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I don't know how to get $$1-x^2+\frac{x^4}{2!}-\cdots.$$

I think it is too complex, if not impossible, to just use the definition of Maclaurin series.


Using the definition: consider the situation that all terms containing $x$ equal to zero, $f(0)=1$, $\,\,f'(0)=0$, $\,\,f''(0)=-2$, $\,\,f'''(0)=0$, $\,\,f^{(4)}(0)=4$, $\,\,f^{(5)}(0)=0$, $\,\,f^{(6)}(0)=16$

so I get the result: $\sum_{n=0}^∞(-2)^n\frac{x^{2n}}{(2n)!}$

Why...

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  • $\begingroup$ It's definitely not impossible! You need just to be able to differentiate and use the maclaurin series formula, but the current answers are quicker $\endgroup$ – danimal Apr 3 '15 at 14:47
  • $\begingroup$ @danimal Could you please make a specific illustration of this method because I always get the wrong answer using this. $\endgroup$ – Samara Apr 3 '15 at 15:07
  • $\begingroup$ ok - see below! $\endgroup$ – danimal Apr 3 '15 at 15:41
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Take the power serie of $e^Y$

$$e^{Y} = \sum_{n=0}^{\infty} \frac{Y^n}{n!}$$

Now replace $Y$ by $-x^2$, and you get

$$e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!}$$

Then you have

$$e^{-x^2} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{n!}$$

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why can't you substitute in $$e^u = 1 + \frac u{1!} + \frac {u^2}{2!} + \frac {u^3}{3!} +\cdots$$ for $u = -x^2$ and get $$e^{-x^2} = 1 - \frac {x^2}{1!} + \frac {x^4}{2!} - \frac {x^6}{3!} +\cdots$$

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Just to add to the other methods here (which are much more efficient and much less tedious) you can also use the Maclaurin formula, which involves differentiating $f(x)$ as many times as needed, finding the values of these derivatives at $x=0$, and then dividing each term by the correct factorial and multiplying by the correct power of $x$.

$$f(x) = e^{-x^2}\implies f(0)=1$$ $$f'(x) = -2xe^{-x^2}\implies f'(0)=0$$ $$f''(x) = -2(1\times e^{-x^2}+(-2x)\times x e^{-x^2})=-2e^{-x^2}(1-2x^2)\implies f''(0)=-2$$ $$f'''(x) = -2(-4xe^{-x^2}+(1-2x^2)(-2xe^{-x^2}))\implies f'''(0)=0$$ etc. The tricky part is doing the differentiation of this function (and I may well have made a mistake as I don't have pen and paper on me!) but I'll leave you to practise your differentiation with the fourth derivative - I strongly recommend it if, as you say, you keep making mistakes doing it this way. Because there will always be an $e^{-x^2}$ factor, it's worth factorising your answer each time as then you only have to apply the product rule once each time.

The Maclaurin series will then be: $$f(x) = f(0) + {f'(0)x\over1!}+ {f''(0)x^2\over2!}+ {f'''(0)x\over3!}+\cdots$$ $$=1+0+{-2x^2\over2!}+0+\cdots$$ $$=1-x^2+\cdots$$

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  • $\begingroup$ I finally get it! But I have to apply product rule as many times as the number of different power or X to get the right answer. $\endgroup$ – Samara Apr 3 '15 at 15:55

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