Question #181556

Obtain a four-term Taylor polynomial approximation valid near x=,0 for each (1+x)^-1/3. Estimate the range of x over which three term polynomials will give two decimal accuracy


1
Expert's answer
2021-04-26T17:47:11-0400

set f(x)=(1+x)1/3f(x)=(1+x)^{-1/3}

for a four term Taylor polynomial centered at x=0

f(x)=(1+x)1/3    f(0)=1f(x)= (1+x)^{-1/3}\implies f(0)=1

f(x)=13(1+x)4/3    f(0)=12f'(x)=-\frac 13 (1+x)^{-4/3}\implies f'(0)=-\frac 12

f(x)=49(1+x)7/3    f(0)=49f''(x)= -\frac 49(1+x)^{-7/3} \implies f''(0)= \frac 49

f(x)=2827(1+x)10/3    f(0)=2827f'''(x)= -\frac {28}{27}(1+x)^{-10/3}\implies f'''(0)= -\frac {28}{27}

now the four term Taylor polynomial approximation to f(x) valid near x=0 is;

f(x)=(1+x)1/3=113x+492!x228273!x3f(x) =(1+x)^{-1/3} = 1-\frac 13x + \frac{\frac 49}{2!}x^2- \frac {\frac {28}{27}}{3!}x^3

=113x+29x21481x3= 1-\frac 13 x + \frac 29 x^2 - \frac {14}{81}x^3

for the range of values of x over which these term polynomials will give a two decimal accuracy we may use the Lagrange error bound;

mxan+1(n+a)!<=m\frac {|x-a|^{n+1}}{(n+a)!}<=  maximum admissible error

where m=maxxIfn+1(x)m= max _{x\isin\Iota}|f^{n+1}(x)|

mx033!<=0.01\therefore m \frac {|x-0|^3}{3!}<= 0.01 or x3<=0.013!m|x|^3<= \frac {0.01*3!}{m}

x<=0.39m3|x|<= \frac {0.39}{\sqrt [3]m}

    0.4<=x<=0.4\implies -0.4<= x<= 0.4 is the range desired.





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