跳转至

多项式反三角函数

Description

给定多项式 f\left(x\right) ,求模 x^{n} 意义下的 \arcsin{f\left(x\right)}, \arccos{f\left(x\right)} \arctan{f\left(x\right)}

Method

仿照求多项式 \ln 的方法,对反三角函数求导再积分可得:

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} x} \arcsin{x} &= \frac{1}{\sqrt{1 - x^{2}}} \\ \arcsin{x} &= \int \frac{1}{\sqrt{1 - x^{2}}} \mathrm{d} x \\ \frac{\mathrm{d}}{\mathrm{d} x} \arccos{x} &= - \frac{1}{\sqrt{1 - x^{2}}} \\ \arccos{x} &= - \int \frac{1}{\sqrt{1 - x^{2}}} \mathrm{d} x \\ \frac{\mathrm{d}}{\mathrm{d} x} \arctan{x} &= \frac{1}{1 + x^{2}} \\ \arctan{x} &= \int \frac{1}{1 + x^{2}} \mathrm{d} x \end{aligned}

那么代入 f\left(x\right) 就有:

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} x} \arcsin{f\left(x\right)} &= \frac{f'\left(x\right)}{\sqrt{1 - f^{2}\left(x\right)}} \\ \arcsin{f\left(x\right)} &= \int \frac{f'\left(x\right)}{\sqrt{1 - f^{2}\left(x\right)}} \mathrm{d} x \\ \frac{\mathrm{d}}{\mathrm{d} x} \arccos{f\left(x\right)} &= - \frac{f'\left(x\right)}{\sqrt{1 - f^{2}\left(x\right)}} \\ \arccos{f\left(x\right)} &= - \int \frac{f'\left(x\right)}{\sqrt{1 - f^{2}\left(x\right)}} \mathrm{d} x \\ \frac{\mathrm{d}}{\mathrm{d} x} \arctan{f\left(x\right)} &= \frac{f'\left(x\right)}{1 + f^{2}\left(x\right)} \\ \arctan{f\left(x\right)} &= \int \frac{f'\left(x\right)}{1 + f^{2}\left(x\right)} \mathrm{d} x \end{aligned}

直接按式子求就可以了。

Code

多项式反三角函数
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
constexpr int maxn = 262144;
constexpr int mod = 998244353;

using i64 = long long;
using poly_t = int[maxn];
using poly = int *const;

inline void derivative(const poly &h, const int n, poly &f) {
  for (int i = 1; i != n; ++i) f[i - 1] = (i64)h[i] * i % mod;
  f[n - 1] = 0;
}

inline void integrate(const poly &h, const int n, poly &f) {
  for (int i = n - 1; i; --i) f[i] = (i64)h[i - 1] * inv[i] % mod;
  f[0] = 0; /* C */
}

void polyarcsin(const poly &h, const int n, poly &f) {
  /* arcsin(f) = ∫ f' / sqrt(1 - f^2) dx  */
  static poly_t arcsin_t;
  const int t = n << 1;
  std::copy(h, h + n, arcsin_t);
  std::fill(arcsin_t + n, arcsin_t + t, 0);

  DFT(arcsin_t, t);
  for (int i = 0; i != t; ++i) arcsin_t[i] = sqr(arcsin_t[i]);
  IDFT(arcsin_t, t);

  arcsin_t[0] = sub(1, arcsin_t[0]);
  for (int i = 1; i != n; ++i)
    arcsin_t[i] = arcsin_t[i] ? mod - arcsin_t[i] : 0;

  polysqrt(arcsin_t, n, f);
  polyinv(f, n, arcsin_t);
  derivative(h, n, f);

  DFT(f, t);
  DFT(arcsin_t, t);
  for (int i = 0; i != t; ++i) arcsin_t[i] = (i64)f[i] * arcsin_t[i] % mod;
  IDFT(arcsin_t, t);

  integrate(arcsin_t, n, f);
}

void polyarccos(const poly &h, const int n, poly &f) {
  /* arccos(f) = - ∫ f' / sqrt(1 - f^2) dx  */
  polyarcsin(h, n, f);
  for (int i = 0; i != n; ++i) f[i] = f[i] ? mod - f[i] : 0;
}

void polyarctan(const poly &h, const int n, poly &f) {
  /* arctan(f) = ∫ f' / (1 + f^2) dx  */
  static poly_t arctan_t;
  const int t = n << 1;
  std::copy(h, h + n, arctan_t);
  std::fill(arctan_t + n, arctan_t + t, 0);

  DFT(arctan_t, t);
  for (int i = 0; i != t; ++i) arctan_t[i] = sqr(arctan_t[i]);
  IDFT(arctan_t, t);

  inc(arctan_t[0], 1);
  std::fill(arctan_t + n, arctan_t + t, 0);

  polyinv(arctan_t, n, f);
  derivative(h, n, arctan_t);

  DFT(f, t);
  DFT(arctan_t, t);
  for (int i = 0; i != t; ++i) arctan_t[i] = (i64)f[i] * arctan_t[i] % mod;
  IDFT(arctan_t, t);

  integrate(arctan_t, n, f);
}

评论