Answer:
Σ(-1)^kx^k for k = 0 to n
Step-by-step explanation:
The nth Maclaurin polynomials for f to be
Pn(x) = f(0) + f'(0)x + f''(0)x²/2! + f"'(0)x³/3! +. ......
The given function is.
f(x) = 1/(1+x)
Differentiate four times with respect to x
f(x) = 1/(1+x)
f'(x) = -1/(1+x)²
f''(x) = 2/(1+x)³
f'''(x) = -6/(1+x)⁴
f''''(x) = 24/(1+x)^5
To calculate with a coefficient of 1
f(0) = 1
f'(0) = -1
f''(0) = 2
f'''(0) = -6
f''''(0) = 24
Findinf Pn(x) for n = 0 to 4.
Po(x) = 1
P1(x) = 1 - x
P2(x) = 1 - x + x²
P3(x) = 1 - x+ x² - x³
P4(x) = 1 - x+ x² - x³+ x⁴
Hence, the nth Maclaurin polynomials is
1 - x+ x² - x³+ x⁴ +.......+(-1)^nx^n
= Σ(-1)^kx^k for k = 0 to n
Since the plot of "The Wife of Bath's Tale" has at its heart a loathly lady who shape-shifts into a beautiful, young damsel, we might expect appearances to be important here. And they are, just not for the reason you might think. For instead of this being a tale about how a knight learns to appreciate people for what's on the inside and that outer appearances don't matter, it's a tale about how a knight learns to give up sovereignty to his wife. That sovereignty includes power over the body. The loathly lady's physical appearance becomes an important symbol of that body, so that, at the end of the tale, when she offers her husband a choice about how he wants her to look, she's in essence offering him control of her body. He grants this control back to her, thus proving his understanding of the doctrine of women's sovereignty in marriage. Medieval stories don't necessarily go in for the whole 'appearances don't mean anything' maxim anyway, as we've seen in the "General Prologue<span>."</span>
Answer:
is this a yes or no question
The midpoint is the average of the endpoints.
((-11+i) + (-4+4i))/2 = -15/2 +5/2i
