Sigmas and Integrals

Before we do any algebra, let’s remove the sigmas and integrals so we don’t have to deal with them later.

Let’s start with the sigma. The first thing to realize is that you can simply substitute $1$ for $h$ inside the limit, as there is no undefined or indeterminate form to deal with. The second thing to realize is that if you expand the sigma, a lot of the terms will cancel out. Keeping this in mind, this is what we get:

$\sum_{x=a}^b (\lim_{h\rightarrow 1} \dfrac{f(x+h)-f(x)}{h}) = \sum_{x=a}^b (f(x+1)-f(x))$ $= (f(a+1)-f(a))+(f(a+2)-f(a+1))+...+(f(b+1)-f(b))$ $= -f(a)+(f(a+1)-f(a+1))+...+(f(b)-f(b))+f(b+1)$ $Q=f(b+1)-f(a)$

Now, let’s go to the integral. The only thing you need to know here is that the limit on the inside gives the derivative of the function, and taking the integral of a derivative provides the difference of the endpoints. This is the basic definition of the integral.

$\int_{a}^{b} (\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \text{d}x) = \int_{a}^{b} \dfrac{\text{d}f}{\text{d}x} \text{d}x$ $R=f(b)-f(a)$

Now, let's simplify a little bit more using the third equation.

$f(x)=(f(x-1))^{2}$ $\text{Let } x=b+1$ $f(b+1)=(f(b))^{2}$

We can substitute this into the result of $Q$ to set up a system of equations in terms of $f(a)$ and $f(b)$ ! Then, we can solve for $f(b)$ . We shouldn’t solve directly for $f(a)$ because there are restrictions on $f(b)$ that we need to deal with in the fourth statement.

$\begin{cases} (f(b))^{2}-f(a)=Q \\ f(b)-f(a)=R \end{cases}$

Subtracting the two equations to eliminate $f(a)$ yields:

$(f(b))^{2}-f(b)=Q-R$

Now, notice that we have a subtraction of $Q$ and $R$ on the right hand side. These two constants are also subtracted in $S$ . Let’s try to construct $S$ on the right hand side.

$(f(b))^{2}-f(b)=Q-R$ $4(f(b))^{2}-4f(b)=4Q-4R$ $4(f(b))^{2}-4f(b)+1=4Q-4R+1$

We’re not going to simplify the right hand side into $S$ yet, for reasons you will soon see. However, we notice that the left hand side is a perfect square polynomial. Using this, we can solve for $f(b)$ .

$4(f(b))^{2}-4f(b)+1=4Q-4R+1$ $(2f(b)-1)^{2}=4Q-4R+1$ $f(b)=\dfrac{1\pm \sqrt{4Q-4R+1}}{2}$

We have two solutions for $f(b)$ ! However, we know from the fourth statement that $f(b)$ needs to be positive. If we use the plus in the plus or minus, we’re guaranteed to get a positive result. However, if we take the minus from the plus or minus, the result could be positive or negative! If it was positive, we would have two answers. Let’s see if the smaller root can be positive.

$\dfrac{1- \sqrt{4Q-4R+1}}{2} \overset{?}{>} 0$ $\sqrt{4Q-4R+1} \overset{?}{<} 1$ $4Q-4R+1 \overset{?}{<} 1$ $Q \overset{?}{<} R$ $f(b+1)-f(a) \overset{?}{<} f(b)-f(a)$ $f(b+1) \overset{\text{X}}{<} f(b)$

We have a contradiction against statement four! Thus, $f(b)$ is only equal to the greater solution. Now that we know the value of $f(b)$ , let’s solve for $f(a)$ !

$f(b)=\dfrac{1+ \sqrt{4Q-4R+1}}{2}$ $R=f(b)-f(a)$ $f(a)=f(b)-R$ $f(a)=\dfrac{1+ \sqrt{4Q-4R+1}}{2}-R$ $\boxed{f(a)=\dfrac{1+S}{2}-R}$

Thus, our answer is $\boxed{\dfrac{1+S}{2}-R}$ .