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Prove that hyperspherical coordinates are a diffeomorphism, derive Jacobian The explicit form for the transformation into hyperspherical coordinates is
$$x_1 = r\sin\theta_1 \sin\theta_2 \dotsb \sin \theta_{n-1} \\ x_2 = r\sin\theta_1 \sin\theta_2 \dotsb \cos \theta_{n-1} \\ x_3 = r\sin\theta_1 \dotsb \cos \theta_{n-2}\\ \vdots \\ x_{n} = r \cos\theta_1$$
for $0 \leq \theta_i \leq \pi \;\;(1\leq i \leq n-2)$ and $0\leq \theta_{n-1} \leq 2\pi$. It has Jacobian $r^{n-1} \sin^{n-2}\theta_1 \sin^{n-3}\theta_2 \dotsb \sin{\theta_{n-2}}.$
I wonder if someone could provide me with a reference for an intuitive explanation as to why this indeed is a diffeomorphism from $\mathbb{R}^n\setminus\{0\} \to \mathbb{R}^n \setminus\{0\}$, and why this is the Jacobian. Or perhaps someone could indicate the idea of a proof. Thanks as always
| I will sketch the proof for the Jacobian by induction on $n$, thus in a different fashion compared to What is the metric tensor on the n-sphere (hypersphere)? (which perhaps is smoother, though).
For $n=2$, we have the transformation law $A_2$
$$
x^1 = \rho\cos\phi\\
x^2 = \rho\sin\phi.
$$
Hence
$$
\frac{\partial(x^1,x^2)}{\partial(\rho,\phi)}=
\left(
\begin{array}[cc]
\ \cos\phi & -\rho \sin\phi\\
\sin\phi & \rho \cos\phi
\end{array}
\right)
$$
and the Jacobian is $J_2=\rho(\cos^2\phi+\sin^2\phi)=\rho$, i.e. $dx^1 dx^2=\rho\, d\rho d\phi$.
The idea of induction works as follows: for $n=3$, instead of using the transformation $A_{3}$, given by
$$
x^1 = \rho \cos\phi \sin\theta\\
x^2 = \rho \sin\phi \sin\theta\\
x^3 = \rho \cos\theta,
$$
directly, one uses the two combined transformations $A_{23}$, given by
$$
z^1 = \rho \cos\phi\\
z^2 = \rho \sin\phi\\
\theta=\theta,
$$
with Jacobian $J_2=\rho$, and, letting $|z|\equiv\sqrt{(z^1)^2+(z^2)^2}=\rho$, $A_{3z}$, given by
$$
x^1 = z^1 \sin\theta\\
x^2 = z^2 \sin\theta\\
x^3 = |z| \cos\theta.
$$
Now,
$$
\frac{\partial(x^1,x^2,x^3)}{\partial(z^1,z^2,\theta)}=
\left(
\begin{array}[ccc]
\ \sin\theta & 0 & z^1\cos\theta\\
0 & \sin\theta & z^2\cos\theta\\
z^1|z|^{-1}\cos\theta & z^2 |z|^{-1}\cos\theta & -|z|\sin\theta
\end{array}
\right)
$$
which gives, expanding with respect to the last line,
$$
J_{3z}=
z^1|z|^{-1}\cos\theta(z^1\cos\theta\sin\theta)+
z^2 |z|^{-1}\cos\theta(z^2\sin\theta\cos\theta)+
|z|\sin\theta \sin^2\theta\\
=|z|\sin\theta
$$
so that, being $A_3=A_{3z}\circ A_{23}$,
$$
J_3 = J_2 \cdot J_{3z} = \rho |z|\sin\theta = \rho^2 \sin\theta,
$$
i.e. $dx^1dx^2dx^3 = \rho^2\sin\theta\, d\rho d\theta d\phi$.
Let $A_n$ be
$$
x^1 = \rho \cos\phi \sin\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3} \sin\theta_{n-2}\\
x^2 = \rho \sin\phi \sin\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3} \sin\theta_{n-2}\\
x^3 = \rho \cos\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3} \sin\theta_{n-2}\\
\ldots\\
x^{n-1}=\rho \cos\theta_{n-3}\sin\theta_{n-2}\\
x^n = \rho \cos\theta_{n-2};
$$
the transformation itself is built recursively: at the $n$th step, multiply the $n-1$ old coordinates by sine of the new angle and add a new coordinate $x^n$ as $\rho$ times the cosine of the new angle.
In the same spirit of the above calculation, let $A_{(n-1)n}$ be
$$
z^1 = \rho \cos\phi \sin\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3}\\
z^2 = \rho \sin\phi \sin\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3}\\
z^3 = \rho \cos\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3}\\
\ldots\\
z^{n-1}=\rho \cos\theta_{n-3}\\
\theta_{n-2} = \theta_{n-2}
$$
which, by induction, will have Jacobian $J_{n-1} = \rho^{n-2}\sin\theta_{1}\sin^2\theta_{2}\ldots\sin^{n-3}\theta_{n-3}$.
Let $A_{nz}$ be
$$
x^1 = z^1 \sin\theta_{n-2}\\
x^2 = z^2 \sin\theta_{n-2}\\
\ldots\\
x^{n-1} = z^{n-1} \sin\theta_{n-2}\\
x^n = |z| \cos\theta_{n-2}.
$$
The Jacobi matrix reads
$$
\frac{\partial(x^1,x^2,\ldots,x^{n-1},x^n)}{\partial(z^1,z^2,\ldots,z^{n-1},\theta_{n-2})}
=
\left(
\begin{array}[ccccc]
\ \sin\theta_{n-2} & 0 & \ldots & 0 & z^1\cos\theta_{n-2}\\
0 & \sin\theta_{n-2} & \ldots & 0 & z^2\cos\theta_{n-2}\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & \ldots & \sin\theta_{n-2} & z^{n-1}\cos\theta_{n-2}\\
z^1|z|^{-1}\cos\theta_{n-2} & z^2|z|^{-1}\cos\theta_{n-2} & \ldots & z^{n-1}|z|^{-1} & -|z|\sin\theta_{n-2}
\end{array}
\right),
$$
where $$|z|\equiv \sqrt{\sum_{k=1}^{n-1}(z^k)^2 }=\rho.$$
Again, the Jacobian for $A_{nz}$ is easily expressed expanding with respect to the last row
$$
J_{nz} = z^1|z|^{-1}\cos\theta_{n-2} z^1\cos\theta_{n-2} \sin^{n-2}\theta_{n-2}\\
+ z^2 |z|^{-1}\cos\theta_{n-2} z^2\cos\theta_{n-2}\sin^{n-2}\theta_{n-2}\\
+ \ldots\\
+ z^{n-1}|z|^{-1}\cos\theta_{n-2} z^{n-1}\cos\theta_{n-2}\sin^{n-2}\theta_{n-2}\\
+ |z| \sin^2\theta_{n-2}\sin^{n-2}\theta_{n-2}\\
= |z| \sin^{n-2}\theta_{n-2}.
$$
Finally $A_{n} = A_{nz}\circ A_{(n-1)n}$ and hence
$$
J_n = J_{n-1}\cdot J_{nz} = \left( \rho^{n-2} \prod_{k=0}^{n-3}\sin^k\theta_{k}\right) |z| \sin^{n-2}\theta_{n-2}\\
=\rho^{n-1} \prod_{k=0}^{n-2} \sin^k\theta_k,
$$
where, for notational convenience $\theta_0\equiv \pi/2$.
This proves the Jacobian formula for any $n$.
If I recall correctly, any transformation with nonsingular Jacobi matrix gives rise to a local diffeomorphism, therefore our derivation above proves that this change of coordinates is good, except at the “north poles” and “south poles” $\theta_k =0,\pi$. To see that this indeed covers the whole $\mathbb{R}^n$, with the exception of such singularities, one can work out explicitly the inversion formulas, expressing the hyperspherical coordinates in terms of $x^1,\ldots,x^n$. This can be done, again, recursively and I think it is nicely given on https://en.wikipedia.org/wiki/N-sphere.
I hope it helps!
| {
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"timestamp": "2023-03-29T00:00:00",
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Help find the MacLaurin series for $\frac{1}{e^x+1}$ What is the MacLaurin series up to $x^4$ for $\frac{1}{e^x+1}$?
My Attempt:
$$\begin{align}
\frac{1}{e^x+1} &=(1+e^x)^{-1} \\
&\approx 1 -e^x+(e^x)^2-(e^x)^3+(e^x)^4 \\
\end{align}
$$
Since
$$
\begin{align}
e^x &\approx \frac{1}{24} \, x^{4} + \frac{1}{6} \, x^{3} + \frac{1}{2} \, x^{2} + x
+ 1 \\
(e^x)^2 &\approx \frac{2}{3} \, x^{4} + \frac{4}{3} \, x^{3} + 2 \, x^{2} + 2 \, x + 1
\\
(e^x)^3 &\approx \frac{27}{8} \, x^{4} + \frac{9}{2} \, x^{3} + \frac{9}{2} \, x^{2} + 3
\, x + 1 \\
(e^x)^4 & \approx \frac{32}{3} \, x^{4} + \frac{32}{3} \, x^{3} + 8 \, x^{2} + 4 \, x + 1
\end{align}
$$
Therefore
$$
1 -e^x+(e^x)^2-(e^x)^3+(e^x)^4 \approx \frac{95}{12} \, x^{4} + \frac{22}{3} \, x^{3} + 5 \, x^{2} + 2 \, x + 1
$$
But the answer is $\frac{1}{48} \, x^{3} - \frac{1}{4} \, x + \frac{1}{2}$
Please advise what did I do wrong and what is the correct way to solve this problem.
| Your initial expansion is an asymptotic series, valid for small values of $e^x$. But when $x$ is small, $e^x \approx 1$, which isn't small at all! If you instead write
$$
\frac{1}{e^x+1}=\frac{1}{2+(e^x-1)}=\frac{1}{2}\cdot\frac{1}{1+\frac{1}{2}(e^x-1)}=\frac{1}{2}\left(1-\frac{1}{2}(e^x-1) +\frac{1}{4}(e^x-1)^2-\frac{1}{8}(e^x-1)^3+\frac{1}{16}(e^x-1)^4+\ldots\right),
$$
and then fill in the expansion for $e^x-1$ where appropriate, you should find the right answer.
| {
"language": "en",
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"source": "stackexchange",
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How to show that $\log (\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2})= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$ If $\log (a +b +c) =\log a+\log b+\log c$ then show that $$\log \left(\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2}\right)= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$$
Trial: If put $a=\frac{2a}{1-a^2}$ and the similar we are done. Can we do this in this way? Otherwise how to do. Please help.
| The expression $\dfrac{2a}{1-a^2}$ should make you think of the tangent function, as in $$\tan(2A)=\frac{2\tan A}{1-(\tan A)^2}$$
Here I'm using capital $A$ to refer to the angle and lower-case $a$ to refer to its tangent: $a=\tan A$.
$\log(a+b+c)=\log a+\log b+\log c$ implies $\log(a+b+c)=\log(abc)$, which then implies $a+b+c=abc$, and following the pattern in capital and lower-case letters above, we have $\tan A+\tan B+\tan C=\tan A\tan B\tan C$.
The identity you're trying to prove is equivalent to
$$
\frac{2a}{1-a^2} + \frac{2b}{1-b^2}+\frac{2c}{1-c^2} = \frac{2a}{1-a^2} \cdot \frac{2b}{1-b^2}\cdot\frac{2c}{1-c^2}
$$
(with lower-case $a,b,c$), so that's the same as
$$
\tan(2A)+\tan(2B)+\tan(2C)=\tan(2A)\tan(2B)\tan(2C).
$$
Now at this point I might not know how to proceed further if I hadn't seen the following at some point in the past. In the first place, the usual formula for the tangent of a sum implies with 30 seconds' more work that
$$
\tan(A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan A\tan C-\tan B\tan C},
$$
and in the second place, if $A+B+C=\pi\text{ radians or }180^\circ\text{ or a half-circle}$ then $\tan(A+B+C)=0$. So the fraction is $0$ and therefore the numerator is $0$ and therefore $\tan A+\tan B+\tan C=\tan A\tan B\tan C$.
So what you're being asked to prove is that if $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ then $\tan(2A)+\tan(2B)+\tan(2C)=\tan(2A)\tan(2B)\tan(2C)$.
We've see that the fact that $A+B+C=\text{a half circle}$ implies the first of these identities only because it implies that $\tan(A+B+C)=0$. So you really just need to show that if $\tan(A+B+C)=0$ then $\tan(2A+2B+2C)=0$. It's not hard to show that that's the same as saying that if $A+B+C$ is an integer multiple of a half-circle, then so is $2A+2B+2C$.
Moral: Don't do this problem if you've forgotten your trigonometry.
However, since the problem as stated doesn't mention trigonometry, I'm wondering if there's another way to do it that avoids that. Probably there is. The tangent function and the identities we used involve a particular way of parametrizing the circle, as $A\mapsto(\cos A,\sin A)$. But the result doesn't seem to be one that should depend on such a choice of parametrization.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Difficult limit evaluation: $\lim_{x\to\infty}(\sqrt{x^2+4x} - x)$ I'm struggling to find the solution of the following:
$$\lim_{x\rightarrow\infty}(\sqrt{x^2+4x} - x)$$
I come to the answer of $0$.
The book has an answer of $4/1$.
The book explains a part of the question briefly. Looking at the brief answer information given I then come to an answer of $2$...
What is the answer?
| Multiplying by the quantity $\sqrt{x^2 + 4x} + x$ on top and bottom leads to
$$\sqrt{x^2 + 4x} - x = \frac{x^2 + 4x - x^2}{\sqrt{x^2 + 4x} + x} = \frac{4x}{\sqrt{x^2 + 4x} + x}$$
Now divide each term by $x$, noting that this becomes $x^2$ under the square root:
$$\frac{4x}{\sqrt{x^2 + 4x} + x} = \frac{\frac{4x}{x}}{\sqrt{\frac{x^2 + 4x}{x^2}} + \frac{x}{x}} = \frac{4}{1 + \sqrt{1 + \frac{4}{x}}}$$
Taking $x \to \infty$, this limit is $2$. There appears to be an error in the book, since Wolfram Alpha also returns $2$.
| {
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Hard contest type trigonometry proof Suppose that real numbers $x, y, z$ satisfy:
$$\frac{\cos x + \cos y + \cos z}{\cos(x + y + z)}
=
\frac{\sin x + \sin y + \sin z}{\sin (x + y + z )}
= p$$
Then prove that:
$$\cos (x + y) + \cos (y + z ) + \cos (x + z) = p$$
I am not even getting where to start? Please help.
| note
$$\cos{(x+y)}=\cos{[(x+y+z)-z]}=\cos{(x+y+z)}\cos{z}+\sin{(x+y+z)}\sin{z}$$
and
$$\cos{(y+z)}=\cos{(x+y+z)}\cos{x}+\sin{(x+y+z)}\sin{x}$$
$$\cos{(z+x)}=\cos{(x+y+z)}\cos{y}+\sin{(x+y+z)}\sin{y}$$
add this three
\begin{align*}
&\cos{(x+y)}+\cos{(y+z)}+\cos{(x+z)}\\
&=(\cos{x}+\cos{y}+\cos{z})\cos{(x+y+z)}+(\sin{x}+\sin{y}+\sin{z})\sin{(x+y+z)}\\
&=p\cos^2{(x+y+z)}+p\sin^2{(x+y+z)}\\
&=p
\end{align*}
| {
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"question_score": "9",
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What is the fastest way to $\pi$? There are many known sequences convergent to $\pi$ with different convergence accelerations. For example both of the following sequences are convergent to $\pi$ when $n$ goes to $\infty$:
(a) $a_n=2^{n+1}\times\sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}}}}$ which the under brace part repeats $n$ times. More clear:
$a_0=2\sqrt{2}$
$a_1=4\sqrt{2-\sqrt{2}}$
$a_2=8\sqrt{2-\sqrt{2+\sqrt{2}}}$
$a_3=16\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}$
$...$
(b) $b_n=4\times\sum_{i=0}^{n}\frac{(-1)^i}{2i+1}$
More clear:
$b_0=4$
$b_1=4(1-\frac{1}{3})$
$b_2=4(1-\frac{1}{3}+\frac{1}{5})$
$b_3=4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7})$
$...$
It seems the first sequence goes to $\pi$ faster than the second one.
Question: What is the fastest known explicit sequence convergent to $\pi$? Please introduce a list of known sequences convergent to $\pi$ and their convergence accelerations.
| The provably fastest convergence, in fractions, is given by the continued fraction:
$$[3;7,15,1,292,1,1,1,2,1,3,1,\ldots]$$
This corresponds to the sequence $$3,\frac{22}{7}, \frac{333}{106}, \frac{355}{113}, \frac{103993}{33102},\ldots$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $I=\int_{0}^{2\pi}\frac{\cos^2\theta}{5+4\sin\theta}d\theta$ I've found by Cauchy's Integral Formula that
$$\int_{\gamma (0,1)}\frac{\Re(z)}{2z-i}dz=-\frac{\pi}{4}i$$ but now not sure how to find $I$ using this? And help much appreciated!
| Sub $z=e^{i \theta}$, $\cos{\theta}=\frac12 (z+z^{-1})$, $\sin{\theta}=\frac{1}{2 i} (z-z^{-1})$. Then $d\theta = -i dz/z$ and we can rewrite the integral as an intgeral over the unit circle:
$$-i \oint_{|z|=1} \frac{dz}{z} \frac{\frac14 (z+z^{-1})^2}{5-i 2 (z-z^{-1})} = \frac14 \oint_{|z|=1} \frac{dz}{z^2} \frac{(z^2+1)^2}{2 z^2+i 5 z-2}$$
You may then apply the residue theorem by finding the zeroes of the denominator and evaluating the residues at the poles inside the unit circle. The answer is then $i 2 \pi$ times the sum of those residues.
The poles are at $z=0$ (double pole), $z_{+}=-2 i$, and $z_-=-i/2$. Of these, only $z=0$ and $z=z_-$ are within the unit circle.
For the double pole at $z=0$, the residue is given by the derivative of $z^2$ times the integrand at $z=0$:
$$\operatorname*{Res}_{z=0} \frac{1}{4 z^2} \frac{(z^2+1)^2}{2 z^2+i 5 z-2} = \frac14 \left [\frac{d}{dz} \frac{(z^2+1)^2}{2 z^2+i 5 z-2} \right ]_{z=0} = -i \frac{5}{16}$$
For the simple pole at $z=z_-$, the residue is given by $z-z_-$ times the integrand evaluated at $z-z_-$. For an integrand of the form $p(z)/q(z)$, the residue may be computed as $p(z_-)/q'(z_-)$ as follows:
$$\operatorname*{Res}_{z=z_-} \frac{1}{4 z^2} \frac{(z^2+1)^2}{2 z^2+i 5 z-2} = \frac{(z_-^2+1)^2}{4 z_-^2 (4 z_-+i 5)} = i \frac{3}{16}$$
The value of the integral is then $i 2 \pi$ times the sum of these residues, or $\pi/4$.
| {
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Show $f(x) = (x^4+x^2+1)/(x^3+1) $ is $O(x)$ How would I find the witnesses $C$ and $k$ such that $f(x)$ is $O(x)$?
What I tried was $$(x^4+x^2+1)/(x^3+1) ≤ (x^4+x^4+x^4)/(x^3+x^3) = (3/2)x $$
for values $x>1$. $C = 3/2, k = 1$
Is this right?
| $$\dfrac{x^4+x^2+1}{x^3+1} = \dfrac{(x^2+x+1)(x^2-x+1)}{(x+1)(x^2-x+1)} = \dfrac{x^2+x+1}{x+1} = x + \dfrac1{x+1}$$
Now you should be able to conclude what you want.
| {
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Find the area enclosed by $\sqrt{(x-2)^2+(y-3)^2} + 2\sqrt{(x-3)^2+(y-1)^2} = 4$ Question: What is the area of the interior of the simple closed curve described by the equation
$\sqrt{(x-2)^2+(y-3)^2} + 2\sqrt{(x-3)^2+(y-1)^2} = 4$?
Comments: I came up with this specific problem myself in response to my earlier question, which I don't think was well posed, or at least it was not clear what I was after: to see how to find the area of the interior of a Jordan curve that is described by an implicit function. To see how this area looks like, I uploaded a picture from WolframAlpha:
As you can see, it is quite egg-like. We can generalize; the object could be called a weighted ellipse [edit: usually called a Cartesian oval] with an equation of the form
$\sqrt{(x-x_1)^2+(y-y_1)^2} + a \cdot \sqrt{(x-x_2)^2+(y-y_2)^2} = k$
where $(x_1,y_1)$ and $(x_2,y_2)$ are the Cartesian coordinates of the focal points of this weighted ellipse and $a$ is a weight (it equals $1$ in the case of an ordinary ellipse). As a bonus question, I would like to see how to find the area of this object.
| Here's the answer using polar coordinates. It turns out that the implicit curve is quadratic in $r$ in polar coordinates which means $r$ can be solved for in terms of the angle $t$ using the quadratic formula. Finally, using $A=\frac{1}{2} \int_0^{2 \pi} r(t)^2 dt$ allows for the area to be computed.
For simplicity, take $x_1=y_1=0$ since we can place the origin anywhere. We start with
$$a \sqrt{x^2+y^2}+\sqrt{(x-x_2)^2+(y-y_2)^2}=k.$$
Notice I swapped where the $a$ occurs. It will make some slightly algebra nicer. Substituting $x=r \cos t$ and $y= r \sin t$ gives
$$\sqrt{(r\cos t-x_2)^2+(r \sin t-y_2)^2}=k-ar$$
and after squaring
$$r^2 - r(2 x_1 \cos t +2 y_2 \sin t+1)+(x_1^2+y_2^2)= k^2-2kar + a^2r^2.$$
With dependencies on $t$, this function has the form
$$r^2-r(A\cos t + B \sin t + C) + D = 0$$
where $A = \frac{2x_1}{1-a^2}$, $B = \frac{2y_2}{1-a^2}$, $C = \frac{1+2k}{1-a^2}$, $D = \frac{x_1^2+y_1^2-k^2}{1-a^2}$. Given our initial parameters, perhaps there are bounds on $A,B,C$ and $D$. I don't care. I'll take the above as the new definition of our curve and find the area in terms of the new parameters making assumptions as I go.
Using the quadratic formula yields
$$ r = \frac{1}{2}\left[(A\cos t + B \sin t + C)\pm \sqrt{(A\cos t + B \sin t + C)^2-4D} \right].$$
We have the form $r(t)=p(t) \pm \sqrt{q(t)}$ for $r$. Assume parameters are chosen so that $q(t)>0$. If we $D<0$ we have one branch, otherwise two curves satisfy the original equation. Then the area of the region enclosed is given by
$$A = \frac{1}{2}\int_0^{2\pi} r(t)^2 dt = T_1 + T_2 + T_3,$$
with $T_1 = \frac{1}{2}\int_0^{2\pi} 2 p(t)^2 dt$, $T_3 = \pm\frac{1}{2}\int_0^{2\pi} p(t) \sqrt{q(t)} dt$, and $T_3 = \frac{1}{2} \int_0^{2\pi} q(t) dt$.
More explicitly, this is
$$T_1 = \frac{1}{8} \int_0^{2\pi} (A \cos t+B \sin t+C)^2 = \frac{1}{8}(\frac{2 \pi A^2}{2} + \frac{2 \pi B^2 }{2} + 2\pi C) = \frac{\pi}{8}(A^2+B^2+C)$$
$$T_2 = \pm \frac{1}{4} \int_0^{2\pi} (A\cos t + B \sin t + C) \sqrt{(A\cos t + B \sin t + C)^2-4D} \; dt$$
$$T_3 = \frac{1}{8} \int_0^{2\pi} \left[(A\cos t + B \sin t + C)^2-4D \right] dt = \frac{\pi}{8}(A^2+B^2+C-4D)$$
As seen, the first and third terms are easily computable. The second requires work. Because $A \cos t + B \sin t = \alpha \sin(t+\delta)$ for $\alpha = \sqrt{A^2+B^2}$ and $\delta$ depending on $A$ and $B$, we can substitute this in the expression for $T_2$ and do a change of variables to get
$$T_2 = \pm\frac{1}{4} \int_0^{2\pi} (\sqrt{A^2+B^2} \sin t + C) \sqrt{(\sqrt{A^2+B^2} \sin t + C)^2-4D} \; dt.$$
I highly doubt the antiderivative of the above can be expressed in terms of elementary functions. Perhaps someone can work this into an expression involving elliptic integrals.
If I get to it, I'll work out the degenerate case of an ellipse. In that situation, my guess is that the $T_2$ term vanishes. I've almost certainly made an error or two--please let me know if I did. However, the overall approach is sound.
| {
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Find all matrices that satisfy $\mathrm B \mathrm A = \mathrm I_2$
Given the matrix $$A=\begin{pmatrix}1&8\\3&5\\2&2\\ \end{pmatrix}$$ find all $2 \times 3$ matrices in $B \in M_{2 \times 3}(\mathbb R)$ with $BA=I_2$.
Here's what I did:
$$\begin{pmatrix}a&b&c\\d&e&f\\ \end{pmatrix} \begin{pmatrix}1&8\\3&5\\2&2\\ \end{pmatrix} = \begin{pmatrix}1&0\\0&1\\ \end{pmatrix}$$
and then multiplying things out:
$$\begin{pmatrix} {a+3b+2c}&{8a+5b+2c}\\{d+3e+2f}&{8d+5e+2f}\\ \end{pmatrix} = \begin{pmatrix}1&0\\0&1\\ \end{pmatrix}$$
So would I just set
$$a+3b+2c=1$$
$$8a+5b+2c=0$$
$$d+3e+2f=0$$
$$8d+5e+2f=1$$
But then this gives $4$ equations in $6$ unknowns, so wouldn't there be infinitely many solutions? Did I do this correctly? Matrices aren't my strong point...
Thanks.
| Since the second column is not a multiple of the first column, the matrix $\pmatrix{1 & 8\cr 3 & 5\cr 2 & 2\cr}$ has rank $2$. That implies that a solution exists. One way to get a solution is to take any two rows of $A$ (check that they are linearly independent) and use the inverse of that $2 \times 2$ matrix for the corresponding columns of $B$, and $0$'s in the other column. Thus if you take the first two rows of $A$, the inverse of $\pmatrix{1 & 8\cr 3 & 5\cr}$ is $\pmatrix{-5/19 & 8/19\cr 3/19 & -1/19\cr}$, corresponding to
$B = \pmatrix{-5/19 & 8/19 & 0\cr 3/19 & -1/19 & 0\cr}$.
For the general solution, write the system using block matrices as
$$ [ B_1 \ b ] \left[\matrix{A_1\cr a^T\cr} \right] = B_1 A_1 + b a^T = I$$
(where $B_1$ and $A_1$ are $2 \times 2$, $b$ is $2 \times 1$ and $a^T$ is $1 \times 2$). We can solve for $B_1$ in terms of $b$: $B_1 = (I - b a^T) A_1^{-1}$. We already have $A_1^{-1} = \pmatrix{-5/19 & 8/19 \cr 3/19 & -1/19 \cr}$, so with $b = \pmatrix{b_1\cr b_2}$ and $a^T = (2, 2)$ we get
$$ B_1 = \pmatrix{1 - 2 b_1 & - 2 b_1\cr -2 b_2 & 1 - 2 b_2\cr} \pmatrix{-5/19 & 8/19 \cr 3/19 & -1/19 \cr} = \pmatrix{ (-5 + 4 b_1)/19 & (8 - 14 b_1)/19\cr
(3 + 4 b_2)/19 & (-1 - 14 b_2)/19\cr}$$
i.e.
$$ B = \pmatrix{ (-5 + 4 b_1)/19 & (8 - 14 b_1)/19 & b_1\cr
(3 + 4 b_2)/19 & (-1 - 14 b_2)/19 & b_2\cr}$$
where $b_1$ and $b_2$ are arbitrary.
| {
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How to find this limit: $A=\lim_{n\to \infty}\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\cdots+\sqrt{\frac{1}{n}}}}}$ Question:
Show that $$A=\lim_{n\to \infty}\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}$$
exists, and find the best estimate limit $A$.
It is easy to show that
$$\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}\le\sqrt{1+\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}}$$
and it is well known that this limit
$$\sqrt{1+\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}}$$ exists.
So
$$A=\lim_{n\to \infty}\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}$$
But can use some math methods to find an approximation to this $A$ by hand?
and I guess maybe this is true:
$$1<A\le (\pi)^{\frac{1}{e}}?$$
By the way: we can prove $A$ is a transcendental number?
Thank you very much!
| Well, you can get very good approximations by computing:
$$\sqrt{a+\sqrt{a+\sqrt{a+...}}}=\frac{1+\sqrt{1+4a}}{2}$$
$$b_1=\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}...}}}=\frac{1+\sqrt{3}}{2}$$
$$b_2=\sqrt{\frac{1}{3}+\sqrt{\frac{1}{3}+\sqrt{\frac{1}{3}...}}}=\frac{1+\sqrt{7/3}}{2}$$
$$b_3=\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}...}}}=\frac{1+\sqrt{2}}{2}$$
Then you use the facts that:
$$A^2-1<b_1$$
$$(A^2-1)^2-\frac{1}{2}<b_2$$
And so on, to calculate upper bounds for $A$ with very good accuracy:
$$A<\sqrt{1+b_1}=1.53819$$
$$A<\sqrt{1+\sqrt{\frac{1}{2}+b_2}}=1.52580$$
$$A<\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+b_3}}}=1.52300$$
$$A<\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\sqrt{\frac{1}{4}+b_4}}}}=1.52224$$
And so on. This is already very close to the accurate expression. It's a rather easy calculation and can be done by hand if needed.
For the lower boundary you can use $b=1$ (it's rather obvious):
$$A>\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\sqrt{\frac{1}{4}+1}}}}=1.51844$$
| {
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proof that a function is decreasing I need to prove that $$\sqrt{3+x^{1/3}}+2\over x-1$$ is a decreasing function in $(1,\infty)$ with the definition of a decreasing function; the problem I have is that this expression is difficult so please I need your help
| $$\sqrt{3+x^{1/3}}+2\over x-1$$
Convince yourself that this function is decreasing on $(1,\infty)$ if and only if (with the substitution $x^3$ instead of $x$) the function
$$\sqrt{3+x}+2\over x^3-1$$
is decreasing. Convince yourself again that this is decreasing on $(1,\infty)$ if and only if (with the substitution $x^2+4x+1$ instead of $x$)
$${\sqrt{x^2+4x+4}+2\over (x^2+4x+1)^3-1} = \frac{1}{x((x^2+4x+1)^2+(x^2+4x+1)+1)}$$
is decreasing. But hey, the final form is obviously decreasing on $(1,\infty)$ (We have used $x^2+4x+4 = (x+2)^2$ and $x^3 - 1 = (x-1)(x^2+x+1)$ for the last step).
| {
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Having A Problem With Chinese Remainder Theorem
Find all solutions, if any, to the system of congruences
$$\begin{align}
x&\equiv 7 \pmod{9}\\
x&\equiv 4 \pmod{12}\\
x&\equiv 16 \pmod{21}
\end{align}$$
Solution: Using the Chinese Remainder theorem, we get that this system is equivalent to the 5 equations:
$$\begin{align}
x&\equiv 7 \pmod{9} \\
x&\equiv 0 \pmod{4} \\
x&\equiv 1 \pmod{3} \\
x&\equiv 2 \pmod{7} \\
x&\equiv 1 \pmod{3} \\
\end{align}$$
The 3rd and 5th equations are superfluous, and the total system has general solution $x\equiv16 \pmod {252}$."
I can't seem to get 16 all I get is 64, why?
I do it like this
$$\begin{align}
x&\equiv a_1M_1y_1+a_2M_2y_2+a_3M_3y_3 \pmod{16}\\
&\equiv 7\cdot28\cdot4 + 0 + 2\cdot36\cdot4 \pmod{16}\\
&\equiv 1072\pmod{16} \\
&\equiv 64 \pmod{16}
\end{align}$$
| I do small ones of these by hand. From the first, we have $x=7+9k$ Taking that mod $4$, we get $x=1+3k \pmod 4$ and we want this to be $\equiv 0$. By inspection, $k=1$ works and we have $x=16 \pmod {36}$. Then $16 \equiv 2 \pmod 7$ and we are done, arriving at the solution $16 \pmod {252}$
You should be working mod $9 \cdot 4 \cdot 7=252,$ not mod $16$. Note that $64 \equiv 0 \pmod {16}$ Your $M_2$ should not be zero, it should be $1 \pmod 4$ and $0 \pmod {9,7}$, which is $-63 \pmod {252}$ Your $M_3$ should be $36 \pmod {252}$ I don't know what your $y_1,y_2,y_3$ are. That multiplication by $4$ is the source of your problem. Then you have the solution is $x\equiv 7\cdot 28+0\cdot (-63)+2\cdot 36=268\equiv 16 \pmod {252}$
| {
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About the Special Solutions Method in general, for solving Ax = 0 [GStrang, P140 3.2.1]
● MIT Lec 7 Course Notes: Letting a different free variable equal 1 and setting the other free variables equal to zero gives us other vectors in the nullspace.
● P133: The nullspace consists of all combinations of the special solutions.
● 38:25 of Lecture 7 : "If I set the free variable to 0 and solve for the pivot variables, I'll get all 0s. No progress." - Prof Strang
I accept the following general method for finding the nullspace and will exemplify with Ex 3.2.1:
$A = \begin{bmatrix}
1 & 2 & 2 & 4 & 6 \\
1 & 2 & 3 & 6 & 9 \\
0 & 0 & 1 & 2 & 3
\end{bmatrix} \implies RREF(A) = \begin{bmatrix}
1 & 2 & 0 & 0 & 0 \\
& & 1 & 2 & 3 \\
& & \mathbf{0} & &
\end{bmatrix}$ Thus, $\mathbf{Ax = 0} \implies$
$x_1 = -2a_2 \\
x_2 = a_2 \\
x_3 = -2a_4 -3a_5 \\
x_4 = a_4 \\
x_5 = a_5
$ $\implies \mathbf{x} = a_2\begin{bmatrix}
-2 \\
1 \\
0 \\
0\\
0 \\
\end{bmatrix} + a_4\begin{bmatrix}
0 \\
0 \\
-2 \\
1 \\
0 \\
\end{bmatrix} +
a_5\begin{bmatrix}
0 \\
0 \\
-3 \\
0 \\
1 \\
\end{bmatrix}. $
$1.$ I can't pinpoint why, but I'm tentative about this method (in the grey box above): for each free variable (there're 3 here), this method sets $1$ for it and sets the other free variables to $0$. How and why does each free variable have one of these special solns? How and why does this function?
$2.$ Since $a_2, a_4, a_5$ are free, each can be any scalar. Say I select $a_2 = a_4 = a_5 = 1$.
Then $x = (-2, 1, -5, 1, 1)$, which is one vector. But it's wrong to infer that $ \ker(A) = \{$ all scalar multiples of this one vector $\}$. How and Why?
This question precedes rank, REF, $\mathbf{Ax = b}$, linear independence, span, basis, dimension, dimensions/theorems of the 4 subspaces, Orthogonality, Determinants, eigenvalues and eigenvectors, and linear transformations. Please omit them from answers.
| From $\begin{pmatrix}1&2&0&0&0\\ 0&0&1&2&3\end{pmatrix}$ no further calculation is required. First fill up the matrix in this way:
$$\begin{pmatrix}1&2&0&0&0\\
\color{red}0&\color{red}{-1}&\color{red}0&\color{red}0&\color{red}0\\
0&0&1&2&3\\
\color{red}0&\color{red}0&\color{red}0&\color{red}{-1}&\color{red}0\\
\color{red}0&\color{red}0&\color{red}0&\color{red}0&\color{red}{-1}
\end{pmatrix}.$$
Sharp eyes provided you'll see that
$$\left\{\begin{pmatrix}\color{green}2\\ \color{green}{-1}\\ \color{green}0\\ \color{green}0\\ \color{green}0\end{pmatrix},\begin{pmatrix}\color{green}0\\ \color{green}0\\ \color{green}2\\ \color{green}{-1}\\ \color{green}0\end{pmatrix},\begin{pmatrix}\color{green}0\\ \color{green}0\\ \color{green}3\\ \color{green}0\\ \color{green}{-1}\end{pmatrix}\right\}$$
is a basis for the kernel of $A$.
| {
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Prove that \frac{2n}{3n+7} converges (Question about my method of proof) Show that the sequence $<\frac{2n}{3n+7}>$ converges. Find its limit.
This is the work I have currently:
$\displaystyle \lim_{n \to \infty} \frac{2n}{3n+7} = \frac{2}{3}$
Let $\epsilon > 0$, and observe that:
\begin{align*}
\left|\frac{2n}{3n+7} - \frac{2}{3}\right| &=
\left|\frac{3(2n) - 2(3n + 7)}{3(3n + 7)}\right|\\
&= \left|\frac{-14}{9n+21}\right|\\
&= \frac{14}{9n+21}
\end{align*}
We want $\displaystyle \frac{14}{9n+21} < \epsilon$. Solving for $\epsilon$ gives us:
\begin{align*}
\displaystyle \frac{14}{9n+21} &< \epsilon\\
14 &< \epsilon(9n + 21)\\
\frac{14}{\epsilon} &< 9n + 21\\
\frac{14}{\epsilon} - 21 &< 9n\\
\frac{14}{9 \cdot \epsilon} - \frac{7}{3} &< n
\end{align*}
So, choose $N$ such that $N = \displaystyle \frac{14}{9 \cdot \epsilon} - \frac{7}{3}$. Then for all $n > N$, we have $\displaystyle \left|\frac{2n}{3n+7} - \frac{2}{3}\right| < \epsilon$.
This is the result I came up with for my homework question. I was wondering if when proving convergence the "scratch work" for finding $N$ is included in the proof. (If there is anything else wrong with my proof please let me know)
| If you just take $$N=\left\lceil\frac{14}{9\epsilon}-\frac73\right\rceil$$ so that $N$ is an integer, then you'll be set. (It actually doesn't matter whether $N$ is an integer, but some texts do specify that $N$ is an integer in their definitions.)
It isn't necessarily required that the "scratch work" be shown in a proof. You can always just pull it out of your hat, and say that if $n>N,$ then $n>\frac{14}{9\epsilon}-\frac73,$ so $9n>\frac{14}\epsilon-21,$ so....
Basically, you'll just be doing the scratchwork in reverse.
| {
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Using the Maclaurin series of $\arctan(x)$ to evaluate $\int_{0}^{\frac{\pi}{2}} \arctan( a \sin x) \, \mathrm dx$ I want to use the Maclaurin series of $\arctan (x)$ to show that
\begin{align}\int_{0}^{\pi/ 2} \arctan (a \sin x) \, \mathrm dx &= 2 \sum_{k=0}^{\infty} \frac{\left(\frac{\sqrt{1 + a^{2}}- 1}{a}\right)^{2k+1}}{(2k+1)^{2}} \\ &= \operatorname{Li}_{2} \left(\frac{\sqrt{1+a^{2}}-1}{a} \right) - \operatorname{Li}_{2} \left(\frac{1-\sqrt{1+a^{2}}}{a} \right).
\end{align}
I guess we should first impose the restriction $ |a| \le 1$.
Then we have
$$ \begin{align} \int_{0}^{\pi/ 2} \arctan (a \sin x) \ dx &= \int_{0}^{\pi /2} \sum_{k=0}^{\infty} (-1)^{k} \, \frac{(a \sin x)^{2k+1}}{2k+1} \, \mathrm dx \\ &= \sum_{k=0}^{\infty} (-1)^{k} \, \frac{a^{2k+1}}{2k+1} \int_{0}^{\pi /2} \sin^{2k+1} (x) \, \mathrm dx \\ &= \sum_{k=0}^{\infty} (-1)^{k} \frac{a^{2k+1}}{(2k+1)^{2}} \frac{2^{2k}}{\binom{2k}{k}}. \end{align}$$
But how do we proceed from here?
| Here is an idea: We prove the following identity instead.
$$ \int_{0}^{\frac{\pi}{2}} \arctan\left(\frac{2a\sin x}{1-a^{2}} \right) \, dx = 2 \sum_{n=0}^{\infty} \frac{a^{2n+1}}{(2n+1)^{2}}. $$
Expanding left-hand side,
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \arctan\left(\frac{2a\sin x}{1-a^{2}} \right) \, dx
&= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{\frac{\pi}{2}} \left(\frac{2a\sin x}{1-a^{2}} \right)^{2n+1} \, dx \\
&= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} (2a)^{2n+1} \int_{0}^{\frac{\pi}{2}} \frac{\sin^{2n+1}x}{(1-a^{2})^{2n+1}} \, dx \\
&= \sum_{n=0}^{\infty} \sum_{j=0}^{\infty} \left( \frac{(-1)^{n}2^{2n+1} }{2n+1} \binom{2n+j}{j} \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}x \, dx\right) a^{2j+2n+1}.
\end{align*}
Plugging $m = n+j$, it follows that
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \arctan\left(\frac{2a\sin x}{1-a^{2}} \right) \, dx
&= \sum_{m=0}^{\infty} \left( \sum_{n=0}^{m} \frac{(-1)^{n}2^{2n+1} }{2n+1} \binom{m+n}{m-n} \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}x \, dx\right) a^{2m+1}.
\end{align*}
Thus plugging the Wallis formula, the problem reduces to showing that
$$ \sum_{n=0}^{m} (-1)^{n} \frac{(m+n)!}{(m-n)!} \left( \frac{n!2^{2n}}{(2n+1)!} \right)^{2} = \frac{1}{(2m+1)^{2}}. $$
But I have no idea how to prove this.
| {
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Partial fraction when $N^r$ and $D^r$ are quadratic and cubic polynomials I need to find the nth derivative of the following function
$$y=\frac {x^2+4x+1}{x^3+2x^2-x-2}$$
The trouble is I don't know how to break a fraction like the above one. How do I break it into partial fractions? Or is there any other way to calculate it's nth derivative(leibnitz theorem?) without breaking it into partial fractions? Thanks in advance.
| As $x^3+2x^2-x-2=x^2(x+2)-1(x+2)=(x+2)(x^2-1)=(x+2)(x-1)(x+1)$
and the order of the numerator is less than that of denominator
Using Partial Fraction Decomposition,
$$\frac {x^2+4x+1}{x^3+2x^2-x-2}=\frac A{x+2}+\frac B{x-1} +\frac C{x+1}$$ where $A,B,C$ are arbitrary constants
Multiply either sides by $x^3+2x^2-x-2$ and compare the constants and the coefficients of the different powers of $x$ to determine $A,B,C$
| {
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Find $(a,b)$ such that in $x^2+ax+b$, whenever $v$ is a root, then $v^2 - 2$ is also a root Find $(a,b)$ such that in $x^2+ax+b$, whenever $v$ is a root, then $v^2 - 2$ is also a root
$a,b$ are real numbers. Roots may or may not be real.
In this question, the aim is to find values of and b ,eg. 2,4.
| In the question as now formulated ("whenever $v$ is a root, then $v^2−2$ is also a root"), the quadratic polynomial serves only to define its set $S$ of roots (one or two of them), so one might as well reason for$~S$. It must be closed under the map $x\mapsto x^2-2$; this can be achieved in several ways.
*
*$S=\{r\}$ where $r^2-2=r$. Then either $r=-1$ or $r=2$, and since $x+ax+b=(x-r)^2$ in this case, one gets $(a,b)=(2,1)$ respectively $(a,b)=(-4,4)$.
*$S=\{r,s\}$ where $r^2-2=r=s^2-2$ and $r\neq s$. Now $r$ has the same two possibilities as above, and $s=-r$. Since $x+ax+b=(x-r)(x-s)=x^2-r^2$ in this case, one gets $(a,b)=(0,1)$ respectively $(a,b)=(0,4)$.
*$S=\{r,s\}$ where $r^2-2=s\neq r=s^2-2$. Now $r$ has to satisfy $r=(r^2-2)^2-2$, which gives the $4$th degree equation $r^4-4r^2-r+1=0$. This looks harder, but we know that the two solutions from 1. will give (unwanted) solutions to this equation, so we can divide $x^4-4x^2-x+1$ without remainder by the polynomial $x^2-x-2$ from 1., giving as quotient the polynomial $x^2+x-1$ for which $r$ has to be a root. Since the same argument holds for $s$, it is clear that $s$ must be the other root, and since $x^2+ax+b=(x-r)(x-s)=x^2+x-1$ in this case one finds a single solution $(a,b)=(1,-1)$ here, even without solving the quadratic equation.
Thus one finds exactly $5$ possible pairs for $(a,b)$, namely $\{(2,1),(-4,4),(0,1),(0,4),(1,-1)\}$.
(Solving the final quadratic in 3., while unnecessary, is not hard: $r=-\frac12-\frac{\sqrt5}2$ and $s=-\frac12+\frac{\sqrt5}2$ is one solution, the negative golden ratio and its conjugate; one can also find the two interchanged.)
| {
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Calculate $\lim_{n\to\infty}\sqrt{n^2+n} - n$. Calculate $\lim_{n\to\infty}\hspace{2 pt}a_n$, where $a_n = \sqrt{n^2+n} - n$ is a sequence of complex numbers.
I performed the ratio test:
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty}\frac{\sqrt{(n+1)^2 + (n+1)} - (n+1)}{\sqrt{n^2+n} - n} = \lim_{n\to\infty}\frac{\sqrt{n^2+3n+2} - (n+1)}{\sqrt{n^2+n} - n}$$
which ultimately leads to, when multiplied by $\frac{\frac{1}{n}}{\frac{1}{n}}$,
$$\lim_{n\to\infty}\frac{\sqrt{1+\frac{3}{n}+\frac{2}{n^2}} - (1+\frac{1}{n})}{\sqrt{1+\frac{1}{n}} - 1}$$
Should I multiply by the complex conjugate? It leads to a rather messy expression. I am stuck.
How can one perform the root test here? I attempted: $$\lim_{n\to\infty}\sqrt[n]{\sqrt{n^2+n} - n} \Longleftrightarrow \lim_{n\to\infty}((n^2 + n)^{\frac{1}{2}} - n)^{\frac{1}{n}}$$
I have also no idea how to proceed.
| $$a_n =\sqrt{n^2+n}-n = \frac{(n^2+n)-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1} \Rightarrow \lim_{n \to \infty} a_n = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/587457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluating Definite Integral $\int_1^2\arcsin\left(\frac{4-3\sqrt{x^2-1}}{5x}\right)dx$ How can I prove that
$$\int_1^2\arcsin\left(\frac{4-3\sqrt{x^2-1}}{5x}\right)dx=\frac{\pi}{3}-2\text{arcsec}(\sqrt{5})+\log(2+\sqrt{3})$$
Can someone help me, please?
| Let $\arcsin\left(\dfrac{4-3\sqrt{x^2-1}}{5x}\right) = t$. We then have
$$\dfrac{4-3\sqrt{x^2-1}}{5x} = \sin(t)\implies(4-5x \sin(t))^2 = 9(x^2-1)$$
This gives us
$$25x^2 \sin^2(t)-9x^2 - 40x \sin(t) + 25 = 0 \implies (4x \sin(t)-5)^2 = (3x \cos(t))^2$$
Hence, now let us set
$$x = \dfrac5{3\cos(t) + 4 \sin(t)} = \sec\left(t+\phi\right)$$
where $\cos(\phi) = \dfrac35$ and $\sin(\phi) = -\dfrac45$.
Hence,
$$\int tdx = tx - \int xdt$$
I trust you can now plug in the appropriate limits for $t$ and obtain the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Showing $x^4-24x^2+4$ is irreducible over $\mathbb{Q}$
I want to show that $x^4-24x^2+4$ is irreducible over $\mathbb{Q}$.
I have tried Eisenstein's criterion using $p=2$ and substituting $x+1$ and $x-1$ but it won't work. Then what I tried was using the rational root test. I got as my possible rational roots $\pm 1, \pm 2, \pm4$. But each one does not work. From here I can consider $x^4-24x^2+4$ over $\mathbb{Z}$. So suppose that $x^4-24x^2+4$ is reducible over $\mathbb{Z}$. Then $$x^4-24x^2+4= (x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+ac+d)x^2+(bc+ad)x+bd.$$ By matching up the coefficients we get $$a+c=0$$ $$b+ac+d=-24$$ $$bd=4$$ $$bc+ad=0$$
Then we see $a+c=0\implies a=-c$. Then $$bc+ad=0\implies bc-cd=0 \implies bc=cd\implies b=d.$$ Since $bd=4$ then $b^2=4\implies b=\pm 2$ and $d=\pm 2$. We get two cases: $b=d=2$ and $b=d=-2$. In the first case $b=d=2$ we get $b+ac+d=-24\implies 4-c^2=-24\implies c^2=28$ (I substituted $a=-c$ into $b+ac+d=-24$), a contradiction since $c$ is an integer. Similarly in the case $b=d=-2$ we have $ b+ac+d=-24\implies -4-c^2=-24\implies c^2=20$. Again a contradiction. Would this be correct?
| Note that you have a monic even polynomial (it is invariant under substituting $x:=-x$) of degree $4$. You can therefore consider two cases for potential factors of the polynomial: factors that are themselves even, and factors that are not; the latter come in pairs whose members are interchanged by the substitution $x:=-x$ (which flips the signs of the odd degree terms). Any non-trivial even proper factor must be of the form $x^2-r$ where $r$ satisfies $r^2-24r+4=0$, but the discriminant $560=16*35$ of that polynomial is not a square in$~\Bbb Z$, so there are no such integer$~r$ (this is the case that the initial answer by Praphulla Koushik checked).
So you are left to check a potential factorisation $x^4-24x^2+4=(x^2+ax+b)(x^2-ax+b)$. This gives equations $b^2=4$ (from the constant term) and $a^2=2b+24$ (from the coefficient of $x^2$). It is easily checked that there are no solutions $a,b\in\Bbb Z$ to this system (neither $24+4$ nor $24-4$ is a square).
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $x$ such that $\sqrt{x+\sqrt{x+7}}\in \mathbb{N}$ Find $x$ such that $$\sqrt{x+\sqrt{x+7}}\in \mathbb{N}$$
I tried many ways: $$\sqrt{x+\sqrt{x+7}}=n$$
$$\sqrt{x+\sqrt{x+7}}^2=n^2$$
$$x+\sqrt{x+7}=n^2$$
then solve for $x$ but didn't do with success.
I think this is the most difficult problem in my lifetime
Also $x$ must be made
of $2$ digit.
Thanks to everybody for helping me understand this problem and its solution!
| We want to find $a \in \mathbb{N}$ such that
$$a = \sqrt{x + \sqrt{x + 7}}.$$
Also implicit in the question was that $x \in \mathbb{N}$, but that wasn't stated, only implied by the statement that $x$ is two-digit.
Let $n = \sqrt{x + 7}$. Then $x = n^2 - 7$. Substituting into the original equation:
$$\begin{align}
a = & \sqrt{n^2 + n - 7}\\
0 = & n^2 + n - 7 - a^2
\end{align}$$
Solving for $n$:
$$n = -\frac{1}{2} \pm \frac{1}{2}\sqrt{4 a^2 + 29}$$
Plugging any $a > 1$ into this equation will yield a valid value for $n$ and therefore $x$, but they will generally be non-integer.
For $n$ and therefore $x$ to be integer, $4 a^2 + 29$ must be a perfect square, and furthermore an odd perfect square. We can rewrite this as $(2a)^2 + 29 = b^2$, where $a, b \in \mathbb{N}$, which means two perfect squares must have a difference of 29, and this only occurs for $(2a)^2 = 14^2$ and $b^2 = 15^2$. Therefore, solving backwards, $a = 7$, $n = 7$, and $x = 42$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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To show inverse of tan x
It quite confuses me. Where do I start?
Please help.
| In fact, we can use this to find,
$$2\tan^{-1}x=\begin{cases} \tan^{-1}\frac{2x}{1-x^2} &\mbox{if } x^2<1\iff -1\le x\le1 \\\pi+ \tan^{-1}\frac{2x}{1-x^2} & \mbox{ else where } \end{cases} $$
If we set $\displaystyle \frac{2x}{1-x^2}=y, x^2y+2x-y=0\ \ \ \ (1)$
$\displaystyle\implies x=\frac{-1\pm\sqrt{1+y^2}}y$
Taking the '-' sign, $\displaystyle x=-\frac{\sqrt{1+y^2}+1}y,$ observe that either $\displaystyle x<-1$ or $\displaystyle x>1$
Taking the '+' sign, $\displaystyle x=\frac{\sqrt{1+y^2}-1}y=\frac y{\sqrt{1+y^2}+1},$ observe that $\displaystyle-1\le x\le 1$
One Observation:
If we set $\displaystyle x=\tan\phi, y=\frac{2x}{1-x^2}=\tan2\phi$
If $x_1,x_2$ be the two roots of $(1),$
$\displaystyle x_1\cdot x_2=\frac{-y}y=-1$ (assuming $y\ne0$) and $\displaystyle x_1+x_2=-\frac2y=-\frac2{\tan2\phi}=-2\cot2\phi$
$\displaystyle x_1=\tan\phi,x_2=-\frac1{x_1}=-\cot\phi$ and $\displaystyle x_1+x_2=\tan\phi-\cot\phi=-2\cdot\frac{1-\tan^2\phi}{2\tan\phi}=-\frac2{\tan2\phi}=-2\cot2\phi$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $x^4+y^4$ and $x^3+y^3$ if $x+y=2$ and $x^2+y^2=8$ Find $x^4+y^4$ if $x+y=2$ and $x^2+y^2=8$
So i started the problem by nothing that $x^2+y^2=(x+y)^2 - 2xy$ but that doesn't help!
I also seen that $x+y=2^1$ and $x^2+y^2=2^3$ so maybe $x^3+y^3=2^5$ and $x^4+y^4=2^7$ but i think this is just coincidence
So how can i solve this problem?
PLEASE i need some help and thanks for all!!
| \begin{eqnarray}
x^4+y^4&=&(x^2+y^2)^2-2x^2y^2=(x^2+y^2)^2-\frac12(2xy)^2=(x^2+y^2)^2-\frac12\left[(x+y)^2-(x^2+y^2)\right]^2\\
&=&8^2-\frac12(2^2-8)^2=56.
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Series, divergent $\sum_{n=1}^\infty\frac{n^4-3n}{n^5+n}$
I want to show, that this series is divergent, but i can not find a divergent minorant...
I tried
$\sum_{n=1}^\infty\frac{n^4-3n}{n^5+n}=\sum_{n=1}^\infty\frac{n^3-3}{n^4+1}\geq\sum_{n=1}^\infty\frac{n^3}{n^4}=\sum_{n=1}^\infty\frac{1}{n}$
But that is wrong, i think. I have to find a minorant and not a majorant.
If someone could help me, that would be great.
Thanks.
| Alternately, You can use the
Limit Comparison Test
$$\sum_{n=n_0}^{\infty} a_n, \sum_{n=n_0}^{\infty} b_n, \lim_{n\to\infty} \frac{a_n}{b_n} \ne 0$$
With $a_n$ and $b_n$ series with positive terms;
If $\sum_{n=n_0}^{\infty} b_n$ converges/diverges, $\sum_{n=n_0}^{\infty} a_n$ converges/diverges respectively.
$$a_n = \frac{n^4 - 3n}{n^5 + n} = \frac{n^4}{n^5} \cdot \frac{1 - 3n/n^4}{1 + n/n^5} \approx \frac{1}{n}$$
($\frac{3n}{n^4}$ and $\frac{n}{n^5}$ $\to 0$ for $n$ very, very large)
Take $b_n = \frac{1}{n}$
$$\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{n^4 - 3n}{n^5 + n} \cdot n = \lim_{n\to\infty} \frac{n^5 - 3n^2}{n^5 + n} = \lim_{n\to\infty} \frac{n^5/n^5 - 3n/n^5}{n^5/n^5 + n/n^5} = \frac{1+0}{1+0} = 1 \ne 0$$
$\sum \frac{1}{n}$ diverges since its a p-series with $p = 1$.
Since all the criteria is met and $\sum b_n (\frac{1}{n})$ diverges, $\sum \frac{n^4 - 3n}{n^5 + n}$ diverges.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving system of linear congruences (3 pairs) So we have the following:
$$2x \equiv 3\pmod {5} \\ 3x \equiv 4\pmod {7} \\ 5x \equiv 7\pmod {11}$$
which reduces to:
$$x \equiv 4\pmod {5} \\ x \equiv 6\pmod {7} \\ x \equiv 8\pmod {11}$$
Now the confusion begins here. At this point, I choose the first two pairs of congruences and equate them, giving:
$$ 5k+4= 7l +6 \\ \\$$
But I'm not sure what to do past this point. I know in essence I need to solve this and pair this new equation with the last one and re-do the steps. It's just past this point I don't know how to solve.
| For solving the system of linear congruence, we need to use the Chinese Remainder Theorem
$$
\left\{\begin{array}{ll}
2 x \equiv 3 & (\bmod 5) \\
3 x \equiv 4 & (\bmod 7) \\
5 x \equiv 7 & (\bmod 11)
\end{array}\right.
\Leftrightarrow \left\{\begin{array}{ll}
x \equiv 4 & (\bmod 5) \\
x \equiv 6 & (\bmod 7) \\
x \equiv 8 & (\bmod 11)
\end{array}\right. $$
First of all, I am going to solve the following system of linear congruence for $A, B$ and $C$
$$
\left\{\begin{array}{ll}
7 \times 11 A \equiv 4 & (\bmod 5) \\
5 \times 11B \equiv 6 & (\bmod 7) \\
5 \times 7 C \equiv 8 & (\bmod 11)
\end{array}\right. \Leftrightarrow \left\{\begin{array}{ll}
77 A \equiv 4 & (\bmod 5) \\
55 B \equiv 6 & (\bmod 7) \\
35 C \equiv 8 & (\bmod 11)
\end{array}\right. $$
$$\Leftrightarrow\left\{\begin{array}{cc}
2 A \equiv 4 & (\bmod 5) \\
-B \equiv 6 & (\bmod 7) \\
2 C=8 & (\bmod 11)
\end{array}\right. \Leftrightarrow \left\{\begin{array}{ll}
A \equiv 2 & (\bmod 5) \\
B \equiv 1 & (\bmod 7) \\
C \equiv 4 & (\bmod 11)
\end{array}\right. $$
Using the Chinese Remainder Theorem, the integer solutions for the system are
$$385k+77\times 2+55\times 1+35\times 444$$
$\text{i.e. }\boxed{385k+349,}\tag*{} $
where $k\in \mathbb Z$.
:|D Wish you enjoy my solution! Your suggestions, comments and alternate solutions are warmly welcome!
| {
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Series $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ I want to check, whether $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ converges or diverges.
I tried to use the Ratio test:
$|\frac{a_{n+1}}{a_n}|= \frac{n+2}{2^{n+1}} \frac{2^n}{n+1} = \frac{1}{2} \frac{n+1+1}{n+1} = \frac{1}{2} (1+ \frac{1}{n+1})$
$\lim\limits_{n \rightarrow \infty}{{(\frac{1}{2}(1+ \frac{1}{n+1})) = \frac{1}{2}}} \leq 1$
So $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ converges.
Could somebody please check my solution?
| The series converges, it's can be simply proved finding it's upper bound or as you've done it by the ratio test. But if you are interested in finding it's actual value here's a way to do it. From the sum of geometric series we have:
$$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$$
Multiply both sides by $r$
$$\sum_{n=0}^{\infty} r^{n+1} = \frac{r}{1-r}$$
Differentiate with respect to $r$. Then we have:
$$\sum_{n=0}^{\infty} (n+1)r^n = \frac{1}{(r-1)^2}$$
Now substitute $r= \frac 12$. We have:
$$\sum_{n=0}^{\infty} \frac{n+1}{2^n} = \frac{1}{(-\frac 12)^2} = 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/595981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Solve$(log_{2}(x+1))^2=4$ $$(log_{2}(x+1))^2=4$$
$$log_{2}(x+1)*log_{2}(x+1)=log_{2}16$$
$$x^{2}+2x-15=0$$
$$(x+1)*(x+1)=16$$
$$x^{2}+2x+1=16$$
$$x^{2}+2x-15=0$$
$$(x+5)(x-3)=0$$
$$x_1=-5; x_2=3$$
The solution is only $x_1=3$. Is this correct?
| Your solution is incorrect. First note that
$$\log_2\left(a^2 \right) \neq \left(\log_2(a) \right)^2$$
In your case, we have
$$\left(\log_2(x+1) \right)^2 = 4 \implies \log_2(x+1) = \pm2 \implies x+1 = 2^{\pm2} \implies x+1 = 4 \text{ or }x+1= \dfrac14$$
Hence,
$$x=3 \text{ or }x=-\dfrac34$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Trigonometric equation: $2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0$ I'm new here, but I need your help so much to solve an equation:
$$2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0$$
I tried a lot like making $2[(\sin^2 x)^3 + (\cos^2 x)^3$
| As $\sin^2x+\cos^2x=1$
Taking square, $\displaystyle1=(\sin^2x+\cos^2x)^2=\sin^4x+\cos^4x+2\sin^2x\cos^2x$
$\displaystyle\implies \sin^4x+\cos^4x=1-2(\sin^2x\cos^2x)\ \ \ \ (1)$
Taking cube, $\displaystyle1=(\sin^2x+\cos^2x)^3=\sin^6x+\cos^6x+3\sin^2x\cos^2x(\sin^2x+\cos^2x)$
$\displaystyle\implies \sin^6x+\cos^6x=1-3(\sin^2x\cos^2x)\ \ \ \ (2)$
Equate the values of $\displaystyle\sin^2x\cos^2x$ from $(1),(2)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\lim_{x \rightarrow 0}\frac{9}{x}\bigl(\frac{3}{(x+3)^3}-\frac{1}{9}\bigr)$ I have to determine the following:
$$\lim_{x \rightarrow 0}\frac{9}{x}\left(\frac{3}{(x+3)^3}-\frac{1}{9}\right)$$
I've got so far:
$$\lim_{x \rightarrow 0}\frac{9}{x}\left(\frac{3}{(x+3)^3}-\frac{1}{9}\right)= \lim_{x \rightarrow 0}\left(\frac{27}{x(x+3)^3}-\frac{1}{x}\right)=\lim_{x \rightarrow 0} \left(\frac{27-(x+3)^3}{x(x+3)^3}\right)=\cdots$$
How to go on? I've got $\frac{\infty}{0}...$
| Just develop the nuumerator and you'll get
$$
\lim_{x \rightarrow 0} \frac{-x^3 - 9x^2 - 27x}{x(x+3)^3}
=\lim_{x \rightarrow 0} \frac{-x^2 - 9x - 27}{(x +3)^3} = -1
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$\displaystyle \lim_{x \to 0}\frac{\ln(1+5x)}{x}$ So have I have to find this limit:
$$\lim_{x \to 0}\frac{\ln(1+5x)}{x}$$
With substitution of $c$ into the function, I get $\lim = \frac{0}{0}$ which is undefined. If this were, say, a polynomial, I'd try to factor it out, but in this case I just don't see a strategy here.
How should I go about this? Any links for extra reading would be especially handy.
Restriction: I can't use the L'Hospital rule.
| $ ln(1+x)= x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+...$
Which for your case leads to
$ ln(1+5x)= 5x-\dfrac{25x^2}{2}+\dfrac{125x^3}{3}-\dfrac{625x^4}{4}+...$
Now,
$ \lim_{x\rightarrow 0}\dfrac{ln(1+5x)}{x}= \dfrac{5x-\dfrac{25x^2}{2}+\dfrac{125x^3}{3}-\dfrac{625x^4}{4}+...}{x}$
$\lim_{x\rightarrow 0}\dfrac{ln(1+5x)}{x}= \dfrac{5x}{x}-\dfrac{25x^2}{2x}+\dfrac{125x^3}{3x}-\dfrac{625x^4}{4x}+...$
$=\lim_{x\rightarrow 0} 5-\dfrac{25x}{2}+\dfrac{125x^2}{3}-\dfrac{625x^3}{4}+...$
$=5$
Edit: how does the series form come about, one (loose) way.
$ \dfrac{d\ ln(1+x)}{dx}=\dfrac{1}{1+x}$
Expanding it in a series form.
$ \dfrac{d\ ln(1+x)}{dx}=\dfrac{1}{1+x} = 1-x+x^2-x^3 ..$
Now to get
$ ln(1+x)$ from $\dfrac{d\ ln(1+x)}{dx}$
, integrate the series term wise.
$\int 1-x+x^2-x^3 ..) dx = x - \dfrac{x^2}{2} + \dfrac{x^3}{3}-\dfrac{x^4}{4}...$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Expansion of polynomial raised to high power Is there an easy way to expand something like (x + x^2 + x^3)^6 ?
Thanks in advance!
| My approach avoids multinomials:
Let's play with it a bit:
$$\begin{align}
(x+x^2+x^3)^6 &= x^6(1+x+x^2)^6\\
&= x^6\frac{((1-x)(1+x+x^2))^6}{(1-x)^6}\\
&= \frac{x^6(1-x^3)^6}{(1-x)^6} \\
\end{align}$$
The numerator and denominator can now be expanded with the binomial theorem, and then long division will simplify the expression. (This is the way I would do it.)
Another approach:
$$\begin{align}
(x+x^2+x^3)^6 &= (x+(x^2+x^3))^6 \\
&= x^6+6x^5(x^2+x^3) + 15x^4(x^2+x^3)^2 \\
& +\, 20x^3(x^2+x^3)^3 + 15x^2(x^2+x^3)^4 + 6x(x^2+x^3)^5 + (x^2+x^3)^6
\end{align}$$
Now, each of the binomials can be expanded as above.
| {
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negative exponent problem $$\sqrt{\frac{1}{3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}}}$$
Does this equal =
$$
\begin{align*}
& \sqrt{3^0 + 3^1 + 3^2 + 3^3 + 3^4} \\
=&\sqrt{1 + 3 + 9 + 27 + 81} \\
=&\sqrt{121} \\
=&11.
\end{align*}
$$
The answer is apparently $\frac{9}{11}$ and I'm not sure what rule of negative exponents I got wrong.
The rule I'm using, incorrectly, is this:
$$\frac{1}{3^{-2}} = 3^2 = 9.$$
| $$\sqrt{\frac1{3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}}}=\sqrt{\frac{3^4}{3^4(3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4})}}$$
$$=\sqrt{\frac{3^4}{3^4+3^3+3^2+3^1+3^0}}=\sqrt{\frac{81}{121}}=\frac9{11}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Algebra Inequality Proof Let $a, b, c, d$ be positive numbers. Prove that $\frac{a+b+c+d}{4} \ge \sqrt{\frac{ab+ac+ad+bc+bd+cd}{6}}$.
I was told to rewrite the sum on the right side in terms of $a^2 + b^2 + c^2 + d^2$ and $a + b + c + d$ but I am unsure how to combine the terms.
| another approach is: square both sides and rearrange, we have:
$\iff3(a^2+b^2+c^2+d^2)\ge 2(ab+ac+ad+bc+bd+cd)$
with $a^2+b^2+c^2\ge ab+bc+ac$ <1>
we also have: $a^2+b^2+d^2\ge ab+bd+ad$<2> and $a^2+c^2+d^2\ge ad+cd+ac$<3>
edit: $b^2+c^2+d^2\ge bc+bd+cd$ <4>
<1>+<2>+<3>+<4> $\implies 3(a^2+b^2+c^2+d^2)\ge 2(ab+ac+ad+bc+bd+cd)$
| {
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Complex number trigonometry problem Use $cos (n\theta)$ = $\frac{z^n +z^{-n}}{2}$ to express
$\cos \theta + \cos 3\theta + \cos5\theta + ... + \cos(2n-1)\theta$ as a geometric series in terms of z. Hence find this sum in terms of $\theta$.
I've tried everything in the world and still can't match that of the final answer. Could I pleas have a slight hint on the right path to follow.
Thanks
| $ cosθ+cos3θ+cos5θ+\cdot \cdot \cdot+cos(2n−1)θ$
=$ \dfrac{z+z^{-1}}{2} + \dfrac{z^3+z^{-3}}{2} + \dfrac{z^5+z^{-5}}{2} + \cdot \cdot \cdot + \dfrac{z^{(2n+1)}+z^{-(2n-1)}}{2} $ $$\\$$ = $ 2^{-1}(z+z^3+z^5+z^7+\cdot \cdot \cdot + z^{2n-1}) + 2^{-1}( z^{-1}+z^{-3}+z^{-5} +\cdot \cdot \cdot + z^{-(2n-1)}) $
Then apply the formula of geometric series .
| {
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What are the odds of getting heads 7 times in a row in 40 tries of flipping a coin? I know if you flip a coin $7$ times, the odds of getting $7$ heads in a row is $1$ in $2^7$ or $1$ in $128$.
But if you flip a coin $40$ times, what are the odds of getting $7$ heads in a row in those $40$ tries? I only want to know the first time there are $7$ heads in a row and not count duplicates. Thanks.
| Let's count the number of ways not to get $7$ heads in a row. We will put together atoms that consist of $0$ to $6$ heads followed by a tail. Any arrangement of heads and tails without $7$ heads in a row, appended with a tail, can be uniquely made up of a number of such atoms.
All arrangements of such atoms appear once somewhere in the sum
$$
\sum_{k=0}^\infty(x+x^2+x^3+x^4+x^5+x^6+x^7)^k
$$
where
$x$ represents $T$
$x^2$ represents $HT$
$x^3$ represents $HHT$
$\vdots$
$x^7$ represents $HHHHHHT$
For example, if we are looking for $HTTHHTTH$, append a $T$ and we get the term for $k=5$ where in the first factor, the $x^2$ ($HT$) was chosen, in the second factor, the $x$ ($T$) was chosen, then $x^3$ ($HHT$), then $x$ ($T$), then $x^2$ (HT), to get $HTTHHTTHT$. Note that the exponent of $x$ matches the number of tosses. To count the number of sequences of $40$ flips that do not contain $7$ consecutive heads, we look at the coefficient of $x^{41}$ in
$$
\begin{align}
\sum_{k=0}^\infty(x+x^2+x^3+x^4+x^5+x^6+x^7)^k
&=\frac1{1-x\frac{x^7-1}{x-1}}\\
&=\frac{1-x}{1-2x+x^8}
\end{align}
$$
The coefficient of $x^{41}$ is $955427104501$. There is a degree $8$ recursion to compute this without dividing polynomials: $c_n=2c_{n-1}-c_{n-8}$, where $c_n$ starts
$$
1,1,2,4,8,16,32,64,\dots
$$
The number of sequences of $40$ flips is $2^{40}$. Therefore, the probability of getting a sequence of $7$ heads in a row in $40$ flips is
$$
1-\frac{955427104501}{2^{40}}=0.131044110526
$$
| {
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How many Positive integer solutions does the equation $x + y + z + w = 15$ have? How many Positive integer solutions does the equation $x + y + z + w = 15$ have?
Attempt:
Let $x = m + 1, y = n + 1, z = o + 1, w = p + 1 $
Then, $ m + 1 + n + 1 + o + 1 + p + 1 = 15$
$ m + n + o + p = 11 $
| Here is my try.
Your equation is $x+y+z+w=(x+y)+(z+w)=15$. First we see $x+y$ and $z+w$ as two unknowns, that is $a+b=15$ and $a,b$ satisfy $2\leq a,b\leq13$. Easily, we can say that there are $12$ positive integer solutions for $a$ and $b$. Then we will see there are how many postive integer solutions for $x+y=a$ and $w+z=b$. We note the number of such solutions as $N(\cdot)$.
If $a=2$, then $b=13$. We see that $a=2=x+y$ has unique $1$ solutions for $x$ and $y$, that is $x=1$ and $y=1$. $b=13=z+w$ has $12$ solutions for $z$ and $w$, that is $z=1,2,\dots,12$ and $w=12,11,\dots,1$. Then there are $N(a=2)*N(b=13)=1*12=12$ solutions for $a=2$ and $b=13$.
Then we do like this, we can make a list of the 12 solutions for $a$ and $b$:
$$N(a=2)=1\Leftrightarrow N(b=13)=12\Leftrightarrow N(a=2,b=13)=1*12$$
$$N(a=3)=2\Leftrightarrow N(b=12)=11\Leftrightarrow N(a=3,b=12)=2*11$$
$$\vdots$$
$$N(a=13)=12\Leftrightarrow N(b=2)=1\Leftrightarrow N(a=3,b=12)=12*1$$
So the number of all the solutions is
$$N(x+y+z+w=15)=\sum_{n=1}^{12}n*(13-n)=364.$$
| {
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Find a direct way to calculate recursive elements (simple problem with matrices) I nearly solved this question, I just need a hand with the last part since it is a bit confusing.
We are given the recursive sequences $\{a_n\}$ and $\{b_n\}$ like this:
$a_1=1$, $b_1=2$
$a_n=a_{n-1}+b_{n-1}$
$b_n=3b_{n-1}-a_{n-1}$
We are asked to find a direct way to find $a_n$ and $b_n$
Here is what I did:
$\begin{pmatrix} a_n \\b_n \end{pmatrix}= \begin{pmatrix} a_{n-1}+b_{n-1} \\ 3b_{n-1}-a_{n-1} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 3 \end{pmatrix} \begin{pmatrix} a_{n-1} \\ b_{n-1} \end{pmatrix} = ... = \begin{pmatrix} 1 & 1 \\ -1 & 3 \end{pmatrix}^{n-1} \begin{pmatrix} a_1 \\ b_1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 3 \end{pmatrix}^{n-1} \begin{pmatrix} 1 \\ 2 \end{pmatrix}$
the problem is that the matrix $\begin{pmatrix} 1 & 1 \\ -1 & 3 \end{pmatrix}$ is not diagonlizable. it's jordan form however is $\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$, so overall, if $B_J$ is a jordan basis:
$(B_J\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}B_J^{-1})^{n-1}\begin{pmatrix} 1 \\ 2 \end{pmatrix} =\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}^{n-1}\begin{pmatrix} 1 \\2 \end{pmatrix} = \begin{pmatrix} a_n \\ b_n \end{pmatrix}$
but what now? How can we find $\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}^{n-1}$??
| Hint: Note that $\begin{pmatrix}2 & 1\\ 0 & 2 \end{pmatrix}=\begin{pmatrix}2 & 0\\ 0 & 2 \end{pmatrix}+\begin{pmatrix}0 & 1\\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix}2 & 0\\ 0 & 2 \end{pmatrix}\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}=\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}\begin{pmatrix}2 & 0\\ 0 & 2 \end{pmatrix}$, therefore you can use the binomial theorem.
Also note that $\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}^2=\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}$.
Alternatively compute the first few powers by hand, then conjecture (and prove by induction) that $$\forall k\in \mathbb N\left(\begin{pmatrix} 2 & 1\\0 & 2\end{pmatrix}^k=\begin{pmatrix} 2^k & k2^{k-1}\\0 & 2^k \end{pmatrix}\right).$$
| {
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If $abc=1$ and $a,b,c$ are positive real numbers, prove that ${1 \over a+b+1} + {1 \over b+c+1} + {1 \over c+a+1} \le 1$.
If $abc=1$ and $a,b,c$ are positive real numbers, prove that $${1 \over a+b+1} + {1 \over b+c+1} + {1 \over c+a+1} \le 1\,.$$
The whole problem is in the title. If you wanna hear what I've tried, well, I've tried multiplying both sides by 3 and then using the homogenic mean. $${3 \over a+b+1} \le \sqrt[3]{{1\over ab}} = \sqrt[3]{c}$$ By adding the inequalities I get $$ {3 \over a+b+1} + {3 \over b+c+1} + {3 \over c+a+1} \le \sqrt[3]a + \sqrt[3]b + \sqrt[3]c$$ And then if I proof that that is less or equal to 3, then I've solved the problem. But the thing is, it's not less or equal to 3 (obviously, because you can think of a situation like $a=354$, $b={1\over 354}$ and $c=1$. Then the sum is a lot bigger than 3).
So everything that I try doesn't work. I'd like to get some ideas. Thanks.
| This answer only assumes that $abc\geq 1$. Make the following substitution $$\sqrt[3]{a}=x,\sqrt[3]{b}=y,\sqrt[3]{c}=z$$ then we have $xyz\geq1$ and we have to prove the following inequality now
$$\frac{1}{1+x^3+y^3}+\frac{1}{1+y^3+z^3}+\frac{1}{1+z^3+x^3} \leq 1 $$
Clearly $$(x^3+y^3)=(x+y)(x^2-xy+y^2)\overset{\text{AM-GM}}{\geq}(x+y)xy$$
We have the following chain of inequalities
$$\frac{1}{1+x^3+y^3}+\frac{1}{1+y^3+z^3}+\frac{1}{1+z^3+x^3} \leq \frac{1}{1+xy(x+y)}+\frac{1}{1+xz(x+z)}+\frac{1}{1+yz(z+y)} \\ \leq \frac{1}{1+\frac{1}{z}(x+y)}+\frac{1}{1+\frac{1}{y}(x+z)}+\frac{1}{1+\frac{1}{x}(z+y)}=1$$
| {
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How can I represent this as a sum? I am solving probability, and I got the need to know what this sum is:
$ \frac{1}{2} \times \frac{1}{3} + \frac{1}{2}\times \frac{2}{3} \times \frac{1}{4} + \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{1}{5} + \cdots$
I tried much with no success. How can I approach such sums during an exam?
| In this case, it is useful to simplify somewhat. We get
$$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5} + \frac{1}{5\cdot 6}+\cdots.$$
This is
$$\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+ \left(\frac{1}{5}-\frac{1}{6}\right)+\cdots.$$
Note the telescoping (almost everything cancels).
Remark: An expression for the sum is $\sum_{1}^{\infty} \frac{k!}{(k+2)!}$.
Telescoping is an idea that comes up moderately often, so it is worth watching out for. On exams, the most common closed from things to look out for a geometric series, the expansion of $e^x$ for some $x$, and telescoping series.
| {
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Evaluating $\lim_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x}$ How can I evaluate $$\lim_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x}$$ without L'Hopital rule. Using L'Hopital rule, it evaluates to 2. Is there a way to do it without using L'Hopital?
| Multiply by the conjugate and use trig identities, factoring appropriately:
\begin{align*}
\lim_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x}
&= \lim_{x\to\frac{\pi}{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x} \cdot \frac{1 + \sqrt{2}\sin x}{1 + \sqrt{2}\sin x} \\
&= \lim_{x\to\frac{\pi}{4}}\frac{(1-\tan x)(1 + \sqrt{2}\sin x)}{1 - 2\sin^2 x} \\
&= \lim_{x\to\frac{\pi}{4}}\frac{(1-\frac{\sin x}{\cos x})(1 + \sqrt{2}\sin x)}{(1 - \sin^2 x) - \sin^2 x} \\
&= \lim_{x\to\frac{\pi}{4}}\frac{(1-\frac{\sin x}{\cos x})(1 + \sqrt{2}\sin x)}{\cos^2 x - \sin^2 x} \cdot \frac{\cos x}{\cos x} \\
&= \lim_{x\to\frac{\pi}{4}}\frac{(\cos x - \sin x)(1 + \sqrt{2}\sin x)}{\cos x(\cos x - \sin x)(\cos x + \sin x)} \\
&= \lim_{x\to\frac{\pi}{4}}\frac{1 + \sqrt{2}\sin x}{\cos x(\cos x + \sin x)} \\
&= \frac{1 + \sqrt{2}\sin \frac{\pi}{4}}{\cos \frac{\pi}{4}(\cos \frac{\pi}{4} + \sin \frac{\pi}{4})} \\
&= \frac{1 + \sqrt{2}(\frac{1}{\sqrt 2})}{\frac{1}{\sqrt 2}(\frac{1}{\sqrt 2} + \frac{1}{\sqrt 2})}
= \frac{1 + 1}{\frac{1}{\sqrt 2}(\frac{2}{\sqrt 2})}
= \frac{2}{2/2} = 2
\end{align*}
| {
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whether $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb Z$ How to check whether $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb Z?$
My guess: If $x^4+x^3+x^2+x+1$ is reducible then $$x^4+x^3+x^2+x+1=f(x)g(x)$$ where $1\le\deg f(x),\deg g(x)\le3$ and $\deg f(x)+\deg g(x)=4$
Checking all the cases for $\deg f(x),\deg g(x)$ is a lengthy exercise. Is there any other way?
| If you do not want to use Eisenstein then :
Suppose it is reducible then you have only two choices (Why ??)
$$(x^4+x^3+x^2+x+1)=(x^2+ax+1)(x^2+bx+1)$$
or $$(x^4+x^3+x^2+x+1)=(x^2+ax-1)(x^2+bx-1)$$
This would not take more than two minutes to verify if $a,b$ are in $\mathbb{Z}$ in this case..
| {
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$F$ is a polynomial, $\deg F = 3$, and $(x^2 - 1)(x^2 - 2) | F(F(x)) - x$. Prove that $F$ exists $F$ is a polynomial, $\deg(F) = 3$, and $(x^2 - 1)(x^2 - 2) | F(F(x)) - x$. Prove that:
a) $F$ exists
b) There are at least 10 such polynomials
What I've tried to do:
$(x^2 - 1)(x^2 - 2) \mid F(F(x)) - x\implies \begin{cases}F(F(1)) = 1 \\ F(F(-1)) = -1 \\ F\left(F\left(\sqrt{2}\right)\right) = \sqrt{2}\\ F\left(F\left(-\sqrt{2}\right)\right) = -\sqrt{2}\end{cases} $
Then I tried to apply interpolation somehow but it didn't help.
Thanks in advance!
| This is a partial answer.
Let $K(x)$ be the factor $\frac{F(F(x))-x}{(x^2-1)(x^2-2)}$.
One can simplify the search significantly by restricting our search of
$F(x)$ to be an odd polynomial in $x$, i.e.
$$F(x) = (ax^2 - b)x,\quad\text{ with }\; a \ne 0$$
There are already 4 pairs of real solutions given by
$$(a,b) = \begin{cases}
\pm (2, 3),\\
\pm (\frac{\sqrt{5}-1}{2}, \sqrt{5} ),\\
\pm (\frac{\sqrt{5}+1}{2}, \sqrt{5} ),\\
\pm (\frac{1}{\sqrt{2}}, \frac{3}{\sqrt{2}} )
\end{cases}
\quad\longrightarrow\quad
\frac{K(x)}{x}
= \begin{cases}
16 x^4- 24x^2+4\\
\\\frac12\left[ (7-3\sqrt{5})x^4 - (9-3\sqrt{5})x^2 + 4\right]\\
\\\frac12\left[ (7+3\sqrt{5})x^4 - (9+3\sqrt{5})x^2 + 4\right]\\
\\\frac14\left[ x^4 - 6x^2 + 7\right]
\end{cases}
$$
If one allow complex coefficients for $F(x)$, there are 4 more complex solutions
and we are done.
$$(a,b) = \pm (\frac{1+i}{2}, \frac{3+i}{2}) \quad\text{ and }\quad \pm (\frac{1-i}{2}, \frac{3-i}{2})$$
If $F(x)$ need to be real, then I don't have any good idea how to get the remaining two real solutions easily.
| {
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Proof of $\frac{1}{\sin{(\frac{\pi}{2x})}}<\frac{2x}{\pi}+1$ I want to show that
\begin{equation}
\frac{1}{\sin{(\frac{\pi}{2x})}}<\frac{2x}{\pi}+1
\end{equation}
for any positive integer $x$.
Seems that it is related to the well-known inequality
\begin{equation}
\frac{1}{\sin{(\frac{\pi}{2x})}}>\frac{2x}{\pi}.
\end{equation}
| $$\sin(x) = \sum_{n=0}^{\infty}{\dfrac{x^{2n+1}(-1)^{n}}{(2n+1)!}}$$ $$\sin(\pi/2x) = \sum_{n=0}^{\infty}{\frac{\pi^{2n+1}(-1)^{n}}{(2x)^{2n+1}(2n+1)!}}$$
The inequality obviously holds for $x=1$ and $x = 2$. For all $x \ge 3$, the above sum is a diminishing alternating series that starts with $\dfrac\pi{2x} - \dfrac{\pi^3}{8x^3}$, and the remainder of the series sums to a positive number. So $\sin{\dfrac{\pi}{2x}} \gt \dfrac\pi{2x} - \dfrac{\pi^3}{8x^3}$
I want to swap a different term for $\frac{\pi^3}{8x^3}$ in my inequality, so I need to prove it is larger than $\frac{\pi^3}{8x^3}$
$\dfrac{\pi^2}{4x^2+2x\pi} -\dfrac{\pi^3}{8x^3} = \dfrac{\pi^2 8x^3 - \pi^3(4x^2+2\pi)}{8x^3(4x^2+2x\pi)} = \pi^2\dfrac{4x^3 - \pi(2x^2+\pi)}{4x^3(4x^2+2x\pi)}$
Since the $\pi^2$ and the denominator are both positive, to determine the sign of the difference of the terms I just need to look at the sign of $4x^3 - \pi(2x^2 + \pi)$ $\gt 4x^3 - 4(2x^2 + 4)$, which is positive for $x \ge 3$.
Now I can say (for $x >= 3$) $\dfrac{\pi^2}{4x^2+2x\pi} \gt \dfrac{\pi^3}{8x^3}$, and $0 > \dfrac{\pi^3}{8x^3} - \dfrac{\pi^2}{4x^2+2x\pi}$, which when added to the inequality above gives $\sin{\dfrac{\pi}{2x}} \gt \dfrac\pi{2x} - \dfrac{\pi^2}{4x^2+2x\pi}$ $= \dfrac{\pi(2x+\pi) - \pi^2}{2x(2x + \pi)} = \dfrac{\pi}{2x+\pi}$
$\sin{\dfrac{\pi}{2x}} \gt \dfrac\pi{2x+\pi}$, and both sides are positive, so $\dfrac1{\sin{\dfrac{\pi}{2x}}} < \dfrac{2x}\pi + 1$
| {
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Prove that $n(n^2 - 1)(n + 2)$ is divisible by $4$ for any integer $n$
Prove that $n(n^2 - 1)(n + 2)$ is divisible by $4$ for any integer $n$
I can not understand how to prove it. Please help me.
| $$\text{Here is a different way out; Recall that } \displaystyle \sum_{k=1}^m k = \dfrac{m(m+1)}2$$
$$\text{Hence, we have }1 + 2 + \cdots +(n-1) = \dfrac{n(n-1)}2 \text{ and } 1 + 2 + \cdots +(n+1) = \dfrac{(n+1)(n+2)}2$$
$$\text{Putting the above two together, we get that}$$
$$\color{red}{\left(1 + 2 + \cdots +(n-1) \right) \left(1 + 2 + \cdots + n + (n+1)\right)} = \color{blue}{\dfrac{n(n^2-1)(n+2)}4}$$
$$\text{ Clearly, the }\color{red}{\text{left side is an integer}}\text{ and hence }\color{blue}{\text{right side is also an integer}}.$$
| {
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Extending Completing the square to expressions of $ (x+y+z)^2$ We all know expressions such as $x^2+14x+49 = (x+7)^2$ because it is easily recognizable as a perfect square. What about a expressions in the form of $x^2+2xy+2yz+2xz+y^2+z^2=(x+y+z)^2$
The question is how do I factor $0.09e^{-2t}+0.24e^{-t}+0.34+0.24e^t+0.09e^{2t}$ into the form of $(x+y+z)^2$?
| Setting $e^t=a,$ we have $\displaystyle\frac{9+24a+34a^2+24a^3+9a^4}{100a^4}$ which can be written as $\displaystyle\frac{(3a^2+ab+3)^2}{(100a^2)^2}$ where $b$ is independent of $a$
$\displaystyle\implies 9+24a+34a^2+24a^3+9a^4=9a^4+a^2b^2+9+6ba^3+6ba+18a^2$
$\displaystyle\implies 9+24a+34a^2+24a^3+9a^4=9a^4+6ba^3+a^2(18+b^2)+6ba+9$
Comparing the coefficients of $a$ or $a^3,\displaystyle24=6b\iff b=4$
Comparing the coefficients of $a^2,\displaystyle34=18+b^2\iff b^2=16\iff b=\pm4$
So, $b=?$
| {
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Function : Find the range of $f(x) = \sin^{-1}x +\tan^{-1}x +\cos^{-1}x$ Problem :
Find the range of $f(x) = \sin^{-1}x +\tan^{-1}x +\cos^{-1}x$
Solution : Since, $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$
Since range of $\tan^{-1}x$ is $ (\frac{-\pi}{2}, \frac{\pi}{2})$
$\therefore, \frac{-\pi}{2} \leq \tan^{-1}x \leq \frac{\pi}{2}$
= $ \frac{-\pi}{2} + \frac{\pi}{2} \leq \tan^{-1}x + \frac{\pi}{2} \leq \frac{\pi}{2} + \frac{\pi}{2}$.
= $0 \leq \tan^{-1}x+ \frac{\pi}{2} \leq \pi $
Is it correct.. please suggest thanks....
| Given $f(x) = \sin^{-1}(x)+\cos^{-1}(x)+\tan^{-1}(x)$
First we will calculate domain of function $f(x)$
function $\sin^{-1}(x)$ is defined in $\displaystyle x\in \left[-1,1\right]$ Similarly function $\cos^{-1}(x)$ is defined in $\displaystyle x\in \left[-1,1\right]$
and function $\tan^{-1}(x)$ is defined in $\displaystyle x\in \left(-\infty,+\infty\right)$
So $f(x) = \sin^{-1}(x)+\cos^{-1}(x)+\tan^{-1}(x)$ is defined in $\displaystyle x\in \left[-1,1\right]$
So $f(x) = \sin^{-1}(x)+\cos^{-1}(x)+\tan^{-1}(x)$ is defined in $$\displaystyle x\in \left[-1,1\right]$$
So $\displaystyle f(x) = \frac{\pi}{2}+\tan^{-1}(x)$
Now $\displaystyle f^{'}(x) = \frac{1}{1+x^2}>0\;\forall x\in [-1,1]$
So $f(x)$ is Strictly Increasing function.
So $\displaystyle f(-1) = \frac{\pi}{2}+\tan^{-1}(-1) = \frac{\pi}{2}-\frac{\pi}{4} = \frac{\pi}{4}$
and $\displaystyle f(+1) = \frac{\pi}{2}+\tan^{-1}(1) = \frac{\pi}{2}+\frac{\pi}{4} = \frac{3\pi}{4}$
So $\displaystyle f(x)\in \left[\frac{\pi}{4}\;,\frac{3\pi}{4}\right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solutions for $\frac{3}{x+1}\le\frac{2}{2x+5}$ Im in search of the solutions for:
$$\frac{3}{x+1}\le\frac{2}{2x+5}$$
So first i tried to combine the two sites:
$$\frac{6x + 15 - 2x + 2}{2x^2 +7x + 5}\le{0}$$
$$\frac{4x + 17}{2x^2 +7x +5}\le{0}$$
My problem is that now i have two solutions for the denominator and i dont know how to continue:
$2x^2+7x+5 = -1 \text{ and } -2.5$
The solution should be: $(-2.5;-1) \cup (-\infty;-3.25)$
| The first problem is that the difference of the two sides is
$$
\begin{align}
\frac{6x+15-2x\color{#C00000}{-}2}{(x+1)(2x+5)}=\frac{4x+\color{#C00000}{13}}{(x+1)(2x+5)}\le0
\end{align}
$$
So there are three points to consider: $-\frac{13}{4}$, $-\frac52$, and $-1$.
To the left of all three points, all three terms, $4x+13$, $x+1$, and $2x+5$, are negative.
Between $-\frac{13}{4}$ and $-\frac52$, only two terms are negative.
Between $-\frac52$ and $-1$, only one term is negative.
To the right of all three points, none of the terms are negative.
Note that all the finite endpoints should be closed, not open, where it doesn't divide by $0$:
$$
\textstyle\left(-\infty,-\frac{13}{4}\right]\cup\left(-\frac52,-1\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
$\omega^2+\omega+1$divides a polynomial The question is
Show that $f(n)=n^5+n^4+1$ is not prime for $n>4$.
The solution is given as
Let $\omega$ be the third root of unity. Then $\omega^2+\omega+1=0$. Since $\omega^5+\omega^4+1=\omega^2+\omega+1$, we see that $\omega^2+\omega+1$ is a *factor of the polynomial. So *$n^2+n+1|n^5+n^4+1$.
Which polynomial are we referring to in the bold typeface above? And how is $n^2+n+1|n^5+n^4+1$ true due to $\omega^5+\omega^4+1=\omega^2+\omega+1$?
| As the 3rd root of identity $\omega=\mathrm{e}^{2\pi i/3}$ satisfies $x^5+x^4+1=0$, then so does its conjugate $\bar\omega=\mathrm{e}^{-2\pi i/3}$, and hence $(x-\omega)(x-\bar\omega)=x^2+x+1$ divides $x^5+x^4+1$.
Note that
$$
x^5+x^4+1=(x^2+x+1)(x^3-x+1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Sum - can I just change the terms? I am looking at an exercise where this sum appears $\sum_{k=0}^{n}\binom{2n+1}{2n+1-k}$, and I saw in my textbook that it should be equal to $\sum_{u=n+1}^{2n+1}\binom{2n+1}{u}$.
I replaced at the first sum $2n+1-k$ with $u$, and I get this sum: $\sum_{u=2n+1}^{n+1}\binom{2n+1}{u}$. How did they find the result? Can I just change the terms $n+1$ and $2n+1$, because $n+1$ is smaller?
| It sometimes helps to write out the terms.
\begin{align*}
\sum_{k=0}^n\binom{2n+1}{2n+1-k} &= \binom{2n+1}{2n+1} + \binom{2n+1}{2n} + \dots + \binom{2n+1}{n+2} + \binom{2n+1}{n+1}\\
\\
\sum_{u=n+1}^{2n+1}\binom{2n+1}{u} &= \binom{2n+1}{n+1} + \binom{2n+1}{n+2} + \dots + \binom{2n+1}{2n} + \binom{2n+1}{2n+1}.
\end{align*}
So the two sums are the same, they just reverse the order of the summands.
To see this via a change of index, let $u = 2n+1-(n-k)$ so that $u = n+1+k$; as $k$ goes from $0$ to $n$, $u$ goes from $n+1+0 = n+1$ to $n+1+n = 2n+1$. You can think of this as a combination of two index changes, first $w = n - k$ reverses the order of the terms, and then $u = 2n+1-w$ shifts the index.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determine the smallest positive value of x(in degrees) for which: $\tan(x+100^{\circ}) = \tan(x+50^{\circ})\tan (x)\tan(x-50^{\circ})$ Determine the smallest positive value of x(in degrees) for which:
$\tan(x+100^{\circ}) = \tan(x+50^{\circ})\tan (x)\tan(x-50^{\circ})$
I tried to apply the formula of $\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$ but that led me nowhere resulting in a huge equation.
Please help.
| I'd like to solve the same sum without skipping steps as done by the previous solver.
$\tan(x+100)=\tan(x+50)\tan x\tan(x-50)$
$\tan(x+100)/\tan(x-50) = \tan(x+50)\tan x$
$\sin(x+100)\cos(x-50)/\cos(x+100)\sin(x-50)=\sin(x+50)\sin x/\cos(x+50)\cos x$
Componendo and Dividendo,
$\sin(x+100)\cos(x-50)+\cos(x+100)\sin(x-50)/
\sin(x+100)\cos(x-50)-\cos(x+100)\sin(x-50)=\sin(x+50)\sin x+\cos(x+50)\cos x/\sin(x+50)\sin x-\cos(x+50)\cos x$
$\sin(x+100+x-50)/\sin(x+100-x+50)=\cos(x+50-x)/-\cos(x+50+x)$
$\sin(2x+50)/\sin 150 = \cos(2x+50)/-$cos50$
$\sin(2x+50)/1/2=-\cos 50/\cos(2x+50) (\sin150 = \sin(180-30) = \sin 30 =1/2)$
$2\sin(2x+50)\cos(2x+50) = -\cos 50$
$\sin(4x+100) = -\sin 40 (\sin 2x = 2\sin x\cos x)$
$\sin(4x+100) = \sin(180+40)$ (3rd quadrant, $\sin x$ is negative)
$4x+100 = 220$
$x=120/4 = 30$
Hope that made it easier :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How prove this inequality $x^2+y^2+z^2+xyz(x+y+z-2)\ge 4$ let $x,y,z\ge 0$,and such
$$xy+yz+xz=xyz+2$$
show that
$$x^2+y^2+z^2+xyz(x+y+z-2)\ge 4$$
my try: let $x+y+z=p,xy+yz+xz=q, xyz=r$
then
$$q=r+2$$
show that
$$p^2-2q+r(p-2)\ge 4$$
then I can't,Thank you
| Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, the condition does not depend on $u$ and we need to prove that $f(u)\geq0$,
where $f(u)=9u^2-6v^2+w^3(3u-2)-4$.
But $f$ is an increasing function, which says that it's enough to prove our inequality
for a minimal value of $u$, which happens for equality case of two variables.
Let $y=x$. Then $z=\frac{2-x^2}{x(2-x)}$ and we need to prove that
$$(x-1)^2(x^2-2)^2(2x+1)\geq0,$$
which is obvious.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/622481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find the exact value of $\tan\left ( \sin^{-1} \left ( \sqrt 2/2 \right )\right )$
Find the exact value of $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )$ without using a calculator.
I started by finding $\sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )=\dfrac{\pi}{4}$
So, $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )=\tan\left( \dfrac{\pi}{4}\right)$.
The answer is $1$. Can you show how to solve $\tan\left( \dfrac{\pi}{4}\right)$ to get $1$? Thank you.
| $\hskip2in$
Using the triangle above...& the fact that $$\tan x = \frac{\text{opp}}{\text{adj}}, \space \tan \left(\frac{\pi}{4}\right)=...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/623703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to find the range of the function: $f(x) = \sqrt{x-1}+2\sqrt{3-x}$ Problem :
Find the range of the function: $f(x) = \sqrt{x-1}+2\sqrt{3-x}$
Solution :
Domain of this function can be determined as :
$x - 1 >0 ; 3-x >0 \Rightarrow x >0 ; x <3 ;$
$\therefore $ domain of $x \in [1,3]$
Now if I put the values of this domain in my function then it gives the following values :
at 1 ; the value of the function is $2\sqrt{2}$
at 2 : the value of the function is $ 1+2 = 3$
at 3 : the value of the function is $2$
Can we say that the maximum value of the function is 3 and minimum value of the function is 2;
Therefore the range of this function is [2,3] but this answer is wrong. please suggest..
Also suggest that how can we use differentiation method to find the range... thanks.
| We get
$$f^\prime (x)=\frac{1}{2\sqrt{x-1}}-\frac{2}{2\sqrt{3-x}}=\frac{\sqrt{3-x}-2\sqrt{x-1}}{2\sqrt{(x-1)(3-x)}}.$$
So,
$$f^\prime(x)\ge0\iff \sqrt{3-x}\ge2\sqrt{x-1}\iff 3-x\ge4(x-1)\iff x\le\frac{7}{5}.$$
Now we know that $f(x)$ is increasing in $1\le x\lt 7/5$ and that $f(x)$ is decreasing in $7/5\lt x\le 3$.
So, we know that the max is $f(7/5)$, and that the min is $\min(f(1),f(3)).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Prove that $1 + \cos\alpha + \cos\beta + \cos\gamma = 0$ If $\alpha + \beta + \gamma = \pi $ and $\tan(\frac{-\alpha + \beta + \gamma}4)\tan(\frac{\alpha - \beta + \gamma}4)\tan(\frac{\alpha + \beta - \gamma}4) = 1$
Then prove that:
$1 + \cos\alpha + \cos\beta + \cos\gamma = 0$.
I have no idea how to go about this.
Please help.
| As $\displaystyle \frac{-\alpha+\beta+\gamma}4=\frac{-\alpha+\pi-\alpha}4=\frac\pi4-\frac\alpha2$
$\displaystyle \tan\frac{-\alpha+\beta+\gamma}4=\tan\left(\frac\pi4-\frac\alpha2\right)=\frac{1-\tan\frac\alpha2}{1+\tan\frac\alpha2}=\frac{\cos\frac\alpha2-\sin\frac\alpha2}{\cos\frac\alpha2+\sin\frac\alpha2}$
$\displaystyle\implies \tan^2\left(\frac{-\alpha+\beta+\gamma}4\right)=\left(\frac{\cos\frac\alpha2-\sin\frac\alpha2}{\cos\frac\alpha2+\sin\frac\alpha2}\right)^2=\frac{1-\sin\alpha}{1+\sin\alpha}$
$\displaystyle\implies \prod\left(\frac{1-\sin\alpha}{1+\sin\alpha}\right)=1$
$\displaystyle\implies\sum\sin\alpha+\prod\sin\alpha=0 $
Now set $\displaystyle\sin\alpha=2\sin\frac{\alpha}2\cos\frac{\alpha}2$ in $\displaystyle\prod\sin\alpha$
and use this for $\sum\sin\alpha$
Cancelling out $\displaystyle\prod\cos\frac{\alpha}2,$ (assuming $\displaystyle\prod\cos\frac{\alpha}2\ne0$ )
we get $\displaystyle4\prod \sin\frac{\alpha}2=-2$
Finally, using this, $\displaystyle\sum \cos A=1+4\prod\sin\frac{\alpha}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
How find this limit $\lim_{x\to 1}\frac{f(2001)-f(2002)}{f(2002)-f(2003)}$ Let $$f(m)=\dfrac{m+1}{\dfrac{m+m+1}{\dfrac{m}{1-x^{m}}-\dfrac{m+1}{1-x^{m+1}}+\dfrac{1}{2}}+\dfrac{m+1+m+2}{\dfrac{m+1}{1-x^{m+1}}-\dfrac{m+2}{1-x^{m+2}}+\dfrac{1}{2}}}$$
Find this limit
$$I=\lim_{x\to 1}\dfrac{f(2001)-f(2002)}{f(2002)-f(2003)}$$
My try: since Use (L'Hôpital's rule） we have
\begin{align*}&\lim_{x\to 1}\left(\dfrac{n}{1-x^n}-\dfrac{n+1}{1-x^{n+1}}\right)\\
&=\lim_{x\to1}\dfrac{n(1-x^{n+1})-(n+1)(1-x^n)}{(1-x^n)(1-x^{n+1})}\\
&=\lim_{x\to 1}\dfrac{n-nx^{n+1}-n-1+(n+1)x^n}{1-x^n-x^{n+1}+x^{2n+1}}\\
&=\lim_{x\to 1}\dfrac{-n(n+1)x^n+n(n+1)x^{n-1}}{-(n+1)x^n-nx^n+（2n+1)x^{2n}}\\
&=\lim_{x\to 1}\dfrac{-n^2(n+1)x^{n-1}+n(n-1)(n+1)x^{n-2}}{-n(n+1)x^{n-1}-n^2x^{n-1}+2n(2n+1)x^{2n-1}}\\
&=\dfrac{n(n+1)[n-1-n]}{-n^2-n-n^2+4n^2+2n}=-\dfrac{1}{2}
\end{align*}
This problem is creat a teacher of China, zhejiang university,it is said that teacher want Floored the arrogance of the students.
then I can't,Thank you
| Let $x = 1-z$, we have
$$\begin{align}
\frac{m}{1-x^m} &= \frac{m}{1-(1-z)^m}\\
& = \frac{m}{1 - \left(1 - mz + \frac{m(m-1)}{2}z^2 - \frac{m(m-1)(m-2)}{6}z^3 + O(z^4)\right)}\\
& = \frac{1}{z - \frac{m-1}{2}z^2 + \frac{(m-1)(m-2)}{6}z^2 + O(z^3)}\\
& =\frac{1}{z}\left[ 1 - \left( -\frac{m-1}{2}z + \frac{(m-1)(m-2)}{6}z^2\right) + \left(\frac{m-1}{2} z\right)^2 + O(z^3)\right]\\
&= \frac{1}{z} + \frac{m-1}{2} + \frac{m^2-1}{12}z + O(z^2)
\end{align}
$$
So
$$\begin{align}
&\frac{m}{1-x^m} - \frac{m+1}{1-x^{m+1}} + \frac12\\
=& \left(\frac{1}{z} + \frac{m-1}{2} + \frac{m^2-1}{12}z\right)
- \left(\frac{1}{z} + \frac{m}{2} + \frac{m^2-2m}{12}z\right) + \frac12 + O(z^2)\\
=&-\frac{2m+1}{12} z + O(z^2)
\end{align}
$$
This gives us
$$f(m) = \frac{m+1}{\frac{2m+1}{-\frac{2m+1}{12}z + O(z^2)} + \frac{2m+3}{-\frac{2m+3}{12}z + O(z^2)}}
= -\frac{m+1}{24} z + O(z^2)
$$
and hence
$$\lim_{x\to 1}\frac{f(2001)-f(2002)}{f(2002)-f(2003)}
= \lim_{z\to 0}\frac{-\frac{2002}{24} z + \frac{2003}{24} z + O(z^2)
}{-\frac{2003}{24} z + \frac{2004}{24} z + O(z^2)} =
\lim_{z\to 0}\frac{\frac{z}{24}+O(z^2)}{\frac{z}{24} + O(z^2)}
= 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Reversion of power series So, I just heard about this method. How does one determine the coefficients, and what is it used for?
For example, given $$ y = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7)$$ reversion would give a series for $x$ in terms of $y$, correct? And then this expansion could be used to expand $y$ in ascending powers of itself?
| I think how you go about inverting the series is to first do this:
$$x = y + \frac{x^3}{6} - \frac{x^5}{120} - O(x^7),$$
so that
$$x = y + \frac{1}{6}(y + \frac{x^3}{6} - \frac{x^5}{120} - O(x^7))^3 -
\frac{1}{120}(y + \frac{x^3}{6} - \frac{x^5}{120} - O(x^7))^5 + O(y^7)$$
Then gather the terms of order less than $y^7$ and discard those $y^7$ and higher. If you still have terms with powers of $x$ that are multiplied by powers of $y$ with order less than $y^7$, substitute the first expression again until you don't.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/625195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all $z\in\Bbb C$ such that $|z+1|+ |z-1|=4$ I'd like to find all points of the complex plane which satisfy
$$|z+1| + |z-1| = 4. $$
I know this is an ellipsis with foci $1$ and $-1$, and I know that the answer is
$$3 x^2+4 y^2 \leq 12,$$
but I can't find a correct way of getting there.
First, I write $z$ as $x + i y$ and square both sides of the equation, then divide by 2 and get
$$x^2+y^2+1+\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =8.$$
Pass $x^2+y^2+1$ to the RHS (right hand side), then
$$\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =7-x^2-y^2.\tag{1} $$
Now, I would have to square both sides of the equation like this,
$$((x-1)^2+y^2)((x+1)^2+y^2) = (7-x^2-y^2)^2, \tag{2}$$
but the problem is that I cannot assure that the RHS is not negative, so there could be a value for $z$ such that (2) is satisfied but not (1), i.e it could exist $z=x + i y$ which satisfies
$$\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =-(7-x^2-y^2)\tag{3} $$
in which case also satisfy (2) but not (1)!
So I would get an incorrect solution.
| An ellipse is defined as the locus of all points,the sum of whose from two given points is constant. Here z is a complex number whose distance from $(1,0)$ and $(-1,0)$ is constant. Hence the locus of z is an ellipse in the complex plane. Hence z will be all those points which lies on the ellipse with focus $(-1,0)$ and $(1,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/626554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Calculation of $\int \sqrt{\tan x+2}dx$ Calculation of $\displaystyle \int\sqrt{\tan x+2}\;dx$
$\bf{My\; solution::}$ Let $\displaystyle \tan x+2 = t^2 $, Then
$$\displaystyle \sec^2 (x)dx = 2tdt\Rightarrow dx = \frac{2t}{1+\tan^2 x}dt = \frac{2t}{1+(t^2-2)^2}dt$$
So Integral convert into $$\displaystyle \int \frac{2t^2}{(t^2-2)^2+1^2}dt$$
Let $\displaystyle (t^2-2) = u\Rightarrow t^2=u+2\;,$ Then $\displaystyle tdt=\frac{1}{2}du$
Now How can i solve after that
please help me
Thanks
| Let
$$I=\int\sqrt{\tan x+2}\;dx,\>\>\>\>\> J=\int\frac1{\sqrt{\tan x+2}}\;dx
$$
and substitute $\tan x= \sqrt5 t^2 -2$ to integrate
\begin{align}
I+\sqrt5 J
= &\ \frac2{\sqrt[4]5}
\int \frac{1+\frac1{t^2}}{t^2+ \frac1{t^2}-\frac4{\sqrt5}}dt
=\frac{1}{\sqrt{\frac{\sqrt5}2-1}}
\tan^{-1}\frac{\sqrt[4]5(t-\frac1t)}{2\sqrt{\frac{\sqrt5}2-1}}\\
I-\sqrt5 J
= &\ \frac2{\sqrt[4]5}
\int \frac{1-\frac1{t^2}}{t^2+ \frac1{t^2}-\frac4{\sqrt5}}dt
=-\frac{1}{\sqrt{\frac{\sqrt5}2+1}}
\coth^{-1}\frac{\sqrt[4]5(t+\frac1t)}{2\sqrt{\frac{\sqrt5}2+1}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/626942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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Finding limit of a recursively defined sequence Let $(x_{n})_{n\geq1}$ be a sequence defined by:
$x_1=1$ and $x_n=n(x_{n+1}-\frac{n+1}{n^2})$. Calculate $\lim_{x\rightarrow \infty }nx_n$.
We can write $x_n=n(x_{n+1}-\frac{n+1}{n^2})$ as $(n+1) x_{n+1} = \frac{n + 1}{n^2}(n x_n + (n+1))$, and by the substitution $y_n=nx_n$ we obtain: $y_{n+1} = \frac{n + 1}{n^2}y_n + \left(1 + \frac{1}{n}\right)^2$. How to go on?
| It is clear that $ x_n> \frac{1}{n}.$
Using mathematical induction is easy to see that $ x_n < \frac{1}{n-4} $ for $n\geq6.$
Proof by mathematical induction:
$x_6=\frac{199}{450}<\frac{1}{2}$
$x_{n+1}= \frac{n+1}{n^2} + \frac{x_n}{n}< \frac{n+1}{n^2}+\frac{1}{n(n-4)}=\frac{n^2-2n-4}{n^2(n-4)}. $
Because $\frac{n^2-2n-4}{n^2(n-4)}<\frac{1}{n-3}<=>-n^2+2n+12<0$ assertion is proved.
It follows that for $n\geq6$ $$\frac{1}{n}<x_n < \frac{1}{n-4} $$
and consequently $\ lim_{n\to\infty} nx_n=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/628113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove Inequality without induction. I showed this inequality by induction. I want other methods to prove it.
$(\frac{2n}{3}+\frac{1}{3})\sqrt{n}\leq \sum_{k=1}^{n}\sqrt{k}\leq (\frac{2n}{3}+\frac{1}{2})\sqrt{n}$
Thank
| For $0\le k\le n-1$, let $R_k$ be the region defined by $k\le x\le k+1$, $\sqrt{x}\le y\le \sqrt{k+1}$,
so its area is given by
$A_k=\sqrt{k+1}\cdot1-\int_{k}^{k+1}\sqrt{x}\;dx=\sqrt{k+1}-\frac{2}{3}((k+1)^{3/2}-k^{3/2})$.
$\textbf{1)}$ Since the graph of $y=\sqrt{x}$ is concave down, $R_k$ is contained in the triangle with vertices $\;\;\;(k,\sqrt{k}), (k,\sqrt{k+1}), (k+1,\sqrt{k+1})$, so
$\;\;\;\sqrt{k+1}-\frac{2}{3}((k+1)^{3/2}-k^{3/2})\le\frac{1}{2}(\sqrt{k+1}-\sqrt{k})$.
Then $\displaystyle \sum_{k=0}^{n-1}[\sqrt{k+1}-\frac{2}{3}((k+1)^{3/2}-k^{3/2})]\le\sum_{k=0}^{n-1}\frac{1}{2}(\sqrt{k+1}-\sqrt{k})$, so
$\displaystyle\sum_{k=0}^{n-1}\sqrt{k+1}-\frac{2}{3}n^{3/2}\le\frac{1}{2}\sqrt{n}$ and therefore
$\displaystyle \sum_{k=1}^{n}\sqrt{k}\le\frac{2}{3}n^{3/2}+\frac{1}{2}\sqrt{n}=\big(\frac{2}{3}n+\frac{1}{2}\big)\sqrt{n}$.
$------------------------------------------$
$\textbf{2)}$ Since $(4k^2+4k+1)k\ge4k^3+4k^2$, $(2k+1)^2k\ge4k^2(k+1)$
$\;\;\;$so $(2k+1)\sqrt{k}\ge2k\sqrt{k+1}\;\;\;$. Then
$\;\;\;2k\sqrt{k}+\sqrt{k}+3\sqrt{k+1}\ge2(k+1)\sqrt{k+1}+\sqrt{k+1}$, so
$\;\;\;\frac{2}{3}k\sqrt{k}+\frac{1}{3}\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}(k+1)\sqrt{k+1}+\frac{1}{3}\sqrt{k+1}$ and
$\;\;\;\sqrt{k+1}-\frac{2}{3}(k+1)\sqrt{k+1}+\frac{2}{3}k\sqrt{k}\ge\frac{1}{3}\sqrt{k+1}-\frac{1}{3}\sqrt{k}$.
Then $\sqrt{k+1}-\frac{2}{3}((k+1)^{3/2}-k^{3/2})\ge\frac{1}{3}(\sqrt{k+1}-\sqrt{k})$, so
$\displaystyle\sum_{k=0}^{n-1}[\sqrt{k+1}-\frac{2}{3}((k+1)^{3/2}-k^{3/2})]\ge\sum_{k=0}^{n-1}\frac{1}{3}(\sqrt{k+1}-\sqrt{k})$.
Then $\displaystyle\sum_{k=0}^{n-1}\sqrt{k+1}-\frac{2}{3}n^{3/2}\ge\frac{1}{3}\sqrt{n}$, and therefore
$\displaystyle\sum_{k=1}^{n}\sqrt{k}\ge\frac{2}{3}n^{3/2}+\frac{1}{3}\sqrt{n}=\big(\frac{2}{3}n+\frac{1}{3}\big)\sqrt{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/628364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How prove this inequality $3a^3b+3ab^3+18a^2b+18ab^2+12a^3+12b^3+40a^2+40b^2+64ab\ge 0$ Let $a,b\in[-1,1]$,
then prove or disprove:
$$f(a,b)=3a^3b+3ab^3+18a^2b+18ab^2+12a^3+12b^3+40a^2+40b^2+64ab\ge 0$$
My try: Since
\begin{align*}
f(a,b)&=3a^3b+3ab^3+18a^2b+18ab^2+12a^3+12b^3+40a^2+40b^2+64ab\\
&=3ab(a^2+b^2)+18ab(a+b)+12(a^3+b^3)+40(a^2+b^2)+64ab
\end{align*}
if
$ab\ge 0$
then
\begin{align*}&f(a,b)\ge 9ab(a^2+b^2)+18ab(a+b)+12(a^3+b^3)+40(a^2+b^2)+52ab\\
&=9ab(a^2+b^2)+18ab(a+b)+12(a+b)(a^2+b^2-ab)+40(a^2+b^2)+52ab\\
&=9ab(a^2+b^2)+6(a+b)(2a^2+2b^2+ab)+40(a^2+b^2)+52ab\\
&\ge0
\end{align*}
but for the other case, $ab\le 0$, I can't proceed.
This problem is a follow up from this.
| for $ab<0$, WLOG,let $a>0,-c=b<0 \to 1 \ge c>0$ so the inequality becomes:
$-3a^3c-3ac^3-18a^2c+18ac^2+12a^3-12c^3+40a^2+40c^2-64ac\ge 0$
$12c^2 \ge 12c^3,3ac^3 \ge 3ac^3,9ac^2+9a^3 \ge 18a^2c,3a^3 \ge 3a^3c,40a^2+28c^2\ge 64ac $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/628832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove the inequality: $\frac{(1+x)^2}{2x^2+(1-x)^2}+\frac{(1+y)^2}{2y^2+(1-y)^2}+\frac{(1+z)^2}{2z^2+(1-z)^2}\leq 8$ Prove that:
$$\frac{(1+x)^2}{2x^2+(1-x)^2}+\frac{(1+y)^2}{2y^2+(1-y)^2}+\frac{(1+z)^2}{2z^2+(1-z)^2}\leq 8$$
subject to the constraints:
$$x,y,z >0$$
and
$$x+y+z=1.$$
| An other proof:
$$\Longleftrightarrow \sum_{cyc}\dfrac{(2a+b+c)^2}{2a^2+(b+c)^2}\le 8$$
since
$$\dfrac{(2a+b+c)^2}{2a^2+(b+c)^2}\le\dfrac{4}{3}\dfrac{4a+b+c}{a+b+c}$$
this is true because
$$\Longleftrightarrow (2a-b-c)^2(5a+b+c)\ge 0$$
so
$$\sum_{cyc}\dfrac{(2a+b+c)^2}{2a^2+(b+c)^2}\le\dfrac{4}{3}\sum_{cyc}\dfrac{4a+b+c}{a+b+c}=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/629811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find the integral: $ \int \frac{2x}{\sqrt {4x-1}}\, \mathrm{d}x\;?$ My problem, is how to find the integral of $$ \int \frac{2x}{\sqrt {4x-1}}\, \mathrm{d}x$$
Can I do this?
$$ \int \frac{\frac{1}{2}\cdot4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$
$$ \frac{1}{2}\int \frac{4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$
$$ \frac{1}{2}\int \frac{4x-1}{\sqrt {4x-1}}+\frac{1}{\sqrt {4x-1}}\, \mathrm{d}x$$
$$ \frac{1}{2}\int \sqrt {4x-1}+\frac{1}{\sqrt {4x-1}}\, \mathrm{d}x$$
then
$$ 4x-1= u/'$$
$$ 4dx= du$$
$$ dx= \frac{1}{4}du$$
insert that in the integral
$$ \frac{1}{2}\int \sqrt {u}+\frac{1}{\sqrt {u}}\cdot\frac{1}{4}\, \mathrm{d}u$$
$$ \frac{1}{8}\int \sqrt {u}+\frac{1}{\sqrt {u}}\, \mathrm{d}u$$
and I get
$$\frac{1}{12}\cdot u\cdot \sqrt{u}+ \frac{1}{4}\cdot \sqrt{u}+C$$
$$\frac{1}{12}\cdot(4x-1)\cdot \sqrt{4x-1}+ \frac{1}{4}\cdot \sqrt{4x-1}+C$$
and that's the correct solution. But can I just rewrite $2x$ as $\frac{1}{2}\cdot4x-1+1$ and then put $\frac{1}{2}$ before the integral? Thank you in advance!
| You can't do in that way, but you can salvage the idea:
$$
\int\frac{2x}{\sqrt{4x-1}}dx=\frac{1}{2}\int\frac{4x}{\sqrt{4x-1}}dx
=\frac{1}{2}\int\frac{4x-1+1}{\sqrt{4x-1}}dx
$$
and go on.
A possibly simpler strategy is doing the substitution
$$
\sqrt{4x-1}=2t
$$
where I use $2t$ because after squaring we get
$$
4x-1=4t^2
$$
so
$$
x=t^2+\frac{1}{4},\quad dx=2t\,dt
$$
and the integral becomes
$$
\int\frac{2t^2+\frac{1}{2}}{t}2t\,dt=\int(4t^2+1)\,dt=\frac{4}{3}t^3+t+c
$$
and the result will follow by substituting back $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/630684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Limit of two sequences
*
*$ \lim_{n\to \infty} \sqrt[n]{3^n+4^n} $ . I think the limit is $4$. I did :
$ \sqrt[n]{3^n+4^n} = 4 \sqrt[n]{(\frac{3}{4}) ^n+1}$ .Am I right?
*$ \lim_{n\to \infty} \frac{1}{1\cdot 4 } + \frac{1}{4\cdot 7} +...+\frac{1}{(3n-2)(3n+1)} $. I know that for each $k$ , this sequence is the sum of $k$ terms, the smallest one is $ \frac{1}{(3n-2)(3n+1)} $, and the largest is $ \frac{1}{1. 4 }$ . But when substituting and trying to use the squeeze thm, I get that the limit should be between $0$ and $\infty$, which gives me nothing.
Thanks in advance.
| 1- Your first limit is correct.
2- As I suggested in the comments, you have to partialize the fractions into two, as such:
$$\frac{1}{(3n+1)(3n-2)}= \frac{1}{3} \left(\dfrac{-1}{(3n+1)}+\dfrac{1}{(3n-2)}\right)$$
Using the fact that it is a telescoping sum, you now have to find the:
$$\sum_{n=1}^{\infty}\left(\frac{-1}{(3n+1)}+\frac{1}{(3n-2)}\right)$$
Listing out the first few terms and the $n$-th term the cancellations become clear,
$$\lim \limits_{n \to \infty} \left[\frac{-1}{4} +\frac{-1}{7}+\frac{-1}{10}...\frac{-1}{3n-2} +\frac{-1}{3n+1} \right] + \left[ 1 + \frac{1}{4} + \frac{1}{7}+...\frac{1}{3n-2} \right]$$
$$=\lim \limits_{n \to \infty} \left(1-\frac{1}{3n+1}\right)=1$$
Therefore, the final sum is $\dfrac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/633522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the minimum value of $x^2+y^2$, where $x,y$ are non-negative integers and $x+y$ is a given positive odd integer Let $k$ be a fixed positive odd integer. Find the minimum value of $x^2+y^2$, where $x,y$ are non-negative integers and $x+y=k$
My approaches:
*
*Since $k$ is odd, $x$ and $y$ have different parity. I consider, $k=2m+1$, so that $x+y=2m+1$.
I also consider $x$ to be even and $y$ to be odd.
So, $2p+2q+1=2m+1$.
Also, $4p^2+4q^2+4q+1+8pq+4p=4m^2+4m+1$ which apparently doesn't lead me anywhere.
*$x+y=k$. Then
$x^2+y^2+2xy=k^2$.
Now I am stuck!
Please help!
| Starting with $x<y$ and $y\neq x+1$ then $\left(x+1\right)^{2}+\left(y-1\right)^{2}<x^{2}+y^{2}$ so then pair
$\left(x,y\right)$ will not provide the minimal value (pair $\left(x+1,y-1\right)$ does 'better'). This shows
that we need $y=x+1$.
If $k=2n+1$ then $x=n$ and $y=n+1$ give the minimum value $n^{2}+\left(n+1\right)^{2}=\frac{1}{2}\left(k^{2}+1\right)$
In situations like this it is a good habit to pick out a 'small' one (p.e. $k=7$) and to investigate the candidates $0+7$, $1+6$,... et cetera. Patterns show up that are useful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/633675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Show that the least squares line must pass through the center of mass My problem:
The point $(\bar x, \bar y)$ is the center of mass for the collection of points in Exercise 7. Show that the least squares line must pass through the center of mass. [Hint: Use a change of variables $z = x - \bar x$ to translate the problem so that the new independent variable has mean 0.]
I have already solved Exercise 7:
Given a collection of points $(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)$, let $\mathbf x = (x_1, x_2, \ldots, x_n)^T$, $\mathbf y = (y_1, y_2, \ldots, y_n)^T$, $\bar x = \frac 1n \sum_1^n x_i$, $\bar y = \frac 1n \sum_1^n y_i$ and let $y = c_0 + c_1 y$ be the linear function that gives the best least squares fit to the points. Show that if $\bar x = 0$, then $c_0 = \bar y$ and $c_1 = \frac {\mathbf x^T \mathbf y}{\mathbf x^T \mathbf x}$.
It is obvious that if $x = \bar x$ then $y = c_0 + c_1x = \bar y + 0 = \bar y$, however the hint suggests that the problem should be solved in another way.
Edit
I have found an answer. It makes use of the following theorem:
If A is an m x n matrix of rank n, the normal equations $ A^T A \mathbf x = A^T \mathbf b$ have a unique solution $ \hat {\mathbf x} = (A^TA)^{-1}A^T \mathbf b$ and $ \hat {\mathbf x} $ is the unique least squares solution of the system $ A \mathbf x = \mathbf b $.
Now let $ \hat {\mathbf x} = \mathbf c = (c_0, c_1)^T, A = \begin{pmatrix}1 & \cdots & 1 \\x_1 & \cdots & x_n \\\end{pmatrix}, \mathbf b = \mathbf y = (y_1, \ldots, y_n)^T $ such that $c = (A^TA)^{-1}A^Ty$, then
$$\begin{pmatrix}c_0\\c_1\\\end{pmatrix} = \begin{pmatrix}n & \sum x_i\\\sum x_i & \sum x_i^2\\\end{pmatrix}^{-1} \begin{pmatrix}\sum y_i\\\sum x_iy_i\\\end{pmatrix} $$
which gives values for $c_0$ and $c_1$. These values should be used in the formula $c_1x + c_0$, which, together with $ x = \bar x = \frac 1n \sum x_i$, indeed results in $ \bar y $.
| Assume we have the linear model
$$y=X\beta$$
where
\begin{align*}
y_{n\times1} =\begin{bmatrix}
y_1\\
y_2\\
\vdots\\
y_n
\end{bmatrix}
\hspace{2cm}
X_{n\times2} = \begin{bmatrix}
1 & x_1\\
1& x_2\\
\vdots & \vdots\\
1 & x_n
\end{bmatrix}
\hspace{2cm}
\beta_{2\times1} = \begin{bmatrix}
b_0\\
b_1
\end{bmatrix}
\end{align*}
and so using linear algebra we have (all of my sums are with respect to $i$ and go to $n$, i.e., $\sum_{i=1}^n$)
\begin{align*}
\beta &= (X'X)^{-1}X'y\\
&=\left( \begin{bmatrix}
1 & 1&\cdots & 1\\
x_1 & x_2 & \cdots & x_n
\end{bmatrix} \begin{bmatrix}
1 & x_1\\
1& x_2\\
\vdots & \vdots\\
1 & x_n
\end{bmatrix}\right)^{-1}
\begin{bmatrix}
1 & 1&\cdots & 1\\
x_1 & x_2 & \cdots & x_n
\end{bmatrix}
\begin{bmatrix}
y_1\\
y_2\\
\vdots\\
y_n
\end{bmatrix}\\
&=\left(\begin{bmatrix}
n&\sum x_i\\
\sum x_i & \sum x_i^2
\end{bmatrix}\right)^{-1}\begin{bmatrix}
\sum y_i\\
\sum x_iy_i
\end{bmatrix}\\
& \hspace{-1.45in}\text{taking the inverse is not hard since it is a }2\times2 \text{ matrix}\\
&=\frac{1}{n\sum x_i^2-\left(\sum x_i\right)^2}\begin{bmatrix}
\sum x_i^2 & -\sum x_i\\
-\sum x_i & n
\end{bmatrix}\begin{bmatrix}
\sum y_i\\
\sum x_iy_i
\end{bmatrix}\\
&=\begin{bmatrix}
\frac{\sum x_i^2\sum y_i-\sum x_i\sum x_iy_i}{n\sum x_i^2-\left(\sum x_i\right)^2}\\
\frac{n\sum x_iy_i-\sum x_i\sum y_i}{n\sum x_i^2-\left(\sum x_i\right)^2}
\end{bmatrix}
\end{align*}
and so,
\begin{align*}
b_0 = \frac{\sum x_i^2\sum y_i-\sum x_i\sum x_iy_i}{n\sum x_i^2-\left(\sum x_i\right)^2}
\end{align*}
and
\begin{align*}
b_1 = \frac{n\sum x_iy_i-\sum x_i\sum y_i}{n\sum x_i^2-\left(\sum x_i\right)^2}
\end{align*}
Now we have $b_0$ and $b_1$ and so for any value of $x$ we can figure out the corresponding $y$ value. The cool thing about the least squares line is that it WILL ALWAYS pass through the point that corresponds to the mean of $x$ and the mean of $y$. Why is that true? Plug in $\bar x$ to $y=b_0+b_1x$ and after some algebra it's easy to see.
Let's plug in $\bar x$ for $x$ in $y=b_0+b_1x$, so
\begin{align*}
b_0+b_1\bar x &=\frac{\sum x_i^2\sum y_i-\sum x_i\sum x_iy_i}{n\sum x_i^2-\left(\sum x_i\right)^2}+\frac{n\sum x_iy_i-\sum x_i\sum y_i}{n\sum x_i^2-\left(\sum x_i\right)^2}\times\frac{1}{n}\sum x\\
&=\frac{\sum x_i^2\sum y_i-\sum x_i\sum x_iy_i}{n\sum x_i^2-\left(\sum x_i\right)^2}+\frac{n\sum x_iy_i\sum x_i-\sum x_i\sum y_i\sum x_i}{n^2\sum x_i^2-\left(\sum x_i\right)^2}\\
&=\frac{\sum x_i^2\sum y_i-\sum x_i\sum x_iy_i}{n\sum x_i^2-\left(\sum x_i\right)^2}+\frac{\sum x_iy_i\sum x_i}{n\sum x_i^2-\left(\sum x_i\right)^2}-\frac{\left(\sum x_i\right)^2\sum y_i}{n^2\sum x_i^2-\left(\sum x_i\right)^2}\\
&=\frac{\sum x_i^2\sum y_i-\sum x_i\sum x_iy_i+\sum x_iy_i\sum x_i}{n\sum x_i^2-\left(\sum x_i\right)^2}-\frac{\left(\sum x_i\right)^2\sum y_i}{n^2\sum x_i^2-\left(\sum x_i\right)^2}\\
&=\frac{\sum x_i^2\sum y_i}{n\sum x_i^2-\left(\sum x_i\right)^2}-\frac{\left(\sum x_i\right)^2\sum y_i}{n^2\sum x_i^2-\left(\sum x_i\right)^2}\\
&=\frac{\sum x_i^2\sum y_i-\left(\sum x_i\right)^2\sum y_i}{n\sum x_i^2-\left(\sum x_i\right)^2}\\
&=\frac{\sum y_i\left(\sum x_i^2-\left(\sum x_i\right)^2\right)}{n\sum x_i^2-\left(\sum x_i\right)^2}\\
&=\frac{1}{n}\sum y_i\\
&\bar y
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/635670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
My work:
$(\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b})=\frac{1}{a+b}$
By squaring both sides, we get,
$\frac{\sin^8 x}{a^2}+\frac{\cos^8 x}{b^2}+2\frac{\sin^4 x \cos^4 x}{ab}=\frac{1}{(a+b)^2}$
$\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
So, now, we have to prove that,
$-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=0$
I cannot do this. Please help!
| My Solution:: Given $$\displaystyle \frac{\sin^4 x}{a}+\frac{\cos^4 x}{b} = \frac{1}{a+b}.$$
Now using the Cauchy-Schwarz inequality we get
$$\displaystyle \frac{(\sin^2 x)^2}{a}+\frac{(\cos^2 x)}{b}\geq \frac{\left(\sin^2 x+\cos^2 x\right)^2}{a+b} = \frac{1}{a+b}$$
and equality holds when $$\displaystyle \frac{\sin^2 x}{a} = \frac{\cos^2 x}{b}.$$
Now using ratio and proportion, we get $$\displaystyle \frac{\sin^2 x}{a} = \frac{\cos^2 x}{b}=\frac{\sin^2 x+\cos^2 x}{a+b}=\frac{1}{a+b}.$$
So $$\displaystyle \sin^2 x = \frac{a}{a+b}$$ and $$\displaystyle \cos^2 x = \frac{b}{a+b}.$$ So we get $$\displaystyle \frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{a^3}{a^2\cdot (a+b)^3}+\frac{b^3}{b^2\cdot (a+b)^3} = \frac{1}{(a+b)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/639223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Evaluating indefinite integrals Evaluate the following indefinite integral.
$ \int { { \sin }^{ 6 } } x\quad dx $
My try :
$ \int { ({ \sin^2x } } )^{ 3 }dx\\ \int { (\frac { 1 }{ 2 } } (1-\cos2x))^{ 3 }dx\\ \int { \frac { 1 }{ 8 } } (1-\cos2x)^{ 3 }dx\\ \frac { 1 }{ 8 } \int { (1-\cos2x)^{ 3 } } dx\\ \frac { 1 }{ 8 } \int { 1-3\cos2x+3\cos^{ 2 } } 2x-\cos^{ 3 }2x\quad dx $
Then i got stuck.
| You're on the right track!
Now do a couple more substitutions:
$$\cos^2 (2x) = \frac{1 + \cos(4x)}{2};$$
$$\cos^3 (2x) = (1 - \sin^2 2x)\cos 2x.$$
So from where you were:
$$I = \frac { 1 }{ 8 } \int ({ 1-3\cos2x+3\cos^{ 2 } } 2x-\cos^{ 3 }2x) dx$$
$$I = \frac { 1 }{ 8 } \int (1 - 3 \cos 2x+\frac{3(1+\cos 4x)}{2} - (1 - \sin^2 2x)(\cos 2x)) dx$$
$$I = \frac { 1 }{ 8 } \int (\frac{5}{2} - 2 \cos 2x + \frac{3}{2}\cos 4x + \sin^2 2x \cos 2x) dx.$$
Can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/641199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
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Two trigonometric eliminations I have a couple of trigonometric elimination questions for you...
$1) a \sin \theta = b \sin (2 \theta)$; $c \cos \theta = d \cos (2 \theta)$
*NOTE: the first equation was incorrect...I had $\cos 2 \theta$ and it should be $\sin 2 \theta$.
I was able to get the answer to $2(abc - a^2 d + 2b^2 d)=0$, but the actual answer is $(c^2 - d^2)(abc - a^2 d + 2b^2 d)=0$. Where does the $(c^2 -d^2)$ come from?
UPDATE: I think I know $(c^2-d^2)$ comes from. If $\theta$ =0, then cos $\theta$ = 1 and in turn cos 2 $\theta$ = 1; plugging these values into the first equation gives us $c = d$. (I already did the second condition where $\frac {a} {2b}$)) If we square the first condition and subtract, that gives us $(c^2-d^2)$; since this also equals $2(abc - a^2 d + 2b^2 d)=0$, we can multiply both sides by $(c^2-d^2)$ to get the final result.
UPDATE #2: Actually, the more I thought about it, this is correct.
$2) \displaystyle x \cos \theta + y \sin \theta = \cos (3 \theta)$; $x \sin \theta - y \cos \theta = 3 \sin (3 \theta)$
I'm not sure where to start...I've tried squaring both equations, multiplying, etc. (The answer for this is $\displaystyle(x^2 + y^2)(x^2+y^2+18)+8x(x^2-3y^2) = 27$.)
UPDATE: I was able to get the following for x and y (thanks to the suggestion!)
$\displaystyle x= 2\cos 2 \theta - \cos 4 \theta$
$\displaystyle y = 2\sin 2 \theta + \sin 4 \theta$
UPDATE 2/16/14: BREAKTHROUGH! Never mind...that answer bombed out.
UPDATE 2/17/14: Not giving up and using the advice below, I finally got $\displaystyle c = \frac {(x-1)^2 + (y^2-4)}{12}$...I'm going to plug this in to the equation for x and see how it goes. If all works well, I should finally get the answer, save all the cleanup work. Never mind.
UPDATE 4/8/17: I reposted question #2 to get a fresh perspective...see Trigonometric elimination (reprise from 2014). If it's a duplicate, my apologies!
Thanks for your help!
| HINT:
$1)$ For the original version:
Divide to find $\tan\theta$
and use $\displaystyle\cos2A=\frac{1-\tan^2A}{1+\tan^2A}$
either $(i)$ in square one of the given equation
or $(ii)$ in $$\left(\frac{b\cos2\theta}b\right)^2+\left(\frac{d\cos2\theta}c\right)^2=\sin^2\theta+\cos^2\theta=\cdots$$
For the edited version
From the first relation, $\displaystyle\sin\theta(a-2b\cos\theta)=0$
If $\displaystyle\sin\theta=0,\cos\theta=\pm1,\cos2\theta=2\cos^2\theta-1=1$
Else $\displaystyle\cos\theta=\frac a{2b}$
Put the values of $\displaystyle\cos\theta,\cos2\theta$ in the second relation, $\displaystyle c\cos\theta=d\cos2\theta$
$2)$
Solving for $x,y$
$\displaystyle x=2\cos2\theta-\cos4\theta=2c-(2c^2-1)\iff 2c^2=2c-1-x\ \ \ \ (A)$ where $c=\cos2\theta$
$\displaystyle y=2\sin2\theta-\sin4\theta=2\sin2\theta(1-\cos2\theta)$
$\displaystyle\implies y^2=4(1-c^2)(1-c)^2\ \ \ \ (B)$
Put the value of $2c^2$ from $(A)$ in $(B)$ to find $c$ in terms of $y^2$
Then, put this value of $c$ in $(A)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/642789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Probability number comes up before another In craps, say I roll a 5 for the place bet. What is the probability that I roll another 5 before rolling a 7?
Is this correct?: $P(5 \text{ before 7}) = P(5) + P(\neg 7 \neg 5, 5) + P(\neg 7 \neg 5, \neg 7 \neg 5, 5) + ....$. This becomes $\frac{4}{36}+\frac{26}{36}\frac{4}{36} + ...\left( \frac{26}{36} \right)^n\frac{4}{36} = \frac{1}{9} \sum \limits_{i=0}^n \left( \frac{26}{36} \right)^i = \frac{1}{9} \frac{36}{10} = \frac{2}{5}$ ?
| Your notation $\neg 7 \neg 5$ confused me a bit and so I composed a long
answer, but now that I understand what you wrote a little better,
yes, what you have written is correct.
When you roll a 5, that becomes your point and then you repeatedly roll the dice until either your point shows up and you win, or you roll a 7 and you lose. Thus, having established a point of 5, your (conditional) win probability is
$$P(5)+P(N,5) + P(N,N,5) + \cdots =
\frac{1}{9} + \frac{13}{18}\times \frac{1}{9}
+ \left(\frac{13}{18}\right)^2\times \frac{1}{9} + \cdots
= \frac{1}{9}\times \frac{1}{1-\frac{13}{18}} = \frac{2}{5}$$
where $N$ is the event that the roll is neither 5 nor 7 (what you
have written as $\neg 7 \neg 5$).
More generally, if $A$ and $B$ are mutually exclusive events,
then on a sequence of independent trials, the probability that
$A$ occurs before $B$ is
$\displaystyle \frac{P(A)}{P(A)+P(B)}.$
See, for example, this answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/643352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$z^2=-3-3i$ solve for $z$ Use De Moivre's theorem to solve the equation $z^2=-3-3i$. (Give your answers in polar form)
Can you please explain why there are two answers? I cannot seem to understand why.
By the way, the answers are
$$z=18^{1/4}\mathrm{cis}(-3\pi/8)$$
$$z=18^{1/4}\mathrm{cis}(5\pi/8)$$
Thanks for the help.
| Notice that $-3 - 3i$ is equivalent to $3\sqrt{2}e^{-\frac{3\pi}{4}i}$. To see how, write $-3-3i$ as $3(-1 - i)$. Look at
$$-1 - i = re^{i\theta}$$
where $\theta$ is the angle and $r$ is the radius. Then
$$\begin{aligned}
-1 &= r\cos(\theta)\\
-1 &= r\sin(\theta)
\end{aligned}$$
With some algebra, we have
$$\begin{aligned}
(r\cos(\theta))^2 + (r\sin(\theta)^2) &= (-1)^2 + (-1)^2\\
r^2(\cos^2(\theta) + \sin^2(\theta)) &= 2\\
r^2 &= 2\\
r &= \sqrt{2}\\
\Rightarrow -1 &= \sqrt{2}\cos(\theta)\\
\Rightarrow -1 &= \sqrt{2}\sin(\theta)\\
\Longrightarrow \theta &= -\dfrac{3\pi}{4}
\end{aligned}$$
So we have $3\sqrt{2}e^{-\frac{3\pi}{4}i}$, which by Euler's identity is equivalent to
$$3\sqrt{2}\left(\cos\left(-\dfrac{3\pi}{4}\right) + i\sin\left(-\dfrac{3\pi}{4}\right)\right)$$
which gives us
$$\begin{aligned}
z^2 &= 3\sqrt{2}\left(\cos\left(-\dfrac{3\pi}{4} \right) - i\sin\left(\dfrac{3\pi}{4}\right)\right)\\
z &= \left(3\sqrt{2}\left(\cos\left(-\dfrac{3\pi}{4} \right) - i\sin\left(\dfrac{3\pi}{4}\right)\right) \right)^{\frac{1}{2}}\\
\end{aligned}$$
with the following solutions found by De Moivre's Theorem:
$$\begin{aligned}
z_1 &= (3\sqrt{2})^{\frac{1}{2}}\left(\cos\left(\dfrac{-\frac{3\pi}{4}}{2} \right) + i\sin\left(-\dfrac{\frac{3\pi}{4}}{2} \right) \right)\\
&= (3\sqrt{2})^{\frac{1}{2}}\left(\cos\left(-\dfrac{3\pi}{8} \right) + i\sin\left( -\dfrac{3\pi}{8}\right)\right)\\
&= (3\sqrt{2})^{\frac{1}{2}}e^{-\frac{3\pi}{8}i}\\
&= (\sqrt{18})^{\frac{1}{2}}e^{-\frac{3\pi}{8}i}\\
&= 18^{\frac{1}{4}}e^{-\frac{3\pi}{8}i}\\
z_2 &= (3\sqrt{2})^{\frac{1}{2}}\left(\cos\left(\dfrac{-\frac{3\pi}{4} + 2\pi}{2} \right) + i\sin\left(-\dfrac{\frac{3\pi}{4} + 2\pi}{2} \right) \right)\\
&= 18^{\frac{1}{4}}e^{\frac{5\pi}{8}i}
\end{aligned}$$
By mathematics2x2life's comment, those solutions are equivalent to the following:
$$\begin{aligned}
z_1 &= 18^{\frac{1}{4}}\mathrm{cis}\left(-\dfrac{3\pi}{8} \right)\\
z_2 &= 18^{\frac{1}{4}}\mathrm{cis}\left(\dfrac{5\pi}{8} \right)
\end{aligned}$$
References
*
*Cis $\theta$: http://math.wikia.com/wiki/Cis_%CE%B8
*De Moivre's Theorem: http://www3.ul.ie/~mlc/support/Loughborough%20website/chap10/10_4.pdf
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/644404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $x\in\mathbb R$, solve $4x^2-40\lfloor x\rfloor+51=0$.
If $x\in\mathbb R$, solve $$4x^2-40\lfloor x\rfloor+51=0$$
where $\lfloor x\rfloor$ denotes the integer part of the number. $\lfloor x\rfloor\le x$ and $\lfloor x\rfloor=x-\{x\}$, where $\{x\}$ marks the fraction part of the number. $0\le\{x\}<1$.
Not sure if all these denotions are conventions that always mean what I've mentioned.
Also, I'm not too sure if $x\in\mathbb R$, since the problem doesn't mention it, but it seems quite obvious that it most likely is that way.
| Since $$x-1\lt \lfloor x\rfloor \le x,$$
we have
$$\begin{align}x-1\lt \frac{4x^2+51}{40}\le x&\iff 4x^2-40x+91\gt0\ \text{and}\ 4x^2-40x+51\le 0\\&\iff 1.5\le x\lt 3.5\ \text{or}\ 6.5\lt x\le 8.5.\end{align}$$
1) When $1.5\le x\lt 2\Rightarrow \lfloor x\rfloor=1$,
$$4x^2-40\times 1+51=0$$
does not have any real solution.
2) When $2\le x\lt 3\Rightarrow \lfloor x\rfloor=2$,
$$4x^2-40\times 2+51=0\Rightarrow x=\pm \sqrt{29}/2\Rightarrow x=\sqrt{29}/2.$$
3) When $3\le x\lt 3.5\Rightarrow \lfloor x\rfloor=3$,
$$4x^2-40\times 3+51=0\Rightarrow x=\pm \sqrt{69}/2.$$
But these don't satisfy $3\le x\lt 3.5.$
4) When $6.5\lt x\lt 7\Rightarrow \lfloor x\rfloor=6$,
$$4x^2-40\times 6+51=0\Rightarrow x=\pm 3\sqrt{21}/2\Rightarrow x=3\sqrt{21}/2.$$
5) When $7\le x\lt 8\Rightarrow \lfloor x\rfloor=7$,
$$4x^2-40\times 7+51=0\Rightarrow x=\pm \sqrt{229}/2\Rightarrow x=\sqrt{229}/2.$$
6) When $8\le x\le 8.5\Rightarrow \lfloor x\rfloor=8$,
$$4x^2-40\times 8+51=0\Rightarrow x=\pm \sqrt{269}/2\Rightarrow x=\sqrt{269}/2.$$
Hence, the answer is
$$x=\frac{\sqrt{29}}{2},\frac{3\sqrt{21}}{2},\frac{\sqrt{229}}{2},\frac{\sqrt{269}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/645024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Solve a system of three equations by rewriting in row-echelon form I tried solving this system of equations and I got what seems to be an inconsistent system. I wanted to post my results here to see if I'm correct.
Here is the original problem:
$$ \left\{ \begin{aligned} 3x_1-2x_2+4x_3&=1 \\ x_1+x_2-2x_3&=3 \\ 2x_1-3x_2+6x_3&=8 \end{aligned} \right. $$
For the first step I interchanged equation one with equation two to give the following:
$$ \left\{ \begin{aligned} x_1+x_2-2x_3&=3 \\ 3x_1-2x_2+4x_3&=1 \\ 2x_1-3x_2+6x_3&=8 \end{aligned} \right. $$
I then added (-3) times equation one to equation two to give a new equation two:
$$ \left\{ \begin{aligned} x_1+x_2-2x_3&=3 \\ -5x_2+10x_3&=-8 \\ 2x_1-3x_2+6x_3&=8 \end{aligned} \right. $$
Next I added (-2) times equation one to equation three to give a new equation three:
$$ \left\{ \begin{aligned} x_1+x_2-2x_3&=3 \\ -5x_2+10x_3&=-8 \\ -5x_2+10x_3&=-2 \end{aligned} \right. $$
So it was at this point that I thought something was wrong, but I kept going. Next I added (-1) times equation two to equation three to give a new equation three:
$$ \left\{ \begin{aligned} x_1+x_2-2x_3&=3 \\ -5x_2+10x_3&=-8 \\ 0=10 \end{aligned} \right. $$
Did I do something wrong, or is the properly solved? We haven't gotten to matrices yet, which I know will be tremendously easier, our teacher wants us to get comfortable with doing it this way first.
| I redid the calculation on the computer, it yields:
$$\begin{bmatrix} 3 & -2 & 4 & 1 \\ 1 & 1 & -2 & 3 \\ 2 & -3 & 6 &
8 \\ \end{bmatrix}
\xrightarrow{R_1 \leftrightarrow R_2}
\begin{bmatrix} 1 & 1 & -2 & 3 \\ 3 & -2 & 4 & 1 \\ 2 & -3 & 6 &
8 \\ \end{bmatrix}
\xrightarrow{R_2 \gets R_2 -3 R_1}
\begin{bmatrix} 1 & 1 & -2 & 3 \\ 0 & -5 & 10 & -8 \\ 2 & -3 & 6 &
8 \\ \end{bmatrix}
\xrightarrow{R_3 \gets R_3 -2 R_1}
\begin{bmatrix} 1 & 1 & -2 & 3 \\ 0 & -5 & 10 & -8 \\ 0 & -5 & 10 &
2 \\ \end{bmatrix}
\xrightarrow{R_3 \gets R_3 -R_2}
\begin{bmatrix} 1 & 1 & -2 & 3 \\ 0 & -5 & 10 & -8 \\ 0 & 0 & 0 &
10 \\ \end{bmatrix}$$
which matches your final answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/646746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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private solution after solving nonhomogenous euler equation
Solve the equalation: $$x^2y''-3xy'+3y=\ln x$$
First I solved the homogeneous part and got $$y_h=c_1x+c_2x^3.$$ Then i wanted using variation of parameters writing that $$c_1' x'+c_2' (x^3)^\prime.$$ I solved the equations for $c_1,c_2$ and got a function which when i substitute in the equation, I get $x^2\ln x$ (not $\ln x$). I verified my answers with MuPaD and seems like the integration was correct. Am I using wrong method?
| You have the correct homogeneous solution: $y_h = c_1 x+c_2x^3$.
Now, let $y_1=x$ and $y_2=x^3$. If we let $y_p=u_1(x)y_1(x) + u_2(x)y_2(x)$, our objective is to find $u_1(x)$ and $u_2(x)$ using Variation of Parameters. When I took differential equations, we used Cramer's rule to show that if $y_p = u_1y_1 + u_2y_2$ was a solution to $y^{\prime\prime}+P(x)y^{\prime}+Q(x)y=f(x)$, then it followed that
$$u_1^{\prime}(x) = \frac{\det\begin{bmatrix}0 & y_2\\ f(x) & y_2^{\prime}\end{bmatrix}}{\det\begin{bmatrix}y_1 & y_2\\ y_1^{\prime} & y_2^{\prime}\end{bmatrix}} = -\frac{y_2 f(x)}{W(y_1,y_2)}\tag{1}$$
and
$$u_2^{\prime}(x) = \frac{\det\begin{bmatrix}y_1 & 0\\ y_1^{\prime} & f(x)\end{bmatrix}}{\det\begin{bmatrix}y_1 & y_2\\ y_1^{\prime} & y_2^{\prime}\end{bmatrix}}= \frac{y_1f(x)}{W(y_1,y_2)}\tag{2}$$
where $W(y_1,y_2)=\det\begin{bmatrix}y_1 & y_2 \\ y_1^{\prime} & y_2^{\prime}\end{bmatrix}$ is the Wronskian of $y_1$ and $y_2$. We also note that we can rewrite our original ODE as follows:
$$x^2y^{\prime\prime}-3xy^{\prime}+3y = \ln x \implies y^{\prime\prime} - \frac{3}{x}y^{\prime}+\frac{3}{x^2}y = \frac{\ln x}{x^2}.$$
Thus, $f(x)=\dfrac{\ln x}{x^2}$. Plugging our $y_1$, $y_2$ and $f(x)$ into $(1)$ and $(2)$ gives us
$$u_1^{\prime}(x) = \frac{\det\begin{bmatrix}0 & x^3\\ \frac{\ln x}{x^2} & 3x^2\end{bmatrix}}{\det\begin{bmatrix}x & x^3\\ 1 & 3x^2\end{bmatrix}} = -\frac{\ln x}{2x^2}\qquad\text{ and } \qquad u_2^{\prime}(x) = \frac{\det\begin{bmatrix}x & 0\\ 1 & \frac{\ln x}{x^2}\end{bmatrix}}{\det\begin{bmatrix}x & x^3\\ 1 & 3x^2\end{bmatrix}} =\frac{\ln x}{2x^4}$$
We now integrate each one to see that $u_1(x)=\dfrac{1+\ln x}{2x}$
$$\begin{aligned}u_1(x) &= -\frac{1}{2}\int \frac{\ln x}{x^2}\,dx \\ &= -\frac{1}{2}\int te^{-t}\,dt;\quad(\text{sub: }t=\ln x)\\ &= -\frac{1}{2}\left[-te^{-t} + \int e^{-t}\,dt\right];\quad(\text{parts: }u=t,\,\,dv = e^{-t}\,dt)\\ &= \frac{1}{2}e^{-t}(1+t)\\ &= \frac{1+\ln x}{2x}\end{aligned}$$
and $u_2(x)=-\dfrac{1+3\ln x}{18x^3}$
$$\begin{aligned}u_2(x) &= \frac{1}{2}\int\frac{\ln x}{x^4}\,dx \\ &= \frac{1}{2}\int te^{-3t}\,dt;\quad (\text{sub: } t=\ln x)\\ &= \frac{1}{2}\left[-\frac{1}{3}te^{-3t} + \frac{1}{3}\int e^{-3t}\,dt\right];\quad(\text{parts: }u=t,\,\,dv = e^{-3t}\,dt)\\ &= -\frac{1}{18}e^{-3t}(3t+1)\\ &= -\frac{1+3\ln x}{18x^3}\end{aligned} $$
Note that I didn't add any constants of integration to $u_1(x)$ and $u_2(x)$; the reason for this is because when we consider the general solution $y = y_h + y_p$, the constants of integration from this process get "absorbed" into the respective arbitrary coefficients of the homogeneous solution.
Therefore, the particular solution you should get is $$\begin{aligned}y_p &= u_1(x)y_1(x) + u_2(x)y_2(x) \\ &= \frac{1+\ln x}{2} - \frac{1+3\ln x}{18} \\ &= \frac{4}{9} +\frac{1}{3}\ln x \\ &= \frac{1}{9}(4+3\ln x)\end{aligned}$$
and hence the general solution is
$$y=c_1x+c_2x^3+\frac{1}{9}(4+3\ln x)$$
which matches the solution given by WolframAlpha.
Alternatively, you could have avoided Variation of Parameters completely by solving this equation using the method of undetermined coefficients; in particular, you could have guessed a particular solution of the form $y_p = A+B\ln x$ and then you could have solved for $A$ and $B$. You can read more about the method of undetermined coefficients for Cauchy-Euler equations here.
I hope this makes sense!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this $P_{n}(x)=\frac{1}{\pi}\int_{0}^{\pi}\frac{dt}{(x-\sqrt{x^2-1}\cos{t})^n}dt$ let $x>1$ and $n\in N$,show that
the function
$$P_{n}(x)=\dfrac{1}{\pi}\int_{0}^{\pi}(x+\sqrt{x^2-1}\cos{t})^ndt$$
is polynomial and the dgree is $n$.and we have
$$P_{n}(x)=\dfrac{1}{\pi}\int_{0}^{\pi}\dfrac{1}{(x-\sqrt{x^2-1}\cos{t})^n}dt$$
maybe this is Legendre Polynomial,But I can't.
| First Integral
Substitute $t\mapsto\frac\pi2-t$, then exploit the oddness of $\sin(t)$:
$$
\begin{align}
\int_0^\pi(x+\sqrt{x^2-1}\cos(t))^n\,\mathrm{d}t
&=\int_{-\pi/2}^{\pi/2}(x+\sqrt{x^2-1}\sin(t))^n\,\mathrm{d}t\\
&=\int_{-\pi/2}^{\pi/2}\sum_{k=0}^n\binom{n}{k}x^{n-k}\left(\sqrt{x^2-1}\sin(t)\right)^k\,\mathrm{d}t\\
&=\int_{-\pi/2}^{\pi/2}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{n-2k}\left(\sqrt{x^2-1}\sin(t)\right)^{2k}\,\mathrm{d}t\\
&=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{n-2k}(x^2-1)^k\frac\pi{4^k}\binom{2k}{k}
\end{align}
$$
and each term of the last sum is a polynomial of degree $n$ with positive lead coefficient.
Second Integral
$$
\begin{align}
\pi I_n(x)
&=\int_0^\pi\frac{\mathrm{d}t}{(x-\sqrt{x^2-1}\cos(t))^{n+1}}\\
&=\frac1{x^{n+1}}\int_{-\pi/2}^{\pi/2}\frac{\mathrm{d}t}{(1-\sqrt{1-1/x^2}\sin(t))^{n+1}}\\
&=\frac1{x^{n+1}}\int_{-\pi/2}^{\pi/2}\sum_{k=0}^\infty\binom{-n-1}{k}\left(-\sqrt{1-1/x^2}\sin(t)\right)^k\,\mathrm{d}t\\
&=\frac1{x^{n+1}}\int_{-\pi/2}^{\pi/2}\sum_{k=0}^\infty\binom{-n-1}{2k}\left((1-1/x^2)\sin^2(t)\right)^k\,\mathrm{d}t\\
&=\frac1{x^{n+1}}\sum_{k=0}^\infty\binom{-n-1}{2k}(1-1/x^2)^k\frac\pi{4^k}\binom{2k}{k}\\
&=\frac\pi{x^{n+1}}\sum_{k=0}^\infty\binom{n+2k}{2k}\sqrt{\frac{1-1/x^2}{4}}^{\,\large2k}\binom{2k}{k}\\
\end{align}
$$
Setting $u=\sqrt{\frac{1-1/x^2}{4}}$, we have $\displaystyle\sum_{n=0}^\infty I_n(x)v^n$ is
$$
\begin{align}
\frac1x\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{n+2k}{2k}\binom{2k}{k}u^{2k}(v/x)^n
&=\frac1x\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{n+2k}{n}\binom{2k}{k}u^{2k}(v/x)^n\\
&=\frac1x\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{-2k-1}{n}\binom{2k}{k}u^{2k}(-v/x)^n\\
&=\frac1{x-v}\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{u}{1-v/x}\right)^{2k}\\
&=\frac1{x-v}\left(1-4\left(\frac{u}{1-v/x}\right)^2\right)^{-1/2}\\
&=\frac1{\sqrt{1-2vx+v^2}}\\
&=\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{v(2x-v)}{4}\right)^k
\end{align}
$$
The terms with $v^n$ appear when $k\le n\le2k$. Thus, $x^k$ appears in $I_n(x)$ when $n/2\le k\le n$. Therefore, $I_n(x)$ is a degree $n$ polynomial and the coefficient of $x^n$ in $I_n(x)$ is $\binom{2n}{n}2^{-n}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.
My attempt: Let
$$\begin{align*}
f_n(x)
&= \frac{\ln\left(1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\left(1-2a\cos x+a^2\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(\displaystyle\frac{1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2}{1-2a\cos x+a^2}\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(1+\dfrac{1}{n}\left(\displaystyle\frac{2a-2\cos x+\frac{1}{n}}{1-2a\cos x+a^2}\right)\right)}{\frac{1}{n}}.
\end{align*}$$
Now it is easy to see that $f_n(x) \to \frac{2a-2\cos x}{1-2a\cos x+a^2}$ as $n \to \infty$. $|f_n(x)|\le \frac{2a+2}{(1-a)^2}$ RHS is integrable so $\lim_{n\to\infty}\int_0^\pi f_n(x)dx = \int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2} dx=I'(a)$. But
$$\int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2}=\int_0^\pi\left(1-\frac{(1-a)^2}{1-2a\cos x+a^2}\right)dx.$$ Consider
$$\int_0^\pi\frac{dx}{1-2a\cos x+a^2}=\int_0^\infty\frac{\frac{dy}{1+t^2}}{1-2a\frac{1-t^2}{1+t^2}+a^2}=\int_0^\infty\frac{dt}{1+t^2-2a(1-t^2)+a^2(1+t^2)}=\int_0^\infty\frac{dt}{(1-a)^2+\left((1+a)t\right)^2}\stackrel{(*)}{=}\frac{1}{(1-a)^2}\int_0^\infty\frac{dt}{1+\left(\frac{1+a}{1-a}t\right)^2}=\frac{1}{(1-a)(1+a)}\int_0^\infty\frac{du}{1+u^2}=\frac{1}{(1-a)(1+a)}\frac{\pi}{2}.$$
So $$I'(a)=\frac{\pi}{2}\left(2-\frac{1-a}{1+a}\right)\Rightarrow I(a)=\frac{\pi}{2}\left(3a-2\ln\left(a+1\right)\right).$$
It looks too easy, is there any crucial lack?
$(*)$ — we have to check $a=1$ here by hand and actually consider $[0,1), (1,\infty)$ but result on these two intervals may differ only by constant - it may be important but in my opinion not crucial for this proof.
| By my post, I had found
$$
\int_{0}^{\pi} \ln (b \cos x+c)=\pi \ln \left(\frac{c+\sqrt{c^{2}-b^{2}}}{2}\right) \tag*{(*)}
$$
for any $c\neq 0$ and $-1\leq \frac{b}{c} \leq 1.$
Putting $b=-2 a$ and $c=a^{2}+1$ into $(*)$, we have
$$
\begin{aligned}
\int_{0}^{\pi} \ln \left(1-2 a \cos x+a^{2}\right) d x =& \pi \ln \left(\frac{a^{2}+1+\sqrt{\left(a^{2}+1\right)^{2}-4 a^{2}}}{2}\right) \\
=& \pi \ln \left(\frac{a^{2}+1+\sqrt{\left(a^{2}-1\right)^{2}}}{2}\right) \\
=& \pi \ln \left(\frac{a^{2}+1+\left|a^{2}-1\right|}{2}\right)
\end{aligned}
$$
We can now conclude that
For any $|a| \leqslant 1$,
$$
I=\pi \ln \left(\frac{\left.a^{2}+1+1-a^{2}\right)}{2}\right)=0
$$
for any $|a|>1$,
$$
I=\pi \ln \left(\frac{a^{2}+1+a^{2}-1}{2}\right)=2 \ln |a|
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/650513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 6
} |
Problem with counting limit. I am obliged to count the
$$\lim_{n\rightarrow\infty} a_n$$
where
$$a_n=\frac{1}{n^2+1}+\frac{2}{n^2+2}+\frac{3}{n^2+3}+\cdots+\frac{n}{n^2+n}.$$
I know that this type of task, I should use squeeze theorem. I should find two sequences one that is smaller and one that is bigger. So I find $\large b_n=\frac{1}{n^2+1}$, $ b_n \le a_n$, but I have trouble in finding the bigger sequence $c_n \ge a_n$. I will be glad for any help.
| We will use the following: $$\sum_{k=1}^n k=\frac{n(n+1)}{2}.$$
Since $n\geq 1$, we have
$$a_{n}=\frac{1}{n^2+1}+\frac{2}{n^2+2}+\frac{3}{n^2+3}+\cdots+\frac{n}{n^2+n}\geq\frac{1}{n^2+n}+\frac{2}{n^2+n}+\frac{3}{n^2+n}+\cdots+\frac{n}{n^2+n}=\frac{n(n+1)}{2(n^2+n)}$$
and
$$a_{n}=\frac{1}{n^2+1}+\frac{2}{n^2+2}+\frac{3}{n^2+3}+\cdots+\frac{n}{n^2+n}\leq\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\cdots+\frac{n}{n^2}=\frac{n(n+1)}{2n^2}.$$
By the squeeze theorem, $\lim_{n\rightarrow\infty} a_n=\frac{1}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Ramanujan's partial fraction decomposition of $\frac{1}{(x^2+a^2)\cdots(x^2+(a+n)^2)}$. \begin{align*}
\frac{1}{(x^2+a^2)\cdots(x^2+(a+n)^2)} &= \frac{2\Gamma(2a)}{\Gamma(n)\Gamma(2a+n)}\left(\frac{a}{x^2+a^2}-\frac{2a}{1!}\frac{n-1}{n+2a}\frac{a+1}{x^2+(a+1)^2}\right. \\
& \qquad + \left.\frac{2a(2a+1)}{2!}\frac{(n-1)(n-2)}{(n+2a)(n+2a+1)}\frac{a+2}{x^2+(a+2)^2}-\cdots\right).
\end{align*}
The preceding was by Ramanujan, appearing in one of his notebooks. How does one prove this?
Especially interesting is motiving the proof: given only the complete fraction on the left, is there a method that makes the right side almost immediately obvious? (Basically, it would be nice if the answers imagined the RHS didn't exist in the above equation).
| Since both the left- and right-hand sides are in terms of $x^2$
we can change variables to $X = x^2$. This is a special case of
the question of finding the partial-fraction expansion
$$
\frac1{(X+A_1)(X+A_2)\cdots(X+A_n)}
= \sum_{i=1}^n \frac{C_i}{X+A_i}
$$
for any distinct $A_1,A_2,\ldots,A_n$.
The easy way to find $C_i$ is to multiply both sides by $X+A_i$
and then to evaluate at $X = -A_i$. On the right side this isolates $C_i$.
On the left side we get the product over $j \neq i$ of $1/(A_j-A_i)$.
So $C_i$ must equal this product.
In the present case, each $A_i$ is $(a+i)^2$,
so $A_j-A_i = (a+j)^2 - (a+i)^2$, which factors further as $(j-i)(2a+j+i)$.
The product of this over $1 \leq j \leq n$ excluding $j=i$ can then be
expressed in various ways in terms of factorials and Gamma functions,
one of which yields Ramanujan's choice.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use induction to prove $2n + 1 \le 2^n$ for $n=3,4,\ldots$ Use induction to prove $2n + 1 \le 2^n$ for $n=3,4,\ldots$
I've plugged $3$ in for $n$
I get $7 \le 8$
then I set $2(n+1) +1 \le 2^{n+1}$
then I'm lost.
| We want to show that if $2k+1\le 2^k$ for a particular integer $k\ge 3$, then $2(k+1)+1\le 2^{k+1}$.
We have $2(k+1)+1=2k+1 +2$. By the induction hypothesis, $2k+1\le 2^k$, and therefore $2k+1+2 \le 2^k+2$.
But if $k\ge 1$, then $2^k+2\le 2^k+2^k=2^{k+1}$, so we are finished.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Closed-Form Solution to Infinite Sum Does the convergent infinite sum
$$
\sum_{n=0}^{\infty} \frac{1}{2^n + 1}
$$
have a closed form solution? Quickly coding this up, the decimal approximation appears to be $1.26449978\ldots$
| (Too long for comment)
For $n \ge 1$,
$$
\frac{1}{1 + 2^n} = -\frac{(-1/2^n)}{1 - (-1/2^n)} = (-1/2^n) + (-1/2^n)^2 + (-1/2^n)^3 + (-2^n)^3 + \cdots
$$
Which is an absolutely convergent series. Thus,
\begin{align*}
\sum_{n=1}^\infty \frac{1}{1 + 2^n}
&= \sum_{n = 1}^\infty \sum_{m = 1}^\infty \frac{(-1)^m}{2^{mn}} \\
&= \sum_{N = 1}^\infty \sum_{d | N} \frac{(-1)^d}{2^N} \\
&= \sum_{N = 1}^\infty \frac{1}{2^N}\left[ \text{# of even divisors of } N- \text{# of odd divisors of } N\right] \\
\end{align*}
Letting $N = 2^k l$ where $l$ is odd, and letting $\sigma_0$ be the number of divisors function, we get
\begin{align*}
\sum_{l \text{ odd }= 1}^\infty \sum_{k = 0}^\infty \frac{1}{2^{2^k l}} \left[ k\sigma_0(l) - \sigma_0(l)\right]
&= \sum_{l \text{ odd }= 1}^\infty \sigma_0(l) \sum_{k = 0}^\infty (k-1) \left(\frac{1}{2^l}\right)^{2^k} \\
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all finite set S such that any $a, b, c \in S$, $ab + bc + ca \in S$ Find all finite set of real numbers S such that: $ab + bc + ca \in S$ with any distinct $a, b, c \in S$
I just can solve the problem when $\exists$ at least $3$ elements $\in S\ge 1$
In another case, I got stuck.
| Any set of cardinality $< 3$ is (trivially) a solution.
There are three-element solutions $\{a, b, c\}$ with $b = \dfrac{a(1-c)}{a+c}$ for any $a$ and $c$ such that $a \ne -c$ and $a, \dfrac{a(1-c)}{a+c}, c$ are distinct.
There are four-element solutions in complex numbers, e.g.
$$ \eqalign{a&=\dfrac{1}{3}+\dfrac{\sqrt {13}}{6}+\dfrac{i}{6}\sqrt {13+2\,\sqrt {13}}\cr b & =\dfrac{1}{3}+\dfrac{
\sqrt {13}}{6}-\dfrac{i}{6}\sqrt {13+2\,\sqrt {13}}\cr c &=\dfrac{1}{3}-\dfrac{\sqrt {13}}{6}-\dfrac{i}{9}\sqrt {13+2\,\sqrt {13}}+\dfrac{i}{18}\sqrt {13}\sqrt {13+2\,\sqrt {13}}\cr
d &=\dfrac{1}{3}-\dfrac{\sqrt {13}}{6}+\dfrac{i}{9}\sqrt {13+2\,\sqrt {13}}-\dfrac{i}{18}\sqrt {13}\sqrt {13+2\,\sqrt {13}}\cr}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
expression for$\sum_{k=0}^{n}(k+1)(k+2)$ what is a closed expression for$\sum_{k=0}^{n}(k+1)(k+2)$? the answer says to look at Generating function $a_k=(k+1)(k+2)$ and derive twice $\dfrac{1}{x}=1+x+x^2+...$ so the solution starts: $\sum_{k=0}^{n}(k+1)(k+2)=[\dfrac{1}{(1-x)}.\dfrac{2}{(1-x)^3}]$but isn't $\sum_{k=0}^{\infty}t^k=\dfrac{1}{(1-t)}$? so how the lhs and rhs in $\sum_{k=0}^{n}(k+1)(k+2)=[\dfrac{1}{(1-x)}.\dfrac{2}{(1-x)^3}]$ are equal?
| Hint. You may want to write
$$ \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} (k+1)(k+2) \right) x^{n} = \frac{2}{(1-x)^{4}}. $$
Here is another solution using the Pascal's triangle:
$$ \sum_{k=0}^{n} (k+1)(k+2) = 2 \sum_{k=0}^{n} \binom{k+2}{2} = 2 \left[ \binom{2}{2} + \cdots + \binom{n+2}{2} \right]. $$
Now we know that $\binom{2}{2} = 1 = \binom{3}{3}$. So by applying the property $\binom{m}{k-1} + \binom{m}{k} = \binom{m+1}{k}$ repeatedly, we observe
\begin{align*}
\binom{2}{2} + \cdots + \binom{n+2}{2}
&= \binom{3}{3} + \binom{3}{2} + \binom{4}{2} + \cdots + \binom{n+2}{2} \\
&= \binom{4}{3} + \binom{4}{2} + \cdots + \binom{n+2}{2} \\
&= \binom{5}{3} + \cdots + \binom{n+2}{2} \\
&\quad \quad \vdots \\
&= \binom{n+2}{3} + \binom{n+2}{2} \\
&= \binom{n+3}{3}.
\end{align*}
So we obtain a special case of Christmas Stocking Theorem:
$$ \sum_{k=0}^{n} (k+1)(k+2) = 2 \sum_{k=0}^{n} \binom{k+2}{2} = 2 \binom{n+3}{3} = \frac{(n+3)(n+2)(n+1)}{3}. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Smallest non-zero eigenvalue of a (0,1) matrix What's the smallest absolute value possible of a non-zero eigenvalue of an $n$ by $n$ square matrix whose entries are either $0$ or $1$ (all operations are over $\mathbb{R}$)?
| For a $2\times 2$ matrix, the smallest is $\left| \dfrac{1-\sqrt{5}}{2}\right|$ from $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ or $\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$.
For a $3\times 3$ matrix, the smallest is $\left|\frac{1}{2}(3-\sqrt{5})\right|$ from $\begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ or any other matrix with all entries $1$ except for a single off (main) diagonal $0$.
Best so far - have more calculations to run and check
For a $4\times 4$ matrix, the smallest is $\left|2-\sqrt{3}\right|$ from $\begin{pmatrix} 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How can we make any integer m>11 using 3's and 5's only?? Is there any general solution of this? using 2 integers, what is the minimum number formed after which we can make any number using those 2 integers?
so it says 3a + 5b = m
now we know 4, 7 do not occur, but how can we say we can make any integer m>11 !!
I know 11 is the minimum, but how did it come,
what if it wasnt 3's and 5's say it was 4's and 5's?
is there any general method for this? I cant find it, please help.
| If you can do $m=12 \to 26$, then you can certainly do them all, since you just repeat adding a multiple of $15$, the LCM of $3$ and $5$.
So
$$12 = 3+3+3+3 \\ 13 = 5+5+3 \\ 14 = 3+3+3+5 \\ 15 = 5+5+5 \\ 16 = 3+3+5+5 \\ 17 = 3+3+3+5 \\ 18 = 3+3+3+3+3+3 \\ 19 = 5+5+3+3+3 \\ 20 = 5+5+5+5 \\ 21 = 3+3+3+3+3+3+3 \\ 22 = 5+5+3+3+3+3 \\ 23 = 5+3+3+3+3+3+3 \\ 24 = 3+3+3+3+3+3+3+3 \\ 25 = 5+5+5+5+5 \\ 26 = 5+3+3+3+3+3+3+3.$$
Now just add three $5$'s or five $3$'s to taste for each group of $15$ natural numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
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$(2^a -1)(2^b -1)=2^{2^c}+1$ has no nonnegative integer solutions $(2^a -1)(2^b -1)=2^{2^c}+1$ is not possible for a,b,c nonnegative integers.
Any solutions using parity
Approach:
$(2^a -1)(2^b -1)=2^{2^c}+1\Rightarrow$
$2^{a+b}-2^a-2^b=2^{2^c}\Rightarrow$
| Assume that there was a solution and wlog that $a<b$. Divide by $2^a$, then you have
$2^b-1-2^{b-a}=2^{2^c-a}$
The left side is clearly odd, the right side is even or a fraction if $2^c<a$. Therefore, there cannot exist such a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\lim_{x\to 0}\frac{1}{1-\cos(x^2)}{\sum_{n=4}^\infty {n^5x^n} }$ I am trying to calculate the following limit:
$$\lim_{x\to 0}\dfrac{1}{1-\cos(x^2)}{\sum_{n=4}^\infty {n^5 x^n} }$$
In general, I don't really know how to deal with Limits of infinite summation, so I tried to turn the series into a function, but I couldn't find the function.
Any help will be appreciated...
| The simplest way is to expand $\cos x^2$ into a Taylor series, then cancel out $x^4$ in the fraction.
See:
$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$$
Therefore $$1-\cos(x^2) = \frac{x^4}{2!} - \frac{x^8}{4!} + \frac{x^{12}}{6!}-\cdots=\\=x^4\left(\frac{1}{2} - \frac{x^4 }{4!}+\frac{x^{12}}{6!} - \cdots\right)$$
As for the numerator, it equals $$4^5x^4 + 5^5 x^5 + 6^5 x^6\cdots=\\=x^4\left(4^5 + 5^5x + 6^5x^2+\dots\right).$$
Writing the fraction allows you to cancel out $x^4$, sending $x$ to $0$ tell you that all the factors stil containing $x$ will equal zero, leaving you with the solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/670838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A closed form for the antiderivative of $\frac 1{\sin^5 x +\cos^5 x} $ Does there exist a closed form expression for $$\int \dfrac {dx}{\sin^5 x +\cos^5 x}? $$
| Computer algebra systems can do this. Maybe this is not needed, since the question (answered by Steven) was only whether closed form exists (not to exhibit it). But just for fun I gave it to Maple...
$$
1/5\,i\sqrt {\sqrt {5}+2}\ln \left( \left( \tan \left( 1/2\,x
\right) \right) ^{2}+ \left( 1/2\,i \left( \sqrt {5}+2 \right) ^{3/2
}+1/2+1/2\,\sqrt {5}-5/2\,i\sqrt {\sqrt {5}+2} \right) \tan \left( 1/2
\,x \right) -1/2\,i \left( \sqrt {5}+2 \right) ^{3/2}+1/2-1/2\,\sqrt {
5}+5/2\,i\sqrt {\sqrt {5}+2} \right)
\\-1/5\,i\sqrt {\sqrt {5}+2}\ln
\left( \left( \tan \left( 1/2\,x \right) \right) ^{2}+ \left( -1/2
\,i \left( \sqrt {5}+2 \right) ^{3/2}+1/2+1/2\,\sqrt {5}+5/2\,i\sqrt {
\sqrt {5}+2} \right) \tan \left( 1/2\,x \right) +1/2\,i \left( \sqrt {
5}+2 \right) ^{3/2}+1/2-1/2\,\sqrt {5}-5/2\,i\sqrt {\sqrt {5}+2}
\right)
\\+1/5\,\sqrt {-2+\sqrt {5}}\ln \left( \left( \tan \left( 1/2
\,x \right) \right) ^{2}+ \left( -1/2\, \left( -2+\sqrt {5} \right) ^
{3/2}+1/2-1/2\,\sqrt {5}-5/2\,\sqrt {-2+\sqrt {5}} \right) \tan
\left( 1/2\,x \right) +1/2\, \left( -2+\sqrt {5} \right) ^{3/2}+1/2+1
/2\,\sqrt {5}+5/2\,\sqrt {-2+\sqrt {5}} \right)
\\-1/5\,\sqrt {-2+\sqrt
{5}}\ln \left( \left( \tan \left( 1/2\,x \right) \right) ^{2}+
\left( 1/2\, \left( -2+\sqrt {5} \right) ^{3/2}+1/2-1/2\,\sqrt {5}+5/
2\,\sqrt {-2+\sqrt {5}} \right) \tan \left( 1/2\,x \right) -1/2\,
\left( -2+\sqrt {5} \right) ^{3/2}+1/2+1/2\,\sqrt {5}-5/2\,\sqrt {-2+
\sqrt {5}} \right)
\\+4/5\,\sqrt {2}{\rm{ arctanh}} \left( 1/4\, \left( 2
\,\tan \left( 1/2\,x \right) -2 \right) \sqrt {2} \right)
$$
In this case, we only had to solve two successive quadratic equations.
| {
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"timestamp": "2023-03-29T00:00:00",
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Volume above cone and below paraboloid. I need to find the volume above the cone $z=\sqrt{x^2+y^2}$ and below the paraboloid $z=2-x^2-y^2$. I thought about using spherical coordinates and finding $p$, which would be (before simplification) :
$$-z=-2+x^2+y^2$$
$$-p \cos(\theta) =-2+p^2 \sin^2 (\theta) \cos^2(\theta)+p^2 \sin^2(\theta) \sin^2(\theta)$$
But I can't seem to be able to isolate $p$ even knowing that $\sin^2(\theta)+\cos^2(\theta)=1$.
Any hint would be greatly appreciated!
Thanks.
| I believe cylindrical coordinates will be the easiest. When setting it up as a triple integral you will have
$$V = \iiint \, dV = \iint\limits_{R} \hspace{-5pt} \int_{\sqrt{x^2+y^2}}^{2 -x^2-y^2} \, dz \, dA = \iint\limits_{R} 2-x^2-y^2 - \sqrt{x^2+y^2} \, dA.$$
These surfaces intersect each other when $x^2+y^2=1$, therefore the projected region in the $xy$ plane is a disk of radius 1. Performing this coordinate change we have
$$
\begin{align}
V & = \iint\limits_{R} 2-x^2-y^2 - \sqrt{x^2+y^2} \, dA \\
& = \int_0^{2\pi} \hspace{-5pt} \int_0^1 (2-r^2 -r) r \, dr \, d \theta \\
& = 2 \pi \int_0^1 2r -r^3 -r^2 \, dr \\
& = 2 \pi \left( 1 - \frac{1}{4} - \frac{1}{3} \right) \\
& = \frac{5 \pi}{6}.
\end{align}
$$
| {
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Given positive real numbers $a, b, c$ with $aI am trying to prove the following:
$$\frac{a}{1+a} < \frac{b}{1+b} + \frac{c}{1+c}$$
given that $a, b, c > 0$ and $a < b+c$. I tried various rearrangements but can't seem to get anywhere with it.
| $$\frac{b}{1+b} + \frac{c}{1+c}>\frac{a}{a+1}$$
$$\frac{b+c+2bc}{1+b+c+bc}>\frac{a}{a+1}$$
Since $a,b,c>0$,
$$ab+ac+2abc+b+c+2bc>a+ab+ac+abc$$
$$abc+2bc+b+c>a$$
| {
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How to evaluate the integral: $I=\int_0^3\frac{x\sqrt{x+1}dx}{x^2+x+1}$ Evaluating this integral: $I=\int_0^3\frac{x\sqrt{x+1}dx}{x^2+x+1}$
I've tried:
Set : $\sqrt{x+1}=t\mapsto dx=2t\,dt \Longrightarrow I=2\int_1^2\frac{t^4-t^2}{t^4-t^2+1}dt=2\int_1^2(1-\frac{1}{t^4-t^2+1})dt$
And then, I set $(t^2-\frac{1}{2})=\frac{\sqrt{3}}{2}\tan v$ ... But I can't continue to solve this integral...
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\begin{align}
I &\equiv \int_{0}^{3}{x\root{x + 1}\,dx \over x^{2} + x + 1}
=\int_{0}^{3}x\root{x + 1}\,
\pars{{1 \over x + 1/2 - \root{3}\ic/2} - {1 \over x + 1/2 + \root{3}\ic/2}}\,
{1 \over \root{3}\ic}\,dx
\\[3mm]&={2\root{3} \over 3}\Im\int_{0}^{3}
{x\ \overbrace{\root{x + 1}}^{\ds{\equiv\ t}} \over x + 1/2 - \root{3}\ic/2}\,\dd x
={2\root{3} \over 3}\Im\int_{1}^{2}
{\pars{t^{2} - 1}t \over \pars{t^{2} - 1} + 1/2 - \root{3}\ic/2}\,2t\,\dd t
\\[3mm]&={4\root{3} \over 3}\Im\int_{1}^{2}
{\pars{t^{2} - 1}t^{2} \over t^{2} + z}\,\dd t
\qquad\mbox{where}\qquad
z \equiv -\,\half - {\root{3} \over 2}\,\ic =\expo{4\pi\ic/3}
\qquad\qquad\qquad\qquad\qquad\pars{1}
\end{align}
$z$ is a root of $z^{2} + z + 1 = 0$.
\begin{align}
I &={4\root{3} \over 3}\Im\int_{0}^{2}
{\bracks{\pars{t^{2} + z} - \pars{z + 1}}\bracks{\pars{t^{2} + z} - z}
\over t^{2} + z}\,\dd t
\\[3mm]&={4\root{3} \over 3}\Im\int_{0}^{2}
{\pars{t^{2} + z}^{2} - \pars{2z + 1}\pars{t^{2} + z}
+\ \overbrace{z\pars{z + 1}}^{\ds{=\ - 1}} \over t^{2} + z}\,\dd t
\\[3mm]&={4\root{3} \over 3}\Im\int_{0}^{2}
\bracks{\pars{t^{2} + z} - \pars{2z + 1} - {1 \over t^{2} + z}}\,\dd t
={4\root{3} \over 3}\pars{{\root{3} \over 2}
- \Im\int_{0}^{2}{\dd t \over t^{2} + z}}
\\[3mm]&=2 - {4\root{3} \over 3}\,
\Im\pars{\expo{-2\pi\ic/3}\int_{0}^{2\expo{2\pi\ic/3}}{\dd t \over t^{2} + 1}}
\end{align}
$$
\mbox{Evaluate the right hand side}:\quad
I=2 - {4\root{3} \over 3}\,
\Im\bracks{\expo{-2\pi\ic/3}\arctan\pars{2\expo{2\pi\ic/3}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/674050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $3^{n+1}+3^n+3^{n-1}$ is divisible by $13$. Prove that $3^{n+1}+3^n+3^{n-1}$ is divisible by $13$ for all positive integral values of $n$.
I tried: $3^n \cdot 3^1+3^n+3^n\cdot\frac{1}{3}$
Then what should I do next? Help please?
| I'm not sure how I would proceed from your path, but here is an alternative:
$$3^{n+1} + 3^n + 3^{n-1}= 3^{n-1}(3^2+3+1)=3^{n-1} \cdot 13$$
So:
$$\frac{3^{n+1} + 3^n + 3^{n-1}}{13}=3^{n-1}$$
And since $n \in \mathbb{N}$:
$$13|(3^{n+1} + 3^n + 3^{n-1})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/674131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Showing that $1-1/2+ \cdots +1/(2n-1)-1/2n$ is equal to $1/(n+1)+1/(n+2)+ \cdots +1/(2n)$
$1-1/2+1/3-1/4+ \cdots +1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+ \cdots +1/2n$
I was asked to prove by mathematical induction the validity of the above equation. It isn't hard to prove that it holds for any arbitrary natural number. But how mathematicians (or anyone) discovered that the left side of that equation equals to the right side, it doesn't seem obvious. I've tried to manipulate them with various means such as multiplying the denominator but I can't observe any pattern. Is it by chance that this equation was discovered? Thanks in advance.
| I don't know how someone first discovered this identity, but here's a clearer way of seeing it:
\begin{align*}
1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots - \frac{1}{2n} &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2n}- 2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} \right)\\
&= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2n}-\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \right)
\end{align*}
This cancels out the first $n$ terms of the sequence, leaving the $(n+1)^\text{st}$ to $2n^\text{th}$ terms, which is the righthand side.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/674363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)= -\frac{\log^33}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}$? Ramanujan gave the following identities for the Dilogarithm function:
$$
\begin{align*}
\operatorname{Li}_2\left(\frac{1}{3}\right)-\frac{1}{6}\operatorname{Li}_2\left(\frac{1}{9}\right) &=\frac{{\pi}^2}{18}-\frac{\log^23}{6} \\
\operatorname{Li}_2\left(-\frac{1}{3}\right)-\frac{1}{3}\operatorname{Li}_2\left(\frac{1}{9}\right) &=-\frac{{\pi}^2}{18}+\frac{1}{6}\log^23
\end{align*}
$$
Now, I was wondering if there are similar identities for the trilogarithm? I found numerically that
$$\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)\stackrel?= -\frac{\log^3 3}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6} \tag{1}$$
*
*I was not able to find equation $(1)$ anywhere in literature. Is it a new result?
*How can we prove $(1)$? I believe that it must be true since it agrees to a lot of decimal places.
| Combining trilogarithm identities 1 and 2, one obtains the formula
\begin{align}
\operatorname{Li}_3\left(\frac{1-z}{1+z}\right)-\operatorname{Li}_3\left(-\frac{1-z}{1+z}\right)=
2\operatorname{Li}_3\left(1-z\right)+2\operatorname{Li}_3\left(\frac{1}{1+z}\right)-
\frac12\operatorname{Li}_3\left(\frac{1}{1-z^2}\right)-\frac74\operatorname{Li}_3\left(1\right)\\
-\frac13\ln^3(z+1)+\frac{\pi^2}{6}\ln(z+1)+\frac{1}{12}\ln^3\left(z^2-1\right)+\frac{\pi^2}{12}\ln(z^2-1).
\end{align}
Now it suffices to set $z=2$ and use that $\operatorname{Li}_3\left(1\right)=\zeta(3)$, $\operatorname{Li}_3\left(-1\right)=-\frac34\zeta(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/676124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
} |
Writing $x^2+y^2+z^2$ as a polynomial combination of $xyz$, $x+y+z$, and $\frac1x+\frac1y+\frac1z$ Can we write $x^2+y^2+z^2$ as a polynomial combination of $xyz$, $x+y+z$, and $\dfrac1x+\dfrac1y+\dfrac1z$?
What about $x^3+y^3+z^3$?
| Let our three expressions be $a$, $b$, and $c$. Note that $xy+yz+zx=ac$.
For $x^2+y^2+z^2$, use the identity
$$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz.$$
For $x^3+y^3+z^3$, use the not difficult to verify identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/676935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Modify the Fibonacci series We know that Fibonacci Numbers start with 0 and next element is 1 and F(n)=F(n-1)+F(n-2) to find nth term where n>=2 and F(0)=0 F(1)=1 .
But what if we suppose the first 2 terms of fibonacci series be a and b then what will be its nth term?
Like if a=1 b=2 then if we want to find 4th term then it will be 5.
| $$F(n) = F(n - 1) + F(n - 2)$$
$$F(n - 1) = F(n - 1)$$
$$
\begin{bmatrix} F(n) \\ F(n - 1) \end{bmatrix} =
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} F(n - 1) \\ F(n - 2)\end{bmatrix}$$
$$
\begin{bmatrix} F(n + 1) \\ F(n) \end{bmatrix} =
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} b \\ a
\end{bmatrix}$$
Now if you want a more closed form than that you can do eigen value decomp on the matrix:
$$
\begin{bmatrix} F(n + 1) \\ F(n) \end{bmatrix} =
\begin{bmatrix} 1 & 1 \\ -\frac 2 {\sqrt{5} - 1} & \frac 2 {\sqrt{5} + 1} \end{bmatrix}
\begin{bmatrix} -\frac{\sqrt{5} - 1} 2 & 0 \\ 0 & \frac{\sqrt{5} + 1} 2 \end{bmatrix} ^n
\begin{bmatrix} 1 & 1 \\ -\frac 2 {\sqrt{5} - 1} & \frac 2 {\sqrt{5} + 1} \end{bmatrix}^{-1}
$$
$$F(n) = \frac {b} {\sqrt{5}} \left(\phi^n - \omega^n\right) + \frac {a} {\sqrt{5}} \left(\phi^{n-1} - \omega^{n - 1}\right)$$
using:
$$\phi = \frac{1 + \sqrt 5} 2$$
$$\omega = \frac{1 - \sqrt 5 } 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/677449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $x+\frac1x=5$ find $x^5+\frac1{x^5}$. If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.
Is there any other way find it?
$$
\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110.
$$
Thanks
| First find out $x^2 + x^{-2}$ and then $x^3 + x^{-3}$ by expanding and then at last find the required answer by taking power $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/678650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 6
} |
Evaluating $\int_{0}^{1}\frac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}\operatorname d\!x$ Find this integral
$$\operatorname I=\int\limits_{0}^{1}\dfrac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}\operatorname d\!x$$
My try: let
$$f(x)=x^4-2x^3+2x^2-x+1$$
I found
$$f(1-x)=(1-x)^4-2(1-x)^3+2(1-x)^2-x+1=x^4-2x^3+2x^2-x+1=f(x)$$
so
$$I=\int_{0}^{1}\dfrac{\arcsin{\sqrt{(1-x)}}}{x^4-2x^3+2x^2-x+1}dx$$
so
$$2I=\int_{0}^{1}\dfrac{\arcsin{\sqrt{x}}+\arcsin{\sqrt{(1-x)}}}{x^4-2x^3+2x^2-x+1}dx$$
then I can't,Thank you very much
| Firstly, as shown in @Ron Gordon's answer, we should eliminate the $\arcsin(\sqrt{x})$ in the numerator. Take the substitution that $\arcsin(\sqrt{x})=t$, then
\begin{equation}
I=\int_0^{\pi/2}\frac{t\sin(2t)}{\sin(t)^8-2\sin(t)^6+2\sin(t)^4-\sin(t)^2+1}dt\\
=\int_0^{\pi/2}\frac{t\sin(2t)}{115/128+1/128\cos(8t)+3/32\cos(4t)}dt\\
\end{equation}
By taking $t'=\pi/2-t$, we can check
\begin{equation}
I=\frac{\pi}{4}\int_0^{\pi/2}\frac{\sin(2t)}{115/128+1/128\cos(8t)+3/32\cos(4t)}dt\\
=\frac{\pi}{4}\int_0^1\frac{1}{x^4-2x^3+2x^2-x+1}dx\\
=\frac{\pi}{4}\int_0^1\frac{1}{(x^2-x+\frac{1}{2})^2+\frac{3}{4}}dx\\
=\frac{\pi}{4}\int_0^1\frac{1}{((x-\frac{1}{2})^2+\frac{1}{4})^2+\frac{3}{4}}dx\\
\stackrel{y=x-1/2}{=}\frac{\pi}{4}\int_{-1/2}^{1/2}\frac{1}{(y^2+\frac{1}{4})^2+\frac{3}{4}}dy
\end{equation}
Also, I do not have some good manner to solve it but to factorize the denominator in complex. We can easy to get the 4 roots of the denominator that
\begin{equation}
x_{1,2}=\pm\frac{1}{2}\sqrt{-1-2i\sqrt{3}}\\
x_{3,4}=\pm\frac{1}{2}\sqrt{-1+2i\sqrt{3}}
\end{equation}
Then, you can factorize the last fraction by solve the following problem:
\begin{equation}
\frac{C_1x+C_2}{x-x_1}+\frac{C_3x+C_4}{x-x_2}+\frac{C_5x+C_6}{x-x_3}+\frac{C_7x+C_8}{x-x_4}=\frac{1}{(y^2+\frac{1}{4})^2+\frac{3}{4}}
\end{equation}
By doing this, you can get the solution of $I$. Since the result is too long I ignore them. The approximation of the result is $0.9095208091$. I think there must be some easier method to deal with the integral of rational fraction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/683454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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"answer_id": 0
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Minimum value of the function $\sqrt{(1+1/m)(1+1/n)}$ If $m, n$ are positive real variables whose sum is a constant $k$, then what is the minimum value of $$\sqrt{\bigg(1 + \frac{1}{m}\bigg)\bigg(1 + \frac{1}{n}\bigg)}$$
| By the Cauchy-Schwarz Inequality, we have:
$$\left(1^2 + \left(\frac{1}{\sqrt{m}}\right)^2\right) \left(1^2 + \left(\frac{1}{\sqrt{n}}\right)^2\right) \ge \left(1 + \frac{1}{\sqrt{mn}}\right)^2\\
\sqrt{\left(1 + \frac{1}{m}\right)\left(1 + \frac{1}{n}\right)} \ge 1 + \frac{1}{\sqrt{mn}}$$
But we also know by the AM-GM inequality that
$$\sqrt{mn} \le \frac{m + n}{2}$$
or
$$\frac{1}{\sqrt{mn}} \ge \frac{2}{m + n}$$
So this gives us
$$\begin{align}\sqrt{\left(1 + \frac{1}{m}\right)\left(1 + \frac{1}{n}\right)} &\ge 1 + \frac{2}{m + n}\\
&= 1 + \frac{2}{k}\end{align}$$
with equality at $m = n = \frac{k}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/684637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $7\sin^2A-3\cos^2A=4$, then $\tan A\neq\pm\frac{1}{\sqrt{3}}$
If $7\sin^2A-3\cos^2A=4$, then we have to show that $\tan A=\pm\dfrac{1}{\sqrt{3}}$
The textbook approaches this by substituting $\sin^2A=1-\cos^2A$ and that method does(not) work.
$7\sin^2A-3\cos^2A=4$
$\implies 7\sin^2A-3(1-\sin^2A)=4$ {contains a sign error)
$\implies \sin^2a=\dfrac{1}{4}$
$\implies \cos^2A=1-\sin^2A=\dfrac{3}{4}$
Then, $\tan^2A=\dfrac{\sin^2A}{\cos^2A}=\dfrac{1}{3}$
$\implies \tan A=\pm\dfrac{1}{\sqrt{3}}$
But what's wrong with mine?(nothing)
$7\sin^2A-3\cos^2A=4(\sin^2A+\cos^2A)$
$\implies 3\sin^2A=7\cos^2A$
$\implies \dfrac{\sin^2A}{\cos^2A}=\dfrac{7}{3}$
$\implies \tan^2A=\dfrac{7}{3}$
$\implies \tan A=\sqrt{\dfrac{7}{3}}$
EDIT:Actually,the textbook(as usual) made a sign error in the second line.Ignore it.My solution is correct.
| The book made a sign error in the second line.My method is correct.
| {
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"url": "https://math.stackexchange.com/questions/687249",
"timestamp": "2023-03-29T00:00:00",
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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.