Mixing
Equality of coefficient of two power series
$begingroup$
How do I prove
If there is an $varepsilon > 0$ such that $sum_{n=0}^{+infty} a_nx^n = sum_{n=0}^{+infty} b_nx^n$ for $-varepsilon < x < varepsilon$, then $a_n = b_n$ for all $n in mathbb{N}$.
I know that the $n$-th coefficient in $f(x) = sum_{n=0}^{+infty} a_nx^n$ is given by $a_n = frac{f^{(n)}(0)}{n!}$.
So what I tried: Assume $sum a_n x^n$ and $sum b_n x^n$ have convergence radius $R_a$ and $R_b$ respectively. Now we choose $0<varepsilon<min{{R_a,R_b}}$: $sum a_n x^n$ and $sum b_n x^n$ are uniformly convergent on $]-varepsilon,varepsilon[$.
Say $f_a(x) = sum_{n=0}^{+infty} a_nx^n = sum_{n=0}^{+infty} b_nx^n = f_b(x)$ $(-varepsilon<x<varepsilon)$. On this interval we get: $a_n = frac{f_a^{(n)}(0)}{n!} = frac{f_b^{(n)}(0)}{n!}$, since $0$ in always an interior point of the interval. The equality follows.
real-analysis power-series
edited Jan 13 at 8:23
Hayk
2,6271214
asked Jan 13 at 8:01
ZacharyZachary
1559
$endgroup$
add a comment |
$begingroup$
How do I prove
If there is an $varepsilon > 0$ such that $sum_{n=0}^{+infty} a_nx^n = sum_{n=0}^{+infty} b_nx^n$ for $-varepsilon < x < varepsilon$, then $a_n = b_n$ for all $n in mathbb{N}$.
I know that the $n$-th coefficient in $f(x) = sum_{n=0}^{+infty} a_nx^n$ is given by $a_n = frac{f^{(n)}(0)}{n!}$.
So what I tried: Assume $sum a_n x^n$ and $sum b_n x^n$ have convergence radius $R_a$ and $R_b$ respectively. Now we choose $0<varepsilon<min{{R_a,R_b}}$: $sum a_n x^n$ and $sum b_n x^n$ are uniformly convergent on $]-varepsilon,varepsilon[$.
Say $f_a(x) = sum_{n=0}^{+infty} a_nx^n = sum_{n=0}^{+infty} b_nx^n = f_b(x)$ $(-varepsilon<x<varepsilon)$. On this interval we get: $a_n = frac{f_a^{(n)}(0)}{n!} = frac{f_b^{(n)}(0)}{n!}$, since $0$ in always an interior point of the interval. The equality follows.
real-analysis power-series
edited Jan 13 at 8:23
Hayk
2,6271214
asked Jan 13 at 8:01
ZacharyZachary
1559
$endgroup$
$begingroup$
How do you know that $f_{a}(x)$ and $f_{b}(x)$ are differentiable, or even smooth?
$endgroup$
– Seewoo Lee
Jan 13 at 8:04
$begingroup$
If $R in mathbb{R}^{+}$, the sum of the power series on $]-R,R[$ is smooth $(C^{infty})$.
$endgroup$
– Zachary
Jan 13 at 8:15
add a comment |
$begingroup$
How do I prove
If there is an $varepsilon > 0$ such that $sum_{n=0}^{+infty} a_nx^n = sum_{n=0}^{+infty} b_nx^n$ for $-varepsilon < x < varepsilon$, then $a_n = b_n$ for all $n in mathbb{N}$.
I know that the $n$-th coefficient in $f(x) = sum_{n=0}^{+infty} a_nx^n$ is given by $a_n = frac{f^{(n)}(0)}{n!}$.
So what I tried: Assume $sum a_n x^n$ and $sum b_n x^n$ have convergence radius $R_a$ and $R_b$ respectively. Now we choose $0<varepsilon<min{{R_a,R_b}}$: $sum a_n x^n$ and $sum b_n x^n$ are uniformly convergent on $]-varepsilon,varepsilon[$.
Say $f_a(x) = sum_{n=0}^{+infty} a_nx^n = sum_{n=0}^{+infty} b_nx^n = f_b(x)$ $(-varepsilon<x<varepsilon)$. On this interval we get: $a_n = frac{f_a^{(n)}(0)}{n!} = frac{f_b^{(n)}(0)}{n!}$, since $0$ in always an interior point of the interval. The equality follows.
real-analysis power-series
edited Jan 13 at 8:23
Hayk
2,6271214
asked Jan 13 at 8:01
ZacharyZachary
1559
$endgroup$
How do I prove
If there is an $varepsilon > 0$ such that $sum_{n=0}^{+infty} a_nx^n = sum_{n=0}^{+infty} b_nx^n$ for $-varepsilon < x < varepsilon$, then $a_n = b_n$ for all $n in mathbb{N}$.
I know that the $n$-th coefficient in $f(x) = sum_{n=0}^{+infty} a_nx^n$ is given by $a_n = frac{f^{(n)}(0)}{n!}$.
So what I tried: Assume $sum a_n x^n$ and $sum b_n x^n$ have convergence radius $R_a$ and $R_b$ respectively. Now we choose $0<varepsilon<min{{R_a,R_b}}$: $sum a_n x^n$ and $sum b_n x^n$ are uniformly convergent on $]-varepsilon,varepsilon[$.
Say $f_a(x) = sum_{n=0}^{+infty} a_nx^n = sum_{n=0}^{+infty} b_nx^n = f_b(x)$ $(-varepsilon<x<varepsilon)$. On this interval we get: $a_n = frac{f_a^{(n)}(0)}{n!} = frac{f_b^{(n)}(0)}{n!}$, since $0$ in always an interior point of the interval. The equality follows.
real-analysis power-series
real-analysis power-series
edited Jan 13 at 8:23
Hayk
2,6271214
asked Jan 13 at 8:01
ZacharyZachary
1559
edited Jan 13 at 8:23
Hayk
2,6271214
asked Jan 13 at 8:01
ZacharyZachary
1559
edited Jan 13 at 8:23
Hayk
2,6271214
edited Jan 13 at 8:23
Hayk
2,6271214
edited Jan 13 at 8:23
Hayk
2,6271214
2,6271214
asked Jan 13 at 8:01
ZacharyZachary
1559
asked Jan 13 at 8:01
ZacharyZachary
1559
asked Jan 13 at 8:01
ZacharyZachary
1559
1559
$begingroup$
How do you know that $f_{a}(x)$ and $f_{b}(x)$ are differentiable, or even smooth?
$endgroup$
– Seewoo Lee
Jan 13 at 8:04
$begingroup$
If $R in mathbb{R}^{+}$, the sum of the power series on $]-R,R[$ is smooth $(C^{infty})$.
$endgroup$
– Zachary
Jan 13 at 8:15
add a comment |
$begingroup$
How do you know that $f_{a}(x)$ and $f_{b}(x)$ are differentiable, or even smooth?
$endgroup$
– Seewoo Lee
Jan 13 at 8:04
$begingroup$
If $R in mathbb{R}^{+}$, the sum of the power series on $]-R,R[$ is smooth $(C^{infty})$.
$endgroup$
– Zachary
Jan 13 at 8:15
$begingroup$
How do you know that $f_{a}(x)$ and $f_{b}(x)$ are differentiable, or even smooth?
$endgroup$
– Seewoo Lee
Jan 13 at 8:04
$begingroup$
How do you know that $f_{a}(x)$ and $f_{b}(x)$ are differentiable, or even smooth?
$endgroup$
– Seewoo Lee
Jan 13 at 8:04
$begingroup$
If $R in mathbb{R}^{+}$, the sum of the power series on $]-R,R[$ is smooth $(C^{infty})$.
$endgroup$
– Zachary
Jan 13 at 8:15
$begingroup$
If $R in mathbb{R}^{+}$, the sum of the power series on $]-R,R[$ is smooth $(C^{infty})$.
$endgroup$
– Zachary
Jan 13 at 8:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assuming the convergence radii of both series are positive (as you do in your question), then one can prove the uniqueness (equality of coefficients) by differentiation.
Namely, setting $x=0$ we get $a_0 = b_0$. In view of absolute convergence of the series, we can differentiate the series term by term, where the resulting series will have the same radius of convergence. Differentiating both series and setting $x=0$ implies $a_1 = b_1$. The general case of $a_n = b_n$ follows by induction on $n$.
answered Jan 13 at 8:17
HaykHayk
2,6271214
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add a comment |
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1 Answer
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1 Answer
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active
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active
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$begingroup$
Assuming the convergence radii of both series are positive (as you do in your question), then one can prove the uniqueness (equality of coefficients) by differentiation.
Namely, setting $x=0$ we get $a_0 = b_0$. In view of absolute convergence of the series, we can differentiate the series term by term, where the resulting series will have the same radius of convergence. Differentiating both series and setting $x=0$ implies $a_1 = b_1$. The general case of $a_n = b_n$ follows by induction on $n$.
answered Jan 13 at 8:17
HaykHayk
2,6271214
$endgroup$
add a comment |
$begingroup$
Assuming the convergence radii of both series are positive (as you do in your question), then one can prove the uniqueness (equality of coefficients) by differentiation.
Namely, setting $x=0$ we get $a_0 = b_0$. In view of absolute convergence of the series, we can differentiate the series term by term, where the resulting series will have the same radius of convergence. Differentiating both series and setting $x=0$ implies $a_1 = b_1$. The general case of $a_n = b_n$ follows by induction on $n$.
answered Jan 13 at 8:17
HaykHayk
2,6271214
$endgroup$
add a comment |
$begingroup$
Assuming the convergence radii of both series are positive (as you do in your question), then one can prove the uniqueness (equality of coefficients) by differentiation.
Namely, setting $x=0$ we get $a_0 = b_0$. In view of absolute convergence of the series, we can differentiate the series term by term, where the resulting series will have the same radius of convergence. Differentiating both series and setting $x=0$ implies $a_1 = b_1$. The general case of $a_n = b_n$ follows by induction on $n$.
answered Jan 13 at 8:17
HaykHayk
2,6271214
$endgroup$
Assuming the convergence radii of both series are positive (as you do in your question), then one can prove the uniqueness (equality of coefficients) by differentiation.
Namely, setting $x=0$ we get $a_0 = b_0$. In view of absolute convergence of the series, we can differentiate the series term by term, where the resulting series will have the same radius of convergence. Differentiating both series and setting $x=0$ implies $a_1 = b_1$. The general case of $a_n = b_n$ follows by induction on $n$.
answered Jan 13 at 8:17
HaykHayk
2,6271214
answered Jan 13 at 8:17
HaykHayk
2,6271214
answered Jan 13 at 8:17
HaykHayk
2,6271214
answered Jan 13 at 8:17
HaykHayk
2,6271214
2,6271214
add a comment |
add a comment |
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$begingroup$
How do you know that $f_{a}(x)$ and $f_{b}(x)$ are differentiable, or even smooth?
$endgroup$
– Seewoo Lee
Jan 13 at 8:04
$begingroup$
If $R in mathbb{R}^{+}$, the sum of the power series on $]-R,R[$ is smooth $(C^{infty})$.
$endgroup$
– Zachary
Jan 13 at 8:15