math question help please`

[Deathslayer7]

I'm helping a friend of a friend with calculus and was wondering if im doing this right.

The question is this:

Use Newton's method to approximate the critical number of the function f(x)= x*sin(x). Use 2 as a first estimate. Perform enough iterations to get a repetition at the ten thousandths digit. Remember use radians.

so newtons method is x_n minus f(x_n)/ f ' (x_n)

so taking the derivative you get

f ' (x) = sin(x) + x*cos(x)

now what I told her to do is take the number 2 and convert to radians and then from there do newtons method.

so 2*pi/180 = pi/90 and that is our first estimate. From there it is just plug and chug right?

[KenderWizard]

I may not be interpreting this correctly, but if you're looking for the critical point of the function, that's where f '(x)=0

Then x_n+1 = x_n - f '(x_n)/f ''(x_n)

You sub in x_0=2, to find x_1, and from there repeat the process.

But, as I say, I might be wrong. I think my Newton-Raphson method for finding critical points is the same as your Newton method for finding critical numbers.

Hope that helps anyway!

[Deathslayer7]

thing is it says to use radians.

[Tirian]

It's up to you whether you start with 2 degrees or 2 radians. I think that 2 radians makes more sense, myself, but you'll probably wind up with the same answer either way.

The thing to keep in mind when they say "use radians" is that you have the radians setting on the calculator when you press the sin button.

[Deathslayer7]

well we got 0 as the answer, which i dont think is right.

the numbers in order for x_n

.03491
.01743
.00882
.00428
.00194
.00097
.00049

thats as far as we went so i believe it goes to 0

[EnderChant]

Looks like you're on the right track.


...Use Newton's method to approximate the critical number of the function f(x)= x*sin(x).
While KenderWizard is right, the critical number of a function is where the first derivative is equal to zero, Newton's method is a way to find the roots of a function, not its critical points. So you're probably looking for the zeroes of the function itself, where ƒ(x) = 0.


...Use 2 as a first estimate. Perform enough iterations to get a repetition at the ten thousandths digit. Remember use radians.
This is just telling you where to begin your approximation. Start at x = 2 radians, and then work from there.


...

so 2*pi/180 = pi/90 and that is our first estimate. From there it is just plug and chug right?

This is kind of weird; I assume they mean 2 radians, or ~114º.

EDIT: No, I think your answer of zero is correct. Take a look at the function :
x*sin(x) (http://www.wolframalpha.com/input/?i=xsin(x))

[Tirian]

Just to be clear, you definitely want 2 to be your starting point and not pi/90, as the answer you'll be looking for is 2.02876. This is the graph (http://www.wolframalpha.com/input/?i=x*sin%28x%29) of the function you're working with, and you would seem to be looking for the local maximum up there and not the local minimum at the origin.

[Deathslayer7]

ok ill her try it again with the starting point x=2

[KenderWizard]

As general advice, if she has notes from her class (I am presuming she isn't doing this for fun, or to annoy you!) it might help to look at them. Often a teacher gives similar examples to tricky questions, or frequent questions, and this one seems kind of tricky. Although I don't know what level she's at, it would be a tricky question for the last or second last school year before college here.

Or, likewise, if you or someone else has notes or a good maths book, you might be able to find similar examples and work out how to deal with that 2 from them.

[Deathslayer7]

i understand how to do it. I was just confused on the wording. but we got it now. Thanks. :smallsmile:

[KenderWizard]

Yay! Well done!

[MarshellaSmith]

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