a)If x (in radians) is small, show that (5 - 2 cos x) / (3 + tan x) is approx = (3+x^2)/(3+x).?

Question

Hence, determine the constants A, B and C such that (5 - 2 cos x) /(3+ tan x) is approx = A + Bx + Cx ^ 2.

Answer 1

Use Taylor approximations.....

5-2cos(x) = 5-2(1-x^2/2...) = 3+x^2
and 3+tan(x)= 3+x.... to the first non-constant term

Then do geometric series on these approximations ...

(3+x^2)/(3+x) = [(3+x^2)/3]* [1/(1+x/3)] = [1+x^2/3]* [1 - x/3 + x^2/9...] = then foil out to get 1-x/3+4/9*x^2

The idea here is that geomteric series is easier then doing "straight" Taylor series on the rational function (quotient rule yuk!) but this "trick" is less useful if you want higher order terms like x^5 or x^6, etc.

Answer 2

If x in radians is small, then x is approx 0, cos x is approx 1 and tan x is approx 0. so, replacing with their approximations,
(5 - 2(1))/(3 + 0) is approx (3 + 0^2)/(3+0) = 3/3 = 1
As for constants A. B, and C, I don't think they would be constants per se, I like math guru's approach, but since I've come this far, let me see if what I have in mind works...
For small values of x (in radians), cos x is approx 1-x and sin s is approx x. Substituting these relations in what you have, and simplifiying algebraically, you get (5-2cosx)/(3+tan x) is approx
x - 1 + 6/(2x + 3).

Answer 3

When x is very small, cos x is very close to 1 - x^2/2, and tan x is very close to x. If you compare the graphs of these functions, you will see that it is so. In calculus this would be called a second order approximation, as we are approximating with linear and quadratic functions.

Then just substitute in the approximations and simplify:

(5 - 2cosx)/(3 + tanx) ~ (5 - 2*(1 - x^2/2))/(3 + x)
~ (3 + x^2)/(3 + x).


Hope this helps,


açafrã[e-mail]

Answer 4

if x is very small, approaching zero, then cosx will approach 1 and tan x will approach 0. If we let cosx = 1 and tanx = 0 the the equation becomes (5-2(1))/(3+0) = 3/3 = 1.

if x is very small then x is about zero. zero squared is zero. adding zero to the top and bottom will not change the outcome which will be about 1.

Say x =.001 the x^2 = 1.0 e-6 so 3.001/3.0000006 = 1.0003 or 1

Answer 5

As x --> 0 sin x --> x and cosx --> 1
Thus, the denominator = 3+tanx = 3 +sinx/cosx --> 3+x/1 = 3+x
and the numerator -->5-2(1)=3
So expression --> 3/(3+x)

Answer 6

#2a. why not tailor? let u(x)=5-2cos(x) and v(x)=3+tan(x), then y(0)=1;
y’(x) = (u’v-v’u)/v^2, where u’=2sin(x), v’=1/(cos(x))^2;
y’(0) = (3*0-1*3)/9 = -1/3;
y’’(x)= (u’’v+u’v’ – v’’u-v’u’) / v^2 – (2/v^3)*(u’v-v’u),
where u’’(0)=2, v’’(0)=0;
y’’(0)= (2*3-0)/9 + (2/27)*3 = 8/9;
y(x)= 1-x/3 +(8/9)x^2/2! = 1 –x/3 +(4/9)x^2
#2b. with no tailor; just take x1=0, x2=-0.01, x3=0.01 and plug into y=(5-2cosx) / (3+tanx); so y(0)=A=1; y(-0.01) = (5-2cos(-0.01)) / (3+tan(-0.01)) = 1.003378; y(+0.01) = (5-2cos(0.01)) / (3+tan(0.01)) = 0.996711
now 2 equations:
1.003378 = 1-0.01B+0.0001C
0.996711 = 1+0.01B+0.0001C
hence C=0.444454; B=-0.33336;