A sequence can be start by the a2term?

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Can i define a sequence which starts with a2. term or i must define the first term a1 as well
 
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You can call the first term anything you want. It is merely a label.
 
erogol said:
Can i define a sequence which starts with a2. term or i must define the first term a1 as well

A sequence of real Nos is a function ,from the natural Nos N to the real Nos R.

But a function from N TO R IS according to the definition of a function a subset of NxR ,such that for all nεN ,there exists a unique xεR ,SUCH that (n,x) BELONGS to the function.

So if you ignore the 1st member ,strictly speaking you go against the definition of the sequence
 
poutsos.A said:
A sequence of real Nos is a function ,from the natural Nos N to the real Nos R.

But a function from N TO R IS according to the definition of a function a subset of NxR ,such that for all nεN ,there exists a unique xεR ,SUCH that (n,x) BELONGS to the function.

So if you ignore the 1st member ,strictly speaking you go against the definition of the sequence

Not really, \{a_n\}_{n=1}^\infty \backslash \{a_1\} is still a sequence, just take the map n \mapsto a_{n+1}. It is perfectly well defined.
 
Focus said:
Not really, \{a_n\}_{n=1}^\infty \backslash \{a_1\} is still a sequence, just take the map n \mapsto a_{n+1}. It is perfectly well defined.

What you have written is a subsequence of the sequence {a_{n}}.So if you start the sequence { a_{n}} from the No 2 ,lets say , the subsequence will start from ,2 as well ,hence violating the definition of the sequence
 
poutsos.A said:
What you have written is a subsequence of the sequence {a_{n}}.So if you start the sequence { a_{n}} from the No 2 ,lets say , the subsequence will start from ,2 as well ,hence violating the definition of the sequence

Did you even read my post? It fits your definition perfectly.
 
Use the definition of the subsequence correctly
 

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