Question

Answer the following:
a) if y = sin2x sin3x, then find yn

(b) if y = sin(m sin-2x) then show that (1 – x?)yn+2 – (2n + 1)xYn+1 –
(n2 – m)yn = 0
| (c) Expand logx in power of (x -- 1) by Taylor's theorem
(3)
(d) Evaluate:
(2*2 =4)
(i), o x logx
. Lt et-e-x-2x
x + 0 x-sinx
(e) Expand the function sinx in power of r in Maclaurin's series.​

Answer 1

Answer:

The function sinx in power of r in Maclaurin's series:

$y^n=\frac{1}{4}\left(5^n \sin \left(\frac{n \pi}{2}+5 x\right)+3^n \sin \left(\frac{n \pi}{2}+3 x\right)-9^n \sin \left(\frac{n \pi}{2}+9 x\right)+\sin \left(\frac{n \pi}{2}+x\right)\right)$

Step-by-step explanation:

Step 1: The Scottish mathematician Colin Maclaurin is honoured with the name of the Maclaurin series, which is a Taylor series extension of a function at zero. Any terms in a maclaurin series expansion must be nonnegative integer powers of the variable.

$\begin{aligned}& \text { I. } y=\frac{1}{2}(2 \sin x \sin 2 x) \sin 3 x=\frac{1}{2}(\cos x-\cos 3 x) \sin 3 x \\& y=\frac{1}{4}(\sin 4 x+\sin 2 x-\sin 6 x) \\& y=\frac{1}{4}\left(4 \sin \left(\frac{\pi}{2}+4 x\right)+2 \sin \left(\frac{\pi}{2}+2 x\right)-6 \sin \left(\frac{\pi}{2}+6 x\right)\right)\end{aligned}$

$\begin{aligned}& \mathrm{y} \frac{1}{4}\left(4^2 \sin (\pi+4 x)+2^2 \sin (\pi+2 x)-6^2 \sin (\pi+6 x)\right) \\& y^n=\frac{1}{4}\left(4^n \sin \left(\frac{n \pi}{2}+4 x\right)+2^n \sin \left(\frac{n \pi}{2}+2 x\right)-6^n \sin \left(\frac{n \pi}{2}+6 x\right)\right)\end{aligned}$

Step 2: A set of polynomial functions can be used to approximate functions using the Taylor and Maclaurin series. In other words, you are combining several smaller functions to create a larger function. For a straightforward illustration, add the following numbers together to get the number 10: 1 + 2 + 3 + 4.

$\begin{aligned}& \text { II. } y=\frac{1}{2}(2 \sin 2 x \sin 3 x \sin 4 x)=\frac{1}{2}(\cos x-\cos 5 x) \sin 4 x \\& y=\frac{1}{4}(2 \cos x \sin 4 x)=\frac{1}{4}(\sin 5 x+\sin 3 x-\sin 9 x+\sin x) \\& y^n=\frac{1}{4}\left(5^n \sin \left(\frac{n \pi}{2}+5 x\right)+3^n \sin \left(\frac{n \pi}{2}+3 x\right)-9^n \sin \left(\frac{n \pi}{2}+9 x\right)+\sin \left(\frac{n \pi}{2}+x\right)\right)\end{aligned}$

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