1.7 Proof Methods

Prove by contradiction that `p^2 - 8q - 11 != 0, " for any " p, q in ZZ`

`"p even " => p^2 - 8q = 11`

even which is a contradiction so p must be odd

THEN

`p = 2k + 1 (k in Z)`

`(2k + 1)^2 = 8q +11`

`4k^2 + 4k + 1 = 8q + 11`

`4k^2 + 4k = 8q + 10`

`2k^2 + 2k = 4q +5`

or equivalent with one side odd and one side even

a contradiction as LHS is even and RHS is odd

therefore, if  `p, q in ZZ " then " p^2 - 8q - 11 != 0`

Use mathematical induction to prove that

`sum_(r=1)^n r/((r+1)!) = 1 - 1/((n+1)!) " for all integers " n >= 1`

let P(n) be the proposition that `sum_(r=1)^n r/((r+1)!) = 1 - 1/((n+1)!) " for all integers " n >= 1`

considering P(1):

LHS = `1/2`  and RHS `= 1/2` and P(1) is true

assume P(k) is true i.e. `sum_(r=1)^n r/((r+1)!) = 1 - 1/((n+1)!)`

considering P(k+1):

`sum_(r=1)^(k+1) r/((r+1)!) = sum_(r=1)^k r/((r+1)!) + (k + 1) / (((k + 1) + 1)!)`

`= 1 - 1/((k+1)!) + (k + 1) / ((k + 2)!) = 1 - ((k + 2) - (k + 1)) / ((k + 2)!)`

`= 1 - 1/((k+2)!) (= 1 - 1/(((k+1) + 1)!))`

P(k+1) is true whenever P(k) is true and P(1) is true, so P(n) is true (for all integers, `n>=1`)

Using mathematical induction and the definition

ⁿC_r = `(n!)/ (r!(n-r)!)`

prove that 

`sum_(r=1)^n`^rC₁ = ^{n + 1} C₂ for all `n in Z^+`

attempt to use a common denominator

`= (k(k+1)!)/(2k!) + (2(k+1)!)/(2k!) (= ((k+2)(k+1)!)/(2k!))`

THEN

`= ((k+2)!)/(2!k!) (= ((k+2)!)/(2!(k+2-2)!))`

= ^k+2C₂

since true for n = 1 , and true for n = k implies true for n = k+ 1, therefore true for all `n in Z^+` 

Let `f(x) = sqrt(1 + x) " for " x > -1`

(a) Show that `f''(x) = - 1 / (4 sqrt((1 + x)^3))`

(b) Use mathematical induction to prove that

`f^((n))(x) = (-1/4)^(n-1) ((2n - 3)!)/((n - 2)!) (1 + x)^(1/2 - n) " for " n in ZZ, n >= 2`

Let `g(x) = e^(mx), m in Q`

Consider the function h defined by `h(x) =``f(x)`x `g(x)`for `x > -1`

It is given that the `x^2` term in the Maclaurin series for `h(x)` has a coefficient of `7/4`

(c) Find the possible values of m.

(a)

attempt to use the chain rule

`f'(x) = 1/2 (1 + x)^(-1/2)`

`f''(x) = -1/4 (1 + x)^(-3/2) = - 1 / (4 sqrt((1 + x)^3))`

(b)

`= (-1/4)^(k-1) ((2k - 3)!)/((k - 2)!) (1 - 2k)/(k - 2) (1 + x)^(1/2 - k)`

`= (-1/2) (-1/4)^(k-1) (2k - 1)/(k - 2) ((2k - 3)!)/((k - 2)!) (1 + x)^(1/2 - k)`

`= (-1/4) (-1/4)^(k-1) ((2k - 1)!)/((k - 1)!) (1 + x)^(1/2 - k)`

`= (-1/4)^k ((2k - 1)!)/((k - 1)!) (1 + x)^(1/2 - k)`

`= (-1/4)^((k+1)-1) ((2(k+1) - 3)!)/(((k+1) - 2)!) (1 + x)^(1/2 - (k+1)) = RHS`

THEN since true for n = 2, and true for n = k + 1 if true for n = k, the statement is true for all `n in Z, n >= 2` by mathematical induction.

(c)

attempt to apply binomial theorem for rational exponents

`f(x) = (1 + x)^(1/2) = 1 + 1/2 x + ((1/2)(1/2 - 1))/(2!) x^2 + ...`

`f(x) = 1 + 1/2 x - 1/8 x^2 + ...`

THEN

`g(x) = 1 + mx + (m^2)/2x^2 + ...`

`h(x) = (1 + 1/2 x - 1/8 x^2 + ...) (1 + mx + m^2/2 x^2 + ...)`

coefficient of `x^2` is `(m^2)/2 + m/2 -1/8`

attempt to set equal to `7/4` and solve

`(m^2)/2 + m/2 -1/8 = 7/4`

`4m^2 + 4m - 15 = 0`

`(2m+5)(2m-3)=0`

`=> m = -5/2 or m = 3/2`

Consider the function `f(x) = (x^2 - x - 12) / (2x - 15), x in RR, x != 15/2`

(a) Find the coordinates where the graph of f crosses the

(i) `x`- axis

(ii) `y`- axis

(b) Write down the equation of the vertical asymptote of the graph of f.

(c) The oblique asymptote of the graph of f can be written as `y = ax +b`where `a, b in Q`

Find the value of a and the value of b.

(d) Sketch the graph of f for 30 `<= x <=` 30, clearly indicating the points of intersection with each axis and any asymptotes.

(e)

(i) Express `1/(f(x)` in partial fractions.

(ii) Hence find the exact value of `int_0^3 1/f(x) dx`, expressing your answer as a single logarithm.

(a)

(i)

attempt to solve `x^2 - x - 12 = 0`

(-3;0) and (4;0)

(ii)

(0;`4/5`)

(b)

`x = 15/2`

(c)

attempts division on

`(x^2 - x - 12) / (2x - 15)`

`x/2 + 13/4 + ...`

`a = 1/2`

`b = 13/4`

`y = x/2 + 13/4`

(d)

two branches with approximately correct shape for `-30<= x <= 30`

their vertical and oblique asymptotes in approximately correct positions with both

branches showing correct asymptotic behaviour to these asymptotes

their axes intercepts in approximately the correct positions

(e)

(i) attempts to split into partial fractions

`(2x - 15) / ((x + 3)(x - 4)) = A / (x + 3) + B / (x - 4)`

`2x - 15 = A(x-4) + B(x+3)`

A = 3

B = -1

`(3/(x+3) - 1/(x-4))`

(ii)

`int_0^3 (3/(x+3) - 1/(x-4)) dx`

attempts to integrate and obtains two terms involving

`"ln" = [3 ln |x + 3| - ln |x - 4|]_0^3 = 3 ln 6 - ln 1 - 3 ln 3 + ln 4 = 3 ln 2 + ln 4 = ln 32`

(a) Solve the inequality `x^2 > 2x +1`

(b) Use mathematical induction to prove that `2^(n+1) > n^2`, for `n in Z, n >=3`

(a)

`x < - 0.414; x > 2.41`

`(x < 1 - sqrt(2); x > 1 + sqrt(2))`

(b)

check for n = 3

16 > 9 so true when n = 3

assume true for `n = k` `2^(k+1) > k^2`

`2^(k+1) > k^2`

prove true for  n = k + 1

`2^(k+2) = 2`x `2^(k+1)`

>`2k^2`

`= k^2 + k^2`

> `k^2 + 2k + 1`from part (a)

which is true for `k >=3`

`= (k+1)^2`

hence if true for n = k true for n = k + 1, true for n = 3 so true for all `n >=3`

Use mathematical induction to prove that `(1-a)^n>1-na`for `{n:n in Z^+, n>=2}`, where a <0 < 1 

(multiply both sides by `(1-a)` (which is positive))

`(1-a)^(k+1)>(1-ka)(1-a)`

`(1-a)^(k+1)> 1- (k+1)a + ka^2`

`(1-a)^(k+1)> 1-(k+1)`

`=> P_k+1`is true (as `ka^2 >0)`

THEN

`P_2`is true and `P_k`is true `=> P_k+1`is true so `P_n`for all `n>=2` (or equivalent)

The cubic equation `x^3 + px^2 + qx + r = 0`where `q, p, r in R`has roots `alpha, beta and gamma`

 (a) By expanding `(x-alpha)(x-beta)(x-gamma)` show that:

`p = -(alpha+beta +gamma)`

q = `alphabeta + betagamma + gammaalpha`

r = - `alphabetagamma`

(b) 

(i) Show that `p^2 - 2q = alpha^2 + beta^2 + gamma^2`

(ii) Hence show that `(alpha - beta)^2 + (beta - gamma)^2 + (gamma - alpha)^2 = 2p^2 - 6q`

(c) Given that `p^2 < 3q deduce that `alpha, beta` and `gamma` cannot all be real.

Consider the equation `x^3 - 7x^2 + qx + 1 = 0`

(d) 

Using the result from part (c), show that when q = 17, this equation has at least one complex root.

Noah believes that if `p^2 > 3q` then `alpha, beta` and `gamma` are all real.

(e)

(i) By varying the value of q in the equation `x^3 - 7x^2 + qx + 1 = 0`, determine the smallest positive integer value of  required to show that Noah is incorrect.

(ii) Explain why the equation will have at least one real root for all values of q.

Now consider polynomial equations of degree 4.

The equation `x^4 + px^3 + qx^2 + rx + s = 0` , where `p, q, r, s in R` , has roots `alpha, beta, gamma` and `delta`.

In a similar way to the cubic equation, it can be shown that:

`p = - (alpha + beta + gamma + delta)`

`q = alphabeta + alphagamma + alphadelta + betagamma + betadelta + gammadelta`

`r = - (alphabetagamma + alphabetadelta + alphagammadelta + betagammadelta)`

`s = alphabetagammadelta`

(f) 

(i) Find an expression for `alpha^2 + beta^2 + gamma ^2 + delta ^2`in terms of p and q.

(ii) Hence state a condition in terms of  and  that would imply `x^4 + px^3 + qx^2 + rx + s = 0` has at least one complex root.

(g) Use your result from part (f) (ii) to show that the equation

`x^4 - 2x^3 + 3x^2 - 4x + 5 = 0` has at least one complex root.

The equation `x^4 - 9x^3 + 24x^2 + 22x - 12 = 0` has one integer root.

(h) 

(i) State what the result in part (f)(ii) tells us when considering this equation `x^4 - 9x^3 + 24x + 22x - 12 = 0`

(ii) Write down the integer root of this equation.

(a)

attempt to expand `(x-alpha)(x-beta)(x-gamma)`

`= (x^2 - (alpha +beta)x + alphabeta)(x- gamma)` OR `(x-alpha )(x^2 - (beta + gamma)x + betagamma)`
`(x^3 + px^2 + qx + r)`

= `x^3 - (alpha + beta + gamma)x^2 + (alphabeta + betagamma + gammaalpha)x - alphabetagamma`

comparing coefficients:

`p = -(alpha+beta +gamma)`

q = `alphabeta + betagamma + gammaalpha`

r = - `alphabetagamma`

 (b)

(i)

`p^2 - 2q = (alpha + beta + gamma)^2 - 2(alpha beta + beta gamma + gamma alpha)`

attempt to expand

`(alpha + beta + gamma)^2`

`= alpha^2 + beta^2 + gamma^2 + 2(alpha beta + beta gamma + gamma alpha) - 2(alpha beta + beta gamma + gamma alpha) " or equivalent" = alpha^2 + beta^2 + gamma^2`

(ii) attempt to write `2p^2 - 6q` in terms of  `alpha, beta, gamma`

`= 2(p^2 - 2q) - 2q`

`= 2(alpha^2 + beta^2 + gamma^2) - 2(alpha beta + beta gamma + gamma alpha)`

`= 2(alpha^2 + beta^2 + gamma^2) - 2(alpha beta + beta gamma + gamma alpha)`

`= (alpha - beta)^2 + (beta - gamma)^2 + (gamma - alpha)^2`

(c)

`p^2 < 3q => 2p^2 - 6q < 0`

`=> (alpha - beta)^2 + (beta - gamma)^2 + (gamma - alpha)^2 < 0`

if all roots were real `(alpha - beta)^2 + (beta - gamma)^2 + (gamma - alpha)^2``>= 0`

(d) `p = (-7)^2 = 49`and `3q = 51`

 so `p^2 < 3q``=>`this equation has at least one complex root.

(e)

(i) 

use of GDC (e.g. graphs or tables); q = 12

(ii) complex roots appear in conjugate pairs (so if complex roots occur the other root will be real OR all 3 roots will be real).

 OR a cubic curve always crosses the x-axis at least one point.

(f)

(i) attempt to expand `(alpha + beta + gamma + delta)^2`

`= alpha^2 + beta^2 + gamma^2 + delta^2 + 2(alpha beta + alpha gamma + alpha delta + beta gamma + beta delta + gamma delta)`

`= alpha^2 + beta^2 + gamma^2 + delta^2 + 2(alpha beta + alpha gamma + alpha delta + beta gamma + beta delta + gamma delta)`

= `p^2 - 2q`

(ii)

`p^2 < 2q`or `p^2 - 2q <0`

hence there is at least one complex root.

(h)

(i)  `p^2 > 2q` (81 > 2 x 24) (so nothing can be deduced)

(ii) -1

attempt to express as a product of a linear and cubic factor `(x+1) (x^3 - 10x^+ 34x - 12)`

since for the cubic, `p^2 < 3q`(100 < 102)

there is at least one complex root 

Use mathematical induction to prove that `(d^n)/(dx^n) (x e^(px)) = p^(n-1) (px + n) e^(px)` for `n in Z^+, p in Q`

n = 1: LHS  = `d/dx (x e^(px)) = x p e^(px) + e^(px) = (px + 1) e^(px)`

RHS = `p^(1-1) (px + 1) e^(px) = p^0 (px + 1) e^(px) = (px + 1) e^(px)`

LHS = RHS so true for n = 1 

assume proposition true for n = k

 `d^k/d^(xk)(xe^(px)) = p^(k-1)(px+k)e^(px)`

We differentiate the assumed result for n = k

`d^(k +1)/ dx^(k+1)(xe^(px))=d/dx(d^k/dx^k(xe^(px)))`

`= d/dx(p^(k-1) (px+k)e^(px))`

`= p^(k-1)(px+k)e^(px)`

`=p^(k-1)((px+k)pe^(px) + e^(px)(p))`

`= p^k (px+k)e^px+ e^(px)(p^k)`

`=p^k(px+k+1)e^(px)`

`= p^((k+1)-1)(px+(k+1))e^(px)`

hence true for n = 1 and n = k true

`=> n = k + 1`true therefore true for all `n in Z^+`

Prove by mathematical induction that  

where n `in Z`, `n >=3`

show true for  n = 3

hence true for n = 3

assume true for n = k:

consider for n = k+1:

= `(k!)/((k-3)!3!) + (k!)/((k-2)!2!)`or any correct expression with a visible common factor.

= `(k!)/(3!(k-3)!)(1+3/(k-2))`or any correct expression with a common denomirator.

= `(k!)/(3!(k-2)!)(k+1)`

`((k+1)!)/ (3!(k-2)!`or equivalent

Result is true for n = 3 . If result is true for k it is true for k + 1. Hence result is true for all `n>=3` . Hence proved by induction.