- #1

- 90

- 0

**Proving - numerical analysis (separation of symbols)**

How do I prove this?

f(x+n) = f(x+n-1) + /\f(x+n-2) + ... + /\^(n-2) f(x+1) + /\^(n-1) f(x) +

/\^n f(x-1)

Last edited:

- Thread starter irony of truth
- Start date

- #1

- 90

- 0

How do I prove this?

f(x+n) = f(x+n-1) + /\f(x+n-2) + ... + /\^(n-2) f(x+1) + /\^(n-1) f(x) +

/\^n f(x-1)

Last edited:

- #2

- 79

- 0

I stumbled upon this problem as well... i can't really see how these two are equivalent without their corresponding binomial coefficients when f(x+n) = E^n f(x) = (1+/\)^n f(x) and (1+/\ )^n is expanded ^^;irony of truth said:Which of these must be true?

f(x+n) = f(x+n-1) + /\f(x+n-2) + ... + /\^(n-2) f(x+1) + /\^(n-1) f(x) +

/\^n f(x-1),

OR

f(x+n) = f(x+n-1) + /\f(x+n-2) + ... + /\^(n-2) f(x+1) + /\^(n-1) f(x) +

/\^n f(x)

?

This was the question given in my homework... I just doubt it because my

professor could have miswritten our assignment.

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 14

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 12

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 578