Taylor Series - Problem Solving Practice Problems Online

确定函数的泰勒级数

$f (x) = \ int _0 ^ 1 \ dfrac {\ sin (x)} {x} dx ~ \文本{集中在}~ x = 0。$

\ sum_ {n = 0} ^ {\ infty} \ dfrac {(1) ^ n} {(n + 2) !｝

\ sum_ {n = 0} ^ {\ infty} \ dfrac {(1) ^ n} {(2 n) !｝

\ sum_ {n = 0} ^ {\ infty} \ dfrac {(1) ^ n} {(2 n + 1) !(2 n + 1)}

\sum_{n=0}^{\infty} \dfrac{(-1)^n}{2n (2n+1)!｝

评估 $\压裂{\π^ 2}{2 !}- \压裂{\π^ 4}{4 !}+ \压裂{\π^ 6}{6 !}- \压裂{\π^ 8}{8 !}+ \ cdots。$

确定函数的泰勒级数

$F (x)= sin x\cos x ~\text{居中}~ x=0。$

\ sum_ {n = 0} ^ {\ infty} \ dfrac {(1) n (2 x) ^ ^ {2 n + 1}} {(2 n + 1) !｝

\ dfrac {1} {2} \ sum_ {n = 0} ^ {\ infty} \ dfrac {(1) ^ n (2 x) ^ {2 n + 1}} {(2 n + 1) !｝

\ sum_ {n = 0} ^ {\ infty} \压裂{x ^ {2 n - 1}} {(2 n - 1)}

2 \ sum_ {n = 0} ^ {\ infty} \ dfrac {(1) ^ n (2 x) ^ {2 n + 1}} {(2 n + 1) !｝

评估

$\ sum_ {n = 1} ^ \ infty \压裂{(1)^ {n + 1}} {n \ cdot 2 ^ n}。$

(e + 1) ^ \压裂{1}{2}

\ ln \ dfrac {1} {2}

\ e ^{\压裂{3}{2}}

\ ln \ dfrac {3} {2}

如果 $f (x) = e ^ {x ^ 2}$ 是什么 $f ^ {(2016)} (0)$ ？

\ dfrac {1 !} {2016 !｝

\ dfrac {1} {1008 !｝

0

\ dfrac {2016 !} {1008 !｝

泰勒级数-解决问题