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Verifying that a function is analytic
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Consider the function $$g(x)=begincases frac1xlnfrac11-xhspace.5cm xneq0,\ 1 hspace1.8cm x=0.endcases$$
Let $f$ be the function defined on $(-infty,1)$ by the integral $f(x)=int_0^xg(u)du=int_0^xfrac1ulnfrac11-udu$. Find the Taylor expansion about 0 of $f(x)$ and use this to confirm that the function $f$ is analytic.
My Taylor expansion is the following: $$0+g(0)(x)+fracg'(0)2!x^2+fracg''(0)3!+cdots$$
which equals: $$f(x)=x+frac12fracx^22!+frac23fracx^33!+frac32fracx^44!$$
Which further equals: $$f(x)=sum_k=1^inftyfracx^kk^2.$$
Now I need to prove that this function is analytic, but I am not quite sure how to do that.
real-analysis proof-writing
asked Jul 26 at 2:16
Peetrius
358111
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up vote
2
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Consider the function $$g(x)=begincases frac1xlnfrac11-xhspace.5cm xneq0,\ 1 hspace1.8cm x=0.endcases$$
Let $f$ be the function defined on $(-infty,1)$ by the integral $f(x)=int_0^xg(u)du=int_0^xfrac1ulnfrac11-udu$. Find the Taylor expansion about 0 of $f(x)$ and use this to confirm that the function $f$ is analytic.
My Taylor expansion is the following: $$0+g(0)(x)+fracg'(0)2!x^2+fracg''(0)3!+cdots$$
which equals: $$f(x)=x+frac12fracx^22!+frac23fracx^33!+frac32fracx^44!$$
Which further equals: $$f(x)=sum_k=1^inftyfracx^kk^2.$$
Now I need to prove that this function is analytic, but I am not quite sure how to do that.
real-analysis proof-writing
asked Jul 26 at 2:16
Peetrius
358111
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the function $$g(x)=begincases frac1xlnfrac11-xhspace.5cm xneq0,\ 1 hspace1.8cm x=0.endcases$$
Let $f$ be the function defined on $(-infty,1)$ by the integral $f(x)=int_0^xg(u)du=int_0^xfrac1ulnfrac11-udu$. Find the Taylor expansion about 0 of $f(x)$ and use this to confirm that the function $f$ is analytic.
My Taylor expansion is the following: $$0+g(0)(x)+fracg'(0)2!x^2+fracg''(0)3!+cdots$$
which equals: $$f(x)=x+frac12fracx^22!+frac23fracx^33!+frac32fracx^44!$$
Which further equals: $$f(x)=sum_k=1^inftyfracx^kk^2.$$
Now I need to prove that this function is analytic, but I am not quite sure how to do that.
real-analysis proof-writing
asked Jul 26 at 2:16
Peetrius
358111
Consider the function $$g(x)=begincases frac1xlnfrac11-xhspace.5cm xneq0,\ 1 hspace1.8cm x=0.endcases$$
Let $f$ be the function defined on $(-infty,1)$ by the integral $f(x)=int_0^xg(u)du=int_0^xfrac1ulnfrac11-udu$. Find the Taylor expansion about 0 of $f(x)$ and use this to confirm that the function $f$ is analytic.
My Taylor expansion is the following: $$0+g(0)(x)+fracg'(0)2!x^2+fracg''(0)3!+cdots$$
which equals: $$f(x)=x+frac12fracx^22!+frac23fracx^33!+frac32fracx^44!$$
Which further equals: $$f(x)=sum_k=1^inftyfracx^kk^2.$$
Now I need to prove that this function is analytic, but I am not quite sure how to do that.
real-analysis proof-writing
asked Jul 26 at 2:16
Peetrius
358111
asked Jul 26 at 2:16
Peetrius
358111
asked Jul 26 at 2:16
Peetrius
358111
asked Jul 26 at 2:16
Peetrius
358111
358111
add a comment |Â
add a comment |Â
3 Answers
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up vote
2
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accepted
Recall that a function is analytic in $x=c$ iff there exists an expansion
$$sum_n=0^infty a_n(x-c)^n$$
holds in a neibourhood about $x=c$. With this definition $lndfrac11-x$ is analytic about $x=0$ and indeed for $|x|<1$
$$lndfrac11-x=sum_n=1^infty dfracx^nn$$
also following holds
$$dfrac1xlndfrac11-x=sum_n=1^infty dfracx^n-1n hspace2cm|x|<1$$
and by integration the convergence interval doesn't change
$$f(x)=int_0^xdfrac1ulndfrac11-udu=int_0^xsum_n=1^infty dfracu^n-1ndu=sum_n=1^infty dfracx^nn^2 hspace2cm|x|<1$$
this expansion shows that $f(x)$ is analytic for $|x|<1$.
answered Jul 26 at 2:48
Nosrati
19.3k41544
Thank you very much! This makes a lot more sense.
– Peetrius
Jul 26 at 2:53
1
You are welcome.
– Nosrati
Jul 26 at 2:53
add a comment |Â
up vote
1
down vote
Since you have $$f(x) = sum_k=1^infty fracx^kk^2$$ does it help that $$fracd^nfdx^n=n!sum_k=1^inftyfracx^k-nk^2$$
See Change of Summation and Differentiation
I wanted to leave a comment but don't currently have enough rep. Also see
Alternative Characterizations of Analytic Functions
answered Jul 26 at 2:46
phandaman
676
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up vote
1
down vote
A function is real-analytic at a point if it is equal to its (convergent) Taylor series in some neighborhood of the point.
You found the Taylor series of $g$ around $x = 0$. If you can show it converges for $x$ in some interval containing $0$ then you are finished (at least as far as analyticity at $0$ is concerned.) Hint: ratio test
Also you could have found the Taylor series by using
$$ln(1/(1-u)) = -ln(1-u) = sum_k=1^infty fracu^kk,$$
dividing by $u$ and integrating termwise:
$$int_0^x frac1u ln left(frac11-u right) , du = sum_k=1^infty int_0^x fracu^k-1k , du = sum_k=1^infty fracx^kk^2$$
answered Jul 26 at 2:45
RRL
43.5k42260
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Recall that a function is analytic in $x=c$ iff there exists an expansion
$$sum_n=0^infty a_n(x-c)^n$$
holds in a neibourhood about $x=c$. With this definition $lndfrac11-x$ is analytic about $x=0$ and indeed for $|x|<1$
$$lndfrac11-x=sum_n=1^infty dfracx^nn$$
also following holds
$$dfrac1xlndfrac11-x=sum_n=1^infty dfracx^n-1n hspace2cm|x|<1$$
and by integration the convergence interval doesn't change
$$f(x)=int_0^xdfrac1ulndfrac11-udu=int_0^xsum_n=1^infty dfracu^n-1ndu=sum_n=1^infty dfracx^nn^2 hspace2cm|x|<1$$
this expansion shows that $f(x)$ is analytic for $|x|<1$.
answered Jul 26 at 2:48
Nosrati
19.3k41544
Thank you very much! This makes a lot more sense.
– Peetrius
Jul 26 at 2:53
1
You are welcome.
– Nosrati
Jul 26 at 2:53
add a comment |Â
up vote
2
down vote
accepted
Recall that a function is analytic in $x=c$ iff there exists an expansion
$$sum_n=0^infty a_n(x-c)^n$$
holds in a neibourhood about $x=c$. With this definition $lndfrac11-x$ is analytic about $x=0$ and indeed for $|x|<1$
$$lndfrac11-x=sum_n=1^infty dfracx^nn$$
also following holds
$$dfrac1xlndfrac11-x=sum_n=1^infty dfracx^n-1n hspace2cm|x|<1$$
and by integration the convergence interval doesn't change
$$f(x)=int_0^xdfrac1ulndfrac11-udu=int_0^xsum_n=1^infty dfracu^n-1ndu=sum_n=1^infty dfracx^nn^2 hspace2cm|x|<1$$
this expansion shows that $f(x)$ is analytic for $|x|<1$.
answered Jul 26 at 2:48
Nosrati
19.3k41544
Thank you very much! This makes a lot more sense.
– Peetrius
Jul 26 at 2:53
1
You are welcome.
– Nosrati
Jul 26 at 2:53
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Recall that a function is analytic in $x=c$ iff there exists an expansion
$$sum_n=0^infty a_n(x-c)^n$$
holds in a neibourhood about $x=c$. With this definition $lndfrac11-x$ is analytic about $x=0$ and indeed for $|x|<1$
$$lndfrac11-x=sum_n=1^infty dfracx^nn$$
also following holds
$$dfrac1xlndfrac11-x=sum_n=1^infty dfracx^n-1n hspace2cm|x|<1$$
and by integration the convergence interval doesn't change
$$f(x)=int_0^xdfrac1ulndfrac11-udu=int_0^xsum_n=1^infty dfracu^n-1ndu=sum_n=1^infty dfracx^nn^2 hspace2cm|x|<1$$
this expansion shows that $f(x)$ is analytic for $|x|<1$.
answered Jul 26 at 2:48
Nosrati
19.3k41544
Recall that a function is analytic in $x=c$ iff there exists an expansion
$$sum_n=0^infty a_n(x-c)^n$$
holds in a neibourhood about $x=c$. With this definition $lndfrac11-x$ is analytic about $x=0$ and indeed for $|x|<1$
$$lndfrac11-x=sum_n=1^infty dfracx^nn$$
also following holds
$$dfrac1xlndfrac11-x=sum_n=1^infty dfracx^n-1n hspace2cm|x|<1$$
and by integration the convergence interval doesn't change
$$f(x)=int_0^xdfrac1ulndfrac11-udu=int_0^xsum_n=1^infty dfracu^n-1ndu=sum_n=1^infty dfracx^nn^2 hspace2cm|x|<1$$
this expansion shows that $f(x)$ is analytic for $|x|<1$.
answered Jul 26 at 2:48
Nosrati
19.3k41544
answered Jul 26 at 2:48
Nosrati
19.3k41544
answered Jul 26 at 2:48
Nosrati
19.3k41544
answered Jul 26 at 2:48
Nosrati
19.3k41544
19.3k41544
Thank you very much! This makes a lot more sense.
– Peetrius
Jul 26 at 2:53
1
You are welcome.
– Nosrati
Jul 26 at 2:53
add a comment |Â
Thank you very much! This makes a lot more sense.
– Peetrius
Jul 26 at 2:53
1
You are welcome.
– Nosrati
Jul 26 at 2:53
Thank you very much! This makes a lot more sense.
– Peetrius
Jul 26 at 2:53
Thank you very much! This makes a lot more sense.
– Peetrius
Jul 26 at 2:53
1
1
You are welcome.
– Nosrati
Jul 26 at 2:53
You are welcome.
– Nosrati
Jul 26 at 2:53
add a comment |Â
up vote
1
down vote
Since you have $$f(x) = sum_k=1^infty fracx^kk^2$$ does it help that $$fracd^nfdx^n=n!sum_k=1^inftyfracx^k-nk^2$$
See Change of Summation and Differentiation
I wanted to leave a comment but don't currently have enough rep. Also see
Alternative Characterizations of Analytic Functions
answered Jul 26 at 2:46
phandaman
676
add a comment |Â
up vote
1
down vote
Since you have $$f(x) = sum_k=1^infty fracx^kk^2$$ does it help that $$fracd^nfdx^n=n!sum_k=1^inftyfracx^k-nk^2$$
See Change of Summation and Differentiation
I wanted to leave a comment but don't currently have enough rep. Also see
Alternative Characterizations of Analytic Functions
answered Jul 26 at 2:46
phandaman
676
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since you have $$f(x) = sum_k=1^infty fracx^kk^2$$ does it help that $$fracd^nfdx^n=n!sum_k=1^inftyfracx^k-nk^2$$
See Change of Summation and Differentiation
I wanted to leave a comment but don't currently have enough rep. Also see
Alternative Characterizations of Analytic Functions
answered Jul 26 at 2:46
phandaman
676
Since you have $$f(x) = sum_k=1^infty fracx^kk^2$$ does it help that $$fracd^nfdx^n=n!sum_k=1^inftyfracx^k-nk^2$$
See Change of Summation and Differentiation
I wanted to leave a comment but don't currently have enough rep. Also see
Alternative Characterizations of Analytic Functions
answered Jul 26 at 2:46
phandaman
676
answered Jul 26 at 2:46
phandaman
676
answered Jul 26 at 2:46
phandaman
676
answered Jul 26 at 2:46
phandaman
676
676
add a comment |Â
add a comment |Â
up vote
1
down vote
A function is real-analytic at a point if it is equal to its (convergent) Taylor series in some neighborhood of the point.
You found the Taylor series of $g$ around $x = 0$. If you can show it converges for $x$ in some interval containing $0$ then you are finished (at least as far as analyticity at $0$ is concerned.) Hint: ratio test
Also you could have found the Taylor series by using
$$ln(1/(1-u)) = -ln(1-u) = sum_k=1^infty fracu^kk,$$
dividing by $u$ and integrating termwise:
$$int_0^x frac1u ln left(frac11-u right) , du = sum_k=1^infty int_0^x fracu^k-1k , du = sum_k=1^infty fracx^kk^2$$
answered Jul 26 at 2:45
RRL
43.5k42260
add a comment |Â
up vote
1
down vote
A function is real-analytic at a point if it is equal to its (convergent) Taylor series in some neighborhood of the point.
You found the Taylor series of $g$ around $x = 0$. If you can show it converges for $x$ in some interval containing $0$ then you are finished (at least as far as analyticity at $0$ is concerned.) Hint: ratio test
Also you could have found the Taylor series by using
$$ln(1/(1-u)) = -ln(1-u) = sum_k=1^infty fracu^kk,$$
dividing by $u$ and integrating termwise:
$$int_0^x frac1u ln left(frac11-u right) , du = sum_k=1^infty int_0^x fracu^k-1k , du = sum_k=1^infty fracx^kk^2$$
answered Jul 26 at 2:45
RRL
43.5k42260
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A function is real-analytic at a point if it is equal to its (convergent) Taylor series in some neighborhood of the point.
You found the Taylor series of $g$ around $x = 0$. If you can show it converges for $x$ in some interval containing $0$ then you are finished (at least as far as analyticity at $0$ is concerned.) Hint: ratio test
Also you could have found the Taylor series by using
$$ln(1/(1-u)) = -ln(1-u) = sum_k=1^infty fracu^kk,$$
dividing by $u$ and integrating termwise:
$$int_0^x frac1u ln left(frac11-u right) , du = sum_k=1^infty int_0^x fracu^k-1k , du = sum_k=1^infty fracx^kk^2$$
answered Jul 26 at 2:45
RRL
43.5k42260
A function is real-analytic at a point if it is equal to its (convergent) Taylor series in some neighborhood of the point.
You found the Taylor series of $g$ around $x = 0$. If you can show it converges for $x$ in some interval containing $0$ then you are finished (at least as far as analyticity at $0$ is concerned.) Hint: ratio test
Also you could have found the Taylor series by using
$$ln(1/(1-u)) = -ln(1-u) = sum_k=1^infty fracu^kk,$$
dividing by $u$ and integrating termwise:
$$int_0^x frac1u ln left(frac11-u right) , du = sum_k=1^infty int_0^x fracu^k-1k , du = sum_k=1^infty fracx^kk^2$$
answered Jul 26 at 2:45
RRL
43.5k42260
edited Jul 26 at 2:50
answered Jul 26 at 2:45
RRL
43.5k42260
answered Jul 26 at 2:45
RRL
43.5k42260
answered Jul 26 at 2:45
RRL
43.5k42260
43.5k42260
add a comment |Â
add a comment |Â
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