Calculus Of One Real
Variable By Pheng Kim Ving |
4.1 |
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Go To Problems
& Solutions
1. Mathematical Induction |
Example 1.1
Solution
EOS
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2. Higher-Order Derivatives |
Let y = f(x)
be a differentiable function. If the derivative y' = f
'(x) is itself differentiable, its derivative is called the
second derivative of y = f(x) and is denoted y'' or f
''(x) or d ^{2}y/dx^{2} or (d ^{2}/dx^{2}) f(x):
_{}
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3. Factorials |
Let
n be a
positive integer. The factorial of n is defined as the product of all integers from 1 thru
to n and is
denoted by
n!, read n factorial, so that n! = 1 Χ 2 Χ 3 Χ ... Χ n. It's a product of n factors, hence the name factorial. The
factorial of
0 is defined to be 1: 0! = 1 (there's a
reason for this definition; it's not needed here). We have:
The factorial expansion is also written in decreasing order of the factors:
n! = n Χ (n 1) Χ (n 2) Χ Χ 2 Χ 1.
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4. Derivatives Of All Orders And
Mathematical Induction |
Example 4.1
Let f(x)
= 1/x. Find enough derivatives of different orders of f
to enable you to guess the general formula for f ^{(}^{n}^{)}(x),
where n is a positive integer. Then use mathematical induction to
prove your guess.
Solution
We have:
_{}
_{}
EOS
Note the use of (1)^{n} to specify the sign. If n
is even then we get the
+ sign; if n is odd then we get the sign. For
(1)^{n}^{+1}, if
n
is even then n + 1 is odd, hence we get the sign; if n is odd then n + 1 is even, hence
we get the +
sign.
Problems & Solutions |
1. Prove the following formula by using mathematical induction:
where n is any positive integer. Evaluate:
1 + 2 + ... + 5,000.
For n = 1 we have:
2. Let f and g be twice-differentiable functions. Prove that ( fg)'' = f ''g + 2 f 'g' + fg''.
( fg)'' = (( fg)')' = ( f 'g + fg' )' = ( f 'g)' + ( fg' )' = ( f ''g + f 'g' ) + ( f 'g' + fg'' ) = f ''g + 2 f 'g' + fg''.
3. Let f(x) = 1/(x + 2).
Calculate enough derivatives of different orders of f
to enable you to guess the general formula
for f ^{(}^{n}^{)}(x), where n is any positive integer.
Then use mathematical induction to prove your guess.
Solution
We have f(x) = (x + 2)^{}^{1}. Then:
_{}
Solution
_{}
5. Let
f(t) = t^{2/3}. Calculate
enough derivatives of different orders of f
to enable you to guess the general formula for
f ^{(}^{n}^{)}(t), where
n is any positive integer. Then use
mathematical induction to prove your guess.
Solution
We have:
_{}
_{}
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