revision notes - c3

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Revision notes: C3 Proofs Proof by direct argument Express the conjecture as a general case by using n. If this is true, the conjecture is true for all cases. A logical step by step approach. For geometrical proofs, draw a clearly labelled diagram. Example 1: Prove that the difference between consecutive square numbers is always an odd number. (n + 1)^2 - n^2 = 2n + 1 which is not divisible by two, so the difference between consecutive square numbers must always be odd. Example 2: Prove that the product of an even number and an odd number is always even Let m and n represent 2 numbers, so 2m will be even and 2n + 1 will be odd. 2m(2n + 1) = 2(m(2n+1)) which is a multiple of 2 so must be even. Proof by exhaustion Test every possible case (exhaust all possibilities) therefore it's best to use when there's a limited number of cases to check. Example 1: Prove that square numbers with two digits are squares of numbers with one digit. Square numbers with 2 digits: 16, 25, 36, 49, 64, 81 Square roots of these numbers are all 1 digit, therefore true: +/- (4, 5, 6, 7, 8, 9) Example 2: prove that 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2 L.h.s R.h.s n = 1 1^3 = 1 1^2 = 1 n = 2 1^3 + 2^3 = 9 (1+2)^2 = 9 n = 3 1^3 + 2^3 + 3^3 = 36 (1+2+3)^2 = 36 n = 4 1^3 + 2^3 + 3^3 + 4^3 = 100 (1+2+3+4)^2 = 100 n = 5 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 225 (1+2+3+4+5)^2 = 225

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Page 1: Revision Notes - C3

Revision notes: C3Proofs

Proof by direct argument

Express the conjecture as a general case by using n. If this is true, the conjecture is true for all cases. A logical step by step approach. For geometrical proofs, draw a clearly labelled diagram.

Example 1: Prove that the difference between consecutive square numbers is always an odd number.

(n + 1)^2 - n^2 = 2n + 1 which is not divisible by two, so the difference between consecutive square numbers must always be odd.

Example 2: Prove that the product of an even number and an odd number is always even

Let m and n represent 2 numbers, so 2m will be even and 2n + 1 will be odd. 2m(2n + 1) = 2(m(2n+1)) which is a multiple of 2 so must be even.

Proof by exhaustion

Test every possible case (exhaust all possibilities) therefore it's best to use when there's a limited number of cases to check.

Example 1: Prove that square numbers with two digits are squares of numbers with one digit.

Square numbers with 2 digits: 16, 25, 36, 49, 64, 81

Square roots of these numbers are all 1 digit, therefore true: +/- (4, 5, 6, 7, 8, 9)

Example 2: prove that 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2

L.h.s R.h.s

n = 1 1^3 = 1 1^2 = 1

n = 2 1^3 + 2^3 = 9 (1+2)^2 = 9

n = 3 1^3 + 2^3 + 3^3 = 36 (1+2+3)^2 = 36

n = 4 1^3 + 2^3 + 3^3 + 4^3 = 100 (1+2+3+4)^2 = 100

n = 5 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 225 (1+2+3+4+5)^2 = 225

Page 2: Revision Notes - C3

Proof by contradiction Prove something cannot possibly be false. Assume something is false, then show that this leads to something that is clearly not true and therefore the conjecture must actually be true. You need to understand definitions e.g. What a rational number is. Example 1:

Example 2: Prove that if x is real, x^2 + 1 is greater or equal to 2xx^2 + 1 is less than 2x therefore x^2 - 2x + 1 is less than 0(X - 1)^2 less than 0. As a square cannot be less than 0, when x is a real number, there must be a contradiction.

Disprove by counter example

Prove something is not true by giving an example when the conjecture is false. Always check whether the conjecture works for negative numbers and fractions.

Example 1:

Example 2:

Page 3: Revision Notes - C3

Natural Logarithms and ExponentialsThe general case of an exponential function is a^xThe graphs of y=a^x and y=loga(x) are inverse functions:

The exponential function e^x or exp(x) where e = 2.718 and lnx where lnx = logex are inverse functions like the graph shows, reflected in the line y = x. For y = e^x, the x-axis is the horizontal asymptote and for y = lnx the y-axis is the vertical asymptote.

Exponential growth: y = Ae^(kt)Exponential decay: y = Ae^(-kt)

Rules:•If lnx = a then x = e^a•lna + lnb = lnab•lna - lnb = ln(a/b)•lna^b = blna•ln1 = 0 •lne^x = xlne = x•e^(lnx) = x for all positive x•ln(e^x) = x for all real values of x•Integrate 1/x = lnx + c•lnx = logex•ln(1/x) = -lnx•ln1 = 0•lne = 1•e^x X e^y = e^(x+y)•e^1 = e• e^x / e^y = e^(x-y)•e^0 = 1•e^-1 = 1/e

Example 1:Solve 2 = e^xx = ln2 = 0.69

Example 2: Solve e^(1 - 3x) = 51 - 3x = ln5x = (1 - ln5)/3 = -0.203

Example 3:Solve ln(2x + 1) - ln(3x) = 3ln((2x + 1)/3x) = 3(2x + 1)/3x = e^3Rearrange to get x = 1/(3e^3 - 2) = 0.017

Page 4: Revision Notes - C3

Functions

Notation: f:x |—> 2x - 1 means f(x) = 2x - 1

Key terms: • Mapping - connects input and output values• Function - A mapping which is one-to-one or many-to-many• A one-to-many or many-to-one mapping is not a function• Domain - input values• Range - output values from domain• Co-domain - all possible output values e.g. f(x) = x^2 gives

f(x) is greater or equal to 0• f(x) - the image of x• Example: f(x) = x^2 + 4x - 1 domain x = any real number(x + 2)^2 - 5 = 0 then the range is f(x) => -5

Composite functions: • y = gf(x) means f followed by g. The range of f becomes the domain of g• fg(x) does not equal gf(x)Example: f(x) = x^2 + 2 g(x) = 1/x fg(x) = f(1/x) = (1/x)^2 + 2

Types of functions: • Periodic f(x) = f(k+x) - the graph has

repeating sections e.g. Cosx• Even f(x) = f(-x) - the graph is

symmetrical about the y-axis• Odd f(-x) = -f(x) - the graph has 180

degrees rotational symmetry about the origin e.g. f(x) = x^3 and f(x) = sinx

Page 5: Revision Notes - C3

Inverse functions: • If a function f maps a number x to a number y, then the inverse function f^-1 maps y to x.

I.e. Swap the x and y values because the inverse is a reflection in the line y = x• The inverse of f(x) is f^-1 (x) • Example: f(x) = 4 - 3x y = 4 - 3x 3x = 4 - y x = (4 - y)/3 f^-1 (x) = (4 - x)/3• A function times it's inverse = x i.e. f^-1 (f(x)) = x• The domain of original = range of inverse• The range of original = domain of inverse• A self-inverse function is one that's inverse equals the original function e.g. f(x) = 1/x

when x doesn't equal 0 but f^-1 (x) also equals 1/x• Only one-to-one functions have an inverse e.g. y = e^x• To find the inverse of a many-to-one you need to restrict the domain• The inverse of a function reverses the effect of a function e.g. If f(x) = x+2, f'(x) = x-2

because subtracting 2 reverses the effect of adding 2

Inverse trig functions:

• The inverse if sinx is sin^-1 (x), but the reciprocal of sinx is 1/sinx• y = sinx is a many-to-one therefore you need to restrict the domain to -90<=x<=90 and

reflect in the line y=x• The inverse domain is the normal range

The modulus function:

• f(x) = |x| the modulus sign means ignore the minus sign. This is the same as f(x) = x when x=> 0 and f(x) = -x when x < 0. When sketching this, it is the same as sketching f(x) and reflecting it in the x-axis. If the domain is x equals any real number, the range is f(x) => 0

• To solve an equation with a modulus sign, square both sides to remove the modulus sign.

• Expressing modulus inequalities: |x - a| < b gives a - b < x < b + a|x - a| > b gives x < a - b and x > a + b

Page 6: Revision Notes - C3

Differentiation

Product rule

Quotient rule

Chain rule

Inverse functions The gradient of f^-1 (x) at (a,b) = 1/gradient at (b,a)

Rate of change:

•dy/dx is +ve when y is increasing and -ve y is decreasing

•dy/dx = 1/(dx/dy)

Page 7: Revision Notes - C3

For rate of change problems write down:

• Any formula that maybe used (area, volume,…) - you will probably need to differentiate this

• Any rate of change given

• The rate of change you have to find (use the chain rule to find the rate of change required)

Differentiating natural logs and exponentials

Implicit differentiation

This is used when it if impossible to rearrange for y = f(x), and hence differentiate so we use the chain rule to differentiate each term in an equation without rearrangement.

The derivative of g(y) with respect to x is g'(y) dy/dx

Parametric differentiation

These give a relationship between variables X and Y in terms of a third variable, a parameter, usually called t or theta.

The parametric equations x = f(t) and y = g(t) can be written as co-ordinaries (f(t), g(t)).

To differentiate parametric functions, differentiate each equation to give dy/dt and dx/dt then do: dy/dx = (dy/dt)/(dx/dt)

The Cartesian equation can be found by eliminating the parameter (t or theta...)

Page 8: Revision Notes - C3

Differentiating sinx, cosx and tanx

Page 9: Revision Notes - C3

Integration

Integrating e^x and lnx

Integration of trig functions

Integration by parts

Integrating partial fractions