Define the function $x(t)$ for $t\ge0$: $$ x(0)=1\\ x'(t)=-x(t/2) $$ I could do a power series from $t=0$ like  (thanks to @JeanMarie for pointing this old question out), but I ultimately want asymptotic bounds on the excursions as $t\to\infty$, so I don't think the power series helps.  Is there any other research on this function?  How quickly do these excursions grow with $t$?
123772
I have the following different equation $$f'(x) = f(x/2)$$ with $f(0)=10$. What type of DE is this, and how would you solve it? It seems $f(x)$ is likely to be some relative of $e^x$, since $f'(x) = f(x)$, which is close, but I don't even know what that type of DE is called with that $f(x/2)$ feature, so I'm not having any luck searching for a tutorial. The best candidate I've found was a &quot;delay differential equation&quot;, but that seems more suited to $f(x-3)$ than $f(x/2)$.
76955
I have figured out a way to show that     $$e=\sum_{i=0}^\infty {1\over i!}$$   I am wanting to formally show that     $$e^x = \sum_{i=0}^\infty {x^i\over i!}$$   I have been looking at power series/Taylor series for a long period of time (absolute convergence) and have seen multiple proofs that I look past because something seems illegitimate with radius of convergence. If someone explains the general proof behind Taylor series/power series absolutely converging so there is no gray area, that would work as well. I will probably ask some questions in the comment area if this is the case. There must be something I am missing.
104500
What would be the closed form of $$\sum_{x=1}^n x \log (x)$$?  I know the summation of $\sum \log x$, but how do I use this series to the $$\sum_{x=1}^n x \log (x)$$?$
50638
Suppose $f$ is positive and increasing on $[0, +\infty)$. Suppose that $F(x)=$$\int_0^x f(t)dt$. Prove that $$\int_0^{+\infty} \frac {1}{f(x)} dx$$ converges if and only if $$\int_0^{+\infty} \frac {x}{F(x)} dx$$ converges.   My thoughts, and attempt at solution -  Since $f$ is increasing, $F(x)=\int_0^xf(t)\,dt\le\int_0^xf(x)\,dt=xf(x).$ We can rearrange that to $\frac1{f(x)}\le\frac{x}{F(x)}.$ Hence if the integral of $\frac{x}{F(x)}$ converges, then so does the integral of $\frac1{f(x)}.$  However, the above working seems incomplete as it bounds the integral from the upper side only. In fact, it shows that if the second integral converges, then the first one may converge, but not vice versa. How do I place a lower bound on the integral, and go about the rest of the solution?   Haven't been able to proceed from here, please help.
121887
Show that there is at least a nonzero function $f$, differentiable on $[0,+\infty)$, satisfying $$f'(x)=2f(2x)-f(x) \qquad \forall x&gt;0 $$   $$M_n:=\int_{0}^{\infty}x^nf(x)dx&lt;\infty \qquad \forall n\in  \mathbb{N} $$   My best idea so far is to assume that the solution is a power series, i.e.  $$ f(x)=\sum_{n=0}^{\infty}a_nx^n\qquad \forall x&gt;0$$ Then the equation becomes $$ \sum_{n=0}^{\infty}na_nx^{n-1}=2\sum_{n=0}^{\infty}a_n2^nx^n-\sum_{n=0}^{\infty}a_nx^n$$ equating all the coefficients of the same degree I get $$na_n=(2^{n}-1)a_{n-1}\qquad \forall n\geq 1$$ So setting $a_0=1$, I get $$a_{n}=\frac{1}{n!}\prod_{k=1}^{n}(2^k-1) \qquad \forall n$$ But does the power series actually converge? Using Hadamard's formula, and that $2^{k}-1\geq 2^{k-1}$, $$ |a_n|^{1/n}\geq\frac{1}{(n!)^{1/n}}\left[2^{n(n-1)/2}\right]^{1/n}\sim\frac{e}{n(2\pi n)^{1/2n}}2^{(n-1)/2}\to \infty$$ so the radius of converge of the series is $0$, so it doesn't actually define a solution on $[0,+\infty)$.
10033
I'd like to determine the function corresponding to the following power series: $$x + \sum_{n=1}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n} \frac{x^{2n+1}}{2n+1}, $$ where $|x|&lt;1$.
68610
What is the closed form of  $$\sum_{k=0}^n \frac{x^k}{k!}$$  as a function of $x$ and $n$?  Knowing that it converges to $e^x$ when $n\to \infty$.
29849
Let $x \in \mathbb R$  Find a closed form of $\int_0^\infty\dfrac{\cos(xt)}{1+t^2}dt$ .   Let me give some context: this is an exercise from an improper integrals course for undergraduates. My teacher was not able to give a valid solution without  resorting to Fourier series.   He actually wanted to prove first that the function is a solution of the ODE $$ y-y'' =0$$ without succeeding in proving it.   Anyway the result sould be $\dfrac{\pi}{2}e^{-x}$  Thanks for any help on this subject.
30826
Let $f$ a continous function defined in the interval $[0,1]$. Assuming: $\forall x \in [0,1]:f(x) &gt; x$ Prove: $\forall x \in [0,1]:f(x) &gt; x + \varepsilon $  I tried to use  and did some Algebra tricks but it didn't bring me to a safe shore :)    What do you suggest? Thanks!
86009
Why is this function continuous over $x$? $$f^{n}(\xi(x))=\frac{-(n!\left(-f(x)+f(x_{0})+f(x_0)(x-x_0)+\ldots+\frac{f^{n-1}(x_0)*(x-x0)^{n-1}}{(n-1)!}\right)}{(x-x_0)^n}$$  I tried solving it via the Heine sequence criteria, but to no avail. I need someone help me with the explanation for this, please! Could the answer be, that since the left side of the equation is the final remainder of a Taylor series and a Taylor series converges to a certain function, then that remainder would ideally converge to zero?
39122
The Problem :  We state the following two definitions of the real exponential function from the . We're interested in showing that the two definitions are valid $($i.e. the defining sequence/series does converge to a unique real number$)$ and that the two definitions are equivalent.   I'm stuck at a couple of points $($which are described in highlighted lines$)$. Any help would be much appreciated. Thank you!     Definition $1$. The exponential function can be defined as the following limit of a sequence $$\exp x := \lim_{n \to \infty} \left({1 + \frac x n}\right)^n$$      Definition $2$. The exponential function can be defined as a  $$\exp x := \sum_{n = 0}^\infty \frac {x^n} {n!}$$   My Progress and two places where I'm stuck : Essentially the solution consists of three parts,  namely validity of definition $1$, validity of definition $2$ and equivalence of the two definitions.     Validity of Definition $1$. I'm stuck here! Can we show that the sequence $(a_n)$ given by $a_n = \left(1+\frac{x}{n}\right)^n$ converges for every $x \in \mathbb{R} ??$ Is it eventually monotone and bounded $??$   Validity of Definition $2$. The  of the power series is $$r=\lim_{n \to \infty}\left| \frac{\frac{1}{n!}}{\frac{1}{(n+1)!}} \right|=\lim_{n \to \infty}\left| \frac{(n+1)!}{n!} \right|=\lim_{n \to \infty}(n+1)=+\infty$$ Thus the infinite series in the right-hand side of Definition $2$ converges to a unique real number for all $x \in \mathbb{R}$. Hence the definition is well-defined.  Equivalence of Definition $1$ and Definition $2$.  There's this  of Definition $1$ $\implies$ Definition $2$ in the Pr$\infty$fWiki page, but it's kind of under construction and I'm not really convinced by it. So I decided to try to take my own shot at it.  For all $n \in \mathbb{N} \cup \{0\},$ let  $$T_n=\left(1+\frac{x}{n} \right)^n, ~S_n=\sum_{k=0}^n \frac{x^k}{k!}$$ We have to show that $\lim_{n \to \infty} T_n = \lim_{n \to \infty} S_n$  Now, \begin{align} T_n &amp;= \left(1+\frac{x}{n}\right)^n\\ &amp;= 1+n\cdot\frac{x}{n}+\frac{n(n-1)}{2!}\cdot\frac{x^2}{n^2}+\cdots +\frac{n(n-1)\cdots 1}{n!}\cdot\frac{x^n}{n^n}\\ &amp;= 1+x+\left(1-\frac{1}{n}\right)\cdot \frac{x^2}{2!}+\cdots +\left(1-\frac{1}{n}\right)\cdots \left(1-\frac{n-1}{n}\right)\cdot \frac{x^n}{n!} \end{align}  Clearly, $$S_n-T_n=\left\{1-\left(1-\frac{1}{n}\right)\right\}\frac{x^2}{2!}+\cdots +\left\{1-\left(1-\frac{1}{n}\right)\cdots \left(1-\frac{n-1}{n}\right)\right\}\cdot \frac{x^n}{n!}\geq 0$$     I'm stuck at this point. Can we show that $S_n-T_n \leq B_n$ such that $B_n \to 0$ as $n \to \infty ??$
47797
Let $f:\mathbb{R}_+ \to \mathbb{R}_+$ be a monotone decreasing function defined on the positive real numbers with $$\int_0^\infty f(x)dx &lt;\infty.$$ Show that $$\lim_{x\to\infty} xf(x)=0.$$   This is my proof: Suppose not. Then there is $\varepsilon$  such that for any $M&gt;0$ there exists $x\geq M$ such that $xf(x)\geq \varepsilon$. So we can construct a sequence $(x_n)$ such that $x_n \to \infty $ and $x_n f(x_n ) \geq \varepsilon$. So $$\frac{\varepsilon}{x_n}\leq f(x_n) \implies \sum_{n\in\mathbb{N}}\frac{\varepsilon}{x_n} \leq  \sum_{n\in\mathbb{N}} f(x_n) \leq \int_0^1 f(x)dx.$$ So we get a contradiction.  I feal like I have the correct idea but some details are wrong.  Any help would be appreciated.
79623
thanks in advance!  Calculate the leading-order asymptotic behaviour of the integral $$I(x) = \int_{0}^{2\pi} (1+t^2) e^{x \cos t} dt \mbox{ as } x \mbox { tends to infinity}$$  So far I know there are two maximas at $0 \mbox{ and }2\pi$ so it can be assumed to be the integral  $$\int_{0}^{\epsilon} (1+t^2) e^{x \cos t} dt + \int_{2\pi-\epsilon}^{2\pi} (1+t^2) e^{x \cos t} dt$$ then using taylor expansions this can be transformed to: $$\int_{0}^{\epsilon} (1+t^2) e^{x (1-\frac{t^2}{2})} dt + \int_{2\pi-\epsilon}^{2\pi} (1+t^2) e^{x (1-\frac{1}{2}(t-2\pi)^2)} dt$$ but where do I go from here? How do I find the leading order term?
43439
Given $$f(x) = \int_{0}^{\phi (x)} g(t) dt$$ How could we find $f&#39;(x)$? Please explain your answer.
106543
I don't know how to prove this example. Let $y(t):\mathbb{R}^+\to \mathbb{R}$ be a nonincreasing function. Assume that there is a constant $A&gt;0$ such that \begin{equation} \int_{s}^{+\infty}y(t)dt \leq Ay(s), \quad 0\leq s&lt;+\infty \end{equation} Then $y(t)\leq y(0)e^{1-\frac{t}{A}}$, for all $t&gt;0$.
85412
How find the asymptotic behavior for $x(n)$  if $x^{x^x} = n$?  I supposed that $x = O(\log\log{n})$ and took logarithm two times.  So I get $x = O(\frac{\log\log{n}}{\log\log\log{n}})$  Is it right? How to improve this estimate?
122021
Does the limit $\lim _{x\to 0} \frac 1x \int_0^x \left|\cos \frac 1t \right| dt$ exists ? If it does then what is the value ?   I don't think even L'Hospital's rule can be applied . Please help . Thanks in advance