Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider a function $f_1$ defined by $f_1(x)=1-x+o(x)$ and $f_1(2x)=f_1(x)^2 + 0$. It's simple to find that $f_1(x)=e^{-x}$ (for example by writing series near $x=0$).

Consider a function $f_2$ defined by $f_2(x)=2-x^2+o(x^2)$ and $f_2(2x)=f_2(x)^2-2$. It can be proven that $f_2(x)=2 \cos(x)$.

Is there any formula (probably with use of special functions) for the generalization of this, i.e. function $f_n$ defined by $f_n(x)=2^{n-1}-x^n+o(x^n)$ and $f_n(2x)=f_n(x)^2-2^{2n-2}+2^{n-1}$?

share|improve this question
    
Last line you probably want $f_n(2x)$ on the LHS. –  Willie Wong Oct 29 '10 at 22:50
    
@Willie Wong Yes, thank you. –  Fiktor Oct 30 '10 at 7:19

1 Answer 1

Assuming Willie Wong's correction, by a method of undetermined coefficients it's not too hard to generate a series for $f_3$, and it begins

$ 4-x^3+(1/56)x^6-(1/14112)x^9+(13/115379712)x^{12}-(53/629973227520)x^{15}+(17413/495415985907548160)x^{18} + \cdots$

but it's not clear what to do with this. For example, the denominators have large-ish prime factors: $495415985907548160 = 2^{14} 3^4 5^1 7^5 13^1 31^1 73^1 151^1$. This seems to suggest that there's not going to be an easy third-order differential equation, whereas your $f_1$ and $f_2$ satisfy first- and second-order differential equations.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.