1
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$$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}$$

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  • $\begingroup$ @EricS. Yes all are scalars and you may assume $r>0$. $\endgroup$ – emcor Dec 16 '15 at 9:56
  • $\begingroup$ If you rewrite both enumerator and denominator into a single fraction, i.e. of the form $X=\frac{\frac{\alpha}{\beta}}{\frac{\gamma}{\delta}}$, you can rewrite that into $X=\frac{\alpha}{\beta}\frac{\delta}{\gamma}$ and a lot of terms will drop $\endgroup$ – Eric S. Dec 16 '15 at 9:58
  • $\begingroup$ @EricS. It would be very nice if you could do that... ; ) $\endgroup$ – emcor Dec 16 '15 at 10:00
  • $\begingroup$ Just be patient. Take lots of paper and multiply top and bottom by common terms. first by $(1+r)^T$. That will cause $\frac {1-c}{(1+4)^{T+1}}$ to reduce to $\frac{1-c}{1+r}$. Then by r. Eventually the whole thing should become manageable. Takes patience though. $\endgroup$ – fleablood Jan 7 '16 at 17:44
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$$\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}= \frac{r^2}{r^2}\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}= \frac{{c}+r^2\frac{1-c}{(1+r)^{T+1}}}{r{c}+r^2\frac{1-c}{(1+r)^T}-r^2}=\\ \frac{(1+r)^{T+1}}{(1+r)^{T+1}}\frac{{c}+r^2\frac{1-c}{(1+r)^{T+1}}}{r{c}+r^2\frac{1-c}{(1+r)^T}-r^2}= \frac{{c}(1+r)^{T+1}+r^2({1-c})}{r{c}(1+r)^{T+1}+r^2({1-c}){(1+r)}-r^2(1+r)^{T+1}}= \frac{{c}(1+r)^{T+1}+r^2({1-c})}{r\left({c}-r\right)(1+r)^{T+1}+r^2({1-c}){(1+r)}}= \frac{{c}(1+r)^{T+1}+r^2({1-c})}{\left(({c}-r)(1+r)^{T}+r({1-c})\right){(1+r)r}} $$ $$%=\frac{c-r^2 (c-1) (r+1)^{-T-1}}{r(c -r)-r^2(c-1) (r+1)^{-T}}$$

would this help?

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  • $\begingroup$ Thank you. Can you please also show the steps how to get there? $\endgroup$ – emcor Dec 16 '15 at 10:03
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$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^{T}}-1}$
$====================$
$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c-r}{r}+\frac{1-c}{(1+r)^{T}}}$
$====================$
$X=\frac{\frac{c(1+r)^{T+1}+(1-c)r^2}{r^2(1+r)^{T+1}}}{\frac{(c-r)(1+r)^{T}+(1-c)r}{r(1+r)^{T}}}$
$====================$
$X=\frac{\frac{c(1+r)^{T+1}+(1-c)r^2}{r(1+r)}}{\frac{(c-r)(1+r)^{T}+(1-c)r}1}$
$====================$
$X=\frac{(c(1+r)^{T+1}+(1-c)r^2)1}{r(1+r)((c-r)(1+r)^{T}+(1-c)r)}$
$====================$
$X=\frac{c(1+r)^{T+1}+(1-c)r^2}{r(1+r)((c-r)(1+r)^{T}+(1-c)r)}$

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