Sequence question

[markraz]

Hello

I have this Sequence...

\(\displaystyle \Large
a_{n+1}={(a_{n})}^4, a_{1}=1\)

is this correct?
\(\displaystyle
a_{1}= (1)^4 = 1\)
\(\displaystyle a_{2}= (1)^4 = 1\)
\(\displaystyle a_{3}= (1)^4 = 1\)
\(\displaystyle a_{4}= (1)^4 = 1\)

Thanks

 

[Deleted member 4993]

Hello

I have this Sequence...

\(\displaystyle \Large
a_{n+1}={(a_{n})}^4, a_{1}=1\)

is this correct?
\(\displaystyle
a_{1}= (1)^4 = 1\)
\(\displaystyle a_{2}= (1)^4 = 1\)
\(\displaystyle a_{3}= (1)^4 = 1\)
\(\displaystyle a_{4}= (1)^4 = 1\)

Thanks

Yes

 

[markraz]

Yes

thanks appreciate it
so then my next question is.... is this just a "book" type question? or would this happen in some real world scenario
since the following would yeild the same sequence too
\(\displaystyle \Large
a_{n+1}={(a_{n})}, a_{1}=1\)
\(\displaystyle \Large
a_{n+1}={(a_{n})}^2, a_{1}=1\)
\(\displaystyle \Large
a_{n+1}={(a_{n})}^3, a_{1}=1\)

So would an explicit formula actually yield this resulting Recurrence Relation?
Hate to ask but what would be an example of an explicit formula that would yield \(\displaystyle a_{n+1}={(a_{n})}^4, a_{1}=1\) ??

Thanks

 

[Deleted member 4993]

thanks appreciate it
so then my next question is.... is this just a "book" type question? or would this happen in some real world scenario
since the following would yeild the same sequence too
\(\displaystyle \Large
a_{n+1}={(a_{n})}, a_{1}=1\)
\(\displaystyle \Large
a_{n+1}={(a_{n})}^2, a_{1}=1\)
\(\displaystyle \Large
a_{n+1}={(a_{n})}^3, a_{1}=1\)

So would an explicit formula actually yield this resulting Recurrence Relation?
Hate to ask but what would be an example of an explicit formula that would yield \(\displaystyle a_{n+1}={(a_{n})}^4, a_{1}=1\) ??

Thanks

explicit formula would be:

a_n = 1