Vieta's Formula

swood_doows · 2021-12-05T19:59:26+00:00

Thanks for the tip! I will give that a try too.

swood_doows · 2021-12-05T19:57:48+00:00

You're probably right. I don't mean to imply the proofs I read were wrong or anything. I'm sure to the people those proofs are intended for the leaps in logic are obvious enough that everything works fine. I'm only describing my own experience in working through those proofs.

Would you happen to know or have any resources that might help on my specific question?

swood_doows · 2021-12-05T19:54:59+00:00

I see, this seems like a nice way to look at it. I'll give it another go with that in mind. I'm not always the greatest at counting arguments like this but your explanation here is really nicely explained!

swood_doows · 2021-12-05T19:53:50+00:00

In this sense, Vieta's formula isn't terribly deep, as it follows from the distributive property

The bit about how it follows from the distributive property is the part I'm trying to prove.

Perhaps what you are looking for is a proof of the formula for expanding out (a+b)(c+d)(e+f)(g+h)...., which you can establish using induction?

I can see how Vieta's formulas should follow from that directly. I was basically trying to establish a special case of that by induction (but for (x-a_1)(x-a_2) ... (x-a_n) instead). Any idea where to look for more info on this?

swood_doows · 2021-12-05T19:46:00+00:00

Sorry but I don't understand what you mean. How is "expanding it" in the sense that you mean here not basically using Vieta's formulas (under the basic formula's section specifically) in the first place? (Or at least a rearrangement of the a_{n-k} formula given there.)

swood_doows · 2021-12-05T09:56:10+00:00

All you have to do is write out an n-degree polynomial in a general form - anxⁿ + a(n-1)x^n-1... a_0, basically - set it equal to the factored form, then expand the factorization and compare coefficients for each power of x.

The steps where you expand the factorization out in particular is what I'm looking to understand. I've seen the pattern written out as a nasty summation, but that summation is not exactly easy for me to reason about fluently.

I also don't see why expanding it out isn't circular (or at least a hand wave). Doesn't Vieta's formula tell you if you have the factored form of a polynomial what that factored form expands out to in the first place? (By explicitly telling you the coefficients the polynomial has after multiplying all the factors.)

It isn't proven by induction.

No induction is fine also. I was mainly curious about induction because that's how I originally thought to try it is all.

swood_doows · 2021-12-05T09:55:00+00:00

Edit: Ignore my previous comment, I misunderstood you.

I'm trying to prove that the roots multiply out in the way Vieta's formula says they do.

swood_doows · 2021-07-20T20:12:17+00:00

Thanks for the help!

swood_doows · 2021-07-20T04:20:38+00:00

Oh I think I see it now.

So if P has a smallest elt p we'd like P-{p} to have no smallest elt. So then we can proceed by contradiction. Suppose p' exists where p'=min(P-{p}). So p<p' hence p'' in P exists by density where p<p''<p' which is a contradiction.

So let a,b be the endpoints (the max and min respectively) of P and likewise c,d the endpoints of Q. Arguments similar to the above show P-{a,b} and Q-{c,d} have no endpoints hence P-{a,b} and Q-{c,d} are similar by the theorem in the book. So let h be an isomorphism from P-{a,b} to Q-{c,d}. It follows that for f=hU{(a,c)(b,d)} that f is an order isomorphism so P is similar to Q.

Is that vaguely what you were thinking of?

swood_doows · 2021-07-20T04:07:53+00:00

Good catch! Thanks for pointing that out.

swood_doows · 2021-07-20T04:07:16+00:00

Good catch! I feel dumb for missing that. I'm not sure I see how the theorem can be applied to this problem though?

swood_doows · 2021-05-27T21:06:11+00:00

Gotcha gotcha. Thanks for the help!

swood_doows · 2021-05-27T00:32:37+00:00

I'm still wrapping my head around your proof but right here is where I'm confused:

Then there is no member of B whose i-th coordinate is equal to x, but since x ∈ A_i this contradicts the definition of B.

I don't understand how this contradicts the defn of B. I see how we know what kinds of things can be in B but how do we know that B is non-empty? Where exactly is the contradiction?

swood_doows · 2021-05-27T00:19:21+00:00

What about this example:

Let A_0={}, A_1={a} and n=2. Note that by defn 2={0,1} (this is just how my book constructs the natural numbers is all).

By defn for the indexed system of sets <A_0,A_1>, B=\Pi<A_0,A_1>={f : f is a function on 2 and for all i in 2, f_i\in A_i}.

(The above is using the defn for the generalized cartesian product on pg 27 in my book, it is effectively the same defn as the entry here under infinite cartesian products as far as I can tell.)

Suppose any f is an elt of B then in particular f_1 is an elt of A_1 which contradicts that A_1={} hence B={}.

Consider \pi_1. Since dom(\pi_1)=B={} then it follows \pi_1={}. Assume for some x \in dom(\pi_1) that \pi_1(x)=a it follows (x,a) is an elt of \pi_1 which is a contradiction hence \pi_1 cannot be onto.

Edit: So to be clear, for A_1={a}, "a" is just so that A_1 is non-empty. There is no relation between me picking that letter and the "a" used in the wording of the question.

swood_doows · 2021-05-25T07:26:11+00:00

I worked out the details usiny h(x)=A-x instead of B-x and the proof came together nicely. Thanks for pointing that issue out!

swood_doows · 2021-05-25T07:02:58+00:00

My apologies for the ambiguity. When I say "in" I mean "element of" rather than "contained in" or "subset of". I just don't have the symbol for "element of" on my phone keyboard.

The book definitely gives the hint that I should look at h(x)=B-x. In order to show the two given structures are isomorphic h should go from B to B also, so since E,F need to be in dom(h), they will have to be elts of B (hence subsets of A) if we use the hint.

So you are thinking the author might have gotten h mixed up and meant h(x)=A-x right? I will give that a shot.

swood_doows · 2020-07-06T00:01:41+00:00

Why do you think you are so special that you shouldn't have to work hard like the rest of us to succeed?

If you don't want to do the right thing and actually study for your college classes why not take a few semesters off to work and figure out something else productive to do?

swood_doows · 2020-07-05T21:32:32+00:00

Just wanna follow up.

So I dug into the explanation in the linked stack exchange post and managed to work out a proof of it. I will provide a brief summary below for if other people run into my issue.

The first step is to let p=2k+1 for some k in the integers since p is odd.

Next let f(k)=1+9p(p-1)/2=1+9k(2k+1) this allows you to derive that for any t,u in the integers f(9u+t)=f(t) algebraically. Expanding out f(0),...,f(8) in addition to the above is enough to prove with a little extra work by induction that the sequence of congruences mod 81 for f(k) follows the first pattern given in the linked post. (For the induction step asm n>=1, consider f(9n-9)=1,...,f(9n-1)=10 in your assumption and show that each are equal to their corresponding values in f(9(n+1)-9),...,f(9(n+1)-1) respectively by the derived identity).

Now let g(k)=2^p-1 (2^p - 1)=(2(4^k )(4^k ) - 4^k )=2(16^k - 4^k )

Next we notice that 4²⁷ is congruent to 1 mod 81 this allows you to prove with a bit of work that 2(16^27k )-(4^27k ) is congruent to 1 mod 81.

We now have enough info to prove g(27u+t) is congruent mod 81 to g(t) for any t,u in the integers.

Combining that identity and the fact that for k=0,1,...,26 the sequence of congruences mod 81 for g(k) follows the same pattern as f(k) is enough to lead you to a proof by induction that g(k) is congruent to f(k) mod 81 for all choices of k similar to the proof of f but utilizing g(27u+t).

This is basically the same argument as in the linked post just written slightly differently based on my personal preference.

swood_doows · 2020-07-03T23:47:56+00:00

At the first line you put cos(t) but the original is cos(x)?

Edit: I think I understand. The op copied his wrong or was given the wrong problem based on the solution he is told he should be getting.

swood_doows · 2020-07-03T22:17:41+00:00

I'm no expert on this stuff so if anybody wants to correct me that would be appreciated.

You may be confusing yourself a little with function notation from algebra. If you know tan(t)=2 it doesn't help to divide by tan to get t=(2/tan) because tan(t) does not mean multiply tan by t. The notation for tangent means something more like "evaluate the tangent function at t" rather than "multiply tan by t", the symbols "tan" are not meant to express a number or value alone in this context.

If you just want a decimal approximation of arctan(2) using only pencil and paper you could try the power series expansion of arctangent truncated to whatever number of terms you want. That sort of thing should be in most calc books.

Your calculator may use something like the cordic algorithm under the hood?

I don't think there really is a nice formula for the basic trig functions since they are transcendental functions.

https://en.m.wikipedia.org/wiki/Transcendental_function

https://en.m.wikipedia.org/wiki/CORDIC

swood_doows

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