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## approximations converging [closed]

what does successive approximations converge mean? can someone give me a headstart/explain the problem in the link exercise 3.1 http://people.math.gatech.edu/~bonetto/teaching/6307-fall09/13.pdf

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## closed as off topic by Andres Caicedo, Will Jagy, Qiaochu Yuan, Martin Brandenburg, Bruno MartelliOct 3 2011 at 7:44

It means:

With $\tilde{T}\hspace{.01 in}$ defined by $\; (\tilde{T}(x))(t) = \displaystyle\int_{\tau}^t f(s,x(s))\;ds \;$, $\;\;$ $(\tilde{T})^0$ defined by $\;(\tilde{T})^0(x) = x\;$,

and for all non-negative integers $n$, $(\tilde{T})^{n+1}$ defined by $\;(\tilde{T})^{n+1}(x) = \tilde{T}\hspace{.01 in}((\tilde{T})^n(x))\;$, $\;\;$

$\displaystyle\lim_{n\to \infty} \; (\tilde{T})^n(x) \;\;$ exists.

The way to do Exercise 3.1 is to use Theorem 3.1's result that $\tilde{T}\hspace{.01 in}$ is a contraction, and follow this proof.

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