Mixing
Function representation of power series convergent at x = a
$begingroup$
SOME BACKGROUND INFO:
Analytic functions may be (locally) represented by a convergent power/Taylor series. The domain is given by the interval in which the power series represents this function. For example, $f(x) = e^x$ has $domf = (-∞,∞)$.
In addition, it has also been said that every power series is the Taylor series of some $C^∞$ function.
My question is thus: suppose we had a power series centered at $a$, whose radius of convergence $R = a$. By the above, it must have a Taylor series representation of some analytic, $C^∞$ function. But, also by the above, the domain of this function must be ${a}$, a collapsed interval. How are these ideas compatible? (I feel as if there is some contradiction: a function defined only at one point cannot be differentiated an infinite number of times, and moreover, there would be an infinite number of functions that could be represented by this $R =a $ power series).
convergence power-series taylor-expansion
asked Jan 27 at 19:03
Julia KimJulia Kim
174
$endgroup$
add a comment |
$begingroup$
SOME BACKGROUND INFO:
Analytic functions may be (locally) represented by a convergent power/Taylor series. The domain is given by the interval in which the power series represents this function. For example, $f(x) = e^x$ has $domf = (-∞,∞)$.
In addition, it has also been said that every power series is the Taylor series of some $C^∞$ function.
My question is thus: suppose we had a power series centered at $a$, whose radius of convergence $R = a$. By the above, it must have a Taylor series representation of some analytic, $C^∞$ function. But, also by the above, the domain of this function must be ${a}$, a collapsed interval. How are these ideas compatible? (I feel as if there is some contradiction: a function defined only at one point cannot be differentiated an infinite number of times, and moreover, there would be an infinite number of functions that could be represented by this $R =a $ power series).
convergence power-series taylor-expansion
asked Jan 27 at 19:03
Julia KimJulia Kim
174
$endgroup$
add a comment |
$begingroup$
SOME BACKGROUND INFO:
Analytic functions may be (locally) represented by a convergent power/Taylor series. The domain is given by the interval in which the power series represents this function. For example, $f(x) = e^x$ has $domf = (-∞,∞)$.
In addition, it has also been said that every power series is the Taylor series of some $C^∞$ function.
My question is thus: suppose we had a power series centered at $a$, whose radius of convergence $R = a$. By the above, it must have a Taylor series representation of some analytic, $C^∞$ function. But, also by the above, the domain of this function must be ${a}$, a collapsed interval. How are these ideas compatible? (I feel as if there is some contradiction: a function defined only at one point cannot be differentiated an infinite number of times, and moreover, there would be an infinite number of functions that could be represented by this $R =a $ power series).
convergence power-series taylor-expansion
asked Jan 27 at 19:03
Julia KimJulia Kim
174
$endgroup$
SOME BACKGROUND INFO:
Analytic functions may be (locally) represented by a convergent power/Taylor series. The domain is given by the interval in which the power series represents this function. For example, $f(x) = e^x$ has $domf = (-∞,∞)$.
In addition, it has also been said that every power series is the Taylor series of some $C^∞$ function.
My question is thus: suppose we had a power series centered at $a$, whose radius of convergence $R = a$. By the above, it must have a Taylor series representation of some analytic, $C^∞$ function. But, also by the above, the domain of this function must be ${a}$, a collapsed interval. How are these ideas compatible? (I feel as if there is some contradiction: a function defined only at one point cannot be differentiated an infinite number of times, and moreover, there would be an infinite number of functions that could be represented by this $R =a $ power series).
convergence power-series taylor-expansion
convergence power-series taylor-expansion
asked Jan 27 at 19:03
Julia KimJulia Kim
174
asked Jan 27 at 19:03
Julia KimJulia Kim
174
asked Jan 27 at 19:03
Julia KimJulia Kim
174
asked Jan 27 at 19:03
Julia KimJulia Kim
174
asked Jan 27 at 19:03
Julia KimJulia Kim
174
174
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Actually, what happens is that every power series with non-zero radius of convergence is the Taylor series of some $C^infty$ function.
On the other hand, if a Taylor series is centered at $a(>0)$ and if its radius of convergence is $a$, then it converges on the interval $(0,2a)$ and its sum defines a $C^infty$ function there.
answered Jan 27 at 19:10
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Much obliged for your help. Another question: If a power series has $R = 0$, then its coefficient terms, $c_n$, can be in function of anything, correct? That is, they cannot be $f^{(n)}(a)/n!$, as per the Taylor series definition, since such an $f(x)$ does not exist.
$endgroup$
– Julia Kim
Jan 29 at 3:16
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 29 at 7:55
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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votes
$begingroup$
Actually, what happens is that every power series with non-zero radius of convergence is the Taylor series of some $C^infty$ function.
On the other hand, if a Taylor series is centered at $a(>0)$ and if its radius of convergence is $a$, then it converges on the interval $(0,2a)$ and its sum defines a $C^infty$ function there.
answered Jan 27 at 19:10
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Much obliged for your help. Another question: If a power series has $R = 0$, then its coefficient terms, $c_n$, can be in function of anything, correct? That is, they cannot be $f^{(n)}(a)/n!$, as per the Taylor series definition, since such an $f(x)$ does not exist.
$endgroup$
– Julia Kim
Jan 29 at 3:16
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 29 at 7:55
add a comment |
$begingroup$
Actually, what happens is that every power series with non-zero radius of convergence is the Taylor series of some $C^infty$ function.
On the other hand, if a Taylor series is centered at $a(>0)$ and if its radius of convergence is $a$, then it converges on the interval $(0,2a)$ and its sum defines a $C^infty$ function there.
answered Jan 27 at 19:10
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Much obliged for your help. Another question: If a power series has $R = 0$, then its coefficient terms, $c_n$, can be in function of anything, correct? That is, they cannot be $f^{(n)}(a)/n!$, as per the Taylor series definition, since such an $f(x)$ does not exist.
$endgroup$
– Julia Kim
Jan 29 at 3:16
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 29 at 7:55
add a comment |
$begingroup$
Actually, what happens is that every power series with non-zero radius of convergence is the Taylor series of some $C^infty$ function.
On the other hand, if a Taylor series is centered at $a(>0)$ and if its radius of convergence is $a$, then it converges on the interval $(0,2a)$ and its sum defines a $C^infty$ function there.
answered Jan 27 at 19:10
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
Actually, what happens is that every power series with non-zero radius of convergence is the Taylor series of some $C^infty$ function.
On the other hand, if a Taylor series is centered at $a(>0)$ and if its radius of convergence is $a$, then it converges on the interval $(0,2a)$ and its sum defines a $C^infty$ function there.
answered Jan 27 at 19:10
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jan 27 at 19:10
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jan 27 at 19:10
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jan 27 at 19:10
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
$begingroup$
Much obliged for your help. Another question: If a power series has $R = 0$, then its coefficient terms, $c_n$, can be in function of anything, correct? That is, they cannot be $f^{(n)}(a)/n!$, as per the Taylor series definition, since such an $f(x)$ does not exist.
$endgroup$
– Julia Kim
Jan 29 at 3:16
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 29 at 7:55
add a comment |
$begingroup$
Much obliged for your help. Another question: If a power series has $R = 0$, then its coefficient terms, $c_n$, can be in function of anything, correct? That is, they cannot be $f^{(n)}(a)/n!$, as per the Taylor series definition, since such an $f(x)$ does not exist.
$endgroup$
– Julia Kim
Jan 29 at 3:16
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 29 at 7:55
$begingroup$
Much obliged for your help. Another question: If a power series has $R = 0$, then its coefficient terms, $c_n$, can be in function of anything, correct? That is, they cannot be $f^{(n)}(a)/n!$, as per the Taylor series definition, since such an $f(x)$ does not exist.
$endgroup$
– Julia Kim
Jan 29 at 3:16
$begingroup$
Much obliged for your help. Another question: If a power series has $R = 0$, then its coefficient terms, $c_n$, can be in function of anything, correct? That is, they cannot be $f^{(n)}(a)/n!$, as per the Taylor series definition, since such an $f(x)$ does not exist.
$endgroup$
– Julia Kim
Jan 29 at 3:16
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 29 at 7:55
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Jan 29 at 7:55
add a comment |
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