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# Directional Derivatives of Homogenous Functions

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

Note by Ethan Robinett
2 years ago

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Wow, Michael Spivak's book is still around? · 2 years ago

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) Staff · 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. · 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. · 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. · 2 years ago

Yeah, I figured after reading ahead a bit that those chapters were what everyone was talking about. Great book though · 2 years ago

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