# Taylor Series Help

1. Jan 31, 2016

### I-aM-Lost

1. The problem statement, all variables and given/known data
Use zero- through third-order Taylor series expansion
f(x) = 6x3 − 3x2 + 4x + 5
Using x0=1 and h =1.

Once I found that the Taylor Series value is 49. I want to be able to check the value. On the board our teacher plugged in a value into the equation to show that the answer is 49. But I do not know how they were able to do that.

2. Relevant equations

Taylor Series

3. The attempt at a solution

zero: 12
first: 12 + 16 = 28
second: 12 + 16 + 15 = 43
third: 12 + 16 + 16 + 6 = 49

I am able to find the value of 49 using Taylor Series so that isn't the problem here. I want to be able to find the actual value of 49 without having to do the Taylor Series that way I can make sure what the correct ending value should be.

2. Jan 31, 2016

### Student100

Questions worded a bit weird can you rephrase a bit what you're wanting to know, do you mean he used $f(2)=6(2)^3-3(2)^2+4(2)+5 = 49$?

3. Jan 31, 2016

### WWGD

I too am confused, since Taylor series are about constructing polynomials to approximate functions. Here you are starting with a polynomial itself.

4. Jan 31, 2016

### Samy_A

Might be an exercise to illustrate that for a polynomial of degree n, the n-th order Taylor series is the polynomial itself.

5. Jan 31, 2016

### HallsofIvy

Staff Emeritus
And the nth order Taylor polynomial of a polynomial of degree m< n is just the part of that polynomial of degree less than or equal to m.

6. Jan 31, 2016

### WWGD

Could be, but I don't see where the numerical values come from.

7. Jan 31, 2016

### Samy_A

He calculates the value of the different Taylor series (order 0 to 3) at $x_0=1$ in x=2.

8. Jan 31, 2016

### WWGD

Good catch! I would have never figured it out.

9. Jan 31, 2016

### HallsofIvy

Staff Emeritus
The easy way to do this is to "shift" the polynomial from powers of x to powers of x- 1. Let y= x- 1. Then x= y+ 1.
$6x^3- 3x^2+ 4x+ 5= 6(y+ 1)^3- 3(y+ 1)^2+ 4(y+ 1)+ 5$. Multiply that out to get the polynomial in y. The "0" order Tayor polynomial is just the constant, the "1" order Taylor polynomial is just the constant and x term, the "2" order polynomial is just up to $x^2$, and the "3" order is the entire polynomial. Apparently those four are to be evaluated at x= 2 (so y= 2+ 1= 3) giving four numerical answers, not just the single "49".

10. Jan 31, 2016

### I-aM-Lost

That's what the teacher wanted us to see that the number of the polynomial would be how many derivatives we would have to take to find the Taylor Series for that polynomial. He then picked some value in order to show that 49 was an answer. I am trying to figure how he decided to plug a certain number in to check that our answer was 49. I know if you plug 2 into the original equation you get 49. But how do you know or what to plug in order to get 49.

11. Jan 31, 2016

### I-aM-Lost

He did the Taylor Series. So we did the 0th,1st,2nd,3rd taylor series and found the value to be 49. But he then decided to plug a value in to check that 49 was correct. I understand if we plug 2 into the original equation we get 49. But why did you decide to plug 2 into the equation.

12. Jan 31, 2016

### WWGD

Look at Samy_A's second post.

13. Jan 31, 2016

### Ray Vickson

Because $x = 1+h$ and $h =1$, so you are looking at $f(2)$.

14. Jan 31, 2016

### Samy_A

I can only guess what he did. But when I compute the Taylor series for $f$ at $x_0=1$, I get:
$T_0(x)=12$
$T_1(x)=12+16(x-1)$
$T_2(x)=12+16(x-1)+15(x-1)²$
$T_3(x)=12+16(x-1)+15(x-1)²+6(x-1)³$

Plug in $x=2(=1+h)$ in these Taylor series, and you get the values from your first post.
As you polynomial $f$ is of degree 3, $f(x)=T_3(x)$. That's why $f(2)=T_3(2)=49$.