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Two problems about infinite processes where smaller and smaller steps are taken and you have to discover what happens in the limit.
Age 16 to 18
Challenge Level
Congratulations to Andrei from Tudor Vianu National College, Bucharest, Romania for this excellent solution.
(1) Plot the graph of the function $y=f(x)$ where $f(x) = \sin x +|\sin x|$. Differentiate the function and say where the derivative is defined and where it is not defined.
I observe that $\sin x$ takes positive values between $2k\pi$ and $(2k+1)\pi$ (for integer $k$), that is in the first two quadrants, and $\sin x$ takes negative values between $(2k+1)\pi$ and $(2k+2)\pi$, that is in the third and fourth quadrants. So
$$\eqalign{ f(x) &= 2\sin x \quad for\ 2k\pi \leq x \leq (2k+1)\pi \cr &= 0 \quad for \ (2k+1)\pi \leq x \leq (2k+2)\pi .}$$
Below is represented the graph of $y = f(x)$:
Now I calculate the first derivative of $f(x)$:
$$\eqalign{ f^\prime (x) &= 2\cos x \quad for\ 2k\pi <
x < (2k+1)\pi \cr &= 0 \quad for\ (2k+1)\pi < x <
(2k+2)\pi}$$
The derivative $f^\prime(x)$ is not defined at the points $x =
k\pi$ for any integer $k$ and it does not have a tangent at these
points.
(2) Now I express the function $f(x) = \sin x + \cos x$ in the
form $f(x)=A\sin (x+\alpha)$, find $A$ and $\alpha$ and plot the
graph of this function. Similarly I express the function $g(x) =
\sin x - \cos x$ in the form $g(x) = B\sin (x +\beta)$ where $-\pi
/2 < \beta < \pi /2$, and plot its graph on the same
axes.
$$\eqalign { f(x) &= \sin x + \cos x = A\sin (x + \alpha)
\cr &= A\sin x\cos\alpha + A\cos x \sin \alpha }$$
As this formula must be valid for any $x$, I obtain: $$A\sin
\alpha = 1 \quad {\rm and} \ A\cos \alpha = 1$$ and hence $\tan
\alpha = 1$, $\alpha = \pi/4$ and $A=\pm \sqrt 2$.
Comment: If $A$ has the meaning of an amplitude, $A$ is
positive, and only the positive solution must be kept. This type of
problem is typical for the composition of oscillations.
Hence the graph of this function is a sine graph with a phase
shift of $\pi/4$, that is $f(x)=0$ when $x=k\pi - \pi/4$, it takes
the value 1 when $x=2k\pi$ and $x=2k\pi + \pi/2$ and the value -1
when $x=(2k+1)\pi$ and $x=2k\pi - \pi/2$, has maximum values
$(2k\pi +\pi/4, \sqrt 2)$ and minimum values $((2k+1)\pi + \pi/4,
-\sqrt 2)$
In a similar manner I write $g(x) = \sin x - \cos x$:
$$\eqalign { f(x) &= \sin x - \cos x = B\sin (x + \beta)
\cr &= B\sin x\cos\beta + B\cos x \sin \beta }$$
So I have $$B\sin \beta = -1 \quad {\rm and} \ B\cos \beta =
1$$ and hence $\tan \beta = -1$, $\beta = -\pi/4$ and $B=\sqrt 2$.
Hence the graph of this function is a sine graph with a phase shift
of $-\pi/4$, that is $f(x)=0$ when $x=k\pi + \pi/4$, it takes the
value 1 when $x=k\pi$ and $x=2k\pi + \pi/2$, has maximum values
$(2k+1)\pi - \pi/4, \sqrt 2)$ and minimum values $(2k\pi - \pi/4,
-\sqrt 2)$.
(3) Now, I calculate the function $f(x) = \sin x + |\cos x|$ and plot the graph.
$$\eqalign{ f(x) &= \sin x + \cos x = \sqrt 2 \sin(x + \pi /4) \ for \ 2k\pi -\pi/2\leq x \leq 2k\pi +\pi/2 \cr &= \sin x - \cos x = \sqrt 2 \sin(x - \pi /4)\ for \ 2k\pi + \pi/2 \leq x \leq 2k\pi + 3\pi/2 .}$$
The derivative is not defined at $x= k\pi + \pi/2$ and for other values of $x$ it is:
$$\eqalign{ f^\prime(x) &= \cos x - \sin x \ {\rm for} \ 2k\pi -\pi/2< x < 2k\pi +\pi/2 \cr &= \cos x + \sin x \ {\rm for} \ 2k\pi + \pi/2 < x < 2k\pi + 3\pi/2 .}$$
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NRICH is part of the family of activities in the Millennium Mathematics Project.
NRICH is part of the family of activities in the Millennium Mathematics Project.