Standard +0.8 This question requires understanding that Taylor series coefficients relate to derivatives via the formula $a_n = f^{(n)}(a)/n!$, then carefully extracting the coefficient of $(x-1)^4$ from the given series (which involves the product $1×3×5×7$ in the denominator and $(-2)^3$ in the numerator), and finally computing $4! × $ (coefficient). While conceptually straightforward for students familiar with Taylor series, the algebraic manipulation and factorial arithmetic require care, making it moderately challenging but still a standard Further Maths technique question.
The Taylor series expansion, about \(x = 1\), of the function \(y\) is
$$y = 1 + \sum_{n=1}^{\infty} \frac{(-2)^{n-1}(x-1)^n}{1 \times 3 \times 5 \times \ldots \times (2n-1)}.$$
Find the value of \(\frac{\text{d}^4 y}{\text{d}x^4}\) when \(x = 1\). [3]
a 4 = 4 ! = 1.3.5.7 from the Taylor series expansion
( )
−2 3 64
⇒ y(4)(1)= ×4 !=−
1.3.5.7 35
ALTERNATIVE
d4y ∞ (−2)n−1n(n−1)(n−2)(n−3)(x−1)n − 4
Diff te . four times to get = ∑
dx4 1.3.5....(2n−1)
n = 4
( )
−2 3×4(3)(2)(1) 64
When x = 1, this is +0=−
Answer
Marks
1.3.5.7 35
M1 A1
A1
M1 A1
A1
[3]
Question 2:
2 | ( )
y(4)(1) −2 3
a 4 = 4 ! = 1.3.5.7 from the Taylor series expansion
( )
−2 3 64
⇒ y(4)(1)= ×4 !=−
1.3.5.7 35
ALTERNATIVE
d4y ∞ (−2)n−1n(n−1)(n−2)(n−3)(x−1)n − 4
Diff te . four times to get = ∑
dx4 1.3.5....(2n−1)
n = 4
( )
−2 3×4(3)(2)(1) 64
When x = 1, this is +0=−
1.3.5.7 35 | M1 A1
A1
M1 A1
A1
[3]
The Taylor series expansion, about $x = 1$, of the function $y$ is
$$y = 1 + \sum_{n=1}^{\infty} \frac{(-2)^{n-1}(x-1)^n}{1 \times 3 \times 5 \times \ldots \times (2n-1)}.$$
Find the value of $\frac{\text{d}^4 y}{\text{d}x^4}$ when $x = 1$. [3]
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2015 Q2 [3]}}