Taylor series leads to two different functions - why?
$begingroup$
Suppose, I want to find a function such that its Taylor series expansion is
$$f(x) = sum_{n=0}^{infty}frac{x^{n+1}}{(n+1)a^n}$$
I could start with $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$
Integrate it, substitute $xrightarrow frac{x}{a}$, multiply by $a$ and get
$$F(x) = -ln|x-1| = sum_{n=0}^{infty}frac{x^{n+1}}{n+1}$$
$$a Fleft(frac{x}{a}right) = -a lnleft|frac{x}{a}-1right| = sum_{n=0}^{infty}frac{x^{n+1}}{(n+1)a^n}$$
On the other hand, I could start with subtituting $x rightarrow frac{x}{a}$ before integration to get
$$frac{a}{a-x} = sum_{n=0}^{infty}frac{x^n}{a^n}$$
and then integrate it to get
$$-aln|x-a| = sum_{n=0}^{infty}frac{x^{n+1}}{(n+1)a^n}$$
As you can see, arguments of $ln$ are not equal. Where did it go wrong?
calculus taylor-expansion
edited May 16 at 9:31
Asaf Karagila♦
312k33446780
asked May 16 at 8:43
persistent_netizenpersistent_netizen
1366
$endgroup$
add a comment |
$begingroup$
Suppose, I want to find a function such that its Taylor series expansion is
$$f(x) = sum_{n=0}^{infty}frac{x^{n+1}}{(n+1)a^n}$$
I could start with $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$
Integrate it, substitute $xrightarrow frac{x}{a}$, multiply by $a$ and get
$$F(x) = -ln|x-1| = sum_{n=0}^{infty}frac{x^{n+1}}{n+1}$$
$$a Fleft(frac{x}{a}right) = -a lnleft|frac{x}{a}-1right| = sum_{n=0}^{infty}frac{x^{n+1}}{(n+1)a^n}$$
On the other hand, I could start with subtituting $x rightarrow frac{x}{a}$ before integration to get
$$frac{a}{a-x} = sum_{n=0}^{infty}frac{x^n}{a^n}$$
and then integrate it to get
$$-aln|x-a| = sum_{n=0}^{infty}frac{x^{n+1}}{(n+1)a^n}$$
As you can see, arguments of $ln$ are not equal. Where did it go wrong?
calculus taylor-expansion
edited May 16 at 9:31
Asaf Karagila♦
312k33446780
asked May 16 at 8:43
persistent_netizenpersistent_netizen
1366
$endgroup$
5
$begingroup$
it is possible that two different functions will have the same derivative. think about $f(x) = 2x^{2} +3, g(x) = 2x^{2} - 4 f'(x)=g'(x) = 4x$
$endgroup$
– Jneven
May 16 at 8:51
5
$begingroup$
It's good that you ask this question. But that's why it always pays to be precise. If you had attempted to define exactly what $F$ was, you would have realized that either you define it as some anti-derivative, or you define it as some specific definite integral, and in both cases you will know that you have to handle the constants that arise correctly.
$endgroup$
– user21820
May 16 at 9:37
add a comment |
$begingroup$
Suppose, I want to find a function such that its Taylor series expansion is
$$f(x) = sum_{n=0}^{infty}frac{x^{n+1}}{(n+1)a^n}$$
I could start with $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$
Integrate it, substitute $xrightarrow frac{x}{a}$, multiply by $a$ and get
$$F(x) = -ln|x-1| = sum_{n=0}^{infty}frac{x^{n+1}}{n+1}$$
$$a Fleft(frac{x}{a}right) = -a lnleft|frac{x}{a}-1right| = sum_{n=0}^{infty}frac{x^{n+1}}{(n+1)a^n}$$
On the other hand, I could start with subtituting $x rightarrow frac{x}{a}$ before integration to get
$$frac{a}{a-x} = sum_{n=0}^{infty}frac{x^n}{a^n}$$
and then integrate it to get
$$-aln|x-a| = sum_{n=0}^{infty}frac{x^{n+1}}{(n+1)a^n}$$
As you can see, arguments of $ln$ are not equal. Where did it go wrong?
calculus taylor-expansion
edited May 16 at 9:31
Asaf Karagila♦
312k33446780
asked May 16 at 8:43
persistent_netizenpersistent_netizen
1366
$endgroup$
Suppose, I want to find a function such that its Taylor series expansion is
$$f(x) = sum_{n=0}^{infty}frac{x^{n+1}}{(n+1)a^n}$$
I could start with $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$
Integrate it, substitute $xrightarrow frac{x}{a}$, multiply by $a$ and get
$$F(x) = -ln|x-1| = sum_{n=0}^{infty}frac{x^{n+1}}{n+1}$$
$$a Fleft(frac{x}{a}right) = -a lnleft|frac{x}{a}-1right| = sum_{n=0}^{infty}frac{x^{n+1}}{(n+1)a^n}$$
On the other hand, I could start with subtituting $x rightarrow frac{x}{a}$ before integration to get
$$frac{a}{a-x} = sum_{n=0}^{infty}frac{x^n}{a^n}$$
and then integrate it to get
$$-aln|x-a| = sum_{n=0}^{infty}frac{x^{n+1}}{(n+1)a^n}$$
As you can see, arguments of $ln$ are not equal. Where did it go wrong?
calculus taylor-expansion
calculus taylor-expansion
edited May 16 at 9:31
Asaf Karagila♦
312k33446780
asked May 16 at 8:43
persistent_netizenpersistent_netizen
1366
edited May 16 at 9:31
Asaf Karagila♦
312k33446780
asked May 16 at 8:43
persistent_netizenpersistent_netizen
1366
edited May 16 at 9:31
Asaf Karagila♦
312k33446780
edited May 16 at 9:31
Asaf Karagila♦
312k33446780
edited May 16 at 9:31
Asaf Karagila♦
312k33446780
312k33446780
asked May 16 at 8:43
persistent_netizenpersistent_netizen
1366
asked May 16 at 8:43
persistent_netizenpersistent_netizen
1366
asked May 16 at 8:43
persistent_netizenpersistent_netizen
1366
1366
5
$begingroup$
it is possible that two different functions will have the same derivative. think about $f(x) = 2x^{2} +3, g(x) = 2x^{2} - 4 f'(x)=g'(x) = 4x$
$endgroup$
– Jneven
May 16 at 8:51
5
$begingroup$
It's good that you ask this question. But that's why it always pays to be precise. If you had attempted to define exactly what $F$ was, you would have realized that either you define it as some anti-derivative, or you define it as some specific definite integral, and in both cases you will know that you have to handle the constants that arise correctly.
$endgroup$
– user21820
May 16 at 9:37
add a comment |
5
$begingroup$
it is possible that two different functions will have the same derivative. think about $f(x) = 2x^{2} +3, g(x) = 2x^{2} - 4 f'(x)=g'(x) = 4x$
$endgroup$
– Jneven
May 16 at 8:51
5
$begingroup$
It's good that you ask this question. But that's why it always pays to be precise. If you had attempted to define exactly what $F$ was, you would have realized that either you define it as some anti-derivative, or you define it as some specific definite integral, and in both cases you will know that you have to handle the constants that arise correctly.
$endgroup$
– user21820
May 16 at 9:37
5
5
$begingroup$
it is possible that two different functions will have the same derivative. think about $f(x) = 2x^{2} +3, g(x) = 2x^{2} - 4 f'(x)=g'(x) = 4x$
$endgroup$
– Jneven
May 16 at 8:51
$begingroup$
it is possible that two different functions will have the same derivative. think about $f(x) = 2x^{2} +3, g(x) = 2x^{2} - 4 f'(x)=g'(x) = 4x$
$endgroup$
– Jneven
May 16 at 8:51
5
5
$begingroup$
It's good that you ask this question. But that's why it always pays to be precise. If you had attempted to define exactly what $F$ was, you would have realized that either you define it as some anti-derivative, or you define it as some specific definite integral, and in both cases you will know that you have to handle the constants that arise correctly.
$endgroup$
– user21820
May 16 at 9:37
$begingroup$
It's good that you ask this question. But that's why it always pays to be precise. If you had attempted to define exactly what $F$ was, you would have realized that either you define it as some anti-derivative, or you define it as some specific definite integral, and in both cases you will know that you have to handle the constants that arise correctly.
$endgroup$
– user21820
May 16 at 9:37
add a comment |
1 Answer
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When you integrate, you should include a constant of integration. What you see here is that when integrating the functions, you get different constants of integration. This is why your answers differ by only a constant, namely $aln a$ (you can see this by use of $log$ rules).
If you take care with the limits or boundary conditions in the integration step, then the answers will agree exactly.
answered May 16 at 8:49
John DoeJohn Doe
13.4k11744
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add a comment |
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$begingroup$
When you integrate, you should include a constant of integration. What you see here is that when integrating the functions, you get different constants of integration. This is why your answers differ by only a constant, namely $aln a$ (you can see this by use of $log$ rules).
If you take care with the limits or boundary conditions in the integration step, then the answers will agree exactly.
answered May 16 at 8:49
John DoeJohn Doe
13.4k11744
$endgroup$
add a comment |
$begingroup$
When you integrate, you should include a constant of integration. What you see here is that when integrating the functions, you get different constants of integration. This is why your answers differ by only a constant, namely $aln a$ (you can see this by use of $log$ rules).
If you take care with the limits or boundary conditions in the integration step, then the answers will agree exactly.
answered May 16 at 8:49
John DoeJohn Doe
13.4k11744
$endgroup$
add a comment |
$begingroup$
When you integrate, you should include a constant of integration. What you see here is that when integrating the functions, you get different constants of integration. This is why your answers differ by only a constant, namely $aln a$ (you can see this by use of $log$ rules).
If you take care with the limits or boundary conditions in the integration step, then the answers will agree exactly.
answered May 16 at 8:49
John DoeJohn Doe
13.4k11744
$endgroup$
When you integrate, you should include a constant of integration. What you see here is that when integrating the functions, you get different constants of integration. This is why your answers differ by only a constant, namely $aln a$ (you can see this by use of $log$ rules).
If you take care with the limits or boundary conditions in the integration step, then the answers will agree exactly.
answered May 16 at 8:49
John DoeJohn Doe
13.4k11744
answered May 16 at 8:49
John DoeJohn Doe
13.4k11744
answered May 16 at 8:49
John DoeJohn Doe
13.4k11744
answered May 16 at 8:49
John DoeJohn Doe
13.4k11744
13.4k11744
add a comment |
add a comment |
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$begingroup$
it is possible that two different functions will have the same derivative. think about $f(x) = 2x^{2} +3, g(x) = 2x^{2} - 4 f'(x)=g'(x) = 4x$
$endgroup$
– Jneven
May 16 at 8:51
5
$begingroup$
It's good that you ask this question. But that's why it always pays to be precise. If you had attempted to define exactly what $F$ was, you would have realized that either you define it as some anti-derivative, or you define it as some specific definite integral, and in both cases you will know that you have to handle the constants that arise correctly.
$endgroup$
– user21820
May 16 at 9:37