Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Show limit of function $\frac{xy(x+y)}{x^2-xy+y^2}$ for $(x,y)\to(0,0)$ Show $$\lim_{(x,y)\to (0,0)} xy \frac{(x+y)}{x^2-xy+y^2}=0$$
If I approach from $y=\pm x$, I get $0$. Is that sufficient?
| No, that is not sufficient. That can only show that if the limit exists, then it should equal $0$. But it doesn't show that the limit exists.
Change to polar coordinate system, so that
$$
\lim_{(x,y)\to (0,0)} \frac{xy(x+y)}{x^2-xy+y^2} = \lim_{r\to 0} \frac{ (r^2 \sin \theta \cos \theta) (r\sin \theta + r\cos \theta)}{r^2-r^2\sin \theta \cos \theta}= \lim_{r\to 0} \frac{r\sin \theta \cos \theta(\sin \theta + \cos \theta)}{1-\sin \theta \cos \theta}
$$
(Check that two limits are equilvalent from the ($\varepsilon$-$\delta$) definition of the limits)
But notice that $|1-\sin \theta \cos \theta|=|1-\sin (2\theta) /2|\ge 1/2$.
Hence $\displaystyle \frac{\sin \theta \cos \theta(\sin \theta + \cos \theta)}{1-\sin \theta \cos \theta}$ is bounded for all $\theta\in \mathbb{R}$, and so
$$
\lim_{r\to 0} \frac{r\sin \theta \cos \theta(\sin \theta + \cos \theta)}{1-\sin \theta \cos \theta}=0
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$f(x) = (\cos x - \sin x) (17 \cos x -7 \sin x) $ $f(x) = (\cos x - \sin x) (17 \cos x -7 \sin x)$
Determine the greatest and least values of $\frac{39}{f(x)+14}$ and state a value of x at which greatest values occurs.
Do I just use a graphing calculator for this? Is there a way I could do this without a graphing calculator?
| HINT:
$$f(x)=17\cos^2x+7\sin^2x-24\sin x\cos x=\dfrac{17(1+\cos2x)+7(1-\cos2x)-24\sin2x}2$$
$$=12+5\cos2x-12\sin2x=12+\sqrt{12^2+5^2}\cos\left(2x+\arctan\dfrac{12}5\right)$$
Now for real $x,-1\le\cos\left(2x+\arctan\dfrac{12}5\right)\le1$
| {
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What type of number is this $\frac{\sqrt2}{2}$? $$\frac{\sqrt{2}}{2}$$
In this monomial, an irrational number is divided by a rational number. However this is not a general case but can any one tell me that when we divide an irrational number or multiply an irrational number or its multiplicative inverse by a rational number then what type of the number we get in output? Either rational or irrational?
| If $p$ is a rational (non-zero) number, and $x$ is irrational, then
$$
px, \quad \frac px, \quad \frac xp, \quad p+x, \quad p-x
$$
are all irrational. This can be proven by contradiction. Assume, for instance, that $px$ is a rational number, call it $q$. That is, set $q = px$. Now divide through by $p$, and we get $$\frac{q}{p} = x$$
The left-hand side is a fraction of two rational numbers, so it is rational. The right-hand side is $x$, so it is irrational. These two can therefore not be equal, and therefore the assumption that $px$ is rational must be false.
The others can be proven the same way. Be careful, however, because for each of the five expressions
$$
xy, \quad \frac xy, \quad \frac yx, \quad x+y, \quad y-x
$$there are some irrational numbers $x, y$ such that makes it rational (although not all at once).
| {
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Integrals with the special functions $Ci(x)$ and $erf(x)$ I'm looking for the solutions of the following two integrals:
$$I_1=\int\limits_0^\infty dx\, e^{-x^2}\text{Ci}(ax)$$
and
$$I_2=\int\limits_0^\infty dx\, e^{-ax}\text{erf}(x)$$
with
$$\text{Ci}(x)=\int\limits_\infty^x\frac{\cos(t)}{t}dt$$
and
$$\text{erf}(x)=\frac{2}{\sqrt{\pi}}\int\limits_0^xe^{-t^2}dt$$
Now I'm not 100% sure what is meant by "the solution of the integrals" since these will probably be not-evaluable. But I'm guessing that the question is to reduce the expression to one single special function in stead of the integral of a special function with an elementary function.
Mathematica yields me the answers:
$$I_1=-\frac{\sqrt{\pi}}{4}\Gamma\left(0,\frac{a^2}{4}\right)$$
and
$$I_2=\exp\left(\frac{a^2}{4}\right)\frac{1-\text{erf}(a/2)}{a}$$
A good first step for evaluating these integrals $I_1$ and $I_2$ seemed to fill in the integral representations of these special functions and try to switch te integrals over $x$ and $t$. However this has not (yet) been a success. I also tried to find a differential equation for these integrals, but also this was not so easy to do. Are there any tips/tricks to evaluate these integrals ?
| For $I_2$, please note that this integral is just the Laplace transform of $\text{Erf}(x)$ w.r.t. to the variable $a$
Furthermore the Laplace transform of an integral $\mathcal{L}(\int f(x) dx)$ is just $\mathcal{L}(f(x))/a$
Using this we find that
$$
I_2=\frac{2}{a\sqrt{\pi}}\int_0^{\infty}e^{- a x}e^{-x^2}dx
$$
Completing the square in the exponent and changing variables $y=\frac{a}{2}+x$
This turns into
$$
I_2=2\frac{ e^{\frac{a^2}{4}}}{a\sqrt{\pi}}\int_{\frac{a}{2}}^{\infty}e^{-y^2}dy
$$
Now using straightforwardly the definition of the Errorfunction we obtain
$$
I_2=\frac{e^{\frac{a^2}{4}}}{a}\left(1-\text{Erf}(a/2)\right)
$$
As promised!
I will have a look on $I_1$ later
| {
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Given angles and area, how to find sides of a spherical triangle? So, given angles and area, how to find the sides of a spherical triangle?
I only know that the angles uniquely determine the sides, but what is the relation?
| For angles A, B, and C and corresponding sides a, b, and c of a spherical triangle:
cos A = -cos B cos C + sin B sin C cos a
cos B = -cos C cos A + sin C sin A cos b
cos C = -cos A cos B + sin A sin B cos c
Just plug in your numbers and solve for the sides.
| {
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Probability Question (Colored Socks)
In a drawer Sandy has 5 pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the 10 socks in the drawer. On Tuesday Sandy selects 2 of the remaining 8 socks at random and on Wednesday two of the remaining 6 socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
She has $5$ different colors: $a, b, c, d, e$.
The number of ways of selecting $2$ socks on Wednesday is:
$$\binom{6}{2}$$
But how do I go further?
| Partial solution
Probability of getting matching socks on day 1 is $\frac19$.
If they were not matching, we have 3 pairs and 2 individual socks left. We must select a 'pairable' sock in our first pick, so $\frac68$ for that, and the other pick must be the other sock of the same pair, so $\frac17$. Therefore, probability of getting matching socks on day 2 (but not day 1) is $\frac89\times(\frac68\times\frac17)=\frac2{21}$
Probability of missing a match on day 1 as well as on day 2 should be $1-(\frac19+\frac2{21})=\frac{50}{63}$
I'm not sure how to get probability of getting a match on day 3, though.
| {
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Fibonacci sequence and the Principle of Mathematical Induction Consider the Fibonacci sequence, $F_n$. Prove that $2 ~\vert~ F_n$ if and only if $3 ~\vert~ n$, using the principle of mathematical induction.
I know that I have to prove two implications here. Looking at the first implication (if $2 ~\vert~ F_n$, then $3 ~\vert~ n$) , I am a little confused as to how my base case would look: is it valid to say that $F_1=1$ is not divisible by $2$ and $n=1$ is not divisible by $3$ making the statement true for $n=1$?
| $$f_{n+3}=f_{n+2}+f_{n+1}=f_{n+1}+f_n+f_{n+1}=2f_{n+1}+f_n$$
Since $2f_{n+1}$ is even, you get $f_{n+3}$ is even if and only if $f_n$ is even.
The statement follows now by induction : Check $P(1), P(2), P(3)$ and prove $P(n) \Rightarrow P(n+3)$.
| {
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Proving divisibility of $a^3 - a$ by $6$ As part of a larger proof, I need to show why $a^3-a$ is always divisible by $6$. I'm having trouble getting started.
| Consider the following options:
*
*$a\equiv0\pmod6 \implies a^3-a\equiv 0-0\equiv6\cdot 0\equiv0\pmod6$
*$a\equiv1\pmod6 \implies a^3-a\equiv 1-1\equiv6\cdot 0\equiv0\pmod6$
*$a\equiv2\pmod6 \implies a^3-a\equiv 8-2\equiv6\cdot 1\equiv0\pmod6$
*$a\equiv3\pmod6 \implies a^3-a\equiv 27-3\equiv6\cdot 4\equiv0\pmod6$
*$a\equiv4\pmod6 \implies a^3-a\equiv 64-4\equiv6\cdot10\equiv0\pmod6$
*$a\equiv5\pmod6 \implies a^3-a\equiv125-5\equiv6\cdot20\equiv0\pmod6$
| {
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Solutions of the Laplace equation How do I find solutions $u=f(r)$ of the two-dimensional Laplace equation $u_{xx}+u_{yy}=0$ that depend only on the radial coordinate $r= \sqrt{x^2+y^2}$
| You should also use the polar coordinate version of the Laplace operator
$$
\Delta u
= {\partial^2 u \over \partial r^2}
+{1 \over r} {\partial u \over \partial r}
+ {1 \over r^2} {\partial^2 u \over \partial \theta^2}
$$
and trying for $u(r, \phi) = f(r)$. This reduces to solving the ODE
$$
0 = g'(r) +{1 \over r} {g}
$$
for $g(r) = f'(r)$.
| {
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Conditional expectation is $L_p$ norm reducing I am trying to follow the proof of
\begin{align*}
E \left(|E(X\mid \mathcal{B} )|^p \right) \le E(|X|^p)
\end{align*}
which goes as follows.
Let $f(t)=|t|^p$ which is convex for $p\ge 1$ then
\begin{align*}
E \left(f(E(X\mid \mathcal{B} )) \right) &\le E \left(E(f(X) |\mathcal{B} ) \right) \text{ this is by Jensen's inequality}\\
&= E(f(X) ) \\
&=E(|X|^p)
\end{align*}
I don't undestand the second step where $E \left(E(f(X) \mid \mathcal{B} ) \right)=E(f(X) ) $. What happend to $\mathcal{B}$?
| Using the definition of conditional expectation, we know that for any random variable $Y$ $$E(E(Y|\mathcal{B})) = \int_{\Omega} E(Y|\mathcal{B}) dP = \int_{\Omega} Y dP = E(Y)$$ This is called the tower property of conditional expectation and intuitively it "gets rid of the $\mathcal{B}." $ Now just apply this to $Y=f(X)$.
| {
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If f is surjective then f is not a right divisor of zero Let $R$ be a ring and $M$ a R-module. For $r\in R$ define $f:M\to M$ by $f(s)=sr$.
Show that $f$ is injective if and only if $r$ is not a right zero divisor.
I have done a similar problem to this in the past, involving surjectivity instead of injectivity and came up with:
Suppose f is surjective and gf=0. To prove f is not a right divisor of zero, we need to show that g=0, i.e. that g(m)=0 for all m. So let m be in M. Since f is surjective, m=f(n) for some n in M. Thus g(m)=g(f(n))=(gf)(n)=0(n)=0.
However I am stuck with how to show this problem using injectivity, and was wondering could anybody provide guidance on this?
| $f$ is injective if and only if
$f(s)=0 \Rightarrow s=0$ if and only if
$sr=0 \Rightarrow s=0$ if and only if
$r$ is not a right zero divisor.
| {
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Linear approximation with two variables The problem I have is this:
Use suitable linear approximation to find the approximate values for given functions
at the points indicated:
$f(x, y) = xe^{y+x^2}$ at $(2.05, -3.92)$
I know how to do linear approximation with just one variable (take the derivative and such), but with two variables (and later on in the assignment, three variables) I'm a bit lost. Do I take partial derivatives and combine then somehow? Can someone guide me through a problem of this type?
Thank you in advance.
| Yes. Denoting the partial derivatives by $f'_x$ and $f'_y$, the formula is:
$$f(x_0+h,y_0+k)=f(x_0,y_0)+f'_x(x_0,y_0)h+f'_y(x_0,y_0)k+o\bigl(\lVert(h,k)\rVert\bigr).$$
| {
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For which $n$ does $n\mid 1^n+2^n+\cdots+n^n$? Find all the natural numbers $n$ such that
$$n\mid 1^n+2^n+\cdots+n^n.$$
We know through Faulhaber's formula, that
$$\sum_{k=1}^{n}k^n=\frac1{n+1}\sum_{k=0}^n\binom{n+1}{k}B_k n^{n+1-k},$$
where $B_k$ is a Bernoulli number. I checked few dozen values of $n$ and it seems that only odd numbers are solutions. Any ideas on how to proceed from here?
| To see tnat odd numbers always work, it is enough to use the formula $a^n+b^n = (a+b)(a^{n-1}-a^{n-2}b+...+b^n)$. Thus, if $n$ is odd, $n | k^n+(n-k)^n$ and therefore it divides the sum $1^n+...+n^n$.
| {
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Expressing as $z(t) = x(t) + iy(t)$ My exercise requires to express the equation of a line connecting two point, $z_1 = -3 +2i$ and $z_2 = 3 - 5i$, as $z(t) = x(t) + iy(t)$.
We know that the equation for a line is $$ Re\left [ (m+i)z + b \right ] = 0$$ where $z = x+iy$
The slope is calculated as $$m = \frac{Im(z_1) - Im(z_2)}{Re(z_1) - Re(z_2)}$$
For $z_1 = -3 + 2i$ and $z_2 = 3 - 5i$, we calculate the slope to be: $m = \frac{7}{-6}$.
$$b =\frac{ [|z_2| - |z_1|]}{2(Re(z_2) - Re(z_1))}$$
$$ = \frac{ \sqrt{34 }- \sqrt{13} }{2(6)}$$
Then the line is $Re\left [ (-\frac{7}{6} + i )z + \frac{ \sqrt{34 }- \sqrt{13} }{2(6)} \right ] = 0$
However, I do not know how to get it into the required form.
| The form your exercise is requesting is often called the "parametric" version of the equation, just to throw the terminology out there.
To find the slope, think in terms of the growth of the line. This is expressed by $z_2-z_1$, the vector between the two points, or $6-7i$, with the calculation done.
All we need to do now to fully define the line segment is add the initial point/position vector, $z_1$ (not that different from with the reals, you may notice). This gives us:
$z(t)=(-3+2i)+t(6-7i).$
| {
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Finding the distance between a point and a parabola with different methods I'm trying to find the shortest distance from point $(3,0)$ to the parabola $y= x^2$ using the method of Lagrange Multipliers (my practice), and by "reducing to unstrained problem in one variable" (assignment). (I think the sheet might have meant unrestrained, but in either case, I don't know how to go about it).
How would I go about this?
For Lagrange Multipliers, I think the set-up (from what I've read) is something like $(x-3, y) = λ(2x, ??)$ But I'm unsure where to get the y coordinate for the lambda side. I then solve for λ. From there on, I have no idea what to do.
I would greatly appreciate if someone could point me in the right direction or walk me through this.
Thank you very much in advance.
| Draw a picture.
For either method, we want to minimize $(x-3)^2+y^2$, subject to $y=x^2$.
Substituting $x^2$ for $y$ in $(x-3)^2+y^2$, we find that we want to minimize $f(x)=(x-3)^2+x^4$. Note that $f'(x)=2(x-3)+4x^3$. The critical points of $f(x)$ are where $2(x-3)+4x^3=0$, or equivalently
$$2x^3+x-3=0.$$
In general cubic equations are unpleasant to solve. However, the above equation has the obvious root $x=1$. And since $f''(x)=3x^2+1\gt 0$, the function $f'(x)$ is increasing, and therefore can only have one zero. By the geometry, there is at least one point on the parabola at minimum distance from $(3,0)$, so we have found it. The minimum distance is the distance from $(3,0)$ to $(1,1)$, which is $\sqrt{5}$.
We now sketch the Lagrange multipliers approach. The Lagrangian is
$$(x-3)^2+y^2-\lambda(y-x^2).$$
Take the partial derivatives, and set them equal to $0$. We get $2(x-3)+2\lambda x=0$ and $2y-\lambda=0$.
So $\lambda=2y$. Substituting in the first equation we get $2(x-3)+4yx=0$. Putting $y=x^2$ we obtain $2(x-3)+4x^3=0$, and we are at an equation we have already dealt with.
| {
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Two complicated limits: $\lim_{x\to 0}\frac{e^{ax}-e^{bx}}{\sin(cx)}$ and $\lim_{x\to 0} x(a^{\frac1x}-1)$ I need to solve these 2 limits ( without using L'Hospital's Rule) , but I can't figure out how to go about them:
Let $a \neq b$, $c \neq 0$.
$$\lim_{x\to 0}\frac{e^{ax}-e^{bx}}{\sin(cx)}$$
Also, let $a>0$, $a \neq 1$.
$$\lim_{x\to 0} x(a^{\frac1x}-1)$$
I don't necessarily need the result, more like understanding the process.
| In short: factor the smaller exponential out. Use l'Hopital's rule on what remains.
One iteration will do: exponentials stay, and $sin(cx)$ will become $c\cos (cx)$, which goes to $1$ instead.
| {
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Finiteness in a Hausdorff space where every open subspace is compact Let be $X$ a Hausdorff space where every open subspace of $X$ is compact. I need to prove that $X$ is finite.
As $X$ is Hausdorff, I have that for every distinct elements of $X$, exist disjoint neighborhoods in $X$ and I try to use the set of all these neighborhoods as an open cover of $X$. I think I need to use that $X$ is compact, and the fact that some pair of sets of this open cover are disjoint to prove the finiteness of $X$ but I could finish the proof.
| This seems to rely on the following two facts about Hausdorff spaces.
*
*All singleton sets are closed. (This holds for the wider class of T1-spaces.)
*All compact subsets are closed.
So given $x \in X$, $X \setminus \{ x \}$ is open, hence compact by assumption, hence closed by the above. Therefore the singleton $\{ x \}$ is open. As all singletons are open, $X$ is discrete. As $X$ is an open subset of itself, it is compact, and now note that the only compact discrete spaces are the finite ones.
| {
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Hankel transform with Bessel functions of the second kind The Hankel transform is defined for Bessel functions of the first kind (see e.g. http://en.wikipedia.org/wiki/Hankel_transform)
I would like to know if it is possible to define a Hankel transform with Hankel functions, or alternatively with Bessel functions of the second kind. It seems like a natural extension of the ordinary Hankel transform, but I have not been able to find any good references.
I know these functions are singular at the origin, but because the Hankel function is in some sense a natural construction, it seems like a reasonable thing to consider at least formally.
If this is possible, I would like to know in what cases it is useful, and if there are certain restrictions on the associated function space.
Thank you.
| Some of the Bessel function class transforms. A general Google, or Google Scholar, search will yield some results linked to publications.
Hankel transform
\begin{align}
f(y) = \int_{0}^{\infty} f(x) \, J_{\nu}(xy) \, \sqrt{xy} \, dx
\end{align}
Y-transform
\begin{align}
f(y) = \int_{0}^{\infty} f(x) \, Y_{\nu}(xy) \, \sqrt{xy} \, dx
\end{align}
K-transform
\begin{align}
f(y) = \int_{0}^{\infty} f(x) \, K_{\nu}(xy) \, \sqrt{xy} \, dx
\end{align}
Kontorovich-Lebedev transform
\begin{align}
f(y) = \int_{0}^{\infty} f(x) \, K_{i x}(y) \, dx
\end{align}
H-transform
\begin{align}
\int_{0}^{\infty} f(x) \, {\bf{H}}_{\nu}(xy) \, \sqrt{xy} \, dx
\end{align}
| {
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Sums of Consecutive Cubes (Trouble Interpreting Question) Show that it is possible to divide the set of the first twelve cubes $\left(1^3,2^3,\ldots,12^3\right)$ into two sets of size six with equal sums.
Any suggestions on what techniques should be used to start the problem?
Also, when the question is phrased like that, are you to find a general case that always satisfies the condition? Or, do they instead want you to find a specific example, since if an example exists then of course the case would be possible.
Thanks!
Edit: I finally found an answer :) $$1^3 + 2^3 + 4^3 + 8^3 + 9^3 + 12^3 = 3^3 + 5^3 + 6^3 + 7^3 + 10^3 + 11^3 = 3042$$
My approach to solving this problem was an extension of David's suggestion below. For six cubes to sum to an even number (3042), there has to either be 0 odd cubes, 2 odd cubes, 4 odd cubes, or 6 odd cubes. There cannot be 3 odd numbers because the sum of 3 odd numbers results in an odd number. The remaining 3 numbers would be even perfect cubes, and their sum would be even.
Thus you would have an even + odd = odd sum, but 3042 is even, not odd. Following this logic, a set of six cubes that sum to 3042 can only have 0 odds, 2 odds, 4 odds, or 6 odds (note that the 0 odd case is the complementary set to the 6 odd case, and the 2 odd case is the complementary set to the 4 odd case).
Checking the 0 or 6 odd case is simple. The sum of all the odd cubes does not equal 3042, so the two sets cannot be composed of all odd or no odds.
Hence one set of six cubes must have 2 odds, and the other set must have 4 odds.
Now it is simply a matter of guess and check, checking all the pairs of odd cubes from
($1^3,3^3$) $\rightarrow$ ($9^3,11^3$). Fortunately, ($1^3,9^3$) works, so we don't have to try too many cases.
Also, if anyone has any other solution method, please let me know :)
| First, you've noticed that each side of the equation has to sum up to $3042$.
Since $11^3+12^3=3059>3042$, each one of them has to be on a different group.
Since $3042$ is even, each group must contain an even number of odd numbers.
So each group must contain $0$ or $2$ or $4$ or $6$ odd numbers.
With $0$ odd numbers on one group, it will definitely sum up to a larger value than the other group.
With $6$ odd numbers on one group, it will definitely sum up to a smaller value than the other group.
So the remaining options are:
*
*"$12$" with $3$ out of $5$ other even numbers and $2$ out of $5$ odd numbers other than "$11$"
*"$12$" with $1$ out of $5$ other even numbers and $4$ out of $5$ odd numbers other than "$11$"
The total number of combinations left to check, is therefore $\binom53\cdot\binom52+\binom51\cdot\binom54=125$.
| {
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Find a polynomial $p(x,y)$ with image all positive real numbers Find a polynomial $p(x,y)$ such that
*
*for each $(x,y) \in \mathbb{R}^2$ we have $p(x,y) > 0$, and
*for each $a>0$ $\exists (x,y)$ : $p(x,y) = a$.
I see that $p(x,y)$ must have constant, but how can choose $p$ that achieves each positive value?
Maybe there is no such polynomial, in which case I want to prove that.
| Hint One way to produce a (real) polynomial $p$ that only takes nonnegative values is to write a sum of squares $$p(x, y) := q(x, y)^2 + r(x, y)^2,$$ and we can ensure that the sum is always positive by insisting that the squared quantities $q(x, y), r(x, y)$ are never simultaneously zero. On the other hand, we can ensure that the $p$ takes on arbitrarily small positive values by choosing
*
*$r$ to be identically zero on some curve $\gamma \subset \mathbb{R}^2$ (for suitable $r$ we can just take $\gamma$ to be the curve defined by $r(x, y) = 0$), and
*$q$ that takes on arbitrarily small positive values along $\gamma$.
(Of course, since $q, r$ cannot be simultaneously zero, $q(\gamma)$ cannot contain its infimum, and hence the curve $\gamma$ cannot be compact.)
Solution One simple noncompact curve is the hyperboloid $x y = 1$, and by construction $$r(x, y) := x y - 1$$ vanishes identically there. On the other hand, we can parameterize one arc of this hyperboloid by $\gamma: s \mapsto (\frac{1}{s}, s)$, $s > 0$. Along this curve, we thus have $$p(\gamma(s)) = p\left(\frac{1}{s}, s\right) = q\left(\frac{1}{s}, s\right)^2 + r\left(\frac{1}{s}, s\right)^2 = q\left(\frac{1}{s}, s\right)^2.$$ Now, $q\left(\frac{1}{s}, s\right)^2$ will take on arbitrarily small positive values---but no nonpositive values---if we take, e.g., $$q(x, y) := x.$$ Thus, by construction $$p(x, y) = q(x, y)^2 + r(x, y)^2 = x^2 + (xy - 1)^2$$ has the desired properties.
Remark Since for any nonconstant one-variable polynomial $u(x)$ we have $\lim_{x \to \infty} u(x) = \infty$ or $\lim_{x \to \infty} u(x) = -\infty$, and likewise for the limits as $x \to -\infty$, the Heine-Borel Theorem implies that $u$ assumes any finite infimum or supremum, that is, the phenomenon in this problem only occurs for real functions of at least two variables.
| {
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Simplify $\sin \arctan x$, assuming $x>0$ Now, naturally I understand that $$\arctan x = \arcsin \frac{x}{\sqrt{x^2+1}}$$ and therefore I should have just $$\sin \arctan x = \frac{x}{\sqrt{x^2 + 1}}.$$ However, I am interested as to the effect of restraining the domain to $\{x > 0\}$, as I believe the function may simplify further. Could someone point me in the right direction?
| It cannot be simplified further. There is nothing simpler than $ x/ \sqrt{ 1 -x^2} $ next in simplicity only to $ \tan^{-1}x $. Draw a triangle to look at the sides $ (\sqrt{1+x^2}, 1,x )$ for yourself. Domain is $ (-\infty, \infty) $.Choose only the positive part if you need only positive argument. Also there is atan2 function or equivalent to make a proper quadrant choice for numerical work for several CAS.
| {
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Suppose that $A$ is a connected subset of a space $X$ and that $A\subseteq B \subseteq \bar A$. Prove $B$ is connected. I think I can prove the closure of $A$, that is $\bar A$, is connected, as there are many other threads on this site. I am then just not sure how to make the jump to show formally that B is connected.
| If $B$ is not connected then non-empty sets $U,V$ exist with $U\cap\overline{V}=\varnothing=\overline{U}\cap V$ and $B=U\cup V$.
Then the sets $U\cap A$ and $V\cap A$ are disjoint, open in $A$ and are covering $A$.
So one of these sets must be empty since $A$ is connected.
If $U\cap A=\varnothing$ then $A\subseteq V$ and consequently $U\subseteq B\subseteq \overline{A}\subseteq\overline{V}$.
Then $\varnothing\neq U\subseteq\overline{V}$ contradicting that $U\cap\overline{V}=\varnothing$.
Likewise $V\cap A=\varnothing$ leads to a contradiction.
| {
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How do I find the value of this weird expression? How can I find the value of the expression $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^...}} $? I wrote a computer program to calculate the value, and the result comes out to be 2 (more precisely 1.999997). Can anyone explain what's happening? Is there any general method to calculate these expressions? I am new to these problems.Thanks in advance!
EDIT
On looking at the answer by Clement C., I thought I could generalize the method to find the value of any expression of the form $\sqrt[n]{n}^{\sqrt[n]{n}^{\sqrt[n]{n}^...}} $. The value should be $n$, but this is not the case. This is the graph for $n<50$.
Any help would be appreciated.
| You can rewrite your expression as $$\sqrt2^{\sqrt2^{...}}=2^{(\frac{1}{2})^{2^{...}}}$$
Clearly multiplying the powers out you end up with $$2^{1^{1^{...}}}=2$$
| {
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Show that if $\gcd(r,s_1) =\gcd(r,s_2) = 1$, then $\gcd(r,s_1s_2) = 1$ Never mind the question. I want to try to solve that on my own. What I want to understand is how this: "Hint. $1 = ar + bs_1,\ 1 = ar + bs_2$" relates to solving it.
I'm a little confused by this statement, especially since it applies to integers.
If we say ar + bs = 1, that says to me that you have different multiples of r and s. And when you add those multiples you get 1. How can this happen unless you are adding fraction together or something, or one expression equals 1 while the other equals 0?
| Integers can be negative, too. You can have $2 \times 5 + ( -3) \times 3= 1$ for example.
| {
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Evaluate $\int x^3(x^2+7)\ dx$ I'm trying to find the indefinite integral of $$\int x^3(x^2+7)\ dx$$ and I've seem to have forgotten how to do it in this case. So if anyone can refresh my memory, I'd appreciate it.
| First expand the integrand.
It simplifies to $x^5+7x^3$. Then we have that
$$\int x^5 + 7x^3=\frac{x^6}{6}+\frac{7x^4}{4}+C$$
This is done by the Power Rule, i.e, $$\int x^n\ dx=\frac{x^{n+1}}{n+1}+C$$
| {
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Orthogonal complement of subspace $W = span(5,1+t)$ I have this subspace of $P_2(\mathbb R)$ and I need to find its orthogonal complemente, using the inner product defined as $$<p(t),q(t)> = \int_o^1 p(t)q(t) dt$$
So I'm assuming the vector $$v = a+bt+ct^2$$ as being the vector such that $$<v,5> = 0\\<v, 1+t> = 0$$
So I did:
$$\int_0^1 (a+bt+ct^2)5dt = 0 \implies \frac{5}{6}(6 a+3 b+2 c) = 0\\\int_0^1 (a+bt+ct^2)(1+t)dt = 0 \implies \frac{1}{12} (18 a+10 b+7 c) = 0$$
So I have two equations:
$$6 a+3 b+2 c=0\\18a+10 b+7 c=0$$
Does it means that my subspace is spanned by what? Do I have to choose $c$ to be a free variable? How do I represent my orthogonal complemente as a subspace? All the $t$'s are gone D:
Thak you so much!
Update:
By fixing c, I've found:
$$a = a, b = -6a$$
Then my polynomial should be:
$$a-6at+ct^2$$
But I don't know what to do for $c$. Could somebody help me?
Update 2:
I've managed to solve for $c$ once I knew $a$ and $b$, so I got: $c=6a$. Then my polynomial should be:
$$a-6at+6at^2$$
But my answer is:
$$-bt^2/6+bt-b$$
| Your computations are correct. Now you want to find one nonzero solution of
$$
\begin{cases}
6a+3b+2c=0\\
18a+10b+7c=0
\end{cases}
$$
Multiply the first equation by $3$ and subtract it from the second, getting
$$
b+c=0
$$
so $b=-c$; then $6a-3c+2c=0$ or $6a=c$. Thus you get a nonzero solution by taking $c=1$ (or any nonzero number). The polynomial you're looking for is $\dfrac{1}{6}-t+t^2$ (or any scalar multiple thereof).
Why just one? The subspace you want the orthogonal complement of has dimension $2$, so the orthogonal complement has dimension $1$; hence a single nonzero vector spans it.
| {
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Proving Any connected subset of R is an Interval Common Proof:
Suppose $S$ is not an interval of $R$.
Then by Interval Defined by Betweenness, $∃x,y∈S$ and $z\in R∖S$ such that $x<z<y$.
Consider the sets $A_1=S∩(−∞,z)$ and $A_2=S∩(z,+∞)$.
Then $A_1,A_2$ are open by definition of the subspace topology on S.
Neither is empty because they contain x and y respectively.
They are disjoint, and their union is S, since z∉S.
Therefore $A_1∣A_2$ is a separation of S.
It follows by definition that S is disconnected.
But why are $A_1,A_2$ open sets?
| Do you know what the subspace topology is? The larger set is $\Bbb R$ with the usual topology. Since $S \subseteq \Bbb R$, we can equip $S$ with the subspace topology, i.e., the topology where each $V$ that is open in S is of the form $V = S \cap U$ with $U \subseteq \Bbb R$ open in $\Bbb R$.
Since $A_{1} = S \cap (-\infty, z)$, and the interval $(-\infty, z)$ is open in $\Bbb R$, then $A_{1}$ is open in $S$ with the subspace topology. Similarly, $A_{2}$ is open in $S$ with the subspace topology.
Please note the following very important fact: we determine whether a set is connected or disconnected based on the open sets in its topology. What I mean is, if $(X, \mathcal{T})$ is a topological space, and $Y \subseteq X$ is a subset, we determine whether $Y$ is connected or not using the sets in the subspace topology of $Y$.
So, the set $[0,1) \cup (2,3]$ is disconnected. Why? Let $A = [0,1)$ and $B = (2,3]$. Neither $A$ nor $B$ are open in $\Bbb R$, but both of them are open in $[0,1) \cup (2,3]$ under the subspace topology of this set. Since they are open in the subspace topology, and clearly both are disjoint and nonempty, and $A \cup B$ is the entire set, then $[0,1) \cup (2,3]$ is disconnected.
| {
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operation to get a diagonal matrix from a vector In many programs you can create diagonal matrix from a vector, like diag function in Matlab and DiagonalMatrix function in Mathematica. I'm wondering whether we can use matrix product (or hadamard product, kronecker product, etc) of a vector and identity matrices to create a diagonal matrix.
Thank you!
| You can use
$$(xe^T) \odot I_n = \mathrm{diag}(x)$$
Where $\odot$ is the hadamard product and $e^T = (1,1,\ldots)\in\mathbb R^n$. The hadamard product basically masks away all off-diagonal elements and $xe^T$ has $x$ as its diagonal.
| {
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What are the irreducible curves on the blow up of $\mathbb{P}^{2}$? On the blow-up of $2$-dimensional complex projective space, $\mathbb{P}^2$, I know that
$$Pic(\mathbb{\tilde{P}^2})=\mathbb{Z}[\tilde{H}]+\mathbb{Z}[E]$$
where $\tilde{H}$ is the blow-up of the hyperplane bundle of $\mathbb{P}^2$, and $E$ is the exceptional divisor on $\tilde{\mathbb{P}}^2$. Note that $[\cdot]$ denotes numerical equivalence class.
I already know that the set $\{[\tilde{H}], [E]\}$ form a basis for $\tilde{\mathbb{P}}^2$. but how do the irreducible curves on the blow-up look like?
| They are either strict transforms of irreducible curves from $\mathbb{P}^2$, or they are the exceptional divisor $E$.
| {
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Euclidean norm of complex vectors I am working on a proof: One has two vectors, $u,v \in \mathbb C^n$, such that $u \cdot v=0$ . I am trying to prove that
$$|u + v|^2 = |u|^2 + |v|^2.$$
I am a little stuck on how to do $u + v$ dotted with the conjugate of $u + v$. Is there anything special I can do with this?
| Here is how.
$$ ||u+v||^2 = \langle u+v,u+v \rangle = \langle u,u \rangle + \langle v,v \rangle + \langle u,v \rangle + \langle v,u \rangle = ||u||^2+||v||^2 +0+0. $$
Note:
$$ \langle u,v \rangle = u \cdot v $$
| {
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If $f$ and $g$ are both uniformly continuous, show that $\max(f, g)$ is uniformly continuous My friend asked me this question and I gave him a sketch of proof. My idea is that to construct a function
$$h = \begin{cases}
f-g & \textrm{if $f \ge g$}\\
0 & \textrm{if $f < g$}
\end{cases}$$
and show that $h$ is uniformly continuous. Then since $\max(f, g) = g + h$, so it is uniformly continuous.
He believed that I oversimplified, and show me this site: http://www.math.unm.edu/~crisp/courses/math402/sol-hw5.pdf
The proof of this statement is on the second page. I completely don't see why the third (and fourth) case should be consider, since there is a theorem that for a continuous function $f$, if $f(x_0) > 0$, then there exist a nbhd $U(x_0)$ of $x_0$ s.t. $f(x) > 0$ for all $x \in U(x_0)$. In other words, I can just consider the nbhd s.t. $f(x) > g(x)$ if $f(x_0) > g(x_0)$. For a larger $\epsilon$, I can simply maintain the $\delta$ and everything is fine.
So my questions are:
*
*Is my sketch correct?
*Why do we need to consider 4 cases as shown in the "answer"?
| It is much simpler if you use the formula (proved here)
$$\max(f, g) = \frac{1}{2}\left(f + g + |f - g| \right)$$ and then we can reason in the following manner. Sum and difference of two uniformly continuous functions is uniformly continuous. Hence both $(f + g)$ and $(f - g)$ are also uniformly continuous. If we note the inequality $||a| - |b|| \leq |a - b|$, then we get that modulus of an uniformly continuous function is also uniformly continuous. Hence $|f - g|$ is uniformly continuous. By sum property $h = (f + g + |f - g|)/2$ is also uniformly continuous.
| {
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To find right cosets of H in G where G= and H=$$ ,where o(G)=10 To find right cosets of H in G where G= and H=$<a^{2}>$ ,where o(G)=10
Since order of $G =10$ , so $a^{10}=e$ .We have
$G= { a,a^{2},a^{3},a^{4},a^{5},a^{6},a^{7},a^{8},a^{9},e}$
and
$H = {a^{2} ,a^{4},a^{6},a^{8},e}$
So $Ha={a^{3},a^{5},a^{7},a^{9},e}$
$Ha^{2} = H$
Is this right way to attempt this
| Almost. The only difference is that the coset need not be a group. The only change to be made is $Ha=\{a,a^3,a^5,a^7,a^9\}$. Then as you've noticed every element is in either $Ha$ or $H$. It is a nice exercise to prove that being in a coset is an equivalence relation and the that cosets partition the group. In this way each element will be in a coset, and only one coset.
| {
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Limit of $\sqrt{25x^{2}+5x}-5x$ as $x\to\infty$ $\hspace{1cm} \displaystyle\lim_{x\to\infty} \left(\sqrt{25x^{2}+5x}-5x\right) $
The correct answer seems to be $\frac12$, whereas I get $0$.
Here's how I do this problem:
$$ \sqrt{25x^{2}+5x}-5x \cdot \frac{\sqrt{25x^{2}+5x}+5x}{\sqrt{25x^{2}+5x}+5x} = \frac{25x^2+5x - 25x^2}{\sqrt{25x^{2}+5x} +5x} = \frac {5x}{\sqrt{25x^{2}+5x}+5x} $$
$\sqrt{25x^{2}+5x}$ yields a bigger value than $5x$ as $x$ becomes a very big number. So the denominator is clearly bigger than numerator. So in this case, shouldn't the answer be $0$?
However, if I keep going and divide both numerator and denominator by $x$ I get:
$$ \frac{5}{ \frac{\sqrt{25x^2+5x}}{x} + 5 }$$
In the denominator, $\frac{\sqrt{25x^2+5x}}{x}$
yields a big number (because top is increasing faster than the bottom), in fact, it goes to infinity as $x$ goes to infinity.
In that case, it's just $5$ divided by something going to infinity, therefore, the answer should be $0$, but it's not, why?
| If we start where you left off:
$$\dfrac{5x}{\sqrt{25x^{2}+5x}+5x}$$
we can factorize the square root by $5x$:
$$\dfrac{5x}{\sqrt{5x(5x+1)}+5x}$$
Take this outside of the root, and re-factorize the denominator:
$$\dfrac{5x}{\sqrt{5x}(\sqrt{5x+1}+\sqrt{5x})}$$
Cancel by $\sqrt{5x}$:
$$\dfrac{\sqrt{5x}}{\sqrt{5x+1}+\sqrt{5x}}$$
The denominator tends to $2\sqrt{5x}$ with large $x$, so the limit is $1/2$.
| {
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Numerical methods to solve nonlinear system of inequalities? I know some methods to solve nonlinear system of equaltites: Relaxation Method, Newton method, nonlinear Jacobi method, nonlinear Seidel method.
Is it exist some analogous method to solve nonlinear systems of inequaltites?
| I'm answering from my question after couple years....
Yes, there are various ways to solve numerically objective plus inequalities.
But what is deeply wrong in question is that word non-linearity should not bring panic, in fact the problem is not in linear/non-linear but in convex/non-convex. In USSR it was realized in 1960, in USA approximately at same years.
And there areas outside math. optimization which are still think that problem is in "non-linearity".
For convex optimization problem there are a bunch of methods (e.g. interior point method, penalty method, projective subgradients).
For non-convex problems precise methods don't exist but there are two ways to handle it:
*
*Forget about non-convexity and convexify problem locally in some way
1.a DCCP(convex concave programming) - represent objective/constraint
functions as convex-convex and then remove concave part or
represent it via affine approximation at point.
1.b Just use convex methods and feed non-convex inequality or objective
1.c Fit convex approximation for objective and constraint via particle
method
1.d Use penalty method and pull containts into objective
1.d Create you own
*Use global methods like branch-and-bound which are leveraged in convex optimization step, but they are usually computanionally very hard (even for small dimensions).
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Does the totient function reach all its values when restricted to odd numbers? This question might be a duplicate, if so, I apologise in advance. It is simple, but answering it is probably harder :)
Is it true, that the $\phi(n)$ function(Euler's totient function) takes on all of it's values, when $n$ is an odd integer?
I obviously tried it for the first some $n$:
$\phi(1) = 1,
\phi(3) = 2, \phi(5) = 4, $ and so on.
It is obvious, that if $n$ is a prime, than $\phi(n) = n-1$, so we cover all the $p-1$ numbers, where $p$ is a prime. I just can't really prove if any number is missing on this list.
The question can be asked in this way too: Is it true, that if we use the totient function with only odd integers, we get all the values from it. I hope you can understand it, if not, just comment below, and I try to answer. :) Thanks for any comments!
| The answer is no. Let $m = 2^kn$, with $k=3$ or $4<k<15$ and $n$ odd. Then $\phi(m) = \phi(2^k)\phi(n)=2^{k-1}\phi(n)$. I claim there is no odd number $x$ with $\phi(x) = \phi(m) = 2^{k-1}\phi(n)$.
Let $x$ be an odd number. It can be written as $\Pi_{i=1}^n p_i^{e_i}$, where because $x$ is odd, none of the $p_i$ is $2$. The formula for the totient given a prime factorization yields $\phi(x) = \Pi_{i=1}^n p_i^{e_i-1}(p_i-1)$.
$p_i^{e_i-1}$ is clearly not a power of two for $p_i \neq 2$, and $(p_i-1)$ is a power of two only for $p_i$ a Fermat prime. The only known such primes are $3,5,17,257, 65537$. This means the only known powers of two that can appear in $\phi(x)$ are $2,4,16,256,65536$, or $2^1,2^2,2^4,2^{16}$. So if $k =4$ or $5<k<17$, $\phi(x)$ cannot equal $\phi(2^kn)$.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How to calculate the area covered by any spherical rectangle? Is there any analytic or generalized formula to calculate area covered by any rectangle having length $l$ & width $b$ each as a great circle arc on a spherical surface with a radius $R$? i.e. How to find the area $A$ of rectangle in terms of length $l$, width $b$ and radius $R$
($A=f(l, b,R)$)?
Note: Spherical rectangle is a quadrilateral having equal opposite sides but non-parallel & all the interior angles are equal in magnitude & each one is greater than $90^\circ$.
| Assume we are working on a sphere of radius $1$, or consider the lengths in radians and the areas in steradians.
Extend the sides of length $l$ until they meet. This results in a triangle with sides
$$
w,\quad\frac\pi2-\frac l2,\quad\frac\pi2-\frac l2
$$
The Spherical Law of Cosines says that
$$
\begin{align}
\cos(A)
&=\frac{\cos\left(\frac\pi2-\frac l2\right)-\cos\left(\frac\pi2-\frac l2\right)\cos(w)}{\sin\left(\frac\pi2-\frac l2\right)\sin(w)}\\
&=\frac{\sin\left(\frac l2\right)}{\cos\left(\frac l2\right)}\frac{1-\cos(w)}{\sin(w)}\\[6pt]
&=\tan\left(\frac l2\right)\tan\left(\vphantom{\frac l2}\frac w2\right)
\end{align}
$$
One quarter of the spherical excess of the rectangle is $D-\frac\pi2$ and
$$
\sin\left(D-\frac\pi2\right)=\tan\left(\frac l2\right)\tan\left(\vphantom{\frac l2}\frac w2\right)
$$
Therefore, the area of the rectangle is
$$
\bbox[5px,border:2px solid #C0A000]{4\sin^{-1}\left(\tan\left(\frac l2\right)\tan\left(\vphantom{\frac l2}\frac w2\right)\right)}
$$
Note that for small $l$ and $w$, this is approximately $lw$; and if $l+w=\pi$ (that is, the rectangle is a great circle), we get an area of $2\pi$ (one half the sphere).
Scaling for a sphere of radius $R$ gives
$$
\bbox[5px,border:2px solid #C0A000]{4R^2\sin^{-1}\left(\tan\left(\frac l{2R}\right)\tan\left(\vphantom{\frac l2}\frac w{2R}\right)\right)}
$$
Note the similarity to the formula for the area of a spherical right triangle with legs $a$ and $b$:
$$
2\tan^{-1}\left(\tan\left(\vphantom{\frac b2}\frac a2\right)\tan\left(\frac b2\right)\right)
$$
or for a sphere of radius $R$,
$$
2R^2\tan^{-1}\left(\tan\left(\vphantom{\frac b2}\frac a{2R}\right)\tan\left(\frac b{2R}\right)\right)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $f(x)$ for $f'(x) = f(x) \ln(f(x))$ and $f(0) = 1$ $$f'(x) = f(x) \ln\big(f(x)\big)$$
$$f(0) = 1, \qquad f(x) > 0$$
I am studying for finals on my own and this exercise is really bothering me because I can't seem to solve it. If I divide by $f(x)$ I get $\ln'f(x) = \ln f(x)$ but then I don't know what to do. Any help is appreciated!
| writing your equations as $$\frac{dy}{dx} = y \ln y $$ you can see that it is separable. so $$\frac{dy}{y\ln y} = dx \to x = \int_1^y \frac{dy}{y \ln y} = \int_1^y \frac{d \,( \ln y)}{\ln y} = \ln (\ln y)\big|_1^y = diverges$$
there is trouble with the initial value $f = 1$ at $x = 0.$
| {
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"timestamp": "2023-03-29T00:00:00",
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surface area of 'cylinder' with the top cut at an angle I don't know what the name for this shape is, so in essence it is a cylinder, radius at base $r$, which has had a wedge of the top cut off at an angle so that rather than a circle the upper face is an ellipse. its height at the top of the slanted ellipse is $h_{max}$, and the height at the bottom is $h_{min}$. the volume was easy to calculate, $\pi r^2 \frac{h_{min}+h_{max}}2$. the surface area is harder: the circle is just $\pi r^2$. I am sure that I could calculate the area of the ellipse, I just haven't got round to it, but the area of the once-rectangle is a challenge, as the upper edge is a wave. I assume it involves trigonometry, but I don't know what the formula is. help please? (apologies if my explanation is not clear)
| sorry, hadn't really thought it through. although it is a wave, it goes above the average height just as much as it goes below it, so it is similar to the volume calculation: $2\pi r \frac{h_{min}+h_{max}}2$. I haven't calculated the surface area of the ellipse yet.
EDIT:
thanks for ellipse formula. semiminor axis $=r$ and semimajor axis $=\sqrt{r^2+\frac{(h_{max}-h_{min})^2}4}$
area of ellipse $=\pi r\sqrt{r^2+\frac{(h_{max}-h_{min})^2}4}$
| {
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Are all convergent sequences bounded and monotone? I know of the monotone convergence theorem, but does this mean that sequences converge only if they are bounded and monotone?
| In general, you can choose some Cauchy sequence $\{a_n\}$ in $\mathbb R$ which is alternating, or "jumps around." Then $\{a_n\}$ is bounded and converges but is not monotone.
For example:
$\{a_n\} = \frac 1n$ for odd $n$, $0$ for even $n$
$\{b_n\} = \frac 1 {n^2} \sin (n) $
| {
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In triangle ABC, ∠ACB = 60. AD and BE are angle bisectors. Prove AE + BD = AB. Here is a diagram if it will be helpful:
| The main idea is to express everything in terms of the triangle sides and then obtain something trivial (or at least easy-to-prove).
For this, just use the property that the angle bisector divides the opposite side in the ratio of the neighbour sides i.e. $\frac{AE}{EC}=\frac{AB}{BC}$.
Then (with the usual notations for triangle sides) you get
$AE=\frac{bc}{a+c}$ and $BD=\frac{ac}{b+c}$ and so we need to prove
$c=\frac{bc}{a+c}+\frac{ac}{b+c}$
which is (after expanding) equivalent to $a^2-ab+b^2=c^2$.
But this is just the Law of Cosines using $\gamma=60^\circ$.
| {
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proving that the set of all english words is countble. This is the question :
Prove that the set of all the words in the English language is countble (the set's cardinality is אo)
A word is defined as a finite sequence of letters in the English language.
I'm not really sure how to start this. I know that a finite union of countble sets is countble and i think this is the way to start.
Thanks in advance !
| There are $26$ letters in the English language.
Consider each letter as one of the digits on base $27$:
*
*$A=1$
*$B=2$
*$C=3$
*$\dots$
*$Z=26$
Then map each word to the corresponding integer on base $27$, for example:
$\text{BAGDAD}=217414_{27}=2\cdot27^5+1\cdot27^4+7\cdot27^3+4\cdot27^2+1\cdot27^1+4\cdot27^0$.
This mapping yields that the cardinality of your set is $\leq|\mathbb{N}|$, hence this set is countable.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find a list of summands and factors adding up to a total? I am neither a mathematician nor do I have an idea on how to write down my problem in accurate mathematic formulas. Please feel free to edit my question into shape and remove this paragraph. Also I am unsure about that tags that apply to this question - problem-solving fits, but maybe there are others?
Given a set of natural numbers: [125, 70, 55] and a total of 250 (All of these numbers may vary).
Now I need to find the integral factors to the numbers that add up to the total like this:
125*a + 70*b + 55*c = 375
Is there a mathematical solution to this problem or do I have to find out the brute force way?
For this example there is more than one valid solution:
*
*a = 3, b = 0, c = 0
*a = 2, b = 1, c = 1
*a = 1, b = 2, c = 2
*a = 0, b = 3, c = 3
*a = 0, b = 1, c = 5
*maybe others...
My goal is it to find all possible solutions - or at least more than one of them.
I already tried some brute force algorithms, but none of them is fast enough to find a solution to this problem for lager numbers.
| Just to make things simpler. I'll assign each number a name:
125: a number
70: b number
55: c number
375:final number
1.Split all the number into their prime factors except for your final number:
5x5x5+2x5x7+11x5=375
*Take the total of their numbers away from your final number (375) and find the prime factors of that.
375-250=125
5x5x5=125
125 will be called: subtracted number
Now there are multiple 'tests' to be done.
a) Do any of the number's prime factors match the prime factors of the subtracted number
Yes, a number matches 125 once.
That collects up, meaning that number needs to occur twice for the numbers to add up to 375. Since we didn't do anything to the other numbers, they remain as value 1.
Note: If the subtracted number contained the prime factors (perfectly) of number a, b or c multiple times, a,b, or c will be equal to however many times they match up (plus 1). For example. If number a's prime factor was 5. It would take 3 a's to equate to the subtracted number
Note 2: If two or three of the numbers (a,b or c)combined match up to the subtracted number's prime factor, do the same thing you did in the test:
-2x5x7+11x5=125
-prime factors of 125=5x5x5
-therefore
5x5x5+5x5x5=375
As we know 5x5x5 matches with the prime factors of our subtracted number. So we increase the numbers b and c (we totalled together) by one. So b is and so is c.
b) If you've finished test a, re-do it except add each number to the subtracted number and give their multiplier (a,b or c) a value of 0. Once you have done that, do it with every possible pair of the three numbers
If you feel I've answered your questioo=n. Please tick the box. ALL of the combinations are below:
a=2, b=1, c=1
a=3, b=0 c=0
a=1, b=2, c=2
a=0., b=3, c=3
| {
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Tychonoff's theorem for $[0,1]^\mathbb{R}$ According to Tychonoff's theorem any uncountable product of compact spaces is compact with respect to product topology.
Then $[0,1]^\mathbb{R}$, the space of all functions defined on $\mathbb{R}$ taking values in $[0,1]$ is compact w.r.t. the product topology.
Consider the function $\delta_0(x)=\max(0,\min(x,1))$ on real numbers and $\delta_t(x)=\delta_0(x-t)$. For $t\to\infty$, it seems that there is no convergent sub-sequence and again it seems that $[0,1]^\mathbb{R}$ is then not compact or sequentially compact?
Could someone point out the problem I have? Thanks.
| The product topology is the topology of pointwise convergence; your functions converge pointwise to the zero function.
You are correct that $[0, 1]^{\mathbb{R}}$ is not sequentially compact; I believe $f_n(x) = |\sin nx|$ is an explicit counterexample, but I haven't checked it carefully. But neither sequential compactness nor compactness imply the other in general (and in particular $[0, 1]^{\mathbb{R}}$ is not metrizable).
| {
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Divergence of $\prod_{n=2}^\infty(1+(-1)^n/\sqrt n)$. Looking looking for a verification of my proof that the above product diverges.
$$\begin{align}
\prod_{n=2}^\infty\left(1+\frac{(-1)^n}{\sqrt n}\right) & =\prod_{n=1}^\infty\left(1+\frac1{\sqrt {2n}}\right)\left(1-\frac1{\sqrt{2n+1}}\right)\\
& =\prod_{n=1}^\infty\left(1-\frac1{\sqrt{n_1}}+\frac1{\sqrt {2n}}-\frac1{\sqrt{2n(2n+1)}}\right)\\
& =\prod_{n=1}{\sqrt{2n(2n+1)}-\sqrt{2n}+\sqrt{2n+1}-1\over\sqrt{2n(2n+1)}}\\
& \ge\prod_{n=1}^\infty{\sqrt{2n(2n+1)+2n+1}-\sqrt{2n}-1\over\sqrt{2n(2n+1)}},\quad\sqrt x+\sqrt y\ge\sqrt{x+y}\\
& = \prod_{n=1}^\infty{2n-\sqrt{2n}\over\sqrt{2n(2n+1)}}\\
& = \prod_{n=1}^\infty{2n-1\over\sqrt{2n+1}}
\end{align}$$
This last product diverges since
$$\lim_{n\to\infty}{2n-1\over\sqrt{2n+1}}=\infty.$$
I'm suspicious because
$$\lim_{n\to\infty}\left(1+{(-1)^n\over\sqrt n}\right)=1.$$
| I would like for some approval to my answer, please.
Show that $$\prod_2^\infty(1+\frac{(-1)^k}{\sqrt k})$$ diverges even though$$\sum_2^\infty \frac{(-1)^k}{\sqrt k}$$ converges.
$\prod_2^\infty(1+\frac{(-1)^k}{\sqrt k})$ diverges if and only if $\sum _2^\infty\log(1+\frac{(-1)^k}{\sqrt k})$ diverges.
Indeed, by Leibnitz' converges test $$\sum_2^\infty \frac{(-1)^k}{\sqrt k}$$ converges. Thus, $\sum _2^\infty\log(1+\frac{(-1)^k}{\sqrt k})$ diverges if and only if $\sum _2^\infty(\log(1+\frac{(-1)^k}{\sqrt k})+\frac{(-1)^k}{\sqrt k})$ diverges.
We observe that for all $k\geq 2$,
$$
|\log(1+\frac{(-1)^k)}{\sqrt k})-\frac{(-1)^k)}{\sqrt k}|\\
=|\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3} + \frac{(-1)^{4k}}{4\sqrt k^4} -\frac{(-1)^{5k}}{5\sqrt k^5}+-...|
$$
But since we know that the last series is convergen (to $|\log(1+\frac{(-1)^k)}{\sqrt k})-\frac{(-1)^k)}{\sqrt k}|$), then it is equal to
$$
=|(\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3}) + (\frac{(-1)^{4k}}{4\sqrt k^4} -\frac{(-1)^{5k}}{5\sqrt k^5})+-...|=:A
$$
but every addend in this series is positive because for all $l\in\mathbb{N}$:
$$
(-1)^{2l}(2l+1)\sqrt k^{2l+1}>(-1)^{2l+1}\cdot 2l\cdot\sqrt k^{2l}
$$
($\sqrt k>1$).
Thus, the previos series $A$ obtains:
$$
A=(\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3}) + (\frac{(-1)^{4k}}{4\sqrt k^4} -\frac{(-1)^{5k}}{5\sqrt k^5})+-...
\\ >\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3}\geq \frac{1}{2k}-\frac{1}{3k^{1.5}}
$$
But because $\sum \frac{1}{2k}$ diverges and $\sum\frac{1}{3k^{1.5}}$ converges then $\sum(\frac{1}{2k}-\frac{1}{3k^{1.5}})$ diverges then $A$ diverges.
| {
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Conditional Expectations Given Sum of I.I.D. Given that $X_1,...Xn$ are all identical independent random variables.
$\mathbb{E}(X_1|\sum_{k=1}^{n}X_k)$ = ?
I am unsure how to proceed on this one. I know the default relation: $\mathbb{E}(X|Y)$ = $\mathbb{E}(X*I_{[Y=y]})\over\mathbb{P}(Y=y))$, where I is an indicator function.
Intuitively, I believe the answer should be the sum of the random variables divided by how many random variables or the average of the sum.
| For the late comers for this question. I think clear solution of the problem might be as follows:
Let $Y = \sum_{i=1}^n X_i$, by using the property of conditional expectation $\mathbb{E}[g(Y)|Y] = g(Y)$ (i.e.$\mathbb{E}[Y|Y] = Y$),
we can write
$$ Y = \mathbb{E}[X_1 + X_2 + \cdots + X_n |Y] = \mathbb{E}[X_1|Y]+ \mathbb{E}[X_2|Y]+ \cdots + \mathbb{E}[X_n|Y] = n\mathbb{E}[X_1|Y],$$
since $X_i$'s are i.i.d., then
$$\mathbb{E}[X_1|Y] = \frac{Y}{n}.$$
| {
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Functions and Sets
Robert, Susan, and Thomas are the sole contestants in a lottery in which two prizes will be awarded. Three tickets with their names on them are placed in a hat. The person whose name is on the first ticket drawn wins prize #1. That ticket is placed back in. Then the second ticket is drawn. You can win both prizes. Let $A = \{1, 2\}$, $B = \{r, s, t\}$ where $R = \text {robert}$, etc
a.) Explain how each possible assignment of prizes to contestants may be thought of as a function from $A$ to $B$ and why $B^A$ may be thought of as representing the set of all such possible assignments.
d.) In terms of assignments of prizes in the lottery, what does it mean to say that an element of $B^A$ is an injection?
My attempt:
a.) So it's essentially
f(1) = r, s, t
f(2) = r, s, t
Because either prize can map to any contestant. But then these aren't functions, are they? Because the same value maps to multiple people (potentially).
And the function A --> B is the set of all possible assignments because it explains how awards can be matched to contestants?
d.) So an injective function is 1 to 1. If an element of A --> B is an injection, that means each value in A (domain) only maps to one value in the codomain (B), and no two values in A map to the same value in B. That is to say, an element of B^A that is an injection means that a player did not win both of the prizes.
Am I vaguely close?
| For a,it is OK for a function to map different inputs to the same output. For functions of the reals, $f(x)=0$ is a fine function-it takes any real as an input and returns $0$. A function just has to return a single value for any input. You haven't done the last sentence of the problem, showing how $B^A$ can represent all assignments.
For d, "each value in A (domain) only maps to one value in the codomain (B)" is required of any function, so is not peculiar to injective functions and could be deleted. The rest of the answer is good.
| {
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Induction proof verificiation P(n) = in a line of n people show that somewhere in the line a woman is directly in front of a man. The first person will always be a woman and the last person in the line will always be a man
I wanted to prove this with induction and I was a little unsure of how to express this problem in the inductive step and whether I had the right cases and if I had shown them correctly.
Base case: n=2:
P(2) holds as a woman (front of line) will be directly in front of the man at the back of line since there are only 2 people.
Inductive hypothesis:
Assume P(k) holds for some integer k >= 2 there will be a woman directly in front of a man somewhere in the line, where the kth person is at the back of the line.
Inductive step:
Assuming P(k) holds, show P(k+1) holds too:
In a group of k+1 people consider the cases: (this is where I am slightly unsure)
(1) There are k males and 1 female
(2) There are k females and 1 male
*
*Case (1):
If there are k males and 1 female, k males will always be behind the 1 female at the front by the rule in the statement and so the 2nd male from the front will be directly behind the female at the front and so there is always at least one case where a female is directly in front of a male while there are k males and 1 female in a group of k+1 people
*
*Case (2):
If there are k females and 1 male, k females will always be in front of the 1 male at the back of the line by the rule of the statement and so the kth person who is a female will be directly in front of the k+1 person who is a male.
So in both cases there will always be at least one case where a female is directly in front of a male in a line of k + 1 people which concludes the inductive step.
As mentioned, I'm unsure if I've included the correct and all cases necessary in the inductive step and if I've explained them correctly for this proof. Could you verify and provide any pointers?
| Your two cases are far from exhausting the possibilities. If $k=10$, for instance, the number of women can be anywhere from $1$ through $10$, and so can the number of men. You need to come up with an argument that covers all possibilities.
Try this: remove the man at the end of the line. Either the new line has a man at the end, in which case you can apply your induction hypothesis to it, or it has a woman at the end. How do you finish the argument in that second case?
| {
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Are the following real numbers constructible? 1) $\sqrt[4]{5+\sqrt2}$
2)$\sqrt[6]{2}$
3) $3/(4+\sqrt13)$
4) $3+\sqrt[5]{8}$
From what I know, a number is constructible if it can be converted in a finite number of steps using only the operations addition, subtraction, multiplication, division, and square roots.
This in mind, I would think #1 is constructible number because it is calculated using addition and square roots, with the 4th being a square root of a square root.
I would say #2 is not constructible as it is not possible to take a third root.
I would say #3 is constructible because it makes use of the addition, division, and square root operations.
Finally, #4 I would say is not constructible because it is not possible to generate the 5th root.
Can anyone weigh in on whether these conjectures are correct/incorrect, or maybe comment on the rationale and whether this method (if correct) could still be used on harder examples?
Thank you in advance!
| I think that
all your answers are correct
for the reasons stated.
The only quibble I might have
is for the non-constructible ones,
where the particular numbers are
non-constructible,
while,
for example,
$\sqrt[3]{8}$
is obviously constructible.
| {
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If $A,B>0$ and $A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$. If $A,B>0$ and $\displaystyle A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$.
$\bf{My\; Try::}$ Given $$\displaystyle A+ B = \frac{\pi}{3}$$ and $A,B>0$.
So we can say $$\displaystyle 0< A,B<\frac{\pi}{3}$$. Now taking $\tan $ on both side, we get
$$\displaystyle \tan(A+B) = \tan \left(\frac{\pi}{3}\right).$$ So $\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B} = \sqrt{3}$.
Now Let $\displaystyle \tan A\cdot \tan B=y\;,$ Then $\displaystyle \tan B = \frac{y}{\tan A}.$
So $$\displaystyle \frac{\tan A+\frac{y}{\tan A}}{1-y}=\sqrt{3}\Rightarrow \tan^2 A+y=\sqrt{3}\tan A-y\sqrt{3}\tan A$$
So equation $$\tan^2 A+\sqrt{3}\left(y-1\right)\tan A+y=0$$
Now for real values of $y\;,$ Given equation has real roots. So its $\bf{Discrimnant>0}$
So $$\displaystyle 3\left(y-1\right)^2-4y\geq 0\Rightarrow 3y^2+3-6y-4y\geq 0$$
So we get $$3y^2-10y+3\geq 0\Rightarrow \displaystyle 3y^2-9y-y+3\geq 0$$
So we get $$\displaystyle y\leq \frac{1}{3}\cup y\geq 3$$, But above we get $\displaystyle 0<A,B<\frac{\pi}{3}$
So We Get $$\bf{\displaystyle y_{Max.} = \left(\tan A \cdot \tan B\right)_{Max} = \frac{1}{3}}.$$
My Question is can we solve above question using $\bf{A.M\geq G.M}$ Inequality or Power Mean equality.
Thanks
| $$
\tan A\tan B=\tan A\tan\big(\frac{\pi}{3}-A\big)=\tan A.\frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}=1-1+\frac{\sqrt{3}\tan A-\tan^2A}{1+\sqrt{3}\tan A}\\
=1+\frac{-1-\sqrt{3}\tan A+\sqrt{3}\tan A-\tan^2A}{1+\sqrt{3}\tan A}=1-\frac{1+\tan^2A}{1+\sqrt{3}\tan A}=1-\frac{\sec^2A}{1+\sqrt{3}\tan A}\\
=1-\frac{2}{2\cos^2A+\sqrt{3}.2\sin A\cos A}=1-\frac{2}{1+2\big[\frac{1}{2}\cos 2A+\frac{\sqrt{3}}{2}\sin 2A\big]}\\
=1-\frac{2}{1+2\sin\big(\frac{\pi}{6}+2A\big)}
$$
$\tan A\tan B$ is maximum $\implies \frac{2}{1+2\sin\big(\frac{\pi}{6}+2A\big)}$ is minimum $\implies \sin\big(\frac{\pi}{6}+2A\big)$ is maximum$\implies \sin\big(\frac{\pi}{6}+2A\big)=1$
$$
(\tan A\tan B)_\text{max}=1-\frac{2}{1+2.1}=1-\frac{2}{3}=\frac{1}{3}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Interior of a set $A$ is dense in $X$ iff $A$ is dense in $X$ I would like to check if the following statement is true:
Interior of a set $A$ is dense in $X$ iff $A$ is dense in $X$
It seems to me that they are equivalent, but I couldn't prove or find a counterexample for it.
Thanks!
| $\mathbb Q$ is dense in $\mathbb R$ but interior of $\mathbb Q$ is empty.
| {
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General versions for quadrics of Pascal and Brianchon theorems I am looking for a generalization to quadrics (with proofs) of Pascal's and Brianchon's theorems. It´s for Three dimensional analytical geometry.
I would be very thankful if you could point me towards one.
| Tried to add this as a comment, but comments not working right now:
The question is an exact duplicate of this one.
But, anyway, start with the classical geometry texts. Salmon, Sommerville, and books like that.
Sommerville, Analytical Geometry of Three Dimensions.
Snyder & Sisam, Analytical Geometry of Space
Salmon, A Treatise on the Analytic Geometry of Three Dimensions
| {
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Subtraction of a number I have a number $x$. If I remove the last digit, I get $y$. Given $x-y$, how can I find $x$?
For example x=34 then y=3 given 34-3=31, I have to find 34.
if x=4298 then y=429 , given 4298-429 = 3869 . how can I find 4298 from given 3869?
| Suppose :
$$x=\sum_{k=0}^na_k10^k $$
then :
$$y=\sum_{k=0}^{n-1}a_{k+1}10^k $$
so :
$$x-y=a_0+\sum_{k=1}^na_k(10^k-10^{k-1}) $$
$$x-y=a_0+\sum_{k=1}^n9a_k10^{k-1}=a_0+\frac{9}{10}\sum_{k=1}^n9a_k10^{k} $$
$$x-y=a_0+\frac{9}{10}(x-a_0)=\frac{9x+a_0}{10}$$
Edit : The first part doesn't give the answer to the question (but the answer follows easily from it), for completude I give the rest of the answer now.
This should give the solution, now we call $f$ the function associating this number to $x$. Suppose now that $x>z$ (with first digit respectively $a_0$ and $c_0$) gives the same number through this function $f$ then you have :
$$9x+a_0=9z+b_0$$
So that :
$$9(x-z)=b_0-a_0=:n$$
The number $n$ must verify $-10<n<10$ from the second term and from the first it sould be a positive multiple of $9$, so the only possibility is $b_0=9$ and $a_0=0$ so that :
$$x\text{ is divisible by } 10 \text{ and } z=x-1 $$
On the other hand if $x=10n$ with $n>0$ then set $z=x-1$ then :
$$f(x)=\frac{9x+0}{10}=9n$$
$$f(z)=\frac{9(x-1)+9}{10}=9n $$
So we see that the set of values of $k$ for wich the equation $f(x)=k$ has 2 solutions is exactly $9\mathbb{N}^*$. In that case one can always recover both values : $10n$ and $10n-1$ (where $n=\frac{k}{9}$).
Suppose finally that :
$$n=9k+r\text{ with }1\leq r\leq 8$$
Then from :
$$f(x)=n$$
we get :
$$90k+10r=9x+a_0 $$
Thus, evaluating modulo $9$ we see that $r=a_0$ and finally :
$$90k=9(x-r)\Rightarrow 10k=x-r\Rightarrow x=10k+r $$
And here there are no ambiguity. To make it complete :
$$f(x)=\frac{90k+9r+r}{10}=9k+r=n $$
So we have proven that $f$ is surjective and essentially give a bijection between classes mod $10$ and classes mod $9$ but sticking classes $0$ and $9$ mod $10$ to the same classe $0$ mod $9$. Just to be precise I am seeing $f:\mathbb{N}^*\rightarrow \mathbb{N}^*$.
| {
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Prove that for $n \gt 6$, there is a number $1 \lt k \lt n/2$ that does not divide $n$ My nine year old asked this question at lunch today: Is there a number that is divisible by everything that is half or less than the number?
I immediately answered, "No. I mean, 6. But not for any number bigger than 6."
So I tried to think why that was true, and my first efforts didn't quite work. I did come up with a proof, but it isn't as elegant as I hoped.
How would you prove this?
| Since $n>6$, we have that $2<3 <n/2$, so $n$ is divisible by $2$ and $3$.
Since $n=2\cdot\frac{n}{2}$ and $\frac{n}{2}-1<\frac{n}{2}$, $\frac{n}{2}-1$ divides $n$. $\frac{n}{2}−1$ must have the next smallest codivisor, which is $3$, so must be equal to $\frac{n}{3}$.
Solving $\frac{n}{2}-1=\frac{n}{3}$ gives $n=6$, a contradiction to $n>6$. Thus, $6$ is the largest such positive integer.
| {
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prime numbers and divisibility question - number theory I have a fairly short and straightforward question I would like to ask. Suppose $p_{1},p_{2},...,p_{n}$ are prime numbers. Suppose $X = p_{1}p_{2}...p_{n} + 1$. If $X$ comes out to be a composite number, then why is it true then that none of the $p_{1},p_{2},...,p_{n}$ divide $X$?
| Formally assume that $p_i$ divides $X=bp_i+1$ so there exists an integer $a$ such that $X=a.p_i$and then $p_i(a-b)=1$ so $p_i$ divides $1$ which is impossible.
| {
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Prove that if $L$ is regular then $f(L)$ is regular Prove that if $L$ is regular then $f(L)$ is regular.
$$f(L)=\{w: \text{every prefix of $w$ of odd length $\in$ L } \}$$
So my attemption is:
Let $M= (Q, \Sigma, \delta, q_0,F) $ will be DFA recognizing $L$
Let construct $M'=(Q', \Sigma,\delta', q_0', F')$ recognizing $f(L)$.
$Q'=Q\times\{1,2\}$
$F'=\{(q,1)|q\in F\}\cup\{(q,2)|q\in F\}$
$q_0'= (q_0, 1)$
$\delta'((q,1),a\in\Sigma)=\begin{cases} (\delta(q,a),2);\delta(q,a)\in F\\ \emptyset; \delta(q,a)\notin F\end{cases} $
$\delta'((q,2), a\in\Sigma)=(\delta(q,a),1)$
What about this solution ?
| You can prove the result without using automata. Let $A$ be the alphabet, and let $$
K = \{ u \in A^* \mid \text{ every prefix of $u$ of odd length is in $L$} \}
$$
Let $K^c$ be the complement of $K$ in $A^*$. Then
\begin{align}
K^c &= \{ u \in A^* \mid \text{ there exists a prefix of $u$ of odd length in $L^c$} \}\\
&= \{ u \in A^* \mid \text{ there exists a prefix of $u$ in $(A^2)^*A \cap L^c$} \}\\
&= \bigl((A^2)^*A \cap L^c\bigr)A^*
\end{align}
Therefore $K = \Bigl(\bigl((A^2)^*A \cap L^c\bigr)A^*\Bigr)^c$ and since regular languages are closed under intersection, complement, product and star, $K$ is regular.
| {
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Proving Gaussian curvature $\leq 0$ on line Given some regular surface $S$ that contains a line $L$, I need to prove that the Gaussian curvature $K\leq 0$ at all points of $L$. I am thinking that if I could show that no points on $L$ can be elliptical $K>0$ then I would be set. But how can I do that?
| Hint: The Gaussian curvature is the product of the principal curvatures. What do the principal curvatures minimize/maximize?
| {
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How many pairs of positive integers $(n, m)$ are there such that $2n+3m=2015$? I know that $m$ must be odd and $m\le671$. Also, $n\le1006$. I can't go any further, any help?
| $$\begin{array} {|cc|} \hline
1 & 2014 \\
{\color{red}{ 2 }} & {\color{blue}{ 2013 }} \\
3 & 2012 \\
{\color{red}{ 4 }} & 2011 \\
5 & {\color{blue}{ 2010 }} \\
{\color{red}{ 6 }} & 2009 \\
7 & 2008 \\
{\color{red}{ 8 }} & {\color{blue}{ 2007 }} \\
9 & 2006 \\
{\color{red}{ 10 }} & 2005 \\
11 & {\color{blue}{ 2004 }} \\
{\color{red}{ 12 }} & 2003 \\
13 & 2002 \\
{\color{red}{ 14 }} & {\color{blue}{ 2001 }} \\
15 & 2000 \\
\vdots & \vdots
\end{array}$$
The ones where both red and blue are highlighted, $2$, $8$, $14$,$\dots$, which are $2$ more than a multiple of $6$. So it's $\left \lfloor \frac {2015 - 2}{6} \right \rfloor + 1 = 336$.
| {
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Big List of examples of recreational finite unbounded games What are some examples of mathematical games that can take an unbounded amount of time (a.k.a. there are starting positions such that for any number $n$, there is a line of play taking $>n$ times) but is finite (every line of play eventually ends.) Also, it would be nice if it were recreational in nature (by this I mean I basically mean its nontrivial, I could conceivably be enjoyed by humans theortically.) One answer per game please, but edit variants into the same answer.
This is a big-list question, so many answers would be appreciated.
| One example is the ring game, which is defined here - in essence, the game (or a slight variant thereon) can be described as:
Start with a Noetherian ring $R$. At each turn, replace $R$ with a quotient thereof. If a player can make no legal moves (i.e. $R$ is a field, thus has no proper quotients).
One can notice that if we play this on $\mathbb Z$, then this is equivalent to the game $*\omega$, since the first move takes it to $\mathbb Z/n\mathbb Z$ which is clearly equivalent to $*k$ where $k$ is the number of (not necessarily distinct) prime factors of $n$. However, as evidenced by certain unanswered questions, the structure of the game can get fairly complicated when we consider more complicated rings. One could generalize to replace "ring" with any algebraic structure they desire - like one could play on groups with no infinite ascending chains of normal subgroups.
| {
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Can I make an assumption about arbitrary numbers in a proof? Assume $ x_1, x_2... x_{10} $ are different numbers and $ y_1, y_2... y_{10} $ are some arbitrary numbers. Prove that there exists some unique polynomial of degree not exceeding 9 such that:
$P(x_1)=y_1,P(x_2)=y_2...P(x_{10})=y_{10}$
Proof to the problem rests on the assumption that there is some polynomial Q(x) of degree not exceeding 9 such that:
$ Q(x_1)=y_1,Q(x_2)=y_2...Q(x_{10})=y_{10}$.
Is it a valid assumption in a proof of this sort?
| Your assumption is incorrect because proving the existence of such a polynomial is an important part of the problem you are working on.
Here is a hint: Suppose you were given the problem with only two pairs $(x_1,y_1)$ and $(x_2,y_2)$. Then how would you find such a polynomial? Think in terms of a straight line. You are given two distinct (since $x_1 \neq x_2$) points so there always exists a straight line that passes through these points and the equation of that line will be polynomial of the required degree.
If you were given $3$ points then a parabola passes through it and so on.
If you are still unable to make progress then look for Lagrange Interpolation.
| {
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How to resolve $x \in A \wedge x \notin A $? Let A and B be two sets. Then
$A \setminus B = \{x: x\in A \wedge x\notin B\}$
$A \setminus B = \{x: x\in A \wedge x\notin A \cap B\}$
How can one prove that two logical statements are equal?
Suppose $x \in A \setminus B$.
Then,
$x \in A \wedge x \notin A \cap B$
$x \in A \wedge x \notin A \wedge x\notin B$
How do I move forward?
| Your implication $\{x \in A \wedge x \notin A \cap B\}\Rightarrow \{x \in A \wedge x \notin A \wedge x\notin B\}$ is not true; By De Morgan's laws it should be$$\{x \in A \wedge x \notin A \cap B\}\Rightarrow \{x \in A \wedge (x \notin A \lor x\notin B)\}$$
Now proof.
You want to prove:
$$\{x: x\in A \wedge x\notin B\}=\{x: x\in A \wedge x\notin A \cap B\}$$
or equivalently:
$$A \setminus B=A \setminus (A\cap B).$$
proof.1.
$$A \setminus (A\cap B)= A\cap (A\cap B)^c = A\cap (A^c\cup B^c)=(A\cap A^c)\cup (A\cap B^c)=(A\cap B^c)= A \setminus B$$
proof.2.
Since $A\cap B \subset B$ we have $A \setminus B\subset A \setminus (A\cap B)$.
on the other hand:
Now let $x\in A \setminus (A\cap B)$. So $\underline{x\in A}$ and $x\notin A\cap B$; which means $\underline {x\notin B} $. Therefore $x\in A \setminus B$. (this direction is just Addem's proof).
| {
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Find a point on the line $-2x + 6y - 2 = 0$ that is closest to the point $(0,4).$ I understand that you would use the distance formula here, but I'm confused as to how you calculate x and y after that. Thanks.
| If $(h,k)$ be any point on on $-2x+6y-2=0\iff x=3y-1$ we have $h=3k-1$
If $d$ is the distance between $(0,4);(3k-1,k)$
$d^2=(3k-1)^2+(k-4)^2=10k^2-14k+17=10\left(k^2-\dfrac75k\right)+17$
$=10\left(k-\dfrac7{10}\right)^2+17-10\cdot\left(\dfrac7{10}\right)^2$
$\ge17-10\cdot\left(\dfrac7{10}\right)^2$
The equality occurs if $k-\dfrac7{10}=0\iff k=\dfrac7{10}$
| {
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What exactly am I being asked in this question? I don't need the answer, just the interpretation.
Write a program that inputs a whole number N and outputs the percentage of relatively prime pairs of numbers a, b in the range 1 to N.
For some reason, I'm having difficulty understanding the question. Do I need to calculate the probability of a and b being co-primes? Do I need to find the gcd? I’m not sure what I'm being asked.
Also, please provide an end answer or an example, so that I have something to check my answer against and know that I'm on the right track. Thank you.
| I think you should interpret the question as finding, for two independent random variables $X,Y$ each uniformly distributed over the set $[n]=\{1,2,\ldots,n\}$, the probability that $\gcd(X,Y)=1$. While other interpretations of the question are possible, this seems the most natural one to me, and in addition it is computationally easiest to handle.
You can do this by two nested loops over $[n]$, but there is a more efficient way if $n$ is large. The value of $\gcd(X,Y)$ is always a value in$~[n]$, and for a given $d\in[n]$ it is easy to find the probability that $\gcd(X,Y)$ is a multiple of$~d$, namely $\bigl(\frac{\lfloor n/d\rfloor}n\bigr)^2$: both $X$ and $Y$ must be multiples of $d$, and there are $\lfloor n/d\rfloor$ such multiples. The constant factor $\frac1{n^2}$ is not very interesting, so define $f(d)=\lfloor n/d\rfloor^2$ for $d\in[n]$; then the number $g(d)$ of pairs $(x,y)\in[n^2]$ with $\gcd(x,y)=d$ satisfies
$$
\sum_{k=1}^{\lfloor n/d\rfloor}g(kd)=f(d).
$$
From this you can solve $g(d)=f(d)-\sum_{k=2}^{\lfloor n/d\rfloor}g(kd)$ by downwards recursion. You can improve this even more by observing that $g(d)$ only depends on the integer value $\lfloor n/d\rfloor$, so only about $2\sqrt n$ values of $g$ need to actually be computed and tabulated. The final value computed $g(1)$ gives your answer $\frac{g(1)}{n^2}$.
As a sample result, for $N=30$ one finds $g(1)=555$. From more sample values, see OEIS:018805.
| {
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Vector bundle of dimension $\leqslant n$ on $n$-connected space is trivial I wonder whether any vector bundle of dimension $\leqslant n$ on an $n$-connected CW-complex is trivial? It seems that,
*
*the complex can be given cellular structure with exactly one 0-dimensional cell and no other cells of dimension $\leqslant n$,
*there is a inner product on the vector bundle by the paracompactness of CW-complexes,
*the corresponding $S^{n-1}$-fiber bundle, by some obstruction theory argument (I am not sure which one, but it should use the fact that all maps $S^{\leqslant (n-1)} \to S^n$ are homotopic), has a section,
*and its orthogonal complement has smaller dimension, so allowing the proof by induction.
For $n=1$ it is true: any covering (with two sheets) over simply-connected space is trivial. But I have not found the general statement anywhere, so it there an error?
| Your proposed argument fails at the third step. It is just not true that if $X$ is $n$-connected then every $S^{n-1}$-bundle over it is trivial.
The desired statement is not even true rationally: the rational homotopy groups of $BO(n)$ are not hard to calculate and in particular for $n \ge 3$,
$$\pi_{4n-4}(BO(2n)) \otimes \mathbb{Q} \cong \mathbb{Q}$$
so it follows that there are countably many nontrivial $2n$-dimensional real vector bundles over $S^{4n-4}$, which moreover can be distinguished by their Pontryagin classes $p_{n-1}$, and this gives an infinite family of counterexamples to the desired statement.
| {
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How to compute Dottie number accurately? Dottie number is root of this equation : $cos \alpha = \alpha$, $\alpha \approx 0.73908513321516064165531208767\dots$.
I wonder how can I compute it ? I have tried to do it with an approximating formula:
$\alpha = \frac{5\pi^2}{\alpha^2 + \pi^2} - 4$
I have solved this equation and i got $\alpha \approx 0.738305\dots$. So , how can i compute it accurately ? Can i use taylor series, etc. ?
| Taylor series of order 2 gives a simple quadratic in $\alpha$:
$$\alpha=1-\alpha^2/2\implies \alpha=0.\color{red}{73}2..$$
Of order 4 gives a bi-quadratic (there's a formula to solve roots of a polynomial of degree less than 5) in $\alpha^2$:
$$\alpha=1-\alpha^2/2+\alpha^4/4\implies 0.\color{red}{739}2..$$
Fairly accurate for practical purposes wherein the correct value is $0.739085...$
| {
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Is $\sum_{n\ge 1} \sin(n^2)/n$ convergent?
Is the series
$$\sum_{n\ge 1} \frac{\sin(n^2)}{n}$$
convergent?
My thoughts so far:
1) This is an alternating series so the integration test does not work here.
2) The Weyl inequality roughly says $$\sum_{n\le N} \sin(n^2)$$ is $O(N^{1/2+\epsilon})$, so the Dirichlet test does not work directly, but one can take $$a_n=n^{-1},b_n=\sum_{k\le n} \sin(k^2)$$
and follow the idea of Dirichlet test. The problem now is that the Weyl bound does not hold for all $N$.
| You are on the right track. The key is to consider partial sums:
$$ S_N = \sum_{n=1}^{N}\frac{\sin(n^2)}{n} $$
then find a good rational approximation of $\pi$ depending on $N$, apply Weyl bound (or Weyl differencing technique) to estimate $\sum_{n=1}^{k}e^{in^2}$ and finish through partial summation.
Details on page $11$ here (it is in Italian, hope you don't mind).
| {
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Integral inequality 5 How can I prove that:
$$8\le \int _3^4\frac{x^2}{x-2}dx\le 9$$
My teacher advised me to find the asymptotes, why? what helps me if I find the asymptotes?
| Direct approach is not so painful:
$$\int_{3}^{4}\frac{x^2}{x-2}\,dx = \int_{1}^{2}\frac{(x+2)^2}{x}\,dx = \int_{1}^{2}\left(x+4+\frac{4}{x}\right)\,dx =\frac{11}{2}+4\log 2 $$
and by the Hermite-Hadamard inequality we have:
$$ \log 2 =\int_{1}^{2}\frac{dx}{x} \in \left(\frac{2}{3},\frac{3}{4}\right) $$
so:
$$\int_{3}^{4}\frac{x^2}{x-2}\,dx \in \left(8+\frac{1}{6},8+\frac{1}{2}\right).$$
| {
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What are the transitive groups of degree $4$? How can I find all of the transitive groups of degree $4$ (i.e. the subgroups $H$ of $S_4$, such that for every $1 \leq i, j \leq 4$ there is $\sigma \in H$, such that $\sigma(i) = j$)? I know that one way of doing this is by brute force, but is there a more clever approach? Thanks in advance!
| We know subgroups of $S_4$ come in only a few different orders: $1,2,3,4,6,8,12,$ and $24$.
If we use the orbit stabilizer theorem, we have that $|H| = |\operatorname{Orb}_H(x)| \cdot |\operatorname{Stab}_H(x)|$; this limits us to only four possible subgroup orders, as $|\operatorname{Orb}_H(x)|$ can only be one thing.
Among those four orders, we can find $5$ non-isomorphic groups fulfilling the criteria. I don't know off-hand how many copies of such groups $S_4$ has offhand, but it shouldn't be too hard to pin that down.
Note that just because two subgroups of $S_4$ may be isomorphic, it doesn't mean that they all (do or don't) act transitively on $\{1,2,3,4\}$. For example, I can only think of one particular Klein four-group in $S_4$ that does, although there are several isomorphic subgroups in $S_4$.
| {
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Inequality and Induction: $\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$ I needed to prove that
$\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$, $\forall n \geq 1$ .
I've atempted by induction.
I proved the case for $n=1$ and assumed it holds for some $n$.
The left-side of the n+1 case is
$\frac{1}{2}.\frac{3}{4}. ... .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2}$.
Using the inductive hypothesis, i could reach the result that
$\frac{1}{2}.\frac{3}{4}. ... .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2} <\frac{2n+1}{\sqrt{2n+1}(2n+2)}=\frac{\sqrt{2n+1}}{(2n+2)}$.
Now, i'm wondering how should i connect it to my goal :
$\frac{1}{\sqrt{2n+2+1}}=\frac{1}{\sqrt{2n+3}} $
I know one way to prove that
$\frac{\sqrt{2n+1}}{(2n+2)}<\frac{1}{\sqrt{2n+3}} $
We just square things, then eventually reach
$ (n+1).(n+3)<(n+2)^2$
Which is easily provable because $3<4$ ....
But I was wondering if there was another way to show that ... perharps a more direct way to show that last bit ... a way that was not so direct and brute as to involve squaring both sides. A way of gradually manipulating the left-side until reaching the inequality with the right side.
Thanks in advance.
| $$\begin{align}\frac{1}{2}.\frac{3}{4}. ... .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2} &<\frac{2n+1}{\sqrt{2n+1}(2n+2)}\\~\\&=\frac{\sqrt{2n+1}}{(2n+2)}\\~\\&=\frac{\sqrt{(2n+1)(2n+3)}}{(2n+2)\sqrt{2n+3}}\\~\\&=\frac{\sqrt{4n^2+8n+3}}{(2n+2)\sqrt{2n+3}}\\~\\&\lt \frac{\sqrt{4n^2+8n+3+\color{blue}{1}}}{(2n+2)\sqrt{2n+3}}\\~\\&= \frac{\sqrt{(2n+2)^2}}{(2n+2)\sqrt{2n+3}}\\~\\&=\frac{1}{\sqrt{2n+3}}\end{align}$$
Alternatively you may also use AM-GM inequality and conclude directly that $\sqrt{(2n+1)(2n+3)}\lt 2n+2$
| {
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How do I use Rouche's theorem here? Suppose I had the polynomial $f(z) = z^5+3z+1$ and I want to find the number of complex roots in the first quadrant.
How would I use Rouche's theorem? or is there a simpler way. I was thinking of comparing it to $z$ because I checked on wolfram that it actually only has 1 root, but this is obviously not a good way to do it in general.
EDIT: I had an idea. Can I compare it to $z^5+1$? We would just have to show that$|f(z)-z^5-1|<|z^5+1|$?
| In order to use Rouché to find the number of roots of $f(z) = z^5 + 3 z + 1$ (counted by multiplicity) in the region inside a simple closed contour $C$, you need to find $g(z)$ such that you know the answer for $g$ and $|f(z) - g(z)|$ is relatively small (specifically $ < |f(z)| + |g(z)|$, in the best version of the theorem) on $C$.
Well, since we can't get the whole first quadrant let's take a sector of it: take $C$ to go on the real axis from $0$ to $R$, then the circular arc $|z|=R$ in the first quadrant from $R$ to $Ri$, then back to $0$ on the positive imaginary axis. I'll take $g(z) = z^5 + 1$: we know its roots, of which one is inside $C$ if $R > 1$.
Now on the arc $|z|=R$,
$$|f(z) - g(z)| = |3 z| = 3 R < R^5 - 1 \le |g(z)| \le |f(z)| + |g(z)|$$
if $R$ is large enough ($R \ge 2$ certainly suffices).
On the positive real axis,
$$|f(z) - g(z)| = 3 z < z^5 + 3 z + 1 = |f(z)| \le |f(z)| + |g(z)|$$
On the positive imaginary axis, writing $z = it$ with $t > 0$ we have
$f(z) = i(t^5 + 3 t) + 1$ so
$$|f(z) - g(z)| = 3 t < \sqrt{(t^5 + 3t)^2 + 1} = |f(z)| \le |f(z)| + |g(z)|$$
So $f$ and $g$ have the same number of roots inside $C$, namely $1$. Taking $R \to +\infty$, we conclude that $f$ has exactly one root in the first quadrant.
| {
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Taylor Expansion of Inverse of Difference of Vectors I am trying to derive the multipole moment of a gravitational potential, but I'm getting stuck on some math I believe. So basically the problem is finding the Taylor Expansion for $$\frac{1}{|\mathbf{x}-\mathbf{x'}|}=\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}.$$ The expansion is supposed to be around $\mathbf{x'}=0$.
I have two questions: 1) Do I take the partial derivatives in terms of x or x' (I'm thinking it should be just x, but I'm not sure)? 2) When I'm supposed to multiply by a factor that is the equivalent of $(x-a)$, what should that be? I was thinking I should just dot with x', but that doesn't give me the correct answer.
| The multivariable Taylor expansion is probably most easily written as
$$ f(\mathbf{x+h}) = \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{i_1,i_2,\dotsc,i_k=1}^n (\partial_{x_{i_1}} \partial_{x_{i_2}} \dotsm \partial_{x_{i_k}} f(\mathbf{x}) ) h_{i_1} h_{i_2} \dotsm h_{i_k} $$
(which keeps all terms of the same order together), so the first few terms are
$$ f(\mathbf{x}) + \mathbf{h} \cdot \nabla f(\mathbf{x}) + \mathbf{h} \cdot (Hf(\mathbf{x}))\mathbf{h} + \dotsb $$
However, probably the simpler way in this case is to write
$$ \lvert \mathbf{x-x'} \rvert^{-1} = (|\mathbf{x}|^2+2\mathbf{x\cdot x'}+|\mathbf{x'}|^2)^{-1/2}. $$
Setting $|\mathbf{x}|=R$, $|\mathbf{x}|=r$, and $\mathbf{x\cdot x'} = 2rR\cos{\theta}$, where $\theta$ is the angle between the vectors, we have
$$ \lvert \mathbf{x-x'} \rvert^{-1} = R^{-1}(1+2r/R \cos{\theta}+r^2/R^2)^{-1/2}. $$
Setting $r/R=s$, we can now expand this as a function of $s$ using the binomial theorem:
$$\begin{align*}
\frac{1}{\lvert \mathbf{x-x'} \rvert} &= \frac{1}{R}(1+2s \cos{\theta}+s^2)^{-1/2} \\
&= \frac{1}{R} \left( 1 - \frac{1}{2}(2s \cos{\theta}+s^2) + \frac{-1}{2}\frac{-3}{2}\frac{1}{2!}(2s \cos{\theta}+s^2)^2 + O(s^3) \right) \\
&= \frac{1}{R} \left( 1 - s\cos{\theta} - \frac{1}{2}s^2 + \frac{3}{2}s^2 \cos^2{\theta} + O(s^3) \right) \\
&= \frac{1}{R} - \frac{s\cos{\theta}}{R} - \frac{1}{2R}(3 \cos^2{\theta}-1)s^2 + O(s^3)
\end{align*}$$
Resubstituting, we get the multipole expansion
$$ \frac{1}{\lvert \mathbf{x-x'} \rvert} = \frac{1}{\lvert \mathbf{x} \rvert} - \frac{\mathbf{x \cdot x'}}{\lvert \mathbf{x} \rvert^2} + \frac{1}{2} \frac{3(\mathbf{x \cdot x'})^2-\lvert \mathbf{x'} \rvert^2}{\lvert \mathbf{x} \rvert^5} + O(\lvert \mathbf{x'} \rvert^3/\lvert \mathbf{x} \rvert^4) $$
To actually answer your questions,
*
*Since $\lvert \mathbf{x -x'} \rvert=\lvert \mathbf{x'-x} \rvert$, here it doesn't actually matter. You're taking derivatives of the function $\lvert \mathbf{x} \rvert^{-1}$ and evaluating them at $\mathbf{x}$.
*You are correct in dotting with $\mathbf{x'}$: notice that the third term in the expansion, for example, is
$$ \sum_{i,j=1}^3 \frac{3x_i x_j- \delta_{ij}}{2\lvert \mathbf{x} \rvert} x'_i x'_j. $$
| {
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If $p$ and $q$ are distinct primes and $a$ be any integer then $a^{pq} -a^q -a^p +a$ is divisible by $pq$. If $p$ and $q$ are distinct primes and $a$ be any integer then $a^{pq} -a^q -a^p +a$ is divisible by $pq$.
Factorising we get $a^{pq} -a^q -a^p +a =a^p(a^q -1) - a(a^{q-1}-1)$ and we know $p \mid a^{p-1}-1$ and $q \mid a^{q-1}-1$.
I can't proceed further with the proof. Please Help!
| For prime $q,$
$$a^{pq}-a^q-a^p+a=[(\underbrace{a^p})^q-(\underbrace{a^p})]-[a^q-a]$$
Now by Fermat's Little Theorem, $b^q\equiv b\pmod q$ where $b$ is any integer
Set $b=a^p, a$
Similarly for prime $p$
Now if $p,q$ both divides $a^{pq}-a^q-a^p+a,$ the later must be divisible by lcm$(p,q)$
| {
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Is my solution correct? Differentiate $y = \left(x + \left(x + \sin^2 x\right)^4\right)^6$
Differentiate $y = \left(x + \left(x + \sin^2 x\right)^4\right)^6$.
I am unsure if I skipped a step in the end. Please take a look at my work. Criticize my strategy, point out obvious math deficiencies, or why I am missing the obvious.
| derivative of $\sin^2(x)$ is $2\sin(x)\cos(x)$, also don't omit parentheses
| {
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How do I find the period of the sine function $y = 20\sin\left[\frac{5 \pi}{2}\left(\frac{x -2}{5}\right)\right] + 100$ Using Desmos I can see the period is $0.8$ but how do I get there?
I understand that the period is $2\pi/$co-efficient of $x$ but the $-2/5$ is throwing me off.
| Shifts are to be neglected, hence, if $T$ is a period, we have
$$
\frac{5\pi}{2}T=2\pi,
$$
which gives $T=0.8$.
| {
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Measurable set and continuous function exercise Let $E \subset \mathbb R^n$ be a measurable set. Show that $f:\mathbb R_{\geq 0} \to \mathbb R$, given by $f(r)=|E \cap B(0,r)|$ is continuous.
So, given a fixed $r_0 \in \mathbb R_{\geq 0}$ and given $\epsilon$, I want to prove there is $\delta>0 : |r_0-x|<\delta \implies |f(r_0)-f(x)|<\epsilon$. We have $$|f(r_0)-f(x)|=||E \cap B(0,r_0)|_e-|E \cap B(0,x)|_e|$$
Since $|E \cap B(0,r)|_e \leq |B(0,r)|_e < \infty$ for any $r$, then $$||E \cap B(0,r_0)|_e-|E \cap B(0,x)|_e| \leq |E \cap B(0,r_0) \triangle E \cap B(0,x_0)|_e$$
Here I got stuck, I still didn't use the fact that $E$ is measurable. Any help is greatly appreciated.
| Let $s \in \Bbb R_{\ge 0}$. For all $r > s$,
$$0 \le f(r) - f(s) = |E\cap [B(0,r)\setminus B(0,s)]| \le |B(0,r) \setminus B(0,s)| = C_n(r^n - s^n),$$
where $C_n$ is a constant depending only on $n$. Since $r^n - s^n \to 0$ as $r \to s$, it follows that $f(r) \to f(s)$ as $r\to s^{+}$. A similar argument shows that $f(r) \to f(s)$ as $r\to s^{-}$. Hence $f$ is continuous at $s$. Since $s$ was arbitrary, $f$ is continuous.
| {
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Vector Analysis - Finding perpendicular vectors Given the vector:
$$
\vec{u}=-7\hat{i}-7\hat{j}+7\hat{k}
$$
I need to find another vector $\vec{v} $ that is parallel to the $xy$ plane and perpendicular to the vector $\vec{u} $ .
How can I do it?
I know that any vector that is perpendicular to $\vec{u}$ must satisfy $\vec{u}\cdot \vec{v} =0$ . But how does this help me with finding such a vector?
| If you have a vector on the Euclidean plane $\vec{v}=(a,b)$ then the vector $\vec{u}=(-b,a)$ is perpendicular to $\vec{v}.$ We can apply the same idea in $3$-dimensions:
If we have a vector $\vec{v}=(a,b,c)$ then the following three vectors $$\vec{u}_1=(-b,a,0),\quad \vec{u}_2=(-c,0,a),\quad \vec{u}_3=(0,-c,b),$$ are perpendicular to $\vec{v}.$ One of these gives you the desired vector.
| {
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elementary deduction on limit of sequence Let $(a_n)$ be a convergent sequence and $M$ a real number such that $a_n ≤ M$ for each $n$. Using the previous question, or otherwise, prove that $\lim_{n\to \infty}a_n≤M$.
I tried the "version" where $a_n > M$ and was able to arrived at a solution but this one seems like a tough nut to crack!
| For every $\epsilon>0$ there is $N$ such that $|\lim_na_n-a_n|<\epsilon$.
Then $\lim_na_n-a_n<\epsilon$, for $n>N$.
We deduce that $\lim_na_n<a_n+\epsilon\leq M+\epsilon$.
Since we have obtained that for all $\epsilon>0$, $\lim_na_n<M+\epsilon$, therefore $\lim_na_n\leq M$.
| {
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Baby Rudin Theorem 9.17 In proof of Theorem 9.17 Rudin writes:
$$
|f(x+h)-f(x)-Ah|\leq \frac{1}{2}|Ah| \tag{23}
$$
It follows that
$$
|f(x+h)-f(x)|\geq \frac{1}{2}|Ah| \tag{24}
$$
I note that $f$ is a $C'$ mapping of a subset $E$ of $R^n$ into $R^m$ and $A = f'(A)$. I don't see how (24) follows from (23).
:
| show |u+2v|<|v|-> |u|>|v|
Played with this in 2 dimensions and its true.
Assume |u|<|v|
Draw the vector 2v and from its tip draw a circle of radius |u|. Then u+2v is a vector from the base of 2v to a point on the circle whose length is always greater than |v|. Contradiction. Therefore |u|>|v|.
In n dimensions you would draw a ball of radius |u| at the tip of 2v and tnen calculate distances using the distance formula.
Note. |u+2v| is the same as |u-2v| because v can be + or -. It's just easier to think about.
| {
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Limit of sequences and integers If $a$ is a non zero real number , $x \ge 1$ is a rational number and $(r_n)$ is a sequence of positive integers such that $\lim _{n \to \infty}ax^n-r_n=0$ , then is it true that $x$ is an integer ?
| According to the Wikipedia article on Pisot–Vijayaraghavan numbers, this is true. In fact, if $x$ is algebraic (not necessarily rational) then it has to be a Pisot number, and in particular an algebraic integer. A rational integer is of course a bona fide integer.
| {
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Counting the number of pairs of $(i,j)$ such that $\operatorname{lcm}(i,j)=n$ Pairs $(i,j)$ such that $\operatorname{lcm}(i,j)=n$
how many pair $(i,j)$ can be formed such that $\operatorname{lcm}(i,j)=n$. here $i\leq n$ and $j\leq n$.
here $n$ is a integer number less than $10^{14}$.
| This can get you started.
Factor $n$ into its prime decomposition:
$$n=p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}$$
All the possible $i$'s where $\operatorname{lcm}(i,j)=n$ are given by
$$i=p_1^{b_1}p_2^{b_2}\dots p_k^{b_k}$$
where $0\le b_r\le a_r$. This gives $a_r+1$ choices for the exponent for $p_r$, and the choices are independent for each $r$, so the number of possible choices for $i$ is
$$(a_1+1)(a_2+1)\ldots (a_k+1)$$
Continue from there. The main difficulty is that the number of possible choices for $j$ depends on the particular $i$.
Show us some of the work you do from here, and we can give you more ideas.
| {
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How to explain to a 14-year-old that $\sqrt{(-3)^2}$ isn't $-3$? I had this problem yesterday. I tried to explain to the kid this: $$\sqrt{(-3)^2} = 3,$$ and he immediately said: "My teacher told us that we can cancel the square with the square root, so it's $$\sqrt{(-3)^2} = -3."$$
He has a lot of problems with maths, and I don't know how I can explain to him this as easily as possible. He still thinks that I lied him.
Thank you.
| I think the kid has misunderstood something his teacher said about cancelling. A math student of any age is bound to misunderstand his teacher at some point or other. When I was in high school, long, long ago, Mr. Jones was fond of saying that such and such equation has no real solutions, equations like, say, $x^2 + 9 = 0$. I thought he meant such an equation has no solutions at all. It wasn't until long after college that I learned about imaginary numbers.
The equation $x^4 - 81 = 0$ has four solutions, anyone who has studied the fundamental theorem of algebra can tell you. If we want to limit ourselves to real solutions, there are still two solutions left. But when we punch up $\root 4 \of {81}$ on a calculator, we want just one answer, and we want that answer to be the same each time, e.g., if it says the answer is $3$ one time and $-3$ another time, we'd think the calculator has a malfunction of some sort.
And so it is with the square root. We want the calculator to say $\sqrt 9 = 3$ every time, no matter how it is that we got the $9$ in the first place, whether by $(-3)^2$ or $3^2$ or any other operation that could possibly give $9$, such as $56 - 47$.
| {
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Proving the implication $\lim_{n\to\infty} a_n a_{n+1} =0 \Rightarrow \lim_{n\to\infty} a_n=0$ Let $\{a_n\}$ be a sequence such that for all $n$, $a_n>0$. I have to prove that if $\lim_{n\to\infty} a_n a_{n+1} =0$ then $\lim_{n\to\infty} a_n=0$. Here's my reasoning: $\forall\varepsilon>0,\exists k$ such that $\forall n>k$ we have $|a_n a_{n+1}|=a_n a_{n+1} < \varepsilon$ and thus $a_n=|a_n|=|a_n-0| < \frac{\varepsilon}{a_{n+1}} \Rightarrow \lim_{n\to\infty} a_n=0$. However I don't feel comfortable with the $a_{n+1}$ on the r.h.s. What do you think?
| As said that's not correct. I think you should better argue by contradiction that $\lim_{n\to\infty}a_n\neq 0$. WLOG the limit exists and $$\lim_{n\to\infty} a_n=\lim_{n\to\infty}a_{n+1}=l>0\Rightarrow \lim a_na_{n+1}=l^2>0$$ in contradiction which means that necessarily $\lim_{n\to\infty}a_n=0$ .
| {
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Finding $\lim\limits_{n \to \infty}(\sin(1)\cdot \sin(2) \cdot \ldots \cdot \sin(n))$ Problem: Find $\lim_{n \to \infty}(\sin(1)\cdot \sin(2) \cdot \ldots \cdot \sin(n))$.
My idea: I suppose it to be $0$, but how might I go about proving this?
| $$\sin(k)\sin(k+1)=\frac{1}{2} [\cos(1)-\cos(2k+1)]$$
Now use the fact that $.5 < \cos(1) < .6$ to conclude that
$$\left| \sin(k) \sin(k+1)\right| < 0.8$$
and
$$\left| \sin(1)\cdot \sin(2) \cdot \ldots \cdot \sin(n) \right| \leq 0.8^{\frac{n}{2}}$$
| {
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implicit differentiation solving for $y\prime$ I'm supposed to implicitly differentiate the following and give the answer in terms of $y\prime$. $$\tan(x-y)={y \over 1+x^2}$$
$${ (1+x^2)y\prime - 2xy \over (1+x^2)^2 }$$
How do I solve for $y \prime$?
Edit:
After a ludicrous amount of algebra, i finally ended up at
$$ { sec^2(x-y)(1+x^2)^2+2xy \over (1+x^2)(1+sec^2(x-y)(1+x^2)) } $$
which apparently is correct.
| Differentiating both sides from
$$\tan (x - y(x)) = \frac{{y(x)}}{{{x^2} + 1}}$$
gives:
$$\frac{1}{{{{\cos }^2}(x - y(x))}} - \frac{{y'(x)}}{{{{\cos }^2}(x - y(x))}} = \frac{{y'(x)}}{{{x^2} + 1}} - \frac{{2xy(x)}}{{{{\left( {{x^2} + 1} \right)}^2}}}$$
which can be solved and simplified for ${y'(x)}$ like so:
$$y'(x) = \frac{{2xy(x)}}{{{x^2} + 1}} + \frac{{2\left( {{x^2} - 2xy(x) + 1} \right)}}{{2{x^2} + \cos (2(x - y(x))) + 3}}$$
(I don't like sec-terms)
| {
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$P( A\triangle B ) = 0 \Rightarrow P(A)=P(B)=P(A\cup B) + P(A \cap B) $
I want to prove the following statement;
$$P( A\triangle B ) = 0 \Rightarrow P(A)=P(B)=P(A\cup B) \color{blue}{=} P(A \cap B) $$
What I did is that
$$P(A\triangle B) =P((A\setminus B ) \cup (B\setminus A))=0$$
$$=P((A\cap B^c)\cup (B\cap A^c))=0$$
$$=P((A\cup B)\cap (A\cup A^c)\cap (B\cup B^c)\cap(B^c \cup B^c))=0$$
$$=P((A\cup B)\cap (B^c\cup A^c))=0$$
$$=P((A\cup B)\setminus (B\cap A))=0$$
Please help me showing this. thank you.
| The probabilities have to look like this, from which everything follows.
| {
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Completing System of Vectors to an Orthogonal Basis I want to complete ((1, -2, 2, -3), (2, -3, 2, 4)) to an orthgonal basis. I honestly don't really know how to do this. This a practice problem to a test, were we haven't done problems related to this. I know that an orthogonal basis has that for basis $(v_1, ... , v_n)$, $v_i\perp v_j$ for $i\ne j$. So I'm assuming I'm working in $R^4$, as the question only specifies "Euclideam Spcae". In terms of actually solving the problem, I'm strugging, thanks in advance!
| here is what you can do. you already have two vectors $u = (1, -2, 2, -3)^T, v = (2,- 2, 3, 4)^T$ that are orthogonal. we will first find four orthogonal vectors; making them of unit length is easier.
what we will do is pick a vector $a$ and find the projection on to the space spanned by $u, v$ and subtract it. we can just start with $a = (1,0,0,0).$ so let $$a = ku + lv + w $$ with $l,k$ are to be chosen so that $u^Tw = v^Tw = 0.$ that is $$k = u^Ta/u^Tu, l = v^Ta/v^Tv \to k = \frac 1{14}, l = \frac 1{33},\\ w =(1,0,0,0)^T - \frac 1{14}((1, -2, 2, -3)^T + \frac1{33}(2,- 2, 3, 4)^T$$
once you have $w,$ or a nice multiple of it, we have $u, v, w$ orthogonal. now you can use another vector hopefully $(0,1,0,0)$ and do the same trick to find the fourth vector.
| {
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Having trouble with the substitution method. I have to evaluate the integral:
I see that with u being equal to $4x$ I will get that $du/dx$ will equal $4$ and lead to $du=4dx$. However I am stuck when I plug in the u values to get $\sec u\cos u\,du$. The problem states to use the subsitution method but it looks like I will have to use the product rule because of $\sec u\cos u\,du$. Is there a way to evaluate this integral without the product rule? Thanks in advance
| i think you can use the change of variable $u = \cos 4x, du = -4 \sin 4x \, dx$ instead.
then $$\int \sec 4x \tan 4x \, dx = \int \frac{\sin 4x}{\cos^2 4x} \, dx
\to -\frac 14\int \frac{du}{u^2} = \frac{1}{4u} = \frac 14\sec4x + C$$
| {
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Convergence of Types Theorem
(Convergence of Types Theorem) Suppose that $F_n(u_nx+v_n) \Rightarrow F(x)$ and $F_n(a_nx+b_n) \Rightarrow G(x)$, where $u_n>0, a_n>0$ and $F$ an $G$ are non-degenerate. Then there exist $a>0$ an $b \in\mathbb R$ such that $a_n/u_n \to a$ and $(b_n-v_n)/u_n \to b$, and $F(ax+b) = G(x).$
In Billingsley's textbook, he proves the above theorem using the following lemmas.
*
*If $F_n \Rightarrow F$, $a_n \to a$ and $b_n \to b$, then $F_n(a_nx +b_n) \Rightarrow F(ax+b)$.
*If $F_n \Rightarrow F$ and $a_n \to \infty$, then $F_n(a_nx) \Rightarrow \Delta(x)$, where $\Delta(x)$ is the degenerate distribution at $x$.
*If $F_n \Rightarrow F$ and $b_n$ is unbounded, then $F_n(x+b_n)$ cannot converge weakly.
*If $F_n(x) \Rightarrow F(x)$ and $F_n(a_nx+b_n) \Rightarrow G(x)$, where $F$ and $G$ are non-degenerate, then $$ 0<\inf_n a_n \leq \sup_n a_n < \infty;\; \sup_n |b_n| < \infty .$$
I have difficulty in understanding the proof of the forth lemma (highlighted parts). The argument in the book is as follows. Suppose that $a_n$ is not bounded above. Arrange by passing to a sub-sequence that $a_n \to \infty$. Then by lemma 2,
$$F_n(a_nx) \Rightarrow \Delta(x).(*)$$
Then since
$$F_n\left[a_n \left(x+\frac{b_n}{a_n}\right)\right] = F_n(a_nx+b_n)\Rightarrow G(x),(**)$$
it follows by lemma 3 that $\frac{b_n}{a_n}$ is bounded. [Note that in lemma 3, we did not have $a_n$ in front of $(x+b_n)$. But we do now. My question is how to use lemma 3 to get the desired boundedness of $b_n/a_n$, please?] By passing to a further sub-sequence, arrange that $b_n/a_n$ converges to some $c$. By $(*)$ and lemma 1,$F_n\left[a_n \left(x+\frac{b_n}{a_n}\right)\right] \Rightarrow \Delta(x+c)$ along this sub-sequence. But $(**)$ now implies that $G$ is degenerate, contrary to hypothesis. Thus $a_n$ is bounded above. If $G_n(x) = F_n(a_nx+b_n)$, then $G_n(x) \Rightarrow G(x)$ and $G_n(a_n^{-1}x - a_n^{-1}b_n) = F_n(x) \Rightarrow F(x)$. The result just proved shows that $a_n^{-1}$ is bounded. Thus $a_n$ is bounded away from $0$ and $\infty$. My question here is how to know $a_n$ is positive, please? How come $a_n$ cannot be negative? Thank you!
| It seems that Lemma 3 is applied with $\widetilde{F_n}(t):=F_n(a_nt)$ instead of $F_n$ (which is allowed by Lemma 2) and the sequence $(b_n/a_n)_{n\geqslant 1}$ instead of $(b_n)_{n\geqslant 1}$.
| {
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Spectrum of a bilateral shift Let $u$ be a bilateral shift on Hilbert space $\ell^2(\Bbb Z)$. As for unilateral shifts, the spectrum of $u$ does not contain any eigenvalue. Also $u$ is unitary, so $\sigma(u) \subset \Bbb S$ ($\Bbb S$ means unit circle). How can I show that $\sigma(u)=\Bbb S$?
| Hint: Let $e_n$ $(n\in \mathbb Z)$ be the standard basis in $\ell^2(\mathbb Z)$. For $\lambda \in {\mathbb T}$, let
$$ \xi_n=\frac{1}{2n+1}\bigl( \lambda^{-n} e_{-n}+\lambda^{-n+1} e_{-n+1}+\cdots+e_0+\cdots+\lambda^{n} e_n\bigr)\qquad (n\in \mathbb Z). $$
Note that $(\xi_n)$ is a sequnce of vectors with norm $1$. Consider
$$ u\xi_n-\lambda \xi_n\quad \text{when}\quad n\to \infty. $$
| {
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If every functional of $f$ is smooth, is $f$ smooth? Let E be a (real or complex) Banach space and suppose $f: \mathbb{R}^n \rightarrow E$ has the property that $\lambda \circ f$ is $C^\infty$ for every bounded linear functional $\lambda \in E^\ast$. Does it follow that $f$ is also $C^\infty$?
I didn't really encounter this question anywhere; it occurred to me while reading about smooth, Banach space valued functions. I remembered that all weakly holomorphic Banach space valued functions are holomorphic and I thought it would be nice to have a similar criterion for smooth functions.
| See Theorem 5.0.3 in these notes by Paul Garrett, where this is shown for weakly smooth $f$ on a closed real interval.
It seems he isn't using anything specific to the fact that $f$ is defined on a dimension $1$ space, so I think the proof generalizes to $\mathbb R^n$.
| {
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In the sequence $1,3,7,15,31\ldots$ each term is $2\cdot\text{immediately preceding term}+1$. What is the $n$-th term? I readily see that it is $2^n-1$, but how can I deduce the $n$-th term from the given pattern i.e. $n$-th term $= 2\cdot(n-1)\text{th term} + 1$ without computation.
| Here is an explicit proof by induction. Given that $a_{n+1}=2a_n+1$ for $n=0,1,...$, with $a_0=1$ prove the proposition $\mathcal P(n):\;\,a_n=2^n-1$ for $n=0,1,...$ .
We are given that $\mathcal P(n)$ holds for $n=0$. Suppose now that we know $\mathcal P(n)$ for $n=k$; that is, $\mathcal P(k):\;\,a_k=2^k-1$. Now $a_{k+1}=2a_k+1=2(2^k-1)+1=2^{k+1}-1$. We have shown that $\mathcal P(k)\Rightarrow\mathcal P(k+1)$. Hence, by induction, $\mathcal P(n)$ holds for all $n=0,1,...$ .
| {
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Find the Critical Points: $f(x) =(x^2-1)^3$ This question probably has more to do with my Algebra skills than Calculus. Nonetheless, can someone explain why the factored "term" is not set to zero (0) [second picture]. Thanks in advance.
| $f(x)=(x^2-1)^3$ is a sixth-degree polynomial with $x=\pm 1$ being triple zeroes.
Moreover, $f(x)$ is an even function, hence $f'(x)$ (that is a fifth-degree polynomial) has double zeroes in $x=\pm 1$ and a simple zero in $x=0$. The critical points of $f(x)$ may occur only at the zeroes of the derivative or at the endpoints of the given interval $I=[-1,5]$, hence to compute $\max_{x\in I}f(x)$ and $\min_{x\in I}f(x)$ it is enough to compute $f(-1),f(0),f(1),f(5)$.
The same approach works for the second exercise, too, by considering $F(t)^2$ in place of $F(t)$, getting a third-degree polynomial.
| {
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Started this problem but can't finish it: Showing pointwise convergence for this summation I know how to start this problem but am having trouble finishing the end of it. Any help would be great! Thanks
We let $g_n: E \rightarrow \mathbb{R}$ be continuous functions for $1 \leq n \leq N$ and we let $a_k^{(n)}$ be $N$ convergent sequences of numbers. Assume lim$_{k\rightarrow \infty}a_k^{(n)} = a_n$. Let $f = \sum_{n=1}^N a_n g_n$.
I want to show that:
$\sum_{n=1}^{N} a_k^{\{n\}} g_n$ converges point-wise to $f$.
My solution so far:
Pointwise convergence definition: A sequence of functions $g_n$ defined on a set M converges pointwise to a function g on M if lim $f_n(x)$ = $f(x)$ as n $\rightarrow$ ∞ holds for all $x$ in M.
I can fix $x$ in $E$ and then look at |$\sum_{n=1}^{N} a_k^{\{n\}}g_n(x)$ - $\sum_{n=1}^{N} a_ng_n(x)$|but from here I am do not know how to break this down to show pointwise convergence.
| Hint: |$\sum_{n=1}^{N} a_k^{\{n\}}g_n(x)$ - $\sum_{n=1}^{N} a_ng_n(x)$| = $|\sum_{n=1}^N (a_k^{\{n\}} - a_n) g_n(x)|$.
Additionally, $|\sum_{n=0}^N x_n| \leq \sum_{n=0}^N |x_n|$ for any $x_n$ (this is the triangle inequality).
Can you take it from here?
| {
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Topology, Showing that two metric spaces are topologically equivalent Can someone verify if this is true?
$X=\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
. $d$
is the standard metric on $\mathbb{R}$
and $d'(x,y)=d\left(\tan(x),\tan(y)\right)$
. We want to show that $\left(X,d\right)$
and $\left(X,d'\right)$
are topologically equivalent. I will just show the case if U
is an open set in $\left(X,d\right)$
then U
is also open in $\left(X,d'\right)$
. Let $x_{0}\in U$
. There exists $0<r\in\mathbb{R}$
such that $B_{r}(x_{0})=\left\{ y\,|\,d(x_{0},y)<r\right\} \subseteq U$
. Since $\tan(x)$
is a monotonic function and continuous, there exists $0<r'\in\mathbb{R}$
such that the the greatest $\delta>0$
that satisfies $d(x_{0},y)<\delta\Rightarrow d(\tan(x_{0}),\tan(y))<r'$
, is smaller than $r$
. Since the function is monotonic $r'$
exists. And therefore $B_{r'}^{'}(x_{0})=\left\{ y\,|\,d^{'}(x_{0},y)=d(\tan(x_{0}),\tan(y))<r'\right\} \subseteq B_{r}(x_{0})\subseteq U$
. Therefore $U$
is open in $\left(X,d'\right)$
.
The part that I am not sure of, justifying the existence of r'
. Is there a way of choosing r'
more specifically ? Thanks!
| I will just comment on this paragraph here:
Since $\tan(x)$ is a monotonic function and continuous, there exists
$0<r′\in\mathbb{R}$ such that the the greatest $\delta>0$ that
satisfies $d(x_0,y)<\delta\Rightarrow d(\tan(x_0),\tan(y))<r'$ , is
smaller than $r$ . Since the function is monotonic $r'$ exists.
I don't know what you need $\tan(x)$ to be monotonic for, and I also doubt that monotonicity makes the argument about "the greatest $\delta>0$..." work. Let me know, if I just fail to understand your idea.
The way I think about it is:
I want to prove existence of an $r'$, such that
$$B'_{r'}(x_0)\subseteq B_r(x_0).$$
This means that whenever $d(\tan(x_0),\tan(y))<r'$, I must have $d(x_0,y)<r$. But this property is continuity of $\tan^{-1}(x)$, the inverse of tangent. Letting $v_0=\tan(x_0)$, and $w=\tan(y)$, I know by continuity of $\tan^{-1}(x)$ that there exists a $\delta>0$, such that
$$d(v_0,w)<\delta\Rightarrow d(\tan^{-1}(v_0),\tan^{-1}(w))<r,$$
which is the same as
$$d(\tan(x_0),\tan(y))<\delta\Rightarrow d(x_0,y)<r.$$
In the above, I can thus choose $r'=\delta$.
| {
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Another optimization problem I am having trouble figuring out a next step in an optimization problem
the question is to find the max and min values of $f(x,y)=\frac{x+y}{2+x^2+y^2}$
I calculated $f_x$ and $f_y$ and set both them equal to zero, and the only possibility you get is x=y. I dont know how else to find it after this. But the back of the book says the answer is a max at $f(1,1)$ and a min at $f(-1,-1)$ but I dont know how?
$$f_x=\frac{-x^2-2xy+y^2+2}{(2+x^2+y^2)^2}$$
$$f_y= \frac{-y^2-2xy+x^2+2}{(2+x^2+y^2)^2}$$
Can anyone see why please?
Thankyou
| Setting $f_x=0$ and $f_y=0$ gives $-x^2-2xy+y^2+2=0$ and $-y^2-2xy+x^2+2=0$, so
subtracting these equations gives $2y^2-2x^2=0, \;\;y^2=x^2,\; $ and so $y=\pm x$.
1) If $y=x$, substituting into the first equation gives $x^2=1$ so $x=\pm 1$.
2) If $y=-x$, substituting into the first equation gives $x^2=-1$, so there is no real solution.
Therefore $(1,1)$ and $(-1,-1)$ are the only critical points.
Since $\;\displaystyle f_{xx}=(2+x^2+y^2)^{-2}(-2x-2y)-4x(-x^2-2xy+y^2+2)(2+x^2+y^2)^{-3}$,
$\;\displaystyle \hspace{.36 in}f_{xy}=(2+x^2+y^2)^{-2}(-2x+2y)-4y(-x^2-2xy+y^2+2)(2+x^2+y^2)^{-3}$,
$\;\displaystyle \hspace{.36 in}f_{yy}=(2+x^2+y^2)^{-2}(-2y-2x)-4y(-y^2-2xy+x^2+2)(2+x^2+y^2)^{-3}$,
A) $\;D=f_{xx}f_{yy}-(f_{xy})^2=(-\frac{1}{4})(-\frac{1}{4})-0^2=\frac{1}{16}>0\;$ and $\;f_{xx}=-\frac{1}{4}<0$ at $(1,1)$,
$\hspace{1.4 in}$so $f$ has a relative maximum at $(1,1)$.
B) $\;D=f_{xx}f_{yy}-(f_{xy})^2=(\frac{1}{4})(\frac{1}{4})-0^2=\frac{1}{16}>0\;$ and $\;f_{xx}=\frac{1}{4}>0$ at $(-1,-1)$,
$\hspace{1.4 in}$so $f$ has a relative minimum at $(-1,-1)$.
| {
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How many ways to get at least one pair in a seven card hand? This was the question and answer I saw:
How many different seven-card hands are there that contain two or more cards of
the same rank?
Solution:
There are C(52,7) total hands. To subtract the ones that don’t have pairs, we observe
that such hands have cards of 7 different ranks, and there are C(13,7) ways to select
those. Then there are 4 choices for each card. So:
C(52,7) – C(13,7) · 4^7
But my solution was 13·C(4,2)·C(50,5), as you choose which rank is the pair (13 ways to do this), then choose which two cards are the pair out of that rank(4 choose 2), and then you choose the remaining 5 cards from the rest of the 50 (50 choose 5). But these two do not give the same number. Am I going wrong somewhere with my alternate solution?
| Consider a hand which has two pairs, for example
$$\clubsuit A,\,\diamondsuit A,\,\heartsuit 10,\diamondsuit10,\,\spadesuit7,\,\spadesuit K,\,\diamondsuit 2.$$
Your method will count this hand twice:
*
*choose ace, then clubs and diamonds, then $\heartsuit 10,\diamondsuit10,\,\spadesuit7,\,\spadesuit K,\,\diamondsuit 2$; or
*choose $10$, then hearts and diamonds, then $\clubsuit A,\,\diamondsuit A,\,\spadesuit7,\,\spadesuit K,\,\diamondsuit 2$.
This is why your answer is wrong. Similarly, there will be hands which contain three pairs, and which you will count three times.
In fact, if you think about it, your method will count some hands up to nine times!
| {
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Application of Differentiation (Doesn't understand) It's given the cubic equation $x^3-12x-5=0$. Show graphically that the iteration $x_{n+1}=\sqrt[3]{12x_n+5}$ should be used to find the most negative root and the positive root, and the iteration $x_{n+1}=\dfrac{x^3_n-5}{12}$ should be used to find the other root.
This is the solution given by book,
$$x_{n+1}=\sqrt[3]{12x_n+5}$$
$$F'(x_n)=\frac{4}{\sqrt[3]{(12x_n+5)^2}}$$
$$F'(-3)=0.41$$
$$F'(-0.5)=4$$
$$F'(3)=0.34$$
I've a problem here. Where the -3,-0.5 and 3 come from?
| The idea to find approximations of roots of $f(x) = 0$ starts with finding an integer $a$ such that a root lies between $a$ and $(a + 1)$. To ensure that this is so we need to guess some integers $a$ such that $f(a)f(a + 1) < 0$ i.e. $f(a)$ and $f(a + 1)$ are of opposite signs. Then by Intermediate Value Theorem there is a root of $f(x)$ between $a$ and $(a + 1)$.
Here $f(x) = x^{3} - 12x - 5$ and clearly we can see that $$f(0) = -5, f(1) = -16, f(2) = -21, f(3) = -14, f(4) = 9$$ so that there is a root in the interval $[3, 4]$. Again checking the negative integers we have $$f(0) = -5, f(-1) = 6$$ so that there is a root in interval $[-1, 0]$. Going to further negative values of $x$ we have $$f(-2) = 9, f(-3) = 4, f(-4) = -21$$ so that the third root lies between in $[-4, -3]$.
Now the method of iteration used here recasts the equation $f(x) = 0$ in the form $x = \phi(x)$ and uses the iteration $x_{n + 1} = \phi(x_{n})$ with suitable starting point $x_{0}$. The suitable starting is normally chosen to be one of the end-points of the interval in which the root lies. Sometimes a midpoint of the interval is also chosen.
Next we need to choose the $\phi(x)$ such that $|\phi'(x)| < k < 1$ (for some fixed number $k$) near the points of iteration $x_{i}$. This is needed to guarantee that the iteration converges to the desired root.
The solution has chosen $\phi(x) = \sqrt[3]{12x + 5}$ for one of the roots namely the one lying in interval $[-4, -3]$ and therefore the derivative $\phi'(-3) = 0.41$ has been calculated. (Your notation uses $F$ instead of $\phi$.) Similarly for the root in interval $[3, 4]$ the value $\phi'(3) = 0.34$ is used. In both the cases we see that the value of $|\phi'(x)|$ is less than $1$ and hence the iterations works.
However for the third root in $[-1, 0]$ it appears that the initial point of iteration is chosen as mid point $-0.5$ of the interval and then $\phi'(-0.5) = 4$ which is much greater than $1$. Choosing initial point as $0$ (or $-1$) would also give $|\phi'(x)| > 1$ so that the form of iteration needs to be changed for the iteration to converge.
For the root in $[-1, 0]$ we can write the equation as $$x = \frac{x^{3} - 5}{12}$$ so that $\phi(x) = (x^{3} - 5)/12$ and $\phi'(x) = x^{2}/4$. Clearly in this case $|\phi'(x)| < 1$ for any value of $x \in [-1, 0]$ and the iteration will converge.
I hope you now understand how we need to choose the form of iteration as well as the starting value for the iteration chosen.
Note: You should also read the textbook to understand why the condition $|\phi'(x)| < k < 1$ is needed to guarantee the convergence of iteration $x_{n + 1} = \phi(x_{n})$ to a root of equation $x = \phi(x)$. It is not very difficult to understand and you can revert back for more details.
| {
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Power Set of a Power Empty Set Find ℙ(ℙ(ℙ(∅))).
I know that ℙ(∅) = {∅}.
Then, ℙ(ℙ(∅)) = {∅, {∅}, {∅,{∅}}?
so, ℙ(ℙ(ℙ(∅))) = {∅,{∅, {∅}, {∅,{∅}}}?
Is it? Will it be ok if someone explain to me this concept?
| I think it's easier if you consider $P(\varnothing)$ to be a one-element set, say $\{1\}$. Then it's pretty clear that $P(P(\varnothing))$ has to be precisely $\{ \varnothing, \{1\}\}$. So $P(P(P(\varnothing))) = \{\varnothing, \{\varnothing\}, \{\{1\}\}, \{\varnothing, \{1\}\}\}$.
| {
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Number of strictly increasing and decreasing 4 digit numbers? I have two questions:
(a) Count the 4 digit numbers whose digits decrease strictly from left to right and;
(b) Count the 4 digit numbers whose digits increase strictly from left to right
I have the answers which are $10 \choose 4$ and $9 \choose 4$ respectively. Can someone explain how to come to these answer? The number of ways to choose 4 elements of 10 or 9 respectively seems to simplistic.
| (a) Since the digits must strictly decrease from left to right, there are 10C4 such numbers. For any selection of four digits from the ten digits, there is exactly one way of arranging those four digits in a strictly decreasing order from left to right. We need not worry about leading zeroes because the order is decreasing, and 0 can never be the left most digit. Hence 10C4.
(b) Here, since the digits must strictly increase from left to right, consider two sub-cases:
(b1) If 0 is not included - Then, there are 9 digits, and for every choice of 4 digits from them, we have exactly one way to arrange them in strictly increasing order. So, there are 9C4 such numbers.
(b2) If 0 is included - Then, 0 will appear as the left most digit, and this will not be a four-digit number. Therefore, there are no such numbers at all.
The answer is therefore, 9C4.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $A$ and $B$ are similar, why does $\text{rank}(A) = \text{rank}(B)$? Suppose $A$ and $B$ are similar matrices over $\mathbb{C}^n$. Why do we have $\text{rank}(A) = \text{rank}(B)$?
| If $A$ and $B$ are similar then $A = PBP^{-1}$ for some invertible matrix $P$. Because $P$ is invertible its null space is trivial, i.e. $nullity(P) = \{0\}$.
Now suppose that $Ax = PBP^{-1}x = 0$. Using the result above $PBP^{-1}x = 0$ if and only if $PBx = 0$; Again $PBx = 0$ if $Bx = 0$. The latter result means that $nullity(A) = nullity(B)$. Using rank-nullity-theorem we can conclude that $rank(A) = rank(B)$.
The proof is taken from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$(\delta,\varepsilon)$ Proof of Limit I wish to prove that $\lim_{x\to 2} {x+1 \over x+2} = {3 \over 4} $.
The $(\delta,\varepsilon)$ limit definition in this case is:
$\forall \epsilon >0, \exists \delta >0$ such that $0<|x-2|<\delta \Rightarrow |{x+1 \over x+2} - {3 \over 4}| < \epsilon.$
Thus, I need to provide a $\delta$, which is a function of $\epsilon$ in order to satisfy the above definition.
I am having a bit of difficulty finding an inequality to continue my work below.
$|{x+1 \over x+2} - {3 \over 4}| = {1 \over 4}|{x-2 \over x+2}|$
| Let $\varepsilon>0$ be given. Consider $|\dfrac{x+1}{x+2}-\dfrac{3}{4}|=|1-\dfrac{1}{x+2}-\dfrac{3}{4}|=|\dfrac{1}{4}-\dfrac{1}{x+2}|$
We want to find a $\delta>0$ such that whenever $|x-2|<\delta$ , the above expression $<\varepsilon$.
So start with $|\dfrac{1}{4}-\dfrac{1}{x+2}|<\varepsilon\iff|\dfrac{x-2}{4(x+2)}|<\varepsilon\iff|x-2|<4|x+2|\varepsilon$
When $|x-2|<\delta$, we can say that $2-\delta<x<2+\delta\implies|x+2|<4+\delta<M$ where $M$ is a uniform upper bound. Thus, we have, $|x-2|<4|x+2|\varepsilon<4M\varepsilon$ i.e. $|x-2|<4M\varepsilon$. Choose $\delta=4M\varepsilon$. Then, we have got our $\delta$.
Now check that this $\delta$ works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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