- #1
brydustin
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There is already an article on physics forums that kind of addresses my issue:
https://www.physicsforums.com/showthread.php?t=107516 and I'm not really satisfied with the wikipedia article.
I am generally confused on what the derivative should be. I'm familiar with Jacobian matrices but am confused by Spivak's definition of a derivative because there is a norm in R^m for the numerator of the limit definition and a norm from R^n in the denominator, so it seems as if the derivative must return a scalar -- unless I am mistaken about the meaning of the norm in this case.
Spivak's definition:
lim h -->0 |f(a+h) - f(a) - λ(h)|/|h| = 0
I'm not really sure if the zeros are vectors or scalars (for example the norm equal to zero?)
On a side note, I'm trying to apply this to solve the problem:
Prove that f:R->R^2 is differentiable at a in R iff f_1 and f_2 are, and then it will follow that
f ' (a) = [ f_1 ' (a) , f_2 ' (a) ]^t
The result seems obvious but having a clear understanding of the definitions would help me prove something. Any help appreciated, thanks
https://www.physicsforums.com/showthread.php?t=107516 and I'm not really satisfied with the wikipedia article.
I am generally confused on what the derivative should be. I'm familiar with Jacobian matrices but am confused by Spivak's definition of a derivative because there is a norm in R^m for the numerator of the limit definition and a norm from R^n in the denominator, so it seems as if the derivative must return a scalar -- unless I am mistaken about the meaning of the norm in this case.
Spivak's definition:
lim h -->0 |f(a+h) - f(a) - λ(h)|/|h| = 0
I'm not really sure if the zeros are vectors or scalars (for example the norm equal to zero?)
On a side note, I'm trying to apply this to solve the problem:
Prove that f:R->R^2 is differentiable at a in R iff f_1 and f_2 are, and then it will follow that
f ' (a) = [ f_1 ' (a) , f_2 ' (a) ]^t
The result seems obvious but having a clear understanding of the definitions would help me prove something. Any help appreciated, thanks