# calculating the square root

Discussion in 'Math' started by suzuki, Apr 10, 2012.

1. ### suzuki Thread Starter Member

Aug 10, 2011
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0
Hi,

i have an equation in the form of

$A^{2} = B^{2}+C \cos(\omega t)$

what is the square root of this equation?

the solutions says

$B+0.5\frac{C(t)}{B}$, but im not sure how to get this.

tia

2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
523
And what might A, B and C be?

3. ### steveb Senior Member

Jul 3, 2008
2,433
469

It looks like that is an approximate solution for the case where B is much greater than C. Note, I'm assuming that C(t)=Ccos(wt).

To see this, square the solution, as follows.

$(B+0.5\frac{C(t)}{B})^2=B^2+C(t)+{{C^2(t)}\over{4B^2}}$

Take the case where B is 10 times greater than C. In this case the order of magnitude of B squared is 100, the order of magnitude of C(t) is 1 (or less) and the order of magnitude of C(t)/(2B) squared is 0.0025 (or less). This means the last term can be neglected in many practical cases, provided the assumption that B>>C is met (note that the double greater than sign >> means "much greater than").

4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
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Perhaps you understand the half a question better than I did Steve.

As a matter of interest if you take the B^2 to the other side and factorise you get

(A-B) (A+B) = Ccos(wt)

t = arccos {(A-B) (A+B) /C }/w

5. ### steveb Senior Member

Jul 3, 2008
2,433
469
It's just a supposition on my part as to what the question really is, but the answer looks like the first two terms of the binomial expansion with power of 1/2.

There is no way that that answer can be exact, and the type of approximation I mentioned is somewhat common in physics and engineering, but typically is unacceptable in a pure math context.

I hope the source of the solution makes it clear that there is an approximation involved.

A good approximation to know by heart is $\sqrt{1+\epsilon}\approx 1+{{\epsilon}\over{2}}$, where it is assumed that ε<<1.

This can be used in the following way, in this case.

$\sqrt{B^2+C(t)}=B\sqrt{1+{{C(t)}\over{B^2}}}\appro B\Bigl(1+{{C(t)}\over{2B^2}}\Bigr)=\Bigl(B+{{C(t)}\over{2B}}\Bigr)$

Last edited: Apr 10, 2012
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6. ### suzuki Thread Starter Member

Aug 10, 2011
119
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EDIT: figured it out, thanks again

Hello steveb,

could you clarify in that last step how you went from

$\sqrt{B^2+C(t)}=B\sqrt{1+{{C(t)}\over{B^2}}}$

it looks like you divided everything under the square root by B^2 but then how did a B appear outside of the square root?

thanks again

Last edited: Apr 11, 2012
7. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
523
It's just a bit of algebra.

$\sqrt {{B^2} + C(t)} = \sqrt {\frac{{{B^2}}}{{{B^2}}}({B^2} + C(t))} = \sqrt {{B^2}({{\frac{B}{{{B^2}}}}^2} + \frac{{C(t)}}{{{B^2}}})} = \sqrt {{B^2}} \sqrt {({{\frac{B}{{{B^2}}}}^2} + \frac{{C(t)}}{{{B^2}}})} = B\sqrt {1 + \frac{{C(t)}}{{{B^2}}}}$

Thanks for the amplification Steve.

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8. ### suzuki Thread Starter Member

Aug 10, 2011
119
0
Hi,

to sort of add onto this problem, I am wondering how I can find the amplitude of a sin wave at multiple frequencies. As an example,

$0.01\sin(\omega t)+0.1\sin(2\omega t)+\sin(3\omega t)$

When you plot this, it gives a single sinusoid, but im not sure how i can calculate the amplitude of that waveform
thanks

As a visual example, i plotted three sinusoids with the same amplitude but varying frequencies. I cant seem to get a grasp of how i would determine the amplitude of the red plot

http://imgur.com/YZq85

Last edited: Apr 20, 2012