lec39.ppt
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The document discusses Taylor and Laurent series expansions. It provides examples of using these expansions to represent functions around points. Taylor series provides a power series representation of an analytic function around a point. Laurent series allows representing functions in annular regions, including points where the function is not analytic, using both positive and negative powers of (z - z0). Examples show deducing Laurent series expansions for simple functions like z4 and 1/z4 around various points, and evaluating coefficients via contour integrals and the residue theorem. The document also gives an example of using a contour integral to compute a Greens function in many-particle physics.
![NPTEL Physics Mathematical Physics - 1
About a point a
4 = ( + )4 = ( + )4
= 4 + 43 + 622 + 43 + 4
Here 0 = 4, 1 = 43, 2 = 62, 3 = 4, 4 = 1
Thus all the coefficients are non zero.
(ii) Do the same as () () for () =
for small | |?
1
. What is the region of validity
4
() = 1
= 1
(+) (+)
1
=
1
= (1+
)
= (1 + )
1
Then,
() = 1
[1
+ (+1)
(
) (+1)(+2)
(
)]
2! 3!
2
=
The region of validity is
|
| < 1
The region of validity has to be specified, as
Otherwise we cannot do the binomial expansion.
For =
= 1
1
= 1 1
4 (+)4 (+)4
=
4(1+
)
4
5 4
= (1 4 + 10 20 + 35 )
1
4 2
2 3
3
4
For small | |, the region of validity is
|
| < 1
Or, | | <
That is for a point inside a circle.
Example 2.
Derive the Laurent series by evaluating the coefficients via the residue
theorem for () = ю where = , the point of expansion is and
= 0 0
the coefficients are given by,
=
1
2 ( )
Joint initiative of IITs and IISc Funded by MHRD Page 50 of 66
(р)р
+ +1
0
Where + must enclose the point of expansion 0.](https://image.slidesharecdn.com/lec39-231023101042-b815e200/85/lec39-ppt-2-320.jpg)
![NPTEL Physics Mathematical Physics - 1
a) Take () = , i.e. no singularity, 0 is finite
b) Take () = 1
i.e. has a singularity at the point of expansion and = 1
р 0
Solution
a) = 0 + ( 0) = 0 + 1 where = 0
Laurent series is valid in an annular region. () is analytic inside the annular region and
on the boundary of the ring spaped region bounded by two concentric circles 1 2 or
radii 1 and 2 respectively (1 > 2). Here the outer boundary may extend upto infinity
(or radius 1) and the inner region can be reduced to a small circle containing the point of
expansion point 0. We shall use the residue theorem to find the coefficients
=
1 (卒 )卒 1
2 (卒 )
= ()
0
+1 2
+ +
=
1
2
[2 (sum of residue of ())]
Now,
=
1
1
(1)!
瑞
р霞0
1
1 [( ) ()]
р 0
Here j = n + 1
Thus,
1 = !
瑞 р介 [(р 0)
1
р霞0
+1
(р )
0
+1]
= 1
!
р霞0
瑞
z
р
Hence it is clear that if > 1, 1 = 0
For = 1
Residue =
1
р介р = 0
2
Joint initiative of IITs and IISc Funded by MHRD Page 51 of 66](https://image.slidesharecdn.com/lec39-231023101042-b815e200/85/lec39-ppt-3-320.jpg)
![NPTEL Physics Mathematical Physics - 1
2
Similarly for = 2, Residue =
integral theorem)
1
р(р )
0 р
+ = 0 (by Cauchys
Thus Residue = 1 for = 1
Residue = 0 for = 0
b) () =
1
, = 1
0
At the point of expression, there is no singularity
1 = (1)!
1 1
р
1 [(р 0)
(р)
(р0)+1]
Here the residue of the function = 1
has a ( + 1)≠ order pole
at 0
So = + 1
Example 3.
In dealing with a many particle system of bosons, the Greens function is written
as,
(0) = 2
(署ゐ1)( +
1
2
0
2)
Where 0 is some characteristic frequency of the boson and ゐ is the Bosonic
Matsubara frequency.
ゐ =
2
where = 0, 賊1, 賊2 and is the inverse temperature.
Compute the Greens function using an appropriate contour.
Solution
Write = ゐ
(0) = 2
(署р1)(2+
1
02)
The integrand on the RHS has two simple poles on the real axis at 賊0 and a
branch cut coming from the first bracket in the denominator. Thus the contour is
as follows-
Joint initiative of IITs and IISc Funded by MHRD Page 52 of 66](https://image.slidesharecdn.com/lec39-231023101042-b815e200/85/lec39-ppt-4-320.jpg)
![NPTEL Physics Mathematical Physics - 1
1 is the contour in the complex w-plane encircling the poles of (署も1)
which lie
1
on the imaginary axis at = 2
.
We can obtain the Greens function in a closed form by deforming the contour 1
to a contour 2 encircle the simple poles 賊0. The integrals over the large arcs
involved in the deformation vanish
Thus,
1
2 (署0 1)(2+0
2)
=
1 1
20 署0 1
[ +
1
1署0
]
=
1
20
≠ ( )
Joint initiative of IITs and IISc Funded by MHRD Page 53 of 66
署0
2](https://image.slidesharecdn.com/lec39-231023101042-b815e200/85/lec39-ppt-5-320.jpg)
![Mathematics of nyquist plot [autosaved] [autosaved]](https://cdn.slidesharecdn.com/ss_thumbnails/mathematicsofnyquistplotautosavedautosaved-150219123133-conversion-gate02-thumbnail.jpg?width=560&fit=bounds)
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lec39.ppt
- 1. NPTEL Physics Mathematical Physics - 1 Lecture 39 TAYLOR & LAURENT SERIES Taylor series: Let = 0 be a point in the region where () is analytic, then there is a unique power series expansion of () given by, () = ( 0) where the coefficients = =0 1 ! ю = | 0 Laurent Series: Let 1 and 2 be circles both centered at 0(2 being the inner circle), and a function () be analytic everywhere in 1 and 2 (including the annular region). Then at any point z, the value of f is given by the convergent power series expansion (called Laurent expansion) of () around 0 () = ( 0) = With, = 1 2 ( ) (р) р 0 +1 Where is any contour lying in and encircling the point 0 Example 1. Use of partial fraction and Binomial Theorem to deduce Laurent-Series (i) Deduce the Laurent series for () = 4 if the expansion point is (a) ) origin, (b) an arbitary point a. a) Let = 0, here 0 = 0 () = ю = 4 = () = 44, 4 = 1 all other coefficients are zero. b) Again () = ю = 4 = Joint initiative of IITs and IISc Funded by MHRD Page 49 of 66
- 2. NPTEL Physics Mathematical Physics - 1 About a point a 4 = ( + )4 = ( + )4 = 4 + 43 + 622 + 43 + 4 Here 0 = 4, 1 = 43, 2 = 62, 3 = 4, 4 = 1 Thus all the coefficients are non zero. (ii) Do the same as () () for () = for small | |? 1 . What is the region of validity 4 () = 1 = 1 (+) (+) 1 = 1 = (1+ ) = (1 + ) 1 Then, () = 1 [1 + (+1) ( ) (+1)(+2) ( )] 2! 3! 2 = The region of validity is | | < 1 The region of validity has to be specified, as Otherwise we cannot do the binomial expansion. For = = 1 1 = 1 1 4 (+)4 (+)4 = 4(1+ ) 4 5 4 = (1 4 + 10 20 + 35 ) 1 4 2 2 3 3 4 For small | |, the region of validity is | | < 1 Or, | | < That is for a point inside a circle. Example 2. Derive the Laurent series by evaluating the coefficients via the residue theorem for () = ю where = , the point of expansion is and = 0 0 the coefficients are given by, = 1 2 ( ) Joint initiative of IITs and IISc Funded by MHRD Page 50 of 66 (р)р + +1 0 Where + must enclose the point of expansion 0.
- 3. NPTEL Physics Mathematical Physics - 1 a) Take () = , i.e. no singularity, 0 is finite b) Take () = 1 i.e. has a singularity at the point of expansion and = 1 р 0 Solution a) = 0 + ( 0) = 0 + 1 where = 0 Laurent series is valid in an annular region. () is analytic inside the annular region and on the boundary of the ring spaped region bounded by two concentric circles 1 2 or radii 1 and 2 respectively (1 > 2). Here the outer boundary may extend upto infinity (or radius 1) and the inner region can be reduced to a small circle containing the point of expansion point 0. We shall use the residue theorem to find the coefficients = 1 (卒 )卒 1 2 (卒 ) = () 0 +1 2 + + = 1 2 [2 (sum of residue of ())] Now, = 1 1 (1)! 瑞 р霞0 1 1 [( ) ()] р 0 Here j = n + 1 Thus, 1 = ! 瑞 р介 [(р 0) 1 р霞0 +1 (р ) 0 +1] = 1 ! р霞0 瑞 z р Hence it is clear that if > 1, 1 = 0 For = 1 Residue = 1 р介р = 0 2 Joint initiative of IITs and IISc Funded by MHRD Page 51 of 66
- 4. NPTEL Physics Mathematical Physics - 1 2 Similarly for = 2, Residue = integral theorem) 1 р(р ) 0 р + = 0 (by Cauchys Thus Residue = 1 for = 1 Residue = 0 for = 0 b) () = 1 , = 1 0 At the point of expression, there is no singularity 1 = (1)! 1 1 р 1 [(р 0) (р) (р0)+1] Here the residue of the function = 1 has a ( + 1)≠ order pole at 0 So = + 1 Example 3. In dealing with a many particle system of bosons, the Greens function is written as, (0) = 2 (署ゐ1)( + 1 2 0 2) Where 0 is some characteristic frequency of the boson and ゐ is the Bosonic Matsubara frequency. ゐ = 2 where = 0, 賊1, 賊2 and is the inverse temperature. Compute the Greens function using an appropriate contour. Solution Write = ゐ (0) = 2 (署р1)(2+ 1 02) The integrand on the RHS has two simple poles on the real axis at 賊0 and a branch cut coming from the first bracket in the denominator. Thus the contour is as follows- Joint initiative of IITs and IISc Funded by MHRD Page 52 of 66
- 5. NPTEL Physics Mathematical Physics - 1 1 is the contour in the complex w-plane encircling the poles of (署も1) which lie 1 on the imaginary axis at = 2 . We can obtain the Greens function in a closed form by deforming the contour 1 to a contour 2 encircle the simple poles 賊0. The integrals over the large arcs involved in the deformation vanish Thus, 1 2 (署0 1)(2+0 2) = 1 1 20 署0 1 [ + 1 1署0 ] = 1 20 ≠ ( ) Joint initiative of IITs and IISc Funded by MHRD Page 53 of 66 署0 2