# Congruence

• May 9th 2009, 09:58 AM
Cairo
Congruence
Can somebody help with this question please?

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).
• May 9th 2009, 10:36 AM
running-gag
Hi

$\prod_{k=1}^{+\infty}e^{\frac{x^k}{k}} = e^{\sum_{k=1}^{+\infty}\frac{x^k}{k}}$

To find $\sum_{k=1}^{+\infty}\frac{x^k}{k}$ let's take the derivative

$\sum_{k=0}^{+\infty}x^k = \frac{1}{1-x}$

Therefore $\sum_{k=1}^{+\infty}\frac{x^k}{k}$ is the antiderivative of $\frac{1}{1-x}$ that is 0 for x=0

$\sum_{k=1}^{+\infty}\frac{x^k}{k} = -\ln|1-x|$

Therefore $\prod_{k=1}^{+\infty}e^{\frac{x^k}{k}} = e^{\sum_{k=1}^{+\infty}\frac{x^k}{k}} = e^{-\ln|1-x|} = \frac{1}{|1-x|}$
• May 9th 2009, 11:51 AM
Cairo
Thanks.

Is anybody able to help with the second part of the question please?
• May 10th 2009, 03:57 AM
running-gag
$e^{x} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} + \cdots$

$e^{\frac{x^2}{2}} = 1 + \frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{8\cdot 3!} + \cdots + \frac{x^{2n}}{2^n\cdot n!} + \cdots$

$e^{\frac{x^3}{3}} = 1 + \frac{x^3}{3} + \frac{x^6}{18} + \frac{x^9}{27\cdot 3!} + \cdots + \frac{x^{3n}}{3^n\cdot n!} + \cdots$

etc

To find the coefficient of x^19 in the power series expansion you need to multiply all these identities and find the coefficient of x^19
• May 10th 2009, 04:32 AM
Cairo
surely there must be an easier way to do this?
• May 10th 2009, 05:21 AM
running-gag
I hope so (Happy)

If you consider each term of the expansion of e^x from x^19 down to 1
- x^19/19! must be multiplied by a constant which is 1 therefore 1/19!
- x^18/18! must be multiplied by something times x, but this is not possible since in the other series are involved terms starting from x², therefore 0
- x^17/17! must be multiplied by something times x², which is x²/2 coming from the expansion of e^(x²/2) therefore 1/(2x17!)
- x^16/16! must be multiplied by something times x^3, which is x^3/3 coming from the expansion of e^(x^3/3) therefore 1/(3x16!)
- x^15/15! must be multiplied by something times x^4, which is x^4/4 coming from the expansion of e^(x^4/4) but also x^4/8 coming from the expansion of e^(x^2/2) therefore (1/4+1/8)/15! = 3/(8x15!)
- x^14/14! must be multiplied by something times x^5, which is x^5/5 coming from the expansion of e^(x^5/5) but also x^5/6 coming from the product of the expansion of e^(x^2/2) and the expansion of e^(x^3/3) therefore (1/5+1/6)/14! = 11/(30x14!)

and so on ... but it becomes more and more difficult (Wondering)
• May 11th 2009, 07:25 PM
Media_Man
You guys are so close...
Yes, this is the ticket. Running-gag has the winning strategy, but don't think for a minute you have to find the actual coefficient. All the problem calls for is showing that it is of the form $\frac{1}{19!}+\frac{1}{19}+\frac{r}{s}$ , for 19 not dividing s.

For reference, call the series expansion $e^{\frac{x^k}{k}}$ by $a_k$, and the coefficient preceding $x^{kn}$ by $a_k(n)$ . For example, $a_1(3)=\frac{1}{3!}$ and $a_3(2)=\frac{1}{18}$ .

Now the coefficient on $x^{19}$ is the sum of all possible products of terms whose exponents sum to $19$, for example, one of which is $a_1(4)*a_2(3)*a_3(3) = \frac{1}{4!}*\frac{1}{2^3 3!}*\frac{1}{3^3 3!}$

(i) $a_1(19)=\frac{1}{19!}$
(ii) $a_{19}(1)=\frac{1}{19}$
(iii) Since $19$ is prime, nowhere else in any coefficient $a_k(n)$ does $19$ appear, for $k<19$ and $n<19$.

Therefore, the sum of all such terms must be of the form $\frac{1}{19!}+\frac{1}{19}+\frac{r}{s}$ , for 19 not dividing s.

QED

EDIT: The coefficient is actually $\frac{1}{19!}+\frac{1}{19}+\frac{31546534213457683 52}{15687942664565435561256}$
• May 12th 2009, 10:19 AM
Cairo
Thanks
How do we deduce that 18! = -1 (mod 19) from all this?