Mixing
Showing a statement is true for all positive integers regarding a complex function
$begingroup$
I am not looking for an answer, for both methods I have one. I'd simply like to check if the statement needs to be proved by induction or simply through rearrangement.
Statement:
For $f(z)=sum_{n=0}^infty z^{2^n}$ show that for all positive integers $k$, $f(z)$ satisfies $f(z)=z+z^2+z^4+...+z^{2^{k-1}}+f(z^{2^{k}})$.
My approach:
begin{align}
z+z^2+z^4+...+z^{2^{k-1}}+f(z^{2^{k}})&=sum_{n=0}^{k-1} z^{2^n}+sum_{n=0}^infty (z^{2^k})^{2^n}\&=sum_{n=0}^{k-1} z^{2^n}+sum_{n=0}^infty z^{2^{k+n}}\&=sum_{n=0}^{k-1} z^{2^n}+sum_{n=k}^infty z^{2^n}\&=sum_{n=0}^infty z^{2^n}\&=f(z)
end{align}
Would this approach be correct? Or should I prove this statement by induction?
complex-analysis power-series
asked Jan 6 at 18:12
Ezhelin900Ezhelin900
152
$endgroup$
add a comment |
$begingroup$
I am not looking for an answer, for both methods I have one. I'd simply like to check if the statement needs to be proved by induction or simply through rearrangement.
Statement:
For $f(z)=sum_{n=0}^infty z^{2^n}$ show that for all positive integers $k$, $f(z)$ satisfies $f(z)=z+z^2+z^4+...+z^{2^{k-1}}+f(z^{2^{k}})$.
My approach:
begin{align}
z+z^2+z^4+...+z^{2^{k-1}}+f(z^{2^{k}})&=sum_{n=0}^{k-1} z^{2^n}+sum_{n=0}^infty (z^{2^k})^{2^n}\&=sum_{n=0}^{k-1} z^{2^n}+sum_{n=0}^infty z^{2^{k+n}}\&=sum_{n=0}^{k-1} z^{2^n}+sum_{n=k}^infty z^{2^n}\&=sum_{n=0}^infty z^{2^n}\&=f(z)
end{align}
Would this approach be correct? Or should I prove this statement by induction?
complex-analysis power-series
asked Jan 6 at 18:12
Ezhelin900Ezhelin900
152
$endgroup$
add a comment |
$begingroup$
I am not looking for an answer, for both methods I have one. I'd simply like to check if the statement needs to be proved by induction or simply through rearrangement.
Statement:
For $f(z)=sum_{n=0}^infty z^{2^n}$ show that for all positive integers $k$, $f(z)$ satisfies $f(z)=z+z^2+z^4+...+z^{2^{k-1}}+f(z^{2^{k}})$.
My approach:
begin{align}
z+z^2+z^4+...+z^{2^{k-1}}+f(z^{2^{k}})&=sum_{n=0}^{k-1} z^{2^n}+sum_{n=0}^infty (z^{2^k})^{2^n}\&=sum_{n=0}^{k-1} z^{2^n}+sum_{n=0}^infty z^{2^{k+n}}\&=sum_{n=0}^{k-1} z^{2^n}+sum_{n=k}^infty z^{2^n}\&=sum_{n=0}^infty z^{2^n}\&=f(z)
end{align}
Would this approach be correct? Or should I prove this statement by induction?
complex-analysis power-series
asked Jan 6 at 18:12
Ezhelin900Ezhelin900
152
$endgroup$
I am not looking for an answer, for both methods I have one. I'd simply like to check if the statement needs to be proved by induction or simply through rearrangement.
Statement:
For $f(z)=sum_{n=0}^infty z^{2^n}$ show that for all positive integers $k$, $f(z)$ satisfies $f(z)=z+z^2+z^4+...+z^{2^{k-1}}+f(z^{2^{k}})$.
My approach:
begin{align}
z+z^2+z^4+...+z^{2^{k-1}}+f(z^{2^{k}})&=sum_{n=0}^{k-1} z^{2^n}+sum_{n=0}^infty (z^{2^k})^{2^n}\&=sum_{n=0}^{k-1} z^{2^n}+sum_{n=0}^infty z^{2^{k+n}}\&=sum_{n=0}^{k-1} z^{2^n}+sum_{n=k}^infty z^{2^n}\&=sum_{n=0}^infty z^{2^n}\&=f(z)
end{align}
Would this approach be correct? Or should I prove this statement by induction?
complex-analysis power-series
complex-analysis power-series
asked Jan 6 at 18:12
Ezhelin900Ezhelin900
152
asked Jan 6 at 18:12
Ezhelin900Ezhelin900
152
asked Jan 6 at 18:12
Ezhelin900Ezhelin900
152
asked Jan 6 at 18:12
Ezhelin900Ezhelin900
152
asked Jan 6 at 18:12
Ezhelin900Ezhelin900
152
152
add a comment |
add a comment |
2 Answers
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Yes, this is correct. Both approaches are suitable
answered Jan 13 at 1:10
piece_and_lovepiece_and_love
283
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add a comment |
$begingroup$
Personally I would have proved it by induction, since "prove this for all positive integers" or whatever often screams that, but your method is valid as well. (And probably the easier of the two, too.)
answered Jan 6 at 18:26
Eevee TrainerEevee Trainer
5,8121936
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, this is correct. Both approaches are suitable
answered Jan 13 at 1:10
piece_and_lovepiece_and_love
283
$endgroup$
add a comment |
$begingroup$
Yes, this is correct. Both approaches are suitable
answered Jan 13 at 1:10
piece_and_lovepiece_and_love
283
$endgroup$
add a comment |
$begingroup$
Yes, this is correct. Both approaches are suitable
answered Jan 13 at 1:10
piece_and_lovepiece_and_love
283
$endgroup$
Yes, this is correct. Both approaches are suitable
answered Jan 13 at 1:10
piece_and_lovepiece_and_love
283
answered Jan 13 at 1:10
piece_and_lovepiece_and_love
283
answered Jan 13 at 1:10
piece_and_lovepiece_and_love
283
answered Jan 13 at 1:10
piece_and_lovepiece_and_love
283
283
add a comment |
add a comment |
$begingroup$
Personally I would have proved it by induction, since "prove this for all positive integers" or whatever often screams that, but your method is valid as well. (And probably the easier of the two, too.)
answered Jan 6 at 18:26
Eevee TrainerEevee Trainer
5,8121936
$endgroup$
add a comment |
$begingroup$
Personally I would have proved it by induction, since "prove this for all positive integers" or whatever often screams that, but your method is valid as well. (And probably the easier of the two, too.)
answered Jan 6 at 18:26
Eevee TrainerEevee Trainer
5,8121936
$endgroup$
add a comment |
$begingroup$
Personally I would have proved it by induction, since "prove this for all positive integers" or whatever often screams that, but your method is valid as well. (And probably the easier of the two, too.)
answered Jan 6 at 18:26
Eevee TrainerEevee Trainer
5,8121936
$endgroup$
Personally I would have proved it by induction, since "prove this for all positive integers" or whatever often screams that, but your method is valid as well. (And probably the easier of the two, too.)
answered Jan 6 at 18:26
Eevee TrainerEevee Trainer
5,8121936
answered Jan 6 at 18:26
Eevee TrainerEevee Trainer
5,8121936
answered Jan 6 at 18:26
Eevee TrainerEevee Trainer
5,8121936
answered Jan 6 at 18:26
Eevee TrainerEevee Trainer
5,8121936
5,8121936
add a comment |
add a comment |
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