Half Range Sine SeriesComputation Of full range fourier seriesFourier Series of The Sine FunctionFourier series with half rangeFourier sine series of $f = cos x$Find the half-range Fourier series expansion of $f(x) = cos(x)$Fourier sine series for $x^3$Half range Fourier sine series of cos(x) on 0 < x < $fracpi2$Find the Fourier series of rectified sine waveFourier Series Expansion for Half-Wave Sine ProblemFourier series of $left|xright|$
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Half Range Sine Series
Computation Of full range fourier seriesFourier Series of The Sine FunctionFourier series with half rangeFourier sine series of $f = cos x$Find the half-range Fourier series expansion of $f(x) = cos(x)$Fourier sine series for $x^3$Half range Fourier sine series of cos(x) on 0 < x < $fracpi2$Find the Fourier series of rectified sine waveFourier Series Expansion for Half-Wave Sine ProblemFourier series of $left|xright|$
$begingroup$
Question:
It is known that $f(x)=(x−4)^2$ for all $xin [0,4]$.
Compute the half range sine series expansion for $f(x)$.
My answer :
Half range series: $p=8$, $l=4$, $a_0=a_n=0$.
$$b_n=frac2Lint_0^Lf(x)sinleft(fracnpi xLright)d(x)=frac24int_0^4(x-4)^2sinleft(fracnpi x4right)d(x)$$
Partial Differentiation
Let $u=(x-4)^2$; $du=2(x-4)dx$; $v=frac-4npicos(fracnpi x4)$
beginalign
b_n&=frac12[frac-4npicos(fracnpi x4)(x-4)^2+frac8npiint(x-4)cos(fracnpi x4)d(x)]|^4_0\
&=frac12[frac-4npicos(fracnpi x4)(x-4)^2+frac8npi[frac4npisin(fracnpi x4)(x-4)-frac4npiintsin(fracnpi x4)d(x)]]|^4_0\
&=frac12[frac-4npicos(fracnpi x4)(x-4)^2$+$frac8npi[frac4npisin(fracnpi x4)(x-4)-frac4npi(frac-4npicosfracnpi x4)]|^4_0\
&=frac12[frac-4npicos(fracnpi x4)(x-4)^2$+$frac8npi(frac16n^2pi^2cosfracnpi x4)]|^4_0
endalign
we know $cosnpi=(-1)^n$
$frac12[(0+frac128(-1)^nn^3pi^3)-(-frac64npi+frac128n^3pi^3)$
$b_n=frac64(-1)^nn^3pi^3-frac64n^3pi^3+frac32npi$
My teacher told me my $b_n$ is wrong and it seems that my working also looks like wrong. Can someone please help me to recalculate my $b_n$ please. Just help me once only guys.If I get this right I would able to get full marks.I would be very very grateful to you guys if you guys able to help me. If can suggest me any ideas or simpler way to solve it if you can.Thanks
calculus real-analysis fourier-series
edited Sep 9 '12 at 8:20
Davide Giraudo
128k17156268
asked Sep 9 '12 at 6:31
DavidDavid
8019
$endgroup$
|
show 2 more comments
$begingroup$
Question:
It is known that $f(x)=(x−4)^2$ for all $xin [0,4]$.
Compute the half range sine series expansion for $f(x)$.
My answer :
Half range series: $p=8$, $l=4$, $a_0=a_n=0$.
$$b_n=frac2Lint_0^Lf(x)sinleft(fracnpi xLright)d(x)=frac24int_0^4(x-4)^2sinleft(fracnpi x4right)d(x)$$
Partial Differentiation
Let $u=(x-4)^2$; $du=2(x-4)dx$; $v=frac-4npicos(fracnpi x4)$
beginalign
b_n&=frac12[frac-4npicos(fracnpi x4)(x-4)^2+frac8npiint(x-4)cos(fracnpi x4)d(x)]|^4_0\
&=frac12[frac-4npicos(fracnpi x4)(x-4)^2+frac8npi[frac4npisin(fracnpi x4)(x-4)-frac4npiintsin(fracnpi x4)d(x)]]|^4_0\
&=frac12[frac-4npicos(fracnpi x4)(x-4)^2$+$frac8npi[frac4npisin(fracnpi x4)(x-4)-frac4npi(frac-4npicosfracnpi x4)]|^4_0\
&=frac12[frac-4npicos(fracnpi x4)(x-4)^2$+$frac8npi(frac16n^2pi^2cosfracnpi x4)]|^4_0
endalign
we know $cosnpi=(-1)^n$
$frac12[(0+frac128(-1)^nn^3pi^3)-(-frac64npi+frac128n^3pi^3)$
$b_n=frac64(-1)^nn^3pi^3-frac64n^3pi^3+frac32npi$
My teacher told me my $b_n$ is wrong and it seems that my working also looks like wrong. Can someone please help me to recalculate my $b_n$ please. Just help me once only guys.If I get this right I would able to get full marks.I would be very very grateful to you guys if you guys able to help me. If can suggest me any ideas or simpler way to solve it if you can.Thanks
calculus real-analysis fourier-series
edited Sep 9 '12 at 8:20
Davide Giraudo
128k17156268
asked Sep 9 '12 at 6:31
DavidDavid
8019
$endgroup$
$begingroup$
Should $b_n$ be the integral from $-L$ to $L$ or the integral from $0$ to $L$ because it switches from the first line to the second line without any multiplicative factor.
$endgroup$
– Euler....IS_ALIVE
Sep 9 '12 at 6:40
$begingroup$
O to L. It alwys like that for Half Range Sine Series
$endgroup$
– David
Sep 9 '12 at 6:48
$begingroup$
But you wrote it as $-L$ to $L$ in your first definition of $b_n$....
$endgroup$
– Euler....IS_ALIVE
Sep 9 '12 at 6:49
$begingroup$
this is the link for reference intmath.com/fourier-series/4-fourier-half-range-functions.php
$endgroup$
– David
Sep 9 '12 at 6:49
$begingroup$
I edited already
$endgroup$
– David
Sep 9 '12 at 6:50
|
show 2 more comments
$begingroup$
Question:
It is known that $f(x)=(x−4)^2$ for all $xin [0,4]$.
Compute the half range sine series expansion for $f(x)$.
My answer :
Half range series: $p=8$, $l=4$, $a_0=a_n=0$.
$$b_n=frac2Lint_0^Lf(x)sinleft(fracnpi xLright)d(x)=frac24int_0^4(x-4)^2sinleft(fracnpi x4right)d(x)$$
Partial Differentiation
Let $u=(x-4)^2$; $du=2(x-4)dx$; $v=frac-4npicos(fracnpi x4)$
beginalign
b_n&=frac12[frac-4npicos(fracnpi x4)(x-4)^2+frac8npiint(x-4)cos(fracnpi x4)d(x)]|^4_0\
&=frac12[frac-4npicos(fracnpi x4)(x-4)^2+frac8npi[frac4npisin(fracnpi x4)(x-4)-frac4npiintsin(fracnpi x4)d(x)]]|^4_0\
&=frac12[frac-4npicos(fracnpi x4)(x-4)^2$+$frac8npi[frac4npisin(fracnpi x4)(x-4)-frac4npi(frac-4npicosfracnpi x4)]|^4_0\
&=frac12[frac-4npicos(fracnpi x4)(x-4)^2$+$frac8npi(frac16n^2pi^2cosfracnpi x4)]|^4_0
endalign
we know $cosnpi=(-1)^n$
$frac12[(0+frac128(-1)^nn^3pi^3)-(-frac64npi+frac128n^3pi^3)$
$b_n=frac64(-1)^nn^3pi^3-frac64n^3pi^3+frac32npi$
My teacher told me my $b_n$ is wrong and it seems that my working also looks like wrong. Can someone please help me to recalculate my $b_n$ please. Just help me once only guys.If I get this right I would able to get full marks.I would be very very grateful to you guys if you guys able to help me. If can suggest me any ideas or simpler way to solve it if you can.Thanks
calculus real-analysis fourier-series
edited Sep 9 '12 at 8:20
Davide Giraudo
128k17156268
asked Sep 9 '12 at 6:31
DavidDavid
8019
$endgroup$
Question:
It is known that $f(x)=(x−4)^2$ for all $xin [0,4]$.
Compute the half range sine series expansion for $f(x)$.
My answer :
Half range series: $p=8$, $l=4$, $a_0=a_n=0$.
$$b_n=frac2Lint_0^Lf(x)sinleft(fracnpi xLright)d(x)=frac24int_0^4(x-4)^2sinleft(fracnpi x4right)d(x)$$
Partial Differentiation
Let $u=(x-4)^2$; $du=2(x-4)dx$; $v=frac-4npicos(fracnpi x4)$
beginalign
b_n&=frac12[frac-4npicos(fracnpi x4)(x-4)^2+frac8npiint(x-4)cos(fracnpi x4)d(x)]|^4_0\
&=frac12[frac-4npicos(fracnpi x4)(x-4)^2+frac8npi[frac4npisin(fracnpi x4)(x-4)-frac4npiintsin(fracnpi x4)d(x)]]|^4_0\
&=frac12[frac-4npicos(fracnpi x4)(x-4)^2$+$frac8npi[frac4npisin(fracnpi x4)(x-4)-frac4npi(frac-4npicosfracnpi x4)]|^4_0\
&=frac12[frac-4npicos(fracnpi x4)(x-4)^2$+$frac8npi(frac16n^2pi^2cosfracnpi x4)]|^4_0
endalign
we know $cosnpi=(-1)^n$
$frac12[(0+frac128(-1)^nn^3pi^3)-(-frac64npi+frac128n^3pi^3)$
$b_n=frac64(-1)^nn^3pi^3-frac64n^3pi^3+frac32npi$
My teacher told me my $b_n$ is wrong and it seems that my working also looks like wrong. Can someone please help me to recalculate my $b_n$ please. Just help me once only guys.If I get this right I would able to get full marks.I would be very very grateful to you guys if you guys able to help me. If can suggest me any ideas or simpler way to solve it if you can.Thanks
calculus real-analysis fourier-series
calculus real-analysis fourier-series
edited Sep 9 '12 at 8:20
Davide Giraudo
128k17156268
asked Sep 9 '12 at 6:31
DavidDavid
8019
edited Sep 9 '12 at 8:20
Davide Giraudo
128k17156268
asked Sep 9 '12 at 6:31
DavidDavid
8019
edited Sep 9 '12 at 8:20
Davide Giraudo
128k17156268
edited Sep 9 '12 at 8:20
Davide Giraudo
128k17156268
edited Sep 9 '12 at 8:20
Davide Giraudo
128k17156268
128k17156268
asked Sep 9 '12 at 6:31
DavidDavid
8019
asked Sep 9 '12 at 6:31
DavidDavid
8019
asked Sep 9 '12 at 6:31
DavidDavid
8019
8019
$begingroup$
Should $b_n$ be the integral from $-L$ to $L$ or the integral from $0$ to $L$ because it switches from the first line to the second line without any multiplicative factor.
$endgroup$
– Euler....IS_ALIVE
Sep 9 '12 at 6:40
$begingroup$
O to L. It alwys like that for Half Range Sine Series
$endgroup$
– David
Sep 9 '12 at 6:48
$begingroup$
But you wrote it as $-L$ to $L$ in your first definition of $b_n$....
$endgroup$
– Euler....IS_ALIVE
Sep 9 '12 at 6:49
$begingroup$
this is the link for reference intmath.com/fourier-series/4-fourier-half-range-functions.php
$endgroup$
– David
Sep 9 '12 at 6:49
$begingroup$
I edited already
$endgroup$
– David
Sep 9 '12 at 6:50
|
show 2 more comments
$begingroup$
Should $b_n$ be the integral from $-L$ to $L$ or the integral from $0$ to $L$ because it switches from the first line to the second line without any multiplicative factor.
$endgroup$
– Euler....IS_ALIVE
Sep 9 '12 at 6:40
$begingroup$
O to L. It alwys like that for Half Range Sine Series
$endgroup$
– David
Sep 9 '12 at 6:48
$begingroup$
But you wrote it as $-L$ to $L$ in your first definition of $b_n$....
$endgroup$
– Euler....IS_ALIVE
Sep 9 '12 at 6:49
$begingroup$
this is the link for reference intmath.com/fourier-series/4-fourier-half-range-functions.php
$endgroup$
– David
Sep 9 '12 at 6:49
$begingroup$
I edited already
$endgroup$
– David
Sep 9 '12 at 6:50
$begingroup$
Should $b_n$ be the integral from $-L$ to $L$ or the integral from $0$ to $L$ because it switches from the first line to the second line without any multiplicative factor.
$endgroup$
– Euler....IS_ALIVE
Sep 9 '12 at 6:40
$begingroup$
Should $b_n$ be the integral from $-L$ to $L$ or the integral from $0$ to $L$ because it switches from the first line to the second line without any multiplicative factor.
$endgroup$
– Euler....IS_ALIVE
Sep 9 '12 at 6:40
$begingroup$
O to L. It alwys like that for Half Range Sine Series
$endgroup$
– David
Sep 9 '12 at 6:48
$begingroup$
O to L. It alwys like that for Half Range Sine Series
$endgroup$
– David
Sep 9 '12 at 6:48
$begingroup$
But you wrote it as $-L$ to $L$ in your first definition of $b_n$....
$endgroup$
– Euler....IS_ALIVE
Sep 9 '12 at 6:49
$begingroup$
But you wrote it as $-L$ to $L$ in your first definition of $b_n$....
$endgroup$
– Euler....IS_ALIVE
Sep 9 '12 at 6:49
$begingroup$
this is the link for reference intmath.com/fourier-series/4-fourier-half-range-functions.php
$endgroup$
– David
Sep 9 '12 at 6:49
$begingroup$
this is the link for reference intmath.com/fourier-series/4-fourier-half-range-functions.php
$endgroup$
– David
Sep 9 '12 at 6:49
$begingroup$
I edited already
$endgroup$
– David
Sep 9 '12 at 6:50
$begingroup$
I edited already
$endgroup$
– David
Sep 9 '12 at 6:50
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The integral defining your $b_n$ is correct, but the final answer is wrong. It differs from mine by $64over npi$.
Here's my work from Mathematica:
Here's graphical verification that we are indeed computing the series correctly (I took the 5th, 10th, and 50th partial sum, respectively):
answered Dec 20 '12 at 7:00
JohnDJohnD
12k32158
$endgroup$
$begingroup$
Hi JohnD. I check out my lecturer he said mine is correct now. Maybe yours is correct too but in different form I think
$endgroup$
– David
May 29 '13 at 5:45
$begingroup$
Yes, they are algebraically equivalent.
$endgroup$
– JohnD
May 29 '13 at 15:06
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
$begingroup$
The integral defining your $b_n$ is correct, but the final answer is wrong. It differs from mine by $64over npi$.
Here's my work from Mathematica:
Here's graphical verification that we are indeed computing the series correctly (I took the 5th, 10th, and 50th partial sum, respectively):
answered Dec 20 '12 at 7:00
JohnDJohnD
12k32158
$endgroup$
$begingroup$
Hi JohnD. I check out my lecturer he said mine is correct now. Maybe yours is correct too but in different form I think
$endgroup$
– David
May 29 '13 at 5:45
$begingroup$
Yes, they are algebraically equivalent.
$endgroup$
– JohnD
May 29 '13 at 15:06
add a comment |
$begingroup$
The integral defining your $b_n$ is correct, but the final answer is wrong. It differs from mine by $64over npi$.
Here's my work from Mathematica:
Here's graphical verification that we are indeed computing the series correctly (I took the 5th, 10th, and 50th partial sum, respectively):
answered Dec 20 '12 at 7:00
JohnDJohnD
12k32158
$endgroup$
$begingroup$
Hi JohnD. I check out my lecturer he said mine is correct now. Maybe yours is correct too but in different form I think
$endgroup$
– David
May 29 '13 at 5:45
$begingroup$
Yes, they are algebraically equivalent.
$endgroup$
– JohnD
May 29 '13 at 15:06
add a comment |
$begingroup$
The integral defining your $b_n$ is correct, but the final answer is wrong. It differs from mine by $64over npi$.
Here's my work from Mathematica:
Here's graphical verification that we are indeed computing the series correctly (I took the 5th, 10th, and 50th partial sum, respectively):
answered Dec 20 '12 at 7:00
JohnDJohnD
12k32158
$endgroup$
The integral defining your $b_n$ is correct, but the final answer is wrong. It differs from mine by $64over npi$.
Here's my work from Mathematica:
Here's graphical verification that we are indeed computing the series correctly (I took the 5th, 10th, and 50th partial sum, respectively):
answered Dec 20 '12 at 7:00
JohnDJohnD
12k32158
answered Dec 20 '12 at 7:00
JohnDJohnD
12k32158
answered Dec 20 '12 at 7:00
JohnDJohnD
12k32158
answered Dec 20 '12 at 7:00
JohnDJohnD
12k32158
12k32158
$begingroup$
Hi JohnD. I check out my lecturer he said mine is correct now. Maybe yours is correct too but in different form I think
$endgroup$
– David
May 29 '13 at 5:45
$begingroup$
Yes, they are algebraically equivalent.
$endgroup$
– JohnD
May 29 '13 at 15:06
add a comment |
$begingroup$
Hi JohnD. I check out my lecturer he said mine is correct now. Maybe yours is correct too but in different form I think
$endgroup$
– David
May 29 '13 at 5:45
$begingroup$
Yes, they are algebraically equivalent.
$endgroup$
– JohnD
May 29 '13 at 15:06
$begingroup$
Hi JohnD. I check out my lecturer he said mine is correct now. Maybe yours is correct too but in different form I think
$endgroup$
– David
May 29 '13 at 5:45
$begingroup$
Hi JohnD. I check out my lecturer he said mine is correct now. Maybe yours is correct too but in different form I think
$endgroup$
– David
May 29 '13 at 5:45
$begingroup$
Yes, they are algebraically equivalent.
$endgroup$
– JohnD
May 29 '13 at 15:06
$begingroup$
Yes, they are algebraically equivalent.
$endgroup$
– JohnD
May 29 '13 at 15:06
add a comment |
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$begingroup$
Should $b_n$ be the integral from $-L$ to $L$ or the integral from $0$ to $L$ because it switches from the first line to the second line without any multiplicative factor.
$endgroup$
– Euler....IS_ALIVE
Sep 9 '12 at 6:40
$begingroup$
O to L. It alwys like that for Half Range Sine Series
$endgroup$
– David
Sep 9 '12 at 6:48
$begingroup$
But you wrote it as $-L$ to $L$ in your first definition of $b_n$....
$endgroup$
– Euler....IS_ALIVE
Sep 9 '12 at 6:49
$begingroup$
this is the link for reference intmath.com/fourier-series/4-fourier-half-range-functions.php
$endgroup$
– David
Sep 9 '12 at 6:49
$begingroup$
I edited already
$endgroup$
– David
Sep 9 '12 at 6:50