P148. Show that there is no real constant c > 0 such that \(\cos\sqrt{x+c}=\cos\sqrt{x}\) for all real numbers \(x\ge 0\).
Solution:
If the given equation holds for some constant c>0 then,
f(x) = \(\cos\sqrt{x}-\cos\sqrt{x+c}=0\) for all \(x\ge 0\)
\(\Rightarrow 2\sin\frac{\sqrt{x+c}+\sqrt{x}}{2}\sin\frac{\sqrt{x+c}-\sqrt{x}}{2}=0\)
Putting x=0, we note
\(\Rightarrow\sin^2\frac{\sqrt{c}}{2}=0\)
As \(c\not=0\)
\(\sqrt{c}=2n\pi\)
\(\Rightarrow c=4n^2\pi^2\)
We put n=1 and x=\(\frac{\pi}{2}\) to note that f(x) is not zero.
Hence no c>0 allows f(x) =0 for all \(x\ge 0\). (proved)
Solution:
If the given equation holds for some constant c>0 then,
f(x) = \(\cos\sqrt{x}-\cos\sqrt{x+c}=0\) for all \(x\ge 0\)
\(\Rightarrow 2\sin\frac{\sqrt{x+c}+\sqrt{x}}{2}\sin\frac{\sqrt{x+c}-\sqrt{x}}{2}=0\)
Putting x=0, we note
\(\Rightarrow\sin^2\frac{\sqrt{c}}{2}=0\)
As \(c\not=0\)
\(\sqrt{c}=2n\pi\)
\(\Rightarrow c=4n^2\pi^2\)
We put n=1 and x=\(\frac{\pi}{2}\) to note that f(x) is not zero.
Hence no c>0 allows f(x) =0 for all \(x\ge 0\). (proved)
No comments:
Post a Comment