Problem on Maclaurin expansion
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I am very confused by the Maclaurin expansion of power functions. For example trying to find the expansion of $(1+x^{2})^{1/6}$ by computing the derivatives and then dividing by the factorial to find the coefficient is very time consuming because the numbers get too big. I assume I can also use the binomial theorem to do this but the exact way eludes me. I would really appreciate if you could help me with this because I am getting very frustrated.
sequences-and-series
edited Dec 7 '18 at 16:58
Leucippus
19.6k102871
asked Dec 6 '18 at 20:41
Michael P.Michael P.
222
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add a comment |
$begingroup$
I am very confused by the Maclaurin expansion of power functions. For example trying to find the expansion of $(1+x^{2})^{1/6}$ by computing the derivatives and then dividing by the factorial to find the coefficient is very time consuming because the numbers get too big. I assume I can also use the binomial theorem to do this but the exact way eludes me. I would really appreciate if you could help me with this because I am getting very frustrated.
sequences-and-series
edited Dec 7 '18 at 16:58
Leucippus
19.6k102871
asked Dec 6 '18 at 20:41
Michael P.Michael P.
222
$endgroup$
add a comment |
$begingroup$
I am very confused by the Maclaurin expansion of power functions. For example trying to find the expansion of $(1+x^{2})^{1/6}$ by computing the derivatives and then dividing by the factorial to find the coefficient is very time consuming because the numbers get too big. I assume I can also use the binomial theorem to do this but the exact way eludes me. I would really appreciate if you could help me with this because I am getting very frustrated.
sequences-and-series
edited Dec 7 '18 at 16:58
Leucippus
19.6k102871
asked Dec 6 '18 at 20:41
Michael P.Michael P.
222
$endgroup$
I am very confused by the Maclaurin expansion of power functions. For example trying to find the expansion of $(1+x^{2})^{1/6}$ by computing the derivatives and then dividing by the factorial to find the coefficient is very time consuming because the numbers get too big. I assume I can also use the binomial theorem to do this but the exact way eludes me. I would really appreciate if you could help me with this because I am getting very frustrated.
sequences-and-series
sequences-and-series
edited Dec 7 '18 at 16:58
Leucippus
19.6k102871
asked Dec 6 '18 at 20:41
Michael P.Michael P.
222
edited Dec 7 '18 at 16:58
Leucippus
19.6k102871
asked Dec 6 '18 at 20:41
Michael P.Michael P.
222
edited Dec 7 '18 at 16:58
Leucippus
19.6k102871
edited Dec 7 '18 at 16:58
Leucippus
19.6k102871
edited Dec 7 '18 at 16:58
Leucippus
19.6k102871
19.6k102871
asked Dec 6 '18 at 20:41
Michael P.Michael P.
222
asked Dec 6 '18 at 20:41
Michael P.Michael P.
222
asked Dec 6 '18 at 20:41
Michael P.Michael P.
222
222
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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The easiest way is to use Newton’s Generalized Binomial Theorem to evaluate the function.$$(1+x)^n=1+nx+frac {n(n-1)}{2!}x^2+frac {n(n-1)(n-2)}{3!}x^3+cdots$$Now replace $x$ with $x^2$ and set $n$ equal to $1/6$.
answered Dec 6 '18 at 20:50
Frank W.Frank W.
3,5121321
$endgroup$
add a comment |
$begingroup$
hint
expand just$$f(X)=(1+X)^frac 16$$
and in the end, replace $X$ by $x^2$.
$$f'(0)=frac 16$$
$$f''(0)=frac 16 (frac 16 -1)$$
....
answered Dec 6 '18 at 20:47
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
$begingroup$
I was confident that this was the way.My problem is mainly on how to expand (1+x)^1/6.
$endgroup$
– Michael P.
Dec 6 '18 at 20:49
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@MichaelP. The derivatives are not difficult to get.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:50
$begingroup$
Isn't $f'(0) = 0$ in the original (with the $x^2$)?
$endgroup$
– Makina
Dec 6 '18 at 20:52
$begingroup$
@Makina $f(x)=(1+x)^frac 16$ not $(1+x^2)^frac 16$.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:54
$begingroup$
Ah ok I see what you did there
$endgroup$
– Makina
Dec 6 '18 at 20:54
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The easiest way is to use Newton’s Generalized Binomial Theorem to evaluate the function.$$(1+x)^n=1+nx+frac {n(n-1)}{2!}x^2+frac {n(n-1)(n-2)}{3!}x^3+cdots$$Now replace $x$ with $x^2$ and set $n$ equal to $1/6$.
answered Dec 6 '18 at 20:50
Frank W.Frank W.
3,5121321
$endgroup$
add a comment |
$begingroup$
The easiest way is to use Newton’s Generalized Binomial Theorem to evaluate the function.$$(1+x)^n=1+nx+frac {n(n-1)}{2!}x^2+frac {n(n-1)(n-2)}{3!}x^3+cdots$$Now replace $x$ with $x^2$ and set $n$ equal to $1/6$.
answered Dec 6 '18 at 20:50
Frank W.Frank W.
3,5121321
$endgroup$
add a comment |
$begingroup$
The easiest way is to use Newton’s Generalized Binomial Theorem to evaluate the function.$$(1+x)^n=1+nx+frac {n(n-1)}{2!}x^2+frac {n(n-1)(n-2)}{3!}x^3+cdots$$Now replace $x$ with $x^2$ and set $n$ equal to $1/6$.
answered Dec 6 '18 at 20:50
Frank W.Frank W.
3,5121321
$endgroup$
The easiest way is to use Newton’s Generalized Binomial Theorem to evaluate the function.$$(1+x)^n=1+nx+frac {n(n-1)}{2!}x^2+frac {n(n-1)(n-2)}{3!}x^3+cdots$$Now replace $x$ with $x^2$ and set $n$ equal to $1/6$.
answered Dec 6 '18 at 20:50
Frank W.Frank W.
3,5121321
answered Dec 6 '18 at 20:50
Frank W.Frank W.
3,5121321
answered Dec 6 '18 at 20:50
Frank W.Frank W.
3,5121321
answered Dec 6 '18 at 20:50
Frank W.Frank W.
3,5121321
3,5121321
add a comment |
add a comment |
$begingroup$
hint
expand just$$f(X)=(1+X)^frac 16$$
and in the end, replace $X$ by $x^2$.
$$f'(0)=frac 16$$
$$f''(0)=frac 16 (frac 16 -1)$$
....
answered Dec 6 '18 at 20:47
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
$begingroup$
I was confident that this was the way.My problem is mainly on how to expand (1+x)^1/6.
$endgroup$
– Michael P.
Dec 6 '18 at 20:49
$begingroup$
@MichaelP. The derivatives are not difficult to get.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:50
$begingroup$
Isn't $f'(0) = 0$ in the original (with the $x^2$)?
$endgroup$
– Makina
Dec 6 '18 at 20:52
$begingroup$
@Makina $f(x)=(1+x)^frac 16$ not $(1+x^2)^frac 16$.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:54
$begingroup$
Ah ok I see what you did there
$endgroup$
– Makina
Dec 6 '18 at 20:54
add a comment |
$begingroup$
hint
expand just$$f(X)=(1+X)^frac 16$$
and in the end, replace $X$ by $x^2$.
$$f'(0)=frac 16$$
$$f''(0)=frac 16 (frac 16 -1)$$
....
answered Dec 6 '18 at 20:47
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
$begingroup$
I was confident that this was the way.My problem is mainly on how to expand (1+x)^1/6.
$endgroup$
– Michael P.
Dec 6 '18 at 20:49
$begingroup$
@MichaelP. The derivatives are not difficult to get.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:50
$begingroup$
Isn't $f'(0) = 0$ in the original (with the $x^2$)?
$endgroup$
– Makina
Dec 6 '18 at 20:52
$begingroup$
@Makina $f(x)=(1+x)^frac 16$ not $(1+x^2)^frac 16$.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:54
$begingroup$
Ah ok I see what you did there
$endgroup$
– Makina
Dec 6 '18 at 20:54
add a comment |
$begingroup$
hint
expand just$$f(X)=(1+X)^frac 16$$
and in the end, replace $X$ by $x^2$.
$$f'(0)=frac 16$$
$$f''(0)=frac 16 (frac 16 -1)$$
....
answered Dec 6 '18 at 20:47
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
hint
expand just$$f(X)=(1+X)^frac 16$$
and in the end, replace $X$ by $x^2$.
$$f'(0)=frac 16$$
$$f''(0)=frac 16 (frac 16 -1)$$
....
answered Dec 6 '18 at 20:47
hamam_Abdallahhamam_Abdallah
38k21634
edited Dec 6 '18 at 20:51
answered Dec 6 '18 at 20:47
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 6 '18 at 20:47
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 6 '18 at 20:47
hamam_Abdallahhamam_Abdallah
38k21634
38k21634
$begingroup$
I was confident that this was the way.My problem is mainly on how to expand (1+x)^1/6.
$endgroup$
– Michael P.
Dec 6 '18 at 20:49
$begingroup$
@MichaelP. The derivatives are not difficult to get.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:50
$begingroup$
Isn't $f'(0) = 0$ in the original (with the $x^2$)?
$endgroup$
– Makina
Dec 6 '18 at 20:52
$begingroup$
@Makina $f(x)=(1+x)^frac 16$ not $(1+x^2)^frac 16$.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:54
$begingroup$
Ah ok I see what you did there
$endgroup$
– Makina
Dec 6 '18 at 20:54
add a comment |
$begingroup$
I was confident that this was the way.My problem is mainly on how to expand (1+x)^1/6.
$endgroup$
– Michael P.
Dec 6 '18 at 20:49
$begingroup$
@MichaelP. The derivatives are not difficult to get.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:50
$begingroup$
Isn't $f'(0) = 0$ in the original (with the $x^2$)?
$endgroup$
– Makina
Dec 6 '18 at 20:52
$begingroup$
@Makina $f(x)=(1+x)^frac 16$ not $(1+x^2)^frac 16$.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:54
$begingroup$
Ah ok I see what you did there
$endgroup$
– Makina
Dec 6 '18 at 20:54
$begingroup$
I was confident that this was the way.My problem is mainly on how to expand (1+x)^1/6.
$endgroup$
– Michael P.
Dec 6 '18 at 20:49
$begingroup$
I was confident that this was the way.My problem is mainly on how to expand (1+x)^1/6.
$endgroup$
– Michael P.
Dec 6 '18 at 20:49
$begingroup$
@MichaelP. The derivatives are not difficult to get.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:50
$begingroup$
@MichaelP. The derivatives are not difficult to get.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:50
$begingroup$
Isn't $f'(0) = 0$ in the original (with the $x^2$)?
$endgroup$
– Makina
Dec 6 '18 at 20:52
$begingroup$
Isn't $f'(0) = 0$ in the original (with the $x^2$)?
$endgroup$
– Makina
Dec 6 '18 at 20:52
$begingroup$
@Makina $f(x)=(1+x)^frac 16$ not $(1+x^2)^frac 16$.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:54
$begingroup$
@Makina $f(x)=(1+x)^frac 16$ not $(1+x^2)^frac 16$.
$endgroup$
– hamam_Abdallah
Dec 6 '18 at 20:54
$begingroup$
Ah ok I see what you did there
$endgroup$
– Makina
Dec 6 '18 at 20:54
$begingroup$
Ah ok I see what you did there
$endgroup$
– Makina
Dec 6 '18 at 20:54
add a comment |
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