Arithmetic-Geometric progression general integral formula

Question

I'm solving Arithmetic-Geometric progression. It's rules: $$ k(x)\in Z, \forall{x}\\ f(0)=k(0)\\ q(x) > 0, \forall{x}\\ f(x)=f(x-1)\frac{q(x)}{q(x-1)}+k(x)-k(x-1) $$ I got the general formula: $$ f(x)=k(x)+q(x)\cdot\lim_{n->0}{\sum_{n=0}^{\frac{x}{d}-1}k(nd)\cdot(\frac{1}{q(nd)}-\frac{1}{q(nd+d)})}\\ f(x)=k(x)+q(x)\cdot\lim_{n->0}{\sum_{n=0}^{\frac{x}{d}-1}\frac{k(nd)\cdot q\prime(nd)}{q(nd)\cdot q(nd+d)}}\\ $$ And I thought it was: $$ f(x)=k(x)+q(x)\cdot\int_{0}^{x}{\frac{k(t)\cdot q\prime(t)}{q^{2}(t)}\cdot dt} $$ But it's not. It's valid for $\forall x, |x|<\epsilon$.
I thought q(nd+d) would be a q(t) under integral. I think it was a mistake.
By the way, I've found another formula for k(x) that works fine (as I think): $$ k(x)=f(x)+\lim_{d->0}{\sum_{n=0}^{\frac{x}{d}-1}{f(nd)\cdot(1-\frac{q(nd+d)}{q(nd)})}}\\ k(x)=f(x)-\lim_{d->0}{\sum_{n=0}^{\frac{x}{d}-1}{\frac{f(nd)\cdot(q(nd+d)-q(nd))\cdot d}{q(nd)\cdot d}}}\\ \lim_{d->0}{\frac{q(t+d)-q(t)}{d}}=q\prime(t)\\ k(x)=f(x)-\lim_{d->0}{\sum_{n=0}^{\frac{x}{d}-1}{\frac{f(nd)\cdot q\prime(nd)\cdot d}{q(nd)}}}\\ k(x)=f(x)-\int_{0}^{x}{\frac{f(t)\cdot q\prime(t)}{q(t)}\cdot dt} $$ Help me find the f(x) formula.

Answer 1

Given two classic functions thus $$ \forall d>0, d \in Z, \forall x\in Z\\ \frac{q(x+d)}{q(x)}=q^{d}, q>0 \\ k(x+d)-k(x)=k\cdot d $$ Now, I'll find the general formula. $$ f(x)=f(x-d)\cdot \frac{q(x)}{q(x-d)}+k(x)-k(x-d)=\\ =q^d\cdot f(x-d)+kd\\ Substitution:\\ f(x)=(f(x-2d)⋅q^{d}+kd)⋅q^{d}+kd=\\ =f(x-2d)⋅q^{2d}+kd(1+q^{d})=\\ =(f(x-3d)⋅q^{d}+kd)⋅q^{2d}+kd(1+q^{d})=\\ =f(x-3d)⋅q^{3d}+kd(1+q^{d}+q^{2d})\\ Generalization:\\ f(x)=q^{x}\cdot k(0)+k\cdot\lim_{d->0}{\sum_{n=0}^{\frac{x}{d}-1}{d\cdot q^{nd}}}\\ \boxed{f(x)=q^{x}\cdot k(0)+k\cdot\int_{0}^{x}{q^tdt}} $$ All the properties of the classic Arithmetic-Geometric progression are met. Prove it yourself. And this is a special case for the formula that I want.

Now, I will substitute k(x) and q(x) given above in my formula: $$ q( x) =q_{0} \cdot q^{x}\\ k( x) =k\cdot x+k_{0}\\ \\ \frac{q( x+d)}{q( x)} =\frac{q_{0} \cdot q^{x+d}}{q_{0} \cdot q^{x}} =q^{d}\\ k( x+d) -k( x) =k\cdot ( x+d) +k_{0} -( k\cdot x+k_{0}) =kd\\ \\ f( x) =k( x) +q( x) \cdot \int _{0}^{x}\frac{k( t) \cdot q\prime ( t)}{q^{2}( t)} dt=\\ =k\cdot x+k_{0} +q_{0} \cdot q^{x} \cdot \int _{0}^{x}\frac{( k\cdot t+k_{0}) \cdot q_{0} \cdot q^{t} \cdot ln( q)}{q_{0}^{2} \cdot q^{2t}} dt=\\ =k\cdot x+k_{0} +q^{x} \cdot ln( q) \cdot \int _{0}^{x}\frac{( k\cdot t+k_{0})}{q^{t}} dt=\\ =k\cdot x+k_{0} +q^{x} \cdot ln( q) \cdot \left( k\cdot \int _{0}^{x}\frac{t}{q^{t}} dt+k_{0} \cdot \int _{0}^{x}\frac{1}{q^{t}} dt\right) =\\ =k\cdot x+k_{0} +q^{x} \cdot ln( q) \cdot \left( k\cdot \frac{q^{x} -x\cdot ln( q) -1}{ln^{2}( q) \cdot q^{x}} +k_{0} \cdot \frac{1-q^{-x}}{ln( q)}\right) =\\ =k\cdot x+k_{0} +q^{x} \cdot \left( k\cdot \frac{q^{x} -x\cdot ln( q) -1}{ln( q) \cdot q^{x}} +k_{0} \cdot \left( 1-q^{-x}\right)\right) =\\ =k\cdot x+k_{0} +k\cdot \frac{q^{x} -x\cdot ln( q) -1}{ln( q)} +k_{0} \cdot q^{x} -k_{0} \ =\\ =k\cdot \frac{q^{x} -1}{ln( q)} +k_{0} \cdot q^{x} =k( 0) \cdot q^{x} +k\cdot \int _{0}^{x} q^{t} dt $$ And I've got the same result. So my formula should be ok.