# Inner Product

## Main Question or Discussion Point

I am working my lonely way through Spivack's "Calculus on Manifolds." (not a registered student anywhere, alas).

On p. 23 is a set of problems involving the inner product. I believe I've got it up to d), which asks for a function f:R->R s.t. f is differentiable but |f| is not differentiable.

I believe that |f| is not differentiable for f(x)=x. (if that's wrong I have to start the book all over).

I'm having trouble seeing how to reach this conclusion from consideration of the inner product. |f|**2 = <f,f>, and for g= <f,f>, g' = <x,1> + <1,x>,
F = g**(1/2) = |f|,

and if this is ok so far, I would expect something to be wrong with the next calculation:

F' =(1/2)g**(-1/2)(g') = (1/2)(x**2)**(-1/2) (2x) = 1.

Ken C

HallsofIvy
Homework Helper
I am working my lonely way through Spivack's "Calculus on Manifolds." (not a registered student anywhere, alas).

On p. 23 is a set of problems involving the inner product. I believe I've got it up to d), which asks for a function f:R->R s.t. f is differentiable but |f| is not differentiable.

I believe that |f| is not differentiable for f(x)=x. (if that's wrong I have to start the book all over).

I'm having trouble seeing how to reach this conclusion from consideration of the inner product. |f|**2 = <f,f>, and for g= <f,f>, g' = <x,1> + <1,x>,
F = g**(1/2) = |f|,

and if this is ok so far, I would expect something to be wrong with the next calculation:

F' =(1/2)g**(-1/2)(g') = (1/2)(x**2)**(-1/2) (2x) = 1.

Ken C
You are correct that f(x)= x is such that |f|= |x| is not differentiable at x= 0.

You define g= <f, f>= <x, x>. Now x, here, is a vector not a number so g'= <x, 1>+ <1, x> is not right. If you are talking about 2 dimensions, then f= <x, y>, g= x2+ y2 so g'= 2x+ 2y a number, not a vector. You seem to be confusing vector values with numerical values.

If $f(\vec{x})= <x, y>$, then $|f(\vec{x})|= \sqrt{x^2+ y^2}$. Also, the derivative of a real function of two variables is basically the gradient vector (strictly speaking, it is the linear transformation, L(v), that corresponds to taking the dot product of the gradient vector with the vector v). Here that is
$$f'(\vec{x})= \nabla f(x,y)= (x^2+ y^2)^{-1/2}<x, y>= \frac{1}{\sqrt{x^2+ y^2}}<x, y>$$
and, because of that fraction, that does not exist at (0,0).

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tiny-tim
Homework Helper
Hi Ken! (have a square-root: √ and a square: ² )
(1/2)(x**2)**(-1/2) (2x) = 1.
No, 2x/2√(x²) = x/|x| ≠ 1. thank you both. If I may, follow up questions:

what rules forbid the usual algebra of exponentials (1/2)(x**2)**(-1/2)(2x) = x(x**2)**(-1/2) = x * x**[2*(-1/2] = x ** (1+ -1) ?

Is this my apparent confusion between one D vector and a number?

tiny-tim
Homework Helper
what rules forbid …
The rule that xm/n is not uniquely defined …

for any integer n, it has n different values. Are you asking why isn't $$\sqrt{x^2} = x$$ but instead |x|?

If that's the case note that $$\sqrt{blah} \geq 0$$ but is x always greater than or equal to zero?

Thank you! So I can't in fact multiply exponents like (x**2)**(1/2) = x without knowing in advance that x=>0. It has been many years since I took algebra.

OK, then, it seems like my initial application of the inner product to showing that |x| is not differentiable is OK, until I misused the algebra of exponents, which, correctly applied, yields F' = x/|x| which is not a unique linear function.