MATLAB: Taylor expansion calculation of exp(x^2)

exponential functiontaylor series

Dear Matlab users and experts,

I am aware that the exponential function is standarized as "exp" in Matlab . However, I need to calculate the function value exp(x^2) adjusting the (N) terms in the power series. Can anyone recommend the correct method to compute the function exp(x^2)?

My approach:

x = -3.0:0.1:3.0;
N = 12;
Taylor_p2 = 0;
for n = 0:N  
    Taylor_p2 = Taylor_p2 + (x.^(2.0.*n))./(factorial(n));  % Taylor_p2 = exp(x^2)
end

isn't giving me the desired value. I am using R2020b Matlab version.

Many thanks in advance.

Bhattarai

Best Answer

The formula for taylor series is correct. Just increase the number of terms.

x = -3.0:0.1:3.0;
N = 12;
Taylor_p2 = 0;
for n = 0:N  
    Taylor_p2 = Taylor_p2 + (x.^(2.0.*n))./(factorial(n));  % Taylor_p2 = exp(x^2)
end
p2 = exp(x.^2);
err = norm(p2-Taylor_p2);
plot(x, p2);
hold on
plot(x, Taylor_p2, '*')

Result

>> err
err =
   9.1544e-05

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MATLAB: Taylor series for ln(x)

As a hint, I might wonder if your instructor expected this problem, perhaps this is why the problem was assigned?

Your formula for the series seems at a quick glance correct. However, you might consider if that series is convergent for x = 2.71828...

That is, will terms that look like

(1.718...)^n/n

ever approach zero? Or, is it possible that those terms do blow up, as you have seen? What does that tell you about the values of x for which this series will be convergent?

Note that there are ways to repair this problem, most notably via transformations. For example, a simple idea is if x > 1, then what is the value of -ln(x)? Consider the transformation

u = 1/x
ln(x) = -ln(u)

Now the terms in your series will look like

(-0.63212)^n/n

Clearly they will approach zero. I'm too lazy to write a loop, so consider this:

N = (1:25)';
S = cumsum((mod(N,2)*2-1).*(u-1).^N./N)
S =
     -0.63212
     -0.83191
      -0.9161
     -0.95602
      -0.9762
     -0.98684
      -0.9926
     -0.99578
     -0.99757
     -0.99859
     -0.99918
     -0.99952
     -0.99971
     -0.99983
      -0.9999
     -0.99994
     -0.99996
     -0.99998
     -0.99999
     -0.99999
           -1
           -1
           -1
           -1
           -1

When you negate the above result, that simply recovers our hoped for value for log(exp(1)).

So when you see a problem with a series like this, consider not only if your formula is correct, but if there are other considerations one must satisfy for that formula to be of any use. Here, convergence radius is what matters.

MATLAB: E^x/x

function plot doesn't work with input type syms

Instead try the following

x=[-10:0.0001:10];
y=exp(x)./x;
plot(x,y);
axis([-10 10 -1e3 1e3])
grid on
title('e^x/x')

and the following doesn't crash

syms x x0;
  y= exp(x-x0)/(x-x0)
  x_tay4=taylor(y, 'order',4)
  y =
  exp(x - x0)/(x - x0)
  x_tay4 =
  (- exp(-x0)/(6*x0) - exp(-x0)/(2*x0^2) - exp(-x0)/x0^3 - exp(-x0)/x0^4)*x^3 + (- exp(-x0)/(2*x0) - exp(-x0)/x0^2 - exp(-x0)/x0^3)*x^2 + (- exp(-x0)/x0 - exp(-x0)/x0^2)*x - exp(-x0)/x0

if you find these lines useful would you please mark my answer as Accepted Answer?

To any other reader, if you find this answer of any help please click on the thumbs-up vote link,

thanks in advance for time and attention

John BG

<mailto:jgb2012@sky.com jgb2012@sky.com>

Best Answer

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