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/sci/ - Science & Math

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6197626 No.6197626[DELETED]  [Reply] [Original]

Do you trust mathematical induction 100%?
Can you really tell that nothing stinks about it?
Where is the rigour?
Pic related as hell , when the graph isnt linear , how the fuck can you use induction , it isnt fucking logical.

This is not a troll thread.

>> No.6197630

>>6197626
induction is a method of prooving something. how is your pic related? post the statement and we will demonstrate how it's valid, if we feel like it.

>> No.6197634

>>6197626
please give your definition of 'mathematical induction'

>> No.6197637

>>6197634

it's when you change the polarity of the taylor expansion so fast that a function is induced in R^n.

>> No.6197641

I don't really understand it

so if the formula works for n = 1 and looks the same for n = k+1 then it's proven?

>> No.6197647

>>6197630
>>6197634
Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers.

>> No.6197650

>>6197630
>>6197634
Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. It is done in two steps. The first step, known as the base case, is to prove the given statement for the first natural number. The second step, known as the inductive step, is to prove that the given statement for any one natural number implies the given statement for the next natural number. From these two steps, mathematical induction is the rule from which we infer that the given statement is established for all natural numbers.

>> No.6197657

>>6197650

so where's the problem? if something proves itself recursively, why should there be an upper boundary where math suddenly breaks for no discernible reason?

>> No.6197655

>>6197630
>>6197634
i can see mathematical induction working on linear statments , but non linear statments? its logically incorrect.

>> No.6197659

>>6197657
see>>6197655

>> No.6197661

>>6197659

why is it logically incorrect? what evidence do you have for this? what lead you to believe this if you have no concrete evidence for this?

>> No.6197664

>>6197661
when the graph between the statment and natural numbers isnt linear , induction loses its rigour

>> No.6197669

>>6197664
Why?
Give an example (based on non-linearity) where the method is not sound.

>> No.6197671

>>6197669
the fact that i cant show any examples doesnt prove the statment that mathematical induction can be applied to non linear statments

>> No.6197674

>>6197671

That may be so but it makes it remarkably difficult to discuss the idea(s) you're trying to address since you seem to be the only one that groks them at any useful level.

>> No.6197673

>>6197626
logically, induction doesn't make much sense.

>> No.6197675

>>6197671
>This is not a troll thread.

Lel, while your latest statement is logically sound it fails to address the underlying issue.
Put forth some reason that induction is flawed or inapplicable or GTFO.

>> No.6197681

Here you go, OP:

http://plato.stanford.edu/entries/induction-problem/

it's a legit problem.

>> No.6197689

>>6197671
Uh... yeah it does! That's exactly what it means you colossal fuckhole!
If no counterexamples exist then it must be legitimate!

>> No.6197691

>>6197681
>has nothing whatever to do with mathematical induction

>> No.6197696

>>6197689
so thats how you prove
a^n+b^n=/=c^n
for any a,b,c and n>2?(a b c n are natural)
you just say if no counterexamples exist then it must be legitimate!

are you retarded? go back to highschool please

>> No.6197697

What the fuck are you talking about OP "non linear statement"

Induction can be proven from the well ordering property of the natural numbers btw: that is that every subset of the natural numbers has a smallest element

>> No.6197698

>>6197691
don't knock it because you don't understand it. it has everything do with it.

>> No.6197700

>>6197698
>>6197681
can you make a tl:dr edition please?

>> No.6197701

>what is the well ordering principle

>> No.6197702
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6197702

>>6197696

>> No.6197710

>>6197701
>>the well ordering principle is true because of induction
>>induction is true because of the well ordering principle

>> No.6197713

>>6197698
That article didn't even mention mathematical induction.
Mathematical induction: for a set <span class="math">S \subseteq \mathbb{N}[/spoiler], if <span class="math">0 \in S[/spoiler], and if <span class="math">n \in S[/spoiler] implies that <span class="math">n+1 \in S[/spoiler], then <span class="math">S = \mathbb{N}[/spoiler].
What exactly does this have to do with induction in the ordinary sense, besides the name?

>> No.6197721

>>6197710
>implying the well ordering principle is not obvious

>> No.6197727

>>6197710
They are equivalent axioms.

Basically, all the other Peano axioms show that the set {0, 1, 2...} (that is, the natural numbers) is a subset of the natural numbers, but you need any axiom that says that there aren't any other natural numbers. Both the well ordering principle and the induction axiom do this.

>> No.6197729

>tfw in last year of BSc in math and don't know how to prove by induction
I didn't have the highschool prerequisites when i started but they didnt check
kek

>> No.6197731

>>6197729
What

>> No.6197734

>>6197729
> applied maths majors

or could be stats

>> No.6197735

>>6197727
>>6197721
i see why they are true , but when the system is non linear (the graph between the natural numbers and the statment we want to prove by induction) , logically the pattern breaks because for each natural number the outcome isnt linear to n+1 or n-1 , so induction breaks because now you have to prove that the fact that the connection isnt linear , it still doesnt break the induction principle

(i know the odds of being right are against me, but i didnt find a good answear to this statment yet)

>> No.6197738

>>6197735
no one gets what you are trying to ask

you can prove stuff like n!> 2^n for all n>3 with induction, which is "non linear".

n=4: 4!=24> 2^4=16

Assume n!> 2^n and n>3
now (n+1)!>(n+1)*2^n> 2^(n+1)

>> No.6197739

>>6197731
what?
>>6197734
Not stats, and the subjects don't differ between pure and applied until the last year so although i will do applied, it hasn't affected me yet.
It hasn't popped up very often though, only like 2 questions in final exams over 2 years so i keep forgetting about it. is this odd?

>> No.6197748

>>6197738
i also see why it works for > or < statments , but not for = statments
(bla+blabla+blablala+(nX+100)=3^n+2)

(when the system is non linear)

>> No.6197758

>>6197739
Yea there aren't many exam questions Ive gotten either that need induction

Still you should probably learn it lol, there's some Khan academy videos I think. Lots of proofs have an induction step. For example the most common proofs of the Binomial Theorem, Jordan Normal Form theorem, Spectral theorem, all use induction. Lots of linear algebra uses induction, because you can use the fact that the dimension of a space is a natural number, so just prove that a statement is true for the vector space {0} and then you can assume its true for any space of any dimension smaller than the one in your proof

>> No.6197763

>>6197735
>non linear
What the fuck are you talking about. Do you even know what induction is?

>> No.6197768

>>6197698
Mathematical induction is actually a form of deductive proof, it's just called "induction" because it proves the general case from the specific case. It is a rigorous, formal proof.

>> No.6197778

>>6197748
Like prove n(n+1)/2=1+2+3+4....n or something?

n(n+1)/2=1+2...n
n(n+1)/2+(n+1)=1+2...n+(n+1)
n(n+1)/2+2(n+1)/2=1+2...n+(n+1)
factorizing gives (n+2)*(n+1)/2=1+2...n+(n+1)

In your original pic are you trying to prove 3^n=3n+1 using induction? Obviously you can't because its wrong but if you tried

n=0: 1=1, so the base case is fine

Assume 3^n=3n+1
Now 3^n*n=(3n+1)*n
3^(n+1)=3n^2+n doesn't equal 3*(n+1)+1 so it fails, you can't get this step to work

>> No.6197777

>>6197748
bla+blabla+blablala+(nX+100)=3^n+2

bla+blabla+blablala+(smallest nX+100)=3^smallest n+2, does it work? great.

bla+blabla+blablala+((n+1)X+100)=3^(n+1)+2, does it work? it's proven!

>> No.6197784

>>6197735
>the graph between the natural numbers and the statment we want to prove by induction
There isn't always a graph of anything, though.
For example, we can show, by induction, that if R is noetherian, then so are R[X], R[X,Y], R[X,Y,Z] ... etc. Now the notions of linearity and nonlinearity stop applying, because each object is a type of ring, not some real number that can sit no a line.

>> No.6197787

>>6197748
>i also see why it works for > or < statments , but not for = statments
Mathematical induction is a general method. The details of the proof will vary depending on the problem. But for instance, say we want to prove that 0 + 1 + 2 + . . . + n = n (n+1)/2 for n in N. The case n = 0 is trivial, because 0 (0+1)/2 = 0.

Now for the induction step. Assuming 0 + 1 + 2 + . . . + k = k (k+1)/2 for some k, prove that 0 + 1 + 2 + . . . + k + (k+1) = (k+1)(K+2)/2.
>1 + 2 + ... + k + (k+1) =
>k(k+1)/2 + (k+1) = (k(k+1) + 2 (k+1))/2 =
>(k+1)(k+2)/2

>> No.6197791

>>6197787
>>6197778
>1+2+3...n=n(n+1)/2 hivemind

On another note /sci/ is a lot nicer a place than people give us credit for look at all these people trying to help OP understand induction

>> No.6197796

>>6197784
OP is an applied-mathfag. Rings might as well not even exist.

>> No.6197803

There is of course an axiom of induction necessary for this proof to work. Specifically, we have to say that if some statement is true for 0 and its truth for k implies its truth for k+1, then it is true for all n. I'm pretty sure this is equivalent to the well-ordering principle, and is a defining characteristic of the natural numbers.

>> No.6197823

Induction is just recursion in programming. Ever written a recursive function that operates on a tree structure, and it magically works?

Same as in math.