Is my proof for the Chain Rule invalid?

[mathwonk]

well you are in good company. as i said, even the great g.h. hardy, and many many other calculus authors since his time, have presented the problem with ratio canceling proof apparently without noticing the special case can be treated easily, but separately, once it is noticed.

but you still have to give a proof that the difference k, in

f(z(x+i)) - f(z(x))/i = z'(x)*f(z(x)) + k, does go to zero even when

z(x+i) = z(x), which is exactly what i was doing in the part where you said you failed to see why i was going to so much trouble. i.e. this is precisely the import of my case 2.[I have been explaining this proof to people for almost 40 years now, even though as i said, it appears correctly in 100 year old books on analysis.]

 

[mathwonk]

i just checked edwards and penney, hass weir thomas, grossman, courant, even joseph kitchens great book, and none of them give this proof.

kitchen gives a two case proof but makes the second case look more complicated. the point is that in the second case the LHS of werg's equation

f(z(x+i)) - f(z(x))/i = z'(x)*f(z(x)) + k, is zero, so it suffices to show that

in that case z'(x) is also zero, which is what i remark as my proof of case 2. i.e. in case 2, k is identically zero. and in case 1, the elementary proof shows it goes to zero.

courant does remark in a footnote that the proof can be carried out by a special argument, but that he prefers to use a proof that does not have two separate cases.

courants proof, i.e. hardy's, is also the one given in , grossman, mattuck, etc ... edwards penney give a proof like that of spivak in edition 4 but the usual hardy proof in edition 6 as i recall.

grossman even declares that a different approach from ours is REQUIRED in case 2, which of course is false. many other books imply the same.

spivaks proof, also given in 4th edition of edwards penney, builds a special function phi, whose definition has two cases, essentially building our two case argument into the definition of this function thus making it look afterwards as if the proof has one case. but the proof there is totally unmotivated, and the two cases are there but hidden.

 

[Werg22]

Hummm I seee... but there's an ambiguity in my mind... I still do not understand what is to be proven with k for the case z(x+i) = z(x). I see it as such: the product of the two ratios is not defined for z(x+i) = z(x), however it's defined for an infinity of i in the neighbourdhood of 0. We need not the product to be always defined: as long as we have enough information on the behaviour of f(z(x+i)) - f(x) /i, we have enough: we know that for a conviniently small i, the absolute value of k gets smaller. This truth is not constrained by the fact that the value of the product is not always defined...

 

[Office_Shredder]

You haven't actually proven it, you've just said something around the lines of

'well, from curves that behave relatively well, it appears as if it should become small'

But either way, you're assuming that df(u(x))/dx = df/du*du/dx in the first place just to make that equality so you can say k becomes small

 

[mathwonk]

i rewrote my post to explain this i hope.

 

[mathwonk]

now that we all aprpeciate the issues, let me try again to give the proof:

we want to show that if z'(x0) exists and f'(z0)) exists where z0 = z(x0),

then also g'(x0) exists where g = foz, and in fact that g'(xo) = f'(z0)z'(x0).

so we want to show the limit of the quotient [f(z(x))-f(z(x0))]/[x-x0]

exists as x-->x0, and is equal to the product f'(z0)z'(x0).

case 1) z'(x0) is not zero. then there is a punctured interval around x0 where z(x) never equals z0.

then the quotient [f(z(x))-f(z(x0))]/[x-x0] equals the product of two quotients

[f(z(x))-f(z(x0))]/[z(x)-z0] . [z(x)-z(x0)]/[x-x0].

and we are done by definition of the derivative and the product rule for lim its.case 2) z'(x0) = 0. this time all we have to show is that the quotient

[f(z(x))-f(z(x0))]/[x-x0] has limit zero as x-->x0, and it suffices to show this separately as x-->x0 through values where z(x) = z0, and those where it does not.

at those x where z(x0) = z0, there is nothing to show, as then the quotient is identically zero.

for values where z(x0) = z0, the argument from case 1 applies again. QED.

 

[Werg22]

You haven't actually proven it, you've just said something around the lines of

'well, from curves that behave relatively well, it appears as if it should become small'

But either way, you're assuming that df(u(x))/dx = df/du*du/dx in the first place just to make that equality so you can say k becomes small

While mathwonk made me realize I did not exactly prove the general case, it's for a totally different reason than whatever you are speaking of.

 

[JonF]

You don't prove any case, you have yet to state anything formal. I mean go back and look at what you responded to my previous question and note how you ignored the part that said "using the formal definition of a limit"

 

[Werg22]

No, it's just that you don't understand.

 

[JonF]

Why don't you try stating the formal definition of a limit and then comparing it to what you said about "keeping the change equal" and then tell me who is failing to understand. But hey, "you don't undertand" is a good fall back position.

 

[Werg22]

How about you check post 32.

 

[prasannapakkiam]

That is it.

Arguments seem to be revolving around whether dy/dx or dy/du can be split up into a fraction. How about we take a look at: f'(x)=lim h--> 0 (f(x+h)-f(x))/h. How did we derive it? We used small values of y and x i.e. dy and dx. Remember that all the equal signs in my proof are the "aprox." signs. The limit in the end ties it all up to a differential. But the limit thing is neccessiarily neccessary...

 

[Werg22]

That is it.

Arguments seem to be revolving around whether dy/dx or dy/du can be split up into a fraction. How about we take a look at: f'(x)=lim h--> 0 (f(x+h)-f(x))/h. How did we derive it? We used small values of y and x i.e. dy and dx. Remember that all the equal signs in my proof are the "aprox." signs. The limit in the end ties it all up to a differential. But the limit thing is neccessiarily neccessary...

This was the argument but it no longer is. The argument now is that the equality of the ratio and the product of the other ratios as well as the equality the limit of the first ratio with the product of the other ratios are doubtful. But as I said, for the case in which there exist a neighborhood in which the functions are monotonic, there's nothing to be doubted.

 

[JonF]

How about you check post 32.

This is not the definition of the limit of a function, and your explanation was far from formal, and didn’t really what exactly it means to take to limits keeping “equal change”. Did you go over the delta epsilon definition of a limit in your class?

 

[Office_Shredder]

That is it.

Arguments seem to be revolving around whether dy/dx or dy/du can be split up into a fraction. How about we take a look at: f'(x)=lim h--> 0 (f(x+h)-f(x))/h. How did we derive it? We used small values of y and x i.e. dy and dx. Remember that all the equal signs in my proof are the "aprox." signs. The limit in the end ties it all up to a differential. But the limit thing is neccessiarily neccessary...

Yes, someone said'hey, if we use this as a definition, we get the slope for a bunch of nice looking curves. So let's extend this, and define it to be the slope for all curves'

Why? Because for all curves, that includes curves that aren't nice. So yes, you can use this intuitive sense of derivative, and if your curve behaves welll, expect to be right. But the point of the definition is that it's the only way to 100% accurately describe the derivative of a function, no matter which function you're using.

Technically, werg has yet to prove that he's even covered all the cases (and as we can see by his initially missing the case where it hits zero infinite times in any interval, it's not obvious that all the cases are covered).

However, by definition using the definition of a derivative hits all the cases

 

[Werg22]

Notice how these [insert you know what here] keep on attacking my proof without actually putting anything forth. Office_Shredder, you are so laughable. As soon as Mathwonk stepped in, you changed your tune. It's pretty obvious now that you really don't know what you are talking about.