Math Problems

[Beren3000]

I didn't get what you did after expanding the log. If you take that "divide by (1-x)" is equivalent to "multiply by (1-x)^-1" Then the expansion works:

(x - 1/2x^2 + 1/3x^3 + ...)/(1-x)
=(x - 1/2x^2 + 1/3x^3 + ...) * (1 + x + x^2 + x^3 + ...)
= x + x^2 + x^3 - 1/2x^2 -1/2x^3 + 1/3x^3
= x + 1/2x^2 + 5/6x^3

And there's your answer.

[Janny]

Ohh...

And I'm supposed to be good at maths...

Thank you ever so much.

[Beren3000]

Originally Posted by Janny

Thank you ever so much.

Don't mention it. And if you come across other interesting problems in that notebook, you know where to post them

[Janny]

Here was the second problem that I was going to post, but fortunately I got it just a while ago. Not that I lost sleep over it. *shifty eyes*

Find an expression for the coefficient x^n in:
ln (2 + x)

Don't you hate the way everything is easy when you know how.

[Beren3000]

Now I want to know the answer...is it:

(-1)^n/ n*2^n

?

[Janny]

Yeah, but I think the book gave it as:

(-2)^-n / n

though.

Next up... I promise it's not every single question I can't do...
What are the conditions for which
x^3 + 3ax^2 + 3bx + 1 = 0
has a repeated root.
If (x - a)^2 is a factor of f(x), then (x - a) is a factor of f'(x). So you eliminate x from f(x) = 0 and f'(x) = 0. Only problem is, how's that done? The book thinks it's too simple to bother with, and I'm not sure what to do with the outcome when I get to it anyway.

Cheers.

[Beren3000]

The answer is (I think) b=a^2
There are 2 ways to prove this:

1st way:
To have repeated roots, you need the expression to be a perfect cube.
Now (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
Apply this to the formula you have, you find that f(x) becomes (x+1)^3 only when b=a^2

2nd way:
A point where roots are repeated is either a turning point or an inflexion point on the curve of the function.
So take d/dx [f(x)]:
f ' (x) = 3x^2 + 6ax + 3b
= 3*(x^2 + 2ax + b)
to have turning points (and inflexion points) you need to have f ' (x) = 0
IOW, x^2 + 2ax + b = 0 but this equation itself must have repeated roots so that the turning "points" are actually at the same point (point of inflexion)
so (from the quadratic formula) the discriminant must equal zero:
4a^2 - 4b = 0, which leads you back to b=a^2

Hope that made sense!

P.S. Nice book

EDIT: Somehow I feel there are other conditions that would yield repeated roots to the equation but what these conditions are is beyond me...

[Janny]

Thanks. I recognised the binomial expansion patterns in f(x) and f'(x), but actually I think your solutions are above the level of the book, especially the second. Your solution, the same as mine, and it makes sense, as it would make the discriminent of f '(x) zero.

I think the actual given solution was in the form "b^2 = 4ac".
(ab - 1)^2 = 4(a-b^2)(b-a^2)

Which implies:
(a - b^2) y^2 + (ab - 1) y + b - a^2 = 0
but I can't work back from there to the original equations.

That bit about points of inflection interests me, I hadn't thought of it like that before. Stupid AS syllabus ignores the existence of inflection points.

[Beren3000]

I can make nothing out of that solution
Are you sure there are no other hints in the book??
By the way, for my 1st method, the perfect cube only works for a=b=1
The 2nd way (because it uses differentiation and therefore doesn't bother with constants) yields the general condition b=a^2, of which a=b=1 is just one solution.

Originally Posted by Janny

Stupid AS syllabus ignores the existence of inflection points.

Hehehe, I'm afraid it doesn't get much better with A2 (if you're studying the Edexcel syllabus, that is)

I have thought of 2 ways to solve the problem but each led me to a dead-end. See if you can make better use of them:

-You have two cases for repeated roots: either all three roots are equal, or there is one root and two equal roots. If all three roots are equal, then a=b=1 as we proved before. So take the case when there's a double root. For this case, you have to have one unrepeated root. By the rational root theorem, one of the roots is +or- 1. So one of the factors is (x+1) or (x-1). After that try long division with each of these factors to get a quadratic factor, then equate its discriminant to zero (in order to have a double root).

-A triple root will create an inflection point, and a double root will create a turning point. See if you can use differentiation to write some equations in a and b...

Hope this helps!

[Janny]

Ah yes. It did mention the Rational Root Theorem, another nugget omitted from the AQA AS. 'Just use trial and error, you'll be fine...'.
I had ruled out the possibility of long division, mainly because I didn't fancy the long division and then the factorisation. Yeah, that's just lazy. I'll have a look.

[Beren3000]

Originally Posted by Janny

It did mention the Rational Root Theorem, another nugget omitted from the AQA AS.

So I take it you don't know this theorem?
Ok, it basically states that if the polynomial

ax^n + bx^(n-1) + .... + z

has rational roots, then the roots are the factors of the expression " + or - z/a " So in the case of the polynomial of the question, we have z = a = 1
So the possible roots are + or - 1

Beren out!

[Janny]

Yeah, it gave a one sentence crash course. I did the division. The remainder of the division (in a and b) can be equated to zero, since the assumption is that the divisor is a factor.
The remainder was subbed into the quotient to get a quadratic equation in only one of either a or b. With (x-1) this quaddie has no real roots, so I presume x-1 is not a factor.
x+1, this will please you, gives a=b, and a = -1/3 v a = 1.

Further problems tomorrow, after I've had a good look at them tonight.

[Janny]

Whole message deleted. Annoyed. Did questions.
With mechanics and parametric equations, position is given by x(t) and y(t). Pythag on x' and y'(t) gives resultant velocity in t.
Pythag on x'' and y''(t) gives resultant acceleration in t.
But differentiating velocity doesn't give acceleration. Why so? Is that because v(t) is somehow actually v(x,y) and needs modification?

Coming soon, mean integration.

[Beren3000]

I think I know what's wrong. Don't apply Pythag. to x' and y', first get the resultant position by applying Pythag. on x(t) and y(t), and then differentiate what you get so that you have the resultant velocity. Differentiate again to get the acceleration (IOW, apply Pythag. before differentiating)

I've got an interesting problem for you:

1600 coconuts are to be divided among a 100 monkeys. They can be divided in any way possible (possibilites where some monkeys receive no coconuts are included). So the problem is: prove that no matter how the coconuts are divided, at least 4 monkeys will have the same number of coconuts.

[Fat middle]

Originally Posted by Beren3000

1600 coconuts are to be divided among a 100 monkeys. They can be divided in any way possible (possibilites where some monkeys receive no coconuts are included). So the problem is: prove that no matter how the coconuts are divided, at least 4 monkeys will have the same number of coconuts.

I'm not sure if this is correct, but I'm putting it in spoiler code just in case someone wants to try it himself.

I'm going to prove it a sensu contrario: if there are no 4 monkeys that aere going to receive the same number of coconuts, then the maximum number of monkeys receivig equal number is 3. In the worst case we would have 3 monkeys with 0 coconuts, 3 m with 1 c, 3m with 2c ... 3 monkeys with 32 coconuts and another monkey with more than 32 coconut, let's say 33. That makes 3x0 + 3x1 + 3x2 + ... + 3x32 + 33. And that is 3 x (0 + 1 + ... 32) +33 = 3 x (16 x 33) + 33 = 1617, which is more coconuts than the number we have, so that last monkey will have to receive less coconuts, and the at least 4 of them will have to have the same number

[Beren3000]

Well, I can't tell you if this is correct, because I don't have access to the "official" solution, but that's how I proved it, too. With a slight difference, however:

Instead of saying that the last one will get 33, I just said that 3 * ( 0+1+2+3+...+32) = 1584 coconuts, so that 16 coconuts are left for the last monkey, so 4 monkeys will have 16 coconuts.

[Beren3000]

Hey, Janny, where's that "mean integration" you promised?

[Janny]

Sorry, I've been on holiday. Right, rotation about axis is coming up, but how would you go about integrating:
(1 + 1/x) ^1/2

As to monkeys... I feel the pigeonhole principle coming on, but I might be animal obsessed. For there to be no sharing of numbers, the fewest number of coconuts required happens when each monkey gets assigned the next smallest number. The first monkey gets 0, the second gets 1 and so on. The minimum coconuts is therefore the sum from 0 to 99.
99(99+1) / 2 = 4950. Woah, huge. And this rises to 5050 if we don't want to be cruel to that poor monkey who gets assigned zero.
So in conclusion, 1600 simply isn't enough coconuts. My advise for bridging deficit of 3350: Lift trade restrictions to China. Or switch to some penut-based foodstuff.

And thank you, I shall look into the parametric differentiation.

[Janny]

I've just suffered from the classic problem... it is remedied in school by the accronym RTBQ. I'll leave that with you.
So yes, 3 . S1 to 32
(3 . 32 . 33) / 2 = 1584
And that last monkey must have himself 16 coconuts too. I wondered why I was flying so far clear of that minimum sum bar.

[Beren3000]

Originally Posted by Janny

Sorry, I've been on holiday. Right, rotation about axis is coming up, but how would you go about integrating:
(1 + 1/x) ^1/2

Aaarghhh! This is a tough one, alright. I can't figure it out... for a while I thought I had the (very long) answer but it turned out that I had made a mistake; now I'm completely stumped Maybe if you tell me the book's answer it would help a little...