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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
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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

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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