Question

Newton's method to find all roots of the equation correct to six decimal places. (Enter your answers as a comma-separated list.)
6 cos x = x + 1

Answer 1

And "Newton's Method" would be... What?

\(f(x) = 6\cos(x) - x - 1\)

The challenge is to find \(f(x) = 0\)

\(f'(x) = -6\sin(x) - 1\)

Okay, now build it!

Answer 2

\[x-((6\cos(x)-x-1) \div (-6\sin(x)-1))\]

isnt that it

Answer 3

@tkhunny

Answer 4

oh your still here cool

Answer 5

Be sure. Is it \(\dfrac{f(x)}{f'(x)}\;or\;\dfrac{f'(x)}{f(x)}\)?

Answer 6

f(x)/f'(x)

Answer 7

i got that part, but then i dont really understand the process that comes after

Answer 8

why should i start off with an initial value of 1, why not 0?

Answer 9

Okay, now we need to know where to start. What do you suggest?

The amplitude of the cosine is 6 and the slope of the line is +1, so I would suspect that ALL roots are in [-7,5], otherwise the linear function is outside the Range of the Cosine. Make sense?

Answer 10

well not really how did you get -7,5

Answer 11

i understand the amp is 6 and slope is 1

Answer 12

It's a linear function, x+1. For what values does that land in [-6,6], where the cosine lives?

Answer 13

oh ok
well wouldnt that be -5,7

Answer 14

No.
-5 + 1 = -4
7 + 1 = 8

Yes
-7 + 1 = -6
5 + 1 = 6

Answer 15

ohhhhh
ok

Answer 16

so do i plug those into my formula

Answer 17

Now, to your question. You do NOT want to start anywhere the derivative is very close to zero. This will shoot you off somewhere and possible never come back. Look what an initial x = 0 does.

0
5
5.904173
6.994311
6.292835
5.070488

It immediately goes past out limit of 5 and stays out there until it goes insane and starts jumping all over.

f'(0) = -1 -- That could be steeper.
f'(1) = -6.049 -- Oh, I like that!

Answer 18

so i have to look at the derivative formula and imagine what number would be far enough away from zero?

Answer 19

because if the derivative is zero its a flat line?

Answer 20

Fair enough. A quick sketch of f'(x) wouldn't hurt.

A derivative of zero is very bad news. That will never work out.

Answer 21

but one is around half way between -7 and 6?

Answer 22

So? Stay away from it. It looks like -6.5, -3, 0, 3, and 6 would be bad places to start. Start somewhere else.

Answer 23

Start at x = 1. What do you get?

Answer 24

1.24181

Answer 25

wait no sorry

Answer 26

Should be 1.205298

Answer 27

got it

Answer 28

so now i assighn that to x in my calculator
??

Answer 29

Yes. And the next iteration...

Answer 30

Note: I'm not rounding intermediate results to 6 decimal places. My results may differ a little.

Answer 31

1.19608

Answer 32

Okay. I got 1.196090, but that's fine. Probably one or two more iterations.

Answer 33

that means keep pushing enter on my calculator?
lol

Answer 34

That is one of the advantages of Newton's Method. It is "Self-Correcting". If we're off by some tiny amount of rounding, the method will just fix the error. It's pretty sweet.

Maybe. It depends on how your function is set up. If you can do that, let's see the next two iterations.

Answer 35

i was supposed to get 3 answers i guess, i guess the only way i could know there were 3 roots would be to graph it?

Answer 36

No, that is not the only way. I like to refer to "Blanket Bombing".

We created a Domain. Why not use all of it? Hunt around a little?

Answer 37

-7 leads to a failure to converge
-6 leads to the middle root
-5 leads to the left-most root
-4 leads to the left-most root
-3 leads to a failure to converge
-2 leads to the middle root
-1 leads to the middle root
0 leads to a failure to converge
1 leads to the right-most root
2 leads to the right-most root
3 leads to the middle root
4 leads to the left-most root
5 leads to a failure to converge

You need to be satisfied that you found all of them. I find behavior around 2, 3, and 4 a little peculiar. I might be feeling a need to investigate around there a little more before I was completely satisfied. Really, though, with this nice little investigation, we keep tripping over the same three roots. Time to call it solved and move on.