(I use the fx notation style here 'cause it's easier than ∂f/∂x.)

My calculus book takes it as a given--or gives it as a given, rather--without any sort of proof even in the appendix.

The limit definition of a derivative, has, however, provided me with almost a proof:

fxy = limΔy→0( (1/Δy)[fx(x,y+Δy,z) - fx(x,y,z)] ) =

limΔy→0{ (1/Δy)[ limΔx→0( (1/Δx)[ f(x+Δx,y+Δy,z) - f(x,y+Δy,z) ] ) - limΔx→0( (1/Δx)[ f(x+Δx,y,z) - f(x,y,z) ] ) ] } =

limΔy→0{ (1/Δy)[ limΔx→0( (1/Δx)[ f(x+Δx,y+Δy,z) - f(x,y+Δy,z) - f(x+Δx,y,z) + f(x,y,z) ] ) ] },

Because (1/Δy) is independent of Δx, it is constant in the eyes of the delta-x limit item; so, whether (1/Δy) is on the inside or outside of the delta-x limit doesn't matter:

(1) fxy = limΔy→0{ limΔx→0( (1/Δx)(1/Δy)[ f(x+Δx,y+Δy,z) - f(x,y+Δy,z) - f(x+Δx,y,z) + f(x,y,z) ] ) }

By identical reasoning,

(2) fyx = limΔx→0{ limΔy→0( (1/Δx)(1/Δy)[ f(x+Δx,y+Δy,z) - f(x,y+Δy,z) - f(x+Δx,y,z) + f(x,y,z) ] ) },

which is virtually identical to (1), and if it can be shown to be true that

limΔx→0{ limΔy→0( [an argument] ) } = limΔy→0{ limΔx→0( [an argument] ) }

for any and all arguments for which an exception like that annoying thing where they or their limits don't exist isn't present, then fxy must equal fxy

It seems intuitively reasonable that if you can cram the "limit" parts up against each other then you can switch the two since nothing in either of them is dependent on the changing variable in the other (since Δx, Δy, and zero are all independent of each other).  As yet, though, I'm not comfortable--I need to know/see why the result of compositions of limits must be the same regardless of order (for limits with independent sub-arguments).

This is going to involve (ε, δ)-definitions, isn't it?

Also, where does the "lim(x,y)→(C,D)" form come from? Those seem like they may be relevant to my predicament.

Views: 591

Replies to This Discussion

The above proof can't work, because f_xy is not always equal to f_yx. You need an additional assumption. For example, if f_xy and f_yx are both continuous functions in an open set, then f_xy=f_yx is true on the set (weaker assumptions are possible).

The usual proof considers the limit as both h and k go to zero of f(a+h,b+h) -f(a,b+h)-f(a+h,b)+f(a,b), and shows that this limit is equal to both f_xy(a,b) and f_yx(a,b). It uses the assumption various partial derivatives are continuous, and the mean value theorem.

Does someone have a link with instructions on how to include subscripts, deltas, and other nice math notation in one's reply? I could write out a proof of f_xy=f_yx (assuming continuity of the derivatives) if I knew how to do that.

RSS

Support Atheist Nexus

Supporting Membership

Nexus on Social Media:

Latest Activity

Qiana-Maieev replied to James Yount's discussion Unique and Interesting Instruments in the group Music Lovers
1 hour ago
Qiana-Maieev commented on Loren Miller's blog post Is Nothing Sacred?
1 hour ago
Qiana-Maieev liked Loren Miller's blog post Is Nothing Sacred?
1 hour ago
Joan Denoo commented on Ruth Anthony-Gardner's photo
1 hour ago
Joan Denoo liked Ruth Anthony-Gardner's photo
1 hour ago
Loren Miller posted a status
"“Sometimes people don't want to hear the truth because they don't want their illusions destroyed.” ― Friedrich Wilhelm Nietzsche"
2 hours ago
Loren Miller commented on Loren Miller's blog post Is Nothing Sacred?
2 hours ago
Bertold Brautigan commented on Loren Miller's blog post Is Nothing Sacred?
2 hours ago

© 2016   Atheist Nexus. All rights reserved. Admin: Richard Haynes.   Powered by

Badges  |  Report an Issue  |  Terms of Service