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Lecture notes Mathematics (MA1203)

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Module
Mathematics (MA1203)

Institution
City University (City)

Lecture notes an sample exams with solutions for a 1st year maths course. Complete notes for all lecture merged into a single PDF. Everything included to get you your first class

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Uploaded on September 8, 2023
Number of pages 61
Written in 2022/2023
Type Lecture notes
Professor(s) Ravi mukherjee
Contains All classes

Comparing coefficients of polynomials

Two real valued polynomials are equal if and only if they
coincide when evaluated at all real numbers. Equivalently, two
real valued polynomials are equal if and only of the coefficients
of terms corresponding to the same powers of x are equal.
Example 1
Find a, b, c, d such that
Supplementary Notes: Rational Functions and Partial
Fractions x3 = ax(x − 1)(x − 2) + bx(x − 1) + cx + d
Expanding we see that
Prof. Markus Linckelmann

School of Engineering and Mathematical Sciences x3 = a(x3 − 3x2 + 2x) + b(x2 − x) + cx + d
Department of Mathematics
City University London = ax3 + (b − 3a)x2 + (2a − b + c)x + d

January 30, 2015 Comparing coefficients on both sides yields a = 1, b − 3a = 0,
2a − b + c = 0 and d = 0. It follows that a = 1, b = 3, c = 1,
d = 0.
Prof. Markus Linckelmann Mathematics for Economics (Post A level)

Rational Functions
A rational function in a variable x is a function of the form
p(x)
q(x) Proper rational functions can be further simplified, using a
where p(x) and q(x) are polynomials with q(x) not identically method called partial fractions.
zero. (That is, there is at least one value of x for which Step 1: Simplify pq to form a sum of a polynomial and a proper
q(x) 6= 0.) For example, the two functions rational function qr .
x2 + 6x + 4 x+7 Step 2: Factorize the polynomial q as a product of linear and
and quadratic polynomials (it is a nontrivial fact that this is always
x2 − 5 3x7 − 2x + 1
are rational functions. Note that the sum, difference or product possible).
of two rational functions is again a rational function. Step 3: (see next side)
A rational function p(x)
q(x) is called proper if the degree of the
numerator p(x) is strictly less than the degree of the
denominator q(x). Otherwise we say that the rational function
is improper. For example the first fraction above is improper
while the second is proper. An improper rational function can
always be written as the sum of a polynomial and a proper
rational function withLinckelmann
Prof. Markus the same denominator q. This (Post
Mathematics for Economics is achieved
A level) Prof. Markus Linckelmann Mathematics for Economics (Post A level)

by dividing p by q with rest.

Step 3: If q has n pairwise distinct factors (which may show up
with some positive power), write the fraction as a sum of n
sums. Those n sums depend on whether the factors are linear,
or quadratic, and their powers with which they appear in the
factorisation of q.
If q(x) has a factor (x − a)n then on the right side we write The constants Ai , Bi in above can be determined by comparing
coefficients, for instance.
A1 A2 An This is, however, not always the quickest way to determine the
+ + ... +
x − a (x − a)2 (x − a)n Ai , Bi . It is sometimes far easier to evaluate both sides at
particular values of the variable x for which ‘most’ terms are
where A1 , A2 ,..,An are real constants. These constants are zero.
determined by comparing coefficients.
If q(x) has a factor (ax2 + bx + c)n then on the right side we
write
A1 x + B1 A2 x + B2 An x + Bn
+ + .... +
ax2 + bx + c (ax2 + bx + c)2 (ax2 + bx + c)n

where the Ai and Bi are constants.

Prof. Markus Linckelmann Mathematics for Economics (Post A level) Prof. Markus Linckelmann Mathematics for Economics (Post A level)

Example 2
Example 3
x+5 A B
= + 2x2 + x − 2 A B C D
(x − 3)(x + 1) x−3 x+1 = + 2+ 3+
x3 (x − 1) x x x x−1
Therefore
Therefore
x + 5 = A(x + 1) + B(x − 3).
We could multiply the right side out and compare coefficients. 2x2 + x − 2 = Ax2 (x − 1) + Bx(x − 1) + C(x − 1) + Dx3
As mentioned above, it is quicker to compare both sides for the
values x = −1 and x = 3. We obtain Substituting x = 0 and x = 1 we obtain C = 2 and D = 1.
Comparing the coefficients of the x3 terms and the x2 terms we
4 = −4B obtain
0=A+D
and
2 = −A + B
Using the values of C and D obtained above we get A = −1 and
8 = 4A
B = 1.
and so A = 2 and B = −1
Prof. Markus Linckelmann Mathematics for Economics (Post A level) Prof. Markus Linckelmann Mathematics for Economics (Post A level)

, Example 5
The following example is an improper rational function.
Example 4 Dividing the numerator by the denominator x3 − x with rest
yields
5x − 12 A Bx + C
= +
(x + 2)(x2 − 2x + 3) x + 2 x2 − 2x + 3
3x3 − x2 + 2 3(x3 − x) − x2 + 3x + 2
this form is because x2 − 2x + 3 cannot be factorized any =
x(x2 − 1) x(x − 1)(x + 1)
further. (This is because x2 − 2x + 3 has no real root.) x2 − 3x − 2
Therefore = 3−
x(x − 1)(x + 1)
5x − 12 = A(x2 − 2x + 3) + (Bx + C)(x + 2) Thus we set
Substituting x = −2 we obtain A = −2. By comparing the x2 − 3x − 2 A B C
= + +
coefficients of the x2 terms (which is zero on the left side) and x(x − 1)(x + 1) x x−1 x+1
constant terms we obtain B = 2 and C = −3.
Multiplying both sides by the denominator yields

x2 − 3x − 2 = A(x − 1)(x + 1) + Bx(x + 1) + Cx(x − 1)

Prof. Markus Linckelmann Mathematics for Economics (Post A level) Prof. Markus Linckelmann Mathematics for Economics (Post A level)

Evaluating at 0, 1, −1 yields A = 2, B = −2 and C = 1.
Therefore we obtain the equality

3x3 − x2 + 2 2 2 1
=3− + −
x(x2 − 1) x x−1 x+1

The key point of this method is that it is possible to integrate
the summands which show up when rewriting a rational
function using partial fractions.

Prof. Markus Linckelmann Mathematics for Economics (Post A level)

, 1. Integration

1.1 Basic theory

Definition.
Let f = f (x) be a real valued function in the variable x defined
on a subset D of R. The set D is called the domain of f . An
1. Integration: Indefinite Integrals antiderivative of f is a differentiable function F having the
same domain as f such that

Prof. Markus Linckelmann d
F =f .
dx
School of Mathematics, Computer Science and Engineering
Department of Mathematics
City University London
Antiderivatives are not unique: if F is an antiderivative of f ,
then for any constant C the function F + C is also an
January 26, 2015 d
antiderivative of f , because dx C = 0.

Prof. Markus Linckelmann Mathematics for Economics (Post A level)

Definition.
One can be more precise: The indefinite integral of a function f (x), denoted by the
symbol
If F is an antiderivative of f , then any other antiderivative of f Z
is equal to F + C for some constant C. f (x)dx

Proof. Let F and G are two antiderivatives of f . Then is the class of all antiderivatives of f ; that is, if F is an
antiderivative of f , then
d d
F = G=f , Z
dx dx f (x) dx = F (x) + C
hence
d where C is an arbitrary constant.
(G − F ) = f − f = 0 ,
dx
The only functions whose derivative is zero are the constant The indefinite integral is only defined up to the undetermined
functions. Therefore G − F = C for some constant C, so G = constant C.
F + C. To keep the terminology simple, in what follows we say
‘integral’ instead of ‘indefinite integral’.

Prof. Markus Linckelmann Mathematics for Economics (Post A level) Prof. Markus Linckelmann Mathematics for Economics (Post A level)

There are obvious extensions of these results, replacing x by
From our standard results for differentiation we deduce the ax + b. For example, for k 6= −1 we have
following integrals, which should be memorised.
(ax + b)k+1
Z
(ax + b)k dx = +C
R a(k + 1)
f (x) f (x) dx
k
x (k 6= −1) xk+1 and
k+1 + C − cos(ax + b)
Z
x−1 ln |x| + C sin(ax + b) dx = + C.
a
ex ex + C
etc. Integration is a linear operation. This means that for
sin x − cos x + C
functions f and g and constants a and b we have
cos x sin x + C
tan x − ln(| cos x|) + C Z Z Z
af + bg dx = a f dx + b g dx.

Prof. Markus Linckelmann Mathematics for Economics (Post A level) Prof. Markus Linckelmann Mathematics for Economics (Post A level)

Example 1.1.1:
For more complicated rational functions it is best to simplify
√
Z
3
Z Z Z
1
x7 + 2 − x dx = x7 dx + 3 x−2 dx − x 2 dx first using partial fractions.
x
Example 1.1.3:
x8 3 2 3
= − − x 2 + C.
8 x 3 1 + 3x2 2 3
Z Z Z
2
dx = − 2
dx + dx
(1 + x) (1 + 3x) (1 + x) 1 + 3x
Example 1.1.2: 2
= + ln(|1 + 3x|) + C.
Z
1 (2x + 3)−3 −1 1+x
dx = +C = + C.
(2x + 3)4 (−3).2 6(2x + 3)3

Prof. Markus Linckelmann Mathematics for Economics (Post A level) Prof. Markus Linckelmann Mathematics for Economics (Post A level)

, Sometimes it is not so easy to spot the integral of a function.
2
Example 1.1.5: 2xex dx.
R
For integrals involving trigonometric functions, we typically use
This does not correspond to one of our standard integrals.
standard identities to simplify the integral.
However, by inspection we can observe that
Example 1.1.4:
2
dex 2
1 x 1 = 2xex
Z Z
sin2 x dx = (1 − cos 2x)dx = − sin 2x + C. dx
2 2 4
using the chain rule, and hence
Z
2 2
2xex dx = ex + C.

We would like to formalise this procedure.

Prof. Markus Linckelmann Mathematics for Economics (Post A level) Prof. Markus Linckelmann Mathematics for Economics (Post A level)

1.2 Method of substitution Writing u = g(x) this becomes

du
Z
Recall the chain rule for differentiation:
f ′ (u) dx = f (u) + C
dx
df (g(x))
= f ′ (g(x))g ′ (x).
dx and so we have
Integrating both sides we obtain
Z Z
f ′ (g(x))g ′ (x) dx = f ′ (u) du
Z
f ′ (g(x))g ′ (x) dx = f (g(x)) + C.
where u = g(x).

Prof. Markus Linckelmann Mathematics for Economics (Post A level) Prof. Markus Linckelmann Mathematics for Economics (Post A level)

Example 1.2.2: Integrate
Example 1.2.1: We return to example 1.1.5, and recalculate Z
3
x2 (x3 + 1) 2 dx.
Z
2
2xex dx.
du
Let u = x3 + 1, so dx = 3x2 . Then
du
Let u = x2 ,
so = 2x. Then
dx 3
u 2 du
Z Z
3
x2 (x3 + 1) 2 dx = dx
du 3 dx
Z Z Z
2 2
2xex dx = eu dx = eu du = eu + C = ex + C. 3
dx Z
u2 2 5 2 5
= du = u 2 + C = (x3 + 1) 2 + C.
3 15 15

Prof. Markus Linckelmann Mathematics for Economics (Post A level) Prof. Markus Linckelmann Mathematics for Economics (Post A level)

1.3 Inverse substitution

In the last section we substituted Example 1.3.1: Integrate
1
Z
f ′ (g(x)) −→ f ′ (u) √ dx.
1+ x
′
g (x) dx −→ du.
√ dx
Let x = u, so x = u2 and du = 2u. Then

Replacing f ′ by h and interchanging the roles of x and u we 1 1
Z Z
√ dx = 2u du
have Z Z 1+ x 1+u
h(g(u))g ′ (u) du = h(x) dx 2
Z
= 2− du
1+u
where x = g(u). Therefore we can substitute √ √
= 2u − 2 ln(|1 + u|) + C = 2 x − 2 ln(1 + x) + C.
h(x) −→ h(g(u))
dx
dx −→ g ′ (u) du = du.
du

Prof. Markus Linckelmann Mathematics for Economics (Post A level) Prof. Markus Linckelmann Mathematics for Economics (Post A level)

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