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1. ## Maths help needed

How do I prove that

2^n > n^3 for n>10 or n=10??? I am suposed to use the induktion principle(this is a straight translation from swedish so might not be the correct english term).

I have done it in steps.

Step 1. 2^10 > 10^3. The difference is valid for n=10.

Step 2. Assume that the difference is true for n=p where p=10 or p>10

2^p > p^3 <=> 2^p - p^3 > 0

Step 3

2^(p+1) - (p+1)^3 = 2*2^p - (p^3 + p^2 + 3p + 1)=
= 2*2^p - p^3 - p^2 - 3p -1 > 0

How do I continue from there?? Or have I made a error in step 3?
Should I replace 2*2^p with 2*p^3?? damn I hate this

2. This thread scared me and reminded me I'm going to college some day.

3. 2+2=4...... F-O-U-R

4. Originally Posted by johan
How do I prove that

2^n > n^3 for n>10 or n=10??? I am suposed to use the induktion principle(this is a straight translation from swedish so might not be the correct english term).

I have done it in steps.

Step 1. 2^10 > 10^3. The difference is valid for n=10.

Step 2. Assume that the difference is true for n=p where p=10 or p>10

2^p > p^3 <=> 2^p - p^3 > 0

Step 3

2^(p+1) - (p+1)^3 = 2*2^p - (p^3 + p^2 + 3p + 1)=
= 2*2^p - p^3 - p^2 - 3p -1 > 0

How do I continue from there?? Or have I made a error in step 3?
Should I replace 2*2^p with 2*p^3?? damn I hate this
ok ok.........i think you are totally making this more complicated then it really is.......first off......does ^ mean multiplication?!?!?........you need to explain all the terms.........

no clue what the induktion principle is................please explain

what Math class are you taking?............

"where p=10 or p>10"....quoting you.........now if P is 10..........then how is P greater then 10......???......10 equals 10.......

im just trying to help you............if i have all the correct terms and know what you are trying to solve...........this problem will be a breeze for me........

5. 2*2^p - p^3 - p^2 - 3p -1 > 0 should be your final answer

Plug in 10 and you should have your proof. If F(P+1) is true, then so is F(P) where n=10, n>10.

I think? Been awhile since I did these. Calc 1 right? Analytic?

6. This post made me so happy i'll never have to take Calc again. God i hated college.

7. I can solve this, but have 2 questions. What level math is this and are you allowed to use L'Hoptial's rule?

8. Originally Posted by bdtr
This post made me so happy i'll never have to take Calc again. God i hated college.
I am laughing at one of my friends who is taking Calc II, and his prof. won't let him use his graphing calculator... only a normal scientific one!!! I feel sorry for him, but glad my math is done!

9. Prove:
2^(n+1) > (n+1)^3 forall n >= 10

Inductive step
2^(n+1) > (n+1)^3 =
(2^n) + (2^n) > (n^3) + (3n^2 + 3n + 1)
By the inductive hypothesis we know that 2^n > n^3 forall n > 10
so it suffices to prove that n^3 > 3n^2 + 3n + 1

Since we know n > 10 then we know that
n^3 > 10n^2 and also
n^3 > 3n^2 + 3n^2 + 3n^2
and since 3n^2 > 3n > 1 we can conclude
n^3 > 3n^2 + 3n + 1
QED

10. I would have never made it through Cal II without a graphing calculator.

Originally Posted by navydevildoc
I am laughing at one of my friends who is taking Calc II, and his prof. won't let him use his graphing calculator... only a normal scientific one!!! I feel sorry for him, but glad my math is done!

11. Calculus is the biggest pain in the ass class i have had thus far..besides Organic Chemistry.

12. Originally Posted by MMC78
Prove:
2^(n+1) > (n+1)^3 forall n >= 10

Inductive step
2^(n+1) > (n+1)^3 =
(2^n) + (2^n) > (n^3) + (3n^2 + 3n + 1)
By the inductive hypothesis we know that 2^n > n^3 forall n > 10
so it suffices to prove that n^3 > 3n^2 + 3n + 1

Since we know n > 10 then we know that
n^3 > 10n^2 and also
n^3 > 3n^2 + 3n^2 + 3n^2
and since 3n^2 > 3n > 1 we can conclude
n^3 > 3n^2 + 3n + 1
QED
that looks spot on to me cause for induction you first have to prove it correct for all "n" then for all "n+1" Fuck do the calc days bring back bad bad memories

13. The worst part is the calcs keep going. I took calc 4 (differential equations) last semester. Hardest class i've ever taken.

14. calc I & II isn't too difficult....the intense algebra in it is what always threw me. but without a graphing calc....ouch.

15. Thanks for reminding me how much I suck at Math

16. www.math.com has tones of calculators thats the only way I passed my math class

17. Thanks everyone.

Yupp Billy its analytic

Thanks for the correct answere MMC78

Danielle, ^ menas...uhhh dont know the english word for it to be honest

but x^2=x*x and x^3=x*x*x and so on. What is the correct english word for it?? It sux that we read the maths on swedish I would prefer it to be on english

And I said "where p=10 or p>10" because the "equals or greater then" isymbol isnt aviable on the keyboard so I had to type it out like that instead.

Cant figure out how to explain the inductive hypothesis on english either

Now I know where to turn when I get stuck in maths expect those kinds of threads alot this weekend because I have a test on monday

Thanks again!

18. No problem. BTW the english word for the ^ operator is exponentiation. 2^n is usually written with the n as a superscript, but it's common when writing on a computer to use ^.

19. Originally Posted by MMC78
No problem. BTW the english word for the ^ operator is exponentiation. 2^n is usually written with the n as a superscript, but it's common when writing on a computer to use ^.
btw just noticed a thing

2^(n+1) > (n+1)^3 =
(2^n) + (2^n) > (n^3) + (3n^2 + 3n + 1)

shouldnt that be (n^3)*(3n^2+3n+1)??

This is how my teacher wanted me to solve it

2^(p+1)-(p+1)^3 = 2*2^p - (p^3+3p^2+3p+1) > 2*p^3 - (p^3+3p^2+3p+1)=p^3-3p^2-3p-1
The above is > or = with
10p^2-3p^2-3p-1=7p^2-3p-1 , p>10 or p=10
That is > or = with
70p-3p-1 , p>10 or p=10
That in turn is > or = with
670-1 , p>10 or p=10
And then we have

2^(p+1)-(p+1)^3 > 660 > 0

So the difference according to the induktion is valid for all n>10 n= 10

Dont know what way to solve it is the best one

20. [I](2^n) + (2^n) > (n^3) + (3n^2 + 3n + 1)
shouldnt that be (n^3)*(3n^2+3n+1)??/[I]

No, that's ok. I just grouped the addition with parenthesis to make the next step clearer. I don't know why your teacher didn't like this proof. It's basically a textbook soultion for a class I taught 2 years ago.