Math Enquiry

[shermaine]

I got some enquiry for the following with regard to maths:

(1) Can we simplified sinh(z)/Z^9, using laurent expansion to 1/Z^9 - (Z - Z^3/3! + Z^5/5! - Z^7/7! + Z^9/9!)?

(2) For lim z --- > 1 [(z^2 + 1) (sin (z)/z(z + 1) (z-3)], is that infinite? or zero?

(3) Can we simplified 2cosdeta to a expression that hv sin?
what's the expression?

 

[shermaine]

I got another question here:
using residue theorem to show that integrate from 0 to 2 pi {3/ (5 + 2 cos deta)} d deta is 6pi/sqr root 21

My working is as followed:
let 2 cos deta = e^jdeta +e^-jdeta
and d deta = dz/jz

Then i subtitue then into {3/ (5 + 2 cos deta)} d deta
then i got 3jz/ (z - za) (z-zb), where za = -0.21 and zb = -4.79

and i found out that za is inside the c: |z| = 1
then I = 2 pi j x res (f,za) = lim z --> -0.21 and i gotten 2 pi j x (0.63j)/4.58

anyone can advise me where i did wrongly?

 

[shermaine]

Got another math issue here:
For lim z -- 4 pi (z + 4 pi) ( z.exp(z)/sin z)
by exp z for 4 pi, is it considered as 0?

 

[steveB]

I'm having trouble understanding some of your notation. I would be easier to help you if you can either scan a hand written document, or if you learn to use the latex equation notation as follows

[latex] x+y=z [/latex]
[latex]e^{j \omega t} [/latex]
[latex]\sin x [/latex] etc.

However, I notice a number of mistakes above. When you ask about the zeros and poles, you are not looking at the correct value of z, for example you need to ask about z=-1 and not z=1 in the first question. Also, that Laurent expansion does not look correct.

 

[shermaine]

Hi steve,

I had retyped that into word format
hopefully it's clearer
Please advise where i did wrongly if u can
Thanks

 

[steveB]

Hi steve,

I had retyped that into word format
hopefully it's clearer
Please advise where i did wrongly if u can
Thanks

Well it was a little clearer, but you should use the equation editor in Word to make it better. Or, the latex notation is good too.

Anyway, I've made some suggestions in the attached document. I went through quickly, so I could have made some mistakes, but I did correct a few of your mistakes and you should be able to figure it out from there.

Good luck!

 

[shermaine]

Hi Steve,

Thanks so much for your advise.
I got one question for (b) and (c)
For (b), how come there is a minus sign for -3j/z2 + 5z + 1)?
For (c), Yes, i mean For sin z = 0, z = 0, z = +/-
, z = +/- 3
, z = +/- 4
(you mean 0, ±p, ±2p, ±3p )
I had amended a bit for (c), but i still blurred with finding residue for it.
Maybe if you can assist me abit further?

 

[steveB]

Hi Steve,

Thanks so much for your advise.
I got one question for (b) and (c)
For (b), how come there is a minus sign for -3j/z2 + 5z + 1)?
For (c), Yes, i mean For sin z = 0, z = 0, z = +/-
, z = +/- 3
, z = +/- 4
(you mean 0, ±p, ±2p, ±3p )
I had amended a bit for (c), but i still blurred with finding residue for it.
Maybe if you can assist me abit further?

I got a negative sign from the j being in the denominator. When you move the j to the numerator it becomes -j. However, like I said, I went quickly, so I could have made a mistake. Go through the calculation slowly and you should get it right.

Also, why are you looking at the point +/-4pi? Isn't 4pi greater than 12 and outside the circle of integration?

Also, why to you need to take the limit of the exponential function. You have already shown that the pole at zero is canceled by a zero at zero. Just use the residue formula at +/-pi, +/-2pi and +/-3pi.

Note that is is very hard to interpret your document. The notation is very confusing. I'm doing my best to interpret it, but I could be misunderstanding something. Anyway, I'm just trying to give you hints that will allow you to get it to the right answer.

 

[shermaine]

I got a negative sign from the j being in the denominator. When you move the j to the numerator it becomes -j. However, like I said, I went quickly, so I could have made a mistake. Go through the calculation slowly and you should get it right.

Also, why are you looking at the point +/-4pi? Isn't 4pi greater than 12 and outside the circle of integration?

Also, why to you need to take the limit of the exponential function. You have already shown that the pole at zero is canceled by a zero at zero. Just use the residue formula at +/-pi, +/-2pi and +/-3pi.

Note that is is very hard to interpret your document. The notation is very confusing. I'm doing my best to interpret it, but I could be misunderstanding something. Anyway, I'm just trying to give you hints that will allow you to get it to the right answer.

Thanks so much! From my understanding,since after taking lim [z.exp z ]/[sin z] = 0/0
z -> 0
then i need to take L'hospital rule, which is lim [z.exp z + exp z]/[sin z] = 1/1 = 1
z -> 0

Is that correct method?
As for the Residue, i am not sure how to express it out for the numerator portion.
Is it for -3pi,

Its as followed:

Res(f(z),-3pi) = lim (z + 3pi) [(z.exp z)/sin z]
z - > -3pi

= lim (z^2 exp z + [-3pi z exp z])
z - > -3pi

= [(3pi)^2 exp 3pi + 3pi (-3pi)exp (-3pi) ] / sin 3pi


May i know what is exp (- 3pi)?
Do i need to differentiate again in residue?
May i know what should be the correct answer for this part?
How do i solve the numberator part?
Plesase advise me a bit more. Thnks.

 

[steveB]

Thanks so much! From my understanding,since after taking lim [z.exp z ]/[sin z] = 0/0
z -> 0
then i need to take L'hospital rule, which is lim [z.exp z + exp z]/[sin z] = 1/1 = 1
z -> 0

Is that correct method?.

This is correct, but I think you have a typo error above. The denominator should be cos z and not sin z.

lim [z.exp z + exp z]/[cosz] = 1/1 = 1

As for the Residue, i am not sure how to express it out for the numerator portion.
Is it for -3pi,

Its as followed:

Res(f(z),-3pi) = lim (z + 3pi) [(z.exp z)/sin z]
z - > -3pi

= lim (z^2 exp z + [-3pi z exp z])
z - > -3pi

= [(3pi)^2 exp 3pi + 3pi (-3pi)exp (-3pi) ] / sin 3pi

.

You really need to be careful when your write mathematical equations. Everything you are writing has mistakes in them. I don't know if these are just typos, or if you just don't understand. Anyway the above is not making sense.

Res(f(z),-3pi) = lim as z-> -3pi ( (z + 3pi).z.exp(z)/sin(z)] )

Note that (z+3pi) goes to zero and sin(z) goes to zero, but in the limit as z goes to - 3pi, (z+3pi)/sin(z) goes to 1.

Hence

Res(f(z),-3pi) = -3pi.exp(-3pi)

 

[shermaine]

Thanks for your reply.
But i was thinking for Res(f(z),-3pi) = lim as z-> -3pi ( (z + 3pi).z.exp(z)/sin(z)] )
(z + 3pi) and sin (z) cant cancelled off, so it remains there.
And thus the equation becomes
( ([-3pi] + 3pi).[-3pi].exp(-3pi)/sin(-3pi)] )
sin(-3pi)] is 0 and ([-3pi] + 3pi) = 0 , so left [-3pi].exp(-3pi)
so can i left the answer in -3pi.exp(-3pi) = -540 exp (-540)
But i got one last question: exp (-540) when keyed into calculator, its a math error.
what answer should i left that be?
Steve, really thanks for all your help

 

[steveB]

so can i left the answer in -3pi.exp(-3pi) = -540 exp (-540)
But i got one last question: exp (-540) when keyed into calculator, its a math error.
what answer should i left that be?

I think you made an error with your calculator: 3pi is not equal to 540

However, you should leave the answer as [latex] -3\pi\ e^{-3\pi} [/latex]

Both [latex] \pi [/latex] and [latex] e [/latex] are irrational numbers and can't be expressed exactly with decimal numbers or fractions.

 

[shermaine]

Hi Steve,

Thanks so much for your help.
I was working on z = -3pi,
Res(f(z),-3pi) = lim z ---> -3pi ( z + 3pi) . [z.exp z / sin z]
since z = -3pi , thus -3pi + 3pi = 0 so 0 multiply anything = 0
I'm not sure if i'm correct and it seem that all the equation for -3pi, 3pi,-2pi is equal to zero.
Could you advise me? Thanks!

 

[steveB]

Hi Steve,

Thanks so much for your help.
I was working on z = -3pi,
Res(f(z),-3pi) = lim z ---> -3pi ( z + 3pi) . [z.exp z / sin z]
since z = -3pi , thus -3pi + 3pi = 0 so 0 multiply anything = 0
I'm not sure if i'm correct and it seem that all the equation for -3pi, 3pi,-2pi is equal to zero.
Could you advise me? Thanks!

I believe you can take the limit as z goes to -3pi. Both the numerator and denominator approach zero, but the ratio is finite when you take the limit. I think if you get a nonzero value for the answer, you will have the right answer. See section 2.3.(4) on page 5 of the attached document.