Assume that $F[h(\xi);x,y]$ be the inverse of $G[h(\xi);x,y]$ in the sense that the following identity is satisfied: \begin{equation} \int dz F[h(\xi);x,z]G[h(\xi);z,y] \equiv \delta(x-y) \end{equation} for any $h(\xi)$. Then one can obtain: \begin{equation} \dfrac{\delta}{\delta{h(z)}}F[h(\zeta);x,y] = -\int d\xi d\eta F[h(\zeta);x,\xi]\dfrac{\delta G[h(\zeta);\xi,\eta]}{\delta{h(z)}}F[h(\zeta);\eta,y] \end{equation} Question:How can I prove/understand this functional derivative?   Note: This functional derivative appears in . [APPENDIX A:(A.7)]
123160
I'm trying to work through a derivation of Hedin's Equations in Effect of Interaction on One-Electron States by Hedin and Lundqvist (1969) and I've come across an identity that is given without much explanation that I'd like to understand.   We start with an equation of motion with single particle Green's functions G:   $\left[ i\frac{\partial}{\partial t_1} - h(1) - \phi(1) - V_H(1)\right] G(1,2) - i\int v(1^{+},3) \frac{\delta G(1,2)}{\delta \phi(3)} d3 = \delta(1,2)$  where $1=(x_1,t_1)$, $h$ is the one-electron kinetic energy and ion interaction, $\phi$ is a small perturbing field that is set to zero at the end, $V_H$ is the Hartree potential, and $v$ is the Coulomb interaction.  They then insert the identity  $\frac{\delta G(1,2)}{\delta\phi(3)} = -\int G(1,4) \frac{\delta G^{-1}(4,5)}{\delta\phi(3)} G(5,2) d(4,5)$  which is then used to define the self energy  $\Sigma(1,2) =  -i \int v(1^+,3) G(1,4)\frac{\delta G^{-1}(4,2)}{\delta\phi(3)} d(3,4)$.  I know this identity is related to the definition of the inverse Green's function  $ \int G(1,3) G^{-1}(3,2) d(3) = \delta(1,2)$  but I can't figure out how to get from this definition to the identity that is inserted into the equation of motion. Any ideas?  Thank you!
79259
I don't understand the task from (b). Is it equivalent to : for every linear functional $f(x_{1}, ..., x_{n})=c_{1}x_{1}+...+c_{n}x_{n}$ on $F^{n}$ which satisfy $c_{1}+...+c_{n}=0$ there exists exactly one and unique functional that belongs to the dual space of $W$?  If it is so, could someone please give me a hint?
29673
I was messing around with the definition of the derivative, trying to work out the formulas for the common functions using limits. I hit a roadblock, however, while trying to find the derivative of $e^x$. The process went something like this:  $$\begin{align} (e^x)' &amp;= \lim_{h \to 0} \frac{e^{x+h}-e^x}{h} \\ &amp;= \lim_{h \to 0} \frac{e^xe^h-e^x}{h} \\ &amp;= \lim_{h \to 0} e^x\frac{e^{h}-1}{h} \\ &amp;= e^x \lim_{h \to 0}\frac{e^h-1}{h} \end{align} $$  I can show that $\lim_{h\to 0} \frac{e^h-1}{h} = 1$ using L'Hôpital's, but it kind of defeats the purpose of working out the derivative, so I want to prove it in some other way. I've been trying, but I can't work anything out. Could someone give a hint?
65576
From the functional equation   $$\text{Li}_{3}(z) + \text{Li}_{3}(1-z)+ \text{Li}_{3} \Big( 1 - \frac{1}{z} \Big) = \zeta(3) + \frac{\ln^{3} (z)}{6}+ \frac{\pi^{2} \ln (z) }{6}- \frac{\ln^{2} (z) \ln(1-z)}{2}$$  one can determine the value of $\text{Li}_{3} (\frac{1}{2} )$ since   $$ \text{Li}_{3}(-1) = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{3}} = -\Big(1 - 2^{-2} \Big) \zeta(3) = - \frac{3}{4} \zeta(3) $$  But how does one derive that functional equation?   A similar functional equation for the dilogarithm can be derived using the integral representation $$ \text{Li}_{2}(z) = - \int_{0}^{z} \frac{\ln(1-t)}{t} \ dt $$  Is there a better way to find the value of $\text{Li}_{3} (\frac{1}{2} )$?
23051
$$f(x,y) = \left\{\begin{array}{cc} \frac{xy}{x^2+y^2} &amp; (x,y)\neq(0,0) \\ f(x,y) = 0 &amp; (x,y)=(0,0) \end{array}\right.$$  In order to verify if this function is differentiable, I tried to prove it by the theorem that says that if $\frac{∂f}{∂x}$ and $\frac{∂f}{∂y}$ exist and are continuous at the point $(x_0,y_0)$, then the function is differentiable at this point. So I did:  $$\frac{\partial f}{\partial x}(0,0) = \lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h} = 0$$ $$\frac{\partial f}{\partial y}(0,0) = \lim_{h\to 0}\frac{f(0,0+h)-f(0,0)}{h} = 0$$  so we have that the partial derivatives at point $(0,0)$ is $0$. Now, if we take the derivative at $(x,y)\neq (0,0)$ and then take the limit of it as $(x,y)\to(0,0)$, we can see if the derivatives are continuous or not. So here it is:  $$\frac{\partial f}{\partial x}(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)}$$  but   $$\lim_{(x,y)\to(0,0)} \frac{y(y^2-x^2)}{(x^2+y^2)} $$ does not exist (by wolfram alpha... but can anybody tell me an easy way to prove this limit does not exist? easier than taking the limit in different directions?), therefore the derivative is not continuous at $(0,0)$, so we can't say $f$ is differentiable at $(0,0)$, but for $(x,y)\neq (0,0)$ the function is continuous, as it is a quotient of continuous functions. So $f$ is at least differentiable at $(x,y)\neq (0,0)$.   Now, to verify differentiability at $(0,0)$ I think we must use the limit definition of differentiablity: A function is differentiable at $(0,0)$ iff: $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = 0$$  Let's calculate this limit:  $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = \\ \lim_{(h,k)\to (0,0)} \frac{\frac{hk}{h^2+k^2}}{\sqrt{h^2+k^2}} = \\ \lim_{(h,k)\to (0,0)} \frac{hk}{(h^2+k^2)\sqrt{h^2+k^2}}$$  which I think, it's a limit that does not exist, therefore the function isn't differentiable at $(0,0)$
65425
This is the question: "Show that the derivative of an even function is odd and that the derivative of an odd function is even. (Write the equation that says f is even, and differentiate both sides, using the chain rule.)"  I already read numerous solutions online.  was the official solution but I didn't quite understand it (particularly, I'm not convinced why exactly $dz/dx=-1$; even though $z=-x$).  Thanks in advance =]
64872
The full question is "Define the operator $\frac{d}{d \overline{z}} = \frac{1}{2}(\frac{d}{dx}+i\frac{d}{dy})$. Show that a  function $f(z)=u(x,y)+iv(x,y)$ satisfies the Cauchy-Riemann equation if and only if $\frac{df}{d\overline{z}}=0$." I have already figured out the $\rightarrow$ direction and only need help with the $\leftarrow$ direction.   So far I have: Let $\frac{df}{d\overline{z}}=0$ and $f(z)=u(x,y)+iv(x,y)$. Then $$0=\frac{df}{d\overline{z}}$$ $$=\frac{1}{2}\frac{d}{dx}(u+iv)+\frac{i}{2}\frac{d}{dy}(u+iv)$$ $$=\frac{1}{2}(u_x-v_y)+\frac{i}{2}(u_y+v_x)$$ And this is where I get stuck. I am not sure where to go from there so any help would be much appreciated. (Also feel free to fix my formatting I know it is atrocious."
14214
Good day, I want to prove the following theorem:  1) For any nonzero vector $ v \in V $, there exists a linear funtional $ f \in V^*$ for wich $f(v) \neq 0 $  I know that if $f$ is a lineal functional then we have 2 posibilities   1)$\dim \ker(f)=\dim V$   2)$\dim \ker(f)=\dim V-1$   I've tried to suppose that, for all $v \neq 0 $ and $f \in V^*$ We have $f(v)=0$,  Then  $Im(f)=0$ for all $f \in V^*$ and $ v \in V$ but then the only linear functional is $f=0$ ($V^*=\{0:V \to F\}$) and from this follows: $\dim \ker(f)=\dim V$.  But 2) suggest the existence of a linear functional which $\dim \ker(f)=\dim V-1$, so ¿is that the contradiction?
71634
I'm trying to find the derivative of $\sin^{-1}(x)$. I know the steps that lead to $\frac{1}{\sqrt{1-x^2}}$, however I don't understand why the following reasoning leads to a wrong answer.  Because  $$\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))} $$  If we plug in for $f(x) = \sin(x)$, and because $\frac{d}{dx}\sin(x) = \cos(x)$ we get  $$\frac{d}{dx}\sin^{-1}(x) = \frac{1}{\cos(\sin^{-1}(x))} $$  Since $\sin(x) = \cos(x-\frac{\pi}{2})$, we can state that $\sin^{-1}(x) = \cos^{-1}(x-\frac{\pi}{2})$. (I suspect this is what is wrong)  Thus,  $$\frac{d}{dx}\sin^{-1}(x) = \frac{1}{\cos(\cos^{-1}(x-\frac{\pi}{2}))}$$  Then, by the definition of inverse function, we have  $$\frac{d}{dx}\sin^{-1}(x) = \frac{1}{x - \frac{\pi}{2}}$$
96135
I saw a proof for the above that went \begin{align*} f(n) &amp;= \sum_{k=1}^{n}\left(\sum_{d\mid (k,n)} \mu(d)e^{\frac{2\pi ik}{n}} \right)\\  &amp;= \sum_{d\mid n} \left( \mu(d) \sum_{k=0}^{(n/d)-1} e^{\frac{2\pi idk}{n}} \right), \end{align*} in which case for $g(n)=\sum_{k=0}^{n-1} e^{\frac{2\pi ik}{n}}$ we have \begin{equation*} f(n)=\sum_{d\mid n} \mu(d) g(n/d) \implies g(n) = \sum_{d\mid n} f(d), \end{equation*} which then proves the statement above by Mobius Inversion once more, since $g$ is $0$ for all $n&gt;1$.   I understand most of the proof; I'm just not sure how we went through the first two steps of transforming $f(n)$ to get to the point where we can apply the inversion. If anyone can explain the first two steps for me I would appreciate it, thanks!
118261
Here is the problem: Find all continuous $f:\mathbb{R}\rightarrow\mathbb{R}$ which satisfy  $$f(x+y)+f(x-y)=2[f(x)+f(y)]\;\;\;(1).$$  Here is my attempt:  Fix $\delta&gt;0$ and let $C=\int_{0}^{\delta}2f(y)dy.$  Then  $$\begin{align*} 2\delta f(x)+C&amp;=\int_{0}^{\delta}2[f(x)+f(y)]dy\\ &amp;=\int_{0}^{\delta}f(x+y)+f(x-y)dy\\ &amp;=\int_{x}^{x+\delta}f(y)dy+\int_{x-\delta}^{x}f(y)dy\\ &amp;=\int_{x-\delta}^{x+\delta}f(y)dy. \end{align*}$$  Now since $f$ is continuous, the last expression is a differentiable function of x and thus the first expression must also be differentiable; hence $f$ is differentiable. By induction, f is infinitely differentiable.  Differentiating (1) first with respect to y, we arrive at: $$f&#39;(x+y)-f&#39;(x-y)=2f&#39;(y)\;\;\;(2).$$   Differentiating once more with respect to x, we have: $f&#39;&#39;(x+y)=f&#39;&#39;(x-y),$ so $f&#39;&#39;$ is constant. It follows that $f(x)= ax^2+bx+c$ are the only potential solutions.  Substituting $x=y=0$ in (1) and (2) implies $f(0)=f&#39;(0)=0$; hence $f(x)=ax^2\;\;\;(a\in\mathbb{R})$.  It is easy to check that all such $f$ are indeed solutions. $\square$  Is my proof correct?
101874
The function   $f(x)=x+\sin(x)$  is easily checked to be a bijection from the reals to itself, and so it has a unique inverse $y\mapsto g(y)$ such that $f\circ g=g\circ f$ are both the identity map.  Now $g$ will almost certainly be a function which is not expressible using "the functions in a high-schooler's toolkit" (by which I guess I mean $\exp$, $\log$, and, if you like, the usual trigonometric functions and their friends like $\sinh$, although of course these can all be of course built from exponentials anyway). For purely recreational reasons (stemming from conversations I've had whilst teaching undergraduates) I'm interested in how one proves this sort of thing.  A few years ago I was interested in a related question, and took the trouble to learn some differential Galois theory. My motivation at the time was learning how to prove things like why $h(t):=\int_0^t e^{x^2} dx$ is not expressible in terms of these calculator-button functions (I'm sure there's a better name for them but I'm afraid I don't know it). I've realised that since then I've forgotten most of what I knew, but furthermore I am also unclear about whether this is the way one is supposed to proceed. Is the idea that I come up with some linear differential equation satisfied by $g$ and then apply some differential Galois theory technique? In fact, one of the many things that I have forgotten is the following: if $F$ is a field equipped with a differential operator $D$, and $E/F$ is the field extension obtained by adding a non-zero root of $Dh=ch$, with $c\in F$, then the Galois group of $E/F$ is solvable, whereas the equation itself might not be, in terms of calculator-button functions, if I can't integrate $c$.   Can a more enlightened soul explain to me how one is supposed to proceed? I wonder whether I am somehow conflating two ideas and the differential Galois theory business is a red herring, but it seemed simpler to ask rather than continuing to flounder around.
23711
I have a problem with derivation of some function. This is related to computing the Poincare Map of Logistic Population model with Periodic harvesting. This function is:  $\phi(t,x_0) = x_0 + \int_{0}^{t} f(s,\phi(s,x_0)) ds $,  where $\phi(t,x_0)$ is a function: $\mathbb{R}  \times \mathbb{R} \rightarrow \mathbb{R}$.  I'm trying to differentiate  equation of this function both sides  with respect to $x_{0}$. In book from this equation comes from the result of this differentiation is:  $\frac{\partial \phi}{\partial x_0} (t,x_0) = 1 + \int_0^{t} \frac{\partial f}{\partial x_0} (s, \phi(s,x_0)) \cdot \frac{\partial \phi}{\partial x_0} (s, x_0)ds$.  I can't understand expression under the integral. In this book it is explained that it is clear from chain rule.Could someone explain to me this transformation in easy way? I will be grateful for your help Best regards
39174
Let $f:\mathbb R^2\to \mathbb R$ defined by $f(x,y)=xy$. Prove $f$ is differentiable.  According to the definition I have to show this $$\frac{\|f((x_0,y_0)+(h_1,h_2)) - f(x_0,y_0)-T(h_1\space  h_2)\|}{\|(h_1,h_2)\|}\to 0$$  But I got stuck in the proof.  \begin{align} &amp; \frac{\|f((x_0,y_0)+(h_1,h_2))-f(x_0,y_0)-T(h_1\space  h_2)\|}{\|(h_1,h_2\|} \\[10pt] = {} &amp; \frac{\|(x_0+h_1)(y_0+h_2)-x_0y_0-(x,y)(h_1\space  h_2)\|}{\sqrt{h_1^2+h_2^2}} \\[10pt] \le {} &amp;\frac{\|x_0h_2+y_0h_1+h_1h_2\|}{\sqrt{h_1^2+h_2^2}}+\frac{\|-(x,y)(h_1\space h_2)\|}{\sqrt{h_1^2+h_2^2}} \end{align} and I think I the first term after the inequality goes to zero, but what happen to the second term?  How can it goes to zero too?
65934
$H$ is a Hilbert space and $B(H)$ is the Banach space of its bounded linear operator on $H$. Assume $A\in$ $B(H)$, $A$ is self-adjoint. Suppose $\lambda_{0}$ is in the resolvent set of $A$. Show that: $||(A-\lambda_{0}I)^{-1}||=\frac{1}{d(\lambda_{0},\sigma(A))}$  Now I can use the knowledge of functional calculus(see $(A-\lambda_{0}I)^{-1}$ as the function of $A$)  to show that $||(A-\lambda_{0}I)^{-1}||\leq \frac{1}{d(\lambda_{0},\sigma(A))}$. But I don't know how to proceed.
72432
I know that the derivative of the inverse function of $f(x)$ is $g'(y) = \frac{1}{f'(x)}$ But how to derive the formula for the second derivative of g(y) knowing that $\left[\frac{1}{f(x)}\right]' = -\frac{f'(x)}{(f(x))^2}$ ?  I just started studying this chapter, so please try to be as simple as possible ;-) Thank you.
83265
Let's define  $$ e^x := \lim_{n\to\infty}\left(1+\frac{x} {n}\right)^n, \forall x\in\Bbb R $$  and  $$ \frac{d} {dx} f(x) := \lim_{\Delta x\to0} \frac{f(x+\Delta x) - f(x)} {\Delta x} $$  Prove that  $$ \frac{d} {dx} e^x = e^x $$  using the definition of $e^x$ and derivation above, without using L'Hôpital's rule or the "" and/or the "inverse function derivative trick".  $$ \left( \frac{d} {dx} f^{-1}(x)= \frac{1} {\left(\frac{d}{d(f^{-1}(x))}f(x)\right)(f^{-1}(x))}\right) $$  Or equivalently prove that the following two definiton of $e$ are identical $$ 1)\space\space\space\space e =\lim_{n\to\infty}(1+\frac{1} {n})^n $$  $$ 2) \space\space\space\space e\in\Bbb R,\space\space(\frac{d} {dx} e^x)(x=0) = 1   $$  What I've got is $$ \frac{d}{dx}e^x=e^x \lim_{\Delta x\to0}\frac{e^{\Delta x} - 1} {\Delta x} = e^x \lim_{\Delta x\to0}\lim_{n\to\infty}\frac{\left(1+\frac{\Delta x}{n}\right)^{n}-1}{\Delta x} =  e^x \lim_{\Delta x\to0}\frac{e^{0+\Delta x}-e^0}{\Delta x}  $$  If i assume that $n\in\Bbb N$ I could use binomial theorem but I didn't got much out of it. Wolframalpha just uses L'Hospital rule to solve it, but I am looking for an elementary solution. What I'm interested in is basically is the equivalence of the two definition of $e$ mentioned above. And I'd like to get a direct proof rather than an indirect(I mean which involves logarithms or the derivatives of invers functions).  I look forward getting your aswers.