• 沒有找到結果。

· K < r\r  Á ž m 2006 c 12 20 F

N/A
N/A
Protected

Academic year: 2021

Share "· K < r\r  Á ž m 2006 c 12 20 F"

Copied!
2
0
0

加載中.... (立即查看全文)

全文

(1)

H ® Œ Æ ê Æ X Á ò

 2  1 1 

2005 / 2006 Æc 1  ÆÏ ‘ § ¶ ¡ êÆ©Û Áòa. A ò Á/ª 4ò ¦ ^  ? 2005 ?

· K < r\r  Á ž m 2006 c 12  20 F

K Ò ˜  n o Ê 8 Ô l Ê › o © ò<

 ©

 ? Æ Ò 6 ¶ .. .. .. .. .. .. .. .. .. .. . C .. .. .. .. .. .. .. .. .. .. . ¾ .. .. .. .. .. .. .. .. .. .. . ‚ .. .. .. .. .. .. .. .. .. .. . `²µ

1. žò? !ÆÒ!6 ¶3Áò†ýC¾‚ .

2. Áò 12 ŒK, ÷© 150 ©, Ážm 180 ©¨ "

˜ ! £ 30 ©¤Þ~µ

1. ޘ‡4:3 [0, 1] «mþȗê.

2. ޘ‡kÄmä:¼ê.

3. ޘ‡ëYØ´˜—ëY¼ê.

4. ޘ‡Œ_Œ‡¼ê, Ù_¼ê،‡.

5. ޘ‡š"Œ‡¼ê, §3,˜:?¿êþ".

6. ޘ‡ Riemann ،ȼê.

7. ޘ‡šK¼ê f (x), §3 [0, +∞) þÈ©Âñ, 4 lim x→+∞ f (x) Ø3.

8. ޘ‡3 [0, 1] × [0, 1] þ½Â¼ê f(x, y), §©OéuCþ x, y ëY, Ø

´ëY¼ê.

9. ޘ‡ ê3, ،‡¼ê.

10. ޘ‡ÂñØýéÂñê‘?ê.

 ! £ 10 ©¤b˜¼ê φ(x) ˜ëYŒ, - f (x) = x 2 · φ(x), OŽ f 00 (0).

n ! £ 10 ©¤ïĘ¼ê y = sin(x 3 ) 4Š: ! ":, ¿xÑúã.

o ! £ 10 ©¤OŽÈ© R

π 2

0 (cos x + cos 2x + · · · + cos 20x) · sin(10x)dx.

Ê ! £ 10 ©¤OŽÈ© R R S (1+x+y) ds

2

, Ù¥ S o¡N x + y + z ≤ 1, x ≥ 0, y ≥ 0,

z ≥ 0 >.­¡.

(2)

H ® Œ Æê Æ X Á ò  2  1 2 

8 ! £ 10 ©¤OŽ­È© R R D |x − y|dxdy, Ù¥ D   x 2 + y 2 ≤ a 2 .

Ô ! £ 10 © ¤  ¼ ê f (x) 3 [0, 1] « m   ë Y Œ ‡, … f (0) = 0, f (1) = 1, f 00 (x) < 0, ∀x ∈ [0, 1]. y² f(x) ≥ x, ∀x ∈ [0, 1].

l! £ 10 ©¤ a n+1 > a n > 0, n = 1, 2, · · ·. y², ?ê P n=1 ( a

n+1

a

n

− 1) Âñ¿

©7‡^‡´?ê P n=1 (1 − a a

n+1n

) Âñ.

Ê ! £ 15 ©¤ f n (x) ´ [0, 1] þšKëY¼ê, é ∀ > 0, ?ê P n=1 f n (x) 3 [0, 1 − ] þ˜—Âñ f(x). e lim x→1

f(x) = a k, y²

(a) P n=1 f n (1) 喦

(b) ?˜Ú§b P n=1 f n (1) = a, - f(x) = ¯

 

 

f (x), x ∈ [0, 1) a, x = 1 K P n=1 f n (x) 3 [0, 1] þ˜—Âñ ¯ f(x).

› ! £ 10 ©¤ f : R 2 → R ëY¼ê. y²3ü‡ØÓ: p, q ¦

f(p) = f (q).

›˜! £ 15 ©¤(1)  f (t) ëY, Áy²

Z ZZ

x

2

+y

2

+z

2

≤1 f (ax + by + cz)dxdydz = π

Z 1

−1 (1 − u 2 )f (ku)du, Ù¥ k = √

a 2 + b 2 + c 2 > 0.

(2) |^ (1) ½†OŽÈ©

Z Z

S ( 1

2 x 2 + xy + xz)dydz

Ù¥ S ´¥¡ (x − α) 2 + (y − β) 2 + (z − γ) 2 = R 2 , …È©´÷¥¡ ý .

› ! £ 10 ©¤ f : R n → R n  C 1 • þŠ¼ê, …÷v^‡

kf(x) − f(y)k ≥ kx − yk, ∀x, y ∈ R n .

ù p, k · k ´ R n þIO‰ê. y² f Œ_, …Ù_N´ C 1 .

參考文獻

相關文件

反之, 有了 parametric equation, 我們可利用這些在 R n 的 direction vectors, 利 用解聯立方程組的方法求出和這些 direction vectors 垂直的 normal vectors,

而利用 row vectors 的方法, 由於可以化為 reduced echelon form, 而 basis 是由此 reduced echelon form 中的 nonzero vectors 所組成, 所以雖然和來的 spanning

We point out that extending the concepts of r-convex and quasi-convex functions to the setting associated with second-order cone, which be- longs to symmetric cones, is not easy

Hence, we have shown the S-duality at the Poisson level for a D3-brane in R-R and NS-NS backgrounds.... Hence, we have shown the S-duality at the Poisson level for a D3-brane in R-R

We compare the results of analytical and numerical studies of lattice 2D quantum gravity, where the internal quantum metric is described by random (dynamical)

 依序填入該學生社團負責人之相關資訊,並於下方

 童書有很豐富的內容,可教的東西很 多,成人可以因應兒童的興趣隨機施

 Compute the resource consumption o f an internal node as follows:. ◦ Find the demanding child with minimum dominant share