P.math induction

👨🏻‍🏫 學術台數學

[匿名]

2009-12-29 3:54 am

1. Given two distinct real numbers α, β.
let α + β = p, αβ = q, where p ≠ 1 and a sequence of real numbers (u2), (u3), (u4)..., (un), ... defined by

(u2) = p-q/(p-1) and (u k+1) = p-q/(u k) (k&gE;2)

show that (un) = {[α^(n+1) - β^(n+1)] - (α^n - β^n)} / {(α^n - β^n ) - [α^(n-1) - β^(n-1)]}

2.a let f(x) be a function defined on [a,b], such that
f(x1) + f(x2) &lE;2 f[(x1+x2)/2], for all (x1),(x2) ε [a,b]

for each positive integer n , consider the statement I(n) :
If (xi) ε [a,b], i = 1,2,3,........n
f(x1) + f(x2) + f(x3) + ......f(xn) &lE;nf[(x1+x2+........xn)/n]

(i) prove by induction that I(2^k) is true for evert positive integer k.
(ii) probe that if I(n) (n&gE;2) is true , then I(n-1) is true.
(iii) prove that I(n) is true for every positive integer n

b. by considering f(x) = sinx defined on [0,π], show that
1/n (sin(θ1) + sin(θ2) + ...+ sin(θn) &lE;sin[(θ1 + θ2 + ....θn)/n]
for 0 &lE;θi &lE;π (i = 1,2,3............n)

更新1:

http://farm3.static.flickr.com/2509/4221713257_eb789ecbbb_b.jpg http://farm3.static.flickr.com/2784/4221712575_84e0d3ebe8_b.jpg 第6和第10條自認好人的請做埋第7條

回答 (3)

用戶9112

2009-12-29 8:16 am

✔ 最佳答案

http://img340.imageshack.us/img340/1386/87580778.png

圖片參考：http://img340.imageshack.us/img340/1386/87580778.png

(2)(a)(i) For n = 2, f(x1) + f(x2) = 2f[(x1+x2)/2] (given) therefore l(2^1) is true
Suppose l(2^k) is true, i.e.
f(x1) + f(x2) + ... + f(x_2^k) <= (2^k)f[(x1+x2 + ... + x_2^k)/(2^k)], and
f(x_2^k+1) + ... + f(x_2^k+2^k) <= (2^k)f[(x_2^k+1 + ... + x_2^k+2^k)/(2^k)]
Add them together =>
f(x1) + f(x2) + .... + f[x_2^(k+1)] <= (2^k){f[(x1+x2 + ... + x_2^k)/(2^k)] + f[(x_2^k+1 + ... + x_2^k+2^k)/(2^k)]}
<= (2^k)(2)f{[x1+x2 + ... + x_2^k)/(2^k) + (x_2^k+1 + ... + x_2^k+2^k)/(2^k)]/2}
= [2^(k+1)]f{[x1+x2+...+x_2^(k+1)]/[2^(k+1)]} => l[2^(k+1)] is true
Hence l(2^n) is true for all positive integers n
(ii) if l(n) is true, then
Define y = (x1 + x2 + ... x_n-1)/(n-1) so that y is also in [a,b]
f(x1) + f(x2) + ... + f(x_n-1) + f(y) <= nf[x1+x2+...+x_n-1 + y)/n]
f(x1) + f(x2) + ... + f(x_n-1) + f(y) <= nf{[(n-1)(x1+x2+...+x_n-1)/(n-1) + (x1 + x2 + ... x_n-1)/(n-1)]/n}
f(x1) + f(x2) + ... + f(x_n-1) + f(y) <= nf[(x1 + x2 + ... x_n-1)/(n-1)]
f(x1) + f(x2) + ... + f(x_n-1) <= (n-1)f[(x1 + x2 + ... x_n-1)/(n-1)]
Hence l(n-1) is true
(iii) f(x1) = f(x1) => l(1) is true.
Assume l(k) is true, where 2^p <= k <= 2^(p+1)
l(k) is true => l(k-1) is true => l(k-2) is true .... => l(2^p) is true
l(2^p) is true => l[2^(p+1)] is true => l[2^(p+1) - 1] is true ... => l(k+1) is true
Hence l(n) is true for all positive integers n.
(b) sinθ1 + sinθ2 = 2sin[(θ1 + θ2)/2]cos[(θ1 - θ2)/2] <= 2sin[(θ1 + θ2)/2]
Hence f(θ) = sinθ satisfies the condition in (a) part where θ in [0,π]
Based on result in (a)(iii), (1/n)(sinθ1 + sinθ2 + ... + sinθn) <= sin[(θ1+θ2+... + θn)/n]
http://img187.imageshack.us/img187/2865/24778303.png
圖片參考：http://img187.imageshack.us/img187/2865/24778303.png

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[匿名]

2009-12-30 5:45 am

I would like to know why and how to define y = (x1 + x2 + ... x_n-1)/(n-1) in question 10

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STEVIE-G™

2009-12-29 4:46 am

can you restate your questions??

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收錄日期: 2021-04-18 18:26:34
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