Asked by kookay
hello i just want a answer in these question:
please. im begging!
I. Use Newton’s method to approximate the real root to 4 decimal places.
1. X^3-3X+1=0
2 . X^3-X-2=0
3. 2X-3SINX=0
please. im begging!
I. Use Newton’s method to approximate the real root to 4 decimal places.
1. X^3-3X+1=0
2 . X^3-X-2=0
3. 2X-3SINX=0
Answers
Answered by
dongo
Well, this is simple, you just have to form the derivative of the presumed functions.
e.g. d/dx[2x-3sinx]=:g(x); [2x-3sinx]=:f(x)
Afterwards, you have to numericaly take the following algorithm as many times as necessary in order to be yielded the wanted result:
x_(n+1)= x_n-f(x_n)/g(x_n)
Bear in mind that this is an iteration, i.e. of repeated application of this algorithms comes to a number whose 4th decimal has not changed concerning the preceding one, you'll be done.
e.g. d/dx[2x-3sinx]=:g(x); [2x-3sinx]=:f(x)
Afterwards, you have to numericaly take the following algorithm as many times as necessary in order to be yielded the wanted result:
x_(n+1)= x_n-f(x_n)/g(x_n)
Bear in mind that this is an iteration, i.e. of repeated application of this algorithms comes to a number whose 4th decimal has not changed concerning the preceding one, you'll be done.
Answered by
kookay
hello dongo. First of all thank you but I'm sorry I can't understand it because the question is from my older sister who has a Flu right now and she ask me to post it on the internet so somebody can answer it. Can you solve it for her? please? She will be very very thankful to you.
Answered by
dongo
There's absolutely no use in doing so. Even though she's ill, I am really convinced about her managing to solve this, supplied with the correct algorithms. (In fact, I do ot understand why it is always about calculating. Proving the applicability of this algorithm would be way better... She can greet her profesor from me...)
Answered by
kookay
she said thanks. i just want to help her. btw thank you very much!
Answered by
jacques
Find the limit. If the limit doesn't exist, explain why. lim x approaches 5 x^2-3x-10/x^2-10x+25
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