2—Infinite Series 572.8 Determine the Taylor series forcoshxandsinhx.
2.9 Working strictly by hand,evaluate^7
√
0. 999. Also√
50.
2.10 Determine the next,x^6 , term in the series expansion of the secant.
2.11 The power series for the tangent is not as neat and simple as for the sine and cosine. You can derive it
by taking successive derivatives as done in the text or you can use your knowledge of the series for the sine and
cosine, and the geometric series.
tanx=sinx
cosx=
x−x^3 /3! +···
1 −x^2 /2! +···=
[
x−x^3 /3! +···][
1 −x^2 /2! +···]− 1
Use the expansion for the geometric series to place all thex^2 ,x^4 , etc. terms into the numerator. Then collect
the like powers to obtain the series at least throughx^5. Ans:x+x^3 /3 + 2x^5 /15 + 17x^7 /315 +···
2.12 What is the series expansion forcscx= 1/sinx? As in the previous problem, use your knowledge of the
sine series and the geometric series to get this one at least throughx^5. Note: the first term inthis series is 1 /x.
Ans: 1 /x+x/6 + 7x^3 /360 + 31x^5 /15120 +···
2.13 The exact relativistic expression for the kinetic energy of an object with positive mass is
K=mc^2[
1
√
1 −v^2 /c^2− 1
]
wherecis the speed of light in vacuum. If the speedvis small compared to the speed of light, find an approximate
expression forKto show that it reduces to the Newtonian expression for the kinetic energy. What is the next term
in the expansion and how large the speedvmust be in order that this correction term is 10% of the Newtonian
expression for the kinetic energy? Ans:v≈ 0. 36 c