Laurent Series Harder Example

1. Formula
   Observation

   $f(z) = \frac{1}{z+z^2}$

$f(z)$

The complex function for which the Laurent series is being found.

Complex numbers excluding $z=0$ and $z=-1$

1. Formula
   Observation

   $\frac{1}{1-w} = \sum_{n=0}^{\infty} w^n$

w

The variable in the standard geometric series formula.

Complex numbers with |w| < 1

1. Formula
   Observation

   $\sum_{n=0}^{\infty} w^n$

n

The index of summation in the geometric series.

Non-negative integers (0, 1, 2, ...)

1. Formula
   Observation

   The function is written as $f(z) = 1/(z+z^2)$.

$f(z)$

The rational function being expanded into a Laurent series.

Complex variable z.

1. Formula
   Observation

   The variable z appears throughout the algebraic manipulations and the region condition.

z

Complex variable in the Laurent series expansion.

Region |$z+1$| > 1.

1. Formula
   Observation

   The geometric series formula uses w: $1/(1-w)$ = sum_{$n=0$}^infty $w^n$ if |w| < 1.

w

Dummy variable in the standard geometric series formula.

Complex number with |w| < 1.

1. Formula
   Observation

   The summation index n runs from 0 to infinity in the geometric series formula.

n

Summation index for the geometric series.

Non-negative integers.

1. Formula
   Observation

   The symbol z appears throughout the derivation as the variable in expressions such as ($z+1$)^{-n}, ($z+1$)^{-k}, and |$1/(z+1)$|<1.

z

Complex variable appearing in the Laurent-series expansion.

Complex numbers; the displayed convergence condition requires |$1/(z+1)$|<1, equivalently |$z+1$|>1.

1. Formula
   Observation

   n is used as the summation index in $\sum_{n=0}^{\infty}$ and in exponents such as ($z+1$)^{-n} and -($n+2$).

n

Integer summation index in the original series.

Nonnegative integers, $n=0$,1,2,$\ldots$

1. Formula
   Observation

   k is introduced by the substitution $k=n+2$ and then used as the new summation index in $\sum_{k=2}^{\infty}(z+1)^{-k}$.

k

Reindexed summation variable after the substitution $k=n+2$.

Integers with $k\ge 2$ in the final series.

1. Formula
   Observation

   $\frac{1}{1-w} = \sum_{n=0}^{\infty} w^n \text{ if }$ |w| < 1

The sum of an infinite geometric series with first term 1 and common ratio w is given by the formula, provided the absolute value of w is strictly less than 1.

1. |w| < 1

1. Audio
   Observation

   We need something to be less than one... we need to write this inequality as an inequality that is less than one.

2. Formula
   Observation

   1 < |$z+1$| $\implies \frac{1}{|z+1|} < 1$

To find a Laurent series in a specific annular region, manipulate the given function algebraically to match the form of a known series expansion (like the geometric series), ensuring the new variable satisfies the convergence condition derived from the region's inequality.

1. Target region must be expressible as an inequality involving a modulus less than 1.

1. Geometric Series Formula

1. Formula
   Observation

   The problem statement asks to find the Laurent series for $f(z) = 1/(z+z^2)$ in |$z+1$| > 1.

The video presents a problem of finding the Laurent series expansion of a rational function around a specific point within a given annular region.

1. Region: |$z+1$| > 1

1. Formula
   Observation

   The formula $1/(1-w)$ = sum_{$n=0$}^infty $w^n$ if |w| < 1 is written on the board and used as the basis for the expansion.

The standard geometric series formula is used to expand terms of the form $1/(1-w)$ into an infinite sum, provided the magnitude of w is less than 1.

1. |w| < 1

1. Formula
   Observation

   The board shows a geometric-series factor $\frac{1}{1-\left(\frac{1}{z+1}\right)}$ rewritten as $\sum_{n=0}^{\infty}\left(\frac{1}{z+1}\right)^n$.

2. Audio
   Observation

   The narrator says, "The infinite sum as n runs from zero to infinity of one over z plus one to the n."

The factor $\frac{1}{1-(1/(z+1))}$ is expanded as an infinite geometric series in powers of $\frac{1}{z+1}$. This is the step that converts the rational expression into a series suitable for Laurent-series manipulation.

1. |$1/(z+1)$|<1

2. equivalently |$z+1$|>1

1. Formula
   Observation

   The board rewrites $\left(\frac{1}{z+1}\right)^n$ as $\frac{1^n}{(z+1)^n}$ and then as $\frac{1}{(z+1)^n}$.

2. Audio
   Observation

   The narrator explains, "one over z plus one to the n, that's really one to the n over z plus one to the n, which is really just one over z plus one to the n."

To simplify the general term, the power n is distributed across numerator and denominator. Since $1^n=1$, the term becomes a single reciprocal power of ($z+1$).

1. n is a nonnegative integer

1. Geometric-series form of the Laurent expansion factor

1. Formula
   Observation

   The board combines the outside factor $\frac{1}{(z+1)^2}$ with the series term $\frac{1}{(z+1)^n}$ to obtain $\sum_{n=0}^{\infty}\frac{1}{(z+1)^{n+2}}$.

2. Audio
   Observation

   The narrator says, "Now we'll distribute. So we end up with the sum as n runs from zero to infinity of one over z plus one to the n plus two."

The constant multiplier $\frac{1}{(z+1)^2}$ is absorbed into each summand by adding exponents, producing a single power ($z+1$)^{-($n+2$)} inside the summation.

1. the series is already in the form obtained from the geometric expansion

1. Distributing the exponent over the fraction

1. Formula
   Observation

   The board rewrites $\frac{1}{(z+1)^{n+2}}$ as ($z+1$)^{-($n+2$)}.

2. Audio
   Observation

   The narrator says, "z plus one, you can bring it upstairs and so you get negative n plus two."

A denominator power is moved to the numerator by changing the sign of the exponent. This prepares the expression for reindexing in standard Laurent-series notation.

1. $z+1\neq 0$

1. Combining the prefactor with the series term

1. Formula
   Observation

   The board writes "Let $k=n+2$" and below it "$n=k-2$".

2. Audio
   Observation

   The narrator says, "maybe let k equal n plus two. And when you do that, n ends up being k minus two."

The summation index is changed by introducing a new variable k equal to the old exponent shift $n+2$. Solving for n gives $n=k-2$, which allows the lower limit and exponent to be rewritten in terms of k.

1. n starts at 0 in the original sum

1. Rewriting a reciprocal power with a negative exponent

1. Formula
   Observation

   The final line on the board is $\sum_{k=2}^{\infty}(z+1)^{-k}$.

2. Audio
   Observation

   The narrator says, "when n is zero, k is two. And then here we get z plus one to the negative k. And that is the final answer."

After substituting $k=n+2$, the original sum starting at $n=0$ becomes a sum starting at $k=2$, and the exponent becomes -k. This is the final simplified Laurent-series form shown in the clip.

1. |$z+1$|>1

2. k runs over integers with $k\ge 2$

1. Index substitution for reindexing the series

1. Formula
   Observation

   The board explicitly states $\left|\frac{1}{z+1}\right|<1$ next to the geometric-series step.

2. Audio
   Observation

   No separate spoken statement of the convergence condition is heard in this clip.

The expansion $\frac{1}{1-(1/(z+1))}=\sum_{n=0}^{\infty}\left(\frac{1}{z+1}\right)^n$ is valid when $\left|\frac{1}{z+1}\right|<1$.

1. The expression is treated as a geometric series with ratio $1/(z+1)$.

For complex z satisfying |$1/(z+1)$|<1.

1. Formula
   Observation

   The board shows Let $k=n+2$, $n=k-2$, and then the final sum $\sum_{k=2}^{\infty}(z+1)^{-k}$.

2. Audio
   Observation

   The narrator says, "when n is zero, k is two."

If $k=n+2$ and n begins at 0, then the new index k begins at 2, so $\sum_{n=0}^{\infty}(z+1)^{-(n+2)}$ becomes $\sum_{k=2}^{\infty}(z+1)^{-k}$.

1. n is an integer index starting at 0

2. k is defined by $k=n+2$

For all integer $n\ge 0$, with corresponding $k\ge 2$.

1. Formula
   Observation

   $f(z) = \frac{1}{z+z^2} = \frac{1}{z(1+z)} = \frac{1}{(z+1-1)(1+z)}$

1. Expression

   $$f(z) = \frac{1}{z+z^2}$$
   Explanation

   Start with the original function.

   Justification

   Given in the problem statement.

   Shown in the video
2. Expression

   $$f(z) = \frac{1}{z(1+z)}$$
   Explanation

   Factor out z from the denominator.

   Justification

   Basic algebraic factoring.

   Shown in the video
3. Expression

   $$f(z) = \frac{1}{(z+1-1)(1+z)}$$
   Explanation

   Add and subtract 1 to the factor z to create the term ($z+1$).

   Justification

   Algebraic identity $z = (z+1) - 1$.

   Shown in the video

The function is rewritten to expose the ($z+1$) term, preparing it for substitution into the geometric series formula.

1. Audio
   Observation

   The speaker narrates the steps: factoring out negative one, rewriting $1+z$ as $z+1$, factoring out $z+1$, and distributing the negative one back.

2. Formula
   Observation

   The board shows the step-by-step algebraic transformation of $f(z)$ into a form suitable for the geometric series.

1. Expression

   $$f(z) = \frac{1}{z+z^2} = \frac{1}{z(1+z)}$$
   Explanation

   Factor the denominator of the original function.

   Justification

   Algebraic factoring.

   Shown in the video
2. Expression

   $$= \frac{1}{(z+1-1)(1+z)}$$
   Explanation

   Rewrite z as ($z+1$)-1 to center the expansion around $z=-1$.

   Justification

   Algebraic manipulation.

   Shown in the video
3. Expression

   $$= \frac{1}{-(1-(z+1))(z+1)}$$
   Explanation

   Factor out -1 from the term ($z+1-1$).

   Justification

   Algebraic factoring.

   Shown in the video
4. Expression

   $$= \frac{-1}{(1-(z+1))(z+1)}$$
   Explanation

   Move the negative sign to the numerator.

   Justification

   Algebraic manipulation.

   Shown in the video
5. Expression

   $$= \frac{-1}{z+1} \cdot \frac{1}{1-(z+1)}$$
   Explanation

   Separate the fraction into two parts.

   Justification

   Algebraic manipulation.

   Shown in the video
6. Expression

   $$= \frac{-1}{(z+1)^2} \cdot \frac{1}{\frac{1}{z+1}-1}$$
   Explanation

   Factor out another ($z+1$) from the first term and adjust the second term.

   Justification

   Algebraic manipulation.

   Shown in the video
7. Expression

   $$= \frac{1}{(z+1)^2} \cdot \frac{1}{1-\frac{1}{z+1}}$$
   Explanation

   Distribute the negative sign into the denominator of the second term.

   Justification

   Algebraic manipulation.

   Shown in the video

The function is successfully rewritten as $\frac{1}{(z+1)^2} \cdot \frac{1}{1-\frac{1}{z+1}}$, which is now in a form where the geometric series formula can be applied with $w = \frac{1}{z+1}$.

1. Formula
   Observation

   The full board sequence shows the transition from $\frac{1}{(z+1)^2}\cdot\frac{1}{1-(1/(z+1))}$ to $\sum_{k=2}^{\infty}(z+1)^{-k}$.

2. Audio
   Observation

   The narration tracks the same sequence: expand as an infinite sum, distribute the power, combine exponents, rewrite with negative exponent, then reindex using $k=n+2$.

1. Expression

   $$\frac{1}{(z+1)^2}\cdot\frac{1}{1-\left(\frac{1}{z+1}\right)}$$
   Explanation

   Start from the displayed product containing the geometric-series factor.

   Justification

   This is the expression already present on the board at the beginning of the clip.

   Shown in the video
2. Expression

   $$\frac{1}{(z+1)^2}\sum_{n=0}^{\infty}\left(\frac{1}{z+1}\right)^n$$
   Explanation

   Replace the geometric factor by its infinite-series expansion.

   Justification

   Use the geometric-series formula under the stated condition |$1/(z+1)$|<1.

   Shown in the video
3. Expression

   $$\frac{1}{(z+1)^2}\sum_{n=0}^{\infty}\frac{1}{(z+1)^n}$$
   Explanation

   Simplify the general term by distributing the exponent over the fraction.

   Justification

   Since $\left(\frac{1}{z+1}\right)^n=\frac{1^n}{(z+1)^n}=\frac{1}{(z+1)^n}$.

   Shown in the video
4. Expression

   $$\sum_{n=0}^{\infty}\frac{1}{(z+1)^{n+2}}$$
   Explanation

   Absorb the outside factor into each summand by adding exponents.

   Justification

   Multiplying $\frac{1}{(z+1)^2}$ by $\frac{1}{(z+1)^n}$ gives $\frac{1}{(z+1)^{n+2}}$.

   Shown in the video
5. Expression

   $$\sum_{n=0}^{\infty}(z+1)^{-(n+2)}$$
   Explanation

   Rewrite the reciprocal power using a negative exponent.

   Justification

   By the rule $\frac{1}{a^m}=a^{-m}$.

   Shown in the video
6. Expression

   $$\sum_{k=2}^{\infty}(z+1)^{-k}$$
   Explanation

   Reindex the sum with $k=n+2$.

   Justification

   When $n=0$, $k=2$; and since $n=k-2$, the exponent -($n+2$) becomes -k.

   Shown in the video

The displayed expression simplifies to the final Laurent-series form $\sum_{k=2}^{\infty}(z+1)^{-k}$.

1. Formula
   Observation

   $\text{Find the Laurent series for } f(z) = \frac{1}{z+z^2} \text{ in } 1$ < |$z+1$|

1. The final series expression is not reached within the 120-second clip.

Find the Laurent series for $f(z) = \frac{1}{z+z^2}$ in the region 1 < |$z+1$|.

1. $f(z) = \frac{1}{z+z^2}$

2. Region: 1 < |$z+1$|

Express $f(z)$ as a Laurent series valid in the given region.

1. Expression

   $$\frac{1}{1-w} = \sum_{n=0}^{\infty} w^n \text{ if } |w| < 1$$
   Explanation

   State the geometric series formula as the target template.

   Justification

   Standard mathematical formula.

   Shown in the video
2. Expression

   $$1 < |z+1| \implies \frac{1}{|z+1|} < 1$$
   Explanation

   Manipulate the region inequality to find a quantity less than 1.

   Justification

   Dividing both sides by |$z+1$| (which is > $1 > 0$).

   Shown in the video
3. Expression

   $$\text{Goal: } \frac{1}{1 - \left(\frac{1}{z+1}\right)}$$
   Explanation

   Identify the desired form based on the geometric series template.

   Justification

   Matching the template with $w = \frac{1}{z+1}$.

   Shown in the video
4. Expression

   $$f(z) = \frac{1}{z(1+z)} = \frac{1}{(z+1-1)(1+z)}$$
   Explanation

   Algebraically manipulate $f(z)$ to introduce the ($z+1$) term.

   Justification

   Factoring and adding/subtracting 1.

   Shown in the video

The setup is complete, but the final series is not written before the clip ends.

The algebraic steps are sound and lead towards the goal of substituting $w = 1/(z+1)$ into the geometric series.

1. Formula
   Observation

   The entire video is dedicated to solving this specific example.

Find the Laurent series for $f(z) = 1/(z+z^2)$ in the region |$z+1$| > 1.

1. $f(z) = \frac{1}{z+z^2}$

2. Region: |$z+1$| > 1

3. Geometric series formula: $\frac{1}{1-w} = \sum_{n=0}^{\infty} w^n \text{ for }$ |w| < 1

Express $f(z)$ as a Laurent series centered at $z=-1$.

1. Expression

   $$\frac{1}{z+z^2} = \frac{1}{z(1+z)} = \frac{1}{(z+1-1)(z+1)}$$
   Explanation

   Factor the denominator and rewrite z to center around $z+1$.

   Justification

   Algebraic manipulation.

   Shown in the video
2. Expression

   $$= \frac{1}{-(1-(z+1))(z+1)} = \frac{-1}{(1-(z+1))(z+1)}$$
   Explanation

   Factor out -1 to match the 1-w form.

   Justification

   Algebraic manipulation.

   Shown in the video
3. Expression

   $$= \frac{-1}{z+1} \cdot \frac{1}{1-(z+1)}$$
   Explanation

   Separate the terms.

   Justification

   Algebraic manipulation.

   Shown in the video
4. Expression

   $$= \frac{-1}{(z+1)^2} \cdot \frac{1}{\frac{1}{z+1}-1}$$
   Explanation

   Factor out ($z+1$) from the first term and divide the second term's denominator by ($z+1$).

   Justification

   Algebraic manipulation.

   Shown in the video
5. Expression

   $$= \frac{1}{(z+1)^2} \cdot \frac{1}{1-\frac{1}{z+1}}$$
   Explanation

   Distribute the negative sign into the denominator.

   Justification

   Algebraic manipulation.

   Shown in the video
6. Expression

   $$\text{Let } w = \frac{1}{z+1}. \text{ Since } |z+1| > 1, |w| < 1.$$
   Explanation

   Identify w for the geometric series and verify the convergence condition.

   Justification

   Given region condition.

   Shown in the video
7. Expression

   $$= \frac{1}{(z+1)^2} \sum_{n=0}^{\infty} \left(\frac{1}{z+1}\right)^n$$
   Explanation

   Apply the geometric series formula.

   Justification

   Geometric series formula.

   Derived from the video
8. Expression

   $$= \sum_{n=0}^{\infty} \frac{1}{(z+1)^{n+2}}$$
   Explanation

   Combine the powers of ($z+1$).

   Justification

   Exponent rules.

   Derived from the video

$\sum_{n=0}^{\infty} \frac{1}{(z+1)^{n+2}}$

The final series is a valid Laurent series in the specified region |$z+1$| > 1, containing only negative powers of ($z+1$).

1. Audio
   Observation

   The narrator calls it "Long problem, long and hard. This is actually a harder one."

2. Formula
   Observation

   The board carries out a complete worked simplification ending in $\sum_{k=2}^{\infty}(z+1)^{-k}$.

1. The original full problem statement before this clip is not shown in the provided segment.

Simplify the displayed Laurent-series expression beginning from $\frac{1}{(z+1)^2}\cdot\frac{1}{1-\left(\frac{1}{z+1}\right)}$ and rewrite the result as a single indexed series.

1. A geometric-series factor $\frac{1}{1-(1/(z+1))}$

2. An outside multiplier $\frac{1}{(z+1)^2}$

3. The convergence condition $\left|\frac{1}{z+1}\right|<1$ is written on the board

Obtain the final reindexed series form.

1. Expression

   $$\frac{1}{(z+1)^2}\sum_{n=0}^{\infty}\left(\frac{1}{z+1}\right)^n$$
   Explanation

   Expand the geometric factor as an infinite sum.

   Justification

   Geometric-series expansion with ratio $1/(z+1)$.

   Shown in the video
2. Expression

   $$\frac{1}{(z+1)^2}\sum_{n=0}^{\infty}\frac{1}{(z+1)^n}$$
   Explanation

   Simplify each term $\left(\frac{1}{z+1}\right)^n$.

   Justification

   Distribute the exponent and use $1^n=1$.

   Shown in the video
3. Expression

   $$\sum_{n=0}^{\infty}\frac{1}{(z+1)^{n+2}}$$
   Explanation

   Combine the outside factor with the summand.

   Justification

   Add the exponents of ($z+1$).

   Shown in the video
4. Expression

   $$\sum_{n=0}^{\infty}(z+1)^{-(n+2)}$$
   Explanation

   Write the summand with a negative exponent.

   Justification

   Move the denominator power to the numerator.

   Shown in the video
5. Expression

   $$\sum_{k=2}^{\infty}(z+1)^{-k}$$
   Explanation

   Reindex using $k=n+2$.

   Justification

   The lower limit shifts from $n=0$ to $k=2$, and the exponent becomes -k.

   Shown in the video

$\sum_{k=2}^{\infty}(z+1)^{-k}$

The board explicitly checks the new lower limit by noting that when $n=0$, $k=2$.

1. Diagram
   Observation

   Problem statement is green, formula is cyan, region manipulation is yellow, target form is orange, and algebraic steps are white/cyan.

1. Text and formulas on a black background

1. Different colors are used to distinguish the problem statement, the general formula, the region analysis, the target form, and the algebraic manipulation.

1. The black background remains constant.

Color coding helps visually separate the logical components of the solution strategy.

1. Animation
   Observation

   The instructor writes the problem statement, the geometric series formula, and the step-by-step algebraic derivation on a digital blackboard.

1. Problem statement

2. Geometric series formula

3. Algebraic steps

1. New equations are written sequentially below the previous ones.

2. Terms are factored and rearranged visually.

1. The core mathematical equivalence is maintained throughout the steps.

The visual progression clearly demonstrates the algebraic strategy required to transform the function into a form applicable to the geometric series formula.

1. Diagram
   Observation

   The entire clip is a black digital whiteboard with cyan handwritten formulas; near the end, the final summation sign and upper limit are written in yellow while the rest remains cyan.

2. Animation
   Observation

   Formulas are added line by line downward: first the geometric-series expansion, then the simplified power, then the combined exponent, then the substitution $k=n+2$, and finally the reindexed sum.

1. cyan handwritten algebraic expressions

2. summation signs with limits

3. the substitution text Let $k=n+2$ and $n=k-2$

4. yellow final summation sign and upper limit

1. A new line $\sum_{n=0}^{\infty}\left(\frac{1}{z+1}\right)^n$ is written below the earlier geometric factor.

2. The term $\left(\frac{1}{z+1}\right)^n$ is rewritten as $\frac{1}{(z+1)^n}$.

3. The outside factor is absorbed to produce $\sum_{n=0}^{\infty}\frac{1}{(z+1)^{n+2}}$.

4. The expression is rewritten as $\sum_{n=0}^{\infty}(z+1)^{-(n+2)}$.

5. The substitution $k=n+2$ and $n=k-2$ is written on the left.

6. The final line $\sum_{k=2}^{\infty}(z+1)^{-k}$ is written, with the summation symbol and infinity in yellow.

1. The variable z and the base ($z+1$) remain the same throughout.

2. The derivation stays on a single black digital canvas with handwritten notation.

The visual progression mirrors the algebraic logic: expand, simplify the summand, combine powers, switch to negative-exponent notation, then reindex the sum.

1. Audio
   Observation

   The speaker says, 'And you can see that it wasn't really necessary to factor out the negative one. So I guess what we can do is we can distribute the negative one through back again.'

Factoring out a negative one early in the process might seem necessary but can lead to extra steps if not handled carefully.

The speaker realizes the initial factoring of -1 was unnecessary and corrects it by distributing the -1 back into the denominator to achieve the desired 1-w form.

1. Formula
   Observation

   The board explicitly computes the new lower limit by stating that when $n=0$, $k=2$.

2. Audio
   Observation

   The narrator says, "Let's see, when n is zero, k is two."

After substituting $k=n+2$, one might incorrectly keep the sum starting at $k=0$.

Because the original index starts at $n=0$, the new index must start at $k=2$; the video explicitly checks this shift before writing the final series.

1. Audio
   Observation

   So there's a formula that we will use... 1 over 1 minus w...

The geometric series formula is the specific tool applied within the general strategy for finding the Laurent series.

1. Audio
   Observation

   The speaker states, 'So now we can use the formula. This is our w...' referring to the geometric series formula.

The geometric series formula is the fundamental tool applied to derive the Laurent series after the algebraic manipulation.

1. Formula
   Observation

   The clip begins with a geometric-series factor and ends with a Laurent-style negative-power series.

2. Audio
   Observation

   The narration treats the geometric expansion as the starting point for the later reindexed series.

The geometric-series expansion is the tool that produces the initial infinite sum, which is then algebraically transformed into the final Laurent-series form.

1. Formula
   Observation

   The final line $\sum_{k=2}^{\infty}(z+1)^{-k}$ is written only after the substitution $k=n+2$ and the relation $n=k-2$ appear on the board.