Reduction Formula $ I_{(m,n)}; =;int x^m(x+a)^ndx $
$begingroup$
Some problem occured in proving the following reduction formula.
$$
\
I_{(m,n)}; =;int x^m(x+a)^ndx; = ; frac{x^m(x+a)^{n+1}}{m+n+1}-frac{ma}{m+n+1}I_{(m-1,n)};;;;;;;;;;;;;;;m,n in N
$$
I have tried by using integration by part,here are my result
$$
begin{align}
I_{(m,n)}; =;frac{x^{m+1}(x+a)^n}{m+1}-frac{n}{m+1}I_{(m+1,n-1)}\
I_{(m,n)}; =;frac{x^{m}(x+a)^{n+1}}{n+1}-frac{m}{n+1}I_{(m-1,n+1)}
end{align}
$$
I have no idea on how to combine the 2 result or my direction of attacking the problem is wrong.
Any help would be appreciated.Thank you.
calculus integration reduction-formula
edited Jan 6 at 23:07
clathratus
5,2291438
asked Dec 7 '12 at 12:24
VulcanVulcan
13118
$endgroup$
add a comment |
$begingroup$
Some problem occured in proving the following reduction formula.
$$
\
I_{(m,n)}; =;int x^m(x+a)^ndx; = ; frac{x^m(x+a)^{n+1}}{m+n+1}-frac{ma}{m+n+1}I_{(m-1,n)};;;;;;;;;;;;;;;m,n in N
$$
I have tried by using integration by part,here are my result
$$
begin{align}
I_{(m,n)}; =;frac{x^{m+1}(x+a)^n}{m+1}-frac{n}{m+1}I_{(m+1,n-1)}\
I_{(m,n)}; =;frac{x^{m}(x+a)^{n+1}}{n+1}-frac{m}{n+1}I_{(m-1,n+1)}
end{align}
$$
I have no idea on how to combine the 2 result or my direction of attacking the problem is wrong.
Any help would be appreciated.Thank you.
calculus integration reduction-formula
edited Jan 6 at 23:07
clathratus
5,2291438
asked Dec 7 '12 at 12:24
VulcanVulcan
13118
$endgroup$
add a comment |
$begingroup$
Some problem occured in proving the following reduction formula.
$$
\
I_{(m,n)}; =;int x^m(x+a)^ndx; = ; frac{x^m(x+a)^{n+1}}{m+n+1}-frac{ma}{m+n+1}I_{(m-1,n)};;;;;;;;;;;;;;;m,n in N
$$
I have tried by using integration by part,here are my result
$$
begin{align}
I_{(m,n)}; =;frac{x^{m+1}(x+a)^n}{m+1}-frac{n}{m+1}I_{(m+1,n-1)}\
I_{(m,n)}; =;frac{x^{m}(x+a)^{n+1}}{n+1}-frac{m}{n+1}I_{(m-1,n+1)}
end{align}
$$
I have no idea on how to combine the 2 result or my direction of attacking the problem is wrong.
Any help would be appreciated.Thank you.
calculus integration reduction-formula
edited Jan 6 at 23:07
clathratus
5,2291438
asked Dec 7 '12 at 12:24
VulcanVulcan
13118
$endgroup$
Some problem occured in proving the following reduction formula.
$$
\
I_{(m,n)}; =;int x^m(x+a)^ndx; = ; frac{x^m(x+a)^{n+1}}{m+n+1}-frac{ma}{m+n+1}I_{(m-1,n)};;;;;;;;;;;;;;;m,n in N
$$
I have tried by using integration by part,here are my result
$$
begin{align}
I_{(m,n)}; =;frac{x^{m+1}(x+a)^n}{m+1}-frac{n}{m+1}I_{(m+1,n-1)}\
I_{(m,n)}; =;frac{x^{m}(x+a)^{n+1}}{n+1}-frac{m}{n+1}I_{(m-1,n+1)}
end{align}
$$
I have no idea on how to combine the 2 result or my direction of attacking the problem is wrong.
Any help would be appreciated.Thank you.
calculus integration reduction-formula
calculus integration reduction-formula
edited Jan 6 at 23:07
clathratus
5,2291438
asked Dec 7 '12 at 12:24
VulcanVulcan
13118
edited Jan 6 at 23:07
clathratus
5,2291438
asked Dec 7 '12 at 12:24
VulcanVulcan
13118
edited Jan 6 at 23:07
clathratus
5,2291438
edited Jan 6 at 23:07
clathratus
5,2291438
edited Jan 6 at 23:07
clathratus
5,2291438
5,2291438
asked Dec 7 '12 at 12:24
VulcanVulcan
13118
asked Dec 7 '12 at 12:24
VulcanVulcan
13118
asked Dec 7 '12 at 12:24
VulcanVulcan
13118
13118
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
$I(m,n)=int x^m(x+a)^ndx=int x^{m-1}(x+a-a)(x+a)^{n}dx=int x^{m-1}(x+a)^{n+1}dx-aint x^{m-1}(x+a)^{n}dx=I(m-1,n+1)-aI(m-1,n)$
Use this with your second reduction formula
answered Dec 7 '12 at 12:35
AmrAmr
14.4k43295
$endgroup$
$begingroup$
why would you think in that way?Is there any hint motivated us to do that?
$endgroup$
– Vulcan
Dec 7 '12 at 12:43
$begingroup$
The reduction fromulae you got will not help you reach your goal, thus its time to try something else
$endgroup$
– Amr
Dec 7 '12 at 12:45
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$I(m,n)=int x^m(x+a)^ndx=int x^{m-1}(x+a-a)(x+a)^{n}dx=int x^{m-1}(x+a)^{n+1}dx-aint x^{m-1}(x+a)^{n}dx=I(m-1,n+1)-aI(m-1,n)$
Use this with your second reduction formula
answered Dec 7 '12 at 12:35
AmrAmr
14.4k43295
$endgroup$
$begingroup$
why would you think in that way?Is there any hint motivated us to do that?
$endgroup$
– Vulcan
Dec 7 '12 at 12:43
$begingroup$
The reduction fromulae you got will not help you reach your goal, thus its time to try something else
$endgroup$
– Amr
Dec 7 '12 at 12:45
add a comment |
$begingroup$
Hint:
$I(m,n)=int x^m(x+a)^ndx=int x^{m-1}(x+a-a)(x+a)^{n}dx=int x^{m-1}(x+a)^{n+1}dx-aint x^{m-1}(x+a)^{n}dx=I(m-1,n+1)-aI(m-1,n)$
Use this with your second reduction formula
answered Dec 7 '12 at 12:35
AmrAmr
14.4k43295
$endgroup$
$begingroup$
why would you think in that way?Is there any hint motivated us to do that?
$endgroup$
– Vulcan
Dec 7 '12 at 12:43
$begingroup$
The reduction fromulae you got will not help you reach your goal, thus its time to try something else
$endgroup$
– Amr
Dec 7 '12 at 12:45
add a comment |
$begingroup$
Hint:
$I(m,n)=int x^m(x+a)^ndx=int x^{m-1}(x+a-a)(x+a)^{n}dx=int x^{m-1}(x+a)^{n+1}dx-aint x^{m-1}(x+a)^{n}dx=I(m-1,n+1)-aI(m-1,n)$
Use this with your second reduction formula
answered Dec 7 '12 at 12:35
AmrAmr
14.4k43295
$endgroup$
Hint:
$I(m,n)=int x^m(x+a)^ndx=int x^{m-1}(x+a-a)(x+a)^{n}dx=int x^{m-1}(x+a)^{n+1}dx-aint x^{m-1}(x+a)^{n}dx=I(m-1,n+1)-aI(m-1,n)$
Use this with your second reduction formula
answered Dec 7 '12 at 12:35
AmrAmr
14.4k43295
answered Dec 7 '12 at 12:35
AmrAmr
14.4k43295
answered Dec 7 '12 at 12:35
AmrAmr
14.4k43295
answered Dec 7 '12 at 12:35
AmrAmr
14.4k43295
14.4k43295
$begingroup$
why would you think in that way?Is there any hint motivated us to do that?
$endgroup$
– Vulcan
Dec 7 '12 at 12:43
$begingroup$
The reduction fromulae you got will not help you reach your goal, thus its time to try something else
$endgroup$
– Amr
Dec 7 '12 at 12:45
add a comment |
$begingroup$
why would you think in that way?Is there any hint motivated us to do that?
$endgroup$
– Vulcan
Dec 7 '12 at 12:43
$begingroup$
The reduction fromulae you got will not help you reach your goal, thus its time to try something else
$endgroup$
– Amr
Dec 7 '12 at 12:45
$begingroup$
why would you think in that way?Is there any hint motivated us to do that?
$endgroup$
– Vulcan
Dec 7 '12 at 12:43
$begingroup$
why would you think in that way?Is there any hint motivated us to do that?
$endgroup$
– Vulcan
Dec 7 '12 at 12:43
$begingroup$
The reduction fromulae you got will not help you reach your goal, thus its time to try something else
$endgroup$
– Amr
Dec 7 '12 at 12:45
$begingroup$
The reduction fromulae you got will not help you reach your goal, thus its time to try something else
$endgroup$
– Amr
Dec 7 '12 at 12:45
add a comment |
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