Question

Let \(f:[a,b]\rightarrow R\) be a continuous function. Let \(\epsilon >0\) be a constant. For \(x \in[a+\epsilon,b-\epsilon]\), define \(g(x):=\frac{1}{2\epsilon} \int_{x-\epsilon}^{x+\epsilon} f\)
a. Show that g is differentiable and find the derivative.
b. Let f be differentiable and fix \(x\in(a,b)\)(let epsilon be small enough 11 years ago

Answer 1

This an application of the fundamental theorem of calculus
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

Answer 2

That would seem to be my initial idea.

Answer 3

I understand the FTC, but I fail to see how it is used to prove that f is differentiable

Answer 4

I think we could use that for part b, because the integral will go to zero while 1/2epsilon will go to infinity, but... yea

Answer 5

should we input g into the limit definition of derivative?

Answer 6

The derivative is
\[
g'(x)=\frac 1 {2 \epsilon} (f(x+\epsilon)- f(x+\epsilon))
\]

Answer 7

The derivative is
\[
g'(x)=\frac 1 {2 \epsilon} (f(x+\epsilon)- f(x-\epsilon))
\]

Answer 8

how did you get that?

Answer 9

since we don't know the variable of integration

Answer 10

ohohohohoh

Answer 11

Variable of integration is some dummy variable like \(t\).

Answer 12

ok, nvm, I see it now

Answer 13

The only problem is the \(\epsilon\) part.

Answer 14

that's just a constant in the first case

Answer 15

Actually, since the epsilon simply shrink the interval, I suppose it isn't a problem at all.

Answer 16

Apply the FTC for the interval \( ( x-\epsilon, \quad c)\) and \( (c , \quad x+\epsilon) \)

Answer 17

where c is in between. \(\epsilon \) is a constant

Answer 18

Okay, but I think we can formalize this better.

Answer 19

I am sorry, I have to go to a meeting. I think, you should be able to continue this problem

Answer 20

Since \(f\) is continuous on \([a,b]\), it is also integrable.

Answer 21

Integrable on \((a,b)\). We have defined \(g(x)\) to have a domain of \((a,b)\) or less.

Answer 22

ok, so would this be a cohesive argument: \[g(x):=\frac{1}{2\epsilon} \int_{x-\epsilon}^{x+\epsilon} f\]\[=\frac{1}{2\epsilon} [F(x+\epsilon)-F(x-\epsilon)]~~by~ the~FTC\]now since we know that F is differentiable since it is the antiderivative of f, we find that \[g'(x)=:=\frac{1}{2\epsilon} [f(x+\epsilon)-f(x-\epsilon)] \]

Answer 23

is that enough?

Answer 24

thanks for the help elias

Answer 25

Differentiable has a definition, but in general if the derivative is continuous, then it is differentiable.

Answer 26

so do I need to add that this shows continuity?

Answer 27

Differentiable means that the limit: \[
\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
\]exists.

Answer 28

but without a function, I don't understand how I would show that here, because I don't know how to apply f(x+h)

Answer 29

I guess I also don't see how things will cancel

Answer 30

All you need is the fundamental theorem of calculus. You are not required to prove it and the function meets the conditions required to apply it. If the FTC was proven in class you are not required to reprove it from first principles.

Answer 31

hmm ok, i guess it just doesn't seem like enough

Answer 32

To use FTC, you have to prove that \(x+\epsilon\) and \(x-\epsilon\) are in \((a,b)\).

Answer 33

It's possible that \(x+\epsilon = (b-\epsilon)+\epsilon = b\)

Answer 34

Wait, I suppose \(x<b-\epsilon\), so we can say \(x+\epsilon < b\)

Answer 35

Anyway, the FTC is enough, just carefully state why the conditions are met. In particular you need \(f\) to be continuous on the interval it is being integrated on. In this case it is.

Answer 36

At the very least, we can say \(g\) is differentiable on \((a+\epsilon,b-\epsilon)\).

Answer 37

hmmm.... so is what I said before enough?

Answer 38

Yeah, I think so. I'm looking at this: http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part

Answer 39

fibo you could use that definition that wio mentioned to find the derivative of g
assume F'=f
\[g'(x)=\lim_{h \rightarrow 0}\frac{g(x+h)-g(x)}{h} \\ =\lim_{h \rightarrow 0}\frac{\frac{1}{2 \epsilon } \int\limits_{x+h-\epsilon}^{x+h+\epsilon}f dx-\frac{1}{2 \epsilon} \int\limits_{x- \epsilon}^{x+\epsilon} f dx }{h} \\ =\frac{1}{2\epsilon} \lim_{h \rightarrow 0}\frac{F(x+\epsilon+h)-F(x-\epsilon+h)-F(x+\epsilon)+F(x-\epsilon)}{h} \\ =\frac{1}{ 2 \epsilon } \lim_{h \rightarrow 0}(\frac{F(x+\epsilon+h)-F(x+\epsilon)}{h}-\frac{F(x-\epsilon+h)-F(x-\epsilon)}{h}) \\=\frac{1}{2 \epsilon}(F'(x+\epsilon)-F'(x-\epsilon)) \\ =\frac{1}{2 \epsilon} (f(x+\epsilon)-f(x-\epsilon))\]
but i know you guys are way passed this point now
and sorry for butting in so late

Answer 40

ok, so for part b, does this work?
since the smallest epsilon can possibly get is 0, we can look at the limit as epsilon goes to zero\[\lim_{\epsilon\rightarrow 0} g'(x)=\lim_{\epsilon\rightarrow 0} \frac{1}{2\epsilon}[f(x+\epsilon)-f(x-\epsilon)]=\]\[\ [\lim_{\epsilon\rightarrow 0} \frac{1}{2\epsilon} ]\lim_{\epsilon\rightarrow 0} f(x+\epsilon)-f(x-epsilon)=\lim_{\epsilon\rightarrow 0} \frac{1}{2\epsilon} *0=0 \]

Answer 41

thanks freckles, I just didn't know how that would work. but I am able to see it now

Answer 42

Let \(y = x-\epsilon\) for a moment. \[
g'(x) = \lim_{\epsilon\to0}\frac{f(y+2\epsilon)-f(y)}{2\epsilon}
\]

Answer 43

it was that second to last step that got me

Answer 44

ok

Answer 45

Let \(h=2\epsilon\).

Answer 46

then we have \[g'(x) = \lim_{\epsilon\to0}\frac{f(y+h)-f(y)}{h}\]

Answer 47

\[f'(y) = \lim_{\epsilon\to0}\frac{f(y+2\epsilon)-f(y)}{2\epsilon}\] ?

Answer 48

which is just the def of the derivative

Answer 49

Maybe I should have still put: \[
\lim_{\epsilon\to 0}g'(x)
\]

Answer 50

yes where
\[\lim_{\epsilon \rightarrow 0}g'(x)=f'(x) \text{ right? }\]

Answer 51

Yeah, the only part that is a bit tedious is we have \(\lim_{\epsilon\to0}\) instead of \(\lim_{2\epsilon\to0}\).

Answer 52

2*0 is still zero though...?

Answer 53

There is some composition of limits rule that would work for it though.

Answer 54

so i take it my explanation isn't enough?

Answer 55

It's good enough I think.

Answer 56

Maybe you could say: \[
\lim_{\epsilon\to 0}2\epsilon = 0
\]

Answer 57

ok, I can add that, but why would it be necessary? I'm not computing that limit to avoid the infinity

Answer 58

There isn't any rule about what level of rigor is necessary.

Answer 59

no, but the more the better in this class