Transcript Power series
ENGG2013 Unit 21 Power Series
Apr, 2011.
Charles Kao
• • Vice-chancellor of CUHK from 1987 to 1996.
Nobel prize laureate in 2009.
K. C. Kao and G. A. Hockham, " Dielectric-fibre surface waveguides for optical frequencies ,"
Proc. IEE,
vol. 133, no. 7, pp.1151
–1158, 1966.
“It is foreseeable that glasses with a bulk loss of about 20 dB/km at around 0.6 micrometer will be obtained, as the iron impurity concentration may be reduced to 1 part per million.” kshum 2
Special functions
From the first paragraph of Prof. Kao’s paper (after abstract), we see • • • J n = nth-order Bessel function of the first kind K n = nth-order modified Bessel function of the second kind.
H (i) = th-order Hankel function of the ith type.
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• • •
J
(x)
There is a parameter called the “order”.
The th-order Bessel function of the first kind – http://en.wikipedia.org/wiki/Bessel_function Two different definitions: – Defined as the solution to the differential equation – Defined by power series: kshum 4
Gamma function
(x)
• • Gamma function is the extension of the factorial function to real integer input.
– http://en.wikipedia.org/wiki/Gamma_function Definition by integral • Property : (1) = 1, and for integer n, (n)=(n – 1)!
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Examples
• The 0-th order Bessel function of the first kind • The first order Bessel function of the first kind kshum 6
INFINITE SERIES
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•
Infinite series
Geometric series – If a = 1 and r= 1/2, – If a = 1 and r = 1 1+1+1+1+1+… diverges – If a = 1 and r = – 1 1 – 1 + 1 – 1 + 1 – 1 + … – If a = 1 and r = 2 1+2+4+8+16+… kshum diverges diverges = 1 8
Formal definition for convergence
• Consider an infinite series – The numbers a i may be real or complex.
• Let S n be the nth partial sum • The infinite series is said to be convergent if there is a number L such that, for every arbitrarily small > 0 , there exists an integer N such that • The number L is called the limit of the infinite series.
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Geometric pictures
Complex infinite series Im Complex plane S 2 S 1 S 0 L Re L Real infinite series L L+ kshum 10
Convergence of geometric series
• If |r|<1, then converges, and the limit is equal to .
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Easy fact
• If the magnitudes of the terms in an infinite series does not approach zero, then the infinite series diverges.
• But the converse is not true.
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Harmonic series
is divergent kshum 13
But
is convergent kshum 14
Terminologies
• • An infinite series z 1 +z 2 +z 3 +… is called absolutely convergent if |z 1 |+|z 2 |+|z 3 |+… is convergent.
An infinite series z 1 +z 2 +z 3 +… is called conditionally convergent if z 1 +z 2 +z 3 +… is convergent, but |z 1 |+|z 2 |+|z 3 |+… is divergent.
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Examples
• • is conditionally convergent.
is absolutely convergent.
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Convergence tests
Some sufficient conditions for convergence.
Let z 1 + z 2 + z 3 + z 4 + … be a given infinite series.
(z 1 , z 2 , z 3 , … are real or complex numbers) 1. If it is absolutely convergent, then it converges.
2. (Comparison test) If we can find a convergent series b 1 + b 2 + b 3 + … with non-negative real terms such that |z i | b i for all i, then z 1 + z 2 + z 3 + z 4 + … converges.
http://en.wikipedia.org/wiki/Comparison_test kshum 17
Convergence tests
3. (Ratio test) If there is a real number q < 1, such that for all i > N (N is some integer), then z 1 + z 2 + z 3 + z 4 + … converges.
If for all i > N , , then it diverges kshum http://en.wikipedia.org/wiki/Ratio_test 18
Convergence tests
4. (Root test) If there is a real number q < 1, such that for all i > N (N is some integer), then z 1 + z 2 + z 3 + z 4 + … converges.
If for all i > N , , then it diverges.
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•
Derivation of the root test from comparison test
Suppose that for all i N. Then for all i N. But is a convergent series (because q<1). Therefore z 1 + z 2 + z 3 + z 4 + … converges by the comparison test.
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Application
• Given a complex number x, apply the ratio test to • The ratio of the (i+1)-st term and the i-th term is Let q be a real number strictly less than 1, say q=0.99. Then, Therefore exp(x) is convergent for all complex number x. kshum 21
Application
• Given a complex number x, apply the root test to • The ratio of the (i+1)-st term and the i-th term is Let q be a real number strictly less than 1, say q=0.99. Then, Therefore exp(x) is convergent for all complex number x. kshum 22
Variations: The limit ratio test
• If an infinite series z 1 + z 2 + z 3 terms nonzero, is such that + … , with all Then 1. The series converges if < 1.
2. The series diverges if > 1.
3. No conclusion if = 1.
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Variations: The limit root test
• If an infinite series z 1 + z 2 + z 3 terms nonzero, is such that + … , with all Then 1. The series converges if < 1.
2. The series diverges if > 1.
3. No conclusion if = 1.
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Application
• Let x be a given complex number. Apply the limit root test to • The nth term is • The nth root of the magnitude of the nth term is kshum 25
•
Useful facts
• Stirling approximation: for all positive integer n, we have J 0 (x) converges for every x kshum 26
POWER SERIES
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General form
• • The input, x, may be real or complex number.
The coefficient of the nth term, a n , may be real or complex number.
http://en.wikipedia.org/wiki/Power_series kshum 28
Approximation by tangent line
2 1.5
1 0.5
0 -0.5
-1 -1.5
-2 -2.5
0 0.2
0.4
0.6
0.8
1 x y = log(x) Tangent line at x=0.6
1.2
1.4
1.6
1.8
2 x = linspace(0.1,2,50); plot(x,log(x),'r',x, log(0.6)+(x-0.6)/0.6,'b') grid on; xlabel('x'); ylabel('y'); legend(‘y = log(x)’, ‘Tangent line at x=0.6‘) kshum 29
Approximation by quadratic
1 y = log(x) Second-order approx at x=0.6
0.5
0 -0.5
-1 -1.5
-2 0.2
0.4
0.6
0.8
x = linspace(0.1,2,50); plot(x,log(x),'r',x, log(0.6)+(x-0.6)/0.6-(x-0.6).^2/0.6^2/2,'b') grid on; xlabel('x'); ylabel('y') legend(‘y = log(x)’, ‘Second-order approx at x=0.6‘) 1 x 1.2
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1.6
1.8
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Third-order
-1 -2 4 3 2 1 0 y = log(x) Third-order approx at x=0.6
-3 0 0.2
0.4
0.6
0.8
1 x 1.2
1.4
1.6
1.8
x = linspace(0.05,2,50); plot(x,log(x),'r',x, log(0.6)+(x-0.6)/0.6-(x-0.6).^2/0.6^2/2+(x-0.6).^3/0.6^3/3,'b') grid on; xlabel('x'); ylabel('y') legend('y = log(x)', ‘Third-order approx at x=0.6') kshum 2 31
Fourth-order
1 0 -1 -2 -3 -4 y = log(x) Fourth-order approx at x=0.6
-5 0 0.2
0.4
0.6
0.8
1 x 1.2
1.4
1.6
1.8
2 x = linspace(0.05,2,50); plot(x,log(x),'r',x, log(0.6)+(x-0.6)/0.6-(x-0.6).^2/0.6^2/2+(x-0.6).^3/0.6^3/3-(x-0.6).^4/0.6^4/4,'b') grid on; xlabel('x'); ylabel('y') legend(‘y = log(x)’, ‘Fourth-order approx at x=0.6‘) kshum 32
Fifth-order
10 8 6 0 -2 4 2 y = log(x) Fifth-order approx at x=0.6
-4 0 0.2
0.4
0.6
0.8
x 1 1.2
1.4
1.6
1.8
2 x = linspace(0.05,2,50); plot(x,log(x),'r',x, log(0.6)+(x-0.6)/0.6-(x-0.6).^2/0.6^2/2+(x-0.6).^3/0.6^3/3-(x-0.6).^4/0.6^4/4+(x-0.6).^5/0.6^5/5,'b') grid on; xlabel('x'); ylabel('y') legend(‘y = log(x)’, ‘Fifth-order approx at x=0.6‘) kshum 33
•
Taylor series
Local approximation by power series.
• Try to approximate a function f(x) near x
0
, by
a 0 + a 1 (x – x 0 ) + a 2 (x – x 0 ) 2 + a 3 (x – x 0 ) 3 + a 4 (x – x 0 ) 4 + …
• •
x 0
is called the centre. When x
0
= 0, it is called Maclaurin series.
a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + a 6 x 6 + …
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Taylor series and Maclaurin series
Brook Taylor English mathematician 1685 —1731 kshum Colin Maclaurin Scottish mathematician 1698 —1746 35
Geometric series
Examples
Exponential function Sine function Cosine function More examples at http://en.wikipedia.org/wiki/Maclaurin_series kshum 36
• • • • –
How to obtain the coefficients
Match the derivatives at x =x
0
Set x = x 0 in f(x) = a
0 +a 1 (x – x 0 )+a 2 (x – x 0 ) 2 +a 3 (x – x 0 ) 3 +...
a 0 = f(x 0 )
Set x = x 0 in f’(x)
a 1 = f’(x 0 )
= a
1 +2a 2 (x – x 0 ) +3a 3 (x – x 0 ) 2 +…
Set x = x 0 in f’’(x) = 2a
2
+6a
3 (x – x 0 ) +12a 4 (x – x 0 ) 2 +…
a 2 = f’’(x 0 )/2
In general, we have a
k = f (k) (x 0 ) / k!
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• • •
Example f(x) = log(x), x
0
=0.6
First-order approx.
log(0.6)+(x – 0.6)/0.6
Second-order approx.
log(0.6)+(x – 0.6)/0.6 – (x – 0.6) 2 /(2· 0.6
2 ) Third-order approx.
log(0.6)+(x–0.6)/0.6 – (x–0.6) 2 /(2· 0.6
2 ) +(x–0.6) 3 /(3· 0.6
3 ) kshum 38
• •
Example: Geometric series
Maclaurin series
1/(1– x) = 1+x+x 2 +x 3 +x 4 +x 5 +x 6 +… Equality holds when |x| < 1
• If we carelessly substitute x=1.1, then L.H.S. of
1/(1– x) = 1+x+x 2 +x 3 +x 4 +x 5 +x 6 +…
is equal to -10, but R.H.S. is not well-defined.
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• • •
Radius of convergence for GS
For the geometric series 1+z+z
2 +z 3
+… , it converges if |z| < 1 , but diverges when |z| > 1
.
We say that the radius of convergence is 1 .
1+z+z 2 +z 3
+… converges inside the unit disc, and diverges outside.
complex plane kshum 40
Convergence of Maclaurin series in general
• If the power series f(x) converges at a point x 0 , then it converges for all x such that |x| < |x 0 | in the complex plane.
Im Re x 0 Proof by comparison test kshum 41
Convergence of Taylor series in general
• If the power series f(x) converges at a point x 0 , then it converges for all x such that |x – c| < |x 0 – c| in the complex plane.
Im c R x 0 Re Proof by comparison test also kshum 42
Region of convergence
• • The region of convergence of a Taylor series with center c is the smallest circle with center c , which contains all the points at which f(x) converges. The radius of the region of convergence is called the radius of convergence of this Taylor series.
Im diverge R c Re kshum 43
•
Examples
: radius of convergence = 1. It converges • • at the point z= –1, but diverges for all |z|>1. exp(z): radius of convergence is , because it converges everywhere.
: radius of convergence is 0, because it diverges everywhere except z=0. kshum 44
Behavior on the circle of convergence
• On the circle of convergence |z-c| = R, a Taylor series may or may not converges.
• All three series z n , z n /n, and z n /n 2 Have the same radius of convergence R=1.
R But z n z n diverges everywhere on |z|=1, /n diverges at z= 1 and converges at z=– 1 , z n /n 2 converges everywhere on |z|=1.
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Summary
• • • Power series is useful in calculating special functions, such as exp(x), sin(x), cos(x), Bessel functions, etc.
The evaluation of Taylor series is limited to the points inside a circle called the region of convergence.
We can determine the radius of convergence by root test, ratio test, etc.
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