Question

find taylor series expansion for : x+3/(x-1)(x-4) at x=2

Answer 1

For a general function f(x) the expansion is _n=0∑^n=∞f⁽ⁿ⁾(a)(x-a)ⁿ/n!, where a is the point of differentiation.

This expands to f(a)+f'(a)(x-a)/1!+f"(a)(x-a)²/2!+... ad infinitum.

When f(x)=(x+3)/((x-1)(x-4)) we need to differentiate f an indefinite number of times.

f(2)=5/((1)(-2))=-2.5.

Let u=x+3 and v=(x-1)(x-4)=x²-5x+4, 

then du/dx=1 and dv/dx=2x-5.

df/dx=(vdu/dx-udv/dx)/v²=(x²-5x+4-(x+3)(2x-5))/(x²-5x+4)²,

f'(x)=df/dx=(x²-5x+4-2x²-x+15)/(x²-5x+4)²=(-x²-6x+19)/(x²-5x+4)².

f'(2)=(-4-12+19)/(4-10+4)²=3/4.

This makes the next term in the series 3(x-2)/4.

Using a similar technique f"(x)=2(x³+9x²-57x+83)/(x²-5x+4)³,

so f"(2)=2(8+36-114+83)/(4-10+4)³=-13/4, making the next term -13(x-2)²/8.

The first three terms are -5/2+3(x-2)-13(x-2)²/8. To get more terms keep differentiating and then plugging in x=2 to find f⁽ⁿ⁾(2).