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Is there an interval notation for complex numbers? Just as $$\{x \in \mathbb{R}: a \leq x \leq b\}$$ can be written in the more-compact form $[a,b],$ is there an analogous notation for $$\{z \in \mathbb{C}:z=x+yi, x \in[a,b], y \in[c,d]\} \quad ?$$
Pictorially, the set of all $z \in \mathbb{C}$ lying in the green area is the set that I'd like to express in a more concise form:
|
Maybe just define something as $[a,b]+[c,d]i$ ?
|
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|
Integration of $1/\sin^3 x$
I need a explanation of this problem:
$$ \int \frac{1}{\sin^3 x}\,dx $$
Change the variable $$ t = \tan (x/2) $$
With use of $\tan$, $\cos$, $\sin$ and $\cot$, only.
So how do I think of this and what are the steps?
Progress
Well I have $\sin x = 2t/(1+t^2)$ and $t = \tan x/2$ gives
$$2 \arctan t = x,\quad \frac{2}{(1+t^2)} dt = dx$$
This all gave me
$$\int \left(\frac{2t}{(1+t^2)}\right)^3 (2/(1+t^2))\,dt$$
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More simple is to change the variable by $t=\cos x$ so $dt=-\sin x dx$ and then
$$\int\frac{dx}{\sin^3 x}=-\int\frac{dt}{(1-t^2)^2}$$
Can you take it frome here?
|
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$f(t) = \cos t^{-1} + \int_t^\infty \frac{1}{\tau^2 + f(\tau)^2} d\tau$ implies the integral is $O(\frac{1}{t})$ The following is a quote from "asymptotic methods in analysis" by de Bruijn (p. 136).
If we know that the real function $f(t)$ satisfies the relation
$$f(t) = \cos t^{-1} + \int_t^\infty \frac{1}{\tau^2 + f(\tau)^2} d\tau \,\,\,\,\,\,\,\,\,(t > 1)$$
then it is easily seen that the integral is $O(t^{-1})$.
The $O$ notation is for $t \to \infty$. If $|f(t)| < t$, I can prove the integral is $O(1)$ by factoring out $1/\tau^2$ from the integrand and expanding the remaining factor using the binomial theorem. However, the author does not specify any extra condition on $f(t)$ at all. How can I prove it generally without the condition that $|f(t)| < t$?
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Since $f$ is real valued, we have, for $0 < t \leqslant \tau$,
$$0 < \tau^2 \leqslant \tau^2 + f(\tau)^2,$$
and therefore
$$0 < \int_t^\infty \frac{d\tau}{\tau^2 + f(\tau)^2} \leqslant \int_t^\infty \frac{d\tau}{\tau^2} = \frac{1}{t}.$$
|
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Number of subsets/open subsets/closed subsets of a metric space. Let $(X,d_1)$ and $(X,d_2)$ be two metric spaces which have the same infinite set $X$, but the different metrics $d_1$ and $d_2$.
Denote the collection of subsets $X$ by $S$, and the collection of all open subsets of $(X,d)$ by $U_{d}$.
Then, is it possible to know $|U_{d_1}|$ and $|U_{d_2}|$, the cardinalities of the collections of all open subsets of $(X,d_1)$ and $(X,d_2)$?
For example, if $X=\mathbb{R}$ and if the metric $d$ on $\mathbb{R}$ is defined to be the discrete metric, then $|U_d|=|S|.$
But, if $X=\mathbb{R}$ and if the distance function $d$ of $X$ is defined to be the usual Euclidean distance function, $|U_d|=c$.
In conculusion, what I'm asking is that is it possible to characterize $|U_d|$ of a metric space $(X,d)$, with $X$ being infinite, in terms of metric function $d$?
Also, is it possible to relate $|S|$ with $|U_d|$ of a metric space $(X,d)$ in terms of metric function $d$?
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Let $X$ be an infinite set.
Note that a topology $\tau$ on X is a collection of subsets of $X$, that is $\tau\subset P(X)$.
We consider the collection $T$ of all topologies on X.
Pick a topology $\tau$ in $T$ one at a time, and let $(X,\tau)$ be the corresponding topological space.
If $(X,\tau)$ is a metrizable space, then let a corresponding metric space be $(X,d)$. Then, $|U_d|$ of the metric space $(X,d)$ is just equal to $|\tau|$. Note that there may be more than one metrics $d$ such that $(X,d)$ is a corresponding metric space to $(X,\tau)$.
If $(X,\tau)$ is not a metrizable space, then there is no metric $d$ such that $\tau$ is the induced topology by $d$, and hence for any metric $d$ on $X$ we will not have $\tau$ as a collection of open sets of a metric space $(X,d)$.
This gives an abstract algorithm to obtain the possible cardinality of a collection of all open sets of a metric space $(X,d)$ from an infinite set $X$.
Lastly, note that $|U_d|\leq |P(X)|$ for any metric $d$ on $X$.
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If a recursive sequence converges, must its inverse be divergent? Suppose I have a recursive sequence $\displaystyle a_{n+1} = \frac{a_{n}}{2}$. Clearly, the sequence converges towards zero. Now, suppose I define an "inverse" sequence $\displaystyle b_{n+1} = 2b_{n}$. While the sequence clearly diverges, could I know that based on the knowledge that $a_{n+1}$ converges? Is there any sequence that converges, whose "inverse" also converges?
.. and could this "inverse" sequence be defined in a more intuitive way?
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Say the sequence $(a_n)$ defined by $a_{n+1} = f(a_n)$ for bijective $f$ converges to $a$.
If $f$ is continuous at $a$, then
$$0 = \lim\limits_{n \to \infty} a_
{n+1}-a_n = \lim\limits_{n \to \infty} f(a_n)-a_n = f(a)-a$$
So $f(a)=a$. But then it also follows that $f^{-1}(a)=a$.
So if the inverse sequence is started with seed value $b_0 = a$, the inverse sequence will converge also.
I'm not $100$% sure what can be said for such convergent recurrences where $f$ is not continuous at the limit.
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How do you calculate the dimensions of the null space and column space of the following matrix?
I understand you are supposed to get the reduced row echelon form, which I did, and this is what I came up with:
1 -2 0 19 -6 0 -37
0 0 1 -6 2 0 6
0 0 0 0 0 1 3
0 0 0 0 0 0 0
From here, I know you're supposed to put it in equations, which I also did, and this is what I got:
x1 – 2x2 + 19x4 – 6x5 – 37x7 = 0
x3 – 6x4 + 2x5 + 6x7 = 0
x6 + 3x7 = 0
x1 = 2x2 – 19x4 + 6x5 + 37x7
x3 = 6x4 – 2x5 – 6x7
x6 = -3x7
From here I know you make the columns, but what I don't know is if I'm supposed to also solve the equations for x2, x4, x5, and x7, and make columns for those as well, which would give me a different dimension for the column space. Do I do that or do I stick with the current equations only and end up with a column space of 4?
Thank you for your help.
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Because the matrix is already in row-echelon form:
*
*The number of leading $1$'s (three) is the rank; in fact, the columns containing leading $1$'s (i.e., the first, third, and sixth columns) form a basis of the column space.
*The number of columns not containing leading $1$'s (four) is the dimension of the null space (a.k.a. the nullity).
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How many combinations from rolling 5 identical dice? Where, for example:
(1,3,1,4,6) is considered the same outcome as (1,1,3,6,4)
How many total outcomes are there?
Edit 1:
My hunch is that there are:
6 outcomes from choosing 1 dice to be missing.
6*5 = 30 outcomes from choosing 1 number to be the same and one missing from the 5 remaining.
6*5*4 = 120 from choosing 1 to be triple, 1 from 5 to be missing, and 1 from 4 to be missing.
6*5 = 30 from choosing 1 to be the quadruple and 1 from 5 to be remaining.
and 6 from choosing 1 to be all the same number.
Which gives 192 in total.
Edit 2:
Thanks JimmyK for correct answer: 252. Obtained by: C(6+6-1, 6-1) = C(10,5)
Can also get the answer by summing these possibilities, of which I missed out loads above:
C(6,1) + C(6,5) + C(6,1)*C(5,3) + C(6,2)*C(4,1) + C(6,1)*C(5,2) + C(6,1)*C(5,1) = 252
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Let $x_i$ be the number of dice with the $i$ face showing. Then, the $x_i$ are non-negative integers with $x_1+x_2+x_3+x_4+x_5+x_6 = 5$. This is now the standard Stars and Bars problem. You can simply use the formula given in that Wikipedia article to get the total number of solutions.
|
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general solution of elastic beam equation For the following equation;
$t^4 \dfrac{d^2u}{dt^2} + \lambda^2 u = 0, \quad \lambda >0, ~ t>0,$
where $u(t)$ is real valued function. by using the change of variables
$t=\dfrac{1}{\tau} , \quad u(t)=\dfrac{v(\tau)}{\tau} ,$
how can we find the general solution of the above equation.
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This shows a method to carry out the change of variable and the change of function, as requested in the wording. Then, you can solve the ODE which is obtained on a well-known form :
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Homology and topological propeties i have this theorem with it's proof but i don't understand the last part
They use this proposition:
My question is Why $\varphi^c\cap U_i$ is closed and pairwise disjoint ? where $\varphi^c=\lbrace x, \varphi(x)\leq c\rbrace$
Please thank you.
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The theorem is applied (see the middle in $6.6$ ) to $X=\varphi^c\cap C$ and $X_i=\varphi^c\cap U_i$ which is a closed subset in $X$.($X_i=X\cap(\cup_{j\neq i}U_j)^c$ where the c's power is the complementary ).
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Guidance or advice with $I=\int_0^{2\pi}\frac{1}{4+\cos t}dt$ Let
$$
\begin{align}
I=\int_0^{2\pi}\frac{1}{4+\cos t}dt
\end{align}
$$
I would like to evaluate this integral using cauchhy's Integral formula, I understand that I have to convert this into a form like $\int_{\gamma}\frac{f(z)}{z-z_0}dz$ .
I tried using $\cos t=\frac{e^{it}+e^{-it}}{2}$, but I didn't get anywhere. I haven't seen an example like this.
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Let $z=e^{it}$, then $dz=ie^{it}\ dt$ or $dt=\dfrac{dz}{iz}$, and $\cos t=\dfrac{e^{it}+e^{-it}}{2}=\dfrac{z+z^{-1}}{2}$.
\begin{align}
\int_0^{2\pi}\frac{1}{4+\cos t}dt&=\oint_C\frac{1}{4+\frac{z+z^{-1}}{2}}\cdot\frac{dz}{iz}\\
&=\frac2i\oint_C\frac{1}{z^2+8z+1}dz,
\end{align}
where $C$ is the circle of unit radius with its center at the origin. The poles are $z_1=-4-\sqrt{15}$ and $z_2=\sqrt{15}-4$, but only $z_2$ lies inside $C$. The rest is finding the residue at $z_2$.
|
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Number of images from $\mathbb{N}$ to {0, 1}.
Are the number of images from $\mathbb{N}$ to {0, 1} countably infinite or uncountably infinite?
I was thinking of counting in base 2 to make a bijection between $\mathbb{N}$ and {0, 1}. So, a table like this one:
Image | 0 1 2 3 ... <- Here is N
-------------------------------
0 | 0 0 0 0 ... <- And here is the image in {0,1}
1 | 1 0 0 0 ...
2 | 0 1 0 0 ...
3 | 1 1 0 0 ...
4 | 0 0 1 0 ...
...
So, right now, in the left column, I can identify every number of $\mathbb{N}$ with an image on the right. Which makes it countably infinite, by definition.
But a co-student, claimed that the prof said it was uncountably infinite. Any idea why?
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You can use that if $A\subseteq \mathbb{N}$ then $f(a)=1$ if $a\in A$ and $f(b)=0$ if $b\notin A$ then $\mid \{0,1\}^\mathbb{N}\mid=\mid P(\mathbb{N})\mid$ by Cantor's theorem, $\mid \mathbb{N}\mid<\mid P(\mathbb{N})\mid$.
|
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What is the probability that the maximum number of shots fired successively from a type A gun is $2$? A gun salute always takes place at the funeral of a military leader who has died in a certain country.
(The $21$ gun salute where $21$ rounds are fired - is the most common for the most senior military leaders and the number of rounds progressively gets smaller as the rank gets lower.)
Lance, who was an air marshal has died and his funeral is there tomorrow. At his funeral, a $14$ gun salute will be held (a total of $14$ rounds will be fired in succession) comprising of $5$ rounds from a Type A gun, $3$ rounds from a Type B gun and the rest from a Type C gun. => $6$ right?
If the guns are fired randomly, what is the probability that the maximum number of shots fired successively from a type A gun is $2$?
Answer: $\frac{121968}{168168} = \frac{66}{91} =0.7253$
I was trying to take the approach of taking the total number of possible ways ($\frac{14!}{{5!}{3!}{6!}} = 168168$) and subtracting the ways where you take all the type A out and then place them back in but in between others in ways that there are more than 2 successive fires.
I got confused because I don't know how to account for the repeated type A after putting them back in in such a fashion.
I had:
$3$ A's in succession and $2$ A's in succession
$3$ A's in succession and $2$ A's not in succession
$4$ A's in succession and $1$ A
$5$ A's in succession
I tried to approach it like this:
I'm just going to show you what I did to attempt to get the numerator of the answer ($121968$) before dividing it by $168168$ (obviously to get the probability).
$168168 - \frac{{9!}{10\choose2}{3!}{2!}{2!}}{5!3!6!} - \frac{{9!}{10\choose3}{3!}{3!}}{5!3!6!} - \frac{{9!}{10\choose2}{4!}{2!}}{5!3!6!} - \frac{{9!}{10\choose1}{5!}}{5!3!6!}$
$= 168168 - 756 - 3024 - 1512 - 840 = 162036$
What am I doing wrong? How do they get $121968$
|
Your expression is mostly right. Take the case "3 A's in succession and 2 A's in succession". We have 2 subcases, either the AAA comes first or the AA comes first. These are equivalent.
Let X denote a B or a C (whether it's B or C, we decide at the end). We have 9 X's to place in 3 slots as shown:
AAA AA
^ ^ ^
Except we must place an X in the middle slot. So we have 8 X's and 2 blocks of A's to arrange; one possible arrangement is (X AAA(X) X X X AA X X X X). Total # of ways to do this is 10C2.
Finally we choose B's and C's among the X's. There are 9C3 ways to do this. So the total # of ways for this case is $2\binom{10}{2}\binom{9}{3}$. The other cases are similar.
|
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Manifolds, charts and coordinates Let's consider the manifold $S^1$
It is well known that we need two charts to cover this manifold.
Nonetheless, we can cover the full space using a single coordinate $\theta$ which is just the angle from the center.
Now, is this a general feature? I mean, is it always possible to have in every manifold a single coordinate set that cover points that are in different charts, just as in $S^1$?
|
Perhaps you're "really" asking whether or not every $n$-manifold $M$ has $\mathbf{R}^{n}$ as a covering space, i.e., whether there's a single (smooth) map $\pi:\mathbf{R}^{n} \to M$ whose restrictions to suitably small subsets define coordinate charts on $M$.
If so, the answer is "no"; even among spheres, only $S^{1}$ has this property.
|
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Simulate repeated rolls of a 7-sided die with a 6-sided die What is the most efficient way to simulate a 7-sided die with a 6-sided die? I've put some thought into it but I'm not sure I get somewhere specifically.
To create a 7-sided die we can use a rejection technique. 3-bits give uniform 1-8 and we need uniform 1-7 which means that we have to reject 1/8 i.e. 12.5% rejection probability.
To create $n * 7$-sided die rolls we need $\lceil log_2( 7^n ) \rceil$ bits. This means that our rejection probability is $p_r(n)=1-\frac{7^n}{2^{\lceil log_2( 7^n ) \rceil}}$.
It turns out that the rejection probability varies wildly but for $n=26$ we get $p_r(26) = 1 - \frac{7^{26}}{2^{\lceil log_2(7^{26}) \rceil}} = 1-\frac{7^{26}}{2^{73}} \approx 0.6\%$ rejection probability which is quite good. This means that we can generate with good odds 26 7-die rolls out of 73 bits.
Similarly, if we throw a fair die $n$ times we get number from $0...(6^n-1)$ which gives us $\lfloor log_2(6^{n}) \rfloor$ bits by rejecting everything which is above $2^{\lfloor log_2(6^{n}) \rfloor}$. Consequently the rejection probability is $p_r(n)=1-\frac{2^{\lfloor log_2( 6^{n} ) \rfloor}}{6^n}$.
Again this varies wildly but for $n = 53$, we get $p_r(53) = 1-\frac{2^{137}}{6^{53}} \approx 0.2\%$ which is excellent. As a result, we can roll the 6-face die 53 times and get ~137 bits.
This means that we get about $\frac{137}{53} * \frac{26}{73} = 0.9207$ 7-face die rolls out of 6-face die rolls which is close to the optimum $\frac{log 7}{log6} = 0.9208$.
Is there a way to get the optimum? Is there an way to find those $n$ numbers as above that minimize errors? Is there relevant theory I could have a look at?
P.S. Relevant python expressions:
min([ (i, round(1000*(1-( 7**i ) / (2**ceil(log(7**i,2)))) )/10) for i in xrange(1,100)], key=lambda x: x[1])
min([ (i, round(1000*(1- ((2**floor(log(6**i,2))) / ( 6**i )) ) )/10) for i in xrange(1,100)], key=lambda x: x[1])
P.S.2 Thanks to @Erick Wong for helping me get the question right with his great comments.
Related question: Is there a way to simulate any $n$-sided die using a fixed set of die types for all $n$?
|
In the long run, just skip the binary conversion altogether and go with some form of arithmetic coding: use the $6$-dice rolls to generate a uniform base-$6$ real number in $[0,1]$ and then extract base-$7$ digits from that as they resolve. For instance:
int rand7()
{
static double a=0, width=7; // persistent state
while ((int)(a+width) != (int)a)
{
width /= 6;
a += (rand6()-1)*width;
}
int n = (int)a;
a -= n;
a *= 7; width *= 7;
return (n+1);
}
A test run of $10000$ outputs usually requires exactly $10861$ dice rolls, and occasionally needs one or two more. Note that the uniformity of this implementation is not exact (even if rand6 is perfect) due to floating-point truncation, but should be pretty good overall.
|
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best approximation polynomial $p_1(x)\in P_1$ for $x^3$ I want to find a best approximation polynomial $p_1(x)\in P_1$ for $f(x)=x^3$ in $[-1,1]$ w.r.t. $||\cdot||_{\infty}$. I want to use Chebyshev polynomial to do that, but I don't know how to hang on.
|
Let $p(x)$ will be polynomial which you find. First note that $f(x)-p(x)$ is polynomial of degree $3$ with leading coefficient $1$. You know that among the polynomials of degree $3$ with leading coefficient $1$
$$w(x)=\frac{1}{4}T_{3}(x)$$
is the one of which the maximal absolute value on the interval $[−1, 1]$ is minimal, see.
So you know that you must have $f(x)-p(x)=\frac{1}{4}T_3(x)$, so $p(x)=x^3-\frac{1}{4}T_3(x)$.
|
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Interval of existence of a certain first-order ODE
Without solving the following initial value problem, determine the interval in which the solution is certain to exist: $$\dfrac{dy}{dx}+(\tan x)y=\sin x, \ \ \ y \left (\frac{\pi}{4} \right )=0.$$
Please help me also solve (if you can) under what conditions the solution of the first-order ordinary differential equation exists?
This question is different is differenr from the previously asked queston in a way that it has two parts furthermore, the equation given is completely different from the previously asked question.
It is sad that none of the repliers have answered the question yet whereas they have accepted that they know answers. I am extremely disappointed with this forum which was suggested to me by my friend. Leaving it right now. Good Bye.
|
Hint: Multiply the entire equation by $\sec{x}$ and note that
$$\frac{d}{dx}\left[\sec{x}\,y(x)\right]=\sec{x}\,y^\prime{(x)}+\sec{x}\tan{x}\,y(x).$$
|
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explicit solution for second order homogeneous linear differential equation If we consider the equation
$(1-x^2)\dfrac{d^2y}{dx^2} -2x \dfrac{dy}{dx}+2y=0, \quad -1<x<1$
how can we find the explicit solution, what should be the method for solution?
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There is one obvious particular solution which is $y=c_1 x$. The second one is much less obvious to me but, using a CAS, I found as general solution $$y=c_1 x+c_2 \left(x \tanh ^{-1}(x)-1\right)$$ I hope and wish this will give you some ideas. As said by achille hui, beside the solution in terms of Legendre polynomials, you can obviously use Frobenius method which leads to the solution I wrote.
|
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How to check for convexity of function that is not everywhere differentiable? I have a question. I have just been introduced to the subject of convex sets and convex functions.
I read this in wikipedia that a practical test for convexity is -
to check whether the 2nd derivative (Hessian matrix) of a continuous differentiable function in the interior of the convex set is non-negative (positive semi-definite).
So how to check for the convexity of functions like $f(x)=|x|$ which is differentiable at all points except at $x=0$ which coincidentally is actually its global minimum?
Thanks all for answering.
|
One option is to check directly that the definition of a convex function is satisfied.
It's useful to know that any norm on $\mathbb R^n$ is a convex function. Proof: If $x,y \in \mathbb R^n$ and $0 \leq \theta \leq 1$, then
\begin{align*}
\| \theta x + (1 - \theta) y \| & \leq \| \theta x \| + \| (1 - \theta) y \| \\
&= \theta \| x \| + (1 - \theta) \| y \|.
\end{align*}
This shows that the definition of a convex function is satisfied.
When $n = 1$, the $2$-norm is just the absolute value function $f(x) = | x |$. This shows that the absolute value function is convex.
A bunch of other techniques for recognizing convex functions are explained in the book Boyd and Vandenberghe (free online).
|
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|
Estimate the discounts... Here is a problem on discount estimation,,
If marked price of an article is $\frac{8}{5}$ times cost price and its selling price is at least $\frac{4}{5}$th the cost price then discount percentage is...
multiple choices are possible...
options:
A) $25$
B) $40$
C) $50$
D) $60$
E) $80$
I am somewhat confused...
Thanks in advance....
|
HINT
Say the cost price is $x$,
the maximum possible discount price = $\frac{8}{5}(x) - \frac{4}{5}(x)$
$\implies $maximum possible discount percent = $\dfrac{\frac{8}{5}(x) - \frac{4}{5}(x)}{\frac{8}{5}(x)}\times 100$
Simplify and see which options are not greater than this percent
|
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|
Show an open set has no minimum Suppose $A$ is an open subset of the real numbers. How to prove $A$ has no minimum?
I guess it should be done with Proof-by-contradiction, but I don't get anywhere at this point.
|
Suppose $m \in A$ is the minimum of A. By definition of open set, you have that $(m-\delta , m + \delta) \subseteq A$ with $\delta > 0$ sufficiently small, then $m- \frac{\delta}{2} \in A$ but this is a contradiction (because we have found an element in A smaller than $m$)
|
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|
Solving integral that contain exponential function and lower incomplete gamma function I have the following integral;
$$y=\int_0^\infty\frac{e^{-xf}}{m+x}\gamma(a,hx)~dx$$
where $f,m,h\in\mathbb{R}^+$ , $a\in\mathbb{N}$ , $\gamma\left(a,h x\right)$ is the lower incomplete gamma function
Can anyone help me how to solve it?
Thank you very much
|
Hint:
$\int_0^\infty\dfrac{e^{-xf}}{m+x}\gamma(a,hx)~dx$
$=\int_0^\infty\dfrac{e^{-fx}}{x+m}\sum\limits_{n=0}^\infty\dfrac{(-1)^nh^{n+a}x^{n+a}}{n!(n+a)}dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nh^{n+a}x^{n+a}e^{-fx}}{n!(n+a)(x+m)}dx$
$=\int_m^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nh^{n+a}(x-m)^{n+a}e^{-f(x-m)}}{n!(n+a)x}d(x-m)$
$=\int_m^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nh^{n+a}e^{fm}(x-m)^{n+a}e^{-fx}}{n!(n+a)x}dx$
$=\int_m^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^{n+a}\dfrac{(-1)^nh^{n+a}e^{fm}C_p^{n+a}(-1)^{n+a-p}m^{n+a-p}x^pe^{-fx}}{n!(n+a)x}dx$
$=\int_m^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^{n+a}\dfrac{(-1)^{p+a}(n+a-1)!h^{n+a}m^{n+a-p}e^{fm}x^{p-1}e^{-fx}}{n!p!(n+a-p)!}dx$
|
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|
Show that $(n+1)^{n+1}>(n+2)^n$ for all positive integers Show that:
$(n+1)^{n+1}>(n+2)^n$ holds for all positive integers
I tried using induction:
for $n=1$ we have 4>3 then for $n+1$ we have to show that $(n+2)^{n+2}>(n+3)^{n+1}$ and here I stuck
|
Since $(1+x)^n \geq 1 + nx, \forall x\geq -1$
$(\dfrac{n+1}{n+2})^n=(1-\dfrac{1}{n+2})^n \geq 1 -\dfrac{n}{n+2} = \dfrac{2}{n+2}$
$(n+1)\dfrac{2}{n+2} >1$
|
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|
Let $J$ be a $k \times k$ jordan block, prove that any matrix which commutes with $J$ is a polynomial in $J$ Let $J$ be a $k \times k$ jordan block, prove that any matrix which commutes with $J$ is a polynomial in $J$.
I appreciate your hints, Thanks
|
2 by 2 only. i really hope you will try the 3 by 3 case by hand, analogous to this:
Aright, eigenvalue $w,$
$$ J =
\left(
\begin{array}{rr}
w & 1 \\
0 & w
\end{array}
\right),
$$
trial
$$ M =
\left(
\begin{array}{rr}
a & b \\
c & d
\end{array}
\right),
$$
Next
$$ JM =
\left(
\begin{array}{rr}
wa + c & wb+d \\
wc & wd
\end{array}
\right),
$$
but
$$ MJ =
\left(
\begin{array}{rr}
wa & wb+a \\
wc & wd+ c
\end{array}
\right).
$$
We find that $JM = MJ$ precisely when
$$ c = 0 \; \; \mbox{AND} \; \; a=d. $$ Notice that $w$ does not appear.
Under these conditions,
$$ M =
\left(
\begin{array}{rr}
a & b \\
0 & a
\end{array}
\right).
$$
That means that $$ M = (a-bw)I + b J, $$ that is a polynomial in $J.$
|
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|
Circumference of separate circle So I have been out of Algebra for a while now. I am trying to help my wife prep for an entrance exam and we ran across this in the practice test:
A concrete walkway 2 feet wide surrounds a circular pool of radius 5 feet.
Find the area of the concrete walkway.
I know that the area of the pool is $153.9384$. ($A = \pi r^2$.)
How do you find the outer circle's square footage though based on that?
No answer please; just the formula. This is a practice test and we want to use the formula not the answer.
|
In this kind of problem, there is an exterior larger shape enclosing an interior smaller shape. Taking the difference between the enclosed areas of the two will give you the area of the border.
So subtract the areas of the "small circle" (just the pool) from the "large circle" (which is basically circle formed by the outer border of the walkway).
This method generalises to any regular shape - e.g. rectangular borders surrounding another rectangle and even irregular shapes where you basically have to be given both areas to work out the area of the border.
|
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|
Showing that the Binary Icosahedral Group (given by a presentation) has order $120$. Say we have a group generated by $a, b$, with the relations $(ab)^2=a^3=b^5$ (note that these are not necessarily equal to the identity). How do I show that the group has $120$ elements? Without having orders given to any of the generators, why can't there be infinitely many elements?
This problem is from Stillwell's "Mathematics and its History," after Section 22.8. I'm first asked to show that adding the relation that all $3$ equal the identity gives the icosahedral group, isomorphic to $A_5$, and then to show that this implies the group without that relation has at least $60$ elements. Then, a parenthetical after that problem says "Harder: Show that it in fact has $120$ elements".
|
Consider the group $G=\langle a, b, c; a^2=b^3=c^5=abc\rangle^{\ast}$. Clearly, $\langle abc\rangle$, which is the same subgroup as $\langle a^2\rangle$, as $\langle b^3\rangle$, and as $\langle c^5\rangle$, is a central subgroup of $G$. As we know from the earlier part of the problem, quotienting out this subgroup gives $A_5$, which has order $60$. Hence, in order to prove the result it is sufficient to prove that $(abc)^{2}=1$ and that $G\not\cong A_5$.
A proof of these facts can be found on the Groupprops website. Their proof shows that $G$ contains a subgroup which is a homomorphic image of the dicyclic group $\langle x, y, z; x^2=y^2=z^5=xyz\rangle$, and the subgroup is embedded such that $x_0y_0z_0=abc$. In a dicyclic group, $(x_0y_0z_0)^2=1$, hence $(abc)^2=1$ as required. The subgroup is given by the following embedding.
$$
\begin{align*}
x_0&\mapsto a^{-1}bab^{-1}a\\
y_0&\mapsto a\\
z_0&\mapsto cac^{-3}
\end{align*}
$$
I will leave you to prove that $G\not\cong A_5$. There are a number of ways of doing this. For example, you could prove that $abc\neq 1$ in $G$, or you could "realise" $G$ as a group of matrices (hint: $\operatorname{SL}_2(5)$) or permutations$^{\dagger}$. Read the wikipedia page to get yourself started.
$^{\ast}$The presentation given in this answer differs from the one you gave. However, proving that $(abc)^2=1$ (which is the purpose of this answer) proves that these presentations define isomorphic groups. Can you see why this is?
$^{\dagger}$Another possible, elegant, way would have been to have found a non-trivial homomorphic image of $G$ which was not $A_5$. However, in reality $G$ has only one proper, non-trivial subgroup so this won't work!
|
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|
Bactracking to find compound interest I'm trying to find what percentage 5000 dollars compounding monthly over 120 months will be if the final sum will be 7000 dollars.
So:
7000=5000(1+r/12)^120
When working backwards to find r I always get the percentage = 0.2% but I did trial and error and know the actual answer is about 3.37%
How do I work backwards to find the correct r?
|
We have
$$1.4=\left(1+\frac{r}{12}\right)^{120}.$$
Now we can use a calculator to find $(1.4)^{1/120}$. I get about $1.0028079$. Subtract $1$, multiply by $12$. I get about $0.0336944$.
Remarks: $1.$ I did this on an ordinary calculator. But it could also have been done by Google. Just type in (1.4)^(1/120). It gives $1.00280787001$.
You can even let Google do the whole calculation. Type in 12*((1.4)^(1/120)-1).
$2.$ Note that we have calculated the nominal annual rate $r$. The effective annual rate is somewhat larger. It is $\left(1+\frac{r}{12}\right)^{12}-1$. In our case, to find the effective annual rate, we do not need to compute $r$. We can simply find $(1.4)^{1/10}-1$. This turns out to be about $0.0342197$. The difference between nominal and effective becomes significantly larger when interest rates are high.
|
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|
How to show the surjectivity of $f(x)=x^5$ on $\mathbb R$? Sasy $f:\mathbb R\to\mathbb R$ define by $f(x)=x^5$
This is definitely injective as $x_1^5=x_2^5 \implies x_1=x_2$
I say it is surjective because for all really $x$ there is all real $y$,
$x \in \mathbb R$ there $y \in \mathbb R$?
|
To show surjectivity, you show that for each $y$, there is an $x$ such that $f(x) = y$. This means you need to find an $x$ : $x^5 = y$.
|
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|
Partial fraction decomposition of type $1/(x^2+k)$ I know that partial fraction of this can be written as:
$$\frac{3x}{(1+x)(2+x)}=\frac{-3}{1+x}+\frac{6}{2+x}$$
Which can be done in these ways:
$$\frac{3x}{(1+x)(2+x)}=\frac{A}{1+x}+\frac{B}{2+x}\implies3x=A(2+x)+B(1+x),\forall\;x$$
And now solving it to get A and B.Also someone told me that write the fraction without that denominator whose coeffecient you want to know and put root of denominator, for e.g.:
$$\text{coefficient of $1/(1+x)$ i.e. A}=\frac{3x}{2+x}|_{x+1=0\implies x=-1}=\frac{-3}{1}$$
I know that this is just the first way disguised in an easy form, now what when I need to decompose:
$$\frac1{(x-3)(x^2-1)}\text{ or }\frac1{(x-3)(x^2+1)}$$
There exist multiple root of first and root to second doesn't exist.
Now what to do, adopt the first method? Is there any easy method for these?
|
These are two different beasts, at least on the reals. First, $$\frac1{(x-3)(x^2-1)}=\frac{a}{x-3}+\frac{b}{x-1}+\frac{c}{x+1},$$ for some suitable real numbers $(a,b,c)$. Second, on the reals, $$\frac1{(x-3)(x^2+1)}=\frac{A}{x-3}+\frac{Bx+C}{x^2+1},$$ for some suitable real numbers $(A,B,C)$. But, on the complex numbers, $$\frac1{(x-3)(x^2+1)}=\frac{D}{x-3}+\frac{E}{x+\mathrm i}+\frac{F}{x-\mathrm i}.$$ for some suitable complex numbers $(D,E,F)$. Naturally, $(D,E,F)$ is related to $(A,B,C)$ since $$D=A,\quad E+F=B,\quad \mathrm i\,(F-E)=C,$$ which is enough to show that $D$ is real and that $F=\bar E$ with $E=\frac12(B+\mathrm i\,C)$ and $F=\frac12(B-\mathrm i\,C)$.
Recall that, in full generality, every polynomial on the reals is the product of some factors $(x-a)^n$ and $(x^2+bx+c)^m$ with $b^2\lt4c$ hence every rational fraction with denominator $$\prod_{k}(x-a_k)^{n_k}\cdot\prod_\ell(x^2+b_\ell x+c_\ell)^{m_\ell}$$ is the sum of a polynomial and of some terms $$\frac{A_{k,p}}{(x-a_k)^p}\qquad\text{and}\qquad\frac{B_{\ell,q}x+C_{\ell,q}}{(x^2+b_kx+c_k)^q},$$ for each $k$ and $\ell$, with $1\leqslant p\leqslant n_k$ and $1\leqslant q\leqslant m_\ell$, respectively.
|
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|
Complex Equations The Equation:
$$ z^{4} -2 z^{3} + 12z^{2} -14z + 35 = 0 $$
has a root with a real part 1, solve the equation.
When it says a real part of 1, does this mean that we could use (z-1) and use polynomial division to extract the other rots? Hence:
$$ (z^{4} -2 z^{3} + 12z^{2} -14z + 35 ) / (z-1) $$
But i don't get the right answer, any other approaches I should try?
|
This is not so different from Lucian's approach, and may (or may not) feel computationally more straightforward.
This equation has real coefficients so that roots come in complex conjugate pairs. We are give that there are roots $z=1\pm bi$. Now let $y=z-1$ so that $z=y+1$ whence $$(y+1)^4-2(y+1)^3+12(y+1)^2-14(y+1)+35=y^4+2y^3+12y^2+8y+32=0$$ has the purely imaginary roots $y=\pm bi$
Putting this in and equating real and imaginary parts gives $$b^4-12b^2+32=0$$ and $$-2b^3+8b=0$$
The first equation factorises as $(b^2-4)(b^2-8)=0$ and the second as $-2b(b^2-4)=0$ so to be consistent we need $b^2=4$, $y=\pm 2i$ and $z=1\pm 2i$.
This gives the factorisation $$(z^2-2z+5)(z^2+7)=0$$ for the original equation, which enables the remaining roots to be found.
|
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|
Converting repeating decimal to fraction How do I convert $0.297$ to a fraction, if the $2$ and $9$ are repeating? The non-repeating number is in the middle, so I am not sure how to proceed from here.
|
Note that parenthesized "phrases" should be considered to repeat ad infinitum.
$.297(29)=.(92)-.632$.
$.(92)={{92}\over{{10}^2}-1}={{92}\over{99}}$.
$.632={{632}\over{1000}}={{79}\over{125}}$.
So, the sought-after fraction $={{92}\over{99}}-{{79}\over{125}}=
{{92·125-79·99}\over{99·125}}={{11500-7821}\over{12375}}={{3679}\over{12375}}$.
|
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|
Simple use of log I am struggling to see how we can go from the first expression to the second:
$$\begin{align}
2\log_3 12 - 4\log_3 6 &= \log_3 \left ( \frac{4^2 \cdot 3^2}{2^4 \cdot 3^4} \right )\\
&= \log_3 (3^{-2}) = -2
\end{align}$$
|
First note that
\[ \log_{b}(x^{n})=n\log_{b}(x) \]
\[ \log_{b}(x)-\log_{b}(y)=\log_{b}\left(\frac{x}{y}\right) \]
Here are the steps
\[
2\log_{3} 12 - 4\log_{3} 6 = 2\log_{3} (4\cdot 3) - 4\log_{3} (2\cdot 3)
\]
\[
=\log_{3} (4\cdot 3)^{2} - \log_{3} (2\cdot 3)^{4} = \log_{3} (4^{2}\cdot 3^{2}) - \log_{3} (2^{4}\cdot 3^{4})
\]
\[
= \log_{3} \frac{4^{2}\cdot 3^{2}}{2^{4}\cdot 3^{4}} = \log_{3} \frac{16}{16\cdot 3^{2}}=\log_{3} \frac{1}{3^{2}}=\log_{3} 3^{-2}=-2\log_{3} 3 = -2
\]
|
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Can I prove this, or hopeless? Deviating too much from mean Can I prove this:
We have a sequence of vectors $\left(X_i(n)\right)$ for $i=1,\ldots,t$, where $n\rightarrow \infty$. $t$ does depend on $n$ and is Chosen such that $1 \ll t \ll n$, for instance, take $t=\log(n)$.
We know that $$X(n):=\sum_{i=1}^t X_i(n).$$
Can we prove that $$X(n)= \left(\sum_{i \ : P(i) } X_i(n) \right)(1+o(1))$$
where $P(i)$ is a predicate with
$$P(i)=1 \text{ iff } X_i(n) = \Omega\left(\frac{X(n)}{t}\right),$$
thus $0$ when $X_i(n)$ is asymptotically smaller than $\frac{X(n)}{t}$.
Stated differently, can we prove that $X(n)$ depends only on the Terms $X_i(n)$ for which the predicate is true?
This seems to be rather intuitive. But I cannot prove it. Would be glad if you could provide a proof.
|
Define
$$ \lambda:=\frac{\sum_{i\not\in P(i)} X_i(n)}{\sum_{i\in P(i)} X_i(n)} $$
and note that $\sum_{i\not\in P(i)} X_i(n)=\mathcal{O}(X(n)/t)$ and $\sum_{i\in P(i)} X_i(n)=\Omega(X(n)/t)$.
Your desired result requires that $\lambda=o(1)$. But this only necessarily follows if $\mathcal{O}(f)/\Omega(f)=o(1)$ for all functions $f$, which isn't true. However, the following (weaker) statement is obviously true
$$X(n)= \left(\sum_{i \ : P(i) } X_i(n) \right)(1+\mathcal{O}(1))$$
The issue is that some of the $X_i(n)$ terms may grow asymptotically to the average $\frac{1}{t}\sum_{i=1}^t X_i(n)$ from below ($\mathcal{O})$ while others do so from above ($\Omega$).
|
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|
What advanced methods in contour integration are there? It is well known how to evaluate a definite integral like
$$
\int_{0}^\infty dx\, R(x),
$$
where $R$ is a rational function, using contour integration around a semicircle or a keyhole.
Most complex analysis books only treat well-known and easy examples like this. What I am looking for is examples of integrals that can be evaluated using contour integration, but require more creative tricks, unusual contours, etc. and are not treated in common textbooks.
Useful answers are applicable not just to one integral, but are somewhat general. Needless to say, answers do not have to include the full computation to be useful.
|
Most interesting examples come from physics, often quantum mechanics. There are interesting methods around asymptotic expansions (when you can't solve an ODE exactly, but still want to know the behavior of the solutions at large values of some parameter, for example), or representation of solutions of ODEs using contour integrals (where you use a Laplace transform, then choose contours to find linearly independent solutions, or at least their asymptotic behavior).
Check out this EdX class for the "next level" in contour integration. No more using just a tame Hankel contour - the class deforms branch cuts to wrap around multiple branch points while preserving steepest descent directions, and uses contours that lie not on one, but multiple Riemann sheets of multi-valued functions! I couldn't find many references that go so deep in using contour integrals, with so many difficult examples. If you master the exercises in this class, you are set for a while. https://www.edx.org/course/complex-analysis-physical-applications-misisx-18-11x
|
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|
Convergence of $\sum \frac{1}{\ln n}(\sqrt{n+1} - \sqrt{n})$ Show that
$$ \sum\frac{1}{\ln n}(\sqrt{n+1} - \sqrt{n})$$
Converges.
I've tried the telescopic property or even write it as
$$\sum \frac{1}{\ln n (\sqrt{n+1}+\sqrt{n})}$$
But didnt help.
Thanks in advance!
|
It diverges. By comparison test,
\begin{align}
\frac{\sqrt{n+1} - \sqrt{n}}{\ln{n}} &= \frac{1}{\ln{n}\left(\sqrt{n+1} + \sqrt{n} \right)} \\
&> \frac{1}{n}
\end{align}
for $n > 100$ (say), and the harmonic series diverges.
|
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|
Quotient set cardinal in $\mathbb{Z}_{12}$ In $\mathbb{Z}_{12}$ define the equivalence relation xRy if $x^2 = y^2$
Then what is the cardinal of the quotient set?
|
Since there are only 12 elements, you can simply check this element by element. Clearly $\bar{0},\bar{1},\bar{2},\bar{3}$ form distinct equivalence classes since their squares are all different. Since $\bar{4}^2=\bar{2}^2, \bar{5}^2 = \bar{1}^2, \bar{6}^2=\bar{0}^2, \bar{7}^2=\bar{1}^2,\bar{8}^2=\bar{2}^2,\bar{9}^2=\bar{3}^2,\bar{10}^2=\bar{2}^2,\bar{11}^2=\bar{1}^2$, we know that this is a complete list. Therefore there are only 4 equivalence classes.
|
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|
Can the winding number be infinite? Let $z$ be a point in the complex plane, and $\gamma$ be a closed curve. Is it possible that
$$n(\gamma,z) = \frac{1}{2\pi i}\int_\gamma \frac{dw}{w-z}$$
becomes unbounded? In other words, is it possible to find a curve $\gamma$ such that it winds around a fixed point infinitely many times?
|
Without loss of generality, we show that the winding number around $0$ is finite (the general proof follows by translating the complex plane).
Remember the alternative definition of the winding number: if $\gamma$ is a closed path, we can choose a continuous choice of argument $\theta:[0,1] \to \mathbb R$ such that $$\frac{\gamma(t)}{|\gamma(t)|}=\mathrm e^{i\theta(t)}$$
The winding number can then be defined as$$w(\gamma, 0)=\frac{1}{2\pi}\big(\theta(1) - \theta(0)\big)$$
But $\theta$ is a continuous function on a compact set so it is bounded. In particular the winding number is finite.
|
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Folland's proof of the Hahn Decomposition. Minor error? Theorem 3.3 of Folland's Real Analysis (ed 2) is the Hahn decomposition theorem. In the proof he assumes that the signed measure $\nu$ he is considering does not take the value $-\infty$. Then he argues:
Let $m$ be the supremum of $\nu(E)$ as $E$ ranges over all positive
sets; thus there is a sequence $\{P_j\}$ of positive sets such that
$\nu(P_j) \to m$. Let $P = \cup_1^\infty P_j$. By Lemma 3.2 and
Proposition 3.1, $P$ is positive and $\nu(P)=m$; in particular, $m < \infty.$
The justification for the claim that $m<\infty$ is not clear to me. Could it be the case that Folland wanted to assume that $\nu$ does not take on the value $\infty$ in which case I think that the rest of the proof would work. Or am I missing something?
|
Yes, you are right. Folland did give an errata for his book which shows that $-\infty$ is a typo in the proof and it should be $+\infty$ instead.
|
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|
Compact opens in sober $T_1$ are closed? I am trying to establish some basic facts about spectral spaces. In relation to this I am looking for a proof of, or a counter example to, the statement that compact open subsets of a sober $T_1$ space are closed.
Recall that a topological space is sober if every closed irreducible subset has a generic point. Whence in a sober $T_1$ space the closed irreducible subsets are precisely the singletons.
Any help is appreciated.
|
You can find counterexamples in some fairly typical sober $T_1$ spaces that are not Hausdorff. For example, consider $\mathbb{N}\cup\{x,y\}$ where every point of $\mathbb{N}$ is isolated and the neighbourhoods of $x$ and $y$ are the cofinite subsets containing $x$ and $y$ respectively. Then $\mathbb{N}\cup\{x\}$ is compact open but not closed.
|
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|
How to find $\int {t^n \, e^{t}}\mathrm dt$? Consider:$$\int {t^n e^{t}}\ \mathrm dt$$
is there any closed formula for this?
W|A gave me this but I don't know what is Gamma function:
$$\int {t^n e^t\ \mathrm dt} = (-t)^{-n}\ t^n\ \Gamma(n+1, -t)+ \text{constant}$$
|
It is easily proved by induction and integration by parts that
$$
\int {t^n e^{t}}\,dt = p_n(t) e^t
$$
where $p_n$ is a polynomial of degree $n$ and
$$
p_{n+1}(t) = t^{n+1}-(n+1)p_n(t)
$$
This will give you the formula mentioned by gammatester.
|
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|
First order ODE: $tx'(x'+2)=x$ $$tx'(x'+2)=x$$
First I multiplied it:
$$t(x')^2+2tx'=x$$
Then differentiated both sides:
$$(x')^2+2tx'x''+2tx''+x'=0$$
substituted $p=x'$ and rewrote it as a multiplication
$$(2p't+p)(p+1)=0$$
So either $(2p't+p)=0$ or $p+1=0$
The first one gives $p=\frac{C}{\sqrt{T}}$
The second one gives $p=-1$. My question is how do I take the antidervative of this in order to get the answer for the actual equation?
|
EDIT:
$x=t(x')^2+2tx'$
$p=x'$
$x=tp^2+2tp$
We differentiate in respect to $t$:
$p=p^2+t2pp'+2p+2tp' \Rightarrow p'(2tp+2t)=(p-p^2-2p) \Rightarrow p'(p+1)2t=-(p^2+p) \Rightarrow p'(p+1)2t=-p(p+1) \Rightarrow p'(p+1)2t+p(p+1)=0 \Rightarrow (p+1)(2tp'+p)=0 \\ \Rightarrow p+1=0 \text{ or } 2tp'+p=0 \\ \Rightarrow p=-1 \text{ or } p'=-\frac{1}{2t}p \\ \Rightarrow p=-1 \text{ or } \frac{p'}{p}=-\frac{1}{2t} \\ \Rightarrow p=-1 \text{ or } \ln{p}=-\frac{1}{2}\ln{t} +c\\ \Rightarrow p=-1 \text{ or } \ln{p}=\ln{t^{-\frac{1}{2}}} +c\\ \Rightarrow p=-1 \text{ or } p= \pm e^c \frac{1}{\sqrt{t}} \\ \Rightarrow x'=-1 \text{ or } x'= \pm e^c\frac{1}{\sqrt{t}} \\ \Rightarrow x(t)=-t+c_1 \text{ or } x(t)=2 C\sqrt{t}+c_2, \text{ where } C= \pm e^c$.
$$$$
*
*$x(t)=-t+c_1 \Rightarrow x'=-1$
Replacing this at the initial equation we get:
$x=t(x')^2+2tx' \Rightarrow -t+c_1=t-2t \Rightarrow -t+c_1=-t \Rightarrow c_1=0$
Therefore, $x(t)=-t$.
$$$$
*
*$x(t)=2 C\sqrt{t}+c_2 \Rightarrow x'(t)=\frac{C}{\sqrt{t}} $
Replacing this at the initial equation we get:
$x=t(x')^2+2tx' \Rightarrow 2 C\sqrt{t}+c_2=t\frac{C^2}{t}+2t\frac{C}{\sqrt{t}} \Rightarrow 2C \sqrt{t}+c_2=C^2+\frac{2Ct}{\sqrt{t}} \\ \Rightarrow 2Ct+c_2\sqrt{t}=C^2\sqrt{t}+2Ct \Rightarrow c_2\sqrt{t}=C^2\sqrt{t} \Rightarrow c_2=C^2$
Therefore, $x(t)=2 C\sqrt{t}+C^2$.
|
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Convergent or Divergent Integral Convergent or Divergent?
$$\int_0^1 \frac {dx}{(x+x^{5})^{1/2}} $$
I have problem with the fact that if we have integration from 0 to a say and a to infinity.
How does this change the way we do the comparison test ?
I have in my textbook: If we have from 0 to a and compare with the function
$$ \frac {1}{(x^{3})} $$ then the exponent of x should be less than 1 for it too be convergent. Which I would say is Divergent. But I am wrong?
|
Set $x=y^2$, then $dx=2ydy$. So the integral becomes:
$$I=\int_0^1 \frac {1}{(x+x^{5})^{1/2}}dx=\int_0^1 \frac {2y}{(y^2+y^{10})^{1/2}}dy=\int_0^1 \frac {2}{(1+y^5)^{1/2}}dy$$
Thus the integral is convergent.
|
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|
Reduction modulo p I am going to begin the Tripos part III at Cambridge in October (after going to a different university for undergrad) and have been preparing by reading some part II lecture notes.
Here is an extract from some Galois theory notes:
'$f(X) = X^4 + 5X^2 − 2X − 3 = X^4 + X^2 + 1 = (X^2 + X + 1)^2$ (mod 2)
$f(X) = X^4 + 5X^2 − 2X − 3 =X^4 + 2X^2 + X = X(X^3 + 2X + 1)$ (mod 3)
So f is irreducible, since f = gh implies deg g = 1 or deg g = 2, which is impossible by
reduction modulo 2 and 3, respectively.'
Could someone please explain the last sentence? Why is deg g=1 or 2 impossible by reduction modulo 2/3?
Thank you in advance.
|
First, note that if a polynomial $f$ is reducible, then you could witness these same factors modulo any prime. For instance, $x^2 - 1$ is reducible, and we see this mod 2: $(x+1)^2$ and mod 3: $(x+1)(x+2)$. Of course, it is possible that an irreducible factor of $f$ were further reduced mod $p$, for instance $x^2+1$ will appear as $(x+1)^2$ mod $2$.
Thus, if $f$ has an irreducible factor of degree $n$ it means that modulo any prime $p$ there must be a subset of the reducible factors such that the degree sum up to $n$.
In this particular case if $f$ were reducible, it would have a factor of degree $2$ or $3$. (Linear factors are easily found using the rational root test.(°)) But no subset of the irreducible factors mod $2$ sums up to $3$; whereas no subset of the degrees of the irreducible factors mod $3$ sums op to $2$. (°°) So this is impossible.
(°) As Mathmo123 writes in his answer: it is also clear from the factorization mod 2 that there could not be a linear factor.
(°°) See the answer by André Nicolas for why the factors of $f$ are indeed irreducible.
|
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How to find $\int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$ $$I=\int x.\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$$
Try 1:
Put $z= \ln(x+\sqrt{1+x^2})$, $\mathrm dz=1/\sqrt{1+x^2}\mathrm dx$
$$I=\int \underbrace{x}_{\mathbb u}\underbrace{z}_{\mathbb v}\mathrm dz=x\int zdz-\int (z^2/2)\mathrm dz\tag{Wrong}$$
Try 2:
Put
$z= x+\sqrt{1+x^2}$
$$\implies x-z
=\sqrt{1+x^2}\implies x^2+z^2-2xz
=1+x^2\implies x
=\frac{z^2-1}{2z}$$
$$\mathrm dz
=\left(1+\frac{x}{\sqrt{1+x^2}}\right)\mathrm dx
=\frac{z\mathrm dx}{x-z}=\frac{-2z^2\mathrm dx}{1+z^2}$$
$$I
=\int\frac{(z^2-1)\ln z}{2z}.\frac{(1+z^2)\mathrm dz}{-2z^2}$$
$$=\int\frac{(z^4-1)\mathrm dz}{4z^3}
=\frac14\int\left(z-\frac1{z^3}\right)\mathrm dz
=z^2/2+2/z^2+C\tag{Wrong}$$
Try 3:
Put
$z
=\sqrt{1+x^2},\mathrm dx
=x/\sqrt{1+x^2}\mathrm dx$
$$I
=\int \ln(x+z)\mathrm dz
=\int \ln(z+\sqrt{z^2-1})\mathrm dz$$
Don't know how to solve this integral.
[Note that if I take $u=z+\sqrt{z^2-1}$, it is $=\sqrt{1+x^2}+\sqrt{1+x^2-1}=x+\sqrt{1+x^2}$; same as first try.]
What's wrong in try 1 & 2, how to further solve try 3 and the best method to solve this question?
Update: Sorry, I don't know hyperbolic/inverse hyperbolic trigonometry.
|
Hint:
$$
\int x\frac{\ln({x+\sqrt{1+x^2})}}{\sqrt{1+x^2}}dx=\int\ln({x+\sqrt{1+x^2})}d\sqrt{1+x^2}
$$
and
$$
(\ln({x+\sqrt{1+x^2})})'=\frac{1}{\sqrt{1+x^2}}.
$$
|
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When can an infinite sum and complex integral be interchanged? Are there some conditions under which the following two are equal?
$$\displaystyle \oint_C \sum f_n(z)= \sum \oint_C f_n(z)$$
In the case of real valued functions, the condition $f_n(z) \geq 0$ suffices (see here). Is there a simple condition like that in the case of complex valued analytic functions with simple/double poles?
Thank you :)
|
Since
$$
\int_C f(z)\,dz = \int_C \operatorname{Re} f(z)\,dz + i\int_C \operatorname{Im} f(z)\,dz,
$$
you can essentially apply the same criteria in the complex case as in the real case. Specifically, you just need the real and imaginary parts of the integrands to be single-signed.
Let $Q_j$, $j=1,2,3,4$, denote the four closed quadrants of the complex plane. If
$$
f_n(z) \in Q_j \quad \text{for all } n \in \mathbb N \text{ and } z \in C
$$
for some $Q_j$ then
$$
\displaystyle \oint_C \sum_n f_n(z)= \sum_n \oint_C f_n(z)
$$
follows from applying the known result for real functions to the real and imaginary parts of the integral.
|
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What is the simplest way to understand $\nabla \cdot (\nabla \times \textbf{A} ) =0$? Of course one can prove it by brute force.
But how to Understand it, in the simplest way?
Another issue is $\nabla \times (\nabla f) =0 $.
|
It's just a way to remember/understand. Think that $\nabla $ is a vector. Then, by definition, $\nabla \times A$ is perpendicular to $\nabla $ and so the product scalar of $\nabla $ and $\nabla \times A$ is nul, therefore
$$\nabla \cdot (\nabla \times A)=0$$
|
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Simplifying nested/complex fractions with variables I have the equation $$x = \frac{y+y}{\frac{y}{70} + \frac{y}{90}} $$ and I need to solve for x. My calculator has already shown me that it's not necessary to know y to solve this equation, but I can't seem to figure it out. This is how I try to solve it:
$$
x = \frac{y+y}{\frac{y}{70} + \frac{y}{90}} = 2y\left(\frac{70}{y} + \frac{90}{y}\right) = 2y\left(\frac{90+70}{y}\right) = 2y\cdot\frac{160}{y} = \frac{320y}{y} = 320
$$
But according to my calculator, this is not correct. The answer should be 78.75, but I don't know why. Any help would be much appreciated.
|
This is my solution:
$$ x=\frac{y+y}{\frac{y}{70}+\frac{y}{90}}={y}\cdot\frac{2}{{y}\cdot\left(\frac{1}{70}+\frac{1}{90}\right)}=\frac{2}{\frac{1}{70}+\frac{1}{90}}=\frac{2\cdot6300}{90+70}=12600/160=\boxed{78.75}$$
|
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How to prove that $\int_{0}^{\infty}{\frac{e^{-nx}}{\sqrt{x}}}\mathrm dx$ exists I am trying to show that the integral $\int_{0}^{\infty}{\frac{e^{-nx}}{\sqrt{x}}}\mathrm dx$ exists ($n$ is a natural number). I tried to use the comparison theorem by bounding from above the integrand by another function which is integrable, but I wasn't successful.
The same question for this integral as well : $\int_{0}^{\infty}x^2e^{-nx}\mathrm dx$
Any help is much appreciated? Thanks!
|
The integrand is less than $1/\sqrt{x}$, so that $0<\int_\varepsilon^1 \frac{e^{-nx}dx}{\sqrt{x}}<\int_0^1 \frac{dx}{\sqrt{x}}=2$, and $\int_0^1 \frac{e^{-nx}dx}{\sqrt{x}}$ exists. As to $\int_1^\infty \frac{e^{-nx}dx}{\sqrt{x}}$, the integrand rapidly decreases to $0$ as $x\rightarrow\infty$, in particular, $0<\int_1^N \frac{e^{-nx}dx}{\sqrt{x}}<\int_1^\infty \frac{dx}{x^2}=1$ is sufficient.
|
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Computing the standard part of $(3-\sqrt{c+2})/(c-7)$ where the standard part of $c$ is $7$ I'm working through Keisler's calculus book based on infinitesimals. The following problem has me a little bit stumped.
Compute the standard part of:
$$\frac{3-\sqrt{c+2}}{c-7}$$
Given that $c\ne7$ and $st(c) = 7$ where $st(x)$ is the standard part function. I know that $$\lim_{c\to7}{\frac{3-\sqrt{c+2}}{c-7}}=-\frac{1}{6}$$ I can't for the life of me though figure out how to work this out using standard parts. I'm probably being dumb. Help?
|
HINT: Try multiplying by $\frac{3+\sqrt{c+2}}{3+\sqrt{c+2}}$
|
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How do you solve for x in this equation? I tried expanding, but I still can't get rid of the exponents to isolate x.
$$\frac{(1+x)^4-1}{x}=4.374616$$
Thank you in advance for your help.
|
I will present two general approaches, starting with the one that works best for this particular case:
Assuming small values of x, we have $(1+x)^4\approx1+4x+6x^2$. Subtracting $1$ and dividing by x, we are left with solving $4+6x=4+\alpha\iff x=\dfrac\alpha4=0.06,~$ where $\alpha=0.374616$.
Let $~f(x)=(1+x)^4.\quad$ Then $~\dfrac{(1+x)^4-1}x=\dfrac{(1+x)^4-(1+0)^4}{x-0}\approx f'(0).\quad$ But $~f'(x)=$
$=4\cdot(1+x)^3.\quad$ So $~4\cdot(1+x)^3\approx a\iff x\approx-1+\sqrt[3]{\dfrac a4}=0.03,~$ which unfortunately is
only half of the true value, $x=0.06.~$ Nevertheless, the reason I chose to present this method as well is because, despite not working very well for this particular case, it works quite well in many other similar situations.
|
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How to prove that $\frac{a+b}{2} \geq \sqrt{ab}$ for $a,b>0$? I am reading a chapter about mathematical proofs. As an example there is:
Prove that: $$(1) \space\space\space\space\space\space\space\space\space\space\space \frac{a+b}{2} \geq \sqrt{ab}$$ for $a,b>0$. There is stated that the thesis for the proof is short multiplication formula: $$(2)\space\space\space\space\space\space\space\space\space(a+b)^2=a^2+2ab+b^2$$ Substracting from the short multiplication $(2)$ formula $4ab$ and using square root yields the proof by implication. (here ends the example in a book)
Please tell me if I am right about it:
*
*we are considering only $a,b \in \Re^+$
*$(1)$ can be transformed into $a-2\sqrt{ab}+b \geq0$ which is an equivalent short multiplication formula $(x-y)^2=x^2-2xy+y^2$
*taking into account assumptions squared number is always equal or greater than zero.
*the implication says that if squared number is greater than zero then $(1)$ is true.
My first question is: is my thinking proper?
My other question is why the authors use $(a+b)^2$ and susbstract $4ab$ instead of starting with $(a-b)^2$. I am also confused about the symbols used - in $(1)$ and $(2)$ authors used $a$ and $b$ - is this just coincidence? Or this would be OK as well if in the first and second $a$ and $b$ were the same numbers?
|
Argument is fine as long as it is clear to you that your are using equivalence and not implication on your way down the bullets. That is for $a, b \in \Bbb R^{+}$
$$ \frac{a+b}{2} \geq \sqrt{ab} \iff a-2\sqrt{ab}+b \geq0 \iff (\sqrt a- \sqrt b)^2 \ge 0 $$
You must use "double arrows". Or "if and only if" connectives. Only then can you justify the leftmost statement from the correctness of the one in the right.
The rightmost equality makes sense due to two reasons. One due to the assumption you mentioned above regarding the non-negativity of a square. And two the square roots of $a$ and $b$ exist since both are positive.
You are right. This is a better way to prove this in my opinion. So I wouldnot worry about it.
As for the notation : Yes the formula would still hold if $a$ and $b$ were both the same number. Toverify just look at your proof. Substitue $a$ for $b$ and see what happens. you will notice the equality in $\geq$ holds. This is in fact a secondary idea of this result. The equality holds if and only if $a = b$. You should try to prove this. Decent exercise.
|
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|
Kirchoff Matrix -Tree Theorem I'm reading a proof of the Kirchoff Matrix -Tree Theorem:
If $G$ is a simple connected graph, $D$ the diagonal matrix with the vertices' degrees and $A$ the adjacency matrix, then in $M = -A+D$ every cofactor is equal to the number of the covering trees of $G$.
I don't understand in the proof the following fact: why all the cofactors of $M$ must be equal ?
|
Another definition of laplacian matrix is M( or L) = $QQ^t$, where $Q$ is the incidence matrix. By cauchy-binet theorem, one can calculate the determinant of $QQ^t$ by considering $Q$ and $Q^t$. So, if we take a cofactor, then it can contain twigs of a tree and since determinant of matrix containing (#nodes-1) and twigs of a tree is +/- 1. So, multiplying two cofactors each from Q and $Q^t$ will result in only +1. Also, for any other cofactor not containing all edges which are twigs of a tree, its determinant goes to 0. As a result we have determinant of cofactor=#(trees).
You can refer to cauchy-binet theorem on wikipedia.
|
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The image of non-reflexive Banach space in its bidual is at large distance from some unit vector in the bidual I am having some difficulties with part of a problem, I am working on.
Let $X$ be a non-reflexive Banach space and let $i: X \to X^{**}$ be the canonical embedding.
Show that for given $\epsilon > 0$ there exists some $x^{**} \in X^{**}$, such that $\|x^{**}\|=1$ and $$\inf_{x \in X} \|x^{**} - i(x)\|_{X^{**}} > 1- \epsilon.$$
I have no idea how to show more than the existence of some $x^{**} \in X^{**}$, such that $\|x^{**}\|=1$ and $$\inf_{x \in X} \|x^{**} - i(x)\|_{X^{**}} > 0,$$ which is by definition of reflexivity and closeness of $i(X)$.
|
Simply use the Riesz lemma as $i(X)$ is norm-closed in $X^{**}$.
|
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Indefinite integral of $\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}$ How do I find $$\int\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}\mathrm dx$$
I used partial fractions by breaking up $x^2 + 2x - 2$ into $(x+1)^2 - 3$ and split it into $(a+b)(a-b)$ but as u can see it's extreme tedious. I was wondering if there is a faster technique to resolve this.
|
HINT:
Write $$\frac{2x^3 + 5x^2 +2x +2)}{(x^2 +2x + 2)(x^2 + 2x - 2)}=\frac{Ax+B}{x^2+2x+2}+\frac{Cx+D}{x^2+2x-2}$$
For the ease of calculation, we can write
$$\frac{Ax+B}{x^2+2x+2}=\frac A2\cdot\frac{\left(\dfrac{d(x^2+2x+2)}{dx}\right)}{x^2+2x+2}+\frac C{(x+1)^2+1^2}.$$
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The number of divisors of any positive number $n$ is $\le 2\sqrt{n}$ How to prove that the number of divisors of any positive number $n$ is $\le 2\sqrt{n}$?
I have started something like below:
$$ n^{\tau(n)/2} = \prod_{d|n} d$$
But not getting ideas on how to take this further and conclude. Any help is appreciated. Thanks!
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There are at most $\sqrt n$ divisors $d\mid n$ with $d\le \sqrt n$, and there are at most $\sqrt n$ divisors with $d>\sqrt n$ as each of them corresponds uniquely to a divisor $\frac nd<\sqrt n$.
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$X$ is contained in an annulus containing a circle in the annulus.Show that $\pi_1(X)$ contains a subgroup isomorphic to $\mathbb{Z}$. In $\mathbb{R^2}$ let $C=\{|x|=2\}$ and $A=\{1<|x|<3\}$, let $X$ be a path connected set such that $C \subset X \subset A$. Show that $\pi_1(X)$ contains a subgroup isomorphic to $\mathbb{Z}$.
Well, it is intuitively obvious but how can one argue rigorously? Can I argue like : $C$ itself can be thought of as a path $\gamma$ in $X$ with $\gamma : [0,1] \to X$ is defined by $\gamma(t)=2e^{2\pi it}$. Since $\gamma$ is not nullhomotopic (My question is : does this fact need proof ? If yes, how should I prove that ?) and hence [$\gamma$] generates an infinite cyclic group which is contained in $\pi_1(X)$... $\blacksquare$
Isn't it too handwavy ? thank you in advance for your helpful comments/answers
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The inclusion $X \hookrightarrow A$ induces a homomorphism on fundamental groups $\pi_1(X) \rightarrow \pi_1(A)$. As you said, pick the element $\gamma$ of $\pi_1(X)$ corresponding to $\{|x| = 2\}$ (i.e., a path traversing this circle once counterclockwise). The image of $\gamma$ in $\pi_1(A)$ represents a generator of $\pi_1(A) \cong \Bbb Z$ (why?), so that $\gamma$ must have infinite order in $\pi_1(X)$ (why?)
(As mentioned in the comments above, this all comes down to having previously computed $\pi_1(S^1)$!)
|
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Show that $ax^2+2hxy+by^2$ is positive definite when $h^2The question asks to "show that the condition for $P(x,y)=ax^2+2hxy+by^2$ ($a$,$b$ and $h$ not all zero) to be positive definite is that $h^2<ab$, and that $P(x,y)$ has the same sign as $a$."
Now I've seen questions similar to this before where it's a two variable quadratic and I'm not too sure how to go about it. Normally with one variable you could just show the discriminant is less than $0$ but since there's two variables I can't use the same process (since I'm not sure how the discriminant is defined for a two variable quadratic). I also tried completing the square but that doesn't seem to make the algebra anything near as simple as the answer (I got a rather complicated fraction with cubes). Is there a way to go about this type of problem?
Thanks.
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You know that $P(x,y)$ and $a$ have the same sign, so $b$ has the same sign as $a$ (consider $P(0,1)$, so you know that $ab \geq 0$. Next:
case 1: $xy\leq 0$
$$P(x,y)=ax^2+2hxy+by^2=ax^2+by^2+2\sqrt{ab}xy-2\sqrt{ab}xy+2hxy= \\ =(\sqrt{|a|}x+\sqrt{|b|}y)^2-2\sqrt{ab}xy+2hxy=(\sqrt{|a|}x+\sqrt{|b|}y)^2+2xy(h-\sqrt{ab})$$
You know that $ab>h^2$, so $h-\sqrt{ab} \leq 0$, so $xy(h-\sqrt{ab}) \geq 0$.
case 2: $xy \geq 0$
$$P(x,y)=ax^2+2hxy+by^2=ax^2+by^2+2\sqrt{ab}xy-2\sqrt{ab}xy+2hxy= \\ =(\sqrt{|a|}x-\sqrt{|b|}y)^2+2\sqrt{ab}xy+2hxy=(\sqrt{|a|}x-\sqrt{|b|}y)^2+2xy(h+\sqrt{ab})$$
You know that $ab>h^2$, so $h+\sqrt{ab} \geq 0$, so $xy(h-\sqrt{ab}) \geq 0$.
|
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Prove that $A \circ B = AB$ if and only if both $A$ and $B$ are diagonal Definition. Hadamard product. Let $A,B \in \mathbb{C}^{m \times n}$. The Hadamard product of $A$ and $B$ is defined by $[A \circ B]_{ij} = [A]_{ij}[B]_{ij}$ for all $i = 1, \dots, m$, $j = 1, \dots, n$.
Remark. See details in this Hadamard product wiki article.
There is the following remark in Million's paper in Chapter 2:
We can relate the Hadamard product with matrix multiplication via
considering diagonal matrices, since $A \circ B = AB$ if and only if
both $A$ and $B$ are diagonal.
So there is a theorem, that $A \circ B = AB$ if and only if both $A$ and $B$ are diagonal, but I don't know how to prove it, and I didn't find it in the literature, because not many books have written in this topic.
Edit. In this theorem probably $m=n$. Million didn't write about it.
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Proof: (assuming $n=m$)
($\Rightarrow$) by assumption, $A$ and $B$ are diagonal. Then $AB = diag(a_{11}b_{11}, a_{22}b_{22}, \dots, a_{nn}b_{nn})$ and $[A\circ B]_{i,j} = [A]_{i,j} [B]_{i,j}$. Since $A$ and $B$ are diagonal, then $[A\circ B]_{i,j} = 0$ for all $i\neq j$, and $[A\circ B]_{i,i} = a_{ii}b_{ii}$ $\Rightarrow$ $[A\circ B]_{i,j} = diag(a_{11}b_{11}, a_{22}b_{22}, \dots, a_{nn}b_{nn})=AB$
($\Leftarrow$) (retracted)
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Proving combinatorial identity with the product of Stirling numbers of the first and second kinds $$ \sum_{k} \left[\begin{array}{c} n\\k \end{array}\right] \left\{\begin{array}{c} k\\m \end{array}\right\} = {n \choose m} \frac{\left( n-1\right)!}{\left(m-1 \right)!}, \quad \text{for } n,m > 0 $$
$ \left[\begin{array}{c} n\\k \end{array}\right] $ is Stirling number of the first kind
$ \left\{\begin{array}{c} k\\m \end{array}\right\} $ Stirling number of the second kind
I don't know how to aproach this. I tried think of combinatorial interpretation and mathematical induction, but I got stuck a the beginning.
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Recall the species of permutations marked by cycle count
which is
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}_{\ge 1}(\mathcal{Z}))$$
so that it has the generating function
$$\exp\left(u \log\frac{1}{1-z}\right)$$
and in particular
$$\sum_{n\ge q} \left[n\atop q\right] \frac{w^n}{n!}
= \frac{1}{q!} \left(\log\frac{1}{1-w}\right)^q.$$
Continuing, recall the species of set partitions marked by the number
of sets which is
$$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
so that it has the generating function
$$\exp\left(u (\exp(z)-1)\right)$$
and in particular
$$\sum_{n\ge q} {n\brace q} \frac{w^n}{n!}
= \frac{1}{q!} \left(\exp(w)-1\right)^q.$$
Introduce the EGF
$$Q(z) = \sum_{n\ge 1} \frac{z^n}{n!}
\sum_{k=1}^n \left[n\atop k\right] {k\brace m}.$$
Interchange the order of summation to get
$$\sum_{k\ge 1} {k\brace m}
\sum_{n\ge k} \left[n\atop k\right] \frac{z^n}{n!}.$$
Now we recognize the inner generating function from the introduction
(partitions into $k$ cycles) so $Q(z)$ simplifies to
$$\sum_{k\ge 1} {k\brace m}
\frac{1}{k!} \left(\log\frac{1}{1-z}\right)^k.$$
This too is a familiar generating function (partitions into $m$ sets)
applied to the logarithmic term so we get
$$Q(z) = \frac{1}{m!}
\left(\exp\log\frac{1}{1-z} - 1 \right)^m
= \frac{1}{m!} \left(\frac{z}{1-z}\right)^m.$$
Finally perform coefficient extraction to obtain
$$n! [z^n] Q(z)
= \frac{n!}{m!} [z^{n-m}] \left(\frac{1}{1-z}\right)^m
= \frac{n!}{m!} {n-m+m-1\choose m-1}
\\= \frac{n!}{m!} {n-1\choose m-1}
= \frac{(n-1)!}{(m-1)!} {n\choose m}.$$
|
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Proof of Incircle A circle is drawn that intersects all three sides of $\triangle PQR$ as shown below. Prove that if AB = CD = EF, then the center of the circle is the incenter of $\triangle PQR$.
Designate the center of the circle $G$.
Thinking about it a bit, we realize if we could prove the incircle of the triangle and the circle given are concentric, we would know that the center of the circle is the incenter, for they have the same center.
If we shrink the circle given down to the incircle, the incircle should intersect the triangle at the midpoints of $AB$, $CD$, and $EF$.
We therefore denote the midpoint of $AB M$, the midpoint of $CD L$, andd the midpoint of $EF N$.
We know the incircle is concentric to the given circle if and only if $NG$ and $LG$ are congruent, because a circle is a circle if and only if the radii are congruent. Thus, we must prove $NG\cong LG$.
When proving lengths equal, congruent triangles are always a good idea. We see that $NGE\cong LGD$ seems likely, and it would also prove our answer if we could prove it.
We know that $GD\cong GE$ because they are both radii of the larger circle, and we also know $DL\cong EN$ because they are both half of two lines that are congruent.
We know $\angle GLD$ and $\angle GNE$ are both right angles, because that is where the smaller circle is tangent to the sides of the triangle. Thus, $NGE\cong LGD$ by $HL$ congruency.
Because congruent parts of congruent triangles are congruent, we know $NG\cong LG$, which proves $G$ is the center of the incircle.
Is this proof sound?
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I would develop your point just using circle symmetry.
Take a circle and an intersecting line. Make the radius of the circle shrink, then by the symmetry of the circle, the chord will shrink symmetrically to the bisecting line (its axis). Always by the circle symmetry, any other chord of same (original) length will shrink in the same way and by the same amount.
So they will altogether become tangent to circle at their center point, i.e. at the respective bisecting line, i.e. at the normal to their central point.
|
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|
What is the expected value of A? The Happy Animals Kennel has 18 cages in a row. They allocate these cages at random to 6 dogs, 6 cats, and 6 pot-bellied pigs (with one animal per cage). All arrangements are equally likely.
Let A be the number of times in the row of cages that two animals of the same species are adjacent. For example, in the arrangement DCPCDPPPDCDPCDCCPD (where D=dog, C=cat, and P=pig), we have A=3.
What is the expected value of A?
(not sure if expectation is the correct tag here)
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The probability that Fido and Rover (two dogs) are in adjacent cages is
$$\frac{17}{\binom{18}2}=\frac19,$$
because there are $\binom{18}2$ choices for the (unordered) pair of cages to put them in, and $17$ pairs of adjacent cages.
To get the expected value, multiply this probability by the number of same-species pairs of animals:
$$\frac19\cdot3\cdot\binom62=5.$$
The expected value is $5$.
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For what $p$ does the series $\sum \frac1{n^p \ln(n)}$ converge "Find the values of $p$ s.t. the following series converges: $\sum_{n=2}^{\infty} \frac{1}{n^p \ln(n)}$"
I am trying to do this problem through using the Integral Test to find the values of $p$. I know that for $p = 0$, the series diverges so I will only be considering values of $p \neq 0$.
The function $f(x) = (x^p \ln(x))^{-1}$ satisfies the criterion of the Integral Test, but I am having a difficult time integrating the function.
We have $$\int_{n=2}^{\infty} \frac{1}{x^p \ln(x)} dx$$ A u-substitution with $u = \ln(x)$ will not help us and neither would setting it to $x^p$. Someone suggested letting $x = e^u$, but...i'm not so sure where we would go with that.
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Comparison test is best here. But here's how to use that substitution idea. Notice how it comes down to comparison in the end.
Using $x=e^u$ we have $dx=e^u du$. Thus $$\int_2^\infty \frac{dx}{x^p \ln x} = \int_{\ln2}^\infty \frac{e^u}{u e^{p u}} du = \int_{\ln2}^\infty e^{(1-p) u} \frac{1}{u} du.$$
It is easy to see that for $p<1$ the integrand is unbounded, so we have divergence. For $p=1$ the antiderivative is known and we again diverge. For $p>1$ we can bound the integrand above by $e^{(1-p)u}$, and the integral of this is easily found to be convergent.
|
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If $(X,\tau)^n$ is Hausdorff, is $(X,\tau)$ also Hausdorff?
If $(X,\tau)^n$ is Hausdorff, is $(X,\tau)$ also Hausdorff?
I know that product of Hausdorff space is Hausdorff, but I want to know if this weaker converse of it is true. Thanks.
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Note that $(X,\tau)$ is homeomorphic to a subspace of $(X,\tau)^n$ by mapping $x \mapsto (x,x_0,x_0,\dots,x_0)$ for some fixed $x_0 \in X$ (ignoring the case $X = \emptyset$ which is even easier). Now use the fact that subspaces of Hausdorff spaces are Hausdorff spaces.
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Is there any proof for this formula $\lim_{n \to \infty} \prod_{k=1}^n \left (1+\frac {kx}{n^2} \right) =e^{x/2}$? Some times ago, In a mathematical problem book I sow that this formula. I don't no whether it is true or not. But now I'm try to prove it. I have no idea how to begin it. Any hint or reference would be appreciate. Thank you.
$$\lim_{n \to \infty} \prod_{k=1}^n \left (1+\frac {kx}{n^2} \right) =e^{x/2}$$
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Notice that $x-x^2/2\leq \ln(1+x) \leq x$ for $x\geq 0$ then : $$\frac{xk }{n^2} \geq \ln\left(1+\frac{ xk}{n^2}\right) \geq \frac{x k}{n^2} -\frac{x^2 k^2}{n^4} $$ Sum form $k=1$ to $n$ : $$\frac{(n+1)x}{2n} \geq \sum_{k=1}^n \ln\left(1+\frac{ xk}{n^2}\right) \geq \frac{(n+1)x}{2n} - \frac{(n+1)(2n+1)x}{6n^3} $$
Then the limit of the middle sum is $x/2$ take the exponential to get that the requested limit is $e^{x/2}$.
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Continuity almost everywhere Let
$$f(x) = \left\{\begin{array}{ll}
1 & \text{if }x\in\mathbb{Q},\\
0 & \text{if }x\notin\mathbb{Q}.
\end{array}\right.$$
$$g(x) = \left\{\begin{array}{ll}
\frac{1}{n} & \text{if }x = \frac{m}{n}\in\mathbb{Q},\\
0 & \text{if }x\notin\mathbb{Q}.
\end{array}\right.$$
I have some difficulties with understanding of continuity almost everywhere.
Honestly, I thought that continuity almost everywhere means that I could take any continuous function and change (or even make them undefined) its values for countable set of points from domain, say, for $\mathbb{Q}$. Based on such thinking, both $f(x), g(x)$ are continuous almost everywhere, because I could get them as a result of change of countable values of $h(x) \equiv 0$
The fact is that only $g(x)$ is continuous almost everywhere, while $f(x)$ is discontinuous. Could you explain what went wrong in my understanding?
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A function $f:X\to Y$ is continuous almost everywhere if it is only discontinuous on a set of measure $0$. Informally, this means that if we choose a random point on the function, the probability that it is continuous is exactly $1$.
For example, any countable set, such as $\mathbb Q$, will be of measure $0$ in an uncountable one, such as $\mathbb R$.
There is an important distinction to make here: a function is continuous almost everywhere if it is continuous on a "large" subset. But it must be continuous on that subset. Your function $g$ is continuous at every irrational point, and hence we can say that it is continuous almost everywhere.
That is not the same as removing a countable set of values from the function, because in doing so, you can damage the continuity of the remaining points. In the case of $f$, you are left with a function that is nowhere continuous!
This only works if you remove a countable set, but make sure that the remaining points are still continuous.
|
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$Ker(T) \subseteq V$ Is A Subspace Let $V,W$ be a vector space over a field $\mathbb F$, and $T$ a linear transformation $T:V \rightarrow W$
$Ker(T) \subseteq V $ to prove that $Ker(T)$ is a subspace can we say that:
by definition $0\in Ker(T)$ and because V is a vector space, and in any vector space there are two trivial subspace (0,V) therefore $Ker(T)$ is a subspace?
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Your reasoning isn't correct. Try the following hints:
To start with, one can show that for any linear map $T$, $T(0) = 0$ and so $0 \in \text{ker }T$ (you should prove this). Thus, since $\text{ker }T$ is non-empty, it suffices to show that it's closed under linear combinations. So suppose $u,v \in \text{ker }T$ and $\lambda, \mu \in F$, and show that $T(\lambda u + \mu v) = 0$ using the fact that $T$ is a linear map.
|
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A number is a perfect square if and only if it has odd number of positive divisors I believe I have the solution to this problem but post it anyway to get feedback and alternate solutions/angles for it.
For all $n \in \mathrm {Z_+}$ prove $n$ is a perfect square if and only if $n$ has odd # of positive divisors.
$\Rightarrow$: If $n$ is a perfect square it must consist of 1+ prime factors each to an even power. If $n$ consists of 1+ prime factors each to an even power it must have an odd # of positive divisors because each even power contributes itself plus $1$(for the $0$ case) to the number of divisors for product $n$.
$\Leftarrow$: If $n$ has an odd number of positive divisors it must consist of 1+ prime factors each to an even power. If $n$ consists of 1+ prime factors each to an even power it must be a perfect square.
Thanks.
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If two integers multiply to equal N, you will add two divisors to your total for that number. This will be true for all values except when the two integers are the same. Divisors will always be added in pairs except in this case. So the only way to get to an odd total, is when the two divisors are equal, or N is a perfect square.
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How do you transpose tensors? We transpose a matrix $A$ by replacing $A_{ij}$ with $A_{ji}$, for all $i$ and $j$.
However, in case $A$ has more than two dimensions (that is, it is a tensor), I don't know how to apply the transpose operation.
*
*If A has dimensions $3\times 3 \times 8$, then what will replace $A_{ijk}$?
*If $A$ has shape $3\times 3\times 8\times 8$, then what will replace $A_{ijkl}$?
|
The operation of taking a transpose is closely related to the concept of symmetry. One paper that addresses this is http://www.iaeng.org/publication/WCE2010/WCE2010_pp1838-1841.pdf. I have been researching $2^m$ dimensional matrices where the indices are zeros and ones. The transpose is found by changing all zeros to ones and ones to zeros.
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Convergence of a sequence of continuous function Len $g_n:\mathbb{R}\rightarrow\mathbb{R}$ continuous with $g_n(x)=0$ for $|x|\geq 1/n$, $g_n(x)\geq 0$, $\int_{-1}^1 g_n(x)\,dx=1$. Consider $f:\mathbb{R}\rightarrow\mathbb{R}$ continuous and:
$$
f_n(x)=\int_{-\infty}^\infty g_n(x-y)f(y)\,dy
$$
I want to show that $f_n$ converges to $f$ pointwise and if $f(x)=0$ for $|x|\geq a$ for some $a>1$ then the convergence is uniform.
For the first part after changing variable $z=x-y$ I get $f_n(x)=\int_{-1}^1 g_n(z)f(x-z)dz$. How should I proceed?
The second part with the uniform convergence follows from Dini's Theorem since f is compactly supported and so fn(x) will be compactly supported also?
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As you have noted $$\int\limits_{ - \infty }^{ + \infty } {{g_n}(x - y)f(y)dy} = \int\limits_{ - \infty }^{ + \infty } {{g_n}(z)f(x - z)dz} = \int\limits_{ - \frac{1}{n}}^{ + \frac{1}{n}} {{g_n}(z)f(x - z)dz} $$Assuming continuity of $f$ and the existence of it's derivative, for large enough $n$ (small enough $z$ in the integrand limits)one could write $f(x - z) = f(x) + O(z)$, so by replacement $$\int\limits_{ - 1}^{ + 1} {{g_n}(z)f(x - z)dz} = f(x)\int\limits_{ - 1}^{ + 1} {{g_n}(z)dz} + \int\limits_{ - 1}^{ + 1} {O(z){g_n}(z)dz} = f(x) + \int\limits_{ - \frac{1}{n}}^{ + \frac{1}{n}} {O(z){g_n}(z)dz} $$Now, it is easy to see that $$\left| {\int\limits_{ - 1}^{ + 1} {{g_n}(z)f(x - z)dz} - f(x)} \right| \le O(\frac{1}{n})\int\limits_{ - \frac{1}{n}}^{ + \frac{1}{n}} {{g_n}(z)dz} = O(\frac{1}{n})$$which implies point-wise convergence.
|
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Analogue of $\zeta(2) = \frac{\pi^2}{6}$ for Dirichlet L-series of $\mathbb{Z}/3\mathbb{Z}$? Consider the two Dirichlet characters of $\mathbb{Z}/3\mathbb{Z}$.
$$
\begin{array}{c|ccr}
& 0 & 1 & 2 \\ \hline
\chi_1 & 0 & 1 & 1 \\
\chi_2 & 0 & 1 & -1
\end{array}
$$
I read the L-functions for these series have special values
*
*$ L(2,\chi_1) \in \pi^2 \sqrt{3}\;\mathbb{Q} $
*$ L(1,\chi_2) \in \pi \sqrt{3}\;\mathbb{Q} $
In other words, these numbers are $\pi^k \times \sqrt{3} \times \text{(rational number)}$. Is there a way to derive this similar to to the famous $\zeta(2) = \tfrac{\pi^2}{6}$ formula?
Here are 14 proofs of $\zeta(2) = \tfrac{\pi^2}{6}$ for reference
|
We have:
$$L(2,\chi_1)=\sum_{j=0}^{+\infty}\left(\frac{1}{(3j+1)^2}+\frac{1}{(3j+2)^2}\right)=-\int_{0}^{1}\frac{(1+x)\log x}{1-x^3}\,dx$$
and integration by parts gives:
$$\begin{eqnarray*}\color{red}{L(2,\chi_1)}&=&-\int_{0}^{1}\frac{\log(1-x)}{x}\,dx+\frac{1}{3}\int_{0}^{1}\frac{\log(1-x^3)}{x}\,dx\\&=&-\frac{8}{9}\int_{0}^{1}\frac{\log x}{1-x}=\frac{8}{9}\zeta(2)=\color{red}{\frac{4\pi^2}{27}.}\end{eqnarray*}$$
With a similar technique:
$$L(1,\chi_2)=\sum_{j=0}^{+\infty}\left(\frac{1}{3j+1}-\frac{1}{3j+2}\right)=\int_{0}^{1}\frac{1-x}{1-x^3}\,dx=\int_{0}^{1}\frac{dx}{1+x+x^2}$$
so:
$$\color{red}{L(1,\chi_2)}=\int_{0}^{1/2}\frac{dx}{x^2+3/4}=\frac{1}{\sqrt{3}}\arctan\sqrt{3}=\color{red}{\frac{\pi}{3\sqrt{3}}.}$$
|
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|
Solution of an equation involving even integers If $x$ is any positive even integer $> 1$. I compute:
$$ c = x + x! $$
Now assume instead $c$ (even integer) is given, and I want to get back the value $x$.
Is it possible to find a simple expression for $x$ ?
(or, if not, a convergent procedure to find it)
|
The $x$ is very small compared to $x!$, so basically you want to invert the Gamma function. See e.g. this MathOverflow question
|
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|
Evaluation of a dilogarithmic integral
Problem. Prove that the following dilogarithmic integral has the indicated value:
$$\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(x)}\operatorname{Li}_2{(x)}}{1-x}\stackrel{?}{=}-11\zeta{(5)}+6\zeta{(3)}\zeta{(2)}.$$
My attempt:
I began by using the polylogarithmic expansion in terms of generalized harmonic numbers,
$$\frac{\operatorname{Li}_r{(x)}}{1-x}=\sum_{n=1}^{\infty}H_{n,r}\,x^n;~~r=2.$$
Then I switched the order of summation and integration and used the substitution $u=-\ln{x}$ to evaluate the integral:
$$\begin{align}
\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(x)}\operatorname{Li}_2{(x)}}{1-x}
&=\int_{0}^{1}\mathrm{d}x\ln^2{(x)}\sum_{n=1}^{\infty}H_{n,2}x^n\\
&=\sum_{n=1}^{\infty}H_{n,2}\int_{0}^{1}\mathrm{d}x\,x^n\ln^2{(x)}\\
&=\sum_{n=1}^{\infty}H_{n,2}\int_{0}^{\infty}\mathrm{d}u\,u^2e^{-(n+1)u}\\
&=\sum_{n=1}^{\infty}H_{n,2}\frac{2}{(n+1)^3}\\
&=2\sum_{n=1}^{\infty}\frac{H_{n,2}}{(n+1)^3}.
\end{align}$$
So I've reduced the integral to an Euler sum, but unfortunately I've never quite got the knack for evaluating Euler sums. How to proceed from here?
|
This integral belongs to a wider class of integrals that can always be reduced to single zeta values and to bi-variate zeta values which in turn -- provided the weight of the zeta function in question is not too big-- can also be reduced to single zeta values. My standard way of doing this integrals is as follows. We need to calculate:
\begin{equation}
{\mathcal I}_{0,2}^{(2)} := \int\limits_0^1 \frac{[\log(1/x)]^2}{2!} \cdot \frac{Li_0(x) Li_2(x)}{x} dx
\end{equation}
which is just one half of your integral. Now we use the following identity:
\begin{equation}
\frac{[\log(1/x)]^2}{2!} = \int\limits_{x < \xi_1 < \xi_2 < 1} \prod\limits_{j=1}^2 \frac{d \xi_j}{\xi_j}
\end{equation}
Inserting this into the above and changing order of integration gives:
\begin{eqnarray}
{\mathcal I}_{0,2}^{(2)} = \int\limits_{0 < \xi_1 < \xi_2 < 1} \frac{1}{\xi_1} \frac{1}{\xi_2} \underbrace{\int\limits_0^{\xi_1} \frac{Li_0(x) Li_2(x)}{x} d x}_{\left[Li_1(\xi_1) Li_2(\xi_2) - \int\limits_0^{\xi_1} \frac{[Li_1(x)]^2}{x} dx\right]} \cdot d\xi_1 d\xi_2
\end{eqnarray}
where we integrated by $x$ using integration by parts. Since two minus zero is even we are left with an irreducible integral that we leave unevaluated for the time being. Now we have:
\begin{eqnarray}
{\mathcal I}_{0,2}^{(2)} &=& \int\limits_0^1 \frac{[\log(1/\xi_1)]^1}{1!} \left( \frac{1}{2} [Li_2(\xi_1)]^2 \right)^{'} d\xi_1 - \int\limits_0^1 \frac{[\log(1/x)]^2}{2!} \cdot \frac{[Li_1(x)]^2}{x} dx \\
&=& \frac{1}{2} \left(
\underbrace{\int\limits_0^1 \frac{[Li_2(\xi)]^2}{\xi} d\xi}_{J_1} -
\underbrace{\int\limits_0^1 [\log(1/\xi)]^2 \cdot \frac{[Li_1(\xi)]^2}{\xi} d\xi}_{J_2}
\right)
\end{eqnarray}
Now from Compute an integral containing a product of powers of logarithms. we have that :
\begin{eqnarray}
J_2&=& -\frac{1}{3} \Psi^{(4)}(1) + 2 \Psi^{(2)}(1) \Psi^{(1)}(1)\\
&=& 8 \zeta(5) - 4 \zeta(3) \zeta(2)
\end{eqnarray}
where $\Psi^{(j)}(1)$ is the polygamma function at unity and $\Psi^{(j)}(1)=(-1)^{j+1} j! \zeta(j+1)$. On the other hand we have:
\begin{eqnarray}
J_1 &=& \sum\limits_{m\ge1,n\ge 1} \frac{1}{m^2} \frac{1}{n^2} \frac{1}{(m+n)}\\
& =& \sum\limits_{m\ge 1} \left(\frac{\zeta(2)}{m^3} - \frac{H_m}{m^4}\right) \\
&=& \zeta(2) \zeta(3) - {\bf H}^{(1)}_4(+1)\\
&=& 2 \zeta(2) \zeta(3) - 3 \zeta(5)
\end{eqnarray}
where in the last line above we used my answer to Calculating alternating Euler sums of odd powers . Therefore:
\begin{equation}
{\mathcal I}^{(2)}_{0,2}= \frac{1}{2} \left(J_1-J_2\right)= \frac{1}{2} \left(-11 \zeta(5)+6 \zeta(3) \zeta(2) \right)
\end{equation}
as expected. Note that exactly the same steps can be performed is we replace the power of the logarithm and the orders of the poly-logarithms by any nonnegative integers. The generic result is then given in An integral involving product of poly-logarithms and a power of a logarithm. .
|
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|
Linearizing systems about critical points. $$\def\q{\begin{pmatrix}}\def\p{\end{pmatrix}}\def\l{\lambda}\def\f{\frac{\sqrt{11}}{2}}$$
Find all the critical points of the following systems and derive the linearised system about each
critical point. Use this to determine the nature of the critical points and whether they are stable
or unstable.
What about say:
$$\q y_1' \\ y_2'\p = \q 2y_1 + y_2\\ -5y_1 + 5 \p$$
Here I have a form I haven't before worked with, being that I can't just look at the eigenvalues to find type of critical point, since I can't put it in the form $\vec{y'} = A\vec{y}$ where $A$ is some coefficient matrix. What do I do here?
Form I have worked with
$$\vec{y'} = \q 2 & 1 \\ -5 & 5 \p \q y_1 \\ y_2 \p$$
$$\q 2-\l &1 \\-5&5-\l \p = 0$$
$$=(2-\l)(5-\l)+5 = 0$$
$$=\l -7\l + 15 = 0$$
$$=\frac{7 \pm \sqrt{(-7)^2 - 60}}{2}$$
$$=\frac{7\pm \sqrt{11}i}{2}$$
$$\l =3.5 \pm \frac{\sqrt{11}}{2}$$
$$\l=3.5 + \f, \l = 3.5 - \frac{\sqrt{11}}{2}$$
Since we have complex conjugate eigenvalues with non-zero real part, we know this is a spiral critical point at the origin. I could then sub in a few vectors to determine if it was asymptotically un/stable.
|
The system is $y'=Ay+u$, where $A= \begin{bmatrix} 2 & 1 \\ -5 & 0 \end{bmatrix}$, and $u = \begin{bmatrix} 0 \\ 5 \end{bmatrix}$.
The critical points are where $y'=0$, this gives $y^* = -A^{-1} u = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$.
The linearised system can be written as $(\upsilon+y^*)' = \upsilon' = A(\upsilon+y^*) +u = A \upsilon$.
Since the eigenvalues of $A$ are $1 \pm 2 i$, it is clear that the linearised system is unstable.
|
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|
Do you paragraph a proof? When writing out a proof of moderate length, i.e. a proof taking less than or equal to 5 A4 papers and with normal spacing (please avoid asking the criterion for "normal"), do you tend to paragraph it or not? Why and Why not? Is there a style guide on such question?
|
Proofs should follow the same rules as any other kind of writing. If the text naturally forms paragraphs then that's how it should be written. Now, usually if you have five pages of text forming a single paragraph that would be poor style, but I suppose it's possible that you could have a five-page-long paragraph if, say, one of the sentences of that paragraph was a several-page-long equation that couldn't be broken up.
|
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|
Why is a complex number plus infinity equal to infinity? Why is $$2 + 3 i + \infty = \infty$$
according to Mathematica and Wolfram Alpha?
Shouldn't it be:
$$2 + 3 i + \infty = \infty + 3 i$$
?
After all:
$$2 + 3 i + 10 = 12 + 3 i$$
and not:
$$2 + 3 i + 10 = 12$$
|
In the context of the complex plane there is a very useful notion of a "point at infinity" but it is in infinity "in every direction". This gives rise to a notion of the Riemann Sphere, where two copies of the complex plane give coordinate charts with transition $z\mapsto 1/z$. Complex functions with poles can then be viewed as functions from the Riemann Sphere to itself satisfying certain niceness conditions.
|
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|
Plotting a polar curve The question is, to generate a polar graph using a graphing utility, and to choose parameter interval so that the complete graph is generated.
$$r=\cos\frac{\theta}{5}$$
To find such an interval, we are looking for smallest number of complete revolutions until value of $r$ begins to repeat. Algebraically,this amounts to
$$\cos\frac{\theta}{5}=\cos\frac{\theta+2n\pi}{5}$$
For this equality to hold,$\frac{2n\pi}{5}$ must be an even multiple of $\pi$,the smallest n for which it occurs is $n=5$.Therefore, the graph will be traced completely in $5$ revolutions ($10\pi$).
But when I draw it the graph is completely traced in $5\pi$, where have I gone wrong?
|
Since the question has been already answered, I post here a visual answer that I hope may help you to understand the problem better:
Cheers!
|
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|
If $f,g$ are entire functions such that $f(g(z))=0, \forall z, $ then $g$ is constant or $f(z) =0, \forall z \ ?$ Let $f,g$ be entire functions such that $f(g(z))=0, \forall z.$ Could anyone advise me on how to prove/disprove: either $g(z)$ is constant or $f(z) =0, \forall z \ ?$
Hints will suffice, thank you.
|
Another way to make it: Assume that $g$ is not constant in particular $g'\neq 0$:
As $f\circ g=0$, then by derivation we get $g'\times (f'\circ g)=0$, but the ring of entires functions ( $\Bbb C$ connected) is an integral domain, it follow that $f'\circ g=0$, another derivation we get $g'\times (f''\circ g)=0$, hence $f''\circ g=0$, by same argument we can show that for all $n\in \Bbb N$; $f^{(n)}\circ g=0$.
Now let $a\in \Bbb C$ and $b=g(a)$, since $f$ is analytic we have :
$$\forall z\in \Bbb C\ \ \ f(z)=\sum_{n=0}^{+\infty}\frac{f^{n}(b)}{n!}(z-b)^n=0$$
because $f^{(n)}(b)=f^{(n)}(g(a))=0$
|
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|
If $p : E \to B$ is a covering map and $U \subset B$ is connected and evenly covered, then $p^{-1}(U)$ as a partition into slices is unique. If I have a covering map $p:E \rightarrow B$ and some connected set $U$, that is evenly covered, then $p^{-1}(U)$ as a partition into slices is unique.
Now, if I assume that $B$ is connected, then I want to show for a particular $b \in B$ with covered surrounding $U$ that $p^{-1}(U)$ has some fixed cardinality $k$ of sets which should then agree with $|p^{-1}(b)|=k$. (This one should then agree with the fibre cardinality of all the other elements in $U$.) But I am not so sure about this. I mean, since the partition into slices does not have to be unique in this case, there does not need to exist a unique partition into slices. Is this problem real or can it easily be resolved?
|
Let $p: E \rightarrow B$ be a covering map. For a fixed finite number $k$, define $A_k = \{ b \in B: |p^{-1}(b)|=k \}$. I claim this set is open and closed in $B$.
To see it is open, let $b \in A_k$ and let $U$ be an evenly covered neighbourhood of $b$. So $p^{-1}[U]= \cup_{i \in I} V_i$, where the $V_i$ are pairwise disjoint, and for every $i$, $p$ restricted to $V_i$ is a homeomorphism between $V_i$ and $U$. Then, as all these restrictions are bijections in particular, $b$ has one pre-image in each $V_i$, so $|I|=k$, and moreover, we know that $|p^{-1}(y)|=k$ as well for every $y \in U$, so that $p \in U \subset A_k$, making $b$ an interior point of $A_k$.
Suppose now that $b \notin A_k$, and we take an evenly covered neighbourhood $U$ of it, with the same notation as before. The same reasoning as the previous paragraph gives that all points in $U$ have the same number of preimages as $b$ has, so if $b \notin A_k$, we also have $U \subset B \setminus A_k$, which shows that all points not in in $A_k$ are also not in its closure, so $A_k$ is closed as well.
Now, as $B$ is connected, the only sets that can be open and closed are $B$ and $\emptyset$, so assuming there is some point with finite fibre, all other points have the same sized fibre as well.
|
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What is the history behind the development of the term "coefficient"? Why are coefficients called "coefficients"?
For example I learned that squaring a number is called "squaring" because it actually refers to "making a square". That's how it was developed.
|<----+-----+---->| 3
squared
+-----+-----+-----+
| | | |
| | | |
| | | |
+-----+-----+-----+
| | | |
| | | |
| | | |
+-----+-----+-----+
| | | |
| | | |
| | | |
+-----+-----+-----+
So what is efficient about a coefficient and why is it a "co" like a "coworker" or "coauthor" or a "coeditor"? I feel like if I understood it's history I might remember what it refers to.
|
The Oxford English Dictionary says:
According to Hutton, Vieta, who died in 1603, and wrote in Latin, introduced coefficiens in this sense.
and general meaning of the word around that time seems to have been
Cooperating to produce a result.
So perhaps the meaning is that the coefficient on $5 x$, namely the numeral $5$, cooperates with the value of $x$ to produce the result.
|
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|
How to solve $\tan x=x^2$ in radians? How to solve $\tan x=x^2$ with $x \in [0, 2\pi]$? I try with trigonometry and many ways but the numerical solutions seems to be difficult.
|
Solving $\tan(x)=x^2$ is the same as finding the intersection of the two curves $y=\tan(x)$ and $y=x^2$.
If you plot these two curves for $x \in [0, 2\pi]$, you will notice that, beside the trivial root $x=0$, they intersect just before $x=\frac{3\pi}{2}$. The function being very stiff, it is then better to search for the solution of $\sin(x)=x^2\cos(x)$ as already mentioned by Jack D'Aurizio.
For getting the solution, there are different ways. First, to get an approximation, you can expand the expression as a Taylor series built at $x=\frac{3\pi}{2}$. This gives, for the second order, $$\sin(x)-x^2\cos(x)=-1-\frac{9}{4} \pi ^2 \left(x-\frac{3 \pi }{2}\right)+\left(\frac{1}{2}-3 \pi \right)
\left(x-\frac{3 \pi }{2}\right)^2+O\left(\left(x-\frac{3 \pi }{2}\right)^3\right)$$ and, solving the quadratic, the solution to be retained is $$x = \frac{-12 \pi +54 \pi ^2+\sqrt{128-768 \pi +324 \pi ^4}}{48 \pi -8}\simeq 4.66651$$ which is not bad since the exact solution is $4.66650$ as already mentioned by Jack D'Aurizio.
To reach accurate solutions, Newton method, starting with a guess $x_0=\frac{3\pi}{2}$ will updtate it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ and the iterates will be $4.667357343$, $4.666499882$, $4.666499563$ which is the solution for ten significant figure.
We could have replace Newton by Halley or Housholder methods; they require higher derivatives but less iterations. I do not think that it could be of major interest for this specific problem.
Just for illustration, Halley method gives as first iterate $$x_1=\frac{3 \pi }{2}-\frac{4 \left(-8+48 \pi +81 \pi ^4\right)}{729 \pi ^6} \simeq 4.666542356$$ while Householder would give $$x_1=\frac{3 \pi }{2}-\frac{4 \left(128+3 \pi \left(3 \pi \left(448-48 \pi ^2+432 \pi
^3+729 \pi ^6\right)-512\right)\right)}{59049 \pi ^{10}} \simeq 4.666501749$$
|
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How can I show why this equation has no complex roots? I've been asked to show why an equation has no complex roots but i'm at a complete loss.
The equation is
$F_{n+2}=F_n$
Where $F_n=(x-1)(x-2)...(x-n)$ and n is a positive integer.
I'd really appreciate if someone could explain how I could go about showing this because I'd really like to understand.
Thanks in advance.
|
$F_{n}=(x-1)(x-2) \cdots (x-n)$
$F_{n+2}=(x-1)(x-2) \cdots (x-n)(x-(n+1))(x-(n+2))$
$F_{n} = F_{n+2}$
$(x-1)(x-2) \cdots (x-n)=(x-1)(x-2) \cdots (x-n)(x-(n+1))(x-(n+2))$
$(x-1)(x-2) \cdots (x-n)[(x-(n+1))(x-(n+2)) - 1] = 0$
$(x-1)(x-2) \cdots (x-n)[(x^2-[(n+2)+(n+1)]x+(n+1)(n+2) - 1] = 0$
$(x-1)(x-2) \cdots (x-n)[x^2-(2n+3)x+n^2+3n+ 1] = 0$
Therefore $x = 1,2,3,...,(n-1),n$ and $x^2-(2n+3)x+n^2+3n+ 1 = 0$
$$x=\frac{(2n+3)\pm \sqrt{(2n+3)^2-4(1)(n^2+3n+1)}}{2(1)}$$
$$x=\frac{(2n+3)\pm \sqrt{5}}{2}$$
$$x=\frac{2n+3+ \sqrt{5}}{2} x=\frac{2n+3- \sqrt{5}}{2}$$
Therefore the roots are not complex
|
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Sum the series $\sum_{n = 1}^{\infty}\{\coth (n\pi x) + x^{2}\coth(n\pi/x)\}/n^{3}$ This sum is from Ramanujan's letters to G. H. Hardy and Ramanujan gives the summation formula as
\begin{align} &\frac{1}{1^{3}}\left(\coth \pi x + x^{2}\coth\frac{\pi}{x}\right) + \frac{1}{2^{3}}\left(\coth 2\pi x + x^{2}\coth\frac{2\pi}{x}\right) \notag\\
&\, \, \, \, \, \, \, \, + \frac{1}{3^{3}}\left(\coth 3\pi x + x^{2}\coth\frac{3\pi}{x}\right) + \cdots\notag\\
&\, \, \, \, \, \, \, \, = \frac{\pi^{3}}{90x}(x^{4} + 5x^{2} + 1)\notag
\end{align}
Since $$\coth x = \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} = \frac{1 + e^{-2x}}{1 - e^{-2x}} = 1 + 2\frac{e^{-2x}}{1 - e^{-2x}}$$the above sum is transformed into $$(1 + x^{2})\sum_{n = 1}^{\infty}\frac{1}{n^{3}} + 2\sum_{n = 1}^{\infty}\frac{e^{-2n\pi x}}{n^{3}(1 - e^{-2n\pi x})} + 2x^{2}\sum_{n = 1}^{\infty}\frac{e^{-2n\pi/x}}{n^{3}(1 - e^{-2n\pi/x})}$$ If we put $q = e^{-\pi x}$ we get sums like $\sum q^{2n}/\{n^{3}(1 - q^{2n})\}$ which I don't know how to sum.
It seems I am going on a wrong track. Please provide some alternative approach.
Update: All the answers given below so far use complex analyis (transforms and residues) to evaluate the sum. I am almost certain that Ramanujan did not evaluate the sum using complex analysis. Perhaps the method by Ramanujan is more like the one explained in this question. Do we have any approach based on real-analysis only?
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Recall the well known Mittag-Leffler expansion of hyperbolic cotangent function (denote $\mathbb{W}=\mathbb{Z}/\{0\}$) :
$$\sum_{m\in\mathbb{W}}\frac{1}{m^2+z^2}=\frac{\pi\coth\pi z}{z}-\frac{1}{z^2}\tag{ML}$$
Hence, your sum is by its symmetry :
$$\begin{align}
S&=\frac{1}{2}\sum_{n \in \mathbb{W}}\{\coth (n\pi x) + x^{2}\coth(n\pi/x)\}/n^{3} \\ \\
&=\frac{1}{2\pi x}\sum_{n \in \mathbb{W}}\left(\frac{1}{n^4}+\sum_{m\in\mathbb{W}}\frac{x^2/n^2}{m^2+n^2x^2}\right)
+\left(\frac{x^4}{n^4}+\sum_{m\in\mathbb{W}}\frac{x^2/n^2}{m^2+n^2/x^2}\right)\tag{1}\\ \\
&=\frac{1}{2\pi x}\left(\zeta(4)+x^4\zeta(4)+\sum_{n,m \in \mathbb{W}^2}\frac{x^2}{n^2}\frac{1}{m^2+n^2x^2}+\frac{x^2}{n^2}\frac{1}{m^2+n^2/x^2}\right)\tag{2}\\ \\
&=\frac{1}{2\pi x}\left(2\zeta(4)+2x^4\zeta(4)+x^2\sum_{n,m\in\mathbb{W}^2}\frac{1}{n^2m^2}\frac{m^2+n^2x^2}{m^2+n^2x^2}\right)\tag{3}\\ \\
&=\frac{1}{2\pi x}\left(2\zeta(4)+2x^4\zeta(4)+4x^2\zeta^2(2)\right)\tag{4}\\ \\
&=\frac{1}{2\pi x}\left(2\frac{\pi^4}{90}+2x^4\frac{\pi^4}{90}+4x^2\frac{\pi^4}{36}\right)\tag{5}\\ \\
&=\frac{\pi^3}{90x}\left(1+x^4+5x^2\right)
\end{align}$$
Explanations
$(1)$ Use the Mittag-Leffler formula (ML) with $z=nx$ and $z=n/x$
$(2,4)$ Recall $\zeta(s)=\sum_{n=1}^{\infty}1/n^s$
$(3)$ In the second sum rename $n \longleftrightarrow m$
$(5)$ Zetas for $s=2$ and $4$ are well known, i.e. $\zeta(2)=\pi^2/6$ and $\zeta(4)=\pi^4/90$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Discontinuous seminorm on Banach space We have known that if $X$ is a Banach space and $\sum_{n=0}^{\infty}x_n$ is an absolutely convergent series in $X$ then $\sum_{n=0}^{\infty}x_n$ is a convergent series. Moreover, we have
$$
(*)\quad \left\|\sum_{n=0}^{\infty}x_n\right\|\leq\sum_{n=0}^{\infty}\|x_n\|
$$
by the continuity of $\|.\|$ on $X$.
My own question is that if we replace $\|.\|$ by a seminorm that is a functional $f$ satisfying
*
*$f(\lambda x)=|\lambda|f(x) \quad \forall \lambda\in\mathbb{R}, \forall x\in X$
*$f(x+y)\leq f(x)+f(y) \quad \forall x,y\in X$
thẹn whether the inequality (*) still holds?
My attemption. We can use Zabreiko's lemma (Zabreiko's Lemma) to give the solution for this question by construct a discontinuous seminorm on a Banach space. But I would like to construct explicit Banach space $X$, discontinuous seminorm $p$, and an absolutely convergent series $\sum_{n=0}^{\infty}x_n$ on $X$ satisfying
$$
p\left(\sum_{n=0}^{\infty}x_n\right)>\sum_{n=0}^{\infty}p(x_n).
$$
Thank you for all helping and comments.
|
There exists a discontinuous functional $f:X\to \mathbb{R}$ take $p(x) =|f(x)|$ . Since $f$ is discontinuous then $\mbox{Ker}(f)$ is dense in $X.$ Take any $v\in X\setminus \mbox{Ker}(f)$ and construct absolutely convergent (with respect to norm) series such that $v=\sum_{k=1}^{\infty} v_k$ then you have $$p(v)>\sum_{k=1}^{\infty} p(v_k) =0.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to express $\log_2 (\sqrt{9} - \sqrt{5})$ in terms of $k=\log_2 (\sqrt{9} + \sqrt{5})$? If $$k=\log_2 (\sqrt{9} + \sqrt{5})$$
express $\log_2 (\sqrt{9} - \sqrt{5})$ in terms of $k$.
|
Adding the logarithms $\log_2{(\sqrt{9}-\sqrt{5})}$ and $\log_2{(\sqrt{9}+\sqrt{5})}$ we get the following:
$$\log_2{(\sqrt{9}-\sqrt{5})}+\log_2{(\sqrt{9}+\sqrt{5})}=\log_2{(\sqrt{9}-\sqrt{5})(\sqrt{9}+\sqrt{5})}=\log_2{(9-5)}=\log_2{4}=\log_2{2^2}=2 \cdot \log_2{2}=2$$
Knowing that $k=\log_2{(\sqrt{9}+\sqrt{5})}$ we have:
$$\log_2{(\sqrt{9}-\sqrt{5})}+\log_2{(\sqrt{9}+\sqrt{5})}=\log_2{(\sqrt{9}-\sqrt{5})}+k$$
Therefore, $$2=\log_2{(\sqrt{9}-\sqrt{5})}+k \Rightarrow \log_2{(\sqrt{9}-\sqrt{5})}=2-k$$
|
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|
Image of centralizer under an isomorphism Suppose we have a group isomorphism $\phi: G\rightarrow K$ between two finite groups and let $H$ a subgroup of $G$. Are there any known facts about the image of the centralizer $C_G(H)$ of $H$ in $G$ under $\phi$?
Also, same question for the special case of $K=G$ i.e. when $\phi$ is an automorphism.
|
Morally, you can view isomorphism as just relabeling the elements of a group without changing any structure. So the best we can hope to say is that $$\phi(C_G(H)) = C_K(\phi(H))$$
Edit: I realize now that "best we can hope to say" is a bit strange. Basically any structure you define in group theoretic terms will carry over precisely under isomorphism.
|
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|
Why is the result of $-2^2 = -4$ but $(-2)^2 =4$? I am really new into math, why is $-2^2 = -4 $ and $(-2)^2 = 4 $?
|
It is true that $-2^2$ is ambiguous (unless you know the convention), because $-(2^2) \ne (-2)^2$. (And, indeed, some calculators or programing languages may do it using their own convention, different from the mathematicians' convention.) The mathematicians' convention is $-2^2 = -(2^2)$. Why? Presumably because we often need to write $-(2^2)$. But if you ever need to write $(-2)^2$ you can just write $2^2$ instead.
|
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|
Question about queues A queue is implemented with a sequence $Q[1\ldots n]$ and two indices $\def\head{\operatorname{head}}\head[Q]$ and $\def\end{\operatorname{end}}\end[Q]$ such that the elements of the queue are the following:
$$Q[\head[Q]], Q[\head[Q]+1], \ldots , Q[\end[Q]-1]$$
Initial $\head[Q]=\end[Q]=1$
When $\head[Q]=\end[Q]+1$ the queue is complete.
*
*Does ”Initial $\head[Q]=\end[Q]=1$” stand when there is one element in the queue?
*Could you explain me what the following means?
”When $\head[Q]=\end[Q]+1$ the queue is complete.”
|
When $head=end$ we consider the queue empty.
If there is 1 element in the queue, the queue looks like:
$$-\ \ - \ - \underset{\stackrel\uparrow{head}}o \underset{\stackrel\uparrow{end}}- \ \ -$$
That means that:
$$head+1 \equiv end\pmod{n}$$
At some point in time the queue might look like:
$$o \underset{\stackrel\uparrow{end}}- - \underset{\stackrel\uparrow{head}}o o \ \ o$$
If we add one more element at the end, we get:
$$o\ \ o \underset{\stackrel\uparrow{end}}- \underset{\stackrel\uparrow{head}}o o\ \ o$$
At this point we cannot add any more elements, because if $head=end$, that means we consider this an empty queue. In other words, the queue is full or complete when:
$$end+1 \equiv head \pmod{n}$$
|
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|
Help Evaluating $\lim_{x\to0}(\frac{\sin{x}}{{x}})^{\frac{1}{x}}$ Does anyone know how to evaluate the following limit?
$\lim_{x\to0}(\frac{\sin{x}}{{x}})^{\frac{1}{x}}$
The answer is 1 , but I want to see a step by step solution if possible.
|
Let $\displaystyle y=(\frac{\sin x}{x})^{\frac{1}{x}}$. Then $\displaystyle\lim_{x\to 0} \ln y=\lim_{x\to 0}\ln \left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=\lim_{x\to 0}\frac{1}{x}\ln\left(\frac{\sin x}{x}\right)=\lim_{x\to 0}\frac{\ln\sin x-\ln x}{x}=^{(LH)}\lim_{x\to 0}\frac{\frac{\cos x}{\sin x}-\frac{1}{x}}{1}$
$\displaystyle=\lim_{x\to 0}\frac{x\cos x-\sin x}{x\sin x}=^{(LH)}\lim_{x\to 0}\frac{-x\sin x+\cos x-\cos x}{x\cos x+\sin x}=^{(LH)}\lim_{x\to 0}\frac{-x\cos x-\sin x}{-x\sin x+\cos x+\cos x}=\frac{0}{2}=0,$
so $\displaystyle\lim_{x\to 0}y=e^0=1$.
|
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|
A linear map that is multiplication by a matrix The problem statement, all given variables and data
Let $T$ be multiplication by matrix $A$:
$$A=
\begin{bmatrix}
1 & -1 & 3 \\
5 & 6 & -4 \\
7 & 4 & 2 \\
\end{bmatrix}
$$
Find the range, kernel, rank and nullity of T.
Attempt at a solution
I can find all of these values except for A. What does the question mean when T is a multiplication by matrix A? How do I get range/kernel etc. of T.. I know how to get them for A.
Thanks.
|
Any matrix $A$ represents a linear function $T$, or more precisely $T_A$, defined by multiplication, that is, by $T(x)=Ax$. Now it makes sense to talk about the range and kernel of $T$. Rank and nullity are simply the dimension of range and kernel of $T$, respectively.
Regarding your question, note that, by definition or a little argument, the rank of any matrix $A$, agrees with the rank of the linear map $T_A$, and the nullity of $A$ is the same as the nullity of $T_A$.
For instance,
$$\ker(T)=\{x\in \mathbb{R}^3\mid T(x)=\mathbf{0}\}=\{x\in \mathbb{R}^3\mid Ax=\mathbf{0}\}.$$
Thus, to find the kernel of $T$, you need to solve the homogeneous system $Ax=\mathbf{0}$.
|
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|
Pólya's urn scheme, proof using conditional probability and induction Problem
An urn contains $B$ blue balls and $R$ red balls. Suppose that one extracts successively $n$ balls at random such that when a ball is chosen, it is returned to the urn again along with $c$ extra balls of the same color. For each $n \in \mathbb N$, we define $R_n=\{\text{the n-th ball extracted is red}\}$, and $B_n=\{\text{the n-th ball extracted is blue}\}.$
Prove that $P(R_n)=\dfrac{R}{R+B}$.
I thought of trying to condition the event $R_n$ to another event in order to use induction. For example, if $n=2$, I can express $$P(R_2)=P(R_2|R_1)P(R_1)+P(R_2|B_1)P(B_1)$$$$=\dfrac{R+c}{R+B+c}\dfrac{R}{R+B}+\dfrac{R}{R+B+c}\dfrac{B}{R+B}$$$$=\dfrac{R}{R+B}.$$
Now, suppose the formula is true for $n$, I want to show it is true for $n+1$.
So, $P(R_{n+1}=P(R_{n+1}|R_n)P(R_n)+P(R_{n+1}|B_n)P(B_n)$$$=P(R_{n+1}|R_n)P(R_n)+P(R_{n+1}|B_n)(1-P(R_n)$$$$=P(R_{n+1}|R_n)\dfrac{R}{R+B}+P(R_{n+1}|B_n)(1-\dfrac{R}{R+B}).$$
I am having some difficulty trying to calculate $P(R_{n+1}|R_n)$ and $P(R_{n+1}|B_n)$. I would appreciate if someone could complete my answer or suggest me how can I finish the proof if what I've done up to now is correct.
|
I would actually still advocate the approach suggested here with a small change in the way it is presented:
$P(R_1)=\frac{R}{R+B}$, now we need to prove that $P(R_n)=P(R_{n+1})$.
$P(R_{n+1})=P(R_{n+1}|R_n)P(R_n)+P(R_{n+1}|B_n)(1-P(R_n))$
$X_n$, the number of red balls in the urn at step $n$, is $P(R_n)T_n$, where $T_n$ is the total number of balls on step $n$ which is deterministic.
$P(R_{n+1}|R_n)=\frac{T_nP(R_n)+c}{T_n+c}$
$P(R_{n+1}|B_n)=\frac{T_nP(R_n)}{T_n+c}$
$P(R_{n+1})=\frac{T_nP(R_n)+c}{T_n+c}P(R_n)+\frac{T_nP(R_n)}{T_n+c}(1-P(R_n))=P(R_n)$.
The approach does not use mathematical expectations, it can be considered as an advantage because this problem is often given to students before the study mathematical expectations.
|
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|
Are all endpoints discontinuous? I learned that something is a limit if the left limit and right limit exist and are equal. But then doesn't this mean that if I have a function on $[a,b]$, that the endpoints $a$ and $b$ are discontinuous because $a$ doesn't have a left limit and $b$ doesn't have a right limit?
|
Given $f : A \subset \Bbb R \to \Bbb R$, we say that $f$ is continuous at $x_0 \in A $ if for every $\epsilon > 0 $, exists $\delta > 0 $ such that: for all $x \in A$ that $|x - x_0| < \delta$ implies $|f(x) - f(x_0)| < \epsilon$.
The points $x$ considered in the last implication must first of all be in the function's domain. The only possibility in the endpoints $a$ and $b$ are going to $a$ from the right, and going to $b$ from the left. It makes no sense otherwise.
In general, when you have a function $f: A \subset \Bbb R \to \Bbb R $, we only consider limits of $x$ going to limit points of $A$, also called points of accumulation.
A point $x_0 \in A$ is said a point of accumulation of $A$ if for all $\epsilon > 0$, we have $(]x_0 - \epsilon, x_0 + \epsilon [ \setminus \{x_0\}) \cap A \neq \varnothing $.
|
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|
Differentiation under the integral sign (one complex variable) Let $u(z), u'(z)$ be complex-analytic functions on an open neighborhood $\Omega \subseteq \mathbb{C}$ of the origin. Also, let $f(X)$ be a complex-analytic function. For $s \in [0,1],$ define
$$g(s,z) = f((1-s)u(z) + su'(z))$$
and
$$h(z) = \int_0^1 g(s,z) ds. $$
How do we compute $h'(z)$?
*
*Is it possible to differentiate under the integral sign? (Most of the results I have seen regarding this assumes that $z$ is also real.) If yes, I would appreciate being pointed to a proof of this result.
*In particular, can we estimate $|h'(z)|$ as follows?
$$ |h'(z)| \leq \int_0^1 |\partial_z g(s,z)| ds. $$
|
Since $g$ is smooth, we see that ${\partial g(s,z) \over \partial z}$ is bounded on $[0,1] \times U$, where $U$ is a bounded neighbourhood of $\Omega$.
The mean value theorem gives $|g(s,z+h)-g(s,z)| \le \left( \sup_{(s,z) \in [0,1] \times U}|{\partial g(s,z) \over \partial z}| \right) | h| $.
Hence ${1 \over h} |g(s,z+h)-g(s,z)| \le \left( \sup_{(s,z) \in [0,1] \times U}|{\partial g(s,z) \over \partial z}| \right)$ and so the left hand side is uniformly bounded. Since $\lim_{ h \to 0} {1 \over h} (g(s,z+h)-g(s,z)) = {\partial g(s,z) \over \partial z}$, the bounded convergence theorem gives
$h'(z) = \lim_{h \to 0} \int_0^1 {1 \over h} (g(s,z+h)-g(s,z)) ds= \int_0^1{\partial g(s,z) \over \partial z} ds $.
The bound $|h'(z)| \le \int_0^1 |{\partial g(s,z) \over \partial z}| ds $ follows.
Addendum: Suppose $|\phi_n(t)| \le K$, $\phi_n(t) \to \phi(t)$ and $A$ has finite measure. Then $|\operatorname{re} \phi_n(t)| \le K$, and
$\operatorname{re} \phi_n(t) \to \operatorname{re} \phi(t)$, so
$\int_A \operatorname{re} \phi_n(t)dt \to \int_A \operatorname{re} \phi(t) dt$. Similarly for the imaginary part. Then linearity of the integral gives
$\int_A \phi_n(t)dt \to \int_A \phi(t) dt$.
In the above case, take any sequence $h_n \to 0$ ($h_n \neq0$), and let
$\phi_n(t) = {1 \over h_n} (g(s,z+h_n)-g(s,z))$.
|
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|
Finding solutions of $y'''-4y''+5y'-2y=-x^2+5x+2$ Find all solutions of $$y'''-4y''+5y'-2y=-x^2+5x+2.$$
I know how to find the solutions of the corresponding homogenous differential equation $y'''-4y''+5y'-2y=0$. I've done that in the following way:
The characteristic equation is $$P(x) = x^3 - 4x^2 +5x-2=(x-1)^2(x-2).$$
So if we put $$f_1:x\mapsto e^x\\f_2:x\mapsto xe^x\\f_3:x\mapsto e^{2x}$$ then the solution basis is $$\{f_1, f_2, f_3\}$$ and all solutions are linear combinations of this basis.
However, I am not sure how to deal with the inhomogenous part of the differential equation.
Please share the general approach with me.
|
Hint
It may not be a very academic solution but since you found the solution for the homogenous part, let us set, because the rhs is a polynomial, that $$y=c_1 e^x+c_2 x e^{x}+c_3 e^{2x}+P(x)$$ Differentiate three times and substitute. You end with $$P'''(x)-4 P''(x)+5 P'(x)-2 P(x)=-x^2+5x+2$$
Try $P(x)=a+bx+x^2+dx^3$ and identify.
|
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|
How to deduce a closed formula given an equivalent recursive one? I know how to prove that a closed formula is equivalent to a recursive one with induction, but what about ways of deducing the closed form initially?
For example:
$$ f(n) = 2 f(n-1) + 1 $$
I know how to use induction to prove that $\forall n \ge1$:
$$ f(n) = 2^n f(0) + 2^n - 1 $$
And I was able to come up with that formula in the first place just by examining it informally. It seems like there should have been a straightforward, formal way to deduce the closed form from the recursive definition in the first place, but I'm blanking.
I know it's not always easy (e.g. Fibonacci) but it seems like it ought to be here. For example, how might I go about procedurally deducing the closed form for:
$$ f(n) = 2*f(n-1) + k $$
Thanks.
|
$$f(n) = 2f(n-1) + k \\ f(n-1) = 2f(n-2) + k \\ f(n-2) = 2f(n-3) + k \\ \dots \\ f(1) = 2f(0) + k$$
$$$$
$$\Rightarrow f(n) = 2f(n-1) + k= \\ 2(2f(n-2) + k)+k= \\ 2^2f(n-2)+(2+1)k=2^2(2f(n-3) + k)+(2+1)k= \\ 2^3f(n-3)+(2^2+2+1)k= \\ \dots \overset{*}{=} \\ 2^nf(0)+(2^{n-1}+2^{n-2}+\dots +2+1)k$$
Therefore, $$f(n)=2^nf(0)+k\sum_{i=0}^{n-1}2^i= \\ 2^nf(0)+k\frac{2^n-1}{2-1}= 2^nf(0)+(2^n-1)k$$
EDIT:
$(*):$ We see that $2^3f(n-3)+(2^2+2+1)k$ is of the form $2^jf(n-j)+(2^{j-1}+2^{j-2}+\dots+2^2+2+1)k$
After some steps we have the following:
$2^nf(n-n)+(2^{n-1}+2^{n-2}+\dots+2^2+2+1)k= \\ 2^nf(0)+(2^{n-1}+2^{n-2}+\dots+2^2+2+1)k$
|
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|
Factoring $x^{4} +1$, using real factoring to the second degree Factoring to the second degree using real numbers
$$x^{4} +1$$
I know that $ x^{4} +1=(x^{2} + i)(x^{2}-i).\;$ But these are complex but I thought using these in some kind of way? Got no where!
And then I tried to guess, two solutions are $\pm (-1)^{1/4},\, $
which gave me $(x-\sqrt{i})(x+\sqrt{i}).\;$ But I have a feeling that I am really off here...
Hmm... Thinking!
What kind of techniques do you use?
|
HINT :$$x^4+1+\color{red}{2x^2}-\color{red}{2x^2}=(x^4+2x^2+1)-2x^2=(x^2+1)^2-(\sqrt 2x)^2$$
|
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|
If $\sin( 2 \theta) = \cos( 3)$ and $\theta \leq 90°$, find $\theta$ Find $\theta\leq90°$ if
$$\sin( 2 \theta) = \cos( 3)$$
I know that $\sin 2\theta = 2\sin\theta\cos \theta$, or alternatively, $\theta = \dfrac{\sin^{-1}(\cos 3)}{2}$.
Can somebody help me?
|
Assuming you mean $3^{\circ}$, here are the steps
$$ \sin 2\theta=\cos 3^{\circ} $$
$$ \arcsin(\sin 2\theta)=\arcsin(\cos 3^{\circ}) $$
$$ 2\theta=\arcsin(\cos 3^{\circ}) $$
$$ \theta=\frac{1}{2}\arcsin(\cos 3^{\circ})=\frac{87^{\circ}}{2}=43.5^{\circ} $$
|
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|
$\ln (2 x-5)>\ln (7-2 x)$ Solve
$$\ln (2 x-5)>\ln (7-2 x)$$
The answer is given as $$3<x<7/2$$
This is what I have done $$\ln (2 x-5)-\ln (7-2 x)>0$$
$$\ln \left(\frac{2 x-5}{7-2 x}\right)>0$$
However I am not able to understand how to get to the answer provided.
|
Hint
If $$\ln \left(\frac{2 x-5}{7-2 x}\right)>0$$ it implies that $$\frac{2 x-5}{7-2 x}>1$$ But take care : the logarithm is such that its argument must be positive.
I am sure that you can take from here.
|
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How to maximize pay with repeated toss of coin repeated toss a coin and you can stop anytime and payoff is just #times you got head divided by total number of throws, how do you maximize your pay.
Does anyone have a clever strategy for this? This was changed from another problem I solved involving flipping over cards from a deck and being paid the fraction of red. Also never saw a satisfying solution
|
So I think that you should stop whenever the ratio of heads to flips is greater than 1/2. Like obviously since there is no maximal number of flips you can "eventually" get this as the ratio so stoping when your ratio is lower is silly but you can't guarantee you will "eventually" get a higher ratio so if you ever do get that (example sequences H or THH) then you should stop.
|
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|
Does validity of contrapostive proofs require the Law of Excluded Middle? I remember that during my first proofs class the hardest thing I had accepting were the logic we had to learn, and it seems I still have questions about.
So I was thinking about why when the contrapositive is proved true then it implies that the original statement was true. The way I've been thinking about it is by considering a statement about an element of some set $A$. Letting $Q(x)$ represent $x\in A$ such that this statement holds true for, so the way I've translated $$Q\rightarrow P\Longleftrightarrow \sim P\rightarrow \sim Q$$
to $$Q(x)\subseteq P(x)\Longleftrightarrow P(x)^{c}\subseteq Q(x)^{c}$$
This makes sense initially to me since it does seem that elements that follow $\sim P$ would be element that don't belong to $P(x)$ (i.e. they belong to $P^{c}(x)$) Thus this would make sense to me because $P(x)\cup P(x)^{c}=A$ since it seems that either P or $\sim P$ must hold for an element.
I guess the main question if this last statement is possible would rely on: is the law of excluded middle always hold? Could you perhaps have a nonsense statement, so its negation is also nonsense, and no possible element is from either. Or perhaps the way I'm thinking about contrapositive statements is wrong?
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In classical logic every statement has a truth value, even if it is nonsense like "if the Moon is made of cheese then the sky is made of rubber" (this one is true because any implication with a false premise is defined to be true). The law of excluded middle is a law of classical logic, so it is always true as an axiom, in particular for any set either $x\in A$ or $x\notin A$ is always true.
There are alternative systems of logic where this law is not adopted, intuitionistic logic for example, but there the interpretation is not that $x\in A$ and $x\notin A$ are both nonsense, but rather that there is no "constructive" way to establish either. See more in Can one prove by contraposition in intuitionistic logic?
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/908984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$ For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$
My try :
Left hand side :
$\begin{align} \sum_{d|p^k}\sigma (d) &= \sigma(p^0) + \sigma(p^1) + \sigma(p^2) + \cdots + \sigma(p^k) \\ &= \dfrac{p^{0+1}-1}{p-1} + \dfrac{p^{1+1}-1}{p-1} + \cdots + \dfrac{p^{k+1}-1}{p-1} \\&= \dfrac{1}{p-1}\left( (p + p^2 + \cdots + p^{k+1}) - (k+1)\right) \\&= \dfrac{1}{p-1}\left(\dfrac{p(p^{k+1}-1)}{p-1}- (k+1)\right) \\
\end{align}$
I'm not sure if I am in right path; this doesn't look simple. Any help ?
|
looking at this intuitively, firstly we note that:
$$
\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d) \\
= \sum_{d|n} d\tau(n/d)
$$
so now we are summing the divisors $d$ of $n$, each divisor being counted with multiplicity $\tau(n/d)$. so you just have to persuade yourself that this multiplicity is appropriate.
but this is evident if we look at a particular $d$, since for any integer $m$ we have
$$
dm | n \Leftarrow \Rightarrow m | n/d
$$
in words there is a 1-1 correspondence between (a) multiples of $d$ which divide $n$, and (b) divisors of $n/d$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/909131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
}
|
Does $G$ always have a subgroup isomorphic to $G/N$? Let $G$ be a group and $N$ a normal subgroup of $G$. Must $G$ contain a subgroup isomorphic to $G/N$? My first guess is no, but by the fundamental theorem of abelian groups it is true for finite abelian groups, so finding a counterexample has been a little tough. The finite case is also interesting for me.
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Let $G$ be the reals $\mathbb{R}$under addition. Let $N$ be the integers $\mathbb{Z}$ under addition. $G/N$ is the group of reals modulo $1$. This has a non-zero element whose square is $1$, the equivalence class of $0.5$. So $G/N$ is not isomorphic to any subgroup of $G$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/909224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
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