Download Motion in a Straight Line

to PHY226
Mathematical Methods for
Physics and Astronomy
Purpose of the course:
To provide the further mathematical
knowledge and skills needed to fully
complete future physics courses
If you missed Monday pick up notes now
Phil Lightfoot, E47, (24533) [email protected]
IQ test time!!!!
If an even function is defined by f(x)=f(-x) and an odd function
is defined by f(x)= -f(-x), are the following even or odd
are these even, odd, or just a mess….?
(a) cos2(x)
(d) ex
(b) sin2x
(e) e-x
(c) sin(x)cos(x)
(f) 0.5(ex + e-x)
How does this observation allow you to short cut integration
across the origin for any odd or even function?
IQ test time!!!!
If an even function is defined by f(x)=f(-x) and an odd function
is defined by f(x)= -f(-x), are the following even or odd
Cos(x)
(a) cos2(x)
Cos2(x)
So an even function x an even function gives an even function
IQ test time!!!!
If an even function is defined by f(x)=f(-x) and an odd function
is defined by f(x)= -f(-x), are the following even or odd
(b) sin2(x)
Sin(x)
Sin2(x)
So an odd function x an odd function gives an even function
IQ test time!!!!
If an even function is defined by f(x)=f(-x) and an odd function
is defined by f(x)= -f(-x), are the following even or odd
(c) sin(x)cos(x)
sin(x)
sin(x)cos(x)
cos(x)
So an odd function x an even function gives an odd function
IQ test time!!!!
If an even function is defined by f(x)=f(-x) and an odd function
is defined by f(x)= -f(-x), are the following even or odd
(d) ex
(e) e-x
exp(x)
exp(-x)
(f) 0.5(ex + e-x)
Single exponentials are neither but adding gives an even
function (coshx) and subtracting gives an odd function.
IQ test time!!!!
So even x even = even
even x odd = odd
odd x odd = even
How does this observation allow you to short cut integration
across the origin for any odd or even function?
x
x
x
0
 even dx  2 even dx
x
 odd dx  0
x
Complex numbers
i  1
Argand diagram
Cartesian
a + ib
Imaginary
r
b
q
a
Real
Complex numbers
i  1
Argand diagram
z = a + ib
Imaginary
r
b
q
a
Real
z* is the complex
conjugate
z* = a - ib
zz*=(a+ib)(a-ib)=a2+b2=r2
Complex numbers
i  1
Argand diagram
Cartesian
a + ib
Imaginary
r
b
q
Polar
a  r cosq
b  r sin q
Real
a
a  ib  r (cos q  i sin q )
so
where
r  a b
2
2
2
b
q  tan  
a
1
Complex numbers
Polar
a  ib  r (cos q  i sin q )
But remember earlier we showed that…
cos q  1 
q
and
So…
2
2!

q
4
4!
 ...
iq
 ...
and i sin q  iq 
3!
3
q2
3
4
i
q
q
iq
e  1  iq  

 ...
2! 3! 4!
z  a  ib  r (cos q  i sin q )  re
iq
Complex numbers
So what’s the point of complex numbers?
SHM / LHO / Schrodinger / Diffusion of gas / Diffusion of heat /
all of optics / standing waves / …
All these rely heavily on a mixture of trigonometry and calculus
iq
e  cos q  i sin q
Adding
iq
e e
cos q 
2
e
 iq
 cos q  i sin q
Subtracting
 iq
eiq  e iq
sin q 
2i
Replacing all the trig with exp makes the calculus trivial!!!!!
Working with complex numbers
Add / subtract
Multiply / divide
Powers
(a  ib )  (c  id )  (a  c)  i(b  d )
iq1
r1e  r2e
z  re
iq
iq 2
 r1r2e
n niq
z r e
n
i (q1 q 2 )
Working with complex numbers
Roots
Example : If
z  9e
i

what is z½?
3
Step 1: write down z in polars with the 2πp bit added on to the
argument.
z  9e


i   2p 
3

Step 2: say how many values of p you’ll need (as many as n) and
write out the rooted expression …..
i

1
  2p 
here n = 2 so I’ll need 2 values of p; p = 0 and p = 1. 2
2 3

z  9e
Step 3: Work it out for each value of p….
1
2
p = 0 z  9e
p=1
1
2
i

  2 0 
2 3

z  9e
 3e
i

  2 
2 3

i

6
 3e
 7 
i

 6 
 3e
 
i   
6

i
 3e e
i
6

 3e 6