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Showing that there exists a solution to an equation and writing its Taylor polynomial.

Write down the equation of the tangent plane and compute the Taylor series of the functionEstimate the degree of a Taylor Polynomial using its Error Termproof that Even powers of an odd function's taylor polynomial vanishFind a polynomial $Q$ of degree $k$ and a remainder function $E$ for $f(x)=frac{1}{1-x}$.Taylor polynomial QuestionProve that $T_n(x^2)$ is taylor polynomial of $f(x^2)$Can the $n^{th}$ order Taylor polynomial ever model its function perfectly?Taylor polynomial and remainderTaylor polynomial and integrationCalculating Taylor Polynomial and its error

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I need to show that for the equation $e^{y}+2y+x=1$, there exists a solution to $y=f(x)$ near $x=y=0$, and then I need to write $f(x)$ as its 3rd degree Taylor polynomial expanded at $x=0$. Is there any hint I could have to figure this out?

multivariable-calculus taylor-expansion implicit-function-theorem

share|cite|improve this question

asked Mar 19 at 3:51

numericalorangenumericalorange

1,879313

$endgroup$

add a comment | 

0

$begingroup$

I need to show that for the equation $e^{y}+2y+x=1$, there exists a solution to $y=f(x)$ near $x=y=0$, and then I need to write $f(x)$ as its 3rd degree Taylor polynomial expanded at $x=0$. Is there any hint I could have to figure this out?

multivariable-calculus taylor-expansion implicit-function-theorem

share|cite|improve this question

asked Mar 19 at 3:51

numericalorangenumericalorange

1,879313

$endgroup$

add a comment | 

$begingroup$

I need to show that for the equation $e^{y}+2y+x=1$, there exists a solution to $y=f(x)$ near $x=y=0$, and then I need to write $f(x)$ as its 3rd degree Taylor polynomial expanded at $x=0$. Is there any hint I could have to figure this out?

multivariable-calculus taylor-expansion implicit-function-theorem

share|cite|improve this question

asked Mar 19 at 3:51

numericalorangenumericalorange

1,879313

$endgroup$

share|cite|improve this question

asked Mar 19 at 3:51

numericalorangenumericalorange

1,879313

share|cite|improve this question

asked Mar 19 at 3:51

numericalorangenumericalorange

1,879313

asked Mar 19 at 3:51

numericalorangenumericalorange

1,879313

asked Mar 19 at 3:51

numericalorangenumericalorange

1,879313

2 Answers
2

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oldest

votes

1

$begingroup$

$$e^y+2y+x=1 tag 1$$
Obviously $(x=0,y=0)$ is solution. Expanding to series around this point :

$$y=a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...$$

$$e^{a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...}+2(a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...)+x-1=0$$

Expanding to series leads to :
$$(3a_1+1)x+(3a_2+frac12 a_1^2)x^2+(3a_3+a_1a_2+frac16 a_1^3)x^3+...=0$$
Then we have to solve
$$3a_1+1=0$$
$$3a_2+frac12 a_1^2=0$$
$$3a_3+a_1a_2+frac16 a_1^3=0$$
$$text{etc.}$$
The result is :
$$a_1=-frac13$$
$$a_2=-frac{1}{54}$$
$$a_3=0$$
$$a_4=frac{1}{8748}$$
$$text{etc.}$$
$$y(x)=-frac13 x -frac{1}{54}x^2+frac{1}{8748}x^4+frac{1}{196830}x^5+...$$
NOTE for information :

This equation can be solved analytically, but it requires a special function.

Of course, this is not the kind of answer expected by the OP.
$$frac12 e^y=-y+frac{1-x}{2}$$
$$(-y+frac{1-x}{2})e^{-y}=frac12$$
$$(-y+frac{1-x}{2})e^{-y+frac{1-x}{2}}=frac12e^{frac{1-x}{2}} tag 2$$
The solution of an equation of this kind :
$$Xe^X=Y$$
cannot be expressed with a finit number of elementary functions. It requires a special function, namely the Lambert W function :
$$X=W(Y)$$
http://mathworld.wolfram.com/LambertW-Function.html

With $quad X=(-y+frac{1-x}{2})quad$ and $quad Y=frac12e^{frac{1-x}{2}}quad$ Eq.$(2)$ becomes $Xe^X=Y$ which solution is :
$$(-y+frac{1-x}{2})=Wleft(frac12e^{frac{1-x}{2}} right)$$
The solution of Eq.$(1)$ is :
$$y=frac{1-x}{2} - Wleft(frac12e^{frac{1-x}{2}}right)$$
Expanding the Lambert W function to series would lead to the same result as above.

share|cite|improve this answer

answered Mar 19 at 9:20

JJacquelinJJacquelin

45.3k21856

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, you really took the time in helping me understand and I am truly grateful. Thank you so much for your help!
  $endgroup$
  – numericalorange
  Mar 21 at 19:09
  

  
  
  
  

add a comment | 

1

$begingroup$

For the first part, this is the implicit function theorem. For the second part, I suggest the following. Differentiate your equation to get
$y'(x)e^{y(x)}+2y'(x)+1=0.$ Then, at $x=0$, knowing that $y(0) = 0$), we get $y'(0)+2y'(0) +1=0.$ Hence, $y'(0) = -frac{1}{3}.$ For $y''(0)$ and $y'''(0)$, same idea.

share|cite|improve this answer

answered Mar 19 at 4:21

PebetoPebeto

5015

$endgroup$

add a comment | 

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1

$begingroup$

$$e^y+2y+x=1 tag 1$$
Obviously $(x=0,y=0)$ is solution. Expanding to series around this point :

$$y=a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...$$

$$e^{a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...}+2(a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...)+x-1=0$$

Expanding to series leads to :
$$(3a_1+1)x+(3a_2+frac12 a_1^2)x^2+(3a_3+a_1a_2+frac16 a_1^3)x^3+...=0$$
Then we have to solve
$$3a_1+1=0$$
$$3a_2+frac12 a_1^2=0$$
$$3a_3+a_1a_2+frac16 a_1^3=0$$
$$text{etc.}$$
The result is :
$$a_1=-frac13$$
$$a_2=-frac{1}{54}$$
$$a_3=0$$
$$a_4=frac{1}{8748}$$
$$text{etc.}$$
$$y(x)=-frac13 x -frac{1}{54}x^2+frac{1}{8748}x^4+frac{1}{196830}x^5+...$$
NOTE for information :

This equation can be solved analytically, but it requires a special function.

Of course, this is not the kind of answer expected by the OP.
$$frac12 e^y=-y+frac{1-x}{2}$$
$$(-y+frac{1-x}{2})e^{-y}=frac12$$
$$(-y+frac{1-x}{2})e^{-y+frac{1-x}{2}}=frac12e^{frac{1-x}{2}} tag 2$$
The solution of an equation of this kind :
$$Xe^X=Y$$
cannot be expressed with a finit number of elementary functions. It requires a special function, namely the Lambert W function :
$$X=W(Y)$$
http://mathworld.wolfram.com/LambertW-Function.html

With $quad X=(-y+frac{1-x}{2})quad$ and $quad Y=frac12e^{frac{1-x}{2}}quad$ Eq.$(2)$ becomes $Xe^X=Y$ which solution is :
$$(-y+frac{1-x}{2})=Wleft(frac12e^{frac{1-x}{2}} right)$$
The solution of Eq.$(1)$ is :
$$y=frac{1-x}{2} - Wleft(frac12e^{frac{1-x}{2}}right)$$
Expanding the Lambert W function to series would lead to the same result as above.

share|cite|improve this answer

answered Mar 19 at 9:20

JJacquelinJJacquelin

45.3k21856

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, you really took the time in helping me understand and I am truly grateful. Thank you so much for your help!
  $endgroup$
  – numericalorange
  Mar 21 at 19:09
  

  
  
  
  

add a comment | 

1

$begingroup$

$$e^y+2y+x=1 tag 1$$
Obviously $(x=0,y=0)$ is solution. Expanding to series around this point :

$$y=a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...$$

$$e^{a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...}+2(a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...)+x-1=0$$

Expanding to series leads to :
$$(3a_1+1)x+(3a_2+frac12 a_1^2)x^2+(3a_3+a_1a_2+frac16 a_1^3)x^3+...=0$$
Then we have to solve
$$3a_1+1=0$$
$$3a_2+frac12 a_1^2=0$$
$$3a_3+a_1a_2+frac16 a_1^3=0$$
$$text{etc.}$$
The result is :
$$a_1=-frac13$$
$$a_2=-frac{1}{54}$$
$$a_3=0$$
$$a_4=frac{1}{8748}$$
$$text{etc.}$$
$$y(x)=-frac13 x -frac{1}{54}x^2+frac{1}{8748}x^4+frac{1}{196830}x^5+...$$
NOTE for information :

This equation can be solved analytically, but it requires a special function.

Of course, this is not the kind of answer expected by the OP.
$$frac12 e^y=-y+frac{1-x}{2}$$
$$(-y+frac{1-x}{2})e^{-y}=frac12$$
$$(-y+frac{1-x}{2})e^{-y+frac{1-x}{2}}=frac12e^{frac{1-x}{2}} tag 2$$
The solution of an equation of this kind :
$$Xe^X=Y$$
cannot be expressed with a finit number of elementary functions. It requires a special function, namely the Lambert W function :
$$X=W(Y)$$
http://mathworld.wolfram.com/LambertW-Function.html

With $quad X=(-y+frac{1-x}{2})quad$ and $quad Y=frac12e^{frac{1-x}{2}}quad$ Eq.$(2)$ becomes $Xe^X=Y$ which solution is :
$$(-y+frac{1-x}{2})=Wleft(frac12e^{frac{1-x}{2}} right)$$
The solution of Eq.$(1)$ is :
$$y=frac{1-x}{2} - Wleft(frac12e^{frac{1-x}{2}}right)$$
Expanding the Lambert W function to series would lead to the same result as above.

share|cite|improve this answer

answered Mar 19 at 9:20

JJacquelinJJacquelin

45.3k21856

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, you really took the time in helping me understand and I am truly grateful. Thank you so much for your help!
  $endgroup$
  – numericalorange
  Mar 21 at 19:09
  

  
  
  
  

add a comment | 

$begingroup$

$$e^y+2y+x=1 tag 1$$
Obviously $(x=0,y=0)$ is solution. Expanding to series around this point :

$$y=a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...$$

$$e^{a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...}+2(a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...)+x-1=0$$

Expanding to series leads to :
$$(3a_1+1)x+(3a_2+frac12 a_1^2)x^2+(3a_3+a_1a_2+frac16 a_1^3)x^3+...=0$$
Then we have to solve
$$3a_1+1=0$$
$$3a_2+frac12 a_1^2=0$$
$$3a_3+a_1a_2+frac16 a_1^3=0$$
$$text{etc.}$$
The result is :
$$a_1=-frac13$$
$$a_2=-frac{1}{54}$$
$$a_3=0$$
$$a_4=frac{1}{8748}$$
$$text{etc.}$$
$$y(x)=-frac13 x -frac{1}{54}x^2+frac{1}{8748}x^4+frac{1}{196830}x^5+...$$
NOTE for information :

This equation can be solved analytically, but it requires a special function.

Of course, this is not the kind of answer expected by the OP.
$$frac12 e^y=-y+frac{1-x}{2}$$
$$(-y+frac{1-x}{2})e^{-y}=frac12$$
$$(-y+frac{1-x}{2})e^{-y+frac{1-x}{2}}=frac12e^{frac{1-x}{2}} tag 2$$
The solution of an equation of this kind :
$$Xe^X=Y$$
cannot be expressed with a finit number of elementary functions. It requires a special function, namely the Lambert W function :
$$X=W(Y)$$
http://mathworld.wolfram.com/LambertW-Function.html

With $quad X=(-y+frac{1-x}{2})quad$ and $quad Y=frac12e^{frac{1-x}{2}}quad$ Eq.$(2)$ becomes $Xe^X=Y$ which solution is :
$$(-y+frac{1-x}{2})=Wleft(frac12e^{frac{1-x}{2}} right)$$
The solution of Eq.$(1)$ is :
$$y=frac{1-x}{2} - Wleft(frac12e^{frac{1-x}{2}}right)$$
Expanding the Lambert W function to series would lead to the same result as above.

share|cite|improve this answer

answered Mar 19 at 9:20

JJacquelinJJacquelin

45.3k21856

$endgroup$

share|cite|improve this answer

answered Mar 19 at 9:20

JJacquelinJJacquelin

45.3k21856

answered Mar 19 at 9:20

JJacquelinJJacquelin

45.3k21856

answered Mar 19 at 9:20

JJacquelinJJacquelin

45.3k21856

1

$begingroup$

For the first part, this is the implicit function theorem. For the second part, I suggest the following. Differentiate your equation to get
$y'(x)e^{y(x)}+2y'(x)+1=0.$ Then, at $x=0$, knowing that $y(0) = 0$), we get $y'(0)+2y'(0) +1=0.$ Hence, $y'(0) = -frac{1}{3}.$ For $y''(0)$ and $y'''(0)$, same idea.

share|cite|improve this answer

answered Mar 19 at 4:21

PebetoPebeto

5015

$endgroup$

add a comment | 

1

$begingroup$

For the first part, this is the implicit function theorem. For the second part, I suggest the following. Differentiate your equation to get
$y'(x)e^{y(x)}+2y'(x)+1=0.$ Then, at $x=0$, knowing that $y(0) = 0$), we get $y'(0)+2y'(0) +1=0.$ Hence, $y'(0) = -frac{1}{3}.$ For $y''(0)$ and $y'''(0)$, same idea.

share|cite|improve this answer

answered Mar 19 at 4:21

PebetoPebeto

5015

$endgroup$

add a comment | 

$begingroup$

For the first part, this is the implicit function theorem. For the second part, I suggest the following. Differentiate your equation to get
$y'(x)e^{y(x)}+2y'(x)+1=0.$ Then, at $x=0$, knowing that $y(0) = 0$), we get $y'(0)+2y'(0) +1=0.$ Hence, $y'(0) = -frac{1}{3}.$ For $y''(0)$ and $y'''(0)$, same idea.

share|cite|improve this answer

answered Mar 19 at 4:21

PebetoPebeto

5015

$endgroup$

share|cite|improve this answer

answered Mar 19 at 4:21

PebetoPebeto

5015

answered Mar 19 at 4:21

PebetoPebeto

5015

answered Mar 19 at 4:21

PebetoPebeto

5015