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The example states: sinh(j*Pi)=cos(Pi)*Sinh(0) +jcosh(x)sin(y)

I am unsure of this relation. I understand why the x terms are zero but don't know the relation to expand sinh. Can anyone shed some light on this for me?

Cheers.

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- Thread starter damo03
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- #1

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The example states: sinh(j*Pi)=cos(Pi)*Sinh(0) +jcosh(x)sin(y)

I am unsure of this relation. I understand why the x terms are zero but don't know the relation to expand sinh. Can anyone shed some light on this for me?

Cheers.

- #2

HallsofIvy

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I recommend you go back and read the example again. What you write has a constant on the left and a function of y on the right. TheyI am reading through a worked example of the Taylor series expansion of Sinh(z) about z=j*Pi

The example states: sinh(j*Pi)=cos(Pi)*Sinh(0) +jcosh(x)sin(y)

Do you mean [itex]sinh(j(y+ \pi))[/itex]?

I am unsure of this relation. I understand why the x terms are zero but don't know the relation to expand sinh. Can anyone shed some light on this for me?

Cheers.

- #3

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The first step showin is to calculate the two non-zero terms in the Taylor series expansion of sinh(z) about z=j*[itex]\pi[/itex]

f(z)=sinh(z) , Z0=j*[itex]\pi[/itex]

Sinh(j*[itex]\pi[/itex])

sinh(x+jy)=cos(y)*sinh(x)+jcosh(x)sin(y) (in this case x=0, y=[itex]\pi[/itex])

It's the above line that I don't understand (how they expand sinh)...

The worked example then goes on to show that sinh(x+jy)=cos(Pi)*sinh(0)+jcosh(0)sin(Pi)

which does indeed =-1*0 + j*1*0 = 0

but that one line expansion of sinh is where i am lost.

- #4

HallsofIvy

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[tex]cos(y)sinh(x)+ i cosh(x)sin(y)= \frac{e^{iy}+ e^{-iy}}{2}\frac{e^x- e^{-x}}{2}+ i\frac{e^x+ e^{-x}}{2}\frac{e^{iy}- e^{-iy}}{2i}[/tex]

The "i" in the numerator and the "i" in the denominator in the second term will cancel. (Sorry about writing "i" instead of "j"- I just can't stop myself!)

The denominators of course will be 4. Multiplying out the numerators,

[tex]e^xe^{iy}+ e^{x}e^{iy}- e^{-x}e^{iy}- e^{-x}e^{-iy}[/tex]

for the first term and

[tex]e^xe^{iy}- e^xe^{-iy}+ e^{-x}e^{iy}- e^{-x}e^{-iy}[/tex]

for the second term.

Now, observe what terms add and what terms cancel.

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