Jump to content

Nice Calculus problem involving Maclaurin series


Recommended Posts

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

Link to comment
Share on other sites

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).

Link to comment
Share on other sites

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]

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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....

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.