Math, asked by nakulsharma55242, 8 months ago

The value of fourier coefficient bn of the fourier sine series f(x)=x^2 for x in (0,π) is
π^2/3
π^2/5
none of these

Answers

Answered by chandanalikhik
0

Step-by-step explanation:

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Answered by jisoo85
0

Answer:

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Step-by-step explanation:

This section explains three Fourier series: sines, cosines, and exponentials eikx.

Square waves (1 or 0 or −1) are great examples, with delta functions in the derivative.

We look at a spike, a step function, and a ramp—and smoother functions too.

Start with sin x. It has period 2π since sin(x + 2π) = sin x. It is an odd function

since sin(−x) = − sin x, and it vanishes at x = 0 and x = π. Every function sin nx

has those three properties, and Fourier looked at infinite combinations of the sines:

Fourier sine series S(x) = b1 sin x + b2 sin 2x + b3 sin 3x + ··· = ∞

n=1

bn sin nx (1)

If the numbers b1, b2,... drop off quickly enough (we are foreshadowing the im-

portance of the decay rate) then the sum S(x) will inherit all three properties:

Periodic S(x + 2π) = S(x) Odd S(−x) = −S(x) S(0) = S(π)=0

200 years ago, Fourier startled the mathematicians in France by suggesting that any

function S(x) with those properties could be expressed as an infinite series of sines.

This idea started an enormous development of Fourier series. Our first step is to

compute from S(x) the number bk that multiplies sin kx.

Suppose S(x) = bn sin nx. Multiply both sides by sin kx. Integrate from 0 to π:

π

0

S(x) sin kx dx =

π

0

b1 sin x sin kx dx + ··· +

π

0

bk sin kx sin kx dx + ··· (2)

On the right side, all integrals are zero except the highlighted one with n = k.

This property of “orthogonality” will dominate the whole chapter. The sines make

90◦ angles in function space, when their inner products are integrals from 0 to π:

Orthogonality π

0

sin nx sin kx dx = 0 if n = k . (3)

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