Solving an equation Picard's way

Introduction: Solving x^2-x-1=0

If we are trying to solve an equation like x^2-x-1=0, we could try x^2=x+1, so that x=1+\frac{1}{x}

And we can substitute the value of x again x=1+\frac{1}{1+\frac{1}{x}}

And again, and again...

x+1+\frac{1}{1+\frac{1}{1+\frac{1}{x}}}

We could, finally, now write

x=1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \ldots}}}}

Picard's way

Hence to solve x^2-x-1=0

We first write it as x=1+\frac{1}{x}

And then we transform it as x_1=1 and x_{n+1}=1+\frac{1}{x_n}

And then we get...

x_1=1; x_2=1+\frac{1}{x_1}; x_3=1+\frac{1}{x_2}; x_4=1+\frac{1}{x_3}

x_1=1; x_2=1+1/1; x_3=1+\frac{1}{1+\frac{1}{1}}; x_4=1+\frac{1}{1+\frac{1}{1+\frac{1}{1}}}

If the sequence \{x_n\}_{n\in\mathbb{N}} converges, an exercise in calculus shows that \displaystyle{\lim_{x\rightarrow \infty} x_n=x} is a solution to the problem.

In this case the sequence converges, other problems like x=1-x lead to non-convergent sequences, and thus they do not lead to solutions...

Now, an interactive demonstration

Solving a differential equation

In a similar way, lets try to solve the initial value problem

\frac{dy}{dx}=y

y(0)=1

This corresponds to y(x)-y(0)=\int_{0}^{x}y(s)ds

Solving, and substituing the value of y(0), we get:

y(x)=\int_{0}^{x}y(s)ds+1

And we can substitute again...

y(s)=\int_{0}^{x}\left(\int_{0}^{s}y(s_1)ds_1+1\right)+sds

Solving a Differential Equation in Picard's way

We can do again the same thing we did for the other equation

y_1=1; the constant function 1, which does satisfy that y(0)=1

y_{n+1}=\int_{0}^{x}y_n(s)ds+1

Now, an interactive demonstration

Notice that this, indeed converges to the power series expansion of e^x, regardless of the initial function!