Let $f : [0, ∞) → \mathbb{R}$ be the solution of the ordinary differential equation  $f'(x)=\frac{1}{x^2 +f(x)^2}$  satisfying $f(0) = 1.$ (You may assume that this solution exists, and is unique.) Prove that $f$ is bounded.  I'm not sure where to start with this question, is there specific theorems that I need to take advantage of? Is it an unimportant fact that this is an ODE?  Any tips would be very helpful!
122394
I am given that $f:[0,\infty)\to \mathbb{R}$ is the unique solution to the ODE: $$f^{\prime}(x) = \frac{1}{x^{2} + (f(x))^{2}}$$ with $f(0)=1$ and I must prove that it is bounded. I have already proven (using the fact that the derivative is always positive and the Mean Value Theorem) that it is monotonically increasing so is bounded below, specifically by $f(0)=1$. However, I am struggling to figure out how to show it is bounded above. One thought I had was to rearrange the ode to get $(f(x))^2$ the subject, however this doesn't seem like a valid method for some reason, I may just be overthinking it though.
109651
Was reading some notes and it states that $f(z) = \frac{1 - e^{iz}}{z^2}$ can be written as $f(x) = \frac{-iz}{z^2} + E(z)$ where $E(z)$ is bounded as $z \rightarrow 0.$ I don't exactly see why. Help is appreciated.
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 $f(x/2)$.
113001
$f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function. Assume that $\lim_{x \rightarrow \pm \infty} f(x)$ exist and are finite. Prove that $f$ is bounded.  So to show that $f$ is bounded, I must show that $\exists M \in \mathbb{R}$ such that $|f(x)| \leq M$ for all $x \in Dom(f)$. Assuming that $f(x)$'s limit exist, then I know that $\forall \epsilon  &gt; 0$, $\exists N$ such that for $n &gt; N$, $|f(n) - L| &lt; \epsilon$. But I am having trouble with figuring out how to connect the two ideas together.
59440
Let $f:[0,1] \rightarrow \mathbb{R}$ be a continuous function. If $f(0)=f(1)$ then exists $x\in [0,1]$ s.t. $f(x) = f(x+\frac{1}{2})$   We have that $f\left(\frac{1}{2}\right)=f(0)$ or $f\left(\frac{1}{2}\right)&gt;f(0)$ or $f\left(\frac{1}{2}\right)&lt;f(0)$.  If $f\left(\frac{1}{2}\right)=f(0)$ then $x = 0$ Solves the problem.  if $f\left(\frac{1}{2}\right)&lt;f(0)$ then by the intermediate value theorem  $$\exists k \in \left[0,\frac{1}{2}\right], f(0)&gt;f(k)&gt;f\left(\frac{1}{2}\right) \implies$$  $$\implies f(1)&gt;f(k)&gt;f\left(\frac{1}{2}\right)\implies \exists t \in \left[\frac{1}{2},1\right] s.t. f(t)=f(k).$$   Now how can I show that exists at least one value of k such that $t = k+\frac{1}{2}$? (The solution is similar if $f(\frac{1}{2})&gt;f(0)$)
16193
Prove that if the function $f$ is defined on the set of positive real numbers, its values are real, and $f$ satisfies the equation $$f\left( \frac{x+y}{2}\right) + f\left(\frac{2xy}{x+y} \right) =f(x)+f(y)$$ for all positive $x,y$, then $$2f(\sqrt{xy})=f(x)+f(y)$$ for every pair $x,y$ of positive numbers.    Source: Miklos Schweitzer Memorial Competition 2001    I can see how the repeated application of the functional equation condition upon itself forms a bound, but how can I formally prove this?
42806
This seems like it should be super simple but I can't figure it out. Prove that if $f(t)$ satisfies $f(t)=f'(t)$, then $f(t+s)=f(t)f(s)$ for all $t,s$ belonging to R without using the fact that $f(t)=c\cdot e^t$. The only hint I'm given is that I can use the fact that two functions satisfying the same ODE $x'=x$ with the same initial condition are equal. Any ideas?
121557
Let $f$ be an entire function, such that $$ \lvert\, f(x+yi)\rvert\le \frac{1}{\lvert y\rvert}, $$ for all $x$ and all $y\ne 0$.  Show that $f$ is bounded.
101332
Suppose $f:[0,\infty)$ $\rightarrow$ R and has continuous $f'(x)$ and $f''(x) $ .  $f(x) $$\rightarrow $0 as $x$ $\rightarrow$ $\infty$. Given $f'(x)\rightarrow b$ as $x \rightarrow \infty$  show that $b=0$  Get stuck in the first place, but I tried to use Mean Value Theorem and then connected it to the limits.  Any hint?
66613
Show that the first order differential equation $y'(x)=\sqrt{|y(x)|}$ with intial value $y(1/2)= 1/16$ has infinitely many solutions on the interval [−1, 1].  My thought were to show that this equation has two solutions (just by ansatz, or just looking at the interval $]0, 1]$ to get $y'(x) = \sqrt{|y(x)|}$) and deduce from that that there must be infinitely many solutions since the solution set forms a vector space. I've got a nagging feeling this isn't the way to go though. Can anybody give me a hint where to look?
99523
The statement is: $f$ is a real fucntion on $\mathbb R$. Then if $f'(x)=f(x)$ and $f(0)=1$, then $f(x)\neq 0.$
120320
Let $f(x)$ be Riemann integrable on $[a,b]$ and satisfy $\inf\{\left|f(x)\right|:a\le x \le b\}=d&gt;0$.  How can we prove that $g(x)=\frac{1}{f(x)}$ is Riemann integrable?  I know from the statement that $f(x)\neq 0$ since its $\inf$ is $d&gt;0$, so $g(x)$ is bounded on $[a,b]$ since $f(x)$ is bounded and there is no division by $0$.    Is there a relationship like $M(f)=m(g)$ and $m(f)=M(g)$, since $g(x)$ is the reciprocal?
101201
The question is :     Let $f : [0,1] \longrightarrow \mathbb {R}$ be defined by    $f(x) = x^{\alpha} \sin {\frac {1} {x^{\beta}}}$ , whenever $x \in (0,1]$ and $=0$ , whenever $x = 0$.  Then find the values of $\alpha$ and $\beta$ for which $f$ becomes a function of bounded variation on $[0,1]$.  Please give me a right way to proceed.Thank you in advance.  I have just obtained the result which is  $\alpha &gt; 2$ and $\beta &lt; \alpha - 2$.Is it correct at all?Please verify it.
77964
Using my right to ask for help once again. I can't solve the following problem for quite a long time. Just have no idea how to do that. Here it is. Let $f(x)$ be bounded on any interval $(1,b), b&gt;1$. Then $\lim_{x\to+\infty}{f(x) \over x}=\lim_{x\to+\infty}(f(x+1)-f(x))$. It is from the Russian book "Lekcii po matematicheskomu analizu" by G.I.Arkhipov, V.A.Sadovnichy and V.N.Chubarikov, 2nd ed., Moscow, 2000, page 676. Please, keep in mind, that times, when I did homework had passed long time ago. Now I'm solving problems for pleasure.
74292
How do I solve $$x'=e^{it}\overline{x}?$$  This is a complex differential equation, but I don't see how to solve it.  Edit: the original ODE is given by $$(x', y')=\begin{pmatrix}\cos t&amp; \sin t\\ \sin t&amp;-\cos t\end{pmatrix}(x,y)$$ I want to show that solutions of this ODE do not remain bounded for all $t$, and the idea was to solve the complex equation.
60141
Solve differential equation:  $x'=x^2- \frac{2}{t^2}$  Maybe is it sth connected with homogeneous equation? I have no idea how to solve it.
65966
Regarding: Let $f:(0, \infty) \to \mathbb{R}$ be a differentiable function for which $f'(x)f'(\frac{1}{x})&lt;0$ for every $0&lt;x \ne 1$. Then $f'(1)=0$. -- So I am trying to prove this, and I said: We know that $f$ is differentiable at $(0, \infty)$, so there exists a $0&lt; \delta &lt; 1$ such that $f$ is differentiable at $[1- \delta, 1+\delta]$ . So there exists a $x_{0} \in [1- \delta, 1+\delta]$ such that $f'(x_{0})f'(\frac{1}{x_{0}})&lt;0$ So by Darboux Theorem, there exists a $x_{0}&lt;c&lt; \frac{1}{x_{0}}$ such that $f'(c)=0$. But now, I do not know how to link $c$ with $1$... Any hints? Thanks! :)