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What is the Taylor expansion of this function? [closed]

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0

$begingroup$

I'm trying to solve this exercise.

Establish if and in what interval the function

beginequation
f(x) =
leftlbrace
beginarrayll
fracarctan xx & mboxif $ xneq 0$ \
1 & mboxif $ x=0$.
endarray
right.
endequation

is expandable in Taylor series of initial point $ x_0 = 0 $ and write the expansion. Finally calculate $$ int_0^1 f(x) dx $$ unless an error of $ 10^-3 $.

Can anyone please help me?

real-analysis integration taylor-expansion

share|cite|improve this question

edited Mar 16 at 10:41

blub

2,417726

asked Mar 16 at 10:38

C. BishopC. Bishop

65

$endgroup$

closed as off-topic by Nosrati, Leucippus, José Carlos Santos, Wouter, Thomas Shelby Mar 19 at 11:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Leucippus, José Carlos Santos, Wouter, Thomas Shelby

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $$frac1xarctan x=frac1xintdfrac11+x^2dx=frac1xintsum_ngeq0(-1)^nx^2ndx$$where $|x|<1$.
  $endgroup$
  – Nosrati
  Mar 16 at 10:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I’m sorry, I don’t understand the last equality
  $endgroup$
  – C. Bishop
  Mar 16 at 11:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nosrati has just substituted the equivalent geometric series in the last step i.e. $frac11+r=1-r+r^2-r^3+...$ where $r=x^2$ in this case.
  $endgroup$
  – James Arathoon
  Mar 16 at 12:07
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm trying to solve this exercise.

Establish if and in what interval the function

beginequation
f(x) =
leftlbrace
beginarrayll
fracarctan xx & mboxif $ xneq 0$ \
1 & mboxif $ x=0$.
endarray
right.
endequation

is expandable in Taylor series of initial point $ x_0 = 0 $ and write the expansion. Finally calculate $$ int_0^1 f(x) dx $$ unless an error of $ 10^-3 $.

Can anyone please help me?

real-analysis integration taylor-expansion

share|cite|improve this question

edited Mar 16 at 10:41

blub

2,417726

asked Mar 16 at 10:38

C. BishopC. Bishop

65

$endgroup$

closed as off-topic by Nosrati, Leucippus, José Carlos Santos, Wouter, Thomas Shelby Mar 19 at 11:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Leucippus, José Carlos Santos, Wouter, Thomas Shelby

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $$frac1xarctan x=frac1xintdfrac11+x^2dx=frac1xintsum_ngeq0(-1)^nx^2ndx$$where $|x|<1$.
  $endgroup$
  – Nosrati
  Mar 16 at 10:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I’m sorry, I don’t understand the last equality
  $endgroup$
  – C. Bishop
  Mar 16 at 11:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nosrati has just substituted the equivalent geometric series in the last step i.e. $frac11+r=1-r+r^2-r^3+...$ where $r=x^2$ in this case.
  $endgroup$
  – James Arathoon
  Mar 16 at 12:07
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to solve this exercise.

Establish if and in what interval the function

beginequation
f(x) =
leftlbrace
beginarrayll
fracarctan xx & mboxif $ xneq 0$ \
1 & mboxif $ x=0$.
endarray
right.
endequation

is expandable in Taylor series of initial point $ x_0 = 0 $ and write the expansion. Finally calculate $$ int_0^1 f(x) dx $$ unless an error of $ 10^-3 $.

Can anyone please help me?

real-analysis integration taylor-expansion

share|cite|improve this question

edited Mar 16 at 10:41

blub

2,417726

asked Mar 16 at 10:38

C. BishopC. Bishop

65

$endgroup$

share|cite|improve this question

edited Mar 16 at 10:41

blub

2,417726

asked Mar 16 at 10:38

C. BishopC. Bishop

65

share|cite|improve this question

edited Mar 16 at 10:41

blub

2,417726

asked Mar 16 at 10:38

C. BishopC. Bishop

65

edited Mar 16 at 10:41

blub

2,417726

edited Mar 16 at 10:41

blub

2,417726

asked Mar 16 at 10:38

C. BishopC. Bishop

65

asked Mar 16 at 10:38

C. BishopC. Bishop

65

2 Answers
2

active

oldest

votes

0

$begingroup$

Hint:

You have the standard Taylor's expansion for $arctan$:
$$arctan x=x-fracx^33+fracx^55-dots+(-1)^nfracx^2n+12n+1+dotsm$$

share|cite|improve this answer

answered Mar 16 at 11:01

BernardBernard

123k741117

$endgroup$

add a comment | 

0

$begingroup$

As already recalled by Bernard, we have
$$arctan x=sum_n=0^infty(-1)^nfracx^2n+12n+1$$ So
$$int_0^1 fracarctan xx=sum_n=0^inftyfrac(-1)^n(2n+1)^2$$ Now, consider that
$$sum_n=0^inftyfrac(-1)^n(2n+1)^2=sum_n=0^pfrac(-1)^n(2n+1)^2+sum_n=p+1^inftyfrac(-1)^n(2n+1)^2$$ and you want to know $p$ such that
$$|R_p|=frac 1 (2p+3)^2 leq epsilon$$ that is to say
$$(2p+3)^2 geq frac 1 epsilonimplies p geq frac12 sqrtepsilon -frac32$$ So, by the end
$$p=leftlceil frac12 sqrtepsilon -frac32rightrceil$$

share|cite|improve this answer

answered Mar 16 at 14:33

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

add a comment | 

0

$begingroup$

Hint:

You have the standard Taylor's expansion for $arctan$:
$$arctan x=x-fracx^33+fracx^55-dots+(-1)^nfracx^2n+12n+1+dotsm$$

share|cite|improve this answer

answered Mar 16 at 11:01

BernardBernard

123k741117

$endgroup$

add a comment | 

0

$begingroup$

Hint:

You have the standard Taylor's expansion for $arctan$:
$$arctan x=x-fracx^33+fracx^55-dots+(-1)^nfracx^2n+12n+1+dotsm$$

share|cite|improve this answer

answered Mar 16 at 11:01

BernardBernard

123k741117

$endgroup$

add a comment | 

$begingroup$

Hint:

You have the standard Taylor's expansion for $arctan$:
$$arctan x=x-fracx^33+fracx^55-dots+(-1)^nfracx^2n+12n+1+dotsm$$

share|cite|improve this answer

answered Mar 16 at 11:01

BernardBernard

123k741117

$endgroup$

share|cite|improve this answer

answered Mar 16 at 11:01

BernardBernard

123k741117

answered Mar 16 at 11:01

BernardBernard

123k741117

answered Mar 16 at 11:01

BernardBernard

123k741117

0

$begingroup$

As already recalled by Bernard, we have
$$arctan x=sum_n=0^infty(-1)^nfracx^2n+12n+1$$ So
$$int_0^1 fracarctan xx=sum_n=0^inftyfrac(-1)^n(2n+1)^2$$ Now, consider that
$$sum_n=0^inftyfrac(-1)^n(2n+1)^2=sum_n=0^pfrac(-1)^n(2n+1)^2+sum_n=p+1^inftyfrac(-1)^n(2n+1)^2$$ and you want to know $p$ such that
$$|R_p|=frac 1 (2p+3)^2 leq epsilon$$ that is to say
$$(2p+3)^2 geq frac 1 epsilonimplies p geq frac12 sqrtepsilon -frac32$$ So, by the end
$$p=leftlceil frac12 sqrtepsilon -frac32rightrceil$$

share|cite|improve this answer

answered Mar 16 at 14:33

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

add a comment | 

0

$begingroup$

As already recalled by Bernard, we have
$$arctan x=sum_n=0^infty(-1)^nfracx^2n+12n+1$$ So
$$int_0^1 fracarctan xx=sum_n=0^inftyfrac(-1)^n(2n+1)^2$$ Now, consider that
$$sum_n=0^inftyfrac(-1)^n(2n+1)^2=sum_n=0^pfrac(-1)^n(2n+1)^2+sum_n=p+1^inftyfrac(-1)^n(2n+1)^2$$ and you want to know $p$ such that
$$|R_p|=frac 1 (2p+3)^2 leq epsilon$$ that is to say
$$(2p+3)^2 geq frac 1 epsilonimplies p geq frac12 sqrtepsilon -frac32$$ So, by the end
$$p=leftlceil frac12 sqrtepsilon -frac32rightrceil$$

share|cite|improve this answer

answered Mar 16 at 14:33

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

add a comment | 

$begingroup$

As already recalled by Bernard, we have
$$arctan x=sum_n=0^infty(-1)^nfracx^2n+12n+1$$ So
$$int_0^1 fracarctan xx=sum_n=0^inftyfrac(-1)^n(2n+1)^2$$ Now, consider that
$$sum_n=0^inftyfrac(-1)^n(2n+1)^2=sum_n=0^pfrac(-1)^n(2n+1)^2+sum_n=p+1^inftyfrac(-1)^n(2n+1)^2$$ and you want to know $p$ such that
$$|R_p|=frac 1 (2p+3)^2 leq epsilon$$ that is to say
$$(2p+3)^2 geq frac 1 epsilonimplies p geq frac12 sqrtepsilon -frac32$$ So, by the end
$$p=leftlceil frac12 sqrtepsilon -frac32rightrceil$$

share|cite|improve this answer

answered Mar 16 at 14:33

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

share|cite|improve this answer

answered Mar 16 at 14:33

Claude LeiboviciClaude Leibovici

125k1158135

answered Mar 16 at 14:33

Claude LeiboviciClaude Leibovici

125k1158135

answered Mar 16 at 14:33

Claude LeiboviciClaude Leibovici

125k1158135