let a,b be members of positive reals s.t a+b1 and f(0)=1 and f(n+1)=af(n)+bf(n-1) Prove this inequality holds true for f(n)

Question

let a,b be members of positive reals s.t a+b>1 and f(0)=1  and f(n+1)=af(n)+bf(n-1)

Prove this inequality holds true for f(n)<- (a=b)^n

amistre64 · Answer

looks like induction ....

amistre64 · Answer

you last line is a little confusing

anonymous · Answer

so I am planning on proving this by strong induction. It is given f(0)=1. so i was thinking that f(1)=a

anonymous · Answer

last line prove that the inequality holds for f(n) less than equal to (a+b)^n

oh i used = instead of +

amistre64 · Answer

f(0+1)=af(0)+bf(0-1)

f(1)=a + b f(-1)

without some value for that f(-1), it would be hard pressed to say f(1) = a

amistre64 · Answer

but the rule is stated as n = Z^+,  so n=0 is taboo eh

amistre64 · Answer

nah, a,b are Z^+ .. no limitations in n that i can tell

amistre64 · Answer

in order for this to make sense (at least to me); the 2 first terms need to be defined upfront since the definition for the nth term relies on 2 prior values.

anonymous · Answer

ok. i was wondering the same thing...
but i wasn't sure if there was something i could do withthe inequality like

anonymous · Answer

f(n+1)=af(n)+bf(n-1)<= a(a+b)^n=b(a+b)^(n-1)

amistre64 · Answer

i vaguely recall seeing something worked like this:

$$f(n+1)=af(n)+bf(n-1)$$
$$f(n+1)-af(n)-bf(n-1)=0$$
$$f^{n+1}-af^n-bf^{n-1}=0$$
$$f^{n-1}(f^{2}-af-b)=0$$

amistre64 · Answer

$$f(1)=\frac{a\pm\sqrt{a^2+4b}}{2}$$
but this is really just a foggy memory :)

anonymous · Answer

hmmm i don't know what to do with that :( but i will look at it a bit

amistre64 · Answer

to see if im recalling things correclty, this looks fibonacci like to me:

$$f(n+1) = f(n) + f(n-1)$$
$$f(n+1) - f(n) - f(n-1)=0$$
$$f^{n+1} - f^n - f^{n-1}=0$$
$$f^{n-1}(f^{2} - f -1)=0$$
$$f(1)=\frac{1\pm\sqrt{5}}{2}$$
murmur ....  the golden rule does have something to do with the limit of the sum of the sequence, but at the moment im clearly missing something

amistre64 · Answer

lol, golden ratio :)  golden rule is something different

anonymous · Answer

:) i see how you got there... how did the f(n+1) become a power of a function?
f^(n+1)-f^n-f^(n-1)=0 jump come from?

amistre64 · Answer

instead of renaming it i just went ahead and used the f as a variable.

amistre64 · Answer

i believe this is what i am trying to recall ... http://www.cimt.plymouth.ac.uk/projects/mepres/alevel/discrete_ch15.pdf