Then for "n\\in \\N"
"=>p(n)+1=0=>p(n)=-1"
Let "q(x)=1,x\\in \\R."
Let "p(x)=|\\sin(\\pi x)|-1"
Then
"\\lambda(x)=|\\sin(\\pi x)|"
Check
"=|\\sin(\\pi x)|=\\lambda(x), x\\in\\R"
"\\lambda(n+\\dfrac{1}{2})=|\\sin(\\pi(n+\\dfrac{1}{2} ))|=\\sin(\\dfrac{\\pi}{2})=1, n\\in\\N"
"\\lambda(x)=q(x)(p(x)+1)=1\\cdot(|\\sin(\\pi x)|-1+1)"
Comments
Dear gmali3, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!
so much late.btw thanks
Leave a comment