## MathSofiya 3 years ago Please help....Please help....Taylor series f(x)=sinx a=1 n=3 $0.8 \le x \le 1.2$ $sinx\approx T_3(x)=sin(1)+\frac{cos(1)}{1}(x-1)-\frac{sin(1)}{2}(x-1)^2-\frac{cos(1)}{6}(x-1)^3$ (b) Use Taylor's Inequality to estimate the accuracy of the approximation $f(x)\approx T_n(x)$ lies in the given interval. $\left|R_3(x)\right|\le \frac{M}{4!} {\left|x-1\right|}^4$ I'm almost there.....

1. MathSofiya

here is an example that me and @smoothmath did http://openstudy.com/study#/updates/500af8b7e4b0549a892f4a6d

2. SmoothMath

Which part is giving you trouble?

3. SmoothMath

It looks like you made some mistakes in constructing the Taylor Series.

4. MathSofiya

where?

5. MathSofiya

i see it

6. SmoothMath

Kind of a lot of mistakes. Let me see you try again.

7. SmoothMath

$\large T_3 = f(a) +\frac{f'(a)}{1!}*(x-a) + \frac{f''(a)}{2!}*(x-a)^2 + \frac{f'''(a)}{3!}*(x-a)^3$

8. MathSofiya

I fixed it

9. SmoothMath

That's just by the definition of the Taylor series.

10. SmoothMath

Much better =)

11. SmoothMath

Okay, that's the first part. Now for the second part, we need to get M.

12. MathSofiya

$f^4(x)=sinx\le M$

13. MathSofiya

$\left|R_3(x)\right|\le \frac{sinx}{4!} {\left|x-1\right|}^4$ shoot me

14. SmoothMath

Hold on. We need to get an actual value of M. Remember how we get that?

15. MathSofiya

it's the absolute value of f^4 (x)

16. MathSofiya

don't give up on me quite yet please..... sin(1) or I have to do something with $x \ge 0.8$

17. SmoothMath

Let's step back, slow down, and understand what it is that we're doing. Just some theory.

18. MathSofiya

$f^{(n+1)}\le f(lower limit)<$

19. MathSofiya

$f^{(n+1)}\le f(lower limit)<M$

20. SmoothMath

Here'es the idea. We've made a taylor polynomial to approximate the function. Now we're trying to make a statement about how accurate the approximation is. Taylor's inequality allows us to do that. Here's how. We look at the derivative one higher than our approximation, and we look at it on the interval that we're interested in. If we can pick out a number, call it M, that the derivative is less than on the whole interval, then that allows us to use that M to limit the error.

21. SmoothMath

The basic statement is: $$\large f^{n+1} \le M$$ Therefore, $$R_n < \text{some thing dependent on M}$$

22. SmoothMath

The lower the M I'm able to pick, the better. Why? Because it allows me to put a lower cieling on the possible error.

23. SmoothMath

ceiling*

24. MathSofiya

yep. So we pick the lowest limit that we're allowed 0.8 in this case

25. SmoothMath

I don't think you're right.

26. SmoothMath

Let me talk about the maximum of a function for a bit.

27. MathSofiya

ok

28. SmoothMath

Okay, I'm thinking just a few examples will help you to see what it is we're doing.

29. SmoothMath

|dw:1342915184406:dw|

30. SmoothMath

Text on right says max=10

31. SmoothMath

|dw:1342915323520:dw|

32. SmoothMath

|dw:1342915451038:dw|

33. SmoothMath

Tell me that you understand. Because I can't make the words to explain this.

34. MathSofiya

I'm really sorry, John. I'm really trying to understand this. Ok here is what I understand. I see that you have drawn three unique functions, and each function was given an interval, and a maximum was chosen. The maximum that you have chosen is the M. The derivative should be less than M on the whole interval.

35. MathSofiya

That M allows us to put a cap on the error

36. MathSofiya

What we're trying to do with M is to put a cap on the error. That's why we need to find an actual value for M.

37. MathSofiya

That value M is greater than the next derivative up.

38. MathSofiya

hold on smooth

39. SmoothMath

Well, let me plot the 4th derivative for us.

40. MathSofiya

no hold on

41. SmoothMath
42. MathSofiya

x=1.2 will give me the ultimate cap. $f^4(x)=sinx\le sin(1.2)<1.932$

43. MathSofiya

ultimate cap for our interval

44. SmoothMath

sin(1.2) is not 1.932.

45. MathSofiya

0.932 :P

46. SmoothMath

That's impossible, since sin oscillates between 1 and -1.

47. SmoothMath

Good good =)

48. MathSofiya

typo dude...chill out

49. MathSofiya

and I didn't look at your wolfram ;P

50. MathSofiya

Victory?

51. SmoothMath

Almost there.

52. MathSofiya

The example is what confused me. They picked 7 instead of 9....but 1/7 is greater than 1/9 that's why they picked 7

53. MathSofiya

remember the interval $7 \le x \le 9$

54. SmoothMath

I do remember. Please realize that the maximum is not always going to be at one of the endpoints.

55. MathSofiya

I understand that now

56. SmoothMath

You really have to look at what the function is. It's going to be different every time and you have to analyze the function to figure out where the upper bound is.

57. MathSofiya

Yes sir.

58. SmoothMath

Okay good =)

59. MathSofiya

$\left|R_3(x)\right|\le \frac{.932}{4!} {\left|1.2-1\right|}^4$

60. MathSofiya

yes?

61. MathSofiya

$\left|R_3(x)\right|\le 0.000054$

62. SmoothMath

Looks good, pal.

63. MathSofiya

Thank ya! You're amazingly patient. Thank you soo much John. :)

64. SmoothMath

My pleasure! =D