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Representing power series as a function - what to do with the constant after integration?
The 2019 Stack Overflow Developer Survey Results Are InWhat does it really mean for the power series of a function to converge?Maclaurin Series with Power in Denominator?Function that Represents Divergent Power Series?Finding the power series of a logarithmic function.Writing $frac11 + w + w^2$ as a power series and finding the ROCPower series expansion of a complex functionPower series representing a rational functionFinding a function corresponding to the complex power seriesFind the sum for the power serieshow to find a function expression of a power series?
$begingroup$
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
edited Apr 7 at 19:10
Leucippus
19.8k102871
asked Apr 7 at 16:49
user3711671user3711671
438
$endgroup$
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
edited Apr 7 at 19:10
Leucippus
19.8k102871
asked Apr 7 at 16:49
user3711671user3711671
438
$endgroup$
$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
Apr 7 at 17:09
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
edited Apr 7 at 19:10
Leucippus
19.8k102871
asked Apr 7 at 16:49
user3711671user3711671
438
$endgroup$
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
power-series
edited Apr 7 at 19:10
Leucippus
19.8k102871
asked Apr 7 at 16:49
user3711671user3711671
438
edited Apr 7 at 19:10
Leucippus
19.8k102871
asked Apr 7 at 16:49
user3711671user3711671
438
edited Apr 7 at 19:10
Leucippus
19.8k102871
edited Apr 7 at 19:10
Leucippus
19.8k102871
edited Apr 7 at 19:10
Leucippus
19.8k102871
19.8k102871
asked Apr 7 at 16:49
user3711671user3711671
438
asked Apr 7 at 16:49
user3711671user3711671
438
asked Apr 7 at 16:49
user3711671user3711671
438
438
$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
Apr 7 at 17:09
add a comment |
$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
Apr 7 at 17:09
$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
Apr 7 at 17:09
$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
Apr 7 at 17:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $$sum_n=0^infty a_n(x-a)^n$$ for some real $a,$ and have (by differentiation and then integration) determined that $$f(x)=C+sum_n=0^infty a_n(x-a)^n$$ 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 $C=f(a)-a_0.$
answered Apr 7 at 17:07
Cameron BuieCameron Buie
86.8k773161
$endgroup$
add a comment |
$begingroup$
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 $C=0$.
answered Apr 7 at 16:56
st.mathst.math
1,143115
$endgroup$
$begingroup$
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
Apr 7 at 17:09
$begingroup$
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
Apr 7 at 17:10
$begingroup$
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
Apr 7 at 17:13
$begingroup$
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
Apr 7 at 17:15
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
Apr 7 at 17:28
|
show 6 more comments
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2 Answers
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active
oldest
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2 Answers
2
active
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$begingroup$
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 $$sum_n=0^infty a_n(x-a)^n$$ for some real $a,$ and have (by differentiation and then integration) determined that $$f(x)=C+sum_n=0^infty a_n(x-a)^n$$ 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 $C=f(a)-a_0.$
answered Apr 7 at 17:07
Cameron BuieCameron Buie
86.8k773161
$endgroup$
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 $$sum_n=0^infty a_n(x-a)^n$$ for some real $a,$ and have (by differentiation and then integration) determined that $$f(x)=C+sum_n=0^infty a_n(x-a)^n$$ 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 $C=f(a)-a_0.$
answered Apr 7 at 17:07
Cameron BuieCameron Buie
86.8k773161
$endgroup$
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 $$sum_n=0^infty a_n(x-a)^n$$ for some real $a,$ and have (by differentiation and then integration) determined that $$f(x)=C+sum_n=0^infty a_n(x-a)^n$$ 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 $C=f(a)-a_0.$
answered Apr 7 at 17:07
Cameron BuieCameron Buie
86.8k773161
$endgroup$
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 $$sum_n=0^infty a_n(x-a)^n$$ for some real $a,$ and have (by differentiation and then integration) determined that $$f(x)=C+sum_n=0^infty a_n(x-a)^n$$ 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 $C=f(a)-a_0.$
answered Apr 7 at 17:07
Cameron BuieCameron Buie
86.8k773161
edited Apr 8 at 17:58
answered Apr 7 at 17:07
Cameron BuieCameron Buie
86.8k773161
answered Apr 7 at 17:07
Cameron BuieCameron Buie
86.8k773161
answered Apr 7 at 17:07
Cameron BuieCameron Buie
86.8k773161
86.8k773161
add a comment |
add a comment |
$begingroup$
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 $C=0$.
answered Apr 7 at 16:56
st.mathst.math
1,143115
$endgroup$
$begingroup$
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
Apr 7 at 17:09
$begingroup$
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
Apr 7 at 17:10
$begingroup$
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
Apr 7 at 17:13
$begingroup$
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
Apr 7 at 17:15
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
Apr 7 at 17:28
|
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 $C=0$.
answered Apr 7 at 16:56
st.mathst.math
1,143115
$endgroup$
$begingroup$
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
Apr 7 at 17:09
$begingroup$
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
Apr 7 at 17:10
$begingroup$
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
Apr 7 at 17:13
$begingroup$
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
Apr 7 at 17:15
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
Apr 7 at 17:28
|
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 $C=0$.
answered Apr 7 at 16:56
st.mathst.math
1,143115
$endgroup$
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 $C=0$.
answered Apr 7 at 16:56
st.mathst.math
1,143115
edited Apr 7 at 17:04
answered Apr 7 at 16:56
st.mathst.math
1,143115
answered Apr 7 at 16:56
st.mathst.math
1,143115
answered Apr 7 at 16:56
st.mathst.math
1,143115
1,143115
$begingroup$
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
Apr 7 at 17:09
$begingroup$
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
Apr 7 at 17:10
$begingroup$
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
Apr 7 at 17:13
$begingroup$
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
Apr 7 at 17:15
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
Apr 7 at 17:28
|
show 6 more comments
$begingroup$
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
Apr 7 at 17:09
$begingroup$
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
Apr 7 at 17:10
$begingroup$
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)?
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– user3711671
Apr 7 at 17:13
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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.
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– st.math
Apr 7 at 17:15
1
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@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.
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– Cameron Buie
Apr 7 at 17:28
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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).
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– user3711671
Apr 7 at 17:09
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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).
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– user3711671
Apr 7 at 17:09
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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 $x=0$.
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– st.math
Apr 7 at 17:10
$begingroup$
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
Apr 7 at 17:10
$begingroup$
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
Apr 7 at 17:13
$begingroup$
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
Apr 7 at 17:13
$begingroup$
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
Apr 7 at 17:15
$begingroup$
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
Apr 7 at 17:15
1
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
Apr 7 at 17:28
$begingroup$
@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
Apr 7 at 17:28
|
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$begingroup$
I was talking about log(x) not being defined at x=0.
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– user3711671
Apr 7 at 17:09