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**lg problem**

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

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.

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

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

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.

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,063

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

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,063

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.

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

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

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

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

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

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.

Why did the vector cross the road?

It wanted to be normal.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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)*

IPBLE: Increasing Performance By Lowering Expectations.

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

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

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

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,063

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. **

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

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

ganesh wrote:

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.

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

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,063

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

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

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

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)*

"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..."

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