# Infinite products

1. Dec 12, 2006

### the dude man

How do you differentiate one? Is it possible? Any formulas?

2. Dec 12, 2006

### AKG

Apply d/dx to the expression.

3. Dec 12, 2006

### hypermorphism

One differentiates functions, not numbers. Do you mean you want to know the derivative of the identity function? Over what domain? Or do you want to know the derivative of the function that always returns 1?

4. Dec 12, 2006

### the dude man

y'(x) = y(x)*M

M=
----
\ f' / f
/
----
Does that work?

Last edited: Dec 12, 2006
5. Dec 12, 2006

### AKG

No, that doesn't work. Change one of the f's to a g, then it should work.

6. Dec 21, 2006

### the dude man

I cant figure this out can someone refer me to proof cause i cant find one.

7. Dec 22, 2006

### the dude man

The derivative of a function in the form of an infinite product.
Anything would be nice.

8. Dec 24, 2006

### the dude man

Bump. .

9. Dec 24, 2006

### matt grime

Please don't bump. You asked how to differentiate an infinite product, then posted an ascii art diagram that indicates an infinite sum, not product. Which is it? If you take time to learn how to post latex here it will help you.

10. Dec 25, 2006

### 3.14159

equation for derivative of n product

A similar question came to me while finding the derivative of a function equal to the product of three differentiable functions here is my generalization to a function equal to an infinite product:

Let $$g_{n} = \prod^n_{k=1}f_{k}$$ and assume that $$f'_n$$ exists. Then by the product rule of differentiation and some clever factoring
$$g'_n = (\prod^n_{k=1}f_{k})\sum^n_{j=1}\frac{f'_j}{f_j}$$
which is even more simply written as
$$g'_{n} = g_{n}\sum^n_{j=1}\frac{f'_j}{f_j}$$.

To see how I get to this generalization I will work out the derivative for
$$g_{3} = \prod^3_{k=1}f_{k}=f_1f_2f_3$$.
First,
$$g'_{3}= f_2f_3f'_1 + f_1(f_2f_3)'= f_2f_3f'_1 + f_1(f_2f'_3+f_3f'_2)= f_2f_3f'_1+f_1f_2f'_3+f_1f_3f'_2$$
Then noticing the relationship between this form and the original equation,
$$g'_3=\frac{g_3}{f_1}f'_1+\frac{g_3}{f_2}f'_2+\frac{g_3}{f_3}f'_3= g_3(\frac{f'_1}{f_1}+\frac{f'_2}{f_2}+\frac{f'_3}{f_3})$$
which is what the general function predicted,
$$g'_3=g_3\sum^3_{j=1}\frac{f'_j}{f_j}$$

I know that this statement is true for specific values of n, but I’m not sure what features this function has when g is an infinite product of differentiable functions. I hope that this helps with the question at hand.

Last edited by a moderator: Dec 25, 2006