calculateA-n-n-1-1-n-n-2-cos-n-andB-n-n-1-1-n-n-2-cos-npi-3-

Question Number 132496 by mathmax by abdo last updated on 14/Feb/21

{calculateA}_{n} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (n) {andB}_{n} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (\frac{n π}{3})

Answered by mnjuly1970 last updated on 14/Feb/21

b y u s i n g d i l o g a r i t h m f u n c t i o n

o r f o u r i e r s e r i e s \dots

f i r s t m e t h o d . .

f (x) = x x \in [- π, π]

f i s a r t i f i c i a l p e r i o d i c f u n c t i o n

o n [- π, π]

f (x) \approx \frac{a_{0}}{2} + \sum_{n = 1} (a_{n} c o s (n x) + b_{n} s i n (n x))

a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) c o s (n x) d x = 0

b_{n} = \frac{1}{π} \int_{- π}^{π} f (x) s i n (n x) d x

= \frac{1}{π} \int_{- π}^{π} x s i n (n x) d x = \frac{2}{π} \int_{0}^{π} x s i n (n x) d x

= \frac{2}{π} {{[- \frac{x}{n} c o s (n x)]}_{0}^{π} + \frac{1}{n} \int_{0}^{π} c o s (n x) d x

2 {(- 1)}^{n + 1} \frac{1}{n}

a_{0} = \frac{1}{π} \int_{- π}^{π} f (x) d x = 0

x \approx \underset{n = 1}{\sum^{\infty}} \frac{2 {(- 1)}^{n + 1}}{n} s i n (n x)

\int x d x \approx 2 \underset{n = 1}{\sum^{\infty}} \int \frac{{(- 1)}^{n + 1}}{n} s i n (n x) d x

\frac{x^{2}}{2} \approx C + 2 \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} c o s (n x)

x = π \Rightarrow \frac{π^{2}}{2} \approx C + 2 \sum_{n = 1} \frac{1}{n^{2}}

C = \frac{π^{2}}{2} - \frac{π^{2}}{3} = \frac{π^{2}}{6}

\frac{x^{2}}{2} = \frac{π^{2}}{6} + 2 \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} c o s (n x)

x = 1 \Rightarrow \frac{1}{2} - \frac{π^{2}}{6} = 2 \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}}

\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} = \frac{1}{4} - \frac{π^{2}}{12} \dots

Commented by mathmax by abdo last updated on 14/Feb/21

thank you sir .

Commented by mnjuly1970 last updated on 15/Feb/21

g r a t e f u l \dots

Answered by mnjuly1970 last updated on 14/Feb/21

s e c o n d m e t h o d

\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} (\frac{e^{i n} + e^{- i n}}{2})

= \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- e^{i})}^{n}}{n^{2}} + \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- e^{- i})}^{n}}{n^{2}}

= \frac{1}{2} ({l i}_{2} (- e^{i}) + {l i}_{2} (- \frac{1}{e^{i}}))

\underset{\frac{- π^{2}}{6} - \frac{1}{2} {l n}^{2} (z)}{\overset{{l i}_{2} (- z) + {l i}_{2} (- \frac{1}{z})}{=}} \frac{1}{2} (\frac{- π^{2}}{6} - \frac{1}{2} {l n}^{2} (e^{i}))

= \frac{- π^{2}}{12} - \frac{1}{4} (i^{2}) = \frac{- π^{2}}{12} + \frac{1}{4} \dots .

Answered by mathmax by abdo last updated on 14/Feb/21

let φ (x) = x, 2 π periodic odd \Rightarrow φ (x) = \sum_{n = 1}^{\infty} a_{n} \sin (nx)

a_{n} = \frac{1}{π} \int_{- π}^{π} x \sin (nx) dx = \frac{2}{π} \int_{0}^{π} xsin (nx) dx \Rightarrow

\frac{π}{2} a_{n} = {[- \frac{x}{n} \cos (nx)]}_{0}^{π} + \int_{0}^{π} \frac{1}{n} \cos (nx) dx = - \frac{π}{n} {(- 1)}^{n} + \frac{1}{n^{2}} {[\sin (nx)]}_{0}^{π}

= - \frac{π}{n} {(- 1)}^{n} \Rightarrow a_{n} = \frac{2}{π} . (- \frac{π}{n}) {(- 1)}^{n} = - \frac{2}{n} {(- 1)}^{n} \Rightarrow

x = - 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \sin (nx) \Rightarrow \frac{x^{2}}{2} = 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (nx) + C

x = 0 \Rightarrow 0 = 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} + c = 2 . (2^{1 - 2} - 1) ξ (2) + C

= 2 (- \frac{1}{2}) \frac{π^{2}}{6} + C = - \frac{π^{2}}{6} + C \Rightarrow C = \frac{π^{2}}{6} \Rightarrow

\frac{x^{2}}{2} = 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (nx) + \frac{π^{2}}{6} \Rightarrow 2 Σ (\dots .) = \frac{x^{2}}{2} - \frac{π^{2}}{6} \Rightarrow

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (nx) = \frac{x^{2}}{4} - \frac{π^{2}}{12}

x = 1 \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (n) = \frac{1}{4} - \frac{π^{2}}{12}

x = \frac{π}{3} \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} \cos (n \frac{π}{3}) = \frac{π^{2}}{36} - \frac{π^{2}}{12} = \frac{π^{2}}{36} - \frac{3 π^{2}}{36} = - \frac{π^{2}}{18}

Commented by mnjuly1970 last updated on 15/Feb/21

v e r y n i c e s i r m a x . .