- #36
Fredrik
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That's not a norm. |f(t)-g(t)| is the absolute value of the real number f(t)-g(t).
Fredrik said:That's not a norm. |f(t)-g(t)| is the absolute value of the real number f(t)-g(t).
Fredrik said:Plot the graphs of the functions
\begin{align}
& x\mapsto x^3+9\\
& x\mapsto x^3+6\\
& x\mapsto x^3\\
& x\mapsto x^3-3
\end{align} and label them 1,2,3,4 respectively. Every g in D is contained in the region between graphs 1 and 4, and goes outside of the region between graphs 2 and 3 at least once.
Fredrik said:[tex]\|f-z\|=\sup_{t\in[0,1]}|f(t)-z(t)|=\sup_{t\in[0,1]}|t^3+3-5t|=3[/tex] The last step is easy since the derivative of f-z doesn't have any zeroes in [0,1]. This means that f-z is either increasing in the entire interval, or decreasing in the entire interval. Since (f-z)(0)=3 and (f-z)(1)=-1, it must be decreasing, and the maximum value is 3.
There seem to be some gaps in you knowledge...bugatti79 said:This is becoming interesting...so the slope if you like is negative ie (0,3) and (1,-1)...but what is the importance of determining its slope?
bugatti79 said:Is knowing what the resulting max value not enough, to determine whether f-z is in A?
Thanks Fredrik :-)
Fredrik said:What you need to determine is the max value in the interval. Since the slope is never zero inside the interval, we know that the function has its max value at one of the endpoints of the interval.
If the slope had been zero somewhere in the interval, then that might have been the point where the function has its max value.
Fredrik said:[tex]\|f-z\|=\sup_{t\in[0,1]}|f(t)-z(t)|=\sup_{t\in[0,1]}|t^3+3-5t|=3[/tex] The last step is easy since the derivative of f-z doesn't have any zeroes in [0,1]. This means that f-z is either increasing in the entire interval, or decreasing in the entire interval. Since (f-z)(0)=3 and (f-z)(1)=-1, it must be decreasing, and the maximum value is 3.
I'm not a fan of the notation f-z=t^3-t^2, because it looks like a function equal to a number. If f-z satisfies [itex](f-z)(t)=t^3-t^2[/itex] for all [itex]t\in[0,1][/itex], then [itex](f-z)'(t)=3t^2-2t[/itex] for all [itex]t\in [0,1][/itex], so [itex](f-z)'(t)=0[/itex] implies [itex]t=2/3[/itex].bugatti79 said:So if I had different function like f-z = t^3-t^2 say, then the derivative of this wrt t and subbing in 0 we would get 0 which would impy the max is at t=0?
Yes, that's right.bugatti79 said:Just one little query regarding the [itex]\|f-z\|=\sup_{t\in[0,1]}|f(t)-z(t)|=\sup_{t\in[0,1]}|t^3+3-5t|=3[/itex]
You put in 0 and you are left with 3. If you put in 1 we get -1. So we take the highest of those two because [itex]sup_{t \in [0,1]} | f-z|=max\{t_1,t_2\}[/itex] or something like this?
Fredrik said:I'm not a fan of the notation f-z=t^3-t^2, because it looks like a function equal to a number. If f-z satisfies [itex](f-z)(t)=t^3-t^2[/itex] for all [itex]t\in[0,1][/itex], then [itex](f-z)'(t)=3t^2-2t[/itex] for all [itex]t\in [0,1][/itex], so [itex](f-z)'(t)=0[/itex] implies [itex]t=2/3[/itex].
I think you called the set D earlier. What we have to do is of course to determine [itex]\|f-g\|[/itex], and this is fairly straightforward. I'll leave that to you.bugatti79 said:but how would we determine whether this new function is in A?
I'm not sure why you would want to know, but in general, if you have two functions F and G, and want to know if one of them is "above" the other, you need to find out if the equation F(x)=G(x) has any solutions. If it doesn't, the graphs never intersect.bugatti79 said:Ok, you have identified that the maximum value lies somewhere at a point corresponding to t=2/3 because this is where the slope is 0. How do we establish whether the function lies above x^3+9 say?
bugatti79 said:Ok, thanks
How would you verify whether another new function like z(x)=5x is an element in D?
According to Wolfram, it looks like this function is with the region...
http://www.wolframalpha.com/input/?i=plot+x%5E3%2C+x%5E3%2B6%2C+x%5E3%2B9%2C+x%5E3-3%2C+5x+between+0+and+1
Fredrik said:[tex]\|f-z\|=\sup_{t\in[0,1]}|f(t)-z(t)|=\sup_{t\in[0,1]}|t^3+3-5t|=3[/tex] The last step is easy since the derivative of f-z doesn't have any zeroes in [0,1]. This means that f-z is either increasing in the entire interval, or decreasing in the entire interval. Since (f-z)(0)=3 and (f-z)(1)=-1, it must be decreasing, and the maximum value is 3.
A function [itex]g\in C[0,1][/itex] (where [itex]C[0,1][/itex] now denotes real-valued continuous functions on [0,1]) is in D if and only if [itex]3\leq\|f-g\|\leq 6[/itex]. So to check if z is in D, you have to find [itex]\|f-z\|[/itex], and I did. [itex]\|f-z\|=3[/itex]. Since [itex]3\leq 3\leq 6[/itex], z is in D.bugatti79 said:Just a question on these 2 post. Do we conclude that the function z(x)=5x is in D because the maximum value of 3 is less than our highest intercept of 9 and greater than -3?
Fredrik said:A function [itex]g\in C[0,1][/itex] (where [itex]C[0,1][/itex] now denotes real-valued continuous functions on [0,1]) is in D if and only if [itex]3\leq\|f-g\|\leq 6[/itex]. So to check if z is in D, you have to find [itex]\|f-z\|[/itex], and I did. [itex]\|f-z\|=3[/itex]. Since [itex]3\leq 3\leq 6[/itex], z is in D.
Fredrik said:A function [itex]g\in C[0,1][/itex] (where [itex]C[0,1][/itex] now denotes real-valued continuous functions on [0,1]) is in D if and only if [itex]3\leq\|f-g\|\leq 6[/itex]. So to check if z is in D, you have to find [itex]\|f-z\|[/itex], and I did. [itex]\|f-z\|=3[/itex]. Since [itex]3\leq 3\leq 6[/itex], z is in D.
f-z has a local maximum at t if and only if (f-z)'(t)=0 and (f-z)''(t)<0. So if (f-z)'(t)≠0 for all t, then we can be sure that f-z doesn't have any local maxima in the interval.bugatti79 said:So we can verify this by checking that the derivative of f-z is not 0. If is WAS 0, then there would be a possibility that the max value could be higher than 6 hence not in D...?
LCKurtz said:I have been reading this thread since my first post which was #2. I am curious bugatti79 about two questions:
1. What course are you currently studying where this topic arose?
2. Have you taken a course in calculus?
Fredrik said:f-z has a local maximum at t if and only if (f-z)'(t)=0 and (f-z)''(t)<0. So if (f-z)'(t)≠0 for all t, then we can be sure that f-z doesn't have any local maxima in the interval.