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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