/sci/ - Science & Math

[Anonymous No.11123427]

Hello nerds, /Biz/ness man here I need your help

i need to create a function to solve for what annual compounding rate of growth (Y) an asset would need to see for it to break even on a linear growth loan (x).

The loan charges 10.5% flat on the original loan, so not compounding. 1000 base, 1100 x=1, 1200 x=2, ect. So 1000 + (105*x) is what I have setup for the left side of the equation. The right side is a bit tricky and probably where I need help, 1000*(1+Y)^x. This should be correct yes? Am I missing anything? That will give me an equation where i can solve for either X or Y on that graph to determine either how much time a certain return would break even over or how much return I would need to breakeven over a specified time right? Is there an easier way to do this equation?

Thank you nerds, buy chainlink and get rich

[Anonymous No.11123477]

>>11123427
>this is your brain on shekels
i refuse to help you in monetary pursuit

[Anonymous No.11123479]

>>11123477
Kek'd and checked.

You should still help me though. U know you wanna

[Anonymous No.11123493]

>>11123479
>checked
this board is glacial, i could quote my own post if i wanted to
in fact, here
>>11123492

[Anonymous No.11123500]

>>11123493
son of a bitch
>>11123499

[Anonymous No.11123504]

>>11123500
fuck this gay earth

[Anonymous No.11123521]

>>11123504
That's funny as fuck not gonna lie. Screenshotted that shit

[Anonymous No.11123557]

>>11123427
P+Prt = P(1+r)^t
1+rt = (1+r)^t
[eqn]1+rt=\sum_{n=0}^{\infty}\prod_{k=1}^{n} \frac{t-k+1}{k} r[/eqn]
[eqn]0=\sum_{n=2}^{\infty}\prod_{k=1}^{n} \frac{t-k+1}{k} r[/eqn]
solutions: r=0, t=0, t=1

[Anonymous No.11123608]

>>11123557
No, he wants 2 different interest rates: P(1+rt) = P(1+s)^t

solve for r
rt = (1+s)^t -1
r= ((1+s)^t -1)/t

solve for s
1 +rt = (1+s)^t
s=(1+rt)^(1/t)-1

solve for t
1 = -rt + (1+s)^t
1/-r = t + (1+s)^t/-r
(1+s)^(-1/r) = (1+s)^t (1+s)^((1+s)^t/-r)
let y=1+s
y^-1/r = y^t y^[(y^t)/-r]
let z=y^t
y^-1/r = z e^[-zln(y)/r]
let x = -zln(y)/r
-y^(-1/r)ln(y)/r = xe^x
x=W(-y^(-1/r)ln(y)/r)
z=-r*W(-y^(-1/r)ln(y)/r)/ln(y)
t=ln(-r*W(-y^(-1/r)ln(y)/r)/ln(y))/ln(y) = ln(-r*W(-(1+s)^(-1/r)ln(1+s)/r)/ln(1+s))/ln(1+s)

There's probably an identity to simply the Lambert W expression but I can't be assed

[Anonymous No.11123651]

>>11123608
[math] t = \log_{1+s}\left(\frac{-r W\left( \frac{-\ln(1+s)} {r \sqrt[r]{1+s}}\right)}{\ln(1+s)}\right) [/math]

[Anonymous No.11123748]

>>11123748

[Anonymous No.11123769]

>>11123651
[math]t = \log_{1+s}\left(\frac{-r W\left( \frac{-\ln(1+s)} {r \sqrt[r]{1+s}}\right)}{\ln(1+s)}\right)
\\
\ln(1+s) t = \ln\left(\frac{-r W\left( \frac{-\ln(1+s)} {r \sqrt[r]{1+s}}\right)}{\ln(1+s)}\right)
\\
\ln(1+s) t = \ln\left(W\left( \frac{-\ln(1+s)} {r \sqrt[r]{1+s}}\right)\right)+\ln\left(\frac{-r}{\ln(1+s)}\right)
\\
\ln(1+s) t = -W\left( \frac{-\ln(1+s)} {r \sqrt[r]{1+s}}\right) +\ln\left(\frac{-\ln(1+s)} {r \sqrt[r]{1+s}}\right)+\ln\left(\frac{-r}{\ln(1+s)}\right)
\\
\ln(1+s) t = -W\left( \frac{-\ln(1+s)} {r \sqrt[r]{1+s}}\right) +\ln\left(\frac{1} {\sqrt[r]{1+s}}\right)
\\
t= \frac{-W\left( \frac{-\ln(1+s)} {r \sqrt[r]{1+s}}\right)}{\ln(1+s)} -\frac{1}{r}
[/math]