Find the upper-bound for the summation, such that:
(using the lowest possible integer)

Question

amistre64 · Answer

it feels like there is something missing in the post

anonymous · Answer

oops

anonymous · Answer

$$\sum_{n=4200}^{x}n ^{1.5} \ge 1.0 \times 10^{8}$$

anonymous · Answer

not sure if possible.  I could solve it if it was something like 5n

amistre64 · Answer

does this have anything to do with a remainder thrm?

anonymous · Answer

I want to solve it for a game.  Putting in random numbers, the closest integer I could get to have the same as close as possible to 100 million is 4563

anonymous · Answer

*starting at 4220 >_>

anonymous · Answer

Wolfram gives me weird zeta functions or harmonic numbers

amistre64 · Answer

10^8, or 10^7?

anonymous · Answer

100 million, so 10^8 (if I did that right)

amistre64 · Answer

$$\int_{4200}^{x}x^{1.5}dx=x^{2.5}/2.5-(4200)^{2.5}/2.5$$

$$x^{2.5}/2.5-(4200)^{2.5}/2.5=10^8$$
$$x^{2.5}-(4200)^{2.5}=2.5*10^8$$
$$x^{2.5}=(4200)^{2.5}+2.5*10^8$$
$$x=((4200)^{2.5}+2.5*10^8$$)^{1/}

amistre64 · Answer

ugh ... firefox crashed at the end of that

amistre64 · Answer

$$x=((4200)^{2.5}+2.5*10^8)^{1/2.5}=4545.ddd$$

anonymous · Answer

I tried integrals, I must have done it wrong.

amistre64 · Answer

http://www.wolframalpha.com/input/?i=sum%28n%3D4200+to+4545%29n^%281.5%29

amistre64 · Answer

letting x=4545 seems to be fair

anonymous · Answer

oh, I see what I did wrong.  Thank you :D

amistre64 · Answer

yw