# Calculate the value of the given function by using the indicated number of terms of the...

## Question:

Calculate the value of the given function by using the indicated number of terms of the appropriate series.

Round to seven decimal places.

{eq}\sqrt[3]{0.85} {/eq} (3 terms)

## Taylor Series Expansion

Although this problem can be solved by various methods like linearization, numerical series expansion etc. To solve this problem we are going to use the Taylor series expansion and calculate the value up to three decimal places.

Using the Taylor series expansion we can write,

{eq}\displaystyle (1-x)^n=1-nx+\frac{n(n-1)}{2!}x^2-.... {/eq}

We have,

{eq}\displaystyle y=\sqrt[3]{0.85} {/eq}

{eq}\displaystyle y=\left ( 1-0.15 \right )^{\frac{1}{3}} {/eq}

So from the information,

{eq}\displaystyle x=-0.15 {/eq}

{eq}\displaystyle n=\frac{1}{3} {/eq}

Using the expansion we can write:

{eq}\displaystyle \left ( 1+(-0.15) \right )^{\frac{1}{3}}=1+\left ( \frac{1}{3} \right )(-0.15)+\left ( \frac{1}{3} \right )\left ( \frac{1}{3}-1 \right )(0.01125)-.... {/eq}

{eq}\displaystyle \left ( 1+(-0.15) \right )^{\frac{1}{3}}=1-0.05-0.0025 {/eq}

{eq}\displaystyle \boxed{\displaystyle \left ( 1+(-0.15) \right )^{\frac{1}{3}}=0.9475\approx 0.948} {/eq}

Thus,

{eq}\displaystyle \boxed{\displaystyle \left ( 0.85 \right )^{\frac{1}{3}}\approx 0.948} {/eq}