Newton's Method For Estimating Roots

[ardentmed]

Alright guys, well I just typed this out and gave an in-depth explanation of my reasoning, but I got a server error upon pressing the post button. So I'll start again, but keep it brief:

Here is the question:

Use Newton's method to estimate \(\displaystyle \sqrt[7]{1000}\) to six decimal places. Explain your choise of starting value.

Use Newton's method to find the roots of \(\displaystyle \, 2\cos(x)\, =\, x^4\) to six decimal places. Explain your choise of starting value.

For the first one, I composed the following function to comply with the given root:

f(x) = x^7 - 1000

And used newton's method with x1=3 because (2187)^1/3 is fairly close to the given root.

I then proceeded to use Newton's method with f'(x) = 7(x^6) and computed the following for x6:

x6 = 2.682695795.

I stopped at x6 because both x5 and x6 respectively had similar answers up until the sixth decimal place.


For the second one, I graphed it and knew that x~+/- 1. So I started with x1=1 for the approximation and stopped at x11, which gave me:

x= +/- 1.01395761 for the positive and negative roots respectively.

Am I on the right track?

Thanks in advance.

 

[Ishuda]

Yes, you are on the right track. However, for the second part I stopped at about x4 after starting at x1=1 [since the function is even, you only have to do one side]

 

[ardentmed]

Yes, you are on the right track. However, for the second part I stopped at about x4 after starting at x1=1 [since the function is even, you only have to do one side]

So is that something I would mention along with the final answer if it was a test question, meaning would it matter if I went ahead and found the other side anyway?

Also, why did you stop at x4?

Thanks in advance.

 

[Ishuda]

So is that something I would mention along with the final answer if it was a test question, meaning would it matter if I went ahead and found the other side anyway?

Also, why did you stop at x4?

Thanks in advance.

First, you could (possibly) save time by just working on the one side, then noting the function was even so that there is another root of the opposite sign. I don't think it would matter if you found both sides anyway and didn't mention the function was even.

I stopped at x4 because it was the same as x3 to 6 decimal places:
____x__________f__________f'_______next x
1.00000000 -0.08060461 5.68294197 1.01418361
1.01418361 0.00132640 5.87072783 1.01395767
1.01395767 0.00000034 5.86770102 1.01395761
1.01395761 0.00000000 5.86770024 1.01395761
1.01395761 0.00000000 5.86770024 1.01395761
1.01395761 0.00000000 5.86770024 1.01395761

 

[ardentmed]

First, you could (possibly) save time by just working on the one side, then noting the function was even so that there is another root of the opposite sign. I don't think it would matter if you found both sides anyway and didn't mention the function was even.

I stopped at x4 because it was the same as x3 to 6 decimal places:
____x__________f__________f'_______next x
1.00000000 -0.08060461 5.68294197 1.01418361
1.01418361 0.00132640 5.87072783 1.01395767
1.01395767 0.00000034 5.86770102 1.01395761
1.01395761 0.00000000 5.86770024 1.01395761
1.01395761 0.00000000 5.86770024 1.01395761
1.01395761 0.00000000 5.86770024 1.01395761

Are you sure? For x4, I got 1.014032912 which differs greatly from the terms before and after it by a few decimal values.