The taylor series is given determine $a_n$ for which x converges to f(x)Taylor Series of ProductsHow to conclude the convergence of the this Taylor Series $sumlimits_n=1^infty a_n x^n$Taylor series QuestionGiven a power series with interval of convergence $(-1,1]$, construct a series with another given interval of convergenceTaylor series of a composed functionIf one series converges than the other doesConvergence of the series $sum_nsqrta_ncdot a_n+1$ given that $sum_na_n$ convergesFor which $n$ the given series convergesDetermine the values of $x$ for which the given power series converges.A power series $sum_n = 0^infty a_nx^n$ such that $sum_n=0^infty a_n= +infty$ but $lim_x to 1 sum_n = 0^infty a_nx^n ne infty$
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The taylor series is given determine $a_n$ for which x converges to f(x)
Taylor Series of ProductsHow to conclude the convergence of the this Taylor Series $sumlimits_n=1^infty a_n x^n$Taylor series QuestionGiven a power series with interval of convergence $(-1,1]$, construct a series with another given interval of convergenceTaylor series of a composed functionIf one series converges than the other doesConvergence of the series $sum_nsqrta_ncdot a_n+1$ given that $sum_na_n$ convergesFor which $n$ the given series convergesDetermine the values of $x$ for which the given power series converges.A power series $sum_n = 0^infty a_nx^n$ such that $sum_n=0^infty a_n= +infty$ but $lim_x to 1 sum_n = 0^infty a_nx^n ne infty$
$begingroup$
The taylor series of the function f(x) = $1-mathrme^-x^2$ around x = 0 is given by $sum_n=0^infty a_nx^n$
determine $a_n$ for all n $geq$ 0 and give for which value of x the series converges to f(x)
I've come to this result $sum_n=1^infty dfracleft(-1right)^n+1x^2nn!$ = $sum_n=0^infty a_nx^n$
Anyone got ideas how to solve this further on?
calculus sequences-and-series summation power-series taylor-expansion
asked Mar 28 at 20:12
J.DoeJ.Doe
63
$endgroup$
|
show 3 more comments
$begingroup$
The taylor series of the function f(x) = $1-mathrme^-x^2$ around x = 0 is given by $sum_n=0^infty a_nx^n$
determine $a_n$ for all n $geq$ 0 and give for which value of x the series converges to f(x)
I've come to this result $sum_n=1^infty dfracleft(-1right)^n+1x^2nn!$ = $sum_n=0^infty a_nx^n$
Anyone got ideas how to solve this further on?
calculus sequences-and-series summation power-series taylor-expansion
asked Mar 28 at 20:12
J.DoeJ.Doe
63
$endgroup$
$begingroup$
You are trying to find radius of convergence right?
$endgroup$
– Nimish
Mar 28 at 20:27
$begingroup$
Unfortunately, you $a_n$'s are wrong.
$endgroup$
– Yves Daoust
Mar 28 at 20:38
$begingroup$
@J.Doe: you just edited, didn't you ?
$endgroup$
– Yves Daoust
Mar 28 at 20:40
$begingroup$
Also fix your $a^n$.
$endgroup$
– Yves Daoust
Mar 28 at 20:42
$begingroup$
No, pay attention. $a^nleftrightarrow a_n$.
$endgroup$
– Yves Daoust
Mar 28 at 20:45
|
show 3 more comments
$begingroup$
The taylor series of the function f(x) = $1-mathrme^-x^2$ around x = 0 is given by $sum_n=0^infty a_nx^n$
determine $a_n$ for all n $geq$ 0 and give for which value of x the series converges to f(x)
I've come to this result $sum_n=1^infty dfracleft(-1right)^n+1x^2nn!$ = $sum_n=0^infty a_nx^n$
Anyone got ideas how to solve this further on?
calculus sequences-and-series summation power-series taylor-expansion
asked Mar 28 at 20:12
J.DoeJ.Doe
63
$endgroup$
The taylor series of the function f(x) = $1-mathrme^-x^2$ around x = 0 is given by $sum_n=0^infty a_nx^n$
determine $a_n$ for all n $geq$ 0 and give for which value of x the series converges to f(x)
I've come to this result $sum_n=1^infty dfracleft(-1right)^n+1x^2nn!$ = $sum_n=0^infty a_nx^n$
Anyone got ideas how to solve this further on?
calculus sequences-and-series summation power-series taylor-expansion
calculus sequences-and-series summation power-series taylor-expansion
asked Mar 28 at 20:12
J.DoeJ.Doe
63
asked Mar 28 at 20:12
J.DoeJ.Doe
63
edited Mar 28 at 20:46
J.Doe
asked Mar 28 at 20:12
J.DoeJ.Doe
63
asked Mar 28 at 20:12
J.DoeJ.Doe
63
asked Mar 28 at 20:12
J.DoeJ.Doe
63
63
$begingroup$
You are trying to find radius of convergence right?
$endgroup$
– Nimish
Mar 28 at 20:27
$begingroup$
Unfortunately, you $a_n$'s are wrong.
$endgroup$
– Yves Daoust
Mar 28 at 20:38
$begingroup$
@J.Doe: you just edited, didn't you ?
$endgroup$
– Yves Daoust
Mar 28 at 20:40
$begingroup$
Also fix your $a^n$.
$endgroup$
– Yves Daoust
Mar 28 at 20:42
$begingroup$
No, pay attention. $a^nleftrightarrow a_n$.
$endgroup$
– Yves Daoust
Mar 28 at 20:45
|
show 3 more comments
$begingroup$
You are trying to find radius of convergence right?
$endgroup$
– Nimish
Mar 28 at 20:27
$begingroup$
Unfortunately, you $a_n$'s are wrong.
$endgroup$
– Yves Daoust
Mar 28 at 20:38
$begingroup$
@J.Doe: you just edited, didn't you ?
$endgroup$
– Yves Daoust
Mar 28 at 20:40
$begingroup$
Also fix your $a^n$.
$endgroup$
– Yves Daoust
Mar 28 at 20:42
$begingroup$
No, pay attention. $a^nleftrightarrow a_n$.
$endgroup$
– Yves Daoust
Mar 28 at 20:45
$begingroup$
You are trying to find radius of convergence right?
$endgroup$
– Nimish
Mar 28 at 20:27
$begingroup$
You are trying to find radius of convergence right?
$endgroup$
– Nimish
Mar 28 at 20:27
$begingroup$
Unfortunately, you $a_n$'s are wrong.
$endgroup$
– Yves Daoust
Mar 28 at 20:38
$begingroup$
Unfortunately, you $a_n$'s are wrong.
$endgroup$
– Yves Daoust
Mar 28 at 20:38
$begingroup$
@J.Doe: you just edited, didn't you ?
$endgroup$
– Yves Daoust
Mar 28 at 20:40
$begingroup$
@J.Doe: you just edited, didn't you ?
$endgroup$
– Yves Daoust
Mar 28 at 20:40
$begingroup$
Also fix your $a^n$.
$endgroup$
– Yves Daoust
Mar 28 at 20:42
$begingroup$
Also fix your $a^n$.
$endgroup$
– Yves Daoust
Mar 28 at 20:42
$begingroup$
No, pay attention. $a^nleftrightarrow a_n$.
$endgroup$
– Yves Daoust
Mar 28 at 20:45
$begingroup$
No, pay attention. $a^nleftrightarrow a_n$.
$endgroup$
– Yves Daoust
Mar 28 at 20:45
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The exponential function has the well-known Taylor series
$$sum_k=0^infty fracx^kk!$$ which converges for all real values.
Substituting $-x^2$ for $x$, you will get an entire series, which is indeed the Taylor series of $e^-x^2$.
You can conclude.
answered Mar 28 at 20:57
Yves DaoustYves Daoust
132k676229
$endgroup$
add a comment |
$begingroup$
It seems that next you want to find the radius of convergence $R$. You can use the ratio test to see that $R = infty$. Consider
$$bigglvert frac(-1)^n+2x^2(n+1)(n+1)! cdot fracn!(-1)^n+1x^2nbiggrvert= fraclvert x rvert^2n+1 to 0 text as n to infty.$$
Edit: Since $R=infty$, it follows that the power series converges (to $f(x)$) for all real $x$.
answered Mar 28 at 20:32
Gary MoonGary Moon
92127
$endgroup$
$begingroup$
I misstated my question i meant you have to find an for which it converges
$endgroup$
– J.Doe
Mar 28 at 20:37
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
The exponential function has the well-known Taylor series
$$sum_k=0^infty fracx^kk!$$ which converges for all real values.
Substituting $-x^2$ for $x$, you will get an entire series, which is indeed the Taylor series of $e^-x^2$.
You can conclude.
answered Mar 28 at 20:57
Yves DaoustYves Daoust
132k676229
$endgroup$
add a comment |
$begingroup$
The exponential function has the well-known Taylor series
$$sum_k=0^infty fracx^kk!$$ which converges for all real values.
Substituting $-x^2$ for $x$, you will get an entire series, which is indeed the Taylor series of $e^-x^2$.
You can conclude.
answered Mar 28 at 20:57
Yves DaoustYves Daoust
132k676229
$endgroup$
add a comment |
$begingroup$
The exponential function has the well-known Taylor series
$$sum_k=0^infty fracx^kk!$$ which converges for all real values.
Substituting $-x^2$ for $x$, you will get an entire series, which is indeed the Taylor series of $e^-x^2$.
You can conclude.
answered Mar 28 at 20:57
Yves DaoustYves Daoust
132k676229
$endgroup$
The exponential function has the well-known Taylor series
$$sum_k=0^infty fracx^kk!$$ which converges for all real values.
Substituting $-x^2$ for $x$, you will get an entire series, which is indeed the Taylor series of $e^-x^2$.
You can conclude.
answered Mar 28 at 20:57
Yves DaoustYves Daoust
132k676229
answered Mar 28 at 20:57
Yves DaoustYves Daoust
132k676229
answered Mar 28 at 20:57
Yves DaoustYves Daoust
132k676229
answered Mar 28 at 20:57
Yves DaoustYves Daoust
132k676229
132k676229
add a comment |
add a comment |
$begingroup$
It seems that next you want to find the radius of convergence $R$. You can use the ratio test to see that $R = infty$. Consider
$$bigglvert frac(-1)^n+2x^2(n+1)(n+1)! cdot fracn!(-1)^n+1x^2nbiggrvert= fraclvert x rvert^2n+1 to 0 text as n to infty.$$
Edit: Since $R=infty$, it follows that the power series converges (to $f(x)$) for all real $x$.
answered Mar 28 at 20:32
Gary MoonGary Moon
92127
$endgroup$
$begingroup$
I misstated my question i meant you have to find an for which it converges
$endgroup$
– J.Doe
Mar 28 at 20:37
add a comment |
$begingroup$
It seems that next you want to find the radius of convergence $R$. You can use the ratio test to see that $R = infty$. Consider
$$bigglvert frac(-1)^n+2x^2(n+1)(n+1)! cdot fracn!(-1)^n+1x^2nbiggrvert= fraclvert x rvert^2n+1 to 0 text as n to infty.$$
Edit: Since $R=infty$, it follows that the power series converges (to $f(x)$) for all real $x$.
answered Mar 28 at 20:32
Gary MoonGary Moon
92127
$endgroup$
$begingroup$
I misstated my question i meant you have to find an for which it converges
$endgroup$
– J.Doe
Mar 28 at 20:37
add a comment |
$begingroup$
It seems that next you want to find the radius of convergence $R$. You can use the ratio test to see that $R = infty$. Consider
$$bigglvert frac(-1)^n+2x^2(n+1)(n+1)! cdot fracn!(-1)^n+1x^2nbiggrvert= fraclvert x rvert^2n+1 to 0 text as n to infty.$$
Edit: Since $R=infty$, it follows that the power series converges (to $f(x)$) for all real $x$.
answered Mar 28 at 20:32
Gary MoonGary Moon
92127
$endgroup$
It seems that next you want to find the radius of convergence $R$. You can use the ratio test to see that $R = infty$. Consider
$$bigglvert frac(-1)^n+2x^2(n+1)(n+1)! cdot fracn!(-1)^n+1x^2nbiggrvert= fraclvert x rvert^2n+1 to 0 text as n to infty.$$
Edit: Since $R=infty$, it follows that the power series converges (to $f(x)$) for all real $x$.
answered Mar 28 at 20:32
Gary MoonGary Moon
92127
edited Mar 28 at 20:59
answered Mar 28 at 20:32
Gary MoonGary Moon
92127
answered Mar 28 at 20:32
Gary MoonGary Moon
92127
answered Mar 28 at 20:32
Gary MoonGary Moon
92127
92127
$begingroup$
I misstated my question i meant you have to find an for which it converges
$endgroup$
– J.Doe
Mar 28 at 20:37
add a comment |
$begingroup$
I misstated my question i meant you have to find an for which it converges
$endgroup$
– J.Doe
Mar 28 at 20:37
$begingroup$
I misstated my question i meant you have to find an for which it converges
$endgroup$
– J.Doe
Mar 28 at 20:37
$begingroup$
I misstated my question i meant you have to find an for which it converges
$endgroup$
– J.Doe
Mar 28 at 20:37
add a comment |
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$begingroup$
You are trying to find radius of convergence right?
$endgroup$
– Nimish
Mar 28 at 20:27
$begingroup$
Unfortunately, you $a_n$'s are wrong.
$endgroup$
– Yves Daoust
Mar 28 at 20:38
$begingroup$
@J.Doe: you just edited, didn't you ?
$endgroup$
– Yves Daoust
Mar 28 at 20:40
$begingroup$
Also fix your $a^n$.
$endgroup$
– Yves Daoust
Mar 28 at 20:42
$begingroup$
No, pay attention. $a^nleftrightarrow a_n$.
$endgroup$
– Yves Daoust
Mar 28 at 20:45