This set of Engineering Mathematics test focuses on “Taylor Mclaurin Series – 2”.
1. Let τa [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 is equal to _______
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
τa1(τa2(…….(f(x))….)) = f(x)
Where a1, a2………….an 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.
a) x+\(\frac{x^3}{3!}+\frac{x^5}{5!}+…..\infty\)
b) 1+\(\frac{x^2}{2!}+\frac{x^4}{4!}+….\infty\)
c) c1 \( \times cosh(x)+c_2\times sinh(x)\)
d) cosh(x) + sinh(x)
View Answer
Explanation: Consider the general form of Mclaurin series for f(x)
f(x)=a0+a1x+a2x2+a3x3+….∞
Then we have
f(0)(0)=f(n+2)(0)
n!an=(n+2)!an+2
an+2=\(\frac{a^n}{(n+1)(n+2)}\)
a2n=\(\frac{a_0}{(2n)!}\)
OR
a2n+1=\(\frac{a_1}{(2n+1)!}\)
One can deduce the general expression for f(x) to be
f(x)=\(a_0 \times (\frac{1}{1} + \frac{x^2}{2!} + …\infty) + a_1 \times (\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}+…\infty)\)
Now writing a0 = c1 and a1 = 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
View Answer
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?
a) \(\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}+…..\infty\)
b) \(\frac{(x-5)}{1!}+\frac{(x-5)^3}{3!}+\frac{(x-5)^5}{5!}+…\infty\)
c) \(1+\frac{x^2}{2!}+\frac{x^4}{4!}+…\infty\)
d) \(-(\frac{(x-5)}{1!}+\frac{(x-5)^3}{3!}+\frac{(x-5)^5}{5!}+…\infty)\)
View Answer
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?
a) g(x)=g(0)+(ex-1)+f(x)-f(0)
b) τ(f(x+tan(45)))=τ45(g(x))
c) g(x)=g(0)+g(1)(0).\(\frac{x}{1!}+g^{(2)}(0).\frac{x^2}{2!}+…\infty\)
d) g(x)=\(g(0)-\sum_{n=1}^{\infty}\frac{x^n \times f^{(1)}(n)}{n!}+(e^x-1)\)
View Answer
Explanation: The general expansion of Mclaurin is given by
g(x)=g(0)+g(1)(0).\(\frac{x}{1!}+g^{(2)}(0).\frac{x^2}{2!}+…\infty\)
We know that f(1) (n) = g(n) (0)
After the graph f(x) is rotated by 45 degrees we have an additional amount of tan45 gets added to slope of g(x)
f(1) (0) + tan(45) = g(n) (0)
After substituting this we have
g(x) = g(0) + (f(1) (1) + tan(45)) \(\times \frac{x}{1!}\) + (f(1) (2) + tan(45)) \(\times \frac{x^2}{2!}+…\infty\)
g(x) = g(0) + \((f^{(1)}(1).\frac{x}{1!} + f^{(1)}(2).\frac{x^2}{2!} + f^{(1)}(3).\frac{x^3}{3!}+…\infty)+(\frac{x}{1!}+\frac{x^2}{2!}+…\infty)\)
g(x)=\(g(0)-\sum_{n=1}^{\infty}\frac{x^n \times f^{(1)}(n)}{n!}+(e^x-1)\)
g(x)=g(0)+(ex-1)+f(x)-f(0)
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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