How Do Series Expansions Relate to Exponential Functions?

  • Thread starter Thread starter e^(i Pi)+1=0
  • Start date Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 1K views
e^(i Pi)+1=0
Messages
246
Reaction score
1
limit as n→∞ of [itex]\frac{(2t)^n}{n!}[/itex] and [itex]\frac{(-t)^n}{n!}[/itex]

Answers are e2t-1 and e-t-1 but I don't know how to work them out, thanks.

edit: btw these are series
 
Physics news on Phys.org
e^(i Pi)+1=0 said:
limit as n→∞ of [itex]\frac{(2t)^n}{n!}[/itex] and [itex]\frac{(-t)^n}{n!}[/itex]

Answers are e2t-1 and e-t-1 but I don't know how to work them out, thanks.

edit: btw these are series


Starting at n=1, it would seem? That would be unusual.