So what you're saying that is if we have k digits ANYWHERE in a repeating interval of k digits, then those k digits repeat? Then we're home free, since we can state that the repeating interval will eventually contain a prime with nk digits with a repeating interval of k digits n times, which is...
I agree that there must be a prime with x digits, whatever x is. Let's say a prime has nk digits. Then let's say m of those digits appear in the first repeating interval containing the prime. So it is structured so that we have m digits/k digits/k digits/k digits/...(n-1 times)/k-m digits. I...
Let's go through some casework.
If no selected light bulbs are defective, there is a (4/5)^8 chance of that happening
If 1 selected light bulb is defective, there is a (4/5)^7*(1/5)*8 chance of that happening.
If 2 selected light bulbs are defective, there is a (4/5)^6*(1/5)^2*8*7...
But that's kind of hand-wavy...just "sketching out" why leaves me unsatisfied, I guess.
And even if the prime is k digits and falls into the repeating interval, does the n<=p<10n really cause a contradiction? Let's assume that the prime of k digits is the highest prime with k digits...then...
Oh! I was always assuming I should use mean value on f, not h. Alright, mean value on h implies that there is a value c on (0,1) s.t. h'(c)=0 i.e. f(c)=0.
I'm stupid.
Well, the argument I made about the nk digits repetition implying that the number is not a prime completely rests on the interval starting at the first digit of the "prime" with nk digits. I completely agree with you that the postulate guarantees a prime with 2k, 3k, 4k digits. But I'm not sure...
Ok, so either the function accumulates zero area everywhere (i.e. is constant at zero) and I'm done, or it accumulates some positive area and the same amount of negative area.
Question, though. Does the accumulation of positive area implies there exists an x on the interval such that f(x)>0...
But my comments in #7 are completely reliant on the prime being k digits and being in the interval. If the prime is, say, 2k digits, then I have no clue where the interval might start. Even if I did, it wouldn't matter. If the interval starts at the first digit of the 2k-digit prime, then the...
How can I exhaust cases where the digits of the prime in question are greater than k?
How can I exhaust cases where the repeating interval contains digits from more than one prime?
Why can I leave off the +C?
h(0)=0 (or C)
h(1)=0 (or C) as well...so no area is accumulated by f from 0 to 1. Thus the areas "cancel out" and either f is constant at 0 or f(0) has an sign opposite that of f(1)...that makes sense, but how can I write that up rigorously?
Hmmm...still stumped. If a prime has 2k digits for example, if the first k digits make up the repeating part, then the last k digits of the prime are just the same as the first k digits. Why can't those k digits repeat forever?
And what if the repeating interval with k digits is made up of...
So I can integrate f and get
h(x)=anxn+1/n+1 + ... + a0*x + C.
I can factor out an x and get x(anxn/n+1 +...+a0)+C.
I'm still left riddled with x-terms and I don't see how the integral can help me find many roots...
Homework Statement
Suppose an/n+1 +...+a0/1=0.
Prove f(x) =anxn +...+a0 has a root between zero and one.
Homework Equations
I'm pretty sure this is induction, but I'm not completely sure.
Mean Value Theorem probably
The Attempt at a Solution
Well f(0)=a0 and f(1)=an + ... +...
Well, if n is equal to the repeating prime with k digits, doesn't the less than or equal to sign make the no next prime question not trivial?
I have absolutely no clue how to apply anything to the scenario where the repeating happens "mid-prime".
And what if the interval contains multiple...
Well, the bottom can, at the least, be equal to 2 (when the square root filth is equal to 0), so the expression has to be less than or equal to just 3D/2, correct?