Proof of the Fundamental Theorem of Calculus.

[Darth Frodo]

Hi all I'm currently working my way through proving the FToC by proving something that is a foundation for it. So I need to prove that;

L(f,P_{1}) ≥ L(f,P) where P\subsetP_{1} i.e where P_{1} is a refinement of P.

So, Let P_{1} = P \cup {c} where c \in [x_{k-1},x_{k}]

Let L' = inf{x|x \in [x_{k-1},c]}

Let L'' = inf{x|x\in} [c,x_{k}]

L = inf{x|x\in [x_{k-1},x_{k}]}

So from this the next line is;

L'(c-x_{k-1}) + L''(x_{k}-c) ≥ L(x_{k}-x_{k-1})

Now this is the line I can't fully grasp. How was this line come up with? I can understand it from a geometrical/graphical/pictorial point of view, but from an analytic point of view I cannot.

So far this is what I have,

L' + L'' ≥ L

And If I multiply by the differences in the x-ordinates I still don't get the same line. Any help would be appreciated.

How exactly did that mystery line happen?

 

[micromass]

I know what you mean, but I'm going to comment on your notation anyway.

Hi all I'm currently working my way through proving the FToC by proving something that is a foundation for it. So I need to prove that;

L(f,P_{1}) ≥ L(f,P) where P\subsetP_{1} i.e where P_{1} is a refinement of P.

If you want to use LaTeX, then you should not just use LaTeX on each individual symbol, but rather on the whole line. So writing something like

L(f,P,[itеx]_{1}[/itеx]) ≥ L(f,P)

it is much handier (and much more aesthetic) to write

[itеx]L(f,P_1)\geq L(f,P)[itеx]

So, Let P_{1} = P \cup {c} where c \in [x_{k-1},x_{k}]

Let L' = inf{x|x \in [x_{k-1},c]}

As you wrote it now, we have that $L^\prime = x_{k-1}$. You should probably have written

L^\prime = inf \{ f(x)~\vert~x \in [x_{k-1},c]\}

Same remark with the others.

Let L'' = inf{x|x\in} [c,x_{k}]

L = inf{x|x\in [x_{k-1},x_{k}]}

So from this the next line is;

L'(c-x_{k-1}) + L''(x_{k}-c) ≥ L(x_{k}-x_{k-1})

As it is now, it is simply not true. Take f(x)=-1 and take x_k=0, $c=1$ and $x_{k+1}=2$. Then $L = L^\prime = L^{\prime\prime} = -1$. But

L^\prime + L^{\prime\prime} = -2 \leq -1 = L

 

[Darth Frodo]

L^{'}(c-x_{k-1}) = (-1)(1-2) = 1

L^{''}(x_{k}-c) = (-1)(0-1) = 1

L = (-1)(x_{k}-x_{k-1}) = -1(-2) = 2

1 + 1 ≥ 2

Does it not hold?

 

[micromass]

L^{'}(c-x_{k-1}) = (-1)(1-2) = 1

L^{''}(x_{k}-c) = (-1)(0-1) = 1

L = (-1)(x_{k}-x_{k-1}) = -1(-2) = 2

1 + 1 ≥ 2

Does it not hold?

Oh, you meant that $L(x_k - x_{k-1})$ is a multiplication. I thought it was part of the notation. Makes sense then.

Here's a hint to prove it

L(x_k - x_{k-1}) = L(x_k-c) + L(c-x_{k-1})

 

[Darth Frodo]

L(x_{k}-x_{k-1}) = L(x_{k}-c) + L(c-x_{k-1})

L ≤ L' and L ≤ L''

\Rightarrow L(x_{k}-x_{k-1}) ≤ L''(x_{k}-c) + L'(c-x_{k-1})

This correct?

 

[micromass]

Yes. Be sure that you know how to prove $L\leq L^\prime$ though.

 

[Darth Frodo]

Hmm, that I don't know but i'll attempt it later and get back to you on it if I don't get it. Thanks for the help! Super as always!

BTW, would this be the kind of exercise one would expect from Spivak?

 

[micromass]

Hmm, that I don't know but i'll attempt it later and get back to you on it if I don't get it. Thanks for the help! Super as always!

BTW, would this be the kind of exercise one would expect from Spivak?

Yes. This is more of an exercise you would encounter in Spivak than in another calculus book. Those exercises tend to be difficult at first, but you get used to them pretty quickly.