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I am trying to develop a nomogram which simultaneously shows the exact Fisher equation $(1+u) = (1+v)(1+w)$ and its linear approximation $u \approx v + w$.

This amounts to finding twelve smooth curves such that the following equations hold:

$$\det \begin{bmatrix}f_1(u) & f_2(u) & f_3(u) \\ g_1(v)& g_2(v) & g_3(v) \\ h_1(w) & h_2(w) & h_3(w)\end{bmatrix} = (1+v)(1+w) - (1+u)$$

$$\det \begin{bmatrix}f_4(u) & f_5(u) & f_6(u) \\ g_1(v)& g_2(v) & g_3(v) \\ h_1(w) & h_2(w) & h_3(w)\end{bmatrix} = v + w - u$$

My question is how to begin thinking about this problem --- how can you decide whether such curves exist, and how might you look for them?

user326210
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Dylan
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1 Answers1

2

For context for people trying to figure out what the question is about - we're discussing this thing: https://en.wikipedia.org/wiki/Fisher_equation

If you mean for them to be shown "all on the same scales" then I don't think this is possible, but if you either:

a) have both nomograms on the same axis-lines (same support lines but all with different scales for the exact and approximate forms, (so different tick marks and axis labels), or

b) allow at least one axis (support line) of the approximate equation to be different from the exact form,

-- then yes.

  1. Note that the most obvious ways to tackle the first (i.e. exact) equation is on the log scale (I'll assume that all the variables exceed -1; interest rates above -100% seems uncontroversial; indeed normally they'd exceed 0)

    That would produce three parallel scales in $log(1+u)$, $log(1+v)$, $log(1+w)$ (but with tick marks labelled in $u, v$ and $w$); on the other side of the same support lines you could place $u, v$ and $w$ as linear scales for the linear approximation, and set the scaling up so that the centres of the $v$ and $w$ scale represented the same value of $v$ and $w$ on both sides (so there'd be a single horizontal line that represented both equations at the same time and you could see the way that the scales diverged away from that)

    That would look roughly like this:

      v v*    u u*    w w*
   |               |
   |       |       |
   |       |       |
   |       |       |
   |       |       |
   |       |       |
   |               |

    where both the left and the right of each line would have its own tick marks and labels; the *-side of each axis-support line would show the scaling for the approximate equation. (It's suggestive to use slightly different colors for the two sides.)

    Constructing this would require dealing with the determinant for the *-variables separately from the non-* ones, but keeping them so that the column representing the abscissae of each axis support-line was the same for both determinants and choosing $w$&$w*$ and $v$&$v*$ scale pairs so they were the same at some convenient chosen point (presumably $0$).

  2. You could set up an $N$ (or "Z") type nomogram for the product in the exact equation, (which has two linear scales on the parallel arms, for $1+u$ and one of the variables in the product, say $(1+w)$), where again it's labelled in $u$, $v$ and $w$, and then set up a third parallel axis (for $v$, say; noting that $(1+w) + (0+v) = (1+u)$ can also be labelled with w,v,u respectively)

    That should look something roughly like this:

    w    v* v   u
|
|10  | / 
|    |/    |0
|    |0    | 
|0  /|     |10
   / |20   | 
           |20

    Here the u and w scales do double-duty, but the v and v* scales are on different lines.

    You can approach this second one by writing a pair of overlapping determinants:

    Let W = 1+w and U = 1+u. Write V = 1+v for the exact equation and v*=v for the approximation.

    So you want to simultaneously write W+v*=U and WV=U; if you write a 3x3 determinant for V W and U and another for W U and v*, but keep W and U the same for both, you can write a 4-line array representing both determinants at the same time:

    V      |  -V    1    0 |
U      ||  0   -W    1 ||
W      ||  U    0    1 ||
v*      |-v*/2 1/2   1  |

    (assuming I didn't screw that up somewhere) and then manipulate rows and columns as you would a 3x3 determinant to produce something in standard form that's reasonably-scaled, keeping in mind that the tick marks and labels will all be in $u,v,w$ (and $v*$).

Glen_b
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