Question

What is Lagrange Error and how do you find the value for it $M$ ?

Answer 1

Hint : We need to know the Taylor series equation in terms of $f\left( x \right)$ around $x = a$ . In that formula next, we put $k = n$ . We need to know the basic equation for ${P_n}\left( x \right)$ and ${R_n}\left( x \right)$ . By using these equations next we would relate the Taylor series equation with the equations of ${P_n}\left( x \right)$ and ${R_n}\left( x \right)$ . Also, we need to substitute the second theorem of the mean in the equation ${R_n}\left( x \right)$ to find the term $M$ .

Complete step by step solution:
Consider the Taylor series of a function $f\left( x \right)$ around $x = a$ ,
$f\left( x \right) = \sum\limits_{k = 0}^\infty {\dfrac{{{f^{\left( k \right)}}\left( a \right)}}{{k!}}{{\left( {x - a} \right)}^k}}$
If we stop the Taylor series at $k = n$ we have,
$f\left( x \right) = {P_n}\left( x \right) + {R_n}\left( x \right)$
Where,
${P_n}\left( x \right) = \sum\limits_{k = 0}^\infty {\dfrac{{{f^{\left( k \right)}}\left( a \right)}}{{k!}}{{\left( {x - a} \right)}^k}}$
And it can be demonstrated that rest can be expressed as,
${R_n}x = \dfrac{1}{{n!}}\int\limits_n^x {{f^{\left( {n + 1} \right)}}} \left( t \right){\left( {x - t} \right)^n}dt$
Applying the second theorem of the mean to this integral we have,
${R_n}\left( x \right) = \dfrac{1}{{\left( {n + 1!} \right)}}{f^{\left( {n + 1} \right)}}\left( \xi \right){\left( {x - a} \right)^{n + 1}}$
Where $\xi$ is a point between $x$ and $a$
Clearly if in the interval delimited by $x$ and $a$ we have,
$\left| {{f^{\left( {n + 1} \right)}}\left( \xi \right)} \right| \prec M$
Then
$\left| {{R_n}\left( x \right)} \right| \leqslant \dfrac{M}{{\left( {n + 1} \right)!}}{\left| {x - a} \right|^{n + 1}}$
So, the correct answer is “ $\left| {{R_n}\left( x \right)} \right| \leqslant \dfrac{M}{{\left( {n + 1} \right)!}}{\left| {x - a} \right|^{n + 1}}$ ”.

Note : Remember the second theorem of the mean to solve these types of questions. Note that $\xi$ is a point between $x$ and $a$ . Also, note that if $\left| {{f^{\left( {n + 1} \right)}}\left( \xi \right)} \right|$ is less than $M$ then $\left| {{R_n}\left( x \right)} \right|$ is also less than or equal to $\dfrac{M}{{\left( {n + 1} \right)!}}{\left| {x - a} \right|^{n + 1}}$ . Also, note that $f\left( x \right)$ is also can be written as the sum of ${P_n}\left( x \right)$ and ${R_n}\left( x \right)$ .