something MATH.......

lovelove · #1

I know this is the wrong forum, but please help me.
It is about a simple looking math problem.

A = 2xy / (x^2 + y^2)
x>0, y>0
What is the maximum of 'A'?

I'm stuck with this. I'd appreciate anyone who can show the solution.

Winter vacation has ended, and the university enterence test is at November 2013.... I sadly have terribly short time for Go.. :cry:

Shaddy · #2

lovelove wrote:

I know this is the wrong forum, but please help me.
It is about a simple looking math problem.

A = 2xy / (x^2 + y^2)
x>0, y>0
What is the maximum of 'A'?

I'm stuck with this. I'd appreciate anyone who can show the solution.

Winter vacation has ended, and the university enterence test is at November 2013.... I sadly have terribly short time for Go.. :cry:

It is 1. There is a common inequality (called arithmetic-geometric mean inequality, but nevermind that) that says that

x^2 + y^2 >= 2xy.

to prove it, you can write x^2 + y^2 - 2xy = (x-y)^2 >= 0

lovelove · #3

Shaddy wrote:

lovelove wrote:

I know this is the wrong forum, but please help me.
It is about a simple looking math problem.

A = 2xy / (x^2 + y^2)
x>0, y>0
What is the maximum of 'A'?

I'm stuck with this. I'd appreciate anyone who can show the solution.

Winter vacation has ended, and the university enterence test is at November 2013.... I sadly have terribly short time for Go.. :cry:

It is 1. There is a common inequality (called arithmetic-geometric mean inequality, but nevermind that) that says that

x^2 + y^2 >= 2xy.

to prove it, you can write x^2 + y^2 - 2xy = (x-y)^2 >= 0

mitsun · #4

For an alternative geometric solution, you could draw a circle and transform to polar coordinates:
x = r sin(t), y = r cos(t), 2xy / (x^2 + r^2) = (some trig manipulation) = sin(2t)

TheBigH · #5

mitsun wrote:

For an alternative geometric solution, you could draw a circle and transform to polar coordinates:
x = r sin(t), y = r cos(t), 2xy / (x^2 + r^2) = (some trig manipulation) = sin(2t)

Ooh, that's elegant. Nice idea!

drmwc · #6

If you've done calculus, a possible thrid approach is to take the partial derivative of A w.r.t. x or y, and set it to 0.

This a lot less elegant than the solutions posted so far, though.

Specifically:

dA/Dy=(2x^3-2xy^2)/(x^2+y^2)^2.

Setting this to 0, and using x>0 and y>0, gets x^2=y^2. Hence x=y at the maximum, where A=1. (There's a little more work to do to show this a maximum not a minimum.)

speedchase · #7

drmwc wrote:

If you've done calculus, a possible thrid approach is to take the partial derivative of A w.r.t. x or y, and set it to 0.

This a lot less elegant than the solutions posted so far, though.

Specifically:

dA/Dy=(2x^3-2xy^2)/(x^2+y^2)^2.

Setting this to 0, and using x>0 and y>0, gets x^2=y^2. Hence x=y at the maximum, where A=1. (There's a little more work to do to show this a maximum not a minimum.)

You also have to show that +/- infinity is greater

lightvector · #8

If you have the right intuition and bits of knowledge, it's possible to quite quickly see the solution to this problem in your head, without even having to write anything down.

The denominator, x^2+y^2, is the squared length of the vector (x,y). And for any particular value of the denominator, that is, for any particular fixed length, how do you choose a vector (x,y) of that length that maximizes the numerator, 2xy? Geometrically, what points on a circle around the origin maximize xy? It should be intuitively obvious that the maximum occurs when x = y.

And when x = y, the whole expression simplifies into 2xx/(x^2+x^2) = 2x^2/(2x^2) = 1, so the maximum is 1 and does not depend on the length of the vector, aside from it needing to be nonzero.

(This is mostly the same as the polar coordinate solution).

lovelove · #9

Real thanks for everyone who replied (and will reply...) to my post. :bow:

And... let me commit my second abuse to this forum.

The (math) problem, below in a italic font, that I'm again in trouble with wants me to find the truth values of propositions about matrix. The original problem is written in Korean, so I may have used some improper mathematical expressions in English.

'A' is a 2x2 matrix, 'O' is a 2x2 zero matrix, 'n' ∈ a set of all natural numbers

(1)
A^n = O → A^2 = O

(2)
A 2x2 matrix 'X' that 'X^2 = A' always exists regardless of the value of 'A'. (Whatever value 'A' has, there must be some kind of a 2x2 matrix 'X' that 'X^2 = A')

(3)
'A' that 'A^2 = ( 1 2 )' exists.
____________( 3 4 )

I'd again, of course, appreciate anyone who can show the solution. :blackeye:

Bill Spight · **#10**

lovelove wrote:

Real thanks for everyone who replied (and will reply...) to my post. :bow:

And... let me commit my second abuse to this forum.

The (math) problem, below in a italic font, that I'm again in trouble with wants me to find the truth values of propositions about matrix. The original problem is written in Korean, so I may have used some improper mathematical expressions in English.

'A' is a 2x2 matrix, 'O' is a 2x2 zero matrix, 'n' ∈ a set of all natural numbers

(1)
A^n = O → A^2 = O

(2)
A 2x2 matrix 'X' that 'X^2 = A' always exists regardless of the value of 'A'. (Whatever value 'A' has, there must be some kind of a 2x2 matrix 'X' that 'X^2 = A')

(3)
'A' that 'A^2 = ( 1 2 )' exists.
____________( 3 4 )

I'd again, of course, appreciate anyone who can show the solution. :blackeye:

To get you started:

Code:

     A = (a11 a12)
         (a21 a22)

What is A^2?

lovelove · **#11**

Bill Spight wrote:

Code:

     A = (a11 a12)
         (a21 a22)

What is A^2?

Code:

A = (a b)
    (c d)
A^2 = (a^2+bc ab+bd)
      (ac+cd d^2+ab)

Bill Spight · **#12**

lovelove wrote:

Bill Spight wrote:

Code:

     A = (a11 a12)
         (a21 a22)

What is A^2?

Code:

A = (a b)
    (c d)
A^2 = (a^2+bc ab+bd)
      (ac+cd d^2+ab)

You have an oops!

OK. What must be true of a, b, c, and d for A^2 to be 0?

lovelove · **#13**

Bill Spight wrote:

lovelove wrote:

Bill Spight wrote:

Code:

     A = (a11 a12)
         (a21 a22)

What is A^2?

Code:

A = (a b)
    (c d)
A^2 = (a^2+bc ab+bd)
      (ac+cd d^2+ab)

You have an oops!

OK. What must be true of a, b, c, and d for A^2 to be 0?

a=b=c=d=0 or something like a=1, b=1, c=-1, d=-1 (some cases when ad-bc = 0)

Bill Spight · **#14**

lovelove wrote:

Bill Spight wrote:

lovelove wrote:

Code:

A = (a b)
    (c d)
A^2 = (a^2+bc ab+bd)
      (ac+cd d^2+ab)

You have an oops!

OK. What must be true of a, b, c, and d for A^2 to be 0?

a=b=c=d=0 or something like a=1, b=1, c=-1, d=-1 (some cases when ad-bc = 0)

Do you see the oops?

Make a general statement about the conditions for A^2 = 0.

I am going to bed now.

Bill Spight · **#15**

A <> 0, A^2 = 0 , where A is a 2x2 matrix.

Write A using 2 real variables.

----

Given: A is a 2x2 matrix.
Prove: A^4 = 0 -> A^2 = 0.

mitsun · **#16**

The three original questions are all essentially the same (as I think Bill was attempting to lead up to). You can Google "matrix square root" to get more information than you want.

The solution to A*A = 0 can be derived with a little algebra (exercise for the reader). Verifying the solution is easy by direct matrix multiplication.

Code:

A = ( a         b )
    ( -a^2/b   -a ) 

For amusement, this is the only matrix with determinant = trace = 0. The determinant property is necessary, but the trace property is not obvious, at least to me.

lovelove · **#17**

Bill Spight wrote:

A <> 0, A^2 = 0 , where A is a 2x2 matrix.

Write A using 2 real variables.

----

Given: A is a 2x2 matrix.
Prove: A^4 = 0 -> A^2 = 0.

Code:

I think I get the first question.

A = ( a b ); 'a, b, c, d' are real numbers; 'n' is a natural number
    ( c d )

If A^n = O, 'A' cannot have an inverse matrix, so the determinant is zero.
using the Caily-Hamilton thing, A^2 = (a+d)A
when n >= 3, A^n = (a+d)^(n-1)A = O, a+d = 0, A^2 = O
when n = 1 or n = 2, A^2 = O

Therefore A^2 = O 

perceval · **#18**

lovelove wrote:

Bill Spight wrote:

A <> 0, A^2 = 0 , where A is a 2x2 matrix.

Write A using 2 real variables.

----

Given: A is a 2x2 matrix.
Prove: A^4 = 0 -> A^2 = 0.

Code:

I think I get the first question.

A = ( a b ); 'a, b, c, d' are real numbers; 'n' is a natural number
    ( c d )

If A^n = O, 'A' cannot have an inverse matrix, so the determinant is zero.
using the Caily-Hamilton thing, A^2 = (a+d)A
when n >= 3, A^n = (a+d)^(n-1)A = O, a+d = 0, A^2 = O
when n = 1 or n = 2, A^2 = O

Therefore A^2 = O 

OK for Q1.

Now for Q2:

you have to either prove the proposed assertions, or find a counter example.

For 2 i suggest finding a counter example ;-)

.
bigger hint:

..........

something MATH.......

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