Question

Can this be simplified? Also my as well check to make sure it is correct. Find the orthogonal projection of f onto g. C[-1,1], f(x)=x, g(x)=e^x

Answer 1

\[<f,g>=\int\limits_{a}^{b}f(x)g(x)dx\]

Answer 2

\[\frac{2e^{-1}}{\frac{1}{2}(e^2-e^{-2})}\]

Answer 3

teacher likes to mess around with equations if he doesn't think they are hard enough

Answer 4

for <f,g> i got 2e^-1

Answer 5

this is only the constant in front o the vector also

Answer 6

so my answer above would have an x at the end

Answer 7

\[\int_{-1}^{1} x*e^xdx\]is it like this?

Answer 8

If it is like that umm then
\[\left| x*e^x - e^x\right|_{-1}^{1}\]
\[e - e - (-e^{-1} - e^{-1})\]
\[-2*e^{-1}\]
\[-\frac{2}{e}\]
maybe I am not sure

Answer 9

i don't belive that is correct

Answer 10

if you use tabular's you get should get (x-1)e^x

Answer 11

which is what you have aboe

Answer 12

\[xe^x-e^2=(x-1)e^x\]

Answer 13

solving that i get\[(1-1)e^x=0e^1-(-1-1)e^{-1}=-2e^{-1}=2e^{-1}\]

Answer 14

you are missing a negative ithink somewhere

Answer 15

oh yeah you are right

Answer 16

so then you have to do the inner product of <g,g> = just switch f(x)g(x) to g(x)g(x)dx

Answer 17

and tell me what you get there

Answer 18

\[\frac{e^{2x}}{2}\] evaluated from -1 to 1

Answer 19

g(x).g(x) = e^(2x)
\[|\frac{e^{2x}}{2}|_{-1}^{1}\]
\[\frac{e^2}{2} - \frac{e^{-2}}{2}\]

Answer 20

is there anyway to simplify that? is there a identity?

Answer 21

\[e^x+e^{-x}/2=\]

Answer 22

either sinh cosh? or no?

Answer 23

simplify that hmm no idea

Answer 24

\[\sinh(2x) or \cosh(2x)?\]

Answer 25

it'd be cosh(2x) but i forget how hyperbolic functions work again?

Answer 26

hmm sorry I haven't even studied hyperbolic functions