How do you divide the series # x-x^3/(3!) +x^5/(5!)-x^7/(7!)+O(x^8) # by #1-x^2/(2!)+x^4/(4!)-x^6/(6!)+O(x^8)#?

1 Answer
Steve M
Oct 16, 2017

# S = x + x^3/3 + (2x^5)/15 + (17x^7)/315 + ... #

Explanation:

We assume that the initial question has been truncated to #O(x^8)#, and we seek:

# S = (x-x^3/(3!) +x^5/(5!)-x^7/(7!)+O(x^8))/(1-x^2/(2!)+x^4/(4!)-x^6/(6!)+O(x^8))#

Let us expand the factorials to get:

# S = (x-x^3/(6) +x^5/(120)-x^7/(5040)+ O(x^9) )/(1-x^2/(2)+x^4/(24)-x^6/(720)+O(x^8))#

Then we can use algebraic long division, working upto #O(x^8)#:

# {: ( , , +x, +1/3x^3,+(2x^5)/15 , +(17x^7)/315 , ), ( , , ――, ――, ――, ――, ), ( 1-x^2/(2)+x^4/(24)-x^6/(720), ")", +x, -x^3/6, +x^5/120, - x^7/5040, ), ( , , +x, -x^3/2, +x^5/24, -x^7/720, -), ( , , ――, ――, ――, ――, ), ( , , , +x^3/3, -x^5/30, +x^7/840, ), ( , , , +x^3/3, -x^5/6, +x^7/72, -), ( , , , , ――, ――, ), ( , , , , (2x^5)/15, -(4x^7)/315, ), ( , , , , (2x^5)/15, -x^7/15,- ), ( , , , , ――, ――, ), ( , , , , , (17x^7)/315, ) :} #

#S = x + x^3/3 + (2x^5)/15 + (17x^7)/315 + ... #

Related questions
Impact of this question
624 views around the world
You can reuse this answer
Creative Commons License