# Indeterminate forms of Taylor series

1. Aug 8, 2013

### fakecop

Can someone please explain how the taylor series would work if x, the given value from the function, is equal to a, the value at which you expand the function?

For example, let's take 1/(1-x) as an example. The taylor series for this with a=0 is Ʃ(n from 0 to infinity) x^n. But if we let x=a=0, then the first term of this sequence is 0^0, or indeterminate.

Am i supposed to take the limit of x^0 as x approaches 0-which is one in this case, or always assume x^0 to be equal to 1, or follow some other formula?

2. Aug 8, 2013

### HallsofIvy

I don't know why you would represent to first term as $$x^0$$ in the first place. The Taylor's series for f(x) at x= a is given by f(x)= f(a)+ f'(a)(x- a)+ (f''(a)/2)(x-a)^2+ ..., etc. When x= a, that is f(a), of course.

3. Aug 8, 2013

### pasmith

You're supposed to understand that using $x^0$ instead of $1$ is a notational convention, so that we can write $\sum_{n=0}^\infty a_nx^n$ instead of $a_0 + \sum_{n=1}^{\infty} a_n x^n$.

4. Aug 8, 2013

### fakecop

Thank you! Now I have another question. When doing term-by-term differentiation, x^0 often appears. As you've said, "x^0" is simply notation, and is equal to 1. The "x^0" in power series is not the same as the actual expression x^0, which is an exponential indeterminate form at x=0.

However, what is really confusing me is that almost all sources I've checked treats x^0 at x=0 to be 1, not indeterminate. For instance, according to the power rule, the derivative of ax^n is equal to n*ax^(n-1), the only exception being n=0. but if n=1, then the derivative of ax^1 would be 1*ax^0. This is clearly wrong as this expression is indeterminate at x=0.

This brings me to the second point-according to the power rule article on wikipedia, the derivative of x^0 is simply 0. This again treats x^0 to be equal to 1, which is wrong.

The third, perhaps most severe inconsistency, is that according to Wolfram Alpha, the function x^0 is continuous, and its derivative is 0, even at x=0! But i'm sure there is a hole at x=0. If x^0 is not even continuous, how can it have a derivative?

5. Aug 8, 2013

### lurflurf

^Again in all those cases x^0 is taken to mean 1. This is different that taking some limit where we need to be careful about such things. It is very handy to write x^0 for 1 in many formula.

6. Aug 8, 2013

### fakecop

You are right, but surely the function f(x) = x^0 is not differentiable at x=0, right? For the derivative to exist at x=0, the right and left hand limits of 0^0 (both equal to 1) must equal to 0^0 itself, which is indeterminate. Thus I don't understand why Wolfram indicated that the derivative would be zero-it shouldn't exist at x=0.

7. Aug 8, 2013

### pasmith

$0^0$ is indeed indeterminate, so $f : x \mapsto x^0$ is only defined for $x \neq 0$. To turn $f$ into a function defined on the whole line one must specify a value for $f(0)$, and the only sensible value is $\lim_{x \to 0} x^0 = \lim_{x \to 0} 1 = 1$. We then have
$$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^0 - 1}{h} = \lim_{h \to 0} \frac{1 - 1}{h} = 0.$$

8. Aug 8, 2013

### lurflurf

x^0 in this context means 1 not whatever you think it should mean. In any case x^0 will be one everywhere except when x=0 where we might assign any of several values or leave it undefined. Choosing x^0=1 makes the most sense and avoid writing a lot of equations with an exception "unless x=0 in which case the equation should have x^a replaced by one." We do not have a deep mathematical question here. We just have a convention for avoiding annoying exceptions.

9. Aug 9, 2013

### HallsofIvy

No, not right! You have been told repeatedly that f(x)= x^0 = 1 and that constant function has derivative 0 for all x.

Every single response (and there have been four) has told you that x^0=1. You say you understand that but then continue to talk about "0^0". There is NO "0^0" involved! It is just 1.
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