# Let f(x)=∑a_n(x^n) for |x|<R. If f(x)=f(-x) for all |x|<R, show that a_n=0 for all odd n.

*print*Print*list*Cite

### 1 Answer

If `f(x)=f(-x)` for all `|x|ltR, f(x)-f(-x)` is a constant 0 function for all `|x|ltR.`

The constant 0 function has a **unique** power series that is ` `

` `

Therefore all the coefficients of the power series of f(x)-f(-x) ` ` are 0.

Now let's find the coefficients of f(x)-f(-x) as an expression of a_n

`f(x)-f(-x)=sum_0^(oo)a_nx^n-sum_0^(oo)a_n(-x)^n`

`f(x)-f(-x)=sum_0^(oo)(a_n-(-1)^na_n)x^n`

Hence` (a_n-(-1)^na_n)=0` for all n.

If n even `(-1)^n=1 ` and the condition becomes `a_n-a_n=0 ` which is true for any even n.

if n odd,` (-1)^n=-1` and the condition becomes `a_n+a_n=0. ` Therefore `a_n=0.`

**In conclusion if** `f(x)=f(-x)` , `a_n=0` **for all odd n**.