Power series for $2^x$ bypassing $e^x$? [on hold]power series for $sqrtz^2-1$ around $z=0$Function whose power series coefficients contain logarithmsPower series convergence in boundary problemExpressing $zinmathbbC[[w]]$ as a power series in $yinmathbbC[[z]]$.Power series solution for ODEIdentity involving $log$ and a power seriesProof of identity principle for power seriesSubtracting two power seriesHow to recast these two exponential of infinite power series as simple power series?Convergence of Power Series and Power Series Solutions of ODE
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Power series for $2^x$ bypassing $e^x$? [on hold]
power series for $sqrtz^2-1$ around $z=0$Function whose power series coefficients contain logarithmsPower series convergence in boundary problemExpressing $zinmathbbC[[w]]$ as a power series in $yinmathbbC[[z]]$.Power series solution for ODEIdentity involving $log$ and a power seriesProof of identity principle for power seriesSubtracting two power seriesHow to recast these two exponential of infinite power series as simple power series?Convergence of Power Series and Power Series Solutions of ODE
$begingroup$
Is it possible to create a power series for $2^x$ that's always convergent, and where the coefficients are all fractional (that is, bypassing the conversion of $2^x$ to $e^xlog2$)?
That is, $2^x=sum_j=0^inftyc_j x^j$, $forall$ $x$, where the $c_j$ are fractional.
Please take kindly to this question, I'm a newbie.
power-series taylor-expansion
asked 2 days ago
JR SousaJR Sousa
1008
$endgroup$
put on hold as off-topic by Eevee Trainer, Alex Provost, Shailesh, Song, Leucippus yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Song, Leucippus
add a comment |
$begingroup$
Is it possible to create a power series for $2^x$ that's always convergent, and where the coefficients are all fractional (that is, bypassing the conversion of $2^x$ to $e^xlog2$)?
That is, $2^x=sum_j=0^inftyc_j x^j$, $forall$ $x$, where the $c_j$ are fractional.
Please take kindly to this question, I'm a newbie.
power-series taylor-expansion
asked 2 days ago
JR SousaJR Sousa
1008
$endgroup$
put on hold as off-topic by Eevee Trainer, Alex Provost, Shailesh, Song, Leucippus yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Song, Leucippus
1
$begingroup$
It's not really clear exactly what you're asking..
$endgroup$
– Antinous
2 days ago
$begingroup$
@Antinous I reworded the post, have a look.
$endgroup$
– JR Sousa
2 days ago
$begingroup$
Such a series is necessarily the Maclaurin series for $2^x,$ and it doesn't have rational coefficients.
$endgroup$
– saulspatz
2 days ago
add a comment |
$begingroup$
Is it possible to create a power series for $2^x$ that's always convergent, and where the coefficients are all fractional (that is, bypassing the conversion of $2^x$ to $e^xlog2$)?
That is, $2^x=sum_j=0^inftyc_j x^j$, $forall$ $x$, where the $c_j$ are fractional.
Please take kindly to this question, I'm a newbie.
power-series taylor-expansion
asked 2 days ago
JR SousaJR Sousa
1008
$endgroup$
Is it possible to create a power series for $2^x$ that's always convergent, and where the coefficients are all fractional (that is, bypassing the conversion of $2^x$ to $e^xlog2$)?
That is, $2^x=sum_j=0^inftyc_j x^j$, $forall$ $x$, where the $c_j$ are fractional.
Please take kindly to this question, I'm a newbie.
power-series taylor-expansion
power-series taylor-expansion
asked 2 days ago
JR SousaJR Sousa
1008
asked 2 days ago
JR SousaJR Sousa
1008
edited 2 days ago
JR Sousa
asked 2 days ago
JR SousaJR Sousa
1008
asked 2 days ago
JR SousaJR Sousa
1008
asked 2 days ago
JR SousaJR Sousa
1008
1008
put on hold as off-topic by Eevee Trainer, Alex Provost, Shailesh, Song, Leucippus yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Song, Leucippus
put on hold as off-topic by Eevee Trainer, Alex Provost, Shailesh, Song, Leucippus yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Song, Leucippus
1
$begingroup$
It's not really clear exactly what you're asking..
$endgroup$
– Antinous
2 days ago
$begingroup$
@Antinous I reworded the post, have a look.
$endgroup$
– JR Sousa
2 days ago
$begingroup$
Such a series is necessarily the Maclaurin series for $2^x,$ and it doesn't have rational coefficients.
$endgroup$
– saulspatz
2 days ago
add a comment |
1
$begingroup$
It's not really clear exactly what you're asking..
$endgroup$
– Antinous
2 days ago
$begingroup$
@Antinous I reworded the post, have a look.
$endgroup$
– JR Sousa
2 days ago
$begingroup$
Such a series is necessarily the Maclaurin series for $2^x,$ and it doesn't have rational coefficients.
$endgroup$
– saulspatz
2 days ago
1
1
$begingroup$
It's not really clear exactly what you're asking..
$endgroup$
– Antinous
2 days ago
$begingroup$
It's not really clear exactly what you're asking..
$endgroup$
– Antinous
2 days ago
$begingroup$
@Antinous I reworded the post, have a look.
$endgroup$
– JR Sousa
2 days ago
$begingroup$
@Antinous I reworded the post, have a look.
$endgroup$
– JR Sousa
2 days ago
$begingroup$
Such a series is necessarily the Maclaurin series for $2^x,$ and it doesn't have rational coefficients.
$endgroup$
– saulspatz
2 days ago
$begingroup$
Such a series is necessarily the Maclaurin series for $2^x,$ and it doesn't have rational coefficients.
$endgroup$
– saulspatz
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, this is not possible. This would only be possible if $ln(2)$ was rational - which it is not. The method of finding this series expansion can be done avoiding the aforementioned identity by evaluating the function at each of its $n$th derivatives.
$$f(0)=2^0=1$$
$$f'(0)=ln(2)times2^0=ln(2)$$
$$...$$
$$f^(n)(0)=ln^n(2)$$
$$f(x)=sum_k=0^infty fracf^(k)(0)x^kk!=sum_k=0^infty fracln^k(2)x^kk!=1+fracln(2)x1!+fracln^2(2)x^22!+...$$
answered 2 days ago
Peter ForemanPeter Foreman
3,6851216
$endgroup$
$begingroup$
I pondered about it and came to the conclusion it's really impossible. Thanks for the insight.
$endgroup$
– JR Sousa
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, this is not possible. This would only be possible if $ln(2)$ was rational - which it is not. The method of finding this series expansion can be done avoiding the aforementioned identity by evaluating the function at each of its $n$th derivatives.
$$f(0)=2^0=1$$
$$f'(0)=ln(2)times2^0=ln(2)$$
$$...$$
$$f^(n)(0)=ln^n(2)$$
$$f(x)=sum_k=0^infty fracf^(k)(0)x^kk!=sum_k=0^infty fracln^k(2)x^kk!=1+fracln(2)x1!+fracln^2(2)x^22!+...$$
answered 2 days ago
Peter ForemanPeter Foreman
3,6851216
$endgroup$
$begingroup$
I pondered about it and came to the conclusion it's really impossible. Thanks for the insight.
$endgroup$
– JR Sousa
2 days ago
add a comment |
$begingroup$
No, this is not possible. This would only be possible if $ln(2)$ was rational - which it is not. The method of finding this series expansion can be done avoiding the aforementioned identity by evaluating the function at each of its $n$th derivatives.
$$f(0)=2^0=1$$
$$f'(0)=ln(2)times2^0=ln(2)$$
$$...$$
$$f^(n)(0)=ln^n(2)$$
$$f(x)=sum_k=0^infty fracf^(k)(0)x^kk!=sum_k=0^infty fracln^k(2)x^kk!=1+fracln(2)x1!+fracln^2(2)x^22!+...$$
answered 2 days ago
Peter ForemanPeter Foreman
3,6851216
$endgroup$
$begingroup$
I pondered about it and came to the conclusion it's really impossible. Thanks for the insight.
$endgroup$
– JR Sousa
2 days ago
add a comment |
$begingroup$
No, this is not possible. This would only be possible if $ln(2)$ was rational - which it is not. The method of finding this series expansion can be done avoiding the aforementioned identity by evaluating the function at each of its $n$th derivatives.
$$f(0)=2^0=1$$
$$f'(0)=ln(2)times2^0=ln(2)$$
$$...$$
$$f^(n)(0)=ln^n(2)$$
$$f(x)=sum_k=0^infty fracf^(k)(0)x^kk!=sum_k=0^infty fracln^k(2)x^kk!=1+fracln(2)x1!+fracln^2(2)x^22!+...$$
answered 2 days ago
Peter ForemanPeter Foreman
3,6851216
$endgroup$
No, this is not possible. This would only be possible if $ln(2)$ was rational - which it is not. The method of finding this series expansion can be done avoiding the aforementioned identity by evaluating the function at each of its $n$th derivatives.
$$f(0)=2^0=1$$
$$f'(0)=ln(2)times2^0=ln(2)$$
$$...$$
$$f^(n)(0)=ln^n(2)$$
$$f(x)=sum_k=0^infty fracf^(k)(0)x^kk!=sum_k=0^infty fracln^k(2)x^kk!=1+fracln(2)x1!+fracln^2(2)x^22!+...$$
answered 2 days ago
Peter ForemanPeter Foreman
3,6851216
answered 2 days ago
Peter ForemanPeter Foreman
3,6851216
answered 2 days ago
Peter ForemanPeter Foreman
3,6851216
answered 2 days ago
Peter ForemanPeter Foreman
3,6851216
3,6851216
$begingroup$
I pondered about it and came to the conclusion it's really impossible. Thanks for the insight.
$endgroup$
– JR Sousa
2 days ago
add a comment |
$begingroup$
I pondered about it and came to the conclusion it's really impossible. Thanks for the insight.
$endgroup$
– JR Sousa
2 days ago
$begingroup$
I pondered about it and came to the conclusion it's really impossible. Thanks for the insight.
$endgroup$
– JR Sousa
2 days ago
$begingroup$
I pondered about it and came to the conclusion it's really impossible. Thanks for the insight.
$endgroup$
– JR Sousa
2 days ago
add a comment |
1
$begingroup$
It's not really clear exactly what you're asking..
$endgroup$
– Antinous
2 days ago
$begingroup$
@Antinous I reworded the post, have a look.
$endgroup$
– JR Sousa
2 days ago
$begingroup$
Such a series is necessarily the Maclaurin series for $2^x,$ and it doesn't have rational coefficients.
$endgroup$
– saulspatz
2 days ago