Come back to be a problem found binomial inversion looks like they forgot.
Forget about the record binomial inversion looks like how to derive (I think fear is this life?)
Binomial inversion
There is a excellent (hao) United States (bei) equation:
\[ f_n = \sum_{i = 0} ^ n (-1)^i \dbinom{n}{i} g_i \Leftrightarrow g_n = \sum_{i = 0} ^ n (-1)^i \dbinom{n}{i} f_i \]
There is also a solid (NaN3) with (bei) of the formula:
\[ f_n = \sum_{i = 0} ^ n \dbinom{n}{i} g_i \Leftrightarrow g_n = \sum_{i = 0} ^ n (-1) ^ {n - i} \dbinom{n}{i} f_i\]
Derivation of a following:
I forget to make the first look like, we assume that all \ [g_ {n} = \ sum_ {i = 0} ^ n t_ {n, i} f_ {i} \]
Into the right side of the equation:
\[ \begin{split} f_{n} &= \sum_{i = 0} ^ n (-1) ^ i \dbinom{n}{i} \sum_{j = 0} ^ {i} t_{i,j} f_j \\ &= \sum_{i = 0} ^nf_i (\sum_{j = i} ^ n (-1) ^j \dbinom{n}{j} t_{j,i}) \\ &= \sum_{i = 0} ^ n f_i ( \sum_{j = 0} ^ {n - i} (-1)^{j + i} \dbinom{n}{j + i} t_{j + i, i} ) \end{split} \]
We want to construct a $ t_ {j, i} $ satisfies \ [[i = n] = \ sum_ {j = 0} ^ {n - i} (-1) ^ {j + i} \ dbinom {n} { j + i} t_ {j + i, i} \]
Consider this expression to represent the left out: \ [[= n-0] = \ sum_ {I} = 0 n-^ (-1) ^ I \ {n-dbinom {I}} \]
Thus \ [[i = n] = \ sum_ {j = 0} ^ {n - i} (-1) ^ j \ dbinom {n - i} {j} \]
$ T_ {j + i, i} $ which should have the form of a number of combinations, but following this formula there is only one combination of numbers, then we accompanied a try.
\[ [i = n] = \sum_{j = 0} ^ {n - i} (-1) ^ j \dbinom{n - i}{j} \dbinom{n}{i}\]
And this can be treated as:
\[ [i = n] = \sum_{j = 0} ^ {n - i} (-1) ^ j \dbinom{j + i}{i} \dbinom{n}{j + i} \]
Than what can be obtained \ [t_ {i, j} = (-1) ^ j \ dbinom {i} {j} \]
Other way
There is a excellent (hao) United States (bei) equation:
\[ f_k = \sum_{i = k} ^ n (-1) ^ i \dbinom{i}{k} g_i \Leftrightarrow g_k = \sum_{i = k} ^ n (-1) ^ i \dbinom{i}{k} f_i\]
There is also a solid (NaN3) with (bei) of the formula:
\[ f_{k} = \sum_{i = k} ^ {n} \dbinom{i}{k} g_{i} \Leftrightarrow g_{k} = \sum_{i = k} ^ n (-1) ^ {i - k} \dbinom{i}{k} f_i\]
Other understanding
cly_none tell a wonderful understanding.
Consider \ [f_ {n} = \ sum_ {i = 0} ^ n \ dbinom {n} {i} g_i \] This equation can be rewritten as:
\[ \frac{f_{n}}{n!} = \sum_{i=0} ^ n \frac{g_i}{i!} \frac{1}{(n-i)!}\]
In the exponential generating function alone \ [f (x) = g (x) * e ^ x \]
Then \ [G (X) = F (X) * E ^ {-} X \] , directly to expand back to complete the binomial inversion.