Donate to Remove ads

Got a credit card? use our Credit Card & Finance Calculators

Thanks to johnstevens77,Bhoddhisatva,scotia,Anonymous,Cornytiv34, for Donating to support the site

Evaluate

cinelli
Lemon Slice
Posts: 550
Joined: November 9th, 2016, 11:33 am
Has thanked: 231 times
Been thanked: 160 times

Evaluate

#216789

Postby cinelli » April 23rd, 2019, 11:46 am

Evaluate
.
------------- -------------
3/ /-- 3/ /--
/ 11 + 4 * V 29 + / 11 - 4 * V 29
V V

I have done my best to indicate cube roots and square roots.
Calculators and computers are not allowed.

Cinelli

UncleEbenezer
The full Lemon
Posts: 10691
Joined: November 4th, 2016, 8:17 pm
Has thanked: 1459 times
Been thanked: 2965 times

Re: Evaluate

#216839

Postby UncleEbenezer » April 23rd, 2019, 3:22 pm

cinelli wrote:Evaluate
.
------------- -------------
3/ /-- 3/ /--
/ 11 + 4 * V 29 + / 11 - 4 * V 29
V V

I have done my best to indicate cube roots and square roots.
Calculators and computers are not allowed.

Cinelli

Took some staring at it, but I think I've got your notation. Perhaps works better as

(11 + 4*29^(1/2))^(1/3) + (11 - 4*29^(1/2))^(1/3)

Looks like a job for the calculator, but since you forbid that,

Let's see if it yields to schoolboy algebra. Breaking it down in simple terms without the encumbrance of numbers, it's
(x+y)^(1/3) + (x-y)^(1/3)
Expanding the cube of that,
(a+b)^3 = a^3 +3*a^2*b + 3*a*b^2 + b^3 = a^3 + 3*a*b*(a+b) + b*3

or

(x+y) + 3*( ((x+y)(x-y))^(1/3) ((x+y)^(1/3)+(x-y)^(1/3)) + (x-y)

= 2x + 3(x^2-y^2)^(1/3) ((x+y)^(1/3)+(x-y)^(1/3))

OK, in numbers, x=11 and y=sqrt(16*29) = sqrt(464). Now x^2 - y^2 = 121 - 464 = -343, whose cube root is -7. Convenient! So we have
22 + 3(-7 ((x+y)^(1/3)+(x-y)^(1/3)))

Now that term we haven't yet evaluated is the original expression come back to haunt us. Let's call it Z. We have
Z^3 = 22 - 3*7*Z
Which has a particularly easy solution, which ought to be a lot more obvious from the original than it is. Ho, hum ..

UncleEbenezer
The full Lemon
Posts: 10691
Joined: November 4th, 2016, 8:17 pm
Has thanked: 1459 times
Been thanked: 2965 times

Re: Evaluate

#216951

Postby UncleEbenezer » April 24th, 2019, 9:52 am

I know it's what's intended, but an incompleteness in my solution is nagging at me. That is to say, I've ignored the alternative values of the cube roots (and indeed the square root, if you don't assume the two instances of sqrt(29) both take the same value - without which the whole solution collapses in a heap and really does need the calculator).

I refer of course not to the cubic equation at the end (where I carefully used the indefinite article to refer to "a solution"), but to the evaluation of an intermediate cube root to its real value. I'm sure that's what was meant, but it bugs me to leave such details as implied.

cinelli
Lemon Slice
Posts: 550
Joined: November 9th, 2016, 11:33 am
Has thanked: 231 times
Been thanked: 160 times

Re: Evaluate

#217778

Postby cinelli » April 27th, 2019, 7:16 pm

UncleEbenezer,

Yes, that was what was intended. You have completed 95% of the solution with just the last little bit remaining.

Cinelli

UncleEbenezer
The full Lemon
Posts: 10691
Joined: November 4th, 2016, 8:17 pm
Has thanked: 1459 times
Been thanked: 2965 times

Re: Evaluate

#217796

Postby UncleEbenezer » April 27th, 2019, 9:13 pm

cinelli wrote:Yes, that was what was intended. You have completed 95% of the solution with just the last little bit remaining.

My late mother enjoyed jigsaws. My dad would sometimes ceremoniously put in the last piece, proclaiming it obviously the most difficult, because all the others were demonstrably easier.

I shall leave the completion to a greater mind than mine.

(Still impressed by how unintuitive it is from the original puzzle).

Gengulphus
Lemon Quarter
Posts: 4255
Joined: November 4th, 2016, 1:17 am
Been thanked: 2628 times

Re: Evaluate

#217896

Postby Gengulphus » April 28th, 2019, 12:09 pm

UncleEbenezer wrote:
cinelli wrote:Yes, that was what was intended. You have completed 95% of the solution with just the last little bit remaining.

My late mother enjoyed jigsaws. My dad would sometimes ceremoniously put in the last piece, proclaiming it obviously the most difficult, because all the others were demonstrably easier.

I shall leave the completion to a greater mind than mine.

(Still impressed by how unintuitive it is from the original puzzle).

Well, I assume that the "last little bit" is just that you've arrived at a cubic that the answer has to satisfy, and observed that that cubic has an obvious solution, but not actually stated what that solution is, nor why the answer isn't instead either of the two other solutions of the cubic. The obvious solution is indeed so obvious that it probably isn't really necessary to state it, though a stickler for completeness might take off a mark for its omission. The fact that the other two solutions aren't applicable is pretty trivial once one has decided that the puzzle implicitly means "evaluate in the real numbers" and not "evaluate in the complex numbers" - I can see a choice of two easy arguments, one based on monotonicity and the other on the standard solution of a quadratic, and there may well be more. And of course, that decision that the puzzle must be about real numbers also resolves your point about which roots are involved, since when one is working in the domain of the real numbers, square roots are non-negative by standard mathematical convention and cube roots are unique.

Gengulphus

cinelli
Lemon Slice
Posts: 550
Joined: November 9th, 2016, 11:33 am
Has thanked: 231 times
Been thanked: 160 times

Re: Evaluate

#219218

Postby cinelli » May 3rd, 2019, 11:28 am

I am probably being pedantic in saying that no one has yet given the answer, so here is my solution.

Let

.      -------------      -------------
3/ /-- 3/ /--
Z = / 11 + 4 * V 29 + / 11 - 4 * V 29
V V

The first thing to say is that Z is real and that the square root and cube roots are real and unique. In cubing both sides of the equation you are turning a linear equation into a cubic so will be introducing extra, spurious, roots. As UncleEbenezer has shown, the resulting cubic simplifies to

Z^3 + 21*Z – 22 = 0

and this has an obvious root, Z = 1, so it is a cubic which we can actually solve. Factorise:

(Z – 1) * (Z^2 + Z + 22) = 0

and note that the quadratic factor has complex roots. Since we know that Z is real, these extra roots can be ignored. Thus we have the remarkable fact that that horrible expression for Z which involves square and cube roots is precisely 1.


Cinelli


Return to “Games, Puzzles and Riddles”

Who is online

Users browsing this forum: No registered users and 3 guests