Banach Fixed Point and Differential Equations

ChemEng1
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Homework Statement


Find the value of x, correct to three decimal places for which: \int^{x}_{0}\frac{t^{2}}{1+t^{2}}dt=\frac{1}{2}.

Homework Equations


Banach's Fixed Point Theorem
Picard's Theorem?

The Attempt at a Solution


I'm not sure where to start with this type of problem.

From other BFPT problems, I will need to show a contraction mapping into itself.

Any pointers would be greatly appreciated.
 
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Banach's fixed point theorem and Picard's method show that this problem has a unique solution. Actually finding the solution doesn't require that.

You could, for example, actually integrate that:
\int_0^x \frac{t^2}{1+ t^2}dt= \int_0^x 1- \frac{1}{t^2+ 1}dt[/itex] <br /> is pretty easy to integrate.
 
\int^{x}_{0}\frac{t^{2}}{1+t^{2}}dt=\int^{x}_{0}(1-\frac{1}{1+t^{2}})dt

Let t=tanθ. Then dt=sec2θdθ.

\int^{x}_{0}(1-\frac{1}{1+t^{2}})dt=\int^{x}_{0}dt-\int^{x}_{0}\frac{1}{sec^{2}\vartheta}sec2θdθ=\int^{x}_{0}dt-\int^{x}_{0}dθ=t-arctan(t) from 0 to x=x-arctan(x).

x-arctan(x)=.5 at x=1.4743.

Are you familiar with any approaches that use BFPT?
 
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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...
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