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di-problem.dvi

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1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =

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I No. sin cos sine, cosine : trigonometric function π : π =.4 : n = 0, ±, ±, sin + nπ = sin cos + nπ = cos : parity sin = sin : odd cos = cos : even.

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欧州特許庁米国特許商標庁との共通特許分類 CPC (Cooperative Patent Classification) 日本パテントデータサービス ( 株 ) 国際部 2019 年 7 月 31 日 CPC 2019.08 版が発効します 原文及び詳細はCPCホームページのCPC Revisions(CPCの改訂 ) をご覧ください https://www.cooperativepatentclassification.org/cpcrevisions/noticeofchanges.html

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(2018) 2018 7 5 f(x) [ 1, 1] 3 3 1 3 f(x) dx c i f(x i ) 1 0 i=1 = 5 ) ( ) 3 ( 9 f + 8 5 9 f(0) + 5 3 9 f 5 1 1 + sin(x) θ ( 1 θ dx = tan 1 + sin x 2 π ) + 1 4 1 3 [a, b] f a, b double G3(double (*f)(),

( z = x 3 y + y ( z = cos(x y ( 8 ( s8.7 y = xe x ( 8 ( s83.8 ( ( + xdx ( cos 3 xdx t = sin x ( 8 ( s84 ( 8 ( s85. C : y = x + 4, l : y = x + a,

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sin cos No. sine, cosine : trigonometric function π : π = 3.4 : n = 0, ±, ±, sin + nπ = sin cos + nπ = cos : parity sin = sin : odd cos = cos : even.

08 No. : No. : No.3 : No.4 : No.5 : No.6 : No.7 : No.8 : No.9 : No.0 : No. : sin cos No. sine, cosine : trigonometric function π : π = 3.4 : n = 0, ±, ±, sin + nπ = sin cos + nπ = cos : parity sin = sin

IS-LM (interest) 100 (net rate of interest) (rate of interest) ( ) = 100 (2.1) (gross rate of interest) ( ) = 100 (2.2)

1 2 2 2 2.1 IS-LM 1 2.2 1 1 (interest) 100 (net rate of interest) (rate of interest) ( ) = 100 (2.1) (gross rate of interest) ( ) = 100 (2.2) 1 1. 2. 1 1 ( ) 2.3. 3 2.3 1 (yield to maturity) (rate of return)

x ( ) x dx = ax

x ( ) x dx = ax 1 dx = a x log x = at + c x(t) = e at C (C = e c ) a > 0 t a < 0 t 0 (at + b ) h dx = lim x(t + h) x(t) h 0 h x(t + h) x(t) h x(t) t x(t + h) x(t) ax(t) h x(t + h) x(t) + ahx(t) 0, h, 2h,

Fourier (a) C, (b) C, (c) f 2 (a), (b) (c) (L 2 ) (a) C x : f(x) = a 0 2 + (a n cos nx + b n sin nx). ( N ) a 0 f(x) = lim N 2 + (a n cos nx + b n sin

( ) 205 6 Fourier f : R C () (2) f(x) = a 0 2 + (a n cos nx + b n sin nx), n= a n = f(x) cos nx dx, b n = π π f(x) sin nx dx a n, b n f Fourier, (3) f Fourier or No. ) 5, Fourier (3) (4) f(x) = c n = n=

IA 2013 : :10722 : 2 : :2 :761 :1 (23-27) : : ( / ) (1 /, ) / e.g. (Taylar ) e x = 1 + x + x xn n! +... sin x = x x3 6 + x5 x2n+1 + (

IA 2013 : :10722 : 2 : :2 :761 :1 23-27) : : 1 1.1 / ) 1 /, ) / e.g. Taylar ) e x = 1 + x + x2 2 +... + xn n! +... sin x = x x3 6 + x5 x2n+1 + 1)n 5! 2n + 1)! 2 2.1 = 1 e.g. 0 = 0.00..., π = 3.14..., 1

数学の基礎訓練I

I 9 6 13 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 3 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............