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Can you convert a taylor series into a power series while avoiding singularities/discontinuities that result...
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Suppose one follows the Taylor series formula for a function, defining a function as $$ sum_{n=0}^{ infty} frac{f^{n}(x)|_{x=a}}{n!}(x-a)^n.$$
Then, suppose you want this function to instead be defined as a power series since that is a little bit simpler.
You might think to simply take $a=0$, except there are plenty of functions like $ln(x)$ or $sqrt{x}$ which don't have a derivative at $x=0$. In such a case, you'd have to pick $a$ to be something else, so how can you convert that function's Taylor series to a power series that results in just $x^n$ instead of $(x-a)^n$?
power-series taylor-expansion
asked Feb 2 at 3:08
Vane VoeVane Voe
396
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add a comment |
$begingroup$
Suppose one follows the Taylor series formula for a function, defining a function as $$ sum_{n=0}^{ infty} frac{f^{n}(x)|_{x=a}}{n!}(x-a)^n.$$
Then, suppose you want this function to instead be defined as a power series since that is a little bit simpler.
You might think to simply take $a=0$, except there are plenty of functions like $ln(x)$ or $sqrt{x}$ which don't have a derivative at $x=0$. In such a case, you'd have to pick $a$ to be something else, so how can you convert that function's Taylor series to a power series that results in just $x^n$ instead of $(x-a)^n$?
power-series taylor-expansion
asked Feb 2 at 3:08
Vane VoeVane Voe
396
$endgroup$
add a comment |
$begingroup$
Suppose one follows the Taylor series formula for a function, defining a function as $$ sum_{n=0}^{ infty} frac{f^{n}(x)|_{x=a}}{n!}(x-a)^n.$$
Then, suppose you want this function to instead be defined as a power series since that is a little bit simpler.
You might think to simply take $a=0$, except there are plenty of functions like $ln(x)$ or $sqrt{x}$ which don't have a derivative at $x=0$. In such a case, you'd have to pick $a$ to be something else, so how can you convert that function's Taylor series to a power series that results in just $x^n$ instead of $(x-a)^n$?
power-series taylor-expansion
asked Feb 2 at 3:08
Vane VoeVane Voe
396
$endgroup$
Suppose one follows the Taylor series formula for a function, defining a function as $$ sum_{n=0}^{ infty} frac{f^{n}(x)|_{x=a}}{n!}(x-a)^n.$$
Then, suppose you want this function to instead be defined as a power series since that is a little bit simpler.
You might think to simply take $a=0$, except there are plenty of functions like $ln(x)$ or $sqrt{x}$ which don't have a derivative at $x=0$. In such a case, you'd have to pick $a$ to be something else, so how can you convert that function's Taylor series to a power series that results in just $x^n$ instead of $(x-a)^n$?
power-series taylor-expansion
power-series taylor-expansion
asked Feb 2 at 3:08
Vane VoeVane Voe
396
asked Feb 2 at 3:08
Vane VoeVane Voe
396
edited Feb 2 at 3:19
Vane Voe
asked Feb 2 at 3:08
Vane VoeVane Voe
396
asked Feb 2 at 3:08
Vane VoeVane Voe
396
asked Feb 2 at 3:08
Vane VoeVane Voe
396
396
add a comment |
add a comment |
1 Answer
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The simple answer is that you can't. If I would write $$ln(x)=sum_{n=0}^infty a_nx^n$$ then obviously I should be able to compute the derivative at $0$. $$frac d{dx} ln x=frac 1x$$ This diverges at $0$. Taking the derivative of the right hand side you get $$sum_{n=1}^infty na_n x^{n-1}$$ In the limit $xto 0$ you get $a_1$, which is finite.
You might ask "but what if I just expand all the $(x-x_0)^n$ terms in the Taylor series?" What will happen is that you get for the all terms $x^n$ coefficients that can be written as $sum_j^infty b_j(n) x_0^j$, but there is no guarantee that these series converge.
answered Feb 2 at 3:27
AndreiAndrei
13.7k21230
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$begingroup$
Isn't there some way to express it as a power series but then just restrict the domain to exclude anything at or less than zero?
$endgroup$
– Vane Voe
Feb 2 at 3:34
$begingroup$
If the two expressions are the same for an infinite number of $x$ points, then the coefficients of $x^n$ must all be the same. I think (just a hunch) that if you write such a series it should diverge everywhere outside of the interval of convergence. But the power series converges at $0$ to $a_0$.
$endgroup$
– Andrei
Feb 2 at 3:47
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Okay...so how would one go about writing a power series that converges everywhere outside of some $a_0$?
$endgroup$
– Vane Voe
Feb 2 at 4:05
$begingroup$
@VaneVoe: I'd say you need Laurent series for that.
$endgroup$
– zipirovich
Feb 2 at 5:25
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So I somehow need to take complex integrals for a purely real function in the real plane? That doesn't quite sound right.
$endgroup$
– Vane Voe
Feb 2 at 5:26
add a comment |
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$begingroup$
The simple answer is that you can't. If I would write $$ln(x)=sum_{n=0}^infty a_nx^n$$ then obviously I should be able to compute the derivative at $0$. $$frac d{dx} ln x=frac 1x$$ This diverges at $0$. Taking the derivative of the right hand side you get $$sum_{n=1}^infty na_n x^{n-1}$$ In the limit $xto 0$ you get $a_1$, which is finite.
You might ask "but what if I just expand all the $(x-x_0)^n$ terms in the Taylor series?" What will happen is that you get for the all terms $x^n$ coefficients that can be written as $sum_j^infty b_j(n) x_0^j$, but there is no guarantee that these series converge.
answered Feb 2 at 3:27
AndreiAndrei
13.7k21230
$endgroup$
$begingroup$
Isn't there some way to express it as a power series but then just restrict the domain to exclude anything at or less than zero?
$endgroup$
– Vane Voe
Feb 2 at 3:34
$begingroup$
If the two expressions are the same for an infinite number of $x$ points, then the coefficients of $x^n$ must all be the same. I think (just a hunch) that if you write such a series it should diverge everywhere outside of the interval of convergence. But the power series converges at $0$ to $a_0$.
$endgroup$
– Andrei
Feb 2 at 3:47
$begingroup$
Okay...so how would one go about writing a power series that converges everywhere outside of some $a_0$?
$endgroup$
– Vane Voe
Feb 2 at 4:05
$begingroup$
@VaneVoe: I'd say you need Laurent series for that.
$endgroup$
– zipirovich
Feb 2 at 5:25
$begingroup$
So I somehow need to take complex integrals for a purely real function in the real plane? That doesn't quite sound right.
$endgroup$
– Vane Voe
Feb 2 at 5:26
add a comment |
$begingroup$
The simple answer is that you can't. If I would write $$ln(x)=sum_{n=0}^infty a_nx^n$$ then obviously I should be able to compute the derivative at $0$. $$frac d{dx} ln x=frac 1x$$ This diverges at $0$. Taking the derivative of the right hand side you get $$sum_{n=1}^infty na_n x^{n-1}$$ In the limit $xto 0$ you get $a_1$, which is finite.
You might ask "but what if I just expand all the $(x-x_0)^n$ terms in the Taylor series?" What will happen is that you get for the all terms $x^n$ coefficients that can be written as $sum_j^infty b_j(n) x_0^j$, but there is no guarantee that these series converge.
answered Feb 2 at 3:27
AndreiAndrei
13.7k21230
$endgroup$
$begingroup$
Isn't there some way to express it as a power series but then just restrict the domain to exclude anything at or less than zero?
$endgroup$
– Vane Voe
Feb 2 at 3:34
$begingroup$
If the two expressions are the same for an infinite number of $x$ points, then the coefficients of $x^n$ must all be the same. I think (just a hunch) that if you write such a series it should diverge everywhere outside of the interval of convergence. But the power series converges at $0$ to $a_0$.
$endgroup$
– Andrei
Feb 2 at 3:47
$begingroup$
Okay...so how would one go about writing a power series that converges everywhere outside of some $a_0$?
$endgroup$
– Vane Voe
Feb 2 at 4:05
$begingroup$
@VaneVoe: I'd say you need Laurent series for that.
$endgroup$
– zipirovich
Feb 2 at 5:25
$begingroup$
So I somehow need to take complex integrals for a purely real function in the real plane? That doesn't quite sound right.
$endgroup$
– Vane Voe
Feb 2 at 5:26
add a comment |
$begingroup$
The simple answer is that you can't. If I would write $$ln(x)=sum_{n=0}^infty a_nx^n$$ then obviously I should be able to compute the derivative at $0$. $$frac d{dx} ln x=frac 1x$$ This diverges at $0$. Taking the derivative of the right hand side you get $$sum_{n=1}^infty na_n x^{n-1}$$ In the limit $xto 0$ you get $a_1$, which is finite.
You might ask "but what if I just expand all the $(x-x_0)^n$ terms in the Taylor series?" What will happen is that you get for the all terms $x^n$ coefficients that can be written as $sum_j^infty b_j(n) x_0^j$, but there is no guarantee that these series converge.
answered Feb 2 at 3:27
AndreiAndrei
13.7k21230
$endgroup$
The simple answer is that you can't. If I would write $$ln(x)=sum_{n=0}^infty a_nx^n$$ then obviously I should be able to compute the derivative at $0$. $$frac d{dx} ln x=frac 1x$$ This diverges at $0$. Taking the derivative of the right hand side you get $$sum_{n=1}^infty na_n x^{n-1}$$ In the limit $xto 0$ you get $a_1$, which is finite.
You might ask "but what if I just expand all the $(x-x_0)^n$ terms in the Taylor series?" What will happen is that you get for the all terms $x^n$ coefficients that can be written as $sum_j^infty b_j(n) x_0^j$, but there is no guarantee that these series converge.
answered Feb 2 at 3:27
AndreiAndrei
13.7k21230
answered Feb 2 at 3:27
AndreiAndrei
13.7k21230
answered Feb 2 at 3:27
AndreiAndrei
13.7k21230
answered Feb 2 at 3:27
AndreiAndrei
13.7k21230
13.7k21230
$begingroup$
Isn't there some way to express it as a power series but then just restrict the domain to exclude anything at or less than zero?
$endgroup$
– Vane Voe
Feb 2 at 3:34
$begingroup$
If the two expressions are the same for an infinite number of $x$ points, then the coefficients of $x^n$ must all be the same. I think (just a hunch) that if you write such a series it should diverge everywhere outside of the interval of convergence. But the power series converges at $0$ to $a_0$.
$endgroup$
– Andrei
Feb 2 at 3:47
$begingroup$
Okay...so how would one go about writing a power series that converges everywhere outside of some $a_0$?
$endgroup$
– Vane Voe
Feb 2 at 4:05
$begingroup$
@VaneVoe: I'd say you need Laurent series for that.
$endgroup$
– zipirovich
Feb 2 at 5:25
$begingroup$
So I somehow need to take complex integrals for a purely real function in the real plane? That doesn't quite sound right.
$endgroup$
– Vane Voe
Feb 2 at 5:26
add a comment |
$begingroup$
Isn't there some way to express it as a power series but then just restrict the domain to exclude anything at or less than zero?
$endgroup$
– Vane Voe
Feb 2 at 3:34
$begingroup$
If the two expressions are the same for an infinite number of $x$ points, then the coefficients of $x^n$ must all be the same. I think (just a hunch) that if you write such a series it should diverge everywhere outside of the interval of convergence. But the power series converges at $0$ to $a_0$.
$endgroup$
– Andrei
Feb 2 at 3:47
$begingroup$
Okay...so how would one go about writing a power series that converges everywhere outside of some $a_0$?
$endgroup$
– Vane Voe
Feb 2 at 4:05
$begingroup$
@VaneVoe: I'd say you need Laurent series for that.
$endgroup$
– zipirovich
Feb 2 at 5:25
$begingroup$
So I somehow need to take complex integrals for a purely real function in the real plane? That doesn't quite sound right.
$endgroup$
– Vane Voe
Feb 2 at 5:26
$begingroup$
Isn't there some way to express it as a power series but then just restrict the domain to exclude anything at or less than zero?
$endgroup$
– Vane Voe
Feb 2 at 3:34
$begingroup$
Isn't there some way to express it as a power series but then just restrict the domain to exclude anything at or less than zero?
$endgroup$
– Vane Voe
Feb 2 at 3:34
$begingroup$
If the two expressions are the same for an infinite number of $x$ points, then the coefficients of $x^n$ must all be the same. I think (just a hunch) that if you write such a series it should diverge everywhere outside of the interval of convergence. But the power series converges at $0$ to $a_0$.
$endgroup$
– Andrei
Feb 2 at 3:47
$begingroup$
If the two expressions are the same for an infinite number of $x$ points, then the coefficients of $x^n$ must all be the same. I think (just a hunch) that if you write such a series it should diverge everywhere outside of the interval of convergence. But the power series converges at $0$ to $a_0$.
$endgroup$
– Andrei
Feb 2 at 3:47
$begingroup$
Okay...so how would one go about writing a power series that converges everywhere outside of some $a_0$?
$endgroup$
– Vane Voe
Feb 2 at 4:05
$begingroup$
Okay...so how would one go about writing a power series that converges everywhere outside of some $a_0$?
$endgroup$
– Vane Voe
Feb 2 at 4:05
$begingroup$
@VaneVoe: I'd say you need Laurent series for that.
$endgroup$
– zipirovich
Feb 2 at 5:25
$begingroup$
@VaneVoe: I'd say you need Laurent series for that.
$endgroup$
– zipirovich
Feb 2 at 5:25
$begingroup$
So I somehow need to take complex integrals for a purely real function in the real plane? That doesn't quite sound right.
$endgroup$
– Vane Voe
Feb 2 at 5:26
$begingroup$
So I somehow need to take complex integrals for a purely real function in the real plane? That doesn't quite sound right.
$endgroup$
– Vane Voe
Feb 2 at 5:26
add a comment |
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