HP Forum Archive 14

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promgramming??(built in func) Message #1 Posted by Harrington on 2 June 2004, 11:35 a.m. This equation dosen't see to work well in my 49G+ I really don understand the underlaying factor for this what I do know is that other simple equations will work jus fine. The equation with the problem is : let assume 1/(x^2+3*x+2) I expect to see -1/(x+2) + (1)/(x+1) or vice versa. This is the simple program in RPN mode << Eter """ALG V}INPUT OBJ-> ->a <<a parfac>> >> all flags are set exact do you see any mistake on my part??

promgramming??(built in func) Message #2 Posted by Veli-Pekka Nousiainen on 3 June 2004, 1:54 a.m., in response to message #1 by Harrington **NOTE ** Capitalization MUST be as shown ** Program Frac: << "Enter Fraction" { "'" ALG V } INPUT OBJ-> -> a << a PARTFRAC >> >> BYTES => #B61Ch 78. key in after ' 1/(X^2+3*X+2 Press ENTER Result: '1/(1+X)-1/(2+X)' {VPN}

Re: promgramming??(built in func) Message #3 Posted by bill platt on 3 June 2004, 1:18 p.m., in response to message #2 by Veli-Pekka Nousiainen Hi VPN, How to modify to work on a 48GX?

Re: promgramming??(built in func) Message #4 Posted by Veli-Pekka Nousiainen on 3 June 2004, 4:33 p.m., in response to message #3 by bill platt Search for ALG48 in the www.hpcalc.org PF command will do the trick. You may need to invest in a 128MB RAM card. I would certainly use a faaaast 49g+ instead. {VPN}

BYTES => #B61Ch 78 always??? Message #5 Posted by Harrington on 3 June 2004, 4:36 p.m., in response to message #4 by Veli-Pekka Nousiainen please give some explination to this fragment. BYTES => #B61Ch 78. Look like an assemble code this might be the winning soulution cuz with out BYTES => #B61Ch 78. The 49G+ will not give an soulution to that problem .... jus explinations and some refrences for reading on BYTES => #B61Ch 78. thanks VPN

BYTES => #B61Ch 78 always??? Message #6 Posted by Veli-Pekka Nousiainen on 3 June 2004, 5:00 p.m., in response to message #5 by Harrington 1) RCL the program on stack level 1 2) spell out the command name BYTES and press [ENTER] 3) you will see the checksum and the used bytes on the stack {VPN}

Thanks VPN(never leave this board) Message #7 Posted by Harrington on 3 June 2004, 6:29 p.m., in response to message #6 by Veli-Pekka Nousiainen you are helpful on this board thanks a lot .

Help help help !!!!! please Message #8 Posted by Harrington on 4 June 2004, 12:57 a.m., in response to message #2 by Veli-Pekka Nousiainen I see all the syntax my question is here : where should I type or key in this BYTEs=> ???Capitalization MUST be as shown ** Program Frac: << "Enter Fraction" { "'" ALG V } INPUT OBJ-> -> a << a PARTFRAC >> >> <=======)I know this is the program BYTES => #B61Ch 78. <===========) now this is my question key in after ' <=======) key what in 1/(x^2+3*x+2) 1/(X^2+3*X+2 Press ENTER Result: '1/(1+X)-1/(2+X)' {VPN} what is happening is that I don know where that supposed to be in program clearify a bit here

How does BYTES => #B61Ch 78 slove this problem Message #9 Posted by Harrington on 4 June 2004, 11:55 p.m., in response to message #2 by Veli-Pekka Nousiainen I get the memory location how does it slove the problem. why should I be worried about this location?

Re: How does BYTES => #B61Ch 78 slove this problem Message #10 Posted by Cameron on 5 June 2004, 1:27 a.m., in response to message #9 by Harrington This isn't my bag, but... It doesn't *solve* the problem. It's not supposed to. And it's not a "memory location". I suspect that you've gone off on a tangent. It's not machine code; it's not tricky programming. It is simply a mechanism for confirming that you have keyed in VPN's program correctly. The first number (#B61Ch) is a checksum of the program you keyed in (in base 16). The second number is the amount of storage consumed by the program. If you get the same numbers as VPN you know you've keyed it in correctly. Try an experiment. VPN stressed that you must match the character case of his listing *exactly*. Re-key the program and change just one character. The program won't work (of course) but when you use BYTES you will get a *different* value to VPN's. That tells you that you made a mistake (although it doesn't show you what the mistake was). I hope that helped. Cameron PS: I know nothing of these new-fangled RPL calculators so I probably can't help you much more. xyzzy Edited: 5 June 2004, 1:29 a.m.

I keyed in these correctly(1/(x^2+3*x+2)) Message #11 Posted by Harrington on 5 June 2004, 3:10 a.m., in response to message #10 by Cameron what is interesting is that hp49G+ was programed under this condtion for built in function of this kind.n(x)/d(x) where n(x)=0 and d(x)=0 are true .cuz it work for any function of this kind x^2+3x+6/(x^3+3*4*x+8) rational function will pass, any thing short will not perhap I have a bug in my calculator ....... I typed this proble over and over . please help see if this work in ya calculator as well mine it is not while I'm posting this again.

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