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1

$begingroup$

This power series $$f(x)=sum_n=1^infty fracx^3n3n$$ when differentiated, loses $3n$ in the denominator, with one manipulation, one can get $$f'(x)=frac1x(1-x^3) $$ using the geometric series sum formula. Since this is $f'(x)$, integration is required so we get $$f(x)=log(x)-frac13log(1-x^3) + C$$
(the log base is probably $e$). The usual step for finding the constant is to find it's value for $x=0$ in the given power series and the new function. This obviously won't work here or in any other case where a function isn't defined for $x=0$. How do I find the constant then?

power-series

share|cite|improve this question

edited 1 hour ago

Leucippus

19.7k102871

asked 3 hours ago

user3711671user3711671

417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was talking about log(x) not being defined at x=0.
  $endgroup$
  – user3711671
  3 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

This power series $$f(x)=sum_n=1^infty fracx^3n3n$$ when differentiated, loses $3n$ in the denominator, with one manipulation, one can get $$f'(x)=frac1x(1-x^3) $$ using the geometric series sum formula. Since this is $f'(x)$, integration is required so we get $$f(x)=log(x)-frac13log(1-x^3) + C$$
(the log base is probably $e$). The usual step for finding the constant is to find it's value for $x=0$ in the given power series and the new function. This obviously won't work here or in any other case where a function isn't defined for $x=0$. How do I find the constant then?

power-series

share|cite|improve this question

edited 1 hour ago

Leucippus

19.7k102871

asked 3 hours ago

user3711671user3711671

417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was talking about log(x) not being defined at x=0.
  $endgroup$
  – user3711671
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This power series $$f(x)=sum_n=1^infty fracx^3n3n$$ when differentiated, loses $3n$ in the denominator, with one manipulation, one can get $$f'(x)=frac1x(1-x^3) $$ using the geometric series sum formula. Since this is $f'(x)$, integration is required so we get $$f(x)=log(x)-frac13log(1-x^3) + C$$
(the log base is probably $e$). The usual step for finding the constant is to find it's value for $x=0$ in the given power series and the new function. This obviously won't work here or in any other case where a function isn't defined for $x=0$. How do I find the constant then?

power-series

share|cite|improve this question

edited 1 hour ago

Leucippus

19.7k102871

asked 3 hours ago

user3711671user3711671

417

$endgroup$

share|cite|improve this question

edited 1 hour ago

Leucippus

19.7k102871

asked 3 hours ago

user3711671user3711671

417

share|cite|improve this question

edited 1 hour ago

Leucippus

19.7k102871

asked 3 hours ago

user3711671user3711671

417

edited 1 hour ago

Leucippus

19.7k102871

edited 1 hour ago

Leucippus

19.7k102871

asked 3 hours ago

user3711671user3711671

417

asked 3 hours ago

user3711671user3711671

417

2 Answers
2

active

oldest

votes

4

$begingroup$

Differentiating $f(x)$:

$$f(x)=fracx^33+fracx^66+cdotsimplies f'(x)=x^2+x^5+cdots=sum_n=1^inftyx^3n-1=fracx^21-x^3$$ for $|x|<1$, which results in $$f(x)=-frac13ln(1-x^3)+C$$ after integration. Your function now is well-defined at $x=0$, and this implies $C=0$.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

st.mathst.math

1,110115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, I've made a mistake at the geometric series, n starts at 1, not 0.My question still remains, what should one do if plugging in x=0 can't be done (because of functions like log(x), 1/x etc)?The interesting thing is that if log(x) remains untouched, C will equal -log(x).
  $endgroup$
  – user3711671
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then plug in another number, e.g. $x=1$. The constant after integration can be "fixed" by moving the graph of the function up or down the $y$-axis. This can be done by using any point; not only $x=0$.
  $endgroup$
  – st.math
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But what about the infinite sum of 1/3n then?Shouldn't I find out what the constant is by inserting the value for which the sum and it's function representation are equal (when x=0 both are equal to zero)?
  $endgroup$
  – user3711671
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can plug in any point where the function is defined. After plugging in $x=0$ into the sum, you get $f(0)=0$. The closed form you obtained after integration ($f(x)=ln(1-x^3)/3+C$) is undetermined by the constant $C$ only, which you can find after plugging in $0$ there, which directly implies $f(0)=0+C$ or $C=0$ (because we know that $f(0)$ has to be $0$ by the sum). You will get the same result $C=0$ after plugging in $x=0.5$, for example.
  $endgroup$
  – st.math
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user3711671: The only kicker is that your value of $x$ must be within the interval of convergence for the power series. The center point is typically the simplest to plug in. I explain a bit further in my answer.
  $endgroup$
  – Cameron Buie
  3 hours ago
  
  

  
  
  
  

 | 
show 6 more comments

2

$begingroup$

You're not quite on track. One way to know that for sure is that your function isn't defined at $x=0,$ but the power series certainly is!

For $|t|<1,$ we have that $$sum_n=0^infty t^n=frac11-t,$$ so $$sum_n=1^infty t^n=frac11-t-1=fract1-t.$$ Thus, we have for $|x|<1$ that $$f'(x)=sum_n=1^infty x^3n-1=frac1xsum_n=1left(x^3right)^n=frac1xcdotfracx^31-x^3=left(1-x^3right)^-1cdot-frac13fracdleft(1-x^3right)dx.$$ This has the antiderivative family $$f(x)=-frac13lnleft(1-x^3right)+C,$$ which is defined at every $x$ in the radius of convergence, allowing you to find your constant in the usual way.

---

More generally, let's suppose that you've used a power series $$sum_n=k^infty a_n(x-a)^n$$ for some real $a$ and some integer $kge 0,$ and have (by differentiation and then integration) determined that $$f(x)+C=sum_n=k^infty a_n(x-a)^n$$ for some constant $C.$ Even without determining the radius of convergence, if you haven't made any errors, then the function is necessarily defined at $x=a,$ so you can use $x=a$ to determine what your value of $C$ will be.

Namely, we will have $C=-f(a)$ if $kge 1,$ and $C=a_0-f(a)$ otherwise.

share|cite|improve this answer

edited 1 hour ago

answered 3 hours ago

Cameron BuieCameron Buie

86.6k773161

$endgroup$

add a comment | 

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4

$begingroup$

Differentiating $f(x)$:

$$f(x)=fracx^33+fracx^66+cdotsimplies f'(x)=x^2+x^5+cdots=sum_n=1^inftyx^3n-1=fracx^21-x^3$$ for $|x|<1$, which results in $$f(x)=-frac13ln(1-x^3)+C$$ after integration. Your function now is well-defined at $x=0$, and this implies $C=0$.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

st.mathst.math

1,110115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, I've made a mistake at the geometric series, n starts at 1, not 0.My question still remains, what should one do if plugging in x=0 can't be done (because of functions like log(x), 1/x etc)?The interesting thing is that if log(x) remains untouched, C will equal -log(x).
  $endgroup$
  – user3711671
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then plug in another number, e.g. $x=1$. The constant after integration can be "fixed" by moving the graph of the function up or down the $y$-axis. This can be done by using any point; not only $x=0$.
  $endgroup$
  – st.math
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But what about the infinite sum of 1/3n then?Shouldn't I find out what the constant is by inserting the value for which the sum and it's function representation are equal (when x=0 both are equal to zero)?
  $endgroup$
  – user3711671
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can plug in any point where the function is defined. After plugging in $x=0$ into the sum, you get $f(0)=0$. The closed form you obtained after integration ($f(x)=ln(1-x^3)/3+C$) is undetermined by the constant $C$ only, which you can find after plugging in $0$ there, which directly implies $f(0)=0+C$ or $C=0$ (because we know that $f(0)$ has to be $0$ by the sum). You will get the same result $C=0$ after plugging in $x=0.5$, for example.
  $endgroup$
  – st.math
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user3711671: The only kicker is that your value of $x$ must be within the interval of convergence for the power series. The center point is typically the simplest to plug in. I explain a bit further in my answer.
  $endgroup$
  – Cameron Buie
  3 hours ago
  
  

  
  
  
  

 | 
show 6 more comments

4

$begingroup$

Differentiating $f(x)$:

$$f(x)=fracx^33+fracx^66+cdotsimplies f'(x)=x^2+x^5+cdots=sum_n=1^inftyx^3n-1=fracx^21-x^3$$ for $|x|<1$, which results in $$f(x)=-frac13ln(1-x^3)+C$$ after integration. Your function now is well-defined at $x=0$, and this implies $C=0$.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

st.mathst.math

1,110115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, I've made a mistake at the geometric series, n starts at 1, not 0.My question still remains, what should one do if plugging in x=0 can't be done (because of functions like log(x), 1/x etc)?The interesting thing is that if log(x) remains untouched, C will equal -log(x).
  $endgroup$
  – user3711671
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then plug in another number, e.g. $x=1$. The constant after integration can be "fixed" by moving the graph of the function up or down the $y$-axis. This can be done by using any point; not only $x=0$.
  $endgroup$
  – st.math
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But what about the infinite sum of 1/3n then?Shouldn't I find out what the constant is by inserting the value for which the sum and it's function representation are equal (when x=0 both are equal to zero)?
  $endgroup$
  – user3711671
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can plug in any point where the function is defined. After plugging in $x=0$ into the sum, you get $f(0)=0$. The closed form you obtained after integration ($f(x)=ln(1-x^3)/3+C$) is undetermined by the constant $C$ only, which you can find after plugging in $0$ there, which directly implies $f(0)=0+C$ or $C=0$ (because we know that $f(0)$ has to be $0$ by the sum). You will get the same result $C=0$ after plugging in $x=0.5$, for example.
  $endgroup$
  – st.math
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user3711671: The only kicker is that your value of $x$ must be within the interval of convergence for the power series. The center point is typically the simplest to plug in. I explain a bit further in my answer.
  $endgroup$
  – Cameron Buie
  3 hours ago
  
  

  
  
  
  

 | 
show 6 more comments

$begingroup$

Differentiating $f(x)$:

$$f(x)=fracx^33+fracx^66+cdotsimplies f'(x)=x^2+x^5+cdots=sum_n=1^inftyx^3n-1=fracx^21-x^3$$ for $|x|<1$, which results in $$f(x)=-frac13ln(1-x^3)+C$$ after integration. Your function now is well-defined at $x=0$, and this implies $C=0$.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

st.mathst.math

1,110115

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

st.mathst.math

1,110115

answered 3 hours ago

st.mathst.math

1,110115

answered 3 hours ago

st.mathst.math

1,110115

2

$begingroup$

You're not quite on track. One way to know that for sure is that your function isn't defined at $x=0,$ but the power series certainly is!

For $|t|<1,$ we have that $$sum_n=0^infty t^n=frac11-t,$$ so $$sum_n=1^infty t^n=frac11-t-1=fract1-t.$$ Thus, we have for $|x|<1$ that $$f'(x)=sum_n=1^infty x^3n-1=frac1xsum_n=1left(x^3right)^n=frac1xcdotfracx^31-x^3=left(1-x^3right)^-1cdot-frac13fracdleft(1-x^3right)dx.$$ This has the antiderivative family $$f(x)=-frac13lnleft(1-x^3right)+C,$$ which is defined at every $x$ in the radius of convergence, allowing you to find your constant in the usual way.

---

More generally, let's suppose that you've used a power series $$sum_n=k^infty a_n(x-a)^n$$ for some real $a$ and some integer $kge 0,$ and have (by differentiation and then integration) determined that $$f(x)+C=sum_n=k^infty a_n(x-a)^n$$ for some constant $C.$ Even without determining the radius of convergence, if you haven't made any errors, then the function is necessarily defined at $x=a,$ so you can use $x=a$ to determine what your value of $C$ will be.

Namely, we will have $C=-f(a)$ if $kge 1,$ and $C=a_0-f(a)$ otherwise.

share|cite|improve this answer

edited 1 hour ago

answered 3 hours ago

Cameron BuieCameron Buie

86.6k773161

$endgroup$

add a comment | 

2

$begingroup$

You're not quite on track. One way to know that for sure is that your function isn't defined at $x=0,$ but the power series certainly is!

For $|t|<1,$ we have that $$sum_n=0^infty t^n=frac11-t,$$ so $$sum_n=1^infty t^n=frac11-t-1=fract1-t.$$ Thus, we have for $|x|<1$ that $$f'(x)=sum_n=1^infty x^3n-1=frac1xsum_n=1left(x^3right)^n=frac1xcdotfracx^31-x^3=left(1-x^3right)^-1cdot-frac13fracdleft(1-x^3right)dx.$$ This has the antiderivative family $$f(x)=-frac13lnleft(1-x^3right)+C,$$ which is defined at every $x$ in the radius of convergence, allowing you to find your constant in the usual way.

---

More generally, let's suppose that you've used a power series $$sum_n=k^infty a_n(x-a)^n$$ for some real $a$ and some integer $kge 0,$ and have (by differentiation and then integration) determined that $$f(x)+C=sum_n=k^infty a_n(x-a)^n$$ for some constant $C.$ Even without determining the radius of convergence, if you haven't made any errors, then the function is necessarily defined at $x=a,$ so you can use $x=a$ to determine what your value of $C$ will be.

Namely, we will have $C=-f(a)$ if $kge 1,$ and $C=a_0-f(a)$ otherwise.

share|cite|improve this answer

edited 1 hour ago

answered 3 hours ago

Cameron BuieCameron Buie

86.6k773161

$endgroup$

add a comment | 

$begingroup$

You're not quite on track. One way to know that for sure is that your function isn't defined at $x=0,$ but the power series certainly is!

For $|t|<1,$ we have that $$sum_n=0^infty t^n=frac11-t,$$ so $$sum_n=1^infty t^n=frac11-t-1=fract1-t.$$ Thus, we have for $|x|<1$ that $$f'(x)=sum_n=1^infty x^3n-1=frac1xsum_n=1left(x^3right)^n=frac1xcdotfracx^31-x^3=left(1-x^3right)^-1cdot-frac13fracdleft(1-x^3right)dx.$$ This has the antiderivative family $$f(x)=-frac13lnleft(1-x^3right)+C,$$ which is defined at every $x$ in the radius of convergence, allowing you to find your constant in the usual way.

---

More generally, let's suppose that you've used a power series $$sum_n=k^infty a_n(x-a)^n$$ for some real $a$ and some integer $kge 0,$ and have (by differentiation and then integration) determined that $$f(x)+C=sum_n=k^infty a_n(x-a)^n$$ for some constant $C.$ Even without determining the radius of convergence, if you haven't made any errors, then the function is necessarily defined at $x=a,$ so you can use $x=a$ to determine what your value of $C$ will be.

Namely, we will have $C=-f(a)$ if $kge 1,$ and $C=a_0-f(a)$ otherwise.

share|cite|improve this answer

edited 1 hour ago

answered 3 hours ago

Cameron BuieCameron Buie

86.6k773161

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 3 hours ago

Cameron BuieCameron Buie

86.6k773161

answered 3 hours ago

Cameron BuieCameron Buie

86.6k773161

answered 3 hours ago

Cameron BuieCameron Buie

86.6k773161