# Pharaoh's Taylor series question from Yahoo Answers

• MHB
• CaptainBlack
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.
CaptainBlack
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​

Last edited:
.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.

## 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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