Sunday, August 26, 2007

graph4

My dear friends,

thank you ever so much for bearing with me. As I suspect is so often the case, it is the mathematician, and not the math, that has gone wrong. Thanks to all your feedback and encouragement, I think I can now call this case closed. For your visual satisfaction, I present below the final graphical solution to this little endeavour:



I don't know exactly where I went wrong with that last graph, but the solution as it stands now should graph as depicted above, and it looks like a set of solutions quite in agreement with what one would expect from the problem as I'd set it out.

Anyway, I really do appreciate not only the conversations we've had, but, more profoundly, that I have friends with whom I can have such conversations. Many thanks again for all your help.

Most fondly,
Paul

Friday, August 24, 2007

One Problem For Another...

All right. At last I know where I went wrong: a little sign change poked its ugly head in there at the last minute, and where it should have read, A=2x^3+x, it read A=2x^3-x. So that helps somewhat. However, that graph looks like this:




As you can see, I am now stuck with the opposite problem; You could give me an x, say, -.25, and my new better function will give you a positive number! I'll let the $20 stand...

Friday, August 17, 2007

A Mathematical Excursion (or, Where I Went Wrong)

o.k. I've taken to "amusing" myself lately by attempting to solve little geometrical problems I've set for myself. On the whole this has been successful, as measured both by the satisfactory completion of said problems (i.e. solving them), and by the "amusement" level associated with the process (e.g., not thinking about my foot, debts, work...).

HOWEVER, I've run into a stumbling block. I've come up with a solution to a problem which just doesn't seem like it could possibly be right. $20 to the first person who can show me where I went wrong. I'll lay out the problem, and then some basic things we'll need to agree on:

I thought it would be fun to see if, given some x, I could find some other number A, such that the line going between (A, 0) and (x, x^2) would be perpendicular to the function f(x)=x^2. A thousand words:

My thinking went something like this:

Axioms:
1. The slope of f(x) at x is equal to the derivative of f(x) evaluated at x.
2. if f(x)=x^n, f'(x)=nx^(n-1) [the "Power Rule"]

3. The slope of a line between two points is the quotient of the difference between y values and x values: Dy/Dx= (y2-y1)/(x2-x1)
4. If two lines are perpendicular, then their slopes are negative reciprocals of each other.

All of this should be, if not obvious, at least readily verifiable. Still with me? Then here's one argument, with justification for each step in brackets[]:

I. f(x)= x^2, therefore f'(x)=2x [stipulated, power rule]
II. The slope of our little green line, then, should be the negative reciprocal of the slope at x, or -1/2x [stipulated, 4]
III. The slope of our little green line should also equal (0-x^2)/(A-x) [stipulated, 3]
IV. Therefore, -1/2x= (0-x^2)/(A-x) [II., III.]

It's not terribly tricky to solve for A, but, for the sake of being explicit, here's how I did it:

1. -(A-x)/2x = (0-x^2) [mult. (A-x)]

2. -(A-x) = 2x(0-x^2) [mult. 2x]

3. -(A-x) = -2x^3 [dist.]

4. (A-x) = 2x^3 [mult. -1}

5. A = (2x^3) - x [sub. x]

Cool. So now we have a function that, when you put in some x, gives you back an A! Can you imagine the plot of this function? I did, and that was when I noticed that something fishy was going on. A thousand more words:


Problem: If (A, 0) is supposed to be the point where the line perpendicular to f(x) intersects the x axis, why would it ever be negative in the first quadrant? I imagine those lines getting closer and closer to vertical as one approaches the origin (f'(0)=0, and the negative reciprocal of this, while undefined, approaches -infinity, something like -1/0), but they would have to have a positive slope to get from anywhere on f(x) in quadrant one to the negative x axis.

So, there you have it. I can't see how to refute the math, but I can't see how it could be right either. For now there is just a hole where reason should be.

Better luck to you,
Paulman

Monday, August 13, 2007

howdyhowdyhowdy

What the heck?! It's been, like, well, way too long since I last posted, and all kinds of things have been going on. I wrecked my heel and have been hobbling around since last Wednesday, but the doctor's say I haven't broken anything, so it's just going to be painful and suck for a few weeks; I miss Kenpo :( I'm also bummed because this means that I can't practise riding the 5' giraffe unicycle I recently acquired. I purchased it for quite a reasonable sum from Joe Lind of Compulsion Cycles. It teases me from my living room. Anyway, hopefully I'll be back to my more physical self soon; in the meantime, I've been laying down the beats for OCEANOVER's upcoming full length album. That's been moving along swimmingly, though there's really never quite enough time. In part because I'm soon to begin my lab work at the U of M. It's all been squared away with the powers that be, and it looks like I'll start Wednesday next. 'Til then I've been trying to learn as much about MRI and DTI (diffusion tensor imaging) specifically as I can. There's a lot to learn though, and it would be easy to spend years on any particular facet. More on this as it comes...
In other news I finished the latest (and, need I say, final) installment of the Harry Potter series. Can I just say, DAMN! I was wrecked for days after I finished it. I haven't yet been to a support group or anything, but that is some series J.K. put together.
I've also been working on some LED lamps.. something's not quite right...

I am STARVING!! What am I doing writing? Dinner, then maybe more. Anyway, I'm still out here, sort of. LOVE!!!!

-p