Taylor series of a function
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I am puzzled with the following problem and I am not able to figure out the answer no matter how hard I try. Let's say we have a function $f(x)$ and we know its Taylor series is of the form $sum_{n=0}^infty a_nx^n$ (center $x_0=0$). We want to find the Taylor series of $f(sqrt[k] x)$ (if it exists). We can write $f(sqrt[k] x)= sum_{n=0}^infty a_nx^{n/k}$. If we differentiate $f(sqrt[k] x)$ and find that it is not differential at $x=0$, does that mean that the Taylor series we obtained is wrong and it does not have one.
sequences-and-series power-series
edited Dec 8 '18 at 13:34
Bernard
119k740113
asked Dec 8 '18 at 13:28
mxaxcmxaxc
976
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add a comment |
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I am puzzled with the following problem and I am not able to figure out the answer no matter how hard I try. Let's say we have a function $f(x)$ and we know its Taylor series is of the form $sum_{n=0}^infty a_nx^n$ (center $x_0=0$). We want to find the Taylor series of $f(sqrt[k] x)$ (if it exists). We can write $f(sqrt[k] x)= sum_{n=0}^infty a_nx^{n/k}$. If we differentiate $f(sqrt[k] x)$ and find that it is not differential at $x=0$, does that mean that the Taylor series we obtained is wrong and it does not have one.
sequences-and-series power-series
edited Dec 8 '18 at 13:34
Bernard
119k740113
asked Dec 8 '18 at 13:28
mxaxcmxaxc
976
$endgroup$
$begingroup$
How do you treat complex numbers, e.g. $x^{1/k}$ for even $k$ and negative $x$? What is the 'Taylor series' of $sqrt{x}$ at $x=0,$ i.e. $f=1, k=2?$
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– gammatester
Dec 8 '18 at 13:40
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You didn't "obtain a Taylor series." In a Taylor series, the exponents are all nonnegative integers.
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– saulspatz
Dec 8 '18 at 14:22
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thanks for your comments, they helped me a lot! :)
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– mxaxc
Dec 8 '18 at 14:32
add a comment |
$begingroup$
I am puzzled with the following problem and I am not able to figure out the answer no matter how hard I try. Let's say we have a function $f(x)$ and we know its Taylor series is of the form $sum_{n=0}^infty a_nx^n$ (center $x_0=0$). We want to find the Taylor series of $f(sqrt[k] x)$ (if it exists). We can write $f(sqrt[k] x)= sum_{n=0}^infty a_nx^{n/k}$. If we differentiate $f(sqrt[k] x)$ and find that it is not differential at $x=0$, does that mean that the Taylor series we obtained is wrong and it does not have one.
sequences-and-series power-series
edited Dec 8 '18 at 13:34
Bernard
119k740113
asked Dec 8 '18 at 13:28
mxaxcmxaxc
976
$endgroup$
I am puzzled with the following problem and I am not able to figure out the answer no matter how hard I try. Let's say we have a function $f(x)$ and we know its Taylor series is of the form $sum_{n=0}^infty a_nx^n$ (center $x_0=0$). We want to find the Taylor series of $f(sqrt[k] x)$ (if it exists). We can write $f(sqrt[k] x)= sum_{n=0}^infty a_nx^{n/k}$. If we differentiate $f(sqrt[k] x)$ and find that it is not differential at $x=0$, does that mean that the Taylor series we obtained is wrong and it does not have one.
sequences-and-series power-series
sequences-and-series power-series
edited Dec 8 '18 at 13:34
Bernard
119k740113
asked Dec 8 '18 at 13:28
mxaxcmxaxc
976
edited Dec 8 '18 at 13:34
Bernard
119k740113
asked Dec 8 '18 at 13:28
mxaxcmxaxc
976
edited Dec 8 '18 at 13:34
Bernard
119k740113
edited Dec 8 '18 at 13:34
Bernard
119k740113
edited Dec 8 '18 at 13:34
Bernard
119k740113
119k740113
asked Dec 8 '18 at 13:28
mxaxcmxaxc
976
asked Dec 8 '18 at 13:28
mxaxcmxaxc
976
asked Dec 8 '18 at 13:28
mxaxcmxaxc
976
976
$begingroup$
How do you treat complex numbers, e.g. $x^{1/k}$ for even $k$ and negative $x$? What is the 'Taylor series' of $sqrt{x}$ at $x=0,$ i.e. $f=1, k=2?$
$endgroup$
– gammatester
Dec 8 '18 at 13:40
$begingroup$
You didn't "obtain a Taylor series." In a Taylor series, the exponents are all nonnegative integers.
$endgroup$
– saulspatz
Dec 8 '18 at 14:22
$begingroup$
thanks for your comments, they helped me a lot! :)
$endgroup$
– mxaxc
Dec 8 '18 at 14:32
add a comment |
$begingroup$
How do you treat complex numbers, e.g. $x^{1/k}$ for even $k$ and negative $x$? What is the 'Taylor series' of $sqrt{x}$ at $x=0,$ i.e. $f=1, k=2?$
$endgroup$
– gammatester
Dec 8 '18 at 13:40
$begingroup$
You didn't "obtain a Taylor series." In a Taylor series, the exponents are all nonnegative integers.
$endgroup$
– saulspatz
Dec 8 '18 at 14:22
$begingroup$
thanks for your comments, they helped me a lot! :)
$endgroup$
– mxaxc
Dec 8 '18 at 14:32
$begingroup$
How do you treat complex numbers, e.g. $x^{1/k}$ for even $k$ and negative $x$? What is the 'Taylor series' of $sqrt{x}$ at $x=0,$ i.e. $f=1, k=2?$
$endgroup$
– gammatester
Dec 8 '18 at 13:40
$begingroup$
How do you treat complex numbers, e.g. $x^{1/k}$ for even $k$ and negative $x$? What is the 'Taylor series' of $sqrt{x}$ at $x=0,$ i.e. $f=1, k=2?$
$endgroup$
– gammatester
Dec 8 '18 at 13:40
$begingroup$
You didn't "obtain a Taylor series." In a Taylor series, the exponents are all nonnegative integers.
$endgroup$
– saulspatz
Dec 8 '18 at 14:22
$begingroup$
You didn't "obtain a Taylor series." In a Taylor series, the exponents are all nonnegative integers.
$endgroup$
– saulspatz
Dec 8 '18 at 14:22
$begingroup$
thanks for your comments, they helped me a lot! :)
$endgroup$
– mxaxc
Dec 8 '18 at 14:32
$begingroup$
thanks for your comments, they helped me a lot! :)
$endgroup$
– mxaxc
Dec 8 '18 at 14:32
add a comment |
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$begingroup$
How do you treat complex numbers, e.g. $x^{1/k}$ for even $k$ and negative $x$? What is the 'Taylor series' of $sqrt{x}$ at $x=0,$ i.e. $f=1, k=2?$
$endgroup$
– gammatester
Dec 8 '18 at 13:40
$begingroup$
You didn't "obtain a Taylor series." In a Taylor series, the exponents are all nonnegative integers.
$endgroup$
– saulspatz
Dec 8 '18 at 14:22
$begingroup$
thanks for your comments, they helped me a lot! :)
$endgroup$
– mxaxc
Dec 8 '18 at 14:32