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How do I compute the Taylor series for $\ln(9x^2-10x+4)$ centered at zero in the form of a summation?

I've tried all the possible methods that I generally use to compute a Taylor series but none of them worked. All the Taylor series that I've done before are able to transform into the summation form because there were always some type of order that I'm able to identify. However, after computing the derivatives for this function centered at zero, there is no order at all.

What are some of the steps that I can take to solve this problem?

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We first remove the constant term in the series, which is $\ln4$: $$\ln(9x^2-10x+4)=\ln4+\ln\left(1+\frac94x^2-\frac52x\right)$$ Now substitute $\frac94x^2-\frac52x$ for $x$ in the Maclaurin series for $\ln(1+x)$ and expand. Since there is no constant term in the substituted expression, the terms contributing to a given power are finite in number: $$\ln\left(1+\frac94x^2-\frac52x\right)=\left(\frac94x^2-\frac52x\right)-\frac{\left(\frac94x^2-\frac52x\right)^2}2+\frac{\left(\frac94x^2-\frac52x\right)^3}3-\cdots$$ $$=-\frac52x-\frac78x^2+\frac5{12}x^3+\cdots$$ Finally $$\ln(9x^2-10x+4)=\ln4-\frac52x-\frac78x^2+\frac5{12}x^3+\cdots$$

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  • $\begingroup$ @Ekidona You can't for this, really. It's just a matter of expansion. $\endgroup$ Commented Apr 22, 2020 at 2:55

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