Finding X-Intercepts for f(x)=(x/2)-sin(x)

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Homework Help Overview

The discussion revolves around finding the x-intercepts of the trigonometric equation f(x)=(x/2)-sin(x). Participants are exploring methods to identify these intercepts, particularly within the constraints of their current coursework, which has not yet introduced Newton's method.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the known intercept at (0,0) and express uncertainty about finding additional roots without using Newton's method. Some suggest that a graphing calculator might be an alternative approach. Others question the necessity of numerical methods and explore the implications of the problem's constraints.

Discussion Status

The conversation is ongoing, with various methods being considered. Some participants have shared their experiences with Newton's method, while others are exploring alternative approaches, such as graphing. There is recognition of the need for a complete problem statement to facilitate collaborative understanding.

Contextual Notes

Participants note that the problem is to be solved within the interval (0, 3π), which adds complexity to the search for roots. There is also mention of the potential oversight regarding the use of graphing calculators in the context of the homework rules.

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Homework Statement


I need to find an x-intercept for this trigonometric equation, f(x)=(x/2)-sin(x). I already know that one of the intercepts in (0,0) because that is a y-intercept. I figure i could use Newton's method but this problem is in a section before Newton's method is introduced.


Homework Equations





The Attempt at a Solution


I figured that the relation (sin(x)/x)=1 could possible be useful at first but now I know it's not. I know it has to be just one or two steps also. Help appreciated.
 
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kevinnn said:

Homework Statement


I need to find an x-intercept for this trigonometric equation, f(x)=(x/2)-sin(x). I already know that one of the intercepts in (0,0) because that is a y-intercept. I figure i could use Newton's method but this problem is in a section before Newton's method is introduced. I figured that the relation (sin(x)/x)=1 could possible be useful at first but now I know it's not. I know it has to be just one or two steps also. Help appreciated.

If you are after an ##x## intercept instead of all of them, you are done because (0,0) is both an x and y intercept. Otherwise there is a root a bit less than 2 you will need Newton's method to find. And the function is odd, so...
 
Sorry I forgot to put the interval, were only suppose to look at (0,3pi). This definitely brings up confusion for me. I thought I could only find the root using Newton's method but that is in a later section. But that is the only way right?
 
kevinnn said:
Sorry I forgot to put the interval, were only suppose to look at (0,3pi). This definitely brings up confusion for me. I thought I could only find the root using Newton's method but that is in a later section. But that is the only way right?

You will need a numerical method of some sort. Maybe you are allowed to use a graphing calculator if you haven't studied Newton's method yet.
 
I assume it must be a graphing calculator problem too. I do know how to use Newton's method,it just didn't make sense for someone that has studied only up to the chapter the problem is on. They must have just left the graphing calculator ok symbol out. But i used Newtons method and got the intercept, only to a couple decimal places though.
 
You told us what you need to do, we need a complete problem statement so we can all work on the same thing.

The title of the thread is "curve sketching", Have you sketched the curve? What tools for that do you have?
 
Ohhh the problem is done and correctly sketched. It was a poor title now that I look at it. That one question was the part of the curve sketching I couldn't do without Newtons method or a calculator, that is what my question was. Sorry.
 
Try this, you have:

[tex]F(x)= \frac x 2- Sin(x)[/tex]

If you want the roots of this set [tex]F(x)=0= \frac x 2- Sin(x)[/tex]

so
[tex]x=2 sin(x)[/tex]

From your sketch you should know the number of roots and their approximate location. Make a rough estimate of a root, plug it into the RHS of the last equation get a new x, repeat several times.
 
Last edited:
I like that. I just used straight forward Newton's method. I will work with this though, it makes for a more complete knowledge of what is going on with a graph.
 
  • #10
Newton's method may converge faster then the above fixed point method, but it does converge after about 10 -12 iterations with an initial guess of .5. An initial guess of 2 took about 6 iterations.
 

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