Mathemagics?

[mmKALLL]

So the matriculation exams are coming up and I'm sharpening up my math skills. However, I ran into a couple of problems I was unable to solve.

13. a) ³√(a) + ³√(b) = p, ³√(ab) = q. Determine a + b as a function of p and q.

16. With which value of n is sum (1 / (√(k+1) + √(k)), k = 1 -> n) equal to 40?

I would greatly appreciate step-by-step solutions. I do not study mathemathics in English, and thus am not too proficient with the conventions and math language, so, if anything needs clarification, I'll be here to answer~

For the first one, I tried all sorts of stuff, mainly altering a + b, but I always ended up with forms which had either a or b somewhere.

As for the second one, I came to a standstill at 1 / (√2 + √1) + 1 / (√3 + √2) + ... + 1 / (√(n+1) + √n) = 40. No idea on how to proceed from that, doesn't seem to be covered in any of the high school literature either so there's some neat trick inside.

Thanks!

Edited February 28, 2013 by Piss Macho Macho Macho

[BBM]

a^(1/3)+ b^(1/3) = p

cube both sides: (x + y)^3 = x^3 + y^3 + 3xy(x + y)

a + b + 3(a^(1/3))(b^(1/3))(a^(1/3) + b^(1/3)) = p^3

(a^(1/3))(b^(1/3)) = q and (a^(1/3) + b^(1/3)) = p

therefore:

a + b + 3qp = p^3

a + b = p^3 - 3qp

Don't have time to look at the second one right now, sorry.

Edited February 28, 2013 by BBM

[mmKALLL]

Oh, right! Thanks a lot!

It didn't occur to me to approach it like that, but it was more or less simple enough, yeah.

Edit: Was able to do the other one. The trick is to multiply the fraction with √(k+1) - √k. should have known, and actually I guessed that too but didn't try

1 / (√(k+1) + √k)

= (√(k+1) - √k) / {[√(k+1) + √k]*[√(k+1) - √k]}

= (√(k+1) - √k) / (k+1 - k)

= √(k+1) - √k

Once opened, we see that the √(k+1) cancels out the √k term of the next sum, as in

(√(1+1) - √1) + (√(2+1) - √2) + (√(3+1) - √3) + ... (√(n+1) - √n)

= -√1 + √(1+1) - √2 + √(2+1) - √3 + ... + √((n-1)+1 - √n + √(n+1)

= √(n+1) - √1

= √(n+1) - 1

And the equation becoming:

√(n+1) - 1 = 40

√(n+1) = 41

n+1 = 41^2

n = 1681 - 1 = 1680.

Perhaps I'll post again if I need more help in the future. Well then!

Edited February 28, 2013 by Piss Macho Macho Macho

[mmKALLL]

Alrighty, after solving a few dozen problems, another tricky one came up.

46a)

Solve y = (e^x - e^-x) / 2 in relation to x. (So essentially, find the inverse function of sinh x)

Any ideas?

[BoxMulder]

Substitute z = e^x. Can you solve for z after doing that?

[mmKALLL]

No, I don't see how it would work out.By multiplying both sides with the 2, we'd get2y = e^x - e^-x,and by substituting e^x = z, it would be2y = z - 1/z,which doesn't really go further as far as I can see. At any rate, given the nature of the question, I'd believe that natural logarithms would be involved at some point, and that there's some way to converge the two terms. I can't figure out how, though.

E: Oh dear, I feel so dumb for not figuring that out on the spot.

Edited March 6, 2013 by Piss Macho Macho Macho

[BoxMulder]

Check that you can rearrange the equation as

z^2-2yz-1 = 0

(You should explain exactly what happened to the 1/z and why you can do whatever needs to be done.) Can you solve for z now?

[Randa]

Are you solving for x or y?

[mmKALLL]

Solving for x. Multiply both sides with z and it becomes a second degree polynom in relation to z.

I feel so dumb for not figuring it out on the spot, the solution struck me right away when I woke up. Oh well, it was around 2 am here at the time.

Thanks for the help, though!

Edited March 6, 2013 by Piss Macho Macho Macho