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Representing power series as a function - what to do with the constant after integration?
What 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 yesterday
Leucippus
19.8k102871
asked yesterday
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 yesterday
Leucippus
19.8k102871
asked yesterday
user3711671user3711671
438
$endgroup$
$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
yesterday
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 yesterday
Leucippus
19.8k102871
asked yesterday
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 yesterday
Leucippus
19.8k102871
asked yesterday
user3711671user3711671
438
edited yesterday
Leucippus
19.8k102871
asked yesterday
user3711671user3711671
438
edited yesterday
Leucippus
19.8k102871
edited yesterday
Leucippus
19.8k102871
edited yesterday
Leucippus
19.8k102871
19.8k102871
asked yesterday
user3711671user3711671
438
asked yesterday
user3711671user3711671
438
asked yesterday
user3711671user3711671
438
438
$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
yesterday
add a comment |
$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
yesterday
$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
yesterday
$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
yesterday
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 yesterday
Cameron BuieCameron Buie
86.7k773161
$endgroup$
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 yesterday
st.mathst.math
1,113115
$endgroup$
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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
yesterday
|
show 6 more comments
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2 Answers
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2 Answers
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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 yesterday
Cameron BuieCameron Buie
86.7k773161
$endgroup$
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 yesterday
Cameron BuieCameron Buie
86.7k773161
$endgroup$
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 yesterday
Cameron BuieCameron Buie
86.7k773161
$endgroup$
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 yesterday
Cameron BuieCameron Buie
86.7k773161
edited 8 hours ago
answered yesterday
Cameron BuieCameron Buie
86.7k773161
answered yesterday
Cameron BuieCameron Buie
86.7k773161
answered yesterday
Cameron BuieCameron Buie
86.7k773161
86.7k773161
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 yesterday
st.mathst.math
1,113115
$endgroup$
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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
yesterday
|
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 yesterday
st.mathst.math
1,113115
$endgroup$
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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
yesterday
|
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 yesterday
st.mathst.math
1,113115
$endgroup$
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 yesterday
st.mathst.math
1,113115
edited yesterday
answered yesterday
st.mathst.math
1,113115
answered yesterday
st.mathst.math
1,113115
answered yesterday
st.mathst.math
1,113115
1,113115
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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
yesterday
|
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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.
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– Cameron Buie
yesterday
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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
yesterday
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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
yesterday
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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
yesterday
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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
yesterday
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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
yesterday
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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
yesterday
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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
yesterday
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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
yesterday
1
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
yesterday
$begingroup$
@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.
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– Cameron Buie
yesterday
|
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$begingroup$
I was talking about log(x) not being defined at x=0.
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– user3711671
yesterday