Bryce Rogers and Taylor Riesland

Bryce Rogers and Taylor Riesland

11/07/16

Professor Ipina

We are asked to prove that the parabola that fits better is given by an even function. To prove that the parabola is an even function, one can examine f(x)=f(-x). If f(x)=f(-x) holds true, then the parabola is even. If f(x)=-f(x), then the function is odd. We chose two random values to plug into

After plugging in two random values for f(x) and f(-x) we noticed above that indeed f(x)=f(-x) proving that the parabola is an even function.

We are asked to find a₂ in order to find the best fit for the semicircle. The equation for the semicircle can be written as, and the equation for the best fit can be written as. To calculate the best fit, we shall apply MacLaurin’s series which entails plugging in x values that increasingly become closer to 0 to find the best fit or a₂.The Maclaurin series is an addition to the Taylor series, named after Scottish mathematician, Colin Maclaurin. Maclaurin was considered to be one of the most influential Scottish mathematicians of the 18th century. The Maclaurin series was named to describe the case in which the variable “a” in the Taylor series equation equals zero(Tweedie 1).

The first series of numbers that will be plugged in for x are .5, and -.5.

when x=-.5, a₂=.536 x=.5, a₂=.536

Because the parabola is an even function, .5 and -.5 yield the same value of .536 for a₂. The next series of numbers to plug in for x consist of .1 and -.1

when x=.1, a₂=.501, x=-.1, a₂=.501

When x=0, a₂ can be any real number, because a₂ is multiplied by the x value. Any number multiplied by 0 is 0.

And lastly x values to plug in entail .01 and -01.

When x=.01, a₂=.5 when x=-.01, a₂=.5

From this we are able to extrapolate that the best fit of the line is when a₂=-.5. The new equation for the parabola is; )

Figure 1: In this figure we can see visually the different lines and how they’re effected for the different a₂ values. The function that is closest to the semi-circle must be the best fit. As shown in figure 1 the best fit is when a₂ is equal to .5.

We were asked to repeat the procedure to find the value of a₄ that gives the “best-fitting” quartic at the top of the semicircle. We decided to use the same three sets of x-values for the intersection points. We found that the value of was .5, so we implemented this into our new quartic equation, which can be re-written as.

. when x=-.5, a₄=.143593539x=.5, a₄=.143593539

when x=.1, a₄=.125628934, x=-.1, a₄=.125628934

When x=.01, a₄=.5 when x=-.01, a₄=.125

From these values, we can conclude that a₄ is approaching .125. The equation can be rewritten as.

Figure 2: Similar to figure 1 are four different lines including the semi-circle and the different functions that are effected by a₄. The function that is closest to the semicircle must be the best fit. From this visualization we can extrapolate that the best fit is when a₄ is equal to .125.

We were next asked repeat the procedure to find the value of that gives the “best-fitting” six order polynomial at the top of the semicircle. We decided to use the same three sets of x-values for the intersection points. We found that the value of was .5 and that a₄ was .125, so we implemented this into our new quartic equation, which was .

When x= -.5, or .5 a₆= .0743

When x=-.1 or .1 a₆= .0629

When x=.01 or.01 a₆=.063

We were able to extrapolate that the best fit for a₆ was .063.

Figure 3: The last visual is for the 3 lines that are effected by a₆ as well as the semicircle. According to the visual the best fit for a₆ is .063.

Finally, we were asked to test what would happen if our semicircle had a radius that was larger than 1. To assess this, we gave our new semicircle a radius of 1.5, and then used the formula to test points on our parabola and improve the fit at the top of the circle.

When x=0.1 or -0.1, = 0.333704529

When x=0, =0

When x=0.5 or -0.5, = 0.343145751

When x=0.01 or -0.01, =0.33333704

From these results, we can extrapolate that the change in the radius of the semicircle caused the best fit for to change from 0.5 to 0.3.

References

Maclaurin Series. (n.d.). Retrieved November 07, 2016, from

Tweedie, C. (1915).A study of the life and writings of Colin MacLaurin. London.