Let 0 < r < 1. Then $$\sup_{x\in[-r,r]}f(x)=f(r)}$$, right? However, the text I'm reading says it's $$f(-r)$$. How could this be? For example, say r = 0.5, then the least upper bound of [-0.5, 0.5] is 0.5, or r, right? I don't see how it could be -r. Thanks for any help.

tiny-tim
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Hi AKBAR! Let 0 < r < 1. Then $$\sup_{x\in[-r,r]}f(x)=f(r)}$$, right?

No.

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? HallsofIvy
Homework Helper
Let 0 < r < 1. Then $$\sup_{x\in[-r,r]}f(x)=f(r)}$$, right? However, the text I'm reading says it's $$f(-r)$$. How could this be? For example, say r = 0.5, then the least upper bound of [-0.5, 0.5] is 0.5, or r, right? I don't see how it could be -r. Thanks for any help.
You are taking the supremum ("least upper bound") of the values of f not x!

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 f(x) on the interval (not the sup of the interval itself) is f(r).

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)= -x2, then f(-r)= f(r)= -r2< 0. The maximum (and so sup) occurs in the middle of the interval. The sup is f(0)= 0.

Ahhhh....that really clears things up. Thank you guys.

The function in question was $$f(x)=\frac{(n-1)!}{(1+x)^n}$$. So $$\sup_{-r\leq x\leq r}\frac{(n-1)!}{(1+x)^n}=\frac{(n-1)!}{(1-r)^n}$$ 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.