Any mathematicians out there?

[Big_B]

I have a problem that I need resolved quickly.

Imagine a function y = f(a,b).

take the first derivative df(a,b)/da and then the derivative of the result to give the cross-partial:
d(df/da)/db

My question is: Do you interpret the cross-partial as a first or second order derivative?

My gut is second order, but it's important so I want a second opinion. Thanks:)

[mdm]

It's irrelevant because the earths gravitational pull counters the derivative altogether.

[OiPaz]

It's a second order derivative.
Do you know, for instance, the Hessian matrix? Its elements are all second order derivatives, and (d^2 f)/dab is an element of that matrix. Besides, in order to obtain d^2/dab you have to derive twice, so it's second order! :-)

[Big_B]

It's irrelevant because the earths gravitational pull counters the derivative altogether.

http://photo.klein-jensen.dk/smilies/Kick.gif ;)

[Big_B]

It's a second order derivative.
Do you know, for instance, the Hessian matrix? Its elements are all second order derivatives, and (d^2 f)/dab is an element of that matrix. Besides, in order to obtain d^2/dab you have to derive twice, so it's second order! :-)

Cool. That's what I was thinking. Thanks! :D

[MattyB]

*crosses his eyes and falls over*

[Claire]

Thank God I didn't have to do maths at Uni! I'm so happy I got transfer credits from high school or I'd flunk the class big time. I didn't understand anything in Big_B's post. LOL

[gmen]

take the first derivative df(a,b)/da and then the derivative of the result to give the cross-partial:
d(df/da)/db
It's certainly a second order derivative... but surely the 'd's should be those curly partial derivative type ones... where do I find them on the keyboard? :lol:

ahh.. the good old character map... ∂²f(a,b)/∂a∂b

Hmmm... ;)

---- Gavin

[Big_B]

Thanks gavin :lol:

Ok, so here's the next stupid question. if the result of the cross partial is >0, does a increase or decrease if I increase b? (and why)

Thanks again, this help is REALLY appreciated!

[Maureen Souza]

Thanks gavin :lol:

Ok, so here's the next stupid question. if the result of the cross partial is >0, does a increase or decrease if I increase b? (and why)

Thanks again, this help is REALLY appreciated!

Are we doing your homework for you :lol: :lol: :lol: ?

[OiPaz]

if the result of the cross partial is >0, does a increase or decrease if I increase b? (and why)
Being a second order derivative, it has no correlation with slope. It gives information about curvature (partial information: you need the whole Hessian matrix).
But probably I have not fully understood your question: if you only increase b, a does not change at all! ;-)

[gmen]

Not quite sure I'm 100% with you (but it's been a long time since Maths was my thing)... but it seems to me that a and b are independent variables, so changing one won't change the other...

---- Gavin

[Big_B]

Thanks for coming back so quickly :)

Sorry, I probably didn't explain very well. Imagine an individual maximising f(a,b) by choosing a. Then some exogenous factor changes b and I'm trying to find out what the optimium response of the individual would be. I.e. how do they change a in response.

Oipaz - Agrree with you about the 2nd order part. That's what's confusing me. I'm researching a paper that appears to be using the cross partial to explain the above, but I can't understand why.

Maureen - it's for work, so a kind of homework I guess ;)

[OiPaz]

OK, let's see if I've understood, now.
So, you have a C^2 function f(a,b), b is fixed to a value b1 and the single-variable function f(a;b1) has a max in a1. So f'(a;b1) is 0 in a=a1, locally positive in a<a1 and locally negative in a>a1.
Now, if someone makes you jump from b1 to b2, slightly larger than b1, what happens? There are three cases:

1) if the max does not move obviously (d^2 f)/(da db) is 0 and the new max is a2=a1. Note that you must not move from a1 to remain in the max;

2) if the max moves to the right you have that f'(a;b2) is 0 in the new max a2>a1 and therefore positive in a1: so f'(a;b1) is an increasing function in a1 (remember that f'(a1;b1)=0) and hence the second order cross derivative in (a1,b1) is positive. Note that you must increase a1 to remain in the max;

3) if the max moves to the left, according to the previous reasoning, the cross derivative is negative in the initial point (a1,b1) and you have to decrease a1 to remain in the max.

Then, in all the three cases, you have to (locally) move a1 according to the sign of the second order cross derivative in order to remain in the max. I must stress that this is valid only locally: if b2>>b1 then you can not say anything about the problem.

Do any of the aforesaid sentences make sense? ;-)

[gmen]

Thanks for coming back so quickly :)

Sorry, I probably didn't explain very well. Imagine an individual maximising f(a,b) by choosing a. Then some exogenous factor changes b and I'm trying to find out what the optimium response of the individual would be. I.e. how do they change a in response.
Sounds like it might be a case for the Lagrangian multiplier which would be a route to solving a constrained maximisation/optimisation problem :rolleyes: ... is this some kind of economic/financial model?

However, I'd have to dig the text books out as my Lagrangian optimisation is somewhat rusty (to say the least) ;) ... but I could be wrong and maybe OiPaz will have a better answer!

---- Gavin

[Ronald S. Jr.]

:shock:

[Scottes]

:shock:
Ditto.

[gmen]

OK, let's see if I've understood, now.
So, you have a C^2 function f(a,b), b is fixed to a value b1 and the single-variable function f(a;b1) has a max in a1. So f'(a;b1) is 0 in a=a1, locally positive in a<a1 and locally negative in a>a1.
Now, if someone makes you jump from b1 to b2, slightly larger than b1, what happens? There are three cases:

1) if the max does not move obviously (d^2 f)/(da db) is 0 and the new max is a2=a1. Note that you must not move from a1 to remain in the max;

2) if the max moves to the right you have that f'(a;b2) is 0 in the new max a2>a1 and therefore positive in a1: so f'(a;b1) is an increasing function in a1 (remember that f'(a1;b1)=0) and hence the second order cross derivative in (a1,b1) is positive. Note that you must increase a1 to remain in the max;

3) if the max moves to the left, according to the previous reasoning, the cross derivative is negative in the initial point (a1,b1) and you have to decrease a1 to remain in the max.

Then, in all the three cases, you have to (locally) move a1 according to the sign of the second order cross derivative in order to remain in the max. I must stress that this is valid only locally: if b2>>b1 then you can not say anything about the problem.

Do any of the aforesaid sentences make sense? ;-):lol: Nice one OiPaz. I think I was over-complicating matters ;)

---- Gavin

[OiPaz]

Thanks. But, reading again my post, I think that probably the opposite approach would be easier: if you apply the definition of derivative you obtain that the second order cross one is the limit for (a2,b2)->(a1,b1) of:

[ f(a2,b2)-f(a2,b1)-f(a1,b2)+f(a1,b1) ] / [ (a2-a1)(b2-b1) ].

Now, you said that b2>b1 so the second factor in the denominator is positive. The sum of the first two addends in the numerator is positive if (a2,b2) is a max and the sum of the last two addends is positive as well if (a1,b1) is a max, so in your case the whole numerator is positive. Then the sign of the derivative is the same of the factor (a2-a1): positive if the max moves to the right, negative otherwise! QED. ;-)



P.s.: as we are speaking of limits, again, the reasoning is valid locally.

[Big_B]

Guys, this is great. Thankyou so much for your help - this looks like it is exactly what I am after. I have to run off to a meeting now so will look through it properly when I get back.

Gavin - yes, it's an economic model of behaviour.

Thanks again :)

[Skip Souza]

I thought that this forum was in English???

[soupdragon]

I had hoped you mods would be educated beyond fundamentals.

[zacker]

.... I have a camera!
-zacker-

[gmen]

.... I have a camera!
-zacker-Some of us have Cameras, some of us have Slide Rules ;)

---- Gavin

[zacker]

...Oooh i have a slide rule also...
I wondered what it was used for though...lol
math was never my strong point... Lunch, recess, art, reading.. In-School suspension, Detention.. thats where I excelled!!
-zacker-

[Jon]

Are we doing your homework for you :lol: :lol: :lol: ?
At least he knew enough to not post it on the Moderators' forum!
:{)#

[Jon]

Some of us have Cameras, some of us have Slide Rules ;)

---- Gavin
I've got both (and can use them), and a couple of HP scientific calculators, but that doesn't mean I understood any of that!

[KevC]

Teehee.
Mathematicians are funny.
There was a bunch of kids in the pasu (physics and astronomy student union) lounge making algebra jokes with pizzas. The kernel of the pizza is the box, the inner product of blah blah blah, something weird about the crust being indeterminant or something. The algebra was far more advanced that what we were required to know for physics. Those pure math kids, wow they always blow my mind =)

[Claire]

My ex once bought a book with various mathematical quizzes in it. Then he'd lie in bed doing them before sleeping. I really need to stop dating engineers/IT men...:rolleyes:


:lol:

[Big_B]

Well i'm an economist, so does that make me better or worse? :lol:

[PacAce]

I'm glad you got that all squared away, Big_Brain (btw, how's Pinky doing?). I was just about to jump in with my 2 cents. :D

(Yeah, right! :o :o :o )

:mrgreen: :lol: :mrgreen:

[Big_B]

I'm glad you got that all squared away, Big_Brain (btw, how's Pinky doing?). I was just about to jump in with my 2 cents. :D

(Yeah, right! :o :o :o )

:mrgreen: :lol: :mrgreen:

:p :lol:

[Big_B]

OK, let's see if I've understood, now.
So, you have a C^2 function f(a,b), b is fixed to a value b1 and the single-variable function f(a;b1) has a max in a1. So f'(a;b1) is 0 in a=a1, locally positive in a<a1 and locally negative in a>a1.
Now, if someone makes you jump from b1 to b2, slightly larger than b1, what happens? There are three cases:

1) if the max does not move obviously (d^2 f)/(da db) is 0 and the new max is a2=a1. Note that you must not move from a1 to remain in the max;

2) if the max moves to the right you have that f'(a;b2) is 0 in the new max a2>a1 and therefore positive in a1: so f'(a;b1) is an increasing function in a1 (remember that f'(a1;b1)=0) and hence the second order cross derivative in (a1,b1) is positive. Note that you must increase a1 to remain in the max;

3) if the max moves to the left, according to the previous reasoning, the cross derivative is negative in the initial point (a1,b1) and you have to decrease a1 to remain in the max.

Then, in all the three cases, you have to (locally) move a1 according to the sign of the second order cross derivative in order to remain in the max. I must stress that this is valid only locally: if b2>>b1 then you can not say anything about the problem.

Do any of the aforesaid sentences make sense? ;-)

I went through this on the train on thursday. It's exactly what I was looking for and makes the intuition really clear. Thanks again :)