Question

20) Use Newton's method to find the negative fourth root of 5 by solving the equation
x^4 - 5 = 0. Start with x1 = -1 and find x2.

Answer 1

\(f(x) = x^{4} - 5\)

\(f'(x) = 4x^{3}\)

\(x_{n+1} = x_{n} - \dfrac{f(x_{n})}{f'(x_{n})}\)

Go!!

Answer 2

You can do it.

You have \(x_{1} = -1\).

I defined \(f(x)\;and\;f'(x)\).

Calculate \(f(x_{1})\;and\;f'(x_{1})\)

Then what?

Answer 3

f'(-1)=-4

Answer 4

Please don't change your mind after you are already correct. No guessing.

\(f(-1) = (-1)^{4} - 5 = 1 - 5 = -4\)

Answer 5

Now, build \(x_{2}\).

Answer 6

how do i

Answer 7

Seriously? If you can't read a formula, you are going to suffer in this course. Go look at my first post, again, and use n = 1.

Answer 8

\(x_{1} = −1 \)

\(f(−1) = −4 \)

\(f′(−1) = −4 \)

\(x 2 =−1−\dfrac{-4}{-4} = \) ??

Use the formula. Don't make it up.

Answer 9

x2=-2

Answer 10

so do i need to do something with -2?

Answer 11

Awesome. Now find \(f(x_{2})\;and\;f'(x_{2})\) and do it all again to find \(x_{3}\).

Answer 12

but is that part of solving the problem?

Answer 13

???

What does this mean to you? \(x_{n+1} = g(x_{n})\)? It is an iterative process. You just keep doing it! That's why one index leads to the next.

Answer 14

f(-2)=-21
f'(-2)=-32
im clueless to how many iteration are needed

Answer 15

Did you miss the part where I said, "You just keep doing it"? It is an iterative process. Each iteration should bring you closer to the solution. You quit when you believe you are close enough. Now go! Iterate!

Answer 16

but i found x2 right? -2

Answer 17

Now go find \(x_{3}\).

Answer 18

im trying to figure what to sub into equation

Answer 19

Why? Substitute \(x_{2}\). What other candidate is there?

Answer 20

-2-1=-2-(-21/-32)

Answer 21

What is that "-2-1" on the left-hand side? That should say \(x_{3}\).

Answer 22

i followed the formula

Answer 23

No. If you had followed the formula, with n = 2, the left hand side is \(x_{3}\).

Also, \(f(-2) = 11 \ne -21\)

Answer 24

-16-5=-21

Answer 25

where is this n coming from?

Answer 26

-2(-2)=4
4(-2)=-8
-8(-2)=16

Answer 27

Is was n = 1 when we started. This leads to 1+1 = 2. This leads to 2+1 = 3. This leads to 3+1 = 4. etc. This is why \(n_{1} = -1\) was supplied in the beginning.

Answer 28

u started using n but it was x1

Answer 29

-53/32

Answer 30

You don't seem to know what an index is.

\(x_{1} = -1\)

\(x_{2} = x_{1} - \dfrac{f(x_{1})}{f'(x_{1})} = -1 - \dfrac{-4}{-4} = -1 - 1 = -2\)

\(x_{3} = x_{2} - \dfrac{f(x_{2})}{f'(x_{2})} = -2 - \dfrac{11}{-32} = -1.65625\)

\(x_{4} = x_{3} - \dfrac{f(x_{3})}{f'(x_{3})} = -1.65625 - \dfrac{2.5249490737915}{-18.1734619140625} = -1.51731394666402\)

Are we seeing it, yet? The index is just a counter. There is no evaluation of the index. It is just counting for us.

Answer 31

Yes! \(x_{3} = -53/32 = -1.65625\)

Answer 32

Now find \(x_{4}\)

Answer 33

well it seems you have given x4 even though all the numbers are confusing

Answer 34

so i have to find the negative fourth root but i have to fin x2