How do I find the period of function f(x) = [sin^3(x/2)]?
A somewhat rigorous analytical solution could be thus. First, consider the auxiliary function[math]g(x) = \sin^3{x}[/math]A good guess for the period of [math]g(x)[/math] would be [math]2\pi[/math], since the period of [math]\sin{x}[/math] is [math]2\pi[/math]. To prove that thereindeed isn't a still smaller number [math]P[/math] for which [math]g(x) = g(x + P), \forall x[/math], we suppose:[math]0 < P < 2\pi[/math]Then, [math]\sin^3{(x + P)} = \sin^3{x}[/math]Taking real cube roots on both sides,[math]\sin{(x + P)} = \sin{x}[/math]But, since we know that no such [math]P[/math] exists, we are sure that [math]2\pi[/math] is the smallest positive number that serves as the period of [math]g(x)[/math].Finally, we note that: [math]\displaystyle f(x) = g\left(\frac{x}{2}\right) = g\left(\frac{x}{2} + 2\pi\right) = g\left(\frac{x + 4\pi}{2}\right)[/math]Hence, [math]f(x) = f(x + 4\pi)[/math]
What is the period of a function f(x)?
[math]f(x) = (\sin^4 x+ \cos^4 x)[/math][math]f(x) = (\sin^2 x+ \cos^2 x)^2 - 2 \sin^2 x \cos^2 x [/math][math]f(x) = 1 - \dfrac{1}{2} 4 \sin^2 x \cos^2 x [/math][math]f(x) = 1 - \dfrac{1}{2} \sin^2 2x [/math][math]f(x) = 1- \dfrac{1}{2} \dfrac{(1- \cos 4x)}{2}[/math][math]f(x) = 1 - \dfrac{1}{4}(1- \cos 4x)[/math][math]f(x) = \dfrac{3+ \cos 4x}{4}[/math]So,Period of [math]\cos x = 2 \pi [/math]Period of [math]\cos4x[/math][math] = \dfrac{2\pi}{4} = \dfrac{\pi}{2}[/math]So, period of f(x) = [math]\dfrac{\pi}{2}[/math]
Is f (x) = 2x + sin(x) onto function?
Yes, [math]f(x)=2x+\sin x[/math] is an onto function.Robby gives a nice proof of the fact that the function is a bijection, but that is overkill.We need only three simple facts, plus the Intermediate Value Theorem, to show that the function is onto. I will simply state them; I leave any necessary proof to the reader.[math]f(x)[/math] is continuous.[math]\displaystyle \lim_{x \to -\infty} f(x) = -\infty[/math].[math]\displaystyle \lim_{x \to \infty} f(x) = \infty[/math].
Can we write every functions as a sum of infinity series like we do for sin(x) ? Or is it only for functions that are infinitely derivable?
A function over the reals is called analytic on an open set [math]D[/math] if for every [math]x_0 \in D[/math] there exists a sequence of real coefficients [math]\{ a_0, a_1, \ldots \}[/math] and an [math]r>0 [/math]such that the open interval [math](a_i-r,a_i+r)[/math] is contained in [math]D[/math] and such that[math]f(x) = \sum^{\infty}_{i=0} a_i (x-x_0)^i[/math]whenever [math]|x-x_0| < r[/math] . Here [math]r[/math] is called the radius of convergence. Analytic functions are equal to their Taylor expansions.However, not all differentiable functions over the reals are analytic. A simple example is[math]f(x) = e^{-\frac{1}{x^2}}[/math]This function has an [math]n[/math]th derivative for every [math]n[/math]. But its Taylor expansion around [math]0[/math] is equal to [math]0,[/math]whereas [math]f(0)[/math] is undefined.
How do you write down the Maclaurin series of expansion for the function f(x)=cos x?
Taylor and Maclaurin SeriesPersonal Website of Kiryl TsishchankaGo here
F(x) = sin 3x + sin 6x, 0 I believe it is just the same as the period of sin(3x) period = 2π/3 if you graph sin3x you get 3 cycles in 2π, for sin6x you get 6 cycles. Add the curves and period = 2π/3 This works because 6 is divisable by 3