f(x)=a1sinx+a2sin2x+...+ansinnx,且满足|f(x)|<=|sinx|

shingrons

f(x)=a1sinx+a2sin2x+...+ansinnx,且满足|f(x)|<=|sinx|
求证：|a1+2a2+3a3+4a4+...+nan|<=1

这题很不错
应该一看就会，不然就是不会

只是找不到初等方法去解

sardinecan

你的方法是将它微分吗？

shingrons

原帖由 shingrons 于 11-11-2006 12:32 PM 发表
f(x)=a1sinx+a2sin2x+...+ansinnx,且满足|f(x)|<=|sinx|
求证：|a1+2a2+3a3+4a4+...+nan|<=1

这题很不错
应该一看就会，不然就是不会

只是找不到初等方法去解

|f'(0)|<=|cos0|
|a1+2a2+3a3+4a4+...+nan|<=1
...
I can't find any other alternative way to prove this
Racking my brain now...
Anyelse can help?

[ 本帖最后由 shingrons 于 15-11-2006 10:16 PM 编辑 ]

dunwan2tellu

原帖由 shingrons 于 14-11-2006 08:59 AM 发表


|f'(0)|<=|sin0|
|a1+2a2+3a3+4a4+...+nan|<=1
...
I can't find any other alternative way to prove this
Racking my brain now...
Anyelse can help?

我不觉得说，如果

f(x) >= g(x)

那么

f'(x) >= g'(x)

example : f(x) = x^2 + 1 , g(x) = 2x

明显 f(x) >= g(x) , 但是 f'(x) = 2x 却不一定大过 g'(x) = 2

sardinecan

if we plot y =|f(x)|  and y =|sinx|               |f(x)|<=|sinx|
We consider when x = 0( this special case)
f (x) = 0
sin x also =0
then |f(0)|<=|sin0| or f(0) = sin 0
At x = o+h (where h is very closed to 0)
the line of |f (o+h)| will not overleap the |sin (0+h)|
so |f' (o )| <= |cos o |  
f'(0)=|a1+2a2+3a3+4a4+...+nan|
(d sin x )/dx, x=0  = cos 0 = 1
so,
|a1+2a2+3a3+4a4+...+nan|<=1
i hope u all can understand...

dunwan2tellu

原帖由 sardinecan 于 14-11-2006 11:52 PM 发表
if we plot y =|f(x)|  and y =|sinx|               |f(x)|<=|sinx|
We consider when x = 0( this special case)
f (x) = 0
sin x also =0
then |f(0)|<=|sin0| or f(0) = sin 0
At x = o+h (where  ...

很有趣的 argument , 是用 |f'(0)| = lim_{h->0}1/h * (|f(0+h)-f(0)|) 吗？

原帖由 shingrons 于 15-11-2006 10:21 PM 发表
|f(x)|<=|sinx|
|f(x)/x|<=|sinx/x|
两边取极限(x趋近0)
得出结果后代x=0

不错的方法

[ 本帖最后由 dunwan2tellu 于 16-11-2006 03:00 PM 编辑 ]

shingrons

|f(x)|<=|sinx|
|f(x)/x|<=|sinx/x|
两边取极限(x趋近0)
得出结果后代x=0

[ 本帖最后由 shingrons 于 15-11-2006 10:22 PM 编辑 ]