Evaluate the limit for a function defined on [0,1]
$begingroup$
The limit is a Riemann sum
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^5}+frac{i}{n^2} right)$$
$delta x=frac{1}{n}$, so I distribute it to the terms to get
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^6}+frac{i}{n^3} right)$$
Now that they have similar denominators I multiply $frac{i}{n^3}cdot(n^3)$ to get
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^6}+frac{2i}{n^6} right)$$
Combining the terms I get $$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{2i^4}{n^6}right)$$
Knowing that $i^2=frac{n(n+1)(2n+1)}{6}$ I split up the $i^4$ into$$lim_{nrightarrowinfty}frac{2}{n^6}sum_{i=1}^nleft(frac{n(n+1)(2n+1)}{6}right)+ left(frac{n(n+1)(2n+1)}{6}right)$$
Am I on the right track?
calculus riemann-sum
asked Dec 10 '18 at 23:16
Eric BrownEric Brown
737
$endgroup$
add a comment |
$begingroup$
The limit is a Riemann sum
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^5}+frac{i}{n^2} right)$$
$delta x=frac{1}{n}$, so I distribute it to the terms to get
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^6}+frac{i}{n^3} right)$$
Now that they have similar denominators I multiply $frac{i}{n^3}cdot(n^3)$ to get
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^6}+frac{2i}{n^6} right)$$
Combining the terms I get $$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{2i^4}{n^6}right)$$
Knowing that $i^2=frac{n(n+1)(2n+1)}{6}$ I split up the $i^4$ into$$lim_{nrightarrowinfty}frac{2}{n^6}sum_{i=1}^nleft(frac{n(n+1)(2n+1)}{6}right)+ left(frac{n(n+1)(2n+1)}{6}right)$$
Am I on the right track?
calculus riemann-sum
asked Dec 10 '18 at 23:16
Eric BrownEric Brown
737
$endgroup$
add a comment |
$begingroup$
The limit is a Riemann sum
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^5}+frac{i}{n^2} right)$$
$delta x=frac{1}{n}$, so I distribute it to the terms to get
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^6}+frac{i}{n^3} right)$$
Now that they have similar denominators I multiply $frac{i}{n^3}cdot(n^3)$ to get
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^6}+frac{2i}{n^6} right)$$
Combining the terms I get $$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{2i^4}{n^6}right)$$
Knowing that $i^2=frac{n(n+1)(2n+1)}{6}$ I split up the $i^4$ into$$lim_{nrightarrowinfty}frac{2}{n^6}sum_{i=1}^nleft(frac{n(n+1)(2n+1)}{6}right)+ left(frac{n(n+1)(2n+1)}{6}right)$$
Am I on the right track?
calculus riemann-sum
asked Dec 10 '18 at 23:16
Eric BrownEric Brown
737
$endgroup$
The limit is a Riemann sum
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^5}+frac{i}{n^2} right)$$
$delta x=frac{1}{n}$, so I distribute it to the terms to get
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^6}+frac{i}{n^3} right)$$
Now that they have similar denominators I multiply $frac{i}{n^3}cdot(n^3)$ to get
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^6}+frac{2i}{n^6} right)$$
Combining the terms I get $$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{2i^4}{n^6}right)$$
Knowing that $i^2=frac{n(n+1)(2n+1)}{6}$ I split up the $i^4$ into$$lim_{nrightarrowinfty}frac{2}{n^6}sum_{i=1}^nleft(frac{n(n+1)(2n+1)}{6}right)+ left(frac{n(n+1)(2n+1)}{6}right)$$
Am I on the right track?
calculus riemann-sum
calculus riemann-sum
asked Dec 10 '18 at 23:16
Eric BrownEric Brown
737
asked Dec 10 '18 at 23:16
Eric BrownEric Brown
737
asked Dec 10 '18 at 23:16
Eric BrownEric Brown
737
asked Dec 10 '18 at 23:16
Eric BrownEric Brown
737
asked Dec 10 '18 at 23:16
Eric BrownEric Brown
737
737
add a comment |
add a comment |
1 Answer
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$begingroup$
We have that by Riemann sum
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^5}+frac{i}{n^2} right)=lim_{nrightarrowinfty}frac1nsum_{i=1}^nleft(frac{i^4}{n^4}+frac{i}{n} right)=int_0^1(x^4+x) dx$$
As an alternative refer to Faulhaber's formula.
answered Dec 10 '18 at 23:18
gimusigimusi
92.8k84494
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We have that by Riemann sum
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^5}+frac{i}{n^2} right)=lim_{nrightarrowinfty}frac1nsum_{i=1}^nleft(frac{i^4}{n^4}+frac{i}{n} right)=int_0^1(x^4+x) dx$$
As an alternative refer to Faulhaber's formula.
answered Dec 10 '18 at 23:18
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
We have that by Riemann sum
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^5}+frac{i}{n^2} right)=lim_{nrightarrowinfty}frac1nsum_{i=1}^nleft(frac{i^4}{n^4}+frac{i}{n} right)=int_0^1(x^4+x) dx$$
As an alternative refer to Faulhaber's formula.
answered Dec 10 '18 at 23:18
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
We have that by Riemann sum
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^5}+frac{i}{n^2} right)=lim_{nrightarrowinfty}frac1nsum_{i=1}^nleft(frac{i^4}{n^4}+frac{i}{n} right)=int_0^1(x^4+x) dx$$
As an alternative refer to Faulhaber's formula.
answered Dec 10 '18 at 23:18
gimusigimusi
92.8k84494
$endgroup$
We have that by Riemann sum
$$lim_{nrightarrowinfty}sum_{i=1}^nleft(frac{i^4}{n^5}+frac{i}{n^2} right)=lim_{nrightarrowinfty}frac1nsum_{i=1}^nleft(frac{i^4}{n^4}+frac{i}{n} right)=int_0^1(x^4+x) dx$$
As an alternative refer to Faulhaber's formula.
answered Dec 10 '18 at 23:18
gimusigimusi
92.8k84494
answered Dec 10 '18 at 23:18
gimusigimusi
92.8k84494
answered Dec 10 '18 at 23:18
gimusigimusi
92.8k84494
answered Dec 10 '18 at 23:18
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
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