2 4 Substituting 2 into gives 2(2) 4. + x + x +...= + + x ! 4! Substituting x= 0 and =1 into (1 + x ) + x = 0, gives 0

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1 Talor Series 6C 1 Differentiating = +, with respect to, gives = 1+ (1) Differentiating (1) gives = () =, = 1 into = +, gives = + (1), so = 1 1 = into (1) gives =1+ = into gives () = () = = Sousing the Talor epansionin the form where =, i.e. ii 1 () = 1+ +! () () = !! 6 Differentiating (1 + ), gives + = (1 + ) = (1) i.e. (1 + ) + + = = an =1 into (1 + ) + =, gives = =, = 1 an = int o (1) gives = 1 So using the Talor epansion in the form ii, () ( 1) = = +!! Pearson Eucation Lt 18. Coping permitte for purchasing institution onl. This material is not copright free. 1

2 Differentiating e, + = gives e + = (1) Differentiating (1) gives + e = () = an = into + e =, gives + 1 =, so = 1 =, = 1 into(1) gives + ( 1) (1) = so = =, into gives () (1) so 1 = () + = = into the Talor series with =, gives () ( 1) ( 1)!! = + 6 = Differentiating + = with respect to gives, i.e = (1) + + = Differentiating (1) gives = (), i.e. + + = =, = 1 an = into gives + + = () = = =, = an 1 into gives = (1) + ( 1) + () =, so = =, =, 1an into gives = = () ( ) ( 1), so + + = = into the Talor series with form ii, gives ( 1) ( ) () 1!!! 1 1 = = Pearson Eucation Lt 18. Coping permitte for purchasing institution onl. This material is not copright free.

3 5 Differentiating gives + = + = + (1) = 1, = 1 an = 1 into + = gives 5 = 1 1 = 1, = 1 = 1 an = 5 into gives = 1 (1) into the form of the Talor series form i, with = 1, gives 6 Differentiating (5) ( 1)!! 5 5 = 1 ( 1) + ( 1) ( 1) + = 1 + ( 1)( 1) + ( 1) + ( 1) + 1, + = + twice with respect to, gives = = =, = 1 an = 1 into + + = 1+ gives = = 1, = 1 an into gives = (1) = = 1, = 1,, into gives 1 = = () = ( ) () (1) So, using the Talor series form ii, = !!! 1 so = a Differentiating ( ) ( 1+ ) + =1+ Differentiating (1) gives (1) 1+ = + with respect to (1) (1+ ) + + = + (1+ ) + (1 ) = () () Pearson Eucation Lt 18. Coping permitte for purchasing institution onl. This material is not copright free.

4 7 b = an = 1 into ( 1+ ) = + gives =1 ( ) = known values into (1) gives ( ) ( )( ) + =1+1 =5 known values into () gives ( ) = =16 So using = !! = = !! 8 Differentiating sin cos + = with respect to, gives sin + cos sin cos + + = (1) or sin +cos sin= =, = 1 1 into sin + cos= gives + = so = =,, = into(1) gives ( ) ( ) ( )( ) + = 1 So + 1= = all values into gives the series solution ( ) = ( )! = Pearson Eucation Lt 18. Coping permitte for purchasing institution onl. This material is not copright free.

5 9 a i Differentiating = with respect to, gives = (1) ii Differentiating (1) gives = So = () b Differentiating () gives = so 6 = () c =, =1, into gives = 1=1, so =1 =, =1, =1 into(1) gives ( )( ) ( ), so 1 1 = = =1, =1, = into () gives ( )( ) ( ), so 1 1 = =8 =1, =1, = an = 8 ( )( ) ( )( ), so = =8 into () gives these values into the form of Talor s series form ii, gives () (8) (8) 7 1 (1) 1!!! 6 = = Pearson Eucation Lt 18. Coping permitte for purchasing institution onl. This material is not copright free. 5

6 1 Differentiating cos + sin + =(1), with respect to, gives cos sin + cos+ sin +6 =, () Differentiating again cos sin sin+ cos +6 +1, = =, = 1 into (1) gives +1 ( ) =, so = =, =1, = into () gives (1)( ) =, so = 11 =, = 1, =, = 11 into() gives + (1)( ) + 6(1)(11) + 1(1)( ), so = 11 () these values into the form of Talor s series form ii, 11 ( 11) gives = 1 + ( ) + + +!! = Ignoring terms in an higher powers, a We consier the ifferential equation: = Differentiating both sies: = + = + = + + = = = i.e. p=, q= 1 Pearson Eucation Lt 18. Coping permitte for purchasing institution onl. This material is not copright free. 6

7 11 b Now use the initial conitions given to fin: ( = 1) = 1 = ( = 1) = + 1 = ( = 1) = = 1 5 ( = 1) = = Plugging this into the Talor epansion for ( ), we see: ( ) = + ( 1) + 1 ( 1) + 1 ( 1)!! ( 1) ( 1) +...! 5! ( ) = + ( 1) + ( 1) + 1 ( 1) ( 1) + ( 1) Pearson Eucation Lt 18. Coping permitte for purchasing institution onl. This material is not copright free. 7

Hyperbolic Functions 6D

Hyperbolic Functions 6D a (sinh cosh b (cosh 5 5sinh 5 c (tanh sech (sinh cosh e f (coth cosech (sech sinh (cosh sinh cosh cosh tanh sech g (e sinh e sinh e + cosh e (cosh sinh h ( cosh cosh + sinh sinh