# Newton's Methodstuck on a simple problem

1. Homework Statement

Find x(sub3) when x(sub1) is 3 and x^3+3x-5 = 0

2. Homework Equations
Mean Value Theoren Equation with Newton's Method.

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

3. 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.
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

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HallsofIvy
Homework Helper
1. Homework Statement

Find x(sub3) when x(sub1) is 3 and x^3+3x-5 = 0

2. Homework Equations
Mean Value Theoren Equation with Newton's Method.

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

3. 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.
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

Halls' - can you check that second iterate again? - I get something more like 1.38

Integral
Staff Emeritus
Gold Member
I get 1.38 for x3 also.

HallsofIvy
Homework Helper
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?

Integral
Staff Emeritus
Gold Member
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.