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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 $lim_x to 1 sum_n = 0^infty a_nx^n ne infty$

-1

0

$begingroup$

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

share|cite|improve this question

edited Mar 28 at 20:46

J.Doe

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

-1

0

$begingroup$

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

share|cite|improve this question

edited Mar 28 at 20:46

J.Doe

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

share|cite|improve this question

edited Mar 28 at 20:46

J.Doe

asked Mar 28 at 20:12

J.DoeJ.Doe

63

$endgroup$

share|cite|improve this question

edited Mar 28 at 20:46

J.Doe

asked Mar 28 at 20:12

J.DoeJ.Doe

63

share|cite|improve this question

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

2 Answers
2

active

oldest

votes

0

$begingroup$

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.

share|cite|improve this answer

answered Mar 28 at 20:57

Yves DaoustYves Daoust

132k676229

$endgroup$

add a comment | 

0

$begingroup$

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 $x$.

share|cite|improve this answer

edited Mar 28 at 20:59

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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0

$begingroup$

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.

share|cite|improve this answer

answered Mar 28 at 20:57

Yves DaoustYves Daoust

132k676229

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Mar 28 at 20:57

Yves DaoustYves Daoust

132k676229

$endgroup$

share|cite|improve this answer

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

0

$begingroup$

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 $x$.

share|cite|improve this answer

edited Mar 28 at 20:59

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 | 

0

$begingroup$

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 $x$.

share|cite|improve this answer

edited Mar 28 at 20:59

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 $x$.

share|cite|improve this answer

edited Mar 28 at 20:59

answered Mar 28 at 20:32

Gary MoonGary Moon

92127

$endgroup$

share|cite|improve this answer

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