Question

Consider the Taylor polynomial T n ( x ) centered at x = 21 for all n for the function f ( x ) = 1 x − 1 , where i is the index of summation. Find the i th term of T n ( x ) . (Express numbers in exact form. Use symbolic notation and fractions where needed. For alternating series, include a factor of the form ( − 1 ) n in your answer.)

Answer 1

Answer:

$\sum_{n=0}^\infty (1/20) (-1)^n(\frac{x-21}{20})^n$

Step-by-step explanation:

Remember the geometric series

$\frac{a}{1-x} = \sum_{n=0}^{\infty} ax^n$

So we can express

$f(x) = \frac{1}{x-1} = \frac{1}{x-1+21-21} \\=\frac{1}{(x-21)+20} \\= \frac{1}{ \frac{20}{20}( (x-21) +20 )} \\= \frac{1/20}{1+(\frac{x-21}{20}) } \\\\= \frac{1/20}{1-(-\frac{x-21}{20}) } \\\\\\= \sum_{n=0}^\infty (1/20) (-\frac{x-21}{20})^n\\= \sum_{n=0}^\infty (1/20) (-1)^n(\frac{x-21}{20})^n$

Answer 2

The ith term of the Taylor polynomial is: $\mathbf{ \sum\limits^{\infty}_0 \frac{1}{20} (-1)^i(\frac{x-21}{20})^i }$

The function is given as:

$\mathbf{f(x) = \frac{1}{x -1}}$

And the Taylor polynomial is centered at: $\mathbf{x = 21}$

Rewrite f(x) as:

$\mathbf{f(x) = \frac{1}{x -1 + 0}}$

Express 0 as 21 - 21

$\mathbf{f(x) = \frac{1}{x -1 + 21 - 21}}$

This gives

$\mathbf{f(x) = \frac{1}{x + 20 - 21}}$

Rewrite as:

$\mathbf{f(x) = \frac{1}{x - 21+ 20 }}$

Multiply f(x) by 1

$\mathbf{f(x) = \frac{1}{x - 21+ 20 } \times 1}$

Express 1 as 20/20

$\mathbf{f(x) = \frac{1}{(x - 21)+ 20 } \times \frac{20}{20}}$

Rewrite as:

$\mathbf{f(x) = \frac{1/20}{((x - 21)+ 20 )\times \frac{1}{20}} }$

 

Expand

$\mathbf{f(x) = \frac{1/20}{\frac{x - 21}{20}+ 1 } }$

Apply associative property

$\mathbf{f(x) = \frac{1/20}{1 + \frac{x - 21}{20} } }$

Express + as --

$\mathbf{f(x) = \frac{1/20}{1 -(- \frac{x - 21}{20} )} }$

Recall that the sum to infinity of a geometric progression is:

$\mathbf{\sum\limits^{\infty}_0 ar^n = \frac{a}{1- r}}$

By comparison:

$\mathbf{a = \frac{1}{20}}$

$\mathbf{r = -\frac{x-21}{20}}$

So, we have:

$\mathbf{\sum\limits^{\infty}_0 \frac{1}{20} \times ( -\frac{x-21}{20})^n = \frac{a}{1- r}}$

Split

$\mathbf{\sum\limits^{\infty}_0 \frac{1}{20} \times (-1)^n \times (\frac{x-21}{20})^n = \frac{a}{1- r}}$

Rewrite as:

$\mathbf{\frac{a}{1- r} = \sum\limits^{\infty}_0 \frac{1}{20} \times (-1)^n \times (\frac{x-21}{20})^n }$

Replace n with i

$\mathbf{T(i) = \sum\limits^{\infty}_0 \frac{1}{20} \times (-1)^i \times (\frac{x-21}{20})^i }$

$\mathbf{T(i) = \sum\limits^{\infty}_0 \frac{1}{20} (-1)^i(\frac{x-21}{20})^i }$

Hence, the ith term is: $\mathbf{ \sum\limits^{\infty}_0 \frac{1}{20} (-1)^i(\frac{x-21}{20})^i }$

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