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Using the Maclaurin expansion formula:Maclaurin formula to find the Maclaurin series for $sin(3x)$, I can get the correct answer by using $x^n$ in the formula above (in the tail-end of the formula).

Similarly, to expand $e^(-x)$, I can get the correct answer by using $x^n$.

But to expand $\frac{1}{(1+2x)}$, I can't get the correct answer by using $x^n$, instead I have to use $(2x)^n$ to get the correct answer.

Why is this so? Shouldn't I have to put $(3x)^n$ into the formula to expand $sin(3x)$, and $(-x)^n$ for $e^(-x)$ as well then? Am I doing something wrong?

Just to be sure, here is my working for all three problems:

No. 1. $f(x) = sin(3x)

f(0)=0; f'(0)=3; f''(0)=0; f'''(0)=-27; f''''(0)=0; f'''''(0)=243$

$sin(3x)= sum of f^(n)(0)/n!*(x^n) = 3x - (27/3!)x^3 + (243/5!)x^5 +...

No. 2. f(x) = e^(-x)

f(0)=1, f'(0)=1, f''(0)=1, f'''(0)=1

e^(-x)= sum of f^(n)(0)/n!*(x^n) = 1 - x + x^2/2! - x^3/3! +...

No. 3. f(x) = 1/(1+2x)

f(0)=1, f'(0)=-1, f''(0)=2, f'''(0)=-6

1/(1+2x)= sum of f^(n)/n!*(x^n) = 1 - x + x^2 - x^3 +... (wrong)

or

1/(1+2x)= sum of f^(n)(0)/n!*(2x^n) = 1 - 2x + 4x^2 - 8x^3 +...$ (correct)

I understand that substituting the X values into the standard Maclaurin series formulas for sin(x), e^x, and 1/(1+x) is easier, but I really rather use the general Maclaurin formula to expand the functions. I'd really appreciate it if someone could tell me what I'm doing wrong here, why I need to use a new X value (i.e. 2x) to expand the polynomial but not for the trigonometric or exponential functions.

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    $\begingroup$ "Shouldn't I have to put $(3x)^n$ into the formula to expand $\sin(3x)$, and $(-x)^n$ for $e^{-x}$ as well then?" Yes, you should. And judging by the sentence "Similarly, to expand $e^{-x}$, I can get the correct answer by using $x^n$.", you did something wrong if you got the right answer. $\endgroup$ – Git Gud Dec 24 '13 at 15:49
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    $\begingroup$ Yes, you should have done that, though it could be these are two different questions: one thing is to express $\;\sin 3x\;$ or $\;e^{-x}\;$ in a power series and you can use the usual ones as those have infinite convergence radius; other question though is to ask to develop that question as power series. Then you have to use $\;3x\;,\;-x\;$ resp. $\endgroup$ – DonAntonio Dec 24 '13 at 15:50
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    $\begingroup$ The procedures are exactly the same for $\frac{1}{1+2x}$ as for the others. Two options are available. (i) Use the basic formula, calculate the derivatives at $x=0$. Perhaps you are differentiating $\frac{1}{1+2x}$ incorrectly, forgetting about the Chain Rule. Or else (ii) use the Maclaurin series for $\frac{1}{1-t}$, substitute $-2x$ for $t$ at the end. $\endgroup$ – André Nicolas Dec 24 '13 at 15:55
  • $\begingroup$ Based on your edit, it looks like your error lay in forgetting the Chain Rule in your calculations. $\endgroup$ – Cameron Buie Dec 24 '13 at 16:12
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In both the first two cases, we can readily obtain the Maclaurin series in two ways:

  • We can start with the Maclaurin series for the basic function--$\sin x,e^x$ respectively--then replace all $x^n$ appropriately--by $(3x)^n$ and $(-x)^n$ respectively.
  • We can explicitly determine all derivatives of the functions and compute the Maclaurin series directly.

If we take the latter approach, you should note (for example) that if $f(x)=\sin(3x)$ and $g(x)=\sin x$, then we will have $f^{(n)}(x)=3^n\cdot g^{(n)}(3x)$ for all $n\ge 0,$ due to the chain rule. Consequently, $$\begin{align}f(x) &= \sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n\\ &= \sum_{n=0}^\infty\frac{g^{(n)}(0)\cdot 3^n}{n!}x^n\\ &= \sum_{n=0}^\infty\frac{g^{(n)}(0)}{n!}(3x)^n\\ &= g(3x),\end{align}$$ as we should expect.

The same ideas work for the function $f(x)=\frac1{1+2x}.$ We could explicitly calculate derivatives of all orders (which is tedious), or (more simply) we could begin with the function $g(x)=\frac1{1-x},$ which has a geometric series as its Maclaurin series, namely $$g(x)=\sum_{n=0}x^n$$ for all $\left|x\right|<1.$ Observing then that $f(x)=g(-2x),$ we then find that $$f(x)=\sum_{n=0}^\infty(-2x)^n=\sum_{n=0}^\infty(-2)^nx^n.$$ I leave it to you to confirm that $f^{(n)}(0)=(-2)^n\cdot n!$ for all $n\ge 0$ (don't forget the chain rule!), so that both approaches once again bear the same fruit. The formula for $f(x),$ though, will hold whenever $\left|-2x\right|<1,$ or equivalently whenever $\left|x\right|<\frac12.$ Aside from that, nothing is different from the first two cases (whose basic and transformed series are everywhere convergent).

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    $\begingroup$ Ah yes, thanks so much! I see now where I went wrong. $\endgroup$ – Harmony Dec 24 '13 at 16:32
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There are two ways of finding the correct series:

  • Find the series for $\sin t$, $\operatorname{e}^t$ and $\frac{1}{1-t}$ and then substitute $t=3x$, $t=-x$ and $t=-2x$, resp.
  • Use $\operatorname{f}(x)=\sin 3x$, $\operatorname{f}(x)=\operatorname{e}^{-x}$ and $\operatorname{f}(x)=\frac{1}{1+2x}$ in the red formula that you posted.

Both approaches will give you exactly the same answers for these three functions.

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