Find the third degree Taylor polynomial centered at \(x=1\) for the function \(f(x) = \sqrt {2x-1}\).

Since we want a Taylor Polynomial centered at \(x=1\), we want to look for a polynomial in powers of \(x-1\). Since we want a third degree Taylor polynomial, this means we are looking for a polynomial of the form:

\(p_3(x) = a_0+a_1x+a_2x^2+a_3x^3\) \(p_3(x) = a_0+a_1(x-1)+a_2(x-1)^2+a_3(x-1)^3\)

The coefficients are found using the requirements:

\(f(1) = p_3(1)\) \(f'(1) = p_3'(1)\) \(f''(1) = p_3''(1)\) \(f'''(1) = p_3'''(1)\) \(f^{(4)}(1) = p_3^{(4)}(1)\) \(f^{(5)}(1) = p_3^{(5)}(1)\)

These requirements are used to establish a formula that relates the coefficients of the Taylor Polynomial to the derivatives of the function that it approximates for each \(k=0,1,2,\ldots ,n\):

\[ a_k = \frac {f^{(k)}(c)}{k!} \]
where \(c\) is the \(x\)-value at which the series is centered. Here, \(c=\answer {1}\).
Complete the table below:
\(k\) \(f^{(k)}(x)\) \(f^{(k)}(1)\) \(a_k = \frac {f^{(k)}(1)}{k!}\)
\(0\) \(\answer {(2x-1)^{1/2}}\) \(\answer {1}\) \(\answer {1}\)
\(1\) \(\answer {(2x-1)^{-1/2}}\) \(\answer {1}\) \(\answer {1}\)
\(2\) \(\answer {-(2x-1)^{-3/2}}\) \(\answer {-1}\) \(\answer {-\frac {1}{2}}\)
\(3\) \(\answer {3(2x-1)^{-5/2}}\) \(\answer {3}\) \(\answer {\frac {1}{2}}\)

Hence, the third degree Taylor polynomial for \(f(x) =\sqrt {2x-1}\) is:

\begin{align*} p_3(x) &=a_0+a_1(x-1)+a_2(x-1)^2+a_3(x-1)^3 \\ &= \answer {1}+\answer {1}(x-1)+\answer {-\frac {1}{2}}(x-1)^2+\answer {\frac {1}{2}}(x-1)^3\\ \end{align*}
The Taylor polynomials are used to approximate the value of a function near the center. As long as we are “close enough" (which we will formalize in a later section), the higher order polynomials should give better approximations. Suppose that we want to approximate \(\sqrt {1.2}\).

The exact value is found using the function, provided we use the correct \(x\)-value. Since \(f(x) = \sqrt {2x-1}\), the \(x\)-value that gives \(\sqrt {1.2}\) is \(x=\answer {1.1}\).

To find this, just set \(2x-1 = 1.2\).

Thus, the exact answer to 6 decimal places is \(\sqrt {1.2} = \answer [tolerance= .000001]{1.095445}\)

We can use the third degree Taylor polynomial to write down the first and second degree Taylor polynomials. We can then substitute \(x=1.1\) into these to approximate \(\sqrt {1.2}\). In fact:

The first degree Taylor polynomial is \(p_1(x) = \answer {1}+\answer {1}(x-1)\) so to 6 decimal places: \(\sqrt {1.2} \approx p_1(1.1) = \answer [tolerance=.000001]{1.100000}\).

The second degree Taylor polynomial is \(p_2(x) = \answer {1}+\answer {1}(x-1)+\answer {-\frac {1}{2}}(x-1)^2\) so to 6 decimal places: \(\sqrt {1.2} \approx p_2(1.1) = \answer [tolerance=.000001]{1.095000}\).

The third degree Taylor polynomial is \(p_3(x) =1+(x-1)-\frac {1}{2}(x-1)^2+\frac {1}{2}(x-1)^3\) so to 6 decimal places: \(\sqrt {1.2} \approx p_3(1.1) = \answer [tolerance=.000001]{1.095500}\).

Do the higher order polynomials provide more accurate results?

Yes No