Working with power series

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Last updated 8:35 AM on 6/13/26
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9 Terms

1
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What changes with the initial n value when you take the derivative of a power series

it becomes 1

2
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What does not change when you take the derivative of aa power series

The radius of convergence

3
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<p>Would the derivative of this function increase the initial n value</p>

Would the derivative of this function increase the initial n value

no, because the 1st term (n=0) of the non-derivative function is X. And the derivative of X is 1, not 0.

4
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When should you use tests to find the radius of convergence and when can you just set the magnitude of X less than 1

If there is only one n in the series equation you can just set the magnitude of x less than 1

If there are multiple n in the series equation you have to use one of the tests to simplify the equation

5
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Where should the c go when solving integrals with a power series

on the left side of the series symbol

<p>on the left side of the series symbol</p>
6
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What must we remove after using integrals to solve for a power series problem

You must remove the “C” by solving for it

7
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How do you solve for the actual value for “C” after using an integral to solve a power series

You sub in any number in the range of the convergence (most likely 0) into both sides of the series and then solve

<p>You sub in any number in the range of the convergence (most likely 0) into both sides of the series and then solve </p>
8
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When do you not have to solve for “C”

when there is an integral symbol on the left side of the equation

<p>when there is an integral symbol on the left side of the equation</p>
9
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How can you find integrals for complicated functions

you set a part of the equation as u and then du is the derivative of u. You then try to organize the u and du to make it match the equation. Then you find the integral of the new equation.

<p>you set a part of the equation as u and then du is the derivative of u. You then try to organize the u and du to make it match the equation. Then you find the integral of the new equation.</p>