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Topic: 4xsquared - 40Floor(x) + 51 = 0 (Read 485 times) |
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ThudnBlunder
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4xsquared - 40Floor(x) + 51 = 0
« on: Jan 14th, 2009, 4:14pm » |
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Find the number of real solutions to 4x2 - 40 x + 51 = 0, where x represents the greatest integer less than x, also known as the integer part of x, and as floor(x).
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teekyman
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Re: 4xsquared - 40Floor(x) + 51 = 0
« Reply #1 on: Jan 14th, 2009, 7:28pm » |
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4
write x = u+5 for convenience,
plugging into the above we get
f(u) = 4u + 40u - 40u- 49 or f(u) = 4u+40{u} - 49 = 0
since {u} goes between 0 and 1, we see that we only need concern ourselves with values of u between -7/2 and 7/2.
by hand, we consider values of the above function at the integers from -4 to 4 - these are easy to compute by hand since the fractional argument is 0 and the function is 4u- 49.
-4: 15
-3: -13
-2: -33
-1: -45
0: -49
1: -45
2: -33
3: -13
4: 15
1. its easy to see that the value of f(-3.00001) is approx f(-3) + 40.
2. Because the function is continuous over intervals of the form [k,k+1) where k is an integer, we can use the IMT to find roots.
3. f(u) is also strictly increasing inside these intervals, as the slope contribution of 4u(at most -32 to 32) is less than that of the contribution from 40{u} (40) in this range. Thus, we can also guarantee the existence of exactly one root in places that we find one.
By observation, we will find roots between x =2 and 3, 6 and 7, 7 and 8, and 8 and 9. Furthermore, by looking at the original eqaution we see that 4xmust be an integer at these roots, so with a little guess and check (with the aid of a for loop) we can find the exact roots, which are x =  29/2 ~ 2.69 , 189/2 ~ 6.87, 229/2 ~ 7.57, 269/2 ~ 8.20
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