December 1st, 2016, 06:42 AM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,282 Thanks: 93  Power Series and Taylor Expansion
f(x)=$\displaystyle \frac{1e^{x}}{x} =1\frac{x}{2!}+\frac{x^{2}}{3!}\frac{x^{3}}{4!}+..... $ Define f(0) = 1 f'(x)=$\displaystyle \frac{xe^{x}(1e^{x})}{x^{2}} $ f'(0)=? f(x) has a valid power series expansion about 0. Yet, there is the theorem: If a function can be represented by a convergent series of powers of x, then the function is the Taylor series expansion about the point x=0. That's my question. 
December 1st, 2016, 07:07 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,688 Thanks: 1522 
f$^{\,'}\!$(0) = 1/2.

December 1st, 2016, 07:57 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,214 Thanks: 2410 Math Focus: Mainly analysis and algebra 
Rather than simply differentiating, you ought to be using the difference quotient $$f'(0) = \lim_{h \to 0} \frac{f(h)  f(0)}{h} = \lim_{h \to 0} \frac{\frac{1  e^{h}}{h}  1}{h} = \lim_{h \to 0} \frac{1  h  e^{h}}{h^2} = \frac12$$ Although it makes no practical difference, it does have the benefit that it only uses values that you have defined. 
December 1st, 2016, 08:57 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 1,781 Thanks: 919 
Given the power series about 0, $f^\prime(0)$ is simply the coefficient of the series term of order 1. By inspection, it's as the others have said: $f^\prime(0)=\dfrac 1 2$. Last edited by skipjack; December 1st, 2016 at 11:13 AM. 
December 1st, 2016, 10:19 AM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,282 Thanks: 93 
You don't differentiate a series to get its Taylor expansion. You differentiate a function to get its Taylor expansion; you don't differentiate the power series of sinx to get the power series for sinx.* L'Hôpital's rule works in OP, as in v8archie's post, for f'(0). I'd rather use it in the actual derivative to be consistent with Taylor's theorem. I assume it works in the higher order derivatives f$\displaystyle ^{(n)}$(0) Thanks for comments. EDIT* But if you just have the convergent power series for an unknown function, then differentiating the series itself does verify it's a Taylor's expansion. Everybody is right, including Taylor. Last edited by skipjack; December 1st, 2016 at 11:16 AM. 
December 1st, 2016, 11:01 AM  #6  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,214 Thanks: 2410 Math Focus: Mainly analysis and algebra  Quote:
It is only when you explicitly defined the function at $x=0$ that anything changes. And only then when the value defined for $f(0)$ is that returned by the power series. Whatever that value may be, the correct way to proceed is then to evaluate the derivative via the definition of the derivative: the difference quotient. There are, after all, functions (that do not have a power series expansion, but do have a derivative. For example: $$g(x)=\begin{cases}x^2\sin\left(\frac1x\right) & (x \ne 0) \\ 0 & (x=0) \end{cases}$$ I suspect that you have omitted some continuity requirement on the function when talking about Taylor's theorem, unless it is hidden in the statement that the function has a Taylor expansion.  
December 1st, 2016, 11:54 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 18,688 Thanks: 1522  Nobody suggested differentiating a series to get its own Taylor expansion, as that wouldn't make sense. However, the original definitions imply f($x$) ≡ $\displaystyle 1\frac{x}{2!}+\frac{x^{2}}{3!}\frac{x^{3}}{4!}+\,... $. It's legitimate to differentiate that to obtain f$\, '\!$($x$) ≡ $\displaystyle \frac{1}{2!}+\frac{x}{3}\frac{x^{2}}{8}+\,...$, and then put $x$ = 0 to obtain f$\,'\!$(0) = 1/2. 
December 1st, 2016, 01:03 PM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,282 Thanks: 93 
Define $\displaystyle f^{(n)}(0) =\lim_{x\rightarrow 0}f^{(n)}(x)$ which exist and are continuous at x=0 for all derivatives by L'Hôpital's rule. Then x=0 belongs to domain of f(x), and obviously the power series is a valid expansion about x=0. romsek had a good point. A convergent power series may define an unknown function, power series solution of an ODE for example. In that case, it is easily shown it is a Taylor series. Last edited by skipjack; December 1st, 2016 at 01:11 PM. 
December 1st, 2016, 01:14 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 18,688 Thanks: 1522  
December 2nd, 2016, 08:01 AM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,282 Thanks: 93  $\displaystyle f(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^3+...\\ f'(x)=a_{1}+2a_{2}x+3a_{3}x^2+...\\ f''(x)=2a_{2}+3\cdot 2a_{3}x+...\\ f'''(x)=3\cdot 2a_{3}+....\\ ...\\ f(0)=a^{0}\\ f'(0)=a_{1}\\ f''(0)=2!a_{2}\\ f'''(0)=3!a_{3}\\ ...\\$ Taylor's (really Maclaurins) expansion of f(x): $\displaystyle f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{f''' (0)}{3!}x^{3}+....$ 

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expansion, power, series, taylor 
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