3
votes
How do I calculate the value of a power series using the radius of convergence of another series?
HINT
Consider the following power series (known as geometric series) which converges whenever $|x| < 1$:
\begin{align*}
\sum_{n=0}^{\infty}x^{n} = 1 + x + x^{2} + x^{3} + \ldots = \frac{1}{1 - x}
\...
- 7,078
3
votes
Is there exist a real continuous function f such that $\int_0^x f(t) dt =mf(x)$ for some real constant m?
We can also try to use the Fundamental Theorem of Calculus here to try and find a solution. We start with the equalty
$$mf(x)=\int_0^x f(t) \mathrm{d}t.$$
Differentiating this then yields
$$mf'(x)=f(x)...
- 2,497
2
votes
Show that $f_n(x)$ converges uniformly to a continuous function on $[0,1 ]$
Use the contraction mapping theorem applied to the space $C[0,1]$ and the mapping $T:C[0,1]\to C[0,1]$ with
$$Tf(x)=-1+\frac{1}{3}\int_0^1\cos\left(3x^2+y^2+1\right)f(y)\,dy.$$
To see that this is a ...
- 4,081
2
votes
Is this condition clearly stated: $f(x)=0 \Rightarrow f(y)\geq(OR \leq) 0$ for all function $f$ in $A$?
It's generally cleaner and clearer to put quantifiers at the front of the expression, and use "for all" instead of "for any". Furthermore, your condition seems to be that if $f(x)$ ...
- 114k
2
votes
Need Help in Reasoning with Extremal Points
Hint : By the triangle inequality, and the fact that $|f(x)| \leq 1$ for every $x$, one has
$$1 = \left| \int_K f(x) d\mu\right| \leq \int_K |f(x)| d\mu \leq \int_K 1 d\mu = \mu(K)=1$$
So you must ...
- 23.8k
1
vote
Problem on studying $\sum_{n\geq 1}\frac{n^{\log{n}}}{\sqrt{n!}}\frac{\tan{n}}{|\tan{n}|+n}$, comparison criterion
Looking at
$\dfrac{n^{\log{n}}}{\sqrt{n!}}\dfrac{\tan{n}}{|\tan{n}|+n}
$,
using the n-th root test,
$\begin{array}\\
\left(\dfrac{n^{\log{n}}}{\sqrt{n!}}\right)^{1/n}
&=\dfrac{n^{\log{n}/n}}{n!^{1/...
- 101k
1
vote
Rigorous definition of $\lim\limits_{{\mathbf{|x|}\to\infty}}\mathbf{f(x)}=\infty$ $\mathbf{f}:M\to\mathbb{R^m}$
$$
\lim\limits_{{\mathbf{|x|}\to\infty}}\mathbf{f(x)}\to\infty
$$
means
For every $A > 0$, there exists $B > 0$ so that:
for all $\mathbf{x} \in \mathbb R^n$, if
$|\mathbf{x}| > B$, then $|\...
- 96.5k
1
vote
Accepted
Rigorous definition of $\lim\limits_{{\mathbf{|x|}\to\infty}}\mathbf{f(x)}=\infty$ $\mathbf{f}:M\to\mathbb{R^m}$
Let $f:M\to\mathbf{R}^m$ be a function from a subset $M$ of $\mathbf{R}^n$ to $\mathbf{R}^m$.
When $M$ is a bounded subset of $\mathbf{R}^n$, the expression $\displaystyle \lim_{|x|\to\infty}f(x)$ ...
- 1,080
1
vote
If $f\in L^1(\mathbb{R}^d)$ is uniformly continuous, do its integrals over spheres of increasing radius decay to zero?
You're right, it's false.
First, consider the function
$$\operatorname{bump}(x;x_0,a)=\begin{cases}\exp\left(\frac{-1}{1-(x-x_0)^2/a^2}\right)&|x-x_0|<a \\ 0 & |x-x_0|\geq a\end{cases}$$
...
- 8,147
1
vote
Proof Check Lemma 2.2.10 in Tao
To prove the existence, it will be easy if we consider the contrapositive of above statement
$\forall a \gt 0 \to \exists b \ (b++ = a)$
$\longleftrightarrow \forall b (b++ \ne a) \to \exists a (a = ...
- 121
1
vote
If $(a_n)\subset[0,\infty)$ is non-increasing and $\sum a_n<\infty$, then $\lim{n a_n} = 0$
https://en.wikipedia.org/wiki/Limit_comparison_test
By contradiction if $\lim\limits_{n\to\infty} n a_n\ne 0 $ then $\lim\limits_{n\to\infty} \frac{a_n}{\frac{1}{n}} \ne 0 $ so $\sum\limits_{n=1}^\...
- 21
1
vote
Accepted
If $a_n$ is a positive sequence and $\lim \limits_{n \to \infty}a_n=0$ then there exists $N>0$ such that $(a_{N+n})$ is decreasing
The counterexample you gave is correct. This is sufficient to solve the problem given. As for what follows..
If we assume $(a_{N+n})$ is decreasing for some $N$, that means $a_{N+n+1} \leqslant a_{N+...
- 6,658
1
vote
Prove $\Bbb E(X)=\int_{\Bbb R}xf(x)\,d\lambda$ in rigorous measure theory
As mentioned in the comment, the integral shouldn't be over $X(\Omega)$. The general change of variables theorem tells you
\begin{align}
\int_{\Omega}X\,dP&=\int_{X^{-1}(\Bbb{R})}(\text{id}_{\Bbb{...
- 34.6k
1
vote
How to prove that $x_n=1+\frac{1}{2!}+\cdots+\frac{1}{n!}$ is a Cauchy sequence?
Let n and m be nonnegative integers (n,m≥1). We have:
|x_(n+m)-x_n |=|(1+1/2!+⋯+1/n!+1/(n+1)!+⋯+1/(n+m)!)-(1+1/2!+⋯+1/n!)|
=|1/(n+1)!+⋯+1/(n+m)!|=1/(n+1)!+⋯+1/(n+m)!≤1/(n+1)^2 +⋯+1/(n+m)^2 =|y_(n+m)-...
1
vote
Sufficient condition for a continuous function to be differentiable
The fact that the Weierstrass function has a special name seems to indicate that not being differentiable is the exception, and your question indicates a similar perception. This is, however, not ...
- 20.4k
1
vote
Accepted
Let $\{E_k\}$ be a sequence of measurable sets in X. Show that $A\in\mathfrak{M}$, where $A$ is the set of all x that lie in infinitely many $E_k$'s.
$$
A =\bigcap_{N=1}^\infty \bigcup_{n\ge N} E_k
$$
By definition, this means any point of $A$ belongs to infinitely many of $E_k$. By the fact that $E_k$ are all measurable, you conclude $A$ is as ...
- 3,857
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