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how to determine the following Taylor series $$x^2-3x+4$$ around the point of development of $x_0$?

Could someone explain that example and provide a general explanation?

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up vote 6 down vote accepted

For polynomials, the Taylor series around $x=0$ is just the polynomial itself. If you want the series around $x=a$, the general formula (for a function $f$ around $x=a$) $$ f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n $$ When $f$ is a polynomial of degree $n$, the ($n+1$)'th derivative equals $0$, so the series is finite.

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Taking Ragnar's approach, we get $$(a^2-3a+4)+(2a-3)(x-a)+(x-a)^2$$ With a polynomial we can also do this directly, by setting $x=y+a$ in the original polynomial to give $(y+a)^2-3(y+a)+4$, which becomes, on expanding as a polynomial in $y=x-a$$$(a^2-3a+4)+(2a-3)y+y^2$$

This method of calculation always works for polynomials. Whether you use it depends on whether you find it convenient, or whether you are doing a problem set or exam which requires the use of a particular method.

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