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Problems in Mathematical Analysis.pdf - pwp.net.ipl.pt

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2689 .<br />

2691 . i_<br />

2694. Solution. 1) 2a = -~ f /(*) cos 2/i* dx<br />

Answers 461<br />

/<br />

slnn/t<br />

4n-l<br />

(* / W cos 2/ix d<br />

-jj-<br />

2 P Ji<br />

H-- \ / (x) cos 2nx dx. If we make the substitution t**-= x <strong>in</strong> the first<br />

Jl<br />

<strong>in</strong>tegral and t = x <strong>in</strong> the second, then, tak<strong>in</strong>g advantage of the assumed<br />

identity /(-^+< )<br />

z<br />

= -/ (^<br />

(n = 0, 1,^2, ...);<br />

n t<br />

2) &, = T / (^) s<strong>in</strong> 2nx dx = -|- f s<strong>in</strong><br />

<<br />

)<br />

will readily be seen that o 2n =0<br />

n<br />

~. T S<strong>in</strong><br />

The same substitution as <strong>in</strong> Case (1), with account taken of the assumed<br />

Identity f (y + /<br />

2697. s<strong>in</strong>h /<br />

[t<br />

2698. > (-1)"<br />

where<br />

) =/ (\^ t ^ leads to the equalities 6 2II = (/i=l, 2, ...).<br />

'<br />

/IJTJC<br />

/cos ~ J<br />

s<strong>in</strong>- cos (2'?<br />

_8_<br />

nx<br />

-16\\-1)"-' -~, 2702.<br />

n=i<br />

_ A V 1 ^ S J 21+ !) n*<br />

n=o<br />

4 1<br />

2 9<br />

^<br />

,(2'H-l)Jix<br />

n<br />

_-!? 1^1<br />

o l<br />

1)2<br />

2'mx .<br />

1<br />

b)<br />

-i-<br />

r s 2 mx

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