- #1
mathwiz1234
- 3
- 0
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)
Cycloid power series. problem from hell.
- Thread starter mathwiz1234
- Start date
-
- Tags
- Power Power series Series
In summary: If so, how do you do that? In summary, the cycloid power series is a series for a function that is defined as the sum of a certain power series for x and another power series for y. The convergence interval for this series is given by the equation y'=-a+acos(theta). However, the series cannot be calculated in parametric form and must be evaluated analytically.
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)
- #2
cristo
Staff Emeritus
Science Advisor
- 8,146
- 74
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!
Welcome to PF!
- #3
mathwiz1234
- 3
- 0
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
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
- #4
wackensack
- 4
- 0
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" :rofl:
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"
This function satisfies the brachistochrone equation:
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.
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" :rofl:
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"
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.
- #5
HallsofIvy
Science Advisor
Homework Helper
- 42,989
- 975
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?mathwiz1234 said:
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)
1. What is a Cycloid power series?
A Cycloid power series is a mathematical function that is used to model the shape of a cycloid, which is a curve traced by a point on the circumference of a circle as it rolls along a straight line.
2. How is a Cycloid power series calculated?
A Cycloid power series is calculated using a power series expansion, where each term in the series is a polynomial function of increasing powers of x. The coefficients of the polynomial are determined using the derivatives of the function at a specific point.
3. What is the significance of the "problem from hell" in relation to Cycloid power series?
The "problem from hell" refers to the difficulty in calculating the coefficients of a Cycloid power series. This is because the derivatives of the function at the specific point can be very complex and difficult to compute, making the calculation of the coefficients a challenging task.
4. What are some real-world applications of Cycloid power series?
Cycloid power series have various applications in physics, engineering, and mathematics. They are used to model the motion of pendulums, rolling objects, and other mechanical systems. They are also used in signal processing, image recognition, and data compression.
5. Are there any limitations to using Cycloid power series?
While Cycloid power series can accurately model certain curves, they have limitations in representing more complex shapes. They also have a limited range of convergence, meaning that they may not provide accurate results outside of a certain range of values.
Similar threads
-
Calculus
-
Calculus
-
Calculus
-
Introductory Physics Homework Help
-
Calculus