Linear regression

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QUESTION
 
1. Let Jn(a) = Z a
0
1
(x
2 + a
2
)
n
dx where a is a real number and n is a non-negative integer. Evaluate
J0(a) and J1(a). As well, for a 6= 0 and n ≥ 1, find a formula for Jn+1(a) in terms of Jn(a). As
part of your solution, state the formula for integration by parts.
2. Suppose f(x) has a continuous second derivative for all x ∈ R. For any x ∈ R, prove
f(x) = f(0) + f
0
(0)x +
Z x
0
(x − s)f
00(s)ds.
3. Provide an example of an indefinite integral that when integrating requires sin(x) as a substitution. As well, evaluate your indefinite integral.
4. Let g be a continuous function. Prove Z b
a
g(x)dx =
Z a+b
2
a
(g(x) + g(a + b − x))dx.
5. Find and explain the errors (if any) in the following solution.
We evaluate Z
1
1 + t
4
dt by first factorizing the denominator as follows. Notice 1 + t
4 = (1 +
t
2
)
2 − 2t
2 and factoring we have
1 + t
4 = (1 + t
2
)
2 − 2t
2 = (1 + t
2 −
√
2t)(1 + t
2 +
√
2t).
Considering the concept of partial fractions we have
1
1 + t
4
=
At + B
t
2 +
√
2t + 1
+
Ct + D
t
2 −
√
2t + 1
(which implies 1 = (At + B)(t
2 −
√
2t + 1) + (Ct + D)(t
2 +
√
2t + 1)) where A, B, C and D are
determined by solving the following system of equations
1 = B + D,
0 = A −
√
2B + C +
√
2D,
0 = −
√
2A + B +
√
2C + D,
0 = A + C.
The solution to this system of equations is A = −C =
1
2
√
2
, B = D =
1
2
. We can now write
1
1 + t
4
as
1
1 + t
4
=
1
2
√
2
t +
1
2
t
2 +
√
2t + 1
+
−
1
2
√
2
t +
1
2
t
2 −
√
2t + 1
=
1
2
√
2
t +
√
2
(t
2 +
√
2t + 1)
−
1
2
√
2
t −
√
2
(t
2 −
√
2t + 1)
=
1
2
√
2
t +
√
2
((t +
√
2
2
)
2 +
1
2
)
−
1
2
√
2
t −
√
2
((t −
√
2
2
)
2 +
1
2
)
 

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