Prove $f^{(n)}$ exists on $mathbb{R}$ and $f^{(n)}(0)=0$ for all $n$, where $f(x)=e^{frac{-1}{x^2}}$ [on...
$f(0)=0$ and $f(x)=e^{frac{-1}{x^2}}$ for $xneq 0$. I need to prove $f^{(n)}$ exists on $mathbb{R}$, $f^{(n)}(0)=0$ for all $n$, and every Taylor Polynomial about $0$ is $0$.
After setting up the Taylor expansion for $f(x)$, I obtain $$sumlimits_{k=0}^{n} dfrac{f^{(k)}(0)}{k!} x^k + dfrac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$ for some $c$ between $0$ and $x$. What would be the best approach from here to accomplish the goal of the proof mentioned in the title and second sentence?
real-analysis polynomials taylor-expansion exponential-function
asked Nov 29 at 3:02
t.perez
538
put on hold as off-topic by Lord_Farin, user10354138, KReiser, Leucippus, mrtaurho 12 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord_Farin, user10354138, KReiser, Leucippus, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
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$f(0)=0$ and $f(x)=e^{frac{-1}{x^2}}$ for $xneq 0$. I need to prove $f^{(n)}$ exists on $mathbb{R}$, $f^{(n)}(0)=0$ for all $n$, and every Taylor Polynomial about $0$ is $0$.
After setting up the Taylor expansion for $f(x)$, I obtain $$sumlimits_{k=0}^{n} dfrac{f^{(k)}(0)}{k!} x^k + dfrac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$ for some $c$ between $0$ and $x$. What would be the best approach from here to accomplish the goal of the proof mentioned in the title and second sentence?
real-analysis polynomials taylor-expansion exponential-function
asked Nov 29 at 3:02
t.perez
538
put on hold as off-topic by Lord_Farin, user10354138, KReiser, Leucippus, mrtaurho 12 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord_Farin, user10354138, KReiser, Leucippus, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Are you asking anything?
– Will M.
Nov 29 at 3:04
@WillM., yes whats in the title. I'll make it more clear in the description
– t.perez
Nov 29 at 3:10
1
Direct computation, nothing else.
– xbh
Nov 29 at 3:22
add a comment |
$f(0)=0$ and $f(x)=e^{frac{-1}{x^2}}$ for $xneq 0$. I need to prove $f^{(n)}$ exists on $mathbb{R}$, $f^{(n)}(0)=0$ for all $n$, and every Taylor Polynomial about $0$ is $0$.
After setting up the Taylor expansion for $f(x)$, I obtain $$sumlimits_{k=0}^{n} dfrac{f^{(k)}(0)}{k!} x^k + dfrac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$ for some $c$ between $0$ and $x$. What would be the best approach from here to accomplish the goal of the proof mentioned in the title and second sentence?
real-analysis polynomials taylor-expansion exponential-function
asked Nov 29 at 3:02
t.perez
538
$f(0)=0$ and $f(x)=e^{frac{-1}{x^2}}$ for $xneq 0$. I need to prove $f^{(n)}$ exists on $mathbb{R}$, $f^{(n)}(0)=0$ for all $n$, and every Taylor Polynomial about $0$ is $0$.
After setting up the Taylor expansion for $f(x)$, I obtain $$sumlimits_{k=0}^{n} dfrac{f^{(k)}(0)}{k!} x^k + dfrac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$ for some $c$ between $0$ and $x$. What would be the best approach from here to accomplish the goal of the proof mentioned in the title and second sentence?
real-analysis polynomials taylor-expansion exponential-function
real-analysis polynomials taylor-expansion exponential-function
asked Nov 29 at 3:02
t.perez
538
asked Nov 29 at 3:02
t.perez
538
edited Nov 29 at 3:11
asked Nov 29 at 3:02
t.perez
538
asked Nov 29 at 3:02
t.perez
538
asked Nov 29 at 3:02
t.perez
538
538
put on hold as off-topic by Lord_Farin, user10354138, KReiser, Leucippus, mrtaurho 12 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord_Farin, user10354138, KReiser, Leucippus, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Lord_Farin, user10354138, KReiser, Leucippus, mrtaurho 12 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord_Farin, user10354138, KReiser, Leucippus, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Are you asking anything?
– Will M.
Nov 29 at 3:04
@WillM., yes whats in the title. I'll make it more clear in the description
– t.perez
Nov 29 at 3:10
1
Direct computation, nothing else.
– xbh
Nov 29 at 3:22
add a comment |
1
Are you asking anything?
– Will M.
Nov 29 at 3:04
@WillM., yes whats in the title. I'll make it more clear in the description
– t.perez
Nov 29 at 3:10
1
Direct computation, nothing else.
– xbh
Nov 29 at 3:22
1
1
Are you asking anything?
– Will M.
Nov 29 at 3:04
Are you asking anything?
– Will M.
Nov 29 at 3:04
@WillM., yes whats in the title. I'll make it more clear in the description
– t.perez
Nov 29 at 3:10
@WillM., yes whats in the title. I'll make it more clear in the description
– t.perez
Nov 29 at 3:10
1
1
Direct computation, nothing else.
– xbh
Nov 29 at 3:22
Direct computation, nothing else.
– xbh
Nov 29 at 3:22
add a comment |
1 Answer
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Prove by induction that there exist polynomials $p_n$ such that $f^{(n)}(x) = p_nleft(dfrac{1}{x}right) e^{-frac{1}{x^2}}$ and conclude. $square$
answered Nov 29 at 3:47
Will M.
2,377314
add a comment |
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Prove by induction that there exist polynomials $p_n$ such that $f^{(n)}(x) = p_nleft(dfrac{1}{x}right) e^{-frac{1}{x^2}}$ and conclude. $square$
answered Nov 29 at 3:47
Will M.
2,377314
add a comment |
Prove by induction that there exist polynomials $p_n$ such that $f^{(n)}(x) = p_nleft(dfrac{1}{x}right) e^{-frac{1}{x^2}}$ and conclude. $square$
answered Nov 29 at 3:47
Will M.
2,377314
add a comment |
Prove by induction that there exist polynomials $p_n$ such that $f^{(n)}(x) = p_nleft(dfrac{1}{x}right) e^{-frac{1}{x^2}}$ and conclude. $square$
answered Nov 29 at 3:47
Will M.
2,377314
Prove by induction that there exist polynomials $p_n$ such that $f^{(n)}(x) = p_nleft(dfrac{1}{x}right) e^{-frac{1}{x^2}}$ and conclude. $square$
answered Nov 29 at 3:47
Will M.
2,377314
answered Nov 29 at 3:47
Will M.
2,377314
answered Nov 29 at 3:47
Will M.
2,377314
answered Nov 29 at 3:47
Will M.
2,377314
2,377314
add a comment |
add a comment |
1
Are you asking anything?
– Will M.
Nov 29 at 3:04
@WillM., yes whats in the title. I'll make it more clear in the description
– t.perez
Nov 29 at 3:10
1
Direct computation, nothing else.
– xbh
Nov 29 at 3:22