1
$\begingroup$

Notation:

Let $$Q_1[f(x)] = \lim _{h \rightarrow 1} \frac{f(x + h) - f(x) }{h}$$

And let

$$Q_1^{-1}\left[f(x)\right] = G(x)|Q_1\left[g(x)\right] = f(x)$$

Consider the equation

$$a_0(x) + a_1(x)f(x) + a_2(x)Q_1\left[f(x)\right] = 0$$

How do we express the general solution to this equation?

A natural first step is to divide the equation by $a_2(x)$ to find:

$$\frac{a_0(x)}{a_2(x)} + \frac{a_1(x)}{a_2(x)}f(x) + Q_1\left[f(x)\right] = 0$$

We can rename the coefficients (being that they are general) to:

$$ b_0(x) + b_1(x) f(x) + Q_1\left[f(x)\right] = 0 $$ Now subtract $b_0(x)$ from both sides: (I am attempting a Duhamel's formula style approach)

$$b_1(x)f(x) + Q_1\left[f(x)\right] = -b_0(x) $$

Now note that for a pair of functions there exists a product rule. Given $a(x)$ and $b(x)$

$$Q_1\left[a(x)b(x)\right] = Q_1\left[a(x)\right]Q_1\left[b(x)\right] + Q_1\left[a(x)\right]b(x) + a(x)Q_1\left[b(x)\right]$$

If we let $b(x) = f(x)$ then we note that we obtain a form of:

$$\left(Q_1\left[a(x)\right] + a(x)\right)Q_1\left[f(x)\right] + Q_1\left[a(x)\right]f(x)$$

At this point it is evident that to solve:

$$b_1(x)f(x) + Q_1\left[f(x)\right] = -b_0(x) $$

We need a function $\lambda(x)$ such that

$$\lambda(x)b_1(x)f(x) + \lambda(x)Q_1\left[f(x)\right] = Q_1\left[f(x)a(x)\right] $$

Therefore:

$$\lambda(x)b_1(x) = Q_1\left[a(x)\right]$$

$$\lambda(x) = Q_1\left[a(x)\right] + a(x)$$

Thus:

$$\lambda(x)b_1(x) + a(x) = \lambda(x)$$

Therefore:

$$\lambda(x) = \frac{a(x)}{1 - b_1(x)}$$

And thus both equations breakdown to:

$$ \frac{b_1(x)}{1 - b_1(x)} a(x) = Q_1\left[a(x)\right] $$

Let $a(x) = 2^{g(x)}$ then it follows

$$Q_1\left[2^{g(x)}\right] = 2^{g(x)}\left(2^{Q_1\left[g(x)\right]} - 1\right) $$

Thus we are merely tasked with solving

$$\left(2^{Q_1\left[g(x)\right]} - 1\right) = \frac{b_1(x)}{1 - b_1(x)} $$

Which has solution

$$g(x) = Q_1^{-1}\left[\log_2\left\lbrace\frac{1}{1 - b_1(x)}\right\rbrace\right] $$

Thus the general solution has the form

$$2^\left({Q_1^{-1}\left[\log_2\left(\frac{1}{1 - b_1(x)}\right)\right]}\right)f(x) = Q_1^{-1}\left[-b_0(x)*\frac{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - b_1(x)}\right)\right]}}{1 - b_1(x)}\right]$$

Therefore

$$f = \frac{Q_1^{-1}\left[-b_0(x)*\frac{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - b_1(x)}\right)\right]}}{1 - b_1(x)}\right]}{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - b_1(x)}\right)\right]}}$$

Simplifying and substituting $\frac{a_0(x)}{a_2(x)}$ for $b_0(x)$ and $\frac{a_1(x)}{a_2(x)}$ for $b_1(x)$

We conclude that the general solution to:

$$a_0(x) + a_1(x)f(x) + a_2(x)Q_1\left[f(x)\right] = 0$$

is:

$$f = \frac{Q_1^{-1}\left[\frac{a_0(x)}{a_2(x)}*\frac{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - \frac{a_1(x)}{a_2(x)}}\right)\right]}}{\frac{a_1(x)}{a_2(x)} - 1}\right]}{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - \frac{a_1(x)}{a_2(x)}}\right)\right]}} $$

I think this is correct. But a second eye is always helpful

$\endgroup$
  • $\begingroup$ *on a side note how do you make equations bigger as well as create left and right parenthesis that engulf the ENTIRE expression between them, resizing if need be? $\endgroup$ – frogeyedpeas Jul 8 '14 at 0:15
  • 1
    $\begingroup$ For parentheses, try \left( and \right) (or \left[ and \right] or \left\{ and \right\}. $\endgroup$ – Mark Fischler Jul 8 '14 at 0:28
  • 1
    $\begingroup$ I added the \left and \right for you but where you have powers of 2, it looks just as confusing (and you can't add any more brackets due to how many are there) $\endgroup$ – Jam Jul 8 '14 at 0:52
  • $\begingroup$ thanks man! I think its an improvement from before $\endgroup$ – frogeyedpeas Jul 8 '14 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.