Any method to verify Talor / Maclaurin series for a function. ( Check my final answer) Ex : e^x (1-x)^-1 = 1 + 2x + 5/2 x^2 +.... or e^x(1-x)^-1 = 1+ 2x + 13/6 x^2 +.... Putting random values and checking using the calc. doesn't work. close even in wrong situations .
Well I got the same as the first one 1+2x+5/2x^2+8/3x^3... and You're supposed to check only with values very close to zero.
e^x = 1 + x + x^2/2! + x^3/3!+ x^4/4! e^x/(1-x): 1 +2x +(2+1/2!)x^2 + ... ------------------------------ 1-x | 1 + x + x^2/2! + x^3/3!+ x^4/4! (1 - x) ------ 2x +x^2/2! (2x -2x^2) ----------- (2+1/2!)x^2 + x^3/3! ((2+1/2!)x^2 -(2+1/2!)x^3) ----------------------- (2+1/2!+1/3!)x^3 or: 1 +2x + (2+1/2!)x^2 + (2+1/2!+1/3!)x^3 +(2+1/2!+1/3!+1/4!)x^4+... 1 +2x + (5/2!)x^2 + (16/3!)x^3 +(65/4!)x^4+...
yeah you can also guess a few terms using their individual series\[e^x = 1 + x + x^2/2! + x^3/3! + x^4/4!\cdots\]\[(1-x)^{-1}=1 + x + x^2 + x^3 + x^4 + \cdots\]
@math&ing001, Tried using very close to zero values: x = 0.001 estimated: got 1.0020025 for the first got 1.002002167 for the second actual: 1.002002503 The first is closer, but the second is too close too. @amistre64, My calculus exam is MCQ so I don't want to waste my time. while the answer is already written in A,B ,C or D @ParthKohli , Yeah , works sometimes,but hell if I forgot a term.
@TrojanPoem that's to be expected. To get really close results you have to go even further with the terms. The more terms you have, the more precise your result will be.
@math&ing001 , maybe It will work that way I will use 0.0000001 as x
Yes, and I was also talking about terms as in e^x (1-x)^-1 = 1 + 2x + 5/2 x^2 + 8/3 x^3 + ax^4 +.... With higher degrees you get better results.
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