AdministratorNovember 13, 2020 at 1:01 pm
Does the function f(x) = 1 + x¹/1! + x²/2! + x³/3! + … + xⁿ/n! have any real roots?
MemberNovember 17, 2020 at 3:18 am
When n is even, It has no real solution. When n is odd, It has exactly one solution. We can prove that by induction. Assume that this is the case when n=2k and n=2k+1. Consider the case when n=2k+2. It is easy to see this curve lies above the x axis, based on differential tests and the fact , f'(n,x)= f(n-1,x). And When n=2k+3. It must have at least one solution. If it has more than one, then by roll’s theorem f(2k+2,x)=0 will have at least one solution, leading to contradiction. And That proves the induction.
AdministratorNovember 17, 2020 at 3:13 pm
That’s a great solution! I will just elaborate/rephrase the induction a bit:)
For even n, you have an even-degree polynomial f(x) such that f(x)=f'(x)+xⁿ/n! > f'(x). Therefore, you can’t have a negative local minimum and thus f(x) > 0 for all x.
For odd n, you have an odd degree polynomial with an always positive derivative and therefore it has a single root.
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