A function that is not a polynomial can often be approximated by a polynomial. For example, for certain x-values, the function f(x) = e^x can be approximated by the fifth degree polynomial.
p(x) = 1+ x + x^2/2 + x^3/6 + x^4/24 + x^5/ 120
a) show that p(1) roughly equals f(1) = e. how good of an estimate is it?
b) calculate p(5). how well does p(5) approximate f(5)?
can you plz explain this to me.....i didnt understand it in my hw.....thank you very much... likes for all. 5 stars for best
p(x) = 1+ x + x^2/2 + x^3/6 + x^4/24 + x^5/ 120
a) show that p(1) roughly equals f(1) = e. how good of an estimate is it?
b) calculate p(5). how well does p(5) approximate f(5)?
can you plz explain this to me.....i didnt understand it in my hw.....thank you very much... likes for all. 5 stars for best
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There's not much to explain.
f(x) = e^x
and
p(x) = 1+ x + x^2/2 + x^3/6 + x^4/24 + x^5/ 120
You are supposed to compare f(1) and p(1), and then compare f(5) and f=p(5)
p(x) is the first 5 terms of the power series representing e^x. If you continue the series, it will be exactly e^x, but it will take a lot of terms to get an accurate value. You can probably guess that the larger the value of x, the worse the approximation will be, since you are missing large powers of the power series if you stop at just x^5.
Here are the values of the two functions, the absolute difference f-p, and the ratio f/p as a percent, to illustrate how good an estimate it is. A spreadsheet like Excel is excellent for studying things like this.
Unfortunately, these numbers are hard to read since Yahoo doesn't preserve tabs, so the numbers are separated by just a single space unless you put in characters like dots.
x.. f..... p..... diff..... ratio
0.00 1.00000 1.00000 0.00000 100.0000%
0.25 1.28403 1.28403 0.00000 100.0000%
0.50 1.64872 1.64870 0.00002 100.0014%
0.75 2.11700 2.11672 0.00028 100.0131%
1 2.71828 2.71667 0.00162 100.0595%
2 7.38906 7.26667 0.12239 101.6843%
3 20.08554 18.40000 1.68554 109.1605%
4 54.59815 42.86667 11.73148 127.3674%
5 148.41316 91.41667 56.99649 162.3480%
10 22026.46579 1477.66667 20548.79913 1…
20 485165195.40979 34887.66667 48513030…
At x=1, the power series is very close, only off by about 0.06%. For x = 5, the correct value is 62% higher than p(x), and it gets worse from there.
f(x) = e^x
and
p(x) = 1+ x + x^2/2 + x^3/6 + x^4/24 + x^5/ 120
You are supposed to compare f(1) and p(1), and then compare f(5) and f=p(5)
p(x) is the first 5 terms of the power series representing e^x. If you continue the series, it will be exactly e^x, but it will take a lot of terms to get an accurate value. You can probably guess that the larger the value of x, the worse the approximation will be, since you are missing large powers of the power series if you stop at just x^5.
Here are the values of the two functions, the absolute difference f-p, and the ratio f/p as a percent, to illustrate how good an estimate it is. A spreadsheet like Excel is excellent for studying things like this.
Unfortunately, these numbers are hard to read since Yahoo doesn't preserve tabs, so the numbers are separated by just a single space unless you put in characters like dots.
x.. f..... p..... diff..... ratio
0.00 1.00000 1.00000 0.00000 100.0000%
0.25 1.28403 1.28403 0.00000 100.0000%
0.50 1.64872 1.64870 0.00002 100.0014%
0.75 2.11700 2.11672 0.00028 100.0131%
1 2.71828 2.71667 0.00162 100.0595%
2 7.38906 7.26667 0.12239 101.6843%
3 20.08554 18.40000 1.68554 109.1605%
4 54.59815 42.86667 11.73148 127.3674%
5 148.41316 91.41667 56.99649 162.3480%
10 22026.46579 1477.66667 20548.79913 1…
20 485165195.40979 34887.66667 48513030…
At x=1, the power series is very close, only off by about 0.06%. For x = 5, the correct value is 62% higher than p(x), and it gets worse from there.