Taylor Polynomial

Taylor polynomial is a polynomial that approximates a function near a specific point, often denoted by a. By taking a finite number of terms from the function’s derivatives, a Taylor polynomial provides a close approximation to the actual function in a neighborhood around that point. 

The Taylor polynomial of degree n for a function f(x) centered at a point a is given by:

${P_{n}\left( x\right) =f\left( a\right) +f’\left( a\right) \left( x-a\right) +\dfrac{f”\left( a\right) }{2!}\left( x-a\right) ^{2}+\ldots +\dfrac{f^{n}\left( a\right) }{a!}\left( x-a\right) ^{n}}$ 

Here,

• fⁿ(a) represents n^th derivative of f(x) at a
• n! is the factorial of n

By including terms up to degree n, the Taylor polynomial provides a finite approximation of f(x) near a. The more terms added, the more accurately P_n(x) represents f(x) around a.

It is widely used in many fields like physics, engineering, and economics to approximate functions, such as potential energy functions, signal processing equations, and utility-cost functions.

Difference Between Taylor Polynomial and Taylor Series

While the Taylor polynomial is a finite sum, the Taylor series takes the process further by adding terms infinitely. In general, the Taylor series can provide a near-exact representation of a function in an interval around a, where the series converges.

The form of the Taylor series for f(x) at a is:

${f\left( x\right) =\sum ^{\infty }_{n=0}\dfrac{f^{n}\left( a\right) }{n!}\left( x-a\right) ^{n}}$

⇒  ${f\left( x\right) =f\left( a\right) +f’\left( a\right) \left( x-a\right) +\dfrac{f”\left( a\right) }{2!}\left( x-a\right) ^{2}+\ldots}$

If a = 0, the Taylor series is known as a Maclauurin series for f(x), given by:

${f\left( x\right) =f\left( 0\right) +f’\left( 0\right) x+\dfrac{f”\left( 0\right) }{2!}x^{2}+\ldots}$

Here,

Maclaurin polynomial (a Taylor polynomial centered at a = 0) is expressed as: ${P_{n}\left( x\right) =f\left( 0\right) +f’\left( 0\right) x+\dfrac{f”\left( 0\right) }{2!}x^{2}+\ldots \dfrac{f^{n}\left( 0\right) }{n!}x^{n}}$

However, it is important to note that not all Taylor series converges to the function they approximate, particularly if evaluated far from the center point a. A Taylor series converges to a function within a specific interval around a, known as the radius of convergence, which varies based on the function.

Taylor Polynomial of Two Variables

In addition to approximating functions of a single variable, Taylor polynomials can also be used to approximate functions of two variables, f(x, y). The Taylor polynomial of degree n for a function f(x,y) centered at a point (a,b) takes into account partial derivatives with respect to both x and y.

The general form of the Taylor polynomial of degree n for a function f(x, y) around (a, b) is given by:

${P_{n}\left( x,y\right) =f\left( a,b\right) +f_{x}\left( a,b\right) \left( x-a\right) +f_{y}\left( a,b\right) \left( y-b\right) +\dfrac{f_{x,x}\left( a,b\right) }{2!}\left( x-a\right) ^{2}+\dfrac{f_{y,y}\left( a,b\right) }{2!}\left( y-b\right) ^{2}+f_{x,y}\left( a,b\right) \left( x-a\right) \left( y-b\right) +\ldots}$

Here,

• f_x and f_y are the 1^st-order partial derivatives with respect to x and y
• f_xx, f_yy, and f_xy are the 2^nd-order partial derivatives

Higher-order terms involve mixed partial derivatives up to the n^th degree.

Approximating

Let us consider a function y = ln(x) and approximate the value of ln(1.2) using P₅(x). 

Since 1 is very near to 1.2, we first find the n^th Taylor polynomial of y = ln(x) at x = 1

Starting with derivatives of ln(x) at x = 1:

${\begin{aligned}f\left( x\right) =\ln \left( x\right) \Rightarrow f\left( 1\right) =0\\ f’\left( x\right) =\dfrac{1}{x}\Rightarrow f’\left( 1\right) =1\\ f”\left( x\right) =-\dfrac{1}{x^{2}}\Rightarrow f”\left( 1\right) =-1\end{aligned}}$

…

${f^{n}\left( x\right) =\dfrac{\left( -1\right) ^{n+1}}{x^{n}}\left( n-1\right) !\Rightarrow f^{n}\left( 1\right) =\left( -1\right) ^{n+1}\left( n-1\right) !}$

Thus, ${P_{n}\left( x\right) =f\left( a\right) +f’\left( a\right) \left( x-a\right) +\dfrac{f”\left( a\right) }{2!}\left( x-a\right) ^{2}+\ldots +\dfrac{f^{n}\left( a\right) }{a!}\left( x-a\right) ^{n}}$

⇒ P_n(x) = ${0+\left( x-1\right) -\dfrac{1}{2}\left( x-1\right) ^{2}+\ldots +\dfrac{\left( -1\right) ^{n+1}}{n}\left( x-1\right) ^{n}}$

Now, the Taylor polynomial P₅(x) for ln(x) by substituting x = 1.2 becomes:

P₅(x) = ${\left( 1.2-1\right) -\dfrac{1}{2}\left( 1.2-1\right) ^{2}+\ldots +\dfrac{\left( -1\right) ^{5+1}}{5}\left( 1.2-1\right) ^{5}}$

⇒ P₅(x) = ${0\cdot 2-\dfrac{1}{2}\left( 0\cdot 2\right) ^{2}+\ldots +\dfrac{1}{5}\left( 0\cdot 2\right) ^{5}}$ = ${\dfrac{34187}{187500}}$ ≈ 0.18233

Also, ln(1.2) ≈ 0.18232

Finding Errors

To determine the accuracy of approximation, we calculate the error bound using Taylor’s Theorem, which states:

If f(x) is a polynomial function that is differentiated n + 1 times on an open interval I containing a, then for each positive integer n and for each x ∈ I,

${f\left( x\right) =f\left( a\right) +f’\left( a\right) \left( x-a\right) +\dfrac{f”\left( a\right) }{2!}\left( x-a\right) ^{2}+\ldots +\dfrac{f^{n}\left( a\right) }{n!}\left( x-a\right) ^{n} +R_{n}\left( x\right)}$

⇒ ${f\left( x\right) =P_{n}\left( x\right) +R_{n}\left( x\right)}$, 

Here, 

P_n(x) is the Taylor polynomial of f(x) at a, and 

${R_{n}\left( x\right) =\dfrac{f^{n+1}\left( c\right) }{\left( n+1\right) !}\left( x-a\right) ^{n+1}}$ 

⇒ ${\left| R_{n}\left( x\right) \right| \leq \dfrac{M}{\left( n+1\right) !}\left| x-a\right| ^{n+1}}$

M = max|f^{n + 1}(x)|

Now, using this theorem, we can find the error bounds for approximating ln(1.2) with P₅(x).

Here, we calculate R₅(x) at x = 1.2 and a = 1 to get the error bounds

${\left| R_{5}\left( 1.2\right) \right| \leq \dfrac{M}{\left( 5+1\right) !}\left| 1.2-1\right| ^{5+1}}$

${\left| R_{5}\left( 1.2\right) \right| \leq \dfrac{M}{\left( 6\right) !}\left| 0.2\right| ^{6}}$

Calculating the value of M, we get

M = max|f^{n + 1}(x)|

f^{n + 1}(x) = ${\dfrac{\left( -1\right) ^{n}n!}{x^{n+1}}}$

⇒ f⁶(x) = ${-\dfrac{5!}{x^{6}}}$

⇒ f⁶(x) = ${-\dfrac{120}{x^{6}}}$ 

⇒ |f⁶(x)| = ${\dfrac{120}{x^{6}}}$

Now, using the interval 1 ≤ c ≤ 1.2, R₅(1.2) = ${\dfrac{\left| f^{1}\left( c\right) \right| }{6!}\left( 1\cdot 2-1\right) ^{6}}$

Since |f⁶(c)| = ${\dfrac{120}{c^{6}}}$, M = 120

Thus, ${\left| R_{5}\left( 1.2\right) \right| \leq \dfrac{120}{\left( 6\right) !}\left| 0.2\right| ^{6}}$

⇒ ${\left| R_{5}\left( 1.2\right) \right| \leq \dfrac{120}{720}\left| 0.2\right| ^{6}}$

⇒ ${\left| R_{5}\left( 1.2\right) \right| \leq \dfrac{120}{720}\cdot 0.000064}$

= ${\dfrac{1}{6}\cdot 0.000064}$

≈ 0.00001067

Thus, the error bound for approximating ln(1.2) with P₅(x) is approximately 0.00001067, which indicates a very close approximation.