Mixing
Square Root (Cauchy Sequence)
So I'm typing on a phone or else I'd use math Jax, but basically take an infinite sequence that is made of the values of the square root function as each successive term. This clearly does not converge to a real number. Yet it is a Cauchy Sequence of real numbers. How can this be? Bare in mind I didn't reach very long for duplicates, and I just learned of Cauchy sequences literally today in class.
cauchy-sequences
asked Nov 20 '18 at 21:25
Christheyankee
83
add a comment |
So I'm typing on a phone or else I'd use math Jax, but basically take an infinite sequence that is made of the values of the square root function as each successive term. This clearly does not converge to a real number. Yet it is a Cauchy Sequence of real numbers. How can this be? Bare in mind I didn't reach very long for duplicates, and I just learned of Cauchy sequences literally today in class.
cauchy-sequences
asked Nov 20 '18 at 21:25
Christheyankee
83
"Infinite sequence that is made of the values of the square root function..." Elaborate on this. So do you mean something like $sqrt{n}$, or $sqrt{1 + {1over n}}$?
– Decaf-Math
Nov 20 '18 at 21:28
I'm not sure I understood the definition of the sequence. You are taking square root of which numbers? Anyway, if you got that a Cauchy sequence of real numbers does not converge to a real number then clearly you did something wrong.
– Mark
Nov 20 '18 at 21:29
1
If you mean the sequence obtained by iterating square root on some iniitial value: $x, sqrt{x}, sqrt{sqrt{x}}, ldots$, then the sequence does converge for non-negative real $x$. (And if asked to prove that this sequence is Cauchy, I'd do it by proving that it converges $ddot{smile}$.)
– Rob Arthan
Nov 20 '18 at 21:31
add a comment |
So I'm typing on a phone or else I'd use math Jax, but basically take an infinite sequence that is made of the values of the square root function as each successive term. This clearly does not converge to a real number. Yet it is a Cauchy Sequence of real numbers. How can this be? Bare in mind I didn't reach very long for duplicates, and I just learned of Cauchy sequences literally today in class.
cauchy-sequences
asked Nov 20 '18 at 21:25
Christheyankee
83
So I'm typing on a phone or else I'd use math Jax, but basically take an infinite sequence that is made of the values of the square root function as each successive term. This clearly does not converge to a real number. Yet it is a Cauchy Sequence of real numbers. How can this be? Bare in mind I didn't reach very long for duplicates, and I just learned of Cauchy sequences literally today in class.
cauchy-sequences
cauchy-sequences
asked Nov 20 '18 at 21:25
Christheyankee
83
asked Nov 20 '18 at 21:25
Christheyankee
83
asked Nov 20 '18 at 21:25
Christheyankee
83
asked Nov 20 '18 at 21:25
Christheyankee
83
asked Nov 20 '18 at 21:25
Christheyankee
83
83
"Infinite sequence that is made of the values of the square root function..." Elaborate on this. So do you mean something like $sqrt{n}$, or $sqrt{1 + {1over n}}$?
– Decaf-Math
Nov 20 '18 at 21:28
I'm not sure I understood the definition of the sequence. You are taking square root of which numbers? Anyway, if you got that a Cauchy sequence of real numbers does not converge to a real number then clearly you did something wrong.
– Mark
Nov 20 '18 at 21:29
1
If you mean the sequence obtained by iterating square root on some iniitial value: $x, sqrt{x}, sqrt{sqrt{x}}, ldots$, then the sequence does converge for non-negative real $x$. (And if asked to prove that this sequence is Cauchy, I'd do it by proving that it converges $ddot{smile}$.)
– Rob Arthan
Nov 20 '18 at 21:31
add a comment |
"Infinite sequence that is made of the values of the square root function..." Elaborate on this. So do you mean something like $sqrt{n}$, or $sqrt{1 + {1over n}}$?
– Decaf-Math
Nov 20 '18 at 21:28
I'm not sure I understood the definition of the sequence. You are taking square root of which numbers? Anyway, if you got that a Cauchy sequence of real numbers does not converge to a real number then clearly you did something wrong.
– Mark
Nov 20 '18 at 21:29
1
If you mean the sequence obtained by iterating square root on some iniitial value: $x, sqrt{x}, sqrt{sqrt{x}}, ldots$, then the sequence does converge for non-negative real $x$. (And if asked to prove that this sequence is Cauchy, I'd do it by proving that it converges $ddot{smile}$.)
– Rob Arthan
Nov 20 '18 at 21:31
"Infinite sequence that is made of the values of the square root function..." Elaborate on this. So do you mean something like $sqrt{n}$, or $sqrt{1 + {1over n}}$?
– Decaf-Math
Nov 20 '18 at 21:28
"Infinite sequence that is made of the values of the square root function..." Elaborate on this. So do you mean something like $sqrt{n}$, or $sqrt{1 + {1over n}}$?
– Decaf-Math
Nov 20 '18 at 21:28
I'm not sure I understood the definition of the sequence. You are taking square root of which numbers? Anyway, if you got that a Cauchy sequence of real numbers does not converge to a real number then clearly you did something wrong.
– Mark
Nov 20 '18 at 21:29
I'm not sure I understood the definition of the sequence. You are taking square root of which numbers? Anyway, if you got that a Cauchy sequence of real numbers does not converge to a real number then clearly you did something wrong.
– Mark
Nov 20 '18 at 21:29
1
1
If you mean the sequence obtained by iterating square root on some iniitial value: $x, sqrt{x}, sqrt{sqrt{x}}, ldots$, then the sequence does converge for non-negative real $x$. (And if asked to prove that this sequence is Cauchy, I'd do it by proving that it converges $ddot{smile}$.)
– Rob Arthan
Nov 20 '18 at 21:31
If you mean the sequence obtained by iterating square root on some iniitial value: $x, sqrt{x}, sqrt{sqrt{x}}, ldots$, then the sequence does converge for non-negative real $x$. (And if asked to prove that this sequence is Cauchy, I'd do it by proving that it converges $ddot{smile}$.)
– Rob Arthan
Nov 20 '18 at 21:31
add a comment |
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"Infinite sequence that is made of the values of the square root function..." Elaborate on this. So do you mean something like $sqrt{n}$, or $sqrt{1 + {1over n}}$?
– Decaf-Math
Nov 20 '18 at 21:28
I'm not sure I understood the definition of the sequence. You are taking square root of which numbers? Anyway, if you got that a Cauchy sequence of real numbers does not converge to a real number then clearly you did something wrong.
– Mark
Nov 20 '18 at 21:29
1
If you mean the sequence obtained by iterating square root on some iniitial value: $x, sqrt{x}, sqrt{sqrt{x}}, ldots$, then the sequence does converge for non-negative real $x$. (And if asked to prove that this sequence is Cauchy, I'd do it by proving that it converges $ddot{smile}$.)
– Rob Arthan
Nov 20 '18 at 21:31