Math

[Native Jovian]

postcount +1

I see what you did there.

[Bleck]

nerdiest thread on the boards

[Bahamut]

I'll take on any (legit) math question.

[The Pezman]

Prove it in a way we can reasonably understand.

[Bahamut]

Prove it in a way we can reasonably understand.

Read this: http://en.wikipedia.org/wiki/Exponential_function#On_the_complex_plane

Basically it boils down to the Taylor series representations for sin, cos, and exp adding up correctly.

[Reaif]

Read this: http://en.wikipedia.org/wiki/Exponential_function#On_the_complex_plane

Basically it boils down to the Taylor series representations for sin, cos, and exp adding up correctly.

Wow Taylor series... It's been a while. Though I am definitely going to have to refresh on it, because I have to take calc 3 next semester, and I hear they use a lot of series' in that class... *Sigh*

But seriously, I am probably screwing myself over next semester. So far I am signed up for Linear Algebra, Calc 3, Engineering Dynamics, and Chemistry (one of the earliest ones). Sounds like a death trap to me.

[Steben]

Read this: http://en.wikipedia.org/wiki/Exponential_function#On_the_complex_plane

Basically it boils down to the Taylor series representations for sin, cos, and exp adding up correctly.

Breaking it down a little more...

Take e^(xi) and turn it into its Taylor series expansion. By computing the i-part of each term (i^0=1, i^1=i, i^2=-1, i^3=-i, etc.), you can then rearrange stuff like above. You get the Taylor expansion of cos(x) plus i times the Taylor expansion of sin(x).

Plug in pi for x and you get cos(pi)+i*sin(pi) = -1 + i*0 = -1.

[Bahamut]

Breaking it down a little more...

Take e^(xi) and turn it into its Taylor series expansion. By computing the i-part of each term (i^0=1, i^1=i, i^2=-1, i^3=-i, etc.), you can then rearrange stuff like above. You get the Taylor expansion of cos(x) plus i times the Taylor expansion of sin(x).

Plug in pi for x and you get cos(pi)+i*sin(pi) = -1 + i*0 = -1.

Yeah, I was too lazy to TeX up the proof...but that's the specifics.

[Red Shadow]

let's try this again:

[angcdixon]

someone help please!!

can you show me how to do these problems:

f(x)= square root of x

g(x)= 1-2x

1) (fog)(x)

2) (gof)(x)

3) (fog)(-4)

4) (gof)(-4)

5) 2g(-1) - 3g(-1)

6) 2f(-9)

the answers are as follows:

1. square root of (1-2x)

2. 1 - (2 sqrt of x)

3. 3

4. 1-4i

5. -3

6. 6i

i just need to know HOW to get the answers!!

please and thank you

[Bahamut]

let's try this again:

Nothing shows up.

someone help please!!

can you show me how to do these problems:

f(x)= square root of x

g(x)= 1-2x

1) (fog)(x)

2) (gof)(x)

3) (fog)(-4)

4) (gof)(-4)

5) 2g(-1) - 3g(-1)

6) 2f(-9)

the answers are as follows:

1. square root of (1-2x)

2. 1 - (2 sqrt of x)

3. 3

4. 1-4i

5. -3

6. 6i

i just need to know HOW to get the answers!!

please and thank you

(f o g)(x) = f(1 - 2x) = (1 - 2x)^(1/2) (square root of 1 - 2x)

(g o f)(x) = g(x^(1/2)) = 1 - 2(x)^(1/2)

3 & 4 you just plug in x = whatever number

2g(-1) - 3g(-1) = -g(-1) = - (1 - 2(-1)) = - ( 1 + 2 ) = -3

2f(-9) = 2 * (-9)^(1/2) = 2 * (9)^(1/2) * (-1)^(1/2) = 2 * 3 * i = 6i

[Red Shadow]

yeah i directed her question here before realizing i could help her with it, which was done

she'll likely ask other stuff that ive decided to forget, however