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1

$begingroup$

I am trying to find the roots of the equation

$$
e^{x} -cos x = 0.
$$

Used the Lambert W function to arrive at
$$
x = W(xcos x),
$$
but I don't know how to proceed from there to get the explicit roots. Any help is much welcome

logarithms

share|cite|improve this question

asked 9 hours ago

DYBnorDYBnor

10466 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you will need a numerical method.
  $endgroup$
  – Dr. Sonnhard Graubner
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sonnhard what numerical method would you suggest?
  $endgroup$
  – DYBnor
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Newton Raphson method
  $endgroup$
  – Dr. Sonnhard Graubner
  8 hours ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

I am trying to find the roots of the equation

$$
e^{x} -cos x = 0.
$$

Used the Lambert W function to arrive at
$$
x = W(xcos x),
$$
but I don't know how to proceed from there to get the explicit roots. Any help is much welcome

logarithms

share|cite|improve this question

asked 9 hours ago

DYBnorDYBnor

10466 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you will need a numerical method.
  $endgroup$
  – Dr. Sonnhard Graubner
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sonnhard what numerical method would you suggest?
  $endgroup$
  – DYBnor
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Newton Raphson method
  $endgroup$
  – Dr. Sonnhard Graubner
  8 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

I am trying to find the roots of the equation

$$
e^{x} -cos x = 0.
$$

Used the Lambert W function to arrive at
$$
x = W(xcos x),
$$
but I don't know how to proceed from there to get the explicit roots. Any help is much welcome

logarithms

share|cite|improve this question

asked 9 hours ago

DYBnorDYBnor

10466 bronze badges

$endgroup$

share|cite|improve this question

asked 9 hours ago

DYBnorDYBnor

10466 bronze badges

share|cite|improve this question

asked 9 hours ago

DYBnorDYBnor

10466 bronze badges

asked 9 hours ago

DYBnorDYBnor

10466 bronze badges

asked 9 hours ago

DYBnorDYBnor

10466 bronze badges

3 Answers
3

active

oldest

votes

2

$begingroup$

Since $-1le cos xle 1$ and $e^x>0$ with $e^x>1, ; x>0$, by IVT we have infinitely many solutions for $xle 0$, one is the trivial $x=0$ the others can by found by numerical methods.

Notably we have $2$ not trivial roots for each pair of intervals with $x<0$ such that $0<cos x<1$ that is for $n=0,1,2,dots$ that is

$$-frac{pi}2 -2pi n pi <x<-2pi n$$

$$-2pi-2pi n<x<-frac32 pi -2pi n$$

Note also that for $|x|$ very large, roots are well approximated by the roots for $cos x=0$ with $x<0$ that is $xapprox -frac{pi}2-npi$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

gimusigimusi

98.1k99 gold badges4747 silver badges9696 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

You could use numerical methods

You could use maclaurin's expansion

$e^x-cos x = x+x^2+frac{x^3}{6}$

Upto 3 degree would be good

Now this becomes a cubic equation
$x=0$

Now this becomes a quadratic equation

$1+x+frac{x^2}{6}$
Roots are -4.725,-1.281

share|cite|improve this answer

answered 8 hours ago

Harshit GuptaHarshit Gupta

6211313 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

A supposed hint is that by using the Mclaurin-series for both the $e^x$ and $cos(x)$, we can arrive at this;

$$x^2sum_{k=0}^{infty}frac{x^{4k}}{(4k+2)!}+frac{1}{2}sinhleft(xright)=0$$

Now take,

$a=sum_{k=0}^{infty}frac{x^{4k}}{(4k+2)!}$

Now we can get different expressions for the same thing;
$$lnleft(-2x^{2}a+sqrt{4x^{4}a^{2}+1}right)-x=0$$
$$e^{x}4x^{2}a+e^{2x}-1=0$$

share|cite|improve this answer

answered 7 hours ago

fishfagfishfag

29077 bronze badges

New contributor

fishfag is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment
 | 

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2

$begingroup$

Since $-1le cos xle 1$ and $e^x>0$ with $e^x>1, ; x>0$, by IVT we have infinitely many solutions for $xle 0$, one is the trivial $x=0$ the others can by found by numerical methods.

Notably we have $2$ not trivial roots for each pair of intervals with $x<0$ such that $0<cos x<1$ that is for $n=0,1,2,dots$ that is

$$-frac{pi}2 -2pi n pi <x<-2pi n$$

$$-2pi-2pi n<x<-frac32 pi -2pi n$$

Note also that for $|x|$ very large, roots are well approximated by the roots for $cos x=0$ with $x<0$ that is $xapprox -frac{pi}2-npi$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

gimusigimusi

98.1k99 gold badges4747 silver badges9696 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

Since $-1le cos xle 1$ and $e^x>0$ with $e^x>1, ; x>0$, by IVT we have infinitely many solutions for $xle 0$, one is the trivial $x=0$ the others can by found by numerical methods.

Notably we have $2$ not trivial roots for each pair of intervals with $x<0$ such that $0<cos x<1$ that is for $n=0,1,2,dots$ that is

$$-frac{pi}2 -2pi n pi <x<-2pi n$$

$$-2pi-2pi n<x<-frac32 pi -2pi n$$

Note also that for $|x|$ very large, roots are well approximated by the roots for $cos x=0$ with $x<0$ that is $xapprox -frac{pi}2-npi$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

gimusigimusi

98.1k99 gold badges4747 silver badges9696 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Since $-1le cos xle 1$ and $e^x>0$ with $e^x>1, ; x>0$, by IVT we have infinitely many solutions for $xle 0$, one is the trivial $x=0$ the others can by found by numerical methods.

Notably we have $2$ not trivial roots for each pair of intervals with $x<0$ such that $0<cos x<1$ that is for $n=0,1,2,dots$ that is

$$-frac{pi}2 -2pi n pi <x<-2pi n$$

$$-2pi-2pi n<x<-frac32 pi -2pi n$$

Note also that for $|x|$ very large, roots are well approximated by the roots for $cos x=0$ with $x<0$ that is $xapprox -frac{pi}2-npi$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

gimusigimusi

98.1k99 gold badges4747 silver badges9696 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

gimusigimusi

98.1k99 gold badges4747 silver badges9696 bronze badges

answered 8 hours ago

gimusigimusi

98.1k99 gold badges4747 silver badges9696 bronze badges

answered 8 hours ago

gimusigimusi

98.1k99 gold badges4747 silver badges9696 bronze badges

0

$begingroup$

You could use numerical methods

You could use maclaurin's expansion

$e^x-cos x = x+x^2+frac{x^3}{6}$

Upto 3 degree would be good

Now this becomes a cubic equation
$x=0$

Now this becomes a quadratic equation

$1+x+frac{x^2}{6}$
Roots are -4.725,-1.281

share|cite|improve this answer

answered 8 hours ago

Harshit GuptaHarshit Gupta

6211313 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

You could use numerical methods

You could use maclaurin's expansion

$e^x-cos x = x+x^2+frac{x^3}{6}$

Upto 3 degree would be good

Now this becomes a cubic equation
$x=0$

Now this becomes a quadratic equation

$1+x+frac{x^2}{6}$
Roots are -4.725,-1.281

share|cite|improve this answer

answered 8 hours ago

Harshit GuptaHarshit Gupta

6211313 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

You could use numerical methods

You could use maclaurin's expansion

$e^x-cos x = x+x^2+frac{x^3}{6}$

Upto 3 degree would be good

Now this becomes a cubic equation
$x=0$

Now this becomes a quadratic equation

$1+x+frac{x^2}{6}$
Roots are -4.725,-1.281

share|cite|improve this answer

answered 8 hours ago

Harshit GuptaHarshit Gupta

6211313 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Harshit GuptaHarshit Gupta

6211313 bronze badges

answered 8 hours ago

Harshit GuptaHarshit Gupta

6211313 bronze badges

answered 8 hours ago

Harshit GuptaHarshit Gupta

6211313 bronze badges

0

$begingroup$

A supposed hint is that by using the Mclaurin-series for both the $e^x$ and $cos(x)$, we can arrive at this;

$$x^2sum_{k=0}^{infty}frac{x^{4k}}{(4k+2)!}+frac{1}{2}sinhleft(xright)=0$$

Now take,

$a=sum_{k=0}^{infty}frac{x^{4k}}{(4k+2)!}$

Now we can get different expressions for the same thing;
$$lnleft(-2x^{2}a+sqrt{4x^{4}a^{2}+1}right)-x=0$$
$$e^{x}4x^{2}a+e^{2x}-1=0$$

share|cite|improve this answer

answered 7 hours ago

fishfagfishfag

29077 bronze badges

New contributor

fishfag is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment
 | 

0

$begingroup$

A supposed hint is that by using the Mclaurin-series for both the $e^x$ and $cos(x)$, we can arrive at this;

$$x^2sum_{k=0}^{infty}frac{x^{4k}}{(4k+2)!}+frac{1}{2}sinhleft(xright)=0$$

Now take,

$a=sum_{k=0}^{infty}frac{x^{4k}}{(4k+2)!}$

Now we can get different expressions for the same thing;
$$lnleft(-2x^{2}a+sqrt{4x^{4}a^{2}+1}right)-x=0$$
$$e^{x}4x^{2}a+e^{2x}-1=0$$

share|cite|improve this answer

answered 7 hours ago

fishfagfishfag

29077 bronze badges

New contributor

fishfag is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment
 | 

$begingroup$

A supposed hint is that by using the Mclaurin-series for both the $e^x$ and $cos(x)$, we can arrive at this;

$$x^2sum_{k=0}^{infty}frac{x^{4k}}{(4k+2)!}+frac{1}{2}sinhleft(xright)=0$$

Now take,

$a=sum_{k=0}^{infty}frac{x^{4k}}{(4k+2)!}$

Now we can get different expressions for the same thing;
$$lnleft(-2x^{2}a+sqrt{4x^{4}a^{2}+1}right)-x=0$$
$$e^{x}4x^{2}a+e^{2x}-1=0$$

share|cite|improve this answer

answered 7 hours ago

fishfagfishfag

29077 bronze badges

New contributor

fishfag is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered 7 hours ago

fishfagfishfag

29077 bronze badges

New contributor

fishfag is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 7 hours ago

fishfagfishfag

29077 bronze badges

New contributor

fishfag is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 7 hours ago

fishfagfishfag

29077 bronze badges