What Is the Solution to the Mathieu-Type Equation with Operator Cos(d/dx)?

[iver]

I have a queation as below

[cos(d/dx)]f(x)

How to solve it? I just konw that the angle is an operator, but the function of x is out of the cosine function.

 

[dx]

They probably mean

\cos ({d/dx}) = 1 - \frac{1}{2!}\frac{d^2}{dx^2} + \frac{1}{4!}\frac{d^4}{dx^4} - ...

 

[HallsofIvy]

Just as I would interpret f(g)(x) to mean f(g(x)), I would interpret cos(d/dx)f(x) to mean cos(df/dx). That would, I believe, give the same thing as dx's Taylor's series interpretation.

 

[g_edgar]

The two suggestions are different. Because the OP says the answer is an operator probably the interpretation of dx is the intended one.

Let's see why they are different. Try the example f(x) = x^2 so that f'(x) = 2x and f''(x) = 2, and all higher derivatives are zero.

Then under the dx method, we get

<br /> f(x) - \frac{1}{2!}\;f&#039;&#039;(x) = x^2 - 1<br />

but under the HallsOfIvy method, we get

<br /> \sin(f&#039;(x)) = \sin(2x)<br />

Not the same.

 

[iver]

Thx~~

But, this term comes from the Mathieu’s equation as below

<br /> \frac{d^{2}f\left(x\right)}{dx^{2}}+\left(\epsilon-\cos\frac{d}{dx}\right)f\left(x\right)=0<br />

If I use series, this equation would be so terrible