Month: February 2018

Series Expansion

| Cosmoshivani

A function can be expanded using Taylor series which is given by,

f(x) = f(a) + \frac{x-a}{1!} f^{'}(a) + \frac{(x-a)^{2}}{2!} f^{''}(a) + \frac{(x-a)^{3}}{3!} f^{'''}(a) + ...

If a=0, it’s called Maclaurin series.

f(x) = f(0) + \frac{x}{1!} f^{'}(0) + \frac{(x)^{2}}{2!} f^{''}(0) + \frac{(x)^{3}}{3!} f^{'''}(0) + ...

Let’s figure out the Maclaurin series for all trigonometric functions,

f(x) = sinx f(0) = 0
f^{'}(x) = cosx f^{'}(0) = 1
f^{''}(x) = -sinx f^{''}(0) = 0
f^{'''}(x) = -cosx f^{'''}(0) = -1

sinx = \frac{x}{1!} - \frac{(x)^{3}}{3!} + ...

f(x) = cosx f(0) = 1
f^{'}(x) = -sinx f^{'}(0) = 0
f^{''}(x) = -cosx f^{''}(0) = -1
f^{'''}(x) = sinx f^{'''}(0) = 0

cosx = 1 - \frac{(x)^{2}}{2!} + ...

f(x) = tanx
f(0) = 0
f^{'}(x) = sec^{2}x = 1+tan^{2}x = 1+f(x)^{2}
f^{'}(0)=1
f^{''}(x) = 2f(x) f^{'}(x)
f^{''}(0) = 0
f^{'''}(x) = 2f^{'}(x)^{2}+2f(x)f^{''}(x)
f^{'''}(0) = 2

tanx = \frac{x}{1!} + \frac{2(x)^{3}}{3!} + ...

f(x) = secx
f(0) = 1
f^{'}(x) = f(x) tanx
f^{'}(0) = 0
f^{''}(x) = f^{'}(x) tanx + f(x) f(x)^{2}
f^{''}(0) = 1
f^{'''}(x) = f{''}(x) tanx + f^{'}(x) f(x)^{2} + f^{'}(x) f(x)^{2} +2 f(x)^{2} f^{'}(x)

secx = 1 + \frac{(x)^{2}}{2!} + ...

f(x) = cotx f(0) = \infty
f(x) = cosecx f(0) = \infty
How can you write series expansion if the value of function is infinite at x=0? I am still trying to figure out a satisfactory answer to this question.

Update:

If a function is not analytic at a point then it cannot be expanded using Taylor series. We have another series expansion for such points and its called Laurent’s expansion.

So, lets evaluate Laurent series expansion of f(z) = cotz at z=0
f(z)=\frac{1}{tanz}
cotz = \frac{1}{z + \frac{z^{3}}{3!} + \frac{2(z)^{5}}{15} + ...}
cotz = \frac{1}{z[1 + \frac{z^{2}}{3!} + \frac{2(z)^{4}}{15} + ...]}

Now we can use the long division method to divide \frac{1}{z} by \frac{1}{1 + \frac{z^{2}}{3!} + \frac{2(z)^{4}}{15} + ...} .

So, we get

cotz = \frac{1}{z} - \frac{z}{3} -\frac{z^{3}}{45} - ...

Similarly we can evaluate Laurent series expansion of cosecz.

f(z) = cosecz at z=0
f(z) = \frac{1}{sinz}
cosecz = \frac{1}{z- \frac {z^{3}}{3!} + \frac{z^{5}}{5!} - ...}
cosecz = \frac{1}{z[1- \frac {z^{2}}{3!} + \frac{z^{4}}{5!} - ...]}

Again using long division method we can divide \frac{1}{z} by \frac{1}{1- \frac {z^{2}}{3!} + \frac{z^{4}}{5!} - ...}

cosecz = \frac{1}{z} + \frac{z}{3!} + \frac{7(z)^{3}}{360} + ...

Thanks to tex.stackexchange.com for helping me write the LATEX notations in this blog.