Mixing
the basis of the natural logarithm
$begingroup$
Euler found out that Euler's number was the basis of the natural logarithm. He started with this equation for infinitely small numbers $w$:
For $a>1$:
$$a^w=1+kw$$
$w$=infinitely small numbers
$a$=base
From $a^w=1+kw$ to this:
$${a}^{jw}={(1+kw)}^{j}$$
$${a}^{jw}={(1+kw)}^{j}=1+frac{j}{1}kw+frac{j(j-1)}{1*2}{k}^{2}{w}^{2}+frac{j(j-1)(j-2)}{1*2*3}{k}^{3}{w}^{3}+...$$
Then he replaced $j$ with $frac{z}{w}$:
$$j=frac{z}{w}$$
$z$ is any finite number, $ω$ is an infinitely small number and $j$ is an infinitely
large number.
Then he replaced $w$ with $frac{z}{j}$:
$${a}^{z}={a}^{jw}={(1+k*frac{z}{j})}^{j}=1+frac{j}{1}k*frac{z}{j}+frac{j(j-1)}{1*2}{k}^{2}{(frac{z}{j})}^{2}+frac{j(j-1)(j-2)}{1*2*3}{k}^{3}{(frac{z}{j})}^{3}+...$$
Since $j$ is infinitely large, you will get this:
$${a}^{z}=1+frac{kz}{1}+frac{{k}^{2}{z}^{2}}{1*2}+frac{{k}^{3}{z}^{3}}{1*2*3}+...$$
Then you replace $z$ and $k$ with $1$ to get the base of natural logarithm:
$${a}=1+frac{1}{1}+frac{1}{1*2}+frac{1}{1*2*3}+...$$
My question is, why did Euler replace $k$ and $z$ with $1$ to get the base of natural logarithm?
He had made also a series for $$log{1+x}=frac{1}{k}(frac{x}{1}-frac{{x}^{2}}{2}+frac{{x}^{3}}{3}-frac{{x}^{4}}{4}+...)$$
You can see that here in this link from page 23 to 25.
Here is the link
reference-request logarithms
edited Jan 29 at 17:19
YuiTo Cheng
2,1863937
asked Sep 23 '18 at 12:19
GiannaGianna
205
$endgroup$
add a comment |
$begingroup$
Euler found out that Euler's number was the basis of the natural logarithm. He started with this equation for infinitely small numbers $w$:
For $a>1$:
$$a^w=1+kw$$
$w$=infinitely small numbers
$a$=base
From $a^w=1+kw$ to this:
$${a}^{jw}={(1+kw)}^{j}$$
$${a}^{jw}={(1+kw)}^{j}=1+frac{j}{1}kw+frac{j(j-1)}{1*2}{k}^{2}{w}^{2}+frac{j(j-1)(j-2)}{1*2*3}{k}^{3}{w}^{3}+...$$
Then he replaced $j$ with $frac{z}{w}$:
$$j=frac{z}{w}$$
$z$ is any finite number, $ω$ is an infinitely small number and $j$ is an infinitely
large number.
Then he replaced $w$ with $frac{z}{j}$:
$${a}^{z}={a}^{jw}={(1+k*frac{z}{j})}^{j}=1+frac{j}{1}k*frac{z}{j}+frac{j(j-1)}{1*2}{k}^{2}{(frac{z}{j})}^{2}+frac{j(j-1)(j-2)}{1*2*3}{k}^{3}{(frac{z}{j})}^{3}+...$$
Since $j$ is infinitely large, you will get this:
$${a}^{z}=1+frac{kz}{1}+frac{{k}^{2}{z}^{2}}{1*2}+frac{{k}^{3}{z}^{3}}{1*2*3}+...$$
Then you replace $z$ and $k$ with $1$ to get the base of natural logarithm:
$${a}=1+frac{1}{1}+frac{1}{1*2}+frac{1}{1*2*3}+...$$
My question is, why did Euler replace $k$ and $z$ with $1$ to get the base of natural logarithm?
He had made also a series for $$log{1+x}=frac{1}{k}(frac{x}{1}-frac{{x}^{2}}{2}+frac{{x}^{3}}{3}-frac{{x}^{4}}{4}+...)$$
You can see that here in this link from page 23 to 25.
Here is the link
reference-request logarithms
edited Jan 29 at 17:19
YuiTo Cheng
2,1863937
asked Sep 23 '18 at 12:19
GiannaGianna
205
$endgroup$
add a comment |
$begingroup$
Euler found out that Euler's number was the basis of the natural logarithm. He started with this equation for infinitely small numbers $w$:
For $a>1$:
$$a^w=1+kw$$
$w$=infinitely small numbers
$a$=base
From $a^w=1+kw$ to this:
$${a}^{jw}={(1+kw)}^{j}$$
$${a}^{jw}={(1+kw)}^{j}=1+frac{j}{1}kw+frac{j(j-1)}{1*2}{k}^{2}{w}^{2}+frac{j(j-1)(j-2)}{1*2*3}{k}^{3}{w}^{3}+...$$
Then he replaced $j$ with $frac{z}{w}$:
$$j=frac{z}{w}$$
$z$ is any finite number, $ω$ is an infinitely small number and $j$ is an infinitely
large number.
Then he replaced $w$ with $frac{z}{j}$:
$${a}^{z}={a}^{jw}={(1+k*frac{z}{j})}^{j}=1+frac{j}{1}k*frac{z}{j}+frac{j(j-1)}{1*2}{k}^{2}{(frac{z}{j})}^{2}+frac{j(j-1)(j-2)}{1*2*3}{k}^{3}{(frac{z}{j})}^{3}+...$$
Since $j$ is infinitely large, you will get this:
$${a}^{z}=1+frac{kz}{1}+frac{{k}^{2}{z}^{2}}{1*2}+frac{{k}^{3}{z}^{3}}{1*2*3}+...$$
Then you replace $z$ and $k$ with $1$ to get the base of natural logarithm:
$${a}=1+frac{1}{1}+frac{1}{1*2}+frac{1}{1*2*3}+...$$
My question is, why did Euler replace $k$ and $z$ with $1$ to get the base of natural logarithm?
He had made also a series for $$log{1+x}=frac{1}{k}(frac{x}{1}-frac{{x}^{2}}{2}+frac{{x}^{3}}{3}-frac{{x}^{4}}{4}+...)$$
You can see that here in this link from page 23 to 25.
Here is the link
reference-request logarithms
edited Jan 29 at 17:19
YuiTo Cheng
2,1863937
asked Sep 23 '18 at 12:19
GiannaGianna
205
$endgroup$
Euler found out that Euler's number was the basis of the natural logarithm. He started with this equation for infinitely small numbers $w$:
For $a>1$:
$$a^w=1+kw$$
$w$=infinitely small numbers
$a$=base
From $a^w=1+kw$ to this:
$${a}^{jw}={(1+kw)}^{j}$$
$${a}^{jw}={(1+kw)}^{j}=1+frac{j}{1}kw+frac{j(j-1)}{1*2}{k}^{2}{w}^{2}+frac{j(j-1)(j-2)}{1*2*3}{k}^{3}{w}^{3}+...$$
Then he replaced $j$ with $frac{z}{w}$:
$$j=frac{z}{w}$$
$z$ is any finite number, $ω$ is an infinitely small number and $j$ is an infinitely
large number.
Then he replaced $w$ with $frac{z}{j}$:
$${a}^{z}={a}^{jw}={(1+k*frac{z}{j})}^{j}=1+frac{j}{1}k*frac{z}{j}+frac{j(j-1)}{1*2}{k}^{2}{(frac{z}{j})}^{2}+frac{j(j-1)(j-2)}{1*2*3}{k}^{3}{(frac{z}{j})}^{3}+...$$
Since $j$ is infinitely large, you will get this:
$${a}^{z}=1+frac{kz}{1}+frac{{k}^{2}{z}^{2}}{1*2}+frac{{k}^{3}{z}^{3}}{1*2*3}+...$$
Then you replace $z$ and $k$ with $1$ to get the base of natural logarithm:
$${a}=1+frac{1}{1}+frac{1}{1*2}+frac{1}{1*2*3}+...$$
My question is, why did Euler replace $k$ and $z$ with $1$ to get the base of natural logarithm?
He had made also a series for $$log{1+x}=frac{1}{k}(frac{x}{1}-frac{{x}^{2}}{2}+frac{{x}^{3}}{3}-frac{{x}^{4}}{4}+...)$$
You can see that here in this link from page 23 to 25.
Here is the link
reference-request logarithms
reference-request logarithms
edited Jan 29 at 17:19
YuiTo Cheng
2,1863937
asked Sep 23 '18 at 12:19
GiannaGianna
205
edited Jan 29 at 17:19
YuiTo Cheng
2,1863937
asked Sep 23 '18 at 12:19
GiannaGianna
205
edited Jan 29 at 17:19
YuiTo Cheng
2,1863937
edited Jan 29 at 17:19
YuiTo Cheng
2,1863937
edited Jan 29 at 17:19
YuiTo Cheng
2,1863937
2,1863937
asked Sep 23 '18 at 12:19
GiannaGianna
205
asked Sep 23 '18 at 12:19
GiannaGianna
205
asked Sep 23 '18 at 12:19
GiannaGianna
205
205
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't claim to really understand Euler's calculation, but here's my take on it. Obviously, to compute $a$ for $a^z,$ you set $z=1,$ so the only question is why Euler sets $k=1.$ The equation $$a^w=1+kw$$ for infinitesimal $w$ means, in standard terminology, $$k= {da^wover dw}|_{w=0}=ln a$$ so if $a$ is the base of natural logarithms, $k=1.$
EDIT
What I mean to say is that $a^w=1+kw$ is the same as $$k={a^w-1over w}$$ for infinitesimal $w$ and this means that $k$ is the derivative of $a^w$ (as a function of $w$) at $w=0$. Now if $a=e,$ we must have $k=1$. I haven't read Euler's calculation, but you say that he was determining the base for natural logarithms, and I take that to mean the number $a$ such that $ln a = 1.$
answered Sep 23 '18 at 12:36
saulspatzsaulspatz
17.1k31435
$endgroup$
$begingroup$
He had made also a series for $$log{1+x}=frac{1}{k}(frac{x}{1}-frac{{x}^{2}}{2}+frac{{x}^{3}}{3}-frac{{x}^{4}}{4}+...)$$
$endgroup$
– Gianna
Sep 23 '18 at 12:44
$begingroup$
But I don't understand this term $$k= {da^wover dw}|_{w=0}=ln a$$
$endgroup$
– Gianna
Sep 23 '18 at 12:46
$begingroup$
Sorry I still don't understand
$endgroup$
– Gianna
Sep 23 '18 at 13:12
$begingroup$
Can you tell me exactly what you don't understand? You know that $${da^wover dw}= a^wln a$$ right?
$endgroup$
– saulspatz
Sep 23 '18 at 13:18
$begingroup$
I don't understand this: $${da^wover dw}= a^wln a$$
$endgroup$
– Gianna
Sep 23 '18 at 13:28
|
show 4 more comments
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't claim to really understand Euler's calculation, but here's my take on it. Obviously, to compute $a$ for $a^z,$ you set $z=1,$ so the only question is why Euler sets $k=1.$ The equation $$a^w=1+kw$$ for infinitesimal $w$ means, in standard terminology, $$k= {da^wover dw}|_{w=0}=ln a$$ so if $a$ is the base of natural logarithms, $k=1.$
EDIT
What I mean to say is that $a^w=1+kw$ is the same as $$k={a^w-1over w}$$ for infinitesimal $w$ and this means that $k$ is the derivative of $a^w$ (as a function of $w$) at $w=0$. Now if $a=e,$ we must have $k=1$. I haven't read Euler's calculation, but you say that he was determining the base for natural logarithms, and I take that to mean the number $a$ such that $ln a = 1.$
answered Sep 23 '18 at 12:36
saulspatzsaulspatz
17.1k31435
$endgroup$
$begingroup$
He had made also a series for $$log{1+x}=frac{1}{k}(frac{x}{1}-frac{{x}^{2}}{2}+frac{{x}^{3}}{3}-frac{{x}^{4}}{4}+...)$$
$endgroup$
– Gianna
Sep 23 '18 at 12:44
$begingroup$
But I don't understand this term $$k= {da^wover dw}|_{w=0}=ln a$$
$endgroup$
– Gianna
Sep 23 '18 at 12:46
$begingroup$
Sorry I still don't understand
$endgroup$
– Gianna
Sep 23 '18 at 13:12
$begingroup$
Can you tell me exactly what you don't understand? You know that $${da^wover dw}= a^wln a$$ right?
$endgroup$
– saulspatz
Sep 23 '18 at 13:18
$begingroup$
I don't understand this: $${da^wover dw}= a^wln a$$
$endgroup$
– Gianna
Sep 23 '18 at 13:28
|
show 4 more comments
$begingroup$
I don't claim to really understand Euler's calculation, but here's my take on it. Obviously, to compute $a$ for $a^z,$ you set $z=1,$ so the only question is why Euler sets $k=1.$ The equation $$a^w=1+kw$$ for infinitesimal $w$ means, in standard terminology, $$k= {da^wover dw}|_{w=0}=ln a$$ so if $a$ is the base of natural logarithms, $k=1.$
EDIT
What I mean to say is that $a^w=1+kw$ is the same as $$k={a^w-1over w}$$ for infinitesimal $w$ and this means that $k$ is the derivative of $a^w$ (as a function of $w$) at $w=0$. Now if $a=e,$ we must have $k=1$. I haven't read Euler's calculation, but you say that he was determining the base for natural logarithms, and I take that to mean the number $a$ such that $ln a = 1.$
answered Sep 23 '18 at 12:36
saulspatzsaulspatz
17.1k31435
$endgroup$
$begingroup$
He had made also a series for $$log{1+x}=frac{1}{k}(frac{x}{1}-frac{{x}^{2}}{2}+frac{{x}^{3}}{3}-frac{{x}^{4}}{4}+...)$$
$endgroup$
– Gianna
Sep 23 '18 at 12:44
$begingroup$
But I don't understand this term $$k= {da^wover dw}|_{w=0}=ln a$$
$endgroup$
– Gianna
Sep 23 '18 at 12:46
$begingroup$
Sorry I still don't understand
$endgroup$
– Gianna
Sep 23 '18 at 13:12
$begingroup$
Can you tell me exactly what you don't understand? You know that $${da^wover dw}= a^wln a$$ right?
$endgroup$
– saulspatz
Sep 23 '18 at 13:18
$begingroup$
I don't understand this: $${da^wover dw}= a^wln a$$
$endgroup$
– Gianna
Sep 23 '18 at 13:28
|
show 4 more comments
$begingroup$
I don't claim to really understand Euler's calculation, but here's my take on it. Obviously, to compute $a$ for $a^z,$ you set $z=1,$ so the only question is why Euler sets $k=1.$ The equation $$a^w=1+kw$$ for infinitesimal $w$ means, in standard terminology, $$k= {da^wover dw}|_{w=0}=ln a$$ so if $a$ is the base of natural logarithms, $k=1.$
EDIT
What I mean to say is that $a^w=1+kw$ is the same as $$k={a^w-1over w}$$ for infinitesimal $w$ and this means that $k$ is the derivative of $a^w$ (as a function of $w$) at $w=0$. Now if $a=e,$ we must have $k=1$. I haven't read Euler's calculation, but you say that he was determining the base for natural logarithms, and I take that to mean the number $a$ such that $ln a = 1.$
answered Sep 23 '18 at 12:36
saulspatzsaulspatz
17.1k31435
$endgroup$
I don't claim to really understand Euler's calculation, but here's my take on it. Obviously, to compute $a$ for $a^z,$ you set $z=1,$ so the only question is why Euler sets $k=1.$ The equation $$a^w=1+kw$$ for infinitesimal $w$ means, in standard terminology, $$k= {da^wover dw}|_{w=0}=ln a$$ so if $a$ is the base of natural logarithms, $k=1.$
EDIT
What I mean to say is that $a^w=1+kw$ is the same as $$k={a^w-1over w}$$ for infinitesimal $w$ and this means that $k$ is the derivative of $a^w$ (as a function of $w$) at $w=0$. Now if $a=e,$ we must have $k=1$. I haven't read Euler's calculation, but you say that he was determining the base for natural logarithms, and I take that to mean the number $a$ such that $ln a = 1.$
answered Sep 23 '18 at 12:36
saulspatzsaulspatz
17.1k31435
edited Sep 23 '18 at 13:01
answered Sep 23 '18 at 12:36
saulspatzsaulspatz
17.1k31435
answered Sep 23 '18 at 12:36
saulspatzsaulspatz
17.1k31435
answered Sep 23 '18 at 12:36
saulspatzsaulspatz
17.1k31435
17.1k31435
$begingroup$
He had made also a series for $$log{1+x}=frac{1}{k}(frac{x}{1}-frac{{x}^{2}}{2}+frac{{x}^{3}}{3}-frac{{x}^{4}}{4}+...)$$
$endgroup$
– Gianna
Sep 23 '18 at 12:44
$begingroup$
But I don't understand this term $$k= {da^wover dw}|_{w=0}=ln a$$
$endgroup$
– Gianna
Sep 23 '18 at 12:46
$begingroup$
Sorry I still don't understand
$endgroup$
– Gianna
Sep 23 '18 at 13:12
$begingroup$
Can you tell me exactly what you don't understand? You know that $${da^wover dw}= a^wln a$$ right?
$endgroup$
– saulspatz
Sep 23 '18 at 13:18
$begingroup$
I don't understand this: $${da^wover dw}= a^wln a$$
$endgroup$
– Gianna
Sep 23 '18 at 13:28
|
show 4 more comments
$begingroup$
He had made also a series for $$log{1+x}=frac{1}{k}(frac{x}{1}-frac{{x}^{2}}{2}+frac{{x}^{3}}{3}-frac{{x}^{4}}{4}+...)$$
$endgroup$
– Gianna
Sep 23 '18 at 12:44
$begingroup$
But I don't understand this term $$k= {da^wover dw}|_{w=0}=ln a$$
$endgroup$
– Gianna
Sep 23 '18 at 12:46
$begingroup$
Sorry I still don't understand
$endgroup$
– Gianna
Sep 23 '18 at 13:12
$begingroup$
Can you tell me exactly what you don't understand? You know that $${da^wover dw}= a^wln a$$ right?
$endgroup$
– saulspatz
Sep 23 '18 at 13:18
$begingroup$
I don't understand this: $${da^wover dw}= a^wln a$$
$endgroup$
– Gianna
Sep 23 '18 at 13:28
$begingroup$
He had made also a series for $$log{1+x}=frac{1}{k}(frac{x}{1}-frac{{x}^{2}}{2}+frac{{x}^{3}}{3}-frac{{x}^{4}}{4}+...)$$
$endgroup$
– Gianna
Sep 23 '18 at 12:44
$begingroup$
He had made also a series for $$log{1+x}=frac{1}{k}(frac{x}{1}-frac{{x}^{2}}{2}+frac{{x}^{3}}{3}-frac{{x}^{4}}{4}+...)$$
$endgroup$
– Gianna
Sep 23 '18 at 12:44
$begingroup$
But I don't understand this term $$k= {da^wover dw}|_{w=0}=ln a$$
$endgroup$
– Gianna
Sep 23 '18 at 12:46
$begingroup$
But I don't understand this term $$k= {da^wover dw}|_{w=0}=ln a$$
$endgroup$
– Gianna
Sep 23 '18 at 12:46
$begingroup$
Sorry I still don't understand
$endgroup$
– Gianna
Sep 23 '18 at 13:12
$begingroup$
Sorry I still don't understand
$endgroup$
– Gianna
Sep 23 '18 at 13:12
$begingroup$
Can you tell me exactly what you don't understand? You know that $${da^wover dw}= a^wln a$$ right?
$endgroup$
– saulspatz
Sep 23 '18 at 13:18
$begingroup$
Can you tell me exactly what you don't understand? You know that $${da^wover dw}= a^wln a$$ right?
$endgroup$
– saulspatz
Sep 23 '18 at 13:18
$begingroup$
I don't understand this: $${da^wover dw}= a^wln a$$
$endgroup$
– Gianna
Sep 23 '18 at 13:28
$begingroup$
I don't understand this: $${da^wover dw}= a^wln a$$
$endgroup$
– Gianna
Sep 23 '18 at 13:28
|
show 4 more comments
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