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Find the terms through degree four of the Maclaurin series of $f(x)$.

$$f(x) = \frac{1}{1+\sin x}$$

My work:

The Maclaurin series for $\sin x$ up to degree $4$ is $x - \frac{x^3}{6} + \frac{x^5}{120}$

The Maclaurin series for $\frac{1}{1+x}$ up to degree $4$ is $1 - x + x^2 - x^3 + x^4$

I substituted $x - \frac{x^3}{6} + \frac{x^5}{120}$ for $x$ in $1 - x + x^2 - x^3 + x^4$

Did I do this right? Plugging this into WolframAlpha, I get this: http://goo.gl/SKddyh

Which doesn't seem like the answer in the text: $1-x+x^2-\frac{5x^3}{6}+\frac{2x^4}{3}$

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  • $\begingroup$ @ABC Thank you for editing! $\endgroup$ – Quaxton Hale Feb 27 '14 at 7:41
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Have a look at this:

Your equation, expanded

As you can see, what you plugged into WolframAlpha was the same equation as the answer in the text.

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You did it right. The answer on WolframAlpha is the same. Try typing "expand" in front on WA.

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You did it right and nothing has to be added to the answers you received.

I do not know if you were obliged to use these steps since there is a direct way of doing this expansion applying the basic rules, that is to say that the Taylor series of $f(x)$ built at $x=0$ just write (up to the fourth degree) $$f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+\frac{1}{6} f^{(3)}(0) x^3+\frac{1}{24} f^{(4)}(0) x^4+O\left(x^5\right)$$ The value of the function and its first four derivatives are $1,-1 ,2,-5,16$ (they are easy to evaluate bcause $sin(0)=0$ and $cos(0)=1$).

For sure, you arrive to your result.

You also have another way : replace in the denominator $sin(x)$ by its Taylor expansion and perform the long division of numerator by denominator.

You have done a good job ! Congratulations.

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Your answer is correct, but there is a somewhat simpler method. You are looking for a polynomial $P=a+bx+cx^2+dx^3+ex^4$ such that $P\times(1+x-\frac{x^3}6)\equiv 1\pmod{x^5}$. This gives (by comparing coefficients of $x^0,x^1,\ldots,x^4$) the equations $$ \begin{align}a&=1\\a+b&=0\\b+c&=0\\-\tfrac16a+c+d&=0\\-\tfrac16b+d+e&=0, \end{align} $$ which you can solve straight away as $a=1$, $b=-1$, $c=1$, $d=-\tfrac56$, $e=\tfrac23$.

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  • $\begingroup$ wish I could give this five +1s; just opened a new way of reasoning about series, for me. Thanks for posting! $\endgroup$ – Rax Adaam Apr 18 '18 at 22:05

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