This is for a personal project of mine, and I am having a hard time conceptualizing this problem.
So let's say that I was able to quantify the price of an asset as the following:
$$P = (3.5 \cdot Q)^{1.3}$$
$P$ represents the price of the asset (in USD for this example's sake), $Q$ represents the current quantity of the asset that has been purchased.
So then let's say that I want to purchase $100 worth of the asset, and currently 5 assets have already been purchased, making the value of the asset currently be at:
$$(3.5 \cdot 5)^{1.3} \approx 41.30$$
But of course, me buying the asset is increasing the price of it, so I can't simply do something like this to get how much of the asset can be purchased:
$$Q = \frac{100}{41.30}$$
Which makes me think that maybe some calculus is required here? I took calculus in college but if I could get some help in explaining this in mathematical terms to get a start on it that would be very helpful.
We have $P = (3.5Q)^{1.3}$, and we want to find how many assets past $5$ we can purchase with $\$100$. If we assume that fractional assets aren't possible, i.e. the number of assets has to be an integer, and that the price changes after each purchase of a single asset, then we can proceed one purchase at a time until we've run out of money:
\begin{array}{c|c|c|c} \text{\$ Available} & \text{Current Assets} & \text{Current Price} & \text{\$ Remaining}\\ \hline \$100 & 5 & (3.5(5))^{1.3} \approx \$41.3 & \$100 - \$41.3 = \$58.70\\ \hline \$58.7 & 6 & (3.5(6))^{1.3}\approx \$52.35 & \$58.70 - \$52.35 = \$6.35\\ \hline \$6.35 & 7 & (3.5(7))^{1.3} \approx \$63.96 & \$6.35 - \$63.96 = \color{red}{-\$57.61} \end{array}
This tells us that the most we can purchase is two additional assets.
If, however, we want to consider the possibility that we can purchase infinitesimally small portions at a time, then we can proceed as follows.
Disclaimer: It's possible I'm 'reinventing the wheel' here, because while I've never seen this particular scenario, it's likely not original, so going through what I'm about to do may be overkill.
If we begin with $5$ assets, and only purchase $1$ asset at a time, then the total amount of money spent on $3$ additional assets can be expressed as follows:
$$(3.5(5))^{1.3} + (3.5(6))^{1.3} + (3.5(7))^{1.3}$$
If we only purchase $0.5$ of an asset at a time, the total amount of money spent on $3$ additional assets can be expressed as: $$0.5(3.5(5))^{1.3} + 0.5(3.5(5.5))^{1.3} + \dotsb + 0.5(3.5(7))^{1.3} + 0.5(3.5(7.5))^{1.3}$$
We can rewrite this expression more compactly as $$(3.5)^{1.3}\sum_{k = 0}^{5}(5+0.5k)^{1.3}(0.5).$$
Note here that we have $5$ as the upper limit on the sum because when purchasing $3$ additional assets, $0.5$ at a time, we need to calculate the price a total of $3/0.5 = 6$ times, but since the first calculation is based on the original price, we need the sum to stop at $\frac{3}{0.5} - 1 = 5$.
In general, then, for a purchase of $x$ additional assets, a fraction of $\Delta x$ at a time, our total expenses can be expressed as $$(3.5)^{1.3}\sum_{k = 0}^{\frac{x}{\Delta x} - 1}(5+k\Delta x)^{1.3}\Delta x.$$
Now, shifting the indices by $n = k+1$, we get the expression
$$(3.5)^{1.3}\sum_{n = 1}^{\frac{x}{\Delta x}}(5+(n-1)\Delta x)^{1.3}\Delta x.$$
We can think of this sum as a Riemann sum, where the interval $[0,x]$ has been partitioned into $x/\Delta x$ sub-intervals, using the points $x_n = (n-1)\Delta x$ in each sub-interval (these corresponding to the left endpoints of each sub-interval here). So, for the function $f(x) = (5+x)^{1.3}$, our expression looks like $$(3.5)^{1.3}\sum_{n = 1}^{\frac{x}{\Delta x}}f(x_{n})\Delta x.$$ Taking the limit as $\Delta x \to 0$, this converges to the limit $$(3.5)^{1.3}\int_{0}^{x}(5+t)^{1.3}\,dt.$$
Notice, however, that purchasing fractional assets continuously in this manner is actually more expensive than purchasing them one at a time as we initially did, because the price is constantly being recalculated. For instance, we found initially that it would cost us $\$93.65$ to purchase two assets one at a time. Using the integral, however, we would calculate $$(3.5)^{1.3}\int_{0}^{2}(5+t)^{1.3}\,dt \approx \$104.88.$$