And generating polynomial function
Polynomial
Dove out open polynomial root (enhanced version).
Here the board put a polynomial Collection
Polynomial multiplication
Back a board just fine.
Taylor and Maclaurin Series expansion
If \ (f (x) \) in \ (x = x0 \) is present at \ (n-\) derivative, then:
\ [\ the begin {align = left} F (X) & = F (X0) + \ FRAC { f ^ 1 (x0)} { 1!} (x-x0) + \ frac {f ^ 2 (x0)} {2!} (x-x0) ^ 2 + ... + \ frac {f ^ n ( x0)} {n!} ( x-x0) ^ n + \ xi \\ & = \ sum_ {i = 0} ^ n \ frac {f ^ i (x0)} {i!} (x-x0) ^ i + \ XI \ End align = left {} \]
\ (\ XI \) is the remainder when \ (n-\) infinity \ (\ XI \) is the high order infinitesimal.
A special case is \ (x0 = 0 \) is the Taylor expansion, called the Maclaurin Series.
\(f(x)=\sum_{i=0}^n\frac{f^i(0)}{i!}x^i\)
More common is the \ (e ^ x \) expansion: \ (E ^ X =. 1 + \ FRAC {X} {. 1!} + \ FRAC {X ^ 2} + \ FRAC {X ^. 3 {2!} 3} {!} + ... \)
Newton iteration
A thing with useful.
We know that any polynomial arithmetic operation can become a \ (F (x) \) and a polynomial \ (B (the X-) \) , meet
\ [F (B (x) ) \ equiv 0 (\ text {mod} \ x ^ n)
\] For example, if an inverse requirements, then \ (F (B (x) ) = A (x) * B (x) -1 \ equiv 0 \)
Consider now when \ (n = 1 \) when it is a good demand.
Consider \ (n-\) extended to \ (2N \) conditions: Let \ (B_ {n} (x ) \) represents a number satisfying \ (n-\) solution.
The \ (F (B_ {2n} (x)) \) in \ (B_n (x) \) at the Taylor expansion, there are:
\ [F. (B_ {2N} (X)) = F. (B_n (X)) + \ frac {F '(B
(x))} {1!} (B_ {2n} (x) -B_n (x)) + ... \] At this time we found no later entries are used, because:
\ [F (B_ {2n}
(x)) \ equiv 0 (\ text {mod} x ^ {n}) \] is then difficult to find, \ (2N B_ {} (X) \) after \ (n- \) term and \ (B_n (x) \) there is no difference, so if it becomes \ ((B_ {2n} -B_n (x)) ^ 2 \) , when it will cover the whole, so in this case must satisfy \ (F (B_ {2n} (x)) \) properties, i.e. \ (\ {text} X ^ n-MOD \ equiv 0 \) .
This time we turned it into the equation:
\ [B_ 2N} {(X) = B_n (X) - \ {F. FRAC (B_n (X))} {F. '(B_n (X))} \]
Note derivative is \ (B_n (x) \) derivation. then it can recursively solved.
Inverse polynomial
\(F(B_n(x))=A(x)*B_n(x)-1 \equiv 0\).
At this point you can get:
\[ \begin{align} B_{2n}(x)&=B_n(x)-\frac{A(x)*B_n(x)-1}{A(x)}\\ &=B_n(x)-(A(x)*B_n(x)-1)B_n(x)\\ &=2B_n(x)-A(x)*B^2_n(x) \end{align} \]
Open polynomial root
\(F(B_n(x))=B_n^2(x)-A(x)\equiv 0\)
At this time can be:
\ [\} B_ the begin align = left {} {2N (X) = & B_n (X) - \ ^ B_n FRAC {2 (X) -A (X)} {2B_n (X)} \\ & = \ frac {1} {2 } (\ frac {B ^ 2_n (x) + A (x)} {B_n (x)}) \\ & = \ frac {1} {2} (B_n (x) + \ frac {A (x)} {B_n (x)}) \ end {align} \]
Need to use the inverse polynomial.
Polynomial derivation
\ ((n-X ^) '= n-*. 1-n-X_ {} \) , the derivative having a linear property, can be directly calculated.
Polynomials
\ (\ X ^ n-int = \ {n-FRAC {+}. 1. 1} ^ {n-X}. 1 + \) , also satisfy the linearity.
Polynomial ln
\[ ln(A(x))=B(x)\\ ln'(A(x))=B'(x)\\ \frac{A'(x)}{A(x)}=B'(x) \]
Direct composite function after derivation derivation + multiplication \ (\ to \) integration can be.
Polynomial exp
See here should be no people do not know \ (exp (x) \) is \ (e ^ x \) it.
\ (F (B_n (the X-)) = \ LN B_n (the X-) -A (the X-) \ equiv 0 \)
\ [\} B_ the begin align = left {} {2N (X) = & B_n (X) - \ {FRAC LN \ B_n (X) -A (X)} {\ FRAC {} {B_n. 1 (X)} } \\ & = B_n (x) -B_n (x) (\ ln B_n (x) -A (x)) \\ & = B_n (x) (1- \ ln B_n (x) + A (x)) \ end {align} \]
to apply the polynomial \ (\ LN \) and a polynomial multiplication.
Fast power polynomial
\ [B (x) = A ^ k (x) \\ \ LN B (x) = k \ LN A (x) \]
Direct access \ (\ LN \) and then do a multiply each coefficient \ (\ exp \) can.
Polynomial division
Given a length \ (n-\) polynomial \ (A (X) \) , a length \ (m \) polynomial \ (B (X) \) , seeking a length \ (nm \) of polynomial \ (C (x) \) and a length of less than \ (nm \) polynomial \ (R & lt (X) \) .
First we define a calculation \ (Reverse \) of \ (A ^ R & lt (X) = X ^ nA (\ FRAC. 1 {X} {}) \) , in fact, equivalent to flip \ (A \) coefficients of the polynomial.
Then this time have:
\ [A (X) = B (X) * C (X) R & lt + (X) \\ A (\ FRAC. 1 {X} {}) = B (\ FRAC. 1 {X} {} ) * C (\ frac {1 } {x}) + R (\ frac {1} {x}) \\ x ^ nA (\ frac {1} {x}) = x ^ mB (\ frac {1} {x}) * x ^ { nm} C (\ frac {1} {x}) + x ^ nR (\ frac {1} {x}) \\ A ^ R (x) = B ^ R (x) * C ^ R (x) +
R ^ R (x) * x ^ {n-m + 1} \] then this formula in \ (\ MOD-X ^ {n-m}. 1 + \) , is:
\ [A ^ R (x) = B ^ R (x) * C ^ R (x) \\ C ^ R (x) = \ frac {A ^ R (x)} {B ^ R (x)} \ ]
is directly applied to the inverse polynomial.
\ (R & lt (X) = A (X) -B (X) * C (X) \) , can be calculated directly.
other
There is also little things did not engage in a later dateThe pigeon
After reading the above you can do polynomial board the
But to use the square root quadratic residue, he is no guarantee that \ (A [0] = 1 \) .
Generating function
Ordinary generating function ( \ (\ text of OGF} {\) )
Consider a series \ (A = <a_0, A_1, A_2, ...> \) , his \ (\ text {OGF} \ ) is \ (a_0 + a_1x + a_2x ^ 2 + ... \)
For example the Fibonacci column \ (\ text {OGF} \ ) is \ (0 + 1x + 1x ^ 2 + 2x ^ 3 + ... \)
Exponential generating function ( \ (\ text of EGF} {\) )
For the number of columns \ (A \) , his \ (\ text {EGF} \ ) is \ (\ sum_ {i = 0 } ^ {\ infty} \ frac {a_i} {i!} X ^ i \)
\ (e ^ x \) is the number of columns \ (<1,1,1,1, ...> \) a \ (of EGF \) .
To be continued.