Re: Taylor Series & Minimum

Hey cliff123.

Hint: Calculate the derivatives at x = -1 (The derivative of e^x is always e^x so this is easy) and expand around (x + 1)^n for each nth term from n = 0 to n = infinity.

Re: Taylor Series & Minimum

Quote:

Originally Posted by

**cliff123** Hi, I have a questions regarding Taylor Series and Minimum topics.

1. With f(x) = (x^2 + 2x + 2) e^x

a. Find the Taylor series at x = -1

b. Find the value of n power from that series.

c. Find the value of f^(10) (-1)

2. A cylinder with an open top and closed bottom is made of glass with 0.5 cm of thickness. Both outer side height and radius of the cylinder is 40cm.

a. Find the volume of the glass material.

b. Find the approx. value of the volume using differential approx.

3. A farmer has a field in a triangle ABC shape. The length of AB and AC are 100 meters and 80 meters respectively and the angle of BAC is 60 degrees. He wants to build a fence on P point on AB side to Q point on AC side to cut the field into two identical areas. If the fence cost $5 / meter, calculate the minimum cost to built it.

This questions are translated from my native language, I'm sorry if there's any errors about how I interpret it in English.

Please help me, I'll be extremely grateful. thanks :)

$\displaystyle \displaystyle \begin{align*} f(x) &= \left(x^2 + 2x + 2 \right) e^x \\ &= \left( x^2 + 2x + 1^2 - 1^2 + 2 \right) e^x \\ &= \left[ \left(x + 1 \right) ^2 + 1 \right] e^x \\ &= \left\{ \left[ x - (-1) \right] ^2 + 1 \right\} e^x \end{align*}$

Now we should try to find the Taylor Series for $\displaystyle \displaystyle \begin{align*} g(x) = e^x \end{align*}$ about x = -1. Note that each derivative of $\displaystyle \displaystyle \begin{align*}e^x \end{align*}$ is $\displaystyle \displaystyle \begin{align*} e^x \end{align*}$, so $\displaystyle \displaystyle \begin{align*} g^{(n)}(-1) = e^{-1} \end{align*}$ for all n. Therefore the Taylor Series is $\displaystyle \displaystyle \begin{align*} g(x) = e^x = \sum_{n = 0}^{\infty}\left\{ \frac{e^{-1}}{n!} \left[ x - (-1) \right] ^n \right\} \end{align*}$. So the Taylor Series for f(x) is

$\displaystyle \displaystyle \begin{align*} f(x) &= \left\{ \left[ x - (-1) \right] ^2 + 1 \right\} \sum_{n = 0}^{\infty} \left\{ \frac{e^{-1}}{n!} \left[ x - (-1) \right] ^n \right\} \end{align*}$

Re: Taylor Series & Minimum

Hi and thanks chiro & Prove It :)

about the other questions, can you help me?