# Math

posted by Jane

Use Newton’s Method to approximate 3^(√7) to four decimal places.
Use x1 = 2 as your seed. Round off intermediate iterates to five decimal place

1. Steve

let f(x) = x^3 - 7
so f(∛7) = 0

See
http://keisan.casio.com/exec/system/1244946907

just enter your function, its derivative, and initial guess.

2. Reiny

the answer I should get is 18.2955

let x = 3^√7
then x^(1/√7) = 3
x^(1/√7) - 3 = 0

let y = x^(1/√7) - 3 = x^.3779645 - 3
dy/dx = (1/√7)^(1/√7 - 1) = .3779645 x^-.62204

iteration expression:

x - (x^.3779645 - 3)/(.3779645x^-.62204)

x = 2 -----> 15.06815
x = 15.06815 --->18.10237..
x = 18.10237.. ---> 18.294858..
x = 18.29456 ----> 8.2955 , which was my calculator answer.

I find this a very strange question.
There is no practical way to do the above calculations without using a scientific calculator.
So why not find that scientific calculator to find the answer in the first place ?

3. Steve

Good job, Reiny. I keep seeing 3^√ being used as cube root, so maybe I was way off.

In any case, the problem is solved.

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