Solving a Power Series Question: X = yt/2(sinh yt + sin yt/cosh yt - cos yt)

In summary, we are trying to calculate X using a power series, with given values for y and t. The equations used for sinh yt and cosh yt are relatively simple and only a few terms are needed to get the desired precision of 5 decimal places. The division parts come from the respective equations for sinh yt and cosh yt, and there may have been a typo in the equation for sin yt.
  • #1
tone999
18
0
another power series question...
I tried it for awhile but it just got out of hand and the amount 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
 
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  • #2
X = yt divided by 2 (sinh yt + sin yt divided by cosh yt - cos yt )
isn't clear. Do you mean
[tex]X= \frac{yt}{2}\frac{sinh yt+ sin yt}{cosh yt- cos yt}[/tex]

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

In either case, the power series for sinh yt+ sin yt and cosh yt- cos yt are relatively simple:
[tex]sinh yt= yt+ (1/6)y^3t^3+ (1/5!)y^5t^5)+... [/tex]
and
[tex]sin yt= yt- (1/6)y^3y^3z+ (1/5!)y^5t^5+...[/tex]
so that
[tex]sinh yt+ sinyt= 2(yt+ (1/5!)y^5t^5+ (1/9!)y^9t^9+ ... [/tex]
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.

[tex]cosh yt= 1+ (1/2)y^2t^2+ (1/4!)y^4t^4+ (1/6!)y^6t^6+ ...[/tex]
[tex]cos yt= 1- (1/2)y^2t^2+ (1/4!)y^4t^4- (1/6!)y^6t^6+ ...[/tex]
so
[tex]cosh yt- cos yt= 2((1/2)y^2t^2+ (1/6!)y^6t^6+ ...[/tex]
where the powers are of the form 4n+ 4. Again, it shouldn't take many terms to get that to 5 decimal places.
 
  • #3
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
 

Related to Solving a Power Series Question: X = yt/2(sinh yt + sin yt/cosh yt - cos yt)

1. What is a power series?

A power series is a mathematical representation of a function as an infinite sum of terms, where each term is a power of the variable. It is commonly used in calculus to approximate functions.

2. How do I solve a power series question?

To solve a power series question, you need to first identify the pattern in the terms of the series. Then, you can use techniques such as substitution, integration, or differentiation to manipulate the series and find a solution.

3. What is the meaning of "X = yt/2(sinh yt + sin yt/cosh yt - cos yt)"?

This equation is a power series representation of a function. The variable "yt" is raised to different powers and multiplied by coefficients to create the terms of the series. The "X =" indicates that the series is equal to the function represented by the series.

4. How do I find the value of "X" in the equation?

To find the value of "X", you need to evaluate the infinite series by either finding a finite number of terms or using a convergence test. Once you have a finite number of terms, you can substitute the value of "yt" into the series and calculate the sum.

5. Can I use power series to solve any type of function?

Power series can be used to approximate any function, but it may not always provide an exact solution. Some functions may require a large number of terms in the series to accurately approximate the function, while others may not have a power series representation at all.

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