
What is Lagrange Error and how do you find the value for it \[M\] ?
Answer
544.2k+ views
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 \[{R_n}\left( x \right)\] .
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 \[{R_n}\left( x \right)\] .
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

