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Suppose I have some polynomial:

$$f(x)=a_0 +a_1x+a_2 x^2+a_3 x^6+a_4x^8$$

How can I construct a power series representation for this polynomial? More generally: How can I construct a power series given ANY polynomial with finite terms?

qmd
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    A polynomial is a power series. – Martin R Jun 28 '19 at 07:45
  • It is a power series. A power series does not necessarily have infinitely many nonzero terms. – lEm Jun 28 '19 at 07:45
  • @MartinR But isn't a power series something of the form $\sum_{n=0}^{\infty} c_n x^n$? – qmd Jun 28 '19 at 07:46
  • Yes, and that's what you have, with $c_0 = a_0$, $c_1 = a_1$, $c_2 = a_2$, $c_3=c_4 = c_5 = 0$, $c_6 = a_3$, $c_7 = 0$, $c_8 = a_4$, and $c_n = 0$ for $n\ge 9$ – Martin R Jun 28 '19 at 07:48
  • @qmd yes. In your case, $\exists N$ so that $c_n=0 \forall n >N$. – Botond Jun 28 '19 at 07:48
  • @MartinR I sort of got that far but in tried to find the radius of convergance for a similar problem with given $a_n$'s and I thought i need to find some sort of "rule" $c_n=f(n)$ and then write out the power series $\sum _{n=0}^{\infty }c_n x^n$ in order to be able to use the root test for example. – qmd Jun 28 '19 at 07:52
  • @InsideOut I am asked to find the radius of convergence for a similar problem with given values for the $a_n's$. I thought I need to write it as a power series ($\sum c_n x^n$) first to find the radius of convergence. – qmd Jun 28 '19 at 07:53
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    If you want to compute it formally: In the case of a polynomial, only finitely many $c_n$ are not zero. Therefore $\limsup_{n \to \infty} |c_n|^{1/n} = 0$, and the radius of convergence is $\infty$. – Martin R Jun 28 '19 at 07:54
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    @InsideOut: I recommend avoiding assigning gender to users of this site when it's not apparent. Our implicit biases cause us to assume that mathematics users are male, and falling into that trap perpetuates the exclusion that women feel in the discipline. – Greg Martin Jun 28 '19 at 07:55
  • @MartinR How can I just say that $\limsup_{n \rightarrow \infty} \lvert c_n \rvert ^\frac{1}{n}=0$ without knowing what $c_n$ is? – qmd Jun 28 '19 at 07:59
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    If the OP is looking for the power series centered in $0$, his question is trivial. But the OP wants a power series centered on an arbitrarily point $x_o\in\Bbb R$ is another matter. @GregMartin Thanks for remind me that. – InsideOut Jun 28 '19 at 07:59
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    You can always define $c_n = \frac{f^{(n)}(x_0)}{n!}$ for a polynomial, and note that $c_n = 0$ for $n > \operatorname{deg}(f)$. – Martin R Jun 28 '19 at 08:16
  • @MartinR That makes sense! Thank you for your help! – qmd Jun 28 '19 at 08:48

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A (real or complex) polynomial in a single variable $x$ $$ f(x) = a_0 + a_1 x + \ldots + a_n x^n $$ it equal to its own power series at $x=0$: With $$ c_j = \begin{cases} a_j & \text{for } 0 \le j \le n\\ 0 & \text{for } j > n \end{cases} $$ we have $$ f(x) = \sum_{j=0}^n a_j x^j = \sum_{j=0}^\infty c_j x^j \, . $$ The series converges for all $x \in \Bbb R$ (or $\Bbb C$), so that the radius of convergence is $\infty$.

For the power series centered at an arbitrary point $x_0$ we have $$ f(x) = \sum_{j=0}^n\frac{f^{(j)}(x_0)}{j!} x^j = \sum_{j=0}^\infty \frac{f^{(j)}(x_0)}{j!} x^j $$ because $f^{(j)} \equiv 0$ for $j > n$.

Martin R
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