# Engineering Mathematics Questions and Answers – Taylor Mclaurin Series – 4

This set of Engineering Mathematics Quiz focuses on “Taylor Mclaurin Series – 4”.

1. The expansion of f(x), about x = a is
a) $$f(a)+\frac{h}{1!} f’ (a)+\frac{h^2}{2!} f” (a)….+\frac{h^n}{n!} f^n (a)$$
b) $$f(a)+\frac{h}{1!} f’ (a)+\frac{h^2}{2!} f” (a)….$$
c) $$hf(a)+\frac{h^2}{1!} f’ (a)+\frac{h^3}{2!} f” (a)…+\frac{h^n}{n!} f^n (a)$$
d) $$hf(a)+\frac{h^2}{1!} f’ (a)+\frac{h^3}{2!} f” (a)…..$$

Explanation: By taylor expansion,
f(a+h) = f(a) + h1! f’ (a) + h22! f (a)…….

2. Find the expansion of ex in terms of x + m, m > 0.
a) $$e^m [1+(x+m)+\frac{(x+m)^2}{2!}+\frac{(x+m)^3}{3!}+….]$$
b) $$e^{-m} [1+(x-m)+\frac{(x-m)^2}{2!}+\frac{(x-m)^3}{3!}+….]$$
c) $$e^m [1+(x-m)+\frac{(x-m)^2}{2!}+\frac{(x-m)^3}{3!}+….]$$
d) $$e^{-m} [1+(x+m)+\frac{(x+m)^2}{2!}+\frac{(x+m)^3}{3!}+….]$$

Explanation: Let, h = x + m = > f(x) = f(h-m) = e(h-m)
By taylor theorem, putting a = -m, we get,
f(a+h) = $$f(a)+\frac{h}{1!} f’ (a)+\frac{h^2}{2!} f” (a)…$$
f(h-m) = $$f(-m)+h/1! f'(-m)+\frac{h^2}{2!} f” (-m)….(1)$$
now, f(-m) = f’(-m) = f’’(-m)=e-m
hence,
f(x)=ex=f(h-m)=$$e^{-m} [1+(x+m)+\frac{(x+m)^2}{2!}+\frac{(x+m)^3}{3!}+….]$$

3. Expand ln(x) in the power of (x-m).
a) ln⁡(m)+$$\frac{h}{m}-\frac{1}{2!} (h/m)^2+\frac{2}{3!} (h/m)^3-……$$
b) ln⁡(m)-$$\frac{h}{m}-\frac{1}{2!} (h/m)^2-\frac{2}{3!} (h/m)^3-……$$
c) ln⁡(m)-$$\frac{1}{2!} (h/m)^2+\frac{2}{3!} (h/m)^4-……$$
d) ln⁡(m)+$$\frac{h}{m}+\frac{2}{3!} (h/m)^3-……$$

Explanation: where, h = x-m
Let, h = x – m => f(x) = f(h+m) = e(h+m)
By taylor theorem, putting a = m , we get,
f(a+h) = $$f(a)+\frac{h}{1!} f’ (a)+\frac{h^2}{2! }f” (a)…$$
f(h-m) = $$f(m)+\frac{h}{1!} f’ (m)+\frac{h^2}{2!} f” (m)$$…….(1)
now,f(m) = ln(m), f’(m)=1/m, f” (m)=-1/m2, f”’ (m)=2/m3,……
hence,
f(x)=ln(x)=f(h+m)=$$ln⁡(m)+\frac{h}{m}-\frac{h^2}{2!}\frac{1}{m^2}+\frac{h^3}{3!} \frac{2}{m^3}-……$$
f(x)=ln(x)=ln⁡(m)+$$\frac{h}{m}-\frac{1}{2!} (h/m)^2+\frac{2}{3!} (h/m)^3-……$$
where, h = x-m

4. Find the value of √10
a) 3.1633
b) 3.1623
c) 3.1632
d) 3.1645

Explanation: Now f(x)=√x,
Hence, f’ (x)=$$\frac{1}{2} x^{-1/2}$$
f” (x)=$$-\frac{1}{4} x^{-3/2}$$
f”’ (x)=$$\frac{3}{8} x^{-5/2}$$
Hence,
f(x+h)=$$\sqrt{x+h}=\sqrt{x}+\frac{h}{2} x^{-1/2}-\frac{h^2}{8} x^{-3/2}+\frac{h^3}{16} x^{-5/2}+….$$
(By Taylor’s expansion)
Putting,
h=1, and x=9 we get,
f(10)=√10=3+1/6-1/216+1/3888+….=3.1623

5. Expand f(x) = 1x about x = 1.
a) 1 – (x-1) + (x-1)2 – (x-1)3 + ….
b) 1 + (x-1) + (x-1)2 + (x-1)3 + ….
c) 1 + (x-1) – (x-1)2 + (x-1)3 + ….
d) 1 – (x+1) + (x+1)2 – (x+1)3 + ….

Explanation: Given f(x) = 1x
Let, x – 1 = h
Hence, x = 1 + h
Hence, f(x) = f(1 + h) = f(1) + h1! f’ (1) + h22! f (1) +h33! f”’ (1)+…
Now, f(1) = 1, f'(1) = -1, f”(1) = 2 ,f”'(1) = -6,…….
Hence, f(1 + h) = 1 – h + h2 – h3+….
hence, 1 – (x-1) + (x-1)2 – (x-1)3 +….
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6. Find the expansion of f(x) = ex1+ex, given ∫f(x)dx = ln⁡(2), for x = 0
a) $$\frac{1}{2}-\frac{x}{4}-\frac{x^3}{48}-…$$
b) $$\frac{1}{2}+\frac{x}{4}-\frac{x^3}{48}+….$$
c) $$\frac{1}{2}+\frac{x}{4}+\frac{x^3}{48}+….$$
d) $$\frac{1}{2}+\frac{x}{4}-\frac{x^3}{48}+….$$

Explanation:
Given,f(x)=$$\frac{e^x}{1+e^x}$$
Differentiating it we get
$$f^1 (x)=ln⁡(1+e^x)+C$$, now putting x=0 we get,c=0
Hence,
$$f^1 (x)=ln⁡(1+e^x)$$
Now,$$f^1 (x)=ln⁡(1+e^x)=ln⁡(2)+\frac{x}{2}+\frac{x^2}{8}-\frac{x^4}{192}+..$$(By,Mclaurin’s expansion)
Hence, Differentiating it we get,
f(x)=$$\frac{1}{2}+\frac{x}{4}-\frac{x^3}{48}+….$$.

7. Find the value of eπ4√2
a) 1.74
b) 1.84
c) 1.94
d) 1.64

Explanation: Let, f(x) = exSin(x), f(0) = 1
Now, the expansion of xSin(x) is $$x^2-\frac{x^3}{3!}+\frac{x^6}{5!}+….$$
Hence, $$e^xSin(x)=e^y=1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+….$$
Hence,
$$e^{xSin(x)}=1+(x^2-\frac{x^4}{3!}+\frac{x^6}{6!}+…..)+\frac{(x^2-\frac{x^4}{3!}+\frac{x^6}{6!}+…..)^2}{2!}$$
$$+\frac{(x^2-\frac{x^4}{3!}+\frac{x^6}{6!}+…..)^3}{6}+….$$
$$e^{xSin(x)}=1+x^2-\frac{x^4}{3!}+\frac{x^6}{5!}+\frac{x^4}{2}-\frac{x^6}{6}+\frac{x^6}{6}+….$$ (we neglect all other other terms by considering the options given)
Hence, $$e^{xSin(x)}=1+x^2+\frac{x^4}{3}+\frac{x^6}{120}+……$$
Putting, x = π/4,
We get,
f(π/4)=eπ/4 Sin(π/4)=eπ/(4√2)=1+(π/4)2+1/3 (π/4)4+….=1+.6168+.1268=1.74

8. Find the value of ln(sin(31o)) if ln(2) = 0.69315
a) -0.653
b) -0.663
c) -0.764
d) -0.662

Explanation: Let, f(x) = ln⁡(sin⁡(x+h))
Then, f(x) = ln⁡(sin⁡(x)), if h=0
f’ (x)=cot⁡(x), f” (x)=-cosec2 (x), f”’ (x)=2cosec2 (x)cot⁡(x)
Hence, by Taylor’s theorem,
f(x+h)=f(x)+hf'(x)+$$\frac{h^2}{2!}$$ f” (x)+$$\frac{h^3}{3!}$$ f”’ (x)+⋯..
Hence, ln⁡(sin⁡(x+h))=ln⁡(sin⁡(x))+h cot⁡(x)-$$\frac{h^2}{2!}$$ cosec2 (x)+$$\frac{h^3}{3!}$$ (2cosec2 (x) cot⁡(x))+⋯.
Now let, x=30o, h=1o;
ln(sin(31o)) = $$ln(sin(30))+\frac{π}{180} cot⁡(\frac{π}{6})-\frac{1}{2!} (\frac{π}{180})^2 cosec^2 (\frac{π}{6})+⋯$$
ln(sin(31o)) = -0.6935+.030231-.000304+0
ln(sin(31o)) = -0.663

9. The expansion of f(x,y), is
a) f(0,0)+$$[x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}-2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+….$$
b) f(0,0)+$$[x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+…$$
c) f(0,0)+$$[x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}-2xy \frac{∂^2 f}{∂x∂y}-y^2 \frac{∂^2 f}{∂y^2}]+…$$
d) f(0,0)-$$[x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]-…$$

Explanation: By taylor expansion,
f(x,y) = f(0,0)+$$[x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+…$$

10. The expansion of f(x, y)=ex Sin(y), is
a) x + xy + ….
b) y + y2 x + ….
c) x + x2 y + ….
d) y + xy + ……..

Explanation: Now, f(x, y)=ex Sin(y), f(0,0) = 0
Therefore,
fx (x,y) = ex Sin(y), hence fx (0,0) = 0

fy (x,y) = ex Cos(y), hence fy (0,0) = 1

fxx (x,y) = ex Sin(y), hence fxx (0,0) = 0

fyy (x,y) = -ex Sin(y), hence fyy (0,0) = 0

fxy (x,y) = ex Cos(y), hence fxy (0,0) = 1
By taylor expansion,
f(x,y) = f(0,0)+$$[x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+…$$
f(x,y) = 0 + 0 + y + $$\frac{1}{2!}$$ [0 + 2xy + 0] +…..
f(x,y) = y + xy + ……..

11. The expansion of f(x, y) = ex ln(1 + y), is
a) f(x,y) = y + xy – y22 +…….
b) f(x,y) = y – xy + y22 -…….
c) f(x,y) = y + x – y22 +……..
d) f(x,y) = x + y – x22 +……..

Explanation: Now, f(x, y) = ex ln(1 + y) , f(0,0) = 0
Therefore,
$$f_x (x,y)=e^x ln(1+y)$$, hence fx (0,0) = 0
$$f_y (x,y)=\frac{e^x}{(1+y)}$$, hence fy (0,0) = 1
$$f_{xx} (x,y)=e^x ln(1+y)$$, hence fxx (0,0) = 0
$$f_{yy} (x,y)=-\frac{e^x}{(1+y)^2}$$, hence fyy (0,0) = -1
$$f_{xy} (x,y)=\frac{e^x}{(1+y)}$$, hence fxy (0,0) = 1
By taylor expansion,
f(x,y) = f(0,0)+$$[x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+1/2! [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+…$$
f(x,y) = y + xy – y22 +…….

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