 #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 $$E=\leftf\left( \frac{1}{2}\right)  P\left( \frac{1}{2}\right)\right=\frac{5}{8}  e^{ \frac{1}{2}}\approx 0.018$$
Method. 2.
Using the formula $$E=\left\frac{f'''(c)(ca)^3}{6}\right$$
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 $$E=\leftf\left( \frac{1}{2}\right)  P\left( \frac{1}{2}\right)\right=\frac{5}{8}  e^{ \frac{1}{2}}\approx 0.018$$
Method. 2.
Using the formula $$E=\left\frac{f'''(c)(ca)^3}{6}\right$$
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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