Math Is Fun Forum

krisper

Member

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 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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mathsyperson

Moderator

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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Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,387

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

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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,387

Re: lg problem

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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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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 Thanks.

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Humankind's inherent sense of right and wrong cannot be biologically explained.

mathsyperson

Moderator

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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Why did the vector cross the road?
It wanted to be normal.

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

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.

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

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krisper

Member

Registered: 2005-12-20

Posts: 19

Re: lg problem

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 I tried the Stirling formula but nothing like this Thanks again.

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Humankind's inherent sense of right and wrong cannot be biologically explained.

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,387

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.

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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

krisper

Member

Registered: 2005-12-20

Posts: 19

Re: lg problem

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

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

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Humankind's inherent sense of right and wrong cannot be biologically explained.

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,387

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

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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Ricky

Moderator

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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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."