Mathematics
Senior High
波線引いてあるところの符号になぜ等しいのかわかりません。
h(z)=(1-z) {log(1-z)-log2}+zlogzにより,
-1
h'(z)=-{log(1-z)-log 2}+(1-z)
1-z
1
+log z +zー
=log z-log(1-2)+log2=log2z-log(1-z)
よって、h'(z)の符号は, 2zー(1ーz)3D32-1の符号に
等しいから,h(z) の
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