MATLAB SUMMATION HELP Please????

So i need to know how i can do this summation? ik its a for loop but when i ran it all i got was zeros.

 e^x=(1+x/1!+x^2/2!+x^3/3!+⋯,)-50<x<50 Eq (1) 
 e^'x=A*(1+x/1!+x^2/2!+x^3/3!+⋯),-50<x<50 Eq (2) 

 For the above equations build a vector named X from -50 to 50 in increment of 0.5. Then calculate the values of ex and e’x. Plot the values of these two functions with the following formats: 

 I got the vector built, but am stuck

[James Tursa]

Please post your code so far and we can comment on it and offer suggestions.

Answer 1

[Roger Stafford]

I notice on your other thread at Answers #142842 you have written:

n = 0:201 
x = [-50:0.5:50] 
xs = sum((x.^n./factorial(n))

As you may have already seen, this will give you trouble because for both ends of that x-range, the value of both x^201 and 201! will be extremely large and overflow to infinity, thereby producing a NaN in taking their ratio.

The way around this is to use a for-loop which runs through the series backwards. You start with 1. On the first trip you compute 1+x/201*1. On the second trip you get 1+x/200*(1+x/201*1). On the third trip you get

 1+x/199*(1+x/200*(1+x/201*1))

And so forth. I think you can see the pattern here. At each step there is first a multiplication of the previous value times x, then a division by an integer that is successively 201, 200, 199, etc., and finally 1 is added. If you follow this method, your computed value of ex will be extremely close to the ideal exp(x) for numbers as extreme as 50 and -50.

You face one additional difficulty. In making a plot, you will find that for the plot to be scaled to fit values for ex with x = 50 on the plot, smaller values of x will give much smaller values for ex and the plot line will look as if those values are all zeros.

I haven't understood the significance of your second equation (2). What is 'A'?