Dekoyl's UMEP Questions

[dekoyl]

Okay.. time to cram for UMEP exams =\

Prove by induction that:


What I got to was, in my inductive steps:
So let be the statement that . If holds, holds.

LHS of

[addikaye03]

1+(1/√2)+(1/√3)+...+(1/√n)≤ 2√n-1

Prove true for n=1, LHS=1/√1=1, RHS=2√1-1=1 therefore LHS≤RHS & true for n=1

Assume true for n=k-1,

1+(1/√2)+(1/√3)+...+(1/√k-1)≤ 2√(k-1) -1

Prove true for n=k

1+(1/√2)+(1/√3)+...+(1/√k-1)+ (1/√k)<=2√k -1

LHS

≤ (2√(k-1)+(1/√k)-1

≤ [2√k(√k-1)+1]/√k  -1

≤ [2√k(√k-1)+1-√k]/√k

≤ [2k(√k-1)-k+√k]/k

≤ [k(2√k+2-1)]/k

≤ 2√k-1

I might have stuffed it up a bit at the end a bit. There's the idea though

[dekoyl]

Thanks That's a nice way to do it. I've come across seemingly different ways to prove by induction but I like your method.

[dekoyl]

Hmm did you make a mistake early on? Because you went from:

to:

[addikaye03]

Hmm did you make a mistake early on? Because you went from:

to:

Yep, that would have been it. Thanks for correcting me. This method is called "strong induction", if you want to learn more about it.

ALSO, if you are thinking of using this method, under my working i would have said, "therefore true for n=k"

Then a final statement like:

Result holds true for n=1 and n=k+1, therefore holds true for all integer n where n>=0, by the process of Mathematical Induction.

NB: The most common mistake in mathematical induction is students believe its an "iterative method", whereby, since true for n=1 and n=k+1 then true for n=1+1=2, 2+1=3,..., n-1+1=n. This is NOT the idea. It's a generalisation that holds true for for all n but some assumption number k. This might help occasionally.

Here are some more Q:

1) Suppose x>0, y>0

Rtp: 4/s ≤ 1/x + 1/y, where s = x + y

2) You are given the following conditions:



for

show (a) for

and deduce (b) for

3) Prove by MI for all integers n≥1 that

(a1 + a2 + ... + an)/n≥ (a1.a2...an)^1/n

Where a1, a2, ..., an > 0

Spoiler

q3. Poyla's method:

 

[dekoyl]

^Thanks
How do we prove something as simple as this? :S This isn't important as the exam's passed already but anyway..

Let . We define equality, addition and multiplication of these rational numbers in the usual way, that is,
 and

Using only the above properties prove that for :

[kamil9876]

let

By property 3:



by property 2:



By distributivity of multiplication of integers with respect to addition:


By property 2:


Now you must prove the the cancellation law, ie: that you can cancel out the f's in first term and d's in the second term. This can be proven with (1) which I will leave as an exercise to you. (also one last application of property 2 at the end)


note: I have assumed certain properties of integers(distributivity, and the last step will use some others as well), which is alright because ussually you learn about integers and prove their properties before dealing with rationals, after all that is how rationals have been defined in this question.

edit: dyslexic mistakes with all these letters :/

[dekoyl]

^Cool. Thanks kamil. I think I did something like that yesterday.. eh hope I get a few marks for my efforts anyway.

[humph]

How did the exam go? Any uber tricky questions?

[addikaye03]

How did the exam go? Any uber tricky questions?

bump, i want to see some harder Q's also, here's a Q i got the the other day, thought it was a good application:

1) When two resistors of resistances z1 and z2 are connected in series, their equivalent resistance z is given by z=z1+z2, and when they are connected in parallel their equivalent resistance z is given by 1/z=1/z1+1/z2.

In an electric circuit that contains two resistors whose resistances are z1=R1+iwL and z2=R2-i/(wC) respectively, find the value of w so that their equivalent resistance is purely real when their resistors are

i) Connected in Series
ii)Connected in Parallel

2)Decompose the integrand into partial fractions with complex linear denominators, hence, find

int. dx/(x^8-1)

* i got the answer with plenty of tedious algebra + noting (x^4-1)(x^4+1) Hence roots of unity of (x^8-1), but want a quicker way*

3) A car of mass M kg, width w metres and centre of mass h metres above the ground travels round a level curve of radius r metres with a constant speed v m/s such that driver's right-hand side is near centre of the curve.

a) By drawing the front view of the car and two normal reactions of the road's surface on the right and the left wheels (neglect length of car), show that the right and left normal reactions respectively are:

M/2rw(rgw-2hv^2) and M/2rw(rgw+2hv^2)

b) hence, show that the car overturns when v^2>= rgw/2h


I suppose these Q would be considered Harder within the Mathematics Extension 2 syllabus (HSC system, NSW)

Cheers

[humph]

Weird, this came up in a proof I'm working on. Well more accurately, I had to show

Didn't prove it using induction though, just by Abel's lemma.