This set of Engineering Mathematics test focuses on “Taylor Mclaurin Series – 2”.

1. Let τ_{a} [ f (x) ] denote the Taylor series of f (x) centered at a then the value of the expression

[ τ199 (τ121 (τ1729 (f (101729))) ]^{3} – [τ342 (f (101729)) ]^{3}

a) 101729

b) 0

c) 1

d) -101729

View Answer

Explanation: The Taylor polynomial of any function is unique at any center. Also observe that Taylor series of any function is some polynomial. Coupling these facts we have

τa_{1}(τa_{2}(…….(f(x))….)) = f(x)

Where a_{1}, a_{2}………….a_{n} are real numbers

Hence the value of the given expression is

= [ τ199(τ121(τ1729(f(101729))) ]^{3} – [τ342(f(101729)) ]^{3}

= [f(101729)]^{3} – [f(101729)]^{3} = 0.

2. Function has the property that f^{(n)} (x) = f^{(n + 2)} (x) : n ≥ 1 : n ∊ N Then which of the following is the expression for f(x) in most general form.

View Answer

Explanation: Consider the general form of Mclaurin series for f(x)

Now writing a

_{0}= c1 and a

_{1}= c2 we have

f(x) = c1 * cosh(x) + c2 * sinh(x).

3. The value of f(1) can be deduced using Taylor series

a) True

b) False

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Explanation: Though the function f(x) is discontinuous at x = 1 we can still find f(1) as the function is continuous and differentiable in the interval. [0, 1 )

4. To find the value of cosh(23) with good accuracy the Taylor series should be centered at

a) 23

b) 22

c) 21

d) Delta (small) interval around 23

View Answer

Explanation: To find the Taylor Series according to the theory we have to find some center which is really close to 23 and has determinate value to find the value of cosh(23) with Taylor series.

5. Find the Taylor Series expansion of sinh (x) centered around 5

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Explanation: We know the Taylor series does not change the polynomial. Hence, be the polynomial centered at 5 or anywhere else would yield the same polynomial. In this case the polynomial centered at 0 has to be equal to the polynomial centered at 5.

6. Let f^{(1)} (n) = g^{(n)} (0) for some functions f(x) and g(x). Now let the coordinate axes having graph f(x) be rotated by 45 degrees (clockwise). Then the corresponding Mclaurin series of transformed g(x) is

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7. Let τa( f(x)) denote the Taylor series of the polynomial f(x) centered at a. Which of the following exactly happens after the Taylor series is formed.

a) τa( f(x)) = f(x)

b) The Taylor series has the effect of scaling GRAPH OF f(x)

c) The Taylor series transforms the origin

d) Scaled up graph obtained by factor of a

View Answer

Explanation: The Taylor series has no effect whatsoever on the nature of the polynomial. This is because the Taylor series is itself a polynomial and hence no change takes place.

τa( f(x)) = f(x).

8. Let τ(x) be the taylor Series centered at 0. Let the Taylor series be approximated to fifth degree polynomial only, which is the interval over which f(x) = sin(x) can be accurately calculated

a) ( -∞, ∞)

b) [ -2π, 2π].

c) [ -4π, 0].

d) It is accurate for any interval

View Answer

Explanation: The function sin(x) is known to have a root at every x = nπ The center is at 0. The Taylor polynomial also has five roots, spread symmetrically around the origin. After the polynomial crosses the roots it goes outward to infinity, But the sine function is bounded.

Hence, the interval where the roots of sine and polynomial approximately coincide is [ -2π, 2π].

9. The Taylor polynomial of degree 6 is approximated for cos(x). Then the interval in which the function can be accurately calculated using Taylor series (center = 80π)

a) [ -3π, 3π].

b) [ 77.5π, 83.5π].

c) [ -2.5π, 2.5π].

d) [ 77π, 83π].

View Answer

Explanation: Given the Taylor polynomial has degree 6 we must have 6 roots.

The symmetry point is at the center of the Taylor polynomial (which is 80π).

The cos(x) function has roots at nπ/2

Hence, the interval over which the roots approximately coincide is [ 77.5π, 83.5π].

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