non-trig fcns w/ infinite numbers of real roots...?
- Thread starter Tania
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Good point, but I thought by itself (1) it was too trivial and not clear how to generalize it, and (2) it would just invite confusion: somehow the function f(x) = 0 must be distinguished from the equation f(x) = 0. The equation 0x = 0 may be the shortest way to express this.tkhunny said:
Thanks for the thoughts.
f(x)=0 is TOO trivial!
I have thought long about this problem (not sure why!) but I just like the idea of there being equations other than trig ones with infinite roots. If no such equation exists does anyone know how to prove that?
What about f(x) = (-1)^x where x is an integer? Not continuous so not really as good as a trig equation.
f(x)=0 is TOO trivial!
I have thought long about this problem (not sure why!) but I just like the idea of there being equations other than trig ones with infinite roots. If no such equation exists does anyone know how to prove that?
What about f(x) = (-1)^x where x is an integer? Not continuous so not really as good as a trig equation.
pka
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Take note that you have ready been given a perfectly good example. However, the function is discontinuous everywhere.
Here is an example with infinitely many zeros and countably many discontinuities.
To understand it you must know some topology of the real number line.
Note: \(\displaystyle (0,1] = \bigcup\limits_{n = 1}^\infty {\left( {\frac{1}{{n + 1}},\frac{1}{n}} \right]} .\)
So:\(\displaystyle x \in (0,1] \Rightarrow \quad \exists (J)\left[ {x \in \left( {\frac{1}{{J + 1}},\frac{1}{J}} \right]} \right]\).
Now define the function.
\(\displaystyle \L f(x) = \left\{ {\begin{array}{rr}
{1,} & {x > 1} \\
{2J(J + 1)\left( {x - \frac{1}{J}} \right) + 1,} & {x \in (0,1]} \\
{ - 1,} & {x \le 0} \\
\end{array}} \right.\)
Here is an example with infinitely many zeros and countably many discontinuities.
To understand it you must know some topology of the real number line.
Note: \(\displaystyle (0,1] = \bigcup\limits_{n = 1}^\infty {\left( {\frac{1}{{n + 1}},\frac{1}{n}} \right]} .\)
So:\(\displaystyle x \in (0,1] \Rightarrow \quad \exists (J)\left[ {x \in \left( {\frac{1}{{J + 1}},\frac{1}{J}} \right]} \right]\).
Now define the function.
\(\displaystyle \L f(x) = \left\{ {\begin{array}{rr}
{1,} & {x > 1} \\
{2J(J + 1)\left( {x - \frac{1}{J}} \right) + 1,} & {x \in (0,1]} \\
{ - 1,} & {x \le 0} \\
\end{array}} \right.\)
pka, since there is always an x equal to to 1/J for some J would \(\displaystyle \,\,f(x)=(x-\frac{1}{J}) \,\,\) have been sufficient in (0,1]? There would be a 1-1 correspondence between N and the zeros of the function.
Also, here's another: \(\displaystyle f(x)=x-\lfloor x \rfloor\).
Also, here's another: \(\displaystyle f(x)=x-\lfloor x \rfloor\).