Friday, August 9, 2019

f(x)=xsinx Find the Maclaurin series for the function.

Maclaurin series  is a special case of Taylor series which is centered at a=0. We follow the formula:
f(x) =sum_(n=0)^oo (f^n(0))/(n!)x^n
or
f(x) = f(0) + (f'(0))/(1!)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3 +(f^4(0))/(4!)x^4 +(f^5(0))/(5!)x^5 +...
To list of f^n(x) up to n=10 , we may apply the Product rule for differentiation: d/(dx) (u*v) = u'*v +u*v '.
f(x) = xsin(x)
f'(x) = xcos(x)+sin(x)
f''(x) = 2cos(x)-xsin(x)
f'''(x) =- xcos(x)-3sin(x)
f^4(x) = xsin(x)-4cos(x)
f^5(x) = xcos(x)+5sin(x)
f^6(x) = 6cos(x)-xsin(x)
f^7(x) = -xcos(x)-7sin(x)
f^8(x) = xsin(x)-8cos(x)
f^9(x) = xcos(x)+9sin(x)
f^(10)(x)= 10*cos(x)-x*sin(x)
Note: d/(dx)x=1 ,  d/(dx) cos(x) =-sin(x) , and d/(dx) sin(x)=cos(x) .
Plug-in x =0, we get:
f(0) = 0*sin(0)
         =0*0
         =0
f'(0) = 0*cos(0)+sin(0)
         =0*1+0
         =0
f''(0) = 2cos(0)-0*sin(0)
           =2*1-0*0
           =2
f'''(0) =- 0*cos(0)-3sin(0)
            =-0*1-3*0
            =0
f^4(0) = 0*sin(0)-4cos(0)
          =0*0 -4*1
          =-4
f^5(0) = 0*cos(0)+5sin(0)
          =0*1+5*0
          =0
f^6(0) = 6cos(0)-0*sin(0)
          =6*1-0*0
          =6
f^7(0) = -0*0cos(0)-7sin(0)
           =-0*1-7*0
           =0
f^8(0) =0*sin(0)-8cos(0)
          =0*0-8*1
          =-8
f^9(0) = 0*cos(0)+9sin(0)
           =0*1+9*0
           =0
f^(10)(0)= 10*cos(0)-0*sin(0)
            =10*1-0*0
            =10
Note: cos(0)=1 and sin(0) =0 .
Plug-in the values in the formula, we get:
f(x) = 0 + 0/(1!)x+2/(2!)x^2+0/(3!)x^3+(-4)/(4!)x^4+0/(5!)x^5
+6/(6!)x^6+ 0/(7!)x^7+(-8)/(8!)x^8+0/(9!)x^9 +10/(10!)x^10+...
 
=0 + 0/(1)x+2/(1*2)x^2+0/(1*2*3)x^3-4/(1*2*3*4)x^4
+ 0/(1*2*3*4*5)x^5 + 6/(1*2*3*4*5*6)x^6+0/(1*2*3*4*5*6*7)x^7
-8/(1*2*3*4*5*6*7*8)x^8 + 0/(1*2*3*4*5*6*7*8*9)x^9 + 10/(1*2*3*4*5*6*7*8*9*10)x^(10)+...
 
=0 + 0+2/2x^2+0/6x^3-4/24x^4 + 0/120x^5 + 6/720x^6
+0/5040x^7 -8/40320x^8 + 0/362880x^9 +10/3628800x^(10)+...
 
=0 + 0+x^2+0-1/6x^4 + 0 + 1/120x^6
+0-1/5040x^8 + 0+1/362880x^(10)+...
 
=x^2 -1/6x^4 + 1/120x^6 -1/5040x^8 +1/362880x^(10)+...
Therefore, the Maclaurin series for the function f(x) =xsin(x) can be expressed as:
xsin(x)=x^2 -1/6x^4 + 1/120x^6 -1/5040x^8 +1/362880x^(10)+...

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