- #1

tompenny

- 15

- 3

- Homework Statement
- Evaluate the Taylor series and find the error at a point

- Relevant Equations
- $$f^{(0)}\left(x\right)=f\left(x\right)=e^{x}$$

And want to approximate it using Taylor at the point $$\frac{1}{\sqrt e} $$

I have the following function

$$f^{(0)}\left(x\right)=f\left(x\right)=e^{x}$$

And want to approximate it using Taylor at the point ##\frac{1}{\sqrt e} ##

I also want to decide (without calculator)whether the error in the approximation is smaller than ##\frac{1}{25} ##

The Taylor polynomial is:

$$f\left(x\right)\approx P\left(x\right) = 1+x+\frac{1}{2}x^{2}$$

I evaluate the function at the point: $$f\left(- \frac{1}{2}\right)=e^{- \frac{1}{2}}$$

I evaluate the series at the point: $$P\left(- \frac{1}{2}\right)=\frac{5}{8}$$

Method. 1.

The error I get is $$E=\left|f\left(- \frac{1}{2}\right) - P\left(- \frac{1}{2}\right)\right|=\frac{5}{8} - e^{- \frac{1}{2}}\approx 0.018$$

Method. 2.

Using the formula $$E=\left|\frac{f'''(c)(c-a)^3}{6}\right|$$

where c is: $$-\frac{1}{2}<c<0$$

I get the error to $$E=\left|\frac{e^{- \frac{1}{2}}(-\frac{1}{2})^3}{6}\right|=\frac{1}{48\sqrt(e)}\approx 0.013$$

So my questions:

1. Which one of my methods is correct(if any)?

2. How do I show that the error is smaller than 1/25 without using a calculator?

Many thanks for any input on this matter:)

$$f^{(0)}\left(x\right)=f\left(x\right)=e^{x}$$

And want to approximate it using Taylor at the point ##\frac{1}{\sqrt e} ##

I also want to decide (without calculator)whether the error in the approximation is smaller than ##\frac{1}{25} ##

The Taylor polynomial is:

$$f\left(x\right)\approx P\left(x\right) = 1+x+\frac{1}{2}x^{2}$$

I evaluate the function at the point: $$f\left(- \frac{1}{2}\right)=e^{- \frac{1}{2}}$$

I evaluate the series at the point: $$P\left(- \frac{1}{2}\right)=\frac{5}{8}$$

Method. 1.

The error I get is $$E=\left|f\left(- \frac{1}{2}\right) - P\left(- \frac{1}{2}\right)\right|=\frac{5}{8} - e^{- \frac{1}{2}}\approx 0.018$$

Method. 2.

Using the formula $$E=\left|\frac{f'''(c)(c-a)^3}{6}\right|$$

where c is: $$-\frac{1}{2}<c<0$$

I get the error to $$E=\left|\frac{e^{- \frac{1}{2}}(-\frac{1}{2})^3}{6}\right|=\frac{1}{48\sqrt(e)}\approx 0.013$$

So my questions:

1. Which one of my methods is correct(if any)?

2. How do I show that the error is smaller than 1/25 without using a calculator?

Many thanks for any input on this matter:)

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