A mathematical expression of polynomial function can be described as below.

*f(x) = a*

_{0}+ a_{1}x + a_{2}x^{2 }+ ... + a_{n}x^{n}**Generating data**

For our test, we need data, and we can generate it as below. You may use your own observation data too.

`peq <- function(x) x^3+2*x^2+5`

` `

x <- seq(-0.99, 1, by = .01) y <- peq(x) + runif(200)

df <- data.frame(x = x, y = y) head(df) x y 1 -0.99 6.635701 2 -0.98 6.290250 3 -0.97 6.063431 4 -0.96 6.632796 5 -0.95 6.634153 6 -0.94 6.896084

Now we have a 'df' data, and we visualize it in a plot. We need to fit this data with the best curve.

plot(df$x, df$y, pch=20, col="gray")

` `

**Building the model**

We build a model with lm() function with a formula. I(x^2) represents x

^{2}in a formula. We can also use poly(x,2) function and it is the same as I(x^2).

> model <- lm(y~x+I(x^3)+I(x^2), data = df) > summary(model) Call: lm(formula = y ~ x + I(x^3) + I(x^2), data = df) Residuals: Min 1Q Median 3Q Max -0.49598082 -0.21488892 -0.01301059 0.18515573 0.58048188 Coefficients: Estimate Std. Error t value (Intercept) 4.3634157 0.1091087 39.99144 x -0.1078152 0.9309088 -0.11582 I(x^3) -0.5925309 1.3905638 -0.42611 I(x^2) 3.6462591 2.1359770 1.70707 Pr(>|t|) (Intercept) < 0.0000000000000002 *** x 0.908039 I(x^3) 0.670983 I(x^2) 0.091042 . --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.2626079 on 96 degrees of freedom Multiple R-squared: 0.9243076, Adjusted R-squared: 0.9219422 F-statistic: 390.7635 on 3 and 96 DF, p-value: < 0.00000000000000022204

Next, we predict the model with our df data.

> pred <- predict(model,data=df)

` `

**Finding the most fitted curve**

Finding the best-fitted curve is important. We check the model with various possible equations. In the below plot, I use a quadratic and cubic formula for curve fitting. The orange line (linear regression) and yellow curve are the wrong choices for this data. The pink curve is close, but the blue curve is the best match for our data trend. Thus, I use the y~x

^{3}+x

^{2}formula to build our polynomial regression model.

You may find the best-fit formula for your data by visualizing them in a plot.

The source code of the above plot is listed below.

windows(width=8, height=6) plot(x=df$x, y=df$y, pch=20, col="grey") lines(df$x, predict(lm(y~x, data=df)), type="l", col="orange1", lwd=2) lines(df$x, predict(lm(y~I(x^2), data=df)), type="l", col="pink1", lwd=2) lines(df$x, predict(lm(y~I(x^3), data=df)), type="l", col="yellow2", lwd=2) lines(df$x, predict(lm(y~poly(x,3)+poly(x,2), data=df)), type="l", col="blue", lwd=2) legend("topleft", legend = c("y~x, - linear","y~x^2", "y~x^3", "y~x^3+x^2"), col = c("orange","pink","yellow","blue"), lty = 1, lwd=3 )

**Plotting the result**

1. Plotting with a plot() function.

pred <- predict(model,data = df) lines(df$x, pred, lwd = 3, col = "blue")

2. Plotting with a ggplot().

Polynomial regression data can be easily fitted and plotted with ggplot().

`library(ggplot2)`

`ggplot(data=df, aes(x,y)) +`

` geom_point() + `

` geom_smooth(method="lm", formula=y~I(x^3)+I(x^2))`

In this post, we have briefly learned how to fit polynomial regression data in R and plot the results with a plot and ggplot functions. The full source code is listed below.

`peq <- function(x) x^3+2*x^2+5`

` `

x <- seq(-0.99, 1, by = .01) y <- peq(x) + runif(200)

df <- data.frame(x = x, y = y) head(df)

plot(df$x, df$y, pch=20, col="gray")

` `

```
model <- lm(y~x+I(x^3)+I(x^2), data = df)
summary(model)
```

` `

`pred <- predict(model,data=df) `

` `

```
windows(width=8, height=6)
plot(x=df$x, y=df$y, pch=20, col="grey")
lines(df$x, predict(lm(y~x, data=df)), type="l", col="orange1", lwd=2)
lines(df$x, predict(lm(y~I(x^2), data=df)), type="l", col="pink1", lwd=2)
lines(df$x, predict(lm(y~I(x^3), data=df)), type="l", col="yellow2", lwd=2)
lines(df$x, predict(lm(y~poly(x,3)+poly(x,2), data=df)), type="l", col="blue", lwd=2)
legend("topleft",
legend = c("y~x, - linear","y~x^2", "y~x^3", "y~x^3+x^2"),
col = c("orange","pink","yellow","blue"),
lty = 1, lwd=3
)
```

` `

```
pred <- predict(model,data = df)
lines(df$x, pred, lwd = 3, col = "blue")
```

```
```

`library(ggplot2) `

`ggplot(data=df, aes(x,y)) + geom_point() `

` + geom_smooth(method="lm", formula=y~I(x^3)+I(x^2))`

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