Simplification of $ sqrt{(1-x^2)}$ to $(1-frac{x^2}{2})$
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While following a proof from an electrical engineering book (Design of Analog CMOS Integrated Circuits, second edition from Behzad Razavi ), I came across a simplification which I found curious. In equations 14.18 to 14.19 they state that the following holds for small values of $x$:
$$
sqrt{1-x^2} approx left(1-frac{x^2}{2}right)
$$
I can see that this appears to be the case after simulating this in matlab but it seems unintuitive to me, and I was wondering if anyone here knows the kind of mathematical terms I can use to find some kind of proof for this (or the proof itself).
calculus approximation
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add a comment |
$begingroup$
While following a proof from an electrical engineering book (Design of Analog CMOS Integrated Circuits, second edition from Behzad Razavi ), I came across a simplification which I found curious. In equations 14.18 to 14.19 they state that the following holds for small values of $x$:
$$
sqrt{1-x^2} approx left(1-frac{x^2}{2}right)
$$
I can see that this appears to be the case after simulating this in matlab but it seems unintuitive to me, and I was wondering if anyone here knows the kind of mathematical terms I can use to find some kind of proof for this (or the proof itself).
calculus approximation
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3
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It's Taylor series of first order - $sqrt{1-x}=1-x+text{O}(x^2)$ or linear approximation.
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– Galc127
Dec 16 '18 at 13:52
add a comment |
$begingroup$
While following a proof from an electrical engineering book (Design of Analog CMOS Integrated Circuits, second edition from Behzad Razavi ), I came across a simplification which I found curious. In equations 14.18 to 14.19 they state that the following holds for small values of $x$:
$$
sqrt{1-x^2} approx left(1-frac{x^2}{2}right)
$$
I can see that this appears to be the case after simulating this in matlab but it seems unintuitive to me, and I was wondering if anyone here knows the kind of mathematical terms I can use to find some kind of proof for this (or the proof itself).
calculus approximation
$endgroup$
While following a proof from an electrical engineering book (Design of Analog CMOS Integrated Circuits, second edition from Behzad Razavi ), I came across a simplification which I found curious. In equations 14.18 to 14.19 they state that the following holds for small values of $x$:
$$
sqrt{1-x^2} approx left(1-frac{x^2}{2}right)
$$
I can see that this appears to be the case after simulating this in matlab but it seems unintuitive to me, and I was wondering if anyone here knows the kind of mathematical terms I can use to find some kind of proof for this (or the proof itself).
calculus approximation
calculus approximation
edited Dec 16 '18 at 14:00
Viktor Glombik
9381527
9381527
asked Dec 16 '18 at 13:48
eosuseosus
82
82
3
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It's Taylor series of first order - $sqrt{1-x}=1-x+text{O}(x^2)$ or linear approximation.
$endgroup$
– Galc127
Dec 16 '18 at 13:52
add a comment |
3
$begingroup$
It's Taylor series of first order - $sqrt{1-x}=1-x+text{O}(x^2)$ or linear approximation.
$endgroup$
– Galc127
Dec 16 '18 at 13:52
3
3
$begingroup$
It's Taylor series of first order - $sqrt{1-x}=1-x+text{O}(x^2)$ or linear approximation.
$endgroup$
– Galc127
Dec 16 '18 at 13:52
$begingroup$
It's Taylor series of first order - $sqrt{1-x}=1-x+text{O}(x^2)$ or linear approximation.
$endgroup$
– Galc127
Dec 16 '18 at 13:52
add a comment |
3 Answers
3
active
oldest
votes
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Term to look for: linear approximation
In general, the best linear approximation for a differentiable function near a point $c$ is
$$
f(x) approx f(c) + f'(c);(x-c)
$$
This is essentially the definition of the derivative. And you should find this in your calculus book soon after the definition of derivative.
Now if $f(x) = sqrt{1-x}$ and $c=0$, we get $f(0)=1$ and $f'(0)=-frac{1}{2}$. So
$$
sqrt{1-x} approx 1 - frac{x}{2}
$$
To get your case, substitute $x^2$ for $x$.
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1
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I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
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– Botond
Dec 16 '18 at 14:04
add a comment |
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Every smooth function can be locally approximated by its tangent (as a consequence of Taylor's theorem).
$$sqrt{1-t}approx 1-frac t2.$$
Hence for small $x$,
$$sqrt{1-x^2}approx 1-frac{x^2}2.$$
The next approximation order is parabolic, corresponding to the "osculatrix parabola" (i.e. same tangent and same curvature)
$$sqrt{1-t}approx 1-frac t2-frac{t^2}8,$$
and
$$sqrt{1-x^2}approx 1-frac{x^2}2-frac{x^4}8,$$
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add a comment |
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Use the Taylor series expansion at $x=0$ to get $sqrt{1-x^2}approx1-frac{x^2}2+o(x^4)$.
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
Term to look for: linear approximation
In general, the best linear approximation for a differentiable function near a point $c$ is
$$
f(x) approx f(c) + f'(c);(x-c)
$$
This is essentially the definition of the derivative. And you should find this in your calculus book soon after the definition of derivative.
Now if $f(x) = sqrt{1-x}$ and $c=0$, we get $f(0)=1$ and $f'(0)=-frac{1}{2}$. So
$$
sqrt{1-x} approx 1 - frac{x}{2}
$$
To get your case, substitute $x^2$ for $x$.
$endgroup$
1
$begingroup$
I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
$endgroup$
– Botond
Dec 16 '18 at 14:04
add a comment |
$begingroup$
Term to look for: linear approximation
In general, the best linear approximation for a differentiable function near a point $c$ is
$$
f(x) approx f(c) + f'(c);(x-c)
$$
This is essentially the definition of the derivative. And you should find this in your calculus book soon after the definition of derivative.
Now if $f(x) = sqrt{1-x}$ and $c=0$, we get $f(0)=1$ and $f'(0)=-frac{1}{2}$. So
$$
sqrt{1-x} approx 1 - frac{x}{2}
$$
To get your case, substitute $x^2$ for $x$.
$endgroup$
1
$begingroup$
I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
$endgroup$
– Botond
Dec 16 '18 at 14:04
add a comment |
$begingroup$
Term to look for: linear approximation
In general, the best linear approximation for a differentiable function near a point $c$ is
$$
f(x) approx f(c) + f'(c);(x-c)
$$
This is essentially the definition of the derivative. And you should find this in your calculus book soon after the definition of derivative.
Now if $f(x) = sqrt{1-x}$ and $c=0$, we get $f(0)=1$ and $f'(0)=-frac{1}{2}$. So
$$
sqrt{1-x} approx 1 - frac{x}{2}
$$
To get your case, substitute $x^2$ for $x$.
$endgroup$
Term to look for: linear approximation
In general, the best linear approximation for a differentiable function near a point $c$ is
$$
f(x) approx f(c) + f'(c);(x-c)
$$
This is essentially the definition of the derivative. And you should find this in your calculus book soon after the definition of derivative.
Now if $f(x) = sqrt{1-x}$ and $c=0$, we get $f(0)=1$ and $f'(0)=-frac{1}{2}$. So
$$
sqrt{1-x} approx 1 - frac{x}{2}
$$
To get your case, substitute $x^2$ for $x$.
edited Dec 16 '18 at 13:58
answered Dec 16 '18 at 13:53
GEdgarGEdgar
62.5k267171
62.5k267171
1
$begingroup$
I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
$endgroup$
– Botond
Dec 16 '18 at 14:04
add a comment |
1
$begingroup$
I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
$endgroup$
– Botond
Dec 16 '18 at 14:04
1
1
$begingroup$
I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
$endgroup$
– Botond
Dec 16 '18 at 14:04
$begingroup$
I think you could go a bit further and show that $(1+x)^r approx 1+rx$.
$endgroup$
– Botond
Dec 16 '18 at 14:04
add a comment |
$begingroup$
Every smooth function can be locally approximated by its tangent (as a consequence of Taylor's theorem).
$$sqrt{1-t}approx 1-frac t2.$$
Hence for small $x$,
$$sqrt{1-x^2}approx 1-frac{x^2}2.$$
The next approximation order is parabolic, corresponding to the "osculatrix parabola" (i.e. same tangent and same curvature)
$$sqrt{1-t}approx 1-frac t2-frac{t^2}8,$$
and
$$sqrt{1-x^2}approx 1-frac{x^2}2-frac{x^4}8,$$
$endgroup$
add a comment |
$begingroup$
Every smooth function can be locally approximated by its tangent (as a consequence of Taylor's theorem).
$$sqrt{1-t}approx 1-frac t2.$$
Hence for small $x$,
$$sqrt{1-x^2}approx 1-frac{x^2}2.$$
The next approximation order is parabolic, corresponding to the "osculatrix parabola" (i.e. same tangent and same curvature)
$$sqrt{1-t}approx 1-frac t2-frac{t^2}8,$$
and
$$sqrt{1-x^2}approx 1-frac{x^2}2-frac{x^4}8,$$
$endgroup$
add a comment |
$begingroup$
Every smooth function can be locally approximated by its tangent (as a consequence of Taylor's theorem).
$$sqrt{1-t}approx 1-frac t2.$$
Hence for small $x$,
$$sqrt{1-x^2}approx 1-frac{x^2}2.$$
The next approximation order is parabolic, corresponding to the "osculatrix parabola" (i.e. same tangent and same curvature)
$$sqrt{1-t}approx 1-frac t2-frac{t^2}8,$$
and
$$sqrt{1-x^2}approx 1-frac{x^2}2-frac{x^4}8,$$
$endgroup$
Every smooth function can be locally approximated by its tangent (as a consequence of Taylor's theorem).
$$sqrt{1-t}approx 1-frac t2.$$
Hence for small $x$,
$$sqrt{1-x^2}approx 1-frac{x^2}2.$$
The next approximation order is parabolic, corresponding to the "osculatrix parabola" (i.e. same tangent and same curvature)
$$sqrt{1-t}approx 1-frac t2-frac{t^2}8,$$
and
$$sqrt{1-x^2}approx 1-frac{x^2}2-frac{x^4}8,$$
edited Dec 16 '18 at 14:10
answered Dec 16 '18 at 14:00
Yves DaoustYves Daoust
128k675227
128k675227
add a comment |
add a comment |
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Use the Taylor series expansion at $x=0$ to get $sqrt{1-x^2}approx1-frac{x^2}2+o(x^4)$.
$endgroup$
add a comment |
$begingroup$
Use the Taylor series expansion at $x=0$ to get $sqrt{1-x^2}approx1-frac{x^2}2+o(x^4)$.
$endgroup$
add a comment |
$begingroup$
Use the Taylor series expansion at $x=0$ to get $sqrt{1-x^2}approx1-frac{x^2}2+o(x^4)$.
$endgroup$
Use the Taylor series expansion at $x=0$ to get $sqrt{1-x^2}approx1-frac{x^2}2+o(x^4)$.
edited Dec 16 '18 at 14:18
answered Dec 16 '18 at 13:59
Chris CusterChris Custer
13.7k3827
13.7k3827
add a comment |
add a comment |
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It's Taylor series of first order - $sqrt{1-x}=1-x+text{O}(x^2)$ or linear approximation.
$endgroup$
– Galc127
Dec 16 '18 at 13:52