Mixing
Solving Maclaurin series
$begingroup$
I have to tackle the following question. My thoughts so far are below it.
(A) By successive differentiation find the first four non-zero terms in the Maclaurin series for $$F(x)=(x+1)ln(1+x)-x$$
(B) Deduce that, for $ngeq2$, the coefficient of $X^n$ in this series is $(-1)^ncdot(1/n(n-1))$
(C) By applying the ratio test, find the radius of convergence for this Maclaurin series
I got $$F(x)=(1/2)x^2-(1/6)x^3+(1/12)x^4-(1/20)x^5$$ for part A.
I can easily find the general coefficient in part B: $$(-1)^n cdot (n－2)!/n! = (-1)^n ·1/n(n-1),$$ but my problem is, what is the right format to deduce?
For part C, I am very confused because after applying the ratio test my result is $1,$ how can I find the radius of convergence?
Help! Thanks!
taylor-expansion euler-maclaurin
edited Jan 17 at 10:15
user376343
3,8483829
asked Jan 17 at 8:13
何雨桓何雨桓
61
$endgroup$
add a comment |
$begingroup$
I have to tackle the following question. My thoughts so far are below it.
(A) By successive differentiation find the first four non-zero terms in the Maclaurin series for $$F(x)=(x+1)ln(1+x)-x$$
(B) Deduce that, for $ngeq2$, the coefficient of $X^n$ in this series is $(-1)^ncdot(1/n(n-1))$
(C) By applying the ratio test, find the radius of convergence for this Maclaurin series
I got $$F(x)=(1/2)x^2-(1/6)x^3+(1/12)x^4-(1/20)x^5$$ for part A.
I can easily find the general coefficient in part B: $$(-1)^n cdot (n－2)!/n! = (-1)^n ·1/n(n-1),$$ but my problem is, what is the right format to deduce?
For part C, I am very confused because after applying the ratio test my result is $1,$ how can I find the radius of convergence?
Help! Thanks!
taylor-expansion euler-maclaurin
edited Jan 17 at 10:15
user376343
3,8483829
asked Jan 17 at 8:13
何雨桓何雨桓
61
$endgroup$
2
$begingroup$
Please don't just write what look like homework questions. Context, and what you've already attempted, and where you're stuck all make for a much better question. As it stands, this question is likely to be downvoted and closed.
$endgroup$
– postmortes
Jan 17 at 8:17
$begingroup$
Sorry I am new to this website, and I’ve already changed my question.
$endgroup$
– 何雨桓
Jan 17 at 8:58
$begingroup$
I've added some formatting to your question -- you might want to take a look to help you with future questions. Thanks for updating, it's much better now
$endgroup$
– postmortes
Jan 17 at 9:43
$begingroup$
Okay. Thank you so much !!!!
$endgroup$
– 何雨桓
Jan 17 at 10:38
add a comment |
$begingroup$
I have to tackle the following question. My thoughts so far are below it.
(A) By successive differentiation find the first four non-zero terms in the Maclaurin series for $$F(x)=(x+1)ln(1+x)-x$$
(B) Deduce that, for $ngeq2$, the coefficient of $X^n$ in this series is $(-1)^ncdot(1/n(n-1))$
(C) By applying the ratio test, find the radius of convergence for this Maclaurin series
I got $$F(x)=(1/2)x^2-(1/6)x^3+(1/12)x^4-(1/20)x^5$$ for part A.
I can easily find the general coefficient in part B: $$(-1)^n cdot (n－2)!/n! = (-1)^n ·1/n(n-1),$$ but my problem is, what is the right format to deduce?
For part C, I am very confused because after applying the ratio test my result is $1,$ how can I find the radius of convergence?
Help! Thanks!
taylor-expansion euler-maclaurin
edited Jan 17 at 10:15
user376343
3,8483829
asked Jan 17 at 8:13
何雨桓何雨桓
61
$endgroup$
I have to tackle the following question. My thoughts so far are below it.
(A) By successive differentiation find the first four non-zero terms in the Maclaurin series for $$F(x)=(x+1)ln(1+x)-x$$
(B) Deduce that, for $ngeq2$, the coefficient of $X^n$ in this series is $(-1)^ncdot(1/n(n-1))$
(C) By applying the ratio test, find the radius of convergence for this Maclaurin series
I got $$F(x)=(1/2)x^2-(1/6)x^3+(1/12)x^4-(1/20)x^5$$ for part A.
I can easily find the general coefficient in part B: $$(-1)^n cdot (n－2)!/n! = (-1)^n ·1/n(n-1),$$ but my problem is, what is the right format to deduce?
For part C, I am very confused because after applying the ratio test my result is $1,$ how can I find the radius of convergence?
Help! Thanks!
taylor-expansion euler-maclaurin
taylor-expansion euler-maclaurin
edited Jan 17 at 10:15
user376343
3,8483829
asked Jan 17 at 8:13
何雨桓何雨桓
61
edited Jan 17 at 10:15
user376343
3,8483829
asked Jan 17 at 8:13
何雨桓何雨桓
61
edited Jan 17 at 10:15
user376343
3,8483829
edited Jan 17 at 10:15
user376343
3,8483829
edited Jan 17 at 10:15
user376343
3,8483829
3,8483829
asked Jan 17 at 8:13
何雨桓何雨桓
61
asked Jan 17 at 8:13
何雨桓何雨桓
61
asked Jan 17 at 8:13
何雨桓何雨桓
61
61
2
$begingroup$
Please don't just write what look like homework questions. Context, and what you've already attempted, and where you're stuck all make for a much better question. As it stands, this question is likely to be downvoted and closed.
$endgroup$
– postmortes
Jan 17 at 8:17
$begingroup$
Sorry I am new to this website, and I’ve already changed my question.
$endgroup$
– 何雨桓
Jan 17 at 8:58
$begingroup$
I've added some formatting to your question -- you might want to take a look to help you with future questions. Thanks for updating, it's much better now
$endgroup$
– postmortes
Jan 17 at 9:43
$begingroup$
Okay. Thank you so much !!!!
$endgroup$
– 何雨桓
Jan 17 at 10:38
add a comment |
2
$begingroup$
Please don't just write what look like homework questions. Context, and what you've already attempted, and where you're stuck all make for a much better question. As it stands, this question is likely to be downvoted and closed.
$endgroup$
– postmortes
Jan 17 at 8:17
$begingroup$
Sorry I am new to this website, and I’ve already changed my question.
$endgroup$
– 何雨桓
Jan 17 at 8:58
$begingroup$
I've added some formatting to your question -- you might want to take a look to help you with future questions. Thanks for updating, it's much better now
$endgroup$
– postmortes
Jan 17 at 9:43
$begingroup$
Okay. Thank you so much !!!!
$endgroup$
– 何雨桓
Jan 17 at 10:38
2
2
$begingroup$
Please don't just write what look like homework questions. Context, and what you've already attempted, and where you're stuck all make for a much better question. As it stands, this question is likely to be downvoted and closed.
$endgroup$
– postmortes
Jan 17 at 8:17
$begingroup$
Please don't just write what look like homework questions. Context, and what you've already attempted, and where you're stuck all make for a much better question. As it stands, this question is likely to be downvoted and closed.
$endgroup$
– postmortes
Jan 17 at 8:17
$begingroup$
Sorry I am new to this website, and I’ve already changed my question.
$endgroup$
– 何雨桓
Jan 17 at 8:58
$begingroup$
Sorry I am new to this website, and I’ve already changed my question.
$endgroup$
– 何雨桓
Jan 17 at 8:58
$begingroup$
I've added some formatting to your question -- you might want to take a look to help you with future questions. Thanks for updating, it's much better now
$endgroup$
– postmortes
Jan 17 at 9:43
$begingroup$
I've added some formatting to your question -- you might want to take a look to help you with future questions. Thanks for updating, it's much better now
$endgroup$
– postmortes
Jan 17 at 9:43
$begingroup$
Okay. Thank you so much !!!!
$endgroup$
– 何雨桓
Jan 17 at 10:38
$begingroup$
Okay. Thank you so much !!!!
$endgroup$
– 何雨桓
Jan 17 at 10:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
$$F'(x)=ln(x+1)$$ and then,
$$F^{(n)}(x)=(-1)^nfrac{(n-2)!}{(x+1)^{n-1}},$$
and
$$F^{(n)}(0)=(-1)^n(n-2)!.$$
Hence the $n^{th}$ coefficient is
$$(-1)^nfrac{(n-2)!}{n!}.$$
Applying the ratio test,
$$lim_{ntoinfty}frac{n(n-1)}{(n+1)n}=1$$ which is the radius of convergence.
Note that for $x=pm1$, the terms decrease like $n^{-2}$ so that the series also converges.
answered Jan 17 at 9:02
Yves DaoustYves Daoust
128k675227
$endgroup$
$begingroup$
I understand. Thank you so much for answering my question!!!!
$endgroup$
– 何雨桓
Jan 17 at 9:25
add a comment |
$begingroup$
Just for fun, another method:
We know the development
$$ln(1+x)=sum_{n=1}^infty(-1)^{n+1}frac{x^n}{n}.$$
Then,
$$(1+x)ln(1+x)=sum_{n=1}^inftyleft((-1)^{n+1}frac{x^n}{n}+(-1)^{n-1}frac{x^{n+1}}{n}right)=x+sum_{n=2}^infty(-1)^{n}left(-frac1{n}+frac{1}{n-1}right)x^n.$$
answered Jan 17 at 9:22
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
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votes
$begingroup$
Hint:
$$F'(x)=ln(x+1)$$ and then,
$$F^{(n)}(x)=(-1)^nfrac{(n-2)!}{(x+1)^{n-1}},$$
and
$$F^{(n)}(0)=(-1)^n(n-2)!.$$
Hence the $n^{th}$ coefficient is
$$(-1)^nfrac{(n-2)!}{n!}.$$
Applying the ratio test,
$$lim_{ntoinfty}frac{n(n-1)}{(n+1)n}=1$$ which is the radius of convergence.
Note that for $x=pm1$, the terms decrease like $n^{-2}$ so that the series also converges.
answered Jan 17 at 9:02
Yves DaoustYves Daoust
128k675227
$endgroup$
$begingroup$
I understand. Thank you so much for answering my question!!!!
$endgroup$
– 何雨桓
Jan 17 at 9:25
add a comment |
$begingroup$
Hint:
$$F'(x)=ln(x+1)$$ and then,
$$F^{(n)}(x)=(-1)^nfrac{(n-2)!}{(x+1)^{n-1}},$$
and
$$F^{(n)}(0)=(-1)^n(n-2)!.$$
Hence the $n^{th}$ coefficient is
$$(-1)^nfrac{(n-2)!}{n!}.$$
Applying the ratio test,
$$lim_{ntoinfty}frac{n(n-1)}{(n+1)n}=1$$ which is the radius of convergence.
Note that for $x=pm1$, the terms decrease like $n^{-2}$ so that the series also converges.
answered Jan 17 at 9:02
Yves DaoustYves Daoust
128k675227
$endgroup$
$begingroup$
I understand. Thank you so much for answering my question!!!!
$endgroup$
– 何雨桓
Jan 17 at 9:25
add a comment |
$begingroup$
Hint:
$$F'(x)=ln(x+1)$$ and then,
$$F^{(n)}(x)=(-1)^nfrac{(n-2)!}{(x+1)^{n-1}},$$
and
$$F^{(n)}(0)=(-1)^n(n-2)!.$$
Hence the $n^{th}$ coefficient is
$$(-1)^nfrac{(n-2)!}{n!}.$$
Applying the ratio test,
$$lim_{ntoinfty}frac{n(n-1)}{(n+1)n}=1$$ which is the radius of convergence.
Note that for $x=pm1$, the terms decrease like $n^{-2}$ so that the series also converges.
answered Jan 17 at 9:02
Yves DaoustYves Daoust
128k675227
$endgroup$
Hint:
$$F'(x)=ln(x+1)$$ and then,
$$F^{(n)}(x)=(-1)^nfrac{(n-2)!}{(x+1)^{n-1}},$$
and
$$F^{(n)}(0)=(-1)^n(n-2)!.$$
Hence the $n^{th}$ coefficient is
$$(-1)^nfrac{(n-2)!}{n!}.$$
Applying the ratio test,
$$lim_{ntoinfty}frac{n(n-1)}{(n+1)n}=1$$ which is the radius of convergence.
Note that for $x=pm1$, the terms decrease like $n^{-2}$ so that the series also converges.
answered Jan 17 at 9:02
Yves DaoustYves Daoust
128k675227
edited Jan 17 at 9:15
answered Jan 17 at 9:02
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 9:02
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 9:02
Yves DaoustYves Daoust
128k675227
128k675227
$begingroup$
I understand. Thank you so much for answering my question!!!!
$endgroup$
– 何雨桓
Jan 17 at 9:25
add a comment |
$begingroup$
I understand. Thank you so much for answering my question!!!!
$endgroup$
– 何雨桓
Jan 17 at 9:25
$begingroup$
I understand. Thank you so much for answering my question!!!!
$endgroup$
– 何雨桓
Jan 17 at 9:25
$begingroup$
I understand. Thank you so much for answering my question!!!!
$endgroup$
– 何雨桓
Jan 17 at 9:25
add a comment |
$begingroup$
Just for fun, another method:
We know the development
$$ln(1+x)=sum_{n=1}^infty(-1)^{n+1}frac{x^n}{n}.$$
Then,
$$(1+x)ln(1+x)=sum_{n=1}^inftyleft((-1)^{n+1}frac{x^n}{n}+(-1)^{n-1}frac{x^{n+1}}{n}right)=x+sum_{n=2}^infty(-1)^{n}left(-frac1{n}+frac{1}{n-1}right)x^n.$$
answered Jan 17 at 9:22
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
Just for fun, another method:
We know the development
$$ln(1+x)=sum_{n=1}^infty(-1)^{n+1}frac{x^n}{n}.$$
Then,
$$(1+x)ln(1+x)=sum_{n=1}^inftyleft((-1)^{n+1}frac{x^n}{n}+(-1)^{n-1}frac{x^{n+1}}{n}right)=x+sum_{n=2}^infty(-1)^{n}left(-frac1{n}+frac{1}{n-1}right)x^n.$$
answered Jan 17 at 9:22
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
Just for fun, another method:
We know the development
$$ln(1+x)=sum_{n=1}^infty(-1)^{n+1}frac{x^n}{n}.$$
Then,
$$(1+x)ln(1+x)=sum_{n=1}^inftyleft((-1)^{n+1}frac{x^n}{n}+(-1)^{n-1}frac{x^{n+1}}{n}right)=x+sum_{n=2}^infty(-1)^{n}left(-frac1{n}+frac{1}{n-1}right)x^n.$$
answered Jan 17 at 9:22
Yves DaoustYves Daoust
128k675227
$endgroup$
Just for fun, another method:
We know the development
$$ln(1+x)=sum_{n=1}^infty(-1)^{n+1}frac{x^n}{n}.$$
Then,
$$(1+x)ln(1+x)=sum_{n=1}^inftyleft((-1)^{n+1}frac{x^n}{n}+(-1)^{n-1}frac{x^{n+1}}{n}right)=x+sum_{n=2}^infty(-1)^{n}left(-frac1{n}+frac{1}{n-1}right)x^n.$$
answered Jan 17 at 9:22
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 9:22
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 9:22
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 9:22
Yves DaoustYves Daoust
128k675227
128k675227
add a comment |
add a comment |
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$begingroup$
Please don't just write what look like homework questions. Context, and what you've already attempted, and where you're stuck all make for a much better question. As it stands, this question is likely to be downvoted and closed.
$endgroup$
– postmortes
Jan 17 at 8:17
$begingroup$
Sorry I am new to this website, and I’ve already changed my question.
$endgroup$
– 何雨桓
Jan 17 at 8:58
$begingroup$
I've added some formatting to your question -- you might want to take a look to help you with future questions. Thanks for updating, it's much better now
$endgroup$
– postmortes
Jan 17 at 9:43
$begingroup$
Okay. Thank you so much !!!!
$endgroup$
– 何雨桓
Jan 17 at 10:38