Mixing
how to find a function expression of a power series?
$begingroup$
I have the following power series: $x+frac{x^2}{2}+frac{x^3}{3}+frac{x^4}{4}+cdots$.
I want to use the sum of the geometric series $sum^infty_limits{k=0}{x^k}=frac{1}{1-x}$ to write a function $f(x)$ for the power series above.
When I look at the series I see that its equal to $sum^infty_limits{k=0}{frac{x^{k+1}}{k+1}}$, which is the integral of the geometric power series.
$$int{sum^infty_{k=0}{x^k}=intfrac{1}{1-x}}implies sum^infty_{k=0}{frac{x^{k+1}}{k+1}}=-ln|1-x|$$
The function expression should be $f(x)=-ln|1-x|$.
Is it correct to do it this way? Am I missing constant values after integrating?
sequences-and-series power-series
edited Jan 25 at 14:40
KM101
6,0901525
asked Jan 25 at 14:32
user3133165user3133165
1928
$endgroup$
|
show 5 more comments
$begingroup$
I have the following power series: $x+frac{x^2}{2}+frac{x^3}{3}+frac{x^4}{4}+cdots$.
I want to use the sum of the geometric series $sum^infty_limits{k=0}{x^k}=frac{1}{1-x}$ to write a function $f(x)$ for the power series above.
When I look at the series I see that its equal to $sum^infty_limits{k=0}{frac{x^{k+1}}{k+1}}$, which is the integral of the geometric power series.
$$int{sum^infty_{k=0}{x^k}=intfrac{1}{1-x}}implies sum^infty_{k=0}{frac{x^{k+1}}{k+1}}=-ln|1-x|$$
The function expression should be $f(x)=-ln|1-x|$.
Is it correct to do it this way? Am I missing constant values after integrating?
sequences-and-series power-series
edited Jan 25 at 14:40
KM101
6,0901525
asked Jan 25 at 14:32
user3133165user3133165
1928
$endgroup$
$begingroup$
I'm trying to relate this to the geometric series
$endgroup$
– user3133165
Jan 25 at 14:36
$begingroup$
If you plug $0$ in the initial power series, you should get the same value with the analytical function.
$endgroup$
– Yves Daoust
Jan 25 at 14:37
$begingroup$
So is this correct? does f(x) equal to the series for all x?
$endgroup$
– user3133165
Jan 25 at 14:39
1
$begingroup$
I haven't looked at your calculations, but yes, differentiating and antidifferentiating are common ways of trying to find closed forms for power seies.
$endgroup$
– Arthur
Jan 25 at 14:39
1
$begingroup$
"Does $f(x)$ equal to the series for all $x$?" No. The series does not converge everywhere, only for $|x|<1$.
$endgroup$
– kccu
Jan 25 at 14:44
|
show 5 more comments
$begingroup$
I have the following power series: $x+frac{x^2}{2}+frac{x^3}{3}+frac{x^4}{4}+cdots$.
I want to use the sum of the geometric series $sum^infty_limits{k=0}{x^k}=frac{1}{1-x}$ to write a function $f(x)$ for the power series above.
When I look at the series I see that its equal to $sum^infty_limits{k=0}{frac{x^{k+1}}{k+1}}$, which is the integral of the geometric power series.
$$int{sum^infty_{k=0}{x^k}=intfrac{1}{1-x}}implies sum^infty_{k=0}{frac{x^{k+1}}{k+1}}=-ln|1-x|$$
The function expression should be $f(x)=-ln|1-x|$.
Is it correct to do it this way? Am I missing constant values after integrating?
sequences-and-series power-series
edited Jan 25 at 14:40
KM101
6,0901525
asked Jan 25 at 14:32
user3133165user3133165
1928
$endgroup$
I have the following power series: $x+frac{x^2}{2}+frac{x^3}{3}+frac{x^4}{4}+cdots$.
I want to use the sum of the geometric series $sum^infty_limits{k=0}{x^k}=frac{1}{1-x}$ to write a function $f(x)$ for the power series above.
When I look at the series I see that its equal to $sum^infty_limits{k=0}{frac{x^{k+1}}{k+1}}$, which is the integral of the geometric power series.
$$int{sum^infty_{k=0}{x^k}=intfrac{1}{1-x}}implies sum^infty_{k=0}{frac{x^{k+1}}{k+1}}=-ln|1-x|$$
The function expression should be $f(x)=-ln|1-x|$.
Is it correct to do it this way? Am I missing constant values after integrating?
sequences-and-series power-series
sequences-and-series power-series
edited Jan 25 at 14:40
KM101
6,0901525
asked Jan 25 at 14:32
user3133165user3133165
1928
edited Jan 25 at 14:40
KM101
6,0901525
asked Jan 25 at 14:32
user3133165user3133165
1928
edited Jan 25 at 14:40
KM101
6,0901525
edited Jan 25 at 14:40
KM101
6,0901525
edited Jan 25 at 14:40
KM101
6,0901525
6,0901525
asked Jan 25 at 14:32
user3133165user3133165
1928
asked Jan 25 at 14:32
user3133165user3133165
1928
asked Jan 25 at 14:32
user3133165user3133165
1928
1928
$begingroup$
I'm trying to relate this to the geometric series
$endgroup$
– user3133165
Jan 25 at 14:36
$begingroup$
If you plug $0$ in the initial power series, you should get the same value with the analytical function.
$endgroup$
– Yves Daoust
Jan 25 at 14:37
$begingroup$
So is this correct? does f(x) equal to the series for all x?
$endgroup$
– user3133165
Jan 25 at 14:39
1
$begingroup$
I haven't looked at your calculations, but yes, differentiating and antidifferentiating are common ways of trying to find closed forms for power seies.
$endgroup$
– Arthur
Jan 25 at 14:39
1
$begingroup$
"Does $f(x)$ equal to the series for all $x$?" No. The series does not converge everywhere, only for $|x|<1$.
$endgroup$
– kccu
Jan 25 at 14:44
|
show 5 more comments
$begingroup$
I'm trying to relate this to the geometric series
$endgroup$
– user3133165
Jan 25 at 14:36
$begingroup$
If you plug $0$ in the initial power series, you should get the same value with the analytical function.
$endgroup$
– Yves Daoust
Jan 25 at 14:37
$begingroup$
So is this correct? does f(x) equal to the series for all x?
$endgroup$
– user3133165
Jan 25 at 14:39
1
$begingroup$
I haven't looked at your calculations, but yes, differentiating and antidifferentiating are common ways of trying to find closed forms for power seies.
$endgroup$
– Arthur
Jan 25 at 14:39
1
$begingroup$
"Does $f(x)$ equal to the series for all $x$?" No. The series does not converge everywhere, only for $|x|<1$.
$endgroup$
– kccu
Jan 25 at 14:44
$begingroup$
I'm trying to relate this to the geometric series
$endgroup$
– user3133165
Jan 25 at 14:36
$begingroup$
I'm trying to relate this to the geometric series
$endgroup$
– user3133165
Jan 25 at 14:36
$begingroup$
If you plug $0$ in the initial power series, you should get the same value with the analytical function.
$endgroup$
– Yves Daoust
Jan 25 at 14:37
$begingroup$
If you plug $0$ in the initial power series, you should get the same value with the analytical function.
$endgroup$
– Yves Daoust
Jan 25 at 14:37
$begingroup$
So is this correct? does f(x) equal to the series for all x?
$endgroup$
– user3133165
Jan 25 at 14:39
$begingroup$
So is this correct? does f(x) equal to the series for all x?
$endgroup$
– user3133165
Jan 25 at 14:39
1
1
$begingroup$
I haven't looked at your calculations, but yes, differentiating and antidifferentiating are common ways of trying to find closed forms for power seies.
$endgroup$
– Arthur
Jan 25 at 14:39
$begingroup$
I haven't looked at your calculations, but yes, differentiating and antidifferentiating are common ways of trying to find closed forms for power seies.
$endgroup$
– Arthur
Jan 25 at 14:39
1
1
$begingroup$
"Does $f(x)$ equal to the series for all $x$?" No. The series does not converge everywhere, only for $|x|<1$.
$endgroup$
– kccu
Jan 25 at 14:44
$begingroup$
"Does $f(x)$ equal to the series for all $x$?" No. The series does not converge everywhere, only for $|x|<1$.
$endgroup$
– kccu
Jan 25 at 14:44
|
show 5 more comments
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$begingroup$
I'm trying to relate this to the geometric series
$endgroup$
– user3133165
Jan 25 at 14:36
$begingroup$
If you plug $0$ in the initial power series, you should get the same value with the analytical function.
$endgroup$
– Yves Daoust
Jan 25 at 14:37
$begingroup$
So is this correct? does f(x) equal to the series for all x?
$endgroup$
– user3133165
Jan 25 at 14:39
1
$begingroup$
I haven't looked at your calculations, but yes, differentiating and antidifferentiating are common ways of trying to find closed forms for power seies.
$endgroup$
– Arthur
Jan 25 at 14:39
1
$begingroup$
"Does $f(x)$ equal to the series for all $x$?" No. The series does not converge everywhere, only for $|x|<1$.
$endgroup$
– kccu
Jan 25 at 14:44