Your argument relies on the radius of convergence of the series being at least Pi, but f(x) is not even defined for x = Pi/2. Compare to the MacLaurin series of the Pi-periodic function tan(x), which has only nonnegative coefficients. Yet, it is indeed true that some coefficients of the MacLaurin series of f(x) must be negative. The MacLaurin series of sin(x) converges for all x, and that of tan(x) converges for |x| < Pi/2. From this, we may deduce that the MacLaurin series of f(x) also converges for |x| < Pi/2. Now, for x just a little less than Pi/2, we have that tan(sin(x)) is close to tan(1), whereas sin(tan(x)) fluctuates between -1 and 1. In particular, there are positive values x < Pi/2 for which f'(x) is negative, which implies that some coefficient of the MacLaurin series must be negative. Correct me if I am wrong. Jakob
It is not true:
the function is 2Pi-periodic, and bounded from above and below (and non-constant)
so f'(x) must be negative for at least one positive value of x, which cannot be the case with only nonnegative coefficients of its series.
(actually not being allowed to use a computer lead me to conjecture that it is not true ;) )
We have In[735]:= # == Normal[Series[#, {x, 0, 23}]] &[Tan[Sin[x]] - Sin[Tan[x]]]
Out[735]= -Sin[Tan[x]] + Tan[Sin[x]] == x^7/30 + 29 x^9/756 + 1913 x^11/75600 + 95 x^13/7392 + 311148869 x^15/54486432000 + 10193207 x^17/4358914560 + 1664108363 x^19/1905468364800 + 2097555460001 x^21/7602818775552000 + 374694625074883 x^23/6690480522485760000
Without using a computer, prove or disprove: The terms of the infinite series are nonnegative.
Christoph