Can I prove there is no real solution except $x=0, x=1$, without using the function $W(x)$?
Can I prove there is no real solution except $x=0, x=1$, without using the function $W(x)$?
And is it possible to do it without using calculus?
$$2^x=x+1.$$
Here is my attempts:
$2^x>0 Rightarrow x+1>0 Rightarrow x>-1$.
Now, I need to look at these intervals.
$$xin (-1,0]; [0,1]; [1, infty)$$
Maybe, it is easy to prove there is no real solution for $x>1$. Because , for $xtoinfty$, we get $2^x>>x$.
Problematic point is , $xin [0,1]$ or $xin [-1,0]$.
For $0<x<1$, we get $2>2^x>1$, which is correct ,because $2>x+1>1$ also true. So, this method doesn't work. I need more rigorous method.
inequality proof-writing exponential-function real-numbers
asked Dec 7 at 22:07
Beginner
848
add a comment |
Can I prove there is no real solution except $x=0, x=1$, without using the function $W(x)$?
And is it possible to do it without using calculus?
$$2^x=x+1.$$
Here is my attempts:
$2^x>0 Rightarrow x+1>0 Rightarrow x>-1$.
Now, I need to look at these intervals.
$$xin (-1,0]; [0,1]; [1, infty)$$
Maybe, it is easy to prove there is no real solution for $x>1$. Because , for $xtoinfty$, we get $2^x>>x$.
Problematic point is , $xin [0,1]$ or $xin [-1,0]$.
For $0<x<1$, we get $2>2^x>1$, which is correct ,because $2>x+1>1$ also true. So, this method doesn't work. I need more rigorous method.
inequality proof-writing exponential-function real-numbers
asked Dec 7 at 22:07
Beginner
848
add a comment |
Can I prove there is no real solution except $x=0, x=1$, without using the function $W(x)$?
And is it possible to do it without using calculus?
$$2^x=x+1.$$
Here is my attempts:
$2^x>0 Rightarrow x+1>0 Rightarrow x>-1$.
Now, I need to look at these intervals.
$$xin (-1,0]; [0,1]; [1, infty)$$
Maybe, it is easy to prove there is no real solution for $x>1$. Because , for $xtoinfty$, we get $2^x>>x$.
Problematic point is , $xin [0,1]$ or $xin [-1,0]$.
For $0<x<1$, we get $2>2^x>1$, which is correct ,because $2>x+1>1$ also true. So, this method doesn't work. I need more rigorous method.
inequality proof-writing exponential-function real-numbers
asked Dec 7 at 22:07
Beginner
848
Can I prove there is no real solution except $x=0, x=1$, without using the function $W(x)$?
And is it possible to do it without using calculus?
$$2^x=x+1.$$
Here is my attempts:
$2^x>0 Rightarrow x+1>0 Rightarrow x>-1$.
Now, I need to look at these intervals.
$$xin (-1,0]; [0,1]; [1, infty)$$
Maybe, it is easy to prove there is no real solution for $x>1$. Because , for $xtoinfty$, we get $2^x>>x$.
Problematic point is , $xin [0,1]$ or $xin [-1,0]$.
For $0<x<1$, we get $2>2^x>1$, which is correct ,because $2>x+1>1$ also true. So, this method doesn't work. I need more rigorous method.
inequality proof-writing exponential-function real-numbers
inequality proof-writing exponential-function real-numbers
asked Dec 7 at 22:07
Beginner
848
asked Dec 7 at 22:07
Beginner
848
edited Dec 7 at 22:53
asked Dec 7 at 22:07
Beginner
848
asked Dec 7 at 22:07
Beginner
848
asked Dec 7 at 22:07
Beginner
848
848
add a comment |
add a comment |
4 Answers
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To prove that no solution exists in $(0,1),$ you may use the series representation for the exponential function (which can be derived without using calculus). If $xin (0,1)$, we have
$$begin{align}
2^x
&=1+frac{ln(2)}{1!}x+frac{ln^2(2)}{2!}x^2+...\
&=1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}x+...bigg)\
&<1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}+...bigg)\
&=1+x\
end{align} $$
Which shows that $2^xlt 1+x$ for $xin (0,1)$.
answered Dec 7 at 22:42
Frpzzd
21.6k839107
add a comment |
With calculus:
Lemma: If a function $f$ is $n$ times differentiable on an interval $I$, and $f$ has $n+1$ distinct zeros on $I$, then the $n$th derivative of $f$ has a zero somewhere in that interval.
Proof sketch: Apply Rolle's theorem repeatedly. There's a zero of $f'$ between each pair of zeros of $f$, for a total of at least $n$, then a zero of $f''$ between each zero of $f'$, and so on.
Now, for this problem? Consider $f(x)=2^x-x-1$. The second derivative is $f''(x)=ln^2(2)cdot 2^x$, which is positive on all of $mathbb{R}$. With no zeros of the second derivative, the contrapositive of the lemma gives us that there are no more than two zeros of $f$ - on all of $mathbb{R}$. We already know of two ($0$ and $1$), so there are no others.
answered Dec 7 at 22:30
jmerry
1,14616
2
In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
– jmerry
Dec 7 at 22:35
add a comment |
Just another way $:quad$ You have already ruled out $x leqslant -1$. For others,
Case $xin (0, 1)$ using Bernoulli's inequality, $(1+1)^x < 1+x$, so there is no solution
Case $x in (-1, 0)cup (1, infty)$, again with Bernoulli's inequality we have $(1+1)^x > 1+x$.
Thus the only remaining points to check are $xin {0, 1}$.
answered Dec 8 at 18:57
Macavity
35.1k52453
add a comment |
Function $xmapsto 2^x$ is strictly convex.
A strictly convex intersects a line in 0, 1 or 2 points.
So there exists at most two solutions of the equation $2^x = x +1$.
answered Dec 7 at 23:01
dallonsi
1187
add a comment |
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4 Answers
4
active
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4 Answers
4
active
oldest
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active
oldest
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active
oldest
votes
To prove that no solution exists in $(0,1),$ you may use the series representation for the exponential function (which can be derived without using calculus). If $xin (0,1)$, we have
$$begin{align}
2^x
&=1+frac{ln(2)}{1!}x+frac{ln^2(2)}{2!}x^2+...\
&=1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}x+...bigg)\
&<1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}+...bigg)\
&=1+x\
end{align} $$
Which shows that $2^xlt 1+x$ for $xin (0,1)$.
answered Dec 7 at 22:42
Frpzzd
21.6k839107
add a comment |
To prove that no solution exists in $(0,1),$ you may use the series representation for the exponential function (which can be derived without using calculus). If $xin (0,1)$, we have
$$begin{align}
2^x
&=1+frac{ln(2)}{1!}x+frac{ln^2(2)}{2!}x^2+...\
&=1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}x+...bigg)\
&<1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}+...bigg)\
&=1+x\
end{align} $$
Which shows that $2^xlt 1+x$ for $xin (0,1)$.
answered Dec 7 at 22:42
Frpzzd
21.6k839107
add a comment |
To prove that no solution exists in $(0,1),$ you may use the series representation for the exponential function (which can be derived without using calculus). If $xin (0,1)$, we have
$$begin{align}
2^x
&=1+frac{ln(2)}{1!}x+frac{ln^2(2)}{2!}x^2+...\
&=1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}x+...bigg)\
&<1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}+...bigg)\
&=1+x\
end{align} $$
Which shows that $2^xlt 1+x$ for $xin (0,1)$.
answered Dec 7 at 22:42
Frpzzd
21.6k839107
To prove that no solution exists in $(0,1),$ you may use the series representation for the exponential function (which can be derived without using calculus). If $xin (0,1)$, we have
$$begin{align}
2^x
&=1+frac{ln(2)}{1!}x+frac{ln^2(2)}{2!}x^2+...\
&=1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}x+...bigg)\
&<1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}+...bigg)\
&=1+x\
end{align} $$
Which shows that $2^xlt 1+x$ for $xin (0,1)$.
answered Dec 7 at 22:42
Frpzzd
21.6k839107
answered Dec 7 at 22:42
Frpzzd
21.6k839107
answered Dec 7 at 22:42
Frpzzd
21.6k839107
answered Dec 7 at 22:42
Frpzzd
21.6k839107
21.6k839107
add a comment |
add a comment |
With calculus:
Lemma: If a function $f$ is $n$ times differentiable on an interval $I$, and $f$ has $n+1$ distinct zeros on $I$, then the $n$th derivative of $f$ has a zero somewhere in that interval.
Proof sketch: Apply Rolle's theorem repeatedly. There's a zero of $f'$ between each pair of zeros of $f$, for a total of at least $n$, then a zero of $f''$ between each zero of $f'$, and so on.
Now, for this problem? Consider $f(x)=2^x-x-1$. The second derivative is $f''(x)=ln^2(2)cdot 2^x$, which is positive on all of $mathbb{R}$. With no zeros of the second derivative, the contrapositive of the lemma gives us that there are no more than two zeros of $f$ - on all of $mathbb{R}$. We already know of two ($0$ and $1$), so there are no others.
answered Dec 7 at 22:30
jmerry
1,14616
2
In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
– jmerry
Dec 7 at 22:35
add a comment |
With calculus:
Lemma: If a function $f$ is $n$ times differentiable on an interval $I$, and $f$ has $n+1$ distinct zeros on $I$, then the $n$th derivative of $f$ has a zero somewhere in that interval.
Proof sketch: Apply Rolle's theorem repeatedly. There's a zero of $f'$ between each pair of zeros of $f$, for a total of at least $n$, then a zero of $f''$ between each zero of $f'$, and so on.
Now, for this problem? Consider $f(x)=2^x-x-1$. The second derivative is $f''(x)=ln^2(2)cdot 2^x$, which is positive on all of $mathbb{R}$. With no zeros of the second derivative, the contrapositive of the lemma gives us that there are no more than two zeros of $f$ - on all of $mathbb{R}$. We already know of two ($0$ and $1$), so there are no others.
answered Dec 7 at 22:30
jmerry
1,14616
2
In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
– jmerry
Dec 7 at 22:35
add a comment |
With calculus:
Lemma: If a function $f$ is $n$ times differentiable on an interval $I$, and $f$ has $n+1$ distinct zeros on $I$, then the $n$th derivative of $f$ has a zero somewhere in that interval.
Proof sketch: Apply Rolle's theorem repeatedly. There's a zero of $f'$ between each pair of zeros of $f$, for a total of at least $n$, then a zero of $f''$ between each zero of $f'$, and so on.
Now, for this problem? Consider $f(x)=2^x-x-1$. The second derivative is $f''(x)=ln^2(2)cdot 2^x$, which is positive on all of $mathbb{R}$. With no zeros of the second derivative, the contrapositive of the lemma gives us that there are no more than two zeros of $f$ - on all of $mathbb{R}$. We already know of two ($0$ and $1$), so there are no others.
answered Dec 7 at 22:30
jmerry
1,14616
With calculus:
Lemma: If a function $f$ is $n$ times differentiable on an interval $I$, and $f$ has $n+1$ distinct zeros on $I$, then the $n$th derivative of $f$ has a zero somewhere in that interval.
Proof sketch: Apply Rolle's theorem repeatedly. There's a zero of $f'$ between each pair of zeros of $f$, for a total of at least $n$, then a zero of $f''$ between each zero of $f'$, and so on.
Now, for this problem? Consider $f(x)=2^x-x-1$. The second derivative is $f''(x)=ln^2(2)cdot 2^x$, which is positive on all of $mathbb{R}$. With no zeros of the second derivative, the contrapositive of the lemma gives us that there are no more than two zeros of $f$ - on all of $mathbb{R}$. We already know of two ($0$ and $1$), so there are no others.
answered Dec 7 at 22:30
jmerry
1,14616
answered Dec 7 at 22:30
jmerry
1,14616
answered Dec 7 at 22:30
jmerry
1,14616
answered Dec 7 at 22:30
jmerry
1,14616
1,14616
2
In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
– jmerry
Dec 7 at 22:35
add a comment |
2
In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
– jmerry
Dec 7 at 22:35
2
2
In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
– jmerry
Dec 7 at 22:35
In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
– jmerry
Dec 7 at 22:35
add a comment |
Just another way $:quad$ You have already ruled out $x leqslant -1$. For others,
Case $xin (0, 1)$ using Bernoulli's inequality, $(1+1)^x < 1+x$, so there is no solution
Case $x in (-1, 0)cup (1, infty)$, again with Bernoulli's inequality we have $(1+1)^x > 1+x$.
Thus the only remaining points to check are $xin {0, 1}$.
answered Dec 8 at 18:57
Macavity
35.1k52453
add a comment |
Just another way $:quad$ You have already ruled out $x leqslant -1$. For others,
Case $xin (0, 1)$ using Bernoulli's inequality, $(1+1)^x < 1+x$, so there is no solution
Case $x in (-1, 0)cup (1, infty)$, again with Bernoulli's inequality we have $(1+1)^x > 1+x$.
Thus the only remaining points to check are $xin {0, 1}$.
answered Dec 8 at 18:57
Macavity
35.1k52453
add a comment |
Just another way $:quad$ You have already ruled out $x leqslant -1$. For others,
Case $xin (0, 1)$ using Bernoulli's inequality, $(1+1)^x < 1+x$, so there is no solution
Case $x in (-1, 0)cup (1, infty)$, again with Bernoulli's inequality we have $(1+1)^x > 1+x$.
Thus the only remaining points to check are $xin {0, 1}$.
answered Dec 8 at 18:57
Macavity
35.1k52453
Just another way $:quad$ You have already ruled out $x leqslant -1$. For others,
Case $xin (0, 1)$ using Bernoulli's inequality, $(1+1)^x < 1+x$, so there is no solution
Case $x in (-1, 0)cup (1, infty)$, again with Bernoulli's inequality we have $(1+1)^x > 1+x$.
Thus the only remaining points to check are $xin {0, 1}$.
answered Dec 8 at 18:57
Macavity
35.1k52453
answered Dec 8 at 18:57
Macavity
35.1k52453
answered Dec 8 at 18:57
Macavity
35.1k52453
answered Dec 8 at 18:57
Macavity
35.1k52453
35.1k52453
add a comment |
add a comment |
Function $xmapsto 2^x$ is strictly convex.
A strictly convex intersects a line in 0, 1 or 2 points.
So there exists at most two solutions of the equation $2^x = x +1$.
answered Dec 7 at 23:01
dallonsi
1187
add a comment |
Function $xmapsto 2^x$ is strictly convex.
A strictly convex intersects a line in 0, 1 or 2 points.
So there exists at most two solutions of the equation $2^x = x +1$.
answered Dec 7 at 23:01
dallonsi
1187
add a comment |
Function $xmapsto 2^x$ is strictly convex.
A strictly convex intersects a line in 0, 1 or 2 points.
So there exists at most two solutions of the equation $2^x = x +1$.
answered Dec 7 at 23:01
dallonsi
1187
Function $xmapsto 2^x$ is strictly convex.
A strictly convex intersects a line in 0, 1 or 2 points.
So there exists at most two solutions of the equation $2^x = x +1$.
answered Dec 7 at 23:01
dallonsi
1187
edited Dec 7 at 23:06
answered Dec 7 at 23:01
dallonsi
1187
answered Dec 7 at 23:01
dallonsi
1187
answered Dec 7 at 23:01
dallonsi
1187
1187
add a comment |
add a comment |
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