Answered

Dear Yousuf,

thank your for contacting us.

Currently, the images are not attached, it is therefore difficult for us to understand the behavior you are facing.

Can you please attach these again (or provide a link where we can download these)?

Best regards,

Olivier

Please see links below:

Equation 1: https://drive.google.com/open?id=10w1T5nYYPpfne2X3TkHbIHSUdtTXwQ80

Equation 2: https://drive.google.com/open?id=1Z4O0vdgs4dP5QLAFnlLVk3gmVeDFXKRT

Regards,

Yousuf

Dear Yousuf,

Thank you for the images.

If I understand well, you are writing t=[v_[i]/µg] and would like to have it recognized as t=[v_[i]/µ]*g?

Currently, our solution doesn't claim to interpret but recognize exactly what is written.

Best regards,

Olivier

Not
quite. I am asking why t=v_[i]/µg=[v_[i]/µg] but t=v_[i]/gµ=[v_[i]/[gµ]]? In
words, when a full fraction is entered with the denominator having a typical
latin character (g in this case) followed by other characters the interpreter
places brackets [] around the denominator (expected behavior). When a non latin
character is the first character in the denominator of a full fraction is not a
typical character (µ in this case) then no brackets are placed around the
denominator (unexpected behavior since order of operations will treat the
characters, beyond the first, as not part of the denominator).

Thus the question is, what is the condition for placing brackets around the
denominator of a full fraction?

Answer

Dear Yousuf,

thank you for the explanation, I could better understand.

Currently, when writing t=v_[i]/µg, the denominator is usually recognized as micro gram, while when writing t=v_[i]/gu, the denominator is recognized as g multiply mu.

The reason is that in our math engine, microgram has more weight than mu multiply g.

If you do not want the micro gram to be recognized, then you can create your own math gramar definition removing micro gram in the latter (see https://developer.myscript.com/docs/interactive-ink/1.3/ios/advanced/build-custom-resources/#math)

Best regards,

Olivier

## yousuf.binazhar@datadynamics-inc.com

Hi,

But the equation transforms correctly to t=[v_[i]/μg] which is usually interpreted as t=[v_[i]/μ]*g.

What is the logic as to why the two are different?

Olivier @MyScriptsaid 8 months agoDear Yousuf,

thank you for the explanation, I could better understand.

Currently, when writing t=v_[i]/µg, the denominator is usually recognized as micro gram, while when writing t=v_[i]/gu, the denominator is recognized as g multiply mu.

The reason is that in our math engine, microgram has more weight than mu multiply g.

If you do not want the micro gram to be recognized, then you can create your own math gramar definition removing micro gram in the latter (see https://developer.myscript.com/docs/interactive-ink/1.3/ios/advanced/build-custom-resources/#math)

Best regards,

Olivier