# Thread: Analysis function

2. ## Re: Analysis function

Ker T = {x \in l^\infinity | T(x) = 0}
= T(x1,x2,...) = 0
= (x2, x3,...)= 0
= xi = 0 for i=2,3,...
= {x | x= (x1, 0,0,...)}
dim ker T = 1

3. ## Re: Analysis function

Thank you very much

4. ## Re: Analysis function

Those are both factoring problems and both use what is known as the FOIL method. (FOIL = First, Outer, Inner, Last). You can look that up on Google to learn more about it. I don't have the words to explain it clearly. It's actually something you learn by practicing; it isn't really all that amenable to just listening to someone tell you how to do it. You have to actually DO it yourself. And honestly? The FOIL method is taught pretty 'early on' these days, like in Algebra 1 classes. I'm kinda surprised you don't know of it.
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Anyways, f(x) = (x^2 - 25) can be factored into [ (x + 5) * (x - 5) ], and the zeros
are x = - 5 and x = 5.

Similarly,

f(x) = x^2 + 4x - 32 = [ (x + 8) * (x - 4) ], and the zeros are x= - 8 and x = 4.