## MathSofiya 3 years ago Approximate f by a Taylor polynomial with degree n at the number a. $f(x)=sinx$ a=1 n=3 $0.8 \le x \le 1.2$

1. MathSofiya

I'm skimming through my book and don't seem to find a concrete answer...what the difference between MacLaurin and Taylor again?

2. MathSofiya

MacLaurin is a type of TAylor?

3. timo86m

maclourin is at 0 tho

4. KingGeorge

A Maclauris series is a Taylor series at $$x=0$$

5. MathSofiya

so I have the equation f=sinx and I'm trying to approximate that line with degree 3 and number 1?

6. KingGeorge

To approximate one with a Taylor series, use the formula $\sum_{i=0}^n \frac{f^{(n)}(a)}{n!}(x-a)^n$

7. KingGeorge

Excuse me, $\sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i$

8. MathSofiya

$\sum_{n=3}^{?} \frac{sinx}{3!} (x-1)^3$ this doesn't look right...where did I go wrong?

9. timo86m

$\sin \left( 1 \right) +\cos \left( 1 \right) \left( x-1 \right) -1/2 \,\sin \left( 1 \right) \left( x-1 \right) ^{2}$

10. KingGeorge

So you have $$a=1$$, and $f^{(0)}=\sin(x)$$f^{(1)}=\cos(x)$$f^{(2)}=-\sin(x)$$f^{(3)}=-\cos(x)$

11. timo86m

oh wait :D are you in radians or degrees sofiya

12. KingGeorge

Use the formula, and we get $\frac{\sin(1)}{0!}(x-1)^0+\frac{\cos(1)}{1!}(x-1)^1-\frac{\sin(1)}{2!}(x-1)^2-\frac{\cos(1)}{3!}(x-1)^3$This can, of course, be simplified a bit.

13. KingGeorge

Be back in a couple minutes.

14. MathSofiya

k

15. timo86m

$\sin \left( a \right) +\cos \left( a \right) \left( x-a \right) -1/2 \,\sin \left( a \right) \left( x-a \right) ^{2}$

16. KingGeorge

Remember we're going to $$n=3$$ and not $$n=2$$.

17. MathSofiya

I'm kinda lost...what's my goal here? to write it in the condensed form $\sum$ or to add or simplify the written out form?

18. KingGeorge

To add and simplify the expanded form.

19. timo86m

king george check inbox

20. timo86m

is that right what i sent you?

21. KingGeorge

One minute.

22. MathSofiya

$\frac{\sin(1)}{0!}(x-1)^0+\frac{\cos(1)}{1!}(x-1)^1-\frac{\sin(1)}{2!}(x-1)^2-\frac{\cos(1)}{3!}(x-1)^3$ = $sin(1)+cos(1)(x-1)-\frac{sin(1)(x-1)^2}{2}-\frac{cos(1)(x-1)^3}{6}$ Like this or in the$\sum$ form?

23. timo86m

sin(a)+cos(a)*(x-a)-(1/2)*sin(a)*(x-a)^2

24. KingGeorge

The first way @MathSofiya. You don't want sigma notation in your answer.

25. timo86m

sin(a)+cos(a)*(x-a)-(1/2)*sin(a)*(x-a)^2-(1/6)*cos(a)*(x-a)^3 with 4th approximation choose a=0 and x-(1/6)*x^3

26. KingGeorge

@timo86m We're told that $$a=1$$

27. MathSofiya

so this way would be correct I guess $sin(1)+cos(1)(x-1)-\frac{sin(1)(x-1)^2}{2}-\frac{cos(1)(x-1)^3}{6}$ and i have to find a way to add this fact to the equation $0.8 \le x \le 1.2$

28. KingGeorge

I have no idea what $$0.8\leq x\leq 1.2$$ is supposed to do to our solution. As far as I'm concerned, that's irrelevant since $$x$$ is a variable. We've just restricted the domain.

29. timo86m

@KingGeorge Taylor is all about knowing a value at a such as 0 90 and such for sin(x) so that you get an appoximation. nobody knows what sin(1) is but we know cos(0) sin(o) and such.

30. KingGeorge

Taylor series works with all $$a$$. Whether we can compute $$\sin(1)$$ by hand or not is irrelevant. If we can't compute it, leave it as $$\sin(1)$$.

31. MathSofiya

ooohhhh can I pick any number between 0.8 and 1.2 for x to test something?

32. KingGeorge

Sure.

33. timo86m

it does work for all a but it is not practical.

34. timo86m

i think it is just f(x)=x-(1/6)*x^3 and the answer is f(1)=5/6

35. MathSofiya

or that might be to estimate the accuracy...the next question reads: "Use Taylors Inequality to estimate the accuracy of the approximation $f(x)\approx T_n(x)$when x lies in the given interval"

36. KingGeorge

That's probably what it's referring to in that case. I can't say I'm familiar with Taylor's Inequality though. @timo86m It specifically asks for $$a=1$$, and not 0.

37. MathSofiya

$\left|R_n(x) \right|=\left|f(x)-T_n(x)\right|$

38. MathSofiya

is the absolute value of the remainder

39. timo86m

a 1 could be what they wanted sin(x) evaluated at. it makes no sense to have cos1 sin1 because it has irrational number.

40. KingGeorge

That's why you don't evaluate it. We merely leave it as $$\sin(1)$$ or $$\cos(1)$$.

41. timo86m

it said approximate

42. KingGeorge

A Taylor series of finite length is by definition an approximation, whether we evaluate $$\sin(1)$$ or not.

43. MathSofiya

if $\left|f^{(n+1)}(x)\right|\le M$ then $\left|R_n(x) \right|\le \frac{M}{(n+1)!}\left| x-a\right|^{n+1}$

44. MathSofiya

There are three possible methods for estimating the size of error: 1. The method above 2. Alternating Series Estimation Theorem...if it's an alternating series 3. Use graphing device

45. MathSofiya

since this is an alternating series we can use method 2 I guess

46. KingGeorge

It says use Taylor's inequality though, so I'm inclined to think method 1 is probably what we want. (as long as that is Taylor's inequality)

47. MathSofiya

sure

48. MathSofiya

It doesn't explain what M is

49. KingGeorge

So do we want to find $$R_n(x)$$?

50. MathSofiya

yes that would be a start |sinx-T_n (x)|

51. MathSofiya

but they also said that $f(x)\approx T_n(x)$

52. KingGeorge

I think we should be using $$x=0.8$$ to get these approximations since $$|\sin(1)-\sin(.8)|>|\sin(1)-\sin(1.2)|$$.

53. KingGeorge

So if we look at $$|\sin(0.8)-T_3(0.8)|$$, we get $$.248284...$$

54. KingGeorge

But I don't think that's all we need.

55. MathSofiya

Are we assuming that f(x) is approximately T_n(x) therefore its sin(x)-sin(x) and in f(1)=sin(1) and T_n(.8)=sin(.8)?

56. KingGeorge

Since $$f(x)\approx T_n(x)$$ we know that $$T_n(0.8)\approx \sin(.8)$$. They aren't exactly the same, but they're about the same.

57. MathSofiya

The above statement makes sense...but when do I know to plug 0.8 in or 1.2 in...I guess I'm missing the bigger picture

58. KingGeorge

I'm really not sure where to go from here. Let me see if I can get someone else to help out with this last part.

59. MathSofiya

I know that's not sin(x) or cosine x...but I'm trying to come as close to a line as possible

60. MathSofiya

ok sounds good.

61. KingGeorge

Current question I'm unable to help on is "Use Taylors Inequality to estimate the accuracy of the approximation $$T_n$$ when $$0.8\leq x\leq 1.2$$?"

62. MathSofiya

Should I post it as a new question?

63. KingGeorge

Might be a good idea. This thread is getting a bit cluttered.

64. MathSofiya

sure