Mixing
Why is $Kcdot f(n) + L cdot g(n) leq (K+L) cdot (f(n) + g(n)) $?
$begingroup$
$Kcdot f(n) + L cdot g(n) leq (K+L) cdot (f(n) + g(n)) $
$n, f(n), g(n), K, L in R^+$
I've seen this inequality a few times in my algorithms course, but I am trying to understand how it works. I would also like to know if it has a name.
An example of someone using it is in the solution to R-3.16
algorithms
asked Jan 29 at 17:24
Cedric MartensCedric Martens
375214
$endgroup$
add a comment |
$begingroup$
$Kcdot f(n) + L cdot g(n) leq (K+L) cdot (f(n) + g(n)) $
$n, f(n), g(n), K, L in R^+$
I've seen this inequality a few times in my algorithms course, but I am trying to understand how it works. I would also like to know if it has a name.
An example of someone using it is in the solution to R-3.16
algorithms
asked Jan 29 at 17:24
Cedric MartensCedric Martens
375214
$endgroup$
1
$begingroup$
Are $f, g$ functions into non-negative numbers, and $K, L$ non-negative numbers? Then simply expand the parentheses on the right hand side. You get the left hand side, plus something non-negative.
$endgroup$
– Mees de Vries
Jan 29 at 17:25
$begingroup$
You are right, it is the case because of the context. Thank you
$endgroup$
– Cedric Martens
Jan 29 at 17:31
add a comment |
$begingroup$
$Kcdot f(n) + L cdot g(n) leq (K+L) cdot (f(n) + g(n)) $
$n, f(n), g(n), K, L in R^+$
I've seen this inequality a few times in my algorithms course, but I am trying to understand how it works. I would also like to know if it has a name.
An example of someone using it is in the solution to R-3.16
algorithms
asked Jan 29 at 17:24
Cedric MartensCedric Martens
375214
$endgroup$
$Kcdot f(n) + L cdot g(n) leq (K+L) cdot (f(n) + g(n)) $
$n, f(n), g(n), K, L in R^+$
I've seen this inequality a few times in my algorithms course, but I am trying to understand how it works. I would also like to know if it has a name.
An example of someone using it is in the solution to R-3.16
algorithms
algorithms
asked Jan 29 at 17:24
Cedric MartensCedric Martens
375214
asked Jan 29 at 17:24
Cedric MartensCedric Martens
375214
edited Jan 29 at 17:33
Cedric Martens
asked Jan 29 at 17:24
Cedric MartensCedric Martens
375214
asked Jan 29 at 17:24
Cedric MartensCedric Martens
375214
asked Jan 29 at 17:24
Cedric MartensCedric Martens
375214
375214
1
$begingroup$
Are $f, g$ functions into non-negative numbers, and $K, L$ non-negative numbers? Then simply expand the parentheses on the right hand side. You get the left hand side, plus something non-negative.
$endgroup$
– Mees de Vries
Jan 29 at 17:25
$begingroup$
You are right, it is the case because of the context. Thank you
$endgroup$
– Cedric Martens
Jan 29 at 17:31
add a comment |
1
$begingroup$
Are $f, g$ functions into non-negative numbers, and $K, L$ non-negative numbers? Then simply expand the parentheses on the right hand side. You get the left hand side, plus something non-negative.
$endgroup$
– Mees de Vries
Jan 29 at 17:25
$begingroup$
You are right, it is the case because of the context. Thank you
$endgroup$
– Cedric Martens
Jan 29 at 17:31
1
1
$begingroup$
Are $f, g$ functions into non-negative numbers, and $K, L$ non-negative numbers? Then simply expand the parentheses on the right hand side. You get the left hand side, plus something non-negative.
$endgroup$
– Mees de Vries
Jan 29 at 17:25
$begingroup$
Are $f, g$ functions into non-negative numbers, and $K, L$ non-negative numbers? Then simply expand the parentheses on the right hand side. You get the left hand side, plus something non-negative.
$endgroup$
– Mees de Vries
Jan 29 at 17:25
$begingroup$
You are right, it is the case because of the context. Thank you
$endgroup$
– Cedric Martens
Jan 29 at 17:31
$begingroup$
You are right, it is the case because of the context. Thank you
$endgroup$
– Cedric Martens
Jan 29 at 17:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $K, L, f(n), g(n)$ are all non-negative numbers, expanding the right hand side of the inequality gives the left hand side plus something non-negative.
answered Jan 29 at 17:32
Mees de VriesMees de Vries
17.6k13060
$endgroup$
add a comment |
$begingroup$
$$(K+L).(f(n) + g(n)) = K.f(n) + L.g(n) + K.g(n) + L.f(n)$$
if $K.g(n) geq 0$ and $L.f(n) geq 0$ then they only increase the value of the RHS which makes it greater than or equal to the first two terms: $K.f(n) + L.g(n)$. Hence:
$$K.f(n) + L.g(n) leq (K+L).(f(n) + g(n)) $$
answered Jan 29 at 17:33
Aloagbaye MomoduAloagbaye Momodu
163
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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$begingroup$
Since $K, L, f(n), g(n)$ are all non-negative numbers, expanding the right hand side of the inequality gives the left hand side plus something non-negative.
answered Jan 29 at 17:32
Mees de VriesMees de Vries
17.6k13060
$endgroup$
add a comment |
$begingroup$
Since $K, L, f(n), g(n)$ are all non-negative numbers, expanding the right hand side of the inequality gives the left hand side plus something non-negative.
answered Jan 29 at 17:32
Mees de VriesMees de Vries
17.6k13060
$endgroup$
add a comment |
$begingroup$
Since $K, L, f(n), g(n)$ are all non-negative numbers, expanding the right hand side of the inequality gives the left hand side plus something non-negative.
answered Jan 29 at 17:32
Mees de VriesMees de Vries
17.6k13060
$endgroup$
Since $K, L, f(n), g(n)$ are all non-negative numbers, expanding the right hand side of the inequality gives the left hand side plus something non-negative.
answered Jan 29 at 17:32
Mees de VriesMees de Vries
17.6k13060
answered Jan 29 at 17:32
Mees de VriesMees de Vries
17.6k13060
answered Jan 29 at 17:32
Mees de VriesMees de Vries
17.6k13060
answered Jan 29 at 17:32
Mees de VriesMees de Vries
17.6k13060
17.6k13060
add a comment |
add a comment |
$begingroup$
$$(K+L).(f(n) + g(n)) = K.f(n) + L.g(n) + K.g(n) + L.f(n)$$
if $K.g(n) geq 0$ and $L.f(n) geq 0$ then they only increase the value of the RHS which makes it greater than or equal to the first two terms: $K.f(n) + L.g(n)$. Hence:
$$K.f(n) + L.g(n) leq (K+L).(f(n) + g(n)) $$
answered Jan 29 at 17:33
Aloagbaye MomoduAloagbaye Momodu
163
$endgroup$
add a comment |
$begingroup$
$$(K+L).(f(n) + g(n)) = K.f(n) + L.g(n) + K.g(n) + L.f(n)$$
if $K.g(n) geq 0$ and $L.f(n) geq 0$ then they only increase the value of the RHS which makes it greater than or equal to the first two terms: $K.f(n) + L.g(n)$. Hence:
$$K.f(n) + L.g(n) leq (K+L).(f(n) + g(n)) $$
answered Jan 29 at 17:33
Aloagbaye MomoduAloagbaye Momodu
163
$endgroup$
add a comment |
$begingroup$
$$(K+L).(f(n) + g(n)) = K.f(n) + L.g(n) + K.g(n) + L.f(n)$$
if $K.g(n) geq 0$ and $L.f(n) geq 0$ then they only increase the value of the RHS which makes it greater than or equal to the first two terms: $K.f(n) + L.g(n)$. Hence:
$$K.f(n) + L.g(n) leq (K+L).(f(n) + g(n)) $$
answered Jan 29 at 17:33
Aloagbaye MomoduAloagbaye Momodu
163
$endgroup$
$$(K+L).(f(n) + g(n)) = K.f(n) + L.g(n) + K.g(n) + L.f(n)$$
if $K.g(n) geq 0$ and $L.f(n) geq 0$ then they only increase the value of the RHS which makes it greater than or equal to the first two terms: $K.f(n) + L.g(n)$. Hence:
$$K.f(n) + L.g(n) leq (K+L).(f(n) + g(n)) $$
answered Jan 29 at 17:33
Aloagbaye MomoduAloagbaye Momodu
163
answered Jan 29 at 17:33
Aloagbaye MomoduAloagbaye Momodu
163
answered Jan 29 at 17:33
Aloagbaye MomoduAloagbaye Momodu
163
answered Jan 29 at 17:33
Aloagbaye MomoduAloagbaye Momodu
163
163
add a comment |
add a comment |
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$begingroup$
Are $f, g$ functions into non-negative numbers, and $K, L$ non-negative numbers? Then simply expand the parentheses on the right hand side. You get the left hand side, plus something non-negative.
$endgroup$
– Mees de Vries
Jan 29 at 17:25
$begingroup$
You are right, it is the case because of the context. Thank you
$endgroup$
– Cedric Martens
Jan 29 at 17:31