Question

Linear approximation:
Verify the given linear approx at a=0. Then determine the values of x for which the linear approx is accurate to within 0.1...
e^x cosx approx 1+x

Answer 1

\[e^xcosx \approx 1+x\]

Answer 2

I know the first part just asks to plug in 0 to verify that the approx is good
The second part i end up with
\[-1.1<e^xcosx-x<1.1\]
everywhere I've looked for help says it must be "solved numerically" but I'm thrown off by all the x's

Answer 3

x should be in radians. Try very small values of x such as +/- 0.01, 0.1, etc and by successive trial and error you can find the domain where the inequality is valid.

Answer 4

ah, so "solve numerically" typically means to find by trial and error? Or is that just in this case?

Answer 5

There are techniques such as Newton-Ralphson method, etc. to find successive approximations. But in numerical analysis, trial and error is an accepted method in finding solutions.

Answer 6

Got it, thanks this helps :)

Answer 7

you are welcome. If you have a functions calculator just input the function and keep trying different x's both negative and positive until you narrow the domain where the inequality is valid up to so many decimal places.

Answer 8

one method I find helpful when using trial and error method is a binary search.
If I am looking for a solution say a root or a domain where an inequality is valid, etc. then I might start with (just an example) x = 0 and x = 100
Then try right in the middle x = 50. Find out if the solution is in 0-50 or 50-100
If it is the first, then try x = 25. Is it to the left or to the right of x = 25. If it is to the right then try x = (25+50)/2 = 37.5, etc. until the solution is narrowed down to however many decimal accuracy you desire.

Answer 9

I'm having a hard time grasping this stuff so I reviewed some examples and made a guide to follow - can you confirm if this is ok:

if\[f(x)\approx L(x)\]
then \[f(x)-0.1<L(x)<f(x)+0.1\]
inputting L(x), f(x)-0.1, and f(x)+0.1 into graphing calculator will show where L(x) intersects either of the other two functions. These intersections mark the interval in which the linear approx is accurate to within 0.1
yes yes?

Answer 10

This is of course only for finding solutions accurate to 0.1
using this guide for my original problem e^(x)cosx approx 1+x, I got
-0.483 < x < 0.416

Answer 11

If graphing calculators are allowed that sounds like a good idea to me.

I am getting -0.763 < x < 0.6072

I did it three ways to find the x coordinates of the intersection points and they all came out the same way although I prefer the first plot.
Plotting y = e^xcos(x) - x -1 ; y = -0.1 ; and y = +0.1
and
Plotting y = e^xcos(x) ; y = x + 1.1; and y = x - 0.9
and
Plotting y = e^xcos(x) + 0.1 ; y = e^xcos(x) - 0.1 ; y = x + 1

I used the website: https://www.desmos.com/calculator

If you are using your calculator make sure the angle is set to radians and not degrees.

Answer 12

Great! Yea I completely forgot to switch over to radians. Thanks again for all your help, you've been very helpful!

Answer 13

you are welcome.