A simple algebra quiz problem

Hi Guys,
  I was attempting to derive an equation for 
the frequency response of a digital filter.  
In my derivation I obtained one term in my 
equation that took the form of what I'll 
call 'y' as given below:
                    *     
  y = x [sqrt(x/x*)]        (1)

where x = re^(jp) is a complex number, 'r' is 
a real-valued scalar magnitude value, and angle p 
is in radians.

In words, Eq. (1) says, "y equals x times the 
conjugate of the square root of the ratio of 
x over x-conjugate."

Here's the quiz problem:
Can you simplify the above expression for 'y' 
so that your simplified 'y' is equal to 
the 'y' in Eq. (1)?

I think I have the correct solution to this quiz 
problem but I thought it was interesting enough 
to plop it down in front of you guys. 

[-Rick-]

On Sun, 07 Sep 2014 16:44:40 -0700, Rick Lyons
<R.Lyons@_BOGUS_ieee.org> wrote:

>
>Hi Guys,
>  I was attempting to derive an equation for 
>the frequency response of a digital filter.  
>In my derivation I obtained one term in my 
>equation that took the form of what I'll 
>call 'y' as given below:
>                    *     
>  y = x [sqrt(x/x*)]        (1)
>
>where x = re^(jp) is a complex number, 'r' is 
>a real-valued scalar magnitude value, and angle p 
>is in radians.
>
>In words, Eq. (1) says, "y equals x times the 
>conjugate of the square root of the ratio of 
>x over x-conjugate."
>
>Here's the quiz problem:
>Can you simplify the above expression for 'y' 
>so that your simplified 'y' is equal to 
>the 'y' in Eq. (1)?
>
>I think I have the correct solution to this quiz 
>problem but I thought it was interesting enough 
>to plop it down in front of you guys. 
>
>[-Rick-]


Just doing it quickly it seems to me to be:

y = sqrt(r)re^jp = (r^1.5)e^jp


Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com

Rick Lyons <R.Lyons@_BOGUS_ieee.org> writes:

> Hi Guys,
>   I was attempting to derive an equation for 
> the frequency response of a digital filter.  
> In my derivation I obtained one term in my 
> equation that took the form of what I'll 
> call 'y' as given below:
>                     *     
>   y = x [sqrt(x/x*)]        (1)
>
> where x = re^(jp) is a complex number, 'r' is 
> a real-valued scalar magnitude value, and angle p 
> is in radians.
>
> In words, Eq. (1) says, "y equals x times the 
> conjugate of the square root of the ratio of 
> x over x-conjugate."
>
> Here's the quiz problem:
> Can you simplify the above expression for 'y' 
> so that your simplified 'y' is equal to 
> the 'y' in Eq. (1)?
>
> I think I have the correct solution to this quiz 
> problem but I thought it was interesting enough 
> to plop it down in front of you guys. 
>
> [-Rick-]

x = r.
-- 
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com

Randy Yates <yates@digitalsignallabs.com> writes:

> Rick Lyons <R.Lyons@_BOGUS_ieee.org> writes:
>
>> Hi Guys,
>>   I was attempting to derive an equation for 
>> the frequency response of a digital filter.  
>> In my derivation I obtained one term in my 
>> equation that took the form of what I'll 
>> call 'y' as given below:
>>                     *     
>>   y = x [sqrt(x/x*)]        (1)
>>
>> where x = re^(jp) is a complex number, 'r' is 
>> a real-valued scalar magnitude value, and angle p 
>> is in radians.
>>
>> In words, Eq. (1) says, "y equals x times the 
>> conjugate of the square root of the ratio of 
>> x over x-conjugate."
>>
>> Here's the quiz problem:
>> Can you simplify the above expression for 'y' 
>> so that your simplified 'y' is equal to 
>> the 'y' in Eq. (1)?
>>
>> I think I have the correct solution to this quiz 
>> problem but I thought it was interesting enough 
>> to plop it down in front of you guys. 
>>
>> [-Rick-]
>
> x = r.

Doh! Sorry.

y = r.
-- 
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com

On Sunday, September 7, 2014 7:53:47 PM UTC-4, Eric Jacobsen wrote:
> On Sun, 07 Sep 2014 16:44:40 -0700, Rick Lyons
> 
> <R.Lyons@_BOGUS_ieee.org> wrote:
> 
> 
> 
> >
> 
> >Hi Guys,
> 
> >  I was attempting to derive an equation for 
> 
> >the frequency response of a digital filter.  
> 
> >In my derivation I obtained one term in my 
> 
> >equation that took the form of what I'll 
> 
> >call 'y' as given below:
> 
> >                    *     
> 
> >  y = x [sqrt(x/x*)]        (1)
> 
> >
> 
> >where x = re^(jp) is a complex number, 'r' is 
> 
> >a real-valued scalar magnitude value, and angle p 
> 
> >is in radians.
> 
> >
> 
> >In words, Eq. (1) says, "y equals x times the 
> 
> >conjugate of the square root of the ratio of 
> 
> >x over x-conjugate."
> 
> >
> 
> >Here's the quiz problem:
> 
> >Can you simplify the above expression for 'y' 
> 
> >so that your simplified 'y' is equal to 
> 
> >the 'y' in Eq. (1)?
> 
> >
> 
> >I think I have the correct solution to this quiz 
> 
> >problem but I thought it was interesting enough 
> 
> >to plop it down in front of you guys. 
> 
> >
> 
> >[-Rick-]
> 
> 
> 
> 
> 
> Just doing it quickly it seems to me to be:
> 
> 
> 
> y = sqrt(r)re^jp = (r^1.5)e^jp
> 
> 
> 
> 
> 
> Eric Jacobsen
> 
> Anchor Hill Communications
> 
> http://www.anchorhill.com

y=(r*exp(jp))*conjg[sqrt( (r*exp(jp)) / (r*exp(-jp)) )]
 =(r*exp(jp))*conjg[sqrt(exp(j2p)]
 =(r*exp(jp))*conjg[+/- exp(jp)]
 =(+/-)(r*exp(jp))*exp(-jp)]
 =(+/-)r

y=(+/-)r


Dirk

Randy Yates <yates@digitalsignallabs.com> writes:

> Randy Yates <yates@digitalsignallabs.com> writes:
>
>> Rick Lyons <R.Lyons@_BOGUS_ieee.org> writes:
>>
>>> Hi Guys,
>>>   I was attempting to derive an equation for 
>>> the frequency response of a digital filter.  
>>> In my derivation I obtained one term in my 
>>> equation that took the form of what I'll 
>>> call 'y' as given below:
>>>                     *     
>>>   y = x [sqrt(x/x*)]        (1)
>>>
>>> where x = re^(jp) is a complex number, 'r' is 
>>> a real-valued scalar magnitude value, and angle p 
>>> is in radians.
>>>
>>> In words, Eq. (1) says, "y equals x times the 
>>> conjugate of the square root of the ratio of 
>>> x over x-conjugate."
>>>
>>> Here's the quiz problem:
>>> Can you simplify the above expression for 'y' 
>>> so that your simplified 'y' is equal to 
>>> the 'y' in Eq. (1)?
>>>
>>> I think I have the correct solution to this quiz 
>>> problem but I thought it was interesting enough 
>>> to plop it down in front of you guys. 
>>>
>>> [-Rick-]
>>
>> x = r.
>
> Doh! Sorry.
>
> y = r.

http://www.digitalsignallabs.com/algebra-quiz-20140907233002.htm
-- 
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com

Am 08.09.2014 05:40, schrieb Randy Yates:

>>>> [-Rick-]
>>>
>>> x = r.
>>
>> Doh! Sorry.
>>
>> y = r.
>
> http://www.digitalsignallabs.com/algebra-quiz-20140907233002.htm
>
Well, the answer there is correct, but the computation is incomplete. 
The major problem here is that the square root is a multi-valued 
function in the complex plane, so the answer could be r or -r, depending 
on whether you pick the lower or the upper sheet of the square root. It 
requires some extra care to justify that the answer is r, and not -r.

For that, do the following: First set \phi to zero, so x is real. By 
convention, the square root of a positive number is positive, so y = r. 
In complex terms, we are on the "upper sheet" of the square root.

Now consider the path z(t) = r * e^{it} for t = 0... and increasing.

Clearly, for t=0 the answer y(t) is r. Now, since the square root is a 
continuous function, and complex conjugation is, and we are away from 
the pole in the square root, y(t) is also a continuous function of t, so 
it cannot jump from +r to -r. Hence, the answer is r all the way, no 
matter what \phi is (and it is never -r).

It is a bit delicate, and unfortunately the above "calculation" misses 
this argument. It's not an algebraic problem anymore due to the multiple 
values of the square root. (Or rather, sqrt is not defined on C, but on 
the double-overlay of C).

Greetings,
	Thomas

On Sunday, September 7, 2014 11:31:48 PM UTC-4, belld...@gmail.com wrote:
> On Sunday, September 7, 2014 7:53:47 PM UTC-4, Eric Jacobsen wrote:
> 
> > On Sun, 07 Sep 2014 16:44:40 -0700, Rick Lyons
> 
> > 
> 
> > <R.Lyons@_BOGUS_ieee.org> wrote:
> 
> > 
> 
> > 
> 
> > 
> 
> > >
> 
> > 
> 
> > >Hi Guys,
> 
> > 
> 
> > >  I was attempting to derive an equation for 
> 
> > 
> 
> > >the frequency response of a digital filter.  
> 
> > 
> 
> > >In my derivation I obtained one term in my 
> 
> > 
> 
> > >equation that took the form of what I'll 
> 
> > 
> 
> > >call 'y' as given below:
> 
> > 
> 
> > >                    *     
> 
> > 
> 
> > >  y = x [sqrt(x/x*)]        (1)
> 
> > 
> 
> > >
> 
> > 
> 
> > >where x = re^(jp) is a complex number, 'r' is 
> 
> > 
> 
> > >a real-valued scalar magnitude value, and angle p 
> 
> > 
> 
> > >is in radians.
> 
> > 
> 
> > >
> 
> > 
> 
> > >In words, Eq. (1) says, "y equals x times the 
> 
> > 
> 
> > >conjugate of the square root of the ratio of 
> 
> > 
> 
> > >x over x-conjugate."
> 
> > 
> 
> > >
> 
> > 
> 
> > >Here's the quiz problem:
> 
> > 
> 
> > >Can you simplify the above expression for 'y' 
> 
> > 
> 
> > >so that your simplified 'y' is equal to 
> 
> > 
> 
> > >the 'y' in Eq. (1)?
> 
> > 
> 
> > >
> 
> > 
> 
> > >I think I have the correct solution to this quiz 
> 
> > 
> 
> > >problem but I thought it was interesting enough 
> 
> > 
> 
> > >to plop it down in front of you guys. 
> 
> > 
> 
> > >
> 
> > 
> 
> > >[-Rick-]
> 
> > 
> 
> > 
> 
> > 
> 
> > 
> 
> > 
> 
> > Just doing it quickly it seems to me to be:
> 
> > 
> 
> > 
> 
> > 
> 
> > y = sqrt(r)re^jp = (r^1.5)e^jp
> 
> > 
> 
> > 
> 
> > 
> 
> > 
> 
> > 
> 
> > Eric Jacobsen
> 
> > 
> 
> > Anchor Hill Communications
> 
> > 
> 
> > http://www.anchorhill.com
> 
> 
> 
> y=(r*exp(jp))*conjg[sqrt( (r*exp(jp)) / (r*exp(-jp)) )]
> 
>  =(r*exp(jp))*conjg[sqrt(exp(j2p)]
> 
>  =(r*exp(jp))*conjg[+/- exp(jp)]
> 
>  =(+/-)(r*exp(jp))*exp(-jp)]
> 
>  =(+/-)r
> 
> 
> 
> y=(+/-)r
> 
> 
> 
> 
> 
> Dirk

An example to consider (MATLAB):

y=(r*exp(j*p))*conj(sqrt( (r*exp(j*p)) / (r*exp(-j*p)) ))

Inputs and Outputs:
Example 1: In: r = 1, In: p = 4.0484, Out: y = -1
Example 2: In: r = 5, In: p = 5.0993, Out: y =  5

Example 1 gives y=-r
Example 2 gives y=+r

While the answer appears independent of 'p' because it seems to factor out, it isn't.  The value of 'p' affects the sign of the output.

Did some more testing.  Appears that for abs(r)>0
y=+r, for p in [0,           .25*(2*pi) ]
y=-r, for p in (.25*(2*pi),  .75*(2*pi) ]
y=+r, for p in (.27*(2*pi), 1.00*(2*pi) ]


Dirk

On Sun, 7 Sep 2014 20:31:48 -0700 (PDT), bellda2005@gmail.com wrote:

>On Sunday, September 7, 2014 7:53:47 PM UTC-4, Eric Jacobsen wrote:
>> On Sun, 07 Sep 2014 16:44:40 -0700, Rick Lyons
>> 
>> <R.Lyons@_BOGUS_ieee.org> wrote:
>> 
>> >Hi Guys,
>> 
>> >  I was attempting to derive an equation for 
>> >the frequency response of a digital filter.  
>> >In my derivation I obtained one term in my 
>> >equation that took the form of what I'll 
>> >call 'y' as given below:
>> >                    *     
>> >  y = x [sqrt(x/x*)]        (1)
>> >
>> >where x = re^(jp) is a complex number, 'r' is 
>> >a real-valued scalar magnitude value, and angle p 
>> >is in radians.
>> >
>> >In words, Eq. (1) says, "y equals x times the 
>> >conjugate of the square root of the ratio of 
>> >x over x-conjugate."
>> >
>> 
>> >Here's the quiz problem:
>> >Can you simplify the above expression for 'y' 
>> >so that your simplified 'y' is equal to 
>> >the 'y' in Eq. (1)?
>> >
>> >I think I have the correct solution to this quiz 
>> >problem but I thought it was interesting enough 
>> >to plop it down in front of you guys. 
>> >
>> >[-Rick-]
>> 
>> 
>> Just doing it quickly it seems to me to be:
>> 
>> y = sqrt(r)re^jp = (r^1.5)e^jp

>
>y=(r*exp(jp))*conjg[sqrt( (r*exp(jp)) / (r*exp(-jp)) )]
> =(r*exp(jp))*conjg[sqrt(exp(j2p)]
> =(r*exp(jp))*conjg[+/- exp(jp)]
> =(+/-)(r*exp(jp))*exp(-jp)]
> =(+/-)r
>
>y=(+/-)r
>
>
>Dirk

Dirk's looks right to me now.

Yesterday in my world r/r = r and other dubious realities, so I didn't
get what Dirk got.   ;)


Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com

On Sunday, September 7, 2014 11:31:48 PM UTC-4, belld...@gmail.com wrote:
> On Sunday, September 7, 2014 7:53:47 PM UTC-4, Eric Jacobsen wrote:
> 
> > On Sun, 07 Sep 2014 16:44:40 -0700, Rick Lyons
> 
> > 
> 
> > <R.Lyons@_BOGUS_ieee.org> wrote:
> 
> > 
> 
> > 
> 
> > 
> 
> > >
> 
> > 
> 
> > >Hi Guys,
> 
> > 
> 
> > >  I was attempting to derive an equation for 
> 
> > 
> 
> > >the frequency response of a digital filter.  
> 
> > 
> 
> > >In my derivation I obtained one term in my 
> 
> > 
> 
> > >equation that took the form of what I'll 
> 
> > 
> 
> > >call 'y' as given below:
> 
> > 
> 
> > >                    *     
> 
> > 
> 
> > >  y = x [sqrt(x/x*)]        (1)
> 
> > 
> 
> > >
> 
> > 
> 
> > >where x = re^(jp) is a complex number, 'r' is 
> 
> > 
> 
> > >a real-valued scalar magnitude value, and angle p 
> 
> > 
> 
> > >is in radians.
> 
> > 
> 
> > >
> 
> > 
> 
> > >In words, Eq. (1) says, "y equals x times the 
> 
> > 
> 
> > >conjugate of the square root of the ratio of 
> 
> > 
> 
> > >x over x-conjugate."
> 
> > 
> 
> > >
> 
> > 
> 
> > >Here's the quiz problem:
> 
> > 
> 
> > >Can you simplify the above expression for 'y' 
> 
> > 
> 
> > >so that your simplified 'y' is equal to 
> 
> > 
> 
> > >the 'y' in Eq. (1)?
> 
> > 
> 
> > >
> 
> > 
> 
> > >I think I have the correct solution to this quiz 
> 
> > 
> 
> > >problem but I thought it was interesting enough 
> 
> > 
> 
> > >to plop it down in front of you guys. 
> 
> > 
> 
> > >
> 
> > 
> 
> > >[-Rick-]
> 
> > 
> 
> > 
> 
> > 
> 
> > 
> 
> > 
> 
> > Just doing it quickly it seems to me to be:
> 
> > 
> 
> > 
> 
> > 
> 
> > y = sqrt(r)re^jp = (r^1.5)e^jp
> 
> > 
> 
> > 
> 
> > 
> 
> > 
> 
> > 
> 
> > Eric Jacobsen
> 
> > 
> 
> > Anchor Hill Communications
> 
> > 
> 
> > http://www.anchorhill.com
> 
> 
> 
> y=(r*exp(jp))*conjg[sqrt( (r*exp(jp)) / (r*exp(-jp)) )]
> 
>  =(r*exp(jp))*conjg[sqrt(exp(j2p)]
> 
>  =(r*exp(jp))*conjg[+/- exp(jp)]
> 
>  =(+/-)(r*exp(jp))*exp(-jp)]
> 
>  =(+/-)r
> 
> 
> 
> y=(+/-)r
> 
> 
> 
> 
> 
> Dirk

An example to consider (MATLAB):

y=(r*exp(j*p))*conj(sqrt( (r*exp(j*p)) / (r*exp(-j*p)) ))

Inputs and Outputs:
Example 1: In: r = 1, In: p = 4.0484, Out: y = -1
Example 2: In: r = 5, In: p = 5.0993, Out: y =  5

Example 1 gives y=-r
Example 2 gives y=+r

While the answer appears independent of 'p' because it seems to factor out, it isn't.  The value of 'p' affects the sign of the output.

Did some more testing.  Appears that for abs(r)>0
y=+r, for p in [0,           .25*(2*pi) ]
y=-r, for p in (.25*(2*pi),  .75*(2*pi) ]
y=+r, for p in (.27*(2*pi), 1.00*(2*pi) ]

Of course this may be related to how sqrt() is defined.

Dirk