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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

ok.

next?

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,066

But, between us I agree with you and TheDude and will stick to the notation you mentioned.

If two processes have a running time of 1111 log(n) and (1 / 111 )n^3 which one is slower? For what n will they be equal? ( here we are using the computer notation for log(n) )

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

hi bobbym

1111log(n) is much better.n^3/111 is slower

i don't know when they are equal.

*Last edited by anonimnystefy (2012-01-25 09:51:45)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,066

You have no idea when they are equal?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

nope.i mean i know i have to solve an equation,but i'm not sure it can be done.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,066

Want to try it?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

yup.how?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,066

Enter this into alpha, just as you see it.

Plot[{1111Log[n,2],n^3/111},{n,0,30}]

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

hi bobbym

i used "solve" instead.i got a real solution- 29.3544.now i'm not sure if i should take 19 or 30,but i'm more on the side of 29.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,066

The plot would have shown you a little more. For n< that value you are better off using the slower process!

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

hi bobbym

i know that.but the big oh gives us results for large enough numbers.i was trying to find the exact value at which they are equal.

btw i looked it up first,but i couldn't see the exact result.

*Last edited by anonimnystefy (2012-01-25 10:15:24)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,066

You can not have n, other than a whole number so 29 and 30 are all you need..

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

next?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,066

One more thing first. Did you see the reason for solving graphically or otherwise?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

hi bobbym

well not really.i know when the latter will be better and when the first will be better.i just need to know when they are equal.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,066

Yes, but the exact point tells you which one is best for particular n's, while Big(O) only tells you for gazillions! If those were two different type sorts and you only had 20 objects to sort the n^3 / 111 is the right one to use.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

hi

well,duh.i'm saying that using the mixed you can do better than using just one.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,066

f(n) = 100n^2, g(n) = n^4

f(10) = 10000, g(10) = 10000

f(50) = 250000, g(50) = 6250000

f(100) = 1,000,000, g(100)= 100,000,000

The above table and a graph show that f(n) grows more slowly than g(n). Would you agree that f(n) = O(g(n))?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

hi

it should be g(50)=6250000.

yes.i would say that f(n) is O(g(n))

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,066

Correct! I have fixed the above question.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

hi bobbym

4 zeros!

next?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,066

I see that my cyber pen ran out of cyber ink.

Order these common functions according to growth rate:

1, n^n, 2^n, log(n), n log(n), n^2, n, n^2

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

hi

by what,complexity?

1,log(n),n,n log(n),n^2 (twice?),2^n,n^n

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,066

Kayrect a mundo! You certainly know your Big(O).

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

well it's not mine.

next?and this should be the last one for today.

*Last edited by anonimnystefy (2012-01-25 10:55:53)*

Here lies the reader who will never open this book. He is forever dead.

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