Daily Archives: September 8, 2019

Finite Difference Approximations of Derivatives

We will derive the finite difference approximations for y'(x) and y''(x).

Let

y'_i, y''_i denotes y'(x_i) and y''(x_i) respectively

and

h = x_{i+1}-x_i = x_i -x_{i-1} > 0.

We prove that

[1] y_i' \approx \frac{y_{i+1} - y_i}{h}

Let x=x_{i+1}, a=x_i, Taylor series

y(x) = y(a) + y'(a) (x-a) +  O((x-a)^2)

is

y(x_{i+1}) = y(x_i) + y'(x_i)(x_{i+1}-x_i)+O((x_{i+1} - x_i)^2)

i.e.,

y_{i+1} = y_i + y'_i h + O(h^2)

Hence,

\frac{y_{i+1} -y_i}{h} = y'_i + O(h)


[2] y_i' \approx \frac{y_{i+1}-y_{i-1}}{2h}

Let x = x_{i+1}, a = x_i, Taylor series

y(x) = y(a) + y'(a) (x-a) + \frac{y''(a)}{2!} (x-a)^2 + O((x-a)^3)

becomes

y(x_{i+1}) = y(x_i) + y'(x_i)(x_{i+1}-x_i) + \frac{y''_i}{2!} (x_{i+1}-x_i)^2 + O((x_{i+1}-x_i)^3)

It follows that

y_{i+1}= y_i + y'_i h + \frac{y''_i}{2!} h^2 + O(h^3)\quad\quad\quad(1)

Similarly, let x=x_{i-1}, a = x_i,

y_{i-1}= y_i + y'_i (x_{i-1}-x_1) + \frac{y''_i}{2!} (x_{i-1}-x_i)^2 + O((x_{x-1}-x_i)^3

Since x_{i-1}-x_i = -(x_i - x_{i-1}) = -h, we have

y_{i-1} = y_i - y'_i h + \frac{y''_i}{2!} + O(h^3)\quad\quad\quad(2)

(1)-(2) \implies

y_{i+1}-y_{i-1} = 2y'_i h + O(h^3)

Therefore,

\frac{y_{i+1}-y_{i-1}}{2h} = y'_i + O(h^2)


[3] y_i'' \approx \frac{y_{i+1}-2y_i+y_{i-1}}{h^2}

Let x = x_{i+1}, a=x_i, Taylor series

y(x) = y(a) + y'(a) (x-a) + \frac{y''(a)}{2!} (x-a)^2 + \frac{y'''(a)}{3!} (x-a)^3 + O((x-a)^4)

becomes

y(x_{i+1}) = y(x_i) + y'(x_i) (x_{i+1}-x_i) + \frac{y''(x_i)}{2!} (x_{i+1} - x_i)^2

+ \frac{y'''(x_i)}{3!}(x_{i+1} - x_i)^3 + O((x_{i+1}-x_{i})^4).

That is,

y_{i+1} = y_i + y_i' h + \frac{y_i''}{2!} h^2 +  \frac{y_i'''}{3!} h^3  + O(h^4)\quad\quad\quad(3)

Similarly, let x = x_{i-1}, a=x_i, we have

y(x_{i-1}) = y(x_i) + y'(x_i) (x_{i-1}-x_i) + \frac{y''(x_i)}{2!} (x_{i-1} - x_i)^2

+ \frac{y'''(x_i)}{3!}(x_{i-1} - x_i)^3 + O((x_{i-1}-x_{i})^4).

i.e.,

y_{i-1} = y_i - y_i' h + \frac{y_i''}{2!} h^2 - \frac{y_i'''}{3!} h^3  + O(h^4)\quad\quad\quad(4)

(3) + (4) \implies

y_{i+1}+y_{i-1} = 2y_i +y_i'' h^2 + O(h^4).

Therefore,

\frac{y_{i+1}-2y_i+y_{i-1}}{h^2} = y_i'' + O(h^2)

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