Newton-Raphson Method [Part E]

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Sakurazaki

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Please don't misunderstand I'm looking for guidance on how to do this problem and the various steps attributed to it. Not trying to be a slacker. I'll be posting work as I go along, also editing the main topic and this post as I progress through the problem.

math1-1.jpg


math2.jpg


For part A this is the work that I have:
math1-2.jpg

dy/dx being the slope of the tangent line at x0y0
Sounds simple enough... but then x1,0 is put in for part b...
Is this correct so far?
 
Sakurazaki said:
... Is this correct so far?
Click to expand...


Hello Sakurazaki:

It looks like a good start for part (A), but it's not finished.

I'm assuming that you used the symbol b in your equation to represent the y-intercept of the tangent line to h(x) at x = x[sub:ibdldkzl]0[/sub:ibdldkzl].

We could express this value in terms of the slope, x[sub:ibdldkzl]0[/sub:ibdldkzl], and y[sub:ibdldkzl]0[/sub:ibdldkzl], but it's less work when applying Newton's Method to leave the equation of this line in point-slope form. The reason for this is that the y-intercept is not important to Newton's Method.

Also, I would suggest using prime notation versus Leibnitz notation (and the vertical evaluation bar) for the slope. The exercise gives you this slope as h'(x[sub:ibdldkzl]0[/sub:ibdldkzl]).

In other words, we have the slope h'(x[sub:ibdldkzl]0[/sub:ibdldkzl]) and the coordinates of a point (x[sub:ibdldkzl]0[/sub:ibdldkzl], y[sub:ibdldkzl]0[/sub:ibdldkzl]), so we can write the equation for this line in point-slope form.

Cheers,

~ Mark :)



PS: Function f in the first paragraph of the exercise looks like a typographical error, to me. I assume, also, that you mean part (B) when you type "part b", as opposed to some statement about the part of the equation that you wrote using the symbol b.
 
Re: Newton-Raphson Method [Part A]

So I would basically solve part a as:

y-yo = h'xo(x-xo)

Correct?

And then for part b I'd assume that I plug in those coordinates to get

0-yo = h'xo(x1-xo)?
 
Sakurazaki said:
So I would basically solve part a as:

y-yo = h'xo(x-xo) ? Yes, but please use proper function notation.

y - yo = h'(xo) (x - xo)

Personally, I would take the typing of math notation one step further.

y - y_o = h'(x_o) * (x - x_o)

Now, having said all of this, I just noticed that they changed from m = h'(x_o) back to m = m for part (A).

Excuse me, but I previously missed this fact. Therefore, the correct result for part (A) follows.

y - y_o = m * (x - x_0)

:roll:

And then for part b I'd assume that I plug in those coordinates ...

This is a good start for part (B), since it carries out the first instruction given, but it's only a start.

Read the second instruction given for part(B), and do that, too.
Click to expand...
 
Re: Newton-Raphson Method [Part A]

Ok, thank you very much.

As for part c...I'm totally lost. I know it's a proof problem somewhat. But not sure what to do. It's referring to line x_2 and I think it wants me to do something similiar to part b's second instruction, however I'm not too sure how I should go about this.

A pointer to get me started off might help.
 
Re: Newton-Raphson Method [Part A]

Sakurazaki said:
Ok, thank you very much.

As for part c...I'm totally lost. I know it's a proof problem somewhat. But not sure what to do. It's referring to line x_2 and I think it wants me to do something similiar to part b's second instruction, however I'm not too sure how I should go about this.

A pointer to get me started off might help.
Click to expand...

Hint:

\(\displaystyle h'(x_1) \, = \, slope \, of \, the \, tangent \, line \, at \, (x_1,f(x_1)) \, = \, \frac{y_1}{x_1 - x_2}\)
 
Re: Newton-Raphson Method [Part C]

D: Well I'm rather stuck on Part C. ....can someone give me the answer and explain it to me? I keep thinking that the x_2=x_1 ....etc (the equation that we're supposed to explain why it works, is actually just the equation for the theorem. (I'm not sure how to phrase it.) So I think I was about to just explain what are the parts of the theorem and how they work to get the answer.... (I'm making no sense.)
 
Re: Newton-Raphson Method [Part C]

Part c:

From the point-slope equation of a line at y=f(x) at the initial approximation, \(\displaystyle y-y_{1}=m(x-x_{1})\),

we have \(\displaystyle y-f(x_{1})=f'(x_{1})(x-x_{1})\)..........[1]

If \(\displaystyle f'(x)\neq 0\), then this line is not parallel to the x-axis and consequently it crosses the x-axis at some point \(\displaystyle (x_{2},0)\).

Sub the coordinates of this point into [1] and solve for \(\displaystyle x_{2}\).
 
Re: Newton-Raphson Method [Part C]

Ah....sorry I still don't understand it. If you could explain it in some other way like step by step, it'd be appreciated.
 
Re: Newton-Raphson Method [Part A]


Hint:

\(\displaystyle h'(x_1) \, = \, slope \, of \, the \, tangent \, line \, at \, (x_1,f(x_1)) \, = \, \frac{y_1}{x_1 - x_2}\)

\(\displaystyle h'(x_1) \, = \, \frac{y_1}{x_1 - x_2}\)

\(\displaystyle h'(x_1) \, = \, \frac{h(x_1)}{x_1 - x_2}\)

\(\displaystyle x_1 - x_2 \, = \, \frac{h(x_1)}{h'(x_1)}\)

\(\displaystyle x_2 \, = \, x_1 \, - \, \frac{h(x_1)}{h'(x_1)}\)

This is simple algebraic manipulation - taught to you somewhere in sixth grade.
 
Re: Newton-Raphson Method [Part C]

Ah, I get it now, I got thrown off and forgot (which I'm sure happens to all of us ^^;) . Thanks.
 
For part E, number 1...

What do they mean by show graphically that if the starting point....a set of convergent iterations is obtained?

Edit: Nvm, I got it. I'm just confused now on the 2nd part of part E, as in how to do it.
 
My interpretation is:

The equation obtained in (c) - need to be shown graphically.

In my opinion, it has already been shown in the accompanying graphics of the problem.
 
This is what I have:
graph1.jpg
Is this correct so far?

The original equation is in red, and the derivative of the equation is in blue. What should I do next?
 
You do not need to plot h'(x).

Study the problem and the accompanying graphics - carefully, slowly and several times over. That is what you would have to do for h(x) = 2 - 1/(x^3).

You'll notice in the graphics of the problem statement - h'(x) was not plotted. They showed the tangent lines (whose slopes are h'(x)).
 
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