The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

Maclaurin Series

oook
so the question asks for the expansion of sin(x+x^3)
using maclaurin series (first two non zero terms)

So I got

= (x+x^3) - 1/6 (x+x^3)^3

but then in the solutions it goes on to get...

= x + 5/6 x^3

and has a note saying x^3 + 0(x^5)

if anyone could explain how this works I would be very grateful, thanks!! :smile:
Reply 1
Avatar for SimonM
SimonM
18
Do you mean (x+x^3)-1/6(x+x^3)^3 ?
Reply 2
Avatar for laura_beth
laura_beth
OP
SimonM
Do you mean (x+x^3)-1/6(x+x^3)^3 ?


yep
sorry typed it out wrong

I got:

(x+x^3) - 1/6 (x+x^3)^3
Reply 3
Avatar for SimonM
SimonM
18
Which expands to:

x + (5 x^3)/6 - x^5/2 - x^7/2 - x^9/6 = x+5/6 x^3 + O(x^5)
Reply 4
Avatar for Beenman500
Beenman500
well, if you expand the brackets (-1/6)(x+x^3)^3 and diregard any power of x^n where n > 3. you get -1/6 * x^3

and if you add that to the original bracket you will get x + (x^3 - 1/6 * x ^3) = x + 5/6 * x^3 as the answers want
Reply 5
Avatar for laura_beth
laura_beth
OP
SimonM
Which expands to:

x + (5 x^3)/6 - x^5/2 - x^7/2 - x^9/6 = x+5/6 x^3 + O(x^5)


ook, get the first bit and the expansion but I'm still unsure how

- x^5/2 - x^7/2 - x^9/6 becomes O(x^5) :/

sorry, probably ridiculously simple...
thanks for the help!
Reply 6
Avatar for laura_beth
laura_beth
OP
Beenman500
well, if you expand the brackets (-1/6)(x+x^3)^3 and diregard any power of x^n where n > 3. you get -1/6 * x^3

and if you add that to the original bracket you will get x + (x^3 - 1/6 * x ^3) = x + 5/6 * x^3 as the answers want


why do you disregard the higher powers? just because you want the first 2 terms?
Reply 7
Avatar for nota bene
nota bene
15
laura_beth
ook, get the first bit and the expansion but I'm still unsure how

- x^5/2 - x^7/2 - x^9/6 becomes O(x^5) :/

sorry, probably ridiculously simple...
thanks for the help!

Well, O(x^5) is by definition including terms of higher order - basically saying 'the largest term that is included here is of order 5'

We want to cut off our expansion to make an approximation; infinitely many terms would make approximating a bit hard!

Related discussions

Latest

Trending

Trending

Articles for you