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
Answer:
perdon no se
Step-by-step explanation:
Step-by-step explanation:
You are just replacing the 'n' with the number in the 'n box'.
So if this is the formula:
then:
B(-3)= 4-(-3/3)
B(-3)= 4-( -1)
B(-3)= 4+1
B(-3)= 5
B(0)= 4-(0/3)
B(0)= 4-0
B(0)= 4
B(3)= 4-(3/3)
B(3)= 4-1
B(3)= 3
I think that it your answer is 12.4 but im not 100% sure
plz mark brainiest
The prove that the equation can be verified using the laws of exponents.
<h3>What is the proof of the equation given; 2^(2x+4)= 16 × 2^(2x)?</h3>
It follows from the task content that the equation given is; 2^(2x+4)= 16 • 2^(2x).
It follows from the laws of indices ; particularly, the product of same base numbers.
The evaluation is therefore as follows;
2^(2x+4)= 16 • 2^(2x)
2^(2x) • 2⁴ = 16 • 2^(2x)
2^(2x) • 16 = 16 • 2^(2x)
Hence, since LHS = RHS, it follows that the expression is mathematically correct.
Read more on laws of exponents;
brainly.com/question/847241
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