Hallo
Für die Aufgabe ergibt sich beispielsweise folgende Rechnung bei Berückischtigung des 6. Grades:
f = e^cosx = 1 +cos x +1/2 cosx^2 +1/6 cosx^3 +1/24 cosx^4 +1/120 cosx^5 ++
Aufgrund der Eulerschen Formeln ergibt sich:
f = e^cosx = 1 +1/4 +1/64 +1/36/64 +1/2(e^ix +e^-ix) +1/16(e^ix +e^-ix) +1/384 (e^ix +e^-ix)
+1/8 (e^i2x +e^-i2x) +1/96 (e^i2x +e^-i2x) +1/48/64 (e^i2x +e^-i2x)
+1/48 (e^i3x +e^-i3x) +1/768 (e^i3x +e^-i3x)
+1/384 (e^i4x +e^-i4x) +1/120/64 (e^i4x +e^-i4x)
+1/3840 (e^i5x +e^-i5x)
+1/720/64 (e^i6x +e^-6x) ++
f = 1 +1/4 +1/64 +1/2304 +(1/2 +1/16 +1/384) (e^ix +e^-ix) +(1/8 +1/96 +1/3072) (e^i2x +e^-i2x) +(1/48 +1/768)(e^i3x +e^-i3x)
+(1/384 +1/7680) (e^i4x +e^-i4x) + 1/3840 (e^i5x +e^-i5x) +1/46080 (e^i6x +e^-i6x) ++
f = 81/64 +1/2304 +(217/192)*1/2(e^ix +e^-ix) +(13/48 +1/1536)1/2(e^i2x +e-i2x) +(17/384)1/2(ei^3x +e^-i3x)
+(1/192 +1/3840)*1/2(e^i4x +e^-i4x) + 1/1920/2(e^i5x +e^-i5x) +1/23040/2(e^i6x +e^-i6x) ++
f = (81/64 +1/2304) +(217/192) cos1x +(13/48 +17/192) cos2x +(17/384) cos3x +(1/192 +1/3840) cos4x + (1/1920) cos5x +(1/23040) cos6x ++
Probe:
f(0) = e ~ (81/64 +217/192 +13/48 +17/384 +1/192 +1/1920) +(1/2304 +1/1536 +1/3840 +1/23040)
= (30*81 +217*10 +13*40 +17*5 +10 +1)/1920 +1/23040(10 +15 +6 +1) = 15648/1920/3 +8/5760 = 15656/5760 = 62624/23040
= 1957/720 = 2.718055556 (Fehler ~ 4/23040 = 0.00013020833333 -> e in [2.717881944..2.718229167])
f(pi/3) e^0.5 = 1.648721271 ~ 81/64 +217/384 -13/96 -17/384 -1/384 +1/3840 = (60*81 +217*10 -13*40 -17*10 -10 +1)/3840 = 6331/3840
= 1.648697917 (Fehler ~ 3/3840 = 0.00078125 -> e^0.5 in [1.647916667..1.649479167])
Liebe Grüße
