Hi
To find let's take the derivative
Therefore is the antiderivative of that is 0 for x=0
Therefore
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).
I hope so
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
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 , for 19 not dividing s.
For reference, call the series expansion by , and the coefficient preceding by . For example, and .
Now the coefficient on is the sum of all possible products of terms whose exponents sum to , for example, one of which is
(i)
(ii)
(iii) Since is prime, nowhere else in any coefficient does appear, for and .
Therefore, the sum of all such terms must be of the form , for 19 not dividing s.
QED
EDIT: The coefficient is actually