zoukankan      html  css  js  c++  java
  • [转]Mathematical Induction --数学归纳法1

    Mathematical Induction

    Mathematical Induction is a special way of proving things. It has only 2 steps:

    • Step 1. Show it is true for the first one
    • Step 2. Show that if any one is true then the next one is true

    Then all are true

    Domino Effect

    Have you heard of the "Domino Effect"?

    • Step 1. The first domino falls
    • Step 2. When any domino falls, the next domino falls

    So ... all dominos will fall!

    That is how Mathematical Induction works.

    In the world of numbers we say:

    • Step 1. Show it is true for n=1
    • Step 2. Show that if n=k is true then n=k+1 is also true

    How to Do it

    Step 1 is usually easy, we just have to prove it is true for n=1

    Step 2 is best done this way:

    • Assume it is true for n=k
    • Prove it is true for n=k+1 (we can use the n=k case as a fact.)

    Step 2 can often be tricky ... because we may need to use imaginative tricks to make it work!

    Like in this example:

    Example: 3n−1 is a multiple of 2

    Is that true? Let us find out.

    1. Show it is true for n=1

    31−1 = 3−1 = 2

    Yes 2 is a multiple of 2. That was easy.

    31−1 is true

     

    2. Assume it is true for n=k

    3k−1 is true

    (Hang on! How do we know that? We don't!
    It is an assumption ... that we treat
    as a fact for the rest of this example)

     

    Now, prove that 3k+1−1 is a multiple of 2

    3k+1 is also 3×3k

    And then split into and

    And each of these are multiples of 2

     

    Because:

    • 2×3k is a multiple of 2 (we are multiplying by 2)
    • 3k−1 is true (we said that in the assumption above)

    So:

    3k+1−1 is true

    DONE!

    Did you see how we used the 3k−1 case as being true, even though we had not proved it? That is OK, because we are relying on the Domino Effect ...

    ... we are asking if any domino falls will the next one fall?

    So we take it as a fact (temporarily) that the "n=k" domino falls (i.e. 3k−1 is true), and see if that means the "n=k+1" domino will also fall.

    Tricks

    I said before that we often need to use imaginative tricks.

    A common trick is to rewrite the n=k+1 case into 2 parts:

    • one part being the n=k case (which is assumed to be true)
    • the other part can then be checked to see if it is also true

    We did that in the example above, and here is another one:

    Example: Adding up Odd Numbers

    1 + 3 + 5 + ... + (2n−1) = n2

    1. Show it is true for n=1

    1 = 12 is True

     

    2. Assume it is true for n=k

    1 + 3 + 5 + ... + (2k−1) = k2 is True
    (An assumption!)

    Now, prove it is true for "k+1"

    1 + 3 + 5 + ... + (2k−1) + (2(k+1)−1) = (k+1)2   ?

    We know that 1 + 3 + 5 + ... + (2k−1) = k2 (the assumption above), so we can do a replacement for all but the last term:

    k2 + (2(k+1)−1) = (k+1)2

    Now expand all terms:

    k2 + 2k + 2 − 1 = k2 + 2k+1

    And simplify:

    k2 + 2k + 1 = k2 + 2k + 1

    They are the same! So it is true.

    So:

    1 + 3 + 5 + ... + (2(k+1)−1) = (k+1)2 is True

    DONE!

    So there you have it!

     
    Search :: Index :: About :: Contact :: Contribute :: Cite This Page :: Privacy

    Copyright © 2014 MathsIsFun.com
     
     
  • 相关阅读:
    树型表的设计 上海
    FTP通讯封装 上海
    线程淡写 上海
    TCP通讯故障 上海
    设计模式引导 上海
    初试Delegate 上海
    c# 扫描端口 上海
    攻读计算机研究生的看法(转载) 上海
    挖掘表字段中的汉字 上海
    新生活运动 上海
  • 原文地址:https://www.cnblogs.com/freebird92/p/5955386.html
Copyright © 2011-2022 走看看