Infinite loop

[user923005]

Re: Infinite loop

On Mar 20, 8:32 pm, "user923005 " <dcor...@connx. comwrote:

/*
Here's my version of your 1/e program:
*/

I call it a 1/e program be cause the divided difference for one unit
step is A(n)/A(n+1) = 1/e (approximately)

You could also call it an e program, because you can get the same
series (asymptotically ) by multiplying the previous term by e.

e.g.
....
9.0000000000000 00000 27145.247361213 536000000
10.000000000000 000000 73788.778437947 738000000

27145.247361213 536 * e =73788.43263101 260340026529804 9315

e.g.
....
99.000000000000 000000
331282515698122 690000000000000 00000000000000. 000000000000000 000
99.500000000000 000000
546192530242544 100000000000000 00000000000000. 000000000000000 000
100.00000000000 0000000
900519242508405 510000000000000 00000000000000. 000000000000000 000

331282515698122 690000000000000 00000000000000 * e =
900519242508405 302330870179174 11

Seems to answer pretty nearly.
Since there are two calls to exp() in there with different args, I
guess it won't be a good way to calculate e, but I hope it has some
interesting properties that you are looking for.

[user923005]

Re: Infinite loop

On Mar 20, 8:47 pm, "user923005 " <dcor...@connx. comwrote:

On Mar 20, 8:32 pm, "user923005 " <dcor...@connx. comwrote:
>

/*
Here's my version of your 1/e program:
*/

>
I call it a 1/e program be cause the divided difference for one unit
step is A(n)/A(n+1) = 1/e (approximately)
>
You could also call it an e program, because you can get the same
series (asymptotically ) by multiplying the previous term by e.
>
e.g.
...
9.0000000000000 00000 27145.247361213 536000000
10.000000000000 000000 73788.778437947 738000000
>
27145.247361213 536 * e =73788.43263101 260340026529804 9315
>
e.g.
...
99.000000000000 000000
331282515698122 690000000000000 00000000000000. 000000000000000 000
99.500000000000 000000
546192530242544 100000000000000 00000000000000. 000000000000000 000
100.00000000000 0000000
900519242508405 510000000000000 00000000000000. 000000000000000 000
>
331282515698122 690000000000000 00000000000000 * e =
900519242508405 302330870179174 11
>
Seems to answer pretty nearly.
Since there are two calls to exp() in there with different args, I
guess it won't be a good way to calculate e, but I hope it has some
interesting properties that you are looking for.

A bit of thought shows why it converges to e per unit step.
exp(x) is [by far] the fastest growing term.
It could be used as an explanation why O(f(n)) needs (primarily) the
dominant term to explain the nature of the algorithm for big
arguments.

[Dave Vandervies]

Re: Infinite loop [OT]

In article <1174448304.561 462.223800@o5g2 000hsb.googlegr oups.com>,
user923005 <dcorbit@connx. comwrote:

>On Mar 19, 6:23 pm, dj3va...@caffei ne.csclub.uwate rloo.ca (Dave
>Vandervies) wrote:
>[snip]

>Exercise for the reader: Does a finitely describable uncountable set
>of discrete elements exist?
>(V guvax gur cbjre frg bs gur frg bs vagrtref dhnyvsvrf.)

>
>Vf gurer n sbezny cebbs gung gur frg bs nyy fhofrgf bs gur vagrtref vf
>hapbhagnoyr?

Yes.
Vg'f npghnyyl fyvtugyl zber trareny; nal frg bs nal pneqvanyvgl unf n
fznyyre pneqvanyvgl guna vgf cbjre frg.
I can dig the proof out of my algebra notes from a year or two ago
if you're interested; this is getting rather far OT, so it's probably
better to email me if you want it. (Using dj3vande at eskimo.com[1]
will make it rather less likely that I'll delete your email without
noticing it while clearing the spam out from the address I post with.)


dave

[1] Protected from Google, not from spammers.

--
Dave Vandervies dj3vande@csclub .uwaterloo.ca
If only... there were people unburdened by the Hippocratic Oath on
ambulances, so they could usefully be fitted with rocket launchers.
--Maarten Wiltink in the scary devil monastery

[Army1987]

Re: Infinite loop

you_idiot_one:

printf("Enter the non-negative starting value\n");
converted = scanf("%lf", &start);
if (converted != 1 || start < 0.0) goto you_idiot_one;

Apart that if, I input some non-numerical data, scanf will try to read it as
a double over and over again, have you ever heard about "while"? C ain't
Commodore BASIC...
I am not one of those structured-programming worshippers who consider it
immoral to ever write "goto", "break", or "return" anywhere else than as the
last statement of a function, but here I just can't see how goto helps...

do {
printf("Enter the non-negative starting value\n");
/* You don't even need printf here, */
/* puts("Enter the non-negative starting value"); would do the job */
converted = scanf("%lf", &start);
} while (converted != 1 || start < 0.0)