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#1 2005-12-20 20:56:04

krisper
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Registered: 2005-12-20
Posts: 19

lg problem

This is a bit of chalange for me and that is why I am posting it here smile The problem says: Prove that
lg(lgX) > 2500, where X equals (2^10000)! . The ! sign means factoriel.  Please if you dont know how to solve it atleast write where you got after trying so we could solve this problem together. Thanks.


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#2 2005-12-20 22:37:46

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: lg problem

I think it would be easier if we transferred the lg bit to the right hand side.

lg(lgX)) > 2500
lgX > 10^2500
X > 10^(10^2500)

(2^10000)! > 10^(10^2500)

These are both extremely large numbers. Most calculators don't even like 80!, so (2^10000)! is huge! Similarly, the right-hand side will have (10^2500)+1 digits!

I can't see how to get the (2^10000)! into a usable form right now, but I'll be back later to ponder over it some more.


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#3 2005-12-20 23:16:31

ganesh
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Registered: 2005-06-28
Posts: 14,886

Re: lg problem

The value of (2^10000)! can be found approximately using the James-Stirling formula,
where n! ~ [√(2*pi*n)]*[(n/e)^n]
2^10,000 contain 3011 digits.
The factorial of this number would be of the order of 10^10^3011 (actually, much greater!).
Log to base 10 of 10^10^3011 is 10^3011
and log of log of 10^10^3011 is 3011, which is greater than 2500!
q.e.d smile


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#4 2005-12-21 16:39:36

ganesh
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Registered: 2005-06-28
Posts: 14,886

Re: lg problem

ganesh wrote:

2^10,000 contain 3011 digits.
The factorial of this number would be of the order of 10^10^3011 (actually, much greater!).

This is true because for any number n greater than 100, n!>>10^n.


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#5 2005-12-23 01:31:51

krisper
Member
Registered: 2005-12-20
Posts: 19

Re: lg problem

"2^10,000 contain 3011 digits" - Can you explain me please how did you find that out smile Thanks.


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#6 2005-12-23 02:50:23

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: lg problem

You work out how many digits a number has by taking log(10) of it and rounding up.

By definition, log(2) 2^10000 = 10000

And by combining that with the rule that log(a)b = log(c)b/log(c)a, you can work it out.

log(2) 2^10000 = [log(10) 2^10000] ÷ [log(10) 2]

∴ log(10) 2^10000 = log(2) 2^10000 * log(10) 2 = 10000 * log(10)2 = 3010.3...

So rounding up gives that 2^10000 has 3011 digits.


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#7 2005-12-23 06:54:18

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: lg problem

2^10000 = 199506311688075838488374216268358508382349683188619245485200894985294388302219\
466319199616840361945978993311294232091242715564913494137811175937859320963239\
578557300467937945267652465512660598955205500869181933115425086084606181046855\
090748660896248880904898948380092539416332578506215683094739025569123880652250\
966438744410467598716269854532228685381616943157756296407628368807607322285350\
916414761839563814589694638994108409605362678210646214273333940365255656495306\
031426802349694003359343166514592977732796657756061725820314079941981796073782\
456837622800373028854872519008344645814546505579296014148339216157345881392570\
953797691192778008269577356744441230620187578363255027283237892707103738028663\
930314281332414016241956716905740614196543423246388012488561473052074319922596\
117962501309928602417083408076059323201612684922884962558413128440615367389514\
871142563151110897455142033138202029316409575964647560104058458415660720449628\
670165150619206310041864222759086709005746064178569519114560550682512504060075\
198422618980592371180544447880729063952425483392219827074044731623767608466130\
337787060398034131971334936546227005631699374555082417809728109832913144035718\
775247685098572769379264332215993998768866608083688378380276432827751722736575\
727447841122943897338108616074232532919748131201976041782819656974758981645312\
584341359598627841301281854062834766490886905210475808826158239619857701224070\
443305830758690393196046034049731565832086721059133009037528234155397453943977\
152574552905102123109473216107534748257407752739863482984983407569379556466386\
218745694992790165721037013644331358172143117913982229838458473344402709641828\
510050729277483645505786345011008529878123894739286995408343461588070439591189\
858151457791771436196987281314594837832020814749821718580113890712282509058268\
174362205774759214176537156877256149045829049924610286300815355833081301019876\
758562343435389554091756234008448875261626435686488335194637203772932400944562\
469232543504006780272738377553764067268986362410374914109667185570507590981002\
467898801782719259533812824219540283027594084489550146766683896979968862416363\
133763939033734558014076367418777110553842257394991101864682196965816514851304\
942223699477147630691554682176828762003627772577237813653316111968112807926694\
818872012986436607685516398605346022978715575179473852463694469230878942659482\
170080511203223654962881690357391213683383935917564187338505109702716139154395\
909915981546544173363116569360311222499379699992267817323580231118626445752991\
357581750081998392362846152498810889602322443621737716180863570154684840586223\
297928538756234865564405369626220189635710288123615675125433383032700290976686\
505685571575055167275188991941297113376901499161813151715440077286505731895574\
509203301853048471138183154073240533190384620840364217637039115506397890007428\
536721962809034779745333204683687958685802379522186291200807428195513179481576\
244482985184615097048880272747215746881315947504097321150804981904558034168269\
49787141316063210686391511681774304792596709376. It has exactly 3011 digits.
smile

Last edited by krassi_holmz (2005-12-23 06:56:53)


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#8 2005-12-25 22:36:15

krisper
Member
Registered: 2005-12-20
Posts: 19

Re: lg problem

ganesh wrote:

2^10,000 contain 3011 digits.
The factorial of this number would be of the order of 10^10^3011 (actually, much greater!).

Sorry again for asking but can you explain how did you got to this - (2^10 000)! = 10^10^3011 smile I tried the Stirling formula but nothing like this sad Thanks again.


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#9 2005-12-25 23:02:42

ganesh
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Registered: 2005-06-28
Posts: 14,886

Re: lg problem

2^10,000 is approximate 10^3011.
We know that for any number greater than 25, n!>10^n.
Therefore, (2^10,000)! would be greater than 10^10^3,011.
You can omit the √(2*pi*n) in the James Stirling formula as it wouldn't make a big difference.
If you are only required to show lg(lgX) > 2500, where X equals (2^10000)!, the James Stirling formula may not be necessary, as the simpler method has been discussed.
Don't be sorry for seeking clarifications to your doubts. smile


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#10 2005-12-25 23:11:55

krisper
Member
Registered: 2005-12-20
Posts: 19

Re: lg problem

ganesh wrote:

We know that for any number greater than 25, n!>10^n.

Is this a theorem or ? smile Thanks very much for your explanation. Happy holidays.


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#11 2005-12-25 23:35:16

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,886

Re: lg problem

No, not a theorem. An observation that can be proved.
By using the calculator (scientific mode), you can see that 25! > 10^25.
Therefater, the LHS of the inequation increases by 26, 27, 28 etc. for 26!, 27!, 28!. On the other hand, the RHS increases only by 10 in each step. Therefore, it can be proved that for any number, n ≥ 25, n!>10^n.
Wish you happy holidays too smile


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#12 2005-12-26 03:23:33

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: lg problem

Proof: This proof is by induction on n.

(Base Case) Let n = 25.  Then 25! > 10^25.  So the base case holds.

(Inductive Assumption) Now let n be an arbitrary number, such that n ≥ 25.  Assume that n! > 10^n.

Since n ≥ 25, n+1 > 10.  Also, since we already know n! > 10^n:

(n+1)*n! > 10*10^n, or rather, (n+1)! > 10^(n+1).

∴By the principle of mathematical induction, n! > 10^n for all n ≥ 25.  QED.

Last edited by Ricky (2005-12-26 03:25:53)


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