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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
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[g(x)=\int\limits_{1}^{x}f(t)dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0Well, you know that this function g exists, so f must have an antiderivative. Just give me a second; I have something else on the go...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, I remember something like that. Now on to how to apply it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's not hard  promise...

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0g'(1)=f(1)=3 and g''(x)=f'(x) so g''(1)=f'(1)=4 and..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[g(x)=\int\limits_{1}^{x}f(t)dt=F(x)F(1) \rightarrow g'(x)=F'(x)=f(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and you continue from there to get your derivatives of g in terms of f

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For your zeroth term, you know that\[g(x)=\int\limits_{1}^{x}3dt \rightarrow g(x)=3x3\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0oh yeah i'm dumb. don't listen to me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0can't I just go from that, to make the polynomial by using the Talyor series?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh I see, Im still missing g(0) and g''' and g'''' right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes. 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
 5 years ago
Best ResponseYou've already chosen the best response.0Scratch that. Im missing G(0)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah. so All I need to find is g(0) and Im good?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you give me several minutes  I have to take care of something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dont quite understand how you got this: g(x)=3x3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0Then you just have to use your g(x)'s in a Taylor series expansion... Bit sketchy...brb

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Forget the 3x3...I'm distracted. brb

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, So g(x) does end up equaling 0. Got it now. Thanks!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, my answer key is different. it says: \[(7/2)+4x(x ^{2}/2)+(2/3)(x1)^{3}(x1)^{4}/12\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Was it wrong to assume g(x) to be g(1)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, you probably have to do some algebra...I just got back...let me do it on paper.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm thinking you should try expanding the first two terms of your Taylor series. See how the cubic and quartic still have the form (x1)^3 and (x1)^4, but there's no (x1) or (x1)^2. I'll do it now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh I see what you mean.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, it's right. Just looks different.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For your final two components, you just have to clean up the numerical factors  they don't have any factorials in the answer key.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah I checked too. meh. that's pretty annoying though. So... In the end it is ok to make g(x)=g(1)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think you're happy with this.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ha. Well to be honest, I didn't quite get the last part.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In 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
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, it just means this:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have a function g(x) defined as\[g(x)=\int\limits_{1}^{x}f(t)dt=F(x)F(1)\]where F is the antiderivative 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
 5 years ago
Best ResponseYou've already chosen the best response.0Take 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
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0You don't use f(1) to find g(x). You use f(1) to find g'(1).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0But I guess that's wrong to say because this g(x) is only an approx. good around the point t=1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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 welldefined 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
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0Ok 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.
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