Total Derivatives of fxns from R^n to R^m

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In summary, Spivak's definition of a derivative is a linear transformation that can be represented as a vector or matrix depending on the values of "n" and "m".
  • #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
 
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  • #2
The derivative is the linear transformation λ in the numerator. It is not the fraction, which is indeed a scalar that approaches 0. In the definition above, you may read it as "The derivative of f at a is the unique linear transformation λ such that the limit as h approaches the 0 vector of ||f(a + h) - f(a) - λ(h)||/||h|| is 0."
 
  • #3
that makes a lot more sense... so it can be a vector value or a matrix depending on the values of "n" and "m"? So if n=1 and m>1 its a row vector? And n>1 and m=1 then its column vector? n=m=1 then its a scalar and it n>1,m>1 then its a matrix?
 
  • #4
Strictly speaking, even in the "Calculus I" sense of a function from R1 to R1 the "derivative" is a linear transformation- it is "[itex]y= mx[/itex]" where m is the "usual" derivative evaluated at that point. Of course, we can represent that transformation by the coefficient df/dx= m.

For a function from R1 to R3, the "vector valued function of a single real variable", f(t)=<f(t), g(t), h(t)>, the derivative is the linear transformation that maps x to <(df/dt)x, (dg/dt)x, (dh/dt)x> which is just the scalar product x<df/dt, dg/dt, dh/dt>. We typically "represent" that linear transformation as the vector <df/dt, dg/dt, dh/dt>.

For a function, f(x,y,z), from R3 to R1, the "scalar valued function of three variables", the derivative is the linear transformation that maps <a, b, c> to [itex](\partial f/\partial x)a+ (\partial f/\partial y)b+ (\partial f/\partial z)c[/itex], the dot product [itex]<\partial f/\partial x, \partial f/\partial y, \partial f/\partial z>\cdot<a, b, c>[/itex] which we can represent by [itex]\nabla f= <\partial f/\partial x, \partial f/\partial y, \partial f/\partial z>[/itex].

Finally, for a function of three variables, which returns a 3-vector, [itex]<f(x, y, z), g(x, y, z), h(x, y, z)>[/itex] is the linear transformation that can be represented by the 3 by 3 matrix
[tex]\begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} & \frac{\partial g}{\partial z} \\ \frac{\partial h}{\partial x} & \frac{\partial h}{\partial y} & \frac{\partial h}{\partial z}\end{bmatrix}[/tex]
 
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  • #5
It is a linear transformation. If you choose a basis for R^n and R^m, then you can determine the image of the basis of R^n under the transformation λ. Due to linearity, if we write e_i as the ith basis vector of R^n in some coordinate system and x^i as the ith component of a vector x in R^n, we can then write
[tex]\lambda(x) = \lambda(x^1e_1 + \cdots + x^ne_n) = x^1\lambda(e_1) + \cdots + x^n\lambda(e_n)[/tex]
The latter sum can be written as a matrix multiplication where λ(e_i) is the ith column of the matrix. However, this is relative to a coordinate system for R^n and R^m. Change those coordinate systems, and the matrix representation of λ will be different. Hence, it is usually preferable to work with λ as a linear transformation when doing theoretical work, with a matrix representation in a given coordinate system.
Halls gives some examples of the differing representations of λ in Cartesian coordinate systems with the standard basis above.
 

FAQ: Total Derivatives of fxns from R^n to R^m

What is a total derivative of a function from R^n to R^m?

A total derivative of a function from R^n to R^m is a generalization of the derivative of a single variable function to a multivariable function. It represents the rate of change of the output of the function with respect to changes in all of the input variables.

How is the total derivative calculated?

The total derivative is calculated using partial derivatives. The partial derivative of the function with respect to each input variable is calculated and then combined to form the total derivative.

What is the difference between a total derivative and a partial derivative?

A partial derivative only considers the rate of change of the output with respect to one input variable, while a total derivative considers the rate of change with respect to all input variables simultaneously.

What is the significance of total derivatives in real-world applications?

Total derivatives have many real-world applications, such as in physics, economics, and engineering. They are used to model and analyze systems with multiple variables, and to optimize functions for maximum or minimum values.

Can total derivatives be negative?

Yes, total derivatives can be negative. This indicates that the output of the function is decreasing with respect to changes in the input variables. It is important to consider the sign of the total derivative when using it to make predictions or optimize functions.

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