Induction

[jetzac]

Show by induction that an >=0 for all n.

Given: a1=2, an+1=an^2+2/2*an

I have found out a2=3/2 and also a3 and a4, this is right.

But I have problems with the induction, as I have never before had this...

I need to make a statment to prove right that would be an >=0 ?

Test for lowest value, 1 And is true.

next step as i have figured it out is: set an as k(or n)
and then k+1 but here I'm lost.

So hopefully someone will help me on my way with this.

 

[daon]

Huh? Is that (a_n+2)/(2*a_n)?

Anyway, a_n >0 implies any squaring, adding. dividing by a_n will give you something positive..

 

[jetzac]

I'm not sure how to put it... but written in another forum they use a[n], it is a and a notation of n down in the bottom right of the a... not as a² but down...

so i know a[1]=2 and a[3]=3/2... (both 2and 3 should go down on the right side os a)

and need to show that a[n] is equal or higher than 0 for all n


I'll try to find a program to write it right...

 

[jetzac]

Know that you have to prove the teorem for 1 first then set it as k and k+1... but I don't understand it...

 

[galactus]

I think that should be

\(\displaystyle a_{n+1}=\frac{(a_{n})^{2}+2}{2a_{n}}, \;\ a_{1}=2, \;\ a_{2}=\frac{3}{2}\)

Which can be written as:

\(\displaystyle a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{2}{a_{n}}\right)\)

Note that this is the interative method of finding \(\displaystyle \sqrt{2}\)

i.e. This sequence's limit converges to \(\displaystyle \sqrt{2}\).

 

[Deleted member 4993]

Using Galactus's expression you can show that a[sub:1y9r3iai]n[/sub:1y9r3iai] and a[sub:1y9r3iai]n+2[/sub:1y9r3iai] must have same sign.

since a[sub:1y9r3iai]1[/sub:1y9r3iai] and a[sub:1y9r3iai]2[/sub:1y9r3iai] are positive - all other numbers (a[sub:1y9r3iai]1+2[/sub:1y9r3iai], a[sub:1y9r3iai]2+2[/sub:1y9r3iai], a[sub:1y9r3iai](1+2)+2[/sub:1y9r3iai], ... etc.) are positive.

.

 

[Deleted member 4993]

Using Galactus's expression you can show that a[sub:2ov2uz4z]n[/sub:2ov2uz4z] * a[sub:2ov2uz4z]n+2[/sub:2ov2uz4z] > 0 ? a[sub:2ov2uz4z]n[/sub:2ov2uz4z] and a[sub:2ov2uz4z]n+2[/sub:2ov2uz4z] must have same sign.

since a[sub:2ov2uz4z]1[/sub:2ov2uz4z] and a[sub:2ov2uz4z]2[/sub:2ov2uz4z] are positive - all other numbers (a[sub:2ov2uz4z]1+2[/sub:2ov2uz4z], a[sub:2ov2uz4z]2+2[/sub:2ov2uz4z], a[sub:2ov2uz4z](1+2)+2[/sub:2ov2uz4z], ... etc.) are positive.

.

 

[jetzac]

thanks for pointing me in the right direction I'll sit down and try again !