# Is there another mathematical way to approach this problem?

$$f(x)-f(x-1) = x-5$$

$$f(16) = 74$$

Compute $$f(1)$$.

So, this is basically a linear function. I, however, calculated it without taking the easier way to approach this problem.

If

$$f(16) - f(15) = 11$$

Then

$$74-f(15) = 11 \implies f(15)=63$$

This will take so long as it seems. Since we're given a linear function, how would we compute it in an other mathematical way? Furthermore, is there a general rule for what you advise?

• integration maybe? – Chase Ryan Taylor Dec 4 '18 at 13:48
• Hint: Try a quadratic $f$. – user10354138 Dec 4 '18 at 13:48
• $f$ is not a linear function – Todor Markov Dec 4 '18 at 13:49
• @ChaseRyanTaylor Why should we integrate it? – Hamilton Dec 4 '18 at 13:50

Hint: Notice that you can write $$f(1)$$ as \begin{align} f(1) &= f(1)-f(2)+f(2) -f(3)+\ldots-f(16)+f(16)\\ &= \sum_{k=2}^{16} (f(k-1)-f(k)) + f(16)\\ &= \sum_{k=2}^{16} (5-k) + 74. \end{align}

• Why does $k=2$? – Hamilton Dec 4 '18 at 13:53
• Because the first term of our sum is $f(1) - f(2)$, which is $f(k-1)-f(k)$ evaluated at $k=2$. – MisterRiemann Dec 4 '18 at 13:55

First, note that $$f$$ isn't linear at all (if it were linear, then $$f(x)-f(x-1)$$ would be constant. However, that difference does suggest something quadratic: its second difference will be constant. Thus, we'll try to find $$a$$, $$b$$, and $$c$$ such that $$f(x) = ax^2 + bx + c$$ satisfies this. For this purpose, we garner some equations relating $$a$$, $$b$$, and $$c$$:

First, $$f(16) = 74$$ tells us that $$256a + 16b + c = 74$$. Second, $$f(x)-f(x-1) = x - 1$$ gives us $$a(x^2 - (x-1)^2) + b(x - (x-1)) = x - 1$$. Simplifying that gives us $$a(2x-1)+b=x-1$$, for all $$x$$. Taking $$x = 0$$ gives us $$b - a = -1$$, $$x = 1$$ give $$a + b = 0$$, so $$b = -a$$, and the $$x = 0$$ case gives $$2a = 1$$, so $$a = \frac{1}{2}$$, and $$b = \frac{-1}{2}$$. Finally, $$f(16)=74$$ gives us $$120 + c = 74$$, so $$c = -46$$. Thus, $$f(x) = \frac{1}{2}x^2 - \frac{1}{2}x - 46$$.

You can quickly go back and check that this satisfies all relevant conditions, if you like (it does), and therefore $$f(1) = \frac{1}{2}(1^2) - \frac{1}{2}(1)-46 = -46$$.

Hint:

Let $$f(x)=g(x)+a+bx+cx^2$$

$$x-5=f(x)-f(x-1)=g(x)-g(x-1)+c(2x-1)+b$$

Set $$2c=1,b-c=-5$$ so that $$T=g(x)=g(x-1)$$

For any integer $$x$$

$$T=\cdots=g(16)=f(16)-a-b(16)-c(16^2)$$

$$f(16)-f(15) = 16-5$$ $$f(15)-f(14) = 15-5$$ $$f(14)-f(13) = 14-5$$ $$\vdots$$ $$f(3)-f(2) = 3-5$$ $$f(2)-f(1) = 2-5$$

So, if we sum this we get $$f(16)-f(1) = (16+15+...+3+2)-15\cdot 5$$ and thus $$f(1) = 75-{15\cdot 17\over 2}+f(16)$$

• I know this way but it makes me confused. – Hamilton Dec 4 '18 at 13:51
• Why.............. – Maria Mazur Dec 4 '18 at 13:51
• Really, its a big trouble for me! :) – Hamilton Dec 4 '18 at 13:52

Hint.

Try

$$f(x) = \frac 12(x-4)(x-5) + C_0$$

The linear recurrence equation can be solved as

$$f(x) = f_h(x)+ f_p(x)\\ f_h(x) = C_1\\ f_p(x) = \frac 12(x-4)(x-5)+C_2$$

with

$$f(16)= 74 \to C_0 = 8$$