Newton's Methodstuck on a simple problem

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Homework Statement



Find x(sub3) when x(sub1) is 3 and x^3+3x-5 = 0
Give your answer to 4 correct decimal places.

Homework Equations


Mean Value Theoren Equation with Newton's Method.

x(sub n+1) = x(sub n) - [(f(x)/f'(x)]


The Attempt at a Solution



Okay, so we have to find x(sub2) before x(sub3).

x(sub2) = 3 - [(3^3 + 3(3) - 5)/(3(3)^2 + 3)

I get x(sub 2) to be ~ 1.966666667

Substituting that in the equation, I keep calculating x(sub3) to be ~1.010865585, which would be 1.0109 when rounded to 4 decimal places.

The homework program (WebAssign) that I am using is telling me that answer is wrong. I thought it might be a rounding error, so I tried 1.0108 and 1.0110, both of which were also wrong.

I don't know what I am doing wrong. I have solved this the same way as the ones prior to it, and they came out correct.

Thanks for any and all help.
 
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syphonation said:

Homework Statement



Find x(sub3) when x(sub1) is 3 and x^3+3x-5 = 0
Give your answer to 4 correct decimal places.

Homework Equations


Mean Value Theoren Equation with Newton's Method.

x(sub n+1) = x(sub n) - [(f(x)/f'(x)]


The Attempt at a Solution



Okay, so we have to find x(sub2) before x(sub3).

x(sub2) = 3 - [(3^3 + 3(3) - 5)/(3(3)^2 + 3)

I get x(sub 2) to be ~ 1.966666667
Okay, that's what I get.

Substituting that in the equation, I keep calculating x(sub3) to be ~1.010865585, which would be 1.0109 when rounded to 4 decimal places.
That's not what I get. Check your arithmetic again. It's more like 2.5.

The homework program (WebAssign) that I am using is telling me that answer is wrong. I thought it might be a rounding error, so I tried 1.0108 and 1.0110, both of which were also wrong.

I don't know what I am doing wrong. I have solved this the same way as the ones prior to it, and they came out correct.

Thanks for any and all help.
 
Halls' - can you check that second iterate again? - I get something more like 1.38
 
I get 1.38 for x3 also.
 
Oh, of course, I subtracted from the original 3 instead of x1! Still be a good idea to check his arithmetic though, wouldn't it?
 
HallsofIvy said:
Oh, of course, I subtracted from the original 3 instead of x1! Still be a good idea to check his arithmetic though, wouldn't it?

Yep, he has a mistake, it is not ~1.01
 
...Still be a good idea to check his arithmetic though, wouldn't it?

Yes, absolutely.
 
Thanks everybody. It did turn out to be ~1.38...

Not sure what I was doing wrong before, but oh well. I guess I will just have to be more careful when test day comes.

thanks again.
 
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