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If I have the equations $f(x)=30c^x$ and $g(x)=g(x-1)+f(x)$, $g(0)=30$ why is $g(x)=30xc^x+30$ not the answer when something like $h(x)=h(x-1)+c$, $h(t)=d$ would make $h(x)=c(x-t)+d$? What I try to do goes as follows $30c^x(x-0)+30$ or $30xc^x+30$ but it does not have the same line or even shape. Here is my graphs desmos.com red=expected, purple=$f(x)$, green=unexpected. What am I doing wrong or is this something that is mathematically incorrect to try and do?

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Simply because $h(x) = d $ is a constant function, you have predetermined the entire future and history of $h(x)$ in $x$. Whereas, $g(0)=30$ only says that at $x=0$, $g(0)=30$, but it's past and future are decided by $g(x) = g(x-1)+f(x)$. A simpler example could be $h_1(x) = 4$ which is a constant function, whereas h_2(0)=4 and h_2(x+1)=h_2(x)+4, $h_2(x)$ is a linearly growing function.

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  • $\begingroup$ This does help except it doesn't answer my initial question. Is there a way to convert the recursive equation to an explicit one (or the other way around) to add them together properly or does the recursive equation depend on the result of the explicit too much? $\endgroup$ – Doshorte Dovencio May 12 at 2:46

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