Match functions a-f with Taylor polynomials A-F (all...

Match functions a-f with Taylor polynomials A-F (all centered at 0). Give reasons for your choices. a. $\sqrt{1+2 x}$ b. $\frac{1}{\sqrt{1+2 x}}$ c. $e^{2 x}$ d. $\frac{1}{1+2 x}$ e. $\frac{1}{(1+2 x)^{3}}$ $\mathbf{f}_{\cdot} \quad e^{-2 \pi}$ A. $p_{2}(x)=1+2 x+2 x^{2}$ B. $p_{2}(x)=1-6 x+24 x^{2}$ C. $p_{2}(x)=1+x-\frac{x^{2}}{2}$ D. $p_{2}(x)=1-2 x+4 x^{2}$ E. $p_{2}(x)=1-x+\frac{3}{2} x^{2}$ $\mathbf{F} . p_{2}(x)=1-2 x+2 x^{2}$

Match functions a-f with Taylor polynomials A-F (all centered at 0). Give reasons for your choices.
a. $\sqrt{1+2 x}$
b. $\frac{1}{\sqrt{1+2 x}}$
c. $e^{2 x}$
d. $\frac{1}{1+2 x}$
e. $\frac{1}{(1+2 x)^{3}}$
$\mathbf{f}_{\cdot} \quad e^{-2 \pi}$
A. $p_{2}(x)=1+2 x+2 x^{2}$
B. $p_{2}(x)=1-6 x+24 x^{2}$
C. $p_{2}(x)=1+x-\frac{x^{2}}{2}$
D. $p_{2}(x)=1-2 x+4 x^{2}$
E. $p_{2}(x)=1-x+\frac{3}{2} x^{2}$
$\mathbf{F} . p_{2}(x)=1-2 x+2 x^{2}$

Key Concepts

Differentiation Techniques

Differentiation techniques are essential for obtaining the terms of a Taylor polynomial. By differentiating the function and evaluating these derivatives at the center, one can find the coefficients of the series. This process is particularly important when dealing with functions that involve compositions or products, as standard derivative rules must be applied correctly.

Coefficient Comparison

When matching functions with their Taylor polynomials, comparing coefficients of corresponding powers of x is a critical technique. By computing or knowing the derivatives at the center point, one can determine the coefficients of the polynomial and then match them with the given approximations.

Taylor Series

A Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. This concept helps in approximating functions by polynomials, where the accuracy of the approximation improves as more terms are included. It is a fundamental tool in calculus for studying local behavior of functions.

Maclaurin Series

A Maclaurin series is a special type of Taylor series that is expanded about the point zero. It is especially useful in approximating functions that are smooth near zero, allowing one to express the function as a power series whose coefficients are determined by the derivatives evaluated at 0.

Transcript

00:01 So you have the following functions, the function a, the function 1, square root of 1 plus 2x, b the function is to the minus 1, square of 1 plus 2x, c the function e to the 2x, d the function 1 plus 2x, and e to the 2x, d the function 1 plus 2x to the minus 1, and e, e, the function 1 plus 2 x to the minus 3 and we have some second degree second degree taylor polynomials on this side so you have 1 plus 2x plus 2x squared 1 minus 6x this is 1 the other one is 1 minus 6x plus 24 x squared um another one is one plus x minus x squared halfs here one is one minus two x plus four x squared and then we also have one minus x plus three halves of times times xx squared and then 1 minus 2 x plus 2x squared so you'd like to match which goes with which which is the taylor polynomial your second degree taylor polynomials so which of these corresponds to which function and that well to first order we see that all of these functions agree so what we need to do would be too much the derivatives so if we differentiate this function a you'd have one half the derivative of a of the function a differentiating that we're gonna have one half of one plus two x to the minus one half times a two by the inner ear so that this at zero is gonna be all just one it's one so if we look over this side the function that has there a derivative one at zero would be only this one so this one should correspond to that one now we differentiate b yeah if we differentiate this side we'll get one minus six times minus x so we'll get one also and now we differentiate the second one the relative of b you have a minus a half times all that to the minus three halves x minus two to the minus three halves and then that times two so that these cancels and then if we evaluate that x equals to zero it gets just minus one that derivative is minus one and all that was easy because the only one that has negative minus one at zero will be this term so that this one should correspond to this one.

05:36 Now we differentiate this, c, the derivative of c is going to be 2, me to the 2 x, evaluate that 0 is equal to 2.

06:01 And the one that has derivative 2 equals to 0 is only this one because of that coefficient...

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