What is the Taylor expansion of $\frac{1}{1-x^4}$ at $x=0$? I was told it is $1 + x^4 + x^8$ + ... but the function's derivatives at $0$ are $0$ so shouldn't the Taylor Series at $0$ just be 1?
Taylor Expansion at x=0 of 1/(1-x^4)
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1The fourth derivative of $\frac{1}{1-x^4}$ at the origin is not zero. – 2017-02-01
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0Be sure to take your derivatives carefully. It would helpful if you included your work so far. – 2017-02-01
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0You both are right. I lost the constant in my fourth derivative and obtained $\dfrac{(-24*65x^4-24*155x^8-24*35x^{12})}{((-1+x^4)^5)}$. Thank you. – 2017-02-01
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Using \begin{align} \frac{1}{1-x} = \sum^\infty_{n=0}x^n \end{align} we see that \begin{align} \frac{1}{1-x^4} = \sum^\infty_{n=0}x^{4n}. \end{align}
Moreover, observe that \begin{align} f^{(4)}(x) = -24 \frac{(1 + 65 x^4 + 155 x^8 + 35 x^{12})}{(-1 + x^4)^5} \end{align} which means \begin{align} f^{(4)}(0)= 24 = 4!. \end{align} Likewise, we can show that other than derivatives which are multiples of 4, the other terms are zeros.
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0Thank you! I was missing a constant in my fourth derivative and I didn't catch the connection to the geometric series at first. That makes it very intuitive! – 2017-02-01