Calculus Cheat Sheet
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Special Limit Theorems
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Special stuff to know when encountering these limits.
lim
a
→
0
sin
α
α
=
1
{\displaystyle \lim _{a\to 0}{\frac {\sin \alpha }{\alpha }}=1}
lim
a
→
0
cos
α
−
1
α
=
0
{\displaystyle \lim _{a\to 0}{\frac {\cos \alpha -1}{\alpha }}=0}
Infinitely Large Demoninator
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lim
x
→
∞
=
any real number
x
any integer
≥
0
=
0
{\displaystyle \lim _{x\to \infty }={\frac {\mbox{any real number}}{x^{{\mbox{any integer}}\geq 0}}}=0}
Euler's Number
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lim
x
→
∞
(
1
+
1
x
)
x
=
e
{\displaystyle \lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=e}
Trigonometric Integrals
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Integrals of trig functions, of course.
∫
sin
x
d
x
=
cos
x
+
C
{\displaystyle \int \sin x\;dx=\cos x+C}
∫
cos
x
d
x
=
sin
x
+
C
{\displaystyle \int \cos x\;dx=\sin x+C}
∫
tan
x
d
x
=
−
ln
|
cos
x
|
+
C
{\displaystyle \int \tan x\;dx=-\ln |\cos x|+C}
Cotangent
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∫
cot
x
d
x
=
ln
|
sin
x
|
+
C
{\displaystyle \int \cot x\;dx=\ln |\sin x|+C}
∫
sec
x
d
x
=
ln
|
sec
x
+
tan
x
|
+
C
{\displaystyle \int \sec x\;dx=\ln |\sec x+\tan x|+C}
Cosecant
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∫
csc
x
d
x
=
−
ln
|
csc
x
+
cot
x
|
+
C
{\displaystyle \int \csc x\;dx=-\ln |\csc x+\cot x|+C}
Inverse Trigonometric Integrals
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Special integrals that can be solved with inverse trig functions.
∫
1
a
2
−
u
2
d
u
=
arcsin
(
u
a
)
+
c
{\displaystyle \int {\frac {1}{\sqrt {a^{2}-u^{2}}}}du=\arcsin \left({\frac {u}{a}}\right)+c}
Arctangent
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∫
−
d
u
=
1
a
arctan
(
u
a
)
+
c
{\displaystyle \int -du={\frac {1}{a}}\arctan \left({\frac {u}{a}}\right)+c}
Arcsecant
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∫
1
u
u
2
−
a
2
=
1
a
arcsec
(
|
u
|
a
)
+
c
{\displaystyle \int {\frac {1}{u{\sqrt {u^{2}-a^{2}}}}}={\frac {1}{a}}\operatorname {arcsec} \left({\frac {|u|}{a}}\right)+c}
Common Series
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The common types of series to recognize.
Geometric Series
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∑
n
=
0
∞
a
r
n
{\displaystyle \sum _{n=0}^{\infty }ar^{n}}
P -Series
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∑
n
=
1
∞
1
n
P
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{P}}}}
Telescoping Series
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∑
n
=
1
∞
(
1
n
−
1
n
+
a
)
{\displaystyle \sum _{n=1}^{\infty }\left({\frac {1}{n}}-{\frac {1}{n+a}}\right)}
Special Series
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These are special series that are good to remember.
Power Series
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∑
n
=
0
∞
a
n
(
x
−
c
)
n
{\displaystyle \sum _{n=0}^{\infty }a_{n}(x-c)^{n}}
Maclaurin Series
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∑
n
=
0
∞
f
(
n
)
(
0
)
x
n
n
!
{\displaystyle \sum _{n=0}^{\infty }{\frac {f^{(n)}(0)x^{n}}{n!}}}
Taylor Series
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∑
n
=
0
∞
f
(
n
)
(
a
)
n
!
(
x
−
a
)
n
{\displaystyle \sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}}
Math Proofs
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Discuss them here .
Prove how
i
i
≈
.20787957635076
{\displaystyle i^{i}\approx .20787957635076}
, a real, irrational number.
It appears to be
e
−
π
2
{\displaystyle e^{-{\frac {\pi }{2}}}}
.
Solution: Euler proved, in 1746, that this has infinitely many values:
i
i
=
e
−
(
π
2
+
2
k
π
)
{\displaystyle i^{i}=e^{-({\frac {\pi }{2}}+2k\pi )}}
. The principal value is k = 0, which provides
e
π
2
{\displaystyle e^{\frac {\pi }{2}}}
.[1]
Prove Euler's Identity, in a way different than the one on the actual Wiki page.
Dividing/Multiplying by Zero
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This is based partially off of a book I am reading.[2]
Using the distributive property, one can tell that
2
×
3
+
2
×
4
=
2
(
3
+
4
)
{\displaystyle 2\times 3+2\times 4=2(3+4)}
. So, one could apply that to zero, allowing
2
×
0
+
2
×
0
=
2
(
0
+
0
)
{\displaystyle 2\times 0+2\times 0=2(0+0)}
. So, we know that
2
×
0
+
2
×
0
=
2
×
0
{\displaystyle 2\times 0+2\times 0=2\times 0}
. Now, if you subtract
2
×
0
{\displaystyle 2\times 0}
from it, you get
2
×
0
=
0
{\displaystyle 2\times 0=0}
.
Division is the undoing of multiplication. So, one could infer that
(
2
×
0
)
0
=
2
{\displaystyle {\frac {(2\times 0)}{0}}=2}
. But you can also tell that
(
3
×
0
)
0
=
3
{\displaystyle {\frac {(3\times 0)}{0}}=3}
, and so on until you realize that
N
×
0
0
=
N
{\displaystyle {\frac {\mathbb {N} \times 0}{0}}=\mathbb {N} }
, which in itself proves that any number times zero is zero (because if you multiply both sides by zero to remove the function, you get
N
×
0
=
0
{\displaystyle \mathbb {N} \times 0=0}
). Since we know that anything times zero is zero,
0
0
=
2
{\displaystyle {\frac {0}{0}}=2}
in the case of
(
2
×
0
)
0
=
2
{\displaystyle {\frac {(2\times 0)}{0}}=2}
. But
0
0
{\displaystyle {\frac {0}{0}}}
must also be 3, 4, et al. So, that would mean that
0
0
=
N
{\displaystyle {\frac {0}{0}}=\mathbb {N} }
. But that is the same as saying
0
0
=
∞
{\displaystyle {\frac {0}{0}}=\infty }
. But if you put, say, 3 on top, you would get infinity also. Since zero plus itself is always zero, regardless of how many times it's done, it would take an infinite amount of zeros to get to three, and then some. Since no matter how many zeros you put into three, you won't get to three, you can say that
N
0
=
∞
{\displaystyle {\frac {\mathbb {N} }{0}}=\infty }
because it would apply for absolutely any number, including 0 and 3. On a side note, though,
0
0
{\displaystyle {\frac {0}{0}}}
is an indeterminate number , as is
∞
∞
{\displaystyle {\frac {\infty }{\infty }}}
.
Inverses
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The inverse of any function of the form
f
(
x
)
=
1
x
|
a
b
|
{\displaystyle f(x)={\frac {1}{x}}^{|{\frac {a}{b}}|}}
is
f
−
1
(
x
)
=
1
x
b
a
{\displaystyle f^{-1}(x)={\frac {1}{x}}^{\frac {b}{a}}}
.
References
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^ Eli Maor (1994) - e : The Story of a Number ISBN 0-691-03390-0
^ Charles Seife (2000) - Zero: The Biography of a Dangerous Idea ISBN 0-670-88457-X