aanpalendmw

2022-07-16

Prove symmetry of natural logarithm

Prove that $f(x)=\mathrm{ln}\sqrt{{x}^{2}+1}$ is symmetrical in $x=0$

$\mathrm{ln}\sqrt{(x-a{)}^{2}+1}=\mathrm{ln}\sqrt{(x+a{)}^{2}+1}$

$\sqrt{(x-a{)}^{2}+1}=\sqrt{(x+a{)}^{2}+1}$

$(x-a{)}^{2}+1=(x+a{)}^{2}+1$

${x}^{2}-2ax+{a}^{2}+1={x}^{2}+2ax+{a}^{2}+1$

$-2ax=2ax$

$-2ax=2ax$

$-x=x$?

I don't know what to do? Is this the proof or did I miss something?

Prove that $f(x)=\mathrm{ln}\sqrt{{x}^{2}+1}$ is symmetrical in $x=0$

$\mathrm{ln}\sqrt{(x-a{)}^{2}+1}=\mathrm{ln}\sqrt{(x+a{)}^{2}+1}$

$\sqrt{(x-a{)}^{2}+1}=\sqrt{(x+a{)}^{2}+1}$

$(x-a{)}^{2}+1=(x+a{)}^{2}+1$

${x}^{2}-2ax+{a}^{2}+1={x}^{2}+2ax+{a}^{2}+1$

$-2ax=2ax$

$-2ax=2ax$

$-x=x$?

I don't know what to do? Is this the proof or did I miss something?

nuramaaji2000fh

Beginner2022-07-17Added 18 answers

If you can prove that your function is an even function that is

$f(-x)=f(x)$

then it is symmetrical about the $y$-axis or the line $x=0$. In your case we have

$f(-x)=\mathrm{ln}(\sqrt{(-x{)}^{2}+1})=\mathrm{ln}(\sqrt{(x{)}^{2}+1})=f(x)$

which proves the symmetry.

$f(-x)=f(x)$

then it is symmetrical about the $y$-axis or the line $x=0$. In your case we have

$f(-x)=\mathrm{ln}(\sqrt{(-x{)}^{2}+1})=\mathrm{ln}(\sqrt{(x{)}^{2}+1})=f(x)$

which proves the symmetry.

Aleah Booth

Beginner2022-07-18Added 5 answers

If a function is symmetrical about a line $x=a$ then it can be said that:

$f(x)=f(2a-x)\mathrm{\forall}x\in \mathbb{R}$

$\mathrm{ln}\sqrt{{x}^{2}+1}=\mathrm{ln}\sqrt{(2a-x{)}^{2}+1}$

By using property $\mathrm{ln}{a}^{b}=b\mathrm{ln}a$

$\mathrm{ln}({x}^{2}+1)=\mathrm{ln}((2a-x{)}^{2}+1)$

Since "$\mathrm{ln}$" is a one-one function:

${x}^{2}=(2a-x{)}^{2}$

Expanding and cancelling:

$4{a}^{2}=4ax$

It gives two solutions:

$a=0,x=a$

Since we want symmetry for all real points we take the solution line $x=a(=0)$

For this question there was an easy way: You could just have shown that

${x}^{2}=(-x{)}^{2}$

For this question there was an easy way: You could just have shown that

${x}^{2}=(-x{)}^{2}$

and the rest follows after.

$f(x)=f(2a-x)\mathrm{\forall}x\in \mathbb{R}$

$\mathrm{ln}\sqrt{{x}^{2}+1}=\mathrm{ln}\sqrt{(2a-x{)}^{2}+1}$

By using property $\mathrm{ln}{a}^{b}=b\mathrm{ln}a$

$\mathrm{ln}({x}^{2}+1)=\mathrm{ln}((2a-x{)}^{2}+1)$

Since "$\mathrm{ln}$" is a one-one function:

${x}^{2}=(2a-x{)}^{2}$

Expanding and cancelling:

$4{a}^{2}=4ax$

It gives two solutions:

$a=0,x=a$

Since we want symmetry for all real points we take the solution line $x=a(=0)$

For this question there was an easy way: You could just have shown that

${x}^{2}=(-x{)}^{2}$

For this question there was an easy way: You could just have shown that

${x}^{2}=(-x{)}^{2}$

and the rest follows after.

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