Question

multiple Choice
6. First two terms in expansion of log cos x by Taylor's theorem in ascending powers of ( x-
(x-4)
is
TC
(B) log
+
...
(A) log}-(x-1)...
to logtz-(x-)...
(D) log3+(x-4)...​

Answer 1

Answer:

Consider a function f that has a power series representation at x=a. Then the series has the form

∞

∑

n=0 cn(x−a)n=c0+c1(x−a)+c2(x−a)2+⋯.

(6.4)

What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series Equation 6.4 is a representation for f at x=a, we certainly want the series to equal f(a) at x=a. Evaluating the series at x=a, we see that

∞

∑

n=0 cn(x−a)n =c0+c1(a−a)+c2(a−a)2+⋯  =c0.  

Thus, the series equals f(a) if the coefficient c0=f(a). In addition, we would like the first derivative of the power series to equal f′(a) at x=a. Differentiating Equation 6.4 term-by-term, we see that

d

dx

 

(  

∞

∑

n=0 cn(x−a)n)=c1+2c2(x−a)+3c3(x−a)2+⋯.

Therefore, at x=a, the derivative is

d

dx

 

(  

∞

∑

n=0 cn(x−a)n) =c1+2c2(a−a)+3c3(a−a)2+⋯  =c1.  

Therefore, the derivative of the series equals f′(a) if the coefficient c1=f′(a). Continuing in this way, we look for coefficients cn such that all the derivatives of the power series Equation 6.4 will agree with all the corresponding derivatives of f at x=a. The second and third derivatives of Equation 6.4 are given by

2

 

(  

∞

∑

n=0 cn(x−a)n)=2c2+3·2c3(x−a)+4·3c4(x−a)2+⋯

and

d3

dx3

 

(  

∞

∑

n=0 cn(x−a)n)=3·2c3+4·3·2c4(x−a)+5·4·3c5(x−a)2+⋯.

Therefore, at x=a, the second and third derivatives

d2

dx2

 

(  

∞

∑

n=0 cn(x−a)n) =2c2+3·2c3(a−a)+4·3c4(a−a)2+⋯  =2c2  

and

d3

dx3

 

(  

∞

∑

n=0 cn(x−a)n) =3·2c3+4·3·2c4(a−a)+5·4·3c5(a−a)2+⋯  

Step-by-step explanation: