dz \bigg | \leq
\frac{a \pi} {a^4 -1}
$$
$$
\int_{\Gamma} \frac{1} {(z^2 +1)^2} dz =
\int_{0}^{\pi} \frac{1} {(a^2 e^{2i \theta} +1)^2} (a i e^{i \theta}) d \theta =
-\frac{a} {a^2 + 1} + \frac{\pi}{2} + tan^{-1} a
,
(a>1)
$$
|
There is a classic bound, best seen by geometric reasoning: $$\left|\int_\Gamma\frac{1}{(z^2+1)^2}\,\mathrm{d}z\right|<\frac{\pi a}{(a^2-1)^2}$$
Which is slightly sharper than yours, $a>1$. The explanation: $\Gamma$ traces a half-circle - the locus of $z^2,\,z\in\Gamma$ traces a full circle (circle wraps around, arguments double...) of radius $a^2$. When you add $1$, you shift this to a circle of radius $a^2$ with centre $1$ - the minimum modulus (distance to the origin!) of any complex number on the circumference is 'clearly' given by that point of this locus on the real axis closest to the origin, which is the point $1-a^2$. The minimum modulus of $z^2+1$ is $a^2-1$, $a>1$, among $z\in\Gamma$.
That means I can bound the integrand from above by: $$\frac{1}{(a^2-1)^2}$$And the inequality follows.
|
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|
How to show that $\int_0^1 (1+t^2)^{\frac 7 2} dt < \frac 7 2 $? I need to show that $\int_0^1 (1+t^2)^{\frac 7 2} dt < \frac 7 2 $. I've checked numerically that this is true, but I haven't been able to prove it.
I've tried trigonometric substitutions. Let $\tan u= t:$
$$\int_0^1 (1+t^2)^{\frac 7 2} dt = \int_0^{\frac{\sqrt 2}{2}} (1+\tan^2 u )^{\frac 9 2} du = \int_0^{\frac{\sqrt 2}{2}} \sec^9 u \ du = \int_0^{\frac{\sqrt 2}{2}} \sec^{10} u \cos u\ du = \int_0^{\frac{\sqrt 2}{2}} \frac {\cos u}{(1-\sin^2 u)^5} du$$
Now let $\sin u = w$. Then:
$$\int_0^{\frac{\sqrt 2}{2}} \frac {\cos u}{(1-\sin^2 u)^5} du = \int_0^{\sin {\frac{\sqrt 2}{2}}} \frac {1}{(1-w^2)^5} dw.$$
This last integral is solvable using partial fraction decomposition, but even after going through all the work required I'm not really sure how to compare it with $\frac 7 2$, because of that $\sin {\frac {\sqrt{2}}{2}}$ term, which is not easy to compare.
|
Since $t^2 \in [0,1],\ $ we may use the Binomial expansion,
$$ \left( 1+ t^2 \right)^{7/2} = 1 + \frac{7}{2}t^2 + \frac{35}{8} t^4 + \frac{35}{16} t^6 + \frac{35}{128} t^8 + (\text{ alternating sequence of decreasing terms with negative leading term }),$$
so,
$$ \left( 1+ t^2 \right)^{7/2} < 1 + \frac{7}{2}t^2 + \frac{35}{8} t^4 + \frac{35}{16} t^6 + \frac{35}{128} t^8\qquad \forall\ t\in(0,1) $$
Unless I've made a calculation error, I get:
$$\ \int_0^1 1 + \frac{7}{2}t^2 + \frac{35}{8} t^4 + \frac{35}{16} t^6 + \frac{35}{128} t^8\ dt < 3.4.$$
|
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|
How do I solve this comparing of coefficient in this differential equation?
Solve $y'' + 2y' = \cos \pi x$
Homogeneous equation: $y= C_1 + C_2e^{-2x}$
Particular solution:
$y_p = A \cos \pi x + B\sin \pi x$
$y'=-A \pi \sin \pi x + B\pi \cos \pi x$
$y'' = -A \pi^2 \cos \pi x - B\pi^2 \sin \pi x$
Substituting back into the equation, $-A \pi^2 \cos \pi x - B\pi^2\sin \pi x + 2(-A\pi\sin \pi x + B\pi \cos \pi x) = \cos \pi x$
$\cos \pi x (-A\pi^2 + 2B\pi) - \sin\pi x(B\pi^2 + 2A\pi)= \cos \pi x$
So,
$ (-A\pi^2 + 2B\pi) =1 $
$(-B\pi^2 - 2A\pi) =0$
How do I solve this to get $y_p = \frac{1}{(4+\pi^2)\pi} (-\pi \cos \pi x + 2 \sin \pi x)$
|
From the second equation
$$A = -\dfrac{B \pi}{2}$$
Substituting into the first
$$B \pi^3 + 4 B \pi = 2$$
Solving for $B$
$$B = \dfrac{2}{\pi(\pi^2+4)}$$
Can you finish it off?
There are other approaches to finding $A$ and $B$.
|
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|
Parametric equation appears to generate circle and violate $x(t)^2+y(t)^2={R+}$ The context of the equations is from plotting the real against the imaginary impedance of an RC circuit however the problem is entirely mathematical.
Given the two equations:
$x(w) = \frac{R}{C^2 R^2 w^2 + 1}$
$y(w) = \frac{C R^2 w}{C^2 R^2 w^2 + 1}$
Where R, C and w are positive reals.
Plotting these equations appears to yeild a arc which follows a circle. Arc shown in green, see https://www.desmos.com/calculator/m4yffxmpn6 for interactive version.
Purple - A circle centered at x=R and r=R
Red/Blue are functions x and y
Following the suggestion of this question I tried to prove this by squaring each component and summing them and seeing if the result is constant.
$x(w)^2 + y(w)^2$
$=\frac{R^2}{((RCw)^2+1)^2} + \frac{R^4 C^2 w^2}{((RCw)^2 + 1)^2}$
$=\frac{R^2}{((RCw)^2 + 1}$
This is still a function of $w$ and so not constant for all $w$ and the curve is not a circluar arc.
Can this be right? It looks so much like a circular arc that I am second guessing myself at this proof being right.
|
Let's assume your parametric equation describes (at least part of) a circle, with unknown center $(a,b)$ and radius $r$. Hence, for all $w$,
$$(x-a)^2+(y-b)^2-r^2=0$$
The first term is
$$\left(\frac{R}{R^2C^2w^2+1}-a\right)^2=\frac{(R-aR^2C^2w^2-a)^2}{(R^2C^2w^2+1)^2}=\frac{a^2R^4C^4w^4+(R-a)^2-2a(R-a)R^2C^2w^2}{(R^2C^2w^2+1)^2}$$
Then
$$\left(\frac{R^2Cw}{R^2C^2w^2+1}-b\right)^2=\frac{(R^2Cw-bR^2C^2w^2-b)^2}{(R^2C^2w^2+1)^2}\\=\frac{b^2R^4C^4w^4+R^4C^2w^2+b^2-2bR^4C^3w^3-2bR^2Cw+2b^2R^2C^2w^2}{(R^2C^2w^2+1)^2}$$
And the last is
$$\frac{-r^2(R^2C^2w^2+1)^2}{(R^2C^2w^2+1)^2}=\frac{-r^2R^4C^4w^4-r^2-2r^2R^2C^2w^2}{(R^2C^2w^2+1)^2}$$
Now, add the terms. Since we managed to get the same denominator, factor it out. When doing the addition, keep $w^k$ terms together.
The equation at the beginning becomes :
$$\frac{1}{(R^2C^2w^2+1)^2}\left[(a^2+b^2-r^2)R^4C^4w^4-2bR^4C^3w^3+(R^4C^2+2b^2R^2C^2-2a(R-a)R^2C^2-2r^2R^2C^2)w^2-2bR^2Cw+(R-a)^2+b^2-r^2\right]=0$$
This has to be true for all $w$, hence the right factor is a polynomial in $w$ with null coefficients.
Therefore:
*
*$a^2+b^2=r^2$
*$b=0$
*$R^2C^2(R^2+2b^2-2a(R-a)-2r^2)=0$ hence $R^2+2b^2-2a(R-a)-2r^2=0$
*$(R-a)^2+b^2-r^2=0$, hence $R^2+a^2+b^2-r^2-2aR=0$
From the first and last we get $R(R-2a)=0$ hence $a=R/2$. Put this and $b=0$ in the first and we get $r=R/2$. We can check with the third:
$$R^2+2b^2-2a(R-a)-2r^2=R^2-\frac{1}{2}R^2-\frac{1}{2}R^2=0$$
So the parametric curve is indeed an arc of the circle of center $(R/2,0)$ and radius $R/2$.
Is it the full circle? No, the circle passes through the origin, but $x$ can't be zero.
Are all other points on the circle reached? Let's see. $y$ is an odd function of $w$, so the curve is symmetric wrt the $x$ axis. For $w\to\infty$, $x\to0$, and for $w=0$, $x=R$. Therefore, all abscissas of points on the circle are reached except $0$. That is, the parametric curve, defined for $w\in\Bbb R$, is exactly the aforementioned circle minus one point at the origin. For $w>0$ you keep only points of the circle such that $y>0$ (upper half circle).
|
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|
$\int \frac{\cos 4x}{4 \sin 2x} dx$ $\int \frac{\cos 4x}{4 \sin 2x} dx$
Let $u=2x$, $dx = 1/2 du$
$\int \frac{\cos 2u}{4 \sin u} \frac{1}{2} du = \frac{1}{8} \int \frac{1-2\sin^2 u}{\sin u}du \frac{1}{8} \int \frac{1}{\sin u} du - \frac{1}{8} \int 2 \sin u$
How do I integrate $\int \frac{1}{\sin u} du$ to get $\ln (\tan x)$ ?
The online calculator told me to use Weierstrass Substitution which I have not learnt before. Is there any other way to solve this ?
|
Hint Rewrite the standard integration formula
$$\int \cot u \,du = - \log\left\vert\csc u + \cot u\right\vert + C$$
using a half-angle identity for cotangent,
$$\csc u + \cot u = \frac{1 + \cos u}{\sin u} = \cot \frac{u}{2} = \cot x .$$
|
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|
Minimize function $\sqrt{x^2-4x+5}+\sqrt{4+x^2}$ Suppose I want to find
$$
\min f(x) = \min(\sqrt{x^2-4x+5}+\sqrt{4+x^2})
$$
I'd start with computing derivative and set it to $0$
$$
f'(x) = \frac{x-2}{\sqrt{x^2-4x+5}} + \frac{x}{\sqrt{4+x^2}}
$$
Then
$$
\frac{x-2}{\sqrt{x^2-4x+5}} + \frac{x}{\sqrt{4+x^2}} = 0
$$
$$
(x-2)\sqrt{4+x^2} + x\sqrt{x^2-4x+5}=0
$$
My first instinct is to rewrite it as
$$
(2-x)\sqrt{4+x^2} = x\sqrt{x^2-4x+5}
$$
and square both sides. The thing is, I get
$$
3x^2 -16x +16=0
$$
with $x_1 = 4/3$ and $x_2 = 4$. So $f'(x_1^-) <0$ and $f'(x_1^+) > 0$ thus in $x = 4/3$ there is a minimum of our function.
Question: is it valid to square both sides of an equation as I did above? Is there a possibility that because of that I lose or introduce a solution that shouldn't exist and might mess everything up?
|
Squaring both sides gives you the solutions to $(2-x)\sqrt{4+x^2} = x\sqrt{x^2-4x+5}$ and to $(2-x)\sqrt{4+x^2} = -x\sqrt{x^2-4x+5}$. So if our possible candidates are $x = 4/3, 4$, we can substitute these values into the equation before squaring.
We find that only $x_1 = 4/3$ satisfies the original equation.
Of course it remains to check that $(x_1, f(x_1))$ is a minimum. Therefore $f(x)$ has exactly one minimum point and no other extrema, as the graph confirms:
|
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|
In $\triangle ABC$, if $AC=4$ and $BC=5$, and $\cos(A-B)=\frac{7}{8}$, find $\cos(C)$ The problem is as the title suggests, in the given figure below, the goal is to find the Cosine of $\angle C$. I tried multiple ways of approaching this, such as with the Law of Sines, area formula, etc but none of them seemed to lead anywhere. My actual approach, which I will post as an answer below, uses the law of Cosines. Please share your own approaches especially if they use a different method!
|
If the Law of cosines works the Law of sines probably works. From $\frac{\sin A}{5}=\frac{\sin B}{4}=k$ and $\cos A\cos B+\sin A\sin B=\cos(A-B)$ we get the equation $$\sqrt{1-25k^2}\sqrt{1-16k^2}+20k^2=\frac{7}{8}$$
and $k=\frac{\sqrt{5}}{8\sqrt{2}}$. Hence, $\sin A=\frac{5\sqrt{5}}{8\sqrt{2}}$, $\sin B=\frac{4\sqrt{5}}{8\sqrt{2}}$, $\cos A=\frac{\sqrt{6}}{16}$, $\cos B=\frac{\sqrt{6}}{4}$ and
$$\cos C=-\cos(A+B)=\sin A\sin B-\cos A\cos B=\frac{5\sqrt{5}}{8\sqrt{2}}\frac{4\sqrt{5}}{8\sqrt{2}}-\frac{\sqrt{6}}{16}\frac{\sqrt{6}}{4}=\frac{25}{32}-\frac{3}{32}=\frac{11}{16.}$$
Note: WA output for Albert Chan's comment. Is this enough? I hope so.
|
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|
Find the solution to a system of equations Initially I'd like to solve the following problem:
Find $x$: $$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}$$
I'm already aware of one approach to solve it, but now I wonder what if I set $y=\sqrt{2+\sqrt{2+x}}$
?
That is, to solve the following system:
$$ \begin{cases} x = \sqrt{ 2+ \sqrt{ 2+y }} \\ y= \sqrt{2+\sqrt{2+x}} \end{cases}$$
|
I've just got an idea.
Let's suppose $x>y$, then $x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt{2+\sqrt{2+y}}<\sqrt{2+\sqrt{2+x}}=y$, which is impossible. When $x<y$ it's the same. Hence $x=y$.
As a result, $x=\sqrt{2+\sqrt2+x}\Rightarrow x^2=2+\sqrt{2+x}\Rightarrow x^4-4x^2+4=x+2\Rightarrow x^4-4x^2-x+2=0\Rightarrow$$\left(x^2+x-1\right)\left(x+1\right)\left(x-2\right)=0\Rightarrow x=2$.
If there's any mistake please tell me.
|
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|
Proof that $\binom{n}{k}\frac{1}{n^k} \le \frac{1}{k!}$ In a Calculus textbook, I was faced with the following
Problem: If $n$ is a natural number with $n \ge 1$, proof that $\binom{n}{k}\frac{1}{n^k} \le \frac{1}{k!}$ for all $k \in \mathbb{N}$.
The book presented a different solution than the one, I came up with. So maybe mine is wrong. My question now is, if my solution is correct and - more importantly - if not, where I went wrong. Thanks in advance for any comments and advice.
Solution: I proof the proposition by induction on $k$. For $k=0$ we have
$$\binom{n}{0}\frac{1}{n^0} = 1 \le \frac{1}{0!},$$
a true statement. Now notice, that if $k \gt n$ we have $\binom{n}{k} = 0$, so the proposition is plainly true. It thus suffices to proof it for $0 \le k \le n$. In this case we have as inductive step
$$\begin{align} \binom{n}{k+1} \frac{1}{n^{k+1}} & = \binom{n}{k} \frac{1}{n^{k}} \cdot \frac{1}{n(k+1)(n-k+1)} \\ & \le \frac{1}{k!} \cdot \frac{1}{n(k+1)(n-k+1)} && \text{by inductive hypothesis} \\ & \le \frac{1}{k!}\frac{1}{k+1} = \frac{1}{(k+1)!} && \text{since $n \ge 1$ and $n-k+1 \ge 1$}\end{align}$$
as required. $\blacksquare$
As additional comment, I have used the following steps to manipulate the binomial coefficient:
$\binom{n}{k+1} = \frac{n!}{(k+1)!(n-k+1)!} = \frac{n!}{(k+1)k!(n-k+1)(n-k)!} = \frac{n!}{k!(n-k)!} \cdot \frac{1}{(k+1)(n-k+1)} = \binom{n}{k} \frac{1}{(k+1)(n-k+1)}$
|
There is a mistake in your derivation: $$\binom{n}{k+1} = \frac{n!}{(k+1)!(n-k\color{red}-1)!} \color{blue}{= \frac{n!}{k!(n-k)!} \cdot \frac{n-k}{k+1} = \binom{n}{k} \cdot \frac{n-k}{k+1}}.$$
So, the final step of your proof should be the following:
\begin{align}
\binom{n}{k+1} \frac{1}{n^{k+1}} & = \binom{n}{k} \frac{1}{n^{k}} \cdot \frac{n-k}{n(k+1)} \\
& \le \frac{1}{k!} \cdot \frac{n-k}{n(k+1)} && \text{by induction hypothesis} \\
& \le \frac{1}{(k+1)!}\frac{n-k}{n} \\
& < \frac{1}{(k+1)!} && \text{since $n-k < n$}
\end{align}
|
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Prove $T(n) = T(\lfloor n/2 \rfloor) * T(\lfloor n/2 \rfloor) \leq 2^n$ I am stuck on the below problem.
Here is what I have so far, and I have highlighted the area that I am not understanding, particularly in the induction step when we start evaluating for k+1.
$T(n) = T(\lfloor \frac{n}{2} \rfloor) * T(\lfloor \frac{n}{2} \rfloor)$
Where $T(1) = 1$.
Show that $T(n) \leq 2^n$ by induction.
To prove that $T(n) \leq 2^n$, given that $T(1) = 1$, you can again use induction. Here's how you can do this for the given recurrence relation:
First, we need to show that the upper bound holds for the base case.
Base Case
In this case, the base case is $n = 1$, and we know that $T(1) = 1$. Since $1 <= 2^1$, the upper bound holds for the base case.
Next, we need to show that if the upper bound holds for a given value of $n$, it must also hold for the next value of $n$. To do this, we can assume that the upper bound holds for some arbitrary value of n and then show that it follows that the upper bound must also hold for $n + 1$.
Induction Step - problem I'm having below
For example, suppose we assume that the upper bound holds for $n = k$. This means that $T(k) <= 2^k$. We can then evaluate $T(k+1)$ as follows:
$T(k+1) = T(\frac{k+1}{2}) * T(\frac{k+1}{2})$
$= T(\frac{k}{2}) * T(\frac{k}{2})$ // WHY ARE THESE TWO THINGS EQUAL?
Since we know that $T(\frac{k}{2}) <= 2^{\frac{k}{2}}$, we can substitute this into the above equation to get:
$T(k+1) \leq 2^{(k/2)} * 2^{(k/2)}$
$= 2^{k/2 + k/2}$
$\leq 2^k$
Since $2^k \leq 2^{k+1}$, the upper bound holds for $n = k + 1$.
Therefore, by induction, we have shown that the upper bound $T(n) <= 2^n$ holds for all values of $n$.
How can we derive that:
$T(k+1) = T(\frac{k+1}{2}) * T(\frac{k+1}{2})$
$= T(\frac{k}{2}) * T(\frac{k}{2})$
|
You've begun well with your base step, but there are a couple of issues with your induction step. First, the problem uses the floor function for the RHS of the function definition, i.e.,
$$T(n) = T\left(\left\lfloor\frac{n}{2}\right\rfloor\right)\times T\left(\left\lfloor\frac{n}{2}\right\rfloor\right) \tag{1}\label{eq1A}$$
but you don't have this. Second, with the floor function being used, note that $T\left(\left\lfloor\frac{k+1}{2}\right\rfloor\right)\times T\left(\left\lfloor\frac{k+1}{2}\right\rfloor\right) = T\left(\left\lfloor\frac{k}{2}\right\rfloor\right)\times T\left(\left\lfloor\frac{k}{2}\right\rfloor\right)$ is only necessarily true in general if $k$ is even since then $\left\lfloor\frac{k+1}{2}\right\rfloor = \left\lfloor\frac{k}{2}\right\rfloor$. Nonetheless, with $T(1) = 1$, note we can prove (e.g., by induction) that $T(n) = 1 \; \forall \; n \ge 1$, so that statement is then always true.
Instead, to more generally support any $0 \le T(1) \le 2$, I suggest using strong induction so that, for some $k \ge 1$, we have all $1 \le n \le k$ satisfying the requested inequality, i.e.,
$$T(n) \le 2^n \tag{2}\label{eq2A}$$
Next, note $\left\lfloor\frac{k+1}{2}\right\rfloor \ge 1$, and consider $k - \left\lfloor\frac{k+1}{2}\right\rfloor$. If $k$ is even, we get $k - \frac{k}{2} = \frac{k}{2} \gt 0$, and if $k$ is odd we then get $k - \frac{k+1}{2} = \frac{2k-(k+1)}{2} = \frac{k-1}{2} \ge 0$. Thus, the arguments in the RHS of \eqref{eq1A} for $T(k+1)$ will always be $\ge 1$ and $\le k$, i.e., \eqref{eq2A} will be true for $n = \left\lfloor\frac{k+1}{2}\right\rfloor$.
There are now $2$ cases to consider. First, $k + 1$ being even gives that
$$\begin{equation}\begin{aligned}
T(k+1) & = T\left(\left\lfloor\frac{k+1}{2}\right\rfloor\right)\times T\left(\left\lfloor\frac{k+1}{2}\right\rfloor\right) \\
& = T\left(\frac{k+1}{2}\right)\times T\left(\frac{k+1}{2}\right) \\
& \le 2^{\frac{k+1}{2}} \times 2^{\frac{k+1}{2}} \\
& = 2^{k+1}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Second, $k + 1$ being odd results in
$$\begin{equation}\begin{aligned}
T(k+1) & = T\left(\left\lfloor\frac{k+1}{2}\right\rfloor\right)\times T\left(\left\lfloor\frac{k+1}{2}\right\rfloor\right) \\
& = T\left(\frac{k}{2}\right)\times T\left(\frac{k}{2}\right) \\
& \le 2^{\frac{k}{2}} \times 2^{\frac{k}{2}} \\
& \lt 2^{k+1}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Therefore, in both cases, we've proven the induction step that if $T(k) \le 2^{k}$, then $T(k+1) \le 2^{k+1}$. Thus, by strong induction, we've shown that $T(n) \le 2^{n}$ (i.e., \eqref{eq2A}) is true for all $n \ge 1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplifying $\frac{1}{x^3+y^3} + \frac{1}{x^6+y^6}$ Simplify: $$\frac{1}{x^3+y^3} + \frac{1}{x^6+y^6}$$
I know $\frac {na}{nb} = \frac {a}{b}$.
But I also know $(\frac{a}{b})^2 = \frac {a^2}{b^2} \neq \frac{a}{b}$
So how do I get the same denominator (go from $x^3+y^3$ to $x^6+y^6$) if I can't just square it, since that would change its value? And even if I did just square it, I would be left with an extra $2x^3y^3$, which is another problem I don't know how to handle. Any guidance on handling these type of problems would be appreciated thanks.
|
How about if you let $x^3 = a$ and $y^3 = b$?
Then $x^3 + y^3 = a + b$ and $x^6 + y^6 = a^2 + b^2$.
|
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|
Is there a better method to factoring large polynomials such as $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$? The question is as such:
Express $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ in irreducible factors $\in \mathbb{R}$
My initial thoughts to this were to use grouping of factors but this method was to no avail. Instead, I took this detour:
$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ is a factor of $x^7 - 1$, that is $x^7 - 1 = (x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)$
Hence, we can factor $x^7 - 1$ instead and just eliminate $x = 1$.
$$\to x^7 - 1 = 0$$
$$\to x^7 = 1 = e^{i2\pi}$$
$$\to x = e^{i\frac{2k\pi}{7}}, k = \{0, 1, 2, 3, 4, 5, 6\}$$
$$\to x^7 - 1 = (x - 1)(x - e^{i\frac{2\pi}{7}})(x - e^{i\frac{4\pi}{7}})(x - e^{i\frac{6\pi}{7}})(x - e^{i\frac{8\pi}{7}})(x - e^{i\frac{10\pi}{7}})(x - e^{i\frac{12\pi}{7}})$$
Finally we have, $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = (x - e^{i\frac{2\pi}{7}})(x - e^{i\frac{4\pi}{7}})(x - e^{i\frac{6\pi}{7}})(x - e^{i\frac{8\pi}{7}})(x - e^{i\frac{10\pi}{7}})(x - e^{i\frac{12\pi}{7}})$.
After a bit more thought, we can obtain real factors by group two factors such that upon multiplication, the angle addition obtains $e^{\frac{14\pi}{7}} = 1$, as the $c$ in $x^2 + bx + c$. This would be like $(x - e^{i\frac{2\pi}{7}})(x - e^{i\frac{12\pi}{7}}) = x^2 - 2\sin(\frac{3\pi}{14})x + 1$ (Using $e^{i\theta} = \cos{\theta} + i\sin{\theta}$).
Of course, all of this is tedious by hand to work out and type here, hence my final output was $$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = (x^2 - 2\sin(\frac{3\pi}{14})x + 1)(x^2 + 2\cos(\frac{\pi}{7})x + 1)(x^2 + 2\sin(\frac{\pi}{14}) + 1)$$.
My Question:
Is this factoring correct, and more importantly is there a more direct/simpler way to factor the polynomial?
|
Another well-known method is to rewrite the polynomial as follows.
$\begin{align}
&1+z+z^2+z^3+z^4+z^5+z^6=\\
&=z^3\left(\dfrac{1}{z^3}+\dfrac{1}{z^2}+\dfrac{1}{z}+1+z+z^2+z^3\right)=\\
&=z^3\left[\left(\dfrac{1}{z^3}+z^3\right)+\left(\dfrac{1}{z^2}+z^2\right)+\left(\dfrac{1}{z}+z\right)+1\right]=\\
&=z^3\left[\left(\dfrac{1}{z}+z\right)^3-3\left(\dfrac{1}{z}+z\right)+\left(\dfrac{1}{z}+z\right)^2-2+\\
+\left(\dfrac{1}{z}+z\right)+1\right]=\\
&=z^3\left[\left(\dfrac{1}{z}+z\right)^3+\left(\dfrac{1}{z}+z\right)^2-2\left(\dfrac{1}{z}+z\right)-1\right].
\end{align}$
By letting $\;y=\dfrac{1}{z}+z\;$ we get that
$\begin{align}
&1+z+z^2+z^3+z^4+z^5+z^6=z^3\left(y^3+y^2-2y-1\right).
\end{align}$
Now consider the real roots of the polynomial $\;y^3+y^2-2y-1\;$, which all belong to the interval $\;\left]-2,2\right[\; .$
|
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|
Prove or disprove the inequality if $a,b,c>0$, $a \geq b+c$. Prove or disprove the inequality
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 7abc$$ if $$a,b,c>0, a \geq b+c.$$
I thought to use this evaluation:
$$a^2b+b^2c+c^2a \geq 3abc.$$
So we have:
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 3abc+3abc=6abc,$$ which is obvious that $$6abc<7abc.$$
Is it right? I'm embarrassed that in my solution I did not have to use the condition that $$a \geq b+c.$$
Any hint would help a lot. thanks!
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Since $$b+c\geq\frac{7bc-b^2-c^2}{2(b+c)},$$ we obtain:
$$\sum_{cyc}(a^2b+a^2c)-7abc=(b+c)a^2+(b^2+c^2-7bc)a+bc(b+c)\geq$$
$$\geq(b+c)^3+(b^2+c^2-7bc)(b+c)+bc(b+c)=2(b+c)(b-c)^2\geq0.$$
|
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|
Solving $3 = x^2 + y^2 + z^2 - xy - yz - zx$ for integer $x$, $y$, $z$ I've been thinking about a solution for the following equation for integers $x, y, z$:
$$3 = x^2 + y^2 + z^2 - xy - yz - zx$$
A possible approach would probably be to transform the original equation to the following one:
$$6 = (x-y)^2 + (y-z)^2 + (x-z)^2$$
But how I do solve this?
|
You have made great Progress to get :
$6 = (x-y)^2 + (y-z)^2 + (x-z)^2$
The Squares $S1,S2,S3$ must be (Case 1) all 3 even or (Case 2) 1 even with 2 odd.
Let $S1 \le S2 \le S3$ , without loss of generality.
Case 1 : This implies that the Squares are (1A) $0,0,6$ (1B) $0,2,4$ (1C) $2,2,2$
We can then see that no Integer Squares give this.
Case 2 : This implies that the Squares are (2A) $0,1,5$ (2B) $0,3,3$ (2C) $1,1,4$ (2D) $1,2,3$
We can then see that no Integer Squares give this , except (2C).
Hence we get Simultaneous Equations to solve.
$x-y=\pm 1$
$y-z=\pm 1$
$z-x=\pm 2$
We have 8 Such Simultaneous Equations which may give Integer Solutions.
Case 2C1 :
$x-y=+ 1$
$y-z=+ 1$
$z-x=+ 2$
No Solution (adding the 3 Equations gives 0 = 4 which will be Inconsistent )
Case 2C2 :
$x-y=+ 1$
$y-z=+ 1$
$z-x=- 2$
This gives $y=x-1$ & $z=x-2$ ( Solution works out )
Case 2C3 :
$x-y=+ 1$
$y-z=- 1$
$z-x=+ 2$
No Solution (adding the 3 Equations gives 0 = 2 which will be Inconsistent )
Case 2C4 :
$x-y=+ 1$
$y-z=- 1$
$z-x=- 2$
No Solution.
Case 2C5 :
$x-y=- 1$
$y-z=+ 1$
$z-x=+ 2$
No Solution.
Case 2C6 :
$x-y=- 1$
$y-z=+ 1$
$z-x=- 2$
No Solution.
Case 2C7 :
$x-y=- 1$
$y-z=- 1$
$z-x=+ 2$
This gives $y=x+1$ & $z=x-2$ ( Solution works out , though it is Equivalent to 2C2 )
Case 2C8 :
$x-y=- 1$
$y-z=- 1$
$z-x=- 2$
No Solution.
Overall Solution Set : $y=x-1$ & $z=x-2$
Examples :
$x=0,y=-1,z=-2$
$x=10,y=9,z=8$
$x=-10,y=-11,z=-12$
$x=1,y=0,z=-1$
|
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|
Getting different values of the integral when integrating with different order I want to determine the volume of the shape given by the following
$$ K = \{(x,y,z): x^2+y^2+z^2 \leq 2 \, \, , x+y>0 \, \, , \, z\leq 1\} $$
I thought that i can integrate with respect to $xy$-plane first by using polar coordinates
$$ \int_{-\sqrt{2}}^1 \biggl( \int_{-\frac{\pi}{4}}^{\frac{3 \pi}{4}} \int_0^{\sqrt{2-z^2}} r \, \, drd\theta \biggr) \, dz \quad=\quad \frac{\pi}{2} \int_{-\sqrt{2}}^1 2-z^2 \, dz \quad=\quad ..... \quad=\quad \frac{\pi}{2} \biggl( \frac{5-2\sqrt{2}}{3} + 2\sqrt{2} \biggr) $$
But if I integrate with respect to $z$ first I get the following
$$ \iint_D \biggl( \int_{-\sqrt{2-(x^2+y^2)}}^1 \, dz \biggr) \, dxdy \quad=\quad \int_{-\frac{\pi}{4}}^{\frac{3 \pi}{4}} \int_0^{\sqrt{2}} \biggl(1+\sqrt{2-r^2} \biggr) r \, \, drd\theta = \, \, ...... \, \, = \quad \frac{\pi}{3} \biggl( 3 + 2 \sqrt{2} \biggr) $$
Which is very confusing since the value of the integral should be the same regardless of the order of integration! Can anyone see where I missed up?
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The first integral is correct. As regards the second integral note that $z=\sqrt{2-x^2-y^2}\leq 1$ for $1\leq x^2+y^2\leq \sqrt{2}$.
Therefore we need to split the $xy$-domain $D$ in two parts:
$$V=\iint_{D_1} \biggl( \int_{-\sqrt{2-(x^2+y^2)}}^1 \, dz \biggr) \, dxdy
+\iint_{D_2} \biggl( \int_{-\sqrt{2-(x^2+y^2)}}^{\sqrt{2-(x^2+y^2)}} \, dz \biggr) \, dxdy $$
where
$D_1=\{(x,y,0): x^2+y^2\leq 1, x+y>0\}$ and $D_2=\{(x,y,0): 1<x^2+y^2\leq 2, x+y>0\}.$
Hence
$$V=\pi\int_{0}^1\Big(1+\sqrt{2-r^2}\Big)r\,dr+\pi\int_{1}^{\sqrt{2}}\Big(2\sqrt{2-r^2}\Big)r\,dr=\frac{(5+4\sqrt{2})\pi}{6}$$
as we expected.
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|
How do we prove $x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0$? Question
How do we prove the following for all $x \in \mathbb{R}$ :
$$x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0 $$
My Progress
We can factorise the left hand side of the desired inequality as follows:
$$x^6+x^5+4x^4-12x^3+4x^2+x+1=(x-1)^2(x^4+3x^3+9x^2+3x+1)$$
However, after this I was unable to make any further progress in deducing the desired inequality.
I appreciate your help
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By using the symmetry in quartic,
\begin{array}
.x^4+3x^3+9x^2+3x+1&=x^2(x^2+3x+\frac{9}{2}+\frac{9}{2}+3x^{-1}+x^{-2})\\
&=x^2\large\left((x+\frac{3}{2})^2+(x^{-1}+\frac{3}{2})^2+\frac{9}{2}\large\right)\geq0
\end{array}
and at $x=0$ it is $1\geq 0$.
|
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|
If $α,β,γ$ are the roots of the equation $f(x)=x^3+qx+r=0$ then find the equation whose roots are, If $α,β,γ$ are the roots of the equation $f(x)=x^3+qx+r=0$ then find the equation whose roots are, $\frac{\beta^2+\gamma^2}{\alpha^2}$,$\frac{\alpha^2+\gamma^2}{\beta^2}$,$\frac{\beta^2+\alpha^2}{\gamma^2}$.
My solution goes like this:
We consider, $a=\frac{\alpha^2+\beta^2}{\gamma^2}$,$b=\frac{\beta^2+\gamma^2}{\alpha^2}$,$c=\frac{\gamma^2+\alpha^2}{\beta^2}$. Now, $$\alpha+\beta+\gamma=0,\alpha\beta+\beta\gamma+\gamma\alpha=q,\gamma\alpha\beta=-r$$ and hence,$\alpha^2+\beta^2+\gamma^2=-2q$. Also, $a=\frac{\alpha^2+\beta^2}{\gamma^2}=\frac{-2q-\gamma^2}{\gamma^2}=\frac{-2q}{\gamma^2}-1$ or $\gamma^2=\frac{-2q}{a+1}$. Also, $$a=\frac{\alpha^2+\beta^2}{\gamma^2}=\frac{\gamma^2-2\alpha\beta}{\gamma^2}=1-\frac{2\alpha\beta\gamma}{\gamma^2\gamma}=1+\frac{2r}{\frac{-2q}{a+1}\gamma}=1-\frac{r(a+1)}{q\gamma}$$ and hence,$\gamma=\frac{r(a+1)}{q(1-a)}$. Also, $a=\frac{\alpha^2+\beta^2}{\gamma^2}=\frac{\gamma^2-2\alpha\beta}{\gamma^2}=1-2\frac{\alpha\beta\gamma}{\gamma^3}=1+\frac{2r}{\gamma^3}$. Thus, $\gamma^3=\frac{2r}{a-1}$. Now, we have, $\gamma^3+q\gamma+r=0$. Thus, $\frac{2r}{a-1}+\frac{r(a+1)}{(1-a)}+r=0$, which implies $a^2-a-2=0$. Thus, the required equation is $x^2-x-2=0$.
Is the above solution correct? If not, then where is it going wrong? I dont get where is the mistake occuring?
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It’s not clear why you have written $$\frac{\gamma^2-2\alpha\beta}{\gamma^2}=\frac{1-2\alpha\beta\gamma}{\gamma^2\gamma}$$ This would mean that $\gamma^3=1$, which is not true.
It should be obvious to you that your answer is incorrect as you have arrived at a quadratic polynomial, not a cubic.
HINT…If you want a straightforward way of doing this question, try the following:
First find a polynomial whose roots are $\alpha^2$, $\beta^2$ and $\gamma^2$ which you can do either by using Vieta’s formulas or by substituting $y=x^2$ i.e. $x=\pm\sqrt{y}$.
Either way, you will get $$y^3+2qy^2+q^2y-r^2=0$$
Now, as you already know, $$\frac{\alpha^2+\beta^2}{\gamma^2}=-\frac{2q}{\gamma^2}-1$$
So now substitute $$z=-\frac{2q}{y}-1\implies y=-\frac{2q}{z+1}$$ into the above polynomial in $y$ and the resulting polynomial in $z$ will have the roots $\frac{\alpha^2+\beta^2}{\gamma^2}$ etc.
I hope this helps.
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|
How to convert Quaternions into Polar form? I would like to know how to write quaternions as polar form. Because I heard that if $A$ and $B$ are elements of $C$, this can be done with the form $A \cdot e^{B \cdot j}$.
But how can I do that?
Can I do it like this
$
\begin{align*}
a + b \cdot i + c \cdot j + d \cdot k &= \sqrt{\left( a + b \cdot i \right)^{2} + \left( c + d \cdot i \right))^{2}} \cdot \exp\left( \tan^{-1}\left( \frac{c + d \cdot i}{a + b \cdot i} \right) \cdot j \right)\\
\end{align*}
$?
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How to convert Quaternions into Polar form?
In general if you have a quaternion $q$ with $ q = q_{w} + q_{x} \cdot \mathrm{i} + q_{y} \cdot \mathrm{j} + q_{z} \cdot \mathrm{k} $, you can write it as $ q = \left| q \right| \cdot \exp\left( \theta \cdot \left( \cos\left( \varphi_{x} \right) \cdot \mathrm{i} + \cos\left( \varphi_{y} \right) \cdot \mathrm{j} + \cos\left( \varphi_{z} \right) \cdot \mathrm{k} \right) \right) $ where $ \left| q \right| = \sqrt{q_{w}^{2} + q_{x}^{2} + q_{y}^{2} + q_{z}^{2}} $.
We already know from studying quaternions in spatial rotation that $ e^{\theta \cdot \left( u_{x} \cdot \mathrm{i} + u_{y} \cdot \mathrm{j} + u_{z} \cdot \mathrm{k} \right)} = \cos\left( \theta \right) + \left( u_{x} \cdot \mathrm{i} + u_{y} \cdot \mathrm{j} + u_{z} \cdot \mathrm{k} \right) \cdot \sin\left( \theta \right) $ so we get $ q = \left| q \right| \cdot \left( \cos\left( \theta \right) + \left( \cos\left( \varphi_{x} \right) \cdot \mathrm{i} + \cos\left( \varphi_{y} \right) \cdot \mathrm{j} + \cos\left( \varphi_{z} \right) \cdot \mathrm{k} \right) \cdot \sin\left( \theta \right) \right) $.
If we compare the ral parts of both sides of the equation of the quaternion (the one in the algebraic form and the one in the polar form), we get:
$$
\begin{align*}
q = q_{w} + q_{x} \cdot \mathrm{i} + q_{y} \cdot \mathrm{j} + q_{z} \cdot \mathrm{k} ~&\wedge ~q = \left| q \right| \cdot \left( \cos\left( \theta \right) + \left( \cos\left( \varphi_{x} \right) \cdot \mathrm{i} + \cos\left( \varphi_{y} \right) \cdot \mathrm{j} + \cos\left( \varphi_{z} \right) \cdot \mathrm{k} \right) \cdot \sin\left( \theta \right) \right)\\
\end{align*}
$$
$$
\begin{align*}
q_{w} + q_{x} \cdot \mathrm{i} + q_{y} \cdot \mathrm{j} + q_{z} \cdot \mathrm{k} &= \left| q \right| \cdot \left( \cos\left( \theta \right) + \left( \cos\left( \varphi_{x} \right) \cdot \mathrm{i} + \cos\left( \varphi_{y} \right) \cdot \mathrm{j} + \cos\left( \varphi_{z} \right) \cdot \mathrm{k} \right) \cdot \sin\left( \theta \right) \right)\\
q_{w} + q_{x} \cdot \mathrm{i} + q_{y} \cdot \mathrm{j} + q_{z} \cdot \mathrm{k} &= \left| q \right| \cdot \cos\left( \theta \right) + \left| q \right| \cdot \left( \cos\left( \varphi_{x} \right) \cdot \mathrm{i} + \cos\left( \varphi_{y} \right) \cdot \mathrm{j} + \cos\left( \varphi_{z} \right) \cdot \mathrm{k} \right) \cdot \sin\left( \theta \right) \quad\mid\quad \Re\left( \right)\\
\Re\left( q_{w} + q_{x} \cdot \mathrm{i} + q_{y} \cdot \mathrm{j} + q_{z} \cdot \mathrm{k} \right) &= \Re\left( \left| q \right| \cdot \cos\left( \theta \right) + \left| q \right| \cdot \left( \cos\left( \varphi_{x} \right) \cdot \mathrm{i} + \cos\left( \varphi_{y} \right) \cdot \mathrm{j} + \cos\left( \varphi_{z} \right) \cdot \mathrm{k} \right) \cdot \sin\left( \theta \right) \right)\\
q_{w} &= \left| q \right| \cdot \cos\left( \theta \right) \quad\mid\quad \div \left| q \right|\\
\frac{q_{w}}{\left| q \right|} &= \cos\left( \theta \right) \quad\mid\quad \arccos\left( \right)\\
\arccos\left( \frac{q_{w}}{\left| q \right|} \right) &= \arccos\left( \cos\left( \theta \right) \right)\\
\arccos\left( \frac{q_{w}}{\left| q \right|} \right) &= \theta\\
\theta &= \arccos\left( \frac{q_{w}}{\left| q \right|} \right)\\
\end{align*}
$$
So all we have to do is $ \varphi_{x} $, $ \varphi_{y} $ and $ \varphi_{z} $ determine.
In fact, we can do this with all components of the quaternions:
$$
\begin{align*}
q_{w} + q_{x} \cdot \mathrm{i} + q_{y} \cdot \mathrm{j} + q_{z} \cdot \mathrm{k} &= \left| q \right| \cdot \left( \cos\left( \theta \right) + \left( \cos\left( \varphi_{x} \right) \cdot \mathrm{i} + \cos\left( \varphi_{y} \right) \cdot \mathrm{j} + \cos\left( \varphi_{z} \right) \cdot \mathrm{k} \right) \cdot \sin\left( \theta \right) \right)\\
q_{w} + q_{x} \cdot \mathrm{i} + q_{y} \cdot \mathrm{j} + q_{z} \cdot \mathrm{k} &= \left| q \right| \cdot \cos\left( \theta \right) + \left| q \right| \cdot \cos\left( \varphi_{x} \right) \cdot \mathrm{i} \cdot \sin\left( \theta \right) + \left| q \right| \cdot \cos\left( \varphi_{y} \right) \cdot \mathrm{j} \cdot \sin\left( \theta \right) + \left| q \right| \cdot \cos\left( \varphi_{z} \right) \cdot \mathrm{k} \cdot \sin\left( \theta \right) \quad\mid\quad \dot\Im\left( \right)\\
\dot\Im\left( q_{w} + q_{x} \cdot \mathrm{i} + q_{y} \cdot \mathrm{j} + q_{z} \cdot \mathrm{k} \right) &= \dot\Im\left( \left| q \right| \cdot \cos\left( \theta \right) + \left| q \right| \cdot \cos\left( \varphi_{x} \right) \cdot \mathrm{i} \cdot \sin\left( \theta \right) + \left| q \right| \cdot \cos\left( \varphi_{y} \right) \cdot \mathrm{j} \cdot \sin\left( \theta \right) + \left| q \right| \cdot \cos\left( \varphi_{z} \right) \cdot \mathrm{k} \cdot \sin\left( \theta \right) \right)\\
q_{a} &= \cos\left( \varphi_{a} \right) \cdot \sqrt{\left| q \right|^{2} - q_{w}^{2}}\\
q_{a} &= \cos\left( \varphi_{a} \right) \cdot \sqrt{\left| q \right|^{2} - q_{w}^{2}} \quad\mid\quad \div \left( \sqrt{\left| q \right|^{2} - q_{w}^{2}} \right)\\
\frac{q_{a}}{\sqrt{\left| q \right|^{2} - q_{w}^{2}}} &= \cos\left( \varphi_{a} \right) \quad\mid\quad \arccos\left( \right)\\
\arccos\left( \frac{q_{a}}{\sqrt{\left| q \right|^{2} - q_{w}^{2}}} \right) &= \arccos\left( \cos\left( \varphi_{a} \right) \right)\\
\arccos\left( \frac{q_{a}}{\sqrt{\left| q \right|^{2} - q_{w}^{2}}} \right) &= \varphi_{a}\\
\varphi_{a} &= \arccos\left( \frac{q_{a}}{\sqrt{\left| q \right|^{2} - q_{w}^{2}}} \right)\\
\end{align*}
$$
Can I do it like this $ a + b \cdot i + c \cdot j + d \cdot k = \sqrt{\left( a + b \cdot \mathrm{i} \right)^{2} + \left( c + d \cdot i \right))^{2}} \cdot \exp\left( \tan^{-1}\left( \frac{c + d \cdot \mathrm{i}}{a + b \cdot i} \right) \cdot \mathrm{j} \right) $?
Nope, because the formula is unfortunately not correct.
You can easily disprove the formula with the simple counterexample $ a = 0 = b $:
$$
\begin{align*}
&\text{counterexample } a = 0 = b\\
&\quad 0 + 0 \cdot i + c \cdot j + d \cdot k = \sqrt{\left( 0 + 0 \cdot \mathrm{i} \right)^{2} + \left( c + d \cdot i \right))^{2}} \cdot \exp\left( \tan^{-1}\left( \frac{c + d \cdot \mathrm{i}}{0 + 0 \cdot i} \right) \cdot \mathrm{j} \right)\\
&\quad 0 + 0 \cdot i + c \cdot j + d \cdot k = \sqrt{\left( 0 + 0 \cdot \mathrm{i} \right)^{2} + \left( c + d \cdot i \right))^{2}} \cdot \exp\left( \tan^{-1}\left( \underbrace{\frac{c + d \cdot \mathrm{i}}{0}}_{\text{Division by } 0 \text{ is not defined!}} \right) \cdot \mathrm{j} \right)\\
\end{align*}
$$
But of course also bring quaternions into the polar form $ A \cdot e^{B \cdot \mathrm{j}} $ (which is discussed here).
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4613868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Tricky integral in a complex plane How to integrate using residues:
$$\oint\limits_{|z|=2}\frac{1}{(z^6-1)(z-3)} dz $$
if the idea probably requires to change the sum of 6 residues to aditive inverse of residue at infinity.
I believe that the sum of 6 residues is:
$$2 \pi i\sum_{i=1}^6 Res_i \frac{1}{(z^6-1)(z-3)}=-2\pi i(Res_\inf \frac {1}{(z^6-1)(z-3)}+Res_3 \frac {1}{(z^6-1)(z-3)})$$
because $${3>|z|=2, \infty>|z|=2.}$$
Since $$\frac {1}{(z^6-1)(z-3)} = \frac {1}{(z^7)(1-\frac{1}{z^6})(1-\frac {3}{z})} = \frac {1}{z^7}+ \frac {3}{z^8}+ ...$$
$$=> -2\pi iRes_\inf \frac {1}{(z^6-1)(z-3)}=0$$
|
$$Res_{\infty}f(z)=Res_{z=0}-\frac{1}{z^2}f(\frac1z)=Res_{z=0}\frac{z^5}{(z^6-1)(1-3z)}=0$$
$$Res_{z=3}f(z)=Res_{z=3}\frac{1}{(z^6-1)(z-3)}=\left.\frac{1}{z^6-1}\right\rvert_{z=3}=\frac{1}{728}.$$
$$I=-2\pi i(Res_{\infty}f(z)+Res_{z=3}f(z))=-\frac{\pi i}{364}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4614189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Does a vector space need a quadratic form in order to define a wedge product? Geometric (Clifford) algebras require the vector space to be endowed with a quadratic form in order to define the geometric product.
Meanwhile, an exterior algebra has the wedge product as its defining product. It is not clear to me if a quadratic form is needed to specify a wedge product. So: does a wedge product need the vector space to be equipped with a quadratic form in order to be defined?
It seems like maybe not. For example, in 2D, the determinant, which can be calculated from a basis independent formula, is related to the definition of the wedge product.
Now, suppose we had a wedge product which was defined without relying on a quadratic form. Could the wedge product be then used to motivate a “dot product” or quadratic form?
For example, lets say that the wedge product was defined without a quadratic form, and we defined the lengths and angle of two vectors via:
$$|a \wedge b|=|a||b|\sin \theta$$
Could this definition of length and angle then be used to define a dot product?
|
The wedge product in geometric algebra is typically defined in one of two ways, the first as the completely antisymmetrized product
$$\mathbf{a} \wedge \mathbf{b} = \frac{1}{{2}} \left( { \mathbf{a} \mathbf{b} - \mathbf{b} \mathbf{a} } \right),$$
and the second, as the grade-2 selection of a product of two vectors
$$\mathbf{a} \wedge \mathbf{b} = {\left\langle{{ \mathbf{a} \mathbf{b} }}\right\rangle}_{2}.$$
In the first case, a metric for the vector space is required for this definition to expand in the usual fashion. In the second case, the definition is really of no use whatsoever without the metric.
Let's look at the antisymmetrized product and see where the metric dependencies are implied. Suppose that we have a two dimensional vector space with basis $\left\{ {\mathbf{e}_1, \mathbf{e}_2} \right\}$, where
$$\begin{aligned} \mathbf{a} &= a^1 \mathbf{e}_1 + a^2 \mathbf{e}_2 \\ \mathbf{b} &= b^1 \mathbf{e}_1 + b^2 \mathbf{e}_2.\end{aligned}$$
If we imposed a metric $ \mathbf{e}_i \cdot \mathbf{e}_j = g_{ij} $, then it gives us a notion of length and orthonormality, but do we need that for the wedge product, if defined as an antisymmetric sum? That antisymmetric sum expands as
$$\begin{aligned}\mathbf{a} \wedge \mathbf{b} &= \frac{1}{{2}} \left( { \mathbf{a} \mathbf{b} - \mathbf{b} \mathbf{a} } \right) \\ &= \frac{1}{{2}} \sum_{i,j = 1}^2 \left( { a^i b^j \mathbf{e}_i \mathbf{e}_j - b^i a^j \mathbf{e}_i \mathbf{e}_j } \right) \\ &= \frac{1}{{2}} \sum_{i \ne j} \left( { a^i b^j \mathbf{e}_i \mathbf{e}_j - b^i a^j \mathbf{e}_i \mathbf{e}_j } \right) + \frac{1}{{2}} \sum_{i = 1}^2 \left( { a^i b^i \mathbf{e}_i \mathbf{e}_i - b^i a^i \mathbf{e}_i \mathbf{e}_i } \right) \\ &= \sum_{i \ne j} a^i b^j \frac{1}{{2}} \left( { \mathbf{e}_i \mathbf{e}_j - \mathbf{e}_j \mathbf{e}_i } \right) \end{aligned}$$
This wedge product definition, even without a metric, allows us to conclude that
$$\begin{aligned} \mathbf{a} \wedge \mathbf{a} &= 0 \\ \mathbf{a} \wedge \mathbf{b} &= -\mathbf{b} \wedge \mathbf{a},\end{aligned}$$
but we don't know how to reduce an expression like
$$\frac{1}{{2}} \left( { \mathbf{e}_i \mathbf{e}_j - \mathbf{e}_j \mathbf{e}_i } \right),$$
nor can we give any meaning to any of $ \left\lVert {\mathbf{a}} \right\rVert $, $\left\lVert {\mathbf{b}} \right\rVert$, $ \left\lVert { \mathbf{a} \wedge \mathbf{b} } \right\rVert $.
In geometric algebra, the metric is usually introduced by way of the contraction axiom
$$ \mathbf{a}^2 = \mathbf{a} \cdot \mathbf{a}.$$
When expanded in coordinates, this brings the metric into the mix explicitly
$$\begin{aligned} \mathbf{a}^2 &= \sum_{i,j = 1}^N \left( { a^i \mathbf{e}_i } \right) \cdot \left( { a^j \mathbf{e}_j } \right) \\ &= \sum_{i,j = 1}^N a^i a^j \mathbf{e}_i \cdot \mathbf{e}_j \\ &= \sum_{i,j = 1}^N a^i a^j g_{ij}.\end{aligned}$$
Given this axiom, we have
$$ \left( { \mathbf{a} + \mathbf{b} } \right)^2 = \mathbf{a}^2 + \mathbf{b}^2 + \mathbf{a} \mathbf{b} + \mathbf{b} \mathbf{a},$$
but also
$$ \left( { \mathbf{a} + \mathbf{b} } \right)^2 = \left( { \mathbf{a} + \mathbf{b} } \right) \cdot \left( { \mathbf{a} + \mathbf{b} } \right) = \mathbf{a}^2 + \mathbf{b}^2 + 2 \mathbf{a} \cdot \mathbf{b},$$
so we are able to see that the symmetric sum is the dot product, that is
$$ \mathbf{a} \cdot \mathbf{b} = \frac{1}{{2}} \left( { \mathbf{a} \mathbf{b} + \mathbf{b} \mathbf{a} } \right),$$
and
$$\mathbf{b} \mathbf{a} = -\mathbf{a} \mathbf{b} + 2 \mathbf{a} \cdot \mathbf{b}.$$
In particular, for $i \ne j$
$$\begin{aligned} \mathbf{e}_i \wedge \mathbf{e}_j &=\frac{1}{{2}} \left( { \mathbf{e}_i \mathbf{e}_j - \mathbf{e}_j \mathbf{e}_i } \right) \\ &=\frac{1}{{2}} \left( { \mathbf{e}_i \mathbf{e}_j - \left( { -\mathbf{e}_i \mathbf{e}_j + 2 \mathbf{e}_i \cdot \mathbf{e}_j } \right) } \right) \\ &=\mathbf{e}_i \mathbf{e}_j - \mathbf{e}_i \cdot \mathbf{e}_j.\end{aligned}$$
Only by virtue of having a metric in play, do we see the notion of grade fall out of the mix in a natural fashion (here, the wedge product of our basis vectors, is observed to be the portion of the vector product that has the scalar part of the product subtracted off.)
|
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"url": "https://math.stackexchange.com/questions/4615329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
$\mathbb{E}[X \mid X \geq n/2]$ when $X \sim \text{Bin}(n, 0.5)$ I'm wondering if there's a closed form for the conditional expectation $\mathbb{E}[X \mid X \geq n/2]$ when $X \sim \text{Bin}(n, 0.5)$ is drawn from the Binomial distribution with parameters $n$ and $0.5$. Thanks!
|
@xzm is correct that
$$
\begin{align}
E[X|X \ge \frac n2] &= \frac{\sum_{k=\lfloor\frac{n+1}2\rfloor}^n k{n\choose k}}{\sum_{k=\lfloor\frac{n+1}2\rfloor}^n {n\choose k}}
\end{align}
$$
but it is possible to obtain this in closed form.
First note that $k{n\choose k}=n{n-1 \choose k-1}$. Then consider odd and even $n$ separately. For $n=2a+1$,
\begin{align}
E[X|X \ge \frac n2] &= \frac{(2a+1)\sum_{k=a+1}^{2a+1} {2a\choose k-1}}{\sum_{k=a+1}^{2a+1} {2a+1\choose k}} &= \frac{(2a+1)\sum_{k=a}^{2a} {2a\choose k}}{\sum_{k=a+1}^{2a+1} {2a+1\choose k}}
\end{align}
For $n$ even the sum of the "upper half" of the Binomial coefficients is $\sum_{k=a}^{2a} {2a\choose k} = \frac{1}{2}\big(2^{2a}-{2a \choose a}\big) +{2a \choose a} = 2^{2a-1} +\frac{1}{2} {2a \choose a}$, whereas for $n$ odd the sum is $\sum_{k=a+1}^{2a+1} {2a+1\choose k} = \frac{1}{2}2^{2a+1} = 2^{2a}$.
Thus
\begin{align}
E[X|X \ge \frac n2] &= \frac{(2a+1)\big(2^{2a-1}+\frac{1}{2} {2a \choose a}\big)}{2^{2a}} &= \frac{(2a+1)\big(2^{2a}+ {2a \choose a}\big)}{2^{2a+1}}.
\end{align}
Similar calculations can be done to give a closed form for the even case. For $n=2a$, the result is
\begin{align}
E[X|X \ge \frac n2] &= \frac{4a\big(2^{2a-2}+ {2a-1 \choose a}\big)}{2^{2a}+{2a \choose a}}.
\end{align}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4616013",
"timestamp": "2023-03-29T00:00:00",
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|
Show that $|\tan x-x|\leq 8x^2$ if $|x|\leq \pi/3$ Show that $|\tan x-x|\leq 8x^2$ if $|x|\leq \pi/3$. I think this is supposed be solved using the maclaurin series.
Let $f(x)=\tan x$ then $f'=1/\cos ^2x$ and $f''=\frac{-2\cos x \sin x}{\cos^4 x}=\frac{-\sin 2x}{\cos^4 x}$. Since $f(0)=0$ and $f'(0)=1$ we have that
$|x +\frac{-\sin 2\theta x}{\cos^4 x\theta}-x|=|\frac{-\sin 2\theta x}{\cos^4 x\theta}|$,where $\theta$ is between $0$ and $1$. Now what is left is to show that $|\frac{-\sin 2\theta x}{\cos^4 x\theta}|\leq 8x^2$ if $|x|\leq \pi/3$.
|
Let $f(x) = \tan x - x$ for all $x$ such that $|x| < \pi/2.$
Then $f(-x) = -f(x)$ for all $x,$ so in order to prove that
$|f(x)| \leqslant 8x^2$ for all $x$ such that $|x| \leqslant \pi/3,$
it is enough to prove it for all $x$ such that
$0 < x \leqslant \pi/3.$
We have $f'(x) = \sec^2 x - 1 > 0$ for all $x$ such that
$0 < x < \pi/2,$ and $f(0) = 0,$ so $f(x) > 0$ for all $x$ such that
$0 < x < \pi/2.$
Therefore it is enough to prove
$$
f(x) < 8x^2 \quad \left(0 < x \leqslant \frac\pi3\right).
$$
The well-known Maclaurin series (both of which converge for all $x$)
\begin{gather*}
\cos x = \sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}, \\
\sin x = \sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}
\end{gather*}
are alternating for all $x$ such that $0 < x \leqslant \sqrt2,$
because the negated ratio of successive terms of the cosine series is
$x^2/((2n+1)(2n+2)) \leqslant 1,$ and the calculation for the sine
series is similar.
Because $\pi/3 < \sqrt2,$ we therefore have $\sin x < x$ and
$\cos x > 1 - x^2/2$ for all $x$ such that $0 < x \leqslant \pi/3.$
Therefore for all such $x,$
$$
f(x) = \frac{\sin x}{\cos x} - x < \frac{x}{1 - x^2/2} - x =
\frac{x\cdot x^2}{2 - x^2} \leqslant \frac{(\pi/3)x^2}{2 - (\pi/3)^2} =
\frac{3\pi x^2}{18 - \pi^2} < \frac{5x^2}4,
$$
because $3\pi < \pi^2 < 10.$
Addendum. I think it is worth doing a little more work,
in order to prove the stronger inequality
$$
|\tan x - x| \leqslant \frac{17x^2}{21} \quad
\left(|x| \leqslant \frac\pi3\right).
$$
Proof.
Suppose $0 < x \leqslant \pi/3.$ Then, taking two more terms in the sine series,
$$
f(x) < \frac{x - x^3/6 + x^5/120}{1 - x^2/2} - x =
\frac{x^3/3 + x^5/120}{1 - x^2/2} = Ax^2,
$$
where
$$
A = \frac{x/3 + x^3/120}{1 - x^2/2} =
\frac{x(1 + x^2/40)}{3(1 - x^2/2)}.
$$
Because $\pi^2 < 10$ and $\pi < 22/7,$
$$
A < \frac{22\cdot(1 + 1/36)}{7\cdot9\cdot(1 - 5/9)} =
\frac{22\cdot37}{7\cdot4\cdot36} = \frac{407}{504} <
\frac{408}{504} = \frac{17}{21}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2}$ ascendingly? How would I order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2} \ $ without approximating the irrational numbers? In fact, I would be interested in knowing a general way to solve such questions if there is one.
What I tried to so far, because they are all positive numbers, is to square $x,y,z$ but, obviously, the rational parts will not be equal so I cannot compare the radicals. Proving that $x<z$ is easy and so is $y<z$, but I'm stuck at $x < y \text{ or } x>y$.
|
A visual demonstration of sorts:
Starting from $ \ x \ = \ \sqrt3 - 1 \ \ , \ $ we have
$$ x \ + \ 1 \ \ = \ \ \sqrt3 \ \ \Rightarrow \ \ x^2 \ + \ 2x \ + \ 1 \ \ = \ \ 3 \ \ . \ $$ So $ \ \sqrt3 - 1 \ $ is the larger of the two zeroes of $ \ x^2 + 2x - 2 \ \ . \ $ As for $ \ x \ = \ \sqrt5 \ - \ \sqrt2 \ \ , \ $ we may write
$$ x \ + \ \sqrt2 \ \ = \ \ \sqrt5 \ \ \Rightarrow \ \ x^2 \ + \ 2 \sqrt2 · x \ + \ 2 \ \ = \ \ 5 \ \ , $$
making $ \ \sqrt5 \ - \ \sqrt2 \ $ the larger of the two zeroes of the polynomial $ \ x^2 + 2 \sqrt2 · x - 3 \ \ . \ $ A graph of the two polynomials (the first in red, the second in blue) indicates the relative positions of these two zeroes.
If one mistrusts a graph (a not unreasonable attitude), we can show that the curve for $ \ x^2 + 2x - 2 \ $ lies "above" that of $ \ x^2 + 2 \sqrt2 · x - 3 \ \ $ [vertex at $ \ (-1 \ , \ -3) \ $ versus $ \ (-\sqrt2 \ , \ -5) \ \ , \ $ "higher" $ \ y-$ intercept; also, $$ x^2 + 2x - 2 \ > \ x^2 + 2 \sqrt2 x - 3 \ \ \Rightarrow \ \ 1 \ > \ 2·(\sqrt2 - 1)·x $$ $$ \Rightarrow \ \ \frac{1}{2·(\sqrt2 - 1)} \ = \ \frac12·(1 + \sqrt2) \ > \ x \ \ . \ ] $$ until the curves intersect at $ \ x \ = \ \frac{1}{2 \sqrt2 \ - \ 1} \ = \ \frac12 + \frac{\sqrt2}{2} \ \ , $ $ y \ = \ \frac32 \sqrt2 - \frac14 \ > \ 0 \ \ . \ $ Since this intersection is "above" the $ \ x-$axis, the positive $ \ x-$intercept of $ \ x^2 + 2x - 2 \ $ is smaller than that of $ \ x^2 + 2 \sqrt2 · x - 3 \ \ . $
|
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"timestamp": "2023-03-29T00:00:00",
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|
Probability of subsets with different constraints Let $U = \{1, 2, \dots, 1000\}$. We pick a subset from U uniformly at random. That is, every subset has an equal chance of showing up.
*
*What is the probability that the empty set is chosen?
*What is the probability that the subset has at most ten elements?
*What is the probability that the subset contains the numbers 500 or 1000?
*What is the probability that the subset does not contain all the multiples of 10?
*What is the probability that the size of the subset is odd?
My solution:
The total number of subsets of a set with size $n$ is $2^n$
*
*$\frac{1}{2^{1000}}$ (only one possibility)
*$\frac{\dbinom{1000}{10} + \dbinom{1000}{9} + \dots + \dbinom{1000}{1} + \dbinom{1000}{0}}{2^{1000}}$
*$(1 - \frac{2^{1000-1}}{2^{1000}}) + (1 - \frac{2^{1000-1}}{2^{1000}})$ (my understanding, if we want the subset to contain a particular number is the same as $1$ minus probability of the subset does not contain a number and itself is equal to a subset of a set of size $1000-1$)
*$\frac{2^{1000-100}}{2^{1000}} = \frac{2^{900}}{2^{1000}}$ (similar to above. there are 100 multiple of $10$ in $1000$)
*$\frac{\dbinom{1000}{999} + \dbinom{1000}{997} + \dots + \dbinom{1000}{1}}{2^{1000}}$
Am I on the right track?
|
1 and 2 - OK.
*$(1 - \frac{2^{1000-1}}{2^{1000}}) + (1 - \frac{2^{1000-1}}{2^{1000}}) = (1 - \frac12) + ( 1 -\frac12) = 1$, but the answer in obviously less than $1$. So no all is good in 4.
In 5 all is OK but there's a closed form.
We know that $(1+x)^n = \sum_{k=0}^{1000} \dbinom{1000}{k}x^k$. Put $x=\pm 1$ and $n=1000$. We got
$$ (1+1)^{1000} = \dbinom{1000}{0} + \dbinom{1000}{1} + \dbinom{1000}{2} + \dbinom{1000}{3} + \dbinom{1000}{4} + \ldots$$
$$ (1 + (-1))^{1000} = \dbinom{1000}{0} - \dbinom{1000}{1} + \dbinom{1000}{2} -\dbinom{1000}{3} + \dbinom{1000}{4} + \ldots$$
Hence $\frac{(1+1)^{1000} - (1 + (-1))^{100} }2 = \dbinom{1000}{1} + \dbinom{1000}{3} + \dots + \dbinom{1000}{999}$ and
$$\frac{\dbinom{1000}{999} + \dbinom{1000}{997} + \dots + \dbinom{1000}{1}}{2^{1000}} = \frac12$$
|
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|
Simplifying $\frac {\cos^4 x}{\cos^2 y}＋\frac {\sin^4 x}{\sin^2 y}＝1$ How do you simplify the following?
$$\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}＝1$$
What I've tried:
$$\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}＝1$$
$$\sin^2(x)+\cos^2(x)=1$$
$$\implies\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}＝\sin^2(x)+\cos^2(x)$$
$$\implies \frac {\cos^4(x)}{\cos^2(y)}－\cos^2(x)＝\sin^2(x)－\frac {\sin^4(x)}{\sin^2(y)}\\~\\\implies \frac {\cos^4(x)－\cos^2(x)\cos^2(y)}{\cos^2(y)}＝\frac {\sin^2(x)\sin^2(y)－\sin^4(x)}{\sin^2(y)}\\~\\\implies \frac {\cos^2(x)}{\cos^2(y)}\left(\cos^2(x)－\cos^2(y)\right)＝\frac {\sin^2(x)}{\sin^2(y)}\left(\sin^2(y)－\sin^2(x)\right)$$
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\begin{eqnarray}\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}&=&1\\
\cos^4x\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\
(1-\sin^2x)^2\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\
(1-2\sin^2x+\sin^4x)\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\
(1-2\sin^2x)\sin^2y+\sin^4x\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\
(1-2\sin^2x)\sin^2y+\sin^4x&=&\sin^2y(1-\sin^2y)\\
\sin^4x&=&\sin^2y(1-\sin^2y)-(1-2\sin^2x)\sin^2y\\
\sin^4x&=&(1-\sin^2y-1+2\sin^2x)\sin^2y\\
\sin^4x&=&2\sin^2x\sin^2y-\sin^4y\\
\sin^4x+\sin^4y&=&2\sin^2x\sin^2y\\
\frac{\sin^2x}{\sin^2y}+\frac{\sin^2y}{\sin^2x}&=&2
\end{eqnarray}
Let $u=\frac{\sin^2x}{\sin^2y}$, then $u+\frac{1}{u}=2$. So $(u-1)^2=0$.
Thus $u=1$ which gives $$\sin^2x=\sin^2y$$
Note that the original expression is undefined when either of $\cos y$ or $\sin y$ is zero whereas the 'equivalent' equation is defined for all $x,y$. If the original equation is set to $a$ instead of $1$ and we allow $a\to1$ we will see why.
The following desmos graph compares the graph for $a=1.2$ (in red) compared to the graph of $\sin^2x=\sin^2y$ (in black).
Here is a link to a graph in motion where $a$ varies in value from 1 to 2.
Desmos dynamic graph
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|
Simplifying partial derivative I have trouble understanding how this simplification was done. Can anyone explain please?
$\frac{\partial V}{\partial M}=\frac{5}{6}\frac{2^{1/3}3^{1/2}}{5^{5/6}}\frac{M^{-1/6}}{p_1^{1/3}p_2^{1/2}}=\frac{5^{1/6}}{2^{2/3}3^{1/2}p_1^{1/3}p_2^{1/2}}M^{-1/6}$
Thanks for your help.
Stan
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Besides some rearrangements, all the simplification comes in the numeric terms through factorisation and applying exponent rules:
$\begin{eqnarray}\frac{5}{6} \frac{2^{1/3} 3^{1/2}}{5^{5/6}} & = & \frac{5}{2 \times 3} \frac{2^{1/3} 3^{1/2}}{5^{5/6}} \\
& = & \frac{5^{6/6}}{5^{1/6}} \frac{2^{1/3}}{2^{3/3}} \frac{3^{1/2}}{3^{2/2}} \\
& = & 5^{5/6} \frac{1}{2^{2/3}} \frac{1}{3^{1/2}}\end{eqnarray}$
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|
Imaginary part of this integral I'm reading a physics paper and there is one step involving the imaginary part of an integral that I cannot reproduce (eq 23). The integral is
$$I=\int_0^1 dx\frac{1}{\Big[xa^2+(1-x)b^2-x(1-x)c^2-i\epsilon\Big]^{3/2}},$$
where all parameters are real and $a,\,b,\,\epsilon>0$. They claim that the imaginary part of this integral is
$$\text{Im}\{I\}=\frac{2\pi}{ab}\Big[\delta(c-a-b)+\delta(c+a+b)\Big].$$
I have no clue how to get that, any help would be appreciated.
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You can just plug this integral into some computer algebra system and get
$$I = \frac{2}{a^4 + b^4 + c^4 - 2(a^2 b^2 + a^2 c^2 + b^2 c^2) + 4i\epsilon c^2}\left[\frac{b^2-a^2-c^2}{\sqrt{a^2-i\epsilon}} + \frac{a^2-b^2-c^2}{\sqrt{b^2-i\epsilon}}\right].$$
Since $a$ and $b$ are nonzero and $\epsilon$ is infinitesimal, the imaginary part is provided by the first factor. Its denominator can be written as
$$[c^2-(a+b)^2][c^2-(a-b)^2]+4i\epsilon c^2.$$
If $c^2 = (a-b)^2$, the terms in square brackets in $I$ cancel each other. In the vicinity of $c^2 = (a+b)^2$
$$I = -2\frac{a+b}{ab}\frac{1}{c^2 - (a+b)^2 + i0},$$ and you get two delta functions by the Sokhotski–Plemelj theorem. Although it seems the prefactor will be $\frac{\pi}{ab}$, not $\frac{2\pi}{ab}$.
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|
One-variable inequality Let $t \in [0;1]$. Prove that
$$\sqrt[3]{{t{{\left( {t + 1} \right)}^2} + 4\left( {t + 1} \right)}} - \sqrt[3]{{12t}} \ge \frac{{\sqrt[3]{{14}}}}{7}\left( {\sqrt[3]{{2{{(t + 7)}^2}(3t + 1)}} - 8\sqrt[3]{t}} \right).$$
Direction 1: I consider two functions $f,g$ as follows
$$f\left( t \right) = \sqrt[3]{{t{{\left( {t + 1} \right)}^2} + 4\left( {t + 1} \right)}} - \sqrt[3]{{12t}},g\left( t \right) = \frac{{\sqrt[3]{{14}}}}{7}\left( {\sqrt[3]{{2{{(t + 7)}^2}(3t + 1)}} - 8\sqrt[3]{t}} \right)$$
However both functions $f,g$ reduce the function (while what I want is for them to be opposite monotonous).
Direction 2: I transform the inequality as follows
$$\begin{array}{l}
{(t - 1)^2}(t + 4) + 3\sqrt[3]{{t{{\left( {t + 1} \right)}^2} + 4\left( {t + 1} \right)}} \cdot \sqrt[3]{{12t}}\left( {\sqrt[3]{{t{{\left( {t + 1} \right)}^2} + 4\left( {t + 1} \right)}} - \sqrt[3]{{12t}}} \right)\\
\ge \frac{4}{{49}}{(t - 1)^2}(3t + 49) + 3\sqrt[3]{{2{{(t + 7)}^2}(3t + 1)}} \cdot 8\sqrt[3]{t}\left( {\sqrt[3]{{2{{(t + 7)}^2}(3t + 1)}} - 8\sqrt[3]{t}} \right).
\end{array}$$
It was too complicated and I stopped.
I have also submitted this question to the AoPS forum at the following link, but no support response yet!
https://artofproblemsolving.com/community/u797276h3001638p26953179
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The desired inequality is written as
$$\sqrt[3]{a} - \sqrt[3]{b} \ge \sqrt[3]{c} - \sqrt[3]{d}$$
or
$$\frac{a - b}{a^{2/3} + a^{1/3}b^{1/3} + b^{2/3}}
\ge \frac{c - d}{c^{2/3} + c^{1/3}d^{1/3} + d^{2/3}} \tag{1}$$
where
$a = t(t + 1)^2 + 4(t + 1)$,
$b = 12t$,
$c = \frac{14}{7^3}\cdot 2(t + 7)^2(3t + 1)$, and
$d = \frac{14}{7^3}\cdot 8^3t$.
We have
\begin{align*}
a - b &= (t + 4)(t - 1)^2 \ge 0, \\
c - d &= \frac{4}{49}(3t + 49)(t - 1)^2 \ge 0,\\
(a - b) - (c - d) &= \frac{37}{49}t(t-1)^2 \ge 0,\\
c - a &= \frac{t(-37t^2+74t+399)}{49} \ge 0, \\
d - b &= \frac{436t}{49} \ge 0.
\end{align*}
Thus, we have
$c \ge a > 0$ and $d \ge b \ge 0$ which results in
$$a^{2/3} + a^{1/3}b^{1/3} + b^{2/3} \le c^{2/3} + c^{1/3}d^{1/3} + d^{2/3}.$$
Also, we have $a - b \ge c - d \ge 0$.
Thus, (1) is true.
We are done.
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|
Please explain how to solve limit. I know the answer but how to explain it? Problem:
$a@b = \frac{a+b}{ab+1}$. Solve limit:
$\lim_{n \to \infty}(2@3@...@n)$.
I've tried to solve this problem by just calculating:
$$2 @ 3 = 0.714$$
$$2 @ 3 @ 4 = 1.222$$
$$2 @ 3 @ 4 @ 5 = 0.875$$
$$2 @ 3 @ 4 @ 5 @ 6 = 1.1$$
I found the pattern. The first number is less than 1, then the next is greater than 1, the next is less than 1, and so on. So the limit must be 1. But how to explain it mathematically?
I've tried to transform this:
$$(n-1)@n = \frac{2n-1}{n^2-n+1}$$
$$n@(n+1) = \frac{2n + 1}{n^2+n+1}$$
But it didn't help me to understand the method how to solve it. I think there should be a simple idea, which I don't see. I appreciate all hints.
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It holds that $\lim x_n = x$ if and only if every subsequence of $x_n$ contains, in turn, a subsequence that converges to $x$.
Check that the subsequence of even partial terms is strictly increasing and of odd partial terms is strictly decreasing.
Take any subsequence of the initial sequence. If it contains odd partial terms infinitely often, then it contains a subsequence converging to $1$. Otherwise, if it eventually contains only even partial terms, it again contains a subsequence converging to $1$. Thus, the initial limit is $1$.
Let's examine odd partial terms, say. It is readily verified that $@$ is associative and
$$ a@b@c = \frac{a+b+c+abc}{ab+bc+ca+1}.$$
My pre-edit answer is redundant. Also, the upper bound of $1+\frac{1}{n}$ does not hold, I miscalculated. We'll have to be more lenient with the bound. We'll accept odd partial terms being $>1$ as given.
It suffices to check that
$$ 2@3@\ldots @2n \leqslant 1+\frac{1}{\sqrt{n}},\quad n\in\mathbb N $$
Base case holds. Suppose $A:= 2@3@\ldots @2n \leqslant 1+\frac{1}{\sqrt{n}}$ for some $n$. Then
$$
\begin{align*}
A@(2n+1)@(2n+2) &= \frac{A+(2n+1)+(2n+2)+A(2n+1)(2n+2)}{A(2n+1)+(2n+1)(2n+2)+A(2n+2)+1} \\
&\leqslant \frac{A+(2n+1)+(2n+2)+A(2n+1)(2n+2)}{(2n+1)+(2n+1)(2n+2)+(2n+2)+1} \\
&= \frac{(4n+3) + A((2n+1)(2n+2)+1)}{(2n+1)+(2n+1)(2n+2)+(2n+2)+1} \\
&\leqslant \frac{(4n+3) + (1+\frac{1}{\sqrt{n}})((2n+1)(2n+2)+1)}{(2n+1)+(2n+1)(2n+2)+(2n+2)+1} \\
&= \frac{4\sqrt{n^3} + 4n^2 + 10n + 6\sqrt{n} + \frac{3}{\sqrt{n}}+6}{4n^2 + 10n+6} \\
&\overset{?}\leqslant 1+\frac{1}{\sqrt{n+1}}
\end{align*}
$$
The last inequality is a matter of direct verification. It suffices to check
$$\frac{4\sqrt{n^3}+6\sqrt{n}+\frac{3}{\sqrt{n}}}{4n^2+10n+6} \leqslant \frac{1}{\sqrt{n+1}}. $$
Note that
$$ \frac{4n^2+6n+3}{\sqrt{\frac{n}{n+1}}} \leqslant 4n^2+10n+6 \Leftrightarrow 8n(n+1)(4n+3)\geqslant 3, $$
the right hand statement is evidently true. Even partial terms can be tackled analogously.
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|
Range of Trigonometric function having square root
Finding range of function
$\displaystyle f(x)=\cos(x)\sin(x)+\cos(x)\sqrt{\sin^2(x)+\sin^2(\alpha)}$
I have use Algebric inequality
$\displaystyle -(a^2+b^2)\leq 2ab\leq (a^2+b^2)$
$\displaystyle (\cos^2(x)+\sin^2(x))\leq 2\cos(x)\sin(x)\leq \cos^2(x)+\sin^2(x)\cdots (1)$
And
$\displaystyle -[\cos^2(x)+\sin^2(x)+\sin^2(\alpha)]\leq 2\cos(x)\sqrt{\sin^2(x)+\sin^2(\alpha)}\leq [\cos^2(x)+\sin^2(x)+\sin^2(\alpha)]\cdots (2)$
Adding $(1)$ and $(2)$
$\displaystyle -(1+1+\sin^2(\alpha))\leq 2\cos(x)[\sin(x)+\sqrt{\sin^2(x)+\sin^2(\alpha)}]\leq (1+1+\sin^2(\alpha))$
$\displaystyle -\bigg(1+\frac{\sin^2(\alpha)}{2}\bigg)\leq f(x)\leq \bigg(1+\frac{\sin^2(\alpha)}{2}\bigg)$
I did not know where my try is wrong.
Please have a look on it
But actual answer is $\displaystyle -\sqrt{1+\sin^2(\alpha)}\leq f(x)\leq \sqrt{1+\sin^2(\alpha)}$
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Let $p=\sin^2\alpha,\; y=\cos 2x,\;$ then
$$\cos x \sin x+\cos x\sqrt{\sin^2x+\sin^2\alpha\mathstrut}
=\dfrac{\pm\sqrt{1-y^2}\pm\sqrt{1-y^2+2p(1+y)\mathstrut}}2 = f(y),$$
$$2f'(y) = \mp\dfrac y{\sqrt{1-y^2}}\pm\dfrac{p-y}{\sqrt{1-y^2+2p(1+y)}}.$$
The set of possible extremes of $f(y)$ corresponds with the expression
$$\left[\begin{align}
&y^2-1=0\\[4pt]
&y^2-2py-2p-1=0\\[4pt]
&y^2(1+2py-y^2+2p)=(p^2-2py+y^2)(1-y^2),
\end{align}\right.,$$
$$\left[\begin{align}
&y=\pm1\\[4pt]
&y=p\pm\sqrt{p^2+2p+1}\\[4pt]
&(2+p)y^2+2y-p=0,
\end{align}\right.$$
$$y_m\in\left\{-1,1, \dfrac p{2+p}\right\},\quad
f(y_m)\in\left\{0, \pm\sqrt{p\mathstrut}, \pm \sqrt{1+p} \right\},$$
$$\mathbb{f(y)\in\left[-\sqrt{1+\sin^2\alpha}, \sqrt{1+\sin^2\alpha}\right]}.$$
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Is it true that if $a, b, x, y>0$, $a+b \ge x+y$ and $ab \le xy$ then $a^n+b^n≥x^n+y^n$ for $n>1$? Let $a, b, x, y>0$, $a+b \ge x+y$ and $ab \le xy$ then $a^n+b^n \ge x^n+y^n$ where $n=2$.
Is it true that if $a, b, x, y>0$, $a+b \ge x+y$ and $ab \le xy$ then $a^n+b^n \ge x^n+y^n$ for $n \ge 1$?
I have just been checked it is true with $1< a, b, x, y \le 1000$ with $n=3,4$ by my computer. I am looking for a proof if this is true inequality.
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Assuming that the inequality holds for $n - 1$ and $n - 2$, we can use induction.
\begin{align}
a^n + b^n &= a^{n - 1}(a + b) - a^{n - 1}b + b^{n - 1}(a + b) - ab^{n - 1} \\
&=(a + b)(a^{n - 1} + b^{n - 1}) - ab(a^{n - 2} + b^{n - 2}) \\
&\geq (x + y)(x^{n - 1} + y^{n - 1}) - xy(a^{n - 2} + b^{n - 2}) \\
&= x^n + y^n + xy(x^{n - 2} + y^{n - 2} - a^{n - 2} - b^{n - 2}) \\
&> x^n + y^n
\end{align}
Indeed, when $n = 1$ we are given the inequality. When $n = 2$, we have
\begin{align}
a^2 + b^2 &= (a + b)^2 - 2ab \\
&\geq (x + y)^2 - 2xy \\
&=x^2 + y^2.
\end{align}
Let me know if there are any mistakes.
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Calculate the following integral $ \int\frac{2x+1}{x^{n+2}(x+1)^{n+2}}\ln\left(\frac{2x^2+2x+1}{x^2(x+1)^2}+\frac7{16}\right)dx$ Hello I am trying to solve a pretty complicated integral. It is a from a set of problems, published in a monthly journal for high school students and they are exercises in preparation for a competition.
So the problem is the following: Calculate the integral $ \int \frac {2x+1}{x^{n+2}(x+1)^{n+2}} \ln\left(\frac {2x^2+2x+1}{x^2(x+1)^2} + \frac{7}{16}\right)dx$
I first tried partial integration, integrating $\frac {2x+1}{x^{n+2}(x+1)^{n+2}}$, it is happily easy to do it and we will end up with $\int \frac {2x+1}{x^{n+2}(x+1)^{n+2}} = \frac {-1}{x^{n+1}(x+1)^{n+1}}$.
I also differentiated $\ln\left(\frac {2x^2+2x+1}{x^2(x+1)^2} + \frac{7}{16}\right)$, ending up with $\frac {-64x^3-96x^2-96x-32}{7x^6+21x^5+71x^3+48x^2+16x}$
Now when I try to solve the integral of the product of these two, I am stuck.
I also thought about some kind of recursive relationship in terms of $n$ but I am not sure about it.
I would happily accept any help in solving this problem. Thanks in advance.
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Integrate by parts:
$$\int \frac{2x+1}{x^{n+2}(x+1)^{n+2}} \ln\left(\frac{2x^2+2x+1}{x^2(x+1)^2} + \frac7{16}\right) \, dx \\
= -\frac1{n+1} \frac{\ln\left(\frac{2x^2+2x+1}{x^2(x+1)^2}+\frac7{16}\right)}{x^{n+1}(x+1)^{n+1}} - \frac{32}{n+1} \int \frac{2x^3+3x^2+3x+1}{x^{n+2}(x+1)^{n+2} \left(7x^4+14x^3+39x^2+32x+16\right)} \, dx$$
Expand into partial fractions:
$$\frac{2x^3+3x^2+3x+1}{7x^4+14x^3+39x^2+32x+16} = \frac18 \left(\frac{2x+1}{x^2+x+4} + \frac{2x+1}{7x^2+7x+4}\right)$$
The remaining integrals are elementary. For instance, substituting
$$x=\frac{\sqrt{4u-15}-1}2 \implies u = x^2+x+4 \implies du=2x+1\,dx$$
leads to
$$\int \frac{2x+1}{x^{n+2}(x+1)^{n+2}(x^2+x+4)} \, dx = \int \frac{du}{\left(\frac{\sqrt{4u-15}-1}2\right)^{n+2} \left(\frac{\sqrt{4u-15}-1}2+1\right)^{n+2} u} = \int \frac{du}{\left(u-4\right)^{n+2}u}$$
and can be further developed with partial fraction expansions.
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Without Calculator find $\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$
Without Calculator find $$\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$$
Where $\left \lfloor x \right \rfloor $ represents floor function.
My Try:
Let $x=2\cos(50^{\circ})+\sqrt{3}$. We have
$$\begin{aligned}
& \cos \left(50^{\circ}\right)<\cos \left(45^{\circ}\right) \\
\Rightarrow \quad & 2 \cos \left(50^{\circ}\right)<\sqrt{2} \\
\Rightarrow \quad & x<\sqrt{3}+\sqrt{2}<3.14
\end{aligned}$$
Now I am struggling hard to prove that $x>3$
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It is known that $\cos(50^o)=\frac{\phi}2$. So, we need to find the floor of $\phi+\sqrt{3}$. $\sqrt{3}$ is less than $2$ since it is less than $\sqrt{4}$. $\phi$ is less than $2$ since $(1+\sqrt{5})/2<(1+\sqrt{9})/2=2$. We could do a similar proof that $\sqrt{3},\phi>1$. So, the answer must be between $\lfloor4\rfloor$ and $\lfloor2\rfloor$. The answer is thus $3$.
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|
Solving for x in logarithmic equation $\log_4(2x) = \frac{1}{2}x^2 - 1$ I am trying to solve for $x$ in the equation $\log_4(2x) = \frac{1}{2}x^2 - 1$. I have tried converting the logarithmic expression to exponential form, but I am not able to isolate $x$ in the resulting equation.
This is what I have tried as of now:
$$\log_4(2x) + \log_4(4) = \frac{1}{2}x^2$$
$$\log_4(8x) = \frac{1}{2}x^2$$
$$2\log_4(8x) = x^2$$
$$\log_4(64x^2) = x^2$$
$$64x^2 = 4^{x^2}$$
after which I am not too sure on how to find x
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$$\log_4{(2x)}=\frac{x^2}{2}-1$$
As stated already in a previous answer, try $x=2$:
$$x=2: \log_4{(2\cdot 2)}=\frac{2^2}{2}-1\Leftrightarrow 1=1$$
So $x=2$ is a solution. So far nothing new (from previous answer).
Next I would transform the equation into an exponential one:
$$\begin{align}\log_4{(2x)}&=\frac{1}{2}(1+\log_2 x)\\ \log_2 x&=x^2-3\\8x&=2^{x^2}\\f(x)&=2^{x^2}-8x\\f(0)&\gt 0, f(1)\lt 0\\f(0)&\gt 0, f(\frac{1}{2})\lt 0,…f(\frac{1}{8})\gt 0, f(\frac{3}{16})\lt 0\end{align}$$
I would continue to chase the root by halving the interval. So far I could tell that the root is located between $\frac{1}{8}$ and $\frac{3}{16}$.
With red is $y=2^{x^2}$ and with blue is $y=8x$. The plot was obtained from desmos.com
|
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|
Determining whether $\frac{1}{2^{2}} + \frac{2}{3{^2}} + ... +\frac{n}{(n+1)^{2}}$ is a Cauchy sequence. My attempt so far: If $n \leq m$, then
$a_{m} - a_{n} = \frac{n+1}{(n+2)^2}+\frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} < \frac{n+1}{(n+2)(n+1)}+\frac{n+2}{(n+3)(n+2)}+...+\frac{m}{(m+1)m} = \frac{1}{(n+2)}+\frac{1}{(n+3)}+...+\frac{1}{(m+1)} < \frac{1}{n}+\frac{1}{n}+...+\frac{1}{m} \leq \frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}$.
A bit stuck here though. The hint in the book said it was not Cauchy.
Edit: If it's not Cauchy, then let me take a new approach..
$a_{m} - a_{n} = \frac{n+1}{(n+2)^2}+\frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} \geq \frac{n+1}{(n+2)^2} + \frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} \geq \frac{1}{(m+1)^2} + \frac{1}{(m+1)^2}+...+\frac{1}{(m+1)^2}$.
The $\frac{1}{(m+1)^2}$ occurs $m-n$ times here.
|
Put $m=2n$ and check that $|a_m-a_n|$ does not go to $0$.
|
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|
Generatingfunctionology Chapter 1 Exercise 1.c Regarding the exercises of the Generatingfunctionology book available at (https://www2.math.upenn.edu/~wilf/DownldGF.html).
In particular Chapter 1, Exercise 1.(c), which asks to find the ordinary power series generating function of
$$a_n = n^2.$$
Following the method explained in the book (Page 8 "The Method"), I define $A(x) = \sum_n a_n x^n$ and arrive (by multiplying by $x^n$ and summing over $n$ both sides) at:
$$(LHS) \quad \sum_n a_n x^n = A(x)$$
$$(RHS) \quad \sum_n n^2 x^n $$$$= \sum_n (x^2 \frac{d^2}{dx^2} + x \frac{d}{dx}) x^n $$$$= (x^2 \frac{d^2}{dx^2} + x \frac{d}{dx}) \sum_n x^n $$$$= (x^2 \frac{d^2}{dx^2} + x \frac{d}{dx}) \frac{1}{1-x}$$
The solution in the book (Page 197) states that I should arrive at $(x D)^2 \frac{1}{1-x}$ which I assume means $(x^2 \frac{d^2}{dx^2})\frac{1}{1-x}$. And it appears I have this $x \frac{d}{dx}$ term, which I shouldn't.
However, $(x^2 \frac{d^2}{dx^2}) x^n \neq n^2 x^n$, actually $(x^2 \frac{d^2}{dx^2}) x^n = (n^2 - n) x^n$, which is exactly why I added the term $x \frac{d}{dx}$ in the first place to remove the $- n x^n$ that shouldn't be there.
Anyway, I'd appreciate any help that you could give me regarding why my solution does not match the book solution.
|
$\left(x^2 \frac{d^2}{dx^2} + x \frac{d}{dx}\right)\left(\frac{1}{1-x}\right)$
gives
$2x^2\left(\frac{1}{1-x}\right)^3+x\left(\frac{1}{1-x}\right)^2$
$(xD)^2$ differentiates in two stages.
$(xD)(xD)\left(\frac{1}{1-x}\right)$
$=(xD)x\left(\frac{1}{1-x}\right)^2$
$=x\left(2x\left(\frac{1}{1-x}\right)^3+\left(\frac{1}{1-x}\right)^2\right)$
so they are equal. ($D=\frac{d}{dx}$, so $xD(xD)=x(xD'+D)$, as we started with).
|
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Evaluate $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$
Evaluate $$\large{\int} \small{\sqrt{\frac{1+x^2}{x^2-x^4}} \space {\large{dx}}}$$
Note that this is a Q&A post and if you have another way of solving this problem, please do present your solution.
|
$$\large{\int} \normalsize{\dfrac{1}{x}\cdot\sqrt{\dfrac{1+x^2}{1-x^2}}} \space \mathrm{dx} \quad \xrightarrow{\large{x \space = \space \tan{u}}} \quad \large{\int} \normalsize{\dfrac{\cos{u}}{\sin{u}}\cdot\sqrt{\dfrac{1+\tan^2{u}}{1-\tan^2{u}}}} \cdot \sec^2{u} \space \mathrm{du} \\ \quad \\$$
$$\require{\cancel} = \large{\int} \normalsize{\dfrac{\cancel{\cos^2{u}}}{\sin{u}}\cdot \dfrac{1}{\cos^\cancel{3}{u}\sqrt{2\cos^2{u} - 1}}} \space \mathrm{du}$$
$$\require{\cancel} = \dfrac{1}{2}\large{\int} \normalsize{\dfrac{1}{\sin{2u} \sqrt{\cos{2u}}}} \space \mathrm{du}$$
$$\require{\cancel} = \dfrac{1}{2}\large{\int} \normalsize{\dfrac{\sin{2u}}{\left(1 - \cos^2{2u}\right) \sqrt{\cos{2u}}}} \space \mathrm{du} \quad \xrightarrow{\large{t \space = \space \cos{2u}}} \quad \dfrac{1}{2}\large{\int} \normalsize{\dfrac{1}{\sqrt{t}\left(1 + t\right)\left(1 - t\right)}} \space \mathrm{dt}$$
$$= -\dfrac{1}{4}\large{\int} \normalsize{\dfrac{1}{1 + \left(\sqrt{t}\right)^2}} \space \mathrm{d\left[\sqrt{t}\space\right]} + \dfrac{1}{8}\large{\int} \normalsize{\dfrac{1}{\sqrt{t} + 1}} - \dfrac{1}{\sqrt{t} -1} \space \mathrm{d\left[\sqrt{t}\space\right]}$$
$$\bbox[5px, border: 2px solid black]{=-\dfrac{1}{4}\tan^{-1}{\sqrt{t}} \space + \space \dfrac{1}{8}\ln{\left|\dfrac{1+\sqrt{t}}{\sqrt{t} - 1}\right|} + C} \quad \quad (1)$$
$\newcommand{\answer}[1]{-\dfrac{1}{4}\tan^{-1}{\sqrt{#1}} \space + \space \dfrac{1}{8}\ln{\left|\dfrac{1 + \sqrt{#1}}{\sqrt{#1} - 1}\right|} + C}$
$x = \tan{u} \longleftrightarrow u = \tan^{-1}{x} \\ \therefore \cos{2u} = \dfrac{1 - x^2}{1 + x^2}$
$$\require{AMScd}\begin{CD}(1) = {-\dfrac{1}{4}\tan^{-1}{\sqrt{\cos{2u}}} \space + \space \dfrac{1}{8}\ln{\left|\dfrac{1+\sqrt{\cos{2u}}}{\sqrt{\cos{2u}} - 1}\right|} + C} \\ @VV \large{\cos{2u} \space = \space \frac{1 - x^2}{1 + x^2}} V \\ \end{CD}$$
$$\bbox[5px, border: 2px solid black]{\answer{\dfrac{1 - x^2}{1 + x^2}}}$$
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|
Given $a + b + c = 20$, find $\max(ab + ac + bc)$ Given that $a + b + c = 20$, what's the maximum possible value for $ab +ac + bc$?
$$\left(a, b, c \in \mathbb{N}\right)$$
I tried following this post, Evaluating max(ab+bc+ac), which lead to:
$ab + bc + ac = 200 - (b - 10)^2 - (c - 10)^2 - bc$, but I couldn't get anywhere afterwards.
Manually trying different numbers lead me to $a = b = 7, c = 6$ with a max result of $133$, but I can't prove it. I assumed for the sum to be the maximum, $a, b, c$ have to be as big as possible, therefore they should be as close to $\frac{20}{3}$ as possible. But is that really the case? If so how can it be proven?
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Here is a Solution :
$a+b+c=20$
$(a+b+c)^2=400$
$a^2+b^2+c^2+2(ab+bc+ca)=400$
$X=(ab+bc+ca)=200-(a^2+b^2+c^2)/2$
We want to maximize X which occurs when $Y=(a^2+b^2+c^2)$ is minimized.
When $a<b$ , we can increase $a$ by $\delta$ while decreasing $b$ by $\delta$ (keeping total sum Constant) , which decreases $Y$.
When $a>b$ , we can decrease $a$ by $\delta$ while increasing $b$ by $\delta$ (keeping total sum Constant) , which decreases $Y$.
Likewise , $b<c$ , $b>c$ , $c<a$ , $c>a$ can all be adjusted to "Equality" to minimize $Y$.
We can not adjust further when we have attained "Equality" for the 3 variables.
In other words , $Y$ is minimum when $a=b=c=20/3$
In that Case , $X=3(20/3)(20/3)=400/3=133.333$ is the maximum , when $a,b,c$ are real numbers. When these are Integers, this value is the upper bound.
We can take nearest Integers $(a,b,c)=(7,7,6)$ to get the Integral maximum $7\times7+7\times6+6\times7=133$.
|
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|
Proof that $3 \mid 10^{n+2} - 2*10^n + 7, \forall n \in \mathbb{Z}^+$. This is what I have so far.
Proof by Induction. Let $n \in \mathbb{Z}^+$ Let $P(n)$ be the statement that $10^{n+2} - 2*10^n + 7$ is divisible by 3.
($\textit{Base Case}$): Let $n = 1$.
$$
10^{1+2} - 2*10^1 + 7 = 1000 - 20 + 7 = 987
$$
$3 \mid 987$ there for $P(1)$ is true.
($\textit{Inductive Step}$): Let $k \in \mathbb{Z}^+$. Suppose $P(k)$ is true. Now we must show that $P(k+1)$ is true.
$$
10^{(k+1) + 2} - 2*10^{k+1} + 7
$$
$$
\Rightarrow 10^{(k+2)+1} - 2*10^{k+1} + 7
$$
$$
\Rightarrow 10^{k+2}(10) - 2*10^{k}(10) - 7
$$
$$
\Rightarrow 10(10^{k+2} - 2*10^{k}) + 7
$$
I don't know how to proceed. I've tried other methods of manipulating the equation and nothing seems to work.
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With the inductive step note that we have:
\begin{align*}
& 10^{k+2} -2\cdot10^{k} + 7 = 3m\quad\text{for some }m\in \mathbb{Z}\\
\implies & 10^{k+2} -2\cdot10^{k} = 3m-7 \qquad(*)
\end{align*}
Now for the case P(k+1):
\begin{align*}
10^{(k+1)+2} -2\cdot10^{k+1} + 7 & = 10\bigg( 10^{k+2} -2\cdot 10^{k}\bigg) + 7\\
& = 10\bigg(3m-7\bigg) + 7 \qquad \text{using }(*)\\
& = 3(10m-21).
\end{align*}
|
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|
Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$
Does it converge? If so, what is its sum?
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In this answer, I used only Bernoulli's inequality to show that
$$
\left(\frac{2n+1}{n+1}\right)^\frac{n}{n+1}
\le\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}
\le\frac{2n+1}{n+1}\tag{1}
$$
The squeeze theorem and $(1)$, show that
$$
\exp\left[\lim\limits_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)\right]=2\tag{2}
$$
That is,
$$
\begin{align}
\lim_{n\to\infty}\left(1-\frac12+\frac13-\frac14+\dots-\frac1{2n}\right)
&=\lim_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)\\[6pt]
&=\log(2)\tag{3}
\end{align}
$$
|
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|
Is there a general formula for solving Quartic (Degree $4$) equations? There is a general formula for solving quadratic equations, namely the Quadratic Formula, or the Sridharacharya Formula:
$$x = \frac{ -b \pm \sqrt{ b^2 - 4ac } }{ 2a } $$
For cubic equations of the form $ax^3+bx^2+cx+d=0$, there is a set of three equations, one for each root.
Is there a general formula for solving equations of the following form [Quartic Equations]?
$$ ax^4 + bx^3 + cx^2 + dx + e = 0 $$
How about for higher degrees? If not, why not?
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Contrary to common opinion, the resolution of the quartic with real coeficients is not so difficult.
I take for granted that depletion and reduction of the equation by a linear change of variable
$$\alpha x^4+\beta x^3+\gamma x^2+\delta x+\epsilon=0\to x^4+px^2+qx+r=0$$ are known.
Then we factor this depleted quadrinomial as
$$(x^2+ax+b)(x^2-ax+c)=x^4+(-a^2+b+c)x^2+a(c-b)x+bc$$ and identify the coefficients,
$$-a^2+b+c=p,\\a(c-b)=q,\\bc=r.$$
We can express $c\pm b$ in terms of $a$, solve for $b,c$ then form the product $4a^2bc$:
$$c-b=\frac qa,\\c+b=p+a^2,$$ hence
$$4a^2bc=\left(a(p+a^2)-q\right)\left(a(p+a^2)+q\right)=a^2(p^2+a^2)^2-q^2=4a^2r.$$
This is a cubic equation in $a^2$, and fortunately it always has a positive solution.
Then after resolution, from $a$ we obtain $b$ and $c$, and the roots of the two quadratic trinomials.
|
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How do I count the subsets of a set whose number of elements is divisible by 3? 4? Let $S$ be a set of size $n$. There is an easy way to count the number of subsets with an even number of elements. Algebraically, it comes from the fact that
$\displaystyle \sum_{k=0}^{n} {n \choose k} = (1 + 1)^n$
while
$\displaystyle \sum_{k=0}^{n} (-1)^k {n \choose k} = (1 - 1)^n$.
It follows that
$\displaystyle \sum_{k=0}^{n/2} {n \choose 2k} = 2^{n-1}$.
A direct combinatorial proof is as follows: fix an element $s \in S$. If a given subset has $s$ in it, add it in; otherwise, take it out. This defines a bijection between the number of subsets with an even number of elements and the number of subsets with an odd number of elements.
The analogous formulas for the subsets with a number of elements divisible by $3$ or $4$ are more complicated, and divide into cases depending on the residue of $n \bmod 6$ and $n \bmod 8$, respectively. The algebraic derivations of these formulas are as follows (with $\omega$ a primitive third root of unity): observe that
$\displaystyle \sum_{k=0}^{n} \omega^k {n \choose k} = (1 + \omega)^n = (-\omega^2)^n$
while
$\displaystyle \sum_{k=0}^{n} \omega^{2k} {n \choose k} = (1 + \omega^2)^n = (-\omega)^n$
and that $1 + \omega^k + \omega^{2k} = 0$ if $k$ is not divisible by $3$ and equals $3$ otherwise. (This is a special case of the discrete Fourier transform.) It follows that
$\displaystyle \sum_{k=0}^{n/3} {n \choose 3k} = \frac{2^n + (-\omega)^n + (-\omega)^{2n}}{3}.$
$-\omega$ and $-\omega^2$ are sixth roots of unity, so this formula splits into six cases (or maybe three). Similar observations about fourth roots of unity show that
$\displaystyle \sum_{k=0}^{n/4} {n \choose 4k} = \frac{2^n + (1+i)^n + (1-i)^n}{4}$
where $1+i = \sqrt{2} e^{ \frac{\pi i}{4} }$ is a scalar multiple of an eighth root of unity, so this formula splits into eight cases (or maybe four).
Question: Does anyone know a direct combinatorial proof of these identities?
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I try to add details with simple calculation:
for example, I show $\binom{n}{1}+\binom{n}{3}+\binom{n}{6}+\cdots=\frac{1}{3}[2^n+2\cos(\frac{n\pi}{3})]$
we know $(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\cdots+\binom{n}{n}x^n$, so if in this identity we put $1,\alpha,\alpha^2$ where $\alpha=\cos\frac{2\pi}{3}+i\sin(\frac{2\pi}{3})$respectively, then we get $$2^n=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots$$, $$(1+\alpha)^n=\binom{n}{0}+\binom{n}{1}\alpha+\binom{n}{2}\alpha^2+\binom{n}{3}\alpha^3+\cdots$$,
$$(1+\alpha^2)^n=\binom{n}{0}+\binom{n}{1}\alpha^2+\binom{n}{2}\alpha^4+\binom{n}{3}\alpha^6+\cdots$$
but if $3\nmid n$, then we can easily see that $1+\alpha^k+\alpha^{2k}=3$ hence
$2^n+(1+\alpha)^n+(1+\alpha^2)^n=3\{\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\cdots\}$ but since $1+\alpha+\alpha^{2}=0$ hence
$1+\alpha=-\alpha^2=\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})$ and also $1+\alpha^2=-\alpha=\cos(\frac{\pi}{3})-i\sin(\frac{\pi}{3})$ hence $$2^n+(1+\alpha)^n+(1+\alpha^2)^n=2^n+2\cos(\frac{n\pi}{3})$$, so we get the desired result.
|
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Help in getting the Quadratic Equation I'm starting a chapter on Functions and they had the steps shown to reach the p-q equation.
$$ x_{1,2} = -\frac{p}{2} \pm\sqrt{\left(\frac{p}{2}\right)^2 - q}$$
So I wanted to do the same with the Quadratic Equation. I'm using the base linear equation
$$ax+by+c = 0.$$
The solution I have so far is as follows:
$$x^2 + \frac{b}{a}x + \frac{c}{a}= 0$$
$$x^2 + \frac{b}{a}x = -\frac{c}{a}$$
$$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$$
$$\left(x + \frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}$$
$$\left(x + \frac{b}{2a}\right) = \pm\sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}$$
$$x = -\frac{b}{2a} \pm\sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}$$
My problem comes from trying to solve the insides of the square root:
$$\sqrt{(\frac{b}{2a})^2 - \frac{c}{a}} = \sqrt{\frac{b^2}{4a2} - \frac{c}{a}}$$
$$= \sqrt{\frac{b^2}{4a^2} - \frac{c}{a} \left(\frac{4a}{4a}\right)} = \sqrt{\frac{b^2 - 4ac}{4a^2}}$$
$$= \sqrt{\frac{b^2 - 4ac}{\left(2a\right)^2}}$$
Then:
$$x_{1,2} = \frac{-\left(\frac{b}{2a}\right) \pm\sqrt{b^2 -4ac}}{2a}$$
but there is still the problem of the -(b/2a) outside of the sqrt.
What am I doing wrong? Also, Tex is awesome; is there a better way to do the 1,2 subscripts than _1,_2?
Answer:
Instead of
$$x_{1,2} = \frac{-\left(\frac{b}{2a}\right) \pm\sqrt{b^2 -4ac}}{2a}.$$
The solution goes
$$\frac{-b}{2a}\pm \sqrt{\frac{b^{2}-4ac}{4a^{2}}}=\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$$
$$= \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$
|
You factored 2a out of the square root and put it in the denominator without factoring it out of -(b/2a).
|
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|
Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$ How does one sum the given series: $$ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{2n+5} + \frac{ 1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{1}{2n+7} + \cdots \ \text{ad inf}$$
Given, such a series, how does one go about solving it. Getting an Integral Representation seems tough for me.
I thought of going along these lines, Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$ but couldn't succeed.
|
Consider the function
$$f(a) = \sum_{k=0}^{\infty} \frac{1\cdot3\cdots(2k-1)}{2\cdot4\cdots(2k)}\cdot a^{2n+2k}$$
$$|a| \leq 1$$
What you want is $$\int_{0}^{1} f(a) da$$
The $k^{th}$ coefficient can be written as
$$\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}{\sin^{2k}x}dx$$
(again Wallis's like product as in the other question).
Thus we have
$$f(a) = \frac{2a^{2n}}{\pi} \int_{0}^{\frac{\pi}{2}} \sum_{k=0}^{\infty} {(a\sin x)^{2k}} dx$$
$$ = \frac{2a^{2n}}{\pi} \int_{0}^{\frac{\pi}{2}} {\frac{1}{1-(a\sin x)^2}}dx$$
Now $$\frac{1}{1-(a\sin x)^2} = \frac{1}{\cos^2 x + (\sqrt{1-a^2}\sin x)^2} = \frac{\sec^2 x}{1+(\sqrt{1-a^2}\tan x)^2}$$ which is the derivative of
$$\frac{\arctan(\sqrt{1-a^2}\tan x)}{\sqrt{1-a^2}}$$ whose integral between $0$ and $\frac{\pi}{2}$ is $$\frac{\pi}{2\sqrt{1-a^2}}$$
Thus we get $$f(a) = \frac{a^{2n}}{\sqrt{1-a^2}}$$
(So in fact, we have shown the identity which Qiaochu uses in his answer).
Hence the required sum is $$\int_{0}^{1} \frac{a^{2n}}{\sqrt{1-a^2}}da$$ which can easily be found by the substitution $a = \sin x$ (like in Qiaochu's answer) and which gives us the result to be
$$\int_{0}^{\frac{\pi}{2}} {\sin^{2n}x}dx$$
Now this integral for positive integer $n$ can easily be seen to be what Mariano got. I am unsure of whether that formula holds for an arbitrary real number $n$.
|
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|
Evaluating the nested radical $ \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots}}} $. How does one prove the following limit?
$$
\lim_{n \to \infty}
\sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots \sqrt{1 + (n - 1) \sqrt{1 + n}}}}}
= 3.
$$
|
This is Ramanujan's famous nested radical.
More information can be found here: http://www.isibang.ac.in/~sury/ramanujanday.pdf
See Also: http://mathworld.wolfram.com/NestedRadical.html (number 26).
Apparently, this is how he came up with it (sorry, no reference for this claim).
Start with
$$3 = \sqrt{9} = \sqrt{1 + 8} = \sqrt{1 + 2 \cdot 4}$$
$$ = \sqrt{1 + 2\sqrt{16}} = \sqrt{1 + 2\sqrt{1 + 3 \cdot 5}}$$
$$ = \sqrt{1 + 2\sqrt{1 + 3 \sqrt{25}}} = \sqrt{1 + 2\sqrt{1 + 3 \sqrt{1 + 4 \cdot 6}}}$$
etc.
|
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|
How to solve this non-linear differential equation? I'm trying to solve this non-linear differential equation using substitution $\dfrac{y}{x}$ to $t$. However, I can't solve this equation.
$$
\text{xy$\prime $} = \left(\frac{y}{x}\right)^3+y
$$
How to solve this equation and what's the general solution?
Thanks.
|
Here is my solution. Is this right?
\begin{align}
\text{xy$\prime $} = \left(\frac{y}{x}\right)^3+y \\ \\
\frac{y}{x}=u, \\
y = \text{ux}, \\
\text{y$\prime $} = \text{u$\prime $x} + u \\ \\
x^2\text{u$\prime $} = u^3 \\
\int \frac{1}{u^3} \, du =\int \frac{1}{x^2} \, dx \\
-\frac{1}{2u^2} = -\frac{1}{x} + c \\
u^{2 }= \frac{1}{\frac{2}{x}+c} \\
y^{2 }= \frac{x^2}{\frac{2}{x}+c} \\
y = x\left(\frac{2}{x}+c\right)^{-\frac{1}{2}}
\end{align}
|
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|
Showing $\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!} $ I am stuck while proving this identity I verified it using induction but like the other two (1) (2), I am seeking a more of a general way (algebraic will be much appreciated)
$$\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!} $$
|
Seems easy is $n$ is odd. Multiplying both sides by $n!$ we have
$$ {n \choose 1} + {n \choose 3} + ... {n \choose n} = \frac{1}{2} \left[ {n \choose 0} + {n \choose 1} + ... {n \choose n} \right] = 2^{n-1}$$
|
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|
The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
|
Another variation. We make use of the following identity (proved at the bottom of this note):
$$\sum_{k=1}^n \cot^2 \left( \frac {2k-1}{2n} \frac{\pi}{2} \right) = 2n^2 – n. \quad (1)$$
Now $1/\theta > \cot \theta > 1/\theta - \theta/3 > 0$ for $0< \theta< \pi/2 < \sqrt{3}$ and so
$$ 1/\theta^2 – 2/3 < \cot^2 \theta < 1/\theta^2. \quad (2)$$
With $\theta_k = (2k-1)\pi/4n,$ summing the inequalities $(2)$ from $k=1$ to $n$ we obtain
$$2n^2 – n < \sum_{k=1}^n \left( \frac{2n}{2k-1}\frac{2}{\pi} \right)^2 < 2n^2 – n + 2n/3.$$
Hence
$$\frac{\pi^2}{16}\frac{2n^2-n}{n^2} < \sum_{k=1}^n \frac{1}{(2k-1)^2} <
\frac{\pi^2}{16}\frac{2n^2-n/3}{n^2}.$$
Taking the limit as $n \rightarrow \infty$ we obtain
$$ \sum_{k=1}^\infty \frac{1}{(2k-1)^2} = \frac{\pi^2}{8},$$
from which the result for $\sum_{k=1}^\infty 1/k^2$ follows easily.
To prove $(1)$ we note that
$$ \cos 2n\theta = \text{Re}(\cos\theta + i \sin\theta)^{2n} =
\sum_{k=0}^n (-1)^k {2n \choose 2k}\cos^{2n-2k}\theta\sin^{2k}\theta.$$
Therefore
$$\frac{\cos 2n\theta}{\sin^{2n}\theta} = \sum_{k=0}^n (-1)^k {2n \choose 2k}\cot^{2n-2k}\theta.$$
And so setting $x = \cot^2\theta$ we note that
$$f(x) = \sum_{k=0}^n (-1)^k {2n \choose 2k}x^{n-k}$$
has roots $x_j = \cot^2 (2j-1)\pi/4n,$ for $j=1,2,\ldots,n,$ from which $(1)$ follows since
${2n \choose 2n-2} = 2n^2-n.$
|
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|
Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$ Using $\text{n}^{\text{th}}$ root of unity
$$\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$$
Prove that
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$
|
Consider $z^n=1$, each root is
$$\xi_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n} = e^{i\frac{2k\pi}{n}}, k=0,1,2,...,n-1 $$
So, we have
$$ z^n -1 = \prod_{k=0}^{n-1}(z-\xi_k)$$
$$\Longrightarrow (z-1)(z^{n-1}+...+z^2+z+1) = (z-\xi_0)\prod_{k=1}^{n-1}(z-\xi_k)$$
$$\Longrightarrow (z-1)(z^{n-1}+...+z^2+z+1) = (z-1)\prod_{k=1}^{n-1}(z-\xi_k)$$
$$\Longrightarrow z^{n-1}+...+z^2+z+1 = \prod_{k=1}^{n-1}(z-\xi_k)$$
By substituting z=1, $$\Longrightarrow n = \prod_{k=1}^{n-1}(1-\xi_k) $$
Next, take the modulus on both sides,
$$ |n| = n = |\prod_{k=1}^{n-1}(1-\xi_k)| = \prod_{k=1}^{n-1}|(1-\xi_k)|$$
$$ 1 - \xi_k = 1-(\cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}) = 2\sin\frac{k\pi}{n}(\sin\frac{k\pi}{n} -i\cos\frac{k\pi}{n})$$
$$ |1 - \xi_k| = 2\sin\frac{k\pi}{n} $$
So,
$$ n = 2^{n-1}\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}$$
$$\prod_{k=1}^{n-1}\sin\frac{k\pi}{n} = \frac{n}{2^{n-1}} $$
|
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|
Eigenvalues of companion matrix of $4x^3 - 3x^2 + 9x - 1$ I want to find all the roots of a polynomial and decided to compute the eigenvalues of its companion matrix.
How do I do that?
For example, if I have this polynomial: $4x^3 - 3x^2 + 9x - 1$, I compute the companion matrix:
$$\begin{bmatrix} 0&0&\frac{3}{4} \\ 1&0&-\frac{9}{4} \\ 0&1&\frac{1}{4} \end{bmatrix}$$
Now how can I find the eigenvalues of this matrix? Thanks in advance.
|
Hey There, so if I am assuming correctly for your case, you want to find eigenvalues for this matrix, which is essentially solving for your roots of the characteristic polynomial of the matrix after doing the determinant operation on it. So to go off from Robert idea, we want to use the equation,
det(A$-\lambda$ I) = $0$ $~~~$(following from this we can plug in the coefficient matrix given).
det(A$-\lambda$ I) =
$\left[\begin{array}{ccc}
0-\lambda & 0 & \dfrac{3}{4} \\
1 & 0-\lambda & -\dfrac{9}{4} \\
0 & 1 & \dfrac{1}{4}-\lambda
\end{array} \right] = 0$, where A is your coefficient matrix and I is the identity matrix.
I = $\left[\begin{array}{ccr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right]$
From here we can now find out the eigenvalues of the matrix A as follows:
$\underline {\text{Evaluation of the determinant expanding it by the minors of column 1:}}$
= $~-\lambda
\left[\begin{array}{cc}
-\lambda & -\dfrac{9}{4} \\
1 & \dfrac{1}{4}-\lambda
\end{array} \right]
-1
\left[\begin{array}{cc}
0 & \dfrac{3}{4} \\
1 & \dfrac{1}{4}-\lambda
\end{array} \right]
+ 0
\left[\begin{array}{rr}
0 & \dfrac{3}{4} \\
-\lambda & -\dfrac{9}{4}
\end{array} \right]
$
$\Rightarrow ~ -\lambda
\left[\begin{array}{c}
\lambda^{2} -\dfrac{1}{4}\lambda + \dfrac{9}{4} \\
\end{array} \right]
-1
\left[\begin{array}{cc}
0 -\dfrac{3}{4} \\
\end{array} \right]
+ ~0
\left[\begin{array}{rr}
0 + \dfrac{3}{4}\lambda \\
\end{array} \right]
$
$\Rightarrow ~$ $-\lambda^3+\dfrac{1}{4}\lambda^2-\dfrac{9}{4}\lambda+\dfrac{3}{4}$, $~$ Hence our characteristic polynomial is now obtained.
$$P(\lambda)=-\lambda^3+\dfrac{1}{4}\lambda^2-\dfrac{9}{4}\lambda+\dfrac{3}{4}$$
If you need assistance on how to find the characteristic polynomial by evaluating the determinant, here is a reference: Computing Determinants
\
After solving this polynomial for its roots (eigenvalues) we get the following:
{$\lambda = (0.329,0.000) ~~ \lambda = (-0.040,-1.508) ~~ \lambda = (-0.040,1.508)$}
I believe all the roots except for $\lambda = 0.329$ are complex conjugate roots. Can someone else please verify that those are all of the roots to this polynomial and that the ones I provided are correct, Thanks. I hope this helps out if this explanation is what you were looking for.
|
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|
Expressions of $\sin \frac{A}{2}$ and $\cos \frac{A}{2}$ in terms of $\sin A$ I am trying to understand the interpretation of $\sin \frac{A}{2}$ and $\cos \frac{A}{2}$ in terms of $\sin A$ from my book, here is how it is given :
We have $ \bigl( \sin \frac{A}{2} + \cos \frac{A}{2} \bigr)^{2} = 1 + \sin A $ and $ \bigl( \sin \frac{A}{2} - \cos \frac{A}{2} \bigr)^{2} = 1 - \sin A $
By adding and subtracting we have $ 2 \cdot \sin \frac{A}{2} = \pm \sqrt{ 1 + \sin A } \pm \sqrt{ 1 - \sin A } $ ---- (1) and $ 2 \cdot \cos \frac{A}{2} = \mp \sqrt{ 1 + \sin A } \mp \sqrt{ 1 - \sin A } $ ---(2)
I have understood upto this far well,
Now they have broke the them into quadrants :
In 1st quadrant :
$$ 2 \cdot \sin \frac{A}{2} = \sqrt{ 1 + \sin A } - \sqrt{ 1 - \sin A } $$
$$ 2 \cdot \cos \frac{A}{2} = \sqrt{ 1 + \sin A } + \sqrt{ 1 - \sin A } $$
In 2nd quadrant :
$$ 2 \cdot \sin \frac{A}{2} = \sqrt{ 1 + \sin A } + \sqrt{ 1 - \sin A } $$
$$ 2 \cdot \cos \frac{A}{2} = \sqrt{ 1 + \sin A } - \sqrt{ 1 - \sin A } $$
In 3rd quadrant :
$$ 2 \cdot \sin \frac{A}{2} = \sqrt{ 1 + \sin A } - \sqrt{ 1 - \sin A } $$
$$ 2 \cdot \cos \frac{A}{2} = \sqrt{ 1 + \sin A } + \sqrt{ 1 - \sin A } $$
In 4th quadrant :
$$ 2 \cdot \sin \frac{A}{2} = - \sqrt{ 1 + \sin A } - \sqrt{ 1 - \sin A } $$
$$ 2 \cdot \cos \frac{A}{2} = - \sqrt{ 1 + \sin A } + \sqrt{ 1 - \sin A } $$
Now, In knew the ALL-SINE-TAN-COSINE rule but still I am not able to figure out how the respective signs are computed in these (above) cases.
|
We have that
$$\cos A=\cos \left(2\cdot\frac{A}{2}\right) = \cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}}$$
and
$$1 = \cos^2{\frac{A}{2}}+\sin^2{\frac{A}{2}}.$$
Therefore
\begin{align*}
1+\cos A & = \cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}} + \cos^2{\frac{A}{2}}+\sin^2{\frac{A}{2}} \\
& = 2\cos^2{\frac{A}{2}} \\
\Rightarrow \frac{1+\cos A}{2} & = \cos^2{\frac{A}{2}} \\
\Rightarrow \cos\frac{A}{2} & = \sqrt{\frac{1+\cos A}{2}}.
\end{align*}
and
\begin{align*}
1-\cos A & = \cos^2{\frac{A}{2}}+\sin^2{\frac{A}{2}} - \cos^2{\frac{A}{2}}+\sin^2{\frac{A}{2}} \\
& = 2\sin^2{\frac{A}{2}} \\
\Rightarrow \frac{1-\cos A}{2} & = \sin^2{\frac{A}{2}} \\
\Rightarrow \sin\frac{A}{2} & = \sqrt{\frac{1-\cos A}{2}}.
\end{align*}
|
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|
Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However, doing the following gets a completely different answer:
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{\sin x}{\sin^2(x)\cos x} dx\\
&=&\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.
\end{eqnarray*}
let $u=\cos x, du=-\sin x dx$; then
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{-1}{(1-u^2)u} du\\
&=&\int \frac{-1}{(1+u)(1-u)u}du\\
&=&\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\
&=&-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C
\end{eqnarray*}
I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?
|
Tangent half-angle substitution
$$\displaystyle \int \frac{1}{\sin x\cos x} dx=\displaystyle \int \frac{(1+t^2)}{t(1-t^2)} dt=\displaystyle \int \frac{1}{t} dt-\displaystyle \int \frac{1}{1-t} dt-\displaystyle \int \frac{1}{1+t} dt$$
|
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|
Calculating the integral $\int_0^\infty \frac{\cos x}{1+x^2}\, \mathrm{d}x$ without using complex analysis Suppose that we do not know anything about the complex analysis (numbers). In this case, how to calculate the following integral in closed form?
$$\int_0^\infty\frac{\cos x}{1+x^2}\,\mathrm{d}x$$
|
The method of brackets, which is an extension of Ramanujan's master theorem, can be used to evaluate this classic integral.
The term bracket refers to the assignment of the symbol $\langle a \rangle$ to the divergent integral $\int_{0}^{\infty} x^{a -1} \, \mathrm{d}x$.
You can read about the method in the following papers:
Definite integrals by the method of brackets. Part 1
The method of brackets. Part 2: Examples and applications
Evaluation of entries in Gradshteyn and Ryzhik employing the
method of brackets
On the Method of Brackets: Rules, Examples, Interpretations and Modifications
The hypergeometric representation of the cosine function is $$ \, _0F_{1} \left(; \frac{1}{2}; - \frac{x^{2}}{4} \right) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \, \frac{\Gamma \left( \frac{1}{2} \right)}{\Gamma \left(n+\frac{1}{2} \right)} \left(\frac{x^{2}}{4} \right)^{n} = \sum_{n=0}^{\infty} \phi_{n} \, \frac{\Gamma \left( \frac{1}{2} \right)}{\Gamma \left(n+\frac{1}{2} \right)} \left(\frac{x^{2}}{4} \right)^{n}. $$
(This representation can be derived from the Maclaurin series of $\cos (x)$ by using the duplication formula for the gamma function).
And according to Rule 3.1 on page 8 of the first paper, the function $ \frac{1}{1+x^{2}}$ is assigned the bracket series $$\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \phi_{k,m} \, x^{2m} \langle k+m+1 \rangle. $$
Therefore, the integral $ \int_{0}^{\infty} \frac{\cos x}{1+x^{2}} \, \mathrm{d}x$ corresponds with the bracket series $$ \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \phi_{k,m,n} \, \frac{\Gamma \left(\frac{1}{2} \right)}{\Gamma \left(n+\frac{1}{2}\right)} \frac{1}{4^{n}} \langle k+m+1\rangle \langle 2m+2n+1 \rangle. \tag{1}$$
To evaluate $(1)$, first let $k$ be a free parameter.
The brackets then vanish if $m=-k-1$ and $n= k + \frac{1}{2}$.
So according to Rule 3.3 in the first paper, the contribution to the integral is
$$\begin{align}\frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} \frac{\Gamma \left(\frac{1}{2} \right)}{\Gamma (k+1 )} \, \frac{\Gamma (k+1) \Gamma \left(-k-\frac{1}{2} \right)}{4^{k+ 1/2}} &= \frac{\sqrt{\pi}}{4} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} \frac{1}{4^{k}} \frac{\pi (-1)^{k-1} }{\Gamma \left(k + \frac{3}{2} \right)} \\ &= - \frac{\pi \sqrt{\pi} }{4} \sum_{k=0}^{\infty} \frac{1}{k!} \frac{1}{4^{k}} \frac{2^{2(k+1)-1} \Gamma(k+1)}{ \sqrt{\pi} \, \Gamma(2k+2)} \\ &= -\frac{\pi}{2} \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} \\ &= -\frac{\pi}{2} \, \sinh (1). \end{align}$$
(I used the reflection formula for the gamma function on the first line, and then I used the duplication formula for the gamma function on the second line.)
Now let $m$ be a free parameter.
The brackets then vanish if $k = -m-1$ and $n=-m -\frac{1}{2}$.
So according to Rule 3.3, the contribution to the integral is $$ \frac{1}{2} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!} \frac{\Gamma \left(\frac{1}{2} \right)}{{\color{red}{\Gamma (-m)}}} \frac{\Gamma(m+1) \Gamma \left(m+ \frac{1}{2} \right)}{4^{-m-1/2}} =0.$$
Finally let $n$ be a free parameter.
The brackets then vanish if $k = - n - \frac{1}{2}$and $m= n - \frac{1}{2}$.
So according to Rule 3.3, the contribution to the integral is
$$\begin{align}\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \frac{\Gamma \left(\frac{1}{2} \right)}{\Gamma (n + \frac{1}{2} )} \, \frac{\Gamma \left(n + \frac{1}{2} \right) \Gamma \left(\frac{1}{2}-n \right) }{4^{n}} &= \frac{\sqrt{\pi}}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \frac{1}{4^{n}} \frac{\pi (-1)^{n}}{\Gamma \left(n+ \frac{1}{2} \right)} \\ &= \frac{\pi \sqrt{\pi}}{2} \sum_{n=0}^{\infty} \frac{1}{n!} \frac{1}{4^{n}} \frac{2^{2(n+1/2)-1} \Gamma(n+1)}{\sqrt{\pi} \, \Gamma(2n+1)} \\ &= \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{1}{(2n)!} \\ &= \frac{\pi}{2} \, \cosh (1). \end{align}$$
Rule 3.4 in the first paper says that we should add these three contributions to get the value of the integral.
Therefore, $$\int_{0}^{\infty} \frac{\cos(x)}{1+x^{2}} \, \mathrm{d}x = - \frac{\pi}{2} \sinh(1) + 0 + \frac{\pi}{2} \cosh(1) = \frac{\pi}{2e}.$$
|
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|
$x^y = y^x$ for integers $x$ and $y$ We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\neq y$) which satisfies the equality $x^y = y^x$?
|
For every integer $n$, $x = y = n$ is a solution. So assume $x \neq y$.
Suppose $n^m = m^n$. Then $n^{1/n} = m^{1/m}$. Now the function $x \mapsto x^{1/x}$ reaches its maximum at $e$, and is otherwise monotone. Thus (assuming $n < m$) we must have $n < e$, i.e. $n = 1$ or $n = 2$.
If $n = 1$ then $n^m = 1$ and so $m = 1$, so it's a trivial solution.
If $n = 2$ then $n^m$ is a power of $2$, and so (since $m > 0$) $m$ must also be a power of $2$, say $m = 2^k$. Then $n^m = 2^{2^k}$ and $m^n = 2^{2k}$, so that $2^k = 2k$ or $2^{k-1} = k$. Now $2^{3-1} > 3$, and so an easy induction shows that $k \leq 2$. If $k = 1$ then $n = m$, and $k = 2$ corresponds to $2^4 = 4^2$.
EDIT: Up till now we considered $n,m>0$. We now go over all other cases. The solution $n = m = 0$ is trivial (whatever value we give to $0^0$).
If $n=0$ and $m \neq 0$ then $n^m = 0$ whereas $m^n = 1$, so this is not a solution.
If $n > 0$ and $m < 0$ then $0 < n^m \leq 1$ whereas $|m^n| \geq 1$. Hence necessarily $n^m = 1$ so that $n = 1$. It follows that $m^1 = 1^m = 1$. In particular, there's no solution with opposite signs.
If $n,m < 0$ then $(-1)^m (-n)^m = n^m = m^n = (-1)^n (-m)^n$, so that $n,m$ must have the same parity. Taking inverses, we get $(-n)^{-m} = (-m)^{-n}$, so that $-n,-m$ is a solution for positive integers. The only non-trivial positive solution $2,4$ yields the only non-trivial negative solution $-2,-4$.
|
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|
How to calculate the expected number of distinct items when drawing pairs? Suppose I have a set $\mathcal{S}$ of $N$ distinct items. Now consider the set $\mathcal{P}$ of all possible pairs that I can draw from $S$. Naturally, $|\mathcal{P}| = \binom{N}{2}$. Now when I draw $k$ items (pairs) from $\mathcal{P}$ with a uniform distribution, what is the expected number of distinct items from $S$ in those $k$ pairs?
P.S.: I also asked this question over at stats, but got no answers so far, so I am trying here. Thanks for your time!
Edit I pick the pairs without replacement.
|
For choosing without replacement, here is an exact answer. Assuming $n \geq 2$, so that there is at least one pair, and $1 \leq k \leq \binom{n}{2}$, so that you're choosing at least one pair but not more than the total number of pairs, the expected value is
$$n - \left(\frac{n^2 - 3n - 2k + 4}{n-1}\right) \frac{\binom{\binom{n}{2} - n + 1}{k-1}}{\binom{\binom{n}{2} - 1}{k-1}}.$$
We can assume that we are choosing pairs in order. Let $X_k$ be the number of distinct items from $S$ through $k$ pairs. Let $Y_i$ be the number of items in the $i$th pair that did not appear in any of the previous pairs. So $X_k = \sum_{i=1}^k Y_i$.
Now, $Y_i$ is either 0, 1, or 2. Since there are $\binom{n}{2} - n + 1$ pairs that do not contain a given item and $\binom{n}{2} - 2n + 3$ pairs that do not contain either of two given items, we have
$$P(Y_i = 1) = \frac{\binom{\binom{n}{2} - n + 1}{i-1} + \binom{\binom{n}{2} - n + 1}{i-1} - 2 \binom{\binom{n}{2} - 2n + 3}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}}$$
and
$$P(Y_i = 2) = \frac{\binom{\binom{n}{2} - 2n + 3}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}}.$$
Thus
$$E[Y_i] = 2\frac{\binom{\binom{n}{2} - n + 1}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}}.$$
It can be proved by induction that
$$\sum_{i=0}^k \frac{\binom{M}{i}}{\binom{N}{i}} = \frac{(N+1)\binom{N}{k} - (M-k)\binom{M}{k}}{(N+1-M)\binom{N}{k}}.$$
Thus
$$E[X_k] = \sum_{i=1}^k E[Y_i] = 2\sum_{i=1}^k \frac{\binom{\binom{n}{2} - n + 1}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}} $$
$$= 2\frac{(\frac{n(n-1)}{2}-1+1)\binom{\binom{n}{2}}{k-1} - (\frac{n(n-1)}{2} - n + 1 - k + 1)\binom{\binom{n}{2} - n + 1}{k-1}}{(\binom{n}{2} - 1+1-\binom{n}{2} + n - 1)\binom{\binom{n}{2} - 1}{k-1}}$$
$$= \frac{n(n-1)\binom{\binom{n}{2}}{k-1} - (n^2 - 3n - 2k + 4)\binom{\binom{n}{2} - n + 1}{k-1}}{(n - 1)\binom{\binom{n}{2} - 1}{k-1}}$$
$$= n -\frac{(n^2 - 3n - 2k + 4)\binom{\binom{n}{2} - n + 1}{k-1}}{(n - 1)\binom{\binom{n}{2} - 1}{k-1}}.$$
|
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|
Another limit task, I've multiplied by the conjugate, now what? The task:
$\lim_{x\to\infty} \sqrt{x^2+1} -x $
I've multiplied with the conjugate expression ($\sqrt{x^2+1} +x$), then I get this
$\lim_{x\to\infty} \frac{1}{\sqrt{x^2+1} +x} $
Is this correct so far? And what would be the next step?
|
We have $$\sqrt{1 + \epsilon} = 1 + \frac{\epsilon}{2} + O(\epsilon^2)$$ using the Taylor expansion. So $$\sqrt{x^2+1} = x\sqrt{1+x^{-2}} = x\left(1 + \frac{1}{2x^2} + O(x^{-4})\right) = x + \frac{1}{2x} + O(x^{-3}).$$ Hence $$\sqrt{x^2+1} - x = \frac{1}{2x} + O(x^{-3}) \longrightarrow 0.$$
|
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|
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$
Possible Duplicate:
On the sequence $x_{n+1} = \sqrt{c+x_n}$
Where does this sequence converge?
$\sqrt{7},\sqrt{7+\sqrt{7}},\sqrt{7+\sqrt{7+\sqrt{7}}}$,...
|
For a proof of convergence,
Define the sequence as
$\displaystyle x_{0} = 0$
$\displaystyle x_{n+1} =\sqrt{7 + x_n}$
Note that $\displaystyle x_n \geq 0 \ \ \forall n$.
Notice that $\displaystyle x^2 - x - 7 = (x-a)(x-b)$ where $\displaystyle a \lt 0$ and $\displaystyle b \gt 0$.
We claim the following:
i) $\displaystyle x_n \lt b \Longrightarrow x_{n+1} \lt b$
ii) $\displaystyle x_n \lt b \Longrightarrow x_{n+1} \gt x_n$
For a proof of i)
We have that
$\displaystyle x_n \lt b = b^2 - 7$ and so $x_n +7 \lt b^2$ and thus by taking square roots $x_{n+1} \lt b$
For a proof of ii)
We have that
$\displaystyle (x_{n+1})^2 - (x_n)^2 = -(x^2_n - x_n -7) = -(x_n-a)(x_n-b) \gt 0$ if $x_n \lt b$.
Thus $\displaystyle \{x_{n}\}$ is monotonically increasing and bounded above and so is convergent.
By setting $L = \sqrt{7+L}$, we can easily see that the limit is $\displaystyle b = \dfrac{1 + \sqrt{29}}{2}$
In fact, we can show that the convergence is linear.
$\displaystyle \dfrac{b-x_{n+1}}{b-x_n} = \dfrac{b^2 - (7+x_n)}{(b+\sqrt{7+x_n})(b-x_n)} = \dfrac{1}{b + x_{n+1}}$
Thus $\displaystyle \lim_{n\to \infty} \dfrac{b-x_{n+1}}{b-x_n} = \dfrac{1}{2b}$.
We can also show something a bit stronger:
Let $\displaystyle t_n = b - x_n$.
The we have shown above that $\displaystyle t_n \gt 0$ and $\displaystyle t_n \lt b^2$
We have that
$\displaystyle b - t_{n+1} = \sqrt{7 + b - t_n} = \sqrt{b^2 - t_n}$
Dividing by $\displaystyle b$ throughout we get
$\displaystyle 1 - \dfrac{t_{n+1}}{b} = \sqrt{1 - \dfrac{t_n}{b^2}}$
Using $\displaystyle 1 - \dfrac{x}{2} \gt \sqrt{1-x} \gt 1 - x \ \ 0 \lt x \lt 1$ we have that
$\displaystyle 1 - \dfrac{t_n}{2b^2} \geq 1 - \dfrac{t_{n+1}}{b} \geq 1 - \dfrac{t_n}{b^2}$
And so
$\displaystyle \dfrac{t_n}{2b} \leq t_{n+1} \leq \dfrac{t_n}{b}$
This gives us that $\displaystyle b - \dfrac{b}{b^n} \leq x_n \leq b - \dfrac{b}{(2b)^n}$
|
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Find the positive real number(s) $c$ such that the graphs of $y=x^c$ and $x=y^c$ intersect (somewhere) at an angle $\frac{\pi}{4}$ Find the positive real number(s) $c$ such that the graphs of $y=x^c$ and $x=y^c$ intersect (somewhere) at an angle $\frac{\pi}{4}$.
A problem from Mark Krusemeyer. Should one be using the symmetry of the graphs along $x=y$ due the functions being inverses of one another? Just a thought.
|
It is possible to use the symmetry as a slight shortcut to Simon's answer: Given the graph and conditions as laid out there, the relevant point of intersection must be at $x=y=1$. Because of the symmetry, the angles formed by the tangent lines to the curves at that point and the line $y=x$ must be $\frac{\pi}{8}$, so the slope of the tangent line to the blue curve (lower for $0< x<1$) at $x=1$ must be $\tan\frac{3\pi}{8}=1+\sqrt{2}$. Now, either $cx^{c-1}|_{x=1}=c=1+\sqrt{2}$ or $c^{-1}x^{1/c-1}|_{x=1}=\frac{1}{c}=1+\sqrt{2}$, so $c=1+\sqrt{2}$ or $c=\frac{1}{1+\sqrt{2}}=-1+\sqrt{2}$.
|
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Can my proof be simplified and is it valid? $n(n+2) = (n+1)^2 - 1$ for all integers I've decided to learn the basics of proofs and here is my first attempt. Could I improve or simplify my proof in any way? Is my formal language correct? Thanks!
Let $n$ be any integer. $$n(n+2) = (n+1)^2 - 1$$ I will prove this identify by induction.
First, check with $n=1$;
$$1 \times 3 = 2^2 + 1 \equiv
3 = 3$$
Inductive step: Assume that the identity is true for n = k;
$$k(k+2) = (k+1)^2 - 1$$
When $n = k + 1$;
$$(k+1)((k+1)+2) = ((k+1)+1)^2 - 1$$
$$\equiv k+1(k+3) = (k+2)^2 - 1$$
Let $n = k + 1$;
$$n(n+2) = (n+1)^2 - 1$$
□
|
Using $a^2 - b^2 = (a-b)(a+b)$ we have $$(n+1)^2 - 1 = (n+1)^2 - 1^2 = (n+1-1)(n+1+1) = n(n+2).$$
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Help solving another integral $\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$ $$\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$$
Complete the square?
$$\int \frac{1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$$
Not sure what do do next, or if I should try something else?
Big help if you can show as "step-by-step" possible as you can.
Thanks in advance!
@Chandru1:
I'm confused at what happens to the 2 that gets factored out when completing the square (the 2 in the denominator) because it looks like Arturo left it out and instead substituted:
$\int \frac {1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$
$u = x+1$
$du =dx$
which changes the integral to $\int \frac {1}{(2(u^2-4))^\frac{3}{2}}$
Would I be best factoring it out as a 1/2 in front of the integral or leaving it in and then using trig. substitution like this?
$u= \sqrt{2}sec(t)$
$du = \sqrt{2}sec(t)*tan(t) dt$
Which changes it to:
$\int \frac {\sqrt{2}sec(t)*tan(t)}{(2(2sec^2(t)-4))^\frac{3}{2}} \ dt$ = $\int \frac {\sqrt{2}sec(t)*tan(t)}{(2(tan^2(t)))^\frac{3}{2}} \ dt$ = $\int \frac {\sqrt{2}sec(t)*tan(t)}{(2tan^3(t))} \ dt$
My $\sqrt{2}$ factors out like yours, but I have no 1/8 so figure I factor it out as a 1/2 earlier like I thought? But then the constant in front of my integral would be $\frac{\sqrt{2}}{2}$ and you have $\frac{\sqrt{2}}{8}$
I know what's wrong, just not how I went wrong.
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Hint. $(2x^{2}+4x-2) = 2(x^{2}+2x +1) -4 = 2(x+1)^{2}-4$. Now subsitute $x+1 = \sqrt{2} \sec{t}$. Then you have $dx = \sqrt{2} \sec{t} \cdot \tan{t} \ dt$. Now, $$2 \cdot \Bigl[(x+1)^{2} -2 \Bigr]= 2 \cdot \Bigl[ 2 \cdot (\sec^{2}{t}-1)\Bigr] = 4 \cdot (\sec^{2}{t}-1)$$.
Now what will you get if you power them by $\frac{3}{2}$. You will get $4^{3/2} \cdot \Bigl[ \tan^{2}{t}\Bigr]^{3/2} = 8 \cdot \tan^{3}{t}$.
And try to evaluate the integral. Simplyfying you should get
\begin{align*}
\int \frac{1}{(2x^{2}+4x-2)^{-\frac{3}{2}}} \ \textrm{dx} &= \frac{\sqrt{2}}{8} \int\frac{\sec{t} \cdot \tan{t}}{\tan^{3}{t}} \ \textrm{dt} \\ &= \frac{\sqrt{2}}{8} \int\frac{\cos{t}}{\sin^{2}{t}} \ \textrm{dt}
\end{align*}
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|
Show that 13 divides $2^{70}+3^{70}$
Show that $13$ divides $2^{70} + 3^{70}$.
My main problem here is trying to figure out how to separate the two elements in the sum, and then use Fermat's little theorem. So how can I separate the two?
Thanks!
|
So by FlT, you know that $2^{12}\equiv 1\pmod{13}$ and $3^{12}\equiv 1\pmod{13}$. So
$$
2^{70}+3^{70}\equiv (2^{12})^5\cdot 2^{10}+(3^{12})^5\cdot 3^{10}\equiv 2^{10}+3^{10}\pmod{13}.
$$
However, again by FlT,
$$
2^{12}\equiv 2^{10}\cdot 2^{2}\equiv 1\pmod{13}\implies 2^{10}\equiv 2^{-2}\pmod{13}.
$$
That is to say, $2^{10}$ is congruent to the multiplicative inverse of $4$ modulo $13$. You can do the same for $3^{10}$, except you will need to find the multiplicative inverse to $9$ modulo $13$. You'll see that the sum of these inverses is $0$ modulo $13$, to get your answer.
|
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|
How to solve $x^3 + 2x + 2 \equiv 0 \pmod{25}$? My attempt was:
$x^3 + 2x + 2 \equiv 0 \pmod{25}$
By inspection, we see that $x \equiv 1 \pmod{5}$. is a solution of $x^3 + 2x + 2 \equiv 0 \pmod{5}$. Let $x = 1 + 5k$, then we have:
$$(1 + 5k)^3 + 2(1 + 5k) + 2 \equiv 0 \pmod{25}$$
$$\Leftrightarrow 125k^3 + 75k^2 + 25k + 5 \equiv 0 \pmod{25}$$
$$\Leftrightarrow 5 \equiv 0 \pmod{25}$$
And I was stuck here :( because k was completely cancelled out, how can we find solution for this equation?
Thanks,
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For this, I think the quickest way is simply to work out the 25 possibilities.
For 0,1,2,...24 you have $x^3+2x+2$ giving 2, 5, 14, 35, 74, 137, 230, 359, 530, 749, 1022, 1355, 1754, 2225, 2774, 3407, 4130, 4949, 5870, 6899, 8042, 9305, 10694, 12215, 13874 equivalent modulo 25 to 2, 5, 14, 10, 24, 12, 5, 9, 5, 24, 22, 5, 4, 0, 24, 7, 5, 24, 20, 24, 17, 5, 19, 15, 24.
So the solution is $x \equiv 13$
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Diophantine equation from Carmichael: $2x^4 - 2y^4 = z^2$ I would to solve the Diophantine equation $2x^4 - 2y^4 = z^2$ by descent. This is an exercise from Carmichael Diophantine Analysis but I cannot do it.
Since the LHS is even $2|z$, let $z = 2z'$ and divide both sides by $2$ to get $$(x-y)(x+y)(x^2+y^2)=2z'^2.$$ Congruence conditions tell us that either x and y are both odd or both even, that tells us both sides are a multiple of $8$ so let $z' = 2z''$ and we have $$\frac{x-y}{2}\frac{x+y}{2}\frac{x^2+y^2}{2}=z''^2.$$ The problem now is I don't know if these numbers on the LHS are coprime so I can't claim they are all squares. I don't see any way to continue so maybe I took a wrong turn earlier.
Any hints about how to deal with this equation?
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If $p$ is a prime and it divides both $x$ and $y$, then it divides $z$: if $p$ is odd, then $p^4|z^2$; if $p=2$, then $32|z^2$. Either way, you can factor out the common prime from all of $x$, $y$, and $z$.
Thus, we may assume that $\gcd(x,y)=1$. Since you also know that $x\equiv y\pmod{2}$, you can reduce to the case where $x$ and $y$ are coprime and both odd.
Then $\gcd(x+y,x-y) = 2$, since any common divisor must divide $(x+y)+(x-y)=2x$ and $(x+y)-(x-y)=2y$, and the gcd of $2x$ and $2y$ is $2$. Therefore, $\frac{x-y}{2}$ and $\frac{x+y}{2}$ are coprime.
Any common divisor of $x-y$ and $x^2+y^2$ must divide $2x^2 = (x-y)(x+y) + (x^2+y^2)$, and must also divide $2y^2 = x^2+y^2 - (x-y)(x+y)$. Hence it must divide $\gcd(2x^2,2y^2) = 2$. Thus, $\gcd(x-y,x^2+y^2) = 2$. By the same argument, $\gcd(x+y,x^2+y^2) = 2$. This means that $\frac{x-y}{2}$ and $\frac {x^2+y^2}{2}$ are coprime, as are $\frac{x+y}{2}$ and $\frac{x^2+y^2}{2}$.
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Prove the reduction formula The question is to "prove the reduction formula"
$$ \int{ \frac{ x^2 }{ \left(a^2 + x^2\right)^n } dx } = \frac{ 1 }{ 2n-2 } \left( -\frac{x}{ \left( a^2+x^2 \right)^{n-1} } + \int{ \frac{dx}{ \left( a^2 + x^2 \right)^{n-1} } } \right) $$
What I got is
Set
$ u = x $
$ du = dx $
$\displaystyle{ dv = \frac{ x }{ \left( a^2 + x^2 \right)^{n} } dx }$
$\displaystyle{ v = \frac{ 1 }{ 2(n+1) \left( a^2 + x^2 \right)^{n+1} } }$
So I got
$$ \frac{ 1 }{ 2n+2 } \left( \frac{x}{ \left( a^2 + x^2 \right)^{n+1}} - \int{ \frac{dx}{ \left( a^2+x^2 \right)^{n+1} } } \right) $$
Which I believe is correct. They are subtracting from n in the integration step and I'm not sure why
|
You went wrong when you integrated $dv$.
You have $dv = x(a^2+x^2)^{-n}\,dx$. When you integrate, you add one to the exponent. But adding one to $-n$ gives $-n+1 = -(n-1)$. So
$$v = \frac{1}{2(-n+1)}(a^2+x^2)^{-n+1} = \frac{1}{2(1-n)(a^2+x^2)^{n-1}}.$$
The minus sign from integration by parts can be cancelled out by switching the sign of $2(1-n)$ to get $2(n-1) = 2n-2$.
If you use the correct value of $v$, I think you will have no trouble establishing the formula.
|
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Coloring of an $1\times n$ board using 4 colors? How can I find the number of ways to color an $1\times n$ board using the colors red, blue, green and orange if:
# of red squares is even
# of green squares is even
We did the tilings of a $1\times n$ board using squares and dominos in class, but I'm not sure how to do the coloring with more than two options
Thanks for any help!
|
I just want to point out that exponential generating functions serve as a good approach here. Let $g(x)$ be the exponential generating function of the sequence $h_0, h_1, h_2,\ldots$, where $h_n$ is the number of acceptable colorings of a $1\times n$ board. So $h_n$ is the coefficient of $x^n$ in $g(x)$.
Then
$$
g(x)=\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)^2\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\cdots\right)^2
$$
where the first squared term comes from the fact that red and green tiles must occurs in even numbers, so we only consider even powers, and the second squared term comes from the fact that orange and blue tiles can occur in any number.
The first series is just the even terms of the series for $e^x$, so bisecting the series into its even terms, one sees $g(x)$ simplifies to
$$
\begin{align*}
g(x) &= \left(\frac{e^x+e^{-x}}{2}\right)^2\cdot(e^x)^2 \\
&= \frac{1}{4}(e^{2x}+2+e^{-2x})\cdot(e^{2x})\\
&= \frac{1}{4}(e^{4x}+2e^{2x}+1)\\
&= \frac{1}{4}+\frac{1}{4}\left(\sum_{n\geq 0}\frac{(4x)^n}{n!}+2\sum_{n\geq 0}\frac{(2x)^n}{n!}\right)\\
&= \frac{1}{4}+\frac{1}{4}\left(\sum_{n\geq 0}\frac{(4^n+2\cdot 2^n)x^n}{n!}\right)
\end{align*}
$$
Since there is a $\frac{1}{4}$ hanging out outside the sum, we calculate $h_0=\frac{1}{4}+\frac{1}{4}(3)=1$. Then for the coefficient of nonconstant terms, we see
$$
h_n=\frac{1}{4}(4^n+2\cdot 2^n)=4^{n-1}+2^{n-1}
$$
as noted in user9325's answer.
|
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|
Trig substitution for a triple integral This problem involves calculating the triple integral of the following fraction, first with respect to $p$:
$$
\int\limits_0^{2\pi} \int\limits_0^\pi \int\limits_0^{2} \frac{p^2\sin(\phi)}{\sqrt{p^2 + 3}} dp d\phi d\theta
$$
This involves trig-substitution (I believe), and I am just hoping for an explanation of how to do it.
|
Update: I don't use a trig-substitution because I don't see which one could work. To integrate $p^{2}/\sqrt{p^{2}+3}$ I applied a technique derived from this more general one.
If $P(x)$ is a polynomial of degree $n\geq 2$, we can find a polynomial $Q(x)$ of degree $n-1$ and a constant $C$ such that
$$\int \frac{P(x)}{\sqrt{ax^{2}+bx+c}}\;\mathrm{d}x=Q(x)\sqrt{ax^{2}+bx+c}+\int \frac{C}{\sqrt{ax^{2}+bx+c}}\;\mathrm{d}x.$$
(Described in Cálculo Integral em $\mathbb{R}$ by M. Olga Baptista.)
EDIT: simplified the exposition. The integral is separable, as already noted by J. M. in a comment. Assuming the limits of integration are as follows, we have
$$\begin{eqnarray*}
I &:&=\int_{\theta =0}^{2\pi }\int_{\phi =0}^{\pi
}\int_{p=0}^{2}\frac{p^{2}\sin (\phi )}{\sqrt{p^{2}+3}}\;\mathrm{d}p\;%
\mathrm{d}\phi \;\mathrm{d}\theta \\&=&\left( \int_{0}^{2\pi }\mathrm{d}\theta \right) \left(
\int_{0}^{\pi }\sin(\phi)\;\mathrm{d}\phi\;\right)\left(\int_{0}^{2}\frac{p^{2}}{\sqrt{%
p^{2}+3}}\;\mathrm{d}p\right) \\
&=&2\pi \left(
\int_{0}^{\pi }\sin(\phi)\;\mathrm{d}\phi\;\right)\left(\int_{0}^{2}\frac{p^{2}}{\sqrt{%
p^{2}+3}}\;\mathrm{d}p\right).
\end{eqnarray*}$$
The antiderivative of $\sin (\phi )$ is $-\cos \left( \phi \right) $; hence
$$I =4\pi \int_{0}^{2}\frac{p^{2}}{\sqrt{p^{2}+3}}\;\mathrm{d}p.$$
For the evaluation of this last integral we are going to apply the following property:
If $P(x)$ is a polynomial of degree $n\geq 2$, we can find a polynomial $Q(x)$ of degree $n-1$ and a constant $C$ such that
$$\int \frac{P(x)}{\sqrt{x^{2}+c}}\;\mathrm{d}x=Q(x)\sqrt{x^{2}+c}+\int \frac{C}{\sqrt{%
x^{2}+c}}\;\mathrm{d}x.$$
In the present case, we have $P(x)=x^{2}$ and $Q(x)$ is of the form $Q(x)=Ax+B$
$$\int \frac{x^{2}}{\sqrt{x^{2}+c}}\;\mathrm{d}x=\left( Ax+B\right) \sqrt{%
x^{2}+c}+\int \frac{C}{\sqrt{x^{2}+c}}\;\mathrm{d}x.$$
Differentiating both sides, we get
$$\begin{eqnarray*}
\frac{x^{2}}{\sqrt{x^{2}+c}} &=&A\sqrt{x^{2}+c}+\frac{x\left( Ax+B\right) }{%
\sqrt{x^{2}+c}}+\frac{C}{\sqrt{x^{2}+c}} \\
&=&\frac{1}{\sqrt{x^{2}+c}}\left[ A\left( x^{2}+c\right) +x\left(
Ax+B\right) +C\right],
\end{eqnarray*}$$
and so
$$x^{2} =A\left( x^{2}+c\right) +x\left( Ax+B\right) +C=2Ax^{2}+Bx+Ac+C.$$
Comparing coefficients, we get $A=1/2,\;B=0,\;C=-c/2$. Hence
$$\begin{eqnarray*}
\int \frac{x^{2}}{\sqrt{x^{2}+c}}\;\mathrm{d}x &=&\frac{1}{2}x\sqrt{x^{2}+c}-%
\frac{c}{2}\int \frac{1}{\sqrt{x^{2}+c}}\;\mathrm{d}x \\
&=&\frac{1}{2}x\sqrt{x^{2}+c}-\frac{c}{2}\ln \left( x+\sqrt{x^{2}+c}\right)
\end{eqnarray*}$$
because
$$\int \frac{1}{\sqrt{x^{2}+c}}\;\mathrm{d}x=\ln \left( x+\sqrt{x^{2}+c}%
\right)+\text{ constant}.$$
Thus
$$\int_{0}^{2}\frac{p^{2}}{\sqrt{p^{2}+3}}\;\mathrm{d}p=\sqrt{7}-\frac{3}{2}%
\ln \left( 2+\sqrt{7}\right) +\frac{3}{2}\ln \left( \sqrt{3}\right).$$
And finally
$$I=4\pi \left( \sqrt{7}-\frac{3}{2}\ln \left( 2+\sqrt{7}\right) +\frac{3}{2}\ln \left( \sqrt{3}\right) \right) .$$
|
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|
Complete induction of $10^n \equiv (-1)^n \pmod{11}$ To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.
It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$
For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$
$n\Rightarrow n+1$
$$\begin{align*}
(-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.
|
You are not setting up your induction very well. You should not start with the equality you want to establish, namely that $(-1)^{n+1}-10^{n+1}$ is a multiple of $11$. Instead, you should start with the Induction Hypothesis, which is that $(-1)^n - 10^n$ is a multiple of $11$.
So: the Inductive Step is to show that if $(-1)^n - 10^n$ is a multiple of $11$, then $(-1)^{n+1} - 10^{n+1}$ is also a multiple of $11$.
Let's write out our Induction Hypothesis: it says that
$$\text{There exists an integer }k\text{ such that }(-1)^n - 10^n = 11k.$$
What we want to prove is that:
$$\text{there exists an integer }\ell\text{ such that }(-1)^{n+1}-10^{n+1}=11\ell.$$
(Note that the multiple may be different, that's why I used a different letter).
So now we can try manipulating the expression we want. One possibility is to use the following identity:
$$a^{n+1}-b^{n+1} = (a-b)(a^n + a^{n-1}b + a^{n-2}b^2 + \cdots + ab^{n-1}+b^n),$$
if you already know this identity.
So we have, with $a=-1$ and $b=10$,
$$
(-1)^{n+1} - 10^{n+1} = \Bigl( (-1) - 10\Bigr)\Bigl( (-1)^n + (-1)^{n-1}(10) + \cdots + (-1)10^{n-1} + 10^n\Bigr).$$
Now notice that you don't even need to use the induction hypothesis to conclude that $(-1)^{n+1}-10^{n+1}$ is a multiple of $11$ (as could be seen in mac's answer).
If you don't know the identity, then you can perform some purely algebraic manipulations. E.g.,
$$\begin{align*}
(-1)^{n+1} - 10^{n+1} &= -1\left( (-1)^n + 10^{n+1}\right)\\
&= -\left( (-1)^n -10^n + 10^n + 10^{n+1}\right)\\
&= -\left( \Bigl((-1)^n - 10^n\Bigr) + 10^n\Bigl(1 + 10\Bigr)\right)\\
&= -\left( 11k + 10^n(11)\right) &\quad&\text{(by the induction hypothesis)}\\
&= -\left( 11(k+10^n)\right)\\
&= 11\left( -(k+10^n)\right),
\end{align*}$$
which gives that $(-1)^{n+1} - 10^{n+1}$ is a multiple of $11$, as desired, from the assumption that $(-1)^n - 10^n$ is a multiple of $11$.
But easier still is to use the following property of congruences:
Proposition. Let $a,b,c,d,k$ be integers. If
$$a\equiv b\pmod{k}\qquad\text{and}\qquad c\equiv d\pmod{k}$$
then $ac\equiv bd\pmod{k}$.
Proof. Since $a\equiv b\pmod{k}$, then $k|a-b$, so $k$ divides any multiple of $a-b$; for example, $k|(a-b)c = ac-bc$. Since $k$ divides $ac-bc$, then $ac\equiv bc\pmod{k}$.
Since $c\equiv d\pmod{k}$, then $k|c-d$, so $k|(c-d)b = cb-db$, hence $bc\equiv bd\pmod{k}$.
Since $ac\equiv bc\pmod{k}$ and $bc\equiv bd\pmod{k}$, then $ac\equiv bd\pmod{k}$. QED
Corollary. If $a_1\equiv b_1\pmod{k}$, $a_2\equiv b_2\pmod{k},\ldots, a_n\equiv b_n\pmod{k}$, then
$$a_1\cdots a_n\equiv b_1\cdots b_n\pmod{k}.$$
Proof. Induction on $n$. QED
(This is where you would want to use induction, rather than the specific case you are looking at).
Corollary. If $a\equiv b\pmod{k}$, then for all positive integers $n$, $a^n\equiv b^n\pmod{k}$.
Proof. Apply previous corollary with $a_i=a$ and $b_i=b$ for all $i$. QED
|
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|
Explaining an algebra step in $ \frac{n^2(n+1)^2}{4} + (n+1)^3 = \frac{(n+1)^2}{4}(n^2+4n+4)$ I have encountered this step in my textbook and I do not understand it, could someone please list the intermediate steps?
$$ \frac{n^2(n+1)^2}{4} + (n+1)^3 = \frac{(n+1)^2}{4}(n^2+4n+4). $$
Thanks,
|
One way to see it is to factor out the $(n+1)^2/4$. Then
$$
\begin{align*}
\frac{n^2(n+1)^2}{4} + (n+1)^3 &=\frac{(n+1)^2}{4}\bigl(n^2+4(n+1)\bigr) \\
&= \frac{(n+1)^2}{4}(n^2+4n+4)
\end{align*}
$$
Since you factor out a $1/4$, you have to keep a $4$ in the numerator in the second term.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
|
A High School Proof:
$$S_n = 1^2 + 2^2 + 3^2 +\dots+ n^2$$
We know,
$$r^3 - 3r^2 + 3r - 1 = (r-1)^3 $$
$$r^3 - (r-1)^3 = 3r^2 - 3r + 1$$
When $r=1, 2, 3,\dots, n$
$$1^3 - 0^3 = 3*1^2 - 3*1 + 1\qquad(1)$$
$$2^3 - 1^3 = 3*2^2 - 3*2 + 1\qquad(2)$$
$$3^3 - 2^3 = 3*3^2 - 3*3 + 1\qquad(3)$$
$$\dots\dots\dots\dots\dots\dots\dots\dots\dots$$
$$n^3 - (n-1)^3 = 3n^2 - 3n + 1\qquad(n)$$
By summing all the equations (from 1 to n) we get,
$$n^3 - 0^3 = 3(1^2 + 2^2 + 3^2 +\dots+ n^2) - 3(1+2+3+\dots+n) + (1+1+1+\dots)$$
$$n^3 = 3S_n - 3\frac{n(n+1)}{2} + n$$
\begin{align}
3S_n & = n^3 + \frac{3n(n+1)}{2} - n\\
& = \frac{2n^3 + 3n^2 + 3n - 2n}{2}\\
& = \frac{2n^3 + 3n^2 + n}{2}\\
& = \frac{n(2n^2 + 3n + 1)}{2}\\
& = \frac{n(2n^2 + 2n + n + 1)}{2}\\
& = \frac{n\{2n(n+1)+1(n+1)\}}{2}\\
& = \frac{n(n+1)(2n+1)}{2}\\
\end{align}
$$\therefore S_n = \frac{n(n+1)(2n+1)}{6}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I resolve a recurrence relation when the characteristic equation has fewer roots than terms? I know how to solve "simple" recurrence relations. For instance, say you have:
$$c_0 = 20$$
$$c_1 = 30$$
$$c_n = 3 c_{n-1} - 2 c_{n-2}$$
We can write the characteristic equation as:
$$3x^{n-1} - 2x^{n-2} = x^n$$
Solving this with $n=2$, we get $x = 1$ or $x = 2$. This lets us write the relation $c_n = \alpha_1 1^n + \alpha_2 2^n$, and we can solve for $\alpha_1$ and $\alpha_2$ with the initial states $c_0$ and $c_1$.
However, this depends on the fact that $3x^{n-1} - 2x^{n-2} = x^n$ has two roots.
Now, I'm stuck on another problem where the characteristic equation has fewer roots than terms.
Say I have this recurrence relation instead:
$$a_0 = 0$$
$$a_1 = 2$$
$$a_2 = −1$$
$$a_n = 9a_{n-1} - 15a_{n-2} - 25a_{n-3}$$
The characteristic equation would be:
$$9x^{n-1} - 15x^{n-2} - 25x^{n-3} = x^n$$
However, solving with $n=3$, we only get two roots: $x=-1$ or $x=5$. There are not enough roots to write a relation in the form of $a_n = \alpha_1 r_1^n + \alpha_2r_2^n + \alpha_3r_3^n$. How do I proceed?
|
Use generating function directly: define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence with no subtractions in indices:
$$
a_{n + 3} = 9 a_{n + 2} - 15 a_{n + 1} - 25 a_n
$$
Multiply by $z^n$ and sum over $n \ge 0$, recognize:
$$
\sum_{n \ge 0} a_{n + k} z^n
= \frac{A(z) - a_0 - a_1 z - \ldots - a_{k - 1} z^{k + 1}}{z^k}
$$
to get:
$$
\frac{A(z) - 2 z + z^2}{z^3}
= 9 \frac{A(z) - 2 z}{z^2}
- 15 \frac{A(z)}{z}
- 25 A(z)
$$
Written as partial fractions:
$$
A(z) = \frac{53}{60 (1 - 5 z)} - \frac{3}{10 (1 - 5 z)^2} - \frac{7}{12 (1 + z)}
$$
The generalized binomial theorem lets you read off coefficients:
\begin{align}
a_n &= \frac{53}{60} \cdot 5^n
- \frac{3}{10} \binom{-2}{n} (-5)^n
- \frac{7}{12} \cdot (-1)^n \\
&= \frac{53}{60} \cdot 5^n
- \frac{3}{10} \binom{n + 2 - 1}{2 - 1} \cdot 5^n
- \frac{7}{12} \cdot (-1)^n \\
&= \frac{(35 - 18 n) \cdot 5^n - 35 \cdot (-1)^n}{60}
\end{align}
Repeat roots give terms including $\binom{-m}{n} = \binom{n + m - 1}{m - 1}$, which is just a polynomial of degree $m - 1$ in $n$.
|
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|
Two problems on number theory I need some ideas (preferable some tricks) for solving these two problems:
Find the largest number $n$ such that $(2004!)!$ is divisible by
$((n!)!)!$
For which integer $n$ is $2^8 + 2^{11} + 2^n$ a perfect square?
For the second one the suggested solution is like this : $ 2^8 + 2^{11} + 2^n = ((2^4)^2 + 2\times2^4\times2^6 + (2^ \frac{n}{2})^2 ) \Rightarrow n=12$
But I can't understand the approach,any ideas?
|
A different way for the second question:
First consider $2^8+2^{11}+2^n$ modulo $3$. Because $2^2\equiv 1\pmod 3$, we have
$$ 2^8 + 2^{11} + 2^n \equiv 2^0+2^1+2^{n\bmod 2} \equiv 2^{n\bmod 2} \bmod 3$$
and since $2^1$ is not a square modulo $3$, $n$ must be even. Similarly, $2^3\equiv 1\bmod 7$, so
$$ 2^8+2^{11} + 2^n \equiv 2^2+2^2 + 2^{n\bmod 3} \equiv 1 + 2^{n\bmod 3} \bmod 7$$
The squares modulo $7$ are $0$, $1$, $2$, and $4$, so $2^{n\bmod 3}$ can only be $1$ and thus $n$ is a multiple of $3$.
Since $n$ is even, $2^n$ is a perfect square, and the next perfect square is $1+2^{n/2+1}+2^n$, which is already too large to be $2^8+2^{11}+2^n$ whenever $n/2+1>11$ -- that is, whenever $n>20$.
Putting it all together, $n$ is a multiple of $6$ that is at most $20$, and we can easily check $n=0, 6, 12, 18$ to find that only $n=12$ works.
|
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|
Limit of $(x+3)^{1 + 1/x} - x^{1 + 1/(x+3)}$ when $x\to \infty$ Okay, this is the last limit I have to solve but it's not that easy ;)
$$\lim_{x\to \infty} (x+3)^{1 + 1/x} - x^{1 + 1/(x+3)}$$
|
I know the following method is not valid for the purpose of this exercise,
but I would like to post it. I computed in the Computer Algebra System included in SWP the following power series expansions:
$$\begin{eqnarray*}
(t^{-1}+3)^{1+t} &=&t^{-1}+\left( 3-\ln t\right) +\left( -\frac{3}{2}+\frac{1
}{2}\left( 3-\ln t\right) ^{2}\right) t+O\left( t^{2}\right) \\
t^{-1-1/(t^{-1}+3)} &=&t^{-1}+\left( -\ln t\right) +\left( 3\ln t+\frac{1}{2}
\ln ^{2}t\right) t+O\left( t^{2}\right)
\end{eqnarray*}$$
and
$$(t^{-1}+3)^{1+t}-t^{-1-1/(t^{-1}+3)}=3+\left( -\frac{3}{2}+\frac{1}{2}\left(
3-\ln t\right) ^{2}-3\ln t-\frac{1}{2}\ln ^{2}t\right) t+O\left(
t^{2}\right).$$
Then I generated the asymptotic power series of $(x+3)^{1+1/x}-x^{1+1/(x+3)}$
by the change of variables $x=1/t$
$$\begin{eqnarray*}
f(x)&=&(x+3)^{1+1/x}-x^{1+1/(x+3)} \\ &=&3+\left( -\frac{3}{2}+\frac{1}{2}\left( 3-\ln
\frac{1}{x}\right) ^{2}-3\ln \frac{1}{x}-\frac{1}{2}\ln ^{2}\frac{1}{x}
\right) \frac{1}{x}+O\left( \frac{1}{x^{2}}\right) \\
&=&3+\left( -\frac{3}{2x}+\frac{1}{2x}\left( 3+\ln x\right) ^{2}+\frac{3\ln x
}{x}+\frac{\ln ^{2}x}{2x}\right) +O\left( \frac{1}{x^{2}}\right) \\
&=&3+\frac{3}{x}+6\frac{\ln x}{x}+\frac{\ln ^{2}x}{x}+O\left( \frac{1}{x^{2}}
\right) \\
&\sim &3,
\end{eqnarray*}$$
and confirmed with the following variant
$$\begin{eqnarray*}
f(x)&=&(x+3)^{1+1/x}-x^{1+1/(x+3)} \\ &=&\frac{\left( x+3\right)
^{1+1/x}x^{-1-1/(x+3)}-1}{x^{-1-1/(x+3)}} \\
&\sim &\frac{1+\frac{3}{x}+\left( -3\ln \frac{1}{x}+3\right) \frac{1}{x^{2}}
+O\left( \frac{1}{x^{3}}\right) -1}{\frac{1}{x}+\left( \ln \frac{1}{x}
\right) \frac{1}{x^{2}}+O\left( \frac{1}{x^{3}}\right) } \\
&\sim &3\frac{x+\ln x+1}{x-\ln x} \\
&\sim &3.
\end{eqnarray*}$$
|
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|
Question about computing a Fourier transform of an integral transform related to fractional Brownian motion I am trying to show an integral transform has a fixed point.
Let $H \in (0,1)$ and consider the following integral transform whose kernel is the density of fractional Brownian motion: $$T_H f(x) = 2 \int_{0}^{\infty} \frac{1}{\sqrt{2 \pi y^{2H}}}\exp\left(\frac{-x^2}{2y^{2H}}\right) f(y) dy.$$
I want to show $T_H$ has a fixed point given by $ \phi(x) = \exp(-c|x|^{\frac{1}{2(1-H)}})$ for some $c \in \mathbb{R}$. That is $T_H \phi (x) = \phi (x)$. Although I do not know if this is true in general I know it is true for the case $H=\frac{1}{2}$.
1) What is the Fourier transform of $T_H \phi(x)$ with respect x? Is it equal to a multiple of the Fourier transform of $\phi(x)$?
The motivation for the problem is is that we know for the $H=\frac{1}{2}$ and $c=2$ then $\phi(x) = e^{-2|x|}$ is a fixed point for $T_{\frac{1}{2}}$. That is the Fourier transform of
$T_{\frac{1}{2}}e^{-2|x|} = 2 \int_{0}^{\infty} \frac{e^{\frac{-x^2}{2y}}}{\sqrt{2 \pi y}} e^{-2y} dy$ is equal to $\frac{4}{k^2+4}$ which is the same as the Fourier transform of $e^{-2|x|}$. This is from a paper titled iterated random walk by L. Turban if anyone is interested in the reference.
|
It is rather easy to verify numerically, that $\phi(x) = \exp(-c \vert x \vert^{\frac{1}{2(1-H)}} )$ does not satisfy $T_H \phi = \phi$.
To this end it is enough to observe that $(T_H \phi_c)(x) = c^{-2 (1-h)^2} (T_H \phi_1)(x c^{2 h(1-h)})$, which is obtained via changing variables $y \to c^{-2(1-h)} y$. This scaling tells you that if $T_H \phi_c = \phi_c$, then $T_H \phi_1 = c^{-2(1-h)^2} \phi_1( x c^{2 h (1-h)} )$.
Thus it suffices to show that functional forms of $T_H \phi$ and $\phi$ are different for your purported function. So we set $c=1$ and change variables into $y \to t^{2(1-h)}$. This gives
$$
(T_H \phi)(x) = 2(1-h) \sqrt{\frac{2}{\pi}} \int_0^\infty t^{2(1-h)^2} \mathrm{e}^{-t} \exp(-\frac{x^2}{2} t^{-4 h (1-h)} ) \frac{\mathrm{d} t}{t}
$$
The integral can be evaluate in terms of Fox H-function. Specifically, in terms of Mellin-Barnes integral:
$$
(T_H \phi)(x) =\sqrt{\frac{2}{\pi}} \frac{z^{2 \rho_h}}{2h} \frac{1}{2 \pi i} \int_{\gamma - i \infty}^{\gamma+i \infty} \left( \Gamma(s) \Gamma( -\rho_h + \frac{s}{\kappa_h} )
z^{-\frac{2s}{\kappa_h }} \right)\mathrm{d}s
$$
where $\gamma$ is an arbitrary $\gamma > \rho_h \kappa_h$ and $z = \frac{x}{\sqrt{2}}$ and we introduced $\kappa_h = 4 h(1-h)$ and $\rho_h = \frac{1-h}{2 h}$.
The Mellin-Barnes integral can be evaluated as the sum over poles of $\Gamma$ functions at $s= -k$ for $k \in \mathbb{Z}^+$ and at $s = \rho_h \kappa_h - k \, \kappa_h$, giving
$$
\begin{eqnarray}
(T_H \phi)(x) = \frac{z^{2 \rho_h} }{2 h} \sqrt{\frac{2}{\pi}} && \sum_{k=0}^\infty (- z^{\frac{2}{\kappa_h}} )^k \frac{\Gamma(-\rho_h - \kappa_h \, k )}{k!} + \\
\sqrt{\frac{4 \rho_h \kappa_h}{\pi}} && \sum_{k=0}^\infty (- z^{ 2 } )^k \frac{\Gamma(\rho_h \kappa_h - k \, \kappa_h )}{k!}
\end{eqnarray}
$$
When $h = \frac{1}{2}$, $\kappa_{1/2} = 1$ and $\rho_{1/2} = \frac{1}{2}$. Then the sums evaluate exactly:
$$
\begin{eqnarray}
T_{1/2} \phi &=& \frac{x}{\sqrt{2}} \sqrt{ \frac{2}{\pi} } \sum_{k=0}^\infty (-\frac{x^2}{2})^k \frac{\Gamma(-\frac{1}{2}-k)}{k!} + \sqrt{ \frac{2}{\pi} } \sum_{k=0}^\infty (-\frac{x^2}{2})^k \frac{\Gamma(1/2-k)}{k!} \\
&=& -\sqrt{2} \sinh( \sqrt{2} x) + \sqrt{2} \cosh( \sqrt{2} x) = \sqrt{2} \, \exp( -\sqrt{2} x)
\end{eqnarray}
$$
It is also possible to evaluate these sums for rational values of $h$, but expressions involve hypergeometric functions and stop being so simple.
Added: Below we construct the $T_H \psi$, where $\psi(x)$ is a general Fox H-function. The class of Fox H-functions is a natural place to search for solutions of $T_H \psi = \psi$ because operator $T_H$ maps a Fox H-function into another Fox H-function. Specifically, let
$$
\begin{eqnarray}
\psi(x) &=& H^{m,n}_{p,q}\left( \left. \begin{array}{c} (a_1, \alpha_1), \ldots, (a_p, \alpha_p) \\ (b_1, \beta_1), \ldots, (b_q, \beta_q)
\end{array} \right\vert x\right) \\ &=& \frac{1}{2 \pi i} \int_{\gamma-i \infty}^{\gamma + i \infty} \frac{ \prod_{k=1}^m \Gamma(1-a_k - \alpha_k s) \prod_{k=1}^n \Gamma(b_k + \beta_k s) }{ \prod_{k=m+1}^p \Gamma(1-a_k - \alpha_k s) \prod_{k=n+1}^q \Gamma(b_k + \beta_k s) } x^{-s} \mathrm{d} s
\end{eqnarray}
$$
Then
$$
(T_H \psi)(x) = \frac{1}{\sqrt{2 \pi}} \frac{1}{h} \left( \frac{x^2}{2} \right)^{\frac{(1-h)}{2h} } H^{m,n+1}_{p,q+1}\left( \left. \begin{array}{c} (a_1, \alpha_1), \ldots, (a_p, \alpha_p) \\ \left( -\frac{1-h}{2h}, \frac{1}{2h}\right), (b_1, \beta_1), \ldots, (b_q, \beta_q)
\end{array} \right\vert \left(\frac{x^2}{2}\right)^{\frac{1}{2h}} \right)
$$
Now let's understand the mechanics of how it works for $h=\frac{1}{2}$. Your function $\phi(x)$ with $h=\frac{1}{2}$ is $\exp(-x) = H^{0,1}_{0,1}(x)$ with $\beta_1=1$ and $b_1 = 0$, so
$$
\begin{eqnarray}
\left. (T_H \phi)(x) \right\vert_{h=\frac{1}{2}} &=&
\frac{2}{\sqrt{2 \pi}} \left( \frac{x^2}{2} \right)^{\frac{1}{2} } H^{0,2}_{0,2}\left( \left. \begin{array}{c} - \\ \left( -\frac{1}{2}, 1\right), (0, 1)
\end{array} \right\vert \frac{x^2}{2} \right) \\
&=& \sqrt{\frac{2}{\pi}} \left( \frac{x^2}{2} \right)^{\frac{1}{2} }
\frac{1}{2 \pi i} \int_{\gamma-i \infty}^{\gamma+i \infty} \Gamma\left( -\frac{1}{2} + s\right)\Gamma(s) \left( \frac{x^2}{2} \right)^{-s} \mathrm{d} s \\
&=& \sqrt{\frac{2}{\pi}} \left( \frac{x^2}{2} \right)^{\frac{1}{2} }
\frac{1}{2 \pi i} \int_{\gamma-i \infty}^{\gamma+i \infty} \sqrt{\pi} \Gamma(2s-1) 4^{1-s} \left( \frac{x^2}{2} \right)^{-s} \mathrm{d} s \left. = \right|_{s \to \frac{s}{2}} \\
&=& 2 x
\frac{1}{2 \pi i} \int_{\gamma-i \infty}^{\gamma+i \infty} \Gamma(s-1) \left( \sqrt{2} x \right)^{-s} \mathrm{d} s \\ &=&
(2 x) \left( \frac{\exp(- \sqrt{2} x)}{\sqrt{2} x} \right) = \sqrt{2} \exp(- \sqrt{2} x)
\end{eqnarray}
$$
Notice that the use of tuplication identity for $\Gamma(x)$ was essential here, this essentially proves that within the family of Fox H-functions the solution to $(T_H \psi)(x) = c_1 \psi(c_2 x)$ for some constants $c_1$ and $c_2$ is only possible for $h=\frac{1}{2}$.
Now, the family of Fox H-functions is big and encompasses all hyper-geometric elementary functions, i.e. those that satisfy a linear differential equations with coefficients polynomial in $x$.
If you would like to find the eigenfunction of $T_h$ operator, I would suggest playing with
quadrature rules. Suppose $(T_H f)(x) = \sum_{i=1}^n \omega_i K(x, y_i) f(y_i)$. Solve the system of equations
$$
\sum_{i=1}^n \omega_i K(x_j, y_i) f(y_i) = f(x_j)
$$
numerically, which would allow you to explore the eigen-function for $h \not= \frac{1}{2}$.
|
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Finding limits of rational functions as $x\to\infty$
Possible Duplicate:
Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials
$$\displaystyle \lim_{x\to\infty}\frac{(2x^2+1)^2}{(x-1)^2(x^2+x)}.$$
I do not know where to start on this, I tried multiplying it out and that didn't help really. It seems very complicated and I know I have to reduce it somehow but everything I do just makes it more complicated.
|
Hint: Divide the numerator and the denominator by the highest power of $x$ that occurs (in this case, $x^4$); distribute it so that it shows you exactly what is going on. For example,
$$\begin{align*}
\frac{1}{x^4}(x-1)^2(x^2+x) &=\frac{1}{x^2}\times\frac{1}{x^2}\times(x-1)^2\times(x^2+x) &&\text{(factor }\frac{1}{x^4}\text{)}\\
&= \frac{1}{x^2}\times(x-1)^2 \times \frac{1}{x^2}\times (x^2+x) &&\text{(reordering)}\\
&= \left(\frac{1}{x^2}(x-1)^2\right)\times\left(\frac{1}{x^2}(x^2+x)\right)&&\text{(associativity)}\\
&= \left(\frac{1}{x}(x-1)\frac{1}{x}(x-1)\right)\times\left(\frac{1}{x^2}(x^2+x)\right)\\
&=\left(\frac{x-1}{x}\times\frac{x-1}{x}\right)\times\left(\frac{x^2+x}{x^2}\right)\\
&=\left(\frac{x-1}{x}\right)^2\left(\frac{x^2+x}{x^2}\right).
\end{align*}$$
Now simplify a bit. Then do the same thing with the numerator, and see what happens as $x\to\infty$.
(Once you have more familiarity with end behavior and limits of rational functions, you'll be able to compute this limit "by eye").
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How to solve this equation: $\frac{\sqrt[5]{x^3 \sqrt{x\sqrt[3]{x^{-2}}}}}{\sqrt[4]{x\sqrt[3]{x}}}=3$ Please, help me to solve this equation:
$$\frac{\sqrt[5]{x^3\sqrt{x\sqrt[3]{x^{-2}}}}}{\sqrt[4]{x\sqrt[3]{x}}}=3$$
I tried to shorten fraction, but I get very weird numbers like
$$\frac{\sqrt[30]{x^{19}}}{\sqrt[3]{x}}=3,$$ and I'm stuck there :(
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If $x\gt 0$, then:
*
*$\sqrt[3]{x^{-2}} = x^{-2/3}$.
*$x\sqrt[3]{x^{-2}} = xx^{-2/3} = x^{1/3}$.
*$\sqrt{x\sqrt[3]{x^{-2}}} = x^{1/6}$.
*$x^3\sqrt{x\sqrt[3]{x^{-2}}} = x^{19/6}$.
*$\sqrt[5]{x^3\sqrt{x\sqrt[3]{x^{-2}}}} = x^{19/30}$.
*$x\sqrt[3]{x} = xx^{1/3} = x^{4/3}$.
*$\sqrt[4]{x\sqrt[3]{x}} = x^{1/3}$.
So the quotient is equal to $x^{19/30}/x^{1/3} = x^{(19/30)-(1/3)} = x^{9/30}$. Your equation is then equivalent to
$$x^{9/30} = 3$$
which can be solved by raising both sides to the $30/9$.
If $x\lt 0$, then you need to throw in a few absolute values, since for example, $\sqrt{x\sqrt[3]{x^{-2}}} = |x|^{1/6}$, instead of $x^{1/6}$.
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|
Proving : $ \bigl(1+\frac{1}{n+1}\bigr)^{n+1} \gt (1+\frac{1}{n})^{n} $ How could we prove that this inequality holds
$$ \left(1+\frac{1}{n+1}\right)^{n+1} \gt \left(1+\frac{1}{n} \right)^{n} $$
where $n \in \mathbb{N}$, I think we could use the AM-GM inequality for this but not getting how?
|
No AM-GM inequality - just simple computation:
$$\begin{align}
\frac{(1+\frac{x}{n+1})^{n+1}}{(1+\frac{x}{n})^n}
&= (1+\frac{x}{n})\left(\frac{1+\frac{x}{n+1}}{1+\frac{x}{n}}\right)^{n+1} \\\\
&= (1+\frac{x}{n})\left(\frac{n(n+1)+nx}{(n+1)(n+x)}\right)^{n+1} \\\\
&= (1+\frac{x}{n})\left(\frac{(n+1)(n+x)-x}{(n+1)(n+x)}\right)^{n+1} \\\\
&= (1+\frac{x}{n})\left(1-\frac{x}{(n+1)(n+x)}\right)^{n+1} \\\\
&> (1+\frac{x}{n})(1-\frac{x}{n+x}) = \frac{n+x}{n} \frac{n}{n+x} = 1.
\end{align}$$
Copied from a previous answer of mine.
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|
Demonstrate inequality $$15\le(3+\sin^2x)(4+\cos^2x)\le16 \mbox{ for any }x \in \mathbb{R}$$
I've wrote everything using $\sin$:
\begin{align*}15\le(3+\sin^2x)(4+1-\sin^2x)\le16&\Rightarrow
15\le(3+\sin^2x)(5-\sin^2x)\le16\\
&\Rightarrow 15\le15+2\sin^2x-\sin^4x\le16 \mid-15\\
&\Rightarrow 0\le2\sin^2x-\sin^4x\le1.
\end{align*}
I've stuck here and I need some help. Thanks.
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Welcome to math.stackexchange Daniel!
Just expand and simplify the middle term as such: $$ (3 + \sin^2 x)(4+\cos^2 x) = 12 + 3\cos^2 x + 4\sin^2 x + \sin^2 x \cos^2 x $$
Now, since $\sin^2 x + \cos^2 x = 1$ we can write that term as $$ 15 + \sin^2x + \sin^2 x\cos^2 x$$
Thus, this reduces our problem to showing
$$ 0 \leq \sin^2 x + \sin^2 x \cos^2 x \leq 1$$
which is easy to see if we write that term as $$ (1-\cos^2 x) + \sin^2 x \cos^2 x = 1+ \cos^2 x(\sin^2x -1) = 1-\cos^4 x.$$
As Srivatsan pointed out, a quicker route is to change into cosines immediately:
$$ (3+\sin^2 x)(4+\cos^2 x) = (3 + (1-\cos^2 x))(4+\cos^2 x) $$
$$ = (4-\cos^2 x)(4+\cos^2 x) = 16 - \cos^4 x$$
which produces the required inequality since $0\leq \cos^4 x \leq 1$.
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|
Find M, since $\log_5 M = 2\log_5 A - \log_5 B+2$ Find M, since $\log_5 M = 2\log_5 A - \log_5 B+2$
I tried this:
The answer is in function of A and B.
$\frac{\log_M M}{\log_M 5} = 2\frac{\log_M A}{\log_M 5} - \frac{\log_M B+2}{\log_M 5}$
$1=2\log_M A - \log_M B+2$
$\log_M A^2 = \log_M B+2$
$A^2=B+2$
$\log_5 M = 2\log_5 A - \log_5 A^2$
$\log_5 M = 2\log_5 A - 2 \log_5 A$
$\log_5 M = 0$
$5^0 = M \implies M=1 $
So I don't find how to get an answer in function of A and B nor there is 1 as answer. What did I do wrong?
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$$\log_5 M = 2 \log_5 A − \log_5 B + 2 = \log_5 A^2 + \log_5 \frac{1}{B} + \log_5 25 = \log_5 \left(\frac{25 A^2}{B}\right) \quad \Longrightarrow \quad M = \frac{25A^2}{B}$$
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|
Inverse Laplace Transform -s domain How can I find the inverse Laplace transforms of the following function?
$$ G\left(s\right)=\frac{2(s+1)}{s(s^2+s+2)} $$
I solved so far. After that, how do I do?
$$ \frac{1}{s}+\frac{1}{s^2+s+2}+\frac{s}{s^2+s+2}=G\left( s \right)$$
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Use:
$$
\mathcal{LT}_s\left( \sin(\alpha x) \mathrm{e}^{-b x} \right) = \int_0^\infty \sin(\alpha x) \mathrm{e}^{-b x} \mathrm{e}^{-s x} \mathrm{d} x = \frac{\alpha}{(s+b)^2 + \alpha^2}
$$
$$
\mathcal{LT}_s\left( \cos(\alpha x) \mathrm{e}^{-b x} \right) = \int_0^\infty \cos(\alpha x) \mathrm{e}^{-b x} \mathrm{e}^{-s x} \mathrm{d} x = \frac{b+s}{(s+b)^2 + \alpha^2}
$$
Completing the square: $s^2+s+2 = \left(s+\frac{1}{2}\right)^2 + \frac{7}{4}$. Therefore, decompose the image of Laplace transform accordingly:
$$
\frac{2 (s+1)}{s
\left(s^2+s+2\right)}=\frac{1}{s}-\frac{s+\frac{1}{2}}{\left(s+\frac{1}{2}\right)
^2+\frac{7}{4}}+\frac{3}{\sqrt{7}}\frac{\sqrt{7}/2}{\left(\left(s+\frac{1}{2}\right)^2+\frac{7}{4}\right)}
$$
Compare with the answer by WolframAlpha.
|
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|
Parametric form of an ellipse given by $ax^2 + by^2 + cxy = d$ If $c = 0$, the parametric form is obviously $x = \sqrt{\frac{d}{a}} \cos(t), y = \sqrt{\frac{d}{b}} \sin(t)$.
When $c \neq 0$ the sine and cosine should be phase shifted from each other. How do I find the angular shift and from there how do I adjust the factors multiplying the sine and cosine?
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If the parametric ellipse coordinates are $\left(x(t),y(t)\right) = (X \cos\varphi \cos(t)-Y \sin\varphi \sin(t), Y \cos\varphi \sin(t)+X \sin\varphi \cos(t)) $
Then the parameters $ X,Y,\varphi $ are
$$ X =\pm \sqrt{\frac{2 d}{a+b+\sqrt{(a-b)^2+c^2}}} $$
$$ Y =\pm \sqrt{\frac{2 d}{a+b-\sqrt{(a-b)^2+c^2}}} $$
$$ \varphi = \frac{1}{2}\tan^{-1}\left(\frac{c}{a-b}\right) $$
Example:
$10 x^2+20 y^2-18 x y = 100$
The coefficients are ($a=10$, $b=20$, $c=-18$, $d=100$)
The parametric coefficients are
$(-1.008311 \cos(t)+3.973667 \sin(t), 1.713639 \cos(t)+2.338119 \sin(t))$
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|
Simplifying fractions How to transform $$\left({1+\sqrt{1-4ab}\over2a}\right)^n-\left({1-\sqrt{1-4ab}\over2a}\right)^n\over \left({1+\sqrt{1-4ab}\over2a}\right)^m-\left({1-\sqrt{1-4ab}\over2a}\right)^m$$ into $$\left({b\over a}\right)^n -1\over \left({b\over a}\right)^m -1$$?
Thanks.
Added: As per comments, there is an additional assumption that $a,b\gt 0$, $b=1-a$, and $b\gt a$.
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With the added information, which is key (otherwise, the two expressions are just not equal in general, as noted by Ross):
If $b=1-a$ and $b\gt a$, then $1-a\gt a$ so $1\gt 2a$. Then
$$1 -4ab = 1-4a(1-a) = 1-4a+4a^2 = (1-2a)^2,$$
and since $1-2a\gt 0$, we have
$$\sqrt{1-4ab} = \sqrt{(1-2a)^2} = |1-2a| = 1-2a.$$
So
$$\begin{align*}
\frac{1+\sqrt{1-4ab}}{2a} &= \frac{1+(1-2a)}{2a} = \frac{2(1-a)}{2a} = \frac{b}{a},\\
\frac{1-\sqrt{1-4ab}}{2a} &= \frac{1-(1-2a)}{2a} = \frac{2a}{2a} = 1.
\end{align*}$$
That means that
$$\begin{align*}
\left({1+\sqrt{1-4ab}\over2a}\right)^n-\left({1-\sqrt{1-4ab}\over2a}\right)^n\over \left({1+\sqrt{1-4ab}\over2a}\right)^m-\left({1-\sqrt{1-4ab}\over2a}\right)^m & =
\frac{\left(\frac{b}{a}\right)^n - 1^n}{\left(\frac{b}{a}\right)^m - 1^m}\\
&=\frac{\left(\frac{b}{a}\right)^n - 1}{\left(\frac{b}{a}\right)^m - 1}.
\end{align*}$$
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.