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## anonymous 5 years ago 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

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1. anonymous

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

2. 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?

3. 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...

4. myininaya

g'(x)=f(x)

5. myininaya

does that help?

6. anonymous

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

7. anonymous

It's not hard - promise...

8. myininaya

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

9. anonymous

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

10. myininaya

oops g''(1)=-1

11. anonymous

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

12. anonymous

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

13. myininaya

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

14. 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.

15. anonymous

lol, don't say that

16. anonymous

What bit?

17. anonymous

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

18. anonymous

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

19. 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

20. anonymous

Scratch that. Im missing G(0)

21. anonymous

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

22. anonymous

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

23. anonymous

pretty much

24. anonymous

no problem. Thanks

25. anonymous

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

26. 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$g(1)=\int\limits_{1}^{1}f(t)dt = F(1)-F(1)=0$

27. anonymous

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

28. anonymous

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

29. anonymous

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

30. anonymous

Yes

31. anonymous

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

32. anonymous

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

33. anonymous

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

34. 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.

35. anonymous

oh I see what you mean.

36. anonymous

Yeah, it's right. Just looks different.

37. 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.

38. anonymous

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

39. 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.

40. anonymous

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

41. anonymous

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

42. anonymous

The g(1) = 0 part?

43. 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.

44. anonymous

Oh, it just means this:

45. 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,$g(1)=\int\limits_{1}^{1}f(t)dt=F(1)-F(1)=0$

46. 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.

47. 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).

48. anonymous

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

49. 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.

50. anonymous

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

51. 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.

52. 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?

53. 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.

54. anonymous

EXACTLY!

55. anonymous

phew

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