Pharaoh's Taylor series question from Yahoo Answers

In summary, the Taylor series expansion method was used to approximate the solution for van der pol's equation. Using the first three non-zero terms, the approximate solutions for y(0.1) and y'(0.1) were calculated, and then used to find the approximate solution at x=0.2.
  • #1
CaptainBlack
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Part 1 of Pharaoh's Taylor series and modified Euler question from Yahoo Answers

consider van der pol's equation
y" - 0.2(1-y^2)y' + y = 0 y(0)=0.1 y'(0)=0.1
1)
You are asked to find the approximate solution for this problem using the Taylor series
method. Your expansion should include the first three non-zero terms and you should
work to six decimal places accuracy. First find the approximate solutions for both y (0.1)
and y’(0.1) using the first three non-zero terms of Taylor series expansion for each
function and then use this information to calculate the approximate solution at x = 0.2.
The Taylor series expansion about \(t=0\) is of the form:
\(y(t)=y(0)+y'(0)t+\frac{y''(0)t^2}{2}+.. \)​
We are given \(y(0)\) and \(y'(0)\) in the initial condition, and so from the equation we have:
\(y''(0) = 0.2(1-(y(0))^2)-y(0)=0.2(1-0.1^2)-0.1=-0.0802\)​
So the Taylor series about \(t=0\) is:
\(y(t)=0.1+0.1t-0.0401t^2+... \)​
and using the first three terms of this we have \(y(0.1)\approx 0.109599\), Also:
\(y'(t)=0.1-0.0802t+...\)​
and so \(y'(0.1) \approx 0.09198\)Now the Taylor expansion about \(t=0.1\) is:

\(y(t)=y(0.1)+(t-0.1)y'(0.1)+\frac{(t-0.1)^2y''(0.1)}{2}+...\)​
where \(y''(0.1)=0.2\left(1-(y(0.1))^2\right)y'(0.1)-y(0.1)=-0.08178796\).So:

\(y(0.2)\approx 0.109599+0.1\times 0.0198-0.1^2\times 0.8178796 = 0.108789 \) to 6 six DP​
 
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  • #2
.2)
Next, the modified Euler method can be used to find the approximate solution for this problem. The modified Euler method is given by:

\(y_{n+1}=y_n+hf(x_n+\frac{h}{2},y_n+\frac{h}{2}f(x_n,y_n))\)

where \(f(x,y)=y"-0.2(1-y^2)y'+y\).

Using \(h=0.1\) and starting with \(y_0=0.1\), we can calculate the values of \(y_1\) and \(y_2\) as follows:

\(y_1\approx 0.1+0.1\left(0.1+\frac{0.1}{2}\left(0.1-0.2(1-0.1^2)0.1+0.1\right)\right)\approx 0.109625\)

\(y_2\approx 0.109625+0.1\left(0.2+\frac{0.1}{2}\left(0.2-0.2(1-0.109625^2)0.2+0.109625\right)\right)\approx 0.10961\)

So, the approximate solution at \(x=0.2\) using the modified Euler method is \(y(0.2)\approx 0.10961\). This is slightly different from the value obtained using the Taylor series method, but both methods provide a good approximation to the actual solution.
 

Related to Pharaoh's Taylor series question from Yahoo Answers

1. What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms, with each term being a polynomial of increasing degree. It is used to approximate a function and can provide a better understanding of the behavior of the function at a specific point.

2. Who was Pharaoh and what does he have to do with a Taylor series?

Pharaoh was the title given to rulers of ancient Egypt. In this context, Pharaoh's Taylor series refers to a hypothetical mathematical problem involving a ruler named Pharaoh and a tailor that has become a popular question on online forums and discussion boards.

3. What is the Pharaoh's Taylor series question from Yahoo Answers?

The Pharaoh's Taylor series question from Yahoo Answers is a mathematical problem that asks how many times a tailor must fold a piece of fabric to reach a certain thickness, assuming the fabric doubles in thickness with each fold.

4. What is the solution to the Pharaoh's Taylor series question?

The solution to the Pharaoh's Taylor series question is infinite, as the fabric will never reach a certain thickness. This is because the series will continue to double with each fold, and there is no limit to the number of times it can be folded.

5. Why is the Pharaoh's Taylor series question considered a paradox?

The Pharaoh's Taylor series question is considered a paradox because it presents a seemingly simple problem with an answer that defies common sense and intuition. It challenges our understanding of exponential growth and the concept of infinity.

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