Anyone ever seen this notation? I've looked in 3 textbooks

>>8996739
It's the function evaluated at 0.

>>8996739
That's just a shorthand way of writing operators.

>>8996842
how would you expand it?

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>>8996966
It simply means each part in the left bracket applied to the part in the right. It is linear.

In the pic: the top line is an example of how you can use the natation for partial derivatives and the bottom line is your example.

I hope that helps.

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>>8996978
What does f^2 mean? f(0) * f(0)?

>>8997412
f^2(0)=f(f(0))

>>8996978
why do you assume [math]f, g[/math] are linear?

>>8997814
If you have an additive structure on the codomain of f and g, you can define f+g by (f+g)(x)=f(x)+g(x). That has nothing to do with linearity.

>>8997835
Actually, now that I think about it, the condition f(0)=0 is redundant. If you have f(ax+by)=af(x)+bf(y), then you have f(0)=f(-0)=-f(0), and that forces f(0)=0.

>>8997814
I mispoke. I have only ever applied this notation to partial derivatives. I assumed it was linear due to the linearity of differentiation. I'm unsure if it's linear in general.

>>8997858
who the fuck would include f(0) = 0 into the definition of linear map?

>>8998056
I have seen f(e)=e' in the def of a group homomorphism too, also redundant

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>>8996739
>>8996978
Horrid notation, yet understandable. Needs tweaks tho.

what does any of this have to do with what OP asked?

>>8997858
>f(0)=f(-0)=-f(0)
Whoa.
f(0)=f(x-x)=f(x)+f(-x)=f(x)-f(x)=0.
That's what I would have done but you wrote it better.

>>8998690
e'
=f^-1(e)f(e)
=f^-1(e)f(ee)
=f^-1(e)f(e)f(e)
=f(e).
Is this correct?

>>9000171
* f(e)^-1 as inverse of f(e).