What is the Taylor series expansion at ##a/2## for finding real functions?

L Navarro H
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Homework Statement
Find the real functions f so: f' (x) + f (a - x) = e^x
a is a constant
Relevant Equations
none
Mod note: Member warned that some effort must be shown.
 
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L Navarro H said:
Homework Statement:: Find the real functions f so: f' (x) + f (a - x) = e^x
a is a constant
Relevant Equations:: none

.
Please post your attempt, per forum rules.
 
haruspex said:
Please post your attempt, per forum rules.

I use the Abel Identity to find the functions, but I'm no sure if it's correct to used it in this problem
my answer is {e^x , - 1/(3*e^(2x))
 
L Navarro H said:
I use the Abel Identity to find the functions, but I'm no sure if it's correct to used it in this problem
my answer is {e^x , - 1/(3*e^(2x))
Are you saying that each of those is a solution to the equation? Doesn’t look that way to me. Shouldn't 'a' figure in the answer?
Please post your working.

Hint: you need a strategy for eliminating references to f(a-x) and its derivatives so that all the references to f are to f(x) and its derivatives.

Btw, sorry for the delay.. for some reason I did not get the alert. Happens sometimes.

Still there @L Navarro H ? If you need a stronger hint, try two things...
1. Write the equation swapping x and a-x.
2. Differentiate the equation.
Can you see how get rid of references to f(a-x)?
 
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have you tried to expand a solution in the Taylor series at the point ##a/2##? That is to look for solutions in the form ##f(x)=\sum_{k=0}^\infty f_k(x-a/2)^k##

upd: but haruspex's idea is better
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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