Requesting solutions to 4U induction questions (1 Viewer)

[NitroBoon]

Hey guys I have some 4U induction questions that i don't know how to solve, if anyone know the solution, please post here thanks.

 

[tommykins]

回复: Requesting solutions to 4U induction questions

A few of them require a previous result, but I'll do them anyways, give me a sec.

 

[knockii]

Isn't the fact that you're having difficulties with the induction questions an indication that 4U might be a bit hard for you?

Instead of asking for the solutions, perhaps ask someone to help you out so that you understand the concepts. It's been said numerous times.... Don't cheat, you're only cheating yourself.

 

[Timothy.Siu]

Isn't the fact that you're having difficulties with the induction questions an indication that 4U might be a bit hard for you?

Instead of asking for the solutions, perhaps ask someone to help you out so that you understand the concepts. It's been said numerous times.... Don't cheat, you're only cheating yourself.

why? are u saying they're easy? and u can learn from solutions and asking for the solution is asking someone to help u out...

 

[Continuum]

Isn't the fact that you're having difficulties with the induction questions an indication that 4U might be a bit hard for you?

Instead of asking for the solutions, perhaps ask someone to help you out so that you understand the concepts. It's been said numerous times.... Don't cheat, you're only cheating yourself.

Says the guy who doesn't even do any math at all. I think you're the one cheating yourself, trying to seem like you know what you're talking about when you actually have no freaking idea. You're laughable.

Oh and looking at the solution of a question is probably the best way of understanding a concept. Reading over math theory and actually seeing it be applied are a whole different ballgame.

 

[NitroBoon]

Its not that 4U is hard for me lol, its that these questions are harder than 4U level

 

[tommykins]

回复: Re: Requesting solutions to 4U induction questions

Isn't the fact that you're having difficulties with the induction questions an indication that 4U might be a bit hard for you?

Instead of asking for the solutions, perhaps ask someone to help you out so that you understand the concepts. It's been said numerous times.... Don't cheat, you're only cheating yourself.

Haha are you serious? I'm struggling with some of them at this very moment.

 

[shaon0]

Its not that 4U is hard for me lol, its that these questions are harder than 4U level

I don't think they are harder than 4Unit level but they are hard none the less.

 

[Continuum]

I don't think they are harder than 4Unit level but they are hard none the less.

If you can judge how hard they are, then obviously you can do them right? Please do them and post up full solutions to every question.

 

[tommykins]

回复: Re: Requesting solutions to 4U induction questions

Dammit, i did these questions before but they had a part i) to do it, will ponder further.

 

[Timothy.Siu]

Re: 回复: Re: Requesting solutions to 4U induction questions

Dammit, i did these questions before but they had a part i) to do it, will ponder further.

lol good luck,
anyways this shuda been posted in the 4unit sections...not many prelims will b able to do this.

 

[Continuum]

Re: 回复: Re: Requesting solutions to 4U induction questions

lol good luck,
anyways this shuda been posted in the 4unit sections...not many prelims will b able to do this.

Probably. The parts that didn't need induction were simple enough, but the ones that did... I got lost half-way in.

 

[bored of sc]

3a could be done by a 2-unit student, I think.

 

[tommykins]

回复: Re: Requesting solutions to 4U induction questions

Yes it can, but the induction is abit of a bitch.

 

[Timothy.Siu]

Re: 回复: Re: Requesting solutions to 4U induction questions

Yes it can, but the induction is abit of a bitch.

lol, hav u thought enough on these questions, u've had all night

 

[tommykins]

回复: Re: 回复: Re: Requesting solutions to 4U induction questions

i've yet to be bothered to obtain paper and pen, slept at 3am cause beast wars was distracting me

 

[lyounamu]

Hey guys I have some 4U induction questions that i don't know how to solve, if anyone know the solution, please post here thanks.

Lol, your sig has questions on them?

 

[jco51]

1. n=1, differentiate y, y' = e^x + y,
n=2, y'' = e^x + y'
...
therefore d^(n)y/dx^n = ne^x + xe^x

2. apparently true for n=1
using the formula sinαsinβ = [cos(α-β) - cos(α+β)] / 2
n=2, (sinx + sin2x)sin(x/2) = [cos(x/2) - cos(3x/2)] /2 + [cos(3x/2) - cos(5x/2)] /2 = [cos(x/2) - cos(5x/2)] /2
thus sinx + sin2x = [sin(3x/2)sinx] / [sin(x/2)]

suppose that it's true for n=k, which is .. (sub n=k in)
when n=k+1, sinx + sin2x + ... + sinkx + sin(k+1)x = [sin((k+1)x/2)sin(kx/2)] / [sin(x/2)] + sin(k+1)x
get a common denominator and use the previous method

3.
1) apparently true for n=1
α + β = 1, αβ = 1, thus α^2 + β^2 = -1

2) [α^(n-1) + β^(n-1)](α+β) = α^n + β^n + α[β^(n-1)] + [α^(n-1)]β = α^n + β^n + αβ[α^(n-2) + β^(n-2)]
therefore A_n = (α+β)A_(n-1) - αβA_(n-2) = A_(n-1) - A_(n-2)

3) true for n=3, 4
assume it's true for n=k-1, k
A_(k+1) = A_k - A_(k-1) = 2cos(kπ/3) - 2cos[(k-1)π/3]
because 2cos[(k-1)π/3] + 2cos[(k+1)π/3] = 4cos(kπ/3)cos(π/3) = 2cos(kπ/3)
therefore 2cos(kπ/3) - 2cos[(k-1)π/3] = 2cos[(k+1)π/3] as req'd

4. n=1, RHS = (1 - cos2θ)/(2sinθ) = (sinθ)^2/sinθ = sinθ = LHS
assume it's true for n=k, therefore.. (...)
when n=k+1, because 2sinθ * [(1 - cos2kθ)/(2sinθ) + sin(2k+1)θ] = 1 - cos2kθ + 2sin(2k+1)θsinθ = 1 - cos2kθ + cos2kθ - cos(2k+2)θ = 1 - cos(2k+2)θ
as req'd

lol