Taylor polynomial question

[sdg612]

Find the Taylor polynomial of degree 10 about x=0 for f(x)=sin2x (show all work)

This is what i have:

M10= f(0)+f$$\hat{}$$1(0)x+f$$\hat{}$$2(0)x$$\hat{}$$2/2!+f$$\hat{}$$3(0)x$$\hat{}$$3/3!+...+f$$\hat{}$$10(0)x$$\hat{}$$10/10!

f(x)=sin2x
f(0)=sin2(0)=0

f$$\hat{}$$1(x)=2cos2x
f$$\hat{}$$1(0)=2cos2(0)=2

f$$\hat{}$$2(x)=-4sin2x
f$$\hat{}$$2(0)=-4sin2(0)=0

f$$\hat{}$$3(x)=-8cos2x
f$$\hat{}$$3(0)=-8cos2(0)=-8

f$$\hat{}$$4(x)=16sin2x
f$$\hat{}$$4(0)=16sin2(0)=0

f$$\hat{}$$5(x)=32cos2x
f$$\hat{}$$5(0)=32cos2(0)=32

f$$\hat{}$$6(x)=64sin2x
f$$\hat{}$$6(0)=64sin2(0)=0

f$$\hat{}$$7(x)=128cos2x
f$$\hat{}$$7(0)=128cos2(0)=128

f$$\hat{}$$8(x)=256sin2x
f$$\hat{}$$8(0)=256sin2(0)=0

f$$\hat{}$$9(x)=412cos2x
f$$\hat{}$$9(0)=412cos2(0)=412

f$$\hat{}$$10(x)=824sin2x
f$$\hat{}$$10(0)=824sin2(0)=0


So, M10=0+2x+0-8x$$\hat{}$$3/3!+0+32x$$\hat{}$$5/5!+0+128x$$\hat{}$$7/7!+0+412x$$\hat{}$$9/9!+0

Therefore, 2x-8x$$\hat{}$$3/3!+32x$$\hat{}$$5/5!+128x$$\hat{}$$7/7!+412x$$\hat{}$$9/9!

 

[sdg612]

Find the Taylor polynomial of degree 10 about x=0 for f(x)=sin2x (show all work)

This is what i have:

M10= f(0)+f$$\hat{}$$1(0)x+f$$\hat{}$$2(0)x$$\hat{}$$2/2!+f$$\hat{}$$3(0)x$$\hat{}$$3/3!+...+f$$\hat{}$$10(0)x$$\hat{}$$10/10!

f(x)=sin2x
f(0)=sin2(0)=0

f$$\hat{}$$1(x)=2cos2x
f$$\hat{}$$1(0)=2cos2(0)=2

f$$\hat{}$$2(x)=-4sin2x
f$$\hat{}$$2(0)=-4sin2(0)=0

f$$\hat{}$$3(x)=-8cos2x
f$$\hat{}$$3(0)=-8cos2(0)=-8

f$$\hat{}$$4(x)=16sin2x
f$$\hat{}$$4(0)=16sin2(0)=0

f$$\hat{}$$5(x)=32cos2x
f$$\hat{}$$5(0)=32cos2(0)=32

f$$\hat{}$$6(x)=64sin2x
f$$\hat{}$$6(0)=64sin2(0)=0

f$$\hat{}$$7(x)=128cos2x
f$$\hat{}$$7(0)=128cos2(0)=128

f$$\hat{}$$8(x)=256sin2x
f$$\hat{}$$8(0)=256sin2(0)=0

f$$\hat{}$$9(x)=412cos2x
f$$\hat{}$$9(0)=412cos2(0)=412

f$$\hat{}$$10(x)=824sin2x
f$$\hat{}$$10(0)=824sin2(0)=0


So, M10=0+2x+0-8x$$\hat{}$$3/3!+0+32x$$\hat{}$$5/5!+0+128x$$\hat{}$$7/7!+0+412x$$\hat{}$$9/9!+0

Therefore, 2x-8x$$\hat{}$$3/3!+32x$$\hat{}$$5/5!+128x$$\hat{}$$7/7!+412x$$\hat{}$$9/9!

does this look right? anyone?

 

[rock.freak667]

well your differentials of cos(2x) are wrong as
$$\frac{d}{dx}(cos2x)=-2sin2x$$

but if I were you, I would just find sinx around x=0 and after you find it, replace all the x's with 2x

 

[colby2152]

It was looking right until the end...

Sixth derivative - you forgot a negative sign
Ninth derivative - 512, not 412
Tenth derivative - the sign probably stays the same after double corrections, 512*2=1024

Other than those small errors, you are golden!

 

[colby2152]

Please see my reply on your other double post.