Is there a solution to this transcendental equation ?

In summary, the equation sinX + cosX = lnX has a unique solution between 1 and √e, but no closed form. Using intermediate value theorem and knowledge of the functions involved, we can show that the solution is approximately 1.8893.
  • #1
Monsterboy
303
96
This is not a homework problem, this equation simply occurred to my mind and my math teacher said such an equation either can't exist or he doesn't know the answer.
sinX ➕ cosX = lnX

I don't know how to start...
 
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  • #2
This looks ugly. To simplify it a bit, notice that the LHS is equal to ##\frac12 \sin (2x)##. In particular, this lies in ##\left[-\frac12,\frac12\right]##, so that any ##x## solving your equation would have to belong to ##\left[\frac1{\sqrt{e}},\sqrt{e}\right]##.

Using some knowledge of the functions ##\sin## and ##\ln## (i.e. knowing where each of them is positive/negative and where each of them is positively/negatively sloped, and using that ##2\in (0,\pi)##), you can use intermediate value theorem to show that the equation has a unique solution ##x^*>0##, and that it satifies ##x^*\in (1, \sqrt{e})##.

As for an explicit solution, good luck.
 
  • #3
Monsterboy said:
This is not a homework problem, this equation simply occurred to my mind and my math teacher said such an equation either can't exist or he doesn't know the answer.
sinX ➕ cosX = lnX

I don't know how to start...
You could have started by drawing a graph of each function. Note that:

##sin(x) + cos(x) = \sqrt{2}sin(x + \frac{\pi}{4})##

Which simplifies things.
 
  • #4
economicsnerd said:
This looks ugly. To simplify it a bit, notice that the LHS is equal to ##\frac12 \sin (2x)##.

##sin(x) + cos(x) \ne \frac12 \sin (2x)##

##sin(x)cos(x) = \frac12 \sin (2x)##
 
  • #5
Was it "+" rather than "*" ?

On my computer, it's showing up as "sin(x) [black square with a white X in it] cos(x)". If it was addition, my bad. Sorry for adding confusion.
 
  • #6
I think it is some special character, something like a large "+" sign. So it is clearly an addition.
This does not change the result - there is a unique solution, but probably no closed form for it.
 
  • #7
with intermediate value method X is something like 1.8893...
 
  • #8
Monsterboy said:
with intermediate value method X is something like 1.8893...

Indeed. Which is the sole numerical solution to sin(x)+cos(x) = ln(x).

It has no closed form, though. Numerical solution is the best you're going to get.
 

Related to Is there a solution to this transcendental equation ?

1. What is a transcendental equation?

A transcendental equation is an equation that contains transcendental functions, such as trigonometric, exponential, or logarithmic functions. These equations cannot be solved algebraically and require numerical methods to find a solution.

2. Why is it difficult to find a solution to a transcendental equation?

Transcendental equations are difficult to solve because they involve functions that cannot be expressed in terms of finite algebraic operations. This means that there is no formula or set of steps that can be used to find a solution, and numerical methods must be used instead.

3. Is there a general method for solving transcendental equations?

No, there is no general method for solving transcendental equations. Each type of transcendental equation requires its own specific method, such as the Newton-Raphson method for finding roots of a function or the Lambert W function for solving exponential and logarithmic equations.

4. Can a transcendental equation have multiple solutions?

Yes, a transcendental equation can have multiple solutions. In fact, most transcendental equations have infinite solutions. For example, the equation sin(x) = 0 has an infinite number of solutions, since sin(x) = 0 for any value of x that is a multiple of pi.

5. Is it possible for a transcendental equation to have no solutions?

Yes, it is possible for a transcendental equation to have no solutions. This can occur when there is no value of the variable that satisfies the equation, such as in the equation e^x = -1, which has no real solutions.

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