not gonna lie the jumpcuts are really throwing me off

if you really feel you need jump cuts to minimize the time, consider slightly fading each audio clip into the other to make the audio sound smoother and more natural. but also just use less jump cuts

Very useful for a particular section in my class, thank you

this is the hardest part of math 3a I think

At 6:44, what is actually happening when you are computing the images of the basis vectors T(1), T(x), and T(x^2)? I am having trouble visualizing what "variable" or "coefficient" the values 1,x,x^2 are being plugged into in the transformation. My book has this exact same example but with x = t and I don't understand how it gets the basis vector images. For instance, if you want T(1) of the transformation defined by T(a_0 + a_1*x + a_2 * x^2), does the 1 go in for the a values? or the x values? or neither, and I'm not visualizing this correctly? Thanks in advance.

@shooseto3932

3 жыл бұрын

You need to re-write "1" in the form "a_0+a_1x+a_2 x^2". This can be accomplished by a_0=1; a_1=1 ;a_2=0 (Actually this is the only way since "a_i" has to be numbers so it CANNOT be something like a_1=1/x.) Then T(1) = T(1+0x + 0x^2 ) = 0 + 2*0x Likewise, T(x) = T(0 + 1x+0x^2)= 1+2*0x = 1 and T(x^2) = T(0+0x+1x^2) = 0+2*(1)x = 2x.

@Chefkevv

3 жыл бұрын

@@shooseto3932 Thanks! I've been studying for my exam tomorrow for 2 days and am feeling much more confident about the material. Seeing your clarification has helped me on this topic!!

@shooseto3932

3 жыл бұрын

@@Chefkevv Good luck!

thankyou sensei scripting helps

why T(1)=0

@shooseto3932

2 жыл бұрын

Basically, T is applying the derivative to the argument, so in this case (1)' = 0. However, if we look at the definition of T, writing out 1 as a linear combination 1=(a_0)1+(a_1)x+(a_2)x^2, we get a_0=1, a_1=0 and a_2=0 so in the output a_1+2a_2x, we get 0.

huhhhh