
#1
Jun408, 03:58 PM

P: 5

Let 0 < r < 1. Then [tex]\sup_{x\in[r,r]}f(x)=f(r)}[/tex], right? However, the text I'm reading says it's [tex]f(r)[/tex]. 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.




#2
Jun408, 04:20 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

Hi AKBAR!
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? 



#3
Jun408, 05:15 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

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)= x^{2}, then f(r)= f(r)= r^{2}< 0. The maximum (and so sup) occurs in the middle of the interval. The sup is f(0)= 0. 



#4
Jun408, 05:24 PM

P: 5

confused about supremum
Ahhhh....that really clears things up. Thank you guys.
The function in question was [tex]f(x)=\frac{(n1)!}{(1+x)^n}[/tex]. So [tex]\sup_{r\leq x\leq r}\frac{(n1)!}{(1+x)^n}=\frac{(n1)!}{(1r)^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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