n=1 (−1)(n− 1)xn+ yn n 3.Since f (x, y

2. Since f (x, y) = ln(1 + x + y + xy) = ln((1 + x)(1 + y)) = ln(1 + x) + ln(1 + y), the Taylor series for f about (0, 0) is

∑∞ n=1

(−1)⁽n− 1)xⁿ+ yⁿ n

3.Since f (x, y) = tan⁻¹(x + xy) = tan⁻¹(ux), where u = y + 1, the Taylor series for f about (0,−1) is

∑∞ n=0

(−1)ⁿ(ux)²ⁿ⁺¹ 2n + 1 =

∑∞ 0

(−1)ⁿx²ⁿ⁺¹(1 + y)²ⁿ⁺¹ 2n + 1 .

5.

f (x, y) = e^x2+y2=

∑∞ n=0

(x²+ y²)ⁿ

n! =

∑∞ n=0

1 n!

∑n j=0

n!

j!(n− j)!x^2jy^2n−2j=

∑∞ n=0

∑n j=0

x^2jy^2n−2j j!(n− j)!. This is the Taylor series for f about(0, 0).

7.let u = x− 2, v = y − 1. Then f (x, y) = 1

2 + x− 2y = 1

2 + (2 + u)− 2(v + 1) = 1

2 + u− 2v = 1 2(1 +^u−2v₂ )

=1

2[1−u− 2v

2 + (u− 2v

2 )²− (u− 2v

2 )³+ ...] = 1 2 −u

4 +v 2+ u²

8 −uv 2 +v²

2 −u³ 16+ ...

The Taylor polynomial of degree 3 for f about (2, 1) is 1

2−x− 2

4 +y− 1

2 +(x− 1)²

8 −(x− 2)(y − 1)

2 +(y− 1)² 2

−(x− 2)³

16 +3(x− 2)²(y− 1)

8 −3(x− 2)(y − 1)²

4 +(y− 1)³

2 .

13. The equation xsiny = y + sinx can be written F (x, y) = 0 where F (x, y) = xsiny− y − sinx. Since F (0, 0) = 0, and F2(0, 0) =−1 ̸= 0, the given equation has a solution of the form y = f (x) where f (0) = 0. Try

y = a1x + a2x²+ a3x³+ a4x⁴+ ....

Then

siny = y−1

6y³+ ... = a1x + a2x²+ a3x³+ a4x⁴+ ..−1

6(a1x + ...)³+ ...

Substituting into the given equation we obtain a1x²+ a2x³+ (a3−1

6a³₁)x⁴+ ... = a1x + a2x²+ a3x³+ a4x⁴+ ... + x−1 6x³+ ...

Comparing coefficients of various powers of x on both sides, we get a₁+ 1 = 0, a₂= a₁, a₃−1

6 = a₂.

1

2

Thus a1= 1, a2=−1, a3=−⁵₆. The required solution is y =−x − x²−⁵₆x³+ ....

16. The coefficient of x²y in the Taylor series for f (x, y) = tan⁻¹(x + y) about (0, 0) is _2!1!¹ f₁₁₂(0, 0) = ¹₂f₁₁₂(0, 0). But tan⁻¹(x + y) = x + y−¹₃(x + y)³+ ... = x + y−¹₃(x³+3x²y+3xy²+y³)+... ,so the coefficient of x²y is−1.Hence f112(0, 0) =−2.