Taylor Series and Picard Method

$y = mx + b$ As we know, **tangent lines are decent approximations** for small intervals. However, they only work for a limited number of points. To approximate larger functions, we use **Taylor series**: $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n$

Problem Example:

• Compute the **Taylor polynomial of degree 2** for the following initial value problem as well as Picard's Method:
  • $\frac{dy}{dx} = x - y$ $y(0) = 1$

Taylor Polynomial Solution Steps:

1. The differential equation is: $\frac{dy}{dx} = x - y$
2. Compute the second derivative for the Taylor polynomial: $\frac{d^2y}{dx^2} = 1 - y' + (-y') * (x - y) = 1 - y' + y * y' - x * y'$
3. Using the initial condition $y(0) = 1$: $\frac{dy}{dx}(0) = 0 - 1 = -1$ $\frac{d^2y}{dx^2}(0) = 1 - (-1) + (1)*(-1) = 1$
4. Approximate the solution using the degree 2 Taylor polynomial: $y(x) \approx 1 + (-1)*(x - 0) + 1/2*(x - 0)^2 = 1 - x + 0.5 x^2$

Picard's Iteration Method:

$y_0(x) = y_0, \\ y_{n+1}(x) = y_0 + \int_{x_0}^{x} f(t, y_n(t)) \, dt$

This iterative method approximates the solution of $dy/dx = f(x,y)$ starting at the initial point $(x_0, y_0)$. Each iteration uses the previous function inside the integral to get closer to the exact solution.

Checking Accuracy:

We can check how good this approximation is by using **Euler's method** with a small step size and comparing the results to the exact solution and the Taylor polynomial. Below is a graph showing:

• Blue: Exact solution $y = x - 1 + 2 e^{-x}$
• Green: Euler's method approximation
• Orange: Taylor polynomial approximation

• Let for example a harder diff eq such as $y'' = 0.1*y' -y +sin(2t)$ and since we know $y(0) = 0$ and $y'(0) = 1$ then we do a simple substitution $v = y'$ and logically $v' = y''$
  • Now we do a substitution $v' = -0.1*v + y + sin(2t)$ and we arrive at two equations: $\begin{cases} y'(t) = v(t) \\ v'(t) = -0.1\,v(t) - y(t) + \sin(2t) \end{cases}$ $y(0) = 0, \quad v(0) = 1$
• Control which iteration of pickard is displayed