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
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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
$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
calculus proof-verification taylor-expansion
edited 2 days ago
Eevee Trainer
7,80621339
7,80621339
asked 2 days ago
MichaelMichael
246
246
add a comment |
add a comment |
1 Answer
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$begingroup$
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.
$endgroup$
$begingroup$
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
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1 Answer
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active
oldest
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1 Answer
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active
oldest
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$begingroup$
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.
$endgroup$
$begingroup$
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.
$endgroup$
$begingroup$
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.
$endgroup$
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.
answered 2 days ago
José Carlos SantosJosé Carlos Santos
167k22132235
167k22132235
$begingroup$
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
$begingroup$
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
$begingroup$
What happened to x is substituting it with 2 and -2 (what I got from interval of convergence)?
$endgroup$
– Michael
2 days ago
$begingroup$
What happened to x is substituting it with 2 and -2 (what I got from interval of convergence)?
$endgroup$
– Michael
2 days ago
1
1
$begingroup$
The series diverges at those two points.
$endgroup$
– José Carlos Santos
2 days ago
$begingroup$
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
$begingroup$
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
$begingroup$
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
1
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
2 days ago
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
2 days ago
|
show 5 more comments
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