Question

Approximate f by a Taylor polynomial with degree n at the number a.
\[f(x)=sinx\]
a=1 n=3
\[0.8 \le x \le 1.2\]

Answer 1

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

Answer 2

MacLaurin is a type of TAylor?

Answer 3

maclourin is at 0 tho

Answer 4

A Maclauris series is a Taylor series at \(x=0\)

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

oh wait :D are you in radians or degrees sofiya

Answer 12

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.

Answer 13

Be back in a couple minutes.

Answer 14

k

Answer 15

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

Answer 16

Remember we're going to \(n=3\) and not \(n=2\).

Answer 17

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?

Answer 18

To add and simplify the expanded form.

Answer 19

king george check inbox

Answer 20

is that right what i sent you?

Answer 21

One minute.

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

@timo86m We're told that \(a=1\)

Answer 27

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
\[0.8 \le x \le 1.2\]

Answer 28

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.

Answer 29

@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.

Answer 30

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 \(\sin(1)\).

Answer 31

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

Answer 32

Sure.

Answer 33

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

Answer 34

i think it is just

f(x)=x-(1/6)*x^3

and the answer is f(1)=5/6

Answer 35

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"

Answer 36

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.

Answer 37

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

Answer 38

is the absolute value of the remainder

Answer 39

a 1 could be what they wanted sin(x) evaluated at.

it makes no sense to have cos1 sin1 because it has irrational number.

Answer 40

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

Answer 41

it said approximate

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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)

Answer 47

sure

Answer 48

It doesn't explain what M is

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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)?

Answer 56

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.

Answer 57

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

Answer 58

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.

Answer 59

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

Answer 60

ok sounds good.

Answer 61

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

Answer 62

Should I post it as a new question?

Answer 63

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

Answer 64

sure