Thread: Newtons Method

Use Newtons method to find the two solutions of e^x=5*x to three decimal places.
Newtons Method: X(n+1) = Xn - f(Xn)/f'(Xn)

I just dont understand how you would set this up. Do you set both the approximations equal to each other to solve or do you solve them individually and keep checking?

Originally Posted by coe236

Use Newtons method to find the two solutions of e^x=5*x to three decimal places.
Newtons Method: X(n+1) = Xn - f(Xn)/f'(Xn)

I just dont understand how you would set this up. Do you set both the approximations equal to each other to solve or do you solve them individually and keep checking?

there are two solutions for x. one is around 2.5 (a little more), the other is between 0.2 and 0.5.

first rewrite:

thus we let

now we guess a root. let's say we're going after the one around 2.5. so we guess 2.5 (that is, we let )

then

you will get a value for

then repeat the same process, using the last value you got as , and keep doing it until you realize the values settling (you always get the same first three decimal places).

do this again for the other root, guess 0.5


below is the graph

Originally Posted by coe236

Use Newtons method to find the two solutions of e^x=5*x to three decimal places.
Newtons Method: X(n+1) = Xn - f(Xn)/f'(Xn)

I just dont understand how you would set this up. Do you set both the approximations equal to each other to solve or do you solve them individually and keep checking?

Hello,

1. make a rough sketch: You'll notice that there are 2 points of intersection thus

2. the equation have 2 solutions.

3. consider the function

then you have:



Now use your calculator:

Code:

start:             0                  2
2. step:       0.25                3.092876765
3. step:       0.25915...          2.70696...
   .
   .
answer:        0.259171101..       2.542641358...

ohh i understand, I kept solving for x and that messed me up. That sounds simple enough. Thanks guys

Originally Posted by coe236

ohh i understand, I kept solving for x and that messed me up. That sounds simple enough. Thanks guys

Hello,

if (and only if) your calculator has a -key then these kinds of problems are simpler than they appear.

With your problem type:

0 = [remark: Now the zero is stored in the ANS]
ANS - ((e^ANS-5*ANS)/(e^ANS-5)) = [now you have the 2nd result]
= [you only hit the -key]
=
...
= until the results don't change

You have finished the complete problem within seconds!

hehe, yeah thats going to be useful method to use on exams. Thankyou =]