# A few taylor expansions - tricks?

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#1
Does anyone know of any tricks that can be used to expand these quickly (ie: without having to differentiate the function lots of times to generate a taylor series the long way), about x = 0.

- tan(x + PI/4)
- ln(sec x)
0
16 years ago
#2
Well if you know the Maclaurin series for tan x then the second one is really easy. Notice that (logsecx)' = tanx, so that to get the Maclaurin series for logsec(x) we can just integrate the series for tan(x) term by term. (If this is for anything other than A-Level you need to justify why you know it converges uniformly)
0
16 years ago
#3
(1)
tan(x + pi/4)
= (cos(x) + sin(x)) / (cos(x) - sin(x))
= 1 + 2sin(x) / (cos(x) - sin(x))
= 1 + 2tan(x) / (1 - tan(x))
= 1 + 2tan(x) + 2tan(x)^2 + 2tan(x)^3 + ... for small x (this is a GP).

Now replace tan(x) by its Taylor series x + x^3 / 3 + 2x^5 / 15 + ... (valid for small x) and expand the powers. You should get

1 + 2x + 2x^2 + 8x^3 / 3 ... .

(2)
ln(1 + x) = x - x^2 / 2 + x^3 / 3 - x^4 / 4 + ... for small x.

So
ln(sec(x))
= -ln(cos(x))
= -ln(1 + (cos(x) - 1))
= -[(cos(x) - 1) - (cos(x) - 1)^2 / 2 + (cos(x) - 1)^3 / 3 - (cos(x) - 1)^4 / 4 + ...] for small x.

You can use the same idea as in (1), replacing (cos(x) - 1) by its Taylor series -x^2 / 2! + x^4 / 4! - x^6 / 6! + ... (valid for small x). You should get

x^2 / 2 + x^4 / 12 + x^6 / 45 ... .
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