Mixing
Proof of an equality
$begingroup$
Show that
$(1+r)(-(1+r)^{-1}-(1+r)^{-2}-...-(1+r)^{-T+1})'_r = frac {1+r}{r^2} (1 - frac{Tr+1}{(1+r)^{T}} )$
sequences-and-series analysis
edited Jan 20 at 19:11
DisintegratingByParts
59.5k42580
asked Jan 20 at 12:07
PhilipPhilip
887
$endgroup$
add a comment |
$begingroup$
Show that
$(1+r)(-(1+r)^{-1}-(1+r)^{-2}-...-(1+r)^{-T+1})'_r = frac {1+r}{r^2} (1 - frac{Tr+1}{(1+r)^{T}} )$
sequences-and-series analysis
edited Jan 20 at 19:11
DisintegratingByParts
59.5k42580
asked Jan 20 at 12:07
PhilipPhilip
887
$endgroup$
$begingroup$
What did you try?
$endgroup$
– rtybase
Jan 20 at 12:19
$begingroup$
$(1+r)(-(1+r)^{-1}-...-(1+r)^{-T+1})'_r=(1+r) (...)'_r$ and here, instead of "..." maybe I can use the geometric progression or something like that. Then, find the derivation of (...) and get the answer? But actually I do not know how to continue this equality.
$endgroup$
– Philip
Jan 20 at 12:29
$begingroup$
Yes, go for it!
$endgroup$
– rtybase
Jan 20 at 12:39
$begingroup$
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
$endgroup$
– Shaun
Jan 20 at 13:22
add a comment |
$begingroup$
Show that
$(1+r)(-(1+r)^{-1}-(1+r)^{-2}-...-(1+r)^{-T+1})'_r = frac {1+r}{r^2} (1 - frac{Tr+1}{(1+r)^{T}} )$
sequences-and-series analysis
edited Jan 20 at 19:11
DisintegratingByParts
59.5k42580
asked Jan 20 at 12:07
PhilipPhilip
887
$endgroup$
Show that
$(1+r)(-(1+r)^{-1}-(1+r)^{-2}-...-(1+r)^{-T+1})'_r = frac {1+r}{r^2} (1 - frac{Tr+1}{(1+r)^{T}} )$
sequences-and-series analysis
sequences-and-series analysis
edited Jan 20 at 19:11
DisintegratingByParts
59.5k42580
asked Jan 20 at 12:07
PhilipPhilip
887
edited Jan 20 at 19:11
DisintegratingByParts
59.5k42580
asked Jan 20 at 12:07
PhilipPhilip
887
edited Jan 20 at 19:11
DisintegratingByParts
59.5k42580
edited Jan 20 at 19:11
DisintegratingByParts
59.5k42580
edited Jan 20 at 19:11
DisintegratingByParts
59.5k42580
59.5k42580
asked Jan 20 at 12:07
PhilipPhilip
887
asked Jan 20 at 12:07
PhilipPhilip
887
asked Jan 20 at 12:07
PhilipPhilip
887
887
$begingroup$
What did you try?
$endgroup$
– rtybase
Jan 20 at 12:19
$begingroup$
$(1+r)(-(1+r)^{-1}-...-(1+r)^{-T+1})'_r=(1+r) (...)'_r$ and here, instead of "..." maybe I can use the geometric progression or something like that. Then, find the derivation of (...) and get the answer? But actually I do not know how to continue this equality.
$endgroup$
– Philip
Jan 20 at 12:29
$begingroup$
Yes, go for it!
$endgroup$
– rtybase
Jan 20 at 12:39
$begingroup$
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
$endgroup$
– Shaun
Jan 20 at 13:22
add a comment |
$begingroup$
What did you try?
$endgroup$
– rtybase
Jan 20 at 12:19
$begingroup$
$(1+r)(-(1+r)^{-1}-...-(1+r)^{-T+1})'_r=(1+r) (...)'_r$ and here, instead of "..." maybe I can use the geometric progression or something like that. Then, find the derivation of (...) and get the answer? But actually I do not know how to continue this equality.
$endgroup$
– Philip
Jan 20 at 12:29
$begingroup$
Yes, go for it!
$endgroup$
– rtybase
Jan 20 at 12:39
$begingroup$
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
$endgroup$
– Shaun
Jan 20 at 13:22
$begingroup$
What did you try?
$endgroup$
– rtybase
Jan 20 at 12:19
$begingroup$
What did you try?
$endgroup$
– rtybase
Jan 20 at 12:19
$begingroup$
$(1+r)(-(1+r)^{-1}-...-(1+r)^{-T+1})'_r=(1+r) (...)'_r$ and here, instead of "..." maybe I can use the geometric progression or something like that. Then, find the derivation of (...) and get the answer? But actually I do not know how to continue this equality.
$endgroup$
– Philip
Jan 20 at 12:29
$begingroup$
$(1+r)(-(1+r)^{-1}-...-(1+r)^{-T+1})'_r=(1+r) (...)'_r$ and here, instead of "..." maybe I can use the geometric progression or something like that. Then, find the derivation of (...) and get the answer? But actually I do not know how to continue this equality.
$endgroup$
– Philip
Jan 20 at 12:29
$begingroup$
Yes, go for it!
$endgroup$
– rtybase
Jan 20 at 12:39
$begingroup$
Yes, go for it!
$endgroup$
– rtybase
Jan 20 at 12:39
$begingroup$
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
$endgroup$
– Shaun
Jan 20 at 13:22
$begingroup$
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
$endgroup$
– Shaun
Jan 20 at 13:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$(1+r)(-(1+r)^{-1}-(1+r)^{-2}-...-(1+r)^{-T+1})'_r$$$$ = -(1+r)((1+r)^{-1}+(1+r)^{-2}+...+(1+r)^{-T+1})'_r$$$$=-(1+r)(frac{(1+r)^{-1}(1-(1+r)^{-T+1})}{1-(1+r)^{-1}})'_r$$$$=-(1+r)(frac{1-(1+r)^{-T+1}}{r})'_r$$$$=-(1+r)(frac{1}{r}-frac{(1+r)^{-T+1}}{r})'_r$$$$=-(1+r)(-frac{1}{r^2}-frac{r cdot (-T+1)(1+r)^{-T} - (1+r)^{-T+1}}{r^2})$$$$=frac{1+r}{r^2}(1+r cdot (-T+1)(1+r)^{-T} - (1+r)^{-T+1})$$$$=frac{1+r}{r^2}(1-frac{rcdot(T-1)+1+r}{(1+r)^{T}})$$$$=frac{1+r}{r^2}(1-frac{rcdot T+1}{(1+r)^{T}})$$
answered Jan 20 at 12:34
Peter ForemanPeter Foreman
2,7421214
$endgroup$
$begingroup$
I just used the fact that $1+x+x^2+...+x^{n-1} = frac{1-x^n}{1-x}$
$endgroup$
– Peter Foreman
Jan 20 at 12:35
add a comment |
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1 Answer
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$begingroup$
$$(1+r)(-(1+r)^{-1}-(1+r)^{-2}-...-(1+r)^{-T+1})'_r$$$$ = -(1+r)((1+r)^{-1}+(1+r)^{-2}+...+(1+r)^{-T+1})'_r$$$$=-(1+r)(frac{(1+r)^{-1}(1-(1+r)^{-T+1})}{1-(1+r)^{-1}})'_r$$$$=-(1+r)(frac{1-(1+r)^{-T+1}}{r})'_r$$$$=-(1+r)(frac{1}{r}-frac{(1+r)^{-T+1}}{r})'_r$$$$=-(1+r)(-frac{1}{r^2}-frac{r cdot (-T+1)(1+r)^{-T} - (1+r)^{-T+1}}{r^2})$$$$=frac{1+r}{r^2}(1+r cdot (-T+1)(1+r)^{-T} - (1+r)^{-T+1})$$$$=frac{1+r}{r^2}(1-frac{rcdot(T-1)+1+r}{(1+r)^{T}})$$$$=frac{1+r}{r^2}(1-frac{rcdot T+1}{(1+r)^{T}})$$
answered Jan 20 at 12:34
Peter ForemanPeter Foreman
2,7421214
$endgroup$
$begingroup$
I just used the fact that $1+x+x^2+...+x^{n-1} = frac{1-x^n}{1-x}$
$endgroup$
– Peter Foreman
Jan 20 at 12:35
add a comment |
$begingroup$
$$(1+r)(-(1+r)^{-1}-(1+r)^{-2}-...-(1+r)^{-T+1})'_r$$$$ = -(1+r)((1+r)^{-1}+(1+r)^{-2}+...+(1+r)^{-T+1})'_r$$$$=-(1+r)(frac{(1+r)^{-1}(1-(1+r)^{-T+1})}{1-(1+r)^{-1}})'_r$$$$=-(1+r)(frac{1-(1+r)^{-T+1}}{r})'_r$$$$=-(1+r)(frac{1}{r}-frac{(1+r)^{-T+1}}{r})'_r$$$$=-(1+r)(-frac{1}{r^2}-frac{r cdot (-T+1)(1+r)^{-T} - (1+r)^{-T+1}}{r^2})$$$$=frac{1+r}{r^2}(1+r cdot (-T+1)(1+r)^{-T} - (1+r)^{-T+1})$$$$=frac{1+r}{r^2}(1-frac{rcdot(T-1)+1+r}{(1+r)^{T}})$$$$=frac{1+r}{r^2}(1-frac{rcdot T+1}{(1+r)^{T}})$$
answered Jan 20 at 12:34
Peter ForemanPeter Foreman
2,7421214
$endgroup$
$begingroup$
I just used the fact that $1+x+x^2+...+x^{n-1} = frac{1-x^n}{1-x}$
$endgroup$
– Peter Foreman
Jan 20 at 12:35
add a comment |
$begingroup$
$$(1+r)(-(1+r)^{-1}-(1+r)^{-2}-...-(1+r)^{-T+1})'_r$$$$ = -(1+r)((1+r)^{-1}+(1+r)^{-2}+...+(1+r)^{-T+1})'_r$$$$=-(1+r)(frac{(1+r)^{-1}(1-(1+r)^{-T+1})}{1-(1+r)^{-1}})'_r$$$$=-(1+r)(frac{1-(1+r)^{-T+1}}{r})'_r$$$$=-(1+r)(frac{1}{r}-frac{(1+r)^{-T+1}}{r})'_r$$$$=-(1+r)(-frac{1}{r^2}-frac{r cdot (-T+1)(1+r)^{-T} - (1+r)^{-T+1}}{r^2})$$$$=frac{1+r}{r^2}(1+r cdot (-T+1)(1+r)^{-T} - (1+r)^{-T+1})$$$$=frac{1+r}{r^2}(1-frac{rcdot(T-1)+1+r}{(1+r)^{T}})$$$$=frac{1+r}{r^2}(1-frac{rcdot T+1}{(1+r)^{T}})$$
answered Jan 20 at 12:34
Peter ForemanPeter Foreman
2,7421214
$endgroup$
$$(1+r)(-(1+r)^{-1}-(1+r)^{-2}-...-(1+r)^{-T+1})'_r$$$$ = -(1+r)((1+r)^{-1}+(1+r)^{-2}+...+(1+r)^{-T+1})'_r$$$$=-(1+r)(frac{(1+r)^{-1}(1-(1+r)^{-T+1})}{1-(1+r)^{-1}})'_r$$$$=-(1+r)(frac{1-(1+r)^{-T+1}}{r})'_r$$$$=-(1+r)(frac{1}{r}-frac{(1+r)^{-T+1}}{r})'_r$$$$=-(1+r)(-frac{1}{r^2}-frac{r cdot (-T+1)(1+r)^{-T} - (1+r)^{-T+1}}{r^2})$$$$=frac{1+r}{r^2}(1+r cdot (-T+1)(1+r)^{-T} - (1+r)^{-T+1})$$$$=frac{1+r}{r^2}(1-frac{rcdot(T-1)+1+r}{(1+r)^{T}})$$$$=frac{1+r}{r^2}(1-frac{rcdot T+1}{(1+r)^{T}})$$
answered Jan 20 at 12:34
Peter ForemanPeter Foreman
2,7421214
answered Jan 20 at 12:34
Peter ForemanPeter Foreman
2,7421214
answered Jan 20 at 12:34
Peter ForemanPeter Foreman
2,7421214
answered Jan 20 at 12:34
Peter ForemanPeter Foreman
2,7421214
2,7421214
$begingroup$
I just used the fact that $1+x+x^2+...+x^{n-1} = frac{1-x^n}{1-x}$
$endgroup$
– Peter Foreman
Jan 20 at 12:35
add a comment |
$begingroup$
I just used the fact that $1+x+x^2+...+x^{n-1} = frac{1-x^n}{1-x}$
$endgroup$
– Peter Foreman
Jan 20 at 12:35
$begingroup$
I just used the fact that $1+x+x^2+...+x^{n-1} = frac{1-x^n}{1-x}$
$endgroup$
– Peter Foreman
Jan 20 at 12:35
$begingroup$
I just used the fact that $1+x+x^2+...+x^{n-1} = frac{1-x^n}{1-x}$
$endgroup$
– Peter Foreman
Jan 20 at 12:35
add a comment |
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$begingroup$
What did you try?
$endgroup$
– rtybase
Jan 20 at 12:19
$begingroup$
$(1+r)(-(1+r)^{-1}-...-(1+r)^{-T+1})'_r=(1+r) (...)'_r$ and here, instead of "..." maybe I can use the geometric progression or something like that. Then, find the derivation of (...) and get the answer? But actually I do not know how to continue this equality.
$endgroup$
– Philip
Jan 20 at 12:29
$begingroup$
Yes, go for it!
$endgroup$
– rtybase
Jan 20 at 12:39
$begingroup$
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
$endgroup$
– Shaun
Jan 20 at 13:22