What is the slope of this line?
f ( x ) = exp ⎝ ⎛ k = 1 ∑ ⌈ x ⌉ k 1 ⎠ ⎞ The graph of the above is drawn in red, where exp ( x ) = e x . The purple line is a linear function of the form y = m x . The value of m is maximized such that the purple line never intersects one of the red segments.
What is ⌊ 1 0 0 0 m ⌋ ?
The answer is 1781.
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4 solutions
Exactly same solution, very elegant sir.
You effectively used abuse of notation when you said the slope was parallel and that got me confused. I reached many equations similar to this, but didn't know of the Euler Mascheroni constant so settled on good old high input approximation (I sort of guessed that this was going to be a limit, so simple calculated a difference between f(n+1)-f(n) for n>> (I settled on n=5))
Very nice solution. Perhaps the one thing worth expanding on would be showing, or at least noting that ∑ k = 1 ⌈ x ⌉ ≥ γ + l n x for all x, so that we can be sure that the line never crosses the graph somewhere earlier on.
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Yes, thank you for pointing this out! The question was originally less specific, so I've updated my solution to match the question.
Why is e^gamma times x equal to e^gamma?
It isn't, the equation of the line produced is y = e γ x , and so the slope is m = e γ .
Used WolframAlpha script
"find (1/10^10)exp(HarmonicNumber[10^10)])"
to obtain 1.7810.... Multiplying with 1000 and using Floor function, Answer=1781.
M = x 2 − x 1 y 2 − y 1 = 2 − 1 e ( 1 + 2 1 ) − e 1 = 1 e 2 3 − e 1 = 1 . 7 6 (2d.p.)
I calculated e^(H_200000)/200000 in wolfram
The sum of reciprocals of positive integers (also known as the harmonic numbers ) can be approximated by the following function: k = 1 ∑ n k 1 ≈ γ + ln ( n ) γ is the Euler-Mascheroni constant , and roughly equals 0 . 5 7 7 2 . This approximation becomes more accurate as n becomes larger.
Since the expression in the question involves ⌈ x ⌉ , which is always greater than or equal to x , we can state that:
exp ⎝ ⎛ k = 1 ∑ ⌈ x ⌉ k 1 ⎠ ⎞ ≥ exp ( γ + ln x ) = e γ + ln x = e γ e ln x = e γ x
And so m = e γ ≈ 1 . 7 8 1