Extension One Revising Game (2 Viewers)

[bored of sc]

Question: ( to keep things going, its 4U tho)
a)If z1=1+i,z2=rt3-i, find the moduli and principal arguments of z1,z2 and z1/z2

a) |z1| = sqrt(2)
arg(z1) = pi/4
|z2| = 2
arg(z2) = -pi/6
z1/z2 = {sqrt(2)cis[pi/4]}/{2cis[-pi/6]}
= [sqrt(2)/2]*cis[5pi/12]

 

[bored of sc]

Derive the quadratic formula by completing the square

Hint: begin with ax²+bx+c = 0

 

[azureus88]

ax^2 + bx + c = 0
a(x^2 + (b/a)x) = -c
a(x^2 + (b/a)x + (b/2a)^2) = -c + (b^2)/4a
a(x + b/2a)^2 = (b^2 - 4ac)/4a
(x + b/2a)^2 = (b^2 - 4ac)/4a^2
(x + b/2a) = +or- [sqrt(b^2 - 4ac)]/2a

x = {-b+or- [sqrt(b^2 - 4ac)]}/2a

 

[addikaye03]

a) |z1| = sqrt(2)
arg(z1) = pi/4
|z2| = 2
arg(z2) = -pi/6
z1/z2 = {sqrt(2)cis[pi/4]}/{2cis[-pi/6]}
= [sqrt(2)/2]*cis[5pi/12]

Part ii) was the focus but welldone

 

[lolokay]

find the limiting sum of 1 - x² + x⁴ - x⁶ + x⁸ ...
given that 0 < or = x < 1

hence find the limiting sum of 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 ...

 

[Trebla]

The answer is π/4, but I'll let others work it out themselves lol. However, some of the answers to this question I have seen are technically flawed, so see if you guys can do it with technical precision (it requires a common Ext2 technique to get it technically precise though lol)

 

[lolokay]

is there anything wrong with

Spoiler

1 - x² + x⁴ - x⁶ + x⁸ ... = 1/(1+x²)
{integral from 0 to 1}[1 - x² + x⁴...]dx = {integral from 0 to 1}1/(1+x²)dx

[x - 1/3x³ + 1/5x⁵ ... ]1,0 = [tan^-1x]1,0
1 - 1/3 + 1/5 - 1/7 ... - 0 = pi/4 - 0

 

[addikaye03]

find the limiting sum of 1 - x² + x⁴ - x⁶ + x⁸ ...
given that 0 < x < 1

hence find the value of 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 ...

1-x^2+x^4-x^6+x^8... [G.P, r=-x^2, a =1]
S=1/(1+x^2)=d/dx(tan^-1x)
therefore, 1-x^2+x^4-x^6+x^8...=d/dx(tan^-1x)
x-x^3/3+x^5/5-x^7/7+x^9/9-x^11/11...+C=tan^-1x+C....( when x=0, both sides = 0 .'. no constant)

By equating 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 ...
tan^-1x=1 therefore pi/4

 

[Trebla]

is there anything wrong with

Spoiler

1 - x² + x⁴ - x⁶ + x⁸ ... = 1/(1+x²)
{integral from 0 to 1}[1 - x² + x⁴...]dx = {integral from 0 to 1}1/(1+x²)dx

[x - 1/3x³ + 1/5x⁵ ... ]1,0 = [tan^-1x]1,0
1 - 1/3 + 1/5 - 1/7 ... - 0 = pi/4 - 0

The fact that you've included 0 and 1 as limits in the integral with respect to x when they are excluded from x in the original question is the technical issue...lol

 

[lolokay]

oh right.. well you can just include 0; that doesn't affect the limiting sum
not too sure about the 1 though.. unless you make the limit 1^- (value of x as x->1), such that it is just below 1, but the sum still approaches the value for x=1.. or something

 

[addikaye03]

does my solution have technical issues too? lol i think it does

QUESTION:If a,b are real positive numbers, Prove (a+b)^2(1/a^2+1/b^2)>=8

 

[lolokay]

does my solution have technical issues too? lol i think it does

QUESTION: Prove (a+b)^2(1/a^2+1/b^2)>=8

(a-b)² >= 0
a² + b² >= 2ab
a/b + b/a >= 2 (dividing by ab which is >0 as a,b >0)
similarly, a²/b² + b²/a² > 2

(a+b)^2(1/a^2+1/b^2)
= (a² + 2ab + b²)(1/a² + 1/b²)
= 1 + a²/b² + 2b/a + 2a/b + b²/a² + 1
= 2 + 2[a/b + b/a] + [b²/a² + a²/b²]
>= 2 + 2(2) + 2
= 8

.'. (a+b)^2(1/a^2+1/b^2)>=8 a,b>0



Another question:
solve for x; (x+1)(x+2)(x+3)(x+4) = 8

 

[Trebla]

I think also, it's more of an issue with the way the question was set up, but anyway since another question is being asked I might as well type up the more technically precise way (and also because I'm bored) lol:

Spoiler

Consider a general geometric series (not necessarily limiting sum)
1 - x² + x⁴ - x⁶ + …… + (–1)ⁿx²ⁿ
a = 1, r = - x² and there are n + 1 terms, so this equals:
1[1 - (- x²)^{n + 1})] / [1 - (- x²)] = [1 - (-1)^{n + 1}x^{2n + 2}] / [1 + x²]
= [1 - (-1)(-1)ⁿx^{2n + 2}] / [1 + x²]
= [1 + (-1)ⁿx^{2n + 2}] / [1 + x²]
1 - x² + x⁴ - x⁶ + …… + (–1)ⁿx²ⁿ = [1 + (-1)ⁿx^{2n + 2}] / [1 + x²]
Integrating both sides with respect to x with limits 0 to 1 (notice there is no restriction on x)
∫₀¹ {1 - x² + x⁴ - x⁶ + …… + (–1)ⁿx²ⁿ}dx = ∫₀¹ (1 + (-1)ⁿx^{2n + 2}) dx / (1 + x²)
[1 - x² + x⁴ - x⁶ + …… + (–1)ⁿx²ⁿ]₀¹ = ∫₀¹ {dx / [1 + x²]} + (-1)ⁿ∫₀¹ (x^{2n + 2}) dx / (1 + x²)
1 - 1/3 + 1/5 - 1/7 + ..... + (–1)ⁿ / (2n + 1) = [tan^-1x]₀¹ + ∫₀¹ (x^{2n + 2}) dx / (1 + x²)
Σⁿ_{k = 0} (–1)^k / (2k + 1) = π/4 + ∫₀¹ (x^{2n + 2}) dx / (1 + x²)

But
x² ≥ 0
x² + 1 ≥ 1
1 / (x² + 1) ≤ 1
=> x^{2n + 2} / (x² + 1) ≤ x^{2n + 2}
Note no change of sign as x^{2n + 2} is always positive for all real x
∫₀¹ x^{2n + 2} dx / (x² + 1) ≤ ∫₀¹ x^{2n + 2} dx
∫₀¹ x^{2n + 2} dx / (x² + 1) ≤ 1 / (2n + 3)
Also, since the integrand is non-negative, in integral is also non-negative hence
0 ≤ ∫₀¹ x^{2n + 2} dx / (x² + 1) ≤ 1 / (2n + 3)
As n --> ∞, 1 / (2n + 3) --> 0, so the ∫₀¹ x^{2n + 2} dx / (x² + 1) --> 0
(known by the rather graphic name as the "squeeze law" lol)
Hence as n --> ∞:
Σ^∞_{k = 0} (–1)^k / (2k + 1) = π/4

*sighs* that took a lot of typing up codes lol...

The inequality set up makes it more of an Extension 2 question. A similar question of this series was asked in a past HSC exam for Extension 2 in 2004 Q8 but it involved an integration reduction formula for sinⁿx and the proof steps required stopped at the inequalities bit though (the explicit series itself was not required for proof in the question).

 

[lolokay]

so you're not able to put a limit in the integal?, ie.

Spoiler

∫₀ⁿ (1/(1+x²))dx as n approaches 1 (approaching from values less than 1)
= tan^-1n which has a limit of pi/4

∫₀ⁿ (1 - x² + x⁴ - ...)dx as n approaches 1
= n - 1/3n³ + ...
but as n approaches 1, n^m for some integer m, also approaches 1
.'. the integral has a limit 1 - 1/3 + 1/5 - 1/7 + ...
and then equate the two limits

+anyone who hasn't read through the solutions, feel free to have a go, or at least think about the original question

 

[Trebla]

so you're not able to put a limit in the integal?, ie.

Spoiler

∫₀ⁿ (1/(1+x²))dx as n approaches 1 (approaching from values less than 1)
= tan^-1n which has a limit of pi/4

∫₀ⁿ (1 - x² + x⁴ - ...)dx as n approaches 1
= n - 1/3n³ + ...
but as n approaches 1, n^m for some integer m, also approaches 1
.'. the integral has a limit 1 - 1/3 + 1/5 - 1/7 + ...
and then equate the two limits

+anyone who hasn't read through the solutions, feel free to have a go, or at least think about the original question

Yeah, that would also work...as opposed to my rather convoluted method lol
though I'm not sure if an integral approaching a limit is still within the scope of the course...haha

 

[bored of sc]

b) If z=(1+i)/(rt3-i), find the smallest postive integer n such that z^n is real and evaluate z^n for this integer n.

z = rt2/2*cis75

cos(75) = cos(45+30)
= cos45*cos30-sin45*sin30
= 1/rt2*rt3/2-1/rt2*1/2
= (rt3-1)/2rt2

sin(75) = sin(45+30)
= sin45*cos30+sin30*cos45
= 1/rt2*rt3/2+1/2*1/rt2
= (rt3+1)/2rt2

z = rt2/2*{[(rt3-1)/2rt2]+[(rt3+1)/2rt2]i}
= 2rt2/4*{[(rt6-rt2)/4]+[(rt6+rt2)/4]i}
= (2rt12-4)/16 + (2rt12+4)/16*i
= (rt3-1)/4 + (rt3+1)/4*i
= (rt3-1+rt3i+i)/4

i⁰ = 1
i¹ = i
i² = -1
i³ = -i
i⁴ = 1
i⁵ = i
i⁶ = -1
i⁷ = -i

No idea where to go to!

 

[bored of sc]

Can you guys post up an easier question from an area like circle geometry or trigonometry?

 

[addikaye03]

z = rt2/2*cis75

cos(75) = cos(45+30)
= cos45*cos30-sin45*sin30
= 1/rt2*rt3/2-1/rt2*1/2
= (rt3-1)/2rt2

sin(75) = sin(45+30)
= sin45*cos30+sin30*cos45
= 1/rt2*rt3/2+1/2*1/rt2
= (rt3+1)/2rt2

z = rt2/2*{[(rt3-1)/2rt2]+[(rt3+1)/2rt2]i}
= 2rt2/4*{[(rt6-rt2)/4]+[(rt6+rt2)/4]i}
= (2rt12-4)/16 + (2rt12+4)/16*i
= (rt3-1)/4 + (rt3+1)/4*i
= (rt3-1+rt3i+i)/4

i⁰ = 1
i¹ = i
i² = -1
i³ = -i
i⁴ = 1
i⁵ = i
i⁶ = -1
i⁷ = -i

No idea where to go to!

think arg (z^n)=narg(z), and umm.. arg(z^n)=kpi, where K is 0, +-1,+-2,...

 

[kurt.physics]

Come on people, were is some difficult circle geometry?

 

[addikaye03]

Come on people, were is some difficult circle geometry?

yer i agree dude, i wouldnt mind some circle geometry.