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July 5

How do I find the inverse

How do I find the inverse of 315773 (mod 91260)

Mod[315773 s, 91260] == 1

or

315773 * s = 1 (mod 91260)

I'm looking for the value of s

Thank you

122.107.135.140 (talk) 02:11, 5 July 2008 (UTC)[reply]

See Extended Euclidean algorithm. Oded (talk) 02:40, 5 July 2008 (UTC)[reply]

How to realize, prove, disprove, and or show the math behind a new type of number. A Transcendental number.

I stumbled across a way to use PI that is very odd or different than known.

I don't know really how to say it so I will just write it all out as best as I can.

If anyone has any opinion on this please help me. I am curious if this number can be calculated. I will start with PI

PI is 3.14...etc so what?

But can this number exist? PI ... infinite <-- 53562951413.1415926535 --> infinite

But it does exist, in my mind. And now yours.

How? Why? Why not? Does anyone care? Opinions?

Can it even fit into known terminology as far as the category it fits into?

What about a number that is infinitely big? WHAT?????

In history, has a number that is infinitely large ever been worked with mathematically?

Is this an integer?

PI.... infinite <-- 53562951413.

Where does this fit into all of it?

These questions are real. And nothing short of a real answer is deserved. NO INSULTS.

If you care the background behind the "BiTransendental" (not a real word, i made it up) number: http://www.pipyramid.com If you think that link is offensive to wikipedia in any way I will remove it, its only there to show the background behind the idea, not to advertise anyone or anything.

My Known BiTranscendental Numbers (many more)

pi = ... 51413.1415 ...

e = ... 8172.718 ...

Euler's constant, gamma = ... 5127750.577215 ...

Liouville's number ... 0001000000000000000001000110.110001000000000000000001000 ...

Chapernowne's number, ... 17161514131211109876543210.1234567891011121314151617 ...

Morse-Thue's number, ... 100101100.01101001 ...

ii = ...6759787020.207879576...

Feigenbaum numbers, e.g. ... 9664.669 ...

My Suggested Addition To Transcendental Numbers (many more)

pi = ...51413.

e = ...7182.

Euler's constant, gamma = ... 5127750.

Liouville's number ... 0001000000000000000001000110.

Chapernowne's number, ...17161514131211109876543210.

Morse-Thue's number, ... 100101100.

ii = ...6759787020.

Feigenbaum numbers, e.g. ... 9664.

Some Known Transcendental Numbers

pi = 3.1415 ...

e = 2.817...

Euler's constant, gamma = 0.577215...

Liouville's number 0.110001000000000000000001000 ...

Chapernowne's number, 0.1234567891011121314151617...

Morse-Thue's number, 0.01101001 ...

ii = 0.207879576...

Feigenbaum numbers, e.g. 4.669 ...

Gravitroid (talk) 07:19, 5 July 2008 (UTC)[reply]

Sounds like you might be interested in transfinite numbers or even surreal numbers. Integers cannot have infinitely many digits to the left of the decimal point though. If there were infinitely many, some operations would lose meaning – very similar to formal Laurent series. In fact, the decimal expansion of a number is a formal Laurent series in 1/10. GromXXVII (talk) 11:43, 5 July 2008 (UTC)[reply]

Makes me wonder, could you have a decimal system analog to formal Laurent series in 10 instead of 1/10, while still keeping the usual metric ? --81.250.16.61 (talk) 16:42, 5 July 2008 (UTC)[reply]

Yes, it would just reverse the digit order. --Tango (talk) 13:03, 6 July 2008 (UTC)[reply]

Hmm that helps, as much as i can understand it. Thats alot of new content for me to absorb...

You said...

Integers cannot have infinitely many digits to the left of the decimal point though. If there were infinitely many, some operations would lose meaning –

But I do have a point to make,(no pun)

Consider this:

This normally happens with infinitely many digits to the right of the decimal point though, some operations would lose meaning. I.E. rounding. And you decide where that happens.

So do I round? To help fix it and fit it into known math?

= ... 62951413.1415926 ...

rounded

=   3951413.141593

I guess I will wait a bit before suggesting new ideas... Let a few more people chime in on this? —Preceding unsigned comment added by Gravitroid (talk • contribs) 12:25, 5 July 2008 (UTC)[reply]

You may also be interested in P-adic numbers. -- Leland McInnes (talk) 13:51, 5 July 2008 (UTC)[reply]

I guess I was thinking of in particular operations common to the study of algebraic structures. Even still consider what rounding is: given a number N, you’re finding a number M that is close to N. In the usual measure of size 3951413.141593 is not at all close to 62951413.1415926, and much less so if you put more digits to the left! So I might actually be tempted to round it to

\infty

instead. In fact in the extended reals it’s probably equal to infinity – as precisely one element is at least as large as every other, and ...62951413.1415926... has that property. Thus

...62951413.1415926...=\infty

(Note that here I was using the assumption that

...62951413.1415926...=\left(3+1\cdot 10+4\cdot 100+...\right)+\left({\frac {1}{10}}+{\frac {4}{100}}+...\right)

)GromXXVII (talk) 14:47, 5 July 2008 (UTC)[reply]

I need to reply to one point that you made here GromXXVII, you said In the usual measure of size 3951413.141593 is not at all close to 62951413.1415926, and much less so if you put more digits to the left! What about this? In the usual measure of size 0.141593 is not at all close to 0.1415926, and much less so if you put more digits to the right! Some may think that is close, because its a SMALL number, when really we are just stepping up or down with the same Magnitude (mathematics)... in opposite directions... ? Gravitroid (talk) 01:31, 6 July 2008 (UTC)[reply]

Well by adding more digits to the left you’re increasing the magnitude of their difference. By adding more digits to the right, your increasing the magnitude of their difference by no more than one order of magnitude larger than the original difference. GromXXVII (talk) 12:15, 6 July 2008 (UTC)[reply]

Uhh, can you think of a way to reword that? With math or an example?Gravitroid (talk) 12:41, 6 July 2008 (UTC)[reply]

Suppose I want to round ...4321.1234... On the left, if I round to 4 digits I get 4321.1234… to 5 digits I get 54321.1234… to 6 digits I get 654321.1234… On the right if I round to 4 digits I get …4321.1234 To 5 digits I get …4321.12345 To 6 digits I get …4321.123456

So on the left the number I get when rounding changes by an order of magnitude each time, while when rounding on the right the numbers are changing by less than 10^-4 each time. I’m not too sure if what I’m doing on the left actually makes any sense though – because I’m applying concepts from the usual measure of size to a string of symbols that should probably have something else. GromXXVII (talk) 16:21, 6 July 2008 (UTC)[reply]

Uhh, heres why rounding on both sides is the same. This number ...54321. large rounded to 5321. NOT 4322. i.e. 54321. - 5321. = 49000 and 54321. - 4322. = 49999 soo.... 49000 < 49999, thus correct rounding of large numbers is ...

This 54321. to 5321. the numbers 54321. , and 5321. are closer to eachother than 54321. and 4322. Its a matter of thinking about HOW size is relative. My brain huts... Because this is totally incorrect, technically. Large rounded would be... This number 54321. large rounded to 9999. but your no longer preserving the structure of your number. The closest preservation of the structure is a rounded answer of 5321. not 9321. Like in the case of small rounding .12345 to .1239 is incorrect. And .12345 to .1235 is correct. And I suppose this is how you work with these numbers? But if thats the case then the number 43210 is rounded to 2210? well no its 3211, lol. (when in rounding you round down if less than 5?) I think my brain is turning to mush with this new number.

;), yes? Gravitroid (talk) 02:38, 7 July 2008 (UTC)[reply]

I was going to link the p-adics, until I saw someone else had. One problem with the numbers described so far is that you haven't told us how arithmetic is to be performed on them. The 10-adics can be created by restricting decimals to finitely many on the right, freeing them to infinitely many on the left, allowing carrying to and from infinity on the left, and otherwise keeping arithmetic the same. To deal with convergence, though, the metric is flipped around and the definition of "small" changes in odd ways. I've seen a paper that described a purely algebraic (ignoring metric properties) extension of the idea to doubly infinite sequences, and I seem to remember that, modulo obvious equivalences, they just got the real numbers back. Food for thought, anyway. Black Carrot (talk) 18:46, 5 July 2008 (UTC)[reply]

Your BT numbers (BT is Bi-Transcendental) is way too cool! But what in the world are you suppose to do with them? As far as I can see, apart from admiring them, they are pretty useless (pun fully intended). You cannot add them in finite time. You cannot multiple them. They don't seems to solve any mathematical puzzles.

But I must say, BT numbers are beautiful. I think of them as sets of (infinite) digits. 122.107.135.140 (talk) 23:26, 5 July 2008 (UTC)[reply]

Isn't there a place for pointless numbers? So that way when we become inter-dimensional beings we can also enjoy our lemonade? I am pretty sure, in time, we will see the point of such a number. The history is told on the pipyramid.com page, so really they aren't even really MY numbers, just like PI isn't someones. BT numbers do have their origin, my brain, but I wouldn't have found them if the retired astrophysicist had not found the number PI to begin with. And if the news didn't carry the story. And if someone didn't take the pictures. Etc... This whole thing is odd. ABSURD! NEW NUMBER FOUND IN CROP CIRCLE?

I pass the torch onto you mathematician, go forth and discover. Gravitroid (talk) 06:47, 6 July 2008 (UTC)[reply]

Mmmh an interesting thing, you might be interested in the omega number a random sequence of 1's and 0's in binary, a web search should explain.87.102.86.73 (talk) 10:20, 6 July 2008 (UTC)[reply]

Actually, π is not an integer [nor are the other infinite numbers (unless it finds a repeated pattern)]. They are irrational numbers. For comprehending irrational numbers, all I can say is that no one will ever know every digit of any of them (though that wouldn't even be neccesary to find the area of a star 10x the size of our sun) so don't put too much stress on yourself to try and figure them all out. However, you can try to think of it as

$\pi ={\frac {C}{d}}$

or as

$\pi \approx {\frac {22}{7}}\approx 3.14$

Earthan Philosopher (talk) 04:55, 10 July 2008 (UTC)[reply]

Miles

I have seen somewhere that the American mile and British mile are of different lengths, with the American mile being roughly 1.8km compared to the British mile of roughly 1.6km. Is this correct? Simply south (talk) 11:13, 5 July 2008 (UTC)[reply]

See Mile. Basically the US and Imperial miles are the same at ~1.6km. Maybe you're thinking of the Nautical mile, which is 1.852km. AndrewWTaylor (talk) 11:51, 5 July 2008 (UTC)[reply]

Exponent/logarithm question

I have an exponential equation: $e^{ax+b}=y$ . I can now express x as $x={\frac {\ln y-b}{a}}$ . Is there any way I can introduce a new variable c to the left-hand side of the equation, so that c affects the equation independently from a and b (i.e. not like $e^{ax+b+c}=y$ , where I can just as easily substitute $b+c$ with a single variable), and still be able to express x in terms of a, b, c and $\ln y$ , other than $e^{ax^{2}+bx+c}=y$ ? JIP | Talk 16:22, 5 July 2008 (UTC)[reply]

How about

ce^{ax+b}=y

? By the way, what on earth is this for? Algebraist 16:27, 5 July 2008 (UTC)[reply]

It's equivalent to

e^{ax+b+\ln c}=y

, which is the same thing as in my

b+c

substitution above. It's for estimating Wikipedia's article growth. JIP | Talk 16:43, 5 July 2008 (UTC)[reply]

Pretty much anything you put in there will add new information. For instance, you could replace x with x^c, or cx^n for n>2. You could add in csin(x) or sin(cx) or sin(c-x), or generall f(x,c) where f is anything besides what you've excluded. Black Carrot (talk) 18:35, 5 July 2008 (UTC)[reply]

No, it won't. You've got an overparametrized model here. Adding c, either as a factor in front, or as a term in the exponent, will not give you any new functions that you haven't already got with a and b. Michael Hardy (talk) 01:16, 6 July 2008 (UTC)[reply]

I get the impression that JIP is looking functions f(x) such that

e^{f(x)}=y

is analytically invertable, which is equivalent to asking when f(x) is analytically invertable, since

x=f^{-1}(ln(y))

. That still leaves a very large set of possible forms. For example, everything of the form

f(x)=ax^{4n}+bx^{3n}+cx^{2n}+dx^{n}+e

for all a, b, c, d, e, and n will have an exact algebraic inverse (though not all give rise to real valued solutions). Dragons flight (talk) 18:48, 5 July 2008 (UTC)[reply]

All of the above suggestions have been in the form

e^{f(x,a,b,c)}=y

. This, of course, automatically means that I can express x in terms of a, b, c and

\ln y

. This is all well and good, but I was wondering if there are possibilities involving changes to the left-hand side itself, not just to the exponent of the exponential term. JIP | Talk 19:50, 5 July 2008 (UTC)[reply]

You are being offly vague. What kind of changes would you like? Dragons flight (talk) 20:17, 5 July 2008 (UTC)[reply]

Maybe

e^{ax+b}+c=y

? Of course, that is the same as

e^{ax+b}=y-c

, but, as I understand, that is not a disadvantage in this case. --Martynas Patasius (talk) 23:43, 5 July 2008 (UTC)[reply]

The answers above are the blind leading the blind. If you write $y=e^{ax+b+c}$ or $y=ce^{ax+b}$ , you won't have any more curves than if you write $y=e^{ax+b}$ . It's a two-parameter family, and if you express it in those ways with three paramaters, it's still a two parameter family with a redundant set of parameters. That means you could change b but then compensate by changing c, so that you get the SAME curve with DIFFERENT values of the parameters. Michael Hardy (talk) 01:18, 6 July 2008 (UTC)[reply]

Michael Hardy is right. Adding c to the exponent, or multiplying the left hand side with c, is a redundant operation. There is an option to make a real change by using

e^{f(x,a,b,c)}=y

where c is a non-redundant parameter of f, but I already know how to do this, because the only difference between the exponential equation and just using f is that

f(x,a,b,c)

equals

\ln y

and not y. I was wondering if there are non-redundant changes that I did not know about. I originally thought about

e^{2ax}+e^{bx}+c=y

but found out I could not express x in terms of a, b, c and

\ln y

. I tried solving it with

e^{x}

and y instead but both my calculator and SciLab had trouble processing the numerical value of

e^{x}

because x is in the order of tens of millions. JIP | Talk 04:57, 6 July 2008 (UTC)[reply]

Given the information given, you might want to consider e^f(x)+e^g(x-t)=y. Reason being that the second part represents the introduction of a 'foreign' language version of wikipedia.. the possibilities are of course endless.87.102.86.73 (talk) 12:11, 6 July 2008 (UTC)[reply]

Also if you think/suspect that the maximum number of articles on wikipedia is finite you might want to consider curves like the Logistic function..87.102.86.73 (talk) 13:16, 6 July 2008 (UTC)[reply]

Unfortunately, I still don't know if it's possible to construct an equation with the left part involving $e^{x}$ in any other form than $e^{f(x)}=y$ , that I could take the logarithm of both sides from, and end up free of exponentiations (is that a word?). I previously thought about an equation in the form of $e^{f(x)}+e^{g(x)}=y$ but apparently I can't take the logarithm of a sum. The idea is to have three parameters, a, b and c, and three reference points 1, 2, and 3, with $x_{n}$ equalling the time since a chosen "epoch" at that reference point, and $y_{n}$ equalling the number of articles at that reference point. With three equations and three unknowns, I can figure out a, b and c, and use that and $y_{4}$ (the target number of articles) to figure out $x_{4}$ (the time when that number will be reached). But this will have to be calculated without using terms in the form of $e^{f(x_{n})}$ because they are simply too large to compute numerically. JIP | Talk 18:12, 6 July 2008 (UTC)[reply]

Why too large? The number of articles is only millions, so the exponentials should evaluate to the same. It strikes me that prehaps you are trying to evaluate e^ax by computing (e^x)^a with a poorly scaled version of x. If you recast x in years, then you wouldn't have any terms larger than ~e⁶ which is very manageable. Doing so would also rescale the meaning of a, but that doesn't seem a problem. I don't see any reason why you would ever need to evaluate the exponent of ten million as you suggest above. Dragons flight (talk) 18:23, 6 July 2008 (UTC)[reply]

I was doing it in seconds, not in years. As you all know, a year is more than 30 million seconds. I guess I could change the scale to days instead of seconds, making the exponentiations much more manageable, and then change it back to seconds once I have solved the equation. I think it won't give me the exact same answer, but it will be on the same scale, and whoever said estimating Wikipedia growth was an exact science anyway? JIP | Talk 18:48, 6 July 2008 (UTC)[reply]

Seconds seems excessively precise for modelling Wikipedia growth! --Tango (talk) 20:49, 6 July 2008 (UTC)[reply]