More generally, with $y=f(x)$ as above, can $f(A+B)$ be expressed in terms of $f(A)$ and $f(B)$? Other things I'd like are a Laurent series for f and, if it were possible, a linear differential equation for f with polynomial coefficients and at most polynomial forcing function. I don't know much about differential equations, but I suspect that last request is impossible. Thanks!
Let $y=f(x)=\cos(x)/\cosh(x)$. Can $f(2x)$ be expressed, nicely, or otherwise, as a function of $y$?
1
$\begingroup$
real-analysis
ordinary-differential-equations
trigonometry
trigonometric-integrals
-
2Try to find $x_1,x_2$ with $f(x_1)=f(x_2)$ but $f(2x_1)\neq f(2x_2)$. Pretty sure this is possible. – 2017-02-16
-
0In particular, there are lots of values $x$ where $f(x)=0$... – 2017-02-16
-
0Thomas, I think I get your clever idea in your first comment, but I didn’t understand the second comment: If f(x)=O=f(w), then doesn’t f(2x) =f(2w)=O? – 2018-10-15
1 Answers
2
In the case where $A=B=x$ $$f(2x) = \frac{\cos^2(x) - \sin^2(x)}{\sinh^2(x)+\cosh^2(x)}=\frac{\cos^2(x)-1/2}{\cosh^2(x)-1/2}$$ We note that this is fairly similar to being $f(x)^2$, and thus $f(x)^2$ is a fantastic approximation for any decently large $x$. We could even split the numerator to get closer to the form the OP desires; however, I am not confident that a simple closed form in terms of $f(x)$ exists
-
0This is an interesting observation, but it doesn't answer the question. – 2017-02-16
-
0@MathematicsStudent1122 True. More of a long comment. Nevertheless, I thought it interesting enough to share. – 2017-02-16