Around the point " x " = 0, we find out that the radius of convergence of this series is \ scriptstyle \ infty meaning that this series converges for all complex numbers.
42.
Another problem of extrapolation is loosely related to the problem of analytic continuation, where ( typically ) a power series representation of a convergence to produce a power series with a larger radius of convergence.
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Hi there guys-I was wondering about determining the convergence of the series \ sum ^ { \ infty } z ^ { n ! }-particularly on the boundary of the radius of convergence.
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From above it is evident that all that is really needed is the radius of convergence of the MacLaurin series be greater than ? " T " ?, the operator norm of " T ".
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To prove ( i ) and ( v ), apply the ratio test and use formula ( 2 ) above to show that whenever " ? " is not a nonnegative integer, the radius of convergence is exactly 1.
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Conversely, for a given holonomic sequence c _ n, the function defined by the above sum is holonomic ( this is true in the sense of formal power series, even if the sum has a zero radius of convergence ).
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For a value such as " t " = 0.000000001 that is significantly smaller than this radius of convergence, the first-order approximation 19.9569 . . . is reasonably close to the root 19.9509 . ..
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Is a formal power series with rational coefficients " a " " n ", which has a non-zero radius of convergence in the complex plane, and within it represents an analytic function that is in fact an algebraic function.
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If the inner radius of convergence is positive, " f " may have infinitely many negative terms but still be regular at " c ", as in the example above, in which case it is represented by a " different"
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And one expects problems when | " t " | is larger than the radius of convergence of this power series, which is given by the smallest value of | " t " | such that the root ? " j " becomes multiple.