Verifying that $sqrt{(R+jomega L)(jomega C)}=0.5frac{R}{sqrt{L/C}}+jomegasqrt{LC}$ for $Rlll omega L$
$begingroup$
So I have the following square root of this two complex numbers and my book provides this:
$$sqrt{(R+jomega L)(jomega C)}=0.5frac{R}{sqrt{frac{L}{C}}}+jomegasqrt{LC}$$
if $$Rlllomega L$$
I have no freaking idea how they do this mathematically. I tried to apply distributive property, which leads to
$$sqrt{jRomega C-omega^2LC}$$
And the second term of my expansion kind of looks like the second term of the expression
$$sqrt{-omega^2LC}=jomegasqrt{LC}$$
But I don't if (a) this is correct and (b) how do I get the first term.
Thanks in advance.
complex-numbers radicals
edited Dec 16 '18 at 18:51
Blue
47.8k870152
asked Dec 16 '18 at 17:37
Granger ObliviateGranger Obliviate
557415
$endgroup$
add a comment |
$begingroup$
So I have the following square root of this two complex numbers and my book provides this:
$$sqrt{(R+jomega L)(jomega C)}=0.5frac{R}{sqrt{frac{L}{C}}}+jomegasqrt{LC}$$
if $$Rlllomega L$$
I have no freaking idea how they do this mathematically. I tried to apply distributive property, which leads to
$$sqrt{jRomega C-omega^2LC}$$
And the second term of my expansion kind of looks like the second term of the expression
$$sqrt{-omega^2LC}=jomegasqrt{LC}$$
But I don't if (a) this is correct and (b) how do I get the first term.
Thanks in advance.
complex-numbers radicals
edited Dec 16 '18 at 18:51
Blue
47.8k870152
asked Dec 16 '18 at 17:37
Granger ObliviateGranger Obliviate
557415
$endgroup$
add a comment |
$begingroup$
So I have the following square root of this two complex numbers and my book provides this:
$$sqrt{(R+jomega L)(jomega C)}=0.5frac{R}{sqrt{frac{L}{C}}}+jomegasqrt{LC}$$
if $$Rlllomega L$$
I have no freaking idea how they do this mathematically. I tried to apply distributive property, which leads to
$$sqrt{jRomega C-omega^2LC}$$
And the second term of my expansion kind of looks like the second term of the expression
$$sqrt{-omega^2LC}=jomegasqrt{LC}$$
But I don't if (a) this is correct and (b) how do I get the first term.
Thanks in advance.
complex-numbers radicals
edited Dec 16 '18 at 18:51
Blue
47.8k870152
asked Dec 16 '18 at 17:37
Granger ObliviateGranger Obliviate
557415
$endgroup$
So I have the following square root of this two complex numbers and my book provides this:
$$sqrt{(R+jomega L)(jomega C)}=0.5frac{R}{sqrt{frac{L}{C}}}+jomegasqrt{LC}$$
if $$Rlllomega L$$
I have no freaking idea how they do this mathematically. I tried to apply distributive property, which leads to
$$sqrt{jRomega C-omega^2LC}$$
And the second term of my expansion kind of looks like the second term of the expression
$$sqrt{-omega^2LC}=jomegasqrt{LC}$$
But I don't if (a) this is correct and (b) how do I get the first term.
Thanks in advance.
complex-numbers radicals
complex-numbers radicals
edited Dec 16 '18 at 18:51
Blue
47.8k870152
asked Dec 16 '18 at 17:37
Granger ObliviateGranger Obliviate
557415
edited Dec 16 '18 at 18:51
Blue
47.8k870152
asked Dec 16 '18 at 17:37
Granger ObliviateGranger Obliviate
557415
edited Dec 16 '18 at 18:51
Blue
47.8k870152
edited Dec 16 '18 at 18:51
Blue
47.8k870152
edited Dec 16 '18 at 18:51
Blue
47.8k870152
47.8k870152
asked Dec 16 '18 at 17:37
Granger ObliviateGranger Obliviate
557415
asked Dec 16 '18 at 17:37
Granger ObliviateGranger Obliviate
557415
asked Dec 16 '18 at 17:37
Granger ObliviateGranger Obliviate
557415
557415
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The basic idea is that $sqrt{1+x} approx 1+frac x2$ when $x ll 1$. The right side is the first two terms of the Taylor series. If you expand the left you have
$$sqrt{(R+jomega L)(jomega C)}=sqrt{Rjomega C-omega^2LC}\
=jomegasqrt{LC}sqrt {Rfrac 1{jomega L}+1}\
approx jomega sqrt{LC}left(1+frac {R}{2jomega L}right)\
=jomega sqrt{LC}+frac R2sqrt{frac {C}{L}}$$
They owe you an approximation sign when they do the Taylor series step.
answered Dec 16 '18 at 18:04
Ross MillikanRoss Millikan
293k23197371
$endgroup$
$begingroup$
"They owe you an approximation sign when they do the Taylor series step." - It looks like it's from a physics book. Allowances must be made... :)
$endgroup$
– Matthias
Dec 16 '18 at 18:20
$begingroup$
@Matthias: Actually the $j$ indicates it is probably engineering. I was a physics student and remember getting the approximation signs, but it was many years ago.
$endgroup$
– Ross Millikan
Dec 16 '18 at 18:23
$begingroup$
Thank you for the clear explanation Ross! I forgot about the Taylor series. It is from an engineering book indeed
$endgroup$
– Granger Obliviate
Dec 16 '18 at 19:58
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
$begingroup$
The basic idea is that $sqrt{1+x} approx 1+frac x2$ when $x ll 1$. The right side is the first two terms of the Taylor series. If you expand the left you have
$$sqrt{(R+jomega L)(jomega C)}=sqrt{Rjomega C-omega^2LC}\
=jomegasqrt{LC}sqrt {Rfrac 1{jomega L}+1}\
approx jomega sqrt{LC}left(1+frac {R}{2jomega L}right)\
=jomega sqrt{LC}+frac R2sqrt{frac {C}{L}}$$
They owe you an approximation sign when they do the Taylor series step.
answered Dec 16 '18 at 18:04
Ross MillikanRoss Millikan
293k23197371
$endgroup$
$begingroup$
"They owe you an approximation sign when they do the Taylor series step." - It looks like it's from a physics book. Allowances must be made... :)
$endgroup$
– Matthias
Dec 16 '18 at 18:20
$begingroup$
@Matthias: Actually the $j$ indicates it is probably engineering. I was a physics student and remember getting the approximation signs, but it was many years ago.
$endgroup$
– Ross Millikan
Dec 16 '18 at 18:23
$begingroup$
Thank you for the clear explanation Ross! I forgot about the Taylor series. It is from an engineering book indeed
$endgroup$
– Granger Obliviate
Dec 16 '18 at 19:58
add a comment |
$begingroup$
The basic idea is that $sqrt{1+x} approx 1+frac x2$ when $x ll 1$. The right side is the first two terms of the Taylor series. If you expand the left you have
$$sqrt{(R+jomega L)(jomega C)}=sqrt{Rjomega C-omega^2LC}\
=jomegasqrt{LC}sqrt {Rfrac 1{jomega L}+1}\
approx jomega sqrt{LC}left(1+frac {R}{2jomega L}right)\
=jomega sqrt{LC}+frac R2sqrt{frac {C}{L}}$$
They owe you an approximation sign when they do the Taylor series step.
answered Dec 16 '18 at 18:04
Ross MillikanRoss Millikan
293k23197371
$endgroup$
$begingroup$
"They owe you an approximation sign when they do the Taylor series step." - It looks like it's from a physics book. Allowances must be made... :)
$endgroup$
– Matthias
Dec 16 '18 at 18:20
$begingroup$
@Matthias: Actually the $j$ indicates it is probably engineering. I was a physics student and remember getting the approximation signs, but it was many years ago.
$endgroup$
– Ross Millikan
Dec 16 '18 at 18:23
$begingroup$
Thank you for the clear explanation Ross! I forgot about the Taylor series. It is from an engineering book indeed
$endgroup$
– Granger Obliviate
Dec 16 '18 at 19:58
add a comment |
$begingroup$
The basic idea is that $sqrt{1+x} approx 1+frac x2$ when $x ll 1$. The right side is the first two terms of the Taylor series. If you expand the left you have
$$sqrt{(R+jomega L)(jomega C)}=sqrt{Rjomega C-omega^2LC}\
=jomegasqrt{LC}sqrt {Rfrac 1{jomega L}+1}\
approx jomega sqrt{LC}left(1+frac {R}{2jomega L}right)\
=jomega sqrt{LC}+frac R2sqrt{frac {C}{L}}$$
They owe you an approximation sign when they do the Taylor series step.
answered Dec 16 '18 at 18:04
Ross MillikanRoss Millikan
293k23197371
$endgroup$
The basic idea is that $sqrt{1+x} approx 1+frac x2$ when $x ll 1$. The right side is the first two terms of the Taylor series. If you expand the left you have
$$sqrt{(R+jomega L)(jomega C)}=sqrt{Rjomega C-omega^2LC}\
=jomegasqrt{LC}sqrt {Rfrac 1{jomega L}+1}\
approx jomega sqrt{LC}left(1+frac {R}{2jomega L}right)\
=jomega sqrt{LC}+frac R2sqrt{frac {C}{L}}$$
They owe you an approximation sign when they do the Taylor series step.
answered Dec 16 '18 at 18:04
Ross MillikanRoss Millikan
293k23197371
answered Dec 16 '18 at 18:04
Ross MillikanRoss Millikan
293k23197371
answered Dec 16 '18 at 18:04
Ross MillikanRoss Millikan
293k23197371
answered Dec 16 '18 at 18:04
Ross MillikanRoss Millikan
293k23197371
293k23197371
$begingroup$
"They owe you an approximation sign when they do the Taylor series step." - It looks like it's from a physics book. Allowances must be made... :)
$endgroup$
– Matthias
Dec 16 '18 at 18:20
$begingroup$
@Matthias: Actually the $j$ indicates it is probably engineering. I was a physics student and remember getting the approximation signs, but it was many years ago.
$endgroup$
– Ross Millikan
Dec 16 '18 at 18:23
$begingroup$
Thank you for the clear explanation Ross! I forgot about the Taylor series. It is from an engineering book indeed
$endgroup$
– Granger Obliviate
Dec 16 '18 at 19:58
add a comment |
$begingroup$
"They owe you an approximation sign when they do the Taylor series step." - It looks like it's from a physics book. Allowances must be made... :)
$endgroup$
– Matthias
Dec 16 '18 at 18:20
$begingroup$
@Matthias: Actually the $j$ indicates it is probably engineering. I was a physics student and remember getting the approximation signs, but it was many years ago.
$endgroup$
– Ross Millikan
Dec 16 '18 at 18:23
$begingroup$
Thank you for the clear explanation Ross! I forgot about the Taylor series. It is from an engineering book indeed
$endgroup$
– Granger Obliviate
Dec 16 '18 at 19:58
$begingroup$
"They owe you an approximation sign when they do the Taylor series step." - It looks like it's from a physics book. Allowances must be made... :)
$endgroup$
– Matthias
Dec 16 '18 at 18:20
$begingroup$
"They owe you an approximation sign when they do the Taylor series step." - It looks like it's from a physics book. Allowances must be made... :)
$endgroup$
– Matthias
Dec 16 '18 at 18:20
$begingroup$
@Matthias: Actually the $j$ indicates it is probably engineering. I was a physics student and remember getting the approximation signs, but it was many years ago.
$endgroup$
– Ross Millikan
Dec 16 '18 at 18:23
$begingroup$
@Matthias: Actually the $j$ indicates it is probably engineering. I was a physics student and remember getting the approximation signs, but it was many years ago.
$endgroup$
– Ross Millikan
Dec 16 '18 at 18:23
$begingroup$
Thank you for the clear explanation Ross! I forgot about the Taylor series. It is from an engineering book indeed
$endgroup$
– Granger Obliviate
Dec 16 '18 at 19:58
$begingroup$
Thank you for the clear explanation Ross! I forgot about the Taylor series. It is from an engineering book indeed
$endgroup$
– Granger Obliviate
Dec 16 '18 at 19:58
add a comment |
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