TITLE: Show that $x(t) = \frac{1}{1+t}$ is a root of $x(t) = 1 - \int_{0}^{t}[x(u)]^2 du $
QUESTION [1 upvotes]: Metric Spaces: Iteration and Application, Exercise 8
Let $X$ be the set of continuous real functions defined on the open interval $(-1, 1)$.  for $x \in X$, let $ f(x) \in X$ be defined by:
$$ (f(x))(t) = 1 - \int_{0}^{t}[x(u)]^2 du $$
Let $x_1(t) = 0$, $x_2 = f(x_1)$, $x_3 = f(x_2)$, etc.
First we are asked to evaluate $x_4$ and $x_5$, and note how they seem be be approaching the function $x \in X$ given by:
$$ x(t) = 1 - t + t^2 - t^3 + \dots = \frac{1}{1+t}$$
Then we are asked to confirm by direct integration that $x(t) = \frac{1}{1+t}$ is a root of the equation:
$$ x(t) = 1 - \int_{0}^{t}[x(u)]^2 du $$
My Question
I don't think I correctly understand what I'm expected to do when asked to confirm the root.  If I substitute $x(t) = \frac{1}{1+t}$ into the equation, I would expect to get 0 when I evaluate
$$ 1 - \int_{0}^{t}\Big[\frac{1}{1+u}\Big]^2 du $$
However, I'm actually calculating $1 - \frac{t}{1+t}$.
What am I doing wrong here?

REPLY [1 votes]: The LHS is $x(t)$, not $0$. So when you substitute for the function $x$, you in fact get $\frac{1}{1 + t} = 1 - \frac t{1 + t}$, which is correct.
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