# Approximate 5th root of 20

• May 21st 2010, 01:36 PM
TsAmE
Approximate 5th root of 20
f(x) = x^5 - 20

f'(x) = 5x^4

x2 = x1 - ( x1^5 - 20 ) / ( 5x1^4 )
x2 = 1 - ( (1)^5 - 20 ) / ( 5(1)^4 )
x2 = 1 - ( -19 / 5 )
x2 = 24 / 5
x2 = 4.08

My answer is wrong and the correct answer is 1.82056420. I used an initial approximation of x1 = 1, as no approximation was given.
• May 21st 2010, 01:59 PM
Also sprach Zarathustra
Quote:

Originally Posted by TsAmE
f(x) = x^5 - 20

f'(x) = 5x^4

x2 = x1 - ( x1^5 - 20 ) / ( 5x1^4 )
x2 = 1 - ( (1)^5 - 20 ) / ( 5(1)^4 )
x2 = 1 - ( -19 / 5 )
x2 = 24 / 5
x2 = 4.08

My answer is wrong and the correct answer is 1.82056420. I used an initial approximation of x1 = 1, as no approximation was given.

You can use Taylor...
x^1/5 ~ 1+(1/5)*(x-1) - (2/25)*(x-1)^2)+(6/125)*(x-1)^3+O((x-1)^4)
You put x=20 and get the needed.
• May 21st 2010, 08:17 PM
TsAmE
I havent learned taylor, I want to do it using newtons method
• May 26th 2010, 02:28 AM
a tutor
Your method is correct. You made a little slip. 24/5=4.8 rather than 4.08. Just keep going.

On a decent calculator you can do this:

1.5 =

ANS - (ANS^5-20)/(5ANS^4) = = = = =

It gives
1.9901..
1.8470..
1.8213..
and pretty soon you will have
1.820564203
• May 26th 2010, 02:56 AM
HallsofIvy
Quote:

Originally Posted by TsAmE
f(x) = x^5 - 20

f'(x) = 5x^4

x2 = x1 - ( x1^5 - 20 ) / ( 5x1^4 )
x2 = 1 - ( (1)^5 - 20 ) / ( 5(1)^4 )
x2 = 1 - ( -19 / 5 )
x2 = 24 / 5
x2 = 4.08

My answer is wrong and the correct answer is 1.82056420. I used an initial approximation of x1 = 1, as no approximation was given.

Well, 24/5= 4.8, not 4.08 but what exactly was the problem? Surely not to do just one iteration? If you keep going, you get successive approximations that converge to 1.82...