求Laurent级数1/(z³+z+2)=∑(z+1)^n c_n的收敛半径与系数c_n

Got it, let's work through this step by step. You were right to start with substituting ( t = z + 1 ) — that's exactly the right move here. Let's break this down into finding the convergence region first, then computing the coefficients ( c_n ).

Let ( t = z + 1 ), so ( z = t - 1 ). Substitute this into the denominator ( z^3 + z + 2 ):
[
\begin{align*}
z^3 + z + 2 &= (t-1)^3 + (t-1) + 2 \
&= t^3 - 3t^2 + 3t - 1 + t - 1 + 2 \
&= t^3 - 3t^2 + 4t \
&= t(t^2 - 3t + 4)
\end{align*}
]
Your original equation becomes:
[
\frac{1}{t(t^2 - 3t + 4)} = \sum_{n=-\infty}^\infty c_n t^n
]

First, find the singularities of the function (where the denominator is zero):

1. ( t = 0 ) (this corresponds to ( z = -1 ))
2. Roots of ( t^2 - 3t + 4 = 0 ): solving this quadratic gives ( t = \frac{3 \pm i\sqrt{7}}{2} ). The modulus of these roots is:
   [
   \left| \frac{3 \pm i\sqrt{7}}{2} \right| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{\sqrt{7}}{2}\right)^2} = \sqrt{\frac{9 + 7}{4}} = 2
   ]

For a Laurent series, the convergence region is a ring between the nearest and farthest singularities from the expansion point (here, ( t = 0 ), i.e., ( z = -1 )). The nearest singularity is at ( |t| = 0 ), and the farthest is at ( |t| = 2 ). So the convergence region is:
[
0 < |t| < 2 \quad \text{or equivalently} \quad 0 < |z+1| < 2
]

If we use the given formula for the outer convergence radius ( R ):
[
R = \frac{1}{\limsup_{n\rightarrow\infty}{\sqrt[n]{|c_n|}}}
]
Since the farthest singularity is at distance 2 from the expansion point, ( R = 2 ) (this matches the singularity-based intuition, which is a quicker way to find it without computing the limsup directly).

First, decompose the rational function into partial fractions:
[
\frac{1}{t(t^2 - 3t + 4)} = \frac{A}{t} + \frac{Bt + C}{t^2 - 3t + 4}
]
Multiply both sides by ( t(t^2 - 3t + 4) ) to get:
[
1 = A(t^2 - 3t + 4) + (Bt + C)t
]
Expand and match coefficients:

• Constant term: ( 4A = 1 \implies A = \frac{1}{4} )
• ( t^2 ) term: ( A + B = 0 \implies B = -\frac{1}{4} )
• ( t ) term: ( -3A + C = 0 \implies C = \frac{3}{4} )

So the decomposition becomes:
[
\frac{1}{4t} + \frac{3 - t}{4(t^2 - 3t + 4)}
]

Now we'll expand each part into power series:

Part 1: ( \frac{1}{4t} ) #

This is already a single term in the Laurent series: ( \frac{1}{4}t^{-1} ). This contributes:

• ( c_{-1} = \frac{1}{4} )
• ( c_n = 0 ) for all ( n \leq -2 )

Part 2: ( \frac{3 - t}{4(t^2 - 3t + 4)} ) #

We need to expand this as a positive power series (since we're in ( |t| < 2 )). Let's denote ( \frac{1}{t^2 - 3t + 4} = \sum_{k=0}^\infty a_k t^k ). Multiply both sides by ( t^2 - 3t + 4 ):
[
1 = (t^2 - 3t + 4)\sum_{k=0}^\infty a_k t^k
]
Expand the left-hand side and match coefficients to get a recurrence relation:

• ( k=0 ): ( 4a_0 = 1 \implies a_0 = \frac{1}{4} )
• ( k=1 ): ( 4a_1 - 3a_0 = 0 \implies a_1 = \frac{3}{16} )
• ( k \geq 2 ): ( 4a_k - 3a_{k-1} + a_{k-2} = 0 \implies a_k = \frac{3a_{k-1} - a_{k-2}}{4} )

Now multiply this series by ( \frac{3 - t}{4} ):
[
\frac{3 - t}{4}\sum_{k=0}^\infty a_k t^k = \frac{3}{4}\sum_{k=0}^\infty a_k t^k - \frac{1}{4}\sum_{k=0}^\infty a_k t^{k+1}
]
Combine the sums to get coefficients for positive ( n ):

• For ( n=0 ): ( c_0 = \frac{3}{4}a_0 = \frac{3}{4} \cdot \frac{1}{4} = \frac{3}{16} )
• For ( n \geq 1 ): ( c_n = \frac{3a_n - a_{n-1}}{4} )

Using the recurrence for ( a_k ), you can compute ( c_n ) recursively for any positive ( n ). For example:

• ( c_1 = \frac{3a_1 - a_0}{4} = \frac{3 \cdot \frac{3}{16} - \frac{1}{4}}{4} = \frac{5}{64} )
• ( c_2 = \frac{3a_2 - a_1}{4} = \frac{3 \cdot \frac{5}{64} - \frac{3}{16}}{4} = \frac{3}{256} )

• Convergence region: ( 0 < |z+1| < 2 ); the outer convergence radius ( R = 2 )
• Coefficients:
  • ( c_n = 0 ) for ( n \leq -2 )
  • ( c_{-1} = \frac{1}{4} )
  • ( c_0 = \frac{3}{16} )
  • ( c_n = \frac{3a_n - a_{n-1}}{4} ) for ( n \geq 1 ), where ( a_k = \frac{3a_{k-1} - a_{k-2}}{4} ), ( a_0 = \frac{1}{4} ), ( a_1 = \frac{3}{16} )

内容的提问来源于stack exchange，提问作者agneau