Finding limit of infinite term

phasair
Messages
1
Reaction score
0

Homework Statement



The problem is not homework, just something that has been bothering me. It's in picture form, in the attachment.
I'm an undergraduate in math, and this type of convergence is a bit unknown to me. What I've seen so far are normal sequences and series, and power series.
This however, is different, because there is no 'first' term, no 'beginning', so I don't really know how to approach this.

Homework Equations



The Attempt at a Solution



The equation I've found that describes this term is:

f(x)^2 = 1 + (x^2)f(x+1)

In this case, we would be looking for f(2). However, that equation doesn't contain enough information to solve it, because there are two free variables.

Any help would be appreciated.
 

Attachments

  • 1345785316964.png
    1345785316964.png
    5 KB · Views: 545
Physics news on Phys.org
You found a functional equation. You only need one particular value of the function. For example, what is f(0)?
Note: You have an identity f(x)=f(-x) here. Use it.
 
Last edited:
Actually, this can be expressed as a recursion, indexed by a single term.Then

you can talk about convergence of the sequence:

You define a base term, and then every other term can be defined in terms of the previous one(s).

a1:=1

a2 is a function of a1.

Now, if you can find general properties of the sequence, you can ( when possible) determine its limit.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top