Question

don't know if doing right? Use the definition to find the Taylor Series (centered at c ) for the function f(x)=sin(x), c=pie/4

Answer 1

ick

Answer 2

ick that dont sound good

Answer 3

you are writing in terms of
\[(x-\frac{\pi}{4})\] right?

Answer 4

first term is
\[\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\]

Answer 5

actually this one is not so bad since you take the derivative of cosine, get -sine, then -cosine, then sine etc etc

Answer 6

so next one the derivative is -sine and
\[-\sin(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}\] so next term is
\[-\frac{\sqrt{2}}{2}x\]

Answer 7

\[\sqrt{2}\]i got cos(\[\pi/4\]=1/ which equals .707

Answer 8

take the derivative of - sine, get - cosine and keep going. it is going to be - - + + - - + + etc and you will always get
\[\frac{\sqrt{2}}{2}\]

Answer 9

just use
\[\frac{\sqrt{2}}{2}\]

Answer 10

and don't forget to divide by n!

Answer 11

1/\[\sqrt{2}\]

Answer 12

so you get
\[\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}\frac{x^2}{2!}+\frac{\sqrt{2}}{2}\frac{x^3}{3!}+\frac{\sqrt{2}}{2}\frac{x^4}{4!}-...\]

Answer 13

you can use
\[\frac{1}{\sqrt{2}}\] if you like

Answer 14

same number.

Answer 15

oh man i wrote it wrong i am sorry let me do it correctly!

Answer 16

\[what i got was sinx=x-(x-c)^3! +(x-c)^5/5!-(x-c)^7/7!\]

Answer 17

x-(x-c)^3/3!

Answer 18

\[\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}(x-\frac{\pi}{4})-\frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^2}{2!}+\frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^3}{3!}+\frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^4}{4!}-...\]

Answer 19

no this one has both even and odd terms

\[\cos(x)\] is even but
\[\cos(x-\frac{\pi}{4})\] is not

Answer 20

oh crap you are doing sine and i am doing cosine. i should just quit. doesn't really matter much though

Answer 21

goes
sine
cosine
-sine
-cosine
sine
cosine
etc

Answer 22

at
\[\frac{\pi}{4}\] you get
\[\frac{1}{\sqrt{2}}\]
\[\frac{1}{\sqrt{2}}\]
\[-\frac{1}{\sqrt{2}}\]
\[-\frac{1}{\sqrt{2}}\]
\[\frac{1}{\sqrt{2}}\]
\[\frac{1}{\sqrt{2}}\]
etc

Answer 23

so your answer is
\[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}(x-\frac{\pi}{4})-\frac{1}{\sqrt{2}}\frac{(x-\frac{\pi}{4})^2}{2}\]
\[-\frac{1}{\sqrt{2}}\frac{(x-\frac{\pi}{4})^3}{3!}+\frac{1}{\sqrt{2}}\frac{(x-\frac{\pi}{4})^4}{4!}\]etc

Answer 24

two pluses followed by two minus and always over n!

Answer 25

ok so I see if I can replace x with -x to see if even or not. I think this is what is confusing me with most problems. I don't know how to tell what will or should end up in my answer

Answer 26

not sure how to tell even odd or both on most questions