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
Answer: b
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_2times sinh(x))
d) cosh(x) + sinh(x)
Answer: c
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
Answer: a
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
Answer: d
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))
Answer: a
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))
Answer: a
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)