Newton's Methodstuck on a simple problem

  • Thread starter syphonation
  • Start date
In summary, the conversation discusses finding the value of x(sub3) using Newton's Method and the Mean Value Theorem when given x(sub1) and a cubic equation. The attempt at a solution involves finding x(sub2) first, which is found to be ~1.966666667. However, there is a mistake in the arithmetic and the correct value is ~1.38. The conversation ends with the individual expressing gratitude for the help and acknowledging the importance of being careful during tests.
  • #1
syphonation
19
0

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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  • #2
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.
 
  • #3
Halls' - can you check that second iterate again? - I get something more like 1.38
 
  • #4
I get 1.38 for x3 also.
 
  • #5
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?
 
  • #6
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
 
  • #7
...Still be a good idea to check his arithmetic though, wouldn't it?

Yes, absolutely.
 
  • #8
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.
 

What is Newton's Method?

Newton's Method is a mathematical algorithm used to find the roots of a function. It involves making an initial guess and then using successive approximations to get closer to the actual root.

How does Newton's Method work?

Newton's Method works by using the slope of a function to estimate where the root of the function might be. It then uses this estimate to make a better approximation, and continues this process until it reaches a satisfactory approximation of the root.

What types of problems can Newton's Method be used for?

Newton's Method can be used to find the roots of any continuous function. It is commonly used in calculus to solve equations that cannot be solved using algebraic methods.

What are some advantages of using Newton's Method?

One advantage of Newton's Method is that it can converge to the root relatively quickly, especially compared to other root finding methods. It is also fairly simple to implement and can be used to find both real and complex roots of a function.

Are there any limitations to using Newton's Method?

Yes, there are some limitations to using Newton's Method. It may not always converge to the correct root, or it may not converge at all if the initial guess is too far from the actual root. Additionally, it requires knowledge of the derivative of the function, which may be difficult to obtain in some cases.

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