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Define the function by the series

[infinity]
f(t)=∑ 7/n5 sin (nπt)
n=1
it turns out we can find derivatives by computing the derivative term-by-term as wel. find the series of the derivative
[infinity]
f'(t)=a0+∑ an cos (nπt) + bn sin (nπt)
n=1
a0=
an=
bn=

User Knitschi
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1 Answer

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a₀ = 0, aₙ = 7/n⁴π, bₙ = 0.

1. Derivative of sin(nπt):

The derivative of sin(nπt) is nπ cos(nπt). This applies to each term in the summation of f(t).

2. Term-by-term differentiation:

Applying the derivative to each term of the series and keeping in mind the constant factor of 7/n⁵:

f'(t) = d/dt [∑ 7/n⁵ sin(nπt)]

= ∑ 7/n⁵ d/dt [sin(nπt)]

= ∑ 7/n⁵ * nπ cos(nπt)

3. Separating coefficients:

The resulting series has the same form as the desired answer with cos(nπt) for each term. We can directly compare coefficients to solve for aₙ and bₙ:

a₀: The constant term comes from differentiating any constant term in the original series, but there is none. Therefore, a₀ = 0.

aₙ: This comes from the coefficient of cos(nπt) in the differentiated series, which is 7/n⁵ * nπ. Therefore, aₙ = 7/n⁴π.

bₙ: Since only cosine terms appear after differentiation, there are no sine terms in the derivative series. Therefore, all bₙ are 0.

Therefore, the coefficients are:

a₀ = 0

aₙ = 7/n⁴π

bₙ = 0

This gives us the complete series for the derivative f'(t):

f'(t) = ∑ (7/n⁴π) * cos(nπt)

User Rcplusplus
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8.4k points