Question

maclaurin series expansion of cos^2x-sin^2x is​

Answer 1

Answer:

$\begin{aligned}& \cos x^2-x \sin 2 x= \\& \sum_{n=0}^{\infty}(-1)^n\left(\frac{x^{4 n}}{(2 n) !}-\frac{2^{1+2 n} x^{(2+2 n)}}{(1+2 n) !}\right)\end{aligned}$

Step-by-step explanation:

Step 1: Given the values of the function's successive derivatives at zero, a Maclaurin series is a power series that enables one to construct an approximation of a function f(x) with input values close to zero. It is equivalent to the function it represents in many practical contexts.

Step 2: The sine function is a case where the Maclaurin series is helpful. The sine function formulation prevents an easy technique of calculating the function's output values at arbitrary input values. However, for x=0, it is simple to determine the values of sin(x) and all of its derivatives. With the aid of the Maclaurin series, it is possible to construct accurate estimates of sin(x) for inputs that are nearly equal to zero but not quite so.

Step 3: A Maclaurin series can be used to calculate an otherwise impossible sum, identify the antiderivative of a complex function, or approximate a function. Polynomial approximations for the function are provided by partial sums of a Maclaurin series.

$\begin{aligned} & \cos x^2=\sum_{n=0}^{\infty}(-1)^n \frac{x^{4 n}}{(2 n) !} \\ & \sin 2 x= \sum_{n=0}^{\infty}(-1)^n \frac{2^{2 n+1} x^{2 n+1}}{(2 n+1) !}\end{aligned}$

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