{"id":570,"date":"2023-05-10T11:02:45","date_gmt":"2023-05-10T11:02:45","guid":{"rendered":"https:\/\/milkyeggs.com\/?p=570"},"modified":"2023-10-30T22:53:48","modified_gmt":"2023-10-30T22:53:48","slug":"exercise-1-6-from-stein-and-shakarchis-fourier-analysis","status":"publish","type":"post","link":"https:\/\/milkyeggs.com\/math\/exercise-1-6-from-stein-and-shakarchis-fourier-analysis\/","title":{"rendered":"Exercise 1.6 from Stein and Shakarchi’s Fourier Analysis"},"content":{"rendered":"\n

The problem is as follows:<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

We have<\/p>\n\n\n\n

  <\/span>   <\/span>\"\begin{align*}g'(t) &= f'(t) \cos ct - c f(t) \sin ct - c^{-1} f''(t) \sin ct - f'(t) \cos ct \\&= - c f(t) \sin ct - c^{-1} f''(t) \sin ct \\&= - \frac{1}{c} \left( \sin ct) (f''(t) + c^2 f(t) \right) \\&= 0\end{align*}\"<\/p><\/p>\n\n\n\n

and<\/p>\n\n\n\n

  <\/span>   <\/span>\"\begin{align*}h'(t) &= f'(t) \sin ct + c f(t) \cos ct + c^{-1} f''(t) \cos ct - f'(t) \sin ct \\&= c f(t) \cos ct + c^{-1} f''(t) \cos ct \\&= \frac{1}{c} \left( \cos ct) (f''(t) + c^2 f(t) \right) \\&= 0.\end{align*}\"<\/p><\/p>\n\n\n\n

Each equality with \"0\" follows from the assumption \"f''(t) + c^2 f(t) = 0\". Therefore \"g'(t) = h'(t) = 0\" and \"g(t), h(t)\" are constant.<\/p>\n\n\n\n

The problem asks for constants \"a\" and \"b\". Could it be that \"g(t)\" and \"h(t)\" are these constants? That would be a natural guess. Let’s just try computing the sum<\/p>\n\n\n\n

  <\/span>   <\/span>\"\begin{align*}g(t) \cos ct + h(t) \sin ct &= f(t) \cos^2 ct + f(t) \sin^2 ct \\&= f(t).\end{align*}\"<\/p>.<\/p>\n\n\n\n

So indeed \"a = g(t)\" and \"b = h(t)\".<\/p>\n","protected":false},"excerpt":{"rendered":"

The problem is as follows: Prove that if \"f\" is a twice continuously differentiable function on \"\mathbb{R}\" which is a solution of the equation \"f''(t) + c^2 f(t) = 0\", then there exist constants \"a\" and \"b\" such that \"f(t) = a \cos ct + b \sin ct\".<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"_links":{"self":[{"href":"https:\/\/milkyeggs.com\/wp-json\/wp\/v2\/posts\/570"}],"collection":[{"href":"https:\/\/milkyeggs.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/milkyeggs.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/milkyeggs.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/milkyeggs.com\/wp-json\/wp\/v2\/comments?post=570"}],"version-history":[{"count":6,"href":"https:\/\/milkyeggs.com\/wp-json\/wp\/v2\/posts\/570\/revisions"}],"predecessor-version":[{"id":578,"href":"https:\/\/milkyeggs.com\/wp-json\/wp\/v2\/posts\/570\/revisions\/578"}],"wp:attachment":[{"href":"https:\/\/milkyeggs.com\/wp-json\/wp\/v2\/media?parent=570"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/milkyeggs.com\/wp-json\/wp\/v2\/categories?post=570"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/milkyeggs.com\/wp-json\/wp\/v2\/tags?post=570"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}