Mixing
Infinitesimal order with Taylor series
$begingroup$
I'm here again with another exercise (after this one) on Taylor series: determine the infinitesimal order for $xto 0$ of the function
$$
f(x)=sqrt{frac{x^4+x^5}{x-sin{x}}} .
$$
The problem is that, after having developed the denominator as $frac{x^3}{6}+o(x^3)$, I have difficulties in determining the Taylor series of the whole function, as $f$ is not differentiable in $x=0$.
Thank you for your help!
limits taylor-expansion infinitesimals
asked Jan 24 at 15:15
Fabio OriFabio Ori
161
$endgroup$
add a comment |
$begingroup$
I'm here again with another exercise (after this one) on Taylor series: determine the infinitesimal order for $xto 0$ of the function
$$
f(x)=sqrt{frac{x^4+x^5}{x-sin{x}}} .
$$
The problem is that, after having developed the denominator as $frac{x^3}{6}+o(x^3)$, I have difficulties in determining the Taylor series of the whole function, as $f$ is not differentiable in $x=0$.
Thank you for your help!
limits taylor-expansion infinitesimals
asked Jan 24 at 15:15
Fabio OriFabio Ori
161
$endgroup$
1
$begingroup$
If you just want the infinitesimal order that is easy: the top is 4, the bottom is 3, so the inside is 1, which makes the square root give 1/2 at the end. If you actually want to expand in (rational, not integer) powers of $x$ then you should factor $sqrt{x}$ out of the square root and then expand what's left.
$endgroup$
– Ian
Jan 24 at 15:40
add a comment |
$begingroup$
I'm here again with another exercise (after this one) on Taylor series: determine the infinitesimal order for $xto 0$ of the function
$$
f(x)=sqrt{frac{x^4+x^5}{x-sin{x}}} .
$$
The problem is that, after having developed the denominator as $frac{x^3}{6}+o(x^3)$, I have difficulties in determining the Taylor series of the whole function, as $f$ is not differentiable in $x=0$.
Thank you for your help!
limits taylor-expansion infinitesimals
asked Jan 24 at 15:15
Fabio OriFabio Ori
161
$endgroup$
I'm here again with another exercise (after this one) on Taylor series: determine the infinitesimal order for $xto 0$ of the function
$$
f(x)=sqrt{frac{x^4+x^5}{x-sin{x}}} .
$$
The problem is that, after having developed the denominator as $frac{x^3}{6}+o(x^3)$, I have difficulties in determining the Taylor series of the whole function, as $f$ is not differentiable in $x=0$.
Thank you for your help!
limits taylor-expansion infinitesimals
limits taylor-expansion infinitesimals
asked Jan 24 at 15:15
Fabio OriFabio Ori
161
asked Jan 24 at 15:15
Fabio OriFabio Ori
161
asked Jan 24 at 15:15
Fabio OriFabio Ori
161
asked Jan 24 at 15:15
Fabio OriFabio Ori
161
asked Jan 24 at 15:15
Fabio OriFabio Ori
161
161
1
$begingroup$
If you just want the infinitesimal order that is easy: the top is 4, the bottom is 3, so the inside is 1, which makes the square root give 1/2 at the end. If you actually want to expand in (rational, not integer) powers of $x$ then you should factor $sqrt{x}$ out of the square root and then expand what's left.
$endgroup$
– Ian
Jan 24 at 15:40
add a comment |
1
$begingroup$
If you just want the infinitesimal order that is easy: the top is 4, the bottom is 3, so the inside is 1, which makes the square root give 1/2 at the end. If you actually want to expand in (rational, not integer) powers of $x$ then you should factor $sqrt{x}$ out of the square root and then expand what's left.
$endgroup$
– Ian
Jan 24 at 15:40
1
1
$begingroup$
If you just want the infinitesimal order that is easy: the top is 4, the bottom is 3, so the inside is 1, which makes the square root give 1/2 at the end. If you actually want to expand in (rational, not integer) powers of $x$ then you should factor $sqrt{x}$ out of the square root and then expand what's left.
$endgroup$
– Ian
Jan 24 at 15:40
$begingroup$
If you just want the infinitesimal order that is easy: the top is 4, the bottom is 3, so the inside is 1, which makes the square root give 1/2 at the end. If you actually want to expand in (rational, not integer) powers of $x$ then you should factor $sqrt{x}$ out of the square root and then expand what's left.
$endgroup$
– Ian
Jan 24 at 15:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Using steps
$$sin(x)=x-frac{x^3}{6}+frac{x^5}{120}-frac{x^7}{5040}+Oleft(x^{9}right)$$
$$x-sin(x)=frac{x^3}{6}-frac{x^5}{120}+frac{x^7}{5040}+Oleft(x^{9}right)$$
$$frac{x^4+x^5}{x-sin{x}}=frac{x^4+x^5}{frac{x^3}{6}-frac{x^5}{120}+frac{x^7}{5040}+Oleft(x^{11}right)}=6 x+6 x^2+frac{3 x^3}{10}+Oleft(x^{4}right)$$
$$sqrt{frac{x^4+x^5}{x-sin{x}}}=sqrt{6} sqrt{x}+sqrt{frac{3}{2}} x^{3/2}+Oleft(x^{5/2}right)$$
answered Jan 25 at 4:35
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Using steps
$$sin(x)=x-frac{x^3}{6}+frac{x^5}{120}-frac{x^7}{5040}+Oleft(x^{9}right)$$
$$x-sin(x)=frac{x^3}{6}-frac{x^5}{120}+frac{x^7}{5040}+Oleft(x^{9}right)$$
$$frac{x^4+x^5}{x-sin{x}}=frac{x^4+x^5}{frac{x^3}{6}-frac{x^5}{120}+frac{x^7}{5040}+Oleft(x^{11}right)}=6 x+6 x^2+frac{3 x^3}{10}+Oleft(x^{4}right)$$
$$sqrt{frac{x^4+x^5}{x-sin{x}}}=sqrt{6} sqrt{x}+sqrt{frac{3}{2}} x^{3/2}+Oleft(x^{5/2}right)$$
answered Jan 25 at 4:35
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
Using steps
$$sin(x)=x-frac{x^3}{6}+frac{x^5}{120}-frac{x^7}{5040}+Oleft(x^{9}right)$$
$$x-sin(x)=frac{x^3}{6}-frac{x^5}{120}+frac{x^7}{5040}+Oleft(x^{9}right)$$
$$frac{x^4+x^5}{x-sin{x}}=frac{x^4+x^5}{frac{x^3}{6}-frac{x^5}{120}+frac{x^7}{5040}+Oleft(x^{11}right)}=6 x+6 x^2+frac{3 x^3}{10}+Oleft(x^{4}right)$$
$$sqrt{frac{x^4+x^5}{x-sin{x}}}=sqrt{6} sqrt{x}+sqrt{frac{3}{2}} x^{3/2}+Oleft(x^{5/2}right)$$
answered Jan 25 at 4:35
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
Using steps
$$sin(x)=x-frac{x^3}{6}+frac{x^5}{120}-frac{x^7}{5040}+Oleft(x^{9}right)$$
$$x-sin(x)=frac{x^3}{6}-frac{x^5}{120}+frac{x^7}{5040}+Oleft(x^{9}right)$$
$$frac{x^4+x^5}{x-sin{x}}=frac{x^4+x^5}{frac{x^3}{6}-frac{x^5}{120}+frac{x^7}{5040}+Oleft(x^{11}right)}=6 x+6 x^2+frac{3 x^3}{10}+Oleft(x^{4}right)$$
$$sqrt{frac{x^4+x^5}{x-sin{x}}}=sqrt{6} sqrt{x}+sqrt{frac{3}{2}} x^{3/2}+Oleft(x^{5/2}right)$$
answered Jan 25 at 4:35
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
Using steps
$$sin(x)=x-frac{x^3}{6}+frac{x^5}{120}-frac{x^7}{5040}+Oleft(x^{9}right)$$
$$x-sin(x)=frac{x^3}{6}-frac{x^5}{120}+frac{x^7}{5040}+Oleft(x^{9}right)$$
$$frac{x^4+x^5}{x-sin{x}}=frac{x^4+x^5}{frac{x^3}{6}-frac{x^5}{120}+frac{x^7}{5040}+Oleft(x^{11}right)}=6 x+6 x^2+frac{3 x^3}{10}+Oleft(x^{4}right)$$
$$sqrt{frac{x^4+x^5}{x-sin{x}}}=sqrt{6} sqrt{x}+sqrt{frac{3}{2}} x^{3/2}+Oleft(x^{5/2}right)$$
answered Jan 25 at 4:35
Claude LeiboviciClaude Leibovici
124k1157135
answered Jan 25 at 4:35
Claude LeiboviciClaude Leibovici
124k1157135
answered Jan 25 at 4:35
Claude LeiboviciClaude Leibovici
124k1157135
answered Jan 25 at 4:35
Claude LeiboviciClaude Leibovici
124k1157135
124k1157135
add a comment |
add a comment |
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$begingroup$
If you just want the infinitesimal order that is easy: the top is 4, the bottom is 3, so the inside is 1, which makes the square root give 1/2 at the end. If you actually want to expand in (rational, not integer) powers of $x$ then you should factor $sqrt{x}$ out of the square root and then expand what's left.
$endgroup$
– Ian
Jan 24 at 15:40