Taylor Series coefficient of x^n

In summary: The coefficient is just a measure of the scaling of that term. In summary, the Taylor series for 0.5x^2[e^x-e^(-x)] around x=0 is ∑[x^(2n+3)/n!] and the coefficient of x^n is given by 1/[(n-3)/2]! except for when n is 1 or even, in which case it is 0.
  • #1
berbek16
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0

Homework Statement


Find the Taylor series for 0.5x^2[e^x-e^(-x)] around x=0. What is the coefficient of x^n?


Homework Equations


e^x=∑x^n/n!


The Attempt at a Solution


I understand how to find the Taylor series for this equation (it being ∑[x^(2n+3)/n!]; x^3+x^5+x^7/2!+...) through manipulation of the Taylor series for e^x; however I'm not quite sure what it means by the "coefficient of x^n?" The answer is apparently 0 when n is 1 or even, and otherwise it is given by 1/[(n-3)/2]!, but how is this found? Thank you so much for any help!
 
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  • #2
The coefficient of ##x^n## is simply the value by which ##x^n## is multiplied when you write out the series. So in your case, assuming your series is correct (which I haven't checked),
$$x^3 + x^5 + \frac{x^7}{2!} + \cdots$$
There is no constant term (##x^0##), so the coefficient of ##x^0## is zero. Similarly, the coefficient for ##x^1## is zero and likewise for all ##x^n## where ##n## is even, since these terms do not appear in the series.

So look at what remains: the coefficient for ##x^3## is ##1##, the coefficient for ##x^5## is also ##1##, and the coefficient for ##x^7## is ##1/(2!) = 1/2##. If you plug ##n=3,5,7## into the formula given in the answer, you can see that it matches these values.

To find the general formula, the question comes down to: what is the coefficient of ##x^n## in your series ##\sum x^{2n+3}/(n!)##?

Hint: try using a different letter instead of ##n## to avoid confusion: ##\sum x^{2m+3}/(m!)##. What is the ##x^n## term of this series?
 
  • #3
The coefficient of ##x^n## is precisely what it means. If your given a value ##n## what is the coefficient for that specific term ##x^n##.
 

What is a Taylor Series coefficient of x^n?

A Taylor Series coefficient of x^n refers to the constant term that multiplies the x^n term in a Taylor series expansion. It represents the slope of the curve at the point x=0.

How is the Taylor Series coefficient of x^n calculated?

The Taylor Series coefficient of x^n can be calculated using the formula: f(n)(0)/n!, where f(n)(0) is the nth derivative of the function f at x=0 and n! is the factorial of n.

What is the significance of the Taylor Series coefficient of x^n?

The Taylor Series coefficient of x^n is significant because it allows us to approximate a function with an infinite series of polynomials. It helps us understand the behavior of a function at different points and can be used to solve differential equations.

How does the Taylor Series coefficient of x^n change as n increases?

As n increases, the Taylor Series coefficient of x^n will decrease in magnitude. This is because the higher order derivatives of a function tend to decrease in magnitude as x approaches 0, resulting in smaller coefficients for higher powers of x.

Can the Taylor Series coefficient of x^n be negative?

Yes, the Taylor Series coefficient of x^n can be negative. This can occur when the function has a point of inflection at x=0, where the slope of the function changes from positive to negative or vice versa.

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