Taylor Series Approximation Help

In summary, the problem statement is to find approximate values for y_1 through y_20 with h=.1 for this Initial Value Problem:y'= cosh(4x^2-2y^2)y(0)=14y_1= (x_0-y_0) = 0-(-1) = 1y_2= y_1 + (x_1+y_1)h + (x_1+y_1+1)h^2/2y_3=y_2 + (x_2+y_2)h + (x_2+y_2+1)h^2/2
  • #1
rail1090
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0

Homework Statement



Use the "Three Term" Taylor's approximation to find approximate values y_1 through y_20 with h=.1 for this Initial Value Problem:
y'= cosh(4x^2-2y^2)
y(0)=14

And write a computer program to do the grunt work approximation



Homework Equations





The Attempt at a Solution



y_n=cosh(4x^2-2y^2)
y''= (cosh(4x^2-2y^2)(4/3x^3-2/3y^3)
y_0=14
y_n+1=y_n+h{y_ncosh(4x_n^2-2y_n^2)+h^2/2[cosh(4x_n^2-2y_n^2)(4/3x_n^3-2/3y_n^3)]
I'm pretty sure I'm doing this wrong, and I also have no idea how to go about writing a computer program to do this work
 
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  • #2
In the original problem statement, is that the second derivative, ie,

y'' = cosh(4x^2-2y^2)

or is it the first derivative?
 
  • #3
wow, my professor changed the IVP to:
y'=xy^3 - cos(x)sin(y)
y(0)= -1
still using h= .1 and looking for y_1 to y_20, as well as writing the program. Apparently the original blew up too quickly.
 
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  • #4
OK. Now to start, what were the various expressions that your prof gave you for approximations to first derivatives. Hint: a two term approx might be

[tex]y'_{+} = \frac{y_{n+1}-y_{n}}{h} + O(h)[/tex]

then again, it might be

[tex]y'_{-} = \frac{y_{n}-y_{n-1}}{h} + O(h)[/tex]

Do you remember how your prof established the three point approx?
 
  • #5
Thanks for the help Theo, but my problem now lies in the programming bit of the problem. I think I've got the steps before it figured out. Please correct me if I'm wrong.

y''= x3(y^2)y' + y^3 -[(-sin(x)siny) + (cos(x)cos(y)y')]
= 3x(y^2)y' + y^3 +[cos(x)cos(y)y' - sin(x)sin(y)]
y'= (0)(-1)^3- [cos(0)sin(-1)] = sin(-1)
y''= 3(0) + (-1)^3 = [(cos(0)cos(-1)sin(-1)) - (sin(0)sin(-1))]
= -1 + sin(-1)
y_1= (x_0-y_0) = 0-(-1) = 1
y_2= y_1 + (x_1+y_1)h + (x_1+y_1+1)h^2/2
= 1 + (.1+1)(.1) + (.1+1+1).005 = 1.1205
y_3=y_2 + (x_2+y_2)h + (x_2+y_2+1)h^2/2
=1.1205 +(.2+1.1205)(.1) + (.2+1.1205+1).005 = 1.2641525
...

You see, now I have no idea how to go about writing a program for this approximation. What sort of program do I use, and where can I find it? Can I just download one? Is it very complicated to program this?
 
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FAQ: Taylor Series Approximation Help

1. What is a Taylor series approximation?

A Taylor series approximation is a mathematical technique used to approximate a function using a series of polynomials. It is based on the idea that any continuous function can be approximated by an infinite series of polynomials.

2. How is a Taylor series approximation calculated?

A Taylor series approximation is calculated using the derivatives of a function evaluated at a specific point. The derivatives are then used to construct a polynomial series that approximates the original function. The more terms included in the series, the more accurate the approximation will be.

3. What are the applications of Taylor series approximation?

Taylor series approximation has various applications in mathematics, physics, and engineering. It is used to approximate functions that are difficult to solve, to evaluate integrals and derivatives, and to model complex physical systems.

4. How accurate is a Taylor series approximation?

The accuracy of a Taylor series approximation depends on the number of terms included in the series. The more terms included, the closer the approximation will be to the original function. However, the approximation may not be accurate for all values of x, as it is only valid in a specific interval around the point of approximation.

5. Can a Taylor series approximation be used for any function?

No, a Taylor series approximation can only be used for functions that are differentiable at the point of approximation. If a function is not differentiable, then the Taylor series will not converge and cannot be used to approximate the function.

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