2-var. induction Q: find n1 for (1+x)^n >1 +n*x + n*x^2

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transgalactic

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n1 is the smallest whole number for which this inequality works :

(1+x)^n >1 +n*x + n*x^2

also i am given that x>0

find n1

and prove this inequality for every n=>n1 by induction.

the base case:

(1+x)^n1 >1+n1*x + n1*x^2

i think its correct because i was told that this inequality works for n1.

n=k step we presume that this equation is true :

equation 1: (1+x)^k >1 +k*x + k*x^2

n=k+1 step we need to prove this equation:
equation 2: (1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2

now i need to multiply equation1 by sum thing
and
do
if a<b<c
then a<c

how to do this thing in this case?
 
Re: two variable induction question..

transgalactic said:
n=k+1 step we need to prove this equation:
equation 2: (1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2

(1+x)^k * (1+x) > (1+x)*(1 +k*x + k*x^2)

(1+x)^k * (1+x) > (1 +k*x + k*x^2) + x + kx^2 + kx^3

(1+x)^k * (1+x) > (1 +k*x + k*x^2) + x + x^2 + (k-1)x^2+ kx^3

(1+x)^k * (1+x) > 1 + (k+1)*x + (k+1)*x^2 + (k-1)x^2+ kx^3

(1+x)^k * (1+x) > 1 + (k+1)*x + (k+1)*x^2
Click to expand...
 
you multiplied equation1 by (1+x) and you got this equation
(1+x)^k * (1+x) > 1 + (k+1)*x + (k+1)*x^2

but in order to prove that equation2 is correct i need to use this equation inside equation 2

how to do that?

you didnt involve equation2 at all.
 
transgalactic said:
you multiplied equation1 by (1+x) and you got this equation
(1+x)^k * (1+x) > 1 + (k+1)*x + (k+1)*x^2

but in order to prove that equation2 is correct i need to use this equation inside equation 2

how to do that?

you didnt involve equation2 at all.

Of course - I didn't.

On top of equation 2 - you wrote " we need to prove this equation:"

and I showed you how to prove it - starting with equation (1)

Please review methods of proof by induction.
Click to expand...
 
this equation comes from equation1

(1+x)^k * (1+x) > 1 + (k+1)*x + (k+1)*x^2

and it differs equation2

i need to transform this equation so it will make equation2 valid

??
 
write your equation 2 along with the equation on the top - tell me how are those different.
 
the changed equation1 is:
A > B
(1+x)^k * (1+x) > 1 + (k+1)*x + (k+1)*x^2


and equation2 is:
C > B
(1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2

the only way it will be ok is if
(1+x)^(k+1)>(1+x)^k * (1+x)

C>A
how this inequality true


??
 
transgalactic said:
the changed equation1 is:

(1+x)^k * (1+x) > 1 + (k+1)*x + (k+1)*x^2


and equation2 is:

(1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2

the only way it will be ok is if
(1+x)^(k+1)>(1+x)^k * (1+x) -------------- why?

Wouldn't equality be sufficient?

how this inequality true
??
Click to expand...
 
i can replace n1 with n

(1+x)^n1 >1 +n1*x + n1*x^2

(1+x)^n1>1+n1*(x+x^2)

how to separate n1 here?

is this the right way??
 
transgalactic said:
i can replace n1 with n

(1+x)^n1 >1 +n1*x + n1*x^2

(1+x)^n1>1+n1*(x+x^2)

how to separate n1 here?

is this the right way??
Click to expand...

If I were to do that - I would try out and check n=1,2,....etc.
 
i was told that when n=n1 then this inequality works
(1+x)^n >1 +n*x + n*x^2

if n=1
(1+x)>1+x+x^2

if n=2

(1+x)^2 >1 +2*x + 2*x^2

if n=3

(1+x)^3>1+3*x+3*x^2

these cases dont work

what to do?
 
transgalactic said:
i was told that when n=n1 then this inequality works
(1+x)^n >1 +n*x + n*x^2

if n=1
(1+x)>1+x+x^2

if n=2

(1+x)^2 >1 +2*x + 2*x^2

if n=3

(1+x)^3>1+3*x+3*x^2

these cases dont work

what to do?
Click to expand...

You need to expand the left hand side so that you can match term to term.

You need to keep on going - till and untill the inequality becomes true.
 
but as i increase "n"
the left hand side goes bigger more then the right side then the previous n

??
 
transgalactic said:
but as i increase "n"
the left hand side goes bigger more then the right side then the previous n

??
Click to expand...
yes?????
 
transgalactic said:
when i did the n=4 and x=2

84>25
i checked n=3

n1>3 ..............................Incorrect ---- n1 = 3

am i correct?
Click to expand...
 
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