Mixing
Evaluate the Riemann sum
$begingroup$
If
$mathrm{f}left(xright) = 2cosleft(xright)$
$ 0 leq x leq 3pi/4$
evaluate the Riemann sum with
$n = 6$,
taking the sample points to be left endpoints. ( Round your answer to six decimal places ).
I've tried this several times now by using
$delta x = pi/8$ and doing:
$$
frac{pi}{8}left[2cosleft(0right) +
2cosleft(frac{pi}{8}right) +
2cosleft(frac{2pi}{8}right) +cdots + 2cosleft(frac{5pi}{8}right)right]
$$
and the answer I keep coming up with is $4.711374$ but my online homework is telling me this is wrong. Can anybody help ?.
Edit: I tried subtracting a $pi/8$ from each value inside the cosine functions and got $3.926437$ can anyone tell me if this is correct ?.
integration definite-integrals riemann-sum
edited Jan 29 at 6:04
Felix Marin
68.9k7109146
asked Nov 8 '16 at 0:48
MaggieMaggie
90210
$endgroup$
add a comment |
$begingroup$
If
$mathrm{f}left(xright) = 2cosleft(xright)$
$ 0 leq x leq 3pi/4$
evaluate the Riemann sum with
$n = 6$,
taking the sample points to be left endpoints. ( Round your answer to six decimal places ).
I've tried this several times now by using
$delta x = pi/8$ and doing:
$$
frac{pi}{8}left[2cosleft(0right) +
2cosleft(frac{pi}{8}right) +
2cosleft(frac{2pi}{8}right) +cdots + 2cosleft(frac{5pi}{8}right)right]
$$
and the answer I keep coming up with is $4.711374$ but my online homework is telling me this is wrong. Can anybody help ?.
Edit: I tried subtracting a $pi/8$ from each value inside the cosine functions and got $3.926437$ can anyone tell me if this is correct ?.
integration definite-integrals riemann-sum
edited Jan 29 at 6:04
Felix Marin
68.9k7109146
asked Nov 8 '16 at 0:48
MaggieMaggie
90210
$endgroup$
2
$begingroup$
Use $cos{(k pi/8)} = operatorname{Re}{e^{i k pi/8}}$ and add up the resulting geometric series.
$endgroup$
– Ron Gordon
Nov 8 '16 at 0:50
$begingroup$
Be sure that your calculator is in radian mode.
$endgroup$
– Matthew Conroy
Nov 8 '16 at 0:59
$begingroup$
Your set-up looks fine. You calculation is off. Right sum $2.06$, Left sum $0.72,$ Trapezoid rule $1.40$ Midpoint rule $1.42.$ Also note that $cos (frac {pi}{2}-x) = - cos (frac {pi}{2}+x)$ So, there is a little bit less calculating necessary than you show.
$endgroup$
– Doug M
Nov 8 '16 at 1:16
$begingroup$
I notice I have my left sum and right sum's backward.
$endgroup$
– Doug M
Nov 8 '16 at 1:36
add a comment |
$begingroup$
If
$mathrm{f}left(xright) = 2cosleft(xright)$
$ 0 leq x leq 3pi/4$
evaluate the Riemann sum with
$n = 6$,
taking the sample points to be left endpoints. ( Round your answer to six decimal places ).
I've tried this several times now by using
$delta x = pi/8$ and doing:
$$
frac{pi}{8}left[2cosleft(0right) +
2cosleft(frac{pi}{8}right) +
2cosleft(frac{2pi}{8}right) +cdots + 2cosleft(frac{5pi}{8}right)right]
$$
and the answer I keep coming up with is $4.711374$ but my online homework is telling me this is wrong. Can anybody help ?.
Edit: I tried subtracting a $pi/8$ from each value inside the cosine functions and got $3.926437$ can anyone tell me if this is correct ?.
integration definite-integrals riemann-sum
edited Jan 29 at 6:04
Felix Marin
68.9k7109146
asked Nov 8 '16 at 0:48
MaggieMaggie
90210
$endgroup$
If
$mathrm{f}left(xright) = 2cosleft(xright)$
$ 0 leq x leq 3pi/4$
evaluate the Riemann sum with
$n = 6$,
taking the sample points to be left endpoints. ( Round your answer to six decimal places ).
I've tried this several times now by using
$delta x = pi/8$ and doing:
$$
frac{pi}{8}left[2cosleft(0right) +
2cosleft(frac{pi}{8}right) +
2cosleft(frac{2pi}{8}right) +cdots + 2cosleft(frac{5pi}{8}right)right]
$$
and the answer I keep coming up with is $4.711374$ but my online homework is telling me this is wrong. Can anybody help ?.
Edit: I tried subtracting a $pi/8$ from each value inside the cosine functions and got $3.926437$ can anyone tell me if this is correct ?.
integration definite-integrals riemann-sum
integration definite-integrals riemann-sum
edited Jan 29 at 6:04
Felix Marin
68.9k7109146
asked Nov 8 '16 at 0:48
MaggieMaggie
90210
edited Jan 29 at 6:04
Felix Marin
68.9k7109146
asked Nov 8 '16 at 0:48
MaggieMaggie
90210
edited Jan 29 at 6:04
Felix Marin
68.9k7109146
edited Jan 29 at 6:04
Felix Marin
68.9k7109146
edited Jan 29 at 6:04
Felix Marin
68.9k7109146
68.9k7109146
asked Nov 8 '16 at 0:48
MaggieMaggie
90210
asked Nov 8 '16 at 0:48
MaggieMaggie
90210
asked Nov 8 '16 at 0:48
MaggieMaggie
90210
90210
2
$begingroup$
Use $cos{(k pi/8)} = operatorname{Re}{e^{i k pi/8}}$ and add up the resulting geometric series.
$endgroup$
– Ron Gordon
Nov 8 '16 at 0:50
$begingroup$
Be sure that your calculator is in radian mode.
$endgroup$
– Matthew Conroy
Nov 8 '16 at 0:59
$begingroup$
Your set-up looks fine. You calculation is off. Right sum $2.06$, Left sum $0.72,$ Trapezoid rule $1.40$ Midpoint rule $1.42.$ Also note that $cos (frac {pi}{2}-x) = - cos (frac {pi}{2}+x)$ So, there is a little bit less calculating necessary than you show.
$endgroup$
– Doug M
Nov 8 '16 at 1:16
$begingroup$
I notice I have my left sum and right sum's backward.
$endgroup$
– Doug M
Nov 8 '16 at 1:36
add a comment |
2
$begingroup$
Use $cos{(k pi/8)} = operatorname{Re}{e^{i k pi/8}}$ and add up the resulting geometric series.
$endgroup$
– Ron Gordon
Nov 8 '16 at 0:50
$begingroup$
Be sure that your calculator is in radian mode.
$endgroup$
– Matthew Conroy
Nov 8 '16 at 0:59
$begingroup$
Your set-up looks fine. You calculation is off. Right sum $2.06$, Left sum $0.72,$ Trapezoid rule $1.40$ Midpoint rule $1.42.$ Also note that $cos (frac {pi}{2}-x) = - cos (frac {pi}{2}+x)$ So, there is a little bit less calculating necessary than you show.
$endgroup$
– Doug M
Nov 8 '16 at 1:16
$begingroup$
I notice I have my left sum and right sum's backward.
$endgroup$
– Doug M
Nov 8 '16 at 1:36
2
2
$begingroup$
Use $cos{(k pi/8)} = operatorname{Re}{e^{i k pi/8}}$ and add up the resulting geometric series.
$endgroup$
– Ron Gordon
Nov 8 '16 at 0:50
$begingroup$
Use $cos{(k pi/8)} = operatorname{Re}{e^{i k pi/8}}$ and add up the resulting geometric series.
$endgroup$
– Ron Gordon
Nov 8 '16 at 0:50
$begingroup$
Be sure that your calculator is in radian mode.
$endgroup$
– Matthew Conroy
Nov 8 '16 at 0:59
$begingroup$
Be sure that your calculator is in radian mode.
$endgroup$
– Matthew Conroy
Nov 8 '16 at 0:59
$begingroup$
Your set-up looks fine. You calculation is off. Right sum $2.06$, Left sum $0.72,$ Trapezoid rule $1.40$ Midpoint rule $1.42.$ Also note that $cos (frac {pi}{2}-x) = - cos (frac {pi}{2}+x)$ So, there is a little bit less calculating necessary than you show.
$endgroup$
– Doug M
Nov 8 '16 at 1:16
$begingroup$
Your set-up looks fine. You calculation is off. Right sum $2.06$, Left sum $0.72,$ Trapezoid rule $1.40$ Midpoint rule $1.42.$ Also note that $cos (frac {pi}{2}-x) = - cos (frac {pi}{2}+x)$ So, there is a little bit less calculating necessary than you show.
$endgroup$
– Doug M
Nov 8 '16 at 1:16
$begingroup$
I notice I have my left sum and right sum's backward.
$endgroup$
– Doug M
Nov 8 '16 at 1:36
$begingroup$
I notice I have my left sum and right sum's backward.
$endgroup$
– Doug M
Nov 8 '16 at 1:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your setup is right.
You might make use of the fact that $cos(frac{pi}{2})=0$ and that
$$
cos left(frac{5pi}{8} right) = -cos left(frac{3pi}{8} right)
$$
to simplify your sum to
$$
frac{pi}{8} left( 2cos (0) + 2cos left( frac{pi}{8} right)
+2 cos left(frac{2pi}{8} right) right)
$$
so you only need to sum three things instead of six, and so you'll have fewer opportunities for error!
answered Nov 8 '16 at 4:28
Matthew ConroyMatthew Conroy
10.4k32836
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Your setup is right.
You might make use of the fact that $cos(frac{pi}{2})=0$ and that
$$
cos left(frac{5pi}{8} right) = -cos left(frac{3pi}{8} right)
$$
to simplify your sum to
$$
frac{pi}{8} left( 2cos (0) + 2cos left( frac{pi}{8} right)
+2 cos left(frac{2pi}{8} right) right)
$$
so you only need to sum three things instead of six, and so you'll have fewer opportunities for error!
answered Nov 8 '16 at 4:28
Matthew ConroyMatthew Conroy
10.4k32836
$endgroup$
add a comment |
$begingroup$
Your setup is right.
You might make use of the fact that $cos(frac{pi}{2})=0$ and that
$$
cos left(frac{5pi}{8} right) = -cos left(frac{3pi}{8} right)
$$
to simplify your sum to
$$
frac{pi}{8} left( 2cos (0) + 2cos left( frac{pi}{8} right)
+2 cos left(frac{2pi}{8} right) right)
$$
so you only need to sum three things instead of six, and so you'll have fewer opportunities for error!
answered Nov 8 '16 at 4:28
Matthew ConroyMatthew Conroy
10.4k32836
$endgroup$
add a comment |
$begingroup$
Your setup is right.
You might make use of the fact that $cos(frac{pi}{2})=0$ and that
$$
cos left(frac{5pi}{8} right) = -cos left(frac{3pi}{8} right)
$$
to simplify your sum to
$$
frac{pi}{8} left( 2cos (0) + 2cos left( frac{pi}{8} right)
+2 cos left(frac{2pi}{8} right) right)
$$
so you only need to sum three things instead of six, and so you'll have fewer opportunities for error!
answered Nov 8 '16 at 4:28
Matthew ConroyMatthew Conroy
10.4k32836
$endgroup$
Your setup is right.
You might make use of the fact that $cos(frac{pi}{2})=0$ and that
$$
cos left(frac{5pi}{8} right) = -cos left(frac{3pi}{8} right)
$$
to simplify your sum to
$$
frac{pi}{8} left( 2cos (0) + 2cos left( frac{pi}{8} right)
+2 cos left(frac{2pi}{8} right) right)
$$
so you only need to sum three things instead of six, and so you'll have fewer opportunities for error!
answered Nov 8 '16 at 4:28
Matthew ConroyMatthew Conroy
10.4k32836
answered Nov 8 '16 at 4:28
Matthew ConroyMatthew Conroy
10.4k32836
answered Nov 8 '16 at 4:28
Matthew ConroyMatthew Conroy
10.4k32836
answered Nov 8 '16 at 4:28
Matthew ConroyMatthew Conroy
10.4k32836
10.4k32836
add a comment |
add a comment |
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2
$begingroup$
Use $cos{(k pi/8)} = operatorname{Re}{e^{i k pi/8}}$ and add up the resulting geometric series.
$endgroup$
– Ron Gordon
Nov 8 '16 at 0:50
$begingroup$
Be sure that your calculator is in radian mode.
$endgroup$
– Matthew Conroy
Nov 8 '16 at 0:59
$begingroup$
Your set-up looks fine. You calculation is off. Right sum $2.06$, Left sum $0.72,$ Trapezoid rule $1.40$ Midpoint rule $1.42.$ Also note that $cos (frac {pi}{2}-x) = - cos (frac {pi}{2}+x)$ So, there is a little bit less calculating necessary than you show.
$endgroup$
– Doug M
Nov 8 '16 at 1:16
$begingroup$
I notice I have my left sum and right sum's backward.
$endgroup$
– Doug M
Nov 8 '16 at 1:36