证明级数(sin nx)/n对于x属于0到2都收敛?
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2021-08-04
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用欧拉公式。e^ix=cosx+isinx
e^-ix=cosx-isinx
(e^inx-e^-inx)/2i=sinnx
原式=(e^inx-e^-inx)/2in
stolz,原早敏液式=e^inx-e^-inx-e^(n-1)ix+e^-i(n-1)x/2i
求和。2i原拿卜式=e^ix(e^nix-1)/[e^ix-1]-e^-ix(e^-inx-1)/(e^-ix-1)-1×[e^inx-1]/(e^ix-1)+1×[e^-inx-1]/(e^-ix-1)=[e^ix(n+1)-e^ix+1-e^inx+e^-inx-1-e^-i(n-1)x+e^ix]/(e^ix-1)=[e^ix(n+1)-e^inx+e^-inx-e^-i(n-1)x]/(e^ix-1)=[e^i(2n+1)x-e^2inx+1-e^ix]/[e^(n+1)ix-e^inx]=cos(2n+1)x+isin(2n+1)x-cos2nx-isin2nx+1-cosx-isinx/cos(n+1)x+isin(陆物n+1)x-cosnx-isinnx=(cos2nx+isin2nx)(cosx-1+isinx)/(cosnx+isinnx)(cosx-1-sinx)=(cos2nx+isin2nx)/(cosnx+isinnx)=e^i2nx/e^inx=e^inx
e^-ix=cosx-isinx
(e^inx-e^-inx)/2i=sinnx
原式=(e^inx-e^-inx)/2in
stolz,原早敏液式=e^inx-e^-inx-e^(n-1)ix+e^-i(n-1)x/2i
求和。2i原拿卜式=e^ix(e^nix-1)/[e^ix-1]-e^-ix(e^-inx-1)/(e^-ix-1)-1×[e^inx-1]/(e^ix-1)+1×[e^-inx-1]/(e^-ix-1)=[e^ix(n+1)-e^ix+1-e^inx+e^-inx-1-e^-i(n-1)x+e^ix]/(e^ix-1)=[e^ix(n+1)-e^inx+e^-inx-e^-i(n-1)x]/(e^ix-1)=[e^i(2n+1)x-e^2inx+1-e^ix]/[e^(n+1)ix-e^inx]=cos(2n+1)x+isin(2n+1)x-cos2nx-isin2nx+1-cosx-isinx/cos(n+1)x+isin(陆物n+1)x-cosnx-isinnx=(cos2nx+isin2nx)(cosx-1+isinx)/(cosnx+isinnx)(cosx-1-sinx)=(cos2nx+isin2nx)/(cosnx+isinnx)=e^i2nx/e^inx=e^inx
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