Calc1231W5T4 - Recurrence Relations

Friday, 24 August 2018

1:05 PM

Untitled picture.png Machine generated alternative text:
1 x3e¯Xdx can be evaluated by three successive applications of integration by parts. An elegant way to perform successive applications of integration by parts is to use 
The integral 13 = 
a reduction formula 
The fist step, much like recusion or induction, is to have a base case. This is typically a simple integral such as: 
Consider the definite integral 
6- 
1 
xne¯X dc 
1 
e dc 
10 
I-I/e 
The following reduction formula is a compact way of writing the result of integration by parts for the integral In 
for n > 1 
Using this reduction formula 
.11 = 10 -e = 
. 12 = 211 ¯ = 
. 13=312 -e = 
Challenge' You may like to derive the reduction formula yourself. 

Ink Drawings















Untitled picture.png Machine generated alternative text:
.10 
.11 = 
Now use the reduction formula 
tann (x) dc, evaluate 
-In(l /sqrt(2)) 
pi/4 - 2/3 
for n > 2 
1 
to evaluate 
.14 
.15 
-1/4 
- In(l isqrt(2)) 









































Untitled picture.png Machine generated alternative text:
The Euler Integral of the first kind, also known as the Beta Function, is defined by 
B(m, n) 
Its values can be computed recursively as follows. First we find 
1 
t(m—l) (1 
Then we use the reduction formula: 
to evaluate 
m+n—l 
B(m, n) 
n—l 
B(m, n — 1) 
m+n—l 
for m > 1 
for n > 1 
Below is a contour plot of the Beta function; against the m and n axis. 

Untitled picture.png Machine generated alternative text:
The Euler Integral of the first kind, also known as the Beta Function, is defined by 
B(m, n) 
Its values can be computed recursively as follows. First we find 
1 
t(m—l) (1 
Then we use the reduction formula: 
to evaluate 
m+n—l 
B(m, n) 
n—l 
B(m, n — 1) 
m+n—l 
for m > 1 
for n > 1 
Below is a contour plot of the Beta function; against the m and n axis. 

Untitled picture.png Machine generated alternative text:
Let's find a recurrence relation for 
x e¯Xdx. 
The base case is 
x e¯Xdc 
The recurrence relation is used to relate In to the base case 10 _ To achieve this we must successively reduce the value of n using integration by parts. 
Assuming n 2 1: 
xne¯Xdx 
where 
Note the strategic choice of u ensures the power is reduced. With these choices of u, t/ we have 
where 
uddx = nIn—1 
Note: the Maple syntax for e 
u t.'dx 
is exp (x)
Untitled picture.png Machine generated alternative text:
Let's find a recurrence relation for 
x e¯Xdx. 
The base case is 
x e¯Xdc 
The recurrence relation is used to relate In to the base case 10 _ To achieve this we must successively reduce the value of n using integration by parts. 
Assuming n 2 1: 
xne¯Xdx 
where 
Note the strategic choice of u ensures the power is reduced. With these choices of u, t/ we have 
where 
uddx = nIn—1 
Note: the Maple syntax for e 
u t.'dx 
is exp (x) 

Untitled picture.png Machine generated alternative text:
Let's find a reduction formula for the indefinite integral 
If n > 2 then we can rewrite this integral as 
Now use integration by parts to get 
secn (x) dc 
n 2 (x) dc 
(1 + tan @))sec 
n 2 (x) dc 
tan (x) sec 
= In—2 + 
tan 
u/ dc 
= [uv] _ 
sec 
with u = tan@) and v' = 2 This gives 
o and 
Hence, without further integration we have the reduction formula 
where g(n) 
1 
n—l 
Use this to evaluate 
Recall the Maple syntax for sin 
n 2 (x) tan@) + 
tan(x) + 2/3 tan(x) 
-2 
0 
14 
is

 

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