# two functions $f(x)$ and $g(x)$

let be two functions $f(x)$ and $g(x)$ with an infinite set of roots $a_{n}$ and $b_{n}$ so $f(a_{n}) =0= g(b_{n})$

also they satisfy the same functional equation $f(1-s)=f(x)$ and $g(s)=g(1-s)$

then we use some numerical method to evaluate the roots and we check that the first 100 roots agree so can we conclude that $f(x)= g(x)$ or at least $f(x)=h(x)g(x)$ they are proportional function

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What if the 101th root is different? – Ben Millwood Sep 1 '12 at 9:59
Apart from a typo, the functional equation simply says that the functions are symmetrical about $s=\frac 1 2$. The only other condition is that they each have a countable set of roots (apparently isolated roots to make better sense of the first 100 roots - there seems to be a natural way of counting them). It is easy to construct functions with these two properties. $f$ and $g$ could be identical for a long finite stretch, but different beyond that - without there being any problem about continuity or differentiability. – Mark Bennet Sep 1 '12 at 10:55
@MarkBennet: for a "first 100" to make sense you only need that the roots are well-ordered. There could be $\omega_1$ of them, although you might struggle to actually list them in that case. – Ben Millwood Sep 1 '12 at 14:21
@BenMillwood - yes, that is right, but the idea of comparing the first 100 roots of two functions without qualification means that two sets of roots need to be well-ordered - and there might be expected to be some natural relationship between those well-orderings, since nothing more is said. There are easy examples where the question makes sense even if the roots are not isolated (roots at $s=1-\frac 1 n$ and $s=\frac 1 n$ being an example). – Mark Bennet Sep 1 '12 at 16:32

$$f = \mathbb 1_{\mathbb R \backslash {\mathbb Q}}$$
Any computer will give you $f = 0$.
Moreover, in your case, $g = \mathbb 1_{{\mathbb Q}}$ gives a counter-example.