Wilsons Theorem?

**Cairo** · May 23rd 2009, 09:53 AM

Hello,

I posted this question some time ago and got a great response which I now understand. However, I am still struggling to understand the leap from (*) to (**). This is Wilsons Theorem right?

Prove that

(e^x)(e^x2/2)(e^x^3/3)...=1+x+x^2+... when |x|<1.

Show that the coefficient of x^19 in the power series expansion on the LHS has the form

1/19! + 1/19 + r/s,

where 19 does not divide s. (*)

Deduce that 18!= -1(mod 19). (**)

Thanks in advance

**Media_Man** · May 23rd 2009, 07:18 PM

Can we assume that if $\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty b_nx^n$ for all values of $|x|<1$ , then $a_n=b_n$ for all $n$ ? Somehow I don't think this is valid, but let's run with it for a moment.

If this is true, then the whole point of the two-part exercise was to make the connection that $\frac{1}{19!}+\frac{1}{19}+\frac{r}{s}=1$ for some $r,s$ where $19\not|s$

Multiplying gives us $1+18!=19!-19!\frac{r}{s}$ . Now reduce the $\frac{r}{s}$ term and because the LHS is a whole number, the RHS must reduce fully. Since $19\not|s$ , 19 will still be a factor in this term after reducing, so it can be written $1+18!=19k$ for some positive integer k, which implies your hypothesis.

Not sure if this quasi-proof holds water, but it is the only connection I see between the two parts of the question.

**ThePerfectHacker** · May 23rd 2009, 08:38 PM

Originally Posted by Media_Man

Can we assume that if $\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty b_nx^n$ for all values of $|x|<1$ , then $a_n=b_n$ for all $n$ ? Somehow I don't think this is valid, but let's run with it for a moment.

Yes. Power series for analytic functions on $(-1,1)$ must be unique. The proof is easy.

**Cairo** · May 23rd 2009, 09:17 PM

Thanks Media Man. Your assumption is valid and it now makes sense.

**Media_Man** · May 24th 2009, 05:11 AM

Yes. Power series for analytic functions on $(-1,1)$ must be unique. The proof is easy.

That is interesting. What is the proof? Does this hold for $x \not\in (-1,1)$ provided the series converges?

**ThePerfectHacker** · May 24th 2009, 08:57 AM

Originally Posted by Media_Man

That is interesting. What is the proof? Does this hold for $x \not\in (-1,1)$ provided the series converges?

Say that, $a_0+a_1x+a_2x^2 + ... = b_0 + b_1 x + b_2 x^2 + ...$ for all $|x|<1$ .
If $x=0$ then we get $a_0 = b_0$ , subtract it off both sides.
That gives us, $a_1 x + a_2 x^2 + .... = b_1 x + b_2 x^2 + ...$ for all $|x|<1$ .
Differenciate and let $x=0$ then we get $a_1 = b_1$ , subtract it off both sides.
That gives us, $a_2 x^2 + ... = b_2 x^2 + ...$ for all $|x|<1$ .
Differenciate twice and let $x=0$ then we get $2a_2 = 2b_2 \implies a_2 = b_2$ .
Now keep on going.

**Media_Man** · May 24th 2009, 09:08 AM

Sensible enough. Why would this proof not work on some other interval of convergence besides (-1,1)? Aren't there Taylor series expansions for functions that converge within other radii?

**ThePerfectHacker** · May 24th 2009, 09:38 AM

Originally Posted by Media_Man

Sensible enough. Why would this proof not work on some other interval of convergence besides (-1,1)? Aren't there Taylor series expansions for functions that converge within other radii?

It works on all other intervals too. I just proved it for that interval because that was the assumption you were using in this problem.

**Media_Man** · May 24th 2009, 10:03 AM

I know I am being picky, but it seems to me that it would be possible to construct two series $f(x)=\sum_{n=0}^\infty a_nx^n$ and $g(x)=\sum_{n=0}^\infty b_nx^n$ such that $f(x)=g(x)$ for $|x|<1$ but $f(x)\neq g(x)$ for $|x|\geq 1$ , both of which converging within, say, $[-2,2]$ .

The preceding proof seems to imply that no such differentiable functions $f$ and $g$ can ever be found.

**ThePerfectHacker** · May 24th 2009, 10:12 AM

Originally Posted by Media_Man

I know I am being picky, but it seems to me that it would be possible to construct two series $f(x)=\sum_{n=0}^\infty a_nx^n$ and $g(x)=\sum_{n=0}^\infty b_nx^n$ such that $f(x)=g(x)$ for $|x|<1$ but $f(x)\neq g(x)$ for $|x|\geq 1$ , both of which converging within, say, $[-2,2]$ .

Once they agree on $(-1,1)$ then it will mean $a_n = b_n$ and so if the series converge all the way up to $(-2,2)$ then definitely $f(x) = g(x)$ on the entire interval since the coefficients are the same.

**Media_Man** · May 24th 2009, 10:59 AM

I think I worked it out in my head now. One of the stipulations of the proof is that the function expressible by the series is infinitely differentiable in the interval of convergence. So defining $h(x)=f(x)-g(x)$ , $h$ could not possibly be equal to zero within some interval $x\in(a-\epsilon, a+\epsilon)$ and nonzero at a point outside this interval. Otherwise the nth derivative of h at $a+\epsilon$ would have a left hand limit of zero and a nonzero right hand limit for some $n$ .

*Not trying to argue with mathematical proof, I'm just trying to get my intuition to catch up to it. Lemma: No infinitely differentiable function can be constant on an interval no matter how small, unless it is constant everywhere.

**ThePerfectHacker** · May 24th 2009, 11:23 AM

Originally Posted by Media_Man

*Not trying to argue with mathematical proof, I'm just trying to get my intuition to catch up to it. Lemma: No infinitely differentiable function can be constant on an interval no matter how small, unless it is constant everywhere.

Careful, let $f(x) = \left\{ \begin{array}{c} 0 \text{ if }x\leq 0 \\ e^{-1/x} \text{ if }x>0 \end{array} \right.$ then $f$ is infinite differenciable with Taylor series identically $0$ centered at $0$ .

The function is identically zero if the interval is contained in the left-hand side of
the number line but it is not identically zero if the line crosses into the right hand side of the number line.

**Media_Man** · May 24th 2009, 11:55 AM

Ah, yes, this is just the kind of function I feared. But alas, let's rephrase again: Lemma: No infinitely differentiable function expressible by a single Taylor series on some interval A can be constant on an interval $B\subset A$ no matter how small, unless it is constant everywhere in A. The point is you can't have two distinct Taylor series that converge at the same value on some interval and converge on different values on some other interval. The above example requires two different Taylor series to fully describe it.

Sure, the values of the functions $g(x)=0$ and $f(x)$ defined above agree on the interval $(-\infty,0]$ and disagree on the interval $(0,\infty)$ but that is because $f(x)$ , though an infinitely differentiable function at every point, cannot be expressed by a Taylor series which converges on $+a$ and $-a$ , for some $a\neq 0$ .

*I may be getting caught up in technicalities, but we have already posed a sound proof here. This is just me making sure my intuition follows suit.

**chiph588@** · May 26th 2009, 07:49 AM

Originally Posted by ThePerfectHacker

Say that, $a_0+a_1x+a_2x^2 + ... = b_0 + b_1 x + b_2 x^2 + ...$ for all $|x|<1$ .
If $x=0$ then we get $a_0 = b_0$ , subtract it off both sides.
That gives us, $a_1 x + a_2 x^2 + .... = b_1 x + b_2 x^2 + ...$ for all $|x|<1$ .
Differenciate and let $x=0$ then we get $a_1 = b_1$ , subtract it off both sides.
That gives us, $a_2 x^2 + ... = b_2 x^2 + ...$ for all $|x|<1$ .
Differenciate twice and let $x=0$ then we get $2a_2 = 2b_2 \implies a_2 = b_2$ .
Now keep on going.

Are we assuming we can differentiate term by term though? If the sum doesn't converge absolutely isn't this invalid?

**ThePerfectHacker** · May 26th 2009, 08:47 AM

Originally Posted by chiph588@

Are we assuming we can differentiate term by term though? If the sum doesn't converge absolutely isn't this invalid?

If you have a power series and the terms $a_n$ are sufficiently bounded, say, $|a_n| \leq n!$ then the series converges absolutely on $(-1,1)$ and so it is okay to differenciate term-by-term.

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