Cycloid power series. problem from hell.

Click For Summary
SUMMARY

The discussion focuses on deriving the cycloid power series, specifically identifying the first non-zero terms, the general term, and the interval of convergence. The cycloid is defined parametrically with equations x = a(θ - sin(θ)) and y = -a + a cos(θ). The series expansion is approximated around t = π, yielding terms such as -1/8 (x - π)² and -1/384 (x - π)⁴. The cycloid function satisfies the brachistochrone equation, indicating its significance in physics and calculus.

PREREQUISITES
  • Understanding of parametric equations and their derivatives
  • Familiarity with Taylor series and power series expansions
  • Knowledge of calculus, specifically differentiation and limits
  • Basic grasp of the brachistochrone problem in physics
NEXT STEPS
  • Study the derivation of Taylor series for parametric functions
  • Learn about the convergence of power series and interval determination
  • Explore the applications of cycloids in physics, particularly in mechanics
  • Investigate the relationship between cycloids and the brachistochrone problem
USEFUL FOR

Mathematicians, physics students, and educators interested in advanced calculus, particularly those focusing on parametric equations and series expansions.

mathwiz1234
Messages
2
Reaction score
0
:devil: Can you handle it?

Find the first non-zero terms, the general term for the cycloid power series, and the interval of convergence for the cycloid power series.

cycloid:
y=-a+acos(theta)
x=a(theta)-asin(theta)
 
Physics news on Phys.org
There seem to be a lot of "limits of death" and "problems from hell" today. I'm afraid you must show some work before anyone assists you on your homework questions. Unfortunately, daring the reader to answer for you doesn't work! Further, you should post in the homework forums next time.

Welcome to PF!
 
My bad. Well, I know that the basics for taylor polynomials. for a cycloid,
I know that f(a)+ f'(a)(x-a)+ f''(a)(x-a)^2... will give me the right
answer but I don't know how to do this in parametric mode or what
really do other than what I've listed below. Thanks!

y=-a+acos(theta)
x=a(theta)-asin(theta)

so, dy/dx=-sin(theta)/1-cos(theta)

I also know that d(dy/dx)/dx=d^2y/dx^2
 
Problem from the sky

Ignoring homotheties, let the following cycloid:
x = t - sin t
y = 1 - cos t,
t in [0, 2pi].

From this, we have y' = dy/dx = (dy/dt) / (dx/dt) = (sin t) / (1 - cos t) = ... = cot (t/2), and then y'' = d(y')/dx = ... = -(1/4) [ cosec (t/2) ]^4, ..., "AND SO ON" :smile:

Maybe it will be interesting calculate the series at t = pi. Doing that and putting y^(m) := (d^m)y /(dx)^m, it is easy to see that y^(2n+1)[pi] = 0, and y^(2n)[pi] < 0, n = 1, 2, 3, ... .

The transcendental "cycloid function" y = cyc(x) is aproximately:
cyc(x) =~
2
-1/8 (x-pi)^2
-1/384 (x-pi)^4
-11/92160 (x-pi)^6
-73/10321920 (x-pi)^8
-887/1857945600 (x-pi)^10
-136883/3923981107200 (x-pi)^12
-7680089/2856658246041600 (x-pi)^14
-26838347/124654178009088000 (x-pi)^16
- ... "AND SO ON" :biggrin:

This function satisfies the brachistochrone equation:

y * (1 + y'^2) = 2 (*)​

However not all curves satisfying this equation are cycloids. To see that, think about inserting a horizontal line segment between two arcs of the same cycloid. It will be a solution of the differential equation (*).

Good luck.
 
mathwiz1234 said:
:devil: Can you handle it?

Find the first non-zero terms, the general term for the cycloid power series, and the interval of convergence for the cycloid power series.

cycloid:
y=-a+acos(theta)
x=a(theta)-asin(theta)
The problem is that the phrase "cycloid power series" makes no sense to me! I know what the power series representation for a function is but what function are you talking about? Do you mean power series representations for x and y separately?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
1
Views
2K