Find the Maclaurin series for e^10x^2
Write in series notation.
This is what I got so far
f(x)=f(0)+(f'(0)/1!)(x)+(f"(0)/2!)(x)^2+(f"'(0)/3!)(x)^3
e^10x^2=1+1x+(1/2!)x^2+(1/3!)x^3
I am stuck, I don't think I am going this right. How would you write the notation
In google paste:
Maclaurin series calculator eMathHelp
When you see list of results click on:
Taylor and Maclaurin (Power) Series Calculator - eMathHelp
When page be open in rectangle Enter a function type your function.
If your function is:
e^(10x^2)
type it like that
In rectangle Order n =
type whatever number you want.
The default value is 5 but you can choose 6, 7, 8 etc.
Then click option: CALCULATE
You will see the solution step by step
You want
∑ (10x^2)^k/k! = 1 + 10x^2 + 50x^4 + ...
To obtain the Maclaurin series for e^(10x^2), we can start by finding the derivatives of e^(10x^2) with respect to x.
Let's denote the nth derivative of e^(10x^2) as f^(n)(x). To find the pattern for the derivatives, we can use the chain rule.
The chain rule states that if we have a function of the form f(g(x)), then the derivative of that composite function is f'(g(x)) times g'(x).
Using this rule, let's find the first few derivatives:
f(x) = e^(10x^2)
f'(x) = d/dx (e^(10x^2))
= 2(10x)e^(10x^2)
= 20xe^(10x^2)
f''(x) = d/dx (20xe^(10x^2))
= 20e^(10x^2) + 20x(20x)e^(10x^2)
= 20e^(10x^2) + 400x^2e^(10x^2)
f'''(x) = d/dx (20e^(10x^2) + 400x^2e^(10x^2))
= 200e^(10x^2) + 400(2x)e^(10x^2) + 400x^2(20x)e^(10x^2)
= 200e^(10x^2) + 800xe^(10x^2) + 8000x^3e^(10x^2)
By observing the pattern in the derivatives, we can conclude that the nth derivative, f^(n)(x), will have a term with a factor of (10x)^n before e^(10x^2). Additionally, the coefficient of (10x)^n will be the factorial of n divided by the factorial of n minus 1.
Now, let's write the Maclaurin series for e^(10x^2) using the formula you provided:
f(x) = f(0) + (f'(0)/1!)(x) + (f''(0)/2!)(x)^2 + (f'''(0)/3!)(x)^3 + ...
From the pattern we observed in the derivatives, we can find the values of f^(n)(0):
f(0) = e^(10(0)^2) = e^0 = 1
f'(0) = 20(0)e^(10(0)^2) = 0
f''(0) = 20e^(10(0)^2) = 20
f'''(0) = 200e^(10(0)^2) = 200
Substituting these values into the series formula, we have:
f(x) = 1 + (0/1!)(x) + (20/2!)(x)^2 + (200/3!)(x)^3 + ...
Simplifying further, we have:
f(x) = 1 + 0 + (10/1)(x)^2 + (40/3)(x)^3 + ...
Therefore, the Maclaurin series for e^(10x^2) in series notation is:
e^(10x^2) = 1 + (10/1)(x)^2 + (40/3)(x)^3 + ...