45. Taylor and Maclaurin Series

Dated: 30-10-2024

One of the early applications of calculus was to approximate numerical values of functions¹ such as \(\sin (x)\) and \(\ln (x)\) etc.
To do this, we would approximate the function¹ using polynomials² and then later approximating the numerical values from that.

Problem

Given a function¹ \(f(x)\) and a point \(a\) over the x axis, find a polynomial² of a specified degree that best approximates the \(f(x)\) over the vicinity of the point \(a\).

Solution

Imagine we have a polynomial²

\[P(x) = c_0 + c_1 \cdot x + c_2 \cdot x^2 + \ldots + c_n \cdot x^n\]

The polynomial² \(P(x)\) has \(n + 1\) coefficients. Therefore, it is reasonable to think that we can impose \(n + 1\) conditions over it.
Assume that our point of interest is \(a = 0\) then to have a high degree of resemblance between \(f(x)\) and \(P(x)\), we find the coefficients of \(P(x)\) such that

\[f^{\prime}(0) = P^{\prime}(0)\]

\[f^{n}(0) = P^{n}(0)\]

Once we start finding some of these derivatives,³ we will get a pattern that is

\[f^n(0) = P^n(0) = n!c_n\]

So the general coefficient \(c_n\) for \(P(0)\) will be

\[c_n = \frac{f^n(0)}{n!}\]

Maclaurin Series

Plugging the \(c_n\) term into our polynomial,² we get

\[P_{n}(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{f'''(0)}{3!}x^{3}+\ldots+\frac{f^{(n)}(0)}{n!}x^{n}\]

This is the case at \(x = 0\)

Taylor Series

If we make the case for \(x = a\) then maclaurin series becomes

\[P_{n}(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+\frac{f'''(a)}{3!}(x-a)^{3}+\ldots+\frac{f^{(n)}(a)}{n!}(x-a)^{n}\]

and if \(f(x)\) has derivatives of all orders of \(a\) then

\[\sum_{i=0}^{\infty}\frac{f^{(i)}(a)}{i!}(x-a)^{i}=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+\ldots+\frac{f^{(i)}(a)}{i!}(x-a)^{i}+\ldots\]

45. Taylor and Maclaurin Series

Problem

Solution

Maclaurin Series

Taylor Series

References