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

- anonymous

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- anonymous

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

- anonymous

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?

- anonymous

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...

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## More answers

- myininaya

g'(x)=f(x)

- myininaya

does that help?

- anonymous

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

- anonymous

It's not hard - promise...

- myininaya

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

- anonymous

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

- myininaya

oops g''(1)=-1

- anonymous

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

- anonymous

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

- myininaya

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

- anonymous

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.

- anonymous

lol, don't say that

- anonymous

What bit?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Scratch that. Im missing G(0)

- anonymous

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

- anonymous

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

- anonymous

pretty much

- anonymous

no problem. Thanks

- anonymous

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

- anonymous

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\[g(1)=\int\limits_{1}^{1}f(t)dt = F(1)-F(1)=0\]

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Yes

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

oh I see what you mean.

- anonymous

Yeah, it's right. Just looks different.

- anonymous

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

- anonymous

yeah I checked too. meh. that's pretty annoying though.
So... In the end it is ok to make g(x)=g(1)?

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

The g(1) = 0 part?

- anonymous

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.

- anonymous

Oh, it just means this:

- anonymous

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

- anonymous

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.

- anonymous

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).

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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.

- anonymous

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?

- anonymous

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.

- anonymous

EXACTLY!

- anonymous

phew

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