University of Florida/Egm6341/s10.team2.niki/HW7

Statement
P.39-1

Solution
We have Verhulst model or the Logistic equation as P.38-3

$$\dot{x}=\frac{dx}{dt}=rx(1-\frac{x}{x_{max}})$$

Separating variables and integrating we have

$$\int_{x_0}^{x}\frac{x_{max}}{x(x_{max}-x)}dx=\int_{t_0=0}^{t}rdt$$

which can be written as

$$\int_{x_0}^{x}\frac{1}{x}dx+\int_{x_0}^{x}\frac{1}{(x_{max}-x)}dx=\int_{t_0=0}^{t}rdt$$

We get

$$log_e(\frac{x}{x_0})-log_e(\frac{x_{max}-x}{x_{max}-x_0})= r(t-0)$$

$$\frac{x(x_{max}-x_0)}{x_0(x_{max}-x)}=e^{rt}$$

Rearranging we have

$$x=\frac{x_{max}x_0e^{rt}}{x_{max}+x_0(e^{rt}-1)}$$

Statement
We have the cosine series expressed as $$f(cos\theta )=\frac{a_0}{2}+\sum_{k=1}^{\infty }a_k cos(k\theta)$$, we need to express teh constants $$a_k$$ as

$$a_k=\frac{2}{\pi }\int_{0}^{\pi}f(cos\theta)cos(k\theta)d\theta$$

Solution
We have the given expression for the cosine series as

$$f(cos\theta )=\frac{a_0}{2}+\sum_{k=1}^{\infty }a_k cos(k\theta)$$

multiplying both sides by $$cos(m\theta)$$ where k is not the same as m and integrating we get,

$$\int_{0}^{2\pi}f(cos\theta )(cos(m\theta))=\int_{0}^{2\pi}\frac{a_0cos(m\theta)}{2}+\int_{0}^{2\pi}\sum_{k=1}^{\infty }a_k cos(k\theta)cos(m\theta)$$

Using the property of orthogonality we know that $$\int_{0}^{2\pi}a_k cos(k\theta)cos(m\theta)$$ exists only when k = m i.e

$$\int_{0}^{2\pi}f(cos\theta )(cos(k\theta))d\theta=\frac{a_0}{2}\int_{0}^{2\pi}cos(m\theta)d\theta+a_k\int_{0}^{2\pi} cos^2(k\theta)d\theta                                    $$

wkt,

$$\int_{0}^{2\pi}cos(m\theta)d\theta=0$$ and

$$\int_{0}^{2\pi}cos^2(k\theta)d\theta = \frac{2\pi-0}{2}=\pi$$

Thus we have by substituting and rearranging terms,

$$a_k=\frac{2}{\pi }\int_{0}^{\pi}f(cos\theta)cos(k\theta)d\theta$$

Author
--Egm6341.s10.team2.niki 14:43, 23 April 2010 (UTC)