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Mathematical induction

  1. Mar 21, 2009 #1
    While learning mathematical induction,an idea occurred to me.
    Mathematical induction is done by proving that the first statement in the infinite sequence of statements is true, and then proving that if any one statement in the infinite sequence of statements is true, then so is the next one.
    Similarly, I can prove a proposition by proving a proposition is true when x=x0 and then proving that if any x can make the proposition true, then so is x+Δx(Δx>0 andΔx→0).
    e.g.:
    prove e^(i m)=cos(m)+i sin(m)(Euler’s formula)
    First, when m=0,obviously,it's true.
    Second, assume it's true when m=x,then when m= x+Δx(Δx>0 andΔx→0),
    e^i(x+∆x) ={d/dx[e^(ix)]}∆x+e^(ix)=e^(ix)+i e^(ix) ∆x=cos(x)+i sin(x)+i[cos(x)+i sin(x) ]∆x=[cos(x)-sin(x)∆x]+i[sin(x)+cos(x)∆x]
    ∵∆x→0,∴cos(∆x)=1,sin(∆x)=∆x;
    ∴cos(x+∆x)=cos(x)cos(∆x)-sin(x)sin(∆x)=cos(x)-sin(x)∆x;
    sin(x+∆x)=sin(x)cos(∆x)+cos(x)sin(∆x)=sin(x)+cos(x)∆x;
    ∴[cos(x)-sin(x)∆x]+i[sin(x)+cos(x)∆x]=cos(x+∆x)+i sin(x+∆x)
    Therefore,if the formula is true for m=x,the formula is also true for m= x+Δx.
    It's easy to prove if the formula is true for m=x,the formula is also true for m=x-Δx
    So the formula is true for any m.
    What do you think of my idea?Please tell me your opinion.
    Note:
    1.I'm not a native English speaker. So there maybe many errors on language.If you don't understand what I'm saying,please tell me and I would correct the mistakes.I would be glad if you point our my errors on language.The reason why I come to Physics Forums is that forums in my country are either unpopular or aimed to college entrance exam,which makes hundreds of thousands of students each province work hard to get a higher score in order to be admitted into a good university,rather than intend for math and and physics fans.
    2.If my idea is wrong,please tell me where the error is.
     
  2. jcsd
  3. Mar 21, 2009 #2
    What do you mean by "Δx→0"?

    It's true for any integer m. Keep in mind that your proof is only of a special case of euler's formula.
     
  4. Mar 21, 2009 #3
    why?
    Δx→0 means that Δx is smaller than any positive number but larger than 0.
    It's true for m=0,then it's true for m=Δx,then it'x true for m=2Δx,then it's true for m=3Δx,...,then it's true for m=(0.01/Δx)×Δx=0.01,then it's true for m=(0.02/Δx)×Δx=0.02,...,then it's true for m=(any x/Δx)×Δx=any x
     
  5. Mar 22, 2009 #4
    The problem with using induction for this is that induction can only possibly prove something for a countably infinite number of cases (integers, etc) and if you are talking about real or complex numbers, you have an uncountable infinity of cases.

    Basically, if you say that delta_x goes to zero in the analysis sense, then you can apply your inductive step as many times as you please and it will never prove anything for a number bigger than zero. Why? 0 + delta_x = 0 as delta_x goes -> 0.

    You can certainly scale the delta_x down to 1, 0.1, 0.01, ..., 0.0000...1, ... but you'll always end up proving Euler's formula for only countably many cases.
     
  6. Mar 22, 2009 #5
    I mean, you should be able to give one proof that will work for all cases. That's the idea. You should be able to write down one proof, and I should be able to verify it in all instances. Your method leaves holes.

    Say you choose delta_x to be 0.0000...1 with 1,000,000,000,000 zeros before the 1 after the decimal. You'd think this gets just about all real numbers, but you'd be wrong. Because between 0 and delta_x you already have infinitly more instances of the problem your theorem says nothing about than all the cases your theorem does cover.
     
  7. Mar 22, 2009 #6
    Even if you "scale down" deltaX I believe the proof still breaks down: As long as deltaX is finite (.1 or .01 or even .0000000001) the proof doesn't work because sin(deltaX) does not exactly equal deltaX and cos(deltaX) does not exactly equal 1.



    But your manipulations are still valuable because they prove Euler's formula in a different way:

    You've got all the ingredients to show that both e^ix and sin(x) + i cos(x) satisfy the same differential equation (f(x+dX) = f(x) +if(x)*dX) and have the same boundary condition (at x=0 they both are 1). Since differential equations like these have unique solutions for a given boundry conditions you've shown that the two functions are the same for all x.
     
  8. Mar 22, 2009 #7
    I see where is the mistake.

    But I have done some work the moment the idea occurred to me to verify whether it's true.I tried to prove x^2<1 when x≥0 using my method.if I succeed,it's obvious that my method is wrong.But I failed.Here is my work:
    First,when x=0,obviously it's true.
    Second,assume it's true for x=m.And then for x=m+Δm:
    (m+Δm)^2=m^2+2mΔm<1+2mΔm.But 1+2mΔm>1.
    It means if it's true for x=m,it's uncertain to be true for x=m+Δm.

    I also tried to prove some other proposition that is obvious wrong and also failed.

    Can you give me a proof of a proposition that is obvious wrong such as 1+1=3 using my method?
     
  9. Mar 22, 2009 #8

    matt grime

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    You're just doing calculus, i.e. exploiting some property of continuity. It's nothing to do with induction.
     
  10. Mar 22, 2009 #9
    Is what you're trying to express: to show f(x)=g(x), show that f(0)=g(0) and f'(x)=g'(x).
     
  11. Mar 22, 2009 #10
    Your limit notation is sloppy.

    What you should be doing is proving it for m and then for (m+dm) as dm->0. So...

    [lim (m+dm)]^2 = lim (m+dm)^2 = lim (m^2 + dm^2 + 2mdm) = lim (m^2) + lim (dm^2) + lim (2mdm) < 1 + 0 + 0 = 1.

    So if it holds for m, so must it hold for m + lim(dm) = lim(m+dm).

    So your very only example can provide the sort of counter-evidence you are asking for.
     
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