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tiny-tim

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No.Let 0 < r < 1. Then [tex]\sup_{x\in[-r,r]}f(x)=f(r)}[/tex], right?

That will be true if f is increasing in [-r,r], but not in most other cases.

What is f? What is the relevance of 0 < r < 1? What is the context?

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HallsofIvy

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You are taking the supremum ("least upper bound") of the values of

For example, if f(x)= x, then f(-r)= -r, f(r)= r and f takes on all values between -r and r. In that case the sup of

But if f(x)= -x, then f(-r)= r, f(r)= -r and now the supremum occurs at -r: f(-r) is the largest value (in fact it is the maximum value).

And it can get more complicated than that. If f(x)= -x

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The function in question was [tex]f(x)=\frac{(n-1)!}{(1+x)^n}[/tex]. So [tex]\sup_{-r\leq x\leq r}\frac{(n-1)!}{(1+x)^n}=\frac{(n-1)!}{(1-r)^n}[/tex] The r comes from the radius of convergence of a Taylor expansion (I'm reading about where T(x) = f(x) ).

Thanks again for the help.

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