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# Worked example: power series from cos(x)

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.E (LO)
,
LIM‑8.E.1 (EK)
,
LIM‑8.F (LO)
,
LIM‑8.F.2 (EK)
Finding a power series to represent x³cos(x²) using the Maclaurin series of cos(x).

## Want to join the conversation?

• At , Sal says we would have had to find the 19th derivative to find the first five, non-zero terms. Why is that the case? I see that the fifth term has x to the 19th power, but by the first derivative we already have x to the 4th power.
• why we can simply put x^3 in the series without effect the convergence (divergence) ?
• Great question! But think about what happens when we multiply some series by x^3. Imagine we have a series: Σ (an)x^n, and you tell me "I have determined that this series converges for x = c, and I'll even tell you that when x = c, the series converges to L." Well, if I multiply the series by x^3, what will happen now when x = c? No problem, the whole thing will certainly converge to (c^3)L. The same type of argument should convince you that if some series diverges when x = c, then x^3 (or x^n for any finite n) times the series will also certainly diverge at x = c.
• my question maybe does not have relevant with this lesson , but pleas help me
(k+1) ! = k! (k+1)
how that's happen?
• if you had (3)! that would be 3*2*1, if you had (3+1)! go decrement the term each time, so it would be (3+1)!= (3+1)*((3+1)-1)*((3+1)-2)*((3+1)-3)= (3+1)*3*2*1 which is the same thing as (3+1)*3!, so just replace 3 with k
• Is there any general prof that the Maclaurin series of f(x)=g(x)*h(x) will be exactly the same as the expression we get if we only substitute the Maclaurin series for g(x) and multiply with h(x)? I am not sure this is intuitive to me.
Thank you!
• In the video Sal managed to represent the function f(x) with a polynomial. But how do we know that that's the same polynomial we would get if we expanded the function using the Maclaurin series? Is there some rule that says there could only be one polynomial expression of a function? Thanks
• Sal started the whole process by basing the polynomial on the MacClaurin expansion of cos x - so it is safe to say that that is the polynomial you would get using the MacClaurin series.
• Something I just noticed about this technique, we are only creating the Maclaurin series based on the sine and cosine equivalents, evaluating them and their series of derivatives at zero and applying the pattern. But why just the trig functions? Why wouldn't we be plugging x=0 for x^3, or even the x^2 in cos(x^2)?
• Maclaurin/Taylor series require that the function be differentiable infinitely many times (not counting a derivative that is just the constant 0). Thus, polynomials cannot be used for creating Maclaurin/Taylor series.

While Maclaurin/Taylor series don't have to be trig functions, they do have to be something that can be differentiated infinitely many times and must be evaluable at the point under consideration.
• Can someone write down the proof for this?
• How do you express this in Sigma notation? I have &Sigma (-1)^n * (x)^?/(n+1)!

I guess a more accurate question would be: How do I express the powers of x as a function?
• I just don't understand how cos(x^2) has the same Maclaurin series as cos(x) by just replacing each (X) with (X^2)
• Perhaps you should try it and see whether you get the series converging on the correct answer.
for x = ½π you should get
cos x = 0
cos (x²) = −0.781211892....
So, check a few terms of Maclaurin and see whether you are converging on those values.