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ÇóÖ¤: n*(1-x^2)*x^(2n)-(2x+x^2)*(1-x^2n)+2x*(1+x)*(1-x^n)ºã´óÓÚ0£¬ÆäÖÐ0<x<1,nΪ´óÓÚ2µÄÕûÊý¡£Çó´óÉñÖ¸½Ì£¡²»Ê¤¸Ð¼¤£¡ÓÉÓÚÊÖ»ú´«²»ÁËͼ£¬Ö»ÄÜÊÖ´ò¡£

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for you, a thousand times over
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zaq123321

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3Â¥: Originally posted by ÂäÒ¶ÎÞÉù=== at 2015-03-24 18:00:34
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Sorry about this. Then should we compute the second derivative, which is negative in (0,1)?
But the calculation looks pretty complex.
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4Â¥2015-03-24 19:48:30
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zaq123321

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ÂäÒ¶ÎÞÉù===: ½ð±Ò+10, ¡ïÓаïÖú, ·Ç³£Ð»Ð»ÄãÄܹØ×¢Õâ¸öÎÊÌ⣡ 2015-03-26 01:42:30
Replace the second x^(2n) with x^n. Combine the second term and third term. Then each term is positive.
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2Â¥2015-03-24 15:56:57
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ľ³æ (ÕýʽдÊÖ)

Dragonborn

ÒýÓûØÌû:
2Â¥: Originally posted by zaq123321 at 2015-03-24 15:56:57
Replace the second x^(2n) with x^n. Combine the second term and third term. Then each term is positive.

Ìæ»»ºóʽ×Ó±»·Å´óÁ˰ɡ­¡­Òª·ÅËõµÄ»°Ó¦¸ÃÒª·ÅС¡­¡­²»¹ý»¹ÊǸÐл£¡

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for you, a thousand times over
3Â¥2015-03-24 18:00:34
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ÂäÒ¶ÎÞÉù===

ľ³æ (ÕýʽдÊÖ)

Dragonborn

ÒýÓûØÌû:
4Â¥: Originally posted by zaq123321 at 2015-03-24 19:48:30
Sorry about this. Then should we compute the second derivative, which is negative in (0,1)?
But the calculation looks pretty complex....

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for you, a thousand times over
5Â¥2015-03-24 20:38:37
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