Prove by Mean Value Theorem $frac{x}{1+x}<ln(1+x)0$
up vote
3
down vote
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Prove for $x>0$
$$
frac{x}{1+x}<ln(1+x)<x
$$
I tried writing $ln(1+x)=ln(1+x)-ln(1)$ and using the MVT for the $(1,1+x)$ interval. I eventually could prove the inequality but how do I have to prove even for $(0,1)$
calculus inequality logarithms
edited Nov 18 at 6:35
Martin Sleziak
44.5k7115268
asked Mar 18 '15 at 17:55
user224677
161
add a comment |
up vote
3
down vote
favorite
Prove for $x>0$
$$
frac{x}{1+x}<ln(1+x)<x
$$
I tried writing $ln(1+x)=ln(1+x)-ln(1)$ and using the MVT for the $(1,1+x)$ interval. I eventually could prove the inequality but how do I have to prove even for $(0,1)$
calculus inequality logarithms
edited Nov 18 at 6:35
Martin Sleziak
44.5k7115268
asked Mar 18 '15 at 17:55
user224677
161
can it be done without Taylor?
– user224677
Mar 18 '15 at 18:08
I know but we haven't quite studied Taylor series so if you could help with some hint, I just don't know how to prove for (0;1) interval because the rest I have done
– user224677
Mar 18 '15 at 18:11
I will think on it; I don't currently have it solved either :)
– graydad
Mar 18 '15 at 18:13
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Prove for $x>0$
$$
frac{x}{1+x}<ln(1+x)<x
$$
I tried writing $ln(1+x)=ln(1+x)-ln(1)$ and using the MVT for the $(1,1+x)$ interval. I eventually could prove the inequality but how do I have to prove even for $(0,1)$
calculus inequality logarithms
edited Nov 18 at 6:35
Martin Sleziak
44.5k7115268
asked Mar 18 '15 at 17:55
user224677
161
Prove for $x>0$
$$
frac{x}{1+x}<ln(1+x)<x
$$
I tried writing $ln(1+x)=ln(1+x)-ln(1)$ and using the MVT for the $(1,1+x)$ interval. I eventually could prove the inequality but how do I have to prove even for $(0,1)$
calculus inequality logarithms
calculus inequality logarithms
edited Nov 18 at 6:35
Martin Sleziak
44.5k7115268
asked Mar 18 '15 at 17:55
user224677
161
edited Nov 18 at 6:35
Martin Sleziak
44.5k7115268
asked Mar 18 '15 at 17:55
user224677
161
edited Nov 18 at 6:35
Martin Sleziak
44.5k7115268
edited Nov 18 at 6:35
Martin Sleziak
44.5k7115268
edited Nov 18 at 6:35
Martin Sleziak
44.5k7115268
44.5k7115268
asked Mar 18 '15 at 17:55
user224677
161
asked Mar 18 '15 at 17:55
user224677
161
asked Mar 18 '15 at 17:55
user224677
161
161
can it be done without Taylor?
– user224677
Mar 18 '15 at 18:08
I know but we haven't quite studied Taylor series so if you could help with some hint, I just don't know how to prove for (0;1) interval because the rest I have done
– user224677
Mar 18 '15 at 18:11
I will think on it; I don't currently have it solved either :)
– graydad
Mar 18 '15 at 18:13
add a comment |
can it be done without Taylor?
– user224677
Mar 18 '15 at 18:08
I know but we haven't quite studied Taylor series so if you could help with some hint, I just don't know how to prove for (0;1) interval because the rest I have done
– user224677
Mar 18 '15 at 18:11
I will think on it; I don't currently have it solved either :)
– graydad
Mar 18 '15 at 18:13
can it be done without Taylor?
– user224677
Mar 18 '15 at 18:08
can it be done without Taylor?
– user224677
Mar 18 '15 at 18:08
I know but we haven't quite studied Taylor series so if you could help with some hint, I just don't know how to prove for (0;1) interval because the rest I have done
– user224677
Mar 18 '15 at 18:11
I know but we haven't quite studied Taylor series so if you could help with some hint, I just don't know how to prove for (0;1) interval because the rest I have done
– user224677
Mar 18 '15 at 18:11
I will think on it; I don't currently have it solved either :)
– graydad
Mar 18 '15 at 18:13
I will think on it; I don't currently have it solved either :)
– graydad
Mar 18 '15 at 18:13
add a comment |
1 Answer
1
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up vote
2
down vote
By the mean value theorem, given $x > 0$, there exists $c in (0,x)$ such that $f(x) - f(0) = f'(c)x$, i.e., $ln(1 + x) = frac{x}{1 + c}$. Since $0 < c < x$, $frac{1}{1 + x} < frac{1}{1 + c} < 1$. Therefore $frac{x}{1 + x} < frac{x}{1 + c} < x$, i.e.,
$$frac{x}{1 + x} < ln(1 + x) < x.$$
answered Mar 18 '15 at 18:15
kobe
34.5k22247
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
By the mean value theorem, given $x > 0$, there exists $c in (0,x)$ such that $f(x) - f(0) = f'(c)x$, i.e., $ln(1 + x) = frac{x}{1 + c}$. Since $0 < c < x$, $frac{1}{1 + x} < frac{1}{1 + c} < 1$. Therefore $frac{x}{1 + x} < frac{x}{1 + c} < x$, i.e.,
$$frac{x}{1 + x} < ln(1 + x) < x.$$
answered Mar 18 '15 at 18:15
kobe
34.5k22247
add a comment |
up vote
2
down vote
By the mean value theorem, given $x > 0$, there exists $c in (0,x)$ such that $f(x) - f(0) = f'(c)x$, i.e., $ln(1 + x) = frac{x}{1 + c}$. Since $0 < c < x$, $frac{1}{1 + x} < frac{1}{1 + c} < 1$. Therefore $frac{x}{1 + x} < frac{x}{1 + c} < x$, i.e.,
$$frac{x}{1 + x} < ln(1 + x) < x.$$
answered Mar 18 '15 at 18:15
kobe
34.5k22247
add a comment |
up vote
2
down vote
up vote
2
down vote
By the mean value theorem, given $x > 0$, there exists $c in (0,x)$ such that $f(x) - f(0) = f'(c)x$, i.e., $ln(1 + x) = frac{x}{1 + c}$. Since $0 < c < x$, $frac{1}{1 + x} < frac{1}{1 + c} < 1$. Therefore $frac{x}{1 + x} < frac{x}{1 + c} < x$, i.e.,
$$frac{x}{1 + x} < ln(1 + x) < x.$$
answered Mar 18 '15 at 18:15
kobe
34.5k22247
By the mean value theorem, given $x > 0$, there exists $c in (0,x)$ such that $f(x) - f(0) = f'(c)x$, i.e., $ln(1 + x) = frac{x}{1 + c}$. Since $0 < c < x$, $frac{1}{1 + x} < frac{1}{1 + c} < 1$. Therefore $frac{x}{1 + x} < frac{x}{1 + c} < x$, i.e.,
$$frac{x}{1 + x} < ln(1 + x) < x.$$
answered Mar 18 '15 at 18:15
kobe
34.5k22247
answered Mar 18 '15 at 18:15
kobe
34.5k22247
answered Mar 18 '15 at 18:15
kobe
34.5k22247
answered Mar 18 '15 at 18:15
kobe
34.5k22247
34.5k22247
add a comment |
add a comment |
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can it be done without Taylor?
– user224677
Mar 18 '15 at 18:08
I know but we haven't quite studied Taylor series so if you could help with some hint, I just don't know how to prove for (0;1) interval because the rest I have done
– user224677
Mar 18 '15 at 18:11
I will think on it; I don't currently have it solved either :)
– graydad
Mar 18 '15 at 18:13