Calucate the coefficient of taylor series
Suppose $f$ is a real valued function with $f(0) = 0$ and $f'(x) = 1/sqrt{1+x^2}$. I want to calculate the coefficient of the $x^7$ term around $0$.
My approch: Although I know the answer if given $$f^{(7)}(0)/7!,$$ I am unable to calculate that far. I was hoping if there is shorter way to get the answer given the fact that this question was previously asked in an exam.
taylor-expansion
edited Nov 28 '18 at 2:28
Tianlalu
3,09621038
asked Nov 28 '18 at 0:54
henceproved
1358
add a comment |
Suppose $f$ is a real valued function with $f(0) = 0$ and $f'(x) = 1/sqrt{1+x^2}$. I want to calculate the coefficient of the $x^7$ term around $0$.
My approch: Although I know the answer if given $$f^{(7)}(0)/7!,$$ I am unable to calculate that far. I was hoping if there is shorter way to get the answer given the fact that this question was previously asked in an exam.
taylor-expansion
edited Nov 28 '18 at 2:28
Tianlalu
3,09621038
asked Nov 28 '18 at 0:54
henceproved
1358
If you know the binomial expansion, then you can get this easily.
– Sean Roberson
Nov 28 '18 at 2:08
@SeanRoberson You want me to expand $(1 + x^2)^{-0.5}$ and check for the coefficient of $x^6$ term i.e. $(^{-.5}C_3) times 6! / 7!)$.
– henceproved
Nov 28 '18 at 2:20
@SeanRoberson But how is binomial related to taylor series?
– henceproved
Nov 28 '18 at 2:24
add a comment |
Suppose $f$ is a real valued function with $f(0) = 0$ and $f'(x) = 1/sqrt{1+x^2}$. I want to calculate the coefficient of the $x^7$ term around $0$.
My approch: Although I know the answer if given $$f^{(7)}(0)/7!,$$ I am unable to calculate that far. I was hoping if there is shorter way to get the answer given the fact that this question was previously asked in an exam.
taylor-expansion
edited Nov 28 '18 at 2:28
Tianlalu
3,09621038
asked Nov 28 '18 at 0:54
henceproved
1358
Suppose $f$ is a real valued function with $f(0) = 0$ and $f'(x) = 1/sqrt{1+x^2}$. I want to calculate the coefficient of the $x^7$ term around $0$.
My approch: Although I know the answer if given $$f^{(7)}(0)/7!,$$ I am unable to calculate that far. I was hoping if there is shorter way to get the answer given the fact that this question was previously asked in an exam.
taylor-expansion
taylor-expansion
edited Nov 28 '18 at 2:28
Tianlalu
3,09621038
asked Nov 28 '18 at 0:54
henceproved
1358
edited Nov 28 '18 at 2:28
Tianlalu
3,09621038
asked Nov 28 '18 at 0:54
henceproved
1358
edited Nov 28 '18 at 2:28
Tianlalu
3,09621038
edited Nov 28 '18 at 2:28
Tianlalu
3,09621038
edited Nov 28 '18 at 2:28
Tianlalu
3,09621038
3,09621038
asked Nov 28 '18 at 0:54
henceproved
1358
asked Nov 28 '18 at 0:54
henceproved
1358
asked Nov 28 '18 at 0:54
henceproved
1358
1358
If you know the binomial expansion, then you can get this easily.
– Sean Roberson
Nov 28 '18 at 2:08
@SeanRoberson You want me to expand $(1 + x^2)^{-0.5}$ and check for the coefficient of $x^6$ term i.e. $(^{-.5}C_3) times 6! / 7!)$.
– henceproved
Nov 28 '18 at 2:20
@SeanRoberson But how is binomial related to taylor series?
– henceproved
Nov 28 '18 at 2:24
add a comment |
If you know the binomial expansion, then you can get this easily.
– Sean Roberson
Nov 28 '18 at 2:08
@SeanRoberson You want me to expand $(1 + x^2)^{-0.5}$ and check for the coefficient of $x^6$ term i.e. $(^{-.5}C_3) times 6! / 7!)$.
– henceproved
Nov 28 '18 at 2:20
@SeanRoberson But how is binomial related to taylor series?
– henceproved
Nov 28 '18 at 2:24
If you know the binomial expansion, then you can get this easily.
– Sean Roberson
Nov 28 '18 at 2:08
If you know the binomial expansion, then you can get this easily.
– Sean Roberson
Nov 28 '18 at 2:08
@SeanRoberson You want me to expand $(1 + x^2)^{-0.5}$ and check for the coefficient of $x^6$ term i.e. $(^{-.5}C_3) times 6! / 7!)$.
– henceproved
Nov 28 '18 at 2:20
@SeanRoberson You want me to expand $(1 + x^2)^{-0.5}$ and check for the coefficient of $x^6$ term i.e. $(^{-.5}C_3) times 6! / 7!)$.
– henceproved
Nov 28 '18 at 2:20
@SeanRoberson But how is binomial related to taylor series?
– henceproved
Nov 28 '18 at 2:24
@SeanRoberson But how is binomial related to taylor series?
– henceproved
Nov 28 '18 at 2:24
add a comment |
1 Answer
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It is not difficult to find the Taylor series of $(1-t)^{alpha}$ around $0$:
$$(1+t)^{alpha}=1+alpha x+frac{alpha(alpha-1)}{2!}x^2+frac{alpha(alpha-1)(alpha-2)}{3!}x^3+cdots$$
Put $t=x^2$, $alpha=-1/2$, the Taylor series of $f'(x)=(1+x^2)^{-frac12}$ around $0$ is given by
$$f'(x)=1-frac12x^2+cdots+frac{-frac12(-frac12-1)(-frac12-2)}{3!}x^6+cdots$$
Take the sixth derivative and put $x=0$,
$$f^{(7)}(0)=left.frac{d^6}{dx^6}f'(x)right|_{x=0}=0+frac{-frac12(-frac12-1)(-frac12-2)}{3!}(6!)+0.$$
answered Nov 28 '18 at 2:28
Tianlalu
3,09621038
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is not difficult to find the Taylor series of $(1-t)^{alpha}$ around $0$:
$$(1+t)^{alpha}=1+alpha x+frac{alpha(alpha-1)}{2!}x^2+frac{alpha(alpha-1)(alpha-2)}{3!}x^3+cdots$$
Put $t=x^2$, $alpha=-1/2$, the Taylor series of $f'(x)=(1+x^2)^{-frac12}$ around $0$ is given by
$$f'(x)=1-frac12x^2+cdots+frac{-frac12(-frac12-1)(-frac12-2)}{3!}x^6+cdots$$
Take the sixth derivative and put $x=0$,
$$f^{(7)}(0)=left.frac{d^6}{dx^6}f'(x)right|_{x=0}=0+frac{-frac12(-frac12-1)(-frac12-2)}{3!}(6!)+0.$$
answered Nov 28 '18 at 2:28
Tianlalu
3,09621038
add a comment |
It is not difficult to find the Taylor series of $(1-t)^{alpha}$ around $0$:
$$(1+t)^{alpha}=1+alpha x+frac{alpha(alpha-1)}{2!}x^2+frac{alpha(alpha-1)(alpha-2)}{3!}x^3+cdots$$
Put $t=x^2$, $alpha=-1/2$, the Taylor series of $f'(x)=(1+x^2)^{-frac12}$ around $0$ is given by
$$f'(x)=1-frac12x^2+cdots+frac{-frac12(-frac12-1)(-frac12-2)}{3!}x^6+cdots$$
Take the sixth derivative and put $x=0$,
$$f^{(7)}(0)=left.frac{d^6}{dx^6}f'(x)right|_{x=0}=0+frac{-frac12(-frac12-1)(-frac12-2)}{3!}(6!)+0.$$
answered Nov 28 '18 at 2:28
Tianlalu
3,09621038
add a comment |
It is not difficult to find the Taylor series of $(1-t)^{alpha}$ around $0$:
$$(1+t)^{alpha}=1+alpha x+frac{alpha(alpha-1)}{2!}x^2+frac{alpha(alpha-1)(alpha-2)}{3!}x^3+cdots$$
Put $t=x^2$, $alpha=-1/2$, the Taylor series of $f'(x)=(1+x^2)^{-frac12}$ around $0$ is given by
$$f'(x)=1-frac12x^2+cdots+frac{-frac12(-frac12-1)(-frac12-2)}{3!}x^6+cdots$$
Take the sixth derivative and put $x=0$,
$$f^{(7)}(0)=left.frac{d^6}{dx^6}f'(x)right|_{x=0}=0+frac{-frac12(-frac12-1)(-frac12-2)}{3!}(6!)+0.$$
answered Nov 28 '18 at 2:28
Tianlalu
3,09621038
It is not difficult to find the Taylor series of $(1-t)^{alpha}$ around $0$:
$$(1+t)^{alpha}=1+alpha x+frac{alpha(alpha-1)}{2!}x^2+frac{alpha(alpha-1)(alpha-2)}{3!}x^3+cdots$$
Put $t=x^2$, $alpha=-1/2$, the Taylor series of $f'(x)=(1+x^2)^{-frac12}$ around $0$ is given by
$$f'(x)=1-frac12x^2+cdots+frac{-frac12(-frac12-1)(-frac12-2)}{3!}x^6+cdots$$
Take the sixth derivative and put $x=0$,
$$f^{(7)}(0)=left.frac{d^6}{dx^6}f'(x)right|_{x=0}=0+frac{-frac12(-frac12-1)(-frac12-2)}{3!}(6!)+0.$$
answered Nov 28 '18 at 2:28
Tianlalu
3,09621038
answered Nov 28 '18 at 2:28
Tianlalu
3,09621038
answered Nov 28 '18 at 2:28
Tianlalu
3,09621038
answered Nov 28 '18 at 2:28
Tianlalu
3,09621038
3,09621038
add a comment |
add a comment |
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If you know the binomial expansion, then you can get this easily.
– Sean Roberson
Nov 28 '18 at 2:08
@SeanRoberson You want me to expand $(1 + x^2)^{-0.5}$ and check for the coefficient of $x^6$ term i.e. $(^{-.5}C_3) times 6! / 7!)$.
– henceproved
Nov 28 '18 at 2:20
@SeanRoberson But how is binomial related to taylor series?
– henceproved
Nov 28 '18 at 2:24