# Taylor polynomial for $f(x)=x^{1/7}$ about $a=1$

Here is the problem:

(a) Determine the Taylor polynomial $$T_2(x)$$ of degree $$2$$ for the function $$f(x)=x^{1/7}$$ centered at $$a=1$$.

(b) Suppose we were to use the approximation $$f(x) \approx T_2(x)$$. Obtain an upper bound on the error of this approximation when $$0.7 \leq x \leq 1.3$$. Give your answer rounded (up) to $$4$$ decimal places.

I'm stuck on (a) currently, I can't figure out the formula to use to get a Taylor series of a certain degree.

• The Taylor polynomial of $f$ degree 2 about $a$ is defined to be $$T_2(x) = f(a) + \frac{f'(a)}{1!} (x-a) + \frac{f''(a)}{2!} (x-a)^2 .$$ You just have to plug in the values of the function and its first and second derivatives in this formula. Commented Dec 15, 2011 at 12:00
• So $T_{n}(x)$ just means it's just the first $n+1$ terms of the series? Commented Dec 15, 2011 at 12:16
• Yes, the first $n+1$ terms give the Taylor polynomial of degree (order) $n$. But note that $T_n(x)$ is the notation this question uses; it might happen that you find some other notation (like, $P_n(x)$) in another place. (On the other hand, the terminology Taylor polynomial is standard.) Commented Dec 15, 2011 at 12:22

To get your polynomial you need to fill in the values of the second degree Taylor polynomial:

$$P\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right) + f''\left( a \right)\frac{{{{\left( {x - a} \right)}^2}}}{{2!}}$$

In general, the Taylor polynomial of $n$th degree around $x=a$ is given by

$${P_n}\left( x \right) = f\left( a \right) + f'\left( a \right)\frac{{x - a}}{{1!}} + \cdots + {f^{\left( {n - 1} \right)}}\left( a \right)\frac{{{{\left( {x - a} \right)}^{n - 1}}}}{{\left( {n - 1} \right)!}} + {f^{\left( n \right)}}\left( a \right)\frac{{{{\left( {x - a} \right)}^n}}}{{n!}}$$

At $x=1$ we have

$f(1) = 1$

$f'(1) = \dfrac{1}{7}$

$f''(1) = -\dfrac{6}{49}$

So we replace

$$P\left( x \right) = 1 + \dfrac{1}{7}\left( {x - 1} \right) -\dfrac{6}{49}\frac{{{{\left( {x - 1} \right)}^2}}}{{2!}}$$

You can see the respective plots here:

For the error, we calculate the remainder of the polynomial, which is given by several formulas, and among one is:

$$R{_n}\left( x \right) = \int\limits_a^x {\frac{{{{\left( {x - t} \right)}^n}}}{{n!}}{f^{\left( {n + 1} \right)}}\left( t \right)dt}$$

If you're not familiar with integration you can use the following theorem:

Suppose $m \leq f^{(n+1)}(x) \leq M$ in some interval $I$ containing $x=a$. Then the error or remainder of degree $n$, $R_n(x)$ satisfies:

$$m\frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}} \leq {R_n}\left( x \right) \leq M\frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}}$$

If $x < a$ the error as to be multiplied by $(-1)^{n+1}$, i.e

$$m\frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}} \leq (-1)^{n+1}{R_n}\left( x \right) \leq M\frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}}$$

In your case you have $0.7 < x < 1.3$ so you can use the theorem and get using $M$ and $m$ as $f'''(0.7)$ and $f'''(1.3)$ since the derivative is monotonically decreasing.

$$0.11\frac{{{{\left( {x - 1} \right)}^3}}}{{3!}} < {R_2}\left( x \right) < 0.63\frac{{{{\left( {x - 1} \right)}^3}}}{{3!}}$$

So using this you can calculate what the error is for example for $x=1.3$

$$0.000495 < {R_2}\left( {1.3} \right) < 0.002835$$

You can see the plots of the cubics here (red and blue), which give upper and lower bounds of the error (green) for $x$ near $1$ (and obviously give $0$ when $x=1$).