Analysis differentiability proof involving sequences
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So I was able to show part (a) using the limit definition of the derivative, but I don't know how to approach part (b). I thought maybe assuming that there did exist such a sequence (yn) but couldn't find a way to produce a contradiction. Help!
real-analysis
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add a comment |
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So I was able to show part (a) using the limit definition of the derivative, but I don't know how to approach part (b). I thought maybe assuming that there did exist such a sequence (yn) but couldn't find a way to produce a contradiction. Help!
real-analysis
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Do you know Taylor's theorem?
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– PhoemueX
Dec 7 '18 at 8:45
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yeah the one that gives a statement about the remainder term?
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– Nick
Dec 7 '18 at 12:21
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Yes. Try to apply that to $f$, developing the power series at $0$. If there does not exist a sequence as in the claim, this would show that $f$ is represented by this power series. Why is this not true?
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– PhoemueX
Dec 8 '18 at 8:27
add a comment |
$begingroup$
So I was able to show part (a) using the limit definition of the derivative, but I don't know how to approach part (b). I thought maybe assuming that there did exist such a sequence (yn) but couldn't find a way to produce a contradiction. Help!
real-analysis
$endgroup$
So I was able to show part (a) using the limit definition of the derivative, but I don't know how to approach part (b). I thought maybe assuming that there did exist such a sequence (yn) but couldn't find a way to produce a contradiction. Help!
real-analysis
real-analysis
asked Dec 7 '18 at 8:18
NickNick
344
344
$begingroup$
Do you know Taylor's theorem?
$endgroup$
– PhoemueX
Dec 7 '18 at 8:45
$begingroup$
yeah the one that gives a statement about the remainder term?
$endgroup$
– Nick
Dec 7 '18 at 12:21
$begingroup$
Yes. Try to apply that to $f$, developing the power series at $0$. If there does not exist a sequence as in the claim, this would show that $f$ is represented by this power series. Why is this not true?
$endgroup$
– PhoemueX
Dec 8 '18 at 8:27
add a comment |
$begingroup$
Do you know Taylor's theorem?
$endgroup$
– PhoemueX
Dec 7 '18 at 8:45
$begingroup$
yeah the one that gives a statement about the remainder term?
$endgroup$
– Nick
Dec 7 '18 at 12:21
$begingroup$
Yes. Try to apply that to $f$, developing the power series at $0$. If there does not exist a sequence as in the claim, this would show that $f$ is represented by this power series. Why is this not true?
$endgroup$
– PhoemueX
Dec 8 '18 at 8:27
$begingroup$
Do you know Taylor's theorem?
$endgroup$
– PhoemueX
Dec 7 '18 at 8:45
$begingroup$
Do you know Taylor's theorem?
$endgroup$
– PhoemueX
Dec 7 '18 at 8:45
$begingroup$
yeah the one that gives a statement about the remainder term?
$endgroup$
– Nick
Dec 7 '18 at 12:21
$begingroup$
yeah the one that gives a statement about the remainder term?
$endgroup$
– Nick
Dec 7 '18 at 12:21
$begingroup$
Yes. Try to apply that to $f$, developing the power series at $0$. If there does not exist a sequence as in the claim, this would show that $f$ is represented by this power series. Why is this not true?
$endgroup$
– PhoemueX
Dec 8 '18 at 8:27
$begingroup$
Yes. Try to apply that to $f$, developing the power series at $0$. If there does not exist a sequence as in the claim, this would show that $f$ is represented by this power series. Why is this not true?
$endgroup$
– PhoemueX
Dec 8 '18 at 8:27
add a comment |
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$begingroup$
Do you know Taylor's theorem?
$endgroup$
– PhoemueX
Dec 7 '18 at 8:45
$begingroup$
yeah the one that gives a statement about the remainder term?
$endgroup$
– Nick
Dec 7 '18 at 12:21
$begingroup$
Yes. Try to apply that to $f$, developing the power series at $0$. If there does not exist a sequence as in the claim, this would show that $f$ is represented by this power series. Why is this not true?
$endgroup$
– PhoemueX
Dec 8 '18 at 8:27