In this problem I have integrated the expression |f'(x) | to get $f(x)$&lt;$x/2$ but after that how to proceed with the problem is clueless.Kindly help.
123909
Let $f:\Bbb R \to \Bbb R$ be differentiable on $\Bbb R$ and $|f'(x)| \leq \frac {1}2$ for all $x\in \Bbb R$. Let $a \in \Bbb R$ and define $x_1 = a, x_{n+1} = f(x_n)$ for all $n \in \Bbb N$. Show that $\{x_n\}$ is a Cauchy sequence. And show that $\exists x \in \Bbb R$ such that $f(x) = x$.
101867
I can't get any of these on my own and I am attempting to do $x^x$ I had it explained to me for about an hour and I still can't do it on my own. I thought I was supposed to make it into $x(\ln x)$ which should be equivalent to the original term. Then the differentiation should be easy for most people from here.  $$1(\ln x)+x(\frac1x) $$ or $$\ln x+1$$  This is of course wrong but I do not know why.
102312
This is question B4 in 1999 Putnam Exam. I would appreciate if somebody could help me with this.      Let $f(x),f'(x),f''(x),f'''(x)&gt;0$ , $f'''(x)$ is a continuous function and $f'''(x)&lt;f(x)$ on $\mathbb{R}$. Then show that   $$f'(x)&lt;2f(x),~ \forall x\in \mathbb{R}.$$
88137
Let $f(x) = \int_x^{x+1} \sin(e^t)dt$. Show that $$ e^x|f(x)| &lt; 2 $$ and that $$ e^xf(x) = \cos(e^x) - e^{-1}\cos(e^{x+1}) + r(x) $$ where $|r(x)| &lt; Ce^{-x}$ for some constant $C$.  I can do the first part easily. Integration by parts gives us $$ f(x) = \frac{\cos e^x}{e^x} - \frac{\cos e^{x+1}}{e^{x+1}} - \int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2}du, $$ so substituting $\cos u = 1$ and $\cos u = -1$ gives \begin{align*} \cos e^x - 1 + \frac{1}{e}\left(1- \cos e^{x+1}\right) &amp;\leq e^xf(x) \leq \cos e^x + 1 - \frac{1}{e}\left(\cos e^{x+1} + 1\right)\\ -2 &amp;&lt; e^xf(x) &lt; 2 \end{align*} So that $e^x|f(x)| &lt; 2$.  I get stuck on the second part, though. From integration by parts, I have $e^xf(x) = \cos e^x - e^{-1}\cos e^{x+1} - e^x\int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2}du$, so it reduces to showing that $$ \left|\int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2}du\right| &lt; Ce^{-2x} $$ for some constant $C$. However, I always end up with a constant for $Ce^{x}$ rather than for $C$.
2385
If $f$ is differentiable on $(a,\infty)$, Show that if $f'(x)\rightarrow L$ as $ x \rightarrow \infty$, $-\infty \leq L \leq \infty $  then  $ f(x)/x \rightarrow L $ as $x \rightarrow \infty$.   Deduce further that if $f(x) \rightarrow M$ as $ x\rightarrow \infty$ , and $ f'(x)\rightarrow L $ as $ x \rightarrow\infty $ then $ L = 0$   For the first part I tried converting $f(x)$ to $F(\frac{1}{x})$ and $x \rightarrow 0$ and then using L'Hospital's rule but I am getting stuck. How do I proceed with this problem ?
46536
I need to show that $\dfrac{2}{\pi}&lt;\dfrac{\sin(x)}{x}&lt;1$ for $0&lt;x&lt;\dfrac{\pi}{2}$.  I know I need to use the mean value theorem, would I just say that since $f$ is continuous in the interval (call it I).  We know $\exists c\in I$ such that $f'(c) = \dfrac{f\left(\dfrac{\pi}{2}\right)-f(0)}{\dfrac{\pi}{2}}$. I don't know where this is going though.  Also, would I need to show $\lim_{x \to 0}\dfrac{\sin(x)}{x} = 1$? Is there a way to do this without L'H Rule?
98417
So we want to find a $\delta&gt;0$   such that for all $2-\delta&lt;x&lt;2+\delta$  , we will have $4-\epsilon&lt;x^{2}&lt;4+\epsilon$ for all $\epsilon&gt;0$  . If we can find a way to express $\delta$  as a function $\delta (\epsilon)$,  $\delta:\mathbb{R}_+ \to \mathbb{R}_+ $ then we will have solved the problem. But I can't see how to relate $\epsilon$ to $\delta$ in this case.  My initial reaction is write $(2-\delta)^2&lt;x^2&lt;(2+\delta)^2$, this makes the two inequalities look related, but I am not sure where to go from here.
81231
This problem is an extension to the simpler problem which deals with $f(x) + f'(x) \to A$ as $x \to \infty$ (see problem 2 on ).     If $f$ is twice continuously differentiable in some interval $(a, \infty)$ and $f(x) + f'(x) + f''(x) \to A$ as $x \to \infty$ then show that $f(x) \to A$ as $x \to \infty$.   However, the approach based on considering sign of $f'(x)$ for large $x$ (which applies to the simpler problem in the blog) does not seem to apply here. Any hints on this problem?  I believe that a similar generalization concerning expression $\sum\limits_{k = 0}^{n}f^{(k)}(x) \to A$ is also true, but I don't have a clue to prove the general result.
22559
Okay I like have no idea how to do this. I tried integration by parts then everything came back uglier. I then put this in wolfram it says "No found in standard mathematical functions"   Is there a easy way to do this?   Maybe like FTOC II?
14252
Could someone advise me on how to determine all entire functions $f$ such that $|f'(z)| \leq |f(z)|, \forall z\ ?$  Hints will suffice, thank you.
102094
let $$f(x)=\int_{x}^{x+1}\sin{(e^t)}dt$$  show that: $$e^x|f(x)|\le 2$$  My idea: let $$e^t=u$$ then $$|f(x)|=|\int_{e^x}^{e^{x+1}}\dfrac{1}{u}d\cos{u}|$$
37731
I'm not entirely sure what I want to ask here, so please bear with me!  I think the explanation they give us in school for how finding the antiderivative by substitution works is: $$\int f(g(t))g&#39;(t)dt=(Fg)(t)=F(g(t))=F(u)=\int f(u)du$$  But I never really understood the equality $F(u)=\int f(u)du$. Why can we behave as if the identity $u=g(t)$ doesn't exist, compute the integral $\int f(u)du$, and then 'substitute' $u$ in the result yielding the correct answer? In a sense, it feels like the variable $u$ switches from being 'meaningful' in $F(u)$ (where it stands for $g(t)$) and being insignificant in $\int f(u)du$, since as I perceive it the "variable name" here is meaningless and we could say $ \int f(u)du = \int f(k)dk = \int f(x)dx = ... $ all the same. Another way to ask the same thing: why $(Fg)(t)=\int f(u)du$ and not, say, $(Fg)(t)=\int f(x)dx$? Where am I getting confused?  Maybe someone can explain this better than my high school teacher? Is there a 'formal' explanation for this? Thank you a lot!
74431
I've been stuck on this question for a long time now and was wondering if anyone could show me how it's done. So far I have done the following: Since $\lim_{x\to \infty}f'(x)=M$ then $\forall \epsilon ,\exists A &gt;0$  s.t. if $x&gt;A$ then $|f'(x)-M|&lt;\epsilon$. So we have that $M-\epsilon&lt;f'(x)&lt;M+\epsilon$. By MVT on $[x,x+1]$ for $x&gt;A$, $\exists c\in (x,x+1)$ s.t. ${f(x+1)-f(x)\over 1}=f'(c)$. Therefore, $M-\epsilon&lt;f(x+1)-f(x)&lt;M+\epsilon$. However, I do not know what to do after this.
14621
For a function $f$ I know that: $$\int{f'(r)dr}=f(r)$$ where $f(r)$ is known. knowing the result of this integral how can i calculate $$\int{(f'(r))^2dr}$$ Is there any relation between these integrals?
89608
Really stuck with this, \begin{cases}  2tx'(t)-1=x^2t \\  x(t=1)=1 \\  \end{cases} I don't understand the variables. What does it mean that $x$ differentiated have $t$ as a function value, like this: $x'(t)$? Don't get it :(
51766
If $f'(c)$ exists then $\lim_{x\to c} f(x)=f(c)$   My attempt is below. Did I do wrong? Since I don't really understand it. I thought $f'(x) = f'(c)$ so $f'(x)-f'(c)=0$. Am I on the right track? Many thanks!  \begin{align} f(x)-f(c)&amp;=\frac{f(x)-f(c)}{x-c}\cdot(x-c) \\ \lim_{x\to c}[f(x)-f(c)]&amp;=\lim_{x\to c}\left[\frac{f(x)-f(c)}{x-c}\cdot(x-c)\right] \\ \lim_{x\to c}[f(x)-f(c)]&amp;=0 \\ \lim_{x\to c} f(x)&amp;=f(c) \end{align}
49670
Show $$\lim_{h \to \ 0} \frac{f(x + 2h) - 2f(x+h) + f(x)}{h^{2}} = f''(x)$$   Proof:  By definition:    $$f'(x) = \lim_{h \to \ 0} \frac{f(x + h) - f(x)}{h}$$  Using this idea it would imply:   $$1)\ \ f''(x) = \lim_{h \to \ 0} \frac{f'(x + h) - f'(x)}{h}$$  As such it is required that I find an expression for $f'(x+h)$. This is where I'm not sure if the step I took is legitimate.  An expression for $f'(x + h)$ is:  $$f'(x+h) =  \lim_{h \to \ 0} \frac{f(x + 2h) - f(x + h)}{h}$$  Combining this with the definition of $f'(x)$ and inserting it into 1) you arrive at: $$\lim_{h \to \ 0} \frac{f(x + 2h) - 2f(x+h) + f(x)}{h^{2}} = f''(x)$$   As required.    Concern: I feel a discomfort with this solution. Even though "mechanically" it worked out, if I am taking the limit as $h \rightarrow 0$ that would mean $x + 2h$ and $x + h$ both go to $x$. But I am attempting to use the idea that $x +2h$ goes to $x + h$. Perhaps it is a notation idea that I need to communicate better, but I feel it is larger than just that.