Which Little "o" to use for Maclaurin expansion?

In summary, you should expand everything to high enough order so that the terms you want in the end are correct and then toss the rest.
  • #1
Anne5632
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2
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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  • #2
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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  • #3
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.
 
  • #4
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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  • #5
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.
 
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1. What is a Maclaurin expansion?

A Maclaurin expansion is a mathematical method used to approximate a given function using a series of terms. It is similar to a Taylor series, but it is centered at 0, making it a special case of the Taylor series.

2. Why is it important to choose the correct "little o" for a Maclaurin expansion?

The "little o" used in a Maclaurin expansion determines the accuracy of the approximation. Choosing the correct "little o" ensures that the approximation is as close to the actual function as possible.

3. How do I know which "little o" to use for a Maclaurin expansion?

The "little o" used in a Maclaurin expansion is typically determined by the highest order term in the function being approximated. It is usually the term with the highest degree in the polynomial.

4. Can I use any "little o" for a Maclaurin expansion?

No, the "little o" used in a Maclaurin expansion must be a function that goes to 0 as the variable approaches 0. This ensures that the approximation is valid for all values of the variable.

5. Are there any limitations to using Maclaurin expansions?

Maclaurin expansions are only valid for functions that are infinitely differentiable at the point 0. Additionally, they may not always converge to the actual function, especially for functions with a limited domain or with discontinuities at 0.

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