âµ[x+1/4]â¥[x]â¥[x-1/2]ï¼ä¸[x+1/4]=[x-1/2]
â´[x+1/4]ï¼[x]ï¼[x-1/2]
å½ä¸ä»
å½0â¤{x}ï¼3/4æ¶ï¼[x+1/4]ï¼[x]ï¼
å½ä¸ä»
å½1/2â¤{x}ï¼1æ¶ï¼[x]ï¼[x-1/2]
â´1/2â¤{x}ï¼3/4æ¶ï¼[x+1/4]=[x-1/2]ã
â´xç解é为{x|n+1/2â¤xï¼n+3/4ï¼nâZ}
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