Question

How do you find the Remainder in the Taylor Series?

Answer 1

Hint : Use the remainder estimation theorem of Taylor Series that is given as if there is a positive constant $M$ such that $\left| {{f^{(n + 1)}}(t)} \right| \leqslant M$ for all $t$ between $x\;{\text{and}}\;a$ , inclusive, then the remainder term represented by ${R_n}(x)$ in Taylor’s Theorem satisfies the inequality
${R_n}(x) \leqslant M\dfrac{{{{\left| {x - a} \right|}^{n + 1}}}}{{(n + 1)!}}$

Complete step-by-step answer :
There are multiple methods to find the Remainder in Taylor Series, we will go through the Remainder estimation theorem of Taylor Series that is the most simplest and common method to find the remainder in Taylor Series. The remainder estimation theorem can be understood as follows:
If there is a positive constant $M$ such that $\left| {{f^{(n + 1)}}(t)} \right| \leqslant M$ is inclusive for all $t$ between $x\;{\text{and}}\;a$ , then the remaining term (represented by ${R_n}(x)$ ) in Taylor's Theorem meets the inequality
${R_n}(x) \leqslant M\dfrac{{{{\left| {x - a} \right|}^{n + 1}}}}{{(n + 1)!}}$
The function ${R_n}(x)$ is the remainder term of the Taylor Series and is defined to be ${R_n}(x) = f(x) - {P_n}(x),\;{\text{where}}\;{P_n}(x)$ is the $nth$ degree Taylor polynomial which is centered at $x = a$
${P_n}(x) = f(a) + f'(a)(x - a) + \dfrac{{f''(a)}}{{2!}}{(x - a)^2} + \dfrac{{f'''(a)}}{{3!}}{(x - a)^3} + .....$
Let us understand it with an example, if
$f(x) = {e^x},\;a = 0\;{\text{and}}\;n = 4$ , we will get
${P_4}(x) = f(0) + f'(0)(x - 0) + \dfrac{{f''(0)}}{{2!}}{(x - 0)^2} + \dfrac{{f'''(0)}}{{3!}}{(x - 0)^3} + \dfrac{{f''''(0)}}{{4!}}{(x - 0)^4} \\\ {P_4}(x) = {e^0} + {e^0}x + \dfrac{{{e^0}}}{{2!}}{x^2} + \dfrac{{{e^0}}}{{3!}}{x^3} + \dfrac{{{e^0}}}{{4!}}{x^4} \\\ {P_4}(x) = 1 + x + \dfrac{{{x^2}}}{{2!}}{x^2} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} \\\$
Also if we take this over the interval $\left[ { - 2,\;2} \right]$ , so that $\left| {x - a} \right| = 2$ then the above inequality will be written as
${R_4}(x) \leqslant \dfrac{{{e^2}{2^{4 + 1}}}}{{(4 + 1)!}} \Rightarrow {R_4}(x) \leqslant \dfrac{{4{e^2}}}{{15}} \approx 1.97$
So we can expect ${P_4}(x)$ to be approximately ${e^x}$ to within the this amount over the interval $\left[ { - 2,\;2} \right]$

Note : ${R_n}(x)$ (remainder of order n) is also known as the error term for the approximation of $f(x)\;{\text{by}}\;{P_n}(x)$ over $I$ . In actual, approximation is a bit better than the error bound or the remainder, the main point is that the remainder estimation theorem gives the assurance.
You can also find the remainder term with the help of a direct formula derived from the Taylor’s formula
${R_n}(x) = \dfrac{{{f^{(n + 1)}}(c){{\left| {x - a} \right|}^{n + 1}}}}{{(n + 1)!}}$ for some $c$ between $a\;{\text{and}}\;x$