Mixing
Approximating $pi$ with arctangent
$begingroup$
Use the fact that $frac{pi}{4} = text{arctangent}(frac{1}{2}) + text{arctangent}(frac{1}{3})$ to determine the number of terms summed to ensure an approximation to $pi$ less than $10^{-3}$.
So far I've got
begin{align*}
pi = 4Bigg(sum_{i=1}^{infty} (-1)^{i+1}frac{1}{2^{2i-1}(2i-1)} + sum_{i=1}^{n} (-1)^{i+1}frac{1}{3^{2i-1}(2i-1)}Bigg)
end{align*}
from the given identity. But I don't know how to proceed.
numerical-methods approximation
asked Jan 23 at 17:53
mXdXmXdX
898
$endgroup$
add a comment |
$begingroup$
Use the fact that $frac{pi}{4} = text{arctangent}(frac{1}{2}) + text{arctangent}(frac{1}{3})$ to determine the number of terms summed to ensure an approximation to $pi$ less than $10^{-3}$.
So far I've got
begin{align*}
pi = 4Bigg(sum_{i=1}^{infty} (-1)^{i+1}frac{1}{2^{2i-1}(2i-1)} + sum_{i=1}^{n} (-1)^{i+1}frac{1}{3^{2i-1}(2i-1)}Bigg)
end{align*}
from the given identity. But I don't know how to proceed.
numerical-methods approximation
asked Jan 23 at 17:53
mXdXmXdX
898
$endgroup$
1
$begingroup$
Welcome to MSE. What you've "got so far" is false; it's only true for sums to infinity. So the real story is $pi = 4(sum^n ldots + E_1(n) + sum^n ldots + E_2(n))$, where $E_1(n)$ and $E_2(n)$ are the "error terms" for those two power series. What you need to do is find a value $n$ large enough that $|E_1(n)| + |E_2(n)| < 10^{-3}$. (You should ask yourself why I put in those absolute-value brackets, too). When you've found the error terms and tried to work with them, if you're still stuck, you can show your work by clicking "edit" below your question; then we can help further.
$endgroup$
– John Hughes
Jan 23 at 18:01
add a comment |
$begingroup$
Use the fact that $frac{pi}{4} = text{arctangent}(frac{1}{2}) + text{arctangent}(frac{1}{3})$ to determine the number of terms summed to ensure an approximation to $pi$ less than $10^{-3}$.
So far I've got
begin{align*}
pi = 4Bigg(sum_{i=1}^{infty} (-1)^{i+1}frac{1}{2^{2i-1}(2i-1)} + sum_{i=1}^{n} (-1)^{i+1}frac{1}{3^{2i-1}(2i-1)}Bigg)
end{align*}
from the given identity. But I don't know how to proceed.
numerical-methods approximation
asked Jan 23 at 17:53
mXdXmXdX
898
$endgroup$
Use the fact that $frac{pi}{4} = text{arctangent}(frac{1}{2}) + text{arctangent}(frac{1}{3})$ to determine the number of terms summed to ensure an approximation to $pi$ less than $10^{-3}$.
So far I've got
begin{align*}
pi = 4Bigg(sum_{i=1}^{infty} (-1)^{i+1}frac{1}{2^{2i-1}(2i-1)} + sum_{i=1}^{n} (-1)^{i+1}frac{1}{3^{2i-1}(2i-1)}Bigg)
end{align*}
from the given identity. But I don't know how to proceed.
numerical-methods approximation
numerical-methods approximation
asked Jan 23 at 17:53
mXdXmXdX
898
asked Jan 23 at 17:53
mXdXmXdX
898
edited Jan 30 at 2:02
mXdX
asked Jan 23 at 17:53
mXdXmXdX
898
asked Jan 23 at 17:53
mXdXmXdX
898
asked Jan 23 at 17:53
mXdXmXdX
898
898
1
$begingroup$
Welcome to MSE. What you've "got so far" is false; it's only true for sums to infinity. So the real story is $pi = 4(sum^n ldots + E_1(n) + sum^n ldots + E_2(n))$, where $E_1(n)$ and $E_2(n)$ are the "error terms" for those two power series. What you need to do is find a value $n$ large enough that $|E_1(n)| + |E_2(n)| < 10^{-3}$. (You should ask yourself why I put in those absolute-value brackets, too). When you've found the error terms and tried to work with them, if you're still stuck, you can show your work by clicking "edit" below your question; then we can help further.
$endgroup$
– John Hughes
Jan 23 at 18:01
add a comment |
1
$begingroup$
Welcome to MSE. What you've "got so far" is false; it's only true for sums to infinity. So the real story is $pi = 4(sum^n ldots + E_1(n) + sum^n ldots + E_2(n))$, where $E_1(n)$ and $E_2(n)$ are the "error terms" for those two power series. What you need to do is find a value $n$ large enough that $|E_1(n)| + |E_2(n)| < 10^{-3}$. (You should ask yourself why I put in those absolute-value brackets, too). When you've found the error terms and tried to work with them, if you're still stuck, you can show your work by clicking "edit" below your question; then we can help further.
$endgroup$
– John Hughes
Jan 23 at 18:01
1
1
$begingroup$
Welcome to MSE. What you've "got so far" is false; it's only true for sums to infinity. So the real story is $pi = 4(sum^n ldots + E_1(n) + sum^n ldots + E_2(n))$, where $E_1(n)$ and $E_2(n)$ are the "error terms" for those two power series. What you need to do is find a value $n$ large enough that $|E_1(n)| + |E_2(n)| < 10^{-3}$. (You should ask yourself why I put in those absolute-value brackets, too). When you've found the error terms and tried to work with them, if you're still stuck, you can show your work by clicking "edit" below your question; then we can help further.
$endgroup$
– John Hughes
Jan 23 at 18:01
$begingroup$
Welcome to MSE. What you've "got so far" is false; it's only true for sums to infinity. So the real story is $pi = 4(sum^n ldots + E_1(n) + sum^n ldots + E_2(n))$, where $E_1(n)$ and $E_2(n)$ are the "error terms" for those two power series. What you need to do is find a value $n$ large enough that $|E_1(n)| + |E_2(n)| < 10^{-3}$. (You should ask yourself why I put in those absolute-value brackets, too). When you've found the error terms and tried to work with them, if you're still stuck, you can show your work by clicking "edit" below your question; then we can help further.
$endgroup$
– John Hughes
Jan 23 at 18:01
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Rewrite the equality as:
begin{array}{}
pi = 4left[sumlimits_{i=0}^{n-1} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)+sumlimits_{i=n}^{infty} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)right]
end{array}
to obtain the inequality to be solved for $n$:
$$
begin{array}{}
4left|sumlimits_{i=n}^{infty} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)right|<epsilon=10^{-3}.tag1
end{array}$$
Observe now that the series (1) is alternating with monotonously decreasing terms, and thus the absolute value of the series is less than that of its first term.
answered Jan 23 at 18:06
useruser
5,31511030
$endgroup$
$begingroup$
Solving that thing for $n$ seemed pretty tedious. I just plugged in values and got $n=4$. Does that seem correct?
$endgroup$
– mXdX
Jan 30 at 2:03
$begingroup$
@mXdX My initial answer was intended as a hint and I am really happy that you have completed the task. Your step is correct. I have added it now in the answer for certainity.
$endgroup$
– user
Jan 30 at 18:46
add a comment |
$begingroup$
The sums are alternating and the terms decreasing in absolute value. This implies that the error is lesser or equal to the first discarded term. The error summing up to $n$ is then lesser or equal to
$$
4Bigl(frac{1}{2^{2n+1}(2,n+1)}+frac{1}{3^{2n+1}(2,n+1)}Bigr).
$$
Now choose $n$ so that the above is $<10^{-3}$.
answered Jan 23 at 18:50
Julián AguirreJulián Aguirre
69.4k24096
$endgroup$
$begingroup$
This is the result I got. From here, it seemed best to just try values for $n$. Looks like it's $n=4$, yes?
$endgroup$
– mXdX
Jan 30 at 2:00
$begingroup$
Yes, that is correct.
$endgroup$
– Julián Aguirre
Jan 30 at 5:55
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%2f3084828%2fapproximating-pi-with-arctangent%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Rewrite the equality as:
begin{array}{}
pi = 4left[sumlimits_{i=0}^{n-1} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)+sumlimits_{i=n}^{infty} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)right]
end{array}
to obtain the inequality to be solved for $n$:
$$
begin{array}{}
4left|sumlimits_{i=n}^{infty} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)right|<epsilon=10^{-3}.tag1
end{array}$$
Observe now that the series (1) is alternating with monotonously decreasing terms, and thus the absolute value of the series is less than that of its first term.
answered Jan 23 at 18:06
useruser
5,31511030
$endgroup$
$begingroup$
Solving that thing for $n$ seemed pretty tedious. I just plugged in values and got $n=4$. Does that seem correct?
$endgroup$
– mXdX
Jan 30 at 2:03
$begingroup$
@mXdX My initial answer was intended as a hint and I am really happy that you have completed the task. Your step is correct. I have added it now in the answer for certainity.
$endgroup$
– user
Jan 30 at 18:46
add a comment |
$begingroup$
Rewrite the equality as:
begin{array}{}
pi = 4left[sumlimits_{i=0}^{n-1} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)+sumlimits_{i=n}^{infty} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)right]
end{array}
to obtain the inequality to be solved for $n$:
$$
begin{array}{}
4left|sumlimits_{i=n}^{infty} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)right|<epsilon=10^{-3}.tag1
end{array}$$
Observe now that the series (1) is alternating with monotonously decreasing terms, and thus the absolute value of the series is less than that of its first term.
answered Jan 23 at 18:06
useruser
5,31511030
$endgroup$
$begingroup$
Solving that thing for $n$ seemed pretty tedious. I just plugged in values and got $n=4$. Does that seem correct?
$endgroup$
– mXdX
Jan 30 at 2:03
$begingroup$
@mXdX My initial answer was intended as a hint and I am really happy that you have completed the task. Your step is correct. I have added it now in the answer for certainity.
$endgroup$
– user
Jan 30 at 18:46
add a comment |
$begingroup$
Rewrite the equality as:
begin{array}{}
pi = 4left[sumlimits_{i=0}^{n-1} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)+sumlimits_{i=n}^{infty} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)right]
end{array}
to obtain the inequality to be solved for $n$:
$$
begin{array}{}
4left|sumlimits_{i=n}^{infty} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)right|<epsilon=10^{-3}.tag1
end{array}$$
Observe now that the series (1) is alternating with monotonously decreasing terms, and thus the absolute value of the series is less than that of its first term.
answered Jan 23 at 18:06
useruser
5,31511030
$endgroup$
Rewrite the equality as:
begin{array}{}
pi = 4left[sumlimits_{i=0}^{n-1} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)+sumlimits_{i=n}^{infty} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)right]
end{array}
to obtain the inequality to be solved for $n$:
$$
begin{array}{}
4left|sumlimits_{i=n}^{infty} (-1)^{i}left(frac{1}{2^{2i+1}(2i+1)} +frac{1}{3^{2i+1}(2i+1)}right)right|<epsilon=10^{-3}.tag1
end{array}$$
Observe now that the series (1) is alternating with monotonously decreasing terms, and thus the absolute value of the series is less than that of its first term.
answered Jan 23 at 18:06
useruser
5,31511030
edited Jan 30 at 18:41
answered Jan 23 at 18:06
useruser
5,31511030
answered Jan 23 at 18:06
useruser
5,31511030
answered Jan 23 at 18:06
useruser
5,31511030
5,31511030
$begingroup$
Solving that thing for $n$ seemed pretty tedious. I just plugged in values and got $n=4$. Does that seem correct?
$endgroup$
– mXdX
Jan 30 at 2:03
$begingroup$
@mXdX My initial answer was intended as a hint and I am really happy that you have completed the task. Your step is correct. I have added it now in the answer for certainity.
$endgroup$
– user
Jan 30 at 18:46
add a comment |
$begingroup$
Solving that thing for $n$ seemed pretty tedious. I just plugged in values and got $n=4$. Does that seem correct?
$endgroup$
– mXdX
Jan 30 at 2:03
$begingroup$
@mXdX My initial answer was intended as a hint and I am really happy that you have completed the task. Your step is correct. I have added it now in the answer for certainity.
$endgroup$
– user
Jan 30 at 18:46
$begingroup$
Solving that thing for $n$ seemed pretty tedious. I just plugged in values and got $n=4$. Does that seem correct?
$endgroup$
– mXdX
Jan 30 at 2:03
$begingroup$
Solving that thing for $n$ seemed pretty tedious. I just plugged in values and got $n=4$. Does that seem correct?
$endgroup$
– mXdX
Jan 30 at 2:03
$begingroup$
@mXdX My initial answer was intended as a hint and I am really happy that you have completed the task. Your step is correct. I have added it now in the answer for certainity.
$endgroup$
– user
Jan 30 at 18:46
$begingroup$
@mXdX My initial answer was intended as a hint and I am really happy that you have completed the task. Your step is correct. I have added it now in the answer for certainity.
$endgroup$
– user
Jan 30 at 18:46
add a comment |
$begingroup$
The sums are alternating and the terms decreasing in absolute value. This implies that the error is lesser or equal to the first discarded term. The error summing up to $n$ is then lesser or equal to
$$
4Bigl(frac{1}{2^{2n+1}(2,n+1)}+frac{1}{3^{2n+1}(2,n+1)}Bigr).
$$
Now choose $n$ so that the above is $<10^{-3}$.
answered Jan 23 at 18:50
Julián AguirreJulián Aguirre
69.4k24096
$endgroup$
$begingroup$
This is the result I got. From here, it seemed best to just try values for $n$. Looks like it's $n=4$, yes?
$endgroup$
– mXdX
Jan 30 at 2:00
$begingroup$
Yes, that is correct.
$endgroup$
– Julián Aguirre
Jan 30 at 5:55
add a comment |
$begingroup$
The sums are alternating and the terms decreasing in absolute value. This implies that the error is lesser or equal to the first discarded term. The error summing up to $n$ is then lesser or equal to
$$
4Bigl(frac{1}{2^{2n+1}(2,n+1)}+frac{1}{3^{2n+1}(2,n+1)}Bigr).
$$
Now choose $n$ so that the above is $<10^{-3}$.
answered Jan 23 at 18:50
Julián AguirreJulián Aguirre
69.4k24096
$endgroup$
$begingroup$
This is the result I got. From here, it seemed best to just try values for $n$. Looks like it's $n=4$, yes?
$endgroup$
– mXdX
Jan 30 at 2:00
$begingroup$
Yes, that is correct.
$endgroup$
– Julián Aguirre
Jan 30 at 5:55
add a comment |
$begingroup$
The sums are alternating and the terms decreasing in absolute value. This implies that the error is lesser or equal to the first discarded term. The error summing up to $n$ is then lesser or equal to
$$
4Bigl(frac{1}{2^{2n+1}(2,n+1)}+frac{1}{3^{2n+1}(2,n+1)}Bigr).
$$
Now choose $n$ so that the above is $<10^{-3}$.
answered Jan 23 at 18:50
Julián AguirreJulián Aguirre
69.4k24096
$endgroup$
The sums are alternating and the terms decreasing in absolute value. This implies that the error is lesser or equal to the first discarded term. The error summing up to $n$ is then lesser or equal to
$$
4Bigl(frac{1}{2^{2n+1}(2,n+1)}+frac{1}{3^{2n+1}(2,n+1)}Bigr).
$$
Now choose $n$ so that the above is $<10^{-3}$.
answered Jan 23 at 18:50
Julián AguirreJulián Aguirre
69.4k24096
answered Jan 23 at 18:50
Julián AguirreJulián Aguirre
69.4k24096
answered Jan 23 at 18:50
Julián AguirreJulián Aguirre
69.4k24096
answered Jan 23 at 18:50
Julián AguirreJulián Aguirre
69.4k24096
69.4k24096
$begingroup$
This is the result I got. From here, it seemed best to just try values for $n$. Looks like it's $n=4$, yes?
$endgroup$
– mXdX
Jan 30 at 2:00
$begingroup$
Yes, that is correct.
$endgroup$
– Julián Aguirre
Jan 30 at 5:55
add a comment |
$begingroup$
This is the result I got. From here, it seemed best to just try values for $n$. Looks like it's $n=4$, yes?
$endgroup$
– mXdX
Jan 30 at 2:00
$begingroup$
Yes, that is correct.
$endgroup$
– Julián Aguirre
Jan 30 at 5:55
$begingroup$
This is the result I got. From here, it seemed best to just try values for $n$. Looks like it's $n=4$, yes?
$endgroup$
– mXdX
Jan 30 at 2:00
$begingroup$
This is the result I got. From here, it seemed best to just try values for $n$. Looks like it's $n=4$, yes?
$endgroup$
– mXdX
Jan 30 at 2:00
$begingroup$
Yes, that is correct.
$endgroup$
– Julián Aguirre
Jan 30 at 5:55
$begingroup$
Yes, that is correct.
$endgroup$
– Julián Aguirre
Jan 30 at 5:55
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%2f3084828%2fapproximating-pi-with-arctangent%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
1
$begingroup$
Welcome to MSE. What you've "got so far" is false; it's only true for sums to infinity. So the real story is $pi = 4(sum^n ldots + E_1(n) + sum^n ldots + E_2(n))$, where $E_1(n)$ and $E_2(n)$ are the "error terms" for those two power series. What you need to do is find a value $n$ large enough that $|E_1(n)| + |E_2(n)| < 10^{-3}$. (You should ask yourself why I put in those absolute-value brackets, too). When you've found the error terms and tried to work with them, if you're still stuck, you can show your work by clicking "edit" below your question; then we can help further.
$endgroup$
– John Hughes
Jan 23 at 18:01