For some peace of mind in a project, I am trying to prove two equations are somewhat equivalent. I have these two equations. $$ i_{1} = \frac{-I_M}{2}\frac{2i_2\left(1+e^{\left(\frac{2i_2R_E}{V_T}\right)}\right)+i_e\left(-1+e^{\left(\frac{2i_2R_E}{V_T}\right)}\right)}{2i_2\left(-1+e^{\left(\frac{2i_2R_E}{V_T}\right)}\right)+i_e\left(1+e^{\left(\frac{2i_2R_E}{V_T}\right)}\right)} $$

$$ i_1 = \frac{I_M}{2}\frac{\left(-1\pm e^{\left(\mp2\frac{i_2R_E + V_T\operatorname{Arctanh}\left[\frac{i_2}{i_e}\right]}{V_T}\right)}\right)}{\left(1\pm e^{\left(\mp 2\frac{i_2R_E + V_T\operatorname{Arctanh}\left[\frac{i_2}{i_e}\right]}{V_T}\right)}\right)} $$

I know these equations are equivalent because their Taylor series coefficients are pretty much the same except for one factor of 2 on the denominator. I have probably made a mistake with the formation of one of the two equations above, but I can't find it, and I think if I convert the top one to a form similar to the bottom one I can 'debug' the error. I also know the Arctanh to exponential identity gives a very similar form to the top one, (seen here 5th one down) but I don't know how to convert between the two with the extra variables in there.

I have tried many times but this is abit beyond me. Does someone want to have a go a converting Eq. 1 to Eq. 2 or vice versa? Even if it's not close, it might give me some insight.

Thanks a lot.

Edit: added missing $V_T$ and minus sign into eq. 2.

  • $\begingroup$ There's not much fancy technique to this: just carefully apply Arctanh definition and expand then simplify. $\endgroup$ – Scaramouche Aug 6 '13 at 0:10
  • $\begingroup$ What happens to the i2 and ie variables in the top equation when you convert it to the Arctanh? $\endgroup$ – T May Aug 6 '13 at 0:13
  • $\begingroup$ The numerator and denominator of the big fraction in your second equation are identical, so that fraction reduces to $1$. This probably isn't what you intend. $\endgroup$ – Blue Aug 6 '13 at 1:08
  • $\begingroup$ Thanks, Edited a missing minus sign into the numerator. $\endgroup$ – T May Aug 6 '13 at 1:15
  • $\begingroup$ There seems to be an error. The equations match (with $\pm = +$ and $\mp = -$ in the second equation), but only if the "$2i_2$" factors in the numerator and denominator of the first equation are replaced by "$i_2$". $\endgroup$ – Blue Aug 6 '13 at 1:41

There are way too many distracting symbols floating around. To simplify, I ignore the "$i_1$"s and "$I_M/2$"s. Also, I define $$a := i_2 \qquad b := i_e \qquad x := \frac{i_2 R_E}{V_T} \qquad y := \operatorname{atanh}\frac{i_2}{i_e} = \operatorname{atanh}\frac{a}{b}$$

Moreover, I assert that, in the second equation, "$\pm$" is "$+$", and "$\mp$" is "$-$"; and, in the first equation, I suppress the factors of "$2$", which I believe are in error.

Then the question becomes $$- \frac{a \left(e^{2x} + 1 \right) + b \left( e^{2x}-1 \right)}{a \left( e^{2x}-1\right) + b \left( e^{2x}+1 \right) } \quad \stackrel{?}{=} \quad -\frac{1-e^{-2(x+y)}}{1+e^{-2(x+y)}}$$ or, more simply, $$\frac{a \cosh x + b \sinh x}{a \sinh x + b \cosh x} \quad \stackrel{?}{=} \quad \frac{\sinh(x+y)}{\cosh(x+y)} = \tanh(x+y)$$ (Here, I multiplied-through the numerator and denominator on the left by $\frac{1}{2}e^{-x}$, and on the right by $\frac{1}{2}e^{(x+y)}$, and then invoked the definitions of $\sinh$ and $\cosh$.)

Now, dividing-through the right by $b\cosh x$, and recalling the definition of $y$, gives

$$\frac{\frac{a}{b} + \tanh x}{\frac{a}{b}\tanh x+1} = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y} \quad \stackrel{\ddot{\smile}}{=} \quad \tanh(x+y)$$

  • $\begingroup$ Thanks for the detailed response! I honestly thought no-one would be bothered helping me. I think I have got my head around it. This means I have neglected to divide some current (i2) from my differential amplifier equations, preceding these two equations. Thank you very much! $\endgroup$ – T May Aug 6 '13 at 5:27
  • $\begingroup$ @TMay: Others may have been waiting to make sure the dust had settled on minor edits to the equations. It's not much fun trying to prove things are equivalent when one is not certain what those things are supposed to be. ;) In any case, I'm glad to have been of service. $\endgroup$ – Blue Aug 6 '13 at 5:34

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