Question

Challenge question....... prove the inequality that exists between

\(\large 2^{2^{2^{2^{...}}}}\) upto n terms, \(\large 4^{4^{4^{4^{...}}}}\) upto (n-1) terms,
\(\large 3^{3^{3^{3^{...}}}}\) upto n terms....

Answer 1

@amistre64
@experimentX
@eliassaab
@mukushla
@mahmit2012

Need your help please......

Answer 2

For ease of discussion we can call this operation tetration, and write 2#n, 3#n, 4#(n-1). http://en.wikipedia.org/wiki/Tetration

Answer 3

ok

Then ........

Answer 4

Well clearly 2#n<3#n, and also 4#(n-1)=(2^2)#(n-1)=2#(2n-2) so 2#n<4#(n-1) if n>2. We also notice that 2#n=4#(n-1) if n=2.

Answer 5

What is that '#'............ ?

Answer 6

That's the symbol I'm using for tetration. I guess the wikipedia article does it this way: \(^n2\), which is new to me, I'd seen mostly # before. Whichever way you want.

Answer 7

It's just a lot easier than writing out the whole exponent chain every time.

Answer 8

# will work

Answer 9

But don't u think that its a bit more theoretical.....

Answer 10

I don't think I know what you mean.

Answer 11

well leave that.....

thanks for this proof..............

any other ....... ?

Answer 12

Via trial and error I have concluded that 3#10>4#9. Not sure how to approach that algebraically tbh.

Answer 13

I'll do the first one ! and you do the rest !
let's define : \[U_{n+1}=2^{U_n} \text{ with }U_0=1\]
and \[V_{n+1}=4^{V_n} \text{ with } V_0=1\]
the question can be translated as the inequality beetween \[V_{n-1} \text{ and } U_n \]
let's have an idea first !
\[U_1=2\text{ } V_0=1\]

Answer 14

\[U_2=2^{2} \text{ and } V_1=4\]
\[U_3=2^{U_2}=2^4 \text{ and } V_2=4^{4}=2^8\]
now ! By induction ! let's assume that \[V_{n-1} > U_{n}\] let's show that :
\[V_n > U_{n+1}\]

Answer 15

\[V_{n-1} > U_{n} \Rightarrow V_{n}=4^{V_{n-1}} > 4^{U_n} > 2^{U_n}=U_{n+1}\]
what we've proved is \[\forall n \ge 3....V_{n-1}>U_n\]
Sorry ! I typed it separately ! I was just afraid to lose all my comment !

Answer 16

its ok @neemo Thanks a lot for giving me another alternative............

Any other method friends.......... ?

Answer 17

Well thanks for your replies everyone