Vectors as Inputs

Metronome

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Jun 12, 2018
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In this video (https://youtu.be/GzBGa1BCXgs?t=267), it is said that the inner function in a composition must be a scalar and can't be a vector. What does this mean? It is common in multivariable calculus to consider functions with multiple inputs, which can be viewed geometrically as a single point or vector. Here (https://youtu.be/qZlBjnC3iro?t=162) is an example of a vector being plugged into a function, in fact. In what sense does plugging a vector into a function not work?
 
I disagree with quite a lot of what he is saying. Whether you have a vector as the "independent variable" of a vector valued function or not depends entirely upon how that vector valued function is defined! There is nothing wrong with defining one vector as a function of another. He also refers to \(\displaystyle \frac{dv}{\alpha}= \frac{dv}{d\f}\frac{df}{dx}\) as "a terrible way of writing the chain rule" on the grounds that you cannot differentiate a function with respect to another function, only with respect to a "variable". I strongly suspect that this person is a physicist, not a mathematician! Any mathematician would agree that a function can a variable and that, indeed, when we talk about the "variable x", we are talking about the function f(x)= x.
 
I disagree with quite a lot of what he is saying. Whether you have a vector as the "independent variable" of a vector valued function or not depends entirely upon how that vector valued function is defined! There is nothing wrong with defining one vector as a function of another. He also refers to \(\displaystyle \frac{dv}{\alpha}= \frac{dv}{d\f}\frac{df}{dx}\) as "a terrible way of writing the chain rule" on the grounds that you cannot differentiate a function with respect to another function, only with respect to a "variable". I strongly suspect that this person is a physicist, not a mathematician! Any mathematician would agree that a function can a variable and that, indeed, when we talk about the "variable x", we are talking about the function f(x)= x.

Thanks for adding this perspective. I thought some of what he said sounded too restrictive. By the way, isn't your final claim circular, since the function's input is also a variable x?
 
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