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How to solve this limit $\lim_{n\to\infty} ((\frac{1}{\sqrt{n^2+1}}) + \cdots + (\frac{1}{\sqrt{n^2+n}}))$ How do I solve this limit $$\lim_{n\to\infty} \left(\frac{1}{\sqrt{n^2+1}} + \cdots + \frac{1}{\sqrt{n^2+n}}\right)\text{ ?}$$
(n goes to plus infinity.)
I tried putting in $n=1,2,3,4,\ldots$ to find some pattern but it's hard to see where it's going.
For example, $n=1$, limit is $\dfrac{1}{\sqrt{2}}$
For example, $n=2$, limit is $\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{6}}$
For example, $n=3$, limit is $\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{6}} + \dfrac{1}{\sqrt{12}}$
For example, $n=4$, limit is $\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{6}} + \dfrac{1}{\sqrt{12}} + \dfrac{1}{\sqrt{20}}$
I'm not exactly sure what this limit is converging to...
| Hint:
$$\frac{n}{\sqrt{n^2+n}}\leq(\frac{1}{\sqrt{n^2+1}} + ... + \frac{1}{\sqrt{n^2+n}}) \leq \frac{n}{\sqrt{n^2+1}}$$
| {
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"url": "https://math.stackexchange.com/questions/959824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove ${_2F_1}\left({{\tfrac16,\tfrac23}\atop{\tfrac56}}\middle|\,\frac{80}{81}\right)=\frac 35 \cdot 5^{1/6} \cdot 3^{2/3}$ I've found the following hypergeometric function value by numerical observation. The identity matches at least for $100$ digits.
$${_2F_1}\left(\begin{array}c\tfrac16,\tfrac23\\\tfrac56\end{array}\middle|\,\frac{80}{81}\right) \stackrel{?}{=} \frac 35 \cdot 5^{1/6} \cdot 3^{2/3}$$
Or using a Pfaff transformation in an equivalent form
$$81^{1/6} \cdot {_2F_1}\left(\begin{array}c\tfrac16,\tfrac16\\\tfrac56\end{array}\middle|\,-80\right) \stackrel{?}{=} \frac 35 \cdot 5^{1/6} \cdot 3^{2/3}$$
How could we prove it?
Other related problem: How could we prove that
$${_2F_1}\left(\begin{array}c\tfrac16,\tfrac23\\\tfrac56\end{array}\middle|\,\frac{80}{81}\right) \stackrel{?}{=} {_2F_1}\left(\begin{array}c\tfrac12,\tfrac56\\\tfrac12\end{array}\middle|\,\frac{4}{9}\right) = {_1F_0}\left(\begin{array}c\tfrac56\\\ - \,\end{array}\middle|\,\frac{4}{9}\right)$$
| The Tito Piezas answer is related to this parameterization for
${}_2F_1(\frac{1}{6},\frac{2}{3};\frac{5}{6};z)$ in terms of modular forms:
$$
{}_2F_1\left(\frac{1}{6},\frac{2}{3};\frac{5}{6};
\frac{j_{3A}}{j_{3A}-108}\right)
= \eta(\tau)\; \eta(3\tau)\;(j_{3A}-108)^{1/6}\;
\Gamma\left(\frac{2}{3}\right)^3
\frac{(-6i\tau+\sqrt{3}-3i)}{2^{7/3}\pi}
\tag{1}$$
where $\eta$ is the Dedekind eta function and $j_{3A}$ is
$$
j_{3A} = \left(\frac{\eta(\tau)^6}{\eta(3\tau)^6}+
\frac{3^3 \eta(3\tau)^6}{\eta(\tau)^6}\right)^2
$$
(Choosing the proper branches of the sixth root and the ${}_2F_1$).
Take $\tau = \frac{1}{2} + i \frac{5}{6}\sqrt{3}$ to get the required result from special values of $\eta$...
$$
\eta\left(\frac{1}{2} + i \frac{5}{6}\sqrt{3}\right)
=2^{-3/2} 3^{3/8} 5^{-5/12} (1+\sqrt5)^{1/2} \pi^{-1}
\Gamma\left(\frac{1}{3}\right)^{3/2} (-1)^{1/24}
$$
and something similar for $\eta(3\tau)$, so
$$
j_{3A}(\tau) = -8640,\qquad
\frac{j_{3A}(\tau)}{j_{3A}(\tau)-108} = \frac{80}{81}
$$
In $(1)$, the product of $\Gamma(1/3)$ and $\Gamma(2/3)$ simplify to something algebraic together with a $\pi$, and the powers of $\pi$ cancel.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x^3(x^2+1)^2(x-1)(x+1)>0$
I need to solve this inequality:
$$x^3(x^2+1)^2(x-1)(x+1)>0$$
Note: I haven't learnt imaginary numbers. Does the $(x^2+1)^2$ affect the inequality?
Thanks
| Hint: You have $(x^2+1)^2\gt0$ over the reals, therefore $$x^3(x^2+1)^2(x-1)(x+1)\gt0\iff x^3(x-1)(x+1)\gt0.$$
| {
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"url": "https://math.stackexchange.com/questions/962217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $A=\pmatrix{1 &0\\-1&1}$, show that $A^2-2A+I_2=0$. Hence find $A^{50}$
If $$A=\pmatrix{1 &0\\-1&1},$$ show that $$A^2-2A+I_2=0,$$ where $I_{2}$ is the $2x2$ Identity matrix. Hence find $A^{50}$.
We have $$A^2-2A+I_2=A(A-2I_3)+I_=\pmatrix{1 &0\\-1&1}\pmatrix{-1 &0\\-1&-1}+I_2
=-I_2+I_2=0.$$
How can I show the second part?
| Since $(A-I_2)^2 = 0$, by binomial theorem, one has
$$\begin{align}A^{50}
&= (I_2 + (A - I_2))^{50}\\
&= I_2 + \binom{50}{1} (A-I_2)^{1} + \binom{50}{2}(A-I_2)^2 + \binom{50}{3}(A-I_2)^3 +\cdots\\
&= I_2 + 50 (A-I_2) + 0 + 0 + \cdots\\
&= \begin{pmatrix}1 & 0\\-50 & 1\end{pmatrix}
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\displaystyle\lim_{x\to -8} \frac {8 − |x|} {8 + x} \ $ How do I solve? \begin{align*}
\lim_{x\to -8^+} \frac {8 − |x|} {8 + x} &=
\lim_{x\to -8^+} \frac {8 − x} {8 + x}\\
&= \frac {8 − 8} {8 + -8} \\ &= 0
\end{align*}
\begin{align*}
\lim_{x\to -8^-} \frac {8 − |x|} {8 + x} &=
\lim_{x\to -8^-} \frac {8 − (-x)} {8 + x} \\
&= \frac {8 + − 8} {8 + -8} \\ &= 0
\end{align*}
However, the answer key says that I am wrong. Where have I failed?
| $$
\lim_{x\rightarrow -8} \frac{8-|x|}{8+x} = \lim_{x<0,\ x\rightarrow
-8} \frac{8+x}{8+x}=1 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is this equality true? Any sources? Is it true that $1 - \frac 12+\frac13-\frac14+\cdots-\frac 1{200}=\frac 1{101}+\frac 1{102}+\cdots+\frac 1{200}$? Where can I find sources for this proof?
| Here is the boring proof by induction. We want to prove that
$$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n} = \frac{1}{n+1} + \cdots + \frac{1}{2n}. $$
Base case: $n = 1$. The left-hand side reads $1-1/2 = 1/2$, and the right-hand side is $1/2$.
Induction step: Suppose
$$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n} = \frac{1}{n+1} + \cdots + \frac{1}{2n}. $$
Then
$$
\begin{align*}
&1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n+1} - \frac{1}{2n+2} \\ =
&1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\ =
&\frac{1}{n+1} + \cdots + \frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\ =
&\frac{1}{n+2} + \cdots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{n+1} - \frac{1}{2n+2} \\ =
&\frac{1}{n+2} + \cdots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}. \end{align*}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this limit (formally) As I was coursing through Spivak's calculus, more like analysis; I found an interesting, questionable example.
Let $\frac{p}{q}$ be in its lowest terms; $p$ and $q$ are integers with no common factors and $q > 0$
If $f(x) = \begin{cases}0 & \text{$x$ is irrational} \\ \frac{1}{q} & x = \frac{p}{q}, 0 < x < 1 \end{cases}$
Show that $f(x)$ approaches $0$ as $x$ approaches $0$
Using the actual definition, not just the rough idea, we must show this is true.
Definition: For every $\epsilon > 0$ there is some $\delta > 0$ such that, for all $x$, if $0 < |x - a| < \delta$, then $|f(x) - l| < \epsilon$ This is if $f$ approaches the limit $l$ near $a$.
So, $a = 0$ and $l = 0$ we dont actually pick a random $\delta$, we just use it analytically.
$0 < |x| < \delta$ because we can choose a $\delta > |x|$ We need to do,
$|f(x)| < \epsilon$
Lets consider $g_1(x) = 0$, where $x$ is irrational first.
$0 < |x| < \delta_1$ then prove $|0| < \epsilon$
It is true since $\epsilon > 0$ part one is done then.
Lets consider $g_2(x) = \frac{1}{q}, x= \frac{p}{q}, 0 < x < 1$
$0 < |x| < \delta_2$ then prove $\frac{1}{q} < \epsilon$
Since $0 < x < 1$, $0 < q < \infty$ and $p < q$ because $0 < x < 1$
How do we go about this then?
Any ideas will be appreciated! Thanks!
| The key step in proving $f(x)$ approaches zero for positive $x$ is that for any $\delta < 1$ and $x \neq 0$,
if $0 < x < \delta$ then
$$
x < \frac{1}{\lfloor \frac{1}{\delta} \rfloor }
$$
where $\lfloor \frac{1}{\delta} \rfloor $ is an integer, being the least integer less than $\frac{1}{\delta}$.
This tells us that if $0 < x < \delta$ then
$ x < \frac{1}{q(x;\delta)}$
with $q(x;\delta)=\lfloor \frac{1}{\delta} \rfloor $. Then $f(x)$ is either $0$ or $\frac{1}{q(x;\delta)}$ or $\frac{1}{k}$ for some $k > q(x;\delta)$. Thus
$$
f(x) \leq 1/q(x;\delta) = \frac{1}{\lfloor \frac{1}{\delta} \rfloor }
$$
So given any $\epsilon > 0$, take
$$
\delta = \frac{1}{\frac{1}{\epsilon} + 1} = \frac{\epsilon}{1+\epsilon}
$$
Then for all $0 < x < \delta$,
$$
f(x) \leq \frac{1}{\lfloor \frac{1}{\delta} \rfloor } =
\frac{1}{\lfloor \frac{1}{\frac{\epsilon}{1+\epsilon}} \rfloor }
$$
Simplifying,
$$f(x) \leq \frac{1}{\lfloor \frac{1+\epsilon}{\epsilon} \rfloor } =
\frac{1}{\lfloor \frac{1}{\epsilon}+1 \rfloor }
< \frac{1}{\frac{1}{\epsilon}} = \epsilon
$$
So for all $0 < x < \delta(\epsilon) = \frac{\epsilon}{1+\epsilon}$,
$$
0 \leq f(x) < \epsilon
$$
and $f(x)$ approaches zero as $x$ approaches zero from above.
You complete the proof by demonstrating similarly that $f(x)$ approaches zero for negative $x$ .
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $f(36)$ given $\frac{f(x)f(y)-f(xy)}{3} = x+y+2$ on $\mathbb R$
Let $f:\mathbb R \to \mathbb R$ be a function with
$$\frac{f(x)f(y)-f(xy)}{3} = x+y+2$$
for all real numbers $x,y$. List all possible values for $f(36)$.
So far I have just been plugging in possible $x$ and $y$.
$$\frac{f(4)f(9)-f(4\cdot9)}{3}=4+9+2$$
So then $f(36)=f(4)f(9)-45$.
$$\frac{f(6)f(6)-f(6\cdot6)}{3}=6+6+2$$
$$\frac{f(2)f(18)-f(2\cdot18)}{3}=2+18+2$$
$$\frac{f(3)f(12)-f(3\cdot12)}{3}=3+12+2$$
$$\frac{f(36)f(1)-f(36\cdot1)}{3}=36+1+2$$
| If you let $x=y=0$ you get $f(0)^2-f(0) = 6$, which has solutions $f(0)=-2$ or $f(0)=3$. In both cases, setting $x=0$ gives
$$f(0)f(y)-f(0) = 3y+6 $$
$$f(y) = \frac{3y+6+f(0)}{f(0)}$$
Inserting $y=36$ and the two possible values of $f(0)$ then gives $f(36)=-56$ or $f(36)=39$.
| {
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"timestamp": "2023-03-29T00:00:00",
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On finding the equilibrium solutions to a system of differential equations I am asked to find all equilibrium solutions to this system of differential equations:
$$\begin{cases} x ' = x^2 + y^2 - 1 \\ y'= x^2 - y^2 \end{cases} $$
and to determine if they are stable, asymptotically stable or unstable.
I do not know how to proceed, I tried switching to polar coordinates to make the system linear in some way but I did not succeed.
Would I need to use a software to solve this?
Context: I am following the course of Arthur Mattuck (MIT opencourseware), but I can't seem to find these type of systems, or equilibrium points.
| This answer posted in response to the modified system, given in (1) and (2) below:
Oy Gevalt! NOW there is work!
Where is Moshe now that we need him to lead us out from under Pharoah's toil?
Well, I ain't no $Moshe$ but I can cut some down some of the work, thus:
The system to be considered is now
$x' = x^2 + y^2 - 1, \tag{1}$
$y' = x^2 - y^2; \tag{2}$
the number of equilibria has jumped from none to four! To see this, note that now $x' = y' = 0$ implies, from (1), (2), that
$x^2 + y^2 = 1, \tag{3}$
$x^2 = y^2; \tag{4}$
using (4) in (3) yields
$2x^2 = 1 \Rightarrow x = \pm \dfrac{\sqrt 2}{2}; \tag{5}$
we see from (4) that $y$ may take the same values; the equilibria occur at the four points
$(x, y) = (\pm \dfrac{\sqrt 2}{2}, \pm \dfrac{\sqrt 2}{2}). \tag{6}$
It may also be seen, geoemtrically, that the equilibria are given by (6), since (3) is the equation of a circle, centered at the origin and of radius $1$, and (4) is the combined equation of the two lines $x \pm y = 0$, since
$x^2 = y^2 \Rightarrow x = \pm y \Rightarrow x \pm y = 0; \tag{7}$
the circle intersects these lines at the specified points (6). In any event, having the equilibria of the system (1)-(2) at hand, the next step is to linearize the equations about these four points, and see what we get. Linearizing requires computation of the Jacobian matrix $J(x, y)$ of the vector field $(x', y')^T = (x^2 + y^2 -1, x^2 - y^2)^T$; we have
$J(x, y) = \begin{bmatrix} \dfrac{\partial x'}{\partial x} & \dfrac{\partial x'}{\partial y} \\ \dfrac{\partial y'}{\partial x} & \dfrac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 2x & 2y \\ 2x & -2y \end{bmatrix} = 2\begin{bmatrix} x & y \\ x & -y \end{bmatrix}, \tag{8}$
and we next must evaluate and eigen-analyze $J(\pm \dfrac{\sqrt{2}}{2}, \pm \dfrac{\sqrt{2}}{2})$ for all four possible combinations of $\pm \dfrac{\sqrt{2}}{2}$, i.e., at all four points $(\pm \dfrac{\sqrt{2}}{2}, \pm \dfrac{\sqrt{2}}{2})$. Well, some the the Oy! Gevalt! can be mollified by lessening the amount of work by realizing that, due to certain symmetries of the problem, there are really only two matrices $J(x, y)$ which need to be considered, not four. This is most easily seen by breaking the situation up into quadrants:
$J(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}) = -J(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}), \; \; \text{for quadrants I, III}, \tag{9}$
$J(-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}) = -J(\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}), \; \; \text{for quadrants II, IV}; \tag{10}$
in this way we only need perform the eigen-analysis on two matrices out of the four $J(\pm \dfrac{\sqrt{2}}{2}, \pm \dfrac{\sqrt{2}}{2})$, so lets start with $J_I = J(\dfrac {\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})$; its characteristic polynomial, call it $p_I(\lambda)$, is
$p_I(\lambda) = \det (\begin{bmatrix} \sqrt{2} - \lambda & \sqrt{2} \\ \sqrt{2} & -\sqrt{2} - \lambda \end{bmatrix}) = \lambda^2 - 4. \tag{11}$
We see from (11) that the eigenvalues of $J_I$ are $\lambda = \pm 2$; since $J_I$ has a positive eigenvalue, the point $(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})$ is unstable; since $J_I$ also has a negative eigenvalue, this point is a saddle; by (9), $J_{III} = -J_I$ also has eigenvalues $\pm 2$, and hence also unstable and a saddle. These facts are borne out by the excellent graphic contributed by Amzoti in his answer. We next turn to $J_{II}$; since we are now in the second quadrant, we have
$J_{II} = J(-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}) = \begin{bmatrix} -\sqrt{2} & \sqrt{2} \\ -\sqrt{2} & -\sqrt{2}\end{bmatrix}, \tag{12}$
$p_{II}(\lambda) = \det(J_{II} - \lambda I) = \det(\begin{bmatrix} -\sqrt{2} - \lambda & \sqrt{2} \\ -\sqrt{2} & -\sqrt{2} - \lambda \end{bmatrix})$
$= (\lambda + \sqrt{2})^2 + 2 = \lambda^2 + 2\sqrt{2} \lambda + 4; \tag{13}$
the zeroes of $p_{II}(\lambda)$ are found via the quadratic formula:
$\lambda = \dfrac{1}{2}(-2\sqrt{2} \pm \sqrt{8 - 16}) = \dfrac{1}{2}(-2\sqrt{2} \pm 2i\sqrt{2}) = -\sqrt{2} \pm i\sqrt{2}; \tag{14}$
we we see that the eigenvalues of $J_{II}$ are not real, but have negative real part; thus $(-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})$ is a stable spiral. Furthermore, since $J_{IV} = -J_{II}$, the eigevalues of $J_{IV}$ are $\lambda = \sqrt{2} \pm i\sqrt{2}$; thus the point $(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2})$ is an unstable spiral. All these computations support and are supported by Amzoti's grahic of the phase portrait of (1)-(2).
The stable spiral point at $(-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})$ is in fact asymptotically stable; this follows from the fact that $\Re(\lambda) < 0$ for each of the eigenvalues of $J_{II}$; this fact is both well-known and well documented, for example in the excellent reference provided by Amzoti in his comment.
There appears to be a discrepancy in the calculation of eigenvalues by Monolinte and myself, but we agree on the qualitative features of the equilibrium points. As ever, abstract analysis is easy but arithmetic proves difficult! Until further notice, I'm standing by my calculations.
Hope this helps! Cheers,
and as ever,
Fiat Lux!!!
| {
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"timestamp": "2023-03-29T00:00:00",
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An Inequality; $a^2+b^2=1$ $a,b$ are tho real numbers such that $a^2+b^2=1$.
To prove that ;
$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{ab+1}\geq\dfrac{3}{1+\cfrac{(a+b)^2}{4}}$$
When I first saw this question, I thought of applying Titu's Lemma, to get
$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{ab+1}\geq\dfrac{9}{1+1+1+a^2+b^2+ab}=\dfrac{9}{3+(a+b)^2-ab}OR=\dfrac{9}{4+ab}$$
Now, from hereI am confused, how to proceed. Can anybody kindly help me over this problem ?
| This is more of a step-by-step version of the trigonometric soln already posted. With $a=\cos t, \; b = \sin t$, the inequality is
$$\frac1{\sin^2t+1}+\frac1{\cos^2t+1}+\frac1{\frac12 \sin 2t+1} \ge \frac{12}{5+\sin 2t}$$
$$LHS = \frac{3}{\frac14\sin^22t + 2}+\frac1{\frac12\sin 2t+1}$$
so with $x = \sin 2t\in [-1, 1]$, the inequality can be written as
$$\frac{12}{8+x^2}+\frac2{2+x}\ge \frac{12}{5+x} \iff (1-x)(4+6x+5x^2) \ge 0$$
which is obvious as $5x^2+4 \ge 4\sqrt5 |x|> 6|x|$.
| {
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Evaluate $\int \frac 1{x^{12}+1} \, dx$
Evaluate $\displaystyle \int \frac 1{x^{12}+1} \, dx$
I tried writing this in partial fractions.
$$\int \frac 1{x^{12}+1} \, dx=\int \frac{1}{[(x^6+1)+\sqrt{2}x^3][(x^6+1)-\sqrt{2}x^3]} \, dx$$
So I did:
\begin{align}
\frac{1}{[(x^6+1)+\sqrt{2}x^3][(x^6+1)-\sqrt{2}x^3]} &= \frac{Ax^5+Bx^4+Cx^3+Dx^2+Ex+F}{[(x^6+1)-\sqrt{2}x^3]}\\&{}\qquad+\frac{A_1x^5+B_1x^4+C_1x^3+D_1x^2+E_1x+F_1}{[(x^6+1)+\sqrt{2}x^3]}\end{align}
After messy work, I found $A=A_1=B=B_1=D=D_1=E=E_1=0$ and
$$C=-\frac 1{2\sqrt{2}},\qquad C_1=\frac 1{2\sqrt{2}},\qquad F=F_1=\frac 12$$
I now get the integral
$$\int \frac{-\frac 1{2\sqrt{2}}x^3+\frac 12}{(x^6+1)-\sqrt{2}x^3}\, dx + \int \frac{\frac 1{2\sqrt{2}}x^3+\frac 12}{(x^6+1)+\sqrt{2}x^3} \, dx$$
But how can I go from here? I am stuck.
| Hint: solve $x^{12}=-1$ in the complex plane, then use partial fractions.
Note that $(-1)^{1/12}=\cos{\left(\frac{2k\pi+\pi}{12}\right)}+i\sin{\left(\frac{2k\pi+\pi}{12}\right)}$, with $k=0,1,2,\cdots,11$
Now the roots are:
$$
\begin{matrix}
x_1=\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{4}i, & x_2=\frac{\sqrt{6}+\sqrt{2}}{4}-\frac{\sqrt{6}-\sqrt{2}}{4}i \\
x_3=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i, & x_4=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i \\
x_5=\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4}i, &x_6=\frac{\sqrt{6}-\sqrt{2}}{4}-\frac{\sqrt{6}+\sqrt{2}}{4}i \\
x_7=\frac{-\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4}i, & x_8=\frac{-\sqrt{6}+\sqrt{2}}{4}-\frac{\sqrt{6}+\sqrt{2}}{4}i \\
x_9=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i, & x_{10}=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i \\
x_{11}=\frac{-\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{4}i, & x_{12}=\frac{-\sqrt{6}-\sqrt{2}}{4}-\frac{\sqrt{6}-\sqrt{2}}{4}i
\end{matrix}
$$
by the factor theorem:
$$x^{12}+1=(x-x_1)(x-x_2)\cdots(x-x_{11})(x-x_{12})$$
further it is known that $(x+a+bi)(x+a-bi)=x^2+2ax+a^2+b^2$,
Then $$x^{12}+1=(x^2+m_1x+n_1)(x^2+m_2x+n_2)\cdots(x^2+m_6x+n_6)$$
Now apply partial fractions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/971973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
proving Fibonacci numbers using mathematical Induction? Can anyone confirm whether my answer is correct, please.
Let suppose we have the following fibonacci numbers as shown: $f(0) = 0, f(1) = 1$, and $f(n) = f(n-1) + f(n-2)$ for $n \geq 2$. Prove that for each $n \geq 0, f(4n)$ is a multiple of $3$
$f_0 = 0,f_1= 1,f_n = f_{n-1}+f_{n-2} \text{ for } n \geq 2$ Prove that for each $n \geq 0, f_{4n}$ is a multiple of $3$.
My solution to this question:
\begin{align}
f_{4(k+1)} & = f_{4k+4} \\
& = f_{4k+3} + f_{4k+2} \\
& = f_{4k+3} + (f_{4k+1} + f_{4k}) \\
& = (f_{4k+2} + f_{4k+1}) + (f_{4k+1} + f_{4k}) \\
& = f_{4k+1} + f_{4k} + f_{4k+1} + f_{4k+1} + f_{4k} \\
& = 3(f_{4k+1}) + 2(f_{4k})
\end{align}
| What you need to prove is that $f_{4(n + 1)}$ is divisible by 3 (or that it has a factor of 3 in it) for all $n \in \mathbb{N}$. You have to prove that the proposition holds for your base case: $f_4$ which it surely does.
Now you assume $f_{4n}$ holds and prove that $f_{4(n + 1)}$ also holds.
To do this, define $f_{4n} = 3m, \hspace{2mm} m \in \mathbb{N}$, (definition of multiple of 3).
Now, we have to construct $f_{4(n + 1)}$, Fibonacci numbers are defined by their predecesors in the form:
$$ f_n = f_{n - 1} + f_{n - 2}$$
We also define $f_{4n - 1} = k, \hspace{2mm} k \in \mathbb{N}$ (this just tells us that it is a natural number, as $F \subset \mathbb{N}$).
We can now procede to construct $f_{4(n + 1)}$ as follows:
$$ f_{4n + 1} = f_{4n} + f_{4n - 1} = 3m + k $$
$$ f_{4n + 2} = f_{4n + 1} + f_{4n} = (3m + k) + (3m) = (6m + k)$$
$$ f_{4n + 3} = f_{4n + 2} + f_{4n + 1} = (6m + k) + (3m + k) = (9m + 2k)$$
$$ f_{4n + 4} = f_{4n + 3} + f_{4n + 2} = (9m + 2k) + (6m + k) = (15m + 3k)$$
Now, we can factor out 3 in the last step:
$$ f_{4n + 4} = 3(5m + k)$$
but $ f_{4n + 4} = f_{4(n + 1)} $, we showed that if the proposition for $f_{4n}$ holds, it also holds for $ f_{4(n + 1)}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Positive values of $x$ that satisfy the inequality $\frac{1}{x}-\frac{1}{x-1}>\frac{1}{x-2}$ Determine the set of positive values of $x$ that satisfy the inequality $$\frac{1}{x}-\frac{1}{x-1}>\frac{1}{x-2}.$$
My attempt:
\begin{align}
\frac{-1}{x(x-1)} & >\frac{1}{(x-2)} \\[0.1in]
\frac{1}{x(1-x)} & >\frac{1}{(x-2)} \\[0.1in]
x(1-x) & <(x-2)
\end{align}
If I put $0.25$ in the original inequality, it works, but not in the last one. What mistake did I make? Please advise.
| If I may suggest, consider $$f(x)=\frac{1}{x}-\frac{1}{x-1}-\frac{1}{x-2}$$ that you want to be positive. Reduce to same denominator and simplify; you then arrive to $$f(x)=\frac{2-x^2}{x(x-1)(x-2)}$$ Multiply the top by the denominator so $$f(x)=\frac{(2-x^2)x(x-1)(x-2)}{\Big(x(x-1)(x-2)\Big)^2}$$ Now, the denominator is positive and so the sign of $f(x)$ is given by the sign of $$(\sqrt 2-x)(\sqrt 2+x)x(x-1)(x-2)$$
I am sure that you can take from here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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coordinate geometry high level problems If the lines $aX^2 + 2hXY + bY^2 =0$ form two sides of a parallelogram and the line $lX + mY =1$ is one diagonal, prove that the equation of other diagonal is $Y(bl – lm) = X(am – hl)$
| Let the pair of straight lines meet the diagonal at $(x_1,y_1)$ and $(x_2,y_2)$, then the other diagonal will pass through $(0,0)$ and $(x_1+x_2,y_1+y_2)$.
$$
\left \{
\begin{align*}
aX^2+2hXY+bY^2 &= 0 \\
\ell X+mY &= 1
\end{align*}
\right.$$
$$\implies
aX^2+2hX\left( \frac{1-\ell X}{m} \right)+
b\left( \frac{1-\ell X}{m} \right)^2=0$$
$$\left( a-\frac{2h\ell}{m}+\frac{b\ell^2}{m^2} \right)X^2+
\left( \frac{2h}{m}-\frac{2b\ell}{m^2} \right)X+\frac{b}{m^2}=0$$
$$x_1+x_2=\frac{2(b\ell-hm)}{am^2-2h\ell m+bl^2}$$
Similarly,
$$y_1+y_2=\frac{2(am-h\ell)}{am^2-2h\ell m+bl^2}$$
Hence the equation for the other diagonal is
$$\frac{Y}{X}=\frac{y_1+y_2}{x_1+x_2}$$
$$(b\ell-hm)Y=(am-h\ell)X$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_0^\infty \frac{dx}{\sqrt{x}[x^2+(1+2\sqrt{2})x+1][1-x+x^2-x^3+...+x^{50}]}$ My brother's friend gave me the following wicked integral with a beautiful result
\begin{equation}
{\Large\int_0^\infty} \frac{dx}{\sqrt{x} \bigg[x^2+\left(1+2\sqrt{2}\right)x+1\bigg] \bigg[1-x+x^2-x^3+\cdots+x^{50}\bigg]}={\large\left(\sqrt{2}-1\right)\pi}
\end{equation}
He claimed the above integral can be generalised to the following form
\begin{equation}
{\Large\int_0^\infty} \frac{dx}{\sqrt{x} \bigg[x^2+ax+1\bigg] \bigg[1-x+x^2-x^3+\cdots+(-x)^{n}\bigg]}=\ldots
\end{equation}
This is a challenging problem. How to prove it and what is the closed-form of the general integral?
| Indeed let
$$
I(n,a)=\int_0^\infty\frac{dx}{\sqrt{x}(1+ax+x^2)(\sum_{k=0}^n(-x)^k)}
$$
The change of variables $x\leftarrow 1/x$ yields
$$
I(n,a)=\int_0^\infty\frac{(-1)^nx^{n+1}dx}{ \sqrt{x}(1+ax+x^2)(\sum_{k=0}^n(-x)^k)}
$$
Thus
$$
2I(n,a)=\int_0^\infty\frac{1+x}{\sqrt{x}(1+ax+x^2)}dx=
2\int_0^\infty\frac{1+t^2}{ 1+at^2+t^4}dt
$$
Or equivalently, setting $u=t-1/t$,
$$
I(n,a)=
\int_{-\infty}^\infty\frac{du}{ 2+a+u^2} =\frac{\pi}{\sqrt{2+a}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/978560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
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Simplify the algebraic expression Can someone please explain to me how the algebraic expression in the picture is simplified. To be more specific, how (1) becomes (2).
*
*$3x^2(6x-4)^4 + x^3(6\times 4\times (6x-4)^3)$
*$3x^2(6x-4)^3(6x-4+x\times 2\times 4)$
*$3x^2(6x-4)^3(14x-4)$
original link: http://i.stack.imgur.com/7rXzS.jpg
| $$3x^2(6x-4)^4+x^3(6\times 4\times (6x-4)^3)$$$$=3x^2(6x-4)^3\times(6x-4)+x^2\times x\times 3\times 2\times 4\times (6x-4)^3$$$$=3x^2(6x-4)^3\times(6x-4)+3x^2(6x-4)^3\times x\times 2\times 4$$$$=3x^2(6x-4)^3((6x-4)+x\times 2\times 4)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Some help with sin and cos I'm having trouble to understand the following equalities in these two equations, i.e. how to apply the addition formulas.
Firstly:
$$ \frac {1- \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} {1+ \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} = \frac {cos^2(\frac x2) - sin^2(\frac x2)} {cos^2(\frac x2) + sin^2(\frac x2)}= cos^2(\frac x2) - sin^2(\frac x2)$$
and secondly:
$$ \frac {2tan(\frac x2)} {1+tan^2(\frac x2)} = 2* \frac {sin(\frac x2)} {cos(\frac x2)(1+\frac {sin^2(\frac x2)}{cos^2(\frac x2)})} = 2* \frac {sin(\frac x2)cos(\frac x2)} {cos^2(\frac x2)+sin^2(\frac x2)}= 2sin(\frac x2)cos(\frac x2) $$
I don't understand all the steps in those equations unfortunately. Can someone help me out here?
| To get the first equality
$$\frac {1- \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} {1+ \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} \underbrace{=}_{(1)} \frac {cos^2(\frac x2) - sin^2(\frac x2)} {cos^2(\frac x2) + sin^2(\frac x2)}\underbrace{=}_{(2)} cos^2(\frac x2) - sin^2(\frac x2)$$
in step $(1)$ you multiply numerator and denominator by $\cos^2\frac x2$ (note that $\frac ab=\frac{ac}{bc}, \forall c\ne 0.$)
In step $(2)$ it is used the equality $\sin^2 t+\cos^2 t=1.$
To get the second equality
$$\frac {2tan(\frac x2)} {1+tan^2(\frac x2)} \underbrace{=}_{(1)} 2\frac {sin(\frac x2)} {cos(\frac x2)(1+\frac {sin^2(\frac x2)}{cos^2(\frac x2)})} \underbrace{=}_{(2)}2\frac {sin(\frac x2)cos(\frac x2)} {cos^2(\frac x2)+sin^2(\frac x2)}\underbrace{=}_{(3)} 2sin(\frac x2)cos(\frac x2)$$ note that:
$(1)$ is just the definition of tangent $\left(\tan t=\frac{\sin t}{\cos t}\right).$ In $(2)$ multiply numerator and denominator by $\cos \frac{x}{2}$ (note that, $\frac{a}{b}=\frac{ac}{bc}, \forall c\ne 0.$) In step $(3)$ it is used again the equality $\sin^2 t+\cos^2 t=1.$
| {
"language": "en",
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how evaluate $\int_0^{\pi}\frac{1}{(a+\cos{\theta})^2}, a>1 $, using residues theorem? how evaluate $\int_0^{\pi}\frac{1}{(a+\cos{\theta})^2}, a>1 $, using residues theorem?
This problem is an exercise book Complex Analysis of Conway.
| Firstly $\int_0^{\pi}\dfrac{1}{(a+\cos{\theta})^2}d\theta = \dfrac{1}{2}\int_0^{2\pi}\dfrac{1}{(a+\cos{\theta})^2}d\theta$.
Write $\cos \theta = \dfrac{e^{i\theta} + e^{-i\theta}}{2} = \dfrac{z+ z^{-1}}{2}$ with $z = e^{i\theta}$, then
$$\int_0^{2\pi}\dfrac{1}{(a+\cos{\theta})^2}d\theta = \int_{|z|= 1} \dfrac{1}{(a+ \dfrac{z + z^{-1}}{2})^2} \dfrac{dz}{iz} = -4i\int_{|z|= 1} \dfrac{z}{(z^2 + 2az + 1)^2}dz$$
Denote the two roots of $z^2 + 2az + 1 = 0$ by $z_1$ and $z_2$ we have $z_1 = \dfrac{-2a + \sqrt{4a^2 - 4}}{2}$ and $z_2 = \dfrac{-2a - \sqrt{4a^2 - 4}}{2}$ and $z_1z_2 = 1$
Since $a>1$, only $z_1 = \sqrt{a^2 -1 } - a$ is inside $|z|=1$, so the residue of $f(z) = \dfrac{z}{(z^2 + 2az + 1)^2}$ at $z_1$ is the value of $\dfrac{d}{dz}\dfrac{z}{(z-z_2)^2}$ taken at $z = z_1$. Then it's easy to get the result by residue theorem
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that: $\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le 3\sqrt[3]{3}$ Given $a,b,c>0$ and $a+b+c=3$. Prove that: $\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le 3\sqrt[3]{3}$
| Let $x = a+2b$, $y = b+2c$, and $z = c + 2a$, then $x+y+z = 3(a+b+c) = 9$, and we have that $f''(x) = -\dfrac{2}{9}\cdot x^{-\frac{5}{3}} < 0$ on $(0,9)$ with $f(x) = \sqrt[3]{x}$. Thus $f$ is concave and we have:
$LHS = f(x) + f(y) + f(z) \leq 3f\left(\frac{x+y+z}{3}\right) = 3f(3) = 3\sqrt[3]{3} = RHS$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Beautiful cyclic inequality Prove that cyclic sum of $\displaystyle \sum_{\text{cyclic}} \dfrac{a^3}{a^2+ab+b^2} \geq \dfrac{a+b+c}{3}$ , if $a, b, c > 0$
I'm really stuck on this one. Tried some stuff involving QM> AM(because the are positive) but can't derive the needed ,can't proceed from it.
| Note that
$$\frac{a^3-b^3}{a^2+ab+b^2}=a-b \implies LHS = \sum_{cyc} \frac{a^3}{a^2+ab+b^2} = \sum_{cyc} \frac{b^3}{a^2+ab+b^2}$$
So we get
$$2LHS = \sum_{cyc} \frac{a^3+b^3}{a^2+ab+b^2}$$
But by rearrangement, $a^3+b^3\ge a^2b+ab^2 \implies 3a^3+3b^3 \ge a^3+2a^2b+2ab^2+b^3 = (a+b)(a^2+ab+b^2)$.
$$\therefore 2LHS = \sum_{cyc} \frac{a^3+b^3}{a^2+ab+b^2} \ge \frac13\sum_{cyc} (a+b)= \frac23(a+b+c)$$
which is what we wanted to show.
| {
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Let $M=\{(x,y,z):z=3x-y\}$. Find the orthogonal projection of a vector $v \in \mathbb{R}^3$ on $M$ and a matrix representation of $p_M$. Let $M=\{(x,y,z):z=3x-y\}$. Find the orthogonal projection of a vector $v \in \mathbb{R}^3$ on $M$ and a matrix representation of $p_M$.
I know that the orthogonal projection of two vectors is $p_vu=\frac{u \cdot v}{\|v\|^2}u$ How would I apply this to an equation.
$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 3 &-1 \end{pmatrix} (\begin{pmatrix} 1 & 0 & 3 \\ 0 & 1 & -1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 3 &-1 \end{pmatrix})^{-1}\begin{pmatrix} 1 & 0 & 3 \\ 0 & 1 & -1 \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 3 &-1 \end{pmatrix} (\begin{pmatrix} 10 & -3 \\ -3 & 2 \end{pmatrix})^{-1}\begin{pmatrix} 1 & 0 & 3 \\ 0 & 1 & -1 \end{pmatrix}=\frac{1}{11}\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 3 &-1 \end{pmatrix}\begin{pmatrix} 2 & 3 \\ 3 & 10 \end{pmatrix}\begin{pmatrix} 1 & 0 & 3 \\ 0 & 1 & -1 \end{pmatrix}=\frac{1}{11}\begin{pmatrix} 2 & 3 \\ 3 & 10 \\ 3 & -11 \end{pmatrix}\begin{pmatrix} 1 & 0 & 3 \\ 0 & 1 & -1 \end{pmatrix}=\frac{1}{11}\begin{pmatrix}2 & 3 & 3 \\ 3 & 10 & -11 \\ 3 & -11 & 20 \end{pmatrix}$
| Hints: find a set of basis $\{x_1,x_2\}$ for $M$ and write $X=[x_1 x_2]$ (each $x_i$ is a column vector and $X$ is $3\times 2$). Then, the matrix representation of $p_M$ is $X(X'X)^{-1}X'$. You need to verify that $p_M$, represented this way, is indeed the orthogonal projection onto $M$.
In general, if $X$ has full column rank, then $X(X'X)^{-1}X'$ projects orthogonally onto the column space of $X$.
Edit: To find a basis, note that each vector in $M$ can be written as
$$
(x,y,3x-y)'=(x,0,3x)'+(0,y,-y)'=x(1,0,3)'+y(0,1,-1)'.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\log_ab+\log_bc+\log_ca\geq1+\log_{ab}bc+\log_{bc}ab$ Prove inequality $$\log_ab+\log_bc+\log_ca\geq1+\log_{ab}bc+\log_{bc}ab$$
for $a>1,b>1,c>1.$
Inequality is interesting because of asymmetry and inhomogeneity and I think the solution might interest someone.
We noted $x=\lg a,y=\lg b, z=\lg c $ and wrote inequality in the form
$$\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\geq 1+\frac{y+z}{x+y}+\frac{x+y}{y+z}$$
for $x>0, y>0, z>0.$
We denote $$ \frac{y}{x}=A,\frac{z}{y}=B, \frac{x}{z}=C$$ with $ABC=1$ and we get
$$A+B+C\geq \frac{A+AB}{1+A}+\frac{1+A}{A+AB}+1.$$
For $C=\frac{1}{AB}$ is obtained
$$\begin{align} A^3B^2+A^3B+AB^3-A^2B^2-2A^2B-2AB+A+1 &\geq0 \\
\implies\quad\quad\quad (A+1)(AB-1)^2+AB(A-B)^2 &\geq0.
\end{align}$$
This is a solution. Has anyone another idea?
| We can use C-S.
Indeed, $\frac{x}{z}+\frac{y}{x}+\frac{z}{y}-1-\frac{x+y}{y+z}-\frac{y+z}{x+y}=\frac{x^2}{xz}+\frac{y^2}{yx}+\frac{z^2}{yz}+\frac{y^2}{y^2}-\left(\frac{x+y}{y+z}+\frac{y+z}{x+y}+2\right)\geq$
$\geq\frac{(x+y+z+y)^2}{xz+yx+yz+y^2}-\left(\frac{x+y}{y+z}+\frac{y+z}{x+y}+2\right)=0$
| {
"language": "en",
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Ladder against a wall. Having a bit of a problem with a question. There is a 4m ladder leaving against a wall. There is a box in between The ladder and wall. The box is a cubic metre.
I have found a quartic to find the length up the wall the ladder will reach. But its proving difficult for me to solve.
$$x^4 +2x^3 -14x^2 +2x +1=0$$
| You can find more about this type of equations under the name reciprocal equation or reciprocal polynomial.
See, for example, also this post Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ (And several of the posts shown there among linked questions.)
In this particular case, you have:
$$
\begin{align}
x^4+2x^3-14x^2+2x+1&=0\\
x^2+2x-14+\frac2x+\frac1{x^2}&=0
\end{align}
$$
If we use the substitution $u=x+\frac1x$, we get $u^2=x^2+2+\frac1{x^2}$ $\Rightarrow$ $x^2+\frac1{x^2}=u^2-2$. So your equation gets to the form
$$
\begin{align}
\left(x^2+\frac1{x^2}\right)+2\left(x+\frac1x\right)-14&=0\\
u^2+2u-16=0\\
(u+1)^2-17=0
\end{align}
$$
which yields
$$u_{1,2}=-1\pm\sqrt{17}.$$
Now we have to solve for each $u$ the equation
$$
\begin{align}
x+\frac1x=u\\
x^2-ux+1=0
\end{align}
$$
which yields
$$x=\frac{u\pm\sqrt{u^2-4}}2.$$
Since we have $u^2=16-2u$, this can be rewritten as
$$x=\frac{u\pm\sqrt{12-2u}}2.$$
So the solutions are
$$
\begin{align}
x=\frac{-1\pm\sqrt{17}+\sqrt{14\mp2\sqrt{17}}}2
\end{align}
$$
Here is what WolframAlpha returns for your equation: (Link)
Note that you can switch there between approximate and exact forms of the result.
| {
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How to find nonnegative solutions of a linear system? I have the following system of $M$ linear equations in $N$ unknowns.
$$
\begin{bmatrix}
3 & 0 & 1 & 0 & -1 & -3 & 2\\
1 & 2 & 0 & 4 & 0 & 0 & -1\\
1 & 1 & 0 & 0 & -1 & -1 & -2\\
0 & 0 & 1 & 0 & -3 & -1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4} \\
x_{5} \\
x_{6} \\
x_{7} \\
\end{bmatrix} =
\begin{bmatrix}
1\\
0\\
0\\
-1\\
\end{bmatrix}$$
Is there any algorithm for finding answers of this equations that ${x_{i} \ge 0}$?
Comment: I just want that $x_i \ge 0$.
It can change to
$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 2/3 & -2/3 & 1/3 & 2/3\\
0 & 1 & 0 & 0 & -5/3 & -1/3 & -7/3 & -2/3 \\
0 & 0 & 1 & 0 & -3 & -1 & 1 & -1 \\
0 & 0 & 0 & 1 & 2/3 & 1/3 & 5/6 & 1/6 \\
\end{bmatrix}
$$
| You can choose three unknown values as variables, for example $x_5$, $x_6$ and $x_7$, then the remaining unknown values become a linear function of these variables. In this case,
$$
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
=\frac{1}{6}\left(
\begin{bmatrix}
4 \\
-4 \\
-6 \\
1
\end{bmatrix}
+
\begin{bmatrix}
-4 & 4 & -2 \\
10 & 2 & 14 \\
18 & 6 & -6 \\
-4 & -2 & -5
\end{bmatrix}
\begin{bmatrix}
x_5 \\
x_6 \\
x_7
\end{bmatrix}
\right)
$$
If you for example want to find which values for $x_5$, $x_6$ and $x_7$ can be used such that $x_3$ and $x_4$ are equal to or greater than zero you will run into a problem. You namely want to $x_5$, $x_6$ and $x_7$ also to be equal to or greater than zero, thus the optimal variable which you would like to increase from zero would need to have the largest positive gain for $x_3$ and the smallest negative gain for $x_4$. This is true for $x_5$, however even this variable will not be able to keep both $x_3$ and $x_4$ positive, namely for $x_5=\frac{1}{4}$ then $x_4=0$ but $x_3=-\frac{1}{4}$and for $x_5=\frac{1}{3}$ then $x_3=0$ but $x_4=-\frac{1}{18}$. Thus is can be concluded that there exists no solution such that $x_i\geq0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/982780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the Laurent series of $f(z) = \frac{1}{z-2} + \frac{1}{z-3}$ for $2 < |z| < 3$ and for $|z| > 3$ Find the Laurent series of $f(z) = \frac{1}{z-2} + \frac{1}{z-3}$ for $2 < |z| < 3$ and for $|z| > 3$.
Is the first step here to notice that
$$ \frac{1}{z-2} + \frac{1}{z-3} = \frac{2z-5}{(z-2)(z-3)}$$
and compute the Laurent series as a single term rather than in two parts?
| Case I. $2<\lvert z\rvert<3$.
$$
\frac{1}{z-2}+\frac{1}{z-3}=\frac{1}{z}\cdot\frac{1}{1-\frac{2}{z}}-\frac{1}{3}\frac{1}{1-\frac{z}{3}}=\sum_{n=1}^\infty\frac{2^{n-1}}{z^n}-\sum_{n=0}^\infty \frac{z^n}{3^{n+1}}.
$$
Case II. $\lvert z\rvert>3$.
$$
\frac{1}{z-2}+\frac{1}{z-3}=\frac{1}{z}\cdot\frac{1}{1-\frac{2}{z}}+
\frac{1}{z}\cdot\frac{1}{1-\frac{3}{z}}=\sum_{n=1}^\infty\frac{2^{n-1}+3^{n-1}}{z^n}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/984772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Sum of $1+2+4+8+....$ I was solving a recurrence problem which had a sequence such as $y = (1+2+4+8+...)\sqrt n$, and I wanted to find what $x = 1+2+4+8+...$ was. So consider $x = 1+2+4+8+...$ as an infinite series.
$$x-1 = 2+4+8+...$$
$$2(x)=2(1+2+4+8+...) = 2+4+8+... = x-1$$
$$2x = x-1 \Rightarrow x=-1$$
What does $x=-1$ represent? Have I made a mistake in my calculations? Of course, we expect that $x=\infty$ since the sum grows by a factor of $2$ with each term.
| Here you manipulate x as if it is finite numbers, so you have $2x$. But in fact $n\cdot\infty = \infty$ for any finite number $n$. And also $n+\infty=\infty$ again for any finite number $n$. So basically your equation is $\infty=\infty$, which have nothing useful.
You might be interested in this proof which shows that:
$1+2+3+...=-1/12$
The proof set $c=1+2+3+...$ and have $4c=4+8+12+...$, then he got $c-4c = 1-2+3-4+...=1/(1+1)^2$, ans so $c=-1/4$, which is absurd.
As said in wikipedia I linked to,
It is dangerous to manipulate infinite series as if they were finite sums.
| {
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"url": "https://math.stackexchange.com/questions/985457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
There are 10 sticks of length 1,..,10. How many triangles can be formed There are 10 distinct sticks of length 1,..,10. How many triangles can be formed? I do not know whether there are some counting tricks for this one.
| given the triangle inequality and a hint of euclidean geometry, a triangle(possibly degenerate) can have sides $a,b,c$ with $a\leq b\leq c$ if and only if $c< a+b$
There are $\binom{10}{3}$ ways to select three integers between $1$ and $10$
It should be clear there are $k-1$ ways to write the number $k$ as a sum of two different positive numbers
Lets count how many of these selections leave the larger number bigger than the sum of the other two classifying on the larger number $c$
$c=4: (4-1)=3$
$c=5:4+3$
$c=6:5+4+3$
In general the number of selections of number that cannot form a triangle where $c$ is the larger number is $3+4+\dots+(c-1)$ This sum is $\frac{(c+2)(c-3)}{2}=\frac{c^2-c-6}{2}$
what we would like is $\sum_{4}^{n}\frac{c^2-c-6}{2}=\frac{1}{2}(\sum_4^nc^2-\sum_n^4c-6(n-3))$
We shall simpify this formula using the formulas for the sum of squares and the gaussian sum to get:
$\frac{1}{2}[(\frac{n(n+1)(2n+1)}{6}-14)-(\frac{n(n+1)}{2}-6)-6(n-3)]$
This can be further simplified by the reader.
All you have to do is change $n$ for $10$ and you should hopefully get the number of triangles up to congruence that can be created using those sticks (some will be degenerate though).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/985930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Confusing algebra rule: why $\frac{7^{n+1}-1}{6} + 7^{n+1} = \frac{7^{n+2}-1}{6}$? Math rule I don't understand.
My discrete math midterm is tomorrow and I'm studying proof styles. I came across a rule (algebra maybe?) I don't quite understand and I was hoping someone could explain it step by step for me.
$$\frac{7^{n+1}-1}{6} + 7^{n+1} = \frac{7^{n+2}-1}{6}$$
I guess I can memorize it, but could someone show me how it works step by step?
Thanks
| This isn't the sort of rule you need to memorize, but you do need know the operations to get from one side of the equation to the other.
$$
\begin{align}
\frac{7^{n+1} - 1}{6} + 7^{n+1}
&= \frac{7^{n+1} - 1}{6} + \frac{6\cdot 7^{n+1}}{6} \\
&= \frac{7^{n+1} - 1 + 6 \cdot 7^{n+1}}{6} \\
&= \frac{7 \cdot 7^{n+1} - 1}{6} \\
&= \frac{7^{n+2} -1}{6}.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/986034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Show that for every integer $n ≥ 1$, $1 + \frac{1}{4} +\frac{1}{9} + · · · + \frac{1}{n^2} ≤ 2 − \frac{1}{n}$
I can just think of trying to prove $\frac{1}{4} +\frac{1}{9} + · · · + \frac{1}{n^2} ≤ \frac{1}{n}$, but remains stuck.
| Hint: Note that
$$\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{n^2} \lt \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{(n-1)\cdot n}.$$
The expression on the right turns out to be a telescoping sum. For $\frac{1}{(i-1)\cdot i}=\frac{1}{i-1}-\frac{1}{i}$.
One could also do an induction version of the above proof, using the fact that $\frac{1}{(n+1)^2}\lt \frac{1}{n}-\frac{1}{n+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/987665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given the matrix $A^k$, how to get $A^{k+1}$? Given: $$A^k = \left(\begin{array}{rr} \cos kx & \sin kx \\ -\sin kx & \cos kx\end{array}\right)$$
$$A^{k+1} \overbrace{=}^? \left(\begin{array}{rr} \cos kx & \sin kx \\ -\sin kx & \cos kx\end{array}\right) \left(\begin{array}{rr} \cos x & \sin x \\ -\sin x & \cos x\end{array}\right) \\ = \left(\begin{array}{rr} \cos kx \cos x-\sin kx \sin x & \cos kx\sin x+\sin kx\cos x \\ -\sin kx \cos x+\cos kx (-\sin x) & \cos kx\cos x-\sin kx \sin x\end{array}\right)\\=\left(\begin{array}{rr} \cos (k+1)x & \sin (k+1)x \\ -\sin (k+1)x & \cos (k+1)x\end{array}\right) $$
How? I really don't understand the final part...
How can for exp.
$$
(-\sin kx)(\cos x) + (\cos kx)(-\sin x)=-\sin (k+1)x
$$
I don't understand much English.
| Hint:
$$
\cos a \cos b - \sin a \sin b = \cos (a+b)\\
\sin a \cos b + \sin b \cos a = \sin (a+b)
$$
to remember this:
if the vector $e_a$ of coordinates $(\cos a, \sin a)$ then
$$ \cos a\cos (-b) + \sin a \sin (-b) =
e_a\cdot e_{-b} = \cos(a-(-b))
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $2\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)}\ge 2ax+2by-ay-bx$? Let $a,b,x,y$ be real numbers , then is it true that
$$2\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)}\ge 2ax+2by-ay-bx?$$ Actually I am trying to prove the triangle inequality for the norm of numbers in $K(\rho)$ , where $\rho$ is the imaginary cube-root of unity , $|a+b\rho|:=\sqrt{(a+b\rho)(a+b \rho^2)}=a^2-ab+b^2$
| By C-S
$$2\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)}=2\sqrt{\left(\left(a-\frac{b}{2}\right)^2+\frac{3}{4}b^2\right)\left(\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2\right)}\geq$$
$$2\left(\left(a-\frac{b}{2}\right)\left(x-\frac{y}{2}\right)+\frac{3}{4}by\right)=2ax-bx-ay+2by$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/989274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Proving that $F(x)$ is a constant This was on a test and i know i was supposed to use 2nd ftoc to prove that $F(x)$ was a constant when $x>0$
$$
F(x) = \int_{0}^{x} \frac{1}{t^2 +1} dt + \int_{0}^{\frac{1}{x}} \frac{1}{t^2 +1} \therefore
$$
knowing the following:
$$
F(x) = \int_{a}^{x} f(t) dt
$$
then
$$
F'(x) = f(x)
$$
I would simply do
$$
F'(x) = \frac{1}{x^2+1} + \frac{1}{\frac{1}{x^2}+1} = \frac{1}{x^2+1}+\frac{x^2}{x^2+1} = 1
$$
But couldn't i do this without knowing the 2nd ftoc?
Example:
Set t = tan(x), which then makes the bounds change
$$
\int_{0}^{\arctan(x)}\cos^2(t)dt + \int_{0}^{\arctan(\frac{1}{x})}\cos^2(t)dt
$$
$$
= \frac{1}{2}(\arctan(x)+\sin(\arctan(x))\cos(\arctan(x)) + \arctan(\frac{1}{x})+\sin(\arctan(\frac{1}{x}))\cos(\arctan(\frac{1}{x}))
$$
then use the arctan identities and knowing $\arctan(x)+\arctan(\frac{1}{x}) = \frac{\pi}{2}$ when $x>0$ and do:
$$
\frac{1}{2}[\frac{\pi}{2} + \frac{1}{\sqrt{x^2+1}} \times\frac{x}{\sqrt{x^2+1}} + \frac{1}{x\sqrt{x^{-2}+1}} \times \frac{1}{\sqrt{x^{-2}+1}}]
$$
which equals:
$$
\frac{1}{2}[\frac{\pi}{2} + \frac{x}{x^2+1} + \frac{1}{\frac{1}{x}+x}]
$$
which is not a constant, Where did i go wrong?
| You forgot to replace the $dt$ with the proper expression in the substitution. With $t = \tan x$, we have $dt = \tan' x\,dx = (1+\tan^2 x)\,dx$, so the integral becomes
$$\int_0^{\arctan x} \frac{1+\tan^2 x}{1+\tan^2 x}\,dx = \int_0^{\arctan x} 1\,dx = \arctan x$$
and analogously for the other integral.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Find derivative of tricky logarithmic functions Find the derivative of $y=(x^{x+1})(x+1)^x$
So this is what I have,
$$\ln y=\ln[(x^{x+1})(x+1)^x]$$
$$= \ln x^{x+1} + \ln(x+1)^x$$
$$\frac{1}{y}y' = (1)(\ln x) + (x+1)\frac{1}{x} + (1)(\ln(x+1)) + (x)\frac{1}{x+1}$$
$$= \ln x + \frac{x+1}{x} + \ln(x+1) + \frac{x}{x+1}$$
$$y' = (x^{x+1})(x+1)^x \left[\ln x + \ln(x+1) + \frac{x+1}{x} + \frac{x}{x+1}\right]$$
Is that the right answer?
| Yes that is the right answer..
| {
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"url": "https://math.stackexchange.com/questions/989907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Cyclic Hexagon Circumradius A cyclic hexagon has side lengths of 2, 2, 7, 7, 11, 11, in that order. Find the length of its circumradius.
Not sure if there is a theorem or formula for this, but I tried dividing it into 30°, 60°, 90° triangles. Is that a possible way to approach the problem? If there is a theorem that can be used for this I would love to know. Any help is greatly appreciated.
| In general would be the solution for
$4 R^3 -(a^2+b^2+c^2) R - a b c = 0$
$P = \sqrt[3]{ \sqrt{4(-12)^3(a^2+b^2+c^2)^3 + 2^8 3^6 a^2 b^2 c^2} + 2^4 3^3 a b c }$
$
R = \frac{P}{12 \sqrt[3]{2}} + \frac{\sqrt[3]{2} (a^{2}+b^{2}+c^{2})}{P}
$
then
$a=2, b=7, c=11$
$4 R^3 - 174 R - 154 = 0$
$R = 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Sum of squares in at least 64 ways? How can I found the smallest positive integer that can be represented as a sum of two positive square integers at least 64 ways? Here for example $5=1^2+2^2=2^2+1^2$ are two different representations of a sum. I found in http://mathworld.wolfram.com/SumofSquaresFunction.html that there is a function $r_2(n)$ but I think it is not suitable as it ignores order and signs. Is $5^3\cdot 13\cdot 17\cdot 29\cdot 37\cdot 41=1215306625$ the smallest integer satisfying the condition?
| The Brahmagupta–Fibonacci identity should get you started. A product of two sums of two squares is a sum of two squares in two different ways. Iterate the process and you can get more than two different ways.
For example, $(1^2+2^2)(2^2+3^2) = 4^2+7^2= 1^2+8^2$.
$$
(4^2+7^2)(1^2+3^2) = 25^2 + 5^2= 19^2 + 17^2 = 650
$$
$$
(1^2+8^2)(1^2+3^2) = 25^2 + 5^2 = 11^2 + 23^2 = 650
$$
One could say this gives you four ways to express $650$, but two of them are the same. The fact that some are the same as each other is a complication that will need to be dealt with.
| {
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"url": "https://math.stackexchange.com/questions/992430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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T= 2x2 matrix with inputs [5/4 -1/4] and [3/4 1/4] across. Set xn=T^n (1,0) for n >=1. Let $T= \begin{bmatrix}\frac{5}{4} & -\frac{1}{4} \\ \frac{3}{4} & \frac{1}{4}\end{bmatrix}$. Set $x_n=T^n \begin{pmatrix}1 \\ 0\end{pmatrix}$ for $n\geq 1$. Need to prove that $(x_n)$ converges and find the limit $y$. Also have to find an explicit $N$ so that $\left\|x_n-y\right\| < (1/2)10^{-100}$ for all $n \geq N$. I am not sure how to solve this question at all. I know that to find an explicit $N$, we can use induction but I don't know how to go about doing that.
| This diagonalizes to
$$ T = SJS^{-1} = \pmatrix{\frac{1}{3} & 1 \\ 1 & 1}\pmatrix{\frac{1}{2} & 0 \\ 0 & 1}\pmatrix{-\frac{3}{2} & \frac{3}{2} \\ \frac{3}{2} & -\frac{1}{2}} .$$
Noting that $T^n = SJ^nS^{-1}$ we then have
$$x_n = \pmatrix{\frac{1}{3} & 1 \\ 1 & 1}\pmatrix{(\frac{1}{2})^n & 0 \\ 0 & 1}\pmatrix{-\frac{3}{2} & \frac{3}{2} \\ \frac{3}{2} & -\frac{1}{2}}\pmatrix{1 \\ 0}.$$
Multiplying $x_n$ out we have
$$ x_n = \pmatrix{-2^{-(n + 1)} + \frac{3}{2} \\ -3\cdot2^{-(n+1)} + \frac{3}{2}}. $$
Thus, noting that $\lim_{n \to \infty} x_n = \pmatrix{\frac{3}{2} \\ \frac{3}{2}}$, we know
$$|| x_n - p || = || \pmatrix{-2^{-(n + 1)} & -3\cdot2^{-(n+1)}}^T || = 2^{-(n - 1)} < \frac{1}{2}10^{-100},$$
assuming that the norm is just the max of the sum of the absolute values of each column. If you are using a different definition of the norm, then change as required.
Using this we now have
$$2^{-(n - 1)} < \frac{1}{2}10^{-100} \iff \ln{2^{-n + 2}} < \ln{10^{-100}} \iff -n < \frac{\ln{10^{-100}}}{\ln 2} - 2 \iff n > -\frac{\ln{10^{-100}}}{\ln 2} + 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate the closed form of the series We know that the closed form of the series
$$\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3),$$
but how to evaluate the following series
$$\sum\limits_{n = 1}^\infty {\frac{{{H_{\left[ {\frac{n}{3}} \right]}}}}{{{n^2}}}{{\left( { - 1} \right)}^{n - 1}}} ,\sum\limits_{n = 1}^\infty {\frac{{{H_{\left[ {\frac{n}{3}} \right]}}}}{{{n^3}}}{{\left( { - 1} \right)}^{n - 1}}} .$$
| $1)$ The case where $\left[ x \right]$ is not considered floor function $\left(\left[ x \right]=x \right)$
$$\sum\limits_{n = 1}^\infty {\frac{{{H_{\left[ {\frac{n}{3}} \right]}}}}{{{n^2}}}{{\left( { - 1} \right)}^{n - 1}}} =$$
$$\frac{1}{12}\log^3(3)-\frac{\pi^2}{36}\log\left(\frac{256}{243}\right)-\frac{7}{24}\zeta(3)-\frac{1}{72}\ln(3)\left(9\ln^2(3)-5\pi^2\right)$$
$$+\operatorname{Li_3}\left(\frac{1}{6}\left(3+i\sqrt{3}\right)\right)+\operatorname{Li_3}\left(\frac{1}{6}\left(3-i\sqrt{3}\right)\right)+$$
$$i\frac{\pi}{6}\left(\operatorname{Li_2}\left(\frac{1}{6}\left(3-i\sqrt{3}\right)\right)-\operatorname{Li_2}\left(\frac{1}{6}\left(3+i\sqrt{3}\right)\right)\right)+i\frac{\pi}{3}\left(\operatorname{Li_2}\left(\frac{1}{4}\left(3+i\sqrt{3}\right)\right)-\operatorname{Li_2}\left(\frac{1}{4}\left(3-i\sqrt{3}\right)\right)\right)$$
$2)$ The case where $\left[ x \right]$ is considered floor function
$$\sum\limits_{n = 1}^\infty {\frac{{{H_{\left[ {\frac{n}{3}} \right]}}}}{{{n^2}}}{{\left( { - 1} \right)}^{n - 1}}} =$$
$$\frac{161}{72}\zeta(3)+\frac{\pi^2}{27}\log(3)+\frac{\pi}{72\sqrt{3}}\left(\underbrace{\psi^{(1)}\left(\frac{2}{3}\right)+\psi^{(1)}\left(\frac{5}{6}\right)-\psi^{(1)}\left(\frac{1}{3}\right)-\psi^{(1)}\left(\frac{1}{6}\right)}_{\displaystyle -36\sqrt3\operatorname{Cl}_2\left(\frac{2\pi}{3}\right)}\right)$$
$$+i\frac{5\pi}{9}\operatorname{Li_2}\left(1-i\frac{\sqrt{3}}{3}\right)-i\frac{5\pi}{9}\operatorname{Li_2}\left(1+i\frac{\sqrt{3}}{3}\right)$$
$$+i\frac{7\pi}{9}\operatorname{Li_2}\left(\frac{1}{6}\left(3+i\sqrt{3}\right)\right)-i\frac{7\pi}{9}\operatorname{Li_2}\left(\frac{1}{6}\left(3-i\sqrt{3}\right)\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Maximum absolute value of polynomial coefficients Suppose we have a polynomial in integer coefficients
$$p = p_0 + p_1 x + p_2 x^2 + \ldots + p_n x^n, p_k \in \mathbb{Z}$$
Now define $M(p)$ as the maximum absolute value of the coefficients of $p$, i.e.
$$M(p) = \max \{|p_k| \: |\: 0 \leq k \leq n\}$$
Is it true that for any polynomial factor $q$ of $p$, $M(q) \leq M(p)$? I'm not sure whether this is obvious and I'm just missing something, or whether my train of thought is just completely off...
| No. There is a famous example of a factor of $x^{105}-1$
\begin{align}
\Phi_{105}(x) = & \; x^{48} + x^{47} + x^{46} - x^{43} - x^{42} - 2 x^{41} - x^{40} - x^{39} + x^{36} + x^{35} + x^{34} \\
& {} + x^{33} + x^{32} + x^{31} - x^{28} - x^{26} - x^{24} - x^{22} - x^{20} + x^{17} + x^{16} + x^{15} \\
& {} + x^{14} + x^{13} + x^{12} - x^9 - x^8 - 2 x^7 - x^6 - x^5 + x^2 + x + 1
\end{align}
See cyclotomic polynomials
Also $x^4+1=(x^2+1)^2-2x^2$ has the factorisation $(x^2+\sqrt 2 x + 1)(x^2-\sqrt 2 x +1)$ which isn't integral, but does suggest that lower degree counterexamples may exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Number of pairs $(A,B)$ with $\gcd(A,B)=B, A \ne B^2$ with $A,B \le n$
How many pairs $(A,B)$ of integers up to $n$ are there such that $\gcd(A,B)=B$, not counting those pairs where $B^2=A$?
If we consider $n = 5$ we have $25$ possible pairs. They are
$$(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(2,5),\dotsc,(5,5)$$
Of them, $8$ pairs satisfy the above condition. My main objective is to find the number of such pairs if $n$ is given, with a efficient method. One of my friends showed me a method but he didn't state the reasoning behind it.
Here, I'm presenting his method for $n=10$.
$$
\begin{align}
& 2 \cdot \left(\left(\left\lfloor\frac{10}{1}\right\rfloor - 1\right) + \left(\left\lfloor \frac{10}{2} \right\rfloor - 2\right) + \left(\left\lfloor \frac{10}{3} \right\rfloor - 3\right)\right) \\
=& 2 \cdot ((10 - 1) + (5 - 2) + (3 - 3)) \\
=& 2 \cdot (9 + 3 + 0) \\
=& 24
\end{align}
$$
Can anyone explain the reasoning behind the above method in details and clearly?
| Observe that $\gcd(a,b) = b$ if and only if $b$ divides $a$, so you want to add the number of multiples of $b$ less or equal than $n$ for every $b \leq n$. To remove the pairs $(b^2,b)$ you just need to subtract the number of perfect squares less or equal than $n$. Putting it together, the formula you're looking for is:
$$
\sum_{b = 1}^n \left\lfloor \frac{n}{b} \right\rfloor
- \left\lfloor \sqrt{n} \right\rfloor
$$
Your friend is probably exploiting some symmetry of this formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If a,b,c $>0$ and a+b+c=1, then find the maximum / minimum value of the following If a,b,c $>0$ and a+b+c=1, then find the maximum / minimum value of the following :
(a) abc
(b) $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
(c) $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})$
Using A.M - GM. inequality on a,b,c :
$A.M \geq G.M
$
$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$
$\Rightarrow \frac{1}{27} \geq abc $
If we use : let a =$\frac{1}{2},b=\frac{1}{3},c=\frac{1}{6}$ can we solve the inequalities by assuming these values somehow. please suggest thanks.
| For (a), you can use the AM-GM inequality
$$
(abc)^{1/3}\leq\frac{1}{3}(a+b+c)=\frac{1}{3}\implies abc\leq\frac{1}{27}.
$$
For (b), use the AM-GM again, twice this time:
$$
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq3\sqrt[3]{abc}\times3\frac{1}{\sqrt[3]{abc}}=9\implies\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 9.
$$
For (c), multiply out the LHS
$$
1+T_1+T_2+T_3\quad\text{where}\quad T_1=\frac{1}{abc},\\
T_2=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right),\quad T_3=\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right).
$$
Using (a) and (b), you have indeed estabished the minimum for $T_1$ and $T_2$ ($27$ and $9$, respectively). For $T_3$, you can use
$$
\frac{1}{3}T_3\geq(ab+bc+ca)\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\geq 9\implies T_3\geq 27.
$$
The first inequality is because $ab+bc+ca\leq\frac{1}{3}(a+b+c)^2$ (this one is true because it is equivalent to $(a-b)^2+(b-c)^2+(c-a)^2\geq 0$) and the second inequality above follows in the same manner as we have done in (b).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to derive an proof for this infinite square root equation? Here is continuous square root, namely:
$\sqrt {1 + a \sqrt {1+b \sqrt {1+c\sqrt {1 +...}}}}$= any integer
Find $a,b,c,d,e,f,...$ in general
Uh, very interesting algebra pre-calculus problem, yet very challenging.
I know part of the answer but doesn't know how to start working on this problem.
The original problem is to prove $\sqrt {1 + 2 \sqrt {1+3 \sqrt {1+4\sqrt {1 +...}}}}$$=3$
However,i am curious on how to prove that we have finite or prove that we have infinite number of answer that satisfy the equation
| Copying from Wikipedia in case of losing the data
Ramanujan posed this problem to the 'Journal of Indian Mathematical Society':
$? = \sqrt{1+2\sqrt{1+3 \sqrt{1+\cdots}}}. \, $
This can be solved by noting a more general formulation:
: $? = \sqrt{ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}}}$
Setting this to F(x) and squaring both sides gives us:$
: $F(x)^2 = ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}}$
Which can be simplified to:
: $F(x)^2 = ax+(n+a)^2 +xF(x+n) $
It can then be shown that:
: $F(x) = {x + n + a}$
So, setting ''a'' =0, ''n'' = 1, and ''x'' = 2:
: $3= \sqrt{1+2\sqrt{1+3 \sqrt{1+\cdots}}}$
Ramanujan stated this radical in his lost notebook
$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}}=\frac{2+\sqrt{5}+\sqrt{15-6\sqrt{5}}}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximal points on an $n$-dimensional ellipsoid For an $n$-dimensional ellipsoid in $\mathbb{R}^n$ centered at $\mathbf{0}_n$, defined by a set of vectors $\mathbf{v}_i$ giving the directions of the principle axes and $\lambda_i$ representing the corresponding magnitudes of each axis (for $i=1,\ldots, n$), is there a straightforward way to work out the maximal $x$-coordinate, for example, on the ellipsoid?
$\textbf{2D case:}$
For the 2D case let $\mathbf{v}_1=\left(x_1,\;y_1\right)^T$ with $\lambda_1=a$ represent the major axis and $\mathbf{v}_2=\left(x_2,\;y_2\right)^T$ with $\lambda_2=b$ represent the minor axis.
Here the geometry is not too complicated. We can define a parametric equation for the x-coordinate of the curve in terms of an angle $t$ as
$$x=a\cos(t)\cos(c)-b\sin(t)\sin(c)$$
which with some basic calc and trig can be shown to be maximized at
$$ x_\text{max} = \frac{a^2\cos(c)^2+b^2-b^2\cos(c)^2}{a\cos(c)\sqrt{\frac{a^2\cos(c)^2+b^2-b^2\cos(c)^2}{a^2\cos(c)^2}}} $$
Similarly a maximal point for $y$ can also be obtained.
$\textbf{General case:}$
For the general case, however, I am stumped. I imagine an analytic result is pretty difficult, so any advice on a numerical approach for finding such a solution in general would also be appreciated.
| First, rewrite the equation of your ellipsoid using matrices:
$$\sum_{i=1}^n \frac{(\vec{x} \cdot \vec{v}_i )^2}{\lambda_i^2} = 1
\quad\iff\quad
\vec{x}^T \Lambda\,\vec{x} = 1
\quad\text{ where }\quad \Lambda = \sum_{i=1}^n\frac{\vec{v}_i \otimes \vec{v}_i}{\lambda_i^2}
$$
where $\otimes$ stands for outer product between two column vectors.
The unit normal vector $\vec{n}$ at a point $\vec{x}$ on the ellipsoid will be in the direction of the gradient:
$$\vec{n}\;\propto\; \vec{\nabla}(\vec{x}^T \Lambda\,\vec{x}) = 2 \Lambda\vec{x}$$
Let $\mu \in \mathbb{R}$ be the number such that
$$\Lambda \vec{x} = \mu \vec{n} \quad\iff\quad \vec{x} = \mu \Lambda^{-1} \vec{n}$$
We have
$$\mu^2 \vec{n}^T \Lambda^{-1} \vec{n} = \vec{x}^T\Lambda\vec{x} = 1
\quad\implies\quad
\mu = \frac{1}{\sqrt{\vec{n}^T\Lambda^{-1}\vec{n}}}
\quad\implies\quad
\vec{x} = \frac{\Lambda^{-1}\vec{n}}{{\sqrt{\vec{n}^T\Lambda^{-1}\vec{n}}}}
$$
Given any direction in the form of a unit vector $\vec{e}$, at the point $\vec{x}$ on the ellipsoid which maximizes $\vec{x}\cdot\vec{e}$, the corresponding normal vector $\vec{n}$ lies in the direction of $\vec{e}$. This means the extend of the ellipsoid
along direction $\vec{e}$ will be given by
$$\vec{e}\cdot\vec{x} = \vec{n}\cdot\vec{x} =
\sqrt{ \vec{n}\Lambda^{-1}\vec{n} } = \sqrt{ \vec{e}\Lambda^{-1}\vec{e} }$$
For example, to obtain the extend of the ellipsoid along the $x$-direction, one just need to compute the inverse matrix $\Lambda^{-1}$ and take the square root of its first diagonal element!
Update
Let us use the $2$-dim example in question as an example.
Instead of using $c$ as the angle between the major- and $x$-axis,
we will use $\theta$ instead.
Redefine $c$ now as $\cos\theta$ and let $s = \sin\theta$. we have
$$
\lambda_1 = a,\;\lambda_2 = b,\;
\vec{v}_1 = \begin{pmatrix} c \\ s\end{pmatrix},\;
\vec{v}_2 = \begin{pmatrix}-s \\ c\end{pmatrix}
$$
The matrix $\Lambda$ is now given by
$$\frac{1}{a^2}
\begin{pmatrix}c \\ s\end{pmatrix} \otimes
\begin{pmatrix}c \\ s\end{pmatrix} +
\frac{1}{b^2}
\begin{pmatrix}-s \\ c\end{pmatrix} \otimes
\begin{pmatrix}-s \\ c\end{pmatrix}
= \frac{1}{a^2}
\begin{pmatrix} c^2 & sc \\ sc & s^2\end{pmatrix} +
\frac{1}{b^2}
\begin{pmatrix} s^2 & -sc\\-sc & c^2\end{pmatrix}
$$
Expand everything out, we find
$$\Lambda = \begin{pmatrix}
\frac{c^2}{a^2} + \frac{s^2}{b^2} &
\frac{sc}{a^2} - \frac{sc}{b^2}\\
\frac{sc}{a^2} - \frac{sc}{b^2} &
\frac{s^2}{a^2} + \frac{c^2}{b^2}
\end{pmatrix}
\quad\implies\quad \Lambda^{-1} = \frac{1}{\Delta}
\begin{pmatrix}
\frac{s^2}{a^2} + \frac{c^2}{b^2} &
\frac{sc}{b^2} - \frac{sc}{a^2}\\
\frac{sc}{b^2} - \frac{sc}{a^2} &
\frac{c^2}{a^2} + \frac{s^2}{b^2}
\end{pmatrix}
$$
where $\Delta = \det\Lambda = \frac{1}{a^2b^2}$.
From this, we can read off the maximal $x$-extend of the ellipse as
$$
x_{max}
= \sqrt{\begin{pmatrix}1 \\ 0\end{pmatrix}^T \Lambda^{-1}\begin{pmatrix}1 \\ 0\end{pmatrix}}
= ab \sqrt{\frac{s^2}{a^2} + \frac{c^2}{b^2}}
= \sqrt{a^2\cos^2\theta + b^2\sin^2\theta}
$$
This is a simplified version of the expression of $x_{max}$ in question.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}$ I want to see OTHER approaches than this one. Make sure they are significantly different and not a direct restatement.
$$\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}\tag{1}$$
$$\sum_{n=1}^x n=\frac{x(x+1)}{2} \; \forall x >0\tag{2}$$
$$\begin{align*}
(1)&\stackrel{(2)}{\iff} \frac{2}{2\cdot 3}+\frac{2}{3\cdot 4}+\frac{2}{4\cdot 5}+\dots+\frac{2}{x(x+1)}=\frac{2011}{2013}\\\\
&\iff 2\left (\frac12 -\frac13+\frac13-\frac14+\frac14-\dots+\frac{1}{x}-\frac{1}{x+1}\right )=\frac{2011}{2013}\\\\
&\iff 1-\frac{2}{x+1}=\frac{2011}{2013}\\\\
&\iff x=2012
\end{align*}$$
| I'm not sure there is much which can be done radically differently. The proof can be recast as an induction - which is what you might do after computing the first few sums by hand as $\frac 13, \frac 12=\frac 24, \frac 35, \frac 23=\frac 46, \frac 57$
The $n^{th}$ term is $\cfrac 2{(n+1)(n+2)}$
We can prove that the sum of the first $n$ terms is $\cfrac n{n+2}$, since this is true for the first term and $$\frac n{n+2}+\frac 2{(n+2)(n+3)}=\frac {n+1}{n+3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Maximum of $\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B$ in triangle What is the maximum value of $$\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B,$$ where $A,B,C$ are angles in a triangle?
We can rewrite as $$-\sin A\sin B\sin(A+B)+\sin B\sin(A+B)\cos A+\sin(A+B)\sin A\cos B$$
Expanding, this becomes
$$-\sin^2 A\sin B\cos B-\sin^2B\sin A\cos A+2\sin B\sin A\cos B\cos A+\cos^2A\sin^2B+\sin^2A\cos^2B$$
| Let $\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B=y$
$$\implies2y=\cos C[\cos(A-B)-\cos(A+B)]+2\sin C[\sin B\cos A+\sin A\cos B]$$
Now $\sin B\cos A+\sin A\cos B=\sin(A+B)=\sin(\pi-C)=\sin C$ and $\cos(A+B)=\cos(\pi-C)=-\cos C$
$$\implies y=\cos C[\cos(A-B)+\cos C]+2\sin C[\sin C]$$
On rearrangement, $$\cos^2C-\cos C\cos(A-B)+2y-2=0$$ which is a Quadratic Equation in $\cos C$
As $C$ is real, $\implies\cos C$ is real, the discriminant, $\cos^2(A-B)-4(2y-2)$ must be non-negative
$\implies 4(2y-2)\le\cos^2(A-B)$ which is $\le1$
$\implies y\le\dfrac98$ the equality occurs if $\cos(A-B)=\pm1$
But as $0<A,B<\pi,$
$(i)\cos(A-B)\ne-1$ and consequently $(ii)\cos(A-B)=1\implies A-B=0$
In that case, $\cos C=\dfrac{\cos(A-B)}2=\dfrac12\implies C=?$
| {
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"timestamp": "2023-03-29T00:00:00",
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How would you show that $\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$? I've recently seen a Highschool problem and I was wondering, how would you show that
$$\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$$
Thank you for your time,
| $$\begin{align}\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} &= \sqrt{\left(\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}}\right)^2} \\
&= \sqrt{14 + 4\sqrt{10} + 14 - 4\sqrt{10} - 2\sqrt{(14 + 4\sqrt{10})(14-4\sqrt{10})}} \\
&= \sqrt{28 - 2\sqrt{14^2 - 4^2\cdot 10}} \\
&= \sqrt{28 - 2\cdot \sqrt{196 - 160}} \\
&= \sqrt{28 - 2\sqrt{36}} \\
&= \sqrt{28 - 12} \\
&= \sqrt{16} \\
&= 4\end{align}$$
In the above, you're applying $(a-b)^2 = a^2 + b^2 - 2ab$ and $(a+b)(a-b) = a^2 - b^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int\frac{1}{1+x^6} \,dx$ I came across following problem
Evaluate $$\int\frac{1}{1+x^6} \,dx$$
When I asked my teacher for hint he said first evaluate
$$\int\frac{1}{1+x^4} \,dx$$
I've tried to factorize $1+x^6$ as
$$1+x^6=(x^2 + 1)(x^4 - x^2 + 1)$$
and then writing
$$I=\int\frac{1}{1+x^6} \,dx=\int\frac{1}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx=\int\frac{1+x^2-x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$
$$I=\int\frac{1}{x^4 - x^2 + 1} \,dx-\int\frac{x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$
However $$x^4-x^2+1=\left(x^2-\frac12\right)^2+\frac{3}{4}$$
But I can't see how it helps
I've also tried to reverse engineer the solution given by Wolfram Alpha
And I need to have terms similar to
$$\frac{x^2-1}{x^4-x^2+1} \quad , \quad \frac{1}{1+x^2} \quad , \quad \frac{1}{(x+c)^2+1}\quad , \quad \frac{1}{(x+c)^2+1}$$ in integrand, How can I transform my cute looking integrand into these huge terms?
Since in exams I will neither have access to WA nor time to reverse engineer the solution moreover it does not seem intuitive,is there any way to solve this problem with some nice tricks or maybe substitutions?
| By partial fraction expansion, $$I=\int{1\over x^6+1}dx=\int{1\over f(x)}={1\over f'(x_1)}\int{dx\over x-x_1}+{1\over f'(x_2)}\int{dx\over x-x_2}\ldots+{1\over f'(x_6)}\int{dx\over x-x_6}=\sum_{k=0}^6\ln(x-x_k)^{1\over f'(x_k)}$$ where $x_k$ are the roots of the denominator polynomial. The six complex roots of −1 are $(-1)^{1/6}=e^{(2k+1)\pi i/6},\;k=0\ldots5$ So the integral is
[Correction/simplification: ${1\over f'(x_k)}={1\over6x_k^5}=-{x_k\over6}$]
$$I={1\over6}\ln\left(\Pi_{k=0}^5\left(x-e^{\frac{(2k+1)\pi}{6}i}\right)^{e^\frac{-5(2k+1)\pi i}{6}}\right)={1\over6}\ln\left(\Pi_{k=0}^5\left(x-e^{\frac{(2k+1)\pi}{6}i}\right)^{-e^\frac{(2k+1)\pi i}{6}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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If the ratio of AM and HM of two positive real numbers (a and b) is m:n. Find a:b. $$AM = \frac{a+b}2\text{ and }HM = \frac{2ab}{a+b}$$
Therefore $AM/HM= (a+b)^2/4ab=m/n$.
I'm not able to proceed further. What do I need to do to get $a/b$? Thanks in advance. :)
Also, the answer is $\dfrac { \sqrt{m} + \sqrt{m-n}}{ \sqrt{m} - \sqrt{m-n}}$ how do I get it?
| $$AM = \frac{a + b}{2}$$
$$HM = \frac{2ab}{a + b}$$
$$\frac{AM}{HM} = \frac{(a + b)^2}{4ab} = \frac{m}{n}$$
Let $m = (a + b)^2$ and let $n = 4ab$. Then $m - n = (a - b)^2$.
Therefore,
$$a + b = \sqrt{m}$$
$$a - b = \sqrt{m - n}.$$
Consequently,
$$a = \frac{\sqrt{m} + \sqrt{m - n}}{2}$$
$$b = \frac{\sqrt{m} - \sqrt{m - n}}{2}.$$
Finally,
$$\frac{a}{b} = \frac{\sqrt{m} + \sqrt{m - n}}{\sqrt{m} - \sqrt{m - n}},$$
as desired.
QED.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
deg(pq)=deg(p)+deg(q)
After following this theorem, I am curious as to when:
$deg(pq) < deg(p) + deg(q)$
Can anyone give me either an example when this is the case or conditions for this to be true?
| When the product of their leading coefficients is 0 (for polynomials in one variable). If there is more than one such coefficient (for multivariate polynomials), then the sum of their pairwise products must be 0.
Example: Ring $\left( {{\mathbb{Z}_6}, + , \cdot } \right)$ and polynomial ring $\left( {{\mathbb{Z}_6}, + , \cdot } \right)\left[ X \right]$. $\left( {3{x^2} + 1} \right)\left( {2{x^3} + 1} \right) = 2{x^3} + 3{x^2} + 1$
Example 2: In $\left( {{\mathbb{Z}_6}, + , \cdot } \right)\left[ {X,Y} \right]$, $\left( {2{x^2}y + 2x{y^2} + 1} \right)\left( {3{x^3} + 3{y^3} + 1} \right) = 0\left( {{x^5}y + {x^2}{y^4} + {x^4}{y^2} + x{y^5}} \right) + 2{x^2}y + 2x{y^2} + 3{x^3} + 3{y^3} + 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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show that $P(X=n) = \frac{4}{n(n+1)(n+2)}$ is pmf, that is show that$\sum^{\infty}_{n=1} P(X = n) = 1$ show that $P(X=n) = \frac{4}{n(n+1)(n+2)}$ is a pmf, that is show that$\sum^{\infty}_{n=1} P(X = n) = 1$
My solution
$\sum^{\infty}_{n=1} P(X = n)$
$ =\sum^{\infty}_{n=1} \frac{4}{n(n+1)(n+2)}$
$ =\sum^{\infty}_{n=1} \frac{1}{n(n+1)} - \frac{1}{n(n+2)}$
$ = 2\sum^{\infty}_{n=1} (\frac{1}{n} - \frac{1}{(n+1)}) - (\frac{1}{n} - \frac{1}{(n+2)})$
$ = 2\sum^{\infty}_{n=1} - \frac{1}{(n+1)} + \frac{1}{(n+2)}$
$ = -1$
What am I doing wrong here?
| The second step should be
$$\sum_{n=1}^\infty \frac{2}{n(n+1)}- \frac{2}{(n+1)(n+2)},$$
and then the sum telescopes so we have $2(\frac{1}{2})=1$. The one half comes from the first part of the first term of $\sum_{n=1}^\infty \frac{1}{n(n+1)}- \frac{1}{(n+1)(n+2)}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}$ Given $a,b,c>0$ such that
$$a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
Prove that
$$a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}.$$
| From the hypothesis, we have
$$abc(a+b+c)\ge ab+bc+ca.$$
Since
$$(ab+bc+ca)^2\ge 3abc(a+b+c)$$
we have
$$abc(a+b+c)\ge ab+bc+ca\ge 3.$$
That means
$$a+b+c\ge \frac3{abc}.$$
It remains to show that
$$a+b+c\ge \frac9{a+b+c},$$
which follows from the obvious
$$\frac1a+\frac1b+\frac1c\ge \frac9{a+b+c}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why is the antiderivative of $\frac{1}{1+x^2}=\tan^{-1}(x)$? My textbook says the antiderivative of $\frac{1}{1+x^2}$ is $\tan^{-1}(x)$.
To confirm this to myself I took the derivative of $\tan^{-1}(x)$ expecting to get $\frac{1}{1+x^2}$ , but instead I ended up with $-\frac{1}{\sin^2(x)}$. So why is $\tan^{-1}(x)$ the antiderivative of $\frac{1}{1+x^2}$ if the derivative of $\tan^{-1}(x)$ is not $\frac{1}{1+x^2}$? Shouldn't the derivative of the antiderivative of a function give you the original function?
| $$
\begin{align}
\arctan(x) &= y\\
x &= \tan(y)\\
\frac{\mathrm d}{\mathrm dx} x &= \frac{\mathrm d}{\mathrm dx} \tan(y)\\
1 &= y' \sec^2(y)\\
y'&=\dfrac{1}{\sec^2(y)}\\
y'&=\dfrac{1}{\tan^2(y)+1}\\
y'&=\dfrac{1}{x^2 + 1}
\end{align}
$$
Because $x = \tan(y)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Why $2^{3^n}=-1 \mod 3^{n+1}?$ Why $2^{3^n}=-1 \mod 3^{n+1}?$
By using the Euler totient function $\varphi$ I have got that
$$
2^{\varphi(3^{n+1})}=2^{3^{n+1}-3^n}=2^{2 \cdot 3^n}=1 \mod 3^{n+1}.
$$
How to prove now that the square root is equal to $-1?$
| We have $$(2^{3^n}+1)(2^{3^n}-1) = 0 \mod{3^{n+1}}$$
then remark that $2^{3^n }- 1 = (3-1)^{3^n} - 1 = (-1)^{3^n} - 1 = -2 = 1 \mod{3}$
so $$2^{3^n}+1 = 0 \mod{3^{n+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find all polynomials with real coefficients that satisfy $(x^2-6x+8)P(x)=(x^2+2x)P(x-2)$ Find all polynomials with real coefficients that satisfy $$(x^2-6x+8)P(x)=(x^2+2x)P(x-2)\forall x\in\Bbb R$$
My work;
$$\frac{P(x)}{P(x-2)}=-\frac{4}{x-2}+\frac{12}{x-4}+1\tag{1}$$
$$\frac{P(x-2)}{P(x)}=-\frac{12}{x+2}+\frac{4}{x}+1\tag{2}$$
I also factorised the two known polynomial which didn't give anything useful. What should I extract from these two ratios?
| The relation is $(x-2)(x-4)P(x)=x(x+2)P(x-2)$ hence $$\frac{P(x)}{(x+2)x^2(x-2)}=\frac{P(x-2)}{x(x-2)^2(x-4)}.$$ Can you conclude?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
find the interval of convergence of Taylor series Represent the function $f(x)= x^{0.5}$ as a power series: $\displaystyle \sum_{n=0}^\infty c_n(x−6)^n$
Got that: $c_0$ = $\sqrt{6}$
$C_1=\dfrac{1}{2\sqrt{6}}$
...
But I couldn't find the interval of convergence. I thought we'd require $|x-6| < 1$.
| Let $f(x) = \sqrt{x}$, and $a = 6$. Apply Taylor expansion formula for $f$ at $x = a$ we have:
$f(x) = f(a) + f'(a)(x-a) + \dfrac{f''(a)}{2!}(x-a)^2 + ....+ \dfrac{f^{(n)}(a)}{n!}(x-a)^n + ... = \sqrt{6} + \dfrac{1}{2\sqrt{6}}(x-6) - \dfrac{1}{48\sqrt{6}}(x-6)^2 + .......+\dfrac{(-1)^{n+1}\cdot \displaystyle \prod_{k=1}^{n-1}(2k-1)\cdot 6^{-(2n-1)/2}}{n!\cdot 2^n}(x-6)^n +......$.
$R = \displaystyle \lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \displaystyle \lim_{n \to \infty} \dfrac{2n-1}{12n}\cdot |x-6| = \dfrac{|x-6|}{6} < 1 \iff |x-6| < 6 \iff -6 < x - 6 < 6 \iff 0 < x < 12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
problematic limit with squares I have problem with calculation such limit : $\displaystyle \lim_{n\to \infty} {\frac{10^{\sqrt{n+1}-\sqrt{n}}-1}{2^{\frac{1}{\sqrt{n}}}-1}}$
I only know the answer from wolfram that it's $\frac{1}{2}+\frac{\ln{5}}{\ln{4}}$
| First of all
$$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}.$$
Now, write $x=1/\sqrt{n}.$ Thus, $\sqrt{n+1}=\frac{\sqrt{x^2+1}}{x}.$
Now, we compute the limit
$$\displaystyle \lim_{n\to \infty} {\frac{10^{\sqrt{n+1}-\sqrt{n}}-1}{2^{\frac{1}{\sqrt{n}}}-1}}=\lim_{x\to 0^+} \frac{10^{\frac{x}{1+\sqrt{x^2+1}}}-1}{2^x-1} \underbrace{=}_{\mathrm{L'Hospital}} \lim_{x\to 0^+} \frac{10^{\frac{x}{1+\sqrt{x^2+1}}} \frac{1}{\sqrt{x^2+1}(1+\sqrt{x^2+1})} \ln 10}{2^x\ln 2}\\=\frac{\frac12 \ln 10}{\ln 2}\underbrace{=}_{\ln 10=\ln (2\cdot 5)=\ln 2+\ln 5}\frac{\frac12 \ln 2+\frac 12 \ln 5}{\ln 2}=\frac12+\frac{\ln 5}{2\ln 2}=\frac12+\frac{\ln 5}{\ln 2^2}=\frac12+\frac{\ln 5}{\ln 4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Verifying $\sec^2x + \tan^2x = (1-\sin^4x)\sec^4x$
Verify: $$\sec^2x + \tan^2x = (1-\sin^4x)\sec^4x$$
My solution:
$$
\begin{align}\sec^2x+\tan^2x&=\frac{1}{\cos^2x}+\frac{\sin^2x}{\cos^2x}\\
&=\frac{1+\sin^2x}{\cos^2x}\\
&=\frac{1+\sin^2x}{\cos^2x}\cdot\frac{1-\sin^2x}{1-\sin^2x}\\
&=\frac{1-\sin^4x}{\cos^2x\cdot\cos^2x}\\
&=\frac{1-\sin^4x}{\cos^4x}\\
&=\frac{1}{\cos^4x}-\frac{\sin^4x}{\cos^4x}\\
&=\sec^4x-\sin^4x\sec^4x\\
&=\sec^4x(1-\sin^4x)\\
\end{align}$$
Is it incorrect to multiply in $1-\sin^2x$ like in the fourth equality?
| we have to show that $\sec(x)^2+\tan(x)^2-(1-\sin(x)^4)\sec(x)^4=0$, the left hand side is equivalent to
$\frac{\sin(x)^4-1}{\cos(x)^4}+\frac{\sin(x)^2+1}{\cos(x)^2}=\frac{(\sin(x)^2-1)(\sin(x)^2+1)+(\sin(x)^2+1)\cos(x)^2}{\cos(x)^4}$
the numerator is equivalent to
$(\sin(x)^2+1)(\sin(x)^2+\cos(x)^2-1)=0$
since $\sin(x)^2+\cos(x)^2=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Example 3.53 in Baby Rudin Here's Example 3.53 in the book Principles of Mathematical Analysis by Walter Rudin, third edition.
Consider the convergent series $$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \ldots$$ and one of its rearrangements $$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} + \ldots$$ in which two positive terms are always followed by one negative. If $s$ is the sum of the original series, then $$s < 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6}.$$ Since $$\frac{1}{4k-3} + \frac{1}{4k-1} - \frac{1}{2k} = \frac{8k-4}{(4k-1)(4k-3)} - \frac{1}{2k} = \frac{2k(8k-4) - (4k-1)(4k-3)}{2k(4k-1)(4k-3)} = \frac{8k-3}{2k(4k-1)(4k-3)} > 0$$ for $k \geq 1$, we see that $$s^\prime_3 < s^\prime_6 < s^\prime_9 < \ldots,$$ where $s^\prime_n$ is the $n$th partial sum of the series after the rearrangement. Hence $$\lim_{n\to\infty}\sup s^\prime_n > s^\prime_3 = \frac{5}{6},$$ so that the rearranged series certainly does not converge to $s$.
Now here's my question:
How to determine, using the machinery developed by Rudin upto this point in the book, if the new (or rearranged) series converges at all? Rudin leaves it to the reader to check that the new series does converge. How to prove this convergence?
I would like to have answers that use only the results that Rudin has discussed so far in the book.
| Hint: Notice that $s_{3k}'$ converge as $k \to \infty$. How far can $s_{3k+1}$ and $s_{3k+2}$ be from $s_{3k}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to prove $0 < a_n < 1$ by induction I know $n \in \mathbb{N}$ and...
$$
a_n = \begin{cases}
0 & \text{ if } n = 0 \\
a_{n-1}^{2} + \frac{1}{4} & \text{ if } n > 0
\end{cases}
$$
*
*Base Case:
$$a_1 = a^2_0 + \frac{1}{4}$$
$$a_1 = 0^2 + \frac{1}{4} = \frac{1}{4}$$
Thus, we have that $0 < a_1 < 1$. So our base case is ok.
*Inductive hypothesis:
Assume $n$ is arbitrary. Suppose
$$0 < a_{n} < 1$$
$$0 < a_{n-1}^{2} + \frac{1}{4} < 1$$
is true, when $n > 1$.
*Inductive step:
Let's prove
$$0 < a_{n+1} < 1$$
$$0 < a_{n}^{2} + \frac{1}{4} < 1$$
is also true when $n > 1$.
My guess is that we have to prove that $a^2_{n}$ has to be less than $\frac{3}{4}$, which otherwise would make $a_{n+1}$ equal or greater than $1$.
So we have $(a_{n-1}^{2} + \frac{1}{4})^2 < \frac{3}{4}$... I don't know if this is correct, and how to continue...
| Ok I'll right it down so that it's clear to you :)
I want to prove the property P: " 0 < $a_n$ < 1 "
I look at the property A: $0 < a_n < \frac{1}{2}$
A => P, that is : If A is true then P is true
I'll prove A using induction (so technically I don't prove P by induction, but by implication).
$ a_o = 0 $ < $\frac{1}{2}$
If $ 0 < a_n < \frac{1}{2} $ , then :
$ a_{n+1} = a_n^2 + \frac{1}{4} $ > $a_n^2 $ > 0
And : $ a_{n+1} = a_n^2 + \frac{1}{4} $ < $ (\frac{1}{2})^2 + \frac{1}{4} = \frac{1}{2} $
Hence you get : 0 < $a_{n+1}$ < $\frac{1}{2}$ : the hypothesis holds for the rank n+1
So you have proven using induction that for every n positive integer you have :
0 < $a_n$ < $\frac{1}{2}$
But since : $\frac{1}{2}$ < 1 , you also have:
0 < $a_n$ < $\frac{1}{2}$ < 1 ie 0 < $a_n$ < 1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that it is the solution of the recurrence I have to show that the solution of the recurrence $$X(1)=1, X(n)=\sum_{i=1}^{n-1}X(i)X(n-i), \text{ for } n>1$$
is $$X(n+1)=\frac{1}{n+1} \binom{2n}{n}$$
I used induction to show that.
I have done the following:
For $n=0$ : $X(1)=1 \checkmark $
We assume that it stands for each $1 \leq k \leq n$:
$$X(k)=\frac{1}{k}\binom{2(k-1)}{k-1} \ \ \ \ \ (*)$$
We want to show that it stands for $n+1$:
$$X(n+1)=\sum_{i=1}^{n} X(i)X(n+1-i)=\sum_{i=1}^{n} \frac{1}{i}\binom{2(i-1)}{i-1}\frac{1}{n+1-i}\binom{2(n-i)}{n-i}=\sum_{i=1}^{n}\frac{1}{i}\frac{(2(i-1))!}{(i-1)!(2(i-1)-(i-1))!}\frac{1}{n+1-i}\binom{(2(n-i))!}{(n-1)!(2(n-i)-(n-i))!}=\sum_{i=1}^{n}\frac{(2(i-1))!}{i!(i-1)!}\frac{(2(n-i))!}{(n-i+1)!(n-i)!}$$
How could I continue??
| These are the Catalan numbers.
This can be done by generating functions which is quite simple.
Set $X_0=0$ and $X_1=1$ and introduce
$$f(w) = \sum_{n\ge 0} X_n w^n.$$
Because $X_0=0$ we can extend the recurrence to
$$X_n = \sum_{q=0}^n X_q X_{n-q}$$
for $n>1.$
This says that, again for $n>1,$
$$[w^n] f(w) = [w^n] f(w)^2.$$
Multiply by $w^n$ and sum over $n>1$ to get
$$\sum_{n\gt 1} w^n [w^n] f(w) =
\sum_{n\gt 1} w^n [w^n] f(w)^2.$$
This is an annihilated coefficient extractor (ACE)
and it simplifies to
$$f(w) - X_1 w - X_0 = f(w)^2 - 2X_0 X_1 w - X_0$$
which is
$$f(w) - w = f(w)^2.$$
Solve the quadratic and pick the proper solution to finally obtain
$$f(w) = \frac{1}{2} - \frac{\sqrt{1-4w}}{2}.$$
We can extract coefficients from this in various ways, by table
lookup or using Lagrange inversion. The Newton binomial series gives
$$[w^n] \sqrt{1-w} =
(-1)^n {1/2 \choose n}
\\ = (-1)^n
\frac{1/2 \times (-1/2) \times (-3/2) \times \cdots
\times }{n!}
= (-1)^n \frac{1}{2^n n!} \times (-1)^{n-1}
\prod_{k=0}^{n-2} (2k+1)
\\ = - \frac{1}{2^n n!} \frac{(2n-3)!}{\prod_{k=1}^{n-2} (2k)}
= - \frac{1}{2^n n!} \frac{(2n-3)!}{2^{n-2} (n-2)!}
= - \frac{1}{2^{2n-1} (n-1)} {2n-2\choose n-2}.$$
The conclusion is that
$$[w^n] f(w) = \frac{1}{2^{2n} (n-1)} 4^n {2n-2\choose n-2}
= \frac{1}{n-1} {2n-2\choose n-2}.$$
Re-write this as
$$\frac{1}{n-1} \frac{n-1}{n} {2n-2\choose n-1}
= \frac{1}{n} {2n-2\choose n-1}$$
which also holds for $n=1.$
The ACE technique is also used at this
MSE link.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculation of $\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\mathrm dx$
Calculate the definite integral
$$
I=\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\;\mathrm dx
$$
given that $a>b>0$.
My Attempt:
If we replace $x$ by $C$, then
$$
I = \int_{0}^{\pi}\frac{\sin^2 C}{a^2+b^2-2ab\cos C}\;\mathrm dC
$$
Now we can use the Cosine Formula ($A+B+C=\pi$). Applying the formula gives
$$
\begin{align}
\cos C &= \frac{a^2+b^2-c^2}{2ab}\\
a^2+b^2-2ab\cos C &= c^2
\end{align}
$$
From here we can use the formula $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$ to transform the integral to
$$
\begin{align}
I &= \int_{0}^{\pi}\frac{\sin^2 C}{c^2}\;\mathrm dC\\
&= \int_{0}^{\pi}\frac{\sin^2A}{a^2}\;\mathrm dC\\
&= \int_{0}^{\pi}\frac{\sin^2 B}{b^2}\;\mathrm dC
\end{align}
$$
Is my process right? If not, how can I calculate the above integral?
| We have
$$\eqalign{I_m(a,b)=
\int_0^\pi\frac{\cos(mx)}{a^2-2ab\cos x+b^2}\,\mathrm dx
&=\left\{\matrix{\frac{\pi}{a^2-b^2}&\hbox{if}&m=0\\\cr
\frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^m&\hbox{if}&m\ne0
}\right.}$$
Proof can be seen here. Hence
\begin{align}
\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\,\mathrm dx&=\frac{1}{2}\int_0^{\pi} \frac{1-\cos2 x}{a^2+b^2-2ab \cos x}\,\mathrm dx\\
&=\frac{1}{2}\left[\frac{\pi}{a^2-b^2}-\frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^2\right]\\
&=\frac{\pi}{2 a^2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Finding the exact amount of a sigma problem? $$\begin{align*}
\sum_{k=1}^\infty \frac{1}{k(k+1)}
\end{align*}$$
This is a telescoping series; therefore I use partial fractions to solve.
$\int_{1}^{∞} (1)/(k+1) $ = $ ((A/k)+(B)/(k+1))$
A= 1
B= -1
After solving for partial fractions I get.
$ (1/k)+(-1/k+1)$|∞
|1
Then I solve by subtracting
$(1/∞)-(1)/(∞+1)-(1/1+-1/1+1) = -1$
The answer is 1 yet I got -1. Some input on where I made a mistake would help.
| You've made this more complicated than perhaps it needs to be. First note that $\displaystyle \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$. Hence
$\displaystyle \sum_{k=1}^N \frac{1}{k(k+1)} = \sum_{k=1}^N \frac{1}{k} - \frac{1}{k+1} $
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - ... + \frac{1}{N} - \frac{1}{N+1} $
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 - \frac{1}{N+1}$
Now take the limit as $N \rightarrow \infty$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that $n^3 > (n+1)^2$ for $n>2$ using mathematical induction Is the following a correct way of showing that $n^3 > (n+1)^2$ for $n>2$ using mathematical induction? Thank you in advance!
$P(n): n^3>(n+1)^2$
$P(3): 3^3>(3+1)^2$
$27 > 16$ (correct)
$P(k): k^3>(k+1)^2$
Assume $P(k)$ is true.
$P(k)$ $\implies$ $P(k+1)$?
$k^3+3k^2+3k+1>(k+1)^2+3k^2+3k+1$
$(k+1)^3>(k+1)^2+3k^2+3k+1$
$(3k-2)(k+1)>0$ $\forall$ $k\in N$
$\implies 3k^2+k-2>0$
$\implies 4k^2+5k+2>k^2+4k+4$
$\implies (k^2+2k+1)+3k^2+3k+1>(k+2)^2$
$\implies (k+1)^2+3k^2+3k+1>(k+2)^2$
Hence $P(k) \implies P(k+1)$
Hence by the principal of mathematical induction $P(n)$ is true $\forall n>2$
| That looks fine, but here is a simpler way to do the inductive step: assuming $k\ge3$ and $k^3>(k+1)^2$, we have
$$\eqalign{(k+1)^3
&=k^3+3k^2+3k+1\cr
&>(k+1)^2+3+2k+0\cr
&=k^2+4k+4\cr
&=(k+2)^2\ .\cr}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Find roots of $f(x)=\left(x^2-1\right)^2\left(x^2-2x+1\right)$ I need to find the roots for the following function:
$$f(x)=\left(x^2-1\right)^2\left(x^2-2x+1\right)$$
I guess I already found one by doing this:
$$ f(x)=(x^2-1)^2(x^2-2x+1)\\
(x^2-1)^2 = 0 | +1\\
x^2 = 1 | √ \\
x = 1 \\
P₀₁(1|0)$$
But I could need some help, to find the second zero digit.
| You want $f(x) = 0$.
$$(x-1)^2(x+1)^2(x-1)^2=(x-1)^4(x+1)^2= 0 $$
Then roots are $x_1=1$ multiplicity $4$ and $x_2=-1$ multiplicity $2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to find the binary expansion of any number in the unit interval [0,1] For each integer $n\geq 1$ and $x\in [0,1]$, define $f_n(x)=x_n$ where $x_n$ is the $n$th binary digit of x. If x is a number with two binary expansions, use the expansion that ends with infinitely many zeroes. Draw the graphs of the first three members of the sequence $\{f_n-1/2\}$.
In fact I don't know how to represent the binary expansion of any number in the unit interval [0,1]. Thanks!
| You can get the binary expansion using the floor function as part of an algorithm that uses theory surrounding the limit,
$\tag 1 \quad a = {\displaystyle \lim _{n \to +\infty} \frac{ \lfloor a 2^n \rfloor}{2^n}},\; \text{ for any } a \in \Bbb R$
EXAMPLE: Write $\frac{3}{14}$ as a binary expansion.
$\text{floor}: \lfloor \frac{3}{14} \rfloor = 0$
$2^{-1}\;\,: \lfloor 2 \times \frac{3}{14} \rfloor = \lfloor 0+\frac{3}{7} \rfloor = 0$
$2^{-2}\;\,: \lfloor 2 \times \frac{3}{7} \rfloor = \lfloor 0 + \frac{6}{7} \rfloor = 0$
$2^{-3}\;\,: \lfloor 2 \times \frac{6}{7} \rfloor = \lfloor 1 + \frac{5}{7} \rfloor = 1$
$2^{-4}\;\,: \lfloor 2 \times \frac{5}{7} \rfloor = \lfloor 1 + \frac{3}{7} \rfloor = 1$
Since the $2^{-5}$-calculation with input $\frac{3}{7}$ can be looped back to the $2^{-2}$-calculation, we can write
$\quad \frac{3}{14} = {0.0\overline{011}}_{\; base=2}$
| {
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Why four roots to this equation: $(7x+1)^{1 \over 3}+(8+x-x^2)^{1 \over 3}+(x^2-8x-1)^{1 \over 3}=2$
$$(7x+1)^{1 \over 3}+(8+x-x^2)^{1 \over 3}+(x^2-8x-1)^{1 \over 3}=2$$
I figured the roots are $0$, $1$, $-1$, and $9$. But why?
| Let: $(7x+1)^{1/3} = a,(x^2-8x-1)^{1/3} = b, (8+x-x^2)^{1/3} = c.$
You have: $$\begin{cases} a+b+c = 2 \\ a^3+b^3+c^3 = 8 \end{cases} \Rightarrow a^3+b^3+c^3 = (a+b+c)^3$$
$$ \Leftrightarrow a^3+b^3 = (a+b+c)^3-c^3 = (a+b)[(a+b+c)^2+c(a+b+c)+c^2)]$$
$$ \Leftrightarrow a+b = 0 \text{ or }a^2-ab+b^2 = (a+b+c)^2+c(a+b+c)+c^2 $$
The first case is easy: $a = -b \Leftrightarrow a^3 = -b^3 \Leftrightarrow 7x+1 = x^2-8x-1 \Leftrightarrow ... $
The second case is equivalent to:
$$ab+c(a+b)+c^2 = 0 \Leftrightarrow (a+c)(b+c) = 0$$
I guess you can finish the rest.
| {
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Calculation of all positive integer $x$ for which $\lfloor \log_{2}(x) \rfloor = \lfloor \log_{3}(x) \rfloor \;,$ Calculation of all positive integer $x$ for which $\displaystyle \lfloor \log_{2}(x) \rfloor = \lfloor \log_{3}(x) \rfloor \;,$
where $\lfloor x \rfloor $ represent floor function of $x$.
$\bf{My\; Try::}$ I have used the fact that $\lfloor x\rfloor = \lfloor y \rfloor\;,$ is possible when $x,y\in \left[k\;,k+1\right)\;,$
where $k\in \mathbb{Z}$ and $\left|x-y\right|<1.$
So $\displaystyle \left|\log_{2}(x)-\log_{3}(x)\right|<1\Rightarrow -1<\log_{2}(x)-\log_{3}(x)<1$
Now how can I calculate it, Help me
thanks
| Let $n$ be an integer such that $\lfloor \log_2 x \rfloor = n = \lfloor \log_3 x \rfloor$. If $n=0$, then $x=1$ is a solution. Suppose that $n\ge 1$. By the definiton of the floor function, we obtain
$$
\begin{cases}
2^n \le x < 2^{n+1}\\
3^n \le x < 3^{n+1}
\end{cases}
$$
It is obvious that $2^n < 2^{n+1} < 3^{n+1}$ and $2^n < 3^n < 3^{n+1}$, but it is uncertain to compare $2^{n+1}$ and $3^n$. If $2^{n+1}\le 3^n$, then there is no solution. Suppose that $2^{n+1}>3^n$, then
\begin{align}
n+1 &> n\log_2 3\\
1+\frac{1}{n} &> \log_2 3\\
\therefore n&< \frac{1}{\log_2 3 -1} =\frac{1}{\log_2 \frac{3}{2}} < \frac{1}{\log_2 \sqrt{2}} =2
\end{align}
Thus $n$ must be $1$, and we obtain $x=3$.
Conclusion: All integer solutions of $\lfloor \log_2 x \rfloor = \lfloor \log_3 x \rfloor$ are $x=1$ and $x=3$.
| {
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} |
Group action and semidirect product Let $m,n \in \mathbb N$, $k$ a field, $X=(k^{n\times m},+)$, and we consider the groups $GL(n,k)$ and $GL(m,k)$. Let $K:=GL(n,k) \times GL(m,k)$. We define
\begin{align*}
K\times X &\to X\\
((A,B),M) &\mapsto AMB^{-1}
\end{align*}
Note that this defines an action of $K$ on $X$. With this action we can consider $X \rtimes K$. Let
$$
G=\left\{\begin{pmatrix}
A & M \\
0 & B \\
\end{pmatrix} : A \in GL(n,k), B \in GL(m,k), M \in X \right\} \, .
$$
Show that $G \cong X \rtimes K$.
I've tried to define the following map
\begin{align*}
\psi:G &\to X \rtimes K\\
\begin{pmatrix}
A & M \\
0 & B \\
\end{pmatrix} &\mapsto (M,(A,B))
\end{align*}
but this is not a morphism since $$\begin{pmatrix}
A & M \\
0 & B \\
\end{pmatrix}\begin{pmatrix}
C & M' \\
0 & D \\
\end{pmatrix}=\begin{pmatrix}
AC & AM'+MD \\
0 & BD \\
\end{pmatrix},$$ so $$\psi\left(\begin{pmatrix}
A & M \\
0 & B \\
\end{pmatrix}\begin{pmatrix}
C & M' \\
0 & D \\
\end{pmatrix}\right)=(AM'+MD,(AC,BD)),$$ and $$\psi\left(\begin{pmatrix}
A & M \\
0 & B \\
\end{pmatrix}\right)\psi\left(\begin{pmatrix}
C & M' \\
0 & D \\
\end{pmatrix}\right)=(M,(A,B)).(M',(C,D))=(M+AM'B^{-1},(AC,BD))$$
Could someone suggest me an appropriate isomorphism between these two groups? Thanks in advance.
| First we identify the subgroups of $G$ corresponding to $X$ and $K$:
\begin{align*}
X &\cong \left\{
\begin{pmatrix}
1 & M\\
0 & 1
\end{pmatrix}
: M \in X \right\} =:\tilde{X}\\
\quad
K &\cong \left\{
\begin{pmatrix}
A & 0\\
0 & B
\end{pmatrix}
: A \in GL_n(k), B \in GL_m(k)\right\} =: \tilde{K}
\end{align*}
where $1$ is the identity matrix of the appropriate size.
Note that
\begin{align*}
\begin{pmatrix}
A & M\\
0 & B
\end{pmatrix}
=
\begin{pmatrix}
1 & M B^{-1}\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
A & 0\\
0 & B
\end{pmatrix} \in \tilde{X} \tilde{K} \, .
\end{align*}
This suggests the map
\begin{align*}
\varphi : G & \to X \rtimes K\\
\begin{pmatrix}
A & M\\
0 & B
\end{pmatrix} & \mapsto (MB^{-1}, (A,B)) \, .
\end{align*}
Can you show that this map is a homomorphism?
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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} |
Number of Triples Between $1$ and $n$ The exact question is:
Let $n$ be a positive integer. Find the number of triples $(a,b,c)$ such that $1\leq a\leq b\leq c\leq n$ .
| Suppose $b=k$, then there are $k$ possible values for $a$ (from $1$ to $k$) and $n-k+1$ values for $c$ (from $k$ to $n$). That makes a total of $k(n-k+1)$ possibilities for each value of $k$ from $1$ to $n$:
$$\begin{align}
\sum_{k=1}^n k(n-k+1) &= (n+1)\sum_{k=1}^n k - \sum_{k=1}^n k^2\\
&= \frac{n(n+1)}{2}(n+1) - \frac{n(n+1)}{6}(2n+1)\\
&= \frac{n(n+1)}{6}(3n+3-2n-1)\\
&= \frac{n(n+1)}{6}(n+2)
\end{align}$$
(For the formulas used to solve the sums, check Faulhaber's formula)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How can I prove $\sqrt{(111...)+(55...)^2}=5...6$ The formula in my question can be illustrated as follow:
$$\sqrt{11+5^2}=6$$
$$\sqrt{111+55^2}=56$$ $$\sqrt{1111+555^2}=556$$ $$\sqrt{11111+5555^2}=5556$$ and so on
How can I prove the general formula
$$\sqrt{\underbrace{11\ldots 1}_{n+1\text{ times}} + (\underbrace{55\ldots 5}_{n\text{ times}})^{2}} = (\underbrace{55\ldots 5}_{n-1\text{ times}}6)^{2}$$
| We have
\begin{align}
\underbrace{\dfrac{10^{n+1}-1}9}_{1111\ldots11} + \underbrace{\left(5\cdot \dfrac{10^n-1}9\right)^2}_{555\ldots55^2} & = \dfrac{9\cdot 10^{n+1} - 9 + 5^2 \cdot 10^{2n} - 5^2\cdot 2 \cdot 10^n + 5^2}{9^2}\\
& = \dfrac{5^2 \cdot 10^{2n} + 4 \cdot 10^{n+1} + 16}{9^2}\\
& = \underbrace{\left(5 \cdot \dfrac{10^n-1}9 + 1\right)^2}_{555\ldots56^2}
\end{align}
| {
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"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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} |
Proving $\int_0^1\frac{\log 2-\log\left({1+\sqrt{1-x^2}}\right)}{x}dx=\frac{\left(\pi^2-12\log^22\right)}{24}$
$$\int_0^1\frac{\log 2-\log\left({1+\sqrt{1-x^2}}\right)}{x}dx=\frac{\left(\pi^2-12\log^22\right)}{24}$$
At first, I think it can be calculated like the following one with differential method.
$$\int_0^1\frac{x-\log\left({x+\sqrt{1-x^2}}\right)}{x}dx$$ $\\$
But, I'm wrong. I try my best to do it with differential method, but failed. Who can help me to prove it? Thank you.
| Denote the integral as $I$ and make the substitution $\sqrt{1-x^2}\mapsto x$, we get
$$
\begin{align}
I&=-\int_0^1\frac{x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\,dx\\
&=\frac{1}{2}\int_0^1\frac{\ln\left(\frac{1+x}{2}\right)}{1+x}\,dx-\frac{1}{2}\int_0^1\frac{\ln\left(\frac{1+x}{2}\right)}{1-x}\,dx\\
&=I_1-I_2
\end{align}
$$
Evaluation of $I_1$
$$
\begin{align}
I_1
&=\frac{1}{2}\int_0^1\frac{\ln\left(\frac{1+x}{2}\right)}{1+x}\,dx\\
&=\frac{1}{2}\int_0^1\frac{\ln\left(1+x\right)}{1+x}\,dx-\frac{1}{2}\int_0^1\frac{\ln2}{1+x}\,dx\\
&=-\frac{\ln^22}{4}
\end{align}
$$
Evaluation of $I_2$
$$
\begin{align}
I_2
&=\frac{1}{2}\int_{1/2}^1\frac{\ln x}{1-x}\,dx\qquad\Rightarrow\qquad\frac{1+x}{2}\mapsto x\\
&=\frac{1}{2}\int_{1/2}^1\sum_{n=1}^\infty x^{n-1}\ln x\,dx\\
&=\frac{1}{2}\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln x\,dx\\
&=\frac{1}{2}\sum_{n=1}^\infty\left[\frac{\ln2}{n\,2^n}+\frac{1}{n^2\,2^n}-\frac{1}{n^2}\right]\\
&=\frac{1}{2}\left[\ln^22+\text{Li}_2\left(\frac{1}{2}\right)-\zeta(2)\right]\\
&=\frac{\ln^22}{4}-\frac{\pi^2}{24}
\end{align}
$$
Thus
$$I=\int_0^1\frac{\ln 2-\ln\left({1+\sqrt{1-x^2}}\right)}{x}dx=\frac{\left(\pi^2-12\ln^22\right)}{24}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 4
} |
To prove in a triangle: $AD^2=AB\cdot AC- BD\cdot CD$ If $AD$ is an angle bisector of $\triangle ABC$ (with $D\in BC$), then we have to prove that:
$$AD^2=AB\cdot AC- BD\cdot CD$$
I have no idea how to do this, can this be proved with simple geometry? :)
| By the bisector theorem,
$$ \frac{BD}{DC}=\frac{BA}{AC}. $$
By Stewart's theorem,
$$ (BD+CD)(AD^2+BD\cdot CD)= CD\cdot AB^2 + BD\cdot AC^2, $$
hence:
$$ AD^2+BD\cdot CD = \frac{CD}{BD+CD}AB^2+\frac{BD}{BD+CD}AC^2$$
$$ AD^2+BD\cdot CD = \frac{AC}{AB+AC}AB^2+\frac{AB}{AB+AC}AC^2$$
$$ AD^2+BD\cdot CD = AB\cdot AC.$$
As an alternative, from the cosine theorem we have:
$$ \cos\widehat{BAD}=\frac{AB^2+AD^2-BD^2}{2\,AB\,AD}=\frac{AC^2+AD^2-CD^2}{2\,AC\,AD}=\cos\widehat{DAC},$$
hence:
$$ AC(AB^2+AD^2-BD^2)=AB(AC^2+AD^2-CD^2) $$
or:
$$ (AC-AB)AD^2 = (AC-AB)AB\,AC+AC\,BD^2-AB\,CD^2.$$
Again, the claim follows from the angle bisector theorem.
| {
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} |
Let $a, b, c$ be positive real numbers such that $a + 2b + 3c = 26$ and $a^2 + b^2 + c^2 = 52$. Find the largest possible value of $a$.
I used the Cauchy Schwarz inequality $(ax+by+cz)^2 \leq (a^2+b^2+c^2)(x^2+y^2+z^2)$ as follows:
$a + 2b + 3c = 26$ is given; adding $a$ to both sides gives $2a + 2b + 3c = 26+a$. Then, we have $x = 2$, $y=2$, $z=3$. Putting this all into the inequality, we get:
$(ax+by+cz)^2 \leq (a^2+b^2+c^2)(x^2+y^2+z^2)$
$(2a+2b+3c)^2 \leq (a^2+b^2+c^2)(2^2+2^2+3^2)$
$(26+a)^2 \leq (52)(17)$
$(26+a)^2 \leq 884$
$26 + a \leq \sqrt{884}$
$a \leq 2\sqrt{221} - 26 \approx 3.73$
However, the listed answer gives $a \leq \frac{26}{7} \approx 3.71$. What am I doing wrong?
| Our conditions give
$$a^2+b^2+\left(\frac{26-a-2b}{3}\right)^2=52$$ or
$$13b^2-2(52-2a)b+10a^2-52a+208=0.$$
Hence, $$(52-2a)^2-13(10a^2-52a+208)\geq0$$ or
$$a(26-7a)\geq0$$ or
$$0\leq a\leq\frac{26}{7}.$$
For $a=\frac{26}{7}$ we obtain $b=\frac{52-2\cdot\frac{26}{7}}{13}=\frac{20}{7}$ and $c=\frac{26-\frac{26}{7}-2\cdot\frac{20}{7}}{3}>0,$
which says that the answer is $\frac{26}{7}$.
Done!
| {
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Solving $3x\equiv 4\pmod 7$ I'm trying to learn about linear congruences of the form ax = b(mod m). In my book, it's written that if $\gcd(a, m) = 1$ then there must exist an integer $a'$ which is an inverse of $a \pmod{m}$. I'm trying to solve this example:
$$3x \equiv 4 \pmod 7$$
First I noticed $\gcd(3, 7) = 1$.
Therefore, there must exist an integer which is the multiplicative inverse of $3 \pmod 7$.
According to Bezout's Theorem, if $\gcd(a, m) = 1$ then there are integers $s$ and $t$ such that $sa+tm=1$
where $s$ is the multiplicative inverse of $a\pmod{m}$.
Using that theorem:
$\begin{align}7 = 3\cdot2 +1\\7 - 3\cdot2 = 1 \\-2\cdot3 + 7 = 1\end{align}$
$s=-2$ in the above equation so $-2$ is the inverse of $3 \pmod{7}$.
The book says that the next step to solve $3x \equiv 4 \pmod{7}$ is to multiply $-2$ on both sides.
By doing that I get:
$\begin{align}-2\cdot3x \equiv -2\cdot4 \pmod 7\\-6x\equiv -8 \pmod 7\end{align}$
What should I do after that?
I am working on this problem for hours.
Thanks :)
| You have arrived at
$$-6x=-8\pmod{7}.$$
Now:
$$-6x=-8\pmod{7} \underbrace{\iff}_{\mathrm{add}\: 7x=0\pmod{7}} 7x-6x=-8\pmod{7}\\ \iff x=-8\pmod{7}\underbrace{=}_{\mathrm{add}\: 14=0\pmod{7}}(2\cdot 7-8)\pmod{7}=6\pmod{7}.$$
| {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Given a matrix, find a matrix that satisfies Let A be a matrix (3x4)
Prove that there does not exists a matrix X that satisfies
$$
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}X = \begin{pmatrix}
1 & 1 & 1 \\
0 & 2 & 0 \\
2 & 1 & 1 \\
\end{pmatrix}
$$
When I try to peform Gaussian elimination to get the reduced form of A, I always get a row of zeroes, e.g:
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}
$$ R_3 - R_1 \to R_3 $$
I get
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}
What can I conclude from the fact that I got a zeroes row?
Does this help solving the problem?
| You have shown that the rank of your $3\times 4$ matrix is only (at most)$~2$ (since only two nonzero rows are left). On the other hand your $3\times 3$ matrix is invertible (easy computation), so it has rank$~3$. A matrix product $AB$ cannot have rank larger than the rank of either $A$ or $B$, which shows that no such matrix $X$ exists.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
What is the value of the constant C? The curve $y=Cx^{\frac{1}{5}}$ (where C is constant) is tangent to the line $y=\frac{x}{20}+\frac{32}{5}\:$ somewhere. What is the value of constant C?
| The curve $y_c(x)=C x^{\frac{1}{5}}$ touches the line
$y_l(x)=\frac{1}{20}x+\frac{32}{5}$ at some point $(x_t,y_t)$,
so we have two conditions:
$y_c(x_t)=y_l(x_t)$
and $\left.\frac{dy_c}{dx}\right|_{x=x_t}=\left.\frac{dy_l}{dx}\right|_{x=x_t}$:
$$\begin{align}
C x_t^{\frac{1}{5}}&=\frac{1}{20}x_t+\frac{32}{5},\ (1) \\
\frac{1}{5}C x_t^{-\frac{4}{5}}&=\frac{1}{20}. \ (2)
\end{align}$$
If we multiply (2) by $x_t$ we'll get
another expression for the left hand size of (1):
$C x^{\frac{1}{5}}=\dfrac{5}{20} x_t$. Now it's trivial to find $x_t$
and then $C$:
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Entropy Solution of Burgers' equation $u_t+(u^2/2)_x=0$ Given the initial data
$$ g(x)= \cases{ 1 & $x< -1$ \\ 0 & $-1 < x< 0$ \\ 2 & $0 < x< 1$ \\ 0 & $1 < x$ \\ } $$
What is the entropy solution of Burgers' equation $u_t+(u^2/2)_x=0$?
| We can first create a solution for small $t$ by simply looking at the behavior of the initial condition. Notice that at at $x=-1$ we have a shock, as we are jumping from $1$ to $0$, a rarefaction at $x=0$ as we go from $0$ to $2$, and finally a shock at $x=1$ as we go from $2$ to $0$.
At $x=-1$, the R-H condition gives $\sigma = \frac{1}{2}$, and at $x=1$, $\sigma = 1$. Thus we can form our solution as follows:
$$
u(x,t) = \left\{
\begin{array}{lc}
1 & x<-1+\frac{1}{2}t \\
0 & -1 + \frac{1}{2}t<x<0 \\
\frac{x}{t} & 0<x<2t \\
2 & 2t < x< 1+t \\
0 & x>1+t
\end{array}
\right.
$$
Notice that we have collisions at $t=2$ and $t=1$, therefore this solution only is valid for $0\leq t\leq 1$. We will have to examine each shock seperately.
For the shock at $t=2$, we have $u_l = \frac{x}{t}$ and $u_{r} = 0$. Denote the speed of this shock according to the R-H condition by
$$ s'(t) = \frac{ \frac{1}{2} \left( \frac{s(t)}{t} \right) ^2}{ \frac{ s(t)} {t}} = \frac{s(t)}{2t}$$
with $s(1)=2$ since that is the value of $x$ at $t=1$. This gives $s(t) = 2\sqrt{t}$.
For the other shock at $t=2$, we have $u_l = -1$ and $u_r = \frac{x}{t}$, so
$$q'(t) = \frac{ \frac{1}{2} - \frac{1}{2} \left( \frac{q(t)}{t} \right)^2 }{1 - \frac{q(t)}{t}} = \frac{1}{2} \left( 1+ \frac{q(t)}{t} \right) $$
with $ q(2) = 0 $ by the same reasoning, so that this gives
$q(t) = t - \sqrt{2t}$. Note that these two shocks in fact will intersect eachother at $t = 6+4\sqrt{2}$, so our solution
for $1 \leq t \leq 2$ and
$2 \leq t \leq 6+4\sqrt{2}$ is
$$
\left\{
\begin{array}{lc}
1 & x<-1+\frac{1}{2}t\\
0 & -1+\frac{1}{2}t < x <0\\
\frac{x}{t} & 0 < x < 2 \sqrt{t}\\
0 & x >2 \sqrt{t}
\end{array}
\right.
$$
and
$$
\left\{
\begin{array}{lc}
1 & x<t-\sqrt{2t}\\
\frac{x}{t} & t-\sqrt{2t} < x < 2 \sqrt{t}\\
0 & x >2 \sqrt{t}
\end{array}
\right.
$$
respectively. Finally, the shock at $t=6+4 \sqrt{2}$ is given by, with $u_l = 1$, $u_r = 0$,
$$ r'(s) = \frac{1}{2}$$
with $r(6+4 \sqrt{2}) = 4+2 \sqrt{2}$. This gives $r(t) = \frac{1}{2} +1$, and so the solution for $t > 6+4 \sqrt{2}$ is given by
$$
\left\{
\begin{array}{lc}
1 & x<1+\frac{1}{2}t\\
0 & x > 1 + \frac{1}{2}t\\
\end{array}
\right.
$$
This gives you a full solution, and a picture of the shocks can be seen here:
You can readily see where each portion of the solution is assigned to in the picture as well.
| {
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Prove $x + \frac{1}{x} \geq 2$ for $x>0$. Proof that $x+\frac{1}{x}\geq2$ for $x>0$
Would this be correct?
$x*(x+\frac{1}{x}\geq2)$
$x^2+1\geq2x$
$x^2-2x+1\geq2x-2x$
$x^2-2x+1\geq0$
Plug in 1 for x:
$(1)^2-2(1)+1\geq0$
$1-2+1\geq0$
$0\geq0$ Therefore, $x+\frac{1}{2}\geq2$ is true for $x>0$
| We know for positive $a,b$, $\frac{ a + b}{2} \geq \sqrt{ab} $. Put $ a = x^2 $ and $b = 1$ and we obtain
$$ x + \frac{1}{x} \geq 2 $$
Added: You can also use calculus: Put $f(x) = x + \frac{1}{x} $. we have $f'(x) = 1 - \frac{1}{x^2} $ with critical values (points where $f'$ vanishes ) : $x=\pm1$. It is easy to see that $x = 1 $ will furnish a global minimum. Hence $f(x) \geq f(1) $ for all $x > 0 $. It follows that
$$ x + \frac{1}{x} \geq 2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Ways to write "50" A really good friend of mine is an elementary school math teacher. He is turning 50, and we want to put a mathematical expression that equals 50 on his birthday cake but goes beyond the typical "order of operations" problems. Some simple examples are
$$e^{\ln{50}}$$
$$100\sin{\frac{\pi}{6}}$$
$$25\sum_{k=0}^\infty \frac{1}{2^k}$$
$$\frac{300}{\pi^2}\sum_{k\in \mathbb{N}}\frac{1}{k^2}$$
What are some other creative ways I can top his cake?
I should note that he is an elementary school teacher. Now he LOVES math, and I can certainly show him a lot of expressions. I don't want them so difficult that it takes a masters degree to solve, but they should certainly be interesting enough to cause him to be wowed. Elementary functions are good, summations are also good, integrals can be explained, so this is the type of expression I'm looking for...
EDIT:: I would make a note that we are talking about a cake here, so use your judgement from here on out. Think of a normal rectangular cake and how big it is. Hence, long strings of numbers, complex integrals, and long summations are not going to work. I appreciate the answers but I need more compact expressions.
| $$50 = 2\cdot(2\varphi - 1)^4$$
where $\varphi$ is the Golden Ratio.
$$50 = \sum_{i=0}^{+\infty} (0.98)^i$$
(Geometric series)
$$50 = \left(\left(\frac{5^5-5}{5}+5^0\right)\cdot\left(5-5^0\right)\right)^{0.5}$$
$$50 = 0.5 \cdot (5+5)^{\frac{5^0}{0.5}}$$
$$50 = 5\cdot\left(\frac{5}{0.5}+5^0\right)-5$$
(Using only the digits "5" and "0")
$$50 = \frac{3^{3!}-3^{3-3^0}}{3^{3-3^0}}-30$$
(Using only the digits "3" and "0")
$$50 = \frac{(10i)^2\log(i^i)}{\pi}$$
(Using imaginary unit $i$)
$$50 = 3 + 47$$
$$50 = 7 + 43$$
$$50 = 13 + 37$$
$$50 = 19 + 31$$
(As sum of two prime numbers)
$$50 = (7+11)\frac{11}{11-7}+\frac{7+11}{11-7}-11+7$$
(Using only the two next prime numbers of $5$)
$$ 50 = 7+3+ (7-3)\cdot(7+3)$$
(Using only the previous and next prime numbers of $5$)
$$50 = 3\cdot(2^3+3^2)-(2\cdot 3)^{3-2}+3+2$$
(Using only the two previous prime numbers of $5$)
$$ 50 = (1^6-2^5+3^4-4^3+5^2-6^1)^2\cdot(4^1 - 3^2 + 2^3 - 1^4)$$
$$ 50 = 3 - (1^9-2^8+3^7-4^6+5^5-6^4+7^3-8^2+9^1)$$
(Using bases/powers in reverse order)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 20,
"answer_id": 9
} |
General Solution for a non-Homogeneous Recurrence
What is the general solution to the recurrence:
$$x(n + 2) = x(n + 1) + x(n) + n - 1$$
where $n\geq 1$ with $x(1) = 0, x(2) = 1$?
It's a question on a practice exam I'm reviewing and I'm not quite sure why my solution is incorrect.
So I went through the characteristic equation and found the constants, and the solution to the homogeneous part of this question is:
$$x(n) = \frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{\sqrt{5}}$$
This is the Fibonacci expression, right? So, what do I do with the non-homogeneous part of the expression? My professor mentioned something about solving them separately but I'm not sure what to do with it now.
| Generating Function Approach
The below recurrence is equivalent to the recurrence in the OP, but has been extended to allow $n=0$:
$$a_{n+2} = a_{n+1} + a_n + n - 1\tag{$\star$}$$
for $n\geq 2$, and $a_0=2, a_1=0$.
Define $A(x) = \sum_{n=0}^\infty a_n x^n$. Then, multiplying $(\star)$ by $x^n$ and summing over $n$, we have: (note: all sums are to be taken as $\sum_{n=0}^\infty$, but I'm leaving off the limits for ease of typing.)
$$\sum_n a_{n+2}x^n = \sum_na_{n+1}x^n+\sum_n a_n x^n + \sum_n(n-1)x^n\tag{$\star\star$}$$
We can rewrite $(\star\star)$ as:
$$\frac{A(x)-2}{x^2}-\frac{A(x)-2}{x} -A(x) = \sum_n (n-1)x^n = \frac{2x-1}{(1-x)^2}$$
Rearranging the above, we have:
$$A(x)\frac{1-x-x^2}{x^2}=\frac{2-2x}{x^2}+\frac{2x-1}{(1-x)^2}$$
Thus,
\begin{align}A(x) &= \frac{2(1-x)}{1-x-x^2} + \frac{2x^3-x^2}{(1-x-x^2)(1-x)^2} \\
&= \frac{5 x^2-6 x+2}{(1-x-x^2)(1-x)^2} \\
&= \frac{x-2}{x^2 + x - 1} -\frac{1}{(x-1)^2} - \frac{1}{x-1}
\end{align}
The Taylor Series for each of the above can be found (albeit the first term takes some work), then the coefficient of $x^n$ is the $n$th term.
| {
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"url": "https://math.stackexchange.com/questions/1064364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solving variables in a matrix for a specific determinant The matrix is as follows:
$$
A =
\begin{pmatrix}
0 & x & 1 & 2 \\
x & 1 & 1 & x \\
1 & x & x & 1 \\
1 & x & 1 & x
\end{pmatrix}
$$
What I want to do is to find all the solutions for the equation: $$\det(A) = 0$$
At first I attempted to simplify it into a polynomial, but ending up with a 4th degree term makes me wonder if there's any easier way of solving this? You can easily see that the rows/columns would be linearly independent if $x$ is equal to $1$. But I'm having a hard time realizing any other solutions this way.
Have you guys got any idea? Any help would be much appreciated!
| $$A =
\begin{vmatrix}
0 & x & 1 & 2 \\
x & 1 & 1 & x \\
1 & x & x & 1 \\
1 & x & 1 & x
\end{vmatrix}\stackrel{R_2-xR_3\,,R_4-R_3}=\begin{vmatrix}
0 & x & 1 & 2 \\
0 & 1-x^2 & 1-x^2 & 0 \\
1 & x & x & 1 \\
0 & 0 & 1-x & x-1
\end{vmatrix}=
$$
$$\begin{vmatrix}
x & 1 & 2 \\
1-x^2 & 1-x^2 & 0 \\
0 & 1-x & x-1
\end{vmatrix}=(1-x^2)(1-x)\begin{vmatrix}
x & 1 & 2 \\
1 & 1 & 0 \\
0 & 1 & \!\!-1
\end{vmatrix}=(1-x)^2(1+x)\left[-x+2+1\right]=$$
$$=-(x-1)^2(x+1)(x-3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Prove: $ \sum\limits_{cyc}\frac{ab}{a^2+b^2}+\frac{1}{4}(\sum\limits_{cyc}\frac{1}{a})\geq\frac{15}{4} $ Let $a,b,c>0$ such that $a+b+c=1$.
Prove: $ \sum\limits_{cyc}\frac{ab}{a^2+b^2}+\frac{1}{4}\sum\limits_{cyc}\frac{1}{a}\geq\frac{15}{4} $
I don't have any idea. You guy have any idea??
| By AM-GM inequality:
$$\frac{ab}{a^2+b^2}= \frac12-\frac{(a-b)^2}{2(a^2+b^2)}\ge \frac12-\frac{(a-b)^2}{4ab}=1-\frac14\left(\frac{a}b+\frac{b}a\right)$$
Also
$$\frac1{4a}= \frac{a+b+c}{4a}=\frac14+\frac14\left(\frac{b}a+\frac{c}a \right)$$
Cyclically summing the above two and adding, we get the required inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $\ln \big( b + \sqrt{b^2 + c^2 + x^2}\,\big)$? I am looking to demonstrate the following result. Any ideas are much appreciated.
$$
\begin{align}\int \ln \left( b + \sqrt{b^2 + c^2 + x^2}\right) dx = &\;x \ln \left( b + \sqrt{b^2 +c^2 +x^2}\right) +b \ln \left(2x + 2\sqrt{ b^2 +c^2 +x^2} \right)\\&\; - c \arctan \left(\frac{ b x} { c \sqrt{b^2 +c^2 +x^2}}\right) + c \arctan \left(\frac{x}{c}\right) -x \end{align}$$
| Since integral to function $f(x)$ is such function $F(x)$ so that $F'(x) = f(x) + C$ for $C \in \mathbb{R}$ you can find derivative of your result and compare it to the original task.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the set of complex numbers $z$ which satisfy: $\left\lvert\frac{z-3}{z+3}\right\rvert=2$ Find the set of complex numbers $z$ which satisfy
$$\left\lvert\frac{z-3}{z+3}\right\rvert=2\text.$$
I need help on that one. Thank you.
| $$\begin{align}
\left\lvert\frac{z-3}{z+3}\right\rvert=2\\
\left\lvert1-\frac{6}{z+3}\right\rvert=2\\
\left\lvert\frac16-\frac{1}{z+3}\right\rvert=\frac13\\
\left\lvert\frac16-Q\right\rvert=\frac13
\end{align}$$
where $Q=\frac{1}{z+3}$. This last equation shows us $Q$ is along a circle of radius $\frac13$ centered at $\frac16$. The reciprocal of a circle in the complex plane is another circle. The circle $Q$ is on has a diameter passing through $-1/6$ and $1/2$ on the real line. So after we take the reciprocal, we have a circle with a diameter on the real line passing through $-6$ and $2$, and so its center is at $-2$ and radius is $4$.
So $z+3$ is on this circle. And then finally, $z$ is on a circle of radius $4$ centered at $-5$. That is, the solution set is the same as for $\left\lvert z+5\right\rvert=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
intial value piecwise linear ODE; slightly wrong answer Where am I going wrong?
Solve the given initial value problem. Use a graphing utility to graph the continuous function y(x).
$\frac{dy}{dx} +2xy=f(x),y(0)=2$ where $f(x)=\left\{
\begin{array}{lr}
x, & 0 \leq x \leq 1\\
0, & x \gt 1
\end{array}
\right.$
I did $\frac{dy}{dx}+2xy=x$ for $0 \leq x \leq 1 $
$P(x) = 2x$ then $\int P(x) dx = x^2$ and $f(x) = x$
$\frac{d}{dx} [ e^{P(x)dx}y]=e^{\int P(x)dx}f(x)$
$\frac{d}{dx}[e^{x^2}y]=e^{x^2}x$
$\int \frac{d}{dx}[e^{x^2}y] dx =\int e^{x^2}x dx$
$e^{x^2}y = \frac{e^{x^2}}{2}+C$
Rewriting $C = \frac{1}{2}+Ce^{-x^2}$
Then for $x>1$ $\frac{dy}{dx}+2xy=0$ and $P(x)= 2x$ then $\int P(x) dx = x^2$ and $f(x)=0$
$\frac{d}{dx} [ e^{P(x)dx}y]=e^{\int P(x)dx}f(x)$
$\frac{d}{dx}[e^{x^2}y]=e^{x^2}0=0$
$\int \frac{d}{dx}[e^{x^2}y] dx= \int 0 dx = C$
rewriting $e^{x^2}y=C$ so $y=Ce^{-x^2}$ and applying $y(0)=2$ gives
$2=C^{-0^2}=C$
So then
$y=\left\{
\begin{array}{lr}
\frac{1}{2}+2e^{-x^2} & -1 \leq x \leq 1 \\
2e^{-x^2} & x > 1 \\
\end{array}
\right.$
But the answer key gives
$y=\left\{
\begin{array}{lr}
\frac{1}{2}+\frac{3}{2}e^{-x^2} & -1 \leq x < 1 \\
(\frac{1}{2} e + \frac{3}{2})e^{-x^2} & x \ge 1 \\
\end{array}
\right.$
| where you went wrong is applying $y(0) = 2$ to the solution that applies for $x>1$. You should have applied it to the solution that applis for $x<1$, giving $\frac{1}{2} + C = 2$. This gives the answer that appears in the key.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Product of numbers in any two cells sharing a side is $2$ In a $3\times 3$ square, every cell has a positive number. The product of numbers in any two cells sharing a side is exactly $2$. What is the minimum sum of all the numbers?
We may color the cells in chessboard fashion, and write $a$ in black and $b$ in white cells, where $ab=2$. The sum of all the numbers is $5a+4b\geq 2\sqrt{20ab}=4\sqrt{10}$. Inequality holds when $5a=4b$ and $ab=2$. But how to show that it suffices to consider this configuration? (It makes sense since it's the extreme configuration, but I'm not sure how to prove it.)
| Let $a_{11} = a$, and $a_{12} = b$, then the condition that adjacent cells product is $2$ gives the following $3\times 3$ matrix:
$\begin{bmatrix} a & b & a\\ b & a & b\\ a & b & a\end{bmatrix}$ with $ab = 2, a > 0, b > 0 \Rightarrow \displaystyle \sum_{i,j=1}^3 a_{ij} = 5a + 4b = 5a + \dfrac{8}{a} = f(a)$.
$f'(a) = 5 - \dfrac{8}{a^2} = 0 \iff a^2 = \dfrac{8}{5} \iff a = \dfrac{2\sqrt{10}}{5} \Rightarrow b = \dfrac{2}{a} = \dfrac{\sqrt{10}}{2} \Rightarrow f_{\text{min}} = 5\left(\dfrac{2\sqrt{10}}{5}\right) + 4\left(\dfrac{\sqrt{10}}{2}\right) = 4\sqrt{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx=1-\frac{\gamma}{2}-\ln2$ Nowadays I encounter an integral which is difficult for me to evaluate it. Please help me to evaluate it. Thank you.
$$\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx=1-\frac{\gamma}{2}-\ln2$$
where $\gamma$ is The Euler–Mascheroni constant.
| A Generalisation:
\begin{align}
\int^\infty_0\frac{x}{(x^2+w^2)(1+e^{2\pi x})}{\rm d}x\tag1
=&\int^\infty_0\frac{xe^{-x}}{(x^2+4\pi^2w^2)(1+e^{-x})}{\rm d}x\\ \tag2
=&\int^\infty_0\frac{x}{x^2+4\pi^2w^2}\left(\sum^\infty_{n=1}(-1)^{n-1}e^{-nx}\right){\rm d}x\\ \tag3
=&\int^\infty_0xe^{-x}\left(\sum^\infty_{n=1}\frac{(-1)^{n-1}}{x^2+4n^2\pi^2w^2}\right){\rm d}x\\
=&\int^\infty_0\frac{e^{-x}}{2x}-\frac{e^{-x}}{4w}\mathrm{csch}\left(\frac{x}{2w}\right){\rm d}x\tag4\\
=&-\frac{1}{2}\int^1_0\frac{1}{\ln{x}}+\frac{x^{1/2w}}{w(1-x^{1/w})}\ {\rm d}x\tag5\\
=&-\frac{1}{2}\int^1_0\frac{x^{w-1}}{\ln{x}}+\frac{x^{w-1/2}}{1-x}{\rm d}x\tag6\\
=&-\frac{1}{2}\int^0_\infty\int^1_0x^{t+w-1}+\frac{x^{t+w-1/2}\ln{x}}{1-x}\ {\rm d}x\ {\rm d}t\tag7\\
=&-\frac{1}{2}\int^0_\infty\frac{1}{t+w}-\psi_1\left(t+w+\frac{1}{2}\right)\ {\rm d}t\tag8\\
=&\boxed{\large{\color{blue}{\displaystyle\frac{1}{2}\psi_0\left(w+\frac{1}{2}\right)-\frac{1}{2}\ln{w}}}}\\
\end{align}
Explanation:
$(1)$: Substituted $x\mapsto\dfrac{x}{2\pi}$.
$(2)$: Expanded $\dfrac{1}{1+e^{-x}}$ as a geometric series.
$(3)$: Substituted $x\mapsto\dfrac{x}{n}$.
$(4)$: One can show that $\displaystyle\sum^\infty_{n=-\infty}\frac{(-1)^n}{x^2+4n^2\pi^2w^2}=\frac{1}{2wx}\mathrm{csch}\left(\frac{x}{2w}\right)$ using the residue theorem.
$(5)$: Substituted $x\mapsto-\ln{x}$.
$(6)$: Substituted $x\mapsto x^w$.
$(7)$: Used the fact that $\displaystyle\frac{1}{\ln{x}}=\int^0_\infty x^{t}\ {\rm d}t$.
$(8)$: Used the integral representation of the polygamma function.
The Integral:
Let $w=1$ to get
$$\int^\infty_0\frac{x}{(x^2+1)(e^{2\pi x}+1)}{\rm d}x=\frac{1}{2}\psi_0\left(\frac32\right)=\boxed{\large{\color{red}{\displaystyle 1-\frac{\gamma}{2}-\ln{2}}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 2
} |
Find all intergers such that $2n^2+1$ divides $n^3+9n-17$ Find all intergers such that $2n^2+1$ divides $n^3+9n-17$.
Answer : $n=(2 \ and \ 5)$
I did it.
As $2n^2+1$ divides $n^3+9n-17$, then $2n^2+1 \leq n^3+9n-17 \implies n \geq 2$
So $n =2$ is solution and doens't exist solution when n<2. How can I do now to find 5 ? Or better, how can you solve this with another good method ?
Thanks
| HINT:
If integer $d$ divides $n^3+9n-17,2n^2+1$
$d$ must divide $2(n^3+9n-17)-n(2n^2+1)=17n-34$
$d$ must divide $17(2n^2+1)-2n(17n-34)=68n+17$
$d$ must divide $68n+17-4(17n-34)=153$
So the necessary condition is $2n^2+1$ must divide $153$
$\implies2n^2+1\le153\iff n^2\le76\iff-9<n<9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074524",
"timestamp": "2023-03-29T00:00:00",
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There is no largest rational number $p$ such that $p^2 < 2$ In Rudin's analysis example 1.1, he tried to show the following
Let $A$ be the set of all positive rationals $p$ such that $p^2<2$ and let $B$ consist of all positive rationals $p$ such that $p^2 > 2$. He show that $A$ contains no largest number and $B$ contains no smallest
More explicitly, for every $p$ in $A$ we can find a rational $q$ in $A$ such that $p < q$ and vice versa
He then shows
$$q = p - \frac{p^2 - 2}{p + 2}$$
$$q^2 - 2 = \frac{2(p^2 - 2)}{(p + 2)^2}$$
and conclude that if $p$ in $A$ then $p^2 - 2 < 0 \rightarrow q^2 < 2$ and $q > p $ and vice versa.
I understand the conclusion, and I understand the hypothesis, but I didn't understand how he got
$$q = p - \frac{p^2 - 2}{p + 2}$$
and
$$q^2 - 2 = \frac{2(p^2 - 2)}{(p + 2)^2}$$
Specifically, I think he is trying to show the following
If $p$ in $A$, let $q$ be a number greater than $p$
$$q = p + \beta$$
now show
$$q^2 - 2 < 0 \Rightarrow q^2 - 2 = p^2 + 2\beta p + \beta^2 - 2$$
I use the quadratic equation to find the threshold for $\beta$, but I can't derive what Rudin give, can someone explain how Rudin got his equations?
| Observe that: $p^2 < 2 \Rightarrow p^2+2p < 2 + 2p \Rightarrow p(p+2) < 2+2p \Rightarrow p < \dfrac{2+2p}{p+2}\in \mathbb{Q}$. He then set $q = \dfrac{2+2p}{p+2}$, then $p < q$ from definition of $q$, and furthermore: $q^2 - 2 = \left(\dfrac{2+2p}{p+2}\right)^2 - 2= \dfrac{4+8p+4p^2}{p^2+4p+4} - 2= \dfrac{2}{(p+2)^2}\cdot (p^2-2) < 0\Rightarrow q^2< 2.$
| {
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"url": "https://math.stackexchange.com/questions/1076841",
"timestamp": "2023-03-29T00:00:00",
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$\int_0^1e^{(ax^2 + bx)}dx$ in terms of erf I'm trying to evaluate the definite integral,
$$\int_0^1e^{(ax^2 + bx)}dx$$
in terms of the function,
$$F(z)=\int_0^ze^{p^2}dp$$
The correct answer that i'm supposed to get is,
$$\int_0^1e^{(ax^2 + bx)}dx=\frac{e^{-\frac{b^2}{4a}}}{\sqrt{a}}[F(\frac{a+b}{\sqrt{a}})-F(\frac{b}{\sqrt{a}})] $$
However i'm getting a slightly different answer with $\frac{1}{2}$ appearing in the arguments to the function F. What I get is the below,
$$\int_0^1e^{(ax^2 + bx)}dx=\frac{e^{-\frac{b^2}{4a}}}{\sqrt{a}}[F(\frac{a+b}{2\sqrt{a}})-F(\frac{b}{2\sqrt{a}})] $$
Can someone advise if i'm doing something wrong?
Many thanks!
| $$
ax^2 + bx = a\left(x+\frac{b}{2a}\right)^{2} - \frac{b^2}{4a}
$$
thus
$$
\mathrm{e}^{ - \frac{b^2}{4a}}\int_0^1 \mathrm{e}^{a\left(x+\frac{b}{2a}\right)^{2}}dx
$$
let $p = \sqrt{a}\left(x+\frac{b}{2a}\right)$
then we have
$$
\int_{\frac{b}{2\sqrt{a}}}^{\sqrt{a}\left(1+\frac{b}{2a}\right)}\mathrm{e}^{p^2}\frac{dp}{\sqrt{a}}
$$
or
$$
\frac{1}{\sqrt{a}}\mathrm{e}^{ - \frac{b^2}{4a}}\left[\int_0^{\sqrt{a}\left(1+\frac{b}{2a}\right)}\mathrm{e}^{p^2}dp -\int_0^{\frac{b}{2\sqrt{a}}}\mathrm{e}^{p^2}dp \right]
$$
set $s = \frac{b}{2}$ and $a = t$
we find
$$
\sqrt{a}\left(1+\frac{b}{2a}\right) = \sqrt{t}\left(1 + \frac{s}{t}\right) = \frac{s+t}{\sqrt{t}}\\
\frac{b}{2\sqrt{a}} = \frac{s}{\sqrt{t}}
$$
and finally
$$
\mathrm{e}^{ - \frac{b^2}{4a}}\frac{1}{\sqrt{t}}\left[\int_0^{\frac{s+t}{\sqrt{t}}}\mathrm{e}^{p^2}dp
-\int_0^{\frac{s}{\sqrt{t}}}\mathrm{e}^{p^2}dp \right]
$$
so your answer is out by a factor of 2 due to not setting your final version of "$x$" as $b/2$. But this is purely a game of trying to match the correct answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove that $f(x) = x^3 + 3x - 1$ is irreducible in $\mathbb Q[X]$
Prove that $f(x) = x^3 + 3x - 1$ is irreducible in $\mathbb Q[X]$.
Let $\theta$ be a root of $f(x)$. Compute $\frac{1}{\theta}$ and $(2 + \theta^2)^{-1} $ in $\mathbb Q[\theta ]$.
\begin{array}{l}
f\left( \theta \right) = \theta ^3 + 3\theta - 1 = 0 \\
\Leftrightarrow \theta \left( {3 + \theta ^2 } \right) = 1 \\
\Leftrightarrow \frac{1}{\theta } = \left( {3 + \theta ^2 } \right);\left( {\theta \ne 0} \right) \\
\end{array}
\begin{array}{l}
\frac{1}{\theta } = 3 + \theta ^2 ;\left( {\theta \ne 0} \right) \\
\Leftrightarrow \frac{1}{\theta } - 1 = 2 + \theta ^2 \\
\Leftrightarrow \left( {\frac{1}{\theta } - 1} \right)^{ - 1} = \left( {2 + \theta ^2 } \right)^{ - 1} \quad ;\left( { \pm \sqrt 2 \notin Q} \right) \\
\Leftrightarrow \left( {2 + \theta ^2 } \right)^{ - 1} = \frac{\theta }{{1 - \theta }}\quad ;\left( {\theta \ne 1} \right) \\
\end{array}
But I can't show that $f$ is irreducible.
| For polynomials of deg $\leq 3$ it is enough to check that there are no rational roots (i.e a linear factor). Applying the Rational Roots Theorem the set of possible roots are $\pm 1$, which clearly $g(\pm 1) \not = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $I(a) = \int_0^{\pi/2} \frac{dx}{1-a\sin x}$ I have a problem with this integral. It seems that solution has to be simple, but I couldn't find out.
$$I(a) = \int_0^{\pi/2} \frac{dx}{1-a\sin x}$$
I tried using integration by parts and differentiating with regard to a, but neither helped.
| Using $u=\sin x$ and $v^2=\frac{1+u}{1-u}$
we have
\begin{eqnarray}
\int_0^{\frac{\pi}{2}}\frac{dx}{1-a\sin x}&=&\int_0^{1}\frac{1}{(1-au)\sqrt{1-u^2}}du\\
&=&\int_0^{1}\frac{1}{(1-au)(1+u)}\sqrt{\frac{1+u}{1-u}}du\\
&=&\int_1^\infty\frac{1}{(1-a\frac{v^2-1}{v^2+1})(1+\frac{v^2-1}{v^2+1})}v\frac{2v}{(1+v^2)^2}dv\\
&=&2\int_1^\infty\frac{1}{(1+a)+(1-a)v^2}dv\\
&=&\frac{2}{1-a}\int_1^\infty\frac{1}{v^2+\frac{1+a}{1-a}}dv\\
&=&\frac{2}{1-a}\sqrt{\frac{1-a}{1+a}}\arctan\left(\sqrt{\frac{1-a}{1+a}}v\right)\bigg|_1^\infty\\
&=&\frac{2}{\sqrt{1-a^2}}\left(\frac{\pi}{2}-\arctan\left(\sqrt{\frac{1-a}{1+a}}\right)\right)
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof that sum of first $n$ cubes is always a perfect square I know that
$$1^3+2^3+3^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2$$
What I would like to know is whether there is a simple proof (that obviously does not use the above info) as to why the sum of the first $n$ cubes always a perfect square.
| Let's prove that using the mathematical induction.
Check the base
$n = 1$
$1^3 = (\frac{1(1 + 1)}{2})^2$
That's true
Suppose that is true for $n - 1$ and prove that is true for $n$.
We know $1^3 + 2^3 + \dots + (n-1)^3 = (\frac{(n-1)n}{2})^2$
Then
$1^3 + 2^3 + \dots + (n-1)^3 + n^3 = (\frac{(n-1)n}{2})^2 + n^3 = \frac{n^2(n-1)^2}{4} + n^3 = \frac{n^4 - 2n^3 + n^2}{4} + n^3 = \frac{n^4 + 2n^3 + n^2}{4} = \frac{n^2(n+1)^2}{4} = (\frac{n(n+1)}{2})^2$
QED
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of $\sum ( \cos \sqrt[3]{n^3 + \sqrt n + 7} - \cos \sqrt[3]{n^3 - 2\sqrt n + 3})$ I have some problem with this example: $$\displaystyle \sum_{n=2}^{\infty}\Bigg(\cos\Big(\sqrt[3]{n^3+\sqrt{n}+7}\Big) -\cos\Big(\sqrt[3]{n^3-2\sqrt{n}+3}\Big)\Bigg)$$
the only idea that crossed my mind is to use that $\cos x-\cos y=-2\sin\big({\frac{x+y}{2}}\big)\sin\big({\frac{x-y}{2}}\big)$ but later I don't know what to do with sines how to compare them or what else I can do with them ?
| Hint. You may write, as $n \to \infty$,
$$ \sqrt[3]{n^3+\sqrt{n}+7}=\left(n^3+\sqrt{n}+7\right)^{1/3}=n\left(1+\frac{\sqrt{n}+7}{n^3}\right)^{1/3}=n+\mathcal{O}_1\left(\frac{1}{n^{3/2}}\right)$$
and
$$ \sqrt[3]{n^3-2\sqrt{n}+3}=\left(n^3-2\sqrt{n}+3\right)^{1/3}=n\left(1+\frac{-2\sqrt{n}+3}{n^3}\right)^{1/3}=n+\mathcal{O}_2\left(\frac{1}{n^{3/2}}\right)$$
thus, using $\displaystyle \cos x-\cos y=-2\sin{\frac{x+y}{2}}\cdot \sin{\frac{x-y}{2}}$, we get
$$\left| \cos\sqrt[3]{n^3+\sqrt{n}+7} -\cos\sqrt[3]{n^3-2\sqrt{n}+3}\right|\leq2 \left|\sin \left(\mathcal{O}_3\left(\frac{1}{n^{3/2}}\right)\right)\right|=\mathcal{O}_4\left(\frac{1}{n^{3/2}}\right)$$ and conclude that the initial series converges as does $\displaystyle \sum \frac{1}{n^{3/2}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $2(x+y)+xy=x^2+y^2$ where $x,y \in \mathbb{Z}$ Solve the equation: $$2(x+y)+xy=x^2+y^2$$
How should I go about solving this? Any guidance appreciated.
Thanks!
| $ 2(x+y) + xy = x^2 + y^2 $
$ 2(x+y) + 3xy = (x+y)^2 $
$ u = x + y, v = xy $
$ 2u + 3v = u^2 $
$ u^2 - 2u - 3v = 0 $
$ \frac{1}{2}u^2 - u - \frac{3}{2}v = 0 $
$ D = 1 + 3v $
$ u = 1 + \sqrt{1 + 3v} $ or $ u = 1 - \sqrt{1 + 3v} $
Let $ \sqrt{1+3v} = k $, hence $ v = \frac{k^2 - 1}{3} $
First case $ u = 1 + \sqrt{1+3v} $
$ x + y = 1 + k $
$ xy = \frac{k^2 - 1}{3} $
$ y = 1 + k - x $
$ 3x(1+k-x) = k^2 - 1 $
$ 3x + 3kx - 3x^2 = k^2 - 1 $
$ 3x^2 - 3(k+1)x + k^2 - 1 = 0 $
$ D = 9(k+1)^2 - 12 (k^2 - 1) = -3k^2 + 18k + 21 = -3(k^2 - 6k - 7) = -3(k+1)(k-7) $
$ -3(k+1)(k-7) >= 0 $, hence $ k \in [-1, 7] $ and $ k \in \mathbb N $
$x = \frac{3(k+1) \pm \sqrt{-3(k+1)(k-7)}}{6}$
Because of this $ k $ may be 1, 5 or 7 for $ \sqrt{-3(k+1)(k-7)} \in \mathbb Z $
$ k = 1 $, hence $ x = 2 $ and $ y = 0 $ OR $ x = 0 $ and $ y = 2 $
Similarly you can regard cases $ k = 5 $ and $ k = 7 $ AND $ u = 1 - \sqrt{1 + 3v} $.
In summary the answer will be $ (0, 0) $, $ (0, 2) $, $ (2, 0) $, $ (2, 4) $, $ (4, 2) $, $ (4, 4) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving inequality $\frac{n}{2} < 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + + \frac{1}{2^{n}-1} < n$ Prove the inequality $\frac{n}{2} < 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + + \frac{1}{2^{n}-1} < n$ where $n\in\mathbb{N}\backslash\{0,1\}$
My work
At first I tried to prove first half by induction
$\frac{n}2+\frac{1}2<1+\frac{1}2+\dots+\frac{1}{2^{n+1}-1}$
Which is $\displaystyle\frac{1}2+\sum_{k=0}^{2^n-1}\frac{1}k<\sum_{k=0}^{2^{n+1}-1}\frac{1}k$
And then I got this $\displaystyle\sum_{k=0}^{2^{n+1}-1}\frac{1}k-\sum_{k=0}^{2^n-1}\frac{1}k=\sum_{k=0}^{2n}\frac{1}{2^n+k}$
And now I see no way to go on $\displaystyle\sum_{k=0}^{2n}\frac{1}{2^n+k}>\frac{1}2$
where $(n>1,n\in\mathbb{N},h=\overline{0,2n})$
My math background is too low so I'm not able to prove this last inequality I've got, maybe someone knows how to?(maybe using limit's definition?) But I think there should be more simplicated solution, which I cannot find for almost 4 hours now. Thank you
| Since $\frac{1}{n}\geq \log\left(1+\frac{1}{n}\right)$, it follows that:
$$\sum_{n=1}^{2^N-1}\frac{1}{n}\geq\sum_{n=1}^{2^N-1}\left(\log(n+1)-\log(n)\right) = \log 2^N = N \log 2.$$
On the other hand, $\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\leq\frac{1}{2n^2}$ gives:
$$\sum_{n=1}^{2^N-1}\frac{1}{n}\leq N \log 2+\frac{1}{2}\sum_{n=1}^{+\infty}\frac{1}{n^2}= N\log 2+\frac{\pi^2}{12}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Find the value of : $\lim_{x \to \infty}( \sqrt{4x^2+5x} - \sqrt{4x^2+x})$ $$\lim_{x \to \infty} \left(\sqrt{4x^2+5x} - \sqrt{4x^2+x}\ \right)$$
I have a lot of approaches, but it seems that I get stuck in all of those unfortunately. So for example I have tried to multiply both numerator and denominator by the conjugate $\left(\sqrt{4x^2+5x} + \sqrt{4x^2+x}\right)$, then I get $\displaystyle \frac{4x}{\sqrt{4x^2+5x} + \sqrt{4x^2+x}}$, but I can conclude nothing out of it.
| Since you already received answers, let me show you another approach you could use. $$A=\sqrt{4x^2+5x} - \sqrt{4x^2+x}=2x \sqrt{1+\frac{5}{4x}}-2x \sqrt{1+\frac{1}{4x}}=2x \Big(\sqrt{1+\frac{5}{4x}}-\sqrt{1+\frac{1}{4x}}\Big)$$ Now, you may be already know that, when $y$ is small compared to $1$ $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Use this twice, replacing $y$ by $\frac{5}{4x}$ for the first radical and by $\frac{1}{4x}$ for the second radical. You will then have $$A=2x \Big(\frac{1}{2 x}-\frac{3}{16 x^2}+\cdots\Big)=1-\frac{3}{8 x}+\cdots$$ which shows the limit and how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Conditional inequalities Let a,b,c be positive real numbers such that $abc=1$. Prove that $$\frac 1{a^3(b+c)}+\frac 1{b^3(c+a)}+\frac 1{c^3(a+b)} \ge \frac 32$$ We can derive the following inequalities from the given equality :-$$ab+bc+ca\ge3$$$$a+b+c\ge 3$$$$\frac 1a+\frac 1b+\frac 1c\ge3$$ None of them help to get a solution. If we simplify the given inequality, we get the following $$(a+b)(b+c)(\frac 2{b^3}-a-c)+(b+c)(c+a)(\frac 2{c^3}-a-b)+(c+a)(a+b)(\frac 2{a^3}-b-c) \ge 0$$ What do i do after this?
| This is IMO 1995, Problem 2.
$\sum \frac{1}{a^3(b+c)}=\sum\frac{1/a^2}{a(b+c)}\ge\sum \frac{(1/a+1/b+1/c)^2}{2(ab+bc+ca)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac32.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve this equation $\frac{1}{x+\sqrt{x^2-1}}=\frac{1}{4x}+\frac{3x}{2x^2+2}$
Solve this equation
$$\frac{1}{x+\sqrt{x^2-1}}=\frac{1}{4x}+\frac{3x}{2x^2+2}$$
This equation has solutions $x=\pm 1$ but I cannot find a solution.
Do you think using an inequality will is the way forward?
| LHS part is equal to $x - \sqrt{x^2-1}$ and RHS is $\frac{1+ 7x^2}{4x^3 +4x}$. Thus,
$$ x - \sqrt{x^2-1} = \frac{1+ 7x^2}{4x^3 +4x} \Rightarrow x^2 -1 = \left(\frac{1+7x^2}{4x^3 +4x} -x \right)^2 \Rightarrow (x^2-1)(4x^3+4x)^2 = (-4x^4 +3x^2+1)^2 \Rightarrow 40x^6 -17x^4 -22x^2 -1 = 0.$$ I think you can solve now.
| {
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"url": "https://math.stackexchange.com/questions/1086527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Summation of series with terms $U_n=\frac{1}{n^2-n+1} -\frac{1}{n^2+n+1}$
Given that $U_n=\dfrac{1}{n^2-n+1} -\dfrac{1}{n^2+n+1}$, find $S_N$= $\sum_{n=N+1}^{2N}U_n$ in terms of $N$. Find a number $M$ such that $S_n<10^{-20}$ for all $N>M$.
I was able to calculate the sum as $\dfrac{1}{N^2+N+1}-\dfrac{1}{4N^2+2N+1}$ using the method of differences. I am having trouble doing the second part. After thinking a while I think that the correct approach is to remove some of the terms while preserving the inequality and then solve for $N$. Two such attempts were $\dfrac{1}{N^2+N+1}-\dfrac{1}{N^2}$ and $\dfrac{1}{N^2+N+1}-\dfrac{1}{N}$ but I don't know how to proceed. Is this the correct way? Any help, including hints, would be appreciated
| You're overthinking it. Just use some crude bounds:
$$\begin{align*}
S_N=\frac{1}{N^2+N+1}-\frac{1}{4N^2+2N+1}&< \frac{1}{N^2+N+1}\\
&< \frac{1}{N^2}.
\end{align*}$$
Therefore $M=10^{10}$ works, by which we mean that for any $N>10^{10}$, $S_N<10^{-20}$:
$$
S_N<\frac{1}{N^2}<\frac{1}{10^{20}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator
It can be calculated that $\frac{555555}{7} = 79365$. What is the remainder of the number $5555\dots5555$ with a thousand $5$'s, when divided by $7$?
I did the following:
$$\begin{array}
& 5 \ \text{mod} \ 7=& &5 \\
55 \ \text{mod} \ 7= & &6 \\
555 \ \text{mod} \ 7= & &2 \\
5555 \ \text{mod} \ 7= & &4 \\
55555 \ \text{mod} \ 7= & &3 \\
555555 \ \text{mod} \ 7= & &0 \\
5555555 \ \text{mod} \ 7= & &5 \\
55555555 \ \text{mod} \ 7= & &6 \\
555555555 \ \text{mod} \ 7= & &2 \\
5555555555 \ \text{mod} \ 7= & &4 \\
\end{array}$$
It can be seen that the cycle is: $\{5,6,2,4,3,0\}$.
$$\begin{array}
& 1 \ \text{number =} &5 \\
7 \ \text{numbers =} &5 \\
13 \ \text{numbers =} &5 \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots & \\
985 \ \text{numbers =} &5 \\
991 \ \text{numbers =} &5 \\
997 \ \text{numbers =} &5 \\
998 \ \text{numbers =} &6 \\
999 \ \text{numbers =} &2 \\
\color{red}{1000} \ \color{red}{\text{numbers =}} &\color{red}{4} \\
\end{array}$$
From here, we can conclude that $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7 = 4$.
However, I wasn't allowed to use a calculator and solved this in about 12 minutes. Another problem was that there was a time limit of about 5 minutes. My question is: Is there an easier and faster way to solve this?
Thanks a lot in advance!
| After noting $555555$ is divisible by $7$, note further that $555555\times 10^r$ is divisible by $7$ for any positive integer $r$. So you can cast out groups of six $5$s starting at the most significant digit, without changing the remainder on division by $7$. This gets rid of $996$ of the $5$s, leaving $5555$. Then $4949$ is obviously divisible by $7$ leaving $606$, and simple division then gives the remainder $4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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