Math Forum

by **vbinterface** » Tue Feb 09, 2010 1:06 am

Hi.

If I have series of 1,2,4,7,11 on X-axis with co-ordinates 0,1,2,3,4 respectively, can I calculate the X-coordinate for any number in the series in one "generalized equation" ?
For example the number 7 in the progression series is located at x=3.
So, if given the number 11, how can it's corresponding x-coordinate be calculated using a single equation?

Thanks in advance.

by **skipjack** » Tue Feb 09, 2010 5:04 am

Yes; x = √(2y - 7/4) - 1/2, where y is the number.

by **vbinterface** » Tue Feb 09, 2010 9:39 pm

Thanks skipjack.
Can you please explain your equation (x = √(2y - 7/4) - 1/2) a little more with an example for "any" number in the series?

by **skipjack** » Tue Feb 09, 2010 10:10 pm

For y = 7, x = √(2(7) - 7/4) - 1/2 = √(49/4) - 1/2 = 7/2 - 1/2 = 3.

For mental evaluation, you might prefer the equivalent equation x = (√(8y - 7) - 1)/2.

by **soroban** » Wed Feb 10, 2010 1:27 am

Hello, vbinterface!

If I have series of 1,2,4,7,11 on X-axis with co-ordinates 0,1,2,3,4 respectively,
can I calculate the x-coordinate for any number in the series with one "generalized equation" ?

For example the number 7 in the progression series is located at x=3.
So, if given the number 11, how can it's corresponding x-coordinate be calculated using a single equation?

$\text{We have this table of values:}$

. . . . $\begin{array}{c|c} x & y \\ \hline \\ \\ 0 & 1 \\ \\ \\ 1 & 2 \\ \\ \\ 2 & 4 \\ \\ \\ 3 & 7 \\ \\ \\ 4 & 11 \\ \\ \\ \vdots & \vdots \end{array}$

$\text{The generating function is: }\;y \;=\;\frac{1}{2}x^2\,+\,\frac{1}{2}x\,+\,1$

$\text{We have: }\:\frac{1}{2}x^2 \,+\,\frac{1}{2}x\,+\,1 \:=\:y \;\;\;\Rightarrow\;\;\;x^2\,+\,x \,+\,(2\,-\,2y) \;=\;0$

$\text{Solve for }x:\;\;x \;=\;\frac{-1 \,\pm\,\sqrt{1^2\,-\,4(2-2y)}}{2} \;=\;\frac{-1\,\pm\,\sqrt{8y\,-\,7}}{2}$

$\text{We will use the }positive\text{ value of }x:\;\;x \;=\;\frac{\sqrt{8y\,-\,7} \,-\,1}{2}$

. . This is equivalent to skipjack's formula.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\text{Suppose we are given: }\:y \,=\,11$

$\text{then: }\:x \:=\:\frac{\sqrt{8(11)\,-\,7}\,-\,1}{2} \:=\:\frac{\sqrt{88\,-\,7} \,-\,1}{2} \:=\:\frac{\sqrt{81}\,-\,1}{2} \:=\: \frac{9\,-\,1}{2} \:=\:\frac{8}{2} \:=\:4$

by **vbinterface** » Thu Feb 11, 2010 2:48 am

Thank you very much skipjack and soroban for your help and detailed explanation. It solved my question.

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