Find the 3 mathematicians which may be revealed, with a little manipulation, from the formulae below. They are well known enough to be in the MacTutor History of Mathematics archive.

$$\frac{\chi }{\csc i} $$
$$\frac{2ar}{a + r}{y}'dx$$
$$\int_{2}^{p} \frac{dt}{ln t}s\chi tz$$

  • $\begingroup$ Is there any reason for adding the images and not using Mathjax? $\endgroup$
    – manshu
    May 9 '16 at 10:25
  • $\begingroup$ @Tom. Do it. it's not like anyone answered so far. $\endgroup$
    – Marius
    May 9 '16 at 11:17
  • 1
    $\begingroup$ Thanks Marius (and Lord of the dark for the Mathjax) - I was able to see how to edit it and fixed now!! Maybe the reason no one answered it was my first one should have the csc i on the denominator. $\endgroup$
    – Tom
    May 9 '16 at 11:21

Here are my guesses

$\frac{\chi}{\csc i}$

Oscar Chisini - the equation is equivalent to $\chi \sin i$ or "chi sin i"

$\frac{2ar}{a+r} y' dx$

GH Hardy - the first bit is the harmonic mean of $a$ and $r$ or $H(a,r)$ and the second bit is $dy$ so $H(a,r) + dy$

$\int_2^p \frac{dt}{\ln t} s \chi tz $

Rudolf Lipschitz - the first part is the offset logarithmic integral of $p$ or $Li(p)$ added to $s$ + "chi" + $t$ + $z$.

  • 1
    $\begingroup$ This answer is even better considering I had ' t' instead of 'ln t' in the denominator of the third one - thanks for pointing out the correction ' Li (t) ' was intended. $\endgroup$
    – Tom
    May 9 '16 at 12:20
  • 1
    $\begingroup$ I think once I read the second half of it, I got it into my head what the answer was and only later realised that you would need $\ln t$ below the line. Really enjoyable puzzle, by the way. $\endgroup$
    – hexomino
    May 9 '16 at 12:25

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