Expressing Riemann sums as integrals
$begingroup$
$$L_2=lim_{n→∞}sum_{k=1}^nfrac{(k-cos^2(k))^4}{n^5}.$$
My teacher said that when brackets at the numarator is expanded the limit of sum except $dfrac{k^4}{n^5}$ equals $0$, so the Riemann sum becomes $limlimits_{n→∞}sumlimits_{k=1}^ndfrac{k^4}{n^5}$. I don't understand this point. Please explain this point. Sorry for my bad English.
limits definite-integrals
edited Dec 21 '18 at 8:11
Saad
19.7k92352
asked Dec 21 '18 at 7:38
1u2i3o41u2i3o4
182
$endgroup$
add a comment |
$begingroup$
$$L_2=lim_{n→∞}sum_{k=1}^nfrac{(k-cos^2(k))^4}{n^5}.$$
My teacher said that when brackets at the numarator is expanded the limit of sum except $dfrac{k^4}{n^5}$ equals $0$, so the Riemann sum becomes $limlimits_{n→∞}sumlimits_{k=1}^ndfrac{k^4}{n^5}$. I don't understand this point. Please explain this point. Sorry for my bad English.
limits definite-integrals
edited Dec 21 '18 at 8:11
Saad
19.7k92352
asked Dec 21 '18 at 7:38
1u2i3o41u2i3o4
182
$endgroup$
add a comment |
$begingroup$
$$L_2=lim_{n→∞}sum_{k=1}^nfrac{(k-cos^2(k))^4}{n^5}.$$
My teacher said that when brackets at the numarator is expanded the limit of sum except $dfrac{k^4}{n^5}$ equals $0$, so the Riemann sum becomes $limlimits_{n→∞}sumlimits_{k=1}^ndfrac{k^4}{n^5}$. I don't understand this point. Please explain this point. Sorry for my bad English.
limits definite-integrals
edited Dec 21 '18 at 8:11
Saad
19.7k92352
asked Dec 21 '18 at 7:38
1u2i3o41u2i3o4
182
$endgroup$
$$L_2=lim_{n→∞}sum_{k=1}^nfrac{(k-cos^2(k))^4}{n^5}.$$
My teacher said that when brackets at the numarator is expanded the limit of sum except $dfrac{k^4}{n^5}$ equals $0$, so the Riemann sum becomes $limlimits_{n→∞}sumlimits_{k=1}^ndfrac{k^4}{n^5}$. I don't understand this point. Please explain this point. Sorry for my bad English.
limits definite-integrals
limits definite-integrals
edited Dec 21 '18 at 8:11
Saad
19.7k92352
asked Dec 21 '18 at 7:38
1u2i3o41u2i3o4
182
edited Dec 21 '18 at 8:11
Saad
19.7k92352
asked Dec 21 '18 at 7:38
1u2i3o41u2i3o4
182
edited Dec 21 '18 at 8:11
Saad
19.7k92352
edited Dec 21 '18 at 8:11
Saad
19.7k92352
edited Dec 21 '18 at 8:11
Saad
19.7k92352
19.7k92352
asked Dec 21 '18 at 7:38
1u2i3o41u2i3o4
182
asked Dec 21 '18 at 7:38
1u2i3o41u2i3o4
182
asked Dec 21 '18 at 7:38
1u2i3o41u2i3o4
182
182
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$sum_{k=1}^{n} k^{j} leq int_1^{n} x^{j} dx =frac {n^{j+1} -1} {j+1}$ and $frac {n^{j+1} -1} {n^{5}} to 0$ as $ n to infty$ if $j <4$. Hence, when you expand $(k-cos^{2}(k))^{4}$ all terms except $k^{4}$ give limit $0$. [Note that $cos^{2} k leq 1$].
answered Dec 21 '18 at 7:50
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
$begingroup$
$$sumfrac{(k-cos^2 k)^4}{n^5}=sum left(frac{k^4-4k^3cos ^2left(kright)+6k^2cos ^4left(kright)-4kcos ^6left(kright)+cos ^8left(kright)}{n^5}right)$$Every term is involving $cos k$ except the term $k^4$. Applying summation and use algebra of limits together with the fact $cos k$ is bounded by $1$ to get your result
answered Dec 21 '18 at 7:58
Chinnapparaj RChinnapparaj R
5,5072928
$endgroup$
add a comment |
$begingroup$
Since
$$
sum_{k=1}^nk^3=frac{n^2(n+1)^2}4
$$
we get
$$
begin{align}
lim_{ntoinfty}sum_{k=1}^nfrac{left(k-cos^2(k)right)^4}{n^5}
&=lim_{ntoinfty}sum_{k=1}^nleft[frac{k^4}{n^5}+O!left(frac{k^3}{n^5}right)right]\
&=lim_{ntoinfty}left[sum_{k=1}^nfrac{k^4}{n^4}frac1n+O!left(frac1nright)right]\
&=int_0^1x^4,mathrm{d}x+0\[6pt]
&=frac15
end{align}
$$
answered Dec 21 '18 at 9:15
robjohn♦robjohn
268k27308633
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
$sum_{k=1}^{n} k^{j} leq int_1^{n} x^{j} dx =frac {n^{j+1} -1} {j+1}$ and $frac {n^{j+1} -1} {n^{5}} to 0$ as $ n to infty$ if $j <4$. Hence, when you expand $(k-cos^{2}(k))^{4}$ all terms except $k^{4}$ give limit $0$. [Note that $cos^{2} k leq 1$].
answered Dec 21 '18 at 7:50
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
$begingroup$
$sum_{k=1}^{n} k^{j} leq int_1^{n} x^{j} dx =frac {n^{j+1} -1} {j+1}$ and $frac {n^{j+1} -1} {n^{5}} to 0$ as $ n to infty$ if $j <4$. Hence, when you expand $(k-cos^{2}(k))^{4}$ all terms except $k^{4}$ give limit $0$. [Note that $cos^{2} k leq 1$].
answered Dec 21 '18 at 7:50
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
$begingroup$
$sum_{k=1}^{n} k^{j} leq int_1^{n} x^{j} dx =frac {n^{j+1} -1} {j+1}$ and $frac {n^{j+1} -1} {n^{5}} to 0$ as $ n to infty$ if $j <4$. Hence, when you expand $(k-cos^{2}(k))^{4}$ all terms except $k^{4}$ give limit $0$. [Note that $cos^{2} k leq 1$].
answered Dec 21 '18 at 7:50
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
$sum_{k=1}^{n} k^{j} leq int_1^{n} x^{j} dx =frac {n^{j+1} -1} {j+1}$ and $frac {n^{j+1} -1} {n^{5}} to 0$ as $ n to infty$ if $j <4$. Hence, when you expand $(k-cos^{2}(k))^{4}$ all terms except $k^{4}$ give limit $0$. [Note that $cos^{2} k leq 1$].
answered Dec 21 '18 at 7:50
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 21 '18 at 7:50
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 21 '18 at 7:50
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 21 '18 at 7:50
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
60.6k42161
add a comment |
add a comment |
$begingroup$
$$sumfrac{(k-cos^2 k)^4}{n^5}=sum left(frac{k^4-4k^3cos ^2left(kright)+6k^2cos ^4left(kright)-4kcos ^6left(kright)+cos ^8left(kright)}{n^5}right)$$Every term is involving $cos k$ except the term $k^4$. Applying summation and use algebra of limits together with the fact $cos k$ is bounded by $1$ to get your result
answered Dec 21 '18 at 7:58
Chinnapparaj RChinnapparaj R
5,5072928
$endgroup$
add a comment |
$begingroup$
$$sumfrac{(k-cos^2 k)^4}{n^5}=sum left(frac{k^4-4k^3cos ^2left(kright)+6k^2cos ^4left(kright)-4kcos ^6left(kright)+cos ^8left(kright)}{n^5}right)$$Every term is involving $cos k$ except the term $k^4$. Applying summation and use algebra of limits together with the fact $cos k$ is bounded by $1$ to get your result
answered Dec 21 '18 at 7:58
Chinnapparaj RChinnapparaj R
5,5072928
$endgroup$
add a comment |
$begingroup$
$$sumfrac{(k-cos^2 k)^4}{n^5}=sum left(frac{k^4-4k^3cos ^2left(kright)+6k^2cos ^4left(kright)-4kcos ^6left(kright)+cos ^8left(kright)}{n^5}right)$$Every term is involving $cos k$ except the term $k^4$. Applying summation and use algebra of limits together with the fact $cos k$ is bounded by $1$ to get your result
answered Dec 21 '18 at 7:58
Chinnapparaj RChinnapparaj R
5,5072928
$endgroup$
$$sumfrac{(k-cos^2 k)^4}{n^5}=sum left(frac{k^4-4k^3cos ^2left(kright)+6k^2cos ^4left(kright)-4kcos ^6left(kright)+cos ^8left(kright)}{n^5}right)$$Every term is involving $cos k$ except the term $k^4$. Applying summation and use algebra of limits together with the fact $cos k$ is bounded by $1$ to get your result
answered Dec 21 '18 at 7:58
Chinnapparaj RChinnapparaj R
5,5072928
answered Dec 21 '18 at 7:58
Chinnapparaj RChinnapparaj R
5,5072928
answered Dec 21 '18 at 7:58
Chinnapparaj RChinnapparaj R
5,5072928
answered Dec 21 '18 at 7:58
Chinnapparaj RChinnapparaj R
5,5072928
5,5072928
add a comment |
add a comment |
$begingroup$
Since
$$
sum_{k=1}^nk^3=frac{n^2(n+1)^2}4
$$
we get
$$
begin{align}
lim_{ntoinfty}sum_{k=1}^nfrac{left(k-cos^2(k)right)^4}{n^5}
&=lim_{ntoinfty}sum_{k=1}^nleft[frac{k^4}{n^5}+O!left(frac{k^3}{n^5}right)right]\
&=lim_{ntoinfty}left[sum_{k=1}^nfrac{k^4}{n^4}frac1n+O!left(frac1nright)right]\
&=int_0^1x^4,mathrm{d}x+0\[6pt]
&=frac15
end{align}
$$
answered Dec 21 '18 at 9:15
robjohn♦robjohn
268k27308633
$endgroup$
add a comment |
$begingroup$
Since
$$
sum_{k=1}^nk^3=frac{n^2(n+1)^2}4
$$
we get
$$
begin{align}
lim_{ntoinfty}sum_{k=1}^nfrac{left(k-cos^2(k)right)^4}{n^5}
&=lim_{ntoinfty}sum_{k=1}^nleft[frac{k^4}{n^5}+O!left(frac{k^3}{n^5}right)right]\
&=lim_{ntoinfty}left[sum_{k=1}^nfrac{k^4}{n^4}frac1n+O!left(frac1nright)right]\
&=int_0^1x^4,mathrm{d}x+0\[6pt]
&=frac15
end{align}
$$
answered Dec 21 '18 at 9:15
robjohn♦robjohn
268k27308633
$endgroup$
add a comment |
$begingroup$
Since
$$
sum_{k=1}^nk^3=frac{n^2(n+1)^2}4
$$
we get
$$
begin{align}
lim_{ntoinfty}sum_{k=1}^nfrac{left(k-cos^2(k)right)^4}{n^5}
&=lim_{ntoinfty}sum_{k=1}^nleft[frac{k^4}{n^5}+O!left(frac{k^3}{n^5}right)right]\
&=lim_{ntoinfty}left[sum_{k=1}^nfrac{k^4}{n^4}frac1n+O!left(frac1nright)right]\
&=int_0^1x^4,mathrm{d}x+0\[6pt]
&=frac15
end{align}
$$
answered Dec 21 '18 at 9:15
robjohn♦robjohn
268k27308633
$endgroup$
Since
$$
sum_{k=1}^nk^3=frac{n^2(n+1)^2}4
$$
we get
$$
begin{align}
lim_{ntoinfty}sum_{k=1}^nfrac{left(k-cos^2(k)right)^4}{n^5}
&=lim_{ntoinfty}sum_{k=1}^nleft[frac{k^4}{n^5}+O!left(frac{k^3}{n^5}right)right]\
&=lim_{ntoinfty}left[sum_{k=1}^nfrac{k^4}{n^4}frac1n+O!left(frac1nright)right]\
&=int_0^1x^4,mathrm{d}x+0\[6pt]
&=frac15
end{align}
$$
answered Dec 21 '18 at 9:15
robjohn♦robjohn
268k27308633
answered Dec 21 '18 at 9:15
robjohn♦robjohn
268k27308633
answered Dec 21 '18 at 9:15
robjohn♦robjohn
268k27308633
answered Dec 21 '18 at 9:15
robjohn♦robjohn
268k27308633
268k27308633
add a comment |
add a comment |
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