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

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

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).

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]

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.

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....

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.

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

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

ack, linear algebra 101

 

Stick to calculus for now

Number theory?

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

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

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

Btw, Calculus help is now UP, yippee

 

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

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.

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

How is group theory related to number thoery?

I think group theory is a part of number theory

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.