Something I don't understand about taylor series

[tamtam402]

Homework Statement

Let's say I'm asked to find the taylor expansion for cot x, at the given point a = π/2.

Homework Equations

The Attempt at a Solution

My first thought would be to take the mc laurin series expansion for cotx, which is:
cot x = 1/x + x/3 - x³/45 ...
and substitute x for z-π/2.

According to my solution manual, that isn't correct. By messing around with trig identities, I solved the problem in the following way:

cot x = cot (π/2 + x - π/2) = -tan(x - π/2)
I took the mc laurin series expansion for tanx, multiplied it by -1, replaced x by z-π/2 and found the right answer.


Why wasn't I allowed to substitute x = z - π/2 in cot x, yet doing the same thing in -tanx gave me the right answer?

 

[vela]

Because cot x, which is what you're being asked to calculate, isn't equal to cot (x-pi/2), which is what you calculated.

 

[tamtam402]

Isn't a taylor series development the development of f(x-a)? I wasn't asked to evaluate cot x for x = pi/2, I was asked to evaluate the Taylor expansion at a = pi/2.

I'm self-studying from Boas book and there's pretty much no explanation given on series. If someone would be kind enough to provide me a link to demistify the difference between Taylor and McLaurin expansions, I'd be very grateful*.

 

[HallsofIvy]

Isn't a taylor series development the development of f(x-a)?

NO, it isnt. A Taylor's series for f(x) at x= a is a power series for f(x) at the point x= a. That is NOT, in general, a power series for f(x-a).

I wasn't asked to evaluate cot x for x = pi/2, I was asked to evaluate the Taylor expansion at a = pi/2.

I'm self-studying from Boas book and there's pretty much no explanation given on series. If someone would be kind enough to provide me a link to demistify the difference between Taylor and McLaurin expansions, I'd be very grateful*.

 

[vela]

I'm self-studying from Boas book and there's pretty much no explanation given on series. If someone would be kind enough to provide me a link to demistify the difference between Taylor and McLaurin expansions, I'd be very grateful*.

You might want to check out a freshman calculus book for review purposes. Boas assumes you are already familiar with the basics of these series. In any case, a MacLaurin series is the Taylor series for a=0. The Taylor series about x=a is simply
$$f(x) = f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots.$$ It's a power series in x-a, not f(x-a).

Here's a specific example of the difference. The MacLaurin series for sin x is
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots.$$ If you substitute, say, $x=z-\pi/2$, you'd get
$$(z-\pi/2) - \frac{(z-\pi/2)^3}{3!} + \frac{(z-\pi/2)^5}{5!} - \cdots = \sin(z-\pi/2) = -\cos z.$$ If you use the formula above to find the series for sine about $x=\pi/2$, you'd actually get
$$\sin x = 1 - \frac{(x-\pi/2)^2}{2!} + \frac{(x-\pi/2)^4}{4!} - \cdots.$$

 

[Ray Vickson]

Homework Statement

Let's say I'm asked to find the taylor expansion for cot x, at the given point a = π/2.

Homework Equations

The Attempt at a Solution

My first thought would be to take the mc laurin series expansion for cotx, which is:
cot x = 1/x + x/3 - x³/45 ...
and substitute x for z-π/2.

According to my solution manual, that isn't correct. By messing around with trig identities, I solved the problem in the following way:

cot x = cot (π/2 + x - π/2) = -tan(x - π/2)
I took the mc laurin series expansion for tanx, multiplied it by -1, replaced x by z-π/2 and found the right answer.


Why wasn't I allowed to substitute x = z - π/2 in cot x, yet doing the same thing in -tanx gave me the right answer?

The Taylor series for cot x at x = π/2 is the Maclaurin series of f(z) = cot(π/2 + z) around z = 0. If you try to get it from the Maclaurin series around x = 0, here is what happens:
$$\cot(\pi/2 + z) = \frac{1}{\pi/2 + z} - \frac{1}{3} (\pi/2 + z) - \frac{1}{45} (\pi/2 + z)^3 - \frac{2}{945} (\pi/2 + z)^5 + \cdots.$$
The constant term (not containing z) in the expansion is
$$c_0 = \frac{2}{\pi}-\frac{1}{3}\frac{\pi}{2} - \frac{1}{45}\frac{\pi^3}{2^3} - \frac{2}{945} \frac{\pi^5}{2^5} - \cdots.$$
Similarly, you can get the coefficient c₁ of z¹, c₂ of z², etc. However, these coefficients themselves are infinite series, so this is not at all useful.

RGV

 

[tamtam402]

Alright, thanks everyone :)