# Need help in maths

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#### Zaknafein

##### New Member
Hi guys, any mathematician can help me with algebras? i cant figure out....

i need to make the equation below into 2(2^k) <=== (as in 2 to the power of k, times 2)

k^3 + 3(k)^2 + 3k + 1

any kind souls can help me out pls? greatly appreciated!! :sweat:

#### blive

##### New Member
Hi guys, any mathematician can help me with algebras? i cant figure out....

i need to make the equation below into 2(2^k) <=== (as in 2 to the power of k, times 2)

k^3 + 3(k)^2 + 3k + 1

any kind souls can help me out pls? greatly appreciated!! :sweat:
Not too sure what you mean. been a long time since I did algrebra. Are you trying to simplify the equation? BTW, what level is this?:sweat:

= 3(2^k)+3(2(2^k))+3(2^k)+1
= 12(2^k) +1 ??

#### Zaknafein

##### New Member
its not really a maths module...
hahah im taking IT actually

what i meant was, i need to simplify that long equation to 2(2^k)

#### Zaknafein

##### New Member
Not too sure what you mean. been a long time since I did algrebra. Are you trying to simplify the equation? BTW, what level is this?:sweat:

= 3(2^k)+3(2(2^k))+3(2^k)+1

= 12(2^k) +1 ??
hmm, i dun really understand how u get the line in bold

#### jnet6

##### Senior Member
Sorry tried, but unable to help.

Is there any example given? in similar form.

#### Zaknafein

##### New Member
thx for the help im actually doing proofing thru induction
i need to proof that (k+1)^3 < 2^(k+1)

so i need to simplify the LHS close to the RHS form

#### blive

##### New Member
sorry, please disregard my post. probably wrong calculations...;(

Just saw your last post. it would be easier if you had posted that in the 1st place. Don't be presumptious in assuming the answer.

#### Zaknafein

##### New Member
oh ic, sorry blive....
i thought i was on the right track...

#### blive

##### New Member
sorry, I tried to help. getting rusty, but what do I know? I am an engineer, and we only use calculator to do simple calculations too!

How about assuming that x = (k+1)? then your equation becomes x^3 <2^x ==>x^2 <2 ===> k+1 < 2 ^(1/2) ==> substitute into original equation??

#### Zaknafein

##### New Member
hmmm... let me try.. haha maybe easier...
thx for suggestion!

#### night86mare

##### Deregistered
Hi guys, any mathematician can help me with algebras? i cant figure out....

i need to make the equation below into 2(2^k) <=== (as in 2 to the power of k, times 2)

k^3 + 3(k)^2 + 3k + 1

any kind souls can help me out pls? greatly appreciated!! :sweat:
hmm, i dun really understand how u get the line in bold

which is it? the two different questions you have posted.. are very different.

induction has a very unique format, if the first one posted is about induction too, then it's a very different story.

#### Zaknafein

##### New Member
thx for the help im actually doing proofing thru induction
i need to proof that (k+1)^3 < 2^(k+1)

so i need to simplify the LHS close to the RHS form
hi night86mare

ok.. to simplify everything, ill just type the actual question out... btw, thx again for all those helping or have helped me

Using induction, prove that (n^3) + 2 < 2^n for all n > 9

#### night86mare

##### Deregistered
let (n^3) + 2 be f(n)

since proof is for n > 9

let n be 10,
f(10) = 1002 < 2^10 = 1024

when n = g (arbitiary value), where g>9
f(g) = g^3 + 2 < 2^g -------- result (1) -- you have to hold this to be true

now take n = g+1, where g>9
f(g+1) = (g+1)^3 + 2 = g^3 + 3g^2 + 3g + 3

from result (1),
2g^3 + 2 < 2*2^g = 2^(g+1) --- result (2)

taking the difference between f(g+1) and result (2),
g^3 - 3g^2 - 3g + 1 has to be greater or equal to 0 for f(g+1) < result (2) [basic algebra rearrangement]
this is true when g > 9
as g^3 - 3g^2 -3g + 1 = (g+1) [(g-2)^2 -3]

if the above is true, and statement is true when n = 10 > 9
then by mathematical induction, the statement is true

btw i did this quickly, and something's not right because if you follow the steps, it seems to suggest that the statement is true for values smaller than n>9 (which isn't true, if you check the results), maybe someone at the jc level who's still familiar with mathematical induction can help.

#### cantaresg

##### New Member
Let me try...

let (n^3) + 2 be f(n)

since proof is for n > 9

let n be 10,
f(10) = 1002 < 2^10 = 1024
Thus f(10) is true.
Assume that f(k) is true, ie (k^3) + 2 < 2^k
To prove f(k+1) is true, ie (k+1)^3 +2 < 2^ (k+1)

From LHS,

(k+1)^3 + 2 = (k^3 + 3k^2 + 3k +1) +2
< 2^k + 3k^2 + 3k +1
= 2^k + k^3 [3/k + 3/ (k^2) + 1/ (k^3)]

Now, for n >9, 3/k + 3/(k^2) + 1/ (k^3) <1

Hence, 2^k + k^3 > 2^k + k^3 [......]

so (k+1)^3 +2 < 2^k + k^3
< 2^k + k^3 + 2
< 2^k + 2^k
< 2 ^ (k+1) (shown)

#### Zaknafein

##### New Member
hmmm, very chiim lehh
anyway, ill try to understand this and thx a lot for the help

#### cantaresg

##### New Member
paiseh.. posted 4 times. think due to server lag

#### cantaresg

##### New Member
Actually is it a necessity to prove by induction? Isn't it easier to sketch two graphs and show that the intersection is at around 9?

#### night86mare

##### Deregistered
Actually is it a necessity to prove by induction? Isn't it easier to sketch two graphs and show that the intersection is at around 9?
yes, you don't even need to sketch graphs, i think logarithm will solve it

but induction is a very weird topic - sometimes you are asked to do it because it is well, part of the syllabus. and if the question asked you to use induction.. then...? :dunno:

#### raptor84

##### Senior Member
Proving by induction is one of the many ways in which to use formal logic to prove theorems. Not much use in real life I guess =p

#### night86mare

##### Deregistered
Proving by induction is one of the many ways in which to use formal logic to prove theorems. Not much use in real life I guess =p
wrong, actually.

most of science is formed on induction. if syndrome x occurs in the past, and happens in the present, therefore we would expect it to extend into the future.

but not many people realise that.

consider things like even your color green. goodman, a philosopher along with hume (who came first), questioned the logic of induction - there is no real logic actually, when it comes to daily life, that much is true. there is not much logic in it. but.. well, it's used.

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