MATLAB SUMMATION HELP Please????
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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
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James Tursa
on 22 Jul 2014
Please post your code so far and we can comment on it and offer suggestions.
Answers (1)
Roger Stafford
on 22 Jul 2014
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'?
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