Fdtxjr

In the late 1940’s to early 1950’s what technology was available that could melt a LOT of ice?

Why don't MCU characters ever seem to have language issues?

Why would a jet engine that runs at temps excess of 2000°C burn when it crashes?

Can you reject a postdoc offer after the PI has paid a large sum for flights/accommodation for your visit?

Why does Deadpool say "You're welcome, Canada," after shooting Ryan Reynolds in the end credits?

Good for you! in Russian

Time travel short story where dinosaur doesn't taste like chicken

Reverse string, can I make it faster?

Space in array system equations

How are such low op-amp input currents possible?

Do items de-spawn in Diablo?

Upside Down Word Puzzle

Should I take out a loan for a friend to invest on my behalf?

How strictly should I take "Candidates must be local"?

Are babies of evil humanoid species inherently evil?

How to create a hard link to an inode (ext4)?

A question on the ultrafilter number

Do Bugbears' arms literally get longer when it's their turn?

Are the terms "stab" and "staccato" synonyms?

Word for a person who has no opinion about whether god exists

Could you please stop shuffling the deck and play already?

Latest web browser compatible with Windows 98

How to pass a string to a command that expects a file?

Low budget alien movie about the Earth being cooked

Solution verification: finding a Maclaurin series for $f$, interval of convergence, and $f^(10)(0)$

Question about Maclaurin Series for $cos x$Taylor and Maclaurin Series for $f(x)=e^x$Maclaurin series for $e^x +2e^-x$Maclaurin series of $sin(2pi x)$Finding interval of convergence for seriesMaclaurin Series expansion intervalMaclaurin Series for a natural logarithmMaclaurin series - Approximation and interval of convergenceMaclaurin series for lnMaclaurin Series from sin(x) to cos(x) using derivative

2

$begingroup$

I have to find Maclaurin series for function $f(x)$ = $2x^2over16+x^4$, it's interval of convergence and $f^(10)(0)$. I managed to calculate Maclaurin series and $10^th$ derivative, but I'm not sure if it's done in proper way and if solution is correct.

---

On determining the series,

$$beginalign
f(x) &= frac2x^216+x^4 \
&= frac2x^216 cdot frac11-frac-x^416 \
&= frac2x^216 cdot sum_i=0^infty left(-fracx^416 right)^n \
&= frac2x^216 cdot sum_i=0^infty frac(-1)^n cdot x^4n16^n \
&= sum_i=0^infty frac(-1)^n cdot x^4n+22^4n+3
endalign$$

for $vert-x^4over16vert<1 implies xin(-2;2)$

---

On determining $f^(10)(0)$,

$$f^(10)(0)cdot frac x^10 10! = fracx^422^43 implies f^(10)(0) = x^32cdot frac10!2^43$$

---

Looking for feedback and opinion if the way I solved it is correct.

calculus proof-verification taylor-expansion

share|cite|improve this question

edited 2 days ago

Eevee Trainer

7,80721339

asked 2 days ago

MichaelMichael

246

$endgroup$

add a comment | 

2

$begingroup$

I have to find Maclaurin series for function $f(x)$ = $2x^2over16+x^4$, it's interval of convergence and $f^(10)(0)$. I managed to calculate Maclaurin series and $10^th$ derivative, but I'm not sure if it's done in proper way and if solution is correct.

---

On determining the series,

$$beginalign
f(x) &= frac2x^216+x^4 \
&= frac2x^216 cdot frac11-frac-x^416 \
&= frac2x^216 cdot sum_i=0^infty left(-fracx^416 right)^n \
&= frac2x^216 cdot sum_i=0^infty frac(-1)^n cdot x^4n16^n \
&= sum_i=0^infty frac(-1)^n cdot x^4n+22^4n+3
endalign$$

for $vert-x^4over16vert<1 implies xin(-2;2)$

---

On determining $f^(10)(0)$,

$$f^(10)(0)cdot frac x^10 10! = fracx^422^43 implies f^(10)(0) = x^32cdot frac10!2^43$$

---

Looking for feedback and opinion if the way I solved it is correct.

calculus proof-verification taylor-expansion

share|cite|improve this question

edited 2 days ago

Eevee Trainer

7,80721339

asked 2 days ago

MichaelMichael

246

$endgroup$

add a comment | 

$begingroup$

I have to find Maclaurin series for function $f(x)$ = $2x^2over16+x^4$, it's interval of convergence and $f^(10)(0)$. I managed to calculate Maclaurin series and $10^th$ derivative, but I'm not sure if it's done in proper way and if solution is correct.

---

On determining the series,

$$beginalign
f(x) &= frac2x^216+x^4 \
&= frac2x^216 cdot frac11-frac-x^416 \
&= frac2x^216 cdot sum_i=0^infty left(-fracx^416 right)^n \
&= frac2x^216 cdot sum_i=0^infty frac(-1)^n cdot x^4n16^n \
&= sum_i=0^infty frac(-1)^n cdot x^4n+22^4n+3
endalign$$

for $vert-x^4over16vert<1 implies xin(-2;2)$

---

On determining $f^(10)(0)$,

$$f^(10)(0)cdot frac x^10 10! = fracx^422^43 implies f^(10)(0) = x^32cdot frac10!2^43$$

---

Looking for feedback and opinion if the way I solved it is correct.

calculus proof-verification taylor-expansion

share|cite|improve this question

edited 2 days ago

Eevee Trainer

7,80721339

asked 2 days ago

MichaelMichael

246

$endgroup$

share|cite|improve this question

edited 2 days ago

Eevee Trainer

7,80721339

asked 2 days ago

MichaelMichael

246

share|cite|improve this question

edited 2 days ago

Eevee Trainer

7,80721339

asked 2 days ago

MichaelMichael

246

edited 2 days ago

Eevee Trainer

7,80721339

edited 2 days ago

Eevee Trainer

7,80721339

asked 2 days ago

MichaelMichael

246

asked 2 days ago

MichaelMichael

246

1 Answer
1

active

oldest

votes

2

$begingroup$

Your computation of the MacLaurin series of $f$ looks right, but remember that $f^(10)(0)$ is a number. You know that$$fracf^(10)(0)10!=frac(-1)^22^11.$$Therefore,$$f^(10)(0)=frac10!2^11.$$

Also, $leftlvertfrac-x^416rightrvert<1$ doesn't just imply that $xin(-2,2)$; it is actually equivalent to it.

share|cite|improve this answer

answered 2 days ago

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What happened to x is substituting it with 2 and -2 (what I got from interval of convergence)?
  $endgroup$
  – Michael
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The series diverges at those two points.
  $endgroup$
  – José Carlos Santos
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I’m ahead with the program and doing it Maclaurin on my own, so sorry for tedious questions. To get 10th derivative you input -2 or 2? Is that the way to calculate it?
  $endgroup$
  – Michael
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To get the $10^textth$ derivative, I put $n=2$, so that $x^4n+2$ becomes $x^10$.
  $endgroup$
  – José Carlos Santos
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  2 days ago
  

  
  
  
  

 | 
show 5 more comments

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3142211%2fsolution-verification-finding-a-maclaurin-series-for-f-interval-of-convergen%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

Your computation of the MacLaurin series of $f$ looks right, but remember that $f^(10)(0)$ is a number. You know that$$fracf^(10)(0)10!=frac(-1)^22^11.$$Therefore,$$f^(10)(0)=frac10!2^11.$$

Also, $leftlvertfrac-x^416rightrvert<1$ doesn't just imply that $xin(-2,2)$; it is actually equivalent to it.

share|cite|improve this answer

answered 2 days ago

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What happened to x is substituting it with 2 and -2 (what I got from interval of convergence)?
  $endgroup$
  – Michael
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The series diverges at those two points.
  $endgroup$
  – José Carlos Santos
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I’m ahead with the program and doing it Maclaurin on my own, so sorry for tedious questions. To get 10th derivative you input -2 or 2? Is that the way to calculate it?
  $endgroup$
  – Michael
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To get the $10^textth$ derivative, I put $n=2$, so that $x^4n+2$ becomes $x^10$.
  $endgroup$
  – José Carlos Santos
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  2 days ago
  

  
  
  
  

 | 
show 5 more comments

2

$begingroup$

Your computation of the MacLaurin series of $f$ looks right, but remember that $f^(10)(0)$ is a number. You know that$$fracf^(10)(0)10!=frac(-1)^22^11.$$Therefore,$$f^(10)(0)=frac10!2^11.$$

Also, $leftlvertfrac-x^416rightrvert<1$ doesn't just imply that $xin(-2,2)$; it is actually equivalent to it.

share|cite|improve this answer

answered 2 days ago

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What happened to x is substituting it with 2 and -2 (what I got from interval of convergence)?
  $endgroup$
  – Michael
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The series diverges at those two points.
  $endgroup$
  – José Carlos Santos
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I’m ahead with the program and doing it Maclaurin on my own, so sorry for tedious questions. To get 10th derivative you input -2 or 2? Is that the way to calculate it?
  $endgroup$
  – Michael
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To get the $10^textth$ derivative, I put $n=2$, so that $x^4n+2$ becomes $x^10$.
  $endgroup$
  – José Carlos Santos
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  2 days ago
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

Your computation of the MacLaurin series of $f$ looks right, but remember that $f^(10)(0)$ is a number. You know that$$fracf^(10)(0)10!=frac(-1)^22^11.$$Therefore,$$f^(10)(0)=frac10!2^11.$$

Also, $leftlvertfrac-x^416rightrvert<1$ doesn't just imply that $xin(-2,2)$; it is actually equivalent to it.

share|cite|improve this answer

answered 2 days ago

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

share|cite|improve this answer

answered 2 days ago

José Carlos SantosJosé Carlos Santos

167k22132235

answered 2 days ago

José Carlos SantosJosé Carlos Santos

167k22132235

answered 2 days ago

José Carlos SantosJosé Carlos Santos

167k22132235