TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1992 question 4 solution
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\displaystyle \text{Consider }\lozenge (\lambda\mu ) = \lambda\lozenge\mu + \mu\lozenge\lambda \text{ [a]} \\
\text{But } \lozenge (\lambda\mu ) = \lambda\lozenge\mu \text{ [d]} \\
\therefore \lambda\lozenge\mu + \mu\lozenge\lambda = \lambda\lozenge\mu \Rightarrow \mu\lozenge\lambda = 0 \Rightarrow \lozenge\lambda = 0.\\ \\
\lozenge x^2 = x\lozenge x + x\lozenge x \text{ [a]} \\
= x.1 + x.1 \text{ [c]} \\
= 2x.\\
\lozenge x^3 = x^2 \lozenge x + x \lozenge x^2 \text{ [a]} \\
= x^2.1 + x.2x \text{ [c]} \\
= 3x^2. \\ \\
\text{Proposition: } \lozenge x^n = nx^{n-1}. \text{ Proceed by induction; assume true for } n \leq k. \\
\lozenge x^k = kx^{k-1} \\
\lozenge x^{k+1} = \lozenge (x^k.x) = x^k\lozenge x + x\lozenge x^k \text{ [a]}
= x^k.1 + x.kx^{k-1}\\
= (k+1)x^k \text{as required}. \\
\therefore \lozenge x^n = nx^{n-1}. \\ \\
h(x) = k_0 + k_1x + k_2x^2 + k_3x^3 + \dots + k_rx^r + \dots\\
\lozenge h(x) = \lozenge (k_0 + k_1x + k_2x^2 + k_3x^3 + \dots + k_rx^r + \dots ) \\
\lozenge h(x) = \lozenge (k_0) + \lozenge (k_1x) + \lozenge (k_2x^2) + \lozenge (k_3x^3) + \dots + \lozenge (k_rx^r) + \dots \text{ [b]} \\
\lozenge h(x) = 0 + k_1\lozenge (x) + k_2\lozenge (x^2) + k_3\lozenge (x^3) + \dots + k_r\lozenge (x^r) + \dots \text{ [d]} \\
\lozenge h(x) = k_1(1) + k_2(2x) + k_3(3x^2) + \dots + k_r(rx^{r-1}) + \dots \\
\therefore \lozenge h(x) = \frac{\text{d}}{\text{d}x} h(x).
Solution by generalebriety.
