Nice Calculus problem involving Maclaurin series

[bloodhound]

Taken from my calculus exam

 

The function f satisfies

[math]f(2x)=xf(x)+1[/math]

 

1)Show that [math]f(0)=1[/math]

2)Show using induction that

[math]2^{n}f^{n}(2x)=nf^{(n-1)}(x)+xf^{n}(x) \forall x \in R[/math]

3)Hence, derive the first three terms of the Maclaurin series of [math]f[/math]

 

 

I managed to do it, but because of 1) and 2) if 3) was given by itself I would think i would be completely lost for some time.

 

edit: [math]f^n[/math] denotes the nth derivative of f

[Dave]

Done (1) - pretty obvious. Working on 2, haven't quite got there yet, but I think I know how to do it

[jordan]

I have absolutly no idea what is going on there, but I was looking over it anyways. Just wondering, what do the last four symbols in number 2 mean (starting with what appears to be an up-side down A).

[Dave]

Read literally the upside-down A means "for all" - the entire thing means that that condition (the f(x) bit) is true for all x in the set of real numbers.

 

btw, to generate a proper real numbers symbol, use \mathbb:

 

[math]\forall x\in\mathbb{R}[/math]

[Dave]

Hmm. I did it, but I'm not sure it's entirely true.

 

We know that [math]f(2x) = xf(x) + 1[/math]. Now say [math]g(x) = xf(x) + 1 - f(2x)[/math]. Then we have that [math]g'(x) = f(x) + xf'(x) - 2f(2x)[/math] by the chain rule (effectively this is implicit differentiation).

 

By differentiating more and more times, it's pretty obvious that you get [math]g^{(n)}(x) = nf^{(n-1)}(x) + xf^{(n)}(x) - 2^n f^{(n)}(x)[/math] as required.

[admiral_ju00]

Dave, I think you should start a Calc 101 section here, and kind of like instead of asking question / recieving answer thing, maybe you guys can start others (like myself) on our way to actually understand Calculus? (Kind of like a teaching/tutoring thing)

 

Or maybe not....

[Dave]

Sounds like a good idea; I'm sure I can draw up a "syllabus" of types

 

Maybe if it kicks off properly I can get blike to create a sub-forum for it.

[jordan]

A daily calc lesson at 9:00 every morning taught by dave. Nice.

[bloodhound]

yes, and if it takes off, we can do others like Linear Mathematics 101, or Mathematical Analysis 101. etc

[Dave]

ack, linear algebra 101

 

Stick to calculus for now

[jordan]

Number theory?

[bloodhound]

I loved Linear algebra. all the vector spaces and subspaces. mind boggling.

I havent done much number theory, except introductory group theory.

[Dave]

Nope, I haven't done any number theory. Maybe next year after I've taken the module

[Dave]

Btw, Calculus help is now UP, yippee

 

(and this is my 2,000th post, woot)

[jordan]

Ok. No need for number theory then. I was interested because I read a book not to long ago that said it would be classified as number theory but never described what number theory was. I liked the book so I assumed I would have to like learning number theory as well.

[bloodhound]

I could put stuff up on group theory if u want.

[jordan]

How is group theory related to number thoery?

[bloodhound]

I think group theory is a part of number theory

[Dave]

Group theory is very interesting because you can use it to define basic addition and multiplication in a rigourous sense; indeed, if you start with set theory you can basically build up the natural numbers, the rationals, reals, etc and all the operations that can be done with them.