# Math Help - complex analysis, inequality contradiction

1. ## complex analysis, inequality contradiction

I realise this is not calculus, yet still I'd appreciate if someone could help me

I have an entire function, f
and I need to show that the following is NOT true for ALL n=1,2,....

|f(n)(0)| >= nn *n!

(ie, the modulus of the nth derivative of f at zero is greater or equal to n to the power of n, times n factorial)

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so, my first attempt was by counter example, I picked a simple polynomial and showed the above expression is not true for all n.
however, I'm not sure if the question just wants a counter example, or a generalised proof for all such entire functions. If so, I'm slightly stumped... the only thought I have in this case is to look at Taylor series, but what will that yield?

2. ## Re: complex analysis, inequality contradiction

You are in the correct path. The negation of a "for all" is "there exists" so to disprove this you only need one counter example. A polynomial would work very well as after taking $n+1$ derivatives of an nth degree polynomial they are all zero! The right hand side is always non negative.

3. ## Re: complex analysis, inequality contradiction

Originally Posted by TheEmptySet
The negation of a "for all" is "there exists" so to disprove this you only need one counter example.
ah, ah, here is where I became unsure about the way to go about this.. it doesn't say "FOR ALL entire functions f .....", it just says "let f be entire, and then for all n we have ....".
so if we pick a different function, we could assume the statement to still be true (even though we know it's not)

do you see a difference? maybe i need to prove it more generally
or am i just making things more complicated..

4. ## Re: complex analysis, inequality contradiction

ah Okay then. Do a google search for Liouville's theorm. The jist of this is that the constants in a taylor expansion are bounded by a polynomial.

Liouville's theorem (complex analysis) - Wikipedia, the free encyclopedia