Linear Approximation

FalconPUNCH! · #1 $Old$ November 1st, 2007, 06:07 PM

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.

TKHunny · #2 $Old$ November 1st, 2007, 06:30 PM

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 $\frac{x^{2}}{2}*(1+\frac{x}{3}+\frac{x^{2}}{12}+\frac{x^{3}}{60}+...)\;<\;0.1$

PK Fire!

FalconPUNCH! · #3 $Old$ November 1st, 2007, 07:15 PM

I still don't understand it. :\

FalconPUNCH! · #4 $Old$ November 1st, 2007, 07:58 PM

Anyone? I really need some help with this and I need to understand it before tomorrow.

topsquark · #5 $Old$ November 1st, 2007, 08:07 PM

Quote:

Originally Posted by FalconPUNCH! $View Post$

Anyone? I really need some help with this and I need to understand it before tomorrow.

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

FalconPUNCH! · #6 $Old$ November 1st, 2007, 08:16 PM

Quote:

Originally Posted by topsquark $View Post$

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.

CaptainBlack · #7 $Old$ November 1st, 2007, 11:27 PM

Quote:

Originally Posted by FalconPUNCH! $View Post$

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