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Newton-Rhapson Formula

Hi,

Please can I have some help with this question

I know the formula is

Xn - (f (x) / f'(x))

I've found f'(x) to be 6x^2 + 2x as required

How do I finish the question please?

So what trouble are you having?
You know the function and it's derivative.
Just sub them in.

Reply 3

M.Johnson2111

I get:

Xn+1 = Xn - ((2x^3 + x^2 - 1) / (6x^2 + 2x))

I know I'm missing the Xn, how do i find it?

Or what am i doing wrong please?

Reply 4

mqb2766

Original post by M.Johnson2111

I get:

Xn+1 = Xn - ((2x^3 + x^2 - 1) / (6x^2 + 2x))

I know I'm missing the Xn, how do i find it?

Or what am i doing wrong please?

Both the function and it's derivative are evaluated at x_n

x_{n+1} = x_n - f(x_n)/f'(x_n)

You know x_n, you use the formula to iterate to x_{n+1}.

Sorry, which is Xn?

Original post by M.Johnson2111

Sorry, which is Xn?

Don't understand what you're asking?
Both the function and it's derivative are evaluated at x_n.
https://en.m.wikipedia.org/wiki/Newton%27s_method

Reply 7

M.Johnson2111

Sorry for the confusion

What is the value for Xn?

Or do I sub Xn into the equation?

Please

Reply 8

mqb2766

Original post by M.Johnson2111

Sorry for the confusion

What is the value for Xn?

Or do I sub Xn into the equation?

Please

Yes, you sub x_n into the function and it's derivative.
You're evaluating both when x=x_n, i.e. when both are evaluated at the current estimate.

(edited 3 years ago)

Reply 9

M.Johnson2111

Original post by mqb2766

Yes, you sub x_n into the function and it's derivative.
You're evaluating both when x=x_n, i.e. when both are evaluated at the correct estimate.

So then

Xn+1 = Xn - ((2Xn^3 + Xn^2 -1) / (6Xn^2 + 2Xn))

Reply 10

mqb2766

Original post by M.Johnson2111

So then

Xn+1 = Xn - ((2Xn^3 + Xn^2 -1) / (6Xn^2 + 2Xn))

Yes, so simplify/combine the right hand side.

Reply 11

DFranklin

Famous Rap by Newton-Rhapson:

Yo, Yo, Here's How We Go,
Take a function f and diff it nice and slow,
Evaluate at a and divide by f prime
Subtract this off to get a new a that's fine.

Reply 12

M.Johnson2111

Original post by mqb2766

Yes, so simplify/combine the right hand side.

How would I do this please?

Reply 13

mqb2766

Original post by M.Johnson2111

How would I do this please?

Put them over the common denominator.
Tbh, you could look at the desired answer and do a bit of method spotting?

Reply 14

M.Johnson2111

Original post by mqb2766

Put them over the common denominator.
Tbh, you could look at the desired answer and do a bit of method spotting?

Do you mean sub x =1?

Reply 15

mqb2766

Original post by M.Johnson2111

Do you mean sub x =1?

No idea what you mean?
I've got the feeling that you're not really understanding something but not sure what.

Reply 16

M.Johnson2111

If i sub x=1 into dy/dx, it proves that it is a stationary point because the dy/dx=0

Reply 17

mqb2766

Original post by M.Johnson2111

If i sub x=1 into dy/dx, it proves that it is a stationary point because the dy/dx=0

The only question I've seen is to prove the x_n -> x_{n+1} formula.
Is this a different question part?

Tbh, it looks like your other thread?

(edited 3 years ago)

Reply 18

the bear

Original post by DFranklin

Famous Rap by Newton-Rhapson:

Yo, Yo, Here's How We Go,
Take a function f and diff it nice and slow,
Evaluate at a and divide by f prime
Subtract this off to get a new a that's fine.