If $I_{m,n} = int_{0}^{pi/2}cos^mxcos nx ~dx$, prove that $I_{m,n} = frac{m(m-1)}{m^2-n^2}I_{m-2,n}$
up vote
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down vote
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If $$I_{m,n} = int_{0}^{pi/2}cos^m x cos nx ~dx$$
prove that
$$I_{m,n} = frac{m(m-1)}{m^2-n^2}I_{m-2,n}$$
I have tried my level best to solve but failed. Please help me out with a detailed solution if possible.This is what I have done...
integration trigonometry trigonometric-integrals
edited Nov 24 at 15:19
Tianlalu
2,8661835
asked Nov 18 at 7:05
Abhishek Ghosh
426
add a comment |
up vote
4
down vote
favorite
If $$I_{m,n} = int_{0}^{pi/2}cos^m x cos nx ~dx$$
prove that
$$I_{m,n} = frac{m(m-1)}{m^2-n^2}I_{m-2,n}$$
I have tried my level best to solve but failed. Please help me out with a detailed solution if possible.This is what I have done...
integration trigonometry trigonometric-integrals
edited Nov 24 at 15:19
Tianlalu
2,8661835
asked Nov 18 at 7:05
Abhishek Ghosh
426
1
You have posted an exact duplicate of a question (now deleted). Re-posting is not the way to call attention to your question, nor to side-step being put "on hold". Instead, you should edit the previous question (which you did) and wait for the community to re-open (which you didn't). Please close this question and un-delete the other. Also, if you can, please upload your image to a service such as Imgur; people are rightfully averse to clicking a link to a Google Drive file that may-or-may-not be an image.
– Blue
Nov 18 at 7:17
But no one is looking into that previous post... I have edited a lot
– Abhishek Ghosh
Nov 18 at 7:19
Unfortunately, Math.SE gets a great deal of traffic. Not every question gets as much attention as we might like. (The same is true of answers. I post a lot of really good ones, and no one notices. ;) Again, re-posting is not the appropriate course of action. That said, showing your work as you did is the right thing to do. (If nothing else, it tells people that you aren't simply trying to have them do your homework for you!) But it's up to the community to vote to re-open.
– Blue
Nov 18 at 7:27
Take a look here: math.stackexchange.com/questions/849826/…
– Robert Z
Nov 18 at 9:25
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
If $$I_{m,n} = int_{0}^{pi/2}cos^m x cos nx ~dx$$
prove that
$$I_{m,n} = frac{m(m-1)}{m^2-n^2}I_{m-2,n}$$
I have tried my level best to solve but failed. Please help me out with a detailed solution if possible.This is what I have done...
integration trigonometry trigonometric-integrals
edited Nov 24 at 15:19
Tianlalu
2,8661835
asked Nov 18 at 7:05
Abhishek Ghosh
426
If $$I_{m,n} = int_{0}^{pi/2}cos^m x cos nx ~dx$$
prove that
$$I_{m,n} = frac{m(m-1)}{m^2-n^2}I_{m-2,n}$$
I have tried my level best to solve but failed. Please help me out with a detailed solution if possible.This is what I have done...
integration trigonometry trigonometric-integrals
integration trigonometry trigonometric-integrals
edited Nov 24 at 15:19
Tianlalu
2,8661835
asked Nov 18 at 7:05
Abhishek Ghosh
426
edited Nov 24 at 15:19
Tianlalu
2,8661835
asked Nov 18 at 7:05
Abhishek Ghosh
426
edited Nov 24 at 15:19
Tianlalu
2,8661835
edited Nov 24 at 15:19
Tianlalu
2,8661835
edited Nov 24 at 15:19
Tianlalu
2,8661835
2,8661835
asked Nov 18 at 7:05
Abhishek Ghosh
426
asked Nov 18 at 7:05
Abhishek Ghosh
426
asked Nov 18 at 7:05
Abhishek Ghosh
426
426
1
You have posted an exact duplicate of a question (now deleted). Re-posting is not the way to call attention to your question, nor to side-step being put "on hold". Instead, you should edit the previous question (which you did) and wait for the community to re-open (which you didn't). Please close this question and un-delete the other. Also, if you can, please upload your image to a service such as Imgur; people are rightfully averse to clicking a link to a Google Drive file that may-or-may-not be an image.
– Blue
Nov 18 at 7:17
But no one is looking into that previous post... I have edited a lot
– Abhishek Ghosh
Nov 18 at 7:19
Unfortunately, Math.SE gets a great deal of traffic. Not every question gets as much attention as we might like. (The same is true of answers. I post a lot of really good ones, and no one notices. ;) Again, re-posting is not the appropriate course of action. That said, showing your work as you did is the right thing to do. (If nothing else, it tells people that you aren't simply trying to have them do your homework for you!) But it's up to the community to vote to re-open.
– Blue
Nov 18 at 7:27
Take a look here: math.stackexchange.com/questions/849826/…
– Robert Z
Nov 18 at 9:25
add a comment |
1
You have posted an exact duplicate of a question (now deleted). Re-posting is not the way to call attention to your question, nor to side-step being put "on hold". Instead, you should edit the previous question (which you did) and wait for the community to re-open (which you didn't). Please close this question and un-delete the other. Also, if you can, please upload your image to a service such as Imgur; people are rightfully averse to clicking a link to a Google Drive file that may-or-may-not be an image.
– Blue
Nov 18 at 7:17
But no one is looking into that previous post... I have edited a lot
– Abhishek Ghosh
Nov 18 at 7:19
Unfortunately, Math.SE gets a great deal of traffic. Not every question gets as much attention as we might like. (The same is true of answers. I post a lot of really good ones, and no one notices. ;) Again, re-posting is not the appropriate course of action. That said, showing your work as you did is the right thing to do. (If nothing else, it tells people that you aren't simply trying to have them do your homework for you!) But it's up to the community to vote to re-open.
– Blue
Nov 18 at 7:27
Take a look here: math.stackexchange.com/questions/849826/…
– Robert Z
Nov 18 at 9:25
1
1
You have posted an exact duplicate of a question (now deleted). Re-posting is not the way to call attention to your question, nor to side-step being put "on hold". Instead, you should edit the previous question (which you did) and wait for the community to re-open (which you didn't). Please close this question and un-delete the other. Also, if you can, please upload your image to a service such as Imgur; people are rightfully averse to clicking a link to a Google Drive file that may-or-may-not be an image.
– Blue
Nov 18 at 7:17
You have posted an exact duplicate of a question (now deleted). Re-posting is not the way to call attention to your question, nor to side-step being put "on hold". Instead, you should edit the previous question (which you did) and wait for the community to re-open (which you didn't). Please close this question and un-delete the other. Also, if you can, please upload your image to a service such as Imgur; people are rightfully averse to clicking a link to a Google Drive file that may-or-may-not be an image.
– Blue
Nov 18 at 7:17
But no one is looking into that previous post... I have edited a lot
– Abhishek Ghosh
Nov 18 at 7:19
But no one is looking into that previous post... I have edited a lot
– Abhishek Ghosh
Nov 18 at 7:19
Unfortunately, Math.SE gets a great deal of traffic. Not every question gets as much attention as we might like. (The same is true of answers. I post a lot of really good ones, and no one notices. ;) Again, re-posting is not the appropriate course of action. That said, showing your work as you did is the right thing to do. (If nothing else, it tells people that you aren't simply trying to have them do your homework for you!) But it's up to the community to vote to re-open.
– Blue
Nov 18 at 7:27
Unfortunately, Math.SE gets a great deal of traffic. Not every question gets as much attention as we might like. (The same is true of answers. I post a lot of really good ones, and no one notices. ;) Again, re-posting is not the appropriate course of action. That said, showing your work as you did is the right thing to do. (If nothing else, it tells people that you aren't simply trying to have them do your homework for you!) But it's up to the community to vote to re-open.
– Blue
Nov 18 at 7:27
Take a look here: math.stackexchange.com/questions/849826/…
– Robert Z
Nov 18 at 9:25
Take a look here: math.stackexchange.com/questions/849826/…
– Robert Z
Nov 18 at 9:25
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Let $f_{m,a}(x)=f_m=e^{ax}cos^mx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-mf_{m-1}sin x.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+mf_{m-1}sin x$, the derivative of $g_{m}$ will cancel $color{blue}{sin x}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{7.3em}+mbig(f'_{m-1}sin xhspace{9.9em}+underbrace{f_{m-1}cos x}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{amf_{m-1}sin x}+color{blue}mbig(color{blue}{af_{m-1}sin x}-(m-1) f_{m-2}color{orange}{sin^2 x}+f_mbig)\
g'_{m}&=(a^2+m)f_m-m(m-1)f_{m-2}color{orange}{(1-cos^2 x)}\
g'_{m}&=(a^2+m^2)f_m-m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$J_{m,a}=int_0^{pi/2}e^{ax}cos^mx~mathrm dx=int_0^{pi/2}f_m~mathrm dx,$$
integrate $(2)$,
$$underbrace{big[g_mbig]_0^{pi/2}}_{=-a}=(a^2+m^2)J_{m,a}-m(m-1)J_{m-2,a}.tag{3}$$
Put $a=in$, notice $Rebig{J_{m,in}big}=I_{m,n}$, by considering the real part of $(3)$, we get the desired result
$$0=(m^2-n^2)I_{m,n}-m(m-1)I_{m-2,n}.$$
answered Nov 18 at 12:17
Tianlalu
2,8661835
Just before equation (2) there is (1-cos²x) how does that term vanish in equation labelled (2) can you please explain? @Tianlalu
– Abhishek Ghosh
Nov 19 at 2:40
@AbhishekGhosh : $$f_{m-2}cos^2 x=f_m.$$
– Tianlalu
Nov 19 at 2:42
Thank you. I got it!
– Abhishek Ghosh
Nov 19 at 2:45
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $f_{m,a}(x)=f_m=e^{ax}cos^mx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-mf_{m-1}sin x.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+mf_{m-1}sin x$, the derivative of $g_{m}$ will cancel $color{blue}{sin x}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{7.3em}+mbig(f'_{m-1}sin xhspace{9.9em}+underbrace{f_{m-1}cos x}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{amf_{m-1}sin x}+color{blue}mbig(color{blue}{af_{m-1}sin x}-(m-1) f_{m-2}color{orange}{sin^2 x}+f_mbig)\
g'_{m}&=(a^2+m)f_m-m(m-1)f_{m-2}color{orange}{(1-cos^2 x)}\
g'_{m}&=(a^2+m^2)f_m-m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$J_{m,a}=int_0^{pi/2}e^{ax}cos^mx~mathrm dx=int_0^{pi/2}f_m~mathrm dx,$$
integrate $(2)$,
$$underbrace{big[g_mbig]_0^{pi/2}}_{=-a}=(a^2+m^2)J_{m,a}-m(m-1)J_{m-2,a}.tag{3}$$
Put $a=in$, notice $Rebig{J_{m,in}big}=I_{m,n}$, by considering the real part of $(3)$, we get the desired result
$$0=(m^2-n^2)I_{m,n}-m(m-1)I_{m-2,n}.$$
answered Nov 18 at 12:17
Tianlalu
2,8661835
Just before equation (2) there is (1-cos²x) how does that term vanish in equation labelled (2) can you please explain? @Tianlalu
– Abhishek Ghosh
Nov 19 at 2:40
@AbhishekGhosh : $$f_{m-2}cos^2 x=f_m.$$
– Tianlalu
Nov 19 at 2:42
Thank you. I got it!
– Abhishek Ghosh
Nov 19 at 2:45
add a comment |
up vote
1
down vote
accepted
Let $f_{m,a}(x)=f_m=e^{ax}cos^mx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-mf_{m-1}sin x.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+mf_{m-1}sin x$, the derivative of $g_{m}$ will cancel $color{blue}{sin x}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{7.3em}+mbig(f'_{m-1}sin xhspace{9.9em}+underbrace{f_{m-1}cos x}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{amf_{m-1}sin x}+color{blue}mbig(color{blue}{af_{m-1}sin x}-(m-1) f_{m-2}color{orange}{sin^2 x}+f_mbig)\
g'_{m}&=(a^2+m)f_m-m(m-1)f_{m-2}color{orange}{(1-cos^2 x)}\
g'_{m}&=(a^2+m^2)f_m-m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$J_{m,a}=int_0^{pi/2}e^{ax}cos^mx~mathrm dx=int_0^{pi/2}f_m~mathrm dx,$$
integrate $(2)$,
$$underbrace{big[g_mbig]_0^{pi/2}}_{=-a}=(a^2+m^2)J_{m,a}-m(m-1)J_{m-2,a}.tag{3}$$
Put $a=in$, notice $Rebig{J_{m,in}big}=I_{m,n}$, by considering the real part of $(3)$, we get the desired result
$$0=(m^2-n^2)I_{m,n}-m(m-1)I_{m-2,n}.$$
answered Nov 18 at 12:17
Tianlalu
2,8661835
Just before equation (2) there is (1-cos²x) how does that term vanish in equation labelled (2) can you please explain? @Tianlalu
– Abhishek Ghosh
Nov 19 at 2:40
@AbhishekGhosh : $$f_{m-2}cos^2 x=f_m.$$
– Tianlalu
Nov 19 at 2:42
Thank you. I got it!
– Abhishek Ghosh
Nov 19 at 2:45
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $f_{m,a}(x)=f_m=e^{ax}cos^mx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-mf_{m-1}sin x.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+mf_{m-1}sin x$, the derivative of $g_{m}$ will cancel $color{blue}{sin x}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{7.3em}+mbig(f'_{m-1}sin xhspace{9.9em}+underbrace{f_{m-1}cos x}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{amf_{m-1}sin x}+color{blue}mbig(color{blue}{af_{m-1}sin x}-(m-1) f_{m-2}color{orange}{sin^2 x}+f_mbig)\
g'_{m}&=(a^2+m)f_m-m(m-1)f_{m-2}color{orange}{(1-cos^2 x)}\
g'_{m}&=(a^2+m^2)f_m-m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$J_{m,a}=int_0^{pi/2}e^{ax}cos^mx~mathrm dx=int_0^{pi/2}f_m~mathrm dx,$$
integrate $(2)$,
$$underbrace{big[g_mbig]_0^{pi/2}}_{=-a}=(a^2+m^2)J_{m,a}-m(m-1)J_{m-2,a}.tag{3}$$
Put $a=in$, notice $Rebig{J_{m,in}big}=I_{m,n}$, by considering the real part of $(3)$, we get the desired result
$$0=(m^2-n^2)I_{m,n}-m(m-1)I_{m-2,n}.$$
answered Nov 18 at 12:17
Tianlalu
2,8661835
Let $f_{m,a}(x)=f_m=e^{ax}cos^mx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-mf_{m-1}sin x.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+mf_{m-1}sin x$, the derivative of $g_{m}$ will cancel $color{blue}{sin x}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{7.3em}+mbig(f'_{m-1}sin xhspace{9.9em}+underbrace{f_{m-1}cos x}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{amf_{m-1}sin x}+color{blue}mbig(color{blue}{af_{m-1}sin x}-(m-1) f_{m-2}color{orange}{sin^2 x}+f_mbig)\
g'_{m}&=(a^2+m)f_m-m(m-1)f_{m-2}color{orange}{(1-cos^2 x)}\
g'_{m}&=(a^2+m^2)f_m-m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$J_{m,a}=int_0^{pi/2}e^{ax}cos^mx~mathrm dx=int_0^{pi/2}f_m~mathrm dx,$$
integrate $(2)$,
$$underbrace{big[g_mbig]_0^{pi/2}}_{=-a}=(a^2+m^2)J_{m,a}-m(m-1)J_{m-2,a}.tag{3}$$
Put $a=in$, notice $Rebig{J_{m,in}big}=I_{m,n}$, by considering the real part of $(3)$, we get the desired result
$$0=(m^2-n^2)I_{m,n}-m(m-1)I_{m-2,n}.$$
answered Nov 18 at 12:17
Tianlalu
2,8661835
answered Nov 18 at 12:17
Tianlalu
2,8661835
answered Nov 18 at 12:17
Tianlalu
2,8661835
answered Nov 18 at 12:17
Tianlalu
2,8661835
2,8661835
Just before equation (2) there is (1-cos²x) how does that term vanish in equation labelled (2) can you please explain? @Tianlalu
– Abhishek Ghosh
Nov 19 at 2:40
@AbhishekGhosh : $$f_{m-2}cos^2 x=f_m.$$
– Tianlalu
Nov 19 at 2:42
Thank you. I got it!
– Abhishek Ghosh
Nov 19 at 2:45
add a comment |
Just before equation (2) there is (1-cos²x) how does that term vanish in equation labelled (2) can you please explain? @Tianlalu
– Abhishek Ghosh
Nov 19 at 2:40
@AbhishekGhosh : $$f_{m-2}cos^2 x=f_m.$$
– Tianlalu
Nov 19 at 2:42
Thank you. I got it!
– Abhishek Ghosh
Nov 19 at 2:45
Just before equation (2) there is (1-cos²x) how does that term vanish in equation labelled (2) can you please explain? @Tianlalu
– Abhishek Ghosh
Nov 19 at 2:40
Just before equation (2) there is (1-cos²x) how does that term vanish in equation labelled (2) can you please explain? @Tianlalu
– Abhishek Ghosh
Nov 19 at 2:40
@AbhishekGhosh : $$f_{m-2}cos^2 x=f_m.$$
– Tianlalu
Nov 19 at 2:42
@AbhishekGhosh : $$f_{m-2}cos^2 x=f_m.$$
– Tianlalu
Nov 19 at 2:42
Thank you. I got it!
– Abhishek Ghosh
Nov 19 at 2:45
Thank you. I got it!
– Abhishek Ghosh
Nov 19 at 2:45
add a comment |
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1
You have posted an exact duplicate of a question (now deleted). Re-posting is not the way to call attention to your question, nor to side-step being put "on hold". Instead, you should edit the previous question (which you did) and wait for the community to re-open (which you didn't). Please close this question and un-delete the other. Also, if you can, please upload your image to a service such as Imgur; people are rightfully averse to clicking a link to a Google Drive file that may-or-may-not be an image.
– Blue
Nov 18 at 7:17
But no one is looking into that previous post... I have edited a lot
– Abhishek Ghosh
Nov 18 at 7:19
Unfortunately, Math.SE gets a great deal of traffic. Not every question gets as much attention as we might like. (The same is true of answers. I post a lot of really good ones, and no one notices. ;) Again, re-posting is not the appropriate course of action. That said, showing your work as you did is the right thing to do. (If nothing else, it tells people that you aren't simply trying to have them do your homework for you!) But it's up to the community to vote to re-open.
– Blue
Nov 18 at 7:27
Take a look here: math.stackexchange.com/questions/849826/…
– Robert Z
Nov 18 at 9:25