Mixing
Integral Method to find upper and lower bounds on the sum
$begingroup$
I got this problem wrong on my exam and I know why I got it wrong but I am not sure how to find the right answer.
This is the question:
Consider$sum_{k=1}^n (3+ 2/k^3)$. Use the integral method to obtain upper and lower bound on the sum. Do not split the summation into $sum_{k=1}^n 3+sum_{k=1}^n ( 2/k^3)$
This is the formula we had to use: Click here to see the approximation integral formula
This function is decreasing and I was able to calculate the lower bound, but I am not sure how to do the upper bound becuase this is what I get $ int_{0}^{n}$(3+2/k^3)dx and after taking the antiderivative. I get 3k - (1/k^2) from 0 to n. so if I take the integral from 0 to n then 1/k^2 becomes undefined. I tried to change the integral from -1 to n but I got it wrong on the exam.
Could someone please tell me how to solve this? Thanks
definite-integrals approximate-integration
asked Apr 2 '17 at 9:35
jenkins342jenkins342
4
$endgroup$
add a comment |
$begingroup$
I got this problem wrong on my exam and I know why I got it wrong but I am not sure how to find the right answer.
This is the question:
Consider$sum_{k=1}^n (3+ 2/k^3)$. Use the integral method to obtain upper and lower bound on the sum. Do not split the summation into $sum_{k=1}^n 3+sum_{k=1}^n ( 2/k^3)$
This is the formula we had to use: Click here to see the approximation integral formula
This function is decreasing and I was able to calculate the lower bound, but I am not sure how to do the upper bound becuase this is what I get $ int_{0}^{n}$(3+2/k^3)dx and after taking the antiderivative. I get 3k - (1/k^2) from 0 to n. so if I take the integral from 0 to n then 1/k^2 becomes undefined. I tried to change the integral from -1 to n but I got it wrong on the exam.
Could someone please tell me how to solve this? Thanks
definite-integrals approximate-integration
asked Apr 2 '17 at 9:35
jenkins342jenkins342
4
$endgroup$
add a comment |
$begingroup$
I got this problem wrong on my exam and I know why I got it wrong but I am not sure how to find the right answer.
This is the question:
Consider$sum_{k=1}^n (3+ 2/k^3)$. Use the integral method to obtain upper and lower bound on the sum. Do not split the summation into $sum_{k=1}^n 3+sum_{k=1}^n ( 2/k^3)$
This is the formula we had to use: Click here to see the approximation integral formula
This function is decreasing and I was able to calculate the lower bound, but I am not sure how to do the upper bound becuase this is what I get $ int_{0}^{n}$(3+2/k^3)dx and after taking the antiderivative. I get 3k - (1/k^2) from 0 to n. so if I take the integral from 0 to n then 1/k^2 becomes undefined. I tried to change the integral from -1 to n but I got it wrong on the exam.
Could someone please tell me how to solve this? Thanks
definite-integrals approximate-integration
asked Apr 2 '17 at 9:35
jenkins342jenkins342
4
$endgroup$
I got this problem wrong on my exam and I know why I got it wrong but I am not sure how to find the right answer.
This is the question:
Consider$sum_{k=1}^n (3+ 2/k^3)$. Use the integral method to obtain upper and lower bound on the sum. Do not split the summation into $sum_{k=1}^n 3+sum_{k=1}^n ( 2/k^3)$
This is the formula we had to use: Click here to see the approximation integral formula
This function is decreasing and I was able to calculate the lower bound, but I am not sure how to do the upper bound becuase this is what I get $ int_{0}^{n}$(3+2/k^3)dx and after taking the antiderivative. I get 3k - (1/k^2) from 0 to n. so if I take the integral from 0 to n then 1/k^2 becomes undefined. I tried to change the integral from -1 to n but I got it wrong on the exam.
Could someone please tell me how to solve this? Thanks
definite-integrals approximate-integration
definite-integrals approximate-integration
asked Apr 2 '17 at 9:35
jenkins342jenkins342
4
asked Apr 2 '17 at 9:35
jenkins342jenkins342
4
asked Apr 2 '17 at 9:35
jenkins342jenkins342
4
asked Apr 2 '17 at 9:35
jenkins342jenkins342
4
asked Apr 2 '17 at 9:35
jenkins342jenkins342
4
4
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint. You may write
$$
begin{align}
sum_{k=1}^nleft(3+frac1{k^3} right)&=left(3+frac1{1^3} right)+sum_{k=color{red}{2}}^nleft(3+frac1{k^3} right)
\\&le left(3+frac1{1^3}right)+int_{color{red}{2-1}}^nleft(3+frac1{x^3} right)dx
end{align}
$$ then getting easily an upper bound.
answered Apr 2 '17 at 9:45
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
$begingroup$
thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
$endgroup$
– jenkins342
Apr 2 '17 at 10:07
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2214219%2fintegral-method-to-find-upper-and-lower-bounds-on-the-sum%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint. You may write
$$
begin{align}
sum_{k=1}^nleft(3+frac1{k^3} right)&=left(3+frac1{1^3} right)+sum_{k=color{red}{2}}^nleft(3+frac1{k^3} right)
\\&le left(3+frac1{1^3}right)+int_{color{red}{2-1}}^nleft(3+frac1{x^3} right)dx
end{align}
$$ then getting easily an upper bound.
answered Apr 2 '17 at 9:45
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
$begingroup$
thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
$endgroup$
– jenkins342
Apr 2 '17 at 10:07
add a comment |
$begingroup$
Hint. You may write
$$
begin{align}
sum_{k=1}^nleft(3+frac1{k^3} right)&=left(3+frac1{1^3} right)+sum_{k=color{red}{2}}^nleft(3+frac1{k^3} right)
\\&le left(3+frac1{1^3}right)+int_{color{red}{2-1}}^nleft(3+frac1{x^3} right)dx
end{align}
$$ then getting easily an upper bound.
answered Apr 2 '17 at 9:45
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
$begingroup$
thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
$endgroup$
– jenkins342
Apr 2 '17 at 10:07
add a comment |
$begingroup$
Hint. You may write
$$
begin{align}
sum_{k=1}^nleft(3+frac1{k^3} right)&=left(3+frac1{1^3} right)+sum_{k=color{red}{2}}^nleft(3+frac1{k^3} right)
\\&le left(3+frac1{1^3}right)+int_{color{red}{2-1}}^nleft(3+frac1{x^3} right)dx
end{align}
$$ then getting easily an upper bound.
answered Apr 2 '17 at 9:45
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
Hint. You may write
$$
begin{align}
sum_{k=1}^nleft(3+frac1{k^3} right)&=left(3+frac1{1^3} right)+sum_{k=color{red}{2}}^nleft(3+frac1{k^3} right)
\\&le left(3+frac1{1^3}right)+int_{color{red}{2-1}}^nleft(3+frac1{x^3} right)dx
end{align}
$$ then getting easily an upper bound.
answered Apr 2 '17 at 9:45
Olivier OloaOlivier Oloa
109k17178294
answered Apr 2 '17 at 9:45
Olivier OloaOlivier Oloa
109k17178294
answered Apr 2 '17 at 9:45
Olivier OloaOlivier Oloa
109k17178294
answered Apr 2 '17 at 9:45
Olivier OloaOlivier Oloa
109k17178294
109k17178294
$begingroup$
thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
$endgroup$
– jenkins342
Apr 2 '17 at 10:07
add a comment |
$begingroup$
thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
$endgroup$
– jenkins342
Apr 2 '17 at 10:07
$begingroup$
thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
$endgroup$
– jenkins342
Apr 2 '17 at 10:07
$begingroup$
thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
$endgroup$
– jenkins342
Apr 2 '17 at 10:07
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2214219%2fintegral-method-to-find-upper-and-lower-bounds-on-the-sum%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
