# Power Series

1. Nov 22, 2006

### tone999

another power series question...
I tried it for awhile but it just got out of hand and the ammount of numbers got unbearable.

Q. The increase in resistance of strip conductors due to eddy currents at power frequencies is given by :

X = yt divided by 2 (sinh yt + sin yt divided by cosh yt - cos yt )

If y = 1.08 and t = 1.3, calculate X, correct to 5 significant figures, using power series.

p.s. i dnt know which font it is to do the mathmatical language style :(

Any help would be greatly appreciated. thanks

2. Nov 22, 2006

### HallsofIvy

X = yt divided by 2 (sinh yt + sin yt divided by cosh yt - cos yt )
isn't clear. Do you mean
$$X= \frac{yt}{2}\frac{sinh yt+ sin yt}{cosh yt- cos yt}$$

(the other possibility is
$$X= \frac{yt}{2(\frac{sinh yt+ sin yt}{cosh yt- cos yt}}$$
which would be more simply written as
$$X= \frac{yt}{2}\frac{cosh yt- cos yt}{sinh yt+ sin yt}$$

In either case, the power series for sinh yt+ sin yt and cosh yt- cos yt are relatively simple:
$$sinh yt= yt+ (1/6)y^3t^3+ (1/5!)y^5t^5)+...$$
and
$$sin yt= yt- (1/6)y^3y^3z+ (1/5!)y^5t^5+...$$
so that
$$sinh yt+ sinyt= 2(yt+ (1/5!)y^5t^5+ (1/9!)y^9t^9+ ...$$
where the powers are of the form 4n+1. Since 1/(4n+1)! goes to 0 rapidly, you shouldn't need many terms to get that to 5 decimal places.

$$cosh yt= 1+ (1/2)y^2t^2+ (1/4!)y^4t^4+ (1/6!)y^6t^6+ ...$$
$$cos yt= 1- (1/2)y^2t^2+ (1/4!)y^4t^4- (1/6!)y^6t^6+ ...$$
so
$$cosh yt- cos yt= 2((1/2)y^2t^2+ (1/6!)y^6t^6+ ...$$
where the powers are of the form 4n+ 4. Again, it shouldn't take many terms to get that to 5 decimal places.

3. Nov 27, 2006

### tone999

hi

hi sorry i took so long 2 come back to this problem.yes the first equation is the one i am using.
theres a few things i didnt understand firstly where do the divison parts come from eg (1/6) (1/2)?
and maybe its just a typo but should the z be there: "(1/6)y^3y^3z" and instead a t^3.
sorry to be so annoying