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

## Comments

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TopNewestWow, Michael Spivak's book is still around? – Michael Mendrin · 2 years ago

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(Though, the class was 10 years ago, so that doesn't preclude the possibility that it is no longer around. LOL) – Calvin Lin Staff · 2 years ago

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– Michael Mendrin · 2 years ago

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.Log in to reply

– Ethan Robinett · 2 years ago

I bought the book after reading that it can be considered a "prerequisite" to Spivak's "Comprehensive Intro to Differential Geometry." If I'm correct, the book is still used in MIT's Analysis II and a few other Analysis classes at other schools. I actually got through most of Rudin's "Principles of Mathematical Analysis" before I picked up this book, and I have to say I like Spivak's general style of presentation much better than Rudin's, though the two books obviously differ on some subject matter. I also read that "calc on manifolds" is an infamously difficult book. I can only reason that this statement is referring to chapters 4 and 5, because I'm almost all the way through chapter 3 and haven't had THAT much difficulty with the material (some difficulty with some of the problems, but nowhere near what I had expected given what I read). I would be interested to hear your opinion on the difficulty of the book.Log in to reply

– Michael Mendrin · 2 years ago

I just now downloaded an online PDF copy of Spivak's book so that I could remember chapters 4 and 5, "Integration on Chains", and "Calculus of Manifold". Yeah, it was pretty tough for me back then, as it did expose me to new material, namely exterior calculus. I have a couple more books on exterior calculus from college days, and I think the other books covered it a bit more clearly for me. If you find it difficult, I don't blame you.Log in to reply

– Ethan Robinett · 2 years ago

Yeah, I figured after reading ahead a bit that those chapters were what everyone was talking about. Great book thoughLog in to reply