Recurrence systems

[Probability]

Please give me if you can a second opinion to the understanding of how these sequences work!

a_1 = 0, a_n+1 = 2a_n+1 (n = 1,2,3...)

So I think a_1 = 0, the "n" value is 1, and the "a" value is 0.

Please advise if I am correct so far, and if so,

a_1+1 = 2 x 0_1+1 = 0

so "a" = 0

I am a bit unsure how to work them out!

 

[Deleted member 4993]

Please give me if you can a second opinion to the understanding of how these sequences work!

a_1 = 0, a_n+1 = 2a_n+1 (n = 1,2,3...)

So I think a_1 = 0, the "n" value is 1, and the "a" value is 0.

Please advise if I am correct so far, and if so,

a_1+1 = 2 x 0_1+1 = 0

so "a" = 0

I am a bit unsure how to work them out!

If your sequence is as you wrote it:

\(\displaystyle a_{n} \ + \ 1 \ = \ 2a_n \ + \ 1\)

It is trivial - all the elements = 0 (as you found out)

However, I think your sequence is:

\(\displaystyle a_{n+1} \ = \ 2a_n \ + \ 1\)

then

\(\displaystyle a_2 \ = 2a_1 \ + \ 1 \ = 1\)

\(\displaystyle a_3 \ = 2a_2 \ + \ 1 \ = 3\)

\(\displaystyle a_4 \ = 2a_3 \ + \ 1 \ = 7\)

and so on.....

 

[Probability]

If your sequence is as you wrote it:

\(\displaystyle a_{n} \ + \ 1 \ = \ 2a_n \ + \ 1\)

It is trivial - all the elements = 0 (as you found out)

However, I think your sequence is:

\(\displaystyle a_{n+1} \ = \ 2a_n \ + \ 1\)

then

\(\displaystyle a_2 \ = 2a_1 \ + \ 1 \ = 1\)

\(\displaystyle a_3 \ = 2a_2 \ + \ 1 \ = 3\)

\(\displaystyle a_4 \ = 2a_3 \ + \ 1 \ = 7\)

and so on.....

You are correct, \(\displaystyle a_{n+1} \ = \ 2a_n \ + \ 1\)

However although I now see this is a multiplication process to solve it, I am still unclear in the correct interpretation of the sequence?

If \(\displaystyle a_1\) = 0, and \(\displaystyle a_n+1 = 2a_n+1\)

If a_1 = 0, then \(\displaystyle a_0+1 = 2(0) + 1 = 1\)

therefore 1 = 1 and not 0, so I am confused?

 

[Deleted member 4993]

You are correct, \(\displaystyle a_{n+1} \ = \ 2a_n \ + \ 1\)

However although I now see this is a multiplication process to solve it, I am still unclear in the correct interpretation of the sequence?

If \(\displaystyle a_1\) = 0, and \(\displaystyle a_n+1 = 2a_n+1\) No it is \(\displaystyle a_{(n+1)} = 2a_n+1\)



If a_1 = 0, then \(\displaystyle a_0+1 = 2(0) + 1 = 1\)

therefore 1 = 1 and not 0, so I am confused?

.

 

[pka]

Please give me if you can a second opinion to the understanding of how these sequences work!
a_1 = 0, a_n+1 = 2a_n+1 (n = 1,2,3...)

Can you show that the general term is:
\(\displaystyle a_n=2^{n-1}-1~?\)

 

[Probability]

Can you show that the general term is:
\(\displaystyle a_n=2^{n-1}-1~?\)

This is what I think I understand.

\(\displaystyle a_n=2^{n-1}-1\)

\(\displaystyle a_1=2^{1-1}-1 = 2(0) - 1= -1\)

So in my example;

\(\displaystyle a_1 = 0, a_{n+1} = 2a_{n+1}\)

\(\displaystyle a_1 = 0, a_{1+1} = 2_{0}{1 + 1} = 2\)

How am I doing?

 

[Deleted member 4993]

This is what I think I understand.

\(\displaystyle a_n=2^{n-1}-1\)

\(\displaystyle a_1=2^{1-1}-1 = 2(0) - 1= -1\) <<< Incorrect \(\displaystyle a_1 \ = \ 2^{1-1} - 1 \ = \ 2^0 - 1 \ = \ = 1 - 1 = 0\)

So in my example;

\(\displaystyle a_1 = 0, a_{n+1} = 2a_{n+1}\)

\(\displaystyle a_1 = 0, a_{1+1} = 2_{0}{1 + 1} = 2\)

How am I doing?

I do not think you understand the solution suggested by pka.

 

[Probability]

I do not think you understand the solution suggested by pka.

While I agree I don't understand the subject to date, I also don't profess to understand where you invented the addional "1" from below?

You wrote;

\(\displaystyle a_1 = 2^{1-1} - 1 = 2^0 - 1 = 1 - 1 = 0\)

If in the example the exponent 1 is subtracted from 1, then on the LHS of the equality sign there is left -1, because \(\displaystyle 2^{1-1} = 2^{0}, as you pointed out\). On the RHS of the equality sign we now have \(\displaystyle 2^0 - 1\)= 1

So if we only have -1 left on the LHS, how then can we invent an additional 1 on the RHS to make 1 - 1 = 0?

Sorry my misunderstandng completely - 2^0 = 1 - 1 = 0