Respuesta :
Answer:
The first four nonzero terms of the Taylor series of [tex]\frac{7}{x + 1}[/tex] around [tex]a=2[/tex] are:
[tex]f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}[/tex]
Step-by-step explanation:
The Taylor series of the function f at a (or about a or centered at a) is given by
[tex]f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k[/tex]
To find the first four nonzero terms of the Taylor series of [tex]\frac{7}{x + 1}[/tex] around [tex]a=2[/tex] you must:
In our case,
[tex]f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k=\sum\limits_{k=0}^{4}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k[/tex]
So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.
- [tex]f^{(0)}\left(x\right)=f\left(x\right)=\frac{7}{x + 1}[/tex]
Evaluate the function at the point: [tex]f\left(2\right)=\frac{7}{3}[/tex]
- [tex]f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(\frac{7}{x + 1}\right)^{\prime}=- \frac{7}{\left(x + 1\right)^{2}}[/tex]
Evaluate the function at the point: [tex]\left(f\left(2\right)\right)^{\prime }=- \frac{7}{9}[/tex]
- [tex]f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(- \frac{7}{\left(x + 1\right)^{2}}\right)^{\prime}=\frac{14}{\left(x + 1\right)^{3}}[/tex]
Evaluate the function at the point: [tex]\left(f\left(2\right)\right)^{\prime \prime }=\frac{14}{27}[/tex]
- [tex]f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(\frac{14}{\left(x + 1\right)^{3}}\right)^{\prime}=- \frac{42}{\left(x + 1\right)^{4}}[/tex]
Evaluate the function at the point: [tex]\left(f\left(2\right)\right)^{\prime \prime \prime }=- \frac{14}{27}[/tex]
- [tex]f^{(4)}\left(x\right)=\left(f^{(3)}\left(x\right)\right)^{\prime}=\left(- \frac{42}{\left(x + 1\right)^{4}}\right)^{\prime}=\frac{168}{\left(x + 1\right)^{5}}[/tex]
Evaluate the function at the point: [tex]\left(f\left(2\right)\right)^{\prime \prime \prime \prime }=\frac{56}{81}[/tex]
Apply the Taylor series definition:
[tex]f\left(x\right)\approx\frac{\frac{7}{3}}{0!}\left(x-\left(2\right)\right)^{0}+\frac{- \frac{7}{9}}{1!}\left(x-\left(2\right)\right)^{1}+\frac{\frac{14}{27}}{2!}\left(x-\left(2\right)\right)^{2}+\frac{- \frac{14}{27}}{3!}\left(x-\left(2\right)\right)^{3}+\frac{\frac{56}{81}}{4!}\left(x-\left(2\right)\right)^{4}[/tex]
The first four nonzero terms of the Taylor series of [tex]\frac{7}{x + 1}[/tex] around [tex]a=2[/tex] are:
[tex]f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}[/tex]
