I try to use Darboux's Thm and IVT but end up stuck. Is using mean-value theorem many time can get the result?
3735
Prove that if $f$ is differentiable on $[a,b]$ and if $f(a)=f(b)=0$ then for any real $\beta$ there is an $x \in (a,b)$ such that $\beta  \cdot f(x)+f'(x)=0$. (Using rolle's theorem)   My attempt:  Using Rolle's theorem we can say that there exists some $c \in (a,b)$, where $f'(c)=0$ . Therefore one factor in the expression $\beta  \cdot f(x)+f'(x)=0$ is $0$ at $c$ but I am unable to prove that the other factor will simultaneously be $0$ at $c$.
75384
Suppose that $f$ is differentiable on a convex domain (open and connected) D and Re($f')&gt;0$ in D. How can we prove that f is injective on D?  I found some answers using the mean value theorem for holomorphic functions, but I am not familiar with that theorem. Actually I want to use the mean value theorem for integrals, which tells us that the value of an differentiable function $f$ at the center $z_{0}$ of the disk $|z-z_{0}|&lt;r$ is equal to the integral mean of its values on the boundary of that disk: $f(z_{0})=\int^{2\pi}_{0}{f(z_{0}+re^{it})dt}$.  Can you help me proving it by the mean value theorem for integrals?
63161
Whats a quick way to work the below (and problems similar) out?   For me to take this derivative it involves a lot of time and boring calculation, there has to be a better way.    This is taken from the solutions to Chris Bishop's Pattern Recognition and Machine learning:
33235
Suppose $f:[0,1] \Rightarrow \mathbb{R}$ is continuous and $\int_0^x f(x)dx = \int_x^1 f(x)dx$. Prove that $f(x) = 0$ for all $x$.  So, I can intuitively see that this is true. My proof mostly makes sense, I think, but I'm not sure if it covers the case where there are negative and positive values in each segment, resulting in a mean value of 0, but still having nonzero values. Can someone tell me how to cover that, or how this does?  Suppose $f(x) \neq 0$ for all x $\in$ [0,1]. By the mean value theorem, there exist some c and d for which $f(c)(x) = f(d)(1-x) = \int_0^x f(x)dx$. Since x $\neq$ (1-x) for all x $\in$ [0,1], it follows that $f(c) = f(d) = 0$. Therefore, $f(x) = 0$ for all $x \in [0,1]$.
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I know how to solve it using Lagrange method of variation of constants, but is there easier way?
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The  that is defined as $Y_{t} = Y_{t-1} + e_t$, where $e_t$ is white noise. Denotes that the current position is the sum of the previous position + an unpredicted term. You can prove that the mean function $\mu_t = 0 $, since $$E(Y_{t}) = E(e_1+ e_2+ ... +e_t) = E(e_1) + E(e_2) +... +E(e_t) = 0 + 0 + \cdots + 0$$ But why is it that the variance increases linearly with time? Does this have something to do with it's not “pure” random, since the new position is very correlated with the previous one? EDIT: Now I have a much better understanding by visualizing a big sample of random walks, and here we can easily observe that the overall variance does increase over time,  and the mean is as expected around zero. Maybe this was trivial after all, since in the very early stages of the time series (compare time = 10 with time = 100) the random walkers have not had yet the time to explore as much.
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$$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$  Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.  The answer is $Θ (n \log \log n)$.  Can anyone arrive at the solution.
27199
I'm trying to solve this mean value theorem problem but confused where to start,     If $0&lt;a&lt;b$ prove that $(1-\frac{a}{b})&lt;\ln\frac{b}{a}&lt;\frac{b}{a}-1$   Can someone please lend me a hand?
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I want to show that $$\frac{x}{1+x}&lt;\log(1+x)&lt;x$$ for all $x&gt;0$ using the mean value theorem. I tried to prove the two inequalities separately.  $$\frac{x}{1+x}&lt;\log(1+x) \Leftrightarrow \frac{x}{1+x} -\log(1+x) &lt;0$$  Let $$f(x) = \frac{x}{1+x} -\log(1+x).$$ Since $$f(0)=0$$ and $$f'(x)= \frac{1}{(1+x)^2}-\frac{1}{1+x}&lt;0$$ for all $x &gt; 0$,  $f(x)&lt;0$ for all $x&gt;0$.  Is this correct so far?  I go on with the second part: Let $f(x) = \log(x+1)$. Choose $a=0$ and $x&gt;0$ so that there is, according to the mean value theorem, an $x_0$ between $a$ and $x$ with   $f'(x_0)=\frac{f(x)-f(a)}{x-a} \Leftrightarrow \frac{1}{x_0+1}=\frac{ \log(x+1)}{x}$.  Since $$x_0&gt;0 \Rightarrow  \frac{1}{x_0+1}&lt;1.$$ $$\Rightarrow 1 &gt; \frac{1}{x_0+1}= \frac{ \log(x+1)}{x} \Rightarrow x&gt; \log(x+1)$$
58278
Let $W(u)$ $(u \geq 0)$ be a Brownian motion on a probability space $(\Omega, \mathscr{F}, \mathbb{P})$. Let $I(T) = \int_0^T W(u) du$.  One can use integration by parts to show that $I(T) = \int_{0}^T (T-u) dW(u)$.  Since the Ito integral of a deterministic integrand is normally distributed, one can use Ito's isometry to show that the variance of $I(T)$ is $\frac{1}{3}T^3$.  Page 112 of Monte Carlo Methods in Finance by Glasserman states that the variance of $I(T)$ is given by  $\mbox{Var}(I(T)) = 2\int_{0}^T \int_{0}^t \mbox{Cov}(W(u),W(t)) du dt = 2 \int_{0}^T \int_{0}^t  u du dt = \frac{1}{3}T^3$.  I do not understand the first part of the equality - is it a consequence of Fubini's theorem?
47039
Just doing this for revision, seems much harder than it should be, should I use $x=\tan u$ ? Any help appreciated.
43089
I have no idea how to do this. Any help would be appreciated. The chapter I'm on is about differentiation and the mean value theorem.  Prove $a^\alpha b^{1-\alpha} \le \alpha a + (1 - \alpha)b, \; a,b &gt; 0,\; 0 &lt; \alpha &lt; 1$
101332
Suppose $f:[0,\infty)$ $\rightarrow$ R and has continuous $f'(x)$ and $f''(x) $ .  $f(x) $$\rightarrow $0 as $x$ $\rightarrow$ $\infty$. Given $f'(x)\rightarrow b$ as $x \rightarrow \infty$  show that $b=0$  Get stuck in the first place, but I tried to use Mean Value Theorem and then connected it to the limits.  Any hint?
108529
I want to prove that if $x$ is a real number and $x&gt;-1$, $$\sqrt{1+x}  \leq 1+\dfrac{x}{2}$$ I'm not sure which one should I choose for $f(x)$. I also don't understand why we need the mean value theorem to prove this.
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What would the normalising  constant be of the following, or atleast an approximation? I would like to avoid sampling.  $$f(\theta)=\exp(-k_1e^{-k_2\theta^2}-\theta^2)\qquad\theta\in(-\infty,\infty), \quad k_1,k_2&gt;0$$  In the event that its not possible, what would be a valid polynomial lower bound on $\log(f(\theta))$, that would take into account $k_1$  For anyone, who is interested, I end up with this expresssion because I place a Gaussian prior over (a squared exponential) kernel parameters, and trying to use variational bayes (hence, a lower bound).
20751
I'm not sure how to go about this. Does it involve using MVT?  I got as far as saying $e = \frac{e^x - e^y}{x-y}$.
71819
This is a more general treatment of the issue posed by  . After deriving the asymptotic distribution of the sample variance, we can apply the Delta method to arrive at the corresponding distribution for the standard deviation.    Let a sample of size $n$ of i.i.d. non-normal random variables $\{X_i\},\;\; i=1,...,n$, with mean $\mu$ and variance $\sigma^2$. Set the sample mean and the sample variance as  $$\bar x = \frac 1n \sum_{i=1}^nX_i,\;\;\; s^2 = \frac 1{n-1} \sum_{i=1}^n(X_i-\bar x)^2$$  We know that  $$E(s^2) = \sigma^2, \;\;\; \operatorname {Var}(s^2) = \frac{1}{n} \left(\mu_4 - \frac{n-3}{n-1}\sigma^4\right)$$  where $\mu_4 = E(X_i -\mu)^4$, and we restrict our attention to distributions for which what moments need to exist and be finite, do exist and are finite.  Does it hold that  $$\sqrt n(s^2 - \sigma^2) \rightarrow_d N\left(0,\mu_4 - \sigma^4\right)\;\; ?$$