Question

f(1)=3, f'(1)=-1, f''(1)=4, f'''(1)=-2
Write the fourth degree Taylor polynomial for g(x)=integral(1 to x) f(t)dt

Answer 1

\[g(x)=\int\limits_{1}^{x}f(t)dt\]

Answer 2

I know how to do this normally, but I am at a loss when there's an integral in it. Do I just find the integral first?

Answer 3

Well, you know that this function g exists, so f must have an anti-derivative. Just give me a second; I have something else on the go...

Answer 4

g'(x)=f(x)

Answer 5

does that help?

Answer 6

Yeah, I remember something like that. Now on to how to apply it.

Answer 7

It's not hard - promise...

Answer 8

g'(1)=f(1)=3 and g''(x)=f'(x) so g''(1)=f'(1)=4 and..

Answer 9

\[g(x)=\int\limits_{1}^{x}f(t)dt=F(x)-F(1) \rightarrow g'(x)=F'(x)=f(x)\]

Answer 10

oops g''(1)=-1

Answer 11

and you continue from there to get your derivatives of g in terms of f

Answer 12

For your zeroth term, you know that\[g(x)=\int\limits_{1}^{x}3dt \rightarrow g(x)=3x-3\]

Answer 13

oh yeah i'm dumb. don't listen to me

Answer 14

ok Wait, I get it up to here:
g'(1)=f(1)=3 and g''(x)=f'(x) so g''(1)=f'(1)=4 and..

But after that Im not sure what you did.

Answer 15

lol, don't say that

Answer 16

What bit?

Answer 17

can't I just go from that, to make the polynomial by using the Talyor series?

Answer 18

oh I see, Im still missing g(0) and g''' and g'''' right?

Answer 19

Yes. You can find your derivative terms easily by taking successive derivatives and matching with your information:

g(x)=f'(x) --> g(1)=f'(1) = -1
g'(x)=f''(x) --> g'(1)=f''(1)=4 etc

Answer 20

Scratch that. Im missing G(0)

Answer 21

yeah. so All I need to find is g(0) and Im good?

Answer 22

Can you give me several minutes - I have to take care of something.

Answer 23

pretty much

Answer 24

no problem. Thanks

Answer 25

I dont quite understand how you got this:
g(x)=3x-3

Answer 26

I made a mistake above (I'm distracted)...I have your answer...like I said, be back in several mins....but basically, it's

g'(x) = f(x)
g''(x) = f'(x)
g'''(x)=f'''(x)
g''''(x)=f'''(x)

SUb x=1 and equate for each

Now, for your zeroth term\[g(1)=\int\limits_{1}^{1}f(t)dt = F(1)-F(1)=0\]

Answer 27

Then you just have to use your g(x)'s in a Taylor series expansion...

Bit sketchy...brb

Answer 28

Forget the 3x-3...I'm distracted.

brb

Answer 29

Ok, So g(x) does end up equaling 0.
Got it now. Thanks!

Answer 30

Yes

Answer 31

Actually, my answer key is different. it says:
\[(-7/2)+4x-(x ^{2}/2)+(2/3)(x-1)^{3}-(x-1)^{4}/12\]

Answer 32

Was it wrong to assume g(x) to be g(1)?

Answer 33

Yeah, you probably have to do some algebra...I just got back...let me do it on paper.

Answer 34

I'm thinking you should try expanding the first two terms of your Taylor series. See how the cubic and quartic still have the form (x-1)^3 and (x-1)^4, but there's no (x-1) or (x-1)^2. I'll do it now.

Answer 35

oh I see what you mean.

Answer 36

Yeah, it's right. Just looks different.

Answer 37

For your final two components, you just have to clean up the numerical factors - they don't have any factorials in the answer key.

Answer 38

yeah I checked too. meh. that's pretty annoying though.

So... In the end it is ok to make g(x)=g(1)?

Answer 39

You have no choice since you're evaluating at that point. You see, your function g was defined in terms of an integral, but it's just like any other function: when you sub. a number in for x in g(x), you do the same in the expression it's equal to, right? In this case, the x in the expression was a limit...and (luckily) it turned out that you were integrating at a single point, which is zero.

Answer 40

I don't think you're happy with this.

Answer 41

ha. Well to be honest, I didn't quite get the last part.

Answer 42

The g(1) = 0 part?

Answer 43

In this case, the x in the expression was a limit...and (luckily) it turned out that you were integrating at a single point, which is zero.

Answer 44

Oh, it just means this:

Answer 45

you have a function g(x) defined as\[g(x)=\int\limits_{1}^{x}f(t)dt=F(x)-F(1)\]where F is the anti-derivative of f.

Like I said, for x=1, this means,\[g(1)=\int\limits_{1}^{1}f(t)dt=F(1)-F(1)=0\]

Answer 46

Take any integral you like, e.g.\[\int\limits_{1}^{x}tdt\]Then the integral is\[\frac{t^2}{2}|_1^x=\frac{x^2}{2}-\frac{1}{2}\]If x=1, then the RHS evaluates to zero.

Answer 47

The problem was that I didnt quite accept the fact that f(1) was the only point they gave us therefore we must use that to find g(x).

Answer 48

You don't use f(1) to find g(x). You use f(1) to find g'(1).

Answer 49

well, what I meant was that we used f(t) at t=1 and it's derivatives at t=1 to find g(x). So basically a specific point to find the equation for a generic function.

Answer 50

But I guess that's wrong to say because this g(x) is only an approx. good around the point t=1.

Answer 51

Well, here's the thing - you didn't use a specific point to find the generic function, you already had the generic function :p. The function was \[g(x)=\int\limits_{1}^{x}f(t)dt\]which is a well-defined function assuming f meets all the necessary conditions for the integral to exist (which is assumed). You're problem came with getting over the presence of another function, f.

Answer 52

If that f(t) was some explicit expression of t, like t^2+1, you would have had no problem, because you would have integrated and taken the limits and you would have had an explicit function for g in terms of x only and you'd be fine with using that to find your Taylor series, right?

Answer 53

Ok that makes a lot of sense. So I just got confused by f(x) havign a 1 inside. so if it was like t^2+1, and x=1, I would have had to find what t^2+1 was at t=1 anyways, but the problem gave it to me already in the form of f(1), f'(1) etc. So as you said, it was a lucky coincidence.

Answer 54

EXACTLY!

Answer 55

phew