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expansion of e^-x/3?

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    the maclaurins series expansion is totally new to me
    I know that
    e^x = 1 + x + x^2/2!.....

    but what would the expansion of e^-x/3 be?

    thank you
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    e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x  ^3}{3!}+...+\dfrac{x^n}{n!} \Rightarrow e^{\frac{-x}{3}}=1+\dfrac{-x}{3}+\dfrac{(\frac{-x}{3})^2}{2!}+\dfrac{(\frac{-x}{3})^3}{3!}+...+\dfrac{(\frac{-x}{3})^n}{n!}={?}
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    Going by the the Maclaurin series equation, you need to find the derivatives of the function you're given. Do you mean, e^(-x/3) or e^(-x)/3? Either way, differentiate it to how many derivatives they ask for (E.g, "find the expansion up to the x^2 term" - the second derivative) and use the formula for it. From memory, I think it's f(x) ~ f(0) + xf'(0) + (x(^2)/2!)f''(0)....
 
 
 
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