Thread: Maths help needed

09252003, 04:18 PM #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

09252003, 04:39 PM #2
This thread scared me and reminded me I'm going to college some day.

09252003, 04:49 PM #3
2+2=4...... FOUR

09252003, 04:55 PM #4Originally Posted by johan
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........

09252003, 06:26 PM #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?

09252003, 06:30 PM #6
This post made me so happy i'll never have to take Calc again. God i hated college.

09252003, 06:35 PM #7
I can solve this, but have 2 questions. What level math is this and are you allowed to use L'Hoptial's rule?

09252003, 06:38 PM #8Originally Posted by bdtr

09252003, 07:03 PM #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

09252003, 07:28 PM #10
I would have never made it through Cal II without a graphing calculator.
Originally Posted by navydevildoc

09252003, 10:12 PM #11
Calculus is the biggest pain in the ass class i have had thus far..besides Organic Chemistry.

09252003, 10:44 PM #12Originally Posted by MMC78

09262003, 12:17 AM #13
The worst part is the calcs keep going. I took calc 4 (differential equations) last semester. Hardest class i've ever taken.

09262003, 12:40 AM #14
calc I & II isn't too difficult....the intense algebra in it is what always threw me. but without a graphing calc....ouch.

09262003, 12:48 AM #15
Thanks for reminding me how much I suck at Math

09262003, 12:50 AM #16
www.math.com has tones of calculators thats the only way I passed my math class

09262003, 01:00 AM #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!

09262003, 03:42 AM #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 ^.

09262003, 08:50 AM #19Originally Posted by MMC78
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^33p^23p1
The above is > or = with
10p^23p^23p1=7p^23p1 , p>10 or p=10
That is > or = with
70p3p1 , p>10 or p=10
That in turn is > or = with
6701 , 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

09262003, 01:56 PM #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.
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