Can linear equations be hard?

Algebra Level 5

Given that

$f\left( x \right) =\left| \left| x \right| -1 \right|$ ${ P }_{ 0 }(x)=f(x)$ ${ P }_{ n+1 }(x)={ f(P }_{ n }(x))$

$A_{k}=\left| \lim _{ n\rightarrow \infty }{ \int _{ 0 }^{ k }{ { P }_{ n }(x) } -0.5dx } \right|$

$A=\max{A_{k}}$

And that $k_{0}$ is the minimum value such that $A=A_{k_{0}}$ , where $k$ ranges over all the positive reals $k>0$ .

Find $\left\lfloor 1000(A+k_{0}) \right\rfloor$

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The answer is 625.

2 solutions

Julian Poon
Feb 18, 2015

Consider the sketch of $f(x)$ . It would be the graph $|x|-1$ and then flipped upwards: hi

Now consider $f(f(x))$

So for $\lim _{ n\rightarrow \infty }{ { P }_{ n } } (x)-0.5$ , the graph would be either

Depending on whether $n$ is odd or even.

It is now easy to see that the maximum of $\left| \lim _{ n\rightarrow \infty }{ \int _{ 0 }^{ k }{ { P }_{ n }(x) } -0.5dx } \right|$ is ${0.5}^{2}/2=0.125=A$ and this occurs in either graph when $x=0.5=k_{0}$

Therefore, $\left\lfloor 1000(A+ k_{0} )\right\rfloor = \boxed{625}$

So, I think you should define $A_k = | \lim \int_0^k \ldots |$ , and then $A = \min A_k$ , and $k_0$ is the minimum value such that $A = A_{k_0}$ .

That will clarify that the $k$ in the integral is fixed as $n$ tends to infinity, instead of being allowed to vary for different $n$ .

Calvin Lin Staff - 6 years, 3 months ago

Or did you mean $A=\max{A_{k}}$ ? I edited it already btw, thanks!

Julian Poon - 6 years, 3 months ago

Akul Agrawal
Oct 27, 2015

Just to elaborate a little

Let's take 2 Cases.

Case(1)

${ A }_{ k }=\left| \frac { \left\lfloor k \right\rfloor }{ 2 } +\frac { { \left\{ k \right\} }^{ 2 } }{ 2 } -\frac { k }{ 2 } \right| \\ \Rightarrow \quad { A }_{ k }=\left| \frac { { \left\{ k \right\} }^{ 2 }-\left\{ k \right\} }{ 2 } \right|$

Case(2)

${ A }_{ k }=\left| \frac { \left\lfloor k \right\rfloor }{ 2 } +\frac { \left\{ k \right\} \left( 2-\left\{ k \right\} \right) }{ 2 } -\frac { k }{ 2 } \right| \\ \Rightarrow \quad { A }_{ k }=\left| \frac { \left\{ k \right\} -{ \left\{ k \right\} }^{ 2 } }{ 2 } \right|$

Note

The equations would remain same if k is odd or even.

Conclusion

$A={ \left| \frac { { \left\{ k \right\} }^{ 2 }-\left\{ k \right\} }{ 2 } \right| }_{ min }=\frac { 1 }{ 8 } \\ at\quad \left\{ k \right\} =\frac { 1 }{ 2 } \quad \Rightarrow { k }_{ min\quad pos }={ k }_{ o }=\frac { 1 }{ 2 }$

Can linear equations be hard?

The answer is 625.

2 solutions

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