Webpage title: Proving a Series Equals Zero: Differentiation Method

[ibmichuco]

Hi all,

In order for the series

a_0 x^0 + a_1 x^1 + a_2 x^2 + a_3 x^3+\cdots

to be equal zero, where x can take any values,say,
from 0 to 1, the coefficients a_i must be all zero.

This sounds reasonable, but I am sure that there is some kind of
prove for this. Most books that I look for did not offer one.

Any suggestion would be appreciated.

Michuco

 

[wisvuze]

EDITED: I realized that this is in the homework section, so I cannot give you the answer, sorry

 

[wisvuze]

But, as a hint, since the a_i's are fixed for ANY x's, fiddle around with setting particular x_k's to be 1 and others to be 0.
What happens when all x's are 1? You get that a_1 + ... .+ a_n = 0 , that is looking good so far, since if the a's are 0 then that will be true for sure. You can find something even more useful though

 

[micromass]

First question is: what does it mean for a series to equal 0? There are definitions for a series to equal something such that the claim in your OP isn't even true!

What you could probably do is to differentiate the series and keep evaluating it in 0... There are some technicalities involved though...

 

[ibmichuco]

EDITED: I realized that this is in the homework section, so I cannot give you the answer, sorry

Sorry, I posted in the wrong section, this is not homework.
It is the standard way that perturbation theory is discussed in
most books. Since it has to do with series, these books just state
the results.

Michuco
PS. MOD, could you move this to the non homework section. Thanks

 

[HallsofIvy]

Are you talking about a finite degree polynomial or a power series? For a finite, say degree n, polynomial, the obvious thing to do is to set x equal to say, 0, -1, 1, 2, -2, etc. for a total of n values. That gives n linear equations, all equal to 0. If you show those equations are independent then the unique solution is all coefficients equal to 0. The simpler way to do it is to set x= 0, then differentiate that sum- if it is a constant (0), then its derivative is also 0, and set x= 0, then differentiate again, then set x= 0, etc. Tht also gives you n equations but n very easy equations.

If this is an infinite series, find the MacLaurin series for the function f(x)= 0, then use the fact that such a power series is unique.

 

[ibmichuco]

First question is: what does it mean for a series to equal 0? There are definitions for a series to equal something such that the claim in your OP isn't even true!

What you could probably do is to differentiate the series and keep evaluating it in 0... There are some technicalities involved though...

This sounds neat! For example, if I expand f(x) as n terms

f(x) = a_0x^0+a_1x^1+a_2x^2+\cdots =0

then I can just differentiate n+1 times and that would leave me with
a_n=0 and in the same way to show that all other coefficients a_i
must also be zero. Is this reasonable?

Regards,

Michuco