# Maths Power SeriesWatch

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#1
How does one

'write down the 1st 4 terms of the power series for cos3x'????

can i use binomial or something?
0
16 years ago
#2
I'm assuming that means it's a Taylor series:

f(a) + f'(a)x + (1/2!)f''(a)x^2 + (1/3!)f'''(a)x^3 + ...

where a is the point around which you're expanding the function. If you pick a = 0, the series is a MacLaurin's series.
0
#3
(Original post by Nylex)
I'm assuming that means it's a Taylor series:

f(a) + f'(a)x + (1/2!)f''(a)x^2 + (1/3!)f'''(a)x^3 + ...

where a is the point around which you're expanding the function. If you pick a = 0, the series is a MacLaurin's series.
so how would i use that for this cos3x?
0
16 years ago
#4
(Original post by Ollie)
so how would i use that for this cos3x?
f(x) = cos3x => f(0) = 1
f'(x) = -3sin3x => f'(0) = 0
f''(x) = -9cos3x => f''(0) = -9
f'''(x) = 27sin3x => f'''(0) = 0
f''''(x) = (3^4)cos3x => f''''(0) = 3^4

I think you will find that you get:

1, 0, -3^2, 0, 3^4, 0, -3^6, 0, 3^8, ...

Sub this into the formula given:

cos3x = 1 - ((3^2)/2!)x^2 + ((3^4)/4!)x^4 - ((3^6)/6!)x^6 + ...
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#5
(Original post by mikesgt2)
f(x) = cos3x => f(0) = 1
f'(x) = -3sin3x => f'(0) = 0
f''(x) = -9cos3x => f''(0) = -9
f'''(x) = 27sin3x => f'''(0) = 0
f''''(x) = (3^4)cos3x => f''''(0) = 3^4

I think you will find that you get:

1, 0, -3^2, 0, 3^4, 0, -3^6, 0, 3^8, ...

Sub this into the formula given:

cos3x = 1 - ((3^2)/2!)x^2 + ((3^4)/4!)x^4 - ((3^6)/6!)x^6 + ...
thanks very much
0
16 years ago
#6
(Original post by Ollie)
thanks very much
There is a proof of maclaurin's theorem here:

0
16 years ago
#7
okay...how advanced is this? AS, A2, Degree? Just interested...looks quite hard. I guess if you haven't learn it yet though then it would, heh.
0
16 years ago
#8
Taylor/MacLaurin's series are covered in Edexcel's P6 (dunno in which module for other boards, but I guess either P5 or P6), so A Level Further Maths. I didn't do A Level Further Maths so I've only done them at uni (1st year Physics), but it doesn't seem that difficult to understand, involves lots of differentiation though.
0
16 years ago
#9
pure 4 on OCR.
0
16 years ago
#10
(Original post by Nylex)
Taylor/MacLaurin's series are covered in Edexcel's P6 (dunno in which module for other boards, but I guess either P5 or P6), so A Level Further Maths. I didn't do A Level Further Maths so I've only done them at uni (1st year Physics), but it doesn't seem that difficult to understand, involves lots of differentiation though.
There are loads of clever methods for doing them - like for tan(x) and arctan(x) etc.

Ben
0
16 years ago
#11
(Original post by Ben.S.)
There are loads of clever methods for doing them - like for tan(x) and arctan(x) etc.

Ben
For arctan x, I know you just use d/dx(arctan x) = 1/1 + x^2 and expand that using the binomial series and integrate each term to get the Taylor series.
0
16 years ago
#12
(Original post by fishpaste)
pure 4 on OCR.
Seems like people who do AQA A do MacLaurin's series in P3.
0
16 years ago
#13
(Original post by Nylex)
Seems like people who do AQA A do MacLaurin's series in P3.
Yeah, but AQA A is weird. P3 is a further module, but P4 is not!

Ben
0
16 years ago
#14
(Original post by Ben.S.)
Yeah, but AQA A is weird. P3 is a further module, but P4 is not!

Ben
Eh?! Strange :/.
0
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