# Math Help - I need some help with taylor polynomials in the worst of ways!

1. ## I need some help with taylor polynomials in the worst of ways!

i have my final coming up very soon, and i am completely lost with taylor polynomials. i cant really grasp what the prof. talks about and the my book is just like a theory book, so i dont really get that either.

anyways, if someone can give me a crash course telling me what they do, how to find them, when to use them, etc...

and also i have these exercises that i obviously cant do, so if someone can walk me through some (preferably after explaining what exactly to do)

obtain the following taylor polynomials...
1. T(subscript=2n+1) (x/(1-x^2))= summation notation from k=0 to n (x^(2k+1))
2. T(subscript=n) (log (1+x))=summation notation from k=1 to n [(-1)^(k+1) (x^k)]/k
3. T(subscript=2n) (sin^2 x)=summation notation from k=1 to n [(-1)^(k+1)][(2^(k-1))/((2k)!)][x^2k]

it says hint=cos 2x= 1- 2sin^2 x

thanks for any and all help!

2. Originally Posted by twostep08
i have my final coming up very soon, and i am completely lost with taylor polynomials. i cant really grasp what the prof. talks about and the my book is just like a theory book, so i dont really get that either.

anyways, if someone can give me a crash course telling me what they do, how to find them, when to use them, etc...

and also i have these exercises that i obviously cant do, so if someone can walk me through some (preferably after explaining what exactly to do)

obtain the following taylor polynomials...
1. T(subscript=2n+1) (x/(1-x^2))= summation notation from k=0 to n (x^(2k+1))
2. T(subscript=n) (log (1+x))=summation notation from k=1 to n [(-1)^(k+1) (x^k)]/k
3. T(subscript=2n) (sin^2 x)=summation notation from k=1 to n [(-1)^(k+1)][(2^(k-1))/((2k)!)][x^2k]

it says hint=cos 2x= 1- 2sin^2 x

thanks for any and all help!
Taylor's theorem - Wikipedia, the free encyclopedia

Essentially a Taylor Polynomial is a polynomial function used to approximate / evaluate a transcendental function.

If you think about it, how do you think a calculator finds the sine of an angle? Or the logarithm of a number? All that a calculator can do is add (and by extension, subtract, multiply, divide and exponentiate).

So it makes sense that when these transcendental functions need evaluating, there must be some combination of addition, subtraction, multiplication, division and exponentiation that can be done.

Some examples:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$

$\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots - \dots$

$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots - \dots$.

3. ok, but where do those numbers come from? and how do i use that for the exercises in my first post?

5. lol yeah, i tried-im still a little shaky on it but its getting better. but then i really have no idea how to apply it to the examples

6. Originally Posted by twostep08
i have my final coming up very soon, and i am completely lost with taylor polynomials. i cant really grasp what the prof. talks about and the my book is just like a theory book, so i dont really get that either.

anyways, if someone can give me a crash course telling me what they do, how to find them, when to use them, etc...

and also i have these exercises that i obviously cant do, so if someone can walk me through some (preferably after explaining what exactly to do)

obtain the following taylor polynomials...
1. T(subscript=2n+1) (x/(1-x^2))= summation notation from k=0 to n (x^(2k+1))
2. T(subscript=n) (log (1+x))=summation notation from k=1 to n [(-1)^(k+1) (x^k)]/k
3. T(subscript=2n) (sin^2 x)=summation notation from k=1 to n [(-1)^(k+1)][(2^(k-1))/((2k)!)][x^2k]

it says hint=cos 2x= 1- 2sin^2 x

thanks for any and all help!

$\frac{x}{1 - x^2} = \sum_{k = 0}^{\infty}x^{2k + 1}$.

I'm going to split it into

$x\cdot \frac{1}{1 - x^2}$.

If you're good with geometric series, you'll know that

$a + ar + ar^2 + ar^3 + \dots = \frac{a}{1 - r}$, provided that $|r| < 1$.

$\frac{1}{1 - x^2}$ is already of this form, with $a = 1$ and $r = x^2$.

So I know that

$\frac{1}{1 - x^2} = 1 + x^2 + x^4 + x^6 + \dots$.

If I multiply everything by $x$...

$\frac{x}{1 - x^2} = x + x^3 + x^5 + x^7 + \dots$

$= \sum_{k = 0}^{\infty}x^{2k + 1}$.

7. Originally Posted by twostep08
i have my final coming up very soon, and i am completely lost with taylor polynomials. i cant really grasp what the prof. talks about and the my book is just like a theory book, so i dont really get that either.

anyways, if someone can give me a crash course telling me what they do, how to find them, when to use them, etc...

and also i have these exercises that i obviously cant do, so if someone can walk me through some (preferably after explaining what exactly to do)

obtain the following taylor polynomials...
1. T(subscript=2n+1) (x/(1-x^2))= summation notation from k=0 to n (x^(2k+1))
2. T(subscript=n) (log (1+x))=summation notation from k=1 to n [(-1)^(k+1) (x^k)]/k
3. T(subscript=2n) (sin^2 x)=summation notation from k=1 to n [(-1)^(k+1)][(2^(k-1))/((2k)!)][x^2k]

it says hint=cos 2x= 1- 2sin^2 x

thanks for any and all help!
2. Assume that $\ln{(1 + x)}$ can be expressed as a polynomial.

Then $\ln{(1 + x)} = a + bx + cx^2 + dx^3 + \dots$.

Letting $x = 0$, we see that

$\ln{1} = a$

Therefore $a = 0$.

Now, we wish to evaluate the other coefficients as well. To do this, we differentiate both sides of the equation - which eliminates the first coefficient and reduces all the powers by 1.

$\frac{1}{1 + x} = b + 2cx + 3dx^2 + 4ex^3 + \dots$.

Letting $x = 0$ we find $b = 1$.

Take the derivative of both sides again...

$-\frac{1}{(1 + x)^2} = 2c + 3\cdot 2 dx + 4 \cdot 3 ex^2 + 5 \cdot 4 fx^3 + \dots$

Let $x = 0$ we find

$c = -\frac{1}{2}$.

Take the derivative of both sides again

$\frac{2}{(1 + x)^3} = 3\cdot 2 d + 4\cdot 3 \cdot 2 ex + 5\cdot 4 \cdot 3 fx^2 + 6 \cdot 5 \cdot 4 gx^3 + \dots$

Let $x = 0$ we find

$d = \frac{2}{3\cdot 2} = \frac{1}{3}$.

Take the derivative of both sides again

$-\frac{3\cdot 2}{(1 + x)^4} = 4\cdot 3 \cdot 2e + 5\cdot 4 \cdot 3 \cdot 2fx + 6\cdot 5 \cdot 4 \cdot 3 gx^2 + \dots$.

Let $x = 0$ we find

$e = - \frac{3\cdot 2}{4\cdot 3 \cdot 2} = -\frac{1}{4}$.

If we continue on like this, we find

$\ln{(1 + x)} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots - \dots$.