I recently bought the book "Calculus on Manifolds" by Michael Spivak and highly recommend it to anyone wishing to learn calculus in a much more general setting. I have not finished the book but I find what I have read to be presented in a very concise, sophisticated manner. The following proof is my answer to one of the exercises in the book, with some notation changed to a more classical form, so that those unfamiliar with the text may understand the proof.

To prove: Suppose \(f: \mathbb{R}^n \rightarrow \mathbb{R} \) is homogenous of degree \(m\) and differentiable. Then:

\(\displaystyle \sum_{i=1}^{n} x_i \frac{\partial f}{\partial x_i}(\vec{x}) = mf(\vec{x})\) \(\forall \vec{x} \in \mathbb{R}^n\)

Proof:

By definition, \(f\) is homogenous of degree \(m\) necessarily implies that: \(f(t\vec{x})=t^mf(\vec{x})\) \(\forall t \in \mathbb{R}\)

Also note that:

\(\displaystyle \sum_{i=1}^{n} x_i \frac{\partial f}{\partial x_i}(\vec{x}) = \nabla f(\vec{x}) \cdot \vec{x}= D_{\vec{x}}f(\vec{x})\)

Where \(D_{\vec{x}}f(\vec{x})\) is the directional derivative in the direction of \(\vec{x}\), defined by:

\(D_{\vec{x}}f(\vec{x}) = \displaystyle \lim_{t \to 0} \frac{f(\vec{x} +t\vec{x})-f(\vec{x})}{t}\)

Then we have:

\(D_{\vec{x}}f(\vec{x}) =\displaystyle \lim_{t \to 0} \frac{f(\vec{x} (t+1))-f(\vec{x})}{t}=\displaystyle \lim_{t \to 0} \frac{(t+1)^mf(\vec{x})-f(\vec{x})}{t}\)

Which can be written as follows using the binomial theorem:

\(D_{\vec{x}}f(\vec{x}) =\displaystyle \lim_{t \to 0} \frac{1}{t} f(\vec{x}) (-1 + \displaystyle \sum_{i=0}^{m} \dbinom{m}{i} t^i)\)

\(\dbinom{m}{0}t^0 = 1\) hence we may write:

\(D_{\vec{x}}f(\vec{x}) =\displaystyle \lim_{t \to 0} \frac{1}{t} f(\vec{x}) ( \displaystyle \sum_{i=1}^{m} \dbinom{m}{i} t^i)\)

And distributing \(\frac{1}{t}\) we have:

\(D_{\vec{x}}f(\vec{x}) =\displaystyle \lim_{t \to 0} f(\vec{x}) ( \displaystyle \sum_{i=1}^{m} \dbinom{m}{i} t^{i-1})\)

Now if we take the limit, notice that \(\forall i \geq 2\), \(\displaystyle \lim_{t \to 0}\dbinom{m}{i} t^{i-1} =0\).

Then we have:

\(D_{\vec{x}}f(\vec{x}) =f(\vec{x})(\dbinom{m}{1})=mf(\vec{x})\)

Recalling that \(D_{\vec{x}}f(\vec{x}) =\displaystyle \sum_{i=1}^{n} x_i \frac{\partial f}{\partial x_i}(\vec{x}) \) immediately leads to:

\(\displaystyle \sum_{i=1}^{n} x_i \frac{\partial f}{\partial x_i}(\vec{x}) =mf(\vec{x})\)

Which completes the proof.

QED

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## Comments

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TopNewestWow, Michael Spivak's book is still around?

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It's not just around, but it is also the textbook that I used in my Honors Analysis class. It is really well written, clear and concise.

(Though, the class was 10 years ago, so that doesn't preclude the possibility that it is no longer around. LOL)

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That book came out in 1965. I had that book in college in 1967, a skinny, dirty paperback. Geez, I never knew that book would become such a mainstay, still going strong after FIFTY years! I still have that book somewhere, if I can only find it.

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