Taylor's Theorem

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GRE Quantitative Reasoning › Taylor's Theorem

Questions 1 - 8

For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms?

$f(x)= \frac{x^{5}}{3}+2$

$f(x)= x^{2}+\sqrt{x}+1$

$f(x)=x+3\sin(x)$

$f(x)=\ln|x+3|$

Explanation

Recall the Maclaurin series formula:

$f(x)=\sum_{n=0}^{\infty}\frac{f^n(0)}{n!}x^n$

$=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...$

Despite being a 5th degree polynomial recall that the Maclaurin series for any polynomial is just the polynomial itself, so this function's Taylor series is identical to itself with two non-zero terms.

The only function that has four or fewer terms is $f(x)= \frac{x^{5}}{3}+2$ as its Maclaurin series is $2+\frac{1}{3}x^5$ .

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

$\left ( -5, 5\right )$

$\left ( \frac{-1}{5}, \frac{1}{5} \right )$

$(-\infty, \infty)$

Explanation

Using the root test

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{(n+1)!(x-5)^{n+1}}{n!(x-5)^n} \right |=\left |x-5 \right |\lim_{n\rightarrow \infty }\left | n+1 \right |$

and

$\lim_{n\rightarrow \infty }\left | n+1 \right |=\infty$ . T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

Write out the first four terms of the Taylor series about for the following function:

Explanation

The Taylor series about x=a of any function is given by the following:

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$

So, we must find the zeroth, first, second, and third derivatives of the function (for n=0, 1, 2, and 3 which makes the first four terms):

$f^{0}(x)=f(x)$

The derivatives were found using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Now, evaluated at x=a=1, and plugging in the correct n where appropriate, we get the following:

$\frac{4(x-1)^0}{0!}+\frac{4(x-1)^1}{1!}+\frac{2(x-1)^2}{2!}+\frac{0(x-1)^3}{3!}$

which when simplified is equal to

Write out the first three terms of the Taylor series for the following function about :

$0+(x-1)+\frac{(x-1)^2}{2}$

Explanation

The general formula for the Taylor series of a given function about x=a is

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$ .

We were asked to find the first three terms, which correspond to n=0, 1, and 2. So first, we need to find the zeroth, first, and second derivative of the given function. The zeroth derivative is just the function itself.

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Now use the above formula to write out the first three terms:

$\frac{(2e^1+2)(x-1)^0}{0!}+\frac{(2e^1)(x-1)^1}{1!}+\frac{(2e^1)(x-1)^2}{2!}$

Simplified, this becomes

Suppose that the derivative of a function, denoted , can be approximated by the third degree Taylor polynomial, centered at :

$f'(x) \approx 1 + (x-2) + \frac{4}{3} (x-2)^2 + \frac{8}{9}(x-2)^3$

If , find the third degree Taylor polynomial for centered at .

$7 + (x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3$

$(x-2) + 2(x-2)^2 + {4}(x-2)^3$

$(x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3 + \frac{8}{27}(x-2)^4$

$7 + x + (x-2)^2 + \frac{4}{9}(x-2)^3$

Explanation

To get , we need to find the antiderivative of by integrating the third degree polynomial term by term.

We only want up to a third degree polynomial, so we can disregard the fourth order term:

$f(x) = c + (x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3$

Since , substitute for the final .

$f(x)=7 + (x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3$

Find the first two terms of the Taylor series about for the following function:

$\infty +2(x-3)$

Explanation

The general formula for the Taylor series about x=a for a function is

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$

First, we must find the zeroth and first derivative of the function.

The zeroth derivative of a function is just the function itself, so we only have to find the first derivative:

The derivative was found using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Now, write the first two terms of the sequence (n=0 and n=1):

$\frac{[2(3)+5](x-3)^0}{0!}+\frac{2(x-3)^1}{1!}=11+2(x-3)$

Let $f(x)=(1+\frac{x}{2})^{1/3}$

Find the the first three terms of the Taylor Series for centered at .

$T_{2}(x)=1+\frac{x}{6}-\frac{x^2}{36}$

$T_{2}(x)=1+\frac{x}{6}+\frac{x^2}{36}$

$T_{2}=0+\frac{x}{6} -\frac{x^2}{18}$

$T_{3}=0+0+0$

$T_3=1+\frac{x}{3}+4x^2$

Explanation

Using the formula of a binomial series centered at 0:

$(1+x)^k =\sum_{n=0}^{\infty } \binom{k}{n}x^n$ ,

where we replace with $\frac{x}{2}$ and $k=\frac{1}{3}$ , we get:

$\sum_{n=0}^{2}\binom{\frac{1}{3}}{n}\left(\frac{x}{2}\right)^n$ for the first 3 terms.

Then, we find the terms where,

$T_2(x)=\binom{\frac{1}{3}}{0}(\frac{x}{2})^0 + \binom{\frac{1}{3}}{1}(\frac{x}{2})^1+\binom{\frac{1}{3}}{2}(\frac{x}{2})^2=1+\frac{x}{6}- \frac{x^2}{36}$

Determine the convergence of the Taylor Series for at where $f(x)=\sum_{n=0}^{\infty }\frac{x^n}{n!}$ .

Absolutely Convergent.

Conditionally Convergent.

Divergent.

Inconclusive.

Does not exist.

Explanation

By the ratio test, the series $\sum_{n=0}^{\infty }\frac{5^n}{n!}$ converges absolutely:

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{5^{n+1}}{(n+1)!} \cdot \frac{n!}{5^n} \right |=\lim_{n\rightarrow \infty }\left | \frac{5}{n+1} \right |=0<1$

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