Maths help needed

**Kärnfysikern** · 09-25-2003, 03:18 PM

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

**Calipso** · 09-25-2003, 03:39 PM

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

***SV-1*** · 09-25-2003, 03:49 PM

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

**Danielle** · 09-25-2003, 03:55 PM

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........

***Billy_Bathgate*** · 09-25-2003, 05:26 PM

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?

**BDTR** · 09-25-2003, 05:30 PM

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

**MMC78** · 09-25-2003, 05:35 PM

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

**navydevildoc** · 09-25-2003, 05:38 PM

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!

**MMC78** · 09-25-2003, 06:03 PM

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

**BDTR** · 09-25-2003, 06:28 PM

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!

**punk_bbuilder** · 09-25-2003, 09:12 PM

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

**Full Intensity** · 09-25-2003, 09:44 PM

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

**Dude-Man** · 09-25-2003, 11:17 PM

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

**symatech** · 09-25-2003, 11:40 PM

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

**mass junkie** · 09-25-2003, 11:48 PM

Thanks for reminding me how much I suck at Math

**wrstlr69sdnl** · 09-25-2003, 11:50 PM

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

**Kärnfysikern** · 09-26-2003, 12:00 AM

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!

**MMC78** · 09-26-2003, 02:42 AM

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 ^.

**Kärnfysikern** · 09-26-2003, 07:50 AM

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

**MMC78** · 09-26-2003, 12:56 PM

[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.