Need help on all steps. Use differentiation to find a power series representation for $$ f(x)=\frac{1}{(1+x)^2} $$
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This function seemed to be pretty much straight forward, but my solution is incorrect. I have two questions: 1. Where did I make a mistake? 2. I learned that there are shortcuts for finding a series (substitution / multiplication / division / differentiation / integration of both sides). Is there something that I can apply here?  $$f(x)=\frac{1}{(1-x)^2}  \qquad  f(0)=1$$ $$f'(x)=\frac{2}{(1-x)^3}  \qquad  f'(0)=2$$  $$f''(x)=\frac{6}{(1-x)^4}  \qquad  f''(0)=6$$  $$f'''(x)=\frac{24}{(1-x)^5}  \qquad  f'''(0)=24$$  $$f''''(x)=\frac{120}{(1-x)^6}  \qquad  f''''(0)=120$$ The series is then $$1+\frac{2x}{2!}+\frac{6x^2}{3!}+\frac{24x^3}{4!}+\frac{120x^4}{5!}$$ and when simplified is  $$1+x+x^2+x^3+x^4$$ But the correct answer is  $$1+2x+3x^2+4x^3+5x^4 $$
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The function $f$ is differentiable twice at x. Prove that:   $$\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$$   Hint: use Peano's remainder (if $f:I\to\mathbb R$ is differentiable $n$ times on $a\in I$ then $R_k(x)=o(|x-a|^k), \ x\to a$).   I just don't see the connection here between the second derivative and the Taylor series which has to do with Peano's remainder...  The remainder is defined to be the difference between the function and polynomial but what does it has to do in this case ?
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How to get $f(x)$, if we know that $f(f(x))=x^2+x$?  Is there an elementary function $f(x)$ that satisfies the equation?
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Let $f$ be a function defined on the interval $(−1, 1)$ such that for all $x, y \in (−1, 1)$, $$f(x + y) = \frac{{f(x) + f(y)}}{{1 - f(x)f(y)}}.$$ Suppose that $f$ is differentiable at $x = 0$, show that $f$ is differentiable on $(−1, 1)$.  Need your kind guidance and help. Thanks in advance!
40332
Let $f(x) = x^2 \sin{\frac{1}{x}}$ for $x\neq 0$ and $f(0) =0$. (a) Use the basic properties of the derivative, and the Chain Rule to show that $f$ is differentiable at each $a\neq 0$ and calculate $f'(a)$. You may use without proof that $\sin$ is differentiable and that $\sin' =\cos$.  Not even sure what this is asking.  (b) Show that $f$ is differentiable at $0$ and that $f'(0) =0$.  $\frac {f(x)-f(0)}{x-0} \to \lim_{x \to 0} x \sin(1/x)$. $x \sin(1/x) \leq |x|$ and $\lim_{x \to 0} |x|=0$. Thus $f(x)$ is differentiable at $0$; moreover $f^{'}(0)=0$.  (c) Show that $f'$ is not continuous at $0$.  $f{'}(x)=x^{2} \cos(1/x) (-x^{-2}) + 2x \sin (1/x)$. In pieces: $\lim_{x \to 0} \cos (1/x)$. $f^{'}(0-)$ nor $f{'}(0+)$ exists as $x \to 0$ $f^{'}(x)$ oscillates infinity between $-1$ and $1$ with ever increase frequency as $x \rightarrow 0$ for any $p&gt;0$ $[-p,0]$, $[-p,p]$ or $[0,p]$ $f$ is not continuous. Question: How to show more rigorously?
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What is the coefficient of $x^{11}$ in the power series expansion of $\frac 1{1-x-x^4}$?    How do I do power series expansions?
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Given $f = f(u(t), t)$, is it possible to find the following derivative? ($f$,$u \in \mathbb{R}^n$, $t \in \mathbb{R}$)  $$ \frac{\partial f}{\partial||u||^2_2} $$  I am aware of the following derivatives:  $$ \frac{\partial||x||^2_2}{\partial x} = \frac{\partial ||x^Tx||_2}{\partial x} = 2x $$  So I can invert it to give me (elementwise division in the RHS)  $$ \frac{\partial x}{\partial||x||^2_2} = \frac{1}{2x} $$  This step does feel like abuse of notation I find it hard to convince myself of its validity.  If it's right though, I could just apply the chain rule to find  $$ \frac{\partial f}{\partial u} \frac{\partial u}{\partial||u||^2_2} $$  However, knowing that both $f$ and $u$ are a function of $t$, I am somehow doubting my own reasoning. Or does that just mean that the derivative will still be a function of $t$ and I have nothing to worry about? Can anyone shed some light?  UPDATE: I'm now doubting the existence of  $$ \frac{\partial u}{\partial ||u||^2_2} $$  because $u = u(||u||_2^2)$ is not injective. Is the inversion step I did above legitimate? Is there another possible approach?
23893
Find the Maclaurin series of $f(x)=\dfrac{1}{1+x+x^2}$ and the radius of convergence of the series.  I can't solve this problem.
47224
Let $a&gt;0$, show that for $x&gt;0$, $1&lt;f(x)&lt;2$, where $$f(x)=\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+a}}+\sqrt{\frac{ax}{ax+8}}$$  I could take the derivative, find the maximum of the function and conclude. But that involves heady algebra and you have to solve a high degree polynomial. Is there an easier way simply using inequalities? Thanks, much appreciated
22219
I am asked to prove $e^{(Y \log(1 + Z))} = (1 + Z)^Y$ using the power series definitions for both $\log(Z)$ and $e^Z$.  Our hint is to find a formula for the nth derivative of the power series expansion of $e^{(Y \log(1 + Z))}$ with respect to Z and then use it to find the coefficient on $Z^n$.  I have found that the formula for the nth derivative is $$\frac{(Y)_n}{(1 + Z)^n}\sum_{n = 0}^\infty\frac{1}{n!}\left[Y \log(1 + Z)\right]^n$$ where $(Y)_n$ is the falling factorial.  I am stuck from here though and cannot figure out how to use this to find the coefficient on $Z_n$.  Any help would be greatly appreciated.  Thanks in advance.
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I need to find the $k$'th derivative of $1/p(x)$, where $p(x)$ is a polynomial and $k\in\mathbb{R}$  It dosen't have to be an explicit formula, an algorithm which finds a formula for some $k$ is fine.
98563
I am not too grounded in differentiation but today, I was posed with a supposedly easy question $w = f(x,y) = x^2 + y^2$ where $x = r\sin\theta $ and $y = r\cos\theta$ requiring the solution to $\partial w / \partial r$ and $\partial w / \partial \theta $. I simply solved the former using the trig identity $\sin^2 \theta + \cos^2 \theta = 1$, resulting to $\partial w / \partial r = 2r$.  However I was told that this solution could not be applied to this question because I should be solving for the total derivative. I could not find any good resource online to explain clearly to me the difference between a normal derivative and a total derivative and why my solution here was wrong. Is there anyone who could explain the difference to me using a practical example? Thanks!
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I do have somewhat of a reasoning for this.   $$S = 1 + x + x^{2} + x^{3} +.. + x^{n} $$   Denoting the derivative of $S$ as $T$   $$T = 1 + 2x + 3x^{2} + 4x^{3} +... + nx^{n-1}$$  $$xT = x + 2x^{2} + 3x^{3} +... + nx^{n}$$  Writing the derivative of this series as $V$  $$V = 1 + 2^{2}x + 3^{2}x^{2} +... + n^{2}x^{n-1}$$  $$xV = x + 2^{2}x^{2} + 3^{2}x^{2} +... + n^{2}x^{n}$$  Which is ${\sum{n^{2}x^{n}}}$  However, I am unable to start from $\sum{n^{2}}$and finish where I started. Can someone give me systematic steps on how to manipulate ${\sum{n^{2}x^{n}}}$into a generating function?   The answers are a little different from what I expected.  Let me explain with $\sum nx^{n}$. This function is not the series of any known function but with one slight manipulation, we can change it.  $$\sum x\frac {d x^{n}}{dx}$$ $$=x \frac{d}{dx} {\sum x^{n}} = x\frac {d}{dx} \frac{1}{1-x}$$ $$=\frac {x}{ (1-x)^2 }$$  I wanted a similar trick for evaluating $\sum n^{2}x^{n}$ $$=\sum nx \frac{d}{dx} x^{n}$$ And, then what ? How can I finish this ? I know how to start at $S$ and finish here, but I'm having trouble doing the algebra of starting with this sum and finishing where I started.
120504
a) Compute the full Fourier series representation of $f(x) = e^{ax}, −π ≤ x &lt; π.$      b) By using the result of a) or otherwise determine the full Fourier series expansion for the function $g(x)=\sinh(x), −π ≤ x &lt; π.$   For part a) this is what I did and I'm pretty sure it's all correct. (Please let me know if it isn't!)  $$\begin{align} a_0\ &amp;=\frac{1}{π}\int_{-π}^{π} f(x)\cos(0x)\;dx\\ &amp;=\frac{1}{π}\int_{-π}^{π} e^{ax}\;dx=\frac{1}{π}\left[\frac{e^{ax}}{a}\right]^π_{-π}=\frac{e^{aπ}-e^{-aπ}}{aπ}\\ a_m\ &amp;=\frac{1}{π}\int_{-π}^{π} e^{ax}\cos(mx)\;dx\\ &amp;=\frac{1}{π}\left[\frac{e^{ax}\sin(mx)}{m}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{ae^{ax}\sin(mx)}{m} dx\\ &amp;=\frac{1}{π}\left[\frac{e^{ax}\sin(mx)}{m}\right]^π_{-π}+\frac{1}{π}\left[\frac{ae^{ax}\cos(mx)}{m^2}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{a^2e^{ax}\cos(mx)}{m^2} dx\\ &amp;=\frac{1}{π}\left[\frac{e^{ax}\sin(mx)}{m}+\frac{ae^{ax}\cos(mx)}{m^2}\right]^π_{-π}-\frac{a^2}{m^2}a_m\\ a_m\left(\frac{m^2+a^2}{m^2}\right)&amp;=\frac{ae^{aπ}(-1)^m-e^{-aπ}(-1)^m}{πm^2}\\ a_m&amp;=\frac{(-1)^m\left(ae^{aπ}-e^{-aπ}\right)}{π\left(m^2+a^2\right)} \end{align}$$  Then $$\begin{align} b_m&amp;=\frac{1}{π}\int_{-π}^{π} e^{ax}sin(mx)\;dx\\ &amp;=\frac{1}{π}\left[\frac{-e^{ax}\cos(mx)}{m}\right]^π_{-π}+\frac{1}{π}\int_{-π}^{π} \frac{ae^{ax}\cos(mx)}{m}\;dx\\ &amp;=\frac{1}{π}\left[\frac{-e^{ax}\cos(mx)}{m}\right]^π_{-π}+\frac{1}{π}\left[\frac{ae^{ax}\sin(mx)}{m^2}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{a^2e^{ax}\sin(mx)}{m^2}\;dx\\ &amp;=\frac{1}{π}\left[\frac{-e^{ax}\cos(mx)}{m}+\frac{ae^{ax}\sin(mx)}{m^2}\right]^π_{-π}-\frac{a^2}{m^2}b_m\\ b_m\left(\frac{m^2+a^2}{m^2}\right)&amp;=\frac{(-1)^{m+1}\left(e^{aπ}-e^{-aπ}\right)}{πm}\\ b_m&amp;=\frac{m(-1)^{m+1}\left(e^{aπ}-e^{-aπ}\right)}{π\left(m^2+a^2\right)} \end{align}$$  so $$\begin{align} f(x)&amp;=\frac{1}{2}a_0+\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx)\\ e^{ax}&amp;=\frac{e^{aπ}-e^{-aπ}}{π}\left[\frac{1}{2a}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2+a^2}(\cos(nx)-n\sin(nx))\right]\\ e^{ax}&amp;=\frac{2}{π}\sinh(aπ)\left[\frac{1}{2a}+\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2+a^2}(\cos(nx)-n\sin(nx))\right] \end{align}$$  but for part b) I wasn't sure of how to get the answer from part a) so instead I did this. I am also unsure if this is right. How do you do part b) using part a) ?  $$\begin{align} g(x)&amp;=\sinh(x)\\ a_0\ &amp;=\frac{1}{π}\int_{-π}^{π}\sinh(x)\cos(0x)\;dx\\ &amp;=\frac{1}{π}\left[\cosh(x)\right]^π_{-π}=0\\ a_m\ &amp;=\frac{1}{π}\int_{-π}^{π}\sinh(x)\cos(mx)\;dx\\ &amp;=\frac{1}{π}\left[\frac{\sinh(x)\sin(mx)}{m}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π}\frac{\cosh(x)\sin(mx)}{m}\;dx\\ &amp;=\frac{1}{π}\left[\frac{sinh(x)\sin(mx)}{m}\right]^π_{-π}+\frac{1}{π}\left[\frac{\cosh(x)\cos(mx)}{m^2}\right]^π_{-π}-\frac{1}{π}\int_{-π}^{π} \frac{\sinh(x)\cos(mx)}{m^2}\\ a_m\left(\frac{m^2+1}{m^2}\right)&amp;=\frac{1}{π}\left[\frac{\sinh(x)sin(mx)}{m}+\frac{\cosh(x)\cos(mx)}{m^2}\right]^π_{-π}=0 \end{align}$$  Then $$\begin{align} b_m&amp;=\frac{1}{π}\int_{-π}^{π}\sinh(x)\sin(mx)\;dx\\ b_m\left(\frac{m^2+a^2}{m^2}\right)&amp;=\left[\frac{-\sinh(x)(-1)^{m}}{πm}\right]^π_{-π}\\ b_m&amp;=\frac{-2m\sinh(π)(-1)^m}{π(m^2+1)} \end{align}$$  so $$\begin{align} g(x)&amp;=\frac{1}{2}a_0+\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx)\\ \sinh(x)&amp;=\frac{-2\sinh(π)}{π}\sum_{n=1}^{\infty}\frac{(-1)^nn\sin(nx)}{n^2+1} \end{align}$$
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Let's say I have this function $f(x)=x$. I want to differentiate with respect to $x^2$. So I want to calculate $\large\frac{df(x)}{dx^2}$. In general, how can I calculate the derivative of a function $f(x)$ with respect to a function $g(x)$, so $\large\frac{df(x)}{dg(x)}$?   (I dont know whether this is a good notation)?
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I am wondering if there are any functions $f(x)$ such that it cannot be expressed as a power series of $x$? This might turn out to be a silly question, but I can't think of one at the moment.   Thanks!
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These days, the standard way to present differential calculus is by introducing the Cauchy-Weierstrass definition of the limit. One then defines the derivative as a limit, proves results like the Leibniz and chain rules, and uses this machinery to differentiate some simple functions such as polynomials. The purpose of my question is to see what creative alternatives people can describe to this approach. The nature of the question is that there is not going to be a single best answer. I have several methods that I've collected which I'll put in as answers to my own question.  It's not reasonable to expect answers to include an entire introductory textbook treatment of differentiation, nor would anyone want to read answers that were that lengthy. A sketch is fine. Lack of rigor is fine. Well known notation and terminology can be assumed. It would be nice to develop things to the point where one can differentiate a polynomial, since that would help to illustrate how your method works and demonstrate that it's usable. For this purpose, it suffices to prove that if $n&gt;0$ is an integer, the derivative of $x^n$ equals $0$ at $0$ and equals $n$ at $1$; the result at other nonzero values of $x$ follows by scaling. Doing this for $n=2$ is fine if the generalization to $n&gt;2$ is obvious.