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

$$f(16) = 74$$

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?

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

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  • $\begingroup$ Why does $k=2$? $\endgroup$ – Hamilton Dec 4 '18 at 13:53
  • $\begingroup$ Because the first term of our sum is $f(1) - f(2)$, which is $f(k-1)-f(k)$ evaluated at $k=2$. $\endgroup$ – MisterRiemann Dec 4 '18 at 13:55
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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 $f(1) = \frac{1}{2}(1^2) - \frac{1}{2}(1)-46 = -46$.

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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 $x$

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

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

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

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

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

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