HP 49G plus, DOMAIN and LIMITS



#17

Hello, I have couple of questions that I just can't figure out, so if someone can help me I would appreciate that a lot.

ok, my first question is on the LIMITS, how would I find the limit of a F(x,y) function ?
for example lim (x^5+4x^3*y-5xy^2)
(x,y)=(5,-2)

I know how to find the limit of normal F(x) function but I can't do this..

and some thing on Domain, i can't do the Domain of f(x,y)..

can someone please help me.

Thank you
Slobodan,


#18

Hi Slobodan,

It's 24 years ago for me since I dropped out of engineering (due to apathy and some stress, not "dumbness" ;-P) so forgive my being rusty. I am not saying I'm not dumb, just that I didn't drop out 'cause I am/was dumb ;-)

Valentin is the maths professor around here (literally,
I think...) so I am sure he will correct me where I am wrong.
His understanding of maths is really cool and he is really helpful.

I am not too worried about making a fool out of myself as the regulars here will know, , so I will say what little I remember
in general terms.

LIMITS:
I'm not gonna tell you the nuts and bolts of it because I have forgotten, BUT:

You just need to do some BASIC READING ON "DOUBLE INTEGRATION".
It's quite involved maths, sort of.

I would google on that to start with. I think there are some equations of (very) different form which are f(x,y), being D.E.'s
That stuff is different, more involved, very useful to engineers, but really made my head spin (memory overload -> "fail" ME201).

Generally the function of one independent variable has a line or curve as a solution or a limit (they're different), but the function of TWO independent variables is a SURFACE in 3-space, so the limit could I think be a surface as well as a curve or line.

So your answer will be often be a non linear equation itself, not
neccessarily an easy solutin like a line or an axis.

There are simple cases, though:

Let's take Space and forget about the third dimension (z).

Say you have a (two dimensional) "gravity" function:
g = F(x,y) = SQRT(1/(x^2+y^2))

So we will make the third axis (z) plot our output or G force.
You would have as an example a black hole with an infinite gravity
force "at" the origin (0,0) with z (= g) = infinity. You can correctly say that the Z axis is a kind of limit, called a "pole".
That function also has a "ZERO" at infinity, the function decays to nearly zero at very large x and y. I visualize this as a circus tent.

Compare that with the simple hyperbolic function y = f(x) = 1/x, where the x and y axes are asymptotes, i.e. limits of the function.

I know I haven't explained much, but you could google on D.E.'s maybe (it's a bit o.t. given the functions you mentioned) and three dimensional calculus (?) Also google or Wiki on "mathematical poles and zeros".

The function DOMAIN is just the PLANE from x=minus infinity and y=minus infinity to x=plus infinity and y=plus infinity
(the "infinite plane")

I am sure Valentin will come to our rescue on this.

Best,

DW


Edited: 3 Oct 2005, 8:56 a.m.


#19

Hi, Don

Don Wallace posted:

"I am sure Valentin will come to our rescue on this."

Thanks for your confidence in my humble abilities, but I understand that what Nikolic actually wants are instructions in how to tackle these advanced subjects with his HP49G+.

In this regard I can be of little help as I don't own any HP48/49 model nor have I ever used one.

Best regards from V.


#20

Hi Valentin,

Yes, I realise that but I also realise (I think) Slobodan is new to 3D calculus, otherwise he would have just gone and done it ;-)

A calculator will just crunch the numbers for you...

DW


#21

Hi again, Don:

Don posted: "A calculator will just crunch the numbers for you..."

Indeed. Let's see if you'd do better than your calculator. For a micro-challenge, try and compute this limit with your HP48/49,
symbolically and numerically. If a symbolic solution is not an
option, see if you can symbolically prove the result yourself:

                 i=(n+1)^2
/ --------- \
| \ |
Limit | \ |
n->Inf | Frac( / Sqrt(i) ) |
| / |
| --------- |
\ i=n^2 /
where, of course, "Inf" is 'positive infinity', "Frac" is 'fractional part', "Sqrt" is 'square root' and the big Sigma
letter means "sum from i=n^2 to i=(n+1)^2".

Try and guess the result before it comes out of the screen !

Best regards from V.


#22

Hi Valentin,

I only have a 41. And I'm nt working in any area where I need
limits, so I'm not motivated to use my head on this too much;
real life presents enough (serious) problems to occupy me...

Thanks for the example though.

I WONDER WHAT SLOBODAN THINKS ABOUT MY EXAMPLES?

DW

Edited: 3 Oct 2005, 6:50 p.m.

#23

After some fiddeling with the syntax, I was at least able to enter the function into my 49G+:

'VA(N)=FP(sigma(I=N^2,(N+1)^2,sqrt I))'
DEF
(substitute sigma and sqrt by the respective symbols). This creates a user function VA which can be evaluted from the VAR menu.

Trying to compute the limit automatically failed: CAS asked for approximate mode and gave up afterwards. But the function VA seems to approximate 1/6 for greater values of N. I'm unable to explain why. Here are some thoughts:

The range of the sum goes over the values between to squared integers, e. g. from 4 to 9 if N=2 or from 100 to 121 if N=10. Since only the fractional part is computed, the first and last summands can be removed from the sum (the square root of a squared integer is an integer.) Therefore, the number of summands is 2*N and the values to be summed up are between sqrt(N^2+1) (>N) and sqrt(N^2+2*N) (<N+1). None of these values can be an integer so all add to the final fraction.

That's the end of my thoughts. I have no idea how a function like FP can be treated analytically...


#24

First I want to say "Thank you" to all the people who tried to help me on my question...

I am acutally not new to domain and limits, I taken caluculs some two years ago in HS and now I'm in CALC III in my sophmore year at university. On most domain and limits, I can find them. Ones that are in 3D are definetly harder but I'm hanging in there (F(x,y,z))

The reason I asked this question was just so I can figure it out faster when midterm comes (Wendsday), because I realized that my calculator can do those funcions(and many other things that I don't know to work on it). So its ok since many people don't use this calculator, I'll stay up all night tonight and tommorow for the test :) Again, I want to say "thank you" to all the people that responed to my question..

I also have another question, I don't know if I should open a new topic for it ? (I promise this is my last question and I won't boather you guys any more. :), its just that I have a midterm on Wendesday and my scholarship is on the line ehehe :-( I must get 3.5 this semseter :-( ))

ok here it is

I am trying to figure out how to use my HP49+ for "solving system of equations with multiple variables" !? if someone knows how to do it will they please post ?

Thank you, and once again thanks for the help

#25

Hi Valentin,

after a few hours of sleep I think I know what's going on with your challenge.

You can look at the sum as an aproximation of the area below a lying parabola y=sqrt x in the interval [n²,(n+1)²]. To get rid of the integer part, simply subtract the area under the straight line crossing the parabola between the points (n²,n) and (n²+2n+1,n+1):

'VA2(N) = integral(0, 2*N+1, sqrt(N^2+X)-X/(2*N+1)-N, X)'
DEF
The limit n->oo of the original sum should be the same as the limit of the above integral.

The exact solution to this integral is 1/6 for any n and so is the limit for n->oo.

Now back to work (or to bed...)


#26

Hi, Marcus:

See ? "Calculators just crunch the numbers for you" but it helps if you previously crunch them up yourself a little bit :-)

Thanks for your interest and keen insights and

Best regards from V.


#27

Uh ?
I don't see how one is so sure that "the area ABOVE the straight line crossing ..." is less than 1.
Hard, dirty direct calculation shows that the sum hac value 2xN^2+3xN+1+1/6+o(1/N^2), as the term in (1/N) vanishes (quite a surprise !).
**THEN** the fractional part is **DEMONSTRATED** to reach 1/6 on infinity.
Please answer my question. Thanks.


#28

GE, let me try to explain my thoughts...

At first, I was just subtracting N from the parabola and I got a "stair" of 2n+1 steps with (total) heights between 0 and 1. If you now look at the summands from both ends (first/last, next to first, next to last) they tend to cancel their fractional part if you add them together. If the summands were lying on the straight line between the endpoints, the effect would be exact (the sum would be exactly one for each pair). Since the parabola always stretches slightly above the straight line, there is some fractional part left.

One consequence is that if you subtract the summands on the straight line you only subtract an integer value which does not effect the final result.

The proof isn't perfect, yet. But since the integral itself evaluates to a value below 1 (1/6), I seem to be correct.

Valentin, any real proof?


#29

Fair enough.
I think you found WHY it goes to 1/6, while calculus only shows WHAT it does (ie go to 1/6). My remark came out of frustration for not finding an elegant proof. There is still something missing from yours, however.
My method was brute force : taylor series to "sufficient" order of sqrt, then summation on each term, being powers of i for i in [n^2, (n+1)^2]. A no brainer, not elegant.
Thank you Valentin for this one, very nice.

As we are quite off topic here and into Math Land, do you know of a method to factor numbers of the form (a + b.sqrt(5)) ? Example : (1+sqrt(5)) = (3-sqrt(5)).(2+sqrt(5))
Thanks.


#30

Hi, GE:

GE posted:

"Do you know of a method to factor numbers of the form (a + b.sqrt(5)) ? Example : (1+sqrt(5)) = (3-sqrt(5)).(2+sqrt(5)) Thanks. "

Surely you're aware that the factorization, if possible, is not unique, right ? For example, for the particular case you gave, we have:

      (1 + sqrt(5)) = (a + b*sqrt(5)) . (c + d*sqrt(5))
with a,b,c,d integer, where your factorization corresponds to the case a = 3, b = -1, c = 2, d = 1. But you can factorize 1 + sqrt(5) using factors of the stated form in infinite ways, these are but a few possibilities:
    a       b       c       d
------------------------------
2 1 3 -1 (your factorization)
7 3 -2 1
9 4 -11 5
29 13 9 -4
38 17 47 -21
123 55 -38 17
161 72 -199 89
so you have, for instance:
  (161 + 72*sqrt(5)) . (-199 + 89*sqrt(5)) = (1 + sqrt(5))
as you may check. The general solution is a simple case of diophantine equations, which, if I'm not wrong, in this particular case reduces to a variation of the generalized Pell's equation. See:

Pell's Equation

I'd like to elaborate but am incredibly busy right now. Nevertheless, for the particular example you gave of factoring
(1+sqrt(5)) and for you to check your future, general results, this small HP-71B program will directly compute (no search) and print the first ten set of coefficients (A,B,C,D) such that (A+B*r)*(C+D*r) = (1+r), where r =sqrt(5).

  100 DISP " A         B           C           D"
110 DISP "--------------------------------------"
120 DESTROY ALL @ STD @ R=SQR(5) @ FOR N=1 TO 5
130 A=IROUND(((2+R)^N+(2-R)^N)/2)
140 B=IROUND(((2+R)^N-(2-R)^N)/(2*R))
150 DISP A,B,FNC(A,B),FND(A,B)
160 A=IROUND(((1+R)*(2+R)^N+(1-R)*(2-R)^N)/2)
170 B=IROUND(((1+R)*(2+R)^N-(1-R)*(2-R)^N)/(2*R))
180 DISP A,B,FNC(A,B),FND(A,B)
190 NEXT N
200 DEF FNC(A,B)=(5*B-A)/(5*B^2-A^2)
210 DEF FND(A,B)=(B-A)/(5*B^2-A^2)
Upon running, it produces:
 A            B            C            D
----------------------------------------------
2 1 3 -1
7 3 -2 1
9 4 -11 5
29 13 9 -4
38 17 47 -21
123 55 -38 17
161 72 -199 89
521 233 161 -72
682 305 843 -377
2207 987 -682 305

which expands on the results given above.

Best regards from V.

Edited: 5 Oct 2005, 11:08 a.m.

#31

Hi Valentin,

thank you for the warm words :-)

I must admit that I had my calculators (HP 49G+ and TI Voyage 200) do the 'dirty work' of computing the integral. In the meantime I did it myself with pencil and paper and the result is defintly 1/6, independent of n.

What is your source for these challenges? Is it just the top of your head?

Marcus


#32

Hi again, Marcus:

Marcus posted:

"thank you for the warm words :-)"

They're well deserved. You've been a regular contributor to my mini- and micro-challenges of all sorts, clearly showing a lot of interest in them and contributing many good ideas.

"Valentin, any real proof? "

Sort of. Using the Euler-Maclaurin formula for sums, we readily
get the asymptotic expression:

k=n
\--- 2 3/2 1 1/2 1 -1/2 1 -3/2
> sqrt(k) = - n + - n + -- n + ---- n + ...
/--- 3 2 24 2880
k=1

and, when applied to our particular sum, we get (asymptotically):
    k=(n+1)^2
\--- 2 7 1 1 1 3n^2+3n+1
> sqrt(k) = 2n + 3n + - - -- ------ - ---- ---------- + ...
/--- 6 24 n(n+1) 2880 (n(n+1))^3
k=n^2
and the fractional part of this expression does tend to 1/6 (as
numerically found), Q.E.D.

"What is your source for these challenges? Is it just the top of your head?"

Not really. I read a lot of math, be they books, newsgroups or whatever. Everytime I find something interesting, I keep a mental note of it and, when appropriate, it may constitute the 'mother lode' for some mini- or micro-challenge, hoping that like-minded people will also be as amused by it as I was in the first place.

Fortunately, math is a *very* amusing discipline and it's actually quite easy to find fascinating quirks and facts aplenty.

  888888888888888888888888888888888888887 = 922166778639871 * 963913371722125383063497

19999999999999999999999999999999999999999999 = 14188750452331121 * 1409567394055769572492644719

66666666666666666666666666666666666666666666666667 = 696005971439957313307 * 95784618814032449036431938481

555555555555554444444444444443333333333333332222222222222222211111111111111 = 3063441154048486369668261625739 * 181350163955772670068231705843686316121284149

Best regards from V.

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