# Linear Approximation

• Nov 1st 2007, 04:07 PM
FalconPUNCH!
Linear Approximation
Verify the given linear approximation at $a = o$. Then determine the values of x for which the linear approximation is accurate to within 0.1

$e^x =^* 1 + x$

*I don't know how to do an approximately sign but it's not supposed to be an equal sign.
• Nov 1st 2007, 04:30 PM
TKHunny
Not real clear what tools can be used...

We have a well-known result: $e^{x}\;=\;1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+...$

So, $e^x\;=\;(1+x)\;+\;\frac{x^{2}}{2}*(1+\frac{x}{3}+\ frac{x^{2}}{12}+\frac{x^{3}}{60}+...)$

Since this is ALWAYS an underestimate, your task is to determine where $\frac{x^{2}}{2}*(1+\frac{x}{3}+\frac{x^{2}}{12}+\f rac{x^{3}}{60}+...)\;<\;0.1$

PK Fire!
• Nov 1st 2007, 05:15 PM
FalconPUNCH!
I still don't understand it. :\
• Nov 1st 2007, 05:58 PM
FalconPUNCH!
Anyone? I really need some help with this and I need to understand it before tomorrow. :confused:
• Nov 1st 2007, 06:07 PM
topsquark
Quote:

Originally Posted by FalconPUNCH!
Anyone? I really need some help with this and I need to understand it before tomorrow. :confused:

Can you give us an example to show us how this is supposed to work? Presumably you went over it in class, or there is an example in your book.

-Dan
• Nov 1st 2007, 06:16 PM
FalconPUNCH!
Quote:

Originally Posted by topsquark
Can you give us an example to show us how this is supposed to work? Presumably you went over it in class, or there is an example in your book.

-Dan

I don't know if this is the exact same thing but it's the only thing I can find in my book that is similar.

Example:

For what values of x is the linear approximation $\sqrt{x + 3} = \frac{7}{4} + \frac {x}{4}$

It's supposed to be an approximately sign not equal

accurate within 0.5? What about accuracy to within 0.1?

Solution:

Accuracy to within 0.5 means that the functions should differ by less than 0.5.

$\sqrt{x + 3} - 0.5 < \frac{7}{4} + \frac{x}{4} < \sqrt{x + 3} + 0.5$

This says that the linear approximation should lie between the curves obtainted by shifting the curve $y = \sqrt{x+3}$ upward and downward by an amount 0.5.

Later it gives me -2.6 < x < 8.6 for the 0.5 and -1.1 < x < 3.9 for the 0.1

I don't understand how they got the answers in the book.
• Nov 1st 2007, 09:27 PM
CaptainBlack
Quote:

Originally Posted by FalconPUNCH!
Verify the given linear approximation at $a = o$. Then determine the values of x for which the linear approximation is accurate to within 0.1

$e^x \approx 1 + x$

*I don't know how to do an approximately sign but it's not supposed to be an equal sign.

Draw a graph of $e^x - (1+x)$ arout $x=0$, the thinterval over which this is $<0.1$ is what you want.

(I would have put absolute values in this except we know that $e^x - (1+x)\ge0$,for all $x \in \mathbb{R}$ )

RonL