93039

a part 要點證?我直覺覺得係conv.
但唔識點寫

[ 本帖最後由 93039 於 2014-4-26 09:49 PM 編輯 ]

hk56740

用comparison test證啦
a_n/n<=a_n
since sum of a_n is convergent
--->sum of a_n/n is convergent

[ 本帖最後由 hk56740 於 2014-4-22 05:42 PM 編輯 ]

93039

係喎...
咁都諗唔到

93039

我唔係好明A PART點解X=0果陣個SERIES係CONV.?

桃子

You can integrate and differentiate convergent power series.

93039

如果要搵可以令到個series converge 既x係咪就咁put x=pi/2 and -pi/2之後話因為係呢兩個pt個series div.
所以係 -pi/2<x<pi/2

[ 本帖最後由 93039 於 2014-4-22 11:29 PM 編輯 ]

桃子

Radius of convergence is unchanged after differentiation or integration.

There is no problem at the point x=0.
ln |sec 0| = ln|1| = 0

93039

咁radius of convergence of  ln lsec xl係咪就係 -pi/2<=x<=pi/2?

cwk7093942

引用:

原帖由 93039 於 2014-4-23 12:55 AM 發表
咁radius of convergence of  ln lsec xl係咪就係 -pi/2

does tan(pi/2) exist ??

the statement said that the series converges at the point |x| = pi/2 , is that T or F ???

please , check the question by yourself

you have the ability to prove that T or F

[ 本帖最後由 cwk7093942 於 2014-4-23 08:01 AM 編輯 ]

cwk7093942

引用:

原帖由 93039 於 2014-4-22 11:15 PM 發表
如果要搵可以令到個series converge 既x係咪就咁put x=pi/2 and -pi/2之後話因為係呢兩個pt個series div.
所以係 -pi/2

乜你本書, 你教授係咁教你代2個數, 就叫搵到個 radius of convergence ??

咁 e^x = 1 + x + x^2/2 + ...

我又代 x = 1 and 2 入去

我話個 radius of convergence 係 1 ≦ x ≦ 2 得唔得 ?

[ 本帖最後由 cwk7093942 於 2014-4-23 08:09 AM 編輯 ]

cwk7093942

引用:

原帖由 93039 於 2014-4-23 12:55 AM 發表
咁radius of convergence of  ln lsec xl係咪就係 -pi/2

你自己係 #6 都知 ln(secx) at |x| = pi/2 係有問題 , 咁你點解仲可以講得出哩句野呢??

引用:

原帖由 桃子 於 2014-4-23 12:40 AM 發表
Radius of convergence is unchanged after differentiation or integration.

There is no problem at the point x=0.
ln |sec 0| = ln|1| = 0

for 93039


你讀到大學, 你自己係有能力去 prove 桃子以上所講...


點解 unchanged ?? 你教授, 你本書, 有教你搵 radius of convergence 方法架嘛...(etc. ratio test , roots test)


點解你唔用哩D方法去搵一搵再去比較呢??? 即使唔識搵, 計唔到, WIKI 都有架嘛, google 都有大把答案俾你選擇 ....






不恥下問係一件好事, 但都要你自己學識搵答案....大學數學, 並唔係咁樣學習架...


我亦唔係叫你收埋自己, 一個人學習, 你唔係我, 你有人教!!!我冇人教...

加油

93039

好啦
我試下自己搵下

cwk7093942

引用:

原帖由 桃子 於 2014-4-23 12:40 AM 發表
Radius of convergence is unchanged after differentiation or integration.

There is no problem at the point x=0.
ln |sec 0| = ln|1| = 0

not always true

by the definition of differentiation

f is continuous on [a, b], but might not differentiable on the end points a and b

same as integration

桃子

Radius of convergence does not tell you the behaviour at the boundary.
Even if two power series have the same radius of convergence, they may behave differently at the boundary.

(In the general sense, the boundary refers to a circle centred at a with radius R in the complex plane. In the simplified case that the series is real and centred at x=0, the boundary is x=R and x=-R.)

[ 本帖最後由 桃子 於 2014-4-23 10:26 PM 編輯 ]

firoazen

好奇一問 依家M2多左convergent divergent 既d野?