Thm 37.1

For 𝑎𝑘𝑥𝑘=𝑓(𝑥) we can prove

  1. 𝑓(𝑥)=𝑘𝑎𝑘𝑥𝑘1
  2. 𝑓(𝑡)𝑑𝑡=𝑎𝑘𝑥𝑘+1𝑘+1

Thm 37.2

For uniform limit lim𝑓𝑛=𝑔 where 𝑔 is continue and lim𝑓𝑛=𝑓 point-wise, we can say 𝑓=𝑔.

Thm 37.3

For lim𝑓𝑛=𝑓 is uniform limit and lim𝑓𝑛=𝑓 in Metric Space lim𝑓𝑛(𝑡)𝑑𝑡=𝑓(𝑡)𝑑𝑡 is uniform limit.

Def 37.4 Cauchy-Hadamard theorem

𝑅=[lim𝑛sup𝑛|𝑎𝑛|]1

Props 37.5

(𝑟<𝑅)𝑓𝑛(𝑥)=𝑛𝑘=0𝑎𝑘𝑥𝑘 are uniformly Cauchy on 𝐶[𝑟,𝑟]

Cor 37.6

(𝑟<𝑅)𝑓𝑛 uniformly converge to 𝑓 on [𝑟,𝑟] and 𝑎𝑘𝑥𝑘 convergent.

Lemma 37.7

lim𝑛sup𝑛|𝑎𝑛|=lim𝑛sup𝑛(𝑘+1)|𝑎𝑛+1|=lim𝑛sup𝑛|𝑎𝑛1|𝑘

Thm 37.8

let (𝑎𝑛) and 𝑅=(lim𝑛sup𝑛|𝑎𝑛|)1
Then on (𝑅,𝑅), 𝑓(𝑥)=𝑘𝑎𝑘𝑥𝑘 is

  1. Well defined
  2. continue
  3. differentiable
  4. integrable

Def 38.1

11𝑥=𝑘=0𝑥𝑘

Remark 38.2

Radius of convergent comes from complex inequality.

Prop 38.3

lim𝑘sup𝑘𝑘=1