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spayn [35]
3 years ago
15

Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of

a. (Enter your answers as a comma-separated list.) f(x) = 7 1 + x , a = 2
Mathematics
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer:

The first four nonzero terms of the Taylor series of \frac{7}{x + 1} around a=2 are:

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}

Step-by-step explanation:

The Taylor series of the function <em>f </em>at <em>a </em>(or about <em>a</em> or centered at <em>a</em>) is given by

f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k

To find the first four nonzero terms of the Taylor series of \frac{7}{x + 1} around a=2 you must:

In our case,

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

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.

  • f^{(0)}\left(x\right)=f\left(x\right)=\frac{7}{x + 1}

Evaluate the function at the point: f\left(2\right)=\frac{7}{3}

  • 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}}

Evaluate the function at the point: \left(f\left(2\right)\right)^{\prime }=- \frac{7}{9}

  • 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}}

Evaluate the function at the point: \left(f\left(2\right)\right)^{\prime \prime }=\frac{14}{27}

  • 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}}

Evaluate the function at the point: \left(f\left(2\right)\right)^{\prime \prime \prime }=- \frac{14}{27}

  • 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}}

Evaluate the function at the point: \left(f\left(2\right)\right)^{\prime \prime \prime \prime }=\frac{56}{81}

Apply the Taylor series definition:

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}

The first four nonzero terms of the Taylor series of \frac{7}{x + 1} around a=2 are:

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}

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