0

Lets say I have colony of 100 bacteria's.

Within 1 year, each of bacteria brings up a "child" and becomes "infertile".

After next year, children of bacteria bring up "children" and becomes "infertile".

Assuming that bacteria is immortal and will stay alive forever.

Which formula can I use for it in WolframAlpha?

I know sequence Plot [(formula), {a, start, end}, but how to form such formula in it?

For you it's walk in the park, that's because you're mathematicians. I am not.

Rik Telner
  • 123
  • You don't need to be a mathematician to solve this problem. Just take out a pencil, and a paper, then sketch on to that paper a 2-colomn chart: The first column is the number of years, and the second is the number of bacteria. The starting year (namely, year 0), you have 100 bacteria, and 1 year later, you have 200 bacteria, and what about 3 years, 4 years? Can you see the pattern now? – user49685 Jan 26 '14 at 13:29
  • I see now. I mean, I thought it would require another function for this. – Rik Telner Jan 26 '14 at 13:33
  • 1
    Mentioning a fact here : This site is for amateur mathematicians (though there are mathematicians here) . Mathoverflow is the site for full-fledged mathematicians . – abkds Jan 26 '14 at 13:39
  • @TrafalgarLaw I saw mindblowing (for me) formula's here. If this is for amateurs. I don't even think about thinking of how MathOverflow is like. – Rik Telner Jan 26 '14 at 13:40

1 Answers1

0

In year $1$, there are $100$ bacteria.

In year $2$, there are $100+100$ bacteria.

In year $3$, there are $200+100$ bacteria.

etc.

Each year we get $100$ more bacteria. So in year $a$ there are $100\cdot a$ bacteria.

If your first year is year $0$, then your formula would be $100 \cdot(a+1)$.

Alex
  • 2,354
  • Oh, wait I thought it was something else. I was thinking it require more complex resolve, but apparently I am just stupid. Gon' accept as soon as possible. – Rik Telner Jan 26 '14 at 13:33