Maclaurin Expansion: Obtaining First 3 Terms

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Homework Help Overview

The discussion revolves around obtaining the first three non-zero terms in the Maclaurin expansion of the function (1 + sin2x). Participants are exploring methods to approach this problem within the context of series expansions in calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to differentiate the function to find the terms but questions the validity of their approach and whether sin2x needs to be transformed for expansion. Some participants suggest using the Maclaurin series for sin(x) and squaring it, while others mention using a trigonometric identity to facilitate the expansion.

Discussion Status

The discussion is active with various approaches being considered. Participants are sharing different methods and questioning the original poster's reasoning without reaching a consensus on a single method.

Contextual Notes

There is an indication that the original poster is uncertain about the expandability of the function and whether their differentiation approach is appropriate. The discussion includes references to potential constraints in the context of a homework assignment.

MisterMan
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Hi, I recently sat my Maths examination and there was a Maclaurin expansion question that I made an attempt at but I think it was wrong, it would be good if I could get help on this, it's too late to be of any real help but it will help me understand where I went wrong:

Obtain the first three non-zero terms in the Maclaurin expansion of (1 + sin2x)

What I done here was, let f(x) = (1 + sin2x) and differentiated until I got three non-zero terms when plugging in x = 0. But I'm not sure if this is "expandable" or whether I need to change sin2x into something that can be expanded.

Did I do the right thing or did I make a mistake?
 
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The first term is 1.
 
What you describe sounds like one of several correct ways to attack the problem. Of course, I cannot tell if you executed the attack correctly.
 
A quicker way (and one probably alluded to by Hurkyl) is to take a couple of terms of the Maclaurin series for sin(x), square them, and then add 1.
 
Or use the trig formula

[tex] \sin^{2} x = \frac{1 - \cos{2 x}}{2}[/tex]

and use the Maclaurin series for the cosine.
 

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