How to calculate complex expression in WolframAlphaHow to figure out how to make WolframAlpha work predictablyQuestion about sum and list in a functionProblem evaluating a complicated integralFailed to use N[%] for a infinite seriesCalculating sum of BesselJ[n, x]Error with Wolfram AlphaNumerical result using summation limitChecking an interesting result for a sum
What checks exist against overuse of presidential pardons in the USA?
What caused the end of cybernetic implants?
Printing a list as "a, b, c." using Python
What is Soda Fountain Etiquette?
Why is there no Disney logo in MCU movies?
What does "-1" represent in the value range for unsigned int and signed int?
Why did Lucius make a deal out of Buckbeak hurting Draco but not about Draco being turned into a ferret?
How can I throw a body?
Why did Starhopper's exhaust plume become brighter just before landing?
Are sweatpants frowned upon on flights?
Is "survival" paracord with fire starter strand dangerous
Why doesn't Starship have four landing legs?
What is the following VRP?
Is it possible for a person to be tricked into becoming a lich?
How to handle inventory and story of a player leaving
Under GDPR, can I give permission once to allow everyone to store and process my data?
Pen test results for web application include a file from a forbidden directory that is not even used or referenced
Fixing a blind bolt hole when the first 2-3 threads are ruined?
How did medieval manors handle population growth? Was there room for more fields to be ploughed?
Can two aircraft stay on the same runway at the same time?
In what language did Túrin converse with Mím?
Cauterizing a wound with metal?
Is there an in-universe explanation given to the senior Imperial Navy Officers as to why Darth Vader serves Emperor Palpatine?
Inspiration for failed idea?
How to calculate complex expression in WolframAlpha
How to figure out how to make WolframAlpha work predictablyQuestion about sum and list in a functionProblem evaluating a complicated integralFailed to use N[%] for a infinite seriesCalculating sum of BesselJ[n, x]Error with Wolfram AlphaNumerical result using summation limitChecking an interesting result for a sum
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I need to evaluate the limit:
$$lim_ntoinftyprod_k=1^infty left(1-fracnleft(fracn+sqrtn^2+42right)^k+fracn+sqrtn^2+42right).$$
I could not type into WolframAlpha and find its value. Can someone help me?
summation wolfram-alpha-queries
asked 8 hours ago
bilgamishbilgamish
61 bronze badge
New contributor
bilgamish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I need to evaluate the limit:
$$lim_ntoinftyprod_k=1^infty left(1-fracnleft(fracn+sqrtn^2+42right)^k+fracn+sqrtn^2+42right).$$
I could not type into WolframAlpha and find its value. Can someone help me?
summation wolfram-alpha-queries
asked 8 hours ago
bilgamishbilgamish
61 bronze badge
New contributor
bilgamish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
The way you could type that into WolframAlpha, or Mathematica isLimit[Product[1-n/(((n+Sqrt[n^2+4])/2)^k+(n+Sqrt[n^2+4])/2),k,1,Infinity],n->Infinity]but it appears that neither of those are able to see a way to give you the limit. They do not appear to solve even the simplerLimit[Product[1-n/(n^k+n),k,1,Infinity],n->Infinity]Perhaps you can think of a way to simplify your problem.
$endgroup$
– Bill
7 hours ago
add a comment |
$begingroup$
I need to evaluate the limit:
$$lim_ntoinftyprod_k=1^infty left(1-fracnleft(fracn+sqrtn^2+42right)^k+fracn+sqrtn^2+42right).$$
I could not type into WolframAlpha and find its value. Can someone help me?
summation wolfram-alpha-queries
asked 8 hours ago
bilgamishbilgamish
61 bronze badge
New contributor
bilgamish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I need to evaluate the limit:
$$lim_ntoinftyprod_k=1^infty left(1-fracnleft(fracn+sqrtn^2+42right)^k+fracn+sqrtn^2+42right).$$
I could not type into WolframAlpha and find its value. Can someone help me?
summation wolfram-alpha-queries
summation wolfram-alpha-queries
asked 8 hours ago
bilgamishbilgamish
61 bronze badge
New contributor
bilgamish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
bilgamishbilgamish
61 bronze badge
New contributor
bilgamish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
bilgamishbilgamish
61 bronze badge
New contributor
bilgamish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
bilgamishbilgamish
61 bronze badge
asked 8 hours ago
bilgamishbilgamish
61 bronze badge
61 bronze badge
New contributor
bilgamish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
bilgamish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
The way you could type that into WolframAlpha, or Mathematica isLimit[Product[1-n/(((n+Sqrt[n^2+4])/2)^k+(n+Sqrt[n^2+4])/2),k,1,Infinity],n->Infinity]but it appears that neither of those are able to see a way to give you the limit. They do not appear to solve even the simplerLimit[Product[1-n/(n^k+n),k,1,Infinity],n->Infinity]Perhaps you can think of a way to simplify your problem.
$endgroup$
– Bill
7 hours ago
add a comment |
1
$begingroup$
The way you could type that into WolframAlpha, or Mathematica isLimit[Product[1-n/(((n+Sqrt[n^2+4])/2)^k+(n+Sqrt[n^2+4])/2),k,1,Infinity],n->Infinity]but it appears that neither of those are able to see a way to give you the limit. They do not appear to solve even the simplerLimit[Product[1-n/(n^k+n),k,1,Infinity],n->Infinity]Perhaps you can think of a way to simplify your problem.
$endgroup$
– Bill
7 hours ago
1
1
$begingroup$
The way you could type that into WolframAlpha, or Mathematica is
Limit[Product[1-n/(((n+Sqrt[n^2+4])/2)^k+(n+Sqrt[n^2+4])/2),k,1,Infinity],n->Infinity] but it appears that neither of those are able to see a way to give you the limit. They do not appear to solve even the simpler Limit[Product[1-n/(n^k+n),k,1,Infinity],n->Infinity] Perhaps you can think of a way to simplify your problem.$endgroup$
– Bill
7 hours ago
$begingroup$
The way you could type that into WolframAlpha, or Mathematica is
Limit[Product[1-n/(((n+Sqrt[n^2+4])/2)^k+(n+Sqrt[n^2+4])/2),k,1,Infinity],n->Infinity] but it appears that neither of those are able to see a way to give you the limit. They do not appear to solve even the simpler Limit[Product[1-n/(n^k+n),k,1,Infinity],n->Infinity] Perhaps you can think of a way to simplify your problem.$endgroup$
– Bill
7 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
With Mathematica I have:
func[n_] := NProduct[1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n + Sqrt[n^2 + 4])/2), k, 1, Infinity]
Table[func[10^n], n, 1, 10]
(*0.454951, 0.49505, 0.4995, 0.49995, 0.499995, 0.5, 0.5, 0.5, 0.5, 0.5 *)
It looks like the limit is: 1/2
answered 7 hours ago
Mariusz IwaniukMariusz Iwaniuk
7,2301 gold badge12 silver badges30 bronze badges
$endgroup$
add a comment |
$begingroup$
Analytically, it is easy to see that the first term in the product has a limiting value of 1/2 and all the other terms have limiting values of 1. My maths is a bit rusty, but I suspect that you can justify changing the order of the two limiting processes.
answered 4 hours ago
mikadomikado
7,3921 gold badge9 silver badges29 bronze badges
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
bilgamish is a new contributor. Be nice, and check out our Code of Conduct.
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%2fmathematica.stackexchange.com%2fquestions%2f204528%2fhow-to-calculate-complex-expression-in-wolframalpha%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
$begingroup$
With Mathematica I have:
func[n_] := NProduct[1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n + Sqrt[n^2 + 4])/2), k, 1, Infinity]
Table[func[10^n], n, 1, 10]
(*0.454951, 0.49505, 0.4995, 0.49995, 0.499995, 0.5, 0.5, 0.5, 0.5, 0.5 *)
It looks like the limit is: 1/2
answered 7 hours ago
Mariusz IwaniukMariusz Iwaniuk
7,2301 gold badge12 silver badges30 bronze badges
$endgroup$
add a comment |
$begingroup$
With Mathematica I have:
func[n_] := NProduct[1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n + Sqrt[n^2 + 4])/2), k, 1, Infinity]
Table[func[10^n], n, 1, 10]
(*0.454951, 0.49505, 0.4995, 0.49995, 0.499995, 0.5, 0.5, 0.5, 0.5, 0.5 *)
It looks like the limit is: 1/2
answered 7 hours ago
Mariusz IwaniukMariusz Iwaniuk
7,2301 gold badge12 silver badges30 bronze badges
$endgroup$
add a comment |
$begingroup$
With Mathematica I have:
func[n_] := NProduct[1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n + Sqrt[n^2 + 4])/2), k, 1, Infinity]
Table[func[10^n], n, 1, 10]
(*0.454951, 0.49505, 0.4995, 0.49995, 0.499995, 0.5, 0.5, 0.5, 0.5, 0.5 *)
It looks like the limit is: 1/2
answered 7 hours ago
Mariusz IwaniukMariusz Iwaniuk
7,2301 gold badge12 silver badges30 bronze badges
$endgroup$
With Mathematica I have:
func[n_] := NProduct[1 - n/(((n + Sqrt[n^2 + 4])/2)^k + (n + Sqrt[n^2 + 4])/2), k, 1, Infinity]
Table[func[10^n], n, 1, 10]
(*0.454951, 0.49505, 0.4995, 0.49995, 0.499995, 0.5, 0.5, 0.5, 0.5, 0.5 *)
It looks like the limit is: 1/2
answered 7 hours ago
Mariusz IwaniukMariusz Iwaniuk
7,2301 gold badge12 silver badges30 bronze badges
answered 7 hours ago
Mariusz IwaniukMariusz Iwaniuk
7,2301 gold badge12 silver badges30 bronze badges
answered 7 hours ago
Mariusz IwaniukMariusz Iwaniuk
7,2301 gold badge12 silver badges30 bronze badges
answered 7 hours ago
Mariusz IwaniukMariusz Iwaniuk
7,2301 gold badge12 silver badges30 bronze badges
7,2301 gold badge12 silver badges30 bronze badges
add a comment |
add a comment |
$begingroup$
Analytically, it is easy to see that the first term in the product has a limiting value of 1/2 and all the other terms have limiting values of 1. My maths is a bit rusty, but I suspect that you can justify changing the order of the two limiting processes.
answered 4 hours ago
mikadomikado
7,3921 gold badge9 silver badges29 bronze badges
$endgroup$
add a comment |
$begingroup$
Analytically, it is easy to see that the first term in the product has a limiting value of 1/2 and all the other terms have limiting values of 1. My maths is a bit rusty, but I suspect that you can justify changing the order of the two limiting processes.
answered 4 hours ago
mikadomikado
7,3921 gold badge9 silver badges29 bronze badges
$endgroup$
add a comment |
$begingroup$
Analytically, it is easy to see that the first term in the product has a limiting value of 1/2 and all the other terms have limiting values of 1. My maths is a bit rusty, but I suspect that you can justify changing the order of the two limiting processes.
answered 4 hours ago
mikadomikado
7,3921 gold badge9 silver badges29 bronze badges
$endgroup$
Analytically, it is easy to see that the first term in the product has a limiting value of 1/2 and all the other terms have limiting values of 1. My maths is a bit rusty, but I suspect that you can justify changing the order of the two limiting processes.
answered 4 hours ago
mikadomikado
7,3921 gold badge9 silver badges29 bronze badges
answered 4 hours ago
mikadomikado
7,3921 gold badge9 silver badges29 bronze badges
answered 4 hours ago
mikadomikado
7,3921 gold badge9 silver badges29 bronze badges
answered 4 hours ago
mikadomikado
7,3921 gold badge9 silver badges29 bronze badges
7,3921 gold badge9 silver badges29 bronze badges
add a comment |
add a comment |
bilgamish is a new contributor. Be nice, and check out our Code of Conduct.
bilgamish is a new contributor. Be nice, and check out our Code of Conduct.
bilgamish is a new contributor. Be nice, and check out our Code of Conduct.
bilgamish is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematica 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%2fmathematica.stackexchange.com%2fquestions%2f204528%2fhow-to-calculate-complex-expression-in-wolframalpha%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
$begingroup$
The way you could type that into WolframAlpha, or Mathematica is
Limit[Product[1-n/(((n+Sqrt[n^2+4])/2)^k+(n+Sqrt[n^2+4])/2),k,1,Infinity],n->Infinity]but it appears that neither of those are able to see a way to give you the limit. They do not appear to solve even the simplerLimit[Product[1-n/(n^k+n),k,1,Infinity],n->Infinity]Perhaps you can think of a way to simplify your problem.$endgroup$
– Bill
7 hours ago