q(x)=0,x∈R Then for n∈N
λ(n)=q(n)(p(n)+1)=0=>
=>p(n)+1=0=>p(n)=−1Let q(x)=1,x∈R.
Let p(x)=∣sin(πx)∣−1
Then
λ(x)=1⋅(∣sin(πx)∣−1+1)
λ(x)=∣sin(πx)∣Check
λ(x+1)=∣sin(π(x+1))∣=∣−sin(πx)∣=
=∣sin(πx)∣=λ(x),x∈R
λ(n)=∣sin(π(n))∣=0,n∈N
λ(n+21)=∣sin(π(n+21))∣=sin(2π)=1,n∈N
λ(x)=q(x)(p(x)+1)=1⋅(∣sin(πx)∣−1+1)
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so much late.btw thanks