Understanding the Binomial Expansion and its Relationship to e^p

  • Thread starter Thread starter thereddevils
  • Start date Start date
  • Tags Tags
    Binomial Expansion
AI Thread Summary
The discussion centers on the relationship between the binomial expansion and the expression for e^p, specifically how the series 1 + p + p^2/2! + p^3/3! + ... converges to e^p. Participants note that this can be understood through Taylor expansions and the definition of e^p as a power series. A suggested approach involves applying the binomial theorem to the limit definition of e^p. Additionally, the importance of the remainder term in Taylor's theorem is highlighted, indicating that it approaches zero as n approaches infinity. The conversation emphasizes the need for a proof to solidify understanding of this mathematical concept.
thereddevils
Messages
436
Reaction score
0
How is

1+p+\frac{p^2}{2!}+\frac{p^3}{3!}+...=e^p ?
 
Physics news on Phys.org
I know it has to do with taylor expansions, but I've never studied this so I can't answer your question. I'd also like to see a proof for this so this is like some pointless post I'm making so I can subscribe to this thread :biggrin:
 
Depends how you define e^p. You could just define it as the power series. However, I'm assuming you are using something like...

<br /> e^p = \lim_{n \rightarrow \infty} (1 + \frac{p}{n})^n<br />

Try using the binomial theorem on the right side, then take the limit.
 
thereddevils said:
How is

1+p+\frac{p^2}{2!}+\frac{p^3}{3!}+...=e^p ?

Mentallic said:
I know it has to do with taylor expansions, but I've never studied this so I can't answer your question. I'd also like to see a proof for this so this is like some pointless post I'm making so I can subscribe to this thread :biggrin:

You just show the remainder upon approximating it with the first n terms goes to zero as n --> infinity. See, for example,

http://en.wikipedia.org/wiki/Taylor's_theorem
 

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
View full post »

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
View full post »

Similar threads

Mathematical Induction with binomial sum
Replies
9
Views
3K
If p is even and q is odd, then ##\binom{p}{q}## is even
Replies
19
Views
3K
Binomial Theorem - determine the term with...
Replies
18
Views
2K
Binomial Expansion: Evaluating Coefficient from two binomials
Replies
7
Views
2K
Find the ##r^{th}## term of ##(a+2x)^n##
Replies
3
Views
2K
Calculating Probability Using Bayes' Formula: Solving for P(urn III | silver)
Replies
6
Views
1K
Find the expected value of a coin flipping game
Replies
7
Views
3K
Show that ## p ## divides ## 2^{(p-1)/2}-1 ##
Replies
1
Views
1K
Conditional Probability + Poisson Distribution
Replies
14
Views
3K
Prove that ## 1^{p-1}+2^{p-1}+3^{p-1}+\dotsb +(p-1)^{p-1}\equiv -1 ##.
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top