My Math Forum Power Series and Taylor Expansion

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 December 1st, 2016, 06:42 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 Power Series and Taylor Expansion f(x)=$\displaystyle \frac{1-e^{-x}}{x} =1-\frac{x}{2!}+\frac{x^{2}}{3!}-\frac{x^{3}}{4!}+.....$ Define f(0) = 1 f'(x)=$\displaystyle \frac{xe^{-x}-(1-e^{-x})}{x^{2}}$ 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: 19,049 Thanks: 1618 f$^{\,'}\!$(0) = -1/2.
 December 1st, 2016, 07:57 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 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. Thanks from zylo
 December 1st, 2016, 08:57 AM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 1,975 Thanks: 1026 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: $f^\prime(0)=-\dfrac 1 2$. Thanks from zylo 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: 1,364 Thanks: 100 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
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 Originally Posted by zylo 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$\displaystyle ^{(n)}$
Note that your original function is not defined at $x=0$. It's a removable singularity which is how we happen to get a Maclaurin series that is not undefined at zero. But speaking strictly, the domain of the function and thus of its power series does not include zero. This means that rhe derivative is also not defined at zero. Again it's a removable singularity so the derivative's power series does return a value at zero, but again, the domain of the derivative and thus of its power series does not include zero, both being undefined at this point

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
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Quote:
 Originally Posted by zylo You don't differentiate a series to get its Taylor expansion.
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: 1,364 Thanks: 100 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
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Quote:
 Originally Posted by zylo In that case, it is easily shown it is a Taylor series.
How?

December 2nd, 2016, 08:01 AM   #10
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Quote:
 Originally Posted by skipjack How?
$\displaystyle f(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^3+...\\ f'(x)=a_{1}+2a_{2}x+3a_{3}x^2+...\\ f''(x)=2a_{2}+3\cdot 2a_{3}x+...\\ f'''(x)=3\cdot 2a_{3}+....\\ ...\\ f(0)=a^{0}\\ f'(0)=a_{1}\\ f''(0)=2!a_{2}\\ f'''(0)=3!a_{3}\\ ...\\$
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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