Colin Series...going off on a tangent!

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abra-cad-abra
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hi were doing macluarin series and taylor expansions, only really maclaurin ones as were looking at derivatives at 0.

Anyhow, we did sinx and cosx but not tanx. And when i asked my teacher, he said its too hard and you wont need to know it.

So could anyone explain how to find the maclaurin series for tanx such that it is in closed form.

Or atleast a good method for working out each term rather than repeated differentiation which i have found to be very tiresome already after the 5th derivative. Is there a recurrence relation between the derivatives?

I understand the closed form involves Bernoulli's function so I assume I would not be able to find the general term with just FP2 knowledge?

Thanks
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davros
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(Original post by abra-cad-abra)
hi were doing macluarin series and taylor expansions, only really maclaurin ones as were looking at derivatives at 0.

Anyhow, we did sinx and cosx but not tanx. And when i asked my teacher, he said its too hard and you wont need to know it.

So could anyone explain how to find the maclaurin series for tanx such that it is in closed form.

Or atleast a good method for working out each term rather than repeated differentiation which i have found to be very tiresome already after the 5th derivative. Is there a recurrence relation between the derivatives?

I understand the closed form involves Bernoulli's function so I assume I would not be able to find the general term with just FP2 knowledge?

Thanks
There are numerous approaches.

If you google "tangent maclaurin series" or "maclaurin series for tan x" you'll find numerous links to different ways
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abra-cad-abra
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(Original post by davros)
There are numerous approaches.

If you google "tangent maclaurin series" or "maclaurin series for tan x" you'll find numerous links to different ways
what way do you think could be approached with my knowledge. Something that a step question could be based off perhaps?
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abra-cad-abra
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(Original post by davros)
There are numerous approaches.

If you google "tangent maclaurin series" or "maclaurin series for tan x" you'll find numerous links to different ways
also if i differentiate a series. is it necessarily the differential of the sigma form of the sum

so if i differentiaed sinx. when looking at it in its maclaurin expansion i can differentiate the general term to get the general term of cosx expansion

is this true for any series? so d/dx SIGMA f(r) (r=0 to r=inf or n) is equivalent to SIGMA f'(r) (r=0 to r=inf or n) ???
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Mr M
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(Original post by abra-cad-abra)
what way do you think could be approached with my knowledge. Something that a step question could be based off perhaps?
http://www.thestudentroom.co.uk/show....php?t=1019661
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davros
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(Original post by abra-cad-abra)
also if i differentiate a series. is it necessarily the differential of the sigma form of the sum

so if i differentiaed sinx. when looking at it in its maclaurin expansion i can differentiate the general term to get the general term of cosx expansion

is this true for any series? so d/dx SIGMA f(r) (r=0 to r=inf or n) is equivalent to SIGMA f'(r) (r=0 to r=inf or n) ???
(Sorry, I'm not normally so lazy with suggestions but I've had a long day and turning in soon, so Google seemed the best answer for suggestions to be honest )

Personally I would just use the repeated differentiation method to get the tangent series but there are other methods such as tan x = sin x / cos x so sin x = (cos x)(tan x) then you can multiply series and compare coefficients (like you do with polynomials!).

You can differentiate a series within its "radius of convergence" so a series like sin x that converges for all values of x can be differentiated to give you the cos x series etc.

You can also go the other way via integration, but obviously that introduces an arbitrary constant so you need a bit more information like the value of the function at 0 to pin down that constant
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Blazy
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(Original post by abra-cad-abra)
Or atleast a good method for working out each term rather than repeated differentiation which i have found to be very tiresome already after the 5th derivative.
You can save a bit of time when you make it to the second derivative...

 f(x) = \tan(x)
 f'(x) = \sec^2(x)
 f''(x) = 2\tan(x)\sec^2(x) = 2f(x)f'(x)

From here on, just keep using product rule, e.g.  f'''(x) = 2[f'(x)]^2 + 2f(x)f''(x) = 2[f'(x)]^2 + 4[f(x)]^2f'(x) etc etc
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ThePerfectScore
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You can't do it with A-level maths.
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Phichi
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(Original post by ThePerfectScore)
You can't do it with A-level maths.
You know knowing PerfectScore


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abra-cad-abra
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(Original post by ThePerfectScore)
You can't do it with A-level maths.
what can you do it with
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