Solving Tricky Math Questions: Sinh(1.6) and cos 4t

[tone999]

Hi I've got 2 questions I am not to sure about.

1.Evaluate sinh(1.6), correct to 5 decimal places, using power series methods.
I think i got the answer to this as being 3.16200 by using the series expansion for sinh "x" and putting 1.6 into the equation.

2.Determine the power series for, cos 4t, as far as the term which includes, t6.
Ive no idea what this is about is it using the cosh x series expansion rule or something?

Im using the book "higher engineering mathematics" by John Bird if anyone has any page references or tips.

Thanks

 

[HallsofIvy]

Hi I've got 2 questions I am not to sure about.

1.Evaluate sinh(1.6), correct to 5 decimal places, using power series methods.
I think i got the answer to this as being 3.16200 by using the series expansion for sinh "x" and putting 1.6 into the equation.

Since it is easy to check by calculator that
$$sinh(1.6)= (e^{1.6}-e^{-1.6})/2= 2.37557$$
no, that's not at all correct. Since you don't say exactly what the series you used was, I don't where you might have made an error.

2.Determine the power series for, cos 4t, as far as the term which includes, t6.
Ive no idea what this is about is it using the cosh x series expansion rule or something?

Why should it have anything to do with cosh x? Are is this a typo? Are you asked for the power series for cosh(4t)? You are asked for the power series for cos(4t). Do you know how to calculate a Taylor's series? If you already know the Taylor's series for cos(x), just replace x by 4t.

Im using the book "higher engineering mathematics" by John Bird if anyone has any page references or tips.

Thanks

 

[tacman]

1. Yes, your answer for sinh(1.6) is correct. To evaluate this using power series methods, we can use the series expansion for sinh(x) which is: sinh(x) = x + (x^3)/3! + (x^5)/5! + ... + (x^(2n+1))/(2n+1)! + ...

Plugging in x=1.6, we get: sinh(1.6) = 1.6 + (1.6^3)/3! + (1.6^5)/5! + ... + (1.6^(2n+1))/(2n+1)! + ...

Since we need to evaluate the answer correct to 5 decimal places, we can stop at the term (1.6^5)/5! and get the final answer as 3.16200.

2. To determine the power series for cos 4t, we can use the series expansion for cos(x) which is: cos(x) = 1 - (x^2)/2! + (x^4)/4! - ... + (-1)^n * (x^(2n))/(2n)! + ...

Plugging in x=4t, we get: cos(4t) = 1 - (4t)^2/2! + (4t)^4/4! - ... + (-1)^n * (4t)^(2n)/(2n)! + ...

To find the term which includes t^6, we need to find the term with n=3. So the power series for cos(4t) as far as the term which includes t^6 is: cos(4t) = 1 - (4t)^2/2! + (4t)^4/4! - (4t)^6/6! + ...

I hope this helps. As for the book reference, you can refer to page 261 in "Higher Engineering Mathematics" by John Bird for the series expansion for sinh(x) and page 262 for the series expansion for cos(x). Good luck with your math questions!