Yesterday, I posted the first three numbers of a number series and left the next three numbers blank. Here’s the answer to yesterday’s teaser.
The completed sequence is:
3, 7, -4, -30, -71, -127
Do the numbers seem unrelated to each other? The formula linking these numbers is:
“ y = -15x2 + 53x – 32 ”
Try it, substitute x=1, 2 and 3 in the above equation and you should get the first three numbers (3, 7, -4) provided in the original question. Put x=4, 5 and 6 to get the next three numbers (-30, -71, -127) in the “sequence”.
If you plot the numbers on a graph, you’ll get a standard U-shaped curve (upside down in this instance) governed by the above quardratic equation.
How does one derive the equation from the first three numbers?
Here are the details for those maths geeks out there:
Let’s assume we have a number sequence where the first three numbers are M, N and P.
We want to end up with a formula that looks like this:
y = M (Exp1) + N(Exp2) + P (Exp3) … … … (1)
Exp1 is a binomial expression that will be non-zero only when x=1, and evaluates to exactly 1 when x=1.
Exp2 is a binomial expression that will be non-zero only when x=2, and evaluates to exactly 1 when x=2.
Exp3 is a binomial expression that will be non-zero only when x=3, and evaluates to exactly 1 when x=3.
When x=1, Exp2 and Exp3 becomes zero, while Exp1=1, so y=M.
When x=2, Exp1 and Exp3 becomes zero, while Exp2=1, so y=N.
When x=3, Exp1 and Exp2 becomes zero, while Exp3=1, so y=P.
Let’s look at how to derive Exp1:
To make Exp1 zero when x=2 and 3, it should contain the factors “(x-2)” and “(x-3)”. We then plug in x=1 in “(x-2)(x-3)” to see what we get. Here, we get “2”. In order to make Exp1 equal to 1, we need to divide “(x-2)(x-3)” by 2.
Hence, Exp1 is “(x-2)(x-3)/2”
We repeat the above to get Exp2:
To make Exp2 zero when x=1 and 3, it should contain the factors “(x-1)” and “(x-3)”. Plugging x=2 in “(x-1)(x-3)”, we get “-1”. In order to make Exp2 equal to 1, we need to divide “(x-1)(x-3)” by “-1”.
Hence, Exp2 is “-(x-2)(x-3)”
We repeat the above yet again to get Exp3:
To make Exp3 zero when x=1 and 2, it should contain the factors “(x-1)” and “(x-2)”. Plugging x=3 in “(x-1)(x-2)”, we get “2” again. In order to make Exp3 equal to 1, we need to divide “(x-1)(x-2)” by “2”.
Hence, Exp3 is “(x-1)(x-2)/2”
Putting the three expressions into Equation (1), we get:
y = M [ ( x – 2 ) ( x – 3 ) / 2 ] + N [ – ( x – 2 ) ( x – 3 ) ] + P [ ( x – 1 ) ( x – 2 ) / 2] … … … (2)
Expanding the above and rearranging, we get:
y = [ ( M – 2N + P ) x2 – ( 5M – 8N + 3P ) x + (6M – 6N + 2P ) ] /2 … … … (3)
Plugging in M = 2, N = 4 and P = 6 in Equation (3), we get the simple formula
y = 2x
Or plugging in M = 1, N = 4 and P = 9 in Equation (3), we get:
y=x2
Back to the original question, plugging in M = 3, N = 7 and P = -4 in Equation (3), we get the formula for the sequence:
y = -15x2 + 53x – 32
Theoretically, you could have plugged in ANY three numbers to continue the sequence. Just repeat the above steps using 6 starting numbers instead of three. You should end up with an equation involving x to the power of 5!
But the idea is there – that you can continue any number sequence with any numbers and claim they’re part of a consistent mathematical progression!