Mixing
Proof: Zeros Localisation Theorem
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I have looked at the suggested related questions before asking.
From a First Course In Mathematical Analysis, David Brannan, page 148.Is the proof wrong?
zeros Localisation Theorem
Let $p(x) = {x^n} + {a_{n - 1}}{x^{n - 1}} + ... + {a_1}x + {a_0}$,$x in R$,be a polynomial.
Then all zeros of p line in $( - M,M)$, where $M = 1 + max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|left| {{a_0}} right|} $
$r(x) = frac{{p(x)}}{{{x^n}}} - 1 = frac{{{a_{n - 1}}}}{x} + ... + frac{{{a_1}}}{{{x^{n - 1}}}} + frac{{{a_0}}}{{{x^n}}},x in R - { 0} $ then, using the triangle inequality for $left| x right| > 1$
$left| {r(x)} right| = left| {frac{{{a_{n - 1}}}}{x} + ... + frac{{{a_1}}}{{{x^{n - 1}}}} + frac{{{a_0}}}{{{x^n}}}} right| le left| {frac{{{a_{n - 1}}}}{x}} right| + ... + left| {frac{{{a_0}}}{{{x^n}}}} right|$
$ le max { left| {{a_{n - 1}}} right|,left| {{a_1}} right|,left| {{a_0}} right|}left( {frac{1}{{left| x right|}} + ... + frac{1}{{{{left| x right|}^{n - 1}}}} + frac{1}{{{{left| x right|}^n}}}} right)$
$ < Mleft( {frac{1}{{left| x right|}} + ... + frac{1}{{{{left| x right|}^{n - 1}}}} + frac{1}{{{{left| x right|}^n}}} + ...} right)$
$ = M$$frac{{{textstyle{1 over {left| x right|}}}}}{{1 - {textstyle{1 over {left| x right|}}}}} = frac{M}{{left| x right| - 1}}$it follows that if $ {x ge M = 1 + max { left| {{a_{n - 1}}} right|,...,left| {{a_0}} right|} } $ then $left| {r(x)} right| < 1$.
But surely, if $left| x right| = M$ then $left| {r(x)} right| < frac{M}{{left| x right| - 1}} = frac{M}{{M - 1}}$ is greater than 1!
real-analysis
edited Jan 8 at 15:36
Ian
67.8k25388
asked Dec 23 '18 at 12:43
MegamaticsMegamatics
194
$endgroup$
add a comment |
$begingroup$
I have looked at the suggested related questions before asking.
From a First Course In Mathematical Analysis, David Brannan, page 148.Is the proof wrong?
zeros Localisation Theorem
Let $p(x) = {x^n} + {a_{n - 1}}{x^{n - 1}} + ... + {a_1}x + {a_0}$,$x in R$,be a polynomial.
Then all zeros of p line in $( - M,M)$, where $M = 1 + max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|left| {{a_0}} right|} $
$r(x) = frac{{p(x)}}{{{x^n}}} - 1 = frac{{{a_{n - 1}}}}{x} + ... + frac{{{a_1}}}{{{x^{n - 1}}}} + frac{{{a_0}}}{{{x^n}}},x in R - { 0} $ then, using the triangle inequality for $left| x right| > 1$
$left| {r(x)} right| = left| {frac{{{a_{n - 1}}}}{x} + ... + frac{{{a_1}}}{{{x^{n - 1}}}} + frac{{{a_0}}}{{{x^n}}}} right| le left| {frac{{{a_{n - 1}}}}{x}} right| + ... + left| {frac{{{a_0}}}{{{x^n}}}} right|$
$ le max { left| {{a_{n - 1}}} right|,left| {{a_1}} right|,left| {{a_0}} right|}left( {frac{1}{{left| x right|}} + ... + frac{1}{{{{left| x right|}^{n - 1}}}} + frac{1}{{{{left| x right|}^n}}}} right)$
$ < Mleft( {frac{1}{{left| x right|}} + ... + frac{1}{{{{left| x right|}^{n - 1}}}} + frac{1}{{{{left| x right|}^n}}} + ...} right)$
$ = M$$frac{{{textstyle{1 over {left| x right|}}}}}{{1 - {textstyle{1 over {left| x right|}}}}} = frac{M}{{left| x right| - 1}}$it follows that if $ {x ge M = 1 + max { left| {{a_{n - 1}}} right|,...,left| {{a_0}} right|} } $ then $left| {r(x)} right| < 1$.
But surely, if $left| x right| = M$ then $left| {r(x)} right| < frac{M}{{left| x right| - 1}} = frac{M}{{M - 1}}$ is greater than 1!
real-analysis
edited Jan 8 at 15:36
Ian
67.8k25388
asked Dec 23 '18 at 12:43
MegamaticsMegamatics
194
$endgroup$
$begingroup$
Please see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 13:01
$begingroup$
Please don't use x for multiplication. If you really must you can use times which renders as $times$, but even this is rarely needed.
$endgroup$
– Ian
Jan 8 at 15:31
add a comment |
$begingroup$
I have looked at the suggested related questions before asking.
From a First Course In Mathematical Analysis, David Brannan, page 148.Is the proof wrong?
zeros Localisation Theorem
Let $p(x) = {x^n} + {a_{n - 1}}{x^{n - 1}} + ... + {a_1}x + {a_0}$,$x in R$,be a polynomial.
Then all zeros of p line in $( - M,M)$, where $M = 1 + max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|left| {{a_0}} right|} $
$r(x) = frac{{p(x)}}{{{x^n}}} - 1 = frac{{{a_{n - 1}}}}{x} + ... + frac{{{a_1}}}{{{x^{n - 1}}}} + frac{{{a_0}}}{{{x^n}}},x in R - { 0} $ then, using the triangle inequality for $left| x right| > 1$
$left| {r(x)} right| = left| {frac{{{a_{n - 1}}}}{x} + ... + frac{{{a_1}}}{{{x^{n - 1}}}} + frac{{{a_0}}}{{{x^n}}}} right| le left| {frac{{{a_{n - 1}}}}{x}} right| + ... + left| {frac{{{a_0}}}{{{x^n}}}} right|$
$ le max { left| {{a_{n - 1}}} right|,left| {{a_1}} right|,left| {{a_0}} right|}left( {frac{1}{{left| x right|}} + ... + frac{1}{{{{left| x right|}^{n - 1}}}} + frac{1}{{{{left| x right|}^n}}}} right)$
$ < Mleft( {frac{1}{{left| x right|}} + ... + frac{1}{{{{left| x right|}^{n - 1}}}} + frac{1}{{{{left| x right|}^n}}} + ...} right)$
$ = M$$frac{{{textstyle{1 over {left| x right|}}}}}{{1 - {textstyle{1 over {left| x right|}}}}} = frac{M}{{left| x right| - 1}}$it follows that if $ {x ge M = 1 + max { left| {{a_{n - 1}}} right|,...,left| {{a_0}} right|} } $ then $left| {r(x)} right| < 1$.
But surely, if $left| x right| = M$ then $left| {r(x)} right| < frac{M}{{left| x right| - 1}} = frac{M}{{M - 1}}$ is greater than 1!
real-analysis
edited Jan 8 at 15:36
Ian
67.8k25388
asked Dec 23 '18 at 12:43
MegamaticsMegamatics
194
$endgroup$
I have looked at the suggested related questions before asking.
From a First Course In Mathematical Analysis, David Brannan, page 148.Is the proof wrong?
zeros Localisation Theorem
Let $p(x) = {x^n} + {a_{n - 1}}{x^{n - 1}} + ... + {a_1}x + {a_0}$,$x in R$,be a polynomial.
Then all zeros of p line in $( - M,M)$, where $M = 1 + max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|left| {{a_0}} right|} $
$r(x) = frac{{p(x)}}{{{x^n}}} - 1 = frac{{{a_{n - 1}}}}{x} + ... + frac{{{a_1}}}{{{x^{n - 1}}}} + frac{{{a_0}}}{{{x^n}}},x in R - { 0} $ then, using the triangle inequality for $left| x right| > 1$
$left| {r(x)} right| = left| {frac{{{a_{n - 1}}}}{x} + ... + frac{{{a_1}}}{{{x^{n - 1}}}} + frac{{{a_0}}}{{{x^n}}}} right| le left| {frac{{{a_{n - 1}}}}{x}} right| + ... + left| {frac{{{a_0}}}{{{x^n}}}} right|$
$ le max { left| {{a_{n - 1}}} right|,left| {{a_1}} right|,left| {{a_0}} right|}left( {frac{1}{{left| x right|}} + ... + frac{1}{{{{left| x right|}^{n - 1}}}} + frac{1}{{{{left| x right|}^n}}}} right)$
$ < Mleft( {frac{1}{{left| x right|}} + ... + frac{1}{{{{left| x right|}^{n - 1}}}} + frac{1}{{{{left| x right|}^n}}} + ...} right)$
$ = M$$frac{{{textstyle{1 over {left| x right|}}}}}{{1 - {textstyle{1 over {left| x right|}}}}} = frac{M}{{left| x right| - 1}}$it follows that if $ {x ge M = 1 + max { left| {{a_{n - 1}}} right|,...,left| {{a_0}} right|} } $ then $left| {r(x)} right| < 1$.
But surely, if $left| x right| = M$ then $left| {r(x)} right| < frac{M}{{left| x right| - 1}} = frac{M}{{M - 1}}$ is greater than 1!
real-analysis
real-analysis
edited Jan 8 at 15:36
Ian
67.8k25388
asked Dec 23 '18 at 12:43
MegamaticsMegamatics
194
edited Jan 8 at 15:36
Ian
67.8k25388
asked Dec 23 '18 at 12:43
MegamaticsMegamatics
194
edited Jan 8 at 15:36
Ian
67.8k25388
edited Jan 8 at 15:36
Ian
67.8k25388
edited Jan 8 at 15:36
Ian
67.8k25388
67.8k25388
asked Dec 23 '18 at 12:43
MegamaticsMegamatics
194
asked Dec 23 '18 at 12:43
MegamaticsMegamatics
194
asked Dec 23 '18 at 12:43
MegamaticsMegamatics
194
194
$begingroup$
Please see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 13:01
$begingroup$
Please don't use x for multiplication. If you really must you can use times which renders as $times$, but even this is rarely needed.
$endgroup$
– Ian
Jan 8 at 15:31
add a comment |
$begingroup$
Please see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 13:01
$begingroup$
Please don't use x for multiplication. If you really must you can use times which renders as $times$, but even this is rarely needed.
$endgroup$
– Ian
Jan 8 at 15:31
$begingroup$
Please see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 13:01
$begingroup$
Please see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 13:01
$begingroup$
Please don't use x for multiplication. If you really must you can use times which renders as $times$, but even this is rarely needed.
$endgroup$
– Ian
Jan 8 at 15:31
$begingroup$
Please don't use x for multiplication. If you really must you can use times which renders as $times$, but even this is rarely needed.
$endgroup$
– Ian
Jan 8 at 15:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your error is that you wrote that the max is less than $M$ but you really should have just denoted it by $M-1$ to keep things tighter. In the end the numerator here should be $M-1$ if $M$ is defined the way you defined it. This is why I don't like the style of injecting this $1+$ into the definition of symbols, I would prefer to define $M$ to be the maximum and use functions of that if need be in the proof (and even in the statement, in this situation).
answered Jan 8 at 15:38
IanIan
67.8k25388
$endgroup$
$begingroup$
I have read your comment. I still do not understand. David Brannan printed this proof. To be clear could you redo the parts that are in error. When you refer to max are you referring to the function max because only mod x is greater or equal M.
$endgroup$
– Megamatics
Jan 8 at 16:15
$begingroup$
@Megamatics The error here is in trying to use $max { |a_0|,dots,|a_n| }<M$. This is true, but it is not sufficiently tight to obtain the desired result. The correct way to proceed would be to simply replace $max { |a_0|,dots,|a_n| }$ by $M-1$ (which is exactly what it is, there is no inequality here). Then you would wind up with $|r(x)|<frac{M-1}{|x|-1}$ which is what you want to see. If $M$ weren't defined with the $1+$ in it to begin with, you would avoid this problem, but the statement would be slightly uglier because the interval would become $[-(M+1),M+1]$.
$endgroup$
– Ian
Jan 8 at 16:39
add a comment |
$begingroup$
I understand now. Because $max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|,left| {{a_0}} right|} le M - 1 < M$. So that
$max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|,left| {{a_0}} right|} < M$ is true but now the tighter inequality is obliterated.
So looking back one line with $M$ in the sum causes a issue. $M - 1$ should have been carried through rather than $M$ to $frac{M}{{left| x right| - 1}}$. So a better bound on ${r(x)}$ is $frac{{M - 1}}{{left| x right| - 1}}$. I.e. $left| {r(x)} right| le frac{{M - 1}}{{left| x right| - 1}}$ then the argument from there is correct.
answered Jan 8 at 19:28
MegamaticsMegamatics
194
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
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active
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votes
$begingroup$
Your error is that you wrote that the max is less than $M$ but you really should have just denoted it by $M-1$ to keep things tighter. In the end the numerator here should be $M-1$ if $M$ is defined the way you defined it. This is why I don't like the style of injecting this $1+$ into the definition of symbols, I would prefer to define $M$ to be the maximum and use functions of that if need be in the proof (and even in the statement, in this situation).
answered Jan 8 at 15:38
IanIan
67.8k25388
$endgroup$
$begingroup$
I have read your comment. I still do not understand. David Brannan printed this proof. To be clear could you redo the parts that are in error. When you refer to max are you referring to the function max because only mod x is greater or equal M.
$endgroup$
– Megamatics
Jan 8 at 16:15
$begingroup$
@Megamatics The error here is in trying to use $max { |a_0|,dots,|a_n| }<M$. This is true, but it is not sufficiently tight to obtain the desired result. The correct way to proceed would be to simply replace $max { |a_0|,dots,|a_n| }$ by $M-1$ (which is exactly what it is, there is no inequality here). Then you would wind up with $|r(x)|<frac{M-1}{|x|-1}$ which is what you want to see. If $M$ weren't defined with the $1+$ in it to begin with, you would avoid this problem, but the statement would be slightly uglier because the interval would become $[-(M+1),M+1]$.
$endgroup$
– Ian
Jan 8 at 16:39
add a comment |
$begingroup$
Your error is that you wrote that the max is less than $M$ but you really should have just denoted it by $M-1$ to keep things tighter. In the end the numerator here should be $M-1$ if $M$ is defined the way you defined it. This is why I don't like the style of injecting this $1+$ into the definition of symbols, I would prefer to define $M$ to be the maximum and use functions of that if need be in the proof (and even in the statement, in this situation).
answered Jan 8 at 15:38
IanIan
67.8k25388
$endgroup$
$begingroup$
I have read your comment. I still do not understand. David Brannan printed this proof. To be clear could you redo the parts that are in error. When you refer to max are you referring to the function max because only mod x is greater or equal M.
$endgroup$
– Megamatics
Jan 8 at 16:15
$begingroup$
@Megamatics The error here is in trying to use $max { |a_0|,dots,|a_n| }<M$. This is true, but it is not sufficiently tight to obtain the desired result. The correct way to proceed would be to simply replace $max { |a_0|,dots,|a_n| }$ by $M-1$ (which is exactly what it is, there is no inequality here). Then you would wind up with $|r(x)|<frac{M-1}{|x|-1}$ which is what you want to see. If $M$ weren't defined with the $1+$ in it to begin with, you would avoid this problem, but the statement would be slightly uglier because the interval would become $[-(M+1),M+1]$.
$endgroup$
– Ian
Jan 8 at 16:39
add a comment |
$begingroup$
Your error is that you wrote that the max is less than $M$ but you really should have just denoted it by $M-1$ to keep things tighter. In the end the numerator here should be $M-1$ if $M$ is defined the way you defined it. This is why I don't like the style of injecting this $1+$ into the definition of symbols, I would prefer to define $M$ to be the maximum and use functions of that if need be in the proof (and even in the statement, in this situation).
answered Jan 8 at 15:38
IanIan
67.8k25388
$endgroup$
Your error is that you wrote that the max is less than $M$ but you really should have just denoted it by $M-1$ to keep things tighter. In the end the numerator here should be $M-1$ if $M$ is defined the way you defined it. This is why I don't like the style of injecting this $1+$ into the definition of symbols, I would prefer to define $M$ to be the maximum and use functions of that if need be in the proof (and even in the statement, in this situation).
answered Jan 8 at 15:38
IanIan
67.8k25388
answered Jan 8 at 15:38
IanIan
67.8k25388
answered Jan 8 at 15:38
IanIan
67.8k25388
answered Jan 8 at 15:38
IanIan
67.8k25388
67.8k25388
$begingroup$
I have read your comment. I still do not understand. David Brannan printed this proof. To be clear could you redo the parts that are in error. When you refer to max are you referring to the function max because only mod x is greater or equal M.
$endgroup$
– Megamatics
Jan 8 at 16:15
$begingroup$
@Megamatics The error here is in trying to use $max { |a_0|,dots,|a_n| }<M$. This is true, but it is not sufficiently tight to obtain the desired result. The correct way to proceed would be to simply replace $max { |a_0|,dots,|a_n| }$ by $M-1$ (which is exactly what it is, there is no inequality here). Then you would wind up with $|r(x)|<frac{M-1}{|x|-1}$ which is what you want to see. If $M$ weren't defined with the $1+$ in it to begin with, you would avoid this problem, but the statement would be slightly uglier because the interval would become $[-(M+1),M+1]$.
$endgroup$
– Ian
Jan 8 at 16:39
add a comment |
$begingroup$
I have read your comment. I still do not understand. David Brannan printed this proof. To be clear could you redo the parts that are in error. When you refer to max are you referring to the function max because only mod x is greater or equal M.
$endgroup$
– Megamatics
Jan 8 at 16:15
$begingroup$
@Megamatics The error here is in trying to use $max { |a_0|,dots,|a_n| }<M$. This is true, but it is not sufficiently tight to obtain the desired result. The correct way to proceed would be to simply replace $max { |a_0|,dots,|a_n| }$ by $M-1$ (which is exactly what it is, there is no inequality here). Then you would wind up with $|r(x)|<frac{M-1}{|x|-1}$ which is what you want to see. If $M$ weren't defined with the $1+$ in it to begin with, you would avoid this problem, but the statement would be slightly uglier because the interval would become $[-(M+1),M+1]$.
$endgroup$
– Ian
Jan 8 at 16:39
$begingroup$
I have read your comment. I still do not understand. David Brannan printed this proof. To be clear could you redo the parts that are in error. When you refer to max are you referring to the function max because only mod x is greater or equal M.
$endgroup$
– Megamatics
Jan 8 at 16:15
$begingroup$
I have read your comment. I still do not understand. David Brannan printed this proof. To be clear could you redo the parts that are in error. When you refer to max are you referring to the function max because only mod x is greater or equal M.
$endgroup$
– Megamatics
Jan 8 at 16:15
$begingroup$
@Megamatics The error here is in trying to use $max { |a_0|,dots,|a_n| }<M$. This is true, but it is not sufficiently tight to obtain the desired result. The correct way to proceed would be to simply replace $max { |a_0|,dots,|a_n| }$ by $M-1$ (which is exactly what it is, there is no inequality here). Then you would wind up with $|r(x)|<frac{M-1}{|x|-1}$ which is what you want to see. If $M$ weren't defined with the $1+$ in it to begin with, you would avoid this problem, but the statement would be slightly uglier because the interval would become $[-(M+1),M+1]$.
$endgroup$
– Ian
Jan 8 at 16:39
$begingroup$
@Megamatics The error here is in trying to use $max { |a_0|,dots,|a_n| }<M$. This is true, but it is not sufficiently tight to obtain the desired result. The correct way to proceed would be to simply replace $max { |a_0|,dots,|a_n| }$ by $M-1$ (which is exactly what it is, there is no inequality here). Then you would wind up with $|r(x)|<frac{M-1}{|x|-1}$ which is what you want to see. If $M$ weren't defined with the $1+$ in it to begin with, you would avoid this problem, but the statement would be slightly uglier because the interval would become $[-(M+1),M+1]$.
$endgroup$
– Ian
Jan 8 at 16:39
add a comment |
$begingroup$
I understand now. Because $max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|,left| {{a_0}} right|} le M - 1 < M$. So that
$max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|,left| {{a_0}} right|} < M$ is true but now the tighter inequality is obliterated.
So looking back one line with $M$ in the sum causes a issue. $M - 1$ should have been carried through rather than $M$ to $frac{M}{{left| x right| - 1}}$. So a better bound on ${r(x)}$ is $frac{{M - 1}}{{left| x right| - 1}}$. I.e. $left| {r(x)} right| le frac{{M - 1}}{{left| x right| - 1}}$ then the argument from there is correct.
answered Jan 8 at 19:28
MegamaticsMegamatics
194
$endgroup$
add a comment |
$begingroup$
I understand now. Because $max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|,left| {{a_0}} right|} le M - 1 < M$. So that
$max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|,left| {{a_0}} right|} < M$ is true but now the tighter inequality is obliterated.
So looking back one line with $M$ in the sum causes a issue. $M - 1$ should have been carried through rather than $M$ to $frac{M}{{left| x right| - 1}}$. So a better bound on ${r(x)}$ is $frac{{M - 1}}{{left| x right| - 1}}$. I.e. $left| {r(x)} right| le frac{{M - 1}}{{left| x right| - 1}}$ then the argument from there is correct.
answered Jan 8 at 19:28
MegamaticsMegamatics
194
$endgroup$
add a comment |
$begingroup$
I understand now. Because $max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|,left| {{a_0}} right|} le M - 1 < M$. So that
$max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|,left| {{a_0}} right|} < M$ is true but now the tighter inequality is obliterated.
So looking back one line with $M$ in the sum causes a issue. $M - 1$ should have been carried through rather than $M$ to $frac{M}{{left| x right| - 1}}$. So a better bound on ${r(x)}$ is $frac{{M - 1}}{{left| x right| - 1}}$. I.e. $left| {r(x)} right| le frac{{M - 1}}{{left| x right| - 1}}$ then the argument from there is correct.
answered Jan 8 at 19:28
MegamaticsMegamatics
194
$endgroup$
I understand now. Because $max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|,left| {{a_0}} right|} le M - 1 < M$. So that
$max { left| {{a_{n - 1}}} right|,...,left| {{a_1}} right|,left| {{a_0}} right|} < M$ is true but now the tighter inequality is obliterated.
So looking back one line with $M$ in the sum causes a issue. $M - 1$ should have been carried through rather than $M$ to $frac{M}{{left| x right| - 1}}$. So a better bound on ${r(x)}$ is $frac{{M - 1}}{{left| x right| - 1}}$. I.e. $left| {r(x)} right| le frac{{M - 1}}{{left| x right| - 1}}$ then the argument from there is correct.
answered Jan 8 at 19:28
MegamaticsMegamatics
194
edited Jan 8 at 23:36
answered Jan 8 at 19:28
MegamaticsMegamatics
194
answered Jan 8 at 19:28
MegamaticsMegamatics
194
answered Jan 8 at 19:28
MegamaticsMegamatics
194
194
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$begingroup$
Please see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 13:01
$begingroup$
Please don't use x for multiplication. If you really must you can use times which renders as $times$, but even this is rarely needed.
$endgroup$
– Ian
Jan 8 at 15:31