Can Factoring Out a Negative One Affect the Convergence of a Taylor Series?

[ronaldor9]

First of all if i have a function with all negative terms is it possible to determine its convergence simply by factoring the negative one, treating the other terms as a positive series determine its convergence then assume that multiplying by the constant negative one will not change its convergence.

Second, one cannot say the taylor series is equal to the function whose taylor series has been determined unless one states explicitly the radius of convergence, true? Can one use the ratio test to determine the interval of convergence or does one have to use the remainder, i.e. Lagrange form or another, and see if it goes to zero as n goes to infinity to determine convergence:
$$R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}.$$

thanks

 

[JazzFusion]

First part: yes, true.

 

[g_edgar]

Use ratio test to determine interval of convergence. But that does not show that the sum of the series is the original function. There is some example with $$\exp(-1/x^2)$$ illustrating this.

 

[ronaldor9]

why is it that even if it converges the sum of the series is not the original function?

 

[HallsofIvy]

The function g edgar gives:
$$f(x)= e^{-1/x^2}$$
if $x\ne 0$, 0 if x= 0 is infinitely differentiable and its n^th-derivative at x= 0 is 0 for all n. That is, its Taylor series about 0 (MacLaurin series) is identically equal to 0. That obviously converges for all x but, just as obviously, it is not equal to f(x) for any x other than 0.

 

[ronaldor9]

hmm, but if you require that x does not equal zero for the taylor series arent the function equal

 

[HallsofIvy]

I have no idea what that means. What do you mean by "require that x does not equal 0 for the Taylor's series"? x= 0 is the only place they are 0. If you mean simply look at the values of the two functions everywhere except at x= 0, you are dropping the only point where they are equal!

 

[ronaldor9]

Oh never mind that, I had to read the past a couple more times but I understand. Thanks for the counterexample.