Which Little "o" to use for Maclaurin expansion?

Anne5632
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Homework Statement
Find McLaren expansion to order 3
Relevant Equations
((2+3x)/(2-x))+(2-3x)sin2x
I have calculated it and got the answer but for the first equation with the division the little o is (x^3), I believe and for the equation being multiplied by sin, little o is (x^4)
For my answer do I add little o(x^4)?
 
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I would say to order 3 means up to and including terms in ##x^3##. I.e. neglecting terms of order ##x^4## and above.
 
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Anne5632 said:
Homework Statement:: Find McLaren expansion to order 3
Relevant Equations:: ((2+3x)/(2-x))+(2-3x)sin2x

I have calculated it and got the answer but for the first equation with the division the little o is (x^3), I believe and for the equation being multiploed by sin, little o is (x^4)
For my answer do I add little o(x^4)?
Generally, you have to play it safe an expand everything to high enough order so that the terms you want in the end are correct and then you toss the rest.

When you have a sum or difference, the order is limited by the term with the lowest order. So in your case, because you expanded the first term to third order, the overall expression you ended up with is only good to third-order. It doesn't matter if you expand the second term to fourth or higher order.

A couple of nitpicks.
  • It's spelled Maclaurin. You have two different spellings, neither correct.
  • ##(2+3x)/(2-x)## is the first term or an expression. It's not an equation as there's no equal sign. Similarly, ##(2-3x)\sin 2x## is not an equation.
 
To address the thread title, keep in mind that
  • ##f(x) = O(g(x))## if ##f/g## is bounded as ##x \rightarrow x_0##
  • ##f(x) = o(g(x))## if ##f/g \rightarrow 0## as ##x \rightarrow x_0##
(whoever invented this notation was clearly taking the proverbial)
 
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Just a couple tips: With ## \frac{1}{2-x} =(1/2)(\frac{1}{1-\frac{x}{2}}) ##, the geometric series approach is the simplest, and you can avoid taking derivatives. Meanwhile the series for the sine function is well known, and the problem is thereby a simple one. I agree with post 2 for what terms to keep.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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