Calculus BC ยท Unit 10
Substitute into a known series, then integrate or differentiate term by term inside the interval of convergence.
Start from a known series
\(e^u=\sum \dfrac{u^n}{n!}\), \(\sin u=u-\dfrac{u^3}{6}+\cdots\), \(\cos u=1-\dfrac{u^2}{2}+\cdots\), and \(\dfrac{1}{1-u}=\sum u^n\) for \(|u|<1\).
Substitute
Replace \(u\) by \(-x^2\), \(2x\), or \(x^2\), then simplify the powers.
Integrate term by term
Inside the interval of convergence, integrate each term and keep the new denominator \(2n+1\) or whatever power appears.
A limit can be the first coefficient
Write the series, cancel the terms that the numerator subtracts, and divide. The constant that remains is the limit.
One substitution into a memorized series.
\(e^u=\sum_{n=0}^{\infty}\dfrac{u^n}{n!}\)
Use \(u=-x^2\).
\(e^{-x^2}=\sum_{n=0}^{\infty}\dfrac{(-x^2)^n}{n!}\)
\(=\sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n}}{n!}\)
Answer: \(\sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n}}{n!}\)
\(\cos u=\sum_{n=0}^{\infty}\dfrac{(-1)^n u^{2n}}{(2n)!}\)
Use \(u=2x\).
\(\cos(2x)=\sum_{n=0}^{\infty}\dfrac{(-1)^n (2x)^{2n}}{(2n)!}\)
\(=\sum_{n=0}^{\infty}\dfrac{(-1)^n 4^n x^{2n}}{(2n)!}\)
Answer: \(\sum_{n=0}^{\infty}\dfrac{(-1)^n 4^n x^{2n}}{(2n)!}\)
\(\dfrac{1}{1-u}=\sum_{n=0}^{\infty} u^n\) for \(|u|<1\)
Use \(u=-x^2\).
\(\dfrac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^n x^{2n}\)
This needs \(|-x^2|<1\), so \(|x|<1\).
Answer: \(\sum_{n=0}^{\infty}(-1)^n x^{2n}\) for \(|x|<1\)
\(\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{x^n}{n}\) for \(-1 The \(n=1,2,3\) terms are \(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}\) Answer: \(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}\)
\(\sin u=u-\dfrac{u^3}{6}+\cdots\)
Use \(u=x^2\).
\(x^2-\dfrac{(x^2)^3}{6}\)
\(x^2-\dfrac{x^6}{6}\)
Answer: \(x^2-\dfrac{x^6}{6}\)
Substitute \(u=3x\) into \(1+u+u^2/2\).
\(1+3x+\dfrac{(3x)^2}{2}\)
\(1+3x+\dfrac{9x^2}{2}\)
Answer: A. \(1+3x+\dfrac{9x^2}{2}\)
Integrate a series, or cancel the first terms before a limit.
\(e^{-t^2}=\sum_{n=0}^{\infty}\dfrac{(-1)^n t^{2n}}{n!}\)
Integrate term by term from 0 to \(x\).
\(\int_0^{x}\dfrac{(-1)^n t^{2n}}{n!}\,dt=\dfrac{(-1)^n x^{2n+1}}{n!(2n+1)}\)
\(\int_0^{x}e^{-t^2}\,dt=\sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n+1}}{n!(2n+1)}\)
Answer: \(\sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n+1}}{n!(2n+1)}\)
\(e^x=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\cdots\)
\(e^x-1-x=\dfrac{x^2}{2}+\dfrac{x^3}{6}+\cdots\)
\(\dfrac{e^x-1-x}{x^2}=\dfrac{1}{2}+\dfrac{x}{6}+\cdots\)
As \(x\to 0\), the limit is \(\dfrac{1}{2}\).
Answer: \(\dfrac{1}{2}\)
Differentiate term by term for \(|x|<1\).
\(\dfrac{1}{(1-x)^2}=\sum_{n=1}^{\infty} n x^{n-1}\)
Answer: \(\sum_{n=1}^{\infty} n x^{n-1}\) for \(|x|<1\)
\(\cos(t^2)=1-\dfrac{t^4}{2}+\dfrac{t^8}{24}-\cdots\)
Integrate from 0 to \(x\).
\(\left[t-\dfrac{t^5}{10}+\dfrac{t^9}{216}\right]_0^{x}\)
\(x-\dfrac{x^5}{10}+\dfrac{x^9}{216}\)
Answer: \(x-\dfrac{x^5}{10}+\dfrac{x^9}{216}\)
\(e^x=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\cdots\)
\(x e^x=x\left(1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\cdots\right)\)
Through degree 3:
\(x+x^2+\dfrac{1}{2}x^3\)
Answer: \(x+x^2+\dfrac{1}{2}x^3\)
The radius is 1, from the ratio test or from the geometric series that produces it.
At \(x=1\), the series is the alternating harmonic series, which converges.
At \(x=-1\), the series is a negative harmonic series, which diverges.
The interval is \((-1,1]\).
Answer: \((-1,1]\)
A limit from a sine series, an arctangent integral, or one coefficient.
\(\sin x=x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\cdots\)
\(\sin x-x=-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\cdots\)
\(\dfrac{\sin x-x}{x^3}=-\dfrac{1}{6}+\dfrac{x^2}{120}-\cdots\)
The limit is \(-\dfrac{1}{6}\).
Answer: \(-\dfrac{1}{6}\)
\(\cos x=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\cdots\)
\(1-\cos x=\dfrac{x^2}{2}-\dfrac{x^4}{24}+\cdots\)
\(\dfrac{1-\cos x}{x^2}=\dfrac{1}{2}-\dfrac{x^2}{24}+\cdots\)
The limit is \(\dfrac{1}{2}\).
Answer: \(\dfrac{1}{2}\)
At \(x=1\), the terms are \((-1)^n\).
Those terms do not go to 0.
The series diverges at \(x=1\), even though the function \(1/(1+x^2)\) is defined there.
Answer: the terms do not go to 0
\(\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots\)
\(x-\dfrac{x^3}{6}+\dfrac{x^5}{120}\)
Answer: \(x-\dfrac{x^3}{6}+\dfrac{x^5}{120}\)
\(\dfrac{1}{1+t^2}=\sum_{n=0}^{\infty}(-1)^n t^{2n}\) for \(|t|<1\)
\(\arctan x=\int_0^{x}\sum_{n=0}^{\infty}(-1)^n t^{2n}\,dt\)
\(=\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{2n+1}\)
Answer: \(\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{2n+1}\) for \(|x|<1\)
The \(n=2\) term of \(e^{-t^2}\) is \(t^4/2!\), which is \(t^4/2\).
\(\int_0^{x}\dfrac{t^4}{2}\,dt=\dfrac{x^5}{10}\)
The coefficient is \(\dfrac{1}{10}\).
Answer: \(\dfrac{1}{10}\)
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