Compute $sumlimits_{n=0}^infty a_nx^n$ if $a_0=3$, $a_1=5$, and $na_n=frac23a_{n-1}-(n-1)a_{n-1}$ for every...
$begingroup$
Assume that $a_0=3$, $a_1=5$, and, for arbitrary $n>1$ , $na_n=frac{2}{3}a_{n-1}-(n-1)a_{n-1}$. Prove that, when $|x|<1$, the series $sumlimits_{n=0}^infty a_nx^n$ converges, and compute its sum.
I tried to let $displaystyle a_n-a_1=sum_{k=2}^{n}frac{5-6k}{3k}a_{n-1}$ , and $displaystyle a_n=(frac{5-3n}{3n})a_{n-1}$
$$
a_{n-1}(frac{5-3n}{3n}-sum_{k=2}^{n}frac{5-6k}{3k})=5
$$
I want to know how to continue it.
Edit: (after reading ideas by @JV.Stalker)
I made the following supplement
$$
S(x)=sum_{n=0}^{infty}a_nx^n\
S'(x)=sum_{n=1}^{infty}na_nx^{n-1}\
sum_{n=2}^{infty}na_nx^n=sum_{n=2}^{infty}frac{2}{3}a_{n-1}x^n-sum_{n=2}^{infty}(n-1)a_{n-1}x^n\
[xS'(x)-5x]=frac{2}{3}x·sum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}-xsum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}\
x[S'(x)-5]=frac{2}{3}x(S(x)-3)-x(xS'(x))\
S'(x)-5=S(x)-3-xS'(x)\
(x+1)S'(x)=frac{2}{3}S(x)+3\
S'(x)-frac{2}{3}frac{1}{x+1}S(x)=frac{3}{x+1}\
S(x)=c(x+1)^{frac{2}{3}}-frac{9}{2}\
S(0)=a_0=3\
c=frac{15}{2}\
S(x)=frac{15}{2}(x+1)^{frac{2}{3}}-frac{9}{2}
$$
sequences-and-series power-series
edited Dec 14 '18 at 11:14
Did
248k23223460
asked Dec 10 '18 at 10:59
Jack throneJack throne
184
$endgroup$
add a comment |
$begingroup$
Assume that $a_0=3$, $a_1=5$, and, for arbitrary $n>1$ , $na_n=frac{2}{3}a_{n-1}-(n-1)a_{n-1}$. Prove that, when $|x|<1$, the series $sumlimits_{n=0}^infty a_nx^n$ converges, and compute its sum.
I tried to let $displaystyle a_n-a_1=sum_{k=2}^{n}frac{5-6k}{3k}a_{n-1}$ , and $displaystyle a_n=(frac{5-3n}{3n})a_{n-1}$
$$
a_{n-1}(frac{5-3n}{3n}-sum_{k=2}^{n}frac{5-6k}{3k})=5
$$
I want to know how to continue it.
Edit: (after reading ideas by @JV.Stalker)
I made the following supplement
$$
S(x)=sum_{n=0}^{infty}a_nx^n\
S'(x)=sum_{n=1}^{infty}na_nx^{n-1}\
sum_{n=2}^{infty}na_nx^n=sum_{n=2}^{infty}frac{2}{3}a_{n-1}x^n-sum_{n=2}^{infty}(n-1)a_{n-1}x^n\
[xS'(x)-5x]=frac{2}{3}x·sum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}-xsum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}\
x[S'(x)-5]=frac{2}{3}x(S(x)-3)-x(xS'(x))\
S'(x)-5=S(x)-3-xS'(x)\
(x+1)S'(x)=frac{2}{3}S(x)+3\
S'(x)-frac{2}{3}frac{1}{x+1}S(x)=frac{3}{x+1}\
S(x)=c(x+1)^{frac{2}{3}}-frac{9}{2}\
S(0)=a_0=3\
c=frac{15}{2}\
S(x)=frac{15}{2}(x+1)^{frac{2}{3}}-frac{9}{2}
$$
sequences-and-series power-series
edited Dec 14 '18 at 11:14
Did
248k23223460
asked Dec 10 '18 at 10:59
Jack throneJack throne
184
$endgroup$
$begingroup$
$a_{n-1}$ twice and two conditions ?
$endgroup$
– Claude Leibovici
Dec 10 '18 at 11:05
$begingroup$
@ClaudeLeibovici I don't know why it's written like this, but it's correct
$endgroup$
– Jack throne
Dec 10 '18 at 11:11
$begingroup$
Did you check the answer below before accepting it?
$endgroup$
– Did
Dec 14 '18 at 11:06
$begingroup$
Maybe I made a mistake as a freshman, but @JV.Stalker really helped me, also thanks very much for your remind and let me know the regulation of the community.
$endgroup$
– Jack throne
Dec 14 '18 at 14:22
add a comment |
$begingroup$
Assume that $a_0=3$, $a_1=5$, and, for arbitrary $n>1$ , $na_n=frac{2}{3}a_{n-1}-(n-1)a_{n-1}$. Prove that, when $|x|<1$, the series $sumlimits_{n=0}^infty a_nx^n$ converges, and compute its sum.
I tried to let $displaystyle a_n-a_1=sum_{k=2}^{n}frac{5-6k}{3k}a_{n-1}$ , and $displaystyle a_n=(frac{5-3n}{3n})a_{n-1}$
$$
a_{n-1}(frac{5-3n}{3n}-sum_{k=2}^{n}frac{5-6k}{3k})=5
$$
I want to know how to continue it.
Edit: (after reading ideas by @JV.Stalker)
I made the following supplement
$$
S(x)=sum_{n=0}^{infty}a_nx^n\
S'(x)=sum_{n=1}^{infty}na_nx^{n-1}\
sum_{n=2}^{infty}na_nx^n=sum_{n=2}^{infty}frac{2}{3}a_{n-1}x^n-sum_{n=2}^{infty}(n-1)a_{n-1}x^n\
[xS'(x)-5x]=frac{2}{3}x·sum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}-xsum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}\
x[S'(x)-5]=frac{2}{3}x(S(x)-3)-x(xS'(x))\
S'(x)-5=S(x)-3-xS'(x)\
(x+1)S'(x)=frac{2}{3}S(x)+3\
S'(x)-frac{2}{3}frac{1}{x+1}S(x)=frac{3}{x+1}\
S(x)=c(x+1)^{frac{2}{3}}-frac{9}{2}\
S(0)=a_0=3\
c=frac{15}{2}\
S(x)=frac{15}{2}(x+1)^{frac{2}{3}}-frac{9}{2}
$$
sequences-and-series power-series
edited Dec 14 '18 at 11:14
Did
248k23223460
asked Dec 10 '18 at 10:59
Jack throneJack throne
184
$endgroup$
Assume that $a_0=3$, $a_1=5$, and, for arbitrary $n>1$ , $na_n=frac{2}{3}a_{n-1}-(n-1)a_{n-1}$. Prove that, when $|x|<1$, the series $sumlimits_{n=0}^infty a_nx^n$ converges, and compute its sum.
I tried to let $displaystyle a_n-a_1=sum_{k=2}^{n}frac{5-6k}{3k}a_{n-1}$ , and $displaystyle a_n=(frac{5-3n}{3n})a_{n-1}$
$$
a_{n-1}(frac{5-3n}{3n}-sum_{k=2}^{n}frac{5-6k}{3k})=5
$$
I want to know how to continue it.
Edit: (after reading ideas by @JV.Stalker)
I made the following supplement
$$
S(x)=sum_{n=0}^{infty}a_nx^n\
S'(x)=sum_{n=1}^{infty}na_nx^{n-1}\
sum_{n=2}^{infty}na_nx^n=sum_{n=2}^{infty}frac{2}{3}a_{n-1}x^n-sum_{n=2}^{infty}(n-1)a_{n-1}x^n\
[xS'(x)-5x]=frac{2}{3}x·sum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}-xsum_{n=2}^{infty}(n-1)a_{n-1}x^{n-1}\
x[S'(x)-5]=frac{2}{3}x(S(x)-3)-x(xS'(x))\
S'(x)-5=S(x)-3-xS'(x)\
(x+1)S'(x)=frac{2}{3}S(x)+3\
S'(x)-frac{2}{3}frac{1}{x+1}S(x)=frac{3}{x+1}\
S(x)=c(x+1)^{frac{2}{3}}-frac{9}{2}\
S(0)=a_0=3\
c=frac{15}{2}\
S(x)=frac{15}{2}(x+1)^{frac{2}{3}}-frac{9}{2}
$$
sequences-and-series power-series
sequences-and-series power-series
edited Dec 14 '18 at 11:14
Did
248k23223460
asked Dec 10 '18 at 10:59
Jack throneJack throne
184
edited Dec 14 '18 at 11:14
Did
248k23223460
asked Dec 10 '18 at 10:59
Jack throneJack throne
184
edited Dec 14 '18 at 11:14
Did
248k23223460
edited Dec 14 '18 at 11:14
Did
248k23223460
edited Dec 14 '18 at 11:14
Did
248k23223460
248k23223460
asked Dec 10 '18 at 10:59
Jack throneJack throne
184
asked Dec 10 '18 at 10:59
Jack throneJack throne
184
asked Dec 10 '18 at 10:59
Jack throneJack throne
184
184
$begingroup$
$a_{n-1}$ twice and two conditions ?
$endgroup$
– Claude Leibovici
Dec 10 '18 at 11:05
$begingroup$
@ClaudeLeibovici I don't know why it's written like this, but it's correct
$endgroup$
– Jack throne
Dec 10 '18 at 11:11
$begingroup$
Did you check the answer below before accepting it?
$endgroup$
– Did
Dec 14 '18 at 11:06
$begingroup$
Maybe I made a mistake as a freshman, but @JV.Stalker really helped me, also thanks very much for your remind and let me know the regulation of the community.
$endgroup$
– Jack throne
Dec 14 '18 at 14:22
add a comment |
$begingroup$
$a_{n-1}$ twice and two conditions ?
$endgroup$
– Claude Leibovici
Dec 10 '18 at 11:05
$begingroup$
@ClaudeLeibovici I don't know why it's written like this, but it's correct
$endgroup$
– Jack throne
Dec 10 '18 at 11:11
$begingroup$
Did you check the answer below before accepting it?
$endgroup$
– Did
Dec 14 '18 at 11:06
$begingroup$
Maybe I made a mistake as a freshman, but @JV.Stalker really helped me, also thanks very much for your remind and let me know the regulation of the community.
$endgroup$
– Jack throne
Dec 14 '18 at 14:22
$begingroup$
$a_{n-1}$ twice and two conditions ?
$endgroup$
– Claude Leibovici
Dec 10 '18 at 11:05
$begingroup$
$a_{n-1}$ twice and two conditions ?
$endgroup$
– Claude Leibovici
Dec 10 '18 at 11:05
$begingroup$
@ClaudeLeibovici I don't know why it's written like this, but it's correct
$endgroup$
– Jack throne
Dec 10 '18 at 11:11
$begingroup$
@ClaudeLeibovici I don't know why it's written like this, but it's correct
$endgroup$
– Jack throne
Dec 10 '18 at 11:11
$begingroup$
Did you check the answer below before accepting it?
$endgroup$
– Did
Dec 14 '18 at 11:06
$begingroup$
Did you check the answer below before accepting it?
$endgroup$
– Did
Dec 14 '18 at 11:06
$begingroup$
Maybe I made a mistake as a freshman, but @JV.Stalker really helped me, also thanks very much for your remind and let me know the regulation of the community.
$endgroup$
– Jack throne
Dec 14 '18 at 14:22
$begingroup$
Maybe I made a mistake as a freshman, but @JV.Stalker really helped me, also thanks very much for your remind and let me know the regulation of the community.
$endgroup$
– Jack throne
Dec 14 '18 at 14:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$na_n=frac{5}{3}a_{n-1}-na_{n-1}$
Multiply by $x^n$ both sides and sum from $n=1$ to $infty$
$sumlimits_{n=1}^infty na_n x^n=frac{5}{3}sumlimits_{n=1}^infty a_{n-1}x^n-sumlimits_{n=1}^infty na_{n-1}x^n$
Reindex of the RHS:
$sumlimits_{n=1}^infty na_nx^n=frac{5}{3}xsumlimits_{n=0}^infty a_{n}x^n-xsumlimits_{n=0}^infty (n+1)a_{n}x^n$
After sorting the eqution:
$sumlimits_{n=1}^infty na_nx^n+xsumlimits_{n=0}^infty na_{n}x^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$
$(1+x)sumlimits_{n=0}^infty na_nx^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$
Use that $nx^{n-1}=frac{dx^n}{dx}$ and let $f(x)=sumlimits_{n=0}^infty a_nx^n$
We have the following differential eqution:
$(x+1)frac{df(x)}{dx}=frac{2}{3}f(x)$
Finally $f(x)=(x+1)^frac{2}{3}+c$
$sumlimits_{n=1}^infty a_n x^n$ is convergent.
answered Dec 10 '18 at 14:24
JV.StalkerJV.Stalker
68339
$endgroup$
$begingroup$
I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
$endgroup$
– JV.Stalker
Dec 10 '18 at 14:30
$begingroup$
I made the following supplement after your method, and Thanks again! @JV.Stalker
$endgroup$
– Jack throne
Dec 13 '18 at 13:46
$begingroup$
@Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
$endgroup$
– JV.Stalker
Dec 13 '18 at 15:30
$begingroup$
"Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
$endgroup$
– Did
Dec 14 '18 at 11:05
$begingroup$
Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
$endgroup$
– Did
Dec 14 '18 at 11:08
|
show 1 more comment
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$begingroup$
$na_n=frac{5}{3}a_{n-1}-na_{n-1}$
Multiply by $x^n$ both sides and sum from $n=1$ to $infty$
$sumlimits_{n=1}^infty na_n x^n=frac{5}{3}sumlimits_{n=1}^infty a_{n-1}x^n-sumlimits_{n=1}^infty na_{n-1}x^n$
Reindex of the RHS:
$sumlimits_{n=1}^infty na_nx^n=frac{5}{3}xsumlimits_{n=0}^infty a_{n}x^n-xsumlimits_{n=0}^infty (n+1)a_{n}x^n$
After sorting the eqution:
$sumlimits_{n=1}^infty na_nx^n+xsumlimits_{n=0}^infty na_{n}x^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$
$(1+x)sumlimits_{n=0}^infty na_nx^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$
Use that $nx^{n-1}=frac{dx^n}{dx}$ and let $f(x)=sumlimits_{n=0}^infty a_nx^n$
We have the following differential eqution:
$(x+1)frac{df(x)}{dx}=frac{2}{3}f(x)$
Finally $f(x)=(x+1)^frac{2}{3}+c$
$sumlimits_{n=1}^infty a_n x^n$ is convergent.
answered Dec 10 '18 at 14:24
JV.StalkerJV.Stalker
68339
$endgroup$
$begingroup$
I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
$endgroup$
– JV.Stalker
Dec 10 '18 at 14:30
$begingroup$
I made the following supplement after your method, and Thanks again! @JV.Stalker
$endgroup$
– Jack throne
Dec 13 '18 at 13:46
$begingroup$
@Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
$endgroup$
– JV.Stalker
Dec 13 '18 at 15:30
$begingroup$
"Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
$endgroup$
– Did
Dec 14 '18 at 11:05
$begingroup$
Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
$endgroup$
– Did
Dec 14 '18 at 11:08
|
show 1 more comment
$begingroup$
$na_n=frac{5}{3}a_{n-1}-na_{n-1}$
Multiply by $x^n$ both sides and sum from $n=1$ to $infty$
$sumlimits_{n=1}^infty na_n x^n=frac{5}{3}sumlimits_{n=1}^infty a_{n-1}x^n-sumlimits_{n=1}^infty na_{n-1}x^n$
Reindex of the RHS:
$sumlimits_{n=1}^infty na_nx^n=frac{5}{3}xsumlimits_{n=0}^infty a_{n}x^n-xsumlimits_{n=0}^infty (n+1)a_{n}x^n$
After sorting the eqution:
$sumlimits_{n=1}^infty na_nx^n+xsumlimits_{n=0}^infty na_{n}x^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$
$(1+x)sumlimits_{n=0}^infty na_nx^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$
Use that $nx^{n-1}=frac{dx^n}{dx}$ and let $f(x)=sumlimits_{n=0}^infty a_nx^n$
We have the following differential eqution:
$(x+1)frac{df(x)}{dx}=frac{2}{3}f(x)$
Finally $f(x)=(x+1)^frac{2}{3}+c$
$sumlimits_{n=1}^infty a_n x^n$ is convergent.
answered Dec 10 '18 at 14:24
JV.StalkerJV.Stalker
68339
$endgroup$
$begingroup$
I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
$endgroup$
– JV.Stalker
Dec 10 '18 at 14:30
$begingroup$
I made the following supplement after your method, and Thanks again! @JV.Stalker
$endgroup$
– Jack throne
Dec 13 '18 at 13:46
$begingroup$
@Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
$endgroup$
– JV.Stalker
Dec 13 '18 at 15:30
$begingroup$
"Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
$endgroup$
– Did
Dec 14 '18 at 11:05
$begingroup$
Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
$endgroup$
– Did
Dec 14 '18 at 11:08
|
show 1 more comment
$begingroup$
$na_n=frac{5}{3}a_{n-1}-na_{n-1}$
Multiply by $x^n$ both sides and sum from $n=1$ to $infty$
$sumlimits_{n=1}^infty na_n x^n=frac{5}{3}sumlimits_{n=1}^infty a_{n-1}x^n-sumlimits_{n=1}^infty na_{n-1}x^n$
Reindex of the RHS:
$sumlimits_{n=1}^infty na_nx^n=frac{5}{3}xsumlimits_{n=0}^infty a_{n}x^n-xsumlimits_{n=0}^infty (n+1)a_{n}x^n$
After sorting the eqution:
$sumlimits_{n=1}^infty na_nx^n+xsumlimits_{n=0}^infty na_{n}x^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$
$(1+x)sumlimits_{n=0}^infty na_nx^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$
Use that $nx^{n-1}=frac{dx^n}{dx}$ and let $f(x)=sumlimits_{n=0}^infty a_nx^n$
We have the following differential eqution:
$(x+1)frac{df(x)}{dx}=frac{2}{3}f(x)$
Finally $f(x)=(x+1)^frac{2}{3}+c$
$sumlimits_{n=1}^infty a_n x^n$ is convergent.
answered Dec 10 '18 at 14:24
JV.StalkerJV.Stalker
68339
$endgroup$
$na_n=frac{5}{3}a_{n-1}-na_{n-1}$
Multiply by $x^n$ both sides and sum from $n=1$ to $infty$
$sumlimits_{n=1}^infty na_n x^n=frac{5}{3}sumlimits_{n=1}^infty a_{n-1}x^n-sumlimits_{n=1}^infty na_{n-1}x^n$
Reindex of the RHS:
$sumlimits_{n=1}^infty na_nx^n=frac{5}{3}xsumlimits_{n=0}^infty a_{n}x^n-xsumlimits_{n=0}^infty (n+1)a_{n}x^n$
After sorting the eqution:
$sumlimits_{n=1}^infty na_nx^n+xsumlimits_{n=0}^infty na_{n}x^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$
$(1+x)sumlimits_{n=0}^infty na_nx^n=frac{2}{3}xsumlimits_{n=0}^infty a_{n}x^n$
Use that $nx^{n-1}=frac{dx^n}{dx}$ and let $f(x)=sumlimits_{n=0}^infty a_nx^n$
We have the following differential eqution:
$(x+1)frac{df(x)}{dx}=frac{2}{3}f(x)$
Finally $f(x)=(x+1)^frac{2}{3}+c$
$sumlimits_{n=1}^infty a_n x^n$ is convergent.
answered Dec 10 '18 at 14:24
JV.StalkerJV.Stalker
68339
answered Dec 10 '18 at 14:24
JV.StalkerJV.Stalker
68339
answered Dec 10 '18 at 14:24
JV.StalkerJV.Stalker
68339
answered Dec 10 '18 at 14:24
JV.StalkerJV.Stalker
68339
68339
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I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
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– JV.Stalker
Dec 10 '18 at 14:30
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I made the following supplement after your method, and Thanks again! @JV.Stalker
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– Jack throne
Dec 13 '18 at 13:46
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@Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
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– JV.Stalker
Dec 13 '18 at 15:30
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"Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
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– Did
Dec 14 '18 at 11:05
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Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
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– Did
Dec 14 '18 at 11:08
|
show 1 more comment
$begingroup$
I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
$endgroup$
– JV.Stalker
Dec 10 '18 at 14:30
$begingroup$
I made the following supplement after your method, and Thanks again! @JV.Stalker
$endgroup$
– Jack throne
Dec 13 '18 at 13:46
$begingroup$
@Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
$endgroup$
– JV.Stalker
Dec 13 '18 at 15:30
$begingroup$
"Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
$endgroup$
– Did
Dec 14 '18 at 11:05
$begingroup$
Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
$endgroup$
– Did
Dec 14 '18 at 11:08
$begingroup$
I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
$endgroup$
– JV.Stalker
Dec 10 '18 at 14:30
$begingroup$
I had two unusual things when I saw the result, the first is that I did not use the values ${a_0}$ and ${a_1}$, the second is that the sum is convergent for every real x.
$endgroup$
– JV.Stalker
Dec 10 '18 at 14:30
$begingroup$
I made the following supplement after your method, and Thanks again! @JV.Stalker
$endgroup$
– Jack throne
Dec 13 '18 at 13:46
$begingroup$
I made the following supplement after your method, and Thanks again! @JV.Stalker
$endgroup$
– Jack throne
Dec 13 '18 at 13:46
$begingroup$
@Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
$endgroup$
– JV.Stalker
Dec 13 '18 at 15:30
$begingroup$
@Jack throne, It was my pleasure, if it was useful please accept my solution. Thank you.
$endgroup$
– JV.Stalker
Dec 13 '18 at 15:30
$begingroup$
"Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
$endgroup$
– Did
Dec 14 '18 at 11:05
$begingroup$
"Finally $f(x)=(x+1)^frac{2}{3}+c$" No, $(x+1)y'(x)=ay(x)$ is not solved by $y(x)=(x+1)^a+c$. "$sumlimits_{n=1}^infty a_n x^n$ is convergent" ?? 1. Why? 2. For which $x$s?
$endgroup$
– Did
Dec 14 '18 at 11:05
$begingroup$
Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
$endgroup$
– Did
Dec 14 '18 at 11:08
$begingroup$
Oh, and $a_1=5ne2=frac23a_0$ hence the expression of $f(x)$ is wrong. "the first is that I did not use the values a0 and a1" Indeed, this signals the solution is wrong. "the second is that the sum is convergent for every real x" It is not.
$endgroup$
– Did
Dec 14 '18 at 11:08
|
show 1 more comment
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$a_{n-1}$ twice and two conditions ?
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– Claude Leibovici
Dec 10 '18 at 11:05
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@ClaudeLeibovici I don't know why it's written like this, but it's correct
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– Jack throne
Dec 10 '18 at 11:11
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Did you check the answer below before accepting it?
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– Did
Dec 14 '18 at 11:06
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Maybe I made a mistake as a freshman, but @JV.Stalker really helped me, also thanks very much for your remind and let me know the regulation of the community.
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– Jack throne
Dec 14 '18 at 14:22