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

  2. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    g'(x)=f(x)

  5. myininaya
    • 5 years ago
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    does that help?

  6. anonymous
    • 5 years ago
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    Yeah, I remember something like that. Now on to how to apply it.

  7. anonymous
    • 5 years ago
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    It's not hard - promise...

  8. myininaya
    • 5 years ago
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    g'(1)=f(1)=3 and g''(x)=f'(x) so g''(1)=f'(1)=4 and..

  9. anonymous
    • 5 years ago
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    \[g(x)=\int\limits_{1}^{x}f(t)dt=F(x)-F(1) \rightarrow g'(x)=F'(x)=f(x)\]

  10. myininaya
    • 5 years ago
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    oops g''(1)=-1

  11. anonymous
    • 5 years ago
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    and you continue from there to get your derivatives of g in terms of f

  12. anonymous
    • 5 years ago
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    For your zeroth term, you know that\[g(x)=\int\limits_{1}^{x}3dt \rightarrow g(x)=3x-3\]

  13. myininaya
    • 5 years ago
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    oh yeah i'm dumb. don't listen to me

  14. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    lol, don't say that

  16. anonymous
    • 5 years ago
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    What bit?

  17. anonymous
    • 5 years ago
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    can't I just go from that, to make the polynomial by using the Talyor series?

  18. anonymous
    • 5 years ago
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    oh I see, Im still missing g(0) and g''' and g'''' right?

  19. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Scratch that. Im missing G(0)

  21. anonymous
    • 5 years ago
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    yeah. so All I need to find is g(0) and Im good?

  22. anonymous
    • 5 years ago
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    Can you give me several minutes - I have to take care of something.

  23. anonymous
    • 5 years ago
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    pretty much

  24. anonymous
    • 5 years ago
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    no problem. Thanks

  25. anonymous
    • 5 years ago
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    I dont quite understand how you got this: g(x)=3x-3

  26. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Then you just have to use your g(x)'s in a Taylor series expansion... Bit sketchy...brb

  28. anonymous
    • 5 years ago
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    Forget the 3x-3...I'm distracted. brb

  29. anonymous
    • 5 years ago
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    Ok, So g(x) does end up equaling 0. Got it now. Thanks!

  30. anonymous
    • 5 years ago
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    Yes

  31. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Was it wrong to assume g(x) to be g(1)?

  33. anonymous
    • 5 years ago
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    Yeah, you probably have to do some algebra...I just got back...let me do it on paper.

  34. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    oh I see what you mean.

  36. anonymous
    • 5 years ago
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    Yeah, it's right. Just looks different.

  37. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    I don't think you're happy with this.

  41. anonymous
    • 5 years ago
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    ha. Well to be honest, I didn't quite get the last part.

  42. anonymous
    • 5 years ago
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    The g(1) = 0 part?

  43. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Oh, it just means this:

  45. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    You don't use f(1) to find g(x). You use f(1) to find g'(1).

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

  51. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    EXACTLY!

  55. anonymous
    • 5 years ago
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    phew

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